New Course

Chemistry
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Vol. I

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Strictly according to new syllabus & examination pattern of CBSE, ISCE and slate boards of Punjab, Haryana,
H.P., J & K, Odisha, Karnataka, Kerala, Nagaland, Manipur, Rajasthan, Jharkhand, Bihar, U.P.,

for F
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Uttarakhand, M.P., Ghhatisgarh, Assam, Gujarat, West Bengal, A.P. etc,
for class XII of+2 stage.
Your
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Dr. S.C. KHETERPAL Dr. S.N. DHAWAN

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M.Sc., Ph D. M.Sc., Ph.D.

Department of Chemistry Department of Chemistry
Institute of Integrated and Honors Studies (IIHS) Kurukshetra University
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Kurukshetra University KURUKSHETRA

KURUKSHETRA
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Find

2023-2024

Pradeep Publications
' j ■ -i* JALANDHAR (INDIA)
 Published by:
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Edition: Forty-fourth
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ISBN : 978-93-95649-27-8
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PRICE : ? 1598.00 (For Vol. I & II]

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 A Few Words From The Authors ...
ABOUT THE FORTY-FOURTH EDITION
Till last academic session (2022-23), Central Board of Secondary Education (CBSE) had not made much
changes in the syllabii for +1 and +2 classes but had changed the pattern of examination to help the

ow
students to feel comfortable in view of the pandemic. But for the current session (2023-24), to reduce the
burden of studies, CBSE has reduced the syllabii. For the +2 students, the number of units to be studied
have been reduced from 16 to 10, i.e., 6 units have been deleted besides a few articles here and there. The
present edition lays full emphasis on the units included in the new syllabus from board examination point

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of view but does not ignore the preparation for competitive examinations as many of these competitive

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examinations have still retained those units left out by CBSE. Therefore, these units have been included in

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the book along with a section laying emphasis on the questions for competitive examinations. However, all
units of the book have been updated by including all good questions from the examinations held upto 2022
(board as well as competitive examinations). The units not included in the CBSE syllabus but have been

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included for competitive examinations have been marked with asterisk (*) and mentioned in the footnote.

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We understand that it makes the book a little voluminous but the important aspects of board examination
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as well as competitiveexaminations cannot be overlooked.
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With addition of new features, the book now contains the following salient features:
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0 Articlewise 'Solved Problems'followed by 'Problems For Practice' with Answers and Hints.
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● NCERT based Multiple Choice Questions and Assertion-Reason Type Questions.

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● ArT/dew/se Conceptual Questions with Answers.

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0 Art/c/eiv/se Very Short Answer, Short Answer and Long Answer Questions, all with Answers and
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Hints.

0 Case-based very short/short questions.

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0 Case-based MCQs and Assertion-Reason Questions.

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0 Questions of NCERT book with Answers and Solutions.

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0 Questions of NCERT Exemplar Problems with Answers and Solutions.

0 Competition Focus consisting of Articlewise up-to-date Multiple Choice Questions of different
typesforNEET,JEE(Main)andJEE (Advanced) with Answers and Explanations,
it is sincerely hoped that the present extensively revised edition of Pradeep'sNew Course Chemistrywill
excellently meet the requirements of examinations of CBSE, ICSE and all State Boards as well as
competition examinations.

We are highly grateful to Mr. Pradeep Jain, our dynamic publisher, who always takes keen interest in
making the book better and better in style which makes its readingmore and more enjoyable.
Suggestions for further improvement of the book are always welcome and appreciated.

Authors
 Contents
Units Pages
*1. Solid state 1/1—1/76
1.1. Generallntroduction 1/1
1/1
1.2. Why do some Substances Exist as Solids ?
1.3. General Characteristics of Solid State 1/2
1/2
1.4. Amorphous and Crystalline Solids

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1.4,1. Points of difference between acrystallinesolid and an amorphous solid
1.4.2. Uses of amorphoussolids

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1.5. ClassificationofCrystallineSolids 1/4
1/7
1.6. Crystal Lattices (or Space Lattices) and Unit Ceils
1.7. AMore Realistic Picture of a Unit Cell (Open Structures and Space-filling Stiuctures) 1/10

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1.8. Calculation of Number ofAtoms in a Unit Cell of aCubic Crystal System 1/11

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1.8.1. Calculation of the contribution of atoms present at different lattice sites
1.8.2. Calculationof number ofAtoms per unit cell
1.9. Close Packing in Crystals (Close Packed Structures)
for 1/14
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1/18
1.10. Calculation of the Spaces Occupied, i. e., Packing Efficiency or Packing Fractions
1.10.1. Packing efficiency in simple cubic lattice
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1.10.2. Packing efficiency in face-centred cubic Structures (cubic close packing)

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1.10.3. Packing efficiency in body centred cubic Structures

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1.11. Sizes of Tetrahedral and Octahedral Voids 1/19

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1.11.1. Derivation of the relationship between the radius (r) of the octahedral void
and the radius (R) of the atoms in close packing.
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1.11.2. Derivation of the relationship between radius (r) of the tetrahedial void and
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the radius(R) ofthe atoms in close packing.

1.12. Formula of the Compound from Number of Voids Filled 1/22
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1.13. Location of Tetrahedral and Octahedral Voids in a Crystal 1/24

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1/25
1.14. Structures of Simple Ionic Compounds
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1/30
1.15. Close Packing in the Ionic Compounds and Occupancy of Voids
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1.16. Radius Ratio Rules 1/30

1/32
1.17. Relationship between the Nearest Neighbour Distance (d). Edge of the Unit Cell (a)
and Radius ofthe Atom (r) for Pure Elements
1.18. Calculations Involving Unit Cell Dimensions 1/34

(Calculation of Density of a Cubic Crystal from its Edge)

1/43
1.19. Imperfections or Defects In Solids
1/49
1.20. Electrical Properties of Solids
1.20.1. Classification of solids on the basis of conductivity
1.20.2. Band theory of metals.
1.20.3. Explanation of behaviour of conductors, insulators and
semiconductors on the basis of band theory.
1.20.4. Effect oftcmperature on electrical conductivity.
1.20.5. Applications ofn-type andp-type semiconductors
1.20.6. Conductivity oftransition metal oxides.
* \’ol included in CBSE syllabus. This chapter has been given only for (tie preparation of competitive examination.
 1.21. Magnetic Properties of Solids 1/52
RevisionAtA Glance 1/55
Competition Focus [NEET/JEE Special] 1/59

2. Solutions 2A—2/146
2.1. General Introduction 271
2.2. Types of Solutions 2/1
2.3. Expressing Concentration of Solutions 2/2
2.4. Solutions of Solids in Liquids 2/14
2.4. J. Solubility ofa solid in a liquid
2.4.2. Factors affecting the solubility of a solid in a liquid
2.5. Solutions ofGascs in Liquids 2/16

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2.5.1. Solubility ofa gas in a liquid
2.5.2. Factors affecting the solubility of a gas in a liquid (including Henry’.s law)
2.6. Solutions of Solids in Solids (Solid Solutions) 2/22
2.7. Vapour Pressure of Liquid Solutions and Raoult’s Law 2/23
2.7.1, Vapour pressure

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2.7.2, Vapourpressuresofliquid-Iiquidsolutions and Raoult’s law
{i.e., Raoult’s law for volatile solutes)

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2.7.3. Vapour pressures of solutions of solids in liquids and Raoult’s law
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(/.e.,Raoull’s law for non-volatilesolutes)

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2.7.4, Raoult’s law as a special case of Hcniy’s law

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2.7,5. Similarities and Dissimilarities between Raoult’s law and Henry’s law
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2.8. Ideal and Non-ideal Solutions 2/31
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2.9. Azeotropic or Constant Boiling Mixtures 2/34
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2.10. Colligative Properties and Determination of Molar Mass 2/35
2.11. RelativeLoweringofVapourPressure 2/35
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2.11.1. General discussion

2.11.2. To conclude that the relative lowering of vapour pressure is a colligative property
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2.11.3. Determination of molecular masses of solutes from lowering of vapour pressure

2.11.4. Measurement of relative lowering of vapour pressure
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2.12. Osmosis and Osmotic Pressure 2/41

2.12.1. General discussion and definitions
2.12.2. Experimental measurement of osmotic pressure
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2.12.3. Expression for the osmotic pressure

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2.12.4. Isotonic solutions

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2.12.5. To conclude that osmotic pressure is a colligative property

2.12.6. Biological importance of osmosis
2.12.7. Explanation of some phenomena on the basis of osmosis
2.12.8. Determination of molecular mass from osmotic pressure measurements
2.13. Elevation of Boiling Point (Ebullioscopy) 2/52
2.13.1. General discussion

2.13.2. Expression for the elevation ofboiling point

2.13.3. Calculation of molecular mass of the solute
2.13.4. Calculation of molal elevation constant fromenthalpy of vaporisation
2.13.5. To conclude that elevation ofboiling point is a colligative property
2.14. Depression of Freezing Point (Cryoscopy) 2/57
2.14.1. General discussion
2.14.2. Expression for the depression of freezing point
2.14.3. Calculation of molal depression constant from enthalpy of fusion
 2.14.4. Calculation of molecular mass of tl\e solute
2.14.5. To conclude that the depression of freezing point is a colligative property
2.14.6. Applications (Importance) ofdepression of freezing point
2.14.7. Rast method
2.15. Abnormal Molar Masses 2/65
Revision AlA Glance 2/75

Multiple Choice Questions 2/78

Assertion-Reason Type Questions 2/82

Conceptual Questions 2/86

Veiy Short Answer Questions [/ mark] 2/89

Short Answer Questions [2 or 3 marks] 2/92

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Long Answer Questions [.5 or more marks] 2/94

Case-Based Ver}’ Short/Short Questions 2/95

Case-Rased MCQs and Assertion-Reason Questions 2/97

NCER T Questions and Exercises with A nswers 2/100
JSCER TExemplar Problems with Answers. Hints and Solutions 2/113

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Competition Focus [NEET/JEESpeciul\ 2/122

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3. Electrochemistry 3/1—3/156
3.1. General Introduction 3/1

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3.1.1. Redox reactions
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3.1.2. Blectrocheniical cells ks
3.2. Electrolytic Cells and Electrolysis 3/3
3.2.1. Definitions ofthe terms used
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3.2.2. Mechanism of electrolysis
3.3. QuantitativeAspects of Electrolysis and Earaday’s Laws 3/6
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3.4. Conductance of Electrolytic Solutions 3/14

3.5. Electrical Resistance and Conductance 3/15
3.6. Specific, Equivalent and MolarConductivities 3/16
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3.6.1. Specific conductivity (or simply called conductivity)

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3.6.2. Equivalent conductivity

3.6.3. Molar conductivity
3.7. Measurement of Electrolytic Conductance, Specific Conductivity, Equivalent 3/19
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Conductivity and Molar Conductivity of Ionic Solutions

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3.8. Variation of Conductance, Specific Conductivity, Equivalent Conductivity 3/25

and Molar Conductivity with Concentration or Dilution

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3.9. Variation ofMolarConductivitywith Concentration (for Strong and Weak Electrolytes) 3/25
3.10. Kohlrausch’sLaw 3/27
3.11. Applications of Kolilrausch's Law 3/30
3.12. GalvanicCells 3/34

3.13. DilTcrence between Galvanic cell and Electrolytic Cell 3/38

3.14. Representation of Galvanic Cell 3/38

3.15. Electrode Potential 3/39
3.16. Measurement of Electrode Potential 3/40
3.17. Cell Potential orEMFofa Cell 3/42
3.18. Electrochemical Series 3/43

3.19. Applications of the Electrochemical Scries 3/43

3.20. Effect of Electrolyte Concentration and Temperature on the Electrode Potential 3/49

(Nemst Eqn. forElectrode Potential)

 3.21. Effect of Electrolyte Concentration and Temperature on the EMF of a Cell 3/50

(Nemst Eqn. for EMF of a Ce! 1)

3.22. Concentration Cells 3/56
3.23. Equil ibrium Constant from Nemst Equation 3/57

3.24. Electrochemical Cell and Gibbs Energy of the Reaction (Relation between AG® and 3/59

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3.25. Predicting Products of Electrolysis from Electrode Potentials 3/64
3.26. Batteries (Commercial Cells) 3/67
3.26.1. Primary batteries or cells

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3.26.2. Secondarybatteriesorcells
3.27. Fuel Cells 3/70

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3.28. Corrosion 3/73
3.29. Hydrogen as a Renewable and Non-Polluting Source of Energy 3/75

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RevisionAtA Glance 3/75

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Multiple Choice Questions 3/80

Assertion-Reason Type Questions 3/84

Conceptual Questions 3/89

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Very Short Answer Questions [/ mark] 3/93

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Short Answer Questions [2 or 3 marks] 3/97

Long Answer Questions [5 or more marks] 3/100

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Case-Based Very Short/Short Questions 3/102

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Case-Based MCQs and Assertion-Reason Questions 3/104

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NCER T Questions and Exercises with Answers 3/107
NCERT Exemplar Problems with Answers, Hints and Solutions 3/115
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Competition Focus [NEET/JEE Special] 3/124
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4. Chemical Kinetics 4/1—4/148

4.1. General 1 ntroduction 4/1
4.2. Classification of Reactions on the Basis of Rates 4/2
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4.3. Rate of Reaction, Its Measurement and Dependence on Various Factors 4/2
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4.3.1. Rate of reaction

4.3.2. Average rate and Instantaneous rate of reaction

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4.3.3. Measurementofthe rate ofreaction

4.3.4. Expressing the rate of reaction in terms of different reactants and products
4.3.5. Factors influencing rate ofa reaction
4.4. Dependence of Rate on Concentration of Reactants 4/12
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(Rate Law Expressions, Rate Constant and Order of a Reaction)

4.5. Units of Rate constant or Specific Reaction Rate for Reactions of Different Orders 4/18
4.6. Distinction between Rate of Reaction and Reaction Rate Constant 4/19

4.7. Molecularity ofa Reaction 4/19

4.8. MechanismofaReaction 4/20
4.9. Distinction between Order and Molecularity of a Reaction 4/24
4.10. Integrated Rate Equations 4/24

4.10.1. Integrated rate equation for zero order reactions

4.10.2. Integrated rate equation for first order reactions
4.10.3. Integrated rate equation for first order gas phase reactions
4.11. Half-life Period of a Reaction 4/30
4.12. Determination of the Rate Law. Rate Constant and Order of Reaction 4/31
4.13. Examples ofthe Reactions of First Order 4/43
4.14. Pseudo First Order Reactions 4/46
4.15. Rate of Radioactive Decay/Disintegration 4/58

Contents Contd. after Periodic Table >

 OLD GROUP NOS.—► rA IIA IIIB IVB VB VIB VIIB

MODERN GROUP NOS. 1 2 3 4 5 6 7

s-BLOCK (ns^-2)
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LON(
GROUPS
"o OF THE PEI
PERIODS
(ns^)
1 1'0079 I ATOMIC NUMBgl^ *26
D H

(Is) Hydrogen e F

low
Is*' (ns2) In
3 8’941 9-102
B LI Be 3d^
A]
Lithium Beryliium
(2s2p)
2s"' 2s2 d-BLOCK {TRANS

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[(n-1)di
11 22-990 12
rF 24-305

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B Na Mg
Sodium Magnesium
(39 3p)
3s
1
3s2 e o e o e

D
19 39-098 20
K
40-078 21
Ca
44-956 22
Sc for 47-88 23
Ti
50-942 24
V
51-996 25
Cr
54-938 26
Mn
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Potassium Calcium Scandium Titanium Vanadium Chromium
(4s 3d 4p) Manganese
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4s 1 4s^ 3d*’ 4s^ 3d2 4s2 3d^ 4s^ 3d® 4s"’ 3d® 4s2
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37 85-408 38 87-62 39 88-906 40 91-224 41 92-906 42 95-94 43 98-906* 44

B
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Rb Sr Y Zr Nb Mo Tc
Rubidium Strontium Yttorium Zirconium Niobium Molybdenum Technetium Rl
(5s 4d 5p)
5s ^ 6s2 4d^ 6s^ 4d2 5s2 4d^ 5s ■’ 4d® 5s^ 4d® 5s^ 4
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55 132-905 56 137-33 57 138-91 72 178-49 73 180-95 74 183-84 75 186-21 76

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B Cs Ba La^ Hf Ta W Re
Cesium Barium Lanthanum Hafnium Tantalum Tungsten Rhenium C
(6s (4f) 5d 6p)
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es"* 6s2 Sd"* 6s2 5d2 6s2 5d®6s2 5d^ 6s2 5d® 6s2 5

226-03 89 227 03* 104 261-1T 105 261-11* 106 263-12* 107 262-12* 101
Re

87 223-02 88
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B Fr Ra Ac# Rf Db Sg Bh
Francium Radium Actinium Rutherfortiium Hi
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Dubnium Seaborgium Bohrium

(7s (5f) 6d 7p)
7s"’ 7s2 ed"* 7s2 6d2 7s2 6d3 7s2 6d"^ 7$2 6d® 7s2 6

-^-REPRESENTATIVE-^
f-BLOCK/IN
ELEMENTS

58 140-12 59 140-91 60 144-24 61 146-92* 62 150-36 63

♦ LANTHANOIDS (4f-series) Ce Pr Nd Pm Sm
Cerium
(4f1-l4 5dC-l 0s2) PraseodyniHjm Neodymium Promethium Samarium Eu

4fl 5d^ 6s2 4f3 od° 6s2 4f^ 5d0 6s2 4fS 5d0 6s2 4f^ 5d0 6s2 4f^
90 232-04* 91 231 04' 92 238-03* 93 237-05* 94 244 ●06* 95
# ACTINOIDS (5f-series) Th Pa U Np Pu

(5f0-14 6c|0-’' 7s2)

Thorium Protactinium Uranium Neptunium Plutonium Arr

5f° 6d2 7s2 5f2 6d' 7s2 5f3 6d^ 7s2 5H 6d"' 7s2 5f6 6d0 7s2 5f

* Atomic masses have not been corrected.

 VIII IB IIB illA IVA VA VIA VIIA 0

9 10 11 12 13 14 15 16 17 18

p-BLOCK (np1“6)
FORM /V

bOIC TABLE <E)

(ns^ np®)
145- RELATIVE ATOMIC MASS
2 4 0026
SYMBOL He
<E> ® ® 0

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Helium
(ne^np^) (ns^ np2) (ns^ np®) (ns^np^) (ns^np®) 1s2
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NAME 5 10-811 6 12-011 7 14-007 8 15-999 9 18-998 20-18

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B C N 0 F Ng
Boron Carbon Nitrogen 0 en Fluorine Neon
)N ELEMENTS) 2s2 2P"* 2s2 2p2 2s2 2p3 2s2 2p5 2s2 2p®

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ns^‘2] 13 26-982 14 28-086 15 30-974 16

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32-066 17 35-453 18 39-948
Al Si P S Ci Ar
1 Aluminium Silicon Phosphorus Sulphur Chlorine Argon
O ® ® 0 3s2 3p1 3S^ 3p 2 3s2 3p3 3s2 3s2 3p5 3s2 3p6
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●845 27 58-933 28 58-71 29 63-546 30 65-39 31 69-723 32 72-61 33 74-922 34 78-98 35 79-904 36 83-80
Co Ni Cu Zn Ga Ge As Br Kr
i
s
Cobalt Nickel Copper Zinc Gallium Germanlun-i Arsenic Selenium Bromine Kryptoii
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ls2 3d^ 4$2
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3cJS 4s2 3d ■’0 4s'’ 3d'’0 4s2 4s^ 4p~’ 4$^ 4p^ 4s^4£^ 4s*^ 4p"* 4s^ 4p^ 4s2 4p6
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1-07 45 102-905 46 106 42 47 107-87 48 112-41 49 114-82 50 118-71 51 121-76 52 I27-60 53 126-90 54 131-29
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Rh Pd Ag Cd in Sn Sb Te I Xe
iium Rhodium Palladium Sliver Cadmium indium Tin Antimony Tellurium Iodine Xenon
>s^ 4d® 5s^ 4d'’0 5s° 4d’’0 5s'’ 4d'”^5s2 5s2 Sp'’ 5s^ 5p^ 5s^
mmEI 5s^ 5p® 5s^ 5p^
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0-23 77 192-217 78 195-08 79 196-97 80

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200-59 81 204-38 82 207-19 83 208-98 84 209-98 85 209-99 86 222-02

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Ir Pt Au Hg Tl Pb Bi Po At Rn
um Iridium Platinum Gold Mercury Thallium Lead
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Bismuth Polonium Astatine Radon

)S 5d^ 6s2 5d^ es"* Sd^’^Bs'’ 5d''0 6s2 6s2 ep'’ 6s2 6p2 6s2 6p3 6s2 6p^ 6s2 6p5 6s2 6p®
nd

265* 109 1* 110 111 272* 112

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277* 113 286* 114 289* 115 288* 116 292* 117 294* 118 294*
Mt Ds Rg Cn Nh FI Me Lv Ts Og
Meitnerium Oarmstadtium Roentgenium
Fi

turn
Copernicium Nihoniiim Flercivlum Mosco\4um Llvermorium Tennessine Oganesson
s2 6d^ 7s2 6d® 7s2 6d'’^7s'’ 6d’’0 7s2 7s2 7p1 7s2 7p2 7s2 7p3 7s2 7p^ 7s2 7pS 7s2 7p6
NOBLE GASES
^TRANSITION ELEMENTS (RARE EARTHS) REPRESENTATIVE ELEMENTS
[(n-2)fO-i4(n-i)dO-i ns2] METALS NON-METALS
1

-96 64 157-25 65 158-93 66 162-50 67 164-93 68 167-26 69 168-93 70 173-04 71 174-97

Gd Tb Dy Ho Er Tm Yb Lu
jm Gadolinium Terbium Dysprosium Erbium Holmium
Thulium Ytterbium Lutetium
6s2 4f 5d^ 6s2 4t^ 5d° 6s2 4f'0 5d®6s2 4f^^5d06s2 4f12 5d06s2 4f^3 5d°6s2 4f^4 5d06s2 4f^4 Sd'' 6s2
06* 96 247-07* 97 247-07* 98 251-08* 99 252-08* 100 257-18* 101 258-10* 102 259-10* 103 262-11*
Cm Bk Cf Es Fm Md No Lr
urn Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
7s2 5f7 6d > 7s2 5f9 6d° 7s2 5f1®6d® 7s2: 5f^'>6d07s2 5fl2 6d®7s2, 5f^36d°7s2 5f1'*6d°7s2 6d1 7s2
 4.16. Activation Energy of Reactions 4/59

4,17. Temperature Dependence of the Rale of Reaction-Effect ofTemperatiire on 4/62

Rate Constant (Arrhenius Equation)

4.18. Effect of Catalyst on the Rate of Reaction 4/70

4.19. Collision Theory ofChemical Reactions 4/70

4.20. Transition State Theory or Theory ofAbsolute Reaction Rates 4/72

Revision At A Glance 4/73

Multiple Choice Questions 4/77

Assertion-Reason Type Questions 4/82

Conceptual Questions 4/86

Very Short Answer Questions [J mark] 4/89

Short Answer Questions {2 or 3 marks] 4/92

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Long Answer Questions [5 or more marks] 4/95

Case-Based Very Short/Short Questions 4/97

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Case-Based MCQs and Assertion-Reason Questions 4/98

NCER T Questions and Exercises with A nswers 4/101

NCER T Exemplar Problems with Answers, Hints and Solutions 4/110

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Competition Focus [NEET/JEE Special] 4/120

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*5. Surface Chemistry 5/1-5/72
5.1. General Introduction
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5/1
5.2. Adsorption
5.2.1. Definitions of the terms u.sed
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5.2.2. Evidence in support of adsorption - adsorption in action

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5.2.3. Mechanismofadsorption
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5.2.4. Adsorption-an exothermicprocess

5.2.5. Entropy change during adsorption and adsorption equilibrium
5.2.6. Distinction between adsorption and absorption
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5.2.7. Examples of adsorption,absorptionand sorption

5.2.8. Positive and negative adsorption
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5.3. l-’actorsAtTectingAdsoiptionofGasesby Solids 5/3

5/5
5.4. Types ofAdsorption
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5.5. FreundlichAdsorptionIsolhenn 5/8

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5.6. Adsorption Isobars 5/10

5.7. Applications of.Adsorption 5/15

5.8. Catalysis 5/16

5.8.1. Definition

5.8.2. Positive and negative catalysis

5.8.3. Promoters and poisons
5.8.4. Types of catalysis (homogeneous and heterogeneous catalysis)
5.8.5. Some importantfeaturesofsolid catalysis(or heterogeneous catalysis)
5.9. Zeolitesas Shape-SelectiveCatalysts 5/20
5/21
5.10. Enzymes as Catalysts
5.10.1. What are enzymes ?
5.10.2. Characteristics of enzymes

* Not included in CBSE syllabus. This chapter has been given onlyfor the preparation of competitive examination.
 5.10.3. Mechanism of enzyme catalysis
5.10.4. Points of difference between enzymes and inorganic catalysts
5.11. Catalysts Used in Industries 5/25
5.12. Colloidal State of Matter 5/26
5.13. Distinction between True Solution, Colloids Solution and Suspensions 5/28
5.14. Dispersed Phase and Dispersion Medium 5/28
5.15. Classification of Colloids 5/28
5.16. Preparation of Colloidal Sols 5/33
5.17. Purification ofColloidal Solutions 5/34
5.18. ProperticsofColloidal Solutions 5/36
5.18.1. Physical properties
5.18.2. Colligativc properties - osmotic pressure
5.18.3. Mechanical properties - brownian movement

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5.18.4. Optical properties - tyndall-elTect
5.18.5. Electrical properties : (a) Stability of colloidal .sols-electrical charge on colloidal particles

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(b) Electrophoresis or cataphoresis (c) Electro-osmosis (d) Coagulation or Flocculation or
Precipitation (hardy-schulze law)
5.19. Protective Action of Lyophilic Colloids and Gold Number 5/43

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5.20. Emulsions 5/45

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5.21. ApplicalionsofColloids 5/47
5.22. Nanomaterials 5/50
5.22.1. Definition
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5.22.2. Why are nanomaterials so important ?
5.22.3. Where are nanomaterials being used?
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5.22.4. Classification of nanomaterials

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5.22.5. Examples of substances used as nanomaterials

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5.22.6. Properties of nanomaterials

5.22.7. Selected applications of nanomaterials
5.22.8. Disadvantages of nanomaterials
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Revision At A Glance 5/51

Competition Focus [NEET/JEE Special] 5/55
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*6, General Principles and Processes of isolation of Elements 6/1—6/36

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6.1. General Introduction 6/1

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6.2. Modes of Occurrence of Elements 6/1

6.3. OccurreneeofMetals — MineralsandOrcs 6/1
6.4. Metallurgical Processes 6/2
6.5. Crushing and Grinding of the Ore 6/3
6.6. Concentration or Benefaction of the Ore (Ore-dressing) 6/3
6.7. Extraction ofCrude Metals from Concentrated Ores 6/6
6.7.1. Conversion of the ore into metal oxide or de-electronatio n of ores
6.7.2. Reduction or electronation of the metal oxide to the free metal
6.8. Thermodynamic Principles ofMeiallurgy 6/9
6.8.1. Ellinghamdiagrams
6.8.2. Limitations of ellingham diagrams
6.9. Theory of Pyromctallurgy 6/10
6.9.1. Effect of temperature on the free energy change (A,G“) of the overall reduction process
* Mot included in CBSE syllabus. This chapter has been given onlyfor the preparationofcompetitiveexamination.
 6.10. Applications ofPyromctallurgy 6/11

6.10.1. Extraction of iron from its oxide

6.10.2. Preparation of wrought iron
6.10.3. Production of steel from pig iron
6.10.4. Extraction ofcopper from cuprous oxide [copper(i)oxide]
6. i 0.5. Extraction of zinc from zinc oxide
6.11. Electrochemical Principles of Metallurgy 6/17

6.11.1. Applications of electrolysis to metallurgy

Extraction ofNon-metals by Oxidation 6/19

6.: -. Extraction of Metals both by Oxidation and Reduction 6/19

6.14. Refining 6/19

6.15. Uses of Aluminium. Copper, Zinc and Iron 6/23

RevLvio/tAtA Glance 6/24
6/27

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Competition Focus [NEET/JEE Special]

7/1—7/124

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*7. P-Block Elements
PART-^I: GROUP 15 ELEMENTS : THE NITROGEN FAMILY
7.1. General Introduction 7/1

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7.2. Occurrence 7/1

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7.3. Electronic Configuration 7/2

7.4. Atomic and Physical Properties 7/2

7.4.1, Atomic properties

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7.4.2. Physical Properties
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7.5. Chemical Properties 7/4
7.5.1. Oxidation states 7.5.2. Trends in chemicu. reactivity
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7.6. Anomalous Properties ofNitrogen 7/11
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7/12
7.7. Dinitrogen, N,
7.7,1. Preparationofdiniirogen 7.7.2. Properties of dinilrogen
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7.7.3. Active nitrogen 7.7.4. Uses of dinitrogen

7.8. Ammonia 7/14
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7.8.1. Preparation of ammonia 7.8.2. Manufacture of ammonia

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7.8.3. Physical properties of ammonia 7.8.4. Structure of ammonia

7.8.5. Chemical properties of ammonia 7.8.6. Usesofammonia
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7.8.7. Tests of ammonia

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7.9. Oxides ofNitrogen 7/17

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7.10. Nitric Acid, HNO] 7/20

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7.10.1, Preparation of nitric ucid 7.10.2. Structure of nitric acid

7.10.3. Physicalproperties of nitric acid 7.10.4. Chemical properties of nitric acid
7.10.5. Ring test for nitrate ion 7.10.6. Uses of nitric acid
7.11. .Allotropic forms of Phosphorus 7/27

7.12. Phosphine, PH, 7/31

7.12.1. Preparation of phosphine 7.12.2. Stmctureofphosphine

7.12.3. Physical properties of phosphine 7.12.4. Chemical properties of phosphine
7/33
7.13. PhosphorusHalides
7.13.1. Phosphorus trichloride,PCI, 7.13.2. Phosphorus pentachloride,PCl5
7.14. Oxides and Oxoacids of Phosphorus 7/35

7.14.1. Some important characteristics of oxoacids

* Not included in CBSE syllabus. This chapter has been given only for the preparation of competitive examination.
 PART-n : GROUP 16 ELEMENTS : THE OXYGEN FAMH.Y
7.15. General Introduction 7/38
7.16. Occurrence 7/38
7.17. Electronic Configuration 7/39
7.18. Atomic and Physical Properties 7/39
7.18.1. Atomic properties 7.18.2. Physical properties
7.19. Chemical Properties 7/42
7.19.1. Oxidation states 7.19.2. Trends in chemical reactivity
7.19.3. Uses of oxygen
7.20. Anomalous Behaviour of Oxygen 7/48
7.21. Dioxygen 7/49

ow
7.21.1. Atomic and physical properties 7.21.2. Chemical properties
7.21.3. Uses of dioxygen
7.22. Oxides 7/52
7.22.1. Simple oxides 7.22.2. Mixed oxides
7.22.3. Trends in acid-base behaviour of oxides in the periodic table

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7.23. Ozone 7/54
7.23.1. Preparation ofozone (trioxygen) 7.23.2. Structure of ozone

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7.23.3. Physical properties of ozone 7.23.4. Chemical properties of ozone
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7.23.5. Uses of ozone (trioxygen) 7.23.6. Tests of ozone

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7.24. Extraction of Sulphur-Frasch Process 7/59
7.24.1. Sulphur-allotropic forms
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7.25. Sulphur Dioxide, SO^ 7/61
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7.26. Oxoacids of Sulphur 7/64
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7.27. Sulphuric Acid, H^SO^ 7/64

7.27.1. Manufacture of sulphuric acid (H^SOJ by contact process

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7.27.2. Physical properties of sulphuric acid

7.27.3. Chemical properties ofsulphuric acid
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7.27.4. Structure of sulphuric acid and sulphates

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7.27.5. Uses of sulphuric acid

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PART-m : GROUP 17 ELEMENTS : THE HALOGEN FAMILY

7.28. General Introduction 7/69
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7.29. Occurrence 7/69

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7.30. Electronic Configuration 7/70

7.31. Atomic and Physical Properties 7/70
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7.31.1. Atomic properties 7.31.2. Physical properties

7.32. Chemical Properties 7/72
7.32.1. Oxidation states
7.32.2. Trends in chemical reactivity ofelements and their compounds
7.33. Anomalous Behaviour of Fluorine 7/80
7.34. Chlorine, CI2 7/81

7.34.1. Preparation of chlorine 7.34.2. Manufacture of chlorine

7.34.3. Physical properties of chlorine 7.34.4. Chemical properties of chlorine
7.34.5. Uses of chlorine
7.35. Hydrogen Chloride, HCl 7/84

7.35.1. Preparation of hydrogen chloride

7.35.2. Physical properties of hydrogen chloride
7.35.3. Chemical properties ofhydrogen chloride
7.35.4. Uses ofhydrochloric acid
 7.36. Oxoacids of Halogens 7/86

7.36.1. Acid strength and oxidising power ofoxoacids of halogens

7.37. Intcrhalogen Compounds 7/88

7.37.1. Preparation of interha logon compounds

7.37.2. Properties of interhaiogen compounds
7.37.3. Structure of intcrhalogen compounds
7.37.4. Uses of intcrhalogen compounds
7.38. Polyhalide Tons 7/91

7.38.1. Propertiesofpolyhalides 7.38.2. Structure of polyhalidcs

7.39. Pseudohalide Ions and Pseudohalogens 7/93
PART-IV : GROUP 18 ELEMENTS : THE NOBLE GASES
7.40. General Introduction 7/93
7.41. Occurrence of Noble Gases 7/93

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7.42. Isolation of Noble Gases 7/94
7.43. Electronic Configuration 7/94

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7.44. Atoniicand Physical Properties 7/94
7.45. Chemical Properties 7/96
7.45.1. Xenon fluorides 7.45.2. Xenon oxides

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7.45.3. Xenon oxyfluoridcs 7.45.4. Structure of xenon compounds

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7.46. Uses of Noble Gases 7/100
Revision At A Glance 7/102

CompeHtion Focus [NEFJ7JEESpecia[\

for 7/109
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8. d-And f-6lock Elements 8/1—8/120
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SECTION I. STUDY OF d-BLOCK ELEMENTS TRANSITION ELEMENTS
8.1. General Introduction — Position in the Periodic Table 8/1
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8.2. Electronic Configuration of d-block (Transition) Elements 8/2

8.3. General Characteristics ofTransilion Elements 8/4

8.4. General Trends in the Properties ofTransilion Elements 8/4

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8.4.1. Atomic and ionic radii 8.4.2. Metallic character

8.4.3. Meltingandboilingpoints 8.4.4. Density

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8.4.5. Ionisation energies or ionisation enthalpies

8.4.6. Standard electrode potentials (E") and chemical reactivity
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8.4.7. Oxidation states

8.4.8. Catalytic properties
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8.4.9. Coloured ions

8.4.10. Magnetic properties
8.4,11. Complex formation
8.4.12. Interstitial compounds
8.4.13. Alloy formation
8.5. Some General Group Trends in the Chemistry ofTransilion Metals (Group 3 to Group 12) and 8/19
TheirUses
SECTION II. SOME IMPOUI ANT COMPOUNDS OFTRANSITION ELEMENTS
8.6. Oxides 8/23

8.7. Salts Containing Oxoanions ofTransilion Metals 8/24

8.7.1. Potassium dichromate (KjCtjO,)

8.7.2. Potassium permanganate
 SECTION III. STUDY OF f-BLOCK ELEMENTS (LANTHANOIDS & ACTINOIDS)
8.8. General Introduction 8/34
8.9. TheLanthanoids 8/34
8.9.1. Electronic configuration of lanthanoids
8.9.2. Oxidation .states oflanthanoids
8.9.3. Atomic and ionic radii oflanthanoids
8.9.4. Some important characteristics oflanthanoids
8.9.5. Uses oflanthanoids
8.10. TheActinoids 8/39

8.10.1. Electronic configurations ofactinoids

8.10.2. Oxidation states of actinoids

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8.10.3. Ionic radii andactinoid contraction
8.10.4. General characteristics of actinoids
8.10.5. Uses ofactinoids
8.11. ComparisonofLanthanoidsandActinoids

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8/41
Revision At A Glance 8/47

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Multiple Choice Questions

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8/52
Assertion-Reason Type Questions

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Conceptual Questions 8/60
Very Short Answer Questions [7 mark] 8/66

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Short Answer Questions [2 or 3 marks] 8/70

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Long Answer Questions f 5 or more marks] 8/73
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Case-Based Very Short/Short Questions 8/77
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Case-Based MCQs and Assertion-Reason Questions 8/78
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NCERTQuestionsandExerciseswith Answers 8/81

NCERTExemplarProblemswithAnswers,HintsandSolutions 8/90
B

Competition Focus [NEET/JEE Special] 8/100

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9, Coordination Compounds 9/1—9/140

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9.1. General Introduction W\

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9.2. Coordination Compounds and Complex Ions 9/1

9.3. Some Important Temis used in Coordination Compounds 9/3

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9.4. Formula Writing and Nomenclature of Coordination Compounds 9/7

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9.4.1. Rules for formula writing of mononuclear coordination compounds

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9.4.2. Rules for naming of mononuclear coordination compounds

9.5. Writing Formuia of a Complex when Name is Given 9/17
9.6. Isomerisation in Coordination Compounds 9/19
9.6.1. Isomerism
9.6.2. Structural isomerism

9.6.3. Stereo isomerism or space isomerism

9.7. Bonding in Coordination Compounds 9/25
9.7.1. Werner’s theory of coordination compounds
9.7.2. Valence bond (VB) theory
9.7.3. Crystal field theory
9.8. Stability ofCoordination Compounds in Solution 9/43
9.9. Organometallics/Organometallic Compounds 9/45
9.9.1. Definitions

9.9.2. Classification of organometallic compounds

9.10. Metal Carbonyls 9/48

9.10.1. Structures of metal carbonyls

9.10.2. Properties of metal carbonyls
9.10.3. Uses of metal carbonyls
9.10.4. Bonding in metal carbonyls
9.11. Importance and Applications of Coordination Compounds/Comp lexes 9/50

9.12. Importance and Applications of Organometallics 9/52

9.13. Some Additional Useful Information About Coordination Compounds 9/53
Revision At A Glance 9/61

Multiple Choice Questions 9/67

Assertion-Reason Type Questions 9/71

Conceptual Questions 9/76

low
Very Short Answer Questions [/ mark] 9/82

ShortAnswerQuestions [2 or 3 marks] 9/85

Long Answer Questions [5 or more marks] 9/89

Case-Based Very Short/Short Questions 9/89

Case-Based MCQs andAssertion-Reason Questions 9/91

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F
NCERT Questions and Exercises with Answers 9/93

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NCERTExemplar Problems with Answers, Hints and Solutions 9/103

Competition Focus [NEET/JEE Special] 9/111

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Log Tables (i)-(iv)
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 *

SOLID STATE

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1.1. GENERAL INTRODUCTION

The different types of substances that are present around us are solids or liquids or gases. However,

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solids are more common than liquids and gases. The main point in which the solids differ from liquids and

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gases is the fact that gases and liquids possess fluidity, i.e., they can flow and hence they are called fluids
whereas solids do not possess fluidity ; instead, they possess rigidity. The reason for fluidity of gases and
liquids lies in the fact that their constituent particles are free to move about whereas rigidity of solids is due

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to the fact that their constituent particles have fixed positions and can only oscillate about their mean positions.
Further, because of rigidity possessed by the solids, they have definite shape and definite volume. Hence, a
solid may be defined as follows :
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A solid is defined as thatform of matter which possesses rigidity and hence possesses a definite
shape and a definite volume.
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Now, our aim will be to know the following about the solids which we shall discuss in this unit:
(/) Why do some substances exist as solids ?
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{ii) What are the general characteristics of solids ?

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(Hi) How can solids be classified into amorphous and crystalline solids ?
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(iv) How can crystalline solids be further classified into different types on the basis of the nature of the
constituent particles and the binding forces operating between them ?
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(v) How different solids are put to different uses on the basis of their properties ?
ivi) How are the properties of the solids related to the structure of the solids ?
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(v/7) How does the correlation between structure and properties of solids help in the discovery of new
solid materials with desired electrical or magnetic properties or biodegradable polymers for packaging,
biocompliant solids for surgical implants etc. ?
(vUi) How are the properties of solids modified due to structural imperfections or by the presence of impurities
in very small amounts ?
1.2. WHY DO SOME SUBSTANCES EXIST AS SOLIDS ?
As already discussed in class XI, whether a substance will exist as a solid or a liquid or a gas depends
upon the net effect of the following two opposing forces :
(0 Intermolecular forces. These are the forces existing among the constituent particles which try to
keep the constituent particles close together.
* Not included in CBSE syllabus. This chapter has been given only for the preparation of competitive examinations.
1/1
1/2 T^nAdee^'a, New Course Chemistry (XII)EEIHI

(h) Thermal energy. This is the energy possessed by the constituent particles due to temperature. This
energy tries to keep the constituent particles apart as it tends to m^e them move faster.
At low temperature, the thermal energy is quite low, but intcrmolccular forces are so strong that the particles
are brought very close together. As a result, the constituent particles occupy fixed positions and are unable to move

except that they can oscillate about their mean positions. Hence, the substance exists in Uie solid state.
1.3. GENERAL CHARACTERISTICS OF SOLID STATE
As explained above, a solid has the following properties :
(/) The constituent particles are very’ closely packed, i.e., intermolecular distances are very small
(if) The constituent particles occupy ifxed positions and can only oscillate about their mean positions.
Because of these properties, the solids possess the following characteristic properties :
(0 They possess rigidity. (//) They have high density.
(///) They possess definite shape and definite volume. (/v) They are incompressible.

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1.4. AMORPHOUS AND CRYSTALLINE SOLIDS

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The various solids are classified into the following two types :
(1) Crystalline solids. All solid elements (metals and non- metals) and compounds exist in this form.

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(2) Amorphous solids, e.g., rubber, glass, pitch, tar, fused silica, plastics, polymers of high molecular

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mass etc.

1.4.1. Points of Difference between a Crystalline solid and an Amorphous solid

for
(i) Arrangement of their constituent particles. In a crystalline solid, the constituent particles (i.e.,
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ions, atoms or molecules) are arranged in a definite geometric pattern in all the three dimensions. The order
is so regular that knowing the arrangement at any one site, that at any other site can be predicted. This is
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called long range order.

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In fact, a crystalline solid consists of a large number of small crystals

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each of which has the same regular pattern or arrangement of particles.

This arrangement repeats itself periodically over the entire crystal.
On the other hand, in an amorphous solid, there is a regular
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arrangement of particles in a small region only. This is called short range

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order. Thus, whereas crystalline solids possess both short range and long
range order, amorphous solids possess only short range order.
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Two dimensional structure of

For example, the structures of crystalline quartz and that of quartz glass (a) quartz and (b) quartz glass
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are shown in Fig. 1.1(a) and (b) respectively.

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Although both the structures look to be identical but there is no long range order in structure (b).
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For this reason, whereas crystalline solids have definite geometrical shapes, amorphous solids have
irregular shape (i.e., no definite geometric shape).
Thus, crystalline and amorphous solids may be defined as follows :
A solid is said to be crystalline if its various constituent particles (i.e. ions, atoms or molecules)
are arranged in a definite geometric pattern in three dimensional space so that there is short
range as well as long range order of the constituent particles. On the other hand, a solid is
said to be amorphous if there is no regular arrangement of its constituent particles or at the
most, there is only a short range order of its constituent particles.
(ii) Melting points. The crystalline substances possess sharp melting points whereas the amorphous
substancesmelt gradually over a temperaturerange.
For this reason, crystalline solids have definite heats of fusion whereas amorphous solids do not have
definite heats of fusion.

(Hi) Isotropy and Anisotropy. In case of amorphous substances, properties like electrical conductivity,
refractive index, thermal expansion etc. are identical in all directions just as in case of gases or liquids. This
 SOLID STATE 1/3

property is called isotropy and the substances showing this property are called isotropic. On the other hand,
in case of crystalline substances, the properties mentioned above have different values in different
directions. This type of behaviour is called anisotropy and the substances FIGURE 1.2
exhibiting this type of behaviour are called anisotropic. Thus, whereas amorphous
substances are isotropic in nature, the crysuilline substances are anisotropic. The
anisotropy exhibited by crystalline substances is obviously due to the fact that
in making measurements in different directions, different types of particles fall
on the way, as shown in Fig. 1.2. >

O'v) Cleavage with a knife. A crystalline solid on being cut with a sharp
Anisortopy exhibited by
knife gives a clean cleavage whereas an amorphous solid undergoes an irregular crystalline substances
breakage as shown in Fig. 1.3.
FIGURE 1.3
(v) Heats of fusion. Crystalline solids have definite
heats of fusion whereas amorphous solids do not have definite
heats of fusion.

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(vi) Cooling curves. Cooling curve of
amorphous solid is smooth without any break, /.e.,

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when a molten amorphous solid is cooled, the
temperature decreases smoothly with time (Fig. (a) A crystalline solid undergoes a clean cleavage
(b) An amorphous solid unde^oes an irregular cut
1.4 a). On the other hand, cooling curve of crystalline

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solid is not smooth, i.e., there is break at the point FIGURE 1.4

(B) when solidification starts and a break at the point

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A

(C) when solidification is complete (Fig. 1.4 b).

In fact, only the crystalline solids are the true
solids whereas amorphous solids are considered to
for Liquid
Solidification
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Q. B iC
be highly supercooled liquids of very high viscosity E
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Solid
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and are called pseudo solids. This is supported by D

the fact that the glass of the window panes of some
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very old houses are found to be thicker at the bottom o Time © Time

than at the top, showing some fluidity possessed by (a) Cooling curve of amorphous solid
glass. (b) Cooling curve of crystalline solid
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Further, it may be pointed out that due to short range order possessed by amorphous solids, small parts
of the amorphous solid may be crystalline while the rest may be non-crystalline. These crystalline parts of the
amorphous solids are called crystallites.
It may also be noted that the window glass of the old buildings looks milky. This is because due to
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heating during the day and cooling at nights, i.e., annealing over a number of years, glass acquires some
crystalline character.
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Thus, the main points of difference between a crystalline solid and an amorphous solid may be summed
up as on next page.

Retain In Memory
Any material can be made amorphous or glassy either by rapidly cooling its melt or fizzing its vapours.
For example, silica (5/O2) crystallizes as quartz in which SiO^ tetrahedra are linked in a regular manner
but on melting and then rapid cooling, it gives glass in which SiO^ tetrahedra are randomly joined to
each othe.r Thus, quartz is crystalline Si02 whereas silica glass is amorphous Si02. Similarly, many
complex materials including metallic alloys have been converted into the glassy form.
1.4.2. Uses of Amorphous Solids
Amorphous solids are very useful materials in our everyday life. A few applications are given below :
(/) The most widely used amorphous solids are the inorganic glasses, which find application in
construction, house-ware, laboratory ware, etc.
(ii) Another well known amorphous solid is rubber which is used in making tyres, shoe soles, etc.
1/4 New Course Chemistry (XII)KSsI9]

TABLE 1.1. Comparison of main characteristics of crystalline and amorphous solids

Crystalline Solid Amorphous Solid

(0 The constituent particles are arranged in a (0 The constituent particles are not arranged in
regular fashion containing short range as any regular fashion ; there may be at the most
well as long range order. some short range order only.
(») They have definite geometric shapes*. («) They have irregular shapes*.
m They have sharp melting points. (m) They melt over a range of temperature.
(iv) They are anisotropic. (iv) They are isotropic.
(V) They undergo a clean cleavage. (V) They undergo an irregular cut.
(vO They have definite heats of fusion. (v/) They do not have definite heats of fusion.

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{v/0 They do not have smooth cooling curves. (v/0 They have smooth cooling curves.
(vm) They are true solids. (v»0 They are pseudo solids or supercooled liquids.

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(in) A large number of plastics which are amorphous solids are being used in articles of everyday life.
(iv) Amorphous silica has been found to be the best material for converting sunlight into electricity (in
photovoltaic cells).

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1.5. CLASSIFIC/.I ION OF CRYSTALLINE SOLIDS

Based upon the nature of the constituent particles and the binding forces present between them, the
crystalline solids are further classified into the following four categories :
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(1) Ionic solids (2) Molecular solids (3) Covalent or Network solids (4) Metallic solids.
Now, let us discuss each of these categories one by one.
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1. Ionic solids. In these crystalline solids, the constituent particles are positive and negative ions (i.e.,
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cations and anions), e.g., Na* and Cl“ ions in case of NaCl, arranged in the three dimensional space. These
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ions are held together by strong coulombic, i.e., electrostatic forces of attraction. Some of their important
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characteristics are below :

(0 Because of strong electrostatic forces of attraction, they have high melting and boiling points.
(ii) They are electrical insulators in the solid state because their ions are not free to move about. However,
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in the aqueous solution or in the molten state, they are good conductors of electricity because ions
become free.
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(Hi) They are soluble in polar solvents but insoluble in non-polar solvents.
(iv) Because of strong electrostatic forces of attraction, the ions are closely packed and hence ionic
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solids are hard. However, they are brittle because their stability depends upon preservation of their
geometric pattern.
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2. Molecular solids. In these solids, the constituent particles are molecules. Depending upon the nature
of molecules, these are further sub-divided into the following three types :
(fl) Non-polar molecular solids. These are those FIGURE 1.5

crystalline solids in which the constituent particles are either Helium Atom-1 Helium Atom-2

atoms like those of noble gases (helium, neon, argon etc.) or Force of ■■’■Ml
attraction
non-polar molecules like H2, CI2, I2. CH4 etc. The forces
operating between them are weak dispersion forces or ■.0
'■ (London force)
London forces (a type of van der Waals forces), as already
Momentary Dipole Induced Dipole
studied in class XI, i.e., momentary dipole-induced dipole
forces as shown in Fig. 1.5 for the case of helium. Weak London dispersion forces.

♦This is because crystalline solids are bound by plane surfaces, called the faces of the crystal. The angles between
the adjacent faces, called interfacial angles are constant for crystals of a particular substance even if we break the
crystal into smaller crystals. On the other hand, amorphous solids do not have well-defined planes. When hammered,
they break in an irregular manner.
SOLID STATE 1/5

Their main characteristics are as follows :

(z) These solids are generally soft because of weak intermolecular forces present in them.
(ii) They are non-conductors of electricity as there are no ions present.
(Hi) As they are soft, they have low melting and boiling points,
(zv) Due to weak intermolecular forces present in them, they are usually gaseous or liquids at room
temperature and pressure.
(b) Polar molecular solids. These are those crystalline solids in which the constituent particles are
polar molecules like HCl, SO2 etc. The forces holding these molecules together are dipole-dipole forces of
attraction (which are again a type of van der Waals forces) as shown in Fig. 1.6.
These intermolecular forces of attraction are comparatively stronger than London dispersion forces.
These solids show the following characteristics :
FIGURE 1.6
(0 They are soft.

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(ii) They are non-conductors of electricity.
(Hi) Their melting and boiling points are comparatively higher
than non-polar molecular solids, though not so high,

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P * P
(zv) As their melting and boiling points are not so high, they
also exist as gases or liquids at room temperature and Dipole-dipole forces of attraction.

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ordinary pressure,

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(c) Hydrogen-bonded molecular solids. In these solids, the constituent particles are such molecules
which contain hydrogen atom linked to a highly electronegative atom small in size such as F, O or N, e.g., in
H2O, NH3, etc. Hence, the intermolecular forces of attraction existing among these molecules are the strong

for
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hydrogen bonds. These solids show the following characteristics :
(z) They exist as volatile liquids or soft solids at room temperatureand ordinary pressure.
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(ii) They are non-conductors of electricity.
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(Hi) Their melting and boiling points are generally higher than the molecular solids of the first two types.
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3. Covalent or Network solids. These are those crystallinesolids FIGURE 1.7

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in which the constituent particles are non-metal atoms linked to the t

154 pm
adjacent atoms by covalent bonds throughout the crystal. As a result, a
network of covalent bonds is formed. Hence, they form giant molecules.
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One of the most common examples of the crystals of this type is that of
A
diamond in which the carbon atoms are linked together by covalent
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bonds to give a three dimensional structure as shown in Fig. 1.7. Another T

very common example is that of silicon carbide (SiC). ● ●
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,● ● ●● .■< ●
nd

The main characteristics of these solids are as under: '

(z) As covalent bonds are strong and directional in nature, these

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Network structure of diamond

solids are very hard and brittle.
(z7) They have extremely high melting points and may even FIGURE 1.8

decompose before melting.

(Hi) They are insulators and do not conduct electricity.
Exception. Graphite is also a covalent solid but it is soft and a good
conductor of electricity. Its exceptional behaviour is due to its typical
structure as shown in Fig. 1.8. In this case, carbon atoms are arranged in
different layers. In each layer, every carbon atom is linked to three
neighbouring carbon atoms. Thus, the fourth electron of each carbon atom
is free to move about. Because of the presence of these free electrons in
different layers, graphite becomes a good conductor of electricity. Further,
the distance between the adjacent layers is greater than carbon-carbon bond z
z z
length. Hence, these layers are not bonded to each other and can easily slip
over each other. For this reason, graphite is soft and acts as a good solid
lubricant. Structure of graphite
1/6 7^Mdee^'4t. New Course Chemistry (XII)BZsmi

4. MetalHc solids. In case of metals, the constiaient ptirtidcs FIGURE 1.9

are positively charged metal ions and free electrons. These are
produced from metal atoms because metal atoms have low ©^ ©^ ©^ ©^ ©^
ionization energy and can easily lose their valence electrons to ©"^ ©^ ©^ ©^ ©^ ©^ ©^ ©^
leave behind positively charged ions (called kernels). These
electrons can easily flow throughout tlie metal crystal like water
©""^ ©"' ©^ ©^ ©^ ©^
in the sea. Hence, we call it a sea of free electroas (Fig. 1.9). ©'' ©"■ "©^ ©"^ ©^ ©^ ©^ ©^
Each metal atom contributes one or more electrons towards this
©^" "©" ©^ ©^ ©^ ©^ ©^
simultaneously
sea of mobile electrons. These mobile electrons are
Metallic solid-Posdve ions in a sea of
attracted by the positive ions and hence hold these positive ions
mobile electrons (Electron sea model)
together.
The force that holds the metal ions together in the crystal is called metallic bond.

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These solids show the following characteristic properties :
(0 They possess high electrical and thermal conductivity. This is because on applying electric current,
electrons start flowing towards positively charged electrode.

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Similarly, heat given to one part is carried by the electrons to all parts of the metal.
{ii) They possess lustre and colour in some cases. This is also explained on the basis of free electrons.

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{Hi) They are highly malleable and ductile.

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This is because unlike ionic crystals, the po.sitions of the positive ions can be altered
without destroying the crystal because of uniform charge distribution provided by freely moving

for
electrons. For this reason, metals can be easily deformed,
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(/v) As the positive ions are closely packed in the crystal, most of the metals possess high melting points
and high densities.
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To sum up, the various types of crystalline solids, their constituent particles occupying the points, the
k
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type of attractive forces, their properties and some examples are given in Table 1.2.
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TABLE 1.2. Different types of crystalline solids

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Type of Constituent Nature of Examples Physical Electrical Melting Other

solid particles binding nature conductivity point charac
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forces teristics
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1. Ionic solids Positive and Couiombic NaCI, LiF, Hard but Insulators in solid High Soluble in
negative ions or Electro MgO, ZnS, brittle state but conduc polar solvents,
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static forces Cap2 tors in molten insoluble in

state or aqueous non-polar
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nd

solution solvents
2. Molecular
solids
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(0 Non-polar Atoms of noble London Ar, H„ Ij, Soft Insulator Very low Low density
gas or non-polar dispersion solid CO, (dry and volatile
molecules forces ice) CHj:
(h) Polar Polar molecules Dipole- HCl. SO, Soft Insulator Low
dipole
attractions

{Hi) Hydrogen Molecules con Hydrogen H,0 (ice), Soft Insulator Low
bonded taining H bonding NH3
linked to F,
OorN
3. Covalent Atoms Covalent C (diamond) Very hard Insulators Very high
or Network bonds SiO, (quartz)
solids SiC,"AlN
C (graphite) Soft Conductor Very
(Exception) high
4. Metallic Positive ions Metallic All metals Hard Conductor in Fairly Malleable
solids in a sea of bonds and alloys solid state as well high and ductile
mobile as molten state and possess
electrons lustre
SOLID STATE 1/7

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
1. Polycrystalline solids. Some solids have a large number
of small crystals each of which has a regular turangcment of its constituentparticlesbut these crystalsthemselves
are arranged in a random manner. Such solids are called polycrystalline solids. Thus, they have structure
between crystalline solids and amorphous solids, e.g., iiiuminium tind steel.
2. Isomorphism and Polymorphism. When two or more crystalline ,solid.s having similar chemical
composition exist in the same crystalline form or structure, the property is called isomorphism, e.g., Na3P04
and Na3As04. On the other hand, when a particular substance exists in more than one crystalline form, the
property is called polymorphism, e.g., calcium carbonate exists in two ciystalline forms called calcite and
aragonite. In case of elements, this property is called 'allotropy'.

Curiosity Questions
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Q. 1. Why is glass of window panes of very old buildings found to be thicker at the bottom
than at the top and why Is It milky ?

F lo
Ans. Glass is an amorphous solid which is a supercooled liquid of high viscosity and hence possesses
fluidity. Due to this property it is thicker at the bottom than at the top. Milkiness of glass is due
to the fact that it undergoes heating during the day and cooling at night, i.e., annealing over a

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number of years. As a result, it acquires some crystalline character.

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Q. 2. What are optical fibers ? What are their advantages over ordinaryglass like that of window
panes ?

for
Ans. Optical fibers are obtained by drawing glass into fibers thinner than even human hair but still
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possess very high mechanical strength. The advantage of optical fibers is that unlike ordinary
glass like that of window panes which stops light in less than one meter, optical fibers transmit
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light over kilometers without any noticeable decrease in intensity. That is why they are being
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used in radio broadcast, television programmes and telecommunication including mobile and
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internet services. They are cheaper than copper coils and take much less space.
J
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1.6. CRYSTAL LATTICES (OR SPACE LATTICES) AND UNIT CELLS

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Space lattice is a regular repealing arrangement of points in space. As it is a regular repeating arrangement, to
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describe the space lattice completely, we can choose a small piul of the lattice, which when repeated in different
directions produces the complete space lattice. This small portion of the lattice is called ‘unit cell’. To understand
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space lattice and unit cell, we shall discuss first two dimensional and then three dimensional lattices.
A. Two Dimensional Lattices. A
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nd

Lattice Unit cell

two dimensional lattice is a regular
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arrangement of points in the plane of

(/) Squiu’e lattice Square
the paper as shown in Fig. 1.10 (a).
We suitably choose four points and (ii) Rectangular lattice Rectangle
connect them to obtain the unit cell. In {Hi) Parallelogram lattice Parallelogram
some cases, the unit cell may have an (/v) Rhombic lattice Rectangular with interior point
interior point also. Such unit cells are (v) Hexagonal lattice Rhombus with an angle of 60“
called centred unit ceils. The unit cell
that does not contain any interior point is called primitive unit cell. There are five types of two dimensional
lattices as shown in Fig. l.IO (a) with their unit cells as shown on the next page.
The complete lattice can be generated by repeatedly moving the unit cell in the direction of its edges by
a distance equal to the cell edge, as shown in Fig. 1.10 (h).
The different types of two-dimensional lattices along with their unit cells are summarised in the table above.
Such two dimensional lattices are often seen in the wall papers and tiled floors.
B. Three Dimensional Crystal Lattices. As mentioned earlier, the constituent particles of a crystalline
solid are arranged in a definite fashion in the three dimensional space. Representing the constituent particles
by points, one such arrangement is shown in Fig. 1.10 (c).
1/8 New Course Chemistry (XII) P2STM1

FIGURE 1.10(a)
O
0 0 0 0 0 0 0 0 0 0 o o 0
o o o e o o o e e
0 0 0 0 0 0 0 0 0 0
O 0 0 0 0 o o
o o o

»0^
0 O 0
607
" “ n °
o o
l90°J
o 0 o
0 o
O 0 o o o 0 0 0 0 0 0 0

O 0 o 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

o o o
Five two dimensional lattices with their unit cells as (i) Square, (ii) Rectangular,
(iii) Parallelogram, (iv) Rectangular with interior point and (v) Rhombus

FIGURE 1.10(b) FIGURE 1.10(c)

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U o o o o I
'C
"I

F lo
I

1
o o o o o o
p,
I

; o..

Generating complete two dimensional lattice by 11

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repeatedly moving the unit cell in the direction of the cell

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edges by distance equal to cell edge Space lattice and unit cell

Such a regular arrangement of the constituent particles (i.e., atoms, ions or molecules) of a
crystal in a three dimensional space is called crystal lattice or space lattice.
for
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Some Characteristics of a Crystal Lattice. These are given below :
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(0 Each point in the crystal lattice erpresents a constituent particle which may be an atom, a molecule or an ion.
{ii) Each point in the lattice is called lattice point or lattice site.
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(///) The points are joined by lines just to represent the geometry of the lattice.
Unit Cell. From the complete space lattice, it is possible to select a smallest three dimensional portion
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(as shown by coloured lines in Fig. 1.10c) which repeats itself in different directions to generate the complete
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space lattice. This is called a ‘unit cell’. Thus, a unit cell may be defined as follows :
The smallest three dimensional portion of a complete space lattice which when repeated over
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and again in different directions produces the complete space lattice is called the unit cell.
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The situation may be compared with that of a thick wall made of bricks where the brick is similar to the
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unit cell and the wall to the complete space lattice.

The unit cell may be considered as the ‘fundamentalbuilding block’ of the crystal lattice because the
lattice can be constructed by stacking the unit cells.
Parameters of a unit cell. A unit cell is characterised by the following parameters which decide the size
and shape of the unit cell:
(z) Its dimensions (lengths) along the three edges a, b and c (as shown in Fig. 1.10c). These edges may
or may not be mutually perpendicular.
(ii) Angles between the edges, i.e., angle a between the edges b and c, angle p between the edges a
and c and angle y between the edges a and b.
Thus, a unit cell is characterised by six parameters, i.e., axial lengths a, b and c and axial angles a,
P and y.
Types of Unit Cells. Based upon the parameters of the unit cells {i.e., lengths a, b and c and the
angles a, P and y), there are seven types of unit cells. These are also called crystal systems or crystal
habits because any crystalline solid must belong to any one of these unit cells. These different types of
crystal systems along with their characteristics and examples are listed in Table 1.3 and are shown in
Fig. 1.11 on page 1/9.
SOLID STATE 1/9

TABLE 1.3. Seven types of unit cells or crystal systems

System Axial Lengths Axial Angles Examples

1. Cubic a = h = c a=P = Y=90® Cu, NaCl, KCl, Alums, Diamond, Zinc blende
2. Tetragonal a = b c a=p = Y = 90° White tin (Sn). Sn02, TiO,, CaS04
3. Orthorhombic a*b*c a=(3 = Y=90‘* KNO3, K2SO4, BaS04, PbC03, CaC03
or Rhombic Rhombic sulphur
4. Monoclinic a^b^ c a = Y = 90® 54 p Na2SO4.10H2O, PbCr04, Monoclinic sulphur
5. Triclinic a^b
CUSO4.5H2O, K2Cr202, H3BO3
6. Rhombohedral a = b = c a = P =s Y ^ 90° NaNOj, ICl, As, Sb, Bi, Calcite (CaC03)
or Trigonal

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7. Hexagonal a = b^c a = P = 90°,Y= 120° Cinnabar (HgS), Ice, Graphite, Mg, Zn, Cd,
ZnO, CdS, Agl, PbU

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FIGURE 1.11

1. CUBIC (3) 2. TETRAGONAL (2)

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4 for
c
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* ^4-
l«
a a

k“
y' ● y
Li
s
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SIMPLE FACE-CENTRED BODY-CENTRED SIMPLE BODY-CENTRED
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3. ORTHORHOMBIC (4)

44171 7471 777

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c
Y

^7
I

1747
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nd

y y
y y
t e. ● C.
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SIMPLE FACE-CENTRED END-CENTRED BODY-CENTRED

4. MONOCLINIC (2) 5. TRICLINIC (1) 6. RHOMBOHEDRAL (1) 7. HEXAGONAL (1)

S ^«
rn m o
c c V □
-i
I I 1 O) c
h*o * O m z

“ -U
□ 2m
p H CO
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SIMPLE END-CENTRED SIMPLE SIMPLE

SIMPLE

Seven types of Unit cells or Crystal systems. Total 14 lattices called Bravais lattices (given in brackets)
Note that cubic is the most symmetrical while triclinic is the most unsymmetrical system.
1/10 ^>uxdecfa.'<^ New Course Chemistry (Xll)SSSm

lypes of Lattices. In the various types of unit cells described above, it was assumed that the particles
are present only at the comers of the unit cells. However, it has been observed that the particles may be
present not only at the comers but may also be present at some other special positions within the unit cell.
Hence, unit cells may broadly divided into the following two categories :
(a) Primitive unit cells.

The unit cells in which the constituent particles are present only at the corners are called simple
lit cells ' or 'primitive unit cells’.
Thus, there are seven types of primitive unit cells as given in Table 1.3.
(b) Non-primitive or centred unit cells.
Those unit cells in which the constituent particles are present not only at the comers of the unit cells
but also at some other positions are called non-primitive unit cells or centt ed unit cells.

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There are three types of non-primitive unit cells* as follows :
. iu ed. When the particles are present not only at the corners but also at the centre of

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each face of the unit cell, it is called face-centred unit cell.
When in addition to the particles at the corners, there are particles at the
centres of any two opposite faces, it is called end-face centred.

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‘ ' When in addition to the particles at the comers, there is one particle present

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at the centre within the body of the unit cell, it is called body-centred unit cell.

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Every crystal system does not have all the four types of unit cells, i.e., simple, face-centred, end-centred
and body-centred. Hence, there are only 14 types of space lattices corresponding to seven crystal system as
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shown in Fig. 1.11.
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The fourteen lattices (shown in Fig. 1.11) corresponding to seven crystal systems are known as
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-.iiices.
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1 / MORE REALISTIC PICTURE OF A UNIT CELL

^EN STRUCTURES AND SPACE-FILLING STRUCTURES)
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It has been explained above that the unit cells are represented by points connected by lines. The points
represent the constituent particles, i.e., atoms, ions or molecules whereas the lines help to visualize the symmetry
of the crystal. These are called open structures {as shown in Fig. 1.11). In actual practice, the particles
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occupy much more space of the crystal lattice and are held together by one or the other type of force. For
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nd

example, a more realistic picture of simple, face-centred and body-centred cubic showing how the panicles
actually pack within the solid is given in Fig. 1.12 (a) with actual particle sizes. These are called space-Hlling
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structures. As mentionedabove, the stmetures as given in Fig. 1.11 are called open structures.In the open
structures, quite often, instead of points, we draw small spheres, the centre of the sphere representing the
centre of the particle as shown in Fig. 1.12 (b).
FIGURE l.lg

SIMPLE FACE-CENTRED BODY-CENTRED SIMPLE FACE-CENTRED BODY-CENTRED

(a) Space-filling structures (b) Open structures of simple, face-centred and body-centred cubic

●Besides these, we also have edge-centred unit cell in which the particles are present not only at the comers but
one particle is present on each edge-centre. Also remember that a cube has 12 edge centres.
SOLID STATE 1/11

It is important to note that the spheres (atoms) at the comers of the face-centred and body-centred cubes
do not touch each other. However, in the body-centred cube, all the spheres at the comers touch the central
sphere. Similarly, in the face-centred cube, each sphere at the comer touches three spheres present at the face-
centres of three adjoining faces.
This type of arrangement extends in the three dimensional space. In general, for any crystal system, the
number of spheres which are touching a particular sphere is called its coordinationnumber.
In ionic crystals, the coordination number may be defined as the number of oppositely charged ions
surrounding a particular ion.

In general, the coordination number of any constituent particle in a crystal is the number of
its nearest neighbours.

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1.8. CALCULATION OF NUMBER OF ATOMS IN A UNIT CELL OF A CUBIC CRYSTAL SYSTEM
As explained earlier, in a crystal lattice, each unit cell is touching a number of other unit cells. Thus, a
particle present at the lattice point may be shared by a number of unit cells. Let us, therefore, first find out the

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poition or the fraction of the particle that belongs to a particular unit cell and then we shall calculate the

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number of particles in the unit cell of different lattices of a cubic crystal system. For simplicity, we shall

F
assume that the constituent particles are atoms.
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1.8.1. Calclation of the contribution of atoms present at different lattice sites

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(/) An atom at the corner is shared by eight unit cells
fo FIGURE 1.13
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[see Fig. 1.13 (a)]. Hence,
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Contribution of each atom present at the comer = —
oo

8 li:
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[as shown in Fig. 1.13 (/?)].

(/i) An atom on the face is shared between two unit cells
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[see Fig. 1.14 (n)]. Hence,

An atom at the lattice
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1 point shared by 8 unit cells

Contribution of each atom on the face = —
2
d

[as shown in Fig. 1.14(/?)j.

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in

FIGURE 1.1^
F

-y.
O ill

07
o o
An atom at the face An atom at the body centre not An atom at the edge
shared by any other unit cell centre shared by 4 unit cells
centre shared by 2 unit cells

{Hi) An atom present within the body of the unit cell is shared by no other unit cell (see Fig. 1.15 (a)]. Hence,
Contribution of each atom within the body = 1 [as shown in Fig. 1.15 (6)].
(/v) An atom present on the edge is shared by four unit cells (see Fig. 1.16). Hence,
1
Contribution of each atom on the edge = -
4
1/12
‘P>teuUe^ ’4. New Course Chemistry (XII) orsTun

1.8.2. Calculation of Number of Atoms per Unit Cell

(a) Number of atoms in a unit cell of Simple (Primitive) Cubic Lattice. A simple cubic lattice has
only eight atoms on the comers. As contribution by each =1/8, therefore, number of atoms present in the unit
1
cell =-x8 = l.
8

(b) Number of atoms in a unit cell of Body Centred Cubic (BCC). This lattice has 8 atoms on the
corners and one atom within the body.
1
Contribution by 8 atoms present on the comers = - x8 = 1
8

Contribution by the atom present within the body = I

Number of atoms present in the unit cell =1 + 1=2.

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(c) Number of atoms in a unit cell of the Face Centred Cubic (FCC). This lattice has 8 atoms on the
comers and 6 atoms on the faces (one on each face).

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Contribution by atoms on the comers =-x8 = 1
8

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1
Contribution by atoms on the faces = —x6 = 3

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2
Number of atoms present in the unit cell = 1 + 3=4.

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Note. The number of particles present per unit cell is called unit cell constant (z). Thus, the general
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expression for calculation of unit cell constant is
N N..I
Nc
s
/
k
z =
Yo
8 2 4 1
oo

Subscripts c,/, e and i represent comer, face centre, edge centre and inside.
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In hexagonal unit cell, comer is shared by 6 unit cells, face is shared by 2 unit cells, edge is shared by 3
unit cells. Hence, for hexagonal unit cell,
r

N N,.I
Nc N.
ou
ad

z = (Refer to Hint 55, page 1/108)

6 2 3 1
Y

Remember that in case of compounds, the number of atoms per unit cell are in the same ratio as the
stoichiometry of the compound. Hence, it helps to predict the formula of the compound.
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nd

FORMULAS USED
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PROBLEMS 1
BASED (0 Contribution by a particle on the comer of a unit cell = —
8
ON
1
Calculation of Contribution by a particle on the face of a unit cell = —
z*

particles per unit Contribution by a particle within the body of a unit cell = 1 ●St
ceil and formula <
f
1
of compound Contribution by a particle on the edge of a unit cell = -

(ii) The ratio of the number of particles A and B present per unit cell
'■V . gives the formula of the compound formed between A and B.
ifc'i irii ●1* ●n

Problem
D Calculate the number of atoms per unit cell present in simple, fee and bcc unit cells.
(Assam Board 2012)
Solution. See calculations given above.
SOLID STATE 1/13

Problem
A compound formed by elements A and B has a cubic structure in which A atoms are
at the comers of the cube and B atoms are at the face centres. Derive the formula of the compound.
Solution. As A atoms are present at the 8 comers of the cube, therefore, number of atoms of A in the unit
cell = -x8 = 1.
8
As B atoms are present at the face centres of the 6 faces of the cube, therefore, number of atoms of B in the

unit cell - —x6 = 3.

2
/. Ratio of atoms A : B = 1 : 3.

Hence, the formula of the compound is AB3.

Problem
A cubic solid is made up of two elements X and Y. Atoms Y are present at the corners
of the cube and atoms X at the body centre. What Is the formula of the compound ? What are the coordination
numbers of X and Y ?

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Solution. As atoms Y are present at the 8 comers of the cube, therefore, number of atoms of Y in the unit cell
I

F lo
=-x8=l.
8
As atoms X are present at the body centre, therefore, number of atoms of X in the unit cell = I
Ratio of atoms X : Y = 1 : 1.

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Fre
Hence, the formula of the compound is XY
Coordination number of each of X and Y=8.
Problem

in which atoms A are at the corners and atoms B are

for
El An ionic compound made up of atoms A and B has a face-centred cubic arrangement
at the face-centres. If one of the atoms is missing from
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the comer, what is the simplest formula of the compound ?
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s
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Solution. No. of atoms of A at the comers = 7 (because one A is missing)

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Contribution atoms of A towards unit cell = 7x - = —

8 8
No. of atoms B at face-centres = 6
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]
.●. Contribution of atom B towards unit cell = 6x — = 3
2

7
dY

Ratio of A:B= — :3 = 7:24 .*. Formula is A7B24

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B Calculate the number of unit cells in 8*1 g of aluminium

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Problem
if it crystallizes in a face-
centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol”^) (CBSE 2017)
Solution. 1 mole of A1 = 27 g = 6 02 x 10^^ atoms
6-02x10^3
No. of A1 atoms present in 8-1 g of A1 = x8-l = 1-806 X 1Q23
27
As face-centred cubic unit cell contains 4 atoms, therefore,
1-806x1023
number of unit cells present = = 4-515 X 1022
4

wmm PPA^iTil^Ei
1. A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms are at the comers
of the cube and X atoms are at the alternate faces. What is the formula of the compound ?
1/14 New Course Chemistry (XIl)EZsIHI

2. Calculate the number of atoms in a cubic based unit cell having one atom on each comer and two atoms on
each body diagonal.
3. A compound made up of elements A and B crystallises in the cubic structure. Atoms A are present on the
comers as well as face centres whereas atoms B are present on the edge centres as well as body centre. What
is the formula of the compound ? Draw the stmeture of its unit cell.
4. If three elements P. Q and R crystallize in a cubic solid lattice with P atoms at the comers, Q atoms at the cube
centre and R atoms at the centre of the faces of the cube, then write the formula of the compound.
5. Sodium crystallizes in a hcc unit cell. Calculate the approximate number of unit cells in 9-2 g of sodium
(Atomic mass of Na = 23) (Assam Board 2013)
ANSWERS
1. XY 2.9 3. AB, structure is same as that of NaCI (See Fig. 1.33 (a), page 1/25)
4. PQR3 5. 1-204 X 10^3

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HINTS FOR DIFFICULT PROBLEMS

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1
1. As there are 8 Y atoms at the comers and contribution of each = -, therefore, no. of Y atoms/unit cell

e
1
= 1. There can be only two X atoms on alternate faces. As contribution of each of them = —, therefore,

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= 8x-

F
no. of X-atoms/unit cell = 2x —= 1. Hence, formula is XY.
2
ur
r
2. There are four body diagonals. Thus, there are 8 atoms within the body of the unit cell which are not shared

by any other unit cell. Contribution by atoms at the comers = 8 x - = 1. fo

ks
Yo
Hence, total atoms/unit cell = 8+1=9.
oo

I I
3. Atoms A/unit cell = 8x- + 6x- = 4
B

(corners) (face-centres)
re

1
Atoms B/unit cell = 12x- + 1=4
u

4
ad
Yo

(edge-centres) (body-centre)
.-. A:B = 4:4=] : 1. So the formula is AB.
4. Atoms P per unit cell = 8 x 1/8 = 1, Atoms Q per unit cell = 1, Atoms R per unit cell = 6 x 1/2 = 3.
d
Re
in

Hence, the formula is PQR3.

5. BCC has 2 atoms per unit cell. Proceed as in Solved Problem 5.
F

■ ^ACi'ING IN CRYSTAL^ (CLOSE PACKED STRUCTURES)

In order to understand the packing of the constituent particles in a crystal, it is assumed that these
particles are hard spheres of identical size (e.g., those of a metal). The packing of these spheres takes place in
such a way that they occupy the maximum available space and there is minimum empty space and hence the
crystal has maximum density. This type of packing is called close packing.
The packing of spheres of equal size may be discussed in the following three steps :
>m -on. There is FIGURE 1.171
only one way in which the spheres can be arranged in
one dimensional close packed structure, i.e., these
spheres should be touching each other in a row as shown
in Fig. 1.17. In this arrangement,each sphere is touching Close packing of spheres in one dimension
two neighbouring spheres. Hence, in this arrangement,
coordination number is 2.
SOLID STATE 1/15

(b) Close-Packing in Two Dimensions. When the rows are stacked over each other, a two dimensional
close-packed structure (called crystal plane) is produced. Tliis stacking can be done in two different ways as follows;
(i) The spheres in the second row may be placed in such a way that they are touching the spheres of the
first row and are exactly above the spheres of the first row, i.e., there is horizontal as well as vertical alignment
of the spheres in the two rows. If the first row is called ‘A’ type row, the second row, being exactly same, will
also be ‘A’ type. Further rows may also be built up exactly in the same way. Hence, we call this iurangement
as AAA... type of arrangement (Fig. 1.18a).
In this arrangement, each sphere FIGURE l.lg
is touching four other spheres. Hence,
in this arrangement, the coordinadon
number is 4. Further, on joining the
centres of the four spheres touching the
particular sphere, a square is formed
as shown in Fig. 1.18 (a). Hence, this
Close packing of spheres in two dimensions

w
packing is caUed square close packing (a) Square close packing (b) Hexagonal close packing
in two dimensions.

F lo
(ii) The spheres in the second row may be placed in the depressions of the first row as shown in Fig. 1.18
(b). This produces a row different from the first row. Hence, if we call first row as ‘A’ type, the second row
may be called as ‘B’ type. When the spheres in the third row are placed in the depressions of the second row,

ee
the third row formed is exactly similar to the first row, i.e., ‘A’ type. The fourth row formed will be again ‘B’

Fr
type and so on. Hence, this arrangement is called as ABAB... type of arrangement. In this arrangement, each
sphere is touching six other spheres. Hence, the coordination number is 6. Further, on joining the centres of

for
the six spheres touching a particular sphere, a hexagon is formed as shown in Fig. 1.18 (/?). Hence, tliis
ur
arrangement is called two dimensional hexagonal close-packing.
The second mode of packing {i.e., hexagonal close packing) is more efficient as more space is occupied
s
ook

by the spheres and empty space is less in this arrangement,

Yo

(c) Close-Packing in Three Dimensions. We have discussed above that two dimensional packing can
eB

be square close packing or hexagonal close packing. Now, let us see how a three dimensional packing can be
obtained from each of these.
our

(0 Three dimensional close packing from two dimensional square close packed layers. Starting
ad

from the square close packed layer, the second layer and all further layers will be built up such that they are
horizontally as well as vertically aligned with each other. Hence, if we call the first layer as ‘A’ type, their
Y

lattice generated will be AAA... type. This will be a simple cubic lattice and its unit cell will be a primitive
Re

cubic unit cell, as shown in Fig. 1.19.

nd

FIGURE 1.19 FIGURE 1.201

Fi

A A

A ^UTI ■ b h

A
A
b

A
A A

Three dimensional packing of square

close packed layers forming simple Representing the two types of triangular
cubic lattice with AAA-arrangement voids in hexagonal close packed layer
1/16 ‘P’ittdee^’4. New Course Chemistry fXTHrosnn

(u) Three dimensional packing from two dimensional hexagonalclose packed layers. We have seen
that for two dimensional packing, a more efficient packing is given by hexagonal close packing. Let us now
consider a three dimensional packing keeping a hexagonal close packed pattern for layers. In the base layer
shown in Fig. 1.20, the spheres are marked as A and the voids between the spheres are marked as ‘a’ and
alternately.
Both 'a' and ‘i>’ voids are triangular in shape. They differ only in the fact that whereas apex of the
triangles of voids 'a' point downwards, those of voids 'h’ point upwards.
When a second layer is placed with spheres vertically aligned with those in the first layer, its voids will
come above the voids in the first layer. This is an inefficient way of filling the space.
When the .second layer is placed in such a way that its spheres find place in the 'a' voids of the first layer, the
'b' voids will be left unoccupied since under this arrangement no sphere can be placed in them [Fig. 1.21].
Now, there are two types of voids in the .second layer. These are marked as voids ‘c’ and voids ‘t/ ’. The

w
voids ‘c’ are ordinary voids which lie above the spheres of the first layer whereas voids ‘‘d' lie on the voids of the
first layer and hence are combinations of two voids, one of the first layer and second of the second layer.
The 'a' and 'b' voids

Flo
FIGURE 1.21
of the first layer are both
triangular while only ‘c’

e
re
voids of the second layer
are triangular. The ‘d'

F
voids of the second layer
are combi- nations of two
ur
or
triangular voids (one each
of first layer and second f
ks
layer) with the vertex of one
Yo
triangle upwards and the
oo

vertex of the other triangle

downwards.
B

A simple triangular
re

void like ‘c’ in a crystal is

surrounded by four spheres
u
ad

Red spheres represent layer A Grey spheres represent layer B

and is called a tetrahedral
Yo

void or a hole [Fig. 1.22. Close packing of two layers with hexagonal close packed base layer
(fl)]. A double triangular FIGURE 1.22
d

void like Vf is surrounded by six spheres and is called octahedral

Re

Lower Layer
in

void [Fig. 1.22. (b)]. Tetrahedral void is so called because this

void is surrounded by four spheres lying at the vertices of a
F

regular tetrahedron and touching each other. Similarly, octahedral

void is so called because it is surrounded by six spheres lying at Lying'
the vertices of a regular octahedron four lying in one plane above at
the centre
touching each other, one lying above the plane and one below
the plane (see lower Fig. 1.22 b).
Tlte voids or holes in crystals are also called interstices.
Now, there are two ways to build up the third layer. o
(/■) When a third layer is placed over the second layer in Two types of voids in crystals
such a way that the spheres cover the tetrahedral or ‘c’ voids, a (a) tetrahedral void (b) octahedral void
three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically
aligned (i.e., the third layer is directly above the first, the fourth above the second layer and so on). Calling
the first layer as layer A and second layer as layer B, the arrangement is called AB AB pattern or hexagonal
close packing (hep) [Fig. l.23a (i) or b (01.
Molybdenum, magnesium, zinc and beryllium crystallise in hep (hexagonal close packing) structure.
SOLID STATE 1/17

FIGURE 1.23

o o

ow
A A

o o

e
Actual view of (i) Hexagonal close A simplified view of (i) Hexagonal close packing

re
Fl
packing (ii) Cubic close packing (AB AB paitern)
(ii) Cubic close packing (ABC ABC pattern)

F
(//) When the third layer is placed over the second layer in such a way that spheres cover the octahedral
ur
r
or Vf voids, a layer different from layers A and B is produced. Let us call it as layer C. Continuing further,

fo
a packing is obtained where the spheres in every fourth layer will be vertically aligned. This pattern of
stacking spheres is called ABCABC pattern or cubic close packing {ccp). It is simihir to face-centred
ks
cubic {fee) packing [Fig. 1.23« (//) or 1.23/? (//)].
Yo
Iron, nickel, copper, silver, gold and aluminium crystallise in ccp (cubic close packing) structures.
oo

Both the above patterns of stacking spheres, though different in form, are equally efficient. They occupy
eB

the maximum possible space which is about 74% of the available volume (the empty space being only 26%).
Hence, they are called close packings.
Further, in both hep and ccp methods of stacking, a sphere is in contact with 6 other spheres in its own
ur

layer. It also touches directly 3 spheres in the layer above and three .spheres in the layer below. Thus, a sphere
ad

has 12 close neighbours. It is said to have a coordination number of 12. As already defined.
Yo

The numberofclosest (or nearest) neighboursofany constituentparticle in the crystallattice is

d

called its coordination number.

Re
in

The common coordination numbers in different types of crystals are 4, 6, 8 and 12. In crystals with
directional bonds, coordination number is lower than that of the crystals with non-directional bonds such as
F

metals and ionic compounds.

To understand as to why they are called hexagonal close packing and cubic close packing, the arrangement
of the layers may be represented as shown in Fig. 1.24.
In addition to these two types of arrangements, a third type of arrangement found in metals is body-
centred cubic (bcc) in which space occupied is about 68%.
Lithium, sodium, potassium, rubidium and caesium crystallize in the bcc structure.
The coordination number of each atom in the bcc structure is 8. Further, it is not as closely packed as the
first two as the empty space is about 32%.
Number of atoms per unit cell in bcp. Look at Fig. l.2A{a), A unit cell has 12 atoms at the comers,
2 atoms at the face-centres and 3 atoms within the body. Each atom at the comer is shared by 6 other unit cells.
1
therefore, contribution of an atom at the comer towards the unit cell is —. Similarly, atom at the face-centre
I 6 ^
makes a contribution of — while atom within the body is not shared by any other unit cell. Hence, in hep
2
1/18 T^uxdeefr'^. New Course Chemistry rxinromwi
FIGURE 1.24!
I

A
I
0 I

\ / I
c
B

B i \ i

w
F lo
I

^ HEXAGONAL CLOSE PACKING (HCP) ^ CUBIC CLOSE PACKING (CCP) = FACE-CENTRED CUBIC (FCC)

ee
(a) ABAB type is hexagonal close packing (b) ABC ABC type is cubic close packing

Fr
1
No. of atoms per unit cell = 12 x — + 2x- + 3 =6
6
(comers)
2
(facecentres)
(wiihin
the body) for
ur
s
1.10. CALCULATION OF THE SPACES OCCUPIED, I.e., Packing Efficiency or Packing Fractions
ook
Yo

We have seen above that in any type of packing, some voids (empty spaces) are always present.
eB

The percentage of the total space filled by the particles is called packing efficiency or the
fraction of the total space filled is called packing fraction. nti-:
our
ad

The calculations for different types of structures are described below ;

1.10.1. Packing Efficiency In Simple Cubic Lattice (Fig. 1.25}

Y

Suppose the edge length of the unit

Re

FIGURE 1.^ FIGURE 1.261

nd

cell = a and radius of the sphere = r

Fi

As spheres are touching each other,

evidently, a = 2 r (Fig. 1.26) /
1
No. of spheres per unit cell = - x8 = 1

4 .3
Volume of the sphere = J ^'
■a- *■

Simple cubic unit cell Top/Bottom/Side view of

Volume of the cube = a^= (2 r)^ = 8 simple cubic unit cell

^Jtr'
= ^ = 0*524 or % occupied = 52*4%
3
.'. Fraction occupied, i.e.. packing fraction
8;-3

*The spheres at the comers are not touching each other.

SOLID STATE 1/19

1.10.2. Packing Efficiency in Face-centred Cubic Structures (Cubic Close Packing) (Fig. 1.27)
As sphere on the face centre is touching the spheres at the comers*, evidently AC = 4 r (Fig. 1.28)
But from right angled triangle ABC,
r 4
ac = Vab2+bc2 =V2 a V2 rt = 4 r or a =

Volume of the unit cell

FIGURE 1.2^ FIGURE 1.^

- a
3 _ f 4 _ 32 , 4- -a- >

No. of spheres in the unit cell (a-

c
= 8x- + 6x- = 4
o
.A®

w
2

4 16
Volume of 4 spheres = 4x— Tir-^ = — nr .3 c.: V'

F lo
3
Top view of face-centred
Fraction occupied, i.e., Face-centred cubic unit cell cubic unit cel!

16TC/-V3 _n^ = 0-74

ee
packing fraction = or % occupied = 74%.

Fr
32r^l^l2~ 6
Packing efficiency of hep and cep structures. Cubic close packing (cep) is same as face-centred cubic

for
packing. Hence, packing efficiency is 74%. Further, cep and hep are equally efficient. Hence, hep also has a
ur
packing efficiency of 74%.
s
1.10.3. Packing Efficiency in Body Centred Cubic Structures
ook
Yo
As the sphere at the body centre touches tlie spheres at the comers*, body diagonal, AD = 4 r (Fig. 1.29)
eB

Further, face diagonal. AC = VaB- +BC2 = ^|a~+c^ = V2 a

4r
and body diagonal, AD = VaC^ -f-CD^ = +a'^ = -Js -yj3a = 4r or a =
r

V3
ad
ou

3 _
4r 64;-3
Volume of the unit cell -a
Y

FIGURE 1.29
3>/3
Re

,.’C
nd

No. of spheres per unit cell =8x--i-l=2

8
Fi

4
Volume of 2 spheres = 2x —jtr^ = -nr^
3 3 ;d
.3
-71/
3 71 V3
Fraction occupied, i.e., packing fraction = 0-68
64 8
Body-centred cubic unit cell
or % occupied = 68%. 3^3
1.11. SIZES OF TETTIAHEDRAL AND OCTAHEDRAL VOIDS

As already discussed, in the close packing structure, (hep or cep), there are two types of voids present in
the crystal: (0 Tetrahedral voids and (//■) Octahedral voids
The radii of the voids in these close packed structures are related to the sizes of the spheres present in
the packing.
*Remember that packing in rhombohedral is .similar to hexagonal close packing with 12 nearest neighbours and
a packing fraction of 0.74.
1/20 ‘P'tadecf.i ^ New Course Chemistry (XII)

If R is the radius of the spheres in the close packed arrangement, then

Radius (r) of the tetrahedral void = 0*225 R
Radius (r) of the octahedral void = 0*414 R

In case of ionic compounds, as usually anions are present in the packing and cations occupy the voids,
hence, we can also write
(0 For cations occupying the tetrahedral voids, = 0-225 r_.
(/7) For cations occupying the octahedral voids, 0-414 r_.

ow
Thus, a tetrahedral void is much smaller than the octahedral void. Further, the maximum size of the
sphere occupying the void is 0-414 R. If the smaller spheres to occupy the voids have size larger than this, the
arrangement will no longer be close packed.
Their size acquires importance in the formation of transition metal hydrides, borides, carbides and

e
nitrides in which the respective non-metal atoms, i.e., H, B, C and N are accommodated in the interstices.
That is why these compounds are called interstitial compounds.

re
Flr
1.11.1. Derivation of the relationship between the radius (r)

F
of the octahedral void and the radius (R) of the atoms in dose packing.
A .sphere fitting into the octahedral void is shown by a small circle (Fig. 1.30). A sphere above and a
ou
sphere below this small sphere have not been shown in the Fig.

sr
Obviously, ABC is a right angled triangle. Applying Pythagoras theorem, FIGURE 1.3^

fo
BC^ = AB^ + AC2

k
oo
(2 R)- = (R + r)2 -I- (R -I- r)- = 2 (R + r)~ or
2
Y
or 2 R2 = (R + r)2 or (■>/2R)2=(R-fr)2 or ■^^2R = R -fr
reB

or ;- = V2R-R = R(V2-1) =R (1-414- 1) = 0-414 R

uY

1.11.2. Derivation of the relationship between radius (r)

of the tetrahedral void and the radius (R) of the atoms in close packing.
ad
do

To simplify calculations, a tetrahedral void may be represented in a cube as shown in Fig. 1.31 below, in
which three spheres A, E and F form the triangular base, the fourth (B) lies at the top and the grey coloured
in

sphere occupies the tetrahedral void.

Re

Suppose the length of the side of the cube = a

F

From irght angled triangle ABC, face diagonal AB = VaC“ + BC“ = sla^+a- = -Jl a

As spheres A and B are actually touching each other in the close

FIGURE 1.31
packing, face diagonal AB = 2 R

Hence, 2R= a or R = a ...(i)

Again from the irght angled triangle ABD,

body diagonal AD = -TaB^TbD^ = .^(V2a)^ = V3 a

But as grey coloured sphere touches other spheres, evidently, body

diagonal AD = 2 (R + r)

2{R +r) = ^a or R-hr = a ...(«)

2

R+r Sal2 _ S
Dividing eqn. {ii) by eqn. (0, we get R a /V2 “ V2
SOLID STATE 1/21

L-A >/3-V2 1-732-1414

or l+L= or = 0-225 or r = 0-225 R.
R V2 R ^|2 V5 1414

Calculation of ideal radius of the cation fitting into the voids. The radius of the telnihedral or octahedral
void can be calculated if the radius of the ions in the packing is known. This gives the ideal radius of the ion
fitting into the void.
Sample Problem A solid B" has NaCI type close packed structure. If the anion has a radius
of 241‘5 pm, what should be the ideal radius of the cation ? Can a cation C"*" having radius of 50 pm be
ntted Into the tetrahedral hole of the crystal A"*" B" ?
Solution. As A'*’ B"ha.s NaCl structure. A'*’ ions will be present in the octahedral voids. Ideal radius of the
cation will be equal to the radius of the octahedral void because in that case, it will touch the anions and the
arrangement will be close packed. Hence,

w
Radius of the octahedral void = i'
A+
= 04I4x K = 0-414 X 241-5 pm = 100*0 pm

Radius of the tetrahedral void =0-225xr

B" = 0-225 X 241-5 pm - 54*3 pm

F lo
As the radius of the cation C*" (50 pm) is smaller than the size of the tetrahedral void, it can be placed into the
tetrahedral void (but not exactly fitted into it).

ee
Fr
for
ur
1. A solid AB has NaCl structure. If the radius of the cation A is 100 pm, what is the radius of the anion B ?
2. A solid AB has NaCl structure. If the radius of cation A'*'is 170 pm. calculate the maximum possible
s
radius of the anion B".
ok
Yo
ANSWERS
o

1. 241 pm 2. 410-6 pm
eB

HINTS FOR DIFFICULT PROBLEMS

r

1. As NaCl has octahedral structure, = 0-414xr

ou
ad

B" ●

2. The formula = 0-414 x gives maximum possible radius of anion for a given cation for close packing.
Y

Summing up the Main Features of the Tetrahedral and Octahedral voids

Re
nd
Fi

Tetrahedral Voids Octahedral Voids

(/) As r = 0-225 R, tetrahedral void is much (0 As r = 0-414 R, the size of the octahedral void
smaller than the size of the spheres in the is smaller than that of the spheres in the packing
packing. but larger than the tetrahedral voids.
(ii) Each tetrahedral void is surrounded by 4 (i7) Each octahedral void is surrounded by 6 spheres.
spheres. Hence, its coordination number is 4. Hence, its coordination number is 6.
(i77) In the hep of cep packing, each sphere is in (i77) As octahedral void is a combination of two voids
contact with three spheres in the layer above of the two layers, number of octahedral voids is
it and three spheres in the layer below it, thus equal to half the number of tetrahedral voids
forming one tetrahedral void above and one and hence equal to the number of spheres in the
tetrahedral void below.Hence, there are two packing.
tetrahedral voids per sphere, i.e., number of
tetrahedral voids is double the number of
spheres in the packing.
 1/22
^na.dee^'^ New Course Chemistry (XII)BSm
1.12. FORMULA OF THE TOMPOUND FROM NUMBER OF VOIDS FILLED

As already discussed, in a close packed structure (hep or cep), there are two types of voids present in the
lattice, i.e., tetrahedral voids and octahedral voids. It is found that
(0 Number of octahedral voids = Number of particles present in the close packing.
(ii) Number of tetrahedral voids = 2 x Number of octahedral voids.
In case of ionic compounds, it is found that the bigger ions (usually anions) are present in the packing
whereas smaller ions (usually cations) occupy the voids.
If the cations are small in size, they may occupy tetrahedral voids (because they are smaller than octahedral
voids). However, if tlie cations are large enough to occupy the tetrahedral voids, they may occupy octahedral
voids. Further, it is not necessary that only tetrahedral or only octahedral voids may be occupied. Also, it is
not necessary that all the voids must be occupied. Only a fraction of the total voids may be occupied.
Knowing the fraction of the voids occupied, the formula of the compound can be calculated or vice-

w
versa. This will become clear from the following solved problems.
Sample Problem Q A compound Is formed by two elements X and Y. Atoms of the element Y (as

Flo
anions) make cep and those of the element X (as cations) occupy all the octahedral voids. What is the
formula of the compound ?

ee
Solution. Suppose number of atoms Y in cep = n

Fr
As number of octahedral voids = No. of atoms in cep .'. No. of octahedral voids = n

As all the octahedral voids are occupied by atoms X, therefore, number of atoms X = n.

for
Ratio ofX:Y = n:n=!:l
ur
Hence, the formula of the compound is XY.
s
Sample Problem Atoms of element B form hep lattice and those of the element A occupy
k
Yo
oo

2/3rd of tetrahedral voids. What is the formula of the compound formed by these elements A and B ?
Solution. Suppose number of atoms B in hep lattice
eB

= n
As number of tetrahedral voids is double the number of atoms in the close packing, therefore, number of
tetrahedral voids = 2n
r

2 An
ou

As atoms A occupy 2/3rd of the tetrahedral voids, therefore, number of atoms A in the lattice = —x2n =
ad

3 3
Y

4/1
.●. Ratio of A : B = : n = l:l = 4:3
3 3
Re
nd

Hence, the formula of the compound A4B3.

Fi

Sample Problem Q In a crystalline solid, anions B are arranged in a cubic close packing. Cations
A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied,
what is the formula of the solid ?

Solution. Suppose the number of anions B = n. Then number of octahedral voids = n

Number of tetrahedral voids = 2 n

As octahedral and tetrahedral voids are equally occupied by cations A and all the octahedral voids are
occupied (given), therefore n cations A are present in octahedral voids and n cations A are present in tetrahedral
voids. In other words, corresponding to n anions B. there axen + n = 2n cations A. Thus, cations A and anions B
are in the ratio 2n : n = 2: 1. Hence, the formula of the solid will be A2B.

Sample Problem E In the mineral, spinel, having the formula MgAl204, oxide ions are arranged
in the cubic close packing, Mg^"*^ ions occupy the tetrahedral voids while Al^ ions occupy the octahedral voids.
(i) What percentage of tetrahedral voids is occupied by Mg 2+ ionss
?

(ii) What percentage of octahedral voids is occupied by Al*^'*' ions ?

SOLID STATE 1/23

2+
Solution. According to the formula MgAl204, if there are 4 oxide ions, there will be 1 Mg ion and
2 Al^'*' ions. But if the 4 ions are in ccp arrangement, there will be 4 octahedral voids and 8 tetrahedral voids.
Thus, 1 ion is present in one of the 8 tetrahedral voids

% of tetrahedral voids occupied by =-xl00=

8
12*5%
Similarly, 2 AP'*' ions are present in two octahedral voids out of 4 available.
1

% of octahedral voids occupied by AI^"^ =-xl00 = 50%

Sample Problem 0 What IS the percent by mass of titanium in rutile, a mineral that contains Ti
and oxygen if structure can be described as a closest packed array of oxide ions with titanium ions in one-
half of the octahedral holes ? What is the oxidation number of Ti ? (Atomic mass of Ti = 48)

w
n
Solution. If no. of oxide ions = n, then octahedral holes = n and hence 0- ions
2

F lo
n
Ti: O = — : « = —; 1 = 1: 2 . Hence, formula of the oxide is TiO^. Molar mass of Ti02 = 48 + 32
2 2
-]
= 80 g mol

ee
48

Fr
%ofTi =—xKX)
80
= 60%. Oxidation state of Ti in Ti02 =+4.

for
ur
s
1. What is the formula of a compound in which the element Y forms ccp lattice and atoms X occupy l/3rd of
ok
Yo
tetrahedral voids. (CBSE 2015, Foreign 2017)
o
eB

2. In comindum, oxide ions are arranged in hexagonal close packing and aluminium ions occupy two-third of
the octahedral voids. What is the formula of corrundum ?

3. In a solid, oxide ions are arranged in ccp. One-sixth of the tetrahedral voids are occupied by the cations A
r

while one-third of the octahedral voids are occupied by the cations B. What is the formula of the compound ?
ou
ad

4. A solid is made up of two elements P and Q. Atoms Q are in ccp arrangement while atoms P occupy all the
Y

tetrahedral sites. What is the formula of the compound ?

5. In chromium (III) chloride, CrClj, chloride ions have cubic close packed arrangement and Cr (III) ions are
Re
nd

present in the octahedral holes. What fraction of the octahedral holes is occupied ? What fraction of total
number of holes is occupied ?
Fi

ANSWERS

1. X2Y3 2. AUO3 3. ABO3 4.P2Q 5. 1/3, 1/9

HINTS FOR DIFFICULT PROBLEMS

1. Suppose no. of atoms Y present in the packing = n

Then, tetrahedral voids = 2 n
2n 2n 2

Atoms X present in the tetrahedral voids = ^ x 2 n = 3

.-. Ratio of X : Y =
3
; n = -:I = 2:3
3

Hence, the formula is X2Y3.

3+ ; 2 2u
2. Suppose oxide ions = n. Then octahedral voids = n. Hence, Ah ions =-x/i = —
3 3
2n
.-. Ratio Al^-": = : n = 2:3 , i.e., formula is ALO3.
3
1/24
^n4idee^'A New Course Chemistry (XI1)C!Z&1SI

3. Suppose 0^ ions = n. Then octahedral voids = « and tetrahedral voids = 2^1. Cations A=-x2« = - and
6 3
1 n
cations B=-x«= —
3 3
n n
Ratio A ; B : 0-- = —: n
= 1:1:3, Le., formula is ABO-,.
3 3

4. Suppose number of atoms Q = n

Then number of tetrahedral sites = 2 n
Number of atoms P = 2 «

Ratio P : Q = 2n : n = 2: 1, i.e., formula is P,Q.

5. See Solved Problem 4 above.

w
1.13. LOCATION OF TETRAHEDRAL AND OCTAHEDRAL VOIDS IN A CRYSTAL

In a close-packed structure (ccp or hep), if there are n spheres (atoms or ions) in the packing, then

F lo
Number of octahedral voids = n

Number of tetrahedral voids = 2 n

ee
For example, in the cubic close packing (ccp), i.e., face-centred cubic (fee) unit ceil, there are 4 atoms

Fr
or ions per unit cell, therefore, there are 4 octahedral voids and 8 tetrahedral voids. These are located at
different positions as shown in Fig. 1.32 and explained on the next page :

for
ur
FIGURE 1.321

o
f
s
o
ook
Yo
.0 o
z
eB

o
< y-' m'*' - 4 o O

» LO*
1 0 ● o
r

^ ' io-
ou
ad

'A
o ●
y
Y

● -o-
(0 (ii) (iii)
Re
nd

Octahedral void at the body centre Octahedral void at the edge 12 Octahedral voids at
(shared by none) centre (shared by 4 unit cells) the edge centres represented by O
Fi

€) k
-4

k
●a
T
;t

/
! 0%
y
'T
'a' ---
(i)
#- -*■
II ● T \
,T- T
● -

Formation of one tetrahedral void Location of all the 8 tetrahedral voids represented by T

(a) Representing (i) Octahedral void at the body-centre (ii) Octahedralvoid at the edge centre
(iii) All the 12 octahedralvoids of ccp unit cell each shared by 4 other unit cells,
(b) (i) Formation of a tetrahedral void
(ii) Representing the location of all the 8 tetrahedral voids on the body diagonals
SOLID STATE 1/25

(«) Octahedral voids. One octahedral void is present at the body centre of the cube and 12 octahedral
voids are present on the centres of the 12 edges of the cube. But each void on the edge centre is shared by 4
unit cells. Hence, its contribution in the unit cell = 1/4. Therefore, the effective number of octahedral voids
in the ccp structure =1 + 12x1/4=1+3=4.
(h) Tetrahedral voids. The 8 tetrahedral voids present in the ccp arise from the fact that there tire 8 spheres
at the comers of the unit cell and each sphere at the corner touches three spheres present on the face-
centres of the three adjoining faces, each giving rise to one tetrahedral void (see Fig. 1.12 on page 1/10). These
tetrahedral voids are found to be present on the body diagonals, two on each body diagonal at one-fourth of
the distance from each end, as shown in Fig. 1.32 (b) (ii).
Thus, in ccp, total number of voids per unit cell = 8 (tetrahedral) + 4 (octahedral) = 12.
Similarly, in hep, total number of voids per unit cell = 12 (tetrahedral) + 6 (octahedral) = 18.

w
1.14. STRUCTURES OF SIMPLE IONIC COMPOUNDS

Simple ionic compounds are the compounds of the type AB or AB2 (or A2B) where A and B represent
the positively and negatively charged ions respectively. It is more difficult to describe the structures of even

F lo
these simple ionic compounds as compared to those of the elements. This is because an element consists of
only one type of atoms, all of which occupy the lattice points, e.g., in case of copper which is face-centred

ee
cubic (FCC), all lattice points are occupied by copper atoms. On the other hand, in case of even the simple

Fr
ionic compounds, the arrangement of both A and B ions has to be described, as given below ;
Structures of the ionic compounds of the type AB. Ionic compounds of the type AB means
compounds having the positively and negatively charged ions in the ratio 1:1. These compounds have any

for
ur
one of the following three types of structures:
1. Rock salt (NaCl) type structure.
s
2. Caesium chloride (CsCl) type structure.
ook
Yo
3. Zinc blende (ZnS) type structure.
The main features of each of these structures are discussed below :
eB

1. Rock salt (NaCl) type structure. The structure of NaCl is as shown in Fig. 1.33.
The main features of this structure are as follows :
r

(/) It has face-centred cubic (fee) arrangement (also called cubic close packing, i.e., cep) in which Cl"
ou
ad

ions occupy the comers and face centres of a cube while Na'*’ ions are present at the body centre and edge
centres. (In fact, Na"^ ions occupy all the octahedral voids).
Y

FIGURE 1^
Re
nd
Fi

O = Na'*’ / 1

● = cr 7/

cr ION
SURROUNDED
OCTAHEDRALLY
BY SIXNa* IONS ^ \

7^ Sf

O
O
Na"" ION SURROUNDED
OCTAHEDRALLY BY SIX Cl" IONS

(a) Structure of NaCl (Rock salt) (b) Showing coordination number of

Na'*’ and CI“ ions
 1/26
New Course Chemistry (XII)EQSXS]

(//) Each Na'*’ ion is surrounded by six Cl ions and each Cl ion is surrounded by six Na'*’ ions. In other
words, the Na'*’ ions as well as Cl“ ions have coordination number of six. i.e., this structure has 6 : 6
coordination.

(///) A unit cell of NaCl consists of four NaCl units, i.e., 4 Na* ions and 4 Cl” ions as calculated below :

8 Cl” ions are present on the corners and 6 Cl” ions on the face centres so that Cl” ions per unit cell
1
= -x8
8
+ —x6 = 4. Similarly, 12 Na^ ions are present on the edges and one within the body so that Na"*" ions

per unit cel! = - x 12 + I = 4.

4

A few examples of the compounds having structure similar to that of NaCl are as follows:

ow
(/) Halides of alkali metals (except those of caesium) and that of ammonium.
(//) Oxides and sulphides of alkaline earth metals (except BeS).
(Hi) Halides of silver (except silver iodide).

e
2. Caesium chloride (CsC!) type structure. The structure of CsCl is as shown in Fig. 1.34.

Fl
re
The main features of this structure are as follows

F
(0 It has body-centred cubic (bcc) arrangement.
ur
or
FIGURE 1.34

Cs^ ION SURROUNDED

sf
Cl” ION SURROUNDED BY
\ 8Cs'*‘lONS
k
Yo
BY 8 Cl” IONS
oo

V
B

\
\ . o
re

\
\
s
—^ _
\r/y

o =Cs*
u

7 '
ad

s
✓
d '
Yo

V
= cr
\
a

X s
d

V
Re
in

O unit cell o 8 unit cells

F

Structure of CsCl

(//) Each Cs'^ ion is surrounded by 8 Cl ions and each Cl” ion is surrounded by 8 Cs” ions, i.e., this
structure has 8 : 8 coordination.

(///) A unit cell of CsCl consists of only one unit of CsCl, i.e.. one Cs'^ ion and one Cl” ion.* This may be
explained as follows :
No. of Cs^ ions per unit cell = I (present within the body)

No. of Cl ions per unit cell z= gx- (from corners) = 1

8

A few examples of compounds having CsCl structure are as follows

CsBr, Csl, CsCN, TlCl, TlBr, Til and TICN.
SOLID STATE 1/27

3. Sphalerite or Zinc blende (ZnS) type structure. The structure of ZnS is as shown in Fig. 1.35.
FIGURE 1.35

ION SURROUNDED
TETRAHEDRALLY BY
X FOUR S^“ IONS

I
o

I
S^“ION SURROUNDED

w
TETRAHEDRALLY BY
1
FOUR IONS

0 = 2n^'^ IONS

F lo
ee
= S^" IONS

Fr
Structure of ZnS (Zinc blende)

The main features of this structure are as follows :

for
ur
(i) The arrangement as possessed by ZnS is called cubic close packing (ccp) in which S^“ ions form a
face centred cubic, i.e., S^~ ions are present at the comers as well as at the centre of each face of the cube.
s
ook

Zn^'*’ ions are present at l/4th of the distance along each body diagonal, one Zn^"'’ ion on each body diagonal
Yo

(In fact, Zn^"*" ions are present in the alternate tetrahedral voids).
eB

(«) Each Zn^'*’ ion is surrounded tetrahedrally by four ions and each is surrounded tetrahedrally
by four Zn^'*' ions. Thus, this structure has 4 : 4 coordination.
our
ad

{Hi) One unit cell of ZnS consists of 4 Zn^'*' ions and 4 S"“ ions, i.e., it has 4 ZnS formula units per unit
cell. This may be explained as follows :
')- 1 i
Y

No. of S*^ ions per unit cell = 8x- (from comers) + 6x— (from face centres) = 1+3 = 4.
Re

8 3
nd

No. of Zn-'*' ions per unit cell = 4 (present with the body of the unit cell)
Fi

(/v) The structure of ZnS is similar to that of diamond in which in place of Zn-"*" and S^“ ions, there are
carbon atoms at each site.

It is important to note that the electronegativity difference between Zn and S is very small (~ 0-9),
therefore, the bond between Zn and S has a large amount of covalent character.
A few examples of ionic compounds having ZnS structures include CuCl, CuBr, Cul, Agl and BeS.
It may be noted that zinc sulphide, in fact, exists in two forms, called zinc blende and wurtzite.
Similarities. (/) Both of them have a close-packed arrangement of S-“ ions.
(ii) In both of them, Zn^'*’ ions occupy the alternate tetrahedral voids (because number of tetrahedral
voids is double the number of S““ ions).
(Hi) Both have 4 : 4 structure.
Differences. They differ only in the fact that zinc blende has cubic close-packed (ccp or fee)
arrangement of S~“ ions whereas wurtzite has hexagonal close-packed (hep) arrangement of S^“ ions. As a
result, unit cell of wurtzite has 6 formula units of ZnS (whereas zinc blende has 4 formula units).
1/28 ^naeie^'^ New Course Chemistry (XII)BZ3SI

Structure of the ionic compounds of the type AB2. These are the ionic compounds having cations and
anions in the ratio I : 2. Most of these compounds have calcium fluorite (CaF2) type structure shown in
Fig. 1.36.

FIGURE 1.36

F~ION SURROUNDED

w
F lo
ION SURROUNDED
BY 8 F" IONS

e
Fre
O =Ca^* IONS F“IONS for
r
. Structure of Cap2 (Fluorite)
You
oks

The main features of this structure are as follows :

eBo

(i) It has cubic close packing {ccp) arrangement in which ions are present at the comers and the
centre of each face of the cube and F" ions occupy all the tetrahedral voids.
ad
our

(«) Each Ca-'*’ ion is surrounded by 8 F^ ions, i.e., has a co-ordination number of 8 whereas each F" ion
is surrounded by 4 Ca^'*’ ions, i.e., has a co-ordination number of 4. Thus, this structure has 8 : 4
co-ordination.

{Hi) Each unit cell contains 4 Ca^'*’ ions and 8 F" ions. This may be explained as follows:
dY
Re

Ca^’’" ions per unit cell = 8x-8 (from corners) + bx-^-2

Fin

(from face-centres) = 1 + 3 = 4.

F“ ions per unit cell = 8 (present within the body)

A few examples of compounds having Cap2 stmcture include BaF2, BaCl2, SrF2, SrCl2. CdF2, PbF2 and
ThO>
Structure of ionic compounds of the type A2B. Some ionic compounds have the structure of the type
A2B, e.g., Na20. Their structure is similar to fluorite structure except that positions of the cations and the
anions are reversed, the smaller cations occupying the position of ¥~ ions whereas the larger anions occupy
the position of Ca^'*' ions. Hence, the compounds of the type A2B are said to possess antifluorite structure.
For example, in Na20, 0^“ ions have a cubic close packed arrangement and Na'*’ ions occupy all the
tetrahedral voids. Each ion is surrounded by 8 Na^ ions and each Na'*’ ion is surrounded by 4 0^~ ions.
Hence, it has 4 ; 8 co-ordination.

Many oxides and sulphides of alkali metals have antifluorite stmcture, e.g., Li20, K2O, Rb20, Na2S,
K2S, Rb-,S, etc.
SOLID STATE 1/29

The above results are summed up in Table 1.4 below :

TABLE 1.4. Main characteristics and examples of some simple ionic solids
Crystal Brief description Examples Coordi Number of formula
Structure nation no. units per unit cell

Type AB
(/) Rock salt It has fee arrangement in which Cl" Halides ofLi.Na, K, Na-^ = 6 4
(NaCl) type ions occupy the comers and face Rb, AgF, AgBr, Cl" = 6 (Na"^ ions = 12x-
centres of a cube while Na"*" ions NH4CI, NH4Br, 4
are present at the edge centres and NH4I, etc. edgeceiitres
body centre. (Refer to Fig. 1.33) + 1 = 4

bodycenlre

w
I
Cr ions = 8x
8

F lo
comers

1
+ 6x

ee
bodycenlre
(/7) Caesium It has the bcc arrangement with CsCl, CsBr, Csl, Cs-" = 8 1

Fr
chloride Cs"*" at the body centre and Cl" ions CsCN, TlCl, TlBr, Cl
(Cs'*’ ions = 1,
(CsCl) type at the corners of a cube or vice Til and TICN
at body centre,
versa. (Refer to Fig. i.34)
for
ur
cr ions = 8x-- =1)
s
comers
ook
Yo
(in) Zinc blende It has cep arrangement in which CuCl. CuBr, Cul, Zn--" = 4 4
(ZnS) type S^" ions form fee and each Zn^"*" Agl, BeS S^- = 4 I
eB

S'" ions = 8x
ion is surrounded tetrahedrally by
four S"~ ions and vice versa. (Refer
comers
to Fig. 1.35) Note. The structure
r

of diamond is similar to that of

ad
ou

+ 6x = 4
ZnS in which in place of Zn^'*' and 2
S"“ ions, there are C-atoms at each face centres
Y

place (Refer to Fig. 1.35) Zn-'*’ ions = 4

(present on the body
Re
nd

diagonal)
Fi

4
Type AB2
(0 Fluorite It has cep arrangement in which BaF2, BaCU, SrF2, Ca^^ = 8
Ca-'*' ions = 8x-
(CaF,) type Ca^"^ ions form fee with each Ca""*" S1CI2, CdF2,‘PbF2 F" = 4
ion surrounded by 8 F“ ions and comers
each F" ion by 4 Ca^'*' ions. F" ions
occupy al! the tetrahedral voids. + 6x- = 4
(Refer to Fig. 1.36) 2
face centres

F" ions = 8 (present

within the body)
(f7) Antifluorite Na-" = 4 4
Here negative ions form the cep Na^O
type (AjB arrangement so that eacli po.sitive 0-- = 8
type) ion is surrounded by 4 negative
ions and each negative ion by 8
positive ions. The cations occupy
half of the tetrahedral voids.
1/30
“Pfutdee^'^. New Course Chemistry fXTTInsTwn

Remember that CsCl lattice is slightly (about \%) more stable than NaCl lattice because in CsCl, each
ion is surrounded by more oppositely charged ions. The question arises then why halides of Li, Na, K and Rb
do not have CsCl type lattice. This is answered by their radius ratio values as small cations cannot accommodate
8 halide ions.

Effect of Temperature and Pressure on Crystal Structure — Conversion of NaCl into CsCl
Structure and Vice-versa. On applying high pressure, NaCl structure having 6 : 6 coordination changes to
CsCl .structure having 8 : 8 coordination. Similarly, CsCl having 8 : 8 coordination on heating to 760 K
change.s to NaCl structure having 6 ; 6 coordination. Thus, increa.se of pressure increases the coordination
number whereas increase of temperature decreases the coordination number.
Pressure
NaCl structure CsCl structure
760 K
(6: 6coordination) (8: Scoordination)

Calculation of number of unit cells in a given mass of an ionic compound. Remembering that 1 mole
of an ionic cojnpound contains Avogadro's number of formula units and also knowing the number of formula
units per unit cell, the number of unit cells in a given mass can be calculated.

w
Sample Problem Calculate the approximate number of unit cells present in 1 g of ideal NaCl
crystals.

F lo
Solution. 1 mole of NaCl = 58-5 g = 6-02 x 10"^ fonnula units.
6-02 xIO-"-^

ee
.●. No. of fonnula units in 1 g of NaCl =
58-5

Fr
As one unit cell of NaCl contains 4 NaCl formula units, therefore, number of unit cells present in 1 g of
_ 6-02x10-3
NaCl
58-5x4 for
= 2-57 X 10^1
ur
1.15. CLOSE PACKING IN THE IONIC COMPOUNDS AND OCCUPANCY OF VOIDS
s
Most of the ionic compounds have close-packed arrangement in which one type of ions (usually larger
ook
Yo

ones) form the close packed arrangement and the other type of ions (smaller ones) occupy the voids. A few
eB

examples are given below :

Compound Ions forming the close-packed arrangement Ions occupying the voids
(0 NaCl Cr ions (fee) Na"^ ions in all octahedral voids
our
ad

(ii) ZnS ions (fee) Zn^"^ ions in alternate tetrahedral voids,

(Hi) Cap2 ions (fee) ions in all tetrahedral voids.
It is further interesting to mention here that
Y
Re

(/) In NaCl, there are 4 Cl“ ions in the unit cell. Therefore, there are 4 octahedral voids, all of which are
nd

occupied by Na'*’ ions. Hence, there are 4 NaCl units per unit cell.
Fi

(//) In ZnS, there are 4 ions in the unit cell. Therefore,there are 8 tetrahedral voids, half of which are
occupied by Zn^"*" ions. Hence, again there 4 ZnS units per unit cell.
(in) In Cap2, there are 4 Ca^"^ ions in the unit cell. Therefore, there are 8 tetrahedral voids all of which
are occupied by F" ions. Hence, there are 4 CaF2 units per unit cell.
1.16. RADIUS RATIO RULES

For the stability of an ionic compound, each cation should be surrounded by maximum number of
anions and vice versa (for maximum electrostatic forces of attraction).
The number of oppositely charged ions surrounding each ion is called its coordinationnumber.
Since ionic bond is non-directional, the arrangement of ions within the ionic crystal is determined by the
sizes of the ions.

The ratio of the radius of the cation to that of the anion is called radius ratio, i.e.,

Radius Ratio = Radius of the cation (r^^

Radius of the anion (r_)
SOLID STATE 1/31

Evidently, greater is the radius ratio, the larger is the size of the cation and hence greater is its coordination
number. The relationships between the radius ratio and the coordination number and the structural arrangement
are called radius ratio rules, and are given in Table 1.5.
TABLE 1.5. Radius ratio rules for AB type structures
Radius Ratio Coordi Structural Structure Examples
nation no. arrangement type

0155 — 0-225 3 Planar triangular B2O3

ow
0-225 0-414 4 Tetrahedral Sphalerite, ZnS CuCI, CuBr, Cul, BaS, HgS
0-414 0-732 6 Octahedral Sodium chloride NaBr, KBr, MgO, MnO, CaO, CaS
(Rock salt) NH4Br*
0-732 — 1 Body-centred cubic Caesium chloride Csl, CsBr, TIBr, TlCl

e
re
Retain In Memory

Frl
In NaCl, for exact fitting of Na'’’ ions in the octahedral voids, the radius ratio rjr_ should be 0-414.

F
Actual ratio is 0-525 which shows that Cl" ions move apart, i.e., they do not touch each other. Same
situation exists for CsCl and ZnS.
ou
r
Sample Problem Q The two ions A'*’ and B have radii 88 and 200 pm respectively. In the close

so
packed crystal of compound AB, predict the coordination number of A***.

_ rCA'*') _ 88pm
kf
oo
Solution. = 0-44
r_ r(B"^ 200pm
Y
It lies in the range 0-414 to 0-732
B

Hence, the coordination number of A'*' = 6.

re

Sample Problem Q Bi^ ions form a close packed structure. If the radius of Br“ ion is 195 pm,
oY

calculate the radius of the cation that just Hts into the tetrahedral hole. Can a cation having a radius of
u

82 pm be slipped into the octahedral hole of the crystal Br~ ?

ad

Solution. (/) Radius of the cation just fitting into the tetrahedral hole= Radius of the tetrahedral hole
d

= 0-225 xr = 0-225 X 195 =43-875 pm

Br
in

(h) For the cation A'*' with radius = 82 pm

Re
F

Radius ratio = 82pm = 0-4205

r
195 pm
As it lies in the range 0414 - 0-732, hence the cation A'*’ can be slipped uito the octahedral hole of the crystal A'*’ Br".
Note. In case (i7), radius of octahedral void in which the cation can be fitted exactly =0414xrBr _
= 0-414 X 195 pm = 80-73 pm
This is the minimum size of the octahedral void in which if the cation is placed, it will touch all the anions
and the anions also touch each other (Refer to Fig. 1.30 on page 1/20). However, if cation bigger than the above
size is slipped into the octahedral void, cation will continue to touch all the anions but anion-anion contact will
vanish. The arrangement remains octahedral upto the maximum size of the octahedral void viz. 0732Xr Br
Hence, in such cases, we apply radius ratio rules to find the range of the void, for a particular arrangement. (Refer
to Problem for Practice 3 below).

*NH4Br is a borderline case between NaCl and CsCI type structure. This is because = 1-43- i-50 A, r
''nh^ Br"

= 1-96 A. If we take r
NH+ = 1-43 A, r^/r_ = 0-729. For higher values of rNH+ , r^/r_ > 0-732.
1/32 New Course Chemistry (X11)BSI9]

1. If the radius of Mg^"^ ion, Cs'*’ ion, 0^ ion, ion and Cl ion are 0-65 A, 1 -69 A, 140 A, 1 ●84A and 1 -81 A
respectively, calculate the coordination numbers of the cations in the crystals of MgS, MgO and CsCl.
2. Predict the close packed structure of an ionic compound A'*’ B“ in which the radius of the cation = 148 pm
and radius of anion = 195 pm. What is the coordinationnumber of the cation ?
3. If the close packed cations in an AB type solid with NaCl structure have a radius of 75 pm, what would be
the maximum and minimum sizes of the anions filling the voids ?
4. A solid A'*’ B" has NaCl type close packed structure. If the anion has a radius of 250 pm, what should be the
ideal radius of the cation ? Can a cation having a radius of 180 pm be slipped into the tetrahedral site of
the crystal B' ? Give reason for your answer.

w
ANSWERS

1.4, 6, 8 2. 8, Body-centred cubic

F lo
3. 102-5 pm, 181-2 pm 4. 103-4 pm. No

HINTS FOR DIFFICULT PROBLEMS

ee
Fr
3. For close packed AB type solid with NaCl structure. = 0-414-0-732
r

for
r
+
75
pm = 102-5 pm ;
ur
Minimum value of r_ =
0-732 0-732
r
+
75
= 181-2 pm.
s
Maximum value of r_ -
ok
0414 0-414
Yo

/y(C+)_ 180 pm
o

4. = 0-72
eB

r_(B“) 250pm
It does not lie in the range 0-225 - 0-414. Hence. C'*’ cannot be slipped into the tetrahedral site.
r
ou
ad

1.17. RELATIONSHIP BETWEEN THE NEAREST NEIGHBOUR DISTANCE {dj, EDGE

OFTHE UNIT CELL (a) AND RADIUS OF THE ATOM (r^ FOR PURE ELEMENTS
Y

1. Simple cubic unit cell (Fig. 1.37). Distance between nearest neighbours (d) = AB = a. For pure
d
Re
nd

elements, radius r = “ (because atoms are touching each other).

^ 1
Fi

2. Face-centred cubic (Fig. 1.38). Distance between nearest neighbours (d) =-

But AC^=AB^+BC^= a~+a~ = 2a^ ^
AC = V2 a

FIGURE 1.371

4=71
.izzp*
A- 2 a
B

Simple cubic unit ceil Face-centred cubic unit cell Body-centred cubic unit cell
 SOLID STATE
1/33

3, Body-centred cubic (Fig. 1.39). Distance between nearest neighbours {d) =~CD

In the right angled triangle ABD, right angled at B, AD = VaB^TbD^

Now, in the right angled triangle ACD, right angled at A,

rw
CD = Vac2+AD2 a
1
●●● d=-CD =
a
and
d
r = —
o

2 2 2 " 4

e
TO SUM UP
A. Relationships between the nearest neighbour B. Relationships between atomic radius, r

e
lo
d

r
distance (d) and the edge (a) of a unit cell (which is = — for crystals ofpure elements) and
A

F
of a cubic crystal the edge (a) of the unit cell of a cubic crystal.

u
oF
Simple Face-centred Body-centred Simple Face-centred Body-centred

rs
d =a
d = -^ d=:H a
a a
V3 a
" 2V2

k
V2
r =
2 2

o
4
= 0-707 a = 0-866 a = 0-3535 a = 0-433 a

o
f
Sample Problem Q Xenon crystallises in the face-centred cubic lattice and the edge of the unit
o
Y
cell is 620 pm. What is the nearest neighbour distance and what is the radius of xenon atom ?
B
rY

Solution. Here, a = 620 pm, d=‘?, r = ?

a 620 d 438-5
For the face-centred cubic d = = 438*5pm ; — = 219’25pm
■J2 1-414 2 2
ue

Sample Problem Q CsCl has bcc arrangement and its unit

od

cell edge length is 400 pm. Calculate the interionic distance in CsCl.
ad

Solution. The bcc arrangement of CsCl is shown in the Fig. 1.40, where
in

black circle is Cs"^ ion and coloured circles are Cl" ions. The aim is to find
half of the body diagonal AE. If the edge of the unit cell is ‘o’, then
Re

CE--<[a- +a^ --sfl

F

AE - -I- =73fl =-s^x400

Interionic distance =-^-AE

2
= V3 x 200 = 346-4 pm CsCl structure

Sample Problem 0 Sodium metal crystallises in body centred cubic lattice with the cell edge,
4-29 A. What is the radius of sodium atom ? What is the length of the body diagonal of the unit cell ?
Solution. For body centred cubic lattice, the relationship between the cell edge (n) and the radius of the atom is
■S 1-732
r = — a = x4-29 = 1.86 A
4 4

Length of the body diagonal = 4r = 4x 1-86A = 7-44 A

Sample Problem 0 In face-centred cubic (fee) crystal lattice, edge length is 400 pm. Find the
diameter of the greatest sphere which can be fitted into the interstitial void without distortion of the lattice.
1/34 New Course Chemistry (XII)CZ£ZS

a 400
Solution. For fee, radius of atom (R) = pm = 141.4 pm
2V2 2V2
As octahedral void is bigger in size than the tetrahedral void, the greatest sphere will fit into octahedral void
Radius of octahedral void (r) = 0-414 R = 0-414 x 141-4 pm = 58-54 pm
Diameter of the greatest sphere fitting into the void = 2 x 58-54 pm = 117-08 pm.

1. If the radius of an atom of an element is 75 pm and the lattice type is body-centred cubic, what is the edge
of the unit cell ?
2. The radius of an atom of an element is 500 pm. If it crystallizes as a face centred cubic lattice, what is the
length of the side of the unit cell ?

w
3. A solid AB has CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of
closest approach between A"*" and B” ions ?

F lo
4. What is the radius of sodium atom if it crystallizes in bcc structure with the cell edge of 400 pm ?
(CBSE Foreign 2017)

ee
ANSWERS

Fr
1. 173-2 pm 2. 1414 pm 3. 349-9 pm 4. 173-2 pra

HINTS FOR DIFFICULT PROBLEMS for

ur
V3 4r 4x75
s
1. For BCC, r=— a or = 173-2 pm .
ook

4 1-732
Yo

a
eB

2. ForFCC, r = or a = 2^r =2x 1-414x500 pm = 1414 pm.

2V2
3. Distance of closest approach is equal to the distance between the nearest neighbours (d). As CsCl has
our
ad

BCC lattice,

, ^|3
d =—a=
1-732
—X 404 pm = 349-9 pm.
Y

2
Re

Sa 1-732
nd

4. For BCC, r = = x400 = 173-2 pm.

4 4
Fi

1.18. CALCULATIONS INVOLVING UNIT CELL DIMENSIONS

(Calculation of Density of a Cubic Crystal from its Edge)
FIGURE 1.41
Knowing the edge of a cubic crystal from X-ray studies and knowing
the type of crystal structure possessed by it so that the number of particles
per unit cell is known, the density of the crystal can be calculated. 1
I
Case I. For cubic crystals of elements. I

Suppose the edge of the unit cell = a pm

Number of atoms present per unit cell = Z
Atomic mass of the element = M
a pm
Mass of the unit cell A cubic unit cell
Density of the unit cell = Volumeof the unit cell

Volume of Uie unit cell = (a pm)-^ = (a x 10“'® cm )^ = x 10“^^ cm^

SOLID STATE
1/35

Mass of the unit cell = Number of atoms in the unit cell x Mass of each atom = Z X m

Atomic mass M
where m - mass of each atom = . Substituting this value,
Avogadro’s number N0

ZxM/N ZxM
0
Density of the unit cell = g/cm^
^ a^xN(jXlO“^®
, i.e.,

where edge, a, is in pm and molar mass, M, is in g mor'

The density of the substance is same as the density of the unit cell.
The above equation involves five parameters. Knowing any four, the fifth can be calculated.
Case II. For cubic crystals of ionic compounds. The fonnula is the same except that now Z is the

ow
number of formula units present in one unit cell and M is fonnula mass.

Retain in Memory
(0 If a is taken in cm and M is molar mass in g mol the above expression becomes

e
Fl
re
ZxM
g/cm^
^ fl^xN 0

F
ur
(») In terms of SI units, M is in kg mol a is in meters, then

or
ZxM sf
^~a^xN 0 kg m ^
k
Yo
oo

{Hi) In case of ionic compounds A'^ B~ having/cc stmcture like NaCl,

Edge (fl) = 2 X Distance between A"^ and B~ ions
B
u re

FORMULAS USED
PROBLEMS
ad
Yo

BASED
ON
(/) For an element, the density of the unit cell and hence the density of a
crystal is given by

I
d

Relation between ZxM

Re
in

density and edge 10-30

of cubic crystals
F

where Z = Number of particles present per unit cell, viz., I for simple, 2
for BCC and 4 for FCC.

M = Atomic mass of the element, a = Edge of the unit cell in pm (Note carefully)
Nq = Avogadro’s number, p = Density of the crystal in g/cm^ (when M is in g mol"*)
{ii) For ionic compounds
Z = No. of formula units in one unit cell, e.g., 4 for NaCl and ZnS, 1 for CsCl, etc.
M = Formula mass (molecular mass) of the compound
a = Edge which is 2 x Distance between Na"^ and Cl" in case of NaCl
ZxM
{Hi) In SI units, p = where M = molar mass in kg mol"', a - edge in metres, p = density in kg m"^
a^xN 0

Remember. For BCC structure of an element (like alkali metals), Z = 2

For BCC structure of ionic compounds (like CsCl), Z = I.
1/36 New Course Chemistry (XIl)KSsIS]

iTYPE I. Calculation of density of a cubic crystal from the edge/nearest neighbour distance/radlus
Problem
D Silver forms ccp lattice and X-ray studies of its crystals show that the edge length
of its unit celi is 408-6 pm. Calculate the density of silver (Atomic mass = 107-9 u).
ZxM
Solution, p =
a^xN 0

For ccp lattice (which is equivalent to fee lattice), Z = 4 atoms/unit cell.

-1
Also we are given that M = 107-9 g mol
-1
-12 4atomsxl07-9gmol - 10-5 g cm ^
a = 408-6 pm = 408-6 x 10 cm
" ^ ” (408-6xl0“'2cm)3(6-022xl023atomsmor^)

w
Problem 0 Sodium has a bcc structure with nearest neighbour distance 365-9 pm. Calculate its
density (Atomic mass of sodium = 23).
S

Flo
distance (^0 is related to the edge (a) d = -;^
a
Solution. For the bcc structure, nearest neighbour

e
■^d

re
or a — — X 365-9 = 422-5 pm
V3 1-732

F
For bcc structure, Z = 2
-1
ZxM 2x23gmol
ur
r
For sodium. M = 23 - 1-51 g/cm^
(422-5xl0-‘‘’cm)3x(6.02xl023mol-’)
fo
0 Gold (atomic mass 197 u, atomic radius = 0-144 nm) crystallizes in a face centred unit
ks
Problem
Yo
cell. Determine the density of gold. (Assam Board 2013}
oo

= 2-yflr = 2 X 1-414 X 0-144 nm = 0-407 nm = 0-407 x 10

a
cm
eB

Solution. For/cc- unit cell, r = or a

2V2
-1
ZxM 4xl97gmol
= 19-4 g cm ^
ur

P =
^xN 0 (0-407 X lO-"^ cm)3 x (6-02 x 10^3 mor‘)
ad

a
Yo

Problem
E3 Gold has a close packed structure which can be reviewed as spheres occupying 0-74 of
the total volume. If the density of gold is 19-3 g/cc, calculate the apparent radius of a gold atom in the solid
d
Re

(Au = 197 amu).

in

Solution. As the packing fraction is 0-74, packing can be hep or fee. Knowing that gold has fee lattice, Z
F

= 4

ZxM 4x197
19-3 = or a = 4-07 X 10“^ cm
^ X N0 6-023x10-^

a 4-07x10"'
For fee. = 1-439 X 10“® cm
^ 2V2 2x1-414
TYPE II. , Calculation of edge/interionic distances/radius from density
Problem 0 CsCl has cubic structure. Its density is 3-99 g cm-^. What is the distance between Cs^
and Cl" ions ? (At. mass of Cs = 133)
Solution. CsCl has BCC structure. It has one formula unit in the unit cell. So Z = 1.
-1
ZxM 3 _
ZxM 1X (133 +35-5) g mol = 70-15 X lO-^*^ cm^
or a
^ a^xN 0 pxNo 3-99 gcm"^x 6-02 xlO^^ mol '
 SOLID STATE 1/37

a = (70-15)'/^ X 10-* cm^ = (70-15)^'^ x 10^ pm (1 pm - 10 cm)

- 4124 X 10^ pm = 412-4 pm
For BCC structure,
a 1-732
Interionic distance (</) = ^2 =^^x412-4
2 = 357pm
^

[To solve (70-15)‘^^ put j; = (70-15)'^^ Then log.r = -los70-15 = -x 1.8460 =0-6153
3 ^ 3
X = Antilog 0-6153 = 4-124].
Problem
B The density of aluminium is 2700 kg m“^. Aluminium crystallises in face-centred
cubic lattice. Calculate the radius of aluminium atom in metres (Atomic mass of A1 = 27)

ow
ZxM
Solution, p = or a
3 _
ZxM 4x0-027 kg mol-'
a^xN 0
pxNjj 2700 kg m“^x6-02x10'^’ mol”*
= 66-4 X 10"^*^ m^ or a = (66-4)*^ x 10“'® m

Putting (66-4)'^^ = .r, log x = -1 ,log 66-4

,,, 1-8222 = 0-6074

e
= x = 4-050 a = 4-05 X 10-'® m
3 ^ 3

Fl
re
a 4-05xl0-'®m

F
For fee, = 1-43 xl(M®
' 2V2"
m
2x1.414
ur
or
Problem
Q The edge length of a unit cell of a metal having molecular mass 75 g/mol is 5 A which
crystallizes in a cubic lattice. If the density is 2 g/cc, then find the radius of the metal atom.
sf (CBSE Sample Paper 2017)
k
Yo
ZxM
PXa^xNq _ 2gcm-^x(5xl0-^cm)^x6Q2xl0^^mol-'
oo

Solution. P - or = 2
M -1
0 75 g mol
B

This shows that the metal has body-centred cubic lattice. For BCC lattice,
re

1-732
r = a = x5A =2-165 A
4 4
u
ad
Yo

TYPE III. Calculation of Avogadro’s Number

Problem Q] Calculate the value of Avogadro’s number from the following data :
d
Re

Density of NaCl = 2-165 g cm“^ Distance between Na+ and Cl" in NaCI = 281 pm. (Pb. Board 2011)
in

Solution. A unit cell of NaCl contains 4 NaCl units, therefore, Z = 4,

F

M = 23 35-5 = 58-5 g mol"', p = 2-165 g cm“^ (Given)

As distance between Na+ and Cl" = 281 pm Edge of the unit cell = 2 x 281 = 562 pm
(Edge is the distance from Na"^ to Na'*'ion or Cl" to Cl" ion. See Fig. 1.33a, page 1/26)
ZxM
Substituting these values in the expression, p =
xN 0

4x58-5g mol"*
2-165 gem"^ =
(562xlO"'®cm)3xN 0 or No = 6-09 X 10^3 mol-^
Problem Q The density of KCI is l-9893gcm"^and the length ofthe side ofthe unit cell is 6-29082 A
as determined by X-ray diffraction. Calculate the value of Avogadro’s number.
Solution. KCI has face-centred cubic structure. Hence, no. of atoms per unit cell, (Z) = 4
Edge of the unit cell (a) = 6-29082 x 10“^ cm

ZxM 4x74-5
P = 1-9893 = or N« = 6-017x10^
a^xN 0 (6-29082X 10-8)3 xN 0
1/38 ‘Pnadecfi-’a, New Course Chemistry (XII)GZslSl

TYPE IV. Calculation of atomic mass and no. of atoms in a given mass

Problem HD An element has a body-centred cubic (bcc) structure with cell edge of 288 pm. The
density of the element is 7‘2 g/cm^. How many atoms are present in 208 g of the element ?
Solution. For the BCC structure, Z = 2
Edge of the unit cell, a = 288 pm. Density of the element, p = 7-2 g/cm^
ZxM
Substituting the values in the expression p =
a^xN 0

ow
2xM I
7-2gcm ^ = or M = 51-8 g mol
(288xl0->0cm)'-^x(602xl023mol-*)
By mole concept, 51-8 g of the element contains = 6-02 x 10^^ atoms
6-02 X 10^3

e
208 g of the element contains = x 208 atoms = 24-17 x 10^^ atoms

re
rFl
51-8

F
Problei I m X-ray diffraction studies show that copper crystallizes in an fee unit cell with cell
edge of 3-608 x 10“* cm . In a separate experiment, copper is determined to have a density of 8-92 g/cm^.

r
Calculate the atomic mass of copper.
ou
pxij^ xNq fo
ZxM
ks
Solution. or M =
Z
0
oo

For fee lattice, Z = 4.

Y
eB

Hence, M =
(8.92g cm“^) (3-608 x 10"^ cm)^ (6 022 x 10-^ atoms mol ’) = 631 g mol -1
4 atoms
r

Atomic mass of copper = 63-1 g mol ^

ou
ad
Y

Problem
iTj An element crystallizes into a structure which may be described by a cubic type of
unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the
d

volume of this unit cell is 24 x 10"^ cm^ and density of element is 7-2 g cm"^, calculate the number of atoms
Re
in

present in 200 g of the element.

F

1
Solution. No. of atoms per unit cell = 8x--t-2 = 3, i.e.. Z = 3
^ 8
Volume of the unit cell = = 24 x 10"“^ cm 3
ZxM

3xM
7-2 = or M = 34-69
(24 xl0"2‘^)x (6023x10^3)
Thus. 34-69 g of the element have atoms = 6-023 x 10^^
23
6-023x10
200 g of the element will have atoms = X 200 = 3-4722 x 10^ atoms
34-69
SOLID STATE 1/39

TYPE V.
Calculation of no. of atoms per unit ceM/type of lattice (Simple/BCC/FCC).
Problem [|§] Density of LI atom is 0*53 g/cm*\ The edge length of Li is 3*5 A. Find out the number
of Li atoms in a unit cell (Nq = 6*023 x 10^, M = 6*94)
ZxM
Solution. The aim is to find Z in the formula p =
0

pXfl^xN^) _ 0-53 g cm~^ x (3-5 x I Q-^cm)^ x (6-023 x 1 q3 mor') = 1-97 = 2.

M -1
6-94 g mol
Problem
EB The density of KBr is 2*75 g cni“^. The length of edge of the unit cell is 654 pm. Predict
the type of cubic lattice to which unit ceU of KBr belongs. (N„ = 6*023 x mol"*, At. mass: K = 39, Br = 80)
ZxM
Solution. For cubic crystals, p =

w
xN
0

F lo
or Z = ~
pXd^xN 0 _ (2-75gcm--^)(654xlQ-'^cm)3 x (6-023x 10-^ moi-^)
= 3-89 = 4
M (39 + 80) g mol '

e
Thus, there are four formula units of KBr present per unit cell. Hence, it has face-centred cubic lattice

Fre
(similar to that of NaCl).
Problem [g The density of copper metal is 8*95 g cm"^. If the radius of copper atom is 127*8 pm,
Cu = 63*54 g mor* and = 6*02 x 10^ mol"*)
for
is the copper unit cell a simple cubic, a body-centred cubic or a face-centred cubic ? (Given At. mass of
(CBSE 2010)
r
You
Solution. If copper atom were simple cubic,
s
ook

a = 2 r = 2 X 127-8 pm = 255-6 pm
eB

ZxM -I
1X 63-54 g mol
Z = 1 ; P = = 6-34 g cm“^
X N
0 (255-6x10 *®cm)^x(6-02x10^^ mol"*)
our
ad

Actual p = 8-95 g cm"^.

Hence, copper atom is not simple cubic.
If copper atom were body-centred,
dY
Re

4r _ 4x127-8
Fin

a =
pm = 295-15 pm
Vs ~ 1-732
ZxM 2 X 63-54 g mol *
Z=2; p=
fl^xN 0 -10 = 8-21 gcm-3
(295-15x10 cm)x6-02xl0^^ mor'
Hence, copper atom is not body-centred.
If copper atom were face-centred
a = 2-V2 r = 2 X I -414 X 127-8 pm = 361-4 pm
ZxM -1
4x63-54g mol
Z=4; p= = 8-94 g cm ^
X N
0 (36l-4x 10"**^ cm)^ x 6-02x 10^3 mol"'
Hence, copper is face-centred cubic.
Note that as radius is given, edge depends upon the nature of the unit cell.
1/40 '4. New Course Chemistry (XII)

I TYPE l.| Calculation of density from the edge/interionic distance/radius

1. Gold crystallizes in a face-centred cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate
the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu.
2. Niobium crystallises in body-centred cubic structure. If the atomic radius is 143-1 pm, calculate the density
of Niobium (Atomic mass = 93 u). (CBSE Sample Paper 2018)
3. The effective radius of an iron atom is 1-42 A. It has rock salt like structure. Calculate its density (Fe = 56 amu).
4. The compound CuCl has ZnS structure and the edge length of its unit cell is 500 pm. Calculate its density
(Atomic masses : Cu = 63, Cl = 35-5. Avogadro’s constant = 6-02 x 10^^ mol'*).

low
5. Copper crystal has a face-centred cubic lattice structure. Atomic radius of copper atom is 128 pm. Calculate
the density of copper. Atomic mass of copper = 63-5. (CBSE 2012)
6. Calculate the density of silver which crystallizes in a face-centred cubic structure. The distance between the
nearest silver atoms in this structure is 287 pm. (Molar mass of Ag = 107-87 g mol"', N^ = 6-02 x 10^^ mol"')
[TYPE 11.1 Calculation (>f edgc/interionic distance/radius from density

ee
density is 3-4 g cm"^. What is the length of the edge of
7. The compound CuCl has ZnS (cubic) structure. Its
rF
Fr
the unit cell ? (At masses Cu = 63-5, Cl = 35-5).
8. Lead (II) sulphide crystal has NaCl structure. What is the distance between Pb^"^ and S^“ in PbS if its density
is 12-7 g cm”^ ? (At. mass of Pb = 207).

r
[TYPE IllT] Calculation of Avogadro’s number
fo
u
9. Iron has a body centred cubic unit cell with a cell edge of 286-65 pm. The density of iron is 7-87 g cm"^. Use
ks
this information to calculate Avogadro’s number (At. mass of Fe = 56 g mol"')
Yo
(CBSE 2009, 2012)
oo

10. The well known mineral fluorite is chemically calcium fluoride. It is well known that in one unit of this
mineral, there are 4 Ca^'*’ ions and 8 F^ ions, and that Ca^^ ions are arranged in a fee lattice. The F" ions fill
B

all the tetrahedral holes in the face-centred cubic lattice of Ca^'*’ ions. The edge of the unit cell is
re

5-46 X 10"® cm in length. The density of the solid is 3-18 g cm"^. Use this information to calculate Avogadro’s
number (Molar mass of CaF2 = 78-08 g mol"'). (CBSE 2010)
u
ad

[TYPE iVTl Calculation of atomic mass/molar mass and no. of atoms in a given mass
Yo

11. An element with density 11-2 g cm"^ forms a f.c.c. lattice with edge length of 4 x 10"^ cm. Calculate the
atomic mass of the element (Given : N^ = 6-022 x 10^^ mol"') (CBSE 2014)
nd
Re

12. An element crystallizes in f.c.c. lattice with a cell edge of 300 pm. The density of the element is 10-8 g cm"^.
Calculate the number of atoms in 108 g of the element. (CBSE Sample Paper 2019)
Fi

I TYPE V. I Calcuiation of number of atoms per unit cell/type of lattice

13. Sodium crystallises in the cubic lattice and the edge of the unit cell is 430 pm. Calculate the number of atoms in
a unit cell. [Atomic mass of Na = 23-0 amu. Density of sodium = 0-9623 g cm"^, N^ = 6-023 x 10^^ mol"']
14. Thallium chloride (TlCl) crystallizes in a cubic lattice whose edge length is found to be 385 pm. If the
density of the solid is found to be 7-0 g cm"^, predict the type of lattice to which the crystals of TlCl belong.
(Atomic mass of T1 = 204, Cl = 35-5)
15. Iron (II) oxide has a cubic structure and each unit cell has side 5 A. If the density of the oxide is 4 g cm"\
calculate the number of Fe^"*" and 0“~ ions present in each unit cell (Molar mass of FeO -12 % mol"',
N^=6-02x 10-3 mol"')
16. An element has atomic mass 93 g mol"' and density 11 -5 g cm"3. If the edge length of its unit cell is 300 pm,
identify the type of unit cell. (CBSE 2017)

Miscellaneous

17. An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this
element contains 2 x 10^ atoms. (CBSE 2016)
SOLID STATE 1/41

18. A crystal of Lead (II) sulphide has NaCl structure. In this crystal the shortest distance between the Pb"'*' ion
and ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide ? Also calculate the
unit cell volume.
19. The unit cube length for LiCl (NaCl structure) is 5-14 A. Assuming anion-anion contact, calculate the ionic
radius for chloride ion.
20. A compound AB crystallises in bcc lattice with the unit cell edge length of 380 pm. Calculate
(/) the distance between oppositely charged ions in the lattice
(//) radius of B" if the radius of A'*' is 190 pm.
21. A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of die metal is 5-96 g cnr\
Find the volume of the unit cell (N^= 6-023 x 10^^ atoms mol"^).
22. Aluminium metal forms a cubic close-packed crystal structure. Its atomic radius is 125 x lO"'" m. (a) Calculate
the length of the side of the unit cell. (/?) How many such unit cells are there in I-OO m^ of aluminium ?
23. A uni-univalent ionic crystal AX is composed of the following radii (arbitrary units):

w
A-^ X-
1-0 2-0

Assuming that ions are hard spheres, predict giving reasons whether the crystal will have .sodium chloride or

F lo
cesium chloride structure. Calculate the volume of the unit cell. (Manipur Board 2011)
24. An element ‘X’ (At. mass = 40 g mol”^) having/cc structure, has unit cell edge length of 400 pm. Calculate
4 g of ‘X’ (N^ = 6-022 x 10^^ mol”'). (CBSE 2018)

ee
the density of ‘X’ and the number of unit cells in

Fr
ANSWERS

1. 19-4 g/cm^, 143-9 pm 2. 8-6 g cm”^ 3. 5-74 g cm"^ 4. 5-24 g/cm^

5. 8-9 g cm"^ 6, 10-71 g cm~^ 7. 578 pm
for 8. 250 pm
ur
9. 6-04 X 10-3 10. 6-033 X 10^3 11. 107-9 g mol"' 12. 14-8 X 10^^ atoms
13. 2, U., BCC 14. BCC 15. Z = 4, i.e., 4 Fe^'*’ and 4 ion
s
ook

18. a =5-94 X 10"® cm, V = =2-096xlO-^2 ^.,^3

Yo
16. BCC 17. 38-4 g cm-3
19. 1-81 A 20. (0 329 pm (it) 139 pm 21. 2-786 x 10-23
eB

22. (a) 354 pm (/j) 2-25 x 10^^ 23. NaCl, 216 (au)^ 24.4-15gcm-3, 1-505x 1022
r

HINTS FOR DIFFICULT PROBLEMS

ad
ou

a
1. For fee lattice, Z = 4. Radius r - = 0-35350.
Y

2V2
Re
nd

s 4r 4x143-1
2. For bcc, r =— a or o = pm = 330-5 pm
4 V3~ 1-732
Fi

ZxM 2x93
= 8-6 g cm 3
“ o3 X N 0 (330-5 X10-10)3 X (6-02 x 1023)
a r 142 pm
3. Rock salt type structure means fee for which r = = 0-3535 o or a = 401-7 pm.
2V2 0-3535 0-3535
For fee, Z = 4.
a
5. For fee, r = or a = l-yjlr = 2 X 1-414 X 128 pm = 362 pm
2V2
ZxM -1
4 X 63-5 g mol
^ " fl3 X N 0 = 8- 9 g cm"3
(362xI0-l0cm)3x6-02xl023moI *
Mass of unit cell 4 X (63-5 g mol"' / 6-02 xip23 mol l)
Alternatively, p = = 8-9 g cm"3
Volume of unit cell (362xl0-'0cm)3
1/42
New Course Chemistry (XlI)BZsISl

6. ForFCC, or a= = 1414 x 287pm = 406 pm. Z = 4, M = 107-87 g mol

-\

V2
ZxM
P =
4x107 ●87gmol-'
0 (406x10 x(6-02xl0^^mol”*) = 10-71 g cm ^
8. Distance between Pb"'*' and S“ ions = — (as it has NaCl structure).
2
9. For BCC unit cell of tlie element Fe, Z = 2
ZxM -I
N 2 X 56 g mol
0 ~ = 6-04 X 10^^ mol-’
xp (286-65xl0"“^cm)3 x(7-87gcin"^)
ZxM

w
11. P= . For element wiih/c.c. lattice, Z = 4

M = pXfl^xN^ _ (1 l-2gcm~^)(4x 10~^cm)^x(6-022xIQ-^mol”')

Flo
-1
Z
= 107-9 g mol
4
12. Calculate the atomic mass, M. Then M grams contain 6-023 x 10^^ atoms. Calculate atoms present in 108

e
g.
14. Calculate Z. It comes out to be 1. This shows that a unit cell of TlCl has one formula unit of TlCl. Hence, it

re
has body-centred cubic lattice (similar to that of CsCl).

F
16. p =
ZxM
pxa^xNf) _ (ll-5gcm”^) (300x10 10
cm)^ (6-023x10^3 mol”') = 2-01 = 2
ur
r
xN M -1
0 93 g mol
Hence, the type of unit cell is BCC.
fo
ks
300
17. Molar mass of the element (M) = X 6-02x10^3 = 90-3 g mol”*
Yo
2x102**
oo

ZxM 4x90-3
Density (p) = - 38-4 g cm 3
B

«3xN 0 (250xl0“‘°)3x(6-02xl0^3)
re

18. Edge 'a' of unit cell = 2 x Distance between Pb^"^ and S^” ions.
5-14
19. Interionic distance of LiCl = = 2-57 A
u
ad

2
Yo

BC = Vab2 + AC2 =^2-51)^ + (2-57)2 ^3.53

d

Radius of Cl" ion = —x 3-63 A = 1-81A .

Re
in

2
3^
20. (/) For bcc. distance between A'*' and = — a,x.e., it is halfofbody diagonal (See Fig. 1.40 on page 1/33)
F

(/V) As the cations and the anions touch each other, r ^ -f r = 329 pm =329-190= I39pm .
A+ 1
R-
21. Mass of the unit cell = Mass of one atom x No. of atoms present in the unit cell
\
50
g x2 [For BCC, Z = 2]
,6-023x1023
Mass 50x2 1
Volume of the unit cell = gx = 2-786 X 10-23 cm3
Density 6-023x1023 5-96 gem ”3
a
22. cep = fee. For fee. or a = 2>^r =2x I4I4X 125x 10-2 m
- 354 X 10 ‘2 m = 354 pm
2■^/2
Volume of one unit cell = (354 x 10-’2 ni)3 = 4-436 x 10”2^ m 3

1
.●. No. of unit cells in 1 m-3 = = 2-25 X 1()28.
4-436x10-29
SOLID STATE 1/43

23. rjr_ = 2/1 = 0-5 which lies in the range 0414 - 0-732. Hence, it has sodium chloride stnaciure. (See Table
1.5, page 1/31)
Edge = 2 + /●_) = 6 cm. Volume = (6 aii)^ =216 {auf
24. Given a = 400 pm. For/cc, Z = 4
ZxM 4x40
= 4-15 g cm ^
6-022xlO-3x{400xl0-’<^)^
40
Mass of one unit cell = x4g
23
6-022x10

Mass of the element 4

No. of unit cells in 4 g of the element = = 1-505x10-2
Mass of one unit cell (40x4)/(6-022x1023)

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1.19. IMPERFECTIONS OR DEFECTS IN SOLIDS

As discussed in the beginning of this unit, in a crystalline solid, there is a regular arrangement of constituent

F lo
pai'ticles. However, this arrangement is generally not found to be perfect. This is because a solid contains a
large number of small crystals and some of these may not have a perfect regular arrangement of the constituent

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particles. This happens when crystallisation takes place at fast or moderate rate because the constituent particles

Fr
may not get sufficient time to arrange themselves in a perfect order. Even if we prepare single crystals by
carrying out the crystallisation at extremely slow rate, these single crystals may not have a perfect arrangement.

for
Any departure from perfectly ordered arrangement of constituent particles in crystal is called
ur
imperfection or defect.
The defects may also arise due to the heat absorbed by the crystals from the surroundings* or due to the
ks
presence of impurities in the ciystals.
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oo

Broadly speaking, there are two types of defects called point defects and line defects.
eB

When the deviations or irregularities exist from the ideal arrangement around a point or an atom in
a crystalline substance, the defect is called point defect However, when the deviation from the ideal
arrangement exists in the entire row of lattice points, the defect is called line defect.
r
ou
ad

It is not possible to explain many properties of solids such as the electrical conductivity and mechanical
strength in terms of structure alone. Imperfections not only modify the properties but also sometimes impart
Y

new properties to the solids.

nd
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Here, we shall discuss only the point defects.

Types of Point Defects. Point defects in a crystal may be classified into the following three types:
Fi

(A) Stoichiometric defects (B) Non-stoichiometric defects (C) Impurity defects

Now, we shall discuss each of these defects one by one.
(A) Stoichiometric Defects. If impeifections in the crystal are such that the ratio between the cations
and anions remains the same as represented by the molecular formula, i.e., stoichiometry of the solid is not
disturbed, the defects are called stoichiometric defects. FIGURE 1.42,

These are also called intrinsic defects or thermodynamic defects. Vacant lattice sites

Basically, these defects are of the following two types;

(/) Vacancy defect. When in a crystalline substance, some of the lattice
sites are vacant, the crystal is said to have vacancy defect (Fig. 1.42). This
generally arises due to absorption of heat from the surroundings (That is why
4 y
it is a thermodynamic defect). It results in the decrease in the density of the
substance (because mass in the same volume of the crystal decreases). Vacancy defect

*There is a perfect ordered arrangement of constituent particles in a crystalline substance only at 0 K (Tliird law
of thermodynamics).
1/44
New Course Chemistry (X1I)B&IS1

(«) Interstitial defect. VJhen some extra constituent particlesare present FIGURE 1.43
in the interstitial sites, the crystal is said to have interstitial defect (Fig. 1.43). Particles in the interstitial sites
This defect results in the increase in the density of the substance (because
mass increases but volume remains the same).
The above two types of defects are generally shown by non-ionic solids.
Ionic solids do not show simple vacancy and interstitial defects. Instead, they
show these defects as Schottky and Frenkel defects, as explained below : 0^0—0—0—o
{Hi) Schottky defect. If in an ionic crystal of the type A*B~, Interstitial defect
equal number of cations and anions are missing from their lattice
FIGURE 1T44^
sites so that the electrical neutrality is maintained, it is called
Schottky defect. The Schottky defect containing one pair of holes
due to missing of one cation and one anion is shown in Fig. 1.44.

w
Thus, basically, it is a vacancy defect.
IVpes of compounds exhibiting Schottky defects. This type
©—0—0—0—0
of defect is shown by highly ionic compounds which have
0—©—©—0—©

F lo
(0 high coordination number, and
(//) small difference in the size of cations and anions
©—©—©—©—©

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A few examples of ionic compounds exhibiting Schottky defect

Fr
Schottky defect
are NaCl, KCI, KBr, AgBr and CsCl.
Effect on density. As the number of ions decreases as a result of this defect, the mass decreases whereas

for
ur
the volume remains the same. Hence, like vacancy defect, the density of the solid decreases.
It is observed that in NaCl crystal at room temperature, there are about 10^^ ions and 10^ Schottky pairs
per cm^. This means that there is one Schottky pair defect per 1 O’® ions. Thus, there are large number of holes
s
ok
present in the crystal. This results in the decrease in the density.
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(/V) Frenkel defect If an ion is missing from its lattice site (causing a vacancy or a hole there) and it
o

occupies the interstitial site, electrical neutrality as well as the stoichiometry of the compound are maintained.
eB

This type of defect is called Frenkel defect. Thus, this defect is a combination of vacancy defect and interstitial
defect. Since cations are usually smaller, it is more common to find the cations occupying interstitial sites.
r

This defect is also called dislocation defect* because smaller ion (usually cation) is dislocated from its
ou
ad

normal site to an interstitial site.

FIGURE 1.^
The Frenkel defect in which one cation is missing from the lattice
Y

site and is occupying the interstitial site is shown in Fig. 1.45.

nd
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lypes of compounds exhibiting Frenkel defects. This type of

defect is present in those compounds which have ©—0—0—0—©
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(0 low coordination number and

(//) large difference in the size of cations and anions ©—®—©—©—©
Frenkel defects are found in silver halides (AgCl, AgBr and Agl)
due to small size of Ag"*" ion. It is also shown by ZnS due to small size
of Zn^"^ ion. They are not found in alkali metal halides as the alkali
metal ions cannot fit into the interstitial sites. Frenkel defect

There are some ionic solids which show both Schottky and Frenkel defects, e.g., AgBr.
*The line defects are also called dislocation.s.
The two main line defects are called edge
dislocations and screw dislocations. If the edge of
an atomic plane terminates within the crystal instead
of passing all the way through, it is called edge
dislocation. However, if parallel atomic planes
normal to the axis are converted into a kind of spiral ^3) Edge dislocation (b) Screw dislocation
(helical) ramp, it is called screw dislocation. (small cubes represent atoms or molecules)
SOLID STATE 1/45

Effect on density. Since no ions are missing from the crystal as a whole, therefore density of the solid
remains unchanged.
Some other consequences of Schottky and Frenkel defects.
(/) Solids having these defects conduct electricity to a .small extent. This is because if an ion moves
from its lattice site to occupy a ‘hole’, it creates a new ‘hole’. In this way, a hole moves across the crystal
which as a result moves the charge in the opposite direction.
(ii) Due to presence of holes, the stability (or the lattice energy) of the crystal decreases
{Hi) In Frenkel defect, similar charges come closer. This results in the increase of dielectric constant of
the crystals.
Difference between Schottky and Frenkel defects. The main points of difference are listed below :

w
TABLE 1.6. Main points of difference between Schotd^ defect and Frenkel defect

Schottky Defect Frenkel Defect

Flo
(0 It is due to equal number of cations and anions (0 It is due to the missing of ions (usually cations)
missing from the lattice sites. from the lattice sites and these occupy the

e
re
interstitial sites.

This results in the decrease in the density of It has no effect on the density of the crystal.

F
the crystal.
This type of defect is found in crystals with
ur
{Hi) This type of defect is found in highly ionic {Hi)

r
fo
compounds with high coordination number and low coordination number and in which the
having cations and anions of similar sizes, e.g., difference in the size of cations and anions is
ks
NaCl, CsCl etc. very large, e.g., silver halides.
Yo
oo

(B) Non-stoichiometric Defects*. I,f as a re.sult of the imperfections in the crystal, the ratio of the
cations to the anions becomes different from that indicated by the ideal chemical formula, the defects are
B

called non-stoichiometric defects.

e

These defects result in either excess of the metal atoms or excess of the non-metal atoms (or deficiency
ur

of the metal atoms). These can occur as follows :—

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(a) Metal excess. This may occur in either of the following two ways :—
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(/) By anion vacancies. A negative ion may be missing from its lattice site, leaving a hole which is
occupied by an electron, thereby maintaining the electrical balance (Fig. 1.46). The sites containing the
d
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electrons thus trapped in the anion vacancies are called F-centres because they are responsible for imparting
in

colour to the crystals (F=Farbe which is a German word for ‘colour’).

F

For example, when NaCl is heated in an atmosphere of Na vapour, FIGURE 1.^

the excess of Na atoms deposit on the surface of NaCl crystal. Cl" ions
then diffuse to the surface where they combine with the Na atoms which B' A* B'

become ionized by losing electrons. These electrons diffuse back into

the crystal and occupy the vacant sites created by the Cl“ ions. These
electrons absorb some energy of the white light, giving yellow colour to
NaCl. Similarly, excess of Li in LiCl makes it pink and excess of K in
KCl makes it violet.

This defect is similar to Schottky defect and is found in crystals

having Schottky defects. Metal excess defect
{H) By the presence of extra cations in the interstitial sites. due to anion vacancy
Metal excess may also be caused by an extra cation occupying the
*Non-stoichiometric substances are sometimes called berthollides after Berthollet who studied such compounds.
 1/46
New Course Chemistry fXTTIrosTWi

intersiiiia! site. Electrical neutrality is maintained by an electron present FIGURE 1.47

in another interstitial site (Fig. 1.47). This defect is similar to Frenkel
defect and is found in crystals having Frenkel defects.
For example, when ZnO is heated, it loses oxygen and turns yellow
due to following reaction : e~

ZnO Zn2++-0-+2i>-
T 2
2

Tlie excess of Zn-'*' ions thus formed get trapped into the vacant
interstial sites and the electrons in the neighbouring interstitial sites. Metal excess defect caused by
extra cation in the interstitial site
Crystals with either type of metal excess defects contain some
tree electrons. Hence, such materials acts as semiconductors,
FIGURE 1.48i
(b) Metal Deficiency. This defect occurs when the metal

w
shows variable valency, i.e., in transition metals. The defect A* B' A B

usually occurs due to the missing of a cation from its lattice site

F lo
and the presence of the cation having higher charge (e.g., + 2 B' A*

instead of + I) in the adjacent lattice site. Examples include

FeO. FeS and NiO. A general case where a metal A shows A- B' A^i B'

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variable valency of + I and + 2 is shown in Fig. 1.48. Two

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monovalent cations are replaced by one divalent cation to B' A* B' A*

maintain electrical neutrality.

Due to metal deficiency, the compounds obtained are non- for Metal deficiency defect due to
missing of a cation of a lower
valency and another lattice site
ur
stoichiometric. For example, it is difficult to prepare ferrous
oxide with the ideal composition, FeO. What we actually obtain occupied by a cation of higher valency
s
ook

is FeQ.ggO or Fe^O with x = 0-93 to 0-96.

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(C) Impurity Defects, These defects arise when foreign atoms are present at the lattice site in place of
eB

host atoms (as shown in Figs. 1.49, 1.50 and 1.51) or at the vacant interstitial sites (e.g. in steel, which is
carbide of iron, C-atoms are present at the interstitial sites in the lattice of Fe atoms). In the former case, we
our

get substitutional solid solutions while in the latter case, we get interstitial solid solutions. The formation of
ad

the former depends upon the electronic structure of the impurity while that of the latter on the size of the
impurity.
dY

Addition of impurities changes the properties of the crystal.

Re

The process of adding impurities to a crystalline substance so as to change its properties is

Fin

called doping.

(/) Introducing impurity defect in ionic solids. In case FIGURE 1.491

of ionic solids, the impurities are introduced by adding impurity
of ions. If the impurity ions are in a different valence state from
that of the host ions, vacancies are created. For example, if molten
NaCl, containing a little SrCU as impurity is allowed to cool, in
the crystals of NaCl formed, at some lattice sites, Na'*’ ions are
substituted by Sr^"^ ion. For every Sr^+ ion thus introduced, two
Na'*' ions are removed to maintain electrical neutrality. One of
these lattice sites is occupied by Sr"'*' ion and the other remains
vacant (Fig. 1.49). These vacancies result in the higher electrical
conductivity of the solid. Similar defect and behaviour is Impurity defect introduced
observed when CdCU is added to AgCl. by subsituting Na"*" ions by Sr2+ ions
SOLID STATE 1/47

(m) Introducing impurity defects in covalent solids. In case of covalent solids such as silicon or
germanium (Group 14 elements) which have 4 valence electrons, the impurities added may be of the elements
which may have more than 4 valence electrons (e.g.. Group 15 elements like P or As which have five valence
electrons) or of the elements which have less than 4 valence electrons {e.g.. Group 13 elements like B. Al or
Ga which have 3 electrons in the valence shell). Thus, the impurities added may be electron rich or electron
deficit. The defects thus introduced in the crystals are called electronic defects*. Each of these is briefiy
described below :

(a) Doping with electron rich impurities. Group 14 element like silicon or germanium has 4 electrons
in the valence shell. Hence, it normally forms four covalent bonds with the neighbouring atoms
(Fig. 1.50a). When it is doped with Group 15 element like P or As, the silicon or germanium atoms at some
lattice sites are substituted by atoms of P or As. Now, as these atoms have 5 electrons in the valence shell,
after forming normal four covalent bonds with the neighbouring silicon atoms, the fifth extra electron is free
and gets delocalized (Fig. 1.50b). These delocalized electrons increase the conductivity of silicon or germanium

w
(Fig. 1.50c). As the increase in conductivity is due to negatively charged electrons, the silicon or germanium
crystals doped with electron rich impurities are called n-type semiconductors.

F lo
FIGURE 1.^
ELECTRON

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EXTRA MOBILE ELECTRON OF ARSENIC

Fr
: Si: Si: Si: Si: : Si: Si: si,,: si:

for
●● ●● ●● ●●
-O':
: Si: Si: Si : Si: ; Si ; Si ; As; Si j
ur
: Si: Si: Si : Si : : Si : Si : Si: Si:
s
ook
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(a) Perfect silicon crystal (b) Doping silicon with arsenic (c) n-type semiconductor
(Flow of electrons)
eB

Doping of silicon with Group 15 elements to produce n-type semiconductors

our

(b) Doping with electron deficit impurities. When Group 14 element like Si or Ge is doped with
ad

Group 13 element like B, Al or Ga, the Si or Ge atom at some lattice sites are substituted by those of B, Al or
Ga. Now, as Group 13 elements have only three valence electrons, they can form three covalent bonds
Y

with the neighbouring silicon atoms. Thus, a hole is created at the site where fourth electron is missing
Re

(Fig. 1.51 b). This is called electron hole or electron vacancy. An electron from the neighbouring atom can
nd

jump to fill up this electron hole but then an electron hole is created at the site from where electron has
Fi

Jumped. As it continues, the electron holes will move in a direction opposite to that of the flow of electrons.
Now, when an electric field is applied, the electrons move towards the positively charged plate and the
electron holes move towards the negatively charged plate as if they carry positive charge (Fig. 1.51 b).
Hence, silicon and germanium doped with electron-deficit impurities are called p-type semiconductors.

^Electronic defects or imperfections may also be present in pure covalent or ionic crystals. This is because in
these crystals, the electrons occupy fully the lowest energy states only at 0 K. For example, in the covalent crystal of
silicon or germanium at 0 K, all the electrons occupy the position of covalent bonds and in an ionic crystal like NaCl,
the electrons are mostly concentrated around the more electronegative element. That is why pure silicon or NaCI at
0 K are bad conductors of electricity. Above 0 K, some of the electrons are likely to occupy higher energy states. For
example, in silicon or germanium, some of the electrons from the covalent bonds get thermally released, creating an
electron deficient site called a ‘hole’. The released electrons are free to move and hence pure silicon above 0 K becomes

a semiconductor. The holes also impart electrical conductivity but the holes in an electric field move in a direction
opposite to that of electrons.
 1/48
“Pn4tdee^'4i. New Course Chemistry (XII)BZSZ9]

FIGURE 1.51
ELECTRON SHIFTS RIGHT TO OCCUPY
ELECTRON HOLE HOLE
HOLE CREATING A HOLE HERE

o
I Si I Si I Si ^ Si * : Si : sUi^i : si :
● ● ft ft ft
+

I Si I Si ^ Si ^ Si * : Si ; Si : Ai : Si :
: Si : Si : Si : si : : Si : Si : Si ; Si :
O O
(a) Perfect silicon crystal (b) Doping (c) p-type semiconductor
silicon with alumimium (Flow of holes like positive charge)
Doping of silicon with group 13 elements to produce p-type
semiconductors with flow of holes like positive charge

w
Calculation of cation vacancies. A metal ion with higher valency replaces more than one metal ion with
lower valency, thereby creating vacancies. Depending upon their charges, number of vacancies can be
calculated.

Flo
Sample Problem Q If NaCl is doped with 10 ^ mol % SrCI 2’ what is the concentration of cation

e
vacancies ?

re
Solution. Doping of NaCl with 10"^ mol% SrCl2 means that 100 moles of NaCl are doped with mol of

rF
StCU.

10-3
ur
mole = 10 ^ mole
1 mole of NaCl is doped with SrCU =

As each Sr^^ ion introduces one cation vacancy, therefore,

100
fo
ks
concentration of cation vacancies
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= ] mol/mol of NaCl
oo

= 10-3 X 6 02 X 10^3 mol-'

B

= 6-02 X 10^® moI-3

re

Sample Problem If Al^^ replaces Na* at the edge centre of NaCl lattice then calculate the
vacancies in I mole NaCU (AIPMT Mains 2008)
u
ad

Solution. 1 mole of NaCl contains 1 mole of Na"* ions, i.e., 6-023 x 10^3 ions.
Yo

NaCl has/c'c arrangement of CC ions and Na^ ions are present at the edge centres and body-centre.As there
1
nd

are 12 edge centres and each edge centre is shared by 4 unit cells, their contribution per unit cell = —x 12 = 3.
Re

4
Fi

Contribution of Na"*" ion at the body-centre = 1.

Thu.s, for every 4 Na'*' ions, the ions present at the edge centres = 3. This means that Na'*' ions which have

been replaced =-x6-023xlO^^ =4-517 x 10^3

4

1 A13+ ion will replace 3 Na'*' ions to maintain electrical neutrality. One vacancy will be occupied by AP+ ion
i
and the remaining 2 will be vacant.
1 2
This means that -rd of these positions will be occupied by AP^ ions and — rd will remain vacant. Hence,
no. of vacancies in 1 mole of NaCl =-^x4-517xl033 =3*01x10^
3

Calculation of percentage of metal ions with variable valency from the formula of a non-
stoichiometric compound. These calculations can be done keeping in mind that the total positive charge on
cations must be equal to total negative charge on anions.
SOLID STATE 1/49

Sample Problem The composition of a sample of wustite is Feo-93®i oo* ''^hat percentage of the
iron is present in the form of Fe (III) ?
Solution. The composition is Feo93 O[.oo instead of FeO because some Fe^'^ ions have been replaced by
Fe^^ ions. Let us first calculate the numl^r of Fe^* and Fe^'*' ions present. The formula Fe 0-93 ^100 implies that
93 Fe atoms are combined with 100 0-atoms. Out of 93 Fe atoms, suppose Fe atoms present as Fe^* = x. Then
Fe^'*’ = 93 -X. As the compound is neutral, total charge on Fe^'*’ and Fe^'*’ ions = total charge on O^" ions. Thus,
3xx + 2(93-j) = 2x 100 or 3a: + 186 - 2jc = 200 or x=14, i.e.. Fe^*= 14.
Hence, Fe2-^=93- 14 = 79.
Thus, out of 93 Fe atoms, Fe present as Fe^'*' = 14
14
% age of Fe present as Fe (III) = —x 100 = 15%
93

w
F lo
Analysis shows that a metal oxide has the empirical formula of M 096^]'00- Calculate the percentage of M^'*’
and M^'*' ions in this crystal. {Pb. Board 2011)

ee
ANSWER

Fr
m2+ = 91-7%, M3+ = 8-3%

1.20. ELECTRICAL PROPERTIES OF SOLIDS

for
ur
1.20.1. Classification of solids on th} basis of conductivity
s
Different types of solids show electrical conductivity which varies from as low as 10“^^ohm”^ m“* to
ook
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10^ ohm"* m”*, i.e., it extends over 27 orders of magnitude. Depending upon the difference in electrical
eB

conductivities, the solids are classified into the following three types:
(i) Conductors. The solids which have conductivities in the range 10"^ to 10^ ohm"* m"* are called
conductors. Metals have conductivities of the order of 10^ ohm"* m"’ and hence are the best conductors of
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ad

electricity.
They are further classified as metallic conductors and electrolytic conductors. In the metallic conductors,
Y

the flow of electricity is due to flow of electrons (electronic conductors) without any chemical change occurring
Re

in the metal. The conductivity of metals depends upon the numberofvalence electronsavailableper atom. In
nd

case of electrolytic conductors like NaCl, KCl, etc., the flow of electricity takes place to a good extent only
Fi

when they are taken in the molten state or in aqueous solution. The flow of electricity is due to flow of ions.
However, in the solid state, they conduct electricity only to a small extent which is due to the presence of
defects (holes, electrons, etc.).
Thus, whereas metals conduct electricity in the solid as well as molten state, electrolytes conduct electricity
only in aqueous solution or molten state.
-20
(ii) Insulators. The solids which have extremely low conductivity, i.e., ranging between 10 to

I0"*0ohm"*m-1‘ are called insulators, e.g., plastics, wood, rubber, sulphur, phosphorus etc.
(Hi) Semiconductors. The solids which have conductivity inbetweenthose of conductorsand insulators,
i.e., ranging from 10"^ to 10"* ohm"* m"* are called semiconductors.
To sum up : Range of conductivity (in ohm"* m"*) is
Conductors Semiconductors Insulators

lO'* to 10"^ 10"^ to 10^ 10-20- 10"*0

1/50
New Course Chemistry (XI1)ESSI9]

1.20.2. Band theory of metals

In case of a metal, the atomic orbitals of the metal atoms are so close in energy that they overlap to forni
a large number of molecular orbitals very close in energy. This set of molecular orbitals is called a band. For
example, the formation of a band by overlap of n 2.v' orbitals of n Li atoms is shown in Fig. 1.52. Remember
that just as two atomic orbitals combine to form two molecular orbitals, n atomic orbitals combine to form n
molecular orbitals.

FIGURE 1.^

ow
Empty orbitals
(Upper half)
J_ JljLJ 1
Overlap of n 2s^ orbitals
from n Li atoms
= * Filled orbitals
» (Lower half)

e
gj (each contg. 2 electrons)

re
rFl
n MO's (2s Energy band of MO's)
(Lifl molecule)

F
2s orbitals of Li atoms combining to form molecular orbitals equal to
the number of Li atoms (2s orbitals) combining together. (Li = Is^, 2s^)

r
ou
fo
ks
In case of lithium, the lower half of molecular orbitals are completely filled while the upper half is
empty. On applying electric current, electrons can move into the vacant levels. Hence, electric current can
oo
flow. However, in case of next metal, beryllium (486 = 1^- 2s^ 2p^), as 25 orbitals are filled, the band formed
from 25 orbitals will be completely filled. Hence, no conduction of electricity should occur. But actually Be
Y
eB

is better conductor than Li. This is explained by suggesting that not only 25 orbitals overlap to form band but
2p orbitals also overlap to form an empty band. The band formed by lower energy valence orbitals (e.g. 2s)
is called valence band and the band formed by slightly higher energy valence orbitals is called conduction
r
ou

band. On applying electric current, electrons can flow from valence band to conduction band. The formation
ad
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of valence band and conduction band in case of Be is shown below in Fig. 1.53.
d

FIGURE 1.531
Re
in

Empty,
F

ortitals

3n MOs
(3n Energy levels)

3n 2p Energy levels 2p Band

from n atoms (Conduction band)

Band overlap
✓

nMOs Fully
filled
{n Energy levels) .
n 2s Energy levels E ^ bitiitals
2s Band ►
from n atoms 11
(Valence band)

Overlap of 2s energy band (valence band) and 2p energy band

(conduction band) of Beryllium. (Be = Is^ 2s^ 2p®)
SOLID STATE 1/51

1.203. Explanation of behaviour of conductors,

insulators and semiconductors on the basis of band theory
If the valence band is partially filled {e.g.. in case of lithium) or it overlaps with the higher energy or
unoccupied conduction band (as in case of be^'llium) then electrons can easily flow under the influence of an
applied electric field. Hence, the metal conducts electricity (Fig. 1.54 a).
If the gap between the filled valence band and the unoccupied conduction band is large,
electrons cannot jump from valence band to conduction band. Hence, the substance has extremely low
conductivity and it behaves as an insulator (Fig. 1.54 b).
If the gap between the valence band and conduction band is small, some electrons may jump from
shows some conductivity and it acts as a semiconductor
valence band to conduction band. Hence, the substance
(Fig. 1.54 c)
FIGURE 1.54]

w
Empty
conduction
band
Empty

F lo
conduction
Empty
conduction - - band
>
band Large energy

T Small energy
0

ee
gap
a: -Empty gap

Fr
UJ valence "
z band
Filled Filled Filled
liJ -Filled valence valence valence
valence
band
band band

for band
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Partially Overlapping
filled valence bands (b) Insulator (c) Semiconductor
band (e.g., Li)
s
(e.g., Be)
ook
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\/

(a) Metals (Conductors)

eB

Distinction between metals, insulators

and semiconductors in terms of band theory
our
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1.20.4. Effect of temperature on electrical conductivity

Electrical conductivity of metals decreases with increase of temperature because on heating, the positive
ions of the metal atoms start vibrating and produce hindrance in the flow of eleemons. The electrical conductivity
Y

of semiconductors increases with increase in temperature because more electrons can jump to the conduction
Re
nd

band. Alternatively, the increase in the conductivity of semiconductors may be explained as given in the
Fi

footnote on page 1/47 i.e., in case of semiconductors like silicon or germanium atom, each atom is linked to
four other atoms by covalent bonds. At 0 K. all the electrons occupy the position of covalent bonds and there
are no free electrons. Hence, at 0 K, Si and Ge are insulators. Above 0 K, some covalent bonds break. As a
result, some electrons are released and holes are created there. Due to free electrons and holes. Si and Ge
show some electrical conductivity and hence become semiconductors. Substances like silicon and germanium
which show this type of behaviour are called intrinsic semiconductors. In fact, all pure substances that show
conductivity similar to that of silicon and germanium are called intrinsic semiconductors.
Increasing the conductivity of intrinsic semiconductors. The conductivity of intrinsic semiconductors
is so low that as such they have no practical use. The conductivity can be increased by doping, as already
discussed, either by adding electron rich impurities or electron deficit impurities. The semiconductors
thus obtained are called n-type and p-type semiconductors,as already discussed. Collectively, they are called
extrinsic semiconductors.

For example, electrical conductivity of silicon and germanium is very low at room temperature. It can
be increased by doping with elements of Group 15 or Group 13 (as already discussed).
 1/52
'Pn^:xdce^'<i' New Course Chemistry (xn)E!ZsZSI

Similar to combination of Group 14 elements with those of Group 15 or Group 13, semiconductors have
been prepared by combination of elements of Group 13 and 15 {e.g., InSb, AlP and GaAs) or Group 12 and
16 {e.g., ZnS, CdS, CdSe and HgTe).
To sum up : Semiconductors are of two types :
(/) Intrinsic semiconductors. These are substances like Si and Ge which are insulators at 0 K but above
0 K {e.g., at room temperature), they become semiconductors.
(//) Extrinsicsemiconductors.These are substancesobtained by doping substances like Si and Ge with
impurities so as to enhance their conductivity.

1.20.5. Applications of n-type and p-type semiconductors

n- and p-type semiconductors are combined suitably to form a number of electronic components, e.g.,
(0 A diode is a combination of p- and n-type semiconductors which is used as a rectifie,r
(ii) The solar cell is an efficient photo-diode used for conversion of light energy into electrical energy.

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(Hi) Transistors are sandwich semiconductors of the type pnp or npn which are used to detect or amplify

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radio or audio signals,
(/v) Gallium arsenide (GaAs) semiconductors have very fast response and have revolutionised the designs
of semiconductor devices.

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1.20.6. Conductivity' of transition metal oxides

F
It is interesting to point out that the variation in conductivity is very large even among similar compounds,
for
e.g., monoxides of transition metals, all of which possess NaCl structure,show such large variationsin electrical
properties. Similar variations are observed among other similar oxides. For example,
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(/) TiO is metallic, MnO, FeO, CuO etc. are insulators whereas VO is metallic or insulating depending
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upon temperature.
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(ii) C1O2 is metallic, Mn02 is insulator whereas VO2 is metallic or insulating depending upon temperature.
(Hi) Re03 is metallic whereas VO3 and Ti03 are metallic or insulating depending upon temperature.
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To sum up : (/) TiO, C1O2 Re03 are metallic.

(ii) MnO, FeO and CuO are insulators.

(Hi) VO, VO2, VO3 and Ti03 change from metallic to insulator at a certain temperature.
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It is interesting to mention that Rhenium oxide, ReO^, has the conducHvity as well as appearance
like that of copper.
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Similar variation is observed among the sulphides of transitionelements.

Curiosity Question
f Q. What are diodes and transistors ? For what purpose are they generaiiy used ?
Ans. A diode is a combination of p- and n-type semiconductors which is used as a rectifier. Transistors
are sandwich semiconductors of the type pnp or npn which are used to detect or amplify radio
or audio signals.
J
1.21. MAGNETIC PROPERTIES OF SOLIDS

The magnetic properties of different materials are studied in terms of their magnetic moments which
arise due to the orbital motion and spinning motion of the electrons. As electron is a charged particle, its
orbital motion produces a small magnetic field along the axis of rotation, as shown in Fig. 1.55 and its
SOLID STATE 1/53

spinning motion produces a small magnetic field along the spin FIGURE 1.^
axis.* Thus, each electron may be considered as a small magnet
having a resultant permanent magnetic moment. As magnetic Orbital magnetic Spin magnetic
moment moment
moment is a vector quantity, the net magnetic moment of an
Electron Direction
electron may be represented by an arrow. Thus, a material may
^of spin
be considered to contain a number of magnetic dipoles (similar
to a bar magnet with north and south poles). Based on the
■A
behaviour in the external magnetic field, the substances are A Electron

divided into different categories as explained below : Atomic

nucleus
Orbit

(i) Diamagnetic substances. Substances which are weakly

repelled by the external magnetic ifeld are called diamagnetic
(a) Magnetic moment of electron due to
substances, e.g., TiO^, H2O, NaCl, benzene, etc. The property orbital motion (b) Magnetic moment of
thus exhibited is called diamagnetism. This property is shown electron due to spin motion

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only by those substances which contain fully-filled orbitals, i.e.,
no unpaired electron is present. The magnetic moment possessed by an electron with spin in one direction is
cancelled by that with spin in the opposite direction.

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(11) Paramagnetic substances. Substances which are attracted by the external magnetic field are called
paramagnetic substances. The property thus exhibited is called paramagnetism. This property is shown by
unpaired electrons, e.g., O2, Cu-*, Fe^'*’ and Cr^"^.

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those substances whose atoms, ions or molecules contain
These substances, however, lose their magnetism in the absence of the magnetic field.

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(ih) Ferromagnetic substances. Substances which show permanent magnetism even in the absence of
the magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co, gadolinium (Gd) and C1O2 show

for
ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. These
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substances are very strongly attracted by a magnetic field. The reason for such a magnetic behaviour by these
substances is that in the solid state, the metal ions of these substances are grouped together into small regions
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called domains. Each domain acts as a tiny magnet having a definite magnetic moment. In an unmagnetised
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piece of any ferromagnetic substance, the domains are randomly oriented such that their magnetic moments
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cancel each other. However, when the substance is placed in a magnetic field, all the domains get oriented in
the direction of the magnetic field, as shown in Fig. 1.56 (a). As a result, the substance has a very high
magnetic moment. This ordering of domains persists even when the external magnetic field is removed.
r

Hence, the substance remains permanently magnetised. In fact, ferromagnetism is a case of large amount of
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paramagnetism. The ferromagnetic material, C1O2, is used to make magnetic tapes used for audio recording,
(tv) Anti-ferromagnetic substances. Substances which FIGURE 1.^
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are expected to possess paramagnetism or ferromagnetism

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on the basis of magnetic moments of the domains but actually

they possess zero net magnetic moment are called anti
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ferromagnetic substances, e.g., MnO. Anti-ferromagnetism is

due to the presence of equal number of domains in the opposite
directions as shown in Fig. 1.56 (Z>).
(v) Ferrimagnetic substances. Substances which are
oCDCDCD©©©
expected to possess large magnetism on the basis of the Alignment of the magnetic moments of
domains in (a) ferromagnetic
magnetic moments of the domains but actually have small net substance (b) antiferromagnetic
magnetic moment are called ferrimagnetic substances, e.g., substance (c) ferrimagnetic substance
magnetite (Fc304), ferrites of the formula M^"^Fe204
where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of domains in opposite direction
resulting in some net magnetic moment, as shown in Fig. 1.56 (c).
It may be noted that all magnetically ordered solids, i.e., ferromagnetic and anti-ferromagnetic solids
change into paramagnetic at high temperature. This is due to randomisation of domains (spins) on heating.
●Magnetic moment is expressed in terms of Bohr Magneton (pg = 9-27 x 10“^^ A m^). Spin magnetic moment of
each electron = ± )ig and orbital magnetic moment = pg where is the magnetic quantum number of the electron.
1/54
New Course Chemistry fXinpzsrwn

Ferrimagnetic substance, Fe304, becomes paramagnetic at 850 K. It may be further pointed out here that each
ferromagnetic substance has a characteristic temperature above which no ferromagnetism is observed. This is
known as curie temperature.
Like electrical conductivity, the magnetic behaviour of the compounds of transition elements with similar
structure may not be same. For example.
(/) TiO, VO and CuO are paramagnetic whereas MnO. FeO, CoO and NiO are antiferromagnetic.
(//) Ti02 is diamagnetic, VO2 is paramagnetic, CrO^ is ferromagnetic whereas M11O2 is antiferromagnetic.
SUPPLetWIENT- YOUR
KNOWLEDGE FOR OOMPETITIOBie

1. Superconductivity. A substance is said to be superconducting when it offers no resistance to the flow of

electricity. The phenomenon was discovered by Kammerlingh Onnes in 1913 when it was found that mercury
becomes superconducting at 4 K. Most of the metals become superconducting at very low temperatures
(2 K - 5 K). Certain organic compounds also become superconducting below 5 K. Such low temperatures
can be attained only with liquid helium which is very expensive.

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Certain alloys of niobium have been found to be superconducting at temperatures as high as 23 K.

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Further researches since 1987 have revealed that many metal oxides become superconducting even at fairly
high temperatures. For example, the oxide TL Ca2 Ba, CU3 0|q has been found to be superconducting at
125 K. Efforts are on to find out the materials which are superconducting at room temperature. Once this is
done, it will revolutionize the fields of electronics, power transmission and levitation transportation, i.e.,

e
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running of trains in air without the rails.
2. Examples of some common magnetic substances
Type of Substance
(0 Paramagnetic
Examples for
Molecules (NO and O,), transition metals (Cr, Mn, Ni, Co, Fe),
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metal ions (Cu^"^, Ni-"^, Fe^"^, Cr^"^) metal oxides (CuO, VO^)
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(ii) Diamagnetic N2, NaCl, Zn. Cd. Cu+, Ti02, H2O

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(Hi) Ferromagnetic Ni, Fe, Co, Gd, C1O2

(/v) Anti-ferromagnetic MnO, Mn203, MnOo
(v) Ferrimagnetic Fc304 and ferrites
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3. Dielectric Properties of Solids. Insulators do not conduct electricity because the electrons present in
them are held tightly by the nuclei. However, when electric field is applied, polarisation takes place because
nuclei are attracted to one side and the electron cloud to the other side. Thus, dipoles are formed. Such
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polar crystals show the following properties ;

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(i) Piezoelectricity. When mechanical stress is applied on such crystals, electricity is produced due to
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displacement of ions. The electricity thus produced is called piezoelectricity and the crystals are called
piezoelectric crystals, e.g., titanates of barium and lead, lead zirconate (PbZr03), ammonium dihydrogen
phosphate (NH4H2PO4) and quartz.
(//) Pyroelectricity. Some piezoelectric crystals when heated produce a small electric current. The electricity
thus produced is called pyroelectricity (pyre means heat).
((//) Ferroelectricity. In some of the piezoelectric crystals, the dipoles are permanently polarized even in
the absence of the electric field. Howeve,r on applying electric field, the direction of polarization changes.
This phenomenon is called ferroelectricity due to analogy with ferromagnetism.Some examples of the
ferroelectric solids are bcirium titanate (BaTi03>, .sodium potassium tartarate (Rochelle salt) and potassium
dihydrogen pho.sphate (KH2PO4). It may be pointed out here that all ferroelectric solids are piezoelectric
hut the reverse is not true,

(iv) Anti-ferroelectricity. hi some crystals, the dipoles align themselves in such a way that alternatively,
they point up and down so that the crystal does not possess any net dipole moment. Such crystals are said
to be antiferroelectric. A typical example of such crystals is lead zirconate (PbZr03).
SOLID STATE 1/55

1. Solid. It is that form of matter which possesses rigidity and hence a definite shape and a definite volume.
2. Cause of existence as solid. If intermolecular forces > thermal energy, substance exists as solid.
3. Classification of solids, (i) Crystalline solids, when the constituent particles are arranged in a definite
geometric pattern in three dimensional space so that there is short range a.s well as long range order of
particles, e.g., all elements and compounds.
(«) Amorphous solids, when there is no regular arrangement of particles or at the most there is only short
range order, e.g., rubber, glass etc.
4. Characteristics of crystalline and amorphous solids.
(/) Crystalline solids : Regular arrangement of particles, definite geometric shapes, sharp melting points,
definite heats of fusion, no smooth cooling curve, anisotropic, undergo clean cleavage.
(«) Amorphous solids i No regular arrangement of particles, in egular shapes, melt over a range, no definite

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heats of fusion, smooth cooling curve, isotropic, undergo irregular cleavage.
5. Uses of amorphoussolids. Glass in household wares, laboratory wares and construction, rubber in tyres

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and shoe-soles,plastics in eveiyday life and amorphoussilica in making photovoltaic cell.
6. Classification of crystalline solids.
(/) Ionic solids, e.g. NaCl, MgO etc.

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(«) Molecular solids, polar, e.g., HCI, SO2 etc. and non-polar, e.g., Ar, Ho, U etc.
(m7) Covalent/Network solids, e.g., diamond, quartz (Si02) etc.

7.
(iv) Metallic solids, e.g., all metals and alloys. for
Crystal lattice/Space lattice. It is a regular arrangement of the constituent particles of a crystal in a three
r
dimensional space.
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8. Unit cell. It is the smallest three dimensional portion of the complete space lattice which when repeated in
different directions produces the complete space lattice.
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9. Parametersof a unit cell. Lengths a, b and c of the three edges and angles a, [5 and 7 between the edges
(a between b and c and so on).
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Types of crystal systems. (/) Cubic (« = /> = c, a = ^ = 7 = 90“) (n) Tetragonal {a = b^c,a = ^= y= 90°)
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10.

(iii) Orthorhombic/Rhombic {a ^ b ^ c, a = ^ = y = 90°) (/v) Monoclinic {a^b^c, a = y= 90° ^ p)

(v) Triclinic (a b ^ c, a ^ ^ ^ y ^ 90°) (v/) Rhombohedral/Trigo nal {a = h = c, a = ^ = y * 90°)
(vn) Hexagonal (a = b^ c, a=^ = 90°, 7 = 120°).
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11. Types of lattices/unit cells, (a) Primitive/Simple- having particles only at the corners
(b) Non-primitive : (0 Face-centred - having particles at comers as well as face centres (//) End-centred
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- having particles at comers and any two opposite faces (Hi) Body-centred - having particles at comers and
one at the body centre.
12. Contribution by particles present at difTcrent positions
1 1
Comer = -. Face-centre = —, Body-centre = 1, Edge-centre = —,
8 2 4

13. Number of particles per unit cell of a cubic crystal

1 1 1
Simple =8x—= 1, Body-centred =8x--t-l = 2. Face-centred =8x- + 6x—= 4
2

14. Close packing in crystals (i) Hexagonal close packing (HCP) viz ABAB.... type, space occupied = 74%,
empty space = 26% (ii) Cubic close packing (CCP) viz ABCABC...typc, space occupied = 74%, empty
space = 26% (Hi) Body-centred cubic (BCC), space occupied = 68%, empty space = 32% (not a close
packing). Also remember that in simple cubic unit cell, .space occupied = 52-4%.

f
1/56
New Course Chemistry (XII)
15. Tetrahedral and Octahedral voids. A simple triangular void surrounded by four spheres is a tetrahedral
void. A double triangular void surrounded by six spheres is an octahedral void.
16. Coordination number. It is the number of closest particles or nearest neighbours of any constituent particle
(anions present around a cation or vice versa in ionic compounds). Coordination number in HCP and CCP
is 12 whereas for BCC, it is 8.
17 Sizes of tetrahedral and octahedral voids. (/) Radius (r) of tetrahedral void = 0-225 x R where R is the
radius of the spheres in close packing («) Radius (r) of octahedral void = 0414 R. For cations in the voids
and anions in the packing, for tetrahedral, = 0-225 r_ and for octahedral, = 0-414 r_.
18. Number of voids filled and formula of a compound
No. of octahedral voids = No. of particles present in close packing

ow
No. of tetrahedral voids = 2 x No. of octahedral voids
1
Assuming n particles of anions B are present in the packing and -rd of octahedral voids are occupied by

e
n
cations A, then ratio ofA:B =-:n = -: \ =1 :3. Hence, formula is AB,.
3 ^

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3

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In ccp, total no. of voids per unit cell = 4 (octahedral) + 8 (tetrahedral) = 12

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In hep, total no. of voids per unit cell = 6 (octahedral) + 12 (tetrahedral) = 18
19. Structures of Simple ionic solids :
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Coord. Formula units
Crystal structure Arrangement of ions

AB Type
kfs No. per unit ceil
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(0 Rock salt (NaCl) type Cr ions = fee arrangement Na+ = 4 4
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Na"*" ions = edge centres and body centre Cl-= 4
B

(zi) Caesium chloride (CsCl) type bee arrangement Cs+=8 1

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cr = comers, Cs'*’ = body centre CI"=8

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{in) Zinc blende (ZnS) type ccp arrangement Zn2+ = 4 4

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S^- = fee, = in tetrahedral voids s2- = 4

d

AB2 Type
(z) Fluorite (Cap2) type Ca^+ = 8
in

ccp arrangement 4
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Ca^"*" = fee, F~ = all tetrahedral voids 1^ = 4

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(») Antifluorite (A^B) type B^ = ccp. A'*' = half tetrahedral voids Na+ = 4 4
02- = 8

o .. .. , r> -●
20. Radius ratio rules. Radius ratio =
Radius of cation (r_,_)
^ +'
Radius of anion (r_)
Radius Ratio Coord. No. Structural arrangement Structure type
(z) 0-155 -0-225 3 Planar triangular
(zi) 0-225 - 0-414 4 Tetrahedral Sphalerite (ZnS)
(Hi) 0414 - 0-732 6 Octahedral Rock salt (NaCl)
(zv) 0-732 - 1 8 Body-centred cubic Caesium chloride (CsCl)
21. Effect of temperature and Pressure on crystal structure
Pres.sure _ _.
NaCl structure ^ CsCl stmeture
760 K
(6 ; 6 coordination) (8 : 8 coordination)
SOUD STATE 1/57

22. Calculation of density of a cubic crystal from its edge,

ZxM
(fl) For cubic crystals o/element, density (p) is given by p =
fl^xN^x
0
10-30
where Z = no. of atoms per unit cell, M = atomic mass of the element, Nq = Avogadro’s number, a = edge of
ZXM
the cube in pm, p = density of the crystal in g cm"3. If edge ‘a’ is in cm, formula used is p =
u^xN 0
In terms of SI units, M is in kg mol"* and ‘a’ is in meters, then p is in kg m“3.
(b) For cubic crystals o/ionic compounds, Z = no. of formula units per unit cell, M = formula mass.
Remember, for ionic compound. A'*' B“ having/cc structure (e.g., Na'*'a-), edge (a) = 2 x Distance between
A'*’ and B“ ions.

23. Impeifections/Defects in Solids. Any departure from perfectly ordered arrangement of constituent particles

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is called defect or imperfection.
(A) Stoichiometric defects. When ratio between cations and anions remains the same.

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(0 Schottky defect When equal number of cations and anions are missing. It occurs in compounds with
high coordination number and small difference in size between cations and anions, e.g., NaCl, KCl, KBr,

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AgBr, CsCl. Density of the solid decreases.

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(ii) Frenkel defect When cations are missing from lattice sites and occupy interstitial sites. It occurs in
compounds with low coordination number and large difference in size of cations and anions, e.g., AgCl,
AgBr, Agl. Thus, AgBr shows both defects. There is no effect on density.
for
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(B) Non-stoichiometric defects. When ratio of cations and anions changes as a result of the defect,
(i) Metal excess : (a) By anion vacancies, e.g., on heating NaCl in presence of Na vapour, some anions
s
(Cr) leave lattice sites which are occupied by electrons. These are called F-centresgiving colour to crystals.
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(b) By presence of extra cations in interstitial sites, e.g., on heating ZnO, O2 gas is lost, Zn^+ ions and e~
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occupy interstitial sites.

(lO Metal deficiency. This occurs where metal shows variable valency, e.g., 2A''’ in lattice sites may be
replaced by in one lattice site and one lattice site remains vacant,
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(ill) Impurity defects. Adding impurities to crystalline solids to change their properties is called doping,
(a) In ionic solids, e.g., SrCl2 into NaCl so that 2 Na*** are replaced by one Sr^'** producing one cation
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vacancy.
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(b) In covalent solids :

nd

— By doping with electron rich impurities, e.g., group 14 elements like Si or Ge with group 15 element
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like R As etc. Free electrons become available which increase conductivity. Such solids are called n-type
semiconductors.

— By doping with electron deficit impurities, e.g., group 14 element like Si or Ge with group 13 element
like B, A1 or Ga. As a result, holes are created. Electrons move to occupy the holes. Thus, holes move
towards negatively charged plate as if they carry positive charge. These are called p-type semiconductors.
24. Conductivities of conductors, insulators and semiconductors. Conductors = 10"* - lO’ ohm"^ m“^
Insulators = - 1(T*® ohm"* m-^ Semiconductors = 1(T® - 10^ ohm"* m“^
25. Band theory. Atomic orbitals of metal atoms overlap to form molecular orbitals (m-o’s). The lower half may be
filled and the upper half may be empty or two sets of molecular orbitals may be formed, the filled m.o’s with
lower energy form valence band and empQr or partially filled iilo’s with higher energy form conduction band.
26. Distinction between metals, insulators and semiconductors on the basis of band theory. Metals have
either only valence band, lower half filled and the upper half empty or there is a partial overlap between
filled valence band and empty conduction band. Insulators have large energy gap between valence band and
conduction band. Semiconductors have a small energy gap between valence band and conduction band.
1/58 New Course Chemistry fxinragnwi

27. Classification of substances on the basis of magnetic properties

(0 Diamagnetic - those which are weakly repelled by the external magnetic field, e.g., Ti02, H2O, NaCl,
benzene etc. They contain fiilly filled orbitals and hence zero magnetic moment.
(h) Paramagnetic- those which are attracted by the external magnetic field, e.g., O2, Cu^"^, Fe^'*', Cr^'*' etc.
Their atoms, ions or molecules contain unpaired electrons and hence a net magnetic moment. They lose
their magnetism in the absence of magnetic field,
(m) Ferromagnetic - those which show permanent magnetism even in the absence of magnetic field, e.g.,
Fe, Ni, Co, Gd, C1O2 etc. This is because when placed in magnetic field, their unpaired electrons (or magnetic
domains) get permanently oriented in one direction,
(rv) Anti-ferromagnetic - those which are expected to possess paramagnetism or ferromagnetism but actually
have zero net magnetic moment, e.g., MnO. This is because of equal number of domains in opposite direction,
(v) Ferrimagnetic - those expected to have large magnetism but actually have small net magnetic moment,
e.g., magnetite (Fe304), ferrites Fe204). This is because of unequal number of domains in opposite

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direction.

28. Curie temperature. It is the temperature above which a ferromagnetic substance shows no ferromagnetism.

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SOLID STATE 1/59

NEET/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to

low
● MCQs in Chemistry for NEET
Pradeep’s Stellar Series.... * MCQs in Chemistry for JEE (Main)
separately available for these examinations.

El Multiple Choice Questions (with one correct Answer)

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F
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I. Amorphous and crystalline solids 5. In a face centred cubic lattice, atoms A occupy
the corner positions and atoms B occupy the face
1. Which of the following exists as covalent crystals centre positions, if one atom of B is missing
in the solid state?
for
from one of the face centred points, the formula
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(a) Phosphorus (h) Iodine of the compound is :
(c) Silicon id) Sulphur (a) AB. ib) A2B3
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{JEE Main 2013) (c) id) A,B (AIEEE2011)

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2. A hard substance melts at high temperature and is 6. The correct option forthe number of body-centred
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an insulator in both solid and in the molten state. unit cells in all 14 types of Bravais lattice unit
This solid is most likely to be cells is
ia) Metallic solid ib) Covalent solid ia)l ib)5
r

(c) Ionic solid id) Molecular solid

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ic)2 id)?> (NEET 2021)

(JEE Main 2021)
3. Choose the correct statement: 111. Close packing in
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(fi) Diamond and graphite have two dimensional crystals and .study of voids
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network
7. If spheres of radius V are arranged in hep
ib) Diamond is covalent and graphite is ionic fashion (AB AB...), the vertical distance between
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(c) Diamond is sp^ hybridised and graphite is sp^ any two consecutive A layers is
hybridized
id) Both diamond and graphite are used as 3
id) 4r^~ ib) 4r.~
lubricants (NEET 2022) \3 V2

II. Crystal lattices, (c) 6r id) rVb

unit cells and their types 8. The pyknometric density of sodium chloride
4. How many unit cells are present in a cube-shaped crystal is 2*165 x 10^ kg m“^ while its X-ray
ideal crystal of NaCl of mass 100 g ? density is 2' 178 x 10^ kg m“^. The fraction of the
[Atomic mass : Na = 23, Cl = 35-5] unoccupied sites in sodium chloride crystal is
(fl) 5-14x10"' (h) 1-28 X 10^' (rt) 5*96 ib) 5-96 X 10"2
-I
ic) 1-71 X 10-' id) 2-57 X 10^' ic) 5*96 X 10 id) 5-96 X 10-3

1. (c) 2. ih) 3. (c) 4. id) 5. (c) 6. id) 7. ia) 8. id)

1/60 T><t<idee^'4r New Course Chemistry (XIl)CSHa

9. The fraction of the total volume occupied by the (c) A2B2O {d) A2BO4
atoms present in a simple cube is (JEE Main 2019)
K n 15. If the unit cell of a mineral has a cubic close
(a) - packed (cep) array of oxygen atoms \vith m
6
fraction of octahedral holes occupied by
n n
aluminium ions and n fraction of tetrahedral
holes occupied by magnesium ions, m and n,
(AIPMT 2007) respectively are
10. The packing efficiency of the 1 1 1
two-dimensional square unit cell ib) l-
2’8
shown in the adjoining fig. is
(a) 39.27% 1 1 1 1
(c)
(i) 68.02% 2’2 Vi
(c) 74.05% (d) 78.54% (UT2010) (JEE Advanced 2015)
11. In a solid ‘AB’ having NaCl structure, ‘A’ atoms

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16. The arrangement of
occupy the comers of the cubic unit cell. If all the X“ ions around A'*'

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face-centred atoms along one of the axes are ion in solid AX is \ /

removed, then the resultant stoichiometry of the given in the fig. (not I
V
I X" I

solid is drawn to scale). If

(a) AB2 ib) A2B the radius of X' is

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(0 A4B3 (d) A3B4. 250 pm, the radius of
A"^ is
12. A compound X^ q/ MQ for
has cubic close n
(a) 104 pm (6) 125 pm
X
packing (cep) arran I
(c) 183 pm (d) 57 pm
r
gement of X. Its unit cr ●
d I
● -o
(JEE Advanced 2013)
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cell structure IS I
17. A metal crystallises in a face centred cubic
shown below. The
structure. If the edge length of the unit cell is 'a\
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empirical fonnula of the closest approach between the two atoms in the
the compound is metallic crystal will be
ia) MX . ib) MX2
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ic) M2X id) M5X14 (IIT2012)

(a) V2 a (.b)
V2
13. A crystal is made up of metal ions ‘Mj’ and ‘M2’
and oxide ions. Oxide ions form a cep lattice ic)2a id) 2V2fl
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structure. The cations ‘Mj’ occupy 50% of the (JEE Main 2017)
octahedral voids and the cations ‘M2’ occupy 18. In a binary compound, atoms of element A form
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12-5% of tetrahedral voids of oxide lattice. The

hep structure and those of element M occupy 2/3
oxidation numbers of ‘Mj’ and ‘M2’ are of the tetrahedral voids of the hep structure. The
respectively formula of the binary compound is :
(fl) + l, + 3 ib) + 3, + 1
ia) M2A3 ib) MA3
ic) + 2,+4 id)+ 4,+ 2
(c) M4A (d) M4A3
(JEE Main 2020) (JEE Main 2021)
14. Element ‘B’ forms cep structure and ‘A’ occupies 19. Right option for the number of tetrahedral and
half of the octahedral voids while oxygen atoms octahedral voids in hexagonal primitive unit cell
occupy all the tetrahedral voids. The structure of IS
the bimetallic oxide is
ia) 8,4 ib) 6, 12
ia) AB2O4 ib) A4B2O (c)2, 1 id)\2,6 (NEET2022)

9. (/») 10. (^/) n. ((/) 12. (/;) 13. (c) 14. («) 15. {«) 16. (a) 17. (b) 18. (rf) 19. (rf)
SOLID STATE 1/61

IV. Structure of simple ionic 2

compounds and Radius ratio niici> j

(a) a (b) 4 a

20. A solid compound XY has NaCl structure. If the V3

(0 - a
radius of the cation is 100 pm, the radius of the
a

anion (Y“) will be

(AIPMT 2014)
(a) 275-1 ib) 322-5 pm
(c) 241-5 pm (d) 165-7 pm VI. Calculations involving
unit cell dimensions
(AIPMT Main 2011)

V. Relation between nearest neighbour 26. A metal has a fee lattice. The edge length of the
distance {d), edge (o) of the unit cell unit cell is 404 pm. The density of the metal is
2-72 g cm“^. The molar mass of the metal is
and radius (r) of atoms of elements
(N^, Avogadro’s
-1
constant = 6-02 x 10^^ mol"’)
-i

w
21. A metal crystallizes with a face-centred cubic (a) 40 g mol (b) 30 g mol
-I -1
lattice. The edge of the unit cell is 408 pm. The (c) 27 g mol (d) 20 g mol
diameter of the metal atom is (AIPMT 2013)

F lo
(a) 288 pm (b) 408 pm 27. Ice crystallises in a hexagonal lattice having a
volume of the unit cell as 132 x 10“^'’ cm^. If
(c) 144 pm (d) 204 pm

ee
density of ice at the given temperature is 0-92 g
(AIPMT PreUm 2012)
cm"^, then number of H2O molecules per unit

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22. An element has a body-centred cubic (bcc) cell is
structure with a ceil edge of 288 pm. The atomic (a) 1 (b) 2
radius is
(c) 3
for {^0 4
ur
V2 4 28. Lithium has a hcc structure. Its density is
(o) — X 288pm (b) -=x288pm 530 kg m"^ and its atomic mass is 6-94 g mol"’.
s
Calculate the edge length of the unit cell of
ook
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4
s lithium metal (N^ = 6-02 x 10^^ mol"’)
■=x288pm (a) 527 pm (b) 264 pm
eB

(<^) id) ~x 288pm

V2
(c) 154 pm {d) 352 pm
(NEET 2020 Phase-1)
(NEET Phase I 2016)
23. A given metal crystallizes out with a cubic
our

29. Iron exhibits bcc structure at room temperature.

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structure having edge length of 361 pm. If there Above 900®C, it transforms to fee structure. The
are four metal atoms in one unit cell, what is the
ratio of the density of iron at room temperature to
Y

radius of one atom ?

that at 900°C (assuming molar mass and atomic
Re

(a) 80 pm {b) 108 pm radii of iron remain constant with temperature) is

nd

(c) 40 pm (d) 127 pm

S 4^l3
Fi

(AIPMT 2015) (a) {b)

3->/2
24. CsCl crystallises in body-centred cubic lattice. If
1
'o' is its edge length then which of the following (c) id) - (NEET 2018)
expressions is correct ? 4-n/2
(a) rCs+ + r
Cl" = %/3fl ib) VII. Imperfections or Defects in solids

(0
3a 30. If NaCl is doped with 10^ mol % of S1CI2, the
''cs-+'cr 2
a
Cl- 2 concentration of cation vacancies will be

(JEE Main 2014) (N^= 6-02x10^3 mol-’)

25. If a is the length of the side of a cube, the distance
(a) 6-02 X 10’^ mol-’ (b) 6-02 x lO’^moI"’
between the body-centred atom and one comer (c) 6-02 X 10'^ mol"’ id) 6-02 x lO’"^ mor'
atom in the cube will be (AIPMT 2007)

20. (f) 21. (a) 22. (J) 23. id) 24. id) 25. [d) 26. (c ) 27. id) 28. id) 29. id) 30. id)
1/62 T^n/xdee^'4. New Course Chemistry (XII)BSm

31. The crystal with metal deficiency defect is (h) Na'*’ ions wil increase by 1 while Cl ions
(a) NaCl (b) FeO will decrease by 1
(c) KCl {d) ZnO (c) Number of Na‘*‘ and Cr ions will remain the
same
(e) LiCI (Kerala PMT 2010)
(d) The crystal structure of NaCl will change.
32. Experimentally it was found that a metal oxide
has formula M,),yj{0. Metal M is present as M”"^ 38. Which of the following statements is not
and M^"*" in its oxide. Fraction of the metal which correct ?
3+
exists as M would be (ft) The fraction of the total volume unoccupied
(fi) 5-08% {b) 7-01% by the atoms in a primitive cell is 048.
(c) 4 08% id) 6-05% (h) Molecular solids are generally volatile,
(JEE Main 2013, AMU Engg. 2015) (c) The number of carbon atoms in a unit cell of
Diamond is 4.
33. Which one of the following compounds shows
both Frenkel as well as Schottky defects ? (d) The number of Bravais lattices in which a

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(fi) AgBr (b) Agl crystal can be categorized is 14.
(AIPMT 2008)
(c) NaCl (d) ZnS

F lo
39. KCl crystallises in the same type of lattice as does
(NEET 2020 Phase-2, JEE Main 2020)
NaCl. Given that r,^/r and
Na-^ cr =0-55
Vni. Electrical and

ee
= 0-74 . Calculate the ratio of the side
V"cr

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magnetic properties of solids
of the unit cell of KCl to that of NaC!
34. Which of the following metal oxides is anti-
fcrromagnctic in nature ? (a) M23
(c) 1414 for (b) 0-891
(d) 0414
ur
(a) MnO-) (b) TiO.
(c) VO. “ (d) Cr02 (e) 1-732 (Kerala PET 2008)
s
40. Which is the incorrect statement ?
35. The energy gap (E^) between valence band and
ook
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conduction band for diamond, silicon and (fl) FeO 0-98 has non-stoichiometric metal
eB

germanium are in the order deficiency defect

(a) (diamond) > E^ (silicon) > E^ (germanium) (b) Density decreases in case of crystals with
(b) Eg (diamond) > Eg (silicon) < Eg (germanium) Schottky’s defect
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(c) E. (diamond) = E„ (silicon) = E„ (germanium)

o n o (c) NaCl (5) is insulator, silicon is semiconductor,
(d) Eg (diamond) > Eg (germanium) > Eg (silicon) silver is conductor, quartz is piezoelectric
crystal
Y

IX. Miscellaneous
(d) Frenkel defect is favoured in those ionic
Re
nd

36. Each rubidium halide crystallising in the NaCl compounds in which the sizes of cations and
type lattice has a unit cell length 0-30 A greater anions are almost equal (NEET 2017)
Fi

41. In the bcc unit cell as

than for corresponding potassium salt (r =
shown, if the radius of
1-33 A) of the same halogen. Hence, ionic radius the central atom is
of Rb"^ is double than that of the
corner atoms, then
o
(a) 1-03 A ib) M8A
(c) 1-48 A (£/) 1-63 A
what is the approximate
packing efficiency ?
37. If the positions of Na'*' and Cl" are interchanged
(a) 75%
in NaCl, having fee arrangement of Cr ions then
in the unit cell of NaCl
ib) 90%
(c) 60%
{a) Na'*’ ions will decrease by 1 while Cl” ions
(d) 45% (JEE Main 2019)
will increase by 1

31. (b) 32. (f) 33. (a) 34. (a) 35, (//) 36. (c) 37. (/;} 38. (<) 39. («) 40. ((ul) 41. (b)
SOLID STATE 1/63

HI Multiple Choice Questions (with One or More than One Correct Answers)

42. Which of the following statements are not true ? 46. Which of the following .statements are correct ?
(fl) An element with BCC structure has two
(a) The coordination number of each type of ion
atoms per unit cell. in CsCl crystal is 8.
(b) An ionic compound A"^ with BCC
(b) A metal that crystallizes in bcc structure has
structure has one AB formula unit per unit coordination number of 12.
cell,
(c) A unit cell of an ionic crystal shares some of
(c) The shape of the octahedral void is its ions with other unit cells.
octahedral.

(d) The edge length of the crystal A'*’ B" is equal (d) The length of the edge of unit cell of NaCI is
to the distance between A'*' and B“ ions. 552 pm (rj,jj,+ = 95 pm, /'c|-= 181 pm).

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43. Which of the following are true ? 47. The correct statement(s) regarding defects in
solids is (are)
(a) In NaCl crystals, Na"*" ions are present in all

F lo
the octahedral voids. (a) Frenkel defects are usually favoured by a
(b) In ZnS (zinc blende), Zn^'*’ ions are present in very small difference in the sizes of the
cation and anion

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alternate tetrahedral voids,

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(c) In Cap2, F“ ions occupy all the tetrahedral (b) Frenkel defect is a dislocation defect
voids.
(c) Trapping of an electron in the lattice leads to
(d) In Na20, ions occupy half the octahedral
for
the formation of F-centre
ur
voids.
(d) Schottky defects have no effect on the
44. Crystal systems in which no two axial lengths are physical properties of solids aiT 2009)
s
equal are
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48. With respect to graphite and diamond, which of

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(a) Tetragonal (b) Orthorhombic the following statement(s) given below is (are)
(c) Monoclinic (d) Triclinic
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correct?

45. A metal has cubic close packed (ccp) (a) Graphite is harder than diamond
arrangement, the layer sequence of which is
(b) Graphite has higher electrical conductivity
r

shown below :
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ou

than diamond

(c) Graphite has higher thermal conductivity

0—®
Y

than diamond

© © ©
Re
nd

(d) Graphite has higher C-C bond order than

diamond (IIT 2012)
©z=0© ©
Fi

z = a/2 49. The CORRECT statement(s) for cubic close

packed (ccp) three-dimensional structure is (are)
©
(fl) The number of neighbours of an atom present
in the topmost layer is 12
(b) The efficiency of the atom packing is 74%
z = 0
(c) The number of octahedral and tetrahedral
voids per atom are 1 and 2 respectively
A face diagonal passes through the centre of
atom 4 and the centre (s) of which other atom (s)7
(d) The unit cell edge length is 2-Jl times the
(a) I (b) 2,5
radius of the atom (JEE Advanced 2016)
(c) 8. 12 (d) 9, 10
ANSWERS
42. ic.d) 1 -
44. [h,i.,d) 45. (b.c.d) 46. (a,c,d) 47. 48. (h,d)
49. ih.c.d)
1/64 “Pn4nxiee^'<t. New Course Chemistry (XIl)EZsZSI

nil Multiple Choice Questions (Based on the given Passage/Compre hension).

Each comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option.
tgompreh'ensiOnlF]
i^,orripi^lTensiofTi|[ By X-ray studies, the (IIT 2008). In hexagonal
systems of crystals, a frequently encountered
packing of atoms in a crystal of gold is found
to be in layers such that starting from any arrangement of atoms is described as a
layer, every fourth layer is found to be hexagonal prism. Here, the top and bottom
exactly identical. The density of gold is of the cell are regular hexagoas and three
atoms are sandwiched inbetween them. A
found to be 19*4 g cm~^ and its atomic mass
is 197 a.m.u. space filling model of this structure called
hexagonal closed packed (HCP) is
50. The coordination number of gold atom in the constituted of a sphere on a flat surface

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crystal is surrounded in the same plane by six
(a) 4 ih) 6 (c) 8 (d) 12 identical spheres as closely as possible. Three

F lo
51. The fraction occupied by gold atoms in the spheres are then placed over the first layer so
crystal is that they touch each other and represent the
second layer. Finally, the second layer is
(a) 0-52 (b) 0*68
covered with a third layer that is identical to

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(c) 0-74 (d) 1-0
the bottom layer in relative position. Assume

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52. The approximate number of unit cells present radius of every sphere to be V’.
in 1 g of gold is
(a) 3-06 X 10^* (b) 1-53x102>
for
55. The number of atoms in this HCP unit cell is
ur
(c) 3-82 X 10^0 (d) 7-64 X (.a) 4 (b) 6 (c) 12 (d) n
53. The length of the edge of the unit cell will be 56. The volume of this HCP unit cell is
s
ook

(a) 407 pm (b) 189 pm

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(a) 24V2r3 (b) 16^2
(c) 814 pm (d) 204 pm
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64 ,
54. Assuming gold atom to be spherical, its radius (c) )2^r^ (d)
will be 3V5'
57. The empty space in this HCP unit cell is
our

(a) 203-5 pm (b) 143-9 pm

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(c) 176-2 pm (d) 287-8 pm (a) 74% (b) 47-6%

(c) 32% id) 26%
09
Y

Matching Type Questions

Re
nd

Match the entries of column I with appropriate entries of column II and choose the correct option
Fi

out of the four options (a), (b), (c), (d) given at the end of each question.
58. Column I (Distribution of particles X and Y) Column II (Formula)
(A) X = At the comers; Y = At face centres (P) XY
(B) X = At the comers and face centres iq) XY2
Y = In all tetrahedral voids

(C) X = At the comers (r) XY3

Y = One on each body diagonal
(D) X = At the comers and face centres is) XY4
'Y = At the edge centres and body centre

(a) A-r, B-q, C-p, D-^ ib) As, B-p, C~r, D~q (c) A-q, B-r, C-p, D-y id) A-r, B-q, C-s, D-p

ANSWERS

50. (d) 51. (c) 52. Ui) S3. (^0 54. (h) 55. ib) 56. ia) 57. (d) 58. (d)
SOLID STATE 1/65

59. Column I (T>pe of crystal) Column 11 (Location of cations/anions)

(A) NaCl (p) Cations-fcc, Anions-all tetrahedral voids
(B) ZnS iq) Anions-fcc, Cations-all tetrahedral voids
(C) CaFj (r) Anions-fcc, Cations-all octahedral voids
(D) Na20 (s) Anions-fcc, Cations-altemate tetrahedral voids

(a) A-r, B-s, C-p, D-q (b) A-q, B-p, C-s, D-r (c) As, B-p, C-r, D-q id) A-r, B-j, C-q, D-p

60. Column I (Position of particles X and Y) Column II (Formula)

(A) Y = in cubic close packing ip) XY
X = in tetrahedral voids

(B) X = in cubic close packing iq) X2Y3

w
Y = equally distributed between octahedral

F lo
and tetrahedral voids

(C) Y = in cubic close packing ir) X2Y

X = in octahedral voids

ee
(D) Y = in hexagonal close packing (s) XY2

Fr
X = in 2/3rd of octahedral voids

(a) A-p, B-q, C-r, D-J ib) A-q, B-p, C-s, D-r for
(c) A-r, B-j, C-p, D-q id) As, B-p, C-r, D-q
ur
s
61. Column t (Defect) Column n (Effect)
ook
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(A) Schottky defect ip) Crystal becomes coloured

eB

(B) Doping silicon with aluminium iq) n-typt semiconductor is formed

(C) Doping silicon with arsenic (r) p-type semiconductor is formed
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(D) Heating NaCl crystal in presence of sodium vapour (j) Density of the crystal decreases

(a) As, B-q, C-r, D-p ib) A-q, B-p, C-r, D-i' (c) A-q, B-5, C-q, D-p id) As, B-r, C-q, D-p
dY
Re

62. Match the following :

Fin

Compound Magnetic property

(A) NaCl ip) Ferrimagnetic
(B) MnO iq) Paramagnetic
(C) C1CI3 (r) Ferromagnetic
(D) Cr02 (.r) Diamagnetic
(E) MgFe204 (/) Antiferromagnetic

(i<) A-p, B-r, C-q, D-t, E-.9 ib) A-t, B-q, C-r, D-p, E-5 ic) A-r, B-f, C-q, D-p, E-i
id) As, B-t, C-q, D-r, E-p ie) As, B-r, C-t, D-q, E-p

(Kerala PMT 2015)

ANSWERS

59. (u) 60. ic) 61. id) 62. (d)

1/66 ‘P>uuieefo-'a, New Course Chemistry (XII)l!ZsI9]

K9 Matrix-Match Type Questions

p q r s

Match the entries of column I with appropriate entries of column II. Each
entry in column I may have one or more than one correct option from
A
'©ii0iOISI
'

column II. If the correct matches are A-p, s ; B-r ; C-p, q ; D-5, then the B

correctly bubbled 4x4 matrix should be as foUows :

!©!l®!f® i®i:
63.

(A)
Column I nVpe «f crystal)
Ionic solids (p)
Column II (Example/Property)
Dry ice
D
®il®ii0 l:

(B) Molecular solids iq) Brass

(C) Covalent solids (r) Generally insulators

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(D) Metallic solids (5) Generally have low melting points
64. Column I (T^’pe of packing) Column II (Metal possessing it/Space occupied)

Flo
(A) Hexagonal cubic packing (hep) (p) Iron

e
(B) Cubic close packing (cep) iq) 52%

re
(C) Body centred cubic (bcc) (r) 68%

F
(D) Simple cubic (s) 74%
65. Match the crystal systems/unit cells mentioned in column 1 with their characteristic features
ur
r
mentioned in column II. Indicate your answer by darkening the appropriate bubbles of the
4x4 matrix given in the ORS.
fo
ks
Column I Column II
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(A) Simple cubic and face-centred cubic (p) have these cell parameters a = b = c and a = P = y
oo

(B) Cubic and rhombohedral (q) are two crystal systems

B

(C) Cubic and tetragonal (r) have only two crystallographic angles of 90°
(D) Hexagonal and monoclinic (5) belong to the same crystal system (IIT 2007)
re

A B C D

VI. Integer Type Questions

u
ad

®®®®
Yo

DIRECTIONS. The answer to each of the following questions is a single 00® O

d

digit integer, ranging from 0 to 9. If the correct answers to the question

©©@©
Re
in

numbers A, B, C and D (say) are 4, 0, 9 and 2 respectively, then the

correct darkening of bubbles should be as shown on the side: ® ®®®
F

66. In hexagonal close packing, the difference in the number of tetrahedral and ® ®®0
octahedral voids per unit cell is
® ®®@
67. NH^ and Br“ ions have ionic radii of 143 pm and 196 pm respectively. The ® ®®®
coordination number of NH^ ion in NH^Br is ©000
68. Iron (II) oxide has a cubic structure and each unit cell has side 5 A. If the
©©©®
density of the oxide is 4 g cm“^, the number of oxide ions present in each unit
cell is (Molar mass of FeO = 72 g mol”*, = 6-02 x 10“^ moi"^) ©©@ @
69. aP"*" ions replace Na"*" ions at the edge centres of NaCl lattice. The number of vacancies in one mole NaCl is
found to be X X 10~^. The value of x approximately is
,NSWEIS3

63. {A-r; B-p,,rs ; C-r : D-q) 64. (/t-.v; B-p.s : C-r ; D-q)
65. (4-/J..VB-p.q : C-q ; D-q,r) 66.(6) 67.(6) 68.(4) 69.(3)
SOLID STATE 1/67

70. The oxide Tl„ Ca2 Bii2 CU3 Ojo is found to be superconductor at 125 K. The value of n is
71. The coordination number of A1 in the crystalline state of AICI3 is (IIT 2009)
72. The number of hexagonalfaces that are present in a truncated octahedron is (IIT 2011)
73. A ciysialline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the
density of the substance in the crystal is 8 g enr^,then the number of atoms present in 256 g of the crystal
is N X lO^'^. The value of N is (JEE Advanced 2017)

VII.
Numerical Value Type Questions Decimal Notation)
For the following question, enter the correct numerical value (in decimal-notation, truncated/rounded-off to the
second decimal place, e.g., 6*25, 7*00, - 0*33, 30*27, - 127*30) using the mouse and the onscreen virtual numeric
keypad in the place designated to enter the answer.
74. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed
from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.

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(0 Remove all the anions (X) except the central one.

F lo
(ii) Replace all the face-centred cations (M) by anions (X)
(Hi) Remove all the corner cations (M)
(iV) Replace the central anion (X) with cation (M)

e
number of anions

Fre
The value of in Z is (JEE Advanced 2018)
number of cations
\

for
75. A certain element crystallizes in a bcc lattice of unit cell edge length 27 A. If the same element under the
same conditions crystallizes in the fee lattice, the edge length of the unit cell in A will be (rounded
r
off to the nearest integer). (JEE Main 2021)
You
oks

[Assume each lattice point has a single atom. Assume ->/3 = 1-73, V2 = 1-41]
eBo

76. Ga (atomic mass 70 u) crystallizes in a hexagonal close packed structure. The total number of voids in
0-581 g of Ga is X 10^' (Round off to the nearest integer) [Given : = 6 023 x 10^^]
(JEE Main 2021)
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77. Metal deficiency defect is shown by FeQ.930 in the crystal. Some Fe^''' cations are missing and loss of
positive charge is compensated by the presence of Fe-^‘*‘ ions. The percentage of Fe^'*' ions in Feo.930 crystals
is (Nearest integer) (JEE Main 2022)
dY
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Vlll. Assertlon>Reason Type Questions

Fin

TYPE I

DIRECTIONS. Each question below contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason).
Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correct
option as under :
(a) Statement 1 is TVue ; Statement-2 is True ; Statement-2 is a correct explanation for statement-1,
(b) Statement 1 is TVue ; Statenient-2 is True ; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statemcnt-1 is True ; Statement-2 is False,
(d) Statement-1 is False ; Statement-2 is True.

78. Statement-1. Covalent crystals have the highest melting point.

Statement-2. Covalent bonds are stronger than ionic bonds.

70.(2) 71.(6) 72.(8) 73.(2) 74.(3.00) 75.(33) 76.(15) 77.(85) 78. (c)
1/68 “PntuUcfi. ^ New Course Chemistry (XII) PPTMl
79. Statement-1.In NaCl crystal, all the octahedralvoids are occupied by Na''' ions.
Statement-2. The number of octahedral voids is equal to the number of Cl“ ions in the packing.
80. Statement-1. The octahedral voids have double the size of the tetrahedral voids in a crystal.
Statement-2. The number of tetrahedral voids is double the number of octahedral voids in a crystal.
81. Statement-1. In any ionic solid, [MX], with Schotiky defects, the number of positive and negative ions are
same.

Statement-2. Equal number of cation and anion vacancies are present.

82. Statement-]. In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are
missing from its unit cells.
Statement-2. In an ionic solid, Frenkel defect arises due to dislocation of cations from its lattice site to
interstitial site maintaining overall electrical neutrality. (NEET 2022)

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TYPE II

DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a

F lo
corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as
(a) If both assertion and reason are true, and reason is the true explanation of the assertion,
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion,

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(c) If assertion is true, but reason is false,

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(d) If both assertion and reason are false.

for
83. Assertion. Graphite is an example of tetragonal crystal system.
ur
Reason. For a tetragonal system, a = /?^c, a=[i = 90®, y = 120®. (AUMS 2006)
84. Assertion. Hexagonal close packing is more closely packed than cubic close packing.
s
Reason. Hexagonal close packing has a coordination number of 12 whereas cubic close packing has a
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coordination number of 8.
85. Assertion. Zinc blende and wurtzite both have/cc arrangement of sulphide ions.
eB

Reason. A unit cell of both has four formula units of ZnS.

86. Assertion. Size of the cation is larger in tetrahedral hole than in an octahedral hole.
r

Reason. The cations occupy more space than anions in crystal packing.
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ad

87. Assertion. In a unit cell of NaCl, all Cl” ions touch Na^ ion as well they touch each other.
Reason. Radius ratio rjr_ in NaCl is 0-414.
Y

88. Assertion. If the length of the unit cell of LiCI having NaCl structure is 5-14 A, the ionic radius of Cl” ion
Re
nd

is 1-82 A.
Reason. Anion-anion contact is retained in LiCI structure because anions constitute the lattice.
Fi

(AUMS 2010)
89. Assertion. The sum of the radii of Na"^ and Cl” ions in NaCl crystal is 281 pm. Hence, edge of the unit cell
is 281 pm.
Reason. Edge of the unit cell is the distance between the centres of Na"*" and Cl” ions touching each other.
90. Assertion. No compound has both Schottky and Frenkel defects
Reason. Both defects change the density of the solid (AUMS 2008)
91. Assertion. When 1-0 mol of NaCl is doped with 10“^ mol SrCl2, the number of cationic sites remaining
vacant is 10“^.
Reason. Each SrCl2 unit produces two cation vacancies. (AUMS 2014)
92. Assertion. Antiferromagnetic substances on heating to high temperature become paramagnetic.
Reason. On heating, randomisation of spins occurs.
EZHEHIi-.
19. ih) 80. W) 81. rti (>■»
' 84. id) 85. id) 86. {d) 87. {d) 88. Ut) 89. {d)
90. id) 91. (d) 92. ia)
SOLID STATE 1/69

For Difficult Questions

Multiple Choice Questions (with one correct Answer)

1. Silicon has covalent network structure whereas
sulphur (Sg), phosphorus (P4) and iodine (I^)
are molecular solids.
Distance between A to B =
ir a

2. If a substance is an insulator in both solid and (calculated tiignometrically)

molten state, it cannot be ionic or metallic. If .●. Distance between two consecutive A layers (d)
the melting point is higher, it cannot be
-4

w
molecular solid. It should be covalent network x2r (va = 2r)
solid.

Flo
3. (fl) is incorrect because diamond has three
= 4r
dimensional network whereas graphite has
two dimensional network.

ee
FIGURE 1.58]
(b) is incorrect because both diamond and

Fr
graphite are covalent,
A LAYER
(c) is a correct statement.

for
ur
(d) is incorrect because only graphite is used as
lubricant.
4. One unit cell has 4 NaCl units, i.e., has mass
/ ●N
s
B LAYER
● /
k
Yo
4x58-5
oo

g
6-02x1023
eB

6-02x1023 A LAYER
.’. No. of unit cells in 1 g =
4x58-5 a sin 60°
a(=2r)
r

= 2-57 X 102‘.
ou
ad

5. Contribution by atoms A on the comers ABAB... type packing (hep)

Y

1
4 = 8x--1 8. Molar volume from pyknometric density
M
Re
nd

Contribution by atoms B on the face centres m3

2-165x103
< 1 5
Fi

=5x-=-
2 2
Molar volume from X-ray density

(Total atoms on face centres = 6. As one atom M

m3
is missing, atoms actually present = 5) 2-178x103
Volume unoccupied
Ratio of A : B = 1: ^=2:5.
2 M I 1
m
3 _ OOI3MXIQ-3
Hence, formula = A2B5. 1031,2-165 2-178 2-165x2-178

6. In 14 types of Bravaislattices, body centredunit .'. Fraction unoccupied

cell is present in cubic, tetragonal and
0-013MX10-3 Mx10“3
orthorhombic crystal systems. Hence, the = 5-96x10-3
2-165x2-178 2-165
number of body centred unit cells present in 14
types of Bravais lattices is 3. 9. In a simple cube, no. of atoms/unit cell
7. Distance between two consecutive A layers (d) 1
= 8x = 1
= A^B->A 8
1/70 ‘P'uidee^'^t New Course Chemistry (X11)E!ZSDB]

Hence, the resultant stoichiometry is A3B4.

For Difficult Questions 1
12. No. of M atoms = 4x- + 1=2
4
4 3
Volume of the atom - —nr (edge centres) (body-centre)
3
1
Volume of the cube = = (2 r)^ = 8 No. of X atoms - 8x- -i- 6x— = 4
2
(corners) (face-centres)

Unit formula = M2X4

ow
Fraction occupied = and empirical formula = MX2
13. In the cep lattice of oxide ions, no. of ions
10. In the two-dimensional packing, per unit cell
Packing fraction
1 1
=8x-+6x-=4

e
Area occupied by circles 8 2
within the square

re
No. of octahedral voids = 4
Area of the square
No. of tetrahedral voids = 8

Frl
F
Ixnr-
1
As 50% octahedral voids are occupied by Mj
a

= Vab2 h-BC^ = +cr =V2 a

ou .●. No. of M| atoms per unit cell = 2

osr
AC As 12-5% of tetrahedralvoids are occupiedby M-,
But AC = 4 r(r = radius of each sphere) 12-5

-J2a = 4r or a = r = 2-yfz r kf
No. of M2 atoms per unit cell = 100
x8 = l
oo
V2
.'. Formula of the compound is (Mj)2 (M2) O4
.●. Packing fraction
Y
.●. Oxidation states of Mj and M2 are -t- 2 and +4,
B

2x7cr^ 7t 3-143
= 0-785,/.e., 78-5%. 14. Suppo.se number of atoms ‘B’ in the cep lattice =
(2V2r)2 4 4 n
re
Y

11. There were 6 A atoms on the face-centres. No. of octahedral voids = n

u

Removing face-centred atoms along one of the No, of tetrahedral voids = 2 n

ad

axes means removal of 2 A atoms.

do

Atoms ‘A’ occupying half of the octahedral voids

Now, no. of A atoms per unit cell n

1 1 2
in

= 8x- -I- 4x- = 3

8 2
Re

Atoms ‘O’ occupying all tlie tetrahedral voids = 2 n

(comers) (face-centred)
F

n 1
No. of B atoms per unit cell .-. RatioA:B:0 = -:n:2n=-:l;2=l:2;4
2 2
1
= 12x- + 1 = 4
4 Hence, the formula is AB2O4.
(edge centred) (body centred) 15. In cep lattice, Z = 4. No. of 0-atoms per unit
A
t
B
cell = 4 (0--).
No. of octahedral voids = 4 and No. of
B tetrahedral voids = 8.
A a
A
B
As m fraction of octahedral voids is occupied by
A
B
Al^'*’ ions, therefore, AP'*' ions present = 4 m.
Similarly, Mg-'*' ions = 8 n. Hence, formula of the
B B mineral is AI4,,, Mg^^ O4. As total charge on the
A
compound is zero, hence
7b
4 m (-1- 3) -i- 8 n (-^ 2) + 4 (- 2) = 0
or 12 /?j -i- 16 u - 8 = 0
SOLID STATE 1/71

.●. Body diagonal, AE = IrCs+ + 2/-

C!"
For Difficult Questions

Substituting the given values of m and n, But body diagonal = a

equation is satisfied only when in = — and cr

2
1 1 1
n = - as 12x- + 16x — 8 = 0
8 2

Cs"
16. Cation A"*" occupies octahedral void formed by
anions X"
Radius of octahedral void = 0-414 x Radius of a

anions the packing

(From right angled ACDE, CE = = -Jl

w
= 0-414 X 250 pm = 103-5 pm = 104 pm a

17. For the/cc structure, nearest neighbour distance From right angled AACE,

Flo
a
(^) =
V2-
AE = ylAC-+CE~ =yla- + 2a~ =yj3a)
18. If atoms A = n, then tetrahedral voids = 2 n. Hence.
2 (r , +r )=^^3 a

ee
+
Cs cr

2 ^ 4

Fr
atoms M = — x2n = — ii
'c,-+'cr=T
or a
3 3
25. Similar to Q. 21 above.

M : A = ^ : 1 = 4 :3 for
ur
ZxM
26. Density, p = -30
19. For a close packed structure containing n atoms, ij^xN^xIO
s
there are n octahedral voids and 2 n tetrahedral
k
4xM
Yo
voids. As hexagonal primitive unit cell contains 6 2-72 =
oo

atoms, therefore, there are 12 tetrahedral voids and (404)-'x(6-02x1023)x 10-30

eB

6 octahedral voids.
or M =
2-72x(404)3 x 6-02x10-3 x lO'^O
20. NaCl has face-centred cubic arrangement of Cl" 4
ions and Na'*’ ions are present in the octahedral
r

-I
= 27 g mol
ou
ad

voids. Hence, for such a solid, radius of cation

ZxM
27. p = . The aim is to find Z.
(
Y

= 0-414 X radius of the anion or— = 0-414 «3xN 0

;●
Re
nd

ZxI8
i.e. = 0-414 X /●_ 0-92 =
(132x10-24) (6-02x1023)
Fi

or r ^ ^ 100 = 241-5 pm.

“ 0-414” 0-414 Zx 18
-1
a 132x6-02x10
21. For fee lattice, r = = 0-3535 a
iS 0-92 X 132 X 6-02 X10"'
= 4-06
or Z =
= 0-3535 X 408 pm = 144 pm 18

Diameter = 2 r = 288 pm. 28. When a is in pm and other quantities in CGS

units, we have
S V3
22. For bcc structure, r = a = — X 288 pm ZxM
4
g/cm3
23. Z = 4 means structure is fee. For fee
361
a
- 127-65 pm rr 127 pm For bcc, Z = 2, M = 6-94 g mor^
^ 2V2 2x1-414 530x1000
24. In body-centred cubic (bcc), oppositely charged p = 530 kg m 3 - = 0-53gcm 3
ions touch each other along the body diagonal.
lx (100)3 cm3
1/72 New Course Chemistry (XIl)E!SZai

2(r
Rb-*^
+ rX_ )-2(r^+r
/ \ x+ x“ ) = 0-30A
For Difficult Questions

or r■ j. ~r . = 0-15 A
2x6-94 Rb+ K+
3 _
a
0-53x6-02xlo2^xl0'^° + 0-I5A = 1-33+ 1-15 = 148 A
Rb-" V
= 4-35x 10'^ = 43-5 X 10^ 37. In NaCl with fee arrangement of Cl” ions (See
a = (43-5)*^3 X 10^ = 3-52 x 10^ pm = 352 pm Fig. 1.33, page 1/25), number of Cl” ion = 14,
Na'*' ions = 13. On interchanging their positions.

ow
ZxM Cl” ions will be 13 and Na"*" ions will be 14.
29. Density (p) =
a^xN 0 38. The number of C-atoms in a unit cell of diamond
is 8 and not 4 as explained below.
As M and a remain constant and Nq is also
constant, p « Z Diamond is a covalent crystal in which each

e
C-atom is sp^ hybridized. Thus, each C-atom is
^bcc _ ^bcc - ^ ^ covalently bonded to four other C-aloms

re
P/< cc ^fcc 4 2 tetrahedrally. These tetrahedra are linked together

Flr
into a three dimensional giant molecule.
30. For each Sr^'*’ ion introduced, one cation vacancy

F
is created because 2 Na"'' ions are removed and The structure of diamond is similar to that of ZnS
ou
one vacant site is occupied by Sr^'*'. Doping with (Fig. 1.35, page 1/27 in which S^”ions form fee
10”^ mol % of S1CI2 means 100 moles of NaCl lattice and Zn^'*’ ions are present on the body

sr
are doped with 10"4 mole of SrCl2- diagonals and occupy alternate tetrahedral voids).

fo
SrCl2 doped per mole of NaCl = 10^/100 In diamond, we have C-atoms in place of Zn^*

k
and S^” ions. Thus, diamond has face-centred
= 10”^ mole = 10”^ x (6 02 x lO^^) Sr^+ ions
cubic structure in which C-atoms are present at
oo
= 6-02x 10*'^Sr2-^ ions the comers as well as face-centres and alternate
Y
Hence, concentration of cation vacancies tetrahedral voids.
reB

= 6-02x lO^’^mol”'
1
31. In FeO, as some Fe may be present as Fe^"*" and Contribution of C-atoms at the comers = 8 x - = 1
8
uY

for charge neutrality 3 Fe^"^ = 2 Fe^'*’. Hence, FeO

Contribution of C-atoms at face-centres
should have metal deficiency defect.
32. The formula MQ.9gO shows that if there were 100 1
ad
do

= 6x-^ = 3
0-atoms present as O^” ions, then there 98 M 2
atoms present as M^'*' and Suppose = x, As four C-atoms are present on the body diagonals
in

then = 98 - jc. As the compound as a whole is

i.e., in the alternate tetrahedral voids, their
neutral.
Re

contribution = 4
Total charge on and M^'*’ = Total charge on
F

Total no. of C-atoms in the unit cell =1+3 + 4

100 02- ions
= 8
xi+ 3) + (98-j:)(+ 2) = 100x2
or 3 a: + 196 - 2 AT = 200 or a: = 4 39. We aim at: (r^^
K + Cl ).

%of M^'*' =^xl00 = 4-08% Given

r
cr
= 0-55 and
r
K

Cl”
= 0-74

34. Mn02 = Antiferromagnetic, Ti02 = Diamagnetic,

VO2 = Paramagnetic, C1O2 = Ferromagnetic. +1 = 1-55 and
il +1 = 1-74
35. (fl) is the correct option because diamond is an r r
cr cr
insulator and non-metallic character decreases
down the group. + r
Cl”
i.e.. = 1-55 ...(/)
36. As RbX crystallies in NaCl type lattice, unit cell r
Cl"
length of RbX = 2 (rRb + rX )
+ r
Cl”
= 1-74
Unit cell length of KX = 2(rK ^ + rX _) and
r
...00
cr
SOLID STATE 1/73

r'N-
Volume occupied by the atoms
For Difficult Questions Packing fraction =
Volume of the unit cell
Dividing (»0 by (/), As bcc contains 1 atom at the body centre and
contribution of atoms at the corners is also
1-74
= M23
1-55 1
^NA-*-''' ^cr = 8x- = l
40. Non-stoichiometric ferrous oxide is
^®0 93-o-98 ^100 ^^^0 98^ which is due -Kr^ +-n{2r)^
to metal deficiency defect. Hence, (a) is incorrect. 3 3
Frenkel defect is favoured in those ionic
Packing fraction = 3
a

compounds in which sizes of cation and anion

are largely different. Hence, (d) is incorrect. -Ti:(r^+8r^)
3 471(9
41. For the given bcc unit cell, if a is the edge length

w
and r is the radius of the atom at the comer so that (2^3 r)3 3x8x3-^/-^
2 r is the radius of the atom at the body centre, 71
= 0906

F lo
then VSfl =2r + 2(2r) 2-73
6r % packing efficiency = 0-906 x 100
yj3a = 6r = 2^r

ee
or or a =
73 = 90% approx.

Fr
D1 Multiple Choice Questions (with One or More than One Correct Answers)

for
ur
42. Octahedral void is so called because it is (a) and (c) are also correct.
surrounded by six atoms. Edge length of A* B“ Only (b) is wrong because for a metal with bcc
s
crystal = 2 x distance between A"*" and B”. structure, coordination number = 8.
ok
Yo
43. Only (d) is not true because in Na20, O^” ions 47. Frenkel defect is also called dislocation defect
o

are present in close packing and Na'*’ ions occupy because smaller ions are dislocated from their
eB

all the tetrahedral voids.

lattice sites into the interstitial sites. Hence, (b) is
44. Only (a) is wrong because in tetragonal system, a
= b^c.ln all other cases a*b^c. correct. Trapping of an electron leads to the
r

formation of F-centre. Hence, (c) is correct.

ou
ad

45.
48. (a) Diamond is harder than graphite. Hence, (a) is
incorrect.
Y

(b) Graphite is a good conductor of electricity as

<e)
Re
nd

each C-atom is attached to three other C-atoms

and 4th electron is free while in diamond, all the
Fi

8 four valencies are satisfied, it is insulator. Thus.

(b) is correct,
(c) Diamond is better thermal conductor than
graphite. This is because thermal conduction is
due to the transfer of thermal vibrations from
Z = 0 forms the bottom layer, Z = a/2 forms
atom to atom which is easier in a compact crystal
second layer above it and Z = a forms the third
like that of diamond. Hence, (c) is incorrect.
layer. Atoms 5, 6, 7, 8, 9 and 14 are present at
face centres. Atoms 1, 2, 3, 4, 10, 11, 12 and 13 (^0 In graphite, C-C bond acquires some double
are present at the comers. Atoms 4,2,5 form one bond character. Hence, it is has higher bond order
face diagonal ; 4, 9, 10 form another face than diamond. Thus, (d) is correct.
diagonal; 4, 8, 12 form one more face diagonal. 49. (a) is incorrect because for any atom in the top
Hence, correct options are b, c and d. most layer, coordination number is not 12 as
46. (d) is correct because length of unit cell there is no layer above the topmost layer.
= 2(rNa-^ + r^j_) = 2(95 + l81)pm (b) is a known fact,
1/74 ‘P’utdeep. ‘<x New Course Chemistry (XII)BSIS

of octahedral voids per atom = 1 and number of

For Difficult Questions
tetrahedral voids per atom = 2.
a
(c) is conect because in ccp (fee), number of (d) For eep (fee), r = or a =^2jlr
atoms per unit cell is 4. Hence, octahedral voids 2V2
= 4 and tetrahedral voids = 8. Therefore, number (Refer to Table on page 1/33)

ii IVIuttipie Choice Questions (Based on the given Passage/Comprehe nsion)

50. The arrangement of layers is ABC ABC type, 56. Refer to Fig. 1.58, page 1/69.
i.e.. it has cubic dose packing arrangement. Volume of the unit cell = Base area x height (h)
Hence, the coordination number is 12. Base area of regular hexagon

w
51. In ccp, the fraction occupied is 0-74. = Area of six equilateral triangles,
each with side a (i.e., 2r)
52. See Problem For Practice 6, page 1/14.

Flo
ZxM ZxM = 6x a
3 _
5.3. p = or a 4
fPxN,,x
0
10-30 pxNqxlO 30 1

e
Area of isosceles triangle =— x Base x Height

re
For ccp, i.e., f.c.c., Z = 4.

F
Hence, «3 =
4x197
—- —1 X a X a sm 60
.£f\o t 2 x—
=—a
^ = —a
2
19-4 X 602x1023x10-30 2 2 2 4
ur
r
= 6-747 X 10"^ = 67-47 X 10^
Height h = 2 X Distance between closest packed
or a = (67-47)
1/3
X 1Q2
fo layers (A and B)
ks
= 4-07x I02pm = 407pm 2
Yo
= 2x — a
\3
oo

a
54. For fee, r = = 0-3535a
2S (By applying rules of Trigonometry)
B

.'. Volume of unit cell

re

= 0-3535x407 pm= 143-9 pm

6x s , 2x,l^alJ = 3V2«3
55. According to the given description, the
I
u

4
ad

arrangement of atoms in HCP will be as shown

Yo

in Fig. 1.58, page 1/103. Effective number of Putting Cl = 2 r. Volume of the unit cell
atoms present in the unit cell of HCP = 3V2 (2r)3 =24V2r3
d
Re

1 1 57. Space occupied by spheres in HCP = 74%

in

= 12 X — (comers) + 3 + 2 x — (face centres) = 6.

Empty space = (100 - 74)% = 26%
F

tvlatching Type Cl

62. NaCl-Diamagnetic ; MnO-Antiferromagnetic ; CrCl3-Paramagneti c ; Cr02-Ferromagnetic ;

MgF'c204-Ferrimagnetic

I ■ 1
''--J— Integer Type Questions
66. In one hep unit cell, atoms in the packing = 6. 68. Z =
pxfl3xN 0
Hence, octahedral voids = 6, tetrahedral voids M
- 12. Difference = 12-6 = 6.
(4gcm 3)(5x10 *^cm)3(6-02x10^3mol ') ^
67.
r(NHj)/r(Br-) = 143/196 = 0-730 which lies 72g mol"*
in the range 0-414 - 0-732. Hence, coordination Thus, there are 4 formula units (FeO) per unit
number = 6.
cell. Hence, O^- ions = 4.
SOLID STATE 1/75

For Difficult Questions

69. Refer to Solved Problem 2. page 1/48.

70. As given molecule is neutral, sum of oxidation
states = 0. Hence,
n (+ 3) + 2 (+ 2) + 2 (+ 2) + 3 (+ 2) + 10 {- 2) = 0
or 3 n + 4 + 4 + 6 - 20 = 0 or 3 « = 6 or n~2. ZxM
73.
71. At low temperature, AICI3 exists as a close
packed lattice of Cl” ions with ions
-1
occupying octaliedral voids. Thus, Al^'*' ions has where a is in pm and M in g mol
6 Cr ions around it. Hence, its coordination

ow
4xM
number is 6. 8 = or M = 76-8
(400)^x6xl0^^xl0"^‘^
72. Truncated octahedra (cubo-octahedra) are the
structures shown by zeolites. A truncated 256
Moles in 256 g — = 3-33
octahedron is an Archimedean solid. It has 14 76-8
faces (8 regular hexagons and 6 square). 36 No. of atoms = 3-33 x 6 x 10-^ = 2 x 10^^

e
edges and 24 vertices.

re
VII.

rFl
F
Numerical Value Type Questions (in Decimal Notation)

r
74. According to the given information, cations are After step (n),
ou
fo
present on the comers and face centres and anions anions (X“) =
ks 1 + 6
are present on the edge centres and body centre. body centre face centres
In MX, their contributions towards the unit cell cations (M^) = 8 + 0
oo
are as under: comers face-centre

After step (Hi),

Y
1
eB

Anions(X)= 12x— + 1 =3+1=4

1 6
4 (body centre) anions (X") = +

body centre facecemres

(edgecentres)
cations (M'*') = 0 + 0
r

1 1
Cations (M'*') = 8x- + 6x- =l-t-3 = 4 comer face centre
ou

8 2
ad
Y

(comers) (faceccntres) After step (iv),

After step (i), anions (X“) left = 4 - 3 = 1 anions (X“) = 0 + 6

body centre face centres
d

After step (i7), cations (M"*") left = 4 - 3 = 1,

cadons (M'*') = 1
Re
in

anions (X“) = 1 -f 3 = 4 body centre 1

As anion at the face centre has contribution = —
F

After step (Hi), cations (M'*’) left =1-1=0 towards the unit cell. ^
After step (iv), anions (X“) left = 4-1=3,
cations (M'*') = 0+1 = 1 .●. Anions (X ) per unit cell in Z = 6 x — = 3

numberof anions 3 As cation at the body centre has contribution= 1

Hence, ratio in Z = - =3
Cations (M"*") per unit cell in Z = 1
numberof cations 1
numberof anions
Alternatively, Ratio inZ =- = 3
numberof cations I
Anions (X“) = 12 + 1
edge centres body centre
V3
75. For bcc, r = — a = x27A
Cations (M'*') = + 6 4 4
comers facecemres
a

After step (/), anions (X ) = I For fee, r=

btxly centre 2V2
cations (M'*’) = 8 + 6
comers facecemres or a ' = 2■^^r = 2V2x -x27 = 32-9 A = 33 A
4
1/76 ‘Pn^xdee^ 4 New Course Chemistry (XIl)CZsIS]
w Total voids = 15 x 10^'
For Difficult Questions
77. Suppose 0^“ ions = 100. in the given sample,
0-581
Fe2+ + Fe^+ = 93
76. No. of moles of Ga =
70 ’ Suppose Fe^'’" = x, then Fe^'*’ = 93 - x.
0-581
For charge balance, 2x + 3 (93 - x) = 200
No. of atoms = X (6-023x1023) = 5x K)21 or 279-x = 200 orx=79
70
79
Octahedral voids = 5 xlO^^ % of Fe^"*" ions = ” ^ 1^0 - 85%
Tetrahedral voids = 2 x 5 x lO^* =: 10 x lO^' 93

mu Assertion-Reason Type Questions

78. Correct Statement-2. Covalent bonds are Correct R. A unit cell of zinc blende has 4
weaker than Ionic bonds. (Higher melting points formula units while that of wurtzite has 6
of covalent crystals are due to net-work structure). formula units of ZnS.

w
79. Correct explanation. Na'*' ions can be placed 86. Correct A. Octahedral hole is larger in size than
only in the octahedral voids. (Tetrahedral voids tetrahedral hole.

F lo
are too small to accommodate them). Correct R. Cations are generally smaller. They
80. Correct Statement-1. Octahedral voids are occupy the voids and hence occupy less space.
larger in size than tetrahedral voids but not 87. Correct A. In a unit cell of NaCl, all Cl" ions

e
double in size. touch Na"*" ion but do not touch each other.

Fre
Statement-2 is correct. Correct R. Actual radius ratio r Jr is 0-525
Na"^ cr
81. Statement-2 is the correct explanation of
Statement-1.
88.
for
and not 0-414.
If anion-anion contact
82. Statement-1 is correct because of Schottky defect is retained, then as
r
(where equal number of anions are also missing) shown in the adjoining
You
oks

or impurity defect (where some host cations are fig.,

replaced by cations of the impurity).
eBo

Interionic distance of
Statement-2 is also correct according to Frenkel a 5.14A
defect but not the correct explanation of LiCl = -
2 2
= 2-57 A
Statement-1.
ad
our

Correct explanation. Electrical neutrality is not BC = Vab2-|-AC2 = 7(2-57)2-I-(2^57)2 = 3.53 A

only due to shifting of some cations to interstitial
1 1 y
sites but may also be due to missing of equal Radius of Cl" ion = — BC = — x 3-63 — 1-8 lA
number of cations and anions from the lattice sites 2 2
Re
dY

or due to impurity defect {e.g., 2 Na'^ ions replaced Hence, R is the correct explanation of A.
by one Mg^'^ ion).
Fin

89. Correct A. The edge of NaCl unit cell is the

83. Correct A. Graphite is an example of hexagonal distance from Na'*' to next Na"*" ion or from Cl" ion
crystal system (Each C-atom is sp^~ hybridized to next Cl" ion.
and is linked to three other C-atoms in a Correct R. Edge of the unit cell of NaCl is
hexagonal planar structure). double the distance between the centre of Na"*"
Correct R. For a tetragonal system, a = b ^ c, and Cl" ions touching each other.
a = P = Y = 90®. For a hexagonal system a = b^c, 90. Correct A. Some compounds may have both
a = p = 90®,y= 120®. Schottky and Frenkel defects (e.g., AgBr).
84. Correct A. Hexagonal dose packing and cubic Correct R. Schottky defect lowers the density
close packing are equally close packed (space while Frenkel defect does not.

occupied = 74%). 91. Correct A. As each SrCl2 unit creates one cation
Correct R. Both have a coordination number of vacancy, number of cation vacancies will be
12. 10"3 mol = 10"3 X (6-02 x 1Q23) = 6-02 x lO^O.
85. Correct A. Zinc blende has fee arrangement of Correct R. Each SrCl2 produces one cation
S2" ions while wurtzite has hexagonal close vacancy.

packing arrangement of S2" ions. 92. R is the correct explanation of A.

 w
SOLUTIONS

Flo
ft iBMHr. r»aTiw»MfK.J‘n»«=Wha»fc» snok

ee
2.1. GENERAL INTRODUCTION

Fr
In our everyday life, we rarely come across pure substances. Most of these are homogeneous mixtures
of two or more pure substances. By homogeneous mixture, we mean that the composition and properties are

for
ur
uniform throughout the mixture. Such homogeneous mixtures are called solutions. Further, a solution may
contain varying amounts of the components. Hence, a solution may be defined as follows :
s
I
k
A solution is a homogeneous mixture of two or more chemically non-reacting substances whose
Yo
oo

composition can be varied within certain limits.

eB

In a homogeneous mixture, all the particles are of molecular size, i.e., up to 10“^ m in diameter and the
different constituents of the mixture cannot be separated by any of the physical methods like filtration, settling
or centrifuging. Every solution is made up of a solvent and one or more solutes. A solution containing only
r
ou
ad

one solute dissolved in a solvent is called binary solution.

A solvent is that component of the solution which is present in larger amount by mass than the other
Y

component, termed as solute.

Re
nd

Solutions in water are called aqueous solutions, and the solutions in which water is not the solvent are
called non-aqueous solutions. The solvents in the non-aqueous solutions are usually benzene, ether, carbon
Fi

tetrachloride, etc.
Importance of solutions. The use of a particular solution in everyday life depends upon its composition.
For example,
(i) Brass is a homogeneous mixture of copper and zinc, German silver is that of copper, zinc and nickel while
bronze is that of copper and tin. They have different properties and hence are put to differentuses.
(«) 1 part per million parts (1 ppm) of fluoride ions in water prevents tooth decay, 1-5 ppm causes the
teeth to become mottled (coloured) and still higher concentration can be used as poison, e.g., for rats
in the form of sodium fluoride.

(Hi) Solutions of intravenous injections should have the same ionic concentration as that of the blood plasma.

2.2. TYPES OF SOLUTIONS

A solution can be solid, liquid or a gas depending upon the physical state of the solvent. The various
types of solutions are listed in Table 2.1.
2/1
2/2 New Course Chemistry (XII)BEIHl

TABLE 2.1. Various types of solutions

! S. No. Solute Solvent Type of Sol. Examples

SOLID SOLUTIONS (Solid Solvent)

1. Solid Solid Solid in Solid Alloys (brass, German silver, bronze. 22 carat gold etc.)
2. Liquid Solid Liquid in Solid Hydrated salts, Amalgam of Hg with Na
3. Gas Solid Gas in Solid Dissolved gases in minerals or in Pd

LIQUID SOLUTIONS (Liquid Solvent)

4. Solid Liquid Solid in Liquid Salt or glucose or sugar or urea solution in water
5. Liquid Liquid Liquid in Liquid Methanol or ethanol in water

w
6. Gas Liquid Gas in Liquid Aerated drinks, O2 in water
GASEOUS SOLUTIONS (Gaseous solvent)

F lo
7. Solid Gas Solid in Gas Iodine vapours in air, camphor in N2 gas
8. Liquid Gas Liquid in Gas Humidity in air, chloroform mixed with N2 gas

ee
9. Gas Gas Gas in Gas Air (O2 + Nj)

Fr
Out of the various types of solutions listed in the table, the most significant types of solutions are those

for
which are in liquid phase, i.e., liquid solutions. Hence, we shall confine ourselves to the study of solutions of
ur
solids, liquids or gases in liquids.
s
2.3. EXPRESSING CONCENTRATION OF SOLUTIONS
ook
Yo

The concentration of a solution is the amount of the solute dissolved in a known amount of the solvent
eB

or solution. The concentration of solution can be expressed in various common ways as discussed below :
1. Percentage ; The percentage of a solution is usually expressed either as ‘percentage by mass’ or
‘percentage by volume’ or ‘mass by volume’*
our
ad

In case of a solid dissolved in a liquid, the percentage by mass means the mass of the solute in grams
present in 100 g of the solution, i.e., it is weight/weight (w/w) and percentage by volume means the mass of
Y

the solute dissolved in 100 ctr? or 100 mL of the solution, i.e., it is weight/volume (w/v). For example, 10%
Re

solution of sodium carbonate by mass means 10 g of Na2C03 are present in 100 g of the solution. Similarly,
nd

10% solution of Na2C03 by volume means 10 g of Na,C03 dissolved in 100 cm^ or 100 mL of the solution.
Fi

In case of a liquid dissolved In another liquid, percentage by mass has the same meaning but percentage
by volume means the volume of the liquid solute in cm^ present in 100 cm^ of the solution, i.e., it is volume/
volume (v/v). For example, 10% solution of ethanol in water by weight means 10 g ethanol are present in 100
g of the solution {i.e., 90g of water) and 10% solution of ethanol in water by volume means lOcm^ of ethanol
are present in 100 cm^ of the solution {i.e., dissolved in 90 cm^ of water).
Thus, the formulae for the calculation of percentage concentration may be expressed as follows :

Mass of the component in the solution

Mass % of a component (vv/vv) = xlOO
Total mass of the solution

Volume of the component

Volume % of a component {v!v) = XlOO
Total volume of the solution

♦When nothing is mentioned, it stand.s for percentage by mass. Further, cm-^ (cc) and mL arc taken as equivalent.
Hence, they are used without any distinction.
 SOLUTIONS 2/3

Mass/volume % of a component {w!v) = Mass of the component in the solution xlOO

Total volume of the solution in mL

2. Strength : The strength of a solution is defined as the amount of the solute in grams present in one
litre (or dm^) of the solution, and hence is expressed in g/litre (or g/din^) and is usually written as g
Thus, the formula for calculation of strength may be written as :
Mass of the solute in grams
Strength of a solution (g L ' or g dm =
Volume of the solution in litres (dm^)
3. Molarity ; Molarity of a solution is defined as the number of moles of the solute dissolved per litre
(ordm^) of solution. It is denoted by ‘M’.

w
Number of moles of solute
Mathematically, M =
Volume of the solution in litres

o
Strength in grams per litre

e
M can be calculated from the strength as : M =

Fl
re
Molar mass of solute in g mol''*

F
w 1000
If ‘w’ gram of the solute are present in V cm^ of a given solution, then M = X
ur Molar mass V

r
fo
Thus, a solution of sulphuric acid having 4-9 grams of it dissolved in 500 cm^ of solution will have its
ks
4-9 g lOOOcm^ L“'
M = X
Yo
molarity. = 01 mol L
98 g mol ● 500 cm^
oo

In case of ionic compounds like NaCl, Na2C03 etc., formality is used in place of molarity. The formality
eB

of a solution is defined as the number of gram formula masses of the solute dissolved per litre of the solution.
It is represented by the symbol F. The term ‘formula mass’ is used in place of molecular mass because ionic
compounds exist as ions and not as molecules. Formula mass is the sum of the atomic masses of the atoms in
ur

the formula of the compound.

ad
Yo

of Na^CO^ contains 04 M Na"^ ions and 0-2 M CO^“

It is important to note that a 0-2 M aqueous solution
ions because in aqueous solution, one mole of Na2C03 gives 2 moles of Na'*' ions and 1 mole of CO|“ ions.
d
Re
in

4. Normality : Normality of a solution is deifned as the number of gram equivalents of the solute
dissolved per litre (dm^) of given solution. It is denoted by ‘N’.
F

Number of gram equivalent of solute

Mathematically, N =
Volume of the solution in litres

Strength in grams per litre

N can be calculated from the strength as : N =
Eq. mass of solute
w 1000
If ‘ w' gram of the solute are present in V cm^ of a given solution, N = X

Eq.mass of the solute V

Thus, a solution of sulphuric acid having 0-49 gram of it dissolved in 250 cm^ of solution will have its
normality.

0-49 g lOOOcrn^ L-i

N = X -1
-1
250 cm^ = 0 04 g eq L (Eq. mass of sulphuric acid = 49).
49geq
2/4 'P'uxdeep. New Course Chemistry (XII)

Equivalent mass. («) Equivalent mass of an element is the mass of the element which combines with or displaces
J-008 parts by mass of hydrogen or 8 parts by tnass of oxygen or 35-S parts by mass of chlorine.
Atomic mass of the element
Eq. mass of an element =
Valency of the element

Mol.massof the acid

Hi) Eq. mass an acid =
Basicity of the acid
Basicity is the number of displaceable ions from one molecule of the acid.
Mol.massof the base
HU) Eq. mass of a base =
Acidity of the base

ow
Acidity is the number of displaceable OH" ions from one molecule of the base
Mol.massof thesalt
(/V) Eq. mass of a salt -
Total positive valency of the metal atoms

e
Formula mass of the ion
(V) Eq. mass of an ion =

re
Fl
Charge on the ion

F
Mol. mass or At. mass
(V/) Eq. mass of an oxidizing/reducing agent =
No. of electrons lost or gained by one molecule of the substance
ur
r
Relationship between normality and molarity of a solution. Suppose a solution is x molar and molecular mass of
the solute is M and its equivalent mass is E. Then
fo
ks
atxM I M
Yo
Molarity of the solution = x mol L ’ = x x M g L * ~—
b
C , i.e.. Normality of the solution = XX —
E
oo

MoLmass
eB

Thus, in general Normality of a solution= Molarity x

Eq.mass
M M
For an acid, — = basicity and for a base, — = acidity.
ur

E E
ad

Hence, Normality of an acid = Molarity x Basicity ;

Yo

Normality of a base = Molarity x Acidity

e.g., Normality of HjSO^ = 2 x Molarity of H2SO4
d

Some common fractions of molarity and normality

Re
in

M M M M
= Centimolar, = Millimolar.
F

1 M = Molar, — = Semimolar, — = Decimolar,

2 10 100 1000

Similar names are used for normality.

5. Molality : Molality of a solution is deifned as the number of moles of the solute dissolved in 1000
grams (1 kg) of the solvent. It is denoted by ‘w’.
Number of moles of the solute Number of moles of the solute
Mathematically, m = or m = xlOOO
Mass of the solvent in kg Mass of the solvent in grams
If ‘a’ gram of the solute are dissolved in ‘b' gram of the solvent, then
a 1000
m — X
Molar mass of the solute b

Thus, a solution of anhydrous sodium carbonate (mol. mass = 106) having 1-325 grams of it dissolved
in 250 grams of water will have its molality,
SOLUTIONS 2/5

-1
1-325 g lOOOgkg
m =
-1
X
= 0-05 mol kg '
106 g mol 250 g

6. Mole Fraction : Mole fraction of a constituent (solute as well as solvent) is the fraction obtained by
dividing number of moles of that constituent by the total number of moles of all the constituents present in
the solution. It is denoted by 'x\
If in a solution, n^ and «2 are the number of moles of solvent and the solute respectively, then
n
1
Mole fraction of solvent in the solution. ^ =
+ «2

"2
Mole fraction of solute in the solution,
«1 + «2

w
If a solution contains 4 moles of alcohol and 6 moles of water, then

F lo
4
Mole fraction of alcohol = = 0-4
4 + 6

e
Fre
and Mole fraction of water - _ = 0-6
4+ 6

for
If a solution contains a number of components (say, i), then mole fractions of different components
will be
r
n
You
1
oks

●^1 = , and so on.

+«2 +nj + ...n-
eBo

The sum of mole fractions of all the constituents of a solution is always equal to unity, i.e.,
+ X2 + Xj+...= 1
ad
our

It may be noted that mole fraction is a dimensionless quantity.

7. Mass Fraction : It is defined as the mass of the given component per unit mass of the solution. If
and Xg denote the mass fractions of the two components A and B respectively when g of one component A
Re

is mixed with Wg g of the second component B in a binary solution, then

dY
Fin

W
B
=
and =

WA+Wg

Evidently, X^+Xg= 1
Mass fraction multiplied by 100 gives mass percentage, e.g., mass percentage of A = x 100.
Like mole fraction, mass fraction is also a dimensionlessquantity.
8. Parts per million parts (ppm): For very dilute solution, i.e., when a very small quantity of a solute is
present in a large quantity of a solution, the concentration of the solute is expressed in terms of ppm. It is defined as
the mass of the solute present in one million (lO^j parts by mass of the solution. Thus, for a solute A,
mass of A
PP^A = xIO^
mass of solution

The pollution of the atmosphere is also reported in ppm but it is expressed in terms of volumes rather
than masses, i.e., the volume of the harmful gas (e.g., SO2) in cm^ present in 10® cm^ of the air.
2/6 “P^uxeUe^'A New Course Chemistry (XII)ESSIMI

To sum up:
Mode of Definition Units
expressing concentration

1. Percentage by mass Mass of solute in g present in 100 g of solution % by mass

(solid in liquid or liquid in liquid) (dimensionless)
2. Percentage by volume (liquid in Volume of solute in inL present in 100 mL of solution % by volume
liquid) (dimensionless)
3. Strength Mass of solute in g present in 1 L of solution gL-'
4. Moiarity Moles of solute present in 1 L of solution mol L“ or M
-1
5. Normality Gram equivalents of solute present in 1 L of solution geqL orN

6. Molality Moles of solute present in 1 kg (1000 g) of solvent mol kg"* or m

w
7. Mole fraction {x) Moles of that component divided by total moles of all Dimensionless
the components
8. Mass fraction Mass of that component divided by total mass of all Dimensionless

F lo
the components
9. Parts per million parts by mass Mass of that component in one million parts of the ppm by mass

ee
mass of all the components
10. Parts per million parts by volume Volume of that component in one million parts by ppm by volume

Fr
volume of all the components
11. Formality (for ionic compounds) No. of gram formula masses of the solute present in F

for
1 L of the solution
ur
Retain in Memory
s
ook
Yo
It may be noted that molality, mole fraction, mass fraction, etc. are preferred to molarity, normality,
etc. This is because the former involve masses of the solute and solvent whereas latter involve
eB

volumes of solutions. Temperature has no elfect on mass but it has significant effect on volumes.
Hence, molality, mole fraction, mass fraction etc. do not change with temperature whereas molarity
r

and normality change with temperature.

ou
ad

Relationship between volume and normality or molarity of a solution (Normality equation and
Y

Molarity equation). If Vj mL of a solution of normality Nj are diluted to volume V2 mL and the normality
of the diluted solution is N2 or if V, mL of a solution of normality N[ react exactly with V2 mL of solution of
Re
nd

normality N2, then we have

Fi

N,xVi=N2XV2
This equation is called normality equation.
Similarly, if V, mL of molarity Mj are diluted to volume V2 mL and the molarity of the diluted solution
is M2, then we have
Ml X Vi = M2V2
This is called molarity equation. However, if reaction between two solutions is studied in terms of
their molarities, fu^t the balanced equation is written. The molarity equation for reaction between two reactants is

MjXVj M2XV2
n
1 ^2

where riy is the number of moles of reactant 1 and ^2 ihe number of moles of reactant 2 in the balanced
equation.
SOLUTIONS 2/7

Retain in Memory
(/) If volume V, of a solution of normality Nj is mixed with volume V2 of another non-reacting
solution of normality N2, then the normality N3 of the final solution can be calculated as follows :
V] + N, V. = N3 (V, + V2) or N3 = (N, V, + N, V^VCVj + V2)
Similarly, if molarities are used, M3 = (M, Vj + M2 V2)/(V, + Vj)
(ii) If a solution of an acid is mixed with the solution of a base, first calculate the number of gram
equivalents present in each of them. The smaller number of gram equivalents will neutralize an
equal number of gram equivalents of the other. Number of gram equivalents of the excess reactant
left divided by the total volume of solution in litres will give the normality of the final solution.
(Hi) As normality of acid = Basicity x Molarity i.e., and normality of base
= Acidity x Molarity, i.e., M^, normality equation becomes

w
Flo
FORMULAS USI

e
NUMERICAL.

re
PROBLEMS No.of molesof solute Icm^ = ImL
BASED
(0 Molarity =
Volumeofsolutioninlitresordm^ [idm-^ =1L

F
ON
No.of molesof solute
ur (ii) Molality =

r
The Calculation Mass of solvent in kg
of Molarity,
(///) Normality = fo
No.of geq.of solute
ks
Molality, Volumeof solution in litres or dm^
Yo
Normality etc. (iv) Mole fraction of a component in the solution
oo

_ No.of molesof that component

B

Totalnumberof molesof all the components

re

Massing Massing
(v) Moles = (vi) No. of g eq. =
Molar mass Eq. mass
u
ad

(vii) The normality equation is N, x Vj = N, x V2

Yo

(For dilution from volume Vj to V2 or reaction between two reactants)

(vm) The molarity equation is M[ x Vj = M2 x V2 (For dilution
d

i*-
from volume V, to V2)
Re
in

M, V, M2 V2
For reaction between two reactants.
F

n
1 ^2
where n, and ^3 are their stoichiometric coefficients in the balanced
equation.
(ix) On mixing two non-reacting solutions, final normality or molarity
is calculated using
Ni Vi + N2 V2 = N3 V3 or M, V, + M2 V2 = M3 (Vi -h V2)
■ \y
On mixing acid and base solutions, first calculate gram equivalents i:\
.s.

neutralised. Final normality = No. of g eq. left/Total volume in L. I-,

Problem
Q Calculate the molality and mole fraction of 2-5 g of ethanoic acid (CHjCOOH) in
75 g of benzene. (NCERT Solved Example, Assam Board 2012)
Solution. Mass of solute (CH3COOH ) = 2-5 g , Mass of solvent (CgHg) = 75 g = 0 075 kg
Molar mass of CH3COOH = 60 g mol"*,
-1
Molar mass of = 78 g mol
2/8 7^>iadee^'4i New Course Chemistry (XlI)BdSI

2-5g
Calculation of molality : Moles of the solute (CH3COOH) = -1
= 00417
60 g mol
Moles of thesolute 0-0417 mol -I
Molality = = 0‘556molkg
Mass of the solvent in kg 0-075 kg
Calculation of mole fraction : Moles of solute (n CH3COOH ) = 0 0417 (calculated above)
75 g
Moles of solvent H ) = = 0-961
6"6 -1
78gmol
n
CH3COOH 00417
Mole fraction of CH3COOH in the solution = = 0-0416

w
n ■¥n 0-0417+0-961
CH3COOH
Problem Q Calculate the molarity and normality of a solution containing 9-8 g of H2SO4 in 250 cm^
of the solution.

o
Solution. Mass of H2SO4 dissolved = 9-8 g, Volume of the solution = 250 cm^ = 0-250 L

e
-1
Calculation of molarity : Molar mass of H,S04 = 98 g mol

re
rFl
Massing _ 9-8g = 0-1 mole
No. of moles of H2SO4 =

F
Molar mass 98gmol ’
No.of moles of thesolute 0-1 mol
Molarity = = 0-4 mol L-' = 0-4M

r
ou
Volume of solution in litres 0-250 L

fo
Calculation of normality : Eq. mass of H^S04 =
ks
Mol. mass of H2SO4 _ 98 = 49
Basicity of H.,S04 2
oo
Massing 9-8
No. of g equivalent of HoS04 = — = 0-2
Y

Eq.mass
B

49

No. of g eq of the solute 0-2 geq

re

Normality =
Volumeof solutionin litres
= 0-8geqL ’ = 0-8 N.
0-250L
ou
Y

Problem Calculate the mole fraction of ethylene glycol (C^Hg02) and water in a solution
ad

containing 20% of C2Hg02 by mass. (NCERT Solved Example)

Solution. 20% of C2H6O2 by mass means that 20 g of C2H6O2 are present in 100 g of the solution, i.e..
d

Mass of solute (C2H^02) = 20 g Mass of solvent (H2O) = 100 - 20 g = 80 g

in
Re

-1 -1
Molar mass of C-,HgO, = 62 g mol Molar mass of H2O = 18 g mol
F

20 80
No. of moles of C2HgOT = —
62
= 0-322 ; No. of moles of H2O = TT
18
= 4-444
n
C2H6O2 0-322
Mole fraction of C2H5O2 in the solution = = 0-068
n 0-322 + 4-444
C2H6O2 ”H20
Mole fraction of H,0 in the solution = 1 - 0-068 = 0-932
Problem
0 Find the molarity and molality of a 15% solution of H2SO4 (density of H2SO4
= 1-020 g cm"^). (Atomic ma.ss : H = 1, O = 16, S = 32 a.m.u.).
Solution. 15 % solution of H2SO4 means 15 g of H-,S04 are present in 100 g of the solution, i.e.,
Mass of H2SO4 dissolved = 15 g ; Mass of the solution = 100 g ;
Density of the solution = 1-02 g/cm^ (Given)
Calculation of molality : Mass of solution = 100 g ; Mass of H2SO4 (solute) = 15 g
85
Mass of water (solvent) = 100 - 15 = 85 g = kg = 0-085 kg
1000
SOLUTIONS 2/9

15g
Molar mass of H2SO4 = 98 g mol 15gH2S04 = = 0153 moles
-1
98 g mol
No. of moles of solute 0-153mol -1
Molality = = 1-8 mol kg = 1-8 ni
Mass of solvent in kg 0-085 kg
Calculation of molarity : 15 g of H2SO4 = 0-153 moles {calculated above)
Mass of solution 100 98-04
Volume of solution = = 98-04 cm^ = L= 0-09804 L
1-02 1000
Density of solution
No. of moles of the solute 0-153mol
Molarity =
0-09804 L
= 1-56 mol L ‘ = 1.56 jyi
Volume of solution in litres

Protalom 0 A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate

low
the mole fraction of each component. (HP Board 2011. Chhatisgarh Board 2011)
Solution. Let us suppose that the total mass of the solution is 100 g. Then
Amount of water = 25 g ; Amount of ethanol = 25 g
Amount of acetic acid = 50 g [ Percentages by mass are given]

ee
25 g
25 g of water = = 1-388 moles [-.● Molar mass of water = 18 g mol *J
ISgmol
-1

rF
Fr
25 g
25 g of ethanol = -1
= 0-543 moles [●/ Molar mass of ethanol {C2H5OH) = 46 g mol ']
46 g mol

r
50g
= 0-833 moles [ fo
Molar mass of acetic acid (CH3COOH) = 60 g mol *]
u
50 g of acetic acid = -1
60 g mol
ks
Yo
1-388 1-388
Mole fraction of water = = 0*503
oo

1-388 + 0-543 + 0-833 2-764

B

0-543 0-833
Mole fraction of ethanol = = 0*196 and mole fraction of acetic acid = = 0*301.
re

2-764 2-764

Problem 0 Calculate the molality of a sulphuric acid solution in which the mole fraction of water
u
ad

is 0*85.
Yo

Solution. Mole fraction of water = 0-85

ti2 = 0-15
Mole fraction of H2SO4 in the solution = 1 - 0-85 = 0-15, i.e., ...(0
nd
Re

+^2
where ^2 is the number of moles of H2SO4 and n, is the number of moles of H2O in the solution.
Fi

Molality of H2SO4 solution means the number of moles of H2SO4 present in 1000 g of H2O. Thus, we have
1000
= 55-55 moles.
to find «2 when w, = 1000 g, i.e., n
1 “
18

Substituting the value of n, in equation (i), we get

!h = 0-15 or «2 = O'/i2 + 8-3325 or 0-85 ^2 = ^'^325
55-55+ «2
8-3325
n
2 ~
0-85
= 9-8moles, /.e.. Molality = 9*8 m.
n
n
1 2—= 1-0-85 = 0-15 ...(»)
Alternatively, = 0-85 ...(0
n, + nj +
0-15 0-15 1000
^2 _ 0-15 «2 n X = 9-8 moles
Dividing (») by (i), we get or or
^ ~ 0-85 18
n
1
0-85 1000/18 0-85

Hence, molality = 9*8 m

1
2/10 ‘P’uxdee^'4. New Course Chemistry (Xll)E!ZslS]

Problem Q What volume of 95 % sulphuric acid (density = 1*85 g/cm^) and what mass of water
must be taken to prepare 1«0 cm^ of 15% solution of sulphuric acid (density = MO g/cm^) ?
Solution. 95 % H2SO4 means that 95 g of H^SO^ are present in 1(X) g of the solution
lOOg
Volume of 100 g of this solution = = 54-05 en"v^
l-85gcm“^
95 g
Molarity of 95% H2SO4 = xlOOOcm^L"' = 17-93 M
98gmol 54-05 cm^
15% H2SO4 means that 15 g H2SO4 are present in 100 g of the solution
lOOg = 90-91 cm^
Volume of 100 g of solution = -1
I-lOgcm

ow
I5g
Molarity of 15% H0SO4 = -1
X xlOOOcm^L-' = 1-68 M
98 g mol 90-91 cm^
Applying molarity equation M,.V, M^x V^
(95% H2SO4) (15% H2SO4)
17-93 xV, = 1-68 X 100 or Vj = 9-37 = 9-4 cm^

e
Fl
re
Mass of 100 cm^ of 15% H2SO4 to be prepared = 100 cm-^ x MO g cm“^ = 110 g
Mass of 9-4 cm^ of 95% H2SO4 = 9-4 cm-^ x 1-85 g cm"^ = 17-4 g

F
Mass of water to be taken = 110 - 17-4 g = 92*6 g
ur
or
Problem 0 Calculate the molarity of water if its density is 1000 kg/m^.
Solution. Molarity of water means number of moles of water in 1 litre of water sf
1 Lofwaler= 1000 cm^= lOOOg (v d- 1000 kg/m^ = 1 g/cm-^)
k
Yo
1000
oo

1000 g H2O = 18
moles = 55-56 moles

Hence, molarity = 55*56 iM.

B
re

Problem The mole fraction of benzene in a solution in toluene is 0*50. Calculate the weight
percent of benzene in the solution.
u

Solution. Suppose the weight percent of benzene in the solution = x. This means that in 100 g solution,
ad
Yo

Mass of benzene = xg ; Mass of toluene = (100 - at) g

Mol. mass of benzene (Cj^H^) = 78 ; Mol. mass of toluene (C^,H5CH3) = 92
d

xll%
Re

= 0-50, = 0-50
in

(ng +W.J.) a:/78 + (100-.v)/92

F

X 78x92
= 0-50
78^92 A: + 78(100-.r) or 92A = 46.v + 3900-39:t or 85 .v= 3900 or .t = 45*9%.

Wo/78
= lf ^
n IV w
Alternatively,
B
= 0-5. i.e..
B
= 0-5 or
B
= 0-5 B ) ^’t Wq.
+M.J- M’g /78 + M’y /92 78 ^ 78 92 21,78 92
2w w- 92
or
B _ B
or
w
B _
or — or ,1 + —1.
HV = ,1 + —
92
78 78 92 78 92 vv 78 tv, 78
B B

or +Wj _ ^ or
vv
B 78
= 0-459 . Hence, wl % = 45*9.
78
vv
B vVb+m’t 170
Problem Calculate the normalityof the solution obtained by mixing
(i) 100 cc of 0*2 N H2SO4 with 50 cc of 0-1 N HCl.
(ii) 100 cc of 0*1 N H2SO4 with 100 cc of 0*2 N NaOH.
(Hi) 100 cc of 0*1 M H2SO4 with 100 cc of 0*1 M NaOH.
SOLUTIONS 2/11

Solution, (i) 100 cc of 0-2 N HoS04 contains H^S04 = 100 x 0-2 meq. = 20 meq.
50 cc of 0 ! N HCl contains HCl = 50 x 01 meq. = 5 meq.
Total meq. present = 20 + 5 = 25
Total volume of the solution after mixing =100 + 50 = 150 cc
25 meq
Normality of the final solution = = 0-167 N.
150cc

Alternatively, N| V, + N2 V3 = N3V3
25
= 0-167 N.
0-2x 100 + 0-] x50 = N3(100 + 50) or 150 N3 = 20 + 5 = 25 or N3 150

iii) lOO cc of O-l N HiS04 = 100 x O-l meq = 10 meq ; 100 cc of 0-2 N NaOH = lOO x 0-2 meq = 20 meq
10 meq of H2SO4 will neutralize 10 meq of NaOH meq of NaOH left after neutralisation = 10 meq

w
10 meq.
Total volume of the solution = 100+ 100 = 200cc Normality of NaOH in the solution = = 0-05N.
200 cc

(m) lOOcc of 01 M H^S04 = 100 x 0-1 x 2 meq = 20 meq ; l(X)cc of 0-1 M NaOH = 100 x 0-1 meq = 10 meq

Flo
10 meq of NaOH will neutralize 10 meq of H2SO4 H2SO4 left after neutralisation = 10 meq

e
lOmeq.
Volume of the solution = 100 + 100 = 200 cc Normality of H2SO4 in the solution = = 0-05N

re
200 cc

F
Retain in Memory
ur
Normality is g eq L“* or number of milliequivalents (meq) per cc of the solution and molarity is mol L
-1

or
or number of millimoles (mmol) per cc of the solution.
No. of meq present in Vcc of .x N solution = V x x f
ks
No. of mmol present in Vcc of x M solution = W xx
Yo
oo

Problem m What is the mole fraction of a solute in 2*5 m aqueous solution ?

B

Solution. 2-5 m aqueous solution means 2-5 moles of the solute are present in 1000 g of water, i.e.,
re

1000
Moles of solute = 2-5, Moles of solvent = = 55-55
18
u
ad

2-5
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Mole fraction of solute in the solution = = 0-043

2-5+ 55-55

Problem [0 A 6-90 M solution of KOH in water contains 30% by mass of KOH. Calculate the
d

density of the KOH solution. (Molar mass of KOH = 56 g mol"*)

Re
in

Solution. 6-90 M KOH solution means 6-90 moles of KOH are present in 1 litre of the solution.
F

But 6-90 moles of KOH = 6-90 x 56 g = 386-4 g

Thus, volume of solution containing 386-4 g KOH = 1000 tnL
30% by mass of KOH solution means that 30 g of KOH are present in 100 g of the solution.
100
Mass of solution containing 386-4 g KOH = x386-4g = 1288g
30
Mass 1288g
Density of solution = = I-288gmL-*.
Volume 1000 mL

Problem m Molarity of H^S04 is 0-8 and its density is 1-06 g/cm^. What will be its concentration
in terms of molality and mole fraction ?
Solution. Molarity of H2SO4 = 0-8 means 0-8 mole H2SO4 are present in I L of the solution
0-8 mole H2SO4 = 0-8 x 98 g = 78-4 g : 1 L H,S04 solution = 1000 x 1-06 g = 1060 g
981-6
Mass of solvent (water) = 1060 - 78-4 g = 981-6 g = moles = 54-53 moles
18
Thus, 981-6 g of water contain 0-8 mole of H2SO4.
2/12 “Pn^uUe^'ii. New Course Chemistry rXTT'^rnwi
08 mol
Molality = xl000gkg-> = 0-815 mol kg-^
981-6 g
Further, the solution contains 0-8 mole of solute in 54-53 moles of solvent.
08
Mole fraction of solute = = 0-014
0-8 + 54-53

Problem One litre of N/2 HCl solution is heated in a beaker. It was observed that when the volume
of the solution was reduced to 600 ml, 3-25 g of HCl is lost Calculate the normality of the new solution.
N 1
Solution. 1 L of — HCl contains HCl ~2^ S = 18-25 g ; HCl now present = 18-25 - 3-25 g = 15 g.
15 1
New volume = 600 ml = 0-600 L ; Normality = molx = 0-685 N .

w
36-5 0-600 L

Flo
e
1. Calculate the molality and mole fraction of the solute in aqueous solution containing 3-0 g of urea (molar

re
mass = 60 g mol"^) per 250 g of water.
2. Calculate the molarity and molality of 20 per cent aqueous ethanol (C2H5OH) solution by volume (density

F
of the solution = 0-960 g per cm^). Assume the solution to be ideal.
3.
ur
A 10 cm^ sample of human urine was found to have 5 milligrams of urea on analysis. Calculate the molarity

r
4.
of the given sample with respect to urea (Mol. mass of urea = 60).
fo
The concentration of H2SO4 in a bottle labelled “cone, sulphuric acid” is 18 M. The solution has a density
ks
of 1-84 g cm“^. What is the mole fraction and weight percentage of H2SO4 in this solution ?
Yo
5. A solution contains 90 g of H2O, 6-4 g of methanol and 18-4 g of glycerol. What is the mole fraction of
oo

glycerol? (Glycerol = CH2OH-CHOH-CH2OH) (AP Board 2012)

B

6. Calculate the molarity and normality of a solution containing 5 g of NaOH in 450 mL Solution.
re

(NCERT Solved Example)

7. A sugar syrup of weight 214-2 g contains 34-2 g of sugar (CJ2H22O1,). Calculate (/) molal concentration
u

(») mole fraction of sugar in the syrup. (Pb. Board 2011, HP Board 2011, Bihar Board 2011)
ad
Yo

8. Concentrated nitric acid used as laboratory reagent is usually 69% by mass of nitric acid. Calculate the
volume of the solution which contains 23 g of HNO3. (The density of the concentrated nitric acid solution
is 1-41 g cm"^).
d
Re

9. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is
in

1-84 gmL-'.
F

10.
How many grams of Na2C03 should be dissolved in 250 g of water to prepare 0-1 m solution?
(AP Board 2012)
11. Calculate the volume of 80% H2SO4 (density = 1-80 g/cc) required to prepare one litre of 20% H^S04
(density = 1-25 g/cc).
12. The solubility of Ba(OH)2. SH^O in water at 288 K is 5-6 g per 100 g of water. What is the molality of the
hydroxide ions in the saturated solution of barium hydroxide at 288 K ? (Atomic masses : Ba= 137,0= 16,
H=l).
13. Calculate the number of moles of methanol in 5 litres of its 2 m solution, if the density of the solution is
0-981 kg L"' (Molar mass of methanol = 32-0 g mol"’).
14. 18 g glucose (molar mass 180 g mol”') is present in 500 cm^ of its aqueous solution. What is the molarity of
the solution ? What additional data is required if the molality of the solution is also required to be calculated ?
15. H2SO4 used in lead storage cell is 38% by mass and has a density of 1-30 g cm”^. Calculate its molarity.
16. If 20-0 cm^ of 1 -0 M CaCl2 and 60-0 cm^ of 0-20 M CaCl2 are mixed, what will be the molarity of the final
solution ?
SOLUTIONS 2/13

17. What volume of 10% (w/v) solution of Na2C03 will be required to neutralise 100 mL of HCl solution
containing 3-65 g of HCl? {Assam Board 2012)
18. Calculate the molality of ethanol solution in which the mole fraction of water is 0-88.
(CBSE Sample Paper 2018)
ANSWERS

1. Molality = 0-2 mol kg ^ Mole fraction = 0 00359 2. 3-48 M, 4-35 m

3. 0-0083 mole/litre 4. 0-81, 95-87 % 5. 0-037 6. 0-278 each
7. (0 0-556 m («) 0-0099 8. 23-6 cm^ 9. 10-43 m 10. 2-65 g 11. 173-6 cc
12. 0-356 m 13. 9-22 moles 14. 0-2 M, Density of solution is required
15. 5-0 M 16. 0-40 M 17. 52-9 mL 18. 7-57 m

HINTS FOR DIFFICULT PROBLEMS

2. Assuming the solution to be ideal,
100 cm^ of solution = 20 cm^ of alcohol + 80 cm^ of water; 100 cm^ of solution = 100 x 0-960 = 96 g
96 g of solution = 20 cm-^ of alcohol + 80 g of water -.- 20 cm^ of alcohol = 96 - 80 = 16 g.

w
3. Concentration of urea = 5 mg in 10 cm^ = 500 mg L"' = 0-50 g = 0-5/60 mol L“' = 0-0083 mol L"'.

F lo
4. Cone. - 18 mol L“*. Solute = 18 moles = 18 x 98 g = 1764 g. Solution = 1000 cm^ = 1840 g.
Solvent = 1840- 1764 = 76 g
18 1764
Mole fraction of H2SO4 = = 0-81 ; Weight % = xl000 = 95-87.
76/18 + 18 1840

ree
5. Mole fraction of glycerol =
18-4/92

18-4/92+90/18 + 6-4/32
for F
0-2

0-2 + 5 + 0-2
0-2

5-4
= 0-037

7. Solute (Cj2H220,,) = 34-2 g, Solution = 214-2 g , Solvent = 214-2 - 34-2 = 180 g = 0-180 kg.
Your

34-2/342mol 34-2/342
ks

Molal cone. = = 0-556 mol kg ● ; Mole fraction = = 0-0099

eBoo

0-180 kg 180/18 + 34-2/342

8. Cone, of HNO3 = 69 g in 100 g of solution = 69 g in 100/1-41 cm^, i.e., 70-92 cm^

ad

70-92
our

23 g HNO3 will be present in solution = x23 = 23-6 cm^.

69

9. 93% H2S04(w/v) = 93 g H2SO4 in 100 cm^ of the solution = 93 g in 184 g of the solution
Re

93/98 mol
Solvent (water) = 184 - 93 = 91 g = 0-091 kg ; Molality = = 10-43m,
Y

0-091 kg
Find

10. 0-1 m solution means 0-1 mole of Na2C03 is present in 1000 g of water
0-1
250 g of water should contain Na2C03 = — mole = 0-025 mole = 0-025 x 106 g = 2-65 g

80/98
xl000 = 14-69 M
11. Molarity of 80% H2SO4 = 100/1-80

20/98
xl000= 2-55M
Molarity of 20% H2SO4 = 100/1-25

MiV, M2V2
(80% H2SO4) (20% H2SO4)
i.e., 14-69 xV,1 =2-55x 1000 or V,1 = 173-6 cm^.
12. Molality of OH~ ions = 2 x molality of Ba(OH)2.
2/14 7^^naxCec^'<^ New Course Chemistry (Xll)E!S2a]

13. 2 m solution of methanol means 2 moles of methanol in 1000 g of the solvent (water).
2 moles of CH,OH = 2 x 32 g = 64 g
1064
Mass of solution = 1000 + 64 = 1064 g = 1-064 kg Volume of solution = = 1-085 L
0-981
Thus. 1-085 L of the solution contain CH3OH = 2 moles
2
— x5 = 9-22 moles
5 L of the solution will contain CH3OH = 1-085
32
= 0-40M.
16. M, V| +M3V2=M3 V3 ; 1 X 20 + 0-2 X 60 = M3 (20 + 60) or M3 = 80
10/53 106
17. 10% Nu2C03 (w/v) solution has normality = xlOOO = 1-89 N Eq. wt. of Na2C03 = — = 53
100 ( 2

ow
3-65/36-5
Normality of HCl solution = XlOOO = IN
100

N,V, = N2V2
(NihCO,) (HCl)

e
1-89 X V, = 1 X 100 or V, = 100/1-89 = 52-9 mL

re
rFl
18. Similar to Solved Problem 6.

F
.UNIONS OF . INLi " JIDS

2.4.1. Solubility of a solid in a liquid

r
ou
fo
When a solid solute is added continuously to a liquid solvent, the solute keeps on dissolving and the
ks
concentration of the solution keeps on increasing. This process is known as dissolution. Ultimately, a stage is
reached when no more solute dissolves at the given temperature. This is because from the solution, the solute
oo
particles keep on colliding on the surface of solid solute particles and get separated out of the solution. The
process is known as crystallisation. No more solute dissolves because the rale of dissolution becomes equal
Y
eB

to rale of crystallisation, i.e., a dynamic equilibrium is reached.

Solute + Solvent ^ - Solution
ur

The solution at this stage is said to be saturated solution. An unsaturated solution is one in which more
solute can be dissolved at the same temperature. The concentration of the saturated solution is called the
ad
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‘solubility’. Thus, solubility may be defined as follows :

The >luhility of a solid in a liquid at any temperature is defined as the maximum amount of the
d

solid (solute) in grams which can dissolve in 100 g of the liquid (solvent) to form the saturated
Re
in

solution at that particular temperature.

F

tor-" -uRf .ubility of a solid in a liquid

The important factors on which the solubility of a solid in a liquid depends are :
(/) Nature of the solute and the solvent («) Temperature
These are briefly explained below :
(/) Nature of the solute and the solvent. In general, a solid dissolves in a liquid which is chemically
similar to it. This is expressed by saying “Like dissolves like”. This statement implies that ionic (polar)
compounds like NaCl dissolve in polar solvents like water and are very little soluble or almost insoluble in
non-polar solvents like benzene, ether, etc. Similarly, non-polar (/.e., covalent or organic) compounds like
naphthalene, anthracene etc. are soluble in non-polar (i.e., covalent or organic) solvents like benzene, ether,
carbon tetrachloride, etc. and are very little soluble in water. For example,
(a) Common salt (an ionic compound) is more soluble in water than sugar (a covalent compound). Their
solubilities in water are 5.3 moles per litre and 3.8 moles per litre respectively.
(b) Iodine (a covalent substance) is more soluble in alcohol or carbon tetrachloride (covalent liquids)
than in water.
 SOLUTIONS 2/15

Explanation for solubility. The reason for the behaviour observed as above may be explained as follows:
For ionic compounds
being dissolved in polar
solvents, the solubility is on
account of the fact that there
are strong electrostatic forces Water molecule
of attraction between the ions (polar molecule)
of the crystal and the polar
solvent molecules ; the
Dissolution of an ionic compound like Na^CI i
in a polar solvent (H2O)
negative ions being attracted

ow
by the positive poles of the solvent molecules and the positive ions by the negative poles of the solvent
molecules. For example, when water is used as the solvent, the situation may be represented as shown in
Fig. 2.1.
Thus, the water molecules pull the ions of the crystal apart and the electrostatic forces of attraction
existing between the ions of the crystal are cut off. Further, the ions are surrounded by the water molecules

e
re
which act as a sheath (or envelope) around the ions and prevent the recombination of the ions. The ions thus
moving freely in the solution are said to be hydrated. It may be mentioned here that whereas energy is

Frl
F
required for the splitting of the ionic compound into ions (called lattice energy), energy is given out when'the
ions get hydrated (called hydration energy). A substance dissolves if hydration energy is greater than the
lattice energy, i.e., H > A lattice H.
ou
or
It may be further mentioned that whereas water is the best polar solvent (having highest dielectric
kfs
constant), liquid ammonia, liquid hydrogen sulphide and liquid sulphur dioxide are also good solvents for
ionic compounds.
oo
For non-polar compounds being dissolved in non-polar solvents, the solubility is due to similar solute-
solute, solute-solvent and solvent-solvent interactions.
Y
B

(«) Effect of temperature on solubility—Solubility curves. On the basis of the effect of temperature
on solubility in water, the various ionic substances are divided into three categories.
re

(a) Those whose solubility increases continuously with increase of temperature. Most of the substances
oYu

like NaN03, KNO3, NaCl, KCl, etc. fall into this category. The reason for this behaviour is that in case of all
ad

such substances, the process of dissolution is endothermic, i.e..

Solute + Solvent -v- Heat - ± Solution
d

Applying Le Chaielier’s principle, as the temperature is increased, equilibrium will shift in a direction in
in
Re

which the heat is absorbed, i.e., in the forward direction; consequently, more of the solute pas.ses into the
solution.
F

{b) Those whose solubility decreases continuously with increase of temperature. There are a few
substances like cerium sulphate, lithium carbonate, sodium carbonate monohydrate (Na-iC03.H20), etc. whose
solubility decreases with increase of temperature. Obviously, it is due to the fact that the process of dissolution
of these substances is exothermic, i.e., it is accompanied by evolution of heat,
(c) Those substances whose solubility does not increase or decrease continuously. There are some
substances which on heating change at a particular temperature from one polymorphic form to another (like
a to P-form as in the case of ammonium nitrate) or from one hydrated form to another (like CaCl2.6H20
CaCl2.4H20' > CaCl2.2H20) or from hydrated to anhydrous form (e.g., Na2SO4.J0H2O >1 Na2S04).
Such substances do not show a continuous increase or decrease of solubility. For example, in case of sodium
sulphate, the solubility First increases upto 32.4“C and then begins to decrease. The temperature at which one
form of the substance changes into another is called transition temperature. Thus, in case of sodium sulphate.
Above 324“C
Na2SO4.10H2O V Below 324°C
i Na2S04
2/16 ‘p>utdec^'^ New Course Chemistry (Xll)ESXSl

When the solubilities are plotted against temperatures, the curves obtained are called solubility curves.
Thus, whereas in case of (a) and (/>), the solubility curves are continuous [Fig. 2.2 (/)], these are disconti
nuous in case of (c) [Fig. 2.2 {»)].
FIGURE 2.2

200 280 P

180
240
160 NH4NO3
q: q:
m iU

5 140 5 200
g g a

O 120 O
O) 160
o o
100 o
o
NaNOs CaCl2.2H20

low
Ul

80 120

CaCl2.4H20
CD 60 m
3 3
60
o O /CaCl2.6H20
40 CO
<0
Na2SO4.10H2O Na2S04

J-
40

ee
20
062(804)3

rF it 4' 40 ""55—S3—735“

Fr
40 60 80 T55
32.40’C
TEMPERATURE ('O . TEMPERATURE CC).^

(I) Continuous solubility curves (II) Discontinuous solubility curves

r
Solubility curves
fo
u
Note. Pressure has very little effect on the solubility of a solid in a liquid because solids and liquids are
ks
highly incompressible.
Yo
oo

T-
: <
OUBPLglWENT YOUR
. i. LEDGE FOR COMPETITIONS
B

1. Nernst Distribution Law. A solute distributes itself between two immiscible solvents in such a way that
the ratio of the concentrations of the solute in the two solvents is constant at constant temperature provided
re

that the solute does not undergo any dissociation or association in any of the solvents. Mathematically,
u
ad

—- = K, d constant called distribution coefficient or partition coefficient.

Yo

C,
2. Extraction of a solute from an aqueous solution. A solvent immiscible with water but in which the solute
is soluble is used for extraction. For the same amount of the solvent, the extraction is more if it is used in a
nd
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number of instalments. In other words, multi-step extraction is more economical than a single step extraction.
Fi

2.5. SOLUTIONS OF GASES IN LIQUIDS

2.5.1. Solubility of a gas in a liquid

Almost all gases are soluble in water though to different extents. The existence of aquatic life in lakes,
rivers, sea etc. is due to dissolution of oxygen gas of the air in water. Some gases are also soluble in solvents
like ethyl alcohol, benzene, etc.
The ^ I dubility of any gas in a particular liquid is the volume of the gas in cc (converted to S..TP.)
that can dissolve in unit volume (1 cc) of the liquid to form the saturated solution at the
temperature of the experiment and under a pressure of one atmosphere.
This method of expressing the concentration is called absorption coefficient of the gas and is usually
represented by a.
Solubility of a gas in a liquid at a particular temperature is also expressed in terms of molarity (moles
of the gas dissolved per litre of the solvent to form the saturated solution, i.e., in terras of mol L"*) or in terms
of mole fraction (jc^) of the gas.

t
 SOLUTIONS 2/17

2.5.2. Factors affecting the solubility of a «as in a liquid

The important factors on which the solubility of a gas in a liquid depends are briefly explained below :
(i) Nature of the gas and the solvent. Gases like hydrogen, oxygen, nitrogen, etc. dissolve in water
only to a small extent whereas gases like CO2, HCl, NH^, etc. are highly soluble. The greater solubility of the
latter gases is due to their reaction with the solvent.
Again oxygen, nitrogen and carbon dioxide are much more soluble in ethyl alcohol than in water at the
same temperature and pressure while H2S and NH3 are more soluble in water than in ethyl alcohol. Evidently,
the greater solubility of gas in a solvent is again due to the chemical similarity between the gas and the
solvent.

(//) Effect of temperature. The soluhility of a gas FIGURE 2.3

decreases with increase in temperature. This is expected because 0.007

on heating the solution of a gas, some gas is usually expelled 03

0.006
out of the solution. The same result also follows in an alternate
o
o

manner as under: 0.005

w
Z E
The dissolution of a gas in a liquid is an exothermic process, ^- 0 004
Le., it is accompanied by evolution of heat. Thus, O It

F lo
— 0.003
00
Gas + Solvent - Solution + Heat V CM
h ^ 0.002
Applying Le Chatelier’s principle, it is evident that increase □ u.

ee
0.001
of temperature would shift the equilibrium in the backward

Fr
direction, i.e., the solubility would decrease. o
CO
0
10 20 30 40 50 60 70
Though O2 gas is more soluble than N2 gas at any
temperature, the solubility of both the gases decreases with
increase of temperature as represented in Fig. 2.3. for TEMPERATURE (“C)
ur
Variation of solubility of O2 and N2
Further, it may be pointed out that though generally the gases with temperature at
s
solubility of a gas in a liquid decreases with increase of
ook

constant pressure of 1 atmosphere

Yo

temperature, there are some exceptions. For example, the

solubility of some sparingly soluble gases, such as hydrogen and inert gases, increases slightly with increase
eB

of temperature especially in the non-aqueous solvents such as hydrocarbons, alcohols and acetone,
iiii) Effect of pressure (Henry’s law). This is the most important factor influencing the solubility of a
our

gas in a liquid at a particular temperature. A little thought clearly reveals that as we compress the gas over the
ad

liquid (i.e., we increase the pressure), the solubility will increase.

This may be explained as follows :
Y

For the solution of a gas in a liquid, consider a system FIGURE 2.4

Re

as shown in Fig. 2.4 (a). The lower pan is the solution and
nd

the upper part is gaseous at pressure p and temperature T. GASEOUS

Fi

Suppose the system is in dynamic equilibrium, i.e., rate of PHASE AT

PRESSURE p,
gaseous particles entering and leaving the solution is the TEMPERATURET
same, which means that rate of dissolution = rate of
evaporation. Now, increase the pressure over the system SOLUTION
as shown in Fig. 2.4 (b). The gas gets compressed to a PHASE
smaller volume. Hence, the number of gaseous particles
per unit volume increases. As a result, the number of o o
gaseous particles striking the surface of the solution and Demonstrating the effect of pressure
hence entering into it also increases, till a new equilibrium on solubility of a gas in a liquid
is reestablished. Thus, on increasing the pressure of the
gas above the solution, the solubility increases.
Quantitatively, the effect of pressure on the solubility of a gas in a liquid was studied by Henry (in 1803)
and is called Henry’s law. It states as follows :
The mass of a gas dissolved in a given volume of the liquid at constant temperature is directly
proportional to the pressure of the gas present in equilibrium with the liquid.
2/18 ‘Pnadeef,t'A New Course Chemistry (XII)E2SZS1

Mathematically, m >=< p or m = Kp ...(0

where m = mass of the gas dissolved in a unit volume of the solvent, p - pressure of the gas in equilibrium
with the solvent, K = constant of proportionality whose value depends upon the nature of the gas, the nature
of the solvent and the temperature.
Henry’s law may also be stated as follows :
The solubility of a gas in a liquid at a particular temperature is directly proportional to the
pressure of the gas in equilibrium with the liquid at that temperature.
Dalton during the same period had concluded independently that ifa mixture ofgases are simultaneously
in equilibrium with the liquid at a particular temperature, the solubility of any gas in the mixture is directly
proportional to the partial pressure of that gas in the mixture.

w
Expressing solubility in terms of mole fraction of the gas in the solution, for a gas A, Henry’s law can be
written as ^a = ^'Pa ...(«)

where x^ is the mole fraction of the gas in the solution, is the partial pressure of the gas above the

o
solution and K' is proportionality constant whose value depends upon the nature of the gas, nature of the

e
solvent and the temperature.

Fl
re
For example, the solubility of pure N., in water at 298 K and a partial pressure of 0-78 atm (which is the

F
patlial pressure of in air at 1 -0 ann) is 5-3 x 10"^ mol L“'. If the partial pressure is doubled to 1 -56 atm, the
solubility of in water is doubled to 1 06 x 10”^ mol (using molarities in place of mole fractions).
ur
r
1

fo
From eqn. (//), Pa^-^^a or ;>a = Kh^a
ks ...m

1
Yo
where = — is called Henry’s constant.
oo
K
This is the most commonly used form of Henry’s law and may be defined as follows :
eB

The partial pressure of a gas in vapour phase (p) is directly proportional to the mole fraction (x)
of the gas in the solution.
ur

Using this expression, the units of Kj^ will be atm or bar (or kbar).
ad

This is a convenient expression for testing the validity of Henry’s law. Plotting mole fractions,
Yo

versus corresponding equilibrium pressures, a straight line plot passing through the origin is obtained
with slope = Kj^ (in atm or bar) as shown in Fig. 2.5 for solubility of HCl gas in cyclohexane at 293 K.
d

Different gases have different values of at the same temperature and in the same solvent. For a few
Re
in

gases in water as the solvent, these values are given in Table 2.2 below :
F

TABLE 2.2. Values of Henry’s

FIGURE 2.5
constant for some gases in water.
5 1-2
Gas Temp. (K) Kh (kbar)
LU
I'O

0-8 O2 293 K 34-86

CO 5
303 K 46-82
0-6
293 K 76-48
si 0-^ N2
30 SLOPE = Kh (atm) 303 K 88-84
cc
0-2
CO
H2 293 K 69-16
3 0-010 0-020 298 K 71-18
O
lU MOLE FRACTION OF HCl IN He 293 K 144-97
SOLUTION IN CYCLOHEXANE
Ar 298 K 40-3
Plot of mole fraction in solution 298 K 1-67
CO2
versus equilibrium pressure CH4 298 K 41-85
 SOLUTIONS 2/19

From these values, the following results may be drawn :

(0 Henry’s constant, K^, is a function of the nature of the gas.
(ii) Greater the value of Kjj, lower is the solubility of the gas at the same partial pressure [according to
eqn. (in)] at a particular temperature.
(in) The value of increases with increase of temperature implying that the solubility decreases with
increase of temperature at the same pressure.
It is for this reason that aquatic species feel more
comfortable in cold water (in which O2 gas dissolved is TABLE 2:3. Values of Henry’s
TOiiiffant for some gases

ow
more) than in warm water (in which O2 dissolved is less). diffdrent solvents at 298 K
Further, as already mentioned, the value of Kjj depends
upon the nature of the gas and the nature of the solvent at Gas Solvent KH(kbar)
the same temperature. The Kjj values of some gases in
H2 Water 7M8

e
different solvents at 298 K are given in Table 2.3.
H2 Benzene 3-67

re
Limitations of Henry’s law : Henry’s law is CO2 Water 1-67
applicable only if the following conditions are satisfied: CO2 Benzene on

F
Frl
(0 The pressure should be low and the temperature CH4 Water 41-85
should be high, i.e., the gas should behave like
ou CH4 Benzene 0-57
an ideal gas.

osr
(ii) The gas should not undergo compound formation with the solvent or association or dissociation in
the solvent.
For example, the law is not applicable in case of dissolution of ammonia in water because it undergoes
compound formation followed by dissociation :
kf
oo
NH3 (g) + H2O » NH4OH (aq)
Y
NH40H(o^) NHJ (aq) + OH-(aq)
B

Sumlarly, the law is not applicable to the dissolution of HCl gas in water because it undergoes dissociation
after dissolution:
re
uY

HCl(g) + a^ >HCl(a^)
HCl (aq) > H+ (aq) + CV (aq)
ad
do

Applications of Henry’s law :

(i) In the production of carbonated beverages. To increase the solubility of CO2 in soft drinks, soda
in

water as well as beer, champagne, etc., the bottles are sealed under high pressure. When the bottle is opened
Re

to air, the partial pressure of CO2 above the solution decreases. As a result, solubility decreases and hence
CO2 bubbles out.
F

(ii) In the deep sea diving. Deep-sea divers (or Scuba divers) depend upon compressed air for their
oxygen supply. According to Henry’s law, solubilities of gases increase with pressure. Thus, both N2 and O2
will dissolve considerably in the blood and other body fluids. Oxygen is used up for metabolism, but due to
high partial pressure and greater solubility, N2 will remain dissolved and will form bubbles when the diver
comes to the atmospheric pressure. These bubbles affect nerve impulses and give rise to a disease called
bends or decompression sickness. To avoid bends and also the toxic effects of high concentration of nitrogen
in the blood, the cylinders used by the divers are filled with air diluted with helium (11-7% helium, 56-2%
nitrogen and 32-1% oxygen).
(Hi) In the function of lungs. When air enters the lungs, partial pressure of oxygen is high. This oxygen
combines with haemoglobin to form oxyhaemoglobin. Partial pressure of O2 in tissues is low. Hence, O2 is
released from oxyhaemoglobin which is utilised for functions of the cells.
(iv) For climbers or people living at hi^ altitudes. At high altitudes, the partial pressure of oxygen is
less than at the ground level. As a result, there is a low concentration of oxygen in the blood and tissues of the
people living at high altitudes or climbers. Consequently, they feel weak and are unable to think properly, a
disease called anoxia.
2/20 “pn^iuCee^’A New Course Chemistry (XII)CSm

NUMERICAL FORMULAS AND UNITS USED

PROBLEMS
BASED Vk -
ON where is the mole fraction of the gas A in the solution, is the
partial pressure of the gas above the solution and is Henry’s constant
Henry’s Law } (in atm or bar or kbar)
Note. If equation = Kp^ ‘s used, units of constant K will be atm"’ or
. bar"’ etc.

Problem Q If N2 gas is bubbled through water at 293 K, how many millimoles of Nj gas would
dissolve in 1 litre of water ? Assume that N2 exertsa partial pressure of 0*987 bar. Given that Henry’s law
constant for N2 at 293 K is 76*48 kbar. (NCERT Solved Example)

w
Solution. According to Henry’s law, Pn, = K,.yx
N2; ●●●
P^2 ^ 0-987 bar = 1-29 X10"^
H
^N2 -

F lo
K.t
H
76480 bar
n n

If n moles of N2 are present in 1 L of water (i.e., 55-5 moles), - (as n « 55-5)

n + 55-5 55-5

e
Fre
n
= ]-29xl0"-‘^ or n = l-29xl0"^ X 55-5 moles = 71-595 x 10“^ moles = 0*716 millimoles
55-5

0 At what partial pressure, oxygen will have a solubility of 0*05 g L"* in water at 293 K ?
Problem
for
Henry’s constant (Ky) for Oj in water at 293 K is 34*86 kbar. Assume the density of the solution to be same
r
as that of the solvent.
You
Solution. Calculation of mole fraction (jTq^)
oks

Mass of I L of solution = 1000 g (*.* i/=lgmL-’)

eBo

Mass of solvent (water) = 1000 g - 0-05 g = lOOOg

lOOOg 0-05 g = 1-56x10 ^mole
= 55-5 moles, n
n
H2O “ O2 " -1
32g mol
our
ad

ISgmol
n
O2
n
1-56x10"-^
= 2-81x10-5
"O2 " n + n n 55-5
O2 H2O “H2O
dY
Re

Calculation of partial pressure. Applying Henry’s law,

Fin

PO2 = K„H xa:^O2 = (34-86 xlO^ bar) x (2-81 x 10"5) = 0*98 bar.
Problem 0 Air contains O2 and N2 in the ratio of 1 : 4. Calculate the ratio of solubilities in terms
of mole fractions of Oj and N2 dissolved in water at atmospheric pressure and room temperature at which
Henry’s constant for O2 and N2 are 3*30 x lO"^ torr and 6*60 x lO’ torr respectively.
Solution. At 1 bar pressure.
4

Partial pressure of Ot(Pq^) = ^x 1 bar = 0-2bar ; Partial pressure of N2 (pj,j^ ^ ~ ^ ^ ^ ~ 0-8 bar
P02
Applying Henry’s law, PO2 ~
or
●^02 -
Kh

or
PN2
P^2 ^N2 “
Kh(N2)
Po. Kh(N,) PQ2„Kh(N2)„ 0 2 bar 6-60x10'^ torr _J_ = 1:2.
X
■*'02 ● ^N2
●^N2 Kh(02) Kh(02) 0-8bar 3-30x10^ torr 2’
SOLUTIONS 2/21

1. The Henry’s law constant for oxygen dissolved in water is 4-34 x 10“^ atm at 25°C. If the partial pressure of
oxygen in air is 0-2 atm under atmospheric conditions, calculate the concentration (in moles per litre) of
dissolved oxygen in water in equilibrium with air at 25°C.
2. The mole fraction of helium in a saturated solution at 20°C is l-2x 10“^. Find the pressure of helium above
the solution. Given Henry’s constant at 20°C = 144-97 kbar.
3. What concentration of nitrogen should be present in a glass of water at room temperature ? Assume a
temperature of 25°C, a total pressure of 1 atmosphere and mole fraction of nitrogen in air as 0-78 [Kj^ for
nitrogen = 8-42 x 10""^ M/mm Hg]. (CBSE 2009)
4. Calculate the solubility of CO, in water at 298 K under 760 mm Hg (K^ for CO2 in water at 298 K is
1-25 X 10^ mm Hg). (CBSE 2020)

w
5. Assume that argon exerts a partial pressure of 6 bar. Calculate the solubility of argon gas in water (Given :
Henry’s law constant for argon dissolved in water, = 40 kbar). (CBSE Sample Paper 2022-23)

F lo
ANSWERS
1.2-55X 10“*molL-‘ 2. 0-174 bar 3.4-99 X lO-^M 4. 3-37 X 10-2 mol L"'
5. Mole fraction = 1-5 x 10“^

ee
Fr
HINTS FOR DIFFICULT PROBLEMS

Po-, 0-2 atm

1. Po. = KoXx
H O2
or
=

K
for
4-34x10'^ atm
= 4-6x10"^
ur
II
n n
O. O,
s
To convert it into molarity. "O2 ■
ook

n n
Yo
HoO
eB

O.
For 1 litre of water, ''h20 = 1000/18 = 55-5 moles. l = 4-6xl0"^’ or nO. = 2-55x10-^ mole
55-5
Hence, molarity = 2-55 xlO"^ mol L“'.
our
ad

2* pHe = ^ -^He “ ^ 44-97 X 10^ bar) (1 -2 x 10"^* ) = 0-174 bar.

3. Pn2 = fraction of N, in air x = 0-78 x 1 atm = 0-78 atm = 0-78 x 760 mm = 592-8 mm
Y

As K is in the units of M (mm)"*, Henry’s law is applied in the form :

Re
nd

Cone, in solution = {<; hPn. = 8-42 X lO""^ M (mm)-i x 592-8 mm = 4-99 x 10“^ M
Fi

Pco^ 760 -i
4. = 608x10
^C02
or
K
H 1-25x10^’

To calculate in terms of moles L proceed as in Hint 1.

6 bar
5. -Pm^ = 1-5x10-^
Pat=^E^-^At or
^Ar K
H
40,000 bar

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Calculation of solubility of a gas at a particular pressure from the known solubility at some other
pressure (at the same temperature). From Henry’s law, m = K p. If is the solubility of a gas at

i = -El
in

pressure P, and m. is its solubility at pressure ?2,

fH2 ?2
2/22 “P^eidee^'^. New Course Chemistry (XlI)EZS19a

Thus, knowing mj al P,, iit P2 can be calculated.

FIGURE 2.6
The equation, m = KP al.so suggests that plot of
pressure, P vs solubility m will be a straight line passing Slope of the line
through the origin with slope = K. Greater the value of (tan 8) ^ K
K. greater is the solubility. For example, Ot is more >
tan 0 j = K (or N2
soluble than N-, at the same temperature and pressure Ian 62 = K for O2
as shown in Fig. 2.6. CQ
D

2. Quantitative effect of temperature on the solubility o

tf)

of a gas in a liquid. It is given by an expression similar

to van* t Hoff equation (giving the effect of temperature
on equilibrium constant) or Clausius-Clapeyron
PRESSURE
equation (giving the effect of pressure on the boiling
point of a liquid). Assuming the gas to be ideal, the Variation of solubility of O2 and N2 gases

w
expression is with pressureat constanttemperature
dine _ AH ...(/)

Flo
JT ~ RT2
where c is the concentrationin moles per litre of the gas in the liquid and AH is the heat of solution of

ee
1 mole of the gas at temperature T.
If AH is supposed to be independent of temperature, then integration of eqn. (/) gives

Fr
ln^ = (K)

for
^1
ur
where c, and C2 are the concentrations of the solution at T, and T2 respectively.
ks
Yo
oo

Curiosity Question
eB

r Q. Why do aquatic species like fish fee! more comfortable in the lakes in winter than in summer ?
Ans. Aquatic species need dissolved oxygen for breathing. As solubility of gases decreases with increase
r

of temperature, less oxygen is available in summer in the lakes. Hence, they feel more comfortable
ou
ad

in winter {low temperature) when the solubility is higher.

I
Y

2.6. SOLUTIONS OF SOLIDS IN SOLIDS (SOLID SOLUTIONS)

Re
nd

Solid solutions are those solutions In which the solvent is solid. However, most common among these
are those in which both solute and solvent are solid. These are classified into two types as follows :
Fi

(1) Substitutional solid solutions. If some atoms, ions or inoleculerof one solid present at the lattice
sites are replaced (substituted) by the atoms, ions or molecules of another similar solid, having similar sizes,
the solids thus obtained are called substitutional solid solutions (Fig. 2.7 (a)). A few common examples of
substitutional solid solutions are given below :
Brass, bronze, monel metal, etc.
FIGURE 2.7
(2) Interstitial solid solutions. If in the lattice of a solid
(called the host), the atoms of some other solid, small in size,
occupy the interstitial sites or voids, the solids obtained are called
interstitial solid solutions (Fig. 2.7 (I?)). One of the most common
example of this type of solid solution is that of tungsten carbide
(WC) in which the tungsten (W) atoms are arranged in a face-
centred cubic lattice and C-atoms occupy the octahedral voids ^2) = Solvent(^= Solute Solvent 0= Solute
(surrounded by 6 W atoms at the vertices of the octahedron). (a) Substitutional solid solution
Similarly, steel is interstitial carbide of iron. (b) Interstitial solid solutions
SOLUTIONS 2/23

2.7. VAPOUR PRESSURE OF LIQUID SOLUTIONS AND RAOULTS LAW

Liquid solutions are the solutions in which the solvent is liquid. The solute can be gas, liquid or solid.
Wc have already discussed about solutions of gases in liquids. Therefore, in this section, our aim is to discus.s
solutions of liquids in liquids and those of solids in liquids. We shall confine our study only to binary solutions,
i.e., solutions having only two components (one solute and one solvent). The main property to be studied
about them is their vapour pressures. We shall, therefore, first discuss what we mean by vapour pressure and
then discuss the vapour pressure of the solution of liquids in liquids, i.e., when both the components are
volatile and the vapour pressure of the solutions of solids in liquids, i.e., when the solute is non-volatile hut
solvent is volatile.

2.7.1. Vapour pressure FIGURE 2.8

If a pure liquid or a solution is taken in an

evacuated vessel placed in a thermostat at a particular

w
temperature and the vessel is connected to a manometer
as shown in Fig. 2.8., the amount of the vapour in the THERMOSTAT- VAPOUR

space above the liquid/solution keeps on increasing and

F lo
ultimately becomes constant (as indicated by the LIQUID
OR
pressure in the manometer). This is because an SOLUTION

equilibrium is attained, i.e., rale of evaporation becomes MANOMETER

ee
equal to the rate of condensation. The pressure

Fr
Concept of measurement of vapour pressure
thus exerted by the vapours is called vapour pressure. of a liquid/solution at a particular temperature
Hence,
for
Vapour pressure of a liquid/solution is the pressure exerted by the vapours in equilibrium with
ur
the liquid/solution at a particular temperature.
s
ook

Factors affecting vapour pressure. The vapour pressure of a liquid depends upon the following factors ;
Yo

(t) Nature of the liquid. Weaker arc the intennolecular forces, greater is the amount of vapour because
eB

more molecules can leave the liquid and come into the vapour phase. For this reason, vapour pressure of
diethyl ether is greater than that of ethyl alcohol.
our

(i7) Temperature. Higher the temperature, greater is the vapour pressure. This is because with increase
ad

in temperature, kinetic energy of the molecules increases and hence more molecules leave the surface of the
liquid and come into the vapour phase.
Y

Quantitatively, the effect of temperature on vapour pressure of a liquid is given by Clausius-CIapeyron

Re

equation, i.e..
nd

P A H PT -T '
iocrjl = .. . 2, _h.
Fi

®P, 2-303 R TjT,

where Pj and P-, are the vapour pressures at temperatures T, and T-, respectively and A vap H is the
enthalpy of vaporisation of the liquid.

2.7.2.Vapour Pressures of Liquid-Liquid Solutions and Raoulfs Law

(i.e., Raoulfs Law for Volatile Solutes)
For a solution of liquid in liquid, as both the components of the solution are volatile, each component
will form vapour above the solution. When equilibrium is reached, each component will exert a vapour
pressure, called its partial pressure whose value depends upon the mole fraction of the component in the
solution and the vapour pressure of that component in the pure state. These studies were made by a French
Chemist, F.M. Raoult in 1886, and he pul forward the following result known after him as Raoult’s law :
In a solution, the vapour pressure of a component at a given temperature is equal to the mole
fraction of that componentin the solution multipliedby the vapour pressure of that component
in the pure state.
2/24 U New Course Chemistry (XII) CSm

Let us consider a mixture of two completely miscible volatile liquids A and B, having the mole fractions
and Xq. Suppose at a certain temperature, their partial vapour pressures are and and the vapour
pressures in the pure state are and According to Raoult’s Law,
Pa = ^aP°a and Pb = -^bP''b ...(/)
If P is the total pressure of the system at the same temperature, then by Dalton’s law of partial pressures,
P=Pa + Pb = ^aP°a + ^bP''b ...(H)
or
P = (1 -Xq) P°^ -^bP B~ B~P ^B'^ p A ...(Hi)
Since p°^ and p°g are constant at a particular temperature, FIGURE 2.9
it is evident from eqn. (Hi) that the total vapour pressure is a
linear function of the mole fraction Xg (or as = 1 - Xg).
Thus, a straight line should be obtained,when P is plotted against

ow
x^ orxg. Such a plot is shown in Fig. 2.9. The lines (I) and (II)
give the plots of partial pressure versus mole fraction and the
line (III) that of the total pressure versus mole fraction.
When = 1, i.e., the liquid is pure A, P = Pa

e
Fl
re
xb = 0 MOLE FRACTION OF B Xb= 1
and when Xg = 1, i.e.. the liquid is pure B P = Pb

F
The relationship between vapour pressure
At any value of Xg (or.r^), the value of the total pressure is and mole fraction of an ideal solution at
given by the equation (//V).
ur constant temeperaturc (P—x diagram)

r
fo
The following points may be noted about such solutions :
(?) Raoult’s law, as explained above, is applicable only if the two components (i.e., the volatile liquids)
ks
form a solution (a homogeneous mixture). The law is not applicable if the two liquids are not completely
Yo
miscible.
oo

(ii) In the P — x diagram described above, x^ and X3 denote the mole fraction of the constituents present
B

in the liquid phase. Since, both the constituents are volatile, the vapour above them will contain both
of them. The vapour will, however, be richer in the more volatile constituent.
re

(///) Liquid pairs that obey the Raoull's law are called ideal solutions. Such pairs are, however, very
u

limited. Some such typical liquid pairs are, benzene + toluene, hexane + heptane, chlorobenzene +
ad
Yo

bromobenzene.

(/V’) The composition of the vapour phase in equilibrium with the solution can be calculated from the
partial pressures of the two components (in the vapour phase). If and represent the mole fractions
d
Re

of the components A and B respectively in the vapour phase, then

in

Pa Pa . Pb _ Pb
F

>’b =
=

Pa Pb ^TotaJ Pa + Pb ^Totai
The above results may also be written as
PA = >’AXPToial and /^B=:VBXPToial
In general, for a solution containing a number of volatile components (liquids), for any component /,

(v) Depending upon the vapour pressures of the pure components, total vapour pressure over the solution
may increase or decrease with increase in the mole fraction of a component. For example, in Fig. 2.9,
the line III slopes upwards from left to right but slopes downwards from right to left.
2.73. Vapour Pressures of Solutions of Solids in Liquids and Raoults Law
(i.e., Raoults Law for Non-volatile Solutes)
As solid solutes are non-volatile, we shall first discuss qualitatively the effect of adding such solutes on
the vapour pressure of the liquid solvent and then discuss the results quantitatively, as derived from Raoult’s
law.
SOLUTIONS 2/25

Effect of adding a non-volatile solute on vapour pressure of a liquid solvent. It is observed that the
presence of a non-volatile solute in a solution reduces the escaping tendency of the solvent molecules into the
vapour phase (because some of the solute particles occupy the position of the solvent molecules on the liquid
surface) and thus lower the vapour pressure of the solvent. This may be seen in a simple way by taking pure
solvent in one beaker and the solution of a non-volatile FIGURE 2.10]
solute in the same solvent in another identical beaker . LESS
MORE O VAPOUR
upto same level and then covering them with evacuated VAPOUR' BELL-JAR

bell-jars as shown in Fig. 2.10 (a) and {b). The level in *

(a) falls more than in (b) when equilibrium is attained. INITIAL ● ♦

UJ

Raoult’s law for solutions of solids in liquids, LEVEL

-- xnooooooooooe
oo^ ooof
>

i.e., for non-volatile solutes. Quantitatively, the effect FINAL <

PURE SOLUTION
of adding a solid (non-volatile) on the lowering of LEVEL
SOLVENT

vapour pressure can be derived from the general

w
o SOLVENT MOLECULES - SOLUTE PARTICLES
definition of Raoult’s law, according to which
Vapour pressure of the solvent in the solution Vapour pressure of (a) pure solvent (b) solution '

F lo
_ Mole fraction of the X
Vapour pressure of ...(/V)
solvent in the solution the pure solvent

ee
Now, if the solute is non-volatile and non-electrolyte (e.g., glucose, urea, etc.), it will not contribute to

Fr
the total vapour pressure of the solution. Thus, the vapour pressure of the solution will be the vapour pressure
due to solvent in the solution only, i.e.,

Vapour pressure _ Vapour pressure of the

for
r
...(V)
solvent in the solution
You
of the solution
s
ook

Combining results (/V) and (v), we have for a solution containing non-volatile solute.
eB

Mole fraction of the Vapour pressure of ...(VI)

Vapour pressure of the solution = X
solvent in solution the pure solvent
our
ad

or in terms of symbols, we can write p, = x^xp

FIGURE 2.11
p p»(V,P.
This can be rewritten in the form —o
= x.I ...(VII) OF PURE
dY

P°
Lli
cc
SOLVENT)
Re

The above expression (p^ = Xj x p°) implies that CO V

Fin

lUH

Q.

i.e., vapour pressure of solution c< mole fraction of the §8 Slope of the line
solvent in the solution. = tan 9 = p®
<
> e
Thus, if mole fraction of the solvent in a solution containing
0.0 0.2 0.4 0.6 0.8 1.0
a non-volatile solute is plotted against the vapour pressure of MOLE FRACTION OF
the solution, a straight line will be obtained as shown in Fig. SOLVENT IN THE SOLUTION

2.11. Slope of the line will be tan Q = p°

Plot of vapour pressure of solution
If the solution contains moles of the solute dissolved in versus mole fraction of the solvent for a
ri] moles of the solvent, we have solution obeying Raoult's law
n
1
Mole fraction of the solvent in solution ●^1 =
«i +

Substituting this value in equation (vii), we get E^ = n,

p° n^ -f- n-y

●w.-
2/26 PfKxdce^'4, New Course Chemistry fXinrasmn

p, n
1 P -Ps «2
Subtracting each side from 1, wc get I ^ = 1— or ...{viii)
P'" «) + nj P «l+«2
P° -P
In this expression, p‘ - expresses the lowering of vapour pressure, — is called relative lowering
P°
'h
of vapour pressure, and represents tlie mole fraction of the solute in the solution. Hence, the expression
/i| + 1I2
{viii) may be expressed in words as follows :
The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal
to the mole fraction of the solute in the solution.

2.7.4. Raoult's law as a special case of Henry's law

According to Raoult’s law, for any volatile component of the solution,

w
Pa = ^a><Pa or
Pa^Pa'x^a
i.e., vapour pressure of the volatile component (solvent or solute) is directly proportional to the mole

F lo
fraction of that component in the solution.
Now. if gas is the solute and liquid is the solvent, then according to Henry’s law,/?^ = i.e., partial
pressure of the volatile component (gas) is directly proportional to the mole fraction of that component (gas)

ee
in the solution.

Fr
Thus, Raoult’s law and Henry's law become identical except that their proportionality constants are
different, being equal to for Raoult’s law and (Henry’s constant) for Henry’s law.
for
It is interesting to point out here that Henry’s law is sometimes obeyed even by liquid solutions. Further,
ur
it is observed that if the solute obeys Henr>'’s law. the solvent obeys Raoult's law. (The components behave so
more ideally if the solution is dilute). However, the reverse is not true.
s
ook

2.7.5. Similarities and Dissiiintcaities between Raoult's law and Henry's law
Yo

Based on the above discussion, these may be summed up as follows ;

eB

Similarities. (/) Both apply to the volatile component of the solution.

(ii) Both state that the vapour pressure of any component in the solution is proportional to the mole
our

fraction of that component in the solution,

ad

Dissimilarities. The two laws differ in the proportionality constant. In Raoult’s law, it is equal to vapour
pressure of pure component (p°) while in Henry’s law. it is equal to an experimentally determined value of
Y

Henry’s constant (K,^).

Re

tt;
nd

sup UR
KNOWLCOj tETITIONS
Fi

1. KonowalofTs rale. At any faed temperature, the vapour phase is always richer in the more volatile component
as compared to the solution phase. In other words, mole fraction of the more volatile component is always
greater in the vapour phase than in the .solution phase. Alternatively, vapour phase is relatively richer in the
component whose addition to the liquid mLxiuiv results in an increase in the total vapour pressure.
2. Relationship between mole fraction of the components of an ideal solution in the liquid phase and
vapour phase. Suppose the two component.s forming an ideal solution are A and B. Further, suppose
that their mole fractions in the liquid phase arc and .Vg while in the vapour phase, these are represented
by and respectively. Then p^ = p^ =
Pa ■-^aPa° ■^aPa°
>A =
■'^aPa'^-^-^bPb ●va (;^a“-/’b°) +V

or Pb° 1
yA ^aPa° Pa" Pa -^a
SOLUTIONS 2/27

Pa1-Pb1]Pa°
o o

or
Pb -Pa
O

●^A V ●
>’A Pa° Pb“ Pb >’a

Hence, plot of 1/a'^ vs 1/v^ will be linear with slope = pp^lp^ and intercept = {p^ - Pa^^P^
3. Relationship between mote fraction of a component in the vapour phase and total vapour pressure
of an ideal solution.
0

Pa ^aPa ^aPa
or x^p^' = Va (/>a” - Pb°)'^a + y’APB°
^- = .PTotal ’^aPa"+^bPb° 0’a‘’-V^-’^a + /’b
o

or
■^A [PA° + (Pb’^ - Pa°) >’a] = >'aPB or

Pa°'^^Pb^~Pa°^>'a
But
Pa _^aPa"
>’A = *Toial
p ^Tolal
Putting the value of we get

Pa" VaPb° /^a°-^(Pb°-Pa°)Va _ Pb°-Pa°

w
X or
>’a - ^Total
p PA" + (PB"-PA"'fyA P.
Total Pa
O

Pb
O

Pb" Pa"pb"

F lo
Hence, plot of I/Pj^ai versus y\ is linear with slope = (pq° - p^^Vpj^ P^ and intercept = \lp^
4. Variation of Vapour Pressure with Temperature. As studied in class +1, in the properties of liquids,
vapour pressure of a liquid increases with temperature as shown in Fig. 2.12.
Quantitatively, the variation of vapour pressure with temperature is given by Ciausius-Clapeyron

ree
equation,

4 (In P) = RT^
AH u
or d (In P) =
AH.
^<rr
for F
dT RT-^

On integration, we get
Your
ks

AH
eBoo

LI
lnP = - + c ,..(/)
RT

where C is constant of integration and AH., is the enthalpy of vaporisation of the liquid. Thus, plot of In
ad

P versus 1/T will be linear with negative slope as shown in Fig. 2.13.
our

FIGURE 2.121
FIGURE 2.13
Re
Y
Find

3
U)
I/I
/
0) /
Q.
3 / In P
O /
Q.
CO \\
>
\
>
Temperature
liT

Variation of vapour pressure Plot of In P versus 1/T

with temperature

If P, is vapour pressure at temperature Tj and P2 that at T2. then from eqn. (/), we have
AH 1 1
vap
log 10
P
i
2 303R T
1 ^2
2/28 New Course Chemistry (XI1)EE2S]

NUMERiCA.I_ FORMULAS U:
I
PROBLEMS
BASED (/) According to Raoult’s law, for an ideal solution of two liquids A
ON and B. Pb Pb
where and /?g are partial pressures of components A and B. and
Raoult’s Law
Pq° are the their vapour pressures in the pure state, .y^ and .tg are the
mole fractions of A and B in the solution,
(/z) Mole fractions of A and B in the vapour phase ( and >’g) can be
calculated from the partial pressures and pg as follows :
P^ Pa Pr
Va = and or = 1 ->’a-
y]i pTotal

w
Pa Pb ^Twai

lo
Problem f] Vapour pressures of chloroform (CHCl,) and dichtoromethanc (CHjClj) at 298 K are

e
200 mm Hg and 415 mm Hg respectively. Calculate

re
(i) the vapour pressure of the solution prepared by mixing 25-5 g of CHClj and 40 g of CH2C12 at

rF
298 K, and (i7) the mole fraction of each component in vapour phase.

F
(NCERT Solved Example, Kerala Board 2012)
Solution. (<) Calculation of vapour pressure of the

r
solution

fo
u
Mass of CHCl3 = 25-5g
ks
Mass of CH2CL = 40 g
Yo
Molar mass of CHCI3 = 12 + 1 + 3 x 35-5 = 119-5 g mol"*
oo

Molar mass of CH^C1^ = 12 + 2 + 2 x 35-5 = 85 g mol"'

eB

25-5 g 40g
Moles of CHCI3 = = 0-213 mole ; Moles of CH^CL = = 0-470 mole
119-5g mol 85 g mol '
ur

0-213
ad

Mole fraction of CHCI3 (-^CHCI3^ “ 0-213 + 0-470 = 0-312

Yo

Mole fraction of CH2CI2 (-’^CH-,CI2^ ~ ^ - 0-312 = 0-688

d
Re
in

^total “ PCHC\^ PcH2C\2 ~ ●'^CHClj ^ P^CHO^ ‘‘■●^CH2Cl2 ^ ^'’cHjCb

F

= 0-312 X 200 + 0-688 x 415 = 62-4 + 285-5 = 347-9 mm

Calculation of mole fraction of each component in vapour phase

= 62-4 mm,
As calculated above, PCHCI3 ^■'cH2Ci2 ^ 285-5 mm, = 347.9 nim

Mole fraction of CHCI3 in the vapour phase (y CHCI3 ) =

/^CHCl3 _ 62-4 mm = 018
347-9 mm
^toial

Mole fraction of CH,Cl2 in the vapour phase (y, CHiClj )=‘->'cHCl3=l-0i8 = 0-82

Note. It is interesting to note that p“ch2C1^ ^ /'”cHCl3 ' shows that CH2CI2 is more volatile than CHCl3.
Also we observe that mole fraction of CH,CL is greater than that of CHCI3 in the vapour phase. Thus, we
conclude that vapourphase is richer in the more volatile component.
SOLUTIONS 2/29

Problem
B Two liquids X and Y on mixing form an ideal solution. At SO^C, the vapour pressure of
the solution containing 3 moles of X and 1 mole of Y is 550 mm Hg. But when 4 moles of X and 1 mole of Y
are mixed, the vapour pressure of the solution thus formed is 560 mm Hg. What would be the vapour
pressure of pure X and pure Y at this temperature ?
Solution. Suppose the vapour pressure of pure X is p°^ and that of pure Y
3
In the first case, mole fraction of X = — = 0-75 and mole fraction of Y = 0-25.
4

By Raoult’s law. ^Total -Px'^Py 550 = 0-75 p\ + 0-25 p\ ...(/)

4
In the second case, mole fraction of X = — = 0-80 and mole fraction of Y = 0-20.
5

low
By Raoult’s law. 560 = 0-80 p\ + 0-20 p\ ...(H)
From eqn. (/), 2200 = 3 + P\ ...{Hi)
From eqn. (ii). 2m = 4p\+p\ ...(rV)
Subtracting eqn. (Hi) from eqn. (/v), we get, p°^ = 600 mm

ee
Substituting in (///), we get, p°y = 2200 - 3 x 600 = 400 mm

F
Fr
Problem
Ethylene dibromide (C2H4Br2) and 1,2-dibromo propane form a series of ideal solutions
over the whole range of composition. At 85®C, the vapour pressures of these two liquids are 173 and 127

for
torr respectively. What would be the mole fraction of ethylene dihromide in a solution at 85“C equilibrated
ur
with 1 : 1 molar mixture in the vapour ?
Solution. Suppose the mole fraction of ethylene dibromide in the liquid phase = x. Then mole fraction of
ks
1, 2-dibromo propane will be = 1 - 'a.
Yo
oo

Vapour pressure of ethylene dibromide = xx p° = xx 173 torr = 173 x torr

eB

Vapour pressure of 1, 2-dibromo propane = (I - a) x 127 torr

As they have 1 : 1 molar ratio in the vapour phase,
127
r

173a = (1-A)x 127 or 173 a = 127 - 127 a or 300a=127 or X = = 0-423.

ou
ad

300
Problem
□ At a given temperature, the vapour pressure in mm of Hg of a solution of two volatile
Y

liquids A and B is given by the equation p - 120 - 80 Xjj (Xg = mole fraction of B)
Calculate the vapour pressures of pure A and B at the same temperature.
nd
Re

Solution. When Xg = 0, we have pure A ●.

p/ = 120 - 80 X 0 = 120 mm
Fi

The given expression can be written as P= 120-80(1 -Xa)

When X_^ = 0, we have pure B Pg° = 120 - 80 (1 - 0) = 40 mm
Alternatively, for pure B, .Vg = 1. Hence, p^ = 120 - 80 x 1 = 40 mm
Problem
At 90"C, the vapour pressure of toluene is 400 mm and that of xylene is 150 mm. What
is the composition of the liquid mixture that will boil at 90"C when the pressure of mixture is 0-5 atm ?
Solution. At the boiling point (90°C),
Vapour pressure of mixture (P,o,ai) = 0-5 atm
760
mm = 380 mm
2

total - P\ ^ ●’^X P"x (T = Toluene, X = Xylene)

- P^J + (1 “ P°X
2/30 New Course Chemistry CXII)E3SIBI

380 = .rT-(400) + (l (v Xj + X^= I)

= 400.r-p+ 150- 150.Vt
230
or
250.tj=230 or Xj - 250 = 0-92

.vj^ = t - 0-92 = W-08

ow
1. The mole Ihiction of ethyl alcohol in its solution with methyl alcohol is 0-80. The vapour pressure of ethyl
alcohol at the temperature of the solution is 40 mm of Hg. What is its vapour pressure in solution if the
solution is ideal ?

2. Benzene and toluene (C^Hg) form a nearly ideal solution at 313 K. The vapour pressures of pure

e
benzene and toluene are 160 mm of Hg and 60 mm of Hg respectively. Calculate the panial pressure of

re
benzene and toluene and the total pressure over the following solutions : (/) containing equal weights of
benzene and toluene, (ii) containing 1 mole of benzene and 4 moles of toluene. (Hi) containing equal molecules

Frl
F
of benzene and toluene.

3. The vapour pressure of a pure liquid A is 40 mm Hg at 310 K. The vapour pressure of this liquid in solution
with liquid B is 32 mm of Hg. Calculate the mole fraction of A in the solution if the mixture obeys Raoult’s
ou
r
law.

so
4. Methanol and ethanol form nearly an ideal solution at 300 K. A solution is made by mixing 32 g methanol
kf
and 23 g ethanol at 300 K. Calculate the partial pressures of its constituents and the total pressure of the
oo
solution. [At 300 K : PcH^OH = ^8’. /^CiHgOH = ]
Y
The vapour pressures of benzene and toluene at 293 K are 75 mm and 22 mm Hg respectively. 23-4 g of
eB

5.
benzene and 64-4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of
benzene in the vapour phase assuming that the vapours are in equilibrium with the liquid mixture at this
ur
oY

temperature.
ad

ANSWERS

1. 32 mm
d

2. (() pg = 86-56 mm of Hg, pj ~ 27-54 mm of Hg, P =pg + py= 114-10 mm of Hg

in
Re

(//) pg = 32 mm of Hg, pj- = 48 mm of Hg, P = pg + = 80 mm of Hg

(Hi) pg = 80 mm of Hg, pj = 30 mm of Hg, P = Pa + p?- = 110 mm of Hg]
F

3. 0-8
4. 7'cH30h =60mm, PC2H5OH =17 mm; Total = 77 mm 5. 0-59

HINTS FOR DIFFICULT PROBLEMS

!● PCiH-iOH ~ -^C2H50H ^ P°C2H50H = 0-80 X 40 mm = 32 mm.

2. (/) Mass of benzene = mass of toluene = m- g
-1
Molar mass of benzene, C^^H^ = 78 g mol"'. Molar mass of toluene, C^Hg = 92 g mol
U’ w
Moles of benzene = — and Moles of toluene = —
78 92

, r 1 ● u
Total no. of moles m the solution =
. ● w 1
H- =
92vv-f78w =
I70vv
78 92 78x92 78x92
 SOLUTIONS 2/31

vv/78 92
Mole fraction of benzene. ■^B = = 0-541
170vv/(78x92) 170

and Mole fraction of toluene. Ay= 1 -0-541 =0-459

o
(i7).Vb= 1/(1 +4)= I/5=0-2,.Xt-={)-8.
.(Hi) /jjj = N/N(), iij = N/N„. Hence, n B = n j = 0-5. (N = No. of moleciilc.s, Nq= Avogadro’s No.)

lo
3.
Pa = -'^A ^ Pa° 32 = .r^ x 40 or = 32/40 = 0 8.

k
4. «
CH3OH
= 32/32 = 1 mol, n — 23/46 - 0-5 mol.
C2II50H

Y
F
5. In the liquid phase. = 23-4/78 < 0-3. = 64-4/92 = 0-7. Hence, x^^ = 0-3, .v^. = 0-7

r
Pb - ^'3 X 75 = 22-5 mm. pj = 0-7 x 22 = 15-4 mm ;

o
Total vapour pressure =22-5 -1- 15-4 = 37-9 mm
Mole fraction of benzene in the vapour phase = 22-5/37-9 = 0-59.

o
Y

o
2.8. IDEAL AND NON-IDEAL SOLUTIONS

On the basis of Raoull’s law. liquid-liquid solutions can be classified into ideal and non-ideal solutions

f
as follows :

B
(1) Ideal solution

dr
An ideal solution is that solution in which each component obeys Raoult ’v law under all conditions
of temperatures and concentrations.

An ideal solution will satisfy the following conditions :

u
n

(/) There will be no change in volume on mixing the two components, i.e., = 0
e

(//) There will no change in enthalpy (i.e., no heal is evolved or absorbed) when the two components are
mixed, i.e., = 0
For example, when we mix 50 cm-^ of benzene with 50 cm^ of toluene, the volume of the solution is
o
i

found to be exactly 100 cm^, i.e., = 0

Further, it is found that = 0
ad F

Hence, the solution obtained is ideal. Thus,

An ideal solution may be defined as that solution in which no volume change and no enthalpy
change take place on mixing the solute and the solvent in any proportion.
Re

At the molecular level, an ideal solution may be defined as follows:

An ideal solution ofthe components A and B is defined as the solution in which the intermolecular
interactions between the components (A—B attractions) are of the same magnitude as the
intermolecular interactions found in the pure components (A A attractions and B—B
attractions).

Practically, no solution is ideal. However, when concentration of solute is very low, the solution behaves
ideally. Therefore, very dilute s lutions are nearly ideal. Substances having similar structures and polarities
form nearly ideal solutions. A w examples are given below :
(/) Benzene -1- Toluene
(//) n-Hexane + w-Heptane
(///) Ethyl bromide -f Ethyl chloride
(/V) Chlorobenzene -1- Bromobenzene
The graphical behaviour of ideal solutions has alreadj^been diA44^^ in Fig. 2.9 (Page 2/24).
- ● W
2/32 “^nxxdeeft-'ii. New Course Chemistry CX1I)E2S!91

(2) Non-ideal solution.

A solution which does not obey RaouWs law is called non-ideal solution.
For such solution. ^ 0. 0. For example, when wc mix sulphuric acid (solute) in
water (solvent) the amount of heat generated is very large and a change in volume is also observed. This is
flue to formation of a non-ideal solution.
In terms of molecular interactions, a non-ideal solution may be dcHned as follows :

A non-idcal solution is that solution in which solute and solvent molecules interact with one
another with a different force than the forces of interaction between the molecules of the pure
components.

Tj'pes of Non-ideal solutions. These are divided into two types as explained below :

low
{a) Non-ideal solution.s showing positive deviations. When a component B is added to another
component A, sometimes the partial pressure of a component A is found to be more than expected on the basis
of Raouli’s law. A similar effect is observed for the other component B in the reversed mixing. The total
vapour pressure for any .solution is thus greater than that corresponding to an ideal solution of the same composition.
Such behaviour of solutions is de.scribed as a positive deviation from Raoult’s law (Fig. 2.14).

ee
The boiling points of such .solutions are relatively lower as FIGURE 2.14

F
compared to those of the pure components (becau.se higher the

Fr
vapour pressure, lowering is the boiling point). For one intermediate
composition, the total vapour pressure of such a solution will be Pb°
the highest and the boiling point will be the lowest. This solution
for
ur
111
tx
acquires the property of boiling at a constant temperature and its CO
co Pa¬
composition remains unchanged. Liquid mixtures which distil
ks
Hi
cc
without any change in composition are called azeotropes or a.
Yo
azeotropic mixtures. In case of solutions showing positive cc
oo

deviations, we get minimum boiling {point} azeotropes. O

eB

CL
<
The positive deviations are exhibited by liquid pairs for >

which the A—B molecular interaction forces are lower than the
A—A or the B—B molecular interaction forces. For example,
r
ou
ad

mixtures of ethanol and cyclohexane (or acetone) show positive Xa = 1 mole fraction xa = 0
deviations. In pure ethanol, a very high fraction of the molecules Xb = 0 Xb = 1
Y

are hydrogen bonded as shown below : A vapour pressure graph showing

5+6- 6+8- 5-1 8- positive deviations (solid lines)
Re
nd

H-0 H-0 ●●●● H-0 from ideal behaviour (dotted lines)

Fi

C2H5 C.Hj
On adding cyclohexane (or acetone), its molecules get in-between the molecules of ethanol, thus breaking
the hydrogen bonds and reducing ethanol-ethanol attractions considerably.
Similarly, when carbon sulphide (CS-,) is added to acetone [(CH3)2C = O]. the solute-solvent dipolar
interactions arc weaker than solute-solute and solvent-solvent interactions. Hence, this solution also shows
positive deviation.
In case of solutions showing positive deviations, a slight increase in volume and absorption of heat
takes place on mixing as expected (i.e., AV and AH both are positive).
To .sum up, for a non-ideal solution showing positive deviation,
P\ > PU > -^B P"^ Pt«U.l > -^A P\ + -^B P%1 (“) = +ve (m) AV^^ = -Fve
A few more examples of non-ideal solutions showing positive deviations are given below :
(/) Acetone -+ Carbon disulphide. (//) Acetone + Ethyl alcohol. (///) Acetone + Benzene,
(/v) Methyl alcohol -i- water (v) Ethyl alcohol + Water. (I’O Carbon tetrachloride + Chloroform
(vii) Carbon tetrachloride + Benzene (vi7i) Carbon tetrachloride -1- Toluene.
 SOLUTIONS 2/33

(h) Non-ideal solutions showing negative deviations. If FIGURE 2.15

for the two components A and B, the forces of interaction between

ACTUAL P
the A and B molecules are more than the A — A and IDEAL P
Pb“
B
B forces of interaction, the escaping tendency of A and B <7
types of molecules from the solution becomes less than from m
q:
D Pa®
the pure liquids. In other words, for any composition of the CO
CO
MIN.
solution, the partial vapour pressure of each component will be LU
a:

less and the total vapour pressure of the solution will also be CL
q:
'Pb
less than that expected from Raoult’s law (Fig. 2.15). The.se D
o
solutions are said to show negative deviations from Raoull’s Q.
<

law. Such solutions have relatively higher boiling points as

>
Pa

compared to those of the pure components (because lower the

vapour pressure, higher is the boiling point). For one intermediate Xb = 0 MOLE FRACTION Xa = 0
composition, the total vapour pressure of the solution will be Xa= 1 xb = 1

w
the least and the boiling point will be the highest. Such a solution
will also distil without any change in composition and provides A vapour pressure graph showing

F lo
an example of another kind of azeotrope. We cal 1 it the maximum negative deviations (solid lines)
from ideal behaviour (dotted lines)
boiling (point) azeotrope.
For example, negative deviation from Raoult’s law is exhibited by a mixture of chloroform (CHCl^) and

ee
acetone, (CH3)2CO. When these are mixed, the hydrogen bonding takes place between the two molecular

Fr
species as shown below due to which the escaping tendency of either of the liquid molecules becomes less.
Consequently, the boiling point of solution increases.
Cl
for
ur
CHj
Cl—C—H ,0 = C
s
CH3
ook
Yo
Cl

Hyeirogen bonding between Chloroform and Acetone

eB

In case of solutions showing negative deviations, a slight decrease in volume and evolution of heal takes
place on mixing as expected (/.c., AV and AH both are negative).
r

To sum up,/or a non-ideal solution showing negative deviation,

ad
ou

(i) P^ < Xj, p\, P^^ < Xit <»> = -ve

(Hi) AV,„i,i„g - -ve
Y

Apart from hydrogen bonding, other weaker intennolecular forces like dipole-dipole attractions may also be
partly responsible for observed deviations from ideal behaviour.
Re
nd

A few more examples of non-idcal solutions showing negative deviations arc given below :
Fi

(/) Chloroform + Benzene. (//) Chloroform + Diethyl ether, (Hi) Acetone + Aniline,
(/v) HCl + Water. (v) HNO, + Water. (v’/) Acetic acid + Pyridine.
Difference between Ideal and Non-ideal solutions. The various points of difference may be summed
up as follows :
Ideal Solutions Non-ideal Solutions

1. The interactions between the components are 1. The interactions between the components are
similar to those in the pure components. different from those of the pure components.
2. There is no volume change and enthalpy change 2. AV ^ 0, AH 0 on mixing the components.
on mixing the components (AV = 0, AH = 0)
3. Each component obeys Raoult's law at ail 3. They do not obey Roult’s law. They show
temperatures and concentrations, i.e.. Pa=-^aP°a positive or negative deviations from Raoult’s
law. i.e„ p^^x^p\
4. Their graphical behaviour is shown in Fig. 2.9. 4. Their graphical behaviour is shown in Figs. 2.12
and 2.13.
5. They do not form azeotropes. 5. They form azeotropes.
2/34 "pfuidee^’^ New Course Chemistry CXI1)BZ3MI
Difference between solutions showiii}* positive deviation and negative deviation from ideal
behaviour. The main points of difference aie summed up in the Table below :

Solutions showing positive deviation Solutions showing negative deviation

I. The interactions between the components are 1. The interactionsbetween the componentsare
less- than in the pure components. greater than the pure components.
2- =+ive. 2. AV = -ve.
nuxing
- -ive.

4* Pa>J^aPa"'Pb>'^bPe’ 4. Pa<XaPa°^Ph<-^bPb°
5. Form minimum boiling azeotropes 5. Form maximum boiling azeotropes.

2.9. AZEOTROPIC OR CONSTANT BOILING MIXTURES

In case of solutions showing positive deviations from Raoult's law, at one of the intermediate

w
compositions, the total vapour pressure is the highest and the boiling point is the lowest. In case of solutions
showing negative deviations from Raoult’s law, for one intermediate composition, the vapour pressure is the

F lo
lowest and boiling point is the highest. For liquid pairs of such intermediate compositions, the composition of
the liquid and vapour phases is the same. So, when liquid mixture of such a composition vaporises without
change in composition, the liquid obtained by the condensation of the vapour also has same composition i.e.

ee
the mixture distils over os if it were a pure licfuid. The composition of such a mixture is, of course, fixed.

Fr
This type of liquid mixture, having a deifnite composition, and boiling like a pure liquid, is
called a constant boiling mixture or an azeotropic mixture or simply an azeotrope.
for
In case of positive deviations we get minimum boiling azeotropes whereas in case of negative deviations
ur
we get maximum boiling azeotropes. Some azeotropic mixtures are given in Table 2.4.
s
TABLE 2.4 Some azeotropic mixtures
ook
Yo

Components Composition by Weight Boiling Points (K)

eB

A B % OF B A B Azeotrope

SOLUTIONS SHOWING POSITIVE DEVIATIONS (Minimum boiling Azeotropes)

our
ad

H.O C.H3OH 95-37 373-00 351-3 351-15

71-69 373-00 370-19 350-72
H2O C3H7OH
Y

67 329-25 319-25 312-25

(CH3).C0 CS.
Re

6-8 334-2 351-3 332-3

CHC13 C2H3OH
nd

SOLUTIONS SHOWING NEGATIVE DEVIATIONS (Maximum boiling Azeotropes)

Fi

H,0 HCl 20-3 373-0 188 383

HoO HI 57-0 373-0 239 400

H^O HNO3 68-0 373-0 359 393-5

71-6 373-0 383 476
H^O HCIO4
It may be noted from the above table that in case of minimum boiling azeotropes, the boiling point of the
azeotrope is less than the boiling point of either of the pure components. For example, a mixture of ethanol
(b. pt. 351-3 K) and water (b. pt. 373 K) containing 95-4% of ethanol fonns an azeotrope with boiling point
351 -) 5 K. Similarly, in case of maximum boiling azeotropes, the boiling point of the azeotrope is higher than
that of either of the pure components. For example, a mixture of nitric acid (b. pt. 359 K) and water (b. pt.
373 K) containing 68% nitric acid fonns an azeotrope with boiling point 393-5 K.
Azeotropic mixtures cannot be separated into their constituents by fractional distillation. For example,
if we try to concentrate a dilute solution of alcohol, we can do so only till it becomes 95-4% because after that
it becomes a constant boiling mixture (azeotrope). The mixture containing 95-4% alcohol and 4-6% water is
called rectified spirit.
 SOLUTIONS 2/35

To obtain pure alcohol (100% alcohol) called absolute

alcohol from rectified spirit, the method used is called azeotropic distillation. For this purpose, rectified
spirit is mixed with a suitable amount of benzene. The mixture is then subjected to fractional distillation.
The following three fractions are obtained at different temperatures.
(0 At 331-8 K, a ternary azeotrope (constant boiling mixture) containing 74-1% benzene, 18 5% alcohol
and 7-4% water by mass is obtained.
(ii) At 341-2 K, a binary azeotrope containing 67-7% benzene and 32-3% alcohol by mass is obtained.
(Hi) At 351 K (last fraction), 100% alcohol is obtained..

Curiosity Question

w
f
Q. Why pure ethyl alcohol cannot be obtained from rectified spirit (95-4% alcohol) even by
fractional distillation ?

Flo
Ans. This is because a mixture of 95-4% alcohol with 4-6% water forms an azeotrope, i.e., a constant

e
boiling mixture.
J

re
F
' ’0. C( jETERMINATIOr'OFMC IASS
We have discussed above that when a non-volatile solute is added to a volatile solvent, lowering of
ur
r
vapour pressure takes place. The decrease in vapour pressure of the solvent depends only on the amount of

fo
the solute dissolved and not on the nature of the solute. For example, we may dissolve 1 mole of sucrose or 1
mole of urea into I kg of water, the lowering of vapour pressure is same at a particular temperature. Similarly,
ks
there are many properties of the solutions, which are connected with the lowering of vapour pressure and
Yo
oo

hence depend only on the amounts of the solutes dissolved. All such properties of the solutions are thus
linked with each other and hence are called colligative properties (From Latin : co means together and ligare
B

means to link). We may thus define colligative properties as follows :

re

Those properties of ideal solutions which depend only on the number of particles of the solute
(molecules or ions) dissolved in a definite amount of the solvent and do not depend on the
u
ad

nature of solute are called ■ operti:'

Yo

The important colligative properties are :

(0 Relative lowering of vapour pressure
d
Re

(ii) Osmotic pressure

in

(Hi) Elevation of boiling point

F

(tv) Depression of freezing point.

Let us discuss each of these properties one by one.
2.1 i . PRESSURE

2..,-i.

This property has already been discussed in Art. 2.7.3 under Raoult’s law for solutions of solids in
liquids. It was explained that when a non-volatile solute is dissolved in a solvent, the vapour pressure of the
solution is lower than that of the pure solvent. Further, it was derived that the relative lowering of vapour
p°~p ni
pressure is given by the equation,
P°
where p° = vapour pressure of the pure solvent.
p^ - vapour pressure of the solution
«2 = number of moles of the solute and
n, = number of moles of the solvent
2/36 Course Chemistry

2.11.2. To conclude that the relative lowering of vapour pressure is a colligative property
This follows directly from the equation given above which shows that relative lowering of vapour
pressure depends only upon the number of moles of the solute dissolved in a definite number of moles of the
solvent and is independent of the nature of the solute.
2.11.3. Determination of molecular masses of solutes from lowering of vapour pressure
p°-p s _ «2
According to Raoult’s law,
P’' /Jj +/1-5
If US g of the solute are dissolved in w, g of the solvent and if M2 and M, are their respective molecular
masses, we have

w
n
2 “
M,
w

Flo
1
and n
I “
M
1

ee
Substituting these values in the above expression, we get

Fr
P -Ps ^2 1^2 ...(f)
P" w
1 /Mj+H'2/M2

or
ur
For a dilute solution, «-> can be neglected in comparison with so that Raoult's law equation becomes
f
ks
P -P s _ ”2
Yo
oo

p n
i
B

P°'P s ~ u'2 /M2

or
re

w.1 /M 1

H'2><Mj
u

P -Ps
ad

or ..Hi)
Yo

P >VjXM2
From expressions (/) and (//), it is clear that if us g of the non-volatile solute are dissolved in w, g of ilie
nd
Re

solvent and the vapour pressure of the pure solvent ip°) and that of the solution (p^) are measured experimentally,
then knowing the molecular mass of the solvent (M |), that of the solute (M,) can be calculated.
Fi

This method is usually not preferred because other methods (namely, elevation in boiling point or
depression in freezing point) give results more easily and accurately.

2.11.4. Measurement of relative lowering FIGURE 2.16

of vapour pressure Air saturated with
Pure dry air is first passed through bulbs vapour of solvent from solution
containing solution and then through bulbs containing DRY
AIR
pure solvent and finally through tubes containing
SOLVENT
anhydrous CaCl2 (if solvent used is water). /

Loss in weight of solution bulbs (w,) SOLUTION

/
oc
vapour pressure of the solution (p^)
Loss in weight of solvent bulbs (^2)
OC V.P. of solvent - V.P. of solution. Passing dry air first through the
Thus, and p^-p^oc >V2. solution and then through the solvent
SOLUTIONS 2/37

This gives ^ W2 Loss in weight of solvent bulbs

P VV| + W2 Total loss in weight of solution
The above method is called Ostwald-Walker dynamic method or transpiration method.
Refer to Solved Problem 5, page 2/39 and 40.
IMPORTANT
Quite often, it is difficult to decide whether the calculations should be done by assuming the solution to be
dilute or not, i.e., ^3 should be neglected in comparison to n,1 or not. The following alternative formula
eliminates this ambiguity.

ow
”2
According to Raoult’s law,
P nj +«2
This may be rewritten as (by taking reciprocals)

e
P° _ ”1 ”2

re
n
= -^ + \ or
P° -1=^ or
P°-(P°-P,)
P°-P

Frl
>h »2
s
P'^-Ps n-y P°-Ps "2

F
Ps
P"~Ps _ «2 ,^2^1 p"-Ps _
ou
or
or or

P°-P.s ^*2 Ps n
1
w,1 /M 1 w,M2 Ps ”'1 ^2
This formula can be used for dilute as well as concentrated solutions.
kfs
oo
Y
eB

NUMERICAL
t, J
PROBLEMS
According to Raoiilt’.s law.
ur

BASED
oY

ON
P -P s — VI.’.) /M., n
2 _ hs/M-> >v.,xM 1
ad

Lowering of 1 P° + fi2 vVj / M| + VV2 / M.) 'h w./M

I I W] XM2
(if the solution is very dilute)
d

Vapour Pressure
„ . P°-p, >h
in

or lor all solutions (dilute or concentrated), =

Re

— = — =-
Ps n
1 VI-,/M i
F

where «j, m', . Mj = no. of moles, mass in g and mol. mass of the solvent
respectively
and ;i2, Mt = no. of moles, ma.ss in g and mol. mass of the solute
respectively.

Prok>lom
fl Vapour pressure of water at 293 K is 17-51 mm. Lowering of vapour pressure of a
sugar solution is 0’0614 mm. Calculate (i) Relative lowering of vapour pressure (ii) Vapour pres.surc of the
.solution and (iii) Mole fraction of water.
Solution. Here, we are given that Vapour pressure of water (p°) = 17-51 mm

Lowering of vapour pressure (p°-Pj) = 0’0614mm

P"-P ●V 0-0614 mm
(/) Relative lowering of vapour pressure = _
= 000351
P° 17-51mm
2/38 'P'uuUe^’A. New Course Chemistry (Xll)SSSlS]

(«) Vapour pressure of the solution (p^)

0-0614 mm ; p° - 17-51 mm
17-51 mm - p^= 0-0614 mm or 17-51 mm - 0-0614 mm = 17*4468 mm
(Hi) To calculate mole fraction of water :

P° -PS ^2 P^-P .V
— = 0-00351
By Raoult’s law — = ^2, mole fraction of solute, i.e., X2 =
«j + n-y P°

Mole fraction of solvent (water), x^ = \ - x^ = I - 0-00351 = 0*99649

Problem
0 The vapour pressure of a 5% aqueous solution of a non-volatile organic substance at
373 K is 745 mm. Calculate the molar mass of the solute. (Jharkhand Board 2012)
Solution. 5% aqueous solution of the solute implies that 5 g of the solute are present in 100 g of the solution

low
i.e.. Mass of solute (^2) = 5 g ; Mass of solution = 100 g
Mass of solvent {w^) = 100 - 5 = 95 g
Further, as the solution is aqueous, it means that the solvent is water and we know that
Vapour pressure of pure water at 373 K (p°) = 760 mm

ee
Vapour pressure of the solution at 373 K (p^) = 745 mm (Given)
rF
Fr
-I
Molar mass of solvent (water), Mj = 18 g mol
Molar mass of solute, M2 = To be calculated

for
Using the formula for dilute solutions, viz..
u
P°~Pi ^ W2/M2 ^2^1
ks
, we get
P° n
1 Wj/Mj Wj M^
Yo
oo

-1
760-745 5gxl8gmol
eB

760 95gxM2
5x18x760
r

g mol ’ = 48 g mol
-1
M2 =
ou

or
ad

15x95
Y

Assuming that the solution is not dilute, we apply the formula

W2 /M2
nd
Re

P° rtj + /I2 Wj / Mj -I- W2 / Mt

Fi

760-745 5/M2
760 95/I8 + 5/M2
5 760 760 760 5 95
^ _5_
T^^1^“3M2
or or

3M2 M2 18
18 M2

or
1
—-5I- — or
I X
745 _ 95
M
2 y
3 18 M2 18

745 18 -1
or
M2 = X — = 47*05gmol
3 95

Note. Usually when the solution is less than 5%, it is taken as dilute and the approximate formula is
employed.
SOLUTIONS 2/39

P°-Ps _ n
2 _ W2/M2 H2M]
Alternatively, applying the formula,
Ps n
1 u’l / M| H'j M,
760 mm - 745 mm 15 5gxl8gmol '
or
or M-> = 47*05 g mol *.

rw
745 mm 745 95gxM-,

Problem The vapour pressure of pure benzene at a certain temperature is 0*850 bar. A non
volatile, non-electrolyte solid weighing 0*5 g is added to 39*0 g of benzene (molar mass 78 g mol"*). The

e
vapour pressure of the solution then is 0*845 bar. What is the molar mass of the solid substance ?
(NCERT Solved Example)

e
-1
Solution. Here, we are given that W2 = 0-5 g, vvj = 39-0 g, Mj = 78 g mol

lo
r
= 0-850 bar, = 0-845 bar

F
u
P"-P, n.
W2/M, ^2^1
Substituting these values in the formula. we get
Wj/Mj vi'i M2'

oF
n
I

rs
0-850 bai-0-845 bar 0-5gx78gmol
0-850 bar
39-0gxM2

k
o
0-5x78 0-850
gmol ' = 170 g mol *

o
M2 =

f
or x
39-0 0-005
o
Y
Problem
B At 298 K, the vapour pressure of water is 23*75 mm Hg. Calculate the vapour pressure
at the same temperature over 5% aqueous solution of urea [(CO{NH2)2].
B
rY

Solution. 5% aqueous solution of urea means that

Mass of solution = 100 g ;
ue

Mass of solute, Le.. urea, w, = 5 g

od

Mass of solvent, /.e., water, Wj =100 - 5 = 95 g ;

Vapour pressure of pure water {p°) at 298 K = 23-75 mm
ad
in

Vapour pressure of urea solution, i.e., =? ;

-1
Molar mass of water (M,) = 18 g mol
-I
Molar mass of urea CO(NH2>2, i.e., M2 = 12 -t- 16 -1- 14 x 2 -t- 2 = 60 g mol
Re
F

Applying Raoult’s law.

p°-p J n
_ "2 _ ^2 /M2 ^2 M
1
X
n
1 w,1 /M 1 M2 w
1

Substituting the values, we get

23-75 - p^ 5 18 C I O
X or 23-75-p=—x-^x
" 60 95
23-75
23-75 60 95

or
23-75-p, = 0-375 or
p, = 23-375
Thus, the vapour pressure of 5% urea solution = 23*375 mm
Problem
H A current of dry air was passed through a solution of 2*5 g of a non-volatile substance
‘X’ in 100 g of water and then through water alone. The loss of weight of the former was 1*25 g and that of
the latter was 0*05 g. Calculate (i) mole fraction of the solute in the solution (ii) molecular weight of the
solute.
2/40 T^tauice^'^ New Course Chemistry (XII)BSm

Solution. Loss in weight of solution « p a

oc
1 -25 g ; Loss in weight of solvent (water) -p^ OC
0-05 g
oc
0-05 +1-25, i.e., p°« l-30g

Hence,
_ 0-05 = 0-0385. le., ^-,=0*0385
P 1-30

Now,
P°~Ps - ^2 _ .
n
1 Wj / M|
2-5/M^ 2-5 _ 100
0-0385 = or
X 0-0385 or M2 = 11-7.

w
100/18 M, “ 18

o
e
1. The vapour pressure of 2-1 % of an aqueous solution of a non-electrolyte at 373 K is 755 mm. Calculate the

re
molar mass of .solute.

2. A solution containing 6 g of benzoic acid in 50 g of ether (C2H5OC2H5) has a vapour pressure of 410 mm of

Frl
F
mercury at 293 K. Given that the vapour pressure of ether at the same temperature is 442 mm of mercury,
calculate the molecular mass of benzoic acid. (Assume that the solution is dilute).
ou
3. The vapour pressure of water is 92 mm at 323 K, 18-1 g of urea are dissolved in 100 g of water. The vapour

sor
pressure is reduced by 5 mm. Calculate the molar mass of urea.
4. Calculate the vapour pressure at 295 K of a 0-1M solution of urea. The density of the solution may be taken
as 1 g/cm^. The vapour pressure of pure water at 295 K is 20 mm.
kf
5. The vapour pressure of an aqueous solution of cane sugar (mol. mass 342) is 756 mm at 373 K. How many
oo
grams of sugar are present in 1000 g of water ?
Y
6. At 25‘’C. the vapour pressure of pure water is 23-76 mm of Hg and that of an aqueous dilute solution of urea
B

is 22-98 mm of Hg. Calculate the molality of this solution ?

7. Vapour pressure of an aqueous solution of glucose is 750 mm of Hg at 373 K. Calculate the molality and
re

mole fraction of solution.

oY
u

8. How much urea (molar mass = 60 g mol"*) should be dissolved in 50 g of water so that its vapour pressure
ad

at room temperature is reduced by 25%. Calculate molality of the solution obtained.

9. At 50°C. the vapour pressure of pure CS, is 854 torr. A solution of 2-0 g of sulphur in 100 g of CSt has
d

vapour pressure of 848-9 torr. Determine the formula of sulphur molecule.

in

10. A 0-2 percent aqueous solution of a non-volatile solute exerts a vapour pressure of 1-004 bar at 100°C. What
Re

is the molar mass of the solute ? (Given ; vapour pressure of pure water at 100°C is 1-013 bar and molar
mass of water is 18 g mol"’)
F

ANSWERS
-1 -1
1. 58-7 g mol 2. 122-65 u 3. 56-7 g mol 4. 19-96 mm 5. 100-5 g 6. 1-8 m
-1
7. x-y = 0-013, molality = 0-73 m 8. 55-556 g, 18-52 m 9. S 8 10. 4-06 g mol

HINTS FOR DIFFICULT PROBLEMS

P^-P W2/M, 442-410 6x74 6x74 X ^2

2 = 122-65
or
M2 =
_

P^ 442 50XM., 50 32

P' -PS —
n~,
3. The solution is not dilute. Apply —
P /?l -I-
4. 0-1 M solution of urea means O-I mol, /.e., 6-0 g of urea are dissolved per litre of solution, i.e., in 1000 g
of solution (-.- density = 1 g/cm^). Hence, W2 = 6-0 g, vv, = 1000 - 6-0 = 994 g, p° = 20 mm (Given),
p^ = To be calculated.
SOLUTIONS 2/41

760-756 Wo/342
5. . Calculate W2-
760
IOOO/I8 + W2/342
1000
6. Calculate . But x-, =■ — . Take n, = moles.
n^ - n. 18

760-750 10 Uj
7.
760
= AT2 or a:
2 “ 760
. But .V
2 = — . To calculate molality, calculate ih when Wj = 1000 g, i.e..
n
I

n = 1000/18 moles

10 1000
Thus, >h ~
760
X
18
- 0-73 , ie., molality = 0-73.

8. If p°= 100 mm, = 75 mm. Hence. -

100-75 W2 / 60 . This gives us = 55-556 g

w
100
50/18 + u’2/60
55-556/60 mol
Molality of the solution = xlOOOgkg'' = 18-52 m

F lo
50g
P°-Ps . 854-848-9 2/M,
9. (Molecular mass of CS, = 12 + 2 x 32 = 76 u)

ee
n w/M 854 100/76
I l I

Fr
^ 76
or X or M, = 254-5 u
854 100
M2
for
molecular mass = .r x 32. Hence, 32 x - 254-5 or x = 8.
ur
If formula of sulphur molecule is
10. Similar to solved problem 2, Page 2/38
s
W2 = 0-2 g, M’l = 99-8 g
ook
Yo

2.12. OSMOSIS AND OSMOTIC PRESSURE

eB

FIGURE 2.17
2.12.1. General Discussion and Definitions
r

Osmosis. Suppose a concentrated solution of ^ WATER OR DIL.

ou
ad

copper sulphate (deep blue in colour) is placed in a -I^-^CuS04 SOL.

beaker and water (or a dilute solution of copper -SOLVENT-
Y

sulphate) is added slowly along the walls of the

,CONC, CUSO4
Re

beaker without much disturbing the concentrated

nd

SOL. (DEEP
s,ouute; BLUE)
copper sulphate solution. The two layers are more
Fi

or less well defined. Now, if the beaker is allowed UNIFORMLY BLUE SOL.
^Initial position
to stand, it is observed that after a few days, the O After a few days
solution in the beaker becomes uniformly blue Diffusion in solution
throughout. This must be obviously due to the fact
that the particles of the solute (Cu^"*" and SO^“ ions) move slowly into the solvent and the molecules of the
solvent (water) move into the copper sulphate solution. In other words, the particles of the solute and solvent
mix spontaneously into each other (Fig. 2.17).
The spontaneous mixing of the particles of the solute (present in the solution) and the solvent
(presentabove the solution)to form a homogeneousmixture is called diffusion,just as the term
is used for the spontaneous mixing of gases to form homogeneous mixtures.

Now, suppose the experiment is performed in a slightly different manner. Suppose the beaker is divided
into two compartments by a semipermeable membrane, i.e., a membrane which allows the solvent molecules
to pass through but not the solute particles. Suppose again that copper sulphate solution is placed in one
2/42 New Course Chemistry (XII)BSm

compartment and water in the other (Fig. 2.18). It is FIGURE 2.18

observed that the level on the solution side begins to
NET FLOW
rise. This must be obviously due to the fact that greater
number of solvent (water) molecules from the solvent
CONCENTRATED
side pass into the solution side through the CUSO4 SOL- h
semipermeable membrane than the number of solvent
molecules going into the solvent from the solution WATER
through the semipermeable membrane. Similarly, if a SEMIPERMEABLE solvent:_;-
MOLECULES--^
OR
MEMBRANE
concentrated solution is separated from a dilute solution DIL. CUSO4 SOL.

by a semipermeable membrane, there is a net flow of

solvent from the dilute solution to the concentrated
solution through the semi-permeable membrane till the
concentration on both sides of the membrane becomes Osmosis

w
equal. FIGURE 2.19

The net spontaneous flow of the solvent

Flo
WATER NaCISOL
molecules from the solvent to the solution
or from a less concentrated solution to a /

ee
more concentrated solution through a
semipermeable membrane is called

Fr
osmosis (Greek : push).
0 SWELLS /O SHRINKS
for
It may be remembered that a solution as such has
ur
only vapour pressure. It develops osmotic pressure only EGG WITHOUT OUTER SHELL
(Solid lines-Before osmosis. Dotted Ilnes-After osmosis)
when it is placed in contact with the pure solvent or
s
less concentrated solution through a semipermeable
k
Demonstrating the phenomenon of osmosis
Yo
membrane.
oo

The phenomenon of osmosis can be demonstrated in the laboratory as shown in Fig. 2.19.
eB

Take two eggs. Remove the outer shells of the eggs by dissolving them in hydrochloric acid. Place one
egg in water and the other in sodium chloride solution. The former swells whereas the latter shrinks. It may be
noted that the membrane beneath the outer shell of an egg acts as a semipermeable membrane.
r
ou
ad

Similarly, if we put grapes in a concentrated solution of NaCl, the grapes shrink but if we put kismis or
grapes into pure water, they swell.
Y

When the membrane permits the solvent molecules to come out through the membrane, it is
Re
nd

called exosmosis (exo-osmosis). When the membrane permits the solvent molecules to enter
inside, it is called endosmosis (endo-osmosis).
Fi

Difference between Diffusion and Osmosis. The main points of difference between diffusion and
osmosis may be summed up as given below :
Osmosis Diffusion

1. In osmosis, a semipermeable membrane is u,sed. 1. In diffusion, no semipermeable membrane is

used.
2. In this process, there is only flow of solvent 2. In this process, the solvent as well as the solute
molecules and that loo through the semi molecules move directly into each other.
permeable membrane.
3. It takes place from lower concentration to higher 3. It takes place from higher concentration to lower
concentration. concentration.
4. It applies to solutions only. 4. It lakes place in gases as well as solutions.
5. It can be stopped or reversed by applying pressure 5. It cannot be slopped or reversed.
on the solution with higher concentration.
SOLUTIONS 2/43

Semipermeable membranes. The semi-permeable membranes (as defined above) are of two types :
(0 Natural semipermeable membranes, e.g., vegetable membranes and animal membranes which are
found just under the outer skin of the animals and plants. The pig’s bladder is the most common
animal membrane used.

(//) Artificial semipermeable membranes. The well known examples of the artificial semipermeable
membranes are parchment paper, cellophane and certain freshly precipitated inorganic substances,
e.g., copper fen ocyanide, silicates of iron, cobalt, nickel, etc. The precipitated substances have to be
supported on some material and this is achieved by preparing the precipitate in the walls of a porous
pot.
Osmotic Pressure. As mentioned above, due to osmosis, there is a net How of solvent from the solvent
to the solution or from the less concentrated solution to the more concentrated solution through the
semipermeable membrane. This flow of the solvent does not
FIGURE 2.20

w
continue indefinitely. For example, consider an inverted thistle
funnel at the mouth of which is tied a semipermeable membrane THISTLE-
(pig’s bladder or cellophane). Suppose the thistle funnel is filled FUNNEL

F lo
with sugar solution and then lowered into distilled water contained SUGAR SOL,

in a beaker (Fig. 2.20). It is observed that the level of the solution

DISTILLED

ee
inside the stem of the thistle funnel starts rising and then after WATER
some time, it becomes constant. The irse of level in the stem of

Fr
the thistle funnel is obviously due to the net flow of solvent into iK:
the solution through the semipermeable membrane. The constancy
in the level shows a state of equilibrium, i.e., as many molecules
for
ur
of the solvent enter into the solution through the semipermeable SEMI PERMEABLE MEMBRANE
membrane, the same number of solvent molecules from the solution
s
(a) Initial State (b) Final State
ook

go into the solvent through the semipermeable membrane in the

Yo
Demonstration of osmotic pressure
same time. The pressure exerted by the column h of the solution
eB

is called osmotic pressure. Thus,

Osmotic pressure tnay be defined as the equilibrium hydrostatic pressure of the column set up
as a result of osmosis.
r
ou
ad

Alternatively, the flow of the solvent into the solution through FIGURE 2.21
the semipermeable membrane can be stopped by applying pressure
Y

PISTON
NARROW
on the solution. To understand this concept, consider the apparatus
shown in Fig. 2.21. It consists of a vessel divided into two L - ^TUBE
Re
nd

compartments by a semipermeable membrane. One of the

Fi

compartments is fitted with a narrow tube and filled with the solvent.
The other compartment is provided with a wider tube which is fitted
with a Ifictionless piston and solution is taken in this compartment. SOLUTION _'SOLVENT „ ■
Due to osmosis, there is a tendency for the net flow of solvent into the
solution through the semipermeable membrane. As a result, the
piston will lend to move outwards and the level in the narrow tube
will fall.
SEMIPERMEABLE MEMBRANE
However, if some suitable pressure is applied on the piston, the
entry of the solvent into the solution can be prevented and the level of Concept of osmotic pressure
solvent in the narrow tube will then remain constant at L.

The minimum excess pressure that has to be applied on the solution to prevent the entry of the
solvent into the solution through the semipermeable membrane is called the osmotic pressure.
It is of interest to mention that if a pressure higher than the osmotic pressure is applied on the solution,
the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. The
process is called erverse osmosis (R.O.).
2/44 New Course Chemistry (X1I)E!SI9]

An important application of reverse osmosis is FIGURE 2.22

in the desalination of sea water, i.e., removal of salts PISTON
● (PRESSURE APPLIED > OSMOTIC PRESSURE)
from sea water. On applying pressure greater than
osmotic pressure, pure water can be made to flow J SEAWATER FRESHWATER
through a semipermeable membrane and this water
I - -. T
can be used for drinking purposes. n
II n
C II i\

The principle of the technique used may be t.

n :●
II 1'

illustrated diagrammatically as shown in Fig. 2.22. OUTLET

FOR
As very high pressures have to be applied for f

w
FRESH

reverse osmosis, the semipermeable generally used SEMIPERMEABLE MEMBRANE WATER

is that of cellulose acetate placed over a suitable

A schematic set-up of a desalination plant
support. Only water molecules can pass through this
membrane but the ions and other impurities present FIGURE 2.^

o
e
in sea water cannot.

re
PRESSURE
2.12.2. Experimental Measurement of Osmotic Pressure M GUN METAL GAUGE

KZTsSOLVENT

Frl
VESSEL

F
A number of methods are available for measurement of
c™
osmotic pressure. The best out of these is Berkeley and Hartley’s SOLUTION
Method.
ou fi mJ

r
This method is based upon applying pressure on the solution T ■;-:SOLVENT

so
which is just sufficient to prevent the entry of the solvent into the
solution through the semipermeable membrane. The apparatus
used by Berkeley and Hartley is shown in Fig. 2.23.
kf ■W_7'
oo
It consists of a porous tube (open at both ends) containing the POROUS TUBE WITH
Y
SEMIPERMEABLE MEMBRANE
semipermeable membrane of copper ferrocyanide. The porous tube
B

is fitted with a reservoir R on one side and a tube T on the other. Berkeley and Hartley's apparatus
re

The porous tube is filled with the pure solvent so that the level in the tube T stands at the mark M. The
oY

porous tube is fitted into an outer vessel made of gun metal. This vessel has a wide tube at the top which is
u

fitted with a frictionless piston and a pressure gauge as shown in Fig. 2.23. The solution under study is taken
ad

in the outer gun metal vessel. As a result of osmosis, the level in the tube T tends to fall. The pressure applied
d

on the solution by means of the piston which keeps the level in the tube T at M is a measure of the osmotic
pressure and can be read directly on the pressure gauge.
in
Re

This method is superior to the other methods because of the following reasons:
F

(0 In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on
the membrane.

(ii) The concentration of the solution does not change because the entry of the solvent into the solution
is prevented by the external pressure.
(Hi) The time taken for the measurement of osmotic pressure is much less in this method as compared to
the other methods.

2.12.3. Expression for the osmotic pressure

Osmotic pressure (ti) of a solution is found to be directly proportional to the molar concentration (C) of
the solution and its temperature (T). Mathematically,
7C« C and 71 o: T TC ec C X T or 7U = R X C X T

where R is a constant (called .solution constant) and its value is found to be same as that of the ‘Gas
constant’. The above equation is usually written as
7t = CRT
SOLUTIONS 2/45

n
But C = -
V
where is the number of moles of the solute and V is the volume of the solution in litres.
n
Hence, 7C = —RT or TtV = «RT
V

This equation is called van't Hojf equation for dilute solutions.

van’t Hoff Theory of Dilute Solutions, van’t Hoff put forward the result that dilute solutions of non-electrolytes
behaved similar to gases. The osmotic pressure was due to bombardment of the solute molecules on the semipermeable
membrane just as pressure of a gas was due to bombardment of molecules on the walls of its container. He put
forward the following laws :
(1) Boyle-van’t Hoff law. According to this law, P « C at constant T (C = molar cone.)
1

w
As C = - where V is the volume of the solution containing 1 mole of the solute,
V

F lo
1
P« or PV = constant or TtV = constant
V

Thus, this result is similar to Boyle's law.

ee
(11) Pressure-Temperature law (Gay Lussac-van’t Hoff law')* According to this law,

Fr
P 71
PocT or = constant or — = constant
T

(ill) Combined Re.sult. Combining the above two laws, for T

ur
V C =
1 'I
P = A- CT = /t-T
oks
Poc CT or
V V j
Yo

or PV = AT or PV = ST or TtV = ST
o
eB

where symbol S, used in place of k, is called solution constant. As value of S is found to be same as that of the gas
constant R, we write PV = RT for 1 mole of the solute or PV = nRT for n mole.s of the solute dissolved in V litres
of the solution
our
ad

This result is the same as ideal gas equation.

Thus, the solute in the dilute solution behaves like a gas and osmotic pressure of the solution is equal to the
pressure which the solute would e.xert if it were a gas at the same temperature and occupied the same volume as that
Y

of the solution. This is called ‘van’t Hoff theory of dilute solutions’.

Re
nd

2.12.4. Isotonic Solutions

Fi

From equation n = CRT, it is evident that if two solutions have the same values of C and T, they must
have the same value of n. In other words, solutions of equimolar concentration at the same temperature must
have the same osmotic pressure.
Such solutions which have the same osmotic pressure at the same temperature are called isotonic
solutions. Thus, two isotonic solutions are also isosmotic.

2.12.5. To conclude that osmotic pressure is a colllgative property

From the equation given above, it is clear that osmotic pressure depends upon the number ‘n’ of moles
of the solute per litre of the solution irrespective of the nature of the solute. Hence, osmotic pressure is a
colligative property.
2.12.6. Biological Importance of Osmosis
The growth of plants and animals depends largely on the phenomenon of osmosis. Plants and animal
bodies are composed of a very large number of microscopic units called cells. Cell contains a fluid (cell sap)
2/46 'P’uuiee^'^ New Course Chemistry (X11)[!S19I

and its wall is composed of a living cytoplasmic membrane which is semipermeable and is responsible for the
phenomenon of osmosis in living organisms. If such a cell comes in contact with water or some dilute solution,
the osmotic pressure of which is less than that of cell sap present in the cell, there will be a tendency of
water to enter into the cell through the FIGURE 2.24
cell wall. This causes rupturing of the
LOW CONCENTRATION HIGH CONCENTRATION
cell, a process known as hemolysis (Fig.
2.24 a). The pressure developed inside t
the cell due to the inflow of water into CELL
CELL-C HIGH. ^ LOW
it is called turgor. On the other hand, WALL : CONCEKlTRATIOI^i
(concentration WALL

if the cell comes in contact with a

sol ution of higher osmotic pressure, the -t CELL

cell would shrink due to going out of CELL-
SAP
SAP

w
water from the cell through the cell
^ Swelling/Rupture of the cell o Shrinking of Ihe cell
wall* (Fig. 2.24 h). This shrinking of (hemolysis) {Plasmolysis/Crenation)
the cell is called plasmolysis or

F lo
Osmosis through the semipermeable membrane (cell walls)
crenation.

Some of the processes regulated by osmosis in plants and animals are as under:

ee
(i) Plants absorb water from the soil through their roots due to osmosis because concentration of cell sap

Fr
inside root hair cells is higher than that of the water present in the soil,
(if) Water assimilated by the plants moves into the body of the plants and reaches even to the top of a tall
for
ur
tree due to osmosis.

(Hi) In animals, water moves into different parts of the body under the effect of the process of osmosis.
s
ook

(iv) Stretching of leaves, flowers, etc. is also controlled by osmosis,

Yo

(v) Osmosis helps in rapid growth of the plants and germination of seeds.
eB

(W) Different movements of plants such as opening and closing of flowers etc. are controlled by osmosis.
(v/7) Bursting of red blood cells when placed in water is also due to osmosis.
r
ou
ad

It is interesting to note that the salt concentration in blood plasma due to different species present in it
is equivalent to 0-9% aqueous solution of NaCl (mass/volume). Hence, we may conclude the following :
Y

(0 A 0-9 % solution of pure NaCl is isotonic with human red blood cells (RBC). Therefore, in this solution,
RBC neither swell nor undergo plasmolysis.
Re
nd

(/■/■) A pure NaCl solution with concentration less than 0-9 % is called hypotonic solution. On placing
Fi

RBC in this solution, they will swell and even burst.

(Hi) A pure NaCl solution with concentration more than 0-9% is called hypertonic solution. On placing
RBC’s in this solution, they shrink due to plasmolysis.
Thus, it is essential that the solutions that flow into the blood stream have the same osmotic pressure as
that of the blooa. A proper osmotic pressure balance inside and outside the cells of the organisms is maintained
by Na'*’ and ions.
In general, if two solutions have same osmotic pressure and hence same molar concentration, they are
called isotonic. If one solution is of lower osmotic pressure, it is called hypotonic with respect to the more
concentrated solution. The more concentrated solution is said to be hypertonic with respect to the dilute
solution.

*It is important to note that the biological cells allow the pas.sage not only lo water but also to the other
selected materials. This helps the entry of the nutrients and the disposal of the waste materials. The transport takes
place from a region of higher concentration to that of lower concentration.
SOLUTIONS 2/47

2.12.7. Explanation of some phenomena on the basis of osmosis

(i) Raw mangoes shrivel (shrink) into pickle when placed in concentrated common salt solution (brine).
This is due to outllow of water through semipermeable membranes of mangoes due to osmosis.
(//) Wilted (withered) flowers revive when placed in fresh wate.r This is due to flow of water into the
withered flowers through their semipermeable membranes due to osmosis.
(Hi) Limped carrots become firm again when placed in wate.r Carrots become limp due to loss of water
to atmosphere. When placed in water, water flows into them due to osmosis and they become firm
again.
(iv) Pre.servation of meat by salting and that offruits by adding sugar. This is also explained by osmosis.
This is because the bacterium which is responsible for spoiling them loses water in the salted meat or
candid fruit due to osmosis. As a result, it shrivels and dies,
(y) People taking a lot of salt or salty food develop swelling or puffiness of their tissues, a disea.’se called

w
edema. This is due to retention of water in the tissue cells and intercellular space on account of
osmosis.

F lo
2.12.8. Determination of Molecular mass from Osmotic pressure Measurements
According to van’t Hoff equation for dilute solutions,

ee
TcV = nRT ...(0

Fr
where n is the osmotic pressure at temperature T for a solution containing n moles of the solute dissolved

for
in V litres of the solution, R is a constant (sometimes called ‘solution constant’).
ur
If W2 grams of the solute are dissolved in V litres of the solution and M2 is the molecular mass of the
solute, then
s
ook
Yo
W2
n =

M2
eB

Substituting this value in equation (/), we get

our
ad

vy^ wRT
TtV - =^RT or
M2 = ...(«)
M^ teV
Y

Thus, measuring the osmotic pressure 7t of a solution containing w grams of the solute in V litres of the
Re
nd

solution, at temperature T, the molecular mass, M2, of the solute can be calculated using equation (//). The
value of the constant R is taken as 0-0821 litre atmosphere per degree per mole when k is expressed in
Fi

atmospheres and T in degrees kelvin.

This method is usually not preferred for the determination of molecular masses because of many
experimental difficulties involved. However, for the determination of molecular masses ofproteins, polymers
and other macromolecules, it is considered to be one of the most suitable methods. For such substances, the
other methods based upon colligative properties (elevation in boiling point, depression in freezing point,
etc.) cannot be used because of the following reasons :
(/) The changes observed ir hese properties are very small., (e.g., 0 00001 K for substances having molar
masses of the order of 10^ g mol"'). On the other hand, osmotic pressures of the order of 10"^ atm can
be easily measured in terms of mm.
(ii) Elevation in boiling point method for biomolecules such as proteins is not suitable because these
molecules arc not stable at higher temperatures whereas the advantage of the osmotic pressure is that it
can be measured at room temperature.
(Hi) The osmotic pressure method has the advantage that it uses molarities instead of molalities.
2/48 ^K<ndeep.’4- New Course Chemistry CXI1)ISSXS

NUMERICAL FORMULAS USED

I
PROBLEMS
BASED (/) The van ’/ Hojf equation for dilute solution is
ON
7CV = ;jRT or 71 = -RT or 7t = CRT

{ }
V
Osmotic Pressure
where n = osmotic pressure in atmospheres
?i = number of moles of the solute present in V litres of the solution
C = molar concentration of the solution .

ow
T = temperature in degrees kelvin
R = 0-082! litre atmosphere per degree per mole (L atm K"' mol"')
= 0-083 L bar mo|-‘
(ii) To calculate the molar mass of the solute.

e
Pul n =
^2
so that we have tcV =
M'
RT or M-, =
W2 RT

re
M, ^2

Frl
where vv = mass of solute in grams dissolved in V litres of the solution

F
Mt = molar mass of the solute
{in) Isotonic solutions are those solutions which have the same osmotic
ou
r
pressure. As ti = CRT, if osmotic pressures (tu) are equal, at the same

so
temperature, concentrations (C) must also be equal.
(/V) If a number of solutes are present in the .solution and
kf
are their individual osmotic pressures, then
oo
Total osmotic pressure = tu, + Jt^ + 713 +...
Note. In all numerical problems on osmotic pressure, percentage means
Y
eB

wt/vol. unless density of the solution is given.

ur

Problem Q (a) Calculate the osmotic pressure of O'Ol M solution of cane-sugar at 300 K
oY

(R = 0'0821 litre atni/degree/mole).

ad

(b) If this solution were placed in a tube of uniform cross-sectional area of 1 cm^ with a semipermeable
membrane at the lower end and this end is dipped in pure water, what will be height of the vertical column
d

developed ? Assume density of the solution as Ig m L~'.

in
Re

Solution, (n) Here, we are given that

C = 0-01 M = 0-01 mole/litre, R = 0-0821 litre atm/degree/mole, T = 300 K
F

Using the relation n = CRT, we get 71 = (0 01 mol L“*) x (0-0821 L atm K"* mol"^) (300 K) = 0-2463 atm.
(/») 71 = hdg
Putting 71 = 0-2463 atm = 0-2463 x l-OI x 10^ N m ^
d = \ ,g mL"’ - 10^ kg m"^, g = 9-81 m s“", we get
0-2463 X 1-01 X 10^ = hx 10^ x 9-81 = 2-54 m.

Problem 0 Calculate the osmotic pressure at 273 K of a S% solution of urea (Mol. mass = 60).
(R = 0-0821 litre atm/degree/moic). (Jharkhand Board 2011, Bihar Board 2012)
Solution. 5% solution of urea means that it contain.s 5 g of urea per 100 cm^ of the solution, i.e.,
VI', = 5 g, V = 100 env^ = — litre = 0-1 litre.,
10

M, = 60 g mol"*, R = 0-0821 L atm K~’ mol"*, T = 273 K

n
71 = -RT =
Vt', I
X —RT =
tv^RT _ (5g)(0-0821LatmK-'mor')(273K)
= 18-68 atm.
V M, V M,V (60gmol-*)(0-lL)
SOLUTIONS 2/49

Problem Calculate the concentration of that solution of sugar which has osmotic pressure of
2*46 atmosphere at 300 K.
Solution. Here, we are given that n = 2-46 atm. T = 300 K, R = 0 0821 L atm mol"’
Using the equation, r = CRT, we have
71 2-46 atm
c =
= 0-1 M approx. = 34*2 g/litre.
RT 0-082IL atm K"' mol"' x300K
(*.● Molar mass of sugar Cj2Ho20y = 342 g mol ’)
Problem
10 200 cm^ of an aqueous solution of a protein contains 1*26 g of the protein. The osmotic
pressure of this solution at 300 K is found to be 2*57 x 10"^ bar. Calculate the molar mass of the protein.
(NCERT Solved Example)
Solution. Here, we are given : Mass of the solute (protein), w, = 1-26 g
Volume of the solution (V) = 200 cm^ = 0-200 L
7C = 2-57 X 10-^ bar, T = 300 K, R = 0-083 L bar K"' mol"'

w
W2 RT
Substituting these values in tlie formula, M2 = 7tV
, we gel

F lo
l-26gx0-083LbarK-' mol"' x300K
M., = = 61039 g mol"!
2-57x10-3 barx0-200L

B A 4 per cent solution of sucrose C,2H220jj is isotonic with 3 per cent solution of an

ree
Problem

unknown organic substance. Calculate the molar mass of the unknown substance.
for F
Solution. Since the two solutions are isotonic, they must have same concentrations in moles/litre.
40
For sucrose solufioii, WQ have Concentration = 4 g/100 cm^ (G/ven) = 40 g/litre = moles/litre
342
Your

(-.- Molar mass of sucrose, C,2H^20|| = 342 g mol ')

ks

For unknown substance, suppose M is the mol. mass. Then

eBoo

30
Concentration = 3 g/100 cm3 = 30 g/Htre =— moles/litre
M
ad

30 _ 40 30x342
our

-1
Thus, we have or M = = 256*5 g mol
M “ 342 40

Problem
B Calculate the osmotic pressure of a solution obtained by mixing 100 cm^ of 1*5 %
solution of urea (mol. mass = 60) and 100 cm^ of 3*42% solution of cane-sugar (mol. mass = 342) at 20"C
Re
Y

(R = 0*082 litre atm/degree/mole) (West Bengal Board 2012)

Find

Solution. After mixing, total volume of the solution = 100 + 100 = 200 cm^
Osmotic pressure due to the urea in the solution:
1-5 g of urea which was present originally in 100 cm^ is now present in 200 cm^, i.e., in the final solution.
7-5
Concentration of urea, C = 1-5 g/200 cm^ = 1-5x5 g/litre = 7-5 g/litre = 60
moles/litre

(-.- Molar mass of urea = 60 g mo! ')

T = 20 + 273 = 293 K. R = 0-082 litre atm/degree/mole
(7-5 molL '
Ti = CRT = (0-082 L atm K ' mol *) x (293 K) = 3-00 atm.
60

Osmotic pressure due to cane-sugar in the solution : In the final solution,

17-10
Concentration of cane sugar, C = 3-42 g/200 cm^ = 3-42 x 5 g/litre = 17-10 g/litre = moles/litre
342

(*.● Molar mass of sugar = 342 g mol ')

2/50 New Course Chemistry (XII)KSSM]

H7-10
n = CRT = mol L-' X (0 082 L atm K"‘ mol"') (293 K) = 1 -20 atm
342

As the two solutes behave independent of each other in the solution.

Total Osmotic pressure = Osmotic pressure of urea + Osmotic pressure of su^ r = 3-00 + 1-20 = 4-20 atm.
Alternatively,as colligalivepropertiesdepend only on the number of moles of the solute and do not depend upon
the nature of the solute, problem can be solved by finding the total number of the moles of die solute as follows :
1-5 342
No. of moles of urea present = — = 0-025 ; No. of moles of cane sugar present = = 0’01
60 342

Total no. of moles of the solute = 0-025 -f 0-01 = 0-035

Total volume of the solution = 100 + 100 cm^ = 200 cm^ = 0-2 L

n 0-035molX0-082ILatmK ’mol *x293K

Applying the relation, rt = —
V
RT, we have Ti- 0-2L
- 4*20 atm.

w
Problem B 10 g of a substance were dissolved in water and the solution was made up to 250 cm^.

F lo
The osmotic pressure of the solution was found to be 8 x 10® N m"^ (pascals) at 288 K. Find the molar mass
of the solute.

Solution. Here, = 10 g, V = 250 cm-^ = 250 x 10~^ m-^ = 2-5 x 10^ m^

ee
TC = 8 X 10® N m-^ R = 8-314 J mol"', T = 288 K

Fr
1V2 RT _ (lOg) (8-314 N m K~’ mol-') (288 K) -t
(IJ = INm) = 119’7 g mol
(8xl05Nm-2)x(2-5xl0-^m^)
for
ur
Problem
Q At 300 K, 36 g of glucose, CgHj20^ present per litre in its solution has an osmotic
pressure of 4*98 bar. If the osmotic pressure of another glucose solution is 1*52 bar at the same temperature,
s
ook

calculate the concentration of the other solution. (CBSE Sample Paper 2017, CBSE 2019)
Yo

Solution. n = CRT (C = molar concentration)

eB

It C 4-98 36/180 36 1-52

1 _ 1
= 0061 M
or
C2 = 180^4-98
^2 ’ 1-52
C2
our

7C,
ad

Problem At 300 K, two solutions of glucose in water of concentrations 0*01 M and 0-001 M are
separated by semipermeable membrane with respect to water. On which solution, the pressure should be
dY

applied to prevent osmosis ? Calculate the magnitude of this applied pressure.

Re

Solution. 71 = CRT
Fin

For 0-01 M solution, Kj = 0-01 x 0-0821 x 300 = 0-2463 atm

For 0-001 M solution, 7C2 = 0-001 x 0-0821 x 300 = 0-02463 atm
As movement of solvent molecules occurs from dilute to concentrated solution, pressure should be applied
on concentrated solution, i.e., on 0-01 M solution to prevent osmo.sis.

Magnitude of external pressure = 0-2463 - 0-0246 = 0-2217 atm

1. An aqueous solution of glucose C^^HpO^ has an osmotic pressure of 2-72 atmospheres at 298 K. How many
moles of glucose were dissolved per litre of the solution ? (R = 0-082 lit. atm. moC’ deg"')
2. A solution of sucrose (molecular mass 342/mol) is prepared by dissolving 68-4 g of it per litre of solution.
What is its osmotic pressure at 300 K ? (R = 0-082 lit. atm deg"' mol"').
SOLUTIONS 2/51

3. Calculate the osmotic pressure of a solution containing 171 g of cane-sugar (molecular mass 342) in 500 g
of water at 300 K (R = 0 082 lit. atm deg"* mol"'). Density of die solution is 1-034 g cm“^.
4. A 5% solution of cane-sugar (m. wt. = 342) is isotonic with 0-877 % solution of urea. Find the molecular
weight of urea.
5. One litre aqueous solution of sucrose (molar mass = 342 g mol"*) weighing 1015 g is found to record an
osmotic pressure of 4-82 atm at 293 K. What is the molality of the sucrose solution ?
(R = 0-0821 atm mol"* K"'}.
6. The osmotic pressure of blood is 8-21 atm at 37“C. How much glucose would be used for an injection that
is at the same osmotic pressure as blood ?
7. The osmotic pressure of 0-200 g of haemoglobin in 20-0 ml of solution is 2-88 torr at 25“C. Calculate the
molecular weight of haemoglobin.

w
8. At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4-98 bar. If osmotic pressure
of the solution is I -52 bar at the same temperature, what would be its concentration ?
9. A solution prepared by dissolving 8-95 mg of a gene fragment in 35-0 mL of water has an osmotic pressure

Flo
of 0-335 torr at 25"C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

e
(CBSE 2011)

re
ANSWERS

F
1. 0-1113 mol 2. 4-92 atm. 3. 2-46 atm. 4. 60u
-1
5. 0-2112 m 6. 58-14gL
ur 7. 64484 u 8. 10-98 g L-

r
9. 14193-3 g mol
fo
-1
ks
HINTS FOR DIFFICULT PROBLEMS
Yo
3. Mass of the solution = 500 + 17-1 g =517-1 g
oo

Volume of the solution = Mass/Density = 517-1/1 034 cm^ = 500 cm^ = 0-5 L
B

vvRT 17-1 g X 0-082 L atm K"' mol" ‘ x 300 K

re

n =
= 2-46 atm .
M^V 342 g mol ‘x0-5L
u

5. K = CRT gives C = 0-2 mol L ' = 0-2 mol in 1015 g solution.

ad
Yo

Hence, mass of solvent = 1015 - (0-2 x 342) = 946-6 g.

0-2 mol
d

Molality = xlOOOgkg-i = 0-2112 m.

Re

946-6 g
in

6. 7C = CRT, i.e.. C = Jt/RT = 8-21 atm/(0-0821 L aim K“* mol"* x 310 K)

F

= 0-323 M = 0-323 x 180 g L"' = 58-14 g L"'.

0-200 1000 10
7. Molar cone, of haemoglobin (C) = X molL"’=—molL-' ; tc = CRT
M 20 M

■ I n
2-88
I.e., atm = molL-' x (0-082 L atm K"' mol"') x (298 K) or M = 64484 g mol
-1

760 ,M
8. 71, = C,RT, 712 = C2RT .-. 7t,/7t2 = Cj/Co
4-98 bar/1-52 bar = (36/180 mol L-'j/Cj or
C2 = 0-061 mol L"‘ = 0-061 x 180 g L"' = 10-98 g L"'

9. M - (8-95xlQ-^g) (0-0821LatmK~'mol"')(298K) = 14193-3 g mol

-1
TCV ~ (0-335/760 atm) (35 xlO-^L)
2/52 ‘Pn<^tiee^'<^ New Course Chemistry fXinvismn

ELEVATION OF BOILING POINT (EBULLIOSCOPY)

General Discussion

The boiling point of a liquid is the temperature at which the vapour pressure of the liquid
becomes equal to the atmospheric pressure.
For example, vapour pressure of water at 100”C (373 K) is 1-013 bar (= 1 atm). This is the reason that

w
water boils at 373 K as at this temperature, its vapour pressure becomes equal to one atmospheric pressure
{i.e., 1-013 bar).
It is found that the boiling point of the solution is always higher FIGURE 2.25

than that of the pure solvent. The increase is called the elevation in

e
B
D
boiling point. The reason for the elevation in boiling point may be UJ
ATMOSPHERIC E
a:

e
explained as follows :— CO
■ PRESSURE ” V F

or
We know that the vapour pressure of the solution is lower than CO

r
liJ

that of the pure solvent and vapour pressure increases with increase
a:
Q.

F
A-
in temperature. Hence, the solution has to be heated more to make Cd

the vapour pressure equal to the atmospheric pressure. o c

oF
ul
Q.

Alternatively, the elevation of boiling point may be explained $ ATb

on the basis of plots of vapourpressureversus temperatureas shown

s
r
in Fig. 2.25. Tb Tb
TEMPERATURE

Vapour pressure of the solvent increases with increase in

ko
temperature, as shown by the curve AB. As at any temperature, vapour Elevation in boiling point

of
pressure of the solution is less than that of the solvent, the curve for
the solution lies below that of the solvent, as shown by the curve CD. The temperatures at which the vapour
o
Y
pressure of the solvent and the solution become equal to the atmospheric pressure are Tf and T^ respectively.
rB

Obviously T^ > T^”. The difference, called the elevation of boiling point, AT^, is given by AT^ ~'^b~
eY

2.13.2. Expression for the elevation of boiling point

The elevation of boiling point depends upon concentration of the solute in a solution and has been found
u

to be related with molality ‘m’ as below :

AT^ = Xm
od
ad

where K;, is called the molal elevation constant or ebullioscopic constant and 'in' is the molality of the
in

solution.

If m = 1. then AT^ = K^,. Hence,

Re

Molal elevation constant may be defined as the elevation of boiling point when the molality of
F

the solution is unity, i.e., 1 mole of the solute is dissolved in I kg (1000 g) of the solvent. The
units ofKjj are, therefore, degree/molality, i.e., KJm or '^Clm or K kg mot~K
The molal elevation constants (ebullioscopic constants) of a few solvents are given below :
Solvent K^(K kg moM) Solvent K^(K kg moM)
Water (H2O) 0-52 Cyclohexane (CgHj2) 2-79

Benzene (C^^H^^) 2-53 Carbon tetrachloride (CCI4) 5-03

Ether (C2H5OC2H5) 2-02

Ethyl ale. (C2H5OH) 1-20

Carbon disulphide (C$2) 2-34 Acetone (CH3COCH3) 1-72

Chloroform (CHCI^) 3-63 Acetic acid (CH3COOH) 2-93

Derivation of the expression ATj, = x m. It may be derived in a simple manner as follows ;

It is evident that greater the lowering in vapour pressure (A p), higher is the elevation in boiling point
(AT,), i.e., AT, Ap.
OC
 SOLUTIONS 2/53

But according to the Raoult’s law, A p OC

x-i, the mole fraction of the solute in the solution
Hence, or
~ kx2
where A: is a constant of proportionality.
lU
But X
2 " = — if the solution is dilute.
«1 +«2

i.e., ^2 =
«2
w.1 /M
AT^=^Mi^ w
1 1

If the mass of solvent, w, = 1 kg, then evidently — = m, molality of the solution. Also, for a given
w

solvent, its molecular mass M, is constant so that /cMj = K^, another constant. Hence, the above result

w
reduces to AT^ = m.

F lo
2.13.3. Calculation of molecular mass of the solute
As molality is the number of moles of the solute dissolved per 1000 g of the solvent, if w, grams of the
W2 1000

ee
solute of molecular mass M2 are dissolved in w, grams of the solvent. 111 - X

Ml

Fr
w
1
Hence, the formula AT^, = x m becomes
1000
for
IOOOxK^xh'2
ur
X

M2 w
or
M2 =
1
AT^ X Wj
s
ook

This formula is often used for the calculation of molecular masses of non-ionic solutes (i.e., non
Yo

electrolytes).
eB

2.13.4. Calculation of molal elevation constant from enthalpy of vaporisation

our

R^o _ MiRTq A...H

ad

vap

1000/,^ 1000A.,„^H
V
M
I
Y

where Tq = boiling point of the liquid (pure solvent)

Re

/,, = latent heat of vaporisation per gram of the solvent

nd

^ ~ latent heat of vaporisation per mole of the solvent

Fi

Mj = molecular mass of the solvent

R = gas constant = 8-314 J K"’ mol"' (if or H are in joules)
= 2 cal deg’'* mol-' (if or A,,;,pH are in calorics)
Sampio Problem Calculate the molal elevation constant of water, it being given that its latent
heat of vaporisation is 2-257 kj/g.
RT?0 8-314 JK-'mol-'X (373K)2
Solution. = = 0-512 K kg mol
1000/
V 1000gkg-'x2257Jg“‘

2.13.5. To conclude that elevation of boiling point is a colligative property

From the expression AT/, = K/, m, it is clear that the elevation of boiling point depends on molality, i.e.,
the number of moles of the solute dissolved in 1000 g of the solvent and not upon the nature of the solute.
Hence, it is a colligative property.
2/54 New Course Chemistry (XII)BZslS]

FORMULAS USED
NUMERICA.I_ I
PROBLEMS
BASED (0 AT^ = .w
ON where AT^ = elevation of the boiling point = T^-T°^
m = molality of the solution {i.e., no. of moles/1000 g of solvent)
Elevation in
= molal elevation constant or ebullioscopic constant of the solvent
Boiling Point in K/w or K kg mol"'
lOOOK^ ^2 1000K^W2
(«) AT^ = or M2 =
Wj b^I.AT^

w
where M2 = molecular mass of the solute,
W2 = mass of the solute in grams, Wj = mass of the solvent in grams
Note. in the above formula is molal elevation constant, i.e., molecular

Flo
elevation constant per 1000 g of the solvent. If Ky, is per 100 g of the
100

e
solvent, the equation is modified to : AT^ =

re
W,M2

F
{Hi) For numerical problems involving two different colligative
properties, the equations for these colligative properties are suitably
ur
r
applied one after the other (see Solved Problem 6).

fo
ks
Q Calculate the molal elevation constant of water, it being given that 0*1 molal aqueous
Yo
Problem
oo

solution of a substance boiled at 100*052‘’C.

B

Solution. Here, we are given that m - 0-1

Boiling point of solution = 100-052"C AT^ = 100 052 - 100 = 0-052T
re

ATj, _ 0^52°C
u

Applying the relationship, AT^ = . m, we get = = 0*52"C/m

ad

0-lm
Yo

Problem
0 The boiling point of benzene is 353*23 K. When 1*80 g of a non-volatile solute was
d

dissolved in 90 g of benzene, the boiling point is raised to 354*11 K. Calculate the molar mass of the solute,
Re
in

for benzene is 2*53 K kg mol”^ (NCERT Solved Example)

F

Solution. Here, we are given Wj = 1-80 g, Wj = 90 g ; AT^ = 354-11 - 353-23 K = 0-88 K

-1
= 2-53 K kg mol
Substituting these values in the formula,

lOOOK^ w. lOOQgkg-^ X 2-53 K kg mor^ x 1-80 g -1

M2 = , we get M2 = = 58 g mol
WlAT^ 90gx0-88K

Problem
0 18 g of glucose, C^Hj20^, is dissolved in 1 kg of water in a saucepan. At what temperature
will the water boil at 1*013 bar pressure ? for water is 0*52 K kg mol"*.
(NCERT Solved Example, CBSE 2013)

♦This is because if 1 mole of the solute is dissolved in 100 g of the solvent instead of 1000 g of the solvent,
1000 ii'2 I00K^W2
elevation in boiling point will be 10 times, i.e., 10 AT^ = xm = x or AT^ =
M2W, vv, M2
SOLUTIONS 2/55

-i
Solution. Here, we are given W2 = 18 g, M’l = I kg = 1000 g, = 0-52 K kg mol
(for glucose, C6Hj20^) = 72 + 12 + 96 = 180 g mol"'
1000 K/, vv2 ^ 1000g kg-' X 0-52K kg mol-' x 18g
AT,= = 0-052 K
vj’i M-, lOOOgxlSOgmol"!
As water boils at 373-15 K at 1-013 bar pressure, therefore, boiling point of solution
= 373-15 0 052 K = 373-202 K.

Problem
Q Calculate the boiling point of a solution containing 0*456 g of camphor (mol. mass =
152) dissolved in 31*4 g of acetone (b.p. = 56-30“C), if the molecular elevation constant per 100 g of acetone
is 17-2"C.

Solution. Here, we have h'2 = 0-456 g, M2= 152, w, =3l-4g. Tq = 56-30"C, K;,= 17-2°C/100g

w
100K/,.u'2 _ 100x17-2x0-456
AT,= = 0-16°C
w,M2 31-4x152

Flo
Boiling point of solution (T^) = + AT,, = 56-30 + 016 = 56-46“C

e
re
Problem 0 A solution containing 0-5126 g naphthalene (mol. mass = 128) in 50-0 g of carbon
tetrachloride yields a boiling point elevation of 0-402" C while a solution of 0-6216 g of an unknown solute

rF
in the same weight of the same solvent gives a boiling point elevation of 0-64T’C. Find the molecular mass
of the unknown solute.
ur
to find the molecular mass. fo
Solution. In this case, the first data is used to find the value of K/,. This value of is used in the second data
ks
Step I. To ifnd K,, for CCf from data on naphthalene solution
Yo
oo

w'2 - 0-5126 g. w
1 = 50 00 g. M2 = 128 g mol"'. AT,, = 0-402"C = 0-402 K*
B

AT,,.>v, M, _ 0-402Kx50gxl28gmol-* = 5-02 K kg mol

-I
re

1000 ^2 1000gkg-'x0-5126g
Step II. To ifnd the mol. moss of unknown solute.
u
ad
Yo

The given data are : k-2 = 0-6216 g, = 50-00 g, AT,, = 0-647°C = 0-647 K
K,, = 5-02 K kg mol"' (calculated above)
d
Re

10(X)K,, 1V2 _ lOOOgkg”'x5-02Kkgmol“'x0-6216g

in

M2 = = 96-46 g mol
-1

w,AT,, 50 gx 0-647 K
F

Problem A solution containing 6 g of a solute dissolved in 250 cm^ of water gave an osmotic
pressure of 4*5 atmosphere at 27"C. Calculate the boiling point of the solution. The molal elevation constant
for water is 0-52" C per 1000 g. (Manipur Board 2012)
Solution. The data on osmotic pressure is : W2 = 6g V = 250 cm^ = 0-25 litre
n - 4-5 atm, T = 27 + 273 = 300 K

R = 0-0821 litre atm/degree/mole

van’t Hotf equation for osmotic pressure is ; TiV = nRT

4-5x0-25
4-5 X 0-25 =« X 0-0821 x 300 or n = = 0-0457 mole
0-0821x300

i.e., 0-0457 mole of the solute are present in 250 mL of water (or 250 g of water).

*AT,, in °C = AT,, in K, as it is a difference of two temperatures.

2/56 New Course Chemistry (XII)DSm

0-0457 mol -I
Molality of the solution = xlOOOgkg = 0-1828 mol kg m = 0-1828
250g

K^ = 0-52°C/1000g (Given)
AT;, = K;, X m = 0-52 x 0-1828 = O-OgS'C
Boiling point of solution (T,,) = + AT;, = lOO^C + 0-095" = 100-095"C
Problem Q Calculate the boiling point of a solution containing 25 g urea (NH2CONH2) and 25 g
thiourea (NH2CSNH2) in 500 g chloroform, CHCI3. The boiling point of pure chloroform is 61*2*'C and
Kj = 3-63 K m-K
25 25
Solution. 25 g urea = — mole = 0-417 mole, 25 g thiourea = — mole = 0-329 mole

w
60 76

Total moles = 0-417 + 0-329 = 0-746

Flo
Mass of solvent = 500 g = 0-500 kg

e
0-746 mol
Molality = = 1-492 m,

re
0-500 kg

F
AT;,= K;, X m = 3-63 x 1-492 = 5-416°
Boiling point of solution = 61-2 + 5-416°C = 66*616 “C.
ur
r
Problem
Rl An aqueous solution of glucose boils at 100*01°C.
fo
The molal elevation constant for
water is 0*5 K kg mol“‘. What is the number of glucose molecules in the solution containing 100 g of water ?
ks
Yo
Solution. AT[j = Kj, x m ; 0-01 = 0-5 x m which gives m = 0-02.
oo

Thus, 1000 g of water contain glucose = 0-02 mole

B

/. 100 g of water contain glucose = 0-002 mole = 0-002 x 6-02 x 10-^ molecules = 1*204 x 10^* molecules.
re

wmm
u
ad
Yo

1. Calculate the molal boiling point constant for chloroform from the fact that its boiling point is 6I-2°C and
0-1 molal solution of an organic substance in chloroform boiled at 61-579°C.
d
Re

2. When 1-80 g of non-volatile compound is dissolved in 25-0 g of acetone, the solution boils at 56-86°C
in

while pure acetone boils at 56-38°C under the same atmospheric pressure. Calculate the molar mass of the
F

compound. The molal elevation constant for acetone is 1-72°.

3. The vapour pressure of an aqueous solution of cane sugar (mol. mass = 342) is 732 mm at 100°C. Calculate
the boiling point of the solution (K;, for water = 0-52°C).
4. On dissolving 3-24 g of sulphur in 40 g of benzene, boiling point of solution was higher than that of
benzene by 0-81 K. K;, value for benzene is 2-53 K kg mol"'. What is the molecular formula of sulphur ?
(Atomic mass of sulphur = 32 g mol"')
5. 0-90 g of a non-electrolytc was dissolved in 87-90 g of benzene. This raised the boiling point of benzene by
0-25°C. If the molar mass of non-electrolyte is 103-0 g mol"', calculate the molal elevation constant for
benzene.

6. A solution of an organic compound is prepared by dissolving 68-4 g in 1000 g of water. Calculate the
molecular mass of the compound and osmotic pressure of the solution at 293 K when elevation of b.pt. is
0-104 and K;, for water is 0-52 K mol"'.
7. A solution prepared by dissolving 1 -25 g of oil of winter green (methyl salicylate) in 99-0 g of benzene has
a boiling point of 80-3rC. Determine the molar mass of this compound (B.P. of pure benzene = 80-10°C
and K;, for benzene = 2-53°C kg mol"'. (CBSE 2010)
SOLUTIONS 2/57

8. A solution of glycerol (€31-1303), molar mass = 92 g mol“', in water was prepared by dissolving some
glycerol in 500 g of water. This solution has a boiling point of 10042°C. What mass of glycerol was
dissolved to make this solution ? for water = 0-512 K kg mor*. (CBSE 2010)

ANSWERS
2. 258 g mol ’
-1
1. 3-79 K kg mol 3. 101-064X 4.S 8
-1
6. 342 g mol"', 4-5 atm.
-1
5. 2-515 K kg mol 7. 152 g mol 8. 37-7 g

HINTS FOR DIFFICULT PROBLEMS

- Ps _ ^2 _ 'h
3. Pq = 760 mm, = 732 mm. First apply (taking u', = 1000 g)
1000/18
/’o n
1
n-y = molality (m)
Apply AT^ = X m.

w
4. Calculate molecular mass of the solute (sulphur). We get M2 = 253. Hence, no. of atoms present in one
molecule = 253/32 = 8, i.e., the formula is S3.

F lo
1000K,,m’2 M2H’,AT^ _ 103gmol"'x87-90gx0-25K = 2-515 K kg mol"'.
5. M2 = or Kf^ =
vv,AT^ 1000m;2 1000gkg-'x0-90g

ee
-1
6. AT^ = K,, X m m = AT,/K^ = 0-104/0-52 = 0-2 mol kg

Fr
68-4
-I
1 mole = - 342g , ic\, Molar mass = 342 g mol
Thus, 0-2 mole = 68-4 g 0-2

for
ur
Mass of solution = I068-4 g. Taking it as a dilute solution, its density =: 1 g cm~\ Hence,
Volume = 1068-4 cm^ = I 0684 L
s
ook
Yo
0-2 mol
Molar concentration = = 0-187 mol L"'
eB

1-0684 L

K = CRT = 0-187 mol L"‘ x 0-0821 L atm K"' mol"' x 293 K = 4-50 atm.
r
ou
ad

100 ^2 WjM2AT^, 500x92x0-42

8. AT,= = 37-7 g
or vv
2 “
M’j Mt 1000 K/, 1000x0-512
Y
Re
nd

2.14. DEPRESSION OF FREEZING POINT (CRYOSCOPY)

Fi

2.14.1. General Discussion

Freezing point of a substance is the temperature at which the soUd and the liquid forms of the substance
are in equilibrium i.e., the solid and the liquid forms of substance have the same vapour pressure. It is
observed that the freezing point of the solution is always lower than that of the pure solvent. The decrease is
called the depression of freezing point. The reason for the depression can be explained as follows ;
We know that the vapour pressure of the .solution is less than that of the pure solvent. As freezing point is the
temperature at which the vapour pressure of the liquid and the solid phase are equal, therefore, for the solution, this
will occur at a lower temperature (because lower the temperature, lower is ilte vapour pressure).
Alternatively, the depression of freezing point may be explained on the basis of plots of vapour pressure
versus temperature (Fig. 2.26) as follows :
On cooling, the vapour pressure of the liquid solvent decreases along the curve AB. At B, the solid starts
appearing and the vapour pressure decreases steeply along the path BC (because solids have lower vapour
pressure). At B, the liquid and the solid solvent are in equilibrium and have the same vapour pressure. Thus,
2/58 “PteieCeefi.'^ New Course Chemistry (XlHrasTWl

B represents the freezing point 'XJ of the pure solvent. As at any FIGURE 2.261
temperature, the vapour pressure of solution is less than that of
the solvent, the curve for the solution lies below that of the solvent. LU
a:

On cooling, it will follow the path DE. At E, the solid appears. CO

CO
Hence, E represents the freezing point T^of the solution. Obviously, UJ
SOLID (frozen:
T^is less than The difference, called the depression of freezing CC
SOLVENT /

point, ATy, is given by ATy= _Ty° - Ty- o c I

e

0. I
AT,
2.14.2. Expression for the depression of freezing point <
>
I

The depression in freezing point depends on the Tf

concentration of the solute in a solution and has been found to be TEMPERATURE

related to molality ‘w’ as below :

Depression in freezing point
ATy^= KyX m

w
where I^is a constant known as molal depression constant or cryoscopic constant of the solvent and'm ’ is the
molality of the solution, i.e., the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent.

F lo
If m = 1, ATy - Kyv Hence,

ee
Molal depression constant may be defined as the depression offreezing point when the molality
of the solution is unity, i.e., one mole of the solute is dissolved in 1000 g (1 kg) of the solvent.

Fr
The units of Kj^are, therefore, degrees/molality, i.e., KJm or K kg mol~K

for
The molal depression constants (cryoscopic constants) of a few solvents are given below :
ur
Solvent K^(K kg moM) Solvent K,(K kg moM)
s
ook
Yo
Water (H,0) 1-86 Ether (C2H5OC2H5) 1-79

Benzene (C^H^) 5-12 Carbon tetrachloride (CCI4) 31-8

eB

Ethyl ale. (C,HgOH) 1-99 Cyclohexane (C^H,.,) 2000

Carbon disulphide (CS2) 3-83 Acetic acid {CH3COOH) 3-90

r

Chloroform (CHCig) 4-70 Camphor 39-70

ou
ad

The expression, ATy= KyX m. can be derived in a simple manner exactly in the same way as in case of
Y

elevation in boiling point.

Re
nd

2.14.3. Calculation of molal depression constant from enthalpy of fusion

Fi

RT„20 M.1 RTi

0
A. H
/
K *.● /
f 1000/ 1000 A H / ~ M
/ fus 1

where T,, = freezing point of the liquid (pure solvent),

/y= latent heat of fusion per gram of the solvent
A,„ H = latent heat of fusion per mole of the solvent,
M, = molecular mass of the solvent
R = gas constant = 8-314 J K"' arol”^ if /yor Ay„^.H are in joules
= 2 cal deg”' mol”' if /y or Ay„^H are in calories.
Sample Protsiem Latent heat of fusion of ice is 1436*3 cal mol”'. Calculate the molal deprc.ssion
constant of water.

Solution. K
M.1 RT^-
0 18g mol”' X 2calK~‘ mol”' x(273K)- -I
/ " = 1-87 K kg mol
I000Ay„,H i 000 g kg-'X1436-3 cal mol”'
SOLUTIONS 2/59

2.14.4. Calculation of molecular mass of the solute

w.
2-X
1000 lOOOxKy xh>2
As before, we may write AT^ = X or
M2 =
M2 w
1
ATr xw 1
/

Thus, the molecular mass of non-ionic solute (i.e., non- electrolyte) can be calculated by studying the
depression in freezing point.
2.14.5. To conclude that the depression of freezing point is a colligative prope
From the expression ATy= Ky-m. it is evident, as before, that the depression in freezing point depends
upon molality of the solution, Le.. the number of moles ofthe solute dissolved in lOOOgof the solvent and not
upon the nature of the solute. Hence, it is a colligative property.
2.14.6. Applications (importance) of depression of freezing point

w
(/) In making antifreeze solutions. The running of a car in sub-zero weather even when the radiator is
full of water (which freezes below 273 K) has been possible due to the fact that depression in freezing point

F lo
of water takes place when appropriate amount of a suitable solute (usually ethylene glycol) called antifreeze
is dissolved in the water.

It is interesting to note that the addition of ethylene glycol in water is also useful in summer as it also

ee
raises the boiling point of water and prevents what antifreeze advertisements call “boil-over”.

Fr
(//) In melting of ice on the roads. In winter or in the hills where it snows heavily, common salt (NaCl)
or calcium chloride (CaCl2) is scattered on the roads to melt ice. This is because salt-ice mixture has very low
for
ur
freezing point. For example, a certain composition of NaCI-ice mixture has a freezing point as low as - 2TC
and that of CaCl2-ice mixture as low as -55“C. Thus, so long as the temperature of the surrounding is above
s
these temperatures,the mixture does not freeze and ice keeps on melting.
ok
Yo
2.14.7. Rast method
o
eB

Camphor has high molar depression constant (39-7°). Hence, it is used as a solvent for finding molecular
masses of organic compounds like naphthalene, anthracene, etc. soluble in camphor in the molten state. The
depression in melting point (which is same as depression in freezing point) is so large that it can be found
r
ou
ad

using ordinary thermometer. The method is called “Rast method.”

Y

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
Re
nd

I. The methods used for the measurement of different colligative properties are as follows ;
Relative lowering of vapour pressure by Ostwald and Walker method, Osmotic pressure by Berkeley
Fi

and Hartley method or Morse and Frazer method, Elevation in boiling point by Landsberger method.
Depression in freezing point by Beckmann method.
2. The experiments on osmosis were carried out first of all by Abbe Nollel in 1748.
3. Persons needing replacement of body fluids or nutrients who cannot be fed orally are administered
solutions by intravenous (I.V.) infusion, meaning slow addition to the veins. To prevent crenation
or hemolysis of red blood cells, the I.V. solutions must be isotonic with the intracellular fluids of the
cells.

4. Variation of Elevation in boiling point or Depression in freezing point with molality

(0ATfc=K^xm or (T^-T;) = K^m or T^=T;-hK^m

Thus, plot of T^ versus molality m will be linear with positive slope (= K^^) as shown in Fig. 2.27
showing that boiling point of solution increases linearly with molality.
2/60 New Course Chemistry (XII)EZSCSl

(ii) ATy= Kyx m or T° -T^ = m or T

/ = T° -Kj.m
Thus, plot of Inversus molality m will be linear with negative slope (equal to K^) as shown in Fig. 2.28
showing that heezing point of solution decreases linearly with molality.
FIGURE 2.27 FIGURE 2.28

c
o
Q.
o>
c

o
m

Molaiity Molality

w
Variation of boiling point with molality Variation of freezing point with molality

F lo
Curiosity Questions
f Q. 1. Why oceans do not freeze? Give two reasons.

e
Fre
Ans. (/) Oceans contain huge amounts of salts dissolved in the water. As a result, freezing point of water
is depressed considerably.
for
(//) Winds blow over the surface of sea water and keep it agitated.
Q. 2. While making Ice-creams in metal or plastic cones, the ice-cream seller puts a mixture of
r
ice and common salt around the cones and not Ice alone. Why?
You
oks

Ans. Ice and common salt form a freezing mixture, i.e., salt added to ice lowers its freezing point to
nearly - 20°C. As a result, the freezing of the contents of the cones becomes faster.
J
eBo

NUMERICAL
FORMULAS Ul
ad
our

PROBLEMS
BASED
(0 ATy= Ky-m where AT^= depression in freezing point = - Ty
ON
m = molality of the solution (i.e., no. of moles of solute per 1000 g
of the solvent)
dY
Re

Depression in
Ky = molal depression constant or cryoscopic constant in KJm or
Freezing Point K kg mor’
Fin

1000 KyW2 1000 K.w,

{//) AT/ " or M2 =
W| M., w,1 AT
/
where M2 = molecular mass of the solute,
W2 = mass of the solute in grams, m'j = mass of the solvent in grams
Note. Ky in the above formula is molal depression constant, i.e.,
molecular depression constant per 1000 g of the solvent. If Ky is
molecular depression constant per 100 g of the solvent, the above formula

is modified to AT
lOOKy h-2
/ "
M’j M2
(Hi) For numerical problems involving two different colligative
properties, the equations for these colligative properties are suitably
applied one after the other. Refer to Solved Problems 4 & 5.
SOLUTIONS 2/61

Problem
Q A solution containing 34*2 g of cane-sugar (C12H22OJ1) dissolved in 500 cm^ of water
froze at - 0-374“C. Calculate the freezing point depression constant of water.
Solution. Here, we have AT, = 0-374 K
/
Concentration of sugar solution = 34-2 g in 500 cm^ = 68-4 g in !000 cm^ = 68-4 g in lOOO g
(●.● density of H2O = I g cnr-^)
68-4

342
molal
(Molar mass of C,2H220| j = 342 g mol"*)
= 0-2m

ATy _ 0-374 K = T87 K m-l

Applying the relationship ATy-= KyX m, Ky — m 0-2m

ow
Problem
0 TOO g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing
constant of benzene is 5-12 K kg moT*. Find the
point of benzene by 0-40 K. The freezing point depression
molar mass of solute. (NCERT Solved Example)
-1
Solution. Here, we are given W2 = 1-00 g, m’| = 50 g, ATy= 0-40 K, Ky= 5-12 K kg mol

e
1000 Ky Wj

re
Fl
Substituting these values in the formula M2 - > we get
ATy

F
1000gkg"*x5-I2Kkgmol *xl0g
M2 =
ur = 256 g mol
-1

r
50gx0-40K

fo
B 45 g of ethylene glycol {C2H(^02) is mixed with 600 g water. Calculate
ks
Problem
Yo
(a) freezing point depression (/>) freezing point of the solution
oo

(Kyfor water = 1*86 K kg moT* ; Atomic masses : C = 12, H = 1, O = 16 amu).

eB

(NCERT Solved Example)

-1
Solution. Here, we are given W2 = 45 g, Wj = 600 g, Ky= 1-86 K kg mol
-1
M2 (for solute C2Hg02) = 24 + 6 + 32 = 62 g mol
ur

lOOOKy W2
ad
Yo

Substituting these values in the formula, ATy = , we get

vVjXMt
d

lOOOgkg"*xl-86Kkgmol ’x45g = 2-25 K

Re

AT
in

/ "
600gx62gmol *
F

Freezing point of pure water = 273-15 K

Freezing point of the solution = Ty“ - ATy= 273-15 - 2-25 K = 270-9 K
Problem
Q A solution of urea in water has a boiling point of 100-128®C. Calculate the freezing
point of the same solution. Molal constants for water Kyand K/, are 1-86®C and 0-512”C respectively.
Solution. Step I. To calculate molality of the solution from boiling point data :
We are given that AT^ = 100-128 - 100 = 0-128°C, = 0-512°C
AT^, _ 0-128 = 0-25
Using the formula, AT^ = where m is the molality, we get m = ~ 0-512
Step II. To calculate the depression in freezing point: We are given that Ky= 1-86°C
m = 0-25 (calculated above)
ATy = KyX m = 1-86 X 0-25 = 0-465“
- AT,= 0°C - 0-465“ = - 0-465”C
Freezing point of the solution (Ty) = Ty /
2/62 “P%^ideefr'^ New Course Chemistry (X1I)S*SIS]
Problem
13 The average osmotic pressure of human blood is ?●? atm at 40"C. (a) What would be
the total concentration of the various solutes in the blood ? (b) Assuming the concentration to be essentially
the same as the molality, find the freezing point of blood (Kj- for water = l-Sb^C).
Solution, (fl) We are given that P = 7-7 atm, T = 40“ C = 40 + 273 = 313 K
R = 0 0821 litre atm/degree/mole
7C 7-7 atm
According to van’t Hoff equation, 7C = CRT C =
RT 00821LatniK-lmol-‘x3l3K
= 0‘30 mole/litre
Taking the molar concentration as equal to molality (Given), we have
m = 0-30 : Ky= !-86“C (Given)
AT^= K/-X m = 1-86 x 0-30 = 0-558“
Freezing point of blood = 0“C - 0-558“ = - 0-558“C

low
Problem 0 A solution containing 2*56 g of sulphur dissolved
in 100 g of naphthalene whose melting
point is 80'1“C gave a freezing point lowering of 0-680”C. Calculate the formula of sulphur (Kf for
naphthalene = 6*8 K/m)
Solution. W2 = 2-56 g, Wj - 100 g ; K^= 6-8 Km-^ AT.= 0-68“C

ee
1000Ky.W2
rF
Fr
Now, M2 = w,I xAT
/

_ 1000gkg-‘x6-8Kkgmo|-'x2-56g

r
Substituting the value of Ky, W2, and ATy, M2
-1
= 256 g mol
fo 100gx0-68K
u
Let molecular formula of sulphur be ; Atomic mass of sulphur = 32
ks
Yo
256
Molecular mass of = ;c x 32 XX 32 = 256 = 8
oo

or X =
32

Therefore, the molecular formula of sulphur is Sg.

B

Q How many grams of sucrose (M. wt. = 342) should be dissolved in

re

Problem
lOOg water in order
to produce a solution with 105*0“C difference between the freezing point and the boiling point ?
u

(Kf = l-86“C/m, Kb = 0-SrC/ra)

ad
Yo

Solution. AT^ = m
B. pt of solution (T^) = 100 + AT^= 100 -f
F. pt. of solution (Ty) = 0 -- ATy = 0 - K^m ; T^ -Ty = (100 + m) - (- Kym)
nd
Re

105= 100 + 0-51 xm + l-86m.

Fi

This gives m = 2-11, i.e., 1000 g of water contain sucrose = 2-11 mole = 2-11 x 342 g
2-11x342
Mass of sucrose to be dissolved in 100 g of water = xl00 = 72g.
1000

Problem The freezing point of a solution containing 50 cm^ of ethylene glycol in 50 g water is
found to be - 34“C. Assuming ideal behaviour, calculate the density of ethylene glycol. (K^for water
= 1-86 K kg mol-i)
M2 X ATy X tV| ^ 62 g mol *x34Kx50g
Solution. W2 = = 56-67 g
lOOOxK
/ 1000gkg-^xl-86Kkgmol-'

= 1*13 g cm ^
V 50 cm^

oblem
A 10% solution (by mass) of sucrose in water has freezing point of 269*15 K. Calculate
the freezing point of 10% glucose solution in water, if freezing point of pure water is 273*15 K. Given :
Molar mass of sucrose = 342 g mol”^ molar mass of glucose = 180 g mol”*). (CBSE 2017)
SOLUTIONS 2/63

Solution. 10% solution of sucrose in water by mass means 10 g of sucrose is present in 100 g of the solution.
/.c..

solute (sucrose) = 10 g, solvent (water) = 90 g

lOOOK.w^
M,=
w.I X AT
/

or K
X X AT^ 342 X 90 X (273-15-269-15)
= 12-312 K kg mol
/ "
lOOOxw, 1000x10

10% of glucose solution in water means = 10, vV| = 90 g

lOOOK
AT
/'‘'2 _ 1000x12-312x10 = 7-6‘
/ “

w
Wj M-, 90x180

Freezing point of solution = 273-15 - 7-6 = 265*55 K

Problem [Q Calculate the amount of ice that will separate

F lo
out when a solution containing 50 g of
ethylene glycol in 200 g of water is cooled to - 9*3“C. (Kyfor H2O = 1*86 K kg mol”*)

ee
AT
1000 h'2
Solution.

Fr
/
,v,Mj
Let us calculate the amount of water (W|) present when ATy= 9-3°

for
ur
1000x1-86x50
9-3 = (Mol. mass of (CH20H)2 = 62)
vv,1 X 62
ks
or
w, = 161-29 g
Yo
oo

Water frozen to ice = 200 - 161-29 = 38*71 g

eB
r
ou
ad

1. Normal freezing point of a solvent is 15°C. A 0-5 molal solution of urea in the above solvent causes a
freezing point depression of two degrees. Calculate the molal depression constant.
Y

2. Calculate the freezing point of a solution containing 60 g of glucose (molar mass = 180 g mol"’) in 250 g of
water (Ky of water = 1 -86 K kg mol"').
nd

(CBSE 2018)
Re

3. Water is used in car radiators. In winter season, ethylene glycol is added to water so that water may not
Fi

freeze. Assuming ethylene glycol to be non-volatile, calculate the minimum amount of ethylene glycol that
must be added to 6-0 kg of water to prevent it from freezing at - 0-30°C. The molal depression constant of
water is 1-86 K/m.

4. Two aqueous solutions, containing respectively 7-5 g of urea (mol. wt. = 60) and 42-75 g substance X in
100 g of water freeze at the same temperature. Calculate the molar mass of X.
5. Pure solvent A has freezing point I6-5°C. On dissolving 0-4 g of B in 200 g of A, the solution freezes at
16-4°C and on dissolving 2-24 g of C in 100 g of A, the solution has freezing point of 16-0°C. If the molar
mass of B is 74 g mol"’,' lat is the molar mass of C ?
6- An aqueous solution freezes at - 0-2°C. What is the molality of the solution ? Determine also (/) elevation in
the boiling point (//) lowering of vapour pressure at 298 K, given that K^= 1-86° kg mol“’, K/, = 0-512° kg
mol"’ and vapour pressure of water at 298 K is 23-756 mm. (MP Board 2012)
7. 68-4 g of sugar (molecular weight = 342) is dissolved in 1000 g of water. What is (a) freezing point
(h) boiling point (c) vapour pressure at 20°C (d) osmotic pressure of the solution at 20°C ? The density of the
solution at 20°C is 1 -024 g cm"^. The vapour pressure of water at 20°C is 17-633 mm. The Ky and values
for water are 1-873° and 0-516° respectively. (Assam Board 2012)
2/64 New Course Chemistry fXinpzsTMi

8. An aqueous solution contains 5% by weight of urea and 10% by weight of glucose. What will be its freezing
point ? [Molal depression constant of water is 1-86°C].
9. Addition of 0-643 g of a compound to 50 ml of benzene (density : 0-879 g/ml) lowers the freezing point
from 5-51®C to 5-03°C. If Ky for benzene is 5-12. calculate the molecular weight of the compound.
(We.st Bengal Board 2012)
10. The temperature at a hill station is - 10*^0. Will it be suitable to add ethylene glycol (mol mass = 62) to water
in the radiator so that the solution is 30% by mass ? (K^-for water = 1-86 K m"’)
11. The molal freezing point depression constant of benzene (C^Hg) is 4-90 K kg mol"'. Selenium exists as a
polymer of the type Se^. When 3-26 g of selenium is dissolved in 226 g of benzene, the observed freezing
point is 0-112®C lower than for pure benzene. Deduce the molecular formula of selenium (At. mass of
Se = 78-8 g mol"*).
12. A solution of an organic compound is prepared by dissolving 34-2 g in 500 g of water. Calculate the molar
mass of the compound and freezing point of the solution. Given that for watei- = 0-52 K mol"' B.pt of
solution = IOO-104'’C. K,-J for water = 1-87 K mol"'.
ANSWERS

w
1.4 K m-' 2. - 2-48“C 3. 60 g 4. 342 g mol
-I

F lo
-1
5. 165-8 g mol 6. Molality = 0-1075, (/) 0-055“C (n) 0 046 mm
7. (a) - 0-375“ (b) 100- 103“C (c) 17-569 mm (d) 4-6 atm 8. 3-03“C
9. 156 u 10. Yes 11. Se 8 12. 342gmol-',-0-374‘’C
HINTS FOR DIFFICULT PROBLEMS

ree
3. Vi
= 6000 g, AT^= 0-30“, Kf= 1-86°C for F
CH^OH
Molar mass of ethylene glycol. (Mo) = 62 g mol
CH^OH
Your
ks

1000K^m’2
eBoo

Substitute in the formula. Mo = . Calculate W2-

H’j ATy
4.
ATj-= K^x m. As ATyis same for the two solutions, their molalities will be equal, i.e.,
ad
our

11 1000 _ 42-75 1000 or M = 342.

60 100 “ M
5. For solute B dissolved in solvent A,
Re
Y

AT
1000 Ky W2 lOOOxK.xO-4
/
Find

-1
/ "
M2
, i.e., 0-1 =
200x74
or Ky= 3-7 K kg mol
1000 1000x3-7x2-24
For solute C dissolved in the same solvent A. M, = = 165-8
w,1 AT 100x0-5
/

Alternatively,
(ATy)e vv
B
X
M'^'X M(|.
(ATy)c
lOOOK lOOOK.n,
6. For (//): AT
y W,
/ "
vvj M^ niM,

Calculate — . Then apply O

. Calculate A p.
r n
1
SOLUTIONS 2/65

7. Apply ATys: Kj in and AT/, = m. Mass of solution = 1000 + 684 = 10684 g.

10684 ,
Volume = cnr^.
1-024

8. Mass of solution = lOt^ g : Mass of urea = 5 g

Mass of glucose = 10 g Mass of water = 100- (5 + 10) = 85 g
1000
Molality of urea in the solution = — x = 0-98m
60 85

10 1000
Molality of glucose in the solution = = 0-65 m
180 85

Total ATy= Ky(m, + ^2) = ! -86 (0-98 + 0-65) = 3-03°C

Freezing point of the solution = - 3-03°C.

w
9. u'2 = 0-643 g, u'l = 50 X 0-879 g = 43-95 g.

F lo
10. ATf=KrXw = 1-86X —X —X1000 = 12-86°C, fe.. water will freeze at - 12-86°C.
^ ■’ 62 70
It will not freeze at - 10°C.

ee
11. Similar to Solved Problem 6.

Fr
-1
12. AT/, = 0-104“, W2 = 34-2 g. w
= 500 g, K/, = 0-52 K mol
1000K/,u’2 _ 1000x0-52x34-2

for
-i
M, = = 342 g moi
ur
500x0-104
h’,xAT/,
ks
34-2 1
ATy =K/-Xm = {l-87)x 342 ^ 500 xlOOO =0-374 F.pt = - 0-374“C.
Yo
oo
eB

2.15. ABNORMAL MOLAR MASSES

The various relations as derived above for the colligativc properties are applicable only to the solutions
r

of rum-electrolytes which do not undergo any dissociation or association in the solution. In case of the aqueous
ou
ad

solutions of electrolytes, i.e., acids (HCl, H7SO4CH3C00H, etc.), inorganic bases (NaOH, KOH, Ca(OH)2,
Y

etc.) and salts (NaCl. KCl, KNO3, BaCl2, etc.) which dissociate completely or to small extent in the solution,
the number of particles in the solution increases. As the value of any colligative property depends upon the
Re
nd

number of particles of the solute, therefore, the experimentallyobserved value of colligative property comes
out to be higher than the theoretically expected value. Further, as molecular mass is inversely proportional to
Fi

the colligative property (Refer to the formulae derived for molecular masses), therefore, the observed molecular
mass comes out to be less than the theoreticallyexpected value. For example, in case of KCl, in the aqueous
solution, each molecule dissociates to give two ions (KCl K* + Cl“). Thus, as the number of particles
becomes double, the observed value of colligative properly is double than expected value and molecular
mass is half of the expected value. Similarly, in case of certain substances, association takes place in the
solution. For example, acetic acid, benzoic acid, etc. when dissolved in benzene* exist as doubly associated
molecules in benzene due to hydrogen bonding which may be representedas :
O H—O
R—C C—R
O—H O

Thus, the number of particles in the solution decreases. As a result, the experimentally observed value
of the colligativepropertyis lower and the molecularmass is higher than the expected value. If the association
*Association generally takes place in the non-aqueous .solvents because in the aqueous solution, the high
dielectric constant of water helps in the dissociation of the associated molecules.
2/66 New Course Chemistry fXinrzsrwn

were complete, the observed value of colligalive property will be half and the molecular mass will be double
than the expected value.
y/hen the molecularmass ofa substance determinedby studying any ofthe colligativeproperties
comes out to be different than the theoretically expected value, the sub ance is said to show
abnormal molecular mass.

Fuither, the various relations for the colligative properties hold good only for dilute solutions which act
as ideal solutions. Thus, even when the solute is non-volatile, non-electrolytic, non-associated and non-
dissociated, deviations from the theoretically expected values of colligative properties and molecular masses
are observed at higher concentrations. This is because at higher concentrations, the molecules are so close to
each other that they exert appreciable intermolecular forces of attraction resulting into the deviations from
ideal behaviour.

To sum up, abnormal molecular masses are observed in any one of the following cases :

w
(1) When the solution is non-ideal, i.e., the solution is not dilute.
(2) When the solute undergoes association in the solution.

F lo
(3) When the solute undergoes dissociation in the solution.
To calculate the extent of association or dissociation, van’t Hoff in 1880 introduced a factor ‘i’ called

ee
van’t HolT factor.

Fr
vaii’t HofT factor is defined as the ratio of the experimental value of the colligative property to
the calculated value of the colligative property, i.e., van^t Hofffactor,
. _ Experimental value of the colligative property for
ur
Observed value of the colligative property
Calculated value of the colligative property Normal value of the same property
s
when the solution behaves ideally
ook
Yo

Further, as the colligative property is inversely proportional to the molecular mass of the solute, we can
eB

also write

. Normal (calculated) molecular mass M

i -
r

Abnormal (Observed) molecular mass M

ad
ou

Modified Expressions of Colligative Properties for Substances undergoing Associations or

Dissociation. For substances undergoing association or dissociation in the solution, the various expressions
Y

for the colligative properties are modified as follows :

Re
nd

-Ps -1
_..
(t) Relative lowering of vapour pressure, — x
2
Fi

iii) Elevation in boiling point, AT^ - / m

(in) Depression in freezing point, ATy s i Ky m
(iv) Osmotic pressure, 7t = i — RT
V

Calculation of Degree of Dissociation or Association*. Using van’t Hoff factor, the extent of
dissociation or association can be calculated as follows :

(a) Calculation of the degree of dissociation.

Degree of dissociation is defined as the fraction of the total substance that undergoes dissociation,
No. of moles dissociated
i.e., Degree of dissociation (a) =
Total number of moles taken

*These are also called apparent degree of dissociation or association.

SOLUTIONS 2/67

Suppose a molecule of an electrolyte when dissolved in a solvent dissociates to give n ions and a is the
degree of dissociation, i.e., in general A » rt| B + «2 C + ....
«[ +
If we start with 1 mole of the .solute, at equilibrium, we have (1 - a) moles of undissociated molecules
and na moles of the ions, so that the total number of moles of ions and the undissociated solute molecules =

1 - a + na = 1 + (« - 1) a
Thus, observed colligative property «= 1 + (n - I) a
and, normal (theoretical) value 1 (as 1 mole of A was taken).
l + (n-l)a /-I
van’t Hoff factor, i = or a =
1 n-1

Norma! (calculated) molecular mass M

c M c -M,o

w
But i =
a =
Observed molecular mass M
V (n-1)

From this equation, the apparent degree of dissociation (a) can be easily calculated.

Flo
ib) Calculation of the degree of association.

ree
Degree of association is deifned as the fraction ofthe total substance which exists in the form of
No. of moles associated

rF
associated molecules, i.e., Degree of association (a) =
Total number of moles taken
ur
Suppose n simple molecules of the solute A associate to form the associated molecule A„ so that we
have the equilibrium : «A V i An
fo
ks
If a is the degree of association and we start with one mole of A, then at equilibrium
Yo
Number of moles of A = 1 - a
oo

Number of moles of A„ = a/n

B

Total number of moles = I - a + aJn.

re

Since the colligative property is proportional to the number of moles of the solute present in solution,
u

l-a + —
ad

n n
Yo

therefore, vanT Hoff factor / = whence a = (l-t)

1 n-1
M
c
Also, i -
nd
Re

M
Q
Fi

where M^. is normal (calculated) molecular mass and is the observed molecular mass.
M M o -M c
a
l-a + - =
M, or 1- ^ = a
a
or = a
1
1— - a
n-1

n M
0
Mo n
Mo n n

a -
M,-M, n

M (n-1)
0

From this equation, the apparent degree of association (a) can be calculated.
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. van’t Hoff factor, i > 1 if there is dissociation of the solute in the solution and / < I if there is association
of the solute in the solution.

2. For 100% dissociation of a solute, van’t Hoff factor, / = number of ions produced from one molecule of
the solute.
2/68 New Course Chemistry fxiltpzsm

NUMERICAL FOHMLH^S USED \

PROBLEMS
BASED (0 van’t Hoff factor,
ON
Obs.colligativeproperty _ Calculated mol. mass (M^)
I Association and I Cal. colligative property Observed mol. mass (M^)
Dissociation (//) If observed colligative property < calculated value or
(or i < 1). there is association.
If observed colligative property > calculated value or
(or I > 1), there is dissociation.

ow
(Hi) Calculation of degree of dissociation/association
Step 1. Calculate van’t Hoff factor (/) using the relation

i -
Observed colligative property Calculated mol. mass
Calculated colligative property Observed mol. mass

e
re
Step 2. Write the dissociation/association equation. Starting with 1 mol

rFl
and assuming a to be the degree of dissociation/association, calculate

F
Actual no.of particles
the actual no. of particles. Then i =

r
Step 3. Equate i from steps 1 and 2 and calculate a.
ou
fo
ks
n The freezing point depression of 0*1 molal solution of acetic acid in benzene is 0*256
oo
Protslem

K. Ky for benzene is 5*12 K kg mol”^. What conclusion can you draw about the molecular state of acetic
Y
eB

acid in benzene ?

Solution. Here, we are given m = 0-1 mol kg"' ;

-1
Ky=5-12Kkg mol
Theoretically calculated value of ATy will be ATy= KyX m = 5-12 K kg mol xO-1 mol kg-'= 0-512 K
-1
ur

Observed (experimental) value of ATy = 0-256 K (Given)

ad
Yo

Thus, the observed value is half of the theoretical value. Since the observed value depends upon the
number of particles actually present, this means that the number of particles actually present is half of the
theoretical value. In other words, the molecules of acetic acid are doubly associated in benzene, i.e., they
d

exist as dimers, (CH3COOH)2.

Re
in

Observed colligative property _ 0-256 _ 1

Alternatively, van’t Hoff factor, i=
F

Calculated colligative property 0-512 2

Calculated mol. mass Calculated mol. mass
Also, t = Observed mol. mass =
Observed mol. mass i

60
But calculated molar mass of CH3COOH = 60 g mol ' Observed molar mass = 1/2
= 120g mol

Thus, the observed molar mass is double of the theoretical value. Hence, acetic acid exists as doubly
associated, i.e., as (CH3COOH)2-
Problem Q 0*5 g KCI was dissolved in 100 g water and the solution originally at 20®C, froze at

- 0*24"C. Calculate the percentage ionization of salt Kj- per 1000 g of water = 1*86®C.
Solution. Observed mol. mass is obtained from the given data, i.e.,
W2 = 0-5 g, = 100 g, M, = 18 g mol"' for H2O ; ATy= 0 - (- 0-24) = 0-24"C
lOOOK.
M, =
lOOOgkg 'xl-86Kkgmol 'x0-5g -1
= 38-75 g mol
ATy X M-’i 0-24Kxl00g
SOLUTIONS 2/69

-1
Calculated (theoretical) mol mass of KCl = 39 + 35-5 = 74-5 g mol
Calculated mol mass 74-5
/ = = 1-92
Observed mol mass 38-75
Now, KCl dissociates as
KCl + cr
Initial moles 1 mole 0 0
Moles after disso. I -a a a

1 + a
Total no. of moles after dissociation = l- a+ a + a= l + a i = ora = i- 1 = 1-92- 1 =0-92
1
Percentage ionization = 0-92 x 100 = 92%.
Problem 0 2*0 g of benzoic acid dissolved in 25-0 g of benzene shows a depression in freezing
point equal to 1*62 K. Molal depression constant (Kj) of benzene is 4-9 K kg mol"*. What is the percentage

w
association of the acid if it forms dimer in the solution ? (NCKRT Solved Example, Pb. Board 2011)
Solution. Mass of solute (benzoic acid), vv2 = 2-0 g ; Mass of solvent (benzene), Wj = 25-0 g

Flo
Observed AT/-=1-62K;
/ Kf = 4-9 K kg mol
Observed molar mass of benzoic acid

ee
1000xK^xw2 ^ lOOOgkg"* x4-9Kkgmol“* x2-0g

Fr
-1
M2 = = 242 g mol
AT^ xiv l-62Kx25-0g
-1
Calculated molar mass of benzoic acid (CgH5COOH) = 72+ 5+12 +32+1 = 122 g mol

for
ur
Calculated mol. mass 122
van’t Hoff factor, / = = 0-504
Observed mol. mass 242
ks
If a is the degree of association of benzoic acid, then we have
Yo
oo

2 Cf^H^COOH V (C^H^COOH)^
Initial moles 1 0
eB

After association 1 -a a/2

a a
r

.●. Total number of moles after association = l- a + — = 1

ou

2 2
ad

. l-a/2
Y

a
= 0-504 or 1--=0-504
1 0
nd
Re

a = (1 - 0-504) X 2 = 0-496 x 2 = 0-992 or % age association = 99*2%

Problom
FI Calculate the boiling point of a one molar aqueous solution (density 1*04 g mL"*) of
Fi

potassium chloride (Kj, for water = 0-52 K kg mol"*. Atomic masse.s : K = 39, Cl = 35 5)
Solution. Concentration of solution = IM ; Density of solution = 1 -04 g mL '
Let us first calculate the molality of the solution.
Amount of solute (KCl) = I mol = 74-5 g ; Volume of solution = IL = 1000 mL

Mass of the solution = 1000 x 1-04 g = 1040 g

Mass of solvent = 1040 - 74-5 g = 965-5 g = 0-9655 kg
No.of molesof the solute Imol
Molality of the solution = = 1-0357 mol kg"* = 1-0357 m
Mass of the solvent in kg 0-9655 kg
KCl dissociates as : KCl ^ K+ + Cl"
Number of particles after dissociation = 2 van’t Hoff factor, i = 2
Now AT/, = / x K^ X m = 2 X 0-52 x 1 -0357 = 1 -078"C
Boiling point of the solution = 100 + 1-078 = 101*078“C
2/70 New Course Chemistry (XII)EZslS]

Problem 0 KI and sucrose solution with 0*1 M concentration have osmotic pressure of 0*465 atm
and 0*245 atm respectively. Find the van’t Hoff of KI and its degree of dissociation. (AIPMT Mains 2008)
Solution. For KI solution, 7t = i CRT ...(/)
0465 = /x 0-1 xRxT

For sucrose solution. 7t = CRT 0-245 = 0-1 x R x T

0-465
Dividing {/) by (//), we gel /= = 1*898
0-245

If a is the degree of dissociation, KI K-" r

Initial 1 mole

After disso. 1 - a a a

Total moles after dissociation = (I-a)-f-a-f-a=l+a

w
i = 1 -f a or a = t - 1 = 1-898 - 1 = 0-898 = 89*8%

Problem
0 0*6 mL of acetic acid (CH3COOH) having a density of 1*06 g mL“* is dissolved in

Flo
1 litre of water. The depression in freezing point observed for this strength of the acid was 0*0205®C. Calculate
the van’t Hoff factor and the dissociation constant of the acid. Kj for water = 1*86 K kg mor^

e
re
(NCERT Solved Example, Pb. Board 2011)

rF
Solution. Calculation of van’t Hoff factor (i)
0-6 mL of acetic acid means solute (^2) = 0-6 x 1-06 g = 0-636 g
ur
1 Litre of water means solvent (wj) = 1000 g ; (ATf)observed = 0'0205<’C
lOOOxKy XIV2 fo
lOOOgmol 'xl-86KkgmoI *x0-636g
ks
Observed molar mass, (M2)observed - w,1 x AT 1000 gx 0-0205 K
Yo
/
oo

1
= 57-7 g mol
B

-1
Calculated molar mass of CH3COOH = 60 g mo!
re

-1
Calculated molar mass 60g mol = 1*04.
van’t Hoff factor (i) = -1
Observed molar mass 57-7 g mol
u
ad
Yo

Calculation of degree of dissociation (a) from van’t Hoff factor (i)

If a is the degree of dissociation of acetic acid, then
CH3COOH ^ CH3COO- -t- H+
nd
Re

Initial moles 1
Fi

Moles at eqm. 1 - a a a, Total = 1 + a

1-i-a
I = = l+ a or a = 1 - i = 1-04 - 1-0 = 0-04
1
Calculation of dissociation constant. If we start with C mol L“' of acetic acid, then
-f
CH3COOH ? - CH3COO- + H
Initial cone. C mol L"*
Cone, at eqm. C - C a Ca Ca

= C(l-a)

[CH3C00"][H+] _ CaCa Ca^

Dissociation constant (K^) =
[CH3COOH] C(l-a) 1-a

0-636gL
-1
(0-0106) (0-04)2 = 1*76 x 10-*'.
But C = 0-636 g L
-1
-i
= 0-0106 mol L-^ .-. K a
60 g mol 1-0-04
SOLUTIONS 2/71

Problem
Wk Assuming complete dissociation of the salts, calculate the molality of sodium chloride
solution whose elevation in boiling point is numerically equal to the depression in freezing point of 0-2 ni
aluminium sulphate solution in water (K^ and Kyfor water are 0*52 and 1*86 K kg mol“® respectively).
Solution, i for NaCl = 2, i for AI-, (804)3 = ^
AT^ [NaCI sol.] = AT,- [Al2(S04)3 sol.] {Given)
/xK^x/»{NaCl) = /xK^x/H[Al2 (504)3]. 2 x 0-52 x= 5 x 1-86 x 0-2 or m = 1-788.
Problem
R1 A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate
the osmotic pressure of the solution. Given solution constant (R) = 8-314 JK“* niol“^
Solution. Molar concentration = 01 mol L“'
0!

10-
^ mol m ^=10^ mol m ^
4-
K4[Fe(CN)6l ^ 4K+ + [Fe(CN)6]

w
Initial I mole 0 0
After 50% disso, 1 -0-5 4x0-5 0-5 Total = 3

F lo
= 0-5 = 2
/ = 3

n = i CRT

ee
= 3 X 10^ X 8-314 X 300 Nm"- = 7-483 x 10^ Nm^^

Fr
Problem
E] Arrange the following solutions in the increasing order of their osmotic pressure

for
(a) 34-2 g/lit sucrose (b) 60 g/Iit urea (NH2CONH2) (c) 90 g/Iit glucose (d) 58-5 g/lit sodium chloride
Give reason in support of your answer
ur
-I
Solution. Molar mass of sucrose (C12H22O11) = 342 g mol
oks

Molar mass of urea (NH2CONH2) = 60 g mol”^ : Molar mass of glucose (CgHi20g) = 180 g mol
Yo

Molar mass of NaCI = 58-5 g mol"^ ;

o
eB

34-2 60
Molar cone, of sucrose = = 0-lM ; Molar cone of urea = — = 1M ;
342 60
90 58-5
our

Molar cone of glucose =

ad

= 0-5M ; Molar cone, of NaCI = = 1M :

180 58-5
However, as NaCI is an electrolyte and one formula unit of NaCI dissociates to give two ions (Na"^ and Cl")
therefore, molar concentration of particles in the solution = 2M. Thus, the order of increasing concentration is
Y
Re

Sucrose < Glucose < Urea < NaCI

nd

(0-lM) (0-5M) (IM) (2M)

Fi

As osmotic pressure (or any colligative property) is directly proportional to the number of particles in the
solution, hence increasing order of osmotic pressure will be : Sucrose < Glucose < Urea < NaCI
Note. If the above solutions are to be arranged in order of their decreasing freezing points, the depression in
freezing points (AT^) will be in the order : Sucrose < Glucose < Urea < NaCI
Thus, sucrose will have minimum depression i.e. its actual freezing point will be maximum. Hence, the
order of decreasing freezing points will be : Sucrose > Glucose > Urea > NaCI

1. The freezing point depression of 01 m NaCI solution is 0-372°C. What conclusion would you draw about
its molecular state ? Ky-for water is 1-86 K kg mol"'.
2. Which of the following solution will have the highest and which will have the lowest freezing point and
why ? (0 O.I M NaCI solution (//) 0.1 M glucose solution {Hi) 0.1 M BaCl2 solution
2/72 New Course Chemistry CX11)CQ]9]

3. Calculate the amount of NaCl which must be added to 100 g of water so that the freezing point is depressed
by 2 K. For water, Ky-= 1-86 K/m.
4. Calculate the van’t Hoff factor of CdS04 (molecular mass 208-4) if the dissolution of 5-21 g of CdS04 in
half litre water gives a depression in freezing point of 0168°C (Kyof water is 1-86 K kg mor')
5. Determine the osmotic pressure of a solution prepared by dissolving 2-5 x 10"" g of K2SO4 in 2 L of water
at 25°C, assuming that it is completely dissociated.
(R = 0-0821 L atm K"' mol"*, Molar mass of K2SO4 = 174 g mol"*) (CBSE 2013)

6. 3-9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1 -62 K. Calculate
the van’t Hoff factor and predict the nature of solute (associated or dissociated).
(Given : Molar mass of benzoic acid = 122 g mol"’, Ky for benzene = 4-9 K kg mol"’) (CBSE 2015)

ow
7. 0-01 m aqueous solution of K3 [Fe(CN)^] freezes at - 0-062°C. What is the apparent percentage of
dissociation ? (Kyfor water = 1-86 K kg mol”’)
8. Phenol associates in benzene to a certain extent to form dimer. A solution containing 2-0 x 10"^ kg of phenol
in 1-0 kg of benzene has its freezing point decreased by 0-69 K. Calculate the degree of association of

e
phenol (K^for benzene = 5-12 K kg mol"*).

re
9. Out of the following three solutions, which has the highest boiling point and why ?

Frl
(0 1 M Glucose (H) 1 M Potassium chloride (///) 1 M Aluminium nitrate

F
10. A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. Calculate the osmotic pressure
of the solution. (R = 8-3141 K~’ mol"*).
ou
11. On a certain hill station, pure water is found to boil at 95°C. How many grams of NaCl must be added to

r
2 kg of water so that it boils at 100°C ?

so
12. Calculate the freezing point depression expected for 0-0711 m aqueous solution of Na-)S04. If this solution
kf
actually freezes at - 0-320°C, what would be the value of van't Hoff factor ? (K^ for water is 1-86 C mol"’).
oo
(CBSE 2009)
13. Calculate the boiling point of a solution containing 0-61 g of benzoic acid in 50 g of carbon disulphide
Y
-1
eB

assuming 84% dimerisation of the acid. The boiling point and K^, of CS2 are 46-2°C and 2-3 K kg mol
respectively.
14. Calculate the freezing point of a solution containing 8-1 g of HBr in 100 g of water, assuming the acid to be 90%
ur

ionized. [Given : Molar mass of Br = 80 g/mol, water = 1-86 K kg/mol] (CBSE Sample Paper 2018)
oY

15. Calculate the freezing point of a solution containing 0-5 g KCI (Molar mass = 74-5 g/mol) dissolved in
ad

100 g water, assuming KCI to be 92% ionized. K^of water = 1-86 K kg/mol. (CBSE Sample Paper 2019)
d

16. The freezing point of a solution containing 5 g of benzoic acid (M = 122 g mol"*) in 35 g of benzene is
depressed by 2-94 K. What is the percentage association of benzoic acid if it forms dimer in solution ?
in
Re

(Ky for benzene = 4-9 K kg mol"*) (CBSE 2020)

F

ANSWERS

1. NaCl is dissociated completely 2. Glucose will have highest and BaCl2 lowest freezing point
3. 3-147 g 4. 1-806 5. 5-27 X 10"^ atm 6. 0-506, associated
7. 77-7% 8.0-734 9.1 M Ai(N03>3 10. 7-483 atm 11. 562-42 g
12. 0-132°C, 2-42 13. 46-333°C 14. 269-47 K 15. - 0-24“C 16. 97-6%

HINTS FOR DIFFICULT PROBLEMS

1. ATy(obs.) = 0-372°C, ATy(cal.) = 1-86 x 0-1 = O-ISOX

As ATy(obs.) > ATy(cal.) this means NaCl is dissociated in solution.
AT,(obs.)
/ 0-372 , No. of particles after dissociation
Further. i = But I =

AT,/ (cal.) " 0-186 No.of molecules taken

This means that each NaCl molecule ionizes to give two panicles, i.e., Na"*" and Cl ions as follows;
-t-aq
NaCl > Na+ + Cl"
SOLUTIONS 2/73

2. Glucose does not dissociate, NaCl dissociates to give two ions, BaCl2 dissociates to give three ions.
Depression in freezing point will be maximum for BaCl2 and minimum for glucose solution. Hence,
BaCl2 solution will have the lowest and glucose solution will have the highest freezing point.
3. NaCl ^Na+ + Cl- / = 2; ATy= i Ky/n, i.e., 2 = 2 x 1-86 x m or m = 0-538
0-538x100
Amount of NaCl to be dissolved in 100 g = x58-5g = 3-147g.
1000

4. 500 cm^ H2O = 500 g, i.e., Wj = 500 g ;

5-21 1000
Molality (m) = = 0-05
208-4^ 500
ATy(calculated) = KyX m = 1-86 x 0-05 = 0-093

ow
. 0-168
= 1-806
ATy (observed) = 0-168“C .-. i =
0-093 (or apply ATy= / I^m)

5. K2SO4 ^ 2 K+ SQ2- .-. / = 3 ; C= X- = 7-18 X 10“^ mol L“^

174 2

e
Fl
re
Jt = i CRT ; Ji = (3) (7-18 X 10-^ mol L-‘) (0-0821 L atm K"* mol"^) (298 K) = 5-27 x atm

F
1000KyW2 _ 1000gkg-^x4-9KkgmoHx3-9g
6. ATy (calculated) z= = 3-197 K
Wj M2
ur 49gxl22 gmoH

or
(Mol. mass of benzoic acid CgH5COOH = 122 g mol *)
sf
ATf (observed) 1-62
k
= 0-506
Yo
i =

ATy (calculated) 3-197

oo

3-9 1
B

Alternatively, molality (m) = x—x 1000 = 0-652;

122 49
re

AT
/ 1-62
= 0-507.
ATy= i Kym
u

or
ad

K.m 4-9x0-652
f
Yo

As I < 1, solute is associated.

7. ATy (observed) = 0-062°C

d
Re

ATy (calculated) = KyX m = 1-86 x 0-01 = 0-0186®C

in

AT^ (observed) o-062

F

i = — = = 3-33
ATy (calculated) 0-0186
K3 [Fe(CN)6l V - 3 K+ -H [Fe(CN)6]3-
Initial 1 mol
After disso. 1 -a 3a a. Total = 1 -I- 3 a

. l-t-3a i-1 3-33-1

or a = = 0-777 = 77-7%.
1 3 3

1000 Ky W2 1000 g kg-» x 5-12 K kg mol-^ x 2-0 x IQ-^ kg -1

8. M2 (observed) = = .148-4 g mol
Wj ATy l-0kgx0-69K

M2 (calculated) for C5H5OH = 94 g mol“^ = 0-633

M2 (observed) 148-4
2A74 New Course Chemistry (XU)BE

2 CgHsOH ;f i (C6H50H)2
Initial 1 mol
After asso. 1 -a a/2 Total =1-0/2

. l-a/2
i =
or 0 = 2(1-0 = 2(1-0-633) = 0-734.
1

10. K4[Fe(CN)6l ^ ± 4K+ + [Fe(CN)6]'^

1-0 4o o. Total = 1 H- 4 o
i= 1 +4o= 1 -»-4x0-5 = 3

Apply 7C = / CRT
AT, 5

ow
11. AT^= 100-95 = 5", AT^ = /K,m or m = = 4-807
/K^ 2x0-52
1000 g should contain NaCl = 4-807 moles = 4-807 x 58-5 g
2000 g should contain NaCl = 2 x 4-807 x 58-5 g = 562-42 g.

e
12. Expected (calculated) ATy= m = 1-86 x 0-0711 = 0-132°C (assuming no dissociation)

re
Fl
Actual (observed) ATy = 0-320"C

F
(ATy)p,, _ 0-320 = 2-42.
i =
ur 0-132

r
13. As in Solved Problem 3, , = 1-_ = 1-
o 0-84
= 1-0-42 = 0-58
fo
ks
2 2
Yo
0-61 1000
oo

AT^ = i K^m = 0-58 x 2-3 x 122

X
50
= 0-1334". Hence, T^ = 46-2 + 0-133 = 46-333"C.
eB

14. HBr ± + Br-

Initial 1 mole 0 0
ur

After 90% disso. 1-0-9 = 01 0-9 0-9 Total = 01+0-9 + 0-9 = 1 -9

ad

/= 1-9
Yo

8*1 1
Molarity of solution = — x — x 1000 = 1 m
8-1 100
d
Re

ATy= I Kym = 1-9 X 1-86 x 1 = 3-53 K

in

Freezing point = 273 - 3-53 = 269-47 K

F

15. KCl > K+ + Cl"

1 mole
1 -o o o Total = 1 + o .*. / = 1 + o
Given o = 0-92 i = 1 + 0-92 = 1-92

fO-5 1
ATy=/Ky/n= 1-92X 1-86 x 1,74-5 100 xlOOO =0-24°
X

ATy = t“ - Ty = 0 - 0-24 = - 0-24"C

16. Similar to Solved Problem 3.
122
M2 = 238-1, / = = 0-512
238-1
o
/ = 1--=(>512 or a = 0-916 = 916%
2
SOLUTIONS 2/75

1, Solution. It is a homogeneous mixture of two or more chemically non-reacting substances whose composition
can be varied within certain limits. If only two components are present, it is called a binaiy soluiion.
2. Solute and Solvent. In a bimuy solution, the component present in smaller amount is called solute while the
other present in larger amount is called solvent. If water is the solvent, solution is called aqueous solution.
3. IVpes of solutions. Depending upon the physical slate of the solute and the solvent, there are 9 types of
solutions viz. (0 Solid in solid (e.g., alloys) (//) Liquid in solid (e.g., hydrated salts) (Hi) Gas in solid (e.g.,
dissolved gases in minerals), (tv) Solid in liquid (e.g., sugar solution), (v) Liquid in liquid (e.g., ethanol in
water), (W) Gas in liquid (e.g., aerated drinks), (vt7) Solid in gas (e.g., iodine vapour in air), (viii) Liquid in
gas (e.g., humidity in air), (ix) Gas in gas (e.g., air).
4. Expressing concentration of solutions

w
(/) Percentage. For a .solid in liquid, it is grams of solute in 100 g of solution (w/w) or grams of solute in
100 mL of solution (w/v).

F lo
Fora liquid in liquid, it is grams of solute in 100 g of solution (w/w) or volume of solute in mL in 100 mL
of the solution (v/v).
(ii) Strength. It is grams of solute in IL or 1 dm^ solution (i.e., g L“* or g dm"^).

ee
(Hi) Molarity (M). It represents moles of the solute in 1 L of the solution (i.e., mol L“').

Fr
(iv) Normality (N). It is gram equivalents of the solute in 1 L of the solution (i.e., g eq L~^). Gram equivalents
= Mass in g/Eq. wt. and Eq. wt. of acid/base = Molecular mass/Basicity or Acidity,
(v) Molality (/n). It represents moles of the solute for
in 1 kg of the solvent (i.e., mol kg"’).
ur
(vi) Mole fraction. Mole fraction of any component in the solution = Moles of that component/Total no. of
moles of all the components. For solution containing «2 moles of solute in «| moles of solvent, mole fraction
s
ook
Yo
of solute (jco) = ti2/(«i + 02) and mole fraction of solvent = n^/(n^ + Hj).
Evidently, .Vj + ^2 = 1.
eB

(vii) Mass fraction. Here, we take masses instead of moles. For a solution containing W2 g of solute in vvj g
of solvent, mass fraction of solute = W2/{wj + W2) and mass fraction of solvent = W]/(W] -1- W2).
our

(v//7) Parts per million parts (ppm). It is the mass of the solute present in million pails (10^) by mass of the
ad

solution or volume of the solute in million parts by volume of the solution.

5. Why is molality or mole fraction preferred ? These involve masses of the solute and solvent which do not
Y

change with temperature.

Re
nd

6. Relationships. For dilution or exact reaction between two solutions, NiVj = N2V2 (normality equation)
or in terms of molarities, for dilution, MjVj = M2V-, (molarity equation) or for exact reaction,
Fi

M]V|//i] = M2V2//J2 («] and «2 represent their moles in the balanced equation). For acid (basicity n^)
reacting with base (acidity /i^), relation is n^M^V^ = n^M/,V^. To calculate normality or molarity of final
solution on mixing two or more solutions, NjV, -1- N2V2 = N3 (Vj + V2) or MjV, + M2V2 = M3 (Vj + V2).
7. Solubility of a solid in liquid. It is the maximum amount of the solute in grams which can dissolve in 100 g
of the solvent to form a saturated solution at that particular temperature.
8. Factors affecting solubility of a solid in a liquid, (i) Nature of the .solute and solvent (Like dis.solves like)
(H) Temperature. If dissolution is endothermic, solubility increases with temperature. If it is exothermic,
solubility decreases with increase of temperature. For substances like Na2SO4.10H2O, CaCl2.6H20 etc.,
the temperature at which trend changes is called transition temperature.
9. Effect of pressure on the solubility of a gas in a liquid (Henry’s Law). The mass of the gas dissolved in
a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas in
equilibrium with the liquid (i.e., m p or m = K p where K is a constant) or for a mixture of ga.ses in
equilibrium with a liquid, the partial pressure of the gas is directly proportional to the mole fraction of the
gas in the solution (i.e., p^ « or p^ = where is called Henry’s constant).
2A76 ’P'Mdee^'^ New Course Chemistry (XU)BZS19I

10. Units of Henry’s constant. As Pf^ = Kjj Xf^ and X/^ is dimensionless, units of Kjj will be same as that of
pressure, i.e., atm or bar or k bar.
11. Characteristics of Henry’s constant. (/) Greater the value of Kfj, lower is the solubility of the gas at the
same partial pressure.
(i7) Kjj increases with increase of temperature implying that solubility decreases with increase of temperature.
12. Limitations of Henry’s law. Henry’s law is valid only
(0 when pressure is low and temperature is high (//) gas does not undergo compound formation with the
solvent or association or dissociation in the solvent.

13. Applications of Henry’s law. (/) In production of carbonated beverages (i7) In deep sea diving («7) In
function of lungs (iV) For climbers or people living at high altitudes where concentration of oxygen is low in
blood and tissues causing a disease called anoxia.
14. Solutions of solids in solids (solid solutions). When some atoms, ions or molecules of one solid present at
its lattice sites are replaced by those of a similar solid having nearly same size, solids obtained are called
substitutional solid solutions, e.g„ brass. If in the lattice of a solid, atoms of some other solid, small in size
occupy interstitial sites, solids obtained are called interstitial solid solutions, e.g., steel (Fe + C).

w
15. Vapour pressure. It is the pressure exerted by vapour present in equilibrium with the solution at a particular

F lo
temperature.
16. Why lowering of vapour pressure occurs on dissolving a non-volatile solute in a liquid solvent ? This
is because some of the molecules of the volatile liquid solvent on the surface are replaced by those of the
non-volatile solute.

e
Fre
17. Vapour pressure of liquid-liquid solutions and Raoult’s law. Vapour pressure of a component in the
solution is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of
that component in the pure state. For a solution of components for
A and B, ^ Pb ~ Pb'
Total vapour pressure, P = Pa+Pb =x^pI+x^pI = d -^b> pI + -^b = (Pb " ^A>^B + ^a-
r
You
plot of p^ vs x^, Pq vs X3 and vs Xg or x^ are linear. Evidently, mole fractions in the vapour phase will
oks
eBo

Pa _ Pa Pb
be Ja =
^total and ^B = ptotal
or
Pa = )'A X Ptotai andpe = yfi x Ptotai-
Pa
18. Vapour pressure of solid-in-liquid solutions (Raoult’s law for non-volatile solutes). Remembering that
ad
our

solute is non-volatile, vapour pressure of solution = vapour pressure of solvent in the solution i.e.,
P'"-Ps_ «2
= Xj X p°. On modifying, it gives = X2
p® r\+n^
Re
dY

p® - Pj is lowering of vapour pressure and (p® - Ps)/p® is called relative lowering of vapour pressure. Hence,
Fin

for a non-volatile solute, relative lowering of vapour pressure is equal to the mole fraction of the solute in
the solution.
19. Ideal and Non-ideal solutions. An ideal solution of components A and B is that solution in which A - B
interactions are of the same magnitude as A - A and B - B interactions or there is no volume change and
enthalpy change on mixing, i.e., AV^^ing = ® ^mixing = 0 or in which each component obeys Raoult’s
law at all temperature and concentrations. If these conditions are not satisfied, the solution is said to be non
ideal.

20. Types of non-ideal solutions.

(0 Non-ideal solutions showing positive deviation from Raoult’s law, i.e., theirp^, pg and are higher
than expected from Raoult’s law. This is because their A - B interactions are wesJcer then A - A or B - B
interactions (e.g., ethyl alcohol + cyclohexane or CS2 + acetone etc.) For such solutions, AVj^jj^jng = + ve.
AHniixi ng = +ve.
(i7) Non-ideal solutions showing negative deviations from Raoult’s law, i.e., their p^, pg and are
lower than expected from Raoult’s law. This is because their A - B interactions are higher than A - A or
B - B interactions (e.g., chloroform + acetone etc.) For such solutions, AV^„g = - ve, AH,jjjjjj„g = - ve.
SOLUTIONS 2/77

21. Azeotropic mixtures/Azeotropes. For solutions showing +ve deviations, there is a composition for which
vapour pressure is maximum and hence boiling point is minimum. Moreover, it forms a constant boiling
mixture. Similarly, for solutions showing - ve deviations, there is a constant boiling mixture with maximum
boiling point. Such constant boiling mixtures are called azeotropes.
22. Colligative properties. These are the properties of ideal solutions which depend only on the number of
particles of the solute dissolved in a definite amount of the solvent and do not depend upon the nature of the
solute. These are

(0 Relative lowering of vapour pressure (ii) Osmotic pressure

(Hi) Elevation of boiling point (/v) Depression of freezing point.
These properties are generally used for calculation of molar masses of solutes.
23. Relative lowering of vapour pressure. By Raoult’s law,
P^'-Ps '2 W2 /M2 _ ”2 _ ^2 ^ ^2
-X2- for dilute solution (as n2«n^)
«l+«2 vvj/Mj + W2/M2 n^ w,/M,

w
W2 = mass of solute, Wj = mass of solvent, Mj - molar mass of solvent, M2 = molar mass of solute,
Pj = V.R of solution, p® = V.P. of pure solvent.

F lo
P'^-Ps _ W2/M2
For dilute as well as concentrated solutions,
Ps Wj/M,

ee
Fr
24. Osmosis. The net spontaneous flow of solvent from solvent to solution or from a less concentrated solution
to a more concentrated solution through a semipermeable membrane is called osmosis.

for
25. Osmotic pressure. It is the equilibrium hydrostatic pressure of the column set up as a result of osmosis or it
is the minimum excess pressure that has to be applied on the solvent to stop the flow of the solvent into the
ur
solution through the semipermeable membrane. It is measured usually by Berkeley and Hartley’s method.
26. Calculation of molar mass from osmotic pressure. Osmotic pressure (tc) oc concentration of solution
s
(C)
ook

and also « temperature (T), i.e., jc« C x T or tc = CRT where R = solution constant = gas constant. Putting
Yo

C = n/V, Jt = («/V) RT or tcV = nRT. Further, putting n = W2/M2, we get M2 = wRT/jcV. Taking w in g, V in
eB

L, K in atm, R = 0*0821 L atm K“^ mol~*, T in K, M2 will be in g mol“^.

27. Isotonic solutions. These are the solutions having same osmotic pressure which is so when they have same
molar concentrations.
our
ad

28. Hypotonic and hypertonic solutions. A solution having lower osmotic pressure or lower molar concentration
than the other is called hypotonic while the one with higher molar concentration and hence higher osmotic
pressure is called hypertonic. Human RBC’s are isotonic with 0*9% solution of NaCl. A solution of NaCl
dY

with concentration < 0*9% is hypotonic with respect to RBC’s. When placed in this solution, RBC’s will
Re

swell and may burst. A solution of NaCl with concentration > 0*9% is hypertonic. In this solution, RBC’s
Fin

will shrink, called plasmolysis.

29. Elevation of boiling point. Solution has lower vapour pressure and hence higher boiling point than pure
solvent. The increase, AT^, = T^ - T^,® is called elevation of boiling point.
30. Relationship between ATj and molality. It is found that AT^ »= m (molality of the solution) or AT^ =
xm
where is called molal elevation constant or ebullioscopic constant. When m = 1, AT^ = K^. Hence,
molal elevation constant is the elevation in boiling point when molality a solution = 1 mol kg“^
Units of = K kg mol"*.
31. Calculation of molar mass from AT^. Putting
1000 K^W2
m = —— X — X1000 in AT^ = x m, we get M2 =
M2 w,1 AT^xWj
32. Depression of freezing point Freezing point of a solution is less than that of the pure solvent. The decrease.

ATy = is called depression of freezing point. Just as in case of elevation in boiling point,
ATy = Kyx m where Kyis called molal depression constant or cryoscopic constant. When m = 1, Ky = ATy.
Units of Ky = K kg mol-l.
2/78 New Course Chemistry (XIl)BZsl91

1000 VV2
Calculation of molar mass from AT^^. Similar to AT^, we have M, =
AT^ X M’j
Rast method. This method is used for compounds soluble in camphor on melting. Camphor is used because
it has high Ky viz 39-7°. Lowering of melting point of the solid solution is found. Then the above formula is
used to calculate M2.
Applications of depression of freezing point : (/) In making antifreeze solutions, e.g., adding ethylene
glycol to water in car radiator at hill stations (ii) In melting of ice on roads by adding NaCl or CaCl2-
Abnormal molecular masses. For substances undergoing association, dissociation etc. in the solution,
molecular mass determined from colligative properties is found to be different from expected value. These
are called abnormal molecular masses.

- Van’t Hoff factor (f)

Observed colligative property Calculated molar mass (M^.)

w
Calculated colligative property Observed molar mass (M^)
If solute undergoes dissociation, / > 1 while for association, / < 1.

F lo
J .
Expressions for colligative properties when association/dissocia tion occurs.
P°-P _ .●

ee
Ji = / - RT
= ix2> AT^ = /K^w, ATy=/Krm, V
P

Fr
QUESTIONS

for
ur
Based on NCERT Book

I. Multiple Choice Questions (a) Low temperature

s
ook

(b) Low atmospheric pressure

Yo
1. What is the mole fraction of benzene in solution
containing 30% by mass in carbon tetrachloride ? (c) High atmospheric pressure
eB

(a) 0-541 ib) 0-459 (cf) Both low temperature and high atmospheric
pressure
(c) 0-514 id) 0-489
7. The value of Henry’s constant is
r

2. An antifreeze solution is prepared from 222-6 g of

ou
ad

ethylene glycol and 200 g of water. What is the (a) greater for gases with higher solubility
molality of tlie solution ? (b) greater for gases with lower solubility
Y

(a) 15-97 m ib) 19-57 m (c) constant for all gases

Re
nd

(c) 17-07 m id) 17-95 m (d) not related to the solubility of gases
3. Calculate the amount of benzoic acid required to 8. Which of the following factors affect the solubility
Fi

prepare 250 mL of 0-15 M solution in methanol of gaseous stale in the fixed volume of the liquid
solvent ?
ia) 4-575 g (h) 5-475 g
(i) Nature of the solute
(c) 4-015 g (d) 5-015g
(ii) Temperature
4. The molarity of 900 g of water is
(///) Pressure
ia) SOM ib) 55-5 M
(n) (0 and (/«) at constant T
(c) 5M {d) cannot be calculated
5. Maximum amount of solid solute that can be
ib) iii) only
(c) (//■) and iiii) only
dissolved in specified amount of given liquid
.solvent does not depend upon id) iiii) only
(a) Temperature ib) Nature of solute 9. The concentration of solution with its vapour
(c) Pressure id) Nature of solvent pressure is related in terms of
ia) mole fraction ib) parts per million
6. Low concentration of oxygen in tissues of people
living at high altitude is due to (c) mass percentage id) molality
SOLUTIONS 2/79

10. Vapour pressure of water at 293 K is 17-535 mm (a) 128 (b) 488
Hg. Calculate the vapour pressure of water at 293 K (c) 32 (d) 64
when 25 g of glucose is dissolved in 450 g of 17. The mass of glucose that should be dissolved in
water
50 g of water in order to produce the same lowering
(a) 15-44 mm Hg (b) 15-43 mm Hg of vapour pressure as produced by dissolving 1 g
(c) 17-44 mm Hg (d) 17-04 mm Hg of urea in the same quantity of water is
11. If two substances A and B have =1:2 (a) 1 g ib) 3g
and have mole fraction in solution 1 :2, then mole ic) 6g id) 8g
fraction of A in vapours is 18. The porous membrane used in Reverse osmosis
(a) 0-33 ib) 0-25 plant is made up by
(c) 0-52 (d) 0-2 (a) Cellulose acetate
12. One component of a solution follows Raoult’s (b) Potassium nitrate

w
law over the entire range 0 < x, < 1. The second (c) Mercuric iodide
component must follow Raoult’s law in the range (d) Starch
when X2 is

F lo
19. An unripen mango placed in a concentrated salt
(a) close to zero (b) close to 1
solution to prepare pickel shrivels because
(c) 0<;c2^0-5 id) 0<JC2< 1 (fl) of endosmosis

ee
13. Which of the following is incorrect for ideal (b) it loses water due to reverse osmosis

Fr
solution ?
(c) it gains water due to reverse osmosis
(«) ^xing = 0
(^) AV^xmg = 0 for
(d) it loses water due to osmosis
ur
20. At a certain temperature, the value of the slope of
(c) does not obey Raoult’s law over entire range the plot of osmotic pressure (ic) against concen
of concentration
s
tration (C in mol L“') of a certain polymer
ook
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(d) A - A, B - B interactions are equal to A - B solution is 29IR. The temperature at which
interactions
osmotic pressure is measured is (R is gas
eB

14. 1 mole of liquid A and 2 moles of liquid B make a constant)

solution having a total vapour pressure of 38 torr. ia) 27 rC ib) 18‘’C
our

The vapour pressures of pure A and pure B are

ad

ic) 564 K id) 18 K

45 torr and 36 torr respectively. The described
solution 21. The unit of ebuUioscopic constant is
-1 -1
id) K kg mol ib) mol kg K
Y

(a) is an ideal solution

ic) kg mol"' K"' id) K mol kg
-]
Re

(b) shows negative deviation

nd

22. 18 g of glucose is dissolved in 1 kg of water. At

(c) is a minimum boiling azeotrope
Fi

what temperature will water boil at 1-013 bar ?

(d) has volume greater than the sum of individual for water is 0-52 K kg mol
-I
volumes
ia) 373-2 K ib) 378-2 K
15. Colligative properties depend upon ic) 381-5 K id) 383-2 K
(a) Nature of the solute particles dissolved in
solution
23. When mercuric iodide is added to the aqueous
solution of potassium iodide
(b) Number of particles of the solute in solution
ia) freezing point is raised
(c) Physical properties of the solute particles ib) freezing point is lowered
dissolved in solution
(c) freezing point does not change
(d) Nature of the solvent particles
id) boiling point does not change
16. The vapour pressure of acetone at 20®C is 185 torr.
24. If molality of the dilute solution is doubled, the
When 1-2 g of a non-volatile was dissolved in
100 g of acetone at 20°C, its vapour pressure was value of the molal depression constant (Ky) will be
183 torr. The molar mass (g mol"') of the (a) doubled ib) halved
substance is ic) tripled id) unchanged
2/80 New Course Chemistry (XII)tEIg

25. Which of the following aqueous solution has the (a) Molecular mass of X is less than molecular
highest boiling point ? mass of Y

ia) 0-1 M KNO3 (h) 01 M Na3P04 (h) Y is undergoing dissociation in water while X
(c) 0-1 MBaCl2 (d) O-I M K2SO4 undergoes no change
26. In comparison to 0 01 M solution of glucose, the (c) X is undergoing dissociation in water
depression in freezing point of 0-01 M MgCl2 (d) Molecular mass of X is greater than the
solution is molecular mass of Y

(a) same (b) twice 34, The van’t Hoff factor for BaCl2 at 0 01 M
(c) three times (d) six times concentration is 1-98. The percentage

ow
dissociation of BaCl2 at this concentration is
27. Solution A contains 7 g/L MgCl2 and solution B
contains 7 g/L of NaCl. At room temperature, the (a) 49 (b) 69
osmotic pressure of (c) 89 (d) 98
(a) solution A is greater than B (e) 100.

e
{b) both have same osmotic pressure 35. The van’t Hoff factor (/) for a dilute aqueous

re
(c) solution B is greater than A solution of the strong electrolyte barium

Frl
(d) can’t determine.
hydroxide is

F
(a)0 ib)\
28. van’t Hoff factors x, v and z for association.
dissociation and no change of solute in the (c)2 (^0 3
ou
36. Which of the following pairs will not form an ideal

or
solution respectively are in the order :
solution ?
(a) x<y <z ib) x>z>y
(c) x<z<y id) x>y>z kfs
(a) Benzene and Toluene
(£>) n-Hexane and n-Heptane
29. Which one of the following electrolytes has the
oo
same value of van’t Hoff factor (/) as that of (c) Ethanol and acetone
Y
A1i(S04)3 (if all are 100% ionized) (d) Bromoethane and Chloroethane (CBSE 2022)
B

(a) K2SO4 ib) K3[Fe(CN)6l 37. An azeotropic solution of two liquids has a boiling
(c) A1(N03)3 id) K4[Fe(CN)6l point higher than either of the two when it:
re

30. Of the following0-10 ni aqueous solutions, which (a) shows a negative deviation from Raoult’s law
oYu

one will exhibit the largest freezing point ib) shows a positive deviation from Raoult’s law
ad

depression ? (c) is saturated

ia) KCl ib) C6H,206
d

id) shows no deviation from Raoult’s law

(c) Al2(S04)3 id) K2SO4 (CBSE 2022)
in
Re

31. The molar mass of the solute sodium hydroxide 38. Pressure does not have any significant effect on
obtained from the measurement of osmotic
F

solubility of solids in liquids because :

pressure of its aqueous solution at 27“C is 25 g
(a) Solids are highly compressible
mol"'. Therefore, its ionization percentage in this
solution is ib) Liquids are highly compressible
ia) 75 ib) 60 (c) Solubility of solid in liquid is directly
proportional to its partial pressure
(c) 80 id) 70
32. 0-6 mL of acetic acid is dissolved in 1 litre of id) Solids and liquids are highly incompressible
water. The value of viin’t Hoff factor is 104. (CBSE 2022)
What will be the degree of dissociation of the 39. Vapour pressure of dilute aqueous solution of
acetic acid ? glucose is 750 mm Hg at 373 K. The mole fraction
of the solute is :
ia) 0-01 ib) 0-02
(c) 0-03 id) 0-04 I 1
ia) ib) —
33. The boiling point of 0-2 mol kg"' solution of X in 7-6 38

water is greater than equimolal solution of Y in 1 1

water. Which one of the following statements is ic) id) — (CBSE 2022)
76 10
true in this case ?
 SOLUTIONS
2/81

40. The freezing point of a 0*2 molal solution of anon- (a) 100 g mol
-I
(b) 50 g mol
-I

electrolyte in water is (K. for water = I -86 K kg (c) 25 g mol

-1
(d) 75 g mor’
mol"^)
(('BSE 2022)
(a) - 0-372T (b)- T86°C
(c) + 0-372°C
47. To increase the solubility of CO2 gas in soft drinks,
{(I) + I -86°C the bottle is sealed under
(CBSE 2022)
(fl) Low pressure (b) High temperature
41. Increasing the temperature of an aqueous solution (c) Constant pressure (d) High pressure
will cause
((’BSE 2022)
(a) Increase in Molarity
48. A solution of a pair of volatile liquids A and B
(b) Increase in Molality shows negative deviation from Raoult’s law. This

ow
(c) Decrease in Molarity is because

(d) Decrease in Molality (CBSE 2022)

(fl) and /»3 > p-Q Xq
42. Which of the following conditions is correct for
an ideal solution ? (b) The intermolecular forces A-A, B-B<A-B

e
= 0 and = 0 (c) Both AH^i^i„g and AV„^i,j„g are positive

re
rFl
(d) All of the above (CBSl’. 2022)
(b) > 0 and AV^^ > 0
49. Which of the following analogies is correct ?

F
(c) AH^i^ < 0 and AV^j^ < 0
(a) Chloroform - acetone : Positive deviation ::
(d) AH^i, > 0 and AV„^;^ < 0 (CBSE 2022)
Ethanol - H2O : Negative deviation

r
43. For determination of molar mass of polymers and
ou
fo
proteins which colligative property is used ? (h) Pp^ > Pp^ x^ ; Henry’s law :: p =
ks ■ x :
(a) Relative lowering in vapour pressure Raoult’s law

ib) Elevation in boiling point (c) Pjota] = Pa Pb ● Non-ideal solution ::

oo

(c) Osmotic pressure PjQfal > Pa Pb ● solution

Y
eB

(d) Depression in freezing point (CBSE 2022) (d) K = CRT : Osmotic pressure :: P > tc : Reverse
osmosis (CBSE 2022)
44. Pure water boils at 373 15 K and nitric acid boils
at 359-15 K. An azeotropic mixture of H^O and 50. Which one of the following pairs will form an ideal
r

solution ?
ou

HNO3 boils at 393-55 K. Distilling the azeotropic

ad
Y

mixture will cause (a) Chloroform and acetone

(a) Pure nitric acid to distil over first (b) Ethanol and acetone
d

(b) Pure water to distil over first (c) n-hexane and n-heptane
Re
in

(c) One of them to distil over with a small amount (d) Phenol and aniline (CBSE 2022)
of the other
51. Which of the following formula represents Raoult’s
F

(d) Both of them to distil over in the same law for a solution containing non-vilatile solute ?
composition as that of the mixture being
distilled (CBSE 2022) /^soluie ~ /^solute ^solute

45. A 5% (by mas.s) solution of glucose (molar mass = W P = Kh ■

180g mol"*) is isotonic with 1% solution (by mass) ^Total ~ ^Solvent
of a substance ‘X’. The molar mass of ’X’ is
/^solute “ ^solveni ’^solveni ((’BSE 2022)
-]
(a) 36 g mol (h) 18 g mol
(c) 72 g mo!
-1
(d) 900 g mo!“’ 52. An azeotropic solution of two liquids has a boiling
point lower than either of the two when it
(CBSE 2022)
(ci) shows a positive deviation from Raoult’s law
46. When 2-5 g of a non-volatile solute was dissolved
(b) shows a negative deviation from Raouit’s law
in 50 mL of water, it gave boiling point elevation
of 0-52°C. The molar mass of the solute is (c) shows no deviation from Raoult’s law

(K^ for water = 0-52 K m"^) id) is saturated (CBSE 202

2/82 New Course Chemistry (X1I)ES191

53. On mixing 20 mL of acetone with 30 niL of II. Assertion-Reason Type Questions

chloroform, the total volume of the solution is
In the questions given below, two statements are
(a) < 50 mL (b) = 50 niL
given one labelled Assertion (A) and the other
(c) > 50 inL id) = 10 mL labelled Reason (R Select the correct answer
(CBSE 2022) to these questions from the codes (a), (6), (c) and
54. An unknown gas ‘X’ is dissolved in water at 2-5 {d) given below :
bar pressure and has mole fraction 0 04 in the (a) Both A and R are correct and R is the correct
solution. The mole fraction of ‘X’ gas when the explanation of A.
pressure on the gas is doubled at the same (h) Both A and R are correct but R is not the correct
temperature is explanation of A.
(a) 0-08 (b) 0 04
(c) A is correct but R is wrong.
(c) 0-02 id) 0-92 (CBSE 2022)
id) A is wrong but R is correct.

w
55. In the adjoining diagram, ‘X’ represents 59. Assertion. Henry’s law constant is the ratio of
pressure of the gas to the mole fraction of the gas

Flo
in the solution.

Reason. Henry’s law constant for a particular gas

ee
in a particular solvent is constant and does not

Fr
depend upon temperature.
Q.
60. Assertion. A mixture of chloroform and acetone
forms a non-ideal solution with positive deviation.
O
a

for
ur
>
Reason. Acetone is more volatile than chloroform.
> Temperature
61. Assertion. In reverse osmosis, pressure higher than
s
(a) Boiling point of the solute osmotic pressure has to be applied.
ok
Yo
ib) Freezing point of the solute Reason. Higher pressure pushes the solution from
Bo

(c) Boiling point of solvent one side of the semipermeable membrane to the
id) Freezing point of solution (CBSE 2022) other side.
re

56. identify the law which is stated as: 62. Assertion. 0-1 M HCl solution has higher osmotic
“For any solution, the partial vapour pressure of pressure than 0-1 M NaCl solution
ou
ad

each volatile component in the solution is directly Reason. Cr ions being common, the small size
proportional to its mole fraction.” H'*’ ions have greater ionic mobility than large size
Y

Na'*’ ions.
(a) Henry’s law ib) Raoult’s law
63. Assertion. Out of various colligative properties,
nd

(c) Dalton’s law id) Gay-Lussac’s law

Re

(CBSE Sample Paper 202L2022) osmotic pressure is used for determination of

Fi

molecular masses of polymers.

57. Solubility of gases in liquids decreases with rise
Reason. Polymer solutions do not possess
in temperature because dissolution is an
constant boiling point or freezing point.
(a) endothermic and reversible process
64. Assertion. Ebullioscopic constant is constant for
ib) exothermic and reversible process
a particular solvent at a particular temperature and
(c) endothermic and irreversible process does not depend upon the solute.
id) exothermic and irreversible process Reason. Elevation in boiling point depends upon
(CBSE Sample Paper 2021-2022) the mass of the solute dissolved in definite mass
58. How much ethyl alcohol must be added to 1 litre of a particular solvent and is independent of the
of water so that the solution will freeze at - 14‘^C ? nature of the solute and solvent.

(Kyfor water = l-86°C/mol) 65. Assertion. Cooking time is reduced in pressure

ia) 7-5 mol ib) 8-5 mol cooker.

(c) 9-5 mol id) 10-5 mol Reason. Boiling point inside the pressure cooker
(CBSE Sample Paper 2021-2022) is raised.
SOLUTIONS 2/83

66. Assertion. Azeotropic mixtures are formed only Reason. Relative lowering in vapour pressure
by non-ideal solutions and they may have boiling depends upon mole fraction of pure solvent.
points either greater than or less than both the (CBSE 202!
components.
71. Assertion. A raw mango placed in a saline solution
Reason. The composition of the vapour phase is
loses water and shrivels into pickle.
same as that of liquid phase of the azeotropic
mixture. Reason. Through the process of reverse osmosis,
raw mangoes shrivel into pickle. (CBSE 20211
67. Assertion. Increasing pressure on pure water
decreases its freezing point. 72. Assertion. Molarity of a solution changes with

ow
temperature.
Reason. Density of water is maximum at 273 K.
68. Assertion. For NaCI solution in water, for the same
Reason. Molarity is a colligative property.
molality, the elevation in boiling point is equal to 73. Assertion. Cryo.scopic constant depends upon the
nature of the solvent.
the depression in freezing point.

e
Reason, van’t Hoff factor/ is fixed forNaCl viz. 2. Reason. Cryoscopic constant is a universal

re
constant.
69. Assertion, van't Hoff factor for benzoic acid in

Frl
F
benzene is less than one. 74. Assertion. One molar barium chloride solution

Reason. Benzoic acid behaves as a weak gives double the elevation in boiling point than
one molar sodium chloride solution.
electrolyte in benzene.
ou
sor
Reason. ions carry double the charge than
70. Assertion. Relative lowering in vapour pressure Na'*' ions.
is a colligaiive property.

ANSWERS kf
oo
I. Multiple Choice Questions
Y
B

I. ib) 2. id) 3. ia) 4. (b) 5. ic) 6. ib) 7. ib) 8. ia) 9. ia) 10.ic)
11. id) 12. id) 13.(c) 14. ib) 15. ib) 16. id) 17. ib) 18.ia) 19. id) 20. ib)
re

21. (fl) 22.(a) 23. («) 24. id) 25. ib) 26.ic) 27.ic) 28. ic) 29. id) 30. (c)
oY
u

31. ib) 32. (^/) 33. (c) 34.ia) 35. id) 36. ic) 37. ia) 38. id) 39. ic) 40. ia)
ad

41. ic) 42. ia) 43.ic) 44. id) 45. (£/) 46. ib) 47. id) 48.ib) 49. id) 50. ic)
51. ic) 52. ia)
d

53. (a) 54. ia) 55. id) 56. ib) 57. ib) 58.ia)
in

II. Assertion-Reason Type Questions

Re

59. ic) 60. id) 61. ic) 62. id) 63. ib) 64. ic) 65.ia) 66. ib) 67. ic) 68. id)
F

69.ic) 70. ic) 71. ic) 72. ic) 73. ic) 74. (^0

For Difficult Questions

I. Multiple Choice Questions 50

Molarity = X1000 = 55-5 M
900
4. Molarity is number of moles per litre
11. In solution, if = x, x^ = 2x
900
900 g H2O = moles = 50 moles and if Pa" = P’ Pb° = 2p
18
PA=xxp, p^ = 2xx2p
900 g H2O = 900 inL of H2O = 4xx p
2/84 'pfuutee^ New Course Chemistry (XII) EZ

Ptotal = 5-tP- Thus, number of ions decreases in the solution.

Mole fraction in vapour phase Hence, depression in freezing point is less, i.e.,
actual freezing point is raised.
=^= = 1 = 0.2 24. Molal depression constant (Ky) depends only on
^ P,oal 5 the solvent and is independent of the molality of
12. When one component follows Raoult’s law, the the solution.
solution is ideal. The second component will also
follow Raoult’s law at all concentrations. 25. Higher the concentration of ions in the solution,
14. According to Raoult’s law, higher is the elevation in boiling point. As 0-1 M
Na3P04 has highest concentration of ions,
X p" = ix45torr =15 torr elevation of boiling point is highest for 0*1 M
Na3P04 and so is the boiling point. Alternatively,

w
2
Pb~^b^ Pb = -><36torr = 24 torr AT^ = I m. For the same value of m, higher is
the value of /, higher is AT^.
Pressure expected by Raoult’s law

Flo
27.7gL-'MgCl2 = ^j^molL->
= 15 + 24 = 39 torr

e
Thus, observed pressure (38 torr) is less than

re
expectedvalue. 7x3
= —molL-‘ = molL ^ of ions

F
Hence, the solution shows negative deviation. 95
95
P°-Ps = 2. = ^
n
ur = 0-22 M

r
16.
/i| Wj/M
185-183 _1-2/M2 fo
7gL-‘ NaCl = M
ks
^ 23 + 35-5
Yo
185 100/58
oo

(Molar mass of CH3COCH3 = 58 g mol"*) 7 7x2

M = molL * of ions
B

1-2x58 1-2x58 18^5 58-5 58-5

or M2 = 100 2
IOOXM2
re

= 0-24 M
-1
= 64-38 g mol
u

As concentration of ions in NaCl solution is

ad
Yo

17. P’^-Ps ^ ”2 ^ ^2^1 greater, NaCl solution (solution B) will have

p° nj w^ M2 greater osmotic pressure.
d

As (p® - Pj)/p” is same in the two cases 28. For association, / < 1, For dissociation, / > 1.
Re
in

For no change, i = 1. Hence, order is jr < z < y.

F

w,1 M
W,M2^ 29.Al2(S04)3 and K4[Fe(CN)6l produce the same
2 y glucose urea
number of ions in the solution (viz. 5). Hence,
W2XI8 _ 1x18 they have the same value of van’t Hoff factor.
or W2 = 3g.
50x180 "50x60
30. ATy = I X X m. As Ky and m are same for all
20. n = CRT. Thus, a plot of Jt vs C will be linear with given solutions
slope = RT. Hence,
RT = 291 R or T = 291 K = (291 - 273)*’C
ATy^oc i and i for KCl = 2 ; for CgHj20g, / = 1 ; for
Al2(S04)3, /■ = 5 and for K2SO4, i = 3.
= 18‘’C.
Hence, Al2(S04)3 will show largest freezing point
23. KI reacts with Hgl2 to form a complex as depression.
ionizes
2KI + Hgl2^K2[Hgl4] » 2K+ + Hgl2-
(2K++2I-) 3 ions 31. i = ^c_f0 = i.6
M 25
4 ions 0
SOLUTIONS 2/85

NaOH ^ Na-*- + OH- 1 1

1 mol 0 0 Molar cone, of X = X xlOOO

1 - a a a Total = 1 + a
Mx 100

i = I + a or a = /-i =0-6 = 60% 10

molL *
32. M
CH3COOH V ^ CUjCOO-+H-^ X
1 mole 0 0
10
1 -a a a
M
—
18
or M X = 36 g mol ’
X
Total = I-f a
t = 1 -f a 1000 X X M', 1000x0-52x2-5
or a = / - 1 = 1 -04 - 1 = 004 46. M2 = 0-52 X 50
AT^x«.
33. For equimolal solutions of two substances, the
= 50 g mol”^ (50 ml of water = 50 g of water)
one that undergoes dissociation produces greater
number of particles (ions) in the solution and 51. As non-volatile solute does not form vapour, total
hence has greater boiling point. vapour pressure of solution will be due to solvent
in the solution.
34. BaCl, V ^ Ba-* -t- 2C1-

w
Initial 1 mole 0 0
53. Acetone-chloroform mixture shows negative
deviation from Raoult’s law therefore,

F lo
After disso. 1 - a a 2 a,
= - ve, i.e., AVjqjjji < 50 mL.
Total no. of particles = 1 -i- 2 a
55. As vapour pressure of solvent is higher than that
f = 1 -f 2 a
of solution, the upper curve is for the solvent. The

ree
i -1 1-98 - 1 0-98 point X lies in the solid solvent curve. The lower
or a = = 0-49 = 49%.
2 2

35. As Ba(OH)2 is a strong electrolyte, it dissociates

2
for F
solution curve meets the solid solvent curve at X.

Hence, X represents the freezing point of the

solution.
completely as
Ba(OH)2 ^ Ba^-^ -H 2 OH 1000 xK^ xm'2
Your
58. AT
f ~
ks

Ions produced = 3. Hence, van’t Hoff factor = 3 M, X u’j

eBoo

39.^ -P .1 "2
= ^2, i.e..
760-750

760
= x^ 14 =
1000 X1-86 xw^ 1000 X1-86 X ri2
n^+ri2 M2 XlOOO 1000
ad
our

10 1
or
■^2 = 14
—^ = n-
760 76 n = 7-5 mol
M, 2
or
2 “
1-86
40. ATy-=Kyxm= 1-86 x0-2 = 0-372
Re

T^=0-0-372 = -0-372“C II. Assertion-Reason Type Questions

Y

41. On increasing the temperature of an aqueous

Find

59. Correct R. Henry’s law constant depends upon

solution, volume of the solution increases but
temperature.
number of moles of the solute dissolved remains
60. Correct A. A mixture of chloroform and acetone
the same. Hence, molarity decreases.
forms a non-ideal solution with negative deviation.
44. On distilling an azeotropic mixture, both the liquids
61. Correct R. On applying higher pressure on the
distil over in the same composition. solution, the solvent will flow from solution into
45. For isotonic solutions
the pure solvent through the semipermeable
Molar concentration of glucose membrane.

= Molar concentration of X 62. Correct A. 0-1 M HCl and 0-1 M NaCl will have

1 the same osmotic pressure because HCl and NaCl

Molar cone, of glucose = x XlOOO dissociate to produce two particles each.
180 100
63. Correct explanation. Due to high molar masses
— mol L ' (Molar mass of glucose of polymers, the elevation in boiling point or
18 depression in freezing point are so low that they
= 180 g mo! ') cannot be measured accurately.
2/86 ^'kuCcc^'a New Course Chemistry CXll)EZsX9]

64. Correct R. It depends upon the nature of the 70. Both A and R are correct and R is the correct
explanation of A.
solvent because K^or values depend upon the
solvent. 71. Correct R. Relative lowering of vapour pressure
66. Correct explanation. The boiling point of the depends upon the mole fraction of the solute in
the solution.
azeotropic mixture may be greater or less
72. Correct R. Through the process of osmosis, raw
depending upon whether solution shows negative
mangoes shrivel into pickle.
or positive deviation from Raoult’s law.
73. Correct R. Molarity is not a colligative property.
67. Correct R. Density of water is maximum at 277 K
74. Correct R. Cryoscopic constant is not a universal
(4"C).
constant. Its value depends upon the solvent.
68. Correct A. AT^ = / K/, m, AT^= i Kj-m. Although / 75. Correct A. BaCU dissociates to produce 3 ions
is same but K^- and are not equal and hence (Ba-'*' + 2 Cr) whereas NaCl dissociates to give

ow
AT,^AT^.. 2 ions fNa"^ + Cl"). Hence, elevation in boiling
69. Correct R. Benzoic acid behaves as a weak point of BaCl2 solution will be 1-5 times that of
electrolyte in water. NaCl.

e
Fl
re
F
1. Types of solutions and Volume of water required to be added
ur
r
expressing their concentration M,Vj

fo
= V2-V,= -V,1
1. When and why is molality preferred over M2
ks
molarity in handling solutions in chemistry ? M 'M,-M2
Yo
1
Ans. Molality is preferred when studies are made -1 V, = V1
oo

M2 M,^
independent of temperature. This is because /
4. How Is the molality of a solution different
eB

molality involves masses which do not change

with temperature. from its molarity ?
2. Which solution has higher concentration, Ans. Molality of a solution is the number of moles
ur

1 molar or 1 molal solution of the same of the solute present in 1 kg of the solvent and
does not change with temperature. Molarity of
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solute? Give reason.

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a solution is the number of moles of the solute

Ans. In aqueous solution, IM solution means 1 mole
of the solute in 1000 cc of the solution whereas present in I litre of the solution and changes
with temperature.
d

1 m solution means 1 mole of the solute in 1000

Re
in

g of water (= 1000 cc of water). Total volume II. Solutions of solids in

of 1 m solution >1000 cc due to presence of liquids and solids in solids
F

extra 1 mole of the solute. Hence, number of

moles/cc in 1 m solution will be less than that 5. Why the solubility of Glauber’s salt
in 1 M solution. In other words, 1 M is more (Na2SO4.10H2O) first increases upto 32‘4’’C
concentrated than I m. and then decreases ?

In non-aqueous solution, 1 M can be greater Ans. Upto 32-4°C, Glauber’s salt remains in the
than, less than or equal to 1 m, depending upon hydrated form whose dissolution is
the density of the solution. endothermic. Hence, its solubility increases with
3. Vj cc of solution having molarity Mj is temperature. Beyond 32-4°C, it changes into
diluted to have molarity M2- Derive anhydrous salt whose dissolution is exothermic.
expression (in terms of Mj, M2 and Vj) for Hence, solubility decreases with increase of
the volume of water required to be added. temperature.
Ans. Suppose the final volume after dilution is V2 6. How are substitutional solids formed ?

MiVi Ans. Substitutional solids are formed when some

Then M]V,=M2V2 or Vj = atoms, ions or molecules of one solid
M2
SOLUTIONS 2/87

occupying the lattice site are replaced by Mole fraction of the volatile liquid.
atoms, ions or molecules of another solid
^1= ’ X2
having similar nature imd nearly same size. or
.^2= I -
in. Solution of gases in liquids _ P.
= 1 - ATj or .r 1
7. How is the solubility of gases in water related P° P°
with their Henry’s constants at the same 13. Derive a relationship between mole fraction
pressure and temperature? and vapour pressure of a component of an
Ans. According to Henry’s law -v^. At ideal solution in the liquid phase and vapour
constant partial pressure, greater the of the phase.
gas, smaller is , i.e., less is the solubility. Ans. Refer to Supplement Your Knowledge, point 2.
8. How does Henry’s constant (K[|) of a gas in page 2/26.
a particular solvent vary with temperature?
V. Ideal and Non-ideal

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Ans. According lo Henry’s law x^. With
increase of temperature, solubility of a gas solutioas and azeotropic mixtures
decreases at the same partial pressure, i.e., 14. Two liquids A and B on mixing produce a

F lo
decreases. Hence, at constant p^, increases. warm solution. Which type of deviationfrom
Thus, increases with increase of temperature. Raoult’s law does it show ?

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9. Why oxygen mixed with helium is used by Ans. Warming up of the solution means that the

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deep sea divers ? process of mixing is exothermic, i.e., AHj^^j„g
Ans. Refer to Application (ii), page 2/19. = -ve. This implies that the solution shows a
10. Give reasons for the following :
for
negative deviation.
ur
(a) Aquatic species are more comfortable in 15. Why does a solution of ethanol and
cold water than warm water. cyclohexane show positive deviation from
s
Raoult’s law ?
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(b) At higher altitudes, people suffer from

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anoxia resulting in inability to think. Ans. On adding cyclohexane, its molecules get
inbetween the molecules of ethanol thus
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(CBSE 2019)
breaking the hydrogen bonds present between
Ans. (fl) Refer to Ans. to Curiosity question on page ethanol molecules and reducing ethanol-ethanol
2/22.
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interactions.
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(b) At high altitude, partial pressure of oxygen

in the air is low. As a result there is low
VI. Relative lowering
of vapour pres.sure
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concentration of oxygen in the blood and

tissues.
16. 2 g each of the solutes A and B (Mol mass of
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nd

IV. Vapour pressure of A > B) are dissolved separately in 20 g each

of the same solvent C. Which will show
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liquid solutions and Raoult’s law

greater lowering of vapour pressure and
11. Why does vapour pressure of a liquid why ?
decrease when a non-volatile solute is added Ans. For dilute solutions,
into it ? (UP Board 2012)
P°-Ps _ ^^2^1 Ap _ h'2M|
Ans. Some liquid molecules at the surface are or

replaced by the molecules of the solute which

P^ tv, M., M2
are non-volatile. For same solvent, p° = constant, Mj = constant.
12. Derive the relationship between relative Also, given that
lowering of vapour pressure and mole W2 = Wa = Wq = 2 g, vvj - Wq = 20 g.
fraction of the volatile liquid.
Hence, . As Ma > Mg, therefore,
P°-Ps ^Pb Ma
Ans. According to Raoult’s law, = (mole
Apg > A Pa* ® show greater lowering
fraction of the non-volatile solute) of vapour pressure.
2/88 New Course Chemistry (XII)ESSm

The solution has to be healed more to make

VII. Osmotic prcs.sure
vapour pressure equal to the external pressure.
17. After removing the outer shell of two eggs in Hence, boiling point is increased.
dil. HCI, one is placed in distilled water and
the other is placed in a saturated solution of
22. Two liquids A and B boil at 145"C and 190X
NaCI. What will you observe and why ? respectively. Which of them has higher
or
vapour pressure at 80"C ?
(Uttarakhand Board 2012)
A peeled egg swells when dipped in water
while shrinks in saturated brine solution, Ans. Liquid A with lower boiling point is more
Why? (lltlarukhand Board 2012) volatile and hence will have higher vapour
Ans. Egg in water will swell while egg in NaCl pressure.
solution will shrink. This is because as a result 23. What freezes out first when a solution of
of osmosis, the net flow of solvent is from less common salt is cooled ?

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concentrated to more concentrated solution.
Ans. Water as ice.
18. What do you expect to happen when Red
Blood Corpuscles (RBC’s) are placed in 24. What is de-icing agent ? How does it work ?

F lo
(/) 1 % NaCI solution Ans. Common salt is called de-icing agent because
(n) 0-5% NaCI solution ? it lowers the freezing point of water to such an

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extent that it does not freeze to form ice. Hence,
Ans. (/) They will shrink due to plasmolysis,

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it is used to clear snow from roads.
(ii) They will swell and may even burst.
This is because RBC’s are isotonic with 0-9% IX. Abnormal molar masses
NaCI solution.
for
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25. Why is freezing point depression of O’l M
VIII. Elevation of boiling point sodium chloride solution nearly twice that of
0*1 M glucose solution ?
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and depression of freezing point
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Ans. NaCI, being an electrolytes, dissociates almost
19. What will happen to the boiling point of a completely to give Na'*' and Cr ions whereas
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solution if the weight of the solute dissolved glucose, being non-electrolyte, does not
is doubled but the weight of solvent taken is dissociate. Hence, the number of particles in O-I
halved ? M NaCI solution is nearly double than in 01 M
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glucose solution. Freezing point depression,

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Ans. Elevation in boiling point will be four times being a colligative property, is therefore, nearly
1000K^;v2 twice for NaCI solution than for glucose solution
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because AT^^ = of same molarity.

M., >V|
26. Explain why equimolar aqueous solutions of
Re
nd

20. Why is camphor preferred as a solvent in sodium chloride and sodium sulphate are not
Fi

i.sotonic ? (Manipur Board 2011)

finding the molecular mass of naphthalene
by Rust method ? Ans. NaCI dissociates to give 2 ions (Na'*' and Cl ).
or Why is camphor preferred in the Na2S04 dissociates to give 3 ions (2 Na"^ and
determination of ATy ? $04“). Thus, equimolar .solutions of NaCI and
Ans. Camphor has a large value of Ky viz, 39-TC so Na2S04 have different concentrations of ions
that depression in melting point is large for a in the solution. As osmotic pressure depends
solution of naphthalene in camphor. upon concentration of particles in the solution,
they have different osmotic pressures.
21. Why boiling point of water is increased on
addition of sodium chloride into it ? 27. The experimentally determined molar mass
for what type of substances is always lower
Ans. When NaCI (or any other non-volatile solute) than the true value when water is used as
is dissolved in water, vapour pressure of water solvent. Explain. Give one example of a such
decreases (because some of the solvent a substance and one example of a substance
molecules on the surface are replaced by the which does not show a large variation from
molecules of the solute which are non-volatile). the true value. (CBSE Sample Paper 2019)
SOLUTIONS 2/89

Ans. In dilute solutions when a substance undergoes X. Miscelianeous

dissociation, number of particles of the solute
increases. As colligative properly depends upon 28. Why melting point of a substance is used as
the number of particles, the value of colligative a criterion for testing the purity of the
property will be higher and as molar mass is substance.
inversely proportional to value of colligative
property, molar mass will be lower, e.g., KCl. Ans. A pure compound has a sharp melting point.
Impurities present, if any, lower the melting
For substances like glucose which do not
dissociate or associate in the solution, does not point of the compound (just as depression in
show a large variation from true value. freezing point takes place).

● Very Short Answer

riic

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1 1 ● Short Answer
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● Long Answer

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VERY SHORT ANSWER QUESTIONS Carrying 1 mark

e
re
I. Types uf solutions and Ans. Molality of a solution involves masses of the

rF
expressing their concentration solute and solvent which do not change with
temperature.
1. How is it that alcohol (ethoxyethane) and
ur
9. Calculate the molality of H2SO4 if the density
water are miscible in all proportions ?
(CBSK 2007) fo
of 10% (w/w) aqueous solution of H2SO4 is
1-84 g cm"^ (Molar mass of H2SO4 = 98 g
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Ans. (0 Both are polar and hence miscible.
mol"*). (Raj. Board 2012)
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(//) They form hydrogen bonds with each other.
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This further increases their miscibility. Ans. 10% (w/w) H^S04 solution means 10 g H2SO4
are present in 100 g of solution, i.e., water
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2. Give an example of a solution containing a

liquid solute in a solid solvent. present = 90 g
Ans. Hydrated salts like CUSO4.5 HiO.
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10/98
Molality = xl0O0 = M3m
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3. Give one example each of solid in gas and 90

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liquid in gas solutions.

Ans. Iodine vapour in air, humidity in air. II, Solutions of solids in
4. Define ‘molality’. (CBSK 2017) liquids und solids in solids
nd
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Ans. Molality of a solution is the moles of the solute 10. What is the effect of temperature on the
dissolved in 1 kg of the solvent.
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solubility of sodium sulphate decahydrate ?

5. What do you mean by saying that the
molality of a solution is 0.1 ? Ans. The solubility first increases upto 32-4°C (called
Ans. It means that 0.1 mol of the solute is dissolved transition temperature) and then decreases.
in 1 kg of the solvent. 11. Define transition temperature in solubility of
6. What is the relation between normality and a solid in a liquid.
molarity of a given solution of H2SO4 ? Ans. The temperature at which the trend of solubility
Ans. Normality= 2 x Molarity. changes (e.g., first increases and then decreases)
7. What is the sum of mole fractions of all the is called the transition temperature.
components in a three component system ? 12. Give one example of an interstitial solid.
(CBSK 201U)
Ans. Tungsten carbide (WC) in which tungsten (W)
Ans. jc, -I- ^2 + X3 = I. atoms are arranged in a face-centred cubic lattice
8. Why does the molality of a solution remain
and C-atoms occupy the octahedral voids
unchanged with temperature ?
(interstitial sites).
(.^ssam Board 2013)
2/90 'Pn^tdee^'a. New Course Chemistry (XII)BSm

Ans. Acetone molecules get inbetween the ethanol

III. Solution of ;tases in liquids
molecules breaking hydrogen bonds thereby
13. State the formula relating pressure of a gas weakening ethanol-ethanol attractions.
with its mole fraction in a liquid solution in Consequently, vapour pressure of ethanol/
contact with it.
solution will be greater than that expected from
Ans. According to Henry’s law, Raoult’s law, i.e., for ideal solution. In other
Partial pressure of a gas above the solution = words, the solution will show positive deviation
K|^ X mole fraction of the gas in the solution from Raoult’s law.
or /^A “ ^ -^A’ Kh ” Henry’s constant.
VI. Relative lowering of vapour
IV. Vapour pressure of pressure - a colligative property
liquid solutions and Raoult’s law
23. Define colligative properties. (CBSE 2017)

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14. Why is the vapour pressure of a solution of
Ans. Refer to page 2/35.
glucose in water lower than that of water ?
24. What is the difference between lowering of
Ans. At a number of sites, non-volatile glucose

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molecules will occupy the surface instead of vapour pressure and relative lowering of
water molecules. Hence, effective surface area vapour pressure ?

e
for evaporation decreases and so the vapour Ans. Lowering of V.P. = p° - p .T*

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pressure is lower. Relative lowering of V.P. = (p® - p^!p°.

F
V. Ideal and Non-ideal 25. Write the mathematical form of Raoult’s law
of relative lowering of vapour pressure.
solutions and azeotropic mixtures
ur
r
15. Under what condition do non-ideal solutions
show negative deviations ? fo
P^-Ps _ ^
(Karnataka Board 2012)
ks
Ans.
Ans. When the new forces of interaction between the
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nj -t- «2
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components are greater than those in the pure

components. VII. Osmotic pressure
B

16. Define azeotropic mixture. (HP Board 2013)

26. What are isotonic solutions ? Give one
re

Ans. Refer to Art. 2.9.

17. Define an ideal solution. (CBSE 2017) example. (CBSE 2014)
u

Ans. Refer to Art. 2.8. Ans. Solutions having equal osmotic pressure are
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18. What type of liquids form ideal solutions ? called isotonic solutions, e.g., 0-1 M glucose
Ans. Liquids having similar structures and polarities. and OT M sucrose solutions.

19. What is the boiling point of an azeotrope of

d

27. What is van’t Hoff equation for dilute

Re

non-ideal solution showing positive

in

solution ?
deviations as compared to the boiling points
Ans. TiV = nRT where n is the no. of moles of solute
F

of Its components ?
Ans. The boiling point of such an azeotrope is lower present in V litres of solution, TZ is osmotic
than that of its components. pressure, T is temperature and R is gas constant
or solution constant.
20. What type of non-ideal solution is formed
when ethyl alcohol is mixed with water ? 28. Why does water from the soil rise to the top
Ans. It forms a non-ideal solution showing positive of a tall tree ?
deviations.
Ans. Due to osmosis through the roots because root
21. When 50 niL of liquid A is mixed with 50 niL
cell walls are made up of semipermeable
of liquid B total volume of the solution after
membrane.
mixing is found to be less than 100 mL. What
type of non-ideal solution is formed ? 29. What happens when blood cells are placed
Ans. As is negative, it forms a non-ideal in pure water ?
solution with negative deviations. Ans. Due to osmosis, water molecules move into
22. What happen when acetone is added to pure blood cells through the cell walls. As a result,
ethanol ? (CBSE 2020) blood cells swell and may even burst.
SOLUTIONS 2/91

30. Define reverse osmosis. Give one use of it. Ans. Greater the number of ions produced on
(CBSE 2011, R^. Board 2012) dissociation , greater is the van’t Hoff factor.
Ans. In osmosis, nel flow of solvent across the Hence, the order is

semipcrmeable membrane is from solvent to 0-1 MC12H22O1, <01 M KC1<0-1 M CaCl2

solution. The flow can be reversed by applying < 0-1 M Al2(S04)3
pressure greater than osmotic pressure on the 38. Give an example of a compound in which
solution. This is called reverse osmosis. It is used
hydrogen bonding results in the formation
for desalination of sea water.
of a dimer.
31. Define osmotic pressure. (CBSE 2007)
Ans. Benzoic acid (CgHjCOOH) in benzene.
Ans. Refer to page 2/43. 39. When is the value of van’t Hoff factor more
32. What happens when a pressure greater than than one ?

ow
osmotic pressure is applied on the solution
Ans. i > 1 when the solute undergoes dissociation in
side separated from the solvent by a the solution.
semipermeabie membrane ?
40. What is the van’t HolT factor for a compound
(CBSE 2016, 2020)
which undergoes tetramerisation in an
Ans. When the pressure applied on the solution

e
organic solvent ?

Fl
re
becomes greater than the osmotic pressure of
Ans. 4 A ■>A4
the solution, then the solvent molecules from

F
I
the solution pass through the semipermeabie 1 mole of A after association gives - mole of
membrane to the solvent side. The process is 4
ur
r
1
called reverse osmosis.
A4, Hence, i = — = 0-25.

VIII. Elevation of boiling 41. Define fo molal elevation constant or

ks
point and depression of freezing point ebullioscopic constant.
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33. Define molal depression constant/cryoscopic Ans. It is equal to elevation in boiling point that takes
constant. place when molality of the solution is unity.
B

Ans. The depression in freezing point that takes place 42. Between 2 M glucose solution and 1 M
e

when the molality of the solution is unity. glucose solution, which one has a lower
ur

34. Give one most important application of the freezing point? (Karnataka Board 2012)
ad

phenomenon of depression in freezing point Ans. Higher the concentration, greater is the
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in everyday life. depression and hence lower is the freezing point.

Ans. In lowering the freezing point of water in the
Thus, 2 M glucose solution will have lower
d

car radiator by adding suitable solutes (ethylene freezing point.

Re
in

glycol) to run it in sub-zero weather. 43. Between 1 M glucose solution and 1 M NaCl
35. What is an antifreeze ? solution which one will have higher boiling
F

point and why ?

Ans. A substance (such as ethylene glycol) which is
added to water to lower its freezing point is
Ans. I M NaCl will have higher boiling point because
NaCl dissociates in the solution and hence the
called an anti-freeze.
number of particles is more and therefore, the
IX. Abnormal molar masses elevation in boiling point is mole. Glucose is a
non-electrolyte and does not dissociate or
36. What would be the value of van’t Hoff factor
associate in the solution.
for dilute aqueous solution of K2SO4?
44. How much molar mass do you expect for
(Uttarakhand 2012)
NaCl using colligative properties ? (Assume
Ans. KoS04 4 2 K+ -H SOj / = 3.
complete dissociation)
37. Arrange the following solutions in increasing Ans. As NaCl dissociates to give 2 ions in the
order of their van’t Hofif factor :
solution, observed colligative property will be
01 M CaCl2, 0-1 M KCl, 0-1 M Al2(S04)3, double than normal value and molar mass will
01 M C12H22OH (R^j. Board 2012) be halved viz. 58-5/2 = 29-25.
2/92 New Course Chemistry (X11)SSI9]

SHORT ANSWER QUESTIONS Carrying 2 or 3 marks

I. Types of solutions and 10. State Raoult’s law. Derive its mathematical

expressing their concentration expression for a solution of a non-volatile solute

in a volatile solvent.
1. To what type of solution an alloy belongs ? Give (J & K Board 2011, MP Board 2012) [Art 2.7.3]
one example of a solution of liquid in solid.
11. Derive the relationship between relative
[Art. 2.2] lowering of vapour pressure and mole fraction
2. Explain the terms 'Mass fraction’ and ‘Mole of the volatile solute.
fraction’ ? [Page 2/5] (CBSE Sample Paper 2017) [Art. 2.7.2]
3. Which out of molality, molarity and mole 12. Vapour pressure of a solution is different from
fraction of a solution will remain unchanged on that of pure solvent
raising the temperature and why ? (0 Name the law which helps us to determine

w
(Bihur Board 2011) [Art. 2.3] partial vapour pressure of a volatile component
in solution.
4. Differentiate between molality and molarity of

F lo
(//) State the above law. (Kerala Board 2012)
a solution. What is the effect of change in
[(0 Raoult’s law («) Art. 2.7]
temperature of a solution on its molality and

ee
molarity ? (CBSE 2009, Jharkhand Board 2011) 13. Define vapour pressure of a liquid. What

Fr
happens to the vapour pressure when {a) a
[Art. 2.3]
volatile solute dissolves in the liquid and ib)
II. Sulutiuns of solids in the dissolved solute is non-volatile. [Art. 2.7]
liquids and solids in solids
for
14. Show that the relative lowering of vapour
ur
pressure for a solution is equal to the mole
5. Discuss the effect of temperature on the fraction of the solute when solvent alone is
s
solubility of solids in liquids. [Art. 2.4]
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volatile, [Art. 2.7]

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6. What are substitutional and interstitial solids ? or What do you understand by Relative lowering
eB

Give two examples of each. [Art. 2.6] of vapour pressure? How is it used to determine
molecular mass of the solute?
111. Solution of gases in liquids
(Uttarakhand Board 2012) [Art. 2.7]
r

7. State Henry’s law correlating the pressure of a

ad
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V. Ideal and Non-ideal

gas and its solubility in a solvent and mention
solutions and azeotropic mixtures
two applications for the law. What helps in
Y

existence of aquatic life ? 15. Non-ideal solutions exhibit either positive or

Re
nd

(CBSE 2008) [Art. 2.5] negative deviations from Raoult’s law. What are
these deviations and why are they caused ?
or State Henry’s law and mention its two important
Fi

Explain with one example of each type.

applications. (Assam Board 2012) [Art. 2.5]
(CBSE 2010) [Art. 2.8]
8. State Henry’s law and explain why are the tanks
used by scruba divers filled with air are diluted or Write five differences in solutions having
positive deviation and solutions having negative
with helium (11 -1% helium, 56-2% nitrogen and deviation. <MP Board 2011)
32-1% oxygen) ?
16. Write three differences between ideal and non
(CBSE Sample Paper 2022-23) [Art. 2.5.2] ideal solutions.

IV. Vapour pressure of (Pb. Board 2011, HP Board 2011,

liquid solutions and Raoult’s law MP Board 2011, CBSE 2017) [Art. 2.8]
17. What are the characteristics of an ideal solution?
9. How can you justify the observation that the
Why do solutions behave ideally only at low
vapour pressure of solution of a non-volatile concentration ? [Art. 2.8]
solute in a given solvent is less than that of the
or Define an ideal solution and write one of its
pure solvent ? Also state the law concerning this
observation.
characteristics. (CBSE 2014) [Art. 2.8]
[Art. 2.7.1. & 2.7.3]
SOLUTIONS 2/93

18. State Raouli’s law. Using the law. how would VI. Relative lowering of vapour
you distinguish between ideal and non-ideal pressure - a colligative property
solutions ? [Art 2.8]
19. What is Raoult’s law ? State the factors res 30. What is a colligative property ? Write down the
ponsible for deviations from this law. Illustrate different types of colligative properties. Show
that relative lowering of vapour pressure is a
deviations with suitable examples. [Art. 2.7]
colligative property. [Art 2.10 & 2.13]
20. What are ideal and non-ideal solutions ? Give
31. What is Raoult’s law ? How can molar mass
reasons for the formation of such solutions. Give of a non-volatile solute be determined with its
one example in each case. help ? (MP Board 2013)
(Raj. Board 2011, Karnataka Board 2012)
VII. Osmotic pressure
[Art. 2.8]
21. What type of deviation (positive or negative) 32. What is osmotic pressure ? Prove that it is a

w
from ideal behaviour will be shown by the colligative property.
solution of cyclohexane and ethanol ? Give (HP Board 2011) [ Art 2.12]
suitable reason. [Art 2.8]

F lo
33. Give four points of difference between osmosis
22. What are non-ideal solutions? What are their and diffusion. [Art 2.12.1]
different types? Explain giving examples. 34. What is the importance of semipcrmcabic

ee
(Chhatisgarh Board 2012t [Art. 2.8] membrane in osmosis? Explain.

Fr
23. What type of azeotropic mixture will be formed (UP Board 2012) [Art. 2.12.8]
by a solution of acetone and chloroform ? Justify 35. Derive van’t Hoff equation for dilute solutions.
on the basis of strength of intermolecular
for [Art 2.12]
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interactions that develop in the solution. 36. (0 What is osmotic pressure ?
(CBSE 2019) (ii) State van’t Hoff-Boylc’s law
s
ook
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24. State Raoult’s law. Write the conditions (Hi) What is an ideal solution ?
necessary for a solution to show ideal behaviour. (Karnataka Board 2012)
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[Art. 2.7 & 2.8] [(/) Page 2/43 («) Page 2/45 (iii) Page 2/31]
25. What are azeotropic mixtures? What are their 37. Define osmotic pressure. How can molar mass
our

different types? Explain with examples. of a substance be determined from the

ad

(Chhatisgarh Board 2012) [Art. 2.9] measurement of osmotic pressure of a

solution ? (Assam Board 2013)
26. State Raoult’s law for a solution containing
Y

volatile liquids. Explain with suitable example

[Page 2/43, 2/47 and Art. 2.12.8]
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38. Define the terms ‘osmosis’ and ‘osmotic

nd

the concept of maximum boiling azeotropes.

pressure’. What is the advantage of using
[Art 2.7 - 2.9]
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osmotic pressure as compared to other

27. What are differences between minimum boiling colligative properties for the determination of
azeotropes and maximum boiling azeotropes ? molar masses of solutes in solutions ?
(Pb. Board 2011) [Art. 2.9] (CBSE 2010) [Art. 2.12]
28. Why constant boiling mixtures behave like a 39. What are colligative properfie.s ? Write the
single component when subjected to colligative property used to find the molecular
distillation ? [Art. 2.9] mass of macromolecules.(CBSEForeign 2017)
29. What is meant by positive deviations from [Art. 2.10, Osmotic pressure]
Raoult’s law ? Give an example. What is the VIII. Elevation of boiling point
sign of A^jj. H for positive deviation ? and depression of freezing point
(CBSE 2015) [Art 2.8]
or Define azeotropes. What type of azeotrope is 40. Why the boiling point of a liquid gets raised on
dissolution of non-volatile solute into it ?
formed by positive deviation from Raoult’s law?
Give an example. (CBSE 2015) [Art. 2.9] [Art. 2.13]
2/94 ^na>tUe^'A New Course Chemistry CXll)EZsX9]

or Explain qualitatively the elevation of boiling IX. Abnormal molar masses

point of solution using Raoult’s law.
(West Bengal Board 2012) [Art. 2.13.1] 47. Why do electrolytes show abnormal molecular
41. Illustrate elevation in boiling point with the help masses ? Name the factors responsible for
of vapour pressure-temperature curve of a abnormality. [Art. 2.15]
solution. Show that elevation in boiling point 48. What is Abnormal Molecular Mass ? Discuss

is a colligative property. [Art. 2.13] its being in Molecular Association/Dissociation.

42. Show graphically how the vapour pressure of a [Art 2.15]
solvent and a solution of a non-volatile solute
49. What is van’t Hoff factor ? How docs it help in
change with temperature ? Show on this graph the determination of degree of association or
the boiling points of the solvent and the .solution. dissociation of a solute in solution ?
Which is higher and why ? [Art. 2.13] [Art. 2.15]
43. Explain why the freezing point of a solvent is 50. What is van’t Hoff factor ? What possible values
lowered on dissolving a non-volatile solute into
can it have if the solute molecules undergo

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It. [Art. 2.14]
(i) Association and
44. Explain how the measurement of depression in
(/7) Dissociation, in solution ?

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freezing point can be used for the determination
of molecular masses of non-volatile solutes. (HP Board 2011) [Art 2.15]
[Art 2.14] 51. Define osmotic pressure. Arrange the foUowing

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solutions in the increasing order of their osmotic
45. With the help of a suitable diagram, show that

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pressure:
the lowering of vapour pressure of a solution
than the pure solvent causes a lowering of (a) 34-2 g/lit sucrose
freezing point for the solution compared to that
of the pure solvent.
(b)
for
60 g/lit urea
ur
[Art. 2.14] (c) 90 g/lit gluco.se
46. An aqueous solution containing urea was found (d) 58-5 g/lit sodium chloride.
oks

to have boiling point more than the normal Give reason in support of your answer.
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boiling point of water (373-13 K). When the
[Art 2.12 and 2.15]
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same solution was cooled, it was found that its

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52. Define
freezing point is less than the normal freezing
point of water (273-13 K). Explain these (0 Abnormal molar mass
observations. [Art. 2.13 and 2.14]
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(ii) van’t Hoff factor. (CBSE 2017) [Art. 2.15]

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LONG ANSWER QUESTIONS Carrying 5 or more marks

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1. (a) Differentiate between molarity and molality

nd

3. (a) State the following : (/) Henry’s law about

of a solution. How does a change in temperature partial pressure of a gas in a mixture,
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influence their values ?

(ii) Raoult’s law in its general fonn in reference
{b) Calculate the freezing point of an aqueous to solutions.
solution containing 10.50 g of MgBrj in 2(K) g (b) What would be the molar mass of a
of water (Molar mass of MgBr2 = 184 g, K^for compound if 6-21 g of it dissolved in 24-0 g of
water = 1.80 K kg mol"*) (CBSE 2011) chloroform form a solution that has a boiling
[(a) Art. 2.3 (b) - 1*59"C (1 = 3)] point of 68-04°C. The boiling point of pure
2. (a) Define the terms osmosis and osmotic chloroform is 6I-7°C and the boiling point
elevation constant for chloroform is 3-63°C/m.
pressure. Is the osmotic pressure of a solution a
colligative property ? Explain. (CBSE 2011)
(b) Calculate the boiling point of a solution [(a) (() Art. 2.5 («) Art. 2.6 (b) 148*1 g mol"*]
prepared by adding 15-0 g of NaCl to 250-0 g 4. (i) What is de-icing agent ? How does it
of water (K^ for water = 0-512 K kg mol"*. function ?
Molar mass of NaCl = 58-44 g) (CBSE 2011) (//) 19-5 g of CH2FCOOH is dissolved in 500 g
[(a) Art. 2.12 (b) 101*05"C (i = 2)j of water. The depression in freezing point of
SOLUTIONS 2/95

water observed is 1-0°C. Calculate the van’t (b) A solution of glycerol (C3Hg03) in water
Hoff factor and dissociation constant of was prepared by dissolving some glycerol in 500
fluoroacetic acid. (Pb. Board 2011) g of water. This solution has a boiling point of
[(/) Ans. to Q. 24, page 2/88 10042°C. What mass of glycerol was dissolved
(ii) Ans. to Q. 2.33, page 2/llOJ to make this solution? (K;, for water = 0-512 K
kg mol *) (CBSE 2012)
5. (0 Why elevation in boiling point is a colligative
property ? [(a) Art. 2.5 & 2.13 (fr) h’2 = 37-7 gl
(ii) Calculate the osmotic pressure in Pascals 8. Define the solubilityof a solid in liquid. Brieily
describe the various factors on which the
exerted by a solution prepared by dissolving
1-0 g of polymer of molar mass 1,85,000 in solubility of a solid in a liquid depends.
450 niL of water at 37®C. (Pb. Board 2011) [Art. 2.4]
[{/) Art. 2.13 (//) Similar to Solved Problem 7, or Define osmotic pressure and describe Berkeley

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page 2/50, Ans. = 30 Pa] and Hartley’s method for tlie determination of
6. (a) Define Azeotropes and explain briefly osmotic pressure.

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minimum boiling azeotrope by taking suitable (J & K Board 2012) [Art. 2.12]
example. 9. State and explain Raoult’s law for (a) volatile
solute (b) non-volatile solute [Art. 2.7]

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(b) The vapour pressures of pure liquids A and
B are 450 mm and 700 mm of Hg respectively or What do you understand by colligative

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at 350 K. Calculate the compositionof the liquid properties of a solution? Explain briefly osmosis
mixture if total vapour pressure is 600 mm of and osmotic pressure.
Hg. Also find the composition of the mixture in
for
(J & K Board 2012) [Art. 2.10 & Art. 2.12]
ur
the vapour phase. 10. (/) When 2-56 g of sulphur was dissolved in
[Ans. (a) Art. 2.8 {b) If is mol fraction of A 100 g of CS->, the freezing point lowered by
s
0-383 K. Calculate the formula of sulphur (S,.)
k
in liquid phase, then
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(Kyfor CSt = 3-83 K kg mor'. Atomic mass of
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(450) -f- (1 - x^) (700) = 600 su iphur = 32 g mol"*).

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or - 0-4, Xjj = 0-6. (ii) Blood cells are isotonic with 0-9% sodium
Hence, = 0*4 x 450 = 180 mm. chloride solution. What happens if we place
blood cells in a .solution containing.
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pp - 0-6 X 700 = 420 mm,

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180 420 (a) 1-2% sodium chloride solution ?

= 0-7
= «-3, 3-p = 600 (b) 0-4% sodium chloride solution.
600
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7. (a) Explain the following ; (CBSE 2016)

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(/) Henry’s law about dissolution of a gas in a [Ans. (i) Sg, similar to solved problem 6,
liquid. page 2/63 (ii) (a) exosmosis takes place
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resulting in shrinking of cells {b) endosmosis

(ii) Boiling point elevation constant for a
takes place resulting in swelling of cells]
solvent.

CASE-BASED VERY SHORT/SHORT QUESTIONS

CASE 1. Liquid solutions are the solutions in temperature is called ‘vapour pressure'. It increases with
which the solvent is liquid. Solution of gases in liquid increase of temperature. The effect was quantitatively
is governed by Henry's law. Cons cring the cases when studied by Clausius and Clapeyron. For binary solutions,
the solutes are liquids or solids, and further considering as a result of detailed studies, a French chemist, F.M.
simple cases when there is only one solute (i.e. studying Raoult in 1886, put forward a law known after him as
binary solutions only), if the solute is liquid, it is volatile Raoult’s law. For liquid in liquid, it states that “the
and if the solute is a solid, it is non-volatile. In either vapour pressure of any component is equal to the mole
case, vapours will be present above the solution taken fraction of that component in the solution multiplied
in a closed vessel. The pressure exerted by the vapours by vapour of that component in the pure state.” In case
in equilibrium with the solution at a particular of solid in liquid, it states that "relative lowering of
2/96 New Course Chemistry (XII)BZs!SI

vapour pressure of a solution containing a non-volatile Based on the above paragraph, answer
solute is equal to the mole fraction of the solute in the questions no. 5 to 8 :
solution." These definitions help us in the calculation 5. Solutions of different molalities of a solute were
of different parameters related to solutions of liquid- prepared and their boiling points determined.
in-liquid and solid-in-Iiquid. Then boiling points were plotted against
Based on the above paragraph, answer molalities. What type of plot will be obtained ?
questions no. 1 to 4 : How does it help to calculate the value of
and ?
1. Two gases A and B have values for Henry’ law
constant as 35 and 77 at 293 K. Which will have 6. The vapour pressure of pure benzene at a certain
greater solubility in water at the same pressure temperature is 0-850 bar. A non-volatile, non
and why ? electrolyte solid weighing 0-5 g is added to
2. Which equation is used to study quantitatively the 39-0 g of benzene (molar mass 78 g mol"^). The
effect of temperature on the vapour pressure of a vapour pressure of the solution then is 0-845 bar.

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What is the molar mass of the solid substance ?
liquid ? Write the equation. What do different
symbols represent ? 7, The vapour pressure of an aqueous solution of
3. At a given temperature, the vapour pressure in cane sugar (mol. mass = 342) is 732 mm at 100°C.

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mm of Hg of a solution of two volatile liquids A Calculate the boiling point of the solution (K^ for
and B is given by the equation water = 0-52“C).

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p = 120 - 80 .rg (.Tg = mole fraction of B) 8. Calculate the amount of ice that will separate out

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when a solution containing 50 g of ethylene glycol
Calculate the vapour pressure of pure A and B at
the same temperature. in 200 g of water is cooled to - 9-3°C. (Ky for
H2O = 1-86 K kg mol"*)

for
4. The vapour pressures of benzene and toluene at
ur
293 K are 75 mm and 22 mm of Hg respectively. CASE 3. Colligative properties depend upon the
23-4 g of benzene and 64-4 g of toluene are mixed. number of particles of the solute present in a definite
amount of the solvent. The calculation of molecular
s
If the two form an ideal solution, calculate the
k
masses of solutes which do not undergo any dissociation
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mole fraction of benzene in the vapour phase
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assuming that the vapours are in equilibrium with or association in the solution is simple. However if a
solute undergoes dissociation or association in the
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the liquid mixture at this temperature.

solution, number of particles increases in the former
CASE 2. There are many properties of the ideal case and decreases in the latter case. For example, if
solutions which do not depend upon the nature of the NaCl is the solute and it dissociates completely, 1 mole
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solute dissolved but depend only on the number of of NaCl will produce 2 moles particles viz. Na'*' and
panicles of the solute present in the solution. Such . . _ . water
Cl" ions (NaCl ^ Na-" + Cl"). Similarly,
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properties of the solution are called “colligative

substances like acetic acid, benzoic acid dimerise in
properties.” For example, when a non-volatile (solid)
solvents like benzene, If dimerisation is complete, 1
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solute is dissolved in a liquid solvent, vapour pressure

mole of benzoic acid will produce 0-5 mole of particles
decreases, boiling point is raised while freezing point
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is depressed. The relative lowering of vapour that takes

viz. dimers (2 CgHgCOOH ^ (CgHsCOOH)^).
place or elevation in boiling point of the solvent or
In such cases, the observedvalue of colligativeproperty
depression In the freezing point that take place are all
is different from the theoretically calculated value.
colligative properties. These properties help us to
Consequently, molecular masses calculated will be
calculate the molecular masses of the solutes. These
different. In other words, ‘abnormal molecular masses’
properties are inter-related to one another. Thus,
will be observed. Further, dissociation or association
knowing the elevation in boiling (or the boiling point
may not be 100%. To account for all such complications,
of the solution), depression in freezing point of that van’t Hoff introduced a factor i known as van’t Hoff
solution (or the freezing point of the solution) can be factor and all the formulae for calculation of colligative
calculated. The study of depression in freezing point
properties have been modified.
has a number of practical applications such as in making
antifreeze solution or melting of ice on roads etc. As Based on the above paragraph, answer
questions no. 9 to 12 :
melting point of a crystalline solid has the same value
as the freezing point of the melt, Rast method is an 9. Arrange the following solutions in order of
application of depression in melting point using a solid decreasing freezing points :
solvent to calculate the molecular mass of a solid solute. 0-1 M Kj Fe(CN)6, 0-1 M BaCl2, 0-1 M NaN03,
0-1 M glucose
SOLUTIONS 2/97

10. What observed molecular mass do you expect for 12. Molality of sodium chloride solution whose
NaCl using colligative properties and why ? elevation in boiling point is numerically equal to
the depression in freezing point of 0-2 m
11. A decimolar solution of potassium ferrocyanide
aluminium sulphate solution in water is close to
is 50% dissociated at 300 K. The osmotic pressure
(assume complete dissociation and take and
of the solution is nearly
Ky for water as 0-52 and 1-86 K kg mol"'
(Given solution constant (R) = 8-314 JK“’ moP') respectively)

ANSWERS

1. Gas A will have greater solubility because can be calculated (Refer to Fig. 2.27 on page
according to Henry’s law expression, 2/60)
greater the value of lower is the value of 6. Refer to Solution to Problem 3, page 2/39.

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at the same value of p^. 7. Refer to Hint 3, page 2/57 (Ans. 10I-064°C)
2. The equation is called Clausius-CIapeyron 8. Refer to Solution to Problem 10, page 2/63.
equation. It is written as

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9. Glucose ^ no dissociation,
P. ^ AH vap ( 1
log V -y NaN03 Na'''-i-N02 (2 particles)

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M > 2-303 R T,1 T
2;
BaCl2 4 Ba-'*' -I- 2 Cr (3 particles),

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where Pj is the vapour pressure at temperature
T], P2 is the vapour pressure at temperature T2 K3[Fe(CN)(,l 3 K+ -t- [Fe(CN)^]^
(4 particles).

for
AV = enthalpy of vaporisation of the liquid,
vap
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R = gas constant Greater the number of particles, greater is the
3. Refer to Solution to Problem 4. page 2/29. depression in freezing point or lower is the actual
s
freezing point. Hence, correct order of freezing
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4. In the liquid phase, Ug = 23-4/78 = 0-3,

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point is
= 64-4/92 = 0-7. Hence, = 0-3, = 0-7
0-1 M glucose > 0-1 M NaNO^ > 0-1 M BaCU >
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Pg = 0-3 X 75 = 22-5 mm, 0-1 M K3[Fe(CN)6l-

Pj = 0-7 X 22 = 15-4 mm ; 10. Expected observed molecular mass of NaCl will
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Total vapour pressure = 22-5 + 15-4 = 37-9 mm

58*5
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Mole fraction of benzene in the vapour phase be half of its normal value, i.e., = 29-25.
2
= 22-5/37-9 = 0-59.
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This is because NaCl dissociates completely in

5.
AT^ = K^xm or T^-T“ the solution to form two particles (Na'*’ and Cl“).
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Hence, observed colligative property value will

orT^=T” + K^m be double and molecular mass will be half.
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Hence, plot of T^ vs w will be linear with slope = 11. Refer to Solution to Problem 8, page 2/71.

and intercept on Tjjj axis =T^- Hence, T^ and 12. Refer to Solution to Problem 7, page 2/71.

CASE-BASED MCQs AND ASSERTION-REASON QUESTIONS

f'ASE 1. There are many phenomena which we solvent molecules can pass through them, bigger
observe in nature or at home. For example, raw mangoes molecules of the solute cannot. This process of flow of
shrivel when placed in brine (salt solution), wilted the solvent through semipermeable membranes is called
flowers revive when placed in fresh water, blood cells ‘osmosis’. If a solvent and a solution are connected

collapse when suspended in saline water etc. if we look through semipermeable membrane, the pressure that has
into these processes, we find one thing common in all, to be applied on the solution to prevent the entry of the
i.e., all these substances are bound by membranes. These solvent into the solution is called ‘osmotic pressure’.
membranes contain microscopic holes or pores. Small Osmotic pressure increases with increase in con-
2/98 New Course Chemistry (X11)C&X9]

centralion and increase in temperature {i.e. n C xTCC

Reason. Measurement of low osmotic pressure
orTt = CRT. where Ris constant, similar to gas constant). is easier than measuring low values of elevation
This expression is applicable only to non-electrolytes. in boiling point or depression in freezing
For electrolytes, the formula is modified. Like other point.
colligalivc properties, osmotic pressure is also a 4. Assertion. 01 M sucrose solution has greater
colligalive property and can be used for calculation of osmotic pressure than 01 M glucose solution.
molar masses of the solutes (especially for proteins, Reason. Molecular mass of sucrose (342) is
polymers etc.). In osmosis, there is a net flow of solvent greater than molecular mass of glucose (180).
from solvent to solution. However, this can be reversed
by applying pressure greater than osmotic pressure on CASE 2. Ill normal life, we raiely come across
the solution. This process is called ‘reverse osmosis’ pure substances. Most of these are mixtures containing

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and is used in the purification of water (e.g. desalination two or more pure substances in varying composition. If
of sea water). Also, it may be remembered that solutions a mixture of two or more chemically non-reacting
having same molar concentration will have same substances is homogeneous in which composition of
osmotic pressure. Such solutions are called ‘isotonic the components can be varied within certain limits, it is
solutions'. called a ‘solution’. The simplest solutions are made up

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of two components called solute and solvent. Depending

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Based on the above paragraph, answer

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upon the physical slate of the solvent (solid, liquid or
questions no. 1 to 4 :
gas), solutions are classified as solid solutions, liquid

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1. Red Blood Cells (RBC’s) are isotonic with solutions and gaseous solutions, the most significant
(a) 01% NaCl solution being liquid solutions. Depending upon the amount of

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solute dissolved in a definite amount of the solvent,
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{b) 0 5% NaCl solution

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(c) 0-9% NaCl solution there are different ways in which the concentration of
ks
(d) 0-7% NaCl solution the solution can be expressed such as percentage,
strength, molarity, molality, normality, mole fraction,
2. When pressure greater than osmotic pressure is
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mass fraction, parts per million parts (ppm) etc. If a
applied on the solution separated by the solvent solution is diluted, the new normality or molarity of
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by a semipermeable membrane the solution can be calculated by applying normality

(tj) solvent molecules will flow from the solution equation or molarity equation. Similarly, if two non
to the solvent
reacting solutions are mixed, the normality or molarity
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ib) solute particles start flowing from solution to of the final solution formed can be calculated by
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the solvent applying similar equations.

(c) solvent molecules from solvent stop flowing Based on the above paragraph answer
d

to the solution questions no. 5 to 8 :

(d) temperature of the solution will rise
in
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5. The hardness of a water sample (in terms of

Choose the correct option out of the four options equivalents of CaC03) containing 10“^ M CaS04
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given below : is: (molar mass of CaS04 = 136 g mol“^)

(fl) Both Assertion (A) and Reason (R) are true and (a) 100 ppm (b) 50 ppm
Reason (R) is the correct explanation of Assertion (c) 10 ppm (d) 90 ppm
(A).
6. 100 cc of 01 M H2SO4 is mixed with 50 cc of
(b) Both Assertion (A) and Reason (R) are true but 0-1 N HCl. The normality of the final solution
reason (R) is not the correct explanation of will be
Assertion (A),
(a) 0-10 N (b) 017N
(c) Assertion (A) is true but Reason (R) is false.
(c) 0-20 N (d) 0-15N
{({) Assertion (A) is false but Reason (R) is true.
Choose the correct option out of the four options
3. Assertion. Out of various colligative properties, given below :
osmotic pressure is considered to be most suitable
(a) Both Assertion (A) and Reason (R) are true and
for determination of molecular masses of
Reason (R) is the correct explanation of
polymers.
Assertion (A).
SOLUTIONS 2/99

(b) Boih Assertion (A) and Reason (R) are true but Reason. 1 molal solution contains I mole of the
reason (R) is not the correct explanation of solute in 1000 g of the solvent whereas 1 molar
Assertion (A), solution contains I mole of the solute in 1 litre of
the solution.
(c) Assertion (A) is true but Reason (R) is false,
(r/) Assertion (A) is false but Reason (R) is true. 8. Assertion. Molarity of a solution changes with
temperature but molality does not.
7. Assertion. One molar solution has higher
Reason. Molarity is the moles of the solute in
concentration than one molal solution of the same
1 L of the solution whereas molality is the moles
solute in water.
of the solute in I kg of the solvent.

ANSWERS

1. (c) 2. (a) 3. (a) 4. (d) 5. (a) 6. (b) 7. (a) 8. (b)

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HINTS/EXPLANATIONS For Difficult Questions

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3. Both A and R are correct and R is the correct 6. Refer to Solution to Problem I0(i), page 2/11.
explanation of A, 7. Both A and R are correct and R is the correct
4. Correct A. 01 M sucrose has same osmotic explanation of A.

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pressure as 0-1 M glucose (both being non 8. Correct explanation. Molarity involves volume

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electrolytes). which changes with temperature but molality
involves masses which do not change with
5. Refer to Hint 11, page 2/136.

for
temperature.
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2/100 New Course Chemisti'y (XII)EZ5iaD

WITH
ANSWERS

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NCERT INTEXT UNSOLVED QUESTIONS & PROBLEMS

Q. 2.1. Calcuiate the mass percentage of benzene (C^H^) and carbon tetracliloride (CCI4) if 22 g of benzene
is dissolved in 122 g of carbon tetrachloride.

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Ans. Mass of solution = Mass of benzene + Mass of carbon tetrachloride = 22 g + 122 g = 144 g

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Mass of benzene 22 g

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Mass percentage of benzene = xl00 = X100= 15-28%

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Mass of solution 144 g
Mass of CCI4 122
Mass percentage of CCI4 = xl00 = xl00 = 84-72%
ou Mass of solution 144

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so
Alternatively, mass percentage of CCI4 = 100 - Mass percentage of benzene = 100 - 15-28 = 84-72%.
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Q. 2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans. 30% of benzene in carbon tetrachloride by mass means that
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Mass of benzene in die solution = 30 g ; Mass of solution = 100 g
-I
Mass of carbon tetrachloride = 100 - 30 g = 70 g ; Molar mass of benzene (C^H^) = 78 g mol
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Molar mass of CCI4 = 12 + 4 x 35-5 = 154 g mol
No. of moles of benzene =
Mikss 30 g = 0-385
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Molar mass 78 g mo! *

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Mass 70 g
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No. of moles of CCI4 = -i

= 0-455
Molar mass 154 g mol
d

Moles of benzene 0-385 0-385

Mole fraction of benzene = = 0-458
in

Total moles in the solution 0-385 + 0-455 0-84

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Mole fraction of CClj = i - 0-458 = 0-542.

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Q. 2.3. Calculate the molarity of each of the folloiving solutions :

(a) 30 g of Co (N03)2.6H20 in 4-3 L of solution (b) 30 niL of 0-5 M H2SO4 diluted to 500 mL.
(Atomic mass of cobalt = 58-7)
-1
Ans. (fl) Molar mass of Co(N03>2.6H,0 = 58-7 + 2 (14 + 48) + 6 x 18 g mol
= 58-7 + 124 + 108 g inoi"' = 290-7 g mol
-1

Mass 30 g
No. of moles of Co(N03)2.6H^O = Molar mass
-I
= 0-1032
290-7 g mol
Volume of solution = 4-3 L

No. of moles of solute 0-1032 mole

Molarity of solution = = 0-024 M
Volume of solution in L 4-3 L

(h) 1000 mL of 0-5 M H2SO4 contain H-»S04 = 0-5 mole

0-5
30 niL of 0-5 M H2SO4 contain H.SOj = 1000
x30 mole = 0-015 mole

Volume of solution = 500 mL = 0-500 L

SOLUTIONS 2/101

No. of moles of solute 0*015 mole

Molarity of solution = = 0-03 M ●
Volume of solution in L 0*500 L
Q. 2.4. Calculate the mass of urea (NH2CONH2) required in making 2*5 kg of 0*25 molal aqueous solution.
Ans. 0*25 molal aqueous solution of urea means that
Moles of urea = 0*25 mole, Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea (NH2CONH2) = 14 + 2 + 12 16 -i-14 + 2 = 60 g mol’^
0*25 mole of urea = 0*25 mole x 60 g mol"^ = 15 g
Total mass of solution = 1(XK) + 15 g = 1015 g = 1*015 kg
Thus, 1*015 kg of solution contain urea = 15 g
2*5 kg of solution will require urea = 15 g
x2*5kg = 37g.
1*015 kg
Q. 2.5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass)
aqueous KI is 1*202 g mL"^.

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Ans. 20% (mass/mass) aqueous KI solution means that
Mass of KI = 20 g. Mass of solution in water = 100 g

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Mass of solvent (water) = 100 - 20 = 80 g = 0*080 kg
(a) Calculation of molality

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20 g
Molar mass of KI = 39 + 127 = 166 g mol"* Moles of KI = = 0*120

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166 g mol“^
No. of moles of KI 0*120 mole
Molality of solution =
Mass of solvent in kg 0*080 kg
for
= 1*5 mol kg"^ ●
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(p) Calculation of molarity
s
Density of solution = 1 *202 g mL“* Volume of solution 100 g
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= 83*2 mL = 0*0832 L
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=

1*202 gmL-l
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Moles of solute 0*120 mole

Molarity of solution = = 1*44M.
Volume of solution in L 0*0832 L
(c) Calculation of mole fraction of KI
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No. of moles of KI = 0*120

Mass of water 80 g
No. of moles of water = = 4*44
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-1
Molar mass of water 18 g mol
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No. of moles of KI 0*120 0*120

nd

Mole fraction of KI = = 0*0263.

Total no. of moles in solution 0*120 + 4*44 4*560
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Q. 2.6. H2S, a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H2S
in water at STP is 0*195 m, calculate Henry’s law constant
Ans. Solubility of H2S gas = 0*195 m = 0*195 mole in 1 kg of the solvent (water)
1000 g
1 kg of the solvent (water) = 1000 g = = 55*55 moles
18 g mor*
0*195 0*195
Mole fraction of H2S gas in the solution (x) = = 0*0035
0*195 + 55*55 55*745
Pressure at STP = 0*987 bar

Applying Henry’s law, = K^ x or K„ _= PH2S ^ 0*987 bar = 282 bar.

0*0035
^H2S
Q. 2.7. Henry’s law constant for CO2 in water is 1*67 x 10* Pa at 298 K. Calculate the quantity of CO2 in
500 mL of soda water when packed under 2*5 atm CO2 pressure at 298 K.
2/102 ‘Pnadeefi-'<». New Course Chemistry (XIl)ESBIS]

Ans. K„= 1-67x10^Pa, Pco, ^ = 2-5x101325 Pa

Applying Henry’s law, Pqq^ = Kj^ ^-^c02
= !!a=i.5i7xl0-3
n
PCO. 2-5x101325 Pa CO2
= ^517xl0"^ i.e..
K
II
1-67x10^ Pa n
H,0 'to-.
n
H2O
500
For 500 niL of soda water, water present 500 mL = 500 g = - 27-78 moles
18

i.e.. n = 27-78 moles

H2O
//
. '‘CO2
= 1-517x10-3 or n,CO^ = 42-14x10“^ mole = 42-14 mmol =42-14xl0^x44g = 1-854 g.
27-78
Q. 2.8. The vapour pressures of pure liquids A and B are 450 and 700 mm Hg at 350 K respectively. Find

w
out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the
composition of the vapour phase.
Ans. Here, pp^ = 450 mm, p^ = 700 mm, p-j-oiai =

F lo
Applying Raoult's law. Pa = -^'a Pa ● Pb ” -'●B ^ Pb*’ - (1 “ -^a) Pb°
^Toial = Pa + Pb = ^aPa + (1 - ●'^a) Pb“ = Pe + (Pa° " Pb“) -^a

ee
Substituting the given values, we get

Fr
100
250a-a= 100 = 0-40
600 = 700 + (450 - 700).Va or or
●^a =
250

Thus, composition of the liquid mixture will be

for
ur
(mole fraction of A) = 0-40, Xy (mole fraction of B) = 1 - 0-40 = 0-60
Pa~^a^Pa ~ ^ - ● 80mm, p^ = xp^° = 0-60 x 700mm = 420 mm
s
ook

180
Yo
Pa
Mole fraction of A in the vapour phase = = 0-30
Pa + Pb 180+420
eB

Mole fraction of B in the vapour phase = 1 - 0-30 = 0-70.

Q. 2.9. Vapour pressure of pure water at 298 K is 23*8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in
850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
r
ad
ou

Ans. Here p° = 23-8 mm

-1
Wj = 50 g, M2 (urea) = 60 g mor' ; w, = 850 g, Mj (water) = 18 g mol
Y

Our aim is to calculate p^ and (jf - p^!p°

Re

P°-P,-.V W2 / Mt 50/60
nd

_ "2
Applying Raoult’s law,
P° n^+n2 Wj/M, + W2/M2 850/18 + 50/60
Fi

0-83 0-83
= 0-017
47-22 + 0-83 48-05
Thus, relative lowering of vapour pressure = 0-017
Substituting p° = 23-8 mm, we get
23-8-p. 23-8-p
= 0-017 or
23-8
^ = 0-017 or 23-8-p, = 0-017 x 23-8 = 0-40
P"
or
p^ = 23-8-0-40 = 23-4 mm
Thus, vapour pressure of water in the solution = 23-4 mm.
Q. 2.10. Boiling point of water at 750 mm Hg is 99*63'’C. How much sucrose is to be added to 500 g of water
such that it boils at 100“C ? Molal elevation constant for water is 0-52 K kg mol"*.
(Assam Board 2012)

Ans. Elevation in boiling point required (AT;,) = 100 - 99-63® = 0-37®

Mass of solvent (water), w, = 500 g
SOLUTIONS 2/103

Molar mass of solvent, M, = 18 g mol"' ; Molar mass of solute, C(2H220,j = 342 g mol
1000 K,^W2
Applying the fonnula, M2 =

_ M2Xvv,xAT^, _ 342g mol"'X500gx0-37 K

or M’.
2 “ = 121-7 g.
lOOOxK^ 1000gkg-lx0-52Kkg mor>
Q. 2.11. Calculate the mass of ascorbic acid {vitamin C, C^H^Og) to be dissolved in 75 g of acetic acid to
lower its melting point by l-ST. Ky^= 3-9 K kg mol"'.
Ans. Lowering in melting point (AT^) = 1-5° ; Mass of solvent (CH3COOH), vv, = 75 g
-1
Molar mass of solvent (CH3COOH), Mj = 60 g mol
Molar mass of solute (Cf,Hg05). M, = 72 + 8 + 96 = 176 g mol
-1

ow
-I
For acetic acid, K^= 3-9 K kg mol
lOOOKy-U'2
Applying the formula, M2 = vv,1 AT
/

M2Xvv2XAT^ (176 g mol"') (75 g) (1-5 K)

e
= 5-077 g

Fl
or
VIS =

re
lOOOxK
/
(1000 g kg"*) (3-9 K kg mol"*)

F
Q. 2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1-0 g of
polymer of molar mass 185,000 in 450 mL of water at 37°C.
ur
or
Ans. 7t = CRT = -RT
V

Here, number of moles of solute dissolved in) =

k sf
1-0 g 1
mol
-1
Yo
185,000 g mol 185,000
oo

V = 450 mL = 0-450 L, T = 37°C = 37 + 273 = 310 K

R = 8-314 kPa L K"* mol"* = 8-314 x 10^ Pa L K"' mol"*
B

Substituting these values, we get

re

1 1
7C = molx x8-3I4xl0'^ Pa L K"' mor'x310K = 30-96 Pa.
185.000 0-45 L
u
ad
Yo

NCERT EXERCISE
Q. 2.1. Define the term solution. How many types of solutions are formed ? Write briefly about each type
d

with an example.
Re
in

Ans. Refer to page 2/1, 2/2.

Q. 2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and
F

the other whose particles are very small. What type of this solid solution is likely to be ?
Ans. Interstitial solid solution.
Q. 2.3. Define the terms : (/) Mole fraction («) Molality (Hi) Molarity (iv) Mass percentage.
Ans. Refer to pages 2/3, 2/4 and 2/5.
Q. 2.4. Concentrated nitric acid used in the laboratory is 68% nitric acid by mass in aqueous solution.
What should be the molarity of such a sample of the acid if the density of the solution is
1-504 g mL"' ?
Ans. 68% nitric acid by mass means that
Mass of nitric acid = 68 g. Mass of solution = 100 g

Molar mass of HNO3 = 63 g mol"'

68 g HNO3 = —
63
mole = 1079 mole, Density of solution = I -504 g arL"'
100
.●. Volume of solution = mL = 66-5 mL = 0-0665 L
1-504
2/104 "Pnadee^’A New Course Chemistry (Xll)E2sISI
Moles of the solute 1-079
Molarity of the solution = M = 16-23 M
Volume of solution in L 0-0665

Q. 2.5. A solution of glucose in water is labelled as 10% w/w. What would be the molality and mole fraction
of each component in the solution ? If the density of the solution is 1-2 g mL“\ then what shall be
the molarity of the solution ? (Pb. Board 2011)
Ans. 10 g glucose is present in 100 g solution, i.e., 90 g of water = 0-090 kg of water
10 90
10 g glucose = mol = 0-0555 mol, 90 g H2O = — = 5 moles
180

0-0555 mol
Molality = = 0-617 m
0 090 kg

w
0-0555
;c (Glucose) =
5+0-0555
= 0-01, .r (HjO) = I - O-OI = 0-99
100 0-0555 mol

o
mL = 83-33 mL = 0-08333 L ; Molarity = = 0-67 M
100 g solution = 0-08333 L
1-2

e
re
Q. 2.6. How many mL of a 0-1 M HCl are required to react completely with 1 g mixture of Na2C03 and

rFl
NaHC03 containing equimolar amounts of the two ?

F
Ans. Step 1. To calculate the number of moles of the components in the mixture.
Suppose Na7C03 present in the mixture = x g NaHC03 present in the mixture = (1 - x) g
-1

r
Molar mass of Na2C03= 2x23+ 12 + 3x 16 = 106 g mol
ou
fo
Molar mass of NaHC03 =23+ 1 + 12 + 3x 16 = 84g mor' ks \-x
Moles of NajCOj in x g = , Moles of NaHC03 in (1 - ^) g =
106 84
oo
As mixture contains equimolar amounts of the two,
Y
eB

106
X _ 1— or 106 - 106.r = 84 a: or a: = g= 0-558 g
106 ~ 84 190

0-558 1-0-558 0-442

r

Thus, moles of Na2C03 = = 0-00526 ; Moles of NaHC03 = = 0-00526

ou

106 84 84
Y
ad

Step 2. To calculate the moles of HCl required.

Na2C03 + 2 HCl + 2 NaCl + H2O + CO2 ; NaHC03 + HCl ^ NaCl + H2O + CO2
d

1 mole of Na^C03 required HCl = 2 moles

Re
in

0-00526 mole of Na^C03 requires HCl = 0-00526 x 2 moles = 0-01052 mole

1 mole of NaHC03 requires HCl = 1 mole
F

0-00526 mole of NaHC03 requires HCl = 0 00526 mole

Total HCl required = 0-01052 + 0-00526 moles = 0-01578 moles
Step 3. To calculate volume of 0-1 M HCl
0-1 mole of 0-1 M HCl are present in 1000 mLofHCl.
1000
0-01578 mole of 0-1 M HCl will be present in HCl = x0-01578 = 157-8 mL
0-1

Q. 2.7. A solution is obtained by mixing 300 g of 25% and 400 g of 40% solution by mass. Calculate the
mass percentage of the resulting solution. ?
Ans. 300 g of 25% solution contains solute = 75 g ; 400 g of 40% solution contains solute = 160 g
Total solute = 160 + 75 = 235 g. Total solution = 300 + 400 = 700 g
235
Mass % of solute in the final solution = X100 =33-5%
700
Mass % of water in the final solution = 100 - 33-5 = 66-5%.
SOLUTIONS 2/105

Q. 2.8. An antifreeze solution is prepared from 222-6 g of ethylene glycol, C2H4( OH )2 and 200 g of water.
Calculate the molality of the solution. If the density of the solution is 1-072 g mL"^ then what shall
be the molarity of the solution ? (CBSE 2007)
Ans. -1
Mass of the solute, C2H4(OH)-> = 222-6 g. Molar mass of C2H4(OH )2 = 62 g mol
Moles of the solute = 222-6 g = 3-59 ; Mass of the solvent = 200 g - 0-200 kg
62 g mol ^
3-59 moles
Molality - = 17-95 mol kg
-I

0-200 kg
Total mass of the solution = 422-6 g
422-6 g 3-59 moles
Volume of the solution = = 394-2 mi = 0-3942 L ; Molarity = = 9-11 mol L"*-
-1
1-072 g ml 0-3942 L

Q. 2.9. A sample of drinking water was found to be severely contaminated with chloroform, CHCI3,
supposed to be carcinogen. The level of contaminationwas 15 ppm (by mass).
(/) Express this in percent by mass, (u) Determine the molality of chloroform in the water sample.

w
Ans. 15 ppm means 15 parts in million (10^) parts by mass in the solution

F lo
15
% by mass = x100 = 15x10-4
10^
Taking 15 g chloroform in 10^ g of the solution, mass of solvent = 10^ g
-I
Molar mass ofCHCl3= 12-f 1 + 3 x 35-5 = 119-5 g mol

ree
15/119-5
Molality =
10^
X1000 = 1-25x10“* m.
for F
Q. 2.10. What role does the molecular interaction play in solution of alcohol and water ?
Ans. There is strong hydrogen bonding in alcohol molecules as well as water molecules. On mixing, the molecular
interactions are weakened. Hence, their solution will show positive deviations from ideal behaviour. As a
Your

result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.
ks
eBoo

Q. 2.11. Why do gases nearly always tend to be less soluble in liquid as the temperature is raised ?
Ans. Dissolution of gas in liquid is an exothermic process (Gas + Solvent ^ - Solution + Heat). As the
ad

temperature is increased, equilibrium shifts backward.

our

Q. 2.12. State Henry’s law and mention some of its important applications. (Assam Board 2012)
Ans. Refer to pages 2/17, 18, 19.

Q. 2.13. The partial pressure of ethane over a saturated solution containing 6-56 x 10~^ g of ethane is 1 bar.
Re

If the solution contains 5-00 x 10~^ g of ethane, then what shall be the partial pressure of the gas ?
Y

Ans. Applying the relationship m = xp

Find

-I
In the first case, 6-56 x 10"" g = K|| x 1 bar or Kj.j = 6-56 x 10"- g bar
5-00x10-2 g
In the second case, 5-00 x 10~- g = (6-56 x 10"^ g bar"*) x p or p = = 0-762 bar.
6-56 X10-2 g bar *
Q. 2.14. What is meant by positive and negative deviations from Raoult’s law and how is the sign of A^| H
related to positive and negative deviations from Raoult’slaw ?
Ans. Refer to pages 2/32, 2/33.
Q. 2.15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1-004 bar at the normal boiling
point of the solvent. What is the molecular mass of the solute ?
Ans. Vapour pressure of pure water at the boiling point (p®) = 1 atm = 1-013 bar
Vapour pressure of solution (p^) = 1 -004 bar ; Ma.ss of solute (W2) = 2 g
Mass of solution = 100 g ; Mass of solvent = 98 g
P"-Ps _ «2 n
2 _ W2/M2 _ W-
2-x
MI
Applying Raoult’s law for dilute solution (being 2%)
—^ + ^2 n
I
w./M
1 1 M2 w
1
2/106 New Course Chemistry CXI1)E&XSI

-1
(1-013-1-004) _ 2g 18 g mol 2x18 1-013 -1
X or
M2 = X g mol * - 41*35 g mol
1-013 bar M2 98 0-009

Q. 2.16. Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquid components
are 105*2 kPa and 46*8 kPa respectively.What will be the vapour pressure of a mixture of 26*0 g of
heptane and 35*0 g of octane ?
-!
Ans. Molar mass of heptane (C7H,^) = 100 g mol“‘; Molar mass of octane (CgH|j^) = 114 g mol
26-Qg 35-0 g
26-0 g heptane = -j
= 0-26 mol - 35-0 g octane - -1
= 0-31 mol
100 g mol 114 g mol
0 26 g
X (heptane) = = 0-456 ; .V (octane) = 1 - 0-456 = 0-544
0-26+ 0-31

ow
p (heptane) = 0-456 x 105-2 kPa = 47-97 kPa : p (octane) = 0-544 x 46-8 kPa = 25-46 kPa
^Total = 47-97 -f 25-46 = 73*43 kPa.

Q. 2.17. The vapour pressure of water is 12*3 kPa at 300 K. Calculate the vapour pressure of 1 molal
solution of a solute in it.
An.s. 1 molal solution means 1 mol of the solute in 1 kg of the solvent (water).

e
Fl
re
Mol fraction of solute = = 0-0177
1+55-5

F
12-3-p s
Now, = .X2»'-^-> = 0-0177 or Pj= 12*08 kPa.
P° 12-3
ur
or
Q. 2.18. Calculate the mass of a non-volatile solute (molar mass 40 g mol ^) which should be dissolved in
114 g octane to reduce its vapour pressure to 80%.
sf
(CBSE Sample Paper 2017)
Ans. Reduction of vapour pressure to 80% means that if p® = 100 mm, then p^ = 80 mm.
k
Yo
Applying complete fonnula
oo

P°~P ●f "2 W2/M2

B

P^ 'h Wj/M| +
re

100-80 W2/4O
(Mol. mass of octane CgHjg = 114 g mol *)
u

100 114/114+vis/40
ad
Yo

20 _ ^2^40 w.
_ ”'2
or or or
W2 = 10 g
100 ~ 1 + w,/40 5 40 40
d
Re
in

Note that complete formula is required because concentration of solution is greater than 5%.
Complete formula can also be applied in the form
F

P°-P, _ ^2-^2 100-80 _ tV2/4Q 1

or or or
W2 = 10 g
Ps Wj/Mj 80 ~ 114/114 4 40

Alternatively, suppose mass of solute dissolved = w g

w
Moles of solute = —g
40

114
Moles of solvent (octane) =
114
= 1 mole (Mol. mass of CgHjg = 114 g mol ^)
I
Mole fraction of solvent =
1 + w/40
For a non-volatile solute,
Vapour pressure of solution = Mole fraction of solvent in the solution x Vapour pressure of pure solvent
p, = x^xp°
SOLUTIONS 2/107

1 100
80 = xIOO or I 4-— =
w
or
H’
-1 1
or w = 10 g
1 + H-/40 40 80 40 4
Q. 2.19. A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of
2'8 kPa at 298 K. Further 18 g of water is then added to the solution, the new vapour pressure
becomes 2*9 kPa at 298 K. Calculate
(0 molar mass of the solute. (h) vapour pressure of water at 298 K.
-1
Ans. (/) Suppose the molar mass of the solute = M g mol
30 90 g
^2 (solute) = —
M
moles, h, (solvent, H^O) = -1
= 5 moles
18 g mol
P°-P, _ , i.e..
p°-2-8 30/M
or 1-
30/M

ow
P^ 5 + 30/M P° 5 + 30/M

2-8 30/M 5 + 30/M-30/M 5 P^ 5 + 30/M

or = 1- or ...(0
P° 5 + 30/M 5 + 30 M 5 + 30/M 2-8 5 M

After adding 18 g of water, n (H^O ), i.e., «| = 6 moles

e
p°-2-9 30/M 2-9 30/M

re
Fl
or 1-
6 + 30/M P^ 6 + 30/M

F
2-9 30/M 6 + 30/M-30/M 6 P^ 6 + 30/M 5
or
ur or = 1+ — ...(//)

r
P’^ 6 + 30/M 6 + 30/M 6 + 30/M 2-9 6 M

2-9 1+6/M
fo r 5 "l ( 6
ks
Dividing eqn. (/) by eqn. (ii), we get or 2-9 1+— =2-8 1+ —
2-8 1+5/M
I M
Yo
oo
14-5 16-8 2-3
or 2-9 + = 2-8 + or = 01 or M = 23 u
M M M
eB

6 29 29
(ii) Putting M = 23 in eqn. (i), we get — = 1+— = — or X2-8 = 3-53kPa.
2-8 23 23 23
ur

Q. 2.20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing
ad

point of a 5% glucose in water if freezing point of pure water is 273*15 K.

Yo

Ans. 5% solution by mass means 5 g of solute is present in 100 g of solution

Mass of solvent (water) = 95 g
d

1000
Re
in

Molality of sugar solution = — X

= 0154 (mol. mass of sugar C,2H^20 11 = 342)
342 95
F

215
ATyfor sugar solution = 273 15 - 271 = 2-15° ; AT^=KyXm K^ 0154
1000
Molality of glucose solution = — X
= 0-292 (mol. mass of glucose C^HpO^ = 180)
180 95

215
.●. ATy(Glucose) = ATjrX/n = 0154
X 0-292 =4-08

.-. Freezing point of glucose solution = 273-15 - 4-08 = 269*07 K.

Q. 2.21. Two elements A and B form compounds having molecular formula AB2 and AB^. When dissolved
in 20 g of benzene (CgH^), 1 g of ABj lowers the freezing point by 2*3 K whereas 1*0 g of AB^ lowers
it by 1*3 K. The molal depression constant for benzene Is 5*1 K kg moP*. Calculate atomic masses
of A and B. I! .nul Board 2012)
1000 Wo
Ans. Applying the formula, M^ =
w,1 X AT
/
2/108 ^/undee^'^ New Course Chemistry (XII)ESm

1000X51X1 1000 x 51x1 -1

M
ABj “
= 110-87 gmor‘ ; M AB4 “
= 196-15 g mol
20x2-3 20x1.3
Suppose atomic masses of A and B are ‘a’ and respectively. Then
-1
Molar mass of AB2 = a+ lb = \ 10-87 g mol .(0
-1
Molar mass of AB4 = a + 4 /? = 196-15 g mol .(«*)
Eqn.(«)-Eqn. (Ogives 2^7 =85-28 or i? = 42-64
Substituting in eqn. (0, we get a+ 2.y 42-64 = 110-87 or a- 25-59
Thus, Atomic mass of A = 25*59 u, Atomic mass of B = 42*64 u
Q. 2.22. At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4*98 bar. If the
osmotic pressure of the solution is 1*52 bar at the same temperature, what would be its

ow
concentration ?

36
Ans. 7t = CRT In the first case, 4-98 = xRx300 = 60R ...(0
180
In the second case. 1-52= CxRx300 .(tf)

e
Dividing («) by (/), we get C = 0*061 M .

re
Q. 2.23. Suggest the most important type of intermolecular interaction in the following pairs :

rFl
(i) n-hexane and n-octane (u) I2 and CCI4 (in) NaC104 and water (iv) methanol and acetone

F
(v) acetonitrile (CH3CN ) and acetone (€311^0).
Ans. (0 Both are non-polar. Hence, intermolecular interactions in them will be London dispersion forces

r
(discussed in class XI).
ou
fo
(ii) Same as (/). ks
(ill) NaC104 gives Na"^ and CIO4 ions in the solution while water is a polar molecule. Hence,
intermolecular interactions in them will be ion-dipole interactions,
oo
(iv) Both are polar molecules. Hence, intermolecular interactions in them will be dipole-dipole interactions,
Y
(v) Same as (iv).
B

Q. 2.24. Based on solute-solvent interactions, arrange the following in order of increasing solubility in
Cyclohexane, KCl, CH3OH, CH3CN
re

n-octane and explain,

Ans. (i) Cyclohexane and n-octane both are non-polar. Hence, they mix completely in all proportions.
ou

(I’O KQ is an ionic compound while n-octane is non-polar. Hence, KCT will not dissolve at all in n-octane.
Y
ad

(in) CH3OH and CH3CN both are polar but CH3CN is less polar than CH3OH. As the solvent is non
polar, CH3CN will dissolve more than CH3OH in n-octane.
d

Thus, the order of solubility will be KCl < CH3OH < CH3CN < Cyclohexane.
in
Re

Q. 2.25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble
in water (i) phenol, (ii) toluene, (iii) formic acid, (iv) ethylene glycol, (v) chloroform, (vi) pentanol.
F

Ans. (0 Partially soluble because phenol has polar -OH group but aromatic phenyl, CgH5~ group.
(ii) Insoluble because toluene is non-polar while water is polar.
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water,
(v) Insoluble because chloroform is an organic liquid.
(vi) Partially soluble because -OH group is polar but the large hydrocarbon part (C5HJ1) is non-polar.
Q. 2.26. If the density of some lake water is 1*25 g mL~^ and contains 92 g of Na*^ ions per kg of water,
calculate the molality of Na*** ions in the lake.
92 g
Ans. No. of moles in 92 g of Na'*' ions = = 4 mole (as atomic mass of Na = 23)
23 g mol"^
As these are present in 1 kg of water, by definition, molality = 4 m.
Q. 2.27. If the solubility product of CuS is 6 x calculate the maximum molarity of CuS in aqueous
solution.
Ans. Maximum molarity of CuS in aqueous solution = Solubility of CuS in mol L“*
SOLUTIONS 2/109

If S is the solubility of CuS in mol L"', then

CuS ^ ± Cu2+
s
+ s2-,K
s
=[Cu2+][S2-] = SxS = S2

82 = 6x10-6 or S = Vbx 10-16 = 2*45x10-* mol L-'.

Q. 2.28. Calculate the mass percentage of aspirin (C9Hg04> in acetonitrile (CH3CN) when 6*5 g of €911304
is dissolved in 450 g of CH3CN.
6-5
Ans.
Mass of aspirin xlOO = 1*424% .
Mass percent of aspirin = xl00 =
6-5 + 450
Mass of aspirin + Mass of acetonitrile
Q. 2.29. Nalorphene (C]9H2|N03), similar to morphine, is used to combat withdrawl symptoms in narcotic
users. Dose of nalorphene generally given is 1*5 mg. Calculate the mass of 1*5 x 10~^ m aqueous
solution required for the above dose.

w
Ans. 1-5 X 10-2 jjj solution means that 1-5 x 10^2 „jole of nalorphene is dissolved in 1 kg of water.
-1
Molar mass of C,9H2iN03 = 19 x 12 + 21 + 14 + 48 = 311 g mol
1-5 X 10-2 mole of C19I4NO3 = 1-5 X 10-2 X 311 g ^ Q467 g = 467 mg

Flo
.*. Mass of solution = 1000 g + 0-467 g = 1000-467 g

e
Thus, for 467 mg of nalorphene, solution required = 1000-467 g

re
1000-467
For 1*5 mg of nalorphene, solution required = xl-5 = 3*21g.

F
467
Q. 2.30. Calculate the amount of benzoic add (C3H5COOH) required for preparing 250 mL of 0*15 M
ur
r
solution in methsmol.
Ans. 0-15 M solution neans that 0*15 mole of benzoic acid
fo
is present in 1 L, i.e., 1000 mL of the solution.
Molar mass of benzoic acid (CgHjCOOH) = 72 + 5 + 12 + 32 + 1 = 122 g mol"’
ks
Yo
.*. 0-15 mole of benzoic acid = 0*15 x 122 g = 18-3 g
oo

Thus, 1000 mL of the solution contain benzoic acid = 18*3 g

eB

18-3
250 mL of the solution will contain benzoic acid = X 250 = 4*575 g.
1000

Q. 231. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic
ur

add and trifluoroacedc acid increases in the order given above. Explain briefly.
ad
Yo

Ans. The depression in freezing points are in the order :

acetic acid < trichloroacetic acid < trifluoroacetic acid
d

/
(Cl
Re
in

Cl-^C-COOH F—C-COOH
(CH3—COOH)
Cl F/
F

Fluorine, being most electronegative, has the highest electron withdrawing inductive effect. Consequently,
trifluoroacetic acid is the strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid
ionizes to the largest extent while acetic acid ionizes to the minimum extent to give ions in their solutions
in water. Greater the ions produced, greater is the depression in freezing point. Hence, the depression in
freezing point is maximum for the fluoroacetic acid and minimum for acetic acid.
Q. 232. Calculate the depression in the freezing point of water when 10 g of CH3CH2CHCICOOH is added
to 250 g of water. = 1*4 x lO"®, K^= 1*86 K kg moH. -1
Ans. Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35-5 + 45 = 122-5 g mol
10
mole = 8-16 XIQ-^ mole
10 g of CH3CH2CHCICOOH = 122-5

816 X10“^ mole

Molality of the solution (m) ~ xlOOOgkg-i =0-3264
250g
If a is the degree of dissociation of CH3CH2CHCICOOH, then
2/110 “Pnauiee^'A New Course Chemistry (XII)BZSi91

CH3CH2CHCICOOH V ± CH3CH2CHCICOO' + H-"

Initial cone. C mol L-’ 0 0

At eqm. C(l-a) Ca Ca

Ca. Ca K 14x10-3
Ka = C a^ or a =
u _
= 0-065
C(l-a) "V C ~ \ 0-3264
To calculate van't Hoff factor :

CH3CH2CHCICOOH V ^ CH3CH2CHCICOO- -I- H-"

Initial moles 1

Moles at eqm. 1 - a a a Total = 1 + a

. 1 + a
i =
I = l + a=l-f0-065 = 1-065 ; AT^= / Ky-m = (1-065) (1-86 ) (0-3264) = 0*65".

w
Q. 2.33. 19-5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point observed
is 1‘0®C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid. Kr for water
is 1*86 K kg mol"*. (Pb. Board 2011)

F lo
Ans. Here, W2 = 19-5 g, w, = 500 g, K^= 1-86 K kg mol^*, (ATyj^,,^ = 1-0“
1000 Ky- VV2 _ (1000 g kg-‘) (1-86 K kg mol"') (19-5 g)

ee
M2 (observed) = = 72-54 g mol-l
>v,1 AT (500 g) (1-0 K)

Fr
/
-1
M2 (calculated) for CH^FCOOH = 14 + 19 -f 45 = 78 g mol

van’t Hoff factor (/) =

<r^2)cal _ 78
= 1-0753.
for
ur
(M2)obs 72-54

Calculation of dissociation constant. Suppose degree of dissociation at the given concentration is a.

s
ook
Yo
Then CH2FCOOH V - CHoFCOO" +
Initial C mol L"' 0 0
eB

At eqm. C(1 - a) C a C a, Total = C (1 + a)

. C(I + a)
i = = l +a or a = /-I = 1-0753-1 =0-0753
r

C
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ou

[CH2FC00-][H^] Cg.Ca Co?

K
a

ICH2FCOOH] C(l-a) 1-a

Y

Taking volume of the solution as 500 mL,

Re
nd

C =
19-5
X
1
xl000 = 0-5 M Ka C g3 _ (0-5) (0-0753)^ = 3-07 X10'^.
Fi

78 500 1-a 1-0-0753

Q. 2.34. Vapour pressure of water at 293 K is 17-535 mm Hg. Calculate the vapour pressure of water at
293 K when 25 g of glucose is dissolved in 450 g of water.
Ans. Here, p° = 17-535 mm, ^2 = 25 g, w, = 450 g
For solute (glucose, C5Hj20j^), M2 = 180 g mol"*, For solvent (H2O), Mj = 18 g mor'
P°-P.v _ "2 P°-Ps _ 'h _ P°
Applying Raoult’s law. or or

/jj + /I2 Ps n
1 w,/M, Ps Wj M2
Substituting the given values, we get
17-535
-1 = 25x18 _ 25 17-535 25 _ 4525 or = 17-535 X
4500
or = 1 + = 17-44 mm.
P. 450x180 ~ 4500 Ps 4500 ~ 4500 4525

Q. 2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4-27 x 10^ mm Hg. Calculate
the solubility of methane in benzene at 298 K under 760 mm Hg.
Ans. Here, Ku = 4-27 x 10^ mm, p = 760 mm
SOLUTIONS 2/111

760 mm
Applying Henry’s law, p = K,.|.v, we have x = — = 1-78x10-3
K
H
4-27x105 mm

i.e., mole fraction of methane in benzene = 1*78 x 10"3.

Q. 2.36. 100 g of liquid A (molar mass 140 g mol"^) was dissolved in 1000 g of liquid B (molar mass
180 g mol"’). The vapour pressure of pure liquid B was found to be 500 torn Calculate the vapour
pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the
solution is 475 torn
100 g
Ans. No. of moles of liquid A (solute) = -1
— mole
140 g mol 7

1000 g 50
No. of moles of liquid B (solvent) = = — mole
180 g mol 9

5/7 5/7 _ 5 63 _ 45
Mole fraction of A in the solution (a:^) = = 0-114
5/7 + 50/9 395/63 " 7^ 395 " 395

w
Mole fraction of B in the solution (.tg) =1-0-114 = 0-886
Also, given = 500 torr

F lo
Applying Raoult’s law. ...(/)
Pb - ^bPb° - 0-886 X 500 = 443 torr
^Toiai -Pa'^Pb

ee
475 - 443

Fr
475 = 0-lI4p/ + 443 or
Pa = 0-114
= 280-7 torr

Substituting this value in eqn. (/), we get = 0-114 X 280-7 torr = 32 torn
Q. 2.37. Vapour pressure of pure acetone and chloroform at 328 K are 741*8 mm Hg and 632-8 mm Hg
for
ur
respectively. Assuming that they form ideal solution over the entire range of composition, plot
Ptotni^ /^chloroform ^nd Pacctone ^ function of ATacetone* ^he experimental data observed for different
s
compositions of mixtures is :
ook
Yo
100 XX acetone 0 11*8 23-4 36-0 50-8 58*2 64-5 72-1

/^accto.>>n’ Hg 0 54-9 110*1 202-4 322*7 405-9 454-1 521-1

eB

/’chloroform/rnm Hg 632-8 548-1 469-4 359-7 257*7 193-6 161-2 120-7

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative
r

deviation from the ideal solution.

ad
ou

Ans. X acetone 0-0 0-118 0-234 0-360 0-508 0-582 0-645 0-721
P^CCloJ^^ Hg 0 54-9 110-1 202-4 322-7 405-9 454-1 521-1
Y

/'chloroform^"!'^ Hg 632-8 548-1 469-4 359-7 257-7 193-6 161-2 120-7

/^ total 632-8 603-0 579-5 562-1 580-4 599-5 615-3 641-8

Re
nd

I
Fi

700 -

l^tolal
-g- 600 -
E
Q> 500 ●
3
X>
10

® 400
0.

- 300-

200-

9®"
100-

0 X X X X X X

0 0-1 0-2 0-3 0-4 0-5 0-6 0-7 0-8

Mole fraction of acetone (Xacetone)

As the plot for dips downwards, hence the solution shows negative deviation from the ideal behaviour.
2/112 New Course Chemistry (XII)ESm

Q. 2.38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures
of pure benzene and toluene at 300 K are 50*71 mm Hg and 32*06 mm Hg respectively. Calculate
the mole fraction of benzenein the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
1
Ans. Molar mass of benzene (CgHg) = 78 g mol“‘, Molar mass of toluene (C^HjCHj) = 92 g mol'
80 g
No. of moles in 80 g of benzene = -1026 mole
78 g mol"*
100 g
No. of moles in 100 g of toluene = -I
= 1-087 mole
92 g mol
1-026 1-026
In the solution, mole fraction of benzene = = 0-486
1-026+1-087 2-113
mole fraction of toluene = 1 - 0-486 = 0-514

/»°Benzene = 50-71 mm, = 32*06 mm

w
Applying Raoult’s law, /’Ben.^ne = ^Benzene /^°Benzene = ^ 50-71 mm = 24-65 mm
PT0ia\ = -'Toluene Aoluenc = 0*514 X 32 06 mm = 16-48 mm

Flo
^Benzene 24-65 24-65
.*. Mole fraction of benzene in the vapour phase = = 0*60.
24-65+16-48 41-13
Pb enzene /^oluene

ee
Q. 2.39. The air Is a mixture of a number of gases. The major components are oxygen and nitrogen with

Fr
approximate proportion of 20% is to 79% by volume at 298 K. The water is In equilibrium
with air at a pressure of 10 atm. At 298 K, If the Henry’s law constants for oxygen and nitrogen are
3*30 X 10^ mm and 6*51 x 10^ mm respectively, calculate the composition of these gases in water.

for
ur
Ans. Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volume.
s
20
.*. Partial pressure of oxygen (p_02 ) = X10 atm = 2 atm = 2 x 760 mm = 1520 mm
ok
Yo
100
Bo

79
Partial pressure of nitrogen )= xlO atm - 7-9 atm = 7-9 x 760 mm - 6004 mm
100
Kh (O2) = 3-30 X 10^ mm, Kpj (N,) = 6-51 x 10^ mm
re

1520 mm
Applying Henry’s law. Pq, = K„xx or = 4*61x10-5
H O2 ^02 -
ou
ad

K
H
3-30x10"^ mm
Y

= K„xx or =
Pj<2 _ 6004 mm = 9*22x10-5.
H N2 K,,
H
6-5lxl0^mm
nd
Re

Q. 2.40. Determine the amount of CaCl2(i = 2*47) dissolved in 2*5 litre of water such that its osmotic pressure
is 0*75 atm at 27“C.
Fi

TCXV 0-75 atm X 2-5 L

Ans. 71 = I CRT = i — RT or « = = 0*0308 mole
V /xRxT 2-47x0-0821 L atm K"* mol-* x300 K
-1
Molar mass of CaCh = 40 + 2 x 35-5 = 111 g mol .*. Amount dis.solved = 0-0308 x 111 g = 3*42 g.
Q. 2.41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of
water at 25*’C, assuming that it is completely dissociated. (Pb. Board 2011)
Ans. K2SO4 dissolved = 25 mg = 0-025 g
Volume of solution = 2 L . T = 25“ C = 298 K
-I
Molar mass of K2SO4 = 2 x 39 + 32 + 4 x 16 = 174 g mol
As K2SO4 dissociates completely as K2SO4-+ 2K'''+S04-, f.e., ions produced = 3, .*. i = 3
. n w I 0-025 g 1
.*. 7i = j CRT = /—RT = /x —x —RT =3x X X 0-0821 L atm K"' mol"' x 298 K
V I
M V 174 g mol 2L

= 5*27 X 10"5 atm.

SOLUTIONS 2/113

I
WITH ANSWERS,
HINTS AND SOLUTIONS

m IT

MULTIPLE CHOICE QUESTIONS-I

1. Which of the following units is useful in relating 7. Considering the formation, breaking and

ow
concentration of solution with its vapour strength of hydrogen bond, predict which of the
pressure ? following mixtures will show a positive
deviation from Raoult’s law ?
(a) mole fraction (b) parts per million
(c) mass percentage (d) molality (a) Methanol and acetone
2. On dissolving sugar in water at room (b) Chloroform and acetone

e
Fl
(c) Nitric acid and water

re
temperature, solution feels cool to touch. Under
which of the following cases dissolution of sugar (d) Phenol and aniline

F
will be most rapid ? 8. Colligative properties depend on
(a) the nature of the solute particles dissolved in
(a) Sugar crystals in cold water
ur
r
solution
(b) Sugar crystals in hot water
(c) Powdered sugar in cold water
fo
(b) the number of solute particles in solution
ks
(d) Powdered sugar in hot water (c) the physical properties of the solute particles
Yo
dissolved in solution
3. At equilibrium the rate of dissolution of a solid
oo

solute in a volatile liquid solvent is (d) the nature of solvent particles

9. Which of the following aqueous solutions
B

(a) less than the rate of crystallisation

should have the highest boiling point ?
(b) greater than the rate of crystallisation
e

(a) l.OMNaOH (b) 10 M N32S04

(c) equal to the rate of crystallisation
ur

(C) I.OMNH4NO3 (d) I.OMKNO3

(d) zero
10. The unit of ebullioscopicconstant Is
ad

4. A breaker contains a solution of substance ‘A’.

Yo

Precipitation of substance ‘A* takes place when

(fl) K kg mol“' or K (molality)”^
small amount of ‘A* is added to the solution. (_h) mol kg K”' or K“’ (molality)
d

The solution is (c) kg mor* or K"‘ (molality)”'

Re
in

(a) saturated (b) supersaturated (d) K mol kg“* or K (molality)

11. In comparison to a 0.01 M solution of glucose,
F

(c) unsaturated {d) concentrated

5. Maximum amount of a solid solute that can be
the depression in freezing point of a 0.01 M
dissolved in a specified amount of a given liquid
MgCl2 solution is
(rt) the same (b) about twice
solvent does not depend upon
(c) about three times (d) about six times
(a) Temperature (b) Nature of solute
12. An unripe mango placed in a concentrated salt
(c) Pressure (d) Nature of solvent
solution to prepare pickle, shrivels because
6. Low concentration of oxygen in the blood and
tissues of people living at high altitude is due
to
(a) it gains water due to osmosis
(b) it loses water due to reverse osmosis
(a) low temperature
(c) it gains water due to reverse osmosis
(b) low atmospheric pressure
(d) it loses water due to osmosis
(c) high atmospheric pressure
13. At a given temperature, osmotic pressure of a
(d) both low temperature and high atmospheric concentrated solution of a substance
pressure
(a) is higher than that at a dilute solution
2/114 T^n^adecfr'^- New Course Chemistry (XII)IS3ZS

(/?) is lower than that of a dilute solution Piston (A) Piston (B)
SPM
(c ) is same as that of a dilute solution
Uf) can not be compared with osmotic pressure of
dilute solution Concentrated
sodium
14. Which of the following statements is false ? Fresh water
chloride
(a) Two different solutions of sucrose of same (A) solution in

molality prepared in different solvents will water (B)

have the same depression in freezing point
(h) water will move from side (B) to side (A) if a
{h) The osmotic pressure of a solution is given by
pressure greater than osmotic pressure is
the equation II = CRT (where C is the molarity
of the solution) applied on piston (B)
(c) water will move from side (B) to side (A) if a
(c) Decreasing order of osmotic pressure for
pressure equal to osmotic pressure is applied
0.01 M aqueous solutions of barium chloride,
on piston (B)
potassium chloride, acetic acid and sucrose is

w
(d) water will move from side (A) to side (B) if
BaCl2 > KCl > CH3COOH > sucrose
pressure equal to osmotic pressure is applied
{d) According to Raoult’s law, the vapour pressure
on piston (A)

F lo
exerted by a volatile component of a solution
20. We have three aqueous solutions of NaCI
is directly proportional to its mole fraction in
the solution labelled as ‘A’, ‘B’ and ‘C’ with concentrations

ee
0.1 M, 0.01 M and 0.001 M, respectively. The
15. The values of van’t Hoff factors for KCl, NaCl

Fr
value of van’t Hoff factor for these solutions
and K2SO4, respectively, are will be in the order
(a) 2, 2 and 2 (/;) 2, 2 and 3
(a) /’a < /b < 'c (b) > /g > Iq
(c) 1, 1 and 2 id) 1, 1 and 1
for (d) < '/b > ‘c
ur
' a ~ 'b “ 'c
16. Which of the following statements is false ? 21. On the basis of information given below mark
(a) Units of atmospheric pressure and osmotic the correct option.
s
ook

pressure are the same

Yo
Information :
(b) In reverse osmosis, solvent molecules move (A) In bromoethane and chloroethane mixture,
eB

through a semipermeable membrane from a intermolecular interactions of A-A and B-B type
region of lower concentration of solute to a are nearly same as A-B type interactions
region of higher concentration
(B) In ethanol and acetone mixture, A-A or B-B type
r

(c) The value of molal depression constant depends

ad
ou

intermolecular interactions are stronger than A-B

on nature of solvent
type interactions
(d) Relative lowering of vapour pressure, is a (C) In chloroform and acetone mixture, A-A or B-B
Y

dimensionless quantity type intermolecular interactions are weaker than

Re

17. Value of Henry’s constant K^j

nd

A-B type interactions

(a) increases with increase in temperature (a) Solution (B) and (C) will follow Raoult’s law
Fi

(b) decreases with increase in temperature {b) Solution (A) will follow Raoult’s law
(c) remains constant (c) Solution (B) will show negative deviation from
{d) first increases, then decreases Raoult’s law

18. The value of Henry’s constant Kf^ is {d) Solution (C) will show positive deviation from
Raoult’s law
(«) greater for gases with higher solubility
(b) greater for gases with lower solubility 22. Two beakers of capacity 500 mL were taken.
One of these beakers, labelled as “A”, was filled
(c) constant for all gases with 400 mL water whereas the breaker labelled
(d) not related to the solubility of gases if,
B” was filled with 400 mL of 2 M solution of
19. Consider the figure and mark the correct option. NaCl. At the same temperature both the beakers
(a) water will move from side (A) to side (B) if a were placed in closed containers of same
pressure lower than osmotic pressure is applied material and same capacity as shown in the
on piston (B) figure.
SOLUTIONS 2/115

^ 1<€ (a) 0.004 {/;) 0.008

A B (c) 0.012 (d) 0.016
25. On the basis of information given below mark
the correct option.
Water NaCI solution
Information : On adding acetone to methanol
some of the hydrogen bonds between methanol
At a given temperature, which of the following molecules break.
statement is correct about the vapour pressure
of pure water and that of NaCI solution. (a) At specific composition, methanol-acetone
mixture will form minimum boiling azeotrope
(a) vapour pressure in container (A) is more than
and will show positive deviation from Raoult's
that in container (B)
law
(b) vapour pressure in container (A) is less than
that in container (B) (h) At specific composition, methanol-acetone
mixture forms maximum boiling azeotrope and
(c) vapour pressure is equal in both the containers

w
will show positive deviation from Raoult's law
(d) vapour pressure in container (B) is twice the
(c) At specific composition methanol-acetone
vapour pressure in container (A)
mixture will form minimum boiling azeotrope

Flo
23. If two liquids A and B form minimum boiling and will show negative deviation from Raoult's
azeotrope at some specific composition, then law

ee
(d) At specific composition methanol-acetone
(a) A-B interactions are stronger than those mixture will form maximum boiling azeotrope

Fr
between A-A or B-B
and will show negative deviation from Raoult’s
(b) vapour pressure of solution increases because law

for
more number of molecules of liquids A and B 26. Kjj value for Ar(g), C02(g), HCHO (g) and
ur
can escape from the solution CH4(g) arc 40.39, 1.67, 1.83 x 10-^ and 0.413
(c) vapour pressure of solution decreases because respectively.
s
less number of molecules of only one of the Arrange these gases in the order of their
k
Yo
liquids escape from the solution increasing solubility,
oo

(d) A-B interactions are weaker than those

(a) HCHO < CH4 < CO2 < Ar
eB

between A-A or B-B

(b) HCHO < CO2 < CH4 < Ar
24. 4L of 0.02 M aqueous solution of NaCI was
(c) Ar < CO2 < CH4 < HCHO
diluted by adding one litre of water. The
r

id) Ar < CH4 < CO2 < HCHO

ou

molality of the resultant solution is

ad
Y

MULTIPLECHOICE QUESTIONS-H
nd
Re

Note : In the following questions, two or more (c) These will form minimum boiling azeotrope
Fi

options may be correct. (d) These will not form ideal solution
27. Which of the following factor (s) affect the 29. Relative lowering of vapour pressure is a
solubility of a gaseous solute in the fixed volume colligative property because
of liquid solvent ? (a) It depends on the concentration of a non
(0 nature of solute (//) temperature electrolyte solute in solution and does not
(Hi) pressure depend on the nature of the solute molecules
(a) (/) and (Hi) at constant T (b) It depends on number of particles of electrolyte
(b) (i) and (if) at constant P solute in solution and does not depend on the
(c) (ii) and (Hi) only (d) (Hi) only nature of the solute particles
28. Intermolecular forces between two benzene mole (c) It depends on the concentration of a non
cules are nearly of same strength as those between electrolyte solute in solution as well as on the
two toluene molecules. For a mixture of benzene nature of the solute molecules

and toluene, which of the following are not true ? (d) It depends on the concentration of an
(^) = zero (&)A,„^.y=zero electrolyte or non-electrolyte solute in solution
as well as on the nature of solute molecules
2/116 ‘pfuidee^'4. New Course Chemistry (XIDESSm
30. Van’t Hoff factor i is given by the expression 34. For a binary ideal liquid solution, the variation
in total vapour pressure versus composition of
Normal molar mass solution is given by which of the curves ?
(a) i =
Abnormal molar mass

Abnormal molar mass

(b) i =
Normal molar mass
(fl) (*)
Observed colligative property P P
(c) i =
Calculated colligative property
^1
Calculated colligative property ^2
id) / =
Observed colligative property
31. Isotonic solutions must have the same

w
(d)
(a) solute (b) density P
(c) elevation in boiling point

F lo
(d) depression in freezing point
32. Which of the following binary mixtures will have

ee
●^2 X2
same composition in liquid and vapour phase ? 35. Colligative properties are observed when

Fr
(a) Benzene - Toluene (b) Water-Nitric acid (a) a non-volatile solid is dissolved in a volatile
(c) Water-Ethanol (d) n-Hexane - n-Heptane liquid

for
33. In isotonic solutions
ur
(b) a non-volatile liquid is dissolved in another
(a) solute and solvent both are same volatile liquid
(b) osmotic pressure is same (c) a gas is dissolved in non-volatile liquid
ks
(c) solute and solvent may or may not be same (d) a volatile liquid is dissolved in another volatile
Yo
oo

(d) solute is always same, solvent may be different liquid

eB

ANSWERS
Multiple Choice Questions -1
r
ou

1. (a) 2.(d) 3.(c) 4,(b) 5. (c) 6.(6) 7. (a) 8.(6)

ad

9.(6) 10. (a)

11.(c) 12. (J) 13.(a) 14.(a) 15.(6) 16. (6) 17.{a) 18. (6) 19.(6) 20.(c)
Y

21.(6) 22. (a) 23. id) 24. (d) 25. (a) 26. (c)
Multiple Choice Questions - II
nd
Re

27. (a, 6) 28. (c. d) 29. (a, 6) 30. (a, c) 31. (c. d) 32. (6, c) 33. (6. c) 34. (a, d)
Fi

35. (a, 6)

HINTS FOR DIFFICULT MULTIPLE CHOICE QUESTIONS

Multiple Choice Questions -1

P°-Ps
1. - ^2 (mole fraction of solute in solution) - Raoult’s law.
P^
2. As solution is cool to touch, dissolution is endothermic. Hence, high temperature will favour dissolution.
Further, powdered sugar has large .surface area and will dissolve faster.
3. At equilibrium, rate of dissolution = rate of crystallisation .

4. In saturated solution, more substance does not dissolve. In a supersaturated solution, substance starts
precipitating out.
6. Body temperature of the human body remains constant. Low concentration of oxygen in the blood at
altitude is due to low atmospheric pressure.
SOLUTIONS
2/117

7. H— O.
. On adding acetone, its molecules get in between the molecules of methanol

CHj CH3 CH3

breaking hydrogen bonds and reducing methanol-methanol attractions.
11. 0 01 M MgCl2 = 0 03 M particle concentration which is three times in comparison to 0-01 M glucose.
Hence, depression will be about three times.
13. n = CRT, i.e., naC
14. Depression in freezing point depends upon the nature of solvent. Hence, (a) is false.
15. KCl ^ K+ + Cr (2 ions), NaCl > Na-^ + Cl" ( 2 ions), K2SO4 > 2 K* + SO4 (3 ions).
17. = Kj^ .v^. At higher temperature, is less. Hence, at the same partial pressure, p^, K^j will be more.

ow
18. Xfi^. Lower the value of greater will be the value of K^j at the same partial pressurep\ and at
the same temperature.
19. (b) represents reverse osmosis.
20. NaCl is a strong electrolyte. It dissociates to give 2 ions at all concentrations. Hence, its van’t Hoff
factor = 2 at all concentrations.

e
re
22. Vapour pressure of a liquid decreases when a non-volatile solute is dissolved in it.

rFl
23. Minimum boiling azeotrope is formed when actual vapour pressure is higher than expected, i.e., solution

F
shows +ve deviation from Raouit’s law which is so when A-B interactions are weaker than A-A or
B-B interactions.

24. M,Vj = M2V2, 0-02 X 4 = M2 X 5 or M2 = 0-016.

r
ou
25. As hydrogen bonds break, vapour pressure will be higher, i.e.. it will show +ve deviation from Raouit’s
law and form a minimum boiling azeotrope.
fo
ks
26. Lower the value of K^, higher is the solubility (Hint 18 above).
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Multiple Choice Questions -II
Y
B

28. They will form ideal solution for which H = 0, V = 0. Hence, (a) and (b) are true. But (c) and
(d) are not true (Ideal solutions do not form azeotropes).
re

31. Isotonic solutions have same molar concentration and hence same elevation in boiling point and same
depression in freezing point.
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32. (b) and (c) form non-ideal solutions and hence azeotropic mixtures which have same composition in
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liquid and vapour phase.

34. (a) and (d) both are possible depending upon which component is more volatile.
d
in
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SHORT ANSWER QUESTIONS

F

36. Components of a binary mixture of two liquids A and B were being separated by distillation. After
some time separation of components stopped and composition of vapour phase became same as that
of liquid phase. Both the components started coming in the distillate. Explain why this happened.
Ans. This shows that liquids A and B on mixing form a non-ideal solution and for a particular composition, they
form an azeotropic mixture.
37. Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases,
while addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point.
Ans. NaCl is a non-volatile solute. On adding NaCl to water, vapour pressure is lowered and hence boiling point
of water increases. Methyl alcohol is more volatile than water. On adding methyl alcohol into water, vapour
pressure of solution is greater than that of water. Hence, boiling point of water decreases.
38. Explain the solubility rule “like dissolves like” in terms of intermolecuiar forces that exist in solutions.
Ans. For one substance to dissolve into another substance, the two substances should have similar intermolecuiar
interactions. This is so only when either both the substances (solute and solvent) are polar or both are non
polar. This rule is called “like dissolves like.”
2/118 ‘Pna.dee^'4. New Course Chemistry (Xll)EZSIS]

39. Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of
temperature, however molarity is a function of temperature. Explain.
Ans. Refer to “Retain in Memory”, page 2/6.
40. What is the significance of Henry’s Law constant Kjj ?
Ans. Higher the value of Henry’s law constant, K^, lower is the solubility of gas in the liquid.
41. Why are aquatic species more comfortable in cold water in comparison to warm water ?
Ans. Refer to Curiosity Question on page 2/22.
42. (a) Explain the following phenomena with the help of Henry’s law.
(i) Painful condition known as bends,
(ii) Feeling of weakness and discomfort in breathing at high altitude,
(b) Why soda water bottle kept at room temperature fizzes on opening ?
Ans. {a) (i), {ii). Refer to Applications (ii) and (iv), page 2/19. (b) Refer to application {/), page 2/19.

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43. Why is the vapour pressure of an aqueous solution of glucose lower than that of water ?
Ans. In pure water, the entire surface is occupied by water molecules which are volatile. On adding glucose,

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some water molecules on the surface are replaced by glucose molecules which are non-volatile. Hence,
vapour pressure is lowered.
44. How does sprinkling of salt help in clearing the snow covered roads in hilly areas ? Explain the

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phenomenon involved in the process.

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Ans. When salt is spread over snow covered roads, depression in freezing point of water takes place. At the
ambient temperature, snow starts melting and it helps in clearing the roads.
45. What is “semi permeable membrane” ? for
ur
Ans. A semipermeable membrane is a continuous sheet or film of suitable material (natural or synthetic) which
has pores such that solvent molecules can pass through but big molecules of the solute cannot.
s
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46. Give an example of a material used for making semipermeable membrane for carrying out reverse
osmosis.
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Ans. Cellulose acetate.

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MATCHING TYPE QUESTIONS

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Note : In the following questions, match the items given in Column 1 and Column II.
47. Match the items given in Column I and Column II.
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Column I Column n
(0 Saturated solution (a) Solution having same osmotic pressure at a given temperature as that of
Fin

given solution.
Hi) Binary solution (h) A solution whose osmotic pressure is less than that of another,
HU) Isotonic solution (c) Solution with two components.
Hv) Hypotonic solution A solution which contains maximum amount of solute that can be dissolved
(d)
in a given amount of solvent at a given temperature.
(V) Solid solution (e) A solution whose osmotic pressure is more than that of another,
(v/) Hypertonic solution (f) A solution in solid phase.
48. Match the items given in Column I with the type of solutions given in Column II.
Column 1 Column II
(i) Soda water (a) A solution of gas in solid
(ii) Sugar solution (b) A solution of gas in gas
(Hi) German silver (c) A solution of solid in liquid
(/v) Air (d) A solution of solid in solid
(v) Hydrogen gas in palladium {e) A solution of gas in liquid
(/) A solution of liquid in solid
SOLUTIONS
2/119

49. Match the laws given in Column I with expressions given in Column II.
Column I Column II
(/) Raoult’s law
(a) AT^=Kym
(ii) Henry’s law (b) k = CRT
(Hi) Elevation of boiling point (c) p = x^p^°+X2Pl
(iv) Depression in freezing point (d) AT/, = K^m
(v) Osmotic pressure (e) p = K^.x
50. Match the terms given in Column I with expressions given in Column II.
Column I Column II

(0
Number of moles of the solute component
Mass percentage (a)
Volume of solution in litres

Number of moles of a component

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(ii) Volume percentage (b)
Total number of moles of all the components

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Volume of the solute component in solution
(Hi) Mole fraction (c) xlOO
Total volume of solution

Mass of the solute component in solution

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(iv) Molality (d) XlOO

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Total mass of the solution

Number of moles of the solute components

(V) Molarity (e)
for
Mass of solvent in kilograms
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ANSWERS
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47. (0 (d), (H) (c), (Hi) (a), (iv) -> (h), (v) -> (/), (vi) (e)
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48. (i) -> (e), (ii) (c), (Hi) -> (d), (iv) -> (h), (v) —> (a)
49. ((') -> (c), (ii) -> (e), (Hi) -> (d), (iv) (a), (v) (b)
50. (i) (d), (H) (c), (///) (b), (iv) (e), (v) -> (a)
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HINTS FOR DIFFICULT QUESTIONS

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48. (f) Soda water = CO2 in water (gas in liquid) (H) Sugar solution = sugar in water (solid in liquid)
Re

(Hi) German silver = an alloy of Cu, Zn and Ni (solid in solid) (/v) Air = a mixture of gases (gas in gas)
nd

(v) Hydrogen gas in palladium = (gas in solid)

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ASSERTION AND REASON TYPE QUESTIONS

Note ; In the following questions, a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices :
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion,
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion and reason both are incorrect statements.
(e) Assertion is wrong statement but reason is correct statement.
51. Assertion : Molarity of a solution in liquid state changes with temperature.
Reason : The volume of a solution changes with change in temperature.
52. Assertion : When methyl alcohol is added to water, boiling point of water increases.
Reason : When a volatile solute is added to a volatile solvent, elevation in boiling point is ob-served.
2/120 7^’uttUefr'A New Course Chemistry (XIDEHn

53. Assertion : When NaCl is added to water, a depression in freezing point is observed.
Reason : The lowering of vapour pressure of a solution causes depression in the freezing point.
54. Assertion : When a solution is separated from the pure solvent by a semipenneable membrane, the solvent
molecules pass through it from pure solvent side to the solution side.
Reason : Diffusion of solvent occurs from a region of high concentration solution to a region of low
concentration solution.

ANSWERS

51.(a) 52. (d) 53. (a) 54.(c)

HINTS FOR DIFFICULT QUESTIONS

52. Correct Assertion. When methyl alcohol is added to water, boiling point of water decreases.

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Correct Reason. When a more volatile solute is added to volatile solvent, vapour pressure of solvent
increases.

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54. Correct Reason. Net flow of solvent occurs from region of low concentration to the region of high
concentration.

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LONG ANSWER QUESTIONS

Fr
55. Define the following modes of expressing the concentration of a solution. Which of these modes are
independent of temperature and why ? for
(ii) V/V (volume percentage)
r
(0 w/w (mass percentage)
You
(Hi) w/V (mass by volume percentage) (iv) ppm (parts per million)
s

(vh) m (Molality)
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(v) X (mole fraction) (vi) M (Molarity)

Ans. Refer to Art. 2.3.
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56. Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction
of components in the following solutions.
(0 CHCI3 (/) and CHjCljll) (ii) NaCl(s) and H2O (/)
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Ans. (0 As both the components are volatile, in the solution

/^CHCl^ ~ ●'^CHCl3 ^ /’CHClj

and
“ ●^CH2Cl2 ^ ^CH2Cl2
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^total or P
soln= ■'^CHClj ^CHClj + = ^CHC\^ ^'CHCl^ + PCH 2CI2
Fin

/^CH2Cl2^ ■'^CHCl^ + ^CH2Cl2

(ii) As solute NaCl is non-volatile, vapour pressure of solvent in solution is equal to vapour pressure of
O

solution, Le., or P^^in "

57. Explain the terms ideal and non-ideal solutions in the light of forces of interactions operating between
molecules in liquid solutions.
Ans. Refer to Art. 2.8.

58. Why is it not possible to obtain pure ethanol by fractional distillation ? What general name is given
to binary mixtures which show deviation from Raoult’s law and whose components cannot be
separated by fractional distillation. How many types of such mixtures are there ?
Ans. Pure ethanol cannot be obtained by fractional distillation because when the mixture attains a composition
of 95.6% alcohol and 4-4% water, it becomes an azeotropic mixture. In general, such mixtures are called
‘azeotropes’. There are two types of such mixtures as follows ;
(0 Minimum boiling azeotropes formed from non-ideal solutions showing +ve deviations.
(ii) Maximum boiling azeotropes formed from non-ideal solutions showing -ve deviations.
SOLUTIONS 2/121

59. When kept in water, raisin swells in size. Name and explain the phenomenon involved with the help
of a diagram. Give three applications of the phenomenon. Semi permeable
Ans. The walls of raisin (partially dried grapes) consist of semipermeable membrane

membrane. When kept in water, they swell as concentration inside

Raisin
raisin is high. The phenomenon is called ‘osmosis’.
Applications. Refer to Art. 2.12.7, page 2/47.
60. Discuss biological and industrial importance of osmosis.
Ans.
Biological importance of osmosis. Refer to An. 2.12.6, pages 2/45, 46. H2O
Industrial importance of osmosis. Refer to Art. 2.12.7. Besides y.

these, reverse osmosis is used for desalination of water.

61. How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable

w
membrane ? Can this egg be inserted into a bottle with a narrow neck without distorting its .shape ?
Explain the proce.ss involved.
Ans. The hard CaC03 layer of the egg can be removed by placing it in HCl acid so that CaC03 layer dissolves.
Yes, it can be inserted into a bottle with narrow neck through the following steps :

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e
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rFl
Egg is now removed
After some
♦ and placed in ●Hypertonic

F
time
hypertonic solution solution
,-0-.
Egg placed in Outer shell Egg with only semi

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HCl acid solution of egg dissolves
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permeable membrane

A A fo After
ks some
Egg is now
Addition of time
placed in a
●4
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hypotonic ^ bottle with
/ solution narrow neck
Y
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Egg regains shape due to osmosis Size of the egg gets reduced as
the egg shrivels due to osmosis
62. Why is the mass determined by measuring a culligative property in case of some solutes abnormal ?
r

Discuss it with the help of van’t HofT factor.

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Ans. Refer to Art. 2.15.

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in
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2/122 New Course Chemistry (Xll)CZslSi

rOLf.' -●i'

L2. l>^

M
i
NEET/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to
● MCQs in Chemistry for MEET . .
Pradeep's Stellar Series....

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● MCQs in Chemistry for JEE (Main)

separately available for these examinations.

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Q Multiple Choice Questions correct Answer)

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(a) 5-55 mL (b) IMOmL
1. Types of solutions and

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(c) 16-65 mL id) 22-20 mL
expressing their concentration
6. How many grams of concentrated nitric acid
1. Which of the following is dependent on
temperature ? for
solution should be used to prepare 250 mL of
ur
2-0 M HNO3? The concentrated nitric acid is
(a) Mobility (b) Molarity 70% HNO3
s
(c) Mole fraction id) Weight percentage (a) 45 0 g cone HNO3 ib) 90 0 g cone HNO3
ook
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(NEET 2017)
(c) 70-0 g cone HNO3 (d) 54-0 g cone HNO3
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2. A 5-2 molal aqueous solution of methyl alcohol, (AIPMT 2013)

CH3OH, is supplied. What is the mole fraction 7. 6-02 X 10^ molecules of urea are present in
of methyl alcohol in the solution ? 100 mL of its solution. The concentration of the
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ia) 0-190 ib) 0-086 solution is

ic) 0-050 (J) 0-100 (AIEEE 2011) io) 0-02 M ib) 0-01 M
3. The molarity of a solution obtained by mixing ic) 0-001 M id) 0-1 M
Y

750 mL of 0-5 M HCl with 250 mL of 2 M HCI

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(AIPMT 2013)
nd

will be
8. To neutralise completely 20 mL of 0-1 M aqueous
(fl) 0-975 M ib) 0-875 M
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solution of phosphorous acid (H3PO3), the volume

(c) 1 -00 M id) M75 M of 0-1 M aqueous KOH solution required is
(JEE Main 2013) (<3) 10 mL ib) 20 mL
4. The density (in g mL“') of a 3-60 M sulphuric (c) 40 mL id) 60 mL
acid solution that is 29% H2SO4 (Molar mass = 9. The volumes of 4 N HCl and 10 N HCl required
98 g mol“^) by mass will be to make 1 litre of 6 N HCl are
ia) 1-45 ib) 1-64 ia) 0-75 litre of 4 N HCl and 0-25 litre of 10 N
ic) 1-88 id) 1-22 (AIEEE 2007) HCl

5. Concentrated aqueous sulphuric acid is 98% ib) 0-25 litre of 4 N HCl and 0-75 litre of 10 N
H2SO4 by mass and has a density of 1-80 g mL"'. HCl

Volume of the acid required to make one litre of (c) 0-67 litre of 4 N HCl and 0-33 litre of 10 N
0-lM H2SO4 solution is HCI

ANSWERS

I. (b) 2. ih) 3. ib) 4. id) 5. (a) 6. (a) 7. {b) 8. (c)

SOLUTIONS 2/123

(d) 0-80 litre of 4 N HCl and 0-20 litre of 10 N (d) Enthalpy of solution can be found using
HCl
Clausius-Clapeyron equation
(e) 0-50 litre of 4 N HCl and 0-50 litre of 10 N 16. Which of the following is not a substitutional
HCl solid ?

10. A person is considered to be suffering from lead (fl) Brass (b) Bronze
poisoning if its concentration in him is more than (c) Steel (d) Monel metal
10 micrograms of lead per decilitre of blood.
Concentration in parts per billion parts is 111. Solution of gases in liquids
(a) 1 (b) 10 17. The mole fraction of a gas dissolved in a solvent
(c) 100 id) 1000 is given by Henry’s law. If the Henry’s law
11. The hardness of a water sample (in terms of constant for a gas in water at 298 K is 5-55 x 10^
equivalents of CaC03) containing 10"^ M CaS04 Torr and the partial pressure of the gas is 200
is : (molar mass of CaS04 = 136 g mol"') Torr, then what is the amount of the gas dissolved
in 10 kg of water ?

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(a) 100 ppm (b) 50 ppm
(c) 10 ppm
(fl) 2 0 X 10-“* mol (b) 2-5 X 10 mol
(d) 90 ppm
(JEE Main 2019)
(c) 3-7 X 10-^ mol id) 1-2 X lO"*^ mol

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12. Calculate the molarity of 3-3 molal solution of KCl 18. The solubility of a gas in water at 300 K under a

whose density is 1-28 g mL

-1 pressure of 100 atmospheres is 4 x 10"^ kg L~‘.

ee
Therefore, the mass of the gas in kg dissolved in
(a) 3-7 M ib) 5-0 M
250 mL of water under a pressure of 250

Fr
(c) 34 M id) 2-5 M atmospheres at 300 K is
(JEE Main 2021) (a) 2-5 X 10-3 ib) 2-0 X 10-3
13. 250 g of D-glucose solution in water contains
for
(c) 1-25 X 10"3 id) 5-Ox 10-3
ur
10-8% of carbon by weight. The molality of the ie) 3x10-3
solution is nearest to (Given : Atomic weights are
s
19. For the solution of gases W, X, Y and Z in water
ook

H = 1 u, C= I2u, 0 = 16u)
Yo
at 298 K, the Henry’s law constants (Kj^) are 0-5,
(a) 1-03 ib) 2 06 2, 35 and 40 kbar respectively. The correct plot
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(c) 3-09 id) 5-40 for the given data is

(JEE Main 2022) Z
Y
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14. In one molal solution that contains 0*5 mole of a

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W
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solute, there is
Partial Partial
(a) 500 mL of solvent ib) 500 g of solvent pressure pressure
Y

(c) 100 mL of solvent id) 1000 g of solvent

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nd

(NEET 2022)
> >●
Mole fraction of H2O
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II. Solutions of solids in Mole fraction of H2O

(a) (b)
liquids and solids in solids
15. Which one of the following statements is not
true?
Partial Partial
(a) Dissolution of all solid solutes in water is pressure pressure
exothermic

ib) Common salt is more soluble in water than

canesugar at the same temperature
Mole fraction of H2O Mole fraction of H2O
(c) Solubility of sodium sulphate decahydrate (c) (cO
crystals first increases upto a certain
temperature and then decreases (JEE Main 2019)

ANSWERS

9. ic) 10.(c) 11.ia) 12.(c) 13. ih) 14.ib) 15.(a) 16.(c) 17. (a) 18.(a) 19. ib)
2/124 “Pmideefi-'A New Course Chemistry CX11)E9S19]
fractions of A and B respectively in the solution
IV. Vapour pressure of and and are the mole fractions of A and B
liquid solutions and RaouU’s law respectively in the vapour phase. A plot of 1/Y^
20. The amount of solute (molar mass 60 g mol"*) along y-axis against l/X^ along .r-axis gives a
that must be added to 180 g of water so that the
straight line. What is the slope of the straight
line ?
vapour pressure of water is lowered by 10% is
(a) (b) Pj^IPb
(a) 30 g (b) 60 g
id) 12 g (c) Pb°-Pa° (^0 Pa~Pb
(c) 120 g
(Kerala PET 2010) (where and p^ are the vapour pressures of
{e) 24 g
the pure components A and B respectively)
21. At 80‘’C, the vapour pressure of pure liquid “A is
(IAS Prelim 2010)
520 mm Hg and that of pure liquid ‘B’ is
1000 mm Hg. If a mixture solution of ‘A’ and ‘B‘ 26. The vapour pressure of a solvent decreases by
boils at 80°C and 1 atm pressure, then amount of 10 mm of mercury when a non-volalile solute
was added to the solvent. The mole fraction of the

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‘A’ in the mixture is (1 atm = 760 mm Hg)
solute in the solution is 0-2. What should be the
(a) 48 mol percent (b) 50 mol percent
mole fraction of the solvent if the decrea.se in

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(c) 52 mol percent (d) 34 mol percent
vapour pressure is to be 20 mm of mercury ?
(AIEEE 2008)
(a) 0-8 (b) 0-6
22. Two liquids X and Y form an ideal solution. The

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(c) 04 (d) 0-4
mixture has a vapour pressure of 400 mm at 300

Fr
K when mixed in the molar ratio of I : I and a 27. Which of the following statements about the
vapour pressure of 350 mm when mixed in the composition of the vapour over an ideal 1 : 1
molar ratio of 1 : 2 at the same temperature. The molar

for
mixture of benzene and toluene
correct ? Assume that the temperature is constant
is
ur
vapour pressures of the two pure liquids X and Y
respectively are at 25°C (Given vapour pressure data at 25'’C,
benzene = 12-8 kPa, toluene = 3-85 kPa)
s
(fl) 250 mm, 550 mm (b) 350 mm, 450 mm
ook
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(c) 350 mm, 700 mm (d) 500 mm, 500 mm {a)- The vapour will contain equal amounts of
benzene and toluene
(Kerala PET 2008)
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(e) 550 mm, 250 mm

23. The relative lowering of vapour pressure of an
(b) Not enough information is given to make a
prediction
aqueous solution containing non-volatile solute
r

is 0 0125. The molality of the solution is (c) The vapour will contain a higher percentage
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of benzene
(a) 0-70 (b) 0-50
(c) 0-60 id) 0-80 (d) The vapour will contain a higher percentage
Y

of toluene (NEET Phase I 2016)

(e) 0-40
Re

28. Two open beakers one containing a solvent and

nd

24. If xj and X2 represent the mole fraction of a the other containing a mixture of that solvent with
component A in the vapour phase and liquid
Fi

a non-volatile solute are together sealed in a

mixture respectively and p/i^ and p^° represent container. Over time :
vapours pressures of pure A and pure B, then total
vapour pressure of the liquid mixture is (a) the volume of the solution and the solvent does
not change
Pa° ●^2 {b) the volume of the solution increases and the
(a) (b)
^2 volume of the solvent decreases

(c) the volume of the solution decreases and the

Pb" Pb° h
ic) id) volume of the solvent increases
X2
(d) the volume of the solution does not change
25. An ideal solution is formed by mixing two and the volume of the solvent decreases.
volatile liquids A and B. X^ and X0 are the mole (JEE Main 2020)

ANSWERS

20.(6) 21.(6) 22. («) 23. (a) 24.(6) 2a5. (a) 26.(6) 27. (c) 28.(6)
SOLUTIONS 2/125

V. Ideal and Non-ideal {a) CH2O ib) C2H4O2

solutions and azeotropic mixtures (c) C^HgOj {d) 0311^03
(Karnataka GET 2011)
29. For an ideal solution, the correct option is
35. A 5-25% solution of a substance is isotonic with
{a) Aj^j^ G = 0 at constant T and P a 1-5% solution of urea (molar mass = 60 g
{b) S = 0 at constant T and P mor*) in the same solvent. If the densities of
(c) Aj^j^ V 0 at constant T and P both the solutions are assumed to be equal to
id) A,„i^ H = 0 at constant T and P (NEET 2019) 1 0 g cm“^, molar mass of the substance will be
-I -1
(rt) 210 0 g mol {b) 90-0 g mol
VI. Colligativc properties and
Relative lowering of vapour pressure (c) 1I5'0 g mol"' (d) 105-0 g mol
(AIEEE 2007)
30. Dry air is passed through a solution containing
36. Insulin (C-)H,q05)„ is dissolved in a suitable
10 g of the solute in 90 g of water and then solvent and the osmotic pressure (k) of solutions
through pure water. The loss in weight of solution of various concentrations (C) is measured at
is 2-5 g and that of pure solvent is 0-05 g.

w
20®C. The slope of the plot of n (atm) versus C
Calculate the molecular weight of the solute.
(in g/cnr^) is found to be 4-65 x 10"^. The
(fl) 50 {b) 180

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molecular weight of insulin is
(c) 100 id) 25 (a) 3-17 X K)*^ (b) 4-17 X lO*^
(e) 51 (c) 5-17 X 10^ id) 6-17 X 10^

ree
31. The mass of glucose that should be dissolved in 37. Osmotic pressure of insulin solution at 298 K is
50 g of water in order to produce the .same lowering found to be 0-0072 atm. Hence, height of the

F
of vapour pressure as produced by dissolving 1 g water column due to this pressure will be (Given
of urea in the same quantity of water is
(«) I g ib) 3 g (a) 7-4 mm
for
density of Hg = 13-6 g niL“')
(/;) 7-4 cm
r
ic) 6g id) 8g (c) 74 cm {d) 760 mm
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oks

32. The vapour pressure of a solution of a non 38. A solution of protein (extracted from crabs) was
volatile electrolyte (A) in solvent (B) is 95% of
prepared by dissolving 0-75 g in 125 cm^ of an
eBo

the vapour pressure of the solvent at the same

aqueous solution. At 4°C, an osmotic pressure
temperature. If molar mass of B is 30% of molar rise of 2-6 mm of the solution was observed.
mass of A. the mass ratio of the solvent and
Then molecular weight of protein is (Assume
our
ad

solute are
density of solution is 1-00 g/cm'-*)
(fl) 0-15 ib) 0-20
(a) 9-4 X 10^ ib) 5-4 X 10^
ic) 4-0 id) 5-7
(c) 5-4x10
10
id) 9-4 X 10'^
dY

33. If 8 g of a non-electrolyte solute is dissolved in

Re

114 g of n-octane to reduce its vapour pressure to

39. The following solutions were prepared by
80%, the molar mass (in g moI“') of the solute is dissolving
Fin

[Given that molar mass of n-octane is 114 g moL^) 10 g of glucose (C6H,20g) in 250 mL of water
ia) 40 (h) 60 (Pi).
(c) 80 id) 20 10 g of urea (CH4N2O) in 250 mL of water (P2),
(NEET 2020 Phase-1) 10 g of sucrose (Ci2H->-)0] j) in 250 mL of water
(P3).
VU. Osmotic pressure The right option for the decreasing order of
34. The empirical formula of a non-electrolyte is osmotic pressure of these solutions is ;
CHoO. A solution containing 3 g L“’ of the (a) P2 > P| > P3 {/;)P, >P2>P3
compound exerts the .same osmotic pressure as (c)P2>P3>Pi id) P3 > P, > ?2
that of 0-05 M glucose solution. The molecular (NEET 2021)
formula of the compound is
ANSWERS

29. id) 30. (c) M.ih) 32. id) 33. (<i) 34. (/;) 35. (d) 36. (c) 37. (/)) 38.(6) 39. («)
2/126 New Course Chemistry (XIDESSBI

(a) 102‘’C (b) m’^c

VIII. Elevation of boiling
(c) 101°C (^/) lOOT
point and depression of freezing point
(NEET Phase I 2016)
40. An aqueous solution of urea is found to boil at 46. In 100 g of naphthalene, 2-423 g of S was
IOO-52‘’C. Given for water is 0-52 K kg dissolved. Melting point of naphthalene
mol”*, the mole fraction of urea in the solution is
= 80-rC. A7f = 0-661°C. = 35-7 cal/g of
(a) 1 (b) 0-5 naphthalene. Molecular formula of sulphur
added is
(c) 0-018 (d) 0-25
(«) S, (h) S4
41. For a dilute solution containing 2-5 g of a non
volatile, non-electrolytic solute in 100 g of water, W S6 i.d) Sg
the elevation in boiling point at 1 atm pressure is 47. K
●/ for water is 1-86 K kg mol"*. If your
2”C. Assuming concentration of the solute is automobile radiator holds 1-0 kg of water, how
much lower than the concentration of the solvent, many grams of ethylene glycol (C2H5O2) must

low
the vapour pressure (mm of Hg) of the solution is you add to get the freezing point of the solution
lowered to - 2-8‘’C?
(take K^ = 0-76 K kg mol”*)
(a) 27 g ib) 72 g
(a) 724 (b) 740
(c) 93 g (d) 39 g (AIEEE 2012)
(c) 736 (d) 718 (IIT 2012)
48. An element X of atomic mass 25-0 exists as X4 in

ee
42. A .solution containing 1-8 g of a compound

F benzene to the extent of 100%. When 10-30 g of

Fr
(empirical formula CH2O) in 40 g of water is saturated solution of X in benzene is added to
observed to freeze at - 0-465"C. The molecular 20-0 g of benzene, the depression in freezing
formula of the compound is (Ky^of water = 1-86 point of the resulting solution is 0-51 K. If Ky for
for
ur
kg K mol”*) benzene is 5-1 K kg mol"*, the solubility of X in
(a) C2H4O2 (b) C3H6 100 g of benzene will be
s
(a) 3-0 g (b) 2-7 g
(C) C4H8O4 (d) CsH.oOg
k
Yo
(c) 0-30 g id) 0-27 g
oo

(e) (Kerala CET 2011)

49. Pure water freezes at 273 K and 1 bar. The
43. A solution containing O-IO g of non-volatile
eB

addition of 34-5 g of ethanol to 500 g of water

solute X (molar mass : 100) in 200 g of benzene changes the freezing point of the solution. Use
depresses the freezing point of benzene by the freezing point depression constant of water as
r

0-25®C while 0-50 g of another non-volatile 2 K kg mol"*. The figures shown below represent
ou
ad

solute Y in 100 g of benzene also depresses the plots of vapour pressure (V.P.) versus temperature
Y

freezing point of benzene by 0-25®C. What is the (T) [Molecular weight of ethanol is 46 g mol”'].
molecular mass of Y ?
Among the following, the option representing
nd

change in the freezing point is

Re

(a) 50 (b) 100

(c) 150 (d) 1000
Fi

(IAS Prelim 2010)

44. A solution of urea (mol. mass 56 g mol”*) boils at (3)=?
q:
lOO-lS^C at the atmospheric pressure. If lyiind :>

for water are 1-86 and 0-512 K kg mol"' 270 273 T/K ►
respectively, the above solution will freeze at
{a) -6-5rC (b) -0-654°C
1
(c) 6-54“C (d) 0-654“C
(6)5
45. At 100°C the vapour pressure of a solution of q:
7>
6-5 g of a solute in 100 g water is 732 mm. If
= 0-52, the boiling point of this solution will be 271 273 T/K ►

ANSWERS

40.(c) 41. (a) 42.(e) 43.(d) 44.(b) 45.(c) 46. (J) 47. (c) 48.(«)
SOLUTIONS 2/127

{a) 9-8% ih) 19-6%

(c) 4-9% ((/) 1-68%
(c) I"'' ■ 54. The freezing point (in °C) of a solution
a;
containing 01 g of K,|Fe(CN)J (Mol. Wt.
270
Water-t-Ethanol
273 T/K ►
329) in 100 g of water (k^,= 1-86 K kg mol"') is
{a) -2-3 X 10-2 (h) -5-7 X 10-2
(c) -5-7 X 10-^ (d) - 1-2 X 10-2
(gl' ■
55. The freezing point depression constant for water
a:
is - l-86°Cm-'. If 5-00 g Na2S04 is dissolved in
45-0 g HoO, the freezing point is changed by
- 3-82“C. Calculate the van’t Hoff factor for
271 273 T/K »
N32S04

low
(JEE Advanced 2017) (a) 0-381 {b) 2-05
50. 0-075 molal solution of sucrose containing 1 kg (c) 2-63 UD 3 -11
water is at a temperature of - 0-4°C. If Ky (H2O) = (AIPMT Prelim 2011)
1-86 K/m, how much ice in g will freeze out ?
56. A 0-1 molal aqueous solution of a weak acid is

ee
(fl)350g {b) 100 g
30% ionized. If Ky for water is 1-86° C/m, the

F
Fr
(c) 180 g {d) 650 g freezing point of the solution will be
(JEE Main 2021) (a) -0-I8°C (h) -0-54°C
51. Two solutions A and B are prepared by dissolving (c) -0-36°C (rf) -0-24°C

for
ur
1 g of non-volatile solutes X and Y respectively (AIPMT xMain 2011)
in 1 kg of water. The ratio of the depression in
57. Benzoic acid undergoes dimerisation in benzene
s
freezing point for A and B is found to be I : 4.
k
solution. The van’t HolT factor (0 is related to the
Yo
The ratio of molar masses of X and Y is
oo

degree of association ‘’x’ of the acid as

(a) 1 : 4 {b) 1 : 0-25
(a) i = il-x) (h) i=(l +x)
eB

(c) 1 : 0-20 {d)\:5

(c) i = (\-x/2) id) l = (l+A-/2).
(JEE Main 2022)
58. The freezing point of benzene decreases by
r
ou

0-45°C when 0-2 g of acetic acid is added to 20 g

ad

IX. Abnormal molar masses

of benzene. If acetic acid associates to form a
Y

52. The correct equation for the degree of association dimer in benzene, percentage as.sociation of acetic
‘a’ of an associating solute, ‘n' molecules of
which undergo association in solution is
acid in benzene will be (Ky for benzene = 5-12 K
nd
Re

kg mol"’)
n(j-l) i(n-l) (o) 76-6% (b) 94-6%
Fi

(a) a= (b) a =
\-n 1-t-/? (c) 64-6% id) 80-4%
i (n +1) i(n + l) (JEE Main 2017)
(c) a = id) a =
\-n n-\ 59. A 0-004 M solution of Na2S04 is isotonic with a
0-010 M solution of glucose at the temperature.
«(!-/)
(e) a =
\-n
(Kerala PMT 2015) The apparent degree of dissociation of Na2S04 is
(a) 25% (b) 50%
53. 1 g of a monobasic acid in ’00 g of water lowers
the freezing point by 0-16- . If 0-2 g of the same (c) 75% id) 85%
acid requires 15 mL of N/10 alkali for complete 60. For a weak monobasic acid, if pK^ = 4, then at a
neutralisation, the degree of dissociation of the concentration of 0-01 M of the acid solution, the
acid is (Kyfor water = 1-86 K kg mol-') van’t Hoff factor is

ANSWERS

49.ia) 50. id) 51.ib) 52.(a) 53.ib) 54.(a) 55.(c) 56.id) 57.(c) 58.ib) 59.(c)
2/128 “PniKiee^'A New Course Chemistry (XII)CS2a]
(a) 1-01 (b) 1-02 («)A {b)B
(c) 110 (d) 1-20 (c) In both A and B (d) Neither in A nor in B
61. The pH of 1 M solution of a weak monobasic acid 67. Pure benzene freezes at 5-3°C. A solution of
(HA) is 2. Then, the van’t Hoff factor is 0-223 g of phenylacetic acid (C(^H5CH2COOH)
(a) 1-01 (b) 1-02 in 44 g of benzene (Ky = 512 K kg mol"^)
freezes at 4-47°C. From this observation, one can
(c) MO id) 1-20
conclude that
62. At a certain Hill station, water boils at 96°C. The
amount of NaCl that should be added to one litre
(a) phenylacetic acid exists as such in benzene
(b) phenylacetic acid undergoes partial
of water so that it boils at lOO^C will be (K^ for ionization in benzene
H2O = 0-52 K/m)
(c) phenylacetic acid undergoes complete
(a) 450 g (b) 225 g ionization in benzene
(c) 125 g id) 250 g (d) phenylacetic acid dimerizes in benzene
63. One molal solution of a complex of cobalt

w
68. Consider separate solutions of 0-500 M C2H5OH
chloride with NHj in water showed an elevation iaq), 0-100 M Mg3(S04)2 (aq), 0-250 M KBr
in boiling point equal to 2-08°. Assuming that the

F lo
(aq) and 0-125 M Na3P04 (aq) at 25“C. Which
complex is completely ionized in the solution, the statement is true about these solutions, assuming
complex is (K^ for water = 0-52 K kg mol”*) all salts to be strong electrolytes ?

ee
(a) [Co(NH3)J CI3 (b) [Co(NH3)5C1] CI2 (fl) 0-500 M C2H5OH (aq) has the highest

Fr
(c) [Co(NH3)4Cl2] Cl (d) None of these osmotic pressure
64. Depression in freezing point of 0-01 m aqueous (b) They all have the same osmotic pressure
acetic acid solution is found to be 0-02046 K.
for
(c) 0-100 M Mg3(P04)^ has the highest osmotic
ur
One molal urea solution freezes at - 1-86“C.
pressure
Assuming molarity equal to molality, pH of
(d) 0-125 M Na3P04 (aq) has the highest
s
acetic acid solution is
ook
Yo

osmotic pressure (JEE Main 2014)

ia) 2 ib) 3
eB

X. Miscellaneous
(c) 3-2 id) 4-2
65. The average osmotic pressure of human blood is 69. For 1 molal aqueous solution of the following
our

7-8 bar at 3TC. What is the concentration of an compounds, which one will show the highest
ad

aqueous NaCl solution that could be used in the freezing point ?

blood stream ?
ia) [Co(H20)6] Cl3
dY

(cr) 0-15mol/L ib) 0-30 mol/L ib) [Co(H20)gCl] CI2.H2O

Re

(c) 0-60 mol/L id) 0-45 mol/L (c) [Co(H20)4Cl2] CI.2 H2O
Fin

66. Solution (A) containing FeCl3 is separated from id) [Co(H20)3Cl3],3 H2O (JEE Main 2018)
solution (B) containing K4Fe(CN)g by a 70. One gram of silver gets distributed between
semipermeablemembrane as shown uelow : 10 cm^ of molten zinc and 100 cm^ of molten
Solution (A) Solution (B) lead at 800"C. The percentage of silver still left
FeCl3 K4Fe(CN)6 in the lead layer is approximately
ia) 2 ib) 5
If FeCl3 on reaction with K4[Fe(CN)g] produces ic) 3 id) 1
blue colour of Fe4[Fe(CN)g]3, the blue colour
(Karnataka CET 2011)
will appear in

ANSWERS

60. (r) 61. («) 62. {/?) 63. (u) 64. (fc) 65. («) 66. id) 67. (</) 68. ib) 69. id) 70. (c)
SOLUTIONS 2/129

m Multiple Choice Questions (with One or More than One Correct Answers)

71. If and are the vapour pressures of the 75. For a solution formed by mixing liquids L and
solvent and its solution respectively and Nj and M, the vapour pressure of L plotted against the
N2 are the mole fractions of the solvent and mole fraction of M in solution is shown in the
solute respectively, then following figure. Here and represent mole
(a) Ps = P‘^N2 fractions of L and M respectively, in the solution.
The correct statement(s) applicable to this system
(b) P'^-Ps = P‘>N2 is (are)
(c) Ps = P‘*N|
(d) (P'^-Ps)/Ps = N,/(Nj + N2).
72. The vapour pressure of a dilute solution of a
solute is not influenced by
Z
pl

w
(a) nature of the solute if it is non-electrolyte
(b) mole fraction of the solute
(c) melting point of the solute

F lo
(d) degree of dissociation of the solute.
73. In the depression of freezing point experiment, it

ee
is found that

Fr
(a) The vapour pressure of the solution is less
than that of pure solvent 0

(b) The vapour pressure of the solution is more

than that of pure solvent for
(a) The point Z represents vapour pressure of
pure M and Raoult’s law is obeyed from
ur
(c) Only solute molecules solidify at the Xl = 0 to Xl = 1
freezing point (b) Attractive intermolecular interactions
s
ook

between L-L in pure liquid L and M-M in

Yo
(d) Only solvent molecules solidify at the
freezing point. pure liquid M are stronger than those between
eB

L~M when mixed in solution

74. Mixture(s) showing positive deviations from
Raoult’s law at 35°C is (are) (c) The point Z represents vapour pressure of
(a) carbon tetrachloride + methanol pure liquid M and Raoult’s law is obeyed
our

when 0
ad

(b) carbon disulphide + acetone

(d) The point Z represents vapour pressure of
(c) benzene + toluene
pure liquid L and Raoult’s law is obeyed when
(d) phenol + aniline (JEE Advanced 2016)
Y

Xt1. —> 1 (JEE Advanced 2017)

Re

mi Multiple
nd

Choice Questions (Based on the given Passage/Comprehens ion)

Fi

Each comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option.

^.omprehension
An ideal solution of two are many solutions which do not obey Raoult’s
liquids is a solution in which each component law. In other words, they show deviations from
obeys Raoult’s law which states that the vapour ideal behaviour which may be p<»sitive or
pressure of any component in the solution negative, However, in either case,
depends on the mole fraction of that component corresponding to a particular composition,
in the solution and the vapour pressure of that they form a constant boiling mixtures called
component in the pure state. However, there azeotropes.

ANSWETO1

71. [h,c) 72. (a.c) 73. Ui,d) 74. (<?,/)) 75. (h.d)
2/130 New Course Chemistry (Xll)S!ZsIS]

76. Which of the following mixture do you expect

Given : Freezing point depression constant
will not show positive deviation from Raoult’s water -1
law? of water (K / ) = 1*86 K kg mol
(a) Benzene-Chlorofonn Freezing point depression constant of
-t
ih) Benzene-Acetone ethanol (Kethanolj - 2-0 K kg mol
(c) Benzene-Ethanol Boiling point elevation constant of water
-1
id) Benzene-Carbon tetrachloride (Kwaler) = 0-52 K kg mol
77. An azeotropic solution of two liquids has boiling Boiling point elevation constant of ethanol
point lower than either of the two liquids when it ethanol -I
) = 1*2 K kg mol
ia) shows no deviations from Raoull’s law
Standard freezing point of water = 273 K
ib) shows a positive deviation from Raoult’s law
Standard freezing point of etlianol = 155*7 K
ic) shows a negative deviation from Raoiilt’s law
Standard boiling point of water = 373 K
(d) is saturated.
Standard boiling point of ethanol = 351*5 K

w
78. A solution has a I : 4 mole ratio of pentane to Vapour pressure of pure water = 32*8 mm Hg
hexane. The vapour pressures of the pure
Vapour pressure of pure ethanol = 40 mm Hg
hydrocarbons at 20”C are 440 mm of Hg for

F lo
-1
pentane and 120 mm of Hg for hexane. The mole Molecular weight of water = 18 g mol
-1
fraction of pentane in the vapour phase would be Molecular weight of ethanol = 46 g mol
In answering the following questions,

ee
ia) 0-200 (b) 0-478
consider the solutions to he ideal dilute

Fr
ic) 0-549 (d) 0-786 solutions and solutes to be non-volatile and
non- dissociative.
fG*om5reh‘gSgi5nlB (IIT 2008). Properties
for
79. The freezing point of solution M is
ur
such as boiling point, freezing point and
vapour pressure of a pure solvent change ia) 268-7 K ib) 268-5 K
when solute molecules arc added to get
s
(c) 234-2 K (d) 150-9 K
ook
Yo
homogeneous solution. These are called
colligative properties. Applications of 80. The vapour pressure of the solution M is
eB

colligative properties are very useful in day- (a) 39-3 mm Hg (b) 36-0 mm Hg
to-day life. One of its examples is the use of (c) 29-5 mm Hg id) 28-8 mm Hg
ethylene glycol and water mixture as anti 81. Water is added to the solution M such that the
our

freezing liquid in the radiator of automobiles.

ad

mole fraction of water in the solution becomes

A solution M is prepared by mixing ethanol 0-9. The boiling point of the solution is
and water. The mole fraction of ethanol in
(a) 380-4 K ib) 376-2 K
Y

the mixture is 0*9.

ic) 375-5 K id) 354-7 K
Re
nd

09
Fi

Matching Type Questions

Match the entries of column I with appropriate entries of column II and choose the correct option out
of the four options given.
82. Column 1 (Solutions mixed) Column U (Normality after mixing)
(A) 100 cc of0-2NH2SO4+ 100 cc of 0-1 N HCl ip) 0-25 N
(B) 100 cc of0-2MH2SO4+ 100 cc of 0-1 M HCl iq) 0-067 N
(C) lOOccofO-1 MH2SO4+lOOccofO-1 MNaOH (/) 0-15 N
(D) 100 cc of 0-1 M HCl + 50 cc of 0-2 N NaOH is) 0-05

ia) A-r, B-^, C-p, D-q ib) A-q, B-p, C-r D-5 (c) A-r, B-p, C-i, D-^ id) A-s, B-r, C-p, D-q

ANSWERS

76. (fl) 77. ib) 78. [h) 79. id) 80. ia) 81. ib) 82. (c)
SOLUTIONS 2/1SI

83. Column I (Siibs(unce) Column II (Solubility)

(A) Li2C03 (P) Increases continuously with increase of temperature
(B) KCl (^) Decreases continuously with increase of temperature
(C) Na2SO4.I0 H2O (r) First increases and then decreases

(D) NH4NO3 is) Increases but not continuously

(a) As, B-r, C-p, D-q (b) A-r, B-j, C-q, D-p (c) A-p, B-q, C-r, D-s (_d) A-q, B-p, C-r, D-5

84. Column 1 (van’t HofT factor) Column II (Behaviour)

(A) (■> 1 (p) There is association.
(B) i < I iq) There is dissociation,
(C) f = 1 (r) Impossible

w
(D) / = 0 (i) No association or dissociation

F lo
(a) A-p, B-q, C-x, D-r (b) As, B-r, C-p, D-q (c) A-r, Bs, C-q, D-p (d) A-q, B-p, Cs, D'f

ee
□ Matrix-Match Type Questions

Fr
Match the entries of column I with appropriate entries of column II. Each entry in column I may

for
have one or more than one correct option from column II. If the correct matches are A-p, s ; B-r ;
ur
C-p, q ; D-s, then the correctly bubbled 4x4 matrix should be as as shown on the side :
s
ook

p q r s
Yo

A 0© o©
eB

©© o©
B :

C 0 ®IO ©1
r
ad
ou

D
®li©il0 ©
Y
Re

85. Column I Column IT

nd

(A) Carbon tetrachloride + Toluene ip) Shows positive deviation from ideal behaviour
Fi

(B) Chloroform + Benzene iq) Shows negative deviation from ideal behaviour
(C) Carbon tetrachloride + Chloroform (r) Mixing is endothermic
(D) Benzene + Toluene (s) Shows ideal behaviour

86. Column I (Solvent) Column II (Value of Kf or K^)

(A) 0-1 M Glucose sol. (P) Lowest freezing point
(B) 0-1 M Sucrose sol. iq) Highest freezing point
iC) 01 M BaCl2SoI. ir) Lowest osmotic pressure
(D) 0-1 M Ca(N03)2 sol. is) Highest osmotic pressure

ANSWERS

83. (r/) 84. Ul} 85. (A-p.r ; B-q ; C-p,r : Ds) 86. (A-q.r ; B-q.r ; C-p.s : D-p.s)
2/132 "P>utdeefr'4. New Course Chemistry (Xll)ISSlSl

VI. Integer Type Questions A B C D

DIRECTIONS. The answer to each of the following questions is a ®© @®

single digit integer, ranging from 0 to 9. If the correct answers to the ®® ®0
question numbers A, B, C and D (say) are 4, 0, 9 and 2 respectively,
(hen the correct darkening of bubbles should be as shown on the side : ®® ®@
87. Number of moles of Na.>C03 that should be dissolved in 4 litres of the solution to
®® ®®
obtain 1 N Na2C03 solution is 0@ @@
88. The molality of a sulphuric acid solution in which the mole fraction of water is
0-86 is
®® ®®
®® ®®

w
89. 100 mL of 1 M H2SO4 are mixed with 2(X) mL of 8 M HCl solution. The
normality of the resulting solution is 0® 00
90. The elevation in boiling point expected for 0-3 m A1^(S04)3 solution will be how 0 0 0®

F lo
many times compared with the elevation in boiling point of 01 m Na2S04
solution ? 00 0®

ee
91. If the compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution.

Fr
Which shows a 2-5 K rise in the boiling point of the solution is molal (K/, = 0-52 K kg mor')
(JEE Main 2021)

for
ur
92. 29-2% (w/w) HCl stock solution has a density of 1 -25 g mL"’. The molecular weight of HCl is 36-5 g mol"'. The
volume (mL) of stock solution required to prepare a 200 mL solution of 0-4 M HCl is (IIT 2012)
ks
93. A compound H^X with molar weight of 80 g is dissolved in a .solvent having density of 0-4 g mL"'. Assuming
Yo
oo

no change in volume upon dissolution, the molality of a 3-2 molar solution is (JEE Advanced 2014)
eB

94. MX-> dissociates into M""*" and X" ions in an aqueous solution with a degree of dissociation (a) of 0-5. The
ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of
freezing point in the ab.sence of ionic dissociation is (JEE Advanced 2014)
r
ou
ad

95. If the freezing point of a 0 01 molal aqueous solution of cobalt (III) chloride-ammonia complex (which
behaves as a strong electrolyte) is - 0 0558®C, the number of chloride(s) in the coordination sphere of the
Y

complex is (K^of water = 1-86 K kg mol"') (JEE Advanced 2015)

nd

96. The mole fraction of a solute in a solution is 0-1. At 298 K, molarity of this solution is same as its molality.
Re

Density of the solution at 298 K is 2 0 g cm"^. The ratio of the molecular weights of the solute and solvent
Fi

MW
solute
, IS (JEE Advanced 2016)
MW
solvent

97. 1'2 mL of acetic acid is dissolved in water to make 2-0 L of solution. The depression in freezing point
observed for this strength of acid is 0 0198°C. The percentage of dissociation of the acid is . (Nearest
integer) (Given : Density of acetic acid is I 02 g mL"', Molar mass of acetic acid is 60 g mol ,K/-(H20) =
-1

1-85 K kg mol"') (JEE Main 2022)

ANSWERS
87.(2) 88.(9) 89.(6) 90.(5) 91.(3) 92.(8) 93.(8)
94.(2) 95.(1) 96.(9) 97.(5)
SOLUTIONS 2/133

VII.
Numerical Value Type Questions Decimal Notation)
For each of the following questions, enter the correct numerical value, (in decimal-notation, truncated/rounded-
off to the second decimal place, e.g., 6-25, 7*00, - 0*33, 30*27, - 127-30) using the mouse and the onscreen virtual
numeric keypad in the place designated to enter the answer.
98. Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary
solution of liquids A and B has vapour pressure 45 Torn At the same temperature, a new solution of A and B
having mole fractions and ATg respectively, has a vapour pressure of 22-5 Torr. The value ofx^Jx^ in the new
solution is

(Given that the vapour pressure of pure liquid A is 20 Torr at temperature T) (JEE Advanced 2018)
99. The plot given in the figure shows P-T curves (where P is the pressure and T is the temperature) for two solvents
X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

w
11 2 V/4

F lo
760

ee
!»:) 1. Solvent X

Fr
X
E / '
2. Solution of NaCl
,//' in solvent X
u
3. Solvent Y

for
3
ur
lA
(/)
4. Solution of NaCl
c-
in solvent Y
k s
Yo
oo

O r- 00
\D sC
eB

r*1 rn n-i

Temperature (K)

On addition of equal number of moles of a non-volatile solute X in equal amount (in kg) of these solvents, the
r

elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerisation
ou
ad

in these solvents. If the degree of dimerisation is 0-7 in solvent Y, the degree of dimerisation in solvent
X is. (JEE Advanced 2018)
Y

100. On dissolving 0-5 g of a non-volatile non-ionic solute in 39 g of benzene, its vapour pressure decreases from
650 mm Hg to 640 mm Hg. The depression of freezing point of benzene (in K) upon addition of the solute
nd
Re

IS.
Fi

(Given data ; Molar mass and the molal freezing point depression constant of benzene are 78 g
mol“' and 5-12 K kg mol“* respectively) (JEE Advanced 2019)
101. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0-05. If the density of the
solution is T2 g cm~^, the molarity of urea solution is
(Given data : Molar masses of urea and water are 60 g mol'' and 18 g mol'' respectively)
(JEE Advanced 2019)
102. A 1 molal K4Fe(CN)g solution has a degree of dissociation 0-4. Its boiling point is equal to that of another
solution which contains 181 weight percent of a non-electrolyte solution A. The molar mass of A is
g/moi (Round off to the nearest integer) (JEE Main 2021)
103. 2-5 g of protein containing only glycine (C2H5NO2) is dissolved in water to make 500 mL of solution. The
osmotic pressure of this solution at 300 K is found to be 5-03 x 10'^ bar. The total number of glycine units
present in the protein is (Given R = 0083 L bar K'' mol'') (JEE Main 2022)

J-* -

98. (19 0(0 99. (0-05) 100. (1-U3) 101. t2 98) 102. t85, 103.(330)
2/134 “Pn^idee^’4. New Course Chemistry CXll)CZs29]

VIII. Assertion-Reason Type Questions

TYPE I

DIRECTIONS. Each question contains STATEMENT-1 (Assertion) and STATEMENT-2

(Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Choose the correct option as under :
(a) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation of Statement-1,
(b) Statement-1 is TVue, Statement-2 is IVue; Statement-2 is NOT a correct explanation for Statenieiit-1.
(c) Statement-1 is True, Statement-2 is False,
(d) Statement-1 is False, Statement-2 is True.

104. Statement-1. If on mixing the two liquids, the solution becomes hot, it implies that it shows negative
deviation from Raoult’s law.

w
Statement-2. Solutions which show negative deviation are accompanied by decrease in volume.
105. Statement-1. If a liquid more volatile than the solvent is added to the solvent, the vapour pressure of the

F lo
solution may increase, i.e., > p®.
Statement-2. In the presence of more volatile liquid solute, Raoull’s law does not hold good.
106. Statement-1. One molar aqueous solution has always higher concentration than one molal solution.

ee
Statemcnt-2. One molar solution contains less solvent than one molal solution.

Fr
107. Statement-1. Vapour pressure of water is less than 1013 bar at 373 K.

for
Statement-2. Water boils at 373 K as the vapour pressure at this temperature becomes equal to atmosphere
pressure.
ur
108. Statement-1. Any concentration of NaCl solution can be injected intravenously as NaCl, being a common
s
table salt, is a harmless chemical.
ook
Yo

Statement-2. 0-9% (mass/volume) NaCl solution is isotonic with the fluid inside the blood cells.
eB

TYPE II

DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a
our
ad

corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as
(a) If both assertion and reason are true, and reason is the true explanation of the assertion,
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion,
Y
Re

(c) If assertion is true, but reason is false,

nd

(d) If both assertion and reason are false.

Fi

109. Assertion. One molar aqueous solution has always higher concentration than one molal.
Reason. The molality of a solution depends upon the density of the solution whereas molarity does not.
(ARMS 2010)
110. Assertion. Greater the value of Henry’s constant of a gas in a particular solvent, greater is the solubility of
the gas at the same pressure and temperature.
Reason. Solubility of a gas is directly proportional to its Henry’s constant at the same pressure and
temperature.
111. Assertion. If a liquid solute more volatile than the solvent is added to a solvent, the vapour pressure of the
solution may increase, i.e., > p°.
Reason. In the presence of the more volatile solute, only the solute will form the vapours ,and solvent will not.

ANSWERS

104. (/;) 105. (c) 106. (o) 107. {d) 108. {d) 109. (/;) 110. {d) 111. (c)
SOLUTIONS 2/135

112. Assertion. If red blood cells were removed from the body and placed in pure water, pressure inside the cells
increases.

Reason. The concentration of the salt content in the cells increases. (AIIMS 2006)
113. Assertion. Addition of a non-volatile solute to a volatile solvent increases the boiling point.
Reason. Addition of a non-volatile solute results in lowering of vapour pressure. (AlIMS 2014)
114. Assertion. The depression in freezing point depends on the amount of the solute dissolved and not on the
nature of the solute or solvent.

Reason. For aqueous solutions of different electrolytes, molal depression constant will have different value.
115. Assertion. 0-1 M solution of glucose has higher increment in the freezing point than 0-1 M solution of urea.
Reason. K^for both has different values.
116. Assertion. The molecular weight of acetic acid determined by depression in freezing point method in
benzene and water is found to be different.

Reason. Water is polar and benzene is non-polar. (AIIMS 2005)

w
117. Assertion. Higher the molal depression constant of the solvent used, higher the freezing point of the
solution.

F lo
Reason. Depression in freezing point does not depend on the nature of the solvent. (AIIMS 2011)

ee
Fr
For Difficult Questions

Multiple Choice Questions (with one correct Answer) for

ur
s
29
1. Molarity of a solution depends upon the volume
ook

Further, 29 g H-,SOj = — mole.

Yo

of the solution which changes with temperature. - ■* 98

All others depend only on the masses of the solute Hence, molarity of the solution
eB

and solvent which do not change with 29/98

X1000 mol L"' = 3-60 mol L ● (Given)
temperature. lOO/d
our
ad

2. 5-2 molal solution means 5-2 moles of the solute

i.e., xl000 = 3-60
(CH3OH) are present in 1000 g of water, i.e., 98 100
1000/18 moles = 55-55 moles of water. 3-60 98
Y

or d = X— = 1-22 gmL
Mole fraction of CH3OH 10 29
Re
nd

5-2
5. 98% H2SO4 by mass means 98 g H^SO^ present
in 100 g of the solution,
Fi

= 0-086 ●
5-2-t-55-55
i.e., 100/1-80 mL = 55-5 mL
3. MjVi -t- M2V2 = M3 (V, + V2) Also 98 g H2SO4 = 1 mole H2SO4
0-5 X 750 + 2 X 250 = M3 (750 + 250) Molarity of 98% H2SO4 solution
1
875 X1000 molL~*
375 + 500= 1000 M3 or M3 = 1000
= 0-875 M 55-5

Now, MjV, = M2V2.

4. 29% H2SO4 by mass means 29 g H2SO4 are
1000
present in 100 g of the solution. If d g mL“’ is the Hence, MxV, = 1000 niLx 0-1M
55-5
100
density of the solution, volume mL ●
or V, =5-55 mL.
d

ANSWERS

112.(c) 113. (a) 114. (d) 115. (d) 116. (a) 117. (d)
2/136 “Pnadee^'4. New Course Chemistry (XII)ISBIM]

12. 3-3 mol are present 1000 g of the solvent

For Difficult Questions
3-3 mol of KCI = 3-3 x 74-5 g = 245-8 g
6. 250 mL of 2-0 M HNO3 will contain HNO3 Mass of solution = 1000 + 245 g = 1245-8 g
1245-8
= -^x250 - 0-5 mol = 0-5 X 63 g = 31 -5 g
Volume of solution =
1-28
= 973-3mL
1000

As concentrated HNO3 is 70%. therefore, concen 3-3

Molarity = xlOOO =3-39 = 3-4 M
100 973-3
trated HNO3 required = x31-5g =45g
70 13. Glucose is C^Hi206 (M = 180)
7. No. of moles of urea present in 100 mL of solution 72
Weight of carbon in glucose = x250 (as it
^6-02x10-° 180

ow
= 10 3 mol
~ 6-02 xlO^^ contains 6 C-aloms)
Molar concentration of urea in the solution = 100g
10-3 As solution contains 10-8% Ccirbon by weight.
xiooo = io-2m = o-oim
100 100

e
Mass of solution = X100 = 925-93 g

Fl
re
8. H3PO3 is dibasic while KOH is monoacidic 10-8
n,M,V,= n,M,V„

F
Molality of solution
2 X 0-1 X 20= 1 X 0-1 X or V^ = 40mL.
(250/180) 250
9. Suppose j: L of 4 N HCI are taken. Then 10 N HCl
ur xl000 = XlOOO

or
(925-93-250) 180x675-93
taken = (1 - .r) L. Applying
N,V, + N2V2=N3V3
14.
sf = 2-06

Molality is the moles of the solute dissolved in

k
4 X a: -I- 10 (1 -x) = 6 X 1
Yo
1000 g of the solvent
oo
4
or 6x = 4 or x= — = 0-67 L . 1 molal solution means 1 mole of the solute
6
eB

10. 10 ^g = lOx 10-^g = 10"^ g aie dissolved in iOOO g of the solvent

I decilitre = 0-1 L = 100 mL 0-5 mole of the solute are dissolved in 500 g of
the .solvent.
Thus, 100 mL of blood contain lead = lO"^ g
ur

15. Dissolution of some solids is exothermic while

1 billion (10^) parts will contain lead
ad

that of some others is endothermic.

Yo

10-3 16. Steel is an interstitial solid in which iron atoms occupy

100
xlO^ = 10^ = 100 parts lattice sites and C-atoms occupy interstitial sites.
d
Re

i.e., concentration = 100 parts per billion parts 17. By Henry’s law, x
in

11. 10"3 M means 10“3 moles of CaSO^ present in I L Pa 2OOT01T

F

of water or X
A “
= 3-6x 10"^
K
H
5-55x10'^ Ton-
Mass of IQ-3 moles of CaS04 = 10"3 x 136 g
n n n
= 0-136g But -
A A _ A

1000/18
Thus. 0-136 g of CaS04 are present in 1 L of 'U ”h,o
water, />., I kg of water or 1000 g of water or
103 g of water 1000
= 3-6xl0“^x
1000
mole
''a = -"a 18
I03 g of water contain CaS04 = 0-136 g 18

= 2-0x 10"^ mole.

10^ g of water will contain CaS04
18. By Henry’s law m = p

10
— xlO =136 g At constant temperature, Kj^ is constant.
136gCaS04s 100gCaCO3 At two different pressures,
Hardness of water in terms of equivalents of _ 4x10-3 kgL-* _ 100
CaC03 = 100 ppm. m' p' m‘ 250
SOLUTIONS 2/137

Molality = 112 when = 1000/18 mole

For Difficult Questions
= 55-55 mole

, 250 ^1 + /J2 1 n
I
or m
x4xl0-^kgL-' 0-0125
= 80 or + 1 = 80
100 «2 »2
10-2 55-55
= 10-2kgL-' = n n
— kg in 250 inL 1 _2l_ = 0-70
4
or
= 79 or ^2 = 79 79
= 2-5x 10-3 kg.
Pa
19. According to Henry’s law. 24. Mole fraction in the vapour phase (.Tj) = —
total
/^gas “ -'^gas ...(0 BuIPa = j:aXPa‘’ = .V2XPa“
or p
gas = Kj[(l-.v^p) - K|^ Kj_j Hence, x, = ~ ^ or .Ptotal _Pa!^

w
total ■^1
Hence, plot of pgas vs 'A
HjO
will be linear with
25. From point 2 of Supplement Your Knowledge,
negative slope. Further, for a fixed value of .v H-,0’ page 2/26,

Flo
Pgas Kf^. Hence, the correct plots arc shown
ec

Pb” 1 Pa°-Pb°
in option (b). (y = mA: + c)

ee
>’a Pa^-'^a Pa”
20. If p°= 100 mm.p^= 100 - 10 = 90 mm

Fr
Hence, plot of I/yA along y-axis versus 1/ata
P°-Ps _ h^_^2 _ vv.
2
X
M
1
along -v-axis will be straight line with slope
P^ w.1 /M 1 M2 w.
= Pb>a“

for
1
ur
100-90 W2 18 26. A p/p^ = X-,. Hence, A p/A p' = X2lX’^, i.e., 10/20
100 60
X
180 = 0-2/X2 or .V2' = 0-4. Hence, x^ = f - 0-4 = 0-6.
ks
27. P Benzene - ●'Benzene
r X P°
Yo
10 Benzene
X 60x10 = 60g .
oo

or ^2 = 100
^Toluene ” ●'■Toluene ^ ^Toluene
eB

21. Ptotal (at 80°C) = 760 mm

For an ideal 1 : 1 mixture of benzene and toluene
Plotal = ^APA'’ + -'^BPB” = -’^APA" +0 “-U)Pb“
1 1
= Pb° + -'^a(Pa°-Pb°)
r

■'’Benzene 2 ’ ‘'toluene 2
ou
ad

1000 + .Ta (520 - 1000) = 760

or 480 Aa = 240 1 1
Y

P = — X P =
- -xl2-8kPa = 64 k Pa
or
Aa = 0-50, i.e., 50 mol percent. Benzene 2 Benzene 2
22. Molar ratio of 1 : 1 means a^ = Xy = 0-5
nd
Re

1 1
P = — X P =
_ Ax3-85kPa= 1-925 k Pa
Ptotal=^xPx‘’ + ^vPY” ‘Toluene 2 ‘Toluene 2
Fi

400 = 0-5 /?x" + 0-5 Py° Thus, the vapour will contain higher percentage
or px” Py° ” ...(/) of benzene as the partial vapour pressure of
1 2 benzene is higher than that of toluene.
Molar ratio of 1 : 2 means, Xy = - , Av = —
X 3 ^ 3 28. Vapour pressure of solvent is higher than that of
the solution. To attain equilibrium over time,
350 =
^Px“ + fpY° vapour pressure of solvent beaker will decrease
and that of the solution beaker will increase, i.e.,
or
Px“ + 2py°= 1050 ...(H)
some vapour from solvent beaker will condense
Eqn. (//) - Eqn. (/) gives py° = 250 arm into the solution beaker. Thus, volume of solution
Putting in (i), we get p^° = 550 mm will increase and that of solvent will decrease.

23. Relative lowering of vapour pressure 29. For an ideal solution, A^^j^ H = 0 at constant T
= mole fraction of the solute in the solution. and P

"2 30. Loss in weight of solution « V.P of solution (p,)

I.e., = 0-0125
Loss in weight of solvent « p° - p^
2/138 'P%«teUep’'<i New Course Chemistry (X[l)C!ZsI93

Empirical formula mass of CH2O = 30

For Difficult Ouestions
Mol. mass 60
.'. n = — = 2
P°~Ps _ Loss in weight of solvent E.F. mass 70

Ps Loss in weight of solution Hence, molecular formula = 2 x CH2O = C0H4O-5

35. Isotonic solutions have the same molar
iv^/Mt w-tM|
concentration. Hence,
IV,/M VV| M2 Molar cone, of the substance = Molar cone, of
0-05 10/Mt 2 urea

90/18 -1
52-5 gL 15gL-'
or M^= 100 M 60gmo]
-1

31. P°-P^ _ ^^2 _ ^2^1 (100 g solution = 100 mL as J = 1 g mL~*)

P° Uj M2 52-5x60

w
M =
15
= 210gmol ’.
As (p° - Pg)/p° is same in the two cases
36. K = C RT. Here, concentration C' is in mol L“'

F lo
w'2 Mj
To convert g cm"^ into mol L”', we have
VW[ M^~ /glucose Wj M2 y urea
C (g cra“^) = -^xlOOO mol Lr'

ee
M2

Fr
VV2XI8 1x18
50x180 “ 50x60 or W2 = 3 g. Thus, C' in the formula should be replaced by this
factor, i.e..
32. = 95% of p°, Le., p^ = 0-95 p° for CxlOOO lOOORT
ur
Mg = 30% of M^, ie.. Mg = 0-30 M^^ n = RT = C

(B = solvent, A = electrolyte)
M2 M2
s
Thus, a plot of 7t vs C will be linear with slope
ook
Yo
Applying complete formula (Refer to page 2/37)
lOOORT
eB

P"-Ps _ n
2 _ W2/M, = ^X M 1 w. M
B = 4-65 X 10 ^ {Given)
X
M2
Ps n
1 vv’j/M, w
I M, w
B
M
A
1000x0-0821x293
our

= 4-65x10-3
ad

p°-0-95 iv 0-05 w
_ "A
xO-30 or ^xO-30 M^
0-95 w 0-95 vv
B B or
M2 = 5-17x 10^
Y

37. P = fidg
Re

vv
B _
0-30x0-95
nd

or = 5-7 But P = 0-0072 atm. Hence, h = 0-0072 x 76 cm

0-05
»’a of Hg column, d = density of Hg = 13-6 g cm“3,
Fi

g = 981 cm s“3
33.
P°~P.s _ ”2 _ '‘VM2 _ Hence, P = 0-0072 x 76 x 13-6 x 981
w./M
n
1 vVjM.,
(for Hg column)
100-80 8x114 100 For water, d=\ g cm-3. Hence, for water column
or M2 = 8 X 20
= 40
P = /ix 1 x981
100 II4XM2
Thus, h x 981 = 0-0072 x 75 x 13-6 x 981
34. Cone, of compound in solution = 3 gL or h = 7-4 cm

= — mol L-' 38. Given h = 2-6 mm - 0-26 cm, V = 125 cm^

M
K - hdg - 0-26 X 1 X 980 dyne/cm^
As it is isotonic with 0-05 M glucose solution,
M =
wRT 0-75x(8-3I4xl0'^)x277
71 V (0-26x980)(125)
= 0-05 or M = 60
M
= 5-4x 103.
SOLUTIONS 2/139

44. AT^=K^xm .-. 0'18 = 0-5I2xm

For Difficult Questions
or w = 0-18/0-512

0-!8
39. Osmotic pressure (jc) = iCRT where C is molar
concentration of the solution and i is van’t Hoff AT^= KyX m = 1-86 X 0-512 = 0-654

factor. As all the compounds are non-electrolytes, Tj-= -0-654“C

i = 1 (same) for all. Thus, greater the C, greater is
K. As volume of .solution in each case is same 45.
P°-Ps n
2
n
2

wJM
and mass dissolved is same, greater the molar n.
1
mass, less is the number of moles and hence less
760 - 732 «■>
is the molar concentration (C). Molar masses are
glucose = 180, urea = 60, sucrose = 342. Hence, 760 100/18
order of molar concentration will be urea >
_ 28 X 100 = 0-2046 mole

ow
glucose > sucrose and, therefore, order of tc will or n
2 “ 760 18
be urea {P2) > glucose (Pj) > sucro,se (P3).
0-2046 mole
40. AT^ = X m. Molality of the solution =
_ AT^ _ 0-52 = 1 100/1000 kg
Hence, molality, m - 2-046 mole kg
-1
0-52

e
Fl
AT^ = X molality = 0-52 x 2-046 = 1-06°

re
Molality = 1 means 1 mole of solute in 1000 g of
solvent.
i.e., T^-T°=l-06°

F
But 1000 g of solvent (water)
or =T° +1-06 = 100 + 1-06°C = 10I-06°C
1000
ur
or
moles = 55-55 moles
18
46. M2 =
sf1000 K^W2 _ 1000 ^2 X
RT«20
1
.●. Mole fraction of urea = = 0-018. w,1 AT
/ VVj ATy- lOOOL
/
k
Yo
1-1-55-55
1000x2423 2x(353-l)^
oo

lOOOK^ Wj 1000 x 0-76 x 2-5

= 256gmol
-1

41. Mj = = 9-5
“ 100x0-661 1000x35-7
WjXAT^
B

100x2
If Sj is the formula, then molecular mass a: x 32
P'^-Ps
re

n
2 ~ = 56 or.^ = 8.
X =—i (asn2«n^)
n
1 47. Refer to Hint to Problem 3, page 2/64
u
ad

VV2/M., 1000 K w.
Yo

f '^2
m',/M I Mj =
vv, AT^
760 _ 2-5/9-5
d

= 0-047 1000x1-86x^2
Re

“ 100/18 62 = or
W2 = 93-3g
in

760
1000x2-8
or 760 - p^ = 760 x 0-047 = 35-7 48. Suppose saturated solution of X in benzene
F

or
p^ = 724-3 mm contains w g of X (present as X4). Hence,
lOOOK^ vv^ 1000x1-86x1-8 benzene present = (10-30 - v»') g
42. M2 = vv.1 X AT 40x0465
= 180
.●. Total benzene present = 20 + (10-30 - vv)
/
= (30-30 - w) g
180
= 6
E.F. mass CH2O = 30 n =
30 lOOOK^ W2 .-. 0-51 =
1000x5-1 x VI'
AT
/ “ (30-30-vv)x 100
.●. Molecular formula = 6 x CH2O = CgH]205 VV] X M 2

lOOOKj, n'2 lOOOK^ W2' CM20fX4 = 25 x4= 100)

43. AT
/ " or 51 (30-3 - w) = 5100 vv or 30-3 - u’= 100 vv
VV] XM2 W,'xM2' 101 w = 30-3
or or w = 0-3 g
w. 0-50 200
1
xlOO Thus, 10-30 g of saturated solution contains 0-3 g
or
M2' = — X
XM2 = 100
X
0-10
vv
1 of X and 10-0 g of benzene.
= 1000. .●. Solubility ofXin 100 g of benzene = 3-0 g
2/140 New Course Chemistry (XII) E*ZSTMl

N
For Difficult Questions 15-1 mL of — alkali neutralise acid = 0-2 g
10
1000 mL of 1 N alkali will neutralise acid
49. As T increases, V.P. increases. Hence, options (c)
and {d) are wrong. 0-2
xlOOOxlO = 13245 g
15-1
ATy= KyX m
34-5/46
Eq. wt. of the acid = 13245
T“ =2x Mol. mass of the acid = 132-45 (as it is
0-5
monobasic)
273-T^=3 or T,.= 270 This is calculated (normal) mol. mass
Hence, only option {a) is correct. Normal mol. mass 132-45
i = = 1-196

50. AT^. =K^xm = K.x-^x

/
Mz
—xlOOO w
1
Observed mol. mass 110-71

HA H^ + A"
1
= K /fXn^ X — X 1000 Initial 1 0 0
vv
1 After disso. 1 - a a a

1000 Total = 1 + a
04 = 1-86 X 0-075 X
(ivj = water present)

F low
vv. ●. /=l+a or a = t- 1 = 0-196 = 19-6%
1

or VV| = 348-75 g 54. K,[Fe(CN)J ^ ^ 3 K* + [Fe(CN)^]^

.*. Water frozen as ice = 1000 - 348-75 = 651-25 i = 4
= 650 g vv, 1
(Refer to Solved Problem 10, page 2/63) AT/= iKftn =jxK-x —X — xlOOO

e
51.
ATy= i K^/n = Ky-m
for Fr 0-1 1
vv
1

(i = 1 for both non-electrolytes) = 4xl-86x — X —XlOOO

AT. (A) 329 100
m 1/M M
A _ X _ Y
= 0-023 = 2-3 X 10-2°CorK
AT,(B)
/
m 1/M M
Your

B Y X
Ty= 0-2-3 X 10-2^ = -2-3 X \0~^°C
s
eBo k

AT4A)
/ 1
But ~ = 025 (Given)
AT,(B) 4 55. ATr=ixK/-xm
/ ●/
= ixKrX
■/
—XlOOO
/ Mz w.
1
ad

M a25
our

Y _ 5 1
or Mx : My = 1 : 0-25 3-82 = / X 1-86 X — X — XlOOO
M 1
X 142 45
52. rt A An (Molar mass of Na2S04 = 142)
Re

1 mole or i = 2-62
a 56. HA —> H* + A"
Find Y

a
I -a Total = 1 - a -1- —
rt n I-a a a

a a 1 1-0-3 0-3 0-3

i = 1 - a+— or 1 -i = a — = a I—
/I n n .-. I = (1 - 0-3)-r 0-3+0-3= 1-3
n~l ATy= I Kym= 1-3 X 1-86x0-1 =0-2418
= a
n Ty= 0 - 0-2418^ = - 0-2418T = - 0-24°C.
57. 2 CgHgCOOH A (C6HgCOOH)2
n
or a = (/-/) = Before asso. 1 mol
(n-\) (l-n)
X
After asso. 1 —.r
53. Mz (mol mass of acid) =
IOOOxK^xk'z 2
w,1 xAT
/ Total =l-.c + - = l--
2 2
1000x1-86x1
= 110-7!
, l-x/2
100x0-168 i = = l-i.
i.e.. Observed molecular mass = 110-71 1 2
SOLUTIONS
2/141
!●' 1 ●:-;-/(
For Difficult Questions
61. pH = 2 means fH'*’] = 10"^ M
HA ^ H-^ + A-
58' AT^(observed) = 0-45<> Initial C mol L ^ 0 0

AT
lOOOK^ h-2 After disso. C-Ca C a C a,
/(calculated) Total = C (i + a)
vv’j X M2
Thus, [H"^] = C a, i.e., 10“^ = 1 x a or a = 10"^
1000x512x0-2 / = 1 + a= 1 +0-01 = I-01
20x60
[M2(CH3C00H) = 60]
62. Required AT^ = 100 - 96 = 4°
= 0-853°
>v^
AT„ = /K,,.u =/K^ ^x —xlOOO
AT M, w
1
/(observed) 0-45
I = = 0-527
Wo

low
AT 0-853 4= 2x0-52x
/ (calculated) l.e.. X xlOOO
58-5 1000
2 CH3COOH ^ ^ (CH3C00H)2 or Wo = 225g (1 LH20= 1000 g)
a
1 -a
2
63. AT^ = / m
2-08 = i X 0-52 X 1 or / = 4

ee
F
a a
Total moles = 1 - a -f — = 1
This means that the complex produces 4 ions in

Fr
2 2
the solution.
a a
1 =1 = 0-527 or — = 0-473 or a = 0-946 Hence, the complex is [Co(NH3)gJCl3.
for
2 2
ur
% association = 94-6% 64. For 1 m urea solution, ATy-= Ky-m gives
59. K (Na2S04) = / CRT = i (0-004) RT Ky= I-86° C/m
s
For acetic acid, Aiy= i Ky-m
ook

TC (Glucose) = CRT = 0-010 RT

Yo
0-02046 = /x 1-86x0-01 or /=1-1
As solutions are isotonic, i (0-004) RT = 0-01 RT.
eB

This gives / = 2-5 CH3COOH ± CH3C00- + H-^

Initial C mol L-‘
Now, Na2S04 ^ ^ 2Na+ + SO^~ After disso. C -C a C a C a.
r

1 mole 0 0
ou
ad

Total = C-fCa = C(l +a)

1-a 2 a a.
. C(l + a)
Y

Total = 1 + 2 a i = = 1 +a or a = / - 1 = 0-1
C
/ = 1 + 2 a
[H-^j = C a = (O-OI) (O-I) = 10“"-^ M
Re
nd

i-l 2-5-1 Hence, pH = 3

or a =
Fi

— =0-75=75%.
2 65. NaCl sol. used should be isotonic with blood
60. pK„ - 4 means for HA = 10“^ stream. For NaCl, / = 2. Tt = / CRT

For weak acid, HA H+ + A- n 7-8 bar

C =
= C
/RT 2x0-083b<irLK-‘ mol"' x3I0K
(Ostwald’s dilution law)
= 0-15 mol L-^
K 10^ 66. As there is only a net flow of solvent from one
a =
a —
= I0"> =0-10
y c y 0-01 solution to the other through the semi-permeable
membraneand there is no flow of solute, the two
HA ^ H^ -b A- solutes will not come in contact with each other.
Initial 1 mole Hence, no colour will appear in A or B.
Moles after 1 - a a a. 67. Observed molecular mass of phenylacetic acid
dissoc. Total = 1 -f a 1000x5-12x0-223
= 312-6
/= 1 +a= 1 -f0-10= 1-10 (5-3-4-47) X 4-4
2/142
New Course Chemistry (XII)Egg:

(c) [Co(H20)4Cl2] C1.2 H2O

For Difficult Questions lCo(H20)4Cl2]-^ + C1-. / = 2
(d) [Co(H20)3Cl3].3 H2O
Calculated molecular mass of CgH.5CH2COOH No ionization, / = 1
= 72 + 5 + 12 + 2 + 12 + 32 + 1 = 136
Thus, [Co(H20)3Cl3l-3 H^O produces minimum
As observed molecular mass is ne;\rly double of number of particles. Hence, it has highest
the theoretical value, it dimerizes in benzene. freezing point.
70. According to Nernst distribution law.
68. Osmotic pressure for electrolytes, 7t = i CRT
Cone. (%) of Ag in molten Zn = K
7U (C2H5OH) = 1 X 0-5 RT = 0-5 RT, Cone. (%) of Ag in molten Pb
TC (Mg3(P04)2> = 5 X 0-100 RT = 0-5 RT
(distribution coeff.)
K (KBr) = 2 X 0-25 RT = 0-5 RT, = 300
n (Na3P04) = 4 X 0-125 RT = 0-5 RT
Thus, all solutions have the same osmotic
If at equilibrium, a: g of Ag is left in 100 cm^ ot
molten lead. Ag that goes into 10 cm^ zinc

w
pressure. = {l-.^)g
69. Maximum freezing point means minimum 10(1-^:)

F lo
= 300 or 10- 10a-=300.v
depression in freezing point. Less the number of X
particles produced, less is the depression in 1
freezing point. We have : or 310.r= 10 or X = — g
31

e
Fre
(a) [CoCHoO)^] CI3 ■>
1

[Co(H20)6l^+ + 3 cr, i = 4 Thus, out of 1 g of Ag, Ag present in Pb layer = ^ g

(b) [Co(H20)5CI] CI2.H2O )
[Co(H20)5C1]2+ + 2 cr. i = 3
for
% of Ag left in the Pb layer = — xlOO = 3%
r
You

n Multiple Choice Questions (with One or More than One Correct Answers)
oks
eBo

74. CCI4 + CH3OH = Positive deviation from Raoult’s law

^CH3
our
ad

CS^+0=C = Positive deviation from Raoult’s law

CH3
Benzene + Toluene = Ideal solution
dY
Re

OH NH2
Fin

= Negative deviation from Raoult’s law

75. Point Z represents the point when mole fraction of M is zero, i.e., for pure L and not for pure M. Hence, (n)
and (c) are wrong. When 1, solution is very dilute {containing very very small amount of M). Hence, it
obeys Raoult’s law, i.e., (d) is correct.
As the given plot shows -fve deviation from Raoult’s law, L-M interactions are weaker
than L-L and M-M interactions. Hence, (b) is correct.

on Multiple Choice Questions (Based on the given Passage/Compreh ension)

76. Benzene-Chloroform system shows negative deviations from Raoult’s law.
77. Lower boiling point means higher vapour pressure than expected which implies positive deviation.
 SOLUTIONS
2/143

For Difficult Questions

ATy= K^(ethanol) x m = 2 0 x 2-4J5 = 4-83 K
Freezing point of the solution (ethanol)
78.
■^pentane ^ -^'hexane
= 0-8 = Tj° (ethanol)-AT^=: 155-7-4-83 = 150-87 K
= 150-9 K
/^pentane “ 0-2 X 440 mm - 88 mm Psol. = P\+ ^iPi = 0-9 X 40 + 0-1 X 32-8 mm
/’hex ane - 0-8 X 120 mm = 96 mm
= 36 + 3-28 = 39-28 mm = 39-3 mm
Mole fraction of pentane in vapour phase
88 81. Now, ethanol will be solute and water will be
= 0-478 solvent.
88 + 96
79. Mole fraction of ethanol in the mixture = 0-9 Molality of ethanol in the solution

ow
i.e., 0-9 mole of ethanol is mixed with 0-1 mole of 0-1
water, i.e., water is the solute and ethanol is the xl000 = 6-17
solvent. 0-9x13

Molality of water in the solution ATft = (water) xm = 0-52 x 6-17 K = 3-2 K

O-I Boiling point of solution

e
xl000 = 2-415
= (water) + AT^ = 373 + 3-2 K = 376-2 K

re
0-9x46

rFl
F
09 Matching Type Questions

r
82. A. 100 cc of 0-2 N H2SO4 = 100 x 0-2 meq C. 100 cc of 0-1 M H2SO4 = 10 mmol = 20 meq
ou
= 20 meq
fo
100 cc of 0-1 M NaOH = 10 mmol = 10 meq
ks
100 cc of 0-1 HCl = 100 X 0-1 meq = 10 meq 10 meq NaOH will neutralize 10 meq of H2SO4
Total meq = 30, Total volume = 200 cc H2SO4 left = 10 meq. Total volume = 200 cc
oo
lOmeq
30 meq Normality = = 0-05N.
Y
Normality = = 0-15N . 200 cc
B

200 cc
D. 100 cc of 0-1 N HCl = 10 meq
B. 100 cc of 0-2 M H2SO4 = 100 x 0-2 mmol
re

50 cc of 0-2 N NaOH = 10 meq

= 20 mmol = 40 meq 10 meq of HCl will completely neutralize 10 meq
ou

of NaOH
Y

1000 cc of 0-1 MHCl = 100 X 0-1 mmol

ad

= 10 m mol = 10 meq NaCl formed in the .solution = 10 meq

Normality of NaCl in the solution
d

50 meq
Normality = = 0-25N. _ 10 meq = 0-067 N.
in

200cc
Re

~ 750cc
F

VI.
Integer Type Questions
87. No. ol moles of Na2C03 required for 1 L of 1 N n
2
= 0-14
solution = 0-5
55-55+ /I2
No. of moles of Na2C03 required for 4 L of 1 or
/i2 = 0-14/72+ 7-777
N solution = 4 X 0-5 = 2.
or
0-86 ti2 = 7-777 or n-, = 9-0.
88. Mole fraction of H2SO4 in ^he solution 89. 100 mL of 1 M H2SO4 = 100 millimoles
= I -0-86 = 0-14 = 200 milli eq
200 mL of 8 M HCl = 1600 millimoles
/»2
= 0-14
= 1600 milli eq
/t| + ti2 Total milli eq = 200 + 1600 = 1800
Total volume = 100 + 200 mL = 300 mL
Molality is /?2 in 1000 g of water, i.e., in n 1
= 1000/18 =55-55 moles 1800
.-. Normality of resulting solution = = 6N .
300
2/144
'a- New Course Chemistry (XU) iMaiai

r !'■● AT 0-0588
For Difficult Questions
/ _ = 3
I =
K ./7J 1-86x0-01
/
90. AT^ = / X
As 3 ions are produced on dissociation of the
For 0-3 m Al2(S04)3, AT^ = 5 x x 0-3 = 1 -5 complex, the molecular formula of the complex
For 0-1 m Na2S04, AT^ = 3 x x 0-1 = 0-3 K^i, is [Co(NH3)5C1] CI2. Thus, only one chloride ion
Thus, AT^ for Al2(S04)3 solution will be 5 limes is present in the coordination sphere.
that for Na9S04 solution. 96. Mole fraction of solute = 0-1 means moles of
91. AB- ^ A-" + B-, f=l-a-t-ot + a= l+ ot solute = 0-1, moles of solvent = 0-9
1-a a a a = 75% = 0-75 Mass of solute = 0-1 x M^oj^e g.
.-. i= 1-75 Mass of solvent = 0-9 x Msoivent 8
Total mass of solution
AT/, = / K/, m
= (0-1 + 0-9 g
2-5 = 1-75 X 0-52 xm
-1
Total volume of solution
or m = 2-75 mol kg

w
Mass 0-1M solute + 0-9M solvent
= 3 mol kg"* (nearest integer)
Density 2x1000

F lo
92. Stock solution of HCl = 29-2% (w/w)
Thus, 29-2 g of HCl are present in 100 g of the Molarity of solution
solution
0-1x2000
molL ’
As density of the solution = 1-25 g mL“',

ree
0-lM + 0-9M
solute solvent

F
100
mL
volume of 100 g of the solution = 200

As molar mass of HCl = 36-5 g mol"*, molarity of

1-25
for0-lM
solute
+ 0-9M solvent

the solution
Molality of solution
r
You
29-2
oks

X xl000 = 10M 0-1x1000 100

-1 _
36-5 100 mol kg
0-9 M solvent
eBo

0-9 M
solvent
M,V, M2V2
(stock solution) (solution required) As molarity = molality
200 100
10 xV| =0-4x200 or Vi = 8mL
ad
our

+ 0-9M solvent 0-9M solvent

93. 3-2 molar solution contains 3-2 moles of H2X in 0-lM
solute
1000 mL of the solution. As there is no change in or 0-1 M solute -I-0-9M solvent = 2x0-9
volume on dissolution, volume of solvent
Re
dY

= 1000 mL. 0-1M

solute
-1 or + 0-9 = 1-8
Mass of solvent = 1000 mL x 0-4 g mL = 400g M
Fin

solvent
= 0-4 kg
M 0-9
solute
3-2 moles or = 9
Molality = = 8 mol kg ' = 8 m M
solvent
0-1
0-4 kg

94. MX2 m2+ + 2X" 97. AT^=/K/-m,

1-a a 2a 1-2x1-02
Moles of solute (acetic acid) =
I = 1 - a -f a + 2a = 1 + 2a 60
But a = 0-5, .-. ( = 1 + 2 X 0-5 = 2 As moles ofsolute are very less, molarity = molality
This is the required ratio. 1-2x1-02
0-0198 = /xl-85x or t=l-05
95. ATy=iKym 60x2
Given m = 0-01 molal, T|-= - 0-0558“C,
-1
Kr= 1-86 K kg mol CH3COOH ■> CH3COO- + H-^
1 - a a a

AT^ = T" - T^ = 0 - (- 0-0558) = 0-0558“ .●./=l+a or a = i-1 = 1-05-I =0-05 or5%

SOLUTIONS
2/145

For Difficult Questions

VII.
Numerical Value Type Questions (in Decimal Notation)
98. In 1st case : = 0-5, = 0-5, P-j- = 45 torr P
Applying Raouit’s law, W.v = > 2
“ -^-A Pa "*■ Pb

45 = 0-5 X 20 -f 0-5 x Pg (ATp, = (0/K^),t«

or 0-5 =45-10 = 35 or
Pb ~ 70 torr
_ (0, (K,)^
(0, (K,)^,

ow
In 2nd case : Pj = 22-5 torr, = ?, Ag = ?
^T~^A Pa+(^--^‘a)Pb 3_(l-P/2) x2
22-5 = AAx20-f(l-AA)x70 0-65
or
22-5 = 20aa + 70-70aa P _ 3x0-65
or = 0-975

e
Fl
47-5 _ 19 2 2

re
or 50 a.A =47-5 or A
A “
50 " 20 P

F
or ^ = 0-025 or P = 0-05
19 2
_ 19/20 = 19
20 20 ■■ 1/20
ur P -Ps _ n
2

r
●% 100.
(/I2 = moles of solute,
P°

fo
+ n.
99. From curves I and 2, elevation in boiling point of «i = moles of solvent)
solvent X = 2°
ks
For NaCI, i = 2 650-640 «2
Yo
(AT,)^ = r(K,),»r or
2 = 2(K^_^/« ...(/) 650 0-5 + n.
oo

From curves 3 and 4, elevation in boiling point of

solvent Y = I® 39
B

tn
(39 g = — mole = 0-5 mole)
fO
re

i=2(KA'« ...(/■/■) 1 "2

Dividing eqn. (0 by eqn! {«). or
or 0-5 -f ii2 = 65 fi2
u

65 0-5 +
ad
Yo

...(f/7)
(K,), 1 or 64 «2 = 0-5 or n — mole
^ 640
d

Also, we are given .: (^'TA_3

Re
in

1
(^T,), 1 Molality of the solution = —-X — xIOOO
640 39
As solute S undergoes dimerisation in solvent X as
F

well as solvent Y
In solvent Y, 2S
■> S2 ATy= KjX m = 5-) 2 X
640
X — X1000
39
Initial 1 mole 0
= 1-0256 = 1-03 K
a
After dimerisation (1 - a)
n
urea
2 101. A urea
n -+■ n
a a
urea
H2O
Total = l- a + — = I —
n
2 2 ao5 = urea

n + 50
a 0-7 1-3 urea

(/), = ■ , 2 2
= 0-65
900
In solvent X, 2S (900 g H2O = moles = 50 moles)
S2 18
Initial 1 mole 0 or n
urea
= 0-05 n urea + 0-05 X 50

P P 2-5
After dimerisation I - p 2
Total = l~^ or 0-95 n urea = 2-5 or n = 2-63
2 urea
0-95
 New Course Chemistry (XIl)EZ5aP
2/146
K4Fe(CN)g ^ 4K++ Fe(CN)J;
For Difficult Questions
1 - a 4 a a

Mass of solution = Mass of urea + Mass of water / = 1 + 4 a = 1 + 4 X 0-4 = 2-6

= 2-63 X 60 + 900 Substituting in eqn. ,/),
= 157-8 + 900= 1057-8 g 181 1
2-6 X 1 = 1 X m = X xlOOO
Mass 1057-8 M 100-18-1
Volume of solution =
Density 1-2
18100
- 881-5 mL 2-6 = or M = 85
81-9M
2-63
XlOOO 103. jc = i CRT (/ = 1)
Molarity of solution - 881-5
= 2-98 mol L" 5-03x10"^ = xl000x0-083x300
M 500
102. AT/, [K4Fe{CN)fi] = AT/, (A)

w
-1
(ixK/, x;n) = (/x K/, x m) M (molar mass of protein) = 24751-5 g mol
-1
[K4Fe(CN)fJ (A) Molar mass of glycine (C2H5NO2) = 75 g mol

F lo
or / X m / X m
24751-5
= 330
[K4Fe(CN)6l (A) No. of glycine units in protein = 75

ee
Fr
VIII. Assertion-Reason Typo Questions

104. Correct explanation. In case of solutions for

111. Correct R. Both solute and solvent will form the
r
showing negative deviation, the forces of vapour but vapour phase will be richer in solute.
You
interaction between the components are stronger 112. Correct R. Water enters into the cells.
s
ook

than those in the pure components. increases the

113. Lowering of vapour pressure
105. Correct statement-2. Raoult’s law is applicable
boiling point as more heal is required to make
eB

to volatile solutes also.

vapour pressure equal to atmospheric pressure.
106. Statement-2 is the correct explanation of Hence, R is the correct explanation of A.
Statement-1.
our
ad

114. Correct A. The depression in freezing point

107. Correct Statement-1. At 373 K, i.e., 100°C, the
vapour pressure of water is equal to 1 atmosphere depends upon the nature of the solvent (because
which is equal to 1-013 bar. different solvents have different value for K/).
dY

Correct R. For aqueous solutions of different

Re

108. Correct Statement-1. Only 0-9% NaCl solution

can be injected intravenously because it is electrolytes, K^is same because solvent is same,
Fin

isotonic with the fluid inside the blood cells, i.e., water.
otherwise they will shrink or swell. 115. Correct A. Both being non-electrolytes with
109. Correct explanation. One molar aqueous same molar concentration will have same
solution contains one mole of the solute in 1000 decrease (not increment) in freezing point.
mL of the solution whereas one molal solution
Correct R. Both being aqueous solution, i.e.,
contains one mole of the solute in 1000 g of the
solvent. Thus, solvent present in one molar solvent is same (water), Ky-will be same for both.
solution is less than 1000 mL. Hence, 116. R is the correct explanation of A. (In water,
concentration is more. benzoic acid undergoes dissociation while in
110. Correct A. Greater the value of Henry’s constant benzene, it undergoes association.)
of a gas in a particular solvent, less is the 117. Correct A. Higher the molal depression constant
solubility of the gas at the same pressure and of the solvent used, higher is the depression in
temperature. freezing point.
Correct R. Solubility of a gas is inversely Correct R. Depression in freezing point depends
proportional to its Henry’s constant at the same upon the nature of the solvent.
pressure and temperature.
 3

w
Flo
ELECTROCHEMISTRY

e
re
F
3.1. GENERAL INTRODUCTION
ur
r
3.1.1. Redox reactions

fo
We studied about Redox reactions in class XI. The main concepts are briefly reviewed below ;
ks
A redox reaction is the reaction which involves both reduction and oxidation, i.e., in which one
Yo
substance (reactant) is reduced while the other substance
oo

(reactant) is oxidized.
As oxidation is a reaction which involves loss of electron and reduction is a reaction which involves
B

gam of electrons, hence m terms of electronic concept, redox reaction is a reaction in which one substance
re

loses electrons, i.e., gels oxidized and the other substance gains electrons, i.e., gets reduced.
For example, consider the reaction
u
ad

Zn (.9) + CUSO4 {aq) ZnS04 (aq) + Cu (5)

Yo

Zn (5) + Cu-+ (aq) Zn^"^ (aq) + Cu (^)

or
...(0
Zn is oxidized to Zn"'*’ ions while Cu^'*’ has been reduced to Cu.
d
Re

Similarly, in the reaction

in

Cu (s) + 2 AgN03 (aq) ■> Cu(N03)2 + 2 Ag U)

F

Cu (s) + 2 Ag-^ ^ Cu^+ (aq) + 2 Ag (s)

or
...(«)
Cu has been oxidized to Cu""*" ions while Ag'*' has been reduced to Ag.
Again in the reaction

Zn (.9) + H2SO4 (aq) > ZnS04 + H2 (g)

or
Zn (j) + 2 H'*’ (aq) > Zn-^ (aq) + H2 (g)
Zn has been oxidized to Zn^'^ whereas H'*’ ions have been reduced to H 2 gas.
_ The substance which gets reduced oxidizes the other substance and is called oxidizing agent or
oxidant while the substance which gets oxidized reduces the other substance and is called reducing agent or
reductant. Thus, oxidizing agent is a substance which gains electrons while reducing agent is a substance
which loses electrons.

3/1
3/2 ‘P>uidee^'4. New Course Chemistry (XlI)EEIBl
A erdox eraction may be considered to be made up two half-reactions, one involving oxidation, U., loss of
electrons and the other involving reduction, fe., gain of electrons. These are called oxidation h^-reaction and
reduction half-reaction. For example, the reaction (i) may be split into two half-reactions as under;
Zn ■> + 2 e" (oxidation half-reaction)
Cu2+ + 2 e' ^ Cu (reduction half-reaction)

Overall reaction : Zn -H Cu-+ ^ Zn^+ + Cu

3.1.2. Electrochemical cells ^

When a redox reaction is allowed to take place in a single beaker, e.g., by dipping a zinc rod in CUSO4
solution, beaker is found to be warmer. This shows that redox reacdons are accompanied by release ot
energy called ‘Chemical energy’. By suitable arrangement, the chemical energy thus produced in a erdox
reaction can be converted into electrical energy. The arrangement is called galvanic cell or voltaic cell.
Conversely a reaction may not be spontaneous. However on supplying electrical energy to it, the
reaction may become spontaneous. For example, no reaction takes place on melting NaCl. But on supplying

w
electrical energy to molten NaCl, Na^ and Cl" ions start moving towards the electrodes. Cl 10ns lose

F lo
electrons to form Cl atoms, i.e., undergo oxidation whereas Na'^ ions gain electrons to form Na atoms, /.e., undergo
reduction. Thus, redox reaction occurs on absorbing electrical energy which was otherwise non-spontaneous.
The flow of current due to movement of ions is called electrolytic conduction. The phenomenon is called
electrolysis is called electrolytic cell.

e
electrolysis and the arrangement used to bring about

Fre
lUeclrochemistO'is defined as that branch ofchemistry which deals with the reUitionship between
electrical energy and chemical changes taking place in redox reactions, i.e., how chemical
for
energy produced in a redox reaction can be converted into electrical energy or how electrical
be used to bring about a redox reaction which is otherwise non-spontaneous.
r
energy can
You
The arrangements used to bring about these conversions are called electrochemical cells. Thus, there
oks

are two principal types of electrochemical cells ;

eBo

(1) Galvanic cells or voltaic cells

(2) Electrolytic cells
Galvanic cells are further classified into two types:
our
ad

(1) Chemical ceils in which electrical energy is produced only due to chemical changes occurring within
the cell and no transfer of matter takes place.
(2) Concentration cells in which electrical energy is produced due to physical change involving transfer
dY
Re

of matter from one part of the cell to the other.

Chemical changes involving production or consumption of electricity are called electrochemical changes.
Fin

Thus, the three main aspects of study in the branch of electrochemistry are :
(1) Electrolytic cells and Electrolysis (2) Electrolytic conduction (3) Galvanic cells/Voltaic cells.
Besides these, the other related topics include studyof batteries, fuel cells and corrosion of metals.
Importance of Hlectroclieniistry.The subject of electrochemistry has great theoretical as well as practical
importance. For example,
(/) A number of metals such as Na Mg, Ca and A1 and a number of chemicals such as NaOH, CI2, F2, etc.
are commercially produced by electrochemical methods.
(ii) Batteries and cells used in various instruments and other devices convert chemical energy into electrical
energy.
(//() The sensory signals sent to the brain through the cells and vice versa and also the communication
among different cells are based on electrochemical phenomena.
Moreover, the reactions carried out electrochemically are generally energy efficient and less polluting,
i.e., they are eco friendly. Hence, their study is gaining importance to develop new technologies.
Now, in this unit, we shall take up the study of different aspects one by one.
ELECTROCHEMISTRY 3/3

3.2. ELECTROLYTIC CELL5 AND ELECTROLYSIS

3.2.1. Definitions of the terms used

Electrolysis may be defined as a process of decomposition of an electrolyte by the passage of

electricity through its aqueous solution or molten (fused) state.

The apparatus used to bring about electrolysis is called FIGURE 3.1

electrolytic cell (Fig. 3.1). It consists of a glass vessel in which the D.C.
SOURCE AMMETER
electrolyte in the form of an aqueous solution or in the molten state is + ■

taken. Two metal rods are dipped into it and are connected to a source ; e
of electricity, i.e., battery. These rods are called electrodes ; the one
connected to the negative pole of the battery is called cathode and
the other connected to the positive pole of the battery is called anode.

w
ANODE CATHODE

3.2.2. Mechanism of electrolysis

The process of electrolysis can be easily explained on the basis

Flo
of ionization theory. According to this theory, whenever an electrolyte
(acid, base or salt) is dissolved in water or is taken in the molten
I

ee
state, the electrolyte dissociates to produce positively and negatively

Fr
charged ions. On passing electric current, the positively charged ions
move towards the cathode and hence are called cations whereas Electrolytic cell

for
negatively charged ions move towards the anode and hence are called anions. On reaching their respective
ur
electrodes, ions lose their charge and become neutral particles/species. The cations accept electrons from the
cathode to become neutral species whereas the anions give electrons to the anode to become neutral species.
s
Thus, oxidation occurs at the anode while reduction takes place at the cathode. The conversion of ions into
k
Yo
oo

neutral species at their re.spective electrodes is called the primary change. The product formed as a result of
primary change may be collected as such or it may undergo a secondary change to form the final products.
eB

Further, when the electrolysis of an electrolyte is carried out in the molten state, the products of electrolysis
are actually the substances obtained from the ions of the electrolyte. But the situation is not so simple when
r

the electrolysis of an electrolyte is carried out in the aqueous solution. This is due to reason that in the
ou
ad

aqueous solution, besides the ions of the electrolyte which are present in large amounts, there are also some
and OH" ions furnished by the slight ionization of water. Consequently, when electric current is passed,
Y

both, the cations of the electrolyte and H'*' ions from water, would move towards the cathode, while the anions
Re
nd

of the electrolyte and OH" ions from water would move towards the anode. In other words, at each electrode,
there are two different ions which can be discharged. Which of these two ions gets actually discharged
Fi

depends
(/) upon their relative discharge potentials (i.e., the potential at which an ion is discharged).
(//) sometimes on the material of the electrode used.
Usually ions with lower discharge potentials are discharged in preference to those which have high
discharge potentials. Let us now discuss the electrolysis of some compounds in the light of the above
mechanism :
2+
(i) Electrolysis of molten lead bromide. In the molten state, PbBr2 exists as Pb and Br" ions. On
passing electricity, Pb^"^ ions move towards cathode and Br" ions move toward anode. On reaching cathode,
each Pb^^ ion picks up two electrons from the cathode, and becomes neutral lead atom (primary change)
Thus, lead is liberated at the cathode.

On reaching anode, each Br" ion gives an electron to the anode and becomes a neutral bromine atom
(primary change). Since bromine atoms are unstable, therefore, two bromine atoms combine to form a
bromine molecule (secondary change). Thus, bromine gas is evolved at the anode.
3/4 New Course Chemistry (XII)BE

The reactions occurring at the two electrodes may be shown as follows :

At cathode: Pb2+ (0 + le- Pb (/) {reduction, primary change)
At anode: Br" (0 - e~ -^●Br {oxidation, primary change)
● Br + ● Br - ^ Br2(g) {secondary change)
Electrolysis
Overall reaction : Pb^"*" (/) + 2Br (/) ^ Pb(/) + Br2(g)
Electrolysis
PbBr2(/) Pb(/) + Br2(g)

ow
or ■>

At cathode At anode

The lead obtained at the cathode is in the molten state,

(ii) Electrolysis of molten sodium chloride. Molten sodium chloride contains Na"^ and Cl“ ions.

e
NaCl(0 ^ ± Na+(/) + Cl-(/)

re
On passing electricity, Na'*’ ions move towards cathode while Cl" ions move towards anode. On reaching

Flr
the cathode, each Na"^ ion picks up one electron from the cathode and becomes a neutral atom. Thus, sodium

F
metal is liberated at the cathode.

ou
On reaching the anode, each Cl" ion gives an electron to the anode and becomes a neutral chlorine atom.
Since chlorine atoms are unstable, therefore, two atoms of chlorine combine to form a chlorine molecule.

sr
Thus, CI2 is evolved at the anode.

fo
The reactions occurring at the two electrodes may be shown as follows :
At cathode: Na"*" + e~ ^ Na
k {reduction, primary change)
oo
At anode: Cl- - e- ^ Cl {oxidation, primary change)
Y
● Cl + ● Cl ^ Cl2(g) {secondary change)
reB

Electrolysis
uY

Overall reaction : 2Na''’ (/) + 2C1- (0 ^2Na (/) + Cl2(g)

Electrolysis
or 2NaCl (0 2Na(/) + Cl2(g)
ad
do

At cathode At anode

The metal obtained at the cathode is in the molten state

in

(iii) Electrolysis of acidulated water. Water is only weakly ionized and hence is a poor conductor of
Re

electricity but the presence of an acid makes it a better conductor. This is because water acidified with
F

sulphuric acid contains three types of ions, i.e., OH" and SOj” produced as follows :
H20(/) ± H+(a^) + OH-(a^);

H2SO4 {aq) ^ 2H+ {aq) + SOj- {aq)

When electric current is passed through acidulated water, ions move towards cathode, OH and
SO4- ions move towards anode. Now, since the discharge potential of OH~ ions is much lower than that of
SOl~ ions, therefore, OH~ ions are discharged at anode while SOj“ ions remain in the solution. Similarly,
H'*’ ions are discharged at the cathode. The reactions, occuring at the two electrodes may be written as :
At cathode: H+ {aq) + e- ^ H- {reduction, primary change)
H- + H ^ H2 {g) {secondary change)
At anode: OH-(a^) - e- —> OH {oxidation, primary change)
4 OH ^ 2H2O + O2 {secondary change)
ELECTROCHEMISTRY 3/5

Thus, to obtain the overall reaction, we have

4 H2O (0 -> 4 H'*' {aq) + 4 OH" {aq) {Ionization of H2O)
4 {aq) + 4 2H2(8) {At Cathode)
4 OH" {aq) 4 2 H2O (/) + O2 (g) + 4 e" {At Anode)
Electrolysis
2H2O (/) ^ 2H2(g) + O2 (5) {Overall reaction)

w
(At cathode) (At anode)
(iv) Electrolysis of an aqueous solution of sodium chloride. Sodium chloride and water ionize as under:
NaCl {aq) ^ Na+(fl^) + Cl"(a^) {almost completely ionized)
H20(/) ± H+ {aq) + OH" {aq) {only slightly ionized)

e
On passing electricity, Na"^ {aq) and H"^ {aq) move towards the cathode while Cl" {aq) and OH" {aq)

ro
re
ions move towards the anode.

At cathode. Both Na"^ {aq) and H'*’ {aq) are present near the cathode. Since the discharge potential of H'^
ions is lower than that of Na"*" ions, therefore,H'*’ ions are discharged in preference to Na"^ ions.

F
Fl
H+ + e" ^ H- {reduction, primary change)

u
H- + H ^ H2(g) {secondary change)
Thus, H2 gas is evolved at the cathode while Na'^ ions remain in the solution.

sr
At anode. Both Cl" and OH" ions are present near the anode. Since the discharge potential of Cl" ions

ko
o
is lower than that of OH" ions, therefore. Cl" ions are discharged in preference to OH" ions.
C\--e-
●C1 + -Cl ^ Cl2(g)
^ -Cl
of {oxidation, primary change)
{secondary change)
o
Y
Thus, CI2 gas is evolved at the anode while OH~ ions remain in the solution.
erB

In nutshell, during electrolysis of aqueous sodium chloride, CI2 is liberated at the anode, H2 is liberated
at the cathode while sodium hydroxide remains in the solution,
uY

(v) Electrolysis of aqueous copper sulphate solution using inert electrodes, e.g., platinum electrodes.
Copper sulphate and water ionize as under :
CUSO4 {aq) ^ Cu^'*’ {aq) + SOj" {aq) {almost completely ionized)
do
ad

H20(/) ^ ± U^{aq) + OH-{aq) {only slightly ionized)

in

On passing electricity, Cu^"^ {aq) and H^ {aq) move towards the cathode while SOj" ions and OH" ions
Re

move towards the anode.

F

At cathode. Both Cu^"^ and H^ ions are present near the cathode. Since the discharge potential of Cu^"^
ions is lower than that of H'*' ions, therefore, ions are discharged in preference to H'*' ions,
Cx?-* + 2e- Cu {primary change)
Thus, copper is deposited on the platinum electrode.
At anode. Both SO^“ and OH" ions are present near anode. Since the discharge potential of OH" ions
is lower than that of SOj" ions, therefore, OH" ions are discharged in preference to SO|" ions.
4 OH" - 4 e" ^ 4 0H {oxidation, primary change)
4 OH ^ 2H2O (/) + O2 (g) {secondary change)
Thus, during electrolysis of aqueous CUSO4 solution, Cu is deposited at the cathode, O2 is liberated at
the anode while the solution contains sulphuric acid
It may be noted here that in the examples {Hi) to (v) discussed above, the electrodes used are those
which are not attacked by ions, i.e., graphite and platinum. However, if the electrodes are attacked by ions,
then the reactions taking place during electrolysis are different. For example, let us consider the electrolysis
of an aqueous solution of copper sulphate using copper electrodes.
3/6 ‘Ptadee^'4^ New Course Chemistry (XII)[SSI9]

(vi) Electrolysis of an aqueous solution of copper sulphate using copper electrodes. Copper sulphate
and water ionize as under:

CuSO^ (aq) —> Cu^'*’ {aq) + SO4 {aq) (almost completely ionized)
np (i) ^ i H-" (fl^) + OH-(a^) (only weakly ionized)
On passing electricity, Cu-"^ and move towards cathode while OH" and SO^“ ions move towards anode.
At cathode. Both Cu^'*’ and H'*’ ions are present near the cathode. But the discharge potential of Cu^"^
ions is lower than that of ions, therefore, Cu^^ ions are discharged in preference to H'*’ ions.
Cu^’^ (aq) + 2 e~ ^ Cu (^) (reduction, primary change)
Thus, copper is deposited on the cathode
At anode. Unlike electrolysis of CuSO^ using platinum electrodes, no ions are liberated here, Instead,
anode itself undergoes oxidation (i.e.. loses electrons) to form Cu-'*’ ions which go into the solution. This is
due to the reason that Cu is more easily oxidised than both OH" and SO^“ ions.

w
Cu (5) Cu^'*' (aq) + 2 (oxidation, primary change)

F lo
Thus, the net result of electrolysis is that copper is depositedat the cathodefrom the solution and an equivalent
amount of copper from the anode dissolves into the solution to form ions. This principle of electrolysis is
widely used in electroplating and purification of less reactive metals such as Pb, Cu, Ag, etc. The impure metal is

ee
made the anode that dissolves on passing current and pure metal is deposited at the cathode.

Fr
Importance of Electrolysis. Electrolysis has wide applications in industries. Some of the important
applicationsare :
(/) Production of hydrogen by electrolysisof water.
for
ur
(//) Electrolyticextraction of metals like Na, K. Mg, Ca, Al. etc. Sodium and potassium are produced by
the electrolysis of their fused chlorides whereas aluminium is produced by electrolysis of molten
s
aluminium oxide mixed with cryolite (discus.sed in unit 6).
ook
Yo

(Hi) Production of chlorine by electrolysis of aqueous NaCl solution.

eB

(iv) Manufacture of heavy water (D2O).

(v) Electroplating and electrolytic refining of metals.
our

3.3. QUANTITATIVE ASPECTS OF ELECTROLYSIS AND FARADAY'S LAWS

ad

The relationship between the amounts of substances liberated at the electrodes during electrolysis and
the amount of current passed was studied by Michael Faraday (1833). He put forward the following two laws
Y

which are called Faraday's Lmws of Electrolysis.

Re
nd

(1) Faraday’ First Law of Electrolysis : It stales that

Fi

The amount of chemical reaction and hence the mass of any substance deposited or liberated al
any electrode is directly proportional to the quantity of electricity passed through the electrolyte
(solution or melt).

Thus, if W gram of the substance is deposited on passing Q coulombs of electricity*, then

W « Q or W = ZQ
where Z is a constant of proportionality and is called electrochemical equivalent of the substance
deposited.
*During Faraday’s time, there were no devices available which could supply constant current. Hence, the
quantity of electricity passed was measured by putting an apparatus called coulonieter (coulomb measurer) in the
circuit which is simply a standard electrolytic cell. Generally, copper or silver coulometers were used and the
quantity of electricity passed was calculated from the mass of copper or silver deposited or consumed in the
coulometer. However, coulometers are no longer used becau.se now we have devices which supply constant current
(I) and quantity of electricity can be calculated using the formula Q = I x r, where Q is in coulombs when I is in
ampere and t in .seconds.
ELECTROCHEMISTRY 3/7

If a current of I amperes is passed for t seconds, then Q = I x r so that W = ZxQ = ZxIxt

Thus, if Q = 1 coulomb or I = 1 ampere and t = 1 second, W = Z. Hence,
Electrochemical equivalent of a substance may be defined as the mass ofthe substancedeposited
when a current of one ampere is passed for one second, Le., a quantity of electricity equal to
one coulomb is passed.
(2) Faraday’s Second Law of Electrolysis : It states that
When the same quantity of electricity is passed through solutions of different electrolytes
connected in series^ the masses of the substances produced at the electrodes are directly
proportional to their equivalent weights.

For example, for AgNOj solution and C11SO4 solution FIGURE 3.2
connected in series, if the same quantity of electricity is passed

w
(Fig. 3.2).
I

Mass of Ag deposited _ Eq.wt.of Ag

Flo
Mass of Cu deposited Eq. wt. of Cu ' >

Quantitative Aspects of Electrolysis.

ee
Consider the electrolysis of molten NaCl, i.e.,

Fr
1
Na+ Cr ^ Na+-CL
2 ^

for
Thus, we have Na"^ + e "> Na
ur
This means that one electron produces one sodium atom. AgNOs SOL. CL1SO4 SOL.

Therefore, passage of one mole of electrons will produce one Electrolytic cells (voltameters)
ks
mole of sodium. connected in series
Yo
oo

Again from the above reaction,

eB

cr >Cl + e" or 2Cr CI2 + 2 e

i.e., 2 moles of electrons produce one mole of CU-
Similarly, looking at the reactions, Cu-^ + le~ 4 Cu and AP'*’ + 3e Al, we find that two moles
r
ou

of electrons produce one mole of Cu whereas three moles of electrons will produce one mole of Al.
ad

The charge carried by one mole of electrons can be obtained by multiplying the charge present on one
Y

electron with Avogadro’s number, i.e., it is equal to (1-6021 x 10"^^ coulombs) x (6-022 x 10-^) = 96487
coulombs. This quantity of electricity is called one faraday. As it is a constant quantity, it is known as
nd
Re

Faraday’s constant and is represented by F. For approximate calculations, the value used is 96500 C mol"'.
Hence, Faraday’s constant, F = 96487 C mol = 96500 C mor*
-1
Fi

Thus, it may be concluded that

If n electrons are involved in the electrode reaction, the passage of nfaradays (i.e., n x 96500 C) of
electricity will liberate one mole of the .substance.
In terms of gram equivalents (old concept), it may be remembered that
One faraday (i.e., 96500 coulombs) of electricity deposits one gram equivalent of the substance.
Atomic weight of the element
This is because equivalent weight of any element =
No.of electrons gained or lost
byoneatom/ionof the element
Two Important Results/Conclusion.s. (/) As one faraday (96500 coulombs) deposits one gram
equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent
weight,
_ Eq.wt.of the substance
I.e.,
~ 96500
3/8 I^ietdeep- <t New Course Chemistry (XII) orsmn

(ii) Knowing the weight of the substance deposited (W gram) on passing a definite quantity of electricity
(Q coulombs), the equivalent weight of the substance can be calculated, i.e.,
W
Eq.wt. = ~x 96500
Q
The quantity of electricity actually passed is calculated from the current and time as follows :
Quantity of electricity in coulombs = Current in amperes x Time in seconds
Thus, knowing the quantity of electricity passed, the amount of substance deposited can be
calculated.

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Faraday’s first law and second law can be combined to give a mathematical relation as follows : —

low
ATr ^ Q r’ Q M Cxr M
W = ZQ=—xQ = —xE = — X— = X—
F F F F z

Z = Electrochemical equivalent; Q = Quantity of electricity passed, E = Eq.wt. of the metal, F = 1 Faraday

M = Atomic mass of the metal ; z = Valency of the metal;

ee
C = Current passed, t = Time for which current is passed.

rF
Fr
2. Electrical energy in joules = Potential difference in volts x Quantity of charge in coulombs
= V (volts) X Q (coulombs) = V (volts) x I (ampere) x t (sec) [v =Q = Ixr]
= Power in watts x t in sec. [ Power in watts = V (volts) x I (ampere)]

r
= watt-sec.

fo
u
From the above relation, we may further conclude that Power in watts = Electrical energy in joules
ks
Time in sec.
Yo
Thus, 1 watt = 1 J s ^
oo

Also as 1 coulomb = 1 ampere x I second I ampere = 1 coulomb sec ^

B

1 joule _ 1 newton x 1 metre

re

1 volt -
Further, as 1 joule = 1 volt x 1 coulomb
1 coulomb 1 ampere x 1 sec.
u
ad
Yo

FORMULAS AND UNIT USED

PROBLEMS
BASED
nd

(/) For the calculation of amount liberated/ deposited by a certain

Re

ON
quantity of the electricity or vice versa, write balanced electrode
Fi

reaction.
Quantitative
Aspects of Number of electrons gives the no. of Faradays required for the no.
Electrolysis of moles of that substance in the balanced equation.
(IF = 96500 coulombs).
(ii) Quantity of electricity in coulombs = Current in amperes x Time in
seconds.

m Power (in watts) = Voltage (in volts) x Current (in amp).

(tv) IF always liberates 1 g eq. of the substance.
(V) Electrochemical equivalent = Weight deposited by 1 coulomb
= Eq. wt. of the substance /96500.
(vO For the flow of same quantity of electricity through different
solutions connected in series, the weights deposited are in the ratio
of their equivalent wts.
ELECTROCHEMISTRY 3/9

Problem Q How much charge Is required for the following reduction ?

(i) 1 mol of to Al. (ii) 1 mol of Cu^^ to Cu. (iii) 1 mol of MnOj to Mn^"^
Solution. (/) The electrode reaction is : + 3 e~ > Al
Quantity of charge required for reduction of 1 mol of AP'*‘= 3 F = 3 x 96500 C = 289500 C.
{ii) The electrode reaction is : + 2 e~ > Cu
Quantity of charge required for reduction of 1 mol of Cu^'^ = 2 faradays = 2 x 96500 C = 193000 C
(iii) The electrode reaction is M11O4 > Mn-'*', i.e.. Mn^'*' + 5 > Mn-"^
Quantity of charge required = 5 F = 5 x 96500 C = 482500 C.
Problem
Q Calculate the charge in coulombs required for the oxidation of :
(i) 2 moles of H,0 to O2 (ii) 1 mol of FeO to Fe^O,.
Solution. (/) The electrode reaction for 1 mol of H2O is :

w
1 1
H2O 0“- -0. + 2e- 2 H-" + 2 e .H2
> H2 + -O2, i.e.,
/ 2 ^
or

F lo
I
or H2O > 2H++-0,+2e-

ee
2 2

Fr
Quantity of electricity required for oxidation of 1 mol of H2O = 2F
or Quantity of electricityrequired for oxidation of 2 moles of H^O = 4 F = 4 x 96500 C = 386000 C
I
for
ur
Fe^"^ + e
^ -2 ^<^2^2
(ii) The electrode reaction for 1 mol of FeO is: FeO I.e., Fe2+
s
Quantity of electricity required = I F = 96500 C.
ok
Yo

H How many coulombs of electricity are required for

o

Problem
eB

(i) Complete oxidation of 90 g of HjO ? (ii) Complete reduction of 100 mL of 0-1 M KMnOj solution ?
90
Solution. (/■) 90 g of H^O = —moles = 5 moles
r
ad
ou

18

As explained in example 2 above, 1 mole of H2O requires electricity = 2 F

Y

5 moles of HiO will require electricity= 5x2F=10F=10x96500 = 965000 C

(ii) 1000 mL of 1 M KM11O4 solution contain KMn04 = 1 mol
Re
nd

100 mL of 01 M KMn04 solution contain KMn04 = X100 X 01 mol = 0 01 mol

Fi

1000

As explained in example 1 above. 1 mol of KMnOj requires electricity = 5 F

O-OI mol of KMn04 will require electricity = 0-01 x 5 F = 0-05 F = 0-05 x 96500 C = 4825 C
Problem 0 A solution of CUSO4 is electrolysed for 10 minutes with a current of 1-5 amperes.
What is the mass of copper deposited at the cathode ? (Molar mass of Cu = 63*5 g mor*)
(NCERT Solved Example)
Solution. Current = 1-5 amperes. Time = 10 min = 10 x 60 s = 600 s
Quantity of electricity passed = Current (amp) x Time (s) = 1.5 x 600 coulombs = 900 C
The reaction occurring at the cathode is : Cu^'*' + 2e " — Cu

Thus, 2 F, i.e., 2 x 96500 C deposit Cu = 1 mol = 63-5 g

63-5
900 C will deposit Cu = x900 = 0-296 g
2x96500

1
3/10 1^'uteCee^'a. New Course Chemistry (XII)EZ5IHI

Problem R l\vo electrolytic cells containing silver nitrate solution and copper sulphate solution
are connected in series. A steady current of 2-5 ampere was passed through them till 1‘078 g of Ag were
deposited.How long did the current flow ? What weight of copper will be deposited ?
(At mass of Ag = 107-8, Cu = 63-5)
Solution. For deposition of Ag, reaction is : Ag"^ + e~ ■> Ag
Thus, 1 mole, i.e., 107-8 g Ag is deposited by 1 F = 96500 C
96500
1-078 g Ag will be deposited by xl-078C = 965C
107-8

Q _ 965
Now Q = I X / = 386s = 6 min 26 s
1 ~ 2-5
For deposition of Cu, the reaction is : Cu-'*' + 2 e > Cu

w
Thus. 2 F = 2 X 96500 C deposit Cu = 1 mole = 63-5 g

63-5

F lo
965 C will deposit Cu = x965 = 0-3175 g
2x96500

Probiem
0 Silver is electro-deposited on a metallic vessel of surface area 800 cm" by passing a

ee
current 0-2 ampere for 3 hours. Calculate the thickness of silver deposited. Given the density of silver as

Fr
10*47 g/cc (Atomic mass of Ag = 107*92 amu)
Solution. Quantity of electricity passed = 0-2 x 3 x 60 x 60 C = 2160 C
Ag-*- -h e~ ^ Ag
for
ur
96500 C deposit Ag = 107-92 g
s
ok

107-92
Yo
2160 C will deposit Ag = 96500
x2160g = 2-4156 g
o
eB

Mass 2-4156
Volume deposited = cc = 0-2307 cc
Density 10-47
r
ad
ou

Volume 0-2307
Thickness deposited = = 2*88x10^ cm
Area 800
Y

Problem B In the electrolysis of acidulated water, it is desired to obtain hydrogen at the rate of
Re
nd

1 cc per second at NTP condition. What should be the current passed ?

Fi

Solution. 2 H'*’ -f 2

Thus. 1 mole of H2, i.e., 22400 cc at NTP requires 2 F = 2 x 96500 coulombs

2x96500
1 cc at NTP requires = xl =8-616C
22400

Q 8-616C
As Q=I X / = 8*616 Ampere
t li

Problem d Find the charge in coulombs on 1 g-ion of N^.

Solution. Charge on one ion of = 3 x 1-6 x 10“*^ coulombs
One g-ion = 6-02 x 10~^ ions
Charge on 1 g-ion of N^“ = 3 x 1-6 x 10"'^ x 6-02 x 10“^
= 2*89 X 10® coulombs

i
 ELECTROCHEMISTRY
3/11

Problem 0 When a current of 0-75 A is passed through CUSO4 solution for 25 min, 0*369 g of
copper is deposited at the cathode. Calculate the atomic mass of copper.
Solution. Quantity of electricity passed = (0-75 A) (25 x 60 s) = 1125 C
Cu2++2e- ■> Cu

Thus. 1 mole of Cu (/.e., gram atomic mass) is deposited by 2F, i.e., 2 x 96500 C
Now,
1125 C deposit Cu = 0-369 g
0-369
2 X 96500 C will deposit Cu = 1125 x2x96500g = 63-3 g
Hence, atomic mass of copper = 63*3

w
Problem [0 Calculate the quantity of electricity that would be required to reduce 12*3 g of
nitrobenzene to aniline if the current efficiency for the process is 50 per cent. If the potential drop across

the cell is 3 volts, how much energy will be consumed ? (AIPMT 200Si

o
Solution. The reduction reaction may be written as ; CgHgN02 + 6H'^ + 6 e~ — 4 Cf,HgNH2 + 3H2O

e
Thus. 1 mol of C5H5NO2, i.e.. 123 g require electricity = 6 x 96500 C = 579000 C

re
12-3 g require electricity = 57900 C

Frl
F
As the current efficiency is 50%, quantity of electricity actually required = 2 x 57900 C = 115800 C
Energy consumed = EMF x Qty. of electricity = 3 x115800 J = 347400 J
Problem [Da curmt of 4 ampere was passed for 1*5 hours through a solution of copper sulphate

or
ou
when 3*2 g of copper was ' .posited. Calculate the current efficiency.
Solution. Cu2++2e- ^Cu
kfs
! mole of Cu (63-5 g) requires charge = 2 F = 2 x 96500 coulombs
oo
2x96500
Y
3-2 g Cu will require charge = x3-2 coulombs = 9726 coulombs
B

63-5
Quantity of electricity actually passed = 4 x 45 x 60 C = 21600 C.
re

9726
Current efficiency = X100 = 45% .
oYu

21600
ad

Problem m How many electrons flow through a metallic wire if

a current of 0*5 A is passed for
2 hours ? (Given 1 F = 96,500 C mol-^)
d

(CnSE 2017)
Solution. Charge llowing through the metallic wire = I x r = 0-5 x 2 x 60 x 60 C = 3600 C
in
Re

But charge flowing = no. of electrons flowing x charge on each electron

F

3600 = n X 1-60 X 10-*^

3600
or n = = 2-25 X 10^2
I-60x10-‘9

Problem jj] Qn passing a current of 1*0 ampere for 16 min and 5 sec through one litre solution of
CUCI2, all copper of the solution was deposited at the cathode. Calculate the strength ot CUCI2 solution
(Molar mass of Cu = 63-5, Faraday constant = 96500 C mol“i).
Solution. Q = I X / = I X 965 = 965 C
Cu2+ + 2e ■> Cu

2 X 96500 C decompose CUCI2 = I mole

1 1
965 C decompose CuC^ = x965 = mole = 0 005 mole
2x96500 200

This was pre.sent in 1 L of CUCI2 solution.

Hence, strength = 0*005 M.
3/12
New Course Chemistry (X11)CSI91

Problem [0 A current is passed through two cells connected in series. The first cell contains
X(m^h(aq) and the second cell contains Y (N03)2 (aq). The relative atomic masses of X and Y are in the
ratio I : 2. What is the ratio of the liberated mass of X to that of Y ?
Solution. Suppose atomic mass of X = a
Then atomic mass of Y = 2 o

2a
Eq. wt. of X - —, Eq. wt. of Y - — - 2 (Valency of X = 3, Valency of Y = 2)

Applying Fiiraclay's second law of electrolysis :

Mass of X Eq.wt.of X _ - =1 :3
Mass of Y Eq.wt.ofY a 3

w
F lo
1. How many grams of chlorine can be produced by the electrolysis of molten NaCl with a current of 100 A

ee
for 15 min ? Also calculate the number of chlorine molecules liberated.

Fr
2. Calculate the mass of silver deposited from silver nitrate solution by a current of 2 amperes flowing for
30 minutes (equivalent mass of silver is 108) (Karnataka Board 2012, Raj. Board 2012, CBSE 2017)

for
3. A current of 10 amp is pas.sed tlirough molten AICI3 for 96-5 seconds. Calcuh.iC the mass of A1 deposited.
ur
(AP Board 2012)

4. How many Faradays/coulombs are required to produce (/) 20 0 g of calcium from molten CaCl2 ?
s
ook

(Hr. Board 2011)

Yo
(//■) 40 0 g of aluminium from molten AI2O3 ?
5. How many hours does it take to reduce 3 mol of Fe^’^ to Fe^"^ with 2 00 A current ? (1 Faraday = 96,500 C
eB

mol-‘,R = 8-314 JK"* moC^).

6. A current of 1 -50 A was passed through an electrolytic cell containing AgN03 solution with inert electrodes.
our

The mass of the silver deposited at cathode was 1-50 g. How long did the current flow? (Atomic mass of
ad

Ag= 108 u,F = 96500 0 (Assam Board 2012, CBSE Compt. 2018)
7. How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing
Y

silver in +1 oxidation state for a period of 8-0 hours at a current of 8-46 Amperes 7 What is the area of the
Re

tray if the thickness of the silver plating is 0-00254 cm 7 Density of silver is 10-5 g/cm^
nd

minutes with a current of 1-5 A. The mass of the metal

Fi

8. A solution of metal salt was electrolysed for 15

deposited was 0-000783 kg. Calculate the equivalent mass of the metal. (Maharashtra Board 2012)

9. 0-3605 g of a metal is deposited on the electrode by passing 1-2 ampere current for 15 minutes through its
salt. Atomic weight of the metal is 96. What will be its valency 7
10. A 100 W, 220 V incande.scent lamp is connected in series with an electrolytic cell containing copper sulphate
solution. What weight of copper will be deposited by a current flowing for 5 hours 7
(at. wt. of Cu = 63-54).
U. (a) Two electrolytic cells containing silver nitrate solution and dilute sulphuric acid solution were connected
in series. A steady current of 2-5 amp was passed through them till 1-078 g of silver was deposited (Ag -
107-8 g mol"'. 1 F = 96500 Cl
(/) How much electricity was consumed ? (li) What wa.s the weight of oxygen gas liberated ?
12. How many moles of mercurj' will be produced by electrolysing 1-0 M Hg(N03)2 solution with a current of
2-00 A for 3 hours ? [Hg(N03)2 = 200-6 g moC^] (CBSE 2011)

f
 ELECTROCHEMiSTRY
3/13

electroplated using an acidic solution containing Cr03 according to the following

Cr03 (aq) + 6 H'*' (aq) + 6 e ^ Cr (s) + 3 H2O (/)

Calculate how many grams of chromium will be electroplated by 24,000 coulombs, How long will it take to
electroplate 1-5 g chromium using 12-5 A current ?
(Atomic mass of Cr = 52 g moC^, 1 F = 96,500 C mol"’) (CBSE 2019)
ANSWERS
1- 0-331 g, 2-80 X 10^’ molecules 2. 4- 029 g 3. 0-09 g
4. (0 1 F or 96500 C (//) 4-44 F or 428460 C 5. 40-2 hrs. 6. 893-5 s
7. l-02x lO^cm^ 8. 56 9.3 10. 2-694 g 11. (0 965 C (») 0-16 g
12. 0-112 mole 13. 1336 s

HINTS FOR DIFFICULT PROBLEMS

w
1. Q-900C
2C1-
CI2 + 2 e-;

F lo
2 X 96500 C liberate CI2 = 1 mole
1
900 C will liberate CI2 = 2x96500 x900 = 4-66 X 10 ^ mole = (4-66 x 10~^) x 71 g = 0-331 g

e
(4-66 X 10 X 6-02 x 10^^ molecules = 2-80 x 10^* molecules

Fre
4. (/) Ca2^ + 2e~ ■> Ca
(ii) AI2O3 2 Al^+ ■f 3Q2-
■I. + 6 e“ for
i - 6 e~

2 A1
r
2 2
5. Fe3+ + e-
> Fe^"*^, i.e., 1 mol requires 1 F. 3 moles will require = 3
You
F = 3 x 96500 C
oks

/ = Q/I = (3 X 96500)/2-0 = 144750 s = 40-2 hrs.

eBo

7.
Calculate mass of Ag deposited. It comes out to be 271-65 g.
271-65
Volume of Ag deposited = = 25-9 cm^
ad
our

10-5

25-9
Area of tray = = 1-02x10^ cm2
0-00254
Re
dY

&783
8. Q = 1 -5 X 900 C = 1350 C. 96500 C deposit 1 g eq. Hence, E = X 96500 = 5597 = 56
1350
Fin

9.
Calculate the wt. deposited by 96500 coulombs. This will be the equivalent wt.

Then valency = At. wt/Eq. wt.

Power 100 5
10. Current =
ampere = — ampere .
Voltage 220 11

11. (0 Ag+ + e -> Ag, I mol (107-8 g) are deposited by 1 F /. 1-078 g will be deposited by 0-01 F = 965 C.
(ii) 2 H2O - -> 4 H'*’ + O', -f- 4 on each electrode
4 F liberate 32 g 03. 0-01 F will liberate O2 = 0-08 g
O2 liberated on both electrodes = 2 x 0-08 g = 0-16 g
12. Hg2+ + 2e- ^ Hg, Q = 2 A X (3 X 3600 s) = 21600 C
2 F = 2 X 96500 C produce Hg = I mole
1
21600 C will produce Hg = x21600mole =0-112 mole
2x96500

[
 New Course Chemistry (XlI)CEIBl
3/14

13. 6 X 96500 C deposit Cr = 1 mole = 52 g

52
24000 C will deposit = X 24000 = 2-15 g
6x96500
6x96500
X 1.5= 16702 C
For depositing 1-5 g Cr, charge required - 52

Q 16702 s = 1336 s
Q =1 X r or f = —I “ 12-5

3.4. CONDUCTANCE OF ELECTROLYTIC SOLUTIONS

Substances which allow electricity to pass through them are known as conductors whereas substances
which do not allow electricity to pass through are known as insulators. Conductors are divided into two
classes :

(/) Those which conduct electricity without undergoing any decomposition. These are called electronic

w
conductors, e.g.. metals, graphite and certain minerals. As the name indicates, the conduction in this

F lo
case is due to the flow of electrons,
(ii) Those which undergo decomposition when current is passed through them. These are known as
electrolytic conductors or electrolytes, e.g., solutions of acids, bases and salts in water, fused salts,

e
etc.

Fre
In this case, the flow of electricity is due to the movement of ions. Hence, electrolytic conductance is
also called ionic conductance.
for
Electrolytes are further classified into two main categories as strong electrolytes and weak electrolytes.
Strong electrolytes are substances like strong acids, e.g.. HCl, HNO3, H2SO4, etc., .strong bases, e.g., NaOH
and KOH, etc., and most of the inorganic salts. These dissociate almost completely in the aqueous solution or
r
You
a large extent. Weak electrolytes are substances like weak
in the molten state and hence conduct electricity to
oks

acids, e.g., CH3COOH, HCN, H2CO3. H3PO4, etc. and weak base.s, e.g., NH4OH, CaCOH)^, Al(OH)3, etc.
eBo

These dissociate to a small extent and hence conduct electricity to a small extent.
The fraction of the total number of molecules of the electrolyte which ionizes in the solution is called
degree of ionization. It is usually represented by a. For strong electrolytes, it is almost equal to 1 and for
our
ad

weak electrolytes, it is always less than 1.

Substances like sugar, urea, etc. the solutions of which do not conduct electricity are known as non¬
electrolytes.
Factors affecting electrolytic conduction. The conductance of the solution of an electrolyte depends
dY
Re

upon the following factors :—

(/) Nature of the electrolyte. Strong electrolytes ionize almost completely in the solution and hence conduct
Fin

electricity to a large extent whereas weak electrolytes ionize to a small extent and hence conduct electricity
also to a small extent.
(//) Size of the ions produced and their .solvation. Greater the size of the ions or greater the solvation of the
ions (i.e.. combination of the ions with the solvent), less is the conductance.
(Hi) Nature of the solvent and its vi.sco.sity. Electrolytes ionize more in a polar solvent. Hence, greater the
polarity of the solvent, greater is the ionization and hence greater is the conductance. Similarly, greater
is the viscosity of a solvent, less is the conductance,
(fv) Concentration of the solution. Higher the concentration of the solution, less is the conduction. This is
because in a weak electrolyte, the ionization is less whereas in a strong electrolyte, the interionic
attractions are high at higher concentrations. With dilution, in case of weak electrolytes, ionization
increases and in case of strong electrolytes, interionic attractions decrease and hence conduction
increases,

(v) Temperature. On increasing the temperature, the dissociation increases and hence the conduction
increases.

1
 ELECTROCHEMISTRY
3/15

Factors affecting metallic or electronic conductance. The electrical conductance

through metals
depends upon the following factors ;
.(:) Nature and structure of,he metal (,V) Number of valence electrons per atom
(w) Temperature. It decreases with increase of temperature (as aiready discussed in unit I)
Difference between metallic conductors and electrolytic conductors. The main points of difference
between metallic conductors and electrolytic conductors may be summed up in Table 3.1.
TABLE 3.1.
Difference between metaUic conductors and electrolytic conductors
Metallic Conductors
Electrolytic Conductors
1.
Flow of electricity takes place without the 1.
Flow of electricity takes place accompanied by
decomposition of the substance.
2.
the decomposition of the substance (electrolyte).
Flow of electricity is due to the flow of electrons 2. Flow of electricity is due to the movement of

w
only, i.e., there is no flow of matter. The ions and hence there is flow of matter.
conductance depends upon the structure and

F lo
density of the metal and the number of valence
electrons per atom.
3. The electrical conduction decreases with 3. The electrical conduction increases with increase

ee
increase of temperature. This is because kernels of temperature. This is generally due to increase

Fr
Stan vibrating which produce hinderance in the in dissociation or decrea.se in the interionic
flow of electrons. attractions.
4.
The resistance offered by metal is also due to
vibrating kernels.
4.
for
The resistance shown by an electrolyte solution
ur
is due to factors like inter-ionic attractions,
viscosity of solvent, etc.
s
ook
Yo

3.5. ELECTRICAL RESISTANCE AND CONDUCTANCE

eB

Every substance offers resistance to the flow of electricity to a small or large extent. The law that gives
the exact value of the resistance is known as Ohm’s law. It states that
If to the ends of a conductor is applied a voltage ‘E’ and a current
our

flows through it then the

ad

resistance ‘R'of the conductor is E/L

Current IS generally measured in amperes, whereas voltage is measured in volts. If one ampere current
Y

flows through a conductor when a voltage of one volt is applied to it, the resistance of the conductor is taken
Re

as 1 ohm (written as 1 Q).

nd
Fi

E Volts
Thus, according to Ohm’s law. R =- or
Ohms =
I
Amperes
In terms of SI base units, .Q = (kg mV(-v^ A^). This is explained below :
Q = _V _ Work per unit charge Work I Fx/ 1 mxaxl kg X m s - X m kgm-
^ — - ^ —
A A
Charge A Ax.v A A^s A2; s2a2
It is seen that like metallic conductors, solutions of all eIectrolyte.s also obey Ohm’s law.
It IS obvious that a substance which offers greater resistance will allow less electricity to flow through it.
1 his result also follows from Ohm’s law according to which 1 l/R. oc

The reciprocal of the electrical resistance is called the conductance. It is usually represented

by G. Thus, G = —
R

It is a common practice to speak of the conductance of a solution rather than its resistance.
3/16
‘PnAeUc^’^ New Course Chemistry (XII)gEM

Units. The units of conductance are :

reciprocal ohms written as ohm"’ or mhos or siemens (S) (IS = I H )
For example, if a solution has a resistance of 10 ohms, it is said to have a conductance of 1/10
ohm"' or 1/10 mhos or 1/10 Q"' or 1/10 siemens.
3.6. SPECIFIC EQUIVALENT AND MOLAR CONDUCTIVITIES
3.6.1. Specific Conductivity (or simply called Conductivity}
It is observed that resistance R of a conductor is
(/) directly proportional to its length {/) and (//) inversely proportional to its area of cross section (a)
or
...(/)
i.e.. R^-
a

where p is a constant of proportionality, called Specific Resistance or Resistivity*. Its value depends
upon the material of the conductor.
If/= 1 cm andu = 1 cm^ R = p. Hence,
Resistivity is defined as the resistance of a conductor whose length is 1 cm and area of cross

F low
section is 1 cm^, ue., it is the resistance ofl cm^ of the conductor or in terms of SI units, it is the
resistance of 1 m^ of the conductor,
“The reciprocal of resistivity is known as specific conductance or simply conductivity”**. It is denoted
by K (kappa). Thus, if K is the conductivity and G is the conductance of the solution, then
for Fre FIGURE 3.3
1 1
R = — and P = -
G K
£
o

l = ix- /wno5 OTiTjrY'irv

Substituting these values in eqn. (i), we get G K a
eBo ks
Your

I
or K= Gx 1 cm
ad

a
Specific conductivity
our

Now, if /= 1 cm and a = 1 sq. cm, then k = G. Hence,

Conductivity of a solution is defined as the conductance of a solution ofl cm length and having
Re

1 sq. cm as the area of cross section. Alternatively, it may be defined as “conductance of one
centimeter cube of the solution of the electrolyte'' (Fig. 3.3) or in terms of SI base units, it is
defined as the conductance of I mf of the conductor.
Find Y

If the volume of the solution is V cm^, the conductivity of such a solution at this dilution V is written as
K^,. Similarly, at concentration C, it is represented as
a (cm^) - ohm cm or Q cm or Q m in SI units
Units. Resistivity (p) = R — = ohms
/ cm

1Q m = 100 Q cm

1 1 -1 -1

Specific conductivity (k) ~ " = ohm"^ cm"' or Q"' cm"' or S cm or S m in SI units.

ohm cm

1 S cm"' = 100 S m"'

*IUPAC recommends the use of the term ‘resistivity’ over ‘specific resistance'.
**IUPAC recommends the use of the term ‘conductivity’ over ‘specific conductance .
 ELECTROCHEMISTRY
3/17

, As conductivity expresses the conductance volume, we compare the conductances of different

m terms of their conductivities. As already discussed in unit I, on the basis of conductivities
dilferenl substances have been divided into three categories - conductors, insulators and semiconductors’
ine conductivities of a few substances of each type are given below (S nr‘).
TABLE 3.2. Conductivities of some materials at 298 K

w
Metallic/Electronic : Copper Silver Gold Iron Sodium Graphite
Conductivity (S m~^): 5-9 X 10-^ 6-2 X 10-^ 4-5 X lO-^ 10 X 1Q3 2-1 X 10"* 1-2 X 10
Insulators :

e
Teflon Glass Diamond Quartz
Conductivity (S m"^): 1-0 X 10-‘^ 10 X 10-1^’ 10x10"'^ 10x10-’^

e
or
Electrolytic conductors: Pure water

r
01 M HCl O IMKCI O lMNaC! 0-1 M HAc O-OlMHAc
Conductivity (S irr^): 3-5 X 10-5 3-91 0-14 0-12

F
0-047 0016
Semiconductors : Si Ge CuO

oF
ul
Conductivity (S m“^): 1-5 X 10"- 2-0 1 X 10“^

rs
Substances having zero resistivity or infinite conductivity are known as superconductors as already
discussed in unit 1.

ko
3.6.2. Equivalent conductivity

of
For comparing the conductances of the solutions of different electrolytes, it is essential that (/) the
volumes of the solutions should be same and (i7) they must contain such definite amounts of the electrolytes
o
Y
which give ions carrying the same total charge. The best choice is. therefore, to consider the solutions having
rB

equa volumes and containing their corresponding gram equivalent weights. The conductance of such a solution
eY

ts called its equivalent conductivity. Hence,

Equivalent conductivity of a solution at a dilution Vis deifned as the conductance ofall the ions
u

produced from one gram equivalent of the electrolyte dissolved in V cm^ of the solution when
d

the distance between the electrodes is one cm and the area of the electrodes is so large that
o
ad

whole of the solution is contained between them. It is represented by (lambda).

in

Relationship between equivalent conductivity and specific conductivity. The equivalent conductivity
of a solution ts usually not found directly but is calculated from the specific conductivity. The relationship
Re

between equivalent conductivity and specific conductivity may be obtained as follows ;

F

Consider a rectangulai- vessel with its two opposite walls one cm apart and made of some metal sheet so
that these act as the electrodes.
Case I. Suppose I cm^ of the solution containing I g eq. of the FIGURE 3.4
electrolyte is taken in the vessel. The conductance of this solution will be 1 cm
then its specific conductivity. Further, if 1 cm^ of the solution
taken contains one gram equivalent of the electrolyte, the conductance of
the solution will be its equivalent conductivity (by definition). Thus, when A frwirv-
1 cm of the solution containing one gram equivalent of the electrolyte is 1 cm3

considered, the equivalent conductivity is equal to its speciifc conductivity.

K>

Case II. Suppose 4 cm^ of the solution containing one gram

equivalent of the electrolyte is taken. The conductance of the solution
will be still equal to its equivalent conductivity at this dilution but now 2 cm
>

there will be four cubes each of volume I cm^ as shown in Fig. 3.4. The Relationship between
conductance of each 1 cm^ of the solution is equal to its specific equivalent conductivity
conductivity so that the total conductance of the solution, i.e., equivalent and specific conductivity
conductivity is four times the .speciifc conductivity.
3/18
4- New Course Chemistry (XII) j^M

In general, if the volume of the solution containing one gram equivalent of the electrolyte is V cm ,
we have; Equivalent conductivity = Specific conductivity x V
A,^ = K,xV
In terms of concentration, if the solution has a concentration of c gram equivalent per litre \.e., c gram
equivalents are present in 1000 cm^ of the solution, then the volume of the solution containing one gram
1000 , 1000
equivalent will be , f.e., V - . Hence, the above expression may be written as
c c

1000 1000

w
A = K. X = K. X
eq c
c Normality
eq

Note carefully that in the above expression, V is the volume of the solution in cm^ containing 1 gram
equivalent of the electrolyte and c is the concentration in gram equivalent/litre.

o
e
3
cm

re
-I
Units : a,..
eq
= k x V - ohms”' cm ' x = ohm"^ cm^ (g eq)“^ or cm^ eq
gram eq.

Frl
or S cm^ eq“* or S eq”* in SI units

F
1S eq”‘ = 10“* S cm^ eq”*
ou
Thus, the above formula along with units of each quantity may be written as :

r
K (S cm”') x (1 OOQcm^L-')

so
(S cm^eq ') =
A
eq
Normality (geqL *) kf
oo
K(Sm ')
In terms of SI units, the formula becomes A (S m^eq ') =
Y
cq
Normality (geqm
B

Note that there is no multiplication with 1000

However, if normality is expressed in g eq L“‘, the above formula becomes
re

K(Sm”’)
oY

(S m^eq”') =
u

A
Ctl
1000 L m”^ X Normality (g eq L ')
ad
d

3.6.3. Molar Conductivity

The molar conductivity of a solution is defined in a manner similar to that of the equivalent conductivity,
in
Re

as under:

Molar conductivity of a solution at a dilution V is the conductance ofall the ions produced from
F

one mole of the electrolyte dissolved in V cm-’ of the solution when the electrodes are one cm
apart and the area of the electrodes is so large that the whole of the solution is contained
between them. It is usually represented by a^. -
the case of equivalent
Relationship between molar conductivity and specific conductivity. As i
in

conductivity, the molar conductivity is related to the specific conductivity as follows;

1000 1000

% = or
A = K,x
m
Molarity

where K is the specific conductivity and V is the volume of the solution containing one mole of the electrolyte
and c is the molar concentration, i.e., mol L”* (or mol dm ^).
will be ohm"* cm- mol"* or
Units; Comparing with the equivalent conductivity, the units of a m

S cm^ mol'* or fl”* cm- mol”* or S m^ mol"* in SI units

1 S m^ mol"* = 10“* S cm^ moU* or 1 S cm^ mol"* = 10-^ S mol"*
 ELECTROCHEMISTRY 3/19

Thus, in this case again, the formula along with units of each quantity may be written as

K(Scm->)x(1000cm3L-i)
(S cm^ mol =
Molarity (mol L *)
k(S m"i)
In terms r;/SI units, the formula becomes A (S m^mol *) -
nt
Molarity (niol m
Note that there is no multiplication with 1000.
However, if molarity is expressed in mol L“’, the above formula becomes
k(S m“')
A,^(S m-mol *) =
lOOOLm ^ X Molarity (mol L“')

Curiosity Question

w
r Q. Teflon coating Is done In a number of items of dally use. Name any tvtfo such items and for

F lo
what purpose this coating is done ?
Ans. Teflon is an insulating material. Now a days, some people go for teflon coating on the body of their
car as it covers the painted surface and protects from scratch, rust etc. Teflon coating is also done

ee
on the inner side of the non-stick pans as it has great chemical inertness and thermal stability
-J

Fr
3.7. MEASUREMENT OF ELECTROLYTIC CONDUCTANCE, SPECIFIC CONDUCTIVITY,
EQUIVALENT CONDUCTIVITY AND MOLAR CONDUCTIVITY OF IONIC SOLUTIONS
for
ur
1. Electrolytic conductance. Conductance is the reciprocal of FIGURE 3.5
resistance. Hence, the conductanee can be obtained by the measurement of B
oks

(VARIABLE) (UNKNOWN)
the resistance and the latter can be found by the principle of Wheatstone
Yo

bridge method shown in Fig. 3.5.

o
eB

It consists of four arms containing the resistances R,, R„ R3 and R^. R 1 C

is the variable resistance and Rj is the unknown resistance. When null point
is obtained, i.e., there is no deflection in the galvanometer G ,
our
ad

R
R3
^2 R4 BATTERY
Y

Knowing the values of R,, R3 and R4, R2 can be calculated. |if^

Re

KEY
nd

In finding the resistance of the solution of an electrolyte, the first

difficulty is that the solution cannot be connected in the bridge like a metallic Principle of D.C.
Fi

Wheatstone bridge method

wire or any other solid conductor. This difficulty is overcome by taking a
special type of vessel such that the solution FIGURE 3.6
is present between the two electrodes. The o 0
cell thus used is called conductivity cell. CONNECTING
WIRES
CONNECTING WIRES

It is made up of pyrex glass and has two STOPPER

SEALED
platinum electrodes at a fixed distance
A
GLASS
apart. A number of designs of the EXPERIMENTAL
TUBES

conductivity cells are available. Two SOLUTION

GLASS VESSEL

^1
— GLASS
simple designs are shown in Fig. 3.6. VESSEL

The second difficulty in the

€
● PLATINZEDPt
\. ELECTRODES
measurement of the resistance (and hence
the conductance) of the solution of an
PLATINZED Pt ELECTRODES ^
electrolyte is that if direct current is used,
it causes electrolysis of the solution. As a Two simple designs of conductivity cells
result, the concentration of the electrolyte (a) Filling type (b) Dipping type
3/20 “P^euiee^'A New Course Chemistry (XlI)E!&ia]

near the electrodes changes and this results in the change in the resistance of the solution. These effects are
called polarization effects. To overcome this problem, an alternating current in the audio frequency range
of 550-5000 cycles per second is used. However, when the alternating current is used, ordinary galvanometer
fails to detect the null point. Hence, the galvanometer is replaced by earphone as the detector. Further, to
mimnuze the polarization effects, the electrodes are coated with platinum black (i.e., finely divided platinum

deposited on the electrodes electrolytically). To sum up, three modifications required in the usual Wheatstone
bridge method are :
(/) Source of direct current is replaced by a source of alternating current which is either an induction coil
or a vacuum lube oscillator (which is quieter and gives more symmetrical current).
(ii) Galvanometer is replaced by an earphone.
(Hi) Electrodes of the cell are coaled freshly with platinum black before use.
The complete assembly for the measurement of the electrolytic conductance is shown in Fig. 3.7.
A suitable value of the resistance R is introduced from the standard resistance box such that when the
sliding contact, Le., the jockey J is moved along the stretched wire, the sound in the earphone is reduced to
minimum at a point somewhere in the middle of the wire AC, say at the point D. Then if X is the resistance of

F low
the electrolytic solution, by Wheatstone bridge principle,

Resistance R _ Resistance of wire AD _ Length AD Length CD

; Resistance, X = Resistance R x
Resistance X Resistance of wire CD Length CD Length AD

Thus, knowing the FIGURE 3.7

e
resistance R and the balance

point D, the resistance X of the

VARIABLE RES.
(STANDARD R '»>
for Fre
CONNECTING WIRES
CONDUCTIVITY

electrolytic solution can be RES. BOX)g '●Jb CELL

[● ilS X [
calculated. Then Conductance,
I 1 Length AD
Your

G=—=—X A
eBo ks

C D
X R Length CD A
J -^JOCKEY C

2. Specific conductivity. PLATINIZED

Pt
ad

Specific conductivity or simply EAR ELECTRODES

our

called conductivity (k) is PHONE

related to the conductance (G)

SOURCE OF ALTERNATING EXPERIMENTAL
according to the expression : CURRENT (OSCILLATOR) SOLUTION
Re

I Apparatus for the measurement of electrolytic conductance

k=Gx-.
Find Y

Thus, the conductivity of a solution can be determined by measuring its conductance and the distance (/)
between the electrodes and the area of cross-section (a) of each of the electrodes. But the measurement of I
and a is neither convenient nor reliable. However, for a particular cell, Ua is constant and this constant is
called cell constant. It is usually represented by G*. Thus,
I
G* = — = cell constant
a

Conductivity (k) = Conductance (G) x Cell constant (G*).

Thus, the cell constant of any particular cell can be found by measuring the conductance of a solution
whose conductivity is known. For this purpose, generally specific conductivities ol KCI solutions are used
which are known accurately at different concentrations and temperatures. For a few concentrations at a
temperature of 298 K, these valves are given below :
ELECTROCHEMISTRY 3/21

TABLE 3.3. Conductivities of KCl solutions at 298 K

Concentration of KCl solution Conductivity (Specific Conductivity)

1 000 M or 1000 mol 01113 S cm-' or 1113 S m''
0-100 M or 100 mol m“^ 0-0129 Sm-' or 1-29 S m"'
0 010 M or 10 mol 0-00141 S cm-' or 0-141 S m"'

Knowing the value of the cell constant, the conductivity of the given solution can be determined by
measuring its conductance and multiplying the value with the cell constant,

ow
/ cm
Units of cell constant, G* = - = cm' (or nr' in SI units)
a cm-

3. Equivalent conductivity. Equivalent conductivity (Ag^) is related to the specific conductivity (tc)
1000 KxlOOO

e
A.,. = KX
according to the equation : eq
Normality (N)

re
Frl
where is the concentration of the solution in gram equivalents per litre (i.e., the nonnality of the solution).

F
Thus, knowing the normality of the solution and the specific conductivity k, a^,^ can be calculated.
4. Molar conductivity. The molar conductivity is related to the specific conductivity according to the
ou
or
1000 KXlOOO
A = KX
relation : m
c kfs
Molarity (M)
where c is the molar concentration. Thus, knowing molar concentration c and the specific conductivity k,
oo
A^ can be calculated.
Y
Conductivity water. For accurate results, the solutions should be prepared in a .specially purified water
B

whose own conductance is very small. Such water is called ‘conductivity water’.
re

Alternatively, to gel the resistance of the electrolyte, the resistance of water used for making the solution
may be determined and then subtracted from the resistance of the solution.
oYu
ad

NUMERiCAI- FORMULAS AND UNIT U

d

PROBLEMS
in

BASED
Re

I R(Q)a(cm^) R(Q)Xfl(m^)
ON (/) R = p- or p = = Q cm or = Qm
a /(cm) /(m)
F

■ Specific,
1 1 ^ 1 /(cm)
(ll) K = - = —X-= ..
X . = Q-' cm
-1
= Sem
-1
Equivalent and p R a R(f2) fl(cm^)
Molar
1 /(m)
Conductivities or X = Cl ' m ' = Sm '
R(Q) f7(m2)
Note. K is called specific conductivity or simply conductivity.
I /(cm) /(m)
(Hi) Cell constant (G*) = — = cm or = m
a a (cm-) fl(m2)
I
(iv) K — G X — = Observed conductance x Cell constant = G x G*
a

= Q 'em ' = S cm ' or O ' m ' = S m ' (in SI units)

K Specific conductivity
(v) Cell constant (G*) = — = Specific conductivity x Resistance
G Conductance
3/22 p>uuUe^'4. New Course Chemistry CXlDBZSm

K(Q“*cm"')xlOOO(cin^L ')
(vO A
= Q-i cm^ (g eq ^) = S cm^ eq ’
eq
Normality, (g eq L ’)
K sr‘ m-l
=Q * (g eq) ' = S m“ eq ’
^eq (Normality) geqm ^
Note carefully. In this formula there is no multiplication with 1000.
K(S m~*)
sSm^eq '
1000 L m"3 X Normality (g eq Lr^)
K:(^2-^cm"*)xl000(cn1^L-') = Q"' cm- mol ' = S cm- mol ’

ow
(vii) A m
Molarity (mol L ^)
K(g-^m-‘)
= Q * m^ mol * = S m^ mol *
Molarity, (mol m"^)

e
Note carefully. In this formula there is no multiplication with 1000.

re
rFl
k(S m~^) = S m^ mol *

F
1000 L m”^ X Molarity (mol L"')
Summing-up the Units of Different Quantities

r
Physical Quantity Commonly used Units SI Units
ou
fo
Resistance ohm (Q) ks ohm (H)
Conductance ohm“' (Q“') siemen (S)
Conductivity ohm^' cm“* cnr’) S m“'
oo
ohm"' cm-eq"' (H~' cm^eq"')
-1
Equivalent conductivity S m- eq
Y

Molar conductivity ohnr' cm^ mol"^ (Q"’ cni^ mol"') S m^ mol"'

eB

-1 -1
Cell constant cm m
r
ou

Probl«m Q If specific conductivity of N/50 KCl solution at 298 K is 0’002765 ohm"' cm ' and
Y
ad

resistance of a cell containing this solution is 100 ohms, calculate the cell constant.
d

Sp.conductivity
Solution. Cell constant = = Specific conductivity x Obs. resistance
in
Re

Obs.conductance

= 0 002765 n-’ cm"' x 100 Q = 0-2765 cm"'

F

Problem Q 0*5 normal solution of a salt placed between two platinum electrodes 2-0 cm. apart
and of area of cross section 4-0 sq. cm. has a resistance of 25 ohms. Calculate the equivalent conductivity of
solution.

Solution. 1st step-Ca/c»/a//wj of speciifc conductivity.

Here, I = 2-0 cm, a = 4-0 cm-. R = 25 ohms

/ 2 cm -I
Conductance G = — — — ohm ' ; Cell constant = — = — cm
R 25 a 4 cm^ 2
1
Sp. conductivity (k) = Observed conductance x Cell constant = —'x
25
—cm
2
' =G02Q 'em ’
1000
2nd Step-Calculation of equivalent conductivity, a...
eq = kx
^eq
Here, c = 0-5 N, k = 0-02 ohm
-1
cm ’ (calculated above)
ELECTROCHEMISTRY 3/23

(0-02n-‘cm-^)x(1000cm^L-^)
= 40 cm^ eq or 40 S cm^eq *
-1

(0-5geqL-i)
Problem The electrical resistance of a column of 0*05 M NaOH solution of diameter 1 cm and
length 50 cm is 5*55 x 10^ ohm. Calculate its resistivity, conductivity and molar conductivity.
(NCERT Solved Example, CBSE 2012)
Solution. (/) Calculation of Resistivity. Electrical resistance of the solution, R = 5-55 x 10^ Q
( <r
Area of cross-section of the column («) = k = 3-14 x 2
cm- = 0-785 cm-
V “ /

Length of the column (/) = 50 cm

ow
/
Applying the formula, R = p —
a

RX(7 (5-55xl03fi)(0-785cm2)
P = = 87-135 £2 cm. i.e.. Resistivity (p) = 87-135 £2 cm
/ 50cm

e
re
I

rFl
1
(ii) Calculation of conductivity. Conductivity (k) = — = 0 01148 S cm-*
p 87-135Qcm

F
(m) Calculation of molar conductivity

r
KxlOOO (0-01148Scm-')(1000cm^L-‘)
ou
= 229-6 S cm^ mor*

fo
Molar conductivity (a,„) =
Molarity (0-05 mol L"*) ks
Problem 0 Resistance of a conductivity cell filled with 0-1 mol L * KCl solution is 100 £2. If the
resistance of the same cell when filled with 0-02 mol L“* KCl solution is 520 £2, calculate the conductivity
oo

and molar conductivity of 0-02 M KCl solution. The conductivity of 0-1 M KCl solution is 1*29 S/m.
Y
B

(NCERT Solved Example, CBSE 2014)

Solution, (i) For 01 M KCl solution. R = 100 £2, K = 1-29 S m '
re

Cell constant = Conductivity x Resistance = 1-29 S m"' x 100 £2 = 129 m"* (S = £2"*) = 1-29 cm"*
ou

-I
Cell constant 129 m
Y
ad

(i7) Conductivity of 0-02 M KCl solution (k) = = 0-248 £2-' m-‘ or 0-248 S m -1
Resistance .520 £2

Concentration of the solution = 0-02 M = 0-02 mot L-' = 0-02 x 10^ mol m"^ = 20 mol m“^
d

0-248Sm-*
in
Re

K
Molar conductivity = — = 0-0124 S m^ mol * = 1-24 x 10"^ S m^ mol *
c
rn
20 mol m ^
F

Cell const. 1-29 cm ’

K = = 0-00248 S cm-’
Alternatively, Resistance 520 £2

K 0-00248 S cm"’ x 1000 cm^ L" ’

A = 124 S cm^ mol ’
0-02 mol L-’
m
c
m

Problem R The conductivity of a solution containing 1 gram of anhydrous BaCl2 in 200 cm^ of
water has been found to be 0-0058 S cm"’. What are the molar conductivity and equivalent conductivity of
the solution ? (At. wt. of Ba = 137 and Cl = 35-5).
-1
Solution. We are given : Conductivity (k) = 0-0058 S cm
-I
Molar mass of BaCU = 137 + 2 x 35-5 = 208 g mol
As 1 gram of BaCl2 is present in 200 cm^ of the solution, therefore, molar concentration (c)
I 1
X X1000 mol L-’ = 0-0240 mol L"'
208 200
3/24 ^>uzeUcfi-’^ New Course Chemistry (XII)B&m

KxlOOO 00058S cm-‘ x lOOOcm^

Molar conductivity, A = 241-67 S cm^ mol-*
m
C
m
0-0240 mol L-*

Mol.Wt 208
= 104
Further, in case of BaCl,, equivalent weight = ^ 2

1 1
Concentration of the solution in gram equivalent per X1000 = 00480
litre (c^^) = 104^200
KXlOOO 00058Scm-*x!0OOcm^L-‘
Equivalent conductivity, a eq = 120-83 S cm- eq“*
0-0480 geqL-

ow
1. Specific conductivity of a 0-12 normal solution of an electrolyte is 0-024 ohm ’em*. Determine its equivalent
conductivity.
2. The resistance and conductivity of a conductivity cell containing 0 001 M KCl solution at 298 K are 1200 Q and

e
1-5 X 10"^ S cm"* respectively. Calculate the cell constant and molar conductivity.

Fl
(CBSE 2022)

re
3. The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms.

F
Calculate the equivalent conductivity of the solution if the electrodes in the cell were 2 cm apart and each
has an area of 3-5 sq. cm.
ur
or
4. A cell with N/50 KCl solution showed a resistance of 550 ohms at 25^. The specific conductivity of N/50
KCl at 25“ C is 0-002768 ohm"’ cm"*. The cell filled with N/10 ZnS04 solution at 25°C shows a resistance
sf
of 72-18 ohms. Find the cell constant and molar conductivity of ZnS04 solution.
k
5. A potential differenceof 20 volts appliedto the ends of a column of M/10 AgN03 solution, 4 cm in diameter and
Yo
12 cm in length gave a current of 0-20 ampere. Calculate the specific and molar conductivities of the solution.
oo

-] -1
6. Calculate the equivalent conductivity of 1 M H2SO4 solution, if its conductivity is 26 x 10"- ohm cm
B

(Atomic weight of sulphur = 32).

The measured resistance of a conductance cell containing 7-5 x 10"^ M solution of KCl at 25°C was 1005
re

7.
ohms. Calculate (a) specific conductance (b) molar conductance of the solution. Cell constant = 1 -25 cm"’.
u

8. The conductivityof 0-2 M solution of KCl at 298 K is 0-0248 S cm”’. Calculate its molar conductivity.
ad
Yo

(CBSE 2007, 2008, 2013 ; Assam Board 2012)

9. The electrical resistance of a column of 0-05 M KOH solution of length 50 cm and area of cross-section
0-625 cm^ is 5 X 10^ ohm. Calculate the resistivity, conductivity and molar conductivity.
d

(CBSE 2020)
Re
in

10. The conductivity of 0-01 M solution of acetic acid at 298 Kis 1-65 x 10^ S cm"’. Calculate molar conductivity
(a„,) of the solution. (CBSE Compt. 2018)
F

11. The molar conductivity of a 1 -5 M solution of an electrolyte is found to be 138-9 S cm^ mor’. Calculate the
conductivity of this solution. (CBSE 2022)
ANSWERS

1. 200 ohm"' cm^ eq"’ 2. 0-18 cm-’, 150 S cm2 mol"’ 3. 23-32 ohm"’ cm^ eq
-1

4. 1-5224 cm"’, 421-8 ohm“‘ cm- mol"’ 5. 9-55 X 10"^ S cm"’, 95-5 S cm^ mol"’
6. 130 ohm"’ cm^ eq"' 7. (a) 0-001244 ohm"’ cm"’ (b) 165-87 ohm"' cm^ mol"’
8. 124 Q-' cm^ mol"’ 9. 62-5 Q cm, 0-016 S cm"’, 320 S cm^ mol"’
10. 16-5 S cm- mor’ 11. 0-208 S cm-’
HINTS FOR DIFFICULT PROBLEMS

1000 1000 cm3 L-’

1. A._ = KX
eq = 0-024n ’em ’ X = 200 Q ' cm^ eq ’.
Normality 0-12 g eq L
-]
ELECTROCHEMISTRY 3/25

1. Cell constant = k x R = 1200 x 1-5 x kH 12“’ cm"’ = 018 cm"'

tcxlOOO (l-5xl0"^)xl000
A = 150 S cm^ mol '
10-3
m
Molarity

/ / 1 2cm
3. k=Gx-=—x-= X
a R a 2450 3-5 cm2

10(X) 2 ' 1000cm3 L-’

= KX X — 0-'cm“’ X = 23-32 0-' cm2
Normality 245 3-5 0-1 mol L '

N M 1000
4. — ZnS04Sol = —ZnSO^sol. and m
= KX
10 20 Molarity ’
£ 20 V
5. By Ohm’s law, R = — = 100VA-‘ =1000
I 0-20A

6. 1 M H2SO4 = 2 N H2SO4.

F low
1 1
7. Specific conductance (k) =●—x cell constant = X1-25 cm-' =0 001244 0"'cm-'
R 10050

KXlOOO 0-0012440-‘cm-'xlOOOcm^L"'
Molar conductance (a,„) = = 165-87 0-‘cm2mol"'.
Molarity 7-5 X10-3 mol L-'
9. Similar to Solved Problem 3.

e
10. A
KXIOOO (1-65x10''Scm ’)xl000cm3L 1 for Fr
= 16-5 S cm2 1
m -i
Molarity 0-01 mol L

KXlOOO A X Molarity 138-9x1-5

11. A or K =
m
= 0-208 S cm-’
Your

m
Molarity 1000 1000
eBo ks

3.8. VARIATION OF CONDUCTANCE, SPECIFIC CONDUCTIVITY, EQUIVALENT

CONDUCTIVITY AND MOLAR CONDUCTIVITY WITH CONCENTRATION OR DILUTION
ad
our

In general, for weak as well as strong electrolytes, electrolytic conductance as well as the equivalent
conductivityand molar conductivityincrease with dilution whereas the speciifc conductivity of an electrolytic
solution decreases with dilution. The reasons for such a behaviour are briefly explained below :
Re

The conductance of a solution is due to the presence of ions in the solution. The greater the number of
ions, the greater is the conductance. As on dilution, more ions are produced in solution, so the conductance
Find Y

should also increase on dilution.

The specific conductivity of an electrolyte falls with dilution because the number of current-carrying
particles, i.e., ions present per centimeter cube of the solution becomes less and less on dilution.
However, the increase of equivalent and molecular conductivity on dilution is due to the fact that these
are the products of specific conductivity and the volume V of the solution containing one gram equivalent or
one gram molecule of the electrolyte. As the decreasing value of specific conductivity is more than compensated
by the increasing value of V, so the values of a,,^^ and a,„ increase with dilution.
3.9. VARIATION OF MOLAR CONDUCTIVITY WITH
CONCENTRATION (FOR STRONG AND WEAK ELECTROLYTES)
As already mentioned in the beginning of this unit, there are two types of electrolytes, namely strong
electrolytes and weak electrolytes. The variation of molar conductivity with concentration (or dilution) for
each of these electrolytes is briefly discussed below :
3/26 ‘Pn*xdec^'<x New Course Chemistry (XII)SSSIS]

Conductance Behaviour of Strong Electrolytes. The molar conductivity of strong electrolytes is found
to vary with concentration according to the equation :

f'm = m
—A -Jc
where A is a constant depending upon the type of the electrolyte*, the nature of the solvent and the temperature,
and is the molar conductivity at infinite dilution, called limiting molar conductivity. This equation is
called Debye Huckel-Onsagerequation and is found to hold good at low concentrations. Thus, if a wi

is plotted against Vc . a linear graph is obtained for low concentrations (with slope = - A) but it is not linear
for higher concentrations, as shown in Fig. 3.8 in which KCl has been taken as an example of a sti'ong
electrolyte.
Further, the curve obtained for a strong electrolyte shows tliat there is only a small increase in conductance

ow
with dilution. This is because a strong electrolyte is completely dissociated in solution and so the number of
ions remains constant. At higher concentrations, the greater inter-ionic attractions retard the motion of ions
and. therefore, the conductance falls with increasing concentrations. With decrease in concentration, i.e.,
with dilution, the ions are far apart and, therefore, the interionic attractions decrease due to which the
conductance increases with dilution and approaches a maximum limiting value at infinite dilution, designated

e
re
Fl
as or A
m '

Conductance Behaviour of Weak

F
FIGURE 3.8
Electrolytes. As a weak electrolyte 160 ● AO(KCI)
dissociates to a much
ur
lesser extent as
KCl

r
140 ●●
compared to a strong electrolyte, therefore,

fo
LU
O-
o,r-
the conductance of a weak electrolyte is Z I
s -■
ks
much lower than that of a strong electrolyte
Yo
at the same concentration. This is shown in QCM
oo

Fig. 3.8 in which acetic acid has been taken o O

0--
as an example of a weak electrolyte. Further, I
eB

the curve obtained for a weak electrolyte

do
shows that there is a very large increase in
conductance with dilution especially near
ur

inifnite dilution. This is because as the

ad
Yo

concentration of the weak electrolyte is

reduced, more of it ionizes. Thus, increase in CONCENTRATION IN (MOL L"1)‘'‘'2
conductance with decrease in concen Variation of molar conductivity of a strong electroIyte(KCl)
d

tration is due to the increase in the number

Re

and weak electrolyte (CH3COOH) with concentration

in

of ions in the solution. However, it does not

F

reach a limiting value.

The experimental results of molar conductivity of KCl and CH3COOH at different concentrations given
below further help to understand the variation more clearly.

Concentration (moi L 01 0-05 001 0005 0-001 0-0005 0

(infinite dilution)
Molar conductivity 129-0 133-4 141-3 143-5 146-9 147-8 149-9
of KCl (S cm^ mor'')
Molar conductivity of 5-2 16-3 49-2 67-7 .390-7

CH3COOH (S cm2 ^0,-1)

*NaCl, BaCl2, MgSO^ are called 1 - I, 2 - I and 2-2 electrolytes respectively depending upon the charges
on cation and anion. For a given solvent and temperature, all electrolytes of a particular type have the same value
for the constant A.
 ELECTROCHEMISTRY 3/27

Further, from the plots shown in Fig. 3.8 it is observed that the plot for a strong electrolyte becomes
linear near high dilutions and thus can be extrapolated to zero concentration to get the value of the molar
conductivity at infinite dilution, Le., a°,„. However, for a weak electrolyte increases steeply on dilution,
especially near low concentrations. Hence, an extrapolation to zero concentration is not possible . as is clear
from the plot for acetic acid in Fig. 3.8. Experimentally also, it is not possible to determine the a“^^ value for
a weak electrolyte because though the dissociation is complete, the concentration of ions per unit volume is
so lov; that the conductivity cannot be measured accurately. The problem was finally solved by Kohlraiisch
who put forward a law known as Kohlrausch law, as discussed in the next section.

Sample Problem The moIar Conductivity of KCl solutions at different concentrations at 298 K
are given below :
c/niol L”‘ a/s cin^mol"* c/mol L ^ a/s cni“ mol *

w
0-000198 148-61 0000521 147-81
0-000309 148-29 0-000989 147-09

Flo
Show that the plot between a and is a straight line. Determine the values of a" and A for KCI.
iNCERT Solved Example)

ee
Solution. Taking the square root of concentration, we obtain :

Fr
c‘/2/{niol A^S cm2 mol“*

for
0-01407 148-61 150.2
ur
0-01758 148-29 149.8

0-02283 147-81
s
..r- 149.4
k
L
Yo
0-03145 147-09 o
oo

e 149.0
CM

A plot of A„, (y-axis) and (ji-axis) is shown in E

eB

« 148.6
the adjoining Fig. : V)I

148.2
It can be seen that it is netirly a straight line. From
r

the intercept (i.e., when = 0). we find that 147.8

ou
ad

A...‘’=
III
150-0 S cm2 mor' 147.4
Y

147.0
and A=-slope=AZ 150-0-147-0
0 .005 .010 .015 .020 .025 .030 .035
Re
nd

A.V 0034
Fi

= 88-23 S cm2 mol'V(mol/L"')’^-.

3,10. KOHLRAUSCH'S LAW

Kohlrausch studied the molar conductivities at infinite dilution a®„ for a number of pairs of strong
electrolytes, each pair having a common cation or anion. Then he calculated the differences of these a° in

values for each pair. A few of these results are given in Table 3.4.
It is observed that the difference between a°,,, values for each pair of sodium and potassium salts
having a common anion was .same, irrespective of what this anion was. Similarly, the difference in the
A°... values for each pair of salts having the same different anions and a common cation was same,
HI

irrespective of what this cation was. In fact, column 3 in the Table 3.4 gives the a° value for ion HI

minus the a° values for Na"^ ion. Similarly, the last column gives
m
a°,„ value for Br~ ion minus a°„, value
for Cl~ ion.
3/28 ^n4sietee^ d. New Course Chemistry (XII)

TABLE 3.4. A°^ values (in ohm"^ cm^ mol"^) of some pairs of electrolytes
(at 298 K) having common ions

Difference Electrolyte A®^at 298 K Difference

Electrolyte A®„at298 K

}
KCl 149-86 KBr 151-92

w
126-45 23-41 KCl 149-86 2-06
NaCl
NaBr 128-51

}
KBr 151-92
128-51 23-41 NaCl 126-45 2-06
NaBr

}
144-96 LiBr 117-09
KNO3

e
121-55 23-41 LiCl 115-03 2-06
NaN03

e
concluded that each ion makes a definite

or
On the basis of the above observations, Kohlrausch

r
contribution to the total molar conductivity of an electrolyte at infinite dilution, irrespective of the nature
of the other ion of the electrolyte. This individual contribution of an ion towards the total molar conductivity

F
of the electrolyte is called molar ionic conductivity. As a result of these studies, Kohlrausch in 1876 put

oF
ul
forward a generalization, known after him as ‘Kohlrausch law’. It states as under:
The limiting molar conductivity of an electrolyte (Le. molar conductivity at infinite dilution) is

rs
the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the

k
number of ions present in one formula unit of the electrolyte.

o
A°„forA,B^ = *X", + :vX«_
of
Mathematically, o
where a° „ is the limiting molar conductivity of the electrolyte, and are the limiting molar conductivities
Y
of the cation (A-'^) and the anion (B*") respectively.
YB

For example. a‘’_ for NaCl = X'’ Na-t- + X° a-

m ’
er

= x° + 2X°
a°^ for BaCl2 Ba
2+
cr
u

A°^forAl2(S04)3 =2X“ + 3X“

502-
od
ad

In terms of equivalent conductivities, Kohlrausch’s law is defined as follows ;—

in

The equivalent conductivity of an electrolyte at infinite dilution is the sum of two values one
depending upon the cation and the other upon the anion, i.e.,
Re
F

TABLE 3.5. Limiting molar

conductivities of some common
where and are called the limiting ionic cations and anions in S cm^ mol-^
conductivities for the cation and the anion respectively.
Cation Xe Anion
The limiting ionic conductivities of some cations and
anions are given in Table 3.5 below : H
+
349-6 OH" 199-1

Calculation of a^^^ from values of ions. A few Na-" 50-1 cr 76-3

examples are given below : K-^ 73-5 Br- 78-1
a"
eq (NaCl) = (Na-^) + X",,, (CD Mg
2-f-
106-0 r 76-8

Ca2+ 119-0 NO3 71-5

(BaClj) = r„, (Ba^n + X”„, (Cr)
Ba2-^ 127-2 SQ2- 160-0

A
%(A1C13)= ^ r,„(Al3-^) + X“„,(Cl-) Sr2+ 118-9 CH3COO- 40-9
ELECTROCHEMISTRY 3/29

1 1
A
% (Al2(S04)3)=- +
2 tSOf)
1 1
In general, a” eq +
(cation) + (anion)
n n

where represents the charge on the cation and n~ represents the charge on the anion (In place of n +

and n~, symbols z+ and z_ are quite often used)

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Relation between molar conductivity (a^) and equivalent conductivity (a^^)

a“
m
a” m
a° or A°
eq eq
V_2.

where v^and v_are stoichiometric numbers of the cations and the anions respectively in one formula unit

w
of the electrolyte (A^^ ) and 4+ and are the charges on the cation and anion respectively.

F lo
As z+ represents the total charge on the cation and v_ z_ represents the total charge on the anion and
v^. = v_ 2_, hence we can write
A°

e
A° m

Fre
eq
Total charge on cation or anion (called ‘n’ factor or valency factor)

For example, for the salt, sodium potassium oxalate,

COONa

COOK
for
, equivalent conductivity (A°g^) is related
r
You
A°
to molar conductivity as, a® eq =
s

(as charge on cations = charge on anion = 2)

ook

Similarly, in case of potash alum, K2S04.Al2(S04)3.24 H2O,

eB

A®
A® m
(as total charge on cations = total charge on anions = 8)
eq
8
our
ad

2. Absolute ionic mobilities and their calculation. The absolute ionic mobility is deifned as the speed of
the ion in cm/sec at infinite dilution under a potential gradient of one volt per cm. (Potential gradient
-1
dY

cms
- cm^s~* volt”*
Re

= Applied EMF/distance between the electrodes). Hence, its units are

volt cm”*
Fin

It is found that the limiting ionic conductance or is directly proportional to the absolute ionic
mobility of that ion («®^ or 11°i.e., X°^« and X® a
« u
a or X®^ = k u°^ and X®^ = k m® a

The value of the constant k is found to be equal to F (i.e. 1 faraday or 96500 coulombs). Hence, we
can write

X®
X®,= Fu®, or u°-c = a
and X® a =Fm®a or
a
F F

Ionic conductance
In general, I> tic mobility =
96500

For example, ionic conductance (X°) for Na+ ion is 50-1 H”* cm^ eq"*. Hence, its ionic mobility
_ 501Q *cm^eq -1

-1 = 519 X 10^0"* cm^Cr* =519 x lO^cm-s"' V”*

96500 Ceq
(v C = AsorCn = AsaButAa = V .-. C n = V s or C”> n”* = V”* s”>)
3/30 7>>uxdeefr’4. New Course Chemistry (Xll)CSiai

3.11. APPLICATIONS OF KOHLRAUSCH'S LAW

1. Calculation of molar conductivity at infinite dilution (a“) for weak electrolytes. As already
mentioned, the molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally,
firstly because the conductance of such a solution is low and secondly because the dissociation of such an
electrolyte is not complete even at very high dilutions. The molar conductivity of such an electrolyte at
infinite dilution can be calculated using Kohlrausch’s law.
Consider the example of acetic acid (CH-,COOH) as the weak electrolyte. By Kohtrausch s law,
a“ (CH3COOH) = rCHXOO"
3

This equation can be arrived at by knowing the molar conductivities at infinite dilution for the strong
electrolytes KCl, CH3COOK and HCl. as by Kohlrausch’s law
a“ (KCl) = (CH3COOK) = ^°cH3COO- , aMHCI) =^'

low
H cr

+ r... )
Hence, we have X® H+
cr K* C|-
CH3COO" CH3COO-
i.e., A° (CH3COOH ) = a'’ (CH3COOK) + A° (HCl) - A° (KCl)

ee
Sample Problem 0 If the molar conductivities at infinite dilution of NaCI, HCl and CH3COONa
rF
Fr
(NaAc) are 126-4,425-9 and 91-0 S cm^ mol"* respectively, what will be that of acetic add (HAc) ?
(NCERT Solved Example, CBSE 2007)

for
Solution. According to Kohlrausch’s law, a® for CH3COOH = X® CH3COO- H
u
A® for NaCl = ,+X® cr = 126-4 Sem^ mol"' ...(/)
ks
Na
Yo
A® for HCl = 425-9 S cm- mol ' ...(«)
o

H cr
Bo

+ X°
A® for CH3COONa “ ^"cHaCOO- Na+ = 91-0 S cm- mol ' ...(m)
re

Adding eqns. (n) and (m) and subtracting (t), we get

+x® +x® -X® = 425-9 + 91-0 - 126-4 S cm^ mol"‘
ou

Na-^
ad

cr CHjCOO" Na Cl"

-1
+X®
Y

X® = 390-5 S cm- mol"’- i.e..

or
CH3COO
A
® for CH3COOH = 390-5 S cm- mol
Sample Problem Calculate a„,® for CaClj and MgS04 from the following data :
nd
Re

X® (S cm2 . ca2^ = 119-0, Mg2+ = 106-0, Cl" = 76-3, SOf = 160-0.

Fi

(NCERT Solved Example)

Solution. By Kohlrausch law, a^® (CaClj) = X‘ Ca 2+ + 2X® cr = 119-0 S cm2 mol"* + 2 x 76-3 S enr mol"*
= (119-0+ 152-6) S cm2 mol"' ^ 271-6 S cm^ mol"‘
+x® = 106-0 S cm2 4. 150 0 S cm2 mol"* _ 266 S cm2 mol"*
A„" (MgSOj) - Mg
2+
S05"

1. Calculate the molar conductance at infinite dilution for acetic acid, given

a“(HC1) =425 n-*cm2mcr‘, a" (NaCl) =188 0"' cm2 mol"*, a" (CH3COONa) = 96 Q"‘cm2 mol
-i

2. The molar conductivity of NH4CI at infinite dilution is 149-7 S cm2 the ionic conductivities of OH" and
Cl" ions are 198 and 76-3 S cm2 respectively. Calculate the molar conductivity of NH4OH at this dilution.
 ELECTROCHEMISTRY 3/31

3. Calculate the molar ionic conductance of AP'*' ions at infinite dilution, given that the molar conductance of

AI2 (504)3 molar ionic conductance of SOj'' ions at infinite dilution are 858 S cm- mol"' and 160 S

rw
cnr mol~^ respectively.
4. The molar conductances of NaOH.NaCl and Bad, at infinite dilution are 2481 x IO-2, 1-265 x and
2-800 X 10 ^ S m- mol"' respectively. Calculate Ba (OH),.
5. Given molar conductivities at Infinite dilution : for Ba (OH), = 517-6 cm^ mol"'
A
for BaCl, = 240-6 U"' cm- mol"', a\ for NH4CI = 129-8 Or' cm^ mol"'. Calculate a"^ for NH4OH.

e
6. Find out the molar conductivity of an aqueous solution of BaCI, at infinite dilution when ionic conductances
of * ^d Cl ion are 127-30 S cm- mol"' and 76-34 S cm- mol"' respectively.

r
(MP Board 2011}

o
7. The Am“ values for NaCl and KCl are 126-5 and 149-9 H"' cm- mol"' respectively. The ionic conductances

lu
F
of Na'*’ at infinite dilution is 50-1 Q"' cm^ mol"'. Calculate the ionic conductance at infinite dilution for
ion.

ANSWERS

F
s
1. 333 cm^ mol"' 2. 271-4 S cm^ moi"' 3. 189 S cm^ mol '
4. 5-232 X lO"^ S mol"' 5. 268-3 £2"' cm^ mol"' 6. 279-99 S cm^ mol"'

o
kr
7. 73-5 £2"' cm^ mol"'

oo
HINTS FOR DIFFICULT PROBLEMS

of
2. A”,,, (NH4OH) = (NH4CI) - I"„, (CI-) + X\, (OH-).
Y
Y
3. a”„, [A1,(S04)3] = 2 K, (Al^-) + 3 X%, (S02-).
B
4. a“„ [Ba (OH),] = a“„, (BaCl2) + 2 a%, (NaOH) - 2 a°,„ (NaCI).
5. A°„,(Ba(OH),) =^°3^,2++2?^'’ OH"
r

...(/)

+ 2X^
e

(BaCl2)
u

2+
...(H)
d

Ba cr

°„,(NH4CI) =^°nhJ
A
+ X^
C\- ...m
o
n
ad

1 1
+X°
a'’^(NH40H) NHJ OH" = eqn. (/■) + eqn. (m) - - eqn. (//)
i

= — X 517-6 -f 129-8 — X 240-6 = 268-3 £2"' cm^ mol"'

F
Re

2 2

6. A‘’„,(BaCI,) = ^° Ba 2+ -»-2>.°
cr
= 127-30 + 2 (76-34) S cm2 = 279-99 S cm2 1

7. X%„ (KCl)-?.%,(NaCi) = ;^“+-A.“^^^=I49-9-I26-5= 23-4£2-'cm2mo|-l

X° K-^. = 234 + NiC
. , = 23-4 + 50-1 = 73-5 £2"' cm2 mol"!

2. Calculation of the Degree of Dissociation. According to Arrhenius theory of electrolytic dissociation,

the increase in the molar conductivity with dilution is entirely due to the increase in the dissociation of the
electrolyte ; the molar conductivity at infinite dilution being maximum because the dissociation is almost
complete. Thus, if a^^ is the molar conductivity of a solution at any concentration c and a°^j the molar'
conductivity at infinite dilution {i.e., zero concentration), we will have at concentration c,
A?m
Degree of dissociation (a) =
A
m

However, this relationship is found to hold good only for weak electrolytes. The value of A°„. m
for the
weak electrolytes can be calculated, using Kohlrausch’s law, as discussed already in the first application.
3/32
New Course Chemistry (XIl)CEIBI

Sample Problem Calculate the degree of dissociation (a) of acetic acid if its molar conductivity
is 39-05 S cm2 Given (H+) = 349-6 S cm^ mori and X" (CH3COO-) = 40-9 S cm^ mor^ (C HSE :017)
A
m
Solution. a =
A°
m

(CH3COOH) = X° {CH3COO-) + V (H-*-) = 40-9 + 349-6 = 390-5 S cm^ mol '

A - 39-05 S cm2 (Given)

39-05
a = = 0-1
390-5

wmm

w
1. If the molar conductivities at infinite dilutionat 293 K for aqueous hydrochloric acid, sodium acetate and
sodium chloride solution are 383-5, 78-4 and 102-0 S cm^ respectively, calculate the molar conductivity of

Flo
acetic acid at this temperature and dilution. If the molar conductivity of acetic acid at some other dilution is
100-0 S cm2 293K, calculate the degree of ionization of acetic acid at this dilution.

e
2. (a) The molar conductivities at infinite dilution of potassium chloride, hydrochloric acid and potassium

re
acetate are 130-1, 379-4 and 95-6 S cm2 respectively. Calculate the value of molar conductivity at

F
infinite dilution for acetic acid. .● yni

(h) If the molar conductivity of a given acetic acid solution is 48-5 S cm2 i 25“C, calculate the degree
ur
or
of dissociation of acetic acid at this temperature.
3. The molar conductivity of acetic acid at infinite dilution is 38712 ' cm2 1 same temperature, but
at a concentration of 1 mole in 1000 litres, it is f
55 Or' cm2 jg % age dissociation of
ks
0-001 M acetic acid ?
Yo
if a,^ and a^ of CH3COOH are 48 S cm2 jjjqJ l
oo

4. Calculate the degree of dissociation (a) of CH3COOH

400 S cm2 mor* respectively.
B

ANSWERS
re

1. A° - 359-9 S cm2, _ Q.278 2. a“= 344-9 ohm~' cm2 . o-=z 0-141 3.14-21% 4.0-12
u
ad
Yo

HINTS FOR DIFFICULT PROBLEMS

2. a“ (CH3COOH) = A° (CH3COOK) + A° (HCl) - a“ (KCl) = 95-6 -f 379-4 - 130-1 = 344-9 S cm2 1

d
Re
in

a£ 48-5
m _
a = = 0-141.
A° m 344-9
F

-I
3. I mol in 1000 litres = 0-001 mol L = 0-001 M.

3. Calculation of dissociation constant of a weak electrolyte. Knowing the degree of dissociation, a

(as calculated above), the dissociation constant (K) of the weak electrolyte like CH3COH or NH4OH at
concentration c of the solution can be calculated using the formula.

ca2
(as illustrated in the solved problems below.)
1-a

Sample Problem Q At 291 K, the molar conductivities at infinite dilution of NH4CI, NaOH and

NaCl are 129*8, 217*4 and 108*9 S cm2 respectively. If the molar conductivity of a centinormal solution of
NH4OH is 9*33 S cm2, jg percentage dissociation of NH4OH at this dilution ? Also calculate the
dissociation constant of NH4OH.
Solution. Here, we are given : A*’ for NH4CI = 129-8 S cm2, A° for NaOH = 217-4 S cm2.
A° for NaCl = 108-9 S cm2
 ELECTROCHEMISTRY 3/33

By Kohlrausch’s law, a° for NH^OH ” NH^^ (NH^Cl) + a” (NaOH) - a“ (NaCI)

= 129-8 + 217-4 - 108-9 = 238-3 S cm^
A^, = 9-33 S cm“ (Given)
A 9-33
Degree of dissociation (a) = — - = 0-0392 or
% age dissociation = 0-0392 x 100 = 3-92%.
A° 238-3
Calculation of dissociation constant

NH4OH NH| + OH-

Initiai cone. c
cox CO CO?
Equilibrium cone. c-co = c{ \ - a) CO CO K =
c(l-a) 1 -a
Putting c = 0-01 N = 0-01 M and a = 0-0392, we get
(0-00(0-0392)^ _ I0“2x(3-92x10-2)2

w
K = = 1-599 X 10"*-'
1-0 0392 0-9608

Sample Problem Q The conductivity of 0-001028 M acetic acid is 4-95 x 10“5 S cm"'. Calculate

F lo
its dissociation constant if a“ for acetic acid is 390-5 S cm- mol"’. (NCERT Solved Example)
Solution. For the given concentration of acetic acid solution,

ee
KxlOOO 4-95 X10-5 S cm"’ X1000 cm^ L" >

Fr
A = 48-15 S cm- mol ’
0-001028 mol L"'
m
C

48-15Scm^mol ’
for
A
m —
0 = = 0-1233
a“ m 390-5 Sem^ mol ’
ur
CH3COOH ~ CH3COO-+H-"
s
Initial cone.
ook

c 0 0
Yo

Equilibrium cone. c-co-c{\ - a) c o c o

eB

K =
caca
CO? _ (0-001028molL-')(0-1233)2 = 1-78 X 10-5 mol L-’
c(I-a) 1-a 1-0-1233
our
ad
Y

Conductivity of 0-00241 M acetic acid solution is 7-896 x 10 5 S cm ’. Calculate its molar conductivity in
Re

this solution. If a„° tor acetic acid be 390-5 S cm^ mol"’, what would be its dissociation constant ?
nd

(CBS!-: 2008)
Fi

ANSWER
32-76 S cm^ mol"’, 1-85 x 10*5
HINT FOR DIFFICULTPROBLEM

KxlOOO 7-896X10"5 S cm"' x lOOOcm-^L"’

A*".
m
= = 32-76 S cm- mol ’
Molarity 0-00241 mol L"'

a =
32-76
= 0-084, K = CO? _ 0-00241 x(0-084)2 = 1-85 X 10-5.
390-5 1-a 1-0-084

4. Calculation of solubility of a sparingly soluble salt. Salts such as AgCl, BaS04, PbS04, etc. which
dissolve to a very small extent in water are called sparingly soluble salts. As they dissolve very little, their
solutions are considered as infinitely dilute. Further, as their solutions are saturated, their concentration is
equal to their solubility. Thus, by determining the specific conductivity (ic) and the molar conductivity (a,„) of
3/34
’a New Course Chemistry (Xll)PSIg!

1000 1000 KXlOOO

such a solution, we have a° = KX = KX or Solubility =
Solubility
III
Molarity A
m

A° = x°
A III
... can be calculated by applying Kohlrausch's law (e.g., for AgCl, III
Ag
+

Sample Problem The conductivity of a saturated solution of AgCl at 288 K is found to be

1-382 X 10“^ S cm“'. Find its solubility. Given ionic conductances of Ag* and Cl“ at infinite dilution are
61-9 S cm- moI"‘ and 76-3 S cm- mol"^ respectively.
Solution. ^°ni (AgCl) - X Ag
+
cr
= 61 -9 + 76-3 = 138-2 S cm- mol"'
KXlOOO 1-382x10-^x1000
Solubility =
A°
= 10-5 mol L-' = 10"5 X 143-5 g L"‘ = 1-435 x 10~^ g L"'.
m 138-2

w
-1
At 291 K, saturated solution of BaS04 was found to have a specific conductivity of 3-648 x 10 ohm

F lo
cm . that of water used being 1 -25 x 10"^ ohm”' cm"'. Ionic conductances of Ba-* and SO4 ions are 110
and 136-6 ohm"' cm- mol"' respectively. Calculate the solubility of BaS04 at 291 K. (At masses ; Ba = 137,
S = 32,0= 16)

e
Fre
ANSWER

2- 266 X 10-5 gL-'

HINT FOR DIFFICULT PROBLEM
for
K X1000 (3-648 - 1-250) 10"^ x 1^ x233gL-' ●
r
K (BaSOj) = K (solution) - k (water). Solubility =
You
oks

A
///
(110+136-6)
eBo

5. Calculation of ionic product of water. It is found that ionic conductances of H* and OH at infinite
dilution are : = 349-8 S cm- mol"' and OH"
= 198-5 S cm- mol"'
ad
our

By Kohlrausch's Law,
X'
H-,0 OH"
= 349-8 + 198-5 = 548-7 S cm^ mol"'
Specific conductance of pure water at 298 K is found to be : k = 5-54 x 10"^ S cm"'.
1000
A°... = KX
Re

Applying the formula,

dY

m
Molarity
Fin

. , KXlOO 5-54x10-8x1000 = 1-01 X 10 ^ g ions L -i

Molarity, i.e., [H*j or [OH J =—^
A 548-3
m

K,,.= [H*] [OH-] = (1-01 X 10"^) X (1-01 X 10"^) FIGURE 3.9

= 1-02 X 10-'** ZINC COPPER
CUSO4 SOL
ROD AgN03SOL. ROD

3.12. GALVANIC CELLS

T
Whenever a redox reaction is allowed to lake
place directly in a single beaker, it is found that the
solution becomes hot. For example, when a zinc rod is
placed in copper sulphate solution (Fig. 3.9.), the
solution is found to be warmer as the reaction proceeds
PARTICLES OF Cu PARTICLES OF Ag
according to the equation
^ ZnS04 (aq) + Cu (s) Redox reactions taking place in a single beaker
Zn (5) + CUSO4 (aq)
ELECTROCHEMISTRY 3/35

Similar results are observed when a rod of copper is placed in silver nitrate solution. The reaction taking
place is as follows :
Cu (j) + 2AgN03 (aq) Cu(N03)2 (aq) + 2 Ag (,?)
Thus, we conclude that whenever a redox reaction takes place directly in a single beake,r chemical
energy in the form of heat is produced. By suitable means, it is possible to bring about the redox reaction
indirectly so as to convert the chemical energy into electrical energy.
A device used to convert chemical energy produced in a redox reaction into electrical energy is
called a galvanic cell or voltaic cell, after the names of Luigi Galvani (1780) and Alessandro
Volta (1800) who were the first to perform experiments on the conversion of chemical energy
into electrical energy.

How a redox reaction can be used to produce electrical energy, leading to the formation of a galvanic
cell, can be understood by taking the above two examples of redox reactions again.

w
1. Redox reaction between Zn and CUSO4: The reaction is
Zn + CUSO4 ^—> ZnSO^ + Cu {.v)

F lo
or it may be written as Zn + Cu-+ ^Zn-+ + Cu
This reaction may be split into two half reactions as under:
^ Zn^+ + le~

ee
Zn (Oxidation half reaction)

Fr
Cu-+ + 2e~ ^ Cu (Reduction half reaction)
The first reaction is called the oxidation half reaction and the second is called the reduction half reaction.

for
The reaction obtained by adding the two half reactions is called the overall reaction.
Thus, if a redox reaction is allowed to take place in such a way that oxidation half reaction takes place
ur
in one beaker and the reduction half reaction in another beaker, the electrons given out by the former will be
s
taken by latter and a current will flow. The two portions of the cell are called half-cells or redox couples. The
ook
Yo
cell is set up as follows :—
A zinc rod is placed in zinc sulphate solution taken in a beaker. A copper rod is placed in copper
eB

sulphate solution taken in another beaker. The two rods are connected by a wire and the two solutions are
connected by a salt bridge (Fig. 3.10).
our

Salt bridge and its function. A salt bridge is a U-shaped tube containingconcentratedsolution of an
ad

inert electrolyte like KCI, KNO3 K^SO^, etc. or solidified solution of such an electrolyte in agar-agar and
gelatine. An inert electrolyte is one who,se ions do not take part in the redox reaction and also do not
Y

react with electrolytes used. The function of the salt bridge is to allow the movement of ions from one
Re

solution to the other without mixing of the two solutions. Thus, whereas electrons flow in the outer circuit in
nd

the wire, the inner circuit is completed by the How of ions from one solution to the other through the salt
Fi

bridge. Moreover, it helps to maintain the electrical neutrality of the solution in the two half cells.
Thus, the main functions of the salt bridge are :
(i) To complete the electrical circuit by allowing the ions to flow from one solution to the other without mixing
of the two solutions.

(ii) To maintain the electrical neutrality of the solutions in the two half-cells.*
Let us see what would happen if the salt bridge were not used in the cell shown in Fig. 3.10.
Electrons given out by the zinc electrode will How to the copper electrode where they will neutralize
some of the Cu“'*' ions of the solution. Thus, SO^“ ions will be left and the solution will acquire a negative
charge. At the same time, Zn”'*' ions produced from zinc plate will enter into ZnS04 solution thus giving it a
positive charge. As a result of accumulation of charges in the two solutions, further flow of electrons will stop
and hence the current stops flowing.
*Another function of the salt bridge is that it prevents liquid- liquid junction potential, i.e., the potential
difference that arises between the two solutions when they are directly in contact with each other (as discussed on
page 3/57).
3/36 “pfusuiee^'^. New Course Chemistry (XII)C&1SI

FIGURE 3.10

VOLTMETER/AMMETER

0
CONVENTIONAL

r
e“
CURRENT
SO42 K-^
© ©
ZINC —: i SALT BRIDGE COPPER
ROD
ROD CONTG. K2SO4

w
Cu2+
ZnS04
*Zn2V CUSO4
/ SOL.
SOL.
SO42-;;

o
ANODE MOVEMENT OF CATIONS CATHODE

Zn-^Zn 2+ +2e~

e
MOVEMENT OF ANIONS; Cu2++2e--^Cu

Fl
4

re
Galvanic cell based on the redox reaction,

F
Zn + CuSO^ ZnS04 + Cu
ur
Some important features of the above cell may be summed up as follows :—

r
fo
(/) The zinc electrodeat which oxidation takes place is called the anode. The copper electrode at which
the reduction takes place is called the cathode.
ks
Yo
Hi) Since electrons are produced at the zinc electrode, this electrode is rich in electrons and pushes the electrons
oo

into the external circuit and hence it is designated as the negative pole. The other electrode, i.e., the copper
electrode is in the need of electrons for reduction of Cu^^ ions into Cu, i.e., this electrode is deficient in
eB

electrons and pulls the electrons from the externa! circuit, therefore it acts as the positive pole.
(Hi) The electrons flow from the negative pole to the positive pole in the external circuit. However,
ur

conventionally this current is said to flow in the opposite direction.

ad

of Zn^'*’ ions in the left beaker. Similarly, the reduction

Yo

(iv) The oxidation of zinc into ions produces excess

of copper ions to copper atoms leaves the excess SO^“ ions in the solution in the right beaker. To
d

maintain electrical neutrality of the solutions in the two beakers, the cations and the anions move
Re

through the salt bridge as shown in Fig. 3.10. This helps to complete the inner circuit, as already mentioned,
in

(v) As copper from copper sulphate solution is deposited on the copper electrode and sulphate ions migrate
F

to the other side, the concentration of copper sulphate solution decreases as the cell operates.
Consequently, the current falls with the passage of time.
(vi) Evidently, the weight of copper rod will increase while that of zinc rod will decrease as the cell works.
(vii) If an increasing external opposing potential is applied to the cell (Fig. 3.11 a), the reaction continues to
take place in the cell so long as is less than When becomes equal to Egg,j, the
reaction stops and the current becomes zero. On further increasing the reaction starts again but
in the opposite direction and the current now flows in the opposite direction (Fig. 3.11 b). The
cell now acts like an electrolytic cell (because the electrical energy is being used to bring about a
reaction which is otherwise non-spontaneous).Now, the electrons will flow from Cu to Zn and current
will flow from Zn to Cu. Further, zinc will be deposited on the zinc electrode and copper will dissolve
from the copper electrode.
The cell based on the above redox reaction is commonly called Daniel! cell. It has an electrical potential
of 1-1 V when concentration of Zn^^ and Cu^'^ ions is unity (1 mol dm"^). Thus, in Daniell cell, above changes
will be observed when <, = and > M V.
ELECTROCHEMISTRY 3/37

FIGURE 3.11

^exi ^ccll ^exi /^'cell ^exi ^ ^cell

■/W\/VVWv^^^vV /VvV'A^AW'VvV .AAA/SAAAWW'-

<2>
I'

<i>
e

Anode Cathode
Current I=0 Cathode Anode
/I +veI
Current
-ve
Zn- Salt — Cu — Cu
-ve
bridge
+ve Zn— Zn—I — Cu

z/t -Vz-z

ZnS04 («) CUSO4 ZnS04 (A) CUSO4 ic)

low
When E ext ^cell Wlien E ext ^cell When E ext > ^’cdl
(0 Electrons flow from Zn rod to Cu rod. (0 No flow of electrons or (/) Electrons flow from Cu to Zn.
Hence, current flows from Cu to Zn. current, Current flows from Zn to Cu.
(/7) Zn dis,solves at anode and Cu (n) No chemical reaction. (//) Zn is deposited at Zn electrode
deposits at Cathode. and Cu dissolves at Cu electrode.

ee
Direction of flow of current on applying external opposing potential when
rF
Fr
(a) £external ^ ^cell ^external = ^cell ^external > ^cel!
FIGURE 3.121
A popular form of this cell is shown in Fig. 3.12. Here, salt

for
VOLTMETER
bridge is replaced by a porous pot through which the SO4’" ions
u
migrate from copper sulphate solution to the zinc sulphate solution. (
ks
e .
2. Redox reaction between copper and silver nitrate
Yo
solution (Fig. 3.13): The reaction is :
o

@ COPPER
Bo

Cu + 2AgN03 4 Cu (N03)2 + 2 Ag VESSEL

♦ POROUS POT
or it may be written as :
re

sol ^CuS04
Cu + 2 Ag+ ^ Cu--" + 2Ag SOL.
sof — ZINC ROD
ou
ad

The two half reactions will be ;

●ZnS04 SOL.
Y

Cu > Cu--^ + 2e~

Daniel! cell
{Oxidation half reaction)
nd

FIGURE 3.13
Re

2 Ag^ +2 ^2Ag VOLTMETER / AMMETER

Fi

{Reduction half reaction) e“

CONVENTIONAL
Hence, the cell may be set up as CURRENT
e~
COPPER
NO3- K-"
shown in Fig. 3.13. o
Here, oxidation takes place at the
e|/ROD ^ SALT BRIDGE SILVER
CONTG. KNO3 ROD
copper electrode, therefore, it acts as
anode or negative ;tole. Reduction takes
place at the silver electrode, therefore, it
AgNOa
acts as cathode or positive pole. The Ag-
■Cu2V Cu(N03)2
-^SOL.
SOL\ NO3
conventional flow of cuirent is from silver

to copper.
ANODE MOVEMENT OF CATIONS CATHODE
In a similar manner, based upon any Cu -►Cu2+ + 2e- MOVEMENT OF ANIONS Ag-" + e ●Ag
redox reaction, a galvanic cell can be
constructed. Galvanic cell based on the redox reaction,
Cu + 2AgN03 Cu (N03)2 + 2 Ag
3/38 New Course Chemistry (Xll)BZsZ9]

Retain in Memory
(f) Electrode on which oxidation occurs is called anode (~ve pole)
(ii) Electrode on which reduction occurs is called cathode (+ve pole)
(Hi) Electrons flow from anode to cathode in the external circuit
(iv) Inner circuit is completed by the flow of ions through the salt bridge.

3.13. DIFFERENCE BETWEEN GALVANIC CELL AND ELECTROLYTIC CELL

The main points of difference between two types of electrochemical cells viz. a galvanic cell and an
electrolytic cell are summed up in Table 3.6.
TABLE 3.6. Difference between Galvanic cell and Electrolytic cell

w
Galvanic Cell Electrolytic Cell
1. It is a device to convert chemical energy into electrical 1. It is a device to convert electrical energy into
energy, i.e., electrical energy is produced as a result chemical energy, i.e., electrical energy is

Flo
of the redox reaction taking place (Refer to Fig. 3.10). supplied to the electrolytic solution to bring
Hence, free energy change (AG) is negative. about the redox reaction (i.e., electrolysis).

e
Hence, free energy change (AG) is positive.

re
2. It is based upon the redox reaction which is 2. The redox reaction is non-spontaneous and takes

F
spontaneous. place only when electrical energy is supplied.
3. Two electrodes are usually set up in two separate 3. Both the electrodes are suspended in the solution
ur
r
beakers. or melt of the electrolyte in the same beaker.
4. The electrolytes taken in the two betikers are different. 4.
fo
Only one electrolyte is taken.
ks
5. The electrodes taken are of different materials. 5. The electrodes taken may be of the same or
different materials.
Yo
oo

6. The electrode on which oxidation takes place is called 6. The electrode which is connected to the negative
the anode (or negative pole) and the electrode on terminal of the battery is called the cathode ;
eB

which reduction takes place is called the cathode (or the cations migrate to it which gain electrons
positive pole) and hence reduction takes place here. The other
electrode is called the anode.
ur

7. To set up this cell, a salt bridge/porous pot is used. 7. No salt bridge is used in this case.
ad
Yo

3.14. REPRESENTATION OF GALVANIC CELL

A galvanic cell is represented in a manner as illustrated below for the Daniell cell.
d
Re

ZniZn-+(c-i)IICu“-'(c2)ICu
in

i.e.. by convention, the electrode on which oxidation takes place is written on the left hand side and the other
F

electrode on which reduction takes place is written on the irght hand side. The electrode on the left hand side
is written by writing the symbol of the metal (or the gas) first followed by the symbol of the ion with its
concentration in brackets. The electrode on the right hand side is written by first writing the ion along with its
concentration in brackets followed by the symbol of the metal (or the gas). Single vertical lines represent the
phase boundaries of the electrodes and the double line represents the salt bridge, c, and 62 represent the
concentrations of Zn-^ ions and Cu"'*’ ions respectively.
In a similar manner, Cu - AgNO^ cell may be represented as : Cu I Cu“'*’ (Cj) 11 Ag"*" (cs) I Ag
To sum up : Meta) I Meta) ion (cone.) Metal ion (cunc.)IMetal
Salt

Anode (Negative pole), Oxidation occurs bridge Cathode (Positive pole), Reduction occurs
Kiectronsllow

Left electrode
4r 4 Right electrode
Conventional
current flows

*To remember. Oxidation and Anode both start with vowels. Reduction and Cathode both start with consonants.
ELECTROCHEMISTRY 3/39

Retain in Memory
In salt bridge, the electrolytes like KCl, KNO3 or NH4NO3 are preferred because their ions have almost
equal transport number, viz., 0-5, i.e., they move with almost the same speed when an electric current
flows through them.
3.15. ELECTRODE POTENTIAL
To understand the
FIGURE 3.14
concept of electrode
potential, consider a METAL ROD(M) METAL ROD(M) METAL ROD (M)
metal rod (M) placed in
contact with its own
ions Then there
is one of the following
three possibilities :
n+
(0 M"'*’ ions may M

w
collide with the
metal rod and n+-

\
n+ +ne M
M
n+
M -►M”*+ne
deflected back

F lo
n+
M
without under
going any change.
(i) Metal ions being deflected (ii) Electrons gained by M
n+
(iii) Electrons lost by M
{ii) M"'*' ions, on without any charge. accumulate on metal rod.

ee
are lost by M. So metal rod
collision with the acquires positive charge, So it acquires negative charge.

Fr
metal rod may (i)M
n+
ions deflected back after colliding without any change
gain electrons
(ii) M”'*' ions gaining electrons to form M (i.e., get reduced)
and change into
metal atoms, i.e., for
(iii) Metal atoms losing electrons to form M"'*’ (i.e., M gets oxidized)
ur
/! +
M ion.'i ore
reduced.
s
ook

M"-" + ne~ ^ M ...(1)

Yo

iiii) Metal atoms of the metal rod may lose electrons and change into M”'*’ ions, i.e., metal atoms get oxidized.
eB

M + ne~ ...(2)
These three possibilities are shown in Fig. 3.14.
our
ad

What actually happens depends upon the relative tendency of the metal or its ions.
If metal ions have relatively higher tendency to get reduced, reaction (1) will occur. Metal ions (M"'*‘)
will gain electrons from the metal rod. As a result, metal rod will develop a positive charge with respect to the
Y

solution and ultimately the following equilibrium will be attained

Re
nd

M"-" + lie- V 4 M

If metal has relatively higher tendency to get oxidized, reaction (2) will occur, the electrons will
Fi

accumulate on the metal rod which will, therefore, develop a negative charge. This in turn may attract some
metal ions from the solution which may change into metal atoms. Ultimately, an equilibrium is reached as
follows :— M ^ ^ M"+4-ne-

Thus, in either case, there is a separation of charges between the metal rod and its ions in the solution.
As a result, a potential difference exists between them.
The electrical potential difference set up between the metal and its ions in the solution is called
electrodepotentialor the electrodepotentialmay he simply defined as the tendency ofan electrode
to lose or gain electrons when it is in contact with solution of its own ions.
The electrode potential is further termed as oxidation potential if oxidation takes place at the electrode
with respect to standard hydrogen electrode and is called reduction potential if reduction takes place at the
electrode with respect to standard hydrogen electrode. If in the half cell, the metal rod is suspended in a
solution of one molar concentration, and the temperature is kept at 298 K, the electrode potential is called
standard electrode potential, represented usually by E“. If a gas is involved, the standard conditions chosen
are 1 atmosphere pressure and 298 K.
3/40 ‘P.ne^dee^'4. New Course Chemistry (X11)S!Q2S]

3.16. MEASUREMENT OF ELECTRODE POTENTIAL

FIGURE 3.15
The absolute value of the electrode potential of a single electrode
(called single electrode potential) cannot be determined because oxidation
half reaction or reduction half reaction cannot take place alone. It can H2GASAT
1 BAR PRESSURE
only be measured by using some electrode as the reference electrode.
The reference electrode used is the standard or normal hydrogen
electrode (S.H.E. or N.H.E.). The design of the hydrogen electrode — CONNECTING
WIRE
(which is in fact a general design for any gas electrode) is shown in Fig. --1 M HCl SOL
3.15. In the standard hydrogen gas electrode, hydrogen gas at I bar
pressure is passed into I M HCl at 298 K in which a foil of platinum PLATINUM FOIL
COATED WITH
coated with platinum black (finely divided platinum) remains immersed. Pt BLACK
It simply acts as an inert electrode through which inflow or outflow of
Hydrogen electrode

ow
electrons takes place.
When in a cell, this electrode acts as the anode, i.e., oxidation takes place, the following reaction takes
place : H2 ^ 2H"- + 2 e-

i.e., some hydrogen gas changes into H'*' ions which go into the solution.

e
When this electrode acts as cathode, i.e., reduction takes place, the following reaction occurs :

re
Fl
2 + 2 e-

F
i.e., some H'*' ions from the solution change into gas. Thus, the electrode is reversible with respect to
ions. This electrode is usually represented as : Pt, H2 (g) I (Cone - c)
ur
The electrode potential of the standard hydrogen electrode is taken as 0 000 at 298 K.

r
fo
To determine the electrode potential of any electrode, a cell is set up using this electrode as one of the
electrodes and the second electrode is the standard hydrogen electrode. The EMF of the cell is measured. As
ks
the EMF of the cell is the sum of the oxidation potential of the electrode where oxidation takes place and the
Yo
reduction potential of the electrode where reduction takes place and one of the electrodes involved is the
oo

standard hydrogen electrode for which the electrode potential is taken as zero, EMF of the cell will directly
eB

give the electrode potential of the electrode under investigation. The direction of the flow of current further
indicates whether oxidation takes place or the reduction takes place on the electrode under investigation
with respect to the hydrogen electrode. Accordingly, the electrode potential is termed as oxidation potential
ur

or reduction potential.
ad
Yo

A reading on the voltmeter will be obtained only if the positive terminal of the voltmeter has been
connected to the positive electrode, i.e., on which reduction occurs and the negative terminal to the negative
electrode, i.e., on which oxidation occur.';. FIGURE 3.16
d
Re

The determination of electrode potential

in

VOLTMETER

may be further illustrated with the help of the

F

following two simple examples :

(/) Determination of standard
electrode potential of Zn/Zn^"^ electrode. A
cell comprising of a zinc electrode immersed O I
SALT BRIDGE

ZINC
in 1 M ZnS04 solution and a standard I ROD
CONTG, KCI ●H2GASAT
hydrogen electrode is set up as shown in f 1 BAR
PRESSURE
Fig. 3.16.
EMF of the above cell comes out to be IMHC1
0-76 volt. Hence, the standard electrode
ZnS04 SOL.
potential of Zn/Zn^* electrode = 0-76 volt. Pt FOIL
Further, the direction of flow of current
indicale.s that oxidation takes place at the zinc Measurement of standard potential of Zn/Zn^'*’
electrode. Hence, 0-76 volt is the standard electrode using standard hydrogen electrode
as the reference electrode
oxidation potential of the Zn/Zn"'*' electrode.
ELECTROCHEMISTRY 3/41

(ii) Determination of standard FIGURE 3.17

electrode potential of Cu/Cu^"^ electrode. A VOLTMETER

cell comprising of copper electrode immersed

in 1 M CUSO4 solution and a standard
hydrogen electrode is set up as shown in
Fig. 3.17. (
0
EMF of the cell comes out to be 0-34 SALT BRIDGE
1©
volt. Hence, the standard electrode potential H2 GAS-*-^ V CONTG. KCI

of Cu/Cu^"^ electrode = 0-34 volt. Further, the AT 1 BAR ]

PRESSURE
direction of flow of current indicates that
CUSO4 SOL.'
reduction takes place at the copper electrode. 1M HCl

Hence, 0-34 volt is the standard reduction COPPER

ROD
potential of Cu/Cu-'^ electrode. Pt FOIL-
ANODE CATHODE
It is important to mention here that it is
common practice to express all the electrode Measurement of standard potential of Cu/Cu^'*'
potentials as reduction potentials. Since the electrode using standard hydrogen electrode

w
reduction half reaction is just the reverse of as the refrence electrode

F lo
the oxidation half reaction, therefore, reduction
potential is obtained from the oxidation potential by simply changing the sign, i.e., for any electrode.
Reduction potential = - Oxidation potential
Thus, for the zinc electrode, Standard Reduction Potential = - 0-76 V

ree
Note. A positive value of the standard electrode potential for copper electrode indicates that Cu-'*’ ions
for F
get reduced more easily than H^ ions under the standard conditions. In other words, Cu cannot be oxidized to
Cu^'*' ions in presence of H"^ ions. Hence, copper does not dissolve in HCl. It may be noted that in case of
nitric acid, copper is oxidized by nitrate ions and not by H'*’ ions.
Your
ks

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
eBoo

1. Different Types of Electrodes. Electrodes are classified into five different types as follows :
Type of Electrode Electrode Reaction
ad
our

(/) Metal-metal ion electrode

e.g.. Zn I Zn-^ Zn^^ (aq) + 2e Zn (s')
Cu I Cu^* Cu^+ (aq) + 2e~
Re

Cu (j)
Y

(«) Amalgam electrodes (generally used for active metals like Na, K, etc.)
Find

2+
e.g.. Zn (Hg)IZn Zn^"*" (aq) + 2 e~ ■ ^ Zn (Hg)
(in) Gas electrodes (using Pt or gold as inert electrode for flow of electrons)
e.g., Pt, H2 (g) I (aq) 2 tr (aq) + 2e- ^H,(g)

Pt, 02(g) \ a-(aq) Cl2(g) + 2e- ^ 2 Cl“ (aq)

Pt, O2 (g) I OH' (aq) 02(g) + 2U20 + 2e- 4 OH"(a^)
(iv) Metal-metal insoluble salKsalt anion electrodes

e.g., Ag, AgCl (.v)IKa (aq) AgCI (s) + e~- Ag (5) + Cl (aq)
Hg (/), Hg2Cl2 (J) I KC! (aq) HgnClj (s) + 2e- -4 2Hg (/) + 2Cl'(afy)
(v) Oxidation-reduction electrodes (prepared by dipping platinum wire in solution of two salts of same
metal with different valencies).

I
3/42 New Course Chemistry (XlI)EZsI9]

Type of Electrode Electrode Reaction

e.g.. Pi, {aq), Fe^'^ {aq) Fe^"^ {aq) + e ● ^ {aq)

Pt, Co^-" {aq), Co“-" {aq) Co^'*’ {aq) + e~ Co^-" {aq)
MnO“ {aq), Mn-"^ {aq), {aq) I Pt MnO_^ {aq) + 8 H"^ {aq) + 5 e
Mn--" {aq) + 4 H,0 (/)

Cr20^ {aq), {aq), H"^ {aq) 1 Pt Cr2Q2- {aq) + 14 {aq) + 6 e"

2 Cr^+ {aq) + 7 H2O (/)
In all the above electrodes, reduction half reaction (as shown above) as well as the reverse reaction
(oxidation half reaction) can lake place depending upon the second electrode combined to form the cell.
Hence, the above electrodes are called reversible electrodes.
2. Use of calomel electrode as the reference electrode. The FIGURE 3.18

use of Normal Hydrogen Electrode (N.H.E.) as reference SALT BRIDGE COPPER

electrode is not convenient due to the following difficulties :

-^LEAD WIRE
(0 Gases are difficult to handle.
(//) It is difficult to maintain the H"^ ion concentration as 1 M

w
and the pressure of the gas at 1 atm.

F lo
{Hi) The electrode is poisoned even by the traces of impurities.
KCLSOL
Hence, it is very common to use calomel electrode as the
reference electrode. Its construction is shown in Fig. 3.18. Its __^.Pt. WIRE
reduction potential depends upon the concentration of KCl

ree
solution used, as given below : PASTE

Cone, of KCl sol.

0-1 N
Reduction potential
+ 0-3338 volt
for F Calomel electrode
Hg

1-0 N + 0-2800 volt

Saturated + 0-2415 volt.
Your
ks

The electrode is represented as Hg (/) I Hg^ CU CO I KCl {aq). The electrode is reversible with respect to
eBoo

Crions, i.e., when it acts as anode, the half-cell reaction is :

2Hg(/) + 2C\-{aq) 4 Hg2 Cl. (5) + 2 e~
and when it acts as cathode, the reverse reaction takes place.
ad
our

Similarly, Ag I AgCl I KCl electrode is another widely used reference electrode. It is prepared by dipping
a silver wire in KCl solution to which a drop of AgN03 is added to form AgCl. At I M concentration of C!"
ions, its E° = 0-2225 V
Re
Y

3.17. CELL POTENTIAL OR EMF OF A CELL

Find

A galvanic cell is made up of two electrodes, i.e., two half cells. One of these electrodes must have a
higher electrode potential (higher tendency to lose electrons) than the other electrode. As a result of this
potential difference, the electrons flow from an electrode at a higher potential to the electrode at a lower
potential.
The difference between the electrode potentials of the two half cells is known cell potential or
\ell voltage. Jt is called electromotive force (emf or EMF) of the cell if no current is drawn from
the cell.

The EMF of the cell depends on (/) the nature of the reactants (//) concentration of the solution in the
two half cells, and {Hi) temperature.
The EMF of a Daniell cell is 1-1 volt at 298 K when the concentration of zinc sulphate and copper
sulphate arc one molar each.
Difference between EMF and Potential difference. EMF of a cell differs from its potential difference
in the important aspects given in Table 3.7.

>
 ELECTROCHEMISTRY
3/43

TABLE 3.7.
Difference between EMF and Potential difference
EMF
Potential Difference
1. EMF is the potential difference between the two 1. Potential difference is the difference between
electrodes of the cell when no current is flowing in the electrode potentials of the two electrodes
the circuit (i.e., in an open circuit). under any condition.
2.
EMF can be measured by potentiometric method* 2.
Potential difference can be measured by a

only so that the EMF sent by the potentiometer simple voltmeter.

ow
source is equal and opposite to that of the cell and
there is no net current flowing in the circuit. It cannot
be measured by a voltmeter which draws current.
3.
EMF is the maximum voltage obtainable from the 3. Potential difference is less than the nuiximum
cell.

e
voltage obtainable from the cell (i.e., EMF of
the cell).

re
4.
EMF is responsible for the flow of steady current 4. Potential difference is not responsible for the
in the circuit.

Flr
flow of steady cuirent in the circuit.

F
3.18. ELECTROCHEMICAL SERIES
ou
The standard electrode potentials of a large number of electrodes have been determined using standard
hydrogen electrode as the reference electrode, for which the electrode potential has been arbitnu'ily fixed as

sr
l„,
zero. Also, as already mentioned, it is a common practice to express the electrode potentials as the reduction

fo
potentials. Since reduction half reaction is just the reverse of oxidation half reaction, the reduction potential

k
of any electrode is obtained from its oxidation potential by simply changing the sign. Further, according to
oo
the latest convention of sign, the electrode at which reduction takes place with respect to standard hydrogen
electrode has reduction potential which is given a positive sign ; the electrode at which o.xidation takes place
Y
with respect to standard hydrogen electrode has a positive oxidation potential or expressed as reduction
reB

potential, it will have a negative sign. The various electrodes have thus been arranged in order of their
increasing values of standard reduction potentials, as shown in Table 3.8. This arrangement is called
uY

electrochemical series. This also includes in itself the activity series of metals because the relative activity
of metals can be obtained by comparing their oxidation potentials.
ad
do

3.19. APPLICATIONS OF THE ELECTROCHEMICAL SERIES

1. To compare the relative oxidizing and reducing powers. As by convention, positive sign is used to
in

represent the reduction potential, this implies that greater is the reduction potential, more easily is the substance
Re

(element or ion) reduced or in other words, stronger oxidizing agent it is. Thus, in Table 3.8, as F2 has the
F

highest reduction potential and Li"*" ion has the lowest reduction potential, this means that F2 is reduced most
easily whereas Li+ ions are reduced with greatest difficulty. In other words, F2 is the strongest oxidizing agent
whereas Li ions are the weakest oxidizing agents. Conversely, writing the reaction in the reverse order, it can
be postulated that lithium (Li) will be oxidized most easily and hence is the best reducing agent whereas
fluoride ions (F^) are oxidized with greatest difficulty and hence are weakest reducing agents.
*Potentiometric method. For no deflection in the galvanometer G, FIGURE 3.19

_ Length AB STANDARD CELL (EMF = Eg)

+

E^ Length AC
C
or g _ Length AC XE,
A
lOCKEY
B

* Length AB +
STRETCHED
G
WIRE
EXPERIMENTAL CELL (EMF = Ex)
3/44
pfuuCee^ New Course Chemistry (XIl)EEIg

2. To compare the relative activities .TABLE 3.8. Standard reduction potentials

of metals. Greater the oxidation potential of at 298 K {Electrochemical Series)
a metal, more easily it can lose electrons and

hence greater is its reactivity. As a result, a Reduction Standard reduction

metal with greater oxidation potential can , half reaction potential E" (in volts)
displace metals with lower oxidation ^
potentials from their salt solution. For j,; Li-" + e- Li -3-05

example, oxidation potentials of Mg, Zn, Fe, K-*-+e^ ■> K -2-93

Cu and Ag are in the order: r- -2-90
Ba-+ + 2e- Ba
i \
Mg > Zn > Fe > Cu > Ag A
Ca-"- + 2e- ^ Ca -2-87
K
Hence, each metal can displace metals Na -2-71
Na+ + e-
on its right from the salt solutions. So their -2-37
reactivity is in the above order. Mg^'^ + 2e“ ^ Mg

w
Al^"- + 3e- 4 A1 - 1-66
3. To calculate the standard EMF of
Mn^"*" + 2e“ Mn - M8
any galvanic cell A galvanic cell is |
based on a reaction which can be split into E 2H2O (/) + 2e" H.fg)

Flo
two half reactions, viz., + 20H^(fl^) -0-83
(/) Oxidation half reaction and + le ■> Zn -0-76

ee
H*
(if) Reduction half reaction + 3e- Cr -0-74 Z

Fr
Z LU
UJ
Standard EMF of the cell o Fe-^ + 2e- ^ Fe -044 O
<
<
Cd^-" + 2e' 4 Cd -040
_ Standard oxidation potential o

for
o
ur
z
~ of the oxidation half reaction z
Cq2+ + 2e- ^ Co -0-28
(0
o
3 Ni^+ + 2e' Ni -0-25 a
Standard reduction potential
s
+ o X
Sn -014
of the reduction half reaction _ UJ + 2er
k
O
Yo
q:
a
oo

Pb^-" + 2e" ^ Pb -0-13

As Oxidation potential = -Reduction ^ o
eB

potential, the above expression may also be « 0-00

z

written in the form : I-

2H"- + 2e- ^ H2 I-
o
o z
Standard EMF of the cell z U1
> Sn2+
r

UJ Sn'^+ 2e- 0-15

ou
ad

O'
_ Standard reduction potential Cu^+ + e- > Cu^ 0-16 (0
CO
“ of the reduction half reaction Cu^-^ + 2e" ^ Cu 0-34 O
Y

O z
z
Cu"^ + e" > Cu 0-52
Standard reduction potential to
CO
<
Re
nd

~ of the oxidationhalf reaction ^ 21- 0-53

<
ui
I2 + 2e- ui

Fe^"^ + e" Fe2^ 0-77 o

Furdier, as in the representation of a cell,
Fi

o z
+ 2e“ ^ 2Hg 0-79
the electrode on which oxidation takes place
(anode) is written on the left hand side and the ^ Ag 0-80
Ag"^ + e“
electrode on which reduction takes place .j Hg^-*- + 2e- ■> Hg 0-86
(cathode) is written on the right hand side (as
^ Hg^
2+

already discussed in Art. 3.14.), the above

2 Hg^-" + 2e- 0-91

■> 2 Br- 1-08

expression may also be written as : Bt2 + 2e'
2C1- 1-36
CI2 + 2e-
Standard EMF of the cell
-I Pt--*- + 2e- -> Pt 1-20
Standard reduction potential
of the right hand side electrode
O2 (g) + 4H'^ (aq) + 4e“
1-23
■i 2H2O
Standard reduction potential !
Au^+ + 3e" ^ Au 140

I
of the left hand side electrode Co^+ + e- ^ Co2+ 1-82
^ 2F" 2-87
i.e.. E” cell = E‘ cathode
-E‘ unodc F2 + 2e-
 ELECTROCHEMISTRY
3/45

The method of calculation is illustrated with the solved numerical problems given below :
Retain In Memory
(/) In a cell involving half reactions like Zn > Zn-*+2e~, Ag* + e~ > Ag
the second half reacUon is multiplied with 2 to cancel the electrons in the overall reaction but the
electrode potential is a fixed quantity and is not multiplied with 2.
(n) Reduction will occur at the electrode having higher reduction potential. Oxidation will occur at the
electrode having lower reduction potential (or higher oxidation potential)
{ill) Oxidation potential of any electrode = - Reduction potential of the .same electrode,
(tv) If we write E°
M"+/M , it represents standard reduction potential (or electrode potential) and if
we write as E.° It represents standard oxidation potential.
Sample
Problem {] Calculate the standard EMF of a cell which involves the following cell reaction
Zn + 2 Ag"*"

w
4 + 2 Ag
Given that E' = 0*76 volt and E' = -0-80 volt
Zn,Zi|2+ Ag,Ag+

F lo
Solution. The cell reaction may be split into two half reactions as:
Zn ^ Zn^+ + 2e~ {Oxidation half reaction)
2 Ag+ + 2e~ ^ 2 Ag {Reduction half reaction)

ee
Here, we are given standard oxidation potentials as E° = 0-76 volt and E' = -0-80 volt

Fr
Zn,Zn-+ Ag,Ag-^
We need oxidation potential of zinc electrode but reduction potential of silver electrode.
Reduction potential of Ag electrode = - Oxidation potential
for
of Ag electrode
= - (- 0-80 volt) = + 0-80 volt
ur
Std. EMF of the cell = Std. oxid. potential of zinc electrode + Std. redn. potential of Ag electrode
s
= + 0-76 + 0-80 volt = 1*56 volts
ook
Yo

Alternatively, as oxidation takes place at the zinc electrode and reduction at the silver electrode, by convention
the cell may be represented as : Zn I Zn-"^ 11 Ag-*-1 Ag
eB

E“ccu = Standard redn. potential of R.H.S. electrode - Standard redn. potential of L.H.S. electrode (E“ aiuxle)
As standard oxidation potential of zinc electrode and silver electrode are given to be 0-76 volt and -0-80
volt, their standard reduction potentials will be - 0-76 volt and + 0-80 volt respectively. Hence,
our
ad

E° cell = 0-80-(-0-76) = 1.56 volts.

Sample Problem Q A ccll is prepared by dipping a copper rod in I M CUSO4 solution and a
Y

nickel rod in I M NISO4 solution. The standard reduction potentials of copper electrode and nickel electrode
Re

are 0*34 volt and - 0-25 volt respectively,

nd

(a) What will be the cell reaction ? {h) What will be the standard EMF of the cell ?
Fi

(c) Which electrode will be positive ? {d) How will the cell be represented ?
Solution, (a) The cell reaction can be :
either
Ni + CUSO4 A NiSOj + Cu, i.e., Ni + Cu 2+ Ni--*- + Cu
or
Cu + NiS04 ^ Cu--*- -1- Ni. i.e., Cu + Ni--" ^ Cu 2+ + Ni
The correct reaction will be the one which gives positive EMF. If the first reaction is the cell reaction we
will have Ni > Ni--" + 2e~ Std. oxdn. pot = - (- 0-25) volt.
Cu- f 2e ^ Cu, Std. redn. pot. - + 0-34 volt
Overall reaction : Ni -hCu--" ^ Ni^-" -I- Cu. Std. EMF = + 0-59 volt.
Thus, EMF is positive for this reaction. Hence, this is the correct cell reaction. (If the other reaction is tried
EMF will come out to be - 0-59 volt)
{b) Standard EMF of the cell as calculated above = 0-59 volt.
(c) Since oxidation takes place at the nickel electrode, therefore, it acts as anode or since electrons are pnxluced at
this electrode, it act.s as a negative pole. Obviously, copper electrode will act as cathode or a po.sitive pole,
{d) By convention, the cell will be represented as : Ni I NiSOj (1 M) 11 CUSO4 (1 M) I Cu.
 'a New Course Chemistry (XlljioaiJ:
3/46

1 For the cell : Zn (s) I ZnS04 (aq) 11 CUSO4 (aq) I Cu (.v), calculate standard cell potential if standard state
reduction electrode potentials for Cu^^/Cu and Zn^^/Zn are + 0-34 V and - 0-76 V respectively.
2. A cell is set up between copper and silver electrodes as ; Cu I Cu"'*’ (aq) 11 Ag {aq) 1 Ag.
If its two half cells work under standard conditions, calculate the e.m.f. of the cell
[Given E“ <E° 1 =+0-34volt, (E“ j) = +0-80voltl
Cu^-"/Cu^^ Ag*/Ag cra
3. The standard EMF of the cell ; Ni I Ni^^ I I Cu^-^ I Cu, is 0-59 volt. The standard electrode potential
(reduction potential) of copper electrode is 0-34 volt. Calculate the standard electrode potential of nickel
electrode.

4. The e.m.f. (E°) of the following cells are Ag I Ag"^ (I M) I Cu-"^ (1 M) 1 Cu : E° = - 0-46 V
Zn I Zn-+(1 M) I Cu--^(1 M) I Cu : E“= + MO V

w
Calculate the e.m.f. of the cell : Zn I Zn-+ (1 M) II Ag+ (1 M) I Ag

F lo
5. The half cell reactions with their oxidation potentials are
(a) Pb (i) - 2e ^ Pb^-" (aq); E%,j = + 0-13 V (/?) Ag (5) - e ^Ag^(ri^);E°^,i = -0-80V
Write the cell reaction and calculate its emf.

ee
6. Two half cells are : AP^ (fl^)/Al and Mg-’^ {aq)/Mg. The reduction potentials of these half cells are

Fr
- 1 -66 V and - 2-36 V respectively. Calculate the cell potential. Write the cell reaction also.
7. Calculate the EMF of the cell containing Chromium and Cadmium electrodes
(Given E° Cr'-’'"/Cr =-0.74V. E° Cd^-^/Cd = -0-40V) for
ur
ANSWERS
s
ook

3. E‘ = -0-25 volt 4. 1-56 V

Yo
1, MO V 2. 0-46 volt Np+,Ni
eB

5. Pb (s) + 2 Ag-*- {aq) — > Pb^-^ {aq) + 2 Ag {s); EMF = 0-93 V

7. 0-34 V
6. 0-70 V, 3 Mg + 2 AP^ -»3Mg“^+2Al
our
ad

HINTS FOR DIFFICULT PROBLEMS

. 0-59 =0-34-E'^ or E° = 0-34-0-59 = -0-25 V
3. P 2+
-E° Np^/Ni Np-*-/Ni
Y

Cu JCu
= +l-l0V
Re

07) E° Cu^+ZCu -E°

= -046V
^°Cu2-"/Cu ^Ag-*-/Ag
nd

Subtract (i) from {ii).

Fi

6. E.M.F. can be positive only if oxidation takes place at Mg-electrode, i.e., the cell reaction is :
3+
3 Mg + 2 A1 ^ 3Mg“'*-+Al
E° cell
= = - '-ea-(-2-36) v_^o^v
4. To predict whether u metal reacts «I'itli acid to give hydrogen ga.s. In order that a metal M (assuming it
to be monovalent) may react with an acid to give gas, following reaction should take place .
I
M +
2 -

which can be split into two half reaction as : M ■¥ M"*- + e {Oxidation half reaction)
1
and H"*- + e -H- {Reduction half reaction)
2 2
 ELECTROCHEMISTRY
3/47

Thus, Ae metal should have the tendency to lose electrons, i.e., undergo oxidauon, with respect to
hydrogen. In other words, the metal should have a negative reduction potential. Thus, all metals lying above
hydrogen in the electrochemical series react with the acid to give hydrogen gas. Further, evidently lower the
reduction potential (i.e., more negative the reduction potential) higher is the reactivity.
Alternatively, as will be seen later, the EMF of the cell comprising of the oxidation half reaction and
reduction ha f reaction, as written above, may be calculated by adding the oxidation half potential and the
reduction half potential. If EMF comes out to be positive, the reaction takes place.
u of a redox rcaelion. To see whether a given redox reaction is feasible or
not, the EMF of the cell based upon the given redox reaction is calculated. For a redox reaction to be
spontaneous, the EMF of the cell must he positive. If the EMF comes out to be negative, the direct reaction
as given, cannot take place ; the reverse reaction may take place. This is illustrated by the numerical problems
given below. '

low
Sample Problem Q Predict whether zinc and silver react with IM sulphuric acid to give out
hydrogen gas or not. Given that the standard reduction potentials of zinc and silver are - 0-76 volt and 0-80
volt respectively.
Solution, (a) To predict reaction of zinc with sulphuric acid : If it reacts, the following reaction should take
place :

ee
Zn + H2SO4

F
> Z11SO4 + H2, I.e. Zn + 2H-"
> Ztr* + H2

Fr
By convention, the cell will be represented as: Zn I Zn-"^ 11 I H.,

Std. EMF of the cell (E' cell ) = Std.redn.pot.of _ Std.redn.pot.of

for
ur
R.H.S.electrode L.H.S. electrode = 0~(~ 0-76) = + 0-76 volt.

Thus, the EMF of the cell comes out to be positive. Hence, the reaction takes place.
s
(/>) To predict the reaction of silver with sulphuric acid : If it reacts, the following reaction should
k
Yo
take
oo

place :
2 Ag + H2SO4
eB

>Ag2S04+ H2, i.e., 2Ag + 2m 4 2Ag-^ + H2

By convention, the cell may be represented as
; Ag[Ag+IIH+IH2
E°
r

cell
Ag^A^. = 0^0-80 =-0-80 volt.
ou
ad

Thus, EMF of the supposed ceil comes out to be negative. Hence, this reaction does not take place.
Y

Sample Problem 0 Can a solution of I M copper sulphate be stored in a vessel made of nickel
metal ? Given that E'
Re
nd

Ni,Ni2+ = +0-25 volt, E“ Cu,Cu2+ = -0*34 volt.

Fi

or

Can a nickel spatula be used to stir a solution of copper sulphate ? Support your answer with a reason.
E'
Ni-/Ni=-0*25V,
Solution. In this problem, we want to see whether the following reaction takes place or not :
N1 + CuSO 4
NiS04 + Cu, i.e., Ni + Cir-^ > Ni2+ + Cu
By convention, the cell may be represented as : Ni I Ni^+ 11 Cu“+1 Cu
^°Ni.Nl2+ =+0-25 volt and E° Qi.Cu 2+ = -0-34 volt.

Hence, the reduction potentials will be

E” = -E°
nE+.ni Ni.Ni-'*’ = “ ^'25 volt and
Cu,Cu
2+ = +0-34 volt

E° cell
“ ^ R.H.S, elccirodc “ ^°L.H.S. electrode “ - (-0-25) = + 0-59 volt
Thus, EMF comes out to be positive. This implies that CUSO4 reacts with nickel. Hence, CuSO, cannot be
stored in nickel vessel. ^
 New Course Chemistry (XII)>aaJil
3/48

Sample Problem 0 Iodine (I2) and bromine (Br,) are added to a solution containing iodide (I )
and bromide (Br") ions. What reaction would occur if the concentration of each species is 1 M ? The electrode
E“ = 0-54 V, E‘ = 1-08 V
potentials for the reactions are : I2/I-
Br-,/Br

Solution. The reaction can be either Br, + 2 1- 42Br-+l2 or 4 21 + Bf2

For 1st reaction, E.M.F. = E° -E° = l-()8-0-54 =0-54 V
Br, / Br" U/I"

For 2nd reaction, E.M.F. = E‘ -E' = 0-54-l-08 = -0'54V

h/r Br-) / Br
4 2 Br + I-).
As E.M.F. is positive for the 1st reaction, hence the cell reaction is Bi^ +2 1'

w
Flo
1. Predict reaction of IN sulphuric acid with : (0 Copper {//) lead (///) iron.
E° Pb-'^.Pb =-013volt, and E“ = -0-44 volt
Given, E° Cu--*.Cu = 0-34 volt : Fc^'^.Fc

ee
2. Can we store ; (o) Copper sulphate solution in zinc vessel ?

Fr
(Hr. Board 2011, HP Board 2011, Assam Board 2012)
(/;) Copper sulphate solution in silver vessel ?
for
ur
(Pb. Board 2011)
(c) Copper sulphate solution in iron vessel ?
Give suitable explanation.
s
=0-80V.EV/Fe=-«-^^-
k
= -0-76V, E‘
Yo
CO
E° = 0-34 V.
Cu--^/Cu ^ Zn-*IZn Ag+/Ag
oo

3. A copper wire is dipped in AgN03 solution kept in beaker A and a silver wire is dipped in a solution of
eB

copper sulphate kept in beaker B. if standard electrode potential for

Cu--^+2e- > Cu is + 0-34 V and for Ag"^ + e > Ag is + 0-80 V,
r

prcdict in which beaker the ions present will get reduced ?

ou
ad

4. Why blue colour of copper sulphate solution gets discharged when zinc rod is dipped in it ? Given
Y

E° =+0-34V, 2-t-
=+0-76V
Cu-*/Cii Zn/Zn
Re
nd

= 0-34 V and E" = 1-36V.

5. Can chlorine gas be stored in a copper cylinder ? Given E° Cu'-^.Cu ci2.cr
Fi

6. Using standard electrode potentials, predict the reaction, if any, that occurs between Fe^’*' (aq) and 1 (aq)
po = 0-77 V ; E” = 0-54V
¥c^*{aq)fFii~*-{aq)
7. Predict whether the following reaction (s) is (are) feasible or not
(/) Fe + Zn
2+
4 Fe2+ + Zn, E°z„ = - 0-76 V. = - 0-44 V
(Chhatisgarh Board 2012)
(//) Zn + 2 Ag^ > Zn^^ + 2 Ag, E“z„ = - 0-76 V, E\^ = - 0-80 V
8. Can a nickel spoon be used to stir a solution of silver nitrate ? Support your answer with reason.
(E° = -0-25V. E° = +0-80V)
Ni^+.Ni Ag'^^.Ag

ANSWERS

1. (i) No (/i) Yes {in) Yes 2. (a) No (b) Yes (c) No 3. In beaker A. Ag'^ ions will be reduced
4.E° ceU = MOV, i.e., +ve. Reaction takes place changing blue CUSO4 to colourless ZnS04
8. No
5. No 6.2Fe^++2T“^2Fe2++l2 7. (0 No in) Yes
 ELECTROCHEMISTRY
3/49

HINTS FOR DIFFICULT PROBLEMS

3. The probable reaction in beaker A will be
Cu + 2 AgN03 > Cu (N03)2 + 2 Ag, i.e., Cu + 2 Ag"*" > Cu^"^ + "’ Ag
E„|, for this reaction = - 0 34 + 0-80 = + 0-46 V U.. it is +ve. Hence, Ag+ ions are reduced to Ag
5. Cu+CU >CuCI,,/.e.. Cu2++2C1-
F° =
^ cell ^ Cu/Cu2+ + E‘ ci-,/2cr = -0-34+ 1-36 V= 1-02 V

i.e.. reaction takes place. Hence, it cannot be stored.

6. 2Fe'^++2r ^2Fe--"+I„ E‘ cell
^ Fe3+/Fe2^ + E°
2J-/I, = 0-77 + (- 0-54) = + 0-23 V.
7. (0 =P

w
+ E°
Zn^VZn = + 0-44 + (- 0-76) = - 0-32 V
2+
Fe/Fe

(/?■) E' ceil = E'* + E‘ = + 0-76+ 0-80= 1-56 V

Zn/Zii 24 Ag^/Ag

Flo
8. The possible reaction is Ni + 2 Ag'*' ^ Ni-+ + 2 Ag
= Ecaihotlc -E
cell nncxlc = 0-80-(- 0-25)= 1-05 V

e
As e.m.f. is positive, reaction is spontaneous.

re
rF
6. To calculate the minimum voltage required to bring about electrolysis of a salt solution.
Sample Problem Calculatethe
minimum voltage required to bring about the electrolysis of 1
ur
M
copper sulphate solution at 25"C. Given that E°
Cu^'^/Cu
= 0-34 V and E°
fo H ,0/hr
= -1-23 V.
ks
Solution. The reactions taking place are ;
Yo
At Anode: Cu-'^(rt^) + 2 e"
oo

^ Cu (A), E° = 0-34 V

^ 2 H^iaq) + ]^0^{g) + 2e- ,'E° = - 1-23 V

B

At Cathode: H.O (/)

re

I
Net Reaction : Cu2+ iaq) + H,0 (/) » Cu (.9) + -02(g) + 2H+(fl^),E° cell = - 0-89 V
u
ad

Hence, minimum voltage required to bring about electrolysis

Yo

will be 0*89 V.
SUPPLEMENT YOUR
ICNOWLEDGE FOR COMPETITIONS
.... . , , Predicting thermal stability of metal oxides. Greater the
nd

oxidation potential of a metal (or lesser the reduction potential), more easily it can lose electrons, i.e.,
Re

^e^er is its electropositive character and hence more stable is its oxide. For example BaO CaO Na,0
Fi

MgO and AI2O3 are very stable (as E°r,j value.s for Ba-+, Ca2+, Na+. Mg^+ and Al^+ are - ’ 2-90, ’ - 2-87,
" ’
- 2-71, - 2-37 and - 1-66 V respectively) whereas Ag20 and HgO are less stable and decompose on heating
^ red Hg^'^are 0-80 and 0-86 V respectively)
Heal
BaO. CaO, Na,0, MgO, AI2O3 -4 No decomposition

> 2Ag + ^02 ;

Heat Heat
Ag20 2 HgO ^ 2Hg + 02

3.20. EFFECT OF ELECTROLYTE CONCENTRATION AND TEMPERATURE

ON THE ELECTRODE POTENTIAL (NERNST EQN. FOR ELECTRODE POTENTIAL)
As already mentioned, the electrode potential of an electrode is said to be standard electrode potential
u It IS measured under standard conditions, i.e., electrolyte concentration is !M and temperature is 298K.
owever, if the electrolyte concentration is different from IM and/or temperature is different from 298 K the
electrode potential has a different value. This value is obtained using Nemst eqn. For this purpose, the electrode
reaction is written as reduction reaction, i.e., in general,
M"++;ic^- ■> M {e.g., Zn-*+2e ●> Zn)
 New Course Chemistry (XII)fcMgai
3/50

RT.In [M]
Then the Nernsl eqn. is applied, viz, E = E'^-
nF

where E = electrode potential under given concentration of M"+ ions and temperature T
E° = standard electrode potential, R = gas constant, T = temperature in K, F = 1 Faraday
II = number of electrons involved in the electrode reaction
For pure solids or liquids or gases at one atmospheric pressure, the molar concentration
2-303 RT
is taken
1
as
RT 1
In = E°- log
unity, e.fi., in the above case [M] or [Zn] = 1 so that E = E° - nF nF [M"+]

w
[M»+l

Putting R = 8-314 JK~* mole"', F = 96500 coulombs and taking T = 298 K, we get
0-0591 1
*
E = E®- log ...(0

o
e
n

re
Remember. While applying Nemst equation, electrode potential is always taken as reduction potential.

Frl
F
3 21. EFFECT OF ELECTROLYTE CONCENTRATION AND
TEMPERATURE ON THE EMF OF A CELL (NERNST EQN. FOR EMF OF A CELL)
ou
Let us consider the example of Daniell cell in which the cell reaction is

r
Zn is) + Cu-"^ (aq) > Zn^* (aq) + Cu (5)

so
Let us apply Nemst equation to calculate the electrode potential of the two electrodes involved.
For Zn (5) 1 Zn--^ (aq) electrode. Zn-'*’ (oq) + 2e ^ Zn (.y) kf
oo
RT
E = E^ In [Zn-"*" {aq)]
Y
2+
Zn“+/Zn Zn / Zn 2F
B

For Cu (i) I Cu^-^ (oq) electrode, Cu^+ {aq) + 2e ^ Cu (.s)

re

RT
= E° ln[Cu2+ {aq)]
^Cu-+/Cti
oY

Cu-+/Cu 2F
u

According to the given reaction, oxidation takes place at zinc electrode and reduction at copper electrode.
ad

Hence, zinc electrode is anode and copper electrode is cathode.

d

- E -E°
^cell “ ^ cathode -E anode 2+
in

~ Cu^''’/Cii Zn /Zn
Re

ln[Zn“"^ («?)]|
RT RT

^Cu-+/Cu ' 2F ln[Cu-‘*‘ {aq)] ^ Zii2+/Z31

F

2F

RT [Cu^^ {aq)] RT [Zn^-^ jag)]

= (E Cu“"^/Ca ^Zn^+ZZii^'*’ 2F "[Cu2+(fli/)]
2F [Zn2+(a^)l
cell

Converting natural logarithm to the base 10 and substituting the values of R and F and taking
0-0591, [Zn-+(«t?)]
T = 298 K, we gel ^cell “ ^°cell 2 ® [Cu2+ {aq)]
ne-

In general, if the cell reaction is aA + bB > xX + yY

RT In
,. lX]-L[Yp-
Then applying Nem.st eqn., we have E^^,, = n F [AlMBf
*
Strictly speaking, activities should be used in place of molar concentrations. Activity is the effective
concentration of ions in the solution. Activity = Activity coefficient x molality, i.e., - Y x in. For very dilute
solution, activity of the solution is very nearly equal to molality.
ELECTROCHEMISTRY 3/51

0-0591 [XJMYF
At 25"C (298 K), Ecel.=E" cell log ...(«)
n [Af [Sf
where n is the number of electrons involved in the cell reaction.

Retain in Memory
In the Nernst equation, the term involving log is the same as the expression for
equilibrium constant when there is a minus sign between the two terms on the right hand side.

NUMERiCAU

ow
L
PROBLEMS
BASED (0 For electrode potential, write electrode reaction as reduction reaction,
ON i.e. M"-" + ne~ ●>M
00591 1
Then by Nernst equation, at 298 K, = E°j^g^ log

e
n

re
Fl
(i'O ForEMFofa cell, write cell reaction, e.g., in general,
aA + bB V- xX+yY

F
Then by Nernst equation at 298 K,
ur
r
00591 [xnY]y
F — F°
log

fo
^Cel! " ^ CeU
n \Ar[Bf
ks
where n is no. of electrons involved in cell reaction.
Yo
For pure solid, pure liquid and any gas at 1 atm pressure, put molar
oo
cone. = 1.
eB

Problem
n Calculate the electrode potential of a copper wire dipped in 0-1 M CUSO4 sedution at
ur

25"C. The standard electrode potential of copper is 0-34 volt. (AP Hoard 2012)
Solution. The electrode reaction written as reduction reaction is
ad
Yo

Cu2+ + 2e ■> Cu so that n = 2

00591, 1
Applying Nernst eqn. we get E = E° -
d

{v [CuJ= 1}
2 [Cu2+]
Re
in

00591, i
= 0-34 - log — = 0-34 - 0 02955 = 0-31045 volt.
F

2 0-1

Problem A zinc rod is dipped in 0-1 M solution of ZnS04. The salt is 95% dissociated at this
dilution at 298K. Calculate the electrode potential /Zn ~
Solution. The electrode reaction written as reduction reaction is : Zn-'*’ + 2 e~ ^ Zn (/j = 2)
00591, 1
Applying Nernst equation, we get E Zir*/Zn ^ Zn^'*’ / Zn 2 °^[Zn2+]
As 0-1 M ZnS04 solution is 95 % dissociated, this means that in the solution,
95
[Zn2+1 = x01M = 0095M
100

0-0591, 1
^Zn2+/Zn = -0-76- — log
0-095
= _ 0-76 - 0-02955 (log 1000 - log 95)
= 0-76 - 0-02955 (3 - 1 -9777) = - 0-76 - 0-03021 = - 0-79021 volt
3/52 “^●usidee^ ’^ New Course Chemistry (X11)C&3S]

Problem
0 Represent the cell in which the following reaction takes place :
Mg(s) + 2 Ag-"( 0*0001 M) (0*130 M) + 2Ag(.v)
Calculate its E cdl . Given that E = - 2*37 V and E‘ = +0*80V
Mr2+,Mk Ag'^.Ag
(NCERT Solved Example, Pb. Board 2011)

Solution. Here, we arc given reduction potentials as E = -2-37V , E ^ = +080V

Mg--^.Mg Ag'^'.Ag
As the emf of the cell must be positive, this can be so only if oxidation takes place at the magnesium
electrode. Hence, the electrode reactions will be
Mg -> Mg-'*’ + 2 e (At anode)
2 Ag-*- + 2 e 4 2 Ag (At cathode)
Thus, the cell may be represented as : Mg I Mg-"^ (0-130 M) II Ag’*’ (0-0001 M) I Ag
Standard emf of the cell will be :

w
cell
= Std. Red. Pot. of R.H.S. electrode - Std. Red. Pot. of L.H.S. electrode = 0-80 - (-2-37) = 3*17 V
The overall reaction is : Mg + 2 Ag-*- ^ ± Mg^'*’ + 2 Ag (n = 2)

F lo
Applying Nemst eqn., we get

0-0591 ,^^[Mg--*-J
F — F°
log
^cell “ ^
^’'[Ag^]-

ee
cell
n

Fr
00591, 0-130 0-0591, 0-130
Ecell = cell —^log (10“^)2 = E”
cell

for
= 3-17 - 0-02955 log (1-30 x 10^) = 3-17 - 0-02955 x (7-1139 ) = 3-17 - 0-21 = 2*96 volt.
ur
Alternatively, this problem may be solved by first calculating the electrode potentials of the two electrodes
s
separately and then calculating the emf from the electrode potentials.
ook
Yo

Problem 0 The EMF of the following cell is found to be 0*20 V at 298 K

eB

Cd I Cd--^ (?) 11 Ni^-*- (2*0 M) I Ni

What is the molar concentration of Cd^ ions in the solution ?
our

(E“ Cd2+/Cd = -0*40V, E = -0*25V)

ad

Solution. The cell reaction is : Cd + Ni^’*’ —> Cd^-*- + Ni, E° ,,„ = -0-25-(-0-40) = 0-15 V
0-0591, [Cd~*]
Y

F — F°
Applying Nemst equation, ^cell “ ^
Re

cell
2 [Ni2+]
nd

0-0591, [Cd2-^]
Fi

0-20 = 0-15- log or

log [Cd--^] - log 2 = - 1-690
2 2

or log [Cd--^] = - 1-690 -f 0-3021 = - 1-3879 or [Cd--^] = antilog 2-6121 = 0*0409 M

Problem
0 At what pH of HCI solution will hydrogen gas electrode show electrode potential of
- 0*118 V ? gas is bubbled at 298 K and 1 atm pressure.
Solution. Writing electrode reaction as reduction reaction. H-" -h (?-

0-0591 1
Applying Nemst equation, = E“
H+/H, 1
log-;-
[H^]
= 0 + 0-0591 log [H+J
or -0-118 = -0-0591 ^H or pH = 2. (-.* /7H=-log [H^J)
Problem
B A galvanic cell is constructed with Ag/Ag’*’ as one electrode and Fe^^’/Fe^ as the second
electrode. Calculate the concentration of Ag’*’ ions at which the E.M.F. of the cell will be zero at equimolar
concentrations of Fe^^ and Fe^ ions. Given E“ = 0*80V, E' Fc^ I Fe-*- =0*77V.
Ag+I Ag
ELECTROCHEMISTRY 3^3

Solution. Given electrode potential values show that EMF of the cell will be positive only if reduction
occurs at silver electrode. Therefore, the cell reaction will be : Fe""^ + Ag'^ > Fe-^"^ + Ag
Hence, the cell may be represented as : Fe^"^ 1 Fe^'*' 11 Ag"^ I Ag
E° CO -po = 0-80 - 0-77 = 0-03 V
cell “
^ Ag+/Ag ^ Fe^+/Fe2+
0-0591. IFc-'^+UAg]
Applying Nemst equation to the above reaction, = E°^^u - log r
1 [Fc^+JlAg+l
But [Fe-+] = [Fe-^^] (Given) and [Ag]=l.

^ceil = E°,^„-0-059 Hog 0 = 0-03 + 0-0591 log [Ag+]

[Ag^]
0-03

low
or
loglAg-^1 = - = -0-5076 = 1-4924 or (Ag"^! = Antilog 14924 =3-1 x10 = 0*31 M
0-0591

Problem H Calculate the potential (emf) of the cell

Cd I Cd^+ (010 M) 1 I (0-20 M) I Pt, (0-5 atm)
(Given E“ for Cd-+/Cd = - 0-403 V, R = 8-314 J moH, F = 96, 500 C mol-')

ee
Solution. The cell reaction is : Cd + 2

F (0-20 M) Cd--"(0-10 M) + H, (0-5 atm)

Fr
E° Cell = E‘
^"cd2+,Cd = 0 - (- 0-403) = 0-403 V

for
Applying Nemst equation to the cell reaction,
ur
2-303 RT [Cd^+JxP* =0-403-
2-303x8-314x298 0-1X 0-5
F = F"
log log
ks
'^ceil ^ cell 2x96500 (0-2)2
/jF
Yo
oo

= 0-403 - 0-003 = 0-400 V

eB
r
ou
ad

1. Calculate the electrode potential of the electrode Zn/Zn^* (cone. = 0-1 M) at 25°C
Given that E° = 0-7618 volt .
Y

Zn.Zn2+
2. Calculate the emf of the cell, Cd I Cd^^ (0-001 M) 11 Fe^'^ (0-6 M) I Fe at 25°C
nd
Re

The standard reduction potential of Cd/Cd^^ and Fe/Fe^'*' electrodes are - 0-403 and - 0-441 volt respectively.
3. Iron and nickel are used to make an electrochemical cell by using a salt bridge to join a half-cell containing
Fi

I-O M solution of Fe^'*’ (aq) in which a strip of iron has been immersed to a second half-cell which contains
1-0 M (aq) solution in which a strip of nickel has been immersed. A voltmeter is connected between the
two metal strips.
(/) In which cell does reduction occur ? (ii) Write the half-cell reactions involved.
(Hi) Which metal is the anode ?
(jv) In which direction are the electrons passing through the voltmeter ?
(w) What would be effect on the voltmeter reading if Fe^^ concentration were increased ?
(vi) What will be the voltmeter reading when the cell reaches equilibrium ?
Given that the standard electrode potentials of Fe^VFe and Ni2'‘‘/Ni electrodes are 0-44 and - 0-25 volt
respectively.
4. Calculate the standard electrode potential of the Ni2'''/Ni electrode if the cell potential of the cell
Ni I Ni^-^ (0-01 M) 11 Cu2^ (0-1 M) I Cu is 0-59 V. Given E‘ Cu2+/Cu = +0-34V .

*Note carefully, pressure of Ho has to be taken into consideration if it is not 1 atmosphere.

3/54 ^MeUc^'4, New Course Chemistry (X1I)S!S2X

5. A voltaic cell is set up at 25°C with the half cells, Ap'*’ (0-001 M) and Ni^'*' (0-50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the
cell potential. (Given E® Ni^^/Ni = -0-25V, E° aP^/AI = -l-66V) (CBSE2011)

6. The measured e.m.f. at 25®C for the cell reaction

Zn (s) + Cu^"^ (1-0 M) . " ● Cu (s) + Zn""^ (0-1 M) is 1-3 volt. Calculate E° for the cell reaction.
7. Calculate the potential of the following cell reaction at 298 K
Sn'^ (1-50 M) + Zn (s) > Sn“+ (0-50 M) + Zn^+ (2-0 M)
The standard potential E° of the cell is 0-89 V. Whether the potential of the cell will increase or decrease, if
the concentration of Sn^ is increased in the cell ? (R = 8-314 JK"' mol"' ; F = 96,500 C mol"')

ow
8. Calculate the potential of a zinc-zinc ion electrode in which the zinc ion activity is 0-001 M
(E° 2-f- = - 0-76 V, R = 8-314 J K"' mol"', F = 96,500 C mol"') (CBSE 2007)
Zn /Zn

9. (a) Calculate the electrode potential of silver electrode dipped in 0-1 M solution of silver nitrate of 298 K
assuming AgNO^ to be completely dissociated. The standard electrode potential of Ag"*" I Ag is 0-80 V at

e
298 K. (Karnataka Board 2012)

re
(h) At what concentration of Ag'*' ions will this electrode have a potential of 0-0 volt ?

Fr l
10. Cu2+ + 2 e" ^ Cu, E^ = + 0-34 V ; Ag-*- -h 1 e" ^ Ag, E° = + 0-80 V

F
(/) Construct a galvanic cell using the above data.
(ii) For what concentration of Ag"*- ions will the emf of the cell be zero at 25°C, if the concentration of Cu^"*-
is 0-01 M? (log 3-919 = 0-593)

or
ou
II. Calculate the potential for half cell containing 0-10 M K^CroOy (aq), 0-20 M Cr^"*^ (aq) and 1-0 x 10^ M

R-*-(a^). The half-cell reaction is Cr20^ (r;^)-i-14H^(«<7)-l-6f kfs 2 Cr^^ (aq) + 7 H^O (/)
and the standard electrode potential is given as E° = 1 -33 V. (CBSE 2011)
oo
12. Calculate the emf of the following cell at 298 K
Y
Fc(,.j I Fc“+ (0-001 M) 11 H-*- (I M) I H2(^, (I bar), Pt^^j (Given E'’^^.,, = -f 0-44 V) (CBSE 2013)
B

13. Calculate the e.m.f. of the following cell at 298 K :

2 Cr (5) -f- 3 Fe-+ (0-1 M) >2 Cp-*- (0-01 M) -f 3 Fe (s)
re

Given : E,°(Cr^+/Cr) “ 0-74 V, E“(Fe-^/Fe) = -0-44V

“ (CBSE 2016)
oYu
ad

14. Write the ceil reaction and calculate the e.m.f. of the following cell at 298 K
Sn (s) I Sn^-^ (0-004 M) I (0-020 M) I H2 (g) (1 bar) I Pt (s)
d

(Given : e'^Sii“-"/Sn
,, = -0-14 V) (CBSE 2018)
in
Re

15. Calculate the emf of the cell ; Zn (s) I Zn-+ (0-01 M) II (0-001 M) Ag-*-1 Ag (.?)
F

Given : E° = -0-76V and E = -t-0-80V

Zn‘-^/Zn Ag+/Ag
(log 2 = 0-3010, log 3 =0-4771. log 10 = 1) (CBSE 2020, 2022)

ANSWERS

1. E‘ =-0-7914volt 2. 0-0441 volt

Zn2+, Zn
2+
3. (/) Nickel half cell (ii) Ni -i-2e- ^ Ni (/i7) Fe is anode (/v) Iron to nickel
(v) Voltmeter reading decreases (vi) V = 0
4. - 0-2205 V 5. 1 -46 V 6. 1-27045 V 7. 0-895 V, Increase 8. - 0-849 V
,-14 M
9. (a) 0-741 V (6) 2-9 X 10 10. 5 X 10"^ M 11. 0-76 V 12. 0-53 V
13. 0-31 V 14. 0-1104 V 15. 1-44 V

HINTS FOR DIFFICULT PROBLEMS

1. Standard reduction potential, viz., E‘ Zn

2+
/Zn
= -0-76I8V.
ELECTROCHEMISTRY 3/55

2. For the cell as represented, = E"p^2+/pg “^"cd^+zcd = “ ~ (0 403) = - 0 038 V

Cell reaction : Cd + Fe2+ (0-6 M) > Cd^^ (0-001 M) + Fe
0-0591, 0-0591, 10-3
^ceU = E® log = -0038- log
cell ~
n ~~[Fe2+] 0-6
= - 0-038 -i- 0-0821 = 0-0441 V.

Alternatively, calculate electrode potentials of L.H.S. & R.H.S. electrodes separately. Then
Ecell = Erhs - E LHS
5. Ni2+ + 2e- 0-0591, 1
^Ni, E =-0-25- log = -0-259 V
Nr^/Ni 0-50

A13 + 3e~ 0-0591, 1

> Al, E = -1-66- log—7 =-1-719 V
Al^+ZAl “ 3 10“3
For EMF to be +ve, oxidation should occur on Al-electrode. Cell reaction : 2 Al + 3 Ni^'*' ^2Al3++ 3Ni
E^g„ = - 0-259 + 1-719 V = 1-46 V.

w
0-0591,. JZn^+] . 1.3-E°
6. E cell = E“cell log
n [Cu2+] ; ^ log^^ or E“ cell = 1-27045 V.

F lo
7. E = E“ ceU “
0-0591, [Sn2+][Zn2+] 0-0591, (0-5) (2-0) = 0-895
log = 0-89- —— log- V.
n [Sn4+] 1-5

If [Sn**^] is increased, the second factor will decrease or E cell will increase.

e
Fre
RT 1 2-303x8-314x298 1
8. E = E“-2-303 log = -0-76 log = - 0-849 V.
«F a
Zn2+
0-0591 1
2x96500 10-3
for
(«) E^Ag+/Ag = E" log On solving, we get =0-7409 V
r
Ag+/Ag 1 [Agn
You
oks

0-0591 1 0-80
=0 0 = 0-80- or log[Ag+] =
V/A* log = -13-5364 = 14-4636
eBo

1 [Ag+] 0-0591

or [Ag+] = 2-9x lO-'^M

10. (/) For E“,.g,j to be +ve, oxidation will occur at copper electrode and reduction at silver electrode. Hence, the
ad
our

cell will be represented as : Cu I Cu2+11 Ag"^ I Ag

(«) Cu + 2 Ag+ > Cu2+ + 2 Ag, n = 2
0-0591, [Cu2+]
^cell “ ^cell “ — o = 0-46V- 0-0591,log 0-1 Calculate [Ag'^J.
dY
Re

■ Hence,
[Ag+]2
[Cr3+]2
Fin

11. E = E"-
0-0591 0-0591 (0-20)2
log = 1-33- ■. los -
n
[Cr202-][H+]14 6 (0-10)(10-^)‘4
= 1-33 - ®log
6
(4 X1052) = 1-33 - 6
(57 + 0-6021) = 1-33-0-57 = 0-76 V

0-0591, [Fe2+] 0-0591, 0-001

12. Fe + 2H+ »Fe2- + H2, E^„=E» cell = 0-44- -—log
2
:r
(1)2
13. E»„„ = = - 0-44 - (- 0-74) = 0-30 V
[Cr3-]2
2-303 RT
Ece»=E° cell log (n = 6)
wF [Fe2+]3
14. At anode : Sn (s) 4 Sn2+ (0-004 M) + 2e~
At cathode : 2 H+ (0-002 M) + 2 e~ — ^ H2 (1 bar)
Net reaction : Sn (^) + 2 H+ (0-002 M) > Sn2+ (0-004 M) + H2 (1 bar)
3/56 7>t^tde^'4. New Course Chemistry (XlOCsm

0-0591. [Sn2+](p^^) -.n'’ _p°

log
^ccll “ ^cdl log '-Sn2+/Sn-' ^ [H^2
0-0591. (0-004) xl
= 8-(-0-14) — log = 0-1104 V
(0-02)2
15. Zn + 2Ag+ >Zn2+ + 2Ag
Kcl\ = 0-80-(-0-76) = 1-56 V
E cell„ = E cell
0-591, lZn2+] = !-56
0-0591.log 10-2 = 1-56- 0-0591,log ,_4
10^
—'“W 2 ® (IO"-^)2 2

= 1-56- 0-0591, ...4

-—log 10^ = 1-56-
0-0591
x4 = l-44V

w
2

3.22. CONCFNTRATION CELLS FIGURE 3.20

F lo
Concentration cells are the cells in which either VOLTMETER

the two electrodes are of the same material hut the

ee
e
concentration of the electrolyte in them is different or ■ CURRENT

Fr
the two electrodes have different concentration hut they
are immersed in the same solution of the electrolyte. The ZINC ZINC
ELECTRODE ELECTRODE
former are called electrolyte concentration cells while

for
ur
the latter are called electrode concentration cells.
The most common concentration cell is the
11--
ks
ZnS04 SOLv
S'

electrolyte concentration cells. For example, (CONC = Ci) - ZnS04 SOL

Yo
H2I H^(C,) 11 H-"(C2) I H2
oo

(CONC = C2)
C2>Ci
and Zn 1 Zn2+ (C,) 11 Zn2+ (C,) I Zn
eB

POROUS PARTITION
One such cell consisting of zinc electrodes is shown
in Fig. 3.20. Concentration cell consisting of zinc electrodes
r
ou
ad

E cell
0 0591 C2
The EMF of a concentration cell at 298 K is given by log
C
Y

n
1

Obviously, EMF of the cell will be positive only if C2> C|.

Re
nd

In such a cell, oxidation takes place on the electrode with lower concentration. Hence, it acts as
Fi

anode (or -ve pole). The reduction takes place on the electrode with higher concentration which, therefore,
acts as cathode (or +ve pole). For example, in the concentration cell consisting of zinc electrodes, the electrode
reactions will be :

Zn (i') ^ Zn2+ (C^) + 2c

Zn2+ (C^) + 2e- Zn (5)

Overall reaction : Zn2+(C.) ^Zn2-" (C|)

Thus, there is no net chemical reaction. It simply involves transfer of electrolytefrom one concentration
to the other.

Applying Nernst equation, at 298 K

0-0591, C, 0-0591. C,
log
n n C

Further, as the cell reaction proceeds, C7 decrea.ses whereas Cj increases till the two become equal, i.e.,
C| = C2. At this stage, = 0 and hence the cell stops working.
ELECTROCHEMISTRY 3/57

Similarly, in a concentration cell of the type Pt, H, (P, atm) I (cone. = C) I H, (P, atm), Pt (called
electrode concentration cell) in which the two electrodes are immersed in the same solution but have different
0-0591, P|
pressures or concentrations. E cell “ Iog;f (P, >P2).
n
P2
Sample Problem Q Calculate the EMF of the following concentration cell at 298 K
Zn 1 ZnSO^ (0-05 M) II ZnS04 (0-5 M) I Zn

Solution. E cell 00591, C-, _ 0-0591.log 0-5 = 002955 V

~ log— - 0-05
n
c,
Sample Problem Q A cell consists of two hydrogen electrodes. The negative electrode is in contact
with a solution having 10"* M H'*' ion concentration. Calculate the concentration of !!■*■ Ions at the positive
electrode, if the einf of the cell is found to be 0-118 V at 298 K.

w
Solution. Here, C] = I0“* M, Ct = To be calculated.
0-0591 C2

F lo
For the given concentration cell. E^^jj = n
log
C

ee
0-118 =
0-0591 C2 C2 C2 = Antilog 2 = 10-
log or log = 2 or

Fr
10"* 10"* I0-*
c,= 10-x 10"*= ia-*ivi

for
ur
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
FIGURE 3.21
s
Liquid Junction Potential. In a concentration cell,
ook
Yo
when two electrolytic solutions are directly in
contact (Fig. 3.21), ions move across the junction
eB

COPPER COPPER
ELECTRODE ELECTRODE
with different speeds. As a result, a potential
difference is set up at the junction. It is called Liquid
our

Junction Potential (UP). The EMF of the cell is

I
ad

thus the sum of oxidation potential of one electrode,

reduction potential of the second electrode and the
CUSO4SOL. C11SO4SOL
Y

liquid junction potential, i.e., E = E, -i- E2 + Ej (CONC = Cl) (CONC = C2)

Re

The type of junction generally applied in the

nd

galvanic cells is the salt bridge which contains salts ' LIQUID
Fi

like KCl in which K'*' and Cl" ions move almost at -JUNCTION

equal speeds and hence Ej is taken as zero. Concept of liquid junction in a concentration cell

3.23. EQUILIBRIUM CONSTANT FROM NERNST EQUATION

Let us consider the cell : Zn (a) I ZnSO^ (aq) i I CuSO^ {aq) I Cu (5)
In this cell, electrons flow from zinc to copper in the external circuit whereas the following processes
take place in the interna! circuit. :
(/) Zinc from the zinc rod passes into the solution as Zn^'*’ ions.
(//) Cu^'*' ions from CUSO4 solution are deposited on tlie Cu plate.
(Hi) SO^“ ions from CUSO4 solution flow to the ZnS04 solution llirough the salt bridge (containing K2SO4
as the electrolyte).
3/58 New Course Chemistry (XII)CZsI91

FIGURE 3.22
As a result, the concentration of CuSO^ solution decreases
whereas that of ZnS04 solution increases. Since electrode <
potentials depend upon the concentrations of the solutions, z
REDUCTION-HALF
therefore, electrode potentials of the two electrodes involved LU
\ REACTION
keep on changing*. Ultimately, a stage comes when the electrode o
0.
● EQM.
potentials of the two electrodes become equal and the current O POINT
stops flowing in the circuit, i.e., EMF of the cell becomes zero O
OXIDATION-HALF
as shown in Fig. 3.22. (This is because the EMF is the difference z>
a REACTION
oi
of the two electrodepotentials).Under these conditions,the cell q:

reaction is said to have attained equilibrium,i.e., we have

PROGRESS OF REACTION

Zn {s) + Cu“+ (aq) Zn^"" (aq) + Cu (5)

Representing how the cell stops working
Thus, the concentrations of Cu^"^ and Zn^'*’ ions at this stage or cell reaction reaches equilibrium

w
are the equilibrium concentrations so that we have
[Zn2+ (aq)] eqift

Flo
= K., where K^. represents equilibrium constant when ^cell ■■
[Cu2+ (aq)] cqm
c

e
Putting these values in the Nernst equation, we get

re
F
RT [Zn2+ iaq)] eqm RT
0 = E‘ cel!
In = E° InK
cell
itF [Cu2+(a^)] eqm
c
ur nF

r
F° -
RT
InK
2-303 RT
logK^ fo
ks
or
^ cell ~ c
nF nF
Yo
oo

0-0591
or E'
logK^ at 298 K
B

cell
It
re

Thus, knowing the standard EMF of the cell, the equilibrium constant can be calculated.
Note that the above expression involves E°j.gu and not E^eu because Eggu at equilibrium = 0.
u
ad

Signifkance of K^. The value of tells about the extent of reaction. For example, the value of for
Yo

the Zn-Cu cell at 298 K is found to be very large (i.e., about 2 x 10^) which shows that the reaction goes
almost to completion.
d
Re

Sample Problem Calculate the equilibrium constant for the reaction

in

Cu (s) + 2Ag+ (aq) , ^ Cu^* (aq) + 2Ag (s),

F

Given that E° = 0-80 V and E° Cu^+/Cu = 0-34V

Ag+/Ag
(NCER’I' Solved Example, Pb. Board 2011, Assam Board 2012)
Solution. The cell may be represented as : Cu I Cu2'*’(a^) 11 Ag*(aq) I Ag
E"cdi = E^rhs - E\hs = 0-80 V - (0-34 V) = 0-46 V
00591
E° cell
n
-logK^
For the given reaction, n = 2, = 0-46 V

0-46 = 00591, ^ 0-46x2

= Antilog 15-5668 = 3-6 x 10^^
—^logK, or log =
0-0591
= 15-5668

*Electrode potential of copper electrode will keep on decreasing while that of zinc electrode will keep on
increasing.
 ELECTROCHEMISTRY 3/59

wmm PPA^TTil^El
1. Calculate the equilibrium constant for the reaction, Zn + Cd^'*’ ^ i Zn^+ + Cd,
if E“ =-0403 V and E° =-0-763 V
Cd^'^/Cd Zn2+/Zn
2. Calculate the equilibrium constant for the reaction, Zn + Cu^"^ ^ i Cu + Zn-+.
Given : E” for Zn^^/Zn = - 0-763 V and for Cu^+ZCu = + 0-34 V ; R = 8-314 J K*' mol-'. F = 96500 C mol
3. Calculate the equilibrium constant for the reaction at 298 K.
4 Br“ + 0^ + 4 H+ > 2 Bf2 + 2 H^O. Given that E° cell = 0-16 V.
4. Calculatethe equilibriumconstant for the reaction,2 Fe^'*' + 31“ ;?= i 2 Fe^'*' + 1J . The standard reduction
potentials in acidic conditions are 0-77 and 0-54 V respectively for Fe^‘*^/Fe and /I" couples.

w
5. Calculate the equilibrium constant for the reaction at 298 K ;
Ni02 + 2 Cl“ + 4 H > CI2 + Ni2+ + 2 HoO if E°ceil = 0-320
+
V.

F lo
6. Calculate the equilibrium constant for the following reaction at 298 K
Cu (5) + CI2 (g) > CUCI2 (aq)

ee
R = 8-314 JK“'mol-‘, E° Cu2+/Cu = 0-34V, E‘ i/2ciwcr -1-36V, F = 96500 C mol" ‘

Fr
ANSWERS

1. 1-52 X 10'2 2. 2-121 X 10^’

for
3. 6-747 X 10'^
ur
4. 6-073 X 10"^ 5. 6-747 X 10'^ 6. 3-295 X 10^
s
HINTS FOR DIFFICULT PROBLEMS
ook
Yo

2. E“cell=E° Cu2+/Cu -E° Zn^-^/Zn = 0-34 - (- 0-0763) V = 1-103 V.

eB

0-0591 0-0591, ^
E”
3. ceil “ logK^, 0-16 = — logK^ or log K^.= 10-8291 or K^= 6-747 X 10"^.
r

n
ou
ad

4.
Cell is ri I3 IIFe3+ I Fe2+. E° cell = 0-77 - 0-54 = 0-23 V. n = 2
Y

5. Ni^-" + 2 e“ > Ni^+ ; 2 Cl“ > CI2 + 2 e n= 2

6.
Re

Anode reaction : Cu (^) > Cu"-^ {aq) + 2 e~ ;

nd

Cathode reaction : CI2 (g) + 2 e“ - > 2C!"(a^)

Fi

Net cell reaction ; Cu (s) + CI2 (g) —+ CuCI-, {aq)

^°cell = £'> cathode -E‘ anode = 1-36-0-34= 1-02 V

0-0591
E°
cel! “
n
logK^, i.e., 1-02 = log or log K^= 34-5178
or
K^= antilog 34-5178 = 3-295 x

3.24. ELECTROCHEMICAL CELL AND GIBBS ENERGY OF

THE REACTION (RELATION BETWEEN DC® AND E®cell)
When a cell reaction takes place, electrical energy is produced. The electrical work thus done by the
system (cell) results in the corresponding decrease in the free energy of the system. Keeping in view the finst
law of thermodynamics,
Electrical work done = Decrease in free energy
3/60 New Course Chemistry CXII)BSXSI

On the other hand, when electrolysis is carried out, electrical energy is supplied which results in the
corresponding increase in the free energy of the system. In this case, as before, the electrical work done on the
system is equal to the increase in free energy of the system.
Thus, if free energy change, AG . of a system comes out to be negative, this means that the electrical
work has been done by the system on the surroundings whereas if the free energy change conies out to be
positive, this would mean that the electrical work has been done on the system by the surroundings.
Taking the ca.se of galvanic cell, the electrical work done is equal to the electrical energy produced which in
turn is equal to the product of the quantity of electricity flowing in the circuit and the EMF of the cell, i.e..
Electrical work done = Electrical energy produced = Quantity of electricity flowing x EMF
Now, for every one mole of electrons transferred in the cell reaction, the quantity of electricity that flows
through the cell is one faraday (IF = 96500 coulombs). Hence, if n moles of electrons are transfened in any cell
reaction, the quantity of electricity flowing = //F faradays. If E^.j.]| represents the EMF of the cell, then

w
Electrical work done = ;i FE cell

Hence, - A^ G = n FEj.p„

F lo
For comparing different cells, standard cell potentials are used which are represented by E°^.^,j. The
corresponding free energy change is called the standard free energy change of the reaction, represented by

ee
A^G°. Hence, we can write

Fr
-A^G“ = « FE' cell ...(0

Also we have already learnt in Art. 3.23 that for a reaction in equilibrium.
for
ur
RT
E° InK
cell
/jF
s
ok
Yo
Substituting this value in eqn. (/), we get
o

RT
eB

-A G® = nF. ...(«)
nF or A,G'’ = -RTln K, or A^ G* = - 2-303 RT log

Hence, knowing E°j.gjj, AG® can be calculated which in turn can be used for the calculation of the
r

equilibrium constant K^.

ou
ad

Fuither, as studied in thermodynamics in class +1, - AG = w m.ix ● Hence, the decrease in free energy is
equal to the maximum work that can be obtained from the cell.
Y

r SUPPLEMENT YOUR
Re
nd

^4fNOWLEDGE FOR COMPETITIONS 1. E^^.,1 is intensive but AG is extensive. Remember that in

the relation, AG = - n FE, E is an intensive quantity (as it depends only on the nature of cell reaction)
Fi

whereas AG is extensive and its value depends upon the value of n. Thus, if we write the cell reaction as
Zn (s) + Cu^'*’ {aq) > Zn-* (aq) + Cu (a), as n = 2, AG = - 2 FE.
But if we write the cell reaction as

2Zn{s) + 2Cu^*(aq) 2 Zn"'*’ (aq) + 2 Cu (s) then as n = 4. AG = - 4 FE.

2. Calculation of electrode potential of a half-cell reaction from other two half-cell reactions. For
example
Given : + 2e~ Cu, E® = 0-337 V ...(0
Cu^+ + e- Cu+, E" = 0-153 V ...(»)
To calculate : Cu"*^ + e~ ^ Cu, E®= ?

We cannot calculate the result by directly subtracting eqn. (/) from eqn. (»). We have to calculate AG° of
each half-reaction (as AG® is extensive property).
3. Relationship between electrical energy and enthalpy change of a cell reaction. This relationship
follows from Gibbs-Helmhottz equation according to which
 ELECTROCHEMISTRY 3/61

a (AG)
AG = AH + T
L AT Jp
where AG = free energy change of the cell reaction and AH = enthalpy change of the ceil reaction
But AG = — n FE

a(-«FE) aE)
-uFE = AH+T «FE = -AH+nFT
ar
or
...(0
Jp - ar Jp
AH aE
or E=- + T —
...no
nF .axjp
aE
— represents the variation of EMF of the cell with temperature and is called temperature coefficient

w
of the EMF, i.e.. 3E _ E2 - E|
aT T2 - T,

F lo
According to eqn. (r), electrical energy produced (n EF) = enthalpy change only when —=0, i.e.,
dT

ee
EMF of the cell almost does not change with temperature,
as

Fr
If — is +ve, i.e. EMF increases with temperature, electrical energy produced > enthalpy change.
oT

aE

for
ur
If Tz: is -ve, i.e. EMF decreases with temperature, electrical energy produced < enthalpy change.
oT
k s
Yo
oo

NUMERICAL
PROBLEMS FORMULAS AND UNIT USED
eB

BASED
ON - AG° = /?F E° cell = 2-303 RT log K = w max
r

Calculation of
ou

F = 96500 coulombs.
ad

AG“ (or
AG° = Joules,
Y

Maximum
Work) and R = 8-314 JK-'mol-
nd
Re

Kf fromE^^ll
Fi

Problem f| (a) Calculate the standard free energy change and maximum work obtainable for the
reaction. Zn (.¥) -4- Cu^* (ag) ± Cu (.9) + Zir* (ag)

[Given =-0-76V, =+0’34V, p = 96500 C mol"*]

(NCERT Solved Example)
(b) Also calculate the eqi .briiim constant for the reaction.
Solution. (fl)The cell ma> oe represented as ; Zn(s) I Zn-"^ (1 M) 11 Cu“^ (1 M) I Cu(5)
E°cell = E°RHS-E' LHS = 0.34 _(_ 0-76 ) = I-10 volt
For ihe given cell reaction, n = 2
AG° = - /jFE‘ = -2 X 96500 Cmol-^x MOV =-212, ,300CVmor'
cell

= - 212. 300 J mol"' (I CV= 1 J) =- 212-300 kJ mor‘.

Thus, the maximum work that can be obtained from the Daniell cell = 212*3 kJ.
3/62 ‘P’uxdecfo.'^ New Course Chemistry CXI1)C&I9]
(/,) AG° = - RT In = - 2-303 RT log
212300
= 37-2074
-2!2300 = -2-303 x8-3I4 x298 xlogK^ or logK^.^ 2-303x8-314x298

K^. = Antilog 37-2074 = 1-6 x

Problem Q Estimate the minimum potential difference needed to reduce AI2O3 at 500"C. The free
energy change for the decomposition reaction
●>
■> — Al + O, is AG = + 960 kj (F = 96500 C mor‘).
3 ^
3
Solution. ARO3 (2 AI'^’*^ + 3 0““) > 2 A1 + -;^02, n = 6 e
0 4

low
-Al.O-, 4-Al-f-O.. n = 4 e
3 ^ ^ 3

Substituting AG = + 960 kJ = -f 960,000 J and /j = 4 in the equation. AG = - /? FE. we gel

960,000 = - 4 X 96500 x E or E = - 2-487 V
Minimum potential difference needed to reduce AI2O3 is 2*487 V.

ee
F
0 The E” values corresponding to the following two reduction electrode processes are :

Fr
Problem

(i) Cu+/Cu = -I- 0-52 V

(ii) Cu2+/Cu*= + 016 V.

for
Formulate the galvanic cell for their combination. What will be the standard cell potential for it ?
ur
Calculate A^ G" for the cell reaction {F = 96500 C mor*) (Uttarakhand Board 2012)
Solution. For EMF to be +ve, oxidation should take place on electrode {ii), i.e., half-cell reactions will be
k s
Yo
Cu^ + e > Cu
oo

Cu^ > Cu-"^ + e~

eB

Overall cell reaction : 2Cu-" ^ Cu + Cu-+

Hence, the cell will be represented as: Cu"^ 1 Cu-"^ 1 1 Cu^ I Cu

r
ou
ad

E° cell = E‘ Red (RHS) - (LHS) = 0-52 - 0-16 = 0*36 V

Y

A^G" = -/t FE’ cell = - I X 96500 C mo|-‘ x 0-36 V = - 34740 CV mol"' = - 34740 J moH.
(1 CV= 1 J)
Re
nd

Problem 0 The zinc/silver oxide cell is used in hearing aids and electric watches. The following
Fi

reactions take place : Zn + 2e~ ; E” = 0*76 V

AgjO + H2O + 2e~ 4 2Ag + 20H-; E” = 0*344 V
{a) What is oxidized and reduced ? (^) Find E” of the cell and AG in joules.
Solution, (a) Zn is oxidized and Ag20 is reduced (as Ag"^ ions change into Ag)
(/>) E’ cdl “ ^"AgiO/Ag Zn/Zn-
2+ (Ox) = 0-344 + 0-76 = M04 V
AG = - n FE’ cell
= -2x96500x M04J = 2*13 X 10^ .T
Problem
B Calculate the standard free energy change taking plac in H2/02fuel cell in which the
following reactions occur :
(/) O2 + 4H’*’ -f 4e A 2 H2O, E“ = 1*229 V
(«)2H2 ^4H*+4€~, £‘■ = 0*000 V
Solution. E° cell = E'^
0,/H-vO
(Red)-HE’ (Ox) = 1-229-1-0 = 1-229 V
H2/H+
AG = - « FE’ cell = - 4 X 96500 X 1 -229 J = - 474*4 kJ
ELECTROCHEMISTRY
3/63

mmm. i c^Ei
2+
1. For the cell. Mg (.v) I Mg (aq) I Ag"*" (aq) I Ag (.r), calculate the equilibrium constant of the cell reaction at
25°C and maxiniuin work that can be obtained by operating the cell.
F® =-2-37V and E‘ = + 0-80V (R = 8'31 J mol-' K"')
Mg2+/Mg

w
Ag+/Ag
2. For the reaction N2 (g) + 3H2 (g) ^ - 2NH3 (g) at 298 K. enthalpy and entropy changes arc - 92-4 kJ
and - 198-2 JK”' respectively. Calculate the equilibrium constant of the reaction (R = 8-314 JK"' mol"').
3. For the equilibrium, 2 H2 (g) + 0.> (g) ^ ± 2 H2O (/) at 25°C, AG° is - 474-78 kJ moC'. Calculate log K

e
for it (R = 8-314 J K"' mo|-*).

e
o
4. The emf (E‘ cell ) of the cell reaction, 3 Sn"^ -f 2 Cr ^ 3 Sn--" + 2 Cr^^ is 0-89 V.

r
r
Calculate AG“ for the reaction (F = 96.500 C mor' and VC = J).

F
5. Calculate the cell e.m.f. at 25°C for the ceil : Mg (s) I Mg-+ (0-01 M) 11 Sn^+ (0-1 M) I Sn (j)

oF
ul
Given £' = -0-136 V. 1 F = 96.500 C mor'
Mg2+/Mg = -2-34V, E° Sn-^/Sn
Calculate the maximum work that can be accompalished by the operation of this ceil.

sr
6. Cr20^-+14H+ + 6e ^ Cr-" + -" + 7H,0. E”= 1-33 V;

ko
3 X f2 r > I2 + 2 e-], E° = - 0-54 V

of
Find out the value ol the equilibrium constant and Gibbs free energy change in the reaction given above.
7. The cell in which the following reaction occurs
o
Y
2 Fe'-*'^ (aq) + 2 r (aq) > 2 Fe-"^ (aq) + ^ (,v)
rB

has E!cell - 0-236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
eY

(Given : 1 F = 96,500 C mol"') (CBSE 2017)

8. Calculate G° and log K^, for the following reaction :
u

Cd-"^ (aq) + Zn (.v) > Zn^"^ (aq) + Cd (.r)

d

= -0-403 V ; E'’
o

Given : E
ad

= - 0-763 V (CBSE 2019)

Cci-+/Cd Zn-'^/Zn
in

ANSWERS

1. Max. work, AG® = 611-81 kJ Eqm. const, K = 1-891 x 10'®’ 2.6-958 X I03
Re

3. 3-41 X 1035.-2-02 X jq5j 4. - 515310 J

F

5. E® cell = 2-50 V. M- = 425-372 kJ 6. K=l-596x 10^®, AG° = 457-41 kJ mor'

max

7. 45-548 kJ mor' 8. 69-48 kJ, 1-58 x 10'-

HINTS FOR DIFFICULT PROBLEMS

2. AG = AH - T AS. Calculate AG.
3. - AG" = 2-303 RT log K, 474780 = 2-303 x 8-314 x 298 log K or log K = 83-2.
4. AG ° = - /( FE® cell = - 6 X 96500 X 0-89 J = - 515310 J.

5. Mg + Sn’-^ >Mg3-" + Sn, E® cel! 0-136-(-2-34) = 2-204 V

F - F‘ 0-0591, [Mg^n
^cell - cell log
[Sn2+1 '
0-0591, 0-01
Ecell =2-204- — log = 2-500 V
0-1
vv
max
= - AG® = n FE® cell = 2 X 96500 X 2-204 = 425372 J

(Note that iv max is related to E° cell and not E,.^|])

3/64
“^teideep.'^ New Course Chemistry (XI1)Il5EIH

6. E°«ii = 1-33 + (- 0-54) V = 0-79 V, n = 6.

7. For the given cell reaction, n = 2. AG° ~-n FE°^.e|].
8. E°eii = - 0-403 - C- 0-763) = 0-36 V
A.. G° = - /? F E cell " 2 X 96500 X 0-36 = - 69480 J

n E°cell _
2 X 0-36
= 12-20
log K^. = 0-0591 0-059!

ow
K,. = antilog (12-20)= 1-58 x I0‘-

3.25. PREDICTING PRODUCTS OF ELECTROLYSIS FROM ELECTRODE POTENTIALS

In Art. 3.2, wc have already discussed about the products of electrolysis on the basis of discharge

e
potentials of the different ions present in the solution. As the reactions occurring at the electrodes are oxidation

re
and reduction reactions, a better method to predict the products of electrolysis of an aqueous solution of an
electrolyte is on the basis of their electrode potentials. A few examples are explained below :

F
Frl
1. Electrolysis of melt/nqueous solution of NaCI. When the electrolysis of an electrolyte is carried out in
the molten state, the products of electrolysis are the substances obtained from the ions of the electrolyte. For example,
ou
electrolysis of molten NaCl gives sodium metal at the cathode and chlorine gas at the anode as follows :

sr
Melting
NaCl (5) — ^ Na^ (/) + Cl" (/)

kfo
At cathode : Na"*" (/) + e~ -> Na (/) {Reduction)
At anode : C\-(l) ^ Cl (g) + (Oxidation)
oo
Cl (g)+Cl (g) ^ Cl. {g)
Y
However, when the electrolysis of a concentrated aqueous solution of the electrolyte is carried out, then
reB

as water is also involved, the products of electrolysis are usually not the same as the substances formed from
the ions of the electrolyte. This is because water can also undergo oxidation or reduction, as follows depending
uY

upon the nature of the ions produced from the electrolyte and the nature of the electrodes used
1
H.O (/) 4—0. (g) -f 2 (aq) + 2e (Oxidation)
ad
do

2H.0 (/) + 2e > H2(g) + 20H“(£ir/) (Reduction)

in

The products of electrolysis of aqueous NaCl solution are chlorine at the anode and hydrogen (and not
Re

sodium) at the cathode. These results are explained as follows:—

F

NaCl (aq) ■> Na"*" (aq) + Cl“ (aq)

Probable reactions at the cathode :

Na'*’ (aq) + e~ 4 Na (5) EVd = -2-71V ...(0

2H2O (0 + 2 c- 4 H. (g) + 20H- (aq)

= - 0-83 V .(ii)
Reduction reaction occurs at the cathode. Now, since the standard reduction potential of water
(- 0.83 volt) is greater than the standard reduction potential of sodium (- 2.71 volt), therefore, reduction of
water takes place easily. As a result, reaction (ii) is the actual reaction taking place at the cathode and not (/).
Probable reactions at the anode :

2Cr (aq) 4 CI2 (g) + 2 E^..

ox
= - 1 -36 V ...(iii)

H.O (/) 4
i02 (g) + 2H-^(aq) + 2e~, ¥f ox 1-23 V ...(/V)

Oxidation reaction occurs at the anode. Now, as the standard oxidation potentials of both the above
reactions are nearly equal [being - 1.36 volt for reaction (in) and - 1.23 volt lor reaction (iv) ; taking the
values of standard reduction potentials with their signs reversed!, therefore, both the reactions have equal
 ELECTROCHEMISTRY 3/65

chances to occur. In tael, reaction (/v) should have slightly more chance. However, the actual reaction taking
place in {he concentrated solution of NaCl is (iU) and not (rV), i.e., CI2 is produced and not O2. This unexpected
result is explained on the basis of the concept of “overvoltage”, i.e., water needs greater voltage for oxidation
to O2 (as it is kinctically slow process) than that needed for oxidation of Cl" ions to CI2. This extra voltage
required is called overvoltage and is explained below in the box.
Thus, the actual reactions taking place at the cathode and the anode and the overall reactions are
as follows :—

w
At cathode :
2H20(/) + 2e- ig) + 20H {aq)
At anode : 2C\-{aq) ^ C\,{g)-¥le-

Overall reaction : 2H2O (/) + 2C1~ iaq) ^ H2 ig) + CU ig) + 2OH- (aq)

e
ro
re
Overvoltage. For metal ions to be deposited on the cathode during electrolysis, the voltage required
is almost the same as the standard electrode potential. For example, standard electrode potential of
Zn~‘^/Zn is - 0-76 V. The potential at which Zn""^ ions are deposited on zinc electrode is almost the same. I

F
Fl
However, in some cases, particularly in case of evolution of gases, e.g., H2 at cathode or O2 at anode, the :

u
potential required is much higher that the standard electrode potential of hydrogen or oxygen electrode. '
This is because kinctically, the evolution of H2 or O2 gas is so slow that at lower voltage, no evolution of

sr
gas seems to occur. Hence, some extra voltage is required than the theoretical value of the standard

ko
electrode potential. This extra voltage required is called overvoltage or bubble voltage as it is observed
o
of
just at the point at which gas bubbles begin to appear. Thus, overvoltage is the difference between the
potential required for the evolution of the gas and its standard reduction potential. For example, standard
reduction potential (E°) of hydrogen electrode using platinum electrode is 0-00 V but using zinc as the
o
Y
cathode, hydrogen overvoltage is 0-70 V, nickel as the cathode, overvoltage is 0-22 V and with iron as the
erB

cathode it is 0-08 V. Thus, the value varies largely with the nature of the metal. Similarly, oxygen overvoltage
also depends upon the nature of metal used as anode. Some results are given in Table below.
uY

Electrode
Platinized platinum Iron Nickel Zinc

Hydrogen overvoltage (V) 000

ad

008
do

0-22 0-70

Oxygen overvoltage (V) 0-25 0-24 006

in
Re

2. Electrolysis of an aqueous solution of CuBrj. The products of electrolysis are copper at the cathode
F

and bromine at the anode. These results are explained below :—

CuBr, (aq) ^ Cu-'*' (aq) + 2 Br“ (aq)
Probable reactions at the cathode :

Cu"'*' (aq) + 2 e Cu (^), Red

= + 0-34 V ...(f)
2 H2O (/) + 2 Ht (g) + 2 OH (aq), Red
po =
“
-0-83 V

As the standard reduction potential of reaction (/) is much greater than that of reaction (ii), therefore,
reaction (/) occurs at the cathode and not (//).
Probable reactions at the anode :

2Br" (aq) ^Bt2(g) + 2e , E°ox="l-08V

H.0 (/)
^0^(g) + m*(aq)+2e-, E° Ox "
= - 1-23 V ...(fv)

As the standard oxidation potential of reaction (Hi) is greater than that of reaction (iV), therefore, reaction
(//7) occurs at the anode.
3/66 ;rw^ 'a. New Course Chemistry (Xll)iaam
Actual reactions occurring at the electrodes and overall reaction :
At cathode : Cu^"*" (ag) + 2 e~ ■) Cu (5)
At anode : 2Br” (ag) ^ Br2(g) + 2e“
Overall reaction : + Bf2 {ag) Cu (5) + Brj (g)
3. Electrolysis of aqueous copper sulphate solution using platinum electrodes. Copper sulphate in
aqueous solution ionizes as :
CUSO4 (ag) {ag) + SOl~ (ag)

ow
Comparing reduction potentials of Cu^'*’ and H2O and the oxidation potentials of SO^ and H2O, the
actual reactions occurring at the electrodes are :
At cathode : Cu“'^(a^) + 2e Cu (5)

e
1
-02(g) + 2H+ {ag) + 2e-

re
At Anode : H.O(/)
2 ^
4. Electrolysis of aqueous copper sulphate solution using copper electrodes. The actual reactions

F
Frl
occurring at the electrodes are as follows :—
At cathode : Cu^'*’ {ag) + 2e~ ■> Cu {s)
At Anode :
ou
Cu (5) 4 Cu^"*" {ag) + 2e~

sr
i.e., copper dissolves from the anode and is deposited at the cathode.

kfo
Conclusion. As reduction occurs at the cathode, therefore, if the cation produced from the electrolyte in
an aqueous solution has higher reduction potential than that of water, the substance liberated at the cathode is
oo
that obtained from the cation of the electrolyte. However, if the reduction potential of the cation is less than
that of water, then H2 gas is liberated at the cathode. Similarly, as oxidation occurs at the anode, therefore, if
Y
the anion produced from the electrolyte in an aqueous solution has higher oxidation potential that that of
reB

water, the substance liberated at the anode is that obtained from the anion of the electrolyte. However, if
oxidation potential of the anion is less than that of water, O2 gas is liberated at the anode.
uY

In general, if more than one substance is present in the electrolytic cell, the substance liberated at the
cathode is one which has highest reduction potential and that liberated at the anode is the one which has the
highest oxidation potential.
ad
do

In view of the above conclusions, some more results are given below :
(/) Na"^, Ctt-'*', Mg^'*’ and Al^"^ have much lower reduction potentials than that of water. Hence, these ions are not
in

reduced in the aqueous solution. Instead, reduction of water occurs giving Ht gas at tlie cathode.
Re

(//) Cu-+ and Ag"^ ions have much higher reduction potential than that of water. Hence, these ions are easily
F

reduced and deposited as Cu and Ag at the cathode.

(///) Br“ and 1“ ions have much higher oxidation potential than that of water. Hence, they are more easily
oxidized in the aqueous solution giving Br2 and I-, respectively at the anode. On the other hand, F“ ions
have much lower oxidation potential than that of water. Hence, F” ions are not oxidized in the aqueous
solution to give F2. Instead, H2O is oxidized to give O2 gas.
SUPPUEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Effect of concentration on the products of electrolysis. In the above discussion, standard electrode
potentials have been used. However, if concentrations of solutions being electrolysed are different
from ! M, electrode potentials calculated from Nemst equation should be used. In such cases, depending
upon concentration, products of electrolysis may be different. For example, during electrolysis of
sulphuric acid, at the anode, the following reactions are possible ;
2 H,0 (/) ^ O2 (g) + 4 H+(a^) + 4 ^r, E‘’=+1-23V
2SOl~{ag) 4 S20|“ {ag)+ 2e~. E° =+ 1-96 V
 ELECTROCHEMISTRY 3/67

If H2S0^ is dilute, first reaction occurs at the anode, i.e., is produced. If H2SO4 is concentrated, the
second reaction occurs.

Similarly, in the electrolysis of aqueous NaCl solution, if the solution is concentrated, Cl, is produced
at the anode but if it is very dilute, O2 is produced at the anode.
2. Order of discharge of ions at the electrodes.
(0 At the cathode, the ions which have reduction potential higher than that of water are discharged in
the order of decreasing reduction potentials, i.e., the ion with higher reduction potential is reduced first
and hence discharged first of all. Remember that ions like Na+. Ca^^ Mg^+, A1 3+ etc. which have
reduction potentials less than that of water are not reduced/di,scharged at the cathode. Instead, H2O is
reduced to give H2. For example, order of discharge of some ions on electrolysis is in the order:
= 0-80 V), Hgf = 0-79 V), Cu^+ (EVa = 0-34 V)

low
Ca^'^ or Mg^"^ ions, if present in the solution, will not be discharged. Instead, H2 gas will be liberated.
(i7) At the anode, the ions which have oxidation potential higher than that of water are discharged in
the order of increasing reduction potentials or decreasing oxidation potentials, i.e., the ion with higher
oxidation potential (or lower reduction potential) will be oxidized first or discharged first of all. For
example, on electrolysis of an aqueous solution containing F", Cl~, Br" and T which have E^^g^ values

ee
2-87, 1-36, 1-08 and 0-53 V respectively, the order of discharge will be : I,, Br,, Cl,
rF
Fr
will not be discharged as F2. Instead, H,0 will be oxidized to liberate O,.

3.26. BATTERIES (COMMERCIAL CELLS)

for
Although every spontaneous redox reaction can be used as the basis of a galvanic cell, yet every such
u
galvanic cell is not suitable for commercial purposes. This is because of the following two reasons :—
ks
(/) Galvanic cells using salt bridges have high internal resistances. As a result, if we try to draw large
Yo
oo

current from them, their voltage drops sharply.

eB

(//) They lack the compactness and ruggedness (rough handling) for portability.
A galvanic cell to be used as a commercialcell must fulfil the following two requirements :
(0 It should have the compactness, lightness and ruggedness for portability.
r
ou
ad

(//) Its voltage should not drop much during use, i.e., drop in voltage should be negligible over the small
interval of time during which it is being put to use.
Y

If a number of cells are connected in series, the arrangement is called a ‘battery'. However, the term
nd

battery is sometimes used even for a single cell. Strictly speaking, such a usage is not correct.
Re

The various batteries or cells may be classified mainly into the following two types :
Fi

(A) Primary batteries or cells

(B) Secondary batteries or cells
Primary cells are those in which the redox reaction occurs only once and the cell becomes dead after
some time, and cannot be used again. Two common examples of this type are dry cell and mercury cell.
Secondary cells are those which can be recharged by passing an electric current through them and hence
can be used over and again. Two well known examples of this type are lead storage battery and nickel-
cadmium storage cell.
A brief description of the examples of each type is given below :
3.26.1, Primary Batteries or Cells
(1) Dry cell. The dry cell, commonly used in transistors and clocks, is a compact form of the Leclanche
cell. It consists of a cylindrical zinc container which acts as the anode. A graphite rod placed in the centre (but
not touching the base) acts as the cathode. The space between the anode and the cathode is so packed that zinc
container is in touch with the paste of NH4CI and ZnCl2 and the graphite rod is surrounded by powdered
3/68 T^uidee^ '4. New Course Chemistry (Xll)KSia]

MpOt and carbon* as shown in Fig. 3.23. The graphite rod is fitted with a metal cap and the cylinder is sealed
at the lop with pitch. The zinc container is covered with cardboard to protect it from the atmosphere.
FIGURE 3.^
CARBON ROD
(+)POLE
..O. . - METAL CAP
(CATHODE) ^

T PITCH SEAL
I
rZINC CONTAINER
I
(ANODE)
I
I CARDBOARD
COVER

I "POROUS PAPER
LINING

w
I —GRAPHITE
●^1
CATHODE
I
Mn02 *" C

Flo
■f-
ZINC
- - CUP
- - MnOo +
PASTE OF NH4CI
+ ZnCl2 (ANODE) CARBON BLACK

ree
(a) (-) POLE (b) + NH4CI PASTE
The dry cell (a) Descriptive diagram

rF
(b) A section of the three dimensionalactual look

The reactions taking place at the electrodes are quite complex. However, they can be represented in a
ur
simple maimer as follows:—
At anode : Zn (t) ^ Zrr* (aq) +2e fo
ks
At cathode : 2Mn02 (^) + 2NH^ {aq) + 2e 4 Mn203 (s) + 2NH, (g) + H.O
Yo
oo

(i.e., Mn has been reduced from oxidation state +4 to +3).

B

The NH3 formed is not liberated as gas but combines immediately with the Zn^"*" ions and the Cl" ions to
re

form complex salt. [Zn(NH3)2Cl2] (diamminedichlorido zinc).

These cells have voltage in the range 1.25 V to 1.50 V. However, they do not have a long life because the
u
ad

acidic NH4CI corrodes the zinc container even when the cell is not in use.
Yo

(2) Mercury cell (Ruben- FIGURE 3.24

Mallory cell or Button cell). This PASTE OF HgO CARBON CATHODE ANODE ANODE CAP
nd

+KOH
is a newer type of diy cell which has
Re

found use in low current devices such — ZINC. \

Fi

as hearing aids and watches. It CONTAINER

ASANODE
consists of zinc container as the («) (6)
anode, a carbon rod as the cathode -POROUS
PAPER GASKET \
and a paste of mercuric oxide mixed SEPARATOR SEPARATOR CELL CAN
with KOH as the electrolyte. A lining CATHODE

of porous paper keeps the electrolyte Mercury Cell (a) Descriptive diagram
separated from the zinc anode (b) A section of the three dimensionalactual look
(Fig. 3.24).
The cell produces electrical energy by the following reacdons :
Al anode : Zn is) + 20H" ZnO (.9) + H,0 (/) + 2e-
At cathode : HgO (j) + H-,0 (/) + 2 e~- ^ Hg (/) + 20H-
Overall reaction : Zn (5) + HgO is) ■> ZnO (S') + Hg (/)

*In .some dry cells, the zinc container is filled with the paste of NH4CI, Mn02 and carbon and the graphite
rod is immersed in the centre.
 ELECTROCHEMISTRY 3/69

As the overall cell reaction does not involve any ion whose concentration may change, therefore, this
cell gives a constant potential of 1.35 V throughout its life. However, the spent cells should be reprocessed
for mercury recovery or treated to prevent mercury or mercury compounds from entering the atmosphere and
causing pollution.
The above cell is also called button cell as it is pellet type, flat in construction and looks like a button in
shape.
3.26.2. Secondary Batteries or Cells
(1) Lead storage battery. This is one of the most common batteries used in the automobiles and invertors.
A 12 V lead storage battery is generally used which consists of 6 cells each producing 2 V. Each cell consists
of a lead anode {or a grid of lead filled with finely divided spongy lead as the anode) and a grid of lead packed
with lead dioxide as the cathode. These electrodes are arranged alternately, separated by thin wooden or fibre

ow
glass sheets and suspended in dilute sulphuric acid (38 % by mass or having a density of 1.30 g cm~^) which
acts as the electrolyte (Fig. 3.25).
FIGURE 3.25!
ANODE

e
e--

Fl
CATHODE ANODE I—CATHODE

re
©

F
ur
r
(o) if>)

fo
'NEGATIVE PLATES:
LEAD GRIDS FILLED
WITH SPONGY
ks
LEAD
Yo
\
oo

LEAD PLATES LEAD GRIDS

POSITIVE PLATES:
OR LEAD GRIDS FILLED WITH DIL. H2SO4 38% SULPHURIC LEAD GRIDS FILLED
FILLED WITH SPONGY LEAD DIOXIDE (PbO?) ACID SOLUTION
B

WITH Pb02
LEAD (NEGATIVE PLATES) (POSITIVE PLATES)
re

Lead storage battery of 6 volts (a) Descriptive diagram (b) A section of the three dimensional actual look
u

To increase the current output of each cell, the cathode plates are joined together and the anode plates
ad
Yo

are also joined together (keeping them in the alternate positions of course), i.e., the cells arc connected in
parallel (anode to anode and cathode to cathode). To have an output voltage of 12 V, six cells are connected
in series.
d
Re

Reactions occuring during discharge and recharging

in

The electrode reactions that occur during discharge, i.e., when the battery is in use (Fig. 3.26a) are as
F

follows :—
At anode : Pb(^) + S02- {aq) ■> PbSO^ (5) + 2 e' {Oxidation)

At cathode : Pb02 (i) + {aq) + 4H+ {aq) + 2e~ ^ PbS04 (.V) + 2H2O {Reduction)
Overall reaction :
Pb {s) + Pb02 (.9) + 4H+ {aq) + 2SO“- {aq) 2 PbS04 (,v) + 2H2O
From the above equations, it is obvious that H-iS04 is used up during the discharge. As a result, the
density of H2SO4 falls. When it falls below 1.20 g cm"^, the battery needs recharging.
During recharging (Fig. 3.26/?), the cell is operated like an electrolytic cell, i.e.. now electrical energy
is supplied to it from an external source. The electrode reactions are the reverse of those that occur during
discharge :
At cathode : PbS04 {s) + 2e~- ^ Pb(i-) + SO^ {aq) {Reduction)
At anode :
PbS04 {s) + 2H2O Pb02 (.v) + SO^“ {aq) + 4H+ {aq) + 2 e {Oxidation)

Overall reaction : 2 PbS04 {s) + 2H2O 4 Pb (.V) + PbO, (.V) +4H+ (fl^) +2SO- - {aq)
3/70 ‘P>uicUcfa-’ii- New Course Chemistry (XIOESSSD

FIGURE 3.261
SOURCE OF
DIRECT CURRENT
METER

CATHODE ▼ ‘ ANODE
ANODE 0 ® CATHODE
©
PbS04
Pb02 s

Pb

H-^ H2O
f (SOME
SO42- &

H2O

w
SO42-)

F lo
0 DISCHARGING 0 RECHARGING
Oxidation at -ve electrode (Anode) Reduction at -ve electrode (Cathode)
Reduction at +ve electrode (Cathode) Oxidation at +ve electrode (Anode)

ee
(in both cases, Oxidation is at Anode and Reduction at Cathode)

Fr
Discharging and Recharging of a lead storage cell

for
Such an operation is possible because the PbS04 formed during discharge is a solid and sticks to the
electrodes. It is, therefore, in a position to either receive or give up electrons during electrolysis.
ur
It is important to note that anode is the electrode on which oxidation occurs, i.e., loss of electrons and
s
cathode is the electrode on which reduction occurs, i.e., gain of electrons. Further, the electrode which acts as
ook
Yo

anode during discharge acts as cathode during charging and vice-versa.

eB

(2) Nickel-Cadmium storage cell (or Nicad Cell) This is FIGURE 3.27
another rechargeable cell which is becoming more and more
popular especially in the calculators. It has a longer life than
our
ad

lead storage cell but is more expensive. It consists of a cadmium

electrode (as anode) and a metal grid containing nickel (IV)
oxide (as cathode) immersed in KOH solution. The electrode
Y

reactions occurring during discharge are as follows:—

Re

.POSITIVE PLATE
nd

At anode :

4 Cd (0H)2 is) + 2 e
Fi

Cd (^) -F 20H- {aq) SEPARATOR

At cathode : /(SOAKED IN KOH)
Ni02 (5) -F 2H2O -F 2 f- ^ Ni(OH)2 is) + 2 OH' (aq)
NEGATIVE PLATE
As in the lead storage cell, the reaction products adhere
to the electrodes. Hence, the reactions can be reversed during
charging. Further, as no gases are produced during discharging
or recharging, the battery can be sealed. Nickel-Cadmium rechargeable cell
The potential of each Ni - Cd cell is approximately 1.4 V.
The cell looks similar to dry cell as shown in Fig. 3.27.

3.27. FUEL CELLS

Fuel cells arc the devices which convert the energy produced during the combustion of fuels like hydrogen,
methane, methanol etc. directly into electrical energy. Such a conversion is possible because combustion
reactions are also redox reactions. One such cell which has been very successful is the hydrogen-o.xygen fuel
ELECTROCHEMISTRY 3/71

cell. This cell was used as the primary source of electrical FIGURE 3.28
energy on the Apollo moon flights. The weight of the fuel ANODE OF CATHODE OF

sufficient for 11 days in space was approximately 200 kg. POROUS

CARBON WATER
POROUS
CARBON
This may be compared with the several tons that would have CONTAINING CONTAINING
SUITABLE SUITABLE
been required for the engine-generator set. Moreover, the CATALYSTS ,9' CATALYSTS

product of combustion, namely, water, was used for drinking

by the astronauts.
The general design ofHT-0^ fuel ceil is shown in Fig.
3.28. It consists of porous carbon electrodes containing
suitable catalysts (generally finely divided platinum and
palladium) incorporated in them. Concentrated KOH or
NaOH solution is placed between the electrodes to act as f
L
the electrolyte. Hydrogen and oxygen gases are bubbled CONCENTRATED AQUEOUS KOH/NaOH
through the porous electrodes into the KOH/NaOH solution.
H2-O2 fuel cell
The following electrode reactions take place :

w
Al anode : 2H, (g) + 4 OH- {aq) ^ 4 H^O (/) -I- 4 e"

F lo
Al cathode : O2 (g) 2H2O (/) + 4 ^ 4 OH- (aq)
Overall reaction : 2 H, (g) -F 0. (g) ^ 2H,0 (/)
Thus, in these cells, the reactants are fed continuously to the electrodes and the products are removed

ree
continuously from the electrolyte compartment.
for F
Advantages of fuel cells. The three main advantages of fuel cells are as under :
(/) Because of the continuous supply, such cells never become dead. Such a cell is usually operated at a
temperature of 70 — 140°C and gives a potential of about 0.9 V.
Your

The theoretical voltage of the cell may be calculated from half- cell reactions as follows :
ks
eBoo

2 ig) + 4 OH- (aq) > 4 H^O (/) + 4e~, E“ = + 0-40 V

O2 (g) -t- 2 H-,0 (/) + 4 e~ > 4 OH" (aq), E° = - 0-83 V
ad

= -F 040 V - 0-83 V) = 1 -23 V

our

(/7) Theoretically, the fuel cells are expected to have an efficiency of 100%. However practically they
give an efficiency of 60 — 70 %. Still they are much superior to the thermal power plants in which fuels are
burnt to produce heat which then changes water into steam to run the turbine. Such a power plant does not
Re

have efficiency of more than 40%.

Y
Find

(Hi) They do not cause any pollution problem unlike thcnnal plants which bum fossil fuels like coal,
gas, oil etc.
Efficiency of the fuel cell. As AH is the heat of combustion and AG is the useful work done, i.e..
AG
electrical energy produced, therefore, the thermodynamic efficiency (ri) of a fuel cell = xlOO. Thus,
AH

for the H2 - O2 fuel cell, theoretically, the value can be calculated as follows :
AG = - // FE° cell = - (2) (96500 C mol-') (1 -23 V) = - 237390 CV mol"'
= - 237390 J mol-' = - 237-390 kJ mol"'
AH =-285-8 kJ mol"'

AG - 237-39 kJ mol-'
ri = xIOO = XlOO = 83%
AH - 285-8 kJ mol-'

I
3/72 New Course Chemistry CXU)CSSHI

Difficulties in the construction of fuel cell. The construction of fuel cells is, however, faced with
certain technical, economic and practical difficulties. A few of these are as follows :—
(/) Providing of contact between the three phases needed in a fuel cell, \.e., the gaseous fuel, the liquid
electrolyte and the solid catalyst.
(ii) The corrosiveness of the electrolytes used.
(///) High cost of the catalysts needed for the electrode reactions {e.g., Pt. Pd, Ag, etc.),
(/v) Problem of handling gaseous fuels at low temperatures or high pressures.

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Hydrocarbon-oxygen fuel cells and methanol-oxygen fuel cells have also been developed. For example,
propane-oxygen fuel cell involve.s the combustion of propane as follows ;
C^Hg (g) + 5 0,_{g) >3 CO2 (g) + 4 H^O (/) ...(0
The two half-cell reactions involved are :

In basic medium :

w
At anode : C3Hg(g)+ 20OH-(fl^) 3COo(g)+ 14H2O(/) + 20e-

F lo
At cathode: 5 O2 (g) 10 H2O (/) + 20 ^ 20 OH-

In acidic medium :

At anode : C3 Hg [g) + 6H,0 (/) ^ 3 CO2 ig) + 20 {aq) + 20 e"

ree
At cathode : 5 O2 (g) + 20 ^ 10 H2O (0
(aq) + 20e~-
for F
The problem with the hydrocarbon-oxygen fuel cells is that their power per unit weight is too low to
make them of practical use in ordinary automobiles. Better catalysts arc needed.
The voltage of the above cell can be calculated by first calculating of the combustion reaction
from values and then applying the reaction A^G“ = - n FE°. Thus, for the combustion reaction (1),
Your

A,G“ = L3 (- 394-4) + 4 (- 237-2)] - [- 23-5J kJ mor' = - 2108 kJ mof*

ks
eBoo

As ri = 20, therefore, E° = -
_ - 2108 kJ mol-* _ 2108000CVmoH
nF 20x96500Cmo)-* “ 20x96500Cmol->
ad
our

= 1-09 V [1 J = 1 CV]
2. Advantage of fuel cells operated at high temperature. Fuel cells are classified according to the
temperature range in which they operate. The advantage of using high temperature is that the catalysts
Re

for the various steps are not so necessary or at least cheaper catalysts can be used.
Y
Find

Curiosity Questions
1 Q. 1. Ignoring the water lost by evaporation, some water has still to be added periodically into
the battery used in an invertor or car. Why ? Why this is not required in the maintenance free
batteries ?
Ans. During discharge, electrolysis of water also takes place. Hence, some water is lost due to electrolysis
of water into and O2. Hence, water has to be added periodically into the acid of the battery to
have desired concentration. Now a days, in the maintenancefree batteries, no water needs to be
added as they use electrodes of Ca/Pb alloy which preventselectrolysisof water.
Q. 2. When a car is running, its battery gets charged. How ?
Ans. In an automobile, when it is running, electrical energy is produced in the alternator which keeps
on charging the battery.
J
1
ELECTROCHEMISTRY 3/73

3.28. CORROSION

As soon as the metals are extracted from their ores, the reverse process begins, i.e., nature tries to
convert them back into the form in which they occur. This is due to the attack of the gases present in the
atmosphere on the surface of the metal converting it into compounds such as oxides, sulphides, carbonates,
sulphates etc.
The process of slowly eating away of the metal due to attack of the atmospheric gases on the
surface of the metal resulting into the formation of compounds such as oxides, sulphides,
carbonates, sulphates etc. is called corrosion.
The most common example of corrosion is the rusting of iron. Rust is hydrated ferric oxide, Fe203aH20.
Some other examples include tarnishing of silver, development of green coating on copper and bronze, etc.
Mechanism or Theory of corro.sion. The theory of corrosion can be explainedby taking the example
of rusting of iron. The theory is called electrochemical theory because it explains the formation of rust on
the basis of the formation of electrochemical cells on the surface of the metal.
The formation of rust on the basis of this theory may be explained in the following steps :
(/) The water vapours on the surface of the metal dissolve CO, and O, from the air. Thus, the surface of

F low
metal is covered with the solution of CO2 in water, Le., carbonic acid (H2CO3)
H,0 + CO, ^ H2CO3 ...(1)
This acts as an electrolytic solution of the cell. The carbonic acid dissociates to a small extent as
follows:—

H2CO3 ± 2H-*-+CO| ...(2)

re
H'*' ions may also be produced by dissolution of other acidic oxides such as SO, etc. present in the
atmosphere.
for F
(ii) Iron in contact with the dissolved O, and presence of ions undergoes oxidation as follows :
Fe Fe^++2e-(E"„^ = 0-44 V) ...(3)
Thus, the sites where the above reaction takes place act as anodes. As a result of the above reaction, iron
Your
s
eBo k

is converted into ferrous (Fe^'*’) ions.

{Hi) The electrons lost by iron migrate through the metal to another site on the surface and reduce
dissolved oxygen in the presence of H"*" ions as follows ;
ad

O2 + 4H-^ + 4 p- ^2H20 (E‘‘,,,,= 1-23 V) ...(4)

our

The sites where the above reaction takes place act as cathodes.
Notice that H'^ ions are involved in the reduction ofO-,. As the concentration of H* ions is lowered (i.e.,
Re

pH is increased), the reduction of oxygen becomes less favourable. It is observed that iron in contact
Y

with a solution whose pH is above 9—10 does not corrode.

Find

Adding equations (3) and (4). the overall reaction of the miniature cell will be
2 Fe (.9) + O, (g) + 4 (aq) ■> 2Fe--" iaq) + 2H,0 {/) ; E°^^„ = 1 -67 V
It may be mentioned here that if wafer is .saline, it helps in the flow of current in the miniature cell and
hence enhances the process of corrosion. It is for this reason that in places where it snows heavily in
winter and salting of roads is done, the autos undergo greater rusting.

(iv) The ferrous ions formed react with the dissolved oxygen or oxygen from the air to form ferric oxide
with further production of ions as follows :—
4Fe-+ + O, + 4H,0 ■> 2F6203 + 8H+
(v) Ferric oxide then undergoes hydration to form rust as follows :—
Fe,03 + .vH,0 Fe,03..YH20
Hydrated ferric oxide (Rust)

I
3/74 New Course Chemistry (Xll)S*ZslSI

h is interesting to note that as cathode is generally the area having the largest amount of dissolved
oxygen, the rust is often formed here. If you look at any rusted article, you will observe that rust is
formed at a site other than the site where pitting has occurred.
It may be noted that the rust is a non-sticking compound, i.e., it does not stick to the surface. It peels off
exposing fresh iron surface for further rusting.
The scheme of rusting is represented diagrammatically FIGURE 3.29

in Fig. 3.29. AIR

Factors which promote corrosion : X^WATER

O2
(/■) Reactivity of the metal. More active metals are readily RUST DEPOSIT
DROPLET=p
corroded.
(Fe203. xH20)\ Fe2*(aq)

*)
v.
(//) Presence of impurities. Presence of impurities in
metals enhances the chances of corrosion. Pure metals CATHODE
IRON

do not corrode, e.g., pure iron does not rust. 02 + 4H* + 4e--^2H20 ANODE -
Fe-*-Fe2* + 2e-
(///) Presence of air and moisture. Air and moisture
accelerate corrosion. Presence of gases like SO2 and

F low
CO2 in air catalyse the process of corrosion. Iron when Corrosion of iron
placed in vacuum does not rust.
(iv) Strains in metals. Corrosion (e.g., rusting of iron) Uikes place rapidly at bends, scratches, nicks and
cuts in the metal,

(v) Presence of electrolytes. Electrolytes, if present, also increase the rate of corrosion. For example, iron

re
rusts faster in saline water than in pure water.
for F
Prevention of corrosion. Prevention of corrosion is very important because enormous damage occurs
to buildings, bridges, ships and all other objects made of metals especially iron. Crores of rupees are spent
every year on replacing the rusted structures. Prevention of corrosion not only saves money but also helps in
preventing accidents such as bridge collapse etc.
Your
s

Some of the methods generally used for prevention of corrosion are briefly explained below :
eBook

1. Barrier Protection. The metal surface is not allowed to come in contact with moisture, oxygen and
carbon dioxide. This can be achieved by the following methods :
ad

(a) The metal surface is coated with paint which keeps it out of contact with air, moisture etc. till the paint
our

layer develops cracks.

(b) By applying film of oil and grease on the surface of the iron tools and machinery, the rusting of iron can
be prevented since it keeps the iron surface out of contact with moisture, oxygen and carbon dioxide,
Re

(c) The iron surface is coated with non-corroding metals such as nickel, chromium, aluminium, etc. (by
Y
Find

electroplating) or tin, zinc, etc. (by dipping the iron article in the molten metal). This again shuts out the
supply of oxygen and water to iron surface.
id) The iron surface is coated with phosphate or other chemicals which give a tough adherent insoluble
film which does not allow air and moisture to come in contact with iron surface.

2. Sacrificial Protection. Sacrificial protection means covering the iron surface with a layer of metal
which is more active (electropositive) than iron and thus prevents the iron from losing electrons. The more
active metal loses electronsin preferenceto iron and converts itself into ionic state. With the passage of time,
the more active metal gets consumed but so long as it is present there, it will protect the iron from rusting and
does not allow even the nearly exposed surface of iron to lose electrons (i.e., undergo oxidation). The metal
which is most often used for covering iron with more active metal is zinc and the process is called galvanisation.
The layer of zinc on the surface of iron, when comes in contact with moisture, oxygen and carbon-dioxide in
air, a protective invisible thin layer of basic zinc carbonate. ZnC03.Zn(0H)T is formed due to which the
galvani.sed iron sheets lose their lustre and also tends to protect it from further corrosion.

\
 ELECTROCHEMISTRY
3/75

Iron can be coated with copper by electro-deposition from a solution of copper sulphate or with tin by
dipping into molten metal. Now, if the coating is broken, iron is exposed and iron being more active than both
copper and tin, is corroded*. Here, iron corrodes more rapidly than it does in the absence of tin.
At times, zinc, magnesium and aluminium powders mixed with paints provide decorative protective
coatings also.
3. Electrical Protection — Cathodic Protection. The iron object to be protected from corrosion !Si
connected to a more active metal either directly or through a wire. The iron object acts as cathode and the
protecting metal acts as anode. The anode is gradually used up due to the oxidation of the metal to its ions
due to loss of electrons. These electrons arc transferred
FIGURE 3.^
through the wire to ions present around the iron EARTH SURFACE
object and thus protect it from rusting. As iron object
MOIST SOIL
acts as cathode, it is called cathodic protection. The
(IONS PRESENT
iron object gets protection from rusting as long as some SERVEAS
of the active metal is present. Metals widely used for ELECTROLYTE)

w
Mg
protecting iron objects from rusting are magnesium, K
zinc and aluminium wriich are called sacrificial anodes.

F lo
Magnesium is oftenly employed in the cathodic ● MAGNESIUM
● IRON PIPE ■
ANODE
protection of iron pipes buried in the moi.st soil, canals, (CATHODE)

storage tanks, etc. Pieces of magnesium are hurried . Mg-..Mg2^..2e- 4H^, .'4e-^ 2H2O

ee
OR ■ .

Fr
along the pipeline and coiinr.ted to it by the wire as
O2 + 2H2O +4e-—►40H“
shown in Fig. 3.30.
Magnesium acts as rjf>de whereas iron pipe acts
as cathode.
for
Electrical protection of
underground iron pipes
r
4. Using anri-ni.st .solutions. These are alkaline phosphate and alkaline chromate solutions. The alkalinity
You
s

prevents availability of hydrogen ions. In addition, phosphate lends to deposit an insoluble protective film of
ook

iron phospha.e on the iron. These solutions are used in car radiators to prevent rusting of iron parts of the
eB

engine.

3.29. HYDROGEN AS A RENEWABLE AND NON-POLLUTING SOURCE OF ENERGY

our
ad

The progress of a country is judged by the energy it consumes in making life more comfortable. The
fossil fuels such as coal, oil, gas etc. are non-renewable and hence may not last for very long. Moreover,
they cause pollution. Hydrogen provides an ideal alternative as on combustion, it fonns only water. It ISi: a
dY

renewable and non-polluting source ot energy. Its production by electrolysis of water and its use in making
Re

fuel cells both involve electrochemical principles. Thus, the future vision of energy is the 'Hydrogen Economy. ’
Fin

©L-.AM^ie-'E
1. Electrochemistry. It is that branch of science which deals with the conversion of chemical energy produced
in a redox reaction into electrical energy {electrochemical cell) and vice versa {electrolytic cell).
2. Electrolysis. It is the process of decomposition of an electrolyte by passing electricity through its aqueous
solution or molten state.

3. Products of electrolysis. In molten stale, the products are directly obtained from the electrolyte . e.g;
Mcll
PbBrj > Pb2+ + 2 Br". Pb2+ + 2e~ ^ Pb (at cathode) and 2 Br" Br2 + 2 e (at anode). Similarly,
NaCl (melt) ■> Na (at cathode) + — CI2 (at anode). In aqueous solution, H2O OH- and

-..ncl E"p^,, ^^=-044V, =-0-14 V, E"^„„ =0-34V

I
3/76 “PnAtCe^’^ New Course Chemistry (XlI)KSsZS]

NaCl (aq) ■> Na"^ {aq) + Cl~ {aq). H'*’ ions have lower discharge potential than Na'^ and OH" ions have
lower discharge potential than Cl". Hence, we get H'*' + -> -^H2 (at cathode) and 2 OH~
1
H2O + —02 + 2e (at anode).
4. Faraday’s laws of electrolysis, (i) First law. Mass of the substance deposited on the electrode is directly
proportional to the quantity of electricity passed (W =c Q or W = ZQ where Z is called electrochemical
equivalent of the substance deposited). Putting Q = I (current in amp.) x t (tine in s), W = ZI r. If Q = 1
coulomb or J = 1 amp and t = I s, W = Z.
(u) Second law. On passing same quantity of electricity through .solutions of different electrolytes, the
masses of the substances deposited are directly proportional to their equivalent weights. (W [/W2 = Ej/E^).
5. Faraday of electricity/charge. 1 Faraday (1 F) = 96,500 coulombs.
6. Calculation of masses deposited. For M + ne' 4 M, n faraday deposit 1 mole of M. Similarly, for
11+

x«- X + ne~, n faraday liberate 1 mole of X.

w
7. Electrical Resistance and conductance. Ohm law states, 7/ to the ends of a conducto,r potential difference
E is applied and current 1flows through it, then E/I = R, called resistance. ’ Its units are ohm or Q. Reciprocal

F lo
of resistance is called conductance (G), i.e.. G = 1/R. Units of G are ohm
-1
or Q"* or mho or Siemen (S).
I S = 1

8. Specific, Equivalent and Molar conductivities. (/) Specific conductivity (k). If / is the length of a conductor

e
Fre
/ /
and a is its area of cross-.section, R OC —
or R = p— where p is called spec fie resistance. Reciprocal of
a a

specific resistance is called specific conductance or for

simply conductivity (k). Putting G = 1/R and k = 1/p,
r
1 MV/^ I
we get — = or K = G X -. If / = 1 cm, fl = I cm^, then k = G. Thus, conductivity is the conductance
You
oks

G a

of a conductor of length I cm and area of cross-section 1 cm"-, i.e. conductance of 1 an^ of the conductor.
eBo

Units of K = an“* or S rm”*.

(h) Equivalent conductivity (a^ ). It is the conductance of a solution containing I g equivalent of the
electrolyte. a,.„
eq = k x V where
V is volume of solution in cm^ containing 1 g eq. of the electrolyte or
our
ad

KXlOOO
A
eq . If K is in S cm"', V in cm^ and nonnality in g eq L ', units of a eq = S cm“ eq '. In SI
Normality
K
dY
Re

units, if K is in S m
-1
and V in m^ or normality in g eq m then a eq and units of a.,„
eq
will be
Normality
S m^ cq-'.
Fin

(Hi) Molar conductivity (a,„). It is the conductance of a solution containing I mole of the electrolyte.
KXlOOO
A = K X V where V is volume of solution in cm^ containing 1 mole of the electrolyte or A
7»
tn
Molarity
If K is in S cm’' and V in cm^ or molarity in mol L ', units of a = S cm^ mor^ In SI units, a,„ = K/molarity
m

= S m"’/(mol m"^) = S m“ mor'.

9. Measurement of k, a^^ and a,,,. By Wheatstone bridge principle, taking .solution in a conductivity cell
fitted with two platinuni electrodes and using alternating current and earphone to detect the null point, we
determine the resistance (R) of the solution. Then G = 1/R. As a^,^^ = Gx lla and for a particular conductivity
ceil, l/a = constant called cell constant (G*) with units cm"' or m"'. we first find cell constant (G*).
Knowing G of the solution, we calculate a^^. Similarly, we find a,,,.
10. Variation of G, k, a^^ and a^ with dilution. G increases with dilution because ions increase, k decreases
with dilution because number of ions per cm^ decrease, a eq and A_. increase with dilution because a = K x V
m

and though k decreases but V increases much more.

I
 ELECTROCHEMISTRY
3/77

11. Variation of with concentration (c) for strong and weak electrolytes. For strong electrolytes,
Afm
~ (Debye-Huckel Onsager eqn.) where A ISi a constant and a^ is molar conductance at
infinite dilution (called limiting molar conductivity). Thus, decreases linearly with Vc . When c = 0,
— ® ^ru ® ^
m
- . Thus, A^ can be determined experimentally.
For weak electrolytes, a^ increases as c decreases but does not reach a constant value even at infinite
dilution. Hence, their a^ cannot be found experimentally.
For strong electrolytes, a^ increases with dilution because inter-ionic attractions decrease but for weak
electrolytes, a^ increases with dilution because dissociation increases.

12. Kohlrausch’s law. For an electrolyte B^, a^ = x: -I- y ^ or a°eq = X°+X,“ where and X.1 are

limiting ionic conductivities of cation and anion respectively. The relation between of an electrolyte

w
and m
of its ions is

F lo
Kq = n (cation) n (anion)
n^ = charge on the cation and n~ = charge on the anion.

ee
Fr
A
Similarly, A®
eq
m

Total charge on cations or anions (n factor)

13. Applications of Kohlrausch’s law for
ur
(0 In calculation of a^ or a°^ for weak electrolytes, e.g..
s
ook
Yo
O — 1®
A + 1® o o

CH3COOH “ \CHjCOO" H+
= A
CHjCOONa ^HCl “ ^NaCl ●
eB

(h) In calculation of degree of dissociation,’ i.e.,’ a = a^m /a®m .

(Hi) In calculation of dissociation constant by using value of a calculated above,
our
ad

(iv) In calculation of solubility of sparingly soluble salts using the relation. Solubility = kx1000/a“ .
14.
Absolute ionic mobility. It is the speed of ion in cm/sec at infinite dilution under a potential difference of
dY

Ionic conductance
one volt per cm. It is related to ionic conductance
Re

as:
Ionic mobility =
96500
Fin

15. Galvanic cell. It is a device used to convert chemical energy produced in a redox reaction into electrical
energy. It involves two half cell reactions, e.g., for Zn + Cu2+ > Zn^+ + Cu, we have Zn ^Zn2+ + 2e-
(oxidation halfreaction) and Cu2+ + 2 > Cu (reduction half reaction). The electrode on which oxidation
half reaction occurs is called anode and the other on which reduction occurs is called cathode. The two half
cells are connected by a salt bridge. Electrons flow in the outer circuit whereas ions flow through the salt
bridge. Galvanic cell based on this redox reaction is called Daniell cell. Such a cell is represented as ●

ZnlZn2+(Ci)IICu2+(C2)ICu
Any redox reaction can be sed to make a cell.
16. Electrode potential. It is le potential difference set up between metal and its ions in the solution or it is
the tendency of an electrode to lose or gain electrons. The tendency to lose electrons is caUed oxidation
potential whereas tendency to gain electrons is called reduction potential. If conditions are 1 M cone, 1 atm
pressure and 25®C (298 K), these are called standard electrode potentials (represented by E®).
17. Measurement of electrode potential. Electrode potentials are measured with respect to Standard/Normal
Hydrogen El^trode (SHE/NHE). It is an electrode of platinum foil dipped in 1 M H+ at 298 K in which H,
gas at 1 atm is passed. Its standard electrode potential is taken as 0 0.

I
3/78
New Course Chemistry (XII)B533B

18 EMF of a cell. It is the potential difference between the two electrodes of the cell when no current is
flowing in the circuit. Under any other condition, it is simply called potential difference.
Electrochemical series. It is an arrangement of all the electrodes in order of their increasing standard
reduction potentials. For electrode on which oxidation occurs with respect to SHE,
Reduction potential = - Oxidation potential.
Applications of Electrochemical series
(0 To compare relative oxidizing/reducing powers : Greater the more easily it is reduced and hence
greater is the oxidizing power.
(i7) To compare relative activities of metals. Greater the tendency to lose electrons, i.e., greater the E , ox ’

greater the activity of the metal, e.g.. Mg > Zn > Fe > Cu > Ag. This is called activity series,
iiii) To calculate = E“,, (anode) + E^d (cathode) = (cathode) - E^d (anomie)
= EVd (RHS electrode) - E^d (LHS electrode).
(iV) To predict spontaneity of any redox reaction : Calculate EMF. If it is +ve, reaction is spontaneous.

w
Effect of concentration and temperature on electrode potential. Write the given electrode reaction as
reduction reaction (M"'*’ + ne ■> M).

F lo
RT.In IM]— RT
= E°- In
Then apply Nerast eqn, viz. E = E° - nF [M"-"l
«F [M"-']

ee
2-303 RT 0-0591 1

Fr
= E°- log = E^- log at 298 K.
nF [M"^] n

Effect of concentration and temperature on EMF of a cell. Write the cell reaction
nc
for
^ Cu-"^ + 2 Ag, n - 2) and then apply Nernst eqn, viz.
r
{a A + bB ■> .V X + y Y, e.g. for Cu -t- 2 Ag'
You
2-303 RT [Xf[Y]>’ 0-0591 [XflYF
s
at 298 K.
= E° log
ook

E .. = E°
cell cell log cell
/iF n
eB

Concentration cell. If two half cells are of the same type differing only in the concentration of ions, it is
called electrolyte concentration cell, e.g., Zn I Zn^'*’ (c,) 11 Zn^'*' (^2) I Zn.
0-0591 c
our

log^ where C2 > C].

ad

For such a cell, E^^n = n c

Equilibrium constant from Nernst eqn. When a cell reaction reaches equilibrium, Ecen = 0. Hence, trom
dY

Nernst eqn. (point 21).

Re

2-303 RT [Xf[Y]>' 2-303 RT

log = E^ logK
Fin

0 = E" ccU
cell
nF [AHB]^ eqm
nF

2-303 RT 0-0591
or E°
cell logK = log K at 298 K.
nF n

25. Gibb’s energy and Cell potential (EMF). In a cell reaction,

Electrical energy produced = Decrease in free energy
/I F E cell = -^G or AG = - n F

For standard conditions. AG° = - F E° ceil

Putting value of E^^^d from point 24, AG“ = - RT In K = - 2-303 RT log K.

Predicting products of electrolysis from electrode potentials. In aqueous solution of an electrolyte on
electrolysis, there is a competition between reduction of cation and that of water (2 H2O + 2e
1
” +
> H2

2 OH“) and that between oxidation of anion and that of water (H2O — O2 + 2 H'*' -I- 2 e ). At cathode,
product obtained is one with higher reduction potential while at anode, product obtained is one with higher
oxidation potential.

f
ELECTROCHEMISTRY
3/79

27. Commercial cells/batteries. These are of two types :

(0 Primary cells which cannot be recharged.
(//) Secondary cells which can be recharged.
28. Primary cells. (/) Dr>' cell : Anode = zinc cylinder,
Cathode = Graphite rod surrounded by Mn02 + C + NH4CI + ZnCl2 acting as electrolyte.
Anode reaction : Zn » Zn^+ + 2 e~.
Cathode reaction : 2Mn02 ^ ^4 + 2e“ > Mn203 + 2 NH3 + H2O. = 1-25 - 1-50 V.
They do not have long life because NH4CI corrodes zinc container.
(11) Mercury cell: Anode = Zn container, Cathode = Carbon rod.
Electrolyte - Paste of HgO + KOH. Cell reaction : Zn + HgO ■> ZnO + Hg
^ceii = 1'35 V (constant). Used in hearing aids and watches.
29. Secondary cells. (0 Lead storage battery. Anode = Lead, Cathode = grid of Pb packed with PbO 2’

w
Electrolyte : H2SO4 (38%). = 2 V.

During discharge, Anode reaction : Pb + SO^”

F lo
^ PbS04 + 2 e (ox.).

Cathode reaction : Pb02 + 804“ + 4H* +2e~ ^ PbS04 + 2 H2O (red.)

ee
During recharging, Cathode reaction : PbS04 + 2e~

Fr
4 Pb + S04~ (red.).
Anode reaction : PbS04 + 2 H2O
(m) Ni-Cd storage cell : Anode = Cd, Cathode = Ni02,
for
Pb02 +S02- + 4H+ + 2e~ (ox.)
Electrolyte = KOH sol. = 14 V.
r
Anode reaction : Cd + 2 OH"
You
> Cd(OH)2 + 2 e".
s

Cathode reaction : Ni02 + 2 H2O + 2 e~ > Ni(OH)2 + 2 OH"

ook

30. Fuel cells, in which energy produced from combustion of a fuel (H2, CH4 etc.) is directly converted into
eB

electrical energy, e.g., H2-O2 cell. Electrolyte = cone. KOH sol. E(.e,i = 1*23 V (theoretical).
Anode reaction : 2 H2 + 4 OH" > 4 H2O + 4 e~.
our

Cathode reaction : O2 + 2 H2O + 4 e~ > 4 OH",

ad

AG -nFE -2 x 96500x1-231
ceil
Efficiency of fuel cell = xl00 = xlOO = xl00 = 83%
AH AH -285800J
dY
Re

31. Corrosion. It is slow eating up of metal surface due to attack of atmospheric gases on it. Mechanism of
corrosion (rusting) is based on the formation of miniature electrochemical cells.
Fin

Step (0 : H2O + CO2 H2CO3 V ± 2H++C02-,

Step (ii): Fe Fe^"*" + 2e (anodic sites).
Step (Hi): O2 + 4 H+ + 4 — > 2 H2O (cathodic sites),
Step (iv) : 4 Fe^+ + O2 + 4 H2O » 2 Fe203 + 8 H+,
Step (v): Fe203 + jc H2O > ^6263. X H2O (Rust).
32. Prevention of corrosion. (/) Barrier protection, by applying paint, oil grease etc. or electroplating with Ni
Cr etc.

(//) Sacrificial protection, by coating with more active metal like Zn (called galvanisation).
(Hi) Electrical (cathodic) protection, by connecting underground iron pipes with a more active metal like
Mg with an electrical wire.
(iv) Using anti-rust solutions like alkaline phosphates or chromates.
3/80 ‘P’uxdecfr '^ New Course Chemistry CXlDBSmi

E- TP^iffT^’^El QUESTIONS
Based on NCERT Book

(a) 96500 C (b) 2 X 96500 C

I. Multiple Choice Questions
(c) 9650 C id) 96-50 C
1. Which one of the following is not a good conductor
of electricity ?
9. During the electrolysis of molten sodium chloride,
the lime required to produce 0-10 mol of chlorine
(a) CH3COONa (b) C2H5OH gas using a current of 3 amperes is
(c) NaCl (d) KOH (a) 55 minutes (h) 110 minutes
2. The number of electrons required to balance the (c) 220 minutes (d) 330 minutes
following equation
10. What changes are observed in the specific
NO" +4H^ + e~ ■> 2 H2O + NO is conductance and molar conductance respectively
on diluting the electrolytic solution ?
(a) 5 (b) 4
(a) Both increase

w
(r) 3 (d) 2
(b) Both decrease

F lo
2+
3. The number of moles of Mn04 reduced to Mn (c) Decrease and increase
by the addition of 7-5 mole electrons in MnO^ is {d) Increase and decrease

(a) 2-5 ib) 5 11. For which of the following compounds, the graph

ree
of molar conductivity and (fnolarity) is obtained
(c) 1-5 id) 7-5
4. When aqueous solution of NaCl is electrolysed
for F
a straight line ?
(a) CsCl ib) NH4OH
(a) CI2 is evolved at cathode (c) CH3COOH id) All of these
(b) H2 is evolved at cathode 12. An increase in equivalent conductance of a strong
Your
ks

(c) Na is deposited at cathode electrolyte with dilution is mainly due to

eBoo

id) Na appears at anode (a) Increase in number of ions

5. What amount of electric charge is required for the (b) Increase in ionic mobility of ions
ad

(c) 100% ionisation of electrolyte at norma!

our

reduction of 1 mole of Cr20^“ into Cr^'*’ ? dilution

ia) 6 F ib) 3F (d) Increase in both, i.e., number of ions and

ionic mobility of ions
Re

(c) 1 F (^/) 4F
13. The equivalent conductance of NaCl at
Y

6. How many grams of Ag will be deposited if 5 F

Find

concentration C and at infinite dilution are and

quantity of electricity is passed through aqueous
solution of AgN03 ? X^ respectively. The correct relationship between
ia) 270 g ib) 540 g
X^ and X^ is given by (where the constant B is
positive)
(c) 180g id)l 135 g
7. Two faradays of electricity are passed through a ia)X^ = X^ + (B) Vc
solution of CUSO4. The mass of copper deposited (b) + (B) C
at the cathode is (at. mass of Cu = 63-5 amu) {c)X^ = X^-(B)C
(0) 2g ib) 127 g
id) X^ = X^~ (B) Vc
(c) 0 g id) 63-5 g 14. The molar conductivity of a 0-5 mol/dm^
solution of AgN03 with electrolytic conductivity
8. When 0-1 mol MnOj' is oxidized, the quantity of 5-76 X 10-’^ S cm-' at 298 K is
of electricity required to completely oxidize (a) 2-88 S cm^/mol {b) 11-52 S cm^/mol
MnO|- to Mn04 is (c) 0-086 S cmVmol (d) 28-8 S cm^/mol
 ELECTROCHEMISTRY
3/81

15. The cell constant of a conductivity cell The strongest oxidizing and reducing agents
(a) changes with change in concentration of the respectively are :
electrolyte (a) CI2 and Br“ (b) Cl2andl2
(b) changes with the nature of the electrolyte (c) F2 and 1“ (d) Br2 and Cl~
(c) changes with increase in temperature of the 21. Standard electrode potential for Sn^'^/Sn^^ couple
electrolyte is + 015 V and that for the CPVCr couple is
(d) remains constant for a cell - 0-74 V. These two couples in their standard state
are connected to make a cell. The cell potential
(NH4OH) is equal to will be

(a) + 1-83 V (b) + M9 V

(a) a; (NH4OH) + (NH4CI) - (HCl) (c) +0-89V (d) +018V

ow
(b) A°^ (NH4CI) + a;, (NaOH) - A»^ (NaCl) 22. Zn gives H2 gas with H2SO4 and HCl but not with
HNO3 because
(c) Al (NH4CI) + Al (NaCl) - Al (NaOH) (a) Zn acts as oxidizing agent when reacts with
HNO3
(d) a“^ (NaOH) + A^ (NaCl) - a“^ (NH4CI)

e
(b) HNO3 is weaker acid than H2SO4 and HCl

Fl
re
17. The limiting molar conductivities a° for NaCl, KBr (c) In electrochemical series, Zn is above
and KCl are 126, 152 and 150 S cm^ mol-‘

F
hydrogen
respectively. The a“ for NaBr is
(^0 NO3 is reduced in preference to hydronium
ur
r
(a) 278 S cm2 mor‘ (b) 176 S cm^ mor>

fo
ion

(c) 128 S cm2 mol-' 302 S cm2 mol"'

23. Aluminium displaces hydrogen from dilute HCl
ks
18. The limiting molar conductivities of HCl, whereas silver does not. The e.m.f. of a cell
Yo
CH3COONa and NaCl are respectively 425, 90 prepared by combining Al/Al^^ and Ag/Ag+ is
oo

and 125 mho cm2 25“C. The molar 2-46 V. The reduction potential of silver electrode
is + 0-80 V. The reduction potential of aluminium
B

conductivity of 01 M CH3COOH solution is

electrode is
7*8 mho cm2 mol”' at the same temperature. The
re

degree of dissociation of 01 M acetic acid (a) + 1-66 V (b) -3-26V

solution at the same temperature is (c) 3-26 V
u

(d) - 1-66 V
ad
Yo

(a) 0-10 (b) 0 02 24. For the reaction Ti, Ti2+ (0-01 M)/Cu2+ (0-01 M)/
(c) 015 (d) 0 20 Cu, the value of E^gji can be increased by
(a) increase in Cu2+
d

19. In a galvanic cell, the salt bridge

Re

(b) increase in Ti2+

in

(a) participates chemically with cell reaction

(c) decrease in Cu2+
F

(b) stops diffusion of ions from one electrode to

(d) None of these
another
25. Which has maximum potential for the half cell
(c) is not necessary for the occurrence of cell reaction : 2 H"^ + 2e~ ^H2?
reaction
(a) 10 M HCl
{d) ensures the mixing of the two electrolytic (b) 10 M NaOH
solutions
(c) Pure water
20. Standard reduction potentials of the half reactions (d) A solution with pH = 4
are given below:
26. The reduction potential of hydrogen half cell will
F2(g) + 2e“ ^2F-{aq); E“ = + 2-85V be negative if:
Cl2(g) + 2e- ->2Cl-(fl^);E"=+ 1-36V (a) p (H2) = 1 atm and [H+] = 10 M
Br2(5) + 2e" ^2Br-(a^);E“ = + 106 V
{b) p (H2) = 2 atm and [H+] = 1 0 M
(c) p (H2) = 2 atm and [H+] = 2-0 M
I2 (s) + 2e~ -^2I-(fl^); E° = + 0-53V
{d) p (H2) = 1 atm and [H+] = 2-0 M
3/82 T^fuuCeefr ^ New Course Chemistry (XII) jg
27. In a cell that utilizes the reaction (a) Reduction occurs at H2 electrode
Zn is) + 2 H+ (aq) ^ {aq) + H2 (g) (/?) H2 is cathode and Cu is anode
addition of H2SO4 to the cathode compartment (c) H2 is anode and Cu is cathode
will (d) Oxidation occurs at copper electrode
(a) lower the E and shift equilibrium to the left 36. The cell reaction of the galvanic cell
(b) lower the E and shift equilibrium to the right Cu (s) I Cu^+ (aq) I Hg2+ (aq) I Hg (/) is
(c) increase the E and shift equilibrium to the right (a) Hg + Cu2+ > Hg2+ + Cu
(d) increase the E and shift equilibrium to the left (b) Hg + Cu2+ > Hg+ + Cu+
(c) Cu + Hg >CuHg
28. A solution of Na2S04 in water is electrolysed using 2+
^Cu2+ + Hg

ow
inert electrodes. The products obtained at the (d) Cu + Hg
cathode and anode are respectively 37. The electrolyte which cannot be used in the salt
bridge in the cell
(a) H2, O2 (b) O2, H2
Cu + 2 AgN03 ^ Cu(N03>2 + 2 Ag is
(c) O2, Na (d) O2, SO2
(a) KNO3 (b) NH4NO3

e
29. What happens during the electrolysis of aqueous (d) All of these can be used
(c) KCl

re
solution of CUSO4 by using platinum electrodes

rFl
38. The electrochemical cell stops working after some
(a) Cu will deposit at cathode

F
time because
(b) Cu will deposit at anode (a) electrode potential of both the electrodes
(c) Oxygen will be released at cathode becomes zero

or
ou
(d) Hydrogen gas will be produced at cathode (b) electrode potentials of both the electrodes
30. The electrolysis of which solution will discharge become equal
ksf
OH' ions in preference to Cl' ions (c) one of the electrodes is eaten away
(d) the cell reaction gets reversed
oo
(a) Molten NaCl (b) Solid NaCl
(c) Dilute NaCl (d) All of these 39. A hypothetical galvanic cell is shown below :
Y
B

53. Which pair of electrolytes could not be ®AIA+(xM) I B+(yM)IB®

distinguished by the products of electrolysis using The emf measured is + 0-20 V. The cell reaction is
re

inert electrodes ? (a)A + B+- ^A+ + B

oYu

(a) 1 M CUSO4 solution, 1 M CUCI2 solution (b)A^ + B- ^ A + B+

ad

(b) 1 M KCl solution, 1 M KI solution (c) A+ + e~ A, B"^ + e~ 4B

(c) 1 M AgN03 solution, 1 M Cu(N03)2 solution (d) The cell reaction cannot be predicted
d

(d) 1 M KCl solution, 1 M NaCl solution 40. Given the electrode potentials
in
Re

(e) 1 M CuBr2 solution, 1 M CUSO4 solution Fe^"*" + e~ 4 Fe2+, E° = 0-771 volt

F

32. Which one of the following metals cannot be I2 + 2 e' ■> r, E“ = 0-536 volt
obtained by electrolysis of aqueous solution of its
salt ? E^j, for the cell reaction
(a) Ag (b) Mg 2 Fe^+ -1-21- ^ 2 Fe^-*- +12 is
(c) Cu (d) Cr (a) (2 X 0-771) - 0-536 = 1-006 V
33. In a lead storage battery, the electrolyte H2SO4 is (b) 0-771 - 0-5 X 0-536 = 0-503 volt
(a) 38% (b) 62% (c) 0-771 - 0-536 = 0-235 volt
(c) 80% (d) 48% (d) 0-536 - 0-771 = - 0-235 volt
34. Galvanisation is applying a coating of 41. The standard e.m.f. of the cell Zn I Zn^"^ II
(a) Cr (b) Cu Ag"*" I Ag is 1 -56 V. If the standard reduction
(c) Zn (d) Pb potential of Ag is 0-8 V, the standard oxidation
potential of zinc is
35. In case of a galvanic cell consisting of Cu and H2
(a) - 0-76 V (b) + 0-76 V
electrodes, which of the following statements is
true ? (c)-2-36 V (d) + 2-36 V
ELECTROCHEMISTRY 3/83

42. Which of the following cannot be stored ? 50. The resistance of ()●! N solution of a salt is found

(a) ZnS04 solution in iron vessel to be 2-5 X 10^ ohms. The equivalent conductance
of the solution in ohm“* cm- will be
(b) FeS04 solution in nickel vessel
(cell constant = M5 cm"’)
(c) MgS04 solution in aluminium vessel
(a) 4-6 ih) 5-6
(d) CUSO4 solution in zinc vessel
(c) 6-6 (d) 7-6
43. The minimum voltage required to bring about the
electrolysis of 1 M copper sulphate solution at 25°C 51. Products of electrolysis of an aqueous solution of
will be AgN03 using silver electrodes will be
(Given: E°Cu^'^/Cu = 0-34V and E? + =-l-23V) (Given E = +0-80V)
Ag'^/Ag
-HjO/H
(a) 1-58 V (b) 0-89 V (a) Ag at cathode, Oo at anode

w
(c) 0-34 V (d) 1-23 V (/?) H2 at cathode, 0, at anode
(c) Ag at cathode, dissolution of Ag from anode
44. If the pressure of hydrogen gas is increased from 1
atm to 100 atm, keeping the hydrogen ion (d) H2 at cathode, dissolution of Ag from anode

Flo
concentration at 1 M, the voltage of the hydrogen 52. What amount of CI2 gas is liberated at the anode if
half cell at 25'’C will be

e
1 amp current is passed for 30 min. through NaCI

re
(a) 0-059 V solution ?
(b) - 0-059 V

F
(c) 0-295 V (d) 0-118 V (rt) 0-66 mole (/?) 0-33 mole

45. The potential of hydrogen electrode in contact with (c) 0-66 g (d) 0-33 g
ur
or
a solution of pY\ 10 is 53. In H2-O2 fuel cell, the reaction occurring at anode
{a) + 0-591 V (/?)-0-591 V
(c) + 0-2955 V
is

f
(a)2H2 + 0,— ^ 2 H2O (/)
ks
Uf) - 0-2995 V
Yo
46. The reduction potential of an electrode can be (b) + OH" — ^H,0
oo

increased by (c) 2 H2 -f 4 OH“ ■> 4 H2O + 4 £●

{a) increasing the area of the electrode
B

(d) 4 OH ^ 2 H,0 -t- 0, + 4 f-

{b) decreasing the temperature 54. Which of the following is correct for spontaneity
re

(c) increasing the temperature of a cell ?

{d) decreasing the concentration of metal ion
u

(a) AG = -ve, E° = +ve

ad
Yo

47. Consider the following galvanic cell (b) AG = +ve, E“ = 0

Zn (s) + 2 Ag-' (0-2 M) > Zn-'' (0-28 M) (c) AG = -ve, E° = 0
+ 2 Ag (5)
d

(d) AG = +ve, E"* = -ve (CBSE 2«20)

Re
in

If E°cell = 2-57 V, then the e.m.f. of the cell at 55. The molar conductivity of CH3COOH at infinite
dilution is 390 S cm^/mol. Using the graph and
F

298 K is :
given information, the molar conductivity of
(a) 2-5 V (b) 1 -5 V
CH3COOK will be
(c) 0-5 V (d) - 0-5 V
48. For the cell reaction Cu (s) + 2 Ag-*" (aq) 4 425

375 HCl
Cu^^ (aq) + 2 Ag (j), E°cn is 0-46 V. The
3 325
equilibrium constant of the reaction is E

(a) 3-92 X I0>2 (b) 3-92 X 10^3

(c) 8-92 X lO”^ (d) 8-92 X 10
10 vj 225
<
49. The standard electrode potential for the Daniell cell 175 KCl

is 1 -1 V. The standard Gibbs energy for the reaction 125

Zn (.9) + Cu"-' (aq) Zn--*- {aq) + Cu (.5) will be
-1 75
(a)-21-23 kJ mol" (/?)-212-3 kJ mol 0 0-1 0-2 0-3
(c) - 10-61 kJ mor' (d) - 106-1 kJ mol
-1
(c/M)
1/2
3/84 New Course Chemistry fXIIlPgTHn

(«) 100 S cm" mol ' (/?) 115 S cm" mol ’ Reason. Copper metal of copper anode is more
(c) 150 S cm^ mor' (d) 125 S cm- mol"’ easily oxidized than other ions in the solution.
(CBSE Sample Paper 2022-23) 61. Assertion. Molar conductance of an electrolyse
increases with dilution.
56. In a lead storage battery
Reason. Ions move faster in dilute solutions.
(a) Pb02 is reduced to PbSO^ at the cathode
62. Assertion. Cell constant is best found by measuring
(h) Pb is oxidized to PbS04 at the anode the distance between the electrodes and dividing
(c) Both electrodes are immersed in the same it by the area of cross-section of the electrode.
aqueous solution of H',S04 Reason. Cell constant = lla where / is the distance
{d) All the above are tioie (CBSE 2020) between the electrodes and a is its area of cross-
57. An electrochemical cell behaves like an electrolytic section.

ow
cell when
63. Assertion. Molar conductivity of a weak
(a) = E^xtcmal (b) E,^|| = 0 electrolyte at infinite dilution cannot be determined
(c) > E^gji ^external ^cell experimentally.
(CBSE 2020) Reason. Kohlrausch law helps to find molar
conductivity of a weak electrolyte at infinite

e
58. Which of the following analogies is correct ?

Fl
re
dilution.
(a) Conductivity : Units S m ' :: Molar
64. Assertion. The molar conductivity of BaCl2 at

F
conductivity : Units S m"^ mol"' infinite dilution is the sum of the limiting ionic
(b) Zn - Cu Cell : Daniell cell Zn Zn cell ; conductivities of Ba"'*’ and Cl" ion.
ur
r
Concentration cell
Reason. According to Kohlrausch law,
(c) Conductivity : increases with dilution :: Molar A®
fo
= X. + X.
ks
eq.
conductivity : decreases with dilution
Yo
65. Assertion. The Daniell ceil becomes dead after
(d) Electrode potential; increases with increase in
oo

some time.
concentration of electrolyte :: EMF of a cell :
Reason. Oxidation potential of zinc anode
B

decreases with increase in concentration of

reactants
decreases and that of copper cathode increases.
re

66. Assertion. All electrochemical cells convert

II. Assertion-Reason Type Questions chemical energy into electrical energy.

u
ad

Reason. All redox reactions are exothermic.

Yo

In the questions given below, two statements are

67. Assertion. Copper sulphate solution can be kept
given one labelled Assertion (A) and the other
in a zinc vessel.
labelled Reason (R). Select the correct answer
d

Reason. The position of zinc is higher than

Re

to these questions from the codes (<z), (/>), (c) and

in

(d) given below : copper in the electrochemical series.

F

(a) Both A and R are correct and R is the correct 68. Assertion. Copper liberates hydrogen from a dilute
explanation of A. solution of hydrochloric acid.
(b) Both A and R are correct but R is not the correct Reason. Copper is below hydrogen in the
electrochemical .series.
explanation of A.
(c) A is correct but R is wrong. 69. Assertion. is stronger oxidizing agent than CI2.
(d) A is wrong but R is correct. Reason. Oxidation potential of F2 is greater than
59. Assertion. The electrolysis of aqueous NaCl that of CI2.
solution gives hydrogen at the cathode and chlorine 70. Assertion. H2-O2 fuel cell gives a constant voltage
at the anode. throughout its life.
Reason. Chlorine has higher oxidation potential Reason. In this cell, H2 reacts with OH" ions, yet
than water. the overall concentration of OH" ions does not
change.
60. Assertion. Electrolysis of CUSO4 solution using
platinum electrodes gives oxygen at the anode but 71. Assertion. When aqueous solution of CUSO4 is
using copper electrodes no oxygen is liberated at electrolysed using copper electrodes, no gas is
the anode. liberated at the anode.
ELECTROCHEMISTRY 3/85

Reason. Cu of the anode is more easily oxidized 74. Assertion. Specific conductivity of an electrolytic
than OH“ ions or SO^“ ions present in the solution. solution increases with dilution.

Reason. With dilution, ionization of a weak

72. Assertion. Equivalent weight of any .substance is
96500 times its electrochemical equivalent. electrolyte increases and hence number of ions
increases.
Reason. Each metal has a fixed value of
electrochemical equivalent. 75. Assertion. During discharge of a lead storage
73. Assertion. Specific conductivity of electrolytic battery, both Pb and PbO-, are reduced to form
solution increa.ses with increase of temperature. PbS04-
Reason. Specific conductivity is the conductance Reason. During recharge, PbS04 is reduced as well
of I cm^ of the solution. as oxidized.

ow
ANSWERS

I. Multiple Choice Questions

l.{b) 2.{c) 3.(c) 4.(b) 5. {a) 6. (b) 7. (d) 8. (c) 9. ib) 10. (c)

e
11. (a) 12. (b) 13. [d) 14. (b) 15. (d) 16. (b) n. (c) 18.ib) 19.(b) 20. (c)

re
Fl
21. (c) 22. (d) 23. (d) 24. (a) 25. (a) 26. (b) 27.(c) 28. ffl) 29.(a) 30. (c)

F
31. (d) 32. (b) 33. (a) 34. (c) 35. (c) 36. (d) 37.(c) 38. (/;) 39.(a) 40. (c)
41. (b) 42. (d) 43. ih) 44. (b) 45. {b) 46. ib) 41. (a) 48. (/;) 49.ib) 50.{a)
ur
r
51. (c) 52. (c) 53. (c) 54. (a) 55.ib) 56. id) 57.(c) 58. ib)

II. Assertion-Reason Type Questions

fo
ks
Yo
59. (c) 60. (a) 61. (h) 62. id) 63. ib) 64. id) 65. (a) 66. id) 67. id) 68. id)
oo

69.(c) 70. (a) 71. (a) 72. ib) 73. ib) 74. id) 75. id)
eB
ur

For Difficult Questions

ad
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5. At cathode e 4 H. H + H
I. Multiple Choice Questions

Cr^O^- + 14 + 6 e' ^ 2 Cr3+ + 7 H-,0

d

2. Charge on L.H.S. = - l +4 = + 3
Re
in

Charge on R.H.S. = 0
Hence, 3 e~ should be added on L.H.S. to balance
Thus, 1 mole of Cr20| requires for reduction
F

the charges. charge = 6 F

2+ 7. Cu^-*- + 2e ^Cu
3. Mn04 + 8 H+ + 5 e- > Mn + 4H2O
Thus, 2 F charge deposit 1 mol of Cu, i.e., 63-5 g.
8. The oxidation reaction is
Thus,5 moles of electrons reduce 1 mole of Mn04
7-5 moles of electrons will reduce MnO^- Mn04 + e
1 1 mol of MnO^ for oxidation requires
Mn04 = - x7-5 = 1-5 moles electricity = 1 F = 96500 C

01 mol of MnO^“ will require electricity

4. NaC! (o<7) ■> Na'*' iaq) +Cr iaq)
= 9650 C
H.O (/) ;; — H"^ iaq) + OH“ iaq) 9. 2 Cl" ■> CI2 ig) + 2e-
As H'*’ ions have lower discharge potential than To produce 1 mol of CI2, charge required
Na"*" ion, H'*' ions are discharged at the cathode = 2 F = 2 X 96500 C
3/86 New Course Chemistry (X11)EZSX91

To produce 0-1 mol of CI2, charge required Ecell=E“ cathode -E“anode

= 2 X 96500 X 0 1 = 19300 C
= E“ -E°
Ag‘*’/Ag aP+/ai
Q = Ixr
^ 19300 sec 19300 . 2-46 = 0-80-E®
A1^+/A1
r-Q- ^
= mm
3x60
= 107-22 min =: 110 min or E" = -1-66 V
Al^+ZAl
11. CsCl is a strong electrolyte while NH4OH and 24. Ti -h Cu2+ > Ti2+ + Cu
CH3COOH are weak electrolytes.
12. A strong electrolyte is completely ionized at all 0-0591 , [Ti2+]
concentrations. Hence, number of ions remains the Ece,. = E”ceU
same. However, on dilution, interionic forces
decrease and hence ionic mobility increases. 25.For2H+ + 2e- )H 2’
Therefore, equivalent conductance increases.

w
0-0591, 1
13. According to Debye-Huckel theory, for a strong E = E“- = 0+
®-®|^log[H-]2
—

electrolyte (like NaCl),

F lo
A, = A^-(B)VC = 0-0591 log [H+] = - 0-0591 pH
(a) [H+] = 1 M, E = 0-0591 log 1 = 0

ee
KxlOOO
14. A m ib) [OH-] = 10°, [H+] = 10-14 pH = 14,

Fr
Molarity
E =-0-0591 X 14 = -0-827 V

(c) [H+] = lO-"^, pH = 7, E = - 0-0591 x 7

(5-76 X10-^Scm-l)(1000cm^L-i)
0-5molL-l
for = -0-413 V
ur
(d) pH = 4, E = - 0 0591 x 4 = - 0-236 V
5-76
s
= 11-52 S cm2 mol-' Thus, potential E is maximum for 1 M HCl.
ook
Yo

0-5
26. For hydrogen electrode.
eB

— A®
17. a“NaBr ” 4. A® —A®
A^aCl ARBr A^ci 1
= 126 + 152 - 150 = 128 S cm^ mor' H+ (aq) + e-
our

o 0 0
ad

18* AcHjCOOH “ A^HjCOONa Ahci ~ Aj^j^q 0-0591

1/2

= 90 + 425 - 125 = 390 mho cm^ mol"' E = E“- log

1 [H+1
Y

m _
7-8
Re

a = — = 0-02
o
1/2
nd

A 390
m

= -0-0591 log (vE‘’ = 0)

Fi

20. Higher the reduction potential, more easily it is [H+]

reduced and hence stronger is the oxidizing agent.
1/2
F2 has highest reduction potential. Hence, F2 is (PuJ
H2
strongest oxidizing agent. Considering the reverse E will be -ve when is -hve.
[H+]
reactions (oxidation reactions), higher the oxidation
potential, more easily it is oxidized and hence 1/2
stronger is the reducing agent. i-e., > [H'*'] which is so in case of (b).

Oxidation potential of l~ ion (- 0-53 V) is highest

out of - 2-85, - 1-36, - 1-06 and - 0-53. Hence, T 00591,
is strongest reducing agent. 27.E,,„=E" cell ^'08 [H+]2
21.E%eii = E° cathode _ po anode
On adding H2SO4, [H'*'] will increase, therefore
= +0-15-(-0-74) = +0-89 V
Ecell will increase. The equilibrium Zn -1- 2 H"^
23. A1 is more reactive than Ag, i.e., cell reaction is V Zn^’*’ -t- H2 will shift towards right on
A1 + 3 Ag+ > Al^+ + 3 Ag. increasing the concentration of H'*' ions.
ELECTROCHEMISTRY 3/87

31. (fl)CuS04 —> Cu + O2, Net reaction :

CuCl2 —> Cu + CI2 1

ib) KCl — »H2 + Cl2,KI > H-> + 1-7

Cu2+ iaq) + H2O (/) > Cu (j) + 2 (5)
(c) AgN03 —>Ag + O2, + 2 H+ (aq), E°Cell = - 0-89 V

Cu(N03)2 » Cu+ O2
Hence, minimum voltage required to bring about
id) KCl- »H2 + Cl2, NaCl- ■> H2 + CI2 electrolysis is 0-89 V.
(e) CuBr -> Cu + Br2, CUSO4 Cu + Oo 1
44. H+ + e- >
Thus, KCl and NaCl solutions give the same
products at anode and cathode. 1/2

34. Galvanisation is applying a coating of zinc metal 0-059 ph;

E = E“- log

w
to prevent corrosion. n

35. In Cu - H2 cell, hydrogen loses electrons (oxidation 0 059 . (100)

1/2
= 0- log = - 0 059 V

Flo
occurs). Hence, H2 acts as anode whereas Cu is 1 1
cathode.
1

ee
36. Cu -> Cu^"^ + 2e~ (Oxidation) 45. For hydrogen electrode, H'*' + e > 2 »2

Fr
Hg2+ + 2 e- -> Hg (Reduction) Applying Nemst equation,
2+
Cu + Hg ^ Cu^-*- + Hg = £”
00591
log
1
^ 1

for
ur
I
37. This is because Cl" ions react with Ag'*’ ions to n [H+]
2
form a precipitate of AgCl.
00591 1
39. ®A/A"*" (x m) reprsents - ve pole (anode) on which
s
= 0- log
ok
1 10-10
Yo
oxidation occurs :

B'*’ (y m)/B® represents +ve pole (cathode) on (pH = 10 means (H+) = lO"'® M)
Bo

= -00591 X 10 = -0-591 V
which reduction occurs. Hence, the cell reaction
46. M”+ + ne~
re

is A + B+ ^ A+ + B.

2-303 RT 1
40. E"Cell =Eox(I2)+e;Red (Fe) E = E"
ou

log
ad

M«+/M M"+/M nF [M"+]

= -0-536 + 0-771 =0-235 V
Y

If T is decreased, magnitude of second factor on

41. Ece„ =E;Red (Ag+ (Ag)) + E°
Ox (Zn/Zn^+) RHS decreases so that E M"+/M
nd

increases.
Re

1-56 = 0-80+E°
Fi

Ox(Zn/Zn^*) 0-0591,log^^
[Zn2+]
47.E,,„=E!cell
E” = 1-56-0-80 = 0-76 V
n IAg-]2
Ox(Zn/Zn2+)
= 2-57 -
0-0591, 0-28
42. Zn is more reactive than Cu. Hence, Zn can displace
Cu from CUSO4 solution, i.e., reaction occurs.
Hence, CUSO4 solution can not be stored in Zn = 2-57 -
0-0591,
vessel. This is not so in the first three cases. ^-log7
43. The reactions taking place are 0-0591
= 2-57 - X 0-8451
At Anode: 2
Cu^'*’ (aq) + 2 e~ + Cu (5), E° = 0-34 V = 2-57 - 0-025 = 2-5 V
At Cathode :
0-0591
1 48* ^;u= n
logK
H2O (/) ^2H^(aq)+- 02(g) + 2e~,
E° = -l-23 V For the given reaction, n = 2, £°g,| = 046 V
3/88 T^'uxcCeefi, '<*. New Course Chemistry (XII) BE

1 1
046 = 00591, ^ X1800
1800 C with liberate Cl, = - x
—^logAT, 2 2 96500
= 0 0093 mol
046x2
or log = = 15-5668 = 00093x71 g
0-0591
= 0-66 g
= Antilog 15-5668 = 3-7 x 10*^
55. (HCl) = 425Scm2 mol"‘
49. AG°=-nFE"CeU

= - 2 X 96500 C mol-' x MO V
a;, (KCl) = 150Scm2 mopi
= - 212,300 CV mol-' a;, (CH3COOK) = a;, (CH3COOH) + A® (KCl)
= - 212300 J mol-' (1 CV = 1 J) -a“(HC1)

low
= - 212-3 kJ mol-' = 390 + 150 - 425 = 115 S cm^ mor*
50. Specific conductivity = Obs. conductivity x 58. (a) is incorrect because units of molar conductivity
Cell const are S m^ mor'.
1 (c) is incorrect because conductivity decreases

ee
xl-5 = 4.6 X 10-^ohm-' cm-' with dilution.
2-5 xlO^
rF {d) is incorrect because EMF of a cell increases

Fr
KxlOOO 4-6x10-^x1000 with increase in concentration of reactants.
^eq Normality 0-1
II. Assertion-Reason Type Questions

r
= 4-6 Q-' cm^ eq-'
fo
59. Correct R. Chlorine has higher reduction potential
u
51. Probable reactions at cathode (Reduction) than water but lower discharge potential.
ks
60. R is the correct explanation of A.
Yo
hg^ + e- >Ag, £,^=0-80V
oo

61. Correct explanation. Molar conductance

increases because with dilution in a weak
2 H2O 2 e- -»H2 + 2 0H- £,^=-0-83V
B

electrolyte, dissociation increases while in a strong

re

Ag'*’ ion has higher reduction potential than H2O. electrolyte, interionic attractions decrease.
Hence, Ag"*" ions are reduced more easily and 62. Correct A. Cell constant is best found by
u
ad

deposited as Ag. measuring the conductance (resistance) of a

Yo

Probable reactions at anode (Oxidation) solution whose conductivity is known (because

Ag (from anode) ^ Ag+ (aq) + e~. cell constant = conductivity/conductance)
63. Correct explanation. The dissociation of weak
nd
Re

=-0-80V
ox electrolyte keeps increasing with dilution and is
not complete even at infinite dilution.
Fi

1
H20(/) > - 02(g)-i-2H+(a^), 64. Correct A. a“ (BaCL) X°Ba 2+ + 2A,“
Q-

65. R is the correct explanation of A.

<x =-1-23V
66. Correct A. Electrolytic cells convert electrical
First reaction has higher oxidation potential. Hence, energy into chemical energy (Remember :
Ag from anode dissolves to form Ag'*’ ions in the Galvanic cells and electrolytic cells are collectively
solution.
called electrochemical cells)
1 67. Correct A. Copper sulphate solution cannot be
52. Cl-
» 2^*2+^ kept in zinc vessel.
68. Correct A. Copper does not liberate hydrogen from
Quantity of electricity passed = 1 x 30 x 60 C dilute hydrochloric acid solution.
= 1800 C
69. Correct R. Reduction potential of F2 is greater than
1 that of CI2.
96500 C liberate Cl, = — mole 70. R is the correct explanationof A.
^ 2
ELECTROCHEMISTRY 3/89

71. Both A and R are correct and R is the coixect therefore, specific conductivity increases witli
explanation of A. temperature.
72. Correct explanation. Electrochemical equivalent 74. Correct A. Specific conductivity of a solution
of a substance is the amount deposited by one decreases with dilution (because number of ions/cc
coulomb of charge. in the solution decreases).
73. Correct explanation. Specific conductivity : 75. Correct A. During discharge of lead storage
Conducance x Cell constant. As conductance of battery, Pb is oxidized to PbS04 while Pb02 is
an electrolytic solution increases with temperature reduced to PbS04.

IVIS

w
I. Electrolysis and Faraday’s laws At Anode :

1. An aqueous solution of copper sulphate is

OH ^0H + ^-,40H 42H2O (0 +02(g)

F lo
electrolysed using platinum electrodes in one Thus, Ht is liberated at the cathode and Ot at
the anode.
case and copper electrodes in another case.

ee
Will the products of electrolysis be same or (ii) AgNO^{s) + aq ■>A$^{aq)+ NO3 (aq);
din^erent ? Give reason.

Fr
Ans. In case of platinum electrodes, the products are H2O V
At cathode. Ag"^ ions have lower discharge
Cu on cathode and O2 gas on anode. This is
because out of Cu"'*' and ions (produced from for
potential than H'*' ions. Hence, Ag^ ions will be
ur
HiO), Cu^'*’ ions have lower discharge potential. deposited on the cathode (Ag^ + e~ ^ Ag)
s
Out of SO^ and OH ions, OH ions have At anode. Ag anode will be attacked by NO3
ook
Yo

lower discharge potential (OH“ ^ OH + e" ions. Hence, Ag anode will dissolve to form Ag'*'
eB

4 OH 4 2 H2O + O,) ions in the .solution (Ag Ag-^ + e )

In case of copper electrodes, product at cathode 3. How much amount of a substance is
is the same, viz., Cu but at anode, out of the deposited by 1 coulomb ? What is it called ?
our
ad

three possible reactions, viz., Ans. One coulomb deposits Eq. wt./96500 gram. It
is called electrochemical equivalent of the
OH' OH + c\ SO^ 4 SO4 + 2 e".
Y

substance.
Cu 4 Cu-+ + 2 C-,
Re

4. One faraday of electricity deposits one mol

nd

Cu is more easily oxidized than OH" and SO|” 1

of Na from the molten salt but ~ mol of Al
Fi

ions. Hence, anode of Cu dissolves into the

solution to form Cu'"^ ions. from an aluminium salt. Why ?
2. State the products of electrolysis obtained on Ans. The reactions at cathode for the deposit of Na
the cathode and the anode in the following and AI are
cases :
Na+ + e- 4 Na and AI^-" + 3 c" 4 Al
(i) A dilute solution of H2SO4 with platinum Thus, 1 faraday deposits 1 mol of Na whereas 3
electrodes
faradays are required for depositing one mol of
(ii) An aqueous solution of AgNOj with silver Al.
electrodes.

II. Conductance, specific,

Ans. (/) H^S04 (aq) 4 2 H+ (aq) + SO^" (aq); equivalent and molar conductivities
H,0 V ^ H^ + OH-
At cathode :
5. Copper is conducting as such while copper
sulphate is conducting only in molten state
H^ + e 4H, H + H 4 H, (g)
or in aqueous solution. Explain.
3/90 New Course Chemistry (XII)EEIHI

Ans. Copper is conducting as such because it contains Ans. The molar conductivity of a solution at infinite
free electrons. CUSO4 is conducting in molten dilution, i.e., when concentration approaches
state or in aqueoussolution because it gives ions zero is called limiting molar conductivity. With
only in the molten state or aqueous solution. decrease in concentration, i.e., with dilution,
6. Which will have greater molar conductivity number of ions per cm^ decreases due to large
increase in volume of the solution. As
and why ?
Sol. A. 1 mol KCI dissolved in 200 cc of the conductivity is the conductance per cm^ of the
solution. solution, hence it decreases,

Sol. B. 1 mol KCI dissolved in 500 ec of the 13. Why A for CH3COOH
m cannot be
solution. determined experimentally ?
Ans. Molar conductivity of weak electrolytes keeps
Ans. Solution B will have greater molar conductivity
on increasing with dilution and does not become
because a„, = k x V. With dilution, k decreases
constant even at very large dilutions.
but V increases much more so that product
increases. 14. X and Y are two electrolytes. On dilution,

w
molar conductivity of X increases 2*5 times
7. Solutions of two electrolytes A and B each while that of 4 increases 25 times. Which of
having a concentration of 0*2 M have

F lo
the two is a weak electrolyte and why ?
conductivities 2 x 10"^ and 4 x 10~* S cm -1
(CBSE 2820)
respectively. Which will offer greater
resistance to the flow of current and why ? Ans. Y is a weak electrolyte because the molar
conductivity of a weak electrolyte increases

e
Fre
^ ^ I / 1 rapidly on dilution whereas it increases slowly
Ans. k = Gx—= — i.e., . Hence. B will
a R a R for strong electrolytes (Refer to Fig. 3.8).
offer greater resistance, for HI. Kohlrausch’s
8. Why alternating current is u.sed in place of law and its applications
r
direct current in measuring the electrolytic
You
15. Write expressionsfor equivalentconductivity
oks

conductance ?
and molar conductivity of Al2(S04)3 at
eBo

Ans. Direct current results in the electrolysis of the infinite dilution in terms of their ionic
electrolytic solution. conductivities.
9. A O’l M aqueous of Na2S04 is diluted by Ans. {A\,S,0^) = X° A] 3+
ad
our

A
adding water. What will happen to the values eq
so^- ’
of its conductance (G), conductivity (k),
molar conductivity (a,„) and equivalent A„,(Al2(S04)3) = 2>.“ A1 3+ +3X°

conductivity (a^) ?
Re
dY

IV. Galvanic cells

Ans. K decreases whereas G , a„, and a^,^^ increase with
Fin

dilution (Refer to Art. 3.8).

16. Why is it necessary to use a salt bridge in a
10. Why in a concentrated solution, a strong Galvanic cell ?
electrolyte shows deviations from Debye- Ans. To complete the inner circuit and to maintain
Huckel-Onsager equation ? electrical neutrality of the electrolytic solutions
of the half-cells.
Ans. In concentrated solution of a strong electrolyte,
the interionic forces of attraction are large. 17. Formulate the galvanic cell in which the
following reaction takes place :
11. How does molar conductivity of a weak
electrolyte vary with its concentration ? Give Zn (s)-i-2Ag’^(a^) Zn"'*' iaq) + 2 Ag (y)
reason also. State (i) Which one of the electrodes is

Ans. Refer to page 3/26. Fig. 3.8. negatively charged ?

12. Define limiting molar conductivity. Why (ii) The reaction taking place at each of its
electrode,
conductivity of an electrolyte solution
decreases with decrease in concentration ? (iii) The carriers of current within this cell.
(OiSE 2015) (CBSE 2008)
ELECTROCHEMISTRY 3/91

Ans. The cell reaction is: 21. On the basis of the standard electrode
Zn (s) + 2 Ag+ iaq)- ^ Zn2+ {aq) + 2 Ag {s) potential values stated for acid solution,
The cell is represented as: predict whether Ti"*^ species may be used to
oxidize Fe“ to Fe^‘*
Zn (5) I Zn^-^ {aq) 11 Ag-^ {aq) I Ag (s)
Reaction E/V
(/) Anode, i.e., zinc electrode is negatively 3+ .
^Ti + 001
charged.
^ Fe--^ ; + 0-77
(ii) At Anode : Zn {iO ^ Zir'*' {aq) + 2 e
(C’BS»
At cathode : Ag'^ (aq) + e > Ag (,y)
{Hi) Within the cell, the current is carried by the Ans. We want to check the spontaneity of the

ow
cations and the anions through the salt bridge. reaction :
In the external circuit, electrons How from zinc Ti^-*- + Fe-+ + Ti3+ 4. Fe^+
to silver and hence conventional current flows
E.M.F. of this reaction = + 001 + (- 0-77)
from silver to zinc.
= -0-76 V
18. In a galvanic cell, the following reaction

e
occurs : As EMF is -ve, the reaction is non-spontaneous,

re
Zn (s) + 2 Ag+ {aq) > Zn^"^ {aq) + 2 Ag (s), />., cannot be used to oxidize Fe” to Fe”'.

Frl
E' cell = + 1-50 V 22. Two half cell reactions of an electrochemical

F
cell are given below :
(a) Is the direction of How of electrons from
zinc to silver or silver to zinc ? MnOj (a?)+ 8H-^ («(?) + >
ou
or
{b) How will the concentration of Zn-'*' ions Mn^+ {aq) + 4 H^O (/), E" = + 1-51 V
and Ag"*" ions be affected when the cell
functions ? (CBSE Foreign 2017)
Ans. (n) Zn loses electrons while Ag'*' ions in Ag/
kfs
Sn2+ {ag) > Sn** {ag) + 2 E" = + 0-15 V
Construct the redox reaction from the two
half cell reactions and predict if this reaction
oo
Ag"*^ electrode gain electrons. Hence, flow of
elections is from Zn to Ag. favours formation of reactants or products
Y
shown in the equation. (CBSE 200'>
B

2+
{b) Zn changes into Zn ions. Hence,
concentration of Zn^'*' ions increases. Ag"*" ions Ans. Multiplying first eqn. by 2 and second eqn. by
re

change into Ag. Hence, concentration of Ag'*' 5 to cancel out the electrons, and adding, the
ions keeps on decreasing. redox reaction will be
oYu
ad

V. Electrode potential, cell 2Mn04+16H-^+5Sn2+ +

potential and electrochemical 2 Mn--" + 5 Sn*^ + 8 H^O

d

series and its applications

Oxidation potential of 2nd reaction = - 0-15 V
in
Re

19. What is the use of platinum foil in the (as the given value is reduction potential).
hydrogen electrode ? Hence, if oxidation occurs on Sn^'*'/Sn'*^
F

Ans. It is used for the inflow and outflow of electrons. electrode, = 1-51 + (- 0'15) V = 1-36 V,
20. An electrochemical cell is made of aluminium i.e., it will be +ve. Hence, reaction will favour
and tin electrodes with their standard formation of products as represented above.
reduction potentials - 1-66 V and 0*14 V VI. Nernst eqn., concentration
respectively. Select the anode and the cell and calculation of equilibrium
cathode, represent the cell and write the cell constant from Nern.st equation
reaction. Find the e.m.f. of the cell.
Ans. For E cell to be positive, the given standard 23. How can the reduction potential of an
reduction potentials show that oxidation will electrode be increased ?
take place on Al-clectrode. Hence. A1 will be Ans. M"'*' + ne~ + M.
anode and Sn will be cathode. The cell will be
RT
represented as : E = E° In
„F
A1 {s) 1 AP-*- {aq) 11 Sn^-^ {aq) I Sn is)
RT
^ Al^+.AI
— CO

cell ^ Sn^+.Sn = E“ ln[M"+]

nF
= 0-14-(- 1-66) = 1-80 V
3/92 ‘P’ui^tcep.'A New Course Chemistry fXinrorwi
Thus, electrode potential can be increased either Ans. Higher the standard reduction potential of a
by increasing metal ion concentration, i.e., .species, more easily it is reduced at the cathode.
[M"^J or temperature T. As Ag'*’ iaq) has greater standard reduction
24. What is the difference between a chemical
potential, therefore, the reaction that will occur
cell and a concentration cell ?
at the cathode is Ag"*" (aq) + e~ Ag (.V).
Ans. A chemical cell is a galvanic cell in which
electrical energy produced is due to chemical 28. Predict the products of electrolysis of an
changes occurring within the cell and no transfer aqueous solution of CuCl2 >vith platinum
of matter lakes place. It involves the use of two electrodes.
different electrodes dipped in solutions of Given : E = + 0-34V,
different electrolytes. Cu2+/Cu
A concentration cell is a galvanic cell in which
E =+l-36V
electrical energy is produced due to physical I

change involving transfer of matter from one -ci2/cr

w
part of the cell to the other. It involves the use
of the same electrodes dipped in solutions of eV =o-oov.
the same electrolyte with different concen

F lo
trations Cor electrodes of different concentration
E”1 = 1*23V (CBSE 2020)
dipped in the same solution of the electrolyte).

ee
-O2/H2O
VII. Relation of Gibbs energy

Fr
Ans. Refer to Example 2, Art. 3.25
(DG“) with cell potential (Ep^jj) and
equilibrium constant (K) CuCl2 {aq) > Cu^-*- {aq) + 2 Cl" {aq)

for H2O (/) > H'*' {aq) + OH“ {aq)

ur
25. What is the free energy change for (a) galvanic
Probable reactions at the cathode :
cell {b) electrolytic cell ? (Pb. Board 2011)
ks
Ans. {a) AG < 0 (as redox reaction is spontaneous) Cu^"^ {aq) + 2e > Cu (5), E Red = +0-34V
Yo
{b) AG > 0 (as redox reaction is non-
oo

spontaneous). 1
H'*' {aq) + e ’ ^Red
eB

VIII. Predicting products of

electrolysis As ElRed for first reaction is higher, first
r

26. Following two eractions can occur at cathode

ou

reaction will occur, i.e., Cu will be deposited

ad

in the electrolysis of aqueous sodium chloride on the cathode.

Y

Na’*' + e~- > Na (s), E “ Red =

“
- 2-71 V Probable eractions at the anode :
2 HjO (/) + 2 e- > H2 (5) + 2 OH- (aq),
^ CI2 {g) + 2 E°^
nd
Re

E'
2 Cr {aq) =-l-36V
Red =
- - 0-83 V

Which reaction takes place preferentially and

Fi

I
why ? H2O (/) ■>
-0^(s)+2HUaq) + 2e-,
Ans. As the standard reduction potential of H2O is
greater than that of Na'*’ ion, reduction of water E°ox
=-1-23 V
lakes place preferentially, i.e.. Ht is liberated at
cathode. As E for second reaction is higher, this
ox

27. Following reactions occur at cathode during reaction will occur at the anode, i.e., will be
the electrolysis of aqueous silver chloride liberated.
solution :

Ag-*" (aq) + e~ ^ Ag (s), E® = -I- 0-80 V ; IX. Some commercial celLs/batteries

1
(aq) + e » - H2 (g), E® = 0 00 V 29. What is the role of ZnCl2 in a dry cell ?
On the basis of their standard reduction
Ans. ZhCIt combines with the NH3 produced to form
electrode potential (E®) values, which reaction
the complex salt [Zn(NH3)2Cl2J as otherwise the
is feasible at the cathode and why ? pressure developed due to NH3 would crack the
seal of the cell.
(CBSE 2015)
ELECTROCHEMISTRY 3/93

30. Which type of cells are rechargeable ? Write down the possible reactions for
Ans. Those cells are rechargeable in which ihe corrosion of zinc occuring at anode, cathode
products formed during discharge are deposited and overall reaction to form a white layer of
on the electrodes and these can be decomposed zinc hydroxide.
to give the original substances when elccti-ical (C’BSE Sample Paper 2022-23)
energy is supplied.
Ans. Anode : Zn (s) > Zn“+ (aq) + 2e~]x2
X. Fuel cells
Cathode : O2 (g) + 2 H2O (/) + 4 e
31. Write down the reactions that occur at the 4 OH~(aq)
anode and cathode of H2-O2 fuel cell and the
overall reaction. Overall : 2 Zn (s) + O2 (g) + 2 H2O (/)
Ans. Refer to page 3/71. 2 Zn“+ (aq) + 4 OH^ (aq)
or 2 Zn (5) + O2 Ig) + 2 H2O (/)
XI. Corrosion

w
2Zn(OH)2(ppt)
32. Give reason :
XII. Miscellaneous

F lo
(0 Rusting of Iron pipe can be prevented by
joining it with a piece of magnesium. 34. Give reasons for the following :
(«) Conductivity of an electrolyte solution (i) Copper displaces silver from silver nitrate

ee
decreases with the decrease In concentration. solution

Fr
Ans. (/) It is due to cathodic protection in which (h) Iron pipes are usually, coated with zinc.
magnesium metal is oxidized in preference to (Karnataka Board 2012)
iron and acts as the anode.

for
Ans. {/) Copper lies above silver in the electro
ur
(ii) It is due to the fact that on dilution, number chemical .series, i.e., copper is more active than
of ions per unit volume decreases. silver.
s
33. Corrosion is an electrocliemical phenomenon.
ook

(//) This is done to protect iron pipes from

Yo
The oxygen in moist air reacts as follows : rusting because zinc is more electropositive than
O2 (g) + 2H20 (/) ^ 4 OH- {aq)
eB

iron.
r

● Very Short Answer

mu
1 1
ou
ad

● Short Answer
● Long Answer
Y
Re

VERY SHORT ANSWER QUESTIONS Carrying 1 mark

nd
Fi

I. Electrolysis and Faraday’s laws 5600 cm^ of 0-> will require charge

1. What is meant by Faraday constant ?

= —^x5600 = I faraday
22400
Ans. Faraday constant. F = 96500 coulombs. It is that 3. What is the effect of temperature on the
quantity of electricity which deposits one gram electrical conduction of (/) metallic conductor
equivalent of the substance or it is the charge («) electrolytic conductor ?
carried by one mole of electrons.
Ans. With increase of temperature, the electrical
2. How many faradays ol' electric charge is conduction of metals decreases whereas that of
required to liberate 56 cm^ of oxygen at electrolytes increases.
STP? (Karnataka Board 2012) 4. How do metallic and ionic substances differ
1
Ans. 0^' ¥ —0.^+2e in conducting electricity ? (CBSE 2009)
Ans. In metallic substances, flow of electricity is due
or 2 ^ 0-) + 4 e to flow of electrons while in ionic substances
Thus, i mole of 0->, i.e., 22400 cm-^ require 4 in the aqueous solution or molten state, the flow
faradays of charge. of electricity is due to flow of ions.
3/94 “Pn^xelcc^'A New Course Chemistry (XII)CZ39D

II. Conductance and specific, 12. >Vhat is the effect of decreasing concentration
(luivalent and molar conductivitic.^ on the molar conductivity of a weak
electrolyte ?
5. Give the relationship between molar Ans. On decreasing concentration {i.e., on dilution),
conductivity and specific conductivity.
the molar conductivity increases.
CHSK 2008, 2011, 2012,
Karnataka Board 2012) 111. Kohlrausch's law and its
1000 applications
Ans. A„
m
= KX , where k is specific conducti
M 13. State Kohlrausch’s law of independent
vity in S cm"* and M is the molar concentration migration of ions.
K lA.s.sam Board 2013. CBSE 2022)
in mol L Am
is in S cm- mol"* or ^ Ans. Refer to Art. 3.10.

w
where icis S m"*, M is in mol m"^, A^,is in S m- 14. What is meant by limiting molar
mor*.
conductivity ? iCBSE 2010)
6. Express the relation among conductivity of Ans. The molar conductivity of a .solution at infinite

Flo
the solution in the cell, the cell constant and dilution is called limiting molar conductivity
resistance of solution in the cell. and is representedby the symbol a° m'

e
re
(CBSE 2011, 2012) IS. Write an expression to relate molar
Ans. Conductivity (k) conductivity of an electrolyte to its degree of

F
dissociation.
1
X Cell constant ●
ur
r
Resistance (R) Ans. a =
in
where = molar conductivity at
7. Give the relationship between equivalent and fo in

given concentration, c and a^^ = molar

ks
molar conductance of a given solution ? conductivity at infinite dilution.
Yo
oo

1000 1000 IV. Galvanic cells

Ans. A III = KX and A,., = KX
eq
Molarity Normality
B

16. What is the direction of flow of conventional

current in a galvanic cell ?
re

a
Normality
Ans. Conventional current flows from cathode to
—

^cq Molarity
anode. It is opposite to the direction of flow of
u
ad

electrons.
8. What are the units of molar conductivity ?
Yo

17. What flows in the internal circuit of a

Ans. Ohm'* cm^ mol"* or S cm^ mol"*.
galvanic cell ?
9. What is the relationship between specific
d

Ans. ions.
Re

conductance and equivalent conductance ?

in

KXlOOO
V. Electrode potential, cell
F

Ans. a
eq , where k is specific potential and electrochemical series
N
and its applications
conductance in S cm"*, N is normality in g eq,
L"', A^^is i S cm^ (g eq)~^. 18. Why is it not possible to measure the single
10. How is cell constant calculated from electrode potential ? (J & K Board 2012)
conductance values ? Ans. Oxidation or reduction cannot take place alone.
Ans. Cell constant = specific conductance/observed Moreover, it is a relative tendency and can be
conductance. measured with respect to a reference electrode
only.
11. Which equation gives the relationship
between equivalent or molar conductance 19. Given that the standard electrode potentials
(E") of metals are :
and concentration of a strong electrolyte ?
K+/K = - 2*93 V, Ag+/Ag = 0-80 V,
Ans. Debye-Huckel-Onsager equation, a = a°- A Vc Cu^+ZCu = 0-34 V, Mg^+/Mg = - 2-37 V,
in which a is the molar conductance at Cr^^/Cr = - 0-74 V, Fe^+/Fe = - 0-44 V
concentration c, a® is the molar conductance at
infinite dilution and A is a constant.
Arrange these metals in an increasing order
of their reducing power. (CBSE 2010)
ELECTROCHEMISTRY 3/95

Ans. Higher the oxidation potential more easily it 25. Write Nernst equation for the reaction
is oxidised and hence stronger is the reducing (0 2Cr (s) + 3Cd-^ + (fl?) ->

power. Therefore order is Ag < Cu < Fe < Mg 2Cr^ (ag) + 3Cd (.v)
<K. (h) 2Cr + 3Fe-^-^ ■> 2Ci^^ + 3Fe
20. Define electrochemical series, RT [Cr3+)’-
or Write two applications of electrochemical Ans. (i) E
6F
In
[Cd++I-
^ n = 6)
series. (UP Board 2012)
RT In
, [Cr3+]2 (« = 6)
Ans. The arrangement of various electrodes in the (ii) E cell = E' cell
6F [Fe++]3

w
decreasing or increasing order of their standard
reduction potentials is called electrochemical 26. Why a galvanic cell cell stops working after
series,
some time ?
Ans. With time, concentrations of the electrolytic
or Two applications of electrochemical series are :
solutions change. Hence, their electrode
(/) To compare oxidizing or reducing powers of

o
e
potentials change. When the electrode potentials
different elements. of the two half-cells become equal, the cell stops

re
(//) To predict the spontaneity of any redox working.

Frl
reaction. 27. What is a concentration cell ?

F
21. Can we store copper sulphate in iron vessel ? Ans. Refer to Art. 3.22, page 3/56.
Why? VII. Relation of Gibbs energy (I)(i
ou
sor
Ans. No, because iron is more reactive than copper.
with cell potential (l\i.||) and
22. Why does zinc react with dilute sulphuric equilibrium constant (K)
acid but copper does not ?
i.Iharkhand Board 2011)
kf
28. How is free energy change of a cell reaction
oo
related to (i) its emf (h) equilibrium constant
Ans. Zinc lies above hydrogen whereas copper lies of the cell reaction ?
Y
below hydrogen in the eleclRx:hemical series.
B

Hence, zinc is more reactive than H'*' whereas Ans. (/) - A^G = oF E^^„ where n is number of
electrons involved in the cell reaction.
copper is less reactive. Therefore, zinc reacts
re

(//) - AG° = RT In K where K is equilibrium

with dilute H2SO4 but copper cannot liberate
oY

constant.
u

H-> gas by reaction with dilute H2SO4.

29. Is free energy change of a cell reaction an
ad

VI. Ncrast etju., concentration cells intensive property or extensive property ?

and calculation of equilibrium
d

Ans. Extensive property.

constant from Nernst equation
in

VTH. Predicting products of

Re

23. Write Nernst equation for single electrode electrolysis

potential.
F

(HP Board 201.^ 1

Ans. For the electrode reaction, ne~ ■> M 30. What is overvoltage ?
Ans. Refer to page 3/65.
2-303 RT [M]
E log 31. Write the product obtained at anode on
M"+/M M^^/M /iF [M"+l electrolysis of concentrated sulphuric acid
using platinum electrodes.
2-303 RT 1
po
log
M"+/M nF [M"+] Ans. 820^“ (a^) (Refer to point 1 of Supplement
24. Write Nernst equation to calculate the cell Your Knowledge, page 3/66).
potential of Mg (s) 1 Mg^'*'(a4f) 11 Ag'*'{fl^) 1 Ag IX. Some commercial cells/batU i 1
Ans. The cell reaction is Mg (i) + 2 {aq)
Mg-"^ (aq) 2 Ag (j), n = 2 32. Name the electrolyte used in (/) dry cell (ii)
Nernst equation is : mercury cell.
Ans. (/) Paste of NH4CI and ZnCl2 in the zinc cylinder
RT [Mg^-^]
Ecell=E°^n "2-303 log and powdered MnO, + C surrounding the
2F [Ag+]2 graphite rod.
3/96 New Course Chemistry (XII)BSI91

(//) Moisi mercuric oxide (HgO) mixed with 43. How does H2- O2 fuel cell operate ?
KOH.
Ans. At anode : 2H2 (g) + 40H" {aq) ^4HoO + 4e
33. What is a primary cell ? Give an example. At cathode :
(CBSE 2008) O, (g) + 2F O ie) + At 40H {aq)
Ans. A primary cell is the cell in which the redox
Overall reaction :
reaction occurs only once and the cell becomes
dead after some time. It cannot used again. 2H2(g) + 02(g) ^ 2H20(/),
34. Give an example of a secondary cell. 44. State two advantages of H2 - O2 fuel cell over
ordinary cell.
(HP Board 2013)
Ans. (0 They do not cause any pollution.
Ans. Lead storage battery.
(/7) They have high efficiency of 60-70%.

w
35. Write the name of a cell used in small
watches.
45. Write the name of the electrolyte used in fuel
(Raj. Board 2011) cell.
Ans. Mercury cell (Rubon-Mallory cell).
Ans. Concentrated aqueous KOH solution.

Flo
36. Write the reaction taking place at the cathode
XI. Corrosion
when the lead storage battery is in use.

e
re
46. What is the overall electrochemical reaction
Ans. Pb02 (j) + SO^ iaq) + 4 {aq) ->r2e ->
taking place in rusting ?

F
PbS04 (s) + 2 H2O (/) Ans. 2 Fe (,v) + Ot (g) + 4 H"*" (aq)
ur 2 Fe-+ (aq) + 2 H,0 (/)

r
37. How much electricity in terms of Faraday is
required to carry out the reduction of one
mole of Pb02 ? fo
47. Out of zinc and tin which one protects iron
better even after cracks and why ?
ks
Ans. From the above equation, 1 mole of PbO->
Ans. Zinc protects better because oxidation potential
Yo
of zinc is greater but that of tin is less than that
oo

requires 2 Faradays. of iron.

38. What is the molarity of sulphuric acid before
B

48. Define corrosion. What is the chemical

discharge ? formula of rust ?
re

38/98 Ans. Corrosion is the slow eating away of the surface

Ans. Molarity = xIOOO =5-02M of the metal due to attack of atmospheric gases.
u
ad

lOO/i-294
The formula of rust is FeiO^. .vH20.
Yo

39. Lead storage battery is considered a 49. Why is chromium used for coating iron ?
secondary cell. Why ? Ans. Chromium is a non-corroding metal which
d

Ans. It is a secondary cell becau.se it can be recharged forms a protective layer on iron.
Re
in

after use. 50. What would happen if the protective tin

40. Write the products of electrolysis when dilute coatingover an iron bucket is brokenin some
F

sulphuric acid is electrolysed using Platinum places ?

electrodes ? Ans. In such a case, iron corrodes faster than it does
Ans. H2 (g) at cathode and O2 (g) at anode. in the absence of tin as oxidation potential of
Fe is greater than that of Sn.
X. Fuel cells
51. Which type of metal can be used in cathodic
41. Which cells were used in the Apollo space protection of iron against rusting ?
program ? What was the product used for ? Ans. A metal which is more electropositive than iron
such as Al, Zn, Mg.
Ans. H2 - Oo fuel cell. The product H2O was used 52. Rusting of iron is quicker in saline water than
for drinking by the astronauts.
in ordinary water. Give reason.
42. Define fuel cell ? (CBSE 2017)
Ans. In saline water, the presence of Na'*' and Cl" ions
Ans. A device in which the heat produced as a result increases the conductance of the solution in
of the combustion of a fuel (like H2 in presence contact with the metal surface. This accelerates
of Oi) is converted into electricity is called a the formation of Fe-^ ion and hence that of msi,
fuel cell. Fe.O^. -V H2O.
ELECTROCHEMISTRY 3/97

SHORT ANSWER QUESTIONS Carrying 2 or 3 marks

I. Electrolysis and Faraday’s laws 111. Kohlrausch's law and v

applications
1. What is electrolysis ? Discuss its mechanism
by taking the electrolysisof lead bromide as an 11. Define Kohlrausch’s law. How can it be used to

example, [Art. 3.2] find the degree of dissociation of a weak

electrolyte ? [Art. 3.10]
or What is electrolysis ? Give the reactions
12. How is the molar conductivity of a weak
occurring at the two electrodes during
electrolyte at infinite dilution determined ?
electrolysis of aqueous solution of sodium [Art. 3.10]
chloride. (Hr. Board 2011)
13. Define molar conductivity. Express the

ow
2. Define electrochemical equivalent. How is it relationship between degree of dissociation of
related to the equivalent weight of the element ? an electrolyte and its molar conductivities.
[Art. 3.3] [Art. 3.11]

II. Conductance and specifk, IV. Galvanic cells

e
Fl
equivalent and molar conductivities

re
14. What is an electrochemical cell ? Describe the
3. List the points of difference between metallic working of Daniell cell ?

F
conductors and electrolytic conductors ? [Art. 3.12]
ur
(HP Board 2011) [Art. 3.4] 15. Explain the function of salt bridge in an

or
electrochemical cell. [Art. 3.12]
4. Define the following and write the formula and sf
unit of each : (a) Conductivity 16. Explain difference between Galvanic cell
(Electrochemical cell) and Electrolytic cell ?
k
(b) Molar conductivity (c) Cell constant
Yo
(Maharashtra Board 2012. J & K Roarr '
oo

(MP Board 2012, .1 & K Board 2012)

[Art. 3.13]
[Art. 3.6]
B

17. Consider the following reaction :

5. Why is it necessary to platinize the electrodes Cu (J-) + 2 Ag-^ (aq)
re

of a conductivity cell before it is used for

2 Ag (.9) + Cu-"^ {aq)
conductance measurement ? [Art. 3.7]
u

(i) Depict the galvanic cell in which the given

ad

6. What is cell constant ? How is it determined ?

Yo

reaction takes place,

[Art. 3.7]
(n) Give the direction of flow of current.
7. What is conductivity water ? How is it obtained ? (Hi) Write the half-cell reactions taking place at
d

[Art. 3.7]
Re

cathode and anode.

in

8. Describe the construction and working of (CBSE Compt. 2018) [Fig. 3.13 page 3/37]
F

calomel electrode. (Maharashtra Board 2012)

[Page 3/42] V. Electrode potential, cell pc V
and electrochemical series "rid i'
9. With the help of a graph explain why it is not
applications
possible to determine a” for a weak electrolyte
by extrapolating the concentration-molar 18. What is understood by a normal hydrogen
conductance curve as for strong electrolyte. electrode ? Give its significance. [Art. 3.16]
[Art. 3.9] 19. What do you understand by the following ?
10. How does molar conductivity vary with (i) negative standard electrode potential
concentration for (/) weak electrolyte (ii) strong (//) positive standard electrode potential.
electrolyte ? Give reasons for these variations [Art. 3.16]
[Art. 3.9] 20. What is meant by reduction electrode potentials
or Define molar conductivity for the solution of of zinc and copper being — 0-76 V and + 0-34 V
an electrolyte. How does it vary with
respectively ? Can an aqueous solution of CUSO4
be stored in a zinc vessel ? Answer with reason.
concentration ? (CBSE 2022)
[Art. 3.16]
3/98 New Course Chemistry fXlltrosT^

21. Define electrode potential, oxidation potential 30. Write Nernst equation for the general
and reduction potential. Why is it not possible electrochemical change of the following type
to determine the absolute value of electrode at 25° C.
potential ? (Karnataka Board 2012) C/A + bB
ne'
^ c C + c/D
[Art. 3.16] (AP Board 2012) [Art. 3.21]
or Explain with example standard electrode 31. What is a concentration cell ? Give one example.
potential. (UP Board 2012) [Art. 3.16) How is the emf of such a cell calculated ?

22. What do you understand by Normal H>'drogcn [Art. 3.22]

Electrode ? Give its construction and working. 32. How Nernst equation can be applied in the
[Art. 3.16] calculation of equilibrium constant of any cell
reaction ? [Art. 3.23]
23. What is the difference between e.m.f. and
potential difference ? VII. Relation of Gibbs energy (DG")

w
(Liflarakhand Board 2012) [Art. 3.17] with cell potential (E^^.,,) and
24. What is an electrochemical series ? How does equilibrium constant (K)

F lo
it help us in predicting whether a redox reaction 33. Derive the relationship between Gibb’s free
is feasible in a given direction or not ? energy change and the cell potential.
[Art. 3.24]

ee
[Art. 3.18 & 3.19]
34. What is free energy and free energy change ?

Fr
25. What is an electrochemical series ? How does
Explain how is free energy related to cell
it help in calculating the e.m.f. of a standard potential ? Also explain its significance in
cell ? [Art. 3.18 & 3.19]

for
predicting the feasibility of the cell reaction.
ur
26. Give the reason why blue colour of copper [Art. 3.24]
sulphate solution is discharged slowly when an 35. Derive relationship between Gibbs energy and
s
iron rod is dipped in it. Given equilibrium constant of a reaction. [Art. 3.24]
ook
Yo

E° = (>34V, E° = -0.44 V VIII. Predicting products of

Cu-+/Cu Fe^-^/Fe
eB

electrolysis
[Art. 3.19]
36. Explain why electrolysis of an aqueous solution
27. The cell reaction as written is spontaneous if
of NaCl gives H2 at cathode and CI2 at anode ?
r

the overall EMF of the cell is positive. Comment

ou
ad

Write overall reaction

on this .statement. [Art. 3.19]
(E = -2-71V; E = -0-83V;
NiC/Na H2O/H2
Y

VI. Nernst eqn., concentration cells E° = +l-36V; E° = 1-23V)

cu/2cr H-"/0,/H20
and calculation of equilibrium constant
Re
nd

[Art. 3.25]
28. Daniell cell is a galvanic cell made of zinc and
Fi

copper electrodes IX. Some commercial ceils/batteries

(0 Write anode and cathode reactions in Daniell
37. Draw a well labelled diagram of “Dry cell".
cell
Give the reactions taking place at the anode and
(//') Nernst equation for the electrode reaction, the cathode. [Art. 3.26]
M"+ + ne- M is 38. Which cell is generally used in the hearing aids ?
Name the material of the anode, cathode and
2-303 RT
E = E‘ log the electrolyte. Write the reactions involved,
M"+/M «F [M"+ 1 or Write the name of the cell which is generally
Derive Nernst equation for Daniell cell. used in hearing aids. Write the reactions taking
place at the anode and the cathode of this cell.
(Kerala Board 2012)
(CBSE 2017)
[(i) Page 3/35 ; (ii) Art. 3.21]
or Write the name of the cell which is generally
29. What is the effect of change in (a) concentration used in invertors. Write the reactions taking
(b) temperature on the electrode potential of a place at the anode and the cathode of this cell.
given half-cell ? [Art. 3.20]
(CBSE 2017)
ELECTROCHEMISTRY 3/99

or Write the name of the cell which is generally Answer the following :
used in transistors. Write the reactions taking (/) Which cell is used in hearing acids ?
place at the anode and the cathode of this cell.
(ii) Which cell was used in Apollo Space
{CBSE 2017) [Art. 3.26] Programme ?
39. What type of battery is lead storage battery ? (Hi) Which cell is used in automobiles and
Write the anode and the cathode reactions and invertors ?
the overall reactions occurring in a lead storage (iv) Which cell does not have long life ?
battery. (CBSE 2011) [Art. 3.26.2] (CUSE 2016) [Art. 3.26, 3.27]
40. Give following information about ‘Nickel- [Ans. (0 mercury cell (ii) Fuel cell
Cadmium storage cell’ : (Hi) lead storage cell (iv) Dry cell]
(/) Material of the cathode (//) Material of the

ow
XI. Corro.sion
anode (Hi) Electrolyte used
(/v) Reactions involved at the anode and cathode
49. What is sacrificial protection of iron from
(v) Approximate voltage of the cell. corrosion ? Give one example. [Art. 3.27]
[Art. 3.26]
50. Explain ‘Iron is galvanised for protecting it from

e
41. What is mercury cell ? Give the electrode rusting’. [Art. 3.27]

Fl
re
reactions. [Art. 3.26] 51. Give a brief account of corrosion and its

F
or What is a Battery ? Describe the mercury cell. mechanism. (Pb. Board 2011.
[Art. 3.26] J & K Board 2012) [Art. 3.27]
ur
or
42. What are primary cells ? How does a dry cell or What is electrochemical theory of rusting ?
function ? [Art. 3.26] Explain.
sf (HP Board 2011)
or What are secondary cells ? Describe the Nickel- 52. What is corrosion ? Explain any four factors
k
Cadmium cell.
Yo
[Art. 3.26] affecting corrosion.
oo

43. Explain discharging and recharging of lead (Chhatlsgiirli Board 2011) [Art. 3.27]
storage battery. (HP Board 2011)
B

53. Describe any two of the techniques used for

[Art. 3.26.2] preventing corrosion of metals. [Art. 3.27]
re

X. Fuel cells 54. Rusting of iron is quicker in saline water than

in ordinary water. Explain. [Art. 3.27]
u
ad

44. What is a fuel cell ? Give the constniction and 55. We can use aluminium in place of zinc for
Yo

working of a fuel cell. [Art. 3.26] cathodic protection of rusting. Comment.

45. Describe the construction of a HtOt fuel cell and [Art. 3.27]
d
Re

the reactions taking place in it. [Art. 3.26.3] 56. What is corrosion ? Give mechanism
in

46. What are the difficulties in the construction of (electrochemical phenomenon) of rusting of
F

a fuel cell ? What are its advantages over the iron. What do you understand by sacrificial,
other cells ? [Art. 3.26] cathodic and barrier protection of corrosion ?
47. Leclanche cell, Lead storage cell and Fuel cell [Art. 3.27]
are galvanic cells having different uses 57. Iron does not rust even if the zinc coating is
(0 Among these Leclanche cell is a primary cell broken in a galvanised iron pipe but rusting
and Lead storage cell is a secondary cell. Write occurs much faster if the tin coating over iron
any two differences between primary cells and is broken. Explain. [Art. 3.27]
secondary cells. Hint. Zinc is more electropositive whereas tin
(ii) What is a fuel cell? is less electropositive than iron.
(Hi) Write the overall cell reaction in H^-Ot XII. Miscellaneous
fuel cell. (Kerulu Board 2012) [Art. 3.26]
48. From the given cells : 58. State reasons for the following :
Lead storage cell, mercury cell, fuel cell and (/) Rusting of iron is said to be an electro
dry cell chemical phenomenon
3/100 'prta.dee^'^ New Course Chemistry CXII)S9San
(/i) For a weak electrolyte, its molar conductance (//) Write overall cell reaction for lead storage
in dilute solutions increases sharply as its battery when the battery is being charged,
concentration in solution is decreased. [(i) Art. 3.9 (h) Art. 3.26.2]
[Art. 3.27 & 3.9] 60. Explain (Definition and Formula)
59. (0 For a weak electrolyte, molar conductance in (a) Kohlrausch law (b) Faraday’s first law of
dilute solution increases sharply as its concen- electrolysis. (MP Board 2012)
tr^ion in solution is decreased. Give reason. [{fl)Art. 3.10 (b) 3.3]

ow
LONG ANSWER QUESTIONS Carrying 5 or more marks

1. (a) The resistance of a conductivity cell 4. (fl) Write the anode and cathode reactions and
containing 0 001 M KCl solution at 298 K is the overall reaction occurring in a lead storage

e
1500 What is the cell constant if the battery.

re
conductivity of 0-001 M KCl solution at 298 K (b) A copper-silver cell is set up. The copper
is 0-146 X 10-^Scm-'? ion concentration is 0-10 M. The concentration

Flr
of silver ions is not known. The cell potential

F
(b) Predict the products of electrolysis of the
when measured was 0-422 V, Determine the
solution of H2SO4 with platinum electrodes.
ou concentration of silver ions in the cell. (Given :
(CBSE 2007)
= -fO-34V)-

sr
E° = +0-80V. E°
[Ans. (a) 0-2I9 cm'‘ (b) Page 3/4 and 3/66] Ag+ / Ag Cu-^/Cu

fo
2. («) Define molar conductivity of a substance and (CBSE 2010)
describe how for weak and strong electrolytes,
k [(a) Page 3/69 (A) Ans. 7-1 x 10"2 M]
oo
molar conductivity changes with concentration of
5. (a) What is a fuel cell ? What is its main
solute. How is such change obtained ?
Y
advantage ?
(fl) A voltaic cell is set up at 25“C with the
reB

(A) What are the reactions occurring at the

following half cells: cathode and anode of a Lechlanche cell ?
Ag^ (0.001 M) I Ag and Cu-+ (0.01 M) I Cu
uY

(c) In the button cell widely used for watches

What would be the voltage of this cell ? (E°(,gu and other devices, the following reaction takes
= 0.46 V) (CBSE 2009) place :
ad
do

m Art. 3.6.3 & Art. 3.9 (A) 0-312 V] Zn (i) + Ag20 (s) -I- H^O (/) 4

3. (a) Corrosion is essentially an electrochemical Zn^"*" (aq) -f 2 Ag (i) + 2 OH"* (aq)

in

phenomenon. Explain the reactions occurring Give the cell representation and determine the
Re

during corrosion of iron kept in an open value of K^. for the above reaction using the
F

atmosphere. following data:

(A) One half-cell in a voltaic cell is constructed Ag20 (s) + H^O (/) + 2 e' 4

from a silver wire dipped in silver nitrate 2 Ag {s) + 2 OH- {aq) (E“ = 0-344 V)
solution of unknown concentration, Its other Zn^-^ (aq) + 2e~ ■> Zn (5) (E“ = - 0-76 V)
half-cell consists of a zinc electrode dipping in [Ans. (a) Art. 3.26.3. (A) Art. 3.26.1
1.0 M solution of Zn(N03)2. A voltage of 1.48 (c) Q. 3.6, page 3/111
V is measured for this cell. Use this information Zn (s) 1 Zn^"^ (aq) 11 Ag20 (s) I OH" (aq)
to calculate the concentration of silver nitrate E‘’ = E' -E* = M04V
solution used. AgjO/Ag Zn^-^/Zn

Zn (s) + AgiO (s) + H2O (/) > Zn^"^ (aq)

= -h0-80V ).
E”zn^^,Zn=-^-^"^’EViAg -I- 2 Ag (s) + 2 OH“ (ag)
(CBSE 2009) 0-059
n = 2, Apply E" = log ,
[(a) Art. 3.27 (A) Similar to Solved Problem 4, n

Page 3/52] [Ans. 4-4 x 10"2 M] K^. = 2-34 X 10^7]

ELECTROCHEMISTRY 3/101

6. (a) State Kohlrausch law of independent (/;) Given are the conductivity and molar
migration of ions. (/?) What is a primary battery? conductivity of NaCl solutions at 298 K at
Give one example, different concentrations ;

(c) Three electrolytic cells A, B and C containing Concentration Conductivity Molar

electrolytes Z11SO4, AgN03 and CuSOj M S cm“*
respectively were connected in series. A steady conductivity
current of 1 -5 A was passed through them. 145 S cm^ mol"*
g of Ag were deposited at the cathode of cell B.
0100 106-74 X 10-^ 106-7
(0 How long did the current How? (//) What
mass of copper and zinc were deposited? 0-05 55-53 X 10^ IIM

(At mass of Cu = 63 u, Zn = 65-3 u, Ag = 08 u) 0-02 23-15 X 10^ 115-8

(Assam Board 2013)

Compare the variation of conductivity and molar
[(a) Art. 3.10 (/>) Art. 3.26 conductivity of NaCl solutions on dilution. Give

w
(c) 863-7 s, Cu = 0-423 g, Zn = 0-438 g] reason,

7. (a) How do you explain with the help of a graph, (c) 0-1 M KCl solution offered a resistance of
the increase in the value of molar conductivity

Flo
100 ohms in a conductivity cell at 298 K. If the
with dilution in case of strong and weak cell constant of the cell is 1-29 cm"*, calculate
electrolyte?

ee
the molar conductivityof KCl solution.
{b) Calculate the e.m.f. of the cell at 298 K in

Fr
(CBSE Sample Paper 2018)
which the following reaction takes place :
Ni (5) + 2 Ag+ (aq) [0-002 M]
[Ans. (a) ^cacij
4

Ni2+ (aq) [0-160 M]2 Ag (s)

for
ur
(Given = 1-05 V) (Assam Board 2013) (b) On dilution, conductivity decrea.ses
[(a) Art. 3.9 (b) 0-914 V] whereas molar conductivity increases. NaCl
s
is a strong electrolyte and is completely
k
8. (fl) State Faraday’s first law of electrolysis. How
Yo
ionized at all dilutions. The decrease in
oo

much charge in terms of Faraday is required for

the reduction of 1 mol of Cu^^ to Cu. conductivity is due to decrease in the number
eB

of ions per cni^ of the solution. Molar

(h) Calculate the emf of the following cell at conductivity is product of conductivity and
298 K :
volume containing 1 mole of the electrolyte.
r

Mg (5)IMg2+(0-l M) I Cu2+(0-01 M)lCu (s) Increase in volume on dilution is much more

ou
ad

[Given = + 2-71 V, 1 F = 96500 C mol-*] than decrease in conductivity. Hence, the

Y

(CBSE 2014) [(a) Art. 3.3, 2 F (b) 2-6805 V] product increases with dilution. Also, molar
9. (a) For the reaction, 2 AgCl (5) + H2 (g) (1 atm) conductivity increases with dilution because
Re
nd

interionic attractions decrease with dilution.

> 2 Ag (s) + 2 H+ (0-1 M) + 2 Cl" (0-1 M)
AG" = - 43600 J at 25"C. Calculate the e.m.f. Cell const. l-29cm-*
Fi

of the cell (log 10“” = - n). (c) Conductivity (k) =

Resistance 100 ohm
(b) Define fuel ceil and write its two advantages. = 1-29 X 10"^ ohm"* cm -1
(CBSE 2018)
Molar conductivity (a,„)
[Ans. (a) AG"=-«FE“
ceU
KXlOOO (l-29xl0-2)(1000)
AG" 43600
or E
Adi “
= 0-226 V Molarity 0-1
-nF -2x96500
- 1-29 X 10^ ohm"* cm" mol"*]
E
0 0-0591. (0-1)- 11. (a) The e.m..f of the following cell at 298 K is
= E log = 0-2851V
cell cell
2 * (1) 0-1745V

(b) Art. 3.27 ] Fe (j)/Fe^+ (0-1 M)/H-" (x M)/H2(g) (1 bar)/Pt(5))

10. («) Applying Kohlrausch law of independent Given E° = _ 0-44 V. Calculate the H+
migration of ions, write the expression to Fe^+ZFe “ . ...
determine the limiting molar conductivity of ion concentration of the solution at the electrode
calcium chloride. when hydrogen is being produced.
3/102 ^^leuUcp-'A New Course Chemistry CX1I)E9SI91

(b) Aqueous solution of copper sulphate and 12. (rt) Calculate the degree of dissociation of
0-0024 M acetic acid if conductivity of this
silver nitrate are electrolysedby 1 ampere cuncnt
solution is 8-0 x 10“-^ S cm”*.
for 10 minutes in separate electrolytic cells. Will
the mass of copper and silver deposited on the Given = 349*6 S cm^ moH,
H
cathode be same or different ? Explain your X = 40-9 S cm- mol”'
answer. (CBSE Sample Paper 2019) CH^COO”
[Ans. (a) Fe + 2 H+ ■> Fe2+ + H, {b) Solutions of two electrolytes ‘A' and ‘B’ are
diluted. The limiting molar conductivity of ‘B’
0-591 [Fe2+] increases while that of ‘A’ increases to a much
E cell = Ecell log
n [H"]2 larger extent comparatively. Which of the two
is a strong electrolyte ? Justify your answer.
r" ¥ri« _ I?"
(CBSE Sample Paper 2019)
cell H^/H^ Fe^'^/Fe
[Ans. (fl) a“ (CH3COOH) = 349-6 + 40-9

w
= 0 - (- 0-44) = 0-44 V
= 390-5 S cm2
0-1745 = 0-44 -
0-0591 , 0-1
log KXlOOO (8-OxlO~^)(1000)
2

Flo
Ai
m
=
C 0-0024
On solving it gives log [H'*'] = - 5 = 33-33 S cm2
or [H+] = 1()-*^M

ee
A^_m 33-33

Fr
ib) The mass of copper and silver deposited a = = 0-085
at the cathode will be different. This is A' 390-5
m

because when the same quantify of electricity (/?) B is a strong electrolyte because limiting

for
ur
is passed through electrolytic solutions molar conductivity of a strong electrolyte
coDvected in scries, the masses dcpo.sited are increases gradually while that of a weak
in the ratio of their equivalent masses]
s
electrolyte increases rapidly with dilution]
k
Yo
oo

CASE-BASED VERY SHORT/SHORT QUESTIONS

eB

CASE 1. Molar conductivity of .strong electrolytes of the weak electrolyte at concentration c can be
is found to vary with concentration according to Debye- calculated. Knowing the value ofe and a, equilibrium
r

Huckel Onsager equation. By plotting molar constant (K^,) of the weak electrolyte can be calculated.
ou
ad

conductance (a„,) versus square root of concentration, Based on the above paragraph, answer
Y

the molar conductance of the strong electrolytes at questions no. 1 to 4 :

1. Write Debye-Huckel Onsager equation in the mo.st
infinite dilution (a®^) called ‘limiting molar
Re
nd

common form. What do different symbols

conductance' can be found out by extrapolation of the signify ?
Fi

linear plot (near high dilutions) to zero concentration. 2. The molar conductances of NaOH, NaCl and
However, in case of weak electrolytes, such an
BaCI-, at infinite dilution are 2-48! x lO”^, 1*265
extrapolation to zero concentration cannot be done as X 10”2 and 2-800 x 10”2 S mol"' respectively.
at high dilutions, the plot for weak electrolyte is almost
Calculate a®„, for Ba(OH>2.
parallel to the molar conductance axis. Hence, to tmd
3. The molar conductivity of 0-025 mol L”* solution
the value of a°III ofwcak electrolytes, studies were made of methanoic acid is 46*15 Q"* cm^ mol”'. What
will be its dissociation constant ?
by Kohlrausch who put forward a law known after his
name as Kohlrausch’s law. Kohlrausch’s law helps to 4. A strong electrolyte is already completely ionized
in the solution. Then why its molar conductivity
calculate the a° values of weak electrolytes from the
increa.ses on dilution ?
A®,
III
values of suitable strong electrolytes. Fuither.
knowing the A^j^ value of the weak electrolyte at a CASE 2. Any cell or battery that we use as a
particular concentration r and also using the a° value III
source of electrical energy is basically a galvanic cell.
of that weak electrolyte, the degree of dissociation (a) But every galvanic cell is not suitable for commercial
ELECTROCHEMISTRY 3/103

use. For a battery to be of practical use. it should be CASE 3. When a D.C. voltage from a battery is
reasonably light, compact and its voltage .should not applied between the two electrodes dipped in the
vary much during its use. Batteries are broadly classified solution or melt of an electrolyte, the ions of the
as “Primary batteries' and ‘Secondary batteries.’ Primary electrolyte move towards the opposite electrodes. The
batteries are those in which the redox reaction occurs process is called as ‘electrolysis’. The process is of great
only once and it becomes dead over a period of time importance in laboratory as well as in chemical industry.
and cannot be reused. Two common examples of this For example, pure copper is obtained from impure
type are dry cell and mercury cell. Secondary batteries copper by this process. Many metals like Na, Mg, A1
are those which can be recharged by passing an electric etc. are produced on a large .scale by this process. If a
cunent through them and hence can be used over and number of ions are present in the solution, the ion with
again. Two well known examples of this type arc lead lowest discharge potential is liberated at the electrode.
Similarly, the products of electrolysis depend upon
storage battery and nickel-cadmium storage cell.
whether the electrodes are attacked by the ions or the
Another source of electrical energy is from fuel cells.
electrodes are inert. Quantitatively, the amounts of the
Their advantage over the thermal plants is that they do
substances liberated at the electrodes during electrolysis
not cause pollution and are comparatively simple to
depend upon the amount of quantity of electricity
operate. The reactants are fed continuously and products passed. These studies were made by Faraday.

w
are removed continuously. They are the devices which Alternatively, the products of electrolysis can also be
convert the energy produced during the combu.stion of predicted by comparing the .standard electrode potentials

F lo
fuel, like hydrogen, methane, methanol etc. directly into of the probable reactions occurring at the electrode.
electrical energy. One such cell which has been very However, in case of liberation of gases at the electrodes,
successful and was used in Apollo moon flights is some overvoltage is required.
H->-02 fuel cell.

e
Based on the above paragraph, answer

Fre
Based on the above paragraph, answer questions no. 9 to 12 :
questions no. 5 to 8 : for
9. What will be the products of electrolysis of CuSO^
5. Why mercury cell gives a constant voltage of solution using (/) copper electrodes (//) platinum
electrodes ?
1-35 V throughout its life ?
r
6. Write the reactions occuring at the anode and 10. What is overvoltage ■?
You
oks

cathode of a lead storage battery during discharge. 11. A current of 4 ampere was passed for 1-5 hours
eBo

through a solution of copper sulphate when

7. Write the reactions occuring at the anode and 3-2 g of copper was deposited. Calculate the
cathode of the H2-O2 fuel cell and the overall current etficicncy.
reaction. What is its voltage ?
ad
our

12. A current is passed through two cells connected

8. The theoretical EMF of H^-02 fuel cell is 1 -23 V in .scries. The first cell contains X (N03)3 (aq)
and the heat of combustion of FU in O2 is and the second cell contains X (N03)2
- 285-8 kJ mol"*. What is the theoretical relative atomic mas.ses of X and Y are in the ratio
Re

1 : 2. Calculate the ratio of mass of X to that of Y

dY

efficiency of the H2-O2 fuel cell ?

liberated.
Fin

ANSWERS

1. Debye-Huckel Onsager equation is = (2-800 X 10"-) + 2(2-481 x 10"-)

a"m = A - 2(1-265 X 10"-) Sm^ mol"'
(2-800 4-962 - 2-530) x 10"^ S m^ mol"*
where = molar conductivity at concentration
= 5-232 X kHs m- mol"'.
c mol L"'
3. A”,(HCOOH)=rHCOO" +r
= molar conductivity at infinite dilution 11

called limiting molar conductivity. = 54-6 H- 349-6

A = constant depending upon the type of = 404-2 S cm^ mol"’
electrolyte.
A^m 46-1
2. a“, Ba(OH)2 = a;;, (BaC!,) + 2 a°, (NaOH) a =
404-2
= 0-114
A
III

- 2 a; (NaCI)

I
3/104 New Course Chemistry (XIl)CZSMl

HCOOH ± HCOO" + 7. At anode : 2 H2 (g) + 4 OH {aq)

Initial cone. c mol L 0 0 4 H2O (/) + 4 e-
Cone, at eqm. c(l - a) ca ca At cathode : (g) + 2 H2O (/) + 4 e
4 OH- iaq)
K
car a ca~ 0-025 X (0-114)^
c(l-a) I-a 1-0-114 Overall reaction : 2 H2 (g) + O2 (g) ■>

= 3-67 X 10^ 2 H2O (/)

Voltage of H2-0-7 fuel cell = 0-9 V
4. At high concentration, ion.s are close to each other
and, therefore, there are strong interionic 8. AG = - /iFE =-(2)(96500Cmo|-*)(l-23 V)
attractions. On dilution, ions go far apart,
therefore, interionic attractions decrease, i.e., ions = - 237390 CV mol-‘ = - 237390 J mol"'
become free to move. Hence, conductance = - 237-390 kJ mol-' (1 CV= I J)
increases.
AG - 237-39 kJmor
xl00 =
5. It involves the reaction between Zn and HgO in = 83%
AH -285-8 kJ mol-'

w
presence of KOH. The OH“ ions used in the
reaction at anode (Zn + 2 OH ^ ZnO + H,0 9. With platinum electrodes : Cu at cathode, O2 at

F lo
anode
+ 2 e~) are reproduced in the reaction at cathode
(HgO + H2O + 2 c ^ Hg + 2 OH-). Thus, With Cu electrodes ; Cu at cathode, no gas at
overall reaction does not involve any ion whose anode

concentration may change. Hence, it gives a For explanation, refer to example (v), page 3/5

e
Fre
constant voltage throughout its life. and example (v/), page 3/6.
6. At anode : Pb is) + SOl~ iaq) for
10. Over voltage is the difference between the
potential required for the evolution of the gas and
PbS04(j) + 2e its standard reduction potential (Refer to page
r
At cathode : PbO^ (s) + SO4 (aq) 3/65).
You
oks

+ 4 H-" (aq) + 2 e~ 11. Refer to Solution to Problem 11, page 3/11.

eBo

-> PbS04 (^) + 2 HjO 12. Refer to Solution to Problem 14, page 3/12.
ad
our

CASE-BASED MCQs AND ASSERTION-REASON QUESTIONS

CASE 1. A number of devices or gadgets require order. The series thus obtained is called ‘electrochemical
electrical energy. One of the sources of this energy is series’. This series helps to calculate the standard EMF
Re
dY

‘galvanic cell’. A galvanic cell is an electrochemical of a cell choosing any two electrodes as well comparing
cell that converts the chemical energy of a spontaneous relative oxidizing and reducing powers or relative
Fin

redox reaction into electrical energy. The most common activities of metals etc. The EMF of the cell or electrode
galvanic cell is Daniell cell based on the redox reaction, potential of the electrodes depends upon the
Zn -f CUSO4 ZnS04 + Cu. The redox reaction concentrations of the electrolytes used in the electrodes
consists of two half reactions, one is oxidation-half and and temperature. These studies were made by Nemsl.
the other is reduction-half. The aiTangemeni is set-up We may have a cell in which the electrodes are same
consisting of electrodes connected by a wire for flow but differ only in the concentrations of the electrolytes
of electrons from one electrode to the other and flow of taken in them. Such cells are called ‘concentration cells’.
ions through a salt bridge from one solution to the other. Based on the above paragraph, answer
The tendency of any electrode to lose or gain electrons questions no. 1 to 4 :
is measured with respect to a standard hydrogen 1. One of the electrodes of a galvanic cell consists
electrode taken as a reference electrode and whose
of Cu/Cu-"*" (IM) electrode. Which of the
electrode potential is taken as zero under standard
following electrodes should be combined with the
conditions (1 M concentration, 1 bar pressure). All these
copper electrode to have maximum standard emf
standard electrode potentials (expressed as standard of the cell ?
reduction potentials) are arranged in the increasing

1
 ELECTROCHEMISTRY 3/105

(Given ; Standard eiecirode potentials : Q

relationship between standard EMF of the cell (Eceii)

Cii^-^/Cu = 0-34 volt. Ag+/Ag = 0-80 volt,
and equilibrium constant (K^.) can be found. Another
Ni-+/Ni = - 0-25 V. Fe“+/Fe = - 0-44 volt.
important aspect is that when electrical energy is
Zn^+/Zn = - 0-74 volt) produced in a cell reaction or electrical work is done,
(c() Ag-'-ZAg (/;) Ni-^/Ni this is at the cost of Gibbs free energy (AG) of the
(c) Fe--^/Fe (J) ,Zn2+/Zn reaction which decreases. Thus, there exits a relationship
2. The pH at which hydrogen electrode will show between decrease in Gibbs energy of a cell reaction
an electrode potential of - 0-118 V when H2 gas (- AGp and the electrical work done (/jFE^.gn). As
is bubbled at 298 K and I atm pressure will be
(fl)2 (b)3
is related to equilibrium constant (K^^^) and E°g|]
is also
related to decrease in free energy of the cell reaction
(r)4 id) 5
(AG“), therefore, AG° is related directly to equilibrium
Choose the correct option out of the four options constant of the cell reaction. Thus, the three quantities,
given below:
(a) Both Assertion (A) <md Reason (R) are true and E^,g|], AG“ and K^. are inter-related with each other.

w
Reason (R) is the correct explanation of Assertion These help in a number of calculations.

F lo
(A). Based on the above paragraph, answer
(b) Both Assertion (A) and Reason (R) are true but questions no. 5 to 8 :
reason (R) is not the correct explanation of 5. Which of the following relationship is not
Assertion (A), correct ?

ree
(c) Assertion (A) is true but Reason (R) is false.
(d) A.s.sertion (A) is false but Reason (R) is true.
3. Assertion. In Cu-Zn galvanic cell, zinc electrode
for F
(^) EceU =
00591
n
logKc

is called a negative pole. (b) - AG° = RT In K^.

Your
Reason. In Cu-Zn galvanic cell, copper electrode (c)-AG“ = nF e;,ii
ks

acts as anode.
eBoo

4. Assertion. The electrode potential (reduction AH OE

id) E=- +T
potential) of any electrode decreases with increase wF 3T yp
in concentration of the electrolytic solution.
ad
our

6. The free energy change for the decomposition

Reason. With increase in concentration, ions are reaction
less free to move about in the solution.

2 AI2O3 ■> -A1 + 0-, is AG = + 960 kJ

Re

CASE 2. When a cell reaction takes place in two 3 ^

Y

half-cells, in one of the electrodes, metal ions from the The potential difference needed to reduce AI2O3
Find

solution are deposited in one of electrode (reduction at SOOT is (F = 96500 C mor')

half reaction occurs) whereas in the other half-cell, metal (a) 4-97 V (b) 3-32 V
from the electrode passed into the .solution as metal ions
(c) 9-95 V (d) 2-49 V
(oxidation half reaction occurs). Thus, concentration
Choose the correct option out of the four options
of metal ions in one of the electrodes keeps on
given below;
decreasing whereas in the other it keeps on increasing.
(a) Both Assertion (A) and Reason (R) are true and
Consequently, the electrode potential of one of the
Reason (R) is the correct explanation of Assertion
electrodes keeps on decreasing whereas that of the other
(A).
electrode keeps on increasing. Ultimately, a stage is
ib) Both Assertion (A) and Reason (R) are true but
reached when the electrode potentials of the two
reason (R) is not the correct explanation of
electrodes become equal and the current stops flowing,
Assertion (A),
i.e., EMF of the ceil becomes zero. The cell reaction is
(c) Assertion (A) is true but Reason (R) is false.
said to have attained equilibrium : Applying Nernst
id) Assertion (A) is false but Reason (R) is true.
equation to the cell reaction in equilibrium, the

i
3/106 New Course Chemistry (X1I)CSX91

8. Assertion. Greater the E”cell of a reaction, greater

7. Assertion. E
cell is intensive property white AG°
is extensive property. is the decrease in free energy of the cell
reaction.
Reason. E°cell does not change if the reaction is
Reason. Decrease in free energy is the maximum
multiplied with an integer whereas AG° changes work that can be obtained from a cell.
if the reaction is multiplied with an integer.

ANSWERS

\.{d) 2. (a) 3. (c) 4. id) 5. (a) 6.{d) 7. (a) S.{b)

HiNTS/EXPLANATIONS For Difficult Questions

1. Cu/Ag cell = 0-80 - 0-34 = 0-46 V, increases, electrode potential increases.

Cu/Ni cell = 0-34 + 0-25 = 0-59 V 0-0591

w
5. e!cell (not E^eii) - logKj.
Cu/Fe cell = 0-34 + 0-44 = 0-78 V, n

F lo
Cu/Zn cell = 0-34 + 0-74 = 1-18 V (At equilibrium, E^^ji = 0)
2. Refer to Solution to Problem 5. page 3/52. 6. Refer to Solution to Problem 2, page 3/62.
7. Both A and R are correct and R is the correct
3. Correct R. In Cu - Zn galvanic cell, copper
explanation of A.

ree
electrode acts as cathode.

4. Correct
0-591
A. Electrode potential (E) =
for F
8. Correct explanation. AG° and E°cell
as AG° =-«FE cell
are related

log [M”'*’]. If concentration of M

H-l-
E“ + ■
n
Your
ks
eBoo
ad
our
Re
Y
Find

1
ELECTROCHEMISTRY 3/107

WITH
ANSWERS

I! ll^Ns

NCERT INTEXT UNSOLVED QUESTIONS & PROBLEMS

Q. 3.1. How w'ould you determine the standard electrode potential of I Mg ?

w
Ans.
We will set up a cell consisting of Mg I MgS04( 1 M) as one electrode (by dipping a magnesium wire in
1 M MgS04 solution) and standard hydrogen electrode Pt. H2 (I atm) I (1 M) as the second electrode

F lo
(simihu" to Fig. 3.16) and measure the EMF of the cell and also note the direction of deflection in the
voltmeter. The direction of deflection shows that electrons flow from magnesium electrode to hydrogen
electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode.

ee
Hence, the cell may be represented as :

Fr
MglMg2-^(l M)1IH+(1 M)IH2, (I atm), Pt
po = po
ceU ^
H+.I/2H2 ^ Mg2+,Mg
for
ur
Put = 0
s
Hence, po = -E°
ook

cell
Yo

Q. 3.2. Can you store copper sulphate solution in a zinc pot ?

eB

Ans. Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows ;
Zn (5) + CUSO4 (aq) ZnS04 (aq) + Cu (5)
r

Thus, zinc reacts with CUSO4 solution. Hence, we cannot store copper sulphate solution in a zinc pot.
ad
ou

Alteniaiively, E%Zn-^'.Zn
2- T = -0-76 V, EVCu^^.Cii
2^ ^ = 0-34 V
Y

To check whether zinc reacts with CUSO4 solution, i.e., whether the following reaction takes place or not
Re
nd

Zn (s) + CUSO4 (aq) 4 ZnS04 (aq) + Cu (.s),

find the EMF of this cell reaction. The cell may be represented as : Zn I Zn"'*’ 11 I Cu
Fi

po = po -E'
^ celt ^ Cu^+.Cu Zn^+.Zn = 0-34 V-(-0-76 V) = M0V
As E°^^[| is positive, the reaction takes place and we cannot store.
Q. 3.3. Consult the table of the standard electrode potentials and suggest three substances that can oxidize
ferrous ions under suitable conditions.
Ans. Oxidation of ferrous ions means that the following reaction should occur ;
Fe^-^ ? Fe^+ +e~; E“ ox = -0-77 V
Only those substances can oxidize Fe^'*' to Fe^'*' which are stronger oxidizing agents and have positive
reduction potentials greater than 0-77 V so that EMF of the cell reaction is positive. This is so for
elements lying below Fe-^'*’/Fe-'^ in the electrochemical series, e.g., Br2, CU and F2.
Q. 3.4. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
1
Ans. For hydrogen electrode, H'*' + e~
3/108 New Course Chemistry (XlI)SZsI9B

a0591 1
Applying Nerast equation, E = E“ 1 log
H+ .-Ho n [H*l
2 ^
00591 1
= 0- log {pH = 10 means [H^ = 10 M}
1 10-10
= - 00591 X 10 = - 0-591 V.

Q. 3^. Calculate the emf of the cell in which the following reaction takes place ;
Ni(s) + 2Ag+(0-002M) Ni2+(0-160 M) + 2Ag(s)
Given that E®cen = 1-05 V

ow
Ans. Applying Nemst equation to the given cell reaction,
0-0591 [Ni2+] 0-0591, 0-160 = 0-0591 ,
^cell = E“ cell n
log
[Ag+]2
= 1-05 V -
2
log 7
(0-002)2
1-05 -
^— log (4x10^)

e
= 1-05 - (4-6021) = 1-05 - 0-14 V = 0-91 V

re
Q. 3.6. The cell in which the following reaction occurs :

Frl
2Fe^'*’(a^) -f 2I~(aq) > 2Fe2+(o^) + l2(s)

F
has = 0-236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant
of the cell reaction.
ou
or
Ans. 2Fe3+-i-2e- >2Fe2+or2r > l2 + 2e"
Hence, for the given cell reaction, n = 2.
A^G® = - nFE®j.eu = - 2 x 96500 x 0-236 J = - 45-55 kj mol"^
kfs
A^G® = -2-303 RT log
oo
-45-55 kJ mol-i
Y
= 7-983
or logK^=- 2-303 RT
eB

2-303x8-314x 10-3 y K"* mol'ix298 K

= Antilog (7-983) = 9-616 x lO’
ur

Q. 3.7. Why does the conductivity of a solution decrease with dilution ?

oY

Ans. Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On
ad

dilution, the number of ions per unit volume decreases. Hence, the conductivity decreases.
Q. 3.8. Suggest a way to determine the a®^ value of water.
d

Ans. a°„(H20) = X®jj^+X“ OH-

in
Re

We find out A®„ (HQ), A®^ (NaOH) and a®^ (NaCl). Then
F

A“;„ (H2O) = A®„ (HCl) -t- A®^ (NaOH) - A®„ (NaCl)

Q. 3.9. The molar conductivity of 0-025 mol L“* methanoic acid is 46-1 S cm2 moi-i. Calculate its degree
of dissociation and dissociation constant Given X° (H"^) = 349-6 S cm2 and A,® (HCOO")
= 54-6 S cm2 mor*.
Ans. A®^ (HCOOH) = X,® (H+) + A.® (HCOO-) = 349-6 + 54-6 S cm2 ^ 404.2 s cm2
Am “ ^ mol“* (Given)
46-1
a = = 0-114
aOm 404-2

HCOOH HCOO- + H+
Initial cone. c mol L-‘
Cone, at eqm. c (1 - a) ca ca

ca.ca c(x? 0-025 X (0-114)2 = 3-67x10-4

c (1 - a) 1-a 1-0-114
ELECTROCHEMISTRY
3/1^09
Q. 3.10. If a current of 0*5 ampere flows through a metallic wire for 2 hours, then how many electrons
flow through the wire ?
Ans. Q (coulombs) = I (ampere) x t (sec) = (0-5 ampere) x (2 x 60 x 60 s) = 3600 C
A flow of 1 F, i.e., 96500 C is equivalent to flow of 1 mole of electrons, i.e., 6 02 x 10^^ electrons
6-02x10^3
3600 C is equivalent to flow of electrons = X3600
96500

= 2*246 x 10^^ electrons

Q. 3.11. Suggest a list of metals that are extracted electrolytically.
Ans. Sodium, calcium, magnesium and aluminium.
Q. 3.12. Consider the reaction :

CrjO^- +14H+ +6c" ^ 2 Cr^ + 7 H2O

What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr20|" ?

w
Ans. From the given reaction, 1 mol of Cr20^“ ions require 6 F = 6 x 96500 C = 579000 C of electricity for

F lo
reduction to

Q. 3.13. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are

involved during recharging.

ree
Ans. A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide

F
(Pb02) and 38% solution of sulphuric acid as electrolyte. When the battery is in use. the following
reactions take place :
At anode ; Pb (5) + SO|“ {aq) ^ PbS04 {s) + 2e~
for ...(/)
r
At cathode : Pb02 (.r) + SO|" {aq) + 4 W*{aq) + 2e"
You
^ PbS04 (^) + 2H20(/) ...(»)
oks

Overall reaction : Pb {s) + Pb02 (s) + 2 H2SO4 {aq) > 2 PbS04 {s) + 2 H,0 (/)
eBo

On charging the battery, the reverse reaction takes place, i.e., PbS04 deposited on the electrodes is
converted back into Pb and Pb02 and H2S04is regenerated.
Reverse of reaction (/) will be reduction and hence will take place at cathode. Reverse of reaction (/V)
our
ad

will be oxidation and hence will take place at anode.

Q. 3.14. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Ans. Methane and methanol.
dY
Re

Q. 3.15. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Ans.
The water layer present on the surface of iron (especially in the rainy season) dissolves acidic oxides of
Fin

the air like CO2, SO2 etc. to form acids which dissociate to give ions

H2O + CO2 > H2CO3 2H++CQ2-

In the presence of ions, iron stans losing electrons at some .spot to form ferrous ions, i.e., its oxidation
takes place. Hence, this spot acts as the anode ;
Fe (.v) » Fe-"^ {aq) + 2 e~
The electrons thus reic; .ed move through the metal to reach another spot where ions and the dissolved
oxygen take up these ectrons and reduction reaction takes place. Hence, this spot acts as the cathode :
C>2 +4 {aq) + 4 ^ 2 H2O (/)
The overall reaction is : 2 Fe (s) + O2 (g) + 4 H"*" {aq) ■>
2 Fe2+ (aq) + 2 H2O (/)
Thus, an electrochemical cell is set up on the surface.
FeiTOus ions are further oxidized by the atmospheric oxygen to ferric ions which combine with water
molecules to form hydrated feme oxide, Fe203..r H,0, which is rust.
3/110 “Pnaxiee^'^ New Course Chemistry (XII)BZsl9]

NCERT EXERCISE

Q. 3.1. Arrange the following metals in the order in which they displace each other from the solution of
their salts : Al, Cu, Fe, Mg and Zn
Ans. Mg, Al, Zn, Fe, Cu, Ag

Q. 3.2. Given the standard electrode potentials K'^/K = - 2-93 V, AgVAg = 0-80 V,
Hg?+/Hg = 0-79 V , Mg2-^/Mg = - 2-37 V, Cr^VCr = - 0-74 V
(Assam Board 2012)
Arrange these metals in their increasing order of reducing power.
Ans. Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus,
increasing order of reducing power will be Ag < Hg < Cr < Mg < K.
Q. 3.3. Depict the galvanic cell in which the reaction Zn (s) + 2 Ag^ {aq) -> Zn^'*‘ (aq) + 2 Ag (s) takes
place. Further, show {/) Which of the electrodes is negatively charged ?

w
(k) The carriers of the current in the cell, (m) Individual reaction at each electrode.
Ans. The set-up will be similar to that shown in Fig. 3.10. The cell will be represented as :

Flo
Zn (s) 1 Zn2+ {aq) 11 Ag+ (aq) 1 Ag (s)
(0 Anode. Le., zinc electrode will be negatively charged.

ee
(i7) The current will flow from silver to copper in the external circuit,

Fr
(in) At Anode : Zn (5) > Zn^"^ (aq) + 2e'
At Cathode : Ag"^ (aq) + e > Ag
Q. 3.4. Calculate the standard cell potentials of galvanic cells in which the following reactons take place :
for
ur
(0 2 Cr (s) + 3 Cd^-^ (aq) > 2 Cr^+ (aq) + 3 Cd (s)
(ii) Fe^+ (aq) + Ag'^ (aq) > Fe^'^ (aq) + Ag (s)
s
= 0-80 V, E” = 0-77 V
k
= -0 40V, E
Yo
Given E' = -0 74 V,E Cd2+,Cd Ag+ Ag Fe^+,Fe2+
Ci^+,Cr
oo

Also calculate A^G® and equilibrium constants of the reactions.

eB

Ans. (0 = - 0-40 V - (- 0-74 V) = -h 0-34 V

A^ G“ = - n FE°^cii = - 6 mol X 96500 C mol"* x 0-34 V -1
= - 196860 CV mol"’ = - 196860 J mol"' = - 196*86 kj mol
r
ou
ad

- A,G° = 2-303 RTlog K

196860 = 2-303 x 8-314 x 298 log K or log K = 34-5014
Y

K = Antilog 34*5014 = 3*192 x 10^

Re
nd

(if) E°,^,| = + 0-80 V - 0-77 V = -f 0-03 V.

A^ G” = - rtF E^eii = - (1 mol) x (96500 C moH) x (0-03 V)
Fi

= - 2895 CV mol"' = - 2895 J mol"’

-1
= - 2*895 kJ mol

A,G" = -2-303 RTlogK

- 2895 = - 2-303 x 8-314 x 298 x log K
or log K = 0-5074 or K = Antilog (0-5074) = 3*22.
Q. 3.5. Write the Nernst equation and the e.m.f. of the following cells at 298 K :
(0 Mg (s)! Mg^+ (0*001 M) 1 1 Cu^+ (0-0001 M) I Cu (s)
(ii) Fe (s) I Fe^^ (0*001 M) 11 (1 M) I H, (g) (1 bar) I Pt (s)
(Hi) Sn (s) 1 Sn^+ (0*050 M) 11 (0-020 M) I (g) (Ibar) I Pt (s)
(iv) Pt (s) I Brj (1) I Br" (0*010 M) I (0-030 M) I Hj (g) (1 bar) I Pt (s)
Given E' = -2*37 V,E Cu2+/Cu = +0*34 V, E” Fe2+,Fe = -0*44 V
Mg2+/Mg
E' = -0*I4 V, E = + 1*08 V .
Sn2+/Sn 1/2 Rrj, Br~
ELECTROCHEMISTRY 3/111

Ans. (0 Cell reaction : Mg + Cu""^ 4 Mg--^ + Cu in = 2)

Nenist eqn. : E cell = E“ 0-0591 [Mg2+]

cell
2 lCu2-^]
E = 0-34-(-2-37)-
0-0591 , 10-3 = 2-71 - 0-02955 = 2-68 V
cell
2 10-^
(//) Cell reaction ; Fe + 2 H'*’ > Fe2++ H, in = 2)

Nemst eqn. : E cell = £'■

00591 , [Fe2+I
cell log

0-0591 , iO-3

ow
Ec,[i=0-(-0-44)- log = 0-44 - X (-3) = 0-44 + 0 0887 = 0-523 V
(D-
iiii) Cell reaction : Sn + 2 —^ Sn2+ + (n = 2)
Nemst eqn. : E cell = E° 0-0591 , [Sn2+] 0-0591 , 0-05
cell
-^log = 0-(-0-14)-
-^log (0-02)2

e
re
rFl
= 0-14 log 125 = 0.14- (2-0969) = 0-078 V

F
(/v) Cell reaction : 2 Br~ + 2 ■> Br, -f H, (Note carefully)
0-059! ,

r
1
Nernst eqn. : F = F“
—7—log [Br-]2[H^]2
ou
^cell ^

fo
cell
ks
0-0591 , 1
Eceii =(0-1-08) log
(0-01)2 (0.03)2
oo
Y

log (]-n 1 xio^) =-1-08-

l»f^(7.(H57) =
eB

= -I-08-
I-08-0-208 =- 1-288 V

Thus, oxidation will occur at the hydrogen electrode and reduction on the Bt2 electrode. £^.^.,1 = 1 -288 V.
r

Q. 3.6. In the button cell widely used in watches and other devices, the following reaction takes place :
ou

Zn (5) + Ag20 (s) + H,0 (0 ^ Zn2+(fl^) + 2Ag(^) +OH"(a^)

ad
Y

Determine E® and \ G“ for the reaction.

Given Zn
> Zn2+ + 2 e~, E° = 0-76 V ; Ag^O + HjO + 2 e" ^ 2 Ag + 2 OH-, E° = 0-344 V.
d

'.'(tarakhami Board .
Re
in

An.s. Refer to Solved Problem 4, page 3/62.

Q. 3.7. Define conductivity and molar conductivity for solution of an electrolyte. Discuss their variation
F

with concentration. j«H)7-

Ans. Conductivity. Refer to Art. 3.6.1.

Molar conductivity. Refer to Art. 3.6.3.
Variation with concentration. Refer to Art. 3.8.
Q. 3.8. The conductivity of 0-20 M solution of KCI at 298 K is 0-0248 S cm“^ Calculate its molar
conductivity.
KxlOOO 0-0248 S cm-l xlOOO cm^ £->
Ans. A MJ = 124 S cm2 I
Molarity 0-20 mol L"i
Q. 3.9. The resistance of a conductivity cell containing 0-001 M KCI solution at 298 K is 1500 ft. What is
the cell constant If conductivity of 0-001 M KCI solution at 298 K is 0-146 x 10"3 S cm"* ?
Ans. Cell constant =
Conductivity
Conductance
= Conductivity x Resistance = (0-146 x 10"3)S cm"' x 1500 Q

= 0-219 cm"'
3/112 “p^utdeep^'^. New Course Chemistry (XlDESSm

Q. 3.10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results
are given below :
0-001 0-010 0-020 0050 O'lOO
Concentration/M
23-15 55-53 106-74
10^ X k/S m ^ 1-237 11-85

Calculate a for all concentrations and draw a plot between a and c*". Find the value of a^.
Ans. 1 S cm * = 100 S m
1S cm '
= I (unit conversion factor)
lOOS m-'

1000 XK
(S cm^ mol *) cl/2
Cone. (M) K (S m"‘) K (S cm"*) A
m
Molarity

w
1000x1-237x10-* 0-0316
10-^ 1-237 X 10-- 1-237 X 10-* = 123-7
10'3

F lo
1000x11-85x10-* 0-100
10-- 11-85 X 10-“ 11-85 X 10-* = 118-5
10-2

ee
1000x23-15x10-* 0-141
23-15 X 10-*

Fr
2x 10"2 23-15 X 10-2
2x10-2
1000x55-53x10-*

for
55-53 X 10-2 = 111-1 0-224
5 X 10-2 55-53 X 10
5x10-2
ur
1000x106-74x10-* 0-316
10-* 106-74 X 10-2 106-74 X 10-* = 106-7
ks
10"
Yo
oo

124.0-
(0.032, 123.7)
eB

122.0-

■} 120.0-
A° = Intercept on the axis = 124.0 S cm2 i 118.0-
,(0.10, 118.5)
r

(on exlrapotation to zero concentration)

ou
ad

ig 116.0- (0.14,115.8)
E
114.0-
Y

« 112.0- (0.22, 111.1)

E
< 110.0-
Re
nd

108.0-
(0.32,106.7)
Fi

106.0
0.0 0.05 0.10 0.15 0.20 0.25 0.30 0.35
4>

Q. 3.11. Conductivity of 0-00241 M acetic acid is 7-896 x lO-^S cm-*. Calculate its molar conductivity and
if a" for acetic acid is 390-5 S cm2 mo\~\ what is its dissociation constant ?
KxlOOO (7-896x10-5 S cm-*)xl000 cm^ L~‘ = 32-76 S cm2
Ans. ~
Molarity 0 00241 mol L-'

A^
m
32-76
= 8-4x10-2
ca2 0-0024 lx (8-4x10-2)2 = 1-86x10-5
a = K
A° 390-5 a
1-a 1-0-084
III

Q. 3.12. Same as Solved Problem 1 on page 3/9.

Q. 3.13. How much electricity in terms of Faraday is required to produce
(0 20-0 g of Ca from molten CaClj («) 40-0 g of Al from molten AljOj ?
Ans. (i)Ca2+4-2^ 4Ca

Thus. 1 mol of Ca. i.e., 40 g of Ca require electricity = 2 F 20 g of Ca will require electricity = 1 F

 ELECTROCHEMISTRY 3/113

(/i) A13++ 3 e- > Al. Thus, 1 mol of Al, Le., 27 g of A1 require electricity = 3 F
40 g of Al will require electricity = — x 40 = 4*44 F.
27
Q. 3.14. Same as Solved Problem 2 on page 3/9.
Q. 3.15. A solution of Ni (N03)2 is electrolysed between platinum electrodes using a current of 5*0 ampere
for 20 minutes. What mass of nickel will be deposited at the cathode ? (At. mass of Ni = 58*7)
Ans. Quantity of electricity passed = (5 A) x (20 x 60 5) = 6000 C
Ni2+ + 2e- ^Ni
Thus, 2 F, Le., 2 x 96500 C deposit Ni = 1 mole, Le., 58*7 g
58-7
6000 C will deposit Ni = X 6000 g = 1-825 g.
2x96500
Q. 3.16. Three electrolytic cells A, B and C containing electrolytes ZnS04» AgNOj and CUSO4 respectively
were connected in series. A steady current of 1-50 ampere was passed through them until 1-45 g

w
of Ag were deposited at the cathode of cell B. How long did the current flow ? What mass of
copper and zinc were deposited ? (At. wts. of Cu = 63-5, Zn = 65-3, Ag = 108)

F lo
Ans. Ag+ + e~ > Ag, Le., 108 g of Ag are deposited by 1 F = 96500 C
96500
1 -45 g of Ag will be deposited by X1-45 C = 1295-6 C
108

ee
Q = I X r or / = Q/1 = 1295-6/1 -50 = 863-7 s = 14 min, 24 sec.

Fr
Cu2++2e- ^ Cu

Le., 2 X 96500 C deposit Cu = 63-5 g for

1295-6 C will deposit Cu =
63-5

2x96500
X1295-6 = 0-426 g
ur
65-3
Zn^*+2e- ^ Zn. .*. Zn deposited = X1295-6 = 0-438 g
s
ook

2x96500
Yo

Q. 3.17. Predict if the reaction between the following is feasible :

(/) Fe^+ (aq) and 1“ (aq)
eB

(U) Ag* (aq) and Cu (ui) Fe^ and Br"^ (aq)

(iv) Ag (s) and Fe^ (aq) (v) Br2 (aq) and Fe^ (aq)
Given standard electrode potentials : E°
our

= 0-541, E“
Cu2+,Cu = 0-34V,
ad

V2i2,r
E“ = 1-09V, E"
l/2Br2,Br- Ag-,Ag=+*-«'V, =
Y

Ans. A reaction feasible if EMF of the cell reaction is +ve

Re
nd

(0 Fe^+ (aq) + I" (aq)

^ Fe2+ (aq) + ^1^ Le., Pt 1121T (aq) 11 Fe^^ (aq) I Fe^^ (aq) I Pt
Fi

E“cell = E‘’ Pe3+ Fe2+

-E“ = 0-77 - 0-54 = 0-23 V (Feasible)
1/212,1-
(ii) Ag* (aq) + Cu > Ag (^) + Cu^+ (aq), Le., Cu I Cu^+ (aq) 11 Ag+ (aq) I Ag
E“
cell = ^“Ag^Ag- ■ ^“cu2^cu = 0 80-0-34 = 0-46 V (Feasible)
(in) Fe^'*' (aq) + Br“ (aq)
^ Fe2+ (aq) + ^Br2, = 0-77 - 1-09 = - 0-32 V (Not feasible)
(iv) Ag (5) + Fe^* (aq) — » Ag+ (aq) + Fe2+ (aq) E^^j, = 0-77 - 0-80 = - 0-03 V (Not feasible)

(V) ^ BT2(aq) + Fe^*(aq) > Br" + Fe^+, E”^^j,= 1-09 - 0-77 = 0-32 V (Feasible)
Q. 3.18. Predict the products of electrolysis of the following :
(0 An aqueous solution of AgN03 with silver electrodes
(ii) An aqueous solution of AgN03 with platinum electrodes
3/114
4. New Course Chemistry (Xll)E

(in) A dilute acqueous solution of H2SO4 with platinum electrodes

(iv) An aqueous solution of CUCI2 with platinum electrodes.
(Given E"
Ag+/Ag = +0-80V,E°^2./cu = '*-®‘^^
Ans. (i) Electrolysis of aqueous solution of AgNOj with silver electrodes.
AgNOj (s) + aq -> Ag+(a^) + N03 (a^)
H2O ^ ^ H++OH-

At Cathode ; Ag+ ions have lower discharge potential than ions. Hence, Ag+ ions will be deposited
as Ag in preference to ions.
Alternatively, we have standard reduction potentials as
Ag+ (aq) + e~ ^ Ag (s), E° = + 0-80 V
1
H+ (aq) + e~ -H2(g),E“ = 0 00V

w
2 ^
As Ag"*" ions have higher standard reduction potential than that of H'*’ ions, hence Ag"*" ions will be

F lo
reduced more easily and deposited as Ag.
At Anode : As Ag anode is attacked by NO3 ions, Ag of the anode will dissolve to form Ag"^ ions in

ee
the solution.

Fr
Ag ■» Ag'*' + e~
Alternatively, out of the three possible oxidation reactions occurring at the anode, i.e.,

Ag -> Ag'*' + e ,2 OH
1
» H,0 + -0» +e for
and NO3 4 NO3 + e~.
ur
2 2 2
Ag has highest oxidation potential. Hence, Ag of anode is oxidized to Ag'*' ions which pass into the
s
ook

solution,
Yo

(ii) Electrolysis of aqueous solution of AgN03 using platinum electrodes.

eB

At Cathode : Same as above.

At Anode : As anode is not attackable, out of OH" and NO3 ions, OH ions have lower discharge
our
ad

potential. Hence, OH" ions will be discharged in preference to NO3 ions, which then decompose to
give out Oj-
OH-(fl^) ^ OH + e-, 4 OH ^2H20(1) +02(g)
Y

(hi) Electrolysis of dilute H2SO4 with platinum electrodes.

Re
nd

H2SO^(aq) ■> 2H+(aq)+SOl~ (aq)

Fi

H2O H-*^ + OH-

At Cathode ; li*+e > H, H + H > H2 (g)
At Anode : OH“ > OH + e, 4 OH > 2 H2O + O2 (g)
Thus, H2 is liberated at the cathode and O2 at the anode,
(iv) Electrolysis of aqueous solution of CUCI2 with platinum electrodes
CUCI2 (s) + aq > 002-*- (aq) + 2 Cl" (aq)
H2O H++OH-
At Cathode ; Cu2'*' ions will be reduced in preference to H'*' ions
Cu2-" + 2 e~ > Cu
At Anode ; d“ ions will be oxidized in preference to OH“ ions
Cl- > Cl + e". Cl + Cl > CI2 (g)
Thus, Cu will be deposited on the cathode and CI2 will be liberated at the anode.
ELECTROCHEMISTRY 3/115

& U m
WITH ANSWERS,
f f HINTS AND SOLUTIONS

ll

MULTIPLE CHOICE QUESTIONS-I

ow
1. Which cell will measure standard electrode 4. The difference between the electrode potentials
potential of copper electrode ? of two electrodes when no current is drawn
(a) Pt (s) I H2 (g, 0-1 bar) I H+ (aq., 1 M) through the cell is called
I Cu^-*- (aq., 1 M) I Cu (a) Cell potential (h) Cell emf

e
(c) Potential difference
(h) Pt is) I ig, 1 bar) I (aq.. 1 M)

re
(d) Cell voltage
II Cu“+ (aq., 2 M) I Cu

Frl
5. Which of the following statement is not correct

F
(c) Pt (s) I H2 (g, 1 bar) I (aq., 1 M) about an inert electrode in a cell ?
I Cu^+ (aq., ,1 M) I Cu (a) It docs not participate in the cell reaction.
(d) Pt (s) I H2 (g, 1 bar) I H+ (aq., 0-1 M)
ou
or
(b) It provides surface cither for oxidation or for
!l Cu--" (aq., I M) I Cu reduction reaction.

2. Electrode potential for Mg electrode varies

according to the equation
kfs
(c) It provide.s surface for conduction of electrons.
(d) It provides surface for redox reaction.
oo
6. An electrochemical cell can behave like an
E _ g© 0059, 1
electrolytic cell when,
Y
|Mg2+]
B

(a) E^.,„ = 0 (h) E,,„ > E cxl

The graph of E vs log is
Mg2*I.Mg (c) > E,,ii (./)E,,„ = E ext
re

7. Which of the statements about solutions of

electrolytes is not correct ?
oYu

(a) Conductivity of solution depends upon size of

ad

ions.

(b) Conductivity depends upon viscosity of

d

solution,
in

log(Mg^*]—^ 2+
Re

loglMg )—P
(c) Conductivity does not depend upon solvation
of ions present in solution.
F

(d) Conductivity of solution increases with

(d)i temperature.
O)

LU
8. Using the data given below find out the
2+
strongest reducing agent.
loglMg ]—^ log[Mg^-"] —►
E® = 1-33V E® = 1-36V
3. Which of the following statement is correct ? CrjOl*- /Cr^ - cij/cr
(a) Ej,g]| and A^G of cell reaction both are extensive E
0
= 1-51V E® = -074 V
properties. Mn04/Mn2+ Cr^+/Cr
(h) E^g]] and A^ G of cell reaction both are intensive («) cr (b) Cr
properties, (c) Cr^+ id) Mn^-"
(c) Eggjj is an intensive property while A,. G of cell 9. Use the data given in Q. 8 and find out which
reaction is an extensive property. of the follow ing is the strongest oxidising agent.
(d) Eggj] is an extensive property while A^ G of (a) Cl (h) Mn2+
cell reaction is an intensive property. (c) MnO- (d)
3/116 ^,uxcUe^’4. New Course Chemistry (Xli)ISS191
10. Using the data given in Q. 8 find out in which 15. While charging the lead storage battery.
option the order of reducing power is correct, {a) PbS04 anode is reduced to Pb.
(a) Cr^+<Cr<Mn2-"<Cr {b) PbS04 cathode is reduced to Pb.
{b) Mn-+ < C|- < C"r'+ < Cr (c) PbS04 cathode is oxidised to Pb.
{d) PbS04 anode is oxidised to Pb02-
(c) Cr^+ < cr < Cr^O^- < MnO^ 0
16. A is equal to
id) Mn2+ < Cr^+ < cr < Cr m(NH^OH)
11. Use the data given in Q. 8 and find out the most + A
0
-A?.(HCl)
stable ion in its reduced form. hi(NH4CI)
{b) Cr^+ -a!>(NaCl)

w
0 0
(a) cr (b) A mCNH^Cl) + A
m(NaOH)
(c) Cr (d) Mn-+
12. Use the data of Q. 8 and find out the most stable + A
0
-A«(NaOH)
»i(NaCl)
oxidised species. 0 0
+ A - A
(a) Cr^^ ^m(NaOH) /H(NaCl) (NH4CI)
(h) Mn04

o
e
17. In the electrolysis of aqueous sodium chloride

re
(c) Cr^O^ (d) Mn-+ solution which of the half cell reaction will
13. The quantity of charge required to obtain one occur at anode ?

Frl
F
mole of aluminium from AI2O3 is
(fl) Na"*" (aq) + e~ > Na (s); = -2-71V
(a) IF (b)6F
(c) 3 F id)2F
ou ib)2H20{l) ^ O2 (g) + 4 iaq) + 4 e' ;

sor
14. The cell constant of a conductivity cell E® „ =1-23V
(a) changes with change of electrolyte
(b) changes with change of concentration of (c) kf(aq) + e
1
=000 V
oo
electrolyte
(c) changes with temperature of electrolyte 0
~CUig) + e- ; E Cell
Y
(d) Cl (aq) = 1-36V
(d) remains constant for a cell
B
re
oY

MULTIPLE CHOICE QUESTIONS-II

u

Note : In the following questions two or more

(Hi) 2SOj (aq) ■> S2OI (aq)+2e ;
ad

than two options may be correct.

d

18. The positive value of the standard electrode E® „ =1-96V

potential of Cu^^/Cu indicates that
in

(a) In dilute sulphuric acid solution, hydrogen will

Re

be reduced at cathode.
(a) this redox couple is a stronger reducing agent
(b) In concentrated sulphuric acid solution, water
F

than the couple.

will be oxidised at anode,
(b) this redox couple is a stronger oxidising agent
than H+/H2. (c) In dilute sulphuric acid solution, water will be
oxidised at anode.
(c) Cu can displace H2 from acid.
(d) Cu cannot displace H2 from acid, (d) In dilute sulphuric acid solution, SO^" ion
will be oxidised to tetratliionate ion at anode.
19. E®CeU for some half cell reactions are given
0
20. E =1TV for Daniell cell. Which of the
below. On the basis of these mark the correct Celt

answer. following expressions are correct description of

state of equilibrium in this cell ?
(1) n*(aq) + e‘ ; E® „ =o*oov
(b)
2-303 RT
log K^=M
(a)V\=K, 2F
(miHjO (/) — ^ O2 (g) + 4 H+ (aq) + 4 ; 2-2
(c) logK^ = (^/)logK,= M
E®„=1-23V 0059
ELECTROCHEMISTRY 3/117

21. Conductivity of an electrolytic solution depends 24. What will happen during the electrolysis of
on.
aqueous solution of CUSO4 in the presence of
(a) nature of electrolyte Cu electrodes ?

(b) concentration of electrolyte (a) Copper will deposit at cathode.

(c) power of AC source (h) Copper will dissolve at anode,
(d) distance between the electrodes (c) Oxygen will be released at anode.
(d) Copper will deposit at anode.
22. a"m H,0 is equal to
25. Conductivity k, is equal to
0 0 0
{a) A /n(HCl) + A
m(NaOH)
- A 1 / G*
/n(NaCl) («) -- ib)
R A R
0 0
ih) A + A -A«m(NaOH)
--
/ntHNOj) /ntNaNOj) I
(C) Am
0 0 0
(C) A -A
(HNO3) »i(NaN03) 26. Molar conductivity of ionic solution depends
0 0 on.
(^0 A

w
ni(NH40H) ■*‘^m{HCl) ^m(NH4CJ) (a) temperature
23. What will happen during the electrolysis of (b) distance between electrodes

F lo
aqueous solution of CUSO4 by using platinum (c) concentration of electrolytes in solution
electrodes ?
(d) surface area of electrodes
(a) Copper will deposit at cathode. 27. For the given cell, Mg I Mg^"*" I Cu^"^ I Cu

e
(b) Copper will deposit at anode,

Fre
(a) Mg is cathode {b) Cu is cathode
(c) Oxygen will be released at anode. (c) The cell reaction is Mg + Cu^'*' ^Mg2+ + Cu
(d) Copper will dissolve at anode. (d) Cu is the oxidising agent
for
r
ANSWERS
You
s

Multiple Choice Questions -1

ook

1. (c) 2. ib) 3. (c) 4. ih) 5. id) 6. (c) 7. (c) 8. ib) 9. (c) 10. ih)
eB

11. (d) 12. ia) 13. (c) 14. id) 15. (a) 16. ib) 17. ib)

Multiple Choice Questions -11

our
ad

18. ib, d) 19. ia. c) 20. ib, c) 21. ia. b) 22. ia, d) 23. ia, c) 24. ia, b) 25. (a. b)
26. ia, c) 27. ib. c)
dY
Re

HINTS FOR DIFFICULT MULTIPLE CHOICE QUESTIONS

Fin

Multiple Choice Questions -1

1. For standard electrode potential, [Cu^'*’] = 1 M.

2. E = E° + log [Mg^"*" ]. Hence, plot of E vs log [Mg^"^] will be linear with positive slope and intercept
= E°, i.e., ib).
3. E^,g[| is intensive while G is extensive (Refer to Supplement Your Knowledge, page 3/60).
5. Inert electrode provides surface for oxidation or reduction reaction but not for redox reaction. Hence,
id) is not correct.
6. When external EMF applied (E^j^^) is greater than E^,gjj, electrons flow from cathode to anode, i.e., like
an electrolytic cell.
7. Greater the solvation of the ions, less is the conductivity. Hence, (c) is not correct.
8. Strongest reducing agent is one which is oxidized most easily, i.e., which has highest oxidation potential.
Given values are reduction potentials. Reversing the sign will give oxidation potentials. Thus, Cr/Cr^"^
3/118 4 New Course Chemistry (Xll)

^ Cr^-"
will have highest oxidation potential (+ 0-74 V). Hence, Cr is oxidized most easily (Cr
+ 3 e~) and, therefore, is the strongest reducing agent.
9. Strongest oxidizing agent is one which is reduced most easily, i.e., which has highest reduction potential
viz. MnO^ (reduced to Mn-'*').
10. As explained in Hint 8, less is the oxidation potential, weaker is the reducing agent. Oxidation potentials
are Cr’’+/Cr,0“" = 1-33V, CI-/CI2 = - 1-36 V. Mn“+/MnO" = -1-51V, Cr/Ci^+ = + 0-74 V. Hence,
order of reducing power will be Mn-'*’ < Cl“ < Cr^'*’ < Cr.
11. ishighestfor MnO^/Mn^'^. This shows that this reduction takes place most easily. Hence, Mn^+
will be most stable ion in the reduced form.

12. E°ox is highest

®
for Cr/Cr^"^. Hence, Cr^'^ ions will be most stable oxidized species,
13. AI2O3 > 2 Al, i.e., 2 + 6 e~ > 2 A1 or A\^* + 3 e~ > Al. Hence to obtain one mole of Al,

w
charge required = 3 F.
15. PbS04 deposited on anode is reduced to Pb. This electrode now acts as cathode during charging (Refer

Flo
to Art. 3.26.2).
17. Oxidation occurs at anode. As {h) has higher oxidation potential than (cr), (b) occurs.

ee
Multiple Choice Questions -11
reduction takes place easily compared to H'^'/H,. Hence, it

Fr
18. Positive value for Cu-'^/Cu indicates that this
is stronger oxidizing agent, i.e., (b) is correct. It also implies that Cu cannot reduce ions of an acid to
H2, i.e.. (cl) is correct.

for
19. In dilute H2SO4, (acidulated water), H"^ ions arc reduced to H2 at cathode, i.e., (a) and H2O is oxidized
ur
to 0-) at anode, i.e., (c). Reaction (r’rV) occurs when H2SO4 is concentrated (Refer to “Supplement Your
Knowledge’ page 3/66 point (1)).
k s
2-303 RT, [Zn-+]
Yo
o

20. For Daniell cell, Zn + Cu“'*' - » Zn^-^ + Cu, E cell = E.cell log
oo

n F ICu2+]
eB

O 2-303 RT
At equilibrium, E(,g|j = 0. Hence, E^,^j[ = 2F
logK^ = M (Given) i.e. (h).
r

2-303 RT 2-2
logK^ = l-l or logK^ = , i.e.,(c).
ou

Putting = 0-059 at 298 K,

ad

F 0-059
.0 . o
Y

22. a! = x . +x . We get this expression from (a) and (d).

m(H20) H OH"

I 1 / 1 /
Re
nd

25. R or R = or K (conductivity) = , i.e.. (a).

K A R A
Fi

/ 1 G*
As — = G* (cell constant), k = —G* = , i.e., (b).
A R R

SHORT ANSWER QUESTIONS

28. Can absolute value of electrode potential of an electrode be measured ?

Ans. No because oxidation or reduction cannot take place alone. Moreover, it has to be determined with respect
to a reference electrode.

29. Can E
cell and A^ G“ for cell reaction ever be equal to zero ?
Ans. No (E^.g]| can be zero but not E°cell ). Further, AG° = FE“^jj . As E°^.,| * 0, therefore, A^ G° also cannot
be zero.

30. Under what condition Is E^,gn = 0 or G = 0 ?

Ans. Ej-gi, = 0 or G = 0 when the cell reaction reaches equilibrium.
ELECTROCHEMISTRY 3/119

31. What does the negative sign in the expression E = - 0*76 V mean ?
Zn^^/Zn

Ans. Negative sign for E°Zn-+/Zii shows that zinc is more reactive than hydrogen. This means that when zinc
electrode is connected to SHE, Zn will be oxidized to Zn^"*" and H'*’ will be reduced to H2-
32. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere
current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on
the cathode be same or different ? F^xplain your answer.
Ans. Masses of Cu and Ag deposited will be different because by the same quantity of electricity, masses deposited
are in the ratio of their equivalent weights (Faraday's second law).
33. Depict the galvanic cell in which the cell reaction is Cu + 2 Ag"*" ^ 2 Ag + Cu-+.

ow
Ans. Cu I Cu^+ I Ag+ 1 Ag.
34. Value of standard electrode potential for the oxidation of Cl" ions is more positive than that of water,
even then in the electrolysis of aqueous sodium chloride, why is Cl" oxidised at anode instead of water ?
Ans. Oxidation of H2O to is kinelically so slow that it needs some overvoltage. Hence. d“ is oxidized
instead of water.

e
re
35. What is electrode potential ? Salt bridge

rFl
Ans. Refer to Art. 3.15. 4

F
^Copper
36. Consider the following diagram in which an plate
electrochemical cell is coupled to an electrolytic
■CL1SO4
cell. What will be the polarity of electrodes *A’

or
ou
/ solution
ZnS04
and ‘B’ in the electrolytic cell ? Zinc plate solution

Ans. In the electrochemical cell, zinc acts as a negative

ksf
pole while copper acts as a positive pole. Hence, (A) Electrode (B) Electrode
oo
electrode ‘A* will have negative polarity while T
●Electrolytic cell
electrode ‘B’ will have positive polarity.
Y
B

37. Why is alternating current used for measuring resistance of an electrolytic solution ?
Ans. Alternating current is used to prevent electrolysis of the electrolytic solution so that concentration of ions
re

in the solution remains constant.

oYu

38. A galvanic cell has electrical potential of 1-1 V. If an opposing potential of M V is applied to this cell,
ad

what will happen to the cell reaction and current flowing through the cell ?
Ans. No current will flow and the reaction stops.
d

39. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed ?
in
Re

Ans.
When aqueous NaCl solution is electrolysed. H2 is liberated at cathode, CI2 at anode and NaOH is formed
in the solution. Hence, pH of the solution will ir.se.
F

40. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why ?
Ans. Overall reaction does not involve any ion whose concentration may change. Hence, it gives constant potential.
41. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The of ‘B’ increases 1’5 times while that of A
increases 25 times. Which of the two is a strong electrolyte ? Ju.stlfy your answer.
Ans. 'B' is a strong electrolyte because on dilution of a strong electrolyte, number of ions remains the same.
Only interionic attraction decreases and therefore increase in is small.
42.
When acidulated water (dil. H2SO4 solution) is electrolysed, will the pH of the solution be affected ?
Justify your answer.
Ans. pH of the solution will not be affected as [H'*’] will remain the same. This is because the amount of ions
reduced is compensated by the H'*' ions produced by oxidation of H->0.
1
At cathode ; 2 H"^ + 2 e ^ H2, At anode : HiO -O,+ 2H-^ + 2e".
2 ^
43. In an aqueous solution how does specific conductivity of electrolytes change with addition of water ?
Ans. Conductivity decreases because number of ions per unit volume decreases.
3/120 "Pt^dee^ New Course Chemistry (XII)
44. Which reference electrode is used to measure the electrode potential of other electrodes ?
Ans. Standard hydrogen electrode (SHE) whose electrode potential is taken as zero.
45. Consider the cell: Cu I Cu^"*" I CI“ I CI2, Pt
Write the reactions that occur at anode and cathode.
Ans. Anode : Cu ■> Cu^"^ + 2 e (Oxidation); Cathode : CI2 + 2 e ■> 2 Cl (Reduction)
46. Write the Nernst equation for the ceil reaction in the Daniel cell. How will the be affected when
concentration of Zn^'*' ions is increased ?
0-059, [Zn-+]
Ans. Zn + Cu2+ Zn-+ + Cu, EceU=E-cell
—

[Cu2-^]
When [Zn^"*"] is increased, E^-gn will decrease.
47. What advantage do the fuel cells have over primary and secondary batteries ?
Ans. Primary batteries contain limited amounts of reactants. These cells become dead when the reactants are
consumed. Secondary batteries can be recharged but take a long time for recharging. Fuel cells can be run

w
continuously so long as the reactants are supplied and the products arc removed continuously.
48. Write the cell reaction of a lead storage battery when it is discharged. How does the density of the

F lo
electrolyte change when the battery is discharged ?
Ans. Pb + Pb02 + 2 H2SO4 > 2 PbS04 + 2 HjO
Density of sulphuric acid solution decreases as water is formed and sulphuric acid is consumed during

ee
discharge.

Fr
49. Why on dilution the am of CH3COOH increases drastically, while that of CH3COONa increases
gradually ?
for
Ans. CH3COOH is a weak electrolyte. On dilution, degree of dissociation increases and hence number of ions
ur
increases and, therefore, increases drastically. On the other hand, CH3COONa is a strong electrolyte.
On dilution, number of ions remains the same. Only interionic attractions decrease and hence a^ increases
s
ook

gradually.
Yo
eB

MATCHING TYPE QUESTIONS

Note : Match the items of Column I and Column II in the following questions.
our
ad

50. Match the terms given in Column I with the units given in Column II.
Column I Column II
Y

(/) A
m {a) S cm”'
Re

-I
(i!) E,e,| (b) m
nd

(m) K (c) S cm^ mol '

Fi

((V) G* id) V
51. Match the terms given in Column I with the items given in Column II.
Column I Column II

(0 A
m (a) intensive property

(») KceU (b) depends on number of ions/volume

(m) K (c) extensive property
(iv) \ ^cell id) increases with dilution
52. Match the items of Column 1 and Column II.
Column I Column II

(1) Lead storage battery (a) maximum efficiency

(ii) Mercury cell (b) prevented by galvanisation
(Hi) Fuel cell (c) gives steady potential
(iv) Rusting id) Pb is anode, Pb02 is cathode
ELECTROCHEMISTRY 3/121

53. Match the items of Column I and Column II.

Column I Column II
(0 K (a) \xt

HO A
III (b)

K
(Hi) a (C)
c

G*
(/V) Q (d)
R
54. Match the items of Column I and Column II.
Column I Column II
(/■) Lechlanche cell (a) cell reaction 2 H-> + O2 ^2H20
07) Ni-Cdcell (/;) does not involve any ion in solution and is used in hearing aids,
(m) Fuel cell (c) rechargeable

w
(iv) Mercury cell (d) reaction at anode, Zn ^ Zn^+ + 2 e"
(e) converts energy of combustion into electrical energy

F lo
55. Match the items of Column I and Column II on the basis of data given below :

= 2-87 V, E" . = -3-5 V, E = 14V, = 109V

1^2/F- Li+/U Au^"^/Au Btj/Br”

ee
(0 F, (a) metal is the strongest reducing agent

Fr
(//) Li {b) metal ion which is the weakest oxidising agent
07/)
(/V)
Au^"-
Br-
(c)
0/)
for
non metal which is the best oxidising agent
unreactive metal
r
(v) Au (e) anion that can be oxidised by
You
(vi) Li+
s
(/) anion which is the weakest reducing agent
ook

(v/7) (g) metal ion which is an oxidising agent

eB

ANSWERS
our
ad

50. (/) (c). (/7) (d), (/77) (a), (iv) {b) 51. 0) -> (d), (ii) -> (a), (/77) -> (b), (iv) -> (c)
52. (j) (d). (ii) -> (r), (/77) (a), (iv) -> (b) 53. 0) (d), (ii) -> (c), (/77) ^ (b), (iv) (a)
54. (/■) -> (d), (ii) -> (c), (i77) (a. e), (/V) ^ (h)
dY
Re

55. (/■) (c), (//) -> (fl), (Hi) -> (g), (iv) (e), (v) -> (d), (vi) (b), (v/7) -¥ (f)
Fin

HINTS FOR DIFFICULT QUESTIONS

55. (/■) F2 has highest reduction potential. Hence, it is best oxidizing agent. Further, it is a non-metal, i.e.,
(i) —> (c).
(//) Li has lowest reduction potential or highest oxidation potential. Hence, it is strongest reducing
agent. It is a metal, i.e., (ii) -> (a).
(t/7) E”
Au^+/Au is +ve. Hence, Au'^'*’ ions can undergo reduction and. therefore, act as an oxidizing
agent, i.e., (Hi) (g).

(/V) E°Au-’^/Au >E°

Br-> /Br
. Hence Br ions can be oxidized by Au^'*', i.e., (iv) (e).

(v) Au is unreactive metal. Hence, (v) —> (d).

(vi) Li'''/Li has lowest reduction potential. Hence, Li'*’ will be weakest oxidizing agent, i.e., (v/) —> (b).
(vii) F“/F2 has low oxidation potential (- 2-87 V). Hence, F" is weakest reducing agent,
i.e., (vii) (f).
3/122 New Course Chemistry (XII)Egggl

ASSERTION AND REASON TYPE QUESTIONS

Note : In the following questions, a statement of assertion followed by a statement of reason is given.
Choose the correct answer out of the following choices :
(a) Both assertion and reason are true and the reason is the correct explanation of assertion.
(b) Both assertion and reason are true and the reason is not the correct explanation of assertion,
(c) Assertion is true but the reason is false.
{d) Both assertion and reason arc false,

ow
(e) Assertion is false but reason is true.
56. Assertion : Cu is less reactive than hydrogen,
Reason : is negative.
Cu2+/Cu

e
57. Assertion : E cell should have a positive value for the cell to function.

re
Reason : Ecathode

Flr
58. Assertion : Conductivity of all electrolytes decreases on dilution.

F
Reason : On dilution, number of ions per unit volume decreases.
59. Assertion : for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
ou
Reason : For weak electrolytes,degree of dissociation increases with dilution of solution.

sr
60. Assertion : Mercury cell does not give steady potential.

kfo
Reason : In the cell reaction, ions {ire not involved in solution.
61. Assertion : Electrolysis of NaCl solution gives chlorine at anode instead of O2.
oo
Rea.son : Formation of oxygen at anode requires overvoltage.
Y
62. Assertion : For measuringresistanceof an ionic .solution, an AC source is used.
reB

Reason : Concentration of ionic solution will change if DC source is used.

63. Assertion : Cuirent stops flowing when E^^jj = 0.
uY

Reason : Equilibriumof the cell reaction is attained,

64. Assertion : E increases with increase in concentration of Ag’*' ions.
Ag^/Ag
ad
do

Reason : E, . , ^ has a positive value.

Ag+/Ag
65. Assertion : Copper sulphate can be stored in zinc vessel.
in

Rea.son : Zinc is less reactive than copper.

Re
F

ANSWERS

56.(c) 57.(c) 58.(a) 59.(«) 60,(e) 61.(«) 62.{a) 63.(a) 64. {b) 65.(J)

HINTS FOR DIFFICULT QUESTIONS

56. Correct Reason. E° 2+ is positive (+ 0-34 V).

Cu /Cii

57. Correct Reason. E cell = E cathode -E anode . Ecell is +ve when E cathode >E anode'

60. Correct Assertion. Mercury cell gives a steady potential.

RT 1
64. Correct explanation. Ag"*" + e > Ag, E Ag"*” / Ag = E“Ag+/Ag log
nF [Ag^]
65. Correct Assertion. Copper sulphate solution cannot be stored in zinc vessel.
Correct Reason. Zinc is more reactive than copper.
ELECTROCHEMISTRY
3/123

LONG ANSWER QUESTIONS

66. Consider the Fig. given beiow and answer the
following questions ; Salt bridge
i
In
(/■) Cell ‘A’ has E cell = 2 V and Cell ‘B’ has
E cell = !●! V. Which of the two cells ‘A’ or ‘B’ plate
will act as an electrolytic cell. Which electrode
reactions will occur in this cell ?
I Q .i
-2n
2+
(aq) I *CuSO/)
solution

I Cell A
(h) If cell ‘A’ hasE cell = 0-5 V and cell ‘B’ has Zinc plate
E
cell = 1*1 V then what will be the reactions at
anode and cathode ? (A) Electrode ■H {B) Electrode
T
Ans. (/) Cell ‘B’ will act as an electrolytic cell as it has

low
Celts
lower emf. Reactions occurring in cell A will be
Zn +2 e (at Zn electrode),
Ci|2+ + 2 e- ■> Cu (at Cu electrode)
Zn plate will act as -ve pole and supply electrons to electrode ‘A’, i.e. reduction of ions present in ceil B
will occur on electrode ‘A’ and oxidation on electrode

ee
'B’.

F
As cell B has higher emf, cell B will act as a source of external cinf and reactions occurring in cell A will

Fr
be reversed, (electrons will now flow from cell B to cell A), i.e.,
Zn^+ Zn (Reduction at cathode)
4 Cu^'*' + 2 c~ (oxidation at anode).
Cu
for
ur
67. Consider the Fig. given below and answer the questions (0 to (vi)
(j) Redraw the diagram to show the direction of electron
ks
flow.
Yo
(«) Is silver plate the anode or cathode ?
oo

(in) What will happen if salt bridge is removed ?

eB

(iv) When will the cell stop functioning ? Salt bridge

(v) How will concentration of Zn^+ ions and Zinc plate Silver plate
r

Ag"^ ions be affected when the cell functions ?

ou
ad

(v/) How will the concentration of ions 2+

Zn (aq) 'tt
and Ag'*’ ions be affected after the cell becomes
Y

‘dead’? .;r
Ag (aq)
Re
nd

Ans. (/) Electrons will flow from Zn to Ag

Zn ■> Zn^'*' + 2e (oxidation, -ve pole)
Fi

Ag-^ + e" Ag (reduction.+ve pole).

(//) As reduction occurs on Ag, it will act as cathode.
(Hi) Current will stop flowing, i.e., cell will stop functioning.
(/V) The cell will stop functioning when the electrode potentials of the two cells become equal so that E cell
becomes zero (Refer to Art. 3.23).
(v) Concentration of Zn^'*’ ions will increase while that of Ag'*' ions will decrease,
(v/) When Ece]| = 0, the cell reaction attains equilibrium. Hence, concentration of Zn^'*' and Ag'*’ ions will
remain constant.

68. What is the relationship between Gibbs free energy of the cell reaction in a galvanic ceil and the emf
of the cell ? When will the maximum work be obtained from a galvanic cell ?
Ans. Refer to Art. 3.24.
3/124
New Course Chemistry (Xll)CZs2ai

■ JSKSSi-: L
^■1
^KK mm
iV k;
f'

I!-)

lilfrul
NEET/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to
● MCQs In Chemistry for NEET
Pradeep's Stellar Series.... ● MCQs In Chemistry for JEE (Main)

w
separately available for these examinations.

F lo
Multiple Choice Questions (with one correct Answer)

(c) 6:3:2 id) 1:2:3

e
I. Electrolysis ana Faraday’s laws

Fre
(e) 3 : 6 : 2 (Kerala PET 2010)
1. What current is to be passed for 0-25 s for 5. 9-65 C electric current is passed through fused
deposition of a certain weight of metal which is for
anhydrous magnesium chloride. The magnesium
metal thus obtained is completely converted into
equal to its electrochemical equivalent ?
r
(a) 4 A ib) 100 A a Grignard reagent. The number of moles of the
You
Grignard reagent obtained is
oks

ic) 200 A id) 2 A

(a) 5xl(r4 ib) 1 X 10-^
eBo

2. In an experiment 0-04 F was passed through (c) 5 X 10-5 (cO 1 X 10-5

400 mi of 1 M solution of NaCl. What would be
(Karnataka CET 2010)
the pH of the solution after the electrolysis ? 6. A current of 2-0 A passed for 5 hours through a
ad
our

ia) 8 ib) 10 molten metal salt deposits 22*2 g of metal (At wt.
(c) 13 (^0 6 = 177). The oxidation state of the metal in the
metal salt is
ie) 9
(a) +1 ib) +2
dY
Re

3. Electrolysis of dilute aqueous NaCl solution was

(c) +3 id) +4.
carried out by passing 10 milli ampere current.
7. Same quantity of electricity was passed through
Fin

The time required to liberate 0-01 mol of H2 gas

at the cathode is (1 Faraday = 96500 C mol-l) solutions of salts of elements A, B and C with
atomic weights 7, 27 and 48 respectively. The
ia) 9-65 X 10'^ sec ib) 19-3 X 10“^ sec masses of A, B and C deposited were 2-1 g, 2-7 g
ic) 28-95 X 10'’ sec id) 38-6 X lO'^sec and 7-2 g respectively. The valencies of A, B and
(IIT 2008) C respectively are
4. One faraday of electricity is passed through ia) 3,2 and 1 ib) 1,2 and 3
molten AI2O3, aqueous solution of CUSO4 and ic) 1,3 and 2 id) 2, 3 and 2
molten NaCl taken in three different electrolytic 8. Al.,03 is reduced by electrolysis at low potentials
cells connected in series. The mole ratio of Al, and high currents. If 4-0 x 10'’ amperes of current
Cu and Na deposited at the respective cathodes is passed through molten Alo03 for 6 hours, what
IS mass of aluminium is produced ? (Assume 100%
(«) 2:3:6 ib) 6:2:3 current efficiency. At. mass of Ai = 27 g mol-')
ANSWERS

1. (a) 2. (c) 3. ib) 4. (c) 5. ic) 6. (c) 7. (c)

ELECTROCHEMISTRY
3/125

(a) 81 X 10^ g (b) 24 X 10‘^g 14. The sequence of ionic mobility or conductivity in
(c) 1-3 X 10-*g (d) 9-0 X lO^g the aqueous solution is
9. The weight of silver (at. wt. = 108) displaced by u (a) K-^ > Na-^ > Rb-" > Cs-"
quantity of electricity which displaces 5600 mL (b) Cs+>Rb-">K-*->Na-^
of O2 at STP will be (c) Rb+>K+>Cs+>Na'-
(a)54g (b) 10-8 g (d) Na*>K-^>Rb*>Cs^
(c) 54-0 g (d) 108-0 g
(AIPMT 2008, JEE Main 2021)
(AIPMT 2014)
15. Resistance of 0-2 M solution of an electrolyte is
10. The number of electrons delivered at the cathode 50 n. The specific conductance of the solution is
during electrolysis by a current of 1 ampere in 60 1-4 S IT)-'. The resistance of 0-5 M solution of
seconds is (charge on the electron= l-60x 10"*^ C) the same electrolyte is 280 Q. The molar
(a) 6 X 10--’ (b) 6 X 102» conductivity of 0-5 M solution of the electrolyte
(c) 3-75 X 1Q20 (d) 7-48 X 10--’ in S m^ mol”^ is
(NEET Phase II 2016) (a) 5 X 10^ ib)5x 10-^

w
11. The number of Faradays (F) required to produce (c)5x 10-^ id) 5x10'-’

F lo
20 g of calcium from molten CaCl2 (Atomic mass (JEE Main 2014)
of Ca = 40 g mol"’) is 16. The molar conductivity of a conductivity cell filled
(a) 2 ib)3 with 10 moles of NaCl in 20 mL solution is a m

ree
(c)4 id) 1 and that of 20 moles of NaCl in 80 mL solution in

(NEET 2020, Phase 1)

12. On electrolysis of dil. sulphuric acid using
for F
another identical cell is a . The conductivities

exhibited by these two cells are same. The

'"2

relationship between a.., and a IS

Platinum (Pt) electrodes, the product obtained at m
'"2
Your
anode will be = 2a
ia) A m^ ib) A m-, = A„. /2
oks

m m
1
(a) Oxygen gas ib) H2S gas = 4a
eBo

(^) ^^2
= A
m m
1
(c) SO2 gas
1
id) Hydrogen gas
(JEE Main 2022)
(NEET 2020, Phase 1)
III. Kohlrausch’s law and it
ad
our

II. Conductance and specific, sapplications

equivalent and molar conductivities
17. Ionic mobility of Ag+ is (X = 5x10“^ ohnr’
13. The resistance of a O-JO M weak acid HA in a Ag+
Re
dY

-K
conductivity cell is 2-0 x 10^ ohm. The cell cm- eq *)
constant of the cell is 0-78 cm“’ and aq of the ia) 5-2 x 10-^ ib) 2-4 X 10-9
Fin

acid is 390 S cm^ mol"’.

(c) 1-52x10-9 id) 8-25 X 10-9
Consider the following statements :
1. pH of the acid solution = 3
18. Degree of dissociation of pure water is
1-9 X 10-9. Molar ionic conductances of and
2. pK^ of the acid = 5 OR- ions at infinite dilution are 200 S cm^
3. Degree of dissociation of the acid = 0 01
moL* and 350 S cm- mol"’ respectively. Molar
Which of the statements given above are conductance of water is
correct ?
ia) 3-8 X 10-^ S cm- moL*
ia) 1 and 2 only ib. 1 and 3 only
ib) 5-7 X lO-^Scm^mor*
(c) 2 and 3 only id) 1, 2 and 3
(c) 9-5 X 10-^ S cm^ mol“‘
(IAS Prelim 2010)
id) 1-045 X 10-6 S cm2 mol-’
ANSWERS
8. ia) 9. id) 10. (C-) 11. id) 12. («) 13. id) U.ib) 15, (h) 16. (a) 17. (a) 18. («/)
 New Course Chemistry (XII) CBiag
3/126

23.
19. A%q for BaCl2, H2SO4 and HCl are a:,, .1:2 and 1

^3 S cm^ eq"‘ respectively. If conductivity of AA/WWW

saturated BaS04 solution is y S cm"', then
Kgp for BaS04 is
106 J.2 109 y3 <i>
{Cl) ib)
2{x^ +X2~‘'2,x^) 8U, +.t2 -2.X3)^ ^ Cu rod
Zn rod
Salt

lO^y 10^y2 Bridge

(C-) id)
2(x^ + -.C., “2^:3) 4(jc, ■¥X2~'2-x-^)'^ IM IM

ZnS04 CUSO4
20. Equivalent conductivity at infinite dilution for Sol Sol

sodium potassium oxalate, (COO")2Na''‘ K‘*‘,

= + 0-34V. 0.76 V
will be (given, molar conductivities of oxalate,
K"^ and Na"^ ions at infinite dilution are 148-2,
Identify the incorrect statement from the options
50-1, 73-5 S cm“ mol"' respectively)

F low
-1
given below for the above cell :
(a) 271-8 S cm- eq (fl) If Egxt = I ● 1 V, no flow of e~ or current occurs
{h) 67-95 S cnr eq"* {b) If Egxt < 1 ● 1 V, Zn dissolves at anode and Cu
-1
(c) 543-6 S cnr eq deposits at cathode
{cf) 135-9 S cm^ eq"' (c) If Egxi > i f V, e" flow from Cu to Zn

e
(West Bengal JEE Engg. 2013)
21. The molar conductivity of 0-007 M acetic acid is
for Fre
{d) If E^xt > M V. Zn dissolves at Zn electrode
and Cu deposits at Cu electrode
20 S cm^ mol"*. What is the dissociation constant (JEE Main 2020)
of acetic acid ? Choose the correct option
V. Electrode potential, cell potential,
Your

I a” ^ = 350 S cm^ mol electrochemical series and its

eBo ks

applications
a” = 50 S cm- mol"'] 24. When measured against a standard calomel
ad

CHjCOO" electrode, an electrode is found to have a

our

(a) 1-75 X 10-^ molL"' standard reduction potential of O-lOO V. If

standard reduction potential of calomel electrode
(b) 2-50 X 10“' mol L"' is -i- 0-244 V and it acts as anode, the standard
Re

(c) 1-75 X 10"^ mol L“' electrode potential of the same electrode against
standard hydrogen electrode will be
{d) 2-50 X 10"^ mol L
-1
(NEET 2021)
Find Y

(a) -0-144 V {b) +0-IOOV

Gutvmiic cell.s (c) - 0-344 V (d) - 0-100 V
25. Quinhydrone electrode is sometimes used to find
22. The reaction taking place in the cell
the pH of a solution. It is based on the following
Pt I H2(^?) 1 HCl(l-OM) I AgCl 1 Agis electrode reaction :
1 atm OH
0
(fl) AgCl + (l/2)H2 ^ Ag -I- H-" + Cl"
(b)
(c)
Ag + H-" + Cr
2Ag^-fH2-
^ AgCl + (l/2) H2
2 Ag + 2
(5) + 2 {aq) + 2e ^ ^ [OJ W
(d) 2Ag + 2H-" ■> 2 Ag'*' + H2
OH
(IAS Prelim 2010) O
Hydroquinone
Quinune

19. id) 21). (</) 21.0 ) 22.ia) 23.id) 24. ih)

ELECTROCHEMISTRY
3/127

Its standard electrode potential is 0-70 V. If in a 30. The standard reduction potentials for Zn-'^/Zn.
particular solution, the electrode potetilial is Ni^+ZNi and Fe^-^/Fe are - 0-76, - 0-23 and
found to be 0-58 V, the pH of the solution is - 0-44 V respectively. The reaction X + Y-*-
(a) 2 (b) 4 X-'*' + Y will be spontaneous when
(c) 6 (d) 8 («) X = Zn, Y = Ni ib) X = Ni. Y = Fe
26. Given E° = -0-74V , (c) X = Ni, Y = Zn
Cr^'^/Cr (d) X = Fe, Y = Zn
E° = 1-51V (AIEEE 2012)
2+
MnO^/Mn 31. A button cell used in watches functions as
follows :
= 1-33V, E“ = 1-36V
CrO?"/Cr^+ o/cr
Zn (s) + Ag20 (5) + HjO (/)
Based on the data given above, strongest 2 Ag (s) + (ag) + 2 OH" (ag)
oxidizing agent will be
The half-cell potentials are

w
(a) MnO^ (b) Cl-
Zn^^ (ag) + 2e 4 Zn (s) ;£●* = - 0-76 V
(c) (f/) Mii2+ Ag20 (i) + H2O (1) + 2 e ■>

Flo
(JEE Main 2013) 2 Ag (s) + 2 OH" (ag); E” = 0-34 V
27. Given -Cli/CI
E? = 1-36 V, Ecr^^/Cr = -0-74V, The cell potential will be

ee
(fl)MOV (/)) 042 V (c) 0-84 V (^0 1-34V
^CriO^-ZCr^-* =1-33V, 2+ =1-51V

Fr
(AIPMT 2013)

Among the following, the strongest reducing agent 32. A solution contains Fe^'*', Fe^'*’ and I" ions. This
solution was treated with iodine at 35°C. E° for

for
is

Fe-'+/Fe2+ is + 0-77 V and E" for I2/2I- is

ur
(fl) (b) cr
2+
0-536 V. The favourable redox reaction is
(C-) Cr (d) Mn
(a) I2 will be reduced to I"
k s
(JEE Main 2017)
Yo
(h) There will be no redox reaction
oo

28. Small quantities of solutions of compounds TX,

TY and TZ are put into separate test tubes (c) r will be oxidised to I2
eB

containing X, Y and Z solutions. TX does not (d) Fe-'*' will be oxidised to Fe-^'*'
react with any of these. TY reacts with both X and (AIPMT Main 2011)
Z. TZ reacts with X. The decreasing order of
r

33. Standard electrode potentials are

ou
ad

oxidation of the anions X" Y", Z~ is :

Fe-'*'/Fe, E" = - 0-44 V
(«) Y",Z-,X- (b) Z", X", Y"
Fe3+yFe2+,
Y

E° = + 0-77 V
(c) Y", X", Z" (d) X" Z",Y-
(DCE 2009)
Fe^"*", Fe-^'*' and Fe blocks are kept together, then
Re
nd

(a) Fe-^'*^ increases (b) Fe-^'*' decreases

29. Which of the following .statements are correct
(c) Fe^-^/Fe^-*- remains unchanged
Fi

concerning redox properties ?

(d) Fe-'*'decreases
(0 A metal M for which E° for the half life
reaction M**'*' + ne“ ; i M is very negative 34. Cr202--hl- ^ I2 + Cr^-'
will be a good reducing agent. E° = 0-79 V, E*" = 1-33 V, E° = 9
cell
Cr20^ 12 ●
(//) The oxidizing power of the halogen decreases
from chlorine to iodine. (a) 0-54 V (b) - 0-054 V
(///) The reducing power of hydrogen halides (c) -I- 0- i 8 V (d) -0-18 V
increases from hydrogen chloride to hydrogen 35. An aqueous solution containing one mole per
iodide
litre of each Cu(N03)2, AgNO^, Hg2(N03)2 and
(a) (i), (ii) and (iVi) (b) (i) and {//) Mg(N03)2 is being electrolysed using inert
(<r) (/) only (d) (ii) and (///) only electrodes. The values of the standard electrode
(e) (Hi) only potentials in volts (reduction potentials) are

25.(0) 26.(0) 27.(0 28. to) 29. (01 30. (o) .31. . t 1 /. .
.34. tot
 New Course Chemistry (XIl)ESSZS
3/128

Ag I Ag+ = 0-80, 2Hg I Hg2+ = +0-79 VI. Nernst eqn.» concentration cells
Cu I Cu“+ = + 0-34, Mg I Mg2+ = - 2-37 and calculation of equilibrium
of constant from Nernst equation
With increasing voltage, the sequence
deposition of metals on cathode will be 39. A hydrogen gas e.ectrode is made by dipping
(a) Ag, Hg, Cu. Mg (b) Mg, Cu, Hg, Ag platinum wire in a solution of HCl of pH = 10 and
(c) Ag, Hg, Cu id) Cu. Hg, Ag by passing hydrogen gas around the platinum
wire at one atm pressure. The oxidation potential
(e) Cu, Hg, Ag, Mg of the electrode would be
36. The EMF of a cell formed by combining a (a) 0-059 V (b) 0-59 V
particular electrode with standard calomel (c) 0-118 V (d) 1-18 V
electrode is found to be 0-344 V and calomel
(AIPMT 2013)
electrode is found to act as cathode. If the same
electrode is combined with standard hydrogen 40. How much will the reduction potential of a
electrode, the EMF of the cell will be (Given hydrogen electrode change when its solution
standard reduction potential, initially at pH = 0 is neutralized to pH = 7?

w
= + 0-244 V) (a) Increases by 0-059 V

F lo
{a) 0-344 V (b) 0-244 V (b) Decreases by 0-059 V
(c) 0-588 V (d) 0-100 V (c) Increases by 0-41 V
37. Given below are the half cell reactions : {d) Decreases by 0-41 V
41. Calculate the reduction potential of a half-cell

e
Mn^-^ + 2 e"

Fre
■> Mil, E° = 1-18 V
consisting of platinum electrode immersed in
2 (Mn^-*- + ->Mn"“+), E° = +l-51 V 2-0 M Fe-+ and 0-02 M Fe^+.
The E“ for 3 Mn~+ ■> Mn + 2 Mn^'*’ will be for
Given E° = 0-771V.
(a) - 0-33 V, the reaction will occur Fe^^/Fe^^
r
{b) - 2-69 V, the reaction will not occur (a) 0-653 V ib) 0-889 V
You
oks

(c) - 2-69 V, the reaction will occur (c) 0-683 V (d) 2-771 V
eBo

(d) - 0-33 V, the reaction will not occur (J & K CET 2010)
(JEE Main 2014) 42. Ej, Ej and E3 are the emf values of the three
38. Given below are half cell reactions : galvanic cells respectively
ad
our

(0 Zn I Zn2+ (I M) I Cu^+ (0-1 M) I Cu

MnO^ + 8 H'*’ + 5 e -> Mn^+ + 4 H2O,
(//) Zn I Zn^+ (1 M) I Cu^+ (1 M) I Cu
£° = -l-510V
(ifi) Zn I Zn^-^ (0-1 M) I Cu^-*- (1 M) I Cu
Re

Which one of the following is true ?

dY

1
-0.+2H* + 2e- ^H20, (a) E2 > E3 > Ej (b) E3 > E2 > Ej
2 2
Fin

e: = 4-1-223V (c) El >E2> E3 (d) Ej > E3 > E2

*-"02/H20 (Karnataka CET 2010, 2011)
Will the permanganate ion, MnO^ liberate O2 43. The cell, Zn I Zn^+ (1 M) 11 Cu^+ (1 M) I Cu, (E^e„
from water in the presence of an acid ? = 1-10 V), was allowed to be completely
discharged at 298 K. The relative concentration
(a) Yes, because E“^u = + 0-287 V
ofZn2+toCu2+, -— IS

(b) No, because = - 0-287 V ([Cu^^lJ

(a) 9-65 X 10^ (b) antilog 24-08
(c) Yes, because e”^j| = + 2-733 V
(c) 37-3 id)
(d) No. because E^^,, = - 2-733 V (NEET 2022) (AIEEE 2007)

ANSWERS

35. (c) 36. (d) 37. (/>) 38. (<i) 39. (h) 40. id) 41. (a) 42. (/;) 43. id)
ELECTROCHEMISTRY 3/129

44. An alloy of Pb-Ag weighing 1 08 g was dissolved

in dilute HNO3 and the volume made to 100 ml. A (a) -0-34 + (6) 0.34 V
silver electrode was dipped in the solution and
the EMF of the cell set up
0-0591 0-0591
Pt (s), H2 (g) I H+ (1 M) 11 Ag-" (aq) I Ag (s) (c) 0-34 + V (d) -0-34- V
2 2
was 0-62 V. If E° cell = 0-80 V, what is the
percentage of Ag in the alloy ? 50. For the calomel electrode, Hg, Hg2Cl2 I Cr (aq),
[At 25“C, 2-303 RT/F = 0-06] electrode potentials measured at different d“ ion
(a) 25 {b) 2-50 concentration are plotted against log [Crj. The
(c) 10 (d) 1
variation is correctly represented by the plot
(e) 5 (Kerala PET 2007)
45. The potential of the cell for the reaction E E

M(s) + 2H-"(1 M) ^H2(g)(l atm) .ia) (b)

+ M^+ (0-IM)

w
log [CP]
is 1-500 V. The standard reduction potential for log [Cn
M-VM (5) couple is

F lo
(a) 0-1470 V (b) 1-470V E
(C) E id)
(c) - 1-47 V (d) none of these

ee
46. Consider the following cell reaction :
log (cn log [crj

Fr
2 Fe (x) + O2 (g) + 4 (aq) > 2 Fe^’*' (aq)
+ 2 H2O (/), E" = 1-67 V 51. The following cell is found to have EMF equal to
At [Fe2+] = 10-3 ^ p = 0-1 atm zero
for
Pt. H2 (x atm) I 0-01 M H+ 11 0-1 M H+ I
r
and pH = 3, the cell potential at 25“C is
You
(a) 1-47 V (b) 1-77 V H2 (v atm). Pi
s
ook

(c) 1-87 V (d) 1-57 V (IIT2011) The ratio jc/y is

47. Value of E (a) 0-01 (b) 0-1
eB

at 298 K would be
H20/H2(latm)Pl
(c) 10 (d) 100
(a) - 0-207 V (b) + 0-207 V
52. The e.m.f. of the cell
(c) -0-414 V
our

(d) + 0-414 V
ad

Zn I Zn2+ (0 01 M) 11 Fe^+ (0-001 M) 1 Fe

48. The pressure of H2 required to make the potential
of H2-electrode zero in pure water at 298 K is at 298 K is 0-2905 volt. Then the value of
,-10 equilibrium constant for the cell reaction is
(a) 10 atm (b) 10^ atm
dY
Re

-14
(c) 10 atm (d) 10“'^ atm 0-32 0-32

(a) e00295 (b) 100 0295

Fin

(AMU Engg. 2015, MEET Phase I 2016)

49. For the electrode Cu/Cu^*, log [Cu^'^] (along 0-26 0-32

X-axis) is plotted against E^^j (along Y-axis). The (c) 1000295 (d) 1000591
plot obtained is shown in the fig. The electrode S3. Given at 25"C,
potential of the half cell Cu I Cu^'^ (0-1 M) will be
lAg(NH3)2r + e“ -> Ag + 2 NH3, E“ = 0-02 V
Ag+ + e' + Ag, E'’ = 0-80V
The order of magnitude of the equilibrium
Ered A constant of the reaction

0.34 V [Ag(NH3)2]-^ ^ ^ Ag+ + 2 NH3 will be

-8
O (a) 10 (b) 10-“’
log [Cu^^l (c) 10-’2 (d) 10-’-^

ANSWERS
44. (d) 45. (c) 46. (d) 47. (c) 48. (c) 49. (a) 50.(a) 51.(a) 52.{h) 53. id)
3/130 ■Pn€tde^p- '4. New Course Chemistry (XlDiz^aigu

^ Hg^ , E“ = + 0-855 V VII. Relation of Gibbs energy

54. 2 Hg
(DG"), cell potential
Hg ^ Hg-+, E° = +0-799 V equilibrium constant (K)
Equilibrium constant for the reaction
58. For the following cell
Hg -I- Hg- > He
++
at ITC is
o2
Zn (j) I ZnS04 {aq) I CUSO4 [aq) I Cu (5)
(a) 89 (b) 82-3 when the concentration of Zn^'*' is 10 times the
(c) 79 (d) none of these concentration of Cu^^ the expression for AG (in
J mo!"') is [F is Faraday constant, R is gas
55. For the following electrochemical cell at 298 K
constant, T is temperature, E° (cell) = M V]
Ft (s) I H2 (g, 1 bar) 1 (aq, 1 M) I M''"" (aq) I (a) 2-303 RT+ 1-1 F (b) I-l F
m2+ (aq) I Ft (.V)
(c) 2-303 RT - 2-2 F (d) - 2-2 F
[M^+ (aq)\ = 10-' (JEE Advanced 2017)
^cell = 0-092 V when
[M-** (aq)]

w
59. Standard free energies of formation (in kJ/moI) at
298 K are - 237-2, - 394-4 and 8-2 for
RT

F lo
Given E^4+/m 2+ = 0-151V. 2-303—
r
= 0-059 V
H,0 (/), CO2 (g) and pentane (g) respectively.
The value of for the pentane-oxygen fuel
The value of a: is
cell is
(^)-l

ee
(a)-2 (a) 1-968 V (b) 2-0968 V

Fr
(c) 1 (d)2 (d) 0-0968 V
(c) 1-0968 V
(JEE Advanced 2016) (AIPMT 2008)
56. In the electrochemical cell : for
60. A fuel cell involves combustion of butane at 1
r
Zn I ZnS04 (0-01 M) I CUSO4 (1-0 M) I Cu, atm and 298 K
You
the e.m.f of this Daniel cell is E|. When the
s
13
ook

concentration of ZnS04 is changed to I -0 M and C^i^]Q(s)+ - 02(g) —>4C02 (g) + 5H2O(0.

that of CUSO4 is changed to 0-01 M, the e.m.f
eB

AG° = - 2746 kJ mol"'

changes to E2. From the following, which one is
the relationship between Ej and E^ ? The E;.cell will be

(b) 1-09 V
our

(a) 0-545 V
ad

RT
Given = 0-059 (c) 0-922 V (d) 0-755 V
’ F
61. For a cell involving one electron, E°^^j = 0-59 V
(b) E| < E2
dY

(a) E, =£2
Re

at 298 K. The equilibrium constant for the cell

(c) E, >£2 (d) E2=0?^E|
Fin

(NEET 2017) 2-303 RT

reaction is (Given that = 0-059 V at
57. Find the emf of the cell in which the following F

reaction takes place at 298 K 298 K)

Ni (.?)-I-2 Ag-*- (O-OOIM) (fl) 1-0x10^'’ (b) 1-0 xlO^
Ni-.2+ (0-001 M) -H 2 Ag (s) (c) 1-0 X 10^ (d) 1-0 X 10"’
[Given that (NEET 2019)

2-303 RT 62. If the E!cell for a given reaction has a negative

E;„ = 1-05V, F
= 0-059 at 298 K]
value, then which of the following gives the
(a) 1-0385 V (h) 1-385 V correct relationships for the values of AG“ and
(c) 0-9615 V (d) 1-05 V(NEET 2022) K ... ?
eq

ANSWERS

54. (c) 55. (d) 56. (c) 57. (c) 58. (c) 59. (c) 60. (b) 61. (d)
ELECTROCHEMISTRY
3/131

{a) AG° > 0 ; K eq < I (ib) AG° > 0 ; K eq > 1

{h) formation of PbS04
ic) AG° < 0 ; K >1 (£0 AG" < 0 ; < ! (c) reduction of Pb^'*’ to Pb
vq

(AIPMT Prelim 2011, NEET Phase II 2016) id) decomposition of Pb at the anode
63. The half-cell reactions for rusting of iron are (AMU Engg. 2015)
1 69. 0 05 F charge is passed through a lead storage
2H++-Oo +2e- 4H2O, E" = + 1-23 V, battery. In the anodic half cell, the amount of lead
2 -

Fe2++2e- sulphate precipitated is (Molar mass of PbS04 =

^ Fe (.9), E" = - 0-44 V 303 g mor')
AG° (in kJ) for the reaction is ia) 22-8 g ib) 15-2 g
ia) -76 (b) - 322 ic) 7-6 g id) 11-4 g
(c) - 122 (£/) - 176 (JEE Main 2019)
64. For the cell reaction
X. Fuel cells
2 Fe^+ {aq) + 21“ {aq) 2 Fe^'*' {aq) + I2 {aq)
E“cell = 0-24 V at 298 K. The standard Gibbs 70. In a fuel cell, methanol is used as fuel and oxy
gen gas is used as an oxidizer. The reaction is

w
energy (A^ G°) of the cell reaction is (Given that 3

F lo
Faraday's constant F = 96.500 C mor') CH3OH (/) + - O2 {g) ■> CO2 (g) + 2 H2O (/)
{a) 23-16 kJ mol“' {h) - 46-32 kJ mol '
At 298 K, standard Gibbs energies of formation
(c) - 23*16 kJ mor' {d) 46-32 kJ mor' for CH3OH (/), H2O (/) and CO-, (g) are - 166-2,
(NEET 2019) - 237-2 and - 394-4 kJ mol“'^ respectively. If

ree
65. For the given cell : standard enthalpy of combustion of methanol is

Cu (5) 1 Cu2-»- (C, M) I Cu-+ (C2 M) I Cu (.9), (n) 80%

for F
- 726 kJ mor'. efficiency of the fuel cell will be
(h) 87%
change in Gibbs energy (AG) is negative, if ; {c) 90% {d) 97%
c,
Your
(AIEEE 2009)
{a) C2 =
V2 {h) C2=V2C|
ks

XI. Corrosion
eBoo

(c) C, =C2 (£0 Cl = 2 C2 71. Rust is a mixture of

(.lEE Main 2020)
(fi) FeO and Fe(OH)2 (b) FeO and Fe(OH)3
ad
our

VIIL Predicting products of electrolysis (f:) Fe203 and Fe(OH)3 (d) Fc304 and Fe(OH)3
66. In the electrolysis of which solution, OH“ ions 72. Zinc can be coated on iron to produce galvanised
are discharged in preference to Cl“ ions ? iron but the reverse is not possible. It is because
Re

(fl) dilute NaCI {h) very dilute NaCI {a) zinc is lighter than iron
Y

(c) fused NaCI {b) zinc has lower melting point than iron
Find

{d) solid NaCI

67. The metal that cannot be obtained by electrolysis (c) zinc has lower negative electrode potential
than iron
of an aqueous solution of its salt is
(a) Cr (b) Ag {d) zinc has higher negative electrode potential
than iron (NEET Pha.se II 2016)
(C-) Ca {d) Cu
(JEE Main 2014) XII. Miscellaneous

IX. Some commercial ce]l.s/l)atterie.s 73. A Daniell cell is set up by dipping a zinc rod
weighing 100 g in 1 litre of I-O M CUSO4
68. In the lead-acid battery during charging, the .solution. How long would the cell run if it
cathode reaction is
delivers a steady current of 1-0 ampere ? (Atomic
{a) formation of PbO-> masses : Cu = 63-5, Zn = 65)

ANSWERS
62. (£7) 63. (fe) 64. (fe) 65. (fe) 66. (fe) 67. (c) 68. (c) 69. (c) 70. (</)
71. (c) 72. (£/)
3/132 ‘p>utdee^ 'a New Course Chemistry (XIl)EBIBl

(a) 82-47 hrs (b) 53-61 hrs

(c) 41-23 hrs (d) 26-80 hrs
A A
74. For the half-cell reaction,

2 H^O + 2e~ -> + 2 OH", E*’ = - 0-8277 V at I I

1

298 K. Auloprotolysis constant of water Volume Volume

calculated from this value will be (P) (Q)
-10 -12
(a) I X 10 (/?) 1 X 10
-13 ,-14
(c) 1 X 10 {(f) 1 X 10
75. A lead storage battery has been used for one A
A
month (30 days) at the rate of one hour per day by I
1
drawing a constant current of 2 amperes. H2SO4

w
I I

consumed by the battery is Volume Volume

(b) 2-24 mole (R) (S)

(a) 1-12 mole

Flo
(c) 3-36 mole {d) 4-48 mole (a) (P) ib) (Q)
76. Given the data at 25®C, (c) (R) {d) (S) (IIT 2011)

e
80. The electrode potentials for Cu^'^ (aq) + e

re
Ag + P 4 Agl+e-;E° = 0-152V
Cii'^ (aq) and Cu"^ (aq) + e~ » Cu (5’) are
Ag ^ Ag-^+e"; E“^-0-800 V

F
■f 0-15 V and 4- 0-50 V respectively. The value of
What is the value of log K^p for Agl ?
ur will be

r
(2-303 RT/F = 0-059 V) ^Cu“+/Cu
(a) -37-83 (b) - 16-13
(d) + 8-612
fo
(a) 0-150 V (b) 0-500 V
ks
(c) -8-12 (c) 0-325 V (d) 0-650 V
Yo
77. The Gibbs energy for the decomposition of (AIPMT 2011)
oo

AI2O3 at 500 “C is as follows : 81. How long (approximate) should water be

B

-1 elecU-olysed by passing through 100 amperes

-AI2O3 — A1 -I- O^, A,.G = 966 kJ mol
3 ^ ' current so that oxygen released can completely
re

3
burn 27-66 g of diborane ?
The potential difference needed for electrolytic
(Atomic weight of B = 10-8 u)
u
ad

reduction of AI2O3 at 500®C is at least

Yo

(a) 6-4 hours (b) 0-8 hour

(a) 2-5 V (b) 5-0 V
(c) 3-2 hours (cf) 1-6 hours
(c) 4-5 V {d) 3-0 V (AIEEE 2010)
(JEE Main 2018)
d

78. For the cell reaction

Re

82. Consider the change in oxidation state of

in

2H2 (g) + 02 (g) ^ 2 H2O (0, bromine corresponding to different emf values
F

as shown in the diagram below :

Ar^298K = - 0-32 kJ K-‘
l●82V 1-5V
4 HBrO
What is the value of Ay-H° (H2O) ? Br04 4
BrOJ
Given : 0-> (g) + 4 H"*" (aq) + 4e~ — 4 2 H2O (0. Br“ 4 Br2 <■ 1-595 V
E°= 1-23 V 1-0652 V

(fl) - 189-71 kJ mol“' {b) - 285 08 kJ mor‘ Then the species under disproportionation is
(c) - 379-42 kJ mol"' (d) - 570-16 kJ mol"' (a) Br ih) BrO-
79. AgN03 (d(i) was added to an aqueous KCl
solution gradually and the conductivity of the (c) BtOJ {d) HBrO
solution was measured. The plot of con
(NEET 2018)
ductance (a) versus the volume of AgN03 is
ANSWERS

73.{b) 74. id) 75. (/j) 76.(/?) 77. («) 78. (h) 79. (f/) 80, (c)
81.(r) 82. id)
ELECTROCHEMISTRY 3/133

83. The standard electrode potential E® and its 84. Calculate the standard cell potential (in V) of the
dE°\ cell in which the following reaction takes place
temperature coefficient tor a cell are 2 V
dT Fe“+ (aq) + Ag+ (aq) > (aq) + Ag (s)
and - 5 X 10~^ V K ’ at 300 K respectively. The Given that
cell reaction is

Zn (j) + Cu^'*’ (aq) 4 (aq) + Cu (5) Ag-^/Ag

= -t V ; E
Fe-'^/Fc
= >’V;

The standard reaction enthalpy (A^ H“) at 300 K

in kJ mor' is E°Fe^-^/Fe
3^ = Z V

(Use R = 8 J K-* mol'^ and F = 96,000 C mor^) (a) x-z (b) x-y
(a) -412-8 (b) -384-0
(c) x + 2y-2z (d) -t + y-z
(c) 206-4 (d) 192-0
(JEE Main 2019) (JEE Main 2019)

w
Dl Multiple Choice Questions (with One or More than One Correct Answers)

F lo
85. Which of the following statements are not
(d) Oxidation of copper anode occurs in the
correct ?
electrolysis of aqueous copper sulphate

e
(a) Same quantity of electricity deposits more of solution using copper electrodes.

Fre
iron from ferric sulphate solution than from 88. In a galvanic cell, the salt bridge
ferrous sulphate solution
(b) Electrochemical equivalent of an element
for
(a) does not participate chemically in tlie cell
reaction
r
can be obtained by dividing its equivalent (b) stops the diffusion of ions from one electrode
You
weight by 96,500 to another
oks

(c) 1 Faraday always liberates 1 mole of the (c) is necessary for the occurrence of the cell
eBo

substance at the electrode reaction

(d) A 60 watt bulb emits 60 Joules of energy per (d) ensures mixing of the two electrolytic
second. solutions (JEE Advanced 2014)
our
ad

86. For the cell, TI I T1+ (0-001 M) I Cu^+ (O-I M) I 89. Some standard electrode potentials at 298 K are
^celf 25°C is 0-83 V. This can be increased given below :

(a) by increasing [Cu^'*’] Pb“+/Pb = - 0-13 V, Ni2+/Ni = - 0-24 V

dY
Re

(b) by increasing [Tl"^] Cd^+/Cd = - 0-40 V, Fe^^/Fe = - 0-44 V

2+
(c) by decreasing [Cu^'*'] To a solution containing 0-001 M of X and
Fin

0-1 M of the metal rods X and Y are inserted

(d) by decreasing [Tl'*’]
(at 298 K) and connected by a conducting wire.
87. Which of the following are correct ? This resulted in dissolution of X. The correct
(a) Electrolysis of dilute NaOH solution gives combination(s) of X and Y, respectively, is (are)
H2 at cathode and O2 at anode. (Given : Gas constant, R = 8-314 JK“' mol"' ;
(b) Electrolysis of sulphuric acid (dilute or concen Faraday constant, F = 96500 C mol"*)
trated) gives H2 at cathode and O2 at anode, (a) Cd and Ni (b) Cd and Fe
(c) Ni and Pb (d) Ni and Fe
(c) Electrolysis of aqueous KF solution gives
fluorine at the anode. (JEE Advanced 2021)

ANSWERS

83. (n) 84.(c) 85. (a,c) 86. (a.d) 87. (a.d) 88. (a.h) 89. ia.b.c)
3/134
New Course Chemistry (XII)BEIHl

nn Multiple Choice Questions (Based on the given Passage/Compreh ension)

Each comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option.

ICdmprehensibniC There are two principal (a) -0-42V (b) -2-20V

types of electrochemical cells. A galvanic cell (c) 0-52 V id) I 04 V.
IS an electrochemical cell that produces 92. The standard reduction potentials for two
electricity as a result of spontaneous reaction reactions are given below :
occurring inside it. An electrolytic cell is an
electrochemical cell in which a non- AgCl (5) + e ^Ag is) + C\-{aq), E° = 0-22V
spontaneous reaction is driven by an Ag* {aq) + e~ Ag (i). E" = 0-80 V
external source of current. Any redox The solubility product of AgCl under standard
reaction may be expressed in terms of two conditions of temperature (298 K) is given by

w
half-reactions which are conceptual
(a) 1-6 X 10-5 {h) 1-5x10-®
reactions showing the loss and gain of

F lo
-10 -10
electrons. Each half reaction has a definite (c) 3-2 X 10 id) 1-5 X 10
value of standard electrode potential. The 93. If hydrogen electrodes dipped in two solutions of
overall reaction is represented by a pH = 3 and pH = 6 are connected by a salt bridge,

e
universally accepted method. Knowing the the emf of the resulting cell is

Fre
standard electrode potentials of the half {b) 0-3 V
(fi) 0-177 V
reactions, the standard EMF of the cell can
be calculated. The standard EMF further
helps in the calculation of free energy
for
(c) 0-052 V (d) 0-104 V
r
change, equilibrium constant of the cell ggomprehensiori] A lead storage battery
You
reaction as well as parameters like solubility consists of a lead anode and a grid of lead
oks

product of a sparingly soluble salt. A cell can packed with lead dioxide as the cathode. The
eBo

also be set up in which the two electrodes electrolyte taken is 39% H2SO4 by mass
may be of the same type, e.g., both may be having a density of 1*294 g mL“^ The
hydrogen electrodes but the concentration of battery holds 3*5 L of the acid. During the
our
ad

ions in the two solutions may be different. discharge of the battery, the density H2SO4
Such cells are called concentration cells.
falls from 1*294 g mL"’ to M39 g niL"^ which
90, The reaction is 20% H2S04by mass.
dY
Re

1
- H2(g) + AgCl(5) ^H+ {aq) + C\-{aq) 94. The reaction occurring at the anode during
Fin

charging is
+ Ag (5)
(a) Pb-*+2e^- ^ Pb
occurs in the galvanic cell
(a) Ag I AgCl (a-) I KCl (aq) I AgN03 (aq) I Ag (h) Pb2++SOj > PbSO 4

(b) Pt I H, (g) I HCl (aq) 1 AgNO, (aq) I Ag (c) Pb ^Pb-++2c-

(c) PtI H,\g) 1 HCl (aq) 1 AgC! (s)\ Ag (d) PbSO^ + 2 H^O ●>

(d) Pt 1 H, ig) I KCl (aq) I AgCl (s) 1 Ag

91. The standard electrode potential (E°) for OC\~/ 2Pb02+4H-"+S05-+2e-
95. Moles of sulphuric acid lost during discharge is
Cl'and Cl respectively are 0-94 V and («) 9-88 (b) 8-88

- 1-36 V. The E" value for OCr/^C^ will be (c) 7-88 (d) 6-88

ANSWERS

90. (c) 91. (c) 92. id) 93. (n) 94, id) 95. (n)
ELECTROCHEMISTRY 3/135

96.
Molarily of the solution after the discharge is 99. For the above cell

(a) 8-136 (b) 4 068 (a) E^eii < 0 ; AG > 0 (b) E^g,, > 0 ; AG < 0
(c) 2-32 id) M6 (c) E^^ii < 0 : AG" > 0 id) E^^ii > 0 ; AG" < 0
97.
The amount of charge in coulombs used up by the 100. If the 0 05 molar solution of M'*’ is replaced by a
battery is nearly 0-0025 molar M'*’ solution, then the magnitude of
the ceil potential would be
(«) 954180 ib) 477090
(a) 35 mV ib) 70 mV
ic) 95418 id) 47709 (c) 140 mV id) 700 mV
98. The number of ampere-hour for which the battery
must have been used is (IIT 2012). The electro
ia) 2650-5 ib) 265-05 chemical cell shown below is a concentration

ic) 26-505
cell: M I (saturated solution of sparingly
id) 2-6505
soluble salt MX2) I (0*001 mol dm"-^) I M

w
The emf of the cell depends on the difference
comprehension' in the concentration of ions in the two
(IIT 2010). The

F lo
electrodes. The emf of the cell at 298 K is
concentration of potassium ions inside a 0 059 V.
biological cell is at least twenty times higher
than the outside. The resulting potential 101. The solubility product (K^^,; moP dm ^) of MX2

ee
difference across the ceil is important in

Fr
at 298 K based on the information available for
several processes juch as transmission of the given concentration cell is
nerve impulses and maintaining the ion
balance. A simple model for such a for
(take 2-303 x R x 298/F = 0-059 V)
ur
ia) 1 X 10-'5 ib) 4x 10
-15
concentration cell involving a metal M is -12 -12
ic) 1 X 10 (i/) 4 X 10
M(s) I M'*' (aq ; 0-05 molar) I M"*" (aq ; 1 molar)
s
ook

102. The value of AG (kJ mol"*) for the given cell is

Yo

I M(s)
For the above electrolytic cell, the magnitude (take 1 F = 96500 CmoH)
eB

of the cell potential I E^^jj I = 70 mV. ia) -5-7 ib) 5-7

(c) 11-4 id) - 11-4
our
ad

09 Matching Type Questions

Y

Match the entries of column I with appropriate entries of column II and choose the correct option
Re

out of the four options (a), (b), (c), (d) given at the end of each question.
nd

103. Column I Column II

Fi

For the Daniell cell using Zn & Cu electrodes

(A) Concentration of copper sulphate solution (P) EMF of the cell increases
is doubled
(B) Concentrations of zinc sulphate solution (d) EMF of the cell decreases
is doubled
(C) Concentrations of both the solutions are (r) EMF of cell becomes equal to standard EMF
doubled
(D) Concentrations of both the solutions are is) No effect on EMF
kept equal

ia) A-p, B-q, C-s, D-r (b) As, B-r, C-p, D-q (c) A~q, B-r, C-^, D-;; (d) A-r, B-s'. C-^, D-p

ANSWERS

96. (c) 97. (a) 98, ib) 99. ih) 100. (c) 101. (h) 102. id) 103. (a)
3/136 ‘PKadee^'^ New Course Chemistry (XII)I!Z5IH1

104. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation
in conductivity of these reactions is given in List II. Match List 1 with List II and select the correct
iinswer using the code given below the lists :
List I List II

P. (C2H5>3N + CH3COOH 1. Conductivity decreases and then increases

X Y

Q. KI (01 M) + AgN03 (0-01 M) 2. Conductivity decreases and then

X Y does not change much
R. CH3COOH + KOH 3. Conductivity increases and then
X Y does not change much
S. NaOH + HI 4. Conductivity does not change much
Y and then increases

low
X

Codes : P Q R s

(a) 3 4 2 I

ib) 4 3 2 1

(c) 2 3 4 1

ee
(d) \ 4 3 2 (JEE Advanced 2013)

rF
Fr
105. The standard reduction potential data at 25°C is given below ;
E° (Fe^+, Fe^-*-) = + 0-77 V, E“ (Fe^^, Fe) = - 0-44 V; E“ (Cu^+, Cu) = + 0-34 V, E“ (Cu-^, Cu) = + 0-52 V

r
E° IO2 (g) + 4 + 4 e" ^2H20]=+1-23V; E” [O2 (g) + 2 H2O + 4 e- 4 OH-] = + 0-40 V
E° (Cr^+, Cr) = - 0-74 V, E" (Cr^+, Cr) = - 0-91 V fo
u
ks
Match E° of the redox pair in List I with values given in List II and select the correct answer using the code
Yo
given below the lists ;
oo

List I List II

E“ (Fe^+, Fe)
B

P. 1. -0-18V

2. -04 V
E°(4 H20^- 4 H-*- + 4 OH")
re

Q.
R. E° (Cu^+ + Cu - -> 2 Cu+) 3. -0-04V
E° Cr^-*-)
u

S. 4. - 0-83 V
ad
Yo

Codes : P Q R S

(a) 4 1 2 3

ih) 2 3 4 1
nd
Re

(c) 1 2 3 4

(d) 3 4 1 2 (JEE Advanced 2013)

Fi

Matrix-Match Type Questions

p q r s

Match the entries of column 1 with appropriate entries of column II. Each 1 r I [

entry in column I may have one or more than one correct option from A 0®O©
column II. If the correct matches are A-p, s ; B-r ; C-p, q ; D-5, then the
correctly bubbled 4x4 matrix should be as follows :
B
©J©IP ©j
106. Column 1 Column II
c ©©0©
(A) Electrode on which oxidation occurs (p) Anode
(B) Electrode on which reduction occurs (q) Cathode
d:
t
®ii©iioiI©
ciz: 'J t 1 f

(C) Electrode connected to negative pole of the battery (r) Negative pole
(D) Electrode connected to the positive pole of the battery (5) Positive pole
ANSWERS

104.{a) 105. (f/) 106. (A-p.r: B-q,s ; C-q ; D-p)

 ELECTROCHEMISTRY 3/137

107. Column I Column II

(A) Copper (/?) Most active metal
(B) Zinc
iq) Least active metal
(C) Silver
(r) Reacts w'ith acid to give gas
(D) Aluminium
(.?) Does not react with acid to give gas
108. Column 1 (Cell) Column II (Electrolyte use '.)
(A) Dry cell ip) Mercuric oxide
(B) Ruben-Mallory cell iq) Zinc chloride
(C) Nicad cel! (r) Potassium hydroxide
(D) - Ot fuel cell (5) Ammonium chloride

VI.
Integer Type Questions

w
A B C D

DIRECTIONS. The answer to each of the following questions is a @ ®®©

F lo
single digit integer, ranging from 0 to 9. If the correct answers to the
question numbers A, B, C and D (say) are 4, 0, 9 and 2 respectively, ® ®o®
then the correct darkening of bubbles should be as shmvn on the side :
® ®®®

ee
109. Three litres of 0-5 M K-,Cr20-7 solution have to be completely reduced in the

Fr
acidic medium. The number of faradays of electricity required will be ® ® ®®
110. In the Mg-Al cell, the number of electrons involved in the cell reaction is ® ® @®
for
ur
111. For the Mg-Ag cell, how many times the difference between the EMF of the cell
and its standard EMF will change if concentration of Mg-'*' ions is
® ©® ®
changed from O-I M to 0 01 M and that of Ag'*’ ions is changed from 0-5 M to ® ©© ©
s
ok

0-25 M ?
Yo

0 0©©
o

112. All the energy released from the reaction X > Y, A^G“ = - 193 kJ mol-' is
eB

used for oxidizing M'*’ as » M^"*" + 2 e", E“ = - 0-25 V. Under standard

conditions, the number of moles of M'^ oxidized when one mole of X is
© ©©©
converted to Y is (F = 96500 C mol"') (.TEE Advanced 2015) ®©®©
r
ad
ou

113. The molar conductivity of a solution of a weak acid HX (0-01 M) is 10 times .smaller than the molar
conductivity of a solution of weak acid HY (0-10 M). If X,°X" = Y" ’ the difference in their /jK^ values.
Y

/?K^ (HX) - (HY) is (consider degree of ionization of both acids to be « 1) (JEE Advanced 2015)
Re
nd

114. The conductance of a 0-0015 M aqueous solution of a weak monobasic acid was detennined by using a
conductivity cell consisting of Pt electrodes. The distance between the electrodes is 120 cm with an area of
Fi

cross-section of I cm^. The conductance of this solution was found to be 5 x 10"^ S. The pW of the solution
is 4. The value of the limiting molar conductivity (a°^) of the weak monobasic acid in aqueous solution is
Z X 10*. S cm^ mor*. The value of Z ISi (JEE Advanced 2017)
115. For the electrochemical cell

Mg is) 1 Mg^+ iaq, 1 M) I Cir+ (aq, 1 M) I Cu (i-)

the standard e.m.f of the cell is 2-70 V at 300 K. When the concentration of Mg^^ is changed to x M, the cel!
potential changes to 2-67 V at 300 K. The value of a: is
F
(Given, —
K
= 11500 K V"' where F is the Faraday constant and R is the gas constant. In (10) = 2-30)

(JEE Advanced 2018)

ANSWERS
107. iA-q.s ; B-r ; C-q.s ; D-p.r) 108. {A-q.s ; B-p.r; C-r ; D-r) 109. (9)
110.(6) 111.(2) 112.(4) 113.(3) 114.(6) 115.(10-00)
3/138 New Course Chemistry (X11)E2SX91

116. The magnitude of the change in oxidizing power of the MnO^/Mii-'*’ couple is r x 10^ V if the H'*’
concentration is decreased from 1 M to 10~^ M at 25*’C (Assume concentration ot MnO^ and Mn" to be
same on change in concentration). The value of .t is (Rounded off to the nearest neighbour)
Given : 2-303 RT/F = 0 059 (JEE Main 2021)

117. 5-0 mmol dm"-^ aqueous solution of KCl has a conductance of 0-55 mS when measured in a ceil of cell
constant 1 -3 cm”^. The molar conductivity of this .solution is in S m- mor’ (Round off to the nearest
integer) (JEE Main 2021)

118. The cell potential for the following cell

Pi I H2 (g) {aq) I Cu--" (0-01 M) 1 Cu (.v)
is 0-576 V at 298 K. The /?H of the solution is (Nearest integer)

2-303 RT

w
(Given ; E°Cu^'^/Cu = 0-34V and = 0-06V ) (JEE Main 2021)
F

F lo
VII. Numerical Value Type Questions Decimal Notation)

ee
For the following question, enter the correct numerical value (in decimal-notation, tnincated/rounded-off to the
second decimal place, e.g., 6-25, 7*00, - 0-33, 30*27, - 127*30) using the mouse and the onscreen virtual numeric

Fr
keypad in the place designated to enter the answer.
119. Consider an electrochemical cell

for
ur
2n+
A (s) I A"+ (aq, 2 M) II B (aq, 1 M) 1 B (s)
The value of AH° for the cell reaction is twice that of AG° at 300 K. If the c.m.f of the cell is zero, the AS° (in
s
ook

JK“^ mol"^) of the cell reaction per mole of B formed at 300 K is

Yo

(Given In (2) = 0-7, R (universal gas constant) = 8-3 mof ^ H, S and G are enthalpy, entropy and Gibbs
eB

energy respectively) (JEE Advanced 2018)

120. Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K.
r

Its cell reaction is

ad
ou

1
H2(g) + -02(g) ^ HoO (/)
Y

The work derived from the cell on the consumption of 1 -0 x 10“^ mol of H2 (g) is used to compress 1 -00 mol
Re
nd

of a monoatomic ideal gas in a thermally insulated container. What is the change on temperature (in K) of the
Fi

ideal gas ?
The standard reduction potentials of the two half cell are given below ;
O2 (g) 4 H'*' (aq) + 4 e ^2H20 (D, E”= 1-23 V
2 H+ (aq) + 2e~ ^ H^, E” = 0-00 V
Use F - 96500 C mol
-1
R = 8-314 Jmol'* K"* (JEE Advanced 2020)

121. The reduction potential (E°. in V) of MnO^ (aq) I Mn (s) is

(Given : E = 1-68V ; E° = 1-21V ; E° = -I-03V)
Mn04 (fl</)/Mn02 (●r) MnO-) (i)/Mn"‘^ (aq) (5)

(JEE Advanced 2022)

ANSWERS

116.(3776) 117.(14) 118.(5) 119. (-11-62) 120.(13-2) 121.(0-77)

ELECTROCHEMISTRY 3/139

Question Stem For Q. Nos. 122,123 (JEE Advanced 2021). At 298 K, the limiting molar conductivityof a
weak monobasic acid is 4 x 10^ S cm" moI~*. At 298 K, for an aqueous solution of the acid, the degree of
dissociation is a and the molar conductivity is y x 10" S cm" mol"*. At 298 K. upon 20 times dilution with
water, the mohu- conductivity of the solution becomes 3 y x 10" S cm^ mol"’.
122. The value of a is

123. The value of y is

'm Assertion-Reason Type Questions

TYPE 1

DIRECTIONS. Each question contains Statcment-1 (Assertion) and Statement-2 (Reason). Each
question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correction
option as under :

w
(a) Statemcnt-1 is TViie, Statement-2 is True ; Statement-2 is a correct explanation of Statenient-1.
(b) Statement-1 is True, Stateinent-2 is True ; Stateinent-2 is NOT a correct explanation of Statement-1,

F lo
(c) Statement-1 is True, Statemeiit-2 is False,
(d) Statement-1 is Fal.se, Statement-2 is 'lYue.

ee
124. Statement-1. Specific conductivity of an electrolytic solution decreases with dilution whereas molar

Fr
conductivity increases wi^'i dilution.

for
Statement-2. Specific ● nductivity is the conductance of a specific amountof the electrolytewhereas molar
conductivityis for 1 rr ale of the electrolyte.
ur
125. Statement-1.Iron is protected from corrosion by connecting magnesiummetal with it.
s
Statement-2. Iron acts as a cathode and magnesium us anode which gradually disappears.
ook
Yo

126. Statement-l. The cel! constant of a conductivity ceil depends upon the nature of the material of the
electrodes.
eB

Statement-2. The electrodes of the cell arc coaled with platinum black to avoid polarisation effects.

127. Statement-1. When acidified zinc sulphate solution is electrolysed between zinc electrodes, it is zinc that is
our

deposited at the cathode and no H, gas is evolved.

ad

Statement-2. The electrode potential of zinc is more negative than that of hydrogen as the overvoltage for
the H2 evolution on zinc is quite large.
Y

128. Statement-1. Electrolysis of an aqueous solution of KI gives I2 at the anode but that of KF gives Ot at the
Re

anode and not F^.

nd

Statement-2. F2 is more reactive than L.

Fi

TYPE II

DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a
correspondingstatementof Reason (R) just below it. Of the statements, mark the correct answer as
(a) If both assertion and reason are true, and reason is the true explanation of the assertion,
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion,
(c) If assertion is true, but reason is false,
(d) If both assertion and reason are false.

129. Assertion. Electrolysis of molten CaH, produces hydrogen gas at anode.

Reason. In CaH2, hydrogen is present in the form of H". (AIIMS 2014)

ANSWERS
122. (0-22) 123. (0-88) 124. (c) 125. (t/) 126. (d) 127. (a)
128.(b) 129. (a)
3/140 T^>uuUefr'A New Course Chemistry CXIl)EZs]Sl

130. Assertion. On dilution, the equivalent as well as molar conductivity of the solution increases.
Reason. With dilution, the number of current carrying particles per cm^ increases. (AlIMS 2009)
131. Assertion. According to Kohlrausch law, the molar conductivity of a strong electrolyte at infinite dilution is
sum of molar conductivities of its ions.
Reason. The current carried by cation and anion is always equal. (AIIMS 2007)

132. Assertion. In the electrolysis of aqueous NaCl. Na is preferentially di.scharged at the mercury cathode
forming sodium amalgam.
Reason. It is due to the fact that hydrogen has high overvoltage at mercury cathode. (AIIMS 2007)

133. Assertion. The cell potential of mercury cell is 1-35 V which remains constant.
Reason. In mercury cell, the electrolyteis a paste of KOH and HgO. (AIIMS 2008)
134. Assertion. As a lead storage batteiy gets discharged,density of the electrolyte present in it decreases.
Reason. Lead and lead dioxide both react with sulphuric acid to fonn lead sulphate. (AIIMS 2011)

w
135. Assertion. Galvanised iron does not rust.

Reason. Zinc has a more negative electrode potential Uian iron.

F lo
ee
For Difficult Questions

Fr
Multiple Choice Questions (with one Correct Answer)
for
ur
1. Electrochemical equivalent is the weight 3. NaCl + aq 4 Na-*- + cr,
s
depositedby 1 coulomb. H2O ^ H+ + OH-
ook
Yo

Q =I X r 1 = I X 0-25 or 1 = 4 A. 1
H* + e~
eB

Electrolysis
2. NaCl + H2O
Thus, 0-5 mole of H2 is liberated by 1 F =
1 1
96500 C
our

NaOH + -H2+-Cl.,
ad

2^2-
0-01 mole of H2 will be liberated by charge
At anode At cathode
96500
400 ml of 1 M NaCl solution contains NaCl x001 = 1930C
dY

0-5
Re

1
X 400 mol =0-4 mol
Q = Ixror , =
Fin

1000
I lOxlO-^A
H2O 1
As Na'*’ + e > Na > NaOH + -H. = 193000 sec = 19-3 x 1 O'* sec
2 ^ 4. Al3++3e-—>A1
1 F produces 1 mol of NaOH C\i^-^ + 2e~ >Cu
.-. 0-04 F produces NaOH = 0 04 mol Na"'" + e~ > Na
Thus, 400 ml of the solution now contain 0 04
mol of NaOH 1 1
Molar concentration of NaOH sol.
Thus, 1 F will be deposit - mole of Al, — mole

0-04 of Cu and 1 mole of Na. Hence, molar ratio

xlOOO =01M .
400 1 1
,-13 : T : 1 = 2 : 3 : 6.
[OH-] = 01 M, [H+] = 10 M pH = 13. 3 2

ANSWERS

130. (c) 131. (d) 132. (a) 133. (b) 134. (a) 135. («)
ELECTROCHEMISTRY 3/141

For Difficult Questions

9. By Faraday’s second law
’ W
5. Mg^-*- + 2e~ ■>Mg O, Eo,
2 F, i.e., 2 X 96500 C deposit Mg = 1 mole
Weight of 5000 mL of O2 at STP
9-65 C will deposit Mg
1 32
x9-65 = 5x10 ^mole X5600 = 8g
2x96500 22400

RX X Mg — ^ RMgX W
Ag _ 108
1 mole of Mg forms G.R. = 1 mole or W^g=108g
5 X 10“^ mole of Mg will form G.R.
= 5 X 10'^ mole. 10. Charge flowing during electrolysis = I x r
= 1 X 60 = 60 coulombs
6. Q = 2 X 5 X 60 X 60 = 36000 C
Thus, 36000 C deposit metal = 22-2 g But charge flowing = number of elecu-ons flowing

w
96500 C will deposit metal X charge on each electron
19
60 = n X 1-60 X 10

F lo
22-2
x96500g = 59-5g
36000 60
or n = = 37-5x 10‘^ = 3-75x 10"®
Thus, Eq. wt. of the metal = 59-5 1-60x10"'^

e
Valency or ox. state = At. wt./Eq. wt.

Fre
11. CaCl2 ^ Ca2+ + 2 Cr,
= 177/59-5 = 3.
Ca^+ -I- 2 e“ ●> Ca
7. Suppose valencies of A, B and C are a, b and c
respectively. Then their equivalent weights will
for
To produce 1 mole of Ca (40 g),
be (At. wt/valency) 7/a, 2Hb and 48/c. When Faradays required = 2
r
same quantity of electricity is passed, masses
You
To produce 20 g of Ca, Faradays required = 1
oks

deposited are in the ratio of their equivalent 12. In H2SO4 solution.

weights, i.e..
eBo

Mass of A Mass of B _ Mass of C H2SO4 ^ ± 2 + SOj-

Eq.wtofA Eq.wtofB Eq.wlofC H2O r ^ H+ + OH-
our
ad

2-1 _ 2-7 _ 7-2 or 0-3 a = 0-1 ef = 0-15c

At cathode :

7/a " 2Hb ~ 48/c 2 H-^ -I- 2 c-

.H2
1 At anode : OH" ions have lower discharge
dY
Re

or )7 a = b = \-5 c. Thus, \f b= 1, then =~

3’ potential than SO4 ions, hence,
Fin

1 _2 4 0H- ^ 2 H2O + O2
1-5 " 3
13. K = Conductance (G) x Cell constant (G*)

Ratio a : b .■ c = - = 1:3:2 I
3 xO-78 =3-9 X KT^ cm
-I

2-0x10^
8. Al^-^ + 3 c- ^ A1

Quantity of electricity passed IOOOk 1000x3-9xl0-^

= (4-0 X 10^ A) X (6 X 60 x 60 .v) A =

Molarity 0-10
= 864 X 10® coulombs
= 3-9 S cm^ mol '
3 F, i.e., 3 X 96500 C produce A1 = 1 mole, i.e.,
27 g A 3-9
= 10“- =0-01
864 X 10® C will produce AI
a =
A 390
0
27
3x96500
x864xl0®g HA ^

= 8-058 X lO^g = 81 X lO'^g. [H+] = C a = (0-10) (10-2) = 10“2 M

3/142 New Course Chemistry (XIOKSSISI

Ionic conductance
For DiffIcuH Questions 17. Ionic mobility =
96,500
/?H = 3
5x10-^

rw
= 518x10“^ cm/sec .
K
[H+][A"1 lO-^xlO"^ = 10-5 96,500
a
[HA] 0-1
a‘:m
pK, = 5. 18. Degree of dissociation (a)

e
14. Smaller the size of an ion, more strongly it is
HI

hydrated and less is the ionic mobility. The sizes a£. =axA HI
0
= (l-9x 10-^) X (200 + 350)

e
HI

of the unhydrated ions increase in the order: S cm^ mol"'

Na"^ < K+ < Rb+ < Cs+

luo
r
= 1-9 X 550 X 10“^ = 1-045 x 10"^ S cm^ mor'.
The sizes of the hydrated ions will be in the

F
order: 19. A% (BaS04) = (BaCl2) + a%^ (H2SO4)
Cs+ < Rb-^ < K+ < Na+. -2 A",^ (HCl)
= ATj + ;C2 - 2 AT3

oF
Hence, their ionic mobilities will be in the order :
Cs-^>Rb+>K+>Na-^ lOOOXK

rs
15. Case I: C = 0-2 M. R = 50 n, k = 1 -4 S m"' A%^(BaS04> = Normality (Solubility in geqL ')

ok
K= Conductance x Cell const. = — x Cell const. lOOOxy
R Solubility in g eq L '

fo
+X2 - 2xj
Cell constant = k x R = (1-4 S m (50 Q)
-1
o lOOOv
Y
= 70m -1
Solubility in mol L
Case II: C = 0-5 M, R = 280 Q, Cell const. = 70 m"' 2(.Yj + .t2 - 2xj)
Y
rB

1 1 -1 ± Ba2++S02-
K = — X Cell constant = x70m BaSO 4
R 280Q

= 0-25 S m"' K^.^, for BaS04 =[Ba^+] fS05’] = sxs = s'^

ue

Molar conductivity (in S m^ mol ')

od

-]2 106
lOOOy
K(Sm ') 2(.r, -f ^2 -2.X3) 4(x^+X2-2x;^)^ ■
ad
in

Molarity in mol m'^

20. A®,„ (Sod. pot. oxalate) = (Na"^) + (K^)
K(Sm-') + X" (oxalate)
Molarity (mol L ')xl000Lm ^ = (50-1 + 73-5 + 148-2) S cm^ mol"'
Re
F

= 271-8 S cm- mol-'

-1
0-25 Sm
A°
0-5molL-'xl000Lm-^
,m
A°
eq
Total charge on cations or anions (n factor)
= 5x10"^Sm^mol '
= .?Zl^
2
= 135.9Scm2eq -1
16. A HI
Kj X1000 _KjX 1000 = 2k
I
Molarity 10 mol
21. A
tn
= 20 Sem^ mol-'
0-02 L

Kj xlOOO A
m
'^CHjCOO-
+ A
H+
A
HI, 20 mol = 4K2
= 50 + 350
0-08 L
= 400 S cm^ mol"'
Given k, = K2 A
A A HI 20 1
a =
niy HI2 or A = 2 A HI A 400 20
2 4 Hlj m
ELECTROCHEMISTRY 3/143

28. TX does not react with any solution. This means

For Difficult Questions
that X" is least easily oxidized.
TY reacts with both X and Z. This means Y is
Ca- 1 ^2
K = Ca2=7xlO“^x oxidized by both.
1 - a V20 (Y- + X ^Y + X-, Y- + Z- 4 Y + Z-)
1
= 7x10"^ x-xl0~- TZ reacts with X only (Z~ + X ■> Z + X").
4 This means Z" is less easily oxidized than
= 1-75 x 10-'-’ moIL-l Y". Thus, decreasing order of oxidation of anions
22. LHS electrode (oxidation half reaction) X", Y", Z- will be ; Y“, Z~, X~
29. (/) High negative E° for M"'^ + lie ± M
^ H-" + e means a high positive value for the reverse reac
H+
tion, M ^ i M + ne . This means M is
RHS electrode (reduction half reaction) oxidized very easily. Hence, it is a good reducing

w
AgCI + e Ag + Cl- agent,
Adding the two half reactions, we get (n) Reduction potentials of halogens are in the

Flo
l order : CI2 > Br2 > l2- Thus, CI2 is reduced most
AgCl + ^H2 ^ Ag + H+ + cr. easily and hence is the best oxidizing agent.

ee
iiii) The size of the halide ions is in the order
23. E“,„=E”Cu^-^/Cu -E = 0-34 - (- 0-76) Cl”< Br“< I-. Greater the size of the halide ion,

Fr
Zn2''-/Zn
more easily it can lose electrons and get oxidized.
= MOV
Thus, I" ions can be oxidized most easily and
If E = M V, no current will flow

for
hence have the greatest reducing power.
ur
If E > 11 V, cell will act as electrolytic cell and 30. The reaction X + Y^^ ^ X2+ + Y will be
the electrons will tlow from Cu to Zn
spontaneous if is -i-ve.
s
If E < 1-1 V, cell acts as galvanic cell so that Zn
k
2+
^ Zn^-" + Ni,
Yo
(a) Zn + Ni
dissolves and Cu deposits
oo

_ pO = -0-23-(-0-76) V
24. Standard electrode potential is fixed quantity for Nr+ZhTi Zn2-*/Zn
eB

a given electrode. Hence, its standard reduction = + 0-53 V

potential against SHE will remain same viz. 2+
(b) Ni + Fe 4 Ni^-^ + Fe,
0100 V (EMF of the cell formed will change).
r
ou

Ecell = ^Fc2-"/Fe = _ 0-44 - (- 0-23) V

ad

25. For the given electrode reaction (reduction

reaction). = -0-21 V
Y

00591, 1 (c) Ni + Zn- 4Ni2+-t-Zn,

E = E°-
2 ®[H+]- Ecell “ ” ^
Re
nd

= E° + 0 0591 log [H-"j = E" - 0-0591 pH (d) Fe -H Zn^-^ ■> Fe-+ + Zn,
Fi

0-58 = 0-70 -0-0591 pH e!cell = E -e! = - 0-76 - (- 0-44) V

Zn-*/Zn Fc-+/Fe
0-12 = - 0-32 V
or pH = = 2-0.
0-059 31. E° cell = E‘ Calhode - E‘ Anode = 0-34 - (- 0-76) V
26. Higher the reduction potential, more easily it is = MOV
reduced and hence stronger is the oxidizing 32. The reaction will take place for which E“j,gu is
2+ : positive.
agent. As reduction potential of MnO^/Mn is

(n) I2 + 2 e > 2 I", = 0-536 V

maximum, hence it is strongest oxidizing agent. Fe--^ ^ Fe^++ <?-, EV =-0-770 V
27. Greater the standard oxidation potential (or lesser
ox

E‘ ceil will be -ve.

is the standard reduction potential), more easily
(b) Redox reaction will occur.
the substance is oxidized and hence stronger is
(c)2l- >12 + 2 e-, E**, = -0-536 V
the reducing agent. As Cr'^^/Cr (i.e., standard ox

Fe^+ -I- e > Fe2+, E" red = + 0-770 V

reduction potential) is lowest, hence, Cr will be
strongest reducing agent. 2Fe^++2I- > 2 Fe^-^ + I,, E° cell = + 0T64 V
3/144 New Course Chemistry (XII)CSIS

As it is +ve, cell reaction will take place, i.e., O2

For Difficult Questions
will be liberated and E cell = + 0-287 V

Thus, E°j.g[j is +ve. Hence, this reaction will 39. For hydrogen electrode, reduction reaction is
occur.
1
(d) also will not occur.
33. For the emf to be +ve, the following half-cell
reactions will occur. 1/2

Fe ^Fe2+-(-2r,E° = + 0-44 V = E“
0-0591,
E”
■^,1
2Fe'-*-"+2e- 4 2 Fe2+, E° = + 0-77 V H+/-H, n
2 ^ 2 ^

ow
Overall reaction : 0-0591 1
= 0- log = -0-0591
Fe + 2 Fe3+ 4 3 Fe^+, E°,,ii= 0-121 V n 10-10
Thus, Fe^'^ will decrea.se. (/7H = 10 means [H"^] = 10
-10
M)
34. In the given reaction, I~ has been oxidized to I2 .●. Oxidation potential = 0-591 V

e
and Cr20^" ions have been reduced to Cr^'*’ I

re
40. H-*- e~ -H-
2 2

Frl
● F° - F® - E“
^ ceil ^ CrjO^- h

F
0-0591, 1
I.e. 0-79 = 1-33 - E° or E°, =0-54V
F
^red
= F°
^ red log——
h h 2 [H^l
ou
or
35. Higher the reduction potential, more easily is the = E“^d +0-0591 log [H+]
metal deposited {M”-*- + ne~ > M). Mg^"*- ions E“red = 0. When pH = 0, [H+] = 10° = 1 M
will not be deposited because H2O is reduced
more easily than Mg^"*- ions.
kfs
E red = 0

When pH = 7, [H-"] = 10"^ M

oo
36. E,,„ = E° cathode -E° anode

0-344 = E° calomel -E° X E^e^j = 0-0591 log lO-"^ (0-0591) (- 7)

Y
= -0-4137
B

0-344 = 0-244 - E°x or E°x = -0-100 V

Thus, E,.gjj decreases by 0-41 V.
As E°„, = 0 , E°(.eii with SHE = 0-100 V.
re

41. Fe^-^ + e~ ■>Fe2+

37. The given cell reaction can be obtained from 0-0591,log [Fe^-^]
oYu

E = F°
half-cell reactions as under : red ^ Fe^+ /Fe2+
ad

1 [Fe^+]
Mn^-*- + 2 e~ 4Mn, E° = -l-18 V
2-0
d

2Mn^+ ->2Mn3+-i-2e- E° = -l-51 V = 0-771-0-0591 log

0-02
in

Overall reaction :
Re

= 0-771 -0-0591 X log 100

3Mn2-^ ^ Mn -H 2 Mn3^ = 0-771 - 0-0591 X 2 = 0-653 V.
F

E° = -l-18 V-i-(- 1-51)

42. Zn -H Cu^-^ ^ Zn2+ -I- Cu
= -2-69 V
AsE° cell is -ve, reaction will not occur.
E =E°
0-0591, [Zn2+]
cell cell
38. We have to check whether the cell formed from
the following two half-cell reactions will have -i-ve
or-ve cell voltage.
Thus, higher the value of [Zn^‘*’]/[Cu^'''], lower is
[Zn^n 1
MnO" -t- 8 H-^ -i- 5 fi- ^ Mn^-" 4- 4 H2O, the emf of the cell. for (0 = 777 = '0,
[Cu2+] 0-1
E = + 1-510 V
Red
1 0-1
= 0-1.
for (li) “ J for (“0 “
^ 02 + 2H+-t2e-, e[
1
H^O ■» = - 1-223 V
Ox “
Hence, E3 > E2 > Ej.
^ Zn^+ -1- Cu
Overall E°^„ = e" E Ox
43. The cell reaction is : Zn -1- Cu^"^
= -f 1-510 -H (- 1-223) V = -t- 0-287 V
When the cell is completely discharged, E^^jj = 0.
ELECTROCHEMISTRY 3/145

47. For water at 298 K, [H+] = IQ-"^ M

For Difficult Quostions
1
Reduction reaction is : H"*" + e
Hence, E
Q ::i,
00591 IZn2+]
log :r-^
2 ^rcu2+]
cell
„l/2 1/2
RT, Ph-, ph;
00591 [Zn2+j E = - — In—=- = -0-059 Hog
i.e.. M0 = log ^ F [H^] [H-^]
2 [Cu2+J
1
[Zn2+I [Zn2+] = 10^7’3. = -0-0591 log = -0-4137 = -0-414 V
or = 37-3 or KT’
[Cu2+] [Cu2+]
48. The reduction reaction for Ho-electrode is

ow
44. The cell reaction is
2 H+ (aq) + 2e- ^ H2 (g)
H2 + 2Ag+ 4 2 H+ + 2 Ag
2-303 RT 1 0-0591,
^cell “ ^°cell log = E‘* — log ^
2F [Ag+]2 [H"]2

e
0-62 = 0-80 + 0-06 log [Ag+]

re
In pure water at 298 K, [H'*'] = lO"^ M
-0-18

Frl
or log [Ag+] = = -3

F
0-06
0 = 0-
0-0591, Ph2
log
[Ag"*"] = antilog (- 3) = 1-0 x 10“^ M (10-2)2
= {I-0x 10-^) X 108 gL-' =0-108 gL-'
ou
r
Amount of Ag present in 100 ml solution

so
= 0-0108 g or log = 0 or = 1 (V log 1=0)

% of Ag =
aoio8
xl00 = l%
kf
10 ●14
10-14
oo
1-08 or = 10'*^ atm
45. The cell is
Y
49. The electrode reaction (reduction reaction) is
eB

M I m2-^ (0-1 M) 11 H+ (I M) I H2 (1 atm) Cu2+ + 2 e~ ^ Cu

0-0591,
EceU=E“ cell ■ ^ log[M-+] 0-0591, _
ur

^ log[Cu-^]
oY

red

1-50 = E° cell 00591 ,

ad

- ^ X log (0-1) Thus, a plot of log [Cu2'*'] vs E,^jj will have

= E° cell + 0-02955 intercept = E°^gj = 0-34 V (Given in the plot)
d

or E** = 1-50-0-02955= 1-47045 V We have to calculate oxidation potential when

in

cell
[Cu2+] =0-1 M
Re

But E’’ cell = E‘

trZ-H-, M^+/M

-2:^ log (0-1)

F

2 E_. =E° ox
ox

i.e.. 147 = 0-E°

m2-^/M

^(-l)=-0.34 +
0-0591
or E° = -0-34
2 2
46. Applying Nemst equation to the given reaction, SO. The half cell reaction of the calomel electrode
0-0591 [Fe2+]2 (written as reduction reaction) is
Eceli=E“ cell log
n
Hg2Cl2(.v) + 2e- ^2Hg (0 + 2 Cl- iaq)

For the given reaction, n = 4 and pH = 3 means 0-0591, _,_^2

^ log[Cl Y
E = E°-
[H+] = 10-2 M

E cell = 1-67 -
0-0591, (10-2)2 or
E = E°-0-0591 log [C1-]
— log
4 0-1X (10-2)4 Thus, plot of E vs log [Cr] is linear with a
negative slope (i.e., E decreases linearly with
00591 , , .
= 1-67 - — logic' = 1.67-0-10= 1-57 V. log [Cl-])
3/146 New Course Chemistry CXI1)CSZ9]

0-056x2
= 1-8983
For Difficult Questions
or !ogK^= 0-059
51. For LHS electrode,
K^,= Antilog 1-8983 = 79-12 79
2 (0-01 M) + 2e~ > H2 {x atm)
55. The reactions occurring in the cell are
0-0591, a-
^U-IS ~ log ^ At anode : ^2 ig) -> 2 H+ (aq) + 2e~
2 (0-01)2
For RHS electrode,
At cathode : (aq) + 2e~ m2-" (aq)
2H-" (0-1 M)-f 2e ~ > H2 (y atm) Overall reaction : H2 (g) + (aq) ■>

E
0-0591, V m2-" (aq) + 2 H-" (aq)
RHS “

= E RHS -E LHS = 0 (Given) E cell = E°cell

0-059,
——log
[m2-"][H-"]2
^ccU
= E RHS [M^n
^LHS

w
log (x X 10"^) = log ()’ X 1Q2) F° = F° -E”
^ cell ^ 2-t-
log A + 4 = log y + 2

F lo
A - = 0-151 -0 = 0-151 V
log A - log y = - 2 or log — = 2
V
0-059
0-092 = 0-151- logdO'x ]2)
i.= 10-2 = 0-01

ee
or
2

Fr
y
or 0-092 = 0-151 -0-0295 log 10'"
52. Zn -f Fe2-" ^ Zn2* + Fe (n = 2)
or 0-0295 log I (F = 0-151 - 0-092 = 0-059
E = E”-
0-0591
log for 0-059
ur
n or log 10-' = = 2
0-0295
s
0-0591. 0-01
10*^ = Antilog 2= 102 a=2
ook

0-2905 = E“- log

Yo

2 0-001
56. The cell reaction is
eB

or E“ = 0-2905 -i- 0-0295 = 0-32 volt

Zn + CUSO4 > ZnS04 + Cu, n = 2
0-0591
E° = logK eq E = E''cell 2-303 RT. [ZnSQ^l
log
our
ad

n
2F
[CuSO^]
0-32
0-0591
0-32 = logK eq K_ = 100-0295 2-303 RT. 0-01
2 eq
In 1st case, E. =E log
Y

1 cell
2F 1
Re

53. iAg(NH3)2]-^-fe- 4 Ag + 2 NH3, E° = 0-02 V

nd

Ag ->Ag+ + ^- E“ = -0-80V 2-303 RT I

In 2nd case, E2 = E”^j, log
Fi

2F 0-01
fAg(NH3)2l^ F ± [Ag+] -I- 2 NH3,
Evidently, Ej > E2.
e;i,=-0-78V
57, For the given cell, applying Nemst equation
0-0591 0-0591
E’’ccU logK -0-78 = logK 0-059 [Ni^-"]
“
n 1 E cell = C- n
log
[Ag-"]2
-0-78
logK = = -13-2
or
0-0591
0-059 (10-3)
= 1-05- log
-14
2 (10-3)2
K = Antilog 14-8 = 6-3x10
54. E° for the given reaction
= 0-855 - 0-799 = 0-056 V = 1-05-^^ log (10)3
= 1-05-0-295 X 3
E° _ 0-0591
cell ~ 2 logK^,
= 1-05-0-0855 =0-9615 V
ELECTROCHEMISTRY 3/147

62. AG® = - hFE® cell -O)

For Difficult Questions
and AG° := - RT In K ...(«)
eq
58. Cell reaction is From (0, if E®^.gjj is negative, AG® will be
Zn + Cu2+. ^ Zn^+ + Cu positive, i.e., AG® > 0
AG = AG® + 2-303 RT log Q From (»), AG® can be positive only if < I.
63. For emf to be +ve, oxidation should occur at iron
[Zn2+] electrode,
= AG® + 2-303 RT log
[Cu2+] ^ceil = 1-23 +0-44 V= 1-67 V
AG ® = - «F = - 2 X 96500 x 1-67 J

w
AG® = -nFE;^„=-2F(M) = -322U.
A AG=-2F(l-l) + 2-303RTlog 10
64. Standard Gibbs of a reaction (A^ G®) = - n F E"cell
= 2-303 RT-2-2F

59. The balanced equation for pentane-oxygen cell For the given reaction, n =2, e:cell = 0-24 V

o
e
reaction will be A, G® == - 2 X 96500 x 0-24 = - 46320 J mol"*

re
CjHj2 8 ©2 ^ 5 CO2 -f 6 H2O, n = 32 = - 46-32 kJ mo!"*

Frl
16Q2-) 65. The given cell is a concentration cell. The EMF

F
(v 16(0)-f32e-
of the cell, as represented, is given by
A^G® = [5 X A^G® (CO2) + 6 A^G® (H2O)]
-[A^G® (C5Hi2) + 8Ay-G® (O2)]
ou 0-0591 ^2

r
^cell “ -log^
= [5 {- 394-5) + 6 (- 237-2)] - [(- 8-2) + 0]

so
n
'“I
= - 1972 - 1423-2 + 8-2 = - 3387 kJ mol"'
A/G® = -nFE®,,„
kf
Eggii will be H-ve only if C2 > Cj
As AG = - n F E^gj|, therefore, AG will be -ve
oo
- 3387000 = - 32 X 96500 x E® cell only when E^gn is +ve. Hence, AG to be negative,
Y
or
E°ceii = I 0968 V C2>C[.
B

-10 -HO 1-3 66. In very dilute solution of NaCl, the following
60. C, Hio(g)-h^02(g) -» reactions take place on electrolysis :
re

44 -A 1
oY

Anode: 2 OH +2e
4 C 02+5H20(/)
u

2 2 2
ad

Change in oxidation number of carbon Cathode : 2 H'*' + 2 e 4H2

= 4(+4)-(- 10) = 26 67. Out of the given metals, Ca is s-block metal and
d

is highly electropositive. It cannot be obtained

in

or — Ol+26e~ ^ 13 02- by electrolysis of an aqueous solution of its salt.

Re

2 ^
68. During charging, the reactions that occur are
(8 02- in 4 CO2 and 5 0^“ in 5 H2O)
F

At cathode :
Thus, cell reaction involves 26 electrons, i.e.,
n = 26 PbS04 (s) + 2e~ ->
Pb(^)-i-S02-(n^)
e!ceil
AG® _-(-2746)xl000 At anode : PbS04 (s) + 2 H2O
= + 1-09 V
nF 26x96500
Pb02 (j) + SO4 {aq) -f 4 H'*' (aq) + 2e
61. The relation between standard e.m.f. of a cell Thus, Pb^'*' ions of PbS04 are reduced to Pb on
(E°g,j) and equilibrium constant (K) of the cell the cathode while PbS04 is oxidized to Pb02 at
reaction at 298 K anode.

0-059 (Remember. Anode is the electrode on which

cell “ log oxidation takes place, i.e., loss of electrons and
n
cathode is the electrode on which reduction takes
0-059
0-59 = log or log Kp = 10 place, i.e., gain of electrons. Further, the
I
electrode which acts as anode during discharge
or
K^. = antilog 10 = 1 x 10*® acts as cathode during charging and vice-versa)
3/148 “PmuCeefr'^. New Course Chemistry (XII)BQ291
N" 75. The reactions occurring during discharge are :
For Difficult Questions
At anode : Pb (.y) + SO^“ {aq) ^ PbS04 {s) + 2e~
69. The anodic half cell reaction is
At cathode : PbO-i (s) + 4 H"*" {aq) + SO^" {aq) + 2e~
> PbS04 (5) + 2 HoO {/)
Pb (s) + SO^” {aq) + > PbS04 {s) + 2 e~
For 2 F current passed, PbS04 deposited = 303 g Overall reaction : Pb (i) + Pb02 (i) + 2 H2SO4
(Molar mass of PbS04 = 303 g mor^) > 2 PbS04 is) + 2 H2O (/)
For 0 05 F current passed, PbS04 deposited Quantity of electricity consumed
303
= (2 A) (30 X 3600 s) = 216000 C
= — X 0-05 g = 7-6 g 2 X 96500 coulombs consume H2SO4 = 2 moles
70. G = - 394-4 + 2 (- 237-2) - {- 166-2) 216(X)0 coulombs will consume H2SO4

w
= - 702-6 kJ mol-^ 1
X 216000 = 2-24 moles .
96500
A^G _ -702-6 xlOO =97%.
Efficiency - 76. Ag-fl"- > AgI + e-,E“ = 0-152 V

Flo
AH “ -726
Ag-*" + e" Ag, E° = 0-800 V
71. Rust is a mixture of Fe203 and Fe(OH)3.

e
Ag-^ -I-1 ^ Agl, = 0-952 V

re
72. Zinc has higher negative electrode potential than
iron, so iron cannot be coaled on zinc. 2-303 RT, ^

rF
At equilibrium, E® =
73. 1 L of 1-0 M CUSO4 solution contains Cu^"^ -y-losK,
= 1 mole
ur
[Agl) 1
But Kc =
100 gZn =
100

65
mole fo [Ag+][n K sp
ks
2-303 RT
Thus, in the cell reaction Zn + Cu^'*’ > Zn^'*' + Cu, 0-952 = - log = - 0-059 log K,sp
Yo
Cu“'*‘ ions are the limiting reagent. The cell will F
oo

run till all the Cu-'*' ions are deposited, i.e., or log = -16-13
1 mole of Cu is deposited. Quantity of electricity
B

4
required for deposition of 1 mole of Cu = 2 F -1
77. -ALO,
- A1 + O2, A^G = -t- 966 kJ mol
re

3 2 3
2 x96500 c
Time, t = = 2 X 96500 s

|x(302-),,'.e„ 2 0^'
u

lA
O2, n = 4
ad

Thus,
Yo

2x96500 AG = - wFE
hr = 53-61 hr
3600 966 X 10^ = - 4 X 96500 x E or E = - 2-50 V
d

74. Autoprotolysis constant of H2O = [H'*'] [OH“] = Thus, minimum potential difference required
Re
in

Redn. half reaction : = 2-50 V.

78. For the given cell reaction involving formation

F

2 H2O + 2e ^ H2 -f 2 OH-,
E° = - 0-8277 V of 2 moles of H2O,
-» 2 + 2e~, E° = 0-0 AG“ = -«FE cell = -4x 96500 X 1-23 J
Ox. half reaction : H2
= - 474780 J = - 474-78 kJ
Cell reaction : 2 H-)0 ^ - 2H-^-f2 0H-,
E” = - 0-8277 V (n = 4 for the given cell reaction)
AG“ = AH“-TAS’’
E = E“-
“|^log[H-l2[OH-l2 or AH" = AG“-TAS"

= - 474-78 kJ + 298 K (- 0-32 kJ K"*)

At equilibrium, E = 0. Hence,
= _ 474-8 - 95-36 kJ = - 570-16 kJ
E° = ^^^log[H^]2
2
[0H-]2 This is enthalpy change for the formation of
2 moles of H2O.
- 0-8277 = 0-0591 log [H+J [OH-]
.'. Enthalpy change for the formation of 1 mole of
= 0-0591 log Kw
570-16
or log K,, = -14orK,,,= lO-l'^ H2O (A^H“) = = -285-08 kJ
ELECTROCHEMISTRY 3/149

12x96500 12x96500
For Difficult Questions or t = s = hrs - 3-2 hrs
100 100x3600

79. \K*{aq) + a-{ag)] + Ag-^(aq)+ l^O^iaq) ■>

82. HBrO
+1 0

^ BI2 . E^BrO/Br, -●■595 V

K"^ (aq) + NO3 (aq) + AgCI (5) +1 +5
Initially the conductance is due to K'*' and Cl” HBrO
-> B1O3, = 1-5V
BrOj/HBrO
ions. On adding AgN03, precipitate of AgCl is
formed. Thus, some Cl~ ions are replaced by ^ceii disproportionation of HBrO,
NOj ions. As both have nearly the same e;ii=e“HBrO/Br, -E”
conductance, the conductance remains almost BiOj/HBrO
constant. After the end point, addition of AgN03 = 1-595- 1-5 =0-095 V = + ve

will give Ag"^ and NOJ ions in the solution. Hence, option (d) Is correct.

w
Hence, conductance will increase rapidly. AH"'!
80. Cu^* + e- ●>Cu+, 83. A, H° = - n F E° + « F T V AT
AGj=-nFE° = -lxFx0-15
= - 2 X 96000 X 2 + 2 X 96000 x 300 (- 5 x 10^)

Flo
Cu-' -t- e- -> Cu, AG2 = - 1 X F X 0-50
= 384000- 28800 J
Cu2+ + 2e“- Cu, AG3 = - 2 X F X E°Cu^+zcu = -412-8 kJ mor'

ee
AG3 = AG I -+■ AG2 84. Given that

Fr
-2F E*" = -0-15 F + (-0-50 F)
Cu-'^/Cu Ag-^ + e — ^Ag,E° = .rV .-. AG" =-F.r ...(0
or -2 E°

for
= -0-15-0-50 = -0-65
Cu^*/Ca Fe^'*’ + 2e~ —>Fe,E° = y V .-. AG"=-2Fy
ur
0-65
or E° = 0-325 V ...Hi)
Cu^"-/Cu 2
s
Fe^-" + 3e~ 4Fe, E° = cV .-. AG3=-3Fz
k
81. B2H6 3 O2 > B2O3 + 3 H2O
Yo
oo

Molar mass of B2H^ = 2 x 10-8 + 6 x I ...(///)

-1
= 27-66 g mol Fe^'*' + Ag
eB

Aim : Fe*'*’ + Ag"*"

27-66 g of B2Hg = 1 mole of B2Hg Adding eqns. (0 and (ii) and subtracting cqn. (///),
It requires 3 moles of O2 for complete burning we get
r
ou
ad

1
H2O > H2 + - O2 Fe^^+Ag-^ >Fc3+ + Ag, AG =-FE“cell (n=l)
Y

AG = AG°+AG2-AG3
- mole of O2 requires 2 faradays
Re
nd

-. -FE° = -Fx-2Fy-(-3Fz)
cell
.●. 3 moles of O2 will require =12 faradays
Fi

12 X 96500 = i X t= 100 x t
or E”cell =x+2y-3z

Dl Multiple Choice Questions (with One or More than One Correct Answers)

85. (a) is wrong because 87. (b) is wrong because electrolysis of concentrated
Fe2+ + 2e~ ^ Fe, Fe^'*’ + 3 e~ ^Fe. H2SO4 does not give 0^ at the anode,
I (c) is wrong because electrolysis of aqueous KF
One faraday from Fe^'*' will deposit — mol of Fe
solution gives O2 at the anode and not F2.
88. Salt bridge does not participate chemically in the
and from Fe^’*, it will deposit - mol of Fe.
cell reaction because the amount of electrolyte in
(c) is wrong because I F liberates one gram the bridge remains unchanged. It keeps the
equivalent of the substance. solutions in the two half-cells electrically neutral.
86. 2 T1 -t- Cu^+ ^ 2 T1+ -I- Cu, It simply allows the passage of ions through it. It
0-0591, [T1+]2 prevents transference or diffusion of ions from
F — F°
^cell “ ^ cell
2 ^[Cu2+] one half-cell to the other.
3/150 “Pn^Kiee^'^. New Course Chemistry CX1I)BZHI91

For Difficult Questions

^ Cd^-*- + Ni = 6'14x 0-551

89. (a)Cd + Ni2+
= 0-71 (+ve)
00591, 0-001
E^^j, =0-40+ (-0-24)- ~^log 0-1
Similarly, test for (fc), (c) and {d).
We get (b) = 0-02 V (+ve), (c) = 0-33 V (+ve),
0-0591
= 0-16 + x2 C^O = -0-14V(-ve).
2

mi Multiple Choice Questions (Based on the given Passage/Compreh ension)

90. H2 has been oxidized and AgCl (Ag'*') has been 94. At anode, oxidation occurs. PbS04 is oxidized to
reduced. Hydrogen electrode on left has to be Pb02 according to reaction (d).
Pt I H21 H-". 95. Mass of the solution before discharge
91. The given half-cell reactions may be written as 3500 mL X 1-294 g mL"’ = 4529 g

w
(0 OCr + H2O + 2e ^ Cl" + 2 OH", Mass of H2SO4 present in this solution

F lo
E°,1 = 0-94 V 39
x4529g = 1766-3 Ig
1 100
(ii) Cl- » -CI2 +«". E“2 = - 1-36 V Mass of the solution after discharge
= 3500 mL X M39 g mL'’ = 3986-5 g

ree
icio+20H“,
(Hi) OCl + H2O + € ■>

E°3 = ?
2 ^ for F
Mass of H2SO4 present in this solution
20
x3986-5g = 797-3 g
Eqn. (Hi) can be obtained by adding eqns. (i) and 100
(ii). Hence, in terms of their free energies Loss in mass of H2SO4 during discharge
Your

(AG “ = - n FE®)
ks

= 1766-31 -797-3 = 969-01 g

eBoo

AG°3 = AG^i + AG°2 Moles of H2SO4 lost during discharge

-1 xFxE°3 = -2xFx(0-94)x 1 x F (x 1-36) 969-01
or
£*’3 = 2 X 0-94 - 1 -36 = 0-52 V. = 9-88786
ad

98
our

92. Subtracting second eqn. from first eqn. we get

96. Moles of H2SO4 present in the solution after
AgCl (s) ^ i Ag'^ (a^) + Cl (aq). 797-3
E° = - 0-58 V = 8-136
discharge - 98
Re

Applying Nemst eqn.,

Y

8-136 moles
a0591 [Ag-^][C1-] .●. Molarity of the solution = = 2-32
Find

E = E°- log 3-5L

1 [AgCl (5)]
97. During the discharge reaction, 4 moles of H'*’
Putting [AgCl (j)] = 1 and at equilibrium E = 0
ions, i.e., 2 moles of H2SO4 require 2 F, i.e., 2 x
E° = 0-0591 log [Ag-^] [Cn = 0-0591 log Ksp 96500 coulombs
or -0-58 = 0-0591 logK sp .●. 9-88786 moles will require
or logK^p =-9-8139 =10-1861 2x96500x9-88786
= 954178 coulombs
K„= 1-535 X 10-^0 2

93. For the given concentration cell, Coulombs 954178

98. Ampere-hour = = 265-05
H2lH+(ci)llH-"(c2)IH2 3600s 3600

0-0591 10-3 99. For the given concentration cell,

^cell “ log — = 0-0591 log
1 10-6
E
2-303 RT ^2
cell log
= 0-0591 X 3 = 0-1773 V. nF C
1
ELECTROCHEMISTRY 3/151

101. The given concentration cell is

For Difficult Questions
MI (Saturated = C,) 11 M^+ (O-OOI M = C2) I M
2-303 RT 1 EMF of concentration cell is
log = +ve , ie., E cell >0.
log ^ _ 0-059 log C2
F 005 2-303 RT
^cell “
For the cell reaction to be spontaneous, AG < 0. /iF C
1
n C
1

100. From the above equation,

0-059 =
0059, 0001
2-303 RT
log
1
= 70 mV
—log^ 1
F 005
10"3
= 70 X 10"^ V = 0-07 V or 2= log or
log i(r^ - log Cj = 2
2-303 RT C,
(1-301) = 0-07 or -3-logCj=2 or logC]=-5

w
F

2-303 RT 0-07
or
C, = 10"5 M
= 0-0538 2+
For the salt MX2, ^^2 ^ + 2X-
or
i M
F 1-301
s

Flo
s ^ 2S
For the ne\v concentration.
= S (2 S)" = 4 = 4 X (10-5)3 ^ 4 iq-15
2-303 RT 1

e
E log 102. AG = - n F £^.^11 = - 2 x 96500 x 0-059 J mor'

re
cell p 0-0025
= - 11387 Jmol-*

rF
= 0-0538 log 400 = 0-0538 x 2-6021 = - H-38Umol-*
= 0-140 V= 140 mV
= -U-4Umol-l
ur
m Matching Type Questions fo
ks
Yo
neutralisation s complete,
oo

103. Reaction occurring in the Daniell cell is

i-
Zn (s) + CUSO4 (aq) -> ZnS04 (aq) + Cu no more (C2H5)3NH will
B

2-303 RT. [ZnS04(fl^)]

re

be formed. Hence, g
EceU=E‘‘ cell tiF
log
conductivity will remain
o
[CuSO^ffl^)]
u

constant. Hence, P -> 3. vol. of (C2H5)3N added

ad

104. P : CH3COOH + (C2H5)3N ->

Yo

weak acid weak base Q : AgN03 + KI > Agl + KNO3

+

Ag+ + NO- +K+ + 1- > AgI+K+ + NO"

d

CH3COO-+ (C2H5)3NH
Re
in

or CH3COO- + H+ +(C2Hg)3N On adding KI to AgN03,

F

-I- Ag'*' ions are replaced by ^

CH3COO-+ (C2H5>3NH K-^ ions, Hence,
conductivity remains -o
CH3COOH is a weak acid. Its conductivity is
almost constant, After o
low. On adding (C2Hg)3N, H"*" ions are replaced +
+ replacement of all Ag
Vol. of KI added
by heavier/bigger (C2Hg)3NH. Hence, ions, KI now added gives
conductivity decreases. But now it becomes a K'*' + 1“ ions. Hence, conductivity keeps
+ increasing. Thus, Q —> 4.
buffer of CH3COOH + (C^Hg)3 NH , therefore,
[H'*'] remains constant but more and more of R : KOH + CH3COOH — ^ CH3COOK -t- H2O
+

(C2H3)3NH will be formed. Hence, K+ + OH- + CH3COOH ->

conductivity keeps increasing. When CH3COO- + K-" + H2O

3/152 ‘PruttUe^'^ New Course Chemistry (XI1)BBI91

or -3E‘’ = -0-77+ 0-88 = + 0 11

For Difficult Questions
Fe^+.Fe
Oil
or E° = - 0 0386 V
Initially, conductivity Fe^-^.Fe 3
decreases due to = - 0-04 V
replacement of OH“
3 Hence, P ^ 3.
ions by CH3COO“ *o

Q:2H20 ^02 + 4H+ + 4e”, E“ = -1-23V

c
o
ions. After complete O

neutralisation of KOH, O2 + 2 H7O + 4e 4 OH”, E“ = + 0-40 V

Vol.ofCHsCOOH added
now CH3COOH added
ionizes very little. Hence, conductivity remains 4H2O > 4 H-" + 4 OH”, E° = - 0-83 V
almost constant. Hence, R —> 2. Hence, Q 4.
S : HI + NaOH ^ Nal + H2O R : Cu^+ + 2e~ -+ Cu, E° = + 0-34 V
H+ + r +Na+ + OH- ^Na+ + r + H20 2Cu- ^ 2 Cu+ + 2 e”, E“ = - 0-52 V

w
As H"*" ions are replaced by
Cu--" + Cu ^2Cu^E® = -0●18 V
Na"*" ions, conductivity ^

F lo
decreases. After complete Hence, R —> 1.
neutralisation of HI, -gC S : Cr^-" + 3 e” ^ Cr. AG°i = - 3 F (- 0-74)
NaOH now added will ^ = 2-22F

ee
give Na"^ + OH” ions. Cr ■> Cr^-" + 2 <?-,

Fr
Vol. of NaOH added
Hence, conductivity will AG“2=-2F(0-91)=-1-82F
keep increasing.
Hence. S ^ 1.
105. P : + e” > Fe^^
Cr^-*- + e~
for >Cr2-", AG° 3 =-1FE° Cr3+,Cr2+
ur
AG°3 = AG°i + AG°2
AG°i = - 1 F X 0-77 = - 0-77 F = 2-22F+ 1-82 F
s
-FE“
Fe^-" + 2 e”
ook

>Fe,
Yo

E° 0-4 V
AG°2 = - 2 F X (- 0-44) = 0-88 F or
Cr^+,Cr2+
eB

Hence, S —> 2.
Fe^-" + 3 e”
>Fe,AG°3 = -3FE°^^3,^^ Note. EMF of a cell can be obtained by adding
our

oxidation potential and reduction potentials of

ad

AG°3 = AG“j + AG°2

two half-cells but electrode potential cannot be
-3FE° = -0-77 F +0-88 F calculated by adding oxidation potential and
Fe^^-.Fe
Y

reduction potential of two half-cell potentials.

Re
nd

M Integer Type Questions

Fi

109. 3 L of 0-5 M K2Cr207 solution will contain 111. Mg + 2 Ag-" > Mg^-" + 2 Ag,
0-5 X 3 = 1-5 moles of K2Cr207. The reduction 0-0591. [Mg^-"]
reaction is E. ,. =E" log
cell cell
n
[Ag+ ]2
Ci!20?" +14H+ +6e” ■>
2 Cr3+ + 7 H2O 0-0591 [Mg2-^]
or E° E log
cell cell

Thus, 1 mole of Cr202 requires 6 F of

n
[Ag-"]2
electricity. 0-0591, 01
In 1st case, E*^ cell -E cell —-—log r-
1-5 mole will require electricity = 9 F. 2 (0-5)2
110. The cell reaction is 0-0591 0-01
In 2nd case, E" cell -E log
3 Mg + 2 AP-" ^ 3 Mg2+ + 2 Al, n = 6. cell 2 (0-25)2
ELECTROCHEMISTRY 3/153

t
From eqns. (/) and (//),
For Difficult Questions

c A \2
a
1 _ 1
m
I 0-011
0-0591 (0-1)2 K C. O-I 10
log ^^2
A
'«2
2 ((0-5)2 ]2
= 0-001 = 10-3
0-0591 0-1
x21og = 2 times. or
logK^ 1 -logK,^2 = log 10-3=-3
2 (0-5)2 a

112. X ^ Y, A^G“ = - 193 kJ moi-' ...(/) or

-logK^^ -(-logK^^) = 3
■> m3+ + 2 e“, E° = - 0-25 V
4-
M

For this reaction, AG° = - m FE

O or pK-pK.. =3
I "2
= - 2 X 96500 X (- 0-25) = 48250 J mor* 114. Cone. C = 0-0015 M, Conductance, G = 5 x ICT^ S
= 48-25 kJ mol-'
1= 120 cm, a = 1 cm2
i.e., 48-25 kJ of energy is used for oxidizing I
mole of M'*' to m3'*‘ / 120

w
k = Gx— = 5x10 2x = 6 X 10 3 S cm *
1
/. No. of moles of M'*' oxidized by 193 kJ of

F lo
a

193 KxlOOO 6x10-3x1000

= 4
energy = A^^m =
48-25
Molarity 0-0015

113. Suppose A„j (HX) is represented by a and

ree
m
\ = 40 S cm2 f^Qj-l
A„, (HY) by .
1
for F
HA

C 0
+ A“

0
Then given that a =— a
m, JO »«2 C-Ca C a C a
Your

[H-^] = C a = 10-^ M (v pH = 4)
ks

HX
H^-fX- K^ =
eBoo

[HX] -4 -4
10 10
a =
K C 0-0015
Representing of HX by a , then as
ad

A^m 10^ 40
our

A
= C «2 and a = m
, we have a =
A® A°
m
0-0015 a”m
HI

^2 40 x 0-0015
Re

A
m
or A® = = 6 X 102$ cm2 mol *
10~^
m
Y

1
K = C,
(I
-(0
Find

A
III
1 115. Mg -> Mg2'‘‘ + 2e
Cu2+ + 2e 4 CU
[H^][Y-]
Similarly, HY H+-fY- Ka
[HY] Mg + Cu2+ — 4 Mg2'*' + Cu, n = 2
K For the given cell.
Representing of HY by <h , we have
RT. 1
E = E ln- = E.0 = 2-70V
\2 0
2F 1
A
nv,
Ka
2=^2 A -(«) When concentration of Mg2'*' is changed to x M,
ni2
300
2-67 =2-70- In-
2x11500 1
X"X“ =x“Y" , ^HX
O O
As = A
HY
300
O 0-03 = Inj: or In at = 2-3 or x= 10
or A. = A 2x11500
m
I m-,
3/154 U New Course Chemistry fxmpzsTMl

117. Specific conductivity (k) = Obsd conductance (G)

For Difficult Questions
X Cell const. (G*)
= (0-55 X 10-^ S) X (1-3 X 10^ m-‘)
116. MnO-+8H++56;- ^ Mn^+ + 4 H2O
= (55 X 10-5 S)x (1-3
Change in oxidizing power means that we have = 55x 1-3 X 10-5 Sm-‘
to find change in the value of reduction potential
(E2 - Ej) when [H'*'] changes from 1 M to 10“^ M. A
K(Sm-‘)
Molarity (mol m 5)
m
In 1st case :

E.1 - E" =
0 059,
— log
_ 55xl-3xlQ-5Sm~^
[H+]*[MnO-] 5 mol m“5

ow
In 2nd case : (5 mmol dm“5 = 5 mol m“5)
= 11 X 1-3 X KT^Sm^mor'
E2 - E° =
0-059
iz£i [Mn^-*-] = 14-3 X 10"5Sm-mol-‘
= 14-3 S m^ moH
As [Mn^"*"] and [MnO^] remain the same

e
Nearest integer = 14

re
rFl
0-0591, 118. H2(g) + Cu2-^(fl^)- ^ 2 iaq) + Cu (n = 2)
E2-E, —^log ^

F
1 E cell.. =E°cell
006 [H^]2
— log
[Cu2+]
^10-4 f

r
0-059,
^x(-32)
ou
= —^log I )
fo
0-576 = 0-34-0-03 log
5
0-01
ks
or
IE2-E,l=^^(32) = xxlO“^ 0-576 - 0-34 = - 0-03 log - 0-03 x 10^
oo
or 0-236 =0-06 pH-0-06
or 0-3776 =.rx 10^
Y
or 0-06 pH = 0-296
eB

or 3776 X 10-^ = xx 10^

0-296
or AT = 3776 or pH = = 4-93 5
0-6
r

VII,
ou

Numerical Value Type Questions (in Decimal Notation)

ad
Y

AG“ = AH°-TAS“
119. For the given cell,
d

A(^) ^ A”"^ (fl^) + ne ] X 2 But AH° = 2 AG“ (Given)

Re
in

iaq) + 2 ne AG*’= 2 AG” - TAS"

F

or TAS” = AG*’
Overall reaction : 2 A (s) + B^"'*' (aq)
AG” -RTln4
^ 2 A"+ (aq) + B or AS” = = -Rln4
T T

E = E”-
RT In
, [A"+]2 = -8-3 X 2x0-7
2n-t-
2nF [B
= -11-62 JK"^ mol-^
(In 4 = In 2^ = 2 In 2 = 2 X 0-7)
0 = E”-
RT ,In (2)2
2nF 1 120. For the given cell reaction, n = 2

RT
ag”=-„fe;,
or E” = In 4 = - 2 X 96500 X 1-23 X x 0-70
2nF
= 166-173 J
RT
AG” = -2/iFE” = -2nF ln4 = -RTln4 w = -AG“ = 166-173 J
2rtF
 ELECTROCHEMISTRY 3/155

A
-■D©
For Difficult Questions
122. a. =
1
m _ yxiO~ ^ —y = a ...(/)
A 4x10^ 4
From thermodynamics (studied in class 11) m

«R(AT)
w = 3yxl0-
«2 = = 3cx,1 = 3a
Y-1 4x10-

For monoatomic gas, T “ = 1-66 Ca^

HA V ^ + A K

ow
a
1 - a
As 1 mol of gas is to be compressed, n = 1
C 0 0
lx8-314xAT
166173 = orAT= 13-20 C(l-a) Ca Ca
1-66-1
As temperature is constant, will be constant.

e
+7 +4 „ -7

121. Mn04+3e 4 Mn02 , E° = 1-68 V C,1 ar C-, a;

re
+4 +2 1 -a 1 -a^
1

Mn02 + 2e £2=1-2 IV

F
Frl
+2
(—
^ r'X
(3a)-
0 Ca- 20; 9/20
Mn^++2e- ■> Mn , £3 =-I-03V
ou 1-a l-3a
or
1-a l-3a

sr
+7 0

MnO; + 7^- ●> Mn, E"

C4 ='>

kfo
- . 9
or
I -a 20 (l-3a)
AG" =AG“+AG. +AG3
oo
or 20 - 60 a = 9 - 9 a or 51 a = 11
-7FE” =-3FE“ -2FE2 -2FE3
Y
II
or a = —= 0-2156 = 0-22
reB

7E; =3Ei+2E2+2E; 51

3xl-68+2xi-2I-2xl-03 .V
uY

E° - = 0-77 V 123. From eqn. (/), a = -^, y = 4a = 0-88

^4 - 7 4 ’

VIII.
ad
do

Assertion-Reason Type Questions

in

124. Correct Statement-2. Specific conductivity is MeK

129. CaH2 Ca2+ + 2 H"
Re

the conductance of I cm^ of the solution whereas

F

molar conductivity is the conductance of a At anode : H' > H -t- e" ;

solution containing 1 mole of the electrolyte. H + H > H2

125. Statement-2 is the correct explanation of Hence, R is the correct explanation of A.
statement-1. 130. Correct R. On dilution, the number of current
126. Correct Statement-1. The cell constant of a carrying particles per cm^ decreases and hence
conductivity cell depends upon the distance specific conductivity decreases, a^.^ and a m

between the electrodes and the area of their cross- increase with dilution because a
e<i
section and not on the material of the electrodes. sp. conductivity x volume of solution containing 1 g
127. Statement-2 is the correct explanation of eq. of the electrolyte and a„, = sp. conductivity
statement-1. X volume of solution containing 1 mole of the
electrolyte and volume increases much more than
128. Correct explanation. T ions have much higher
decrease in specific conductivity.
oxidation potential than that of water while F"
ions have much lower oxidation potential than 131. Correct A. The molar conductivity of any
that of water, electrolyte (weak or strong) at infinite dilution is

k
3/156 “Pn^xeUcfi-'^i. New Course Chemistry (Xll)EZsl9]

ion whose concentration may change (OH" ions

For Difficult Questions
consumed in one half reaction are produced in
sum of molar conductivities of the ions the second half reaction).

multiplied by the respective number of ions in 134. R is the correct explanation of A because
one formula unit. sulphuric acid is consumed due to formation of
Correct R. The current carried by cations and lead sulphate and hence density decreases.
anions is not equal. 135. R is the correct explanationof A (More negative
132. R is the correct explanation of A. electrode potential of zinc means higher
133. Correct explanation. The cell reaction, oxidation potential. Hence, zinc is more easily
Zn + HgO ^ ZnO + Hg does not involve any oxidized than iron).

o w re
e
rFl
F
r
ou
fo
ks
oo
Y
B
re
ou
Y
ad
d
in
Re
F
 w
Flo
CHEMICAL KINETICS

e
re
F
ur
r
4.1. GENERAL INTRODUCTION

fo
There are following three main aspects related to the study of reactions:
ks
1. Whether the reaction at all takes place or not, i.e., spontaneityof a reaction (as discussed in terms of
Yo
oo

Gibbs energy change in thermodynamics).

2. If it is spontaneous, then to what extent the reaction takes place before equilibrium is attained (as
B

studied under Chemical Equilibrium).

re

3. Rates or speeds of reactions, which the branch of thermodynamics is unable to answer.

For example, thermodynamics predicts that conversion of diamond to graphite is feasible but the process
u
ad
Yo

is so slow that the change is not perceptible at all. Hence, most people think that diamond is for ever.
Besides studying the rates of reactions, we are also interested to know the factors by which rates of
reactions can be altered. The study of the rates of reactions also helps us to understand the pathway from
d
Re

reactants to products, called ‘Mechanismof reaction’.

in
F

The branch of chemistry which deals with the study of the speeds or the rates of chemical
reactions, the factors affecting the rates of the reactions and the mechanism by which the
reactions proceed is known as chemical kinetics*

In this unit, our main aim will be to discuss what we mean by ‘Rate of Reaction’, ‘Factors on which
the rate of reaction depends’ and ‘Mechanism of reaction’. Also, the study of the effect of concentration
on the rate of reaction gives rise to a new term called ‘Order of Reaction’ which will also be discussed
in detail.

Further, to explain the mechanism of reactions at the molecular level, we shall discuss about a theory
involving collisions of molecules, called ‘collision theory’ of reaction rates.

*The word ‘kinetics’ is derived from the Greek word ‘kinesis’ which means ‘movement’.

4/1
4/2 “Pn^uCee^ 'a New Course Chemistry (XII)C&19I

4.2. CLASSIFICATION OF REACTIONS ON THE BASIS OF RATES

Based upon their rates, the various chemical reactions are classified into three types as follows :
(/) Very fast reactions, i.e., which take place instantaneously (time taken is generally 10"*^ to 10”*^ second),
e.g., ionic reactions like
AgN03 {aq) + NaCl iaq) ^ AgCl {s) + NaN03 {aq)
BaCl2(a^) + H2S04{a^) ^ BaS04 {s) + 2 HCl {aq)
NaOH {aq) + HCl {aq) ^ Na+Cl-{a<?) + H20(/)
(//) Very slow reactions, i.e., which may take days or months, e.g., rusting of iron
(///) Reactions which are neither very slow nor very fast but take place at moderate speeds.
A few examples included in this category are :

w
{a) Hydrolysis of ethyl acetate : CH3COOC2H5 + H2O ^ CH3COOH + C2H5OH
Ethyl acetate Acetic acid Ethyl alcohol

Flo
H+
{b) Hydrolysis of sucrose : CpH220ii + H2O ^ ^6^12^6

e e
Sucrose Glucose Fructose

Fr
Heal
(c) Decomposition of ammonium nitrite : NH4NO2 ^ 2H2O + N2

r
ur
1
{d) Decomposition of hydrogen peroxide : H2O2
fo » H,0 + -0.
2 2 2
ks
The main category of reactions for the kinetic study have been the reactions belonging to
Yo
category {Hi), i.e., which are neither very slow nor very fast.
oo

Reasons for the difference in rates. An obvious reason is that a reaction involves the breaking and
eB

making of bonds. Since different bonds require different amounts of energy for breaking and different amounts
of energies are evolved when different kinds of new bonds are formed, the rates of different reactions are
different. The instantaneous nature of the ionic reactions is due to the fact that these do not involve any
ur
ad

breaking of bonds (as ions are already pre.sent in the aqueous solutions).
Yo

Another reason for the difference in the rates of different reactions in terms of ‘activation energy’ will be
discussed later in this unit.
nd
Re

4.3. RATE OF REACTION, ITS MEASUREMENT AND DEPENDENCE ON VARIOUS FACTORS

Fi

4.3.1. Rate of Reaction

The rate of a reaction means the speed with which the reaction takes place. Just as the speed of a vehicle
is expressed in terms of distance travelled per unit time, the speed or the rate of a chemical reaction is
expressed either in terms of decrease in the concentration of a reactant per unit time or increase in the
concentration of a product per unit time. Hence, the rate of a reaction may be defined as follows :
The rate of reaction is the change in the concentration of any one of the reactants or products
per unit time

Thus, for the hypothetical reaction R ^ P,

Decrease in the concentration of a reactant R
Rate of reaction =
Time interval

or
Increase in the concentration of a product P
Time interval
CHEMICAL KINETICS 4/3

Thus, if [R]] and [Pj] represent the molar concentrations of the reactant and the product respectively at
time and [R2] and [P2] are their respective concentrations at time t^, then changes in concentrations of the
reactant and product will be : A [R] = R2 - Rj and A [P] = P2 - Pi and time interval is : A r = ?2 “
The rate of reaction in terms of reactant or product will be written as

Rate of reaction = -
A[R1 . A[P]
= +
At

SigniHcance of negative and positive sign. It may be emphasised that the rate of reaction is always
positive. The minus sign along with the first term is used simply to show that the concentration of the

ow
reactant is decreasing, i.e., A [/?] is negative so that (-) x (-) = +. The plus sign along with the second
term is used to show that the concentration of the product is increasing, i.e., A [P] is -¥ve.
For example, consider the reaction :
PClg > PCI3 + Cl2

e
re
Rate of reaction = - A[PCIg] = + A[PCl3]_^A[Cl2l

Frl
At At At

F
In general, for any reaction of the type : A + B ^C + D,

A [A]
ou A[B] A[C] A[D1

r
Rate of reaction = - = + = +

so
At At At At

kf
Units of the Rate of Reaction. Since concentration is usually expressed in moles/litre and the time is
taken in seconds or minutes, the unit of the rate of reaction is moles litre”’ sec”’ (mol L”’ s”’) or moles litre”’
oo
min”’ (mol L"' min”’).
Y
In case of gaseous reactions, pressures are used in place of molar concentrations. As pressures are
eB

expressed in atmospheres, therefore, the units of the rate of reaction are atm min”’ or atm s”‘, etc. The
relationship between partial pressure of a gas in a reaction mixture and its molar concentration(c) follows
ur
oY

from the relationship : pW = nRT or ^ “ "y ^ ...(0

ad

No. of moles of the gas

d

where partial pressure, p = X Total pressure

Total no. of moles
in
Re
F

From eqn. (/), — = —.rt i.e.. Rate in atm s ’ = (Rate in mol L ’ s ’) x RT

dt dt

4.3.2. Average Rate and Instantaneous Rate of Reaction

A difficulty arises in stating the rate of reaction as above. This is because according to the Law of Mass
Action, the rate of reaction depends upon the molar concentrations of reactants which keep on decreasing
with the passage of time (while those of the products keep on increasing).Therefore,the rate of reaction does
not remain constant throughout. Hence, the rate of reaction as defined above is the 'average rate of reaction'
during the time interval chosen.
Thus, the expressions for the rate of reaction written above represent the average rate of reaction, i.e..

A[R] . A[P]
Average rate of reaction (r^^) = - = +
At At

Ax
In general, rav
At
4/4 ^,n>zdeefi-'a, New Course Chemistry CX11)ES2S

To know the rate of reaction at any instant of time during the course of a reaction, we introduce the term
‘instantaneous rate of reaction’ which may be defined as follows : —
The instantaneous rate of reaction, i.e., rate of reaction at any instant of time is the rate of
change of concentration (i.e. change of concentration per unit time) offl y one of the reactants
or products at that particular instant of time.

To express the instantaneous rate of reaction, a small interval of time (dt) is chosen at that particular
instant of time during which the rate of reaction is supposed to be almost constant. In fact, instantaneous rate
is the average rate during the smallest possible interval of time (i.e., when At approaches zero). Suppose the
small change in concentration is dx in the small interval of time dt. Then the rate of reaction at that instant is

given by —. Thus, Instantaneous rate (r■^J^^^) = —

ow
dt dt
^/[R] . ^[P]
For example, for the reaction R inst
dt
= +
dt

AfR] or +. A[P]
In fact. 'i«st as A / > 0

e
At At

Fl
re
Retain in Memory

F
Symbol A is used for larger change, i.e., for average rate whereas symbol'd' is used for small
ur
or
change, i.e., for instantaneous rate.

4.3.3. Measurement of the Rate of Reaction

k sf
Yo
In order to measure the rate of a reaction, the progress of the reaction is followed by studying the
oo

concentration of one of the reactants or products at different intervals of time. The most common practice to
do so is to withdraw small amount of the reaction mixture (2 cm^ or 5 cm^) at different intervals of lime, cool
B

it down immediately to nearly 0°C to arrest the reaction (called freezing the reaction) and then find out the
re

concentration of the reactant or the product by suitable method usually by titration against a suitable reagent.
However, this method is not preferred when some observable property like volume, pressure, optical rotation
u
ad

etc. changes with time and can be observed directly at different intervals of time without stopping the reaction.
Yo

It is important to mention that except concentration, all other factors (like temperature etc.) which affect the
rate of the reaction are kept constant during the kinetic study of the reaction.
d

If concentration of one of the reactants is studied at different intervals of time and these values are
Re
in

plotted against the corresponding times, a graph of the type shown in Fig. 4.1 (a) is obtained.
Calculation of instantaneous rate of reactions. To know the rate of the reaction at any time t, a
F

tangent is drawn to the curve at the point corresponding to that time (Fig. 4.1 (a)) and it is extended on either
side so as to cut the axes, say at the points A and B. Then
Rate of reaction = Change in the concentration
Time
Ax OA
= Slope of the tangent
At OB

Thus, the slope of the tangent gives the rate of reaction.

For example, from Fig. 4.1 (a) in the present case, at time t = 10 minutes, Ax = OA = 0-04 mole and
A / = OB = 20 minutes = 20 x 60 = 1200 seconds, therefore, rate of reaction at the end of 10 minutes will be
0-04/1200 = 3-33 x 10*^ mol L“* s"’.
Calculation of the average rate of reaction. To calculate the average rate of reaction between any two
instants of time say /j and t2, the corresponding concentrations .Tj and X2 are noted from the graph as shown
in Fig. 4.1 (b). Then
CHEMtCAL KINETICS 4/5

FIGURE 4.1
ZA 0,0 A 0,05
<

^ y 0.04-jsA _ d[R)_
n “ 0,04
'^inst = - Slope

It
dl
<K 0,03-
'\{Rl I Ar

||ll Jo^
VV

CO
z lU 0,02- X
O -JI
pO
Ax = 0,04
go,,-, At
MOLE r _A[Pl >^-><1
<2 0,01- LLI Q. 'I
'av
F Z \

i‘inst'^= Slope
O < I
18O-0i I
ID
o
0.001-°i-^ o I
o
_ 5 10 15 20 25 H<2 TIME=t>
i
10 15 20 25

o
At = 20 MIN, 0 TIME(INMlNUTES)c ©
TIME (IN MINUTES)«=5>

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(a) Calculation of the instantaneous rate of a reaction from plot of concentration of a reactant vs time
(b) Calculation of the average rate of reaction (c) Calculation of instantaneous rate and
average rate of a reaction from a plot of concentration of a product vs time

F lo
A[R]
Average rate of reaction = -
At

ee
For example, from the Fig. 4.1 (b) between the time interval 5 to 15 minutes,

Fr
0012-003 0-018
Average rate = - - 0-0018 mol L ^ min ’

for
15-5 10
ur
Calculation of rates from plot of concentration of a productversus time. If concentrationof one of
the products is plotted against time, the type of curve obtained is shown in Fig. 4.1 (c). The average rate of
ks
Yo
reaction between times and i2 is found by noting the concentrations x^ and X2 corresponding to these times
oo

as shown in the lower part of the Fig. 4.1. (c) and then calculating the average rate, as done above. The rate
eB

of reaction at any instant of time is calculated, as before, from the slope of the tangent at that point as shown
in the upper part of Fig. 4.1 (c), i.e.,
r

OA _ Ax
ou
ad

Rate = = Slope of the tangent

OB ~ At
Y

The expression for the average rate will be ; Average rate of reaction = ,^1^ = ^^
nd
Re

Af ?2 -?i
Fi

Significance of negative and positive slopes and rates of reaction. From Figs. 4.1 (a) and (c), it may
be noted that in the plot of concentration of reactant versus time, the tangent at any instant oftime has a
negative slope whereas in the plot of concentration of product versus lime, the tangent has a positive
d[R] d[P]
slope. Hence, we have Ptnsi = - Slope of the tangent or r-insl = +
= Slope of the tangent
dt dt

4.3.4. Expressing the Rate of Reaction in terms of different Reactants and Products.
In the example discussed earlier, i.e., PCI5 PCI3 + CI2, 1 mole of PCI5 dissociates to form 1
mole of PCI3 and 1 mole of CI2, i-e., the stoichiometric coefficient of each reactant and product is the same.
For such a case, the rate of reaction will be same whether we express it in terms of decrease in the concentration
of PCI5 or increase in the concentration of PCI3 and CI2. However, a difficulty arises, if the stoichiometric
coefficients of a reaction, i.e., coefficients of reactants and products in the balanced equation are not same.
For example, consider the reaction, 2N,05 ^ 4NO^ + O^.
4/6 New Course Chemistiy C^)ISSI91

^[N^Og]
For this reaction. Rate of decomposition (disappearance) of N2O5 = - dt

dlNO^]
Rate of formation of NO2 = + dt

Rate of formation of O2 = +£I2i]

dt

However, the rates as expressed above are not equal. This is because when 2 moles of N2O5 decompose,
4 moles of NO2 and 1 mole of O2 are produced in the same time. Thus, the rate of decomposition of N2O5 is
twice as fast as that of formation of O2 and the rate of formation of NO2 is four times as fast as that of O2.
Hence, to get identical value of the rate of reaction in terms of any reactant or product, the rates as expressed
above are further divided by the stoichiometric coefficients. Thus, for the above reaction,

w
\d[N^5]_ M[N02l_ I fr[02]

F lo
Rate of reaction = -
2 dt ^4 dt dt

Similarly, for the reaction, N2 + 3H2 ^2NH3

ee
1 dENHg]

Fr
Rate of reaction = - = + -
dt 3 dt 2 dt

or for the reaction 5 Br~ (aq) + BrOj (aq) + 6 H'*’ (aq) for 3Br2(o9) + 3H20(/)
ur
1 fr[Br-] _ ^/[BrOj] _ 1 J[H-»-]
oks

Rate = -
Yo

5 dt dt 6 dt 3 dt
o
eB

Note that in aqueous solutions, the rate of reaction is not expressed in terms of change of concentration
of water because the change is very small (negligible), e.g., say from 55-5 mol L“* to 5549 mol L”^
our
ad

In general, for the reaction, aA + bB ^ X X + y Y, the rate expressions may be written in any
one of the following ways:
Y

ld[Y]
ld[B] , 1 ^[X]
Re

ld[A\
nd

Rate = - = +—
a dt b dt X dt y dt
Fi

where d [A], d [B] represent small decrease in the concentrations of A and B respectively and d pC] and d [Y]
represent small increase in the concentrations of X and Y respectively in the small interval of time dt.
To sum up
(z) The rate of a reaction is equal to the change in concentration divided by the time interval in which
that change has taken place,
(ii) The rate of reaction can be expressed in terms of any reactant or product.
(Hi) As concentration of reactant decreases, a negative sign is used to express the rate of reaction in
terms of reactants. As concentration ofproduct increases, a positive sign is used in terms ofproducts,
(iv) To make the different expressionsfor the rate to be equivalent, the rate expressions are divided by the
stoichiometric coefficients present in the balanced chemical equation.
(v) The rate of reaction at any instant of time is equal to the slope of the tangent to the curve between
concentration of any reactant or product versus time at that instant of time.
CHEMICAL KINETICS 4/7

PROBLEMS
o a
BASED For the reaction, a A + bB > xX + yY
ON
Rate of reaction = - 1 d[A] ^ ^ ^ +1^111
Rates of a dt b dt X dt y dt
Reactions
d[A] d[B]
Rate of disappearance of = - Rate of disappearance of B = -
dt ' dt

d[X] ti[Yl
Rate of formation of X = Rate of formation of Y =

ow
dt dt
Note carefully that reciprocals of coefficients are written in the expressions for rate of reaction but
not with rate of disappearance of a reactant or formation of a product. Knowing the rate of reaction,
the rate of disappearance of a reactant or formation of a product can be calculated or vice versa.
d[A]
For example. Rate of disappearance of A, - = a X Rate of reaction

e
dt

re
rFl
-d[A] 1 d[A]
Conversely, knowing rate of disappearance of A, , Rate of reaction = -

F
dt a dt
d[X]
Similarly, Rale of formation of X, = X X Rate of reaction
dt

r
ou
^[X] 1 d[X]
Conversely, knowing rate of formation of X,
fo
Rate of reaction = + —
dt X dt
ks
oo

Protslem ^ Express the rate of the following reaction in terms of difi*erent reactants and products :
Y
eB

4NH3(g) + 5 02(g) »4NO(g)+ 6H20(g)

If the rate of formation of NO is 3*6 x 10"^ mol L"* s"^, calculate (i) the rate of disappearance of NH3
(i!) rate of formation of H2O. (Assam d J0I2)
r
ou

I ^[H^O]
ad
Y

Solution. Rate of reaction = = +-

4 dt 5 dt 4 dt 6 dt

Rate of disappearance of NH3 = Rate of formation of NO = 3*6 x 10”^ mol L"^ s~^
d
Re
in

Rate of formation of HoO = —

- X Rate of formation of NO
^ 4
F

^f[H,0] 3
i.e., - —X (3-6 X10 ^mol L ' s"’) = 5*4 x 10 ^ mol s“*
dt 2

Alternatively,
1 fr[H20] _ 1 tf[NO]
-
6 d[NO]
dt 4 dt
=> Rate of formation of H2O, dt ~A dt
Prokslem
The reaction, 2 N2O5 (g) ^ ± 4 NO2 (g) + O2 (g) was studied in a closed vessel
It was found that the concentration of NO2 increases by 2*0 x 10~^ mol L~^ in five seconds. Calculate
(i) the rate of reaction (ii) the rate of change of concentration of N2O5.

I
Solution. (/) Rate of reaction = — fr[NQ2l . But ^/[NQ^] _ 2-0xlQ-^molL-‘ = 4x 10-3 mol L-ig-'
4 dt dt 5s

1
Rale of reaction = — x4x 10“3 = 10 ^ mol L * s *
4
4/8 ^nnuCcep-’^ New Course Chemistry (X11)ISSX9]

1
(//) Rate of change of cone, of N2O5 = — — X Rale of formation of NO,
dt 2 ^

1 ^[NO^]
— X (4 X 10~^ mol L“’ s“*) = 2 x 10"^ mol L~* s
-1

~ 2 di ~ 2

Problem 1^ For an elementary reaction, 2A + B ——> 3 C, the rate of appearance of C at time ‘t’
is 1*3 X mol L“* s"^ Calculate at this time
(i) rate of the reaction (ii) rate of disappearance of A.

ow
1 i/[A] I t/[C]
Solution. Rate = -
2 di di ~ 3 di '
<nc]
Given : = 1-3x10-^ molL-'s-*
di

e
re
Rate of reaction = - = - x 1-3 x 10 mol L ’ s * = 4*33 x 10"^ mol L * s ^
3 dt 3

Frl
F
(/[A] _ 2^[C] -xl-3xl0^molL-ls-> =8-66x 10-= mol L"^ s"*
Rate of disappearance of A = -
dt ~ 3 dt 3
ou
r
Problem El For the decomposition of dinitrogen pentoxide at 200"C,

so
1
N205(g) > N204(g) + -02(g),
jL
kf
oo
if the initial pressure is 114 mm and after 25 minutes of the reaction, total pressure of the gaseous
mixture is 133 mm, calculate the average rate of reaction in (a) atm mln"^ (b) mol L"‘ s“*.
Y
B

Solution. {0' If p is the decrease in pressure of N2O5 in 25 minutes, then

1
re

N205(g) > N2O4 ig) + - O2 (g)

oY
u

Initial pressure 114 mm 0 0

ad

1
d

After 25 min 114-p P

2^
in

1
Re

I
Total pressure =114-p + p + —p =114 + — p = 133mm (Given)
F

p = 2 X (133 - 114) = 38 mm
38
i.e., Decrease in pressure in 25 min = 38 mm = atm
760

38/760atm
Average rate of reaction = = 0*002 atm min ^
25 min

(/j) Applying pV = nRT, i.e., — =

V RT

38/760 atm
Decrease in molar concentration = = 0 0012875 mol L"*
0 0821L atm K"’ mol”* x 473 K

00012875 mol L-i

Average rate of reaction = = 8*58 x 10-^ mol L”* 8"^
25x60s
CHEMICAL KINETICS 4/9

Problem
B From the concentrations of C4H9CI (butyl chloride) at different times given below,
calculate the average rate of reaction, C4H9CI + H2O ●> C4H9OH + HCl, during different intervals
of time
-1 -I
t/s [C4H9CI ]/moI L t/s [C4H9CI ]/mol L
0 0-100 300 0-0549

50 0-0905 400 0-0439

100 00820 500 0-0335

150 0-0741 700 0-0210

200 0-0671 800 0-017 (NCERT Solved Example)

ow
-{[C4H9CIJ,,-[C4H9CI]/,}
Solution. Average rate of reaction in the interval /j to ^2
^2-^1
-1
Time/s [C4H9Cl]/mol L Time interval Average Rate (mol L”’ s"’)

e
re
0 0100

rFl
-1
-(0 0905-0-100) mol L

F
50 0-0905 0-50 = 1-96x10'^
(50-0)s

-(0-0820-0-0905) mol L-' = 1-70x10^

or
ou
100 0-0820 50-100
(l()0-50)s
ksf
-(10-0741-0-0820) molL
-I
150 0-0741 100-150 = 1-58x10’^
(150-100)s
oo

-(0-067-0-074) mol L-' = 1-40x10-^

Y
B

200 0-067 150-200

(200-150)s
re

-1
-(0-0549-0-067) mol L
300 0-0549 200-300 = 1-22x10^ & so on.
(300-200) s
oYu
ad

Problem
B The decomposition of N2O5 in CCI4 solution at 318 K has been studied by monitoring
the concentration of N2O5 in the solution. Initially the concentration of N2O5 is 2*33 M and after 184 minutes, it
d

is reduced to 2*08 M. The reaction takes place according to the equation : 2 N2O5 > 4 N02-f O2
in
Re

Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of
production of NO2 during this period ? (NCERT Solved Example)
F

-I
Solution. Average Rate = -12 Ar
1 (208 - 2-33) mol f = 6-79 X 10“^ mol L * min *
2 1-84 min

6-79x10“^ molL-^ 1 min

X = 113 X 10-5 L-i s-i
imn 60s

6-79 X10-^ mol L-' 60min

X
= 4*07 X 10-^ mol L"' hr"*
min 1 hr

Rate - -
1 A[Np2] _ lA[N205l = 6-79 X lO”^ mol L ' min ' (calculated above)
4 At 2 At

A[NP2] = 4 X 6-79 X 10“^ mol L ’ min' - 2*72 X 10"5 mol L"^ min
s.,-1
Rate of production of NO2, Ar
4/10 'P’uxtUef '^ New Course Chemistry (XII) EEan

1. Nitrogen dioxide (NO2) reacts with fluorine (F2) to form nitryl fluoride {NO2F).
2N02(g) + F2(g) — ^ 2N02F(g)
Write the rate of reaction in terms of

(/) rate of formation of NO2F (ii) rate of disappearance of NO, (Hi) rate of disappearance of F2
2. Express the relationship between the rate of productionof water and the rate of disappearanceof oxygen in
the reaction: 2 H2+ O2 ^ 2H20
3. For the reaction, 4 NH3 (g) + 5 O2 (g) ■> 4NO (g) + 6 H2O (g), if the rate expression in terms of disappearance

w
A[NH3] , write the rate expression in terms of concentrations
of NH3 is - of Oj and HjO.
At

4. A + 2B 3 C + 2 D. The rate of disappearance of B is 1 x 10 ^ mol lit ^ sec K What will be (0 Rate of

Flo
the reaction (ii) Rate of change in concentration of A and C ?

e
5. For the reaction 2 N2O5 (g) > 4 NO2 (g) + 0, (g), the rate of formation of NO2 (g) is 2-8 x 10 ^ Ms '.

re
Calculate the rate of disappearance of N2O5 (g). aiSE 2018)

F
> 2 NH3 (g), the rate of formation of NH3 is 3-6 x 10“^ mol s
-1
6. In the given reaction N2 (g) + 3 H2 (g)
calculate
ur
or
(0 rate of reaction, and
(ii) rate of disapperance of H2 (g) f CBSK 2022)
ks
-1
7. In the reaction, 2 N2O5 (g)I > 4 NO2 (g) + O2 (g), the concentration of N2O5 decreases from 0-5 mol L
Yo
to 0-4 mol in 10 minutes. Calculate the average rate of this reaction and rate of production of NO2
oo

during this period. (CBSE 2022)

B

ANSWERS
re

1 d[N02F] 1 d[NO^] d[¥^]

1.(0 +- (ii) -- ^ (Hi) -
u

2 dt dt dt dt 2 dt
ad
Yo

3. Rate = - 4A[02]^4A[H20]
5 At 6 At
d

4. (0 0-5 X 10“^ mol lit”' sec"*

Re
in

^[A] ^^[C]
(ii) - = 0-5 X 10“^ mol lit"* sec"*, + = 1-5 X 10"^ mol lit * sec *]
F

dt dt

5. l-4x 10"^ Ms"* 6. (0 1-8 X lO"* mol L * j * (ii) 54 x 10"^ mol L * ^ ’

7. 5 X 10"^ mol L"' .r*, 0-02 mol L"' r*

HINTS FOR DIFFICULT PROBLEMS

3.
1 A[NH3l 1 A[P2] = +—
1 A[H20]
4 At 5 At 6 At

5.
1 ^[N^Og] _ 1 J[NO,]
2 dt A dt

d[Np^]_ id[NO^]_ -x(2-8xl0“^)

1 = 1 4x 10"^ Ms"*
dt 2 dt ~ '>\
CHEMICAL KINETICS 4/11

6. Rate of reaction = (/[NH3]

dt ~ 3 dt ~2 dt

Given, J[NH3] = 3-6 X 10"^ mol L ’ s *

dt

Rate of reaction = — x 3-6 x 10“^ = 1-8 x 10^ mol L * s '

2

Rate of disappearance of = _d\^^3d[l^ =2 x3-6x 10~^ = 54x 10^molL"' s

-1

dt 2 dt 2

7. Similar to Solved Example 6.

c- ‘ Reactic;’

The rate of any particular reaction depends upon the following factors :—
I. Nature of the reactants. Consider the following two reactions :

F low
2 NO (g) +02(g) ^2N02 ig) ●fast
2 CO (g) +02(g) ^2CO^(g) slow

These reactions appear to be similar but the first is fast while the second is slow. This is because different
amounts of energies are required for breaking of different bonds and different amounts of energies are released
in the formation of different bonds.

re
2. Concentration of the reactants. Greater are the concentrations of the reactants, faster is the reaction.
for F
Conversely, as the concentrations of the reactants decrease, the rate of reaction also decreases.
3. Temperature. The rate of reaction increases with increase of temperature. In most of the cases, the
rate of reaction becomes nearly double for 10° rise of temperature. In some cases, reactions do not take place
at room temperature but take place at higher temperature.
Your
s
eBo k

4. Presence ofcatalyst. A catalyst generally increases the speed of a reaction without itself being consumed
in the reaction. In case of reversible reactions, a catalyst helps to attain the equilibrium quickly without
disturbing the state of equilibrium.
ad

5. Surface area of the reactants. For a reaction involving a solid reactant or catalyst, the smaller is the
our

particle size, i.e., greater is the surface area, the faster is the reaction.
6. Presence of radiation. Some reactions do not take place in the dark but take place in the presence of
light, e.g., H2 + CU > 2 HCl. Such reactions are called “photochemical reactions.” The absorption of
Re

quantum energy of light facilitates the breaking of bonds of the reactants. Further, the rate increases with
Y

increase in the intensity of light.

Find

Now, we shall discuss the quantitative effect of concentration and temperature and the qualitative effect
of catalyst on the rate of reaction. The study of quantitative effect of concentradon on the rate of reaction
leads to the introduction of a number of new terms such as order of reaction, specific reaction rate (rate
constant) and molecularity of a reaction. It also leads to the study of mechanism of the reactions. These
different aspects are discussed in the following sections.

Curiosity Question
f Q. Why do pieces of wood burn faster than a log of wood of the same mass ?
Ans. Pieces of wood have larger surface area than the log of wood of the same mass. Greater the
surface area, faster is the reaction.
I
4/12 ^>uutee^ 4 New Course Chemistry (XII) com
4.4. DEPENDENCE OF RATE ON CONCENTRATION OF REACTANTS
(RATE LAW EXPRESSIONS, RATE CONSTANT AND ORDER OF A REACTION)
Rate law and Rate constant. As a chemical reaction proceeds, the FIGURE 4.2

concentrations of the reactants keep on decreasing while those of the

products keep on increasing, as shown graphically in Fig. 4.2. However, PRODUCT

the rate of reaction is also found to decrease. This shows that the rate of <
a:
H
reaction is directly related to the concentration of reactants.
mO
The effect of concentration on the rate of reaction was first studied o
REACTANT

by Guldberg and Waage in 1867. They put forward the following result, I
called the Law of Mass Action. TIME (IN MINUTES).

For a general reaction, « A + /; B Products Variation of concentration of

Rate « [A]'" [B]^ or Rate = k [A]" [B]^ reactants and products with time
However, experimentally, it is observed that the rate of this reaction may not depend upon all the 'a

w
concentration terms of A and all the 'h' concentration terms of B. Suppose, experimentally the rate of the
reaction is found to depend upon ‘a' concentration terms of A and *P’ concentration terms of B. Then we may

F lo
wnte :

Rate OC
[A]« [B]P or Rate = it[Af [B]P ...(0

ee
where [A] and [B] are the molar concentrations of A and B respectively and A: is a constant called

Fr
velocity constant or rate constant. The above expression is called Rate law. Thus, rate law is the expression
which expresses the rate of reaction in terms of molar concentrations of the reactants with each term raised

for
to some powe,r which may or may not be same as the stoichiometric coefficient of that reactant in the
ur
balanced chemical equation. This also leads us to conclude that the rate law for any reaction cannot be
predicted simply by looking at the balanced chemical equation, i.e., theoretically but has to be determined
s
experimentally. If all concentrations are taken as unity, i.e., [A] = [B] = 1 mole/litre, then Rate = k. Hence,
ook
Yo

Rate constant may be defined as the rate of the reaction when the molar concentration of each
eB

reactant is taken as unity. That is why the rate constant is also called speciifc reaction rate.
Characteristics of rate constant. Some important characteristics of the rate constant are as follows :
r

(/) Rate constant is a measure of the rate of reaction. Greater is the value of the rate constant, faster is the
ad
ou

reaction.

(//) Each reaction has a deifnite value of the rate constant at a particular temperature.
Y

(Hi) The value of the rate constant for the same reaction changes with temperature.
Re
nd

(iv) The value of the rate constant of a reaction does not depend upon the concentrations of the reactants.
(v) The units of the rate constant depend upon the order of reaction (as will be discussedlater).
Fi

Dlfferenco ijcI ween Rate law and Law of mass action. Rate Law differs from the Law of Mass Action
in the fact that whereas the Rate Law states that the rate of reaction depends upon the concentration terms on
which the rate of reaction actually depends, as observed experimentally, the Law of Mass Action is simply
based upon the stoichiometry of the equation. For example, in the above case.
According to Rate Law. Rate = ife [A]“ [B]P ;
b
According to Law of Mass Action, Rate = k [AJ^* [B]
Order of reaction. It is defined as follows :

The sum of the exponents (powers) to which the molar concentrations in the rate law equation
are raised to express the observed rate of the reaction is called the order of reaction.
For example, in the above case, order of reaction = a + p. Depending upon whether <2 + P is equal to
0, 1, 2 or 3, the reactions are said to be of zero order, 1st order, 2nd order and 3rd order respectively, a and
p are called the orders of the reaction with respect to A and B respectively.
CHEMICAL KINETICS 4/13

For example, consider the reaction :

5 Br“ {aq) + BrOj {aq) + 6 («^) ^ 3 Br2 {aq) + 3 H2O (/)
Experimentally, it is observed that the rate law for this reaction is : Rate = k [Br" ] [BrOj ] [H"*"
Thus, order with respect to Br“, BrOj and H'*' ion are I, 1 and 2 respectively and the overall order of
reaction = 1 + 1 + 2 = 4.

Some Common Examples of Reactions of Different Orders.

/iv
(/) Photosynthesis of HCl H2 (g) + C\2 (5) ■> 2HC1

Experimentally, it is found that the rate of reaction is independent of the concentration of and C^.
Hence, it is a reaction of the zero order. Evidently, for such reactions. Rate = k
(ii) Dissociation of A^2^5 2N2O5 4NO2 + O,

w
Experimentally, it is found that; Rate of reaction «= [N2O5]
i.e., it depends only on one concentration term. Hence, the reaction is said to be of first order.
{Hi) Combination between NO2 and F2 to form NO2F 2NO2 (g) + p2 (g) >2N02F (g)

Flo
Experimentally, it is found that: Rate of reaction ->= [NO2] [F2]

ee
i.e., it depends on one concentration term of NO2 and one concentration term of F2. Hence, the order of
reaction is 2, i.e., it is a reaction of the second order.

Fr
Further, it may be mentioned here that the order of a reaction may nor always be a whole number. For
example, at 723K, acetaldehyde decomposes as :

for
ur
723K
CH3CHO ^ CH4 + CO
1.5
ICH3CHOJ . Hence, the order of reaction is 1.5.
Experimentally, it is found that: Rate of reaction
ks
Yo
Some More Examples of Reactions of DilTcrent Orders,
oo

(fl) Reactions of zero order : Some enzyme catalysed reactions and reactions occurring on metal
eB

surfaces are reactions of zero order. For example,

(i) Decomposition of ammonia on a hot platinum surface at high pressure
r

Pt
2NH3 ^ N2 + 3 H2
ou
ad

1130K

The rate law equation is : Rate = k = k*

Y

{ii) Thermal decomposition of HI on gold surface

nd
Re

Au
2 HI H2+I2
Fi

In these reactions, the concentration of the gaseous reactant on the metal surface remains constant
because as the reactants react off on the surface, more molecules are adsorbed on the surface. Hence, rate
remains constant throughout.
*The exact expression for the rate for decomposition of NH3 on Pt surface is given by
Rate = where k^ and k2 are constants
\ + k2[^nfi
When pressure (concentratiorx) of ammonia, i.e., [NHfi is very small, /.'o [NH3I can be neglected in comparison
to 1 so that the above equation becomes
Rate = jtj [NH3], i.e., it is a reaction of first order.
When pressure (concentration) of ammonia is high, i.e., [NH3] is very large, 1 can be neglected in comparison
to k-, [NH3] so that the equation becomes
Rate =
_ -i-
k^ = k ie., it is a reaction of zero order.
k2[^Hfi *2
4/14 T^^iadec/i^ <t New Course Chemistry (XII)IS2SSm
0) Reactions of first order:
(0 Decomposition of ammonium nitrite : NH4NO2 - ^N2 + 2H20
The rate law equation is : Rate = k [NH4NO2]
1
(if) Decomposition of hydrogen peroxide : H2O2 » H2O + -O2
The rate law equation is : Rate = k [H2O2]
{Hi) Decomposition of SO2CI2: SO2CI2 > SO2 + CI2
The rate law equation is : Rate = k [SO2CI2]
(iv) Hydrogenation of ethene : C2H4 (g) + H2 (g) » (g)
The rate law equation is : Rate = k [C2H4]

ow
(v) Decomposition of 1^20^ : 2 N2O5 (g) > 4 NO2 (g) + O2 (g)
The rate law equation is : Rate = k [N2O5]
60(«:
(vO Decomposition of N2O : 2N2O 2 N2 + O2

e
Fl
The rate law equation is : Rate = k [N2O]

re
F
(vh) Natural and artificial decay of unstable nuclei, e.g..
ur
Rate = k [Ra]

or
(c) Reactions of second order :
(i) Dissociation of HI: 2HI H2 + I2
sf
The rate law equation is ; Rate = k [HI]^
k
Yo
oo

(«) Saponification of ethyl acetate : CH3COOC2H5 + NaOH > CH3COONa + C2H5OH

or
CH3COOC2H5 + OH- >CH3C00- + C2H50H
eB

The rate law equation is : Rate = k [CH3COOC2H5I [NaOH]

{Hi) Reaction between NO and O3: NO + O3 > NO2 + O2
ur

The rate law equation is : Rate = k [NO] [O3]

ad
Yo

(d) Reactions of third order:

(0 Combination ofNO and O2 to form NO2 : 2NO + O2 > 2NO2
d

The rate law equation is : Rate = k [NO]^ [O2]

Re
in

(«) Combination of NO and Br2 to form NOBr: 2 NO + Br2 ^ 2NOBr

F

The rate law equation is : Rate = k [NO]^ [Br2]

{e) Reactions of fractional order
(/) Reaction between H2 and Br2 to form HBr : H2 + Br2 ^2HBr

The rate law equation is : Rate = k [H2] [Br2]^^, order = li2

(i7) Combination between CO and C/2 : CO + CI2 ^COCl2
1
The rate law equation is : Rate = k [CO]^ [Cy
1/2
, order =2-
2

{Hi) Dissociation of COCI2: COCI2 >CO + Cl2

The rate law equation is : Rate = k [COCl2]^^, order = ll2

CHEMICAL KINETICS 4/15

Sample Problem |] For a homogeneous decomposition of N2O5 into NO2 and O 2’

2N205(^) ^ 4NO2 (g) + O2 (g), rate = = *[N205]

At

Find out the order with respect to N2O5.

Solution. As rate - k [N2O5], i.e., rate depends only on one concentration term of N,Oc, therefore, order
v.r.t. N205= 1. - ^
Sample Problem Calculate the overall order of a reaction which has the rate expression :
(a) Rate = k [A]*^ (b) Rate = k [Ap [B]"* (NCERT Solved Example)

Solution. . Order w.r.t. A = 1/2, Order w.r.t. B = 3/2 ; Overall order =1 + 1 = 2

w
2 2

3 1
' t>) Order w.r.t. A = 3/2, Order w.r.t. B = - 1 ; Overall order = —

Flo
2 2

Sample Problem)^ The rate law for a reaction is found to be : Rate = k[N02l [I“] [H+]^

e
re
How would the rate of reaction change when (i) Concentration of is doubled

F
(ii) Concentration of I~ U halved (iii) Concentration of each of NOj, I" and are tripled?
ur
or
Solution. Suppose inii* ily the concentrations are :

[NOj] = a mol L“‘, [!"] = b mol and [H^] - c mol L“'

f Rate = kabc^
ks
Yo
(0 New [H+] = 2 c .-. New Rate-kab{2c)- = A kabc^ = 4 times
oo

(h) New 1“ ss 1 1
New Rate = ka-c^ = — kabc^. i.e., rate of reaction is halved.
B

2 2 2
re

(»V)New [NOj] = 3 a, [I'] =3b, [H+] = 3 c New Rate = k (3 a) (3 b) (3 c)2 = 81 abc^ = 81 itme.s.
u
ad

Sample Problem 0 The rate of a gaseous reaction is

Yo

halved when the volume of the vessel IS

i
Dubled. What is the order of reaction ?
Solution. Suppose the reaction is :
d
Re
in

A ■> Products, and order of reaction is n. Then

Rate = k [A]" or r = ka'‘ ...(0
F

When volume of vessel is doubled, molar concentration is halved. As rate becomes half, therefore,

...(ii)
2 2j
Dividing (i) by (ii), we get 2 = 2” or n = 1

Sample Problem NH3 decomposes as : 2 NH3 N2 + 3 H2

^[NHa] diN^]
dt
= ft,[NH3],+ dt
= At2[NH3],+ = *3(NH3]
dt

Derive the ratio between ki and

Solution. The rate of reaction is given by —1 =+ =+ii2y

2 dt dt "^3 dt
4/16 ‘Pn.eidec^li'^ New Course Chemistry (XIl)BZsIS]

Substituting the given values, we get ifc,[NH3l = = ^fc3[NH3]

1 1
or ~ ht — h-y — IC'i

Thus, if *2 = 1. then = 2 and = 3, i.e., ratio : 'A2 : *3 = 2 : 1 : 3. i.e., ratio A, : /:2 : A3 = 2 : 1 : 3.

Sample Problem 0 For the reaction A ^ B, the rate of reaction becomes three times when th
concentration of A is increased by nine times. What is the order of reaction ? (CIJSE 2018 (C]

Solution. Suppose order of reaction is 'n\ Then

Rate (r) = k [A]"
Initially, r = An" ...(1

When concentration is increased 9 times,

w
3r=A(9 af
Dividing eqn. (ii) by eqn. (0

F lo
I
3=l?i^ = 9^ =32"
n
2n = 1 or n =—
2
or 0-5
a

ee
Fr
for
ur
1. For the reaction 2 X > X2, the rate of reaction becomes 27 times when the concentration of X is increase
three times. What is the order of the reaction ?
s

second order in reactant B. How is the rate of reactic

ook

2. A reaction is of first order in reactant A and of

Yo

affected when (0 concentration of B alone is increased to three times (ii) the concentration of A as well as
eB

is doubled ? (CBSE 20H

3. Show by using rate law, how much rate of the reaction, 2NO + O2 ■> 2NO2, will change if the voluir
of the reaction vessel is reduced to one third of its initial value ?
our
ad

ANSWERS

1.3 2. (i) 9 times («) 8 times 3. 27 times

Y

HINTS FOR DIFFICULT PROBLEMS

Re
nd

1. (0/- = A[Xr, 07)27r=A[3Xr Dividing (//) by (i). 27 = 3' or 3^ = 3" or n = 3

Fi

2. r = k [A] [B]2 = kab^ (i) r = ka{3b)^ = 9 kab'^ (ii) r = k{2a)(2 bf = 8 kab^.

3. Initially, rate = k [NO]^ [Ojl -kc-Pb (say).
When volume is reduced to one-third, the concentration of each reactant will become three times.

New rate = A (3 d)^ (3 h) = 27 A b.

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Rate law equation for reversible reaction. For example.

^1 1 ^[HI]
H2+I2 V i 2HI ; Rate = -
2 dl
= Ai [HjJ [l2]-A2lHI]2
*2

Aj = rate constant for forward reaction, A2 = rate constant for backward reaction
CHEMICAL KINETICS
4/17

2. Rate law equation for reactions involving parallei/side reactions. For example,
OH

NO2

Q
OH
r C2H4 + KBr + H2O
Ethylene

00 [Q
o-Nifrophenol
(0 C2H5Br + KOH + HNO
Ethyl bromide
C2H5OH +KBr Phenol
Ethyl alcohol

w
^-Nitrophenol
B (90%) ... Main reaction

Flo
In general, A

e
re
*2
■►0(10%) ■ ■ Side reaction

F
Rate =
-d{A]
= /:,[A]+/:2lA] = (*,+yt2)[A],
ur
or
dt

% Yield of B = and % Yield of C = *2 f

ks
/tj + k2 ^ ^2
Yo
oo

The variation of the concentrations of A, B and C with time may

be represented graphically as shown in Fig. 4.3.
B

3. Consecutive reactions. The reactions which take place in steps,

re

i.e., through an intermediate are called consecutive reactions. In

cone.

general, for a reaction (A —> C) taking place in two steps,

u

each of
ad

first order, we may write

Yo

A » B ^> C
Variation of concentration with time
d

(intermediate)
Re

for reactant A, main reaction

in

Each step has its own reaction rate as well as rate constant. forming B and side reaction
Starting with pure A, its concentration will decrease with time, forming C
F

concentration of B will increase with time and reach a maxunum FIGURE 4.41
and then decrease or decay to form more and more of C whose
Maximum C
concentration will increase with time (Fig. 4.4)
Examples. (/) Radioactive decay A -4 B ^ C.
/ \
/ \
/

(h) Decomposition of ethylene oxide /

/ \
\
' / \ B
cone. \

(CH2)20 > CH3CHO —^ CH4 + CO /

/
\

4. Dependence of rate on the species other than reactants.

For example. -A
Time

CH3COCH3 + I2 - CH3COCH2I + HI; Representing change of concentration

of reactant A, intermediate B and
Rate — — dU^] product C with time for a
dt = k [CH3COCH3] (H^l
consecutive reaction
4/18 7>n4zde€p. U New Course Chemistry (XIl)>:^ai3d

5. Negative order w.r.t. a reaclant/product.* For example,

2O3 ■> 3O2 ; Rate = - 2
lrf[03]^ JO3P - k [03!^ [02]“^ Thus, order w.r.t. O2 is - 1.
[O2J
dt

Similarly, for the following reaction in alkaline medium ;

OC!" + r
OH"
> or + Cl" ; Rate = =k ^ = k [!"] [OC1-] [OH-] -i

dt [OH-]
Thus, order w.r.t. OH ions is-1.
substance indicates that the rate of reaction decreases as the concentration of

ow
The negative order w.r.t. a
that substance increases.

4.5. UNITS OF RATt CONSTANT OR SPECIFIC

REACTION RATE FOR REACTIONS OF DIFFERENT ORDERS

e
re
dx
In general, for a reaction of nth order, Rate, —
dt
= k (cone.)

Frl
F
1 1 1
7 ^
k = —X
1 cone.
X
X
rt-1
= (cone.)*"" time ^
dt (cone.)' time (cone.)' time (cone.)
ou
Putting n = 0, 1,2, 3 etc., units of rate constant for zero order, 1 st order, 2nd order, 3rd order, etc. can be

sor
found. These are given below :
Order of reaction
kf
Units of Rate constant (k)
(mol L"‘)'"® s"‘ = mol L"^ s'^
oo
Zero (n - 0)
First order (n = 1) (molL"i)’-‘s"i = s"^
Y
(molL-*)'-2 s"^ = (mol L"')“* s"‘ = L mol ^ s"^
B

Second order (n = 2)
Third order (« = 3) (mol L-')'-3 s"‘ = (mol L"‘)'^ s“‘ = mor^ s"^
re

In case of gaseous reactions, instead of molar concentration, we have pressures in atmospheres. Hence,
oY
u

the units of rate constant are as follows :

ad

1 1
k = = atm*“" time"^
d

time atm” *
Units of k
in

Order of reaction
Re

Zero (n = 0) (atm)'"® time"' = atm time"', e.g.. atm s“'

F

-1
1st(« = 1) (atm)'"' time"' = time"', e.g., s
2nd (n = 2) (atm)*"^ time"' = atm"' time"', e.g., atm"' s"'
3rd (n = 3) (atm)'"^ time"' == atm"^ time"', e.g., atm"^ s"'
Sample Problem Identify the reaction order from each of the following rate constant:
(i) k = 2*3 X 10"'' litre mol"' sec"' (ii) k = 3-1 X 10-^ sec"'
sec"' (NCERT Solved Example)
(iii) k = 9 3 X lO"* mol litre
Solution. On the basis of units (and not the values) of the rate constant,
■ i) is a reaction of 2nd order, (ii) is a reaction of 1st order, and (Hi) is a reaction of zero order.
*The mechanism is (/) O3 ^ ± O, + 0 (fast) (//) O + O3- 4 2 02(slow)
From slow step. Rale = k [O3] 10]
[0.][0]
From eqn. (/). K
eqm or lO] = K,^^ [031/102]
103]
Hence, Rate = k [O3] Keqm [03]/[02l = ^' [03p[02]-'.
 CHEMICAL KINETICS 4/19

1. The rate constant of a reaction is 3 x 10^/? K What is the order of reaction ?

2. In the reaction A ■> B. the value of the rate constant was found to be 1 -0 x 10"^ mol"’ L s"‘. What is the
order of the reaction ? How will the catalyst affect the value of the rate constant ?

w
3. The rate of reaction A + B ^ Products is given by Rate = k [A]'^ [B]-. What are the units of the rate
constant ?
4.
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration
of the reactant is reduced to half ? What is the unit of rate constant of such a reaction ? (CBSK 2011)

e
5. What are the units of rate constant for zero order and first order reactions ? (j & K Board 2011)

re
o
ANSWERS

r
1. 1st order
2. 2nd order, catalyst increases the value of the rate constant
3. mol s"’ 4. Rate reduces to l/4th, units of jt = L mol"* s“*

F
5. mol L ^ s“', s"^

oF
ul
HINTS FOR DIFFICULT PROBLEMS

k= _ mol L~^ .Y~‘

sr
3.
= (mol L ') '■“* = mol
s s"*.
[A]'^“[B]2 (molL''*)*^2(j^Q]L-ij2

ko
4,6. DISTINCTION BETWEEN RATE OF REACTION AND REACTION
of RATE CONSTANT
o
Y
Rate of Reaction Reaction Rate Constant
B

1. Rate of reaction is the change in concentration of 1. It is a constant of proportionality in the rate law
Y

a reactant or product per unit time. equation and is equal to the rate of reaction when the
er

molar concentration of each of the reactants is unity.

2. The rate of reaction at any instant of time depends
u

2. The rate constant is constant for a particular

upon the molar concentrations of the reactants at reaction at a particular temperature and does not
d
o
ad

that time.
depend upon the concentrationsof the reactants.
3. Its units are always mol litre"’ time"' 3. Its units depend upon the order of reaction
in

(discussed in Art. 4.5.).

4.7. MOLECULARITY OF A REACTION
Re
F

We have already discussed tiiat the rate of reaction may not depend upon the stoichiometric coefficients of
the reactants in the balanced chemical equation. This means that a balanced chemical equation does not give a true
picture of how a reaction takes place. Thus, it is very rare that tlie reaction may be completed in one step.
The reactions taking place in one step are called elementary reactions.
The reactions which do not take place in one step are called complex reactions. The different steps in
which the complex reaction takes place is called the mechanism of the reaction. Each step of the mechanism
of the reaction is an elementary reaction.
Now, according to the ‘Collision Theory’ in order that a reaction may take place, the atoms, ions or
molecules of the reactants must come close together simultaneously and collide with one another.
The number of atoms, ions or molecules that must collide with one another simultaneously so

as to result into a chemical reaction is called the molecularity of the reaction.

In case of simple reactions (also called elementary reactions), all the atoms, ions or molecules of the
reactants present in the balanced chemical reaction can come together simultaneously to collide with each
other. Thus, such reactions take place in one step. Hence, the molecularity is simply the sum of the molecules
4/20 7>>unU<f New Course Chemistry (Xll)tEIBI

of the different reactants as re presented by the balanced chemical equation. A few examples are given
below :

(i) Decomposition of O2F2 ● balanced equation is : O2F2 ^ O2 + F2

Hence, the molecularity of the reaction is 1 and the reaction is called Uni lolecular.
(n) Dissociation of HI: The balanced equation is : 2HI H2 +12
Hence, the molecularity is 2 and the reaction is called Bimolecular.

w
^2NO.
iiii) Reaction between NO and O2 : The balanced equation is ; 2NO + O2
Hence, the molecularity is 3 and the reaction is called Termolecular.
It may be noted that for elementary reactions, the order of reaction is same as its molecularity and order

e
with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical

re
ro
equation.
However, there are many reactions in which all the atoms, ions or molecules as represented in the
balanced chemical reaction may not come together to collide simultaneously. For example, in the reaction,

F
Ful
5 Br- {aq) + BrOj {aq) + 6 H"" {aq) 3 Bt2 {aq) + 3 H2O (/),
it is impossible for all the 12 ions of the reactantsto come together simultaneously to collide. Such reactions

sr
are called complex reactions. These reactions are supposed to take place in a sequence of a number of steps.
Each of these step reactions is called an elementary reaction. Molecularity can be deifned onlyfor an elementary

ko
o
reaction. It has no meaning for the complex reaction.
of
Further, the different elementary reactions of a complex reaction {i.e., sequence of steps in which the
complex reaction takes place) do not take place at the same rate. The niP"
overall rate of the reaction depends
o
Y
upon the slowest step. Hence, the slowest step is called the 1 ●
erB

The situation is similar to that of a cycle factory. The different sections of the factory manufacture
etc. The overall number of cycles manufactured per day
uY

different parts like handles, paddles, rims, seats

depends upon the slowest working section of the factory.
Thus, it is obvious that the tnolecularity of a reaction must always be a whole number whereas the order of
a reaction can be zero orfractional also (as already mentioned). Though generally in a complex reaction, the order
o
ad
d

of reaction is equal to the molecularity of the slowest step, yet the order of any reaction (elementary or complex)
has always to be determined experimentally to find out the exact power of the molar concentration for each
in

reactant on which the rate of reaction actually depends.

As stated earlier, the order of an elementary reaction is equal to its molecularity. However, if the conditions
Re
F

of pressure or concentration are changed, the order may not be equal to molecularity because in that case, the
reaction may not take place in one step, i.e., it may not remain an elementary reaction but may become a
complex reaction. For example, a unimolecular reaction is of first order at high pressure but at low pressure,
it becomes of second order.

A.. :HANif \REAC

As mentioned above, in case of complex reactions, i.e., reactions involving a large number of molecules
of the reactants according to the balanced equation, the chances for all the molecules to come together and
collide are rare. Hence, in such cases, the reactions are supposed to take place in a number of steps.
A series of step reactions or elementary reactions proposed to account for the overall reaction is
called the 'J

The writing of step reactions is usually based upon experimental evidence, e.g., the detection of the
presence of some short lived intermediates etc. However, a complete certainty of the steps is very rarely
possible. But one thing is certain, i.e., slowest step (called the rate determining step) must involve the molecules
on which the rate of reaction actually depends as observed experimentally and written in the Rate Law
CHEMICAL KINETICS 4/21

equation. Further, the sum of the step reactions must satisfy the overall stoichiometry of the reaction. The
bllowing examples will illustrate the point more clearly:
(0 Thermal decomposition of dfnitrogen pentoxide. The balanced equation may be written as :
2NA ^ 4NO2 + Oo
However, according to the Rate Law, it is found that Rate of reaction = k fN^Oj]
Hence, the above reaction must be split into steps in such a way that the slowest step should involve only
tne
molecule of N2O5. Thus, the probable mechanism is as follows:
Slow Fast
Step 1 ; N2O5 ^ NO2 + NO3 ; Step 2 : N2O5 + NO3 3 + O9

ow
Both the above steps are elementai-y reactions. The slow step is unimolecular whereas the fast step is
imolecular.

Hence, the above reaction is a unimolecular reaction or a reaction of the ifrst order.
Substances like NO3 which are believed to be formed during the course of reaction but they do not

e
ppear in the final products are called intermediates.

re
Frl
(//) Combination of NOj and F2 to form NO2F : The reaction may be represented as :

F
2 NO2 + F2 ^ 2 NO2F
Experimentally, it is found that: Rate of reaction - k [NO2] [F2]
ou
or
i.e., the rate law equation involves only one molecule of NO2 and one molecule of F2. Hence, the
robable mechanism is as follows : kfs
Slow Fast
Step 1: NO2 + F2 » NO2F + F ; Step 2 : NO, + F > NO2F
oo

Here, the slow step is bimolecular. Hence, the above reaction must be a biniolecular reaction
Y
or a
’action of the second order.
B

(Hi) Reaction between NOj and CO to form COj and NO : NO, + CO > CO, + NO
re

Experimentally, it is found that the rate expression below 500K is : Rate of reaction = k [NO2]"
oYu

i.e., the rate of reaction does not depend upon the concentration of CO but depends upon [NO,]l
ad

ence, the slowest step must involve two molecules of NO2. The proposed mechanism is as follows:
Slow
d

Fast
Step 1: NO2 + NO2 ^ NO + NO3 ; Step 2 : NO3 + CO > CO2 + NO2
in

This mechanism is confirmed by the fact that the intermediate NO3 is found to be present in small amounts.
Re

The above reaction is thus again a bimolecular reaction or a reaction of the second order,
F

(iv) Decomposition of hypochlorite. Hypochlorite decomposes to form chlorate and chloride as follows :
3C10-
» C10J + 2C1'
Experimentally, the rate law equation is found to be : Rate = k [CIO"]^
Hence, the probable mechanism is as follows :
Slow Fast
Step 1: CIO- + CIO" > CIO2+CI- ; step 2: CIO2+CIO- ^ cioj+cr
It may be pointed out that having known the rate expression, the writing of mechanism is based upon
lagmation. Out of a number of probable mechanisms, the best is considered to be the one which is supported
some experimental evidence.
Note. The rate of reaction may depend not only on the reactants but may also depend upon the substances
esent
as catalyst. For example, the decomposition of hydrogen peroxide is catalysed by iodide ion in an
caline medium. The complete reaction may be represented as follows :
4/22 ‘P>uxd€e^'^ New Course Chemistry (XII)I!EIH]
r
2H2O2 > 2 H2O + O2
alkaline medium

The rate law equation for this reaction is found to be :

Rate = -
2 dt
= ife[H202][I-]

Thus, order wxt. H2O2 - 1 and order w.r.t. I" = 1, and overall order of reaction is 2.
place in steps and the slow step should involve on
The rate law equation suggests that the reaction takes
H2O2 molecule and one I" ion. The evidence suggests that the reaction takes place in two steps as follows :
Fast
Slow
Step 1: H2O2 + I > H2O + 10- ; Step 2 : H2O2 + lO H2O + I- + O;
Hypoiodite ion

w
Flo
1. Nitric oxide, NO reacts with oxygen to produce nitrogen dioxide: 2N0 (g) + O2 (g) ^ 2NO2 (8)

ee
k

Fr
What is the predicted rale law, if the mechanism is : NO + O2 V ± NO3 (fast)
*1

for
NO3 + NO NO2 + NO2 (slow)
ur
2. For the reaction, NO2 (g) + CO (g) 4
CO2 (g) + NO (g), the experimentally determined rate expressio
below 400 K is: rate'= k [NO,l^. What mechanism can be proposed for this reaction ?
k s
For the chemical reaction, 4 HBr + O2 4 2 H2O + 2 Br,, Rate = k [HBr] [O2].
Yo
3.
oo

What is the probable mechanism of the reaction ?

eB

4. Nitric oxide reacts with hydrogen to give nitrogen and water (2 NO + 2 H2 ■> Nj + 2 HoO). The kinetu
of this reaction is explained by the following steps:
(ii) H2O2 + H2 ^ 2 H^O (fast)
r

(0 2 NO + H2 N2 + H2O2 (slow)
ou
ad

(CBSK 200
What is the predicted rate law ?
For the reaction at 500 K, NO2 (g) + CO (g) COo (g) + NO (g), the proposed mechanism is as belov
Y

5.

(0 NO2 + NO2 -> NO + NO3 (slow) (ii) NO3 + CO CO2 + NO2 (fast)
Re
nd

What is the rate law for the reaction ?

The possible mechanism for the reaction : 2 H2 + 2 NO ^ N2 + 2 H2O is
Fi

6.

4 N2O + H2O (slow) (Hi) N2O + H2 Nt + H-,0 (fast)

(0 2 NO V i N2O2 (/ON2O2+H2
What is (0 the rate law for the reaction (ii) the order of the reaction ?
7. Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions
2 H2O2 2 H2O + O2
This reaction takes place in two steps as given below :
Step-I. H2O2 + r > H2O + 10“ (slow)
Step-II. H2O2 + 10- > H2O + r + O2 (fast)
(a) Write the rate law expression and determine the order of reaction w.r.t. H202-
(b) What is the molecularity of each individual step ?
(c) Write the overall order of reaction.
(CBSE 201
(d) Out of step I and II, which one is the rate determining step ?
CHEMICAL KINETICS 4/23

ANSWERS

1. Rale = k [NO]^ [O2] 2. See Example («7) on page 4/20

3. HBr + O2 > HOOBr ...(slow), HOOBr + HBr ■> 2 HOBr ...(fast),
HOBr + HBr > H2O + Br,] x 2 ...(fast)
4. Rate = k [NO]- [H2] 5. Rate = k [NO-,]-
6. Rate = k [NOJ-[H2], Order = 3
7. (a) Rate = k [H2O2I [I~], order w.r.t. H^O-i = 1.
(b) Molecularily : Step-1 = 2, Step-II = 2.
(c) Overall order = 2. (d) Step I is the rate determining step.

ow
HINTS FOR DIFFICULT PROBLEMS

1. From the slow step, Rate = [NO3] [NO] ...(0

[N03J

e
From fast step, eqm. const, k = ...(«●)

re
[N0][0,j

Substituting the value of [NO3] from (//) in (/), we

rFl
F
get:
Rate = k' (NOj^ [O^j
5. From the slow step, Rate = k [NO2]-.

r
ou
sfo
6. From (//), Rate = k [N2O2] [H,j
[N2O2]
k
From (0, K
eq or [N2O2J = [N0J2
[N0]2
oo

Rate = k. [NO]- [H,] = k' [NO]- [H2]

Y
B
re

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
ou
Y

Kinetic study of reaction of iodide ion with hydrogen peroxide at room temperature. Iodide ions are
ad

oxidized by H2O2 in presence of an acid as follows:

H2O2 + 2 r + 2 H-^ ^12 + 2 H2O ...{/)
d

The reaction is carried out in presence of a small amount of thiosulphate (Na2S203) and starch solution.
in
Re

The iodine liberated combines with thiosulphate as follows :

F

+ StOj' ^ S4O2- +21- ...(H)

When all S2O3 ions react away with l2, further liberated combines with starch to give a blue
colour

l2 + Starch ^ Blue colour ...(Hi)

If reaction (/) is fast, S2O3 ions will react away quickly and hence blue colour will appear quickly. In
other words, rate of reaction is inversely proportional to the time taken for blue colour to appear (i.e.,
rateoe
]/!). Thus, starling with same concentration of H-,0-) and adding same amount of S.,Oj" but different
concentrations of iodide, rate of reaction with respect to T ions can be studied. Likewise, rale of reaction
with respect to H2O2 can be studied. It is found that rate « [H20->] [L].
The set of reactions (/) to (Hi) is called a ‘clock reaction’ as rate can be easily timed by sudden appearance
of blue colour. Reaction (/) is called the main reaction, reaction (//) is called monitor reaction whereas
reaction (Hi) is called indicator reaction. Further, reaction (i) is slow and hence the rate deiennining step.
4/24 New Course Chemistry (XlUESESQBl

4.9. DISTINCTION BETWEEN ORDER AND MOLECULARITY OF A REACTION

The main points of difference between the order and the molecularity of a reaction may be summarized
as follows :

Order of Reaction Molecularity of Reaction

1. It is the sum of the concentration terms on which 1. It is the number of atoms, ions or molecules that
the rate of reaction actually depends or it is the must collide with one another simultaneously so
sum of the exponents of the molar concentrations as to result into a chemical reaction.
in the rate law equation.
2. It need not be a whole number, i.e., it can be 2. It is always a whole number. It cannot be zero or
fractional as well as zero. fractional.

It can be calculated by simply adding the

w
3. It can be determined experimentally only and 3.
cannot be calculated. molecules of the slowest step.
4. Order is applicable to elementary as well as 4. Molecularity is applicable only to elementary
reactions. The overall molecularity of a complex

Flo
complex reactions.
reaction has no significance. It is only the slowest
step whose molecularity has significance for the

e
re
overall reaction. Molecularity of the slowest step
is the same as the overall order of reaction.

rF
5. Even the order of a simple reaction may not be 5. For simple reactions, the molecularity can
equal to the number of molecules of the reactants usually be obtained from the stoichiometry of
ur
as seen from the balanced equation.
fo
the equation.

It may be mentioned here that the reactions of higher order are very rare. It is because according to
ks
collision theory, for a reaction to take place, the molecules must come together .simultaneously and collide
Yo
oo

with one another. Greater the number of molecules, lesser are the chances for the molecules to come together
simultaneously and collide.
B

Retain in Memory
re

1. Dissociation of potassium chlorate to form potassium perchlorate (4 KCIO3 ■>3 KCIO4+KCI) is

u

an example of a reaction of 4th order.

ad
Yo

2. Order of a reaction can change with change in conditions such as temperature, pressure, etc. but
molecularity of a reaction is invariable.
d
Re

4.10. INTEGRATED RATE EQUATIONS

in

We have already discussed that for the general reaction, a fs + bB cC + dD,

F

^[R]
Rate = - = k[Af-[B#
dt
This form of the equation is called differential rate equation. This form is not convenient to determine
the rate law and hence the order of reaction. This is because the instantaneous rate has to be determined from
the slope of the tangent at time t in the plot of concentration versus time. This measurement is not only
inconvenient but also lacks accuracy quite often.
To overcome the above difficulty, we integrate the differential equation for the reaction of any order.
This gives us an equation relating directly the experimental data, i.e., time, concentrations at different times
and rate constant.
The integrated rate equations thus obtained are different for reactions of different order. We shall derive
here only for reactions of zero order and first order.
4.10.1. Integrated Rate Equation for Zero Order Reactions
Consider the general reaction : A Products
dW
If it is a reaction of zero order. Rale = - = k[Af'=k or d[A]=-kdt
dt
CHEMICAL KINETICS 4/25

Integrating both sides, we get: [A] = - kt + I ...(0

where I is a constant of integration. At r =0, |A] = [A] 0 [A)o = I
Substituting this value of I in eqn. (/), we get [A] =-kt + [A] 0 ...OV)

or kt=[A\ 0 -[A] or =

This is the expression for rate constant for reactions of zero order.
Some Important Characteristics of Reactions of Zero Order :
(0 Any reaction of zero order must obey eqn. (ii). As it is an equation of a straight line (y =mx + c), the
plot of [Aj versus t will be a straight line with slope = - k and intercept on the concentration axis = [A]q,
as shown in Fig. 4.5.
Also, as rate is independent of concentration, plot of rate versus concentration will be a straight line
parallel to concentration axis (Fig. 4.6).

w
FIGURE 4.5 FIGURE 4.6 FIGURE 4.7
<.

F lo
SLOPE OF THE
z [AJo LINE = -k
o
OR ^c = -SLOPE
OFTHELINE
CC ■

ee
lU
t

Fr
UJ
<
o (T

o
o

TIME, t CONCENTRATION , for INITIAL CONC,, [A]q

ur
Plot of [A] versus t for Plot of Rate versus concentration Plot of half-life (tj/2) versus initial
reactions of zero order for reactions of zero order concentration for a zero order reaction
s
ook
Yo

(ii) Half-life period. Half-life period {tj/2) is the time in which half of the substance has reacted.
eB

[A] 0
This implies that when [Aj = ,t = t Ml-
2
our

[A] 0 [An]
ad

1 [A] 0
Substituting these values in eqn. (Hi), we get t^2 = ~ uA]q ^ 2k
, i.e., t 1/2 “
2k
...(/V)

Thus, half-life period of a zero order reaction is directly proportional to initial concentration, i.e.
Y

1
Re

t 1/2 oc
[A]q. Hence, a plot of t^^ versus [A]q will be a straight line passing through the origin and slope =
nd

, as
2k
shown in Fig. 4.7
Fi

Molar cone. molL *

(Hi) Units of Rate constant (k). From eqn. (Hi), k = - molL ^ time"^ ●
Time time

SUPPLEMENT YOUR
KNOWLEBGE FOR COMPET»T10t*f

1. If the reactant ‘A’ is gaseous and pressures are measured instead of molar concentrations, then for reactions
of zero order,
I P.
0
/: = -[Po-P^] and / 1/2 “
2k
where Pg is initial pressure of A and P^ is the pressure of A after the t.
Further. Pq - Pa = P* decrease in pressure of A in time t. Hence, we can also write

k =— or p = kt or p OC
t, i.e., decrease in pressure of the gaseous reactant A « time.
r
4/26 ^xAdee^'4. New Course Chemistry (Xll)Qsl9]

2. The time taken for the zero order reaction to complete can be calculated as follows :
When the reaction is complete, [A] = 0
[Alo
k = -[A] 0 or t ...(V/)
t 100% “ ^
3. For 3/4th of the reaction to complete, t =

[A] = [A]p ~^[A]q = -^[Aly. Hence, FIGURE 4.8

3 [A] 0
...(w7)

ow
t At the end
3/4 “ 4 k of the reaction

Dividing cqn. (vii) by eqn. (/v), we get

t
3/4 ^3[A]q/4^_3- = l-5

e
0)

t
1/2 [Alo/2/: 2 n;

re
a:

rFl
Hence, for a zero order reaction, t
3/4
= 1-5/ 1/2* 0

F
Time >

4. A zero order reaction proceeds with constant rate with time but at
Plot of rate versus time for a zero
the end of the reaction, rate of reaction will drop to zero as no order reaction

or
ou
reactant is left {Fig. 4.8).
ksf
Sample Problem Q During the decomposition of a gas on the surface of a solid catalyst, the
pressure of the gas at different times was observed to be as follows :
oo

t/s 0 100 200 300

Y

3-40 X 10^ 2-60 X 10-^

B

p/Pa 5-00 X 10^ 4-20 X 10^

Calculate order, rate constant and half-life period of this reaction.
re

Solution. Let us first calculate rates of reaction during different intervals of time
oYu

lnter\>al Rate of reaction

ad

0-100 5
(4-20-5-00)xl(}3pa = 8 Pas-*
d

100s
in
Re

(3-40-4-20) xl(>* Pa = 8 Pas-*

100-200
100s
F

(2-60 - 340) X10^ Pa = 8 Pas-*

200-300 .t
100s

As the rale of reaction remains con.stant throughout, therefore, the reaction is of zero order.
Alternatively, let us test it by integrated rate equation
1
t (sec) /t = -[Py
l
- PJ (for zero order reaction)
1
100 k = (5 00 - 4-20) X 10^ Pa = 8 Pa s~*
100s

1
200 k = (5-00-340) X 10^Pa=8Pas-‘
200 s

I
300 k = (5-00 - 2-60) X 10^ Pa = 8 Pa s-‘
300s
CHEMICAL KINETICS 4/27

As k comes out to be constant, the reaction is of zero order.

-1
Rate constant, /t = 8 Pa s

[Pq] _ 5 00x10^ Pa
Half-life period, tyy = 2 k 2x8Pas-'
= 312-5 s.

Ft
Sample Problem Q The decomposition of NHj on platinum > N2(g)
surface, 2 NH3 (g)
+3 (g), is zero order with k = 2-5 x mol L“* .s"^ What are the rates of production of Nj and H2 ?
(CBSE 2007)

Rale = —
1 ^[NH3] 4/[N2l 1 r/[H,l
Solution. 2 NH3 2 di
= +
dt
= -I--
3 di

ow
For zero order reaction, rate = k
I ^[NH3J_^[N^J _ 1 dtH^] = 2-5 X 10-* mol L-‘ s"‘
2 dt ~ dt 3 dt

^[N,]
Rate of production of Nt = = 2-5 X 10~* mol L"^ s"*

e
dt

re
d[H^]

rFl
= 3 X (2-5 X 10-^ mol L"' s'*) = 7-5 x 10"* mol L"' s"*.

F
Rate of production of Ht =
dt

Sample Problem E] The rate constant of a reaction of zero order in A is 0-0030 mol L * s~^ How

r
ou
long will it take for the initial concentration of A to fall from 0-10 M to 0*075 M ?

fo
(CBSE 2010)
Solution. For a zero order reaction.
ks
/=- [AL-[A] (010-0-075) M =8-3s
oo
k^ 0-003 Ms“‘
Y
eB

4.10.2. Integrated Rate Equation for First Order Reactions

A reaction is said to be of the ifrst order if the rate of the reaction depends upon one concentration term
only. Thus, we may have
ur

For the reaction : A ^ Products, Rate of reaction «= [A],

ad
Yo

For the reaction : 2A ^ Products. Rate of reaction «= [A] only.

For the reaction : A -1- B Products, Rate of reaction « [A] or [B] only.
d

Let us consider the simplest case, viz.. A -> Products.

Re
in

Suppose we start with a moles per litre of the reactant A. After time r, suppose x moles per litre of it have
F

decomposed. Therefore, the concentration of A after time t = (a-x) moles per litre. Then according to Law
dx dx
of Mass Action, Rale of reaction <x (a-a:), i.e., ~ oc (a - x) or ~ = k{a-x) ...(0
dt dt
where k is called the rate constant or the specific reaction rate for the reaction of the first order.
The expression for the rate constant k may be derived as follows :
dx
Equation (/) may be rewritten in the form = kdt Mi)
a- X

dx
Integrating equation (//), we get a-x
h dt, i,e„ — In {a -x)~kt +1 ...(iiV)

where I is a constant of integration.

In the beginning, when t = 0, x = 0 (as no substance has decomposed/reactedat the stait of the reaction).
Putting these values in equation (///), we get - In (a -'0) = i x 0 -i-1 or - In a = l .Mv)
4/28 "PnaxCee^ d New Course Chemistry (XII) PlSTMl

Substituting this value of I in equation (Hi), we get

a
-\n (a-x) = kt + (- In a) or /:? = In fl - In (rt-;c) or kt = ]n ...(V)
a - X

1 a 2-303 a
or k = - \n or log ...(vO
t a-x t a-x

Equation (vi) is sometimes written in another form which is obtained as follows:

If the initial concentration is [A]g and the concentration after time t is [A], then putting a = [A]q and

ow
(a-x) = [A], equation (v/) becomes
,
k =
2-303,lOg-TTT
[AJo (V«)
t [AJ
[A] 0

e
Further, putting a = (A]q and {a - .v) = [A] in eqn. (v), we get kt = In ...(VHl)
[A]

re
which can be written in the exponential form as

Frl
F
[A] 0 [A] -kt
- gkt or = e or
[A3 = [AJoe ...(a)
[A] [A] 0
ou
sor
Expression for the rate constant for reactions of first order from concentrations of the reactant at
two different times. As derived above, for reactions of first order.

kf [from eqn. (vn/)]

oo
t [A]
Y
or
kt = In [A]q - In [A] or
In [A] =- kt+\n [A]q
B

At r = r, if [A] = [A], and at t - t2, [A] = [A]2, then

re

In [A]i=-/:ri + ln[A]„
oY
u

In [A2] = -^^2 + In [A]q ...{xi)

ad

[A] 1
Subtracting eqn. (xi) from eqn. (x), we get, In [A] 1 In f A]2 = -/:!] - (- kl2) or In — k (^2 tj)
d

[A]2
in

k =
1
In
[A], or k =
2-303 [A] I
log
Re

or

(^2-^1) tA]2 [A]2

F

Some Imp* ‘ sties of First Order Reaction.s :

(/) Any reaction of the ifrst order must obey equations (vi), (vii) and (ix). This may be tested in any one
of the following ways :
(a) Substitution method. Starting with a known concentration ‘a’ or [A]q, the concentration of the reactant
(a - x) or [A] at different intervals of time may be noted. For every value of t, the corresponding value of
(a-x), i.e., [A] may be substituted in equation (vi) or (vii). If the values of k thus obtained are nearly constant
(within the experimental error), the reaction is of the first order.
(b) Graphical method. Equation (vii) may be written as
[A] 0
1 - log = log [AJo-log [A]
2-303 [A]

or log[A] = - t + log [A](, ...(xii)

2-303 ^
CHEMICAL KINETICS 4/29

This is the equation of a straight line (>) = mx + c). Thus, if FIGURE 4.9

log [A] or log (a-x) values are plotted against time the graph obtained
should be a straight line if the reaction is of the first order (Fig. 4.9).
The intercept made on the v-axis would be ‘log lA]y’ and the slope of <

k ] k o

the line would be equal to , i.e.y slope = - . From this, the o

2-303 ) 2-303
value of k can be calculated. t

[A] 0 [A] 0 riot of log [A] vs t

Further, from eqn. (v/7), log -— t. Thus, if we plot lo O

[A] 2-303 "[A]

FIGURE 4.10
k
versus r, a straight line graph will be obtained with slope = . as shown

ow
2-303
in Fig. 4.10.
(it) Half-life Period. The rime taken for any fraction of the reaction to
complete is independent of the initial concentration. For example, let us test
the truth of this statement for half of the reaction to complete. Equation (v/)

e
Fl
re
may be written as TIME, t

Plot of log {[A] Q / [A]} vs time

F
2-303, a
for a first order reaction
t = — log ...{xiii)
a — x
ur
r
When half of the reaction is completed, x = all. Representing the time taken for half of the reaction to
be completed by t^/2, equation {xiii) becomes fo
ks
Yo
2-303 0*693
a 2-303, ^ 0-693
hn-~
oo
, i.e.,
hn - k a
a —
2
eB

Thus, 'a' does not appear in this equation which shows that tyi is independent of a. Similarly, it can be
seen that t^fj, ^2/3 etc. will also be independent of the initial concentration. Also, it may be seen that ty2 is
ur

inversely proportional to the rate constant k.

ad
Yo

(Hi) The units of k are independent of the units in which the concentrations are expressed. This is
obvious from equation (v/) because in this equation, k depends upon the ratio of two concentrations, viz.,
d

a
and the lime t. Thus, so long as both a and (a - x) are expressed in the same concentration units, the
Re
in

a-x

value of k is not affected. The units of k, therefore, depend only upon the units of time't'. Depending upon
F

whether t is expressed in seconds, minutes or hours, k would be in sec"*, min"’ or hr"' respectively.
4.103. Integrated Rate Equation For First Order Gas Phase Reactions
Consider the general first order gas phase reaction :

A(^) >B(^)-*-C(g)
Suppose the initial pressure of A = Pq atm.
After time t, suppose the pressure of A decreases by p atm.
Now. as 1 mole of A decomposes to give 1 mole of B and 1 mole of C. pressure of B and C will increase
by p each. Hence, we have
A(g) ^ B (g) + C (g)
Initial pressure : Pq atm 0 0

Pressure after time, t: (Po-/^) p atm p atm

4/30 ‘Pn.ddee^'A. New Course Chemistry fXll^Pgmn

Total pressure of the reaction mixture after time t,

P,= (Pq - p) + p + p = Pq + P -dim ●●● /' = P,-Pq
Pressure of A after time T (P^) = Pq - p = p^ - (P^ - P^) = 2 P,) - P^
But initial pressure of A (Pq) oc initial cone, of A, i.e., [A] 0
and Pressure of A after time t (P^) «= cone, of A at time t, i.e., lA]
2-303, [A] 0
Substituting these values in the first order rate equation, k = log , we get

w
t [A]

2*303 P
0
k = log
2Po-P/

e
For example, azoisopropane decomposes as :

ro
re
{CH3)2Chn = nch(CH3)2 ao — ->N2 (g) + C6H,4 (g)
or sulphuryl chloride (SO-iCl-)) decomposes as

F
SO2CI2 (g) ^ S02 (g) + C12 (g)

Fl
u
Expression for k will be same as derived above.
For Sample Problems, refer to Problem 4.20. page 4/106 and Problem 4.21, page 4/106.

sr
4.11. HALF-LIFE PERIOD OF A REACTION

ko
o
The time taken for half of the reaction to complete, i.e., the time in which the concentration of
a reactant is reduced to half of its original value is called half-life period of the reaction.
of
o
Y
General expression for half life period of a reaction of nth order. As already discus.sed,
erB

0
for zero order, t 1/2 [Aq], for 1st order, /j/2 [AoJ
oc OC

Similarly for 2nd order, t^p « [Ay]“^ and so on.

uY

1 - «
1
In general, for a reaction of nth order, t 1/2 OC
fAo] or /
1/2
OC

[Aof-'
1
ad
do

For a gaseous reaction of nth order, t 1/2 OC

H-1
P6
in

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS
Re
F

1. General expression for the time taken for nth fraction of a reaction of 1st order to complete.
2-303 a
Putting X = — and t= in the first order eqn., t = — log , we get
n k a- X

2-303 a 2-303 1
2*303. n
t
\/n - log — log or t
1/n = —log
k a k 1-1/n n-1
a —
n

Thus, the time taken for any fraction of the 1st order reaction to complete is independent of the initial
concentration. This method can be used to lest a reaction of first order. The method is called ‘fractional
life method.’

2. Some useful relationships between times for different fractions of reaction of first order to complete
%4 or /75% = 2/j/2> ^93-75% = ^ fi/2' ^99-9% = 1® ^1/2-
3. General expression for the time taken for the reactant to reduce to nth fraction.

Here, (a~x) = -
a 2-303 a 2*303,
t = — log or t = ——logn.
n k a! n k
CHEMICAL KINETICS 4/31

Sample Problem D The rate constant of a first order reaction is 60 s ^ How much time will
it take to reduce the initial concentration of the reactant to its 1/16th value ?
Atf=0 [Aol
2-303, a 2-303
Solution. / = — log Iogl6 = 4*62xl0-2s.
k «/l6 60s"i
4. Amount of the substance left after n half-lives.
Atl-t-i/2 (Alo
(One half-life)
Amount of the substance left after one half-life =
[A]o 2

ow
Amount of the substance left after two half lives = 1 [Alo _ fA] 0
—

Atf = 2fi/2 1 [Alo [Alo

2 2 22
(Two half-lives) 2 2 ^
1 [AJo [A](,
Amount of the substance left after three half-lives = — x -

e
2 22 2^

re
[Alo AU = nfi/2 [Alo
In general, Amount of the substance left after n half lives = (/I half-lives) 2"
2"

F
Frl
Sample Problem The half-life period of a first order reaction is 60 minutes. What
percentage will be left after 240 minutes ? ●

Solution. No. of half-lives =

240
ou
- = 4, i.e., /I = 4

sr
60

kfo
Amount left after 4 half-lives = [A]q _ [A] 0
= 00625 of [Alo = 6-25%.
2^ 16
oo
5. Average life. In case of a radioactive substance, the whole of it never disintegrates. In other words, time
required for complete disintegration is infinity. Hence, we talk of average life. As faster is the
Y
disintegration, smaller is the average life, therefore, average life (x) is taken as the reciprocal of
reB

1 1/2
disintegration constant (k), i.e., x = = l-44xr
1/2 ●
k ~ 0-693
uY

In general, average life of a first order reaction {also called natural life time) is the time in which
the concentration of the reactant, A reduces to I/e of the original concentration, [Alg, i-e..
ad
do

r -
1 [A]q .1
= — ln„ e = — ●
1
av
k " [Alo/e k ^ k
in
Re

4.12. DETERMINATION OF THE RATE LAW, RATE CONSTANT AND ORDER OF REACTION
F

The different experimental methods used for the determination of the rate law, rate constant and order of
reaction are briefly described below :—
1. Graphical Method (Using DlfTerential Rate Law Equations).This method can be used when there
is only one reactant. For example, consider the reaction : n A ^ Products

If this reaction is of zero order,

FIGURE 4.11
o ( ^ d[A]
= k [Ap = k ...(/) ZERO ORDER ■>RDEr.
n dt . I

Thus, rate is independent of concentration. Hence, t

plot of Rate vs [A] will be a horizontal line (Fig. 4.11 (())● $
If the reaction is of first order.
Rate oc [A] or Rale = k [A] ...(H)
(i) [A] (ii) [A] (iii) (A)2
Thus, plot of rate vs. molar concentration of the Differential rate law method for
reactant would be linear (Fig. 4.11 (//)) and equation testing the order of a reaction graphically
iii) will be the ‘rate law’

*
4/32 U New Course Chemistry (Xll)BZs!9]

If the above reaction is of the second order. Rate « [A]^ or Rate = k [A]~,
Thus, plot of rate vs [A]" would be linear (Fig. 4.11 (///)) andeqn. (/n) would be the ‘rate law’ and so on.
can be measured by the method already explained in Art.
The rate of reaction at different instants of time
4.3.3 (ie.. by noting the concentration of any suitable reactant or product at different times and then plotting the
concentrations vs. time and finding out the slopes of the tangents corresponding to different times).
The method is illustrated below, taking the example of the decomposition of N2O5. The reaction is :
2N205(g) ^4N02(g) + 02(g)
The progress of this reaction can be studied by measuring the increase in pressure of the reaction mixture
at different times. From these values, the partial pressure of N9O5 at different times can be calculated from

w
which the molar concentration of N2O5 can be calculated. The molar concentrations of N.>05 are then plotted
against time. A graph of the type .shown in Fig. 4.12 is obtained.
The rates at different times are obtained by measuring the slopes of the tangents corresponding to these
times (or the approximate values can be obtained by dividing the change in concentration by the time interval).
The data for one such experiment is given in Table 4.1.

o
e
Data for the decomposition of N20g

re
TABLE 4.1.
FIGURE 4.12

Frl
[N2O5] (mol L-n RATE (mol L""* mln"^)

F
0,01^- TIME (min)
^ 0.014f' 0 1.72 X I0-“
-1

i 0.010-
ou 1.23 X 10-2 34 X 10-^

r
10
z
0.84 X 10-2 25 X 10-5

so
% 0.006- 20
o
0.62x 10-2 18 X 10-5
CN

z 0.002}-
30
40 0.46 X
kf
10-2 13 X 10-5
oo
20 40 60 80 100 50 0.35 X 10-2 10 X 10-5
TIME IN MIN —►
Y
60 0.26 X l(r2 8x 10-5
B

Plot of [N2O5] in moles/litre 70 0.29 X 10-2 6 X 10-5

versus time in minutes*
80 0.14 X 10-2 4x 10-5
re
oY

Now. graphs are plotted for (/) Rate vs [N2O5], and (//) Rate vs [N205]2. These are shown in Fig. 4.13.
u
ad

FIGURE 4.13
d

4X10-4
in
Re

3 X 10-4'
^ 3X10^
s
F

2 X 10-4 2X10-4-
UJ
1 X10-4 1 X lO-*-
a:
2 4 6 8 10 12 14 X 10-5
e o INzOs]^ IN MOL^ L"^
Plot of (a) Rate vs [N2O5] and (b) Rate vs [N205]2
The fact that the plot of Rate vs [N2O5] is linear but that of Rate vs [N205]2 is non-linear shows that
Rate « [N2O5] and not [N205]2. Thus, the rate law is : Rate = k [N^05]
This further shows that the order of reaction is 1, though the coefficient of N2O5 in the balanced equation
is 2. The rate constant k is evidently equal to the slope of the line or it can be calculated as :
k = Rate/(N205]
*From this plot, we may note that concentration of N2O5 does not decrease to zero even after a very long
time (as the curve does not meet the time axis). Thus, the rate of reaction may decrease to a very small value but
does not become zero even after a long time.

a
CHEMICAL KINETICS 4/33

Sample Problem D The experimental data for the decomposition of N2O5 [2 NjOj ^4N02 +
O2] in gas phase at 318 K are given below :
t/s 0 400 800 1200 1600 2000 2400 2800 3200

10^ X [NjOgl/mol L-‘ 1*63 1*36 1-14 0-93 0-78 0-64 0-53 0-43 0-35
(i) Plot [N2O5] against time, (ii) Find the half-life period for the reaction,
(iii) Draw a graph between log [N20g] and t. (iv) What is rate law ?
(v) Calculate the rate constant, (vi) Calculate the half>iife period from k and compare it with (ii).
Solution.

t(s) 0 400 800 1200 1600 2000 2400 2600 3200

[N2O5] (mol L"’) 1-63 1-36 M4 0-93 0-78 0-64 0-53 0-43 0-35] X 10"-
log [N2O5] -1-79 -1-87 -1-94 -2-03 -2-11 -2-19 -2-28 -2-37 -2-46

ow
(/) Plot of [NjOg] vs time (ii) Inilial cone, of N2O5 = 1-63 x 10"“ M,
Half of this concentration = 0-815 x 10"“ M

Time corresponding to this concentration

= 1440 s. Hence, r,/2 = 1440 s.

e
Fl
re
{Hi) Plot of log [N20g] vs time

F
-1’80®
ur -1-90 -

or
- 2 00 -
'13
O
sf
^-2-10
k
Yo
O1
oo

2 - 2-20
B

-2-30
re

-2-40

i I
-2-50
u

0 400 800 1200 1600 2000 2400 2800 3200

ad

TlME/(s) TlME/(s)
Yo

(/»') As plot of log [N20.;] vs time is a straight line, hence it is a reaction of first order, i.e., rate law is:
Rate = k [N2O5].
d
Re
in

k
(\’) Slope of the line = -
2-303
F

-2-46-(-1-79) 0-67 k 0-67

Slope = i.e..
3200-0 3200 ’ 2-303 3200

0-67 0-693 0-693

or k = X 2-303 = 4'82xl0^molL-‘s- -1 (w) = 1438 s.
3200 4-82x10"^

Sample Problem Q The rate of formation of a dimer in a second order dimerisation reaction is
9*5 X 10"^ mol L“* s'* at 0*01 mol L"‘ monomer concentration. Calculate the rate constant.

Solution. If the monomer is represented by A, then the reaction is : 2A ^(A),

As the reaction is of second order, the rate of reaction will be given by : Rate = k [A]“

1^[A1
Rate of reaction = - = + = 9-5 X 10”^ mol L"’ s'*
2 dt dt

9-5 X 10"^ mol L"^ s"^ = k (0-01 mol L"’)^ = /c x 10“^ (mol L"‘)^ or k- 0*95 L mol"* s"^

1 a
4/34 ^nadeefi. <j- New Course Chemistry (XII) PZSTin

Sample Problem B The following data were obtained during the first order thermal
decomposition of N2O5 (^) at a constant volume : 2 N2O5 (g) > 2 N2O4 (g) + O2 (g)
S. No. Time (sec) Total pressure (atm)
1 0 0-5

2 100 0*512

Calculate the rate constant. (CBSE Sample Paper 2018)

Solution. 2 N2O5 (g) ^ 2N204(g) + 02(g)
Initial pressure Pq 0 0

Pressure after time t (Pq -2 p) 2p P

ow
Total pressure after time / (P,) = (Pq - 2 />) + 2 p + /; = Pq + p or p = P, - Pq

Pressure of N2O5 after time ^ *^N205 ^ = Pq “ ^ P = Pq “ ~ ^ Pq “ ^ Pf

Thus, [N205lo « Pq . [N205],-3Pq-2P,

e
2-303 P. 2-303 0-5

Fl
0

re
k = log log
t
3Pq-2P, 100 3x0-5-2(0-512)

F
2-303 0-5 _ 2-303 2-303
log log 1-05 = (0-0212) =4*88 X 10-*s-‘
100 0476 ~ 100
ur 100

or
sf
wmm
k
Yo
oo

1. The decomposition of N2O5 in carbon tetrachloride solution was studied

B

N2O5 (solution) ^ 2NO2 (solution) +

re

The reaction has been found to be first order and the rate constant is found to be 6.2 x 10^ s~'. Calculate the
rate of reaction when (a) [N^O^] = 1.25 mol and (b) IN2O5] = 0.25 mol L"*.
u
ad

(c) What concentration of N-iOg would give a rate of 2.4 x 10“^ mol L“* s”* ?
Yo

2. For the reaction : 2 A + B -f C -» A2B + C, the rate law has been determined to be
Rate = k [A] [B]2 with A = 2.0 x mor- s"*
d
Re

For this reaction determine the initial rate of the reaction with [A] = 0.1 mol L"', [B] = 0.2 mol L"'. [C] = 0.8
in

mol L"’. Determine the rate after 0.04 mol L”^ of A has reacted.
F

3. Following data are obtained for the reaction

1
N2O5 » 2 NO, + - 0,
i 2 ^
t/s 0 300 600

[NjOjJ/mol L
-I
1-6 X 10"- 0-8 X 10-2 0-4 X 10-2

(a) Show that it follows first order reaction

(b) Calculate the half-life
(Given log 2 = 0-3010, log 4 = 0-6021) (CBSE 2017)

ANSWERS

1. (a) 7.75 X 10-^ mol L ' s ^ (d) 1.55 x lO"^ mol L ^ s ^ (c) cone. = 3.87 mol L *
2. 8 X 10-^ mol L-‘ s~\ 3-9 x 10"^ mol s'* 3. f,;2 “ 300 s

a 1
CHEMICAL KINETICS 4/35

HINTS FOR DIFFICULT PROBLEMS

1. (a) Rate = A [N2O5I = (6-2 x 10"^ s”*) (1-25 mol L'^) = 7-75 x 10*^ mol L"‘s"'
(/7) Rate = (6-2 x 10^ s”') (0-25 mol L“') = 1-55 x IQ-^ mol L~’ s~'
(c)24x lO'^-^molL-'s-' = (6-2 x s"*) [N2O5] or [N2O5I = 3-87 mol f'.
2. Initial rale = (2.0 X 10“^ mor-L^s“*) x 0.1 mol L“’x (0.2 mol L“*)^ = 8 x 10~‘^ mol L“’s“*
After 0.04 mol L“‘ of A has reacted. [Aj = 0.1 - 0.04 = 0.06 mol L“'
[B] = 0.2-0.02 = 0.18molL-’
Then rate = (2 x 10“^) x (0.06) x (0-18)^ = 3.9 x 10“^ mol L“' s““'.
2-303.log ['^2^5)0
3. fN20s]o = l-6x 10--M, k =
/
[N2O5]

w
At r = 300 s, k =
2-303 1-6x10-^ _ 2-303 log2 =2-31 X 10-^
300 0-8x10-2 " 300

F lo
2-303 1-6x10-2 2-303
At r = 600 s, k = xlog log 4 =2-31 X 10-^
600 0-4x10-2 600

ee
Fr
As k comes out to be constant, it is a reaction of 1st order.

0-693 0-693
t
1/2“ ^ 2-31x10-^
= 300s
for
ur
or 1-6 X 10-2 Q.g jq-2 means half of the substance has reacted in 300 s. Hence, t\/2 = 300 s.
s
ook

2. Initial Rate Method. This method can be used irrespective of the number of reactants involved, e.g.
Yo

consider the reaction.

eB

A + ti2B + n^C ^ Products

The method consists in finding the initial rate of reaction taking known concentrations of the different
reactants (A, B, C).* Now, the concentrationof one of the reactants is changed (say that of A) taking the
our
ad

concentrations of other reactants (B and C) same as before. The initial rate of the reaction is determined
again. This gives the rate expression with respect to A and hence the order w.r.t A. The experiment is repealed
by changing the concentration of B and taking the same concentrations of A and C and finally changing the
Y

concentration of C and taking the same concentrations of A and B. These will give rate expressions w.r.t. B
Re
nd

and C and hence the orders w.r.t. B and C respectively. Combining the different rale expressions, the overall
rate expression and hence the overall order can be obtained.
Fi

Suppose it is observed as follows :

(/) Keeping the concentrations of B and C constant, if concentration of A is doubled, the rate of reaction
becomes four times or if the concentration of A is trebled, the rate of reaction becomes 9 times. This
means that
Rate OC
[A]2, i.e., order w.r.t A is 2
(i7) Keeping the concentrations of A and C constant, if concentration of B is doubled, the rate of reaction is
also doubled. This means that

Rale [B|, i.e., order w.r.t. B is 1

(in) Keeping the concentrations of A and B constant, if concentration of C is doubled, the rale of reaction
remains unaffected.This means that rate is independentof the concentration of C, i.e. order w.r.t. C is
zero.

*The initial rate of the reaction with respect to a particular reactant can be found from the slope of the
tangent at r = 0 to the graph between concentration vs time.
4/36 New Course Chemistry (Xll)CSlSi

Hence, the overall rate law expression will be

Rate = /t[A]2[B] [Cf Overall order of reaction = 2 + 1 + 0 = 3

Alrernalively, if order with respect to A, B and C are a, P and y. then we have :

Rate = ^[Af [B]P[Cr
Starting with two different initial concentrations of A, keeping concentrations of B and C constant, the
initial rates of the reaction are determined, then

ter ...(0

(ro)2=/:[Ao]“ [B]P [C]Y ...(«)

a

('b)l _[Ao]f [Ao]i

or log = a log
^'*0^2 [Ao]? [AoJ.

w
a =
log[(ro),/(ro)2]

F lo
log{[Ao], /[Aq]^}
This is the order with respect to A. Similarly, order with respect to B and C can be determined. On

ee
adding, we get the overall order of the reaction (= a + p + y)-

Fr
Alternatively, for a reaction of nth order having one reactant A only, Rate = k [A]" = kC
Starting with two different initial concentrations Cj and C2 and assuming that the corresponding initial
rates are rj and r^, we have for
ur
r^=kC'‘■, r2=-tq
s
ook
Yo
C C.1 .r
r, I
or nlog = Iog-J- or n =
\og(r^/r^)
C2 ^2 logfCj/C^)
eB

'2 "2
Rate at any time can be found form the plot of concentration versus time and slope of the tangent on the
curve at any time gives the rate of reaction at that time (r = dCldt). As it involves differentials of
r
ad

concentration with time to express the rates, the method is called van’t Hoff differential method,
ou

Sam
pie Problem a The following rate data were obtained at 303K for the following reaction :
Y

2A + B ^ C + D

[B]/mol L-^ Initial rate of formation of D/mol L“^ min"*

Re

-1
Experiment [A]/mol L
nd

I 0.1 0.1 6.0 X 10-3

Fi

II 0.3 0.2 7.2 X 10-2

III 0.3 0.4 2.88 X 10-*
IV 0.4 0.1 2.4 X 10-2
What is the rate law ? What is the order with respect to each reactant and the overall order ? Also
calculate the rate constant and write its units. (CBSE Foreign 2017)
Solution. From experiments I and IV, it may be noted that [B] is same but [A] has been made four times, the
rate of reaction has also become four times. This means that w.r.t. A,

Rate « [A] ...(/)

From experiments II and III, it may be noted that [A] is kept same and [B] has been doubled, the rate of
reaction has become four times. This means that w.r.t. B,

Rate [B]2 ...(H)

Combining (1) and (11), we get the rate law for the given reaction as : Rate = k [A][B]2
Thus, order w.r.t. A = 1, order w.r.t. B = 2 and overall order of the reaction =1+2 = 3
The rate constant and its units can be calculated from the data of each experiment using the expression.
CHEMICAL KINETICS 4/37

^ Rate molL~^min~^ = mol"2 L2 min”‘

" [A][B]2 ” (molL-l)(molL-l)2
Expt. k (mor^ l2 min“‘) Expt. k (mol"^ \? min"*)
60x10-2 2-88x10-*
I = 6-0 m = 6-0
aix (0-1)2 0-3 X (0-4)2
7-2x10-2 2-4x10-2
n = 6-0 IV = 6-0
0-3x(a2)2 0-4 X (0-1)2

Rate constant, k = 6.0 moH l} min-*

Alternatively, suppose order w.r.t. A is a and w.r.t B is p. Then the rate law will be Rate = k [A]“ [B]P

w
Our aim is to find a, p and k.
Substitutingthe values of expts I to IV, we get
(Rate)^^p, 1 = 6.0 X KT^ = k (0.1)“ (O.l)P

F lo
.(0
(Rate)expt 2 = 7.2 X 10~2 = k (0.3)“ (0.2)P ...(«)

ee
(Rate)„p, 3 = 2.88 x KT* = Jt (0.3)“ (0.4)P (Hi)

(Rate),,p,4 = 2.4 X 10"2 = k (0.4)“ (O.l)P

Fr
(Rate) exptl _ 6.0x10-2 _ cf(0.1)«(0.1)P 1 ^ (0.1)“ ^ r if or a = 1
4“(0.4)«"t4,
or
2.4x10-2 *(0.4)“(0.1)P
for
ur
(Rate)expt4
(Rate) expt 2 _ 7.2x10-2 _fc(0.3)“(0.2)P 1 _ (0.2)P _ f 1
ks
(Rate)expt 3 "2.88x10-* " jfc(0.3)“(0.4)P 4"(0.4)P"t2j
or
Yo
oo

lt= fl- ^P or P = 2
eB

or

.’. Rate law expression is : Rate = k [A] [B]2. k can be calculated as in the first method.
r

ISaSipl^^oBiem^ In a reaction between A and B, the initial rate of reaction was measured for
ou
ad

different initial concentrations of A and B as given below :

Y

A/mol L-* 0-20 0-20 0*40

B/mol L-* 0-30 0-10 005
nd
Re

rg/mol L“* s"* 5 07 X 10-^ 5 07 X 10-® 716 X 10-®

Fi

What is the order of reaction with respect to A and B ?

Solution. ro = [A]“[B]P
(ro), = 5-07 X 10-2= (0-20)“ (0-30)P .(0
(ro)2 = 5-07 X 10-2 ^ (0-20)“ (O-IO)P ,(ii)
(ro)3 = 7-16 X 10-2 = (0-40)“ (0-05)P .(lii)

(^o)l ^0-30 ^P = (3)P .-. p=0

(^0^2 ,0-10,

(/b)3 ^ m C 040 0-05

1-412 = 2“ - = 2“ (.*. P = 0)
(ro)2 5-07 " to-20
or
/ \ 0-10 J 12
log 1-412 = a log 2 or a = 0-1523/0-3010 = 0-5.

(or directly 1-412 = = 2*^2 2“ = 2*^2 or a = 1/2)

Thus, order w.r.L A = 0*5, order w.r.t. B = 0
4/38 “Pn^ieUe^'4. New Course Chemistry (XII)BSm

Sample Problem B The rates of a reaction starting with initial concentrations of 2 x 10~'^ M
and 1 X 10"^ M are equal to 2*40 x 10"^ M s"^ and 0-60 x 10“* M s~^ respectively. Calculate the order of
the reaction with respect to the reactant and also the rate constant.
Solution. If n is the order of reaction

Rate = k [Aq]” = kC’

For two different initial concentrations, we have

(C I \"
c
I
log—= « log
C2 2 C2
log{;j//2) log(2-4xl0-^/0-6xl0^)
=^=2

w
or n —

log(Cj/C2) log(2xl0-^/lxl0-3) log 2

Hence, order of reaction = 2

Flo
Rate - k [Af,]-

e
Rate 24x10-^ molL-*s-'

re
k =
= 0*6 X 10^ L mol-* S-*
[Ao]2 (2xlO-^molL“‘)2

F
ur
or
k sf
Yo
1. For a gaseous reaction 2A + ^ 2AB, the following rate data were obtained at 300 K
oo

Rote of disappearance 0/82 Concentration

(mol Hr' min"')

eB

[A] [B4
(0 1.8 X 10-^ 0.015 0.15

(fi) 1.08 X 10-2 0.09 0.15

ur

(Hi) 5.4 X 10-2 0.015 0.45

ad
Yo

Calculate the rate constant for the reaction and rate of formation of AB when [A] is 0.02 and [B,] is 0.04
mol lir' at 300 K.
d

2. For the reaction 2 N-)05 (g) > 4 NOt (g) + O') (g), the following results have been obtained :
Re
in

-1
S. No. [N-)05] mol L Rate of disappearance of
N2O5, mol L"' min"'
F

1 113 X 10-- 34 X I0"5

2 0-84 X 10"- 25 X 10-5
3 0-62 X 10-2 18 X 10-5
(«) Calculate the order of reaction
(/;) Write rate law
(c) Calculate rate constant of the reaction. (Assam Board 2U13)
3. For the thermal decomposition of acetaldehyde, CH3CHO (5) ^CH^ (g) + CO (g).
the following rate data were obtained :
Experiment Initio! pressure (torr) Initial rate of increase in total pressure (torr)
1 300 0.61 (r,)
2 200 0.27 (r^)
Predict the order of reaction.
CHEMICAL KINETICS 4/39

4. The initial rate of reaction A + 5 B + 6 C > 3 L + 3 M has been determined by measuring the rate of
disappearance of A under the following conditions :
Expt. No. [Alo/M [B]o/M [C]o/M Initial rate/M min~^
1 0-02 0-02 0-02 2-08 X 10-3
2 001 0-02 0-02 1-04 X 10-3
3 0-02 0-04 0-02 4-16 X 10-3
4 0-02 0-02 0-04 8-32 X 10-3
Determine the order of reaction with respect to each reactant and overall order of the reaction. What is the
rate constant ? Calculate the initial rate of change in concentration of B and L.
5. For the reaction A + B ■> products, the following initial rates were obtained at various given initial

w
concentrations

-1
S. No. [A] mol L [B] mol L-^ Initial rate M/s

1 01 01 005

o
e
2 0-2 0-1 0-10

re
3 0-1 0-2 0-05

Frl
Determine the half-life period.

F
(CBSE Sample Paper 2019)
ANSWERS

1. A: = 0.8 litre mol ’ min ' and required rate of formation = 1.28 x 10“3 mol lit"' min"*

or
ou
2. Order = 1, Rate = k [N20g], A: = 1-5 x IQ-^min-* 3.2

4. Rate = k [Aq] [Bq] [Cq]^, Overall order = 1 1 2 = 4, = 1-3 x 10“^ M-^ min-^
kfs
d midt = - 1 -04 X 10-2 jyi ^ = 5.24 x 10"3 M mhr^ 5. 1-386 s
oo
Y
HINTS FOR DIFFICULT PROBLEMS
B

1. From (0 and (//), when fB^J is kept constant and [AJ is made 6 times, the rate also becomes six times. Thus,
re

Rate «
[A]. Further, from (/) and (///), when [A] is kept constant and [B2] is made three times, rate also
becomes three times. Thus, Rate [B2]. Hence, rate = k [A] [B2]. Put the values of [A] and [B^] and
oYu

OC
ad

calculate k. It comes out to be 0-8 litre mol"' min . When [A] = 0.02 M and 82= 0.04 M,

Rate of disappearance of B2 = /: [AJ [82] = 0-8 x 0 02 x 0-04 mol L

-1
d

Rale of formation of AB = 2 x Rate of disappearance of B2

in
Re

[N2Q5]i
2. Rate = /t[N205]". .-. . Substitute given values and calculate n.
F

R2 .[N205]2j
1
Rate of reaction =
— X Rate of disappearance of N,Og

1
When [N2O5] = M 3 X 10-2 mol ; Rate = - X 34 X 10-3 = 17 X 10-3 L-i min

Rate 17 X10 ^mol L ’ min *

k =
= 1-5 X 10-2
M3xlO“2molL“'
mm
[N2O5]
3. If n is the order. Rate (r) = k [CH3CHO]
\n

. Solve for n.
0-27 200
4/40 ‘Pna„eUefi.'<i. New Course Chemistry (X11)ES!91

4. Rate (r(,) = k [Ao]“ lB„jP [C()]V Calculate (r„),/(ro)2, (^o)4^(^o)i

I ^[B] 1 d[L]
Initial rate = - = +- = 2-08 X 10"-^ M niiir' {Given)
5 (h 3 dt

^/IBl
Initial change in cone, of B, i.e., = - 5 X (2-08 X 10“^) = - 1-04 X 10 - M min '
dl

d{L]
= 3 X (2-08 X 10'^) = 6-24 x 10*^ M min
-I
Initial change in cone, of L, i.e., dt

5. Calculate the rate law expression as in solved examples. We gel Rale = /: [A]' [B]^’, i.e., it is a first order
reaction

Rate 0-693 0-693

= 0-5s~‘’ = l-386s

ow
k = t
[A] 1/2- ^ 0-5

3. Integrated Rate Law Method. This is the most common method for studying the kinetics of a
chemical reaction. For example, consider the reaction : n A » Products

If we Stan with a moles/litre of A and in time t, x moles/litre have reacted so that the concentration after

e
Fl
re
time / is (fl-.v) moles/litre, then

F
dx,
if the reaction is of first order, — = k {a - x)
dt
ur
or
and if the reaction is of second order, — - k{a- x)- and so on.
dt sf
These differential equations can be integrated to get expressions for the rate constants. These are given
k
Yo
below for zero, first and second order reactions:
oo

For zero order.

1
For 1 st order. ,
k =
2-303,log —^
[Aq]
k = -{[A^]-[A]},
B

t t [A]
re

, *

For 2nd order. 1

k = -< J 1_
t[[A] [Alo
u
ad
Yo

The advantage of the integrated method is that these integrated forms of equations contain the
concentration of a reactant at different times and hence can be solved to find the value of k from the data of
the run of one experiment only and need not start with different initial concentrations. Moreover, they can be
d
Re

used to find the time for any fraction of the reaction to complete.
in

The above expressions can be used to test the order of reaction by either of the following two methods :
F

0) Substitution method : The results

FIGURE 4.141
of the experiment are substituted in 2ND ORDER
ZERO ORDER 1ST ORDER
the expressions for rate constant of
different orders one by one. The k SLOPE =;
-k
2.303 1
expression which gives a constant [A]
SLOPE = k
log [A]
value of the rate constant k decides
the order of the reaction. The ;intercept=[A]o iNTERCEPT= LOG [AJo
[■INTERCEPT= 1/[A]o
limitation of this method is, t t t

however, that it is a hit and trial

Graphical method for testing order
method. This method along with of reaction using integrated equations
examples is explained in Art. 4.13
for the reactions of the first order.

1 1
♦When [A] = [Aj{/2, / = t^(2■ Substituting, we get t
1/2 “
k[A] 0
or
ka
CHEMICAL KINETICS 4/41

(ii) Graphical method. The above expressions can also be tested graphically by rearranging them in suitable
forms as follows ;

For zero order,

[A] =- kt + [A]q, so that plot of [A] vs t will be linear with slope = -k and intercept = [A] 0-
For 1st order.

k
log[A] = - r— + log [A]q so that plot of log [A] vs t will be linear with slope = _
2-303

and intercept = log [A] 0

For 2nd order,

1 1 1
= kt + so that plot of vs t will be linear with slope = k and intercept =

w
[A] [A] 0 [A] [A] 0

4. Half-life method. As already discussed in Art. 4.11. the half-life period of a reaction of wth order is

Flo
related to initial concentration of the reactant as : t 1/2 oc
. The order ‘/f can be found by either of the
H-1
[Aq]

ee
following two methods :

Fr
(a) Substitution method. Starting with two different initial concentrations and finding their half lives,
the order of reaction (n) can be calculated as ;
n—1

for
ur
H-1
^^1/2^1 _ 1^Aq]2 = .
[Aq].
(^l/2>2 EA,,]-' [A(),l| ●●● log (^1/2)1 - log (^1/2)2 = (« - 1) {log [Agio- log [Aq], }
ks
Yo
oo

or n = l-^
log(/j/2>l~^Qg(^l/2^2
eB

●o8[A()]2 -logEAfl],

Sample Problem The half life period of a substance is 50 minutes at a certain initial
r
ou

concentration. When the concentration is reduced to one half of its initial concentration, the half life
ad

period is found to be 25 minutes. Calculate the order of reaction.

Y

Solution. Suppose the initial concentration in ±e first case is a mol L~K Then
[Aq] 1 = a, (t 1/2)1 = 50 minutes
Re
nd

a
Fi

tAol2= 2’ {^1/2)2 = 25 minutes

n-l rt-i

We know that for a reaction of nth order. ^1/2

OC (h/2)j _ [AoJ2 [{Aq]21
[A,,]"-' UAoU
n-l
(h/2)l {AqIi
yi-1 \«-I
50 (a/2 2 fl 2
Substituting the values, we get or
25 a 1 2j 1

or 1 - /I = 1 or n = 0. Hence, the reaction is of zero-order.

(b) Graphical method. Graphically, half-life method can be used to test the order of reaction as follows :

1 I
t oc or t = K
1/2 n-l 1/2 «-l
{Aq] [AqI
where K is a constant of proportionality (not the rate constant).
4/42 ‘P,nadee/t.’^ New Course Chemistry (XII)

Thus, for zero order, = K [A]q. Hence, plot of [Alo be linear passing through the origin
and having slope = K*.
For 1st order, r,/2 “ K [AqI® = K. Thus, is independent of initial concentration. Hence, plot of r,/2 vs
[A]q will be a horizontal line.
FIGURE 4.15

ZERO ORDER 1ST ORDER 2ND ORDER

(Alo

ow
Testing order of a reaction graphically by half-life method
1
For 2nd order, . Hence, plot of vs 1/[A]q will be linear passing through the origin and
[A] 0

e
having slope = K.

Fl
re
Sample Problem Predict the order of the reaction in the given plots :

F
(0 («●)
ur
or
*1/2
k sf (CBSE Foreign 2017)
Yo
[Rio—►
oo

Solution. (/) 1st order (//) Zero order

B

1
This is because t 1/2 OC
[Refer to Art. 4.11]
re

H-l
[RI 0
u

*Exact values of the slopes in terms of rate constants can be found by using tlie expressions for half-life (fj/2) in
ad
Yo

terms of rate constant k as follow.s:

For zero order reaction : From page 4/25

[A] 0
d
Re

r
1/2 “
in

2k
1
F

Hence, plot of /j/2 vs [A]q will be linear passing through the origin and having slope =
2k
For 1st order reaction : From page 4/29

_ 0-693 = con.stant
'1/2- j
Hence, the plot of vs [A]q is independent of initial concentration, i.e., it is a horizontal line.
For 2nd order reaction : From page 4/40
1 1 1
it =-
t[[A] [A] 0
[Al. 1 I

When / = t|/2, [A] = —^ . Substituting and solving, we get ty2 = k[A] 0

1
Hence, plot of /j/2 vs is linear with slope = — .
[A] 0 ^ k
CHEMICAL KINETICS 4/43

5. Ostwald’s isolation method. This method is used when there are a number of reactants. All reactants
except one are taken in large excess so that the change in their concentrations is negligible and the rate of
reaction depends only on the reactant taken in smaller amount. Thus, for the reaction,
flA + /?B + cC Products.

if order with respect to A, B and C are a, P and y respectively, then according to Rate law,
Rate = /: [A]“ [B]P[C]T'
If B and C are in large excess, their concentrations remain almost constant. These can be included in the
rate constant. Hence, we can write
a
Rate = [A]
Thus, order with respect to A is found out. Similarly, order with respect to B and C can be determined
and hence the overall order of the reaction can be determined.

SUPPLEMENT YOUR

w
KNOWLEDGE FOR COMPETITIONS

Expression for degree of dissociation at time t:

F lo
X
For a first order reaction involving the dissociation of a reactant, degree of dissociation = — .
a

ee
1 a a a a- X
From expression for rate constant, /: = - In , i.e. In = kt or

Fr
or =

t a-X a — X a- X a

£ = e-kt ,
or I
a
or — = I - e~^, i e.. Degree of dissociation = 1 - e
a
for at any time t.
ur
4.13. EXAMPLES OF THE REACTIONS OF FIRST ORDER
s
ook
Yo

1. Decomposition of Nitrogen Pentoxide. The compound, nitrogen pentoxide, is a volatile solid which
decomposes in the gaseous state as well as in the form of its solution in an inert solvent like carbon tetrachloride,
eB

chlorofonn etc. according to the equation :

1
N2O5 > N20,+-02
our
ad
Y
Re

2NO2
nd

When the reaction is carried out in the solution, N^04 and NO2 remain in the solution and the volume of
Fi

oxygen gas collected is noted at different intervals of time. It is obvious that

Volume of oxygen gas collected at any time (V,) «><: Amount o/A^2^5 decompose (x)
Volume of oxygen gas collected at any time (V^ )}c«: Amount of A^2^5 decomposec (x)
i.e., X Vj

1
Volume of oxygen gas collected at inifnite time (V^)
(which is done by heating the reaction vessel)
OC
Amount of /^2^5 initially taken (a)

i.e., a OC V,
Substituting these values in the first order equation viz.
, 2.303 a 2.303 V
k = log , we get k= log
t a-x t V.. -V.I

The constancy in the value of k proves the reaction to be of the first order.
4/44 'P>ta,deefi^'4i. New Course Chemistry (XI1)E&Z91

Sample Problem Q From the following data for the decomposition of N2O5 in carbon
tetrachloride solution at 321 K, show that the reaction is of the first order and calculate the rate
constant.

Time (in minutes) 10 15 20 25 00

Vol. of O, evolved (in cm*^) 6.30 8.95 11.40 13.50 34.75

Solution. If the reaction is of the first order, it must obey the equation :
V
, 2.303 a 2.303
k = log log
t a- X t V -V
I

In the present case, we are given that V^= 34.75 cm^

The value of k at each instant can be calculated as follows :

ow
2.303 V
k = log
t (inin) V, v„-v, t V.-V.I

2.303 . 34.75 -1
10 6.30 34.75 - 6.30 = 28.45 k = log = 0.01997 min

e
10 min 28.45

re
rFl
2.303 , 34.75 - 0.01985min"’
15 8.95 34.75 - 8.95 = 25.80 k = ^^log

F
15min 25.80

2.303 34.75
k = log = 0.01987 min"’

r
20 11.40 34.75 - 11.40 = 23.35
ou
20min 23.35

25 13.50 34.75- 13.50 = 21.25

k k = sfo2.303 , 34.75 = 0.01967 min-’
r^^log
25 mm 21.25
oo
Since the value of k comes to be nearly constant, hence given reaction is of the first order. The average value
Y
of the rate constant = 0.01984 minute"’
B

Sample Problem Q The decomposition of NjOg in CCI4 solution follows the first order rate law.
re

The concentrations of measured at different intervals are given below :

ou

Time in seconds (t) 0 80 160 410 600 1130 1720

Y
ad

5-0 4-8 40 3-4 2*4 1-6

[NjOg] mol/L 5-5

Calculate the rate constant at t = 410 s and t = 1130 s. What do these results show ?
d

2-303 a 2-303, [N205l(,

in

Solution. According to first order rate law, k = log

Re

log
t a-x t
[N2O5],
F

,
k =
2-303 5-5
= 7-768xl0"^s"'
At / - 410 s, log
4105 4-0

2-303, 5-5
At/= 1130 s. k = log — = 7-341xl0-‘’s"’
11305 2-4

These results show that the rate constant is nearly constant at different times.

1. The following data were reported for the decomposition of N2O5 in CCI4 at 303 K.
Time (in minutes) 120 160 200 240 00

Vol. of O2 (in cc) 37.70 45.85 52.67 58.34 84.35

Show that the reaction is of the first order and calculate the rale constant.
CHEMICAL KINETICS 4/45

2. The rate of decomposition of N2O5 in CCI4 solution has been studied at 318 K and the following results
have been obtained :
//min 0 135 342 683 1693
c/M 2-08 1-91 1-67 1-35 0-57

Find the order of the reaction and calculate its rate constant. What is the half-life period ?
[Hint. As rate does not remain constant with time, it is not a zero order reaction. Hence, we try integrated

first order equation, i.c., k = 2-303,

t c
ANSWERS

1. 0.0049 minute ' 2. order = 1. /: = 6-31 x 10^ min"*, / 1/2 = 1 098 X 10^ min

2. Decomposition of Animoniuni Nitrite. An aqueous solution of ammonium nitrite at any moderate

low
temperature decomposes according to the equation
NH4NO2 {aq) ^ 2H2O + N2
The volume of nitrogen gas collected is noted at different intervals of time and the result calculated as
in the case of decomposition of N^Oj.

ee
3. Decomposition of Hydrogen Peroxide. The decomposition of hydrogen peroxide in aqueous solution
rF
Fr
{catalysed by the presence of finely divided platinum) takes place according to the equation :

H2O2 ^ H.O-1--O.,
2 2 2

r
fo
The kinetics of this reaction may be studied either by the same method as done earlier (/.e., by collecting
u
the oxygen gas produced and noting its volume at different intervals of time) or by making use of the fact that
ks
H2O2 solution can be titrated against KMnO^ solution. Thus, by withdrawing equal amounts of the solution
Yo
(usually 5 cc) at regular intervals of time and titrating against the same KMn04 solution, the amount of H2O2
oo

present can be found every time. It is obvious that/or the same volume of the reaction solution withdrawn,
B

Volume of KMnO^ solution used before the commen OC

^Initial concentration of ^^2^2
re

cement of the reaction, i.e.,at zero time (Vq)

i.e.. aoeV,0
u
ad
Yo

H
Volume of KMnO^ solution used Amount of H2O-, present
i.e., (o-A')«V I
at any instant of time t {V^) at that instant i.e. {a — x)
Substituting these values in the first order equation, we get
nd
Re

2.303 a
303, Vy
Fi

k = log or k = — log—
i a - X

The decomposition of hydrogen peroxide, as tested by this equation, is found to be of the first order.
Sample Problem From the following data, show that the decomposition of hydrogen peroxide is
a reaction of the first order :
t 0 10 20
X 46.1 29.8 19.3
where t is the time in minutes and x is the volume of standard KMn04 .solution in cm^ required for titrating
the same volume of the reaction mixture.

Solution. Volume of KMn04 solution used OC

amount of present. Hence, if the given reaction is of the
2.303 a 2.303 V0
k = log log
first order, it must obey the equation : t a- X t V
t

(Note that the symbol x in the numerical problem is in fact a- x, i.e., V,)
In the present case, Vq = 46.1 env^
4/46 New Course Chemistry (XII)

The value of k at each instant can be calculated as follows:

2.303, Vq
r {min) V
r
k =
t log^ I

2.303 , 46.1 = 0.0436 min *

10 29.8 k = log
lOmin 29.8

2.303 , 46.1 = 0.0435 min-‘

r-—^log

ow
20 19.3 k =
20 min 19.3

Thus, the value of k comes out to be nearly constant. Hence, it is a reaction of the first order.

e
re
The catalysed decomposition of H2O2 in aqueous solution is followed by removing equal volume samples at

Flr
various time intervals and titrating them with KMn04 to determine the undecomposed H2O2. The results

F
thus obtained are :

Time (seconds)
cm-^ of KMnO^
ou
0
22.8
600
13.8
1200
8.2

sr
Show that the reaction is mono-molecular. What is the value of the specific reaction rate ?

fo
k
ANSWER

8.45 X 10^ s-‘

oo
Y
4.14. PSEUDO FIRST ORDER REACTIONS
reB

Consider the general reaction, A + B Products, in which order with respect to each reactant is
1 so that the overall order of reaction is 2. The rate law equation is :
uY

Rate = k [A] [B]

However, if one of the reactants is present in large excess, e.g., one of the reactants may be solvent, its
ad
do

concentration will remain almost constant. In that case, the rate of reaction will depend only on the concentration
of the reactant present in smaller amount. Thus, the reaction becomes a first order reaction.
in

Such reactions which are not truly ofthe first order but under certain conditions become reactions
Re

of the first order are called pseudo-ifrst order reactions.

F

Examples:
(/) Acid-catalysed hydrolysis of ethyl acetate :

CH3COOC2H5 + H2O ^CHjCOOH + C2H5OH

Ethyl acetate Acetic acid Ethyl alcohol
(ii) Acid-catalysed inversion of cane-sugar :

C]2Ht)Oj] -I- H2O ^ C6H[20g -I- C6H12O6

Sucrose Glucose Fructose

Both the above reactions are bimolecular but are found to be of the first order, as experimentally it is
observed that

For the first reaction, rate of reaction [CH3COOC2H5] only

OC

and/or the second reaction, Rate of reaction [C12H22O]]] only.

oc

The reason for such a behaviour is obvious from the fact that water is present in such a large excess that
its concentration remains almost constant during the reaction.
 CHEMICAL KINETICS 4/47

The kinetics of the above reactions have been studied as follows :

1. Hydrolysis of Ethyl acetate

CH3COOC2H5 + H2O ^ CH3COOH + C2H5OH

In this reaction, acetic acid is one of the products, the amount of which can be found by titration against
standard NaOH solution. But being an acid-catalysed reaction, the acid present originally as catalyst, also
reacts with NaOH solution. Hence, a little careful thought reveals that for the same volume of reaction mixture
withdrawn at different times\
Volume of NaOH solution used in OC
Amount of acid present only as catalyst
the beginning, i.e., at zero time (Vq) {as no CH^COOH is produced at t = Q) ...(/)

Volume of NaOH solution used] J Amount of acid present as catalyst

at any instant of time / (V^) * J ] + amount of CH-^COOH produced ...(«)

w
Combining results (/) and (//), we find that

F lo
Amount of CH^^COOH produced]
at any instant of time J
oc
(V,-Vo) ...{Hi)

ee
But amount of CH^COOH produced OC Amount of CH^COOCr^H^

Fr
at any instant of time that has reacted (x)
Hence, X«:(V,- Vo) ...{iv)

Further, Volume of haOH solution used after the reaction for

ur
Amount of acid present as ...(V)
has taken place for a long time, say 24 hours or so, OC
catalyst + Max. amount
called infinite time (V^) *
s
of CH^COOH produced
ook
Yo

Combining results (;) and (v), we find tliat

eB

Max. amount of CHfCOOH produced «= (V^ - Vq)

But Max. amount of CH^COOH produced Initial concentration of CH^COOC-,H^ {a)
oc
...(W)
our

Hence, «-(V„-Vq)
ad

From equations {iv) and (v/), we have

(a-x)-(V„-Vo)-(V,-Vo) (fl-x) « (V^-V,) ...{vii)
Y

or

Substituting the values of a and {a - x) from equations (w) and {vii) in the first order equation, we get
Re
nd

, 2.303 a
, 2J03 V -V
Fi

k = log 0
or k = log~
t a - X
t « V -V
(

Sample Problem Q 1.0 ml of ethyl acetate was added to 25 ml of N/2 HCl. 2 ml of the mixture
were withdrawn from time to time during the progress of the hydrolysis of the ester and titrated against
standard NaOH solution. The amount of NaOH required for titration at various intervals is given below :
Time (min) 0 20 75 119 183 OO

NaOH used (ml) 20.24 21.73 25.20 27.60 30.22 43.95

The value at«time was obtained by completing the hydrolysis on boiling. Show that it is a reaction of
the first order and find the average value of the velocity constant.
*The reading at any time r is taken by placing the aliquot (small amount of the reaction mixture) in ice to
arrest the reaction.

** The reading is taken by either leaving the reaction mixture overnight or by healing/boiiing the reaction
mixture and then cooling to room temperature.
4/48 New Course Chemistry (Xll)ISSIS]

Solution. As explained above, if the given reaction is of the first order, it must obey the equation
V -V
, 2.303, a 2.303
log
0
k= log
l a — x t V^-V I
V -Vn0 = 43.95-20.24 = 23.71 ml
In the present case, Vq = 20.24 ml, = 43.95 ml
The value of k at different instants can be calculated as follows :
2.303 V -V,0
I (min) VI k = log
I V ~V.i

2.303 23.71
20 21.73 43.95 —21.73 =22.22 k = log = 0.00324 min -1
20 min 22.22

2.303 . 23.71 -1
75 25.20 43.95 —25.20= 18.75 k = log = 0.00313 min

w
75 min 18.75

2.303 23.71
119 27.60 43.95 — 27.60= 16.35 k = —log = 0.00312min-'

Flo
119min 16.35

2.303 23.71
log = 0.00299 min"'

ee
183 30.22 43.95 — 30.22 = 13.73 k =
183min ''13.73

Fr
Since the value of k comes out to be nearly constant, it is a reaction of the first order. The mean value of
Jt = 0.00312 min'*.

for
Hydrolysis of methyl acetate in aqueous solution f as been studied by titrating
ur
Sample Problem
the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given
s
below :
k
Yo
f/min 0 30 60 90
oo

C/mol L-* 0-8500 0-8004 0-7538 0-7096

eB

Show that it follows a pseudo first order reaction as the concentration of HjO remains nearly constant
(55 mol L“^) during the course of the reaction. What is the value of k' in this equation ?
(NCERT Solved Example)
r

Rate = k' [CH3COOCH3] [HjO]

ou
ad

Solution. For the given reaction, rate = k' [CH3COOCH3I [H2OI

But as [HoOJ remains constant, therefore, we put k' [H^O] =kso that rate = k (CH3COOCH3]
Y

, 2.303 Cn
Now, it is a pseudo first order reaction. Hence, for the given data, k = —-— tog-^
Re
nd
Fi

We are given Cq = 0-8500 mol L '

303, Co
^.og-
-I k =
//min C,/mol
f
L

2-303 0-8500
0-8004 /t = = 2-004x10 ^ min *
30 30min 0-8004

2-303 0-8500
60 0-7538 k = — log = 2-002x10"^ min"*
60 min 0-7538

2-303 0-8500
k = log = 2-005 X10-3 min"*
90 0-7096
90 min 0-7096
-1
Average value of k= 2-004 x 10 ^ min
Thus, k'[R-,0]=k or /t'[55 mol L"'] = 2-004 x KT^ miir*
_ 2-004 X10 3 min
-1
= 3-64 X 10"® L mol * min
“ 55molL->

i
 CHEMICAL KINETICS 4/49

Methyl acetate was subjected to hydrolysis in N—HCl at 298 K. 5 cm^ of the mixture are withdrawn at
N
different intervals and titrated with about NaOH. The following results were obtained :
Time (mts) 0 25 40 61 OQ

N 19.24 24.20 26.60 29.50 42.1

where N = vol. of alkali used in cm^. Show that the reaction is of first order.

2. Inversion of Cane Sugar (Sucrose). The hydrolysis of sucrose in presence of a mineral acid takes
place according to the equation :

^12^22^11 + HoO

w
CgH|20g + CgHj20g
Sucro.se Glucose Fructose
(dextro-rotatory) (dextro-rotatory)

F lo
(laevo-rolatory)
+ 66.5° + 52,5° -92°

(laevo - rotatory)

ee
An important characteristic of the reaction is that sucrose is dextro-rotatory whereas the products-

Fr
glucose and fructose are dextro-rotatory and laevo-rotatory respectively. Further, the laevo-rotation of fructose
is more (being - 92°) than the dextro-rotation of glucose (being + 52.5°) so that the mixture as a whole is
for
laevo-rotatory. Thus, on hydrolysis, the dextro-rotatory sucrose gradually changes into the laevo-rotatory
r
mixture. It is for this reason that the reaction is called ‘inversion of sucrose.’
You
The kinetics of the above reaction is studied by noting the angle of rotation at different intervals of time
s
ook

with the help of a polarimeter. Suppose

Reading of the polarimeter at zero time = Tq
eB

Reading of the polarimeter at any time t = r^

Reading of the polarimeter at infinite time = (i.e., after 24 hours or more)
our
ad

It is evident that the reading at zero time will be positive and

would decrease with the passage of time, pass through zero and FIGURE 4.16
ultimately become negative. r, .● ri
dY

\
0 /
A little careful consideration shows that
Re

\ /
s

Angle of rotation at any instant of time,

s
s
Fin

i.e., (fq - r^) oc amount of sucrose hydrolysed (at), —90

V 90 —

I.e.,
(^0 - ...(0
Angle of rotation at infinite time, 180

i.e., fry - r^) cc the initial concentration of sucrose (a)

I.e., a
{^0 - O ...(«) Readings of the polarimeter
From results (/) and {ii), we have
(a-j:)°^(ro- rj-(ry-r,) I.e., {a-x)o^ {r, - rj ...{Hi)
Substituting the values of a and {a-x) from equations («) and {Hi) in the first order equation,
,
k =
2.303 a 2.303
log , we get k = log
t a-x t

The applicability of this equation for the inversion of sucrose was first shown by Wilhelmy (1850). It
should be noted that the actual value of k, of course, depends upon the concentration of H* ions.
4/50 New Course Chemistry CXII)E!S1S1

Sample Problem The inversion of cane sugar was studied in 1 N HCl at 298 K. The following
polarimetric readings were obtained at different intervals of time :
0.0 7.18 18.00 27.05 CO
Time (minutes)
+ 24.09 + 21.41 + 17.74 + 15.00 -10.74
Reading (degree)
Show that the inversion of cane sugar is a unimolecular reaction.
Solution. In the present case. r„ = +24.09, 10.74 ^0 - + 24.09 - (-10.74) = 34.83
The value of k at different instants may be calculated as under:
, 2.303, 'b - r

t (min) r
/
r. -
f
r k= log
t — r

ow
2.303 34.83 -1
7.18 + 21.41 21.41 (_ 10.74) = 32.15 k = — log = 0.0116min
7.18mm 32.15

2.303 34.83
k = — log = O.OI118min-'
18.00 + 17.74 17.74 _(_I0.74) = 28.48
IS.OOmin 28.48

e
re
2.303 34.83
k = log = 0.01119 min“
+ 15.00 15.00- (-10.74) = 25.74

Frl
27.05
27.05 min 25.74

F
The constancy in the value of k proves that the reaction is of the first order.
ou
or
A 20% solution of cane-sugar having dextro-rotation of 34.50
kfs
is inverted by 0.5 N lactic acid at 298 K. The
rotations determined polarlmetrically are as follows :
oo
0 14.55 111.36 oo
t (minutes)
Y

31.10 13.98 - 10.77

B

Rotation 34.50

Show that the inversion of sugar is unimolecular reaction.

re
oYu

FORMULAS USED
ad

PROBLEMS I
BASED 2.303 .● a 2.303, [A] 0
log
d

ON (0 k = log
t a-x t [A1
in

where k = specific reaction rate in the units of time" ,

Re

Reactions of the
First Order a or [A]q = initial concentration of the reactant
F

(a - x) or [A] = concentration of A at time i

t = time (in minutes or seconds) in which the amount x of the substance
has reacted or decompo.sed, a and x must be expressed in the same units.
(//) The values of a and xor(a- x) have to be calculated carefully depending upon the nature of the
reaction and the data given.
2.303 [A],
(Hi) For concentrations [A], and [AJ^ at two different times t^ and /2- k - log
^2 ~ [A],
2.303, a 0.693
(iv) For half-life period, t = t^2> = ^^2. The formula (/) becomes iy2 - — log
k a~all k

(v) For any other fraction of the reaction to complete like l/3rd, l/4th, etc.
t = t 1/3’ t‘1/4 , etc. X = a/3, a/4, etc.
2.303, n

In general, time taken for mh fraction of a reaction of 1st order to complete is : —^log n -1
i
Mn ~
Aq
(vi) Amount of the substance left after n half-lives = .
 CHEMICAL KINETICS 4/51

Probtem Q At 373 K, the half-life period for the thermal decomposition of is 4.6 sec and is
independent of the initial pressure of N2O5. Calculate the specific rate constant at this temperature.
Solution. Since the half life period is independent of the initial pressure, this shows that the reaction is of
ihc ifrst order.

IT ● r . ^ . 0-693 0.693 0.693

hor a reaction of the first order, we know that ^1/2 =
h
or k = = 0.1507 s*
/
1/2
4.6s

w
Problem
0 A first order reaction is found to have a rate constant, it = 5-5 x Find the half-
life of the reaction.
(NCERT Solved Example)
0.693 0.693
Solution. For a first order reaction, ^1/2 = — = 1*26 X 10*3 s
5.5xlO-l‘*s-'

e
ro
Problem 0 A first order reaction is 40% complete in 50

re
minutes. Calculate the value of the rate
constant. In what time will the reaction be 80% complete ?

F
2.303 a
Solution. (/) For the first order reaction k =

Fl
log

u
t a-X

40
When X =
a = 0.4 a, / = 50 minutes (given)

sr
100

ko
2.303
k =
a
o 2.303 a
= 0.010216 min-*
log or k = log
50 a-0.4 a

(/i) t = ?, when reaction is 80% complete, ie., .r= 0.8 a

50 min 0.6
of
; k = 0.010216 min ’ {calculated above)
o
Y
2.303, a 2.303 a 2.303 1
erB

t =
— log log log = 157,58 min
k a-x 0.010216min-l a 0.8a 0.0l0216min-* 0.2
uY

Problem Q Show that in case of a first order reaction, the time required for 99.9% of the reaction
to take place is about ten times than that required for half the reaction. (NCERT Solved Example)
2-303
ad
do

a 2-303 2-303
Solution. For reaction of first order. ^\I2 ~ — log
k ^a a
log 2 = ^=^(0-3010)
—
k k
in

2-303
Re

99.9% — log -
a
2-303. , 2-303
t.
log 10 3 = x3 .- *99.9% _ 3
F

a-0-999 a fc
= 10
k
h/2 0-3010

1^1
Problem
0 The reaction, SO^CIj —-> SOj + CI2, is a first order reaction with k, = 2.2 x 10"^ sec"*
at 575 K. What percentage of SO2CI2 will get decomposed in 90 minutes when the reaction is carried out at
575 K ?
(Assam Board 2012)
2.303
Solution. Since the reaction is of the first order (given), k = log
a

t a-x

Here, we are given that k=2.2x 10"^ sec -1 ; / = 90 minutes = 90 x 60 = 5400 sec.
2.303 a a
2.2xI0-5s-* = log or log = 0.0516 or
a

5400 s a -X a — x
= anlilog (0.0516)= 1.127.
a-x

or
a = 1.127 a- 1.127X or 0.127 a = I.127x
-t 0.127
= 0.113 or % decomposed = 0-113 x 100 = 11*3%.
a 1.127
4/52 New Course Chemistry CXll)CZsI91

Problem 0 The initial concentration of NjOj in the first order reaction, N2O5 (g) »2N02 ig)
+ 1/2 O2 (g), was 1-24 X 10"^ mol L"’ at 318 K. The concentration of N2O5 after 60 minutes was 0-20 x 10"^
mol L”*". Calculate the rate constant of the reaction at 318 K. (NCERT Solved Example)

2-303.log 2-303 [N205Jo _ 2-303 1-24x10-2 n,

Solution, k = log
[A] - r 60 min 0-2 X10-2 mol L-‘
/
[N.OgJ,
2-303 2-303
log 6-2 min * = x0-7924min * = 0*0304 min ‘
60 60

Problem Q A first order reaction has a specific reaction rate of 10 ^ sec"*. How much time will it
take for 10 g of the reactant to reduce to 2.5 g ? Given log 2 = 0.301, log 4 = 0.6021, log 6 = 0.778.

ow
2.303, a
Solution. We know that for a first order reaction, t = log
k a- X

Here, initial concentration, a = 10 g and concentration left after lime t sec. = 2.5 g, i.e„ {a-x) = 2.5 g
Specific reaction constant k= 10"^ sec"*

e
2.303 10 2.303 2.303

Fl
— x0.6021s

re
Time required for the reactant to reduce to 2.5 g =
10"^^xlog
s 2.5 10" ^ X log 4 = 10-

F
= 1386.6 s. [Given log 4 = 0.6021]
Problem 0 The decomposition of N2OS (g), i.e., N2OS ig)
ur 4 NO7 ig) + O2 ig) is a first order

r
reaction with a rate constant of 5 x lO"^ sec"* at 4S"C. If initial concentration of NjOj is 0-25 M, calculate its
concentration after 2 min. Also calculate half life for the decomposition of N2O5 (g).
fo
ks
Solution. it = 5x 10^sec"‘,a = 0-25M, ia-x) = 2
Yo
t - 2 min = 120 sec
oo

2-303 a 2-303. 0-25 0-25

= 0-026
B

For a first order reaction, k = log 5x10-^ = log or log

t a-x 120 ia-x)
re

0-25 0-25
= Antilog 0-026 = 1 -062 or (a-x) = M =0*235 M
u

a-x 1-062
ad
Yo

0-693 0-693
t - 1386 sec = 23 min 6 sec.
‘^2- k 5xl0“^sec *
d
Re

Problem El The rate constant for an isomerisation reaction, A ^ B is 4*5 X 10"^ min"*. If the
in

initial concentration of A is 1 M, calculate the rate of reaction after 1 h.

F

2-303. [A] 0
Solution. Rate constant in min"* shows that it is a reaction of 1st order. Hence, k- log
t [A]

2-303 1
or 4-5x10"^ min ’ = — log or log [A] =-0-1172 = 1-8828 or [A] = Antilog 1-8828
60 mm [A]
[A] = 0-7635 mol L"*. This is the concentration after 1 h.
Rate after Wi = [A] = 4-5 x 10"^ x 0-7635 mol L“' min"* = 3-44 x 10"^ mol L"* min"*.
Problem
m The following data were obtained during the first order thermal decomposition of
N2O5 (g) at constant volume : 2N20s(g) ^2N20^(g) + 02(g)
S. No. Time/s Total pressure (atm)
1 0 0*5

2 100 0*512

Calculate the rate constant. (NCERT Solved Example)

 CHEMICAL KINETICS 4/53

Solution.
2 N20g (g) ^ 2N204(g) + O2C?)
Initial : 0*5 atm
After 100 s : 0-5 -2/7 2/? P

^Toiai ~ - 2/y) + 2p + p = 0-5 + /? = 0-512 atm (Given) /. p = 0-012 atm

Thus, initial pressure of N20g (Pq) = 0-5 atm
Pressure of N2O5 after 100 s ) = 0-5 - 2 x 0-012 = 0-476 atm

k =
2-303
log-
^Pn20,)o 2-303 0-5 atm
log = 4-98x 10^s-‘.
i 100s 0-476 atm
N2O5

w
Problem [fl At 373 K, a gaseoas reaction A ■» 2B C is observed to be of first order. On starting
with pure A, it was found that at the end of 10 minutes, the total pressure of the system was 176 nim of
mercury and after a long time, when dissociation of A was complete, it was 270 mm. From these data,

o
calculate (i) the initial pressure of A (ii) the pressure of A at the end of 10 minutes, (iii) the rate constant.

e
Solution. Suppose initial pressure of A = P mm.

re
Decrease in the pressure of A after time r = p mm.

Frl
F
A ^ 2B -f C Total pressure
Initial pressure P 0 0 P
Pressures after time i P-p P P + 2p

or
ou
Final pressures 0 2P P 3P
(i) Final pressure = 270 mm (Given)
Ui) Pressure after 10 minutes = 176 mm (Given)
3P = 270 or P= 90 mm kfs
P + 2p = 176 or 90 2p = 176 or p = 43 mm
oo
/. Presssure of A after 10 min = P - p = 90 - 43 = 47 mm
(iii) a "X P and x »= p
Y
B

,
k =
2-303 a 2-303
, P
log
re

t a - X t
oYu

2-303 90 2-303 Qf)

k =
log— = 6-496 X 10-2
ad

or
log min
lOmin "'90-43 lOmin “47

Problem [Q The decomposition of CI2O7 at 400 K in the

d

gas phase to CI2 and Oj is a first order

reaction.
in
Re

(/) After 50 seconds at 400 K, the pressure of CI2O7 falls from 0-062 to 0-044 atm. Calculate the rate
F

constant.

(ii) Calculate the pressure of CI2O7 after 100 sec of decomposition at this temperature.
2-303, Pn
Solution, a « Po and (a —x) OC
P,. Hence, k = log^ (P, = pressure of CI2O7 at time 0
t
Pr

2-303, 0-062 atm

(0 i = 50 sec, Pq - 0-062 atm, P^ = 0-044 atm. k = log = 6-86xl0*--^s“J
50 j 0-044 atm

(ii) r = 100 sec, P,= ?, A: = 6-86 x 10"^ s'

2-303 0-062 atm

6-86x10-35-1 = log
100.?
P.
. 0-062 0-062
log — = 0-2979 or
-— = Antilog 0-2979 = 1 -986 or P, = 0-0312 atm
4/54 ‘p’uuUcfi-’^ New Course Chemistry (XII)CEIS

Problem [0 The rate of a first order reaction is 0-04 mol L"‘ s"* at 10 minutes and 0*03 nioi L"* s ‘ at
20 minutes after initiation. Find the half life of the reaction.
Solution. For a first order reaction, A ^ Products, for concentration of the reactant at two different times,
2-303, [A],
k = log—^
[AJ2
(rate) 1 _ [A] I
But as rate = k [A], therefore
(rate)2 \M2
2-303. (rate) 1 2-303 0-04
= 2-88 X 10-2 niin-i
Hence, k = log log
^2-^1 (rate), ~ (20-10) min "0-03

w
0-693 0-693
/ - 24-06 min = 1443*6 s.
1/2 “
k 2-88xl0-2min"‘
m One-fourth of a first order reaction is completed in 32 minutes. What is the half-life

o
Problem

e
period of this reaction? (HP Board 2011)

re
rFl
2-303 a 2-303 a 2-303 .log —mm
4 . -I = 0-009 min"'
Solution, k = log log
^3

F
/ a-x 32 min a~al4 32

0-693 0-693
t = 77 min.

r
1/2 -
k 0-009
ou
fo
Note. In the various examples and problems on the reactions of first order discussed above, it may be observed
ks
that the most common experimental methods for determination of rate constant include :
(/) Concentration change method in which the concentration of one of the reactants or products is measured
oo

at different limes.
Y

gaseous product is noted at different times, e.g., in the

eB

(;/) Volumetric method in which the volume of some

decomposition of N20g or HjO,.
(Hi) Optical rotation method in which the angle of rotation, i.e., polarimetric readings are noted at different
r

times, e.g., in the inversion of cane sugar.

ou

(iv) Pressure change method (for gaseous reactions) in which the pressure of the reaction mixture is noted at
Y
ad

different times, e.g., in the solved example 10 and 11 above or Problem for Practice 16 below.
d
Re
in
F

1. A first order reaction is found to have a rate constant k = 7.39 x 10 ^ sec ^. Find the half life of this reaction
(log 2 = 0.3010).
2. Show that for a first order reaction, lime required for completion of 99% of reaction is twice the lime
required for completion of 90% reaction. (CBSE 2019)

3. The half life period of a first order reaction is 60 minutes. What percentage of the reactant will be left behind
after 120 minutes ?
4. It was found that the solution of cane-sugar in water was hydrolysed to tlie extent of 25% in 60 minutes.
Calculate the time taken for the sugar to be 50% hydrolysed, assuming that the reaction is of the first order.
5. Decomposition of a gas is of first order. It takes 80 minutes for 80 % of the gas to be decomposed when its
initial concentration is 8 xl0"2 mole/litre. Calculate the specific reaction rate.
6. The data for the conversion of compound A into its isomeride B are as follows :
Time in hr: 0 1 2 3 4 00

49-3 35-6 25-8 18-5 13-8 4-8

%age of A:
Show that this is a first order reaction.
CHEMICAL KINETICS 4/55

7. The following data were obtained during the catalysed decomposition of N2O at I ] 73 K :
Time (min) 30 53 63 80 100 120

%age decomposed 32 50 57 65 73 78

Show that the reaction is of the first order and calculate velocity constant of the reaction.
8.
Find the two third life (^2/3) of a first order reaction in which k = 5.48 x lO"*"^ sec^',
(log 3 = 0.4771, log 2 = 0.3010)
9. A first order reaction has rate constant of /: = 1.15 x 10"^ s“’. How long will it take for 6 g of reactant to
reduce to 3 g ? (Hr. Board 2011)
10. For a first order reaction, calculate the ratio between the time taken to complete three fourth of the reaction
and the time taken to complete half of the reaction.
or

If the rate constant of a first order reaction at a certain temperature is 1-5 x 10 * s * and fj and ^2 are the

w
respectivetimes for 50% and 75% completion of the reaction, determine the ratio of ?2 to
(West Bengal Board 2012)
n. For a first order reaction, it takes 5 minutes for the initial concentration of 0-6 mol to become 0-4 mol

Flo
L"'. How long in all will it lake for the initial concentration to become 0-3 mol L“* ?

e
12. A first order reaction is 75% complete in 60 minutes. Find the half-life of this reaction.

re
13. In a reaction. 5g ethyl acetate Is hydrolysed per litre in the presence of dil HCl in 300 minutes. If the reaction

F
is of first order and the initial concentration of ethyl acetate is 22g/L, calculate the rate constant of the
reaction,
ur
r
14. A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its ty2 value.

15. fo
A first order reaction takes 30 minutes for 75% decomposition. Calculate ty2
(CBSE 2008, Hr. Board 2011)
ks
Yo
[Given : log 2 = 0-3, log 3 = 048, log 4 = 0-6, log 5 = 0-7] (CBSE 2022)
oo

16. In a particular reduction process, the concentration of a solution that is initially 0-24 M is reduced to 0-12 M
in 10 hours and 0-06 M in 20 hours. What is the rate constant of this reaction ? f issam Board 2012>
B

17. The following rate data were obtained for the thermal decomposition of (g)
re

2N20g (g) ^ + 02(g)

Time (sec) 0 50
u
ad
Yo

Total pressure (atm) 0-2 0-25

Calculate the reaction rate when the total pressure is 0-28 atm
18. The half-life period of a first order reaction is 600 s. What percent of A remains after 30 minutes ?
d
Re
in

19. 50% of a reaction is completed in 16 minutes. What fraction of the reaction would occur in 32 minutes ?
20. The decomposition of a compound is found to follow a first order rate law. If it takes 15 minutes for 20
F

percent of original material to react, calculate (i) specific rate constant (ii) the time at which 10 percent of
the original material remains unreacted, (iii) the time it takes for the next 20 percent of the reactant left to
react after the first 15 minutes.
21. A first order reaction is 15% complete in 20 minutes. How long will it take to be 60% complete. (CBSE 2007)
22. The rale constant for a first order reaction is 60s"'. How much time will it take to reduce the concentration
of the reactant to 1/10th of its initial value ? (CBSE 2007)

23. Rate constant of a first order reaction, A Product is 0-016 min"’. Calculate the time required for 80% of
the reaction to be completed. (Karnntnka Board 2012)
24. A first order reaction has a rate constant of 0-0051 min"'. If we begin with 0-10 M concentration of the
reactant, what concentration of the reactant will remain in the solution after 3 hours ? (CBSE 2009)
25. A first order reaction has A: = 1 -5 x 10"^ per second at 240°C. If the reaction is allowed to run for 10 hours,
what percentage of initial concentration would have changed to products ? What is the half-life period of
this reaction ?
4/56 New Course Chemistry (X11)C!2S[9]

26. The thermal decomposition of HCO2H is a first order reaction with a rate constant of 24 x 10“^ s“' at a
certain temperature. Calculate how long will it lake for three-fourths of initial quantity of HCO-,H to
decompose (log 0-25 = - 0-6021). (CBSE 2011)
27. The rate constant of a first order reaction is 60 s ’. How much time it will take to reduce 75% of its original
concentration? (Bihar Board 2012)
28. The thermal decomposition of a compound is of first order. If 50% of the compound is decomposed in 120
minutes, how long will it take for 90% of the compound to decompose ?
(HP Board 2011, Chhatisgarh Board 2012)
29. A first order reaction takes 20 minutes for 25% decomposition. Calculate the lime when 75% of the reaction
will be completed.

ow
(Given : log 2 = 0-3010, log 3 = 0-4771, log 4 = 0-6021) (CBSE 2017)
30. The following data were obtained during the first order thermal decomposition of SO2CI2 at a constant
volume

SO2CI2 (g) ^S02(g) + Cl2 (g)

e
Experiment TIme/s Total pressure/atm

re
Frl
1 0 0-4

F
2 100 0-7

Calculate the rate constant.

ou
or
(Given : log 4 = 0-6021, log 2 = 0-3010) (CBSE 2014)
ANSWERS kfs
1. 9-38 X 103 s 5. 0.02012 min-*
oo
3. 25% 4. 145 minutes
8. 2 X 10’3 sec. 9. 869-6 s 10. 2 11. 8.54 min
Y
eB

12. 30 minutes 13. 8-6 X loomin'* 14.77-86 min 15. 15 min

16. 6-93 X 10-2 hr-* or 1-9 X IQ-^s-* 17. Rate = 5-54 X IQ-^atm s"* 18. 12-5%
19. 3/4 or 75% 20. (0 0-01488 min-* («) 154-77 min (m) 15 minutes 21. 112-7 min
ur
oY

22. 0-0384 s 23. l(X)-6 min 24. 0-04 M 25. 5-2%, 128-3 hrs
ad

26. 9 min 38 s 27. 0-023 s 28. 398-8 min 29. 96-3 min 30. 1-3 X 10-2 S-’
d

HINTS FOR DIFFICULT PROBLEMS

in
Re

2.303 a
1. For a first order reaction, k - log For t = X = all
F

/ a-x

2.303
2.303, a 2.303
x0.3010=9.38x103s
a- all ~7.39xl0-5s-*
3. 120 min = Two half lives, i.e., n-1
1 1 [A] 0
Amount left =
-[A]o=^[AJo=^ = 25% of [A] 0-

2-303 a 2-303
5. k =
80 min
log
fl-0-80n 80
log5 = 0-02012 min"’.
6. a OC
49-3, (a-x) ec 35-6 at r = 1 hr and so on. Substitute in 1st order eqn. k comes out to be constant.
2-303 a 2-303 1
7. % decomposed means—X100 k=~ log log
a t a-x t l-xia

2-303
8 , 2-303, a
Iog3 = 2xl0*^
*2/3=-^>og o-2fl/3
s.
5-48xl0"*'*s-*
CHEMICAL KINETICS 4/57

10. k =
2-303, a
——log
k a-x

2-303
For 3/4 of the reaction to occur, t = (a-x) = a-3 a/4 = a/4 <3/4= — log 4

2-303, a
2-303, _
For half of a reaction to occur, t = ty/2, {a-x) = a-al2 = all. <1/2 = — = ——log2
k
all

Hence,
ht4 _ _ 0-6021 = 2.
hlA ” “ 0-3010
. 2-303, 0-6
11. a - 0-6 mol L~*, {a-x) = 0-4 mol L■^ t = 5 min ; k = —^log = 0-081 lmin“^
5mm 0-4

0-693

w
0-6 mol L-' ^ 0-3 mol L ' means t 1/2- Hence, t 1/2 ~
= 8-54 min.
0-081 lmin“*
13. a = llgL-\x = 5gL-Kt= 300 min.

Flo
2-303 2-303 a
0-693 0-693
a
- 77-86min.
14. * = log log = 0-0089 min“* ; ^1/2 ~ 0-0089 min“^
t a-x 40min a-0-30fl

e
re
0-693
15. k =
2-303
log 100 ^ 2-303 log 4 = 0-046 min
-1 .
’ ^1/2 ~ = 15-0 min

F
-I
30 min 100-75“ 30 0-046 min

Alternatively, 75% is completed in two half-lives, i.e., 2xty2 = 30 min.

ur
r
16. As tij2 is independent of initial concentration, it is a reaction of 1st order. hours,
k = 0-693/fi;2 = 0-693/10 hr = 6-93 x 10-2 j^-i fo
0-693/(10 x 3600) s"* = 1-9 x lOr^ s~K
ks
17. lN20s(g) ^ 1^20^ ig) + 02(g)
Yo
Initial 0-2 atm
oo

P
After 50 s. (0-2-p) P p/2. Total =0-2-1-^
eB

Atr = 50s, 0-2 + -^2 = 0-25 or p = 0-10

ur

Thus, a oc 0-2, a-xoc (0-2 -p), Le., (0-20- 0-10), i.e., 0-10
ad
Yo

2-303, 0-2
k = log — = 0-01386s-l.
50 s 0-1
d
Re

When total pressure = 0-28 atm, 0-2 -J- — = 0-28 or p = 0-16

in

2
F

Pressure of N2O5 at that time = 0-20 - 0-16 = 0-04 atm

Rate = cf X P, = 0-01386 s-i X 0-04 atm = 5-54 x lO"^ atm s"‘.
N2O5
1 1
18. 600 ^ = 10 min, 300 min = 3 half lives, i.e., n-3. Hence, % of A left = — [A]q = = 12-5% .
2-303 a 2-303 a
20. (i) k = log log = 0-01488 min-i
t a-x 15min a -0-20fl

2-303 a 2-303
(») t = log log 10 = 154-77 min.
k 0-10 a 0-01488

(ill) In first 15 minutes, 20% has reacted. .*. Amount left unreacted = 80% of a = 0-80 a
This is now initial concentration. For next 20% of this amount to react means x = 20% of 0-80 a = 0-16 a
2-303 0-80a 2-303 0-80
t- log log
k 0-80 a-0-16 fl 0-01488 min-i 0-64

= 154-77 min x log 1-25 = 154-77 x 0-0969 min = 15 minutes

 4/58 New Course Chemistry fXinrosran

2-303 100 2-303 X 0-0706 .

21. k = log min
20 min 100-15 20

2-303, 100 2-303x20 100 20

f - ——log log X 0-3979 =112-7 min
k 100-60 2-303x0-0706 40 ~ 0-0706
2-303, a 2-303 a 2-303
22. r = — log log log 10 = 00384 s
a-x 60s Cl/10 60s
2-303 a 2-303 a 2-303
23. t = — log log log 5 = 100-6 min
a~x 0-016min * ''ci-0-8a 0-016

2-303. [A] 0 2-303 0-10

24. k = log '● 0-0051 min
-1 _
log
t [A] ● 3x 60 min [A]
0-10 0-10 0-10

w
or
log = 0-3986 or = Antilog (0-3986) = 2-5 or [A] = = 0-04M
[A] [A] 2-5

2-303

F lo
a a
25. l-5xl0-^s"‘ = log or log = 0-0234
10x60x60s a-x a-x

a X 0-055

ee
or - = 1-055 or Cl = 1-055 a - 1-055 .v or 1-055-r = 0-055 a or = 0-052 = 5-2%
o-x 1-055

Fr
a

0-693 4-62x105
t
1/2 ~
= 4-62xlO\s = hr = 128-3hrs
1-5x10-6 ^-1 60x60

for
ur
2-303 a 2-303 2-303
26. x = —a.
- log
4
t =
k CJ-3/4C1 2-4x10
-^log4s =
2-4x10
3j(0-602I)s = 578s = 9 min 38 s
s
ook

0-693 0-693
Yo
/
or s = 288-75 s, =2 X
'^2- k 2-4x10-5
eB

2-303 a 2-303
27. t = log log4 = 0-023 s
k Cl-0-75 rt 60s-'
r

0-693 . 2-303 a
ad

log
ou

28. ij/2 = 120 min. k - 120

min t =
k <1-0-9 Cl ●
29. Proceed as in Solved Problem 3.
Y

2-303 a 2-303 4
k = log log - = 0-0144 min
Re
nd

20 a -0-25 Cl 20

2-303. 2-303
Fi

Cl
i = — log -log4 = 96-3 min
a-0-15a 0-0144
30. Similar to Solved Problem 10.

4.15. RATE OF RADIOACTIVE DECAY/DISINTEGRATION

There are many elements like uranium, thorium etc. which spontaneously emit a, P paiticles or y-
radiation. This phenomenon is called natural radioactivity. On the other hand, certain elements are not
radioactive as such but on bombardment with subatomic particles like neutron, proton etc. produce isotopes
which are radioactive. This phenomenon is called artificial radioactivity.
The emission of particles by any radioactive element or radioactive isotope is called radioactive decay
or radioactive disintegration. It is observed that the rate of disintegration or decay at any time depends upon
the number of atoms of the radioactive element/isolope present at that time. Hence, radioactive disintegration/
decay follows first order kinetics. Thus, if Nq atoms of the radioactive element are present initially and N
atoms are present after time, /, then
CHEMICAL KINETICS 4/59

-dN -dN
Rate of disintegration at time t OC N or = kN
dt dt

where k is called disintegration constant or decay constant.*

As in case of chemical reactions of first order, the above equation on solving gives

2-303 N0 2-303 a
k = log or k = log
t N t a — X

element and ‘a’ is the amount that has disintegrated in lime

where ‘a’ is the initial amount of the radioactive
7’. When half of the substance disintegrates. N = N(,/2 (or a = all) and / = t i/o.
2-303 , 0-693
Half - life period (ty2) = — Iog2 = ——
k k

w
Thus, half-life period of a radioactive element is independent of the initial amount oj the element.
Further, whole of the radioisotope never disintegrates, i.e., total life of any radioisotope is infinity.

F lo
Hence, we talk of average life (also called natural life). Slower the disintegration, i.e., smaller is the
disintegration constant, k, greater is the average life. Hence,

ee
1 t 1/2
Average life (x) = - = i-44x/ 1/2

Fr
k 0-693

Sample Problem Q After 24 hours, only 0-125 g out of the initial quantity of 1 g of a radioactive
for (CBSE Sample Paper 2017)
ur
isotope remains behind. What is its half-life period ?
Solution. Here, we are given a = \ g, « - a = 0-125 g, / = 24 hours
s
ook
Yo
, 2-303, a 2-303 1 0-693 0-693
k = log log .-1 = 8-Ohours
= 0-0866 hr
t a-A 24 hr 0-125 '^2- k 00866 hr-'
eB

Sample Problem Half-life period of a radioactive element is 100 seconds. Calculate the
disintegration constant and average life. How much time will it take to lose its activity by 90% ?
our
ad

1 1
, 0-693 0-693
= 0-00693S-' = 144-3s
Solution. k = : Average life (t) = —
t 100s ic 000693s-'
Y

1/2
Re

N 2-303, N0 2-303
nd

100 _
Loss of 90% activity means
0 _ = 10 ; / = ——log logl0 = 332-3s
N 100-90 k N 0-00693S-'
Fi

It is important to note that the rate of radioactive disintegration is independent of the external factors
like temperature, pressure etc. However, the rates of chemical reactions are affected by temperature, presence
of catalyst etc.
Therefore, our aim now is to discuss the effect of temperature and the effect of adding catalyst on the
rate of reaction. The study of both these factors is ba.sed on the concept of ‘activation energy’. Hence, in the
next section, we shall first take up a brief explanation of the term ‘activation energy’ of reactions.

4.16. ACTIVATION ENERGY OF REACTIONS

According to collision theory (as discussed later in Art. 4.19), a reaction takes place because the reactant
molecules collide with each other. However, every collision is not an effective collision, i.e., does not result
into a chemical reaction. For the collision to be effective, the colliding molecules must have energy more than
a particular value.
*Rate of disintegration/decay is also called activity of the element. Thus, activity = /:N.

I
 4/60 New Course Chemistry fXIltrowi

The minimum energy which the colliding molecules must have in order that the collision between
them may be effective is called threshold energy.
At room temperature, most of the reactant molecules have energy less than the threshold value.
However, if energy is supplied in the form of heat, light, etc., the reactant molecules absorb this energy
and their energy becomes equal to or greater than threshold value. Hence, they start reacting and change
into products.

The minimum extra amount of energy absorbed by the reactant molecules so that their energy
becomes equal to threshold value is called activation energy. In other words, activation energy
is the difference between the threshold energy and the average kinetic energy of the reactant
molecules, i.e.,

Activation energy = Threshold energy — Average kinetic energy of the reactants

w
Evidently, less is the activation energy, faster is the reaction or greater is the activation energy, slower
is the reaction. For example, the reaction,
FIGURE 4.17

F lo
2NO + O2 ■> 2 NO2 is fast reaction because of low ENERGY BARRIER

activation energy whereas the reaction, (ACTIVATED COMPLEX)

t \

ee
2CO + O., 2 CO2, though appears to be similar, is a slow >● THRESHOLD ACTIVATION
O ENERGY / ■'ENERGY
Eg

Fr
reaction because of high activation energy. cc
LU

.Er
Reaction profile (Path followed by a reaction) and concept LU

of Activated Complex (TVansition state). In order that the reactants

for
< REACTANTS
●ENERGY
ur
AH
EVOLVED
may change into products, they have to cross an energy barrier lU
I- Ep
O
(corresponding to threshold energy) as shown in Fig. 4.17. This CL
s
proPTOts
ook

Fig. has been obtained by plotting potential energy vs reaction

Yo

REACTION COORDINATE
coordinate. Reaction coordinate represents the reaction profile, i.e.,
eB

progress of reaction from reactants to products accompanied by- Reaction profile of

an exothermic reaction
changes in potential energy.
our

It is believed that when the reactant molecules absorb energy, their bonds are loosened and new loose
ad

bonds are formed between them. The intermediate thus formed is called an activated complex or transition
state complex. It is unstable and immediately dissociates to form the stable products.
Y
Re

A—A + B—B > A A 2AB

nd

(Reactants) (Products)
Fi

b' B
(Activated complex)

0-74 A 0-77 A
e.g., H H H H H H
>
2-66 A > +
1-60 A
I 2-95 A
I- I I
Reactant molecules
Activated complex Product molecules

It may be noted that when the activated complex dissociates to form the products, it is accompanied by
release of energy. But the enthalpy change (AH) during the reaction at a particular temperature remains
constant because it depends only on the nature of the reactants and the products as shown in the Fig. 4.17.
Also, from the Fig. 4.17, it may be noted that
Threshold energy = Activation energy + Energy possessed by the reactants

i
CHEMICAL KINETICS 4/61

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

Activation energy diagram for two step reaction. Suppose the reaction A ■> C takes place in two
steps;
Step(i) Step(/t)
A > B > c

To see whether step (/) or step (ii) is the rale determining step (slow step), we study the activation energy
separately for each step. Thus, we may write the two steps as :
E, E2
Step (i) : A A* ^ B

ow
E3 E4
Step (ii): B _i_> B* —^ C
Their activation energies are shown on the arrows. If A C is an exothennicreaction,the two possibilities
arc shown in Fig. 4.18.

e
re
Frl
F
ou
or
o
kfs
oo
Y
B

In (n), Ej > E3, i.e., A changing to A* requires more activation than B changing to B* or in other words,
re

activation energy of step (i) is greater than that of step (/i), therefore step (i) is slow, i.e., rate determining
oYu

step.
ad

In (b), E, < E3, i.e., A changing to A* requires less activation energy than B changing to B* or activation
energy of step (i) is lower than that of step (ii) therefore step (ii) is the rate determining step.
d
in
Re
F

Curiosity Question
f Q. Carbon of the coal can combine with oxygen of the air to form CO, CO2 etc. {i.e., AG is -ve
and the reaction Is thermodynamically feasible). Then why It does not happen unless a
flame is applied to start combustion, though O2 is available in abundance ?
Ans. This is because the activation energy for the combustion reaction is very high. On applying flame
to the fuel, the part of the fuel and air In contact with the flame absorb heat which provides the
necessary activation energy and the combustion starts. The heat liberated provides activation
energy to the remaining fuel and hence the combustion continues. Thus, fuels which are
thermodynamically unstable are kinetically stable (i.e., from chemical kinetics point of view). In
other words, the stability of the fuels is due to the existence of energy barriers between the reactants
and the products. These energy barriers have helped in the existence of life on the earth because
if these barriers were not present, all the fuels on this earth would have burnt away by themselves.
 4/62 New Course Chemistry fxmrosm

4.17. TEMPERATURE DEPENDENCE OF THE RATE OF REACTION-

EFFECT OF TEMPER/TURE ON RATE CONSTANT (ARRHENIUS EQUATION)
Temperature has a marked effect on the rate of reaction. For most of the reactions, the rate of reaction
becomes nearly double or even more for 10®rise of temperature. The effect of temperature is usually expressed
in terms of temperature coefficient which is defined by the equation ;
Rate constant at T +10° (308 K)
Temperature coefficient (n) =
Rate constant at T° (298 K)

ow
For most of the reactions, 2 < h < 3
Explanation. As already explained, for a collision to be etfective,
FIGURE 4.19
the colliding molecules must have energy more than threshold energy. At a
particular temperature T, all the reacting molecules do not have the same w T

e
kinetic energy. However, fractions of molecules having a particular kinetic D T+ 10°

re
o
/
energies at a particular temperature remains constant. At a particular UJ

o
temperature, it fractions of molecules are plotted versus corresponding

Flr
F
kinetic energies, a graph of the type shown in Fig. 4.19 is obtained (called o

Maxwell’s distribution of energies). The peak of the curve represents the o

ou
kinetic energy possessed by the maximum fraction of molecules and is
called most probable kinetic energy. If the point "a" represents threshold
O
<
a:
r MOST
PROBABLE
KINETIC d

sr
ENERGY

energy, the shaded area abed represents the fraction of molecules having b

fo
energy greater than threshold value. KINETIC ENERGY

k
When the temperature is increased to T+I0°, the curve shifts as Distribution of energy at
temperatures T and T+ 10° K
oo
shown in the Fig. 4.19. Now, the fraction of molecules having kinetic
energy greater than threshold value is represented by the shaded area abef which is almost double than the
Y
area abed. Thus, the inerea.se in the rate of reaetion with inerease in temperature Ls mainly due to inerease in
reB

the number of effeetive eollisions.

Remember that the increase in the rate of reaction with rise in temperature is not due to the increase in
uY

the total number of collisions but due to increase in the total number of effective collisions.
Also note that with increase of temperature, peak shifts forward but downward. This means that with
ad

increase of temperature,
do

(/) most probable kinetic energy increases,

in

(//) the fraction of molecules possessing most probable kinetic energy decreases.
Re

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k, was
F

proposed by Arrhenius (1889). The equation, called Arrhenius equation, is usually written in the form
-E_/RT
jfe = Ae ...(I)
where the pre-exponential factor A is a constant and is called frequency factor (because it gives the frequency
of binary collisions of the reacting molecules per second per litre), is the energy of activation, R is gas
constant and T is the absolute temperature. The two quantities ‘A’ and ‘E^’ are collectively called Arrhenius
parameters. The factor, gives the fraction of molecules (N^/N^) having energy equal to or greater
than the activation energy, E^ where Ng represents the number of molecules with energy E and Nj represents
the total number of molecules.

The energy of activation (E„) is an important quantity as it is characteristic of the reaction. Using the
above equation, its value can be calculated as follows :

Taking logarithm of both sides of equation (i), we get

E
In /: = In A - ...(H)
RT
CHEMICAL KINETICS 4/63

If the value of the rate constant at temperatures T, and T2 iire and Ao respectively, then we have
E
a
In k,I = In A -
RT,

Ea
and In = In A - ...{iv)
RT2
Subtracting equation (///) from equation (/V), we get
E..a ( E_a \ Ea E.0
In ^2 “ In =- or
a
L_±l = 5«f Vili
RT2 RT
I RT,1 RT, k
1 R T,l T-, R TjT^
" j

w
Ea
*2 T,-T,
or ●og-^ = ...(V)
2303 R T,T.
h 1-2 )

F lo
Thus, knowing the values of the rate constants k^ and at two different temperatures T, and T2, the
value of E^ can be calculated. Alternatively, knowing the rate constant at any one temperature, its value at

ee
another temperature can be calculated provided the value of E^, is known for that reaction.

Fr
To test the validity of Arrhenius equation, let us reconsider equation (//). It may be written as
Ea
In A’ = - + In A ...(Vi)
for
ur
RT

FIGURE 4.20
Eo
s
or log A = - + logA ...(vii) -1.0 I
ook
Yo
2.303 RT
-2.0 ●

This equation is of the form, y = nix + c, i.e., the equation of a

eB

-3.0-

straight line, Thus, if a plot of log A is a straight line, the validity of -4.0
o
r

the equation is confirmed. This is actually found to be so as shown for a

o
ad
ou

-* -5.0 ■
typical plot (dissociation of HI) in Fig. 4.20.
Further, from equation (vii), we will have 1. 1.6 1.8)x10-3
Y

1/T
E
Re

Atypical plot of log k vs 1/T

nd

Slope of the line = -

a

2.303 R
Fi

Thus, measuring the slope of the line, the value of E^, can be calculated.
Note. According to eqn. (vi), if In A is plotted against I/T, slope of the line will be = - E^R
Explanation of the increase of rate constant and rate of reaction with increase of temperature
from Arrhenius equation (assuming activation energy remains constant)
E E -E../RT .
T increases
a
decreases ■>
a
increases > e a increases ■> Rate constant (A)
RT RT
and rate increase.

Explanation of the increase of rate constant and rate of reaction with decrease in activation at the
same temperature and concentration of reactants
E E -E../RT .
a a
E.a decreases decreases ■> increa.ses > e a increases
RT RT
Rale constant (A) and rate increase.
4/64 New Course Chemistry (X11)ESBI9]

■m
n iMENT YOUR
ISR COMPETITIONS

1. Eqn. (/■) shows that as T increases, k increases exponentially as shown in

Fig. 4.21.
2. The factor in the Arrhenius equation is called Boltzmann factor. It
represents the fraction of molecules (N^/N-j-) having energy equal to or greater
than E where Ng represents number of molecules with energy E and Nj k

represents the total number of molecules.

3. As the exponential factor is dimensionless (E^/RT = J morVj mol"'), the pre
exponential factor A has the same units as that of rate constant, e.g., for first
order reaction, k has the units sec*. That is why A is called frequency factor. Exponential increase
of rate constant
4. An example of a reaction showing a small negative temperature coefficient,
i.e., in which the rate of reaction decreases with increase in temperature is : with temperature

w
N0 + -0, ^N02.
2 2
d\nk g
5. Arrhenius equation can also be written in the form ss —● This is obtained by differentiating

F lo
eqn. {ii) w.r.t. T. i/T RT2
6. Calculation of the ratio of rate constants of a reaction at particular temperature in the absence of

ee
catalyst and in the presence of catalyst. As the activation energies are different in the two cases

Fr
(generally lower in the presence of catalyst unless it is a negative catalyst), let us represent them as E^neat
and Ej.jjj. The corresponding rate constant may be represented as k unc and k^^^. According to Arrhenius
equation
for
ur
k
uncat
= A e ^uncai^^^

-Egaf )/RT _ ^AE/RT

s
cat uncat
ook

cat
k
Yo
uncat FIGURE 4:^
k AE k AE
eB

cat cat
or In or 2-3031og-
k RT k RT
uncat uncat
r

k AE
ad
ou

cat
or - Antilog logA-
k 2-303 RT
uncal
Y

Refer to Problems for Practice 7 and 8 page 4/67.

7. Different reactions will have different slopes for plots of
Re

1
nd

log k versus 1/T. Greater slope means lesser value of E^, which T

means greater increase in the rate of reaction (or rate constant)

Fi

for the same increase in temperature or greater sensitivity to Plots of log i r versus ^ for two
temperature change. Thus, in the Fig. 4.22 reaction II has greater different reactions
slope and hence is more sensitive to temperature change.
8. Photochemical Reactions. Reactions which do not take place in dark but take place only in presence of
light are called photochemical reactions, e.g.,
Light
H2(^) + Cl2(g) ^ 2HCI(g)
Light
6CO2(g) + 6H20(g) Chlorophyll
■>
^6^12^6
Glucose
From air

Some substances do not react even in the presence of light but react if another substance is also present.
Such a substance which helps to bring about a photochemical reaction is called a photosensitizerand
the phenomenon is called photosensitization, e.g., green coloured pigment, chlorophyll, in plants acts
as a photosensitizer.
CHEMICAL KINETICS 4/65

NUMERICAL I FORMULAS USED I

PROBLEMS
Ea
BASED (i) log A = - + log A
ON 2.303 RT

where k = rale constant, = energy of activation in joules

Arrhenius
R = gas constant = 8.314 J K“* mor', T = temperature in K.,
Equation
A = frequency factor

^2
(//) log -A =
k 2.303R T,T,
r2

where efj = rate constant at temperature T,, k2 = rate constant at temperature T,.

w
{Hi) Fraction of molecules having energy equal to or greater than activation energy E^^ is given by
Ea

F lo
X = or
log a: = -
2.303 RT

ee
Problsm
Q The rate constant of a reaction is 1.2 x 10“^ sec"* at 30" C and 2.1 x 10“*^ sec"* at 40" C.

Fr
Calculate the energy of activation of the reaction.
Solution. Here, we are given that: it, = 1.2 X 10"^ sec -1 T, = 30 + 273 = 303 K
it-, = 2.1 X 10 ^ sec -1
for 12 = 40 + 273 = 313 K
r
Substituting these values in the equation :
You
s
k E
T2-TO 2.1xl0“^s"* E 313K-303K
ook

a a
log we get log X
it
1
2-303 R
T,T2 1.2x10"3s"* 2.303x8.314JK"*raol"* 303KX313K
eB

2.1 E 10
log - a ^ ir
1.2 2.303x8.314 303x313(1 mol"*)
our
ad

-i
This on solving gives E„ = 44126.31 mol"* = 44-13 kj mol
Problem B The rate of a particular reaction doubles when temperature changes from 27 C to
dY

37"C. Calculate the energy of activation of such a reaction.

Re

Solution. Here, we are given that

Fin

When Ti = 27"C = 300 K, itj = it (say). When T2 = 37"C = 310 K, k2=2k

Substitutingthese values in the equation :
*2
Iog-^ =
E..
a fT2-Tj^ 2it E
Cl
310K-300K
2.303 R
, we get, log X
^1 T.T
1^2 )
k 2.303x8.3141K"i mol"‘ 300KX310K

E 10 E 10
a a
or
log 2 = X or log 2 = X
2.303x8.314 300x310(1 mol"*) 2.303x8.314 300x310(1 mol"*)
This on solving gives 'a
- 53598-6 1 mol"' = 53-6 kJ mol"*

Retain in Memory
For a large number of reactions, activation energy (E^) at room temperature is about 54 kl mol"'.
U.sing this value if we calculate k2/k^ when temperature is increased by 10®, from
300 K to 310 K, we get k2/k^ = 2. This means ^2 = 2 A:[. This proves that rate of reaction is doubled for
10® rise of temperature.
4/66 ‘P'utdee^’A New Course Chemistry (XII)C2S1S]

Problem 1^ The activation energy of a reaction is 94.14 kj mol"' and the value of rate constant at
313 K is 1.8 X 10"' sec"'. Calculate the frequency factor, A.
Solution. Here, we are given that = 94.14 kJ mol"' = 94140 J mor*
T = 313K, 1.8 X 10-5 sec
-I

E
a
Substituting the values in the equation : log/: = - 2.303 RT
+ logA

E 94140JmoI-'
or log A = \ogk +
a
= log{1.8xl0"5 s"') +
2.303 RT 2.303x8.314JK"' mol"' x313K
= (log 1.8) - 5 + 15.7082 = 0.2553 - 5 + 15.7082 = 10.9635
A = antilog (10.9635) = 9.194 x 10‘" s"^

ow
Problem j^] The first order rate constant for the decomposition of ethyl iodide by the reaction ;
C2H5I (g) ^C2H4(g) + Hl (g)
at 600 K is 1-60 x 10"® s"'. Its energy of activation is 209 kj/mol. Calculate the rate constant of the reaction
at 700 K. (NCERT Solved Example) (CBSE 2007)

e
re
Tj = 600 K, /t, = 1-60X 10"5
-1

rFl
Solution. Here, s

T2 = 700 K, *2 = ?, E^, = 209 kJ mol"'

F
Applying Arrhenius equation, log—-
kI
E. flT,
2-303 R
1

r
ou
log k2 = log k^ +
E
a T.-T^ = log(l-60xl0“5) + fo 209000 J mol"' r 700-600
or
ks
2-303 R T.T
1^2
2-303x8-314JK"'mor' 600x700
oo

= (- 5 + 0-2041) + 2-5989 = -2-197 = 3-8030

Y
eB

*2 = Antilog 3-8030 = 6-353 x 10"^ s

-1

Problem Rate constant k of a reaction varies with temperature according to the equation
ur

Ea 1
logk - constant -
ad

2-303 RT
Yo

where is the energy of activation for the reaction. When a graph is plotted for log k versus 1/T, a straight
line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction. State the units
d

(R = 8-314 JK-* mol"^)

Re
in

Solution. Slope of the line = - = -6670K

F

2-303 R
or
E^ = 2-303 X 8-314 (JK"' mol"') x 6670 K = 127711-4 J mol"*.
Problem The rate constant of a reaction increases by 5% when the temperature of the reaction
is increased from 300 to 301 K whereas equilibrium constant increases only by 2%. Calculate the activation
energy for the forward as well as backward reaction.

Solution. According to Arrhenius equation, log — =

E»,/ T2-T1
k.1 2-303 R T,T2

If /;] = /: at 300 K, then at 301 K, k2 = k + = \05k

100

E
, \-05k «./ /'301K-300K
k 2-303x8-314JK-l'moi-'1 300Kx301K
(log 1-05 ) X 2-303 x (8-314 JK"' mol"') x 300 x 301 (K) = 36654 J mol"' = 36-65 kJ mol -1
CHEMICAL KINETICS 4/67

According to van’t Hoff equation (giving the effect of temperature on equilibrium constant),

log
AH° T2-Tj '
K, 2-303R T.T
1^2

If
K = K, at 300 K, then at 301 K. =K+ ^K = 102 K
100

1-02 K AH° 30IK-300K^

K 2-303x8-314JK-'mol-' 300Kx301K
-I
or AH ° = log 1-02 x 2-303 x 8-314 x 300 x 301 J mol REACTION COORDINATE

ow
= 14869 J mor' = 14-87 kJ mol-*
ThUvS, the reaction is endothermic. For such a reaction,
AH ° = Activation energy for forward reaction-Activation energy for backward reaction = y- ^
E
a, b
= E AH° = 36-65 - 14-87 kJ mol"* = 21-78 kJ mol'^

e
Q At 27®C in the presence of a catalyst, the activation energy of a reaction is lowered by

re
Problem

2 kcal. Calculate by how much the rate of reaction will increase ?

Frl
E

F
Solution. In the absence of catalyst, suppose rate constant = k. Then log A: = log A
(t
...(/)
2-303 RT
In the presence of catalyst, suppose rate constant = k'.
ou
or
Now, activation energy = E^^ - 2 (if E^^ is in kcal mo|-')
2 kcal mol *
log k' = log A -
E.a -2
2-303 RT
or log A-' = log A
Ea
;
2-303 RT
kfs 2-203 RT
oo
2 kcal mop*
Subtracting eqn (/) from eqn (//). we get log A'-log A =
2-303 RT
Y
eB

2 kcal mol '

1
log — = = 1 -4474 (-.- R = 2 X 10"^ kcal K ' mol *)
A 2-303(2x 10-3 kcal K"* moI-‘)(300K)
k'
ur

— = Antilog 1-4474 = 28 or k' = 28 A, i.e., the rate of reaction will increase 28 times.
oY

or
k
ad

Problem A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in
the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation
d

energy of the reaction if the catalyst lowers the activation energy by 20 kJ mol"*.
in

Solution. Suppose activation energy in the absence of catalyst is E„ and that in the presence of catalyst, it is
Re

-E../500R
E^. Then A = A^ =A ^■*'****^ (as A is same in both the cases).
F

Ea E.
Hence,
500 R 400 R
or
e.=|e a

But E„-E^=20 kJ or Ea = 20 or
5 "
= 20 or
E,,= 100 kJ mol-*.
Problem H The rate constants of a reaction at 500 K and 700 K are 0-02 s“* and 0*07 s“* respectively.
Calculate the values of E„a and A. (NCERT Solved Example)

Solution, log
A0 E
a fT2-T,
A, 2-303 R
\ ' ^ J

0-07 E 700 K-500 K E 200 K

a a
log X : log 3-5 = X
002 2-303x8-314JK-*moP* 700KX500K ' 2-303X 8-314 JK-* moP* 700 K x 500K
0-5441X 2-303 x 8-314 x 700 x 500
or E
a
JmoP* = 18231-4 J moP* = 18-23 kJ moP*
200
4/68 New Course Chemistry (XlI)BZsl9]

E Ea
log A = log /: +
a
Further, log k= — + !ogA or
2-303 R 2-303 RT

Substituting T = 500 K,k= 0-02 s"', we get

-1
18231-4Jmol
log A = log 0-02 + = -log 50+ 124095
2-303x8-314JK“‘ moI"^ x500K
= - 1-699 + 1-904 = 0-205

A = Antilog (0-205) = 1-603

Problem
m Two reactions of the same order have equal pre exponential factors but their activation
energies differ by 24-9 kj mol"*. Calculate the ratio between the rate constants of these reactions at 27®C

w
(Gas constant, R = 8-314 JK"* mol"*). (CBSE Sample Paper 2018)
-E./RT
Solution. According to Arrhenius equation k = Ae a

-E.a /RT -E.. /RT

k, = Ae 1
, k2= Ae (A = same given)

o
e
re
-E_ /RT
^^(E,^-E,,)/RT _ ^(24-9xl000)/(8-314) (300) _ ^9-983
^2 e

Frl
«2

F
k
k ,In—I-
k, 9-983
In-l- = 9-983 or =
2-303
= 4-3348 7*- = Antilog (4-3348) = 2-162 x 10^
ki
ou
r
Problem m The values of the rate constant for the decomposition of HI into H2 and I2 at different

so
temperatures are given below :
T/K 633 667
kf 710 738
oo
10“**/!^-* 0-19 1-00 8-31 25-1
Y
Draw a graph between In k against 1/T and calculate the values of Arrhenius parameters.
B

Solution. From the given data, we have

T(K) 633 667 710 738
re
oY

1-58 X 10"^ 1-50 X 10"^ 1-41 X 10-3 1-36 X 10"3

u

T
ad

/:(M"* r*) 0-19 X 10^= 1-9 X 10"^ 1-00 X 10-^ 8-31 X 10^ 25-1 X 10^ = 2-51 X 10-3
In k - 10-87 - 9-21 - 7-09 - 5-99
d

(= 2-303 log k)
in

Graph of In k vs 1/T. The plot obtained is as shown in the Fig.

Re

Slope of the line = ^^ = - 20-62 x 10^ K

F

E
a
From Arrhenius eqn., Slope =- R
(for plot of In/: of 1/T)
-6 T
Ey = - Slope X R
-7
= 20-62 X 103 K X (8-314 JK’* mor‘)
= 171-4 kJ mol-* -8“

A
E E -9--
a a
Further, In /: = In A - or In A = In /: +
RT RT ^-10--
Substituting T = 633 K. k = 0-19 x 10"^ s"*, -11 -●
i.e. In A: = - 10-87, we get
-12 “● .
171400
In A = -10-87 + = -10-87 + 32-57 = 21-70
-13 I I I I I I I I I I I I I I
8-314x633
1.30 1.34 1.38 1.42 1.46 1.50 1.54 1.58
or A = 2-65 X 10’M“‘s-* (1/Tx 103)=>
CHEMICAL KINETICS 4/69

1. The rate constants of a reaction are ! x 10 ^ sec ’ and 2 x 10“^ sec ' at 27"C and 37°C respectively. Calculate
the activation energy of the reaction. (Karnataka c-m .
2. The rate of a particular reaction quadruples when the temperature changes from 293Kto313 K. Calculate the
energy of activation for such a reaction assuming that it does not change with temperature.
3. The rate of a reaction triples when temperature changes from 50 to 100“ C. Calculate the energy of activation
for such a reaction (R = 8.314 J K ' mor')
4. For a reaction, the energy of activation is zero. What is the value of rate constant at 3(X) K, if
= 1 -6 X 10^ s”' at 280 K ? [R = 8 31 JK”' mol”' ] (Jharkhand Board Z ‘
The activation energy of a first order reaction at 3(X) K is 60 kJ mor'. In the presence of a catalyst, the activation

w
5.
energy is lowered to 50 kJ mol”' at the same temperature. How many times the rate of reaction will change ?
6. Given that the temperature coefficient for saponification of ethyl acetate by NaOH is 1-75. Calculate the
activation energy of the reaction.

o
7. The activation energy of a reaction is 75-2 kJ mol”' in the absence of a catalyst and 50-14 kJ mol”' with a

e
catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds

re
at 25T?(R = 8-314JK-' mol-').
8. The rate constants of a reaction at 700 K and 760 K

rFl are 0 011 M”' s”' and 0105 M“'s”' respectively.

F
Calculate the values of Arrhenius parameters.
9. A 1st order reaction is 50% complete in 30 minutes at 2TC and in 10 min at 47®C. Calculate

r
(0 rate constant for the reaction at 27°C and 47“C
ou
(») energy of activation for the reaction.
fo
(A.ssani Board 2012. ■
ks
10. The rate constant for the first order decomposition of Ni05 is given by the following equation
k = (2-5xl0'‘’s-')e”(250ooK)n- '
oo

Calculate for this reaction and rate constant if its half-life period be 300 minutes.
● (
Y
eB

ANSWERS

1.53.6 kJ mol ' 2. 52.86 kJ mol”' 3. 22.01 kJ mol-' 4. 1-6 X 10^ s-'
r

5. It will increase 55 times 6. 42-70 kJ mol”' 7. 24660 times

ou
Y
ad

8. = 166-3 kJ mol-', A = 2-824 x I0'°

9. k2Yx: = 2-31 x 10”^ min”', k^-joQ = 6-93 x 10"^ min"', = 43-85 kJ mo!"*
10. 207-85 kJ mol-', 2-31 x 10-^ :.,-l
d

mm
Re
in

HINTS FOR DIFFICULT PROBLEMS

F

2. k^ = 4 or k-y/k^ = 4.

4. E^ = 0 .-. log
k
0 or ^2 = 11 or*,
- I, = k1
*1 I

5. Proceed as in Solved Problem 7.

k 308 K
6. Temp, coeff. = — = 1-75, i.e., Tj = 298 K, Tj = 308 K
^ k 298 K
7. Similar to Solved Problem 7.

0-693 0-693
9. k = 2-31x10 2 min ' *47"C “ = 6-93x10-2 min"'
2TC - 10 min
30 min

6-93x10-2 Ea 20 _ (log3)2-303x8-314x3Q0x320 = 43-85 kJ mor'

log or E a
2-31x10-2 2-303x8-3141,300x320 20
4/70 .‘p'utdt-e^'A New Course Chemistry (X11)B2S!S
-E7RT
10. k = A e

E 25000
U —
or E„a = 25000 x R = 25000 x 8-314 j = 207850 J = 207-85 kJ
RT T

0-693 0-693
k = — = 0-00231 min-' =2-31 x lO-"** min
/ 300 min
1/2

4.18. EFFECT OF CATALYST ON THE RATE OF REACTION

A catalyst is a substance which increases the speed of a reaction without itself undergoing any chemical
change. For example, the rate of thermal decomposition of potassium chlorate is accelerated considerably in
the presence of mangane.se dioxide (Mn02) as catalyst:
Heat
2 KCIO3 MnO->
^ 2 KCl + 3

w
The action of a catalyst can be explained on the basis of “intermediate complex formation theory”.
According to this theory', it is believed that the reactants first combine with the catalyst to form an intermediate
complex which is short-lived tmd decomposes to form the products and regenerating the catalyst. Tlius, in general,

F lo
A -H B -I- C - A. B FIGURE 4.23
Catalyst ACTIVATION ENERGY
WITHOUT CATALYST

ee
Reactants ●c
/ ACTIVATION ENERGY

Fr
Intermediate complex WITH CATALYST
A '
'REACTION PATH
A-B + C
\ WITHOUT CATALYST
Prodiicl Catalyst
(Regenerated)
for
>-
a
cc
LU
REACTION PATH
WITH CATALYST
ur
Z \
/ \
LU

The inteimediate complex formed has much lower potential < \

\

energy thtm the intermediate complex formed between the reactants i- REACTANTS Iah
s
ook
Yo

ill the absence of the catalyst, as shown in Fig. 4.23.

LU
I-
O PRODUCTS
Thus, the presence of catalyst lowers the potential energy Q.
eB

barrier and the reaction follows a new alternate pathway which

requires less activation energy. We know that lower the activation REACTION COORDINATE

energy, faster is the reaction because more reactant molecules can

our

Effect of catalyst on activation energy

ad

cross the energy barrier and change into products.

Some Important Characteristics of a Catalyst and a Catalysed Reaction
Y

(/) A small amount of the catalyst is sufficient to bring about a lai-ge change in the rate of reaction.
Re

(/7) A catalyst cannot initiate a reaction. Itcan only accelerate the rate of reaction. In other words, it catalyses
nd

a spontaneous reaction and does not catalyse non-spontaneous reactions.

Fi

(Hi) A catalyst does not alter the free energy change (AG) of a reaction.
(iv) A catalyst does not alter the enthalpy change (AH) of the reaction,
(v) A catalyst nikes part in tlie reaction but is regenerated without any chemical change at the end of the reaction,
(v/) For a reversible reaction, it accelerates the speed of forward reaction as well as that of backward
reaction to the same extent. Hence, it does not disturb the equilibrium, i.e.. does not change the
equilibrium constant of the reaction but simply helps to attain that equilibrium quickly.
4.19. COLLISION THEORY OF CHEMICAL REACTIONS

Arrhenius equation (k = A ) significantly explains the effect of temperature on the rate of reaction
and hence the rale constant, k. It lakes inio consideraiion ihe frequency factor, A, and the activation energy,
E^, of the reaction. However, a much belter explanation about the rates of reaction is given by ‘collision
theory’ pul forward by Max Trautz and William Lewis in 1916-18. It is based upon kinetic theory of gases.
According to this theory, it is assumed that the reactant molecules are hard spheres and reaction between
ihem occurs only when they collide with each other.
CHEMICAL KINETICS 4/71

The number of collisions that take place per second per unit volume of the reaction mixture is
called collision frequency.
At ordinary temperature and pressure, the value of collision frequency is so high (10“^ to 10^® in a
gaseous reaction) that if all the collisions were effective, i.e., result into chemical reaction, the reaction
should be completed in the fraction of a second. However, in actual practice, this is not so. This is explained
as follows :

For a collision to be effective, the colliding molecules must have energy more than a particular value
called threshold energy. Thus, at ordinary temperature and pressure, most of the reactant molecules may not
possess energy equal to or greater than tlmeshold value.
For an elementary molecular reaction, A + B > Products, ifZ AB is the collision frequency and/is the
fraction of molecules whose collisions are effective, then evidently.

ow
dx
Rate of reaction = _ ...(f)
dt
“ ^AB
Further, according to kinetic theory of gases, the fraction of molecules having energy more than a
particular value, E at temperature T is given by Boltzmann factor, viz.,

e
-E/RT

re
f=e ...(«)

rFl
Now, as the rate of a chemical reaction depends upon the activation energy, E^, eqn. (//) may be written

F
as
/ = ^-e/rt
Substituting this value in eqn. (/), we get Rate = Z AB ^-E/RT

r
ou
fo
As rale of reaction is directly related to rate constant, k, we can also write k = ...(///)
ks
This equation predicts the values of rate constants fairly accurately for reactions involving atomic species
or simple molecules. However, if reactant molecules are complex, significant deviations are observed. This is
oo

because even if the colliding molecules have energy greater than threshold energy, they may not have proper
Y

orientation at the time of collision and no breaking of bonds in the reactant molecules and formation of new
B

bonds to form product molecules may occur. Thus, for a collision to be effective, the colliding molecules
re

must have energy greater than threshold energy and also must have proper orientations. For example, the
reaction between bromomethane and an alkali may or may not lead to the formation of methanol depending
ou

upon whether they have a proper orientation or not at the time of collision as shown in Fig. 4.24 below :
Y
ad

FIGURE 4.24

CH3Br + OH CH3OH + Br
d

H
H. H
in
Re

Proper -6 +5 -5
>HO +H- C —8r—►HO C- Br ►HO C; H + Br“
Orientation
F

H
-6 H H
H H
H- ;C — Br + "OH Intermediate Methanol
H
H
Improper
Bromomethane H- “OH- No products
Orientation
H
(Repel)

Proper and improper orientations of colliding molecules (bromomethane and an alkali)

This may be further illustrated by taking the example of the following reaction :
N02(g) + N02(g) >N204 (g)
Two probable cases are shown in Fig. 4.25 on the next page :
Fig. 4.25(a) represents a proper orientation of the colliding molecules resulting into the fonnation of
N2O4. Fig. 4.25(b) represents improper orientation at the time of collision and hence no new product is
formed.
4/72 J^n^uUefr'4. New Course Chemistry fXTntrown
To account for the orientation factor for the collision to be effective, another factor, P, called orientation
factor or steric factor or probability factor has to be taken into consideration. Hence eqn. (Hi) is modified
to the form :

k = PZ AB ...(IV)

FIGURE 4.25

o
o . O o
Molecules Bond

tN ; (N ■●V
approach formed
O o O o

NO2 * NO2 COLLISION N2O4

ow
o o
o
●’s Molecules Molecules
N ; O O
I
approach separate
O O
0

e
NO2 + NO2 COLLISION N02 N02

re
rFl
Collision between two NO2 molecules (a) Proper orientation forming N2O4

F
(b) Improper orientation-no new molecule formed

Thus, according to collision theory, the rate of a reaction depends upon the number of effective collisions

r
ou
which in turn depends upon the following two factors :

fo
(/) Energy factor, i.e., the colliding molecules must have energy greater than the threshold energy.
ks
(ii) Orientation factor, i.e., the colliding molecules must have proper orientations at the time of collision.
oo
-E../RT
Comparing eqn. (/v) with Arrhenius equation viz. k = Ae Cl
we observe that the Arrhenius parameter
Y
eB

A, also called frequency factor is mathematically given as A = PxZ AB

Limitations of Collision Theory. Though collision theory is able to explain the results of a number of
reactions, yet it does not give satisfactory results in .some cases. This is because it assumes atoms or molecules
r

to be hard spheres and does not take into account their structural aspects. Some other theories like ‘Transition
ou
ad
Y

State Theory’ etc. have been put forward to explain the reaction rates. Their discussion is beyond +2 level.
However, a brief idea of ‘transition state theory’ is given below.
d

4.20. TRANSITION STATE THEORY OR THEORY OF ABSOLUTE REACTION RATES

Re
in

This theory was put forward by Eyring in 1935. It is based upon the following fundamental postulates:
(0 Before the reacting molecules change into products, they come close together and form an intermediate
F

called ‘activated complex’ which has higher energy than both the reactants and the products (as shown
in Fig. 4.17, page 4/60). It is for this reason that the theory is called ‘Transition state theory’.
(ii) The activated complex is assumed to be a normal molecule except that it is unstable with respect to one
special mode of vibration which leads to its dissociation into products. In other words, activated complex
has a transient existence.

(Hi) The activated complex is in equilibrium with the reacting molecules. The special mode of vibration is
converted into translational degree of freedom along the reaction coordinate and as a result, the bonds
of the activated complex are broken leading to the formation of products. Thus, we may represent the
situation as follows :

A +B ^ X* Products
Reactants Acti vated complex

By making use of some fundamental properties of molecules, it was shown by Eyring that rate constant,
k is given by
CHEMICAL KINETICS 4/73

RT
k = K*
Nh

[X*]
where equilibrium constant. K* =
lAIIBJ
Further, as AG* = -RTlnK* or _ ^,-AG*/RT and AG* = AH* - TAS*,
RT
Combining these equations, we get, k = ^AS*/R ^-AH*/RT ...(Eyring equation)
Nh
-E./RT
Comparing with the equation derived from collision theory, viz. k = PZ^g a , we get
PZ
AB
_ ^AS’"/R (●.● AH* E,,)
Thus, transition state theory is better in the sense that whereas in collision theory, correction factor P

w
was introduced arbitrarily, transition state theory justifies its inclusion in terms of entropy of activation, AS*.

z5\Tf ^

Flo
Chemical kinetics. It is that branch of chemistry which deals with the study of rates of reactions, factors

ee
1.
affecting them and the mechanism of reactions.

Fr
2. Rate of reaction. It is the rate of change of concentration, i.e., change in the concentration of a reactant or
A(R] . A[P]
a product per unit time. For the reaction R —> P, Rate =
for
- or +
ur
At Al
3. Significance of -ve and +ve sign. Rate of reaction is always +ve. As [RJ decreases with time, A [R] =-ve
so that (-) X (-) = +ve. As [PJ increases with time, A [P] = +ve.
k s
Cone.
Yo
4. Units of rate of reaction. For reactions in solutions. Rate = - mol L ’ s * etc.
oo

Time
eB

Pressure
-1
For gaseous reactions. Rate = - atm s ‘ etc.
Time
5. Average rate and Instantaneous rate. Rate of reaction does not remain constant throughout because amount
r

of reactant decreases with time while that of product increases with time. Average rate of reaction means
ou
ad

average value during a large time interval. It is represented by Ax/At. Instantaneous rate of reaction means
the rate of a reaction at a particular instant of time during a very small time interval. It is represented by dxidt.
Y

6. Measurement of rate of reaction. Concentration of a suitable reactant or product is measured at different

Re
nd

times and then plotted against time. Slope of the tangent to the curve at the particular instant of time gives
instantaneous rale of reaction at that particular instant. In case of plot of the reactant, slope of tangent is -ve
Fi

(rins, = - d [R]/c/r). In case of plot of product, slope of tangent is +ve = ■¥ d [P]/f/f).

7. Expressing rate of reaction in terms of different reactants/prod ucts. Concentrations of different reactants
and products may not change by the same amount in the same interval of time. But the rate of reaction has
a unique value. Hence, for a reaction, a A + £> B 4 a: X + y Y,

Rate = - ld[A]^ l(j[B]^ ^ lt/[X]_ ^ 1^[Y]

a dt b di X dt y dt '
8. Factors affecting rate of reaction
(0 Nature of reactants, e.g., 2 NO + O2 > 2 NO2 and 2 CO + O2 > 2 CO2 are similar but former is fast
but latter is slow.

{ii) Concentration of reactants. Greater the concentration, faster is the reaction.

(Hi) Temperature. Generally, rate becomes double for every 10° rise of temperature,
(/v) Presence of cataly.st. It generally increases the speed of reaction,
(v) Surface area. Greater the surface area of reactants, faster is the reaction,
(v/) Presence of light. Photochemical reactions take place in presence of light.
4/74 T^fuuieefa.’^, New Course Chemistry (XIl)l!ZsZSI
9. Rate law and Order of reaction. For the reaction a A + /> B Products, rate may not depend upon all a
terms of A or all terms of B. Experimentally, suppose it is found that rate depends upon a terms of A and
p terms of B. Thus, rate oc [A]“ [B]P or rate = k [A]® [B]P. This equation is called rate law equation.
Constant k is called rate constant or speciflc reaction rate. If [A] = [B] = 1 mol L“^ rate = k. Hence, rate
constant is the rate of reaction when the concentration of each reactant is unity. The sum of powers of molar
concentrations of the reactants in the rate law equation, i.e. a + P w called order of reaction.
10. Difference between Rate law and Law of mass action. Law of mass action is a theoretical law, i.e.,
rate = k [A]** [B]*. Rate law is an experimental law, i.e., rate = k [A]“ [B]P where a and P can be found only
experimentally.
11. Examples of reactions of different order :

w
hv Pt
(0 Zero order ; H2 + CI2 ^ 2 HCl, 2 NH3 1130K
» N2 + 3 H2

(ii) First order : 2 N2O5 - ^ 4 NO2 + O2, Rate — k [N2O3]

H2O + i O2, Rate = k [H2O2]

o
H2O2

e
re
SO2CI2 ^ SO2 + CI2, Rate — k [SO2CI2]

Frl
(ill) Second order : 2 HI > H2 +12, Rate = k [HI]^

F
2 NO2 + F2 » 2 NO2F, Rate = k [NO2] [F2]
(iv) Third order : 2 NO + O3 — 2 NO2 + O2, Rate = k [NO]^ [O3]
ou
CH4 + CO, Rate = k [CH3CHO]^ ^ order = 1*5

or
(v) Fractional order : CH3CHO
H2 + Br2 2 HBr, Rate = k [H2] [Br2]‘^, order = 1-5
12.
CO + CI2 ^ COCI2, Rate = k [CO]^ [Cl2]'^2, order = 2-5
Units of rate constant of reactions of different orders.
kfs
oo
dx cone 1 1 1
Rate, — = A: [cone]". Hence k =
Y
X
n-l
dt Time [cone]" Time [cone]
eB

cone
For zero order (n = 0), k = = molLT* s“* or atm s“*
Time
ur
oY

1
For 1st order (n = 1), k = = s ^ or min ^ etc.
ad

Time
1 1
d

For 2nd order (n = 2), k = =Lmol * s ^ or atm * s ^ and so on.

Time cone
in
Re

13. Molecularity of reaction. All the atoms or ions or molecules of the reactants may not collide simultaneously.
The number of atoms or ions or molecules that collide simultaneously with one another so as to result into
F

a chemical reaction is called the molecularity of the reaction. Thus, generally the reaction takes place in a
number of steps. Such reactions are called complex reactions. The slowest step is the rate determining
step. Writing of steps for the overall reaction is called mechanism of the reaction. The different step
reactions are called elementary reactions or an elementary reaction is the reaction taking place in one step.
Molecularity is defined only for an elementary reaction. It has no meaning for complex reaction. For example,
for the reaction 2 NO2 + F2 > 2 NO2F, Rate = k [NO2] [F2]. Probable mechanism is
slow fast
(/) NO2 + F2 » NO2F + F (ii) NO2 + F ^ NO2F. Hence, it is bimolecular.
14. Half-life period of a reaction (f|/2)< It is the time in which half of the reactant reacts.
15. Expressions for rate constant and half-life period for reactions of different orders.
d[A]
(a) Zero order reactions. A Products, Rate = -
dt
= k [A]® = k. On integrating and rearranging.
1
k = -t {[A]q - [A]} or [A] = [A]q - kt. Thus, a plot of [A] vs t will be linear with -ve slope, i.e., - k.
[A] 0 P
0
Putting t = r,/2 when [A] = [A]q/2, ^1/2 = 2k
or
2k
for gaseous reactions.
CHEMICAL KINETICS 4/75

{b) First order reactions. A > Products, Rate = = k[A]. On integrating and rearranging,
dt

(r = iln
t [A]
or [A]
or [A] = [A]o e-*'.
If ‘fl’ is the initial amount of A and x is the amount reacted in time t, we can write

a 2-303 a
k=-\n log
t a-x t a-x

Rewriting as kt = In [A]q - In [A] or In [A] = - ^/ + In [A]q, plot of In A vs r will be linear with -ve slope
k
= - k or rewriting as log [A] = -
2*303
r + log[A]Q, plot of log [A] vs t will be linear with -ve slope
k .. . [A] 0 _ kt [A] 0
or rewnting as log plot of log vs t will be linear passing through the origin
2-303 ^ [A] 2-303 ’ [A]

with +ve slope = ——.

2-303
2-303 ^ 0-693

F low
Putting t = ti/2 when [A] = [AIq/2, t^f2 =
Thus, half-life period of a 1st order reaction is independent of initial concentration.
16. General expression for half-life period. For zero order reaction, t^/2 [Aq]. For 1st order reaction,
hn
oc
[Aq]®. For 2nd order reaction, /j/2 [Aq]”* and so on. In general, for reaction of nth order,
tl/2 l/[Ao]”“* or ^1/2 [Aq]^"”.

e
[A 1
17. Amount left after n half-lives. After one half-life, amount left = — .
for Fr
2

[A 1
After two half lives amount left = . In general, after n half-lives, amount left = —^.
2 2 2^ 2"
Your
s

18. Determination of order of reaction.

eBo k

(*) Graphical method. For nth order, rate = k [A]". For n = 0, rate = k. Hence, plot of Rate vs (A) will be a
horizontal line. In other cases, plot of Rate vs [A]” will be linear with slope = k.
> Products, taking suitable concentrations of A, B and C,
ad

(U) Initial rate method. For a A + bB + cC

our

rate is determined. Now concentration of A is changed (say doubled), keeping that of B and C constant.
Effect on rate of reaction is found again. If order with respect to A, B and C are a, p and y respectively, then

('b)i=«Aol“[V[CoF and (r„)j=fc[A„]“[B/[CF

Re

Dividing, we get (ro)|/(ro)2 = {[Ao]i/[A()]2}“. Thus, a can be found. Similarly, we can find p and y. Overall
Y

order = a + P -f y.
Find

(iii) Integrated rate law method. Substitute the concentrations at different times in the expression for rate
constant (k) and find which expression gives constant value of k.
(iv) Half-life method. r|/2 [Aq]*"”- Starting with two different initial concentrations, find ty2- Then
(H/2)Ah/2h = {[Ao1i/[Ao]2)^“”- Take log and calculate n.
19. Modified expressions for rate constant for some common reactions.
cci
4 , 2-303,
V.. oo

(0N2O5 , k = log —
t
-V, 00

(V, = volume of O2 collected in time t, = volume of O2 collected at «> time).

1 , 2-303, Vq
(«) H2O2 2’
2 2
(Taking same volume of reaction mixture, Vq = volume of KMn04 used for titration at zero time and V, is
that used at time t)

1
4/76 New Course Chemistry (XII)QE

20. Pseudo first order reactions. These are those reactions which are not truly of 1st order but under certain
conditions become reactions of 1st order, e.g.,
(0 Acid hydrolysis of ethyl acetate
H+
CH3COOC2H5 + H2O ^ CH3COOH + C2H5OH, Rate oc [CH3COOC2H5] as H2O is in excess
(«) Acid catalysed inversion of cane-sugar
H+
C12H22O11 + H2O ^6^12^6 + C6H12O6, Rate [C|2i^22®ii^
Sucrose Glucose Fructose

(dextro-rotatory) (dextro-rotatory) (laevo-rotatory)

leavo-rotatory
21. Rate of radioactive disintegration/Decay. It follows first order kinetics.
2-303 N0 2-303 a
Disintegration constant (k) = log or
log
t N t a-X

(Nq — initial no. of atoms, N = atoms present at time tor a = initial amount, x - amount disintegrated in time t).
Half-life (fj/2> = 0-693/A: and average life (x) = 1/A = t|/2/0-693.

w
22. Threshold energy and Activation energy of reactions. The minimum energy which the colliding molecules

F lo
must have so that the collision is effective (i.c., results into eraction) is called threshold energy (Ep. The minimum
extra energy absorbed by the reactant molecules so that their energy becomes equal to threshold energy is
called activation energy (E„). Thus, - Eregctants- Lower the activation energy, faster is the reaction.
23. Effect of temperature on rate of reaction (Arrhenius equation). A = A

ree
A = pre-exponential factor (frequency factor)
Eg = activation energy of the reaction, T = temperature in K, R = gas constant
for F
E E
a
Taking logarithm. In A = In A - or log A = log A -
a

RT
Your
2-303 RT
ks

Thus, plot of In A vs 1/T is linear with -ve slope = - E^R or plot of log A vs 1/T is linear with -ve slope
eBoo

= -Ey2-303R.
24. Effect of catalyst on the rate of reaction. Presence of catalyst lowers the activation eneigy barrier and hence the
reaction becomes faster (Fig. 4.23). It does not affect the value of AG or AH. For a reversible reaction, it
ad
our

affects the forward and the backward reaction to the same extent and hence equilibrium is not disturbed.
25.
Collision theory of chemical reactions. Reaction takes place because the molecules collide. However, all
collisions are not effective. Effective collisions depend upon the following two factors :
(0 Energy factor, i.e., colliding molecules must have energy greater than threshold value,
Re
Y

(/i) Orientation factor, i.e., colliding molecules must have proper orientation at the time of collision.
Find

Combining the effect of both factors, we have A = PZ^

= collision frequency between molecules of A and B
P = orientation factor/steric factor/probability factor.
26. Transition state theory or Theory of absolute reaction rates. According to this theory (put forward by
Eyring), before changing into products, the reactant molecules first combine to form an intermediate called
activated complex which is unstable with respect to a special mode of vibrationthat is converted into
translational degree of freedom and hence it dissociates to form the products, i.e.

A + B V X Products
Reactants Activated
complex
By using fundamental properties of molecules, it was derived by Eyring that
RT
A = g&S*/R g-AH*/RT (Eyring equation)
Nh
AS* and AH* represent entropy of activation and enthalpy of activation respectively.

\
CHEMICAL KINETICS 4/77

QUESTIONS

Bssed on Book

I. Multiple Choice Questions 5. The rale law for a rcaclion between the substances

A and B is given by
1. For the reaction, NO2 ig) + CO (g) ^ NO (g)
Rale = k fA]" [B]
III

+ CO2 (g), the correct expression for the rate of

reaction is On doubling the concentration of A and halving
the conceniralion of B. Ihe ratio of the new rate to
d [NO,1 d\CO,]
the earlier rate of the reaction will be
(a) Rate =- (/?) Rate = -
dt dt
(a) m + n (h) (n - m)
d[m^]-d[co]

low
(c) Rate =
dl (c) 2*"""'^ (d) 'j{m+n)

d [CO] 6. A hypothetical reaction 2 p + q ^ + /● has rate

(d) Rate =
dt constant as 20 x 10“^ mol lir* sec"'. The order
of the reaction is
2. Which of the following expression is correct for

ee
the rate of reaction given below ? (a) unpredictable (b) zero

rF
Fr
(f) one (d) two
5 Br“ (aq) + 5 BrOJ (aq) + 6 H’*' (aq) ■>

7. The rate law' for the chemical reaction.

3 Bf2 {aq) + 3 H2O (/)

r
2 NO2CI > 2 NO2 + Cl2 is r = /: [NO2CI].
A[Brl^_5 A[in fo
u
(a) - Which of the following is rate controlling step ?
Ai At
ks
(a) 2NO2 + CI >N02C1
Yo
A[Br"]_ 5 A[H-*-] (/j) NO2CI + CI >N02 + Cl2
oo

(b) -
At 6 Ar
(c) NO2CI is) > NO2CI (g)
B

A[Br"] 6 A[H+] (d) 2NO2CI >2N02 + Cl2

(c) -
re

At 5 At 8. Higher order (> 3) reactions are rare due to

(n) Low probability of simultaneous collision of
A[Br]^_6 A[H^
u
ad

(cl) - all the reacting species

Yo

At At

3. The rate of formation of SO3 in the reaction (h) Increase in entropy and activation energy as
1 more molecules are involved
2 SO2 + O2 r- - 2 SO3 is 100 kg min
nd
Re

(a) The rate of disappearance of SO^ is 50 kg min

-1 (c) shifting of equilibrium towards reactants due
to elastic collisions
Fi

{b) The rate of disappearance of O2 is 50 kg min“

(rf) loss of active species on collision
(c) The rate of appearance of SO2 cannot be
increased by adding catalyst as the reaction is 9. When initial concentration of reactant is doubled
in equilibrium in a reaction, the half-life period is not affected.
{d) The overall order of reaction is three The order of reaction is

4. In a catalytic experiment involving Haber’s (fl) second (b) zero

process, N2 (g) + 3 H2 (g) ^ 2 NH3 (g), (c) first
the rale of reaction was measured as : Rate
(d) more than zero but less than first
_ d[NH^] = 2-0 X 10^ M s"'. If there were no 10. The reaction 3 A B -I- C would be a zero order
dt reaction when
side reactions, what was the rate of reaction (a) The rate of reaction is proportional to square
expressed in terms of ? of concentration of A
(a) 1 X 10-^ M s-' (b) 4 X 1(H M s“‘ (/?) The rate of reaction remains the same at any
(c) 5 X 10-3 g (d) 1 X !0"3 M s"' concentration of A

4
4/78 New Course Chemistry CX11)ESSSI

(f) The rate of reaction doubles if concentration 16. In the reaction, P + Q ^ R + S, the time taken
of B is increased to double for 75% reaction of P is twice the lime taken for

{d) The rate remains unchanged at any 50% of the reaction of P. The concentration of Q
concentration of B and C varies with the reaction time as shown in the figure.
11. The rate law for the reaction RCl + NaOH

ROH + NaCI is given by r= k [RCI]. The rate of

reaction is

(a) doubled by doubling the concentration of

NaOH

(h) is halved by reducing conentration of RCl by

one half

ow
(c) is increased by decreasing the temperature
The overall order of the reaction is
(d) is unaffected by change in temperature
12. For a reaction, (a) 2 (h)3
> 4 NO2 + O-,, rate and
rate constant are 1-02 x 10^ and 3-4 x 10“^ sec -1 (c)0 id) 1
respectively. Then concentration of No05 (in mol

e
17. A first order reaction is carried out starting with

Fl
L~') at that time will be

re
10 mol L~' of the reactant. It is 40% complete in
(a) 1-732 (b) 3 one hour. If the same reaction is earned out with

F
(c) 1-02x10-^ id) 3-4 X 10-^ an initial concentration of 5 mol L"', the percentage
13. The rate constant of the reaction A
ur of the reaction that is completed in one hour will

or
^ B is
be
0-6 X 10^ mole per litre per second. If the sf
concentration of A is 5 M. then concentration of B (a) 40% (b) 80%
after 20 minutes is
k
(c) 20% (d) 60%
Yo
oo

(a) 0-36 M (b) 0-72 M

18. The time taken for 10% completion of a first order
(c) 108 M (d) 3-60 M reaction is 20 mins. Then for 19% completion, the
B

reaction will take

14. The half-life of a substance in a certain enzyme-
re

catalysed reaction is 138 s. The time required for (a) 40 mins {b) 60 arins
the concentration of the substance to fall from
(c) 30 mins id) 50 mins
u

1-28 mg L“* to 0-04 mg L‘'*, is

ad
Yo

19. For the reaction A -t- 2 B ■> C, the reaction rate

(a) 414 s (b) 552 s is doubled if the concentration of A is doubled.
(c) 690 s (d) 276 s The rate is increased by four times when
d
Re

IS. The plot between concentration versus time for a concentrations of both A and B are increased by
in

zero order reaction is represented by four times. The order of reaction is

F

(a) 3 (b) 0
(c) 1 id) 2
20. When initial concentration of a reactant is doubled
c
in a reaction, the half-life period is not affected.
The order of reaction is
o 4> o t

{a) second
(b) more than zero but less than first
(c) zero
{d) first

21. The adjoining graph shows how T,/2 (half-life) of

a reactant R changes with the initial reactant
concentration a 0
CHEMICAL KINETICS 4/79

A (a) 25 g (b) 50 g
(c) 12-5 g (d) 6-25 g
T 28. The ratio tyg : t^p for the first order reaction is
1/2
(a) 3 (b) 5
(c) 2 (d) 7
Man=>
0 29. A first order reaction has a half-life period of 34-65
seconds. Its rate constant is
The order of the reaction will be

(a) 0 (b) 1
(a) 0-2 X 10-2 sec-* (b) 4 x lO"- sec-'
(c) 20 sec- (d) 2x10^ sec -1
(c) 2 (d) 3
30. The rate of reaction increases with increase in
22. For the elemental^ reaction M N, the rate of

ow
temperature because
disappearance of M increases by a factor of 8 upon
(a) Activation energy of the reactant molecules
doubling the concentration of M. The order of decreases
reaction with respect to M is
(h) Kinetic energy of the product molecules
(fl)4 (b)3 increases

e
re
(c)2 (d) 1 (c) The fraction of the reacting molecules
23. When initial concentration of the reactant is

rFl possessing an energy equal to the activation

F
doubled, the half-life period of a zero order reaction energy is more
(d) The collisions between the molecules decrease
(<3) is halved (b) is doubled

r
31. Which of the following is not a characteristic of a
ou
fo
(c) is tripled (d) remains unchanged
catalyst ?
ks
24. In the reaction A + B ^ Product, if B is taken in (a) It changes the equilibriumconstant
excess, then it is an example of (/;) It alters the reaction path
oo

(a) second order reaction (c) It increases the rale of reaction

Y
eB

(/?) zero order reaction (d) It does not affect the Gibbs energy
(c) pseudo first order reaction 32. For which of the following is the temperature
coefficient maximum ?
ur

{d) first order reaction

(a) A ^ B, = 50 kJ
ad

25. The half-life period of a radioactive decay of

Yo

is 5730 years. An archeological artifact containing

(b) P ^ Q. E„ = 40 kJ
(c) X -> Y, = 60 kJ
wood had only 80% of the found in the living
d

tree. The rate constant of the reaction is id) W ^ Z, E^, = 80 U

Re
in

2-303 . 100 2-303 100

(a) log (b) log 33. The plot of log k \s — helps to calculate
F

5730 80 5730 100-80

2-303 80 2-303 20 (fl) Energy of activation

(c) log (d) log (b) Rate constant of the reaction
5730 100 5730 100-20
(c) Order of the reaction
3
26. The ratio of the time period for — th of the reaction id) Energy of activation as well as frequency factor
34. In a reversible reaction, the activation energies of
of first order to complete to that required for half the forward and the backward reaction are equal.
to the reaction is: In such reactions

(fl) 4 : 3 ib) 3:2 (a) AH = 0 ib) AS = 0

(c) 2 : I id) \:2 (c) order is zero id) AH ?£ 0, AS 0

35. The rate of a chemical reaction doubles for every
27. Half-life of the substance ‘A’ following first order
10° rise in temperature. If the temperature is raised
kinetics is 5 days. Starting with 100 g of A, amount
by 50°C, the rate of reaction increases by about
left after 15 days is
4/80 Vnadee^’^ New Course Chemistry (XII)EZ
(a) 10 times (b) 24 times
(d) 2
(c) 32 times (d) 64 times
36. The rate of a reaction doubles when its temperature 42. The rate constant of a reaction is 2-3 x 10“^ L
changes from 300 K to 310 K. Activation energy mor* s“*. The order of reaction is
of such a reaction will be
(a) zero (b) one
-1
(a) 60*5 kJ mol (b) 53-6 kJ mol-^ (c) two (d) three
-1
(c) 48-6 kJ mol (d) 58-5 kJ mol-^ 43. For the reaction system

w
37. For an endothermic reaction, where AH represents 2N0(g) + 02(g) >2N02(g)
the enthalpy of the reactionin kJ/mol, the minimum
the volume is suddenly reduced to half its value
value for the energy of activation will be
by increasing the pressure on it. If the reaction is
(a) .less than AH (b) zero
of first order with respect to O2 and second order

e
(c) more than AH (d) equal to AH. with respect to NO, the rate of reaction will be

ro
re
38. The rate of a reaction is expressed in different (a) diminish to one-eight of its initial value
ways as follows:
(b) increase to eight times of its initial value

F
= _,ld[A]_ d[B] (c) increase to four times its initial value

Fl
2 dt 3 dt 4 dt dt ●

u
(d) diminish to one-fourth of its initial value
The reaction is
44. In the first order reaction, 75% of the reactant

sr
(a) 4A + B->2C+3D disappeared in 1-386 h. Calculate the rate

ko
(b) B + 3D^4A + 2C constant of reaction,
o
(c) A + B->C+D (a) 3-6 X 10-3 s-l 2-8 x lO"^ s"‘
39. For the reaction
(d) B + D A +C
of
(c) 17-2 X 10-3 s"i 1-8 X 10-3 s-‘
o
1 45. A first order reaction is half-completed in 45
Y
N2O5 ig) >2N02(g)+- 02(g), minutes. How long does it need for 99-9% of the
erB

reaction to be completed ?
the value of rate of disappearance of N2O5 is
uY

given as 6-25 x 10~3 mol L-* s-^ The rate of (a) 20 hours (/>) 10 hours
formation of NO2 and O2 is given respectively as: 1
{a) 1-25 X 10-2 mol L"‘ s-‘ and 6-25 x 10-3 (c) 7- hours (d) 5 hours
ad

mol L-* S-*

do

46. For a first order reaction at 27®C, the ratio of time

{b) 6-25 X 10-3 njoi L-l s-l 5.25 ^ 10"3
required for 75% completion to 25% completion
in

mol L“* S-* of reaction is

(c) 1-25 X 10-2 mol L-* s-^ and 3-125 x 10-3
Re

(fl) 3-0 (b) 2-303

F

mol L-^ s-^ (c) 4-8 (d) 0-477

{d) 6-25 X IQ-3 mol L-* s-* and 3-125 x 10-3 47. In a first-order reaction A B, if k is rate
mol L-* S-* constant and initial concentration of the reactant A

40. 100 cm3 of 1 M CH3COOH was mixed with is 0-5 M, then the half-life is:
100 cm3 of 2 M CH3OH to fonn an ester. The In 2 0-693
change in the initial rate if each solution is diluted (.) ib)
0-5*
with equal volume of water would be
(a) 4 times {b) 0-25 times log 2 log 2
(0- (d)
(c) 2 times {d) 0-5 times
41. The order of a reaction with rate equal to 48. For which order of reaction, a straight line is
is obtained on plotting half-life period (/1/2) against
initial concentration (a)?
1 (a) 0 (b) 1
(a) 1 (^)
(c) 2 (d) 3

A
CHEMICAL KINETICS 4/81

49. When concentration of reactant in reaction (b) Catalyst decreases the rate of backward
^ B is increased by 8 times, the rate reaction so that the rate of forward reaction

increases only 2 times. The order of the reaction increases

would be
(c) Catalyst decreases the enthalpy change of the
reaction
ia) 2 (b) 1/3
(c) 4 (d) 1/2 (d) Catalyst provides an alternative path of lower
activation energy.
50. The following data are obtained for the reaction,
56. For a first order reaction, the slope of the plot of
XxY ^ Products
In [R] vs time gives
Expt [Xo]/mol [VyJ/mol rate/mol s“* +k
1 0-25 0-25 1-0 X 10-6 (a) + k ib)

ow
2-303
2 0-50 0-25 4-0 X 10-6
-k
3 0-25 0-50 8 0 X 10-6 (c)-k (d)
2-303
The overall order of the reaction is
(CBSE 2020)

e
ia) 2 ib) 4
57. The half-life period of a zero order reaction is equal

re
ic) 3 id) 5

rFl to

F
51. An endothermic reaction A ^ B has an
0-693 2k
activation energy of 15 kcal/mole and the ia) (b)
k

r
enthalpy change of reaction is 5 kcal/mole. The
ou
activation energy for the reaction B
fo
^ A is
2-303 [Rnl
ks
(a) 20 kcal/mole (b) 15 kcal/mole ic) id)
k 2k
(c) 10 kcal/mole (d) zero
oo

52. The increase in reaction rate as a result of

(CBSE 2020)
Y

58. For a zero order reaction, the slope of the plot of

eB

temperature rise from 10° to 100° is

[R] vs time is
ia) 112 (b) 512
ur

ic) 400 id) 614 ib)-k

2-303
ad

53. The Arrhenius equation expressing the effect of

Yo

temperature on the rate constant of the reaction is 2-303 IRn]

id)
E 2k
d

(a) k = ib) k =
Re

RT
in

(CBSE 2020)
E 59. Which of the following analogies is correct ?
F

(c) k = log^ id) k = Ae~^d^'^

RT In 2
ia) First order reaction : ^j/2 = :: Zero order
k
54. The rate of a reaction doubles when its
[A] 0
temperature changes from 300 K to 310 K. reaction :
Activation energy of such a reaction will be : '1/2- j
(R = 8-314 J K-l mol-^ and log 2 = 0-301) ib) Zero order reaction : Units ofk = mol L”' s“*
-I -1 Second order reaction : Units of k = L mol"’ s“*
ia) 58-5 kJ mol ib) 60-5 kJ mol
(c) 53-6 kJ mol
-1
id) 48-6 kJ mol
-1 (c) Acid catalysed hydrolysis of ethyl acetate :
Second order reaction :: Base catalysed
55. Which of the following explains the increase of
hydrolysis of ethyl acetate: First order reaction
reaction rate by a catalyst ?
id) 2 N2O5 4 NO2 + O2 : First order reaction
(a) Catalyst provides the necessary energy to the
colliding molecules to cross the barrier :: 2 HI ^ H2 + I2 : First order reaction
4/82 New Course Chemistry (X1I)ESSI9]

11. Assertion-Reason Type Questions 68. Assertion. The time taken for 2 g of uranium to
decay to 1 g is double the time taken for 1 g uranium
For questions below, two statements are given to decay to 0-5 g.
one labelled Assertion (A) and the other labelled
Reason (R). Select the correct answer to these Reason. Rate of disintegration is directly
questions from the codes (o), (Z>), (c) and (</) proportional to the amount of radioactive isotope.
given below : 69. Assertion. In the presence of catalyst, enthalpy
(«) Both A and R are true and R is the correct change of the reaction decreases.
explanation of the assertion. Reason. In the presence of catalyst, activation
(h) Both A and R are true but R is not the correct energy of the reaction decreases.
explanation of the assertion, 70. Assertion. According to collision theoiy, rate of
(c) A is true but R is false, reaction does not depend on the total number of
collisions.
(r/) A is false but R is true.
Reason. The overall rate of reaction depends upon

w
60. Assertion. The reaction NO2 + CO ^ COo
+ NO is a reaction of second order. the number of effective collisions.

Reason. Rate law expression for the above reaction 71. A.ssertion. Rate of a reaction can be positive

F lo
as
is : Rale = k [NO^J [CO]. well as negative.
61. Assertion. Order of reaction is never negative with

ee
^[A]
respect to a reactant or product. Reason. In terms of reactant (A), rate = -
dt

Fr
Reason. The rate of reaction is always positive.
62. Assertion. Order of a reaction can be fractional ^/[P]
but molecularity is never fractional.
for
whereas in terms of a product (P), rate = + dt
ur
Reason. Order of reaction does not depend upon 72. Assertion. Rate constant of the reaction
the stoichiometric coefficients of the balanced
N2 + 3 H2 ^ 2 NH3 and 2 N2 + 6 H2 -> 4 NH3 is
s
equation.
ook
Yo
same at 25°C.
63. Assertion. Rate constant of a zero order reaction
Reason. Equilibrium constant of the reaction
eB

has the same units as those of the rate of reaction.

N7 + 3 —> 2 NH3 is half of the equilibrium
Reason. For a zero order reaction. Rate = Rate
constant.
constant for 2 N2 + 6 H2 —> 4 NH3 at 25®C.
r

73. Assertion. Order of reaction can be fractional but

ad
ou

64. Assertion. 50% of a reaction is completed in 50

molecularity is always a whole number.
sec, therefore, 75% of the reaction will be
Reason. Order of reaction can be found
completed in 75 sec.
Y

Reason. The rate constant of a zero order reaction experimentally only whereas molecularity of a
Re
nd

reaction can be written from the balanced chemical

depends upon time as k =-[ [A]^ - [A]}.
t ^ equation.
Fi

65. Assertion. For a first order reaction, the

concentration of the reactant decreases 74. Assertion. Half-life period of a zero order reaction
doubles if the initial concentration is doubled.
exponentially with time.
Reason. Rate of a zero order reaction doubles if
Reason. Rate of reaction at any time depends upon
the concentration of the reactant at that time. the initial concentration is doubled.

66. Assertion. For the reaction RCl + NaOH (aq) 75. Assertion. The presence of catalyst lowers the
ROH + NaCl, the rate of reaction is reduced to activation energy of a reaction.
half on reducing the concentration of RCl to half. Reason. The reactants combine with the catalyst
Reason. The rate of the reaction is represented by to form an intermediate complex.
k [RCl], i.e., it is a first order reaction. 76. Assertion. The plot of log k versus 1/T is always
67. Assertion. Greater the half-life period, faster is the linear and has a negative slope.
reaction. Reason. Arrhenius equation giving the effect of
Reason. Half-life period depends upon initial temperature on the rate constant k involves an
concentration as well as order of reaction. exponential factor involving activation energy.
CHEMICAL KINETICS 4/83

77. Assertion. Hydrolysis of an ester follows first order 78. Assertion. For complex reactions, molecularity and
kinetics. order are not same.
Reason. Concentration of water remain remains
Reason. Order of reaction may be zero.
nearly constant during the course of the reaction.
(CBSE 2020)
(CBSE 2020)

ANSWERS
I. Multiple Choice Questions
1. (a) 2. (b) 3. (b) 4. (a) 5. ic) 6. ib) 7. ic) 8. ia) 9. ic) 10.ib)
II. (b) 12. (b) 13. (/?) 14. (£●) 15. ib) 16. (cO 17. (n) 18.ia) 19. (r) 20. id)
21. (c) 22. (b) 23. (b) 24. (c) 25. (a) 26. (c) 27.(c) 28. (a) 29. id) 30. ic)
31. {a) 32. (d) 33. (d) 34. (a) 35.ic) 36. ib) 37.(c) 38. ib) 39. {(■) 40. ih)
41. (a) 42. (c) 43. (b) 44. (b) 45.(c) 46.(c) 47.(a) 48.(n) 49. ib) 50. id)
51. (c) 52. {b) 53. (rf) 54. (c) 55. id) 56. (c) 57. id) 58.ib) 59. ib)

w
II. Assertion-Reason Type Questions

F lo
60.(c) 61,id) 62.(a) 63.(a) 64.(c) 65. (6) 66. ic) 67. (i/) 68. (i/) 69. (t/)
70. (c) 71. id) 72. (c) 73. (a) 74.ic) 75. (n) 76. (/>) 77. (ti) 78.ih)

e
Fre
for
For Difficult Questions
r
I. Multiple Choice Questions
You
oks

9.r 1/2 oc
1/2 is independent of [Aq] when
[An]
-1 A[Br~] -1 A[H^1
eBo

2. n = 1
5 At 6 At
12. The given reaction is a reaction of 1st order
A[Br"] 5 AjH-*-] Rate = k [N2O5]
our
ad

or
At 6 At Rate 1-02x10-^
[N2O5] = k - 3 mol L'
ld[NH^]_ \d[U,J d[N,\ 34 X 10-5
4. Rate = —
2 dt 3 dt dt 13. The units mol L”* s”* of the rate constant show
dY
Re

that it is a reaction of zero order. For a zero order

reaction, x = kt.
Fin

dt 2 dt
Amount of A reacted (.v)

lx(2xl0-^Ms-‘)= lO^Ms-'. = (0-6 X 10“^ mol L-' s-') (20 x 60 s)

= 0-72 mol L-'
-I
5. Earlier rate = k o" /?'” [B] formed = [A] reacted = 0-72 mol L
[A1 0
(bT 14. Amount left after n half lives =
New rate =A:(2a) — 2'
V2
1-28
New rate 2“ a" /j"' 2-'" 004 = or 2« = 32 = 25 . n =5
2'
Earlier rate tj” b"'
Time taken = 5 half lives = 5 x 138 s = 690 s.
_ 2« 2^m —

8. Higher order > 3 for reactions is rare because there 15. For zero order reaction, k = - {[A]q - [A]}
is low probability of molecules to come together or [A] =-kt+ [A]q. Thus, plot of [A] vs t is linear
simultaneously and collide. with -ve slope (= - k).
4/84 ‘^'uieCeefi-'A New Course Chemistry (XIOBZSSQ

16. With respect to P, as is twice that of /j/2. 36. Refer to Solved Problem 2, page 4/65.
therefore, order = 1. With respect to Q, from the 37. E„a > AH.
graph, order = 0. Hence, overall order = 1.
17. For a first order reaction, time taken for a definite
fraction to complete is independent of initial >
O
concentration. In other words, in same time, same a
UJ
% age of reaction will be completed. z
UJ

2-303. a
18. / = ——log 2
k a — X

20 =
2-303. a 38. Minus signs are for reactants and positive signs
— log
a —O-lOfl for products. Dividing numbers are the

w
coefficients. Hence, B and D are the reactants and

20
2-303,
=—
100 A and C are the products.
or log ...(/)
k 90
d[Np^] _ ^ 1 ^[NO^J ^,^[0,]

Flo
39. Rate = -
2-303 a dl 2 dl dt
In 2nd case, t = ^log

ee
k a-0-19 a
Given
-d[Np^] = 6-25 X 10“^ mol L“* s"'

Fr
dt
\2
2-303, 100 2-303 10
=— log — log — -(«) Rate of formation of NO2
k 81 k ^ 9

for
[NO,]
ur
Eqn. (n)/Eqn. (i) gives dt dt
t
= 2x6-25 X 10-3 L-l 5-1
s
= 2 or r = 40 min.
ok
20
= 12-50 X 10-3 jjjoi L-l -I
Yo
s

19. Rate, r=k [A]« [B]* -(0 = 1-25 X 10-2 mol L-' s"'
Bo

2r=[2 Af [B]^ -m
Rate of formation of 0-,
4r=[4 A]“ [4B]^
re

_ d[0^] _ 1 d[Np^]
Dividing eqn. (n) by eqn. (/), 2 = 2^ a= 1
dt 2 dt
ou

Dividing eqn. (h) by eqn. (i),

ad

1
4 _ 4«7 46 _ 4 ^ 4* _ ^b+l = - X 6-25 X 10 3 mol L * s“*
Y

b-¥\ = l or b = 0
= 3-125 X 10-3 mol L~^
Hence, overall order = 1.
nd
Re

20. Half-life period of a first order reaction is 40. CH3COOH + CH3OH > CH3COOCH3
1 M 2M
Fi

independent of initial concentration of the reactant + H2O

Rate (ri) = k [CH3COOH] [CH3OH] = /: (1) (2)
_ 0-693 = 2k
'>'2- k
When each solution is diluted with equal volume
1 1 of water, concentration of each is halved. Now,
21. ti;2 OC
; For n = 2, t1/2 oc —
rate
n—1
a a

22. Initially, rate r = it [M]“ = ...(/)

V2=k -
1
2'|_^
On doubling the concentration of M
a
.2 j\ 2j~2
8r=it(2 a) -.(«) 1
Dividing eqn. (//) by eqn. (/), 21c “4
- = 0-25 or r2=0-25r]
8 = 2“ or 2“ = 23 or a = 3
[A] 0 41. Rate = k C5f2 ^-1/2 ,
23. For a zero order reaction, = '. Thus, if (A]q
2k
I
Overall order = - + = 1
is doubled, ty^ Is also doubled. 2
CHEMICAL KINETICS 4/85

42. Based on the units of rate constant, order of

51.
reaction = 2.

43. Rate = k [NO]“ [O2].

5=10
Initially, rate = ka^ b
If volume is reduced to half, concentrations are >,
00
doubled so that
U

5
New rate = k{2 a)^ {2b) = %ka^ b i.e. 8 times A
2-303 a
44. it = log
1-386x60x60 a-0-15 a

2-303 Progress of Reaction

log4 =2-8x 10-^ 'i-'
1-386x60x60

w
52. Increase in steps of 10° has been made 9 times.
45. k =
0-693 . 2-303 a
Hence, rate of reaction should increase 2^ times,
mm log
45 t
99-9% a - 0-999 a i.e., 512 times.

Flo
-E../RT
2-303x45 53. Arrhenius equation is k = Ae
or .t
99-9% ~ log 10^ =448 min
0-693

ee
54. ^'2 E
{2
I 1

Fr
k 2-303 R T ^2
= 7 — hours. 1
2
E
46. For 75% completion, log 2 =
[Cl 1_

for
2-303x8-3141^300 310
ur
2-303. a 2-303
t
75% “ — log log 4 E 10
k a-0-15 a k or 0-3010 =
a
X
ks
2-303x8-314 300x310
For 25% completion,
Yo
or £.. = 53-6 kJ mol-'
oo

1:

2-303, a 303, 4 55. Catalyst provides an alternative path of lower

^25% “ ^
eB

a -0-25 a activation energy.

1 . [A] 0
. _ log 4 0-6021 56. For 1st order reaction, /: = - In
r

t log(4/3) 0-6021-0-4771
t [A]
ou

25%
ad

or kt = In [A]q - In [A] or In [A] = In [A]q - ki

0-6021
Y

= 4-8 Hence, plot of In [A] vs t will be linear with slope

0-125 = -k
nd
Re

47. Half-life of a first order reaction does not depend 57. Refer to page 4/25.
upon initial concentration. It is equal to In 2!k. 58. For a zero order reaction.
Fi

1 1
48. r
1/2
oc
H-1 when n = 0, /j/2 [Aq]- k = - {[Alo-[A]}or[A]=-/:t + [A]o
tA)] t

Hence, plot of [A] vst r will be linear with slope

49. (/) r=ka" (ii) 2 r=k (8 a)". Dividing (ii) by (/), 2 = -k
= 8" or 2^" = 2 or 3 n = 1 or n = 1/3.
59. (a) is not correct because for zero order reaction,
50. From expt. 1 and 2, [Yg] is kept constant, [Xg] is
doubled, rate becomes 4 times. [A] 0
1/2 "
2k
Hence, rate « [Xg]^. From expt. I and 3, [Xg] is
kept constant, [Yg] is doubled, rate becomes 8 (c) is not correct because acid catalysed hydrolysis
times. of ethyl acetate is a first order reaction whereas
Hence, rate « [Yg]^. Hence, overall reactions is : base catalysed hydrolysis of ethyl acetate is a
second order reaction.
Rate = k tXg]2 rYg]3
Overall order = 2 -f 3 = 5. (d) is not correct because 2 HI —> H2 + I2 is a
reaction of second order.
4/86 “PnaxCec^'A New Course Chemistry (XlI)ESZsIS]
II. Assertion-Reason Type Questions 68. Correct A. The lime taken for half of the substance
is same as it is a reaction of 1 st order.
60. Correct R. Rate law expression for the reaction
NO2 + CO > CO2 + NO is : Rate = k [NO,]^ 69. Correct A. In tlie presence of catalyst, there is no
change in the value of enthalpy change (AH).
61. Correct A. Order of reaction can be negative with
respect to a product (e.g., in case of 2 O3 3O2, 70. Correct R. The overall rate of reaction depends
order w.r.t. to O2 is - I, page 4/18). upon the number of effective collisions as well as
the orientation factor.
62. R is the correct explanation of A.
71. Correct A. Rate of reaction is always positive.
63. R is the correct explanation of A because for a zero
order reaction, rate = k [conc]*^ = k. 72. Correct R. If equilibrium constant of the reaction
64. Correct R. Rale constant of any reaction is 2 N2 + 6 H 2 ^ - 4 NH3 is K, then equilibrium
constant al constant temperature and does not constant for the reaction N2 + 3 H2 v '*■

w
change with time or concentration. However,
Asseilion-l is correct because 2 NH3 is Vk .
73. Both A and R are correct and R is the correct
I
-{[A]o-[A]}

Flo
i =
explanation of A.
74. Correct R. Rate of a zero order reaction is

e
I [A| 0 lAl 0 independent of initial concentration.

re
L_
For 50%, /5o% = 2 2k 75. Both A and R are true and R is the correct

F
- 50 sec explanation of A.
ur 76. Correct explanation. Arrhenius equation is

or
For 75%, / 75% - E
f
k = Ae
-E../RT
SO that In k = In A . Hence,
ks
RT
Yo
E
oo

* 4 2it 2 I a
plot of In k\s ~
T
has slope = - R
65. Correct explanation. |A] = LA]Q^'“^^
B

77. Both A and R are correct and R is the correct

66. Correct R. The rate of reaction is represented by k
re

IRCl] [NaOH], i.e., it is a second order reaction. explanation of A.

67. Correct A. Greater the half-life period, slower is 78. Correct explanation. Order of reaction may be
u
ad

the reaction. zero but molecularity is never zero.

Yo
d
Re
in
F

I. General introduction 2. Why in general a reaction does not proceed

and rates of reactions with a uniform rate throughout or why
instantaneous rate is preferred over average
I. For a reaction A -> B, the rate of reaction rate?
dA dB
can be denoted by — or + . State the An.s. This is because the rate of reaction at any lime
dt dt depends upon the concentrations of the
significance of plus and minus signs in this reactants at that time which keep on decreasing
case. with time.

Ans. Rate of reaction is always positive. Minus sign 3. Why boiling of an egg or cooking of rice in an
indicates decrease in the concentration of the open vessel takes more time at a hill station ?
reactant, i.e., d [Aj is -ve so that (-) x (-) = Ans. At a hill station, due to higher altitude,
+ve. Plus sign indicates increase in the atmospheric pressure is low. Hence, water boils
at a lower temperature. At a lower temperature,
concentration of the product with lime during
the rate of reaction decreases. Hence, it takes
the reaction, i.e., d [B] is -i-ve.
longer time.
CHEMICAL KINETICS 4/87

4. A reaction proceeds with a uniform rate Ans. For an elementary reaction, order of reaction
throughout. What do you conclude? or Is should be equal to molecularity and further
there any reaction whose rate does not molecularity should be integral. For the given
decrease with time ? reaction, order of reaction = 1 + 1/2 = 3/2. Since
Ans. Reaction is of zero order. molecularly cannot be fractional, therefore, for
5. When is the rate of reaction equal to specific the given reaction, order is not equal to
reaction rate ? moleculity. Hence, given reaction cannot be an
(Manipiir Board 2012)
elementary reaction.
Ans. Rate = k [A]“ [B]^ etc. 10. The rate expression for the reaction,
Rate = k when either concentration of each
(CH3)3C-C1 + OH" ■¥
reactant is 1 mol or order with respect to
each reactant is zero, i.e., the reaction is of zero (CH3)3C - oh + Cl"
order. is Rate = k [(CH3)3C-CI]. Propose the
mechanism for the reaction.
II. Rate law, order of reaction,

w
rate constant and its units
Ans. The rate of reaction depends on the slowest
step. Hence, we shall have
6. What is the order of reaction whose rate slow

F lo
constant has the same units as the rate of Step 1 : (CH3)3C - Cl ■>

reaction ? (CH3)3C+ + Cr ;

ee
Ans. Zero order. fast
Step 2 : + OH"
7. For a reaction, the graph of the rate of

Fr
reaction agaimst molar concentration of the (CH3)3C - OH
reactant is as shown. What is the order of
IV. Integrated rate equations,
the reaction? (Karnataka Board 2012)
for
half-life, determination of rate law,
ur
rate const, and order
s
t K
ok
Yo
11. Consider the reaction A ^ P. The change
p
in concentration of A with time is shown in
o

(0
(T
eB

the plot.

y.
[A]=t>
r

An.s. Zero order.

ou
ad

8. A reaction is first order in X and second c

o
order in Y.
Y

ns

(z) How is the rate affected on increasing the C

concentration of Y three times ? 8

Re
nd

c
o
(zz) How is the rate affected if the concen o
Time, t=t>
Fi

tration of both X and Y are doubled ?

(CBSE 2020) (z) Predict the order of the reaction
Ans. Rate = k [X] [Y]“ (z*z‘) Derive the expression for the time
If [Y] is increased three times, new rate = k [X] required for the completion of the reaction.
[3 Y]- = 9 /: [X] [Y]^, i.e., increases a times.
If both X and Y are doubled.
(iii) What does the slope of the curve indicate.

Rate = k [2 X] [3 Y]^ = 18 it [X] [Y]^, (CBSE 2019)

i.e.. increases 18 times. Ans. (z) Order = zero (See Fig. 4.5, page 4/25)
[A]
III. Molecularity, (ii) I = — (See page 4/26).
mechanism and difference k

between order and molecularity (iii) Slope of the curve indicates that
concentration of the reactant decreases linearly
9. For a reaction, the rate law is : Rate = A: =
with time and slope = - k (Refer to Fig. 4.5,
[A] [B]^. Can this reaction be an elementary
reaction ? page 4/25).
4/88 ^nadee^% New Course Chemistry CX11)B&X9]

12. A reaction is 50% complete in 2 hours and Ans. The former has lower aclivation energy than
75% complete in 4 hours. What is the order the latter reaction. Lower the activation energy,
of reaction ? faster is the reaction.
Ans. ^ hours. As 75% completion lakes two 17. Can a reaction have zero activation energy ?
half lives, this shows that is independent
of initial concentration. Hence, it is a reaction Ans. If E^, = 0, then according to Arrhenius equation,
of first order. k - = A, i e.. Rate constant =
13. For the reaction :
collision frequency. This means every collision
Pt
results into a chemical reaction which cannot
2NH3 (g) Nj (^) + 3 H2 Rate = k
be true. Hence, 0.
(i) Write the order and moleciilarity of this
reaction. 18. Can a reaction have negative activation
(//) Write the units of/:. (CBSE 2016) energy ?
Ans. (f) Rate = k shows that the rate of reaction is -E../RT
independent of the concentrations of the A11.S. According to Arrhenius eqn., k = Ae

w
reactants. Hence, it is a reaction of zero order. If were negative, the exponential factor can
As two molecules of the reactant are present in be written as t?'’ where x = E^RT. If jc < < 1, we

F lo
the reaction, molecularity is two. have
07) Units of k will be same as that of the rale of
.2 .3
reaction viz. mol L"’ s“‘.
= 1 + a: + — + — +... = I + .r

ee
2! 3!
V. First order reactions,

Fr
pseudo first order reactions (Neglecting higher powers of jc)
and radioactive disintegration
.●.
for = I + E„/RT
& = A{1 + EyRT)./.tr.,/:>> A which
r
14. What are the units of the rate constant of a
pseudo unimolecular reaction ? is impossible. Hence, E^ cannot be negative.
You
s
ook

k' min
19. What is the fraction of molecules having
Ans. k' = k [H2OJ .● energy equal to or greater than activation
■ [H2O] molL-‘
eB

= L mol-* nun energy, ? What is this quantity called ?

Thus, it has the units of a second order reaction. -E./RT
Ans. It is equal to e at temperature T. This
our

15. The rate constant for a .second order reaction

ad

quantity is called ‘Boltzmann factor’.

2-303 b{a~x)
is A: = log 20. What is the value of rate constant at
l{a-b) (b-x)
extremely high temperature (= «>) ? Is this
dY

where a and h arc initial concentrations of

rate constant feasible ? Why or why not ?
Re

the two reactants A and B involved. If one of

Fin

the reactants is present in excess, it becomes Ans. At T = 00

, k = Ae = Ae*^ = A , i.e..
pseudo unimolecular. Explain how ?
Rate constant = collision frequency which is
Ans. Suppose B is in excess so that b » a or x.
not feasible as explained in Q. 16.
Neglecting a and x in comparison to b, the
2-303 a 21. Why equilibrium constant of a reaction does
equation reduces to k b = k' = log not change in the presence of a catalyst ?
t a- X

which is same as for reactions of 1st order. Ans. The catalyst increases the speed of forward
reaction as well as backward reaction lo the
VI. Activation energy, effect of
same extent. Hence, equilibrium is not
temp, on rate (Arrhenius eqn.)
disturbed but is attained quickly.
and effect of catalyst on rate
22. What is the effect of adding catalyst on the
16. The reaction 2 NO (g) + O2 (g) > 2 NO2 (g) free energy change (AG) of a reaction ?
and 2 CO (g) + O2 (g) > 2 CO2 (g) look to Ans. No change in AG occurs in the presence of
be similar. Yet the former Is faster than the
catalyst (as free energy of the reactants and
latter at the same temperature. Explain
products remains the same).
why ?
CHEMICAL KINETICS 4/69

23. The rate of a reaction depends upon the I

temperatureand quantitativelyexpressedas Plot of log ^ vs — will be linear with slope

E
(/) If a graph is plotted between log k and u

1/T, write the expression for the slope of the 2-303 R ●

graph.
(//) Lower the activation energy, faster is the
(«) At different conditions, and E„ are
°2 reaction and hence greater is the rate constant.
the activation energies of the two reactions. Hence, > k2-
If a = 40 J/moI and = 80 J/mol, which
1
V!i. ●'ollision theory
of the two has larger value of the rate
constant ? {CBSE Sample Paper 2019) 24. In some cases, it is found that a large number
E
of colliding molecules have energy more than
a
Ans. (/) In = In A - threshold value, yet the reaction is slow.

w
RT
Why ?
E

F lo
or log k = log A -
(I
Ans. This is because of improper orientation of the
2-303RT colliding molecules at the time of collision.

ee
Very Short Answer

Fr
Short Answer
r«'
● Long Answer

VERY SHORT ANSWER QUESTIONS

for Carrying 1 mark
r
You
I. General introduction Ans. (/) The rate of reaction will be doubled, (ii) No
s
ook

effect on rate.
and rates of reactions
4. Express the rate of the following reaction in
eB

1. Write expression for rate of reaction in terms terms of disappearance of hydrogen in the
of each reactant and product for the reaction reaction
N2+3H2 >2NHy
our
ad

3H2(g) + N2(g) ^2NH3(g)

Ans. ^[N2]_ lf/[H2]_ I ^[NH3] (CBSE 2007)
dt ~ 3 dt ~ 2 dt
1^[H^
dY

2. For the straight reaction, Ans. Rate - -

Re

3 dt
X2 (g) + 2Y2 (g) > 2 XYj (g), write the
Fin

rate equation in terms of disappearance 5. For the reaction

ofYj
N2 (g) + 3 H2 (g) >2 NH3 (g), if
ld[Y^]
Ans. Rate = -
2 dt
= 'A[X,][Y,]2. A [NH3]/A/ = 4 X I0“* mol L“^ s"^ what is the
value of-A[H2]/A/?
Rate of disappearance of
d[Y^] Ans. Rate = -
I A[H^]_ 1 A[NH3]
Y. = 2*[X2l(Y2]- 3 ~ 2 At
dt )
3. The rate law for the reaction :
Ester + H-^ ■¥ Acid -f Alcohol is : or
A[H^]_3A[NH3]
At 2 At

— = it [Ester] [H+f
dt
= -x4xl0-* molL“U-‘
What would be the effect on the rate if 2
(i) concentration of the ester is doubled ?
= 6 X 10“* mol L * s
{ii) concentration of H’*' is doubled ?
4/90 ^>uxdee^'4. New Course Chemistry CXII)BZS[9;

6. Write the names of four factors affecting the Ans. Units of rate = mol L * s '
rate of a reaction. (Raj. Board 2011)
Rate mol L“’ s“*
Ans. (i) Concentration of reactants (n) Temperature Units of/: =
(Hi) Surface area of reactants (zv) Presence of [A]^ [B] (mol L“^)^ (mol L*)^
catalyst. - mol"^ s“*.

II. Rate law, order of reaction, 15. The decomposition reaction of ammonia gas
rate constant and its units on platinum surface has a rate constant =
2'5 X 10"^ moi L“^ s“^ What is the order of
7. Define specific reaction rate or rate constant. the reaction ?
Ans. Specific reaction rate is the rate of reaction Ans. Zero order (on the basis of units of k)
when the molar concentration of each of the
16. The rate constant of a reaction is 3 x 10^
reactants is unity.
mm . What is the order of reaction?
8. Define : Rate of reaction and rate constant.

w
(Assam Board 2012)
(CBSE 2010)
Ans. On the basis of units of rate constant, order = 1.
Ans. Rate of reaction is defined as the change in the

Flo
concentration of any one of the reactants or in. Molecularity,
products per unit time. Rate constant of a mechanism and difference
reaction is equal to the rate of reaction

ee
betwec'i order and molecularity
when concentration of each reactant is taken

Fr
as 1 mol L"'. 17. Define ‘molecularity of a reaction’.
9. What is the difference between Rate law and (Pb. Board 2011)
Law of Mass Action?

for
Ans. See definition on page 4/19.
ur
Ans. Rate law is an experimental law whereas law 18. Give one example of a reaction where order
of mass action is a theoretical law based on the and molecularity are equal.
k s
balanced chemical reaction. Ans. Order and molecularity are same for an
Yo
oo

10. Give an example of a reaction having elementary reaction, i.e., reaction taking place
fractional order. in one step, e.g., for the reaction,
eB

Ans. Decomposition of acetaldehyde (order = 1.5) 2 HI ig) ^H2(s) + h isl

723 K order = molecularity = 2.
r

CH3CHO > CH4+CO, 19. What is the rate determining step of a

ou
ad

1-5
Rate = k (CH3CHO] reaction ? (Chhatisgarh Board 2011)
Y

11. The rate law for the decomposition of N2O5 Ans. The slowest step is the rale determining step.
is : rate = k [N2O5]. What is the significance 20. What is meant by an elementary reaction ?
Re
nd

of k in this equation ? Ans. A reaction which takes place in one step is

Ans. k represents the rate constant or specific called an elementary reaction. For example,
Fi

reaction rate for the reaction and is equal to the reaction between H2 and I2 to from 2 HI is an
rate of reaction when concentration of N2O.; is elementary reaction. Different steps of a
1 mol L"'. complex reaction are also elementary reactions.
12. Define the term ‘order of reaction’ for 21. What is the molecularity of the reaction.
chemical reactions. (CBSE 2008, 2010 ;
1
Pb. Board 2011 ; AP Board 2012) Cl
> 2 ^'2(g>?
Ans. Refer to page 4/12.
Ans. One.
13. Give an example of first order reaction.
(HP Board 2013)
22. For the reaction NO2 + CO > CO2 + NO,
the rate law is : Rate = k [N02l^. Propose
Ans. 2 N-,05 4 NO^ + 0-), Rate = k [N.,05]. the probable mechanism of this reaction,
14. Rate of a reaction is given by the equation :
slow
Rate = k [A]2 [B] Ans. NO2 + NO2 > NO + NO3;
What are the units for the rate and the rate fast
constant for this reaction ? NO3 + CO > CO2 + NO2
CHEMICAL KINETICS 4/91

23. For a reaction : A + H2O B, Rate« A]. 30. Write expression for halMife in case of a
What is Its (i) Molecularity («) Order of reaction between hydrogen and chlorine to
reaction ? form hydrochloric acid gas.
Ans. (/) pseudo-unimolecular (//) order = 1. Ans. Given reaction is a reaction of zero order.

24. State any one condition under which a Hence, /j/2 = [A]q/2 k (page 4/25).
bimolecular reaction may be kinetically of 31. Draw a schematic graph showing how the
first order.
rate of a first order reaction changes with
Ans. A bimolecular reaction may become kinetically change in concentration of the reactant.
of first order if one of the reactants is present Ans. As for a first order reaction, rate molar
in excess.
concentration, graph between Rale vs Molar,
cone, will be a straight line passing through the
rV. Integrated rate equations,
half-life, determinationof rate law, origin. (Similar to Fig. 4.13a, page 4/32).
rate const and order 32. In the reaction aA + products, if
concentration of A is doubled (keeping B

w
25. The reaction A + B ^ C has zero order. What constant) the initial rate becomes four times
is the rate equation ? and if B is doubled (keeping A constant), the

F lo
rate becomes double. What is the rate law
dx
Ans. Rate= — = k[Af^[Bf = k {rate const.) equation and order of reaction?
dt (Manipur Board 2012)

e
Ans. Rate = k [AJ- [BJ, order = 2+1=3

Fre
26. A substance with initial concentration ‘a’
follows zero order kinetics. In how much 33. Give an example of a zero order reaction and
time will the reaction go to completion ? for
write an expression for its velocity constant.
(Jharkhand Board 2011, UP Board 2012)
dx
r
Ans. — = A: or dx = kdf x = kt+ I. hv
dt Ans. H2 + CI2 2 HCI
You
oks

When / = 0, -A = 0. .-. I = 0. or 2 NH3 N2 + 3 H2.

eBo

Hence, x = kt or t = xlk. For completion,

x = a i = a/k Rate constant, -A=-|[A]q-[A]},
ad

27. How is half-life period (/|/2) of a first order

our

34. Write the expression showing the change of

reaction related to its rate constant (A)?
concentration with time in the exponential
(MP Board 2011, J & K Board 2012) form for reactions of first order.

Ans. [A] = [A]q e~ where [A]q is initial

Re

In 2 2-303, ,, 0-693
dY

Ans. f\/2- k concentration, [A] is concentration at time t and

Fin

k is rate constant.
28. How is half-life period related to initial
concentration for a second order reaction ? 35. If half-life period of a reaction is inversely
proportional to initial concentration of the
-I ;
1 reactant, what is the order of reaction ?
Ans. ti/2^ ”● For n = 2, ,l.e..ty2
oc

a Ans. Two. This is because for reaction of nth order.

Thus, half-life period of a second order reaction 1
is inversely proportional to the initial
t
1/2
OC
n-1 (Art. 4.11, page 4/30). Hence,
concentration of the reactant.
fAoJ

29 For a reaction R —> P, half-life (ti/2) is t

1/2
OC when n=2.
observed to be independent of concentration [A] 0
of reactants. What is the order of reaction ? 36. Three-fourth of a first order reaction is
(CBSE 2017) completed in 32 minutes. What is the half-
life period of this reaction ?
0-693
Ans. 1st order t Ans. 16 minutes.
1/2
k J
4/92 ‘Pna.eUefr'a, New Course Chemistry (XII)SSE3MI

V. First on r reactions, 42. The activation energy of a reaction is zero.

Will the rate constant of the reaction depend
pseudo tlrst i der reactions
:::. i radiuact! disintegration upon temperature ? Give reason.
Ans. No. When = 0. A = = A. Thus, k
37. What is pseudo first order reaction? Give
an example of it. HP Board 2011, 2013) becomes independent of T.
Ans. Pseudo first order reaction - Refer to Art. 4.14. 43. The rate of a reaction at 30®C is x, then what
page 4/46. will be the rate of the same reaction at bO^C ?

H-^ Ans. For every 10° rise of temperature, the rate of

Example. CH3COOC2H5 + H2O ●>
lx

CH3COOH + C2H5OH reaction is doubled. Hence, 30° ^ 40°

38. Radioactive disintegration is a first order 4x

4 50° 4 60°, i.e., rate of reaction will
reaction. Explain why. (Raj. Board 2011)
become 8 times.
Ans. The rate of disintegration of a radioactive

w
substance is directly proportional to the amount 44. The plot of In k versus 1/T for a reaction has
of the substance present. Hence, it is a reaction a slope equal to - 4500 K. What is the

F lo
of first order.
activation energy of the reaction ?
Activation energy, effect ox Ans. The plot of In k vs 1/T has slope

ee
tc np. on rate (Arrhenius eqn.) E

Fr
and effect of catalyst on rate R
= - 4500 K

39. In the Arrhenius equation k = A exp (- E/RT), Hence, = 4500 K x 8-314 JK"' mor'
A may be termed as rate constant at
for = 37413 Jmol"^
ur
Ans. Infinite temperature.
= 37-4 kJ mol-’
40. Define activation energy of a reaction.
s
ook

iHP Board 20
VII, Collision theory
Yo

Ans. Refer to page 4/60.

45. What are the two necessary conditions for a
eB

41. Define temperature coefficient of a reaction.

collision between two molecules to be
(Karnataka Board 2012,
effective collision?
Chhatisgarh Board 2012)
our
ad

Ans. Temperature coefficient of a reaction is the ratio Ans. At the time of collision, (/) they should have
of the rate constant of the reaction at 308 K to energy more than threshold value, and (n) they
that at 298 K. Its value lies between 2 and 3.
should have proper orientation.
Y
Re
nd

SHORT ANSWER QUESTIONS Carrying 2 or 3 marks

Fi

1. General introduction II. Rate law, order of reaction,

and rates of reactions rate constant and its units

1. What is meant by reaction rate ? Give its 4. Define the following :

symbolic expression and units for the reaction (/) Order of a reaction (//) Molecularity of a
reaction (iii) Rate law (iV) Law of mass action
CO(g) + N02(g) ^C02(g) + NO(g)
[Art. 4.4]
[Art. 4.3]
5. What is meant by ‘rate constant, k’ of a
2. Show graphically the average and instantaneous reaction ? If the concentration be expressed in
rate of a reaction. (Hr. Board 2011) [Art. 4.3.3] mol units and time in seconds, what would
3. Express the relationship between the rate of be the units for k (/) for a zero order reaction
(h) for a first order reaction,
production of HCl and the rate of disappearance
of H2 in the reaction : H2 + CI2 ^2HC1 (CBSK 2008, Manipur Board 2012)
[Art. 4.4 & Art. 4.5]
[Art. 4.3.4]
CHEMICAL KINETICS 4/93

or What is order of reaction? Write units of the

rate constant k for zero order, first order and
III. Molecularity,
second order reaction.
meclii) tisni and ditTerence
(MP Board 2012)
[Art. 4.4 & 4.5] between . dcr and nudecularity
6. The gas phase decomposition of acetaldehyde, 10. Comment on the statement that molecularity
CH3CHO (g) ^CH4(g) + CO (i-), of the slowest step in a reaction gives the overall
at 680 K is observed to follow the rate molecularity of the reaction. [Art. 4.7]
expression : 11. What is meant by the rate controlling step in a
3/:
Rate = -d fCH3CHO]/^// = k [CH3CHO] reaction ? [Art. 4.7]
If the rate of decomposition is followed by 12. The rate expressions of some reactions are
monitoring the partial pressure of acetaldehyde, given below :
we can express the rate as :
3/2
(a) 2N,Og ^ 4NOt + O2, Rate = k [N2O5]
~^CH3CH0 ^ 1*CH3CHo1 (./;) 2NO, + Ft > 2NO2F,

low
If the pressure is measured in atmospheres and Rate = k [NO2] [F2]
the time in minutes, then
(c)NOt + CO ^C02 + N0, Rate = /;[N02l^
(a) What are the units of the rate of reaction ?
Propose the probable mechanism of each of the
{h) What are the units of the rate constant k ? above reactions. [Art. 4.8]

ee
[Art. 4.3 & 4.5]
13. Nitric oxide, NO, reacts with oxygen to produce
7. Write any four differences between ‘rale of

F
Fr
reaction’ and ‘rate constant’ of a reaction. nitrogen dioxide ;
(CBSE 2011, MP Board 2011, 2NO {g) + 02 (g) ^2N02(g)
Chhatisgarh Board 2011) [Art. 4.6]
for
The rate law for this reaction is :
ur
8. Answer the following questions (Do any two): Rate = i[N0]2[02].
{a) Identify the order of reaction from the Propose a mechanism for the above reaction.
ks
following units of its rate constant :
Yo
[Art. 4.8]
mol“^ s“*.
oo

14, Give any four differences between order of a

{b) The conversion of molecule A to B follows
eB

second order kinetics. If concentration of A is reaction and its molecularity ?

increased to three times, how will it affect the (Chhatisgarh Board 2011,
rate of formation of B ? MP Board 2011, CBSE 2014) [Art. 4.9]
r
ou
ad

(c) Write the expression of integrated rate 15. Explain with suitable example, how the
equation for zero order reaction. (CBSE 2022) molecularity of a reaction is different from the
Y

[(a) 2nd order, page 4/18 (b) Rate = k [A]^. If order of a reaction ?
[A] is increased 3 times. Rate will become
(UP Board 2012) [Art. 4.9]
nd
Re

9 times and so rate of formation will become

9 times (c) Art. 4.10.1] or Define and explain the term ‘molecularity of a
Fi

9. Observe the graph shown in the figure and reaction' with suitable example.
answer the following questions : (Maharashtra Board 20121 [Art. 4.7]

IV. Integrated rate equations,

half-life, determination of rate law,
rate con.st. and order

16. Derive the relationship between rate constant

CO
o and half-life period of a first order reaction.
(Karnataka Board 2012) [Art. 4.10.1]
0 Time 17. Show that for the reactions of first order
(fl) What is the order of reaction ? (/) Half-life period is independent of initial
concentrations.
(b) What is the slope of the curve ?
(c) Write the relationship between k and Iy2 (») Units of rate constant do not depend upon
the units of concentration. [Art. 4.10.2]
(half-life period). (CBSE 2022) [Art. 4.10.2]
4/94 New Course Chemistry (XII)KSE19]

18. Derive the equation for the rate constant of a 24. What do you mean by molecularity and order
first order reaction and show that the time of a reaction ? Explain why their values are
required for the completion of half of the first different in the reaction ;
order reaction is independent of initial CH3COOC2H5 + H2O
concentration.
(Bihar Board 2012, UP Board 2012,
CH3COOH + CoHgOH.
[Art. 4.7 & 4.14]
Assam Board 2012) [Art. 4.10.2]
19. Rate of reaction is the change in concentration 25. Define half-life period and average life. Derive
relation between them. {,] & K Board 2011)
of any one of the reactants or any one of the
products in unit lime [Art. 4.15]
(/) Express the rate of the following reaction in H

terms of reactants and products: 26. For the reaction C12H22O11 + H2O ■>

2 HI -> H, +12 ^6^12^6 ^6^12^6 ’ '^rite

(ii) If the expression for the above reaction is,

w
(/■) Rate of reaction expression (//) rate law
rate = k [HI]^, what is the order of the reaction. equation (Hi) molecularity (/V) order of reaction
(Hi) Are the order and molecularity of the above

Flo
reaction same? Give reason. [Ans. (/) Rate = -
dt
(Kerala Board 2012)
rf[C^Hj206 (glucose)]

ee
id[m_d[^_d\^
(/)-- = +
dt

Fr
2 dt dt dt

d (fructose)]
= -I-
(«)2(/K)No,Art4.7

for
dt
ur
(ii) Rate = k [C|2H220j|] (Hi) Molecularity = 2
20. Derive integrated rate equation for rate constant
(iV) order = 1, Art 4.14]
for a first order reactants. What would be units
s
k
Yo
of the first order rate constant, if the VI. Activation energy, effect of
oo

concentration is expressed in moles per litre and temp, on rate (Arrhenius eqn.)
time in seconds ?
eB

and effect of catalyst on rate

(Pb. Board 2011, Hr. Board 2011) [Art. 4.10.2]
21. (a) How does the rate of reaction depend upon 27. Define the term Activation energy. Why
r

the concentration of reactants? different reactions proceed at different speeds ?

ou
ad

(J & K Board 2012) (J & K Board 2012) [Art. 4.16]

(b) What is rate constant? Write ‘two or Define energy of activation. Draw diagram of
Y

applications’ of rate law. energy profile to show the influence of a

positive catalyst on the energy of activation of
Re
nd

(Maharashtra Board 2012)

a reaction. (Karnataka Board 2012)
22. Define half-life period. How is it related to the
Fi

order of a reaction? (Karnataka Board 2012) [Art. 4.16 & 4.18]

[Art. 4.11] 28. The reaction 2H2(g) + 02(g) > 2H20(g) is
feasible. How is that hydrogen and oxygen
V. First order reactions, mixture allowed to stand at room temperature
pseudo first order reactions shows no formation of water at all ?

and radioactive disintegration [Art. 4.16]

23. What are Pseudounimolecular reactions ?

Hint. Because of high activation energy at room
temperature.
Explain with the help of a suitable example.
29. State the role of activated complex in a reaction
(J & K Board 2011) [Art. 4.14]
and state its relation with activation energy.
or H'*' catalysed hydrolysis of ester [Art. 4.16]

(CH3CO2C2H5 -I- H2O Hint. Activation energy = Energy of activated

complex — Average energy of reactants
CH3CO2H -t- C2H5OH)
30. What is the effect of temperature on the rate of
is a pseudo-uniraolecular reaction. Explain.
reaction ? Explain giving reasons. [Art. 4.17]
(West Bengal Board 2012)
CHEMICAL KINETICS 4/95

31. Write Arrhenius equation showing the effect increase in temperature may be enough to
of temperature on the reaction rate. What do double the rate of reaction.
different symbols signify? How does it help in [Art. 4.3.5 & 4.17]
the calculation of activation energy of a
37. Explain the effect of (a) Concentration of the
reaction ? [Art. 4.17]
reactants {b) Presence of a catalyst on the rate
32. What is activation energy ? How is the rate of a chemical reaction. (Hr. Board 2011)
constant of a reaction related to its activation
[Art. 4.3.5 & 4.18]
energy ? (CBSE 2007) [Art. 4.16 & 4.17]
38. Define the following terms : (/) Collision
33. How is rate consttint of a reaction affected by
frequency
temperature ? Write the name of this relation.
How will you calculate activation energy of a (/7) Half-life period of a reaction.

ow
reaction with the help of this relation 7 [Hr. Board 2011) [(/) Art. 4.19 («) Art. 4.11]
[Art. 4.17] 39. Define the following terms : (/) Order of
reaction
34. Briefly explain the effect of adding catalyst on
the rate of a reaction. [Art. 4.18] (ii) Threshold energy. (Bihar Board 2011)

e
[(i) Art. 4.4 {«) Art. 4.19]
VII. Collision theory

re
40. Define the following terms ;

rFl
35. Write a short note on 'collision theory of
(/) Activation energy (i/) Temperature

F
reaction rates’.
coefficient (m) Half-life period of a reaction.
Vlll. Miscellaneous (.\ssain Board 2012)

or
ou
36. Explain : (i) One gram of pulverised wood [(/) Art. 4.16 (h) Art. 4.17 (/») Art. 4.11]
bums faster than a one gram piece of wood.
ksf
41. Write short notes on :

{ii) An increase of 10 K in temperature rarely (0 Activation energy {ii) Arrhenius equation.

oo
doubles the kinetic energy of particles but this (MP Board 2013) [(/) Art. 4.16 (ii) Art. 4.17]
Y
B

LONG ANSWER QUESTIONS Carrying 5 or more marks

re

1. {a) Derive the general form of the expression 2. (a) Explain the following terms : (/) Rate of a
oYu

for the half- life of first order reaction. reaction, (h) Activation energy of a reaction.
ad

(b) The decomposition of NH3 on platinum (b) The decomposition of phosphine, PH3,
surface is a zero order reaction. What are the
proceeds according to the following equation :
d

rates of production of N2 and H, if /: = 2-5 x

lO^mol-'Ls-l? (CBSE 2008) 4 PH3 (g) > P4 (g) + 6 H2 (g)
in
Re

It is found that the reaction follows the

A[N2] following rate equation : Rate = k LPH3]
[(a) Art 4.10.2 (b)
F

At
= 2-5 X 10-^ mol L-*
The half-life of PH3 is 37-9 s at 120”C.
(0 How much time is required for 3/4th of PH3
A[H2] = 7*5xir^mol L-^s-*] to decompose, {ii} What fractional of the
At original sample of PH3 remains behind after
I minute ? (CBSE 2010)
or {a) List the factors on which the rate of a
chemical reaction depends. [Ans. (a) (i) Art 4.3.1. (ii) Art. 4.16 (b)
{b) The half-life for the decay of radioactive 2-303 a
(i) / = log = 75-82 s
is 5730 years. An archaeological artifact 0-693// 3
1/2 a —a
containing wood has only 80% of the 4
activity as found in living trees. Calculate the
age of the artifact. (ii) log
[A]q _ kt _ (0-693/37-9)(60) = 0-4763

(CBSE 2008) [(a) Art. 4.3.5 (b) Art. 4.15, [A] " 2-303 2-303

2-303 100 [AyiA] - Antilog 0-4763 = 1-2994

t = log = 1845 years ] or [A]/[A]o = 1/1-2994 = 0-77 = 77%]
(0-693/5730) 80
4/96 New Course Chemistry (Xn)EZsZ91

3. (iz) Illustrate graphically the effect of catalyst

t/s 0 30 60
on activation energy.
-1
(b) Catalysts have no effect on the equilibrium [CH3COOCH3]/molL 0-60 0-30 0-15

constant. Why ?
(/) Show that it follows pseudo first order
(c) The decomposition of A into product has reaction as the concentration of water remains
value of k as 4-5 x 10^ s“' at lO^C and constant.
activation energy is 60 kJ mol"’. Calculate the (//) Calculate the average rate of reaction
temperature at which the value of k will be

w
between the time interval 30 to 60 seconds.
1-5 X 10*^ s-’.
tCBSE 201S)
[(fl) Art. 4.18 (/») Art. 4.18 (c) 24“C]
2-303. FAl 0
4. (n) A reaction is second order in A and first [Hint, (i) For 1st order, k = log
order in B t [A1

e
(/) Write the differential rate equation Here [A]q = 0-60 mol L“^. At / = 30 s,

ro
re
[A] = 0-30. At / = 60 s, [A] = 0-15. Calculate k
(ii) How is the rate affected on increasing the in each case. It comes out to be constant
concentration of A three times?
(= 0-231 s-i).

F
(Hi) How is the rate affected when the

Fl
concentration of both A and B are doubled? 0-30-0-15
(ii) Average rate =

u
(b) A first order reaction lakes 40 minutes for '2-^1 60-30
30% decomposition. Calculate t 1/2 for this
= 5 X 10-3 mol L-l s-‘]

sr
reaction.
or (a) For a reaction A + B ■»P, the rate is given

ko
(Given log 1428 = 0-1548)
o
(CBSE 2013) by Rale = k [A] [B]^
[Ans. (i) -
d\

dt
= k[kf [B]
of
(0 How is the rate of reaction affected if the
concentration of B is doubled ?
o
(//) Rate increases 8 times (Hi) 77-87 min]
Y
(ii) What is the overall order of reaction if A is
or (a) For a first order reaction show that time present in large excess ?
erB

required for 99% completion is twice the time (b) A first order reaction takes 30 minutes for
uY

required for the completion of 90% of the 50% completion. Calculate the time required
reaction.
for 90%. Completion of this reaction (log 2 =
(b) Rate constant 'k' of a reaction varies with 0-310). (CBSE 2015)
temperature T’ according to the equation : [Ans. (a) (/) 4 times (ii) 2
ad
do

E 1 0-693 . -1
log k = log A - (b)k = nun
in

30
2-303 Rl^T )
where is tlte activation energy. When a graph 2-303, 100
Re

^90% “ —k Jog 100-90

= 99-69 min]
F

is plotted for log k vs 1/T, a straight line with a

.slope of - 4250 K is obtained. Calculate 7. (a) Draw the plot of In k vs 1/T for a chemical
for the reaction (R = 8-314 JK"’ mol"’) reaction. What does the intercept represent ?
(CBSE 2013)
What is the relation between slope and E^ ?
(b) A first order reaction takes 30 minutes for
[(fl) Similar to Problem 4, page 4/51
20% decomposition. Calculate fj/2 (log 2 =
(b) Similar to Problem 5, page 4/66] 0-3010). (CBSE 2019)
5. (a) Discuss the difference between order and
E
molecularity of a chemical reaction. [Ans. (a) In ^ = In A -
a
. Plot is linear as
RT
(h) Discuss the concept of activation energy in
a chemical reaction and explain the relation shown in Fig. 4JJ0, page 4/63. Intercept = In A,
between reaction rale and temperature. Slope = - Eg/R.
(West Bengal Board 2013) 2-303 100
(b)k = log = 7-4 X 10"3 miir^
[(a) Art. 4.9 (b) Art. 4.16, 4.17] 30 100-20

6. For the hydrolysis of methyl acetate in aqueous 0-693

solution the following results were obtained : t
V2 - —;— = 93-65 min]
k
CHEMICAL KINETICS 4/97

CASE-BASED VERY SHORT/SHORT QUESTIONS

CASE 1. We have maximum number of reactions CASE 2. For a reaction of the type oA
of first order. The question arises how we test Products, if n is the order of the reaction, the rale law
experimentally that it is a reaction of first order. The general expression is written as
integrated expression for the reaction of first order d[A]
Rate - - = A(A]-
dt
,
k =
2-303 a
log is modified based on tlie nature
For reactions of zero order, n = 0. Likewise, for
t a-x)
reactions of 1st, 2nd, 3rd order etc. n = 1. 2, 3 etc. The
of reaction. For example, in the decomposition of N2O5 expression thus written are the differential equations.
1 For many reasons, these are not convenient equations
in CCI4 solution. No05^N204(2N02) + -0.> , for solving problems. We integrate these equations to
2 ~)
get suitable expressions which make the task of solving

w
O2 gas is collected at different times and the formula problems easier. For reactions of different orders
for k is modified accordingly. Further, there are many (0, 1, 2), we get expression for the rate constant k in
terms of time and the initial concentration of the

F lo
reactions which are not truly of first order but under
certain conditions, they become reactions of first order. reactant. Using the experimental data, we can confirm
These lu-e called pseudo-first order reactions, e.g. acid- the order of reaction. Further, from these expressions,

ee
catalysed hydrolysis of ethyl acetate or inversion of
it is useful to know the half-life period of a reaction,

Fr
i.e., the time in which half of the substance is reacted.
cane-sugar (sucrose). Radioactive disintegration is also
A general expression is obtained for half-life period in
a reaction of first order and the expressions for rate terms of initial concentration of the reactant and order
constant (here called disintegration constant or decay
for
of reaction. Using these expressions, the experimental
ur
constant) or for the half-life or average life are same as data of any reaction can be used to determine the rale
for other reactions of first order. law, rate constant and order of reaction.
s
ok

Based on the above paragraph, answer Based on the above paragraph, answer
Yo
questions no. 1 to 4 : questions no. 5 to 8 :
o

1. For a zero order reaction, calculate the ratio of the

eB

5. For the decomposition of N2O5 in CCI4 solution,

3
times taken for — th (75%) of the reaction to if V, represents volumes of O2 gas at different
times and represents the final volume of O2
r

complete and half (50%) of the reaction to collected when decomposition is complete, what
ou
ad

complete. expression will be written for k in terms of

2. What is the order of reaction if a straight line plot and V^.
Y

is obtained when half-life period (/1/2) of a 6. How does hydrolysis of ethyl acetate in acidic
medium differ from that in the alkaline medium
Re

reaction is plotted against 1/[Aq] where [A]q is

nd

the initial concentration of the reactant ? from the point of view of chemical kinetics ?
Fi

7. One fourth of a reaction is completed in

3. On doubling the concentration of the reactant,
32 minutes. What is the half life period of the
half-life period of the reaction is also doubled. reaction ?
What is the order of reaction ?
8. Half-life period of a radioactive element is
4. For the reaction 2A + B C + D, you are given 100 seconds. How much time will it take to lose
the following data its activity by 90% ?
●1 -1
Experi [AJ/moI L- lB]/mol L Initial rate of
CASE 3. According to collision theory, a reaction
ment iurniation of
takes place because the reactant molecules collide with
D/mul min”*
each other. But every collision is not an effective
I O.I 0.1 6.0 X 10”^ collision, i.e., does not result into a chemical reaction.

0.3 7.2 X 10-2

For a collision to be effective, the colliding molecules
II 0.2
must have energy more than a particular value. This
III 0.3 0.4 2.88 X 10"'
value is called ‘threshold energy’. Thus, reactant
IV 0.4 0.1 2.4 X 10-2 molecules have to absorb a certain amount of energy to
What is the overall order of reaction ? reach the threshold value. This extra energy absorbed
4/98 ‘Pnadee^ '<t New Course Chemistry (XII) PZgnin

is called ‘activation energy’. This quantity is important 10. Which of the following property/properties
in deciding the effect of temperature on the rate of changes/change with temperature and why ?
reaction. Quantitatively, the studies were made by AG, AH, E^,. K
Arrhenius. He put forward an equation giving the effect
11. The rate constant of a particular reaction doubles
of temperature on the rate constant of a reaction. This
when temperature changes from 27"C to 37°C.
equation is called ‘Arrhenius equation’. It is generally
observed that for every 10® rise in temperature, the rate The energy of activation of the reaction is
-1 -1
of reaction is nearly doubled. Besides temperature, («) 43-6 kJ mol (b) 13-6 kJ mol
another factor affecting the rate of reaction is the (c) 63-6 kJ mofr’ (d) 73-6 kJ mol
-1

presence of a catalyst. A catalyst lowers the activation 12. For a particular reaction, the rate constant (/:) were
energy barrier and hence the rate becomes faster.
found at different temperatures (T). A graph was
Based on the above paragraph, answer plotted between log k vs 1/T. The slope of the
questions no. 9 to 12 :
linear plot was found to be - 6670 K. The
9. The rate constant of a first order reaction is 10~^
activation energy of the reaction is

w
min"^ at 27°C. The temperature coefficient of this (a) 117-7 kJ mol
-1
(b) 127-7 kJ mol
-1
reaction is 2. What will be the rate constant of
-j -1
(c) 137-7 kJ mol id) 197-7 kJ mol

F lo
this reaction at 17°C ?

ANSWERS

ee
Fr
1. For a zero order reaction, V
2-.303
5. k = — log — (Refer to page 4/43)
[A,] ^ 3 [Aol t ® V

for
k and [
1/2 -
Ik 3/4-4 ^
ur
6. In acid medium, hydrolysis of ethyl acetate is a
(Refer to point 3 of Supplement Your Knowledge,
reaction of first order whereas in alkaline medium,
page 4/26)
s
it is of second order.
ook
Yo
t
3/4 liioj X
2k
1=1.5 7. 77 min. Refer to Solution to Problem 14, page
I
1/2 4 k [Aq] 2 4/54.
eB

1
2. Second order because /1/2 cc 8. 332-3 seconds. Refer to Solution to Sample
»I-I
[Aq] Problem 2, page 4/59.
r
ad
ou

1 9. Suppose rate constant at 17°C = k

Of ^1/2
ec
only when n = 2
tAo] Then as rate is doubled for 10®C rise in
temperature, rate at 27®C = 2 k - 10“^ min"*
Y

3. Zero order because for a zero order reaction

(Given). Hence,
[Aq] .
Re
nd

'1/2
OC

2k
● i e., ti/2« [A]q
min * = 5 X 10^ min *
Fi

4. From I and IV, [B] is kept constant, [A] is made 4 2

times, rate also becomes 4 times, therefore, order
w.r.t. [A] = 1
10. Addition of catalyst lowers the activation energy
barrier. Hence, activation energy is lowered.
From II and III, [A] is kept constant, [B] is
doubled, rate becomes 4 times, therefore, order 11. Refer to Solution to Problem 2, page 4/65.
w.r.t. B = 2 12. Refer to Solution to Problem 5, page 4/66.
Hence, overall order = 3

CASE-BASED MCQs AND ASSERTION-REASON QUESTIONS

CASE 1. From experimental studies, it has been may be completed in one step. The reactions taking
found that the rate of reaction may not depend upon place in one step are called ‘elementary reactions’. The
the stoichiometric coefficients of the reactants in the reactions which do not take place in one step are called
balanced chemical equation. This means that a balanced ‘complex reactions.’ The different steps in which the
chemical equation docs not give a true picture of how complex reaction takes place is called the ‘mechanism
reaction takes place. Thus, it is very rare that the reaction of the reaction’. Each step of the mechanism of the
CHEMICAL KINETICS 4/99

reaction is an elementary reaction. According to {b) Both Assertion (A) and Reason (R) are true
‘collision theory*, in order that the reaction may take but reason (R) is not the correct explanation
place, the atoms, ions or molecules of the reactants must of Assertion (A),
come together simultaneously and collide with one (c) Assertion (A) is true but Reason (R) is false.
another. The number of aton>s, ions or molecules that
id) Assertion (A) is false but Reason (R) is true.
must collide with one another simultaneously so as to
result into a chemical reaction is called the ‘molecularity 3. Assertion. Reactions of higher order (> 3) are
of the reaction’. Further, the different elementary rare.

reactions of a complex reaction do not take place at the Reason. Larger number of collisions take place
same rate. The overall rate of the reaction depends upon due to large number of molecules involved.
the ‘slowest .step’. Hence, the slowest step is called the
4. Assertion. For the reaction (g) + Bt2 (g) — 4

‘rate determining step’. A .series of step reactions or 2 HBr (g), molecularity is 2 but order is not 2.

ow
elementary reactions proposed to account for the overall
reaction is called the mechanism of the reaction. Reason. The rate of reaction does not depend
upon the concentration of Br-,.
Based on the above paragraph, answer
questions no. 1 to 4 : CASE 2. Besides the feasibility and extent of
1. The following mechanism has been proposed for reaction, it is equally important to know the rate at which

e
the reaction takes place, the factors controlling the rate

re
rFl
the reaction of NO with Br2 to fonn NOBr
of a chemical reaction. For example, which parameters

F
NO (^) + Bf2 (g) ± NOBr2 (g) determine as to how rapidly food gets spoiled ? How to
NOBr2 (g) + NO (g) ^ 2 NOBr (g) design the rapidly setting material for dental filling ?
What controls the rate at which fuel burns in an auto

r
If the second step is the rate determining step, the
ou
fo
engine ? Thermodynamics predicts only the feasibility
order of the reaction with-respect to NO (g) is
of a reaction. It can only tell diamond can change into
ks
(a) 3 ib) 2
graphite but in reality the conversion is so slow that the
(c) 1 (d) 0 change is not perceptible at all. Chemical kinetics is
oo

2. The reaction of hydrogen and iodine monochlo the branch of chemistry which helps to study the rates
Y

of reactions, conditions by which the reaction rates can

eB

ride is given as
be altered and effect of factors like temperature, pressure
H -t-2ICl
2(«) (g) ^2 + *2(s) and catalyst on the rate of reaction. There are two ways
ur

This reaction is of first order with respect to to express the rale of reaction viz average rate over the
H2{^) and ICl(^y The following mechanisms time or instantaneous rate (at a particular instant of
ad
Yo

were proposed: time). The rate of reaction can be expressed in teniis of

Mechanism A: any reactant or product. As rate of reaction should have
d

the same value in terms of any reactant or product, the

H2(,) + 2IC1 (g)
Re

change in concentration per unit time has to be divided

in

Mechanism B: by its stoichiometric coefficient in the balanced

F

H + ICI chemical equation.

2(5) (5) >HCI(,, + HI (5),; show
HI + IC1
Based on the above paragraph, answer
(5) (5) > HCl,^j -1- l2(g) ; fast questions no. 5 to 8 :
Which of the
above mechanism(s) can be 5. Rate of reaction is
consistent with the given information about the (a) always positive
reaction?
(b) positive or negative depending upon whether
(a) A only {b) B only we express in terms of reactant or product
(c) A and B both id) Neither A nor B (c) dependent upon the stoichiometric coefficient
of the reactant or product
Choose the correct option out of the four options (d) constant throughout the reaction
given below :
6. The reaction 2 N20g (g) ^ - 4 NO2 (g) + O2 (g)
(a) Both Assertion (A) and Reason (R) are true
was studied in a closed vessel. It was found that
and Reason (R) is the correct explanation of
the concentration of NO2 increa.ses by 2 0 x 10"^
Assertion (A). mol L"* in five seconds. The rate of reaction is
4/100 ^>uxdee/t’A New Course Chemistry CXlI)l!ZsZSI

(tj)4 0x 10-* molL-l s-> (c) Assertion (A) is true but Reason (R) is false.
(h) 2x10-3 mol L-‘ s -1 (d) Assertion (A) is false but Reason (R) is true.
(c) 10“3 mol L“* s“* 7. Assertion. For the reaction A -> 2 B, rate of
reaction in terms o*‘ B is double than that in terms
(d) 10^ mol L-' s-'
of A.
Choose the correct option out of the four options
Reason. In the reaction A —> 2 B, if 1 mole of A
given below:
reacts, 2 moles of B are produced in the same time.
(a) Both Assertion (A) and Reason (R) are true
8. Assertion. Rate of reaction changes with
and Reason (R) is the correct explanation of
temperature but rate constant does not.
Assertion (A).
Reason. Rate of reaction depends upon the
(h) Both Assertion (A) and Reason (R) are true
concentrations of the reactants but rate constant
but reason fR) is not the correct explanation
does not depend upon the concentrations of the
of Assertion (A). reactants.

w
ANSWERS

F lo
l.(b) 2.(0) 3.(c) 4.(c) 5.(0) 6.(c) 7. (d) 8. (d)

ee
HINTS/EXPLANATIONS For Difficult Questions

Fr
1. Rate = k [NOBr2] [NO] ...(0
for
4. Correct R. For the reaction H2 (g) + Br2 (g)
ur
[NOBrJ » 2 HBr (g), rate = k [H2] Hence,
From step I, AT eq
~[NO][Br3] 1
s
order = 1 —.
ook

2
Yo
or
[NOBr2] = [NO] [Br2]
Substituting in eqn. (i), we get 5. Rate of reaction is always positive. In terms of
eB

Rate = k fNO]^ [Br2] = k' [NO]^ [Br2] product also (-) x (-) = + ve. Refer to box on
Hence, order with respect to NO is 2. page 4/3.
r

2. As the given reaction is first order with respect to 6. Refer to Solution to Problem 2, page 4/7.
ad
ou

H2 and ICl, the slow step should involve 7. Correct A. Rate of reaction is same whether we
1 molecule of H2 and 1 molecule of ICl. Hence,
express it in terms of a reactant or in terms of a
Y

B is the correct mechanism.

product.
3. Correct R. Higher order > 3 for reactions is rare
Re
nd

8. Correct A. Both rate of reaction as well as rate

because there is low probability of simultaneous
collision of all the reacting species. constant change with change of temperature.
Fi
CHEMICAL KINETICS 4/101

1 WITH
ANSWERS

;&>■!
1|
I 9>

NCERT INTEXT UNSOLVED QUESTIONS & PROBLEMS

w
Q. 4.1. For the reaction R ■> P, the concentration of a reactant changes from 0*03 M to 0-02 M in 25
minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Flo
Ans. Average rate = -
AIRJ [R]2-[R|]_ 0-02 M-0-030 M -001 M
= 4 X 10“^ M min ^
At 25 min 25 min
ti fj

ee
-001 M
= 6-66x10'^ Ms-‘

Fr
or
25x60 s

Q. 4.2. In a reaction, 2 A ■> Products, the concentration of A decreases from 0*5 mol L * to 0*4 mol L * in
10 minutes. Calculate the rate during this interval.
for
ur
Ans. Average rate = - \_ A [A] _ _J_ [A]2-[A,J _ _\_ 0-4 M-0-5 M 1 -0-1 M
= 5 X 10-^ M min-'.
2 At ~
s
2 ^2 ^1 2 10 min 2 10 min
k
Yo
oo

Q. 4.3. For a reaction, A + B ^ Product, the rate law is given by : r = A: [A]'^^ [B]^
eB

What is the order of the reaction ?

Ans. Order of reaction = — + 2 = 2 — or 2’5.

r

2 2
ou
ad

Q. 4.4. The conversion of the molecules X to Y follows second order kinetics. If the concentration X is
increased to three times, how will it affect the rate of formation of Y ?
Y

Ans. For the reaction, X ■> Y, as it follows second order kinetics, the rate law equation will be
Re
nd

Rate = k [X]^ = ka'^ {if [X] = a mol L"'}

If concentration of X is increased three times, now, [Xj = 3 a mol L"'
Fi

Rate = /:(3a)2=9irfl2
Thus, the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.
Q. 4.5. A first order reaction has a rate constant 1*15 x 10"^ s“^ How long will 5 g of this reactant take to
reduce to 3 g ?
Ans. Here, [A]q= 5 g, [A] = 3 g, /t = 1-15 x 10"^ s"'. As the reaction is of 1st order,
2-303, [A] 0 2-303
t
log
[A]
_
1-15x10-^ s"'
log ^
3g
= 2-00 X10^ (log 1-667) s
= 2-0 X 10^ X 0-2219 s = 443-8 s = 444 s.
Q. 4.6. Time required to decompose SO2CI2 to half of its initial amount is 60 minutes. If the decomposition
is a first order reaction, calculate the rate constant of the reaction.
0-693 0-693
Ans. For a first order reaction, k = = 1*155x10 2 min '
60 min
h/2
4/102 New Course Chemistry (XlI)ESSia]

0-693
or = 1-925 X10^ s-'-
60 x 60s

Q. 4.7. What will be the effect of temperature on rate constant ?

Ans. Rate constant of a reaction is nearly doubled with rise in temperature by 10°. The exact dependence of
-E./RT
the rate constant on temperature is given by Arrhenius equation, k = Ae where A is called

frequency factor and is the activation energy of the reaction.

Q. 4.8. The rate of the chemical reaction doubles for an increase of 10 K from 298 K. Calculate E^.
Ans. Refer to Solved Problem 2, page 4/65.
Q. 4.9. The activation energy for the reaction, 2 HI (g) » (g) + I2 (g), is 209-5 kj mol"* at 581 K.
Calculate the fraction of molecules of reactants having energy equal to or greater than activation
energy.

n
Ans. Fraction of molecules having energy equal to or greater than activation energy = x = = ^-E./RT
N

w
E E
In -t = - or log x = -
RT 2-303 RT

F lo
209-5 xlO^Jmor*
or
log =- = -18-8323
2-303x8-314 JK-‘mol-lx58I K

e
-19

Fre
a: = Antilog (- 18-8323 ) = Antilog 19-1677 = 1-471 x 10

NCERT EXERCISE for

Q. 4.1. From the rate expressions for the following reactions, determine their order of reaction and the
r
dimensions of the rate constants :
You

^ NjO (g) + NO2 (g); Rate = k [NO]^

oks

(/) 3 NO (5)
eBo

(it) H2O2 (ag) ■¥ 3 l~(aq) -f 2 > 2H2O (/) +15 ; Rate = k [H2O2I [I"]
3/2
(Hi) CH3CHO (g) —> CH4 (g) + CO (g); Rate = k [CH3CHO]
(iv) C2H5CI (g)- > C2H4 (g) + HCl (g); Rate = k [C2H5CI].
ad
our

Rate mol L ’ s ’
Ans. (0 Order = 2, Dimen.sions of /: = = L mol * s ^
[NO]" (mol L“1)2
{//) Order = 2, Dimensions of /: = Same as in (i)
dY
Re

3 Rate mol L“*s“*

= mol“*^^s ^
Fin

(///) Order — —, Dimensions of k =

[CH3CH0]3/2 (mol L"‘)3/2
Rate mol s ’
(/v) Order = 1, Dimensions of k = = s"'.
[C2H5CI] mol L ’
Q. 4.2. For the reaction 2 A B > A2B, rate = k [A] [Bj^ with k = 2-0 x 10^ mol"^ 8"^ Calculate the
initial rate of the reaction when [A] = O-I mol L"* and [B] = 0-2 mol L"*. Calculate the rate of
reaction after [A] is reduced to 0-06 mol L"^.
Ans. Initial rale = [A] [Bp=(2-0 x KT* moF^ s'") (0-1 mol L^*) (0-2 mol L ')-= 8 x lO"® mol L“'
When [A] is reduced from 0-10 mol L“‘ to 0-06 mol L"', i.e., 0-04 mol L“^ of A has reacted, B reacted
1
xO-04 mol L-i = 0-02 mol L"'
2

Hence, new [BJ = 0-2 - 0-02 = 0-18 mol L"'.

Now, rate = (2-0 x 10“^ mol-^ L" s"') (0-06 mol L"') (0-18 mol L~^)- = 3-89 x 10"^ mol s‘*.
CHEMICAL KINETICS 4/103

Q. 4.3. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of
production of N2 and if = 2*5 x 10^ mol”^ L s“^ ?
Ans. Sample Problem 2, Page 4/27.
Q. 4.4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction
rate is given by Rate = k pressure is measured in bar and time in minutes,
then what are the units of the rate and rate constants ?
Ans. In terms of pressures, units of rate = bar min“^
Rate bar min *
Units of ^ = = bar ^^^in *
[P,
i"CH30CH3
F bar^/2
Q. 4.5. Mention the factors that effect the rate of a chemical reaction.
Ans. Refer to Art. 4.3.4.

w
Q. 4.6. A reaction Is second order with respect to a reactant. How is the rate of reaction affected if the
concentration of the reactant is (/) doubled (it) reduced to 1/2 ? (CBSE2009, 2012)

F lo
Ans. Rate = k [Al“ = k a~.
If [Aj =2 a. rate = k{2 a)^ = A k = A times.

ee
1 a 1
= —ka^ = — th

Fr
If [A] = -fl. rate = k
2 2 4 4

Q. 4.7. What is the effect of temperature on the rate constant of a reaction ? How can this temperature
effect on the rate constant be represented quantitatively ?
for
ur
Ans. The rate constant of a reaction increases with increase of temperature and becomes nearly double for
every 10° irse of temperature. The effect can be represented quantitatively by Arrhenius equation,
s
ok
Yo
o

where represents the activation energy of the reaction and A represents the frequency factor.
eB

Q. 4.8. In a pseudo first order hydrolysis of an ester in water, the following results were obtained :
t/s 0 30 60 90
r

[Ester]/moI L"’ 0-55 0-31 0-17

ou

0-085
ad

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds,
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Y

Ans. (/) Average rate during the interval 30-60 sec.

Re
nd

C^-Ci _0-3l-0-17 _014 mol L-' s"' = 4-67 x 10“^ mol L"' s"‘
Fi

^2 h 60-30 "lo”

iU) k' = 2-303,log [Aq] .

111 which [A()] = 0-55 M
t [A]

2-303. 0-55
t = 30 sec, k' = log = l-9Ixl0“-s-'
30 s 0-31

2-303. 0-55
/ = 60 sec. ic' = log = l-96xl0“2s-l
60 j 0-17

2-303 0-55
t = 90 sec, k' = log = 2-07x10-2$''
90 J 0-085

1-91+ 1-96+2-07
Average k' = xlO-2 = 1-98 X 10*2s-'.
3
4/104 New Course Chemistry (XII) DE

Q. 4.9. A reaction is first order in A and second order in B.

(0 Write differential rate equation.
(«) How is the rate affected on increasing the concentration of B three times ?
(Hi) How is the rate affected when concentration of both A and B is doubled ?

Ans. (/) — = it [A] [B]2 .

dt

(ii) Rate = k ab^

If [B] is tripled, Rate = ka (3 b)^ = 9 k ab^ = 9 times.
(Hi) If both [A] and [B] are doubled, Rate = k(2a) (2 b)^ = Sk ab^ = 8 times.
Q. 4.10. Same as Sample Problem 2, page 4/37.

ow
Q. 4.11. Same as Sample Problem 1, page 4/36.
Q. 4.12. The reaction between A and B is first order w.r.t A and zero order w.r.t B. Fill in the blanks in the
following table :
Experiment [A]/moI L
-1
[B]/mol L"' Initial rate/mol L"' mm

e
I 0.1 0.1 2.0 X

Fl
re
H 0.2 4.0 X 10-2

F
HI 0.4 0.4
IV 0.2 2.0 X 10-2
ur
or
Ans. The rate expression will be : Rate = /: [A]^ [B]® = k [A]
For expt. I, 2-0 x 1(T2 = mol min"^ = k (0-1 M) or k = 0*2 min‘^ sf
For expt. II, 4 0 x lQ-2 mol L-' min-* = 0*2 min“* [A] or [A] = 0*2 mol L-'
k
Yo
For expt. Ill, Rate = (0-2 min"*) (04 mol L-*) = 0*08 mol L“' min-^
oo

For expt. IV, 2 0 x 10“2 mol L“* min-* - 0-2 min-* [A] or [A] = 0*1 mol L-^
B

Q. 4.13. Calculate the half life of a first order reaction from their rate constants given below :
(a) 200 s-2 (b) 2 min-^ (c) 4 year"^
re

0-693
u

Ans. Half-life period of a first order reaction is given by ?|/2 = —■

ad

Jc
Yo

0-693
(a) /,/2 = 200 S-* = 0346X10-2 s = 3-46 X ir^ s
d
Re
in

0-693
= 0*346 min
(b) tj/2 - 2 min *
F

0-693
(c) tj/2 - -1
= 0*173 year
4 year
Q. 4.14. The half-life for radioactive decay of is 5730 y. An archaelogical artifact contained wood that
had only 80% of the found in living tree. Estimate the age of the sample.
Ans. Radioactive decay follows first order kinetics.
0-693 0-693 -1
Decay constant (k) = yr
5730
hn

?:?2?log^^ =
2-303 100
t = log
k [A] (0-693/5730 yr-*) 80
2-303x5730
X 0-0969 yr = 1845 years
0-693
 CHEMICAL KINETICS 4/105

Q. 4.15. Same as Sample Problem 1, page 4/33.

Q. 4.16. Same as Sample Problem 1, page 4/31.
Q. 4.17. During nuclear explosion, one of the products is ^Sr with half life of 28-1 years. If 1 pg of ^Sr was
absorbed in bones of a newly bom baby instead of calcium, how much of it will remain after
10 years and 60 years if it is not lost metabolically ?
Ans. As radioactive disintegrations follow first order kinetics,
0-693 0-693
Decay constant of ^Sr (k) = = 2-466x10-2 y-l
h/2 28-1 y
To calculate the amount left after 10 years,
a = 1 pg, / = 10 years, k = 2-466 x 10"2 y~^, (a-x) = l

w
,
k =
2-303 a
log
t a-x

2-303 1

Flo
2-466x10-2 = log
10 (a-x) or log (a-jc) = -0-1071

e
re
or
(a - x) = Antilog = 1-8929 = 0*7814 pg

F
To Calculate the amount left after 60 years.
2-303 1
ur
r
2-466x10-2 = log or log (a - x) = - 0-6425
or
60 a-x

(a - x) = Antilog T-3575 = 0*2278 pg. fo

ks
Q. 4.18. For a first order reaction, show that the time required for 99% completion of a first order reaction
Yo
is twice the time required for the completion of 90%.
oo

2-303, a
B

Ans. For first order reaction, t - -—log

k a-x
re

99% completion means that x = 99% of a = 0-99 a

u
ad

2-303, a
2-303,loglO^
,_2 2-303
^99% “ = —— = 2x
Yo

a -0-99 a k k

90% completion means that x = 90% of a = 0-90 a

d
Re

2-303 2-303 2-303

in

a
^90% - —
a-0-99 a
F

^99% ^ f 2 X 2-303 ^ K 2-303 ^ = 2 orWo-2x^90%-

Wo V * k

Q. 4.19. A first order reaction takes 40 min for 30% decomposition. Calculate t VV
Ans. 30% decomposition means that x = 30% of a = 0-30 a

2-303 a 2-303 a 2-303 10 .

As reaction is of 1st order, k = log log
-1

Xlogy
mm
t a-x 40 min a-0-30 a 40

2-303
X0-1549 min”* = 8-918xlO'^min”*
40

For a 1st order reaction, t _ 0-693 0-693

= 77*7 min.
8-918x10-3 min-*
4/106 T^'McCeefi. New Course Chemistry (Xll)i^MB

Q. 4.20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are
obtained :
t (sec) P (mm of Hg)
0 350
360 540
720 630

Calculate the rate constant.

Ans. (CH3)2CH N = N CH(CH3)2 (g) 4 N2(g)+
0 0
Initial pressure Po

ow
After time t Pq-P P P

Total pressure after time t (P,) = (Pq - p) + P + /^ = Pq + P P = P/ ~ Pq

a oc Pq and (a - jr) « Pq - p or substituting the value of p,
fl -r oc Pq- (P,- Pq), I.e.,{a-x)^?Q- P,

e
re
As decomposition of azoisopropane is a first order reaction,
Po

Frl
, 2-303, a 2-303

F
k = log log
t a-x t
2Po-P,
2-303
ou 35-0
2-303 ^ _ 2-303 (0-3400) = 2-175x10-3 s-'

r
When t = 360 sec, k = log
360 s ° 2x35-0-54-0 360 s 16 “ 360 s

so
When t = 720 sec, k =
2-303
720 s
log
35-0

2x35-0-63-0
2-303
720 s
log5 =
kf2-303
720
(0-6990) = 2-235 X10-3 s-‘
oo
Y
2-175 + 2-235
Average value of /c = xl0-3s-3 =2-20x10-3 S-*.
B

Q. 4.21. The following data were obtained during the first order thermal decomposition of SO2CI2 at a
re

constant volume : SO2CI2 (g) > SO2 (g) + CI2 (g)

oY
u

Experiment Time/s Total pressure/atm

ad

1 0 0-5

2 100 0-6
d

Calculate the rate of reaction when total pressure is 0*65 atm.

in
Re

Ans. SO2CI2 (g) > SO2 (g) + CI2 (g)

F

2-303 R0
Proceeding exactly similar to Q. 4.20, k = log
t
2Po-P,

2-303 0-5 2-303

When t = 100 s, k = log log (1-25)
100s 2x0-5-06 100s

2-303
(0-0969) = 2-2316 X10-3 s-‘
100s

When P, = 0-65 atm, Le., Pq + p = 0-65 atm

p = 0-65 - Po = 0-65 - 0-50 = 0-15 atm

Pressure of SO2CI2 at time tip^Q^ch ^ -0-15 atm = 0-35 atm

Rate at that time =kx = (2-2316 X 10-3 s-i) (0-35 atm) = 7*8 x lO"^ atm s“^
CHEMICAL KINETICS 4/107

Q. 4.22. The rate constant for the decomposition of N2O5 at various temperatures is given below :
T/“C 0 20 40 60 80
105 X k/s-i 0*0787 1*70 25*7 178 2140
Draw a graph between In k and 1/T and calculate the values of A and E^. Predict the rate constant
at 30"C and 50”C.

Ans. Proceed as in Solved Problem 11, page 4/68.

Q. 4.23. The rate constant for the decomposition of a hydrocarbon is 2*418 x 10“^ s at 546 K. If the
energy of activation is 179*9 kj/mol, what will be the value of pre-exponential factor ?
Ans. Here, k = 2-418 x IQ-^ s"’, = 179-9 kJ mol"*,

ow
T = 546 K.
According to Arrhenius equation.
E Ea
k = Ae or In it = In A - —^ or log k = log A -
RT 2-303 RT

e
re
E 179-9 kJmol-*
or log A = log /: +
a

2-303 RT
= log (2-418 X10-5 s-') +
2-303 X 8-314 X10-3 kJ K~^ mol-‘ x 546 K

F
Frl
= (-5 + 0-3834) S-* + 17-2081 = 12-5924 s"'
A = Antilog (12-5924) s“* = 3*912 x 10^^ s“^
or

Q. 4.24. Consider a certain reaction A

ou
> Products with A: = 2*0 x 10 ^ s ^ Calculate the concentration of

osr
A remaining after 100 s if the initial concentration of A is 1*0 mol

Ans. The units of k show that the reaction is of first order. Hence, k =
kf 2-303. [A] 0
log
oo
t [A]
Y
2-303 1-0 mol L '
2-Ox 10-2 S-* =
B

log or log [A] = - 0-8684

100s [A]

[A] = Antilog (- 0-8684) = Antilog (1-1316) = 0*1354 mol

re
uY

Q. 4.25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law
with h/2 = 3*00 hours. What fraction of the sample of sucrose remains after 8 hours ?
ad
do

2-303. [A] 0
Ans. As sucrose decomposes according to first order rate law, k = log
t [A]
in

The aim is to find [A]/[AJq

Re
F

, 0-693 0-693 ,
As t 1/2 = 3-0 hour, .'. k = = = 0-231 hr-*
3 hr
h/2

Hence, 0-231 hr-* = 2-303,log—

[A] 0 [AT
or log ^ = 0-8024
8 hr [A] [A]

[A] 0 [A] _ 1
or
= Antilog (0-8024) = 6-345 or = 0*158.
[A] [AJo 6-345
Q. 4.26. The decomposition of a hydrocarbon follows the equation k = (4*5 x 10^^ s“*) ^ Calculate E^.
Ans. According to Arrhenius equation, k =

. _ 28000 K
RT T
or
E„ = 28000 K X R = 28000 K x 8-314 JK-* mor* = 232*79 kJ mol~K
4/108 ● <2 New Course Chemistry (XII) ESE

Q. 4.27. The rate constant for the first order decomposition of H2O2 is given by the following equation :
log k = 14-34 -1-25 X 10^ K/T
Calculate for this reaction and at what temperature will its half*period be 256 minutes ?

Ans. According to Arrhenius equation,

E E.
or In it = In A - — or log = log A -
RT 2-303 RT

1-25x104K
Comparing with the given equation
’ 2-303 RT T

or = 2-303 R x 1-25 x 10^ K = 2-303 x (8-314 JK"' mol'*) x 1-25 x lO'^ K = 239-34 kj moH
0-693
When ti/2 - 256 min, k = = 4-51X10-5 s-l
256 x60 s

Substituting this value in the given equation,

F low
1-25x104k l-25xl0'^K
log (4-51x10-5) = 14-34- T
,ie., (-5-f0-6542) = 14-34-
T

1-25x104 K
or = 18-6858 orT = 669K.
T

re
Q. 4.28. The decomposition of A into products has value of A: as 4-5
x 10^ s~^ at lOX and energy of activation
60 kJ mol"^. At what temperature would k be 1-5 x 10** s“^ ? for F
Ans. it, = 4-5 X 105 s~K T, = 10 + 273 K = 283 K ; it2 = 1-5 x 10^ s-*, T2 = ?, = 60 kJ mor*
Applying Arrhenius equation.
Your

1 ^2 E
T2-T.1
s
eBook

log-^ = 2-303 R
1,T,T2
1-5x104 60000 J mol-* 'T2-283'
ad

log
our

4-5x105 2-303x8-314 JK-‘mol-l 283 T,^

V y

'T2-283' 0-5228 T2-283

Re

or log 3-333 = 3133-63 or

3133-63 283 T2 or 0-0472 T2 = T2-283
2831,^
Y

V /
Find

283
= 297 K = 297 -273”C = 24“C.
or
0-9528 T2 = 283 or 7^ = 0-9528

Q. 4.29. The time required for 10% completion of a first order reaction at 298 K is equal to that required
for its 25% completion at 308 K. If the value of A is 4 x 10^® s“^, calculate k at 318 K and E^.
2-303 a 2-303, 10 2-303 0-1055 0-1055
Ans. ^298^^ - log log- = (0-0458) = or =
t
1
fl-O-lOflt h h h *298K

0-2879 0-2879
2-303 a 2-303, 4 2-303
^2 =
^08K “
h
log
a-0-25 a h logj = h
(0-125) =
h
or
*308K

But/ 1 = u. Hence, —
0-1055 _ 0-2879 or
^308K = 2-7289
^ k 298K ^308K ^298K
CHEMICAL KINETICS 4/109

Now, applying Arrhenius equation, log -

^308K _ T2-T,1
k 298K 2-303 R
V T,T2
' ^ J

log (2-7289) =
^ (308-298) K
2-303x8-314 JK"* mol”' ^ 298 Kx308 K
E 10
0-4360 = a

2-303x8-314 298x308
or E^ = 76-623 kjmol-^

ow
Calculation of k at 318 K

E 7623 JK-* mol-l

log k = log A- = log (4x10^0)-
a

2-303 RT 2-303 X 8-314 JK~^ mol"! x 318 K

= 10-6021 - 12-5843 =- 1-9822

e
re
or k = Antilog (- 1-9822) = Antilog (2-0178) = l-042x 10"2s"^.
Q. 430. Same as Problem 2 For Practice, page 4/68.

F
Frl
ou
osr
kf
oo
Y
B
re
uY
ad
do
in
Re
F
4/110 New Course Chemistry (Xll)l!Zs]Sl

1 WITH ANSWERS,
HINTS AND SOLUTIONS

'4

fisi

MULTIPLE CHOICE QUESTIONS-I

1. The role of a catalyst is to change The initial pressure of the system before
(a) Gibbs energy of reaction decomposition of A was . After lapse of time

w
f/j) enthalpy of reaction ‘r’, total pressure of the system increased by x
(c) activation energy of reaction units and became ‘p/. The rate constant k for
the reaction is given as
(rf) equilibrium constant

Flo
2. In the presence of a catalyst, the heat evolved 2.303
or absorbed during the reaction
ia) k =
t
log—^
Pi~x

ee
(a) increases (b) decreases
2.303

Fr
(c) remains unchanged Pi
(b) k = log
(d) may increase or decrease t
^Pi-Pr
3. Activation energy of a chemical reaction can

for
2.303
ur
be determined by (c) * =
(«) determining the rate constant at standard
t
2Pi+P,
temperature
s
2.303
k
(b) determining the rate constantvS at two (d) ^ = log
Yo
oo

temperatures
/
Pi+x
(c) determining probability of collision 6. According to Arrhenius equation, rate constant
eB

(d) using catalyst k is equal to A e

-E./RT
. Which of the following
4. Consider Fig. below and mark the correct option.
1
r

options represents the graph of In k vs

9
Activated complex
ou
ad

T
Y

A El

t
Re
nd

Products
O! In k
3? E2
Fi

C
LU

Reactants
(a) 1/T—► (b) 1/T—►
Reaction coordinate +■

(cj) Activation energy of forward reaction is r

E] + E-, and product is less stable than reactant. In k in k
{b) Activation energy of forward reaction is
Ej + E-) and product is more stable than
reactant, ic) 1/T (d)
(c) Activation energy of both forward and 7. Consider the Arrhenius equation given below
backward reaction is E| + E2 and reactant is and mark the correct option.
more stable than product. -E_/RT
k = \ e
(d) Activation energy of backward reaction is E]
and product is more stable than reactant. (ci) Rate constant increases exponentially with
5. Consider a first order gas phase decomposition increasing activation energy and decreasing
reaction given below ; A(g) 9 B(g) + C(g) temperature
CHEMICAL KINETICS 4/111

(b) Rate constant decreases exponentially with 11. Which of the following statements is correct ?
increasing activation energy and decreasing (<i) The rate of a reaction decreases with passage
temperature
of time as the concentration of reactants
(c) Rate constant increases exponentially with decreases
decreasing activation energy and decreasing (h) The rate of a reaction is same at any time during
temperature the reaction
(d) Rate constant increases exponentially with (c) The rate of a reaction is independent of
decreasing activation energy and increasing temperature change
temperature
(d) The rate of a reaction decreases with increase

ow
8. A graph of volume
Vs in concentration of reactant(s)
of hydrogen V4
released vs time 12. Which of the following expressions is correct
for the reaction for the rate of reaction given below ?
between zinc and
5Br“ (aq) + BrO^ (aq) + 6H"^(aq) 4 3Br2(aq)

e
dil. HCI is given in

re
Fig. below. On the + 3H20tD
basis of this, mark
(o)

F
Frl
the correct option. Ar At

A|Br~J _6A[H-*-|
0
20 30 40
ou50 (b)
A! ~ 5 At

osr
ia) Average rale upto .Js is
V
.4 -''2 (C)
A[Br~] ^ 5 A[H~^J
At 6 At
40

V. kf A[Br“] _^A|m]
oo
^2
(h) Average rate upto 40 seconds is — (d)
At
40-30 At
Y
13. Which of the following graphs represents
B

(c) Average rate upto 40 seconds is ~

^3 exothermic reaction ?
40
re

Activated complex Activated complex

uY

V' -Vj
(d) Average rate up to 40 seconds is
40-20
9. Which of the following statements Is not correct
ad
do

about order of a reaction ?

pi p
(a) The order of a reaction can be a fractional 03 Reactants
0)
Products
in

c
c
number. UJ
ILi
Re

Products Reactants
(h) Order of a reaction is experimentally (/) Reaction coordinate —► (/i) Reaction coordinate
F

determined quantity,
Activated complex
(c) The order of a reaction is always equal to the
sum of the stoichiometric coefficients of
reactants in the balanced chemical equation for
a reaction. cn
03

(d) The order of a reaction is the sum of the powers

c
LIJ
Reactants Products
of molar concentration of the reactants in the
rate law expression.
{Hi) Reaction coordinate
10. Consider the graph given in Fig. in Q. 8. Which
of the following options does not show (a) (0 only (b) {it) only
instantaneous rate of reaction at 40th second ?
(c) (Hi) only (d) (/) and (ii)
^4-^2 14. Rate law for the reaction A + 2B C is found
(a) (b) to be
50-30 50-30 Rate = K 1.A] [B]
Concentration of reactant *B’ is doubled,
(c)
^3-^2 keeping the concentration of ‘A’ constant, the
(d)
40-30 40-20 value of rate constant will be
4/112 New Course Chemistry (XlI)CBgg

(a) the same (b) doubled (a) It catalyses the forward and backward reaction
to the same extent
(c) quadrupled (d) halved
15. Which of the following statements is incorrect {b) It alters AG of the reaction
about the collison theory of chemical reaction ? (c) It is a substance that does not change the
(fl) It considers reacting molecules or atoms to be equilibrium constant of a reaction
hard spheres and ignores their structural {d) It provides an alternate mechanism by reducing
features activation energy between reactants and
(b) Number of effective collisions determines the products.
rate of reaction 19. The value of rate constant of a pseudo first
order reaction
(c) Collision of atoms or molecules possessing
sufficient threshold energy results into the (a) depends on the concentration of reactants
product formation present in small amount
(J) Molecules should collide with sufficient (/;) depends on the concentration of reactants
threshold energy and proper orientation for the present in excess
collision to be effective. (c) is independent of the concentration of reactants

w
16. A first order reaction is 50% completed in (d) depends only on temperature.
1.26x 10*'* s. How much time would it take for 20. Consider the reaction A ^ B. The

F lo
100% completion ? concentration of both the reactants and the
(a) 1.26 X I0‘^ s (b) 2.52 X lO'-* s products varies exponentially with time. Which
(c) 2.52 X 10^^ s (d) infinite of the following figures correctly describes the

ee
17. Compounds ‘A’ and ‘B’ react according to the change in concentration of reactants and

Fr
following chemical equation : products with time ?
A (g) + 2 B (g) ^ 2C (g)
Concentration of either ‘A’ or ‘B’ were changed
t for [B]
ur
keeping the concentrations of one of the c
[B]
reactants constant and rates were measured as o o
s
a function of initial concentration. Following
CO
ook
Yo
c

results were obtained. Choose the correct option s

<3>
O
c
for the rate equations for this reaction.
c

[A]
eB

o o
u [A] O

Exp. Initial Initial Initial rate of

concentration concentration formation of
our

of [A]/mol L
-1
of [B]/mol L~* [Cl/mol L-* S-*
ad

1. 0,30 0.30 0.10

2. 0.30 0.60 0.40 c

Y

3. 0.60 0.30 0.20 CO

Re
nd

(a) Rate = /: [A ^ [B1 (b) Rate = k [A] [B]-

<u

(c) Rate = k fA] [B] {d) Rate = k [A]^ [B]

0
Fi

Time ♦
18. Which of the following statement is not correct
for the catalyst ? (d)

MULTIPLE CHOICE QUESTIONS-II

Note : In the following questions, two or more 22. Which of the following statements are
options may be correct. applicable to a balanced chemical equation of
21. Rate law can not be determined from balanced an elementary reaction ?
chemical equation if (a) Order is same as molecularity
(«) reverse reaction is involved (b) Order is less than the molecularity
(h) it is an elementary reaction
(r) Order is greater than the molecularity
(c) it is a sequence of elementary reactions
(d) Molecularity can never be zero,
(d) any of the reactants is in excess.
CHEMICAL KINETICS 4/113

23. In any unimolecular reaction (d) most probable kinetic energy decreases at
(a) only one reacting species is involved in the higher temperatures.
rate determining step 28. In the graph showing Maxwell Boltzman
(b) the order and the molecularity of slowest step distribution of energy,
are equal to one (a) area under the curve must not change with
(c) the molecularity of the reaction is one and order increase in temperature
is zero
(b) area under the curve increases with increase in
(d) both molecularity and order of the reaction are temperature
one.
(c) area under the curve decreases with increase
24. For a complex reaction in temperature
(a) order of overall reaction is same as molecularity (d) with increase in temperature curve broadens
of the slowest step and shifts to the right hand side.

w
(b) order of overall reaction is less than the 29. Which of the following statements are in
molecularity of the slowest step accordance with the Arrhenius equation ?
(c) order of overall reaction is greater than (a) Rate of a reaction increases with increase in

F lo
molecularity of the slowest step temperature

(d) molecularity of the slowest step is never zero (b) Rate of a reaction increases with decrease in

ee
or non integer. activation energy

Fr
25. At high pressure, the following reaction is of (c) Rate constant decreases exponentially with
zero order. increase in temperature
1130K
for
(d) Rate of reaction decreases with decrease in
ur
2NH3(g)
Platinumcatalyst ^2 ^ H2 (g) activation energy.
30. Mark the incorrect statements,
s
Which of the following options are correct for
ook
Yo
this reaction ? (a) Catalyst provides an alternative pathway to
reaction mechanism
(a) Rate of reaction = Rate constant
eB

(b) Catalyst raises the activation energy

(b) Rate of the reaction depends on concentration
of ammonia (c) Catalyst lowers the activation energy
r
ad

(d) Catalyst alters enthalpy change of the reaction.

ou

(c) Rate of decomposition of ammonia will remain

constant until ammonia disappears completely 31. Which of the following graphs is correct for a
zero order reaction ?
Y

(d) Further increase in pressure will change the

rate of reaction.
Re
nd

26. During decomposition of an activated complex t

Fi

<D

(a) energy is always released s

C

c
(b) energy is always absorbed o
§£
■s
(c) energy does not change
C ««-
<t>
01 <3°
(d) reactants may be formed Time ►
27. According to Maxwell Boltzmann distribution (b)
of energy,
(a) the fraction of molecule ^ith most probable
kinetic energy dec eases at higher
temperatures
(b) the fraction of molecules with most probable
kinetic energy increases at higher temperatures
(c) most probable kinetic energy increases at Time ►
higher temperatures (d)
 New Course Chemistry (XII)BETBl
4/1J4
32. Which of the following graphs is correct for a first order reaction ?

t k
slope = 2.303
Q.

C
o o
ni
(5 tr: q:
O
5 c o>
0)
o
c
o
u

[Rlo [Rio Time—►

Time ►

{<>) (b) (c) id)

ow
ANSWERS

Multiple Choice Questions -1

6. (a) 7. (c/) 8. (c) 9. (c) 10.{h)
l.(c) 2.(c) 3.{b) 4. (a) 5. (/;)
17. (b) 18. ib) 19. (fc) 20.(b)

e
11.(a) 12.(c) 13.(a) 14.(a) 15.(c) 16. (f/)

re
Multiple Choice Questions - II

rFl
24. (fl.rf) 25. (a,c,d) 26. («,(/) 27. (o.c) 28. (a.^/)

F
21. (a,c,d) 22. (a.t/) 23.

29. (a,b) 30. (b,d) 31. (a.i/) 32. ia,d)

or
ou
HINTS FOR DIFFICULT MULTIPLE CHOICE QUESTIONS ksf
Multiple Choice Questions -1
1. In the presence of catalyst, activation energy barrier is lowered.
oo

2. Refer to Fig. 4.23, page 4/70.

Y
B

3. Knowing ky and /c2 at T, and T2, can be determined (Arrhenius equation).

4. (forward) = E, + E,. Product is less stable because it has higher energy.
re

5. Refer to Art. 4.10.3, page 4/29.

oYu

E 1
/. In^ = InA- — Hence graph of In k — will be linear with negative slope and
ad

-E./RT
6. k = Ae
RT' ® T
intercept = In A, i.e., (a).
d

7. k = Ae
-E./RT
. As Ey increases, - E,/RT decreases and so does k or conversely as E^, decrea.ses. k will
in
Re

increase exponentially. As T increases, E^j/RT decreases or - E^RT increases and so does k.

F

Increase in volume _ V3
8. Average rate at 40s =
Time 40

9. Order of reaction may or may not be equal to sum of stoichiometric coefficients. Hence, (c) is not correct.
in volume during small interval of time close to 40 sec.
10. Instantaneous rate at 40th sec. will be change
viz. 40-30. 50-40, 50-30 or 40-20 .sec. Volume change during 50-30 sec. interval = V5 - V2 and
not V4 - Vo. Hence, {b) is not correct.
AlBr-] A[BrQ-] 5A[H^J
12. Rate = -
I A(Br-| A[BrO;l _ 1 A[H^J _. c

6 A/ At 6 A/
5 At A/

13. For exothermic reaction. Energy of reactants > Energy of products, i.e., in (/) only.
14. Rate constant of a reaction does not depend upon concentrations of the reactants.
15. (c) is incorrect because formation of product depends not only on energy but also on proper orientation
at the time of collision.
16. Whole of the substance never reacts because in every half life, 50% of the substance reacts. Hence, time
taken for 100% completion of a reaction is infinite.
CHEMICAL KINETICS
4/115

17. From expts. 1 and 3, [B] = constant, [A] = doubled, rate is doubled.
Hence, Rate oc [AJ.
From expts. 1 and 2, [A| = constant, [B] = doubled, rate quadrupled.
Hence, Rale « [B]*.
Combining. Rate = k [A] [B]^.
18. Presence of catalyst docs not change AG or AH value. Hence, (b) is not correct.
19. The value of rate constant of a pseudo first order reaction depends not only on temperature but also on
concentration of reactant present in excess (See Sample Problem 2, page 4/48). Hence, (b) is correct.
20. Concentration of reactant. [A] decreases exponentially whereas that of product, [B] increases
exponentially with time. Hence, (b) is the correct option. ●
Multiple Choice Questions - II
25. For zero order reaction, Rate = k [A]*^ = k. Hence, (a) is correct. Rate of zero order reaction is independent
of concentration of reactants. Hence, (b) is wrong.
For the same reason, (c) is correct.

F low
Rate of decomposition of a gas depends upon pressure. Hence, (</) is correct.
26. Activated complex has higher energy. When it decomposes, energy is always released and it may give
products or reactants back.
27. Refer to Fig. 4.19, page 4/62.
28. Area under the curve must remain the same because sum of all fractions must be equal to 1, i.e., total

re
probability is one. Hence. («) is correct. Also (cf) is correct and (b), (c) are wrong.
for F
31. Out of (a) and (r), (a) is correct because in zero order reaction, rate remains constant with time and falls to
zero when whole of reactant has reacted (Refer to Fig. 4.5, page 4/25). Out of (b) and (d), (d) is correct.
32. (a) is correct because ty2 of a 1st order reaction is independent of initial concentration, (b) and (c) are

wrong, (d) is correct (Refer to Fig. 4.10, page 4/29)

Your
s
eBook

SHORT ANSWER QUESTIONS

33. State a condition under which a bimolecular reaction is kinetically first order reaction.
ad
our

Ans. A bimolecular reaction becomes first order reaction when one of the reactants is in excess.
34. Write the rate equation for the reaction 2 A + B ^ C if the order of the reaction is zero.
Ans. Rate = k [A]0 [Bj“ = k.
Re

35. How can you determine the rate law of the following reaction ?
2 NO ig) + 02ig) >2 NO2 (g)
Y
Find

Ans. The rate law can be determined by 'initial rate method’ (page 4/35). Keeping the concentration of one of
the reactants constant and changing the concentration of the other, the effect on the rate of reaction is
determined. For example, for the given reaction,
(0 Keeping [O2I constant, if [NO] is doubled, rate is found to become four times. This shows that
Rate oc [N0]2.
(//) Keeping [NO] constant, if [02] is doubled, rate is al.so found to become double. This shows that
Rate « [0->].
Hence, overall rate law will be ; Rate = k [NO]- [O2].
36. For which type of reactions, order and molecularity have the same value ?
Ans. For elementary reactions, order = molecularity.
37. In a reaction, if the concentration of reactant A is tripled, the rate of reaction becomes twenty seven
times. What is the order of the reaction ?

Ans. r,=k
I [A]“ = ka^\r2 = k (3 a)“ ; = 3“ = 27 (Given) = 3^ a = 3.
'i
4/116 7^'u^eieefi^ New Course Chemistry (XII) BE

38. Derive an expression to calculate time required for completion of zero order reaction.
[A] 0
Ans. t 100% ~ (Refer to page 4/26).
k
39. For a reaction A + B ^ Products, the rate law is : Rate = k [A] Can the reaction be an
elementary reaction ? Explain.
Ans. No, the given reaction cannot be an elementary reaction. This is because the elementary step of the slowest
step involves 1 molecule of A and 3/2 molecules of B, i.e., fractional.
40. For a certain reaction, large fraction of molecules has energy more than the threshold energy, yet the
rate of reaction is very slow. Why ?
Ans. This is because colliding molecules may not be having proper orientation for collision to be effective.
41. For a zero order reaction, will the molecularity be equal to zero ? Explain.

ow
A
Ans. No because molecularity can never be zero or fractional.
42. For a general reaction A > B, plot of concentration of A vs <
time is given in Fig. Answer the following questions on the basis o

of this graph, 6

e
(i) What is the order of the reaction? («) What is the slope of the curve? o

re
(iii) What are the units of rate constant ?

rFl
■>

Ans. (/) This type of plot is obtained for reactions of zero order (Fig. 4.5, page 4/25).

F
(//) Slope of the line = -k (Fig. 4.5, page 4/25).
{iii) Units of k for reactions of zero order = mol L“* s“* (Page 4/18).

r
43. The reaction between H2 (g) and O2 (g) is highly feasible yet aUowing the g^es to stand at room
ou
fo
temperature in the same vessel does not lead to the formation of water. Explain.
ks
Ans. This is because activation energy of the reaction is very high at room temperature.
44. Why does the rate of a reaction increase with rise in temperature ?
oo
Ans. At higher temperature, average energy of the reactant molecules is greater. Hence, larger fractions of colliding
molecules can cross the energy barrier and, therefore, the rate is faster.
Y
B

45. Oxygen is avaUable in plenty in air yet fuels do not burn by themselves at room temperature. Explain.
Ans. The activation energy for the combustion of fuels is very high at room temperature. Therefore, they do not
re

bum by themselves.
46. Why is the probability of reaction with molecularity higher than three very rare ?
ou
Y

molecules to collide simultaneously is very small.

ad

Ans. This is because the probability of more than three

47. Why does the rate of any reaction generally decreases during the course of the reaction ?
Ans. The rate of reaction depends upon concentrations of reactants. As the reaction proceeds, reactants are
d

consumed. Hence, their concentration decreases and so the rate of reaction decreases.
in
Re

48. Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with
the help of one example.
F

Ans. Thermodynamically, the conversion of diamond to graphite is highly feasible because energy is released in
this conversion or combustion of fuels is also feasible. But these reactions do not occur at room temperature
because their activation energy is very high.
49. Why in the redox titration of KMn04 vs oxalic acid, we heat oxalic acid solution before starting the
titration ?
Ans. The reaction between KMn04 and oxalic acid is very slow at room temperature. On heating, the rate
becomes faster.
50. Why can’t molecularity of any reaction be equal to zero ?
Ans. Molecularity of a reaction means the number of molecules of the reactants taking place in an elementary
reaction. Since at least one molecule must be present, molecularity is at least one.
51. Why molecularity is applicable only for elementary reactions and order is applicable for elementary
as well as complex reactions ?
Ans. A complex reaction takes place through a number of elementary reactions. Each elementary reaction may
have different molecularity. Therefore, it is meaningless to talk of molecularity of the overall complex
reaction. On the other hand, order of reaction depends upon the molecularity of the slowest step. Hence, it
is not meaningless to talk of the order of a complex reaction.
CHEMICAL KINETICS 4/117

52. Why we cannot determine the order of a reaction by taking into considerationthe balanced chemical
equation ?
Ans. The rate of reaction may not depend upon all the molecules of a reactant present in the balanced chemical
equation. It has to be determined experimentally that on what power of molar concentration of that reactant,
the rate depends. Summing up the powers of molar concentrations of different reactants as determined
experimentally gives the overall order of reaction.

MATCHING TYPE QUESTIONS

Note: In the following questions, match the items of Column I with appropriate item given in Column II.

ow
53. Match the graph given in Column I with the order of reaction given in Column II. More than one
item in Column I may link to the same item of Column II.
Column I Column II

e
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t

Frl
F
(0
<D
CD
a:
ou
or
Concentration ♦

kfs
oo
Y
B

(«) (a) 1st order

CD
cc
re
oYu

Concentration *■
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d
in

c
Re

ra
(Hi) (b) Zero order
F

s
c
o
O

Time >

o
(iV) c
o
O
O)
o

Time
4/118
New Course Chemistry (XII)KElgl

II
54. Match the statements given in Column I and Column
Column I Column II

(a) cannot be fraction or zero

(/) Catalyst alters the rate of reaction
(//) Molccularily ib) proper orientation is not there always
(//7) Second half life of first order reaction (c) by lowering the activation energy
i-:„“ / RT (d) is same as the first
(IV) c

(V) Energetically favourable reactions are (e) total probability is one

sometimes slow
(vi) Area under the Maxwell Boltzmann curve (/) refers to the fraction of molecules with energy
is constant equal to or greater than activation energy

ow
55. Match the items of Column I and Column II.
Column I Column II

(/) Diamond ia) short interval of time

(ii) Instantaneous rate ib) ordinarily rate of conversion is

e
imperceptible

Fl
re
iiii) Average rate (.c) long duration of time

F
56. Match the items of Column 1 and Column II.
Column I Column II
ur
or
(/) Mathematical expression for rate of reaction (fl) rate constant

ib) rate law

(ii) Rale of reaction for zero order reaction is equal to sf
(//;) Units of rate constant for zero order reaction is same ic) order of slowest step
k
Yo
as that of
oo

(cf) rate of a reaction

(nO Order of a complex reaction is determined by
B

ANSWERS
e

53. (/) (fl), (i7)

(b), {Hi} (hi (iv) ^ (a)
ur

54. ((■) (c). {ii){a), {Hi) -> ((/), (iv) -> (/), (v) ^ {h), {vi) -> (e)
ad

55. (0 ^ (/?), (i7) -> («), {Hi) (c)

Yo

56. (i) ^ (/;), (//) ^ {a), {Hi) (r/), (iv) ^ (c)

d
Re

ASSERTION AND REASON TYPE QUESTIONS

in

Note ; In the following questions, a statement of assertion followed by a statement of reason is given.
F

Choose the correct answer out of the following choices :

{a) Both assertion and reason are correct and the reason is correct explanation of assertion.
(b) Both assertion and reason are correct but reason does not explain assertion.
{O .\ssertion is correct but reason is incorrect.
id) Both assertion and reason are incorrect.
{e) Assertion is incorrect but reason is correct.
57. Assertion : Order of the reaction can be zero or fractional.
Reason : We cannot determine order from balanced chemical equation.
58. Assertion : Order and molecularity are same.
Reason : Order is determined experimentally and molecularity is the sum of the stoichiometric coefficients
of rate dclcnnining elementary step.
59. Assertion : The enthalpy of reaction remains constant in the presence of a catalyst.
Reason : A catalyst participating in the reaction forms different activated complex and lowers down the
activation energy but the difference in energy of reactant and product remains the same.
CHEMICAL KINETICS 4/119

60. Assertion : All collisions of reactant molecules lead to product formation.

Reason : Only those collisions in which molecules have correct orientation and sufficient kinetic energy
lead to compound formation.
61. Assertion : Rate constants determined from Arrhenius equation are fairly accurate for simple as well as
complex molecules.
Reason ; Reactant molecules undergo chemical change irrespective of their orientation during collision.

ANSWERS

57. {b) 58. (e) 59. (a) 60. (e) 61. (c)

w
HINTS FOR DIFFICULT QUESTIONS
57. Correct explanation. Rate of reaction may not depend upon the concentration of a reactant or may
depend upon fractional power of molar concentration.

o
58. Correct Assertion. Order may or may not be equal to molecularity.
60. Correction Assertion. All collisions of reactant molecules do not lead to product formation.

e
re
61. Correct Reason. Arrhenius equation is applicable to collisions between simple as well as complex

rFl
molecules.

F
r
LONG ANSWER QUESTIONS
ou
62.
fo
All energetically effective collisions do not result in a chemical change. Explain with the help of an
ks
example.
Ans. Refer to page 4/71.
oo

63. What happens to most probable kinetic energy and the energy of activation with increase in
Y
eB

temperature ?
Ans. Refer to page 4/62 and Fig. 4.19.
64. Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction.
r

Ans. Refer to page 4/70 and Fig. 4.23.

ou
Y
ad

65. Explain the difference between instantaneous rate of a reaction and average rate of a reaction.
Ans. Refer to Art. 4.3.2, page 4/3.
66. With the help of an example, explain what is meant by pseudo first order reaction.
d

Ans. Refer to Art. 4.14, page 4/46.

Re
in
F
4/120 New Course Chemistry CXII)iSsX9]

> mff Kn.

i
NEET/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to
● MCQs in Chemistry for MEET
Pradeep's Stellar Series.... ● MCQs in Chemistry for JEE (Main)

w
separately available for these examinations.

F lo
Multiple Choice Questions (with one correct Answer)

(a) 1-8 X I0“2 (b) 1-2 X 10"^

e
L General introduction

Fre
and rates of reactions (c) 4 X 10-2 (d) 3-6 X 10^2
1. For the chemical reaction, for
4. 100 cm^ of 1 M CH3COOH was mixed with
(J & K CET 2009)

N2(5) + 3H2 (g) ^ ^ 2 NH3 (g), 100 cm^ of 2 M CH3OH to form an ester. The
r
the correct equation is change in the initial rate if each solution is diluted
You
oks

with equal volume of water would be

(a) 3
eBo

dt dt (a) 4 times (b) 0-25 times

(c) 2 times (d) 0-5 times
-
1 djH^]
-
1 ^[NH3] (Karnataka CET 2015)
dt 2 dt
ad
our

5. The rate of a gaseous reaction is generally

(c) - d[N^]^ d[NH^] expressed in terms of
dP
. If it were expressed in
dt dt dt
terms of change in number of moles per unit time
1 ^[NH3]
dY
Re

id) - (NEET 2019) dn

dt 2 dt or in terms of change in molar
\ dt
Fin

2. The decomposition of N2O5 in CCl4 at 318 K is dC

studied by monitoring the concentration of N2O5 concentration per unit time , which of the
in the solution. Initially, the concentration of N2O5 K dt )
is 2-4 mol L“* and after 200 minutes, it is reduced following relationship will hold good ?
to 2-00 mol L-*. What is the rate of production of dC dn d?
NO2 during this period in mol L“* min"’ ? id)
dt dt dt
(a) 4x10-3 {b) 2 X 10“2
(c) 1 X 10-3 (d) 10^
ib)
(e) 5 X IQ-3 (Kerala PMT 2015) dt ~ ) RTl,i//
3. In the synthesis of ammonia from nitrogen and
hydrogen gases, if 6 x 10-2 hydrogen (c)
dt dt RT dt
disappears in 10 minutes, the number of moles of
ammonia formed in 0-3 minutes is (d) None of these

ANSWERS

1.(^0 2.{a) X{h) 4. (/;) 5. (/>)

CHEMICAL KINETICS 4/121

^1 10. For the reaction

6. In the reversible reaction, 2NO -) t
- N2O4,
*2 2 H2 (s) + 2 NO ig) > N, (S) + 2 H.O (g)
the rate of disappearance of NO2 is equal to the observed rate expression is, rate = k fNO]^ [H2I.
The rate expression for the reverse reaction is
ia)
h (a) [Nj] [H20]2 ih) kf, LN2] [H2O]
(c) ki^ [N,] [H20]2/[H2l
ib) 2^l[N02]2-2^:2[N204]
(c) 2 LN02]2 - k^ [N2O4] (d) kf, [N2I [H20]2 [NO] (JEE Main 2020)
(^0 (2*i-/:2)[N02J (AMU Engg. 2015) II. Rate law, order of reaction,
rate constant and its units
7. Consider the decomposition of N.,05 as
1 11. For the reaction A + B > C + D. doubling the
N2O5 » 2NO2+-O2 concentration of both the reactants increases the
reaction rate by 8 times and doubling the
The rate of reaction is given by

w
concentration of only B simply doubles the
reaction rate. The rate law is given as
1/2
(a) r = ib) r = HA][B]2

F lo
dt 2 dt dt
(c) r = /:rAl2fB] (d) r = k [A] [B]
Therefore, -
d[Np^] (AIPMT Prelim 2012)
dt
= *,(N205]

ee
12. If rate = k [H'*’]" and it becomes 100 times when

Fr
J[N02] the pH changes from 2 to 1, then the order of
+
dt
= 2/:i[N205J = i:j'[N205l reaction is
(a) 0 ib) 1 (c) 2 (d) 3
+
dlO,] 1
= -ki[Np^] = ki" [N2O5I for
ur
dt III. Molecularity,
nicchuni.sm and dilTerencc
Choose the correct option
s
between order and molecularity
ook

(a) kj =ki' = k/' (b) k( =2 ki' = ky

Yo

(c) 2 k, = k,'= 4 k," (d)4k,=2ki' = k,

tf

13. A hypothetical reaction, A2 + B2 ^ 2 AB,

eB

8. The rate constant is numerically the same for three follows the mechanism as given below :
reactions of first, second and third order A2 ± A + A (Fast)
^
respectively. Which one of the following is true
our

A+ B, > AB + B (Slow)
ad

for the rate of these three reactions if concentration

A + B > AB (Fast)
of the reactant is same and greater than 1 M?
The overall order of reaction is
(a) r,=r2 = rj (b) ry> r2> ^3
Y

(a) zero (b) 1

(c) r, < ^2 <
Re

1
nd

(d) There can be no definite order (0 1- id) 2

Fi

9. For the reaction,

rv. Integrated rate equations,
Ag^ + 2NH3 [Ag(NH3)2]^ halt-life, determinationof rate law,
the net rate of reaction is given by rate const, and order

^ = 2 X 10^ [Ag+] [NH3]2 - 1 X 10-2 [Ag(NH3)2]^ 14. The rate of the reaction A
initial concentration of 3-24 x 10"2m is nine times
^ Products, at the

Then which of the following statement/s is/are its rate at another initial concentration of
correct ?
1 -2 X 10“^ M. The order of the reaction is
(a) Rate constant for forward reaction = 2 x 10^
1 3 3 2
(b) Rate constant for backward reaction = 1 x 10“2
(«) 3 (^) 3 (c) 2 (^) 3
(c) Equilibrium constant of the reaction = 2 x 10^
1
(d) All the above (Kerala CET 2011)

ANSWERS

6. (/;) 7. (r) 8. (c) 9. id) 10. (r) 11. (c) 12. (c) 13. (c) 14. id)
4/122 New Course Chemistry (XI1)SSS3SI

15. The time for half life period of a certain reaction (a) 10 (b) 1000
A ■> Products is 1 hour. When the initial con (c) 100 (d) IQ6
centration of the reactant ‘A’, is 2-0 mol L"’, how 20. The following data is obtained during the first
much time does it take for its concentration to order thermal decomposition of
come from 0-50 to 0-25 mol L"* if it is a zero 2A(g) ^B(g) + C(s)
order reaction ?
at constant volume and temperature
(a) 0-25 h ib) Ih
S. No. Time Total pressure in pascals
(c) 4h id) 0-5 h
16. The figure below depicts the change in the I. At the end of 300

concentration of the species A and B for the 10 minutes

reaction A —> B, as a function of time. The point 2. After completion 200
of intersection of the two curves represents
The rate constant in min * is

w
(a) 0-0693 ib) 6-93
(c) 0-00693 id) 69-3
(Karnataka CET 2010)

F lo
21. Rate constant of a reaction is 0-0693 min”’.
Starting with 10 mol L“’, rate of reaction after 10

ee
minutes will be
-1

Fr
-1
(a) 0-0693 M min (b) 0-0693 X 2-5 M min
ia) /|/4 (c) 0-0693 X 5 M min”'
ib) I 1/3
for
id) 0-0693 X 10 M min”'
ur
ha 22. The initial rates of reaction 3 A -t- 2 B -f C
Products, at different initial concentrations are
id) The method cannot be used to predict
s
fractional life time given below :
ook
Yo

17. Half-lives of a first order and a zero order reaction Initial rate, Ms”’ tA]o, M IBlo. M [CJo, M
eB

are same. Then the ratio of tlie initial rates of the 5-0 X 10”^ 0-010 0-005 0-010
first order reaction to that of zero order reaction is 5-0 X 10”^ 0-010 0-005 0-015
1 I-O X 10-2 O-OlO 0-010 0-010
our

ia) ib) 2x0-693

ad

0-693 1-25 X 10”2 0-005 0-005 O-OIO

The order with respect to the reactants A, B and C

2
(c) 0-693 id) are respectively
Y

0-693
(a) 3,2,0 ib) 3, 2, 1
Re
nd

ie) 6-93 (Kerala PET 2010) (r) 2,2,0 id) 2, 2, I

18. For a first order reaction, the time taken to reduce (e) 2, 1,0 (Kerala PMT 2011)
Fi

I 23. At 500 K, the half-life period of a gaseous reaction

the Initial concentration by a factor of — is 20
at the initial pressure of 80 kPa is 350 sec. When
minutes. The time required to reduce initial the pressure is 40 kPa, the half-life period is
concentration by a factor of 1/16 is 175 sec. The order of the reaction is

id) 20 min ib) 10 min ia) zero ib) one

(c) 80 min id) 40 min (c) two id) three
ie) 5 min (Kerala PMT 2011) ie) half (Kerala PMT 2007)
19. The rate constant of a second order reaction, 24. A reaction P Q is completed 25% in 25 min,
2A > Products, is 10”^ lit mol”' min”'. The 50% completed in 25 min if [P] is halved, 25%
initial concentrationof the reactant is 10”^ mol completed in 50 min if [P] is doubled. The order
lit”'. What is the half-life (in min) ? of reaction is

ANSWERS

15. ia) 16. (c) 17. ih) 18. id) 19. id) 20. («) 21. (c) 22. ie) 23. ia)
CHEMICAL KINETICS 4/123

(a) 1 (b) 2 29. If the rate constant for a first order reaction is

(c) 0 id) 3 the time (0 required for the completion of 99% of

25. t 1/4 can be taken as the time taken for the the reaction is given by
(rt) / = 2-303/k ib) t = 0-693/k
3
concentration of a reactant to drop to — of its (c) ; = 6-909/k id) /=4-606/k
4
initial value. If the rale constant for a first order (MEET 2019)
reaction is K, then can be written as 30. The half-life of a reaction is halved as the initial
ib) 0-29/K concentration of the reactant is doubled. The order
ia) 010/K
of reaction is
(c) 0-69/K id) 0-75/K
id) 0-5 ib) 1
26. A plot of log /j/2
(c) 2 id) 0
versus log Cq is given
in the adjoining fig. ! CSJ
31. A substance undergoes first
B

The conclusion that

03
order decomposition. The A
can be drawn from this O
decomposition follows two

w
graph is parallel first order reactions as
log Cq
= i-26 X 10^ s^' , k2 = 3-8 X 10“^ s“'

F lo
id) Order = 1, ^j/2 - ka The percentage distribution of B and C are
2-303 (a) 75% B and 25% C (fc) 80% B and 20% C
ib) Order = 1, /]/2 ~ piog2 (c) 60% B and 40% C id) 90% B and 10% C

e
k

Fre
1
ie) 76-83% B and 23-17% C
(c) Order = 0, h/2 ~ 32. For a second order reaction, 2 A > Products,
2ka for
a plot of log /[/2 vs log a (where a is the initial
1 concentration) will give an intercept equal to
r
id) Order = 2. ^i/2 - "" which of the following ?
You
a
oks

27. For a reaction A -f- B Products, rate law is ia) l/k ib) log (1/2 A)
eBo

d[A] (c) log (I/*) id) log* (SCRA2007)

dt
= k [A]q . The concentration of A left after 33. Under the same reaction conditions, initial
1 concentration of 1-386 mol dm“^ of a substance
time t when / = — is
ad

becomes half in 40 seconds and 20 seconds

our

k
through first order and zero order kinetics respec
[A] 0
ia) ib) [A]f^x e tively. Ratio (Aj/Icq) of the rate constants for first
order (kj) and zero order ik^) of the reactions is
e
dY
Re

[A] 0 1 ia) 0-5 mol"* dm^ ib) 1-0 mol dm ^

ic) id) [A] 0 (c) 1-5 mol dm"^ id) 2-0 mol"* dm^
Fin

e-

(IIT 2008)
28. Decomposition of H2O2 follows a first order
reaction. In fifty minutes, the concentration of 34. In the catalysed decomposition of benzene
diazonium chloride.
H2O2 decreases from 0-5 to 0-125 M in one such

<0>
decomposition. When the concentration of H2O2 A/Cu
reaches 0-05 M, the rate of formation of 0-? will N2CI Cl N2.
be
half-life period is found to be independent of the
ia) 6-93 X 10"~ mol min"* initial concentration of the reactant. After 10
ib) 6-93 X 10“^ mol L"* min"* minutes, the volume of N2 gas collected is 10 L
ic) 2-66 L min
-1
atSTP and after the reaction is complete, it is 50 L. Hence,
id) 1 -34 X 10“^ mol min"’ (JEE Main 2016)
rate constant of the reaction (in min"’) is
ANSWERS

24. ic) 25. ih) 26. ih) 27. ia) 28. ib) 29. (rf) 30. (c) 31. (e) 32. id) 33. ia)
4/124 “PnAdee^’^ New Course Chemistry (XII)CSm

2-303, lO 2-303 50 40. At 300°C, the half-life for the decomposition of

(a) log — (b) log AB2 is 200 s and is independent of the initial
10 50 10 50-10
concentration of AB2. The time required for 80%
2-303. 50 2-303 10 of the AB2 to decompose is
(c) log (cT) log
10 10 10 50-10 (Given : log 2 = 0-30, log 3 = 0-48 )
35. Kinetics of the reaction A (g) ^ 2 B (g) -f (a) 2(X) s (h) 323 s
C (g) is followed by measuring the total pressure (c) 467 s (d) 532 s
at different times. It is given that (JEE Main 2022)
Initial pressure of A = 0-5 atm. 41. The given graph is a representation of kinetics of
Total pressure of A after 2 hours = 0-7 atm. a reaction

Rate constant of the reaction =1x10"^ s“’

What is the rale of reaction -

d[A\
when the total Constant temperature T

w
dt
pressure is 0-7 atm ?
(a) 2-0 X 10^ atm s“^ (b) 4-0 x 10^ atm s“*

F lo
(c) 5-0 X 10^ atm s“* (d) 7-0 x 10^ atm s“* X

(IAS Prelim 2010)

The y and x axes for zero and first order reactions,

ee
36. The rate of a first order reaction is 0-04 mol s"*
respectively are

Fr
at 10 seconds and 0-03 mol L"^ s“* at 20 seconds
after initiation of the reaction. The half-life
(fl) zero order (y = concentration and x = time)
first order (y = tj/2 and x = concentration)
period of the reaction is
for
ur
(a) 44-1 s (b) 54-1 s (b) zero order (y = concentration and x = time)
first order (y = rate constant and
(c) 24-1 s id) 34-1 s
s
X = concentration)
ook

(NEET Phase I 2016)

Yo
(c) zero order (y = rate and x = concentration)
37. At 518®C. the rate of decomposition of a sample
first order (y = tj/2 and x = concentration)
eB

of gaseous acetaldehyde, initially at a pressure of

363 torr, was 1-00 torr s“^ when 5% had reacted {d) zero order (y = rate and x = concentration)
and 0-5 torr s“* when 33% had reacted. The order first order (y = rate and x = rj/2) (NEET 2022)
r

of the reaction is
ad
ou

V. First order reactions,

(a) 2 ib) 3 pseudo first order reactions
Y

(c) 1 (d) 0 (JEE Main 2018) and radioactive disintegration

38. Decomposition of X exhibits a rate constant of
Re

42. For a first order reaction, (average life time),

nd

0-05 |i, g/year. How many years are required for t and t75% are in the order :
50%
the decomposition of 5 |i g of X into 2-5 |i g ?
Fi

ia) 50 ib) 25
ia) i^Q<t < t75 av ib) tjo < /75 < r av

(c) 20 id) 40
(^) ^av < ^50 hs id) r^,, = tjo < ^75
43. In the study of inversion of sucro.se in presence of
(JEE Main 2019)
acid, if rQ, r, and represent the polarimetric
39. For a first order reaction, the time required for the readings at times 0, t and «> respectively, then at
completion of 90% reaction is ^x' time the half-
the 50% inversion, which of the following
life of the reaction. The value of ‘a’ is
relationship will hold good ?
(Given In 10 = 2-303 and log 2 = 0-3010) 1
(a) 1-12 ib) 2-43 ia) r, = /y + ,r (b) +

(c) 3-32 id) 33-31

(c) r,= rQ-r, (^0
(JEE Main 2022)
ANSWERS

34. ib) 35. ib) 36. (c) 37. («) 38. (n) 39. (<●) 40. (/;) 41. (c) 42. ia) 43. ib)
CHEMICAL KINETICS 4/125

-I ~!
44. Number of natural life times required for a («) 5 kJ mol (b) - 5 kJ mol
first order reaction to complete 99-9% is (c) 149 kJ mol
-I
(d) - 149 U mol-'
(a) 6-93 {b) 2-31 50. When a catalyst increases the rate of a chemical
(c) 9-2 {d) infinite reaction, the rate constant
45. Half-life period of a first order reaction is 100 min. (a) remains constant
After 144-3 min, concentration of reactant is
(b) increases
reduced to ... of the original concentration
(c) decreases
(fl) 40% {b) 30%
(d) may increase or decrease depending on the
(c) Me (f/) Me-
order of reaction (AIPMT 2013)
VI. Activation energy, effect of 51. For a reaction taking place in three steps, the rate
temp, on rate (Arrhenius eqn.) constants are ^2 ^^d The overall rate
and effect of catalyst on rate

w
k k
constant is k = ^ . If the energy of activation
46. The rate constant of a first order reaction becomes
h
six times when the temperature is raised from

F lo
350 K to 400 K. Calculate the activation energy values for the first, second and third stages are
of the reaction (R = 8-314 JK"' moU') 40, 50 and 60 kJ mol"' respectively, then the
overall energy of activation in kJ mol"' is

ee
-1
(a) 4-17 kJ mol-' (b) 41-7 kJ mol

Fr
(a) 30 ib) 150
(c) 417-0 kJ mol
-1
(d) 417-0 U mol-'
(c) 50 id) 60
(AUMS 2014)
47. A reactant (A) forms two products :
for (Kerala CEE 2008, AMU Engg. 2014)
ur
52. The rate constants k^ and kn for two different
k1
reactions are lO'^ . ^Q\5 ^-lOoo/T
A » B, Activation Energy £
s
«i
respectively. The temperature at which cfj = k2 is
ook
Yo

2000
A ^ C, Activation Energy E ^2 (a) 1000 K (b) K
eB

2-303
If E„ =2E then k^ and k^ are related as 1000
«2
(c) 2000 K (d) - K
r

Ea,/RT
2-303
ou
ad

(a) = (b) k2= k^e 1

(AIPMT 2008)

(c) (d) 53. For a first order reaction A ■> P, the temperature
Y

(T) dependent rate constant (k) was found to follow

48. In the presence of a catalyst, the activation energy
Re
nd

of a reaction is lowered by 2 kcal at 2TC. The

rate of reaction will increase by the equation log k = — (2000) — + 6-0 ●
Fi

(a) 2 times {b) 14 times

The pre-exponential factor A and the activation
(c) 28 times (d) 20 times
energy E^, respectively, are
49. H20^ is formed in the upper atmosphere through (a) 1-0 X 10^ S-* and 9-2 kJ mor'
the following mechanism
(/)) 6-Os-' and 16-6 kJ mol"'
H2O + (O) > 2 OH > H2O2
(c) 1-0 X 10"' s"' and 16-6 kJ mol”'
The overall enthalpy change and activation energy -I
for the forward reaction are 72 kJ mol"' and 77 kJ (d) 1-0 X 10^ s“' and 38-3 kJ mol (IIT 2009)

mol-' respectively. The activation energy for the 54. Plots showing the variation of the rate constant
decomposition of H2O2 to give back H2O and (O) (k) with temperature (T) are given below. The plot
will be that follows Arrhenius equation is

ANSWERS

44.(a) 45.(c) 46.{b) 47.(d) 48. (c) 49.(a) 50.(h) 51.(a) 52.id) 53.{d)
4/126 New Course Chemistry fXTI^gfZSgwn

58. For a reaction A ^ B, enthalpy of reaction is

/ - 4-2 kJ mor* and enthalpy of activation is 9-6 kj
mor^ The correct potential energy profile for the
/ reaction is shown in the option.

T ► T ►

(a) (b) (a)

t t
PE (/>) PE

Reaction Progress Reaction Progress

T ► T ►

(c) (d) (c)

t id)
T

w
PE PE

(IIT 2010)
55. The activation energy for a reaction at temperature

F lo
T K was found to be 2-303 RT J mor*. The ratio Reaction Progress Reaction Progress^
of the rate constant to Arrhenius factor is
(NEET 2021)

ee
-1
(a) 10 (b) 10“2

Fr
I
(c) 2 X 10-2 (d) 2 X 10-2 59. The slope of the Arrhenius plot In
V
vs —
TJ
of first
(Karnataka GET 2010, 2011) order reaction is - 5 x 10^ K. The value of of
56. Two reactions Rj and R2 have identical pre for
the reaction is:
ur
exponential factors. Activation energy of R 1 Choose the correct option for your answer
exceeds that of R2 by 10 kJ mor^ If /tj and k2 are [Given R = 8*314 JK ’ mol *]
s
ook

the rate constants for reactions Rj and R2

Yo

(a) 41*5 kJ mol-* (b) 83-0 U mol-*

respectively at 300 K, then In (k2/k^) is equal to (c) 166 kJ mol
1
(tO - 83 kJ mol
-1
eB

(R = 8-314 J mole-* K"*)

(NEET 2021)
(a) 6 (b) 4
60. The rate constant (k) of a reaction is measured at
our

(c) 8 (d) 12 different temperatures (T) and the data are plotted
ad

(JEE Main 2017) in the given figure. The activation energy of the
57. For a reaction, consider the plot of In k versus 1/T reaction in kJ mor* is (R is gas constant)
Y

given in the figure. If the rate constant of this

Re

reaction at 400 K is 10-^ s'*, then the rate constant

nd

at 500 K is
Fi

In k 4606 K

irr
(a)R (b)2R
(a) 2 X 10-^ S-* (b) 10-^ S-* 2 1
(c) TT (d) TT
(c) 10-^ S-* (d) 4 X 10^ S-* R R

(JEE Main 2019) (JEE Main 2020)

ANSWERS

54. (a) 55. (a) 56. (b) 57. (b) 58. (i») 59. (a) 60. (b)
CHEMICAL KINETICS 4/127

VII. Collision theory 66. The decomposition of PH3 on tungsten at low

pressure is a first order reaction. It is because the
61. In collision theory of chemical reaction, Z^3 (a) rate is independent of surface coverage
represents
{b) rate of decomposition is very low
(fl) the fraction of the molecules with energies
greater than (c) rate is proportional to the surface coverage
ib) the collision frequency of reactants A and B (^0 rate is inversely proportional to the surface
coverage (NEET Phase II 2016)
(c) steric factor
67. The correct difference between first and second
id) the fraction of molecules with energies equal
order reaction is that
to E, (NEET 2020, Phase 2)
{a) the rate of a first-order reaction does not
VIII. Miscellaneous
depend on reactant concentrations ; the rate of
62. One mole of N2O4 gas at 300 K is kept in a closed a second-order reaction does depend on
container at 1 aim. It is heated to 600 K when 20% reactant concentrations

w
by mass of N2O4 decomposes to NO2 (g). The {b) the half-life of a first order reaction does not
resultant pressure in the container would be

F lo
depend on [A]q ; the half-life of a second-
(a) 1 -2 atm ih) 2-4 atm order reaction does depend on [A] 0
(c) 2-0 atm id) 1 -0 atm (c) a first-order reaction can be catalysed ; a

ee
63. Consider the following statements : second order reaction cannot be catalysed

Fr
(i) increase in concentration of reactant increases id) the rate of a first-order l eaction does depend
the rale of zero order reaction, on reactant concentrations ; the rate of a
(ii) rale constant k is equal to collision frequency
for
second-order reaction does not depend on
r
reactant concentrations (NEET 2018)
AifE^ = 0.
You
(iii) rate constant k is equal to collision frequency k
s
1
k
^ C, if
ook

68. For the reaction scheme, A ^ B

AifE, — 00^
all the reactions are of 1st order and if the rate of
(iv) In /c vs T is a straight line,
eB

formation of B is set to be zero, then the

(v) In vs 1/T is a straight line.
concentration of B is given by
Correct statements are :
our
ad

(fl) (0 and (iv) ib) iii) and (v) ^k

ia) iky + k^) [A] ib) -L [A]
(c) iiii) and (iv) id) iii) and (i/i) k-,- y
dY

ie) ii) and (v) (Kerala PET 2010)

Re

64. The oxidation of a certain metal is found to obey

ic) kyk^lA] id) iky-k2)[A]
(JEE Main 2019)
Fin

the equation A^ = a/ -1-13, where A is the thickness

of the oxide film at time t, a and p are constants. 69. In the reaction x A ■>yB,
The order of this reaction is
i/[A] d[B]
ia) 0 ib) 1 log - = log -1-0-3010
dt d!
(c) -1 id) 2
65. The initial rate of hydrolysis of methyl acetate ‘A’ and ‘B’ respectively can be
(1 M) by a weak acid (HA, 1 M) is 1/lOOth of that of (a) C2H4 and C4Hg
a strong acid (HX, 1 M) at 25°C. The K„ of HA is ib) n-Butane and iso-Butane
(£7) 1 X 10-^ ib) 1 X 10-5
(c) N204andN02
ic) 1 X 10"^ id) 1 X 10-3
id) C2H2 and C2H(-j (JEE Main 2019)
(JEE Advanced 2013)

ANSWERS

61. ib) 62. ib) 63. ib) 64. (c) 65. (a) 66. (c) 67. ih) 68. (/;) 69. (u)
4/128 ^uzdee^'^ New Course Chemistry (XlI)S!2sISI

^ Multiple Choice Questions (with One or More than One Correct Answers)
70. In acidic medium, the rate of reaction between 74. The increase in rate constant of a chemical reaction

BrOj and Br‘ is given by the expression with increasing temperature is (are) due to the fact
(s) that
J[BrOj] {a) the number of collisions among the reactant
di
= it[BrOj] IBr-] [H*]^
molecules increases with increasing
It means temperature
(a) Rate constant of the reaction depends upon {b) the activation energy of the reaction decreases
the concentration of H"*" ions. with increasing temperature
(b) Rate of reaction is independent of the (c) the concentration of the reactant molecules
concentration of the acid added, increases with increasing temperature
(c) The change in pH of the solution will affect {<]) the number of reactant molecules acquiring

w
the rate of reaction, the activation energy increases with increasing
(r/) Doubling the concentration of ions will temperature (West Bengal JEE 2015)
increase the reaction rate by 4 times. 75. According to Arrhenius equation,

F lo
71. The rate constant of a reaction is given by (a) a high activation energy usually implies a fast
k = 2-\x I0"’exp (-2700/RT) reaction

ee
It means that
(b) rate constant increases with increase in

Fr
(a) log k vs. 1/T will be a straight line with slope temperature. This is due to greater number of
27(X) collisions whose energy exceeds the
2-303 R
for
activation energy
ur
(c) higher the magnitude of activation energy,
{b) log k vs. 1/T will be a straight line with stronger is the temperature dependence of the
10
intercept on log k axis = 2-1 x 10
s
rate constant
ook
Yo
(c) The number of effective collisions are (d) the pre-exponential factor is a measure of the
2-1 X 10 cm ^ sec *
rate at which collisions occur, irrespective of
eB

(d) Half life of the reaction increases with increa.se their energy (JEE Advanced 2016)
of temperature. 76. In a bimolecular reaction, the steric factor P was
72. In a hypothetical reaction X Y, the activation
experimentally determined to be 4-5. The correct
r
ad
ou

energy for the forward and the backward reaction oplion(s) among the following is (are)
are 15 and 9 kJ mor’ respectively. The potential
energy of X is 10 kJ mol"’. Then (a) The activation energy of the reaction is
Y

unaffected by the value of the steric factor

(a) Threshold energy of the reaction is 25 kJ
Re

(b) Experimentally determined value of

nd

(h) The potential energy of Y is 16 kJ

frequency factor is higher than that predicted
(c) Heat of reaction is 6 kJ
Fi

by Arrhenius equation
(d) The reaction is endothermic.
73. For the first order reaction
(c) The value of frequency factor predicted by
Arrhenius equation is higher than that
2 N2O5 (g) ^ 4 NO2 (g) + O2 (g) determined experimentally
(«) the concentration of the reactant decreases
(d) Since P = 4-5, the reaction will not proceed
exponentially with time
unless an effective catalyst is added
(b) the half-life of the reaction decreases with
(J£E Advanced 2017)
increasing temperature
(c) the half-life of the reaction depends on the 77. For a first order reaction, A (g) ^2B(g)-nC(g)
initial concentration of the reactant at constant volume and 300 K, the total pressure
(d) the reaction proceeds to 99-6% completion in
at the beginning (/ = 0) and at time t are Py and
Pp respectively. Initially only A is present with
eight half-life duration (IIT 2011)
ANSWERS

70. U-.d) 71. {a.h) 72. {a.b.c.d) 73. {a,b,d) 74. {a.d) 75. {b.c.d) 76. {u.h)
CHEMICAL KINETICS 4/129

concentration [A],, and /j/3 is the time required 78. For the reaction 2 X + Y
k
■> P, the rate of
for the partial pressure of A to reach l/3rd of its
initial value. The correct option (5) is (are) . dP
(Assume that all gases behave as ideal gases)
reaction is = k [X]. Two moles of X are mixed
dt

(«) A (h) A with one mole of Y to make 1 -0 L of solution. At

50 s, 0-5 mole of Y is left in the reaction mixture.
6/3 The correct statement(s) about the reaction is (are)
o

(Use In 2 = 0-693)
c

(«) The rate constant, k, of the reaction is

>
Time 13-86 X 10"^ s~'
[A]o
(C) A id) A (b) Half-life of X is 50 s
c

(c) At 50 s, = 13-86 X 10"^ mol L’' s

CS
a.
I

w
c
o o di
o
c
OL
id) At 100 s, = 3-46 X 10-3 ,„oi L-l

F lo
S
>
Time
dt
[A]o
(JEE Advanced 2021)
(JEE Advanced 2018)

ee
Qa}

Fr
Multiple Choice Questions (Based on the given Passage/Comprehe nsion)

for
Each comprehension given below is followed by some multiple choice questions. Each question has
ur
one correct option. Choose the correct option.

|[|E.ofnpr!eh'ensioni|^ 80. Which of the following plot will be linear ?

s
Arrhenius studied the
ook
Yo

(fl) In k versus T with -ve slope

effect of temperature on the rate of a
ib) k versus 1/T with -ve slope
eB

reaction and postulated that rate constant

(c) In k versus 1/T with -ve slope
varies with temperature exponentially as
{d) In k versus 1/T with -i-ve slope.
k = . For most of the reactions,it
our
ad

81. If the rate of reaction grows 15-6 times on increa

was found that the temperature coeHIcient sing the temperature by 30 K, the temperature
of the reaction lies between 2 to 3. The coefficient of the reaction will be nearly
Y

method is generally used for finding the ia) 2 ib) 2-5

Re

activation energy of a reaction. Keeping

nd

(c) 3-0 id) 3-5

temperature constant, the effect of catalyst 82. If the rate of reaction doubles for 10° rise of
Fi

on the activation energy has also been temperature from 290 K to 300 K, the activation
studied by studying how much the rate of energy of the reaction will be approximately
reaction changes in the presence of catalyst. (fl) 40 kJ mol ih) SOkJmol
-1

In most of the cases, it is observed that -1 -1

(c) 60 kJ mol (d) 70 kJ mol
catalyst lowers the activation energy
83. If X is the fraction of molecules having energy
barrier and increases the rate of reaction.
greater than E.,, it will be given by
79. The pre-exponential factor in the Arrhenus E
id) x = -~ ih) lnAr = —
equation of a second order reaction ha.s the units RT RT
-1
(a) mol L ■ s ib) L mol"' s ●1

-1 (c) .Y = id) Any of these.

(c) s id) dimemsionless

77. Ui.d) 78. ia.h.c.d) 79. [h) 80. (c) 81. (a) 82. ih) 83. (h)
4/130 New Course Chemistry (XII)BSm

lEOmpreiTenslonlH
84. The value of n is
The progress of the ia) 1 (b) 2
reaction A ^ /iB with time is
(c) 3 id) 1-5
represented in the Fig. below :
85. The equilibrium constant K will be
A
(a) 2
- 0.5 (b) 1-2
(c) 0-5
o
2 0.3
(d) 6-67
86. The initial rate of conversion of A will be
o
z
(a) 0-1 mol L^* hr"^
8 0.1
(b) 0-2 mol L-* hr-^
7
1 3 5
(c) 04 mol L“* hr“'

w
TIME/HOUR c=t>
(d) 0-8 mol L-* hr-1

F lo
09 Matching Type Questions

ee
Fr
Match the entries of column i with appropriate entries of column 11 and choose the correct option
out of the four options (a), (b), (c), (d) given at the end of each question.
87. Column I (Reaction)
for
Column II (Order)
ur
(A) 2 HI
H2 +12 ip) 0
s
(B) 2NH3 —> N2 + 3 H2 iq) 1
ook
Yo

1
eB

(C) 2H202 2 H2O + O2 (D I5

(D) COC12 ^ CO + CI2 is) 2
our
ad

id) A-q, B-p, C-r, D-s ib) A-r, B-p, C-s, D-q (c) A-s, C-r, D-p id) As, B-p, C-q, D-r
dY

88. Column I Column II

Re

1
Fin

(A) Zero order reactions ip) t

1/2
OC

[A] 0

(B) First order reactions iq) ^100%

(C) Second order reactions (r) Involves at least two reactants
-ki
(D) Pseudounimolecular reactions is) [A] = [AJo^

id) As, B-q, C-p, D-r ib) A-p, B-r, C-q, D-^ (c) A-q, B-j, C-p, D-r id) A-r, B-p, C-5, D-^

89. Match rate expression in List>I for the decomposition of X with the corresponding profiles
provided in List-ll. and k are constants having appropriate units

ANSWERS

84. (/;) 85. ib) 86. (a) 87. id) 88. (c)
CHEMICAL KINETICS 4/131

LIST-I LIST-II

I. Rate =
k[X]
(P)
X^+[X]
under all possible initial concentrations
ofX
4^
/
X

Initial cone, of X

U. Rate -
*[X]
(Q)
X^+[X] fS

w
where initial concentrations of X are

much less than X^.

F lo
4I.

Initial cone, of X

ee
Fr
*[X]
in. Rate = (R)
X^+[X]
where initial concentrations of X is for
ur
much higher than X^
s
ook
Yo
eB

Initial cone, of X

TV. Rate =
k[Xf
(S)
our
ad

X,+[X]
where initial concentration of X is

much higher than X^

Y
Re
nd
Fi

Time

(T)

Time

(a) I-P; II-Q; III-S; IV-T (b) I-R; II-S; ffl-S; IV-T
(c) I-P; II-Q; ill-Q; IV-R (d) I-R; II-S; IH-Q; IV-R (JEE Advanced 2022)

ANSWERS

89. (a)
4/132 ‘pftetdeep^'4. New Course Chemistry (Xll)S!ZsIS]

p q r s

Matrix-Match Type Questions

f

A I

Match the entries of column 1 with appropriate entries of column II. Each
entry in column I may have one or more than one correct option fror
B
|©i ® :0| ©
column II. If the correct matches are A-p, s ; B-r ; C-p, q ; D-s, then the
correctly bubbled 4x4 matrix should be as follows : '=ii©ii©©'!®!
90. Column I (Units/Property) Column II (Quantity)
D
©I!©i©@!^
-1 ..-1
(A) mol L s (/;) Zero order rate constant
(B) L mol”' s"' {q) First order rale constant
(C) Does not depend upon initial concentration (r) Second order rate constant

w
(D) Affected by temperature (.9) Rate of reaction

91. Column I Column II

Flo
(A) Rate constant of an exothermic reaction (p) increases with increase of temperature.
(B) Rate constant of an endothermic reaction (<7) decreases with increase of temperature,

e
(C) Equilibrium constant of an exothermic reaction (r) . not affected by adding catalyst.

re
(D) Equilibrium constant of an endothermic reaction (.y) not affected by change of initial concentration.

F
A B C D

VI. Integer Type Questions

®®®©
ur
r
DIRECTIONS. The answer to each of the following questions is a fo ooo©
ks
single digit integer, ranging from 0 to 9. If the correct answers to the
©©®©
Yo
question numbers A, B, C and D (say) are 4, 0, 9 and 2 respectively,
oo

then the correct darkening of bubbles should be as shown on the side : ® ®©©
eB

92. The order of the reaction

© © © ®
5Br (ci^) + BrOj (a^) + 6 H'*'(a^) ^ 3 Br2 {aq) + 7 H2O (/) is © ©©©
ur

93. The half-life period of a 3rd order reaction is found to be I hour when we ® © ©©
ad
Yo

.start with a concentration of 0-3 mol L“' of the reactant. If we start with
0-1 mol L"', the half-life period in hours will be ©©©©
94. In the reaction a A + /? B ■> Products, if only concentration of A is doubled ® ® ©©
d
Re

the rate is quadrupled but if further the vessel is reduced to half the volume,
in

the rate becomes 8 times. The overall order of reaction is

F

95. Starting with an initial pressure of 5 atm of azoisopropane, 40% of it decomposes into nitrogen and hexane
vapours in one hour. The pressure in atm exerted by the mixture in atm at this time will be
96. The concentration of R in the reaction R ^ P was measured as a function of time and the following data is
obtained :

[R] (molar) 1-0 0-75 0-40 0-10

I (min.) 0-0 005 0-12 0-18

The order of the reaction is (IIT 2010)

97. A and B decompose via first order kinetics with half-lives 54-0 min and 18-0 min respectively. Starting from
an equimolar non-reactive mixture of A and B. the time taken for the concentration of A to become 16 times
that of B is min (Round off to the nearest integer). (JEE Main 2021)

90. {A-p.s ; B-r : C-p.q.r ; D-p.q..rs) 91. {A-q..s: B-p.s ; C-q.ns : D-p..rs)
92.(4) 93.(9) 94.(3) 95.(7) 96.(0) 97.(108)
CHEMICAL KINETICS 4/133

98. If the activation energy of a reaction is 80*9 kJ mol the fraction of molecules at 700 K having enough
energy to react to form products is e~'^. The value of x is
(Round off to the nearst integer)
[Use R= 8-31 J K"' mol"'] (JEE Main 2021)

99. Sucrose hydrolyses in acid solution into glucose and fructose following first order kinetics with half-life of
3-33 h at 25°C. After 9 h, the fraction of the sucrose remaining is/. Tlie value of/is X 10"" (Round off
to the nearest integer) [Assume In 10 = 2-303, In 2 = 0-693] (JEE Main 2021)

> C -J- >’4 D, concentration of C changes from 10 mmol dm ^ to

100. For a given chemical reaction, yj A + y-, B
20 mmol dm"^ in 10 seconds. Rate of appearance of D is 1 -5 times the rate of disappearance of B which is
twice the rate of disappearance of A. The rate of appearance of D has been experimentally determined to be
9 mmol dm"^ s"‘. Therefore, the rate of reaction is mmol dm ^ s (Nearest integer)

low
(JEE Main 2022)
101. At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55-5 kPa was
340 s. When the pressure was 27-6 kPa, the half-life was found to be 170 s. The order of the reaction is
(integer answer) (JEE Main 2022)

ee
VII.
Numerical Value Type Questions

F Decimal Notation)

Fr
For the following question, enter the correct numerical value, (in decimal-notation, truncated/rounded-off

for
ur
to the second decimal place, c.g., 6’25, 7*00, - 0*33, 30’27, - 127*30) using the mouse and the onscreen
virtual numeric keypad in the place designated to enter the answer.
s
102. Consider the following reversible reaction A (g) ■+■ B (g) v AB (g)
k
Yo
The activation energy of the backward reaction exceeds that of the forward reaction by 2 RT (in J mol"*). If
oo

the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value
eB

of AG° (in J mol"’) for the reaction at 300 K is

(Given : In (2) = 0-7, RT = 2500 J mol"' at 300 K and G is Gibbs energy) (JEE Advanced 2018)
r

103. Consider the kinetic data given in the following table for the reaction, A -h B + C Product
ou
ad

Experiment [A] [B] [C] Rate of reaction

Y

(mol dm ^) (mol dm"^) (mol dm"-^) (mol dm"^ s“^)

6-0 X 10"5
nd

1 0-2 O-I O-I

Re

2 0-2 0-2 0-1 6-0 X 10"^

Fi

3 0-2 O-I 0-2 1-2 X 10"^

4 0-3 0-1 0-1 9-0 X lO"*-^

The rate of the reaction for [A] = 0-15 mol dm [B] = 0-25 mol dm ^ and [C] = 0-15 mol dm ^ is found to
be Y X 10 ●’ mol dm s The value of Y is (JEE Advanced 2019)
A

104. The decomposition reaction 2 N-)05 (g) » 2 N->04 (g) ●¥ O2 (g) is started in a closed cylinder under
isothermalisochoric condition at an initial pressure of 1 atm. After Y x 10'^ s, the pressure inside the cylinder
is found to be 1 -45 atm. If the rate constant of the reaction is 5 x 10"^ s"', assuming ideal gas behaviour the
value of Y is (JEE Advanced 2019)

ANSWERS

98.(14) 99.(16) 100.(1) 101.(0) 102.(-8500) 103. (6-75) 104. (4-60)

4/134 New Course Chemistry (XII)BSm

VIII. Assertion-Reason Type Questions

TYPEl

DIRECTIONS. Each question given below contains Statement-1 (Assertion) and Statement-2 (Reason).
Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correct
option as under :
(a) Statement-1 is IVue, Statement-2 is IVue ; Statement-1 is a correct explanation of Statement-1,
(b) Statement-1 is True, Statement-2 is IVue ; Statement-2 is NOT the correct explanation of Statement-1,
(c) Statement-1 is True, Statement-2 is False,
(d) Statement-1 is False, Statement-2 is IVue.

105. Statement-1. The dissociation of NH3 on hot platinum surface may be reaction of zero order or first order.
Statement-2. Pressure of the gas affects the order of reaction.

w
106. Statement-1. In the reaction, N2 + 3 > 2 NH3, the rate of reaction is different in terms of N2, H2 and
NH3.

F lo
Statement-2. Rate of disappearance of N2 and H-, and rate of formation of NH3 are not equal to each other.
107. Statement-1. Rate constant of a zero order reaction has same units as the rate of reaction.

ee
Statement-2. Rate constant of a zero order reaction does not depend upon the units of concentration.

Fr
108. Statement-1. In a zero order reaction, if concentration of the reactant is doubled, half-life period is also
doubled.

Statement-2. The total time taken for a zero order for

reaction to complete is double of the half-life period.
ur
109. Statenient-1. The rate constant of a pseudounimolecular reaction has the units of the rate constant of a
second order reaction.
s
ook

Statement-2. A pseudounimolecular reaction is a reaction of second order in which one of the reactant is
Yo

present in large excess.

eB

TYPE II

DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a
our
ad

corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as
(a) If both assertion and reason are true and reason is the true explanation of the assertion,
(b) If both assertion and reason are true but reason is not the true explanation of the assertion,
dY
Re

(c) If assertion is true, but reason is false. (d) If both assertion and reason are false.
Fin

110. A.ssertion. In any reaction, the rate of disappearance of a reactant is same as the rate of reaction.
Reason. Both rate of reaction and rate of disappearance of a reactant represent decrease in the concentration
of a reactant per unit time.
111. Assertion. The molecularity of the reaction H2+ Br2 -> 2 HBr is 2.
Reason. The order of the reaction is 3/2. (AIIMS 2004)
112. Assertion. The order of a reaction can have fractional value.
Reason. The order of reaction cannot be written from the balanced equation of the reaction. (AIIMS 2008)
113. Assertion. Half-life of a reaction is independent of the initial concentration of the reactant.
Reason. Half-life of a reaction is independent of the order of reaction.
114. Assertion. The hydrolysis of methyl acetate by dilute HCl is a pseudo first order reaction.
Reason. HCl acts as a catalyst for the hydrolysis. (AIIMS 2007)

ANSWERS

105.(a) 106.(d) 107. (c) 108.(b) 109.(n) 110. (d) 111. (h) 112. (h) 113. (d) 114. (h)
CHEMICAL KINETICS 4/135

115. Assertion. Average life of a radioactive element is tliat period in which 63% of it is decayed.
Reason. Average life t = 1-44 tj/2- (AIIMS 2007)
116. Assertion. By 10° rise of temperature, the rate of reaction increases by 100% to 200%.
Reason. By 10° rise of temperature, the collision frequency increases by 100 to 200%.
117. Assertion. A catalyst increases the rate of a reaction.
Reason. In presence of a catalyst, the activation energy of the reaction decreases. (AIIMS 2011)
118. Assertion. According to transition state theory, for the formation of an activated complex, one of the vibrational
degree of freedom is converted into a translational degree of freedom.
Reason. Energy of the activated complex is higher than the energy of the reactant molecules.
(AIIMS 2006)

m Multiple Choice Questions (Based on Practical Chemistry)

w
119. In the study of oxidation of iodide ions by (c) blue colour will appear only on heating the

F lo
hydrogen peroxide in presence of an acid and reaction mixture
starch as a clock reaction, (d) blue colour will appear after a very long time
(a) thiosulphate added should be in large amount

ee
121. In the study of oxidation of iodide ions by
(b) thiosulphate added should be in small amount hydrogen peroxide in presence of dilute H2SO4,

Fr
(c) amount of thiosulphate added has no thiosulphate and starch as a clock reaction, if
importance concentration of iodide ions is increased

for
(d) thiosulphate is added to speed up the (a) less lime will be taken for blue colour to
ur
reaction appear

120. In the study of oxidation of iodide ions by (b) more time will be taken for blue colour to
ks
hydrogen peroxide in presence of dilute H2SO4 appear
Yo
and starch as a clock reaction, if excess of (c) same time will be taken for blue colour to
oo

thiosulphate is added appear

eB

(n) blue colour will appear immediately (d) any one of the above can happen depending
(b) blue colour will not appear at all upon the amount of H2SO4 present.
r
ou
ad
Y

For Difficult Questions

nd
Re

Multiple Choice Questions correct Answer)

Fi

1. For the given chemical reaction

-d[Np^]
Rate of reaction Rate of disappearance of N-)05 dt

d[N^J _ 1 c/[H,]_ 1 r/[NH3] (2-4-2-0)

di ~ 3 dr ~ 2 dt mol L
20 min
2. 2 N2O5 4 NO2 + O2
0-4
1 [d[NO,] mol L ' min ^
Rate of reaction = - = +— 200
2 dt 4 dt
= 2 X 10"^ mol L"’ min -1
...(/)

ANSWERS

IIS.(M) 116. (c) 117. (d) 118. (a) 119. {/;) 120. (A) 121.(0)
4/136 “P>uuCec^'4i New Course Chemistry (Xll)iZEIB]

For Difficult Questions 6. Rate of reaction = - 1 ^[NO^]

2 dt

From eqn. (/), = ^1 [N02]2-^2 [N2O4]

d[NO^ Rate of disappearance of NO^
Rate of production of NO2 dt dlNO^]
dt
= 2/:, IN2O4].
= -4x-
I ^/[N^Og]
2 dt
7.

ow
= - 2 X (- 2 X ! 0'^) mol L“' min“’ dt 2 dt dt
= 4 X 10"'^ mol L'* min”’
3. N2 + 3H2 >2NH3 A-,[N205] = -/:,'[N,05] = 2^j"|N205l
1
d[N^]_ 1 d[U^] _ 1 ^[NH3]

e
//

Rate = - or L =~k/ = 2k or 2k.I = k/ = 4k,” .

dt 3 dr ~ 2 dt 1 2 ’ 1 I

re
Rate of disappearance of H2,
8. r^=k [A], r2 = k [A]“, r^ = k [A]^
If [A] > 1 M, T3 > T2 > T| or rj < ^2 < r3

Flr
F
_ 6xl0”^mol 9. Net reaction = Rate of forward reaction - Rate of
i.e.. backward reaction
dt lOmin

= 6 X 10”'^ mol min”’

ou = kf [Reactants] - k^, [Products]

sr
Comparing with the given equation,
Rate of formation of NH3,

kfo
kf=2x\[P, k^=\x 10-2
^[NHg] _ 2 ^[H2] Eqm. const. = k^lk^ = (2 x 102)/(10”2) = 2 X 10'^.
i.e.,
3l dt J
oo
dt 10. 2H2(g) + 2NO(g) ± N2(g) + 2H20(g)
Equilibrium constant,
Y
2
— X 6 X10”^ mol min”’
reB

3 K ...(0
[H2]^[N0]2
eq
= 4 X 10”2 mol min”’ h
uY

Now, if dt = 0-3 min, or

NH3 formed = (4 x 10”^ mol min”’) (0-3 min) At equilibrium.

= l-2x 10”^ mol
ad

rate of forward reaction = rate of backward

do

reaction
4. CH3COOH+CH3OH ^ CH3COOCH3 + H2O
IM 2M
But rate of forward reaction = k^ [NO]^ [H2I
in

Rate (ri) = k [CH3COOH] [CH3OH] (Given)

Re

= k(\)(2) = 2k Rate of backward reaction = kj- [NO]^ [H2]

F

When each solution is diluted with equal volume = kyK eq ■ [NO]2 [Hj]
of water, concentration of each is halved. Now,
[N2HH2OI2
~x[NO]2[H2l
rale
= *^x
[H2]2[N0]
'2 = =*fi2 \(22 k .
2k
- = 0-25
4
V y V /
2
[N2][H20]2
= k^x
or
r2 = 0-25 rj [H2I
n 11. (/) r = ka^b^ 0V)8r=i:(2a)“(2 h)^
5. C = -
V dt V dt (Ui)2r = ka^(2 b)^

Eqn. (/7)/Eqn. (in) gives ^ = 2“

n
As PV = n RT, i.e., P = —RT = CRT
V
or 2“ = 4 = 2^ or a = 2.
p
or C = — Eqn. (i7i)/Eqn. (/) gives 2 = 2^ or p = 1. Hence,
RT dt " i/r [ RT ^ RT dt
rate = k [A]* [B],
CHEMICAL KINETICS 4/137

I
For Difficult Questions 18. Time taken to reduce by factor of — = 20 min.
1
12. At pH = 2, i.e., 1H+] = \0~^ M Time taken to reduce by factor of ~,
r = it (10-2)"
At pH= 10“' M i.e., ty2 = 10 min
t I t /
1/2 a 1/2 a 1/2 a 1/2 a
lOOr 10-"
100r=/t(10-'f or 100= 10
n a >—
2
>—
4
»—
16
r 10-2«
Hence, n = 2. Total lime required = 4 x /jp = 4 x 10 min = 40 min
19. Half-life period of a second order reaction
13- As rate depends upon the slow step.
1 I
Rate = k [A] IB2] /
1/2 -
Rate constant x Initial cone, ka
[A]2
For the equilibrium, A2 ^ - 2 A, K,, = I
IA2]

w
10“^ L mol"' min"' X 10"^ molL"'
or
[A] = 7kja^ = 10^ min
Ratc = *:Ky2[A,]l'2[B,l =nA,l''2[B,l

F lo
20. 2A(g) -> B(5) + C{s)
1 Initial Po 0 0
Hence, order =—4-1 = 1
2 After time / Pq - 2 p 0

ee
2 P

14. 9r=it(3-24x lO'^) n

Pressure after time f, P, = Pq - 2 /? -t- p = Pq - p

Fr
...(/)
r = it (1-2 X 10“Y ...Hi) After completion 0 P(/2, Total = P,/2
Dividing (/) by (/i).
for
Thus, pQ - p = 300 pascals, -^ = 200 pascals
P,
ur
V'
3-24x10-2 u
9 = or 9 = (27)
1-2x10"^
s
Pq = 400 pascals. p = 100 pascals
ook
Yo

or (3)2 = (33)" = (3)3". « Pq, xoc 2 p

eB

Hence, 3 n = 2 or n = 2/3. P.
2-303 a 2-303 0
IS, For a zero order reaction, it = log log-
t a — X t
0 -2p
_ 2 0molL~'
our
ad

2r 2xlhr
= 1 -0 mol L ‘ hr ^ 2-303 400 2-303
1/2 log x0-.3010min
10 400-200 10
1 1
= 0-0693 min *.
Y

t = - {[A.] - [A]} = -(0-50-0-25) = 0-25 hr

k ^ 1-0
Re

21. To calculate concentration after 10 mm :

nd

16. At the point of intersection, [A] = [B], i.e., half of

the reactant has reacted. Hence, it represents t 1/2- , 2-303, [A] 0 2-303, 10
Fi

k = log 0-0693 = log —

17. t 1/2 for 1st order reaction =
0-693 t [A] 10 ^lA]
it
10 10
or log = 0-3010 = 2
[A] 0 or

tyj for zero order reaction = [A] [A]

2 k'
or [A] = 5 mol L-'
Initial rate for 1st order reaction (rj) = /r [A]q Rate of reaction after 10 min = k [A]
Initial rate for zero order reaction {kq) = kf = 0-0693 X 5 mo! L“* min“‘
''1 k[A] 0
22. Suppose order w.r.t. A, B and C are a, p and 7
it' respectively. Then
'b
0-693 _ [A1 0 5-0 X 10-3 = (0-010)" (0-005)P (0-010)7 ...{/)
But (Given)
it " 2k' 5-0 X 10-3 ^ (O-OIO)" (0-005)P (0-015)7 ...(//)
1-0 X 10-2 ^ (0.010)“(0 010)P (0-010)7
*[A] 0 'i
= 2x0-693, i.e., -i- = 2x0-693. 1-25 X 10-3 = (0-005)" (0-005)P (0-010)7 _{iv)
it' 'b
4/138 'a New Course Chemistry (XII)

27. The given rate law shows that it is a reaction of

For Difficult Questions 1st order. For 1st order reactions, [A] = [A]q
Dividing (i) by («), 1 [Alo
When f = -, [Al = [AJoe '^=[A]qC
-1 _

^Q-OIQ \Y /2 y "2 h e
1 = or Y=0
0015
3J 28. Decomposition of H2O2 takes place as :
1
Dividing (///) by (ii),
» H-0 + -0,
2 22

2=(2)Pf|
10x10-2
= 2P
5-0x10" j = (2)P v-’ or Decrease of concentration of H2O2 from 0-5
to 0-125 in 50 minutes means two half lives
or 2P = 2 or (3=1. (0-5 0-25 ->0-125)
Dividing (/) by C/v), 4 = (2)“ or a = 2 i.e. 2x/ 1/2 = 50 min or t 1/2 = 25 min

.●. Orders w.r.t. A, B and C are 2, 1,0. For a 1 st order reaction,

w
23. When initial pressure is halved, the half-life period
is also halved. This shows that t 1/2 oc initial j. _ Q-693 _ 0-693 min
. -1

t 25
pressure. This is so for reactions of zero order.

F lo
1/2

24. By hit and trial method, if reaction is of zero order. Rate of reaction = Rate of decomposition of
1

ee
k=-[[A]fi-[A]]
t H2O2, i.e., = /c[H202]

Fr
dt
1 25
Case I. k = —x = 10-2 0-693
25 100
for
25
X 0-05 = 1-386 X 10"^ mol L * min'*
ur
1 1

Case n. ^ ^ ^ 2 ^ ^ halved) ^[0,] U[H,0^1

s
Rate of formation of O2 =
ook

dt 2 dt
Yo

1 1 50
= 10-2
"25^2^100
eB

= - X 1-386 X 10"^
1
Casein. *= —x2{[Alo-[A]} = 6-93 X 10^ mol L"' min"’
our
ad

(as [P] is doubled) 29. For 1 St order reaction,

1 25 2-303 a 2-303 100
= —x2x = 10-2 — log = —— log
Y

t =
50 100 k a- X k 100-99
Re
nd

As k comes out to be constant, hence the given 2-303 2-303 . ,/^ 4-606
reaction is of zero order. log 1Q2 = — x2 1ogl0 = ——
Fi

k k k
2-303 2-303
25. t 1/4 ~ log
a
,log-4
n-1
K (3/4)a K
30. / 1/2 OC
1
n—1 ’
^^1/2)1 ^ ^Aq]^ * _ [[Aq]2
[[AoliJ
n-1
2-303x0-125 0-29 [Aq] ^^1/2^2 ^AqIi
K K
sn-l
I (2a
26. t 1/2 OC
(Cq)' " or /j/2 = " (K = constant) or 2 = (2)
n-1

t/2 a
-■- log hl2 = log K + (1 - n) log Cq (y = nu + c)
or /I - 1 = 1 o r n = 2.
Thus, plot of log /j/2 vs log Cq will be linear with
slope = I - n. But as graph is parallel to log Cq *1 1-26x10-^
axis, slope = 0, Le., 1 - /i = 0 or n = 1. Hence, it is 31. %ofB = xlOO
a reaction of 1 st order and for 1 st order reactions, fe] + kj 1-26x10'^+3-8x10-5

2-303, - 1-26x10^
t
1/2 “ — log2. XlOO = 76-83%
K 10-^(1-26 + 0-38)
CHEMICAL KINETICS 4/139

.'. Pressure of A after 2 hours = 0-5 - 0-1 = 04 atm

For Difficult Questions
Rate constant in shows that it is a reaction of
I St order.
32. For a second order reaction, 2 A ^ Products,
Rate after 2 hours = /: [A] = /: P
1 1 1
k=- = (I0“^ s"') (04 atm) =4-0 x 10^ atm s“*.
t (A1 [A] 0 36. For a first order reaction, A Products, and
for concentration of the reactant at two different
times.
2-303 . [A1
4 k = log (Refer to page 4/28)
{^2 ^1) [Al2
'0Qti/2 . log k Also for first order reaction, rate a [A]. Hence.

ow
2-303 (Rate) I
k = log
h h (Rate)2
logac=t> /
2-303 0-04
log = 0-0287 s-‘
[A] 0 (20-10) VO-03

e
When t -1 1/2 , fA] =

Fl
re
2 ■ 0-693 0-693
/ = 24-14s
1/2- ^ 0-0287 s“*

F
1 J1 1_ 1
Hence, ^1/2 = 7 37. Suppose order of reaction = x. Then
k [A] OJ
0 A: [A] 0
ur
or
1st case : Rate = A: [CH^CHO]-"
log r,/2 = - log * - log [A]q = - log - log a sf 1 = k (363 X 0-95Y
Plot of log ty/2 vs log a will be linear with intercept (95% CH3CHO is present)
k
Yo
= log k. or 1 = k (344-85>^ ...(0
oo

33. For 1 st order reaction,

2nd case : 0-5 = k (363 x 0-67>*
0-693 0-0693 0-693
B

t or k,I = (V 67% CH3CHO is present)

k t 40 s
1/2 0-5 = k (243-2\y ...(H)
re

or

For zero order reaction. Dividing eqn. (/) by eqn. (if).

u

[A] 0 [A]g _ 1-386 mol dm ^

ad

or k 0 344-85 V
Yo

t
1/2 “ “
2k It 2x20s
1/2
0-5 1^243-21 )
fc, _ 0-693 X 2x20s
2 = (\-4\AY or x =2
d

- 0-5 mol ^ dm^ or

Re

40s l-386moIdm“^ 38. The units of the rate constant show that it is a
in

reaction of zero order. Half-life for a reaction of

34. As half-life period is independent of initial
F

concentration, hence it is a reaction of 1 st order. zero order is given by

2-303 a
[A] 0 _
t
k = log 1/2 “ = 50 years
t a— X 2k 2 X 0-05 p g/year
But V OC
a and V, oc x. 0-693 0-693
39. V2= — or A: =
i
2-303 50 1/2
Hence, k = log
10 50-10 2-303 100 2-303 / 1/2
35.
t.
90% - log
A(g) > 2B(g) -f C(g) (0-693//J/2) '' 100-90 0-693
Initial 0-5 atm
= 3-32r 1/2
After 2 hrs 0-5 - p 2p P
40. As ty2 is independent of initial concentration, it
Total pressure after 2 hours = (0-5 -p) + 2 p + p is a reaction of 1st order. Hence,
= 0-5+2 p 0-693 0-693
0-5 + 2p = 0-7 or p = 0-1 atm k = s-1
t 200
1/2
4/140 New Course Chemistry (XII)BZS9]

46. According to Arrhenius equation,

For Difficult Questions

T -T 'i
^'2 E
a ‘2 M
2-303 100 log — =
k 2-303R T T
/
80% “ log = 462-2 log 5 1 ‘1 ^2
0-693/200 100-80

= 462-2 X 0-699 6 E 400 - 350

a
log - =
= 323 s 1 2-303x8-314 350x400

41. For zero order reaction, rate = k [A]° or rate = k

E 50
(constant).

ow
a
0-7782 = X
2-303x8-314 350x400
Hence, plot of rate (j-axis) vs concentration (a-
axis) will give a horizontal line. 0-7782 x 2-303 x 8-314 x 350 x 400
or E
u
0-693 . 50
For 1 St order reaction, t1/2 -, i.e.. t|/2 of a

e
k = 4172051 = 41-72 kJ

re
1 St order reaction is independent of concentration. -E..Cl /RT -E.. /RT
47. k.I =A,e 1
j k^ - ^2
Hence, plot of (.v-axis) vs concentration (x- 1

Frl
F
axis) will be a horizontal line. k
42. As explained on page 4/31,
1
^^(E„^-E„,)/RT ^^(2E„,-E.,)/RT
A.
1
ou
sor
/
1/2
^av = = 1-44/ 50% (●.● E
k 0-693// 1/2 0-693

Also, we know that /75% = 2x/50%-

or k^-=k^Ae kf E. /RT
oo
Hence, /gy^^ < / av
< /
75%- -E,/RT
43. Rotation in time t = 48. it, =Ae a
Y
it2 = Ae-<E<i-2)/RT^A^-E„/RT^2/RT
B

Rotation in time ■» = (rQ - r^) « a

50% inversion means x = all
^_g2/RT ln^= ^
re

or
Y

k k RT
] i
u

i.e.
or 2 Tq - 2 r, = rg - r.
2 Putting R = 2 X 10"^ kcal K-‘ mor^
ad
do

and T = 300 K, we get

1
or 2 r,f = rn0 + r or
'i =r('b + '-=o)
2
k 2
log-^ = = 1-4473
in

44. /99-9 = lOx/ 1/2 *1 2-303x2x10-3x300

Re

(Refer to Solved Problem 4, page 4/51) ^2

F

^
k
= 28 or it2 = 28xit,
0-693 0-693 'l
= lOx
k ●■■'1/2- ^ (or put e = 2-7 and calculate)
49. Activation energy for backward reaction
= 77 -72 = 5 kJ mol-'.
= 10 X 0-693 X /av <n-
k )
= 6-93 X /av’ i.e., 6-93 natural life limes. t 'V
A I 3^Eb=?
1 /
t

77 kJ
^ H2O2
_ '1/2
45. /av = 1 -443 / 1/2 I 72 kJ
k 0-693 o I
0: I
I
liJ I
= 1 -443 X 100 min = 144-3 min i / y
UJ

As explained on page 4/31, average life is the time H20-*-0

in which the concentration is reduced to Ue of PROGRESS OF REACTION
the original concentration.
CHEMICAL KINETICS 4/141

2-303 RT
For Difficult Questions
2-303 log k = 2-303 log A - RT

50. Rate = k fReactantJ or log /: = log A - 1

In the presence of catalyst, at the same concen k k
-- = 10-*.
or log— = -1 or
tration of the reactant, as rate increases, k increases. A. A
-E../RT
Alternatively, k=Ae . In presence of
a -E.. /RT -E., /RT
“2
56. k. = Ae
catalyst at the same temperature, decreases. 1 , k^ = Ae
Hence, k increases.

ow
I

A, *2
^2
51. k = k
1
*3
E.a -E a
iOxlO^
^^-E,/RT^^-E2/RT or In -=■ =
! 2 _
= 4

e
Ae"^T = k RT 8-314x300
A^-E3/rt I

re
-E
^-E^T =^(-Ei-E,+ E3)/RT 57. Arrhenius equation is \nk =
a
-h In A

Flr
or
RT

F
or
E -E| -E2 +E3 Thus, plot of In k vs 1/T is linear with slope
RT RT
ou -E
a

sr
R
or
E = E| 4- E2 - E3 = 40 50 - 60
= 30 kJ mor*

fo
k^ E 1
ln-^ =

k
a

52. *1 = 10*6 ^2=1015 e'

1000/T
k
1
R ,T
1 T2J
When k^ = k2, 10’^ e-2000/r ^ jqIS ^-iooo/t
oo
or IQ g-2()00/T _ g-lOOO/T *2 1 1
Y
2-303 log = 4606
Taking natural logarithm of both sides, we get 10“^ 400 500
reB

2000 -1000 2000 -1000

InlO- or 2-303 -
kj _ 4606 1
uY

T T T T log
10"^ 2-303 ^ 2000
1000 1000
or
T
= 2-303 or T =
2-303
K or
log k2 - log 10'^= 1 or log k2 + 5 = 1
ad
do

or
log /:-, = - 4 or ^2 = ●
2000
53. Given log/: = 6- 58. As A,H = Hg - and enthalpy of reaction is
in

T negative, it can be so only when > Hg i.e. PE

Ea of A is greater than PE of B.
Re

Comparing with log k = log A -

2-303 RT This is so only in diagram (b). In such cases,
F

E
a
activation energy of backward reaction (E/,) >
= 2000
log A = 6, i.e., A= lO^s ’ and 2-303 R activation energy of forward reaction (Ey). This
is again so in diagram (b) only.
or
E„ = 2000 X 2-303 x 8-314 J mo|-‘
= 38294 J mol-* = 38-3 kJ mol"* In the present case, Ey= 9-6 kJ mol"*,
= 9-6 4-4-2 = 13-8 kj mol”*
-e./RT A
54. k = Ae a
. Thus, as T increases, 59. According to Arrhenius equation.
E /RT
e "
-E./RT
k = Ae a

the exponential factor in the denominator decreases

or Ae
-E_/RT
a increases. Hence, k increases or In A = In A 4- In e~^°'
expionenlially as T increases. (Refer to page 4/64). E E a f \ '\
a
or In /t = In A = InA-
E
a RT R l,T
55. 2-303 log cf = 2-303 log A-
RT
E
a
Putting E^ = 2-303 RT (Given), we get Hence, slope of the plot of \n kvs —
T
= R
4/142 New Course Chemistry fXlD^gmn

^lA]
For Difficult Questions represents the rate of increase of thickness
di
E
or -5xlO-^K = - a of oxide film with time.
8-314JK"‘mol
65. Rate w.r.l. weak acid, R, = /: [ester]
or E„ = (5x 10^ K) (8-314 JK-' mor') Rate w.r.t. strong acid, R2 = ^ Lister]
= 41-57 X 10^ Jmo!-'
= 41-5 kJ mol-‘ R, _[H+1 WA _
(Given)
-E../RT R2 [H^] SA 100
60. k = Ae

Ea
In/: = In A- [H-^] SA _ 1
RT ●●● fH"]wA = 100 100
xlM =0-01M

w
E
a l(p = 10-- M
In/: = In A- X
RxlO^ T
HA H-" + A"

Flo
(Multiplying numerator and nominator by 10^) Initial C mol L“
EO

e
10
At eqm. C- C a C a C a

re
Slope of the graph RxlO^ 5
= C(1 - a)

rF
E„=2Rx 10-M = 2RkJ
a
[H+] = C a or 10-“ = 1 X a or a = 10"“
61. In collision theory of chemical reaction, Z^g
ur
represents the collision frequency of the reactants Ca.Ca Ca^
A and B
K
fo a
C(l-a) 1-a
-Ca2
ks
62. N2O4 ^ 2NOo
= 1 X (10-2)2= 10^
Yo
Initial moles 1
oo

Moles after dissociation 1 -0-2 0-4 66. The decomposition of PH3 on tungsten at low
B

= 0-8 pressure is a first order reaction because rate is

Total moles after dissociation = 0-8 -1- 0-4 =1-2 proportional to the surface coverage.
re

Initial temp. = 300 K 67. The half-life of a first-order reaction is

u

PjV = /![ RTi or 1 X V = 1 X R X 300 ...(0 independent of [A]q whereas half-life of a second
ad
Yo

Temp, after dissociation = 600 K order reaction is inversely proportional to [A]q.

68. Rate of formation of B = Rate of formation of B
^‘2^2 ~ ”2 Of ?2 X V = 1 -2 X R X 600 ...(i7)
from A - Rate of decomposition of B to C
d

Dividingeqn. {ii) by eqn. (/),

Re
in

1-2 X 600 ^/IB]

^2 = = 2-4atm. i.e.. = k^[A]-k^[B]
F

300 dt

63. Rate of a zero order reaction is independent of

^/fB]
concentration. Hence, (/) is wrong. Setting - 0 means 0 = [A] - /.'2 [B]
dt
-E../RT
k = Ae a

k
When E^ = 0, k = A. When = oo,k^A. or
k2lB] = k^\A] or [B]=-L[A]
k.
Plot of In k vs 1/T is a straight line.
Hence, only («) and (v) are correct. 69. The given expression can be written as
64. a2 = ar + ^ [^A] ^[B]
Dilferentiating this equation, we get log - = log + log2
dt . dt

2 AdA = a.dt dA _ a a -1
or
dt ~ 2A = ^|AJ [^/B]
log = log 2X
Hence, order of reaction = - 1. dt dt
CHEMICAL KINETICS 4/143

For Difficult Questions

or [dA] ^ X f/[BJ ...(//)
df V dt
[dB]
or = 2x ...(0 Comparing eqns. (/) and (//).
dt dt
.X
For .1" A
V

This is so for the reaction 2 C^H4

.V dt '> dt

01 Multiple Choice Questions (with One or More than One Correct Answers)

ow
70. (a) is wrong because rale constant is constant at Amount reacted = 0-996 of [AJo, i.e. 99-6% of
constant temperature, (b) is wrong because rate [A]y. Hence, (d) is correct.
depends upon (c) is correct because change 74. Both (a) and (d) are the correct reasons.
in pH means change in [H'*'], {d) is correct because

e
75. (a) is incorrect because high activation energy

re
rate «= | H'*’]-
implies a slow reaction (/;), (c) and (d) are correct
71. (c) is wrong because frequency factor gives total

rFl and follow from Arrhenius equation, viz.,

F
number of collisions and not effective collisions
-E./RT
cm"^ sec"’. k = Ae a

r
(d) is wrong because half-life of the reaction
ou
-E./RT

fo
76. Actual k = PAe a
decreases with increase of temperature (as reaction ks
becomes faster).
-E./RT
Theoretical k = Ae a
oo

72. is independent of steric factor. Hence,

J
Y
A t option (a) is correct.
B

9 kJ
k
actual
re

0
a: 15 kJ Further, P= -
LU
z
? Y T (a) k
theoretical
ou

UJ (c)
Y
ad

(b) (actual means observed or experimental value,

10 kJ
X theoretical means predicted by Arrhenius
▼ ^
equation)
d

PROGRESS OF REACTION =t>

As P = 4-5,ic actual >k theoretical
in
Re

This can be so when A actual > A theoretical

All (a) to (d) are correct.
F

Hence, option (b) is correct.

73. [A] = [A]q e~^'. Hence, (a) is correct. 77. A (g) ■» 2 B (g) -f C (g),
0-693 V constant, T = 300 K
hn = . As k increases with increasing / = 0
k Po
temperature, therefore, /[/2 decreases. Hence, t -1 R.0 -
2P,0 4P,0 2P,0
(b) is correct, 1/3
3 ) 3 3

(c) is incorrect because t^/2 of a first order reaction R

_ ^0
is independent of initial concentration.
3

[A] 0 t -1 Po- X 2x X
Amount left after 8 half-lives =
28 .-. ?.f = ?n-x+2x^>l■x
0 = ?[^^^2x
0 or 2x = P,-P
t 0

1 Po
of [A]y = 0-004 of [A]y
256 k(?Q-x)
4/144 ^%/xdee^'^ New Course Chemistry (XlI)CZSISl
78. 2X + Y > P
For Difficult Questions
/ = 0 2 0

/ = 50 2 - 2 X 0-5 0-5
Po = 1
or k (P,-Pq)
Po- , 1 4/X lN cl?
Rate = - — =k [X] iGlvm)
1 2R0 2 (h d{ (It
= — In
k 2Po-P,+Po At 50 s, half of Y has reacted, therefore t 1/2 = 50 s.

1 2P,0 In 50 s, 1 mole of X reacts, therefore, ty2 ^

= — In
k 3P.-P, = 50s
0 /

0-693 0-693

w
2R0 k = s-' = 13-86 X 10''-^ s"‘
or ki = In t 50
3Pn-R
0 /
1/2

or
/t/ = In 2 Pq - In (3 Pq - P,) 1 dX cIP

Flo
At 50 s, rate = - — = k\X]
or
In (3 Po - Pf) = - /:/ -f 2 In Py 2 dt dt

e
Thus, graph between In (3 Pq - P,) vs is a = (13-86 X 10”^ s"l) 1 mol L"‘

re
straight line with -ve slope.
= 13-86 X 10-3 mol s"’

F
So, (a) is correct.
At 100 s (two half-lives), amount of Y left
1
^ = -in3 = 0-25 mol L-‘
ur
/,y3 - In

or
k P./3
0
k
r/Y d?
Thus, t 1/3 is independent of initial concentration. f
Rate - — = k[Y]
ks
dt dt
So ib) is wrong. As rate constant is a constant
Yo
quantity and independent of initial = (13-86 X ir3)(0-25)
oo

concentration, so graph (d) is correct. = 3-46x 10-3 mol L"’ s"'

B

[m Multiple Choice Questions (Based on the given Passage/Comprehens ion)

u re
ad

79. The factor A has the same units as those of k. For 84. From time t - I hour to / = 5 hour, A / = 4 hr
Yo

a second order reaction, k has the units of L Decrease in cone, of A = 0-5 - 0-3 = 0-2 mol L"’
mol"^ s“^ (For a first order reaction, it has the Increase in cone, of B = 0-6 - 0-2 = 0-4 mol L”'.
units of s"’)
d

Thus, in the same lime inler\'al, increase in cone

Re
in

Ea of B is double than decrease in the cone of A.

80. In A: = In A - . Hence, plot of In k versus 1/T Hence, n = 2.
F

RT
85. At equilibrium, there is no further change in cone.
is linear with slope = - EyR. Hence, concentrations at equilibrium are
81. If n is the temperature coefficient, after 10 K rise, [A] = 0-3 inol L“’, [B] = 0-6 mol L“’
rate constant will increase to 2 /i, after next 10 K
rise to 4 n , nd after further 10 K rise to 8 n. Hence,
Eqm. const. K = [B]3 ^ (0-6)3 = 1-2 mol L
8 /I = 15-6 or n = 1-95 = 2. [A1 0-3
82. On calculation, comes out to be nearly 86. From / = 0 to / = I hour, for A,
50kJmol-'.
A .r = 0-6 - 0-5 = 0-1 mol L"’
83. Fraction of molecules having energy greater than
A.r
Initial rate of conversion of A =
E^ at temperature j = (Boltzmann At

E 0-1 molL"'
a
factor), i.e., x = ● Hence, In x = - = 0-1 mol L ’ hr '
RT I hr
 CHEMICAL KINETICS 4/145

For Difficult Questions

09 Matching Type Questions

89. II Rate = ^IX] _k[X] = 1 St order III -> Rate = *[X]

= k = zero order
X^+[XJ X s
X^+[X]

IV ^ Rate =
*[xp
= k [X] = 1st order
X^ + [X]

VI.
Integer Type Questions

w
92. Rate law equation for the given reaction is

F lo
95. (CH3)2CHN = NCH(CH3)2 (g)
N2 (5) + C6H,4 (g)
Rate = ^{Br-][Br03][H-"]2 Initial pressure 5 atm 0 0

ee
Hence, overall order of the reaction is 4.

Fr
40
After 1 hour 5 x5 2 atm 2 atm
1 100
93. t1/2 = k
[A] 0
n-1 ■
for= 5-2 atm = 3 atm
ur
Total pressure after one hour = 3 + 2 + 2 = 7 atm.
1 96. Rate of reaction in the interval 0 0 to 0 05 min
s
For 3rd order reaction, r i/2 = k
ook

[AJo'
Yo

1-0-0-75
= 5 mol L * min
0-05
eB

1 Rate of reaction in the interval 0-05 to 012 min

\=k or k^Q-09.
f0-3)2 0-75-0-40
our

= 5 mol L ’ min ’
ad

012-005
When [A]() = 0-1 mol L“',
Rate of reaction in the interval 012 to 018 min
Y

i,/2 = (0-09) = 9 hours. 0-40-010

(0-1)2
Re

= 5 mol L ' min

nd

018-012
94. r = k [A]“ [B]P = ...(0 Thus, rate of reaction remains constant
Fi

4 r=*(2fl)“ (fc)P ...(//) irrespective of the change of concentration, i.e.,

it is independent of concentration. Hence, it is a
Now, if the volume of the vessel is reduced to half, reaction of zero order.
concentration of both the reactants will be
doubled. Alternatively, for zero order reaction, k= — and
t
k comes out to be constant.
Hence, 8 r = /: (4 (2 bf ...{Hi)
97. Suppose after time t min, concentration of A,
Dividing eqn. (i7) by eqn. O. [A], = 16 times the concentration of B, i.e.,
4 = 2“ or a = 2 [A],= 16[B1 t
Dividing eqn. {Hi) by eqn. {ii), t
Half lives of A in time i = —
8 = 2“ X 2P = 2^ X 54
or 2P = 2
I
Half lives of B in time
or
p=l. 18
Hence, overall order = 3.

ft
4/146 ^●uuieep. 'a New Course Chemistry (XII)
r

For Difficult Questions

dC 20-10 10 ,
— =1 mmol dm ^ s ‘
, j -3 -1
dt 10 10

[B] 0
and [B], = 2//18 dD -f/B (-dk
= 2x
■■ 2''^ = 1-5 X
dt dt } \ dt
As [A], = 16 [B], and [A]q = [BJq
1 16 (ID _ (-d^ A = ,.5
or 2-/754 _ 24-/718 But
2/754 2'/** dt
>’2'. dt J " >’2

w
or ± = 4-± or - / = 216 - 3 / dP _y4 f-dA I± = 2
54 18 dt y\ V dt .Vl
2 7 = 216 or t= 108 min

o
e
or

dD
>k^=2kxl or 2^ = 9

re
98. Fraction of molecules able to cross energy barrier = 9 =
dt >^3 dt >■3 >'3

Frl
-E./RT
= e-^

F
a
= e

Thus,y4= 1-5 y2=2y, =9^3

E.(/ 80-9x1000
_
= 13-91 = 14
X =
ou
r
RT 8-31x700 Rate of reaction
y^ dt dt

so
9 _ 9
99. No. of half-lives in 9 /i =
3-33 10/3
= 2-7
kf 1
= — X 9 mmol dm ^ s *
oo
9

Co C
Y
C or
f
= = X (say) = 1 mmol dm * s '
B

/
●22-7 C
0
1
101. For a gaseous reaction, 7j/2
re

log X = - 2-7 log 2 = - 2-7 X 0-301 = - 0-8127

oc

pn—1
0
oY
u

= 1-1873
ad

nn-1

;c = Antilog (1-1873) =0-16= 16 x 10"^ ^72^1 ^ ^^0 h _ ^^0^2

d

(^172)2 (P5-‘), (Pq)i

in

_l_(dA zif^’
Re

100. - N/I-l
yi < dt yiV^^ - 340 (21-6
\n-\
1
2‘ = -- _ Ql—n
F

or

170 55-5 12
1 fdC\ 1 (ID
or 1 - 71 = I or n =0
>'3 dt y4y^‘

VII. Numerical Value Type Questions (in Decimal Notation)

102. A(g) + B(g) ± AB (g)

Given: (E^)^ - (E„)^= 2 RT and A^.= 4A^ Equilibrium constant,

Now, Rate constant for forward reaction.
K
kj^ A
kf-^f
eg
h \
and Rate constant for backward reaction. _ 4 g2 RT7RT := 4 g2

A
CHEMICAL KINETICS 4/147

Now, at the given concentrations

For Difficult Questions
Rate = k [A] [CJ = (3 x IQ-^) (0-15) (0-15)
= 6-75 X 10"^ mol dm“^ s“^
Now, AG“ = -RT!nK
eq
104. The units of the rate constant show that it is a
= - 2500 In (4 eh
reaction of 1st order. Suppose after time /, decrease
= -2500 (In 4 +In e~) in pressure of N2O5 is 2 p atm
= -2500(1-4 + 2) 2N2O5 + 2 N2O4 + O2
= - 2500 X 3-4 Initial pressure 1 atm 0 0
After time / l-2p 2p
= - 8500 J mor’ P

103. Rate = k [A]'" [B]^ [CY Total pressure after time / (P,):
l-2p + 2p + p = (I+p) atm
By expt no. 1 and 2, =0
1 + p = 1 -45 (Given) or p = 0-45 atm
By expt no. 1 and 3, c = 1 As given reaction is of 1st order.

w
By expt no. 1 and 4, <2 = 1 2-303 1 2-303 1
[[B] and [C] are constant. [A] is made 1-5 times,
/ = — log log

F lo
k l-2p 5x10 1-2x0-45
rate also becomes 1 -5 times]
2-303 1
Rate = it [A]' [B]*^ [C]’ = [A] [C] Yx1Q3 = log — = 4-60x103

ee
5x10^ ''0-1
Also, from expt I, 6 0 x 10”^= k (0-2) (01)

Fr
or t = 3 X 10-3 Y = 4-60

VIII. for
ur
Assertion-Reason Type Questiont>
s
105. Refer to Footnote on page 4/13. At low pressure,
ook

112. Correct explanation. The order of reaction

Yo

it is first order reaction while at high pressure, it depends upon the power to which the
eB

is a zero order reaction.

concentration is raised to express the rate of
106. Correct Statement-1. The rate of reaction is same reaction as found experimentally.
in terms of different reactants and products. 113. Correct A. Half-life of a first order reaction only
our
ad

107. Correct Statement-2, k = conc./time = mol is independent of the initial concentration of the
L"^ i.e., it depends upon units of reactant.

concentration.
Y

1
Re

108. Correct explanation. For zero order reaction, Correct R. t 1/2 oc

i.e. depends upon the
nd

n-1
^1/2 a = all k). [An]
Fi

109. Statement-2 is the correct explanation of order n of the reaction.

Statement-1 (k' = k [H2O] so that 114. Correct explanation. During hydrolysis of
k = /:7[H20] = time-Vmol L"' = L mor’ time”*) methyl acetate by dil HCl, water is taken in large
110. Correct A. Rate of reaction may or may not be excess and hence, its concentration does not
equal to rate of disappearance of a reactant. It change during hydrolysis.
depends upon the stoichiometric coefficient of the 115. R is the correct explanation of A.
reactant in the balanced chemical equation.
116. Correct R. By 10“ rise of temperature, the
Correct R. Rate of disappearance represents collision frequency increases by 2 to 3%.
decrease in concentration of a reactant per unit
117. Correct A. A catalyst may increase or decrease
time but rate may or may not. the rate of a reaction.
111. Correct explanation. Molecularity is two because
two molecules of the reactants are involved in the
Correct R. A negative catalyst does not decrease
the activation energy of a reaction.
given elementary reaction.
4/148 New Course Chemistry (XlI)EZsZ91

For Difficult Questions

118. A + B 4 X* ^ Products decomposes such that one vibrational degree of

freedom is converted into a translational degree
The activated complex X* has higher energy than
reactants. It is a special molecule which of freedom along the reaction coordinate.

IQ Multiple Choice Questions (Based on Practical Chemistry)

119. Thiosulphate added should be in small amount so that it reacts away with iodine and the iodine further
liberated can give blue colour with starch.
120. If thiosulphate is present in excess, all the iodine liberated will react with the thiosulphate and no blue colour
will appear at all.

w
121. On increasing the concentration of T ions, the rate of reaction will increa.se. After reaction with thiosulphate,
less time will be taken for blue colour to appear.

F lo
ee
Fr
for
ur
s
ook
Yo
eB
r
ad
ou
Y
Re
nd
Fi
 *

SURFACE CHEMISTRY

w
Flo
5.1. GENERAL INTRODUCTION

ee
Surface chemistry is that branch of chemistry which deals with the study of the phenomena

Fr
occurring at the surface or interface, i.e., at the boundary separating two hulk phases.

The two bulk phases can be pure compounds or solutions. The interface is represented by putting a hyphen
for
ur
or a slash between the two bulk phases involved, e.g., solid-liquid or solid/liquid. No interface exists between
gases as they are completely miscible. It is an important branch of chemistry as a number of phenomena occur
at the interface, e.g., dissolution, crystallisation, con-osion, heterogeneous catalysis, electrode processes, etc.
k s
The results about surface studies can be satisfactory only if we take a real clean surface. Ultra clean
Yo
oo

surfaces of metals are obtained when a high vacuum of the order of ICH to 10"^ pascal is applied on them. To
eB

keep the surfaces of the solid materials clean so that they are not covered by the gases of the air (O2, etc.),
these solid materials must be kept stored in vacuum.
In this unit, we aim at discussing briefly three important topics related to surface chemistry, viz.,
r

adsorption, catalysis and colloids including emulsions and gels.

ou
ad

5.2. ADSORPTION
Y

5.2.1. Definitions of the terms used

Re
nd

It is observed that there are certain solids like charcoal, silica etc. which when added into a closed vessel
containing some gas or to the aqueous solution of a solute {e.g., acetic acid solution), the molecules of the gas
Fi

or the solute from the solution are attracted towards the surface of the solid and then retained on the surface.
As a result, the concentration ol the gas or the solute on the surface of the solid increases. A similar phenomenon
is observed when a liquid is added into a closed vessel containing a gas. This is known as “adsorption”.
Hence, it may be defined as follows :
The phenomenon of attracting and retaining the molecules of a substance on the surface of a
liquid or a solid resulting into a higher concentration of the molecules on the surface is called
adsorption. The substance thus adsorbed on the surface is called the adsorbate and the substance
on which it is adsorbed is called adsorbent. The reverse process, i.e., removal of the adsorbed
substance from the surface is called desorption (which can be brought about by heating or
reducing the pressure). The adsorption of gases on the surface of metals is called

* Not included in CBSE syllabus. This chapter has been given only for the preparation of competitive examinations.
5/1
5/2 New Course Chemistry (XI1)ESB!9]

As greater is the surface area of the adsorbent, greater is the adsorption, therefore, finely divided
metals or substances having porous structure, e.g., charcoal, silica gel, alumina gel, clay, etc. act as excellent
adsorbents.

'..2.2. Evidence in support of adsorption-Adsorption in action

The following experiments and results show that adsorption occurs :
(/) To the solution of an organic dye such as methylene blue, add animal charcoal and stir. It is observed

w
that the intensity of the colour in the solution decreases showing that some amount of the dye has been
adsorbed.
(//) Introduce finely divided solid into a closed vessel containing a gas at low pressure. The pressure of the
gas is found to decrease showing that some amount of the gas has been adsorbed.

e
(///) When aqueous solution of raw sugar (which has yellowish brown colour) is shaken with animal charcoal

re
ro
and then filtered, the filtrate is colourless which gives white crystals on crystallisation. This is because
the colouring substances are adsorbed by animal charcoal,
(iv) If silica gel is placed in a closed vessel containing moist air, the air becomes dry after some time. This

F
is because the water molecules are adsorbed on the surface of silica gel.

Ful
Thus, we observe that adsorption of gases or liquid molecules or the solid solute molecules from the
solution occurs on the surfaces of the solid adsorbents.

sr
:-Ii.B. Mechanism of Adsorption FIGURE 5.1

ko
o
The situation existing at the surface of a liquid or a solid is
different from that in the interior. For example, a molecule in
the interior of a liquid is completely surrounded by other
of O
o
^3-
I I
Y
molecules on all sides and hence the intermolecular forces of — Ni-Ni— ^
erB

attraction are exerted equally in all directions [Fig. 5.1. (a)]. — Ni —Ni— ^
However, a molecule at the surl'ace of a liquid is surrounded by O
uY

larger number of molecules in the liquid phase and fewer Molecules at the surface
molecules in the vapour phase, i.e., in the space above the liquid experiencing a net inward force of
surface. As a result, these molecules lying at the surface, attraction in case of (a) liquid
(b) solid (c) metal with free valencies
o
ad
d

experience some net inward force of attraction which causes

surface tension. Similar inward forces of attraction exist at the surface of a solid [Fig. 5.1. (/?)]. Alternatively,
in

in case of certain solids such as transition metals (like Ni), there are unutilized free valencies at the surface
[Fig. 5.1(c)].
Re

Because of the unbalanced (or residual) inward forces of attraction or free valencies at the surface,
F

liquids and solids have the propeny to attract and retain the molecules of a gas or a dissolved substance onto
their surfaces with which they come in contact.

^. A(isorptlon**an exothermic process

When adsorption lakes place, the residual forces on the surface of the adsorbent decrease. In other
words, surface energy decreases. This appears in the form of heat called heat of adsorption. Hence, adsorption
always negative.
is invariably an exothermic process, /.e., AH^j^Q^piion
;.2.b. Entropy change during adsorption and Adsorption equilibrium.
As the molecules of the adsorbate are held on the surface of the solid adsorbent, entropy decreases, i.e.,
AS is also negative. As AG = AH - TAS, hence for the process of adsorption to occur, AG must be
negative. As AS is negative, AG can be negative only if AH is negative and AH > TAS in magnitude. This is
true in the beginning. However, as the adsorption proceeds, AH keeps on decreasing and TAS keeps
on increasing till ultimately AH becomes equal to TAS so that AG = 0. This state is called adsorption

equilibrium.
 SURFACE CHEMISTRY 5/3

5.2.6. Distinction between Adsorption and Absorption FIGURE 5.2

The term adso)-ption differs from the term absorption in the Z
WATER VAPOUR
fact that whereas the former refers to the attraction and retention 7 ANHYDROUS
SILICAGEL CALCIUM
of the molecules of a substance on the surface only, the latter CHLORIDE
involves passing of the substance through the surface into the bulk
of the liquid or the solid [Fig. 5.2]. In other words, in absorption
the substance gets uniformly distributed throughout the liquid or ADSORPTION ON ABSORPTION
the solid. THE SURFACE INTO THE INTERIOR
(NO ADSORPTION)
In a number of cases, both adsorption and absorption take
place simultaneously. For such processes, simply the term sorption Illustrating the process of
is used. adsorption and absorption

ow
Thus, in adsorption whereas the concentration is different at the surface than in the bulk, in absorption,
the concentration is same throughout. Moreover, whereas adsorption is fast in the beginning and then the rate
decreases till equilibrium is attained, absorption takes place at uniform speed.
Thus, the main points of difference between adsorption and absorption may be summed up as follows :

e
Adsorption Absorption

re
1. It is a surface phenomenon, i.e., it occurs only

rFl
at 1. It is a bulk phenomenon, i.e., occurs throughout

F
the surface of the adsorbent. the body of the material.
2. In this phenomenon, the concentration on the surface 2. In this phenomenon, the concentration is same

r
of adsorbent is different from that in the bulk. throughout the material.
ou
3. Its rate is high in the beginning and then decreases 3.
fo
Its rate remains same throughout the process.
till equilibrium is attained.
ks
oo
5.2.7. Examples of adsorption, absorption and sorption
Y

(0 If silica gel is placed in a vessel containing water vapours, the latter are adsorbed on the former. On the
eB

other hand, if anhydrous CaCl2 is kept in place of silica gel, absorption takes place as the water vapours
are uniformly distributed in CaCl2 to form hydrated calcium chloride (CaCl2.2H^O). See Fig. 5.2.
r

(«) Ammonia gas placed in contact with charcoal gets adsorbed on the charcoal whereas ammonia gas
ou

placed in contact with water gets absorbed into water, giving NH4OH solution of uniform concentration.
ad
Y

{Hi) Dyes get adsorbed as well as absorbed in the cotton fibres, i.e., sorption takes place.

5.2.8. Positive and Negative Adsorption

d

In case of adsorption by solids from tlie solutions, mostly the solute is adsorbed on the surface of the
Re
in

solid adsorbent so that the concentration of solute on the surface of the adsorbent is greater than in the bulk.
F

This is called positive adsorption. However, in some cases, the solvent from the solution may be adsorbed by
the adsorbent so that the concentration of the solution increases as compared with the initial concentration.
This is called negative adsorption. For example, when a concentrated solution of KCl is shaken with blood
charcoal, it shows positive adsorption but with a dilute solution of KCl, it shows negative adsorption.
To sum up :
When the concentration of the adsorbate is more on the surface of the adsorbent than in the
bulk, it is called positive adsorption. If the concentration of the adsorbate increases in the bulk
after adsorption, it is called negative adsorption.

5.3. FACTORS AFFECTING ADSORPTION OF CASES BY SOLIDS

Almost all solids adsorb gases to some extent. However, the exact amount of a gas adsorbed depends
upon a number of factors, as briefly explained below :
(i) Nature and Surface area of the adsorbent. It is observed that the same gas is adsorbed to different
extents by different solids at the same temperature. Further, as may be expected, the greater the surface area
5/4 ^titdee^'4. New Course Chemistry CXI1)E&19]

of theadsorbeni, greater is the volume of the gas adsorbed. It is for this reason that substances like charcoal and
silica gel are excellent adsorbents because they have highly porous structures and hence large surface areas.
For the same reason, finely divided substances have larger adsorption power than when they are present
in the compact form.
Since the surface area of adsorbents cannot always be determined readily, the common practice is to
express the gas adsorbed per gram of the adsorbent (The surface area per gram of the adsorbent is called
speciifc surface area).
(ii) Nature of the ga.s being adsorbed. Different gases are adsorbed to different extents by the same
adsorbent at the same temperature. This is illustrated in Table 5.1 in which the volumes of different gases
(reduced to N.T.P. conditions) adsorbed by one gram of charcoal at 288 K have been given. The critical
temperatures of the gases are also included in the Table.
TABLE 5.1. Volumes of gases at N.T.R, adsorbed by Ig of charcoal at 288 K

w
Gas H2 N2 CO CH4 CO2 HCl NH3 SO2

F lo
9-3 16-2 48 72 181 380
Volume adsorbed (mL) 4-7 8-0

190 304 324 406 430

Critical temp (K) 33 126 134

ee
Critical temperature increases

Fr
Ease of liquefaction increases
Adsorption increases
for ■>
ur
From Table 5.1, it may be seen that higher the critical temperature* of a gas, greater is the amount of
that gas adsorbed. In other words, a gas which is more easily liquefiable or is more soluble in water is more
s
readily adsorbed. This relationship is not surprising as the critical temperature of a gas is related to the
ook
Yo

intermolecular forces which is an important factor for adsorption too. Higher the critical temperature, more
eB

easily the gas can be liquefied, i.e., greater are the intermolecular forces of attraction. For such a gas, greater
will be the attraction of the gas molecules on the surface of the adsorbent and hence greater will be the
adsorption,
our
ad

(ill) Temperature. Studying the adsorption of any particular gas by some particular adsorbent, it is
obser\'ed that the adsorption decreases with increase of temperature and vice versa. For example, one gram of
charcoal adsorbs about 10 mLof N2at 273 K, 20 mL at 244 K and 45 mLat 195 K. The decrease of adsorption
Y

with increase of temperature may be explained as follows :

Re
nd

Like any other equilibrium, adsorption is a process involving a true equilibrium. The two opposing
processes involved are condensation {i.e., adsorption) of the gas molecules on the surface of the solid and
Fi

evaporation (i.e., desorption) of the gas molecules from the surface of the solid into the gaseous phase. Moreover,
the process of condensation (or adsorption) is exothermic so that the equilibrium may be represented as :
Condensation
Gas {Adsorbate) + Solid {Adsorbent) ^ ± Gas adsorbed on solid + Heat
Evaporation
Applying Le Chatelier’s principle, it can be seen that increase of temperature decreases the adsoiption
and vice versa.

The amount of heat evolved when one mole of the gas is adsorbed on the adsorbent is called the
heat of adsorption.
(iv) Pre.s.sure. At constant temperature, the adsorption of a gas increases with increase of pressure. It is
observed that at low temperature, the adsorption of a gas increases very rapidly as the pressure is increased
from small values.

*Critical temperature of a gas is defined as the minimum temperature above which the gas cannot be liquefied
however high pressure we may apply.
 SURFACE CHEMISTRY 5/5

The variation of adsorption with pressure at different FIGURE 5.3

constant temperatures is shown in Fig. 5.3 for the adsorption 50 + 9^.
of N2 gas by 1 g of charcoal. These curvc.s indicate that at a
fo
CD

fixed pressure, there is a decrease in physical adsorption with

q:
oO
_I 40 O

increase in temperature, E O o3>

.oG
? 30 ■ o

(v) Activation of the soiid adsorbent. It means Q

03
oa
OoC
increasing the adsorbing power of an adsorbent. This is usually
LJJ I
CD
c.c 20 ■■
done by increasing the surface area (or the specific area) of o

the adsorbent which can be achieved in any of the following

CO
D 10-

ow
<
ways:
+ +
(a) By making the surface of the adsorbent rough, e.g., by 20 40 60
mechanical rubbing or by chemical action or by PRESSURE IN cm OF Hg
depositing finely dispersed metals on the suri'ace of the
Variation of adsorption of N2 on charcoal with

e
adsorbent by electroplatin O
C*'

pressure at different constant temperatures

re
{b) By subdividing the adsorbent into smaller pieces or
grains. No doubt, this method increases the surface area but it has a practical limitation, that is, if the

Flr
adsorbent is broken into too fine particles that it becomes almost powder, then the penetration of the

F
gas
becomes difficult and this will obstruct adsorption,
(c) By removing the gases already adsorbed, e.g., charcoal is activated by heating in superheated steam or
ou
in vacuum at a temperature between 623 to 1273 K.

sr
5.4. TYPES OF ADSORPTION

fo
k
An experimental study of the adsorption of gases on solids shows that there are two main types of
adsorption. These are briefly explained below :
oo
1. Physical ad.sorption or van der Waals adsorption or Physisorption. When a gas is held (adsorbed)
Y
on the surface of a solid by van der Waals forces (which are weak inlermolecular forces of attraction)
reB

without resulting into the formation of any chemical bond between the adsorbate and the adsorbent, it is
called 'physical adsorption' or ‘van der Waals adsorption' or 'physisorption'.
Characteristics of physical adsorption or physisorption. Some important characteristics of
uY

physisorption are briefly explained below ;

(0 Physical change. As explained above, it is simply a physical change as no chemical bond is formed.
ad
do

That is why it is called physical adsorption or physisorption.

(ii) Non-specific in nature. As van der Waals forces are universal, on every solid adsorbent, every gas is
in

adsorbed, only the extent of adsorption may vary. In other words, the surface of the solid adsorbent
does not have any preference for a particular gas for this type of adsoi-ption. Hence, it is non-specific in
Re

nature.
F

(Hi) Nature of adsorbate. As already explained, easily liquefiable gases, i.e., having high critical temperature
are adsorbed more strongly because they have stronger van der Waals forces of attraction.
(iv) Reversible nature. Physisorption is a reversible process represented as follows :
Gas + Solid - Gas/Solid + Heat
(Adsorbate) (Adsorbent)
Applying Le Chatelier’s principle, when we increase the pressure or decrease the temperature, equilibrium
shifts forward, i.e., adsorption increases. On reversing the conditions, the adsorption will decrease,
(v) Low enthalpy of adsorption. As physisorption involves only van der Waals forces of attraction and no .

chemical change, the process is exothermic but the enthalpy of adsorption is quite low (20-40 kJ mol"*).
(vi) Low activation energy/No activation energy. As this type of adsorption involves no chemical reaction,
no activation energy or very low activation energy is required for the adsorption to occur.
(v/7) Multimolecular layers of adsorption. As layers of the gas can be adsorbed one over the other by van
der Waals forces, multimolecular layers are formed under high pressure.
(viii) Surface area of the adsorbent. The adsorption increases with increase in surface area of the adsorbent.
5/6 'P’uuUe^'iir New Course Chemistry (XII)snsiHI

2. Chemical adsorption or Chemisorption or Langmuir adsorption. 'When a gas is held on to the

surface of a solid by forces similar to those of a chemical bond (which may be covalent or ionic in nature),
the type of adsorption is called chemical adsorption or chemisorption. This type of adsorption results into
the formation of what is called a ‘surface compound’.
Characteristics of chemisorption. Some important characteristics of chemisorption are briefly explained
below :

(0 Chemical change. It involves llie formation of a chemical bond between the adsorbate and the adsorbent.
Hi) Highly specific nature. This type of adsorption is highly specific in nature because it occurs only if
there is a chance of chemical bond formation between the adsorbate and the adsorbent. For example,
oxygen will be chemisorbed only on those solid adsorbents (e.g., carbon or metals) with which it can
form oxides or hydrogen will be chemisorbed on those solid adsorbents with which it can form hydrides
(e.g., on transition metals).
iiii) Nature of adsorbate. As explained above, the adsorbate should be such that it could form chemical

w
bond with the adsorbent.
(iv) Irreversible nature. As it involves the formation of a compound, the efforts to free the adsorbed gas

F lo
will release the compound instead of the free gas. For example, oxygen adsorbed on tungsten or carbon
is liberated as tungsten oxide or carbon monoxide or carbon dioxide. Hence, it is irreversible in nature.

ee
(V) High activation energy. As chemisorption involves a chemical reaction, it is slow at low temperature
because like other chemical reactions, it requires high energy of activation. Hence, like most of the

Fr
chemical changes, chemisorption increases with increase of temperature. For this reason, a gas may be
physically adsorbed at low temperature but chemisorbed at higher temperature. For example, it happens
for
in case of adsorption of hydrogen on nickel. When chemisorj^tion takes place by raising the temperature,
ur
i.e., by supplying activation energy, the process is called ‘activated adsorption’.
(v/) High Heat of adsorption. Chemisorption is also exothermic and is accompanied with the evolution of
s
ok

large amount of heat (200-400 kJ mol"') which is of the same magnitude as those involved in chemical
Yo

reactions.
o
eB

(vn) Unimolecular layer of adsorption. As chemical bond can be formed only with the layer of molecules
coming in direct contact with the surface of the adsorbent, hence this type of adsorption is unimolecular.
(viii) Surface area of the adsorbent. Like physical adsorption, chemisorption also increases with increase
r
ou
ad

in surface area of the adsorbent.

The main points of difference between physical adsorption and chemisorption are summed up in Table 5.2
Y

below:

Distinction between physical adsorption and chemisorption

Re

TABLE 5.2,
nd
Fi

Physical Adsorption Chemisorption

1. The forces operating in these cases aie weak 1. The forces operating in these cases are similar to
van der Waals forces. those of a chemical bond.
2. The heats of adsorption are low, viz., about 2. The heats of adsorption are high, viz., about
20-40 kJ mol"'. 40-400 kJ mol-'.
3. No compound formation takes place in the.se 3. Surface compounds are formed.
cases.

4. The process is reversible, i.e., desorption of 4. The process is irreversible. Efforts to free the
the gas occurs by increasingthe temperature adsorbed gas does not give back the adsorbed gas
or decreasing the pressure. but gives some definite compound.
5. It does not require any a activation energy. 5. It requires activation energy.
6. This type of adsorption usually takes place 6. This type of adsorption first increases with increase
at low temperature and decreases with of temperature. The effect is called activated
increase of temperature. adsorption.
SURFACE CHEMISTRY 5/7

7. It is not specific in nature, i.e., all gases are 7. It is specific in nature and occurs only when there
adsorbed on all solids to some extent. is some possibility of compound formation between
the gas being adsorbed and the solid adsorbent.
8. The amount of the gas adsorbed is related to 8. There is no such correlation.
the ease of liquefaction of the gas.
9. It forms multimolecular layer. 9. It forms unimolecular layer.
ccrrrrrrrc)
ADSORBENT

ADSORBENT [
10. It increases with increase in the surface area 10. It also increases with increase in the surface area
of the adsorbent. of the adsorbent.
11. It increases with increase of pressure. II. It also increases with increase of pressure. Decrease

w
Decrease of pressure causes desorption of pressure does not cause desorption as explained
above.

Flo
● *1
SUPPLEMENT YOUR '
KNOWLEDGE FOR COMPETITIONS

1. Mechanism of chemisorption. Taking the example of adsorption of H2 on the surface of platinum, as

ee
shown in the fig. below, hydrogen molecules are first attracted towards the surface by weak van der

Fr
Waals forces and hence adsorbed on it due to the presence of unbalanced attractive forces or free valencies
on the surface of the solid (physical adsorption). The adsorbed molecules then dissociate into atoms
which are then chemisorbed and hence held strongly.

for
ur
H—H
H—H H H
ks
i
Yo
Pt Pt Pt Pt Pt Pt
oo

Physical Chemical

adsorption adsorption
eB

Pt Pt Pt Pt Pt Pt

When the solid adsorbent is broken into pieces, the number of unbalanced forces (free valencies) increases
r

and hence adsorption increases. For example.

ou
ad

I i
Y

●--Pt Pt Broken Pt —Pt

+
nd

into two pieces

Re

Pt Pt Pt Pt
Fi

Free valencies = 8 Free valencies = 12

2. Why heat of adsorption is greater for chemisorption than physisorption ? As adsorption involves
attraction of the molecules of the adsorbate on the surface of the adsorbent, surface energy decreases and
appears as heat. That is why heat of adsorption is negative, i.e., adsorption is always exothermic. Further,
physical adsorption involves weak forces of attraction (van der Waals forces), heat evolved is less whereas
chemical adsorption involves chemical reaction, heat evolved is much higher.
3. Why heat of adsorption is negative-An alternative explanation. As the molecules of the adsorbate are
held on the surface of the solid adsorbent, entropy decreases, i.e., AS is negative. As AG = AH - TAS and
for the process to occur AG must be negative which is possible only if AH is negative (and greater in
magnitude than TAS)
Exceptions. In some cases, chemisorption may be endothermic. For example, chemisorption of H2 as H
atoms on the surface of glass is endothermic.But in this case, as H2 (g) > 2 H (g), the entropy change
(AS) is positive and sufficient to dominate over the small positive value of AH. Similarly, when highly
hydrated solutes are adsorbed on the surface of solid adsorbents, they are accompanied by large increase
of entropy (due to release of water molecules on adsorption) which overcomes the effect of small positive
enthalpy change.
5/8 'a New Course Chemistry (XII)

4. Adsorption of nitrogen on iron. This is an example where the gas is physiosorbed at low temperature
and chemisorbed at high temperature. At 83 K, nitrogen is physiosorbed on iron surface as which
decreases with increase of temperature and at room temperature there is almost no adsorption. At 773K
and above, there is chemisorption of nitrogen as N atoms.
5. Competing adsorption. Different adsorbates are adsorbed to different extents on the same adsorbent
under the same conditions of temperature and pressure,
//fl mixture of gases is allowed to come in contact with a particular adsorbent, the more strongly
adsorbahle adsorbate is adsorbed to a greater extent irrespective of its amount present.
Similarly it is observed that
A more strongly adsorable adsorbate can displace a weakly adsorbed substance from the surface of the
adsorbent.

For example, charcoal used in the mask contains 0-,, N2 etc. of the air already adsorbed in it. But when it
comes in contact with the poisonous gases like CI2, SO2, CH4 etc. which are more strongly adsorbable on
charcoal, Oo, etc. are displaced by these gases.

5.5. FREUNDi lCH ADSORPTION ISOTHERM

F low
The extent of adsorption on a given surface generally increases with increase in pressure (for gases) and
concentration (for solution) at constant temperature. At low temperatures, the physical adsorption of a gas
increases very rapidly as the pressure rises. When the temperature is high, the increase in adsorption is
relatively less.
To understand the effect of pressure on adsorption of a gas. we consider adsorption as an equilibrium

e
process. When the adsorbent and the adsorbate are enclosed in a closed vessel, after an initial decrease in the
for Fre
pressure of the gas, gas pressure as well as the amount of gas adsorbed reach constant values. This is because
after the equilibrium is attained, rate of adsorption becomes equal to the rate of desorption.
The amount of gas adsorbed depends on the surface area of the adsorbent or on its mass if the adsorbent
is taken in the form of powder.
Your

The extent of adsorption is usually expressed as xhn, where m is the mass of the adsorbent and x is the
eBo ks

mass of the adsorbate when adsorption equilibrium is reached. Thus, x/m is the mass adsorbed per gram of the
adsorbent.

The surface area per gram of the adsorbetlt (which may be in the form of powder or porous mass) is
ad
our

called its speciifc surface area. It is expressed in square metres per gram. Highly active solids with large
surface area (several hundred square meters per gram) are used as adsorbents.

A graph between the amount of the gas adsorbed per gram of the adsorbent (x/m) and the
Re

equilibrium pressure of the adsorbate at constant temperature is called the adsorption isotherm.
Find Y

The simplest type of adsorption isothenn is shown in FIGURE 5.4

Fig. 5.4. At a value of of equilibrium pressure, x/rn reaches INCONSTANT B

its maximum value and then it remains constant even though JL= kPo
m

the pressure P is increased. This is the saturation state and P^ 1

is the .saturation pressure. This type of adsorption isotherm is
t
A=kP"
observed when the adsorbate forms a uniform molecular layer m

A
on the surface of the adsorbent. X
m
A relationship between the amount adsorbed (x/m) and SATURATION

the equilibrium pressure (P) can be obtained as follows : 4- = kp PRESSURE, Ps

/
m

At low values of P, the graph is nearly straight and sloping

(shown by OA in Fig. 5.4). This is represented by the following 0 r
equation : Variation of x/m with increase in
pressure at constant temperature
— O. P — = constant xP' ...(/)
m
or
(Freundlich adsorption isotherm)
rn

1
 SURFACE CHEMISTRY 5/9

At high pressure, xhn becomes independent of the values of P (beyond point B in Fig. 5.4). In this range
of pressure,

-oc pO or — - constant xP^’ ...(//)

m m

In the intermediate range of pressure (shown by AB in Fig. 5.4), xhn will depend on P raised to powers
between I and 0, i.e., fractions. For a small range of pressure values, we can write :

pl/« -=kV^>

ow
m
or
...{in)
m

where n is a positive integer and n and k are constants depending upon the nature of the adsorbate and adsorbent
at a particular temperature. The factor Mn has values between 0 and 1 (probable range is 01 to 0-5).
This relationship was originally put forward by Freundlich in 1909 and is known as Freundlich

e
adsorption isotherm.

re
To test the validity of this equation, taking logaiithms of both sides of eqn. {in), we get

Flr
1

F
log-=logk + - log P ..m
m n
ou
A graph between log xhn against log P should, therefore, give a straight
line with slope equal to Un and ordinate intercept equal to log k (Fig. 5.5).
FIGURE 5.5

sr
fo
Limitations of Freundlich adsorption isotherm. The experimental
1
k
values, when plotted, however, show some deviation from linearity, specially x|E
at high pressures. Hence, the relation is considered as an approximate one and
oo
holds good over a limited range of pressure. Moreover, concept of Freundlich INTERCEPT=LOGk
Y
adsorption isotherm is purely empirical and it considers adsorption to be LOGP
reB

multimolecular. Hence, it is applicable only to physical adsorption.

Adsorption from solutions. Solid surfaces can also adsorb solutes from Linear graph between
log xhn versus log P.
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the solutions. A few examples are given below :

(/) When a solution of acetic acid is shaken with charcoal, some amount of acetic acid from the solution is
adsorbed on the charcoal and hence the concentration of acid in the solution decreases.
ad
do

{ii) When litmus solution is shaken with charcoal, it becomes colourless because the dye of the litmus
solution is adsorbed by charcoal.
in

(Hi) When the colourless Mg(OH)2 is precipitated in the presence of magneson reagent (a blue coloured
Re

dye), it acquires blue colour because the dye is adsorbed on the solid precipitate.
F

Factors affecting adsorption from solutions. The adsorption from solutions by solid adsorbents is
found to depend upon the following factors :
(/) Nature of the adsorbateand the adsorbent,
(ii) Temperature. The adsorption decreases with increase of temperature.
(///) Surface area of the adsorbent. Greater the surface area of the adsorbent, greater is the adsorption,
(/v) Concentration of the solute in the solution.
The isotherm for the adsorption of solutes from solutions (by the solid adsorbents) is found to be similar
to that shown in Fig. 5.4. Hence, the relationship between — (mass of the solute adsorbed per gram of the
m

adsorbent) and the equilibrium concentration, C, of the solute in the solution is also similar, i.e..

Taking logarithms of both sides of the equation, we get log — = log 'A + —logC
m n
5/10 7^n<^eicc^'4■ New Course Chemistry (X11)E9SIS]

r . . . I
This equation implies that a plot of log — against log C should be a straight line with slope —
n
and
m

intercept log k (similar to that shown in Fig. 5.5). This is found to be so over small ranges of concentration.
Experimental verification. Take four stoppered bottles. Put charcoal (say, 2 g) in each bottle. Add
acetic acid solution of different concentrations in the different bottles. Shake them well. Now, find out the
concentration of acetic acid in each bottle. It is observed that in each bottle, concentration of acetic acid
solution is found to be less than the original concentration. These are equilibrium concentrations (C). The
difference of the original concentration and the equilibrium concentration after shaking with charcoal gives
the amount (a:) of acetic acid adsorbed, x/ni values of different bottles can be plotted against log C.
The equation for adsorption from solutions is found to give better results than for adsorption of gases by
solids.

w
5.6. ADSORPTION ISOBARS

As already discussed, adsorption is a case of dynamic equilibrium in which forward process (adsorption)

F lo
is exothermic while backward process (desorption) is endothermic. Thus, applying Le Chalelier’s principle,
increase of temperature will favour the backward process, i.e., adsorption decreases.

ee
A graph drawn between the amount of the gas adsorbed per gram of the adsorbent (x/m) and

Fr
temperature T at a constant equilibrium pressure of adsorbate gas is known as adsorption
isobar.

for
ur
Adsorption isobars of physical adsorption and FIGURE 5.6

chemical adsorption show an important difference

s
[Fig. 5.6 (a) and 5.6 (£»)] and this difference is helpful
ook
Yo
in distinguishing these two types of adsorption. P = CONSTANT
t P = CONSTANT

The physical adsorption isobar shows a

eB

m m

decrease in x/m throughout with rise in temperature.

The chemisorption isobar shows an initial increase
o o
r

with temperature and then the expected decrease. t t

ou
ad

The initial increase is because of the fact that the (a) Physical adsorption isobar,
heat supplied acts as activation energy required in (b) Chemisorption isobar
Y

chemisorption (like chemical reactions).

Re
nd

Sample Problem Q 50 mL of 1 M oxalic add (molecular mass = 126) is shaken with 0*5 g of
wood charcoal. The final concentration of the solution after adsorption is 0-5 M. Calculate the amount of
Fi

oxalic acid adsorbed per gram of charcoal.

Solution. Calculation of initial amount of oxalic acid in 50 mL solution. I M oxalic acid solution means
1 mole of oxalic acid (126 g) present in 1000 mL solution.
126
50 mL of 1 M solution will contain oxalic acid = x50 = 6-6 g
1000

Calculation of amount of oxalic acid in 50 mL solution after adsorption. Concentration of solution after
adsorption = 0-5 M. Thus, 10(X) mL of the solution contain oxalic acid = 0-5 mol = 63 g.
63
50 mL of the solution will contain oxalic acid = x50 = 3-15g
1000

Calculation of amount adsorbed per gram

Amount of oxalic add adsorbed by 0-5 g charcoal = 6-30 - 3-15 g = 3-15 g
.-. Amount of oxalic acid adsorbed per gram of charcoal = 6*30 g
 SURFACE CHEMISTRY 5/11

During the adsorption of acetic acid vapours on the surface of 1 g of animal

charcoal, the following observations were recorded where jr represents mass of acetic acid vapour adsorbed

Otlservatldn il

rw
x(g) 0-726 0-438

P(atm) 0-576 0-210

Calculate the values of the constants n and k of Freundlich adsorption isotherm.

e
Solution. According to Freundlich adsorption isotherm

r
luo
1
X . „n X 1

F
— = k? or log—= log^ + -logP
m m n

0-726

oF
1
For observation I: log = log A: -f—log 0-576 ..(0
1

rs
n

0-438 1

ok
For observation n : log = log A:-f-log 0-210 m)
1 n

Subtracting eqn. (ii) from eqn. (0

fo
, 0-726 1, 0-576
,og_ = -log o
Y
n 0-210
Y
rB
1 1 0-438
or
log (1-657) = -n log (2-742) or 0-219 = -n (0-438)
0-219 ^
or n =

Substituting the values of observation I and n = 2, in the equation

due

one
, 0-726
no

= A:(0-576)^^ = ifc(0-759) = 0-956 = 1

ad

or
1 0-759

Applying Freundlich adsorption isotherm, calculate the amount of acetic

i

acid adsorbed by 1 kg of blood charcoal at 25®C from a 5% vinegar solution (mass/volume). Given that if
the concentration is expressed in molarity (mol dm*^), x/m is mass of the solute adsorbed per gram of
Re
F

adsorbent, then k = 0-160 and n = 2-32.

Solution. According to Freundlich adsorption isotherm, — = k C*/” -(0

m

5% vinegar (acetic acid solution) means 5 g of acetic acid are present in 100 mL of the solution

Molar mass of acetic acid (CH3COOH) = 60 g mol“* 5 g of acetic acid = — mol

60
1
1000 mL of the solution will contain acetic acid = — x— xlOOO = 0-837 mol
60 100
Le., Concentration of the solution (C) = 0-837 mol L“*

Substituting the values in eqn. (/), we get - = 0-160x(0-837)*/2-32

m

1
log- = log(ai60)-h log (0-837) = - 0-7959 + 0-435 (- 0-0773) = -0-8295 = 1-1705
m 2-32
5/12 ‘PnacCeeft '<*. New Course Chemistry (XII) BE

-1
— = Antilog (i-1705)=0 1481 g (g charcoal)
m

Amount adsorbed by 1 kg (1000 g) of charcoal = 148*1 g.

Sample Problem Assuming that van’t Hoff type equation can be used to determine the
temperature dependence of the amount of the gas adsorbed on the surface of a soUd, calculate the enthalpy
of adsorption, for Nj at 1 atm. Given that 155 cm^ of the gas measured at STP is adsorbed by 1 g of
charcoal at 88 K and 15 cm^ at 273 K.
Solution, van’t Hoff type equation for adsorption of a gas at two different temperatures can be written as
V
ln^ = ^adsrj__11 V AH
ads 1 1 ^
log-^ =-
T2 tJ
or
R
V
1 V
1 2-303 R V % T,J
n

w
155
log
15 2-303 X 8-3141^88 273^

F lo
AH 185
ads
(2-1903-1-1761) = -
2-303x8-314 88x273
= - 2521-7 J mol-‘ = - 2*52 kj moH

ee
or AH

Fr
s^mbib ProbJemiB A one-litre vessel contained a gas at 27®C. 6 g of charcoal was introduced
into it. The pressure of the gas fell down from 700 mm to 400 mm. Calculate the volume of the gas (at S.T.P.)
adsorbed per gram of charcoal. Density of charcoal sample used was 1*5 g cm“^
for
ur
Solution. Volume of the vessel = 1000 cm^
6g
s
Volume of charcoal present in the vessel = = 4 cm^
ook

1-5 g cm“^
Yo

Volume of the gas initially at 2TC, 700 mm pressure = 1000 - 4 = 996 cm^
eB

Volume of the gas at 400 mm and 27“C is again equal to the volume of the vessel excluding that of charcoal,
Le., 996 cm^. Let us calculate equivalent volume of the gas at 700 mm at the same temperature
PiVj = P2V2 i.e. 400 X 996 = 700 x V2 or V2 = 569-1 cc
r
ou
ad

Volume of the gas adsorbed at 27*’C, 700 mm pressure = 996 - 569-1 = 426-9 cm^
426-9 3 -1
cm^g = 71-1 cm^g-l
Y

Volume adsorbed per gram of charcoal =

6
Re
nd

PiVi P9V2 . 700x71-8 760 XV2

Converting this volume to S.T.R, we get = -z:— , i.e., 300 273
or
V2 = 59-6 cm3
T2
Fi

■■sampib~PrbbieSilti! 1 g of charcoal adsorbs 100 ml of 0-5 M CH3COOH to form a monolayer

and thereby the molarity of CH3COOH reduces to 0-49 M. Calculate the surface area of the charcoal
adsorbed by each molecule of acetic acid. Surface area of charcoal = 3*01 x 10 m g.
Solution. 100 ml of 0-5 M CH3COOH contains CH3COOH = 0-05 mole
After adsorption, CH3COOH present = 0-049 mole
Acetic acid adsorbed by 1 g charcoal = (0-05 - 0-049) mole = 0-001 mole = 6-02 x lO^o molecules
Surface area of 1 g of charcoal = 3-01 x 10^ m^
3-01x102 m2
Surface area of charcoal adsorbed by each molecule = = 5x10-1’m2
6-02x1020
In an adsorption experiment, a graph between log {x/m) versus log P was
found to be linear with a slope of 45®. The intercept on the log (x/m) axis was found to be 0*3010. Calculate
the amount of the gas adsorbed per gram of charcoal under a pressure of 0*5 atmosphere.
(Manipur Board 2012)
 SURFACE CHEMISTRY
5/13

Solution. According to Freundlich equation, —=k P'/« or log — = logi + - log P

m m n

X
1
Plot of —
m
versus log P is linear with slope = - and intercept = log k. Thus,
n

- = tan9 = tan 45°= I or n= 1

n

log k = 0-3010 or
k = Antilog 0-3010 = 2
At P = 0-5 atm, ~ = k P'/" =2x(0-5)' = 10
m

Sample Problem Q The volume of nitrogen gas (measured at S.T.P.) required to cover a

sample of silica gel with a mono-molecular layer is 129 cm^ of gel. Calculate the surface area per gram
of the gel if each nitrogen molecule occupies 16*2 x 10’^^ m^.

w
Solution.
22400 cm3 § j p contain = 6-022 x 10^3 molecules

Flo
6-022x1023x129
129 cm3 of N2 at S.T.P. will contain = = 3-468x102’ molecules
22400

ee
Area occupied by a single molecule = 16-2 x I0"20m2
Area occupied by 3-468 x 102> molecules of N2= (16-2 x 10-20) x (3-468 x 102’) m2 = 561-8 m2

Fr
Surface area per gram of gel = 561*8 m2.

for
ur
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Langmuir Adsorption Isotherm

k s
Yo
Assuming adsorption to be unimblecular, Langmuir deduced a relationship theoretically (called Langmuir
oo

adsorption isotherm) which is superior to that of Freundlich in a number of ways as given below :
eB

(0 The results obtained from this isotherm are in better agreement with the experimental values.
(z7) It is applicable over a wider range of pressure.
r

{Hi) Freundlich isotherm followed as a special case for intermediate pressure.

ou
ad

The relationship is as follows :

Y

X nP

m 1+6P
...(/)
Re
nd

where a and b are constants.

Fi

This equation is known as Langmuir adsorption isotherm. The constants a and b depend upon the nature
of the gas adsorbate, the nature of the solid adsorbent and the temperature. Their values can be found
from the experimental data.
To verify the validity of Langmuir adsorption equation, both sides of the equation may first be divided by
x! m a
P so that we get ...(«)
P

P \ b^
Taking the reciprocal of both sides, we get =-+-P ...(m)
xhn a a

p
As a and b are constants, a plot of versus
P should be a straight line with slope equal to b!a and
xhn

intercept on tlie y-axis equal to 1/a. This is actually found to be so as shown in Fig. 5.7 (a) for the
adsorption of N2 on mica at 90 K.

»
5/14 ‘P>tad€efr’4. New Course Chemistry (XII)iiiaiJI
FIGURE 5.7

1000 -

800

b
600
a

£. 1 m
Intercept = — 1
x/m 400 d X
Slope = —

200 Intercept = —
3

0 6 12 18 24 30 36 1/P

low
P

(a) I-angmuir adsorption isotherm (b) Testing Langmuir adsorption

for adsorption of N2 on mica at 90 K equation by plotting m/x versus 1/P

Alternatively, taking reciprocal of both sides of eqn. (iV), we have

ee
1+/>P
L ^
m

rF m
...(iv)

Fr
or + -
X a? X aV a

Thus, a plot of mJx versus 1/P will be a straight line with to bta (Fig. 5.1b)
slope equal to Ha and intercept equal aC

r
X

For adsorption from solution, the Langmuir equation (i) is written in the form
fo m \+bC
u
where C is the equilibrium concentration of the solute in the solution. It can be tested in a manner similar
ks
C m I
Yo
to that of adsorption of gases, i.e., by plotting vs. C or — vs
oo

xlm x C

Freundlich adsorption isotherm as a special case of Langmuir adsorption isotherm.

B

At low pressures, 6 P « 1 so that eqn. (1) becomes

re

-=kP‘
- = ^P. i.e.. or
u

m m m
ad

At high pressures, b P » 1 so that eqn. (i) becomes

Yo

a
X aP _ a = it = cfp0 k = — = constant
m bP~ b b
nd
Re

Hence, for intermediate pressure, we must have — = kP^^"

Fi

where I/n has a value lying between 0 and I which is Freundlich adsorption isotherm.
FIGURE 5.8
Note that Freundlich adsorption is applicable to physical
adsorption only whereas Langmuir adsorption isotherm is
applicable to physisorption as well as chemisorption.
2. Adsorption isostere. It was discussed in this unit that
adsorption increases with increase of pressure and decreases 3

with increase of temperature. Thus, if temperature is increased n

to a certain value (so that the adsorption decreases), then

o>
a.
E
pressure will have to be increased in order to have the same o>

amount of adsorption, i.e., to keep the adsorption constant. The Pressure

implies that for the same amount of adsorption the plot
between pressure versus temperature will be linear. Plot of temp. vs. pressure for a given
The plot of temperature versus pressure for a given amount of amount of adsorption (adsorption
isostere)
adsorption is called adsorption isostere (Fig. 5.8).

t
 SURFACE CHEMISTRY
5/15

5.7. APPLICATIONS OF ADSORPTION

Adsorption finds extensive applications both in research laboratory and in industry. A few applications
are bnefly described below :
fO In preserving vacuum. In Dewar flasks, activated charcoal is placed between the walls of the flask so
that any gas which enters into the annular space either due to glass imperfection or diffusion through
glass is adsorbed. ®
{//) In gas masks. All gas masks are devices containing suitable adsorbent so that the poisonous gases
present in the atmosphere are preferentially adsorbed and the air for breathing is purified.
(in) In clarification of sugar. Sugar is decolorised by treating sugar solution with charcoal powder. The
latter adsorbs the undesirable colours present.
(iV) In chromatographic analysis. The selective adsorption of certain substances from a solution by a
particular solid adsorbent has helped to develop a technique for the separation of the components of

w
the mixture. This technique is called chromatographic analysis. For example, in column chromatography,
a long and wide vertical tube is filled with a suitable adsorbent and the solution of the mixture poured
from the top and then collected one by one from the bottom,

Flo
(v) In heterogeneous catalysis. The action of certain solids as catalysts is best explained in terms of
adsorption. The theory is called adsorption theory. According to this theory, the gaseous reactants are

ee
adsorbed on the surface of the solid catalyst. As a result, the concentration of the reactants increases on

Fr
the surface and hence the rate of reaction increases, e.g., in the manufacture of ammonia using iron as
catalyst, in the manufacture of sulphuric acid by contact process using V2O5 as catalyst and use of
finely divided nickel in the hydrogenation of oils. The theory is also able to explain the greater efficiency

for
ur
of a catalyst in the finely divided state, the action of catalytic poisons etc.
(W) In adsorption indicators. Various dyes, which owe their use to adsorption, have been introduced as
indicators particularly in precipitation titrations. For example, KBr is easily titrated with AgNO-, using
s
ok
eosin as an indicator. ^
Yo

(vn) In softening of hard water. The use of ion exchangers for softening of hard water (as already discussed
Bo

in class +1) is based upon the principle of competing adsorption just as in chromatography,
(vm) In removing moisture from air in the storage of delicate instruments. Such instruments which may
re

be harmed by contact with the moist air, are kept out of contact with moisture using silica gel. Silica gel
is the most commonly used dehumidizer, i.e., to adsorb humidity or moisture from the air.
ou
ad

iix) In the separation of inert gases. Different inert

gases are adsorbed to different extents at different
Y

temperatures on coconut charcoal. This forms the basis of their separation from a mixture.
(j:) In froth floatation process. When sulphide ore is shaken with pine oil and water, the ore particles are
nd
Re

adsorbed on ^e froth that floats and the gangue panicles (silica,

earthy matter etc.) settle down in the tank.
The method is known as froth floatation process for concentration of sulphide ores (discussed in unit 6).
Fi

(xi) In paint industry. The paint should not contain dissolved gases as otherwise the paint does not adhere
well to the surface to be painted and thus will have poor covering power. The dissolved gases are,
therefore, removed by suitable adsorbents during manufacture. Further, all surfaces are covered with
layers of gaseous, liquid or solid films. These have to be removed before the paint is applied. This is
done by suitable liquids which dissolve these films. Such liquids are called wetting agents. The use of
spirit as welting agent in furniture polishing is well known.
(xii) In dyeing. In dyeing, clothes are first kept dipped in a mordant (generally, alum) and then in the
solution of the dye. The mordant is first adsorbed on the clothes and then the dye is adsorbed on the
mordant. In other words, mordant adheres well both to fibre and the dye. If this is not done, uneven
dying would take place.
(xiii) In water conservation. In countries like Australia where there is acute scarcity of water during summer,
layer of stearic acid etc. is sprayed over the lakes and other water reservoirs. It is adsorbed on the
surface of water thereby minimising the loss of water by evaporation,
(.r/v) In curing diseases. Some drugs can adsorb the germs on them and hence kill them and save us from diseases.
5/16 7>niicUe^ 4 New Course Chemistry (XII) EEM

5.8. CATALYSIS

5.8.1. Definition
U is found that the rates of a number of chemical reactions are altered simply by the presence of certain
foreign substances. These studies were first made by Berzelius in 1835. He suggested the name catalyst for
such foreign substances. Thus,
Catalyst is a substance which can change the speed of a chemical reaction without itself
undergoing any change in mass and chemical composition at the end of the reaction and the

ow
phenomenon is known as catalysis.
For example, when potassium chlorate is heated strongly, it decomposes to give dioxygen. The
decomposition takes place at a high temperature in the range 653-873 K.
2 KCIO3 > 2 KCI + 3 O2

e
However, when a small amount of manganese dioxide (Mn02) is added, the decomposition not only

re
becomes faster but takes place at a much lower temperature, i.e., in the range 473-633 K. Moreover, Mn02 is
found to have same mass and composition at the end of reaction. Thus, Mn02 acts as a catalyst for the above

Flr
F
reaction.

5.8.2. Positive and Negative Catalysis

ou
If a catalyst increases (accelerates) the speed of a reaction, it is called a positive catalyst and the

sr
phenomenon is called positive catalysis. On the other hand, if a catalyst decreases (retards) the

fo
speed of a reaction, it is called a negative catalyst and the phenomenon is called negative
catalysis.
k
oo
For example, oxidation of SO2 to SO3 in FIGURE 5.9
Y
presence of NO (lead chamber process) or in .In presence
reB

presence of V2O5 (contact process) are In absence

of catalyst -V
\
1
of negative
catalyst
examples of positive catalysis. Decomposition >.
^ In presence g Vj^ In absence
of H-,02 in presence of glycerol or acetanilide OJ
uY

of positive V y of catalyst
or phosphoric acid and oxidation of chloroform LU catalyst m
in presence of alcohol are examples of negative Reactants
■a

catalysis. However, the term catalyst used

ad
do

Products
without any prefix implies positive catalyst. Q. CL

0 Reaction Coordinate —► (5 Reaction Coordinate—►

It may also be noted that whereas a positive
in

catalyst lowers the potential energy barrier, i.e., (a) Positive catalyst lowers the potential energy bajrier.
Re

lowers the activation energy of the reaction, a (b) Negative catalyst raises the potential energy barrier
F

negative catalyst raises the potential energy

barrier, i.e., increases the activation energy
(Fig. 5.9).
5.8.3. Promoters and Poisons
Certain substances like arsenic, CO, etc. if present, lower the activity of the catalyst. These substances
are called catalytic poisons.
This seems to be due to the fact that these poisons are preferentially adsorbed at the active sites of the
catalyst as compared to the reactants.
A substance, if present along with the catalyst, enhances the activity of the catalyst, it is called a catalytic
promoter, e.g., molybdenum acts as a promoter for iron catalyst in the manufacture of ammonia by Haber s
process.
This seems to be due to the fact that the promoter increases the roughness of the catalyst surface. As a

result, free valences increase. Hence, adsorption increases and so does the rate.
5.8.4. Catalysis (Homogeneous and Heterogeneous Catalysis)
There are two types of catalysis : (1) Homogeneous Catalysis (2) Heterogeneous Catalysis.
 SURFACE CHEMISTRY 5/17

(1) Homogeneous Catalysis. If the catalyst is present in the same phase as the reactants, it is called a
homogeneous catalyst and this type of catalysis is called homogeneous catalysis. Two common examples of
this type of catalysis are as follows :
(0 Oxidation of sulphur dioxide to sulphur trioxide in presence of nitric oxide as catalyst (in lead
chamber process for manufacture of H2SO4).
NO(g)
2S02(g) + 02(^) > 2803(g)
Here, all substances are present in the gaseous phase.
Similarly, oxidation of CO by O2 takes place in presence of NO as catalyst.
NO
2CO(g) + 02(g) > 2C02(g)
(i7) Decomposition of ozone in presence ofNO or Cl atoms as catalyst.

low
NO/Cl
O3 + o > 2O2
{iii) Decomposition of hydrogen peroxide in presence iodide ion as catalyst
I-
2 H2O2 > 2 H2O + O2

e
(iv) Hydrolysis of ethyl acetate or sucrose solution in presence of dilute sulphuric acid.

re
rF
F
CH3COOC2H5 (/) + H2O (0 > CH3COOH (0 + C2H5OH (/)
Ethyl acetate Acetic acid Ethyl alcohol

r
H+(o9)I

fo
u
^12^22^11 + ^^2^ (0 ^ ^6^12^6 (^9) +
Sucrose
ks
Glucose Fructose
Yo
Here, all substances are present in the liquid phase,
oo

(v) Preparation of diethyl ether from ethyl alcohol using concentrated H^O^ as catalyst
Conc.H2S04 (/)
B

2 C2H5OH (/) C2H50C2H5(/) + H20(0

re

Ethyl alcohol Diethyl ether

Here again, all substances are present in the liquid phase
u
ad

Theory of Homogeneous Catalysis. .(Intermediate compound formation theory). According to modem

Yo

views, a catalyst enters into chemical combination with one or more of the reactants forming an intermediate
compound which then decomposes or combines with one of the reactants to produce the product and the
d

catalyst is regenerated. The involvement of the catalyst in the reaction lowers the free energy of activation
Re
in

and hence accelerates the speed of the reaction. In other words, a new more efficient path is followed, e.g.,
(i) Oxidation of SO2 to SO3 in presence of NO takes
F

place as follows :
02(g) +2NO(g)- ^ 2N02(g)
Reactant Catalyst Intermediate

S02(g) + N02(g) - > 803(g) + NO(g)

Reactant Intermediate Product Regenerated catalyst
(ii) Decomposition of H2O2 in presence of 1“ ion takes place as follows :
H2O2 + r > H2O +10" (hypoiodite ion) ; lO" + H2O2 ● ■> H2O + O2 + I
(iii) Decomposition of ozone in presence of NO takes place as follows :
NO + O3 NO2 + O2 i NO2 + O »N0 + 02
(tv) Formation of diethyl ether from ethyl alcohol using concentrated sulphuric acid takes place as follows ●
C2H5OH (/) + H28O4 (/) > C2H5H8O4 (aq) + H2O (0
Reactant Catalyst Intermediate

C2H50H(/) + C2H5H804(o^) » C2H5 —O —CjHjW + H2S04(a«)

Reactant Intermediate
(Diethyl ether) Product Regenerated catalyst
5/18 “pfusUec^ New Course Chemistry

(2) Heterogeneous Catalysis, If the catalyst is present in a different phase than that of the reactants, it
is called a heterogeneous catalyst and this type of catalysis is called heterogeneous catalysis.
The catalyst in heterogeneous catalysis is generally solid and the reactants are mostly gases and sometimes
liquids. In heterogeneous catalysis, the reaction starts at the surface of the solid catalyst. That is why it is also
known as surface catalysis.
Some of the important examples of heterogeneous catalysis which are of commercial or technological
importance are given below :
(i) Manufacture of ammonia from N2 and H2 by using iron as catalyst (Haber s process)
Fe(.T)
N2 (g) + 3 H2 (g) — » 2NH3(g)
Here, reactants are gaseous whereas catalyst is solid.
((7) Synthesis of methyl alcohol from CO and H2 using a mixture of Cu, ZnO and Ci20^,
as catalyst.

w
Cu(i) +
CO(g) + 2H2(g) ZllO(5) +
^ CH3OH
Cf203 (i’)

F lo
Here, again reactants are gaseous and catalyst is a mixture of solids
(Hi) Manufacture of sulphuric add by the oxidation of SO 2 to SOj using platinised asbe.stos or V2O5 as

ee
catalyst (contact process).

Fr
Pt(j)orV205 (.0
S02(g) + 02(g) > 2 SO3 (g)

for
(iv) In the manufacture of nitric add, oxidation of ammonia to nitric oxide using platinum as catalyst
ur
(Ostwald process).
Pt
s
4NH3 + 5O2 ^ 4NO + 6H2O
ook
Yo
800"C

(v) Hydrogenation of unsaturated hydrocarbon in presence ofifnely divided Ni, Pd or Pt, e.g., ethylene
eB

(CH2 = CH2) to ethane (CH^ - CH^) and conversion

Ni(j)
of vegetable oil into margarine (imitation butter) or ghee.
Vegetable oil (/) + H2 (g) > Vegetable ghee (5)
Here, one of the reactants is a liquid, the other is gaseous whereas catalyst is solid,
r
ou
ad

(vi) Polymerisation of ethylene using 1TCI4 and trialkyl aluminium as catalyst (Ziegler - Natta catalyst).
Y

TiCl

« CH2 = CH2(g) U -fCH2-CH2^

+ R3AI
Re
nd

The catalyst is prepared by dissolving (C2Hg)3Al and TiCl4 in a hydrocarbon solvent which react
Fi

exothermically to form a brown solid.

(vii) Oxidation of CO and unburnt petrol (hydro- carbons) to CO2 and reduction of NO to N2 in the
automobile exhaust gases by the catalytic converter (consisting of transition metals and their oxides) fitted
in the automobile exhaust system (Fig. 5.10). FIGURE 5.10
Air Intake Catalyst bed
Catalyst
2 CO + Oo > 2CO2 ;
Catalyst 1 polluting^
2 NO > N2 + O2 Exhaust
V
gases from gases
Catalyst compressor

Hydrocarbons ^ CO2 "I" H2O Catalytic converter

(unburnt petrol) O2

(vm) Cracking of hydrocarbons in the presence of hydrogen by zeolite catalyst,

(ix) Production of hydrocarbons from CO and H2 using iron or cobalt as catalyst. (Fischer - Tropsch
process).
SURFACE CHEMISTRY 5/19

Theory of Heterogeneous Catalysis. {Adsorption theory). According lo the old adsorption theory of
heterogeneous catalysis, it was believed that the reactants in the gaseous state or from the solutions are
adsorbed on the surface of the catalyst. As a result, the concentrationof the reactant molecules on the surface
increases and hence the rate of reaction increases. Further, as adsorption is always exothermic, the heat
released further helps to speed up the reaction.
Another theory of heterogeneous catalysis, as already discussed in unit 4, is the intermediate compound
formation theory. This theory suggests that the reactants (A and B) first combine with the catalyst (C) to
form an intermediate complex which decomposes to form the products and regenerates the catalyst.
The ‘modern adsorption theory’ is a combination of both the above theories. According to this theory,
there are free valencies on the surface of solid catalysts (generally metals) and the mechanism of catalysis
involves the following five steps :
{/) Diffusion of reactant molecules towards the surface of the catalyst (Fig. 5.11 u).
(//) Adsorption of the reactant molecules on the surface of the catalyst by forming loose bonds with the

low
catalyst due to presence of free valencies (Fig. 5.11 b).
{Hi) Occurrence of a chemical reaction between the reactants and the catalyst forming an intermediate, as
shown in Fig. 5.11 (c).
{iv) Desorption of the product molecule from the surface due to lack of its affinity for the catalyst surface,

ee
thereby making the surface free for fresh adsorption of reactant molecules (Fig. 5.11 {d)).
rF
Fr
(v) Diffusion of product molecules away from the surface of the catalyst.
FIGURE 5.11

A+B (Reactant A— B A (Product molecule)

▼ Molecules)

for ' Diffuses away

u
Diffusion Desorption
ks
of reactant of product
Yo
molecules molecules
o

towards
Adsorption of Occurrence of
Bo

Catalyst surface catalyst surface reactant molecules chemical reaction Catalyst

with free valencies regenerated
forming loose bonds forming an intemediate
o o o o
re

Systematic mechanism of modern adsorption theory of catalysis

ou
ad

Advantages of modern adsorption theory. This theory can explain the following :
Y

(/) Small quantity of the catalyst is sufficient because the catalyst is regenerated again and again.
(//) The catalyst takes part in the reaction but is produced back unchanged in mass and chemical composition
nd
Re

at the end of the reaction.

{Hi) Catalytic poisons, if present, are preferentially adsorbed on the catalyst surface, thereby hindering the
Fi

adsorption of the reactant molecules. Hence, they slow down the activity of the catalyst.
This theory, however, cannot explain satisfactorily the action of catalytic promoters.
5.8.5. Some Important Features of Solid Catalysts (or Heterogeneous Catalysis)
Some important aspects of heterogeneous catalysts are briefly described below :
(1) Activity. By activity of the catalyst we mean its capacity to increase the speed of the chemical
10
reaction. In certain cases, the extent to which speed of the reaction i,s increased is as high as 10 times.
Combination of H2 and O-, in the presence of platinum (catalyst) to form water with explosive violence is an
excellent example of catalytic activity. In the absence of the catalyst, platinum, H2 and O2 do not combine
and can be stored as such for an indefinite period.
Pt
2H2 + O2 ^ 2 H2O
The activity depends upon the extent of chemisorption. The adsorption should be reasonably strong but
not so strong that the adsorbed molecules become immobile and no space is available for other reactants to
get adsorbed.
5/20 'P’utdee^’^ New Course Chemistry (XII)GSm

(2) Selectivity. By .selectivity of a catalyst we mean its ability’ to direct the reaction to form particular
products excluding others. For example,
(/) CO and H2 react to form different products in presence of different catalysts as follows :
Cu Cu/Zn0-Cr203
CO ig) + (g) ^ HCHO(g) ; CO(g) + 2U2(g) » CH3OH (g)
Ni
CO ig) + 3H2ig) ^ CH4(g) + H20(g)
H H
{//) Acetylene on reaction with H2 in presence of
Pt
platinum as catalyst gives ethane while in the presence ●►H—C—C—H
of Lindlar’s catalyst (palladium + BaS04 poisoned with
quinoline or sulphur) gives ethylene. H H
Ethane
(3) Specificity. Action of a catalyst is highly specific H—C=C—H + H2-

low
(selective) in nature, i.e., a given substance can act as a Acetylene H H
(Ethyne)
catalyst only in a particular reaction and not for all the Lindlar’s
► H—C=C—H
reactions. It means a substance which acts as a catalyst catalyst
in one reaction may fail to catalyse other reaction. Ethylene
{Pd+BaSOi
(Elhene)
+ quinoline)

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Curiosity Question

F
Fr
f Q. What is done to reduce pollution by carbon monoxide or nitric oxide etc. formed during
combustion of fuels in automobiles ?

for
ur
Ans. Catalytic convertors consisting of transition metals and their oxides are fitted into the automobile exhaust
system. They help in oxidation of CO to CO2 and reduction of NO to N2 which are harmless gases.
J
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5.9. ZEOLITES* AS SHAPE-SELECTIVE CATALYSTS
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Zeolites are aluminosilicates with the general formula, [(A102)x (Si02)y] - z H2O, where n is the
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n+
charge on the metal cation M , which is usually Na"*", or and z is the number of water mole*
cules of hydration which is highly variable. They are microporous three dimensional net work silicates in
which some silicon atoms are replaced by aluminium atoms giving Al-O-Si frame work. They have honey
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comb like structure.** They are found in nature as well as synthesised in the laboratory or industry. They
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form an important class of oxide catalysis. Zeolites to be used as catalysts are heated in vacuum so that the
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water of hydration is lost. As a result, zeolite becomes porous, i.e., the cavities in the cage-like structure
which were occupied by the water molecules become vacant. The size of the pores generally varies between
nd
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260 pm and 740 pm. Thus, only those molecules can be adsorbed in these pores whose size is small enough
to enter these cavities and also leave easily. It will not adsorb those molecules which are too big to enter.
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Thus, zeolites act as selective adsorbents and hence as ‘molecular sieves’. For example, the zeolite, sodium
alumino silicate, can adsorb straight chain hydrocarbonsand not branchedchain or aromatic ones.
Consequently, the reaction taking place in zeolites depends upon the size and shape of reactant and
product molecules as compared to those of the pores and cavities of the zeolite. That is why these types of
reactions are called ‘shape-selective catalysis’.
5.12
*The word ‘zeolite’ in Greek language means ‘boiling stone’.
0= Silicon or
It has been so named because water trapped in the pores boils off Aluminium
when zeolite stone is heated.
O = Oxygen

**Structure of zeolites consists of truncated octahedra (cubo-

octahedra) as the building blocks. This structure is called (5-cage or
sodalitc cage (Fig. 5.12).

Cavities in zeolites
SURFACE CHEMISTRY 5/21

Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons
and isomerization. An important zeolite catalyst used in the petroleum industry is ZSM—5* (Zeolite Sieve of
Molecular Porosity 5). It converts alcohols directly into gasoline (petrol) by first dehydrating them so that a
mixture of hydrocturbons is formed.
ZSM-5
Alcohols 7-> Hydrocarbons
Dehydralion
It may be mentioned that hydrated zeolites are used as *ion- exchangers' in .softening of hard water.

5.10. ENZYMES AS CATALYSTS

5.10.1. What are enzymes ?

All biological reactions are catalysed by special catalysts called enzymes. Thus, enzymes are deifned as
bio-chemicalcatalysts. Chemically,all enzymesare globular proteins (which are complex nitrogenous compounds)

w
with high molar ma.^s rangingfrom 15,000 to 1,000,000 g mol~^ and form colloidal solution in wate.r
Every biological reaction requires a different kind of enzyme. Since there is a large number of such
biological reactions, therefore, there is a large number of enzymes functioning in a living system. A typical

Flo
cell, on the average, contains about 3000 different kinds of enzymes, each catalysing a different reaction.
Enzymes are produced by living plants and animals. Many of them have been obtained in the pure

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crystalline state from living cells. However, the first enzyme was synthesised in the laboratory in 1969.

Fr
Enzymes are vital for biological processes. Without them, the life processes would be very slow and
sluggish. For example, if there were no enzymes in our digestive system, it would take us 50 years to digest
a single meal.
for
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Examples of a few enzyme catalysed reactions are given below :
s
S. No. Reaction Enzyme Source
k
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(0 Inversion of cane-sugar Invertase Yeast

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Iiivertase
^12^22^11 ^2^ ^ ^6^12^6 ^6^12^6
Sucrose Glucose Fructose
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(ii) Conversion of glucose into ethyl alcohol Zymase Yeast

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Zymase
C6H,206(«^) > 2 C2H5OH (aq) + 2 CO, ig)
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Glucose

(Hi) Conversion of starch into maltose Diastase Malt

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nd

Diasta.se
2(C6H,o05)„(fl«7) + nH20 (/) > n C12H22O1, (aq)
Fi

Starch Maltose

(fv) Conversion of maltose into glucose Maltase Yeast

Maltase
C,2H220,i(a^) + H20 (/) > 2 CgHi205 (aq)
Maltose Glucose

(V) Manufacture of acetic acid from ethyl alcohol klycoderma aceti Old vinegar
CH3CH2OH (/) + O, (g) > CH3COOH (0 + H,0 (0
(V/) Decomposition of urea into NH3 and CO, Urease Soyabean
___ Urease
NH2CONH2 (aq) + H2O (/) ^ 2NH3(g) + CO,(g)
Urea

(V//) Conversion of proteins into peptides in the stomach Pepsin Stomach

(viii) Conversion of proteins into amino acids by hydrolysis in intestines Trypsin Intestines

(ix) Conversion of milk into curd Lactic bacilli Curd

*Its formula is H, [(AlO,)^ (SiO^Ig^..,] . 16 H2O.

5/22 ‘P%ad^e^'A New Course Chemistry (XII)ESsISl

5.10.2. Characteristics of Enzymes

Some important characteristics of enzymes are given below :
(i) Specificity. Each enzyme catalyses only one chemical reaction. For example, the enzyme urease
hydrolyses urea to NH3 and CO-) but it does not catalyse the hydrolysis of N-methylurea which is similar in
constitution to urea.
Urease
NH2CONH2 + H2O ^ 2NH3 + CO2
Urea

Urease
CH3NHCONH2 + H2O ^ No action

N-Methylurea
Similarly, the enzyme invertase hydrolyses sucrose to glucose and fructose but does not hydrolyse maltose
to glucose though both are hydrolytic reactions. For the hydrolysis of maltose, the enzyme maltase is used.
Invertase
C12H22O11 + H2O » C6H12O5 + C5Hi20g

w
Sucrose Glucose Fructose

Mallasc

F lo
^12^22^11 + H2O
Maltose Gluco.se

(ii) Efficiency. Enzymes are very efficient catalysts. They speed up the rate of a reaction by factors of

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iipto 10^^.* To get an idea about the tremendous efficiency of enzymes, let us consider the hydrolysis of amides.

Fr
Amide bonds are chemically very stable. That is why, hydrolysis of amides in the laboratory requires the
heating of amides with an alkali for a few hours. In contrast, proteins present in our food, which also contain
for
amide (peptide) bonds, arc easily hydrolysed in our body by different enzymes to individual a-amino acids.
ur
(Hi) Small quantity. Only small amounts of enzymes can be highly efficient. This is due to the reason
that like chemical catalysts, enzymes are also regenerated after their catalytic activity but their rate of
s

regeneration is very fast, of the order of about 1 million times per minute. For example, the enzyme renin which
ook
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is used for coagulation of milk to make cheese coagulates over a million times its own weight of milk protein.
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Similarly, the enzyme carbonic anhydrase, present in the red cells of the blood, catalyses the reversible
reaction involving the decomposition of carbonic acid to CO2 and H2O.
Carbonic ajihydrasc .
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H^CO 3 CO2 + H2O

Carbonic hydrase
Under ideal conditions, a single molecule of carbonic anhydrase can decompose 36 million molecules
of carbonic acid to CO2 and HoO in one minute. The reverse reaction, i.e., the conversion of CO2 and H^O to
dY
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form carbonic acid is catalysed by the enzyme carbonic hydrase. This enzyme is responsible for maintaining
the CO2 level in the body fluids and tissues,
Fin

.13
(iv) Optimum temperature and pH.
Enzyme catalysed reactions have maximum
rate at physiological pH of around 7-4 and
human body temperature of 37"C {3}Q K)
under one atmospheric pressure**. Under
these conditions, most of the chemical
reactions do not occur at appreciable rales if
ordinary laboratory catalysts are used. In fact,
as the temperature or pH is increased, the rate TEMPERATURE (°C]

rises to a maximum (at 37® C or /?H = 74)

and then falls off. Variation of enzyme activity with temperature and pH

*Like other catalysts, enzymes also accelerate a particular reaction by lowering its energy of activation.
**The optimum temperature range for enzymatic activity is 298-310 K whereas optimum pH range is 5-7.
SURFACE CHEMISTRY 5/23

(v) Enzyme activators (co-enzymes). The activity of certain enzymes is increased in the presence of
certain substances, called co-enzymes. It has been observed that if a protein contains a small amount of
vitamin as the non-protein part, its activity is enhanced considerably. The activators are generally metal ions
such as Na'*’, Mn-^, Cu^'*’, Co-'*’ etc. These metal ions are weakly bonded to the enzyme molecules and
increase their catalytic activity. For example, the enzyme, amylase, in pre.sence of NaCl, which provides Na'*’
ion, shows a very high catalytic activity,
(vi) Enzyme inhibitors and poisons. Just as in the case of catalysts, the activity of enzyme is slowed
down in the presence of certain substances. Such substances are called inhibitors or poisons. They act by
combining with the active functional group thereby reducing or completely destroying the catalytic activity
of the enzymes. The use of many drugs is on account of their action as enzyme inhibitors in our body.

5.10.3. Mechanism of Enzyme Catalysis

low
The most accepted mechanism of enzyme catalysed reaction is known as Lock and Key mechanism. It
briefly explained below ;
Enzymes are highly specific in their action. The specificity of the enzymes is due to the presence of
some specific regions, called the active sites, on their surface. These active sites are associated with some

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functional groups such as - NH2 - COOH, - OH, - SH etc. which form weak bonds such as H-bonds, van der
rF
Fr
Waals attraction etc., with the substrate (reactant) molecules. The shape of the active site of any given enzyme
is like a cavity such that only a specific substrate can fit into it, in the same way as one key can fit into a

r
particular lock. This specific binding leads to the formation of an enzyme-substrate complex which accounts
for the high specificity of enzyme-catalysed reactions. fo
u
Once the proper orientation has been achieved, substrate molecules react to form the products in two
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steps as shown in Fig. 5.14. Since the product molecules do not have any affinity for the enzyme surface, they
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at once leave the enzyme surface making room for the fresh substrate molecules to bind to the active sites.
B

The whole sequence is diagrammatically represented in Fig. 5.14.

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FIGURE 5.14
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V_yl
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Enzyme Substarate Enzyme-Subslarate Enzyme-Product Enzyme Products

Catalyst (Recatants) Complex Association
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(Like a lock) (Like a key)

Lock and key mechanism of enzyme action

Thus, the enzyme catalysed reactions lake place in two steps as follows :
Step 1. Formation of enzyme-substrate complex.
E + S - ES (Fast and Reversible)
Enzyme Substrate Enzyme - Substrate complex

Step 2. Dissociation of enzyme-substrate complex to form the products.

ES ■> [EP] E + P (Slow and Rate determining)
Enzyme Sub Enzyme-Product Enzyme Product
strate complex association (Regenerated)
The rate of formation of product depends upon the concentration of ES.
5/24 New Course Chemistry CX1I)ESSI9]

5.10.4. Points of Difference between Enzymes and Inorganic Catalysts

rw
Enzymes Inorganic Catalysts
1. They are proteins with high molar mass. 1. They are simple ions or molecules with low
molar mass.
2. They catalyse biological reactions. 2. They are used to catalyse non-biological reactions.
3. They are highly specific in nature. 3. They are not so specific in nature.

e
4. Their efficiency is very high. 4. Their efficiency is generally not so high.
5. Their efficiency is greatly affected by change

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5. Their efficiency is generally not much affected

o
in temperature or pH. by change in temperature or pH.

lu
F
6. Their action is explained by lock and key type 6. Their mechanism is explained either by
mechanism. intermediate compound formation theory or by
adsorption theory.

F
s
o
Curiosity Question

kr
r Q. Why food in our body Is digested so fast ?

oo
Ans. The digestion of food in our body takes place through reactions catalysed by enzymes present in
our body. These biochemical catalysts are highly efficient. They speed up the rate of reaction by

of
factors as high 10^*^. Moreover, their very small quantity is needed as they are regenerated very
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fast of the order of about 1 million times per minute.. I
B

SUPPLEMENT ; i.
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KNOWLEDGE FOR C0 riTIONS

Kinetics of Enzyme catalysed Reactions (Michaelis-Menten equation)

de
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The enzyme catalysed reaction lakes place in two steps as follows :

o
n

E + S ES
ad

T 7 E + P
Enzyme Substrate *1 Enzyme substrate Producl
complex
i

The rate (velocity) of this reaction is given by

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d[S] [Eq][$] ...(/)

V = -
= k2\ES] = k
dt 2k +[S]
This equation is called Michaelis-Menten equation. In this equation,
[S] = Concentration of the substrate (reactant)
[Ey] = Total concentration of the enzyme (free + combined)
K,„ = Constant called Michaclis constant
The velocity is maximum (= when [6'] is very large, i.e., [S] » K so that in eqn. (/),
III can be
neglected in comparison to [SJ and hence, we have
V
= *2iEoi m ...{it)
i.e.. it is independent of the concentrationof the substrate and hence is a reaction of zero order with
respect to .substrate.
V [S]
From eqn. (/) and (n),
V
m K,„+[S]
SURFACE CHEMISTRY 5/25

V 1
When [S] = Km'
V 2
m

I
or V = —V
2

Thus, Michaelis constani is equal to that concentration of the substrate at which the rate of enzyme
catalysed reaction falls to one-half of the maximum value.

ow
Three cases arise as explained below :
{/) If substrate concentration [5], is very high, the reaction is of zero order with respect to substrate as
explained above.
(«) If substrate concentration [5] is low, [S] can be neglected in comparison to K,„ in eqn. (i)- Then we

e
have

re
v = - d[S] ^ k^[Eo][S]
dt ~ K

F
Frl
m

^[S]
or v = -
ou
dt
oc
[EqHS] ...(//)

sr
kfo
(Hi) If enzyme concentration [£q] is kept constant, then from eqn. (ii) above, the reaction becomes of the
first order.
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5.11. CATALYSTS USED IN INDUSTRIES
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To accelerate the rates of reactions so as to get maximum yields of the products in minimum time,
catalysts are very often used in the chemical industries along with other suitable conditions. A few common
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examples are given in Table 5.3 below :

TABLE 5.3. Some Industrial Catalytic Processes

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do

Process Reactions with catalyst and other conditions

in
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Finely divided iron (catalyst)

1. Haber’s process for the N. (g) + 3 H2 (g) 2 NH3 ig)
F

manufacture of ammonia Molybdenum (Promoter)

200 bar pressure
723-773 K temp.

Platinised
2. Ostwald’s process for the 4NH3 (g) + 5 02(g) > 4 NO (g) + 6 H2O (g)
manufacture of nitric acid asbestos (catalyst)

573 K
2 NO (g) +02(g) ^ 2 NO2 (g)
4N02(g) + 2H20 (/) +02(g) ^ 4 HNO3 (/)
3. Contact process for the Platinisedasbeslos ^
2S02(g) + 02 (g) 2 SO3 (g)
manufacture of sulphuric ^ or\^0^7cataiyst)
acid
673-723K + HiO(/)
SO3 (g) + H2S04 (/) H2S207(/) > 2 H2SO4 {aq)
Oleum

*
5/26 '4. New Course Chemistry (XII) vrsrm

4. Fe-) O3 (Catalyst)
Bosch’s process for the CO + H, + H^O(g) > C0,(g) + 2H,(g)
manufacture of hydrogen. +Cft03 (promoter)
Watergas 673-87.1 K

CUCI2
5. Deacon’s process for the 4HCl(g) + O.ig) > 2H,0(/) + 2Cl,(g)
manufacture of chlorine 773K

Zi)0 +Cr^03
6. Synthesis of methanol CO (g) + 2H2(g) 200 bar
^ CH30H(/)
432 K

Finely divided Ni
7. Hydrogenation of Oil (O + H.lg) 4 Vanaspati ghee (^)
423-473K
vegetable oils
High pressure

low
PdCl2
8. Wacker process for CH2 = CH2 + H2O t+CuCb) > CH3CH0+Pd+2HC1
preparation of aldehydes Acetaldehyde
and ketones from alkcnes
PdCh
^ CH.COCH. + Pd + 2HCl
CH3-CH = CH2 + H2O (+CuCh) 3 3

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Acetone

F (Refer to Unit 12)

Fr
SUPPLEMENT YOUR

for
KNOWLEDGE FOR COMPETITIONS
ur
1. Autocatalysis. During a chemical reaction, if one of the products formed acts as a catalyst, the
phenomenon is called aulocaialysis, e.g., in the titration of oxalic acid solution with KMn04 solution in
ks
presence of dilute H2SO4, the colour of KMn04 finst fades slowly and then faster due to the formation of
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Mn^"*" ions which act as autocatalysl.

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2 KMn04 + 3 H2SO4 + 5 H2C2O4 > K2SO4 + 2 MnS04 + 8 H2O + 10 CO2

Similarly, decomposition of nitroglycerine gives oxides of nitrogen which then catalyse the reaction.
Ester hydrolysis is slow in the beginning but becomes faster after some time because acid produced acts
r
ou
ad

as an autocalalyst (RCOOR' + H2O > RCOOH + R'OH). Rate of reaction = k [RCOOR'] [RCOOH]
2. Induced catalysis. When a reaction affects the rate of another reaction which may not otherwise occur
Y

under the same conditions, the phenomenon is called induced catalysis, e.g., sodium arsenite is not
oxidized by air but if air is passed through a solution containing sodium arsenite and sodium sulphite,
Re
nd

both undergo oxidation. Thus, oxidation of sodium arsenite has been induced by oxidation of sodium
Fi

sulphite.

5.12, COLLOIDAL STATE OF MATTER

Thomas Graham (1861) studied the rates of diffusion of solutions of different substances through
parchment membrane or animal membrane (a semipenneable membrane) and as a result of his experiments,
he divided the substances into two classes - (/) crystalloids and (ii) colloids. The substances like common
salt, sugar, urea, etc. which can be obtained in the crystallineform and in the dissolved state diffuse readily
through parchment membrane were termed as crystalloids, and the substances like starch, gum, glue, gelatine,
albumin, silicic acid etc. which are non-crystalline in nature and in the dissolved state do not diffuse or
diffuse at a very slow rate through the parchment membrane were given the name colloids (Greek : glue
like). Solutions formed by crystalloids are called true solutions while those formed by colloids are called
colloidal solutions.

Graham’s classification of substances into crystalloids and colloids was soon proved to be wrong. A
crystalloid was found to behave as a colloid under different set of conditions and vice versa. For example,
 SURFACE CHEMISTRY 5/27

sodium chloride behaves as a crystalloid when dissolved in water but behaves as a colloid when dissolved
in benzene.

It has since then been realised that a clear cut difference does not exist between crystalloids and
colloids.

The non-diffusibility of colloids and diffusibility of crystalloids through the animal or vegetable membrane
was shown to be due to the difference in the size of particles of crystalloids and colloids. The colloids form
bigger ptulicles which cannot pass through the membrane. It is, therefore, preferable to speak of a substance
in ‘colloidal state' rather than calling it a crystalloid or a colloid. The particles in a colloidal solution do not
settle down on standing. However, if the size of the particles is still larger, these settle down on standing. Such
a system is called suspension. The particle size limits are given in the Table below :

Size (diameters) of Particles*

w
'fVue Solutions Colloidal Solutions Suspensions
Less than 10“^ m or I nm Between 10 ^ and 10"^ m or 1 nm to More than 10"^ m or
1000nm i.e., 10 A—10000 A

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i.e., < 10 A 1000 nm i.e., > 10000 A

Thus, the size of the colloidal particles is intermediate between that of particles of true solution and

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those of suspension. These sizes are such that particles of true solution can pass through parchment membrane

Fr
as well as filter paper whereas those of colloidal solution cannot pass through parchmentmembranebut can
pass through filter paper and those of suspension can pass neither through parchment membrane nor through
filter paper.
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Colloidal state of matter is, therefore, a state in which the size of the particles is such
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(1 to 1000 nm) that they can pass through filter paper but not through animal or vegetable
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membrane.
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Thus, every substance can be brought into the colloidal state by adopting suitable methods.

*10-10 m = 1 A. Sometimes sizes are also expressed in terms of milli-micron (mji). 1 /«l-i = 10 micron
r

= 10~^ m = 10 A = 1 nanometer (1 nm).

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It is important to note that due to their small size, colloidal particles have large specific surface area (surface
Y

area per unit mass). Thus, if a particle of radius 1 cm (Volume = — 7t cm^ and surface area = 4tc =; 12 cm^) is
3
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broken into colloidal particles of radius 1000 A, i.e., 10“^ cm (so that each has volume = —
- jixlO”’^ cm^ and
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surface area = 4ti x 10"*^ cm^ s= 12x10"'^ cm-), number of colloidal particles = (4/37i:)/(4/37t x 10“*^) = lO’^.
Their total surface area = (12 x 10'^®) x lO’^ = 12 x 10^ cm-. A number of properties of the colloids are due to
their large surface area.
To understand the above concept in a more simple way, consider a cube of 1 cm side (though colloidal
particles are not cubical).
Total surface area of the cube = 6 = 6 cm-. Now, if the cube is divided into equal sized 0^^ cubes, these
small cubes will have size equal to large colloidal particles. Their total surface area can be calculated as follows

Volume of the large cube = = 1 cm^

.●. Volume of each small cube = cm^ = 10 cm
3

I0'2
As volume is (side) A therefore, side of each small cube = 10”^ cm
Surface area of each small cube = 6 = 6 x 10"*^ cm-
.●. Total surface area of all the 10*- cubes = (6 x 10“^) x 10*^ cm^ = 60,000 cm-
5/28 New Course Chemistry CXII)S!Zsl9Q

5.13. DISTINCTION BETWEEN TRUE SOLUTIONS, COLLOIDS SOLUTION AND SUSPENSIONS

The distinction between characteristics of true solutions, colloidal solutions and suspension is given in
Table 5.4.

TABLE 5.4. Characteristics of true solutions, colloidal solutions and suspensions

S. No. Property True Solutions Colloidal Solutions Suspensions

1. Nature Homogeneous Heterogeneous Heterogeneous
2. Particle size Less than 10"^ m or Between 10"^to 10“^’mor I nmto More titan 10"^’ in or
(diameters) 1 nm (i.e., < 10 A) 1000 nm(i.e., 10 A to 10000 A). 1000 mil (i.e.. >10000 A)
3. Filtrability Pass through ordinary Pass through ordinary filter paper Do not pas.s through filter
filter paper as well as but not through animal membrane. paper and animal
animal membrane. membrane.

4. Settling Do not settle. Do not settle. Settle on standing.

5. Visibility Particles are invisible. Scattering of light by the particles Particles are visible to
is observed under ultra-micro- naked eye or under a

F low
scope. microscope.
6. Diffusion Diffuse quickly. Diffu.se slowly. Do not diffuse.
7. Appearance Clear and transparent Translucent. Opaque.

A simple example to understand a true solution, a suspension and a colloidal solution is to consider the
for Fre
solution of sugar in water and that of water in which mud is stirred. The former is a true solution but in case
of the latter, so long as the stirring is continued, the mud remains in suspension but as soon as the stirring is
stopped, the larger particles fall to the bottom of the vessel. However, smaller panicles remain in suspension
for a longer lime (say for several days). This part of water containing minute mud particles which cannot be
seen even under a microscope and pass through the filter paper is a colloidal solution.
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5.14. DISPERSED PHASE AND DISPERSION MEDIUM

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Just as in true solution, the substance dissolved is called the solute and the medium in which it is
our

dissolved is called the solvent, similarly in a colloidal system, the terms solute and solvent are replaced by the
terms dispersed phase and dispersion medium respectively. Thus, dispersed phase means the substance
distributed in the dispersion medium in the form of colloidal particles and the dispersion medium means the
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medium in which the substance is dispersed in the form of colloidal particles. The colloidal system thus
obtained is .sometimes called a colloidal dispersion or simply called colloid. Evidently, unlike a true solution
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which is homogeneous, a colloidal system is heterogeneous consisting of two phases — the dispersed phase
and the dispersion medium.

5.15. CLASSIFICATION OF COLLOIDS

Colloids are classified in three different ways as follows :

(A) Ba.scd on physical state of dispersed phase and dispersion medium. Depending upon whether
the dispersed phase and the dispersion medium are solids, liquids or gases, eight types of colloidal systems
are possible. A gas mixed with another gas forms a homogeneous mixture and not a colloidal system. The
examples of the various types along with their typical names are listed in Table 5.5.
Out of the various types of colloidal systems listed in the Table 5.5, the most common arc sols (solids in
liquids), gels (liquids in solids) and emulsions (liquids in liquids). However, in the present unit, we shall take
up a detailed discussion of the ‘sols’ and ‘emulsions’ only. Further, it may be mentioned that depending upon
the dispersion medium, the sols are given special names as follows :
 SURFACE CHEMISTRY 5/29

TABLE 5.5. Types of colloidal systems (or colloidal dispersions)

S. No. Dispersed Dispersion Name Examples
Phase Medium

1. Solid Solid Solid sol Some coloured glasses, gem stones

2. Solid Liquid Sol Some paints, cell fluids, muddy water
3. Solid Gas Aerosol Smoke, dust

4. Liquid Solid Gel Cheese, butter, jellies

5. Liquid Liquid Emulsion Milk, hair cream
6. Liquid Gas Aerosol Fog, mist, cloud, insecticide sprays

low
7. Gas Solid Solid foam Pumice stone, foam rubber
8. Gas Liquid Foam Froth, whipped cream, soap lather

Dispersion medium Name of the sol

Water

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Aquasol or Hydrosol
Alcohol

F AIcosol

Fr
Benzene Benzosol
Gases Aerosol

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(B) Based on nature of interaction between dispersed phase and dispersion medium. On this basis,
colloidal sols are divided into two categories, namely, lyophilic and lyophobic. If water is the dispersion
ks
medium, the terms used are hydrophilic and hydrophobic.
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1. Lyophilic colloids. The word lyophilic means liquid-loving.

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Substances like gum, gelatine, starch, rubber etc. which when mixed with a suitable liquid as
the dispersion medium directly form the colloidal sol are called lyophilic and the sols thus
obtained are called lyophilic sols.
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As they form the colloidal sol directly, they are also called intrinsic colloids. An important characteristic
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of these sols is that if the dispersed phase is separated from the dispersion medium (say, by evaporation), the
sol can be made again by simply remixing with the dispersion medium and shaking. That is why these sols are
nd
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also called reversible sols. Further, these sols are quite stable and cannot be easily precipitated.
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2. Lyophobic colloids. The word ‘lyophobic’ means liquid-hating.

Substances like metals, their sulphides etc. when simply mixed with the dispersion medium do
not form the colloidal sol. Their colloidal sols can be prepared only by special methods (as
discussed later). Such substances arc called lyophobic and the sols formed by them are called
lyophobic sols.

As their colloidal sols have to be prepared by indirect methods, they are also called extrinsiccolloids.
These sols are readily precipit' ed (or coagulated) and hence are not stable. Further, once precipitated, they
do not give back the colloidal sol by simple addition of the dispersion medium. Hence, these sols are also
called irreversible sols.

The greater stability of the lyophilic colloidal sols than the lyophobic colloidal sols is due to the fact that
the former are highly hydrated in the solution.
The essential points of difference between the lyophilic sols and lyophobic sols are given in
Table 5.6.
5/30 'a New Course Chemistry (XII)

TABLE 5.6. Points of difference between lyophilic and lyophobic sols

S. No. Property Lyophilic Sols Lyophobic Sols

1. Ease of preparation Prepared easily by directly mixing with Cannot be prepared directly. Prepared
the liquid dispersion medium. by special methods only.
2. Stability They are quite stable and are not easily They are easily precipitated by addition
precipitated or coagulated. of a small amount of a suitable
electrolyte.
3. Hydration They are highly hydrated They are not much hydrated.
4. Reversible and They are reversible in nature, i.e., once They are irreversible in nature, i.e.,
irreversible nature precipitated can refonn the colloidal once precipitated cannot form the

ow
sol by simply remixing with the colloidal sol by simple addition of the
dispersion medium. dispersion medium.
5. Nature of substances These sols are usually formed by the These sols are usually formed by the
organic substances like starch, gum. inorganic materials like metals, their
proteins etc. sulphides etc.

e
Fl
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6. Viscosity Their viscosity is much higher than that Their viscosity is almost the same as
of the medium. that of the medium.

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7. Surface tension Their surface tension is usually lower Their surface tension is nearly same as
that of the dispersion medium.
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than that of the dispersion medium.

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(C) Based on the type of particles of the dispersed phase. (Multimolecular, Macromolecular and
ks
Associated Colloids)
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Depending upon how the different substances may have size in the range of the colloids, the various
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types of colloids or colloidal dispersions may be divided into the following three categories:
(1) Multimolecular colloids. When on dispersion of a substance in the dispersion medium, a large
B

number of atoms or smaller molecules of the substance (with diameters less than I nin) aggregate together
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to form species having size in the colloidal range, the species thus formed are called multimolecular colloids.
For example, a gold sol may contain particles of various sizes having several atoms. Sulphur sol consists of
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particles containing a thousand or so of Sg sulphur molecules. These are held together by van der Waals
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forces.

(2) Macromolecular colloids. When certain substances having big size molecules, called
d

macromolecules, having large molecular masses are dissolved in a suitable Hcfuid, they form a .solution in
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in

which the molecules of the sub,stance, i.e., the dispersed particles have size in the colloidal range. Such
substances are called macromolecular colloids.
F

These macromolecular substances are usually polymers with very high molecular masses. Examples of
naturally occurring macromolecules are starch, cellulose, proteins, enzymes and gelatine. Examples of man
made macromolecules are polyethylene, nylon, polystyrene, synthetic rubber, etc.
As these molecules have large sizes and have dimensions comparable to those of colloidal particles,
their dispersions are called macromolecular colloids. Their colloidal solutions are quite stable and resemble
true solutions in many respects.
(3) Associated colloids — Micelles. The substances which when dissolved in a medium at low
concentrations behave as normal, strong electrolytes but at higher concentrations exhibit colloidal state
properties due to the formation of aggregated particles are called associated colloids. The aggregated
particles thus formed are called micelles. The formation of micelles takes places only above a particular
temperature called Kraft temperature (T^ and above a particular concentration called Critical Micelle
Concentration (CMC). The most common example of associated colloids is that of surface active agents
such as soaps and synthetic detergents. For soaps, the CMC is 10"^- 10“^ mol L"*. Each micelle contains at
least 100 molecules.
 SURFACE CHEMISTRY 5/31

Mechanism of micelle formation. Let us take the example of soap solution. Soap is sodium salt of
higher fatty acid and may be represented as RCOONa, e.g., sodium stearate, viz., CH3 (CHi)i6COO"Na‘^ or
sodium palmilate, viz., CH3 (CH2)i4COO"Na^. When dissolved in water, it dissociates into RCOO" and Na'*'
ions. The RCOO“ions, however, consist of two parts, i.e., non-polar long hydrocarbon chain R, called the
tail, which is hydrophobic (water repelling) and the polar group COO", called the head, which is hydrophilic,
i.e., water loving (Fig. 5.15).*
FIGURE 5.15

/N /X /X /X /X /X /X /X /X
CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 0"Na*
SODIUM STEARATE (Ci7H35C00“Na*) /'O

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/
I

/ \CH2/ \CH2/ \CH2/ \CH2/ \CH2/ \CH2/ \CH2/XCH2;^v^ /

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I

OJ'

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HYDROPHOBIC TAIL HYDROPHILIC
V, HEAD .J
— STEARATE ION

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2 4 6 10 12 14 16
HYDROPHILIC HEAD

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1 3 5 7 9 11 13 15 17
HYDROPHOBIC TAIL

Hydrophobic tail and hydrophilic head of stearate ion for

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The RCOO" ions, are therefore, present on
s
FIGURE 5.16
the surface with their COO" groups in water and
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STEARATE
the hydrocarbon chains R staying away from it ION ● ►; WATER

and remain at the surface (Fig. 5.16 (n)).

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COO^
However, at higher concentration, these ions do PART +

not remain on the surface but are pulled into the

r

WATER-
bulk of the solution. As a result, at higher
ad
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concentration, the RCOO“ ions form an

aggregate of spherical shape with their o
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hydrocarbon chains (represented by wavy lines)

(a) Stearate ions present on the surface of water
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pointing towards the centre and the COO" part

nd

(represented by a small circle) outwards on the at low concentration of soap solution,

(b) Stearate ions pulled together to form
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surface of the sphere (Fig. 5.16 (/?)). An aggregate a spherical ionic micelle inside the bulk of
thus formed is known as “ionic micelle”. Each
water at high concentration (CMC.)
of these micelles may contain upto 100 ions.
The Cleansing Action of Soaps - An application of ionic micelles formation. Suppose some grease
or oil is sticking on the surface of a cloth as shown in Fig. 5.17 (a). When it comes in contact with soap
solution, the stearate ions arrange themselves around it in such a way that hydrophobic parts of the stearate
ions are in the oil (or grease) and the hydrophilic parts project outside the grease droplet as shown in Fig.
5.17 (b). As hydrophilic partis polar, these polar groups can interact with the water molecules present around
the oil droplet. As a result, the oil droplet is pulled away from the surface of the cloth into water to form ionic
micelle which is then washed away with the excess of water. In fact, the stearate ions of soap molecules help
in making a stable emulsion of oil with water which is washed away with the excess of water. It may be noted
that a sheath of negative charge is formed around the oil globule (as shown in Fig. 5.17 (c)) which prevents
them to come together and form aggregate.
*Molecules like these are called amphipathie. They have both polar and non-polar ends and are big enough
for each end to display its own solubility behaviour.
5/32 “pfiaidee^'^ New Course Chemistry (XlI)CZsMl

FIGURE 5.17

SOAP SOLUTION WATER

GREASE
DROPLET

CLOTH

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(a) Grease or oil on surface of cloth (b) Stearate ions arranged around the oil droplet
(c) Ionic micelle formed surrounded by sheath of negative charge

Similarly, in case of detergents, e.g., sodium lauryl sulphate, viz, CH3(CH2)|[ OSO3 Na'*'. the polar

w
group is OSOj along with the long hydrocarbon chain. It is an example of an anionic detergent as anions
associate together to form an ionic micelle, similar to that of soap. A well known example of a cationic
detergent forming an associated colloid is that of cetyl trimethyl ammonium bromide, CH3(CH2)i5(CH3)3N‘‘'Br“.

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The mechanism of micelle formation in these cases is similar to that of soap.

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TABLE 5.7. Comparison of some important characteristics of muitimolecular,

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macromolectilar and associated colloids

Multimolecular colloids Macromolecular colloids Associated colloids

or
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1. They are formed by the
sf
They are molecules of large size, They are formed by of aggregation
aggregation of a large number of e.g.. polymers like rubber, nylon, of a large number of ions in con
k
atoms or molecules which starch, proteins, etc. centrated solution, e.^.. soap sol.
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generally have diameters less than

I nm, e.g., sols of gold, sulphur, etc.
B

2. Their molecular masses are not They have high molecular masses. Their molecular masses are
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very high. generally high.

3. Their atoms or molecules are held Due to long chain, the van der Higher is the concentration,
u
ad

together by weak van der Waals Waals forces holding them are greater are the van der Waals
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forces. comparatively stronger. forces.

SUPPLEMENT YOUR
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KNOWLEDGE FOR COMPETITIONS FIGURE 5.18

Molar
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1. The CMC value of the surfactants depends both on the size of the conductivity
hydrophobic chain (tail) and size of the hydrophilic domain (head).
The two contributions are counteracting with a lower CMC for a m.’
larger hydrophobic chain and a higher CMC for a larger hydrophilic Surface
domain. For sodium alkyl sulphates, tension

log (CMC) = A - B (No. of C-atoms) D.

O
0-
where A and B are constants. w
u
Osmotic
2. Variation of some physical properties close to critical micelle pressure
concentration (CMC). The CMC is detected by noting a pronounced
a.

discontinuity in physical properties of the solution, particularly the

molar conductivity. The typical variation of some physical properties
of an aqueous solution of sodium dodecylsulphate close to CMC is
shown in the adjoining Fig. 5.18.
Concentration of surfactant
 SURFACE CHEMISTRY 5/33

5.16. PREPARATION OF COLLOIDAL SOLS

As discussed earlier, the lyophilic sols can be prepared directly by mixing the substance with the dispersion
medium. For example, colloidal sols of stiuch, gelatine, gum arabic, soaps etc. are prepared by simply dissolving
these substances in warm water. Similarly, a colloidal sol of cellulose nitrate is obtained by dissolving it in a
mixture of ethyl alcohol and ether. The product (4% solution) obtained is called collodion. However, the
lyophobic sols cannot be prepared directly. Hence, the following two types of methods are employed for
obtaining the lyophobic sols.
FIGURE 5.19
A. Dispersion or Disintegration methods. These SUSPENSION
methods involve the breaking of the bigger particles to
colloidal size. The methods generally employed for this i ●HOLLOW SHAFT
DRIVING BELT
purpose are briefly described below :
(1) Mechanical disintegration. The mechanical
disintegration is carried out in a colloid mill or hall mill or DISCHARGE f DISCHARGE
ultrasonic disintegrator. A ‘colloid mill’ (Fig. 5.19)

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consists of two steel discs with a little gap in-between and
capable of rotating in the opposite directions at high speed STEEL DISCS

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(7000 revolution per minute). A suspension of the substance ROTATING IN
OPPOSITE DIRECTIONS
in water is introduced into the mill. The size of suspension Colloid Mill
particles is reduced to that of colloidal size.

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(2) Electro-disintegration {Bredig's method). This method

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is ciitploycd for obtaining colloidal solutions of metals like gold, FIGURE 5.20

silver, platinumetc. An electric arc is struck between two metallic METAL

electrodes suspended in a trough of water. The intense heat of the ELECTRODES

arc converts the metal into vapours which are condensed for
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ELECTRIC
immediately in the ice cold water bath resulting in the formation ARC
of particles of colloidal size (Fig. 5.20).
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ICE
(3) Peptization.
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A
Peptization is a process of converting afresh
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precipitate into colloidal particles by shaking it with DISPERSION

the dispersion medium in the presence of a small MEDIUM

amount of a suitable electrolyte. The electrolyte added Bredig's method

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is called peptizing agent.

A few examples of sols obtained by peptization are given below :
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(i) A reddish brown coloured colloidal solution is obtained by adding small quantity of ferric chloride
solution to the freshly precipitated ferric hydroxide.
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nd

(//) A precipitate of silver iodide can be peptized by shaking with a dilute solution of silver nitrate or KI.
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(///) On adding insufficient quantity of very dilute HCI solution to the freshly precipitated aluminium
hydroxide, a sol of aluminium hydroxide is obtained.
Cause of peptization. As the electrolyte FIGURE 5.21

is added to a freshly precipitated substance, +

+
+ +
+
+

the particles of the precipitate preferentially +FeCl3 + + + +

Fe{OH)3U Fe(OH)3. ^ + Fe(OH)3 + + Fe(OH)3 + -f

adsorb one particular type of ions of the + + + +

electrolyte (ions common with the precipitated

+ + + +
+ +
FRESHLY
ADSORPTION OF Fe^'^IONS
substance). As a result, they get dispersed due PRECIPITATED
Fe(OH)3 HITTINGAND
to electrostatic repulsions. This gives particles ,, BREAKING
of colloidal size. An example of peptization
of freshly precipitated ferric hydroxide with
ferric chloride solution is shown in Fig. 5.21. + * + +

COLLOIDAL PARTICLES
Freshly prepared precipitates are preferred as OF Fe(OH)3
they are easily peptized. Preparation of colloidal sol by peptization
5/34 7^>uxdecfi^’>^ New Course Chemistry (XII)BZ39]

H. Cimdcnsatlon or Aggregation methods. These methods involve the joining together of a large
number of smaller particles to form particles of colloidal size. The methods generally employed for this
puipose are as follows :
(1) By chemical reactions
(/) By double decomposition : When H^S is passed through a dilute solution of arsenious oxide in water,
a colloidal solution of arsenious sulphide is obtained.
AS2O3 + 3H,S AS2S3 + 3H2O
(Colloidal solution)

(//) By reduction : Gold, silver, platinum, etc. are obtained in colloidal form by reduction of very dilute
solutions of their salts with a suitable reducing agents, e.g.,
2AUCI3 + 3SnCU ^ 2Au + 3SnCl4
Gold sol

Gold sol can also be obtained by reduction of AuClj solution with formaldehyde and tannic acid or with

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hydrazine :
2AUCI3 + 3HCHO + 3H2O ^ 2Au + 3HCOOH + 6HC1

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Formaldehyde Gold sol Formic acid

or
4AUCI3 + 3NH2NH2 ^ 4Au +3N2 + 12HC1

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Hydrazine Gold sol

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(Hi) By oxidation : Sulphur is obtained in the colloidal fonn when H2S is bubbled through the solution
of an oxidising agent like nitric acid, bromine water, etc.
H2S + Bt2 4 S + 2HBr H2S + 2HNO3
;
for
4 2H2O + 2NO2 + S
(/v) By hydrolysis : Fe(OH)3 and Al(OH)3 sols are obtained by boiling solutions of their corresponding
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chlorides, e.g..
s
FCCI3 + 3H2O Fe{OH)3 + 3HCI
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4
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(Colloidal sol)
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(2) By exchange of solvent. Substances like sulphur and phosphorous are fairly soluble in alcohol but
less soluble in water. If their alcoholic solutions are poured in water, colloidal solutions of sulphur and
phosphorus are obtained.
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(3) By condensing vapours of a substance into the solvent. Colloidal solutions of sulphur and mercury
in water are prepared by passing their vapours in cold water containing a little stabilising agent like ammonium
nitrate.
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(4) By excessive cooling. Colloidal solution of ice in an organic solvent like ether is obtained by freezing
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a mixture of the solvent and water.

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SUFPLEIIAENT YOUR
faetWt.gDGE FOR COMPETITIONS
Preparation of colloidal sol of mercury in water. Ultrasonic vibrations, /.e.. vibrations having frequency
more than that of the sound are allowed to hit a beaker containing mercury and water. As a result, mercury
changes into vapours which are dispersed in water to fonn the colloidal sol. This method is called ultrasonic
dispersion method.

5.1/. PURIFICATION OF COLLOIDAL SOLUTIONS

When a colloidal solution is prepared, quite often it contains excessive amounts of electrolytes which
arc crystalloidal in nature and other soluble impurities. Whereas the presence of trace amounts of the electrolytes
is essential for the stability of the colloidal solution, the presence of large amounts results in coagulation of
the colloidal sol. Hence, it is essential to reduce their amount to the minimum required level.

The process of reducing the impurities of the electrolytes to the minimum required level is
known as puriifcation of the colloidal solution.
 SURFACE CHEMISTRY 5/35

The following two methods are used for the purification of colloidal solutions.
1. Dialysis
The process of separating the particles of colloids from those of crystalloids by diffusion of the
mixture through a parchmentor an animal membraneis known as dialysis.
The separation of crystalloids from the colloids is based upon the principle that the particles of the
crystalloids pass through animal membrane (bladder) or parchment paper or cellophane sheet whereas those
of the colloids do not.

The apparatus used is called a dialyser and is shown in Fig. 5.22.

A cellophane sheet or parchment paper is turned into a bag with a funnel tied in the mouth of the bag, as
shown in Fig. 5.22 for the addition of the impure sol. The impure sol is filled into the bag which is then
suspended into a vessel containing distilled water. After some time, all the crystalloids in solution pass out
leaving the colloidal sol behind in the bag. The distilled water is renewed frequently to avoid accumulation of
the crystalloids as otherwise they may start diffusing back into the bag. The above process can be quickened

w
if an electric field is applied around the membrane (the process is then called Electro-dialysis).
Note. The most important application of dialysis is in the purification of blood in the artificial kidney

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machine as shown in Fig. 5.23. The dialysis membrane permits small particles of the excess ions and waste
products to pass through whereas colloid-sized particles such as haemoglobin do not pass through the

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membrane.

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FIGURE 5.22 FIGURE 5.23

. = CRYSTALLAOIDS = COLLOIDS BLOOD CELLOPHANE

ADDITION OF
A IMPURE SOL A for
BLOOD FROM
ARTERY
TO
VEINS
TUBING OR
OTHER DIALYSIS
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MEMBRANE
WATER
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DISTILLED ESSENTIAL
WATER ^ ■ '-IONS".. -WASHING
FUNNEL
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SOLUTION

SOLUTION
OF
●●EXCESSIONS -'-J
CRYSTALLOIDS
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& WASTE PRODUCTS ' -"

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V -

CELLOPHANE
OR Purification of blood by dialysis
PARCHMENT BAG
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Eicctro-dialyser
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2. Ultraflltratioii. Ordinary filter paper is permeable to colloidal dispersions (pores being of colloidal
size) as well as to true solutions, and thus cannot be employed for filtering sols. However, a filter paper
treated with collodion or gelatin solution followed by hardening by dipping in formaldehyde solution serves
the purpose. The collodion solution generally used is 4% solution of nitrocellulose in a mixture of alcohol
and ether. Due to this treatment, the pore size of the filter paper is reduced. The filter paper so obtained is
called ultrafilter and the filtration device using such a filter paper is called ultrafiltration. By using the
impregnating solutions of different concentrations, a series of graded ultrafilters can be obtained. With such
ultrafilters, solute impurities of different sizes can be effectively removed. The sol is poured over the ultrafilter
which permits solution of electrolytes to pass through but retains the colloidalpaiticlcs in the form of slime.
Slime in contact with water disperses spontaneously to form a colloidal system.
Other filter media commonly used are unglazed porcelain and finely sintered glass. Since under ordinary
conditions, filtration proceeds very slowly, pressure or suction has to be applied to speed up the process.
3. Ultra-centrifugation. In this method, the impure sol is taken in a tube which is placed in an
ultracentrifuge. In this machine, the tube is rotated at a very high speed. As a result, the colloidal particles
5/36 New Course Chemistry (XII)CZslSl

settle down at the bottom of the tube whereas the crystalloids and other soluble impurities remain in the
solution. This solution is decanted off and the colloid particles are remixed with the dispersion medium to
give the pure colloidal sol.

Curiosity Question
f Q. Why some patients have to undergo dialysis ? How does it help ?
Ans. Patients suffering from kidney damage due to high glucose in their blood (diabetic patients) have
to undergo dialysis. Kidney filters the blood and purifies it. When it stops functioning, blood has
got to be purified in an artificial kidney machine called dialyser. It takes blood from artery and after
purification, i.e., separation of excess ions and waste products sends back to veins.
J
5.18. PROPERTIES OF COLLOIDAL SOLLITIONS

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5.18.1. Physical Properties
(a) Heterogeneous character. Colloidal sols form heterogeneous mixture consisting of particles of

F lo
dispersed phase and the dispersion medium. The phenomena of Tyndall effect, electrophoresis and electro
osmosis (discussed later in this section) confirm heterogeneity of colloidal systems.
(b) Stability. Colloidal sols are quite stable. Only a few colloidal particles of companilively larger size

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may settle but very slowly,

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(c) Filtrability. Ordinary filter paper cannot be used for removing the dispersed phase because size of
pores of filter paper is bigger than the size of colloidal particles which can easily pass through the pores of the
for
ordinary filter paper. Animal membrane or parchment paper does not allow the colloidal particles to pass
ur
through it. This forms the basis of separating the particles of the colloid from those of the crystalloids in the
process called dialysis, discussed earlier.
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(d) Visibility and colour. The particles in colloidal solution are not visible to naked eye or under
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ordinary microscope. However, as discussed later, they scatter the light falling on them. The colour of the
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colloidal solution depends upon the wavelength of the light scattered by the colloidal particles which in turn
depends upon the size and nature of the particles. For example, in a gold sol, if the particles are very fine, the
colour is red, but with growing size of the particles, the colour changes to purple, tlien blue and finally golden.
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ad

The colour also depends upon the manner in which the light is observed. For example, for the light focussed
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on a mixture of milk and water, if reflected light is viewed, it is blue but if transmitted light is viewed, it is red.
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5.18.2. Colllgative Properties-Osmotic pressure

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There are four colligative propenies-osmotic pressure, elevation in boiling point, depression in freezing
nd

point and relative lowering of vapour pressure. These properties depend upon the number of moles and hence
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on the number of particles present. Colloidal particles have very high average molecular masses, and hence
the number of moles present in solution will be extremely small. Thus, the value for any of the colligative
properties for a particular substance will be smaller as compared to its value when it is a part of true solution.
However, some colloids have measurable osmotic pressures which have been determined with a reasonable
degree of accuracy. Hence, measurement of osmotic pressure has been used to determine the average molecular
masses of colloidal particles.
5.18.3. Mechanical Properties-Brownlan movement
When viewed through an ullramicroscope, colloidal particles
are seen continuously moving in a zig-zag way (Fig. 5.24a). Robert
Brown, in 1827, observedsuch a movementof pollen grains suspended
in water and hence it is called Brownian movement. Thus,
o
● RESULTANT
Brownian movement may be defined as continuous FORCE

zig-Zfig movement of the colloidal particles in a colloidal soL Brownian movement

 SURFACE CHEMISTRY 5/37

Brownian movement does not depend upon the nature of the colloid but depends upon the size of the
colloidal particles and viscosity of the sol. Smaller the size and lesser the viscosity, faster is the movement of
the particles.
Cause of Brownian movement. The reason for Brownian movement is based on the fact that the
molecules of dispersion medium due to their kinetic motion strike against the colloidal particles (dispersed
phase) from all sides with different forces. The resultant force causes them to move (Fig. 5.24/;). However,
colloidal particles being comparatively heavier, move with a slower speed.
Importance of Brownian movement.
(/) Brownian movement opposes the force of gravity and does not allow the colloidal particles to settle
down. Thus, it is responsible for the stability of the colloidal solution. (The stability is also explained on the
basis of electrical charge, as will be discussed later).
(//) It has also helped in the determination of Avogadro’s number.
5.18.4. Optical Properties-Tyndall-effect
If a strong converging beam of light is passed through a colloidal solution placed in a dark room, the

w
path of beam gets illuminated with a bluish light when viewed at right angles to the direction of the passage
of light. The path of the light becomes visible due to scattering of light by the colloidal particles. The

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phenomenon was observed by Tyndall in 1869 and is called Tyndall effect. Thus,
Tyndall effect may be deifned as the scattering of light by the colloidal particles present in a
colloidal sol.

ee
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The illuminated path of beam is called l^ndall cone (Fig. 5.25). The phenomenon is also observed
when a beam of light is projected in a cinema hall tmd it becomes visible due to the scattering by colloidal
dust particles in the air of the room. for
ur
A simple experiment as shown in Fig. 5.26 may be performed in the laboratory (in a dark room) to see
the scattering of light by a colloidal solution and distinguish between a true solution (free from any dust
s
particles) and a colloidal solution. The path of light is visible in the beaker containing colloidal sol (due to
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scattering) but no path is visible in the beaker containing true solution (as there is no scattering).
It may be noted that Tyndall effect is observed only when the followingtwo conditionsare satisfied :
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(/) The diameter of the dispersed particles is not much smaller as compared to the wavelength of the light
used, and
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ad

(//) There is a large difference in the refractive indices of the dispersed phase and the dispersion medium.
Importance of Tyndall effect. Based on Tindall effect, Zsigmondy, in 1903, devised an instrument
called ultramicroscope. The set-up is same as shown in Fig. 5.25. An intense beam of light (e.g., from an arc
Y

lamp) is focussed on the colloidal solution taken in a glass vessel and then observed in the microscope at right
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angles to the beam. The light scattered by each individual colloidal particle appears as a bright star against a
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dark background. Thus, number of colloidal particles can be counted. Knowing the volume of the solution,
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number of particles per unit volume can be determined. This further helps to find out the average mass of the
FIGURE 5.26

FIGURE 5.25
MICROSCOPE
SLIT

TORCH
0
SCREEN
TRUE SOLUTION COLLOIDAL SOL
(NO SCATTERING (SCATTERING
OF LIGHT) OF LIGHT)
TYNDALL CONTAINING
LAMP
CONE SOL
Demonstration of Tyndall effect and
distinction between a true solution and
Tyndall effect a colloidal solution
5/38 New Course Chemistry (X1I)E!S19I

particles. Note that in the ultramicroscope, we do not see tiie actual particles but only the light scattered by
them. Thus, ultramicroscope gives no information about the size and shape of the colloidal particles.
The importance of Tyndall effect also lies in the fact that it helps to confirm the heterogeneous nature of
the colloidal solutions.

5.18.5. Electrical Properties

(fl) Stability of colloidal sols - Electrical charge on colloidal particles. The stability of a colloidal
solution is due to the fact that the colloidal particles in the sol are electrically charged. The particles, therefore,
repel one another and do not coalesce (come close together) to form large non-colloidal particles. Colloidal
particles carry either positive or negative charge. All the dispersed particles in a colloidal solution carry the
same charge while the dispersion medium has an equal and opposite charge. For example, arsenious sulphide
particles are negatively charged whereas the dispersion medium (water) is positively charged. Ferric hydroxide
particles arc positively charged whereas the dispersion medium (water) is negatively charged.
Using water as the dispersionmedium, the charge on the particles of some common sols is as given in
Table 5.8.

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TABLE 5.8. Charge on particles of some common sols

F lo
Negatively Charged Positively Charged

(/) Metallic particles {e.g., Cu, Ag, Au .sols.). (i) Metal hydroxides like Fe(OH)3, AUOHij.
Cr(OH)3, Ca(OH)2

ee
(//) Meta! sulphides like As-^S^, CdS, etc. (ii) Hydrated metallic oxides, e.g., AUO3. x H2O,

Fr
Fc203. -y H-,0, Cr^O^. X H^O etc.
(Hi) Starch, gum, gelatin, clay, charcoal, silicic acid sols, (Hi) Oxides like TiOi, etc.
(tv) Acidic dyes like congo red, eosin sols for
(/v) Haemoglobin
ur
(v) Basic dyes like methylene blue, prussian blue etc.
s
Origin of electrical charge on colloidal particles. The various theories for the origin of electrical
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charge on the colloidal particles are as follows:

eB

(/) Frictional electrification caused by the mutual rubbing of the colloidal particles with molecules of the
dispersion medium due to movement of the colloidal particles in the dispersion medium.
(//) Electron capture by particles from air and during electro-dispersion in Bredig’s arc method.
our
ad

(Hi) Preferential adsorption of ions from solutions. An ionic colloid adsorbs ions common to its own lattice
during the preparation of the colloidal sol. For example, if colloidal sol of Agl is prepared by adding KI
solution to AgN03 solution till Kl is in slight excess, iodide ions (U) will be adsorbed on the surface of
Y

Agl particles thereby giving them a negative charge :

Re

FIGURE 5.27
nd

Agl -H 1- ^ Agl : I
Fi

(from Kl) Negative sol.

On the other hand, if the colloidal sol of Agl is prepared

by adding AgN03 solution to KI solution till AgN03 is in
slight excess. Ag"*" ions will be adsorbed thereby giving
positive charge to the colloidal particles :
^ Agl:Ag+ Preferential adsorption of common
Agl -1- Ag+ - ions from the solution
(from AgNO^) Positive .sol

In either case, the left out ions will remain in the dispersion medium (e.g., K'*' ions in the first case and
NO3 ions in the second case) thereby giving equal and opposite charge to the dispersion medium.
Similarly, if ferric chloride is added to excess of hot water, a positively charged sol of hydrated ferric
oxide is formed due to adsorption of Fe-^"^ ions. However, if ferric chloride is added to sodium hydroxide
solution, a negatively charged sol is obtained due to adsorption of OH" ions.
SURFACE CHEMISTRY 5/39

FeClj + H-,0 (excess) ^ Feo03. .x: H^O i Fe-^-"

Positively charged sol

FeCl3 + NaOH ^ FC2O3 . .x: H2O i OH-

Negatively charged sol
(iv) Dissociation of molecules followed by aggregation of ions. For example, in case of soap, the RCOO“
groups gel dissociated from Na'*’ ions and have a tendency to aggregate into a cluster carrying negative
charge, as already explained. The particles of a dye have a tendency to dissociate to form aggregates
carrying positive or negative charge depending upon its composition,
(v) Dissociation of the molecular electrolytes FIGURE 5.28
adsorbed on the surface of particles. For
example, H2S molecules get adsorbed on Ag I r ;K’-
i Ag I I Ag’’ NO3
colloidal particles of sulphide (e.g., As-,S3) Ag 1 i-|k» ' Ag I Ag'^ NO3
during precipitation. By dissociation of H^S,

w
Ag L 1- ! K’- Ag 1 ' Ag'^ 1 NO3
ions are lost thereby giving a negative
charge to colloidal particles. Feme hydroxide / \ / \

F lo
FIXED MOBILE/ FIXED MOBILE/
sol particles are positive due to self LAYER DIFFUSED LAYER LAYER DIFFUSED LAYER

dissociation ; OH" ions are lost to the solution (a) When Kl is in excess (b) When AgNO^ is in excess
giving positive charge to particles.

ee
Formation of electrical double layer
Electrokinetic or zeta potential. Out of the

Fr
various theories of the origin of charge discussed
FIGURE 5.29
above, preferential adsorption of ions is considered
as the best explanation. When one type of ions of for
ur
the electrolyte are adsorbed on the surface of the
colloidal particles, it forms a ‘fixed layer’. It
s
attracts the counter ions from the medium forming
ook
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a second layer which is mobile and is called

‘diffused layer’ as shown in Fig. 5.28.
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Alternatively, these may be represented as

shown in Fig. 5.29.
our

The double layer of opposite charges thus

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formedis called Helmholtzelectricaldouble layer. Alternate representation of electrical double layer

As a result, a difference of potential exists between
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the fixed layer and the diffused layer. This potential FIGURE 5.3d1
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difference is known as electrokinetic potential or

nd

+ COLLOIDAL SOL _
zeta (^) potential. 1 - >
Fi

(b) Electrophoresis or cataphoresis. The

existence of the electrical charge (positive or negative)
can be shown by the process of ‘Electrophoresis’ also
called ‘Cataphoresis’ which involves the movement of
colloidal particles towards one or the other electrode
4
when placed under the influence of an electric field y V
«
y
(Fig. 5.30).
The movement of colloidal particles under
the influence of an electric field is called
electrophoresis or cataphoresis.

As soon as the colloidal particles reach the & ©V.

oppositely charged electrode, they get neutralised and
hence coagulated. (a) Before electrophoresis
Electrophoresis can be used to find out the nature (b) After Electrophoresis (For negatively
of the charge that the colloidal particles carry. charged colloidal particles)
5/40 ^>n<xeCee^k.'^ New Course Chemistry rxil^rosTWi

Earlier ihe term used was ‘cataphoresis’ because most of the colloidal sols studied at that time contained
positively charged colloidal particles and thus migrated towards cathode. However, now the term electro
phoresis is preferred because colloidal pju-ticles may migrate towards either of the electrode, i.e., anode or
cathode depending upon the charge on the colloidal particles,
(c) Electro-osmosis. On placing FIGURE 5.31
a colloidal solution under the influence
of an electric field, the particles of the WATER \ COLLOIDAL WATER
y SOL
dispersion medium (which are also
F
electrically charged) move towards V :>.N-
T
oppositely charged electrode, provided
the colloidal particles are not allowed
to move as shown in Fig. 5.31. This
SEMIPERMEABLE MEMBRANE
phenomenon is called electro-osmosis.
Electro-osmosis

w
Thus,

F lo
Electro nusis wtay be defined as a phenomenon in which the molecules of the dispersion
medium are allowed to move under the influence of an electricfield whereas colloidalparticles
are not allowed to move.

e
Fre
This phenomenon was observed by Reuss (1809) and Porret (1816).
{d) Coagulation or Flocculation or Precipitation :
for
Coagulation or precipitation is a process of aggregating together the colloidal particles so as to
change them into large sized particles which ultimately settle as a precipitate.
r
You
oks

Coagulation is generally brought about by the addition of electrolytes*. When an electrolyte is added
to a colloidal solution, the particles of the sol take up the ions which are oppositely charged and thus get
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neutralised. The ion responsible for neutralization of charge on the colloidal particles is called the coagulating
ion or flocculating ion. The neutral particles then start accumulating to form particles of larger size which
settle down.
our
ad

When a small amount of the electrolyte is added, i.e., when the concentration of the electrolyte added
is low, the process is called flocculation. It can be reversed on shaking. However, at higher concentration,
coagulation takes place and the process cannot be reversed simply by shaking.
dY
Re

The minimum amount of an electrolyte (millimoles) that must be added to one litre of a colloidal
Fin

solution so as to bring about complete coagulation is called the coagulation or flocculation or

precipitation "uluc of the electrolyte. Thus, smaller is the coagulation value of an electrolyte,
greater is its coagulating or precipitating power.

Hardy Schulze Law. The quantity of the electrolyte which is required to coagulate a definite amount of
a colloidal solution depends upon the valency of the coagulating ion (ion having a charge opposite to that of
the colloidal particles). This observation of Hardy and Schulze is known as Hardy Schulze Law, the main
points of which may be stated as follows :

(i) The effective ions of the electrolyte in bringing about coagulationare those which carry
charge opposite to that of the colloidal particles. These ions are called coagulating ions or
flocculating ions
(ii) Greater is the valency of the coagulating or the flocculating ion, greater is its power to bring
about coagulation.

*Note that the presence of a very small amount of a suitable electrolyte is es,sential for the stability of the sol.
SURFACE CHEMISTRY 5/41

Thus, for coagulation of negatively charged arsenious sulphide sol., trivalent cations (Al^'*’) are far more
effective than divalent (Ba^'*’) cations which in turn are more effective than monovalent (Na'*') cations. Similarly,
for coagulation of positively charged ferric hydroxide sol, tetravalent [Fe(CN)g]'*“ anions are more effective
than trivalent anions (PO^“) which are more effective than divalent (SO^“) anions which in turn are more
effective than monovalent (Cl“) anions.
The coagulation or precipitation values of a few electrolytes are given below :
For coagulation of a negatively charged sol (AS2S3)
Electrolyte NaCl KCl HCl MgCl2 BaCl2 AICI3
Cation Na+ K+ Mg
2+
Ba2+ AP+
Coagulation value 52 49-5 30-8 0-72 0-69 0093

w
For coagulation of a positively charged sol, Fe(OH)3
Electrolyte KBr HCl KNO3 K2SO4 K2C2O4 K3[Fe(CN)6]

Flo
Anion Br- ci- NOJ S02- C2O2- [Fe(CN)6l
3-

e
Coagulation value 138 132 132 0-210 0-238 0-096

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As coagulating power is inversely proportional to coagulation/flocculation value, to compare the relative

F
coagulating powers of two electrolytes for the same colloidal sol, we have
Coagulating power of electrolyte 1 Coagulation value of electrolyte 2
ur
r
Coagulating power of electrolyte 2 Coagulation value of electrolyte 1
For example, for coagulation of negative AS2S3 sol, fo
ks
Coagulating power of AICI3 Coagulation value of NaCl _ 52
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= 559.
oo

Coagulating power of NaCl Coagulation value of AICI3 0-093

B

Thus, AICI3 has 559 times more coagulating power than NaCl.
re

For a negatively charged sol, like that of AS2S3 the coagulating powers of the cations are in the order :
Al3+>Ba2+>Na+.
u
ad

For a positively charged sol like that of Fe(OH)3, the coagulating powers of the anions are in the order :
Yo

[Fe(CN)j]‘^ > PO^ >S02- >C|-

d

Coagulation can also be caused by the following methods :

Re
in

(0 By electrophoresis. In electrophoresis, the particles of dispersed phase move towards oppositely charged
electrode and get neutralised. If the process is continued for sufficient time, these neutral particles
F

unite and grow in size and settle down,

(ii) By mutual precipitation. Mutual precipitation is a process in which oppositely charged sols are mixed
in proper proportions to neutralise the charges of each other causing coagulation of both the sols. For
example, if positively charged ferric hydroxide and negatively charged arsenious sulphide sols are
mixed, both the sols get coagulated,
(iii) By prolonged dialysis. The stability of a colloidal sol is due to presence of a small amount of the
electrolyte. On prolonged dialysis, the electrolyte is completely removed. As a result, the colloidal sol
becomes unstable and gets coagulated,
(iv) By heating (boiling) or cooling. In some cases, heating the sol results into coagulation, e.g., coagulation
of butter. When a sol is boiled, the adsorbed layer is disturbed because the number of collisions on
them by the molecules of the dispersion medium increases. Consequently, the charge on the particles
decreases. Hence, the stability decreases leading to their settling down as a precipitate.
Similarly, in some cases, cooling the sol results in coagulation, e.g., coagulation of milk, i.e., on cooling
milk, fats start floating on the surface.
5/42 'a New Course Chemistry (XII)

Comparison of stability or coagulation of lyophobk and lyophilic colloids. Lyophobic sols are less
stable and hence more easily coagulated than lyophilic colloids. This is because the stability of lyophilic sols
is due to two factors :

(/) Same charge on all the colloidal particles.

(//) Solvation of the colloidal particles.
Thus, to bring about coagulation, both these factors have to removed. This is done
(/) by adding electrolyte.
(//) by adding suitable solvent.
When a solvent like alcohol or acetone is added to a hydrophilic sol, the dehydration of the colloidal
particles occurs. Now, only a small amount of the electrolyte can bring about coagulation.
The stability of lyophobic sol is only due to charge. This factor can be removed by adding only electrolyte.

w
Hence, they can be easily coagulated.

PROBLEMS

Flo
BASED
IVIETHOD USED
ON I

ee
Calculation of It involves the calculation of the amount of electrolyte in millimoles

Fr
required to bring about complete coagulation of one litre of the colloidal
Coagulation/ solution.
Flocculation

for
ur
Value
k s
Problem
For the coagulation of 100 ml of arscnious sulphide sol, 5 ml of 1 M NaCl is required.
Yo
oo

What is the flocculation value of NaCI ?

eB

1
Solution. 5 ml of 1 M NaCl contains NaCl = x5 moles = 5 millimoles
1000

Thus, 100 ml of AS2S3 sol require NaCl for complete coagulation = 5 millimoles
r
ou
ad

1 L, i.e., 1000 ml of the sol require NaCl for completecoagulation= 50 millimoles

By definition, flocculation value of NaCl = 50
Y

SUPPLEMENT YOUR
Re
nd

KNOWLEDGE FOR COMPETITIONS

Fi

1. Positive charge on Sn02 solution in acidic and negative charge in basic medium. Sn02 is amphoteric
in nature. It reacts with acid, e.g., with HCl to form SnCl4 in the solution, the common Sn'*'*’ ions are
adsorbed on the surface of Sn02 particles thereby giving them a positive charge

Sn02 + 4HCI > SnCl4 + 2 H2O

SnO^ + Sn^+ ●>
[Sn02J : Sn'*+
Positively charged colloidal particles
Sn02 reacts with a base, e.g., NaOH to form sodium stannate (Na2Sn03) in the solution. The slannate
ions are adsorbed on the surface of SnO^ particles giving them a negative charge.

Sn02 + 2 NaOH > Na2Sn03 + H2O

Sodium stannate

Sn02 + SnO| [Sn02l: SnO^”

Negatively charged colloidal particles

1
 SURFACE CHEMISTRY 5/43

2. Isoelectric point of a colloid. Lyophobic sols have a specific charge due to adsorption of either positive
or negative ions. Some lyophilic sols also have a specific charge. However in case of certain lyophilic sols,
particularly proteins, the sign of charge depends upon the pH of the solution. Above a certain pH
(characteristic of each sol), the panicles are negatively charged while below this pH, they have a positive
charge. At that particular pH, the particles are uncharged and, therefore, do not migrate under the influence
of ‘electric field’. This pH is called the isoelectric point of the colloid. For example, for gelatin it is at
pH of 4-7. Usually it is not at a definite pH but over a pH range, e.^.. for casein from human milk, it is
4-1 - 4-7, and for haemoglobin, it is 4-3-5-3.

Curiosity Questions
f

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Q. 1. Why do we see a beam coming from projector to screen in a cinema hail ?
Ans. This is due to scattering of light coming from the projector by the particles of the dust present in

F lo
the air and falling In the path of the beam of fight.
Q. 2. Why the sun looks red at the time of setting ? Explain on the basis of colloidal properties.

ee
Ans. At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer

Fr
distance through the atmosphere. As a result, blue part of the light is scattered away by the dust
particles in the atmosphere.Hence, the red part is visible. I
for
ur
5.19. PROTECTIVE ACTION OF LYOPHILIC COLLOIDS AND GOLD NUMBER

It has already been explained that lyophobic sols like tho.se of metals
s
FIGURE 5.32
ook
Yo
(Au, Ag, etc.) are unstable and are easily precipitated by addition of LYOPHILIC PARTICLES

electrolytes. However, it is oKserved that the addition of certain lyophilic (PROTECTING PARTICLES)
eB

colloids like gums, soaps, gelatine, etc. to lyophobic colloids (like a metal
sol) render lyophobic colloids difficult to coagulate by the addition of
our

electrolytes*. The process is known as 'protection' and the lyophilic

ad

colloids are tenned as Protective colloids. It is believed that the protective

action of the lyophilic colloids is due to the covering up of the particles

Y

of the lyophobic colloid by those of the lyophilic colloid (Fig. 5.32). LYOPHOBIC PARTICLES
Re

(PARTICLES BEING PROTECTED)

nd

However, this explanation does not seem to be fully correct because

Protection of colloids
the particles of the protecting substance have almost the same size as
Fi

those of the substance being protected. Thus, the exact mechanism of

protection is not cleai'.
To compare the protective action of different lyophilic colloids, Zsigmondy (in 1901) introduced a term
called Gold number. It is defined as follows :

Gold number of a protective colloid is the minimum mass of it in milligrams which must be
added to 10 mL of a standard red gold sol (containing 0.0053 to 0.0058 per cent gold) so that no
coagulation of the gold sol (i.e., the change of colour from red to blue) takes place when
1 mL of 10% sodium chloride solution is rapidly added to it.

Evidently, smaller the gold number of a protective colloid, the greater is it.s protective action. The gold
numbers of a few protective colloids are listed in Table 5.9.

*Moreover, now the sol of the lyophobic colloid is no longer irreversible but becomes reversible, i.e.. can be
evaporated to dryne.ss and remade by simply adding the dispersion medium.

I
5/44 New Course Chemistry (XI1)CZ£ZM]

TABLE 5.9, Gold numbers of a few protective colloids

Sol Gold Numbers Reciprocal

Gelatine 0005 — 0-01 200— 100

Casein 0-01 — 0-02 100 — 50

Haemoglobin 0-03 — 0-07 33—14

Albumen 01 — 0-2 10 — 5

Gum arabic 0-15 —0-25 7 — 4

Potato starch 20 — 25 0-5 — 0-4

The reciprocals of gold numbers are included in the above table as they give directly a comparison of
the protective action.

w
PROBUEMS

F lo
BASED
ON
It involves the calculation of the weight of the protective colloid in
Calculation of milligrams to be added to 10 mL of gold sol to prevent its coagulation

ee
on addition of 1 mL of 10% NaCl solution.
Gold Number

Fr
for
The coagulation of 100 mL of a colloidal sol of gold is completely prevented by addition
r
of 0*25 g of starch to it before adding 10 mL of 10% NaCl solution. Find out the gold number of starch.
You
s

Solution. Starch added to 100 mL of gold sol to completely prevent coagulation by 10 mL of 10% NaCl
ook

solution = 0-25 g = 250 mg. This is 10 times the amount required for coagulation of 10 mL of gold sol by 1 mL of
eB

10% NaCl solution.

Starch required to be added to 10 mL of gold sol to completely prevent coagulation by 1 mL of 10% NaCl sol
= 25 mg
our
ad

By definition, gold number of starch = 25.

Note that to test 100 mL of gold sol, 10 mL of 10% NaCl solution is required as for 10 mL of gold sol, 1 mL
of 10% NaCl solution is required.
dY
Re
Fin

50 mL of -Standard gold sol needs 0 05 mg of gelatine for its protection from coagulation. Calculate gold
number of gelatine.
ANSWER

001

WOUR
IMPETITIONS

Congo Rubin Number. Instead of gold number, the protective action of a lyophilic colloid is sometimes
expressed in terms of Congo rubin number, as suggested by Osiwald. It is defined as the minimum amount
of the protective colloid in milligrams that prevents the colour change of 100 ml of 0-01% congo rubin dye
to which 0-16 g equivalent of KCl is added.

i
 SURFACE CHEMISTRY 5/45

5.20. EMULSIONS

An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion
medium are liquids. (The two liquids involved are otherwise immiscible).
Method of preparation. An emulsion is prepared by shaking stronglythe mixtureof the two liquids or
by passing the mixture through a colloid mill, called the homogenizer. The emulsions thus prepared from the
pure liquids are usually not stable and the two liquids separate out on standing. To get a stable emulsion,
small quantities of certain other substances are added during preparation. The substances thus added to
.stabilize the emulsions are called emulsifiers or emulsifying agents. The substances commonly used as
emulsifying agents are soaps of various kinds, long chain sulphonic acids or lyophilic colloids (proteins,
gum, agar etc.).
Role of emulsifier. The role of emulsifier can be explained taking the example of soap as an emulsifier.
Soaps are sodium or potassium salts of higher fatty acids, e.g., sodium palmitate (Cj5H3jCOONa). sodium

w
stearate (C|7H35COONa), etc. A molecule of soap consists of two parts, the hydrocarbon pan (Ci^H 31 " ●
C17H35 - , etc.) which is soluble in oil and the polar group (COO'Na'*') which is soluble in water

F lo
R COO-Na-"

Hydrophobic part Hydrophilic part

(Soluble in oil) (Soluble in water) FIGURE 5.33

ee
Thus, if a drop of oil is surrounded by soap solution, CIRCLE (0)

Fr
WATER
REPRESENTS
the R - part of the soap remains in the oil and the COO“Na'*‘ POLAR GROUP (COO“Na*)
part remains in water as shown in Fig. 5.33. Thus, soap
molecules are concentrated over the surface of the drop of for OIL: WAVY LINE (uvw<)
REPRESENTS
ur
oil. As a result, the interfacial tension between oil and water NON-POLAR

decreases and hence they are intermixed into each other to GROUP (R-)
s
ook

form the emulsion. Thus, micelles are formed (compare with

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Fig. 5.16, page 5/31). Role of soap as an emulsiner

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Some other commonly used stabilizing agents are proteins, gums and agar.
Types of Emulsions. If the tenn “oil” is used for any liquid FIGURE 5.34]
our

which is immiscible with water and is capable of forming an

ad

OIL DROPLETS WATER DROPLETS

emulsion with water, then the various types of emulsions may be
classified into two types (Fig. 5.34). These are :
dY

(/) Emulsion of oil-in-water (o/w) in which oil is the dispersed

Re

phase and water is the dispersion medium.

Fin

For example, milk is an emulsion of liquid fat dispersed in

water. Another well known example is that of vanishing \
cream.
WATER© O OIL
Types of emulsion
(//) Emulsion of water-in-oil (w/o) in which water is the
(a) oil in water (b) water in oil
dispersed phase and oil is the dispersion medium.
For example, cod liver oil is an emulsion of water in oil, i.e., water is the dispersed phase and oil is the
dispersion medium. Two other common examples of this type are butter and cold cream.
Evidently, the type of em Ision depends upon the relative amounts of the two liquids. If water is in
excess, it is oil-in-water emuls i. If oil is in excess, it is water-in-oil emulsion. The type of emulsion formed
also depends upon the nature of the emulsifying agent. For example, the presence of soluble soaps a,s the
emulsifying agents generally favours the formation of emulsions of oil-in-water whereas the presence of
insoluble soaps (containing non-alkali metal atoms) favours the formation of emulsions of waler-in-oil.
Some very commonly used emulsifying agents for o/w emulsions are : proteins, gums, natural and
synthetic soaps whereas for w/o emulsions are : heavy metal salts of fatty acids, long chain alcohols, lamp
black, etc.
5/46 New Course Chemistry (XIl)CZ£l9]

To test the type of emulsion. The type to which the given emulsion belongs can be tested experimentally
by any one of the following methods:
(/) Microscopic method. To a small amount of the given emulsion, a few drops of water are added. The
water will mix completely if the emulsion is oil-in-water type. Likewise, for the water-in-oil emulsion,
the miscibility with a few drops of oil can be tested. The miscibility can be seen under a microscope.
(//) Conductance method. This method is based upon adding a small amount of an electrolyte to the emulsion.
If the conductance increases, the emulsion is oil-in-walertype and if there is no significant change, it is
water-in-oil type.
(///) Dye method. A small amount of an oil-soluble dye is added to the emulsion. If it is water-in-oil type, it
becomes deeply coloured, otherwise it remains colourless.
Properties of emulsions. (0 Emulsions exhibit all the properties like Tyndall effect, Brownian movement.
Electrophoresis, Coagulation on addition of electrolytes (containing multivalent positive ions, as the dispersed
particles, i.e., globules are usually negatively charged) as shown by colloidal sols.

w
(ii) Emulsions can be separated into their constituent liquids by boiling, freezing, centrifuging, electrostatic
precipitation by adding large amounts of the electrolytes to precipitate out the dispersed phase or by

F lo
chemical destruction of the emulsifying agent. The separation of cream from milk is a well known
example of centrifuging.

ee
The process of separation of the constituent liquids of an emulsion is called demulsiflcation.

Fr
{Hi) Emulsions can be diluted by adding any amount of the dispersion medium, i.e., water for o/w emulsions
and oils for w/o emulsions. However, if dispersed phase is added into the emulsion, it forms a separate
layer. for
ur
Applications of emulsions. Emulsions are useful in a number of ways. A few of the applications are
given below :
s
ook

(/) In the metallurgical processes, the concentration of ore by froth floatation process is based upon the
Yo

treatment of the powdered ore with oil emulsion. The valuable particles of the ore form foam which
eB

comes to the surface and is skimmed off.

{ii) Asphalt emulsified in water is used for building roads without the necessity of melting the asphalt.
(///) Milk which is an important constituent of our diet is an emulsion of liquid fats in water.
our
ad

(/V) Several oily drugs are prepared in the form of emulsions,

(v) Certain disinfectantssuch as detlol and lysol give emulsions of oil-in-water type on mixing with water.
Y

(vi) The cleansing action of ordinary soap for washing clothes, crockery, etc. is based upon the formation of
oil-in-water emulsion.
Re
nd

{vii) The digestion of fats in the intestines takes place by the process of emulsification. A small amount of
the fat reacts with the alkaline solution present in the intestines to form a sodium soap. This soap causes
Fi

the emulsificationof the rest of the fat thereby making the function of the digestive enzymes easier in
carrying out the metabolic processes.
Harmful effects of emulsions. The formation of emulsions is sometimes harmful. For example, petroleum
forms emulsion with water. Thus, in the petroleum wells, we get the emulsion sometimes instead of petroleum
alone. Likewise, in the areas where small amounts of petroleum are present, the well water becomes unfit for
use.

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

Gels. There are some sols that have a high concentration of dispersed solid and change spontaneously into
semi solid form (jelly-like) on cooling. These are known as gels and the process is known as gelation. For
example, gelatine dissolves in warm water forming a colloidal solution which when cooled sets to a jelly.
Other examples include gum arable, agar, silicic acid, ferric hydroxide, cheese, etc.
The colloidal system constitutingthe liquid as the dispersed phase and the solid as the dispersion ,,
medium is known as gel.

k
 SURFACE CHEMISTRY 5/47

Gels are formed by the interlocking of the particles of solid dispersion medium in the form of a loose
frame work inside which liquid (dispersed phase) is contained.
When the gel is allowed to stand for a long time, it shrinks and loses the entire liquid held by it. This
shrinking of gel is termed as syneresis or weeping.
Classification of gels. Gels are classified into elastic and non-elastic gels. The main points of difference
between the two are as follows :

Elastic Gels Non-Elastic Gels

(0 These are those gels which possess the property (0 These are those gels which do not possess the
of elasticity,, i.e., they change to solid mass on property of elasticity, i.e., they change into
dehydration which can again be converted into solid mass on dehydration which becomes rigid
gel by addition of water followed by heating and and cannot be converted into the original form
cooling. by heating with water.
(ii) When placed in contact with water,, they absorb («) They do not show the phenomenon of
water and swell. This property is called imbibition,

w
imbibition.

(Hi) Examples include gelatine, agar agar, starch, etc. (Hi) Mo.st common example is silicic acid.

F lo
Thixotropy. Some gels like that of gelatine on mechanical shaking lose their semi-solid gel-like character
and change into a liquid sol. This sol on standing again changes into gel. This reversible phenomenon
involving gel-sol transformation is known as thixotropy.

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Curiosity Question for
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Q. Vanishing cream and cold cream both are emulsions. Then what Is the difference between
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the two ?
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Ans. Vanishing cream is an emulsion of oii-in-water whereas cold cream is an emulsion of water-in-oii.
The latter is used for dry skin as the base (dispersion medium) in this cream is oil.
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5.21. APPLICATIONS OF COLLOIDS

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Colloids play a very significant role in nature and in our daily life. .Some of the important applications
of colloids are discussed below :

A. Applications of colloids in everyday life and explanation of certain phenomena :

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(/) Medicines. Most of the medicines used are in the colloidal form because in this form, they are more
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easily assimilated due to large surface area and hence are more effective. For example,
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(a) Argyrol is a silver sol used as an eye lotion, (b) Colloidal antimony is used for curing kalazar.
(c) Colloidal gold is used as an intramuscular injection, (d) Milk of magnesia, which is an emulsion, is
used as an antacid for reducing acidity in the stomach.
(ii) Food articles. A number of food articles that we eat arc colloidal in nature, e.g.. milk, butter, ice
creams, fruit juices, halwa, etc.
(Hi) Fog, mist and cloud. In winters, at night, the moisture of the air condenses on the surface of dust
particles forming fine droplets. These droplets are colloidal in size and hence continue to float in the air
in the form of fog or mist.*
Clouds are aerosols consisting of small droplets of water suspended in the air. In die upper almosplicre
where the temperature is low, they condense together to form bigger drops which come down in the
form of rain. Rain is also caused when two oppositely charged clouds meet each other.
*If smoke is present in the atmosphere (due to burning of coal, petroleum etc. or husk by the farmers), the
fog condenses on the carbon particles. This combination of smoke and fog is called smug. It is highly injurious to
health and leads to asthma, lung or throat cancer etc.
5/48 'Pn^deef.i'a. New Course Chemistry (X11)EZS19]

(/v) Artificial rain. Artificial rain can be caused by spraying electrified or oppositely charged colloidal
dust or sand particles over a cloud. The colloidal water particles present in the cloud will get neutralized
and coagulate to form bigger water drops causing artificial rain. Artificial rain can also be caused by
throwing common salt or silver iodide on the clouds, as it is an electrolyte and brings about coagulation
of water particles,
(v) Formation of Delta. River water contains charged FIGURE 5.35
colloidal particles of clay, sand and many other

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●'●RiVeR-*'- SAND & CLAY
materials. Sea water is a very big store-house of a I..; I»;.:
PARTICLES
variety of electrolytes dissolved in it. As soon as river /A: I COAGULATED
BY
water comes in contact with sea water, the electrolytes '/>//./ SALTS OF

present in sea water coagulate the suspended colloidal ● SEAWATER

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'//////●_,

particles which ultimately settle down at the point of ■■n m ?v'.\

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contact. Thus, the level of the river bed rises. As a
M r ' M (

or
result, water adopts a different course and della is

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formed in due course of time (Fig. 5.35). It is so called

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-iSEAr
because the heap formed has a shape similar to the
Formation of Delta

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Greek letter A. In other words, delta is formed due to

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silting of the estuary {i.e. deposit of sediment at the
mouth of the river),

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(v/) Blue colour of the sky. This is due to the scattering of light by colloidal dust particles present in air

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(Tyndall effect). This is explained by Rayleigh scattering according to which if diameter of particles
is smaller than the wavelength of the incident radiation, intensity of scattered radiation \/X^. As blue
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of
colour of the white sunlight has minimum wavelength, it shows more intense scattering. Hence, sky
looks blue.
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Similarly, sea water looks blue due to scattering of light by the colloidal impurities present in sea water,
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(v//) Tail of comets. It is seen as a Tyndall cone due to the scattering of light by the liny solid particles left
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by the comet in its path.

iyiii) Blood. It is a colloidal solution of an albuminoid and the bleeding stops on applying ferric chloride
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solution due to coagulation of blood forming a clot.

(ix) Cleansing action of soap. Soap solution is colloidal in nature. It removes the dirt particles either by
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adsorption or by emulsifying the greasy matter sticking to the cloth, as already explained in Fig. 5.32.
Smoke screen. In warfare, smoke screens are used which are nothing but colloidal dispersion of certain
in

(.y)
substances {e.g., titanium oxide or SiCl4) in the air.
B. Industrial applications of collold.s :
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(/') Sewage disposal. Colloidal particles of dirt, mud, etc. carry electric charge. Hence, when sewage
water is passed through the plates kept at a high potential, the colloidal particles are coagulated due to
electrophoresis and the suspended matter gets removed.
(//) Purification of water. The precipitation of colloidal impurities present in water can be done by adding
certain electrolytes like alum etc. The negatively charged colloidal particles of impurities gel coagulated
by the Ap'*’ ions and settle down and pure water can be decanted off.
(///■) Smoke precipitation. Smoke particles are actually electrically charged colloidal particles of carbon in
the air. Precipitation of smoke particles is carried out by Cottrell precipitator which is based on the
principle of electrophoresis. Smoke is allowed to pass through a chamber having a number of metal
plates attached to a metal wire connected to a source of high potential (as shown in the Fig. 5.36).
Charged particles of smoke get attracted by oppositely charged electrode, get precipitated after losing
their charge and the hot air passes out through the chimney. The dust particles are also removed in this
process. Thus, the nuisance of smoke in big industrial cities can be avoided,
(/v) Photography. A colloidal solution of silver bromide in gelatine is applied on glass plates or celluloid
films or paper to form sensitive plates in photography.
SURFACE CHEMISTRY 5/49

(v) Rubber Industry. Latex is a colloidal solution of FIGURE 5.36

negatively charged rubber-particles. From latex, PARTICLE FREE
HIGH D.C. VOLTAGE
rubber can be obtained by coagulation. Rubber- (2250,000 VOLTS)
WASTE GASES

plated articles are prepared by depositing

negatively charged rubber-particles over the article PLATE
ELECTRODES
to be rubber-plated by making that article an anode
in a rubber-plating bath, GASES N.
(v/) Tanning/leather industry. The process of CARRYING
DUST
hardening of leather is known as tanning. Tannin, OR
CHIMNEY

which is obtained from plants, is a mixture of CARBON

PARTICLES
derivatives of polyhydroxy benzoic acids. It
contains negatively charged colloidal particles.
Animal hides are also colloidal in nature and

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contain positively charged particles. When they are
soaked in tannin, their mutual coagulation takes
place and leather becomes htud. Chromium salts

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are also used in place of tannin. CARBON OR DUST
“ PARTICLES REMOVED
(vii) Colloidal industrial products. A number of

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industrial products which we use in our everyday EARTH

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life are colloidal in nature, e.g., inks, paints, Cottrell precipitator
lubricants, synthetic plastics, rubber, cement etc.

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SUPPLEMENT YOUR
KNOWLEOGE FOR COMPETITIONS
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1. Colloidal solution of graphite in water is called “aqua dag’’ while that in oil is called “oil dag”.
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2. Colloidal solution of gold in water is called ‘Purple of cassius”.
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3. Colloidal Fe (OH)3 is given to a person having poisoning due to arsenic as the former adsorbs the
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latter and then can be vomitted out.

4. Some General Problems on Colloids :
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P. 1. What is the surface area of a cube having an edge length of 1 cm ? What would be the total
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surface area of the same material if it were subdivided into colloidal size cubes each having an
edge length of 10“’ cm ?
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Sol. Surface area of a cube = 6/^ = 6 x (1 cm)~ - 6 cm^

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Volume of cube of edge length 1 cm = P = 1 cm^

Volume of cube of edge length 10“’ cm = (10“’)^ = 10“^’ cm^
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1
.●. No. of smaller cubes in big cube = = 102'
10-21

Surface area of each smaller cube = 6/2 = 6 (10“’)2 = 6 x 10“1‘^ cm2

.●. Total surface area of all the smaller cubes = (102l) x (6 x 10-1“^) cm2 _ 5 x lO’ cm2
P 2. A particle of suspension of radius 1 mm is broken to form colloidal particles of radius 1000 A.
How many times will be the total surface of the colloidal particles as compared to the surface
area of the particle of suspension.
4 7 4 T 1 4
Volume of particle of suspension of radius 1 mm = ^ ^~ -71(0-1)^ cm2 =—
Sol. TcxlO
3 3

= 4 -s
Volume of particle of radius 1000 A (10“^ cm) = -^(10 =-nxl0-‘5 cm 3
3
5/50 ‘Pn<td€cfi.'4i. New Course Chemistry CXIl)EZsI9I

(4/3)71X10“^ = 10‘-
Number of colloidal panicles formed = -15
(4/3)71x10

Surface area of each colloidal particle = 4 Tcr^ = 4 tc (10"^)^ cm- = 4 7C x 10"*® cm-
Total surface area of 10*^ particles = (4 tc x 10"*®) x 10*^ = 4 ;c x 10^ cm^
Surface area of particle of suspension = 4 7i/- = 4 jc (OT)- cm- = 4 tc x 10~" cm-

Total surface area of colloidal particles _ 4 7tx 10- = 10*^

Surface area of particleof suspension 47txl0“^

5.22. NANOMATERIALS

5.22.1. Definition

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which at least one dimension is less than approximately
Nanomaterials are defined as a set of substances in

100 nm which is approximately 100,000 limes smaller than the diameter of a human hair.
5.22.2. Why are nanomaterials so important?
Nanomaterials are important or of great interest because at this scale, they possess unique optical,

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magnetic, electrical and other properties. These properties have great impact in electronics, medicines and

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other fields.

5.22.3. Where are nanomaterials being used?

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Nanomaterials are being used in a number of commercial consumer products such as stain-resistant and
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wrinklefree textiles, cosmetics, sun screens, electronics, paints and varnishes. A few examples are given
below :
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(/) There are novel UV-blocking coalings on glass bottles which protect beverages from damage by sunlight.
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(//) Using butyl rubber nano-clay composites, there are longer lasting tennis balls.
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(Hi) Nanoscale titanium dioxide is finding applications in cosmetics, sun-block creams and .self-cleaning
windows,
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(/v) Nanoscale silica is being used as a filler in a range of products including cosmetics, dental fillings and
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optica! fibres.
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5.22.4. Classification of nanomaterials

Nanomaterials are classified as (/) Zero dimensional (OD) spheres and clusters, (ii) one dimensional
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(1D) nanofibers,wires and rods, (Hi) two dimensional(2D) films, plates and network.s and (/V) three dimensional
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(3D) nanomalerials.

5.22.5. Examples of substances used as nanomaterials

DilTerent substances w'hich are being used as nanomaterials include gold, carbon (Buckminster fullerene),
metals, metal oxides and alloys.
5.22.6. Properties of nanomaterials
Nanomaterials have structural features inbetween those of atoms and bulk materials. However, their

properties are significantly different from those of atoms and bulk materials. This is mainly due to the nanometer
size of the materials which renders them the following special characteristics ;
(/) large fraction of surface atoms
(//) high surface energy
(///) spatial confinement (due to extremely large surface area to volume ratio)
(iv) reduced imperfections
SURFACE CHEMISTRY 5/51

A few properties are discussed below :

(1) Optical properties. Applications based on optical properties include optical detector, laser, sensor,
imaging, phosphor, display, solar cell, photocatalysis, photoelectrochemistry and biomedicine.
(2) Electrical properties. The use of nanomaterials in this field is based on their special photoconductivity
and electrical conductivity.
(3) Mechanical properties. The special mechanical properties imparted by nanomaterials include
polymers filled with nanoparticles or nanorods or nanotubes.
(4) Magnetic properties. Non-ferromagnetic bulk materials exhibit ferromagnetic-likebehaviourwhen
prepared in nano range. For example, bulk gold and platinum are non-magnelic but at the nano size they are magnetic.
5.22.7. Selected applications of nanomaterials
(/) In fuel cells (as nanomaterials provide more active catalyst due to large surface area to provide maximum
contact of catalyst, reactant gases and electrolyte).

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(ii) In catalysis (as nanomaierial counterparts provide large surface area).
(Hi) As phosphor for high definition TV (as resolution is improved due to reduction in the size of pixels),
(tv) Next-generation computer chips (as they provide miniaturization whereby circuits such as transistors,

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resistors and capacitators are reduced in size),
(v) Elimination of pollutants (as they are highly active catalysts to convert toxic gases such as CO, NO

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etc. to harmless gases in automobile exhausts).

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5.22.8. Disadvantages of nanomaterials
(i) Instability of the particles. They are prone to attack and undergo transformation.
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(ii) Highly explosive. This is due to their extremely large surface area coming in direct contact with oxygen.
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(///) Impurity. They take up impurities from the surroundings very easily. Hence, synthesis of pure
nanomaterials is very difficult.
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(iV) Biologically harmful. They cause irritation and have been found to be cacinogenic. If inhaled, they arc
entrapped in the lungs and in no way can be expelled out of the body,
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(v) Difficulty in synthesis, isolation and application. This is due to their extremely small size.
(vi) Recycling and disposal. Nothing concrete has been done in this direction so far.
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ZiX CSik/VNIOE
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1. Surface chemistry. It is that branch of chemistry which deals with the phenomena occurring at the surface
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or interface separating two bulk phases.

2. Adsorption, Adsorbate and Adsorbent. The phenomenon of attracting and retaining the molecules of a
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substance on the surface resulting into higher concentration on the surface is called adsorption. The substance
adsorbed on the surface is called adsorbate and the substance tm which it is adsorbed is called adsorbent.
The reverse process, /.e., removal of adsorbed substance from the surface is called desorption. The adsorption
of a gas on the surface of a metal is called occlusion.
3. Cause of adsorption. In case of liquid and solid absorbents, the adsorption is due to the unbalanced forces
acting inwards. In case of metals, it is due to presence of free valencies on metal surface.
4. Enthalpy & Entropy changes during adsorption. As molecules tire held on the surface, entropy decreases,
i.e., AS = - ve. As adsorption is a spontaneous process, AG = - ve. As AG = AH - TAS and AS is - ve, AG
can be - ve only if AH is - ve. Hence, AG, AH and AS all arc - ve.
5. Difference between Adsorption and Absorption. Adsorption is a surface phenomena whereas absorption
is a bulk phenomenon, i.e., occurs throughout the body of the material. For example, placed in a closed
vessel, water vapour are adsorbed on silica but absorbed in CaCl->.
6. Positive and Negative adsorption. In case of solutions, if solute is adsorbed on the adsorbent, it is called
positive adsorption. However, if solvent is adsorbed on the adsorbent, it is called negative adsorption.
5/52 “Pfutdeefa-'A New Course Chemistry (XII)CE
7. Factors affecting adsorption of gases by solids
(i) Nature and surface area of adsorbent, e.g., charcoal and silica have different adsorbing powers. Further,
greater the surface area, greater is the adsorption,
(li) Nature of the gas being adsorbed, e.g., adsorption on charcoal is in the order : H2 < N2 < CO < CH4
< CO2 < HC^l < NH3 < SO2. Greater the ease of liquefaction of a gas (or higher the critical temperature),
greateris the adsorption,
(m) Temperature : Adsorption decreases with increase of temperature. This is because adsorption is
exothermic.

Gas (adsorbate) + Solid (adsorbent) ^ - Gas adsorbed on solid + Heat

The heat released when 1 mole of gas is adsorbed is called heat of adsorption,
(iv) Pressure : Adsorption increases with increase of pressure at constant temperature. The effect is greater
at lower temperature,

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(v) Activation of solid adsorbent: This is done by increasing the adsorbing surface area by subdividing the
solid adsorbent or removing the gases already adsorbed.
8. Types of adsorption

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(i) Physical adsorption or physisorption. In this case, the gas molecules of the adsorbate are held on the
surface of the adsorbent by van der Waals forces. It is a physical change, reversible and non-specific in

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nature. It has low heat of adsorption (20-40 kJ mol"*) and decreases with increase of temperature. It is

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multimolecular.

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(ii) Chemical adsorption or chemisorption. In this case, gas the is held to the surface of solid adsorbent as
a result of chemical reaction forming surface compounds. It is chemical change, irreversible and specific in
nature. It has high heat of adsorption (40-400 kJ mol"^) and first increases and then decreases with increase
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or
of temperature. It is unimolecula.r
9. f
Freundlich adsorption isotherm. It is the plot of gas adsorbed per gram of adsorbent (x/m) versus equilibrium
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pressure at constant temperature (Fig. 5.4). Mathematically, — = where n is a positive integer (so that
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X 1
1/n lies between 0 and 1) and n and k are constants. Taking logarithm, log — = logk + —log P. Hence, plot
B

m n
X j
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of log — vs log P is linear with slope = — and intercept = log k.

nt n

X JC 1 X
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Similarly, for adsorption of solutes from solutions, — = kC^^'' or log — = log /: +—log C. Plot of log— vs
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m m n m

log C is linear.
10. Adsorption isobars. These are the plots of x/m vs temperature at constant pressure. In physisorption, x/m
d

decreases throughout as temperature is increased. In chemisorption, x/m fust increases and then decreases
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with increase of temperature. This is because initially, the heat absorbed acts as activation energy.
11. Applications of Adsorption. (0 In preserving vacuum (U) In gas masks (Hi) In decolorizing sugar (iv) In
F

chromatography (v) In heterogeneous catalysis (vi) In adsorption indicators (vii) In softening hard water
(viii) In separation of inert gases (ix) In removing moisture of the air present around certain electronic
instruments or commodities.
12. Catalysis. A substance which changes the speed of a reaction without itself undergoing any change in mass
and chemical composition is called catalyst and the phenomenon is called catalysis. A catalyst is generally
used to speed up a reaction (called positive catalyst or simply catalyst). If it is used to slow down a reaction
(e.g., glycerol in the decomposition of H2O2), it is called negative catalyst.
13. Typesof catalysis(i) Homogeneouscatalysiswhere catalyst is present in the same phase as reactants, e.g.,
no (8)
2SO2 (g) +02(g) > 2 SO3 (g).
(ii) Heterogeneous catalysis where catalyst is present in a different phase than that of reactants, e.g..
Fe(5) Co (s)
N2 (g) + 3 H2 (g) > 2 NH3 (g) (Haber’s process) or CO (g) + H2 (g) ^ Hydrocarbons
TiCl
(Fischer-Tropsch process) or n CH2 = CH2 (g) » -(-CH2 — CH2^ (Zeigler-Natta catalyst).
+R3AI
SURFACE CHEMISTRY 5/53

14. Zeolites as shape-selective catalysts. Zeolites are alumino-silicates having three-dimensional honey comb
like network structure and containing water molecules of hydration. To use them as catalyst, they are heated
so that water of hydration is lost. As a result, cavities become vacant and they become porous. The size of
the pores is in the range 260-740 pm. Thus, only molecules smaller than the size of the pores can be absorbed
here and not the bigger ones. Hence, they act as molecular sieves. An important zeolite used in petroleum
industry is ZSM-5. It is used to convert alcohols into hydrocarbons.
15. Enzymes as catalysts. Enzymes are bio-chemical catalysts which are globular proteins and form
macromolecular colloidal solution in water. They are obtained from living plants and animals. Their important
characteristics are : (i) Specificity, i.e., every biological reaction requires a different enzyme. («) Efficiency,
i.e., they speed up the reaction by a factor upto 10^®. (Hi) Small quantity because they are regenerated at the
rate of 1 million times per minute, (/v) Optimum temperature and pH. Their activity is maximum at
physiological pH of about 7.4 and human body temperature of 3TC.

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16. Mechanism of enzyme catalysis (Lock and key mechanism). Only a particular reactant (substrate) can
combine with a particular enzyme (just as a particular key can fit into a particular lock) to form enzyme-
substrate complex which then decomposes to form the product and regenerate the enzyme.

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17. Autocatalysis. If one of the products acts as a catalyst, it is called autocatalysis, e.g., in the reaction of
oxalic acid with acidified KMn04 solution, Mn^'*' ions produced act as autocatalyst. The reaction speeds up

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with time.

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18. Colloidal state of matter. All substance are classified into three categories, viz. (/) crystalloids (ii) colloids

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and (Hi) suspensions. When shaken with water, crystalloids can pass through filter paper (FP) as well as
animal membrane (AM) or parchment paper (PP), colloids can pass through FP but not through AM/PP
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whereas suspension can neither pass through FP, nor through AM/PP. The reason for such a behaviour is the
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size of their particles. Depending upon size, same substance may act as colloid or suspension. Hence, a
substance is said to be in colloidal state when size of particles is 1 nm-1000 nm. If size is < 1 nm, it forms
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true solution. If size > 1000 nm, it forms suspension.
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19. Dispersed phase and Dispersion medium. Just as in true solution, we have solute and solvent, in colloidal
solution (called colloidal sol), the substance dispersed is called dispersed phase and the medium in which it
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is dispersed is called dispersion medium.

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20. Classification of colloids

(i) Based on physical state of the dispersed phase and dispersion medium. Solid in solid (solid sol, e.g.,
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coloured glass). Solid in liquid (sol, e.g., muddy water). Solid in gas (aerosol, e.g., smoke). Liquid in solid
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(gel, e.g., butter). Liquid in liquid (emulsion, e.g., milk). Liquid in gas (aerosol, e.g., clouds or fog). Gas in
solid (solid foam, e.g., foam rubber) and Gas in liquid (Foam, e.g., soap lather).
d

(ii) Based on nature of interaction between dispersed phase and dispersion medium. These are of two
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types:
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(a) Lyophilic colloids which directly form the colloidal sol when shaken with the dispersion medium, e.g.,
gum, gelatine, starch etc. They are also called intrinsic colloids. They form reversible sols. They are highly
hydrated and are quite stable. (4% solution of cellulose nitrate a mixture of ethyl alcohol and ether is called
collodion),
(b) Lyophobic colloids which do not form colloidal sols directly with the dispersion medium, e.g., metals,
their sulphides etc. They are also called extrinsic colloids. They form irreversible sols. They are not much
hydrated and are less stable and easily precipitated.
(Hi) Based on the type of particles of the dispersed phase. These are of three types :
(a) Multimolecular colloids. Their particles consist of aggregates of molecules, e.g., Sg, gold etc.
(b) Macromolecular colloids. Their particles are macromolecules, e.g., starch, proteins, rubber, gelatine etc.
(c) Associated colloids. They are formed by electrolytes like soaps and detergents above a particular
concentration and a particular temperature. Usually ions aggregate to form micelles. The required concentration
is called critical micellisation concentration (CMC) viz. 1(H - 10“^ mol L“* and temperature is called
Kraft temperature. The cleansing action of soap is due to formation of such ionic micelles.
5/54 New Course Chemistiy (XU) BE

21. Preparation of colloidal sols

(a) Dispersion/Disintegration methods : (i) Mechanical disintegration (colloid mill) (//) Electro
disintegration (Bredig’s arc method) («t) Peptization, i.e„ shaking a freshly precipitated substance with a
suitable electrolyte, e.g., Fe(OH)3 ppt with FeCl3 solution.
(b) Condensation/Aggregation methods : (/) By chemical reaction, e.g., double decomposition (AS2O3 +
3 H2S > AS2S3 + 3 H2O) or reduction (e.g., 2 AUCI3 + 3 SnCl2 >2 Au + 3 SnC^) or oxidation (e.g.,
H2S + Br2 > 2 HBr + S) or hydrolysis (e.g., FeCl3 + 3 H2O > Fe(OH)3 + 3 HCl)
(«) By exchange of solvent, e.g. adding water into solution of sulphur in alcohol.
22. Purification of colloidal solutions - Dialysis. It is the process of separating the particles of the colloids
from those of the crystalloids by diffusing the mixture through a parchment or animal membrane. If electric
field is applied to speed up the process, it is called electro-dialysis.
23. Properties of colloidal solutions
(i) Heterogeneous character, i.e., they consist of dispersed phase and dispersion medium.

low
(ii) Brownian movement, i.e., zig-zig movement of colloidal particles in a colloidal sol.
(Hi) Tyndall effect, i.e., scattering of light by colloidal particles,
(/v) Electrical charge. All colloidal particles in a colloidal sol carry the same charge. Hence, they repel each
other and do not aggregate, (cause of stability). Charge is generally due to adsorption of common ions (e.g.,
Agl + r > Agl ; r or Agl + Ag'^ > Agl ● Ag"*").

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(v) Electrophoresis/cataphoresis. It is the movement of colloidal particles under the influence of an electric field,

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(v/) Electro-osmosis. It is the movement of the molecules of the dispersion medium under the influence of
an electric field when colloidal particles are not allowed to move.
(vii) Coagulation/Flocculation. It is the process of aggregating colloidal particles together to form large
for
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particles which settle down. This is usually done by adding electrolytes. The minimum amount of the electrolyte
(millimoles) that must be added to one litre of a colloidal solution so as to bring about complete coagulation
is called coagulation or flocculation value of the electrolyte. Greater the valency of the oppositely charged
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ions of the electrolyte, greater is its coagulating power. This is called Hardy-Schulze rule. Coagulation
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also occurs on electrophoresis or prolonged dialysis or mixing two oppositely charged colloidal sols.
24. Protective action of lyophilic colloids and Gold number. Addition of ‘lyophilic colloid' into a lyophobic
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colloid protects the latter from coagulation. Hence, lyophilic colloids act as protective colloids. The minimum
mass of the protective colloid in milligrams that must be added to 10 mL of a standard red gold sol so that
no coagulation occurs when 1 mL of 10% NaCl solution is rapidly added to it is called the gold number of
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the lyophilic colloid (protective colloid).

25. Emulsions. It is a colloid dispersion of liquid in liquid. The substance added to stabilize it is called emulsifying
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agent/emulsifier. Generally, soaps are used as emulsifiers. The liquid immiscible with water is termed as oil.
We thus have oil-in-water emulsions, e.g., milk or vanishing cream or water-in-oil emulsion, e.g., butter or
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nd

cold cream.

26. Applications of colloids : (/) Artiifcial rain, i.e., by throwing oppositely charged sand or common salt on
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the clouds to bring about coagulation of water particles.

(ii) Formation of delta, i.e., formation of delta shaped heap of sand, clay etc. where river falls into sea due
to coagulation of sand/clay particles by NaCl present in sea water.
(Hi) Blue colour of sky which is due to scattering of light by colloidal dust particles,
(iv) Puriifcation of water by coagulation of mud particles on adding alum,
(v) Smoke precipitation (Cottrell precipitator) by coagulating the carbon particles of the smoke on metal
plates connected to source of high potential before allowing to pass through the chimney.
(vi) Rubber industry to coagulate rubber from latex.
27. Nanomaterials. These are a set of substances in which at least one dimension is less than approximately
100 nm which is approximately 100,000 times smaller than the diameter of a human hair. Because of their
small size, they have large fraction of surface atoms, high surface energy, spatial confinement and reduced
imperfections. Because of these features, they possess special optical, electrical, mechanical and magnetic
properties. They are being used in fuel cells, cat^ysis, phosphor for high definition (HD) TV, next generation
computer chips and elimination of pollutants. Their disadvantages include their instability, highly explosive
nature, impurity and biologically harmful.
SURFACE CHEMISTRY 5/55

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NEET/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to
● MCQs in Chemistry for NEET
Pradeep's Stellar Series.... ● MCQs in Chemistry for JEE (Main)

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separately available for these examinations.

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Multiple Choice Questions (with one correct Answer)

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{d) At 83 K, nitrogen is adsorbed as atoms.
1. Adsorption

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4. Which of the following curves is in accordance
1. Adsorption is accompanied by with Freundlich adsorption isotherm ?
(a) decrease in enthalpy and increase in entropy for
(6) increase in enthalpy and increase in entropy
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(c) decrease in enthalpy and decrease in entropy
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ia) c:
ib)
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id) no change in enthalpy and entropy D)

O
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O)
(Kerala PMT 2010, Karnataka CET 2010) o

P—► log/J—►
2. Given :
Gas Hz CH4 CO2 SO2
our
ad

Critical Temp/K 33 190 304 630 c

On the basis of the data given above predict which (c) O) id)
S

of the following gases shows least adsorption on o

cn
dY

o
Re

a definite amount of charcoal ?

P—► log/7 ►
(a) H2 ib) CH4
Fin

(Karnataka CET 2015)

(c) SO2 id) CO2
5. For a gas adsorbed on a particular adsorbent at
(JEE Main 2019)
3. Although nitrogen does not adsorb on surface at 0°C, the plot of log — versus log P where P is in
17]
room temperature, it adsorbs on the same surface
at 83 K. Which one of the following statement is atm has a slope and intercept as shown in the Fig.
correct ?

(a) At 83 K, there is formation of monomolecular

layer.
45‘>
(b) At 83 K, there is formation of multimolccular log^ "S
layer. 0,301
(C) At 83 K, nitrogen molecules are held by
chemical bonds. log P

ANSWERS

1. (c') 2. (i/) 3. (h) 4. (d)

i
5/56 ‘P%<!idee^'4. New Course Chemistry (XII)CSm

The mass of the gas adsorbed by 10 g of the (c) 2 and 3 id) 2 and 4
adsorbent at 0-2 atm is
{e) 2 and 5 (Kerala PMT 2007)
Ka) 2g (b) 4 g 9. What is the equation form of Langmuir isotherm
(f) 6g (^0 8g under high pressure ?
6. In Frcundlich adsorption isotherm of a gas shown
a
X
ia) - = ib) - = flP
in the figure, — oc
P'. The value of 'a' from the m b m
m

figure will be
1
(d)
in a.P m a

10. 3 g of activated charcoal was added to 50 mL of

. X
log — acetic acid solution (0 06 N) in a flask. After an
m
hour it was filtered and the strength of the filtrate
was found to be 0-042 N. The amount of acetic
acid absorbed (per gram of charcoal) is
(a) 42 mg (b) 54 mg

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log P
(c) 18 mg ((/) 36 mg

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2 1
(«)- (JEE Main 2015)
11. Which one of the following characteristics is
3
(0 - (d) log- associated with adsorption ?

ree
m
(a) AG and AH are negative but AS is positive

7. Langmuir adsorption isotherm is deduced using

(JEE Main 2019)
for F
(b) AG and AS are negative but AH is positive
(c) AG is negative but AH and AS are positive
the assumption
(d) AG, AH and AS all are negative
(a) The adsorbed molecules interact with each
(MEET Phase I 2016, JEE Main 2021)
Your

other
ks

12. Adsorption of a gas follows Freundlich adsorption

eBoo

(b) The adsorption takes place in multi layers

isotherm. If x is the mass of the gas adsorbed on
(c) The adsorption sites are equivalent in their
ability to adsorb the particles mass ni of the adsorbent, the correct plot of —
in
ad
our

(d) The heat of adsorption varies with the versus p is

coverage (AIPMT 2007)
270 K 200 K
8- Which among the following statements are correct X
250 K 250 K
Re

with respect to adsorption of gases on a solid ? m

200 K 270 K
Y

1. The extent of adsorption is equal to k P" (o) ib)

Find

according to Freundlich isotherm.

2. The extent of adsorption is equal to k P'^"
according to Freundlich isotherm.
3. The extent of adsorption is equal to (1+/? P)/ 270 K 200 K

a P according to Langmuir isotherm. 250 K 250 K

200 K 270 K
4. The extent of adsorption is equal to
a P/(\+h P) according to Langmuir isotherm.
5. Freundlich adsorption isotherm fails at low
pressure.
(«) I and 3 (b) 1 and 4 (JEE Main 2020)

ANSWERS
5. (h) 6. (n) 7. {(●) 8. UO 9. (a) 10. (c) 11. (</) 12. ,(h)

i
SURFACE CHEWHSTRY 5/57

II. Catalysis III. Colloids

Cu 19. The volume of a colloidal particle, V^, as compared

13. CO(g)+H2(g) ^ X
to the volume of a solute particle in a true solution,
Cu/ZnO-CfjO^ V,., could be
CO (^) + H2 (g) > Y
V V
Ni
(a) Tf =1 (b) c ^jq23
CO (5) + H, (8) ^ Z
V
X, Y and Z respectively are
V V
(a) CH3OH, HCHO. CH4 (c) id) -f «10^
V V
{/?) HCHO, CH3OH. CH4
(c) CH4, CH3OH, HCHO (AIIMS 2015)

(d) HCHO, CH4, CH3OH 20. The dispersed phase and dispersion medium in
soap lather are respectively
14. Decomposition of H^Ot can be prevented in
presence of (a) gas and liquid (b) liquid and gas

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(a) glycerol (/>) acetanilide (c) solid and gas (d) solid and liquid

(c) phosphoric acid (d) all of these (e) gas and solid (Kerala PET 2010)

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21. Fog is a colloidal solution of
15. An example of autocatalysis is
(a) solid in gas ib) gas in gas
(a) oxidation of NO to NOo
(c) liquid in gas (d) gas in liquid
(b) oxidation of SOt to SO3

e
Fre
(NEET Phase 1 2016)
(c) decomposition of KCIO3 to KCl and O2 22. Which one of the following is correctly
(d) oxidation of oxalic acid by acidified KMn04
(Karnataka CET 2006)
matched ? for
(a) Emulsion-smoke (b) Gel-butter
16. Hydrolysis of protein in the stomach and intestine
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(c) Aerosol-hair cream (d) Sol-whipped cream
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takes place due to presence of the enzymes (e) Foam-mist (Kerala PMT 2011)
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(a) trypsin and pepsin respectively 23. Stability of lyophilic colloids is due to
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{b) pepsin and trypsin respectively (a) same charge on all the colloidal particles
(c) trypsin in both cases (b) solvation of the colloidal particles
(d) pepsin in both cases (c) both (a) and (b)
ad
our

17. Given below, catalyst and corresponding process/ id) the fact that they are organic substances
reaction are matched. The mismatch is
24. Which of the following colloids cannot be easily
(a) lRhCl(PPh3)2] ; hydrogenation coagulated ?
Re
dY

(b) TiCl4 + AI(C^Hg)3 : polymerization ia) Multimolecular colloids

(c) V20g : Haber-Bosch process ib) Irreversible colloids
Fin

(d) nickel : hydrogenation (c) Lyophobic colloids

18. Which one of the following statements is not id) Macromolecular colloids
correct ? (Karnataka CET 2015)
(a) Catalyst does not initiate any reaction 25. Among the following, surfactant that will form
ib) The value of the equilibrium constant is micelles in aqueous solution at the lowest molar
changed in the presence of a catalyst in the concentration at ambient conditions is
reaction at equilibrium ia) (CH3)(CH2>,5N^(CH3)3Br-
(c) Enzymes catalyse many bio-chemical
reactions ib) CH3(CH2)ijOSOjNa+
(d) Coenzymes increase the catalytic activity of (c) CH3(CH06COO-Na+
enzyme (NEET 2017) id) CH3(CH2),,N^(CH3)3Br- (IIT 2008)

13.(b) 14. id) 15.id) 16. ib) 17.(c) 18.ib) 19.id) 20.(a) 21.(c)
22. ib) 23. ic) 24. id) 25. ia)

I
5/58 New Course Chemistry (XI1)S!SZ9]

26. Wiieii an excess of a very dilute aqueous solution The isoelectric point of the colloidal sol will be
of K1 is added to a very dilute aqueous solution of (a) 4-2 (h) 4-8
silver nitrate, the colloidal particles of silver iodide (c) 7-0 {(I) 5-20
are associated with which of the following
31. Glutamic acid,
Helmholtz double layer ?
H2N-CH(CH2CH2C00H)-C00H has
(a) AgI:Ag+ir ib) AgljK+iNOj (a-COOH) = 2-2, (a - NH3) = 9-8
(c) AgliNOjiAg^ (d) AgiiriK+ and = (R group COOH) = 4-3. The
27. Which mixture of the solutions will lead to the
isoelectric point of glutamic acid is
formation of negatively charged colloidal (Agljr
(a) 5-9 (h) 7-0
sol. ?
(c) 10-2 (d) 3-2
(fl) 50 niL of 0-1 M AgN03 + 50 mL of 0-1 M KI
32. Which of the following electrolyte will have
(h) 50 mL of 0-1 M AgN03 + 50 mL of 1 -5 M KI
maximum flocculation value for Fe(OH)3 sol ?
(c) 50 mL of 1 M AgN03 + 50 mL of 0-2 M KI

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(fl) NaCl (b) Na^S
(d) 50 mL of 2 M AgN03 + 50 mL of 1-5 M KI
(C) (NH4)3P04 {d) K2SO4

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(NEET 2019)
28. The Tyndall effect is observed only when the 33. Among the electrolytes Na2S04, CaCl2, Al2(S04)3
following conditions are satisfied : and NH4CI. the most effective coagulating agent
(0 The diameter of the dispersed particles is much for Sb2S3 sol is

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Fre
smaller than the wavelength of the light used. {«) Na2S04 (h) CaCU
(tV) The diameter of the dispersed particles is not (c) Al2(S04)3 id) NH4CI (IIT 2009)
much smaller than the wavelength of the light
used.
for
34. The coagulation of 200 mL of a positive colloid
took place when 0-73 g HCl was added to it
without changing the volume much. The
r
(fVO The refractive indices of the dispersed phase
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flocculation value of HCl for the colloid is
and dispersion medium are almost similar in
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magnitude. (a) 0-365 (b) 36-5

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(tV) The refractive indices of the dispersed phase (c) 100 (d) 150
and dispersion medium differ greatly in (e) 200
magnitude 35. Ferric chloride is applied to stop bleeding due to
ad
our

(a) (/) and (//) {b) (//) and (Hi) a cut because :
3-f
(c) (0 and (/»') (d) (/7)and(/V) (a) Fe ion coagulates blood which is a
(JEE Main 2017) negatively charged sol.
Re

{b) Fe^'^ ion coagulates blood which is a positively

dY

29. Intensity of the scattered light depends upon the

difference of which of the following property of charged sol.
Fin

the dispersed phase and the dispersion medium ? (c) Cl" ion coagulates blood which is a positively
(a) densities (b) viscosities charged sol.
(c) surface tension (d) refractive indices id) Cl" ion coagulates blood which is a negatively
30. The speeds of colloidal particles in a colloidal sol charged sol.
at different pH values during electrophoresis are 36. Molar conductivity (A,„) of aqueous solution of
given below : sodium stearate, which behaves as a strong
pH 4-20 4-56 5-20 5-65 6-30 7-0 electrolyte, is recorded at various concentrations
Speed + 0-50 +0-18 -0-25 -0-65 -0-90 - 1-25 (C) of sodium stearate. Which one of the following
plots provides the correct representation of micelle
(p ms"‘) formation in the solution ?
(opposite signs indicate opposite direction of
(Critical micelle concentration (CMC) is marked
travel)
with an arrow in the figures)
ANSWERS

26. id) 27. ih) 28. id) 29. (d) 30. ib) 31. (<I) 32. (a) 33. (c) 34. (c) 35. (a)

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 SURFACE CHEMISTRY 5/59

(«) t ib) I 42. Which one of the following statements is FALSE

CMC
for hydrophilic sols ?
(a) These sols are reversible in nature
{b) The sols cannot be easily coagulated
dc dc
(c) They do not require electrolytes for stability
(d) Their viscosity is of the order of that of water
ic) t (d) .. {JEE Main 2021)
43. The charges on the colloidal CdS sol and Ti02 sol
are respectively
(a) positive and negative
dc
>
dc
(b) negative and negative
(c) negative and positive
(JEE Advanced 2019)
(d) positive and positive

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(JEE Main 2021)
37. Which one of the following has minimum gold
number ? 44. . Which condition is required to show the Tindall
effect by colloidal particles ? The refractive indices

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(a) starch (b) sodium oleate
(c) gum arabic
of the dispersed phase and dispersion medium
(d) gelatin
(a) differ greatly in magnitude
(Bihar CECE 2010)

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38. On addition of one mL solution of 10 % NaCl to (b) are exactly same in magnitude

Fr
(c) do not differ much in magnitude
10 mL gold sol in the presence of 0 0250 g of
(d) none of these (JEE Main 2021)
starch, the coagulation is just prevented. Starch
has the following gold number :
for
45. Measuring zeta potential is useful in determining
ur
ia) 0-025 (b) 0-25 which property of colloidal solution ?
(c) 2-5 id) 25. (a) Solubility
s
39. Gold number of gum arabic is 0-15. The amount (b) Stability of the colloidal particles
ook
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of gum arabic required to protect 100 mL of red (c) Size of the colloidal particles
gold sol from coagulation by 10 mL of 10% NaCl
eB

(d) Viscosity (NEET 2020)

solution is
46. A mixture of gases O2, H-) and CO are taken in a
(a) 015 millimoles (b) 015 mg closed vessel containing charcoal. The graph that
our

(c) 1-5 millimoles

ad

(d) 1-5 mg represents the correct behaviour of pressure with

40. The coagulation values in millimoles per litre of time is

the electrolytes used for the coagulation of As^Oj

Y

are given below :

Re

P
I (NaCl) = 52, 11 (BaCb) = 0-69, P
nd

III (MgS04) = 0-22 (0) (/>)

Fi

The correct order of their coagulating power is

(a) III > II > I (b) m > I > II >

(c) I > II > III Time Time

(^0II > I > III
(NEET Phase II 2016)
41. On which of the following properties does the P P
coagulating power of an ion depend ?
(a) The magnitude of the charge on the ion (c) id)
(b) Size of the ion alone
(c) Both magnitude and sign of the charge on the ion
Time Time
(d) The sign of charge on the ion alone
(NEET 2018) (JEE Main 2020)

isi SWERS
36. (a) 37. (d) 3S.(d) 39. (d) 40. (a) 41. (o) 42. id) 43. (c) 44. («)
45. (b) 46, (c)

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5/60 ‘P'uidee^’A New Course Chemistry (XII)BZsIS]
B. Micelle formation is an endothermic process
47. Tyndall effect is observed when :
{a) The diameter of dispersed particles is similar C. The entropy change is positive
to the wavelength of light used D. The entropy change is negative
(/?)The refractive index of dispersed phase is (fl) A and D only (h) A and C only
greater than that of the dispersion medium (c) B and C only (d) B and D only
(c) The diameter of the dispersed particles is much (JEE Main 2022)
smaller than the wavelength of the light used
IV. Miscellaneous
(c/) The diameter of the dispersed particles is much
larger than the wavelength of the light used 51. Select the wrong statement:
(JEE Main 2020) (a) If a very small amount of AICI3 is added to
48. A sample of red ink (a colloidal suspension) is gold sol, coagulation occurs but if a large
prepared by mixing cosine dye, egg white, HCHO quantity of AICI3 is added, there is no
and water. The component which ensures stability coagulation.

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of the ink sample is (h) Organic ions are more strongly adsorbed on
(«) HCHO (b) water charged surfaces in comparison to inorganic

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ions,
(c) Eosine dye id) Egg white
(JEE Main 2020) (c) Both emulsifier and peptising agents stabilise
colloids but their actions are different.

ee
49. Using very little soap while washing clothes does (d) Colloidal solutions are thermodynamically

Fr
not serve the purpose of cleaning of clothes, stable. (AMU Engg. 2009)
because
52. 0-27 g of a long chain fatty acid was dissolved in
(fl) soap particles remain floating in water as ions
for
100 mL of hexane as solvent. 10 mL of this
ur
(h) the hydrophobic part of the soap is not able to solution was added dropwise over a round glass
take away grease plate. Now the solvent (hexane) is allowed to
s
(c) the micelles are not formed due to evaporate off so that only fatty acid remains. The
ook
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concentration of soap below its CMC value distance from the centre to the edge of round plate
is 10 cm. Density of fatty acid is 0-9 g cnr^.
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(f/)colloidal structure of soap in water is

completely disturbed (JEE Main 2022) Calculate the height of the layer (Take 7t = 3)
50. For micelle formation, which of the following ia) 10-*m ib) 10-2 m
our

id) 10-^ m
ad

statement are correct ? (c) 10 * m

A. Micelle formation is an exothermic process (JEE Main 2019)
Y

in Multiple Choice Questions (with One or More than One Correct Answers)
Re
nd

54. Choose the correct reason (s) for the stability of

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53. The correct slatement(s) pertaining to the

adsorption of a gas on a solid surface is (are) lyophobic colloidal particles
(a) Adsorption is always exothermic ia) Preferential adsorption of ions on their surface
from the solution
ib) Physisorption may transform into
chemisorption at high temperature ib) Preferential adsorption of solvent on the
surface from the solution
(c) Physisorption increases with increasing
temperature but chemisorption decreases with (c) Attraction between different particles having
increasing temperature opposite charges on their surface
id) Chemisorption is more exothermic than id) Potential difference between the fixed layer
physisorption, however, it is very slow due to and the diffused layer of opposite charges
(IIT 2011) around the colloidal particles (IIT 2012)
higher energy of activation
ANSWERS

47. ia) 4S.id) 49.(c) 50.(c) 51. ia) 52. id)

53. ia.b.d) 54. iaM)

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 SURFACE CHEMISTRY
5/61

55. The given graphs/data I, II, III and IV represent 57. The correct statement(s) about surface properties
general trends observed for different physisorption is (are)
and chemisorption processes under mild
(«) The critical temperature of ethane and
conditions of temperature and pressure. Which of
the following choice (s) about 1, II, 111 and IV is
nitrogen are 563 K and 126 K respectively.
The adsorption of ethane will be more than
(are) correct
that of nitrogen of same amount of activated
P constant charcoal at a given temperature
P constant {h) Cloud is an emulsion type of colloid in
03 ^
X-

2-e
0)
which liquid is dispersed phase and gas is
2-e
c
a
o
«
C o
3 W
dispersion medium
o ^
9 ^
I (c) Adsorption is accompanied by decrease in
♦ II
enthalpy and decrease in entropy of the
T T system

id) Brownian motion of colloidal particles does

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</3 200 K >s
not depend on the size of the particles but
O)
depends on viscosity of the solution

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03^
'S 43
^250 K
“●e
(JEE Advanced 2017)
LO
c o
□ “ ●=0
Distance of molecule
E "
<
from the surface 58. The correct stalement(s) related to colloids is (are)
o

{a) The process of precipitating colloidal sol by

e
0. -1
= 150 kJ mol

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IV
III
P
an electrolyte is called peptisation.
{b) Colloidal solution freezes at high er
(a)
{h)
I is physisorption and II is chemisorption
I is physisorption and III is chemisorption
for
temperature than the true solution at the same
concentration
r
(c) IV is chemisorption and II is chemisorption (c) Surfactants form micelle above critical
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{d) IV is chemisorption and III is chemi.sorptio
oks

n
micelle concentration (CMC). CMC depends
(IIT 2012) upon temperature
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56. When O2 is adsorbed on a metallic surface, (J) Micelles are macromolecular colloids
electron transfer occurs from the metal to 03. The (JEE Advanced 2021)
true staiement(s) regarding this adsorption is (are) 59. The correct option (s) related to adsorption is (are)
our
ad

(a) O2 is physisorbed
(a) Chemisorption results in a unimolecular layer
{b) heat is released
ib) The entholpy change during physisorption is
(c) occupancy of of O2 is increased in the range 100 to 140 kJ mol"*
dY
Re

(c) Chemi.sorption is an endothermic process

{d) bond length of O2 is increased
Fin

(JEE Advanced 2015) id) Lowering the temperature favours physisorp

tion processes (JEE Advanced 2022)

Q]] Multiple Choice Questions (Based on the given Passage/Comprehens ion)

Each comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option,
(SSgiGiygGiai^SGD ill
U‘ A chen . was studying the solution and with concentrated KCI solution.
phenomenon of adsorption by putting blood He also made a detailed study of the
charcoal in KCI solution. He observed adsorption of gases on solid adsorbents. He
difference in behaviour with dilute KCI observed that at the same temperature,

ANSWERS
55. (o.f) 56. {h,c,d) 57. Ut.c) 58. {b,v) 59. {a.d)

\
5/62 “pmicUep. '4. New Course Chemistry (XII)jg5Igl
m 1
different amounts of gases like NHj, SOj, (c) Plot of — versus P
X
CO2, HCl were adsorbed by the same amount
of the adsorbent. He further studied the effect (d) Plot of log — versus log P
of temperature on adsorption and observed X

that in some cases, absorption showed a

regular trend while in some other cases, the rg.om^Wensi5nlFI Earlier the term ‘Colloids’
trend was not regular. Quantitative studies was used for a category of substances.
on adsorption have been made by Freundlich
However, later, the term colloidal state of
and Langmuir. They put forward mathe matter was preferred. Colloidal dispersions
matical expressions relating x/m (mass of the have been classified into different types
adsorbate adsorbed per gram of the depending upon the physical state of the
adsorbent) with equilibrium pressure,p, if the dispersed phase and the dispersion medium
adsorbate were a gas and with equilibrium or the nature of interactions between them
concentration, C, if the adsorbate were a the nature of the colloidal particles. They

w
or

solute from an aqueous solution. They are prepared in the industry or in the
authenticated their expressions by suitable laboratory by a number of methods and then

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plots. purified. Their properties have also been
studied in detail. Hardy and Schulze made a
60. Which of Ihe following result the chemist must substantial contribution in studying the

e
have observed about his studies with KCl

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coagulation of the colloids. The protective
solution ?
action of lyophilic colloids was studied by
(rt) Dilute KCl solution shows no adsorption Zsigmondy and he introduced a term, called
for
whereas concentrated KCl shows adsorption. ‘gold number’.
r
(/?) Concentrated KCl solution shows no
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adsorption whereas dilute KCl solution 63. Which of the following does not form a lyophilic
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adsorption, colloid ?
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(c) Dilute KCl solution shows positive adsorption (a) Rubber dissolved in benzene
whereas concentrated KCl solution shows
(b) White of the egg dissolved into water
negative adsorption.
(c) Common salt added into benzene
our
ad

{d) Concentrated KCl solution shows positive id) Stannous chloride solution added to gold
adsorption whereas dilute KCl solution shows chloride solution,
negative adsorption.
64. Critical micelle concentration (CMC) of soap
61. The correct order of adsorption of the gases studied
dY
Re

solutions lies in the range

will be
(fl) 10-^-10-5 M (b) 10-5-10^ M
Fin

ia) NH3>S02>C02>HC1 (c) 10^- 10-5 M (d) 10-5-10-2 M

(b) CO2 > SO2 > NH3 > HCl 65. In the experiment on electro-osmosis, in which of
(c) S02>NH3>HC!>C02 the following the level of the dispersion medium
((/) HC1>S02>NH3>C02 will fall on the cathode side ?
62. Which of the following plot will not be linear ? (a) Gold sol (b) Starch sol
(c) Fe(OH)3Sol (d) AS2S3 sol
(a) Plot of log — versus? 66. Which of the following has minimum gold
m

number ?
ib) Plot of — versus P (a) Polato starch (b) Gum arabic
x! m
(c) Gelatine (d) Albumen

60. (</i 61.1. 62. 63. u/i 64.(c) 65.(c) 66. (f)

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SURFACE CHEMISTRY
5/63

m Matching Type Questions

Match the entries of column I with appropriate entries of column II and choose the correct option out
of the four options (a), (b), (c), (d). given at the end of each question,
67. Column I oluniii II
(A) Fog (p) Gel
(B) Milk (q) Foam
(C) Cheese (r) Emulsion
(D) Soap lather (5) Aerosol

(a) A-q, B-5, C-r, D-^ (b) A^s, B-r, C-p. D-^ (c) A-q, B-s, C-r, D~p (d) A-r, B-s, C~q, D-p
68.
Match the following correctly
Catalyst Industrial product
(A) V2O5 (0 High density polythene

F low
(B) Zeigler-Natta O'O Polyacrylonitrile
(C) Peroxide
iiii) NH3
(D) Finely divided Fe (iv) H2SO4
(a) A-(iv), C-(/0, D-(//7) (b) A~(iv), B-(m), D-(/) (c) A-(/77), B-(0, C-(/7), D-(/v)

e
(d) A-iiv), B-(ii), C-(i), D-(di)

e
(e) A-(iv), C-(m), D-(//) for Fr (Kerala PMT 2011)
69.
Match the catalysts to the correct processes
Catalyst Procc
(A) TiCl4 (z) Wacker process
Your

(B) PdClj
s

(z7) Zeigler-Natta polymerisation

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(C) CUCI2 (z77) Contact process

fD) V2O5 (zV) Deacon’s process
ad

(n) (A)-(f7), (B)-{i77), (C)-OV), (D)-(/)

our

(b) (A)-(ii), {B)-(z), (C)-(z7), (D)-(zV)

(c) (A)-(/77), (B)-07), (C)-(iV). (D)-(O (d) (A)-(z7), (B)-(O, (C)-(zV), (D)-(z77)
Re

(JEE Main 2015)

70. Match List-I with LisMI
Y
Find

Usi-I

(A) Deacon’s process (i) ZSM—5

(B) Contact Process
(zz) CUCI2
(C) Cracking of hydrocarbons (///) Ni particles
(D) Hydrogenation of vegetable oils (zV) V2O5
(a) (A).(z), (B)-(/z7), (C)-(z7), (D)-(zv) (b) (A)-(zV), (B)-(z7), (C)-(z), (D)-((77)
(c) (A)-(z7), (B)-(z'v), (C)-(z), (D)-(/z7) (d) (A)-(z77), (B)-(/), (C)-(zV), (D)-(z7)

(JEE Main 2021)

67. (/;) 68. (.'i 6M, !-/ Tir. ,

5/64 ‘P>uidcefi^'4. New Course Chemistry (Xll)ISElS

71. Match List-I with List-II

List-I List-II

(A) Lyophilic colloid (j) Liquid - liquid colloid

(B) Emulsion (//) Protective colloii'
(C) Positively charged (Hi) FeCl3 + NaOH colloid
(D) Negatively charged (iv) FeCl3 + hot water colloid
{a) (A)-(//), (BHO, (CHfV), (DH»0 (b) {A)-{ni), (B)-(O. (C)-(/v). (D)-(/7)
(c) (AH/0, (B)-(O. (CHiii), (D)-(/v) (d) (AHi/O, (B)-(iO. (C)-(O- (D)-(tv)
(JEE Main 2022)

Matrix-Match Type Questions

w
Match the entries of column I with appropriate entries of column II. Each entry in column I may

F lo
have one or more than one correct option from column II. If the correct matches are A-p, s ; B-r, C-
P, q ; D-s, then the correctly bubbled 4x4 matrix should be as shown on the side :
s

ee
P P

©ji©iiOii®'

Fr
A'

B:
^ ©ijQllO!©' for
©ietoil©!:
ur
C

: © @! Oi © '
s
D;
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■-■-II
eB

72. Column I Column II

(A) Physisorption (p) High heat of activation

(q) Multimolecular
our

(B) Chemisorption
ad

(C) Activated adsorption (r) High temperature is required

(D) Desorption (s) Low pressure is required
Y
Re

Column II
nd

73. Column I
(p) Lyophilic
Fi

(A) Gold sol

(B) Starch sol (q) Associated
(C) Soap sol (r) Multimolecular
(D) Gelatin sol (,s) Macromolecular

74. Column I Column II

(A) Conversion of proteins into amino acids (p) Shape selective catalysts
(B) Conversion of alcohols into gasoline (q) Enzymatic catalysis
(C) Polymerisation of ethylene (r) Zcigler-Nattacatalyst
(D) Manufacture of margarine (.y) Heterogeneous catalysis

ANSWERS
71. (a) 72. (A-Q ; B-p.r ; C-p.r : D-s) 73. (A-r : B-p,s : C-q ; D-p.s)
74. (A-q : B'P,s : C-r,s ; D-s)
 SURFACE CHEMISTRY
5/65

VI.
Integer Type Questions A B c n

DIRECT IONS. The answer to each of the following questions is a single digit
®®®®
integer, ranging from 0 to 9. If the correct answers to the question numbers ®®O®
A, B, C and D (say) are 4,0,9 and 2 respectively, then the correct darkening
of bubbles should be as shown on the side : ®@ ©®
75. Water vapour were introduced into a vessel containing the following substances : ®®®®
silica, alumina, quick lime, charcoal, calcium chloride, phosphorus pentoxide, ®0 ® @
calcium carbonate, powdered cellulose, kieselguhr, Fuller’s earth.
The number of cases of adsorption is ® ®®©
76. The number of free valencies available for adsorption if four Pt atoms are linked
®® ®®

low
together by covalent bonds is
77. If 772 niL of SOt gas at STP is adsorbed on 2 g of charcoal at an equilibrium 0 0 0®
pressure of 16 atmospheres and the value of the constant
equation is 0-48, the value of the constant ‘n’ will be
'k' in the Freundlich
®®®®
78. How many of the following are aerosols ? ®®®®

ee
rF
Fog, froth, .soap lather, smoke, clouds, mist, foam rubber, dust, insecticide spray, hair cream

Fr
79. How many of the following are negatively charged sols ?
Silver sol, CrCOH)^ sol, AS2S3 sol, starch sol, silicic acid sol, haemoglobin, congo red dye, prussian blue

r
gum, clay, charcoal.
fo
80. When 2(K) mL of 0-2 M acetic acid is shaken with 0-6 g of wood charcoal, the fmal concentration of acetic
u
acid after adsorption is 0-1 M. The mass of acetic acid adsorbed per gram of carbon ISi
ks
g-
Yo
(JEE Main 2022)
oo

VII.
Numerical Value Type Questions Decimal Notation)
B
re

For the following question, enter the correct numerical value, (in decimal-notation, truncated/rounded-off to the
second decimal place, e.g., 6-25, 7 00, - 0-33, 30-27, - 127-30) using the mouse and the onscreen virtual numeric
u
ad

keypad in the place designated to enter the answer.

Yo

81. Emulsification of 10 mL of oil in water produces 24 x 10**^ droplets If the surface tension of the oil-water
interface is 0-03 Jm and area of each droplet is 12-5 x 10"'^ the energy spent (in joules) in the formation of
oil droplets is
nd
Re

82. 3-12 g of oxygen is adsorbed on 1-2 g of platinum metal. The value of oxygen adsorbed per gram of the
adsorbent at 1 atm and 300 K in L is [R = 0-0821 L atm K“' mol"']
Fi

(JEE Main 2021)

VIII.
Assertion-Reason Type Questions
TYPE I

DIRECTIONS. Each question contains Statement-1 (Assertion) and Statement-2 (Reason). Each question
has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correct option as under :
(a) Statement-1 is TYue, Statement-2 is TVue ; Statement-2 is a correct explanation of Statenient-1.
(b) Statement-I is IVue, Statement-2 is True ; Statement-2 is NOT a correct explanation of Statement-1,
(c) Statement-I is True, Statement-2 is False,
(d) Statement-1 is False, Statement-2 is True.

ANSWERS
75.(7) 76. (8) 77. (4) 78. (6) 79. (8) 80.(2)
81. 90-00 82. (1-2)
 New Course Chemistry (XIpEsIBl
5/66

83. Statement-1. Small quantity of soap is used to prepare a stable emulsion.

Statenient-2. Soap lowers the interfacial tension between oil and water.
84. Statement-1. For arsenic sulphide sol, BaCU has higher coagulation value than NaCl.
Statement-2. Higher the valency of the oppositely charged ion of the electrolyte added, higher is the
coagulating power of the electrolyte.
85. Statement-1. Micelles are formed by surfactant molecules above the critical micellar concentration (CMC)
Statement-2. The conductivity of a solution having sufficient molecules decreases sharply at the CMC.
(IIT 2007)

86. Statement-1. Lyophilic colloids are more stable than lyophobic colloids.
Statement-2. In lyophobic system, the dispersed particles are more solvated than in lyophilic system.
(IAS Prelim 2010)

87. Statement 1. In the coagulation of a negative sol, the flocculating power ot the three given ions is in the

w
order ; > Ba“+ > Na-^
Statement 2. In the coagulation of a positive sol, the flocculating power of the three given salts is in the

F lo
order : NaCl > Na.S04 > Na3P04
TYPE II

ee
DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a

Fr
corresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as
(a) If both assertion and reason are true, and reason is the true explanation of the assertion,
for
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion,
ur
(c) If assertion is true, but reason is false,
s
fd) If both assertion and reason are false.
ook
Yo

88. Assertion. Emulsions of oil in water a-e unstable and sometimes they separate into two layers on standing.
eB

Reason. For stabisation of an emulsion, excess of electrolyte is added. (JEE Main 2021)

89. Assertion. Activated charcoal adsorbs SO^ more efficiently than CH4.
r

Reason. Gases with lower critical temperature are readily adsorbed by activated charcoal.
ad
ou

(JEE Main 2022)

90. Assertion. Langmuir adsorption is a single layer phenomenon.

Y

Reason. It is due to van der Waals forces.

Re
nd

91. Aasertion. Physical adsorption of molecules on the surface requires activation energy.
Fi

Reason. Because the bonds of adsorbed molecules are broken.

92. Assertion. The catalytic convertor in the car’s exhaust system converts polluting exhaust gases into
non-

toxic gases.
Reason. Catalytic convertor contains a mixture of transition metals and their oxides embedded in the inner
(AHMS 2011)
support.
93. Assertion. Alcohols are dehydrated to hydrocarbons in the presence of acidic zeolites.
Reason. Zeolites are porous catalysts.
94. Assertion. Activity of an enzyme is pH-dependeni.
Reason. Change in pH affects the solubility of the enzyme in water.
95. Assertion. The micelle fonned by sodium stearate in water has -COO“ groups at the surface.
Reason. Surface tension of water is reduced by addition of stearate.
ANSWERS
83. («) M.id) SS.lh) 86. (c) 87. (c) 88. (c) 89. (c) 90. (c) 91. id)
92.(fl) 93.{b) 94.(b) 95.(n)
 SURFACE CHEMISTRY 5/67

96. Assertion. The conversion of fresh precipitate to colloidal state is called peptization.
Reason, It is caused by addition of common ions. (AIIMS 2007)
97. Assertion. In chemisorption, adsorptions keeps on increasing with temperature.
Reason. Heat keeps on providing more and more of activation energy. (AIIMS 2011)
98. Assertion. Isoelectric point is the pH at which colloidal particles can move towards either of the electrodes.
Reason. At the isoelectric point, colloidal particles may carry positive or negative charge.

m Multiple Choice Questions (Based on Practical Chemistry);

99. Which one of the following will form a (b) treating ferric chloride with dilute ammonium
hydrophobic sol ? hydroxide solution
(a) Gum (b) Starch (c) adding ferric chloride solution to boiling water
(c) Egg albumin (d) Ferric hydroxide (d) any one of the above methods

w
100. Which one of the following will form a hydrophilic 102. Sol of egg albumin is prepared by
sol ?
(a) mixing the yolk and white of the egg by

F lo
{a) Egg albumin (b) Ferric hydroxide beating and then adding it into water
(c) Arsenious sulphide (d) Gold (b) adding only the yolk of the egg into water

ee
101. Ferric hydroxide sol is prepared by (c) adding only the white of the egg into cold

Fr
water
(a) treating ferric chloride with dilute sodium
hydroxide solution {(f) adding the white of the egg into hot water
for
r
You
s
ook

For Difficult Questions

eB

Multiple Choice Questions (with one correct Answer)

our

1. Adsorption is accompanied by decrease in

ad

5. - = k
enthalpy (AH = -ve, and decrease in entropy m
(AS = -ve). Refer to page 5/2 and 5/7)
dY

2. Smaller the value of critical temperature of a ga.s, log — = log A: + - log P (y = mx + c)

Re

less is the adsorption {i.e., adsorption oc T^) m n

Fin

Hence, least gas adsorped is Ho. I

Plot of log — vs log P is linear with slope = —
3. At 83 K, there is physisorption and hence m n

multimolecular. and intercept = log k. Thus,

4. According to Freundlich adsorption isotherm, 1
X \ X - = taii45°=l or« = 1
log —
m
= log A + — log p . Hence, plot of log — vs
n

n m
and log A-= 0-3010 :.k=2

log p will be linear with positive slope =— and X

1/1
n Hence, at 0-2 atm, — = 2 x (0-2)
m

● X
= 2x0-2 = 0-4
intercept on y-axis log—axis equal to log A.
m
Mass of gas adsorbed by 10 g of adsorbent = 4 g

ANSWEI

96. (/?) 91. {d) 9%.(d) 99. (d) 100. (</) lOl. (c) 102. (t)
5/68 New Course Chemistry CXIl)K!ZsIS]

12. From Frcundlich adsorption isotherm

For Difficult Questions

— ocp (At low pressure)

m
6. According to Freundlich equation
a: 1 — ozP° (At high pressure)
— « P" (when fl = —) m

m n
On increasing the temperature, physical adsorption
X
or log — = log k + a log P decreases.
m in
13. X, Y and Z will be HCHO, CH3OH and CH4
Thus, plot of log — vs log P is linear with slope respectively.
m

- a
14. All the given substances act as negative catalyst
for decomposition of H2O2.
Here, slope = ^^ = —. Hence, a = — 15. In {d), ions produced in the reaction act as
x^-x^ 3 3 autocaialyst.

w
7. Langmuir adsorption isotherm is based on the 16. Pepsin in stomach and trypsin in intestine.
assumption that every adsorption site is equivalent 17. Haber-Bosch process is used for the synthesis of

F lo
and that the ability of a particle to bind there is ammonia. The best catalyst is finely divided iron
independent of whether nearby sites are occupied along with molybdenum or oxides of K and A1 as

ee
or not. promoters.

Fr
18. As catalyst decreases the activation energy of the
I//I
8. According to Freundlich isotherm, — = ^ P forward as well the backward reaction by the same
m
amount, therefore, it increases the speed of forward
for
as well backward reaction to the same extent. As
ur
aP
According to Langmuir isotherm, — = K = kf/k(^, therefore, value of the equilibrium
in l + bP
constant (K) remains unaffected by the presence
s
ok
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X aP of catalyst at a particular temperature.
9. According to Langmuir isotherm, — =
o

\+bP 19. For true solutions, the diameter range of the

eB

particles is I A to 10 A and for colloidal particles,

aP a
it is 10 A to 1000 A.
At high pressure, bP>>]. Hence, —
r

in bP b
^ (4/3)7cr3 fr f 10
ad
ou

= 10^
10. millieq of acetic acid present initially “ (4/3) Tcr^ r
< ^ y
1 )
Y

= 0 06 X 50 = 3
20. Soap lather is a dispersion of gas in liquid.
millieq of acetic acid left after adsorption
Re

21. Fog is an example of aerosol in which dispersed

nd

= 0-042 x 50 = 2-1
phase is liquid and dispersion medium is gas.
Fi

.-. millieq of acetic acid adsorbed by 3 g of 23. The stability of lyophilic colloids is both due to
charcoal
same charge and solvation.
= 3-21 =0-9 meq - 0-9 x 60 mg = 54 mg 24. Macromolecular colloids are quite stable and
Amount of acetic acid adsorbed per gram of resemble true solution in many respects.
54 25. Longer the carbon chain, i.e., greater is the size of
charcoal = — = 18 mg the hydrophobic chain (tail), less is the solubility
3
in water and greater is the tendency of surfactant
11. As the molecules of the adsorbate are held on the molecules to associate to form micelle, /.e., lower
surface of the solid adsorbent, entropy decreases, is the cmc. Hence (a) with the longest carbon chain
i.e., AS is -ve. has the lowest cmc.
As adsorption is a spontaneous process, [In fact, log (cmc) = A - B (C#)]
AG = -ve.
26. As excess of K1 has been added, I" ions are
Now, as AG = AH - TAS, and TAS is -ve, AG can adsorbed on Agl forming a ifxed layer (and giving
be -ve only if AH is -ve. it a negative charge). It then attracts the counter
SURFACE CHEMISTRY 5/69

A1^(S04)3. The order is

For Difficult Questions

Al^-" > Ca--" > Na^ > NH|.

ions (K'^) from the medium forming a second layer
{diffused layer). 34. 200 niL of the sol require = 0-73 g HCI
27. 50 mL of 1 -5 M K1 will contain greater number of 0-73
moles of KI than the number of moles of AgNO^ mol = 0 02 mol = 20 mmol.
36-5
in 50 mL of 0-1 M AgN03-
.●. 1000 mL (1 L) of the sol will require
Hence, when KI solution is added to AgN03
solution, first all Ag'*’ ions will be precipitated as 20
xl000 = 100mmol.
Agl. Further addition of KI will produce K"^ and 200
I" ions. The common L ions will be adsorbed on
Agl forming a negative collodial sol. of [Agl] T. 35. Fe-^"^ ions coagulate the soluble proteins (present
in blood plasma) which are negatively charged.
28. For Tyndall effect, refractive indices of dispersed (Remember that when a blood vessel breaks,
phase and dispersion medium must differ

w
bleeding stops by itself after sometime. It is due
significantly. Secondly, size of the dispersed
to clustering together of platelets (blood particles)
particles should not differ much from the

F lo
to form a plug).
wavelength of light used.
36. As the concentration of sodium stearate increases,
29. Intensity of scattered light depends upon the
the molar conductivity decreases. But when it
difference of refractive indices of the dispersed

ee
increases beyond CMC, stearate ions start
phase and the dispersion medium.

Fr
associating together to form micelles. As a result,
30. Isoelectric point of a colloidal sol is the pH at the molar conductivity starts decreasing abruptly.
which colloidal particles carry no charge. Above
this pH, colloidal particles carry positive or
for
These changes are correctly represented in plot (<a).
38. Amount of starch in mg that prevents coagulation
r
negative charge and below this pH, the charge is by 1 ml of 10% NaCl solution = 0-025 x 1000
You
reversed. Thus, reversal lakes place between 4 56
s

= 25. Hence, gold number = 25.

ook

and 5-20.
39. By definition, amount required for 10 mL of gold
eB

On neutralization
sol = 0-15 mg
31. H3N-CH-COOH ■>
^ I 1st ionization
.●. Amount required for 100 mL of gold sol
CH2CH2COOH = 0-15 X 10= 1-5 mg
our
ad

+
Cationic form
H.N-CH-COO-
^ I (Note that for 10 mL of gold sol, 1 mL of 10%
NaCl solution is used. For 100 mL of gold sol,
CH2CH2COOH
10 mL of 10% NaCl sol. is required).
dY
Re

+
> H.N-CH-COO" 1
2nd ionization
I
40. Coagulating power
Fin

Coagulation value
CH2CH2COO- Hence, the order is III > II > I.
41. The coagulating power of an ion depends upon the
pK„, + pK «2
Isoelectric point (pi) = magnitude of the charge on the ion.
2
45. Zeta potential is useful in determining stability of
2-2 + 4-3 the colloidal particles.
= 3-25
2 46. As H2, O2 and CO get adsorbed on the surface of
charcoal, the pressure decreases. So options (a)
32. Fe(OH)3 is positively charged. Hence, Cl" ions and (d) are ruled out. After some time, almost all
with minimum opposite charge will have maxi the surface .sites are occupied, the pressure remains
mum flocculation value (minimum coagulating constant as it reaches a state of equilibrium. Hence,
power). plot (c) is correct.
33. Sb2S3 is a negatively charged sol. The most 47. Tyndall effect is observed when diameter of the
effective coagulating agent will be electrolyte with dispersed particles is similar to the wavelength of
highest positive charge on the cation, vz., light used.
5/70 7^>usuiec^ ^ New Course Chemistry (XII)

Area of the round plate = Ttr^

For Difficult Questions
= 3x(10)2
= 300 cm^
48. Red ink is a colloidal sol. So it can be stabilised
by material like natural gum or egg white/albumen. Volume of the fatty acid
Height of the layer =
51. Coagulation occurs even when large quantity of Area of the plate
AICI3 is added.
52. Mass of fatty acid in 10 mL of solution = 0 027 g 0-03
cm
-1
Density of fatty acid = 0-9 g mL 300

= 10"^ cm
a027 g
.-. Volume of fatly acid - = 003 cm^ = 10-6
0-9 g cm ^
m

Dl Multiple Choice Questions (with One or More than One Correct Answers)

w
53. Only (c) is incorrect because physisorption 57. Higher the critical temperature, higher will be the

F lo
decreases with increasing temperature and extent of adsorption. Cloud is an aerosol, Emulsions
chemisorption increases initially with increasing are colloidal systems of liquid in liquid. For
temperature.
adsorption. AH = -ve, AS = -ve.

ee
Brownian movement of colloidal particles depends
54. Stability of lyophobic colloids is due to (a) and

Fr
on size of the particles.
id).
55. In graphs I and III, the amount of adsorption 58. (fl) Process of precipitating colloidal sol by using
decreases with increase of temperature and for
an electrolyte is called ‘coagulation’ and not
peptisation.
ur
increases with increase of pressure. Hence, they
represent physisorption. (b) As molar mass of colloidal solution is much
s
In graph II, amount of adsorption increases with higher than true solution, (AT^)j.o| < (ATy) true'
ook
Yo

increase of temperature. Hence, it represents Hence, freezing point of sols > freezing point
solution of true solution. So option (b) is correct,
eB

chemisorption.
Graph IV shows the formation of chemical bond. (c) Micelles are formed at or above CMC and
Hence, it again represents chemisorption. above craft temperature. Hence, (c) is correct.
our

{d) Micelles are associated colloids and not

ad

Thus, (a) and (c) are correct while {b) and (d) are
incorrect. macromolecular colloids.

56. When electron transfer occurs from metal to O2, 59. (a) Chemisorption results in unimolecular layer
Y

chemical bond is formed. Hence, it is while physisorption results in multimolecular

Re

chemisorption and not physisorption. Hence (a) layer. So (a) is correct.

nd

is not true. As it is chemisorption, heat is released. (b) Both physisorption and chemisorption are exo
Fi

I
Hence, (b) is true. When O2 takes up one electron, thermic but AH for physisorption is 20-40 kJ mol
it becomes O2 . According to molecular orbital and for chemisorption, it is 40—400 kJ mol-’. So
(b) is not correct,
theory, new electron will enter into T^p. As a
(c) Chemisorption is exothermic.
result, bond order decreases. Hence, bond length
increases. Hence, (c) and (d) arc ime. (d) is correct.

QH Multiple Choice Questions (Based on the given Passage/Compreh ension)

60. Solute is adsorbed from concentrated KCl solution 62. All except (d) are linear
whereas solvent (water) is adsorbed from dilute
KCl solution.
63. SnCU -I- AuCl^ react to form gold sol which is a
lyophobic sol.
61. The critical temperatures of the given gases are in
64. CMC of soaps lies in the range 10“^*- 10'^
the order : SO2 > NH3 > HCl > CO2. Hence, their mol L"’.
adsorption is in the same order.
SURFACE CHEMISTRY 5/71

For Difficult Questions

65. Level on cathode side will fall when dispersion medium moves towards anode, i.e., dispersion medium
carries -ve charge or colloidal particles carry +ve charge, viz, Fe(OH)3.
66. Gelatine has minimum gold number and hence highest protective power.

‘ Matching Type Questions

69. TiCl4 - Zeigler-Nalta polymerisation, PdCL - Wacker process (Refer to unit 12)
CuCl2 - Deacon’s process, V2O5 - Contact process

ow
70. Deacon’s process is used for the manufactureof chlorine gas.

VI.
Integer Type f questions

e
re
75. Except quick lime (CaO), calcium chloride 0-96 = 048 X (16)
1/71
or 2 = (16)
l/;j

(CaCU) and phosphorus pentoxide (P2O5),

rFl n = 4.

F
which combine with water and show absorption,
78. Aerosols are dispersions of solid in gas or liquid
all the remaining 7 substances are adsorbents for
in gas, Smoke and dust are aerosols of solid in

r
water.
gas while fog, mist, cloud and insecticide spray
ou
fo
76.
are aerosols of liquid in gas.
ks
Pt—Pt
79. Silver sol, AS2S3 sol, starch sol, silicic acid sol,
Congo red dye, gum, clay and charcoal are
oo
Pt—Pt
negatively charged sols.
Y
eB

0-2
Thus, there are 8 free valencies. 80. Initial moles of acetic acid = x200
1000
-1
77. Molar mass of SO2 = 64 g mol = 004 mole
r

Molar volume at STP - 22400 mL

ou

Final moles of acetic acid after adsorption

ad
Y

772 mL of SO2 at STP will have mass

0-1
64 X 200 = 002 mole
X 772 = 1-92 g 1000
d

22400
Acetic acid adsorbed = 004 - 002 = 002 mole
Re
in

X 1-92
= 0-96 = 002 x 60g= 1-2 g
m 2
F

1-2
.●. Mass adsorbed per gram of charcoal = — =
But — = k P‘^" m 0-6
m
= 20g

VII.
Numerical Value Type Questions (in Decimal Notation)
3-12
81. Total droplets = 2 4 x lO’® 82. Moles of O2 = = 00975
32
Total area = Total no. of droplets x Area of each drop
nRT 00975 X 0082-300
= (24x 10‘8)x(12-5 X lO”’''’) Volume of O2 =
P I
= 12-5 X 2 4 X 10^ m^
= 2-3985 L = 24 L
Total energy consumed in the formation of
droplets = (003 J m“^) x (12-5 x 24 x 10^ m^) Volume of O2 adsorbed per g of platinum =
24

= 90J 2
= L2g
5/72 “P^uzdee^’^ New Course Chemistry (XII)EHHI

For Difficult Questions

A.asertion-Reason Type Questions

84. Correct Statement 1. For -vely charged AS2S3 91. Correct A. Activation energy is needed for
sol, BaClo has lower coagulationvalue (or higher chemical adsorption and not for physical
coagulating power) than NaCl. adsorption.
85. The formationof micelleslakes place above CMC. Correct R. Breaking of bonds also takes place
Hence, statement-1 is True. Each micelle contains only in chemical adsorption.
at least 100 molecules. Hence, conductivity of the 92. R is the correct explanation of A.
solution decreases sharply at the CMC. Hence
93. Correct explanation. Both assertion and reason
statement-2 is also Tine but it is NOT the correct

w
are correct but dehydration is not due to porous
explanation of statement-1.
nature but due to the acidic group present in
86. Correct Statement-2. In hyophilic system, the zeolites.

F lo
dispersed particles are more solvated than in
94. Activity of enzyme increases with increase in /?H
hyophobic system.
and is maximum of = 74 and then falls off.

ee
87. For a negatively charged colloid, greater the charge
Hence, assertion is correct.
on the cation, greater is its flocculating power

Fr
(Hardy-Schulze rule). Correct explanation. With change in pH, the
Hence Statement 1 is correct.
extent of protonation of the bases changes.

for
96. Correct explanation. It is caused by adsorption
ur
For a positively charged colloid, greater the charge
of common ions on the surface of the precipitate
on the negative ion, greater is its flocculating
resulting into repulsion and fragmentation.
power, i.e., the order would be Na3P04 > Na2S04
s
ook

> NaCl. Hence, Statement-2 is incorrect. 97. Correct A. In chemisorption, adsorption first
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88. Correct R. On adding excess of electrolyte, increases with temperature and then decreases.
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coagulation occurs and emulsion is further Correct R. Heat absorbed initially provides
destabilized. activation energy for the reaction between
adsorbate and adsorbent.
89. Correct R. Gases with higher critical temperature
r
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are adsorbed more. SO2 being more polar, has 98. Correct A. Isoelectric point is the pH at which
higher critical temperature and hence is adsorbed the colloidal particles do not move towards
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more. cathode or anode.

90. Correct R. The single layer formed may be due Correct R. At the isoelectric point, colloidal
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nd

to van der Waals adsorption or chemisorption. particles do not carry any charge.
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 *

GENERAL PRINCIPLES

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AND PROCESSES OF

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ISOLATION OF ELEMENTS

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F
6.1. GENERAL INTRODUCTION
ur
r
Elements are the basic units of all types of matter in this universe. How do these elements occur in the
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earth’s crust (lithosphere) atmosphere and the sea (hydrosphere) and how are these elements obtained from
them ? These are some of the basic questions which we shall discuss in this unit. Since 80% of the elements
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known are metals, therefore, in this unit, our main emphasis shall be on the extraction/isolation and purification
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of metals.
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6.2. MODES OF OCCURRENCL JF ELEMENTS

B

Elements are found to occur in nature either in the free state (also called the native state) or in the
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combined state, i.e., in form of their compounds. This is mainly due to the reason that different elements have
different chemical reactivities,
u
ad

(i) Native state. Elements which are not attacked by moisture, oxygen and carbon dioxide of the air, occur
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in the native state. For example, carbon, sulphur, gold, platinum, noble gases, etc.
(ii) Combined state. Elements which are readily attacked by moisture, oxygen and carbon dioxide of the
d
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air, occur in the combined state in form of their compounds called minerals. In the combined state,
in

non-metals are found in the reduced form, i.e., X“ (where X = F, Cl, Br, I) while metals are found in the
F

oxidised form, i.e., oxides (e.g., Fc203, AI2O3, Sn02, Mn02).

Further, most of the metal halides being soluble in water are found in lakes and seas while majority of
the metal oxides and sulphides being insoluble arc found in rocks.
6.3. OCCURRENCE OF METALS - MINERALS AND ORES
Majority of the metals occur in the crust of the earth in the combined state in form of compounds called
minerals. Thus,

The naturally occurring chemical substances in form of which the metals occur in the earth
along with impurities are called minerals.
The relative abundance of these metals in the earth’s crust, however, varies from one metal to the other
(Fig. 6.1). For example, aluminium is the third most abundant element by mass (approx. 8-3%) in the earth’s
* Not included in CBSE syllabus. This chapter has been given only for the preparation of competitive examinations.
6/1
6/2 New Course Chemistry (XII)CZsI9]

FIGURE 6.1
crust after oxygen (46-6%) and silicon {211%). Many
gemstones are impure forms of alumina (AI2O3). For Aluminium 8.3
Iron 4.7
example, the impurity present in ruby is Cr while sapphire Calcium 3.6
Other
contains Co as the impurity. Iron is the second most Non-Metals
Sodium 2.8
'.Silicoiy ' Potassium 2.6
abundant (approx. 4-7% by mass) metal in the earth’s crust. Hydrogen 0.1
k 27.7# Metals Magnesium 2
It is a very important metal because it forms a large number Phosphorus 0.1
Others < 0.1
Titanium 0.4
Manganese 0.1
of compounds which have a variety of uses. Iron is also Oxygen
46.6
Others <0.1

one of the essential elements in the biological systems {i.e.,

haemoglobin in blood). Some important minerals of Percentage distribution (by mass) of the
aluminium, iron, copper and zinc are given in Table 6.1. most abundant elements in the earth's crust

TABLE 6.1. Principal Minerals of Some Important Metals

w
Metai Mineral

Aluminium Bauxite AlO^ (OH )3_2j. where 0 < .r < 1

F lo
Kaolinite (a form of clay) [Al2(0H)4Si205] or AI2O3.2 Si02.2 H2O
Iron Haematite Fe,b3
Magnetite Fe304

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FeC03

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Siderite
Iron pyrites or Fool’s gold FeS-)
Copper Copper pyrites
Malachite
CuFe$2
for
CuC03-Cu(0H)2
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Cuprite CU2O
s
Copper glance or Chalcocite CU2S
ook
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Zinc Zinc blende or Sphalerite ZnS

Calamine ZnC03
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Zincite ZnO

From the above table, it is clear that a metal may be found in form of a number of minerals. But every
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mineral is not suitable for the extraction of thatmetal. This is because it may contain either low concentration
of metal, or may contain impurities which may be difficult to remove or the process of extraction from that
mineral may not be chemically feasible or commercially viable. Thus,
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The mineral from which the metal is conveniently and economically extracted is called an ore.
nd

Thus, all ores are minerals but all minerals are not ores.
Fi

For example, iron is found in the crust of the earth as oxides, carbonates and sulphides. Out of these
minerals of iron, oxides of iron are employed for extraction of the metal because they are most abundant and
do not produce polluting gases (like SO^ that is produced in case of iron pyrites). Thus, oxides of iron are
called ores of iron. Similarly, aluminium occurs in earth’s crust in form of two minerals, i.e., bauxite (AI2O3.2
H2O) and clay (AI2O3.2 Si02-2 H2O). Out of these two minerals, aluminium can be conveniently and
economically extracted from bauxite while no easy and cheap method is so far available for extraction of
aluminium from clay. Therefore, the ore of aluminium is bauxite. For copper and zinc, any of the minerals
listed in table 6.1 may be employed depending upon their availability and other factors.
6.4. METALLURGICAL PROCESSES

The process of extracting metals from their ores is called metallurgy.

The process actually employed for the extraction of a metal depends upon its physical and chemical
properties and the impurities associated with it. Since different metals have different physical and chemical
properties, they have different methods of extraction. Therefore, a single, universal method cannot be applied
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/3

for the extraction of all the metals. Still there are some procedures and processes which are common to the
metallurgies of various metals. These are termed as metallurgical operations.
Some important metallurgical operations are :
(/) Crushing and grinding of the ore
(//) Concentration or benefaction of the ore.
(Hi) Extraction of crude metal from the concentrated ore.
(iv) Reifning of the crude metal.
These are briefly described below :
6.5. CRUSHING AND GRINDING OF THE ORE

The ores usually occur in nature in form of big lumps. These FIGURE 6.2
lumps are broken into small pieces with the help of crushers or

w
grinders. This process is called crushing. These small pieces are
then reduced to a fine powder with the help of stamp mill or hall
mill. This process is called pulverisation.

Flo
The crushed ore is fed into a battery of stamp mills working
in series. Each stamp mill (Fig. 6.2) consists of a heavy stamp or M ROTATING

e
CAM SHAFT

re
bolt (weighing over thousand pounds) which rises through a height
of about 20 cm by rotating cam shaft and falls under gravity 100

F
times per minute on the die. The pulverised ore is carried off by a
ur
current of water through the screen provided at the other end. The

r
CRUSHED
coarser particles, if any, xe retained back and pulverised again.
A ball mill, on the other hand, consists of a big steel lank foORE
ks
containing a few steel balls or a few pieces of hard stone like flint in it.
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The steel tank is fed with crushed ore and water, and rotated. Steel
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balls fall on ilie crushed ore and reduce it to a powder. STAMP I

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6.6. CONCENTRATION OR SCREEN/

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BENEFACTION OF THE ORE (ORE-DRESSING) ✓

Ores as they are obtained from earth’s crust are never pure.
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ad
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They are usually associated with earthy and silicious impurities (in
addition to the impurities of other minerals) called gangue or
Stamp mill
matrix. These impurities have to be removed from the ore before
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the extraction of the metal is actually done.

in

The removal of unwanted earthy and silicious impurities (i.e., gangue or matrix) from the ore
F

is called ore-dressing or concentration of ores and the process used to concentrate an ore is
called the benefaction process.

The finely ground ore is concentrated by the following methods :

1. Hand picking, in case the impurities are quite distinct from the ore so that these may be differentiated
by naked eye, these may be separated by hand picking.
2. Hydraulic washing or Lcvigation or Gravity separation. The process by which lighter earthy
particles arefreedfrom the heavier ore particles by washing with water is called levigation. For this purpose,
the powdered ore is either agitated with water or washed with a running stream of water. Tlie lighter gangue
particles are washed away while the heavier ore particles settle down. Two common methods of gravity
separation are described below :
(i) Hydraulic classifier. In this method, the finely powdered ore is dropped through a hopper into a
conical reservoircalled hydraulicclassifier,from the top (Fig. 6.3). A powerful stream of water is introduced
from the bottom of the reservoir. The lighter gangue particles are carried up by the current of water and pass
6/4 New Course Chemistry (XII)ESm

out along with water through the outlet provided near FIGURE 6.3,
the top. The heavier ore particles get collected at the POWDERED ORE
base of the cone. The conical shape of the re.servoir
helps in reducing the velocity of water and thus 1
prevents the ore particles being carried away along
wiili the stream of water.

Hi) Wilfley table. The other method of gravity SUSPENSION OF

separation is Wilfley table, li is a wooden table (Fig. ORE PARTICLES-
GANGUE
6.4) having slanting floor with riffles or grooves or IN WATER

cleats fixed on it.

The pulverised ore is placed on the trays
provided at one end of the table and a stream of water
under pressure is passed over it. A rocking motion is

w
given to the table. The lighter gangue particles are WATER-

carried away by the stream of water while the heavy

ore particles get deposited in the grooves. The ore

F lo
CONCENTRATED ORE
particles deposited in the grooves move towards one
side of the table as a result of rocking motion given
Hydraulic classiRer

ee
to the table and are finally collected.

Fr
This method is used when the ore

particles are heavier than the earthy or FIGURE 6.41

rocky gangue particles. The oxide ores
for PULVERISED ORE
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such as those of iron (haematite), tin 1 ::a
(tinstone) and native ores of Au, Ag, etc.
s
— WATER
are usually concentrated by this method. 7i
ook
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Electromagiieth <'.ntiun.
This method of concentration is mi.i
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y.

employed when either the ore or the GROOVES

impurities associated with it are mag

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netic in nature. For example, chromite,

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■>/-E-V.rV7
(Fe0-Cr^03 = FeCr20^)—an ore of i
chromium, magnetite (Fe304)—an ore
.mm
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of iron and pyrolusite (MnO^)—an ore GANGUE

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of manganese being magnetic are

nd

CONCENTRATED ORE
separated from non-magnetic silicious
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gangue by this method. Similarly, ^Ifley table

tinstone or cassiterite (Sn02), an ore of
tin being non-magnetic can be separated from magnetic
impurities like those of tungstates of iron and manganese FIGURE 6.5

(FeW04, MnW04), which are generally associated with it, ELECTROMAGNETIC FINELY

by this method. ROLLER POWDERED

ORE
In this method, tlie powdered ore is dropped over a
conveyer belt moving around two rollers—one of which
has an electromagnet in it. As the ore particles roll over the NON

belt, the magnetic particles are attracted by the magnetic MAGNETIC-

PARTICLES MAGNETIC PARTICLES
roller. As a resulL two heaps are fonned separately. The heap
collected below the magnetic roller contains the magnetic
particles while the heap formed away from the magnetic
roller contains the non-magnetic impurities (Fig. 6.5). Electromagnetic separation
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/5

Iti case of tinstone, the tungstates being magnetic, fall in a heap under the magnetic roller while the ore
particles, i.e., Sn02 fall in a separate heap away from the magnetic roller.
4. Electrostatic separation. This method is used for concentration/separation of ores which are good
conductors of electricity from those which are poor conductorsof electricity. The method is based upon the
principle that when an electrostatic field is applied, the ore particles which are good conductor of electricity
get electrically charged and hence are repelled by the electrode having the same charge and thus are thrown
away. This method is used for concentration/separation of PbS and ZnS ores occurring together in nature.
The mixture of ores is powdered and is dropped over a roller which is subjected to electrostatic field. PbS
being a good conductor immediately gets charged and is thrown away from the roller whereas ZnS being a
poor conductor does not get charged and hence fails vertically from the roller.

ow
5. Froth floatation. This method is widely used
FIGURE 6.6
for the concentration of sulphide ores such as zinc blende
(ZnS), copper pyrites (CuFeSj), galena (PbS), etc. This ROTATING PADDLE
method is based upon the fact that the sujface of sulphide AIR
ORE

e
ores is preferentially wetted by oils while that ofgangue i'o: FROTH

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is preferentially welted by water. I

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.PULP OF
The ore is crushed into a fine powder and mixed

F
ORE + OIL
with water to form a suspension in a lank (Fig. 6.6). To CS 3 PADDLE DRAWS
this suspension are added collectors (e.g. pine oil, INAiRAND
xanthates, i.e., sodium ethyl xanlhate* and potassium
ou
or
STIRS THE PULP

ethyl xanlhate, and fatty acids) which enhance the non Froth floatation process
wettability of the ore particles and froth-stabilizers (e.g.,
cresols and aniline) which stabilize the froth.
kfs
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The suspension is violently agitated by the rotating paddle which draws in air causing frothing.
During this process, the ore particles which are preferentially wetted by the oil become lighter and thus
Y
B

rise to the surface along with the froth while the gangue particles which are preferentially welted by water
become heavier and thus settle down at the bottom of the tank. The froth is skimmed off. It is allowed to
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collapse and finally dried to get the concentrated ore.

oYu

Separation of two sulphide ores. If the mineral to be concentrated consists of sulphides of two metals,
ad

then by adjusting the proportion of oil to water, it is often possible to separate one sulphide from the other.
Sometimes additional reagents called froth depressants are used to prevent one type of sulphideore particles
d

from forming the froth with air bubbles. For example, sodium cyanide is used as a depressant to separate lead
in

sulphide (PbS) ore from zinc sulphide (ZnS) ore. This is due to the reason that NaCN forms a soluble zinc
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complex. Na2 [Zn(CN)4] which goes into solution thereby preventing it from forming the froth. Under these
F

conditions, only PbS forms a froth and hence can be separated from ZnS ore.
4 NaCN + ZnS Na2 lZn(CN)4l -F Na-)S
Depressant Ore Sod. teiracyanozincute (II)
6. Leaching. This process consists in treating the powdered ore with a suitable reagent (such as acids,
bases or other chemicals) which can selectively dissolve the ore but not the impurities. The following examples
will illustrate the procedure.
(a) Leaching of bauxite, ’'’iirc alumina (AI2O3) is obtained from the bauxite ore by Baeyer’s process.
The bauxite ore usually conta . impurities of iron oxide (Fc203), silica (Si02) and titanium oxide (TiO,).
The powdered ore is heated with a concentrated (45%) solution ofNaOH at 473-523 K and 35-36 bar pressure.
Under these conditions, alumina (A1203) dissolves forming sodium meta-aluminate and silica (SiO-,) as sodium
silicate, leaving behind the impurities

* CH3CH2—O— (sodium ethyl xanlhate)

S"Na+
6/6 ‘P'Ktdee^'A New Course Chemistry (XII)BZSDHI
473-523 K

AI203-2H20 (s) + 2 NaOH (aq) + H2O (/) 35-36 bar pressure

4
2 Na[Al(0H)4] (a^)
Bauxite (45%) Sod. mela-aluminate

473-523 K
Si02 (s) + 2 NaOH (aq) » Na2Si03 (aq) + H2O {/)
Silica Sod. silicate

The resulting solution is filtered {to remove undissolved iinpurities), cooled and its pH adjusted
downward either by dilution or by neutralization with CO2. Thereafter, the solution is seeded with freshly
prepared samples of hydrated alumina when hydrated alumina gets precipitated leaving sodium silicate in the
solution.

2Na[Al(OH)4] (a^)2 CO2 (g) ^ AI2O3 . r H2O (s)i + 2 NaHCOj (aq)

Hydrated alumina, obtained as above, is filtered, washed and finally heated to about 1473 K to get pure
alumina (AI2O3).

w
I473K

AI2O3 . x H2O (s) AI2O3 (5) + X H^O (g)

F lo
(b) Leaching of silver and gold ores. Leaching is also used for extracting precious metals like silver
and gold by converting these metals or their ores into their soluble complexes. This method is also called
Mac Arthur Forest cyanide process.

ee
In this process, the finely powdered native silver or gold is treated with a dilute solution (0.5 %) of

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sodium or potassium cyanide while a current of air is continuously passed. As a result, silver and gold get
oxidised which then combine with CN” ions forming their respective soluble complex cyanides while the
impurities remain unaffected which are filtered off. Thus ; for
ur
4 M + 8 CN- + 2 H2O + O2 ■ 4 [M(CN)2]- + 4 OH-
(air) Soluble complex
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ook
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(z) For silver, 4 Ag + 8 NaCN + 2 H2O + O2 ■ ^4 Na [Ag(CN)2] + 4 NaOH

If instead of native silver ore, silver glance or argentite (Ag2S) is used for leaching of silver, the initial
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reaction forming the soluble silver complex is reversible. However, the current of air used oxidises largely
Na2S (formed in the reaction) to Na2S04 thereby pushing the equilibrium in the forward direction.
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Ag2S + 4NaCN ^ 2 Na[Ag(CN)2l + Na2S

Sod. dicyanoargentale (I)
(Soluble complex)
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4 Na2S + 2 H2O -1- 5 O2 (air) 2Na2S04 + 4Na0H + 2S

Re
nd

(/() For gold, 4 Au + 8 KCN + 2 H2O + O2 ^ 4 K [Au(CN)2] + 4 KOH

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Pot. dicyanoaurate (I)

(Soluble complex)

6.7. EXTRACTION OF CRUDE METALS FROM CONCENTRATED ORES

The process used to obtain metals in free state from concentrated ores is called extraction.
Since many metals can be easily obtained from their respective oxides by reduction (electronation),
therefore, the extraction of metals actually involves the following two chemical processes :
1. Conversion of the ore into metal oxide, i.e., de-electronation of ore.
2. Reduction or electronation of the metal oxide to the free metal.
6.7.1. Conversion of the Ore Into Metal Oxide or De-electronatlon of Ores

Metals are usually present in ores as hydrated oxides (hydroxides), carbonates and sulphides. Depending
upon the nature of the minerals present in the ores, the following two methods are used for conversion of ores
into their respective oxides.
GENERAL PRiNCiPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/7

(a) Calcination.

Calcination is the process of converting an ore into its oxide by heating it strongly below its
melting point either in absence or limited supply of air.

This method is commonly used to convert metal carbonates and hydroxides to their respective oxides.
During the process of calcination, the following chemical changes occur :
(0 Moisture is driven out.
(») Volatile impurities ofS, As and P are removed as their volatile oxides.
m Water is removed from hydrated oxides and hydroxide ores.
Heat Heal
AI2O3.2 H2O AI2O3 + 2 H2O FC'>03.3 H2O

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Bauxite Alumina Limonite Ferric oxide

iiv) Carbonate ores are converted into their respective oxides by loss of carbon dioxide.
Heat Heat
CaC03 CaO + CO2T; CaC03.MgC03 CaO + MgO + 2 CO21
Limestone Calcium oxide Dolomite

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Fl
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Heat Heat
CuC03.Cu(0H)2 ^ 2 CuO + H2O T + CO2 T ; ZnC03 ZnO + CO2 T

F
Malachite Calamine

(V) It makes the ore porous and hence easily workable in subsequent stages. Calcination is usually carried
ur
r
out in a reverberatory furnace (Fig. 6.4).

fo
The concentrated ore is placed on the hearth of the furnace and heated by flames deflected from the roof,
ks
(b) Roasting.
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Roasting is the process of converting an ore into its metallic oxide by heating strongly below its
melting point in excess of air.
B

This process is commonly used for sulphide ores. The following changes occur during roasting :
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(0 Moisture is removed.
u

in) Organic matter is destroyed.

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iiii) Non-metallic impurities like that of sulphu,r phosphorus and arsenicare oxidisedand are removedas
volatile gases.
d

So + 8 Ot -> 8SO2T {Sulphur dioxide)

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in

P4 + 5O2 -> P4O10T {Phosphorus pentoxide)

■> 2 AS2O3 T
F

4 As + 3 Ot {Arsenous oxide)
(/V) Ores are generally converted into metallic oxides.
2 ZnS +3 O2 ^ 2 ZnO + 2SO2T ; 2 PbS +3 O2 ■>
2 PbO +2 SO2 T
Zinc sulphide Zinc oxide Lead sulphide Lead oxide

2 Cu-jS + 3 O2 2 CU2O + 2 SO2 T

Cuprous sulphide Cuprous oxide
The SO2 thus produced is used for manufacture of H^S04.
(v) It makes the ore porous and hence easily workable in subsequent stages.
Like calcination, roasting is also carried out in a reverberatoryfurnace* (Fig. 6.7). During roasting air
vents are kept open while during calcination, air vents are either partially or completely closed.

♦This is a kind of furnace in which fuel does not come in direct contact with the charge. The flames are
directed from the roof of the furnace on to the charge and air supply can be controlled by vents (Fig. 6.7). This
furnace can be used both for oxidation and reduction purposes.
6/8 ‘P>t<xdce^'A New Course Chemistry (XII)EZs39DI

Flux. If the calcined or the roasted ore still contains non- FIGURE 6.7

fusibie impurities of earthy matter, an additional substance

CO CHARGE HOOPER
3 CO
^
called the flux is usually added during the reduction process. 1
I
<
O
AIR VENTS

Thus, I yj
z>
I
I /

Flux is a substance that chemically combines with I

L ^
gangite (earthy impurities) which may still be I
□ □
present in the roasted or the calcined ore to form 1
HOT GASES
an easily fusible material called the lag. AND FLAMES

●: :CHARGE
Flux + Gangue ^ Slag
The slag thus formed melts at the temperature of the LINING FUEL
being lighter
furnace. It is insoluble in the molten metal and also
floats over the surface of the molten metal from where it can
Reverberatory fiirnace
be skimmed off from time to time.

w
Types of fluxes. Depending upon the nature of the impurities present in the ore, fluxes are classified

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into the following two types* :
(i) Acidic fluxes. For basic impurities like lime or metallic oxides {CaO, FeO, MnO, etc.) present in the
ore, acidic fluxes like silica (Si02) and borax (NajB^^O-^.IO H2O), etc. are used.

e
CaO Si02 ^ CaSi03

Fre
+

FeO + S1O2 ^ FeSi03

(Basic impurities) (Acidic flux) for
{Fusible slag)

(ii) Basic fluxes. For acidic impurities like silica (Si02), pho.'iphorus pentoxide (P^Oiq), etc. present
r
in the ore, basic fluxes like limestone (CaCOj), magnesite (MgCOj), haematite (Fe^O^), etc. are
You
oks

used.

Si02 +
CaC03 - CaSi03 + CO.t
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Si02 +
MgC03 - MgSi03 + CO,t
(Acidic impurilie.s) (Basic flux) (Fusible slag)
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6.7.2. Reduction or Electronation of the Metal Oxide to the Free Metal

The roasted or the calcined ore is then reduced to the metal by using a suitable reducing agent. The
choice of the reducing agent, however, depends upon the reactivity of the metal. If the metal to be extracted
dY
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is very reactive like the alkali metals (Na, K, etc.), alkaline earth metals (Ca, Mg, etc.) and aluminium, then
the reduction of the oxide can be done only by the electrolytic method. The oxides of less reactive metals
Fin

such as zinc, iron, copper, lead, tin, manganese, chromium, etc. can be reduced by a number of reducing
agents such as carbon (coke), carbon monoxide or even another metal. When carbon is used as the reducing
agent, it combines with oxygen of the metal oxide to form carbon monoxide.
M..0 + vC ^ xM + vCO
,i-“y
Metal oxide Carbon Metal Carbon monoxide

The process usually involves heating the metal oxide with a suitable reducing agent. Thus,

The process of extracting the metal by heating the metal oxide with a suitable reducing agent is
called pyromctallurgy.

*Besides acidic and basis fluxes, a third type of fluxes are called neutral fluxes. These fluxes are added to
decrease the melting point and make the ore conducting in the electrolytic cell. Some examples of neutral fluxes
are ; KF, CaF^, Na3AlFg (cryolite), etc.
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/9

It is observed that some metals are easily reduced, (/.<?., reduction occurs at low temperatures) but others
are reduced with difficulty, {i.e., reduction occurs at high temperatures). To predict which element will be the
most suitable reducing agent for a given metal oxide and what would be the optimum temperature at which
reduction will proceed smoothly, the basic concepts of thermodynamics are quite useful as explained below :
6.8. THERMODYNAMIC PRINCIPLES OF METALLURGY
For any process, Gibbs free energy change (AG) is given by the equation,
AG = AH - TAS ...(/)
where AH is the enthalpy change and AS is the entropy change and T is the absolute temperature. The
free energy change is also related to the equilibrium constant K of the reactant-product system at temperature
T by the following equation.
AG° = - RT In K ...(n)
If AG° is -ve, then K will be positive. This means that the reaction will proceed towards products. From

w
these facts, we can draw the following conclusions:
1. The criterion of feasibility of a reaction at any temperature is that the AG of the reaction must be

F lo
negative. Thus, if AS is +ve in Eq. (/), on increasing the temperature (T), the value of TAS will increase and
when TAS > AH, AG will be -ve and the reaction will proceed in the forward direction.
2. A reaction with AG positive can still be made to occur by coupling it with another reaction having

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large negative AG so that the net AG of the two reactions is negative.

Fr
Such coupling reactions can be easily understood in terms of Ellingham diagram which plots standard
free energy of formation of oxides {i.e., A^G°) per mole of O2 as a function of absolute temperature.
6.8.1. Ellingham Diagrams for
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Ellingham diagram normally FIGURE 6.8
s
consists of plots of A^G® vs T for
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formation of oxides of elements (Fig.

6.8). Similar diagrams can also be
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constructed for sulphides and halides of

elements. These diagrams help us in -100
our

predicting the feasibility of thermal

ad

-200
reduction of an ore.
-300 ,2Fe°
Let us study the AG° during the
Y

formation of a metal oxide. ^00

o
Re

Zx M (s) + 02(g) > 2 M_^.0(.v) o

nd

-500

In this reaction, dioxygen-a gas has o

Fi

£ -600
been used up and metal oxide- —>

-700
a solid is formed. Since gases have higher %
<
entropy than liquids and solids, therefore, -800

during this reaction, AS becomes -900

negative. Thus, if temperature is
-1000
increased, TAS becomes more and more
negative. Since TAS is subtracted in Eq. -1100
(i), therefore, AG becomes less and less
-1200
negative. In other words, AG increases I 630K
u I I I 1 1
with increase in temperature (nonnally, ox 400X 800X 1200X 1600X 2000X
AG decreases with increase in 273 K 673 K 1073 K 1473 K 1873 K 2273 K

temperature). In other words. AyG° T

lines have +vc slopes for most of the Ellingham diagram - vs T plots for formation
of oxides of some metals. The small circles on the
reactions involving the formation of metal
oxides, M^O (s) as shown in Fig. 6.8. curves indicate melting and boiling points of metals.
6/10 ^>ui<Ceeib. ^ New Course Chemistry (XlOSSSm

Each plot is a straight line except when some change in phase (solid —> liquid or liquid gas) takes
place. The temperature at which such a change occurs, is indicated by an increase in the slope on the +ve side.
For example, in the Zn-ZnO plot, at the boiling point of zinc (1180 K), there is an abrupt increase in the
+ve slope of the curve. Similarly, the +ve slope of the Mg-MgO curve increases abruptly at the boiling point
of Mg at 1363 K. The small circles on the curve indicate melting points (circle on the left) and boiling points
(circle on the right) of the metals.

6.8.2. Limitations of Ellingham Diagrams

1. Kinetics of reduction. Ellingham diagrams are based on thermodynamic concepts, therefore, these
diagrams simply suggest whether the reduction of a given metal oxide with a particular reducing agent is
possible or not. It, however, does not tell anything about the kinetics of the reduction process, i.e., how fast

w
will be the reduction process.
2. Reactant/product equilibrium. The interpretation of AG° is based upon K (i.e., AG° = - RT In K).
Thus, it is presumed that reactants and products are in equilibrium.

o
M.,0 + xM + AO

e
ox

But, this is not always true because the reactant/product may be solid.

re
rFl
However, it nicely explains why the reactions are slow when every species (reactanl/product, reducing/

F
oxidising agent) is in the solid state and smooth when the oxide ore melts. It is interesting to note that AH and
AS values for any chemical reaction remain nearly constant even on changing temperature. Therefore, the
only dominant variable in Eq. (/), is T. However, AS depends much upon the physical state of the compound.

r
ou
fo
Since entropy depends upon the degree of disorder or randomness in the sy.stem, it will increase if a compound
ks
melts (.9 1) or vaporises (/ - g) because the degree of molecular randomness increases on changing the
phase from solid to liquid or from liquid to gas.
oo

6.9. THEORY OF PYROMETALLURCY

Y
B

III the light of the thermodynamic principles and Ellingham diagram, let us first discuss the reduction of
a metal oxide with carbon.
re

The process of extracting a metal by reduction of its oxide with carbon (in form of coke, charcoal
ou

or carbon monoxide) is called smelting.

Y
ad

Now we know that during reduction, the oxide of the metal decomposes, i.e..
d

Reduction : M^.O (^) > X M (s or /) + 1/2 O2 (g) ...(0

in

and the reducing agent takes away the oxygen given by the metal oxide, i.e.,
Re

Oxidation : C{s) + m 02(g) >CO (g) ...(n)

F

The role of the reducing agent here is to provide a large negative AG® to make the sum of the AG® of the
above two reactions (i.e., reduction of the metal oxide and oxidation of the reducing agent) negative. Now
Eq. (0 is actually reverse of Eq. (Hi) representing the oxidation of the metal.
I
A:M(5or O + -O2 (g) > Mp (s) ; AG® ...(Hi)

However, if, instead of partial oxidation of C to CO, complete oxidation of C to CO2 occurs, the oxidation
of the reducing agent may be represented as follows :
1 1 1 1

-C(r) + -02(g) >-C02(g); ^AG (C,CO^)

...(iv)

If, instead of C, CO is used as the reducing agent, the oxidation of the reducing agent may be represented
as follows:

I
C0(g) + -02(g) > CO2 (g); aG® (CO.COo) ...(V)
 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/11

On subtracting Eq. (Hi) from each of the three equations, (it), (iv) and (v), we have,
M,0 (s) + C (j) ■>xM (.9 or /) + CO (g) ...(V/)
1 1
M^O(i)+-C(5) > ^M(^or/) + -C02(^) ...(vii)

M^O (j) + CO (g) > xM(sovl) + CO2 (g) ...(viii)

These three equations (vi), (vii) and (viii) describe the actual reduction of the metal oxide, M^O to the
free metal, M. The values for these reactions, in general, can be obtained by subtraction of the
corresponding AyG° values from the Ellingham diagram (Fig. 6.8). If these values are -ve, the reduction is
feasible otherwise not.

Alternatively, without going into the calculations for A^G°, we can still predict the feasibility of a reduction
process simply by looking at the Ellingham diagram. Metals for which free energy of formation (AyG°) of
their oxides is more negative can reduce those metals for which the free energy of formation
(Ay^G°) of their respective oxides is less negative. In other words, at any given temperature, any element

w
(metal) will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free
energy change (A^ G°) for the combined redox reaction will be negative by an amount equal to the difference

F lo
between the free energy of formation (AjG°) of the two oxides at that temperature. Further, greater the
difference, easier is the reduction. For example. Mg can reduce AI2O3, ZnO, FeO and CU2O more readily
than A1 can reduce ZnO, FeO and CU2O. Similarly, A1 can reduce ZnO, FeO and CU2O more readily

ee
than Zn reduces FeO and C112O. Thus, the relative tendency of these metals to act as reducing agents is

Fr
Mg > A1 > Zn > Fe > Cu.
However, in case of less reactive metals like silver
for
and mercury. AyG° becomes positive at higher
temperatures. This suggests that both silver oxide (Ag,0) and mercury oxide (HgO) are unstable and hence
ur
decompose at high temperatures to liberate the conesponding metal.
s
Heal 630 K
ook

2Ag20(.9) ^ 4Ag(s) + 02(g); 2 HgO (.9) ^ 2Hg(/) + 02(g)

Yo

6.9.1. Effect of Temperature on the Free Energy Change {D^G®) of the overall Reduction Process
eB

We know that for any process,AG = AH - TAS

Since on increasing the temperature, the values of AH and AS nearly remain constant, therefore, the
our
ad

value of A^G° becomes more negative. This means that if a particular reduction process docs not occur at a
lower temperature, it may occur at a higher temperature. The only thing we have to do is to select the temperature
in such a way that the A^G" of the overall redox reaction becomes - ve.
Y

In the Ellingham diagram (Fig. 6.8), this temperature is indicated by the intersection of the two curves
Re
nd

(i.e., curve for formation of M^O and the curve for the oxidation of the reducing agent. At the point of
intersectionof the two curves, AG° = 0. Below this temperature, the AG° is -ve and hence the oxide is stable.
Fi

Above this temperature, AG® is +ve and hence the oxide is unstable, i.e., it would give metal and oxygen. For
example, the temperature at the intersection point ‘A’ of the (Al, AI2O3) and (Mg, MgO) curves is approx.
1623 K. Therefore, below 1623 K, MgO is stable and hence Mg can reduce AI2O3 to Al but above 1623 K,
AI2O3 is more stable and hence Al can reduce MgO to Mg.
6.10. APPLICATIONS OF PYROMETALLURCY
To illustrate the utility of Ellingham diagrams in pyronietallurgy, let us discuss the extraction of iron,
copper and zinc from their respective oxides.
6.10.1. Extraction of Iron from Its Oxides
The chief ores of iron are :

(0 Haematite, Fe203 (red oxide of iron). (ii) Limonile, Fe203.3 H2O (hydrated oxide of iron).
(Hi) Magnetite, Fe304 (magnetic oxide of iron), (iv) Siderite or Spathic ore, FeC03-
(v) Iron pyrites (fool’s gold), FeS.,.
6/12 ‘Pxa.dce^'^ New Course Chemistry CXII)BZs!91

Iron is usually extracted from the oxide ore, i.e., haematite. It involves the following steps :
1. Concentration. The ore is crushed in jaw crushers and the crushed ore is concentrated by gravity
separation process in which the crushed ore is washed in a stream of water when lighter sand and clay
particles are washed away while the heavier ore particles settle down. In case of sulphide ore (iron pyrites),
concentration is carried out by froth floatation process.
2. Calcination. The concentrated ore is then calcined, i.e., heated strongly in the presence of a limited
supply of air in a reverberatory furnace. During calcination, the following changes occur,
(i) Moisture is removed.
{//) Impurities of sulphur, phosphorus and arsenic escape as their volatile oxides.
Sjj + 8 O2 > 8 SO2 T ; P4 + 5 O2 > P40iot; 2 As + 5 O, > AS2O5 T
{Hi) Ferrous oxide is oxidised to ferric o.xide thereby preventing the loss of iron as slag during smelting
4 FeO (s) + O2 ig) > 2 F6203 (a)
(fv) The ore becomes porous and hence is more suitable for reduction to the metallic state.

w
In case of carbonate ore (siderite), during calcination, it is converted into ferric oxide.
FeC03 is) > FeO (s) + CO. (g); 4 FeO (s) + O. (g) > 2 Fe.03 (s)

F lo
However, in case of sulphide ore (iron pyrites), conversion to oxide is carried out by roasting.
4 FeS. (s) +110. (g) > 2 Fe.03 i.s) + 8 SO2 (g)

ee
3. Smelting. The calcined ore is reduced with carbon, i.e., smelted in a blast furnace (Fig. 6.9). It is

Fr
a tall cylindrical furnace made up of steel and lined inside with fire bricks. It is slightly narrow at the top and
again at the bottom. This facilitates the proper flow of materials. The furnace is provided with a double cup

for
and cone arrangement which helps to feed the charge from the top without letting any gases from inside to
escape. Near the top, furnace is also provided with an outlet for waste gases. At the base, the furnace is
ur
provided with
(/) small pipes called tuyeres through which a blast of hot air is admitted.
oks
Yo

(//) a tapping hole for withdrawing molten iron.

o

{Hi) a slag hole for withdrawing slag.

eB

The charge consisting of calcined ore (8 parts), coke (4 parts) and limestone (1 part) is introduced into
the furnace from the top through cup and cone arrangement. At the same time, a blast of hot air pre-heated to
our
ad

FIGURE 6.9

Furnace Double
charge cup
Y

(ore + coke) and cone

+ limestone arrangement
Re

I
nd

Furnace gases
Ore loses
containing 25% CO
Fi

moisture and ,
——^ used to heat
becomes more
porous
finn K incorpino air blast
523 K
1— Fire bricks Reduction begins
Slag formation ●SFejOals) + CO(g) ^ 2Fe304(s)+ C02(g)
CaC03(s) —^ CaO(s) C02(ff) A /Steel shell Fe304(s) + 4CO(g) ^ 3Fe(s) + 4C02(g)
Ca0(s)+Si02(s) -»CaSi03 900 K
F6203(s) .+ CO(g) 2FeO(s) + C02(g)
(Slag) —, 1123 K Reduction completed
Melting of slag FeO(s) + C(s) -> Fe(s) + CO(g)
and iron 1423 K >1073K,
Fe203(s) + 3C(s) 2Fa(&7) + 3CO(g)
Fluid slag and ■*' 1673 K Coke burns to form CO2,
liquid Fe trickle 2170 K =» which on passing up the
down into hearth ^ furnace through more
hot coke, is reduced to CO
C(s) + 02(g) -* COglg)
Tuyeres Hot air blast ^02(g) + C(s) ^ 2Cb(g)
I I I T FeO(s) + C(s) Fe(/) + CO(g)
Molten pig iron .4r" Slag

Blast furnace for the manufacture of pig iron

 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/13

about 1000 K is blown into the furnace through tuyeres. The added coke serves both as a fuel as well as a
reducing agent while added limestone acts as the basic flux.
The temperature near the top of the furnace is around 523 K while near the bottom it is around 2170 K.
(a) Theory of reduction process
Thermodynamics helps us to understand how coke reduces iron oxide to iron. In fact, iron oxide is
reduced to iron mainly by CO though perhaps some reduction by C also takes place.
FeO (5) + C (s) ^Fe (s/O + CO ig) ...(0
This redox reaction can be divided into the following two half reactions—one involving reduction and
the other oxidation :
1
Reduction : FeO (5) > Fe(5) + -02 (FeO, Fe) ...(H)

Oxidation : C(5) + ^02 (g) > CO (g) ; AG° (C, CO) ...(in)

w
The net free energy change of these two combined
FIGURE 6.10
reactions is

F lo
-100-
AG“ -I-AG“
(C, CO) = A,G°
(FeO. Fe)
C'J
-200-
or AG® -AG° = A^G" ...(/V)
o
(C. CO) (Fe. FeO)

ee
-300-

Obviously, the resultant reaction, i.e., Eq. (/), will oFft+O^ -» 2FeQ.

Fr
^00--
take place only when the A^G is -ve. E
-500-●
Let us now find out the approximate temperature at
which this reduction occurs. For this purpose,
2
<
for
-600-
ur
consider the Ellingham diagram as shown in Fig. 6.10. 273 K
673 K 1073 K 1473 K
At approximately 1073 K or above, the C, CO line is
s
ook

Temperature
Yo
much below the Fe, FeO line. This means that,
AG°pg pgo > AG®c, CO hence A^G® is -ve. Ellingham diagram for formation of FeO
eB

In other words, at 1073 K or above, coke will reduce from Fe, CO from C and CO2 from CO
FeO Fe and itself will be oxidised to CO.
our

In contrast, at temperatures below 1073 K, the CO, CO2 line lies below Fe, FeO line. Therefore, below
ad

1073 K, CO reduces the oxides of iron, i.e., Fc203, Fe304, etc.

Thus, in the blast furnace, reduction of iron oxides takes place at different temperatures. Since the air
Y

passes through in few seconds, the individual reactions do not reach equilibrium,
Re

(b) Reactions taking place in the furnace. The following reactions

nd

occur in the blast furnace,

(i) Zone of combustion. Near the tuyeres, coke bums to form carbon dioxide.
Fi

C (.v) -F O2 (g) > CO2 (g); AH = - 393-3 kJ

Since the reaction is exothermic, lot of heat is produced and the temperature here is around 2170 K.
(ii) Zone of heat absorption. This is lower part of the furnace and the temperature here is between
1423-1673 K. As CO2 foraied near tuyeres moves up, it meets the descending charge. The coke present in the
charge reduces CO2 to CO.
CO2 (g) + C (5) ^2CO(g) ; AH = -h 163-2 kJ
Since this reaction is endothermic, therefore, the temperature gradually falls to 1423 K.
(iii) Zone of slag formation. It is the middle part of the furnace. The temperature here is around
1123 K. In this region, limestone decomposes to form CaO and CO2. The CaO thus formed acts as a flux and
combines with silica (present as an impurity) to form fusible calcium silicate slag
1123K
CaC03 (j) ■>
CaO (s) + CO2 (g); AH = -^ 179-9 kJ
1123K
CaO (j) + Si02(s) ^ CaSi03 (.v)
Calcium silicate (slag)
6/14 ‘PnattUe^'4. New Course Chemistry (XII)EZsZSl

(iv) Zone of reduction. This is the upper pan of the furnace. The temperature here is between
500-900 K. Here, the ores are reduced to Fe by CO.

3Fe203(.?) + C0 ig) 573-673K 2 Fe304 (i) + CO, (^)

773-873K
Fe304(^) +4CO(^) > 3 Fe (j) + 4 CO, (5)
m-813K
Fe,O3(.0 + CO ig) > 2 FeO (j)-h CO, (g)
In the middle pan of the furnace, the temperature is between 900- 1500 K. At temperatures above 1073
K, reduction of FeO formed in the upper part of the furnace occurs to Fe by carbon.
>1073K

ow
FeO (5) + C (s) » Fe (.v/7) -f- CO (g)
Further, direct reduction of iron ores, (i.e. haematite, magnetite, etc.), left unreduced around 873 K,
occurs completely to iron by carbon above 1073 K.
>1073K
P^2^3 (-j) + 3 C (^) » 2 Fe is/l) + 3 CO (g)

e
(v) Zone effusion. This is the lower part of the furnace. Temperature here is in between 1423-1673 K.

Fl
re
In this region, spongy iron melts and dissolves some C, S*. P*, Si*, Mn, etc. CaSi03 slag also melts in this

F
region. Both the molten slag and the molten iron trickle down into hearth where they form two separate
layers. The molten CaSi03 slag being lighter forms the upper layer while molten iron being heavier forms the
ur
r
lower layer. The two liquids are periodically tapped off. The iron thus obtained from the furnace contains about

fo
4% carbon and many impurities (e.g., S, P, Si, Mn) in smaller amount. This is called pig iron and is cast into
variety of shapes. Cast iron is different from pig iron and is made by melting pig iron with scrap iron and
ks
coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.
Yo
oo

6.10.2. Preparation of Wrought Iron

B

Wrought iron is the purest form of commercial iron. It contains about 0-2—0-5% carbon besides traces
of P and Si in the form of slag. Carbon in wrought iron is present partly as graphite and partly as cementite,
e

Fe3C. Wrought iron is ductile, soft and malleable. The presence of slag in it makes it lough and resistant
ur

towards rusting and corrosion. It is, therefore, used to make chains, anchors, bolts, nails and railway carnage
ad

couplings. It can be easily magnetised and hence it is used to make magnets in electric cranes and dynamos.
Yo

Wrought iron is prepared from cast iron by decreasing the carbon content and oxidising the impurities
(S, P, Si, Mn, etc.) in a reverberatory furnace lined inside with haematite. The haematite oxidises C to CO, S
d
Re

to SO2, Si to SiO,, P to P4O10 and Mn to MnO.

in

Fc203 (j) + 3 C (s) -> 2 Fe (5) + 3 CO (g)

F

2 Fc203 (^) + 3 S (s) ^ 4 Fe (.0 + 3 SO2 (g)

CO and SO, thus formed escape whereas manganous oxide (MnO) and silica (Si02) combine to form
slag.
MnO(^) + Si02(.y) MnSi03 (0
Manganous silicate (.«/tig)
Similarly, phosphorus pentoxide combines with haematite to form ferric phosphate slag.
2Fe,03(i) + P40,„(g) 4
4FePO4(0
Ferric phosphate (slag)
Sometimes, limestone is added as a flux. This also helps to remove Si02 as CaSi03 slag.
After the reduction, the metal is removed from the furnace and is freed from the slag by passing through
rollers.

*S, P* and Si* are produced from sulphates, phosphates and silicates respectively present in the ore by
reduction with carbon.
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/15

6.10.3. Production of Steel from Pig Iron

Pig iron is converted into steel as follows:
(i) The carbon content from 3-4% present in pig iron is reduced to 0-05-1-5% in steel.
(ii) The impurities of S, P, Si, Mn, etc. present in pig iron are removed either by oxidizing them to volatile
oxides or through slag formation.
(Hi) Alloying elements such as Cr, Ni, Mn, V, Mo and W are added to obtain steel of desirable properties.
Although many processes such as Bessemer process, Thomas process, Siemens open hearth process, Seimens
electric arc furnace, etc. have been developed for production of steel but by far the most widely used process for
making steel today is Basic Oxygen Process (BOP). The furnace is charged with molten pig iron and lime and
pure O2 is blown over the surface of the metal at a great speed through water-cooled retractable lances. The O2
penetrates through the metal and oxidises the impurities rapidly.
2C + O2 ->2 CO 2 Fe + O2- ->2 FeO
2 FeO + Si 2 Fe + Si02 FeO + Mn - 4 Fe + MnO

FeO + SiO-) FeSi03 MnO + Si02 MnSi03

w
Slag Slag
P4 + 5 O2 ^ P4O10 6 CaO + P^OjQ —> 2 Ca3(PO^)'>

F lo
Slag
When all the impurities are removed, the required alloying elements are added to produce steel of desirable
properties. The advantages of using 0-> rather than air are :

e
Fre
(0 The process is fast and can produce about 300 tonnes of steel in about 40 minutes.
(ii) It gives purer product and the surface is free from nitrides.

6.10.4. Extraction of Copper from Cuprous Oxide [Copper (I) Oxide]

for
In the Ellingham diagram shown in Fig. 6.11, the FIGURE 6.11
r
(Cu, CU2O) curve is almost at the top while (C, CO) and
You
-100- 0
oks

(CO, CO2) lines lie much below it particularly in the -200- 6,Oi
^0%
rO^
eBo

o ■ - ^
temperature range 500-600 K. This means that it is very o
-300---, >0'
easy to reduce cuprous oxide to metallic copper by I
o
- oC’
C+O2 —^ CO2
heating with coke. But most of the ores of copper are E -400-
ad

O
our

●?c *0,
sulphides. Therefore, the sulphide ores (after -500-●
2 *Oa.
concentration by froth floatation process) are first roasted <
-600-. 1180 Kl 11270 K
in a reverberatory furnace to convert them into oxides.
I T I T I I
2 CU2S -H 3 O2 > 2 CU2O + 2 SO2
Re

273 K
dY

500 K 1000 K 1500 K 2000 K

The oxides can then be reduced to metallic copper Temperature

Fin

using coke as the reducing agent.

Ellingham diagram showing the formation
CU2O + C >2Cu + CO of Cu20 from Cu, ZnO from Zn, CO from C
However, in actual process, the sulphide ore (i.e., and CO2 from C and CO.
copper pyrites) after concentration by froth floatation
process is roasted in a reverberatory furnace when copper pyrites is converted into a mixture of FeS and CU2S
which, in turn, are partially oxidised.
2 CuFeS2 + O2 > CU2S H- 2 FeS + SO2
Copper pyrites
2 FeS -1-3 O2 2 FeO + 2 SO-,; 2 CU2S -t- 3 O2 2 CU2O + 2 SO2
Since iron is more reactive than copper, FeS is preferentially oxidised to FeO than CujS to CU2O. If at
all any Cu^O is formed, it combines with FeS and is changed back to CujS
CU2O -I- FeS ^ Cu-,S -I- FeO
Thus, the roasted ore mainly contains CU2S and FeO along with some unreacted FeS.

;
6/16 ‘P'uxcUe^'A New Course Chemistry (XlI)CZsISl
The roasted ore is then mixed with silica (Flux) and some powdered coke (to check the oxidation of FeO to
Fe203) and heated strongly in a blast furnace. This process is called smelting. During smelting, FeO combines
with silica to form fusible ferrous silicate slag.
FeO + Si02 FeSi03
Silica Ferrous silicate (slag)
At the temperature of the furnace, the entire mass melts and two layers of molten mass are fonned. The slag
being lighter makes the upper layer which can be withdrawn from the slag hole from time to time. The lower
molten layer is called copper matte. It chiefly consists of C112S and some unchanged FeS.
Recovery of Copper from Matte FIGURE 6.12

To obtain pure copper from matte, the molten matte is FLAME

transferred to a bessemer converter (Fig. 6.12) which is a

v\
HOTAIR
pear shaped furnace made up of steel and lined inside with &SAND
silica. It is mounted on a horizontal axel and can be tilted in
any position. It is fitted with small pipes called tuyeres EI\
MOLTE^
through which a blast of hot air and fine sand is admitted. MATTE

During the process of bessemerization, any sulphur,

w
MOLTEN
COPPER
arsenic and antimony still present as impurity in matte escape

F lo
as their respective volatile oxides while FeS is oxidised to
Bessemer converter for Copper
FeO which combines with silica to form FeSi03 slag.
2 FeS + 3 O2 ^ 2 FeO + 2 SO2 ; FeO + Si02 FeSiOj (slag)

ree
The slag thus formed melts and floats on the top of the molten mass and is removed. When whole of iron
for F
has been removed as slag, some of the cuprous sulphide undergoes oxidation to form cuprous oxide which
then reacts with cuprous sulphide to form copper metal.
2 CuoS + 3 Oo ^ 2 CU2O + 2 SO2 ; 2 CU2O + CU2S ^ 6 Cu + SO2
Copper metal so produced falls below tuyeres and thus escapes the oxidising action of the blast. After
Your
ks

the completion of the reaction, the converter is tilted and the molten metal is poured in sand moulds. As the
eBoo

metal cools, dissolved SO2 escapes. Some of the gas bubbles are, however, entrapped during solidification
giving blister like appearance to the metal. The impure metal (containing about \% impurity) thus obtained
is, therefore, called blister copper. Blister copper is finally purified by electrolytic refining as discussed in
ad
our

sec. 6.14, page 6/20.

6.10.5. Extraction of Zinc from Zinc Oxide
Re

Refer to Ellingham diagram as shown in Fig. 6.11, page 6/15. It is evident from the diagram that
Y

intersection of (Zn, ZnO) and (C, CO) curves lies at a higher temperature than that of (Cu, CU2O) and
Find

(C and CO) curves. Therefore, reduction of ZnO with coke is carried out at a higher temperature than that of
CU2O.
Further, all the three curves representing the oxidation of carbon [(i.e., (C, CO) ; (C, CO2) and (CO,
CO2)] lie above the oxidation curve of Zn till the boiling point of zinc (i.e., 1180 K) is reached. Therefore,
above 1180 K, A^G° for the formation of CO decreases while that for the formation of ZnO increases very
rapidly and intersects the (C, CO2) curve at 1270 K.
In other words, above 1270 K, A^G° for ZnO is higher than that of CO2 and CO from carbon. Therefore,
above 1270 K, A^G° for the reduction of ZnO by carbon is negative and hence ZnO is easily reduced by coke
above 1270 K. For the purpose of reduction, ZnO is made into brickettes with coke and clay and heated
above 1270 K usually around 1673 K so that the reduction process essentially goes to completion.
It may, however, be noted here that AyG° of CO2 from CO is always higher than that of ZnO. Therefore,
CO cannot be used for reduction of ZnO to Zn.

Since the boiling point of zinc is low (1180 K), the metal is distilled off and collected by rapid chilling.

\
 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/17

Curiosity Question
f Q. Usually a broken iron article is taken to the welding shop for repair. But it is difficult to take
broken girders or the cracked railway track to the industry for repair. Can the welding
equipment be brought to the site for repair ? Name the process, Illustrate the principle of its
working and name some other practical applications of this process.
Ans. The equipment used for on the spot welding is called FIGURE 6.13
aluminothermy. In this process, a mixture of aluminium Mg RIBBON
FLUORSPAR
powder and ferric oxide called thermite is ignited in a closed
crucible by inserting a burning magnesium ribbon into the Mg POWDER
ignition mixture consisting of magnesium powder and barium + Ba02
peroxide as shown in the Fig. 6.13. The ignition powder burns
Al POWDER
to produce a large amount of heat needed to initiate the
+ F6203

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reaction. Since A^G® of AI2O3 is much more negative than
that of FegOg (Fig. 6.8, page 6/9), therefore, A1 reduces FegOg
to metallic iron. PLUG

Flo
Aluminothennic process
2 Al (s) + FegOg (s) —> AI2O3 (s) + Fe {1}
This reaction is highly exothermic. As a result, Fe is produced in the molten state and thus can be

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used to weld heavy machinery of all kinds, such as girders, railway tracks, etc.

Fr
Besides Fe203, Al can also be used to reduce oxides of other metals such as CrgOg, Mn304, etc.
CfgOg + 2 Al > ^^2^3 ^ ® MngO^ + 8 Al > 4 AlgOg + 9 Mn
J
or
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f
In general, the process of reduction of a metal oxide to the metal with the help of aluminium powder
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as the reducing agent is called aluminothermy or Goldschmidt thermite process.
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6.11. ELECTROCHEMICAL PRINCIPLES OF METALLURGY

B

We have discussed above how the principles of thermodynamics have been useful
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(0 in the choice of a suitable reducing agent for a given metal oxide and also
(//) in the selection of an optimum temperature at which reduction will proceed smoothly.
u
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The same principles can also be applied in the reduction of metal ions in solution or molten state. These
reductions are usually carried out either by electrolysis or by adding a suitable element as the reducing agent.
The process of electrolysis has been used to carry out the reduction of molten metal salts. The electrochemical
nd
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principles of this method {i.e., electrolysis) can be understood in terms of the following equation,
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AG° = - ;i FE‘ ...(0

where n is the number of electrons involved in the reduction process, E° is the standard electrode potential of
the redox couple (M/M"'*') present in the system. More reactive metals have large negative values of electrode
potentials and hence are difficult to reduce. If the difference in E® values of two redox couples is positive, and
consequently the AG® in Eq. (/) is negative, then the more reactive metal will displace the less reactive metal
from the solution. In other words, less reactive metal will come out of the solution and the more reactive
metal will go into the solution. For example,
(aq) + Fe (5) ■> Cu (5) + Fe^-^ (aq)
More reactive metal Less reactive metal

In simple electrolysis, the M""*" ions are discharged at the cathode (negative electrode) and deposited there.
M"'*'(«^) + ne ^ M(.t)
Depending upon the reactivity of the metal produced, the materials of the electrodes are selected.
Sometimes, a flux is added for mtiking the molten mass more conducting.
Electrolytic method is used for the extraction of active metals like Na, Mg, Ca, Al, etc.
6/18 ^●tendec^ 'a. New Course Chemistry (XII) BQm

6.11.1. Applications of Electrolysis to Metallurgy

The process of extraction of metals by electrolysis of their fused salts is called electrometallurgy.
In this process, electrons serve as the reducing agent.

(a) Extraction of Aluminium from Alumina FIGURE 6.14

Fused alumina (Al-,03) is a bad conductor of Carbon Anode

electricity. Therefore, cryolite (Na3AlF(-,) and fluorspar Iron I P -Copper -r^ n- -

(CaFo) are added to purified alumina which not only Tank I Clamp ' ...Powdered
make alumina a good conductor of electricity but also I f T Coke

reduce the melting point of the mix to around 1140 K. Carbon lining
■ (Cathode)
This process of obtaining aluminium by : Molten AI2O3 + Na3AIFg
electrolysis of a mixture of purified Outlet for

alumina and cryolite is called Hall and /aluminium

Heroiilt process.
Molten aluminium

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The electrolysis of the molten mass is carried out Electrolytic cell for the extraction

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in an electrolytic cell or tank (Fig. 6.I4) made of iron of Aluminium from puriHed alumina
using carbon electrodes. The molten electrolyte
is covered with a layer of powdered coke to prevent oxidation and loss of heal due to radiation. The
temperature of the bath is maintained around 1173 K.

ree
The reactions taking place during the electrolysis are :
Cathode : (melt) + 3 e~ > Al (t) for F
Anode : C(.v) + 0^“ (melt) > CO (g) + 2e^; C(s) + 2 0^~ (melt) > CO2 (g) + 4 e~
For each kg of aluminium produced, 0-5 kg of carbon anode is burnt away. Because of this, the anodes
have to be replaced from time to time,
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at the high temperature (1173 K) of the tank. The

The aluminium metal liberated at the cathode melts
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molten metal being heavier than the molten electrolyte, sinks to the bottom of the tank from where it is
withdrawn periodically through the lapping hole. The metal obtained by this process is about 99-95% pure.
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(b) Extraction of Copper from Low Grade Ores and Scraps

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Copper is extracted by hydrometallurgy (as explained below) from low grade ores. The low grade ores
are leached by treating with an acid or bacteria when copper metal goes into solution as Cu^'*' ions. The
solution containing Cu^''' ions is then treated with scrap iron or H2 gas.
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Cu^"^ (aq) + Fe (s) > Cu (s) + Fe^"^ (aq) ; Cu^’*' (aq) + H2 (g)
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> Cu (s) -t- 2 (aq)

Find

Since E“ of F“’^/Fe (- 0-44 V) or that of H'^/H2 (0-0 V) redox couple is lower than that of Cu^VCu
(+ 0-34 V), therefore, Fe or H2 can displace Cu from Cu^"^ ions.

Curiosity Question
f Q. Iron articles usually rust within few years. But since 400 A.D. the Delhi iron pillar Near Kutab
Minar is existing without any rust or sign of decay. Explain how ?
Ans. Iron rusts due to the formation of hydrated ferric oxide (Fe203.xH20) which keeps on peeling off,
exposing fresh surface to air and moisture for rusting. On the other hand, magnetite (Fe304) is
another oxide of iron which sticks to the surface of metal for centuries. Thus. Delhi iron pillar does
not rust since a thin film of magnetite has been applied on its surface by applying a mixture of
different salts, heating and quenching.
J
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/19

6.12. EXTRACTION OF NON-METALS BY OXIDATION

So far we have discussed extraction of metals by reduction. But non-metals occur in the combined state
in the reduced form. Therefore, they are generally extracted or isolated by oxidation of their compounds.
For example,
Lsolation of chlorine. A very common example of extraction based on oxidation is the isolation of
chlorine from brine (chlorine is abundant in sea water as common salt).
2 Cl- («f7) + 2H20 (/) ^2 0H-(oq) + H2(g) + CU(g)
Since oxidation cannot be carried out by ordinary chemical methods, therefore, it is accomplished by
electrolysis. The AG® for this reaction is -i- 422 kJ. The minimum potential difference required for oxidation
can be calculated as follows : AG° = ~ n FE®

or 422 X 1000 J = - 2 X 96500 x E‘ or E® = -2-18 V = -2-2V

However, due to overvoltage*, actual electrolysis requires an external e.m.f. which is greater than 2-2 V.

w
During electrolysis, CI2 is liberated at the anode and H, at the cathode while NaOH is obtained In the
solution. In other words, H2and NaOH are the by products of this electrolysis. Electrolyis of molten NaCl

F lo
can also be carried out but in Uiis case Na metal is liberated at the cathode and CI2 at the anode.

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6.13. EXTRACTION OF METALS BOTH BY OXIDATION AND REDUCTION

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Extraction of gold and silver. We have already discussed on page 6/6 that extraction of gold and
silver involves leaching of' e metals present in the ore with CN- ions. This is also an oxidation reaction

for
because during the leachin, process, Ag is oxidised to Ag* and Au to Au* which then combine with CN" ions
ur
to form their respective soluble complexes.
4 Ag is) + 8 CN- iaq) + 2 H2O (/) + Oj (g) ^4[Ag(CN)2]-(n^) + 4 0H-(fl^?)
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Silver Soluble complex

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4 Au is) + 8 CN" («r/) + 2 H.O (/) + Oj ig) ^ 4 [Au(CN)2r iaq) + 4 OH" iaq)
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Gold Soluble complex

The metals are then recovered from these complexes by reduction or displacement method using a
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more electropositive zinc metal.

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2[Ag(CN)2l-(^r^) + Zn is) ^2Agis)+ [Zn(CN)4]2-(fl^)

2 [Au(CN)2]- iaq) + Zn (5) ^ 2 Au (i) + [Zn(CN)4l^“ iaq)
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In these displacement reactions, zinc acts as the reducing agent.

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nd

From the above discussion, we conclude that like extraction of copper from low grade copper ores, gold
and silver are also extracted from these ores by hydrometallurgy. Thus,
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The process of extraction of metals by dissolving the ore in a suitable reagent followed by
precipitation or displacement of the metal by a more reactive or more electropositive metal is
called hydrometallurgy.

6.14. REFINING

The metals obtained by any of the methods discussed above are still impure and hence are called crude
metals. The impurities generally present in the crude metals are : (/) Other metals—produced by simultaneous
reduction of their respective oxides originally present in the ore. iii) Non-metals—like silicon and phosphorus
formed by reduction in the furnace, iiii) Unreacted oxides and sulphides of the metals, and (/V) Substances
taken up in the furnace, e.g., residual slag, flux, etc.
*Liberation of gases requires some extra voltage than the theoretical value of standard electrode potential.
This extra voltage required is called overvoltage or bubble voltage.
6/20 “pputdee^'^. New Course Chemistry (XIl)CZsISI

The crude metals are, therefore, purified or refined. The method actually used for purification depends
upon the nature of the metal and the nature of the impurities to be removed.
The process of purifying the crude metals is called refining.
Some common methods used for refining of metals are discussed below ;
1. Distillation Process. This method is employed for purification of volatile metals like zinc, mercury,
cadmium, etc. The impure metal is heated in an iron retort and the vapours are condensed in separate receivers.
The pure metal distils over leaving behind the non-volatile (having higher boiling points) impurities in the
retort.
FIGURE 6.15]
2. Liquation Process. This method is used for IMPURITIES
purification of such metals as contain impurities which are
less fusible than the metals them.selves, i.e., the melting points
#
of the metals are lower than those of the impurities. The IMPURE
METAL
crude metal is heated in an inert atmosphere of carbon /A/G PURE

monoxide on the sloping hearth of a reverberatory furnace METAL

(Fig. 6.15) when the metal melts and flows down into the
receiver leaving the infusible impurities on the hearth. Metals

F low
Liquation process
like tin and lead are purified by this method.
3. Electrolytic refining. A large number of metals such as copper, silver, gold, lead, nickel, chromium,
zinc, aluminium, etc., are refined by this method. FIGURE 6.16
In this method, the impure metal is converted into Bf TERY
a block which forms the anode while cathode is made

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PURE

up of a strip of the same metal (Fig. 6.16) in the pure

for Fr
impure;
II- COPPER
AS
form. These electrodes are suspended in an electrolyte COPPERj
AS
CATHODE
which is the solution of a soluble salt of the metal ANODE

usually a double salt of the metal. When electric current

is passed, metal ions from the electrolyte are deposited
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at the cathode in the form of pure metal while an ACIDIFIED

I I - COPPER
equivalent amount of metal dissolves from the anode ANODE
SULPHATE
and goes into the electrolyte solution as metal ions, i.e., MUD
SOLUTION
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Anode : M(5) ■> (ag) + ne~

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Electrolytic-refining of copper
Cathode : (aq) + ne ^ M (s)
The net result of electrolytic refining is the transfer of pure metal from the anode to the cathode. The
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voltage applied for electrolysis is such that the impurities of more basic metals (more electropositive metals)
remain in the solution as ions whereas impurities of the less basic metals (less electropositive metals) settle
Find Y

down under the anode as anode mud or anode sludge.

In case of electrolytic refining of copper, crude copper metal, i.e., blister copper is made the anode, a
thin sheet of pure copper is made the cathode while copper sulphate solution acidified with sulphuric acid is
taken as the electrolyte. The net result of electrolysis is the transfer of copper in pure form from the anode to
the cathode.
Anode : Cu (j) ■> Cu-‘*‘ (aq) + 2 e~
Cathode : Cu^-^ (aq) + 2e~ -> Cu (i)
The impurities of iron, nickel, zinc and cobalt present in blister copper being more electropositive pass
into solution as soluble sulphates while the impurities of antimony, selenium, tellurium, silver, gold and
platinum being less electropositive are not affected by CUSO4- H^S04 solution and hence settle down under
the anode as anode mud or anode sludge. The recovery of precious metals such as silver, gold and platinum
from the anode mud more than compensates for the high cost of electrolyticrefining.Copperthus obtained is
99-95-99-99% pure.
Zinc can also be refined by this method.
 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/21

4. Zone Refining (Fractional Crystallization)

This method is extremely useful for producing metals of very high purity, e.g., germanium, silicon,
boron, gallium and indium, etc.
The method is based upon the principle that the impurities are more soluble in the molten state
(melt) than in the solid state of the metal.
In other words, when the melt of an impure metal is allowed to cool, the pure metal crystallises out
while the impurities remain in the melt.
In this method, the impure FIGURE 6.17
metal is converted into a bar
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which is heated at one end with
a moving circular heater (Fig. NOBLE GAS ATMOSPHERE
6.17) so that this end melts and
forms a molten zone or the IMPURE METAL
.i

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melt. As the heater is slowly ^TXX^ >●
MOLTEN ZONE
moved along the length of the METAL ROD
CONTAINING INDUCTION c01L°IRECTION OF MOVEMENT

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rod, the pure metal crystallises IMPURITIES MOVING HEATERS OF MOLTEN ZONE AND
IMPURITIES
out of the melt whereas the
Zone-refining of Metals
impurities pass into the adjacent

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molten zone. This process is repeated several times till the impurities are completely driven to one end of the

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rod which is then cut off and discarded. The process is usually carried out in an inert atmosphere to prevent
the oxidation of the metal. Semiconductors like silicon, germanium and gallium are purified by this method.
for
The elements thus obtained are of high purity. For example, germanium prepared by this method contains as
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little as 7-10 ppm of impurities.
5. Vapour phase refining. In this method, the crude metal is freed from impurities by first converting
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it into a suitable volatile compound by heating it with a specific reagent at a lower temperature and then.
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decomposing the volatile compound at some higher temperature to give the pure metal. Thus, the two
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requirements are :
(i) the metal should form a volatile compound with a suitable reagent,
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(ii) the volatile compound should be easily decomposable so that the recovery is easy.
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This method is illustrated by the following two processes :

(a) Mond process. It is used for refining of nickel. When impure nickel is heated in a current of CO at
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330-350 K, it forms volatile nickel tetracarbonyl complex leaving behind the impurities. The nickel
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tetracarbonyl thus obtained is then heated to a higher temperature(450-470 K) when it undergoes thermal
decomposition giving pure nickel.
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330-350K 450-470K
Ni(r) +4CO(g) Ni(CO)4(g) 4 Ni(^) +4CO(g)
Impure nickel Nickel tetracarbonyl Pure nickel

(b) van Arkel method. This method is very useful for preparing ultra-pure metals by removing all the
oxygen and nitrogen present in the form of impurity in certain metals such as zirconium and titanium which
are used in space technology. In this method, crude zirconium is heated in an evacuated vessel with iodine at
870 K. The covalent volatile z conium tetraiodide thus formed is separated. It is then decomposed by heating
over a tungsten filament at 2075 K to give pure zirconium.
870 K 2075 K
Zr (s) + 2 I2 (g) ^ Zrl4(g) Tungsten filament
^ Zr(5) +212(g)
Impure zirconium Pure zirconium

523 K 1700K

Similarly, Ti(^) + 212(5) ^ Til4(g) ■>

Ti(5) + 212(g)
Impure titanium Pure titanium

{
6/22 New Course Chemistry (X11)B&!9]

6. Chromatographic methods. Chromatography is the most modem and versatile method for sepjiration,
purification and testing the purity of elements and their compounds.
The method is based upon the principle that the different components of a mixture are adsorbed
to different extents on an adsorbent
Cliromatography is essentially a physical technique. It consists of two phases, tlie stationiuy phase and
the mobile phase. The stationary phase can be eitlier a solid {such as alumina, silica gel) or tightly bound
liquid on a solid support (such as paper in which the liquid water is held by the solid cellulose). The mobile
phase, on the other hand, can be a liquid, gas or a super critical fluid such as CO2. Depending upon the
physical state of the stationary phase and the moving phase and also on the process of passage of the moving
phase, chromatography can be of several types such as column chromatography, partition chromatography,
gas chromatography, etc.
Column chromatography. Column chromatography is one of the simplest chromatographic techniques
and is widely used.
In column chromatography,an adsorbent such as alumina (AUO3), silica gel or some ion exchange resin

w
is packed in a column (Fig. 6.18). This forms the stationary phase. The mixture to be separated is dissolved in

F lo
a suitable solvent (mobile phase) and applied to the top of the column.
Differentcomponents ofthe mixture are adsorbed to different extents depending upon their polarity.

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Later the adsorbed

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FIGURE 6.18
components are extracted
(eluted) from the column with
a suitable solvent (eluent).
The component which is
for FLOWMETER

DETECTOR—'
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more strongly adsorbed on
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the column takes longer time
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to travel through the column

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COLUMN
than a component which is SOLVENT — SAND
weakly adsorbed. Thus, the {MOBILE PHASE)
various components of the MIXTURE OF-^
our

A OVEN
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COMPOUNDS ADSORBENT >-

mixture are separated as they (A+B+O+SAND B
travel through the adsorbent ADSORBENT-

(stationary phase). (STATIONARY C -«

PHASE)
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Suppose a mixture of
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GLASS WOOL

three different components :0

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(A, B and C) is applied to the

column for separation. SOLVENT
FLOW INJECTOR
TANK
Further, suppose that l_l
component A is most strongly o PUMP 0
LABORATORY METHOD INDUSTRIAL METHOD
adsorbed, compo-nent B is
moderately adsorbed while Apparatus used in column chromatography-
component C is only weakly (a) Laboratory method (b) Industrial method
adsorbed. Upon elution, the
three components begin to separate and form three different coloured bands (A, B and C as shown in Fig.
6.18a) if the mixture is coloured. The component C which is weakly adsorbed travels fastest down the column,
followed by component B while component A is eluted last of all. These are collected in different flasks.
Evaporation of the solvent from each flask gives the desired component.
If. however, the mixture is colourless, the column is extracted with a suitable solvent or a mixture of
solvents and several small fractions (20-25 cm-^ volume) are collected in different flasks. With the help of
suitable physical or chemical methods, the flasks containing the same component are identified and mixed

i
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/23

together. Evaporation of the solvent from the fractions thus combined gives the desired component. In this
way, the different components of the mixture tue separated.
This technkjue is especially suitable for such elements which are available only in minute quantities
and the impurities are not very much different in chemical properties from the elements to be puriifed.
Lanthanoids (rare earth elements) are purified by this technique using ion-exchange as the adsorbent.
The industrial method of chromatography is shown in Fig. 6.18/?.
6.15. USES OF ALUMINIUM, COPPER, ZINC AND IRON
(a) Uses of Aluminium
(/) Aluminium foils are used for wrapping fine articles like photographic films, pharmaceutical products,
cigarettes, chocolates, sweets, etc.

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(//) Fine dust of aluminium is used for making aluminium paints and lacquers. For example, aluminium
powder mixed with linseed oil shines like silver and is called silver paint.
(Hi) Aluminium powder, being highly reactive, is used as a reducing agent in aluminothemiic process for
the extraction of chromium and manganese from their oxides,

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(iv) Being light and good conductor of electricity (on weight for weight basis, A1 conducts twice as Cu),

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aluminium is used for making transmission cables and for winding the moving coils of dynamos or
motors.

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(V) Aluminium is a cheap metal which resists corrosion. Therefore, it is used for making household utensils,
cans for drinks, tubes for tooth-paste, picture frames, trays, etc. It is used in buildings for making

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angles for doors, windows, etc.
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(vi) Aluminium powder is used for flash light bulbs in indoor photography.
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(vii) Alloys of aluminium, being light, are very useful as listed below :
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Alloy Composition Important properties Uses

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1. Aluminium bronze A1 95%, Cu 5% Light strong alloy with Coins, utensils, jewellery,
golden lustre, resistant picture frames, etc.
to corrosion
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2. Magnalium A1 95% Mg 5% Light, tough and strong Light instruments, balance

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beams, pressure cookers, etc.

3. Duralumin A1 95% Cu 4% Light, tough, ductile, Making aeroplanes,
d

Mg 0-5% Mn 0-5% resistant to corrosive automobile parts, pressure

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in

action cookers, etc.

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(b) Uses of Copper

(/) Copper being a good conductor of electricity is extensively used for making electric cables and other
electrical appliances.
(ii) Copper being a good conductor of heat is used for making utensils, steam pipes, kettles, evaporating.
pans, calorimeters, etc.
(///) It is used in several alloys which are even tougher than the metal itself. Some important alloys are :

Alloy Composition Uses

1. Brass Cu 60%, Zn 40% For making utensils, condenser tubes, wires,

parts of machinery, etc.
2. Bronze Cu 80%, Zn 10%, Sn 10% For making cooking utensils, statues, coins,
etc.

3. German silver Cu 25-30%. Zn 25-30%, For making silver wires, resistance wires.
Ni 40-50% etc.
6/24 New Course Chemistry CX1I)E!SIS]

(c) Uses of Zinc

(/) Zinc is used for galvanising iron to protect it from corrosion.
(«) Zinc plates and rods are used in batteries and dry cells.
(Hi) Zn dust is used as a reducing agent in the manufacture of dye-stuffs, pai.its, extraction of gold and
silver by the cyanide process, etc.
(?v) Zinc is used in making some useful alloys as mentioned under uses of copper listed above.
(d) Uses of Iron
(0 Cast iron, which is the most important form of iron, is used for casting stoves, railway sleepers, gutter
pipes, toys, etc.
(ii) It is used in the manufacture of wrought iron and steel. Wrought iron being tough and resistant to
rusting, is used for making anchors, wires, bolts, nails, railway carriage couplings, and agricultural

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implements.
(Hi) Steel finds number of uses. Alloy steel is obtained when other metals are added to it. Some alloy steels

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and their composition, properties and uses are listed below :

Aiioy steei Composition Properties Uses

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1. Stainless steel Fe73%, Crl8%, Resists corrosion For making utensils, cutlery,

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Ni 8% and carbon 1% cycles, automobiles, pens.
etc.

2. Nickel steel Fe 96-98%, Ni 2-4%

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Resistant to corrosion, For making cables,
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hard and elastic automobiles and
aeroplane parts.
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3. Chrome steel Fe 96-98%, Cr 2-4% High tensile strength For making axles, ball
bearings, cutting tools and
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crushing machines.
4. Invar Fe 64%, Ni 36% Practically no coefficient For making watches, meter
of expansion scales, pendulum rods, etc.
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5. Alnico Fe 60%, A1 12%, Highly magnetic For making permanent

Ni 20%, Co 8% magnets
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1. Minerals. The natural materials in which the metals occur in the earth are called minerals.
2.
Ore. The mineral from which the metal is conveniently and economically extracted is called an ore.
3. All ores are minerals but all minerals are not ores, e.g., haematite, siderite and iron pyrites are minerals of
iron but only haematite is an ore of iron. Similarly, bauxite and clays are minerals of aluminium but only
bauxite is an ore of aluminium.
4.
Gangue or Matrix. The unwanted earthy and silicons impurities, e.g., sand, clay, rock, etc. associated with
the ore is called gangue or matrix.
5.
Ore dressing. The removal of gangue or matrix from the ore is called ore dressing or concentration of
ores.

6.
The most common ores are oxides, carbonates, sulphides, silicates, sulphates and halides of the elements.
7.
Benefaction. The process used to concentrate an ore is called benefaction process, e.g., levigation, froth
floatation, magnetic separation, electrostatic separation, leaching, etc.
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION 0*= ELEMENTS 6/25

8.
Levigation. The process of removing lighter earthy particles from the heavier ore particles by washing with
water in a hydraulic classifier or Wilfley table is called levigation or gravity separation.
For example, the oxide ores of iron {haematite) and tin (tinstone) and native ores of Ag, Au, etc. are
concentrated by levigation.
9. Froth floatation is widely used for the concentration of sulphide ores, i.e., zinc blende (ZnS), copper
pyrites (CuFeS2>, galena (PbS), etc. It is based upon preferential wetting of ore particles by oil (pine oil) and
gangue particles by water.
10. Electromagnetic separation. This method of concentration is employed when either the ore or the impurities
associated with it are magnetic in nature. For example, chromite (magnetic) - an ore of chromium is separated
from non-magnetic silicious impurities. Likewise tin stone or casseterite (non-magnetic), an ore of tin is
separated from magnetic impurities of tungstates of iron and manganese by this method.
11.

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Electrostatic separation. This method is used for the separation of ores which are good conductors of
electricity (/.e., PbS) from ores which are poor conductors of electricity (Le., ZnS) when these ores occur
together in nature. The process of levigation is based upon the difference in specific gravities of the ore and

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the gangue particles.
12. Leaching involves treatment of an ore with a suitable reagent which can selectively dissolve the ore while

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impurities remain insoluble. For example, bauxite which contains impurities of Fe203 and silica is treated

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with a hot solution of NaOH which dissolves alumina as sodium meta-aluminate leaving behind the impurities
(Si02) which are filtered off.
13. Extraction. The process used to obtain metals in the ftee state from concentrated ores is called extraction.

for
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The extraction of metals from concentrated ores actually involves the following two steps.
(0 Conversion of the ore into metal oxide, i.e., de-electronatio n of ore.
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(ii) Reduction or electronation of the metal oxide to the free metal.
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14. Conversion of the concentrated ore into metal oxide involves the following two processes,
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(a) Calcination. This process is used to convert the carbonate and hydrated oxide ores to their respective
oxides by heating strongly below their melting points either in absence or in limited supply of air. Non-
metallic impurities of S, P, As, etc. are removed as their volatile oxides.
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(b) Roasting. This process is used to convert sulphide ores into their respective oxides by heating strongly
in presence of excess of air. Non-metallic impurities of As, S and P are removed as their volatile oxides.
Y

15. Reduction. Oxide ores obtained after calcination and roasting are reduced to free metals by different reducing
agents such as C, CO or even by some active metals such as Na, K, Mg, Al, etc.
nd
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16. Choice of reducing agents. The choice of a suitable reducing agent is made on the basis of thermodynamic
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principles. All those metals which have more negative Gibbs energies of formation of their oxides can
reduce the oxides of other metals whose Gibbs energies of formation are less negative.
17.
EUingham diagrams. The plots of variation of AyG® vs T for the formation of metal oxides from metals are
called EUingham diagrams.
18.
EUingham diagrams can be used to determine the optimum temperature at which a particular metal can
reduce the metal oxide of other metal. For example, below 1623 K, (Fig. 6.8, page 6/9), Mg can reduce
AI2O3 to Al but above 1623 K, it is Al which can reduce MgO to Mg.
19.
Choice of better reducing agents. EUingham diagrams can also be used to determine which of the two is a
better reducing agent at a particular temperature. For example, below 983 K, CO is a more effective
reducing agent than C but above 983 K, it is C which is a better reducing agent than CO (refer to Fig. 6.19,
page 6/47).
20. Smelting is the process of extraction of a metal from its oxide by reduction with carbon (coke), e.g., Zn
from ZnO, Sn from Sn02, Pb from PbO and Fe from Fe203. Alkali metals, alkaline earth metals, aluminium,
etc. cannot be prepared by reduction of their respective oxides with carbon.
6/26 7^h4x<Ue^’4. New Course Chemistry (XII)BSm
21. Flux is a substance which combines with gangue which may still be present in the calcined or roasted ore to
form an easily fusible material called the slag
Flux + Gangue > Slag (fusible)
22. Acidic flux. When the ore is contaminated with basicimpurities like those of FeO, MgO, CaO, MnO, acidic
fluxes such as silica (Si02) and borax (Na2B4O7.10 H2O) are used
FeO + Si02 > FeSi03 (slog)
23. Basic fluxes. When the ore is contaminated with acidic impurities like those of Si02, P4O10, etc. basic
fluxes like CaO and MgO are used.
SiO-) + CaO 4 CaSi03 (slag)
24. Alumino-thermiteprocess. It involves reduction of metal oxides such as Fc203, Cr203, Mn304, etc. to

ow
metals with aluminium. A mixture of metal oxide and aluminium powder is called thermite while a mixture
of magnesium powder and BaO.> is called ignition mixture.
25. Auto-reduction. The sulphide ores of certain less electropositiv e metals like those of Hg, Pb, Cu, etc. when
heated in air bring about the conversionof a part of the sulphide ore into oxide or sulphate which then reacts
with the remaining sulphide to give the metal and SO2. Ao external reducing agent is used in this process.

e
Fl
re
2 Cu-)S 4- 3 O') ■> 2 CU2O + 2 SO2
2 Cu,0 -1- Cu->S 4 6 Cu + SOt

F
26. Electrometallurgy. It is the process of extraction of metals by electrolysis of their fused salts. Active metals
ur
such as alkali metals, alkaline earth metals, aluminium, etc. are obtained by this method. For example, in

or
Hall and Heroult process. A1 is obtained by electrolysis of a mixture of fused purified alumina and cryolite.
sf
Similarly, in Down’s process, sodium is obtained by electrolysis of fused NaCl.
Hydrometallurgy. It is the process of dissolving the ore in a suitable chemical reagent followed by extraction
k
27.
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of the metal either by electrolysis or precipitation of the metal by a more electropositive metal. Ag and Au
oo

are extracted by this process.

B

4 Ag + 8 NaOH + O2 + 2 H^O ^ 4 Na [Ag(CN>2] + 4 NaOH

4 Na2lZn(CN)4] + 2 Ag i
e

2 Na[Ag(CN)2l + Zn
ur

28. Refining. The process of purifying the crude metals is called refining.
ad

29. Liquation. Metals like Sn and Pb are purified by this process since the melting points of these metals are
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lower than those of the impurities.

30. Distillation is used for purificationof volatile metals like Zn, Cd and Hg.
d
Re

31. Electrolytic refining. Metals like Cu, Ag. Au, Ni, Cr, Al, etc. are purified by this method. In this method,
in

the impure metal is made the anode, a strip of pure metal is made the cathode while electrolyte consists of a
F

solution of a soluble salt of die metal.

32. Zone refining involves the purification of metals by fractional crystallization. Semiconductors like silicon,
gemianium and gallium are purified by this method.
33. Van Arkel method. Titanium and zirconium are purified by this method.
500 K 1700 K

Ti(s) 212(5) ^ Til4(g) > Ti(^) + 212(g)

Impure Pure

34. Chromatography. It can be used for separation of such elements which are available only in minute
quantities and the impurities as.wciated with them are not vety much different in properties from the elements
to be puriifed. For example, lanthanoids are purified by chromatographic technique using ion exchange as
adsorbent.
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/27

- I X'.

ISI

ia
to
NEET/JEE
SPECIAU

For ultimate preparation of this unit for competitive examinations, students should refer to

w
● MCQs in Chemistry for MEET
Pradeep's Stellar Series. w MCQs in Chemistry for JEE (Main)
separately available for these examinations.

F lo
D Multiple Choice Questions (with one correct Answer)

ee
Fr
I. Occurrence and
5. Which of the following ore is concentratedusing
group 1 cyanide salt ?
concentration of Metals
(fl) Sphalerite
for (b) Malachite
ur
1. Identify the incorrect statement (c) Calamine (d) Siderite
(a) Gangue is an ore contaminated with undesired (JEE Main 2021)
s
materials
ook
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6. Sulphide ores of metals are usually concentrated
(b) The scientific and technological process used by Froth Floatation process. Which one of the
for isolation of the metal from its ore is known
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following sulphide ores offers an exception and

as metallurgy
is concentrated by chemical leaching ?
(c) Materials are naturally occurring chemical
(a) Sphalerite (b) Argentite
r

substances in the earth’s cnist

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(c) Galena (d) Copper pyrite

(d) Ores arc minerals that may contain a metal
(AIPMT 2007)
{NEET 2021)
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7. In the cyanide extraction process of silver from

2. Which one is malachite from the following ;
argentite ore, the oxidising and reducing agents
Re
nd

(fl) CuC03.Cu(0H)2 (h) CuFeS2 used are

(f) Cu(OH)2 (r/)Fe304 (NEET 2019) (a) O2 and CO respectively
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3. Calamine, malachite, magnetite and cryolite (b) O2 and Zn dust respectively

respectively are
(c) HNO3 and Zn dust respectively
ia) ZnS04, Cu(OH)2, Fe304, Na3AlF^, (d) HNO3 and CO respectively (IIT 2012)
(b) ZnC03, CUCO3. Fe^Oj, 8. Which of the reactions is suitable for concentrating
(c) ZnS04, CUCO3, Fc203, AIF3 ore by leaching process.
(d) ZnCOi. CUCO3, Cu(OH>2, Fe304, Na3AlF^ (a) 2 CU2S + O2 ^ 2 CU2O + 2 SO2
(JEE Advanced 2019) (.b) F6304 + CO ■> 3 FeO + CO2
4. Sulphide ores arc common for metals (c) AI2O3 + 2 NaOH + 3 H2O ^2Na[Al(OH)J
(a) Ag, Cu and Pb (h) Ag, Cu and Sn (d) AI2O3 + 6 Mg 4 6 MgO + 4 Al
(c) Ag, Mg and Pb id) Al, Cu and Pb (JEE Main 2022)

l.(w) 2. (a) 3. id) 4. («) S.ia) 6. (/;) 1. ih) 8. (c)

6/28 'PnauUep.’A New Course Chemistry (XII)B3Siai

9. AUO3 was leached with alkali to get X. The

solution of X on passing of gas Y, forms Z, X. Y -600-

and Z respectively are -800-

(fl) X = AI(0H)3, Y = SO2, Z = AI2O3..V H,0 L

o -1000- A + O2 —^ AO2
(/;) Na[Al(OH)4|, Y = SO2, Z = AI2O3
?O -1200-
(c) Na[Al(OH)4], Y = CO2, Z = Al203^ H2O <3
B + ©2 —* BO2

(i/) X = A1(0H)3, Y = CO2, Z = AljOg 200

T
400
T
600 800 1000 1200
T
1400 1600
(JEE Main 2021)
T(»C)
n. Extraction of Metals (a) < I200°C
{b)> 1200°Cbut< 1400X
10. The pair that does not require calcination, is
(c) < 1400'’C
ia) ZnO and Fc203.a: H-^O
id) > I400“C (JEE Main 2020)

w
ib) ZnC03 and CaO
16. For a reaction, 4 M (5) -i- n O2 > 2 M20„ (5)
(c) ZnO and MgO
the free energy change is plotted as a function of

F lo
id) Fe203 and CaC03.MgC03 temperature. The temperature below which the
(JEE Main 2019) oxide is stable could be infeired from as the point
11. The process that involves the removal of sulphur at which

ee
from the ores is
(a) free energy change shows a change from

Fr
(a) leaching ib) smelting negative to positive value
(c) refining id) roasting (h) the slope changes from positive to negative
(JEE Main 2021) for
(c) the slope changes from negative to positive
ur
12. Ellingham diagram is a graphical representation (d) the slope changes from positive to zero
of
(JEE Main 2020)
s
ook

ia) AG vsT ib) (AG-TAS) vsT

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17. Which of the following reduction reaction cannot

(c) AHvsT id) AG V.9 P be carried out with coke ?
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(JEE Main 2021) (n) Fe203 > Fe (b) ZnO-Zn

13. AG° vs T plot in the Ellingham’s diagram slopes (c) AI2O3 > A1 id) CU2O > Cu
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downward for the reaction (JEE Main 2021)

ad

18. Among the reactions A-D, the reaction(s) that

(a) Mg + -02- MgO does/do not occur in the blast furnace during the
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extraction of iron is/are

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(^) 2Ag-Hi02 ^ Ag20 I. CaO -t- SiO-> > CaSi03

nd

II. 3 Fe203 + CO > 2 Fe304 + CO2

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III. FeO -I- SiO-> > FeSi03

(c) C + -0.— 4 CO
2 2 IV. FeO >Fe+I/2 O2
(a) (in) and aV) (b) (I)
1
(d) CO+ 2.0, 4 CO, (AIIMS 2015) (c) (I) and (IV) id) (IV) (JEE Main 2020)
2 ^
19. Identify the correct statement from the following :
14. Considering Ellingham diagram, which of the
ia) Blister copper has blistered appearance due
following metals can be used to reduce alumina ?
to evolution of CO2
(n) Fe ib) Zn
ih) Vapour phase refining is carried out for nickel
(c) Mg (^0 Cu (NEET 2018) by van Arkcl method
15. According to the following diagram, A reduces (c) Pig iron can be moulded into a variety of
BO2 when the temperature is shapes
ANSWERS

9.1c ) 10. (c) 11. (rf) 12. (n) 13. (r) 14. (c) IS. id) 16. (a) 17. (c) 18. (a)
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/29

id) Wrought iron is impure iron with 4% carbon III. Refining of Metals
(NEET 2020. Phase 2)
20. In the extraction of copper from its sulphide ore, 24. The method of zone refining of metals is based
the metal is finally obtained by the reduction of on the principle of
cuprous oxide with {a) greater mobility of the pure metal than that of
(a) carbon monoxide (/?) copper (I) .sulphide the impurity.
(c) sulphur dioxide (d) iron (II) sulphide (b) higher melting point of the impurity than that
(RE-AIPMT 2015) of the pure metal,
21. In the metallurgical extraction of copper following (c) greater noble character of the solid metal than
reaction is used : that of the impurity.
FeO + SiO^ FeSi03 (d) greater solubility of the impurity in the molten
FeO and FeSi03 respectively are slate than in the solid.

(a) Gangue and flux (b) Flux and slag 25. The refining method used when the metal and the
(c) Slag and flux (d) Gangue and slag impurities have low and high melting temperatures
(JEE Main 2022) respectively, is

w
22. In the Hall-Heroult process, aluminium is formed (a) distillation (b) liquation

F lo
at the cathode. The cathode is made out of
(c) zone refining (d) vapour phase refining
(a) platinum (b) pure aluminium (JEE Main 2020)
(c) copper (<●/) carbon 26. Which of the following methods can be used to
(JEE Main 2019) obtain highly pure metal which is liquid at room

e
Fre
23. In the context of the Hall-Heroult process for the temperature ?
extraction of Al, which of the following statements (a) Zone refining (b) Electrolysis
is false ?
(a) aP"*" is reduced at the cathode to form Al
for
(c) Chromatography (d) Distillation
(NEET 2021)
(b) Na3AlFg serves as the electrolyte
r
27. The metal that can be purified economically by
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(c) CO and CO2 are produced in this process
oks

fractional distillation is
(d) AI2O3 is mixed with Cap2 which lowers the
(a)Fe (b) Ni
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melting point of the mixture and brings

conductivity (c) Cu id) Zn (JEE Main 2021)

01 Multiple Choice Questions (with One or More than One Correct Answers)
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28. Copper is purified by electrolytic refining of blister 30. Which of the following statement(s) is (are) true
copper. The correct statement(s) about this process for the extraction of aluminium from bauxite ?
dY

is (are) (a) Hydrated alumina precipitates, when CO2 is

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(a) impure Cu strip is used as cathode bubbled through a solution of sodium

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aluminate
ib) acidified CUSO4 is used as electrolyte
(c) pure Cu deposits at cathode (b) Addition of Na3AlFg lowers the melting point
of alumina
((/) impurities settle as anode mud
(JEE Advanced 2015)
(c) CO-, is evolved at the anode during electrolysis
(d) The cathode is a steel vessel with a lining of
29. The cyanide process of gold extraction involves carbon (JEE Advanced 2020)
leaching out gold from its ore with CN“ in the
31. The electrochemical extraction of alumina from
presence of Q in water to form R. Subsequently R bauxite ore involves
is treated with T to obtain Au and Z. Choose the
correct options (a) the reaction of AI2O3 with coke (C) at a
temperature > 2500°C
(a) R is [Au(CN)4]- (b) T is Zn
(b) the neutralization of aluminate solution by
(c) Q is O2 (d) Z is [Zn(CN)4]2- passing CO2 gas to precipitate hydrated
(JEE Advanced 2019) alumina (AI2O3.3 H-^O)
ANSWERS
19.(c) 20. (ft) 21.fr/) 22.id) 23. (ft) 24. (d) 25. (ft)
26. id) 27. (d) 28. (h.c.d) 29. (h.c.d) 30. (o.h.c.d)
6/30 ‘P'uuCee^’a^ New Course Chemistry (XlI)ESSm

(c) the dissolution of alumina in hot aqueous (/;) in the extraction process of copper from
NaOH copper pyrites, silica is added to produce
copper silicate
id) the electrolysis of AI2O3 mixed with Na^AlF^
(c) partial oxidation of sulphide ore of copper by
to give A1 and CO2 ' (JEE Advanced 2022) roasting followed by self-reductionproduces
32. The correct statement(s) related to metal extraction blister copper
process is (are) {d) in cyanide process, zinc powder is utilized to
(a) a mixture of PbS and PbO undergoes self precipitate gold from Na[Au(CN)2l.
reduction to produce Pb and SO2 (JEE Advanced 2021)

[in Multiple Choice Questions (Based on the given Passage/Comprehens ion)

The comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option.

w
Copper is extracted from
tgdmpreii^n^onlF] Copper is the most noble
copper pyrites. After roasting, the ore is mixed of first row transition metals and occurs in

F lo
with silica and coke and then smelted in a blast small deposits in several countries. Ores of
furnace. The matte obtained from the blast copper include chalcanthite (CUSO4.5 H2O),
furnace is charged into a silica lined converter. atacamite (Cu2Cl(OH)3), cuprite (CU2O),

ee
Some silica is also added and a hot air blast is copper glance (CU2S) and malachite

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blown into the mixture to obtain blister copper (Cu2(0H)2C03). However, 80% of the world
which is purified by electrorefining. copper production comes from the ore

33. Coke is added during smelting

for
chalcopyrite (CuFeS2)* The extraction of
copper from chalcopyrite involves partial
ur
(a) to reduce FeO to Fe roasting, removal of iron and self-reduction.
s
(h) to reduce CuiO to Cu (UT Paper I, 2010)
ook
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(c) to check the oxidation of FeO to Fe->03 35. Partial roasting of chalcopyrite produces
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(d) to check the oxidation of CU2O to CuO

(a) CU2S and FeO (b) CU2O and FeO
34. The chemical composition of the slag formed
(c) CuS and Fc203 (d) CU2O and Fe',03
during smelting process is
36. Iron is removed from chalcopyriteas
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(a) CUS1O3 (a) FeO (b) FeS

{b) FeSi03 (c) Fe203 (d) FeSi03
(c) CaSi03 37. In self-reduction, the reducing species is
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(d) Cu20.Si02 (a) S (b) 0-- (c) S^- (d) SO2

Re
nd

09 Matching Type Questions

Fi

Match the entries of column I with appropriate entries of column II and choose the correct option out
of the four options (a), (b), (c) (<f) given at the end of each question.
38. Column I Column II

(A) Hematite (/■) AI2O3.2 H2O

(B) Bauxite
(//) Fe203
(C) Magnetite (Hi) CuC03-Cu(0H)2
(D) Malachite (iv) Fc304
(a) A-/7, C-/, D-jV (b) A-iv, B-/, C~H, D-Hi (c) A-/. B-iii, C-ii, D-/v (d) A-ii, B-i, C-zv, D-m

(JEE Main 2021)

ANSWERS

31. (b.c.d) 32. Ui,c.d) 33, (c) 34. {/;) 35. (a) 36. (d) 37. (c) 38. (d)
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/31

39. Column I Column II

(A) Concentration of gold ore (/) Aniline

(B) Leaching of alumina (n) NaOH
(C) Froth stabilizer {in) SOt
(D) Blister copper (iv) NaCN

(a) A-tv, B-//, C-n'f, D-/ (b) A-fv. B-ii, C-i, D-iii
(c) A-iii, B-ii, C-i, D-iv (d) A-ii, B-iv, C-iii, D-i (JEE Main 2022)

40. Column I Column II

(A) Zone refining (p) Titanium

(B) Monds’s process (q) Lead
(C) Liquation (r) Nickel

low
(D) van-Arkel (,v) Germanium

(fl) A-r, B-p, C-i, D-^ {b) As, B-r, C-q, D-p (c) A-i, B-^, C-r, D-/j {d) A-q, B-i, C-y^, D-r

ee
F
Matrix-Match Type Questions

Fr
Match the entries of column I with appropriate entries of column II. Each entry in column I may

for
have one or more than one correct option from column II. If the correct matches are A-p, s ; B-r ;
ur
C-p, q ; D-s, then the correctly bubbled 4x4 matrix should be as follows :
s
p q r s
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1

A, &
ii©i!©ii®
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c i©j|©!|0|i©j!
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D
i©!i®i©:,©:
;
L I.. —,i L
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41. Column I Column II

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nd

(A) PbS > PbO (P) Roasting

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(B) CaC03 > CaO (Q) Calcination

(C) ZnS > Zn (r) Carbon reduction

(D) CU2S 4 Cu (s) Self reduction (IIT Paper II 2008)

42. Column I Column II

(A) Carbonate (P) Siderite

(B) Sulphide Malachite

(C) Hydroxide (r) Bauxite

(D) Oxide (s) Calamine

(0 Argentite (JEE Advanced 2015)

ANSWERS

39. (/>) 40. (h) 41. {A-p ; B-q ; C-p.r : D-p.s) 42. {A-p,q.s ; B-t; C-q,r ; D-r)
6/32 'P’uzeUefi-'^- New Course Chemistry (XII)ISSIMI

m Integer Type Questions

A B C D
DIRECTIONS. The answer to each of the following questions Is a single digit
integer, ranging from 0 to 9. If the correct answers to the question numbers © ©©®
A, B, C and D (say) are 4, 0, 9 and 2 respectively, then the correct darkening
of bubbles should be as follows : ©©O©
43. Amongst the following, total number of metals which occur in the native state in ©©©©
the earth's crust are : Fe, Zn, Na, An, Ni, Sb, Sn, Pt, Hg.
44. Amongst the following, oxide ores are ; calamine, fool’s gold, cuprite, zincite,
© © ©©
chalcocite, haematite, bauxite, magnetite, cassiterite. 0 000
45. Amongst the following, the total number of ores which may be concentrated by
froth floatation process is : haematite, bauxite, galena, copper pyrites, sphalerite, ©©®®
cassiterite, calamine, argentite, chalocite.
© ©©©
46. Amongst the following, the ores which are roasted to convert them into their

w
corresponding metal oxides are : ©0® ©
alumina, zinc blende, iron pyrites, copper pyrites, calamine, galena, chalcocite. © ©®®

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47. How many of the following metals can be refined by vapour phase refining ?
Zr, Zn, Cd, Hg. Ni. Co, Pt, Fe, Ti.
®©®©

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48. Among the following ores : Bauxite, Sideritc, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite,

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Sphalerite, Limonite, Cryolite, the number of principal ores of iron is (JEE Main 2022)

VII.
Numerical Value Type Questions (>n Decimal Notation) for
r
For the following que.stion, enter the correct numerical value (in decimal notation, truncated/rounded-olT
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to the second decimal place ; e.g., 625, 7*00, - 0*33, - *30, 30-27, - 127*30) using the mouse and the on*
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ook

screen virtual numeric keypad in the place designated to enter the answer.
49. Galena (an ore) is partially oxidi.sed by passing air through it at high temperature. After some time, the
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passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self
reduction. The weight (in kg) of Pb produced per kg of O2 consumed is (JEE Advanced 2018)
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Assertion-Reason Type Questions

TYPE I
dY
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DIRECTIONS. Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each
Fin

question has four choices (a), (b), (c) and (d) of which ONLY ONE is correct. Choose the correct option
as under

(a) Statement-1 is True, Statement-2 is TViie ; Statement-2 is correct explanation of Statement-1,

(b) Statement-1 is True, Statement-2 is True ; Statement-2 is not a correct explanation of Statement-1,
(c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True.
50. Statement-1. During the reduction of ZnO to Zn, C is more efficient than CO.
Statement-2. The standard free energy of formation of CO2 from CO is always higher than that of ZnO.
51. Statement-1.Metals of high purity are obtained by zone refining.
Statement-2. Impurities are more soluble in the melt than in pure metal.
52. Statement-1. Nickel is purified by reacting it with CO.
Statement-2. Impurities present in nickel form volatile compounds.
,.ITISWERS
43.(2) 44. (6) 45. (4) 46. (5) 47. (3) 48. (4) 49. (6-47 kg)
50.(b) 51. (<i) 52. (f)
GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/33

53. Statement-1. According to Ellingham diagram any metal oxide with higher AG“ is more stable than the one
with lower AG”.
Statement-2. The metal involved in the formation of oxide placed lower in the Ellingham diagram can
reduces the oxide of metal placed higher in the diagram. (JEE Main 2022)
TYPE II

DIRECTIONS. Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each
question has four choices {a), (b), (c) and (d) of which ONLY ONE is correct. Choose the correct option
as under:

(a) Both statement-1 and statenient-2 arc true, 0) Both statement-1 and statement-2 are false.
(c) Statement-1 is true but statement-2 is false. id) Statement-1 is false but statement-2 is true.

54. Statement-1. During electrolytic refining, blister copper deposits precious metals.
Statement-2. In the process of obtaining pure copper by electrolysis method, copper blister is used to make
the anode. (JEE Main 2022)

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55. Statement-1. Leaching of gold with cyanide ion in absence of air/02 cyano complex of Au(IlI).
Statement-2. Zinc is oxidised during displacement reaction carried out for gold extraction.

F lo
(JEE Main 2022)
56. Statement-1. Pig iron is obtained by healing cast iron with scrap iron.

ee
Statement-2. Pig iron has a relatively lower carbon content than that of cast iron. (JEE Main 2022)

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TYPE III

DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a
for
correspondingstatement of Reason (R) just below it. Of the statements,mark the correct answer as
ur
(a) If both assertion and reason arc true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion,
s
ook

(c) If assertion Is true, but reason is false. id) If both assertion and reason are false
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57. Assertion. Gold and platinum occur in native stale.

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Reason. Gold and platinum are not attacked by moisture, oxygen and carbon dioxide of the atmosphere.
58. Assertion. Carbonate and hydroxide ores are concentrated by froth floatation process.
Reason. In froth floatation process, mineral oil is used because it preferentially wets the gangue ptulicles.
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(AIIMS 2015)
59. Assertion. Ag and Au are extracted by leaching their ores with a dilute solution of NaCN.
Reason. Impurities associated with these ores dissolve in NaCN.
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60. Assertion. The reduction of metal oxide is easier if the metal formed is in liquid state than solid state.
nd

Reason. The value of AG” becomes more on negative side as the entropy is higher in the liquid than in solid
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state. (JEE Main 2022)

61. Assertion. Magnesium can reduce AI2O3 at a temperature of I350°C, while above 1350°C aluminium can
reduce MgO.
Reason. The melting and boiling points of magnesium are lower Ilian those of aluminium. (JEE Main 2022)
62. Assertion. Aluminothermy is used for extraction of chromium from chromium oxide.
Reason. Alumina has a high melting point.
63. Assertion. AI is obtained by high temperature reduction of alumina with carbon.
Reason. Alumina reacts with carbon to form aluminium carbide which decomposes at high temperature to
form AI while carbon is oxidised to CO.
64. Assertion. Ti can be purified by van Arkel process.
Reason. Til4 is a volatile compound which decomposes at a high temperature.
ANSWERS
53. (d) 54. ia) 55. (d) 56. (h) 57. (a) 58. id) 59. (c) 60. («) 61. 0)
62. 0) 63. id) 64. (a)
6/34 New Course Chemistry (XIOISSQBI

For Difficult Questions

\ IViultiple Choice Questions (with one correct Answer)

4. Silver glance (Ag->S), Copper pyrites (CuFeS2) tmd Mg > A1 > Cr > Zn. Thus, Mg can reduce A1^03
galena (PbS). to Al.

ow
5. Sphalerite (ZnS) can be dissolved in group 1 3 Mg + AI2O3 ■> 2 Al + 3 MgO
cyanide, le., NaCN or KCN. 15. A will reduce the oxide of B only above 1400°C.
ZnS + 4 NaCN Thus, option {d) is correct.
^ Na2[Zn(CN)4] + Na2S
Sohiblecomplex 16. In the Ellingham diagram, the temperature below

e
which the oxide is stable, is the point at which
6. Argentite (Ag2S) being a low grade ore is

re
free energy change shows a change from negative
preferably extracted by leaching with NaCN. to positive.

Frl
F
7. O2 oxidises Na2S formed in the reaction to 18. Reactions (111) and (IV) do not occur in the blast
Na2S04 and Zn reduces Ag"*" to metallic Ag. furnace.

13. For the reaction. C + 1/2 O2

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^ CO, the AS 19. Pig iron (impure form) is obtained directly from

r
increases. Therefore, as the temperature increases the blast furnace and can be moulded into variety

so
TAS increases and hence AG (AH - TAS) of shapes.
decreases. In other words, the slope of the curve
for formation of CO decreases. However, for all
kf
20. Copper (I) sulphide reduces CU2O to metallic Cu
oo
2 Cu->0 + Cu-)S 6 Cu + SO2
other oxides, it increases, (refer to Fig. 6.8, page
Y
6/9). 21. FeO is a gtmgue while FiS03 is a slag.
B

14. Any metal oxide whose A G° value is lower is more 23. Na3AlFg does not serve as electrolyte. Instead
molten alumina serves as the electrolyte.
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stable than the metal oxide with higher A G°. This

implies that the metal oxide which is placed higher 25. In liquation method of refining, low melting metal
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u

in the Ellingham diagram can be reduced by the flows down while the high melting impurities
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metal involved in the formation of metal oxide remain on the hearth.

placed lower in the diagram.

d

26. The best method to purify liquid metals is

distillation.
From the diagram (Fig. 6.8 page 6/9), the tendency
in
Re

of the four metals (Fe, Zn, Mg, Cu) to act as 27. Being low melting, Zn is most economically
reducing agents follows the order: purified by distillation.
F

m IViultiple Choice Questions (with One or More than One Correct Answers)

28. Impure Cu strip is not used as the cathode but is Thus, R is [Au(CN)4l is wrong while all other
used as the anode. All other options are correct. options are correct.
30. All statements are correct.
29. 4AU + 8CN- + 2H2O+ O2 4[Au(CN)2]"
Q R 31. In electrochemical reduction, such a high
temperature (2500°C) is not needed. Thus option
+ 4 0H"
(a) is wrong while all other options are correct.
32. Silica is added to remove FeO (produced from
2[Au(CN),J- + Zn ■> 2 Au + [Zn(CN)4]-“
R
T
Z
CuFeS2) to FeSi03 as slag.
 GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS 6/35

For Difficult Questions

DQ Multiple Choice Questions (Based on the given Passage/Comprehens ion)

35. During partial roasting, CuFeS2 is first converted 36. The partially roasted ore is mixed with silica (flux)
into a mixture of CU2S and FeS. Further, since iron and little coke and heated in a blast furnace when
is more reactive than Cu, therefore. FeS is FeO present as impurity is removed as ferrous
preferentially oxidised to FeO. Thus, partial silicate slag.
roasting of chalcopyrite produces Cu^S and FeO.
FeO + Si02 FeSiO, i
2 CuFeS2 + O2 CU2S + 2 FeS + SO2 37. During self-reduction, Cu-)S, i.e., S~~ acts as a

Chalcopyrite reducing species and hence reduces C112O to

copper metal.

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A
2 FeS + 3 O2 ^ 2 FeO + 2 SO2
Thus, option (a) is correct. S +2 Cu^O ^ 4 Cu + SO2.

F lo
K9 Matrix-Match Type Questions

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42. (A-/),q,s): Carbonate ores arc (B-/): t (sulphite); AgoS
p (Siderite): FeC03 (C-q,r): Hydroxide ion is present in

q (Malachite) ; CuC03-Cu(0H)2 :
for
q (Malachite) : CuC03.Cu(0H)2 :
ur
i (Calantine): ZnC03 r (Bauxite): AI2O3.2 H->0 or A10^.(0H)3_2^ where
0<jf< 1.
s
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VI. Integer Type Questions

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43. 2 (two, i.e., Au and Pt). low. It is better concentrated by leaching with
our
ad

NaCN.
44. 6 (six ; cuprite, zincite, haematite, bauxite,
magnetite, cassiterite). 46. 5 (five ; zinc blende, iron pyrites, copper pyrites,
45. 4 (four, i.e., galena, copper pyrites, sphalerite, galena, chalocite).
Y

47. 3 (three, i.e., Zr, Ni, Ti).

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chalcocite). Please note that although argentite

nd

(Ag-)S) is a sulphide ore, it is not concentrated by 48. 4 (four, i.e., Siderite, Haematite, Limonite and
froth floatation process since its silver content is
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Magnetite).

VII.
Numerical Value Type Questions (in Decimal Notation)

49. Oxidation of galena (PbS) by O2 occurs according From {Hi),

to the following equation. 3 moles of O2 produce Pb = 3 moles
2 PbS + 3 Oo ^ 2 PbO + 2 SO2 ...(0 Now 1 mole of O-j = 32 g
Self reduction of PbO by PbS occurs as follows : and 1 mole of Pb = 207 g
2PbO+ PbS 4 3 Pb + SO2 ...(//) 3 X 32 kg of O2 produce Pb = 3 x 207 kg
Adding equations (1) and {Hi), we gel, .. ... 3x207
3 PbS + 3 O2 ^ 3 Pb +3 0, ...{Hi) or 1 kg of O2 will produce Pb = = 6-47 kg
3x32

I
6/36 New Course Chemistry fxmpraMi

For Difficult Questions

VIIIJ Assertion-Reason Type Questions

52. Correct statement-2. Nickel reacts with CO to 59. Correct reason. Ag and Au dissolve in NaCN
form volatile Ni(CO)4. solution to form their soluble complexes.
53. Statement-1 is false because metal oxide with 61. Both A and R are correct but R is not the correct
lower AG" more stable than the one with higher explanation ol‘ A.
AG".
62. Correct explanation. G® of AI2O3 is more
54. Both Statement-1 and Statement-2 are true.
negative than that of CroO^.
55. Statement-1 is false while Statement-2 is true.
63. Correct assertion. A1 is extracted by high
56. Both Statement-1 and Statement-2 are false.
temperature electrolytic reduction of AI2O3

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58. Correct assertion. Sulphide ores are concentrated to which fluorspar and cryolite have been
by froth floatation process. added.

F lo
Correct reason. In froth floatation process, pine Correct reason. AI2O3 reacts with C at high tempe
oil preferentially wets the ore particles. ratures to form AI4C3 which does not decompose.

ee
Fr
for
ur
s
ook
Yo
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our
ad
Y
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nd
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I
 p-BLOCK ELEMENTS

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In class XI, we have studied that the elements of group 13-18 constitute p-block elements. Their valence
shell electronic configuration is ns^np^~^ where n = 2-6 (except helium which has 1^^configuration). Like other

F lo
elements, the properties of p-block elements are also greatly influenced by variation in their atomic size, ionisation
enthalpy, electron gain enthalpy and electronegativity. The absence of ^/-orbitals in the elements of second period
and presence of d- and/or/-orbitals in heavier elements (starting from third period onwards) have significant effects

e
on the properties of these elements. Further, the p-block elements can be either metals, metalloids or non-metals.

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This leads to the diversification in the chemistry of these elements.
We have already discussed the chemistry of the elements of groups 13 and 14 in class XI. Therefore, in
for
this unit, we shall discuss the chemistry of the elements of groups 15-18.
PART —I: GROUP 15 ELEMENTS : THE NITROGEN FAMILY
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7.1. GENERAL INTRODUCTION

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The elements of nitrogen family, i.e., group 15 of the periodic table are : nitrogen (N), phosphorus (P),
arsenic (As), antimony (Sb) and bismuth (Bi). Collectively, the group 15 elements are called 'pnicogens' and their
compounds as 'pniconides'. The name is derived from the Greek word 'pnicomigs' meaning 'suffocation'.
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our

In this section, we will discuss the main group trends and the chemistry of nitrogen and phosphorus and
their compounds.
7.2. OCCURRENCE
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Nitrogen. The fin>t member of this group, i.e., niu-ogen occurs as a diatomic ga.s, N,. It makes about 78% by
volume of the atmosphere. Despite this, nitrogen is not a very abundant element in the etirth’s crust. It is only the
Fin

thirty-third most abundant element by mass in the earth’s crust (~ 9 ppm). It mainly occurs as nitrates, i.e.,
(Chile saltpetre) and KNO3 (Indian saltpetre). Nitrogen is the essential constituent of proteins, amino acids and
nucleic acids which regulate the growth and control the hereditary effects in living beings.
Phosphorus. It is a very reactive element and hence it does not occur free in nature. However, it is the
eleventh most abundant (~ 1120 ppm) clement by mass in the earth’s crust.
It occurs in the minerals of the apatite family. Ca<^(PO)4 . CaX, (X = F, Cl or OH), e.g.,
(i) Fluoroapatite, 3Ca3t''04)i.CapT (//) Chloroapatite, 3Ca3(P04)2.CaCl2
(in) Hydroxyapatite. 3C PO4 )2.Ca(OH)-,
These minerals are the major components of the phosphate rocks.
Large deposits of phosphate rocks are found in North Africa and North America. In India, commercially
viable phosphate rocks are found in Rajasthan.
* Not included in CBSE syllabus. This chapter ha.s been given only for the preparation of competitive examinations.
7/1

1
7/2 7^>%cideefi.'<x New Course Chemistry (XII)SSSMl

Phosphorus is an essential constituent of animal and plant matter. It is present in bones as well as in
living cells. About 60% of our bones and teeth are Ca3(P04)2 or fluoroapatite. Phosphoproteins are present in
milk and eggs. It also occurs in nucleic acids (DNA and RNA) which control the hereditary effects in human
beings and in ATP (adenosine triphosphate) and adenosine diphosphate (ADP) which are of vital importance
for production of energy in cells.
Arsenic, antimony and bismuth. The elements As, Sb and Bi are not very abundant. They mainly occur as
sulphides (/>., arsenopyrites, FeAsS ; stibnite, Sb2S3 and bismuth glance, 81283) as traces in other ores.
73. ELECTRONIC CONFIGURATION

The electronic configurations of the elements of group 15 are given in Table 7.1.
The general valence shell electronic configuration of the elements of group 15 is up^, where n = 2 to
6. The three electrons in p-orbitals are distributed as p[, Py, in accordance with Hund’s rule.

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TABLE 7.1. Electronic Configuration of Elements of Group 15

Element
Atomic Electronic configuration

Flo
number Complete With noble gas core

ls-2s^2p^ IHeJ 2s-2p^

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Nitrogen (N) 7

ls-2s~2pV3p^ INeJ 3^-3p^

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Phosphorus (P) 15

Arsenic (As) 33 ls^2s~2p^3s^3p^3d^°4s^4p^ [Ar] 4s^ 4p^

Antimony (Sb) \s^2s^2p^3s^3p^3d^^4sMpHd'°5s^5p^
for IKr] 4d^'^5s^5p^
51
ur
Bismuth (Bi) 83 ls'2s-2p^3s'^3p^3d'Hs^4pHd'Hf-^5s^5p^5d'%s^6p^ [Xe] 4/^5d'%s^6p^
k s
7.4. ATOMIC AND PHYSICAL PROPERTIES
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Some of the important atomic and physical properties of the elements of nitrogen family or group 15
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elements are summarised in Table 7.2.

TABLE 7.2. Some Atomic and Physical Properties of Group 15 Elements

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Elements
Property
Y

N P As Sb Bi

Atomic number 7 15 33 51 83
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nd

-I
Atomic mass/g mol 14-01 30-97 74-92 121-76 208-98
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Ionisation enthalpy A,H 1402 1012 947 834 703

-1
(A,- H)/kJ mol A,H2 2856 1903 1798 1595 1610

A,H3 4577 2910 2736 2443 2466

Electronegativity 3-0 2-1 2-0 1-9 1-9

Covalent radius/pm 70 no 121 141 148

Ionic radius/pm 171 (N-^^) 212 (P^) 222 (As-^) 76 (Sb^^) 103 (Bi^-")
Melting point/K 63 317-1 1089 904 544

(white P) (grey a-foi m

at 38-6 atm)
Boiling point/K 77-2 554 888 1860 1837

(while P) (sublimation
temperature)
Density/g cm"^ at 298 K 0-879 1-823 5-778 6-697 9-808
at 63 K (grey a-form)

/
 p»BLOCK ELEMENTS 7/3

7.4.1. Atomic Properties

Some atomic properties of the elements of group 15 are discussed below :
1. Atomic and ionic radii, (a) The atomic (covalent) and ionic radii (in a particular oxidation state) of the
elements of nitrogen family (group 15) are smaller than the corresponding elements of carbon family (group 14).
Explanation. This is because on moving from left to right, i.e., from group 14 to 15 in a given period, the
nuclear charge increases while the new electron enters the same shell. Fuither, the electrons in the same shell do not
screen each other. Therefore, the effective nuclear charge increases and hence the electrons are more strongly
attracted towards the nucleus. This results in decrease in covalent radii. Same is true of ionic radii.
(b) On moving down the group, the covalent and ionic radii (in a particular oxidation state) increase
with increase in atomic numbe.r There is considerable increase in covalent radius from N to .P Howeve,r
from A.9 to Bi, only a small increase is observed.
Explanation. Down the group, the covalent radii increase primarily due to the addition of a new principal
energy shell in each succeeding element. The considerable increase in covalent radius from N to P is not only
due to the addition of a new energy shell but is also due to strong shielding effect of the s- and /j-electrons

w
present in the inner shells. However, the small increase in covalent radii from As to Bi is due to the poor
shielding of the valence electrons by the d- and/or f-electrons present in the inner shells of these heavier

F lo
elements. As a result, the effective nuclear charge increases which reduces the effect of addition of a new
energy shell to some extent. Consequently, the increase in covalent radius is small from As to Bi. Same is true
of ionic radii.

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2. Ionization enthalpy. The ionization enthalpies of the elements ofgroup 15 are much higher than those of

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the corresponding elements of group 14. Down the group, the ionization enthalpiesdecrease regularly.
Explanation. Because of increased nuclear charge, reduced atomic radii and stable half-filled electronic
for
configurations, the valence electrons of group 15 elements are strongly attracted by the nucleus and hence
ur
they have less tendency to lose electrons. As a result, ionization enthalpies of the elements of the nitrogen
family are much higher as compared to the corresponding elements of the carbon family. But the decrease in
s
the values of ionization energies as we move down the group is due to gradual increase in the atomic size
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which reduces the force of attraction on the electrons by the nucleus.

As expected, successive ionization enthalpies of these elements increase in the order :
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A,. Hj < A. < A- H3 (Table 7.2)

3. Electronegativity. Group 15 elements are more electronegative than group 14 elements. Electronegativity
our

of elements of group 15 .shows a gradual decrea.se on moving down the group from N to Bi.
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Explanation. Due to smaller atomic size and smaller number of electrons needed to attain noble gas
configuration the elements of group 15 are more electronegative as compared to those of group 14. The
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electronegativity of group 15 elements, however, decreases down the group due to a corresponding increase
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in atomic radii from N to Bi. But the decrease is not regular.

nd

7.4.2. Physical Properties

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Some important physical properties of the elements of group 15 are discussed below :
1. Metallic character. The elements of group 15 are less metallic than the correspondingelements of
group 14. Howeve,r on moving down the group, the metallic character increases.
Explanation. Due to increased nuclear charge and higher electronegativity, the elements of group 15
are less metallic than the corresponding elements of group 14. On movingdown the group, the electronegativity
decreases. As a result, valence electrons are lost more readily and hence the metallic character increases.
Thus, first two elements of this group (N and P) are non-metals,the next two (As and Sb) are metalloids,
while Bi is a typical metal. In other words, the metallic character increases from N to Bi.
N, P As, Sb Bi
non-metaLs metalloids metal

Metallic chaiacter increases -»

2. Melting and boiling points. The melting points of group 15 elements ifrst increase from nitrogen to
arsenic and then decrease to antimony and bismuth. The boiling points, however, increase regularly as we
move from N to Bi.
7/4 7^>u^eU^'4l^ New Course Chemistry (X11)BSI9]

Explanation. The melting points increase down the group from N to As due to increase in their atomic
size. Thereafter, even though the atomic size increases but the melting points of Sb and Bi decrease. This
unexpected decrease in the melting points of antimony and bismuth is because of their tendency to form three
covalent bonds instead of five covalent bonds due to inert pair effect. This results in weakening the attraction
among their atoms thereby lowering their melting points. Because of larger size of atoms, bismuth has still
weaker interatomic forces than antimony and thus has still lower melting point.
The boiling points increase down the group from N to Bi due to an increase in their atomic size.
3. Density. The density of the elements of group 15 increases regularly from top to bottom, i.e.,from N
to Bi as usual.

4. Allotropy. Except nitrogen and bismuth, all the elements of this group show allotropy.

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Pho.sphorus exists in three allotropic forms, i.e., white, red and black phosphorus.
Arsenic and antimony exist in two allotropic forms, i.e., yellow and grey.
7.5. CHEMICAL PROPERTIES

Some important chemical properties of the elements of group 15 are discussed below :

e
7.5.1. Oxidation States

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(i) Negative oxidation states. All the elements of this group have five electrons in the valence shell (ns^

F
np^) and thus require three more electrons to acquire the nearest noble gas configuration. Although gain of
three electrons from more metallic elements to form M^" ions requires large amount of energy yet it lakes
place only with nitrogen becau.se it is the smallest and the most electronegative element of this group. Thus,

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nitrogen forms (nitride) ion and shows an oxidation state of -3 in nitrides of some highly electropositive
metals such as Mg3N^, CajN,, etc. Other elements of this group form covalent compounds even with metals
ksf
and show a formal oxidation state of -3. For example, calcium phosphide (Ca^Pj), sodium aisenide (Na3As),
zinc stibide (Zn^Sb^) and magnesium bismuthide (Mg3Bi2). Howeve,r the tendency of these elements to show
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-3 oxidation .state decreases as we move down the group from N to Bi due to a gradual decrease in their
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electronegativity and ionization enthalpy.

B

Besides -3, N and P also show oxidation states of -2 in hydrazine (NH2NH2) and diphosphine (P2H4)
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respectively. Nitrogen also shows an oxidation state of-1 in hydroxylamine (NH-,OH) but phosphorus does
not.
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(n) Positive oxidation states. All the elements of this group show positive oxidation states of +3 and
+5. However, on moving down the group, the stability of +5 oxidation state decreases while that of +3
oxidation state increases due to inert pair effect. Thus, +5 oxidation state of Bi is less stable than +5
d

oxidation .state ofSb. The only stable compound of bismuth having +5 oxidation state is BiFg.
in
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Further due to large amount of energy needed to lose all the five valence electrons, M^'*' ions cannot be
formed, In other words, all the compounds of group 15 elements having +5 oxidation state (i.e. PFg, PCI5,
F

SbFg, BiFj) are essentially covalent. However, elements of this group form both ionic {i.e., BiF3, SbFj) and
cov^ent compounds (i.e. NCI3, PCI3, ASCI3, SbCl3) in +3 oxidation state. In nut shell, the covalent character
decreases in the order: N > P > As > Sb > Bi

It may, however, be pointed out here that nitrogen because of its small size, high electronegativity and strong
tendency to form pn-pK multiple bonds, shows all the oxidation states from -3 to +5 as shown below :
Compound NH3 NH2NH2 NH2OH N2 N2O NO N2O3 NO2 or N2O4 N2O5
Oxidation state -3 -2 -1 0 +1 +2 +3 +4 +5

(Hi) Maximum covalency. Since nitrogen does not possess ^/-orbitals in its valence shell {n = 2), therefore,
it can show a maximum covalency of 4 and that too when it donates its lone pair of electrons [NH^, R^N'^ ].
In other words, nitrogen cannot extend its covalency beyond 4. That is why nitrogen does not form NF5 or
NCI5. On the other hand, phosphorous and all other elements have empty ^/-orbitals and can utilize all their
valence orbitals to exhibit covalency of 5 or 6, e.g., PCI5, ASF5, [PF^]", [SbF^]“, etc.
(iv) Disproportionation. All the oxidation states of nitrogen from +1 to +4 show disproportionation in
acidic medium. For example,
p-BLOCK ELEMENTS 7/5

+3 +5 +2

3HNO2 > HNO3 + 2NO + H^O

Nitrous acid Nitric acid Nitric oxide

Similarly, in case of phosphorus, all the intermediate oxidation stales from -3 to +5 show
disproportionation both in acidic and basic media. For example.
+3 +5 -3
A

4H3PO3 - 3H3PO4 +
PH3
Phosphorus acid Phosphoric acid Phosphine
However, due to inert pair effect as the stability of +3 oxidation slates increases from As to Bi, their
tendency to undergo disproportionation decreases.
7.5.2. Trends in Chemical Reactivity
Some important trends in the chemical reactivity of some of the compounds of these elements are
discussed below :

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1. Reactivity towards Hydrogen (Formation of Hydrides)
All the elements of group 15 fonn volatile trihydrides of the formula EH3 where E = N, P, As, Sb, Bi.

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These are named as

NH3 PH3 ASH3 SbH3 BiH3

Ammonia Phosphine Arsine Stibine Bismuthine

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Some of the properties of these hydrides are shown in Table 7.3.
TABLE 7.3. Properties of Hydrides of Group 15 Elements

Property NH3 PH3

for
ASH3 SbHg BIH3
r
Melting point/K
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195-2 139-5 156-7 185
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Boiling point/K 238-5 185-5 210-6 254-6 290

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E-H distance/pm 101-7 141-9 151-9 170-7

HEH angle (°) 107-8 93-6 91-8 91-3

-1
A^°/kJ mol -46-1 13-4 66-4 145-1 278
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Adiss H° (E-H)/U mol-' 389 322 297 255

The lighter elements also form hydrides of the formula E2H4 viz. (hydrazine), P0H4 and As,H4.
Nitrogen also forms a special hydride of the formula HN3. It is called hydrazoic acid.
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Preparation, (i) By hydrolysis of binary metal compounds. The hydrides of group 15 are prepared
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from their corresponding binary metal compounds such as Mg3N2, Ca3P2, Zn3As2, Mg3Sb2 or Mg3Bi2 by
hydrolysis with water or dilute acids. Thus
Mg3N2 + 6H2O >3Mg(OH)2+ 2NH3 ; Ca3P2 + 6H2O > 3Ca(OH)2 + 2PH3
Mag. nitride Ammonia Cal. phosphide Phosphine
Zn3As2 + 6HC1 ^ 3ZnCl2 + 2ASH3; Mg3Sb2 + 6HC1 — 3MgCl2 + 2SbH3
Zinc arsenide Arsine Mag. stibide Stibine

Mg3Bi2 + 6HC1 ^ 3MgCl2 + 2BiH3

Mag. bismuthide Bismuthine

(li) By reduction of trichlorides. The trichlorides of these elements except that of bismuth upon reduction
with Zn/acid or LiAlH4 give the corresponding hydrides.
ECI3 + 3UAIH4 ■> EH3 + 3 LiCl + 3 AIH3 (E = N, P, As, Sb)
Structure. The valence shell electronic configuration of these elements is ns^np^^ np^ np' and all these
elemenLs undergo .iy^^-hybridization.
7/6 New Course Chemistry (XIl)CZsI9]

Three of the four FIGURE 7.1

LONE PAIR
hybridized orbitals overlap with \s-
orbitals of three hydrogen atoms to
fomi three E-H, o-bonds while the
fourth contains the lone pair of
\
electrons (Fig. 7.1a). E
\
According to VSEPR theory, N
H
lone pair-bond pair repulsion is \
H
107.8° N

larger than bond pair-bond pair o o

Bond pairs are Bond pairs are away
repulsion. As a result, the near the N atom, greater from the P atom, lesser
temihedral shape of the molecule repulsion, larger angle repulsion, smaller angle
gets distorted and the bond angle
becomes less than 109"-28'
Pyramidal structure of (a) EH3 molecules, bond angle and
bond length in (b) NH3 molecule and (c) PH3 molecule.
(tetrahedral angle). Thus, all the
hydrides of group 15 elements have pyramidal structures and their bond angles are found to decrease in the
following order:

F low
NH3 PH3 ASH3 SbH3 BiH3
107-8“ 936“ 91-8' 91-3“ 90“

Explanation. This decrease in bond angles can be explained on the basis of the size and electronegativity
of the central atom. As we move down the group, the size of the central atom goes on increasing and its
electronegativity goes on decreasing. As a result, the bond pairs of electrons tend to lie away and away from

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the central atom as we move from NH3 to BiH3. In other words, the force of repulsion between the adjacent
for F
bond pairs is maximum in NH3 and minimum in BiH3. Consequently, the bond angle is maximum in NH3
(Fig. 7.16) and minimum in BiHj.
FIGURE 7.2
Properties of hydrides, (i) Boiling points. The boiling points
254.6 SbH3(254.6)
increase regularly as we move from PHj to BlHj. Howeve,r the
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boiling point of NHj is higher thanihose of PH^and AsHj. (Fig. 7.2) g-

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1- 238.5
NH3(238.5)
NH3 PH3 ASH3 SbH3 BiH3
O
238-5 K 185-5 K 210-6 K 254-6 290 K Q.

§ 210.6 ^AsH3(210-6)
ad

This variation in boiling points can be explained on the basis

our

of (0 hydrogen bonding (ii) van der Waals forces of attraction. O

□a
185.5
PH3(186.5K)
The electronegativity of N (3-0) is much higher than that of H
(,2-i). As a result, N—H bonds are polar and hence NH3 undergoes
Re

17 34 50 78 100128 150
intermoleculai' H-bonding as shown below : MOLECULAR MASS
Y
Find

H. H. - H.
Plot of boiling points of the
5- 5- 5-
5-^
H N
5-^
-H N
5-^
H N
hydrides of group 15 elements
versus their molecular masses
H H H

In contrast, the electronegativities of P (2-1), As (2-0), Sb (1-9) and Bi (1-^) arc close to H (2-1) and
hence PH3, ASH3, SbH3 and BiH3 do not form H-bonds. Thus, the abnonnally high b.p. of NH^ is due to
intermolecular H-bonding. V -5-;
Further, as we move from PH3 to BiH3, the molecular mass increases. As a result, the van der Waals
foices of attraction increase and the boiling points increase regularly from PH3 to BiH3. Howeve,r the b.p. of
NHj is higher than those ofPHj and AsHj hut lower than those ofShH^ and BiH3. This is probably due to the
reason that the increase in the boiling points of SbH3 and BiH3 because of higher van der Waals forces of
attraction more than compensates tlie increase in b.p. of NH3 because of H-bondijng. Thus, in a nutshell, the
boiling points of hydrides of group 15 elements follow the order :
PHj < AsH:; < NH, < SbH^ < BiHj.
 p-BLOCK ELEMENTS 7/7

(ii) Melting points. Because of intermolecular H-bomling, FIGURE 7.3

NH3 has the highest melting point. The other hydrides of this group 195.2 NH3 (195.2)
do not form H-bonds and hence their melting points are lower than
that of NH3, However, as we move down the group from PH3 to 2 185.0
O
SbHg (185.0)
SbH3, the molecular size increases and hence the melting points d.

increase regularly due to a corresponding increase in their van der

C3
Z 156.7
ASH3 (156,7)
Waals forces of attraction (Fig. 7.3). Thus, the melting points of the O

hydrides of group 15 elements follow the sequence : £0 139.5

PH3 (139.5)
PH3 < AsHj < SbH3 < NH3 17 34 50 78100128150
139-5 156-7 K 185 K 195-2 K MOLECULAR MASS

(Hi) Solubility. NH, forms H-bonds with H2O while PH3 and Plot of melting points of the
other hydrides do not. Therefore, NH3 is soluble fn H2O while PH3 hydrides of group 15 elements
and other hydrides are insoluble in water. versus their molecular masses,

w
(iv) Thermal stability. The thermal stability of the trihydrides of group 15 elements decreases as we
move from NH3 to BiH3. Thus, NH3 is most stable, the stability of PH3, ASH3 and SbH3 decreases progressively

F lo
while BiH3 is the least stable. It is so unstable that it has a half-life of only 20 minutes at ordinary temperatures.
Explanation. As we move down the group the size of the central atom increases and hence its tendency

ee
to form covalent bond with comparatively small H-atom decreases. In other words, the M-H bond strength
decreases and hence the thermal stability goes on decreasing as we move from NH3 to BiHj. Thus, the

Fr
thermal stability decreases in the order: NH3 > PH3 > ASH3 > SbH3 > BiH3.
NH3
>
<r
UJ

for
(v) Reducing character : The reducing character of the
hydrides of group 15 elements increases in the order
ur
d:
o UJ :
PH3
NH3 < PH3 < ASH3 < SbH3 < BJH3
~U} CO CO
m UJ q: UJ u
< CO <C0
s
AsHg s<
Explanation : As we move from NH3 to BiH3, the thermal
ook

Uuj < UJ
Yo
<0
S LU zO stability of the hydrides decreases. In order words, their tendency
SbH3 o2
O UJ
o:d
to liberate hydrogen increases and hence their reducing character
eB

UJ 3
<
increases from NH3 to BiH3. Thus. NH3 being most stable is
X D
m
BiHa H UJ
01
not a good reducing agent.* The other hydrides being unstable
act as good reducing agents,
r
ad
ou

(vi) Basic nature. All the hydrides of group 15 elements possess a lone pair of electrons on the central
atom and thus behave as Lewis bases. The basic character, however, decreases as we move down the group
from NH3 to BiH,. Thus, NH3 is distinctly basic and forms salts with strong and weak acids. PH3 is a very
Y

weak base (may even be regarded as neutral) and thus forms salts only with anhydrous hydrogen chloride gas.
Re
nd

ASH3, SbH3 and BiH3 are not at all basic.

Explanation. The decrease in basic character of the hydrides of group 15 elements may be explained in
Fi

terms of electron density on the central atom. As the size of the central atom increases, the lone pair of
electrons occupies a larger volume. In other words, the electron density on the central atom decreases and
consequently its tendency to donate a pair of electrons decreases and hence the basic strength decreases as we
move from NH3 to BiH3.
Alternatively, the basic character of these hydrides can be explained in terms of strength of the additional
element-hydrogen bond formed when the hydride accepts a proton.

: EH3 +^H+ > EH4

On moving down the group, the size of the element increases and the strength of the E-H bond decreases.
Consequently, the tendency of the hydride to accept a proton decreases and hence the basic strength decreases
down the group.
♦However, at high temperatures, NH3 reduces CuO to Cu.
A

3 CuO -L 2 NH3 ■> 3 Cu + N2 -t- 3 H.,0

7/8
New Course Chemistry (XII)jg

2. Reactivity towards Oxygen (Formation of Oxides)

Nitrogen has a strong tendency to form p7t-p7t multiple bonds
between N and O atoms, whereas other
elements of this group do not. Consequently N forms a number of oxides which have no P, As, Sb or Bi
analogues. Actually, nitrogen forms five oxides with oxidation states ranging from +1 to +5 while other
elements form oxides only in +3 and +5 oxidation states as shown in Table 7.4.
[TABLE 7.4: Oxides of Group 15 Elements
Oxides of Oxides of Oxides of Oxides of Oxides of
Oxidation state
Arsenic Antimony Bismuth
of element Nitrogen Phosphorus
+1 N2O

w
Nitrous oxide

+2 NO
Nitric oxide

o
+3 N2O3 P4O6 AS4O6 SbA BI2O3

e
Arsenic trioxide Antimony trioxide Bismuth
Dinitrogen trioxide Phosphorus trioxide

re
rFl
trioxide

F
+4 N2O4 P4O8
Dinitrogen tetroxide Phosphorus tetroxide

r
orN02
ou
Nitrogen dioxide
fo
ks
+5 N2O5 P4^10
Dinitrogen pentoxide Phosphorus pentoxide Arsenic pentoxide Antimony pentoxide
oo

In general, oxides of non-metals are acidic, those of metalloids are amphoteric while those of metals
Y
eB

are basic. Further, greater the electronegativity of the element, more acidic is the oxide. Among the oxides of
the same element, higher the oxidation state of the element, more is its acidic strength.
r

In the light of the above facts, the following conclusions can be drawn :
You

in the order: N2O <NO< N2OJ < N2O4 < ^^2^5

ad

(i) Acidic strength of oxides of nitrogen increases

NjO and NO are, however, neutral.
(if) Acidic strength of trioxides follows the order: ^20^ > P40^ > As^O^ > Sb/)^
d
Re

In fact, As40g and Sb40g are amphoteric while Bi203 is basic in nature.
in

(Hi) Acidic strength ofpentoxides follows the order : N205> P40jq> As40jq> Sb40jo
F

3. Reactivity towards Halogens (Formation of Halides)

The elements of group 15 form two types of halides, viz., trihalides and pentahalides as shown in Tables
7.5(a) and (b).

TABLE 7.5. llr^alidi^ (EXg) of 15 pieiBy^

E Flourine Chlorine Bromiiie Iodine

N NF3 NCI3 NBrj.NHj NI3.NH3

P PF3 PCI3 PBr3 PI3
As ASF3 ASCI3 AsBr3 ASI3
Sb SbF3 SbCl3 SbBr3 SW3
Bi BiFa BiCl3 BiBr3 Bila
 p-BLOCK ELEMENTS 7/9

TABLE 7.5.
(b) Pentahalides (EXg) of Group 15 Elements
E Fluorine Chlorine Bromine iodine

N
P
PF5 PCI5 PBrj
As
AsFj AsCIj
Sb
ShF, SbClg
Bi
BiF^
(a) Trihalides. AIJ the elements of group 15 form trihaiides FIGURE 7.4
of the general formula, EX3. All these trihalides F
E = N, P, As, Sb or Bi and X = F. Cl, Br or I) are known.

w
£
●Lone Pair Q.

Structure. Like ammonia, trihalides have pyramidal CO F

structures, i.e., the central element is ip^-hybridized. Three of

Flo
the four jp^-orbitals form a-bonds with halogen atoms while
the fourth jp^-orbital contains the lone pair of electrons (Fig / 96’3® F*^
120®
■F

ee
7.4 a). f'
The bond angles of the trihalides of an element decrease

Fr
(a)PF3 (b)PF5 .
as the electronegativity of the halogen increases, i.e., (Pyramidal) (Trigonal bipyramidal)
PF3 < PCI3 < PBrj < PI3
for
Structures of (a) PF3 and (b) PF5
ur
(96-3°) (100-4°) (lOr) (102°)
The reason bring that as the electronegativity of halogen increases, the bond pairs move away from the
ks
central P atom. As a result, bond pair-bond pair repulsions decrease and hence the bond angles also decrease.
Yo
Properties. (0 These trihalides are predominantly covalent with the ionic character increasing down
oo

the group. Thus, Bip3 is predominantly ionic while other halides of Bi, i.e., BiCl3, BiBr3, etc. and Sbp3 arc
eB

partly covalent and partly ionic.

(ii) Of all the trihalides, trihalides of N are the least stable. NF3 is, however, stable. NCI3 is explosive
while NBr3 and NI3 are known only as their unstable ammoniates, i.e., NBr3.NH3 and NI3.NH3. The nitrogen
r

triodide ammoniate is stable only in the moist state. In the dry state, it explodes with noise when struck
ou
ad

liberating vapours of I2. Thus, it is a mild and harmless explosive.

Y

8 NI3.NH3 5 N2 + 9 I2 + 6 NH4I
The instability of NCl^, NBrj and Nl^ is because of the weakness of N—X bond due to large difference
nd
Re

in the size of N and X atoms. Further since the difference in size of N (75 pm) and F (72 pm) is small,
^ ^ bond is quite strong. As a result, NF3 is an exothermic compound. This accounts for the stability of
Fi

NF3. Being stable, NF3 does not undergo hydrolysis with water, dilute acids or alkalies.
(Hi) The trihalides readily undergo hydrolysis but the products of hydrolysis depend up the nature of the
element. For example,
NCI3 + 3 H2O 4 NH3 -f 3 HOCl ; PCI3 + 3 H2O — ^ H3PO3 + 3 HCI
AsCl3+ 3H2O — ^ H3ASO3 + 3 HCI
In contrast, the trichlorides of Bi and Sb are only partly and reversibly hydrolysed to give HCI and
oxychloride of the corresponding metal, i.e.,
SbClj + H2O V ^ SbO+Cl- +2 HCI ; B1C13 + H20 ± BiO-^Cr + 2 HCI
Antimony Bismuth
oxychloride oxychloride
In accordance with Le Chatelier principle, the addition of excess of HCI suppresses the hydrolysis by
shifting the equilibrium to the left.
From the above discussion, it follows that the ease of hydrolysis of trihalides follows the order :
NCI3 > PCI3 > ASCI3 > SbCl3 > BiCl3.
7/10 New Course Chemistry (XII) i^M

(iv) The trihalides of P, As and Sb (especiaUy the fluorides and chlorides) due to the presence of
^/-orbitals, accept a pair of electrons and thus behave as Lewis acids.
PF3 + F2 ^ PF5 PCI3 + CI2 - ^ PCI5
SbF3 + 2F- [SbF5]2- SbClj + 2 Cl" [SbCls]^-
The Lewis acid strength, however, decreases in the order :
PC/3 > AsClj > SbClj and PF^ > PCI3 > PBr^ > PI^.
In contrast, the trihajides of nitrogen due to the absence of J-orbitals and presence of a lone pair of
electrons, behave as Lewis bases. However, NF3 has a little tendency to donate a pair of electrons because of
the high electronegativity of E The Lewis base strength of the other trihalides increases as the electronegativity
of halogen decreases, i.e., NF3 < NCI3 < NBr3 < NI3.
(b) Pentahalides. P, As and Sb form pentahalides of the general formula EX5 due to the presence of

w
vacant </-orbitals in their respective valence shells. N does not form pentahalides due to the absence of
^/-orbitals in its valence shell.
Actually, pentahalides are thermally less stable than trihalides because of the following two reasons :

Flo
(/) As we move down the group, the stability of +3 oxidation state increases while that of +5 oxidation
state decreases due to inert pair effect

e
re
(ii) As the size of the halogen increases from F to I, the strength of the phosphorus-halogen bond decreases
and the steric hindrance increases.

F
The combined effect of these two factors makes pentabromides and pentaiodides particularly unstable.
ur
In the light of the above two reasons, the pentahalides of group 15 are fewer in number as compared to

or
trihalides. For example, pentafluorides of P, As, Sb and Bi are all known, pentachlorides of only P, As and Sb are
sf
known, pentabromide of only phosphorus, i.e., PBrj is known while pentaiodides of none of these elements is
stable. In other words, Bi forms only pentafluoride. The non-existence of pentachloride, pentabromide and
k
Yo
pentaiodide of Bi is probably due to the strongly o^dising properties of Bi^ due to inert pair effect
oo

Preparation. The pentahalides are prepared as follows :

eB

3 PCI5 + 5 ASF3 > 3 PF5 + 5 ASCI3 ; PCI3 + CI2 (in CCI4) > PCI5
AS4O10 + 10 F2 > 4 AsFs + 5 O2 ; Sb40jo + 10 F2 > 4 SbFj + 5 O2
ur

Structure. AU the pentahalides have trigonal bipyramid geometry, i.e., the central atom is sp^d hybridized.
ad

The three halogen atoms occupy equatorial positions while the other two occupy axial positions, Since the
Yo

three equatorial E—X bonds are repelled by two electron pairs but the two axial E—X bonds are repelled by
three electron pairs, therefore, axial bonds are usually longer than the equatorial bonds. The structure of
d

pentahalides is illustrated by the structure of PF5 as shown in Fig. 7.4 b.

Re
in

Properties. (/) As stated above, the pentahalides are thermally less stable than their corresponding
trihalides, i.e., PCI5 > PCI3 + CI2. That is why PCI5 behaves as a good chlorinating agent.
F

PCI5 is covalent in the vapour state but exists as [PCl4]+ [PCy- in the crystalline state ; the ions have
tetrahedral and octahedral structures respectively. In the solid state, PBrj exists as [PBr4]'*^Br while PI5
exists as [Pl4]'^ and 1“ in solution.
(//) All the pentahalides act as Lewis acids due to the presence of vacant ^/-orbitals, the central atom can
accept a pair of electrons thereby expanding its coordination number to 6, i.e.,
MX5 + X- > [MXg]-
As a result, the hybridization of the central atomchanges from sp^d to sp^tf.
(ill) PF5 does not undergo hydrolysis since P-F bond is stronger than P - O bond. All other pentahalides
of P, however, undergo hydrolysis to yield H3PO4.
4. Reactivity towards Metals (Formation of Binary Compounds)
All the elements of group 15 form binary compounds with metals in the -3 oxidation state. For example,
Ca3N2 (calcium nitride), Ca3P2 (calcium phosphide), Na3As (sodium arsenide), Zu3Sb2 (zinc stibide) and
Mg3Bi2 (magnesium bismuthide).
 p-BLOCK ELEMENTS 7/11

7.6. ANOMALOUS PROPERTIES OF NITROGEN

Nitrogen, the first member of group 15 elements, shows anomalous behaviour and differs from rest of
the members of its family. The main reasons for this difference are as follows :
(/) exceptionally small atomic size (//) high electronegativity
{Hi) high ionisation enthalpy

rw
(iv) absence of d-orbitals in its valence shell.
Some important properties in which nitrogen differs from rest of the members of its group are as follows :
1. p7C-p7t multiple bonds. Nitrogen because of its small size and high electronegativity forms
pK - pTC- multiple bonds with itself and with other elements having small size and high electronegativity (e.g.,

e
C, O). Thus, nitrogen exists as a diatomic molecule with a triple bond (N = N, one <5- and two Tt-bonds)
between the two atoms. These N, molecules are held together by weak van tier Waals forces of attraction

e
which can be easily broken by the collisions of the molecules at room temperature. Therefore. N., is a gas at

lo
r
room temperature.
The other elements of this group do not form pn-pti multiple bonds because their atomic orbitals

F
are so

u
large and diffused that they cannot have effective overlapping. Thus, phosphorus, arsenic and antimony do
not form pn-p% multiple bonds. Instead they prefer to form single bonds as P—P, As—As and Sb—Sb while

oF
bismuth forms metallic bonds in the elemental state. Actually, phosphorus, arsenic and antimony exist as

rs
discrete tetraatomic tetrahedral molecules such as P^ (Fig. 7.6, page 7/29), AS4, Sb4, etc. containing E—E
single bonds.

ok
Due to bigger size, the forces of attraction holding the tetraatomic molecules of P4, AS4, Sb4, etc. are
quite strong and hence cannot be broken by the collisions of the molecules at room temperature. Therefore,

o
f
P4, As^, Sb^, all are solids at room temperature.
2. Catenation. The elements of group 15 also show the property of catenation {self linking of atoms)
o
Y
but to a much smaller extent than carbon. The reason being that the E—E bond strength of these elements is
B
rY

much lower than that of C—C bond.

Bond C—C N—N P—P As—As
Bond strength (kJ mol”*) 347 159 213 1474
ue

Among the elements of group 15, phosphorus has the maximum tendency for catenation forming cyclic
as well open chain compounds consisting of many phosphorus atoms. Nitrogen has little tendency for catenation
od

since N—N single bond is very weak due to large interelectronic repulsions between the lone pairs of electrons
ad

present on the N-atoms of N—N bond having small bond length. Nitrogen can form chains containing up to
in

three N-atoms, e.g., hydrazoic acid, N3H or azide ion. NJ ion.

Re

Hydrazoic acid : H — N = N = Nr; Azide ion :

F

-;N=N=Nr
3. Reactivity. Due to the presence of a triple bond between the two N-atoms, the bond length is very
small (109 pm) and hence bond dissociation energy is very large (9414 kJ mol"‘). As a result, Nj is inert and
unreactive in its elemental state.
Since P—^P single bond is much weaker (213 kJ moI“^) than N s N triple bond, therefore, phosphorus is
much more reactive than nitrogen. The reactivity of other elements decreases down the group from As to Bi.
4. Maximum covalency. N is an element of the second period {n = 2). Therefore, it has only s- and
p- orbitals and no fr-orbitals in the valence shell. As a result, nitrogen can show a maximum covalency of 4
and that too when it donates its lone pair of electrons ( NH4, R4N'‘‘). In other words, nitrogen cannot expand
its covalency beyond 4. That is why nitrogen does not form pentahalides (NF5, NCI5, etc.). On the other hand,
phosphorus and other members of group 15 have empty t/-orbitals and hence can exhibit a covalency of either
5 or 6, e.g., PCI5, AsF^, [PF^r, ISbFJ-.
5. -pTC-muUiple bonds. Due to absence of c/-orbitais in the valence shell, nitrogen cannot form
dti-pn multiple bonds. That is why nitrogen does not form compounds of the type, R3N = O and R3N = CH2.
In contrast, due to the presence of vacant ^/-orbitals in the valence shell, phosphorus and other heavier elements
7/12 New Course Chemistry (XlI)tgalBl

of group 15 readily form c/tu - p-K multiple bonds. That is why phosphorus forms compounds of the type R3P
= O. R3P = CH2 (R may be any alkyl group), O = PX3 and RN = PX3 (X = F, Cl, or Br). Phosphorus and
arsenic can also form dK-pn bonds with transition elements when their compounds like P(C2H5)3
(triethylphosphine) and As(C^H5)3 (triphenylarsine) act as ligands.
6. Non-metallic character. On account of high electronegativity, N is the most non-metallic element of
this group, followed by P. In contrast. As and Sb are metalloids while Bi is a metal.
7. Formation of trinegative ion. Oue to small size and high electronegativity N can form trinegative
ions. This tendency is, however, less in P but absent in other elements.
8. Nature of hydrides. Hydride of nitrogen (NH3) is stable while the hydrides of other elements are not
slable. Vunher due to small size and high electronegativity, NH3 undergoes H-bonding while other hydrides
of tliis group do not.
9. Nature of oxides. Nitrogen forms five oxides {N2O, NO, N2O3, NO^ or N2O4 and N2O5) which are
monomeric. Phosphoru.s forms three oxides which are dimeric (P4O6, P40g, and P40](,). Arsenic and antimony
form only two dimeric oxides (AS4O6, AS4O10 and Sh_P(^, Sb40io) while Bi forms only one monomeric oxide
(Bi203).

w
10. Na.'ure of trihalidcs. Except NF3, the trihalides of nitrogen (NCI3, NBr3 and NI3) are unstable and
explosive. The trihalides of other elements are, however, stable.

F lo
7.7. DiNfTROGEN,
Nitrogen was discovered by Daniel Rutherford in 1772. It exists as a diatomic gas (N2) in the elemental

e
Fre
state. Therefore, it is also called dinitrogcn.

7.7.1. Preparation of DInitrogen

for
(«) Commercial preparation of dinitrogen from air. Dinitrogen is prepared commercially from air by
liquefaction and fractional distillation. When liquid air is allowed to distil, dinitrogen having lower boiling
r
point (77-2 K) distils over first leaving behind liquid dioxygen (b.p. 90 K).
You
oks

(b) In the laboratory. Dinitrogen can be prepared by the following three methods:
(/) By thermal decomposition of ammonium nitrite. When an equimolar aqueous soludon of ammonium
eBo

chloride and sodium nitrite is heated, ammonium nitrite is first formed as a result of double decomposition
reaction. It being unstable, decomposes immediately to form dinitrogen gas.
NH4CI (aq) + NaN02 (aq) > NH4NO2 (aq) + NaCl (aq)
our
ad

Amm. nitrite

Heat
NH4NO2 (aq) > N2(g)+2H2O(0
dY
Re

Small amounts of NO and HNO3 are also formed in this reaction. These impurities can be removed by
passing the gas through aqueous sulphuric acid containing potassium dichromate.
Fin

(//) From ammonium dichromate. Dinitrogen can also be prepared by thermal decomposition of
ammonium dichromate.
Heat
(NH4)2Cr207 ^ N2 + 4H2O + Cr203
Amm. dichromate Chromic oxide

Ammonium dichromate may be prepared in situ by heating an equimolar mixture of potassium dichromate
and ammonium chloride.
Heat

K2Cr207 + 2 NH4CI > (NH4)2Cr207 + 2 KCl

(Hi) By thermal decomposition of sodium or barium azide. Very pure dinitrogen is obtained by
thermal decomposition of sodium or barium azide.
573K Heat

2 NaN3 - ^ 2 Na + 3 Nj ; Ba(N3)2 - Ba + 3 N2
Sodium azide Barium azide

Thermal decomposition ofNaN^ is used to inflate the air bags used for safety devices in some cars.
 p-BLOCK ELEMENTS 7/13

Curiosity Question
Q. How do the airbags installed in the dashboard of your car work ?
Ans. When accident occurs, the sensor provided in the car detects the collision. It sends an electrical
signal which overheats and ignites sodium azide (NaNg) placed in the airbag. Sodium azide is a
fast burning fuel that produces large amounts of Ng gas which goes through filters and fills the nylon
airbag. The airbag then hits your head and protects you from injury.
2NaNg >2Na + 3N2
After your head hits the nitrogen filled bag, it deflates by releasing the gas through tiny holes. The
cloud of smoke that fills the vehicle is actually talcum powder or corn starch. The powder prevents
the bag from sticking to itself, while it is folded inside the car. The Ng gas that is released is absolutely
harmless because Ng constitutes about 79% of the air that we inhale. One just needs to open the

w
doors or windows for the gas and powder to escape.
J
7.7.2. Properties of Dinitrogen

Flo
{a) Physical properties, (i) Dinitrogcn is a colourless, tasteless, non-toxic gas. But animals die in it for
want of oxygen.

ee
{//) U has two stable isotopes : and

Fr
(m) It is very slightly soluble in water (23-2 cnr^ per litre of water at 273 K and I bar pressure).
(iv) It has low freezing (63-2 K) and boiling point (77-2 K).

for
(v) It is adsorbed by activated charcoal.
ur
(b) Chemical properties. Dinitrogen is rather inert and unreactive at room temperature due to its high
bond dissociation enthalpy (941-4 kJ mor'). Reactivity, however, increases with rise in temperature. Some
s
important chemical properties of dinitrogen are discussed below :
ok
Yo
1. Action of litmus. It is neutral towards litmus.
Bo

2. Action of active metals. Dinitrogen is neither combustible nor a supporter of combustion. However,
many active metals when heated at high temperatures, keep on burning in an atmosphere of dinitrogen forming
re

their respective ionic nitrides.

Heat Heat
ou
ad

6 Li + Ng ■>
2 LigN ; 3 Mg + Ng MggNg
Lithium nitride Magnesium nitride
Y

Heat Heat
3 Ca + Ng ■>
CagNg 2 A1 + Ng 2 AIN
nd
Re

Calcium nitride Aluminium nitride

3. Action of non-metals. Dinitrogen also combines with non-metals as discus.sed below :

Fi

(i) Action of dihydrogen. When a mixture of dinitrogen and dihydrogen is healed to about 700 K under
a pressure of 200 atmospheres, in presence of iron oxide as catalyst and molybdenum as promoter, ammonia
is formed.

Ng (g) -H 3 Hg (g) ^ ~ 2 NHg (s)

This reaction forms the basis of Haber's process for the manufacture of ammonia,
(ii) Action of dioxygen. Dinitrogen and dioxygen combine to form nitric oxide when the mixture is
healed to 2273—3273 K in an electric arc.

Ng + Og ^ ± 2ND
Nitric oxide

This reaction forms the basis of manufacture of nitric acid by Birkland and Eyde process.
4. Action of alumina in presence of coke. Aluminium nitride is formed.
1273K
AlgOg -1- 3C-1-N2 2 AIN -1-3CO
Alumina Aluminium nitride
7/14 “p>uidee^'^ New Course Chemistry (XII)BSm

5. Action of calcium carbide. Calcium cyanamide is fonned.

1273 K

CaC2 + N2 > CaCNj + C

Cal. carbide Cal. cyanamide
Calcium cyanamide reacts with water to form ammonia.
CaCN2 + 3 H2O ^ CaC03 + 2 NH3
Cal. cyanamide
Therefore, it is used as a fertilizer under the name nitrolim (CaCN2 + C).

7.73. Active Nitrogen

It can be made by passing an electric spark through N2 gas at a low pressure (about 2 mm). This forms
atomic nitrogen and the process is associated with yellow-pink after glow. Active nitrogen is very reactive
and reacts with a number of elements and breaks many stable molecules.

w
HC^CH + 2N ^ NsC—C = N + H2
Acetylene Active nitrogen Cyanogen

Flo
7.7.4. Uses of Dinitrogen
{/) It is used in the manufacture of nitric acid, ammonia, calcium cyanamide, etc.

ee
(//) Dinitrogen provides an inert atmosphere in iron and steel industry. It also acts as an inert diluent in
reactive chemicals

Fr
{Hi) Dinitrogen is used in filling electric bulbs to reduce the rate of volatilisation of the tungsten filament.
(iv) Dinitrogen gas-filled thermometers are used for measuring high temperatures,

for
(v) Liquid dinitrogen is used as a refrigerant to preserve biological materials, food items and in cryosurgery.
ur
7.8. AMMONIA
s
Ammonia is present in small quantities in air and soil where it is formed by the decay of nitrogenous
k
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organic matter, e.g., urea

NH2CONH2 + 2H2O ^ (NH4)2C03- 2 NH3 + HjO + CO2
eB

Urea Amm. carbonate Ammonia

7.8.1. Preparation of Ammonia

r
ou
ad

(1) By heating ammonium salts with a strong base. On a small scale, ammonia is prepared by heating
ammonium salts with a strong base.
Y

Heat
(NH4)2$04 + 2NaOH ■>
2 NH3 + 2 H2O + Na2S04
Re
nd

Amm. sulphate Ammonia

Heat
Fi

NH4CI -h KOH > NH3 + H2O + K.C1

In the laboratory, ammonia is prepared by heating a mixture of slaked lime and ammonium chloride.
Heat
2 NH4CI + Ca(OH)2 > 2 NH3 + 2 H2O + CaCl2
Slaked lime

(2) By the action of water on metal nitrides

Mg3N2 + 6 H2O > 3 Mg(OH)2 + 2 NH3 ; AIN -h 3H2O 4 A1(0H)3 + NH3
Aluminium nitride
Mag. nitride
Drying of ammonia gas. The moisture present in ammonia may be removed by passing it through a
glass lower packed with quick lime, CaO. It cannot be dried by :
(/) Cone. H2SO4 since it reacts with it forming ammonium sulphate.
2 NH3 + H2SO4 > (NH4)2S04
(//) Anhydrous calcium chloride since it forms a complex having the composition, CaCl2.8NH3.
{Hi) Phosphorus pentoxide since it reacts to form ammonium phosphate.
P2O5 + 3 H2O > 2 H3PO4; 3 NH3 4- H3PO4 > (NH4)3P04
p-BLOCK ELEMENTS 7/15

7.8.2. Manufacture of Ammonia

On a commercial scale, ammonia is manufactured by Haber’s process.

N2(g) + 3H2(g) ± 2 NH3 (g); = - 92.4 kJ mol-‘

This reaction is reversible, exothermic and occurs with decrease in volume. Therefore, according to
Chatelier’s principle, the favourable conditions for the manufacture of ammonia are :
(/) Low temperature. Since the forward reaction is exothermic, low temperature will favour the formation
of ammonia. However, at low temperatures, the rate of the reaction will be slow. The optimum temperature
for the reaction has been found to be around 700 K.

(//) High pressure. Since the forward reaction occurs FIGURE 7.5

with decrease in volume, high pressure will favour H2 sv

the formation of ammonia. The reaction is usually ^ Pump
carried out at a pressure of about 200 x 10^ Pa or

w
N2-^ Compressor
200 atmospheres. 20 MPa
N2 H2^
I
(/;/) Catalyst. The rate of reaction is fairly low around
Catalyst
700 K. It is increased by using iron oxide as catalyst

Flo
At
Iron oxide
700 K
with small amounts of K2O and AI2O3. Sometimes, AI2O3 +K2O
molybdenum is used as a promoter (which

e
re
increases the efficiency of the catalyst). (N2 + H2 + NH3) Liquid NH3
The (low chart for production of ammonia is shown

F
Flow chart for the manufacture
in Fig. 7.5. of ammonia.
ur
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7.8.3. Physical Properties of Ammonia
fo
1. Ammonia is a cohurless gas with a characteristic pungent smell called the ammoniacal smell. It
causes tears in the eyes.
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2. It is lighter than air (density = 0-68 g/enr^)
Yo
oo

3. It is extremely soluble in water; one volume of water can dissolve about 1000 volumes of the gas at
eB

273 K.

4. Ammonia can be easily liquefied by cooling under pressure. Liquid ammonia boils at 239-6 K and freezes
at 198-4 K. Like water, ammonia is also a.ssociated in solid and liquid states through hydrogen bonding. This
ur

accounts for its higher melting and boiling points than expected on the basis of its molecular ma.s.s.
ad

5. When vapourized, liquid ammonia cau.ses intense cooling.

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7.8.4. Structure of Ammonia

The nitrogen atom in NH3 is sp^-hybridised. The three of the four ,vp^-orbitals form three N—H ct-
d
Re

bonds while the fourth contains a lone pair of electrons. Since the lone pair-bond pair repulsions ai-e stronger
in

than bond pair-bond pair repulsions, therefore, the H—N—H bond angle decreases from 109° 28' to 107-8°.
F

As a result, molecule has pyramidal geometry with N—H bond length of 101-7 pm and bond angle oj
lOV-S". (Fig. 7.1/7, page 7/8).
7.8.5. Chemical Properties of Ammonia
1. Basic nature. Ammonia is highly soluble in water. Its aqueous solution is weakly basic due to the
formation of OH" ions.

NH3(g)-pH20 (/) ^ ± NH4OH {aq) V i NH4(«^)-f-OH-(a^)

Being basic, it turns moist red litmus blue and neutralises acids in the dry stale as well as in aqueous
solutions forming their corresponding salts.
NH3-+- HCl > NH4CI ; 2 NH4OH-I-H2SO4 (NH4)2$04-P 2 H2O
2. Reaction with heavy metal salt solutions. Ammonium hydroxide reacts with many metallic salts
and precipitates them as hydroxides.
FeCl3 iaq) + 3 NH4OH (aq) » Fe(OH)3 (5) 4. -P 3 NH4CI (oq)
Ferric chloride Brown ppt.
7/16 7^nadeefi>'^ New Course Chemistry (XII)EE

AICI3 (aq) + 3 NH4OH iaq) ■>

A1(0H)3 (s)i +3 NH4CI iaq)
Aluminium chloride Gelatinous white ppt.
CrCl3(fl^) + 3NH40H(a^) Cr(OH)3 (s)i + 3 NH4CI iaq)
Chromium chloride Green ppt.
This property is made use of in precipitating these metals as their hydroxides in the group III of qualitative
analysis.
3. As a Lewis base. Due to the presence of a lone pair of elections on the nitrogen atom, ammonia acts as a
Lewis base. Consequently, it can easily donate its electron pair to form co-ordinate bond with electron-deficient
molecules (such as BF3) or transition metal cations having vacant J-orbitals to form complexes. For example,
NH3 + BF3 >H3N->BF3 ; Ag+(a^) + 2NH3(fl^) > [Ag(NH3)2r (o^)

w
Cu2+ (aq) + 4 NH3 (aq) > [Cu(NH^)4\^* (aq); Cd^^ (aq) + 4 NH3 (aq) > [Cd(NH3)4]2+ (aq)
Thus, ammonia acts as a ligand.
Due to the formation of complex ions, white ppt. of silver chloride dissolves in excess of ammonia to form a
complex compound [Ag(NH3)2]Cl, diamminesilver (I) chloride. Similarly, copper sulphate dissolves in excess of

o
ammonia to form soluble deep blue coloured complex [Cu(NH3)4]S04, tetraamminecopper (II) sulphate.

e
re
AgCl is) + 2 NH4OH iaq) > [Ag(NH3)2]Cl iaq) + 2H2O (/)

rFl Diamminesilver (I) chloride

F
CUSO4 iaq) + 4 NH4OH iaq) [Cu(NH3)4] SO4 (09) + 4H2OH)
Tetraamminecopper (II) sulphate

r
These reactions are used as tests for these cations in qualitative analysis.
ou
fo
4. Oxidation, (a) When ammonia is passed through a solution of calcium hypochlorite (bleaching
ks
powder), bromine water or passed over heated copper oxide, it is oxidised to dinitrogen gas.
4 NH3 + 3 Ca(OCl)2 > 2 N2 + 3 CaCl2 + 6 H2O
oo

8NH3 + 3Br2 > N2 + 6NH4Br

Y
eB

Heat
2 NH3 + 3 CuO ^ 3CU + 3H2O + N2
ib) When ammonia, mixed with an excess of ai,r is passed over Pt/Rh gauze at 500 K under a pressure
r

of 9 ba,r it is oxidised to nitric oxide.

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Pt/Rh gauze
Y
ad

4NH3 + 5O2 ^ 4NO + 6H2O

500K,9bar
This reaction forms the basis for manufacture of nitric acid by Ostwald’s process.
d

5. Action of halogens. Halogens react with ammonia as follows :

in
Re

With excess NH3: 8 NH3 + 3 CI2 ^ 6NH4CI + N2

F

>^^th excess CI2: NH3 + 3CI2 ^ NCI3 + 3HC1

With Br2 : 8 NH3 -I- 3 Br2 ^ 6NH4Br + N2

With I2 : 2NH3-h3l2 ■ 4 NH3.NI3 -t- 3HI
Nitrogen tri-iodide
ammoniate {brown ppt.)
When NH3.NI3, in the dry state, is either rubbed or struck against a surface, it explodes with noise
liberating vapours of iodine.
8 NH3.NI3 ^ 5N2 + 9I2 + 6NH4I
Thus, it is a mild and harmless explosive.
6. Reaction with carbon dioxide. WTien gaseous CO2 is reacted with liquid NH3 at 453—473 K under
a pressure of 220 atmospheres, it first forms ammonium carbamate which subsequently decomposes to give
urea.
453-473K
2 NH3 -I- CO2 220atm
^ [NH2COONH4] NH2CONH2 + H2O
Amm. carbamate Urea
p-BLOCK ELEMENTS 7/17

Urea has a very high nitrogen content (46.66%). It is widely used as a fertilizer since it slowly decomposes
in soil to give NH3 and CO2.
NH2CONH2 + H2O ^ 2 NH3 + CO2
7. Action of metals. Ammonia when passed over molten sodium or potassium metal at 575 K form
corresponding amides with the liberation of dihydrogen.
575 K 575K

2 Na + 2 NH3 ^ 2 NaNH2 + H2 ; 2 K + 2 NH3 ^ 2 KNH2 + H2

Sodamide Pot. amide

8. Reaction with Nessier’s reagent. With Nessler’s reagent (an alkaline solution of K2Hgl4), ammonia
or ammonium salts give a brown precipitate due to the formation of iodide of Millon's base.
2 K2Hgl4 + NH3 + 3 KOH ^ H2N — Hg — O — Hg — I + 7 KI + 2 HjO
Nessler’s reagent Iodide of Millon’s base {brown ppt.)
9. Reaction with sodium hypochlorite. When a large excess of aqueous ammonia is treated with
sodium hypochlorite in presence of glue or gelatine, it gets oxidised to hydrazine.

w
2 NH3 + NaOCl ^ NH2.NH2 + NaCl + H2O
Hydrazine

F lo
Actually the reaction proceeds in two steps :
(0 NH3 + NaOCl ^ NaOH + NH2CI {Fast reaction)
Chloramine

ee
(/ONH2CI + 2NH3 ^ NH2.NH2 + NH4CI (Slow reaction)

Fr
Glue or gelatine catalyses the slow reaction and prevents the oxidation of hydrazine to nitrogen
N2H4 + 2 NH2CI > N2 + 2 NH4CI
for
10. Liquid ammonia as a solvent Just like water, ammonia also undergoes self-ionisation in liquid stale.
ur
2NH 3
^ NHj + NH2 ; {cf 2 H2O H30-" + OH-)
s

Therefore, liquid ammonia is used for dissolving many polar compounds and also for carrying out many
ook
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reactions in the non- aqueous medium.

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7.8.6. Uses of Ammonia

Some important uses of ammonia are :
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(0 In the formation of various nitrogenous fertilizers such as urea, ammonium nitrate, ammonium sulphate,
ad

ammonium phosphate, calcium ammonium nitrate (CAN), etc.

(ii) In the manufacture of nitric acid by the Ostwald’s process and in the manufacture of sodium carbonate
dY

by the Solvay’s process.

Re

(Hi) Liquid ammonia is used as a refrigerant in ice-factories and cold storages,

(/v) As a cleansing agent for removing grease, etc.
Fin

(v) As a laboratory reagent.

7.8.7. Tests of Ammonia
(0 It turns moist red litmus paper blue and moist turmeric paper brown.
(ii) It gives brown precipitate with Nessler’s reagent
{Hi) With a drop of HCl, it produces dense white fumes of ammonium chloride,
(/v) With copper sulphate solution, it gives a deep blue solution,
(v) It gives a yellow precip ite with chloroplatinic acid.
H2PtCl6 + 2NH3 (NH4)2PtCl6
Chloroplatinic acid Amm. chloroplatinate {yellow ppt.)

7.9. GOES OF NITROGEN

Nitrogen forms a number of oxides in which the oxidation stale varies from +1 to +5. The names,
formulae, common methods of preparation, physical appearance and chemical nature of these oxides are
given in Table 7.6.
7/18 T^'Kutee/b-'A New Course Chemistry (XIl)B!ZSQn

TABLE 7.6. Oxides of Nitrogen

Name Formula Oxidation Common methods of Physical appearance and

state of preparation chemical nature
nitrogen

473-533K
1. Dinitrogen NoO 4- 1 (/) It is a colourless neutral gas
(0 NH4NO3
oxide or with a sweet odour
N2O -i- 2 H2O
Nitrogen (I)
473 K
oxide or
(//●) HNO2 + NH2OH (//) N2O -I- NaNHo ■ ■>

ow
Laughing gas -Sodamide
Hydroxylamine
Aq NaN3 -1-H2O
^ N2O-I-2H2O Sod. azide

(m)HN02 + HN3 873 K

e
m 2 NjO (g)

re
Hydrazoic acid

rFl Aq 2N2(g) + 02 (g)

F
> N20-HN2 + H20
2. Nitrogen NO + 2 (/) 2 NaN02 + 2 FeS04 (0 It is a colourless neutral gas

r
monoxide or
4- 3 H2SO4 Fe2(S04)3
ou
fo
1373-1473K
Nitrogen(II)
oxide or 4- 2 NaHS04 4- 2 HjO 07) 2 NO (g)
ks
4-2 NO
Nitric oxide N2 (g) 4- O2 (g)
(ii) 3 Cu -H 8 HNO3 (dil) 077) Being very reactive due to the
oo

3 Cu(N03)2 + 2 NO presence of an odd electron

Y

(paramagnetic), it instantly reacts

eB

4-4H2O
with O2 to form NO2 and with CI2
(m) 4 NH3 4- 5 O2
Pi
to form nitrosyl chloride (NOCl)
r

4 NO 4- 6 H2O 2NO4-O2 >2N02

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HOOK
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2 NO 4- CU > 2 NOCl
ad

(/v)2HN02 4-2I-4-2H-^
> 2 NO 4-12 4- 2 H2O
d

<250K
in
Re

3. Dinitrogen N2O3 4-3 2 NO 4- N2O4 4

(/■) Blue liquid/solid, acidic in
trioxide or
(anhydride of 2N2O3
F

nature
Nitrogen nitrous acid)
sesquioxide
N2O3 -h H2O ^2HN02
Nitrogen(III) >250K
oxide (//) N2O3 ^ NO 4-NO,
673 K

4. Nitrogen NO, 4-4 (/) 2 Pb(N03)2 (0 NO2 is a brown gas, acidic in

dioxide or
2 PbO 4 NO2 4- O2
nature while N2O4 is a colourless
Nitrogen solid/liquid, acidic in nature.
(ii) Cii + 4 HNO3 (cone.)
peroxide or (i7) Being paramagnetic, NOj
Nitrogen(IV) > Cu(N03)2 4- 2 NO2 reacts with F2 and CI2 forming
oxide + 2 H,0 nitryl fluoride (NO2F) and nitryl
chloride (NO7CI) respectively.
2NO24-F2 4 2 NO,F
2 NO2 4- CI2 2 NO,Cl
(/77) It is acidic in nature
p-BLOCK ELEMENTS 7/19

Name Formula Oxidation Common methods of Physical appearance and

state of preparation chemical nature
nitrogen

Cold
Dinitrogen N2O4 + 4 2NO2 ^
■»
N2O4 (or 2 NO2) + H,0
Heal
tetroxide or (Brown) HNO, + HNO3
(mixed N2O4
Nitrogen (IV) anhydride of (Colourless)
oxide
HNO2 and
HNO3)
5. Dinitrogen N2O5 -1-5 4 HNO3 -I- P40,o (i) Exists as colourless solid
pentoxide or below 273 K. Above this temp
(anhydride of 4 HPO3 -I- 2 N2O5
erature, it starts decomposing

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Nitrogen (V) nitric acid)
oxide N2O5 ^ NO2 + NO O2
(ii) It is acidic in nature.

F lo
N205-HH20 >2HN03

ee
Lewis dot structures, main resonance structures and bond parameters of oxides of nitrogen are given in

Fr
Table 7.7.
TABLE 7.7. Structures of Oxides of Nitrogen

for
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Other Characteristic
Formula Resonance structure Bond Parameters
Properties
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ok
Linear molecuie, has
-A-J?
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+ 113pm 119pm low dipole moment
;n=n=!=o:'«-^:n=n—o: O
N2O
o

(0-17D).
eB

:n=o:
A +
115pm (Linear)
Odd electron molecule
NO ●n=o:-<-^'N=o:
Dimer of nitric oxide paramagnetic in the
r

0= N—N=0 gaseous state,

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ad

or diamagnetic In liquid
and solid states due to
N O
238 pm dimerisation
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115 pm
O N
Re
nd

It exists in two different

Fi

N2O3
0: P forms, symmetric and
105^
+ asymmetric. These may
or
N N |g'^114pm be interconverted by
ON
\^/
O
NO
/ irradiation with light of
V. :o
0121pm appropriate waveiength.
Symmetric The N-N bond in
form Asymmetric form (Planar)
asymmetric form is very
weak as compared with
N-N bond in hydrazine.

NO2 N N Odd electron molecule,

●4 paramagnetic,
:p. o: :o Q- dimerises on cooling to
Resonance hybrid form N2O4
(Angular)
7/20 New Course Chemistry (XII)ESSm

Other Characteristic
Formula Resonance structure Bond Parameters
Properties

(i) NO2-N2O4 system

O. is a strong oxidising
o: o:
175pm agent. It oxidises HCI
N2O4
*
135 ;n n:
N N to Ci2 and CO to CO2
O' ''o 2N02+4HCI^
'6:
Resonance hybrid 2NOCI + Ci2 + 2H2O
(Planar) NO2 + CO NO + CO
(ii) Diamagnetic,
dissociates on heating
to form NO2

O Solid N2O5 exists as

w
N205 X/“v
.-cy' \
\ / \Aj
o'^
:n
112

N^134" 0
NO2 NO3 and hence
is cailed nitronium
nitrate
:o

Flo
:o 0: :o Resonance hybrid
(Planar)
it acts as a strong

ee
oxidising agent.
N205+Na^NaN03

Fr
+ NO2
N2O5 + i2 l2^5 ^2

for
ur
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. N2O is used as an anaesthetic agent by dentists for minor operations.

s
ok
Yo
2. Although nitric oxide (NO) is very reactive and harmful yet it occurs in traces in biological systems. It
acts as a neurotransmitter and helps in controlling blood pressure by relaxing blood vessels.
Bo

3. N2O4 is used as an oxidiser for rocket fuels in missiles and space vehicles.
4. Nitric oxide (NO) readily forms complexes with transition metals called nitrosyls. Two important nitrosyl
re

complexes are : [Fe(H20)5N0]^''' (which is responsible for the colour in the brown ring test for nitrates)
and sodium nitroprusside Na2[Fe(CN)5N0].2H20. (which is used for testing
ou

ions.
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5. Nitric oxide acts both as an oxidising as well as a reducing agent while all other nitrogen oxides act as
Y

oxidising agents only.

Reducing agent: 6 KMn04 + 9 H2SO4 + 10 NO ^ 3K2SO4 + 6MnS04 + IOHNO3 + 4H2O
nd
Re

3 I2 + 2 NO + 4 H2O ■^61-+ 2NO“ + 8

Fi

Oxidising agent: SO2 + 2 NO + HjO -> H2SO4 + N2O

H2S + 2 NO S + N2O + HjO

7.10. NITRIC ACID, HNO3

Nitrogen forms three oxoacids such as H.,N202 (hyponitrous acid), HNO2 (nitrous acid) and HNO3
(nitric acid). Their structures are :
OH O O
■N=N' H—O—N=0 H—O—N or H—O—N
HO Nitrous acid
N
o O-
Hyponitrous acid Nitric acid Nitric acid

Amongst these, HNO3 is the most important.

7.10.1. Preparation of Nitric Acid
In the laboratory, nitric acid is prepared by heating KNO3 or NaN03 and concentrated sulphuric acid
in a glass retort.
p-BLOCK ELEMENTS 7/21

Heat
NaN03 + H2SO4 ■> NaHS04 + HNO3
On commercial scale, it is mainly prepared by Ostwald’s process. This method is based upon the
catalytic oxidation of ammonia by atmospheric oxygen.
Pt/Rh gauze catalyst
4NH3(g) + 502(g) ^ 4N0(g) + 6H20(g)
500K,9bar

Nitric oxide thus formed readily combines with oxygen to give NO2.
2 NO (g) + O2 (g) ^ 2 NO2 (g)
Nitrogen dioxide thus formed dissolves in water to form HNO3.
3 NO2 (g) + H.O {/) ^ 2 HNO3 (fl^) + NO (g)
HNO3 is concentrated by distillation to give
The nitric oxide thus formed is recycled and the aqueous
68% HNO3 by mass. Funher concentration to 98% can be achieved by dehydration with concentrated H2SO4.
7.10.2. Structure of Nitric acid

w
Spectroscopic studies have shown that in the gaseous stale, HNO3 exists as a planar molecule with the

F lo
bond angle and bond length as shown.
102°
<V?
N 130°

ee
140-6pm

Fr
O
Gaseous nitric acid

Actually nitric acid is a resonance hybrid of the following two structures : for
ur
o: .0:
s
+ +
ook
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HO N ► HO N

X...o: V.
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0:

Resonance structures of HNO3

our

Like nitric acid, nitrate (NOJ) ion is a resonance hybrid of the following three resonance structures;
ad

:o: :o: :b 0
I
Y

+
I r\i20°
N ■4- > N
Re
nd

o: 0 o:
o o
Fi

Resonance
Resonating structures of NO3 ion hybrid
As a result of resonance, all the three N—O bond lengths in NOJ ion are equal and are 218 pm long.
7.10.3. Physical Properties of Nitric Acid
1. Pure nitric acid is a colourless fuming liquid with pungent odour. However, impure nitric acid is
yellow due to the presence of dissolved oxides of nitrogen.
2. Pure nitric acid freezes at 231-4 K and boils at 355-6 K.
3. Laboratory grade nitric acid is a constant boiling (394 K) mixture, (i.e., azeotrope) containing 68%
HNO3 and 32% HjO by mass. It has a specific gravity of 1-504.
4. Fuming nitric acid is pure nitric acid containing oxides of nitrogen dissolved in it.
7.10.4. Chemical Properties of Nitric Acid
1. Decomposition. It decomposes on healing giving nitrogen dioxide, oxygen and water.
Heat
4 HNO3 (aq) > 4 NO2 (g) + O2 (g) + 2 H2O (0
7/22 7\eidee^'4. New Course Chemistry (XII)CEiai

2. Acidic nature. Nitric acid is a strong monobasic acid. In aqueous solution, it ionizes as :
HNO3 (fl^) + HoO (/) > H3O+ {aq) + NOJ {qq)
Being a monobasic acid, it reacts with metallic oxides, hydroxides, carbonates and bicarbonates giving
only one series of salts called nitrates.
CaO + 2 HNO3 > Ca(N03>2 + H2O ;
NaHC03 + HNO3 > NaN03 + CO2 + H2O
Na2C03 + 2 HNO3 > 2 NaN03 + CO2 + H2O
3. Oxidising agent. Nitric acid is a very strong oxidising agent since it readily gives nasent oxygen both
in the concentrated as well as in the dilute form.

2 HNO3 (cone.) > H2O + 2 NO2 + [O]

ow
2 HNO3 idil) > H2O + 2 NO + 3 [O]
Therefore, nitric acid oxidises many non-metals and compounds.
I. Oxidation of non-metals. Dilute nitric acid has no action on non-metals. However, cone, nitric acid
oxidises many non-metals such as carbon, sulphu,r iodine and metalloids like arsenic, antimony, etc. to their
corresponding oxy-acids while nitric acid itself is reduced to nitrogen dioxide. For example,

e
re
rFl
(0 Carbon is oxidised to carbonic dioxide {CO^.
2 HNO3 > H2O + 2 NO2 + [O] X 2

F
C + 2[0] >C02

r
C + 4HNO3 >C02 + 2H20 + 4N02

fo
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(//) Sulphur is oxidised to sulphuric acid (H2SO4).
ks
2 HNO3 > H2O + 2 NO2 + [O] X 24
Sg + 24 [O] + 8 H2O > 8 H2SO4
oo

Sg + 48 HNO3 > 8 H2SO4 + 48 NO2 + 16 H2O

Y
eB

(Hi) Iodine is oxidized to iodic acid (HIO^).

2 HNO3 ^^20 + 2 NO2 + [O] X 5
ur

l2 + 5[0] >1205
ad

I2O5 + H2O >2HI03

Yo

I2 + 10 HNO3 >2 HIO3 + 10 NO2 + 4 H2O

d

(/v) Phosphorus is oxidised to orthophosphoric acid (H^PO^)

Re
in

2HNO3 H2O + 2 NO2 + [O] X 10

F

P4+IO [O]- ^4^10

P4010 ^ ^2^' ■^4H3P04
P4 + 20 HNO3 > 4 H3PO4 + 20 NO2 + 4 H2O
(v) Arsenic is oxidised to arsenic acid (H^AsO^)
2 HNO3 > H2O + 2 NO2 + [O] X 5
2 As + 5 [O] > AS2O5
AS2O3 + 3 H2O' ■ ') 2 H3ASO4
2 As + 10 HNO3 2 H3ASO4 + 2 H2O + 10 NO2
or
AS + 5HN03- H3ASO4 + H2O + 5 NO2
Arsenic acid

Similarly, Sb + 5 HNO3 ^ H3Sb04 + H2O + 5 NO2

Antimonic acid
p-BLOCK ELEMENTS 7/23

n. Oxidation of compounds. Dilute as well as concentrated nitric acid oxidises a number of compounds.
For example,
(/) Hydrogen sulphide is oxidised to sulphur
(a) With cone. HNO3 : 2HNO3 -4 H2O + 2 NO2 + [O]
H2S + [O] - ^H20 + S
H2S + 2HNO3 2 H2O + 2 NO2 + S i
(b) With dil. HNO3 : 2HNO3 H2O + 2 NO + 3 [O]
H2S + [0]- ^H20 + S]x3
3 H2S + 2 HNO3 ■» 4 H2O + 2 NO + 3 S 4
(//) Sulphur dioxide is oxidised to sulphuric acid

low
(a) With cone. HNO3 : 2HN03 H2O + 2 NO2 + [O]
S02 + H20 + [0] ^H2S04
S02 + 2HN03- ^H2S04 + 2N02
(b) With dil. HNO3 ; 2HN03 -4 H2O + 2 NO + 3 [O]

ee
SO2 + H2O + [O] H2SO4] X 3
rF
Fr
3 SO2 + 2 HNO3 + 2 H2O ^ 3H2SO4 + 2 NO
iiii) Ferrous sulphate is oxidised to ferric sulphate
(a) With cone. HNO3 : 2HNO3 H2O + 2 NO2 + [O]
for
u
2 FeS04 + H2SO4 + [O] ■¥ Fe2(S04)3 + H2O
ks
Yo
2 FeS04 + 2 HNO3 + H2SO4 - ^ Fe2(S04)3 + 2 H2O + 2 NO2
o

(b) With dil. HNO3 : 2HNO3 -4 H2O + 2 NO + 3 [O]

Bo

2 FeS04 + H2SO4 + [O] 4Fe2(S04)3 + H20]x3

re

6 FeS04 + 2 HNO3 + 3 H2SO4 4 3 Fe2(S04)3 + 4 H2O + 2 NO

Nitric oxide thus evolved combines with ferrous sulphate to form a dark brown addition compound,
ou
ad

FeSO4.NO, nitroso-ferrous sulphate. This reaction forms the basis of the ring test for nitrates.
Y

4. Action on metals. With the exception of gold and platinum (noble metals), nitric acid attacks all
metals forming a variety of products. On die basis of reactivity, metals have been divided into the following
nd
Re

three categories :
I. Reaction with metals which are more electropositive than hydrogen (Na, K, Ca, Mg, Al, Mn, Zn, Cr,
Fi

Cd, Fe, Co, Ni, Sn, Pb, Sb, etc). In this case, nascenthydrogen is liberated which further reduces nitric acid giving
a variety of reduction products such as NO2, NO, NH3, NH4NO3 and N2O as shown below :
Metal + HNO3 4 Metal nitrate + H

HNO3 + H > Reduction product + H2O

+H +2H +5H HNO A
HNO3 4 NO2 4 NO 4 NH3 U NH4NO3 4 N2O
-H2O -H2O -H2O -2H2O
The stage to which the reduction actually occurs, however, depends upon the following factors :
(a) nature of the metal, {b) concentration of the acid and (c) temperature. In general, more the dilution
greater is the extent of reduction. For example,
(i) Very dilute nitric acid. Magnesium and manganese are the only metals which produce hydrogen
with very dilute (1—2%) nitric acid.
Mg + 2HN03 4 Mg(N03)2 + H2
Mn + 2 HNO3 4 Mn(N03)2 + H2
7/24 “Pfutdee^'^, New Course Chemistry (XII)GZ

(ii) Cold dilute nitric acid. More active metals like magnesium, zinc, tin and iron react with cold dil.
HNO3 to form ammonium nitrate.
Zn + 2HN03 ■^Zn(N03)2 + 2H]x4
HNO3 + 8H ^ NH3 + 3H2O
NH3 + HNO3 ■» NH4NO3
4 Zn + 10 HNO3 > 4 Zn(N03)2 + 3 H2O + NH4NO3
Amm. nitrate

Similarly, 4 Sn + 10 HNO3 —^ 4 Sn(N03)2 + 3 H2O + NH4NO3

4 Fe + 10 HNO3 —^ 4 Fe(N03)2 + 3 H2O + NH4NO3
Lead, under similar conditions, gives nitric oxide instead of ammonium nitrate.

ow
Pb + 2 HNO3 > Pb(NOi3)2 + 2H] X 3
HNO3 + 3H > NO + 2H2O] X 2
3 Pb + 8 HNO3 ^3Pb(N03)2 + 4H20+ 2 NO

e
Nitric oxide

re
rFl
(iii) With hot dilute nitric acid. With hot dilute nitric acid, ammonium nitrate thus formed undergoes
decomposition to form nitrous oxide (N2O).

F
Zn + 2 HNO3 ■ ^ Zn(N03>2 + 2 H] X 4
HNO3 + 8 H 3H2O + NH3

r
ou
NH3 + HNO3 -4 NH4NO3
fo
ks
NH4NO3 N2O + 2H2O
4 Zn + 10 HNO3 ^ 4 Zn(N03)2 + 5 H2O + N2O
oo

Nitrous oxide
Y
eB

Similarly, 4 Mg + 10 HNO3 ■> 4 Mg(N03)2 + 5 H2O + N2O

(iv) With cone, nitric acid. With cone. HNO3, metals like zinc, magnesium, bismuth, lead, etc. form
nitrogen dioxide (NO2).
ur

Zn + 2HN03 ^Zn(N03)2 + 2H
ad
Yo

HNO3 + H ->H20 + N02]x2

Zn + 4HN03 >Zn(N03)2 + 2H20+ 2NO2
d
Re

Nitrogen dioxide
in

Similarly, Mg + 4 HNO3 > Mg(N03)2 + 2 H2O + 2 NO2

F

Bi + 6 HNO3 > Bi(N03)3 + 3 H2O + 3 NO2

Pb + 4 HNO3 ^ Pb(N03>2 + 2 H2O + 2 NO2
Exception. Tm, however, eracts with cone. HNO3 to form meta-stannic acid (H2Sn03) and nitrogen dioxide.
2 HNO3 > H2O + 2 NO2 + O] X 2
Sn + 2[0] + H2O > H2Sn03
Sn + 4HNO3 > H2Sn03 or Sn02.H20 + 4 NO2 + H2O
Meta-stannic acid

Passivity. When dipped in cone. HNO3, metals like iron, chromium, nickel and aluminium lose their
normal activity and become passive. The passivity of these metals is due to the formation of a thin protective
layer of the metal oxide on the surface of the metal which prevents further action.
For example, in case of iron, a protective layer of ferrosoferric oxide, Fe0.Fe203 or Fe304 is formed on
the surface of iron

3 Fe + 8 HNO3 Fe0.Fe203 + 8 NO2 + 4 H2O

p-BLOCK ELEMENTS 7/25

U. Reaction with metals which are less electropositive than hydrogen (Cu, Ag, Hg, etc.)-
In this case, hydrogen is not liberated. Instead nitric acid oxidises these metals to their corresponding
oxides while it itself is reduced to form either brown fumes of NO2 or colourless vapours of NO.
The metal oxide thus formed reacts further with excess of nitric acid to form the corresponding metal
nitrate and water. For example.
Metal + HNO3 > Metal oxide + NO2 or NO + H2O
Metal oxide + HNO3 > Metal nitrate + H2O
(0 With cone. HNO^, nitrogen dioxide (NO2) is formed.
2 HNO3 > H2O + 2 NO2 + [O]
Cu + [O] > CuO
CuO + 2HNO3 > Cu(N03)2 + H2O

w
CU + 4HNO3 - ^ Cu(N03)2 + 2H2O + 2NO2
Similarly, Ag + 2HN03 -^AgN03 + H20 + N02

Flo
Hg + 4HN03 -4 Hg(N03)2 + 2 H2O + 2 NO2
(ii) With dil HNO^, nitric oxide is evolved.

ee
2HNO3 H2O + 2 NO + 3 [O]

Fr
Cu + [0]- CuO ] X 3
CUO + 2HNO3- Cu(N03>2 + H2O] X 3

for
ur
3 Cu + 8 HNO3 ^ 3 Cu(N03)2 + 4 H2O + 2 no
Similarly, 3 Ag + 4 HNO3 ^ 3 AgN03 + 2 H2O + NO
ks
6 Hg + 8 HNO3 3 Hg2(N03)2 + 4 H2O + 2 NO
Yo
III. Reaction with noble metals. Noble metals like gold and platinum do not react with cone. HNO3.
oo

However, these metals dissolve in aqua regia (3 parts of cone. HCl + 1 part cone. HNO3) forming their
eB

respective chlorides.
HCl first reacts with HNO3 to produce nascent chlorine which then reacts with noble metals forming
r

their respective chlorides.

ou
ad

3HCI + HNO3 NOCl + 2H2O + 2 Cl

Y

Nitrosyl chloride Nascent chlorine

Aqua regia

AU + 3C1 > AUCI3 ; Pt + 4C1 > PtCl4

nd
Re

Auric chloride Platinic chloride

Fi

These chlorides subsequently dissolve in excess of HCl forming their corresponding soluble complexes.
Thus,
AUCI3 + HCl H[AuCl4] ; PtCl4 + 2 HCl H2[PtCl6]
Auric chloride Aurochloric acid Platinic chloride Chloroplatinic acid
5. Action of organic compounds
(i) Oxidation. Many organic compounds are oxidised by cone. HNO3. For example, cane sugar (sucrose)
on oxidation gives oxalic acid.
COOH
C12H22O11 + 36HNO3 ■» 6 I + 36 NO2 + 23 H2O
Sucrose COOH
Oxalic acid

(ii) Nitration. A mixture of cone. HNO3 and cone. H2SO4 (also called nitrating mixture) is used for
introducing one or more - NO2 (nitro) groups into the benzene ring. This process is called nitration. For
example.
7/26 ^nadee^’ii' New Course Chemistry (X11)B2S19]

NO-, CH3 CH3

Cone. Cone. O2N. NO2
H2SO4
I + HNO3 ♦
* H2O ; + 3HNO3 H2SO4 ^ + 3H2O
330 K A
(Cone.) (Cone.)
Benzene Nitrobenzene Toluene

NO2
2,4, 6-TinitroloIuene
(T.N.T.)

CH2-OH CH2-O-NO2
Cone. H2SO4
CH-OH + 3HNO3 CH -0-N02 + 3 H2O
(Co»jc.)

ow
CH2-OH CH2-O-NO2
Glycerol Trmitrogiycerol
Similarly phenol reacts with a mixture of cone. HNO3 + cone. H2SO4 to form 2, 4, 6-trinitrophenoI
(picric acid), though in low yield.

e
Both T.N.T. and trinitroglycerol (or trinitroglycerine or simply nitroglycerine) are used as explosives.

Fl
re
Dynamite, the well-known explosive which is widely used for shooting oil wells, building roads, dams and
tunnels in rocks, is a mixture of glyceryl trinitrate and glyceryl dinitrate absorbed over kieselguhr (a kind of

F
porous earth).
ur
Nitric acid also reacts with proteins giving a yellow compound called xantfioprotein. It is because of

or
this reason that nitric acid turns the skin as well as wool yellow.

7.10.5. Ring Test for Nitrate ion f

ks
In qualitative analysis, the presence of nitrate ion is detected by ring test. In this lest, a freshly prepared
Yo
oo

solution of ferrous sulphate is added to the aqueous solution of a nitrate. Pure cone. H2SO4 is then added
carefully dropwise along the walls of the test tube. The appearance of a dark brown ring at the junction of the
B

two layers indicates the presence of a nitrate ion.

e

The brown irng test for nitrates depends upon the ability of Fe“‘*‘ ions to reduce nitrate ions to nitric
oxide which then reacts with Fe^'*’ ions to form brown coloured complex
ur

NO (g) + 3 Fe^-" (aq) + 2 H2O (1)

ad

NO- (aq) + 3 Fe--" (aq) + 4 (aq) 4

Yo

Nitric oxide

[Fe(H20)6]2+ (aq) + NO (g) ^ [Fe^(H20)5N0+] (aq) + H2O (1)

d

Pentaaquanitrosoniumiroii (I)
Re
in

(Brown complex)
The colour of the complex is actually due to charge transfe,r i.e., an electron is transferred from NO to
F

Fe^-" ion. As a result, this complex formally contains iron in the +1 state and NO'^ (nitrosonium ion).
Consequently, the correct name of this complex is pentaaquanitro soniumiron (I) and not pentaaquanitrosyliron
(II), [Fe(H20)5N0]2^.
7.10.6. Uses of Nitric Acid
Some important uses of nitric acid are given below :
(i) In the manufacture of ammonium nitrate and basic calcium nitrate [Ca0.Ca(N03)2] which are used as
fertilizers. NH4NO3 is also used as an explosive, since on strong heating (above 573 K), or with a
detonator, it decomposes rapidly. The solid has almost zero volume but on heating, it produces seven
volumes of gas which causes the explosion.
>573K

2 NH4NO3 (S) ^ 2N2(g) + 02(g) + 4H20(g)

(ii) In the manufacture of explosives and pyrotechnics such as gun cotton, nitroglycerine, trinitrotoluene
(T.N.T.), picric acid, etc. Nitroglycerine (commercial name sorbitrate) in 5-10 mg doses is used to cure
Angina Pectoris (severe chest pain).
p-BLOCK ELEMENTS 7/27

(Hi) In the preparation of nitro compounds which are used as perfumes, dyes and medicines,
(iv) In the manufacture of artificial silk,
(v) For purification of gold and silver,
(vi) For pickling (cleaning) of stainless steel and etching of metals,
(vii) As an oxidiser in rocket fuels.
(viii) As a reagent in the laboratory.
SUPPLEMENT YOUR
KNOWLEDGE FOR COMP^ITIONS

Properties of nitrous add versus nitric acid. Although nitric acid acts only as an oxidising agent,
nitrous acid (HNO2) acts both as an oxidising as well as a reducing agent as shown below :
(a) Reducing properties. Since HNO^ can be easily oxidised to HNO3, therefore, it acts a reducing agent.

ow
For example,
(0 It reduces acidiifed KMnO^ to manganese salts
2 KMn04 + 3 H2SO4 > K2SO4 + 2 MnS04 + 3 H^O + 5 (O)
HNO2 + O > HNO3] X 5

e
2 KMn04 + 3 H2SO4 + 5 HNO2 > K2SO4 + 2 MnS04 + 5 HNO3 + 3 H2O

re
or
2 MnO^ + 6 + 5 NO”

rFl
> 2 Mn^-*- + 5 NOJ + 3 H2O

F
(ii) It reduces acidified K2Cr20y to chromium salts
K2Cr202 + 4 H2SO4 + 3 HNO2 > K2SO4 + Cr2(S04>3 + 3HNO3 + 4H2O

r
ou
or + 8H+ + 3 NO- >2
fo
+ 3 NO" + 4 H2O
ks
(///) It reduces I2 and Br2 to HI and HBr respectively
-^2HI + HN03 ; HN02 + Br2 + H20 > 2 HBr + HNO3
oo
HNO, + I2 + H2O
(iv) It reduces H2O2 to H2O ; H2O2 + HNOt > H2O + HNO3
Y
eB

(b) Oxidising properties. Since it can be easily reduced to NO, therefore, it acts a good oxidising agent.
For example,
(i) It oxidises H2S toS : H2S + 2 HNO^ > 2 H2O + 2 NO + S
r
You

(ii) It oxidises iodide ion to iodine : 2 KI + 2 HNO2 > 2 KOH + It + 2 NO

ad

(Hi) It oxidises SO2 to H2SO4: SO2 + 2 HNOt > H2SO4 + 2 NO

(iv) It oxidises stannous chloride (SnC/,) to stannic chloride (SnCl^):
d

SnCl2 + 2 HCl + 2 HNO2 ^ SnCl4 + 2 NO + 2 H2O

Re
in

(v) It oxidises ferrous sulphate to ferric sulphate ;

2 FeS04 + H2SO4 + 2 HNO2 ^ Fe2(S04)3 + 2 NO + 2 H2O
F

7.n. ALLOTROPIC FORMS OF PHOSPHORUS

Phosphorus exists in many allotropic forms. Out of these white, red

and black are important.
1. White phosphorus. It is obtained when phosphate rock is heated
with coke and sand in an electric furnace at 1773 K
1773 K

2 Ca3(P04)2 + 6 Si02 6 CaSi03 + P4O1Q

1773K

P4O10+ IOC ^ P4+ 10 CO

Structure. It exists as P4 units. The four s'p-^-hybridized phosphorus
atoms lie at the corners of a regular tetrahedron with ZPPP = 60'’ as shown Structure of
in Fig. 7.6. Each phosphorus atom is linked to three other P-atoms by covalent white phosphorus
bonds so that each P-atom completes it octet.
7/28 T^cideefr’^ New Course Chemistry (XU)BE
On exposure to light, white phosphorus turns yellow, therefore, it is also called yellow phosphorus.
Some important properties of white or yellow phosphorus are given below :
(i) Physical state, colour and smell. It is a soft, translucent waxy white solid with garlic odour. Being
soft, it can be cut with a knife,
(ii) Poisonous nature. It is very poisonous. Persons working with white phosphorus develop a disease
known as Phossy jaw in which jaw bones decay,
(iii) Melting point and Boiling point The various P4 molecules of white phosphorus are held together
only by weak van der Waals’ forces of attraction and hence the melting point (317 K) and boiling point
(553 K) of white phosphorus are quite low.
(iv) Solubility. It is insoluble in water but is soluble in organic solvents such as CS2, alcohol and ether,
(v) Stability. The P4 molecules are stable upto 1070 K even in the vapour phase. But at temperatures
higher than 1070 K, P4 molecules dissociate to give P2 molecules.
>1070K

w
P4^ - 2P2
(vi) Action of air. Since PPP angle is much smaller (60®) than the normal tetrahedral angle
(109® - 28'), therefore, P4 molecules are highly strained. Due to this angle strain, white phosphorus is very

F lo
reactive. It readily catches fire in air with a greenish glow which is visible in the dark. This phenomenon is
called chemiluminescence or phosphorescence. When it bums in air, it produces dense white fumes of

ee
phosphoms pentoxide (P4OJ0).

Fr
P4 (5)+ 5 02(g) >P40io(^)
Because of its very low ignition temperature (303 K), it is always kept under water,
for
(vii) Reaction with halogens. White phosphorus spontaneously catches fire in chlorine forming
ur
phosphorus trichloride (PCI3) and phosphoms pentachloiide (PCI5).
P4 + 6CI2 >4PCl3 P4+IOCI2 4 PCI5
s
ook

(viii) Reaction with sulphur. White phosphoms combines with sulphur with explosive violence forming
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a number of sulphides such as P2S3, P2S5, P4S3, P4S7 and P4S10 depending upon the relative amounts of the
eB

reactants.

8P4 + 3S8 8 P4S3

P4S3 is the most stable sulphide. It is used in 'Strike anywhere Matches’. Matches contain P4S3, KCIO3,
r
ad
ou

fillers and glue. The friction between the match and the sand paper on the side of the match box initiates a
violent reaction between P4S3 and KCIO3. This generates sufficient heat which causes match stick to burst
Y

onto flame.

(ix) Reaction with metals. Phosphoms combines with number of metals such as Na, K, Mg, Ca, Ag,
Re
nd

Cu, etc. forming their respective phosphides.

Fi

12Na + P4 4 Na3P ; 6Mg + P4 2Mg3P2

Sod. phosphide Mag. phosphide
12Ag + P4 4 Ag3P ; 6 Ca + P4 2 Ca3P2
Silver phosphide Cal. phosphide
The reaction with sodium and potassium is very vigorous.
(x) Reaction with aqueous alkalies. When white phosphoms is heated with sodium hydroxide solution
in an inert atmosphere, phosphine (PH3) is formed.
A. CO2
P4 + 3 NaOH + 3 H2O PH3 + 3 NaH2P02
Phosphine Sod. hypophosphite
It is an example of a disproportionation reaction in which oxidation state of phosphoms decreases from
0 in P4 to -3 in PH3 while it increases from 0 in P4 to +1 in NaH2P02.
(xi) Reducing properties. Since white phosphoms can be easily oxidised, it acts as a strong reducing
agent. It reduces HNO3 to NO2 and H2SO4 to SO2.
p-BLOCK ELEMENTS 7/29

P4 + 20 HNO3 ● ^ 20 NO2 + 4 H3PO4 + 4 H2O

P4 + 10 H2SO4 ■4 10 SO2 + 4 H3PO4 + 4 H2O
It also reduces solutions of copper, silver and gold salts to their corresponding metals. For example,
P4 + 3 CUSO4 + 6 H2O ^ CU3P2 i + 2 H3PO3 + 3 H2SO4
CU3P2 + 5 CUSO4 + 8 H2O ^ 8 Cu -I +5 H2SO4 + 2 H3PO4
Similarly, P4 + 2O AgN03+ I6H2O 20 Ag i +4 H3PO4 + 20 HNO3
2. Red phosphorus. It is obtained by heating white phosphorus at 573 K out of contact with air in an
inert atmosphere (CO2 or coal gas) for several hours.
573 K
P4(^) ^ P4W
C02or coalgas
White phosphorus Red phosphoms

low
Like white phosphorus, red FIGURE 7.7
phosphorus also exists as P4 tetraliedra but
these are joined together through covalent
bonds to give a polymeric structure as
shown in Fig. 7.7.

e
re
Because of polymeric structure, its
melting point (883 K) is much higher than
rF
F
Structure of red phosphorus.
that of white phosphorus (317 K).
Some important characteristics of red phosphorus are :

or
u
(0 It is a hard crystalline odourless solid with iron grey lustre.
(ii) It is non-poisonous in nature. f
ks
Yo
(in) It is insoluble in water as well in organic solvents such as CS2, alcohol and ether.
oo

(iv) It is a relatively stable allotrope of phosphorus at room temperature. Its ignition temperature (543 K) is
much higher than that of white phosphorus (303K). As a result, it does not catch fire easily,
B

(v) It sublimes on heating giving vapours which are the same as given by white phosphorus. When these
re

vapours are condensed, white phosphorus is obtained. This gives us a simple method of reconverting
red phosphorus into white phosphorus.
u
ad

(vi) It is denser (216 g cm“^) than white phosphorus (1-84 g cm“^) and is a bad conductor of electricity.
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(vii) Being polymeric in nature, it is less reactive than white phosphorus.

(viii) It bums in oxygen at 565 K to yield phosphorus pentoxide.
d
Re
in

565 K
P4 (s) + 5 O2 (g)
F

(ix) Being less reactive than white phosphorus, it reacts with halogens, sulphur and alkali metals only when
heated forming their corresponding salts.
A A

P4 + 6 CI2 ^ 4PCI3 P4+IOCI2 > 4 PCI 5

453 K A

8 P4 + 3 Sg ^ 8 P4S3 P4+ 12Na ^ 4Na3p

Inert atmosphere

(x) It does not react with caustic alkalies. This property is made use in separating red phosphorus from
white phosphorus.
Comparison of properties of white and red phosphorus are given in Table 7.8 on next page.
3. Black phosphorus. It has two forms : a-black phosphorus and p-black phosphorus. a-Black
phosphorus is formed when red phosphorus is heated in a sealed tube at 803 K.
803 K
Red phosphorus » a-BIack phosphorus
Sealed tube
7/30 ‘P'uuiee^’4- New Course Chemistry (XI1)I!SIM]

TABLE 7.8. Comparison between White and Red Phosphorus

Sr. No. Property White Phosphorus Red Phosphorus

1. Physical state Soft waxy solid, can be cut with Hard, crystalline solid
knife

2. Colour White when pure, turns yellow on Red

exposure to light
3. Physiological action Poisonous
Non-poisonous
4. Molecular formula Consists of discrete tetrahedra
Consists of P4 tetrahedra joined
(Fig. 7.6, page 7/29) together through covalent bonds to
give a polymeric structure

ow
(Fig. 7.7, page 7/31)
5. Odour Garlic Odourless
6. Density I’84 g cm"^ 2-16 g cm”^
7. Ignition temperature 303 K. kept under water 543 K
8. Phosphorescence Burns in air with greenish glow Does not glow

e
which is visible in dark

Fl
re
9. Solubility Insoluble in H2O but soluble in Insoluble both in H2O and CS2

F
CS2. undergoes disproportionation
10. Reaction with NaOH Forms NaH2P02 with evolution of
ur No reaction

r
PH3
n. Action of CI2
burns in CI2 forming PCI3 andfo
Being more reactive spontaneously Being less reactive reacts with CI2
ks
only on healing forming first PCI3
PCI5 and then PCI5
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oo
12. Action of air Reacts at r.t. to form P4O10 Forms P40j{, only on heating
13. Reaction with metals Reacts with Na, K, Mg, Ca, Ag, Cu, No reaction
eB

etc. forming their respective

phosphides
14. Reducing prop>erties Reduces solutions of copper, silver No reaction
ur

and gold salts to their respective

ad

metals
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15.
Action of cone. HNO3 Forms H3PO4 evolving NO2 Forms H3PO4 evolving NOo
d

It can be sublimed in air and has opaque monoclinic or rhombohedral crystals. It is a very stable allotrope
Re
in

of phosphorus and does not oxidise in air until heated very strongly. It does not conduct electricity.
p-Black phosphorus (orthorhombic) is prepared by heating white phosphorus at 473 K under a very
F

high pressure (4000—12000 atm.) in an inert atmosphere.

473 K
White phosphorus > p-Black phosphorus FIGURE 7.8
4000-12000 aim pressure I
P-Black phosphorus has a layered structure in which each \p p p p
phosphorus atom is covalently bonded to three neighbouring phosphorus I 990 I I I I
atoms as shown in Fig. 7.8. The P—P—P angles are of 99° and P—P
distance is 218 pm. I I
The adjacent layers are held 368 pm apart. The atoms within a layer
I I
are more strongly bound than the atoms in adjacent layers. This gives p-
black phosphorus graphite like structure.
!
Some important characteristics of black phosphorus are :
Layered structure of
(0 It is a highly polymerized form of phosphorus and has black metallic P-black phosphorus
lustre.
p-BLOCK ELEMENTS 7/31

(h) It has a sharp m.p. of 860 K (587®C). Its specific gravity is 2-69.
(in) Like graphite, it is a fairly good conductor of electricity.
(iV) It is thermodynaniically, the most stable allotrope of phosphorus and does not bum in air even upto 673 K.
Comparison of reactivity of allotropic forms of phosphorus. The three allotropic forms of phosphorus
differ widely in their chemical reactivity. White phosphorus is the most reactive while black phosphorus is
the least reactive. Therefore, white phosphorus is stored under water to protect it from air while red and black
phosphorus are stable in air.

7.12. PHOSPHINE, PHg

Phosphine (PH3) is also known as hydrogen phosphide or phosphoretted hydrogen.
7.12.1. Preparation of Phosphine
Phosphine may be prepared by the following methods :

w
1. From metal phosphides. By the action of water or dilute HCl on metal phosphides.
Ca3P2 + 6 H2O > 3 Ca(OH)2 + 2 PH3 T ; Ca3P2 + 6 HCl > 3 CaCl2 + 2 PH3 T

F lo
Calcium phosphide Phosphine
AlP + 3H2O 4
A1(0H)3 + PH3 T

ee
Aluminium phosphide
2. From phosphorous acid. By heating phosphorus acid.

Fr
473 K

4 H3PO3 ■>
3 H3PO4 + PH3 T
for
ur
Phosphorous acid Phosphoric acid Phosphine
3. From phosphorus trichloride by reduction with lithium aluminium hydride or lithium hydride
ks
<273K
Yo
4 PCI3 + 3 LiAlH4 ^ 4 PH3 + 3 LiCl + 3 AICI3
oo

Elher
eB

Wann
PCI3 + 3LiH PH3 + 3 LiCl
4. From phosphonium iodide. Very pure phosphine can be prepared by heating phosphonium iodide
r

with a cone, solution of caustic alkali.

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ad

PH4I + NaOH > Nal + H2O + PH3 T

Phosphonium iodide Phosphine
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5. Laboratory method of preparation. In the laboratory, it is prepared by heating white phosphoms

Re
nd

with concentrated caustic alkali solution in an inert atmosphere of oil gas or CO^.

P4 (s) + 3 NaOH {aq) + 3 H2O (/) > 3NaH2P02(«9) + PH3(g)T

Fi

Sod. hypophosphiie Phosphine

It may be noted that pure phosphine is not inflammable but the gas produced usually catches fire in air
due to llie presence of impurity of diphosphine (P2H4)* which is highly inflammable. Therefore, a current of
carbon dioxide or coal gas or oil gas is passed through the flask to displace air.
The vapours of PH3 thus produced are passed into water through a delivery tube. Each bubble of gas
when it comes in contact with air, undergoes combustion and forms a ring of smoke.These irngs are called
vortex rings or philosopher’s rings. These are due to the formation of phosphorus pentoxide (P4OJ0).
To prepare pure phosphine, the gas is passed through a U-tube placed in a freezing mixture. Diphosphine
being a volatile liquid condenses while pure phosphine passes over. Tlie impure gas may also be purified by treat
ing it with hydrogen iodide followed by heating the phosphonium iodide thus foimed with caustic soda solution.
Heat

PH3 -H HI > PH4I ; PH4l + NaOH ^ PH3 + NaI + H20

*This is the origin of the flickering light called will-o-the-wisp, which is sometimes seen in marshes.
7/32 “pruideefi.'^ New Course Chemistry (XII)BZS19]

7.12.2. Structure of Phosphine

P in PH3, like N in NH3, involves 5/>^-hybridisation. Three of the four 5p^-orbilals overlap with b-
orbitals of hydrogen atoms to form three P-H, o-bonds while the fourth contains the lone pair of electrons
(Fig. 7.1(c), page 7/8). Thus, like NH3, phosphine is also pyramidal. The HPH bond angle in PH3 is, however,
much lower (93-6°) than that in NH3 (107-8") due to less repulsions between bond pairs. Further, due to
bigger size of P than N, P—H bond length is longer (142 pm) than N—H bond length (101-7 pm).
7.12.3. Physical Properties of Phosphine
(/) Phosphine is a colourless gas possessing unpleasant odour similar to that of garlic or rotten fish and is
highly poisonous.
(//) Unlike NH3, it is only slightly soluble in water; the aqueous solution being neutral. However, it is more
soluble in CS2 and other organic solvents.

ow
(Hi) Its vapour density is 17, i.e., little heavier than air.
(/v) On cooling to 185-5 K, phosphine condenses to a liquid and on cooling to 139-5K, it solidifies.
7.12.4. Chemical Properties of Phosphine

e
(/) Basic nature. Phosphine is feebly basic and forms salts with mineral acids under anhydrous conditions.

re
rFl
PH3 + HX ph;x- (X = Cl, Br, I)

F
Phosphine Phosphonium halide
(//) Combustibility. Pure PH3 is stable in air but catches fire when heated to 423 K.

r
423 K
ou
fo
PH3 + 2 O2 » H3PO4
It explodes in contact with traces of oxidising agents like HNO3, CI2 or Br2 vaporus.
ks
(Hi) Decomposition. When heated to 713 K. in absence of air, it decomposes into its elements, i.e., P4
oo

and H2. 713K

4PH3 Absence of air ^ P4-f6H,

Y
eB

The solution of phosphine in water decomposes in presence of light giving red phosphorus and H2.
hv
P4 (5) + 6 H2 (g)
ur

4PH3 (aq) 4

Red phosphorus
ad
Yo

(/v) Action of chlorine. PH3 bums in CI2 to form phosphorus trichloride or pentachloride
PH3 + 3 CI2 > PCI3 + 3 HCl ; PH3 + 4 CI2 PCI5 -b 3 HCl
d

Phosphorus Phosphorus
Re
in

trichloride pentachloride
(v) Action of nitrous and nitric oxides. PH3 reduces N2O and NO to N2
F

PH3 + 4N2O >H3P04-f4N2 ; PH3+4NO >H3P04 + 2N2

(v/) Reaction with metallic salt solutions. When PH3 is bubbled through the aqueous solutions of
copper, and mercury salts, the precipitates of corresponding phosphides are formed.
3 CUSO4 + 2 PH3 CU3P2 i + 3 H2SO4
Copper sulphate Copper phosphide

3 HgCl2 + 2 PH3 - Hg3P2 i + 6 HCl

Mercuric chloride Mercuric phosphide
With silver nitrate, first a complex of silver phosphide with silver nitrate is produced. This is subsequently
reduced in presence of water to metallic silver which appears as a black ppt.
PH3 -f 6 AgN03 Ag3P.3 AgN03 + 3 HNO3
Inlermediate complex

Ag3P.3AgN03 + 3 H2O 6 Ag i -1- 3 HNO3 + H3PO 3

Black ppt.
 p-BLOCK ELEMENTS 7/33

Uses. (/) As Holme’s signals in deep seas and oceans for signalling danger points to steamers. Containers
containing a mixture of calcium phosphide and calcium carbide are pierced and thrown into the sea. In
contact with water, a mixture of phosphine and acetylene gases is produced. Phosphine frequently contains
traces of highly inflammable disphosphine P2H4 which catches fire spontaneously. This ignites acetylene
which bums with a luminous flame and thus serves as a signal to the approaching ship.
(ii) For the production of smoke screens. Shells containing calcium phosphide are exploded by warships.
Calcium phosphide reacts with water producing phosphine which burns in air to give clouds of P4OKJ which
act as smoke screens.

7.13. PHOSPHORUS HALIDES

Phosphorus forms two types of halides, PX3 (X = F, Cl, Br, I) and PX5 (X = F, Cl, Br).
7.13.1. Phosphorus It'Ichlorlde, PCI3

low
Preparation. It is prepared by the action of thionyl chloride on white phosphoms.
P4 + 8 SOCI2 - 4PCI3 + 4 SO2 + 2 S2CI2
Thionyl chloride Phosphorus trichloride Sulphur monochloride
In the laboratory, it is prepared by heating white phosphorus in a current of dry chlorine when phosphorus
trichloride foraied distils over

ee
P4 is) + 3 CI2 (g)
rF
^ 4 PCI3 {/)
FIGURE 7.9

Fr
Structure. Like in PH3, P in PCI3 is sp^-hybridized. Three of its sp^-
orbitals overlap with p-orbitals of three chlorine atoms to form three P—Cl,
a-bonds while the fourth orbital contains the lone pair of electrons. Therefore,
like PH3, PCI3 is also pyramidal (Fig. 7.9). The CIPCl bond angle in PCI3 is, for
u
however, greater (100-4") than in PH3 (93-6'’) due to steric crowding of the
ks
two Cl atoms. As expected, the P—Cl bond is much longer (204 pm) than
Yo
o

P—H bond (142 pm) due to bigger size of the Cl atom. Structure of
Bo

Physical Properties. It is a colourless pungent smelling liquid which phosphorus trichloride.

boils at 347 K and solidifies at 161 K. It fumes in moist air.
re

Chemical Properties, (i) Action of water. It reacts violently with water to produce phosphorous acid
(H3PO3) and hydrochloric acid.
ou
ad

PCI3 + 3 H.O > H3PO3 + 3 HCl

Y

(ii) Action of cone. H2SO4. It reacts with cone. H2SO4 forming chlorosulphonic acid.
2 HO — SO2 — OH + PCI3 > HO —SO2 —Cl + HPO3 + SO2 + 2HCI
nd
Re

Sulphuric acid Chlorosulphonic acid

(iii) Action of Cl2, Oj and S
Fi

PCI3 + CI2 4 PCI5

2 PCI3 + O2 ? 2 POCI3 ; PCI3 + S PSCI3
Phosphorus oxychloride Thiophosphoms oxychloride
(iv) It acts as a reducing agent
PCI3 + SO3 )■ POCI3 + SO2 ; PCI3 + SO2CI2 — ^ PCI5 +SO2
Sulphuryl chloride
3PCI3 + SOCI2- ’ POCi3 + PSCI3 + PCI5 ; 3 PCI3 + S2CI2 ^ PCI5 + 2 PSCI3
(v) Reaction with Grigna. d reagents. PCI3 reacts with Grignard reagents to form substituted phosphines.
For example.
PCI3 + 3C6H5MgCl - P(C6Hg)3 + 3 MgCl2
Phenylmag. chloride Triphenylphosphine
(vi) Action of finely divided metals. On heating finely divided metals react to form metal chlorides.
A A
12 Ag + 4PCl3 ^ 12AgCl + P4 ; 6 Na + PCI3 ^ 3 NaCl + NagP
7/34 7^n4uUe^'<i- New Course Chemistry (Xll)CSsia]

Uses. Phosphorus trichloride is widely used as an important reagent in organic chemistry for replacing
hydroxyl groups by chlorine atoms in organic reactions.
(/) 3 CH3CH2OH + PCI3 3 CH3CH2CI + H3PO3
Ethanol Chloroethane

(ii) 3 CH3COOH + PCI3 3 CH3COCI + H3PO3

Acetic acid Acetyl chloride
(iii) 3 C(^H5S020H + PCI3 3 CfiH5S02Cl + H3PO3
Benzenesulphonic acid Benzenesulphonyl chloride

7.13.2. Phosphorus Pentachioride, PCI5

Preparation. (0 Phosphorus pentachioride is prepared by the reaction of white phosphorus with excess
of dry chlorine or by the action of dry chlorine on phosphorus trichloride
P4 (5) + 10 CI2 (g) > 4 PCI5 (s) ; PCI3 (/) + CI2 (g) > PCI5 {s)
(ii) It can also be prepared by the action of sulphuryl chloride (SO2CI2) on white phosphorus
P4 is) + 10 SO2CI2 (0 >4 PCI5 is) + 10 SO2 ig)
Structure. In PClg, phosphorus undergoes 5p^rf-hybridizaton and

F low
has trigonal bipyramidal geometry (Fig. 7.10) in gaseous and liquid states.
It has three equatorial (e), P—Cl bonds and two axial (a), P—Cl bonds.
Since two axial P—Cl bonds are repelled by three bond pairs while three
equatorial bonds are repelled by two bond pairs, therefore, axial bonds
Cl
are longer (240 pm) than equatorial bonds (202 pm).

e
Physical Properties. Phosphorus pentachioride is a pale yellow
for Fre 3 240 pm

C!
crystalline solid with a characteristic pungent smell. In the solid state, it
Structure of
exists as an ionic solid, [PCIJ"^ [PCl^]" in which the cation, [PC^]"^ is phosphorus pentachioride.
tetrahedral while the anion, [PCl^]" is octahedral.
Chemical Properties, (i) Dissociation. When heated, it sublimes but decomposes on stronger heating
Your
eBo ks

into PCI3 an CI2.

PCI5 PCI3 + CI2
ad

(ii) Action of moisture, air and water. In moist air, it undergoes hydrolysis to first form POCI3 and
our

then finally phosphoric acid

PCI3 + H2O > POCI3 + 2 HCl POCI3 + 3 H2O ^ H3PO4 + 3 HCl
However, with excess of water, it reacts violently to form H3PO4 and HCl
Re

PClg + 4 H2O {excess) > H3PO4 + 5 HCl

Find Y

(iii) Reaction with compounds containing hydroxyl groups. It reacts with compounds containing
hydroxyl groups to give the corresponding chloro compounds in which each hydroxyl group is replaced by a
chlorine atom. For example.
CH3COOH + PClg -> CH3COCI + POCI3 + HCl
Acetic acid Acetyl chloride
CH3CH2OH + PClg -4 CH3CH2CI + POCI3 + HCl
Ethyl alcohol Ethyl chloride
HO —SO2 —OH + 2 PClg- ^ Cl—SO2—Cl + 2 POCI3 + 2 HCl
Sulphuric acid Sulphuryl chloride
(iv) Reaction with metals. Finely divided metals, on heating with PClg, give corresponding chlorides.
Zn + PClg > ZnCl2 + PCI3 ;
2 Ag + PClg > 2 AgCl + PClg
Sn + 2 PClg
> SnCl4 + 2 PCI3
(v) Reaction with phosphorus pentoxide, phosphorus pentasulphide and sulphur dioxide. The
reactions occur as follows ;
 p-BLOCK ELEMENTS 7/35

6PCI5 + P4O10 ■>

10 POCI3 6PCI5 + P4S10 - ■>
IOPSCI3
Phosphorus Phosphorus Phosphorus
oxychloride pentasulphide thiochloride

PCI5 + SO2 > POCI3 + SOCI2

Thionyl chloride
(vi) Reaction with chloride ion acceptors. With chloride ion acceptors like boron trichloride and
niobium tetrachloride, it forms addition compounds containing tetrahedral [PCI4] species. +

PCI5 + BCI3 > fPCl4]-^ [BCI4]- ; PCI5 + NbCl4 > [PCl4]+ [NbCljr
(vii) Reaction with potassium fluoride. With KF, PCI5 forms potassium phosphorus hexafluoride,
K^PFfi]-.
PCI5 + 6 KF ^ K+ [PF^]- + 5 KCl
In this reaction, PCI5 acts as a fluoride ion acceptor and hence acts as a lewis acid,

ow
(viii) Reduction. PCI5 reacts with H2 to form PCI3.
PCI5 + H, ^ PCI3 + 2 HCl
In this reaction, PCI5 acts as an oxidising agent and thus oxidises H, to HCl.
7.14. OXIDES AND OXOACIDS OF PHOSPHORUS

e
Fl
re
1. Oxides of phosphorus. Phosphorus forms two oxides : phosphorus (III) oxide, P4O6 and phosphorus
(V) oxide, P4O10. Due to large size and lower electronegativity, phosphorus does not form pK-pK multiple

F
bonds. As a result, oxides of phosphorus have cage structures and exist as dimers,
ur
(i) Preparation. Phosphorus (III) oxide, P4O6 is obtained by burning white phosphorus in a limited

r
P40g P4 + 5 O2 (exce5’.9) fo
supply of air while phosphorus (V) oxide, P4O10 is obtained by burning white phosphorus in excess of air
P4 + 3 O2 (limited) ^4^10
ks
(ii) Structure. Each phosphorus atom in
Yo
FIGURE 7.11
oo

P4O6 and P40]o lie at the corners of a

0 ^
tetrahedron (just as P4 units in elemental
B

CO

phosphorus). In P4O6, each P is covalently ■D

re

bonded to three O atoms and each O is bonded ■P'

O
to two P atoms. Thus, the six O atoms lie along O
127°
o 0
u

1230
the edges of the tetrahedron forming six
ad

O
O
Yo

P—O—P single bonds (Fig. l.Wa). However, 05

O

in P4O1Q, each P also forms a double bond (or T3

3
coordinate bond) by sharing (or donating) its
d

o
Re

—p
o
lone pair of electrons with an O atom (Fig.
in

/
o
o
7.11/7). O 166 pm o ●p,
F

(Hi) Properties. Both P4O6 and P4O10 are o O

0
acidic oxides which dissolve in water to form
phosphonic acid (orthophosphorous acid, Structures of (a) Phosphorus trioxide (P4O5)
H3PO3) and phosphoric acid (orthophosphoric (b) Phosphorus pentoxide (F40j^q).
acid. H3PO4) respectively.
P4O6 + 6 H2O > 4 H3PO3 P4010 6 H2O ^ 4 H3PO4
Because of its great affinity for water, P4O1Q acts as a powerful dehydrating agent. As such, it extracts
water from many inorganic and organic compounds.
2H2SO4 +P40,o > 2SO3 +4HPO3; 4HNO3 + P4O10 ^ 2N2O5 + 4HPO3
4HCIO4 +P40,o > 2CI2O7 +4HPO3; 2CH3CONH2 + P40,o ^ 2 CH3CN + 4 HPO3
Perchloric acid Chlorine (VII) oxide Acetamide Acetonitrile

2. Oxoacids of phosphorus. Phosphorus forms a number of oxoacids. Some of the important oxoacids of
phosphorus along with their formulas and oxidation state (O.S.) of phosphorus, methods of preparation and
basicity, and the number and type of characteristic bonds present in their structures are given in Table 7.9.
7/36 “pitadee^'^^ New Course Chemistry (XII)BZ

TABLE 7.9. Some Oxoacids of Phosphorus

Name Formula OSwOf P Characteristic BasiPity Methods of preparation
bonds and
their number

+ 1 One P—OH, two Monobasic 2 P4 + 3 Ba(OH)2 + 6 H2O

1. Hyphos- H3PO2
P—H, one P = O
phorous acid 2 PH3 + 3 Ba(H2P02)2
(Phosphinic Ba(H2P02)2 + H2SO4
acid)
BaS04 + 2 H3PO,
2. Orthophos- H3PO3 + 3 Two P—OH, one Dibasic P4O6 + 6H2O 4 H3PO3
phorous acid P—H, one P = O PCI3 + 3H2O H3PO3 + 3 HCl
{Phosphonic
acid)

w
3. Pyrophos- H4P2O5 + 3 Two P—OH, two Dibasic 5 H3PO3 + PCI3 ^ 3 H4P2O5
phorous acid P—H, two P = O + 3 HCl

F lo
and one P—O—^P

4. Hypo- + 4 Four P—OH, two Tetrabasic 2 P (red) + 4 NaOCl+2 H2O

H4P2O6
phosphoric P = O, one P—P
H4P2O6 + 4 NaCl
acid

ee
5. Orthophos- H3PO4 + 5 Three P—OH, Tribasic ^4^10 ^ H2O — ^ 4 H3PO4

Fr
phoric acid one P = O
523 K
6. Pyrophos- H4P2O7 +5 Four P—OH, Tetrabasic
2H3PO4 ^ H4P2O7 + H2O
phoric acid two P = O, for
ur
one P—O—^P
-1-5
7. Metaphos- HPO3
s
phoric acid A, sealed tube
ook
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{exists in (HP03>3 -1-5 Three P—OH, Tribasic 3 H3PO3 3 Br2

polymeric three P = O, (HP03>3-l-6HBr
eB

form), e.g., three P- ■P

cyclotrimeta-
phosphoric
our
ad

acid
or linear One P = O, Monobasic
(HPO3), + 5
polymeta- one P—OH,
Y

phosphoric one P—O— P

Re

acid
nd

8. Peroxomono- H3PO5 -^●5 TwoO—H, Tribasic P4O10 + 4 H2O2 (30%) + 2 H2O

Fi

phosphoric one P = O, > 4 H3PO5

acid one P—O—O—H

9. Peroxodiphos H4P2O8 -h5 Four O—H, Tetrabasic Electrolysis of a mixture of

-phoric acid two P = O, K2HPO4 + KF
one P—O—O—P

The structures of these oxoacids are given below :

Hypophosphorous Orthophosphorous Pyrophosphorous acid, Hypophosphoric acid,

acid or Phosphinic acid or Phosphonic
H4P2O6 {Tetrabasic)
acid, H3PO2 acid, H3PO3 {Dibasic) H4P2O5 {Dibasic)
(MonoBasic)
 ●{sjiqdsoqd usSojp/Cq uiniposip) P-^ (sirq^soqd uafojpXqip uinipos) ^Od^H^N ‘sq^s
JO sauas omj suijoj aouaq puB sdnojS hO surejuoo q aouis oisvqip s] '^od^H ‘^(^Od^H)’-a
JO ^Od^H^^N ‘'2M ‘saqqdsoqdodXq pa||BD sqes jo sauas auo Xjuo suijoj aauaq pun ‘dnojS HO“
auo SBq \\ ‘asuBoaq Disnqouoiu si ^Od^H ‘^idurexa joj -ppo aip/o Kipisoq dqj sdpp9p judsaud sdnouS
HOP Jaqunm ayj ‘snqx aiqBziuoi jou si j oj Xjjoajip paquiy uiojb h inq ayqBsiuoi si dnoj§ hO ^MX (^)
DH 3 + ●'Od^H + t ^TD^3h < O^H + ^Od‘^H + ^IO§H Z
^ONH Z + ^Od'^H +1 3V S + ^Od'^H + ^ON§V Z
*'OS^H + *’Od‘-H +1 ^3 < O^H + ^Od^H + ^OSnD
mz + ^Od^H < 0"H + ‘^Od^H +
o^H e + ^od^-H e + ^osm z + <— ^od^H e + *'os"h e + ^o^mi z
●oja ‘^inajaui ‘jaAps ‘jaddoa jo sqBS pUB ‘^j P^UiPP^ saonpaj q ‘^Od^H om} oj pajodwoo sv puoq
H—d Kjuo suwjuoo ji asnoDdq ^Od^H ua^oaM qSnoqj juaSv Supnpaj b sb sjdb osyB ^Od^H

ow
3U3ZU?g apuoiqo uiniiiozRipauszuag

IDHZ + ^HZ + ^Od^H + ^H^DZ < 0^H3 + Wh +

●sauajB oj sqBS uiniuozBipauajB aonpaj oi Xjisiuiaqa oiubSjo ui pasn si ij
I3H P + ^Od^H + ^HZ <r
O^H Z + ^Od^U + ^13§H Z

e
^ONH P + *’Od^H + t 3v t <- O^H Z + "Od^H + ^ON^V P

re
rFl
■§H OJ ^I3§H PUR J9AIP oipBiaiu
OJ ^ON^V saonpaj snqj puB spuoq h—d oa\j suibjuod q sb iua§B Suionpaj poo§ n s| (^Od^H) PP^’

F
snojoqdsoqdodXq ‘aydurexa joj saiyadojd Suionpaj guojys OABq spuoq h—d uibiuod qoiqM spioy (<^)
ppc ouoqdsoqdoquo suiqdsoqj piOR snojoqdsoqdoquo

r
ou
t> f
od^He ^Hd ^Od'^H P
fo
+ <r
IB3H £+
£+ e-
ks
‘ayduiBxa joj 'saiBys uoiiBpixo jaMOj puB
jaqSiq ui spunoduioo pyaiX oj uonoeaj uopBuoiiJodojdsip oSjapun ajBis uoijBpixo £+ ui spiaeoxo aqx (a/)
oo

●qioq you mq spuoq (^O^d^'H d~d JO (^Od^H ‘^Od^H “^■3) H—d JOMIP ‘spuoq HQ—d
Y

uouBpixo SBq siuoqdsoqd qaiqM ui spioBOXo aqx (/»)

B

puB o = d OJ uoijippB UI ‘UIBJUOO ‘g+ UBqj ssay ajBjs

●puoq HO—d 900 js’Bay jb puB q = d 9uo uibjuoo spioB asaqj yyy (//)
re

●sdnojS JO suiojB jaqjo jnoj Xq papunojjns XyyBJpaqBJjaj sy siuoqdsoqd ‘spiOBoxo yyB uy {/)
ou

sppeoxo io s^|JS|J^peJell^ juepodiui aujos 'VVVL

Y
ad

’●‘(^OdH) ‘ppe ouoqdsoqdejaujAiod jeaun ^(^OdH) ‘PPe ouoqdsoqdBjaujujoio^O

d
in
Re
F

(o!SBqej}9±)
{oiseqejjei) 8o^d*'H (o/sequi) Sod^H‘PPe ‘ppe ouoqdsoqdojAd P/sequi) t'od^H‘Pi^e
'ppe ojjoqdsoqdjpoxojdd ojjoqdsoqdouoLuoxojad -^O^d^H oijoqdsoqdoqpo

HO-O.

© © © ©
le/i siNswana xooia-d
7/38 ‘Pxiutee^’^ New Course Chemistry (XII)i:^aiai
H3PO4, on the other hand, is tribasic because it contains three OH groups and hence forms three series
of salts, viz., NaH2p04 (sodium dihydrogen phosphate), Na2HP04 (disodium hydrogen phosphate)
and Na3P04 (trisodium phosphate or simply sodium phosphate).
(v/0 Metaphosphoric acid does not exist as a monomer but exists either as a cyclic trimer in form of
cyclotrimetaphosphoric acid or as a linear polymer in form of polymetaphosphoric acid.
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS Polyphosphates are used as water softeners in detergents.
For example, sodium tripolyphosphate NajPjOjo is one of the most commonly used polyphosphate. It softens
2+
water by forming soluble complexes with Ca?'* and Mg ions present in hard water. However, excessive
use of polyphosphates as water softeners is discouraged by environmentalists since it leads to water pollution

as follows. The phosphates in domestic waste water pass through sewage disposal systems into rivers and
lakes. There they nourish bacteria, which grow excessively and deplete water of dissolved oxygen thus

w
killing fish and other small aquatic animals. Therefore, in U.S.A, the use of phosphates in laundry detergents
has been banned in some states.

Flo
PART—11

GROUP 16 ELEMENTS : THE QXYGEN FAMILY

ee
7.15. GENERAL INTRODUCTION

Fr
Group 16 of the periodic table consists of five elements viz., Oxygen (O), Sulphur (S), Selenium (Se),
Tellurium (Te) and Polonium (Po). The elements of this group are commonly known as the oxygen family
after the name of its first member. These are alsocalled chaicogens (pronounced as kat-ke-jens) (meaning
for
ur
ore forming elements) because many metal oxides occur as oxides and sulphides.
Like elements of groups 14 and 15, the elements of this group also show a gradual increase in metallic
s
character on moving down the group. Thus, oxygen and sulphur are typical non-metals, selenium and tellurium,
ok
Yo
though essentially non-metallic, show some metallic character as well and hence may be called as metalloids.
The last member of this group, i.e., polonium is, however, markedly metallic. It is also radioactive with very
Bo

short half-life.
re

7.16. OCCURRENCE
Oxygen is the most abundant of all the elements. It occurs in the free state as dioxygen (O2) and makes
ou
ad

up 2046% by volume and 23% by mass of the atmosphere. Most of the O2 present in the atmosphere is
produced by photosynthesis in which chlorophyll, the green colouring matter of the plants, absorbs energy
Y

from the sun to convert CO, and HjO of the atmosphere into glucose and O2.
/IV
nd
Re

6 CO2 + 6 H2O CgHjoOg + 6 O2

Chlorophyll
Fi

It also occurs in fonn of ozone (O3), an allotrope of oxygen, in the upper atmosphere which protects us
from the harmful radiations of the sun.
Oxygen makes up 46-6% by weight of the earth’s crust where it mainly occurs as silicate minerals. It
also makes up 89% by weight of water in oceans.
Sulphur, on the other hand, occurs less abundantly. It is the sixteenth most abundant element and constitutes
0 03 to 0-1% by mass of earth’s crust.
In the combined state, it primarily occurs as sulphates such as gypsum, CaS04.2 H^O ; epsom salt,
MgS04.7 H-,0 ; baryte, BaS04 and as sulphides such as galena, PbS ; zinc blende, ZnS and copper pyrites,
CuFeS..
It also occurs in the free stale in the volcanic areas of Sicily, Italy, Japan, America, Russia and Iceland.
Small deposits of sulphui' are found in Kangra in Himachal Pradesh and Baluchistan in Pakistan.
Another major .source of sulphur is H^S present in natural gas and crude oil. Sulphur also exists in combined
state in living matter and is a constituent of many organic materials such as eggs, proteins (amino acids, cystine,
cysteine and methionine contain sulphur), enzymes, garlic, onion, mustard, hair and wool.
Selenium and tellurium are less abundant than sulphur (0-05 ppm in earth’s crust for Se ; 0-002 ppm for
Te) and occur as selenides and tellurides in sulphide ores. The principal source of Se and Te is the anode mud
p-BLOCK ELEMENTS 7/39

or the anode slime deposited during electrolytic refining of Cu. Polonium is even less abundant (0-001 ppm)
in the earth’s crust where it occurs as decay product in thorium and uranium minerals.
7.17. ELECTRONIC CONFIGURATION
The elements of group 16 have six electrons in the valence shell and hence their general outer electronic

arranged in three />orbilals as p\, p\ in accordance with

configuration is iisvip^. The four /^-electrons are
Hund’s rule. The names, electronic configurations and common oxidation states of group 16 elements are
given in Table 7.10.
TABLE 7.10. Electronic Configuration of Elements of Group 16
Atomic Electronic configuration
Eiements Oxidation states
number Complete With noble gas core

low
Oxygen (O) ls~ 2s^ 2p* [He] 2a2 2// -2,- l,+l,+2
Sulphur(S) 16 2s- 2p^ 3s~ Sp"^ [Ne] 3^2 3p‘^ - 2, +2, +4, +6
Selenium (Se) 34 U- 2s- 2/ 3s- 3p^ [ArJ 3d'^4s-4p
4
- 2, +2. +4, +6
4s- 4p^

ee
Tellurium (Te) 52 Is- 2s- 2/ 3a- 3p^ 3cl'^ [KrJ 4d'^5s-5p^ - 2, +2, +4, +6

rF
Fr
4s^ V 5.y2 Sp'*
Polonium (Po) 84 Is^ 2.y2 2p^ 3a- 3p^ 3d 10 [Xe]4f'^5d^^6s-6p-^ +2, +4, +6
4s^4p^4d'^4f'"^ 5s‘^5p^

r
6^2 6p"^
fo
u
ks
7.18. ATOMIC AND PHYSICAL PROPERTIES
Yo
oo

Some of the important atomic and physical properties of the elements of group 16 are listed in Table 7.11.
B

TABLE 7.11. Some Atomic and Physical Properties of Group 16 Elements

re

Elements
Property
0 S
u

Se Te Po
ad
Yo

Atomic number 8 16 34 52 84
-1
Atomic ma.ss/g mol 1600 32-06 78-96 127-60 210-00
nd

Covalent radius (single 66 104 117 137 146

Re

bond)/pm
Fi

Ionic radius (E2~)/pm 140 184 198 221 230 (app. value)
Ionisation enthalpy, Aj H 1 1314-0 1000 941 869 813
-1
(A,- H ) kJ mol A,H2 3388 2251 2045 1790

Electronegativity 3-50 2-44 2-48 2-01 1-76

Electron gain enthalpy - 141 -200 - 195 - 190 - 174

H)/kJ mo|-‘
Densily/g cm"^ at 298 K 1-32 2-06 4-19 6-25 9-4
(at m.p.) (rhombic (hexagonal
sulphur) grey)
Melting poini/K 55 393 490 725 520

(monoclinic
form, 673 K)
Boiling point/K 90 718 958 1260 1235
Oxidation states -2.- 1.+1.+2 - 2, +2, +4, +6 - 2, +2, 44. 46 -2,+2.44,46 42, 44
7/40 New Course Chemistry (XII)BEU

7.18.1. Atomic Properties

Some important atomic properties of group 16 elements are discussed below :
1. Atomic and ionic radii
The atomic radii of the elements of group 16 are smaller than those of the corresponding elements of
group 15. Further, as expected, the atomic radii of the elements of oxygen family increase down the group.
Explanation. The smaller radii of group 16 elements compared to group 15 elements are due to the
increased nuclear charge of the group 16 elements which results in greater attraction of the electrons by the
nucleus. The increase in the atomic radii of group 16 elements down the group is primarily due to increase in
the number of electron shells.

2. Ionization enthalpy. The first ionization enthalpies (y) of the elements ofgroup 16 are unexpectedly
lower while their second ionization enthalpies (A,.7/2) are higher than those of the corresponding elements
of group 15.
Explanation. The first ionization enthalpy (A-//y) of group 16 elements is lower than those of group 15
elements despite their smaller atomic radii and higher nuclear charge. This is due to the relatively symmetrical

w
and more stable electronic configuration of the elements of group 15 as compared to those of the elements of
group 16 (oxygen family), e.g..

F lo
Group 15 element N(Z = 7) U'^2s^2p\2p'^lp\ {more symmetrical, more stable)

Group 16 element 0(Z = 8) 1j2 2s^2p^2p'2pl {less symmetrical, less stable)

e
A y e

Fre
The A,-H2 values of the members of the oxygen family are, however, higher than those of the nitrogen
family because after the removal of first electron, the second electron has to be removed from a more

A.Hj
for
symmetrical exactly half-filled stable electronic configuration of the unipositive ion as explained below :
A.H2
N(l5- 2,2 2p\2p\2p[) N+ (Is^ 2s^2p\2p\2pl) > N-^l\s'^2s~2p^X Ip^lp^)
r
-1 -1 y z
1402 kJmol 2856kJniol
You
{More symmetrical, {Less symmetrical,
s
ook

more stable) less stable)

eB

A,H A, H

0{ls~2s^2pl2p'^2p'^) y 2p*y 2pl) 0^*(.U-^2^-2p'2p' 2p«)

l314kJ moH X z 3388kJmor‘ i
{Less symmetrical, {More symmetrical,
our
ad

less stable) more stable)

As aspected, the first and second ionization enthalpy values of the elements of group 16 decrease down
the group due to increasing atomic size and increasing shielding effect of the inner electrons.
dY

3. Electron gain enthalpy.* The elements of group 16 have two electrons less than the nearest noble
Re

gas configuration. Therefore, they have a high tendency to accept two additional electrons and hence have
Fin

large negative electron gain enthalpies next only to the halogens. The electron gain enthalpy of oxygen is,
howeve,r least negative in this group. This is due to its small size. As a result of which, the electron-electron
repulsions in the relatively small 2p-subshell are comparatively large and hence the incoming electrons are
not accepted with the same ease as in case of other elements of this group.
4. Electronegativity. The elements of group 16 have higher values of electronegativity than the
corresponding elements of group 15. Actually, oxygen is the second most electronegative element (EN =
3-5), the first being fluorine (EN = 4-0).
Explanation. This is due to (/) .smaller atomic size of group 16 elements as compared to those of group
15 elements and (//) elements of group 16 need only electrons while those of group 15 need three electrons
to attain the stable noble gas configuration.
The electronegativity of sulphur is, howeve,r much lower than that of oxygen. This is probably due to an
unexpected increase in the size of sulphur (107 pm) as compared to that of oxygen (66 pm). Thereafte,r the
electronegativity decreases slowly but regularly from selenium to polonium as the size of atom increases
slowly but regularly.
*Electron gain enthalpy is the negative of electron affinity.
p-BLOCK ELEMENTS 7/41

Further since after oxygen there is a steep drop in electronegativity of S, Se. Te and Po, therefore, their
compounds have less ionic chai-acter as compared to those of oxygen.
7.18.2. Physical Properties
Some physical properties of the elements of group 16 are discussed below :
1. Non-mctallic/metallic character. Because of high ionization enthalpy values, the elements of group

rw
16 are less metallic. Howeve,r as we move down the group, the ionization enthalpy decreases and hence the
metallic character increases.

Thus, oxygen is the most non-metallic element of group' 16. Sulphur is also a typical non-metal and is an
insulator. Se and Te are metalloids and hence are semiconductors . Polonium is, however, metallic in nature

e
but is radioactive with a short half life (13-8 days).
2. Melting and boiling points. The melting points, boiling points and densities increase regularly as

e
ulo
we go down the group upto tellurium. Howeve,r the melting and boiling points of polonium are lower than

r
those of tellurium.

F
Explanation. As we move down the group, the atomic size increase. As a result, van dcr Waals forces of
attraction among their atoms also increase and hence melting and boiling points regularly increase from O to

oF
Te. However, due to maximum number of intervening d- {i.e., 3d, 4d and 5d) and /- (i.e., 4/)-eIectrons, the
inert pair effect is maximum in polonium. Consequently, tlie .v-valence electrons in polonium are less available
as compared to those in tellurium. As a result, van der Waals forces of attraction will be weaker in Po than in

rs
Te and therefore, m.p. and b.p. of Po will be lower than that of Te.

k
o
3. Elemental state. Oxygen exists as a diatomic gas at room temperature while other elements (S, Se,
and Te) exist as octaatomic solids.
Explanation. Due to small size and high electronegativity,
oxygen atom forms pK - pn double bond with other oxygen atom of FIGURE 7.12
o
Y
to form 0 = 0 molecule. The intermolecular forces of attraction
B

between oxygen molecules are weak van der Waals forces and
Y

hence oxygen exists as a diatomic gas at room temperature.

r

However, the rest of elements of this group, do not form p

ue

n-pn multiple bonds due to their larger size and hence do not
exist as diatomic (Mj) molecules. Instead they prefer to form
Puckered S-membered ring
d

single bonds and have complex structures. For example, S, Se

o

structure of sulphur molecule (S3).

ad

and Te exist as octa-atomic molecules {Sg, Se^ and having

in

puckered 8-membered crown shaped rings. (Fig. 7.12).

4. Multiple bonding. As discussed above, oxygen forms pn - pn double bonds but other elements do
not. However, S and other elements of this group possess J-orbitals and hence form pK - dti multiple bonds.
Re
F

To obtain effective pn - drc-overlap, the size of d-orbilal must be similar to that of /7-orbital. Thus, sulphur
forms stronger pn - c/rc-bonds than the larger elements in this group. In other words, the tendency of these
elements to form pn - dn multiple bonds decreases from S to Se. For example, S = C = S is moderately stable,
Se = C = Se undergoes decomposition readily while Te = C = Te is unstable.
5. Catenation. Because of stronger S—S bonds as compared to O—O bonds, sulphur has a stronger
tendency for catenation than oxygen.
Explanation. Due to small size, the lone pairs of electrons on the oxygen atoms repel the bond pair of
the O—O bond to a greater extent than the lone pairs of electrons on the sulphur atoms in S—S bond. As a
result, S—S bond is much stronger (213 kJ mor^) than O—O bond (138 kJ mor') and hence sulphur has a
much stronger tendency for catenation than oxygen. Further, as the size of the atom increases down the group
from S to Po, the strength of the element-element bond decreases and hence tendency for catenation decreases
accordingly.
Thus, chains of sulphur atoms are present in polysulphides, polysulphanes,H— —H and polysulphuric
acids, HO3S — — SO3H. Similarly, oxygen forms polyoxides such as H-,Oo
H2S2, Hydrogen disulphide H—S—S—H

H2S3, Hydrogen trisulphide H—S—S—S—H

7/42 "P^uLcCec^'^. New Course Chemistry (X11)E!SJ9]

H2S4, Hydrogen tetrasulphide H—S—S—S—S—H

Hydrogen peroxide H_0—O—H

The S—S bond is very important in biological systems and occurs in proteins and enzymes.
6. Allotropy. All the elements of this group show allotropy. For example, oxygen exists in two non-
metallic forms, viz., dioxygen (0->) and ozone or trioxygen (O3). Sulphur has several allotropic forms as
discussed in Art. 7.24.1, pages 7/60-7/61.
Selenium has eight allotropic forms, of which three are red monoclimc forms containing Seg rings. The
thermodynamically most stable form is grey hexagonal metallic selenium which consists of polymeric helical
chains. The common form of the element is the amorphous black selenium. Grey selenium is the only allotrope
of selenium which conducts electricity. Tellurium has only one crystalline form which is silvery white and
semi- metallic. Like grey selenium, it also consists of polymeric helical chains. Polonium exists in two metallic
forms, i.e. a-form which is cubic and the ^-form which is rhombohedral. Thus, there is marked decrease in
the number of allotropic forms from S to Se to Te.

w
SUPPLEMENT YOUR
KNOWLEDCE FOR COMPETITIONS Photosensitive elements. The grey form of selenium
(metallic) and tellurium consist of parallel chains held by weak metallic bonds. In the presence of light, the

F lo
weak metallic bonds are excited and as a result, the number of free electrons increases and so does the
conductivity. Thus, these elements conduct electricity significantly only in presence of light. That is why Se
and Te are called photosensitive elements.

ee
Further since conduction increases with intensity of light, selenium is particularly used for measuring the

Fr
intensity of light.

7.19. CHEMICAL PROPERTIES

for
ur
Some important chemical properties of the elements of group 16 are discus.sed below :
s
7.19.1. Oxidation States
ook
Yo

Since all the elements of this group have ns~ np"^ configuration in their valence shell (outermost shell),
they can attain noble gas configuration viz. ns- np^ either by gaining or by sharing electrons. These elements,
eB

therefore, show two types of oxidation states,

(i) Negative oxidation states. Oxygen the first element of this group, has high electronegativity. Therefore,
our

it preferably completes its octet by gaining of electrons. As a result, all metal oxides are ionic and contain ions
ad

in which oxygen shows an oxidation state of -2. In addition, oxygen shows an oxidation state of-1 in peroxides
such as H20^, - 1/2 in superoxides such as KO^, zero in and O3, +1 in 0-,p2 and + 2 in OFj.
Y

Since the electronegativities decrease as we move down the group, the tendency of these elements to
Re

show - 2 oxidation state decreases from sulphur to polonium. Hence, there is much less probability of the
nd

formation of dinegative ions in case of S, Se and Te. The least electronegative element, polonium, in fact,
Fi

does not exhibit negative oxidation state at all. Rather it shows positive oxidation states only,
(ii) Positive oxidation states. Oxygen does not show positive oxidation states except in O2F2 and OFj.
The other elements of this group show positive oxidation states of +2, +4 and +6 due to promotion of electrons
to vacant r/-orbitals as explained in tiie diagram shown below ;
2s 2p
Oxygen atom in There are no d-orbitals
the ground state ti n t t Excitation not easy.
3s 3p 3d
Sulphur atom in Two unpaired electrons account
the ground state t; ti t t for an oxidation state of + 2.
Sulphur atom in the Four unpaired electrons account
first excited stale t; t t t t for an oxidation state of + 4

Sulphur atom in the Six unpaired electrons account

second excited state t t t t t t for an oxidation state of + 6
p-BLOCK ELEMENTS 7/43

In the ground state, these elements have two unpaired electrons and hence can form two bonds. This
explains their +2 oxidation state. In the ifrst excited state, one of the paired p-electron goes to the vacant d-
orbital of the same shell, thus making four unpaired electrons available for chemical bonding. This accounts
for their +4 oxidationstate. On further excitation, one of the 5-eleclrons also gets promoted to ^/-orbital, thus
making six unpaired electrons available for bond formation. This explains their +6 oxidation state. However,
due to inert pair effect, the stability of +6 oxidation state decreases down the group. Thus, + 6 oxidation
state is most stable in case of S and least stable in case of Po.

In general, the compounds of S, Se, Te and Po with oxygen arc tercovalent (+4 oxidation state). These
+ 4 compounds show both oxidising and reducing properties.
Since fluorine is the strongest oxidising agent, therefore, an element shows its maximum oxidation state
in its compound with fluorine. Thus, the compounds of S, Se, Te and Po with fluorine show an oxidation state
of + 6.

low
These compounds in +6 oxidations .state show only oxidising properties.
The behaviour of oxygen is different, since it does not have <i-orbitals in its valence shell. Therefore, in
case of oxygen, the 2p-electrons on excitation have to go to 3.9-orbital. But, since too much energy is required
to excite an electron to a higher shell (in this case from K to L-shell), the electrons in oxygen do not get

ee
unpaired. Therefore, oxygen behaves as a divalent element only.
rF
Fr
The bonding in higher oxidation states, particuhnly in +4 and +6 oxidation states, is primarily covalent.
7.19.2. Trends in Chemical Reactivity

for
Oxygen is most reactive of all the elements of group 16. Its reactivity is only slightly less than the most
u
reactive elements, halogens. S is also very reactive particularly at high temperatures which helps in the
s
cleavage of S—S bonds pre.scnt in Sg molecules. However, as we move down the group, the reactivity decreases,
ok
Yo
i.e., O > S > Se > Te > Po.
Bo

Some important trends in the chemical reactivity of group 16 elements are discussed below :
1. Reactivity towards Hydrogen (Formation of Hydrides). All the elements of group 16 foim hydrides
re

of the general formula, H,E where E = O, S, Se, Te, Po, i.e., H2O, H^S, H2Se, HiTe and H2P0.
Preparation. The hydrides of S, Se and Te are prepared by the action of acids on metal sulphides,
ou
ad

selenides and tellurides respectively. You are already familiar with the Kipp’s apparatus used for generation
Y

of H2S in the laboratory by the action of dil. H,S04 on FeS.

FeS (5) + 2 H30"'' (aq) > Fe“'*' (aq) + H^S (g) + 2 H2O (/)
nd
Re

Structure. All these hydrides have angular shape (Fig. 7.13) involving .sp^-hybridization of the central
atom. The bond angles, however, decrease from H,0 to H.,Te as shown below :
Fi

H2O H2S H.Se H2Te

104-5" 92-1" 91" 90"

Explanation. Due to stronger lone pair-lone pair than bond pair FIGURE 7.13
LONE,
bond pair repulsions, the bond angle in water decreases from the PAIRS
tetrahedral value of 109-28' to 104-5". As we move down the group
from O to Te, the size of the central atom goes on increasing and its
electronegativity goes on decreasing. As a result, the position of the BOND BOND
two bond pairs shifts away and away from the central atom as we PAIR PAIR

move from H^O to H^Te. Consequently, the repulsions between the

bond pairs decreases as we move from H-,0 to H.,Te and, therefore, H H

the bond angle decreases in the same order ; Structure of H2O molecule.
H2O > H2S > H2Se > H2Te
Oxygen forms another important hydride, H^O-).
7/44 New Course Chemistry (XIl)BZsXSl

Some properties of hydrides are given in Table 7.12.

TABLE 7.12, Properties of Hydrides of Group 16 Elements
Property HjO H2S H2Se H2Te
m.p./K 273 188 208 222

b.p./K 373 213 232 269

H-E distance/pm 96 134 146 169

HEH angle (“) 104-5 92-1 91 90

H/ H7kJ mol-' -286 -20 73 100

H ( H - E)/kJ mol-' 463 347 276 238
Dissociation constant in 1-8 X 10-'^ 1-3 X 10-’ 1-3 X 10^ 2-3 X iO"-'

w
(aqueous solution at 298 K)

These properties are briefly discussed below :

lo
(/) Physical state, colour, odour, etc. The hydride of O, i.e., H2O is a colourless, odourless liquid while
the hydrides of all the other elements are unpleasant, foul smelling, poisonous gases.

e
re
(ii) Volatility. All the hydrides of the elements of group 16 are volatile. Their volatility, however, first

rF
increases from H2O to H2S and then decreases from H2S to H2Te (Fig. 7.14). In other words, the boiling point

F
first decreases and then increases as shown below :
FIGURE 7.14
H2O H2S H^Se H2Te H2P0 373

r
373 K 213 K 232 K 269 K 309"-2 K
fo
u
Explanation. The abnormally high boiling point of H2O is
ks
t 323
due to strong intermoleciilar H-bonding. Since all other elements
Yo
O
have much lower electronegativity than O, they do not undergo Q.
oo
o
H-bonding. However, since the size of the atom increases regularly ? 273
from O to Te, therefore, van der Waals forces of attraction increase O
B

CD

with increase in the molecular size and hence the boiling point 213
increases gradually from H-,S to H2P0. Thus, the overall trend in
e

volatility is : > H2Se > H2Te > H-,Po > H2O.

ur

50 100 150
{in) Acidic character. The hydrides of group 16 elements act
ad

MOLECULAR MASS
Yo

as weak, diprolic or dibasic acids and thus dissociate in aqueous

solutions in two stages to give H"^ ions. Plot of boiling points of hydrides
of group 16 elements versus
H2E + aq ^ ± H-" (flcy)-t-HE-(c7r/)
d

their molecular masses.

Re

± HHaq) + E^-{aq)
in

HE" + aq ^
Their acidic strength, however, increases down the group, from H^O to H->Te, i.e., H^O < H^S < H2Se <
F

H2Te as is evident from the values of their dissociation constants (K^)

Hydride H2O H,S H-,Se H2Te
K a
I-OxlO-'^* 1-3x10-^ 1-3x10^ 2-3 X 10"^ H
Explanation. This increase in acid strength can be easily E These bonds break on heating
H
explained on the basis of their bond dissociation energy. As the ^ (E = O, S, Se, Te, etc).
atomic size increases down the group, the bond length increases H2O C£
and hence the bond strength decreases. Consequently, the CC
UJ
LU
H
cleavage of E—H bond (E = O, S, Se, Te, etc.) becomes easier. ~(T>
K
OCO
O

H2S CD UJ <(/>
As a result, the tendency to release hydrogen as proton increases, 5CO
< LU
CC CO
a: 01

i.e., acid strength increases down the group. Thus, H2O is least CO UJ < <
o 5
HjSe
acidic while H2P0 is most acidic. <0
Suj
occ

(iv) Thermal stability. The thermal stability of the UJ

Q- 0 =
HsTe O 3
hydrides decreases from H2O to H-,Te. Thus, water dissociates
I
< o
UJ

at about 2073—2273 K, H^S at 673—873 K, H2Se at about q:

423 K while H^Te decomposes even at ordinaiy temperature.

p-BLOCK ELEMENTS 7/45

ThivS is because as the size of the atom E in H2E increases, the bond H—E becomes weaker and thus breaks
on healing,
(v) Reducing character. Hydrides of all elements of this group except that of oxygen. Le.. water are
reducing agents. Their reducing character, however, increases from H2S to H2Te. This is due to the decrease
in their thermal stability. In other words, as the thermal stability decreases, the reducing character increases.
2. Reactivity towards Oxygen (Formation of Oxides)
All the elements of this group form Elements Dioxide (EO2) Trioxide (EO3)
two types of oxides, i.e., EO2 and EO3
as given below. O O3
s S02 S03
(/') Dioxides. Sulphur, selenium and
Se Se02 SeOs
tellurium when burnt in air form dioxides

ow
Te Te02 Te03
of the formula EO^-
Po P0O2
Sjj (5) + 8 02(5) ^8S02(g)
Sulphur dioxide exists as discrete molecules even in the solid state. These molecules are held together
by weak van der Waals forces of attraction. Therefore, SO2 is a gas at room temperature. In SO2, S is sp--

e
Fl
re
hybridized. Two of the three i7?^-orbitals form two o-bonds with oxygen atoms while the third contains the
lone pair of electrons.

F
35 3d
S in the first
tl
ur t t t t

r
excited state

ip^-Hybridization,
Forms
fo
Forms

pK - dn
ks
pn -pn
hybrid orbilals form two
double bond double bond
Yo
S-0, o-bonds
with O with O
oo

Sulphur is now left with one half-filled /7-orbital and one half filled r/-orbital. These form one pK - pn
double bond and one p k - d n double bond with oxygen atoms. Thus, SO2 has bent (angular) structure.
B

Because of bent structure, SO-, molecule has a dipole moment of 1.63 D.

e

The actual bond angle is 119-5° Despite the fact that two 7C-bonds are formed due to overlap of different
ur

orbitals, the two S—O bonds are, however, equal (143 pm) because of resonance between two structures
ad

(land II) as shown in Fig. 7.15.

Yo

Selenium dioxide is a white crystalline solid. In the gaseous state, it exists as discrete monomeric molecules
having structure (Fig. 7.16a) similar to that of SO2. However, in the solid state, it has a non-planar polymeric
d

structure (Fig. I.l6h) consisting of infinite chains. Te02 and P0O2 are non-volatile crystalline ionic solids
Re
in

and each one of these exists in two crystalline forms. Te02 in the solid state has a layered structure consisting
of Te04 units.
F

The different structures of the dioxides of group 16 elements arise due to the reason that the tendency of
these elements to form pn- pn multiple bond decreases as the atomic size increases down the group.
Propertie.s.
(fl) Acid-base character. The dioxides also differ from one another in their reaction with water. SO2
dis.solves giving sulphurous acid (H2SO3) which exists only in solution and cannot be isolated.
FIGURE 7.16
● O Sp2-0RBITAL

Structures of Se02 (a) in the gas phase

Resonance structures of S02- (b) linear chain form of solid Se02-
7/46 New Coiirse Chemistry (XII)S!ZsIS]

SeOo dissolves giving selenious acid (H2Se03) which can be isolated in the crystalline state. Te02 is
almost in.soluble in water. However, it dissolves both in alkalies to form tellurites and in acids to form basic
salts. Thus, Te02 is amphoteric in nature. Similarly, P0O2 is also amphoteric though more basic in character
than TeOi. Thus, the acidic character of the dioxides of group 16 elements decreases as we move down the
group from S to Po.
{b) Reducing-oxidising properties. Since +6 oxidation state of S is more stable than +4, therefore,
SOi acts as a reducing agent. Further, since the stability of +6 oxidation decreases from S to Te, therefore, the
reducing character of the dioxides decreases while their oxidising character increases. Thus, 3>C>2 on
oxidising agent, FIGURE 7.17
(ii) TrioxIdes. All the
o o O O
elements of this group form
p7t-dn pn-pir 142pm
trioxides of the formula EO3. pTt-d:i
S

w
O3 and SO3 are gases. Like
in SO2, sulphur in SO3 is also sp~- 0
hybridized. The three 5/?“-orbitals 0 o o o o o

Flo
of sulphur overlap with /^-orbitals 1 II 111

of oxygen to form three S—O, Resonance structures of SO3.

a-bonds. The sulphur atom (in the

ee
second excited state) is now left with one p-and two ^/-orbitals which overlap with /?-orbitals of oxygen to

Fr
form three 7t-bonds. However, due to resonance between the three canonical structures (I, II and III), all the
three S—O bond lengths are equal (Fig. 7.17).

for
ur
S in the second
excited state t t t t t t
s
7 Fomis Fonn
,\7? -Hybridization,
k
Yo
one pK - pn, two pK - (Ik,
hybrid orbitals fomi three
oo

double bond double bonds

S-0, CT-bonds with O with 0
eB

FIGURE 7.18
The molecules of SO3 are held by weak
van der Waals forces of attraction. Therefore, 0
O o 0
o
r

SO3 exists as a triangular planar gaseous

ou
ad

Vv
molecule at room temperature (Fig. 7.18fl). ..-s
O'
// // //
●O' O'
Due to trigonal planar structure, the three
Y

o o o 0
S-0 dipoles cancel one another. Therefore, the Planar triangular Linear chain of 803(5)
structure of 803(9)
Re
nd

net dipole moment of SO3 is zero. 0

0

However, in the solid state, SO3 exists

Fi

in several modifications of which a cyclic

trimer (Fig. 7.18/7) or a linear polymeric chain
o
/X/
structure (Fig. 7.18c) are important. Se03, on
O' o
o Se
o
/% O

0X3
o o
the other hand, exists as a cyclic tetramer, o
X /
8ev,
8640,2 in the solid state (Fig. 7.18</).
Solid TeOj has a network structure in Cyclictrimerof 803(5) Cyclic tetramer of solid 8003
which TeO^ octahedra share all vertices.
All the trioxides are acidic in nature. Structures of (a) gaseous S03 (b) cyclic trimer of
solid S03 (c) linear chain form of solid S03
3. Reactivity towards Halogens
(d) cyclic tetramer of solid Se03.
(Formation of Haiides).
The elements of group 16 form a number of halides in the oxidation states +1, +2, +4 and +6. The well
characterised chalcogen halides are listed in Table 7.13.
p-BLOCK ELEMENTS 7/47

TABLE 7.13. Some Important Halides of Elements of Group 16

Types of halide Oxygen Sulphur Selenium Tellurium Polonium

1. Monohalides 0,F, S2F2, Se2Cl2, Se^Br,

S2CI2, S2Br2
2. Dihalides OFo, CUO, SF2, SCI2 TeCl-?, TeBf2 P0CI2, PoBr2
Br^O
3. Tetrahaiides SF4, SCI4 Sep4, Tep4, TeCl4, P0CI4, PoBr^,
SeCl4, SeBr4 TeBr4, Te^ P0I4
4. Hexahalides SF6 SeF, TeF^

ow
The stability of the halides in any particular oxidation stale decreases in the order: F > C! > Br > J. The
highest oxidation state is realised only in case of fluorides, i.e., SFj-,.
(/) Hexahalides. Only fluorine forms hexafluorides. Their relative stability decreases in the order :
SF,>SeF,>TeF,.

e
SF^ is a colourless, odourless and non-toxic gas at room temperature. It is thermally stable and chemically

Fl
re
inert. Its chemical inertness is due to the reasonthat the six F atoms protect the sulphur atom from attack by

F
the reagents to such an extent that even thermodynamically most favourable reactions like hydrolysis do not
occu.r However, as the size of the central atom increases, steric hindrance decreases and hence the hydrolysis
ur
or
becomes easy.
Alternatively, as the size of the atom increases, the electronegativity decreases and the bond polarities
sf
increase leading to increase in reactivity or decrease in stability down the group. Thus, the order of hydrolysis
k
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of hexafluorides is : SF6<SeF6<TeFf,.
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In other words, TeF^ undergoes hydrolysis readily.

eB

TeFg + 6 H2O > HJeOg + 6 HF

Structure. All hexafluorides have octahedral structures. For example, S in SF^^ is 5/)V"-hybridized and
hence it has octahedral structure (Fig. 7.19 a) with FSF angle = 90° and S—F bond distance = 155 pm.
ur

({/) Tetrahaiides. FIGURE 7.19

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Amongst tetrahaiides, £
Q.
F
letrafluorides are the
d

most stable. SF4 is a gas,

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in

SeF4 is a liquid while Cl

Tep4 is a solid. These
/l99pm
F

fluorides have sp^d- S

205pm /
r S

hybridization and, thus, 104°

have trigonal bipyramid o

structures in which one
of the equatorial posi (a) Octahedral structure of SFg (b) Trigonal bipyrc midai
tions is occupied by a strucuture of SF4 (c) Bent strucuture of SCI2 (d) Structure of S2CI2
lone pair of electrons.
This lone pair repels the axial bond pairs thereby decreasing the angle from 180° to 173° in SF4 (Fig. 7.19 b).
This geometry is also called see-saw geometry.
The tetrahaiides can act both as Lewis bases due to the presence of a lone pair of electrons and Lewis
acids because the central atom can extend its co-ordination number to six.

SF4 -H BF3 - 4 F4S ^ BF3 SF4 4- 2F- > [SF^]2-

Lewis base Lewis acid
7/48 New Course Chemistry (XII)ISSISI

Further, unlike SF^ which does not undergo hydrolysis because the six F atoms protect the sulphur
atom from attack by water due to steric hindrance, SF^ readily undergoes hydrolysis because the four F
atoms cannot protect the S atom from attack by wate.r
SF4 + 2 HP > 4 HF + SO2 SFg + H.O ^ No action

(m) Dihalides. All elements except selenium form dihalides. These dihalides are formed by sp^-
hybridization. However, due to the presence of two lone pairs of electrons they have bent structures like that
of H2O. The structure of SCU is shown in Fig. 7.8(c).
However, in SCI2, the bond angle is little smaller (103°) as compared to that in H^O (104-5°). This is
due to the reason that S is less electronegative than O. As a result, the bond pairs of the two S—Cl bonds lie
away from the S atom in SCI2 as compared to those of O—H bonds in H2O. Consequently, bond pair-bond
pair repulsions decrease and hence the bond angle decreases to 103° in SCI2 from 104-5° in H-,0 (Fig.
7.8(c)).
(/v) Monohalides. The well known monohalides are dimeric in nature. Examples are SiF-,, S->CU, S->Br^,

w
Se2Cl2 and Sc2Br2. These dimeric halides undergo disproportionat ion. For example,

F lo
+1 +4 0

2Se2Cl2 SeCl^ + 3Se

Their structures are similar to H2O2 as shown in Fig. !.%{d).

ee
Fr
Uses. (0 Because of its inertness and good dielectric properties, SF^ is used as a gaseous insulator in
high voltage generators and switch gears.

for
(«) Both SF4 and SeF4 are used as fluorinating agents for conversion of - COOH into - CF3 and
ur
C = O and P = O groups into CF2 and PF2 groups respectively.
7.20. ANOMALOUS BEh WIOUR OF OXYGEN
s
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Oxygen, the first element of Group 16, like carbon (Group 14) and nitrogen (Group 15) differs
considerably from the rest of the elements of Group 16 because of the following inherent characteristics of
eB

oxygen:
(/) Small size, (ii) Higher electronegativity and (Hi) Non-availability of d-orbitals.
our

The main points of difference between oxygen and the remaining members of Group 16 are given
ad

below :

(/) Physical state. Oxygen is a gas at room temperature while all other members of this group are solids,
Y

(ii) Atomicity. Oxygen molecule is diatomic (O2) while the molecules of the other elements are more
Re
nd

complex, c.g., sulphur and selenium have octaatomic molecules (i.e., Sg and Scg) with puckered irng structures.
(Hi) Non-metallic/metallic character. Being highly electronegative (3-5), oxygen is a typical non-
Fi

metal. Sulphur (EN = 2-5) is also non-metallicbut other members of this group exhibit metallic character also.
(iv) Oxidation states. Being most electronegative element of the group, oxygen usually shows an
oxidation state of-2 in most of its compounds (except in peroxides, 0-,p2 and OF-,). Sulphur also shows an
oxidation state of-2 to some extent. Other elements of this group, however, do not show negative oxidation
states. Further, oxygen does not have tZ-orbilals. That is why it cannot expand its octet and hence cannot show
positive oxidation states. However, other elements of this group have vacant d- orbitals in the valence shell
and hence can show positive oxidation states of +2. +4 and +6.
Please note that oxygen also shows positive oxidation state of+1 in O2F2 and + 2 in OF2. This is due to
the reason that F is more electronegative than O and not due to the reason that O can expand its octet,
(v) Nature of compounds. Due to high electronegativity, oxygen is more ionic in its compounds. The
dinegative anion (0^“) is very common. The dinegative anions of other elements (i.e., S~~, Se^", Te“”) are less
common. In other words, compounds of oxygen are ionic as well as covalent while those of sulphur and
other members of this group are mostly covalent.
p-BLOCK ELEMENTS 7/49

(v/) Multiple bonds. Due to small size and high electronegativit y, oxygen forms pK - pn multiple
bonds with elements having similar size, i.e., carbon, nitrogen (C = O, C = N, C s N), etc. This tendency of
forming multiple bonds is not shown by other elements of this group mainly due to bigger size and lower
electronegativity.
(vii) Hydrides. Due to high electronegativity, oxygen forms H-bonds. For example, the hydride of
oxygen (H2O) forms intermolecular H-bonds while hydrides of other elements of this group (H->S, H^Se, etc.)
do not. As a result, H,0 is a liquid at room temperature while hydrides of other elements are gases.
(v//f) Magnetic nature. Due to the presence of two unpaired electrons, dioxygen (O2) is paramagnetic
in gaseous, liquid and solid states. The other elements of this group are, however, diamagnetic.
7.21. DIOXYGEN

Preparation. Dioxygen is prepared by the following methods :

1. By decomposition of oxygen-rich compounds. Certain compounds containing large amounts of

w
oxygen such as KCIO,, KMn04, KNO3, etc., give dioxygen on strong heating.
Heal Heal

F lo
2 KMn04 KoMn04 4- MnO^ + O2 ; 2 KCIO3 > 2 KCl + 3O2
Pot. permanganate Pol. manganaie Pol. chlorate Pot. chloride

ee
Heat Heat
2KNO3 > 2 KNO2 + O2 ; 2 BaO, - 2BaO + O2

Fr
Pot. nitrate Pot. nitrite Barium peroxide Baiium oxide

Laboratory method of preparation (a) Thermal decomposition of potassium chlorate. In the

for
laboratory, dioxygen is produced by heating a mixture of potassium chlorate and manganese dioxide (catalyst)
ur
in the ratio 4 : 1 in a hard glass test tube at 420 K.
MnOo
ks
2 KCIO3 (s) ^ 2KC1 (5) 4 3 02(g)
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420 K
oo

(b) By the action of water on sodium peroxide. Dioxygen can also be prepared in the laboratory by
eB

the action of water on sodium peroxide.

2 Na202 (s) 4 2 H2O (/) ■» 4 NaOH (aq) 4 O2 (g)
Dioxygen can also be prepared by the action of acidified potassium permanganate on sodium peroxide.
r
ou
ad

2 KMn04 4 3 H2SO4- ^ K2SO4 4 2 MnS04 4 3 H2O 4 5 [O]

Na202 4 H2SO4 -> Na2S04 4 H2O2] X 5
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H2O2 4 [O] - H2O 4 O2] X 5

Re
nd

2KMn04 4 5Na202 4 8H2SO4 4 K2SO4 4 2M11SO4 4 5 Na2S04 4 8H2O 4 5O2

Fi

2. By thermal decomposition of oxides of metals low in the electrochemical series and higher
oxides of some metals. The oxides of certain heavy metals such as Hg, Pb, Ag, etc. on heating decompose to
liberate dioxygen.
Heat Heat
2HgO > 2Hg 4 O2 ; 2 Pb304 6PbO 4 0-)
Mercuric oxide Mercury Red lead Litharge
Heat Heat

2 Ag20 - > 4 Ag 4 O2 ; 3MnO, Mn^O^ 4 O,

Silver oxide Silver Manganese dioxide Trimanganic tetroxide
Manufacture. The main jurces for large scale preparation of dioxygen are (/) air and (//) water,
(i) From liquid air. The most economical method for commercial preparation of dioxygen involves
liquefaction of air (after removing CO2 und water vapours) followed by fractional distillation of the liquid
air thus obtained. During this process, dinitrogen (N2) with lower boiling point (77 K) distils over in the
gaseous form leaving behind dioxygen with higher boiling point (90 K) in the liquid slate,
(ii) From water. Dioxygen can also be prepared by the electrolysis of water containing a small amount
of a mineral acid or an alkali.
7/50 New Course Chemistry (Xll)CZsJ9]

Electrolysis
2H2O (/) > 2H2(g) +02(g)
Dioxygen is collected at the anode while dihydrogen is liberated at the cathode.
7.21.1. Atomic and Physical Properties
I. Physical properties.
(/) Dioxygen is a colourless, odourless and tasteless gas.
(U) It is slightly soluble in water and its solubility is about 3-08 cm^ in lOO cm^ of water at 298 K and one
atmosphere pressure. Even this small amount of dissolved dioxygen is sufficient to sustain marine and
aquatic life.
(Hi) It liquefies at 90 K and freezes at 55 K.
(iV) It has three stable isotopes : *^0, ^ZO and 'fo.

low
(v) Dioxygen is paramagnetic in spite of having even number of electrons.
7.21.2. Chemical Properties
The bond dissociation energy of dioxygen is high (4934 kJ mol"') and hence the reactions of dioxygen
require initiation by external heating. However, when the reaction starts, it continues on its own. This is due

ee
to the reason that the chemical reactions of dioxygen are exothermic and the heat liberated during the reaction
is sufficient to carry on the reactions.

F
Fr
Some important chemical properties are discussed below :
1. Action of litmus. Dioxygen is neutral to litmus.
for
ur
2. Supporter of combustion. Dioxygen is not combustible but is a supporter of combustion. When a
glowing wooden splinter is brought in an atmosphere of dioxygen, it bursts out into a flame.
ks
3. Oxidation. Dioxygen is a powerful oxidising agent and hence can oxidise metals, non-metals and
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other compounds to their respective oxides,
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(a) Reaction with metals. Dioxygen directly combines with most of the metals (except noble metals
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such as gold and platinum) to form their respective oxides. For example,
(/) Active metals like sodium, calcium, etc. react at room temperature forming their respective oxides.
r

4Na (j) + O2 (g) ^2Na20(^) ; 2 Ca (5) +02(g) ^ 2 CaO (i)

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ad

However, sodium also reacts with dioxygen at 575 K to form sodium peroxide.
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575 K
2Na (5) +02(g)> Na202 (j)
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nd

(ii) Magnesium bums in dioxygen to fonn magnesium oxide.

Heal
Fi

2 Mg (^) +02(g) » 2 MgO (5)

(/i7) Aluminium and iron. When heated in air form their respective oxides.
Heat Heat
4A1 (5) +302(g) > 2 AI2O3 (s) ; 4 Fe (5) + 3 O2 (g) > 2 Fe203 (i)
(b) Reaction with non-metals. Dioxygen combines with many non-metals except noble gases forming
their corresponding oxides. For example. 1073K
(/) With dihydrogen : 2 H2 (g) + O^ (g) > 2H20(g)
or Electric discharge
Water

3300 K
(ii) With dinitrogen : N2(g) + 02(g) ^ 2 NO (g)
Nitric oxide

This reaction occurs in the atmosphere, when lightning occurs in the sky.
Heat
(///) With sulphur: S (j) + O2 (g) SO2 (g)
Sulphur dioxide
p-BLOCK ELEMENTS 7/51

Heat Heat
(iv) With carbon: 2C(5) + 02(g) 4
2CO(g) ; C(s)+ 02(g) > C02(g)
(limited) Carbon monoxide (excess) Carbon dioxide

Heat
(v) With phosphorus : P4 (5) + 3 O2 (g) 4 P406(5) ; P4(5)+ 502(g) ^ ^4^10
(limited) Phosphorus (excess) Phosphorus
trioxide pentoxide
(c) Reaction with compounds
(i) With sulphur dioxide. Dioxygen reacts with sulphur dioxide at 720 K under a pressure of 2
atmospheres and in presence of platinum or vanadium pentoxide as catalyst to form sulphur trioxide
720K,2atm
2S02(g) + 02(g) ^ 2503(g)
PtorV205
This reaction forms the basis of Contact process for the manufacture of sulphuric acid,
(li) With ammonia. Dioxygen oxidises ammonia to nitric oxide in presence of Pt/Rh gauze as catalyst

w
at 500 K under a pressure of 9 bar.
1100K,Pt
4 NH3(g) +502(g) > 4N0(g) + 6H20(g)

F lo
This reaction forms the basis of Ostwald process for the manufacture of nitric acid.
(iii) Hydrogen chloride. Dioxygen oxidises hydrogen chloride to chlorine in presence of cupric chloride

ee
as catalyst at 700 K.

Fr
700K,CuCl2
4HCl(g) + 02(g) ^ 2H20(g) + Cl2(g)
This reaction forms the basis of Decon's process for the manufacture of chlorine.
for
(iV) With carbon disulphide. Carbon disulphide bums in dioxygen to form carbon dioxide and sulphur
ur
dioxide.
s
Heat
ook
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C52(g) + 3 02(g) > C02(g) +2 502(g)

(v) With metal sulphides. Many metal sulphides such as Zn5, Hg5, etc. react with dioxygen at high
eB

temperature to form metal oxides and sulphur dioxide.

Heat
2Zn5(5) + 3 02(g) > 2 ZnO (5)+ 2 502(g)
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ad

Heat
2Hg5(5) + 3 02(g) > 2 HgO (5)+ 2 502(g)
(vi) With hydrocarbons. Both saturated and unsaturated hydrocarbons bum in excess of air or dioxygen
Y

to form carbon dioxide and water.

Re
nd

CH4(g) +202(g) CO2 (g) + 2 H2O (g); = - 890 kJ mol"'

Fi

Methane

CH2 = CH2(g) +302(g) ■ 2002 (g) + 2H2O (g); H” = - 1411 kJ mol-'

Ethylene
2CHsCH (g) + 502(g)- ^ 4CO2 (g) + 2H2O (g); A^ = - 1300 kJ mol"'
Acetylene
These reactions are called combustion reactions and are highly exothermic in nature. That is why
hydrocarbons are used as fuels.
4. Action of silent electric discharge. Under the action of silent electric discharge, dry dioxygen gets
converted into trioxygen, i.e., ozone which is an allotropic form of oxygen.
Silent electric discharge
3O2 » 2O3
Dioxygen Ozone

5. Respiration. When air is inhaled, dioxygen is carried by the haemoglobin of the blood to various
tissues of the body where it is used in the slow oxidation of biomolecules such as glucose, fats and oils, etc.
The energy produced during this oxidation sustains life while CO2 and H2O produced are exhaled.
7/52 New Course Chemistry (XI1)|SSI9D|

7.213. Uses of Dioxygen

(/) Dioxygen is used in oxy-hydrogen and oxy-acetylene torches which are used for cutting and welding of
metals.

(//) It is used in metallurgical processes to remove the impurities of those n nals and non-metals which
form volatile oxides.

{in) It is used for artiifcial respiration in hospitals during surgery, by mountaineers and pilots at high
altitudes and by divers in deep-water diving,
(/v) Liquid dioxygen is used as a rocket fuel. For example, the combustion of hydrazine in liquid dioxygen
provides the tremendousthrust in rockets,
(v) It is used in the manufacture of a large number of compounds such as phenol, ethylene oxide, sulphur
dioxide, sulphuric acid, nitric acid, chlorine, etc.

low
(v/) It is also used in large scale production of TiO^ and synthesis gas (CO + H2).
7.22. OXIDES

Most of the elements in the periodic table, metals as well as non-metals, form binary compounds with

e
oxygen. These binary compoundsof metals and non-metals with oxygen are called oxides. In these oxides,

re
the oxidation state of oxygen is always -2. Some elements {e.g., Li, Mg, Al, Zn, etc.) form only one oxide

rF
while many others (e.g., Fe, Cu, N, S, P, etc.) form more than one oxide. The nature of bonding in these

F
oxides may be either ionic or covalent.
Oxides can be either simple (i.e., MgO, AI2O3, etc.) or mixed (i.e., Pb304, Fe304, etc.).

or
u
7.22.1. Simple Oxides f
ks
Simple oxides are best classified on the basis of their acid-base properties. These are classified into the
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following four categories : (/) Basic oxides, (ii) Acidic oxides, {in) Amphotericoxides and (iv) Neutral oxides
oo

(0 Basic oxides. An oxide is said to be basic if it combines with water to form a base. Alkali metals,
B

alkaline earth metals (except beryllium) and tran.sition metals generally form basic oxides. For example,
re

Na20 (s') + H2O (/) > 2NaOH {oq) ; MgO (5) + H2O (/) > Mg(OH)2 {aq)
Fe203 {s) + 3H2O (/) > 2Fe(OH)3 {aq)
u
ad

The basic oxides react with acids to form salts and

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water.

Na20 (5) -I- 2 HCl {aq) > 2NaCl {aq) + H2O (/)
d

Fe203 (5) + 3 H2SO4 {aq) > Fc2(S04)3 {aq) + 3H2O (0

Re
in

(«) Acidic oxides. An oxide which combines with water to form an acid is called an acidic oxide. The
oxides of non- metals are acidic in nature. For example,
F

CO2 (g) -H H.O (/) > H2CO3 {aq) ; SO2 (g) + H2O (/) ^ H2SO3 {aq)
Carbonic acid Sulphurous acid
^4^10 6H2O (/) ^ 4H3P04(ary) ; N2O5 (/) + H2O (/) ^ 2 HNO3 {aq)
Phosphoric acid Nitric acid

CI2O7 (g) + H2O (/) ^ 2HCl04(a^)

Perchloric acid

These acidic oxides react with bases to form salts and water. For example,
SO, (g) -1- 2NaOH {aq) Na2S03 {aq) + H2O (/)
P4O10 is) + 12 NaOH {aq) -> 4Na3P04 {aq) + 6H2O (/)
Besides, oxides of non-metals, the oxides of some metals in higher oxidation states are also acidic in
nature. For example, Mn20y, Cr03, V20g, etc.
Mn^O^ + H,0 ■>
2 HMn04 ', CrO^ + H,0 ^ H2C1O4
Permaganic acid Chromic acid
 p-BLOCK ELEMENTS 7/53

{Hi) Amphoteric oxides. Oxides which react with both acids and bases are called amphoteric oxides.
These are generally formed by elements that are on the border line between metals and non-metals, i.e.,
elements (Al, Zn, Sn, Pb, etc.) which lie in the centre of the periodic table. For example,
AI2O3 is) + 6HC1 {aq) + 9 H^O (/) ^ 2[Al(H20)f,P+(a<?) + 6Cl-(a?)
(Basic) Hexaaquaatuminium (III) ion
AI2O3 (i) + 6NaOH (aq) + 3H2O (/) ^ 2N^[A\(OH)^]{aq)
{Aciiiic) Sod. hexahydroxoaluminate (III)
Other examples of amphoteric oxides are : BeO, ZnO, PbO, SnO and Sn02.
(/v) Neutral oxides. Oxides which neither react with acids nor with bases are called neutral oxides.
For example, H^O, N2O, NO. CO, etc.
7.22.2. Mixed Oxides

w
Metal oxides which consist of two simple metal oxides with the metal in different oxidation states are
■ called mixed oxides. These mixed oxides show the properties of both the metal oxides simultaneously. For
example,

Flo
Pb304 -F 4 HNO3 Pb02 +
2 Pb(N03>2 + 2 H2O
Red lead (from lead dioxide) (from lead oxide)

e
Thus, red lead is considered to be a mixture of two oxides - lead dioxide (Pb02) and lead oxide (PbO)

re
and is given the formula, PbO-,.2 PbO.

F
Similarly, magnetic oxide or ferrosoferric oxide (Fe304) is considered to be a mixture of ferric oxide
(Fe-,03) and ferrous oxide (FeO) and is given the formula, Fe203.Fe0. Consequently, it gives a mixture of
ur
r
ferric and ferrous salts on treatment with acids.
Fe203.Fe0 + 8HC1 ^ 2 FeClj + FeCl2 + 4 H2O fo
ks
Likewise, trimanganese tetroxide, Mn304 (2 Mn0.Mn02) is considered to be a mixed oxide of manganous
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oxide (MnO) and manganese dioxide (MnO-,).
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7.22.3. TVends in Acid-Base Behaviour of Oxides In the Periodic l^bie

B

(/) On moving from left to right in a given period in the periodic table, the nature of the oxides shows a
re

gradual variation from strongly basic through amphoteric to strongly acidic. For example,
Na20 MgO AI2O3 Si02 ^4^10 SO2 CI2O7
u
ad

Strongly basic Basic Amphoteric Weakly acidic Moderately acidic Strongly acidic Very strongly acidic
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(//) Amongst .9-block elements, on moving down the group, the basic character of the oxides increases.
For example
d

Li^O Na20 K2O Rb20, CS2O

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in

Weakly basic Basic Strongly basic Very strongly basic

F

(Hi) Among p-block elements, on moving down the group, the acidic character of the oxides decreases
while the basic character increases. For example, in group 13, the acidic character of oxides decreases and
basic character increases as shown below :

B2O3 AI2O3 In-,03 TI2O3

Acidic Amphoteric Amphoteric Basic Strongly basic
(/V) Among non-metallic oxides, the acidic character increases as the oxidation state of the non-metal
increases. For example.
4l 4-2 4-3 44 +5

N2O NO N2O3 N2O4 N2O5

Neutral Neutral Acidic Strongly acidic Very strongly acidic
(i') Among i/-block elements, the ba.sic character decreases and the acidic character increases with increase
in the oxidation state of the element. For example,
42 48/3 43 44 47
Mn,0 Mn.O
MnO 3 ""4 Mn.,03 Mn02 2 ""7
Basic Amphoteric Amphoteric Amphoteric Acidic
7/54 'PxcieCee^'4. New Course Chemistry fXH^Fgmn
SUPPLEMENT YOUR ^
KNOWLEPQE FOR COMPETITiONS Types of oxides. On the basis of oxygen content, the oxides
can be classified into the following types ;
1- Normal oxides. These oxides contain oxygen atoms as required by the normal oxidation number of-2 of
oxygen. For example, Na20, H2O, MgO, AI2O3, CO2, etc.
2. Polyoxides. These oxides contain oxygen atoms more than the normal oxidation number of -2 of oxygen
atom. These are of the following two types :
(i) Peroxides. In peroxides, the two oxygen atoms are linked to each other by a single covalent bond and
each oxygen atom has an oxidation number of -1. In other words, all peroxides contain a peroxide ion

ow
(0^“) having the structure, ~:0 or

For example, H2O.,, Na20„ Ba02, etc. The metallic peroxides on treatment with dilute mineral acids give
hydrogen peroxide i.e., BaO, + H,SO^ ^ BaS04 + H2O2

e
(ii) Dioxides.Oxides which have the same oxygencontent as peroxides but do not produce H2O2 on treatment

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with dilute mineral acids are called dioxides. They do not contain peroxide, i.e., O^" units but each oxygen

Flr
atom in dioxides has an oxidation number of -2. For example, MnO^ (0 = Mn = O),

F
PbO, (O = Pb = O). CO2 (O = C = O), NO2 (O = N O), etc.
ou
(iii) Superoxides. These oxides contain O2 (-0 —O:~) units in which each O atom has an oxidation
number of-1/2. All superoxides contain odd number of electrons (i.e., 13) and hence are paramagnetic in

sr
nature. For example, KO2, Rb02, CSO2, etc.

fo
3. Suboxides.These oxides contain less oxygen atoms as required by the normal oxidationnumber of-2 of
oxygen. For example, nitrous oxide (N^O), carbon suboxide (C3O2), etc.
k
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4. Mixed oxides. Already discussedin Art. 7.22.2, page 7/53.
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7.23. OZONE
reB

Elemental oxygen exists in two molecular allotropic forms, i.e., O2 and O3. The triatomic molecule, i.e.,
O3 is called trioxygen or ozone.
uY

Ozone is present in the upper atmosphere called stratosphere (about 20 km from the surface of the earth)
where it is formed by the action of ultraviolet (UV) radiations on dioxygen.
ad
do

UV light
30. (g) > 2O3 (g)
Ozone absorbs ultraviolet radiations (220-290 nm) and thus protects the earth and its inhabitants from
in

the harmful radiations of the sun. If there were no ozone layer, more ultraviolet radiations will reach the
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surface of the earth causing much damage to plant and animal life.
F

Ozone layer in the stratosphere is depleted by (a) nitric oxide, (b) atomic oxygen and (c) reactive
hydroxyl radicals which are also present in the atmosphere due to bio-mass burning.
(fl) 03(^) + N0 (g) ^N02(g) +02(g) ; (b) 0-,(g) + 0(g)- ^ O2 (g) + O2 (g)
(c) 03(g) + H0- (g) ^0.(g) + H00(g) ; HOO- (g) + O (g) -^HO-(g)+ 02(g)
Flights of supersonic aeroplanes, nuclear explosions and lightning release NO into the atmosphere.
Ozone layer is also depleted by man-made chlorofluorocarbons (CFC’s) which have been used since
long as aerosol propellants and as refrigerants (CF2CI., freon). Chlorofluorocarbons are long lived molecules
and diffuse into the stratosphere where they undergo photochemical dissociation to produce chlorine atoms.
These chlorine atoms react with ozone thereby decreasing the concentration of ozone at a rate much faster
than its formation from O2.
The chlorine radicals, however, do not combine to fonn CI2 molecule, because they need a three-body
collision to dissipate the energy and such collisions are extremely rare in the upper atmosphere.
hv
a^cF^ig) ^ Cl(g) + -CClF2aO ; ● Cl (g) + O3 (g) > CIO- (g) -h O2 (g)
slow
CIO ● > Cl- + ● 0 ; C10-Cg)+ -0(g) >-Cl(g) + 02(g)
 p-BLOCK ELEMENTS 7/55

Volcanic eruptions also release chlorine atoms into the atmosphere. Due to the hazardous effects of
CFC’s on the ozone layer, U.S.A. has banned the use of CFC’s in spray cans.
Three scientists namely P.J. Crutzen, M.J. Molina and F.S. Rowland received 1995 Nobel Prize in
Chemistry for their pioneering work in this area. As a result of their work, there is a world-wide concern about
the depletion of the ozone layer in the stratosphere and serious efforts are being made to find alternatives for
CFC’s and to limit the emissions of oxides of nitrogen (NO)^ into the stratosphere.
7.23.1. Preparation of Ozone (TVioxygen)
Ozone is obtained when silent electric discharge is passed through pure, cold and dry dioxygen in a
specially designed apparatus called the ozoniser. During this reaction, conversion of O, to ozone is 10% and
the product is called ozonised oxygen.
Silent electric discharge

w
302(g) r i 2O3 (g); (298 K) = + 142 kj mol
-1

Since the formation of ozone from oxygen is endothermic, therefore, it is necessary to use a silent

F lo
electric discharge (sparkless electric discharge) in its preparation. A silent electric discharge produces less
heat and thus prevents the decomposition of ozone back to oxygen.

ee
Ozone is prepared in the laboratory* by the FIGURE 7.20

Fr
following two types of ozonizers ; (/) Siemens Pure and
Ozonizer and (ii) Brodie’s Ozonize.r The description Dry O2—^
of Siemen’s ozonizer is given in Fig. 7.20.
for Tin Foil
ur
It consists of two co-axial glass tubes sealed at
-T
one end. The inner surface of the inner tube and the Tin Foil
s
outer surface of the outer tube are covered with tin
ook
Yo
Ozonised
foils. The inner and outer coatings of tin foil are Oxygen Tin Foil
eB

connected to the terminals of an induction coil. One ■►Induction ●4—

Coil
● end of the coaxial tubes has an inlet for pure and dry
dioxygen and the other end has an outlet for the Seimen’s Ozoniser
r
ad
ou

ozonised oxygen.
A slow stream of pure, dry and cold dioxygen is passed through the annular space between the concentric
Y

tubes and subjected to the silent electric discharge when some of the dioxygen gets converted into ozone. The
product which actually comes out of the outlet is a mixture of
Re

and O3. It is called ozonised oxygen and

nd

contains about 10% ozone. If concentration of ozone greater than 10% is required, a battery of ozonizers is
Fi

used.

Preparation of pure ozone. In order to prepare pure ozone, the ozonised oxygen is cooled with
liquid air when ozone (b.p. 161-2 K) condenses in preference to oxygen (b.p. 90-2 K). The liquid
ozone thus formed still contains some dissolved dioxygen which can be easily removed by fractional
distillation.

7.23.2. Structure of Ozone

The central oxygen atom in ozone is .vp--hybridized containing a lone

pair of electrons. As a result, ozone has an angular structure with a bond
angle of 117° as shown in Fig. 7.21.
It is actually a resonance hybrid of the following two resonating
Structure of ozone.
structures (I and II) as shown in Fig. 7.22.
*Iiicluded only in the syllabus of ISC.
7/56 New Course Chemistry (XII)ESIS

FIGURE 7.2^

Resonating structures II Resonance hybrid

Resonance structures of ozone.

Because of resonance, both the oxygen-oxygen bonds have partial double bond
character. In other words, both the oxygen-oxygen bonds are equal (128 pm) and lie in between those of
oxygen-oxygen double bond length of 121 pm (in O2) and oxygen-oxygen single bond length of 148 pm (in
H.O.).

7.23,3. Physical Properties of Ozone

w
(f) Ozone is a pale blue gas having a strong characteristic smell.

F lo
In small concentration, it is harmless. However, if the concentration rises above 100 parts per million
(ppm), breathing becomes uncomfortable resulting in headache and nausea.
(//) When cooled in liquid air, pure ozone condenses to a deep blue liquid (b.p. 161 K) and then to a violet-

e
Fre
black solid (m.p. 80 K).
(Hi) It is about 1 -67 limes heavier than air because its vapour density (V.D.) is 24 while that of air is 144.
for
(/v) It is slightly soluble in water but readily dissolves in organic solvents such as turpentine oil, cinnamon
oil, carbon tetrachloride, glacial acetic acid, etc.
r
(v) Ozone is diamagnetic while dioxygen is paramagnetic.
You
oks

7.23.4. Chemical Properties of Ozone

eBo

1. Neutral character. Ozone (trioxygen) is neutral to litmus.

2. Decomposition. Ozone is thermodynamically unstable as compared to oxygen since its decomposition
into oxygen results in liberation of heat (AH is -ve) and an increase in entropy (AS is +ve). These two factors
our
ad

reinforce each other, resulting in large negative Gibbs free energy change (AG) for the decomposition of
ozone to oxygen. Therefore, high concentration of ozone can be dangerously explosive.
Even at ordinary temperature, it decomposes slowly to give oxygen. However, when heated to 473 K or
dY

in presence of a catalyst such as manganese dioxide, platinum or cupric oxide, the decomposition takes place
Re

rapidly.
473Kor catalyst
Fin

2O3 (g) ^ 302(g)

Ozone Dioxygen
3. Oxidising agent. Ozone is a powerful oxidising agent, second only to F2 bat much stronger than
dioxygen. This is due to the reason that ozone has higher energy content than dioxygen and hence decomposes
to give dioxygen and atomic oxygen.
03(g) 4
02(^) + 0(g)
Ozone Dioxygen Atomic oxygen
The atomic oxygen thus liberated brings about the oxidation while molecular oxygen is set free. Some
important oxidation reactions of ozone are discussed below :
(a) Oxidation of compounds
(i) Ozone oxidises black lead sulphide to white lead sulphate.
O3 (g) O2 (g) + O (g)l X 4
PbS (5) + 4 O (g) PbSp4 is)
PbS (5) + 4 03 (g) ^PbS04 (i')+4 02 (g)
p-BLOCK ELEJWENTS_ 7/57

Similarly, ozone oxidises sulphides of Cu, Zn and Cd to their respective sulphates.

CuS (s) + 4 O3 (g) ^CuS04(5)+4 0.(g)
ZnS (5)+4 03(g) ^ ZnSO^ (5) + 4 O2 (g)
CdS (s) + 4 O3 (g) ->CdS04 (5)+ 4 0, (g)
(//) Ozone oxidises halogen acids to corresponding halogens
O3 (g) -> O2
2HC1 iaq) + O (g) ^a(g) + H20 (/)
2HC1 (aq) + O3 (g) ^Cl2 (g) + H2 0(/) +02(g)
Similarly, HBr is oxidised to Br2 and HI to I,.
(Hi) Ozone oxidises nitrites to nitrates
O3 (i') ^0,(g) + 0(g)
KNO2 +o — + KNO3 {aq)

w
KNO2 {aq) + O3 (g) 4 KNO3 {aq) + O2 (g)

F lo
(/v') Ozone oxidises moist potassium iodide to iodine
O3 (g) + O2 (g) + O (g)
2KI (fl^) + H20 (/) + 0(g)- ^ 2K0H {aq) + I, {s)

e
Fre
2K1 {aq) + H2O (/) + O3 (g) ■> 2K0H {aq) + U (s) + O2 (g)
for
When ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9-2)
iodine is liberated quantitatively which can be titrated against a standard solution of sodium thiosulphate.
This reaction is used for quantitative estimation of ozone gas.
r
You
However, if the oxidation is carried out in the acidic
medium, I" is oxidised to IJ
oks

3 r {aq) + O3 (g) + 2 H+ {aq) ^ I- {aq) + 0^{g) + H20{l)

eBo

(v) Ozone oxidises acidified ferrous salts to ferric salts.

0,(S) ^0,(g) + 0(g)
ad
our

2FeS04 {aq) + H2SO4 {aq) + O (g) 4Fe2( 504)3 («^) + H20 (/)
2FeS04 + H2SO4 (aq) + O3 (g) ^ Fe2(S04)3 {aq) + H2O (/) + O2 (g)
(vO Ozone oxidises potassium ferrocyanide to potassium ferricyanide
Re
dY

0^(g) + 02(g) + 0(g)

Fin

2K4[Fe(CN)J {aq) + H^O (/) + O (g) ^ 2K3[Fe(CN)J (aq) + 2K0H (aq)

2K4[Fe(CN)6l (aq) + HjO (/) + O3 (g) ^ 2K3 [Fe(CN)6l {aq) + 2 KOH (aq) + O2 (g)
{vii) O.zone oxidises potassium inanganate (green) to potassium permanganate (pink-violet)
0,(g) + O2 (g) + O (g)
2K2Mn04 {aq) + H2O (/) + O (g) ■> 2 KMn04 (aq) + 2 KOH (aq)
2K,Mn04 (aq) + H,0 (/) + O3 (g) ■> 2KMn04 (aq) + 2KOH (aq) + O2 (g)
(viii) Ozone oxidises solid powdered KOH at temperatures below 263 K to potassium ozonide
0^(g) + 02(g) + 0(g)]x5
2 KOH (s) + 5 0(g) ^ 2 KO3 (s) + H,0 (/)
2 KOH (s) + 5 O3 (g) ■>2K03(i) + 5 02 (g) + H20 (/)
Potassium ozonide is an orange coloured solid and contains the paramagnetic ozonide (OJ) ion.
7/58 New Course Chemistry (XII)BSm

(ix) Ozone also oxidises cyanides to cyanaies and nitrogen dioxide to dinitrogen penloxide
CN- + O3 ^ OCN- + O2
2 NO2 + O3 4 N20<; + O2
(b) Oxidation of non-metals. Ozone oxidises non-metals like iodine, sulphur and phosphorus to their
corresponding oxyacids. For example,
(0 Ozone oxidises moist iodine to iodic acid (HIO^)
0,(g) 4 O2 ig) + O (5)] X 5
Uis) + 50 (g) 4 I2O5 (5)
I2O5 Csl + H.O (/) -4 2HIO3 (aq)

12(^)4503 (g) + H20 (/) 4 2HIO3 (aq) + 50, (g)

(//) Ozone oxidises moist sulphur to sulphuric acid (H2SO^).

low
03(S)' 4 02(g) + 0(g)]x24
Sg (5)+ 24 0(g) ^8S03(g)
503(g)+ H2O (/) 4 H2SO4 (aq)] X 8
S^(s) +24 0^(g) + %U,0(l) ^^H,S0^(aq) + 240, (g)

ee
F
(Hi) Ozone oxidises moist phosphorus to phosphoric acid (H;jP04).

Fr
03(g) 4 O2 (g) + O (g)] X 10
P4(^)+10(O) (g)
for
ur
P40m(A) + 6H20 (/) 4 4H3PO4 (aq)
P4(5)+ IOO3 (g) + 6H20 (/) 4 4H3PO4 («(/)+ 1002(g)
ks
Yo
(c) Oxidation of metalloids. Like non-metals, ozone also oxidises metalloids like arsenic and anti
oo

mony to their corresponding oxy-acids. For example,

eB

(/) Ozone oxidises moist arsenic to arsenic acid (H^sO^)

03(g) 4 02(g) + 0(g)]x5
r

2 As (s) + 50 (g) 4 AS2O5 (s)

ou
ad

AS2O5 (s) + 3HjO (/) -4 2H3ASO4 (aq)

Y

2As (,v) + 503(g) + 3H20 (/) 4 2H3ASO4 (aq) + 5O2 (g)

(ii) Ozone oxidises moist antimony to antimonic acid (H^SbO^)
Re
nd

03(g) 402(g) + o‘(g)]x5

Fi

2Sb (s) + 50 (g) 4 Sb.O^ (s)

Sb205 (.v) + 3H2O (1) 2H3Sb04 (aq)
2Sb (.v) + 5O3 (g) + 3H.0 (/) 4 2H3Sb04 (aq) + 5O2 (g)
(d) Oxidation of metals. Ozone also oxidises certain metals like silver, mercury, copper, etc. to their
corresponding oxides. For example.
(/) Silver is oxidised to silver oxide (Ag-,0)
03(g) — 4 02(g) + 0(g)
2Ag (5) + 0(g) 4 Ag,0 (s)

2Ag (.V) 403(g) 4 Ag20 (4) + O2 (g)

(//) Mercury is oxidised to mercurous oxide (Hg,0)
03(g) 4 O2 (g) 4 O (g)
2Hg (/) 4 O (g) 4 Hg20 (5)
2Hg (/) + 03(g) 4 Hg20 (.v) 4 O2 (g)
p-BLOCK ELEMENTS 7/59

The mercurous oxide thus formed dissolves in mercury which loses its meniscus and starts slicking to
the glass. This is called tailing of mercury. The meniscus can, however, be restored by shaking it with water
which dissolves mercurous oxide,
(e) Exceptional behaviour. In all the oxidiition reactions discussed above, the oxidation is brought about by
the atomic oxygen and dioxygen is always evolved. Howeve,r during the oxidation 0J SO2 to SOj and SnCh in
presence of cone. HCl to SnCl^, the ozone is utilized as a whole and dioxygen is not evolved. Thus,
(0 3S02(g) + 03(g) ^ 3803(g)
Sulphur dioxide Sulphur trioxide
Hi) 3 SnCl2 (y) + 6HC1 (c'o/ic.) + O3 (g) ^ SnC\^(acj) + 3H20(/)
Stannous chloride Stannic chloride

(f) Bleaching action. Because of its oxidising action, ozone acts as a mild bleaching agent and bleaches
vegetable colouring matter.
Vegetable colouring matter + O3 ^ Colourless oxidised matter + O^

w
As such ozone bleaches indigo, ivoi7, litmus, delicate fabrics like silk. wool. etc.
4. Action of peroxides. Ozone reacts with peroxides such as hydrogen peroxide and bm ium peroxide
liberating dioxygen.

F lo
H2O2 {aq) + O3 (g) ^ H2O (/) + 2O2 (g); Ba02 H) + O3 (g) ^ BaO (s) + 2 O2 (g)
5. Formation of ozonides - addition reaction

ee
When ozone is bubbed through the solution of an alkene in an inert solvent such as CH2CU, CHCI3,

Fr
CCI4, etc. at 196 K, it adds across the double bond forming an ozonide. O,
CH2 .CH2 O CH
. O3
196 K
> O
for
- O3
196K
> HC CH
ur
CCI4 CCI4
CH2 CH
CH2—O Acetylene
s
Ethylene Ethylene ozonide O, O O
ook
Yo
Acetylene ozonide
196 K
CH3—CH=CH2 + O3 CH3—CH CH,
eB

CCI4
Propylene

O o
Propylene ozonide
r
ad
ou

7.23.5. Uses of Ozone (Trioxygen)

(/) Because of its oxidising nature, it is used as a disinfectant and as a germicide for sterilizing water.
Y

(//) It is used for bleaching delicate fabrics, oils, starch, ivory, flour, starch, etc.
Re

(in) It is used for purifying air in crowded places such as cinema halls, underground railway stations, tunnels,
nd

mines, slaughter houses, etc.

Fi

(iv) It is used as an oxidising agent in the manufacture of artificial silk, synthetic camphor, potassium
permanganate, etc.
(v) It is used for delecting the position of double bond in unsalurated organic compounds.
7.23.6. Tests of Ozone
(/) It turns alcoholic solution of benzidine brown.
(ii) In presence of O3, mercury loses meniscus and starts sticking to glass (tailing of mercury)
(Hi) It turns moist starch-iodide paper blue,
(/v) Ozone does not reduce acidilled solution of KMn04 and K^Cr207.
7.24. EXTRACTION OF SULPHUR-FRASCH PROCESS'
This process is also called Louisiana Proces.s, since it is employed in Louisiana State of America. It is
particularly suitable when the sulphur deposits are found at a depth of 600-1000 ft covered with hard layers
of rocks, clay imd quick sand which prevent the ordinary mining operations.
^Included only in the syllabus of ISC.
7/60 New Course Chemistry (XII)BEiai

Three concentric tubes are introduced into the sulphur FIGURE 7.23
deposits as shown in Fig. 7.23.
Super heated water at 450 K and under pressure is
forced down the outer most pipe which melts the sulphur £ E
CD
(m.p. 385-8 K) thereby making a pool of liquid sulphur.
CD
Upper CP CD Grest-
CO CO
Hot compressed air under a pressure of 25-30 atmosphere ■u T3
E £ V
is blown down the innemiost pipe. The super heated water CP CP
f < CD
:ayef
CD O O
Sand-:
V

and air melt the sulphur forming an emulsion with air. This (P
I
o
T
U. (P
I

emulsion of molten sulphur and air bubbles are forced up

■i, 3 D
(P sz 0)
a. o. Q. O.
and pumped out through the central tube by suction pumps. Lime Sindle
3
CO CO
3
^ CO
CO
3
Layer

ow
The sulphur-air emulsion is collected in large wooden + +

vats where it solidifies. The solidified sulphur is broken up < <

into small pieces. By this process, 99-5% pure sulphur is

obtained. The crude sulphur thus obtained is generally
purified by distillation.

e
7.24.1. Sulphur-Allotropic Forms

re
Sulphur exists in several allolropic forms which may

Frl
F
be classified into the following three categories as discussed
below :

1. Homocyclk species containing 6-20 sulphur

ou
r
atoms. Several allotropic forms of sulphur containing - -Pool’df Liquid Sulphur —

so
6-20 sulphur atoms per ring have been synthesized in the
past two decades. Two important allotropes of this category kf
are yellow rhombic (a-sulphur) and monoclinic Extration of Sulphur
oo
(P-sulphur). The stable fonn at room temperature is rhombic
sulphur which gets converted into monoclinic sulphur when
Y
heated above 369 K.
eB

The method of preparation and properties of these two allotropes are described below :
(a) Rhombic or Orthorhombic sulphur or Octahedral sulphur or a-Sulphur.
ur
oY

Preparation. It is prepared by slowly evaporating the solution of roll sulphur in CS2 in a china dish
when octahedral crystals of rhombic sulphur appear.
ad

Properties. (/) This is the most stable form of sulphur at room temperature while all other varieties of
d

sulphur gradually change into this foim on standing.

in

(//) It is insoluble in water but dissolvesto some extent in benzene, alcohol and ether. It is readily soluble
Re

in CS2. It also dissolves in boiling concentrated solution of sodium sulphite to form sodium thiosulphate.
F

Boil
Na2S03 -I- S — ^ Na2S-50-^
(///) It has little thermal and electrical conductivity,
(ii’) Its specific gravity is 2 06 g cm"'^ and melts at 385-8 K.
(b) Monoclinic sulphur or (3-Sulphur
Preparation. It is prepared by melting sulphur in a china dish. The molten sulphur is allowed to cool till
a crust is formed. Two holes are made in the crust and the remaining liquid is poured out. On removing the
crust, needle-shaped crystals of monoclinic sulphur separate out.
Properties. (/) It is dull yellow in colour and is soluble in CS2 but insoluble in H2O.
(//') It has a specific gravity of I -98 g cm"^ and melts at 393 K.
(///) It is stable only above 369 K. Below this temperature, it slowly changes into rhombic sulphur. Thus, at
369 K, both the varieties of sulphur co-exist and this temperatureis called the transition temperature.
> 369K
Rhombic sulphur ^ < 369K
- Monoclinic sulphur
 p-BLOCK ELEMENTS 7/61

Structure. Both orthorhombic and monoclinic FIGURE 7.24

forms of sulphur arc molecular solids which consist s.
s. .s. s
■S
of puckered {non-planar) rings having crown ●S' ■S' 'S'

shape. The two forms, however, differ in the manner ■S s. ●S.

of packing of the molecules in the crystal lattice. (Fig. 'S' 'S'

S'
■S'
'S
,s,
7.24 a and 7.24 b). S' ●s s.
●s.
●s
S' ■S' ●S' 'S'
(c) Cyclo-S(^ form. Another important allotropic ■ S. ●S,
fonn of this category is cyclo-S^^ (Engel’s sulphur or S' ■s ■S' ●s

t-sulphur) in which the six-membered ring adopts s-

■ S.
s
●s.

the chair form as shown in Fig. 7.25. ●S' ■S' ■S'

●s.
2. Different chain polymers known as catena S
●S'
●s
'S'
sulphur. The most important aliotrope of this
●s

category is plastic sulphur or Ar-sulphur. o o

Its method of preparation and properties are Packing of 8-membered ring structure of

w
discussed below :
(a) orthorhombic sulphur (b) monoclinic
Preparation. It is obtained by pouring molten sulphur in the crystal lattice.

F lo
sulphur into cold water when a soft rubber like mass
called plastic sulphur is obtained. S
FIGURE 7.25

Propertie.s. (/) It is an amorphous form of sulphur. It is soft and

ee
elastic like rubber in the beginning but hardens on standing and cooling

Fr
and gradually changes into rhombic sulphur. s 205-7 pm 3
(//) It has rubber like transparent yellow threads and is insoluble in V 102'2®

C$2 and H-,0. for s ●S'

ur
(in) It does not have a sharp melting point. s

(iv) Its .specific gravity is 1-95 g cm"^.

s
Structure of cyclo-Sg
ook
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Plastic sulphur is regarded as a super-cooled liquid, i.e., a liquid allotropic form of sulphur
which because of rapid cooling below its melting point has no time to
eB

settle in a crystalline form. FIGURE 7.26

Structure. Plastic sulphur consists of zig-zag chains (Fig. 7.26)

our

and sometimes Sg and other rings.

ad

3. Unstable small molecules S„ (n = 2 - 5). These molecules

exist in different concentrations in liquid sulphur at higher
Y

temperatures and in sulphur vapours. ●S:

Re

St .species predominate at about 1(X)0 K. Like dioxygen (O2),

nd

:s:
$2 is paramagnetic and blue coloured and presumably has similar
Fi

bonding.

7.25. SULPHUR DIOXIDE, SOj

Preparation. Sulphur dioxide is formed together with a little Zig-zag chains of sulphur
(6-8%) sulphur trioxide when sulphur is burnt in air or oxygen atoms in plastic sulphur.
Sg (s) + 8 Ot (g) ^8S02(g)
In the laboratory, SO2 is prepared by treating a sulphite with dilute sulphuric acid,
(s) + H2SO4 (aq) ^ SOt (g) + NH2SO4 (aq) + H2O (/)
or
SO|“ (aq) + 2 H"*" (aq) - ^SOt (g) + H20 (0
It may also be prepared in the laboratory by heating copper turnings with cone. HTSO4.
Heat
Cu (s) + 2 H2SO4 (/) > CUSO4 (aq) + SO2 (g) + 2 H2O (/)
In hidiistry, SO2 is obtained as a by product of roasting of sulphide ores such as iron pyrites.
7/62 New Course Chemistry CXlI)CSm

4FeS2(s)+ 11 02(g) ^2Fe203 (^) + 8S02(g)

The gas is dried, liquefied under pressure and stored in steel cylinders.
Properties. 1. CoUmr, smell and solubility. Sulphur dioxide is a colourless gas with pungent smell and is
highly soluble in water. It liquefies at room temperature under a pressure of 2 atmospheres and boils at 263 K.
2. Acidic nature. Sulphur dioxide dissolves in water to form sulphurous acid
SO2 (g) + H2O (/) > H2SO3 {aq)
Therefore, SOt is regarded as anhydride of sulphurous acid. Its aqueous solution is acidic and turns
blue litmus red.
(0 Being acidic, it reacts readily with sodium hydroxide solution, forming sodium sulphite, which then
reacts with more sulphur dioxide to fonn sodium bisulphite (sodium hydrogen sulphite).
2 NaOH (aq) + SOo (g) ^ Na2S03 (aq) + H^O {/)
Sod. sulphite

w
Na2SOj (aq) + SO, (g) + H^O (/) > 2 NaHS03 (aq)
Sod. bisulphite
(ii) Being acidic, it also decomposes carbonates and bicarbonales evolving CO2 gas.

Flo
Na2C03 (aq) + 2 SO, (aq) + H,0 (/) > 2 NaHS03 (aq) + CO2 (g)

e
(Hi) Action of lime water. Like CO2, it turns lime water milky due to the formation of insoluble calcium

re
sulphite. However, if SO2 is passed for a longer time, milkiness disappears due to the formation of soluble
sodium bisulphite.

F
Ca(OH), + SO, ■>
CaSOj + H2O
ur (Milkiness)

r
CaS03 + SO, + H,0
fo
Ca(HS03), + H2O
Cal. bisulphite (.soluble)
ks
3. Non-supporter of combustion. SO, gas is neither combustible nor a supporter of combustion.
Yo
However, certain substances like heated carbon, lighted magnesium irbbon or heated potassium metal keep
oo

on burning. C + SO, CO2 + S

B

2Mg + SO2 ■> 2 MgO + S ; 4 K + 3 SO2 ■¥ K2SO3 + K2S2O3

This is due to the reason that high heal of the reaction dissociates SO2 to S and O, and the O2 thus
re

produced helps combustion.

u

4. Reducing properties. In presence of moisture, SO2 acts as a good reducing agent. Its reducing
ad
Yo

character is due to the evolution of nascent hydrogen

SO2+2H2O H2SO4+ 2 [H]
Some important reducing properties of SO2 are discussed below :
d
Re
in

(/) SO2 reduces halogens to halogen acids

Cl, + H,0 2 HCl + [O]
F

S0,+ H.,0 + [6] ^ H2SO4

Cl2+S0, + 2H,0 ^ 2 HCl + H2SO4
(ii) It decolourises pink violet colour of acidiifed KMnO^ solution
2 KMn04 + 3 H2SO4 4 K2SO4 + 2 MnS04 + 3 H2O + 5 [O]
S0, + H20+[0] H2SO4] X 5
2 KMn04 + 5 SO2 + 2 H2O ^ K2SO4+ 2 MnS04 + 2 H2SO4
(Pink violet) (Colourle.ss)
(Hi) It turns orange coloured potassium dichromate solution green
K2Cr207 + 4 H2SO4 K2SO4 + Cr2(S04>3 + 4 H2O + 3 [O]
SO, + H,0 + [O] 4 H2SO4] X 3
K,Cr,07 + H2SO4+ 3 SO2 ^ K2SO4+ Cr2(S04)3 + H,0
(Orange) (Green)
p-BLOCK ELEMENTS 7/63

(/V) It reduces ferric to ferrous salts

SO2+2H2O ^H2S04+2[H]
Fe2(S04)3 + 2[Hl ^ 2 FeS04 + H2SO4
S02+Fe2(S04)3+2H20 ^ 2 FeS04+ 2 H2SO4
(v) It reduces acidiifed potassium iodate to iodine
SO2 + 2 H2O -4 H2SO4+ 2 HI X 5
KIO3 + H2SO4 4 KHSO4 + HTO3] X 2
2HIO,+
3
10 H -4 l2 + 6 H2O

2 KIO3 + 5 SO2 + 4 H2O -42KHSO4+3H2SO4+I2

Pb02+S02
(vi) It reduces lead dioxide to lead sulphate >PbS04
5. Bleaching action. SO, is a mild bleaching agent. Its bleaching action is due to reduction of the
vegetable colouring matter by nascent hydrogen liberated in the presence ot moisture.

w
SO, + 2 H,0 - 4 H2SO4 + 2 [H]
Vegetable colouring matter + [H] Colourless vegetable matter

F lo
The bleaching by SO, is, however, temporary. When the bleached article is exposed to air, it regains its
colour due to oxidation.

6. Oxidising properties. It acts as a mild oxidising agent particularly when it reacts with strong reducing

ee
Fr
agents.
(/) It oxidises H2S to S : 2 H2S + SO2 4 2 HjO + 3 S
(ii) Active metals are oxidised : 3 Mg + SO2 4 2 MgO + MgS for
ur
4 K + 3 SO, > K2SO3 + K,S,03
3 Fe + SO, 4 2 FeO + FeS
s
ook

(Hi) CO is oxidised to CO2: 2 CO + SO2 4 2 CO2 + S

Yo

(iv) In presence of HCl, stannous and mercurous salts are oxidised

eB

2SnCl2 + S02+4HCl 4 2 SnCl4+ 2 H2O + S

2 Hg2Cl, + SO, + 4 HCl 4 4HgCl, + 2H,0 + S
our
ad

7. Addition reactions. (/) It combines with O2 in presence of platinised asbestos at 723 K or in presence
of V,05 at 773 K to form sulphur trioxide
2 SO2 + O, 4 2 SO3
Y

(ii) It reacts with CI2 in pre.sence of charcoal as catalyst to form sulphuryl chlorite
Re
nd

churcoiil
SO2 + Cl, 4 SO,CI,
Fi

Sulphuryl chloride
{///) It combines with sodium peroxide to form sodium sulphate.
Na202 + SO2 4 N32S04
Structure. The structure of SO2 has already been discussed in Fig. 7.15 on page 7/45.
8. Reaction with PCI5, it gives thionyl chloride, PCI5 + SO2 4 SOCl, + POCI3
Uses. SO2 is used (/) in the manufacture of sulphuric acid, calcium hydrogen sulphite (industrial chemical)
and sodium metabisulphite which is used as a preservative for jams, jellies and squashes.
(i7) in the refining of petroleum and bleaching of sugarcane juice.
(Hi) as a bleaching agent for delicate articles like wool and silk,
(/v) as a disinfectant for killing germs, fungi and moulds,
(v) as an anticlilor for removing excess of CI2 from bleached articles.
(vi) liquid SO, is used as a refrigerant and as a solvent to dissolve number of organic and inorganic chemicals.
7/64 “Px^xdee^’^. New Course Chemistry (XlI)ESsSD
7.26. OXOACIDS OF SULPHUR
FIGURE 7.27
Sulphur forms a number
of oxoacids such as H2SO3,
H2S2O3, H2S2O4, H2S2O5,
H.Sp^ix = 2- 5), H2SO4,
H2S2O7, H2SO5 and H2S20^.
Some of these acids are
toOHJ
unstable and hence cannot be H2SO3 (+4) H2SO4 (+6) H2S2O3 (+6, —2)
Sulphurous acid Sulphuric acid Thiosuiphuric acid H2SO5 (+6)
isolated. They are known only Peroxomonosuiphuric
in aqueous solutions or in form add or Caro's acid

of their salts. The structures

along with the oxidation state

ow
of the sulphur atom in some
important oxoacids are given in
Fig. 7.27.
Out of all these oxoacids, H2S2O8 {+6) H2S2O6 (+6) H2S2O7 (+6)

e
Peroxodisulphuric acid Dithionic acid Pyrosulphuric acid (oleum)
sulphuric acid is the most

re
or Marshall's acid

important.

rFl
Structures of some important oxoacids of sulphur.

F
7.27. SULPHURIC ACID (h2S04)

r
7.27.1. Manufacture of Sulphuric add (H2SO4) by Contact Process
ou
fo
These days sulphuric acid is mostly prepared by the Contact process. The acid produced by this method
ks
is free from arsenic impurities and hence can be safely used for the preparation of edible products.
Theory. The manufacture of H2SO4 by the Contact process involves the following steps:
oo

1. Production of SO2 by burning sulphur or roasting iron pyrites.

Y
eB

Sg + 80-) ^8S02 ; 4FeS2 + IIO2 ■> 2Fe203 + 8SO2

Iron pyrites
2. Catalytic oxidation of sulphur dioxide by air to give .sulphur trioxide.
ur

2SO2 (^) + O2 (g) 2S03(g);A^H'’ =-196.6 kJmoI-'

ad
Yo

This step is the key step in the manufacture of H2SO4 . In accordance with Le Chatelier’sprinciple, the
favourable conditions for the maximum yield of SO3 are :
d

(/) High pressure. Since the forward reaction proceeds with decrease in number of moles, therefore, high
Re
in

pressure will favour the reaction. In actual practice, a pressure of about 2 bars is used. This is because
gases are acidic and corrosion of the plant occurs at high pressures.
F

{//) Low temperature. Since the forward reaction is exothermic, therefore, low temperature will favour the
reaction. However, rate of the reaction decreases with decrease in temperature. Therefore, the reaction
is carried out at an optimum temperature of 720 K.
(///) Use of a catalyst. To increase the rate of reaction at low temperature, a catalyst is to be used. The
common! used catalyst is platinum, or divanadium pentoxide (V-^05). Since platinum is quite costly
and is easily poisoned by arsenic impurities usually present in SO2, therefore, these days, divanadium
pentoxide is employed because it is not only cheaper but is also not easily poisoned.
{iv) Purity of gases. To prevent poisoning of catalyst, the gases must be free from the impurities of As.,03,
dust particles and moisture.
(v) Excess of oxygen. To have maximum yield of SO3, O2 is used in excess,
(v/) Absorption of sulphur trioxide in 98% .sulphuric acid to form oleum.
SO3 + H2SO4 > H2S20y {oleum)
(vii) Dilution of oleum to get sulphuric acid of desired concentration.
H2S2O7 + H2O > 2H,SO 4
p-BLOCK ELEMENTS 7/65

In industry, the two steps, i.e., absorption of SO3 in H2SO4 to form oleum and its subsequent dilution
with water are carried out simultaneously to make the process a continuous one and also to reduce the cost.
The sulphuric acid obtained by the Contact process is generally of 96-98% purity.
Plant and its working. The plant used in the Contact process is shown in Fig. 7.28.
FIGURE 7.28
WATER CONC. H2SO4
SPRAY SPRAY CONC. H2SO4

IMPURE
i DRY
i.
>S02 + 02f—y—I
m\\
~T~] SO;, /m\

o 0 oe V2O5. 0
0000
O O 00 c.**.
SULPHUR- O O O o M t ft
e QUARTZ
e 0 o o g
c
o 00 o 0 <i
O O O o

ow
A
O 00 0
O O ® o PREHEATER 1
o o o < \
a ooo ■ » > «
o O 00
O 00 o
»cVA
O O oO
o oo o
AIR o o e<

1
> o oOO
OO o c
CATALYTIC
T hi WASTE ^ASTE OLEUM

e
CONVERTER
SULPHUR] ARSENIC PURIFIER
burner/ / I WATER ACID (H2S2O7)

re
CONTAINING

rFl
DUST WASHING AND DRYING GELATINOUS HYDRATED
PRECIPITATOR COOLING TOWER TOWER FERRIC OXIDE

F
Contact process for the manufacture of sulphuric acid.

r
The plant consists of the following parts:
ou
fo
1. Pyrite or sulphur burners. Here, sulphur dioxide is produced by roasting of pyrites or by burning
ks
sulphur.
2. Purifying unit. It consists of the following parts :
oo

(/) Dust remove.r Dust is removed either by blowing steam or by using Cottrell electrical precipitator.
(//) Scrubber or wmber Soluble impurities are removed by washing gases with water.
Y
eB

(///) Drie.r A spray of cone. H2SO4 is used for drying of gases,

(/v) Arsenic puriife.r Gelatinous Fe(0H)3 is used to absorb impurities.
r

3. Catalytic converter. The pure gases coming from the testing box are preheated to 720 Kina preheater.
ou

The gases are then passed over a catalyst consisting either of platinized asbestos or V2O5. Under these
ad
Y

conditions, SO2 gets oxidised to SO3 and the heat produced in the reaction is used for heating the gases in the
preheater.
d

4. Ab.sorption tower. SO3 from the catalytic converter is introduced at the base of the absorption tower
from the top of which HnS04 (98%) is showered. As SO3 moves above, it is absorbed by H2SO4 forming
Re
in

oleum.
F

SO3 + H2SO4 > H2S2O7 (oleum)

Oleum is then diluted with water to get H-,S04 of desired concentration.
H2S2O2 + H2O > 2 H2SO4
It may be FIGURE 7.29
notedthat sulphur COOLING
trioxide is not SO2 DUST ARSENIC TESTING
BURNERS AND
CHAMBER PURIFIER ^ BOX
directly absorbed DRYING

in water to form IMPURE SO2

“ PURE
sulphuric acid CONC. H2SO4
because it forms a ,,S02
H2O ABSORP SO3
dense fog of 2 H2SO4 H2S2O7 ^ TION ^
CATALYTIC PRE
CONVERTER HEATER
sulphuric acid Oleum TOWER

which does not

Flow-sheet diagram of contact process
conden.se easily.
A llow-sheet diagram of Contact process is given in Fig. 7.29.

4
7/66 '<t New Course Chemistry (XII) LV«JWl

7.27.2. Physical Properties of Sulphuric Acid

(/) Pure concentrated sulphuric acid is a colourless synipy liquid. It is also known as oil of vitriol.
(//■) Concentrated sulphuric acid (98-3%) has a specific gravity of 1-84 at 298 K. It boils at 590 K and
freezes at 283 K.
FIGURE 7.30
(Hi) It is highly corrosive and
O
produces bums on the skin, — HO

(/v) Sulphuric acid has a high boiling \ / HYDROGEN \ ^ \/

HYDROGEN
point (590 K) and a high y BONDS / V BONDS
viscosity because its molecules HO O — HO
'O
are associated due to inter-
H-Bonding in H2SO4.

ow
molecular hydrogen bonding as
shown in Fig. 7.30.
(y) It has a strong affinity for water. Tlius, when sulphuric acid dissolves in water, a large amount of heat is
produced which may spurt the acid out of the container. Hence, cai'e must be taken while preparing
dilute sulphuric acid from concentrated sulphuric acid (which is 98-3% and is 18M or 36N). Therefore,

e
concentrated H2SO^ is always diluted by adding the acid slowly into water with constant stirring and

re
not by adding water to the acid.

Frl
F
7.27.3. Chemical Properties of Sulphuric Acid
The important chemical properties of H7SO4 are discussed under the following categories :

or
ou
1. Acidic character. In aqueous solution, H,S04 ionizes in two steps ;
H2SO4 (aq) > (aq) + HSO^ (aq) K, = 1 X 10^
kfs
HSO; (aq) >nyaq)+sol- K2= 1.2 X 10-2
oo
Thus, H2SO4 acts as a strong dibasic acid and forms two series of salts - normal sulphates such as
Y

sodium sulphate (NaoS04) and acid sulphates or hydrogen sulphates or bisulphates such as sodium bisulphate
B

(NaHS04).
re

(/) Being an acid, it neutralizes alkalies to form salts and water and decomposes carbonates and
bicarbonates with the evolution of carbon dioxide gas.
oYu

NaOH + H2SO4 > NaHS04 + H2O : 2 NaOH + H2SO4 ^ Na2S04 -I- 2H2O
ad

Sod. hydrogen sulphate Sod. sulphate

2 NaHC03 H2SO4 > Na2S04 + 2 CO2 + 2 H.O ; Na2C03 + H2SO4
d

^ NaHS04 + CO2 + H2O

(ii) Being an acid, it reacts with active metals like Zn and Mg to evolve gas.
in
Re

Zn + H1SO4 (dilute) > ZnS04 + Mg + H2SO4 (dilute) ^ MgS04 + H2

F

2. Dehydrating agent. Concentrated H2SO4 has a strong affinity for water and hence it acts as a strong
dehydrating agent. Some important examples are :
Conc.H2S04
(i) Charring of .sugar : ^12^22^11 12c 4- llHjO
Sugar Sugar charcoal
(ii) Dehydration offormic acid atid oxalic acid
Coiic.H.,SO. COOH
Conc.HTS04
HCOOH H2O + CO ; I -V H2O + CO + CO2
Formic acid COOH
Oxalic acid

(Hi) It removes water of crystallization from hydrated salts. For example.

Conc.H^S04
CUSO4.5 H2O CUSO4 -h 5 H,0
Hydrated copper sulphate {blue) Anhydrous copper sulphate (white)
p-BLOCK ELEMENTS 7/67

(/v) Action on ethyl alcohol

C011C.H2SO4
(a) CH3CH2—OH 443 K
CH2 = CH2 + H20 {Intramolecular dehydration)
Ethyl alcohol Ethylene
Conc.H2S04
{b) 2 CH3CH2OH 410K ^ CH3CH2OCH2CH3 + H2O {Intermolecular dehydration)
Ethyl alcohol Diethyl ether
(v) Drying ofgases. Many gases such as O2, N2, CI2, HCl, etc. which do not react with cone. H2SO4 can
be dried by passing them through a bubbler containing cone. H2SO4.
3. Oxidising agent Hot concentrated sulphuric acid is a moderately strong oxidising agent since it on
decomposition gives nascent oxygen. Its oxidising power, however, lies between nitric acid and phosphoric acid.
Heat

low
H2SO4 ^ H2O + SO2 + O
This nascent oxygen brings about oxidation of a number of non- metals and metals.
(0 It oxidises carbon to carbon dioxide.
H2SO4 ^ H2O + SO2 + O] X 2

e
C + 20 ^C02

re
rF
C -I- 2 H2SO4 — ■> 2 SO2 + CO2 + 2 H2O

F
(«) It oxidises sulphur to sulphur dioxide
H2SO4- -^H20 + S02 + 0] X 16

r
S«+ 16 0 — ^8 802
fo
u
8

88+I6H28O4 ^24 802+ I6H2O

ks
Yo
(Hi) It oxidises phosphorus to phosphoric acid
oo

H2804- ^H20-i-802 + 0]x 10

B

P4+ 10 O ^*4^10
P4O10 + 6 H2O -4 4 H3PO4
e
ur

P4+ 10 H28O4 44H3PO4+ IO8O2 + 4H2O

ad

(tv) It oxidises metals such as Cu, Pb, Hg, Ag, etc. first to their oxides and then to their corresponding
Yo

sulphates liberating sulphur dioxide gas.

H28O4 4 H2O + 8O2 + O
d
Re

Cu + 0 4 CuO
in

CuO + H28O4 4 CU8O4 + H2O

F

Cu + 2 H28O4 ^Cu804+802 + 2H20

>^^th Zn, which is a stronger reducing agent than Cu, the reduction of 8O2 occurs further to give 8 or H28.
4. Action on salts. Because of its low volatility cone. H28O4 can be used to manufacture more volatile
acids from their corresponding salts.
(/) With dilute acid. Carbonates, bicarbonates, sulphites, sulphides, thiosulphates and nitrites are
decomposed by dil. H28O4 at room temperature. For example,
Na2C03 + H28O4 ^ [Na2804 + H2CO3] ^ Na2804 + CO2 + H2O
Strong acid Weak acid

NaHC03 + H2804 -4[NaH804+H2C03] - ^ NaH804 + C02 + H20

Na28 + H28O4 ^ Na2804 + H28
Na2803 + H28O4 4 [Na2804 + H28O3] — 4 Na2804 + 8O2 + H2O
Na28203 + H2SO4 4[Na2804 + H2803 + 8] > Na2804 + 8O2 + H2O + 8
2NaN02 + H2804 4 [Na2804 + 2HNO2] — ■4 Na2804 + NO + NO2 + H2O
7/68 New Course Chemistry (XII)CSm

(//) Wiih hoi concentrated acid. Fluorides, chlorides, nitrates, acetates, oxalates and tartrates are
decomposed by hot cone. HiS04 liberating their corresponding acids. For example,
CaF, + H2SO4 ^ CaS04 + 2HF NaCl + H2SO4 > NaHSO.4 + HCI

KNO3 + H2SO4 —>KHS04+HN03 ; 2 CH3COOK + H^S04 — > K2SO4 + 2 CH3COOH

COONa COOH

COONa
+ H2SO4 4
COOH
+ Na2SO^

In case of bromides and iodides, the halogen acid first liberated being a reducing agent reduces HoS04
to SO^ while it itself is oxidised to the corresponding free halogen. For example,
NaBr 4- H^SO4 4 NaHS04 4- HBrl x 2
H2SO4 4 H^O + SOt + O

w
2HBr4-0 HoO 4- Br,

2 NaBr 4- 3 H2SO4 - -4 2 NaHS04 4- SO, + Br, 4- 2 H,0

F lo
Similarly, 2 Nal + 3 H2SO4 - 4 2 NaHS04 + SO,"4-1, + 2 H,o”
5. Precipitation reactions. When the aqueous solutions of barium, strontium, calcium and lead salts

ee
are treated with dil. H,S04, white precipitates of their corresponding metal sulphates are formed.
BaCl,4- H,s6 4

Fr
4 BaS04 4- 2HC1 ; CaCl, + H2SO4 4 CaS04 4- 2 HCI
(White ppt.) (White ppt.)
(CH3COO)2Pb 4- H2SO4 4 PbS04 + 2 CH3COOH
for
ur
Lead aceuitc (While ppt.)
6. Miscellaneous reactions. (/) Sulphonation. Concentrated sulphuric acid and oleum react with aromatic
s
compounds to form sulphonic acids.
ook

330 K
Yo
H2SO4 4
C6H5SO3H + H2O
Benzene (cone.) Benzenesulphonic acid
eB

(ii) Action of pota.ssium ferrocyanide. Carbon monoxide (CO) is formed

Heat
K4[Fe(CN)(,] + 6 H2SO4 + 6 H2O 4 2 K2SO4 4- FeS04 4- 3 (NH4)2S04 + 6 CO T
r
ou
ad

Pot. ferrocyanide
{Hi) Reaction with .sulphur trioxide. SO3 dissolves in cone. H2SO4 forming oleum.
Y

H,S04 4- SO3 4 H2S2O2 (oleum)

(iv) Reaction with PCl^. Depending upon the amount of PCI5 used, one or both the hydroxyl groups of
Re
nd

H,S04 can be replaced by chlorine atoms. For example.

HO —SO2 —OH + PCI5
Fi

^ Cl —SO2 —OH + POCI3 + HCI

Sulphuric acid Chloro.sulphonic acid
HO — SO, — OH + 2 PCI5 -4 Cl — SO, — Cl 4- 2 POCI3 + 2 HCI
Sulphuric acid Sulphury! chloride
(v) Reaction with KCIO When heated with cone. FIGURE 7.31
H2SO4, KCIO3 gives chlorine dioxide (CIO2) with
_2-
explosion. O

Heat
3 KCIO3 + 3 H2SO4 4

3 KHSO4 4- HCIO4 + 2 CIO2 4- H2O OH

7.27.4. Structure of Sulphuric Acid and Sulphates HO

Both sulphuric acid and sulphate ion have o O

tetrahedral structure, (Fig. 7.31) in which sulphur is Tetrahedral structures of

4’/7^-hybridized with an oxidation slate of 4- 6 (a) sulphuric acid and (b) sulphate ion.
p-BLOCK ELEMENTS 7/69

Since in sulphate
ion. all the S—O bond
lengths are equal,/.e. 149
pm, therefore, it is a
resonance hybrid of four
canonical structures (1,11,
111 and IV) as shown in
Fig. 7.32 :
2_
Resonance structures of SO^ ion.
7.27.5. Uses of Sulphuric Acid
Sulphuric acid has a vast variety of applications both in industiy as well as in laboratory The industrial

w
strength of a nation largely depends on the amount of sulphuric acid it produces and consumes. Therefore, it
is rightly called King of Chemicals. Some important commercial uses of sulphuric acid are :

F lo
(i) In chemical indu.stry. Sulphuric acid is employed in the manufacture of hundreds of other chemicals
such as hydrochloric acid, nitric acid, phosphoric acid, sulphates and bisulphates, diethyl ether, etc.
(ii) In fertilizer industry. Suli^huric acid is used to manufacture fertilizers like ammonium sulphate and

ee
super phosphate of lime,

Fr
(iii) In dyes, drugs, paints, pigments and detergents. It is used directly or indirectly in the manufacture
of number of chemicals such as dyes, drugs, paints, pigments and detergents,
for
(iv) In petroleum refining. Crude petroleum is treated with sulphuric acid to remove unwanted sulphur
ur
and o±er tarry compounds,
(v) In the manufactureof explosives. A mixture of sulphuric acid and nitric acid called nitrating mixture
oks
Yo
is used for nitration of organic compounds. This nitration process is used for the manufacture of a large
number of explosives like dynamite (a mixture of glyceryl trinitrate and glyceryl dinitrate absorbed
o
eB

over Kieselguhr — a porous earth), gun cotton, {nitrocellulose) T.N.T. (2, 4, 6-trinitrotoluene), picric
acid {2,4,6-trinitropbenol), etc.
(vi) In metallurgy. A number of metals like copper and silver are extracted from their ores using sulphuric
our
ad

acid,

(vii) In pickling. Sulphuric acid is used for cleaning (pickling) the surface of metals before enamelling,
Y

electroplating and galvanising,

Re

(viii) In lead storage batteries as electrolyte,

nd

(ix) As laboratory reagent. It is widely used in laboratory as a drying and dehydrating agent.
Fi

PART->III. GROUP 17 ELEMENTS : THE HALOGEN FAMILY

7.28. GENERAL INTRODUCTION

Group 17 of the periodic table consists of five elements, i.e., fluorine, chlorine, bromine, iodine and
astatine. They are all non-metallic elements and are collectively named as halogens. The name halogen
(Greek, halo = sea salts, genes = producing) meaning sea salt formers was given to them by Schweigger in
1811 because the salts (chlorides, bromides and iodides) of the first three elements occur in sea water. Of all
the halogens, fluorine is the most reactive and hence is also called super halogen. Astatine, however, is
radioactive and hence occurs in nature only in traces. Therefore, not much is known about its chemistry,
though it might be expected to be a metalloid.
7.29. OCCURRENCE

Halogens are highly eractive elements and hence do not occur in \hefree state or native stale. They mainly
occur in the combined slate in fomi of their halide (X~) salts although iodine also occurs as iodate (lOj).
7/70 ‘PxeuCee^'^i- New Course Chemistry (XI1)CE]9]

(a) Fluorine occurs as insoluble fluorides to the extent of 0.07% in earth’s crust. The chief minerals of
fluorine are : (/) Fluorspar, CaF,, (ii) Cryolite, Na3AlFfj and (Hi) Fluoroapatiie. CaF2.3Ca3(P04)2.
Small amounts of fluorides are also present in soil, river water, plants, bones and teeth of animals,
(b) Chlorine is the most abundant of all the halogens and occurs as chlorides to the extent of 0.14% in
earth’s crust. The chief sources of chlorine are sea water (2.8% by mass), salt wells and salt beds. In salt beds,
it mainly occurs as : (/) Sodium chloride (Rock salt). NaCl, (ii) Carnallite KCl.MgCl2.6H2O, and
(Hi) Calcium chloride, CaCl2.
(c) Bromine occurs as bromides to the extent of 2.5 x 10“^% in earth’s crust. It mainly occurs in sea
water and salt lakes as bromides of alkali and alkaline earth metals, i.e., NaBr, KBr and MgBr2.
(d) Iodine occurs to the extent of 8 x 10"^% in earth’s crust. It mainly occurs in (0 sea weeds (0.5% by
mass) as alkali metal iodides, (ii) Caliche (crude chile saltpetre) which is mainly sodium nitrate containing
(0.2% by mass) of iodine as sodium iodate (Nal03).
730. ELECTRONIC CONFIGURATION

w
The elements of group 17 have seven electrons in the valence shell and hence their general electronic
configuration is ns^iip^, i.e., ns~ np^ np^~, np^, where n = 2 - 6. Thus, they contain one electron less than the

F lo
nearest inert gas configuration. The names, electronic configuration and common oxidation states of group
17 elements are given in Table 7.14.

ee
TABLE 7.14. Electronic Configuration and Oxidation States of Halogens

Fr
Elements Atomic Electronic configuration Oxidation states
Number Complete for
With noble gas core
ur
Fluorine (F) 9 I 2s- 2p^ [He] 2s- Ip^ -1
s
Chlorine (Cl) 17 \s^ 2s^ 2/’ 3.9^ 3p^ [Ne] 3^^ -l,+l,+3,+5, +7
ook
Yo

Bromine (Br) 35 l.r 2s- 2/’ 3s- 3p^ 3d^^ 4s~ 4p^ [Ar] -I,+l,+3,+5, +7
eB

Iodine (1) 53 ls~ 2s- 2p^ 3s- 3p^ 3d''^ [Krl Ad^^5s-5p^ -l.+l,+3, +5, +7
4s- 4p^ 4r/"’ 5.9“ 5p^
Astatine (At) 85 Is- 2.9- 2p^’ 3.9^ 3p^ 3(/'0 [Xe] 4/ '■» 5d'^ 6!t 6p^
our
ad

4.92 4/4^/'‘’4/
14

5.9- 5/ 5rf‘" 6.9- 6p^

Y
Re

7.31. ATOMIC AND PHYSICAL PROPERTIES

nd

Some of the important atomic and physical properties of the elements of group 17 are listed in table 7.15
Fi

and physical properties of molecules are given in Table 7.16.

TABLE 7.15. Some Atomic Properties of Group 17 Elements (Halogens)
Property F Cl Br I At
radioactive

Atomic number 9 17 35 53 85
-I
Atomic mass/g mol 1900 35-45 79-90 126-90 210

Covalent radius/pm 64 99 114 133

Ionic radius (X")/pm 133 184 196 220

-I
Ionisation enthalpy/kj inol 1680 1256 1142 1008

Electronegativity (Pauling scale) 4-0 3-2 3-0 2-7 2-2

-1
Electron gain enthalpy/kJ mol -333 -349 -325 -296
A
Hy<i. H(X-)/kJ mol-’ 515 381 347 305
p-BLOCK ELEMENTS 7/71

TABLE 7.16. Some Physical Properties of Molecular Halogens

Fa CI2 Bra *2
Melting point/K 54-4 172-0 265-8 386-6

Boiling point/K 84-9 239-0 332-5 458-2

Density/g cm"^ 1-51 for liquid 1-66 for liquid 3-19 for liquid 4-94 for liquid
at 85 K at 203 K at 293 K at 273 K

Distance X—X/pm 143 199 228 266

X^ig) >2XU)/(kJ mol-*) 158-8 242-6 192-8 151-1

EW 2-87 1-36 1-09 0-54

7.31.1. Atomic Properties

Some important atomic properties of the elements of group 17 are discussed below :

w
1. Atomic and ionic radii. Halogens have the smallest atomic radii in their respective periods. This is
due to the reason that atomic radii decrease as the nuclear charge increases as we move from left to right in a

F lo
period.
2, Ionization enthalpy. The ionization enthalpies of halogens are very high next only to the inert gas
(noble gas) in each period. This implies that halogens have very little tendency to lose electrons. However, on

ee
moving down the group, the ionization enthalpies progressively decrease as the size of the halogen increases

Fr
from F to I.

3. Electron gain enthalpy.* Halogens have one electron less than the nearest noble gas configuration.
for
Therefore, they have a strong tendency to accept an additional electron and hence halogens have large
ur
negative electron gain enthalpies.
Further, as we move down the group, the electron gain enthalpies become less and less negative as the
s
ook

size of the halogen increases. However, electron gain enthalpy of fluorine is less negative than that of chlorine.
Yo

This is due to the small size of fluorine atom. As a result of which strong electron-electron repulsions are
eB

present in the relatively compact 2p-orbitals of fluorine. Thus, amongst halogens, chlorine has the most
negative electron gain enthalpy. In nut shell, electron gain enthalpies become less and less negative as we
move from Cl (-349) F (-333) Br (-325) ^ I (-296)
our
ad

4. Electronegativity. Due to small size and higher nuclear charge, each halogen has the highest
electronegativity in its period. Thus, halogens are highly electronegative elements. Fluorine has been assigned
the highest electronegativity of 4.0. As we move down the group from fluorine to iodine, electronegativity
Y

decreases due to a corresponding increase in the size of the atom.

Re
nd

7.31.2. Physical Properties

Fi

Some important physical properties of elements of group 17 are discussed below :

1. Atomicity. Each halogen has one electron less than the nearest inert gas. As a result, halogens are
very reactive elements. They readily share their single unpaired electron with other atoms to form covalent
bonds. Thus, all halogens exist as diatomic molecules.
2. Physical state - Melting and boiling points. As discussed above, all halogens exist as diatomic
molecules in the elemental state. The different molecules of a halogen are held together by weak van der Waals
forces of attraction. The strength of these van der Waals forces increases as the size of the halogen increases
from fluorine to iodine. As a result, F2 and CI2 are gases at room temperature, Br2 is a liquid whereas is a
solid. Further due to increasing van der Waals forces of attraction between the molecules, as we move down
the group, the melting points and boiling points increase with increase in atomic number from fluorine to
iodine.

3. Non-metallic character. Due to their high ionization enthalpies and high electronegativities, halogens
are non-metallic in nature. However, the non-metallic character decreases gradually as we move down the

*Electron gain enthalpy is the negative of electron affinity.

7/72 New Course Chemistry (XII)CZsl9]

group from fluorine to iodine. In fad, iodine is a solid and possesses metallic lustre. It also forms a number
of compounds such as ICl, ICI3, etc. which form positive ions such as I'*' and respectively.
4. Enthalpy of dissociation. Enthalpy of dissociation decreases as the bond distance increases from F2
to I2 due to a corresponding increase in the size of the atom as we move down the group from F to I. The F—
F bond dissociation enthalpy is, however, smaller than that of Cl—Cl and even smaller than that of Br—Br.
This is due to the reason that F atom is very small and hence the electron-electron repulsions between the lone
pairs of electrons are very large. Thus, the bond dissociation enthalpy follows the sequence :
CI2 > Br2 > F2 > l2-
5. Colour. All the halogens are coloured. The colour is due to the fact that their molecules absorb light
in the visible region as a result of which their electrons are excited to higher energy levels while the remaining
light is transmitted. The colour of halogens is actually the colour of this transmitted light, i.e., halogens have
complementary colours
Halogen Fluorine Chlorine Bromine Iodine
Colour Pale yellow Greenish yellow Reddish brown Deep violet
The amount of energy required for the excitation (i.e. excitation energy) decreases progressively from
fluorine to iodine as the size of the atom increases. Conversely, energy of the transmitted light increases

F low
progressively and hence, there is a progressive deepening of colour of the halogens from fluorine to iodine.
Fluorine absorbs violet light (higher excitation energy) and hence appears pale yellow while iodine absorbs
yellow and green light (lower excitation energy) and hence appears deep violet. Similarly, we can account for
greenish yellow colour of chlorine and orange red colour of bromine.
6. Solubility. F2 and CI2 react with water. Br, and I2 are sparingly soluble in water but are soluble in
various organic solvents such as chloroform, carbon tetrachloride, carbon disulphide and hydrocarbons such
for Fre
as hexane, benzene, etc., to give coloured solutions.
However, I2 dissolves in an aqueous solution of KI due to the formation of the complex, KI3
KI + I2 ^ KI3, i.e., K+ [I—I <- I]*
7. Nature of bonds. In order to attain noble gas configuration, a halogen atom either
Your
eBo ks

(/) gains an electron from an electropositive element forming the halide ion X" as in Na'^'Cl" (ionic
compounds) or
(//) shares an electron with another atom to form a covalent bond as in halogen molecule, X-X and hydrogen
ad
our

halide molecule, H-X.

Halogens thus form ionic as well as covalent compounds. The halides of highly electropositive metals
are ionic, those of weakly electropositive metals and non-metals are covalent.
Re

Fluorine, because of its highest electronegativity, forms ionic compounds even with metals like Hg
(group 12), Bi (group 13) and Sn (group 14), etc., which otherwise form covalent halides.
Find Y

Thus, in nutshell, as the electronegativity of the halogens decreases from fluorine to iodine, their tendency
to form ionic bonds decreases while the tendency to form covalent bond increases down the group from
fluorine to iodine.

7.32. CHEMICAL PROPERTIES

Some chemical properties of group 17 elements are discussed below :

7.32.1. Oxidation States

Fluorine imially shows a negative oxidation state of - 1 except in HOF where it shows an oxidation
state of+ 1. The other halogens, in addition to a negative oxidation state of - 1 and positive oxidation state
of+i, also show higher positive oxidation states of +3, -i-5 and +7.
All the halogens have 7 electrons in their respective valence shells and thus need one more electron to
acquire the stable electronic configuration of the nearest noble gas. This is possible either by gaining or
sharing one electron with other elements :
 7/73
p-BLOCK ELEMENTS

(0 If they accept an electron from highly electropositive element, such as Na. K, Ca, etc., they form
uninegative X“ions and thus show an oxidation state of - 1. They can also show an oxidation state of
- 1 by shtu-ing its unpaired electron with less electronegative elements than themselves such as H, B,
Al, C. N. P. S, etc. Thus, all halogens show an oxidation state of - 1 by accepting an electron from
highly electropositive elements (Na, K, Ca, etc.) or by sharing an electron with less electronegative
elements than themselves (H, B, Al, C, N, ,P S, etc.),
(ii) Alternatively, halogens can also share their unpaired electron with more electronegative elements and
thus can also show an oxidation state of + 1.

For example. Cl, Br and I show an oxidation state of +1 by sharing its unpaired electron with more
electronegative elements such as F as in interhalogen compounds (CIF, BrF and IF) or with compounds of O
such as in oxoacids (HOCl, HOBr and HOI). In principle, F should not show positive oxidation stales since
it is the most electronegative element known. In spite of this, F shows an oxidation state of +I in highly

w
unstable HOF molecule.

Another aspect in which fluorine differs from other halogens is that it does not have d-orbitals while all
other halogens have vacant d-orbitals in their respective valence shells. As a result, one, iwo and even three

lo
electrons can be excited from np and ns orbitals to ;ic/-orbitals thereby making 3, 5 or 7 half-filled orbitals

e
available for bond formation as shown below :

re
nd
Ground state
ns

rF np

F
(Oxidation state = i.+i) n n it t
First excited state

r
(Oxidation state =+ 3) ti n t t t
fo
u
Second excited state
t; t t t t t
ks
(Oxidation state = + 5)
Yo
Third excited state
t t t t t t t
oo

(Oxidation state = + 7)
B

Thus, besides - 1 and + 1 oxidation states. Cl, Br and I also show other positive oxidation slates of -t-3,
+5 and +7 in their oxides, oxoacids and interhalogen compounds.
e

In addition to the above positive oxidation states, Cl also shows oxidation states of +4 (in CIO2) and -1- 6
ur

(in Cl20g and CIO3) while Br shows and oxidation state of +4 in B1O2
ad
Yo

From the above discussion, itfollows that fluorine does not show higher positive oxidation states of +3,
+5 and +7 while other halogens do.
d

732.2. Trends In Chemical Reactivity of Elements and their Compounds

Re
in

All the halogens are highly reactive elements due to the following two reasons :
F

(i) Low bond dissociation energy (X2 > 2 X). Halogens have low bond dissociation energy as compared
to common molecules like 0„ N-,, etc. Therefore, they readily dissociate into atoms and react with
other substances readily.
(ii) High negative electron gain enthalpies. Because of high negative electron gain enthalpies, halogens
have a very strong tendency to gain one electron and thus are very reactive.
They react with metals and non-metals to form halides. However, as we move down the group, the
reactivity decreases in the order; F2 > CU > Br^ > h- Some important trends in the chemical reactivity of
these elements and their compounds are discussed below :
1. Oxidising Power.
Since all the halogens have a strong tendency to accept an electron, they act as strong oxidising agents.
However, their oxidising power decreases from F, to I2. Since F2 is the strongest oxidising agent amongst halogens,
it will oxidise all other halide ions to the corresponding halogen in solution or in the dry state.
F^ + 2X- ^2F"-hX2 (X- = C1-, Br- l~)
Similarly, CI2 will oxidise Br“ and I" ions while Br2 will oxidise only 1“ ions from their solutions.
7/74 New Course Chemistry (XI1)BZSI9]

CU + 2X~ 4 2Cr + X2(X-=Br-, n

Bf2 + 21- 2Br" + It
In general, a halogen of lower atomic number oxidises halide ions of higher atomic numbe.r
The oxidizing power of halogens can be compared on the basis of their standard electrode potentials
(Table 7.16, page 7/71) as follows :
p2 + 2e-- ^ 2 F- ; E°= + 2-87 V ; Cl2 + 2e-- 4 2C1-;E° = +1-36V
Br2 + 2 e~ -^2 Br" ; E° = + 1-09 V ; I2 + 2 e- -4 21-; E° = + 0-54 V

The electrode potential of F2/F- couple is the maximum while that of is the minimum, i.e., F2 is
reduced most easily but I, is reduced least easily. In other words, F2 is the strongest oxidising agent while I2
is the weakest oxidising agent. Conversely, 1~ ion is the strongest reducing agent while F ~ ion is the weakest
reducing agent.

w
The relative oxidising power of halogens can further be illustrated by their reactions with water. F2
oxidises H2O to O2 and O3 whereas CI2 and Br2 react with water to form corresponding hydrohalic and
hypohalous acids

Flo
2F2(g) + 2H20 (/) ^4 {aq) + 4 _{aq) + O, (g)
3F2(g) + 3H20 (/) —> 6 H'*’ {aq) + 6 F {aq) + O3 (g)

e
re
Cl2(g) + H20 (/)- 4 HCl {aq) + HOCl {aq)
Hypochlorous acid

F
Br2(g) + H20 (/) ■> HBr {aq) -t- HOBr {aq)
ur
or
Hypobromous acid
The reaction of U with H2O is non-spontaneous. In fact, I~ ions can be oxidised by O2 in acidic medium
which is just the reverse of reaction observed with F2. f
ks
41-(«g) + 02(g) + 4H^(fl^) ^ 2 I2 (s) + 2 H2O (/)
Yo
oo

IMT wuir
ITION9
(a) Comparison of oxidising powers of Fj and CI2. Please
B

remember that electron gain enthalpy of fluorine (- 333 kJ mol"^) is less negative than that of chlorine
re

(- 349 kJ nior^) but still it is the strongest oxidising agent. This is because of its low bond dissociation
energy {158-8 kJ mor') and high heat of hydration (515 kJ mol-’) as compared of those of chlorine for
u

which values are 242-6 and 381 kJ mol"’ respectively.

ad
Yo

{b) Comparison of oxidising powers of Br2 and I2. Br2 is a stronger oxidising agent than I2 can be easily
demonstrated by its action on sodium thiosulphate whereas Br2 oxidises it to NaHS04 where oxidation state
of S is -f 6 while I2 oxidises it to Na2S40g where oxidation number of S is 2-5.
d
Re

-1-6
in

2-0

Na.,S202 + 4 Br2 + 5 H^O ^ 2NaHS04 + 8 HBr

F

Sod. ihiosulphaic Sod. sulphate

2-0 2-5

+ I, ^ 2Na2S40g + 2 Nal
Colourless
Sod. Ihiosulpiiale Violet Sod. letraihionace

The reaction of 12 with Na2S20^ is used as a test for I2 since during this reaction violet 12 is reduced to
colourless Nal.

Another useful test for iodine is that it gives blue colour with starch solution.
(c) Both CI2 and Br2 are stronger oxidising agents than I2 can be easily demonstrated by their action of
sodium thiosulphate. Whereas CI2 and Br2 oxidise Na2S203 to NaHS04 where the oxidation stale (O.S.) of
S increases from + 2 in Na2S203 to + 6 in NaHS04 while I2 oxidises it to Na2S40Q where the O.S. of S
increases from + 2 in Na2S203 to only + 2-5 in Na2S40g
-*-2 +6

Na2S203 + 4CI2 + 5H2O ^ 2NaHS04

Sod. thiosulphate
 7/75
p-BLOCK ELEMENTS

+2 + 6

+ 4 Bf2 + 5 H2O ^ 2NaHS04 + 8HBr

+ 2 2-5

2Na2S203 + I2 ^ Na2S406 + 2Nal

Violet Sod. tctrathionalc Colourless

(d) Reducing action of 1“ ion. As stated above, I' ion is a weak reducing agent. This can be demonstrated by
its action with CUSO4 where 1“ ion reduces Cu^ to Cu"*" ions which combine with 1“ ions to form colourless
CU2I2 and r ions are oxidised to violet coloured I2.
+ 2 + I

2C\iSO^(.aq) +4Nnl{aq) > Cu.,!.,'!' + I2 (-0 + 2 N32S04 (aq)

Blue Colourless Violet

2. Reaclivil owards vdrogt'P.

w
All the halogens combine with hydrogen to form hydrogen halides (HX).

F lo
^2HX
H2 + X2
The reactivity of halogens towards hydrogen decreases from fluorine to iodine. For example, fluorine
combines with hydrogen violently even in the dark, chlorine reacts in diffused sunlight, bromine reacts

ee
with hydrogen only on heating while iodine reacts with hydrogen on heating in presence of platinum as

Fr
catalyst.
(a) Preparation of HF and HCl. HF and HCl are prepared in the laboratory as well as in industry by
heating fluorides and chlorides respectively with concentrated sulphuric acid for
ur
Heat
CaF2 (5) + H2SO4 (Cone.) > CaS04(.r) + 2HF(g)
s
ook
Yo
Heat

2NaCl (j') + H2SO4(Cone.) > Na2S04 (aq) + 2HCI (g)

eB

(b) Preparation of HBr and HI. Unlike HF and HCl, HBr and HI cannot be prepared by the action of
cone. H2SO4 on a bromide or an iodide. The reason being that both HBr and HI evolved are moderately
strong reducing agents and hence they reduce sulphuric acid to sulphur dioxide and are themselves oxidised
our
ad

to Br2 and l2 respectively.

2NaBr + H2SO4 Na2S04 + 2HBr
Y

H2SO4 -^H20 + S02+0

Re

2HBr + 0- ●> H2O + Br2

nd
Fi

2NaBr + 2H2SO4 -> Na,S04 + SO, + Bt2 + 2H3O

Similarly, 2NaI + 2H2SO4 Na2S04 + SO2 + I2 + 2H2O
Therefore, these acids are usually prepared by heating the bromide or the iodide with a non-oxidising
acid such as phosphoric acid.
Heat Heat

3 NaBr + H3PO4 » Na3P04 3HBr

; 3 Nal + H3PO4 > Na3P04 + 3Hl
Commercially, HBr is prepared by the direct reaction of H2 and Br2 at around 570 K in presence of Pt
as catalyst
Pt/a,sbe.stos
H2 (g) + Br2 (g) > 2HBr(g)
570 K

On the other hand, HI is produced by the reaction of 1, either with H2S or with hydrazine (N2H4).
I2 0) + H2S (g) 2 HI (g) + S is)
573 K
2 I2 (s) + N2H4 (aq) ^ 4 HI (aq) + Nj (g)
7/76 New Course Chemistry (XIl)C2SX9[i

(c) Properties of Hydrogen Halides

Some impoitant characteristics of hydrogen halides are discussed below :
(/■) Physical state. Hydrogen fluoride is a low boiling liquid {b.p. 293 K) whereas HBr, HCl and HI are
gases. The anomalous behaviour of HF is due to the presence of intermolecular hydrogen bonding in
its molecules. Consequently, it exists as an associated molecule. (HF)^. In the solid stale, (HF)^ has the
zig-zag structure as shown :

'H

F'

(//) Melting points and boiling points. Amongst hydrogen halides, the boiling point of HF is highest, i.e.,
293 K due to extensive intennolecuiar hydrogen bonding. The other hydrogen halides do not show

w
hydrogen bonding because of the lower electronegativity of the halogens in them. Therefore, the boiling
points of HCl, HBr and HI are much lower than that of HE However, as we move from HCl to HI, their
boiling points show a regular increase due to a corresponding increase in the magnitude of van der

F lo
Waals’ forces of attraction as the size of the halogen increases.
Like boiling point, tlie melting point of HF is higher than that of HCl due to intermolecular hydrogen

ee
bonding. The melting points of other halogen acids show a gradual increase from HCl to HI as the size

Fr
of the halogen atom increases.
Some important physical properties of hydrogen halides are given in Tabic 7.17.

for
ur
TABLE 7.17. Some Physical Properties of Hydrogen Halides
PROPERTY HF HCl HBr HI
ks
Yo
Melting point (K) 190 159 185 222
oo

Boiling point (K) 293 189 206 238

eB

Bond length, (H—X)/pm 91.7 127.4 141.4 160.9

'^diss H“/kJ mol-’ 574 432 363 295

r

Dipole moment }i/D 1-86 Ml 0-79 0-38

ou
ad

Acid dissociation constant/pK II

3-2 -7 0 -9-5 -100
Y

(Hi) Dipole moments. As the electronegativity of the halogen decreases, the dipole moment of the hydrogen
nd

halides decreases in the same order, i.e. HF (1-86 D) > HCl (Ml D) > HBr (0-79 D) > HI (0-38 D)
Re

(/V) Percentageof ionic character increases in the order: HI < HBr < HCl < HF. As the electronegativity
Fi

of the halogen decreases, the dipole moment of the hydrogen halides and hence the ionic character
decreases in the same order,
(v) Bond length iiicrease.s in the order t HF < HCl < HBr < HI. As the size of the halogen atom increases,
the bond length increases in the same order, i.e. H—I (160-9 pm) > HBr (141-4 pm) >
HCl (127-4 pm) > HF (91-7 pm)
(vi) Bond strength increases in the order : HI < HBr < HCl < HF
Bond strengths or bond dissociation energie.s are inversely proportional to the bond lengths, i.e. shorter
the bond length, higher is the bond strength. Since the bond lengths increase in the order:
HF < HCl < HBr < HI,
therefore, bond dissociation energies increase in the reverse order, i.e, HI < HBr < HCl < HF.
(vii) Thermal stability increases in the order : HI < HBr < HCl < HF
Thermal stability is directly proportional to the bond dissociation energy. Since bond dissociation
energy of HF is the highest and that of HI is the least, therefore, HF is the most stable halogen acid
while HI is the least stable halogen acid. In other words, the order of stability of the halogen acids
increases in the order : HI < HBr < HCl < HF.
 7/77
p-BLOCK ELEMENTS

{viii) Acid strength increases in the order : HF < HCl < HBr < HI
The strength of tm acid depends upon its degree of ionization which, in turn, depends upon the bond strength.
HX {aq) - («.?)+ X'iaq)
Thus, higher the bond dissociation energy, lower is the degree of ionization and hence weaker is the
acid. Since the bond dissociation energies of the halogen acids increase in the order: HI < HBr < HCl
< HF, therefore, strength of the acids increases in the reverse direction, i.e., HF < HCl < HBr < HI.
Since lower the value stronger is the acid, therefore, their values decrease in the order : HF
{3-2) > HCl (-7-0) > HBr (-9-5) > HI (-10-0). Another reason for low acidity of HF as compared to
other halogen acids is the strong H-bonding of F" ion to H30''’ as compared to other halide ions.
Reducing power increases in the order : HF < HCl < HBr < HI
The reducing power of a halogen acid depends upon the ease witli which it decomposes to give H2 and X2.
2HX ^ H., +
This, in turn, depends upon the bond dissociation energy. Thus, greater the bond dissociation energy,

w
more stable is the halogen acid and hence weaker is the reducing agent. Since the bond dissociation
energies of the halogen acids increase in the order: HI < HBr < HCl < HF.

F lo
therefore, reducing power of these acids increases in the reverse order ; HF < HCl < HBr < HI.
U) Corrosive nature. All the halogen acids, in particular, HF are corrosive and hence extreme care should
be taken while working with them. Out of all the hydrohalic acids, HF attacks glass and hence it is used

ee
for etching glass and manufacture of glass shell for television tubes.

Fr
3. Reactivity towards Oxygen.
Halogens form many binary compounds with oxygen but most of them are unstable. Some important
oxides of halogens along with their oxidation states are given in Table 7.18. for
ur
TABLE 7.18. Compounds of Halogens with Oxygen
oks
Yo
Oxidation state Fluorine Chlorine Bromine iodine
o
eB

-1 OFo.
2’ 0.,F.,

+1 CI2O Br20
our

+4 CIO2 BrO, I2O4

ad

+5 ^2^5
+6 CI2O6, CIO3
Y

+7 CI2O7
Re
nd

I. Oxides of fluorine, (a) Preparation. Fluorine forms two binary compounds with oxygen, i.e., OF2
Fi

and O2F2. However, only OF2 is thermally stable at 298 K while O2F2 is highly unstable and decomposes into
its elements even at 178 K. They are called as oxygen fluorides rather than oxides of fluorine since the
electronegativity of F is higher than that of O. Oxygen difluoride (OF2) is prepared by passing F2 through 2%
NaOH solution while O2F2 (dioxygen difluoride) is obtained by passing electric discharge through a mixture
of F2 and O2 under low pressure and at liquid air temperature.
2 F2 + 2 NaOH ^ 2 NaF + H^O + OF2
(2%) Oxygen difluoride

Electric discharge
F2 + O') ■>
liq.air temp., low pressure
Dioxygen difluoride
OF, dissolves in water and gives a neutral solution, therefore, it is not an acid anhydride. It dissolves in
NaOH to give sodium fluoride and dioxygen
2 NaOH + OF2 4 2 NaF + H,0 + O.,
7/78 7^'uieieefi. '<2- New Course Chemistry (MI) BE

w
(b) Structure. The structure of OF2 is similar to that of H2O while that of O2F2 is similar to that of H20_^
involving jp^-hybridization of 0-atoms. However, due to greater electronegativity of fluorine than oxygen,
the bond pairs in OF2 lie nearer to the fluorine atom.

r
103® ^ 104-5®
F F H

In contrast, in H2O, due to greater electronegativity of O than H, the bond pairs lie nearer the O atom.
Consequently, bond pair-bond pair repulsions in H2O are stronger than bond pair-bond pair repulsions in OF2

u
and hence F—O—F bond angle is little shorter (103®) than H—O—bond angle (104-5®).

o
F
Further since, F is much more electronegative than O, therefore, it attracts the lone pairs of electrons on the O
atom towards itself. Therefore, lone pair-lone repulsion between the two O atoms in O-^ bond is much lower in
O2F2 than in H2O2. In other words, the O—O bond length in O2F2 is much shorter (122 pm) than that in H2O2 (148

s
pm). However, the O—^F bond lengths in O2F2 are much longer (158 pm) than in OF2 (141 pm).

Fo
\
F.

k
l
r
,_122pm _
\H %

o
F

Y
(c) Uses. Both OF2 and O2F2 are strong oxidising and fluorinating agents. O2F2 oxidises plutonium (Pu)

f
o
Y
to PuFg and this reaction is used in removing Pu as PuFg from spent nuclear fuel.
O2F2 also oxidises H2S to SFg
B
H2S + 4O2F2 4SFg-F2HF-F402
Being a strong oxidising agent OF2 has been used as a rocket fuel,
r

n. Oxides of other halogens. Chlorine, bromine and iodine form oxides in which the oxidation states
d
u

of halogens range from +1 to +7. Chlorine forms the largest number of oxides while iodine forms the least. In
e

these binary compounds, the bonds are mainly covalent due to small difference in electronegativity between
the halogens and oxygen ; the bond polarity, however, increases as we move from Cl to I. The stability of
on

oxides of iodine is greater than those of chlorine while bromine oxides are the least stable. Iodine-oxygen
ad

bond is stable due to greater polarity of the bond while the stability of the chlorine-oxygen bond is due to
i

multiple bond formation involving d-orbitals of the chlorine atom. Bromine being in between lacks both
these characteristics. Thus, the stability of oxides of halogens decreases in the order: I> Cl > B.r Furthe,r
F

the higher oxides of halogens tend to be more stable than the lower ones.
Re

(a) Preparation of chlorine oxides. Since the oxides of chlorine are strongly endothermic compounds
and have large positive free energies of formation, they are not prepared by direct reactions between CI2 and
O2. Therefore, they are usually prepared by indirect methods. For example,
(0 Dichlorine monoxide (CI2O) is best prepared by heating freshly prepared yellow mercuric oxide with
CI2 diluted with dry air at 573 K. 573 K
2Cl2+2HgO ^ HgCl2.Hg0-FCl20(g)
(«) Chlorine dioxide {ClOf) is prepared in the laboratory by the reduction of sodium chlorate with oxalic
acid at 363 K.
H20.363K
2 NaClOj + 2 (C00H)2 ■> 2 CIO2 + 2 CO2 + (COONa)2 + 2 H2O
It can also be prepared by reduction of NaC103 with SO2
2 NaC103 {aq) -F SO2 ig) > 2 CIO2 {aq) + Na2S04 iqq)
m Dichlorine hexaoxide (Cl20g) is prepared by ozonolysis of CIO2.
2 CIO2 -F 2 O3 > Cl20g -F 2 O2
 p-BLOCK ELEMENTS 7/79

(iv) Dichlorine heptoxide is the anhydride of perchloric acid and is conveniently prepared by careful
dehydration of HCIO4 with or with H3PO4 at 263 K followed by distillation at 238 K under a
pressure of 1 mm Hg.
^4®10 orHjPO
2 HCIO4 263 K
^ CI2O7 + H2O
(b) Structure of chlorine oxides. The O atom in CI2O is sp^-hybridized and its structure is similar to
that of OF2. However, due to steric crowding of the two Cl atoms, the Cl—O—Cl bond angle is 111° as
compared to 103° in OF2.
In CIO2, the central Cl atom is 5p^-hybridized with O—Cl—O angle of 118°. Both the Cl—O bonds
have equal (149 pm) bond lengths and are quite shorter than those in CI2O (171 pm). Therefore Cl—O bond,
has appreciable double bond character due to pn- dn bonding. The molecule is paramagnetic since it has

w
one odd electron in a p-orbital.

Flo
111°
Cl Cl
o

e
structure of OCI2 structure of CIO2

re
Odd electron molecules often dimerize in order to pair the electrons but CIO2 does not. This is probably

F
due to the reason that odd electron is delocalized. In contrast, the odd electron on N in NO is localized and
hence NO readily dimerises to form (NO)2-
ur
r
CljOg is a solid. It exists in equilibrium with the monomer, CIO3, which is a liquid. The structure of
fo
neither the liquid nor the solid is known. Both are diamagnetic and have no unpaired electrons. Possible
structures of CI2O6 are :
ks
Yo
O O O O
oo

0-^Cl—Cl^O or XI Cl
o O o o o
B

Spectroscopic data show that the molecule of CI2O7 consists of two

re

CIO3 groups linked by an oxygen atom as shown below :

(c) Uses ofchlorine oxides. Chlorine oxides (CI2O, CIO2, C^Og and CI2O7) are highly reactive oxidising
u
ad

agents and tend to explode. CIO2 is used as a bleaching agent for paper pulp, textiles and in water treatment.
Yo
d
Re
in
F

in. Bromine oxides. The bromine oxides, Br20, Br02, B1O3 are the least stable halogen oxides (middle
row anomally) and exist only at low temperatures. They are very powerful oxidising agents. Their structures
are similar to those of chlorine oxides having comparable molecular formulae.
IV. Iodine oxides. Iodine forms three oxides, I2O4,12O5 and I4O9. Out of these, I2O5 is the most stable.
It is the only trrue oxide of iodine while others are regarded as iodates of tripositive iodine, i.e., I2O4 is

probably IO'*’IOj and I4O9 is probably I^'*’(IOj)2.

(a) Preparation. I2O5 is obtained by dehydrating iodic acid at 473 K
473 K
2HIO3 I2O5 H2O
Therefore, it is the anhydride of iodic acid. It dissolves in water to produce iodic acid.
It is a strong oxidising agent and oxidises CO to CO2 rapidly and quantitatively at room temperature,
liberating I2 which can be titrated against a standard solution of Na2S203 using starch as an indicator.
7/80 New Course Chemistry (X1I)B3SSBI

Room temp.
I2O5 + 5 CO ^ I2 + 5 CO2
Therefore, I2O5 is used for detection and estimation of CO in the atmosphere and other gaseous
mixtures.

(h) Structure. I2O5 consists of two IO3 pyramidal units joined

through a common oxygen atom. Each iodine has two I —> O coordinate
bonds, one I—O covalent bond and a lone pair of electrons.
4. Reactivity towards Metals.
Because of high oxidising power, halogens combine directly w'ith most metals to form their corresponding
halides. For example, Br2 reacts with magnesium to form magnesium bromide.
Mg (^) -I- Br2 (/) MgBi2 (.v)
As expected, the ionic character of the M—X bond, decrea.scs in the order :
M—F > M—Cl > M—Br > M—1

w
This is due to a corresponding decrease in the electronegativity of the halogen from F to I. However, if
the metal exhibits more than one oxidation states, the halide in the higher oxidation state will he more

F lo
covalent than the one in the lower oxidation stale. For example, SnC^, PbCl4, SbCl^, UFg are more covalent
than SnCl2, PbCl2, SbCl3 and UF^ respectively.
5. Reactivity of Halogens towards other Halogens.

ee
Halogens react with each other to form a number of compounds called interhalogens of the type

Fr
XK, XX', XX' and XX^ where X is larger size halogen (or less electronegative halogen) and X'is smaller
for
size halogen (or more electronegative halogen). These are discussed in Art. 7.23, pages 7/51-7/53.
ur
7.33. ANOMALOUS BEHAVIOUR OF FLUORINE
s
ook

Among p-block elements, the first element of each group differs significantly from rest of its members.
Yo

Fluorine, the first member of the halogen family also shows a remarkable different behaviour from rest of the
eB

halogens. For example, ionisation enthalpy, electronegativity, and electrode potential of fluorine are higher
while electron gain enthalpy and bond dissociation enthalpy are lower than the trends set by other halogens.
The main reasons for this anomalous behaviour of fluorine are :
our
ad

(/) Its small atomic size, (ii) Highest electronegativity. {Hi) Low bond dissociation enthalpy of F2 molecule
and (iv) Absence of d-orbitals in the valence shell.
Y

Some points of difference between fluorine and rest of the halogens are as follows :
Re

1. Reactivity. Fluorine is most reactive of all the halogens. This is due to the reason that fluorine has a
nd

much higher electrode potential than those of other halogens.

Fi

2. Oxidation state. Since it is the most electronegative element, it mostly shows a negative oxidation
state of - I, except +1 in HOF. Further, it does not show any higher oxidation slates (positive or negative)
because of the absence of cZ-orbitals in its valence shell. Other members, however, show both negative oxidation
state of- 1 and different positive oxidation states of + 1, +3, +5 and +7.
Since fluorine usually does not show positive oxidation slates, therefore, F2 does not undergo
disproportionation reactions in the alkaline medium while other halogens do.
Cold

X2 + 2 NaOH » NaX + NaOX + H2O (where X = Cl, Br or I)

Hot

3 X2 + 6 NaOH ^ 5 NaX + NaX03 -i- 3 H2O (where X = Cl, Br or I)

3. Hydrogen bonding. Hydrogen bonding (due to small atomic size and high electronegativity of fluorine)
is a distinct phenomenon observed in some fluorine compounds.
(/) HF is a liquid with a boiling point of 293 K while other halogen halides are gases under ordinary
conditions with low boiling points.
 p-BLOCK ELEMENTS 7/81

(//) HF is a weak acid as compared lo other halogen acids which are strong and highly ionised. This is due to
higher dissociation energy of H—F bond and molecular association due to hydrogen bonding in HF
(m) Due to hydrogen bonding, HF can form acid salts of the type KHF2, i.e.. [K+ (F* H—F)] while
other halogen acids do not form such compounds.
4. Nature of compounds. Because of its highest electronegativity , fluorine has highest tendency for
forming ionic compounds and thus fluorides have the maximum ionic chitracter. For example, AIF3 is ionic
whereas other halides of aluminium are covalent.
S. Oxidising power. Fluorine has the highest electrode potential. Thus, it is reduced most easily and
hence acts as the strongest oxidising agent among the halogens. It brings about highest oxidation state of
other elements with which it combines. For example, with sulphur it gives SF^j and with osmium it gives
OsFg. Other halogens do not always bring about the highest oxidation state of the element.
It can oxidise most of the other elements, including some of the noble gases (Kr, Xe).

ow
6. Reactivity of halogen acids. Although weak, HF is the most reactive of all halogen acids. It can
attack glass because of its ability to displace oxygen from silicates to form fluorosilicates.
Na2Si03 + 6HF
^ Na2SiF^+ 3H2O
(Simplified formula of glass)

e
That is why HF is used for etching of glass, in marking thermometers, burettes, etc.

Fl
re
Other halogen acids, i.e., HCl, HBr and HI do not give this reaction.

F
Due to the above special properties offluorine, it is usually called as super halogen.
7.34. CHLORINE, CLj
ur
r
Chlorine was discovered in I774 by Scheele by heating HCl with Mn02- In 1810, Davy established iLs
fo
elementary nature and suggested the name chlorine on the basis of its colour (Greek, chloros - greenish
ks
yellow).
Yo
oo

7.34.1. Preparation of Chlorine

It can be prepared by healing concentrated hydrochloric acid with an oxidising agent such as manganese
eB

dioxide, potassium permanganate, potassium dichromate, lead dioxide, red lead, bleaching powder , ozone.
etc.

(/) With manganese dioxide : Mn02 + 2 HCl

ur

MnCl2 + H7O + O
2 HCl + O -> CU + H7O
ad
Yo

Mn02 + 4 HCl ^ MnCl2 + 2 H2O + CI2

However, in the laboratory, CU is prepared by heating a mixture of sodium chloride, manganese dioxide
d

and cone. H2SO4. The HCl produced in situ is oxidised by MnO^ as shown above
Re
in

NaCl + H2SO4 NaHS04 + HCl) x 2

F

Mn02 + H7SO4 -> MnS04 + H7O + O

2HC1 + 0- ^ H7O + CI2
2 NaCl + 3 H2SO4 + MnOo » 2 NaHS04 + MnS04 + 2 H^O + CU
(ii) With potassium pennanganate : 2 KMn04 + 16 HCl ^ 2 KCl + 2 MnCU + 8 H7O + 5 CU
{Hi) With potassium dichromate : K7Cr20y + 14 HCl - ^ 2 KCl + 2 CrCl3+ 7 H^O + 3 CU "
(zV) With bleaching powder : CaOCl2 ^ HCl -> CaCU + H7O + CI2
(v) With lead dioxide : PbOo + 4 HCl PbCU + 2 H2O + CI2
(vi) With red lead : Pb304 + 8 HCl -^3PbCl2 + 4H20 + Cl2
(vii) With ozone : 2 HCl + O3 ■>H20 + 02+Cl2
In all these reactions, HCl acts a reducing agent.
7.34.2. Manufacture of Chlorine
(0 Deacon’s process. In this process, hydrogen chloride gas is oxidised by atmospheric oxygen in the
presence of CUCI2 as catalyst at 723 K
7/82 New Course Chemistry (XII)

CuC4,723K
4 HCl + O2 > 2 CI2 + 2 H2O
(/O Electrolytic process. Chlorine is obtained by the electrolysis of a concentrated solution of NaCl
(brine), when CI2 is liberated at the anode and the sodium metal liberated at the cathode reacts with H2O to
form NaOH and H2.
At anode : ci- Cl + e 2 Cl ^CU
At cathode : Na"^ + e Na 2 Na + 2 H2O ^ 2 NaOH + H2
It is also obtained as a by product during manufacture of sodium by electrolysis of fused NaCl in
Down’s process.

w
It is a greenish yellow gas with pungent and suffocating odour. It is about 2-5 times heavier than air. It
can be liquefied easily into greenish yellow liquid which boils at 239 K. It is soluble in water.

lo
1. Reaction with metals, non-metals and metalloids. Chlorine reacts with a number of metals, non-

e
metals and metalloids to form their corresponding chlorides. For example,

re
2 Na + CI2 ^NaCl 2 AI + 3CI2 >2AlCl3

rF
F
2 Fe + 3 CI2> 2 FeClg P4 + 6CI2 >4PCl3
Sg + 4 CI2> 4 S2CI2 2 AS-H3C12 >2 ASCI3

r
2. Oxidising properties. When CI2 is dissolved in water it forms chlorine water. On standing, chlorine
fo
ou
water loses its yellow colour due to the formation of HCl and HCIO. Hypochiorous acid (HOCl) thus
oxygen which is responsible for oxidising properties
ks
formed, being unstable, decomposition to give nascent
of chlorine
oo

CI2 + H2O ^ [HCl + HOCl] 2 HCl -H [O]

Nascent oxygen
Y
eB

Thus, in presence of moisture or in aqueous solution, CI2 acts as a powerful oxidising agent. Some of
these oxidising properties are discussed below :
ur

CI2 + H2O 2 HCl + O

(0 It oxidises acidified ferrous to ferric salts :
^ Fe2(S04)3 -I- H2O
ad

2 FeS04 + H2SO4 O
Yo

2 FeS04 + H2SO4 + CI2 -> Fe2(S04)3 -I- 2 HCl

d

(//) It oxidises sulphites to sulphates : Na2S03 + CI2 + H2O ^ Na2S04 -H 2 HCl

Re
in

(Hi) It oxidises thiosulphates to bisulphates ; No2S203 + 4 CI2+ 5 H2O ^ 2 NaHS04 -t- 8 HCl
Sod. thiosulphate
F

(iv) It oxidises sulphur dioxide to sulphuric acid in aqueous solution

SO2 + 2 H2O + CI2 > H2SO4 -I- 2 HCl
(v) It oxidises moist iodine to iodic acid : I2+6H2O + 5CI2 >2HI03+ lOHCl
Iodine Iodic acid

water
^ 2 HCl S
(v/) It oxidises H2S to S in presence of water: H2S + CU
(vii) It oxidises nitrites to nitrates : NaN02+Cl2 + H20 ^ NaN03 -I- 2 HCl
(viii) It oxidises sodium arsenite to sodium arsenate
Na3As03 + H2O + CI2 ^ Na3As04 + 2 HCl
Sod. arsenite Sod. arsenate

3. Bleaching action. Dry CI2 does not act as a bleaching agent. But in presence of moisture or in
aqueous solution, CI2 acts as a powerful bleaching agent and bleaches vegetable or organic matter
CI2 + H2O ■> 2 HCl + [O] (Nascent oxygen)
Coloured substance -1- O Colourless substance
 p-BLOCK ELEMENTS 7/83

The bleaching action of CI2 is due to oxidation of coloured substances to colourless substances by
nascent oxygen. Since the bleaching action of CI2 is due to oxidation and that of SO^ is due to reduction,
therefore, bleaching effect of O2 is permanent while that ofS02 is temporary.
4. Affinity for hydrogen. Chlorine has great affinity for hydrogen. Therefore, it reacts with compounds
containing hydrogen to form HCl.
Difrused sunlight
H2+CI2 > 2 HCl
CioH,6 + 8CI2 > 16 HCl +I0C
Turpentine oil
5. Reaction with ammonia. With excess ammonia, chlorine gives nitrogen and ammonium chloride
whereas with excess chlorine, nitrogen trichloride (explosive) is formed.
8 NH3 + 3 CI2 > 6 NH4CI + N2 ; NH3 + 3 CI2 ^ NCI3 + 3 HCl
(exce.i.s) (excess)

w
6. Reaction with cold and hot alkalies. With cold dilute alkalies, a mixture of chloride and hypochlorite
is formed
0 -1 +1
Cold
Cl2 + 2NaOH(c///.) ^ NaCl + NaOCl +H2O

Flo
Sod. hypochlorite
Cold

e
2Cl2+2Ca(OH)2 (dil.) > CaCl2 + Ca(OCl)2 + 2 H2O

re
Cal. hypochlorite
However, with hot and concentrated alkalies, a mixture

F
of chloride and chlorate is formed.
0 Hot -1 +5
ur
3 CI2 + 6 NaOH (cone.)

r
^ 5 NaCl + NaC103 + 3 H2O
0 -1 +5
fo
Sod. chlorate
ks
Hot
6CI2 + 6 Ca(OH)2 (cone.) 5CaCl2 +Ca(C103)2 +6H2O
Yo
oo

Cal. chlorate

During these reactions, chlorine is simultaneously reduced to chloride ion (Cl") and is oxidised to either
B

hypochlorite ion (OCl ) or to chlorate ion (CIO3). Such reactions which involve simultaneous oxidation-
re

reduction are called disproportionation reactions.

u
ad
Yo

1. Disproportionation reactions. Like CI2, Br2 and I2 also undergo disproportionation reactions with
cold and hot alkalies.
d

Cold
Re

Br2 -I- 2 NaOH (dilute) > NaBr -1- NaOBr + H2O

in

Hot
3 Br2 -H 6 NaOH (cone.)
F

> 5 NaBr + NaBr03 -1- 3 H2O

Strictly speaking, Fj also undergoes disproportionation reaction but only at very low temperatures. For
example.
233K
F2 + H2O (ice) ^ ^ HOF -H HF

Since HOF is extremely unstable, for all practical purposes, it is said that F2 does not undergo
disproportionation reactions.
Further since F2 does not show higher positive oxidation states of + 3, + 5 and +7, therefore, F2 does
not react with NaOH to produce NaFOj.
3 Fj -H 6 NaOH X > 5 NaF -t- NaF03 -H 3 HjO
Instead, it reacts with a very dilute solution of NaOH (2%) to form OF2 as stated on page 7/42.
2. Disproportionation of halates. Halates on heating undergo disproportionation to form perhalates and
halides.

Heal Heat
For example, 4 KCIO3 ■ ■>
3KCIO4 + KCl ; 4KIO3 3 KIO4 + KI
Pot. chlorate Pot. perchlorate Pot. iodate Pot. periodate
7/84 New Course Chemistry (Xll)E!ZsX9]

7. Reaction with dry slaked lime. Chlorine reacts with dry slaked lime to form bleaching powder
2 Ca(OH)2 + 2 CI2 Ca{OCl)3 + CaCl2 + 2 H2O
Slaked lime Cal. hypochlorite
The actual composition of bleaching powder is Ca(OCl)2.CaCl2-Ca( OH)- 2 H2O.
8. Substitution reactions. Chlorine reacts with saturated hydrocarbons to form substitution products.
hv
CH 4 + CI2 CH3C! + HCl
Methane Methyl chloride
(Suluraled hydrocarbon) {Substirution product)
9. Addition reactions. Chlorine reacts with unsaturated hydrocarbons to form addition products.
Room temp.

ow
CH2 = CH2 + CI2 *■ CICH2—CH2CI
Ethcne I, 2-Dichloroethane
(Umaturated hydrocarbon) {Addition product)

C1-, also reacts with many oxides of non-metals to form addition products.
SO2+CI2 >S02Cl2 (Sulphuryl chloride)

e
2 NO + Cl2 > 2 NOCI {Nitrosyl chloride)

re
rFl
CO + CI2 >COCl2 (Carbonyl chloride or phosgene)

F
7.34.5. Uses of Chlorine

r
It is used
ou
fo
(/) for bleaching wood pulp (required for manufacture of paper and rayon), cotton and textiles.
ks
(ii) in the metallurgy (extraction) of gold and platinum.
(Hi) in the manufacture of dyes, drugs and organic compounds such as CHCI3, CCI4, DDT, refrigerants
oo

(CCloFo, freon), and bleaching powder,

Y

(/v) in the preparation of poisonous gases such as phosgene (COCI2), tear gas (CCI3NO2), mustard gas
eB

(CICH2CH2SCH2CH2CI), etc. Mustard gas was used by Germany in World War I.

CH2 Cl Cl CH2CI CH2CI
ur

2 II + S
CH2 CH2—s —CH2
ad

s—s
Yo

Ethylene Sulphur Mustard gas

monochloride
d

(v) in sterilising drinking water.

Re
in

7.35. HYDROGEN CHLORIDE, HCL

F

Glauber prepared hydrogen chloride in 1648 by heating common salt with concentrated sulphuric acid.
Davy in 1810 established its true nature and showed that it is a compound of hydrogen and chlorine.
7.35.1. Preparation of Hydrogen Chloride
(f) Hydrogen chloride is manufactured by the ‘salt cake’ method. In this method, cone. H2SO4 is added
to rock salt (NaCl). The reaction is endothermic and is performed in two stages at different temperatures.
In the ifrst stage, solid NaCl was reacted with H2SO4 at 420 K when it got coated with insoluble
NaHS04. This prevented further reaction and hence the name ‘salt cake’ was given to this method.
In the second stage, the mixture was heated to about 823 K when further reaction with H2SO4 occurred
and Na2S04 was formed (Kraft process). This, by product, Na2S04 was sold, mostly for paper making.
420 K 823 K

NaCl -I- H2SO4 4 NaHS04 + HCl ; NaCl 4- NaHSO 4 4 Na2S04 + HCl

(ii) Large amounts of impure HCl are obtained in recent years as a by product from the heavy organic
chemical industry. For example, HCl is produced in the conversion of 1, 2-dichloroethane to vinyl chloride
and in the manufacture of chlorinated ethanes and chlorinated fluorocarbons (used as refrigerants).
 p-BLOCK ELEMENTS 7/85

Heat
CICH2 — CH2CI ■> CH2 = CHC1 + HCl
1.2-Dichloroethane Vinyl chloride
(Hi) Highly pure HCl is made by direct combination of H, and CI2.
In the laboratory, HCl can be prepared by heating a mixture of sodium chloride and cone. H.,S04.
420 K
NaCl + H,SO 4 > NaHS04 + HCl
But it is conveniently prepared by heating NH4CI with cone. H2SO4. NH4CI is prefeired even though it
is costlier than NaCl because NH4HSO4 is soluble and the reaction does not stop at the half way stage.
Heat
2 NH4CI + H2SO4 > 2 HCl + (NH4)2S04
Drying of HCl gas. It can be dried by passing through cone. H,S04.
7.35,2. Physical Properties of Hydrogen Chloride
It is a colourless pungent smelling gas. It is easily liquefied to form a colourless liquid (b.p. 189 K) and

w
freezes to a colourless solid (f.p. 150 K). It is extremely soluble in water and its aqueous solution is called
hydrochloric acid.

F lo
7.35.3. Chemical Properties of Hydrogen Chloride
1. Acidic nature. It ionizes in aqueous solution.

ee
HCl (g) + H2O (/) > H3O+ (aq) + Cl- (aq) ; = lO"^

Fr
The high value of dissociation constant (K^) indicates that it is a strong acid in water. It reacts with NH,
and gives white fumes of NH4CI
NH3 + HCl -> NH4CI for
ur
2. Reaction with metals. Active metals like Zn, Mg, Fe, Al, etc. react with HCl to produce the
corresponding metal chlorides with the liberation of H2 gas.
s
ook

Zn + 2 HCl 4 ZnCl2 + H2
Yo

Mg + 2 HCl
^ MgCl2 + H2
Fe + 2 HCl > FeCl2 + H2 2 Al + 6 HCl
> 2 AICI3 + 3 H2
eB

3. Reaction with oxidising agents. Concentrated hydrochloric acid reacts with strong oxidising agents
like KMn04, MnOj, K2Cr207, Pb02, etc. to yield chlorine gas. For details refer to sec. 7.20.1, page 7/46 for
our

preparation of CI2 gas.

ad

4. Reaction with nitric acid. V/hen three parts of cone. HCl and one part of cone. HNO3 are mixed,
aqua regia is formed (Refer to page 7/25). It dissolves noble metals like gold and platinum due to the formation
Y

of their complex chlorides. For example,

Re

Au +4H-"+ NO" +4Cr ^ [AuCy- +NO + 2H2O

nd

Gold Tetrachloroaurate(iri)
Fi

3Pt + 16 m + 4NOJ + 18 Cl-

Platinum
> 3[PtCy2- + 4NO + 8H2O
Hexachloroplatinate (IV)
5. Reaction with salts of weak acids. Hydrochloric acid decomposes salts of weaker acids such as

carbonates, hydrogen carbonates, sulphides, phosphides, sulphites, thiosulphates, nitrites, etc.

Na2C03 + 2 HCl » 2 NaCl + H2O + CO2 NaHC03+ HCl > NaCl + H2O + CO2
Na2S + 2 HCl > 2 NaCl + H2S Na2S03 + 2 HCl > 2 NaCl + H2O + SO2
Na2$203 + 2 HCl ^ 2 NaCl + SO2 + S + H2O ; 2 NaN02 2 HCl > 2 NaCl + H2O + NO2 + NO
6. Precipitation reactions. HCl forms insoluble chlorides with soluble salts of silver, lead and mercurous

AgN03 + HCl ^ AgCii + HNO3

{White ppt.)
Pb(N03)2 + 2 HCl -4 PbCl, i + 2 HNO3; Hg2(N03>2 + 2 HCl 4 Hg2Cl2 i + 2 HNO3
{White ppt.) {White ppt.)
7/86 New Course Chemistry (XIDBZsiai

7.35.4. Uses of Hydrochloric acid

It is used

(/) in the manufacture of metal chlorides, chlorine, NH4CI and glucose (by hydrolysis of com starch).
(ii) in pickling or cleaning iron sheets (by removing oxide layer from their surfaces) before subjecting
them to tin plating or galvanisation.
(//7) in dyeing, calicoprinting, tanning and sugar industry.
(iv) for the preparation of aqua regia which is used to dissolve noble metals,
(v) for extracting glue from bones and purifying bone black,
(w) in medicine and as a laboratory reagent.
7.36. OXOACIDS OF HALOGENS
Due to high electronegativity and small size, fluorine forms only one oxoacid, HOF known as
fluoric (I) acid or hypofluorous acid. Chlorine, bromine and iodine form four series of oxoacids with
formula HOX [halic (I) acid or hypohalous acid], HOXO [halic (III) acid or halous acid], HOXO2 [halic

w
(V) acid or halic acid] and HOXO3 [halic (VII) acid or perhalic acid], although many of these are known
only in solution or as salts. The names of some important oxoacids along with their oxidation states are

F lo
given in Table 7.19.
TABLE 7.19. Some Oxoacids of Halogens

e
Name/Oxidation Fluorine Chlorine Bromine Iodine

Fre
state

Halic(l) acid HOF HOCl for HOBr HOI

(Hypohalous acid) (Hypofluorous Hypochlorous Hypobromous Hypoiodous

r
acid acid acid
acid)
You
oks

Halic (III) acid HOCIO or HCIO2

eBo

(Halous acid) Chlorous acid

Halic(V) acid HOCIO2 or HCIO3 HOB1O2 or HBr03 HOIO2 or HIO3

Chloric acid Bromic acid Iodic acid
(Halic acid)
ad
our

Halic(VIi) acid HOC103 or HCIO4 H0Br03 or HB1O4 HOIO3 or HIO4

Perchloric acid Perbromic acid Periodic acid
(Perhalic acid)

The structures of oxoacids of chlorine are given in Fig. 7.33.

dY
Re

In all these oxoacids, Cl is i-p-^-hybridized. FIGURE 7.33

Fin

7.36.1. Acid strength and oxidising power

of oxoacids of halogens
1. (a) Acid strength of hypohalous acids.
Acid strength of oxoacids of different halogens Cl Cl Cl Cl

having same oxidation state decreases with the o. (O. o o.

0
increase in atomic number, or decrease in o.
HYPOCHLOROUS CHLOROUS CHLORIC PERCHLORIC
electronegativity of the halogen, e.g.. ACID ACID ACID ACID
+I +1 +1
Oxoacids of chlorine.
HOCl > HOBr > HOI

Explanation. This can be explained on the basis of electronegativity of the halogen atom. Oxygen is
more electronegative than the halogen (Cl, Br or I), therefore, it attracts the electrons of the oxygen-halogen
bond towiirds itself. Now, as the electionegativity of the halogen decreases from Cl to I, the shared pair of
electrons of the oxygen-halogen bond moves closer and closer towards the oxygen atom. Consequently,
electron-density on the oxygen atom increases from Cl to I. In other words, O atom in H—O—Cl has the
p-BLOCK ELEMENTS 7/87

lowest electron density while it has the highest in H—O—I. Consequently, O in H—O—Cl attracts the
electrons of O—H bond towards itself most strongly followed by in H—O—Br and least strongly in
H—O—As a result, O—breaks most readily in H—OCl and least easily in H—01. In other words, the
acid strength of the hypohalous acids decreases in the order:
H-^0--e-Cl>H-^0 —^Br>H^—O—«-I

(b) Oxidising power of hypohalous acids. As the size of the halogen increases, the thermal stability of
the O—X bond increases and Ae oxidising power of the hypohalous acid or the hypohalite ion decreases.
Thus, hypochlorites are stronger oxidising agents than hypoiodites. Surprisingly, hypobromite ions and

ow
hypochlorite ions are equally strong oxidising agents. This is supported by their electrode potentials :
0Cl- + 2H+ + e" >1/2C12 + H20 ; E‘> = +1-61V
0Br- + 2H+ + e" >l/2Br2 + H20 ; E" = + 1-60V
Or + 2H+ + e- >1/2I2 + H20 ; E‘’ = +1-44V

e
2. (a) Acid strength of perhalic acids. As the electronegativity of the halogen decreases, the tendency

re
of the XO3 group to withdraw electrons of the O—H bond towards itself decreases and hence the acid
strength of the perhalic acid decreases in the same order, i.e., HCIO4 > HB1O4 > HIO4.

Flr
F
(b) Oxidising power of perhalates. Perhalates are strong oxidising agents, their oxidising power
decreases in the order : BrO^ > IO4 > CIO4 ou
This order can be explained on the basis of their electrode potentials for the following reactions, i.e.

sr
B1O4 + 2 H+ + 2 e" > BtO~ + H2O ; E“ = 1-74 V

fo
k
lO" + 2 H+ + 2 e- > lO~ + H2O ; E° = 1-65 V
oo
CIO4 + 2 H+ + 2 e- > CIO- + H2O ; E° = M9 V.
Y
Although among perhalates, BrO” is the strongest oxidising agent, yet it is a weaker oxidising agent
reB

than F2. It is because of this reason that perbromates and perbromic acid can be obtained by oxidation of
bromates (BrOj) by F2 in alkaline solution.
uY

BrO" + F2 + 2 OH- > BrO" + 2+ H2O

3. Acid strength of oxoacids of the same halogen in different oxidation states. Acid strength of
ad
do

oxoacids of the same halogen increases with increase in oxidation number of the halogen, e.g..
+7 +5 +3 +1
Acid
HCIO4 > HCIO3 > HCIO2 > HCIO
in

P^a -10 -1-2 20 7-5

Re

Explanation. This can be explained on the basis of the relative stability of the anion (or the conjugate
F

base) left after removal of a proton.

HC10„ + H20 V - H3O+ + CIO;
Oxoacid Anion
(n=l-4) {Conjugate base)
Now greater the number of oxygen atoms in the anion (conjugate base), greater will be the dispersal of
the negative charge through pK-dn back bonding and hence greater will be the stability of the anion.
O
I
I
I
:‘cr* Cl Cl
/ \ i\ i\
0 — Ci:
/ \
, //// v\ V! \
S, // // N\ // n \
o oj \p o 0_ o d o
STABILITY OF THE ION INCREASES
7/88 New Course Chemistry fXIUPZsun

Since ihe stability of the anions decreases in the order : CIO^ > CIO^ > CIO2 > CIO .
therefore, their acid strength also decreases in the same order : HCIO4 > HCIO3 > HC102 > HCIO
Further, as the stability of the anion increases, its oxidising power decreases accordingly. In other
words, the oxidising power of the oxoacids of chlorine increases in the reverse order of their acid strength,
i.e.. HCIO4 < HCIO3 < HCIO2 < HCIO
SUPPLEMENT YO.UR
KNOWLEDGE FOR COMPETITIONS

Hypofluorous acid (HOF) was prepared in 1971 by fiuorination of ice.

233 K
F2 + H2O (ice) V ^ HOF + HF

w
HOF thus prepared is immediatelyremoved from the reaction mixture otherwiseit will be decomposedby
HF, F2 or H2O.
In HOF, F has an oxidation state of +1. The oxygen atom in HOF like that in HOCl is -hybridized.
However, due to higher electronegativity of F over Cl, the bond pair-bond pair repulsions of H—O and

o
0—F bonds in HOF are weaker than those of H—O and O—Cl bonds in HOCl.

e
As a result, HOF bond angle is little smaller (97°) than HOCl

re
angle (103°) as shown. As expected due to bigger size of Cl

rFl
over F, the 0—Cl bond is much longer than the O—F bond.

F
But the O—H bonds in both these molecules have approx,
the same length.
HOF is highly unstable and rapidly decomposes at room

or
ou
temperature to form HF and O2 with a half life of 9.30
minutes. ksf
2 HOF -> 2 HF + O2
It reacts with H2O to form HF and H2O2; HOF + H2O ■> H2O2 + HF
oo
Y

7.37. INTERHALOGEN COMPOUNDS

B

Halogens react with each other to form a number of compounds called interhalogen compounds. Their
re

general formula is XX' where X is a less electronegative halogen (halogen of larger size) while X' is a more
oYu

electronegative halogen (halogen of smaller size) and n is its number.

ad

Types. Depending upon the oxidation state

of halogen X (i.e., -i-l, +3, -i-5 or +7), these are xx' xx' xx' xx:7
d

divided into the following four types :

in

Nomenclature. The interhalogen CIF, BrF CIF3, BrF3 CIF5

Re

compounds are named as halogen halides. The BrCI IF3 BrFj

F

halogen with the positive oxidation state (i.e. ICl, IBr, IF ICI3 (I.Qg) IF7
X) is named as such while the halogen with the
negative oxidation state (i.e. X') is named as
halide. For example,
BrCI ICI3 IF 7
Bromine chloride Iodine trichloride Iodine heptafluoride

7.37.1. Preparation of Interhalogen Compounds

All interhalogen compounds are prepared either by direct combination between the halogens or by the
action of a halogen on a lower interhalogen. The product formed, however, depends upon the conditions.
473 K 573 K
CI2 (g) + F2 (g) 2ClF(g) ; CI2 (g) + 3 F2 (g) 2 CIF3 (g)
{equal volumes) Chlorine monofluoride (excess) Chlorine trifluoride

475-575K
ClF(g) + F2(g) ^ ClF3(g) l2(s) + Cl2(0 2 ICI (.?)
(equimolar) Iodine monochloride
 7/89
p-BLOCK ELEI|^EN_TS
Bf2 (/) + 3 F2 (g) - ->
2 BrF3 (/) ; Br, (/) + 5 F2 (g) ^ 2 BrFj {!)
{diluted with Bromine trifliioride (excess) Bromine pentafluoride
228 K 293 K
12(g) +P2(g) 2IF(g) I2 (5) + 5 F2 (g) 2 IF5 (g)
(in CClfF soln.) Iodine monofiiioride Iodine peniafluoride
(unstable)
523-573K
I2 (8) + 7 F, (g) 2IF7 (g)
Iodine hcptafluoridc

7.37.2. Properties of Interhalogen Compounds

Some properties of interhalogen compounds are given in Table 7.20.
TABLE 7.20. Some Properties of Interhalogen Compounds

w
Type Formula Physical state and colour Structure

Linear

Flo
XX' CIF Colourless gas
BrF Pale brown gas Linear

e
IF (very unstable) Detected spectroscopically Linear

re
BrCl (pure solid is known at r.t.) Gas Linear

F
ICI Ruby red solid (a-form) Linear

Brown red solid (p-form)

ur
r
IBr Black solid Linear

XX3 CIF3 Colourless gas

fo Bent T-shaped
ks
Yo
Brp3 Yellow green liquid Bent T-shaped
oo

IF3 Yellow powder Bent T-shaped

B

ICI3 (dimerises as Cl-bridged dimer, LClg) Orange solid Bent T-shaped

re

XX5 IF5 Colourless gas but solid below 77 K Square pyramidal

BrFg Colourless liquid Square pyramidal

u
ad
Yo

CIF5 Colourless liquid Square pyramidal

XX'2 IF7 Colourless gas Pentagonal bipyramidal

d
Re
in

The physical and chemical properties of interhalogen compounds are described below ;
F

(/) In interhalogens, the larger halogen always serves as the central atom.

(ii) The higher interhalogen compounds such as XX^ and XX^ are formed by large atoms such as Br and
I associated with small atoms such as F. This is due to the reason that it is possible to pack more small
atoms around a large atom.
(///) All interhalogen compounds are essentially covalent compounds because of small electronegativity
differences,

(/v) All interhalogen compounds have paired electrons and hence are diamagnetic in nature,
(v) The physical properties of inierhalogen compounds are intermediate between those of constituent
halogens except their melting points and boiling points are little higher due to some polarity associated
with these molecules. Further, their melting points and boiling points increase as the difference in
electronegativity increases or the polarity increases.
(vi) They are either volatile liquids or solids except CIF, BrF, CIF3, IF5 and IF^ which are gases at room
temperature.
7/90 New Course Chemistry (XII)I!ZsXSI

(v») Reactivity. Interhalogen compounds are generally more reactive than the halogens (except F2) since
the X — X' bond between two dissimilar electronegative elements is weaker than the bond between
two similar atoms, i.e., X—X or X' — X'. This is due to the reason that overlapping of orbitals of two
dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.
The sequence of reactivity of various interhalogen compounds decreases in the order.
CIF3 > BrF5 > IF7 > CiF > Brp3 > IF5 > BrF > IF3 > IF.
It is evident from the above sequence that, for a given XF„, the sequence follows the order :
Cl > Br > I, i.e., CIF3 > BrF3 > IF3, and for a given halogen the reactivity of XF„ decreases with
decrease in the value of n, i.e., BrFj > BrF3 > BrF.
(viii) Thermal stability. Thennal stability decreases as the size difference or the electronegativity difference
between the two halogen atoms decreases. In other words, the stability of the intcrhalogen compound

w
increases as the size of the central atom increases.

In general, the thermal stability decreases in the order :

Thermal stability IF > BrF > CIF > ICl > IBr > BrCl

Flo
Electronegativity difference 1-5 1-2 1 0 0-5 0-3 0-2

e
Thus, IF is more stable than CIF3 and CIF is more stable than IBr.

re
(/x) Ionization. Interhalogen compounds often ionize in solution or in the liquid state :

rF
2IC1^^ r+ICl; ; 2 ICl 3
± ICIJ+ICI-
ur
fo
When fused ICl is subjected to electrolysis, U is liberated at the cathode while both I2 and CI2 are
liberated at the anode. In contrast, when ICI3 is subjected to electrolysis, l2 and CI2 both are liberated at both
ks
the electrodes,
Yo
oo

(x) Hydrolysis. The reactions of interhalogens are simihu’ to those of halogens. For example, both on
hydrolysis give halide and oxohalide ion. However, in case of interhalogen compound, tlie oxohalide ion is
B

always formed from the larger halogen present. For example,

re

ICI + H2O >HC1 + HOI ; BrFg + 3 H2O ^ 5 HF + HBrOj

Hypoiodous acid Bromic acid
u
ad

(x/) Reaction with alkali metal halides

Yo

Interhalogens react with alkali metal halides to form polyhalides. For example.
ICl + KCl > K-*- [ICI2]- ; ICl + KBr > K+ [BrCll”
d
Re
in

ICI3 + KCI > K+ [ICl4]- ; IFg + CsF > Cs+ [IF^]-

(x/7) Flurinating agents. Interhalogen compounds of llourine ai’e good fluorinating agents and hence
F

fluorinate metals, metal oxides and metal halides. For example,

3 UO2 + 4 BrFj > 3 UF4 + 2 Br^ + 3 O2

UF4 + CIF3 > UFg + CIF FIGURE 7.34

7.37.3. Structure of Interhalogen Compounds X' X'

(/) Structure of XX' type interhalogens. As expected, interhalogens

X' X'
having single covalent bonds such as ICl. IBr, etc. have linear structures.
● ^

:’x

(//) Structure of XX-^ type interhalogens. All interhalogen compounds

of the type XX3 (i.e., IF3, BrFg, CIF3, etc.) involve sp^d hybridization of X' X'

the central atom X and hence have trigonal hipyramidal geometries or are Bonding and structure
T-shaped molecules as shown in Fig. 7.34. of XX'3 molecules
 7/91
p-BLOCK ELEMENTS

ns np nd
Electronic structure of X
(Ground state) ti tut t
Electronic structure of X
(First excited state) ti tl t t t
-v/7^i/-Hyl)ridization-trigonal bipyramid
with two positions occupied by lone pairs

(Hi) Structure of XX^ type imerhcilogens. All interhalogen X’

FIGURE 7.35
X-

compounds of the type XX^ involve sp^d~ hybridization of the X’

central halogen atom X and hence have octahedral (or square X'
X-^
pyramidal) geometry with one position occupied by a lone pair as X

w
shown in Fig. 7.35 (a).
ns np nd
'\/P X’
X'

Electronic
© O X’

Flo
structure of X t; t t t t
Bonding and Structures of
(in the second

e
(a) XX'5 Molecules

re
excited state) .v/)^i/^-Hybridlzation-oclahcdraI geometry
with one position occupied by a lone pair, (b) XX'y Molecules

F
(/v) Structure of XX-j' type interhalogens. All interhalogen compounds of the type XX^ involve sp^d^
ur
r
hybridization of the central halogen atom X and thus have pentagonal bipyramid geometry as shown in Fig.
7.35 (b).
ns np
fo nd
ks
Electronic structure of X
t t t t t
Yo
(in the third excited state)
oo

sp^iP - Hybridization-pentagonal bipyramid geometry

eB

On the basis of VSEPR theory, shapes of inierhalogen compounds can also be deduced as follows :
ur

Molecule Total no. of electron pairs No. of bond pairs No. of lone pairs Shape
ad
Yo

xx' 1±1 = 5 3 (three X-X' bonds) 5-3 = 2 T-shaped

2
d

7 +5
Re

XX' = 6 5 (five X-X' bonds) 6-5= 1 Distorted octahedral

in

2
or square pyramidal
F

7 +7
XX' = 7 7 (seven X-X' bonds) 7-7 = 0 Pentagonal bipyramid
2

737.4. Uses of Interhalogen Compounds

(0 Interhalogcn compounds lure used as non-aqueous solvents,
(ji) Interhalogen compounds of fluorine are very useful tluorinaiing agents. For example, CIF3 and BrF3
are useful in making gaseous UFg which is useful in making enriched fuel.
Further, they are useful in separating the fission products from spent fuel rods. Pu and most of the
fission products form non-volatile tetrafluorides like PUF4 while U forms volatile UFg.
3 CIF3 + U + Up6+ 3 CIF ; 4 CIF3 + 3 Pu + 3 PUF4 + 2 CI2
738. POLYHALIDE IONS
Halogens or interhalogens combine with halide ions to fomi polyhalide ions. The most common example
of polyhalide formation has been provided by the increased solubility of I2 in water in the presence of KI due
to the formation of triiodide ion, Ir
3
 7/92 'P’utdee^'a. New Course Chemistry (XI1)BS2SI

I2CV) + I {aq) > {aq)

In contrast, tribromide (Br^) is much less stable than . A few compounds of unstable Cl^ ion are
known but no compounds containing F^" ionareknown. Similarly, more complex ions such as Ij (pentaiodide),
(heplaiodide) and 1^ (enneaiodide) have also been prepared.
Besides these, many polyhalides are known which contain two or three different halogens. For example,
ICI + KCl ? K+ [ICI2]- ; ICI3 + KCI > [ICI4]-
IF5 + CsF > Cs-" [IFg]- ; ICI + KBr > [BrlCl]-
7.38.1. Properties of Polyhalldes
(/) Polyhalides are stable crystalline ionic compounds soluble in water. Their aqueous solutions conduct
electricity.
(//■) On heating, polyhalides decompose. The products of decomposition, i.e., which halogen remains attached

w
to the metal depends upon the lattice energy of the products. Since the lattice energy of the alkali metal
halides is higher for the smaller ions, therefore, during decomposition smaller halogen remains attached

F lo
to the metal.
FIGURE 7.36

Cs[l3] —^ Csl + I2 ; Rb[ICl2] RbCl + ICI

ee
Fr
7.38.2. Structure of Polyhalldes
The trihalides, i.e, Ij (I—I <- I“) and ICI“ (Ci—I ^ Cl") have one
covalent and one coordinate bond. For covalent bond — a half filled orbital for ■ -j.
ur
is needed and for a coordinate bond — an empty orbital is needed. Here,
s
half-lllled orbital of iodine is 5 p and the empty orbital is 5 d. Thus, iodine
ook
Yo

undergoes ^p^tZ-hybridization. In other words, both I" and ICI" have linear
Structure of I3 ion.
eB

structures involving trigonal bipyramidal geometry with three equatorial

positions occupied by lone pairs as shown in Fig. 7.36.
r
ad
ou

55 5p 5d
Electronic configuration of
iodine atom (ground state) ti tut t
Y

Structure of iodine having

Re

tl ti ti t»
A:
nd

formed one covalent bond (^^) and

Fi

one coordinate bond (t^) in IJ. 5p^i/-Hybridization

Five electron pairs—trigonal bipyramid with
three positions occupied by lone pairs
FIGURE 7.37

Similarly, in the pentahalides, i.e., I^ , [1014]" and [BrF4]", each

central atom has three covalent and one coordinate bond. Therefore, we
need three half-filled and one empty orbital for bonding. To achieve this, Cl Cl

one of the filled 5p-orbital electron gets promoted to 5J-orbital. Therefore,

iodine in [ICl4]“ undergoes .?p^i/^-hybridization. In other words, IICI4]"
has octahedral geometry with two positions occupied by lone pairs of Cl- ●Cl

electrons or square pyramidal as shown in Fig. 7.37.

Structure of ICI4 ion.

 7/93
p-BLOCK ELEMENTS

5s SP 5d
Electronic configuration of iodine
atom (first excited state state) Ti ti t t t
Structure of iodine having formed
three covalent bonds and one
ti t; t; t; n
coordinate bond (f i ii [IC^JT .yp i/ -Hybridization
Six electron pairs—octahedral shape with two positions
occupied by lone pairs

7.39. PSEUDOHALIDE IONS AND PSEUDOHALOGENS

Ions which consist of two or more electronegative atoms of which at least one is nitrogen and have
properties similar to those of halide ions are called pseudohalide ions. Just as halide ions have corresponding
diatomic molecules, pseudohalide ions also have corresponding dimeric molecules. These are called

w
pseudohalogens and show properties similar to those of halogens. A few examples of pseudohalide ions and
pseudohalogens are given below :
Pseudohalogens

F lo
P.seudohalides

Cyanide, CN" Cyanogen, (CN)2

ee
Thiocyanate, SCN" Thiocyanogen, (SCN)2

Fr
Cyanate, OCN" Oxycyanogen, (OCN)2

Some other pseudohalide ions are: NCN-" (cyanamide ion), ONC' (fulminate ion), NJ (azide ion), etc.

for
ur
PART—IV. GROUP 18 ELEMENTS : THE NOBLE GASES
s
7.40. GENERAL INTRODUCTION
ook
Yo

The group 18 of the periodic table consists of six monoatomic gases, i.e., helium (He), neon (Ne), argon
(Ar), krypton (Kr), xenon (Xe) and radon (Rn). Except radon, all other gases are present in the atmosphere in
eB

very small quantities and hence they are known as rare gases of the atmosphere. These gases are also sometimes
referred to as aerogens (present is the air).
r

All these gases do not show any chemical reactivity at ordinary temperatures and hence they were
ou
ad

earlier called inert gases. However, later a number of compounds of these gases, particularly those of xenon
and krypton, have been prepared. This suggests that these gases are not completely inert. Consequently, these
Y

gases are now called noble gases on analogy with noble metals like gold and platinum which show reluctance
to react rather than complete inertness.
Re
nd

7.41, OCCURRENCE OF NOBLE GASES

Fi

Except radon which is a radioactive element, all other noble gases occur in the elemental state in the
atmosphere. Their total percentage in dry air is about 1% by volume of which argon is the major component.
Helium is the second most abundant element in the universe (23% as compared to 76% hydrogen) although
its terreslial abundance is very low. Helium is also present in the natural gas to an extent of 2—7% . Helium
and sometimes neon are also present, in small quantities, in various radioactive minerals such as clevite,
monazite. pitchblende, etc. Helium, neon and argon are also present in trace amounts in some spring waters.
The relative abundance of different noble gases in the atmosphere are given below :
Element Abundance in Air Element Abundance in Air
(% by volume) (% by volume)

Helium (He) 5.24 X 10-^ Krypton (Kr) 1.14 X 10^

Neon (Ne) 1.82 X 10“^ Xenon (Xe) 8.7 X 10-^
Argon (Ar) 0.934 Radon (Rn) Trace
 7/94 New Course Chemistry (XlI)CZSCSl

Retain in Memory,
1. Lord Rayleigh and William Ramsay made the first discovery of the noble gas, argon in 1894.
2. J.N. Lockyer first discovered helium in the sohir spectrum during a total solar eclipse on 18lh
August, 1868. Later on in 1895 it was discoveredon the earth by William Ramsay.

7.42. ISOLATION OF NOBLE GASES

Helium. The main commercial source of helium is natural gas which mainly contains hydrocarbons
(mostly methane along with small turiounis of hydrocarbons up to six carbon atoms) along with varying
amounts of carbon dioxide, nitrogen, hydrogen sulphide and helium (2-7%). The natural gas is compressed
to about 100 atm and cooled to 73 K. Under these conditions, helium does not liquefy while all other gases
gel liquefied. About 99% pure helium is prepared by this method.

w
Neon, argon, krypton and xenon are obtained by fractionation of liquid air. Fractional distillation of
liquid air gives O2, and the mixture of noble gases. The individual noble gases are then separated by
adsorption over coconut charcoal which adsorbs different gases at different temperatures.

Flo
Radon is obtained as the decay product of radium. It has a half life of 3-82 days.
226

e
Ra

re
7.43. ELECTRONIC CONFIGURATION

F
The general outer electronic configuration of noble gases is ns^ np^. Helium, however, has I5-
ur
configuration. In all tlicse gases, all the orbitals which are occupied by the electrons are completely filled.

r
fo
This imparts stability to the atoms of these gases. As a result of this stable arrangement of electrons, the atoms
of these gases neither have any tendency to gain electrons nor to lose electrons. Consequently, these gases are
ks
almost chemically ineil. The electronic configuration of noble gases is given in Table 7.21.
Yo
oo

TABLE 7.21. Electronic Configuration of Noble Gases

B

Name Atomic Number Electronic Configuration

re

Complete With noble gas core

h'2
u

Helium (He) 2
ad
Yo

Neon (Ne) 10 \s^ 2s^2p^ [He] 2^2 2/

Argon (Ar) 18 I.S- 2s~ 2/ 3^2 3p^ [Ne] 3i-2 3p^
d

Krypton (Kr) 36 1 ,v- 2s~ 2/ 3s^ 3p^ 3(/*0 4^- 4p^ [At] 4j2 4p^
Re
in

Xenon (Xe) 54 l.s-2 2s^ Ip^ 3^2 3^6 3^10 4^2 4^^6 4^10 5^2 5^6 [Kr] 4d^^ 5s- 5/
F

Radon (Rn) 86 h-2 2j2 Ipf' 3^2 3p^ 4^2 4p^ 4t/‘0 4/ [Xe] 4/5^10 652 6/
5i'2 5p^ 5^/“* 6^2 6p^.

7.44. ATOMIC AND PHYSICAL PROPERTIES

Some important atomic and physical properties of noble gases are discussed below :
1. Monoatomic nature. A// the noble gases are monoatomic, colourless and odourless. Their monoatomic
nature is due to the stable outer electronic configuration (ns^ np^) of their atoms. As a result, they do not enter
into chemical combination even amongst themselves. Their monoatomic nature is further supported by the
following facts :
(/) The ratio of their specific or molar heats at constant pressure and constant volume, i.e., /Cy is 1.67.
(ii) 22.4 litres of each gas at NTP weighs equal to the atomic mass of the gas in grams.
2. Atomic radii. The atomic radii of noble gases are by far the largest in their respective periods. This
is due to the reason that noble gases (because they do not form molecules) have only van der Waals radii
while others have covalent radii, van der Waals radii, by definition, are larger than covalent radii. As we
 p-BLOCK ELEMENTS 7/95

move down the group from He to Rn, the atomic radii further increase (Table 7.22) primarily due to the
increase in the number of shells.

TABLE 7.22. Atomic and Physical Properties of Group 18 Elements (Noble Gases)
Property He Ne Ar Kr Xe Rn

Atomic number 2 10 18 36 54 86
-1
Atomic mass/g mol 4-00 20-18 39-95 83-80 131-30 222-00
Atomic radius/pm 120 160 190 200 220
Ionization enthalpy/kJ 2372 2080 1520 1351 1170 1037
mor*
Electron gain 48 116 96 96 77 68
-1
enthalpy/kJ mol
Density (at STP)/g 1-8 X 10-^ 9-0 X 10-^ 1-8 X 10-3 3-7 X 10-3 5-9 X 10-3 9-7 X 10-3

w
-3
cm

Melting point/K 24-6 83-8 115-9 161-3 202

F lo
Boiling point/K 4-2 27-1 87-2 119-7 165-0 211

3. Ionisation enthalpy. The ionization enthalpies of noble gases are the highest in their respective,

ee
periods due to their stable electronic configurations. However, as we move down the group from He to Rn,

Fr
the ionization enthalpies decrease due to a corresponding increase in their atomic radii and shielding effect of
the inner electrons.

for
4. Electron gain enthalpy. Noble gases have completely filled subshells. As a result, there is no vacant
ur
room in their valence orbitals and hence the additionalelectron has to be placed in an orbital of next higher
shell. In other words, energy has to be supplied to add an additional electron and hence, the electron gain
s
ook

enthalpy of noble gases is positive*. Further, as we move down the group, the size of the atom increases.
Yo

As a result, its tendency to accept an additional electron decreases and hence the electron gain enthalpy
eB

decreases from Ne to Rn. However, because of its small size. He has much higher tendency to accept an
additional electron and hence its electron gain enthalpy is the least positive of all the noble gases. Surprisingly,
the electron gain enthalpy of Kr is the same as that of Ar. The reason for this exceptional behaviour is not
r
ad

known. Thus, the electron gain enthalpy of noble gases decreases in the order: Ne > Ar = Kr > Xe > Rn > He
ou

5. Melting and boiling points. The melting and boiling points of noble gases are very low. This is due
to the reason that the atoms of these elements are held together by weak van der Waals forces (dispersion
Y

forces) of attraction both in the liquid as well as in the solid states. Further, as we move down the group from
Re
nd

He to Rn, the melting and boiling points of these elements show a regular increase due to a corresponding
increase in the magnitude of their van der Waals forces of attraction as the size of the atom increases.
Fi

Helium, however, has the lowest boiling point (4-2 K) of all the known substances.
6. Ease of liquefaction. The ease of liquefaction of a gas depends upon the magnitude of the attractive
forces present in its atoms or molecules. Since the atoms of noble gases are held together by weak van der
Waals forces of attraction, these gases cannot be easily liquefied. However, as the atomic size increases, the
magnitude of their van der Waals forces of attraction increases and hence ease of liquefaction increases as
we move down the group from He to Xe.
7. Solubility in water. These gases are slightly soluble in water. The solubility, in general, increases
from He to Rn.
Explanation. Water is a polar molecule. Therefore, when it comes near a noble gas, it induces dipole in
the noble gas by distorting or polarizing its otherwise symmetrical electron cloud as shown in Fig. 7.38. The
induced dipole thus set up in the noble gas then interacts with the dipole of polar water molecule. As a result
of this dipole-induced dipole interaction, the noble gas dissolves in water. As the size of the noble gas
♦Earlier electron affinity of the noble gas was taken to be zero.
7/96 ‘P%<xdee^'4- New Course Chemistry fXinrosTwn
FIGURE 7.38
increases, the extent of polarization increases.
Consequently, the magnitude of the dipole-induced POLAR WATER POLAR WATER
NOBLE GAS
dipole interaction increases and hence the solubility MOLECULE MOLECULE

of the noble gas in water increases from He to Rn.

8. Enthalpy of fusion and enthalpy of s-
o
vaporization. The enthalpy of fusion and enthalpy
6*
of vaporization of noble gases are very low due to H

weak interatomic forces of attraction in the liquefied A

DISTORTED
state. The values of these properties are found to ELECTRON CLOUD
SYMMETRICAL
ELECTRON CLOUD
increase from He to Rn due to an increase in the
magnitude of the interatomic forces as the atomic size Polarization of a noble gas
of the noble gas increases. by polar water molecules.
9. Adsorption over charcoal. Except He, all other noble gases are adsorbed by coconut charcoal at low
temperatures. The extent of adsorption increases as the atomic size of the noble gas increases.

w
SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

F lo
1. Helium has the unusual property of diffusing through most commonly used laboratory materials such
as rubber, glass or plastics.

ee
2. Clathrates of noble gases. Noble gases form a number of compounds in which the gases are entrapped

Fr
within cavities of crystal lattices of certain organic and inorganic compounds. Such compounds are
called clathrates or cage compounds. These are also sometimes referred to as host-guest compounds.
The organic or inorganic compound with cavities is called the host while the gas atoms entrapped in it
for
are called the guest. Only Ar, Kr and Xe form clathrates since their sizes match the size of cavities
ur
present in most of the organic and inorganic compounds. As such, He and Ne do not form clathrates
since they are too small to be entrappedwithin the cavities of organic and inorganic compounds.
s
ook
Yo
3. Hydrates and deuterates of noble gases. Hydrates of noble gases are formed by compressing the
gases with H,0. Similarly, deuterates are formed by compressing the gases with D2O. The hydrates
eB

and deuterates of the heavier noble gases are known to be more stable. Thus, Xe.6H20 is the most
stable hydrate of noble gases. The cause of hydrate formation is the dipole-induced dipole interaction
between noble gases and water. Since the dipole-induced dipole interaction increases down the group
r
ad
ou

from He to Rn, therefore, the solubility of the noble gases in water increases from He to Rn.
4. Interstitial compounds. These compounds are formed when small atoms occupy positions in the
interstices of the metal lattice. Only He forms such compounds since the atomic size of He is the
Y

smallest amongst the noble gases and matches the size of the interstices available in the lattice of most
Re
nd

of the heavy metals.

Fi

Curiosity Question
f Q. it is often difficult to store radioactive isotopes of Kr and Xe. Suggest a method to store them.
Ans. During nuclear reactions, radioactive isotopes of Kr and Xe are produced which find large number of
applications in medicine and industry. For this purpose, these radioactive isotopes must be stored. A
useful method to store them is to make their hydrates by compressing them in water. The hydrates
thus formed are stable and are kept at low temperatures and used as and when required.
J
7.45. CHEMICAL PROPERTIES
The noble gases are chemically inert due to the following reasons.
(/) The noble gases have completely filled ns~np^ electronic configuration in their valence shells,
(ii) The noble gases have high ionization enthalpies.
(Hi) The electron gain enthalpies of noble gases are positive.
p-BLOCK ELEMENTS 7/97

Therefore, noble gases neither have a tendency to lose nor to gain electrons and hence do not enter into
chemical combination. Before 1962. it was thought that noble gases do not combine at all and hence no
compounds of noble gases were possible.
However, in 1962, NeU Bartleit observed that platinum hexafluoride (PtF^) — a powerful oxidising
agent reacts with dioxygen to yield an ionic solid, dioxygenylhexafluoroplatinate (V), O:^ [PtFg]“.
02(g)+ PtF6(g) ^ OJ [PtFgl-Fv)
In this reaction, has been oxidised to by PtFg.
Since the first ionization enthalpy of xenon (1170 kJ mor*), is fairly close to that of 0-, molecule (1175
kj mol ’), Bartlett thought that PtFg should also oxidise Xe to Xe'*’. Thus, when Xe and PtF^ were mixed, a
rapid reaction occurred and a red solid with the formula, Xe'^fPtF^]" was obtained.
278 K
Xe + PtFg ^ Xe-^[PtFJ
After this discovery, a large number of xenon compounds (Table 7.23) mainly with most electronegative

w
elements like fluorine and oxygen have been prepared,

F lo
TABLE 7.23. Some Important Stable Compounds of Xenon
4-2 oxidation state ●f 4 oxidation state + 6 oxidation state + 8 oxidation state

ee
Xep2. XeF4, XeFg, Xe04,

Fr
Xenon difluoride Xenon tetrafluoride Xenon hexafluoride Xenon telroxide

XeOp2,
Xenon oxydifluoride
XeOp4,
for
Xenon oxytetrafluoride
Xe03p2
Xenon trioxydifluoride
ur
Xe02F2,
s
Xenon dioxydifluoride
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Xe03,
eB

Xenon trioxide

The compounds of krypton are fewer. Only krypton difluoridc (Krp2) has been studied in detail. The
our

compounds of radon (e.g., Rnp2) has not been isolated but has only been identified by radiotracer technique.
ad

No true compounds of Ar, Ne or He are yet known.

7.45.1. Xenon Fluorides
Y

Preparation. Xenon reacts directly with fluorine under appropriate conditions to form three binary
Re
nd

fluorides Xep2 {xenon difluoride), Xep4 (xenon tetrajhioride) and XeF^ (xenon hexafluoride).
Fi

673 K 873 K
Xe (g) + p2 (g) I bar
4 XeF2(.y) ; Xe(g) + 2F2(g) 4 XeF4(.v)
7 bar
(Xenon in excess) (1:5 ratio)
573 K
Xe (g) + 3 p2 (g) 60-70 bar
4 XeF6(5)
(! : 20 ratio)

Xep2 can also be preparet’ by iiradiaiing a mixture of xenon and fluorine with sunlight or light from a
high pressure mercury arc lam
XeFg can also be prepared by the interaction of Xep4 and O2F2 at 143 K.
I43K „
Xep4 + OoF-) > XeF^ + 02
Properties. (/) All the fluorides are colourless crystalline solids and sublime readily at 298 K.
(//) The lower fluorides form higher fluorides when heated with F^ under pressure.
(Hi) All fluorides are extremely strong oxidising agents. They react quantitatively with H2 as follows.
7/98 “Pnctdee^'A New Course Chemistry (XIl)EZsl9l
4 HF + Xe
XeF^ + ^ 2 HF + Xe Xep4 + 2 H2
XeFg + 3H2 ^ 6 HF + Xe

They oxidise CF to Ci2 and 1“ to I2 and Ce (III) to Ce (IV)

XeF2 + 2 HCl ^ 2 HF + Xe + CU ; Xep4 + 4 KI ^ 4 KF + Xe + 2Io

+ Xep2 + Ce2(S04 )3 ^ 2 Ce(S04)2 + Xe + F2

(/v) They all act as strong fluorinating agents.
XeF4 + 2 SF4 ^ Xe + 2 SFg ; XeF4 + Pt ^ Xe + PtF4
Xep2 + 2 NO ■ ^ Xe + 2 NOF

(v) Xenon fluorides can simultaneously oxidise and fluorinate the heteroatom in an organic compound
without attacking the alkyl or aryl groups.
CH3I + Xep2 ^ CH3IF2 + Xe ; C6H5l + XeF2 — ^C6H5lF2 + Xe

w
(C6H5)2S + XeF2 ->(C5H5)2SF2 + Xe
(v/) Xenon fluorides react with fluoride ion acceptors to form cationic species and with fluoride ion
donors to form fluoroanions.

Flo
Xep2 + PF5 ^ [ XeF [ PF^]- XeF4 + SbF5 ^ [XeF3l^ [ SbF^j-

e
XeF^+ MF ^ M"*" [XeF^]" where M = Na, K, Rb or Cs

re
(v/7) Reaction with water. They are readily hydrolysed by even traces of water. However, they differ in

F
their reactivity with water and undergo complete hydrolysis to give different products.
XeF2 is soluble in water but undergoes slow hydrolysis evolving O2. Hydrolysis is, however, more
ur
r
rapid with alkalies.
2 XeF2 (s) + 2 H2O (/) fo
>2 Xe (g) + 4 HF (aq) + O2 (g)
XeF4 reacts violently with water and undergoes disproportionatio n giving xenon and xenon trioxide
ks
(Xe03) which is a highly explosive solid.
Yo
oo

6 XeF4 (^) + 12 H2O (0 > 4 Xe (g) + 2 XeOj (s) + 24 HF (aq) + 3 O2 (g)

XeFg also reacts violently with water, but slow hydrolysis by atmospheric moisture gives highly explosive
B

solid, Xe03
re

XeFg (s) + 3 H2O (/) > Xe03 (s) + 6 HF (aq)

With two equivalents of H2O, Xe02F2 is obtained.
u
ad

XeFg + 2 H2O > Xe02F2 + 4 HF

Yo

and with one equivalent of H2O, XeOp4 is obtained.

XeFg + H2O > XeOp4 + 2 HF
d
Re

7.45.2. Xenon Oxides

in

Preparation. Xenon forms two oxides, i.e., xenon trioxide (Xe03) and xenon tetroxide (Xe04). We
F

have discussed above that complete hydrolysis of XeFg gives Xe03. Xenon tetroxide (Xe04) is, however,
obtained by the action of anhydrous or cone. H2SO4 on barium perxenate.
Ba2[XeOg] + 2 H2SO4 > Xe04 + 2 BaS04 + 2 H2O
Properties. (0 Xe03 is a colourless explosive solid and acts as a powerful oxidising agent. It oxidises
Pu^'*’ to Pu*^ in the presence of H'*' ions.
3 Pu2+ + Xe03 + 6 H-" > 3 Pu'^'*’ + Xe + 3 H2O
(«) Xe03 reacts with aqueous alkali to form the hydrogen xenate ion, KXeO^ , which slowly undergoes
disproportionation to give xenon and perxenate ion, XeOg' in which Xe has an oxidation state of +8.
Xe03 + OH" HXeO" ; 2HXe04+20H~ > ^^^5” + + O2 + 2 H2O
Hydrogen xenate ion Perxenate ion

Solutions of perxenates are yellow and are powerful oxidising agents.

(Hi) Xe04 is not as stable as Xe03 and decomposes to give xenon and oxygen.
Xe04 > Xe + 2 O2
p-BLOCK ELEMENTS 7/99

7.45.3. Xenon Oxyfluorides

Xenon forms number of oxyfluorides such as XeOp2, (xenon oxydifluoride), Xe02p2 (xenon
dioxydifluoride), XeOp4 (xenon oxytetrafluoride), XeOP^ (xenon oxyhexafluoride), Xe03p2 (xenon
trioxydifluoride), Xe02p4 (^^^on dioxytetrafluoride) etc.
Preparation. Partial hydrolysis of Xep4 gives XeOp2 and that of XeP^ yields XeOp4 and Xe02p2-
XeP4 + H20 >XeOp2 + 2HF ; XePg + H2O - ^ XeOp4 + 2 HP
XepQ + 2 H2O — ^ Xe02p2 + 4 HP
7.45.4. Structure of Xenon Compounds
Both VSEPR theory and concept of hybridization are applied to predict the molecular geometries of
xenon compounds as discussed below.

ow
There are two, four and six Xe—P covalent bonds in Xep2, Xep4 and XePg respectively. Depending on
the number of Xe—F covalent bonds to be formed in each case, the requisite number of electrons of the 5p-
orbital of the valence shell of Xe get unpaired and promoted to the vacant 5f/-orbitals followed by hybridization.
According to VSEPR theory, the shape of the molecule is predicted by the total number of election pairs
(lone pairs + bond pairs) in the valence shell of the central Xe atom as shown in Table 7.24.

e
Fl
re
TABLE 7.24. Hybridisation and Shapes of Xenon Fluorides

F
Molecule Total no. of No. of bond No. of lone pairs Hybridisation Shape
electron pairs pairs of electrons of electrons
ur
or
8 + 2 sf
XeFo
2
2 (two X—F bonds) 5-3=2 sp^d Linear
k
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8 +4 ,
oo

XeF4 2
= 6 4 (four X—F bonds) 6-4 = 2 sp^d~ Square planar
B

8 + 6
= 7
XeFg 6 (six X—F bonds) 7-6= I sp\fi Distorted octahedral
re

The shapes of a number of xenon fluorides and oxyfluorides are given in Fig. 7.39.
u
ad
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FIGURE 7.39

0 ©
d
Re
in
F

©
XeFj Xe03 X©Or2
r(Xe-F) - 200 pm r(X&-F)=195pm DISTORTED OCTAHEDRAL PYRAMIDAL T-SHAPED
LINEAR SQUARE PLANAR

©
XeOF4 Xe02F2
TETRAHEDRAL SQUARE PYRAMIDAL TRIGONAL XeOaFj
BIRYRAMIDAL TRIGONAL BIPYRAMIDAL Xe02F4
OCTAHEDRAL

Shapes of compounds of xenon.

7/100 ‘P'uidee^'4. New Course Chemistry (XlOEZSm

The geometry of XeFg has not yet been confirmed. It probably has a distorted octahedral structure in
which all the six positions are occupied by fluorine atoms and the lone pair is present at the corner of one of
the triangular faces (Fig. 7.39).
The geometries and shapes of xenon oxides and oxyfluoridcs can be beti 'r explained on the basis of
concept of hybridization (Table 7.25).
TABLE 7.25. Hybridization and Shapes of Some Xenon Oxides and Oxyfluorides
Molecule Hybridization Geometry Shape

Xe03 sp^ Tetrahedral Pyramidal

Tetrahedral Tetrahedral
Xe04 sp-

XeOF, sp^d Triogonal bipyramidal T-shaped

XeOp4 sp^d^ Square pyramidal Square pyramidal
sp^d Trigonal bipyramid Distorted trigonal bipyramid

w
Xe02p2
Xe03p2 sp^d Trigonal bipyramid Trigonal bipyramid

F lo
Xe02p4 sp^d^ Octahedral Octahedral

Since in all these compounds, each oxygen atom forms a double bond with xenon, therefore, total

e
number of half-filled orbitals needed for bond formation = number of monovalent atoms + twice the number

Fre
of oxygen atoms. The requisite number of half-filled orbitals are then obtained by promotion of 5p and even
55-electrons to 5t/-orbitals. Now, in order to make 7C-bonds with the O atom, 5d orbitals equal to the number
for
of O atoms are left out while the remaining orbitals are hybridized. The type of hybridization thus obtained
determines the geometry of the molecule. As an illustration, consider the molecule of Xe02F2.
r
You
Total number of half filled orbitals needed = 2 x 2 (O atoms) -i- 1 x 2 (F atoms) = 6
oks

Now promotion of three electrons, one from each of the three 5p-orbitals to 5(i-subshell gives six half
eBo

filled orbitals.

5s 5d

ti n It ti
our
ad

Xenon atom (ground state)

Xenon atom (in the third tl t t t t t t

excited state)
dY
Re

Form two rc-bonds

5p^(/-Hybridization with 0-atoms.
Fin

Two of the 5c(-orbitals are kept as such to form two TC-bonds with oxygen atoms. The remaining five
orbitals undergo 5p^<7-hybridization to give five sp^d- orbitals. Two of these five orbitals form two a-
bonds with F-atoms, another two form two o-bonds with 0-atoms while the fifth one contains a lone pair
of electrons. The two unhybridized half filled 5r/-orbitals overlap with half-filled, p-orbitals of
0-atoms to form two 7C-bonds. As a result, the molecule of Xe02p2 has bipyramidal geometry with one
of the equatorial positions occupied by a lone pair. Consequently, the molecule has trigonal bipyramidal
shape (Fig. 7.39). Similarly, the shapes of all the other oxides and oxyfluorides of xenon (as shown in
Fig. 7.39) can be deduced.
7.46. USES OF NOBLE GASES

1. Uses of Helium

(0 The chief use of helium is iniflling of balloons which aie employed for meterological observations.
Although lifting power of helium is 8% less than that of hydrogen, yet it is preferred because it is non-
inflammable.
p-BLOCK ELEMENTS 7/101

(//) Liquid helium (b.p. 4.2 K) is used as a cryogenic agent for carrying out various experiments at low
temperatures. It is used to produce and sustain powerful superconducting magnets which form essential
part of modern NMR spectrometers and Magnetic Resonance Imaging (MRI) systems used for clinical
diagnosis.
(///) Because of very low solubility of helium in blood, an oxygen-helium mixture is used/or artiifcial
respiration in deep sea diving instead of air because nitrogen present in air dissolves in blood under
high pressure when sea diver goes into deep sea. When he comes to the surface, nitrogen bubbles out of
the blood due to sudden release of pressure causing pain. This disease in called "‘bends",
(iv) Oxygen-helium mixture is also used in the treatment of asthma.
{v) Helium is also used /or creating inert atmosphere during welding of Mg and A1 which are easily oxidizable.
(vi) Helium is chemically inactive and does not become radioactive. Hence it is used in gas cooled atomic
reactors as a heat transfer agent.
2. Uses of Neon

w
(0 Neon is mainly used in discharge tubes and fluorescent lamps for advertising purposes. Neon when
used in a discharge tube produces an orange red glow which can be seen at long distances even in mists

F lo
and fogs. However, when neon is mixed with other gases and used in discharge tubes, glows of different
colours known as neon signs are produced. These are widely used for advertising purposes.

ee
Neon bulbs are also used in botanical gtu-dens and green houses.

Fr
(//) Neon has a remarkable property of carrying extremely high currents even under high voltage. Therefore,
neon is used in safety devices for protecting electrical instruments such as voltmeters, relays and rectifiers
from high voltage.
for
ur
(Hi) It is used in hecon light as safety signal for air navigators since its light has fog penetration power, i.e.,
light cannot be stopped by fog.
s
ook

(iv) It is used for filling sodium vapour lamps.

Yo

3. Uses of Argon
eB

(/) Argon is used mainly to provide an inert atmosphere in high temperature metallurgical processes such
as arc welding of metals and alloys.
our

(//) In the laboratory, it is used for handling substances which are air sensitive.
ad

(Hi) Pure argon is used as a carrier gas in gas chromatography.

(iv) It is used in iflling incandescent andflouorescent lamps where its presence retards the sublimation of
Y

the filament and thus increases the life of the lamp,

Re
nd

(v) It is mixed with neon in "neon .signs’’ for obtaining lights of different colours.
Fi

4. Uses of Krypton
(/) Krypton is used in high efficiency miners’ cap lamps.
(ii) Krypton-85, one of the isotopes of Kr, is used in electric tubes for regulating voltage and for testing
leaks. Krypton-85 is also used for measuring thickness of metal and plastic sheeting.
5. Uses of Xenon

(/) Ktypton and xenon are more efficient than argon in gas iflled lamps because of their lower thermal
conductivities but due to their scarcity and high cost they are used to a much smaller extent.
(ii) A mixture of krypton and xenon has also been used in some tubes for high speed photography.
6. Uses of Radon

(/) Being radioactive, radon is used in radioactive research,

(ii) It is used for normal treatment of cancer and other malignant growths.
(Hi) It is used in X-ray photography for the detection of flaws in metals and other solids.
7/102 'PnzuCeefi-'<i. New Course Chemistry CXI1)ESBI9]

1. The p-block elements consist of elements of groups 13-18 having np^~^ as valence shell electronic
configuration where /i = 2 - 6.

I. Group 15 Elements
2. Group 15 consists of elements : N, P, As, Sb and Bi having ns^ as valence shell electronic

configuration where n = 2 - 6. These elements are also called pnicomigs.

3. Nitrogen forms pn-pn multiple bonds and exists as a diatomic molecule while phosphorus forms only single
bonds and exists as tetratomic (P4) molecules.
4. Nitrogen shows 9 oxidation stales (- 3, - 2, - 1, 0, + I + 2, + 3, + 4, + 5), phosphorus shows 5 oxidation

w
states (-3,-2,0, + 3, + 5) while all the remaining elements show 2 oxidation states of+ 3 and +5. However,
due to inert pair effect, the stability of + 5 oxidation state decreases while that of + 3 oxidation state

F lo
increases from As to Bi.

5. Nitrogen differs from other elements of this group due to its (/) small size, (it) high electronegativity,

ee
{Hi) absence of ^/-orbitals in the valence shell and (/v) tendency to form pK-pK multiple bonds with itself and
other electronegativeelements (O or C).

Fr
6. The hydrides of group 15 elements show the following trends ;
Bond angle : NH3 > PH3 > ASH3 > SbH3 > BiH3-
for
ur
Ba.sic character : NH3 > PH3 while ASH3, SbH3 and BiH3 are not at all basic
Boiling point: PH3 < ASH3 < NH3 < SbH3 < BiH3
s
ok
Yo
Melting point: PH3 < ASH3 < SbH3 < NH3
o

Thermal stability : NH3 > PH3 > ASH3 > SbH3 > BiH3
eB

Reducing character : NH3 < PH3 < ASH3 < SbH3 < BiH3
Solubility : NH3 forms H-bonds with H^O and hence is soluble in H2O while all other hydrides do not form
r
ad
ou

H-bonds with H2O and hence are insoluble.

7. Nitrogen forms five oxides : N^O. NO, N-,03. NO2 or N2O4 and N2O5. Their acidic strength increases in the
order : N2O < NO < N2O3 < N0O4 < N205‘
Y

8. N,0 is called laughing gas and is obtained by thermal decomposition of NH4NO3.

Re
nd

NH4N03-^ N2O + 2H2O

Fi

9, NO is an odd electron (Le., 11) species which is paramagnetic in the gaseous state but diamagnetic (due to
dimerization) both in the liquid and solid states.
10. 'IVihalidcs. All the elements of group 15 form trihalides having pyramidal geometry. (/) Amongst trihalides
of N, NF3 is an exothermic compound due to strong N-F bonds and does not undergo hydrolysis at all.
Other trihalides of N (i.e., NCl3, NBr3 and NI3) are endothermic compounds and readily undergo hydrolysis
giving NH3 and the corresponding hypophalous acid (NX3 + 3H2O > NH3 + .3 HOX where X = Cl, Br
or I). The trichlorides of P and As give HCl and the corresponding H3PO3 and H3ASO3 acids. In contrast,
trichlorides of Sb and Bi are only partly and reversibly hydrolysed to give HCl and oxychloride of the
^ SbOCI + 2 HCl ; BiC^ + H2O ^ i BiOCl + 2HCl
corresponding metal, i.e.. SbCl3 + H20 T
(/) Thus, the ease of hydrolysis of trihalidies follows the order : NCI3 > PCI3 > ASCI3 > SbCl3 > BiCl3.
(//■) The trihalides of N act as Lewis bases. Their Lewis base strength increases in the order :
NF3 < NCI3 < NBr3 < NI3
p-BLOCK ELEMENTS 7/103

11. Pentahalides. (/) P, As and Sb form pentahalides due to the presence of vacant d-orbitals. However, N does
not form pentahalides due to the absence of cf-orbitals. All the pentahalides have trigonal bipyramid geometry,
(ii) Since as we move down the group the stability of + 5 oxidation state decreases while that of + 3 oxidation
state increases due to inert pair effect, therefore, pentahalides are less stable than the corresponding trihalides.
Further, as the size of the halogen increases from F to I, the stability of the pentahalides decreases. Thus, the
pentafluorides of P, As, Sb and Bi are all known; pentachlorides of only P, As and Sb are known; pentabroraide
of only P, i.e., PBr5 is known while pentaiodides of none of elements is known.
(Hi) PCI5 is covalent in the vapour phase but exists as [PC14]'*' [PCl^]" in the crystalline state.
12. Nitrogen gas. It is prepared in the laboratory by thermal decomposition of either NH4NO2 (NH4CI + NaN02>
or (NH4)2Cr207 (NH4CI + K2Cr207). It is rather inert and unreactive at room temperature due to its high
bond dissociation enthalpy (941.4 kJ mol"^). Reactivity, however, increases with rise in temperature.
13. Ammonia. NH3 is manufactured by Haber’s process. In this process, a mixture of N2 and H2 in the ratio

w
1 : 3 is passed over heated iron oxide at 700 K as catalyst and Mo as promoter.
(0 NH3 is a base. Its aqueous solution, i.e., NH4OH reacts with iron, aluminium and chromium salts to give

Flo
reddish brown ppt, of Fe(OH)3, gelatinous white ppt. of Al(OH)3 and dirty green ppt. of Cr(OH)3 respectively.
(ii) NH3 acts as a reducing agent and thus reduces heated CuO to Cu metal while it itself is oxidised to N2 gas.

ee
(Hi) Sodium hypochlorite (NaOCl) oxidises NH3 to NH2NH2 (hydrazine).

Fr
(iv) NH3 acts as a ligand and forms complexes with </-block elements, e.g., AgCl dissolves in NH3 to form
the complex, [Ag(NH3)2]Cl and CUSO4 dissolves in excess of NH3 to form the blue coloured complex,

for
[Cu(NH3)4]S04.
ur
(v) Nessler’s reagent is an alkaline solution of K2Hgl4. It reacts with NH3 or ammonium salts to give a
brown ppt. of iodide of Millon’s base (H2N.HgO.HgI). This reaction is used as a test for NHj salts.
ks
Yo
14. Oxoacids of nitrogen. Nitrogen forms two oxoacids, HNO2 and HNO3. Whereas HNO3 acts as a strong
oo

oxidising agent, HNO2 behaves both as a reducing as well as an oxidising agent.

eB

(a) HNO3 is manufactured by Ostwald’s process. NH3 is first oxidised to NO with O2 in presence of Pt/Rh
gauze as catalyst at 500 K under a pressure of 9 bar. NO thus obtained is further oxidised to NO2 by O2 at
r

310 K. Dissolution of NO2 in H2O gives HNO3 (Art. 7.10.1, page 7/21).
ou
ad

(b) HNO3 acts as a strong oxidising agent both in the concentrated as well as in the dilute form.
Y

(0 Metals which are more electropositive than hydrogen react with HNO3 to first liberate nascent hydrogen
which then reduces HNO3 to NO2, NO, NH3, NH4NO3 and N2O depening upon the nature of the metal,
nd
Re

concentration of the acid used and the temperature.

(ii) Metals such as Cu, Ag, Hg, etc. which are less electropositive than hydrogen react with cone. HNO3
Fi

forming NO2 and the corresponding metal nitrate. With dil. HNO3, metal nitrate and NO are formed.
(Hi) Noble metals like gold and platinum do not react with cone. HNO3. However, these metals dissolve in
aqua regia (1 part cone. HNO3 + 3 part cone. HCl) forming their respective chlorides. The high reactivity
of aqua regia is due to nascent chlorine which is formed by the action of cone. HNO3 on cone. HCl.
HNO3 + 3 HCl > NOCl + 2 H2O + 2 [Cl]
(iv) Passivity. Metals like Fe, Cr, Ni and A1 become passive, i.e., lose their normal activity when dipped in
cone. HNO3. This passivity is due to the formation of a thin protective layer of the metal oxide (i.e., Fe304,
Cr203, NiO, AI2O3 on the surface of the metal),
(v) Cone. HNO3 oxidises canesugar to oxalic acid.
15. Brown ring test. In qualitative analysis, nitrate ion is detected by brown ring test. The brown colour of the
ring is due to the formation of [Fe(H20)5N0]^''’ in which the oxidation state of Fe is +1.
16. Phosphorus exists in three allotropic forms, i.e., white (or yellow), red and black phosphorus. Out of these,
black phosphorus is least reactive.
7/104 New Course Chemistry fXn~>rosTMi

17. White phosphorus is obtained by heating Ca3(P04)2 with coke and sand in an electric furnace at 1770 K.
1770K
2 Ca3(P04)2 + 6 SiO, + 10 C > 6CaSi03 + P4+ loco
18. White phosphorus exists as P4 units in which the four P atoms lie at the comers of a regular tetrahedron with
ZPPP = 60°.

19. The P4 units of white phosphorus are held together by weak van der Walls forces of attraction. As a result,
its ignition temperature (303 K) is very low and easily catches fire in air. It is, therefore, kept under water.
20. Red P is obtained by heating white P at 540—570 K in an inert atmosphere (CO2 or coal gas). Like white
phosphorus, it also consists of P4 tetrahedra but these are joined together through covalent bonds to give
polymeric structure.
21. White P is soluble in CS2 but red P is insoluble. When white P is heated with a solution of NaOH in an inert

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atmosphere, it undergoes disproportionation to fomi PH3 (- 3) and NaH^PO-) (+ 1).
P4 (s) + 3 NaOH {aq) + 3 H2O (/) > 3 NaH2P02 (aq) + PH3 (g)
22. PH3 reacts with aqueous solutions of CUSO4, HgCl-) and AgN03 to form their corresponding metal phosphides.
23. Phosphorus reacts with required amount of CI2 gas to form PCI3 and PClg respectively. PCI3 is also formed

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when while P reacts with SOCI2 and PClg is formed when it reacts with SO2CL.

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Both PCI3 and PClg react with compounds containing hydroxyl groups to form the corresponding chloro

F
compounds.
24.
Phosphorus forms two oxides P4O6 and P40,q. In P4O6, there are 12 P—0 bonds while in P4OJ0, there are
ur
or
12 P—O and 4 P = O bonds.
25. P40,o is a powerful dehydrating agent. As such it dehydrates H^S04 to SO3, HNO3 to N2O5 and HCIO4 to
sf
CI2O7.
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26. Phosphorus forms five oxoacids, i.e., hypophosphorus acid (HgPO^), phosphorus acid (H3PO3),
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orthophosphoric acid (H3PO4), pyrophosphoric acid (H4P2O7) and metaphoshoric acid (HPO3).
B

27. H3PO2 is a monobasic acid and behaves as a strong reducing agent due to the presence of two P—H bonds
in its molecule. It reduces AgN03 to Ag, HgCl2 to Hg and arenediazonium salts to arenes.
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28. H3PO3 is a dibasic acid. It is a weaker reducing agent tlian H3PO-) since it has only one P—H bond while
ur

HgPO^ has two.

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29. H3PO4 is a trihasic acid. It does not act as a reducing agent since it has no P—H bond. With ammonium
molybdate, it gives yellow ppt. of ammonium phosphomolybdate.This reaction is used as a test for phosphate
radical.
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30. HPOj is a monobasic acid. It usually exists as a polymer, (HPO3),, having either linear or ring structures.
in

The irng stmeture is called cyclotrimetaphosphoric acid, (HPOgjg.

F

II. Group 16 Elements

31. Group 16 consists of elements : O, S, Se, Te and Po having ns^ np^ np[ as the valence shell electronic
configuration where n = 2 - 6. Group 16 elements are also called, chaicogens.
32. Oxygen exists as a diatomic gas at room temperature while other elements (S, Se and Te) exist as octaatomic
solids having puckered 8-membered irng structures.
33. Oxygen shows oxidation states of - 2, - 1, + 1 and + 2 while all other elements show oxidation states of
- 2, + 4 and + 6. The stability of - 2 oxidation state decreases from S to Po as the electronegativity of the
element decreases down the group. Further, due to inert pair effect, the stability of + 6 oxidation state
decreases while that of + 4 oxidation state increases down the group. Thus, + 6 oxidation state is most stable
in case of S, i.e., (SFg) and least stable in case of Po.
34.
Hydrides of group 16 elements show the followingtrends :
Physical state : Due to extensive intermolecuiar H-bonding, H2O is a liquid while ail other hydrides are
unpleasant, foul smelling, poisonous gases.
p-BLOCK ELEMENTS 7/105

Boiling point: H2O has the highest boiling point due to intermolecular H-bonding while the boiling points
of other hydrides increase as the atomic mass of the element increases, i.e., H2S < H2Se < H2Te « H2O
VolatiUty : H2S > H2Se > H2Te > H2P0 > H2O or H2O < H2P0 < H2Te < H2Se < H2S
Bond angle : H2O > H2S > H2Se > H2Te
Acidic character : H2O < H2S < H2Se < H2Te
Thermal stability : H2O > H2S > H2Se > H2Te
Reducing power : H2Te > H2Se > H2S > H2O
35. Sulphur forms two oxides : SO2 and SO3. In both these oxides, S is sp^-hybridized ; both the S-O bonds in
SO2 and all the three S-O bonds in SO3 have equal length due to resonance. However, SO2 has a finite
dipole moment while SO3 has zero dipole moment.

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36. SO2 may be regarded as an anhydride of sulphurous acid (H2SO3) while SO3 is regarded as an anhydride of
sulphuric acid (H2SO4).
37. SO2 acts both as a reducing as well as an oxidising agent while SO3 acts only as an oxidising agent.
38. The elements of group 16 form a number of monohalides, dihalides, tetrahalides and hexahalides. Of these,

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tetrahalides and hexahalides of S are very important. Whereas SFg is octahedral while SF4 is trigonal

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bipyramidal. Further, SF4 acts as a Lewis base and easily undergoes hydrolysis but SFg does not undergo

F
hydrolysis.
39. Oxygen exists, is two allotropic forms, i.e., O2 and O3.
ur
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(0 O2 is prepared by thermal decomposition of oxygen rich compounds such as KMn04, KCIO3, KNO3,
Ba02, etc. and metal oxides such as HgO, Pb304, Ag20, Mn02, etc.
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(ii) In the laboratory, O2 is, however, prepared either by heating KCIO3 in presence of Mn02as catalyst or by
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the action of either H2O or acidified KMn04 on Na202.
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(Hi) O3 is present in the upper atmosphere where it is formed by the action of UV radiations on O2. O3
B

protects us from the harmful UV radiations which cause skin cancer. Ozone layer in the stratrosphere is
depleting due to NO released by supersonic aircrafts and due to chlorine radicals produced by decomposition
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of chlorofluorocarbons (CFC’s), Le., freon which is increasingly being used in aerosols and as a refrigerant,
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(iv) Ozone is a powerful oxidising agent, much stronger than oxygen. In most of the oxidation reactions, O2
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is always liberated but during the oxidation of SO2 to SO3 and SnCl2 to SnCl4 by O3, O2 is not liberated,
(v) O3 oxidises Hg to Hg20 which dissolves in Hg. As a result, Hg starts sticking to the glass. This is called
d

tailing of mercury,
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(vi) O3 is used as a disinfectant and as a germicide. It is also used for purifying air in crowded places such as
cinema halls, tunnels and underground railway stations,
F

(vit) O2 is paramagnetic but O3 is diamagnetic. The central oxygen atom in O3 is sp^-hybridized with
ZOOO = 117® and O—O bond length = 128 pm.
40. Sulphur exists in three allotropic forms, i.e., rhombic, monoclinic and plastic sulphur. Rhombic sulphur is
however, the most stable form of sulphur.
(0 Both rhombic and monoclinic sulphur consist of 8-membered puckered irngs but differ from each other
in the manner of packing of the molecules in the crystal lattice.
(ii) Plastic sulphur consists of long zig-zag chains of sulphur atoms obtained by opening of Sg irngs.
(Hi) Both rhombic and monoclinic sulphur are soluble in CS2 but plastic sulphur is insoluble.
41. Sulphur forms a number of oxoacids such as H2SO3, H2SO4, H2S2O3, H2SO5, H2S20g, H2S2O5 and H2S2O7.
Of these, H2SO4 is very important and is called king of chemicals because of its large number of applications.
42.
H2SO4 is manufactured by contact process. In this method, SO2 obtained by burning S or iron pyrites
(FeS2) is oxidised to SO3 in presence of Pt or V2O5 as catalyst. V2O5 is, however, preferred since it is much
cheaper than Pt and is also not poisoned by arsenic impurities usually present in SO2.
7/106 New Course Chemistry fXinpzsTMi

SO3 thus obtained is absorbed in 98% H-)S04 to give oleum (H',8907) which on dilution with calculated
amount of water gives H2SO4 of desired concentration.
Cone. H2SO4 containing dissolved SO^ is called fuming sulphuric acid or oleum.
43. H2SO4 has high b.p. due to H-bonding. It acts as a .strong dibasic acid. Metals more electropositive than
hydrogen react with dil. H2SO4 to evolve H2 gas but less electropositive metals on heating with cone.
H2SO4 evolve SOt.
(0 H2SO4 acts as an oxidising agent though weaker than HNO3.

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(//) H2SO4 liberates HCl from chlorides, H2S from sulphides, HNO3 from nitrates, CO^ from carbonates
and bicarbonates and CO from K4[Fe(CN)(^].
(Hi) H2SO4 behaves as a .strong dehydrating agent and as such dehydrates sugar to sugar charcoal, formic
acid to CO, oxalic acid to CO + CO2 and ethyl alcohol to ethylene.

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111. Group 17 Elements

44. Group 17 consists of elements ; F. Cl, Br. I and At having ns~ np\ np^ np\ where n = 2 - 6 as the valence

F
Fl
shell electronic configuradon. These elements are collectively called as halogens.

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45. (/) All halogens uniformly show an oxidation state of - 1. Fluorine shows only one positive oxidation state
of + 1 in HOF while all other halogens show positive oxidation states of + 1, + 3, + 5 and + 7.

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(//) Cl has the most negative electron gain enthalpy while the negative electron gain enthalpies of other

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halogens decrea.se in the order :
o F>Br>l.

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(Hi) The colour of the halogens deepens down the group : F2 (pale yellow), CU (greenish yellow), bromine
(reddish brown) and iodine (deep violet).
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(iv) Enthalpy of dissociation of halogens decrease in the order :
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CI2 > Br2, > F2 > G

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(v) Oxidising power of halogens dccrea.scs in the order ; F^ > CI2 > Br2 > l2-
Thus, F2 oxidises all other halide ions to halogens :

F^ + 2 X- ^2F- + X, (X = CI. Br. I)

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CI2 will oxidise Br“ ions to Br2 and I~ ions to U while Br-) will oxidise only 1“ ions to I-,,
(vi) The reducing power of halide ions decreases in the order : 1“ > Br" > Cl" > F"
in

(vii) CU, Br2 and I2 show disproportionation reactions since these halogens show three oxidation states ;
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one lower than 0, i.e., - 1 and one higher that 0, i.e., + 1, + 3 or + 5. For example,
F

0 + 1
Cold
Cl., +2NaOH ■> NaCl + NaOCl + H->0

0 + 5
Hoi
3 Cl, + 6 NaOH 5 NaCl + NaC103 + 3 H2O
In contrast, F2 does not show the.se disproportionation reactions since F2 does not show positive oxidation
states of + 3 and + 5. It. however, reacts with dil. NaOH solution forming OF-)
2F, + 2NaOH ^ 2 NaF + H2O + OF2
and with cone. NaOH, it gives O2
2 F2 + 4 NaOH ^ 4 NaF + 2 H2O + O2
The only true dispropoitionation reaction of F2 is its reaction with ice at low temperature.
0 233K -1 + i

F, + H2O (ice) T ± UF + HOF

 7/107
p-BLOCK ELEMENTS

46. Hydrogen halides. All the halogens react with hydrogen forming hydrogen halides. The order of reactivity
being : F2 > CIt > Br^ > I->. HF and HCi can be prepared in the laboratory by the action of cone. H2SO4 on
CaF2 and NaCf respectively. However. HBr and HI cannot be prepared by the action of cone. H2SO4 on
NaBr and Nal respectively. The reason being that HBr and HI are moderately strong reducing agents and
hence reduce H2SO4 to SOi while they themselves are oxidised to Bt2 and U respectively.
{/) Because of strong intcrmolecular H-bonds. HF is a liquid while all other halogen acids arc gases.
((7) Boiling points : HF > HI > HBr > HCI
(Hi) Melting points : HI > HF > HBr > HCI
(iV) Dipole moments : HF > HCI > HBr > HI
(v) Bond lengths : HI > HBr > HC! > HF
(vi) Thermal .stability : HF > HCI > HBr > HI
(w7) Acid strength : HI > HBr > HCI > HF

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(viii) Reducing power : HI > HBr > HCI > HF.
47. Preparation halogens. (/) F-> was first prepared by Moissan by electrolysis of a molten mixture ol KHF2 +

F lo
HF (1:5) in a vessel made up of monel metal (Ni-Cu-Fc alloy) using carbon electrodes. Recently it has been
prepared chemically by reacting pota.ssium hexafluoromanganate (IV), K2MnF^ with the Lewis acid antimony
pentafluoride, Sbp5.

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(/;) C!->, Br2 and I2 can be prepared in the laboratory by healing a mixture of NaCl + Mn02, NaBr + Mn02,

F
KI + MnOo with cone. H2SO4.
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48. Oxoacids. Ft does not form oxoacids while all other halogens form oxoacids : HOX, HXO2, HXO^, HXO4
(X = Cl, Br. I).
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(i) Acid strength : HCIO4 > HCIO^ > HCIO2 > HCIO and HOCl > HOBr > HOI
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((7) Oxidising power : HOCl =: HOBr > HOI and BrO^ > IO4 > CIO^
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49. Interhalogen compounds. Halogens react with each other to form interhalogen compounds of four types,
i.e., XX' (CIF, BrF, BrCI, ICl, IBr, etc.) XX'3 (CIF,. BrF3. IF3. ICI3, etc.), XX'5 (CIF5. B1F5, IF5, etc.) and
XX'7 (IF7).
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On the basis of VSEPR theory, XX'3 molecules have T-shaped, XX'5 molecules have square pyramidal
while XX'7 molecules have pentagonal bipyramid geometry.
50. Polyhalides. Due to the absence of J-orbitals. fluorine does not form polyhalides while all other halogens
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do. For example. I2 dissolves in KI to form IJ ion.

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51. Pseudohalides (CN“, SCN", OCN") possess properties similar to halide ions and contain at least one N
atom. The corresponding dimeric molecules such as (CN)2, (OCN)2, (SCN)2 are called pseudohalogens.

IV. Group 18 Elements

52. Group 18 consi.sts of elements : He. Ne, Ar, Kr, Xe and Rn having ns^ np^ (except He whose electronic
configuration is i.v-) where n = 2 - 6 as the valence shell electronic configuration. All these elements are
also called noble gases or inert gases or the rare gases of the atmosphere or simply aerogens.
53. (0 Most abundant noble gas : Ar (~ 1%) by volume of the atmosphere.
(//) Melting and boiling points : He < Ne < Ar < Kr < Xe < Rn.
(iii) Ionization enthalpy : He > Ne > Ar > Kr > Xe > Rn
(iv) Electron gain enthalpy : Ne (+ 116) > Ar (+ 96) = Kr (+ 96) > Xc (+ 77) > Rn (+ 68) > He (+ 48)
(v) Polarizability, Ease of liquefaction and Solubility in H2O : He < Ne < Ar < Kr < Xe < Rn.
(v/) Clathrate formation : He and Ne do not form clathrate compounds while all other noble gases do.
7/108 New Course Chemistry (XII)EZ

54. Neil Bartlett observed that PtFg oxidises O2 to give

Ptl^ and Xe to give Xe'*’PtI^ because the ionization
enthalpy of O2 (1175 kJ mol"*) and Xe (1170 kJ moH) are very close.
55. The more polarizable noble gases such as Kr and Xe form compounds with strong oxidising agents such as

oxygen and fluorine.

56. Xe forms three fluorides (Xep2, Xep4, XePg), two oxides (Xe03, Xe04) and a large number of oxyfluorides
(XeOp2, XeOp4, Xe02p2, Xe03p2, etc.) TTieir shapes and geometries can be easily predicted on basis of
hybridization and VSEPR theory : Xep2, .{sp^d, linear), XP4 {sp^d^, square planar), XPg {sp^S, distorted
octahedral), Xe03 {sp^, pyramidal), Xe04 {sp^, tetrahedral), etc.
57. Liquid helium (b.p. 4-2 K) is used as cryogenic agent for carrying out various experiments at low temperatures.

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Oxygen-helium mixture is used for artificial respiration. Neon is mainly used in discharge tubes and in
fluorescent lamps for advertising purposes. Argon is used mainly to create inert atmosphere in high
temperature metallurgical processes. It is also used in filling incandescent and fluorescent lamps.

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p-BLOCK ELEMENTS 7/109

Competition
it

■ m
NE^/JEE
SPECIAL

For ultimate preparation of this unit for competitive examinations, students should refer to
● MCQs in Chemistry for MEET
Pradeep's Stellar Series.... ● MCQs in Chemistry for JEE (Main)

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separately available for these examinations.

F lo
Multiple Choice Questions (with one correct Answer)

(c) + 2 (d) + 5

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I. Group 15 Elements

Fre
(JEE Main 2020)
1. Nitrogen gas is obtained by thermal decomposition 5. On heating compound (A) gives a gas (B) which
of for
is a constituent of air. This gas when treated with
(a) Ba(N03)2 (h) BafNj). in the presence of a catalyst gives another gas
((5) which is basic in nature. (A) should not be
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(c) NaNO^ (f/) NaNOj
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(JEE Main 2022) (a) Pb{N03)2 ib) (NH_j)2Cr207

(c) NH4NO2 (d) NaNj
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2. The compound that does not produce nitrogen gas

(JEE Main 2021)
by thermal decomposition is
(b) (NH4)2Cr207 6. The hydrolysis of NCI3 by H2O produces
(a) Ba(N3)2
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(c) NH4NO2 id) (NH4)2S04 ia) NH20HandHCl (b) NH2NH2 and HCl
(JEE Main 2018) (c) NH4OH + HOCl (r/) NH2CI and HOCI
7. A set that represents the pair of neutral oxides of
3. Which ordering of compounds is according to
nitrogen is
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decreasing order of the oxidation state of nitrogen?

(a) NO and N^O (b) NO and NO2
ia) HNO3, NO. NH4CI, N2
Fin

(c) N2O and NO2 id) N2O and N2O3

(b) HNO3. NO, N2, NH4CI (JEE Main 2021)
(c) HNO3. NH4CI, NO, N2 8. The correct order of the acidic nature of oxides is
(d) NO. HNO3. NH4CI, Nt in the order

(ilT Paper I, 2012 ; NEET 2018) (a) NO < N2O < N2O3 < NO2 < N2O5
4. On heating lead (11) nitrate gives a brown gas (A). (b) N2O < NO < N2O3 < NO2 < N2O5
The gas on cooling gives a colourless solid/liquid (c) N2O5 < NO2 < N2O3 < NO < N2O
(B). (B) on healing with NO changes to a blue (d) N^Og < N2O3 < NO2 < NO < N2O
solid (C). The oxidation number of nitrogen in 9. The brown ring lest for nitrates depends upon
solid (C) is (a) the reduction of nitrate to nitric oxide
(a)+ 3 ib) + 4 ib) oxidation of nitric oxide to nitrogen dioxide

1. {b) 2. {d) 3. ib) 4. [a) 5. (a) 6. (c) 7. («) 8. (fc)

7/110 ^>uiuUe^'4. New Course Chemistry (XII)EEIMI

(c) reduction of ferrous sulphate to iron 14. Sulphury! chloride (SO2CI2) reacts with white
(d) oxidising action of sulphuric acid phosphorus (P4) to give
2+ ;
10. lFe{H20)5N0] is a complex formed during the (a) PClg,S02 (b) PCI3, SOCI2
brown irng lest for NOJ ion. In this complex, (c) PClg, SO2. S2CI2 (d) POCI3, SO2, S2CI2
(West Bengal JEE 2015)
(a) there are three unpaired electrons so that its
magnetic moment is 3-87 B.M. 15. White phosphorus reacts with thionyl chloride to
give
{/;) NO transfers its electron to so that iron
exists as Fe (I) and NO as NO"^ («) PCI5, SO2 and S2CI2
(f) the colour is because of charge transfer ih) PCI3, SO2 and S2CI2
(d) all of the above statements are correct (c) PCI3, SO2 and CI2
(AIIMS 2015) (d) PClg, SO2, CI2 (JEE Main 2022)

11. Match List-1 with List-II 16. Identify the incorrect statement related to PClg
from the following :

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List-I List-II (a) PClg molecule is non-reactive
(Name of the 0x0 acid) (Oxidation state {b) Three equitorial P-Cl bonds make an angle

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ofP) of 120® with each other

(c) Two axial P-d21 bonds make an angle of 180°

(A) Hypophosphorus acid (/) + 5

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with each other
(B) Orthophosphoric acid (ii) + 4

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(</) Axial P-Cl bonds are longer than equatorial
(C) Hypophosphoric acid (in) + 3 P-Cl bonds (NEET 2019)
(D) Orthophosphorus acid ((V) 2
for
17. What will be the resultant products formed when
the phosphorus halide PBrg splits up ?
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(V) + 1
(«) [PBr4]’^ and Br“ (b) [PBr^J"* and LPBr4]’*'
(a) (A) - (/v), (B) - (v), (C) - (ii), (D) - (Hi)
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(c) [PBr4]-^ id) IPBr^]-
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(b) (A) - (/v), (B) - (i), (C) - (ii), (D) - (Hi)

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(c) (A) - (v), (B) - ((V), (C) - (i7), (D) - (Hi) II. Group 16 Elements
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(d) (A) - (v), (B) - (/). (C) - (;7), (D) - (Hi) 18. Which is the correct thermal stability order for H2E
(JEE Main 2021) (E = O, S, Se. Te and Po)
12. The reaction of white phosphorus with aqueous («) H2Se < H2Te < H2P0 < H2O < H2S
r
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NaOH gives phosphine along with another (b) H2S < H2O < H2SC < H^Te < H2P0
compound. The reaction type ; the oxidation stales (c) H2O < H2S < H2Se < H2Te < H2P0
Y

of phosphorus in phosphine and the other product (d) H2P0 < H2Te < H2Se < H2S < H2O
are respectively
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(NEET 2019)
nd

(a) redox reaction ; - 3 and - 5

19. Reaction of an inorganic sulphite X with dilute
Fi

(h) redox reaction ; -1- 3 and + 5

H2SO4 generates compound Y. Reaction of Y with
(f) disproportionation reaction ; - 3 and - 5 NaOH gives X. Further, the reaction of X with Y
(d) disproportionation reaction ; - 3 and + 1 and water affords compound Z. Y and Z
(IIT Paper II, 2012) respectively are ;
13. Which is the correct statement for the given acids ? (a) SO, and NaHS03 (b) S and Na,S03
(a) Phosphinic acid is a monoproiic acid while (c) SO, and Na2S03 (d) SO3 and NaHSOg
phosphonic acid is a diprotic acid (JEE Main 2020)
(h) Phosphinic acid is a diprotic acid while 20. In which pair of ions both the species contain S-S
phosphonic acid is a monoprotic acid bond ?
(c) Both are diprotic acids
(d) Both are triprotic acids (NEET Pha.se-12016) (a) S4O2-, S2O5- (b) s,o^-, S2O2-
ANSWERS
9. Ui) 10.(d) II. id) 12. (d) 13.la) 14. (a) 15. (b) 16. (a) 17.(a) 18. (d)
19. («)
p-BLOCK ELEMENTS 7/111

(c) S^Ol~,Sp^f (d) sp^~, spj- NaOH

(hoi (iiul coiic.)
+ CI2 ^ (A) + Side products
(NEET 2017)
Ca(OH), +CI2 (B) + Side products
21, Which of the following oxo acids of sulphur (dry)
containssulphurin two differentoxidationstates?
(a) NaC103 Ca{C103)2
(a) H2S2O3 {b) H2S2O6
(b) NaOCI and €3(0103)2
(c) H2S2O2 {d) H2S-)Og
(c) NaCIO, and Ca(OCI)2
(JEE Main 2022)
((/) NaOCI and Ca(OCi)2 (.lEE Main 2020)
in. Group 17 Elements 28. Total number of lone pairs of electron in ion
22. Which of the following orders is correct for the IS

bond dissociation enthalpy of halogen molecules ? (a) 3 (b) 6

(a) Br2 > I2 > Ft ^ F2 > CI2 > Br-, > I, (c) 9 (d) 12
(c) U > Bf2 > CI2 > F2 id) Cl, > Br, > F2 > I2 (JEE Main 2018)

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(NEET Phase-I 2016 ; JEE Main 2021) 29. The correct statement for the molecule, CSI3, is
23. Which among the following factors is most (a) it contains Cs^, 1“ and lattice I2 molecules

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important in making fluorine the strongest (b) it is a covalent molecule
oxidising agent ?
(c) it contains Cs^ and IJ ions
(a) electron affinity (b) ionization energy

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(d) it contains Cs^'*’ and 1“ ions

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(c) hydration energy
(JEE Main 2014)
(d) bond dissociation energy
30. In the structure of CIF3, the number of lone pairs
24. The shape of O2F2 is similar to that of
(«) C2F, (b) H2O2
for
of electrons on central atom ‘Cl’ is
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(a) one (b) two
(c) H2F2 id) C,H,
(c) four (d) three (NEET 2018)
s
25. Match the following :
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31. Match the interhalogcn compounds of colunin-1

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(A) Pure nitrogen (/) Chlorine with the geometry in column-Il and assign the
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(B) Haber process (ii) Sulphuric acid correct code.

(C) Contact process (m) Ammonia Column I Column II

(D) Deacon’s process (/v) Sodium azide or (A) XX' (0 T-shape

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Barium azide
(B) XX' (ii) Pentagonal bipyramidal
Which of the following is the correct option?
(C) XX' (f77) Linear
(A) (B) (C) (D)
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(D) xx;
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(a) (/v) (Hi) (ii) (i) (iv) Square pyramidal

(b) (i) (ii) (Hi) (iv) (v) Tetrahedral
Fin

(c) 07) Ov) (i) (Hi) Code :

(d) (Hi) (iv) (H) (i) (NEET 2019) A B C D

26. Among the following oxoacids, the correct (a) (Hi) (i) (/v) (H)
decreasing order of acid strength is (b) (v) (iv) (Hi) (H)
(a) HCIO, > HCIO4 > HCIO3 > HOC! (c) (iv) (Hi) (H) (i)
(b) HOCI > HCIO2 > HCIO3 > HOCI4 (d) (Hi) (iv) (i) (ii) (NEET 2017)
(c) HCIO4 > HOCI > HCIO2 > HCIO3 32. The correct statement about ICI5 and ICI4 is
(d) HCIO4 > HCIO3 > HCIO2 > HOC! (fl) both are isostructural
(JEE Main 2014, NEET Phase-I 2016)
27. In following reactions, products (A) and (B)
(b) ICI5 is square pyramidal and ICI4 square
respectively, are planar

ANSWERS
20. (fl) 21. (a) 22. (d) 23. (c) 24. (b) 25. (a) 26. (d) 27. (c) 28. (r) 29. (c)
30. (b) 31. (a)
7/112 “Pn^eideefi.'^ New Course Chemistry (XIl)KSSQO!

(c) ICI5 is trigonal bipyramidal and ICI4 is (c) T-shape and linear
octahedral {d) square planar and trigonal bipyramidal
{d) ICI5 is square pyramidal and ICI4 is (AIIMS 2015)
tetrahedral (JEE Main 2019) 38. The correct geometry of hybridization for XCF4 is
33. Match List-I with List-II {a) Octahedral, sp^d^
List-I List-II
ib) trigonal bipyramidal, sp^d
(A) PCl^ (/) Square pyramidal (c) planar triangle, sp^d^
(B) SF^ (//) Trigonal planar
{d) square planar, sp^d~ (NEET Phase-II 2016)
(C) BrFg (iVi) Octahedral
39. Match the compounds given in column I with the
(D) BF3 (/v) Trigonal bipyramid
hybridization and shape given in column II and
Choose the correct answer from the options given mark the correct option.
below :
Column 1 Column II
(a) (A) - (iv), (B) - (i). (C) - (ii), (D) - (Hi)
(A) XeF^ {/) Distorted octahedral

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(b) (A) - (iv), (B) - (Hi), (C) - (0, (D) - (H)
(B) Xe03 (ii) Square planar
(c) (A) - (H), (B) - (iv), (C) - (Hi), (D) - (0

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(C) XeOF4 (Hi) Pyramidal
(d) (A) - (/), (B) - (Hi), (C) - (iv), (D) - (i)
(NEET 2021) (D) XeF4 (/v) Square pyramidal
Code :
34. The reaction of HCIO3 with HCl gives a

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A B C D
paramagnetic gas, which upon reaction with O3

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produces (a) (iv) (Hi) (i) (H)
(fl) CI2O (b) CIO2 (b) (iv) (i) (H) (Hi)
(c) CljOf, (d) CI2O7 (c) for(i) (Hi) (iv) (ii)
r
(JEE Advanced 2022) (d) (i) (H) (iv) (Hi)
You
35. Noble gases are named because of their inertness (NEET Phase-I 2016, 2019)
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towards reactivity. Identify an incorrect statement 40. Match the compounds of Xe in column I with the
about them. molecular structure in column II.
eB

(a) Noble gases are sparingly soluble in water Column I Column II

(b) Noble gases have very high melting and (A) Xep2 (i) Square planar
our

boiling points
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(B) XeF4 (H) Linear

(c) Noble gases have weak dispersion forces. (Hi) Square pyramidal
(C) Xe03
(d) Noble gases have large positive values of (iv) Pyramidal
(NEET 2021)
(D) XeOF4
electron gain enthalpy
dY

(a) (A) - (ii). (B) - (i), (C) - (iv), (D) - (Hi)

Re

IV. Group 18 Elements-Noble Gases (b) (A) - (ii), (B) - (i), (C) - (Hi), (D) - (iv)
Fin

36. What is the hybridization and geometry of the (c) (A) - (ii), (B) - (iv), (C) - (Hi), (D) ^ (i)
given species ? The species are Xep2 and ICI2 (d) (A) - (ii), (B) - (Hi), (C) - (i), (D) - (iv)
(a) sp^ d and trigonal brpyramid (NEET 2020)

(b) sp^ d^ and square planar 41. The shape/structure of [XeFjl and Xe03p2
(c) sp^ d and linear respectively are
(d) sp^ and irregular tetrahedron (a) pentagonal planar and trigonal bipyramid
(J & K CET 2018) (b) trigonal bipyramidand pentagonalplanin'
37. The shapes of SF4 and Xep2 respectively are (c) octahedral and square pyramidal
(a) trigonal bipyramidal and trigonal bipyramidal (d) trigonal bipyramidal and pentagonal planar
(ib) see-saw and linear (JEE Main 2020)

ANSWERS

32. (/;) 33. ((;) 34. (c) 35. [h) 36. (c) 37. (b) 38. (d) 39. (c) 40. (a) 41. (a)
p-BLOCK ELEMENTS 7/113

42. In which of the following pairs both the species 44. Xenon hexafluoride on partial hydrolysis produces
are not isostructural ? compounds ‘X’ and ‘Y’ and the oxidation state of
(fl) Diamond, silicon carbide Xe are respectively
(/?) NH3, PH3 (a) Xe02F2 (+ 6) and XeO, (+ 4)
(c) XeP4. XeO^ (b) XeOF^ (+ 6) and Xe02F2 {+ 6)
(f) XeOF4 (+ 6) and Xe03 (+ 6)
(d) SiCl4, PClJ {RE-AIPMT 2015)
(d) XcOt (+ 4) and Xe03 (+ 6)
43. Which of the following reactions is an example (JEE Main (Online) 2018)
of a redox reaction ?
45. Which one of the following reactions of xenon
(a) XeFfi + FI2O — XeOp4 + 2 HF compounds are noi feasible ?
ib) XeF^ + 2 H2O —> Xe02F2 + 4 HF {«) 3 XeF4 + 6 H^O
(c) Xep4 + O2F2 - —^ XeFf; + O7 2Xe + Xe03+ I2HF+ 1.5 O2

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(d) XeF2 + PFg — ^ [XeF]+PF^ (b) 2XeF2 + 2H20 2 Xe + 4 HF + O7
(c) XeFg + RbF ^ Rb [XeF^l
(JEE Main 2017)
{d) XeOg + 6 HF

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XeFg + 3 HoO

m Multiple Choice Questions (with One or More than One Correct Answers)

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46. Which of the following pairs of ions is (c) N2O4 (d) N2O5
isoelectronic as well as isostructural ?
(IIT Paper II 2009)
(a) COj-, NO" (b) CIO3, CO| for
51. The correct statement(s) about the oxoacids,
ur
HCIO4 and HCIO, is(are)
(c) sq2-, NOJ id) CIOJ, SOf- (a) the conjugate base of HCIO4 is weaker base
s
(NEET Phase-II 2016)
ook

than H7O
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47. The compound(s) which generate(s) Nt gas upon (/;) the central atom in both HCIO4 and HCIO is
thermal decomposition below 300°C is (are)
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sp^ hybridized
(a) NH4NO3 (b) (NH4)2Cr707 (c) HCIO4 is formed in the reaction between CI7
(c) Ba(N3)2 (d) Mg3N2 and H2O
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(JEE Advanced 2018)

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(d) HCIO4 is more acidic than HCIO because of

48. Based on the compounds of group 15 elements, the resonance stabilization of its anion
the correct state(s) is (are) (JEE Advanced 2017)
Y

(a) Bi20^ is more basic than N2O5 52. The correct statenient(s) regarding (/) HCIO,
Re

(b) NF3 is more covalent than BiFg

nd

(//') HCIO2, (///) HCIO3 and (/v) HCIO4, is (arc)

(c) PH3 boils at lower temperature than NH3 (a) The number of Cl = O bonds in (//) and {Hi)
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(d) the N-N single bond is stronger than P-P together is two
single bond (JEE Advanced 2018) (h) The total number of lone pairs of electrons on
49. The nitrogen containing compound produced in Cl in Hi) and (Hi) together is three
the reaction of HNO3 with P4O1Q (c) The hybridization of Cl in (/v) as sp^
(a) can be prepared by reaction of P4 and HNO3 (d) Amongst (i) to (/v), the strongest acid is (/)
(b) is diamagnetic (c) contains N~N bond
(JEE Advanced 2015)
(d) reacts with Na-metal producing brown gas 53. The compound with two lone pairs of electrons
LE Advanced 2016) on the central atom is (are)
50. The nitrogen oxide(s) that ^ ●mains N—N bond(s) (a) BrFg (b) CIF3
is (are)
(c) XeF4 (d) SF4
(a) N2O (b) N2O3 (JEE Advanced 2016)

ANSWERS
42. (r) 43. (c) 44. (b) 45. (d) 46. (a.d) 47. (b,c) 48. (a.b.c)
49. (b,d) 50. (a.b.c) 51. (a,b.d) 52. (b,c) 53. (b,c)
7/114 ‘pKzdee^ 4- New Course Chemistry (XII)gElBl
54. Each of the following options contains a set of 55. With respect to hypochlorite, chlorate and
four molecules. Identify the option(s) where all perchlorate ions, choose the correct statement(s).
four molecules possess permanent dipole moment (a) Hypochlorite ion is the strongest conjugate
base
at room temperature.
(b) The molecular shape of only chlorate is
{«) NO2, NH3, POCI3, CH3CI iniluenced by the lone pair of electrons of Cl
ib) BF3,03,SF6,XeF6 (c) The hypochlorite and chlorate ions dispro
(c) BeCl3, CO2, BCI3, CHCI3 portionate to give rise to identical set of ions
(d) SO2, C^HgCl, H2Se, BrFj (d) The hypochlorite ion oxidises sulphite ion
(JEE Advanced 2020)
(JEE Advanced 2019)

Multiple Choice Questions (Based on the given Passage/Comprehens ion)

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The comprehension given below is followed by some multiple choice questions. Each question has
one correct option. Choose the correct option.

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DUpon heating KCIO3 in
interhalogens to form polyhalide ions
consisting either of the same halogen or of two
the presence of catalytic amount of Mn02, a or three different halogens. Besides these, a

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gas W is formed. Excess amount of W reacts few other anions are known, which do not

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with white phosphorus to give X. The reaction contain any of the halogen atoms but behave
of X with pure HNO3 gives Y and Z. like halide ions. These anions are called

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pseudohalides and consist of two or more
ur
56. Y and Z are, respectively atoms of which one is always a nitrogen atom.
(fl) N2O5 and HPO3 (h) N2O3 and H3PO4 58. Which one of the following is not a pseudohalide?
s
(c) N2O4 and H3PO3 id) N2O4 and HPO3
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(b) RCOO"
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(a) CNO"
57. W and X are, respectively (c) OCN- (d) NNN-.
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(a) O2 and P4O6 ib) O2 and P40jq 59. The isoelectronic pair is
(c) O3 and P4O6 id) O3 and P4O10 (a) CI2O.ICI- ib) ic\-;,c\o^
(JEE Advanced 2017)
r

ic) n^,i3 id) C10-,CII^

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Halogens combine with 60. The correct order of pseudohalide, poly halide and
Y

each other to form interhalogen compounds inlerhalogen are

(XX',XX',XX5 and XX^). Halide ions (a) Brl2,OCN-,IFg ib) IF5, Brl2 , OCN"
Re
nd

often react with molecules of halogens or (c) OCN-, IF5, Bri; id) OCN-, Bri; , IFg
Fi

Iviatching Type Questions

Match the entries of Column I with appropriate entries of Column II and choose the correct option
out of the four options (ct), ib), ic) and (rf) given at the end of each question.
61. Column I Column II

(A) PCIg ip) Anguliu

(B) IF7 iq) Pyramidal
CC) H3O+ (/■) Trigonal bipyramidal
(D) CIO2 (^) Pentagonal bipyramidal
(a) A-r, B-i, C-q, D-p (b) A-r, B-q, C-p, D-^ (c) A-p, B-s, C-q, D-r id) A-s, B-p, C-r, D-q

54. ui.tii 56. : 57.i/n 58. (/>) 59. id) 60. Id) 61.(a)
p-BLOCK ELEMENTS 7/115
62. Column I - «)luinii II

(A) Helium ip) Ionization energy comparable to Ot

(B) Argon (q) Provides inert atmosphere in metallurgy
(C) Neon (r) Cryogenic
(D) Xenon (s) Advertising sign

(a) A-q, B'S, C-p, D-r (h) A-q, B-p, C-i, D-r (c) A-r, B-q, C~s, D-p (d) A-r, C-q, D-p
63. Column ] ! (dtirnn II

(A) SFg ip) Undergoes disproportionation in alkaline

medium

(B) H3PO2 iq) Liberates CI2 from KCI

(C) P4 (r) Does not undergo hydrolysis by water
(D) F, (s) Reduces diazonium salts

(a) A-r, B-5, C-q, D-p ib) A-r, B-s, C-p, D-q (c) As, B-q, C-r, D-p id) A-q, B-s, C-p, D-r

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64.
The unbalanced chemical reactions given in List I show missing reagent or condition(?) which

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are provided in List II. Match List I with List II and select the correct answer using the code
given below the lists.
List I M.S: II

ree
?
P. PbOo + H2SO4 > PhS04 + O2 1. NO

Q- Na2S203 + H,0 —> NaHSO 4

7

2.
for F
7
R.
N2H4 N2 + other product 3. Wtirm
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S. XeF2 Xe + other product 4. CI2

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P Q R S
ia) 4 2 3 1

(b) 3 9 I 4
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(c) 1 4 2 3

id) 3 4 2 1 (HT Advanced 2013)

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01
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Matrix-Match Type Questions

Fin

p q r s
1 I 1
Match the entries of column I with appropriate entries of column II. Each
entry in column I may have one or more than one correct option from * ©il@l!OIG
column n. If the correct matches are A-p, s ; B-r, C-p, q ; D-s, then the
correctly bubbled 4x4 matrix should be as follows :
B
©ll@ilO!©
65. Column 1 Column II i©li@!iOi ©i
(A) SO2 ip) Basic D
©II© I© s

(B) H2SO4
J L
iq) Acidic
(C) HNO 3 (r) Reducing
(D) NH3 (i) Oxidising

62.(r) 63. (/') 64. ul) 65. -A-'f /■'

7/116 ^n/nctecfr'^ New Course Chemistry (XI1)CS3S]

66. Column I Column II

(A) H3PO3 — (;>) One of the products acts as a reducing agent.

A
(B) PCI3+H2O Kci) One of the products is a tribasic non-reducing acid.
(C) NO-,+ H-,0 ■> (/●) Dehydration
A
(D) HNO3 -H P4OJ0 (.v) In one of the products, central atom has -i-5
oxidation state.

67. Column I (Reaction) Column II (Product)

(A) Cu + dil. HNO3 (P) NO

(B) Cu + cone. HNO3 Ui) NO,
Zn + dil. HNO3 (/■) N2O

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(C)
(D) Zn + cone. HNO3 (^) Cu(N03)2
(r) Zn(N03)2 (IIT Paper II, 2009)

VI. Integer Type Questions

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F
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A B C D
DIRECTIONS. The answer to each of the following questions is a single
digit integer, ranging from 0 to 9. If the correct answers to the question
numbers A, B, C and D (say) are 4, 0, 9 and 2 respectively, then the
® @®®
for
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correct darkening of bubbles should be as follows : ® 000
68. The maximum number of oxidation states which nitrogen can show in its compounds @ ®®®
k s
® ®®®
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IS.
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69. The total number of lone pairs of electrons in N2O3 is (JEE Advanced 2015)
0 0 00
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70. Amongst the following, the metals which become passive when dipped in cone.
HNO3 are Sn, Pb. Fe, Cr, Zn, Ni, Hg, Al, Cu. ® ®©®
71. What is the basicity of hypophosphorus acid ?
r

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72. X is a pale yellow solid. It hydrolyses to POCI3 in moist air and finally gets converted
into phosphoric acid. X exists as an ionic solid. The total number of atoms present 0 ®00
Y

in its cation is.

® ®®®
nd

73. Among the following, the number of compounds that can react with PCI5 to give
Re

(IIT Paper II, 2011)

® ®®®
POCI3 is : 0,, CO2, SO2, H2O, H2SO4, P40iy.
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74. The difference in the oxidation number of two types of sulphur atoms in Na2S40^ is (IIT Paper I, 2011)
75. The total number of diprotic acids among the following is
H3PO4, H2SO4, H3PO3, H2CO3. H2S2O7, H3BO3, H3PO2, H2Cr04, H2SO3 (IIT Paper II, 2010)
76. Among the following, the number of elements showing only one non-zero oxidation state is
O. Cl. F, N, P. Sn, Tl, Na. Ti (IIT Paper II, 2010)
77. What is the oxidation state of chlorine in HCIO4 ?
78. The ratio of number of dii-pTi bonds in Xe04 and SO3 is.
79. Amongst the following, the maximum number of isoelectronic molecules/ions are
XeOj, BrO" . CIF, Xep2, OF2, Xep4, ICI4 , CIO", IBr2" .
ANSWERS

66. {A-p,q,s ; B-p : C-p,s ; D-.rs) 67. (A-p,s ; B-q,s; C-,rt: D-q.t) 68.(9) 69.(8) 70.(4)
71. (1) 72.(5) 73.(6) 74.(5) 75.(6) 76.(1) 77.(7) 78.(2) 79.(3)
p-BLOCK ELEMENTS 7/117

80. Among the triatomic molecuies/ions, BeClj, N", N^O. NO+ , O3, SCI2. ICl”, l~ and Xep2, the total
number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution
from the c/-orbitaI (s) is (JEE Advanced 2015)
81. The sum of the number of lone pairs of electrons on each central atom in the following species is
[TeBrg]2-, [BrF2]+, SNF3 and [XeF3J-
(Atomic numbers : N = 7. F = 9, S = 16, Br = 35, Te = 52, Xe = 54) (JEE Advanced 2017)
82. The total number of compounds having at least one bridging 0x0 group among the molecules given below i
IS

N2O3, N2O5. P4O6, P4O7, H4P2O5, H3P3O10, H2S2O3, H.S,0 (JEE Advanced 2018)
83. Consider PF^, BrF^, PCI3, SF^^, ICI4 , CIF3 and IFg. Among these moIecule(s)/ion(s) having sp^c'P hybridization
IS
(JEE Main 2022)
VII.
Numerical Value Type Questions Decimal Notation)

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For the following question, enter the correct numerical value (in decimal notation, truncated/rounded off
to the second decimal place ; e.g., 6-25, 7-00, - 0-33, - 0-30, 30-27, - 127-30) using the mouse and the on

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screen virtual numeric keypad in the place designated to enter the answer.
84. A mixture of 0-1 mole of Nad and Mn02 was heated with excess of cone. H2SO4. The CI2 thus obtained
was passed through a concentrated aqueous solution of Kl. The amount of I2 (in g) produced is

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VIII.
Assertion-Reason Type Questions
TYPE I for
ur
DIRECTIONS. Each question contains Statement-1 (Assertion) and Statement-2 (Reason). Each question
s
has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correct option as
ook
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under :
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(a) Statement-1 is TVue, Statement-2 is True ; Statement-2 is a correct explanationfor Statement-1,

(b) Statement-! is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statenient-1.
(c) Statement-1 is True, Statement-2 Is False,
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(d) Statement-I Is False, Statement-2 is True.

85. Statement-1. Although NaH2P02 contains two H-atoms, it is not an acid salt.
Statement-2. It contains two ionisable hydrogens.
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86. Statement-I. Sulphur exhibits paramagnetic behaviour in the vapour stale.

nd

Statement-2. In vapour state, sulphur partly exists as S2 molecules which have two unpaired electrons in
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antibonding k* orbital.
87. Statement-1. Salts of CIO3 and CIO4 are well known but those of FOJ and FO4 are unknown.
Statement-2. F is more electronegative than 0 while C! is less electronegativethan O.
88. Statement-1. CIF is more reactive than F2.
Statement-2. The F—F bond is weaker than Cl—F bond.
89. Statement-1. ICl on hydrolysis gies HI and HOCl.
Statement-2. Water can attack iodine more readily.
90. Statement-1. The pentavalent oxides of group 15 elements, E2O5, are less acidic than trivalem oxides, E2O3
of the same element.

Statement-2. The acidic character of group 15 elements, E2O3 decreases down the group.
(JEE Main 2022)
ANSWERS
80. (4) 81. (6) 82. (5 or 6) 83. (4) 84.(12-7) 85. (c) 86. (a)
87. (b) 88. (d) 89. (d) 90. (d)
7/118 New Course Chemistry (XlI)iS2lS]

TYPE H

DIRECTIONS. Each question contains Statement-1 (Assertion) and Statenient-2 (Reason). Each question
has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Choose the correct option as under :
(a) Both Statement-1 and Statement-2 are true, (b) Both Statement-1 and Statement-2 are false.
(c) Statement-1 is true but Statement-2 is false. (d) Statement-1 is false but Statenient-2 is true.

91. Statenient-1. Acid strength increases in the order given as HF « HCl « HBr« HI.
Statement-2. As the size of the elements F, Cl, Br. I increases down the group, the bond strength of HF, HCl,
HBr and HI decreases. (NEET 2022)
92. Statement-1. The boiling points of the following hydrides of group 16 elements increase in the order :
H2O < H2S < H2Se < H2Te.
Statement-2. The boiling points of these hydrides increase with increase in molar mass. (NEET 2022)

TYPE III

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DIRECTIONS. In each of the following questions, a statement of Assertion (A) is given followed by a
CO rresponding statement of Reason (R) just below it. Of the statements, mark the correct answer as

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(a) If both assertion and reason are true, and reason is the correct explanation of the assertion,
(b) If both assertion and reason are true, but reason is not the correct explanation of the assertion,
(c) If a.ssertion is true, but reason is false,

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(d) If both assertion and reason are false.

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93. Assertion. Ozone is destroyed by CFCs in the upper atmosphere.
for
Reason. Ozone holes increase the amount of UV radiation reaching the earth. (JEE Main 2019)
ur
94. Assertion. Nitrogen and oxygen are the main components in the atmo.sphere but these do not react to form
oxides of nitrogen.
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Reason. The reaction between nitrogen and oxygen requires high temperature. {JEE Main 2015)
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95. Assertion. Nitrogen is less reactive than molecular oxygen.

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Reason. The bond length of N2 is shorter than that of oxygen. (AIIMS 2007)
96. Assertion. HNO3 renders iron passive.
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Reason. Iron reacts with HNO3 to form ferric nitrate.

97. Assertion. When NO reacts with FeS04, a brown coloured complex is formed.
Reason. In the complex, the coordination number of Fe is 6. (AIIMS 2009)
Y
Re

98. Assertion. NF3 is a weak ligand than N(CH3)3.

nd

Reason. NF3 ionises to give F” ion in aqueous solution.

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99. Assertion. On cooling, the brown colour of nitrogen dioxide disappears.

Reason. On cooling, NO-, undergoes dimerisation resulting in the pairing of odd electron of NO^
(AIIMS 2013)

100. Assertion. Elementary phosphorus exists in three principal allotropic forms, /.e., white (yellow), red (or
violet) and black.
Reason. Of the three forms, white phosphorus is the most important and most reactive. (AUMS 2015)

101. Assertion. H3PO3 is a dibasic acid.

Reason. There are two H atoms directly attached to P. (AIIMS 2007)

102. Assertion. H—S—H bond angle in H2S is closer to 90° but H—0—H bond angle in H2O is 104-5°.
Reason. Ip-lp repulsion is stronger in H2S than in H2O. (AIIMS 2007)
ANSWERS
91. (o) 92. (/;) 93. (b) 94. (a) 95. {b) 96. (c) 97. (b) 98. (c) 99. (a)
100. ia) 101. (c) 102. (c)
p-BLOCK ELEMENTS 7/119

103. Assertion. F—F bond in F2 molecule is weak.

Reason. F atom is small in size. (AIIMS 2007)

104. Assertion. SiF^ is known but SiCl^ is not.

Reason. Size of F is small and its lone pair of electrons interact with cf-orbitals of Si strongly.
105. Assertion. HOF bond angle is higher than HOCl bond angle in HOX.
Reason. Oxygen is more electronegative than halogens. {AIIMS 2010)
106. Assertion. The ion exists in the solid state and also in liquid state but not in aqueous solution.

Reason. The magnitude of hydrogen bonds in between HF—HF molecule is weaker than that in between HF
and H2O molecules. (AIIMS 2010)
107. Assertion. HCIO4 is a stronger acid than HCIO3.
Reason. Oxidation state of Cl in HCIO4 is +7 and in HCIO3 it is +5.

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108. Assertion. Noble gases have positive electron gain enthalpy.
Reason. Noble gases have stable closed shell electronic configuration. (AIIMS 2010)

F lo
109. Assertion. Xenon forms fluorides.
Reason. Because 5d orbitals are available for valence shell expansion.

ee
110. Assertion. Fluorine forms one 0x0 acid.

Fr
Reason. Fluorine has smallest size amongs halogens and is highly electronegative. (JEE Main 2022)
111. Assertion. ICl is more reactive than I2.
Reason. I-Cl bond is weaker than I-l bond.
for (NEET 2022)
ur
s
ook
Yo

For Difficult Questions

eB

a Multiple Choice Questions (with one correct Answer)

our
ad

5. Compound (A) can be either (NH4)2Cr202,

1. Ba(N3)2 Ba + 3 N2
NH4NO2 or NaNj because on heating these give
Y

2. (NH4)2S04 2 NH3 + H2SO4 N2 (B) gas (a constituent of air) which when heated
Re

with H2 gas gives NH3 (C) gas which is basic in

nd

5 +2 0 -3
nature.
3. HNO3, NO. N2, NH^Cl
Fi

Compound (A), however cannot be Pb(N03)2

4. 2Pb(N03)2 2PbO+ 4NO2 + O2 because on heating it gives NO2 and O2
(A)
2 Pb(N03)2 2 PbO + 4 NO2 + O,
Cool 6. N does not have i/-orbitals to accommodate the
2NO2 {Dimerisea) electrons donated by O of H^O. therefore, attack
(A) Colourless(B)
of H2O occurs on C! atom which has f/-orbitals to
Cool +3 accommodatethe extra electronsdonatedby H,0.
N2O4 + 2 NO 2N2O3 Consequently, Cl-0 bond is formed leading to the
Bluesolid (C)
formation of HOCl and NH3
The O.N. of N in N2O3 is +3. NCI3 + 3 H2O ^ NH3 + 3 HOCl
ANSWERI

103.(1?) 104. (a) 105. (d) 106. (a) 107. (/?) 108. (n) 109. (/?) 110. (a) 111. (n)
7/120 ^xtxdccfr'^ New Course Chemistry (XIl)CZs!9]

For Difficult Questions

15. P4+ 8SOCI2 ^ 4 PCI3 + 4 SO2 + S2CI2
Thionyl chloride

NH3 thus produced combines with H^O to form 16. PCI5 molecule is quite reactive.
NH4OH. Thus, the products of hydrolysis are 17. [PBr^^ and Br“
NH4OH and HOCl.
18. As the size of the element (E) increases, the E-H
7. NO and N^O are neutral oxides.
bond becomes weaker and thus breaks on heating.
8. Acidic nature increases as theO.S, of N increases :
Therefore, the thermal stability decreases from
4l +2 +3 44 +5
H2O to H-,Po. In other words, option (cl) is correct,
N2O < NO < N2O3 < NO, < N2O5 . i.e., U2?o< H2Te < H2Se < H^S < H^O.
9. Reduction of NO^ to NO. Refer to Ring Test,
19. Na,S03 +H2SO4 4 N32S04 + SO2 + H2O
page 7/26. (X) (Y)

w
+
10. :N = 0: 4 : N = 0 : +e
2 NaOH + SO2 4 Na2SOj + H2O
24- (X)
Fe-"" + e 4Fe-"

Flo
3cl'^ Na,S03 + SO2 +H2O 4 2NaHS03
(Y) (Z)
> t; t; t t t
+ e (X)
ti t t t

ee
Thus, option (a) is correct.

Fr
4 unpaired electrons 3 unpaired electrons

.-. Magnetic moment of Fe"^ = ^n{n+2) BM 20. Both 540^” and 5203“ contain S-S bonds.

for
ur
= -n^^BM = 3-87BM O O S

All statements are correct. "O—S—S—O"

k s
11. The structures of oxyacids of P are :
Yo
0"
oo

0 0
o o
S406 S20^
eB

R
21. In H2S2O3, sulphur exhibits two different.
H OH HO OH
r

-2
H OH
ou

S
ad

Hypophosphorus acid (A) Orthopliosphoric acid (B)

O.S = + 1 O.S = + 5 46
HO—S—OH oxidation states of + 6 and - 2,
Y

0 o O
0
Re
nd

P P
22. CI2 > Bf2 > Ft > I2. Refer to Table 7.16 and page
Fi

HO OH H OH
7/72.
HO OH OH
Hypophosphoric acid (C) Orthophosphorus acid (D) 23. Although both hydration energy and bond
O.S = + 4 O.S = + 3
dissociation energy make F2 a strong oxidising
Thu.s, option (d) is correct. agent but the effect of hydration energy is more
0 ■H -3
important.
12. P^+3NaOH+3H20 4 3NaH2P02+ PH3 24. H2O2. Refer to page 7/78.
It hadisproportionation reaction. In this reaction, 26. As the oxidation number of Cl decreases from +7
oxidation state of P increases from 0 in P4 to + 1 in HCIO4 to +5 in HCIO3 to +3 in HCIO2 and +1
in NaH2POT and decreases from 0 in P4 to - 3 in in HOCl, its tendency to attract the electrons of
PH3. the O-H bond decreases. As a result, the tendency
13. Phosphinic acid, H3PO2 (monoprotic), of the O-H bond to dissociate to give H"^ ions
phosphonic acid, H3PO3 (diprotic), decreases and hence the acid strength decreases
14. Refer to page 7/34. accordingly in the order : HCIO4 > HCIO3 >
P4(^)+ 10SO2Cl2(/) 4 4 PCIj (5) + 10 SOt U) HCIO2 > HOCl.
p-BLOCK ELEMENTS 7/121

41. Shape of Xe¥7

For Difficult Questions
No. of electrons in the valence shell of the central

27. 6NaOH + 3 CU atom (Xe) = 8

■> NaC103 ^
(hot and cone.) (A) No. of electrons provided by five F atoms
+ 3 H-yO
=5x1=5
2CafOH)2 +2CI2 4 Ca(OCl)2 + CaCl2
(dry) Charge on the central atom = 1
+ 2 H^O
.●. Total number of electron pairs around the central
28. : I-1:^:1 atom = ilAii = 7
2
29. It contains Cs'*’ and I3 ions. No. of bond pairs = 5 (●.● there are 5 Xe-Ir
30. Refer to Fig. 7.34. page 7/90. Like Brp3, CIF3 ha,s (T-bonds)
two lone pairs of electrons.

w
.●. No. of lone pairs = 7-5 = 2
31. Refer to Art. 7.37.3, page 7/90-7/91. Thus, according to VSEPR theory a molecule with
XX' is linear, XX^ is T-shaped, 5 bond pairs and two lone pairs must be

F lo
pentagonal plana.r
XX5 is square pyramidal and XX7, is pentagonal Shape of Xe03F,

ee
bipyramid. Thus, option (a) is correct. No. of electrons in the valence .shell of the central

Fr
32. Structure of ICI5 is square pyramidal (refer to Fig. atom (Xe) = 8
7.35(a), page 7/91) and iCl^ is square planar No. of electrons provided by two F atoms = 2
(refer to Fig. 7.37, page 7/92)
for
Total number of electron pairs around the central
ur
33. (A) PCI5 is trigonal bipyrainid (iv), atom _8-h2_g
2
(B) SFg is octahedral (Hi),
s
According to VSEPR theory, double bonds do not
ook
Yo
(C) BrFj is square pyramidal (i), contribute any electron while single bonds
(D) BF3 is trigonal planar (ii). contribute one electron each towards the total
eB

Thus, option (b) is correct. number of electrons in the valence shell of the

central atom. However, both double and single

34. 2HC103-1-2HC1 > 2CIO2 -t-CI2 -1-2H20
r

bonds contribute one electron pair.

ad
ou

(Paramagnetic gas)
.●. Total number of electron pairs contributed by 2
2 CIO2 -1- 2 O3 > CI2O6 + 2 O2 single bonds of F and three double bonds of O.
Y

Thus, option (c) is correct. Thus, = 2 -h 3 = 5

Re

Thus, no. of lone pairs = 5-5 = 0

nd

35. Noble gases have weak dispersion forces, hence

they have low melting and boiling points. Now according to VSERP theory, a molecule with
Fi

Thus, option (b) is incorrect. 5 bonds and no lone pair must be trigonal
bipyraniidal.
36. Refer to Fig. 7.36, page 7/92, ICI^ is sp^d Thus, option (a) is correct.
hybridized and linear. Similarly, Xep2 is sp^d- 42. Xep4 is square planar while Xe04 is tetrahedral.
hybridized and linear (Fig. 7.39, page 7/99). +4 + 1 -ft) 0

37. SF4 (sp^d, see-saw) and Xep2 (sp^d, linear). 43. Xep4 + O2F2 A XeF^-h02 is a redox
38. Xep4 (Xe is 5/7'^£/^-hybridized) and has square reaction.

planar geometry. Refer to Table 7.24, page 7/99. 44. Hydrolysis with one equivalent of H2O gives
39. Refer to Table 7.24 and Fig. 7.39, page 7/99 and
XeOF4 and with two equivalents of H^O, Xe07p2
is obtained. In both the.se compounds O.S. of Xe
Table 7.25, page 7/100. is + 6.

XeF, Xe03 XeF4 45. The reaction, Xe03 + 6 HF ■> XeFg -I- 3 H2O is
sp^d^ sp^d^ sp^d^- not feasible since the products actually react to
Distorted Pyramidal Square Square
octahedral pyramidal planar give the reactants back.
7/122 New Course Chemistry (Xll)l!ZsI91

For Difficult Questions

01 Multiple Choice Questions (with One or More than One Correct Answers)
{d) Statement {d) is wrong because strongest acid
46. COj” ion has 32 electrons (6 + 3 x 8 + 2) and C is (iv) and not (/)●
is sp^ hybridized. NOj has 32 electrons and N is S3. BrFj has one lone pair (refer to Fig. 7.35, page 7/91.
also 5/)^-hybridized. Thus, CO|“ and NOJ are CIF3 has two lone pairs (refer to Fig. 7.34,
isoelectronic as well as isostructural, i.e., option page 7/90), XeF4 has two lone pairs (refer to Fig.
(a) is correct. 7.39, page 7/99) and SF4 has only one lone pair
(refer to Fig. 7.19(b), page 7/47. Thus, options (b)
ClOj ionhas42eleclrons(17 + 3x8-i- l)andCIis

ow
and (c) are correct.
-hybridized, while SO3” also has 42 electrons 54. Symmetrical molecules like BF3, SFg, BeCl2 and
and S is -hybridized (refer to Q. 41, page 7/122). CO2 have zero dipole moment even though they
contain polar bonds. Unsymmetrical molecules
"nius, CIOj and SO3" are also isoelectronic and like POCI3 (tetrahedral), CH3C! (tetrahedral),
isostructural, i.e., option (d) is correct.

e
XeFg (distorted octahedral), NH3 (pyramidal), SO2

Fl
re
48. Refer to page 7/11. N-N single bond is weaker (bent), H2Se (V-shaped), NO2 (V-shaped), O3
(159 kJ mol-*) than P-P single bond (213 kJ (bent), BrF^ (square pyramidal) have some finite

F
mol"*). dipole moment.
49. 4 HNO3 + P40,o ^ 2N2O5 + 4HPO3 55. A weakest acid has strongest conjugate base. Thus,
ur
or
N2O5 -I- 2 Na ^ Na20 -I- 2 NO2 option (a) is correct. Since chlorate ion has one
lone pair of electrons, it changes its shape from
51. CI2 reacts with H2O to form HCl + HCIO. As such f
ks
tetrahedral to pyramidal. Thus, option (b) is
it does not give HCIO4 and hence option (c) is
Yo
correct.
wrong. All other options are correct.
oo

52. Refer to Fig. 7.33, page 7/86. Being an oxidising agent, OCl“ oxides SO^“ to
B

{a) Statement (a) is wrong because the number of SO|” . Thus, option (d) is correct.
Cl = O bonds in (//) and (m) is 1 + 2 = 3.
re

Option (c) is wrong since CIO and CIO3 ions

(b) Statement (b) is correct because the lone pairs
give different disproportionation products.
u

of electrons on Cl in (ii) and (Hi) is 2 -1- 1 = 3.

ad

-> 2 cr + CIOJ
Yo

3 CIO-
(c) Statement (c) is also correct since hybridization
of Cl in (iv) is sp^. In fact, it is sp^ in all the four 4CIO3 ^C\-+ 3C10-
acids.
d

Thus, option (a), (b) and (d) are correct.

Re
in

DU Multiple Choice Questions (Based on the given Passage/Comprehens ion)

F

56,
P4 + 5O2 ^4^10 59. Both CIO2 (17 2 X 8 -i- 1) and CIP^- (17 + 2 x
w X
9-1) have 34 electrons.
4HN03 + P40,o ^ 2N205 + 4HP03 60. OCN” (pseudohalide),
Y z

58. RCOO does not contain a N atom and hence is

Brl2 (polyhalide),
not a pseudohalide. IF5 (interhalogen).

m Matching Type Questions

warm (3) +212(2)
64. P : 2 Pb02 + 2 H2SO4 -* 2PbS04 + 02 R: N2H4 N2 + 4 HI
+ 2H2O +2NO(l)
+ 4CI2 (4) S: XeF, Xe + 2NOF
Q : Na2S203 + 5 H2O 2 NaHS04 + 8 HCl
p-BLOCK ELEMENTS 7/123

For Difficult Questions

ra Matrix-Match Type Questions

65. A (SO2) - q (basic), r (reducing), s (oxidising) ; B (H2SO4) - q (acidic), s (oxidising) ; C (HNO3) - q

(acidic), s (oxidising); D (NH3) - p (basic), s (reducing).

VI.
Integer rpe Questions

74. Structure of Nu2S40^ is Thus, there are only four molecules/ions (/-/V)
O 0 where the hybridization of the central atom does

w
, r II 0 0 II .r not have contribution from c/-orbitals.
Na-"”0—S—S—S—S— O-Na-"
81. Compound Lone pair of electrons

F lo
O O
Tep2- 1
Difference in O.N = 5-0 = 5

75. Diprotic acids have two replaceable H-atoms. For Brf^- 2

ee
example, H2SO4, H3PO3, H2CO3, H2S2O7,

Fr
SNF3 0
H2C1O4 and H-,S03.
3
76. Non-zero oxidation states of O are -1,-2, +1 Xel^
and +2 ; that of Cl are - 1,+1, -t-3, +5 and +7 ; that for Total = 6
ur
of F is - 1 and + 1 ; that of P is - 3, - 2, +1, +3 and
-f5 : that of Sn is +2 and +4 ; that of Tl is +1 and 0
s
o o
ook

+3 ; that of Na is +1 and that Ti is +1 and +4.

Yo
82. 1. ON NO or N-N
Thus, only Na can exist in one non-zero oxidation -o
eB

state.

0
77. The oxidation state of Cl in HCIO4 is +7.
78. Xe04 (refer to Fig. 7.39, page 7/99) has four dK-pn 2. O2N NO2
our
ad

bonds while SO3 (refer to Fig. 7.17, Page 7/46) has 3. P4O6 (Fig. 7.11(a), page 7/35)
two dK-pK bonds. Thus, the ratio is 4 : 2 or 2 : 1.
4. P4O7 (P4O6 + one P = O)
Y

79. 3. CIF (17 + 9 = 26), OF2 (8 + 2 x 9 = 26), CIQ-

(17 + 8+1 =26).
Re

O O
nd

80. Molecule/ion Hybridization Shape

Fi

(0 BeCl2 sp Linear
5. H
(ii) N3 sp Linear OH OH

iiii) N2O sp Linear H4P2O5

(iv) NO+ sp Linear

O O O

(V) O3 sp^ Bent

P P P
(Vi) SCI2 sp^ Bent
6. HO O O OH
(vii) ICI2 sp^d Linear HO OH OH
H5P3O10
(vm) l~ sp^d Linear

(ix) XeF2 sp^d Linetu- 83. BrF5, SFg, ICI4 and IF5 have hybridization.

i
7/124 ‘P.na^tieefr’4. New Course Chemistry CXIl)CZs291

O^JCVCTf
For Oifficult Questions

VII.
Numeric al Value Type Questions (>ri Decimal Notation)
84. The composite equation of the reaction is 1
= — X — = — mole
2 NaCl + MnOT + 2 H2SO4 + 2 KI — -> 2 10 20
Na2S04 + MnS04 + 2 H2O + 2 iCCl + I2
But mol. wt. of U = 254 g mol"'
From the above equation,
1 1
2 mole of NaCl produce U = I mole
— mole of I2 - 254x—
20
= 12*7 g
0-1 mole of NaCl will produce I2 20

VIII.
Assertion-Reason Type Questions

w
85. Correct statement>2. NaH2P02 does not have any 97. Correct explanation. The brown colour is due to

F lo
ionizable hydrogens. formation of charge transfer complex between NO
87. Correct explanation. F cannot show positive and Fe-'*' ions.

ee
oxidation states of +5 and +7 and hence FOJ and 98. Correct reason. The lone pair of electrons on the

Fr
FO4 are unknown. N atom is strongly attracted by the three strongly
88. Correct statement-1. F, is more reactive than CIF. electronegative F atoms. As a result, it is not
89. Correct statement-1. ICl on hydrolysis gives HCl for
available for donation and hence NF3 is a weak
ligand.
ur
and HOI.

90. Correct statement-1. The pentavalent oxides, 101. Correct reason. There are two H atoms directly
s
E2O5 are more acidic than the corresponding attached to O, /.c., it has two ionizable H atoms.
ook
Yo

irivalent oxides, E2O3. 103. Correct explanation. Due to small size, the lone
92. Correct statement-1.The boiling points of the pairs of electrons on F-atoms repel the bond pair
eB

followinghydrides of group 16 elements increase between F-aloms. Therefore, F—F bond in F2

molecule is weak.
in the order : H2S < H^Se < H2Te < H2O -
our
ad

2I3K 232K 269 K 373K 105. Correct assertion. HOF bond angle is smaller
Correct statement-2. The boiling points of these than that of HOCl bond angle.
hydrides decrease from H2O to H-,S and then Correct reason. Oxygen is more electronegative
Y

increase abruptly with increase in molar mass. than all halogens except fluorine.
Re
nd

93. Correct expIanation-1. CFCs convert O3 to 0^. 107. Correct explanation. Due to higher (+ 7) O.S. of
As a result, O3 holes increase and hence the Cl in HCIO4 than in HCIO3 (-1- 5). the CIO3 part
Fi

amount of UV radiations reaching the earth of HCIO4 pulls the electrons of the O—H more
increases.
strongly than C102 part in HCIO3. As a result, the
95. Correct explanation. Because of die presence of
O—H bond in HCIO4 can break more easily to
a triple bond, bond dissociation energy of nitrogen
is much higher than that of oxygen. liberate a proton than in HCIO3 and hence HCIO4
is more acidic than HCIO3.
96. Correct reason. HNO3 reacts with Fe to form a
thin impervious layer of ferrosoferric oxide 109. Correctexplanation.The ionizationenergy of F-,
is much higher than that of Xe and 5d orbitals are
(Fe0.Fc203 = Fc304) on the surface of iron which
protects it from further action. available for valence shell expansion.

1
 w
d- AND f- BLOCK ELEMENTS

Flo
ee
SECTION I. STUDY OF d-BLOCK ELEMENTS (TRANSITION ELEMENTS I

Fr
8.1. GENERAL INTRODUCTION - POSITION IN THE PERIODIC TABLE

for
Refer to the Periodic Table given inside the title cover of the book. The elements which lie in- between
ur
s and p block elements in the long form of periodic table (/.e., elements of groups 3-12) are called transition
elements. These elements have their properties which are intermediate between those of ^ and p block elements
ks
and represent a change from the most electropositive 5-block elements to the least electropositive p- block
Yo
oo

elements, thus acting as transition elements in accordance with their name. Further, as they are metallic in
nature, the name transition metals is often used for these elements.
eB

These elements are called (^-block elements since in them, 3d, Ad, 5d and 6r/ subshells are incomplete
and the last electron enters the {n-\)d orbital, i.e., penultimate (last but one) shell. Their general electronic
r

conifguration is
ou
ad

(«-l) d^-^^ ns
1-2
where n is the outermost shell.
Y

Classification of d-BIock Elements. Unlike s or p block elements which are usually discussed as
columns or groups, c/-block elements are better discussed by classifying them into horizontal series. In the
nd
Re

periodic table, there are four main transition series of elements corresponding to filling of 3d, Ad, 5d and 6d
sublevels in 4th, 5th, 6th and 7th periods.
Fi

(0 First transition series or 3d series corresponding to iflling of 3d sublevel consists of the following
10 elements of 4th period : Sc (At. No. = 21) Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn (At. No. = 30)
(ii) Second transition series or 4d series corresponding to iflling of 4d sublevel consists of the following
10 elements of 5th period : Y (At. No. = 39) Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag and Cd (At. No. = 48)
(Hi) Third transition series or 5d series corresponding to iflling ofSd sublevel consists of the following 10
elements of 6th period.
La (At. No. 57) ; Hf (At. No. = 72), Ta, W, Re, Os, Ir, Pt„ Au and Hg (At. No. = 80)
(iv Fourth transition series or 6d series corresponding to the iflling of 6d sublevel starts with Actinium
(At. No. = 89) followed by elements with atomic no. 104 onwards.
The study of transition elements is important because precious metals such as silver, gold and platinum
as well as industrially important metals such as iron, copper and titanium are transition elements. Besides

8/1
8/2 T^n^tdeefit'4. New Course Chemistry fxmrasTMi

these, many industrially useful compounds such as KMn04, K2Cr207, etc. are also the compounds of transition
metals. Hence, in addition to the study of trends in the general properties of ii-block elements, our aim in this
unit will be to study the preparation and properties of some important compounds.
8.2. ELECTRONIC CONFIGURATION OF D-BLOCK fTRANSlTlON) ELEMENTS
As discussed in Atomic Structure In +1 class, the sequence of filling of energy levels of multi-electron
atoms is in the order : l^, 2^, 2p, 3.v, 3/?, 45, 3^/, 4/7, 55, Ad, 5p, 65,4/, 5d, 6p, Is, etc.
In the ground state, the electron occupies the available orbital of lowest energy.
The orbital electronic configuration of argon (At. No. Z = 18) is l5^ 25^ 2p^ 3s^ 3p^, but in potassium
(Z= 19), the 19th electron does not enter 3d because it is higher in energy as compared with 45 orbital.
Therefore, the 19th electron goes into 45 orbital giving potassium an electronic configuration of 15^ 25^ 2p^
3s^ 3p^ 45^ Similarly, calcium (Z = 20) has the configuration, l5^ 2s^ 2p^ 3s^ 3p^ As-. In calcium, the 45
orbital is complete and the next higher in energy is the 3d sub-shell which starts filling up atZ = 21. Scandium
(Z=21) is, therefore, assigned the electronic configuration of l5^ 2s^ 2p^ 3.5“ 3p^ 3d^ As-.
The filling of d orbitals continues till the 3d series ends up at zinc (Z = 30) with the configuration :

w
l5“ 25- 352 3p^ 3d 45“.
In transition elements (leaving a few exceptions), the number of electrons in their outermost subshell

F lo
remains two while their penultimate shell of electrons is being expanded from 8 to 18 electrons due to addition
of electrons in the c/-subshell.

The electronic configurations of transition metals are listed in Table 8.1.

ee
TABLE 8.1. Electronic configurations of transition metals

Fr
1. First (3df) Transition Series (Sc-Zn)

At. No. 21 22 23 24 25
for
26 27 28 29 30
r
You
Element Sc Ti V Cr Mn Fe Co Ni Cu Zn
s
ook

E.C. 3d'As^ 3d2 4^2 3^/3 4^2 3^54^1 3^5 4_y2 2d^4s^ 3d’’As^ 3d^A.s^ 3d 45' 3d45^
eB

2. Second (4d) Transition Series (Y-Cd)

At. No. 39 40 41 42 43 44 45 46 47 48
our
ad

Element Y Zr Nb Mo Tc* Ru Rh Pd Ag Cd

E.C. Ad ^ 5s^ Ad'^ 5s^ Ad*5s^ 4d^ 55* Ad^ 5s^ Ad'^ 5s^ Ad^ 5s^ 4d55° 4d 55Md 55^
dY
Re

3. Third (5d) Transition Series (La-Hg)

Fin

At. No. 57 72 73 74 75 76 77 78 79 80

Element La Hf Ta W Re Os Ir Pt Au Hg
E.C. 5d * 65^ 5d ^ 65^ 5d ^ 65^ 5d 6s^ 5d ^ 65^ 5d ^65^ 5d ^ 65^ 5d ^ 65* 5d 65* 5d 65^

4. Fourth (6d) Transition Series

AL No. 89 104 105 106 107 108 109 110 111 112

Element Ac Ku Ha Sg Bh Hs Mt Uun Uuu Uub

(Unh) (Uns) (Uno) (Une)

E.C. 6d^ls^ ed^ls^ (id^ls^ (>d^lP- 6d^7r td''1^ 6d^ls^ 6d 75' 6d75^

*Technetium (Tc) is a synthetic transition metal which has been made artificially. This has been so named because
the word technetium means artificial.
d- AND f-BLOCK ELEMENTS 8/3

These electronic configurations have the following characteristics :

(0 An inner core of electrons with noble gas configuration.
(//) («—\)d orbitals are filled up progressively with electrons.
{Hi) Most of the members have 2 electrons in the outermost orbital, i.e., ns. Some of the members, i.e., Cr, Cu,
Nb, Mo, Ag, Au, etc. have only one electron in ns orbital whereas Pd has no election in the /i5-orbital.
(/v) In lanthanum (Z=57), one electron goes to 5d orbital before filling of 4/ orbital (an exception from
aufbau order).

Fundamental difference in the electronic configuration of Representative Elements and Transition

Elements. In the representative elements (s- and p-block elements), the valence electrons are present
only in the outermost shell while in the transition elements, the valence electrons are present in the

ow
outermost shell as well as d-orbitals of penultimate shell.

Exceptional configuration of Cr and Cu. The exceptions observed in the first series are in case of electronic
configurations of chromium (Z = 24) and copper (Z = 29). It may be noted that unlike other elements, chromium
and copper have a single electron in the 4s-ort)jtal. This is due to the gain of additional stability by the atom by

e
either having half-filled configuration {i.e., containing 5 electrons in the^/-sublevel) or completely filled configuration,

Fl
re
{i.e., containing 10 electrons in the c/-sublevel). Tfie 3J-level in case of chromium gets exactly half-filled with

F
configuration ~id^ 4i’ and that in case of copper, it gets completely filled with configuration
3d 4^'. This can be explained on the basis of exchange energy (as discussed in +1 class).
ur
or
Thus, the electronic configuration of chromium (Z = 24) and copper (Z = 29) are Is- 2s^ 2p^ 3s^ 3p^
3d ^ 4^* and 1^^ Zr 2p^ .'i-P 3p^ 3d 4s’ respectively.
sf
In second and subsequent series, we come across more irregularities than in the first transition series.
k
Yo
This may be due to the significant role of tire nuclear-electron and electron-electron forces.
oo

Definition of Transilion Elements. Strictly speaking, transition elements are those elements which
B

have partly filled {n—l)d subshell in their elementary state.

re

Let us examine the electronic configuration of copper atom, cuprous ion and cupric ion.
Cu (Z = 29) = is^ 2s^ 2p^ 3s'^ 3p^ 3d 45*
u
ad
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Cu’’’ = 1 s^ 2p 2p^ 3s^ 3p^ 3d

10

(Cuprous ion)
d

Cu2+ = \s^ 2P 2p^ 3^2 3d ^

Re
in

(Cupric ion)
F

Obviously, copper in the ground state or as cuprous ion cannot be considered as transition element and
rightly so it does not exhibit the characteristic properties of the transition elements. However, cupric ion has
incomplete d-sublevel, i.e., d^ electronic configuration and hence is a transition metal ion.
A transition element may thus be deifned as the element whose atom in ground state or ion in
one of the common oxidation states, has incomplete d-subshell (partly filled), Le., having electrons
between 1 to 9.

This definition excludes zinc, cadmium and mercury from the transition elements because they do not
have partly filled (or incomplete) tZ-subshell in their atomic state or their common oxidation state {i.e., Zn^'*’.
Cd^'*’, Hg^'*’). The i/-subshell configurations of these atoms or their dipositive ions are 3d ’®, Ad and 5d
They do not show properties of transition elements to any appreciable extent except for their ability to form
complexes. However, being the end members of the three transition series, they are generally studied with the
rf-block elements.
 8/4 New Course Chemistry (XII)CSm

83. GENERAL CHARACTERISTICS OF TRANSITION ELEMENTS

(1) Except for mercury which is a liquid, all transition elements have typical metallic structure and show typical
metallic properties such as conductivity, malleability and ductility, lustre, high tensile strength etc.
(2) Their atomic radii are in-between those of s- and p-block elements. In a series, they decrease with
increase in atomic number but the decrease is small after midway.

(3) They have high melting and boiling points, high enthalpies of vaporisation, high enthalpies of
atomisation and high enthalpies of hydration of their ions. These properties depend upon the strength
of the metallic bond in them.

(4) Their first ionization energies are higher than those of .v-block elements and less than those of
/)-block elements.
(5) They are electropositive in nature.
(6) They show variable oxidation states.

w
(7) The stability of any oxidation state or the tendency for an transition metal ion to act as oxidizing or
reducing agent depends upon its electrode potential.

F lo
(8) A number of these transition metals and their compounds show catalytic properties.
(9) Most of the transition elements form coloured compounds.

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(10) Their compounds are generally paramagnetic in nature.

Fr
(II) They have a great tendency to form complexes.

for
(12) They fonn interstitial compounds with elements like H, C, B and N.
ur
(13) They form alloys.
Exceptional Behaviour of Zinc, Cadmium and Mercury. As already mentioned, due to completely
s
ook

filled d^^ configuration, Zn, Cd and Hg show exceptional

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behaviour as compared to the other members of

their respective series. Taking the example of first transition series, zinc differs from the other members in the
eB

following characteristics :
(/) It has much higher atomic radius and ionization enthalpies,
our
ad

(/f) Its compounds are while while those of the other members are coloured.
(///) It does not show variable oxidation slates unlike other members.
Y

(iV) lt.s compounds (containing Zn^'*' ions) are diamagnetic while those of the other members are
Re

paramagnetic,
nd

(v) It forms lesser number of complexes than other members.

Fi

8.4. GENERAL TRENDS IN THE PROPERTIES OF TRANSITION ELEMENTS

The </-block (transition) elements have similar ns- orbital electron configuration in the outermost energy
shell, n, and they differ from one another only in the number of electrons in the cZ-orbitals (subshell) of the
penultimate, {n — /), shell. Since only the penultimate shell of electrons of all these elements is expanding,
they may have resemblance in their physical and chemical properties. Thus, the (/-block elements are hard
metals and are good conductors of heat and electricity. They have high melting and boiling points and form
alloys with other metals.
Further, the (/-orbitals of the transition elements project to the periphery of the atom more than the
s and />orbitals. Hence, the (/-electrons affect the properties of the elements to a great extent. Thus, ions with
similar (/" configuration (n = 1 - 9) have similar magnetic and electronic properties. In other words, there are
some similarities in the elements of the same group. However, there are greater horizontal similarities in the
properties of transition elements in addition to the trends in the properties.
 d- AND f>BLOCK ELEMENTS 8/5

The similarities and trends in some of the properties of the transition elements along the horizontal row
and in the vertical group are briefly described below.
8.4.1. Atomic and Ionic Radii

Atomic Radii. (/) The atomic radii of the transition metals lie in-between those of s- and p-block
elements.

iii) Generally the atomic radii of d-block elements in a series decrease with increase in atomic number
but the decrease in atomic size is small after midway. Thi.s may be seen in the first transition series, where
the atomic radii of the metals decrease from scandium to chromium but after that it remains almost same
(Table 8.2).

TABLE 8.2. Atomic radii of d-block elements in pm

Sc Ti V Cr Mil Fc Co Ni Cu Zn
162 147 134 127 126 126 125 124 124 138

w
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
180 160 146 139 136 1.34 134 137 144 154

F lo
Ta W Re Os Ir Pt Au Hg
146 139 137 135 136 138 144 157

e
14 Lanthanoid Lanthanoid

Fre
elements (Z=58-71) contraction
for
Explanation. In the beginning, the atomic radius decreases with the increase in atomic number as the nuclear
charge increases whereas the shielding effect of J-electron is small. After midway, as the electrons enter the last but
r
one (penultimate) shell, the added c/-electron shields (screens) the outermost electron. Hence, with the increase in
You
oks

the </-electrons, screening effect increases. This counterbalances the increased nuclear charge due to increase in
atomic number. As a result, the atomic radii remain practically same after chromium. For example, in Fe, the two
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opposing tendencies almost counterbalance and there is no change in the size from Mn to Fe.
Same type of behaviour is found in second and third transition series (Table 8.2).
ad
our

(///) At the end of the period, there is a slight increase in the atomic radii.
Expiaiiation. Near the end of series, the increased electron-electron repulsions between added electrons
in the same orbitals are greater than the attractive forces due to the increased nuclear charge. This results in
Re
dY

the expansion of the electron cloud and thus the atomic radius increases,
(fv) The atomic radii increase down the group. This
Fin

FIGURE 8.1
means that the atomic radii of second series are larger than 5d series
1901187 > 190
those of first transition series. But the atomic radii ofthe second
4d series
and third transition series are almost the same. Il80i ● 180
c
series
(Fig. 8.1.) ^ 170 170

Explanation. Since in the atoms of the second transition I 160 ,1

>8 157,
160

■154
O
series, the number of shells increases, their atomic radii are g 150 146
138 150
139
larger than those of the elemen s of the first transition series. O
5 140
147
J37i35l^ 144 140
T37 138
The atomic radii of the elements of the second and third 1
130 ! 130
12f>*^126 125 124 128
transition metals are nearly same due to lanthanide 120
126
-J 120
contraction (or also called lanthanoid contraction) discussed TRANSITION METALS

later. Trends in atomic radii

of transition elements
Ionic radii. (0 The trend followed by the ionic radii is
the same as that followed by atomic radii.

»
8/6 New Course Chemistry fXII~>rosTWl

(ii) Ionic radii of transition metals are different in different oxidation states. In general, the ionic radii
decrease with increase in oxidation state. Thus, the ionic size of cations are smaller than those of
cations. This is because the ionic radius decreases with increase in effective nuclear charge. However, the
ionic radii of cations in the same oxidation state decrease with the increase in atomic number as shown in
Table 8.3 below :

TABLE 8.3. Ionic radii of first transition series (in pm)

Element Sc Ti V Cr Mn Fe Co Ni Cu Zn

M2+ 81 91 84 88 80 76 74 72 72 74

76 74 69 66 64 63

iiii) The ionic radii of the transition metals are smaller than those of the representative elements (main
group elements, i.e., s and p-block elements) belonging to the same period. It may be remembered that

w
whereas s- block and «i-block elements form positive ions, /j-block elements mostly form negative ions.
8.4.2. Metallic character

F lo
Except for mercury which is a liquid, all the transition elements have typical metallic structure (hep, cep or
bcc). They exhibit all the characteristics of metals. For example, they are hard, lustrous, meallable and ductile,
have high melting and boiling points, high thermal and electrical conductivity and high tensile strength.

e
Fre
Explanation. Transition elements have relatively low ionization energies and have one or two electrons
in their outermost energy level or ns^). As a result, metallic bonds are formed. Hence, they behave as
for
metals. The unpaired d electrons also result in the formation of metallic bonds. Thus, in the formation of
metallic bonds, both ns and (n - 1) d electrons participate. Greater the number of unpaired d electrons,
r
stronger is the bonding due to the overlapping of unpaired electrons between different metal atoms.
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oks

Cr, Mo and W have maximum number of unpaired d electrons and are, therefore, hard metals whereas Zn, Cd
and Hg are not very hard metals due to the absence of unpaired electrons.
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The heavier transition metals show stronger tendency to FIGURE 8.2

4
form strong M-M bonds than their lighter congeners. For /
THIRD TRANSITION
SERIES
W
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ad

example, in the dimeric unit ReXI?”,

1 O
there is a strong
Re
quadruple bond between Re and Re (one a-bond, two 7i-bonds Ta, SECOND
TRANSITION
and one 5-bond)*. <0
3--
Mo
/ SERIES
Ru
dY

Nb ''
Re

o
f Jr
8.4.3. Melting and Bolling Points P /
Tc
5 Hf' / \
Fin

The transition metals have very high melting and boiling 0. Cr

\
\ PI
points. The melting points of the transition metals rise to a Rhs
\

maximum and then fall as the atomic number increases. Fe Co

\

Ti
Manganese and technetium have abnormally low melting points NI

(Fig. 8.2). FIRST

TRANSITION
y
Mn
SERIES
Au
Explanation. Strong metallic bonds between the atoms of 1
Ag
these elements are responsible for the high melting and boiling TRANSITION METALS

points. This is clear from their high enthalpies of atomization. Trends in melting points of
transition metals
(i.e., heat required to break the metal lattice to getfree atoms).
*Two square planar ReCl4 units are linked through Re-Re bond. Each Re uses s, p^ and d x~-v~
, , orbitals
to form four a-bonds with Cl atoms. The four bonds between Re and Re consist of one a-bond by overlap of p,
and d 2 orbitals along the axis (z-axis), two Ji-bonds by overlap of d^ and dy^ orbitals and one 6-orbital by
overlap of d^ orbitals.

i
d- AND f-BLOCK ELEMENTS 8/7

TABLE 8.4. Enthalpies of atomization (A^H") of first transition series (in kJ mol"^)
Sc Ti V Cr Mn Fe Co Ni Cu Zn

326 473 515 397 281 416 425 430 339 126

The metals of the second {Ad) and third (5^ series have
greater enthalpies of atomisation than the corresponding elements FIGURE 8.3

of the first series (Fig. 8.3). This is due to more frequent metal- 800

metal bonding present in them. This also accounts for the fact THIRD (5d)
SERIES
that there occurs much more frequent metal-metal bonding in 700

compounds of heavy transition elements (3rd series). 600 SECOND (4d)

As already explained above, the strength of the metallic I
o
SERIES

bond depends upon the number of unpaired i/-electrons (half- E 500i FIRST (3d)

w
SERIES
filled f/-orbilals). Greater is the number of unpaired electrons
^to 400
(half-filled orbitals), stronger is the metallic bonding. Because <3

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of the stronger metallic bonding, these elements have high 300
melting and boiling points.
200'
In a particular seiies, the metallic strength increases upto

ee
the middle with increasing number of unpaired electrons, i.e.,

Fr
100
upto d ^ configuration {e.g.. Sc has 1, Ti has 2, V has 3 and Cr ATOMIC NUMBER

has 5 unpaired electrons). After Cr, the number of unpaired

electrons goes on decreasing {e.g.y Fe has 4, Co has 3 unpaired for
Trends in enthalpies of atomisation
ur
of transition metals
electrons and so on). Accordingly, the melting points decrease
after middle (Cr) because of increasing pairing of electrons.
s
ook
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Note that tungsten (W) has the highest melting point among the d-block elements. {Fig. 8.2)
The dip in the melting point at Mn in the first transition series and at Tc in the second transition series
eB

and somewhat in Re in the third transition series can be explained on the basis that they have exactly half-
filled (/-orbitals. As a result, in each case, the electronic configuration is stable, i.e., electrons are held tightly
r

by the nucleus so that the delocalisation is less and the metallic bond is much weaker than that preceding
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ou

element.

As there are no unpaired electrons in Zn, Cd and Hg, they are soft and have low melting points. Hg is
Y

liquid at ordinary temperature with melting point of 234 K.

Re
nd

Note that mercury is the only metal which is liquid at room temperature.
Fi

8.4.4. Density
As we move along a transition series from left to irght, the atomic radii decrease due to increase in
nuclear charge. Hence, the atomic volume decreases. At the same time, atomic mass increases. Hence, the
density increases. For the first transition series, these values are given in Table 8.5 below. The last element,
zinc, is an exception, having large atomic volume and hence lower density.
TABLE 8.5. Density of elements of first transition series (in g/cm^)
Element Sc Ti V Cr Mn Fe Co Ni Cu Zn

Density (g/cm^) 343 4 1 607 7-19 7-21 7-8 8-7 8-9 8-9 7-1

Remember that among the d-block elements, iridium has the highest density (22-6J g cm~^) whereas
scandium has the lowest density (3-43 g cm~^}. Osmium has slightly lesser density (22-59 g cm~^) than
iridium. Thus, osmium and iridium have almost the same density.
8/8 New Course Chemistry fXII~)rosw

8.4.5. Ionisation energies or Ionisation enthalpies

(/) The ifrst ionisation enthalpies of d-block elements lie between s-block and p-block elements. They
are higher than those of s-block elements and are lesser than those of p-block elements.
The ionisation enthalpy gradually increases with increase in atomic mint, er along a given transition
series though .some irregularities are obseiyed (Table 8.6).

TABLE 8.6. Ionisation enthalpies of first transition series (in kJ mol~^)

Element Sc Ti V Cr Mn Fc Co Ni Cii Zn

1st Ionization enthalpy 631 656 650 653 717 762 758 736 745 906

2nd Ionization enthalpy 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734

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3rd Ionization enthalpy 2393 2657 2833 2990 3260 2962 3243 3402 3556 3829

Explanation. The increasing ionization enthalpies are due to increased nuclear charge with increase in

Flo
atomic number which reduces the size of the atom making the removal of outer electron difficult.

ee
The irregular trend in the first ionisation enthalpy of 2>d metals is due to the fact that the removal of the
electron alters the relative energies of 45 and 3d orbitals. Thus, there is a reorganisation energy accompanying

Fr
ionisation. This results into the release of exchange energy which increases as the number of electrons increases
in the d'^ configuration and also from the transference of s- electrons into i/-orbitals. Cr has low first ionization

for
ur
energy because loss of one electron gives stable configuration (3d ^). Zn has veiy high ionization energy
because electron has to be removed from 45 orbital of the stable configuration {3d 45“).
s
(//) In a given scries, the difference in the ionisation enthalpies between any two successive d-
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block elements is very much less than the difference in case of successive s-block or p-block elements
Bo

(Table 8.6).
Explanation. The addition of d electrons in last but one [(« — 1) or penultimate] shell with increase in
re

atomic number provides a screening effect and thus shields the outer s electrons from inward nuclear pull.
Thus, the effects of increased nuclear charge and addition of d electrons tend to oppose each other. This is
ou
ad

due to their counter effects that the ionisation enthalpies show a little variation on moving along a period of
Y

c/-block elements.

(///) The first ionization enthalpy of Zn, Cd and Hg are, howeve,r very high because of the fully iflled
nd
Re

(n-l) d ns^ configuration.

Fi

(iV) Although second and third ionization enthalpies also, in general, increase along a period, but the
magnitude of increase for the successive elements is much higher. Moreover, exceptions are observed from
the regular increase as under :
(a) There is a fall in 2nd ionization enthalpy from Cr to Mn and from Cu to Zn. This is because after the
removal of 1st electron, Cr and Cu acquire a stable configuration (d^ and d and the removal of 2nd
electron is very difficult. Thus, the order for IEt is
23V < 24Cr > and < 29^^ ^ 30^*^-
(b) Third ionization enthalpy of Mn is very high because the 3rd electron has to be removed from the
stable half-filled 3d orbital. ('j^Mn is 3d ^ 45^).
(c) Third ionization enthalpy for change from Fe“'*' to (25^® = 3 d^ A s^) is very small because loss
of 3rd electron gives the stable configuration of 3d^. Hence, IE3 for 26pe < < IE3 for ^jMn.
(v) The high values of 3rd ionization enthalpies for Cu, Ni and Zn explain why they show a maximum
oxidation state of+2.
d- AND f-BLOCK ELEMENTS 8/9

(vi) The first ionisation enthalpies of 5d

900 -
elements are higher as compared to those of 3d and
4d elements (Fig. 8.4). This is because the weak
shielding of nucleus by 4/ electrons in 5d elements
results in greater effective nucletir charge acting on 800 -
the outer valence electrons.

The ionization enthalpies of 3d and 4d metals O

are irregular.
E 700 -
8.4.6. Standard electrode potentials lU

(E°) and chemical reactivity

ow
The magnitude of ionization enthalpy gives a
clue to the amount of enthalpy required to raise TRANSITION ELEMENTS
the metal to a particular oxidation state in a
compound. In other words, thermodynamic Trends in ionization enthalpies of transition
metals. The figure illustrates that the first

e
stability of the compounds of transition elements

re
can he evaluated in terms of the magnitude of
ionisation enthalpies of the metals —smaller the

rFl
ionization enthalpies of 5d metals are
higher than those of 3d and 4d metals.

F
ionisation enthalpy of the metal, stabler is its compound. For example, nickel (II) compounds are more
stable than platinum (II) whereas platinum (IV) compounds are more stable than nickel (IV). For example,

r
fo
ou
K2PtClg is a well known compound and corresponding nickel salt is unknown. This is evident from the
following table in which the first four ionisation enthalpies of both nickel and platinum are given.
ks
In solution, the stability of the
Element -> Ni Pt
oo

compounds depends upon electrode potentials

rather than ionization enthalpies. Electrode
Y

(IE, + IE2) k.1 mol

-1
2-49 X lO"^ 2-66 X 10^
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potential values depend upon factors such as

enthalpy of sublimation (or atomisation) of the (IE3 + IE4) W mol
-1
8-80 X 10^ 6-70 X 10^
metal, the ionisation enthalpy and the
ur

Total (kj mol-^) 11-29 X 10^ 9-36 X 10^

hydration enthalpy, i.e..
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^sub H
M(j) M (g), (A^yb^ enthalpy of sublimation/atomisation)
d

A:H
Re
in

M(g) > M"*" (g) + e , (Aj H is ionisation enthalpy)

F

\ydH
M+ (g) + aq > M'*' (aq), (A|,yd H is enthalpy of hydration)
The total energy. Ay H, for the process involving sublimation, ionsiation and hydration simultaneously,
/.e., for the process, M (5) > M'^ (aq) + e~, will be the sum of the three types of enthalpies, i.e.,
ATH = A,,,H+A,H + A,^dH.
Thus, Ay H, is the total enthalpy change when solid metal, Ay H
U(s) (aq)
M is brought in the aqueous medium in the form of monovalent I III
^sub ^ H
ion, M"*" (aq). ,,

II
The above steps may be represented by a cyclic process U(g) > M^(g)
A;H
(called Bom-Haber cycle) as shown on the irght hand side :
The electrode potentials are a measure of Ay H. Hence, quantitatively, the stability of the transition
metal ions in different oxidation states can be determined on the basis of the electrode potential data. The
lower the electrode potential, i.e., more negative the standard reduction potential of the electrode, more
stable is the oxidation state of the transition metal in the aqueous medium.
8/10 New Course Chemistry (X1I)BZSI91

values (M-'''/M) as well as for for first row transition metals are given below :

Element Ti V Cr Mn Fe Co Ni Cu Zn

E® (M2+/M) in volts (V) -1-63 -118 -0-91 -M8 -0-44 -0-28 -0-25 + 0-34 -0-76

E** (M^+/m2+) in volts (V) -0-37 -0-26 -041 +1-57 + 0-77 + 1-97

The observed values of E° and the calculated values are compared in Fig. 8.5.
The thermochemical parameters used in calculation of E° values for M (s)/M'*‘^ (aq) are given in
Table 8.7 below for the first row transition elements.

TABLE 8.7. Thermochemic£d data (kj mol~^) for FIGURE 8.5

First Row Transition Elements 0-5
ni
●jz
= 0
NdH(M^)

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Element (M) o

orA,„bH“ 4)
●o
-0-5

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Ti 469 661 1310 - 1866 ■B _1
Q>
0>
V 515 648 1370 - 1895 ■2-1-5
<Q
Cr 398 653 1590 - 1925 T3

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C

3 -2 1
Mn 279 716 1510 - 1862 10

Fr
Tl V Cr Mn Fe Co Ni Cu Zn
Fe 418 762 1560 - 1998 Observed values^ Calculated values*

Co
Ni
427
431
757

736
1640

1750
-2079

-2121 for Comparison of the observed

values and calculated values
ur
Cu 339 745 1960 -2121 for standard electrode potentials (E°)
for the elements from Ti to Zn.
s
Zn 130 908 1730 -2059
ook
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The above values lead to the following results ;

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TVends in the Standard Electrode Potentials

(/) There is no regular trend in the E° (M^'^/M) values. This is because their ionization enthalpies
(lEj + IE2) and sublimation enthalpies do not show any regular trend.
our
ad

(//■) The general trend towards less negative E° values along the series is due to the general increase in the
sum of first and second ionization enthalpies.
Y

(Hi) Copper shows a unique behaviour in the series as it is the only metal having positive value for E°. This
Re

explains why it does not liberate H2 gas from acids. It reacts only with the oxidizing acids (HNO3 and
nd

H2SO4) which are reduced. The reason for positive E° value for copper is that the sum of enthalpies of
Fi

sublimation and ionization is not balanced by hydration enthalpy.

(iv) The values of E° for Mn, Ni and Zn are more negative than expected from the general trend. This is due
to greater stability of half-filled </-subshell (d^) in Mn^'*', and completely filled J-subshell in
Zn-'*'. The exceptional behaviour of Ni towards E° value from the regular trend is due to its high
negative enthalpy of hydration.
Trends in the Standard Electrode Potentials
(/) A very low value for E° (Sc-^'^/Sc*''') (not given in the Table above) reflects the stability of Sc^'*’ ion
which has a noble gas configuration.
(ii) The highest value for Zn is on account of very high stability of Zn^"*" ion with configuration. It is
difficult to remove an electron from it to change it into +3 state.
(Hi) The comparatively high value of E° (Mn-^'^/Mn^^) shows that Mn^"^ is very stable which is on account
of stable configuration of
(iv) The comparatively low value of E° (Fe*^''’/Fe^'*’) is on account of extra stability of Fe^'*' (d^), i.e., low
third ionization enthalpy of Fe.
 d- AND f-BLOCK ELEMENTS 8/11

(v) The comparatively low value for V is on account of the stability of V^'*’ ion due to its half-filled 2g
configuration (discussed in unit 9).
Chemical Reactivity and E“ Values. The transition metals vary very widely in their chcmieal reactivity.
Some of them are highly electropositive and dissolve in mineral acids whereas a few of them are ‘noble’, i.e.,
they do not react with simple acids. Some results of chemical reactivity of transition metals as related to their
E° values are given below :
(0 The metals of the first transition series (except copper) are relatively more reactive than the other
series. Thus, they are oxidized by ions though the actual rate is slow, Ti and V are passive to
dilute non-oxidizing acids at room temperature.
(ii) As already explained, less negative E° values for along the series indicate a decreasing tendency
to form divalent cations,

(m) More negative E° values than expected for Mn, Ni and Zn show greater stability for Mn-"^ and

w
Zn2+.
(iv) E° values for the redox couple indicate that and Co^'*‘ ions are the strongest oxidizing

F lo
agents in aqueous solution whereas Ti^"*", V-'*'and Cr^'*' are strongest reducing agents and can liberate
hydrogen from a dilute acid, e.g., 2 Cr-"^ (aq) + 2 (aq) 2Cr^*(aq) + H2(g')

ee
8.4.7. Oxidation states

Fr
All transition elements, except the first and last member of tiie series, exhibit a number of oxidation
states as shown in the Table 8.8.

for
ur
TABLE 8.8. Commonly shown oxidation states of transition metals
(very rare oxidation states are given in parentheses and
most common ones are enclosed in boxes for first transition series)
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1st Series : Sc Ti V Cr Mn Fe Co Ni Cu Zn
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+ 1

(+2) +2 +2 +2 +2 +2 +2 +2 +2
r
ou
ad

+3 +3 +3 -^3 +3 +3 +3 (+3)
Y

+4 +4 +4 +4 +4 +4 (+4)
Re
nd

+5 +5 (+5)
Fi

+6 +6 (+6)

+7

2nd Series : Y Zr Nb Mo Tc Ru Rh Pd Ag Cd

+ 1

(+2) +2 +2 +2 +2 +2 (+2) +2
+3 (+3) (+3) +3 +3 +3 (+3) (+3)
+4 (+4) +4 (+4) +4 +4 +4
+5 +5 (+5) (+5)
+6 (+6) (+6)
+7 (+7)
(+8)
8/12 ‘P>ieuiee^'4, New Cotirse Chemistry (XII)BI

3rd Series: La Hf Ta W Re Os It Ft Ait Hg

(+1) +1 +1

+2 (+2) +2 +2 +2 +2
(+2)
(+3) (+3) +3 +3 +3 (+3) +3
+3 (+3)
+4 (+4) +4 +4 +4 +4 +4

+5 +5 +5 (+5)
+6 (+6) +6 (+6) (+6)
+7
+8

Explanation. The transition elements have their valence electrons in two different sets of orbitals, i.e.,
(n - 1) d and ns. Since there is very little difference in the energies of these orbitals, both energy levels can be
used for bond formation. In simple compounds, we may presume that the two electrons from ns orbital of a

w
transition element are used to give an oxidation state of +2 and the {n—l)d electrons remain unaffected. But
the higher oxidation states like +3, +4, +5, +6 and +7 correspond to the use of all 45 and 3d electrons in the

F lo
transition series of elements. In the excited state, the (n-l)d electrons become bonding and thus give variable
oxidation states to the atoms.

From the study of the oxidation states, we observe the following points :

e
(/) The most common oxidation state of the first row transition metals is +2 except in the case of scandium

Fre
(which has +3); +2 state is due to the loss of two ns^ electrons. It follows from this that d orbitals are
more stable than the 5-orbital after scandium. for
(//) Mostly ionic bonds are formed in +2 and +3 oxidation states. In compounds of higher oxidation states,
the bonds formed are mostly covalent since they are formed by sharing of d- electrons. For example, all
r
You

the bonds between manganese and oxygen are covalent in MnO^ (permanganate ion).
oks

(Hi) The elements which show the greatest number of oxidation states occur in or near the middle of the
eBo

series. For example, Mn shows all the oxidation states from +2 to +7. The lesser number of oxidation
states on the extreme ends are either due to too few electrons to lose or share (e.g.. Sc, Ti) or too many
^/-electrons so that fewer orbitals are available to share electrons with others (e.g., Cu, Zn). The
ad
our

highest oxidation state shown by any transition metal is +8 (by osmium or in a very few compounds by
ruthenium),
(/v) The maximum oxidation states of reasonable stability in the first transition series is equal to the
Re
dY

sum of 5 and d electrons upto Mn followed by an abrupt decrease in the stability of higher oxidation
Fin

states,

(v) The variability of oxidation states of transition elements, which is due to incompletely filled ^/-orbitals,
arises in such a way that successive oxidation states differ by unity, e.g., V™, V^, This is unlike
non-transition elements where oxidation states normally dilfer by two units, e.g., Pb^, Pb^; Sn^, Sn^
etc.

(vi) In a group of d-block elements, the higher oxidation states are more stable for heavier elements. For
example, in group 6, Mo (VI) and W (VI) are found to be more stable than Cr (VI). For this reason, Cr
(VI) in the form of dichromate is a stronger oxidizing agent in acidic medium whereas M0O3 and WO3
are not. This is again unlike p-block elements where lower oxidation states are more stable for heavier
members due to inert pair effect.
(vii) Transition metals also show low oxidation states in some compounds or complexes having ligands like
CO which not only form sigma bonds with the metal atom but also have n acceptor character (discussed
in unit 9). For example, copper in CuCl is in +1 state whereas nickel in nickel tetracarbonyl, Ni(CO)4
or iron in Fe(CO)5 is in the zero oxidation state.
d- AND f-BLOCK ELEMENTS 8/13

(viii) The oxidation state of a metal in a solvent depends on the nature of the solvent. The metal in a

particular oxidation state may undergo oxidation or reduction in the solvent under appropriate conditions.
For example, Cu"*" is unstable in water as it may undergo oxidation whereas Cu^'*' is stable in water.
Similarly, Fe^'*’ is unstable in aerated water (water in which air is dissolved) as it undergoes oxidation
in it.

(ix) If the element exists in more than one oxidation state, their relative stabilities can be known from the
standard electrode potential data. For example, in case of copper, we have
CW^(aq) +e~ - ^ Cu(.v), =0-52V
Cu2+ (aq) + 2 e~ > cu(^), e;Red = 0-34 V
Thus, Cu'*' is reduced more easily and hence is less stable than Cu^’*’.
(jc) The oxidation state of a transition metal depends on the nature of the combining atoms. The compounds
of metals with fluorine and oxygen exhibit the highest oxidation states as fluorine and oxygen are small

w
in size and the most electronegative elements.
IVends in the stability of higher oxidation states of transition metals halides. The stable halides

F lo
of the transition metals of the first transition series in different oxidation states are given in Table 8.9
below :

ee
TABLE 8.9. Stable halides of 3d series in different oxidation states.

Fr
(X = F ■> I. Exceptions are given ih brackets)
iOxidiatilon ■n V Cr Mn Fe Co Ni Cu Zn
No.
for
ur
+6
CrFg
s
+5
VFs CrFj
ook
Yo
+4 TiX4 VX4 CrX4 MnF4
(F^, cr, Br-)
eB

+3 TiX3 VX3 CrX3 Mnp3 FeX3 C0F3

(I^, cr, Br-)
r

+2 TiX2 VX2 CrX2 Mnp2 FeX2 C0X2 NiX2 CuX2

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ou

(Cr, Br- I-) (Cl-,Br,I-) (T, C1-)

+1 CuX
Y

(Cr, Br- I-)

Re
nd

Some of the results observed from this table and their explanation is given below :
Fi

(1) The highest oxidation numbers along the series vary fiom +4 for Ti (in TK4, tetrahalides) to +5 for V (VF5)
to +6 for Cr (in CrFg), then to +7 for Mn (not in simple halides but in Mn03F). Beyond Mn, no metal shows
an oxidation number of more than +3, which is only in the trihalides, FeX3 and C0F3. The highest oxidation
state is generally most stable with fluorine. This is due to either higher lattice energy, e.g., in C0F3 or higher
bond enthalpies for the higher covalent compounds, e.g., in VF5 and CrF^.
(h) Although V directly shows an oxidation state of +5 only in VF5, other halides, however, undergo
hydrolysis to give oxohalides, VOX3 in which oxidation state of V is +5.
(in) In the low oxidation states, fluorides are unstable, e.g., in TiX2, VX2, CuX, X = Cl", Br“ r but not F~.
However, copper (II) halides are known including Cup2 but except iodide. Cu^"*" ions oxidize iodide to
iodine and C11I2 itself is reduced to CU2I2 :
2Cu^+ + 41- ■» Cu2l2(^) + I2
(2CUI2) Cuprous iodide
Copper (II) iodide (More stable)
(Less stable)
8/14 New Course Chemistry (XIl)BZsl91

(/y) Many copper (I) compounds are unstable in aqueous solution and undergo disproportionation* as
follows ;
2Cu+ ●> Cu2+ + Cu
The greater stability of Cu-+ {aq) than Cu"^ {aq) is due to much more negative enthalpy of hydration for
Cu^+ {aq) than Cu+ {aq). This more than compensates for the high second ionization enthalpy of Cu.
TVends in the stability of higher oxidation state.s of transition metal oxides. Oxygen stabilizes the
highest oxidation states more than fluorine. The oxides of first transition series in different oxidation states
are given in Table 8.10 below :
TABLE 8.10. Oxides of metals of first transition series in different
oxidation states (*Mixed oxides)

Oxidation Sc Ti V Cr Mn Fe Co Ni Cu Zn

w
No.

(Gr3) (Gr4) (Gr5) (Gr6) (Gr7) (Gr8) (Gr9) (Gr 10) (Gr 11) (Gr 12)

F lo
+7 Mh207
+6 Cr03

ee
+5 V2O5

Fr
+4 TiOj V2O4 C1O2 Mn02
+3 SC2O3 Ti203 V203 Mnz03 FC')03
Mn30; Fe30; for
C03O4
ur
+2 TiO VO (CrO) MnO FeO CoO NiO CuO ZnO
s
+1 Cu-,0
ook
Yo
+1 Cu^O
eB

Some results observed from the above table are given below ;
(/) The highest oxidation state in the oxides is same as that of group number upto group 7 (SC2O3 to Mn207).
r

{ii) Beyond group 7, the maximum oxidation is +3 (in Fe203), although in the ferrates, (Fe04)^“,
ou
ad

formed in the alkaline medium, Fe has oxidation state of +6. But these readily decompose to Fe203 and
O..
Y

{iii) Besides the oxides, higher oxidation states are also found in the oxocations and oxoanions
Re
nd

e.g., V (V) in VOj, V (IV) in and Ti (IV) in TiO^^ and Mn (VII) in MnO~, Cr (VI) in CrOj
Fi

and Cr20““ imd V (V) in VO^” (vandate ion),

(/v) Oxygen stabilizes the highest oxidation slate even more than fluorine, e.g., O O
highest fluoride of Mn is MnF4 whereas highest oxide is Mu207. The reason
for this is the ability of oxygen to form multiple bonds with the metal atoms.
^Mn
For example, the structure of Mn207 is shown on the right hand side.
Each Mn is telrahedrally surrounded by four 0-atoms including a
o
-7 \/
o o
0

Mn—O—Mn bridge. Similarly, [M04l"“ ions exist having a tetrahedral (Mn207)

stmcture where M = V (V), Cr (VI), Mn (V). Mn (VI) and Mn (VII).

*Disproportionation reactions. These are those reactions in which the same substance undergoes oxidation
as well as reduction, i.e., oxidation state of an element of the reactant increases in one product and decreases in
another product.
 d- AND f-BLOCK ELEMENTS 8/15

8.4.8. Catalytic properties

Many of the transition metals and their compounds are used as catalysts e.g.,
(i) Vanadium as V2O5 in the manufacture of H2SO4 by contact process.
(t7) Cobalt as Cobalt-Thorium in the Fischer - Tropsch proce.ss in the synthesis of gasoline (petrol).
(in) Nickel in the finely divided state in the hydrogenation reactions, e.g., hydrogenation of oils into fats.
(iV) Platinum as spongy platinum in the manufacture of H->S04 by contact process and in the oxidation of
ammonia to nitric acid by Ostwald's process,
(v) Iron in the synthesis of ammonia by Haber’s process. Molybdenum is used as a promoter.
Explanation. The different reasons as to why transition elements act as catalysts are outlined
below :

w
(/) The catalytic properties of the transition elements are probably due to the presence of unpaired
electrons in their incomplete d-orbitals and hence posse.ss the capacity to absorb and re- emit wide range of

F lo
energies. This makes the required energy of activation available.
(//) In some cases, the transition element (i.e.. catalyst) with varying oxidation states may form

ee
intermediate compounds with one of the reactants (Intermediate compoundformation theory or Activated

Fr
complex theory of catalysis). These intermediates provide a new path with lower activation energy.
Further, they get decomposed on reaction with the other reactant regenerating the catalyst (i.e., transition
metal).
for
ur
X Catalyst
+ ■>
X.Catalyst
One reactant Transition metal Intermediate unstable compound
s
ook
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(Activated complex)
X.Catalyst + Y XY + Catalyst
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Other reactant Product

How the change in oxidation state of the transition metal helps the reaction through the formation of
intermediate may be seen from the reaction between SO2 and O2 to form SO3 in presence of V2O5.
r
ou
ad

V20g + SO2 V2O4 + SO3

Divanadium tetraoxide
Y

2 V2O4 + O2 ^ 2V2O5
Re
nd

In the above reaction, vanadium changes its oxidation stale from + 5 to + 4 and again to + 5.
Another example may be quoted of the reaction between iodide and persulphate ions in the presence of
Fi

Fe (III) as catalyst.
Fe3+
2I-+S2OI- ^ I2+2SO2-
The reaction is believed to take place as follows :
2Fe^+ + 21- — 2Fe2+ + I2
2Fe2++S20|- — + 2Fe3++2S02-

(Hi) In a number of cases, the transition metal provides a suitable large surface area with free valencies
on which the reactants are adsorbed. As a result, the concentration of the reactants on the surface of the
catalyst increases. Hence, the rate of reaction increases (Adsorption theory)
8.4.9. Coloured Ions

Most of the transition metal compounds (ionic as well as covalent) are coloured both in the solid stale
and in aqueous solution in contrast to the compounds of s and p- block elements.
8/16 'P'taulee^'4. New Course Chemistry (XII)C&m

Explanation. Colour is due to the presence of

incomplete <rf-subshell. Further, when the anions or the
ligands approach the transition metal ions, their (i-orbitals
do not remain degenerate. They split into two sets, one ^xv ^yz ^xz
f/-orbitais
consisting of lower energy orbitals (^2^ which includes ^2g
dy, and d^ and the other consisting of higher energy
orbitals {e S which includes d X i_ y 2 d^^^ )● called crystal field splitting (discussed in detail in
unit 9).
Thus, the electrons can jump from lower energy i/-orbilals to higher energy i/-orbitals. The required
amount of energy to do this is obtained by absorption of light of a particular wavelength in the region of
visible light. One or more electrons are promoted from a lower to higher level within the same t/-subshell.
The energy required to promote such an electron is very small. Radiations of light corresponding to such
small amount of energy are available within the visible region of light. The transition metal ions have a

w
property to absorb such radiations from the visible light and appear coloured due to emission of the remainder
coloured light. If a substance absorbs wavelength corresponding to red light, the transmitted light will

F lo
as

consist of wavelengths corresponding to the other colours especially greenish blue colour and the substance
will appear greenish blue in colour.
V B

e
The observed colour and the colour absorbed by the .substance / V = Violet

Fre
' B = Blue
are called complementary colours. Complementary colours can R ■ p G = Green
be identified using Munsell colour wheel as shown on right hand Y = Yellow

side ; for O
.
\
Y
O = Orange
R = Red
r
Opposite colours ai’e complementary.
You
Thus, Cu^"^ salts look blue due to the absorption of the red wavelength*. Anhydrous cobalt (II) compounds
s
ook

also absorb red light and appear blue. A complex of titanium [Ti(H20)^]^’*’ is purple in colour. It can be
eB

explained on the basis of crystal field theory as follows :

In [Ti(H20)g]^‘^, Ti has d ^ configuration and this electron is present in t2g orbital in ground state as
shown below. On absorption of yellow wavelength, the electron is excited to the orbital.
our
ad

.\bsorption
>,
C3> ofYcllow
e
g t
dY

Wave
Re

LU
t
Length 2g
Fin

Ground State Excited State

Since yellow wavelength is absorbed from visible region of light, the blue and red light will be transmitted
and solution of (Ti(H-,0)gl^''' will appear purple which is the mixed effect of blue and red colours.
The wavelength of the light absorbed depends upon the nature of the ligand. In aqueous solution in which
water molecules are the ligands, the colours of some metal ions are given in Table 8.11 on the next page,
Sc^'*' and Ti'^'*' have completely empty i/-orbitals and are colourless. Cu"^, Ag'*', Au"*", Zn^"^, Cd^"*" and
Hg^'^ have completely filled r/-orbitals and there are no vacant r/-orbitals for promotion of electrons. Hence,
they are also colourless.
Note, s- and p-block elements do not have a partially filled rZ-subshell. Hence, there cannot be any
d-d transition. The energy to promote an s orp-electron to a higher energy level is much greater and corresponds
to UV light being absorbed. Therefore, the compounds of s- and p-block elements are white and not coloured.

*Red and greenish blue are said to be complemenuiry colours. Complementary colours are those which
when mixed together produce white light.
 d- AND f-BLOCK ELEMENTS 8/17

8.4.10. Magnetic properties

On the basis of behaviour in a magnetic field, substances arc classified as paramagnetic, diamagnetic
and ferromagnetic. Those substances which are attracted by the applied magnetic
field are called paramagnetic whereas those which are repelled by the magnetic field are called diamagnetic*.
Substances which are very strongly attracted by the applied field are called ferromagnetic. Thus,
ferromagnetism is an extreme case of paramagnetism.
Paramagnetism is a property due to the presence of
unpaired electrons. In case of transition metals, as they contain TABLE 8.11. Colours of some

unpaired electrons in the (n— 1) d-orhifals, most of the transition of the hydrated first row

w
transition metal ions
metal ions and their compounds are paramagnetic As the
number of unpaired electrons increases from one to five, the
ion Configuration Colour
magnetic moment and hence paramagnetic character also
increases. Those transition elements which have paired electrons Sc^* 3c/
0
Colourless

o
e
are diamagnetic. Ti4+ 3c/
0
Colourless

re
The magnetic moment of an electron is partly due to its Tj3+ 3c/ I
Purple
orbital motion and partly due to its spin motion. Since in y4+ 3c/
I
Blue

Frl
F
transition metal ions, the unpaired electrons are present in the V3+ 3c/2 Green
outer shells, hence in such cases, spin contribution is much more V2* 3c/3 Violet
ou
important than the orbital contribution. Hence, magnetic moment Cr^-^ 3c/ Violet

sor
is calculated from ‘spin only formula’, viz., Mn^+ 3c/4 Violet
3c/4 Blue
P = t/«(« + 2) B.M. kf
Mn2+ 3c/5 Pink
oo
where n is number of unpaired electrons and B.M. stand for pg3+ 3c/5 Yellow
Bohr magneton. 3c/^ Green
Y
B

Thus, the magnetic moment of a substance varies only with Co3+ 3c/6 Blue

the number of unpaired electrons present in it. The calculated Cq2* 3c/’ Pink
re

values as well as the observed values (for hydrated ions in Ni2+ 3c/S Green
oY

solution or in the solid state) for ions of the first transition series Cu2+ 3c/^
u

Blue
are given in Table 8.12 on the next page. Zn--" 3c/
lu
Colourless
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d

*The paramagnetic or diamagnetic nature of a substance can be found by weighing the substance, first in
in

the air and then after suspending it

Re

between the poles of magnetic field.

F

If the sample weighs more in the

magnetic field, it shows that it is
attracted by the magnetic field and
hence it is paramagnetic. If it weighs
less in the magneticfield, it shows that
it is repelled by the magnetic field and
hence it is diamagnetic. A similar type o
I © i e
of balance is used for measurement of N s N S

magnetic susceptibility of a substance.

The sample is taken in a thin tube. This (a) Weighing in the absence of magnetic field
magnetic balance is known as Gouy’s (b) Sample attracted by the magnetic field-paramagnetic
balance. (c) Sample repelled by the magnetic field-diamagnetic

eh
**The magnetic moment is expressed in Bohr magnetons (B.M.). 1 B.M. = 4tc me
= 9-27 X A m2
where h is the Planck’s constant, e is the electronic charge, c is the velocity of light and m is the mass of electron.
8/18 7^>uidce^ <t New Course Chemistry (XII)

TABLE 8.12. Magnetic moments of some of the transition metal ions

ion Eiectronic Number of unpaired Magnetic moment (Bohr Magneton)
configuration eiectrons Calculated Observed

0
Sc^-" 0 0 0
1
Tj3+ 3J 1 1-73 1-75

Ti
2+
3J2 2 2-84 2-76

V-
2+
3d^ 3 3-87 3-86

ow
Cr2+ 3d-^ 4 4-90 4-80

Mn-+ 3d^ 5 5-92 5-96

Fe-+ 3d^ 4 4-90 5-3 — 5-5

Co^+ 3d'^ 3 3-87 4-4 — 5-2

e
Ni-+ 3^i 2 2-84 2-9 — 3-4

re
Cu--*- 3^^ 1 1-73 1-8 — 2-2

Frl
10 0 0
Zn-+ 3d 0

F
8.4.11. Complex Formation
ou
Transition metal ions form a large number of complex compounds. Complex compounds are those

r
which have a metal ion linked to a number of negative ions (anions) or neutral molecules having lone pairs of

so
electrons. These ions or molecules are called ligands. They donate lone pairs of eiectrons to the central
transition metal ion forming coordinate bonds (Refer to unit 9). kf
oo
A few examples are given below :
[Fe(CN)6]^-, [Fe(CN)6]'^, [Cu(NH3)4]2^-, [Zn(NH3)4]2+, [Ni(CN>4]2- and [PtC^l^-
Y
B

Such complex compounds are not formed by s- and /7-block elements.

Explanation. The transition elements form complexes because of the following reasons :
re

(/) Comparatively smaller size of their metal ions.

oY
u

(ii) Their high ionic charges. (Because of (0 and (ii), they have large charge/size ratio)
ad

(Hi) Availability of vacant <i-orbitals so that these orbitals can accept lone pairs of electrons donated by the
ligands.
d

FIGURE 8.7
TRAPPED INTERSTITIAL
8.4.12. Interstitial Compounds
in

ATOMS VOIDS
Re

The transition metals form a large number of interstitial compounds

F

in which small atoms such as hydrogen, carbon, boron and nitrogen occupy
the empty spaces (interstitial sites) in their lattices (Fig. 8.7).
They are represented by formulae like TiC, TiH2, Mn4N, Fe3H, Fe3C*
etc. However, actually they are non-stoichiometric materials, e.g., TiHj.-y,
VH 056 etc. and tlie bonds present in them are neither typically ionic nor Formation of

covalent. Some of their important characteristics are as follows : interstitial compounds.

(/) They are very hard and rigid, e.g., steel which is an interstitial compound of Fe and C is quite hard.
Similarly, some borides are as hard as diamond.
(ii) They have high melting points which are higher than those of the pure metals.
(in) They show conductivity like that of the pure metal,
(tv) They acquire chemical inertness.
Note. The adsorption of H2 by transition metals such as Pt, Pd, Ni etc. is called occlusion. It is due to
interstitial hydride formation.
*Fe3C is called cementite. It is an important constituent of steel.
d- AND f-BLOCK ELEMENTS 8/19

FIGURE 8.8
8.4.13. Alloy formation
ATOMS OF METAL M
Alloys are homogeneous solid solutions of two or more metals obtained by
melting the components and then cooling the melt. These are formed by metals
whose atomic radii differ by not more than 15% so that the atoms of one metal can
easily take up the positions in the crystal lattice of the other (Fig. 8.8).
Now, as transition metals have similar atomic radii and other characteristics,
hence they form alloys very readily. Alloys are generally harder, have higher melting ATOMS OF METAL M'

points and more resistant to corrosion than the individual metals. Formation of alloys.

The most commonly used are \ht ferrous alloys The metals chromium, vanadium, molybdenum, tungsten
and manganese are used in the formation of alloy steels and stainless steels. Some alloys of transition metals with
non-transition metals are also very common, e.g., brass (Cu + Zn) and bronze (Cu -i- Sn)
8.5. SOME GENERAL GROUP TRENDS IN THE CHEMISTRY OF

w
TRANSITION METALS (CROUP 3 TO GROUP 12) AND THEIR USES
Group 3 Elements (Scandium Group)

F lo
Elements Present: Scandium (Sc), Ytterium (Y), Lanthanum (La) and Actinium (Ac).
General valence shell configuration : (n - \) d^ 4.v- (« = 4 to 7)

ee
Characteristics : (/) They show an oxidation state of +3 in their compounds.

Fr
(//) Their compounds are c jlouriess and diamagnetic.
(Hi) Due to electropositi' e nature, they are reactive metals. Their reactivity increases with increased size.
(iv) They tarnish in air tnd bum in oxygen, giving the oxides, M2O3. for
ur
(v) They react with water to evolve H2 gas e.g., 2 La + 6 H2O — 2 La (OH)^ + 3 H2.
(vi) Their oxides and hydroxides are basic in nature.
s
ook

(vz7) Their atomic and ionic radii (M^"*" ions) increase from Sc to Ac.
Yo

(viii) Their tendency to form complexes decreases down the group due to increase in the size of the ions.
eB

Uses. The oxides of ytterium in the pure state are used in hearing aids, electronic and computer memory
cells.
our

Group 4 Elements (Titanium Group)

ad

Elements Present: Titanium (Ti), Zirconium (Zr) and Hafnium (HD-

General valence shell configuration : (n - 1) d^ 4r^ (n = 4 to 6).
Y

Characteristics : (/) They exhibit most stable oxidation state of +4 in their compounds (e.g., TiC^,
Re

Z1CI4, HfCl4 or Ti02, ZrO^ and Hf02). Titanium also shows an oxidation state of +3 (e.g., in TiCl3).
nd

(ii) The compounds in +4 state in the anhydrous state are covalent and tetrahedral. They are diamagnetic
Fi

and colourless. Titanium compounds in +3 state are paramagnetic and coloured.

(Hi) They are hard metals with high melting and boiling points,
(iv) These elements are unreactive at room temperature. However, on heating with hydrogen, carbon, nitrogen,
boron etc., they form interstitial compounds which are hard and refractory (not damaged by heating to
high temperatures). The best solvent for Ti is Hf.
(v) Due to lanthanide contraction, Zr and Hf have nearly same atomic and ionic radii (Zr = 160 pm, Hf =
159 pm ; Zr“^+ = 79 pm, HD*"^ = 78 pm). Hence, they show similar properties, e.g.,
(a) Both Zr and Hf are silvery white metals, (b) They occur together in minerals.
Note. Titanium is quite abundant in nature and mainly occurs in the form of oxide ores as rutile (Ti02)
and ilmenite (FeTi03).
Uses, (i) When alloyed with aluminium or tin, titanium has high strength. It is used in making of aircraft
frames and jet engines. Due to its resistance to corrosion, Ti is also used in making pipes, vessels for chemicals
etc. Ti02 is a brilliant white pigment used in paints.
(ii) Zirconium oxide is used as a refractory material.
 8/20
'P’uziieep. '4. New Course Chemistry (XII) ilwi
(iroup 5 Elements (Vanadium Group)
Elements Present: Vanadium (V), Niobium (Nb) and Tantalum (Ta).
General valence shell configuration : (n - \) ns^ {n = A - 6).
Characteristics : (/) Most stable oxidation state of this group is +5. In this state, the compounds are
diamagnetic and colourless except V2O5 which is red orange. This is due to crystal defects. Vanadium also
shows an oxidation of +3 and +4 (e.g., in V2O4, VCI3, VCI4).
(//) These metals are unreactive at room temperature. On heating with O2, they form oxides of the type
M2O5. Vanadium also forms V2O4.
(Hi) Due to lanthanide contraction, Nb and Ta have nearly the same size and hence many common properties.
(iv) They form interstitial compounds which are hard and refractory,
(v) Their tendency to form complexes decreases down the group.

w
Uses. (/) Vanadium is added to steel to increase its strength and toughness.
(/7) Niobium alloys are used in making jet engines.
(///) As tantalum is resistant to corrosion, it is used in making analytical weights and in surgery for bone

lo
nails.

e
(iv) V2O5 is used as catalyst for many reactions.

re
Group 6 Elements (Chromium Group)

rF
F
Elements Present: Chromium (Cr), Molybdenum (Mo) and Tungsten (W).
General valence shell configuration :(«-!) d'^ns^ or(n-l)d^ ns^ (n = 4 to 6).

r
Characteristics : (0 These elements show variable oxidation states. The most important oxidation
fo
u
states of Cr are +3 and +6 and those for Mo and W are +5 and +6. In carbonyls, they show an oxidation state
ks
of zero, e.g., in (Cr(CO)5].
Yo
(//) They are hard silvery white metals with high melting and boiling points. Tungsten is a metal with
oo

highest melting point.

B

Uses. (0 Chromium being resistant to corrosion is used in the chrome plating of iron articles.
(ii) All these metals form excellent alloys. For example, chromium is used to make ferrous alloys like
e

stainless steel, chrome steel etc. Mo and W form very hard alloys with steel which are used in making
ur

cutting tools.
ad
Yo

(Hi) These metals and their compounds are used as catalysts in a number of reactions.
(iv) Molybdenum is used in X-ray tubes,
d

(v) Tungsten is used for making filaments of electric bulbs,

Re
in

(v/) K2Cr207 is an important compound of Cr used as an oxidizing agent.

Note. These elements are relatively more abundant, chromium being the most abundant among these
F

metals. Their common ores are : chromite (FeCr204) for Cr. molybdenite (M0S2) for Mo and wolframite
(FeW04.MnW04) for W.
(iroup 7 Elements (Manganese Group)
Elements Present: Manganese (Mn), Technetium (Tc) and Rhenium (Re).
General valence shell configuration : (n- \) d^ ns^ (n = 4 to 6).
Characteristics : (1) These elements show an oxidation state of 0 to +7. For Mn, the important oxidation
states are -J-2, -1-4 and +7 and for Tc and Re, these are -1-4 and +7. The stability of higher oxidation state
increases with increase in atomic number.
(H) Both Mn and Re form a number of oxides. The basic nature of the oxides decreases as the oxidation
state increases, e.g.,

MnO Mn203 Mn02 MnOj Mn207

Basic Amphoieric Acidic
d- AND f-BLOCK ELEMENTS 8/21

(Hi) Due to lanthanide contraction, Tc and Re have similar atomic radii and hence similar characteristics.
(iv) They have high melting and boiling points.
Uses. (/) Manganese as such does not have any use but it used in the manufacture of alloys such as
ferromanganese (Fe + Mn) and manganese bronze (Mn + Cu + Zn).
(H) KMnO^ is an imporlanl useful compound of Mn.
(//7) MnO-, is used as a catalyst.
(iv) Rhenium is used in making electronic filaments, high temperature thermocouples and in flash bulbs.
Note. Among the elements of this group, only Mn is quite abundant. It occurs in the lorm of oxide ore,
pyrolusite (MnOo).
Group 8, 9 and 10 Elements
Elements Present and their Electronic Configurations : The elements of these groups have more
horizontal similarity than vertical similarity. Hence, they are grouped into three triads as follows :

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Group 8 Group 9 Group 10

First triad : Iron (Fe) Cobalt (Co) Nickel (Ni)

F lo
Valence shell E.C.: 4.S- 3^/■^ 45^ 4,r

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Second triad : Ruthenium (Ru) Rhodium (Rh) Palladium (Pd)

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Valence shell E.C.: Ad'' 5^' 4r/® 55^ Ad 5s^

Third triad : Osmium (Os) Iridium (Ir) for Platinum (Pt)

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Valence shell E.C,; 5d^ 6s- Sd'' 5d^ 6.V*
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The elements of the first triad are called ferrous metals whereas elements of the second and third triads
are called platinum metals.
eB

Characteristics of Ferrous Metals : (0 Iron and cobalt exhibit oxidation states of +2 and +3 in their
compounds whereas nickel generally shows an oxidation state of +2.
(//) They have high melting and boiling points.
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(//7) All the three are ferromagnetic.

(iv) Cobalt and nickel are not affected by air under ordinal^ conditions whereas iron undergoes rusting.
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(v) They react with dilute H2SO4 and dilute HNO3 but with cone. HNO3, they become passive.
Re

(vi) All of them combine with CO to form volatile carbonyls.

nd

(vii) Nickel forms complexes with coordination number 4 or 6 whereas iron and cobalt form complexes
Fi

with coordination number 6 only.

(viii) They form coloured compounds. For example, ferrous compounds are green, ferric compounds are
yellow, nickel compounds are green and cobalt compounds are pink.
Uses of Ferrous Metals. (/) All the three metals are used as catalysts especially in the reactions involving
addition of hydrogen.
(i7) Iron is the most abundant transition metal or second most abundant metal (next only to AI) in the
earth’s crust, ft is mainly obtained from its oxide ore, haematite (Fe20-^). It is used as such or converted into
steel (by adding carbon and m^ Is like Cr, Mn and Ni). Both iron and steel are most important construction
materials.

Characteristics of Platinum Metals : (/) They show a large number of oxidation states. The highest
oxidation state, +8 is shown by ruthenium and osmium in oxides, fluorides etc. However, lower oxidation
states are more common.

(ii) All of them form complexes.

(z7i) They are much less abundant than ferrous metals.
Uses of Platinum Metals. All of them are used as catalysts in industry.
8/22
New Course Chemistry fXinrosmn

Group U Elements (Coinage Metals)

Elements Present: Copper (Cu), Silver (Ag) and Gold (Au).
General valence shell E.C.: (n - 1) ‘ (n = 4 - 6).
The elements of this group are called coinage metals because in the early times, they were used in
making coins.
Characteristics : (i) Copper shows oxidation states of+1 and +2, +2 being more stable than +1. That
is why soluble cuprous salts in aqueous solution change into cupric salts (2 Cu"*" > Cu-'*' + Cu, a
disproportionation reaction). The insoluble cuprous salts like CU2I2, Cu2(CNS)2 etc. are comparatively stable.
Silver mainly shows an oxidation state of +1 whereas gold shows oxidation states of + l and +3, +3 being
more stable.

(//) They have a strong tendency to fonn complexes.

(/«) They lie below hydrogen in the electrochemical series (/.<?., they have positive standard electrode
potentials). Hence, they do not react with dilute acids to liberate H2 gas.

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(iv) They readily form alloys with each other as well as with many other metals. They also form amalgams
with mercury,

F lo
(v) Their compounds in +1 state are colourless whereas in higher oxidation states, they are coloured.
Uses. Silver and gold are used in jewellery. Silver halides (especially silver bromide) is used in
photography.

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Group 12 Elements (Zinc Group)

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Elements Present: Zinc (Zn), Cadmium (Cd) and Mercury (Hg).
General valence shell E.C.: (n - 1) J (n = 4 to 6).
Characteristics : In this group, although all of them have
for
configuration, yet zinc and cadmium are
ur
similar whereas mercury is somewhat different from them. For example,
(/) Zinc and cadmium are solids whereas mercury is liquid.
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(») Zinc and cadmium show an oxidation state of +2 only whereas mercury shows oxidation states of + 1
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and +2.
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{Hi) Zinc and cadmium dissolve in dil. HCl and dil. H2SO4 with evolution of H2 but mercury is not affected.
In fact, due to their completely filled d configuration, they show very little characteristics of transition
metals.
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Uses. (0 Zinc is mainly being used to prevent corrosion of iron by depositing a thin film of zinc on iron
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surface (a process called galvanisation).

(«) Zinc is used in making alloys like brass (Cu + Zn) and bronze (Cu + Zn + Sn).
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{Hi) Zinc is also used in several types of batteries.

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nd

{iv) Mercury is used in thermometers, barometers, gas pressure regulators and as electrode etc.
Fi

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. All transition elements are d-block elements but all d-block elements are not transition elements. For
example, elements of Group 12, namely Zn, Cd and Hg, though belong to fr-block but are not considered
as transition elements because they contain completely filled {n—1) ^/-orbitals and do not show properties
like other transition elements except the formation of complexes.
2. Elements of Group 3 namely Sc, Y, La and Ac, also differ from the transition elements in the facts that
they do not show variable oxidation state, Le., they show oxidation state of + 3 only, their compounds are
diamagnetic and colourless while those of other transition elements are paramagnetic and coloured. That
is why elements of Group 3 and Group 12 are called ‘non-typical transition elements' while the rest are
called ‘typical transition elements’.
3. The highest oxidation state shown by any transition element is + 8, viz., Ru in 4 J-series and Os in 5i/-series.
4. Isomorphism is most common among transition metal compounds, e.g., FeS04.7 H2O and ZnS04.7H-,0.
This is due to similarity in the size of their metal ions.
5. The transition metal which is most abundant and used very largely is ‘iron’.
d- AND f-BLOCK ELEMENTS 8/23

6. Both Au and Pt are noble metals and dissolve only in aqua regia (cone. HCl + cone. HNO3 in ratio of
3 : 1) due to formation of H2PtClg and HAUCI4 whereas silver reacts with cone. HNO3 and cone. H2SO4
but is not attacked by aqua regia.
7. Mercury is the only transition metal which is liquid at room temperature. This is due to the very high
ionization enthalpy making it difficult for electrons to participate in metallic bonding.

Curiosity Questions
r Q. 1. Which element/elements is/are alloyed with aluminium or tin in making air craft frames and
jet engines and why ?
Ans. Titanium and niobium are used for alloying with aluminium or tin in making aircraft frames and jet
engines. This is because the alloys obtained have high strength.

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Q. 2. Which element Is used in making bone nails for surgery and why ?
Ans. Tantalum is used because it is resistant to corrosion.

Q. 3. Which element is used for making filaments of electric bulbs and why ?

Flo
Ans. Tungsten Is used in making filaments of electric bulbs because it has very high melting point.
Q. 4. Which element is generaliy used in the X-ray tube for production of X-rays and why ?

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Ans. Molybdenum is used because It is a heavy element. When cathode rays hit this element, X-rays

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are produced.
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for
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SECTION II. SOME IMPORTANT COMPOUNDS OF TRANSITION ELEMENTS
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8.6. OXIDES
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(/) The metals of the first transition series form oxides with oxygen at high temperature.
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(ii) The oxides are formed in the oxidation, stales +1 to +7 (as given in Table 8.10, page 8/14).
(///) The highest oxidation state in the oxides of any transition metal is equal to its group number, e.g„ 7 in
re

Mn-,07. Beyond group 7, no higher oxides of iron above Fc203 are known. Some metals in higher
oxidation state stabilize by forming oxocations, e.g., as VO^, as V02+ and Ti‘^ as 1TO-*.
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(iv) All the meiais except scandium form the oxides with the formula MO which are ionic in nature. As the
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oxidation number of the metal increases, ionic character decreases, e.g., Mn207 is a covalent green oil.
Even C1O3 and V2O5 have low melting points.
nd
Re

+2 +8/3 +3 +4 +7

MnO Mn304 MnaOs Mn02 Mn207

Fi

IONIC CHARACTER DECREASES

(v) In general, the oxides in the lower oxidation states of the metals are basic and in their higher oxidation
states, they are acidic whereas in the intermediate oxidation state, the oxides are amphoteric.
For example, the behaviour of the oxides of manganese may be represented as follows:
+2 +8/3 +3 +4 +7

MnO Mn304 Mn203 Mn02 Mn207

Basic Amphoteric Amphoteric Amphoteric Acidic

ACIDIC CHARACTER INCREASES

Thus, Mn207 dissolves in water to give the acid HMn04.

 8/24 'PxncUeilt-'A New Course Chemistry (X11)E!2SZS]

In case of vanadium, there is a gradual change from the basic lo less basic V2O4 and to amphoteric
V2O5 though it is mainly acidic. Thus, V2O4 dissolves in acids to give VO-'*’ salts whereas V->05 reacts with
alkalies as well as acids to give VO^“ and VO4 respectively.
Similarly, CrO is basic, Cr203 is amphoteric while Cr03 is acidic.
Thus, C1O3 dissolves in water lo give the acids H,Cr04 and H2Cr207.
8.7. SALTS CONTAINING OXOANIONS OF TRANSITION METALS

8.7.1. Potassium Dichromate (K2Cr202)

ow
Preparation. It is prepared from the ore called chromite or fenvchrome or chrome iron, Fe0.Cr^03.
The various steps involved are as follows :
(0 Preparation of sodium chromate. The ore is finely powdered, mixed with sodium carbonate and
quick lime and then roasted, i.e., heated to redness in a reverberatory furnace with free exposure to air when
sodium chromate (yellow in colour) is formed and carbon dioxide is evolved. Quick lime keeps the mass

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porous and thus facilitates oxidation.

re
4Fe0.Cr203 + O2 ■> 2Fe,03 + 4Cr,03
4Na2C03 + 2Cr203 + 30,

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4Na,Cr04 + 4CO2] x 2

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4Fe0.Cr203 + 8Na2C03 + 7O2 8Na,Cr04 + 2Fe,03 + 8CO2
Chromite ore
ou Sod. chromate Ferric oxide

sor
After the reaction, the roasted mass is extracted with water when sodium chromate is completely dissolved
while ferric oxide is left behind. Ferric oxide is separated out by filtration.
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(i7) Conversion of sodium chromate into sodium dichromate. The filtrate containing sodium chromate
oo
solution is treated with concentrated sulphuric acid when sodium chromate is converted into sodium dichromate.
2Na2Cr04 + H2SO4 ■> Na2Cr,07 + Na,S04 + H,0
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B

Sod. chromate (Cone.) Sod. dichromate

Sodium sulphate being less soluble crystallizes out as decahydrale, Na,S04.10H,0 and is removed.
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The clear solution is then evaporated in iron pans to a specific gravity of 1-7 when a further crop of sodium
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sulphate is formed. It is removed and the solution is cooled when orange crystals of sodium dichromate,
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Na2Cr207.2H20 separate on standing.

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{Hi) Conversion of sodium dichromate into potassium dichromate. Hot concentrated solution of sodium
d

dichromate is treated with calculated amount of potassium chloride when potassium dichromate, being much
less soluble than sodium salt, crystallizes out on cooling as orange crystals.
in
Re

Na2Cr207 + 2KC1 K,Cr207 + 2NaCl

F

Sodium dichromate Potassium dichromate

Properties. I. Colour and Melting point. It forms orange crystals which melt at 669 K.
2. Solubility. It is moderately soluble in cold water but freely soluble in hot water.
3. Action of heat. When heated to a white heat, it decomposes with the evolution of oxygen.
4K2Cr20y 4K2Ci04 + 2Cr203 + 3O2
4. Action of alkalies. When an alkali is added to an orange red solution of dichromate, a yellow solution
results due to the formation of chromate.

K2Cr207 + 2KOH ^ 2K2C1O4 + H2O

Pot. dichromate Pot. chromate

or
Cr202-+20H" 2Cr02-+H20
On acidifying, the colour again changes to orange red due to the reformation of dichromale.
2 K2Cr04 + H2SO4 ■4 K2Cr,07 + K2SO4 + H,0
or 2Cr02-+2H+ ^ Cr20^“ + H2O
d- AND f-BLOCK ELEMENTS 8/25

This interconversion is explained by the fact that in dichromate solution, the 0x2^^ ions are invariably
in equilibrium with CrO^“ ions at pH = 4, i.e.,
pH=4
Cr^O^” + H2O V 2Cr02- + 2H-"
Orange red Yellow
(dichromale) (chromate)
On adding an alkali {le., increasing the pH of the solution), the ions are used up and according to the law
of chemical equilibrium, the reaction proceeds in the forward direction producing yellow chromate solution. On
the other hand, when an acid is added {i.e., pH of the solution is decreased), the concentration of H'*' ions is
increased and the reaction proceeds in the backward direction producing an orange red dichromate solution.

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Retain in Memory
The oxidation state of chromium in chromate and dichromate is same, i.e., +6.
Further, Cr(V) undergoes disproportionation in acidic medium as follows :

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V VI ill

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2Q03-+8H+ > Cr0|- + Cr3++4H20

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5. Action of concentrated sulphuric acid
(a) In cold, red crystals of chromic anhydride (chromium trioxide) are formed.
K2Cr207 + 2H2SO4
ou 4 2Cr03 + 2KHSO4 + H^O

sor
(b) On heating the mixture, oxygen is evolved.
2X2^307 + 8H2SO4 4 2K2SO4 + 2Cr2(S04 )3 + 8H2O + 3O2
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6. Oxidising properties. It is a powerful oxidising agent. In the presence of dilute sulphuric acid, one
molecule of potassium dichromate furnishes 3 atoms of available oxygen as indicated by the equation :
oo
K2Cr207 + 4H2S04- ■> K2SO4 + Cr2(S04)3 + 4H2O + 30
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Cr202-+14H+ +6e~ 2Cr^^ + 7H2O
B

or ■>

In other words, Cr20^“ ion takes up electrons and hence acts as an oxidising agent.
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Thus, Cr,07“ ion gains 6 electrons or in the above reaction, Cr (+6) is reduced to Cr (+3) so that total
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decrease in oxidation state of Cr207~ ion is 6.

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C * f 294
d

Eq. wt. of K2Cr207 = = 49

6 6
in

It is on this account that it is used in titrations for which one sixth of the molecular weight of K2Cr207
Re

dissolved in a litre of water would give one normal (IN) solution (containing one gram equivalent of K2Cr207).
F

Some of the oxidizing properties of K2Cr207 in presence of dilute H2SO4 are given below :
(/) It liberates l2from Kl
K2Cr207 + 4H2SO4 4 K2SO4 + Cr2(S04 )3 + 4H2O + 30
6KI + 3H2SO4 + 30 -4 3K2SO4 + 3I2 + 3H2O
K2Cr207 + 7H2SO4 + 6KI ■> 4K'5S04 + Cr2(S04 )3 + 3I2 + 7H2O
or
Cr202- + I4H-^ + 61- 2Cr^-^ + 7H2O + 3I2
iii) It oxidises ferrous salts to ferric salts
+ 4H2SO4 4 K2SO4 + Cr2(S04)3 + 4H2O + 30
2FeS04 + H2SO4 + O -> Fe2(S04)3 + H2O] X 3
K2Cr207 + 7H2SO4+ 6FeS04- ^ K2SO4 + Cr2(S04)3 + 3F62(S04)3 + 2H2O
or
Cr202- + 6Fe2+ + 14H^ ->2Cr3-"+6Fe^-^+7H20
The above two reactions are used in the estimation of iodine and ferrous iron in volumeliic analysis.
 8/26 l^uieice^'4. New Course Chemistry (XII)SSSIS]
(///) It oxidises to sulphur
K2Cr207 + 4H2SO4 ^ K2SO4 + Cr2(S04)3 + 4H2O + 30
H^S + 0 ^H,0 + S]x3

K2Q-207 + 4H2SO4 + 3H2S ■ ^ K2SO4 + Cr2(S04)3 + 7H2O + 3S

or
Cr202- + 8H-^ + 3H2S -> 2Cr^* + 7 H-,0 + 3S
(iv) It oxidises sulphites to sulphates and thiosulphates to sulphates and sulphur
^ K2SO4 + ^2(804)3 + 4H2O + 30
N32S03 + O Na2S04l X 3
K2Cr207 + 4H2SO4 + 3N32S03 - K2SO4 + ^2(804)3 + 4H2O + 3NE2S04
or
Cr2Q2-+8H++3S02- — ^ 2Cr3+ + 3S02'+4H20

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Similarly, Cr20^“+8H++3S20j“- 2 + 3SQ2- + 3 S + 4 H2O
Thiosulphaic

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(v) It oxidises nitrites to nitrates
K2Cr207 + 4 H2SO4 -4 K2SO4 + 019(804)3 + 4 H2O + 3 (O)

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NaN02 + {0) ^ NaN03l X 3

Fr
KoCr^O^ + 4 H2SO4 + 3 NaN02 K2SO4 + Cr2(S04 )3 + 3 NaNOj + 4 H2O
Cr202- +8H+ +3NO2 ^ 2 Cr^-" + 3 NOj + 4 H2O
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or
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(i^z) It oxidises halogen acids to halogen
+ 14 HCI ^ 2KC1 + 2CrCl3 + 7H2O + 3CI2
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(w7) It oxidises SO2 to sulphuric acid
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K2Ct2P-, + 4H9SO4 K9SO4 + Cr2(S04)3 + 4H2O + 30

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SO2 + O + H2O ^ H2SO4] X 3

K2Cr207 + H2SO4 + 3SO2 -> K9SO4 + Cr2(S04)3 + 3H2O
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or
Cr^O^- +2H+ + 3S02- 012(804)3 + H2O
Reduction of dichromate by sulphur dioxide can be used to prepare chrome alums.
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iviU) It oxidises stannous salts to stannic salts

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nd

0r2O2- + I4H++3Sn2+ ■>2 0r3+ + 3Sn^ + 7H20

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(zx) It oxidises ethyl alcohol to acetaldehyde and acetic acid.

K90r207 + 4H2SO4 4 K2SO4 + 0r2(SO4)3 + 4H2O + 30
OH3OH2OH + O OH3OHO + H2O ; CH3CHO +0 ^ OH3OOOH
Ethyl alcohol Acetaldehyde Acetaldehyde Acetic acid

7. Chromyl chloride test {Reaction with a chloride and cone, sulphuric acid). When heated with
concentrated hydrochloric acid or with a chloride and strong sulphuric acid, reddish brown vapours of chromyl
chloride are obtained.

K20r207 + 2H2SO4 -4 2KHSO4 + 2O1O3 + H2O

KOI + H2SO4 -4 KHSO4 + HOI] X 4
2O1O3 + 4H01 4 2O1O2OI2 + 2H2O
K20r9O7 + 4K01 + 6H2SO4 4 20rO20l2 + 6KHSO4 + 3H2O
Chromyl chloride
(Red vapour)
d- AND f-BLOCK ELEMENTS 8/27

This reaction is used in the detection of chloride ions in qualitative analysis.

However, it may be mentioned here that chlorides of Ag, Hg, Pb and Sn do not give this test.
8. Reaction with hydrogen peroxide. Acidified K^Cr^O-^ solution reacts with H,0, to give a deep blue
solution due to the formation of peroxo compound, CrO (0,)t.

Cr^O^- + 2 + 4 H2O2 2 Cr05 + 5 H^O

The blue colour fades away gradually due to the decomposition of CrOj into Cr^'*’ ions and oxygen.
O O
The structure of CrOj is in which Cr is in +6 oxidation state.
O O
O

Uses, (f) In volumetric analysis, it is used as a primary standard for the estimation of Fe^'*’ (ferrous ions)

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and 1~ (iodides) in redox titrations.
In medical field, estimation of Fe^^ ion is needed in the patient suffering from anaemia to know

F lo
the deficiency of iron. However, it is important to remember that though Na2Cr20j and K2Cr20y are
equally good oxidizing agents but Na2Cr-,Oj is not used in the estimation of Fe^* ions because it is
deliquescent. Moreove,r it has a strong corrosive action on skin and is highly poisonous as well as carcinogenic

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in nature.

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(ii) In industry, it is used (a) in chrome tanning in leather industry, (b) in tlie preparation of chrome alum
K2S04.Cr2(S04)3.24H20 and other industrially important compounds such as Cr20:;, CtO^, CrOiC^, K2C1O4,
CrClj etc. (c) in calico printing and dyeing, (d) in photography and in hardening gelatine film.
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(Hi) In organic chemistiy, it is used as an oxidising agent.
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Note. Both sodium and potassium dichromates are strong oxidizing agents. The sodium salt has a greater
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solubility in water and is extensively used as an oxidising agent in organic chemistry.

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Structures of chromate and dichromate ions.

FIGURE 8.9
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12- t2-
,0. o
/Os /Os o 0
/ \ / \ / \
/ \ / \ / \
163 pm
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/ \ / \ / \
/
Cr '
/ \
or
Cr
/ \ 1
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nd

/ \
/ \ .9
"Oc- O" o 0 .0.
''O O
o ^ o o
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0 O

0Chromate ion-Tetrahedral structure 0Dichromate ion-Two tetrahedra sharing one oxygen atom at one corner

(a) Structure of chromate ion (b) Structure of dichromate ion.

In Cr04 ion, Cr is .^p^ hybridised and all the Cr—O bonds are equivalent. In Cr202 ion, the two
Cr—O bonds which share an oxygen atom at the common vertex of two tetrahedral units lu-e longer than the
other six equivalent Cr—O bonds.
Retain in Memory
It is important to know that both in CrO^" and Cr^O^' ions, Cr (VI) has efi configuration.
Hence, yellow colour of CrO\~ and orange colour of Cr20^~ are not due to d-d transition but
due to charge transfe,r i.e., momentary transfer of charge from 0-atom to metal atom thereby
changing 0"~ ion momentarily to O" ion and reducing the oxidation state of Cr from +6 to +5.
8/28 ‘pfuxcU^^’^ New Course Chemistry (XII)CZslSl

Curiosity Question
r Q. Which solution is used by police to test that a person is drunk ? How is it done ?
Ans. Acidified potassium dichromate solution is used which has orange colour. The person is asked to
breathe into the solution taken in a test tube. If the person has consumed alcohol, the orange
colour will change into green colour due to oxidation of alcohol and reduction of acidified potassium
dichromate to green coloured chromium sulphate.
J
8.7.2. Potassium Permanganate
Preparation. On a large scale, it is prepared from the mineral, pyrolusite, MnO-,. The preparation
involves the following two steps :
(/) Conversion of Mn02 into potassium manganate. The finely powdered pyrolusite mineral is fused

low
with potassium hydroxide or potassium carbonate in the presence of air or oxidising agent such as potassium
nitrate or potassium chlorate when green coloured potassium manganate is formed.
2Mn02 + 4KOH + O2 ->
2K2Mn04 + 2H2O
Potassium manganate

2Mn02 + 2K2CO3 + O2 - ^ 2K2Mn04 + 2 CO2

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Mn02 + 2KOH + KNO3

F^ K2Mn04 + KNO2 + H2O

Fr
3Mn02 + 6KOH + KCIO3 ^ 3K2Mn04 + KCl + 3H2O
Potassium manganate thus formed undergoes disproportionation in the neutral or acidic solution as

for
ur
follows, if allowed to stand for some time :
VI VII IV

3MnOj- +4H 2Mn04 -1- M11O2 + 2H2O

ks
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Manganate ion Permanganate ion
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(//) Oxidation of potassium manganate to potassium permanganate,

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{a) Chemical oxidation. The fused mass is extracted with water and the solution after filtration is converted
into potassium permanganate by bubbling carbon dioxide, chlorine or ozonised oxygen through the green solution.
^ 2KMn04 + Mn02 -I + 2K2CO3
r

3K2Mn04 2CO2
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ad

2K2Mn04 + Cl2- >2KMn04+2KCi

2K2Mn04 + H2O + O3 > 2KMn04 + 2KOH + O2
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The carbon dioxide process is uneconomical as one third of the original manganate is reconverted to
nd
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manganese dioxide. However, this process has the advantage that the potassium carbonate formed as a by
product can be used for the oxidative fusion of manganese dioxide. In the chlorine process, potassium chloride
Fi

obtained as a by-product is lost.

(b) Electrolytic oxidation. Now a days, for manufacturing potassium permanganate commercially, the
method of electrolytic oxidation is preferred. The alkaline manganate solution obtained in step (i) is electrolysed
between iron electrodes separated by diaphragm. The reactions taking place are as follows :

KiMn04 V ^ 2K-^+ MnO“-

H2O ^ H-" + OH-

At Anode: MnOj- Mn04 +e

At cathode : H-" -H e" ^ H ; 2H ^ H2
Thus, manganate ions are oxidized to pennanganate at the anode and hydrogen gas is liberated at the
cathode.

After the oxidation is complete, the solution is filtered and evaporated under controlled conditions to
obtain the crystals of potassium permanganate.
d- AND f-BLOCK ELEMENTS 8/29

Laboratory Preparation. In the laboratory, potassium permanganate is prepared by oxidation of

manganese (II) ion salt by peroxodisulphate
2Mn2++ SSjOf- + 8H2O 2Mn04 +10SO2-+16H+
Peroxodisulphate Permanganate
Properties.
1. Colour. Potassium permanganate exists as deep purple black prisms with a greenish lustre which
become dull in air due to superficial reduction.
2. Solubility. It is moderately soluble in water at room temperature and it is more soluble in hot water.
3. Action of heat. When heated to 513 K, it readily decomposes giving oxygen.
2KMn04 K2Mn04 + Mn02 + O2
Pot. manganate

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At red heat, potassium manganate formed decomposes into potassium manganite (K2Mn03) and oxygen.
2K2Mn04 > 2K2Mn03 + O2
4. Action of cone. H2SO4. With well cooled cone H2SO4, it gives Mn207 which on warming decomposes
to Mn02.

e
Fl
2KMn04 + 2H2S04 ■^Mn207 + 2KHS04+H20

re
2Mn20y - ^ 4Mn02 + 3O2

F
With warm cone. H2SO4, O2 gas is given out.
4KMn04 + 6H2S04
ur
^ 2K2SO4 + 4MnS04 + 6H2O + 5O2

r
fo
5. Action of heat in current of hydrogen. When heated in a current of H2, solid KMn04 gives KOH,
MnO and water vapours.
ks
2KMn04+5H2 2KOH + 2MnO + 4H2O
Yo
6. Oxidising property. Potassium permanganate is a powerful oxidising agent. The actual course of
oo

reaction depends on the use of the permanganate in (a) neutral {b) alkaline or (c) acidic solution.
B

(a) In neutral solution. Potassium permanganate acts as a moderate oxidising agent in neutral aqueous
solution because of the reaction :
re

2KMn04 + H20 -» 2KOH + 2Mn02 + 30

u
ad

or
Mn04 + 2H2O + 3 e~- ^ Mn02 + 40H-
Yo

Thus, Mn04 ion gains 3 electrons. Also, according to the above equation, in neutral medium, from two
moles of permanganate, three oxygen atoms are available. In fact, during the course of reaction, the alkali
d
Re

generated renders the medium alkaline even when we start with neutral solutions. Hence, Eq. wt. of KMn04
in

in neutral or weakly alkaline medium

F

Mol.wt. 158
= 52-67
3 3

Some oxidizing properties of KMn04 in the neutral medium. These are given below :
(i) It oxidises hot manganous sulphate to manganese dioxide.
2KMn04 + H2O ^2K0H-f2Mn02+30
3MnS04+3H2O + 30 4 3Mn02 + 3H2S04
2K0H + H2SO4 ->K2S04 + 2H20
2KMn04 + 3MnS04 + 2H2O ^ K2SO4 + 2H2SO4 + 5Mn02
Presence of ZnS04 or ZnO catalyses the oxidation.
(«) It oxidises sodium thiosulphate to sodium sulphate.
8KMn04 + 3Na2S203 + H2O — ^ 3K2SO4 -f 8Mn02 + 3Na2S04 + 2K0H
(Hi) It oxidises hydrogen sulphide to sulphur.
2KMn04 + 4H2S — ^2MnS + S + K2S04 + 4H20
8/30 New Course Chemistty CXlI)ISEISly

MnO| (manganate) ion is produced.

(b) In alkaline solution. In strongly alkaline solution,
2KMn04 + 2K0H 4 2K2Mn04 + H2O + O or
MnO]j“ + e~ MnOj“
Mol. wt. 158
Eq. wt. of KMn04 = = 158
1 1
Potassium manganate is also further reduced to Mn02 when a reducing agent is present.
K2Mn04 + H20 4Mn02 + 2K0H + 0
or
MnOj- + 2H2O + 2 C-- ^Mn02 + 40H-

ow
So the complete reaction is : 2KMn04 + H20- ^2Mn02 + 2K0H + 30
or
MnO^ + 2H2O + 3 e~ —>Mn02 + 40H-
which is the same as that for neutral medium. Hence, equivalent weight of KMn04 in weakly alkaline

e
medium is same as that in the neutral medium, viz., 52-67

re
Some oxidizing properties of KMn04 in the alkaline medium. These are given below :
(/) It oxidises potassium iodide to potassium iodate.

F
Frl
2KMn04-f H2O + KI ^ 2Mn02 + 2KOH + KIO3
Pot. iodate

or I- + 60H-
ou ^ lOj + 3H2O + 6 e-

osr
In this case, iodine is not liberated unlike the case of acidic medium.
Similar reaction takes place with KBr.
kf
(«) It oxidises olefinic compounds to glycols, i.e., when an olefmic compound is shaken with alkaline KMn04,
oo
pink colour of KMn04 is discharged.
Y
CH. CH.OH
B

II ' + H2O + (O)

Alkaline
^ I '
CH2 KMn04 CH2OH
re
uY

Ethylene Ethylene glycol

Alkaline KMn04 used for this test is known as Baeyer’s reagent. It is used for oxidation of a number of
ad

organic compounds,
do

(c) In acidic medium. Potassium permanganate in the presence of dil. sulphuric acid, Le., in acidic
medium, acts as a strong oxidising agent because of the reaction
in

2KMn04 +3H2SO4 ^ K2SO4 -1- 2MnS04 + 3H2O + 50

Re

Mn04 +8H-^+5e- ■^Mn2++4H20.

F

or

Since in the above reaction, MnO^ ion gains 5 electrons or five atoms of oxygen are available from two
molecules of KMn04. Hence,

Eq. wt. of KMn04 = 1^8

5
= 31-6

Some oxidizing properties of KMn04 in the acidic medium. These are given below :
(r) It oxidises H2S to S.
2KMn04+3H2S04 K2SO4 + 2MnS04 + 2H2O + 50
H2S -1- O ^ H2O + S] X 5
2KMn04 + 3H2S04-f 5H2S -4 K2SO4 + 2MnS04 + 3H2O + 5 S
or
2Mn04 + 16H+-1-5S2- - ^2Mn2+-i-8H20-i-5S
d- AND f-BLOCK ELEMENTS 8/31

(//) It oxidises sulphur dioxide to sulphuric acid.

2KMn04 + 3H2SO4 ^ K2SO4 + MnS04 + 3H2O + 50
SOt + H-)0 + O H2SO4IX5
2KMn04 + 5 SO2 + 2H2O > K2SO4+ 2MnS04+ 2H2SO4
or 2Mn07 + 5 SO, + 2H,0 ^ 5S0^
4
+2Mn-^ + 4H+

(Hi) It oxidises nitriles (NO2 ) to nitrates (NO^ ), ursenites (AsO^ ) to arseniaies (AsO\ } and
sulphites and thiosulphates to sulphates.

ow
2KMn04 + 3H,S04 4 K2SO4 + 2MnS04 + 3H,0 + 50
KNO2 + O ■4 KNO3] X 5 ~
2KMn04 + 3H2SO4 + SKNO, 4 K2SO4 + 2MnS04 + 3H,0 + 5KNO3

e
or
2Mn04+6H++5N02 4
2Mn~+ +3H2O+5NOJ

re
Similar equations can be written for the oxidation of arsenite to arsenate, sulphites and thiosulphates to
sulphate.

Frl
F
2MnO-+6H-"+5AsO;^- 2Mn^+ +3H20 + 5As0^"
4

4 2Mn-+ + 3H20 + 5SO^-

2Mn04+6H++5S0“- -
ou
sor
2+
2Mn0- + 6H++5S^05^ 4 2Mn + 3H.,0 + 5S0““+5S
(iv) It oxidises oxalates or oxalic acid to CO2
2KMn04 + 3H2SO4 ^4 kf
K2SO4 + 2MnS04 + 3H2O + 50
oo
C2H2O4 + O 4H2O + 2CO2JX5
Y
2KMn04 + 3H2SO4 + 5C2H2O4 - 4 K2SO4 + 2MnS04 + 8H2O + 10 CO,
B

or
2Mn04 + 16H-*-+5C20^- 2Mn--" + 8H,0 + lOCO,
re
oY

(v) It oxidises ferrous sulphate to ferric sulphate (i.e., ferrous salt to ferric salt).
u

2KMn04 + 3H,S04 4 K2SO4 + 2MnS04 4- 3H,0 + 50

-4 Fe,(S04)3 + H,0] X 5 "
ad

2FeS04 + H2SO4 4- O
d

2KMn04 4-8H2S04 4- 10FeSO4 > K2SO4 + 2MnS04 4- 5Fe2(S04)3 4- 8H2O

in

or 2MnO- 4- 16H-^4- lOFe

2+
4 2Mn-^ + 8H,0+ 10Fe^+
Re

(vi) It oxidises ^2^2 ^2- becau.se acidified KMn04 is a stronger oxidising agent than
F

H2O2.
2KMn04 4- 3H2SO4 4 K2S04 4- 2MnS04 4- 3H2O + 50
H2O2 -4 H,0 4- 01 X 5
04-0 -> 0,1 X 5

2KMn04 4- 3H2SO4 4- 5H2O, - 4 K2SO4 + 2MnS04 4- 8H,0 4- 5O2

or
2MnO“ 4- 6H+ 4- 5H,02 -4 2Mn2+ 4- 8H2O + 5O2
(v/0 It oxidises potassium iodide to iodine
2KMn04 4-3H2S04- 4 K2SO4 + 2MnS04 4- 3H,0 4- 50
2KI 4- H2SO4- -4 K,S04 4-2HI] X 5
2H1 + O - 4 H,b +12] X 5
2KMn04 4-8H2S04+ lOKI -4 K2SO4 4. 2MnS04 4- 8H2O + 51,
or 2MnO“ + 16H-^4- 10 r- 4 2Mn--" 4-8H,0 4-512
8/32 New Course Chemistry fXIl~>ro5Tm

{viii) It oxidises HX (where X = Cl, B,r 1) to X-^

2KMn04 + 3H2S04- ^ K2SO4 + 2MnS04 + 3H2O + 50
2HX + 0- ^ Hp + X2] X 5
2KMn04 + 3H2SO4 + 10 HX - ^ K2SO4 + 2MnS04 + 8H2O + 5X2
or 2MnO- + 16H++ 10X“ -> 2Mn2-" + 8H2O + 5X2
(ix) It oxidises ethyl alcohol to acetaldehyde
2KMn04 + 3H2S04- K.SO4 + 2MnS04 + 3H2O + 50
CH3CH2OH + O - ^ CH3CHO + H2O] X 5
2KM11O4 + 3H2SO4 + 5CH3CH2OH ^ K2SO4 + 2MnS04 + 5CH3CHO + 8H2O.

Retain in Memory

w
From the different reactions discussed above, we find that when KMn04 acts as an oxidizing

Flo
agent, itself it is reduced to MnO^’ or Mn02 or Mn"'*’. Representing the reductions by half
reactions, we have

e
MnO^ ,E° = + 0-56 V

re
Mn04 + e 4

F
Mn04; + 4 + 3 e" ^ Mn02 + 2 H2O, E“ = + 1-69 V
Mn04 + 8 H++ 5 ^Mn2+ + 4HP,E° = + 1-52V
ur
or
In these reactions, we notice that ion concentration plays an important part in influencing the

kinetics is affected by H'*' ion concentration. At

f
reaction. The reduction potential values can explain the occurrenceof different reactions but the
= 1 M, permanganate should oxidize water
ks
for H2O = - 1 -23 V) but the reaction is extremely slow unless either Mn^"^ ions are present or
Yo
oo

temperature is raised.
B

Uses. (/) It is often used in volumetric analysis for the estimationof ferrous salts, oxalates,iodides and
hydrogen peroxide. However, it is not a primary standard because it is difficult to obtain it in the pure state and free
re

from traces of Mn02. It is, therefore, always first standardised with a standard solution of oxalic acid.
Remember that volumetric titrations involving KMnO^ are carried out only in presence of dilute
u
ad
Yo

//25O4 but not in the presence ofHCl or HNOy This is because oxygen produced from KMn04 + dil. H2SO4 is
used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the
reducing agent. In case HCl is used, the oxygen produced from KMn04 partly used up to oxidize HCl to
d

chlorine and in case HNP3 is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent.
Re
in

{ii) It is used as a strong oxidising agent in the laboratory as well as in industry. It is a favourite and
effective oxidising agent used in organic synthesis. Alkaline potassium permanganate is used for testing
F

unsaturation in organic chemistry and is known as Baeyer’s reagent.

(Hi) Potassium permanganate is also widely used as a disinfectant and germicide. A very dilute solution of
permanganate is used for washing wounds and gargling for mouth sore. It is also used for purifying
water of stinking wells. FIGURE 8.1^
(iv) Because of its strong oxidizing power, it is also used o 0“
for bleaching of wool, cotton, silk and other textile /
4/\ / //\
fibres and also for decolourisation of oils. \
\
/ \
\
/ /
/ \ / \
Structure of permanganate and nianganate ions. /
/ \
\ /
/ \
\

Manganese in MnO“ undergoes sp^ hybridisation and /

/
/ \
\
/
/
/ \
\
\
hence four oxygen atoms are arranged tetrahedrally around Mri' \
Mrv
',0 “0
manganese (Fig. 8.10). Also remember that in
r-'-l/
N
N

(a) (b)
permanganate as well as manganate ion (MnO“"), '0
O' ^
o
7t-bonding takes place by overlap of /^-orbital of oxygen (a) Structure of permanganate ion.
with ^/-orbital of manganese. (b) Structure of manganate ion
d> AND f-BLOCK ELEMENTS 8/33

Retain in Memory

As in the case of CrO^ and Cr,0^ ions, Mn in MnO^ is in +7 state with ^ configuration. The
deep purple colour of KMnO^ is not due to d-d transition but due to charge transfer from O to
Mn, i.e., from Ligand to Metal (L —> M) reducing the oxidation state of Mn from +7 to +6
momentarily.

SUPPLEMENT YOUR
KNOWLEDGE FOR COMPETITIONS

1. Mohr salt, FeS04.(NH4),S04.6H,0 is used as a primary standard in KMn04 titrations but FeS04.7 H2O
cannot be used. This is because FeS04.7 H-,0 contains some Fe^'*' ions due to aerial oxidation but Mohr
2+ ;
salt contains only Fe ions.

2. (0 Tests for and Fe^ ions. Ferric salts react with potassium ferrocyanide to give blue ppt. or coloration

w
due to the formation of ferric ferrocyanide or Prussian blue
4FeCl3 + 3K4[Fe(CN)6] ^ Fe4[Fe(CN)6l3 + 12 KCl

F lo
Ferric ferrocyanide
m n

ee
or
FeCl3 + K4[Fe(CN)6] ^ KFe[Fe(CN)g] +3KC1

Fr
Potassium ferric ferrocyanide
(Prussian blue)
for
(ii) Test for Fe^'*’ ions. Ferric ions react with ammonium or potassium sulphocyanide (also called
ur
thiocyanate) to give blood red coloration due to the formation of ferric sulphocyanide or thiocyanate.
oks

FeCl3 + 3 KCNS ■>

Fe(CNS)3 + 3KCI
Yo

Ferric thiocyanate
o

{Blood red coloration)

eB

3. The species [CuCy^ exists but [€014]^ does not. The reason being that 1“ ion, being a stronger reducing
agent than Cl" ion, reduces to Cu'*’. As a result, CUI2 is converted into Cul and hence [€014]^" does
our
ad

not exist.

2 Cul-, 2 Cul +12

4. Like iodide ions, cyanide ions also act as a reducing agent. When potassium cyanide is added to a
Y

solution of copper sulphate, CN" ions first reduce Cu^^ to Cu"^ which then combines with CN" ions to
Re
nd

form insoluble cuprous cyanide which subsequently dissolves in excess of KCN to give the complex
compound, potassium tetracyanocuprate (I)
Fi

2Cu2-*- + 2CN- ^ 2 Cu-*- + (CN)2

Cyanogen
Cu-*- + CN" ^ CuCN J-
CuCN + 3 KCN ■»
K3[Cu(CN)4]
Pot. tetracyanocuprate (I)
{Soluble complex)
5. Decomposition of ammonium dichromate (Volcano experiment). When ammonium dichromate is heated
slightly, it decomposes to form Cr203 (green solid), N2 and H2O. Once initiated, a large amount of heal is
produced and the decomposition continues at high speed.
(NH4)2Cr207 ^ Cr203 + N2 + H2O
Green .solid

The large volumes of N2 and H^O vapour blow the green Cr203 into the air and it looks like a valcano.
8/34 New Course Chemistry fXinrosTWi

SECTION III. STUDY OF f-BLOCK ELEMENTS (LANTHANOIDS & ACTINOIDS)

8.8. GENERAL INTRODUCTION

The elements in which the last electron (also called differentiating electron) enters the ante-penultimate
energy level, /.£?., (« — 2)/-orbitals are called f- block elements. These elements have been termed as/-block
elements since the last electron enters one of the /- orbitals. These elements have also been called inner
transition elements. This is because the last electron in them enters into ante-penultimate shell viz. (n — 2)
y-orbitals, i.e., inner to the penultimate energy level and they form a transition series within the transition
series (^-block elements). Their general electronic configuration is :
1-14
(n-2)f in -l)d

Thus, they have three incomplete shells, viz., (n-2), (ri-l) and nth.
Classification of f-block elements. Depending upon whether the last electron enters a 4/-orbital or a

w
5/-orbital, the/- block elements have been divided into two series as follows :
(/) Lanthanoids. The elements in which the last electron enters one of the 4/-orbitals are called 4/-

F lo
block elements or first inner transition series. These are also called lanthanides or lanthanons or
lanthanoids because they come immediately after lanthanum. Earlier, these 14 elements (Z = 58 to 71) were
called rare earths.

ee
(/7) Actinoids. The elements in which the last electron enters one of the 5/-orbitals are called 5f-block

Fr
elements or second inner transition series. These 14 elements (Z = 90 to 103) are also called actinides or
actinons or actinoids because they come immediately after actinium.

for
Lanthanum, though a d-block element, is included in the lanthanoid series because it closely resembles
ur
lanthanoids. Similarly, actinium is also included in the actinoid series. Furthe,r the study of lanthanoids is
comparatively easier because they show only one stable oxidation state. On the other hand, the chemistry of
s
actinoids is much more complicated partly because they show a wide range of oxidation states and partly
ok
Yo
because they are radioactive.
o

Now. we shall discuss each of these series one by one.

eB

8.9. THE LANTHANOIDS

r

8.9.1. Electronic Configuration of Lanthanoids

ou
ad

The members of this series along with their electronic configurations are given in Tables 8.13. Their
Y

common oxidation states and the electronic configurations in these oxidation states are also included in the
table.
Re
nd

TABLE 8.13. Electronic configuration of lanthanum and lanthanoids

Fi

1-14
(General symbol = Ln) (General E.C. = [Xe] 4/ 5 d 6 s2)

Name of the Symbol At. No. (Z) Electronic Oxidation E.C. outside [Xe] core
element (Ln) configuration states Ln^+ Ln**^

Lanthanum La 57 [Xe] 5d^ 652 +3 5d‘ 4/

0

Cerium Ce 58 [Xe] 4f5d^6s^ +3, +4 4/2 4/

1
4/
0

Praseodymium Pr 59 [Xe] 4/3 5d^ 6.?2 +3, +4 4/3 4/2 4/

Neodymium Nd 60 [Xe] 4/'» 5d^ 6^2 +2, +3, 44 4/4 4/3 4/2
Promethium* Pm 61 [Xe] 4/^ 5d^6s- 4-3 4/^ 4y4
Samarium Sm 62 [Xe] 4f^5d^6s^ 4-2, +3 4f(> 4/5

*Promethium is the only synthetic (man-made) radioactive lanthanoid.

d- AND f-BLOCK ELEMENTS 8/35

Europium Eu 63 [Xe\4f5d^6s- +2, +3 4/' 4y6

Af 5d
\
Gadolinium Gd 64 [Xe] 4/^5(/‘ 6^-2 +3

4f
8
Terbium Tb 65 [XeJ 4/^5j0 6s^ +3, +4 4r 4/
4f 4/
10 8
Dysprosium Dy 66 [Xe] 4/“>5rf” 6.t2 +3, +4 4/
' i]
4/
0
Holmium Ho 67 [Xel 4/“ 5t/" 6.S- +3 4J
12 II
Erbium Er 68 [Xe1 4/‘“5rf“ 6.v^ +3 4/ 4/
Thulium Till 69 [Xel 4/'^5rf**6r +2, +3 4/
●13
4/12
●14 13
Ytterbium Yb 70 [Xej 4/-^ 5rf" 6^^ +2, +3 4y 4/
14
4/'-^5(/‘
1
Lutetium Lu 71 [Xel 4f'"^5d' 6s- +3 4/
In the electronic configurations of lanthanoids given in the Tables 8.13, the following points are note-worthy.
(/) Lanthanoids have the electronic configuration with 6i’^ common but with variable occupancy of 4/
level. However, the electronic configurations of all the iriposilive ions (the most stable oxidation state

w
of all the lanthanoids) aie of the form 4f" {n = I to 14) with increasing atomic number.
(//■) The electronic configuration of cerium (Z = 58) is 4/^ 5d^^ 6s^.

F lo
(Hi) The electronic configuration of europium (Z = 63) is 4/^ 6^" and that of gadolinum (Z = 64) is
4/^ 5d' and this is explained on the basis of extra stability of the half-filled orbitals in their cores.

ee
(iv) The electronic configuration of ytterbium (Z = 70) is 4/“^ 6s- and that of lutetium (Z=71) is 5r/’

Fr
6s~. This is also explair.ed on the basis of extra stability of the completely iflled orbitals in their cores.
8.9.2. Oxidation States of Lanthanoids
for
ur
The typical oxidation state of the lanthanoids is +3. The oxidation state of +2 and +4 are exhibited by
some of the elements. These are shown by those elements which by losing 2 or 4 electrons acquire a stable
s
configuration of/®,or/^'*. e.g., Eu is [Xe] 4/^ Yb^-" is [Xe] 4/
2+
Ce'^'" is [Xe] 4/" and Tb*^-" is
ook
Yo

[Xe] 4/^. Each case tends to revert to the more stable oxidation state of +3 by loss or gain of an electron. That
is why Sm^'*’, Eu"'*’ and Yb"'*' ions in solutions are good reducing agents and aqueous solution of Ce**'*' and
eB

Tb^'*' are good oxidising agents.

The E® value for Ce^/Ce^'^ is + 1 -74 V which suggests that it can oxidize water. However, the reaction
r
ad
ou

rate is very slow and hence Ce (IV) is used as a good analytical reagent.
Exceptions. Looking at Table 8.13, it may be noticed that some elements show an oxidation state of +2
Y

or +4, even though their ions do not have /® and or / configuration, e.g., Pr'^'*' (4/'), Nd^'*’ (4/'^).
Re

Nd^-^ (4/ \ Sm2+ (4/ ^’), Dy'^^ (4f etc.

nd
Fi

Retain in Memory

(i) Lanthanoids show limited number of oxidation states because the energy gap between 4/
and 5 d subshells is large.
(/7) Comparing with tran.sition metals, which show many different oxidation states, the obvious
reason is that in case of transition elements, the r/-electrons are present in the
(n - 1) fr-subshell which can easily participate in bond formation but in case of lanthanoids, the/-
electrons are present in the deeper {n - 2)/-subshell which cannot participate in bond formation.
8.9.3. Atomic and Ionic Radii of Lanthanoids
In lanthanoid series, with increasing atomic number, there is a progressive decrease in the atomic as
well as ionic radii of trivalent ions from La^'*' to Lu^"^.

This regular decrease (contraction) in the atomic and ionic radii of lanthanoids with increasing
atomic number is known as lanthanoid contraction.
 8/36 New Course Chemistry fXincTTSTWi

The values of the radii of lanthanoids and their tripositive ions are given in Table 8.14 below. The
variation in the trend is represented graphically in Fig. 8.11.
FIGURE 8.11
TABLE 8.14. Atomic and Ionic radii
(pm) of lanthanum and lanthanoids 110

Element Atomic Radii (Ln) Ionic Radii (Ln^"^)

Ui 187 106
Ce 183 103 E 100
Q.
Pr 182 101 c/3

Nd 181 99 Q

ow
Pm 181 98
o
Sm 180 96
90
Eu 199 95 o

Gd 180 94
Tb 178 92

e
Dy 177 91

re
Ho 176 90

rFl
57 59 61 63 65 67 69 71
Er 175 89

F
ATOMIC NUMBER
Till 174 88
Yb 173 87 Variation of ionic radii

r
Lu 172 86 of trivalent lanthanoids
ou
fo
It may be noted that the decrease in atomic radii is not quite regular wherea'.-' for the ionic radii of Ln^'*’
ks
ions, it is quite regular.
oo
Cause of lanthanoid contraction. As we move along the lanthanoid series, the nuclear charge increases
by one unit at each successive element. The new electron is added into the same subshell (viz., 4/). As a
Y
B

result, the attraction on the electrons by the nucleus increases and this tends to decrease the size. Further, as
the new electron is added into the /- subshell, there is imperfect shielding of one electron by another in this
re

subshell due to the shapes of these/-orbitals. This imperfect shielding is unable to counterbalance the effect
of the increased nuclear charge. Hence, the net result is a contraction in the size though the decrease is very
ou
Y
ad

small. It is interesting to note that in lanthanoids, the decrease in the atomic radius for 14 elements [Ce (58)
to Lu (71)] is only 11 pm (from 183 to 172 pm). Similarly, decrease in ionic radii from Ce-^'*’ to Lu^'*' is only
17 pm (103 to 86 pm)
d

The contraction is similar to that observed in any transition series. The cause is also similar, just as in
in
Re

the transition .series where the contraction is due

to imperfect shielding of one d electron by another. But the
shielding of one 4/electron by another is less than one d electron by another with increase in nuclear charge
F

along the series.

Consequences of lanthanoid contraction.
(0 Difficulty in .separation of lanthanoids. Since the change in ionic radii (size of the ions) in lanthanoids
is very small, their chemical properties are similar. This makes the separation of the lanthanoids in the pure
state difficult. However, lanthanoid contraction results in slight difference in the size of the lanthanoids
which results in the differences in properties like solubility, complex ion formation, hydration, basic character
of their hydroxides, etc. These differences enable the separation of individual lanthanoid elements by ion
exchange methods.*
(//) Similarity in size of elements belonging to same group of second and third transition series. In
general, we know that the size of the atoms increases down a group. However, whereas the size of the atom of
ai^ element lying in the second transition series is larger than that of the atom of the element lying in the same
*Ion exchange method is based on the fact that hydrated radii of lanthanoid ions increase with increasing
atomic number. Thus, when a solution containing several lanthanoid ions passes slowly through the column of
cation exchange resin, the heavier members will come through first.
d- AND f'BLOCK ELEMENTS 8/37

group of the first transition series (as expected), the size of any atom of the third transition series (after
lanthanum) is nearly same as that of the atom of the element lying in the same group of the second transition
series. A few elements of the 1st. 2nd and 3rd transition series along with their atomic radii are given below :

Group No. —> 3 4 5

1st Transition series 2jSc (144 pm) 22Ti (132 pm) 23V (122 pm)
2nd Transition series 39Y (180 pm) 4oZr (160 pm) 41 Nb (146 pm)

3rd Transition series LANTHANOIDS 72Hf (159 pm) 73Ta (146 pm)
3yLa (187 pm)
(58 — 71)
The similarity in size of the atoms of the elements belonging to the same group of the 2nd and 3rd
transition series (after lanthanum) (c.g. = rj^ etc.) is evidently due to the effect of lanthanoid

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contraction.
3+
(Hi) Effect on the basic strength of hydroxides. As the size of the lanthanoid ions decrease.s from La
to the covalent character of the hydroxides increases and hence the basic strength decreases. Thus,

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La(OH)3 is most basic whereas Lu(OH)3 is least basic.
8.9.4. Some Important Characteristics of Lanthanoids

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1. Silvery appearance and Softness. All the lanthanoids are silvery white soft metals and tarnish

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rapidly in the air. Their hardness increases with increasing atomic number. Samarium (Sm) is exceptionally
hard like steel.

for
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2. Melting points. They have high melting points in the range 1000 to 1200 K except samarium which
has a very high melting point of 1623 K.
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3. Electrical and Thermal conductivity. All of them have typical metallic structure and are good
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conductorsof heal and electricity.
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4. Density and other properties. They have high densities which lie in the range 6-77 to
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9-74 g cm"^. Density and other properties vary smoothly with increasing atomic number except for Eu and
Yb and occasionally for Sm and Tm.
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5. Colour. The lanthanoids are silvery white metals. However, most of the trivalcnt metal ions are
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coloured, both in the solid state and in aqueous solution, This is due to the partly filled/-orbitals which permit
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f-f transition.

Note. It is interesting to observe that lanthanoid ions with xf electrons have a similar colour to those with
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(14-j:)/electrons, e.g., La-^'^ and Lu^'*' are colourless, Sm^"*" and Dy^'*’ are yellow and Eu^"*" and Tb^"^ are
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pink.

6. Magnetic behaviour. Paramagnetism is shown by the positive ions of lanthanoids except La^'*'
(lanthanum ion, / ^) and Lu^'^ (lutetium ion, / ^‘^). This property of the lanthanoids is due to presence of
unpaired electrons in the incomplete 4/subshell.
Lanthanoids differ from transition elements in the fact that their magnetic moments do not obey ‘spin
only’ formula, viz., (n-r2) B.M. where n is the number of unpaired electrons. This is because
in case of transition element.' is quenched by the electric field of the environment
the orbital contribution
but in case of lanthanoids, 4/orbitals lie loo deep to be quenched. Hence, their magnetic moment is
calculated by considering spin as well as orbital contribution, i.e.,

= ^4S(S+1) + L(L+1) B.M.

where S is spin quantum number and L is orbital quantum number.
8/38 'pfuxdee^'A New Course Chemistry (XII)CZSI9]

7. Ionization enthalpies. The first ionization enthalpies of lanthanoids are around 600 kJ mol~* and the
second about 1200 kJ mol"’ which are comparable with those of calcium. The variation of the third ionization
enthalpies shows that just as in case of 3^/ transition series, the loss of the first two electrons is accompanied
by exchange enthalpy. Further, the loss of third electron is easier, i.e., third ionization enthalpy is low if it
leads to stable empty, half-filled or completely filled configuration, as indicated by the abnormally low third
ionization enthalpies of La, Gd and Lu.
8. Electropositive character. They arc highly electropositive because of their low ionization enthalpies.
9. Standard electrode potentials. Their standard reduction potentials, Le., E° values for the half reaction,
(aq) + 3 e- ■> M (s), lie in the range of - 2-2 to - 2-4 V, the only exception being europium (Eu) for
which the E“ value is - 2 0 V.

10. Reducing agents. They readily lose electrons and are thus good reducing agents.
11. Complex formation. The lanthanoids do not have much tendency to form complexes due to low

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chinge density because of their large size. The tendency to form complexes and their stability increases with
increasing atomic number.

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12. Chemical behaviour. The first few members of the series are quite reactive, almost like calcium.
However, with increa.sing atomic number, their l^ehaviour becomes similar to that of aluminium. A few properties

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are given below :

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(/) They combine with H2 on gentle heating. When heated with carbon, they form carbides. On burning in
the presence of halogens, they form halides.
(//) They react with dilute acids to liberate H2 gas.
for
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(///) They form oxides and hydroxides of the type M2O3 and M (OH)3 which are basic like alkaline earth
metal oxides and hydroxides.
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Representing the lanthanoids by the general symbol Ln, the general reactions may be represented as
shown given below ;
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Halogen
► LnXj
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C,2773 K
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■►Ln3C, Ln2C3 & LnC,

Dilute acids
Ln
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(Lanthanoid)
■►Liberate H2 gas
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Bum in 0-.
nd

^Ln203
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Healed with N->

LnN

Healed with S
Ln2S3
H2O
■►Ln (0H)3 + H2
8.9.5. Uses of Lanthanoids

1. Lanthanoids do not find any use in the pure state. The most important use of lanthanoids is in the
production of alloy steels to improve the strength and workability of steel. A well known alloy is mischmetal
which has the following composition :
Lanthanoid metals 95%

(about 50% Ce, 40% La and the rest other lanthanoids)

Iron 5%

S. C, Ca and A1 = traces
d- AND f-BLOCK ELEMENTS 8/39

The maximum amount of mischmetal is used in making a magnesium based alloy {i.e., Mg mixed with
about 3% mischmetal to increase the strength of Mg). It is a pyrophoric alloy {i.e., an alloy which emits
spark when struck, i.e., takes up fire easily) and is used in making bullets, shells and lighter flints.
2. Their oxides (e.g., La203) are used in glass industry, for polishing glass and for making optical
glasses (such as Crooke’s lenses) as they give protection against UV light and as phosphor for television
screens and similar fluorescing surfaces.
Mixed oxides of lanthanoids are used as catalysts in petroleum cracking.
3. Because of their paramagnetic and ferromagnetic properties, their compounds imc used in making
magnetic and electronic devices.

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4. Ceric sulphate is a well known oxidizing agent in volumetric analysis.
5. Recently, lanthanoids have attracted considerable attention because of their use in lasen. Thus,
neodymium oxide dissolved in selenium oxychloride is one of the most powerful liquid laser known so far.

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Curiosity Questions
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Q. 1. Which alloy is generally used in making bullets, shells and lighter flints ? What is its

F
composition? With which other element it is generally alloyed and why ?
Ans. The alloy used is known as mischmetal. For details refer to Art. 8.9.5 above.
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Q. 2. Which compound is generally used in making optical glasses such as Crooke’s lenses

sor
which protect eyes against UV light ?
Ans. Lanthanum oxide (18203). I
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8.10. THE ACTINOIDS
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As mentioned earlier, the ability of actinoids to exist in different oxidation states has made their chemistry
B

more complex. Moreover, most of these elements are radioactive and the study of their chemistry in the
laboratory is difficult. The earlier members of the series have relatively long half- lives, the later ones have
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half-lives ranging from a day to 3 minutes (for lawrencium, Z = 103). The later members have, therefore,
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been prepared only in nanogram quantities.

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8.10.1. Electronic Configurations of Actinoids

d

The members of this scries along with their clecuonic configurations lue given in Table 8.15. Theii'common
in

oxidation states and the electronic configurations in these oxidation states are included in the Table.
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TABLE 8.15. I Electronic configuration of actinium and actinoids

F

(General E.C. = [Rn] 5 / 6 7 s^)

Name of the Symbol At. No. (Z) Electronic Oxidation states E.C. outside [Rn] core
element (M) configuration
-0
Actinium Ac 89 IRnl 6d' Is- +3
●0
Thorium Th 90 |Rn] 6d-lr (+3), +4 5/
Protactinium Pa 91 [Rn] 5/-6f/' ls~ +3, +4, +5 5/2 5/
Uranium U 92 [Rn] 5/3 td'ls- +3, +4, +5, +6 5/3 5/^-
Neptunium Np 93 [Rn] 5/^6r/' 7^2 +3, +4, +5, +6, +7 5/^ 5/^
Plutonium Pu 94 [Rn] 5f^6d^ls~ +3, +4, +5. +6, +7 5/^ 5/**
Americium Am 95 [Rn] 5fHd'^ls^ +3, (+4), +5, +6 5/6 5/'
Curium Cm 96 [Rn] 5f6d' 7s^ +3,(+4) 5/2 5/6
Berklium Bk 97 [Rn] 5/9 6f/'’ Is- +3, 44 5/« 5/2
8/40 ‘P>i4xdee^'A New Course Chemistry (XII)E!Z&iai

Californium Cf 98 [Rn] 5/‘^6f/^7^- +3 5/’ 5/

Einsteinium Es 99 [Rn] 5/'* 6^/” Is^ +3 5/
● 10
5/^
Fermium Fm 100 [RnJ 5/ ‘2 Is^ +3 5/
11 .
5/
10

Mendelevium Md 101 [Rn] 5/^i^6(/0 7.r +3 5/

12
5/
11

13
Nobelium No 102 [Rn] Is- +3 5/ 5/1^-
Lawrencium Lr 103 [Rn] 5/‘●^6^/' Ir +3 5/
14
5/
13

Note. Oxidation slates given in brackets su-e less stable.

From this table, the following points may be noted :
(/) All the actinoids have common 7^“ configuration and variable occupancy of 5/and 6r/ subshells.
(i7) The 14 electrons are being added into 5/, except in thorium (Z = 90) but filling of 5/continues again

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after thorium till 5/orbitals are complete at Z = 103 (Lawrencium).
iiii) The irregularities in the electronic configurations of actinoids like those in the lanthanoids,are related
to
the stabilities of/^,/^ and/ configurations. For example, the configurations of Am and Cm are
[Rn] 5/^6</®7.s- and [Rn] 5/^6r/‘ 7s- re.spective]y.
(iv) Berkelium (Bk, Z = 97) also shows an exceptional configuration of [Rn] 5/^ 6d^ 7.v^ which is neither

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half-filled nor completely filled,

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(v) Though 4/and 5/orbitals have similar shapes but 5/is less deeply buried than 4/ Hence, 5/electrons

F
can participate in bonding to a far greater extent.
Since now some isotopes of many of the actinoids are available in good amounts, their chemistry has
been studied by tracer techniques.
ur
or
'."0.2. Oxidation States of Actinoids sf
The dominant oxidation state of these elements is +3 (similar to lanthanoids). Besides +3 state, actinoids
k
also exhibit an oxidation state of +4. Some actinoids
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show still higher oxidation states. The maximum oxidation
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state first increases upto the middle of the series and then decreases, e.g., it increases from + 4 for Th to +5,
+6, and +7 for Pa. U and Np but decreases in the succeeding elements.
B

The actinoids resemble lanthanoids in having more compounds in +3 state than in the +4 state. However,
the compounds in the +3 and +4 state tend to undergo hydrolysis.
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Further, it may be noted that in case of actinoids. the distribution of oxidation states is so uneven that it
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is of no use to discuss their chemistry in terms of their oxidation states.

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lit

Unlike lanthanoids, actinoids show a large number of oxidation states. This is because of very
d
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small energy gap between 5/ 6d and 7s subshells. Hence, all their electrons can take part in bond
in

formation.
F

R.in.3. Ionic Radii and Actinoid Contraction

The actinoids show actinoid contraction (very much like lanthanoid contraction) due to poor shielding
effect of the 5/-electrons. As a result, the radii of the atoms or ions of these metals decrease regularly across
the series. The contraction is greater from element to element in this series due to poorer shielding by 5/
electrons. This is because 5/orbitals extend in space beyond 6^ and 6p orbitals whereas 4f orbitals are hurried
deep inside the atom.
The ionic radii of actinoids (M^"^ and M"*"^) are given in Table 8.16 below :
TABLE 8.16. Ionic radii (pm) of actinium and actinoids in +3 and +4 oxidation state

Element Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Ionic Radii 111 103 101 100 99 99 98 98

Ionic Radii (M^) 99 96 93 92 90 89 88 87 86

d- AND f-BLOCK ELEMENTS 8/41

8.10.4. General Characteristics of Actinoids

1. Silvery appearance. Like lanthanoids, actinoids are metals with silvery appearance.
2. Structural variability. As they have much greater irregularities in their metallic radii than lanthanoids,
they show great structural variability.
3. Colour. These metals are silvery white. However, actinoid cations are generally coloured. The colour
of the cation depends upon the number of 5/-electrons. The cations containing no 5/-electron or having seven
5/-electrons (i.e., exactly half-filled/-sub.shell) are colourless. The cations containing 2 to 6 electrons in the
5/-subsheil are coloured both in the crystalline state as well as in aqueous solution. The colour arises due to
/—/transition, e.g.. Ac^^(5f^) = colourless, = Red. Np^+CS/*^) = Blue, Pu^+(5/^) = Violet,
Am^'^(5/'’) = Pink, Cm^'^(5/^) = Colourless, Th^"^ = Colourless and so on.
4. Melting and Boiling points. The actinoids like lanthanoids have high melting and boiling points.
However, they do not show any regular trend with rise in atomic number.
5. Density. All the actinoids except thorium and amercium have high densities.
6. Ionization enthalpies. The actinoids have lower ionization enthalpies than lanthanoids because 5/is
less penetrating than 4/and hence is more effectively shielded from the nuclear charge.

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7. Electropositive character. All the known actinoid metals are highly electropositive. They resemble

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the elements of lanthanoid series in this respect.
8. Magnetic behaviour. Like lanthanoids, the actinoid elements are strongly paramagnetic.
The variation in magnetic susceptibility of actinoids with the increasing number of unpaired electrons is
similar to that of lanthanoids but the values are higher for the actinoids than for the lanthanoids.

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9. Reducing agents. All the actinoids are strong reducing agents.
for F
10. Radioactivity. Ail the actinoid elements are radioactive. First few members have relatively long hiilf-
lives. However, the remaining members have half-lives ranging from a few days to few minutes (e.g., 3 min for Lr)
11. Complex Formation. Actinoids have higher tendency to form complexes than lanthanoids. This is
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due to higher chtirge and smaller size of their ions. Most of the complexes are formed by halides of actinoids
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with alkali metal halides.

12. Chemical behaviour. They are highly reactive metals especially in the finely divided state. A few
properties are given below :
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(/) They react with boiling water to give a mixture of oxide and hydride.
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(ii) They combine with most of the non-metals at moderate temperature.

(Hi) All these metals are attacked by hydrochloric acid but the effect of nitric acid is very small due to the
formation of a protective oxide layer on their surface.
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(iv) Alkalies have no action on them.

Find

8.10.5. Uses of Actinoids

The three most important actinoids which find uses as such or in the form of their compounds are
thorium, uranium and plutonium. A few uses of these elements are as follows:
1. Uses of thorium. It is used in atomic reactors and in the treatment of cancer. Its salts are used in
making incandescent gas mantles.
salts are used in glass industry (for imparting green
2. Uses of uranium. It is used as a nuclear fuel. Its
colour), textile industry, ceramic industry and in medicines.
3. Uses of plutonium. It is used as a fuel for atomic reactors as well as for making atomic bombs.

8.11. COMPARISON OF LANTHANOIDS AND ACTINOIDS

Similarities. As both lanthanoids and actinoids involve filling of /-orbitals, they show similarities in
many respects as follows:
(i) Both show mainly an oxidation state of -i-3.
8/42 “P-itidecfr New Course Chemistry fXlllrosTun

(i7) Both are electropositive and very reactive.

(Hi) Both exhibit magnetic and spectral properties.
(iv) Actinoids exhibit actinoid contraction like lanthanoid contraction shown by lanthanoids.
Differences. They show differences in some of their characteristics as follows :
Lanthanoids Actinoids

(/) Besides +3 oxidation state, they show +2 (0 Besides +3 oxidation state, they show higher
and +4 oxidation states only in few cases. oxidation states of +4, +5, +6, +7 also.
(ii) Most of their ions are colourless. (ii) Most of their ions are coloured.
m They have less tendency towards complex m They have greater tendency towards complex
formation. formation.
(iv) Lanthanoid oxides tind hydroxides are less basic. (iv) Actinoid oxides and hydroxides are more basic.

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(V) Do not form oxocation (V) Fonn oxocations e.g. U02'*’,Pu02''’ ^^2
(vi) Except promethium, they are non-radioaclive (Vi) They are all radioactive.

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(vii) Their magnetic properties can be explained (vii) Their magnetic properties cannot be explained
easily. easily, as they are more complex.

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Retain in Memory
1. The most commonly occurring lanthanoid is ‘cerium’ which constitutes about 3 x ICH % of earth’s crust.

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2. The most common mineral containing lanthanoids is ‘Monazite sand'. It is mainly lanthanoid
ordiophosphate.
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8.12. SOME ADDITIONAL USEFUL INFORMATION ABOUT D- AND f^BLOCK ELEMENTS

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1. Some Common .\lloys of Transition Metals

Alloy Percentage Composition Uses
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A. ALLOYS OF COPPER

Brass Cu = 80%, Zn = 20% For making utensils, parts of machinery

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(screws, nuts, bolts etc.), condenser tubes etc.

0 Bronze Cu = 90%, Sn = 10% For making statues, coins, medals, ship’s
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propeller etc.
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3. Gun metal Cu = 90%, Sn = 6-8%, Zn = 2-4% For making gun barrels

4. Bell metal Cu = 80%, Sn = 20% For making bells and gongs
5. Monel metal Cu = 30%, Ni = 67%. Fe and For making corrosion resistant pumps,
Mil = 3% containers for strong acids
6. German silver Cu = 60%, Zn = 20%, Ni = 20% For making resistance wires, utensils and
ornaments

7. Phosphor bronze Cu = 95%, Sn = 4-8%, P = 0-2% For making springs and eleclric switches
B. ALLOYS OF SILVER

1. Coinage silver Ag = 90%, Cu= 10% For making silver coins

2. Silver solder Ag = 63%, Cu = 30%, Zn = 7% For soldering
3. Dental alloy Ag = 33%, Hg = 52%, Sn = 12-5%, For filling of teeth
Cu = 2%. Zn = 0-5%
d- AND f-BLOCK ELEMENTS 8/43

Aiioy Percentage Composition Uses

C. ALLOYS OF IRON

1. Steel Fe = 99 - 99-8%, C = 0-2 - 1% For construction of machinery, knives, razors

etc.

2. Stainless steel Fe = 73%, Cr = 18%, Ni = 8%, For making cutlery, utensils, surgical
C= 1% instruments, blades etc.
3. Nickel steel Fe = 96-98%, Ni = 2-4% For making automobile and aeroplane parts,
armour plates, gears and drilling machines
4. Chrome steel Fe = 98%, Cr = 2% For making axels, ball bearings, files and
cutting tools

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5. Alnico Fe = 60%. Ni = 20%, A1 = 12%, For making permanent magnets
Co = 8%

2. Some Common Alloys of Non-transition elements

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Aiioy Composition Uses

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1. Duralumin A1 = 95%, Cu = 4%, Mg = 0-5%, For making automobile and aeroplane parts,

F
Mn = 0-5% pressure cooker etc.
2. Magnalium A1 = 95%, Mg = 5% For making light weight instruments and
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balance beams

3. Aluminium bronze Al = 95%, Cu = 5%

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For making coins, picture frames and cheap
jewellery
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4. Solder Pb = 50%, Sn = 50% For joining electrical wires together
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5. Pb = 60%, Sb = 30%, Sn = 10% Making type tor printing as it gives a sharp cast
B

3. Some Important Compounds of IVansition Metals and their Special names

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S. No. Compound Formula Special name

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1. Ferrous sulphate FeS04.7HoO Green vitriol

2. Ferrous ammonium sulphate FeS04. (NH4)2S04.6H2O Mohr sail

d

3. Copper (II) sulphate pentahydrate CUSO4. 5H2O Blue vitriol

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in

4. Basic copper carbonates CUCO3. Cu(OH)2 Malachite (green)

F

2 CuCO,. Cu(OH)2 Azurite (deep blue)

5. Basic copper (II) acetate Cu (CH3C00)2 . Cu{OH)2 Verdigris
6. Copper sulphate + slaked lime CUSO4 + Ca(OH)2 Bordeaux mixture (fungicide)
7. Silver nitrate AgNOj Lunar caustic

8. Zinc oxide ZnO Philosphcr’s woo!

9. Zinc sulphide -1- Barium sulphate ZnS + BaS04 Lithopone (white pigment)
10. Titanium oxide + Barium sulphate TiO^ + BaSO 4 Titanox

11. Zinc sulphate ZnS04.7H2O White vitriol

12. Mercury (I) chloride Hg2Cl2 Calomel

13. Lead (II) oxide PbO Litharge

14. Trilead teiraoixde Pb304 Red lead (Sandhur)
8/44 New Course Chemistry (X1I)C2SZS]

4. Some Common Reagents containing Compounds of Transition Metals

Reagent Composition and Use

1. Bacyer’s reagent A dilute solution of alkaline KMn04 used for testing unsaturation
2. Fehling solution A mixture of CUSO4, NaOH and Roschelle’s salt (sodium potassium
tartarate) used to distinguish aldehydes and ketones
3. Tollen's reagent An ammoniacal solution of silver nitrate used in silver mirror test for

aldehydes
4. Benedict’s reagent A solution of CUSO4, sodium citrate and Na2C03 used to distinguish
between aldehydes and ketones and to detect sugar in diabetic patients.
5. Luca’s reagent A mixture of cone. HCl and anhydrous ZnCI^ used to distinguish between

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primary, secondary and tertiary alcohols.
6. Fenton’s reagent A mixture of FeS04 and H2O2 used to oxidize alcohols to aldehydes

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7. Etard’s reagent Chromyl chloride (CrO^Cl2) dissolved in CCI4 used to convert toluene to
benzaldehyde

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Nesseler’s reagent An alkaline solution of K2Hgl4 used for testing NH^ ion and NH3

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9. Millon’s reagent A solution of mercuric and mercurous nitrate used to detect the presence
of soluble proteins.

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10. Schweitzer’s reagent A blue coloured solution of the complex [Cu(NH3)4]S04 used for
dissolving cellulose in the manufacture of artificial silk.
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5. Some Common Catalysts containing Transition metals or Their Compounds
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Catalyst Composition and Use

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I.
Vanadium pentoxide (V2O5) Used for oxidation of SO-i to SOo in contact process for manufacture of
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H2SO4
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2. Lindlar’s catalyst Palladium (Pd) deposited over BaS04 along with a small amount of S or
quinoline, used in the reaction of acid chloride with H2 to obtain aldehydes
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(called Rosenmund’s reduction)

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nd

3. Wilkinson’s catalyst A complex of rhodium chloride with triphenyl phosphine having the
formula [RhCl(PPh3)3] used for the selective hydrogenation of alkenes
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(hydrogenation of double bond at the end of the chain).

4. Zeigler-Natta catalyst Trialkyl aluminium (R3AI) mixed with TiCl4 used for low temperature
polymerisation of alkenes
5. Adams catalyst Pt + PtO used for reductions

6. Raney nickel Finely divided nickel used for low temperature (room temperature)
hydrogenation of oils (unsaturated) into fats (saturated)

6. Some Typical Problems Based on Reactions of KMn04 and K2Cr207

P* L Pyrolusite on heating with KOH in the presence of air gives dark green compound (A). The solution
of (A) on treatment with H2SO4 gives a purple coloured compound (B) which gives the following
reactions :

(0 Alkaline solution of B on reaction with KI changes it into compound (C).

(«) The colour of the compound (B) disappears on treatment with acidic solution of FeS04.
d- AND f-BLOCK ELEMENTS 8/45

(iii) On reaction with cone. H2SO4, compound (B) gives another compound (D) which can decompose
to produce compound (E) along with oxygen gas. Identify compounds (A) to (E) and write balanced
chemical equations involved in each case.
Sol. Chemical equation for the formation of A :

2Mn02 +4KOH + O2 2K2Mn04 + 2H2O

Pyrolusite (A)
Potassium manganate

Chemical equation for the formation of B from A :

3 K2Mn04 + 2 H2SO4 4
2 KMn04 + Mn02 + 2 K2SO4 + 2 H2O
(B)
Potassium permanganate

low
Chemical equations of reactions of compound B :

(0 2 KMn04 + H2O + KI >2Mn02 + 2K0H + KIO3

(C)
Potassium iodate

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iii) 2 KMn04 + 10 FeS04 + 8 H2SO4 > K2SO4 + 2 MnS04 + 5 Fe2(S04)3 + 8 H2O
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{Hi) 2 KMn04 + H2SO4 > Mn207 + K2SO4 + H2O
(B) (D)

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2Mn207 ^ 4Mn02 + 3 O2
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(D) (E)
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P. 2. When an orange crystalline compound (A) was heated with common salt and concentrated sulphuric
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acid, an orange red gas (B) was evolved. Acidified solution of the compound (A) reacts with H2O2 to
oo

produce a deep blue solution due to the formation of compound (C). The gas (B) on passing through
B

NaOH solution gave a solution (D). The solution on reacting with an aqueous solution of lead acetate
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gave a yellow precipitate of (E).

Name the compounds A, B, C, D (present in the solution) and E (present as ppt).
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Sol. K2Cr207 + 4 NaCl + 6 H2SO4 > 2 KHSO4 + 4 NaHS04 + 2 Cr02Cl2 + 3 H2O

(A) (B)
Potassium dichromate Chromyl chloride
nd
Re

K2Cr207 + H2SO4 + 4 H2O2 2Cr05 +K2SO4 + 5H2O

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(C)
Peroxo compound, Cr0(02)2

Cr02Cl2 +4 NaOH Na2Ci04 + 2 NaCl + 2 H2O

(B) (D)
Sodium chromate

Na2Cr04 + (CH3COO)2Pb PbCr04 + 2 CHjCOONa

(E)
Lead chromate
(Yellow ppt.)

P. 3. When a white crystalline compound X is heated with K2Cr207 and concentrated H2SO4, a reddish
brown gas A is evolved. On passing A into caustic soda solution, a yellow coloured solution of B is
obtained. Neutralizing the solution B with acetic acid and on subsequent addition of lead acetate, a
yellow precipitate C is obtained. When X is heated with NaOH solution, a colourless gas is evolved
and on passing this gas into K2Hgl4 solution, a reddish brovm precipitate D is formed. Identify A, B,
C, D and X. Write the equations of the reactions involved.
8/46 New Course Chemistry (Xli)BSXSI

Hg
Sol. X = NH4CI, A = Cr02CU, B = Na2Cr04, C = PbCr04. D = o; NH2 I, NH2^g^
Hg
(Iodide of Million’s base)*
The chemical reactions involved are as follows ;

4NH4CI + KnCr^O^ + 6H2SO4 > 2Cr02Ci, + 4NH4HSO4 + 2KHSO4 + SH^O

Reddish brown

Cr02Cl2+2Na0H > Na2Cr04 + 2HC1

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Yellow sol.

Na2Cr04 + Pb(CH3COO)2 > PbCr04 + 2 CHjCOONa

Yellow ppt.
NH4CI + NaOH > NaCl + NH3 -t- H2O

e
2K2Hgl4 + NH3 + 3KOH > NHjHgO Hgl + 7KI + 2H2O.

re
P. 4. A mixed oxide of iron and chromium, Fe0.Cr203, is fused with sodium carbonate in presence of air
to form a yellow coloured compound (A). On acidification, the compound (A) forms an orange

Frl
F
coloured compound (B) which is a strong oxidizing agent.
(0 Identify the compounds (A) and (B)
ou
(ii) Write balanced chemical equations for each step.

osr
Sol.
4 FeO . Cr203 + O2 > 2 Fc203 + 4 Cr203
4 Na2C03 + 2 Cr203 + 3 O2 > 4 Na2Cr04 + 4 CO2J x 2
4 FeO . Cr-,03 + 8 Na2C03 + 7 4 8 Na^Cr04 + 2 kf + 8 CO.,
oo
(A)
Sodium chromate
Y
B

2 Na2Cr04 + H2SO4 ^ Na2Cr203 + Na2S04 + H2O

(B)
Sodium dichromaie
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Y

P. 5. (a) A blackish brown coloured solid ‘A’ when fu.sed with alkali metal hydroxides in presence of air,
u

produces a dark green coloured compound ‘B’, which on electrolytic oxidation in alkaline medium
ad

gives a dark purple coloured compound C. Identify A, B and C and write the reactions involved,
do

(b) What happcn.s when an acidic solution of the green compound (B) is allowed to stand for some
time ? Give the equation involved. What is this type of reaction called ?
in

Fuse
Re

Sol. (a) M11O2 + 4 KOH + 0, ■>

2 K2MnO 4 + 2 H,0
Pyrolusite
F

Potassium manganate
(A) Blackish brown (B) - Green coloured

Alkaline
2 K2Mn04 + H2O + (O) medium 2 KMn04 + 2 KOH
(B) Potassium permanganate
(C) - Purple coloured
or MnO-r
4 Mn04 + e~
{b) When acidic solution of green compound (B), i.e., potassium manganate is allowed to stand for some
time, it disproporlionates to give permanganate as follows :
3 MnOj- + 4H+ 4 2 Mn04 + MnO, + 2 H^O
This reaction is called disproportionationreaction.
Hg.
*Mlllon’s base is O'
NH2OH . The iodide is obtained by replacing OH group by iodine atom.
Ha
d- AND f-BLOCK ELEMENTS 8/47

P. 6. Calculate the number of moles of KMn04 needed to react completely with one mole of ferrous
oxalate in acidic medium.
Sol. 2 KMn04 + 3 H2SO4 ■) K2SO4 + 2 MnS04 + 3 H2O + 5 (O)] x 3
FeC')04 + H-)S04 FeS04 + H2C2O4] X 10
2 FeS04 + H2SO4 + (O) Fe2(S04)3 + H2O] X 5
H2C2O4 + (O) 2 CO-, + H-,0] X 10

6 KMn04 + 24 H2SO4 + 10 FeC204 3 K2SO4 + 6 MnS04 + 5 Fc2(S04)3 + 20 CO2 + 24 H2O

Thus, 10 moles of ferrous oxalate react with 6 moles of KMn04.

1 mole of ferrous oxalate will react with KMn04 = —

10
mole = - mole
5

w
F lo
1. Transition elements. Elements belonging to i/-block lying between .v and /^-blocks are called transition
elements or transition metals. They are .so called because their properties lie between s and /?-block. They
have the general electronic configuration : (n - 1)
2. IVansition series : 1st transition series (3d series) for n = 4 (21SC — 3oZn)

ree
2nd transition scries
3rd transition series
for F
(4d series) for n = 5 (39Y — 4«Cd)
(5d series) for n = 6 (57La, 72Hf — 8oHg)
4th transition series (6d series) for n = l (ggAc, )Q4Ku onwards)
3. Exceptional configurations of Cr and Cu in 1st transition series :
Your

24Cr : Expected E.C. = [Ar]'^ 3d^ 4s\ Actual E.C. = [Ar]*^ 4d^ 4i‘
ks
eBoo

29CU ; Expected E.C. = [Ar]'^ 3c^ 4s\ Actual E.C. = [Arj'® 3t/"^ 4s'
This is because exactly half-filled (3d^) and fully-filled (3d'^) configurations are more stable.
4. Transition elements in terms of electronic configuration. These are those elements in which atoms or
ad
our

ions in their common oxidation states have incompletely filled fr-subshell (having electrons 1-9). For this
reason, Zn, Cd and Hg are not considered as transition elements because their ions, Zn-'*’, Cd^'*’ and Hg^"^

have completely filled r/-subshell (3J*®, 4d"^, 5d'^ respectively).

Re

5. General Trends in Properties of Transition Elements

Y

(1) Atomic and Ionic radii. Atomic radii decrease in the series with increase in atomic number because
Find

nuclear charge increases. After midway, decrease is small because of increased shielding effect of ^-electrons.
Down the group, atomic radii increase. However, radii of 2nd and 3rd transition series metals are nearly
same due to lanthanoid contraction after lanthanum (Z = 57). Ionic radii follow the same trend. M^'*’ ions are
smaller than M^'*' ions as ions have greater effective nuclear charge,
(it) Metallic character. All transition elements are typical metals due to their low ionization energies and
form metallic bonds. Both ns and (n - \ ) d electrons participate. Greater the number of unpaired electrons,
stronger axe the metallic bonds. That is why Cr, Mo and W are hai'd whereas Zn, Cd and Hg are soft (Hg is
liquid).
(hi) Melting and Boiling points. They have high melting and boiling points. In any series, melting points
first increase, rise to a maximum and then decrease. This is because number of unpaired electrons first
increase, rise to a maximum and then decrease and so are metallic bonds. Thus, Cr, Mo and W have maximum
melting points in their respective series (W > Mo > Cr). W has highest tn.pt. Enthalpies of atomization (to
break metal into free atoms) also follow the same trend. M.pt. of 2sMn < 26pe or 43Tc < 44RU is due to
presence of half-filled rZ-subshells of Mn and Tc and hence are stable and form weak metallic bonds.

1
8/48 New Course Chemistry (XII)E2

(*V) Densities. Densities increase along a series because atomic size decreases whereas atomic mass increases,
(v) Ionization enthalpies. Their ionization enthalpies are higher than 5-block and less than p-block elements.
They generally increase along the series as nuclear charge increases. The first ionization enthalpies of Zn,
Cd and Hg are very high because of fully filled configurations. lE^ of 24Cr > 2sMn and 29CU > 3gZn because
after removal of one electron, Cr (I) is and Cu (I) is 3d^^, i.e. stable. IE3 of Mn is very high because
Mn (II) is 3f/^. IE3 of 26pe is very small because it leads to stable 3^.
(v/) Standard electrode potentials (E®). In solid state, stability of oxidation state can be predicted from
ionization enthalpies. For example, lEj + IE2 for Ni < Pt. Hence, Ni (II) compounds are more stable
lEj + IE2 + IE3 + IE4 for Ni > Pt. Hence Pt (IV) compounds are more stable. In aqueous solution, stability
depends upon electrode potentials, related as AHjoj^j = AH (sublimation) + AH (ionization) + AH (hydration)
viz. M (5) > M (g) > M+ (g) > M+ iaq).
E® (M^'VM) do not show regular trend because lEj + IE2 and sublimation enthalpies do not show regular
trend. In 3d series, only Cu has +ve E® (M^‘*'/M) because the sum A^^jj H + A,- H is not balanced by A|j H.
That is why it does not react with acids to give H2 gas.

w
E° (M^'^/M^'*') studies show that E° (Sc^''’/Sc2‘^) is very low showing stability of Sc^'*’. E® (Mn^'VMn^^) is
high because of stable for Mn^"*". E® (Fe^’^/Fe^'*’) is low due to extra stability of Fe^'*' (cfi). For these
reasons, Mn^+ and Co^"^ are strong oxidizing agents in aqueous solution whereas

F lo
and Cr^’*’ are
strongest reducing agents,
(vii) Oxidation states. The first element of each series (Sc, Y, La) shows an oxidation state of +3. The last

ee
element of each series (Zn, Cd, Hg) shows an oxidation state of +2 (Hg also shows +1). All other elements
show a number of oxidation states, +2 being most common (due to nP). Other oxidation states are due to

Fr
participation of (n - 1) in addition to ns electrons. Lowest oxidation state is +1 (shown by Cu, Ag, Au,

for
Hg). Highest oxidation state is +8 (shown by Os). Highest oxidation states are shown in compounds with
oxygen and fluorine. Higher oxidation states are also found in oxocations and oxoanions e.g. Mn (VII) in
ur
MnO^ and Cr (VI) in CrO^" and CrjO^". Oxygen stabilizes higher oxidation state more than fluorine.
s
ook
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IV vn

e.g., Mn forms MnF^ but it forms Mu202 . This is because oxygen forms multiple bonds.
eB

Many Cu (I) compounds are unstable in aqueous solution and undergo disproportionation (i.e., same
substance undergoes oxidation as well as reduction). Thus, 2 Cu'*' > Cu^'*' + Cu.
(viti) Catalytic properties. Most of the transition metals and their compounds are used as catalyst. This is
our
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mainly due to their variable oxidation states. Reactant reacts with catalyst in one oxidation state to form an
intermediate in another oxidation state which reacts with another reactant to form product and regenerate
the catalyst (intermediate compound formation theory). Alternatively,transitionmetals have a number of
Y

free valencies on their surface and hence can adsorb the reactant so that concentration on the surface increases
Re
nd

and so does the rate (adsorption theory),

(ix) Coloured ions. Most of the transition metal compounds are coloured in solid state and in aqueous
Fi

solution. This is due to presence of incomplete </-subshell. In presence of anion (or ligand), t/-subshell splits
into two sets, one with lower energy called t2g (includes d^, dy^, d^) and the other with higher energy called
Cg (includes d^ ^ and d ^). Electron in t^g absorb particular wavelength of visible light and jump to eg.
Remainder colour is emitted. For example, Cu^"*" salts look blue as they absorb red colour or Tp'*’ complexes
are purple because they absorb yellow colour. Sc^''" and Tl^'*’ compounds are colourless because they have
empty J-orbitals. Cu"*”, Ag"*", Au'*', Zn^"*", Cd^"*", Hg^+ compounds are also colourless because they have
completely filled c?-orbitals.
(xr) Magnetic properties. Compounds of transition metals are generally paramagnetic (attracted by external
magnetic field) or some of them are diamagnetic (repelled by extemd magnetic field) or ferromagnetic
(strongly attracted by external magnetic field). Paramagnetism is due to presence of unpaired electrons
while diamagnetism is due to presence of no unpaired electron. Greater the number of unpaired electrons,
greater is the paramagnetism (Magnetic moment, P- = -^n (n + 2) B.M. called spin only formula where n -
no. of unpaired electrons).
d- AND f-BLOCK ELEMENTS 8/49

(jo) Complex formation. Transition metals form a number of complexes, e.g. K4 [Fe(CN)g], [Cu(NH3)4] SO4
etc. This is due to availability of vacant ^-orbitals and large charge/size ratio of their ions,
(xw) Interstitial compounds. They form interstitial compounds like HC, Fe3H, Fe3C (cementite) etc. in
which small atoms like H, C, B and N occupy interstitial sites in their lattices.
(xtiV) Alloy formation. They form alloys with metals whose atomic radii do not differ by more than 15% so
that some of the host atoms of the lattice sites are easily replaced by the other metal atoms, e.g. brass
(Cu + Zn) and bronze (Cu + Sn).
6. Group-wise study of transition metals

Name of the Group Elements General E.C. Common ox. Important

present cliaracteristics

(i) Group 3 elements 21 Sc, 39Y, 57La, («-l) d\ AP- +3 Reactive metals. Tarnish in air.

(Scandium group) Ac (n = 4-7) React with H2O to evolve

low
89
H2.
Hard metals. Unreactive at
(ii) Group 4 elements 2zTi, 4oZr, 72Hf (n-1) S AP- +4
room temp. Form refractory
(Titanium group) (n = 4 - 6)
materials with H, C, N, B. Zr
and Hf have nearly same size

ee
due to lanthanoid contraction

F and hence occur together.

Fr
{Hi) Group 5 elements 23V, 4iNb, 73Ta (n-1) ^ Ap- +5 Unreactive at room temp. On
(Vanadium group) (n = 4-6) heating in O2, form M2O5 Nb

for
and Ta have nearly same size
ur
and many commonproperties.
(iv) Group 6 elements 24Cr, 42M0, 74W (n-1) ^ ns^, Cr = +3, +6 Hard, silvery white metals. W
ks
1
Mo, W = +5, +6
has highest m.pt. used for
(Chromium group) (n-1) ns
Yo
making filaments of bulbs.
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Form oxides whose basic

(v) Group 7 elements 25Mn, 43TC, 75Re (n-1) nP- Oto+7
eB

character decreases as ox. state

(Manganese group) increases. Tc and Re have
similar size.
r

(vi) Groups 8,9 and 10 26^®’ 2gNi Ferrous metals Fe, Co = +2, +3 All are ferromagnetic.
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ad

elements Ni = +2 Combine with CO to form

volatile carbonyls.
Y

(consist of 3 triads)
44RU, 4sRh, 46 pdl Platinum metals No. of ox. states Much less abundant than
ferrous metals.
nd
Re

760s, 77lr, 7gPt Highest = +8

(Ru, Os)
Fi

(vii) Group 11 elements 29CU, 47Ag, 79AU (n-1) d'O ns

1
Cu = +l,+2 They lie below H in
electrochemical series and do
(Coinage metals) (n = 4 - 6) Ag = +1
Au = +1, +3
not react with acids to give H2.
(vii7) Group 12 elements 30^^» 48^^^* 80^S (n-1) ns^ Zn = +2 Zn and Cd are similar (being
(Zinc group) (n = 4 - 6) Cd = +2 solids) but Hg is different
(being liquid). Zn and Cd
Hg = +1, +2 dissolve in dil. HCl/dil. H2SO4
to give H2 but Hg does not.

7. Oxides. They form oxides with oxidation states +1 to +7. Their ionic character decreases and acidic character
increases as oxidation state increases, e.g..
+2 +3 +4 +7
+ 8/3

MnO ^^304 Mn203 Mn02 Mn20.y

Basic Amphoteric Acidic

Ionic ^ Covalent
8/50 New Course Chemistry (XII)CE

8. Potassium dichromate (K2Cr207)

Preparation. It is prepared from chromite ore (Fe0.Cr203) in three steps :
(0 Fusing with Na2C03 and quick lime and extracting with water to get Na2CK)4 (4 Fe0.Cr203 + 8 Na2C03
+ 7 O2 > 8 Na2Cr04 + 2 Fe203 + 8 CO2).
(«) Treating with cone. H2SO4 to convert Na2Cr04 to Na2Cr207 and separating out less soluble Na2S04 by
fractional crystallisation (2 Na2Ci04 + H2SO4 > Na2Cr207 + Na2S04 + H2O).
(in) Treating with KCl to convert Na2Cr207 to K2Cr207 and separating out less soluble orange crystals of
K2Cr207 by fractional crystallisation (Na2Cr207 + 2 KCl > K2Cr207 + 2 NaCl).

ow
White heat
Properties. (/) 4 K2Cr207 ^ 4 K2C1O4 + 2 Cr203 + 3 ©2
(/0K2Cr2O7+ 2KOH 2K2C1O4 + H2O
2K2Cr04+ H2SO4 —> K2Cr207 + K2SO4 + H2O

e
Base ,

re
Thus, Cr202- + H20 Acid
2Cr02- + 2H+

(Orange)

Flr
(Yellow)

F
Cold
(in) K2CT20-J + 2 H2SO4 » 2 C1O3 + 2 KHSO4 + H2O
ou
Heal
2 K2Cr207 + 8 H2SO4 ■ > 2 K2SO4 + 2 Cr2 (804)3 + 8 H2O + 3 ©2

sr
(iv) Oxidizing properties ;

fo
Mol. wt.

k
K2Cr207 + 4 H2SO4— ^ K2SO4 + Cr2 (804)3 + 4 H2O + 3 (O), Eq. wt. = —-—
oo
(dil)
2 KI + H28O4 + (O) K28O4 + H2O +12
Y
2Fe804+H28©4 + (0) Fc2(804)3 + H2O
reB

H28 + (O) ■> H2O + 8

Na28©3 + (O) ■> Na28©4
uY

NaN©2 + (O) ^NaN©3

(O)
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CH3CH2OH + (O) ^ CH3CHO ^ CH3COOH

-H2O
(v) Chromyl chloride test
in

Heat
K2Cr2©7 (s) + 4 KCl (s) + 6 H28O4 4
2Cr02Cl2 + 6 KH8O4 + 3 H2O
Re

(ebne.)
F

Chromyl chloride
(Reddish brown vapour)

(vi) With H2O2. Cr20|- + 2 H+ + 4 H2O; ^ 2 CrOg + 5 H2O

(Deep blue)

Reason for colour of CrO^" and Cr207~. Cr in both has oxidation state + 6 and hence configuration.
Hence, colour is not due to d-d transition but due to charge transfer (from 0-atom to Cr changing 0^“ to 0“
and Cr (VI) to Cr (V)).
9. Potassium permanganate (KMn04)
Preparation. It is prepared from the mineral pyrolusite (Mn02) in two steps :
(0 Conversion of Mn©2 into K2Mn©4 (potassium manganate) by fusing with KOH or K2CO3 in presence of
air or oxidizing agent like KN63, KCIO3 etc.
2 Mn©2 + 4 KOH + ©2 ^ 2 K2Mn©4 + 2 H2O
or
3 Mn©2 + 6 KOH + KCIO3 ^ 3 K2Mn©4 + KCl + 3 H2O
d- AND f-BLOCK ELEMENTS 8/51

(ii) Oxidation of K2Mn04 to KMn04 chemically with CO2, CI2, O3 etc. or electrolytically (by electrolysis)
3 K2Mn04 + 2 CO2 > 2 KMn04 + Mn02 i+ 2 K2CO3
or
K2Mn04 2 K+ + MnOj", H2O ^ H+ + OH"

MnO^" > Mn04 + e~ (at anode) and H"'' + e~ > ^ ^2 cathode))

513K
Properties : (i) 2 KMn04 (5) » K2Mn04 + Mn02+O2
Red heat
2 K2Mn04 (5) - 2 K2Mn03 + O2
(Pot. manganite)

ow
Cold

(ii) 2 KMn04 + 2 H2SO4 (cone.) ^ Mn207 + 2 KHSO4 + H2O

Warm
4 KMn04 + 6 H2SO4 (cone.) > 2 K2SO4 + 4 MnS04 + 6 H2O + 5 O2
(Hi) Oxidizing properties
Mol. wt.

e
Neutral solution : 2 KMn04 + H2O > 2 KOH + 2 Mn02 + 3 (O), Eq. wt. =

re
rFl
Mol. wt.

F
Alkaline solution : 2 KMn04 + 2 KOH ^ 2 K2Mn04 + H2O + (O), Eq. wt. = 1

In presence of reducing agent, K2Mn04 +H2O 4 Mn02 + 2 KOH + (O)

r
ou
Mol. wt.
Complete reaction : 2 KMn04 + H2O k sfo
> Mn02 + 2 KOH + 3 (O), Eq. wt. =
Alkaline KMn04 is called Baeyer’s reagent, it oxidizes olefinic compound to glycols, e.g..
oo

> I
Y
CH2 = CH2 + H20 + (0)
B

OH OH
„ _ Mol. wt.
re

Acidic medium: 2 KMn04 + 3 H2SO4 (dil.) > K2SO4 + 2 MnS04 + 3 H2O + 5 (O), Eq. wt.
H2S + (O) H2O + S
ou
Y
ad

S02 + H20 + (0) H2SO4

KN02 + (0) ^KN03
d

C2H2O4 (oxalic acid) + (O) 2 CO2 + H2O

in
Re

2FeS04 + H2S04 + (0) -> Fe2(S04)3 + H2O

2KI + H2S04 + (0)- K2SO4 + H2O +12
F

2HX + (0)- ■» H2O + X2

CH3CH2OH + (O) > CH3CHO + H2O
Why KMn04 cannot be used in presence of HCl or HNO3 in ittrations ? Oxygen produced from KMn04
+ HCl not only oxidizes the reducing agent but is also partly used for oxidation of HCl to CI2. HNO3 itself
is an oxidizing agent.

Reason for colour of MnO^. Mn in MnO^ is in oxidation state + 7 with d^ configuration. Hence, colour
is not due to d—d transition but due to charge transfer (from O to Mn) changing Mn from +7 to +6.
1-14
10. Lanthanoids and Actinoids (^-block elements/inner transition elements). General E.C. = (n - 2) f
(n - 1) nP-. Lanthanoids (58 - 71, La - Lu) are elements in which 4/orbitals are progressively filled.
Actinoids (90 - 103, Ac - Lr) are elements in which 5/orbitals are progressively filled. Lanthanoids
(58 - 71) are also called rare earths. Actinoids are radioactive and show many oxidation states and hence
their study is complicated.
11. Lanthanoids (58 - 71, La - Lu) E.C. = [Xe] 4/ 5d^^ 6s^
8/52 New Course Chemistry (Xll)KSSlS]
12. Oxidation states of lanthanoids. Typical is + 3 but + 2 and + 4 are also found. However, they lend to
change to + 3. That is why Eu^"^ and Yb^'^ are reducing agents while Ce*^^ and Tb'*^ are oxidizing
agents. Due to large energy gap between 4/ and 5d, they show limited number of oxidation stales.
13. Atomic and Ionic radii of lanthanoids - Eanthanoid contraction. The small regular decrease in atomic
and ionic radii of lanthanoids is called lanthanoid contraction. The small net decrea.se is due to increase in
nuclear charge which outweighs the imperfect shielding by /-electrons. Due to this contraction, atomic radii
of elements of the 3rd transition series (after La) are nearly same as those of 2nd series in the same group.
14.
Some characteristics of lanthanoids. (/) They are silvery white metals. Mo.st of the trivalent metal ions are

w
coloured due to / -/transition. Those with x/-electrons have same colour as with (14 - .y)/-electrons.
iii) Except La^'*' (f and Lu^"^ (/'‘^), all show paramagnetism due to presence of unpaired electrons. Magnetic
moment, = .^4S(S-h 1)+L(L-l-1) B.M. where S = spin quantum number and L = orbital quantum

e
number.

ro
re
(Hi) They form basic hydroxides. As size decreases from La^’*' to Lu^+, the covalent character of hydroxides
increases and hence basic strength decreases. Thus, La(OH)3 is most basic while Lu(OH)3 is least basic.

F
15. Uses of lanthanoids. They are used only as alloys. The most common being misch metal (Ln = 95%, Fe =

Fl
5%, S, C, Ca, Al= traces). It is used in making Mg based alloy. It is a pyrophoric alloy (emits spark when

u
struck), used in making bullets, shells, lighter flints etc.
16. Actiiioids. (89 - 103, Ac - Lr) E.C. = [Rn] 5/’-*“^ 6d^^ ls~

sr
17. Oxidation states of actinoids. Like lanthanoids, most common is + 3. They also show oxidation slate of

ko
o
+ 4, + 5, -F 6 and + 7, e.g., in Th, Pa, U and Np respectively. They show a large number of oxidation states

18.
because of very small energy gap between 5/, 6d and Is subshells.
of
Ionic radii and Actinoid contraction. It is similar to lanthanoid contraction but contraction is greater due
o
to poor shielding effect of 5/electrons. Further, 5/orbitals extend in space beyond 6.v and 6p orbitals whereas
Y
4/orbitals are buried deep.
erB

19. Some characteristics of actinoids. They are silvery white metals. Their cations are coloured due to/-/
transition. Ac^"*" (5/^), (5/^) and Th'^"^ (5/*^) are colourless. They are strongly paramagnetic. They
uY

have higher tendency to form complexes than lanthanoids due to higher charge and smaller size of their ions.
20. Uses of actinoids. Th is used in atomic reactor and treatment of cancer. U and Pu are used as fuel in nuclear
reactor.
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do
in

QUESTIONS
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Based on NCERT Book

F

I. Multiple Choice Questions (c) They are much harder than the pure metal
1. Which of the following ion has the electronic (d) They have higher melting points than the pure
metal
configuration 3cfi ?
(a) Ni
3+
(b) Mn^-^ 4. Among the following transition metals, the
maximum number of oxidation .states is exhibited
(c) Fe^-^ (d) Co^-^
by
2. The colour of the light absorbed by an aqueous
(g) Mn (Z = 25) (b) Fe(Z = 26)
solution of CUSO4 is
(a) orange-red (b) blue-green (c) Cr(Z = 24) (d) Ti (Z = 22)
(c) yellow (d) violet 5. Of the following outer electronic configurations
3. Which of the following statements about the of first transition series atoms, the highest oxidation
interstitial compounds is incorrect? state is achieved by which one of them ?
(a) They retain metallic conductivity (a) (n - I) d^ ns^ (b) (n - 1) d^ ns-
(b) They are chemically reactive (c) (n - I) d^ 115- {d) {n-\)d^ ns
d- AND f-BLOCK ELEMENTS 8/53

6. The magnetic moment of (NiCl4)^ is (a) CrS04 ib) Cr2(S04)3

(a) 1-82 BM (b) 546 BM
(c) Ct02- id) CiO-
(c) 2-82 BM id) 141 BM
16. In acidic medium, equivalent wt. of K2Cr207 is
7. The stability of ferric ion is due to
(M = molecular wt.)
(a) half-filled ^/-orbitals
(fl) M ib) Mil
{b) half-filled/-orbitals
(c) M/3 id) M/6
(c) completely filled fr-orbitals
17. In dichromate dianion
(d) completely filled/-orbitals
8. Among the following pairs of ions, the lower id) 4 Cr—O bonds are equivalent
oxidation state in aqueous solution is more stable ib) 6 Cr—O bonds are equivalent
than the other, in (c) All Cr—O bonds are equivalent

ow
id) T1+,T13+ ib) Cu+,Cu2+ id) All Cr—O bonds are non-equivalent
id) Cr^+, Cr^+ id) v2+,VQ2+
9. A transition metal exists in its highest oxidation 18. In acidic medium, one mole of Mn04 accepts how
state. It is expected to behave as many moles of electrons in a redox reaction ?

e
(c) a chelating agent (a) 1 Qy) 3

Fl
re
ib) a central metal in a coordination compound ic) 5 .id) 6

F
(c) an oxidizing agent 19. The number of moles of KMn04 that will be
(^0 a reducing agent needed to react with one mole of iodide ion in
ur
or
10. Magnetic moment of 2-84 B.M. is given by : alkaline solution is
(At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27) (a) 2 sf ib) 2/5
ia) Cr2+ ib) Co2+ id) 1
ic) 1/5
k
id) tP+
Yo
ic) Ni2+
20. Name the gas that can readily decolorise acidifed
oo

11. The basic character of the transition metal

monoxides follow the order
KMn04 solution
eB

ia) CO2 ib) SO2

ia) CrO > VO > FeO > TiO
ib) TiO > FeO > VO > CrO ic) NO2 id) P2O5
ur

21. In chromite ore, the oxidation number of iron and

ic) TiO > VO > CrO > FeO
chromium are respectively
ad
Yo

id) VO > CrO > TiO > FeO

ia) +3,+2 ib) +3,+6
12. The product of the oxidation of r with MnO^ in id) +2,+3
ic) +2,+6
d

alkaline medium is
22. When hydrogen peroxide is added to acidified
Re
in

ia) lOj ib) I2 potassium dichromate, a blue colour is produced

F

due to formation of
ic) lOr id) IO4 ia) C1O3 ib) Cr^O^
13. The formula of pyrolusite is
ia) Mu304 ib) Mn02 ic) CrOs id) CtO}-
ic) MnO id) Mu207 23. Which of the following compounds will not give
14. Which of the following statements is correct when positive chromyl chloride test ?
SO2 is passed through acidified K2Cr207 solution ? ia) CUCI2 ib) HgCl2
ia) Green Cr2(S04)3 is fo. ed +

ib) The solution turns blue (c) ZnCl2 id) C6H5NH3CI-

(c) The solution is decolourised 24. Which of the following does not give oxygen on
id) SO2 is reduced heating?
15. Acidified K2Cr207 solution turns green when ia) KCIO3 ib) Zn(C103)2
Nu2S03 is added to it. This is due to the formation ic) K2^^2^7 id) (NH4)2Cr207
of
8/54 New Course Chemistry (XII)EZ

25. Identify the product and its colour when Mn02 is (b) The +3 oxidation state of cerium is more stable
fused with solid KOH in the presence of O2 than +4 oxidation state,

(a) KMn04, purple (i?) K2Mn04, dark green (c) The 44 oxidation state of cerium is not known
(c) MnO, colourless {d) Mn203, brown in solutions.

(e) Mn02, black (d) Cerium (IV) acts as an oxidizing agent.

26. In acidic medium, MnO^~ 35. Which of the following exhibits only + 3 oxidation
state?
(a) disproportionates to Mn02 and MnO^
(a) Ac (b) Pa
(b) is oxidized to MnO^ (c) U (d) Th
(c) is reduced to Mn02 (^0 is reduced to Mn^^ 36. Lanthanide contraction is observed in
27. A student accidently added cone. H2SO4 to (a) Gd (b) At

w
potassium permanganate and it exploded due to
the formation of an explosive which is (c) Xe (d) Ac
(a) MnO (b) Mn203 37. Which of the following is not an actinoid ?

Flo
(c) Mn205 (d) Mn207 (a) Curium (Z = 96)
28. The electronic configuration of gadolinium (At.

e
(b) Californium (Z = 98)
No. = 64) is

re
(c) Uranium (Z = 92)
(a) 4y® 5 efi 6 (b) 4/ 5d^6s^

F
(d) Terbium (Z = 65)
{c) AfSd^ 6 P- (d)
38. Among the following actinide pairs, the maximum
29. Which one of the following shows maximum
ur
r
number of oxidation states in its compounds ? oxidation states is shown by
(a) Eu {b) U fo
(a) U and Np ib) Np and Pu
ks
(c) Gd {d) Am (c) Pu and Th {d) U and Pa
Yo
30. Lanthanoid contraction is due to increase in : 39. Prussian blue is
oo

{a) Effective shielding of/-electrons {a) K3[Fe(CN)6l ib) K4[Fe(CN)6l

B

{b) Atomic number (c) KFe[Fe(CN)6] (d) Fe4 [Fe(CN)6l

re

(c) Effective nuclear change 40. Which one of the following reactions involves
{d) Atomic radius disproportionation?
u
ad

31. Which of the following is the most common (a) 2H2S04 4-CU- > CUSO4 + 2 H2O + SO2
Yo

oxidation state among the lanthanoids ?

{a) +3
(h) AS2O3 + 3H2S > AS2S3 + 3 H2O
(b) +4
(c) 2KOH-HC12- KCl KOCl 4- H2O
d

(c) 4-2 id) 4-5

Re
in

32. Which of the following statements is not correct ? {d) Ca3P2-f6H20 >3Ca(OH)2 4-2PH3
(e)4NH3-<-02 4 2N2-I-6H20
F

(fl) La(OH)3 is less basic than Lu(OH)3

{b) La is actually an element of transition series 41. Which one of the following is not a transition
rather than lanthanoids element ?

(c) Atomic radius of Zr and Hf is same id) Fe ib) Sn

(d) In lanthanoid series, the ionic radius of Lu^'*' (c) Cu id) Hg
is smallest
42. Which of the following has maximum magnetic
33. Which of the following pair have same size ? moment ?
(fl) Fe2+,Ni2+ ib) Zr^+,Ti4+ id) Fe2+ ib) Mn2+
(c) Zr^+, Hf^+ id) Zn2+,Hf^
(c) Cu+ id) Cr2+
34. Cerium (Z = 58) is an important member of the
lanthanoids. Which of the following statements 43. In the first transition series, the highest b.p. and
about cerium is incorrect 7 m.p. is of
(a) The common oxidation states of cerium are 43 id) Cr ib) V
and +4. (c) Ni id) Fe
d- AND f-BLOCK ELEMENTS 8/55

44. The correct order of decreasing second ionisation (a) Solution in test tube becomes green and lime
enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is water turns milky
(a) Mn>Cr>Ti>V (b) Ti>V>Cr>Mn (b) Solution in test tube is colourless and lime
(c) Cr>Mn>V>Ti (d) V>Mn>Cr>Ti water turns milky
45. The E® value for Mn^'^/Mn^'*’ couple is much more (c) Solution in test tube becomes green and lime
water remains clear
positive than for Cr^VCr^"^ or Fe^’‘‘/Fe^'^ because
(a) Mn^"^ is much more stable than Mn^'*’ (d) Solution in test tube remains clear and lime
water also remains clear.
(b) Mn^"*" has 3d^ configuration which is exactly
half-filled and hence is very stable 53. Gaseous product obtained on thermal
(c) the third ionization enthalpy of Mn will be decomposition of (NH4)2Cr207 is
very high (a) NH3 (b) N2
(d) all the three above (c) O2 (d) NO
46. Which of the following ions has a magnetic 54. Volumetric titrations using acidified KMn04 are
carried out in presence of

w
moment of 5-93 B.M.?

(At. No. V = 23, Cr = 24, Mn = 25, Fe = 26) (a) dilute H2SO4 (h) dilute HCl

F lo
(a) Mn^'*' (b) Fe2+ (c) dilute HNO3
(c) id) V3+ (d) both (a) and (b) but not (c)
47. Which of the following ions will exhibit colour in 55. Mn02 when fused with KOH and oxidised in air,

e
gives a dark green compound X. In acidic

Fre
aqueous solutions ?
(a) Sc3+(Z = 21) (b) La3+(Z = 57) solution, X undergoes disproportionation to give
an intense purple compound Y and Mn02. The
(c) tP+(Z = 22) id) Lu3+(Z = 71)
48. Ammonia is a Lewis base. It forms complexes
for
compounds X and Y respectively are
(a) K2Mn04 and KMn04
r
with cations. Which one of the following cations
(b) Mn207 and KMn04
You
does not form complex with ammonia ?
oks

(a) Ag+ ib) Cu-^ (c) K2Mn04 and Mn207

o

(d) KMn04 and K2Mn04

eB

(c) Cd-^ id) Pb++

49. The catalyst used for olefin polymerization is 56. The number of moles of Br2 produced when two
moles of potassium permanganate are heated with
(a) Ziegler-Natta catalyst
our

excess potassium bromide in aqueous acid

ad

(b) Wilkinson’s catalyst medium is

(c) Raney nickel catalyst (a) 2 ib) 3
(d) Merrifield resin
dY

(c) 4 id) 5
Re

50. The basic character of the transition metal

57. In following reaction
monoxides follows the order
Fin

(a) TiO > VO > CrO > FeO

yMn04 +xH++ C2O5- ^yMn++ + 2C02
(b) VO > CrO TiO > FeO + -H2O,
2 2
(c) CrO > VO > FeO > TiO
X and y are
(d) TiO > FeO > VO > CrO
(a) 2 and 16 ib) 16 and 2
(Atomic no. Ti = 22, V = 23, Cr = 24, Fe = 26)
(c) 8 and 16 id) 5 and 2
51. C1O3 dissolves in aqueous NaOH to give 58. Which of the following is not an actinoid ?
(a) Cr202- (b) CrOj- (a) Am ib) Cm
(c) Fm id) Tm
ic) Cr(OH)3 id) Cr(OH)2
52. A mixture of salts (Na2S03 + K2Cr207) in a test 59. The outer electronic configuration of Gd (At. No.
64) is
tube is treated with dil. H2SO4 and resulting gas is
passed through lime water. Which of the (a) 4/3 5d^ 6^2 ib) 4/8 5652
following observations is correct about this test? (c) 4f^5d^6s^ id) 4f 5d^ 6^2
8/56 New Course Chemistry (XII)SSSIS]

60. Most common oxidation states shown by cerium Reason. In the 31/transition series, first the number
are
of unpaired electrons increases and then decreases.
{a) +2,+ 4 ib) +3,+ 4
68. Assertion. E° has a positive value.

rw
Cu-'^/Cu
(c) +3,+ 5 id) +2,+ 3
61. Which of the following trivalent ion has the Reason. The sum of enthalpies of sublimation and
largest ionic radus in the lanthanide series ? ionization balances the hydration energy.
(n) La-’-" ib) Ce^-" 69. Assertion. The maximum oxidation state shown
U
(c) Pim id) Lu^-" by osmium is + 6.

e
62. Reason of lanthanoid contraction is Reason. Osmium has the outer configuration
(fl) negligible screening effect of/-orbitals 5d^ 5^ in which one <i-orbitai contains a paired

r
o
electron.
ib) increasing nuclear charge

F
lu
(c) decreasing nuclear charge 70. Assertion. Cup2, CuCl2, CuBr2 all exist but Cul^
does not.
id) decreasing screening effect
Reason. Iodide ion is too big in size.

oF
63. Which of the following analogies is correct ?
71. Assertion. Cu^’*’ and V^'*' salts are both blue.

rs
(a) Cadmium : transition metal; : Mercury : Not a
transition metal Reason. Cu-'*' and contain the same number

ok
of unpaired electrons.
ib) In third transition series W : highest melting
point: : Au : lowest melting point 72. Assertion. Equivalent mass of KMn04 is same in

fo
the neutral as well as alkaline medium.
(c) Standard electrode potentials (M^-"/M)
Cr = + ve : : Cu = - ve Reason. The product formed in both cases is
o
Y
id) Colours of the hydrated ion : Mn02.
Y
Ti*^-" = coloured : : Ti’-" = colourless 73. Assertion.Copper (II) iodide is not known.
rB

Reason. Cu^"*" oxidizes I” to iodine.

II. Assertion-Reason Type Questions
74. Assertion. Separation of Zr and Hf is difficult.
For questions below, two statements are given
ue

one labelled Assertion (A) and the other labelled Reason. Zr and Hf lie in the same group of the
d

Reason (R). Select the correct answer to these periodic table.

questions from the codes (a), ib), (c) and id) 75. Assertion. Actinoids show larger number of
o

given below :
ad
n

oxidation states than lanthanoids.

id) Both A and R are true and R is the correct
explanation of the assertion. Reason. Highest oxidation state shown by
i

lanthanoids is + 4 while that of actinoids is + 7.

ib) Both A and R are true but R is not the correct
F
Re

explanation of the assertion, 76. Assertion. Transition elements exhibit higher

(c) A is true but R is false. enthalpies of atomisation.
id) A is false but R is true. Reason. Transition metals contain a large number
64. Assertion. Mercury is not considered as a transition of unpaired electrons.
element. 77. Assertion. Cr^'*’ is reducing whereas Mn^"*" is
Reason. Mercury is liquid. oxidizing while both have d‘^ configuration.
65. Assertion. Manganese shows a maximum Reason. E° value for Cr^’^’/Cr^'*' is positive whereas
oxidation state of +5.
E° value for Mn^'^/Mn^'^ is negative.
Reason. Manganese has 5 electrons in the 3d
78. Assertion. Mercury is not considered as a transition
subshell.
element.
66. Assertion. Ce’-" is used as an oxidizing agent in
volumetric analysis. Reason. Mercury is liquid at room temperature.
Reason. Ce’-" has the tendency to change to Ce'’^. 79. Assertion. Mn^"^ compounds are more stable than
67. Assertion. First ionization enthalpies of the 3d Fe^"^ compounds towards oxidation to + 3 state.
transition series first increase to a maximum and Reason. Mn has lower atomic number (25) than
then decrease. Fe (26).
d- AND f-BLOCK ELEMENTS 8/57

80. Assertion. Ti'*+ and both form coloured 82. Assertion. In aqueious solution, Eu^'*' and
compounds. Yb-"*" are good reducing agents while Ce^'*' and
are good oxidizing agents.
Reason. Colour of the compound is due to presence
Reason. Some of the oxidation states shown by
of unpaired electrons in the zZ-subshell.
lanthanoids are + 2, + 3, and + 4.
81. Assertion. Lanthanoids show limited number of
83. Assertion. Transition metals have low melting
oxidation states whereas actinoids show a large
points.
number of oxidation stales.
Reason. The involvement of greater number of
Reason. Lanthanoids have lower atomic numbers
(n - 1) frand electrons in the interatomic metallic
than actinoids. (CBSE 2020)
bonding.

ANSWERS

w
I. Multiple Choice Questions
l.(d) 2. ia) 3. (b) 4. (a) 5. (b) 6. (c) 7. {a) 8. (a) 9. (c) 10.(c)

Flo
11. (c) 12. (a) 13.(b) 14. (a) 15. ib) 16. id) 17. ib) 18.(r) 19.(«) 20. (b)
21. (d) 22.(c) 23.(b) 24. id) 25. (b) 26. ia) 27. id) 28. (b) 29. id) 30.(c)

e
re
31. (a) 32. (a) 33. (c) 34. (c) 35.(a) 36. ia) 37. id) 38. (b) 39. (c) 40. (c)
41. ih) 42. (h) 43. (a) 44. (c) 45. (d) 46. ia) 47.(c) 48. (d) 49. (a) 50. (o)

F
51. ib) 52. (c) S3, (h) 54. (fl) 55.(a) 56. id) 57. ib) 58. (/;) 59. id) 60. (b)
61. (a) 62.(a) 63. (b)
ur
II. Assertion-Reason Type Questions
f or
ks
64.ib) 65. id) 66. (c) 67. id) 68. (c) 69. id) 70. ib) 71. (fl) 72. ia) 73. ia)
Yo
oo

74. ib) 75.ib) 76.(b) 77. (c) 78. ib) 79. (fe) 80. id) 81. ih) 82. ib) 83. id)
B

A A O
re

For Difficult Questions

u
ad
Yo

I. Multiple Choice Questions 7. Fe (Z = 26) = 3d^ 4s^

2. The aqueous solution of CUSO4 absorbs orange- Fe^-^ = 3d^. Thus, Fe^+ ion is quite stable due
d
Re

to half-filled ^/-orbitals.
red colour of light (The complementary colour,
in

blue, is reflected). 8. Tl'^ is more stable than TP"^ due to inert pair effect
F

(T1 belongs to Group 13 and is the last element of

3. Interstitial compounds are chemically inert and not
the Group).
reactive. Hence, ib) is incorrect.
9. In the highest oxidation state, it can undergo
6. In [NiC^]^", Ni is present as Ni^'*’ reduction and hence acts as oxidizing agent, e.g., in
Ni = [Ar]‘^3(i^ 4^^ KMn04, Mn is in +7 oxidation state. It is one of the
.-. Ni2+ = [Ar]'^3i/^ strongest oxidizing agent in solution.

10. Magnetic moment, p = in + 2)

3^*= n n n t t
)i = 2-84 BM when n - 2, i.e., there are 2
It has 2 unpaired electrons. unpaired electrons.
Magnetic moment Cr^"^ = 3d“^ (unpaired electrons = 4)
Co^‘*‘ = 3d'^ (unpaired electrons = 3)
= -fnin + 2) BM = V2x4 = ^ Ni'"^ = 3d^ (unpaired electrons = 2)
= 2-82 BM 3+
Ti = 3rf* (unpaired electrons = 1)
8/58 New Course Chemistry (XII)BEfflO

20. SO2 readily decolourises acidified KMn04 41. Sn is a p-block element and not a transition
solution (Refer to reaction (i7), page 8/31). element.

21. Chromite ore is FeCr204 42. The outer electronic configuration of the given
elements are as follows :
Ox. state of Fe = +2, Cr = +3.
Fe^+ = 3d^
22. Blue colour is due to the formation of CrOj as
follows :
“It 1 1 1 1 (four unpaired electrons)
Cr20^- +2H-^ + 4H202 ^2Cr05 + 5H20 Mn2+ = 3rf5

23. HgCl2 being covalent does not give chromyl 1 “I 'I 1 1 (five unpaired electrons)

ow
chloride test.
Cu+ = 3d‘0

24. (NH4)2Cr207 " ■■) Cr20j + N2 + 4 H2O 1 11 U U 11 unpaired electrons)

25. 2 Mn02 + 4 KOH + O2 Cr3+ = 3d3

e
2 K2Mn04 + 2 H2O 1 1 1 (three unpairedelectrons)

re
Pot. manganate
(Green) Thus, Mn^'*’ has the maximum number of

Frl
F
unpaired electrons and hence has maximum
26. MnO^" disproportionates in acidic medium as magnetic moment.
follows : 43. Cr has highest m.p. and b.p. due to maximum
ou
or
number of unpaired electrons which results in
3MnQ2- +4H+ strong metallic bonding.

2 Mn04 + Mn02 + 2 H2O

kfs
44. The second ionization enthalpy generally
increases form Ti to Mn but Cr has higher value
oo
than Mn. This is because after the removal of first
27. Mn207 is formed which decomposes explosively
Y
on heating (See reaction4, page 8/29). electron, Cr acquires a stable configuration of
eB

33. 2x^ and Hf*'*' have similar ionic radius due to

3d^ and the removal of second electron is very
lanthanoid contraction.
difficult. Thus, the order for IE2 is
ur

Cr > Mn > V > Ti.

34. Ce"*^ exists in solution though it tends to change
oY

to more stable Ce^'*’. 45. The large positive E® value for Mn^'''/Mn^‘^ shows
ad

that Mn^^ is much more stable than Mn^"*". This is

35. Ac (6d* 7^“) shows only + 3 oxidation state. due to the fact that Mn^"*" has 3d^ configuration
d

34. Gadolinium (Gd) is a member of lanthanide series. which is half-filled and hence, is very stable.
in

36. Actinoids are elements from Z = 89 to 103. Thus, Thus, the third ionization enthalpy of Mn will be
Re

very high. In fact, this is the reason that + 3 state

Terbium (Tb, Z = 65) is not an actinoid but a
F

lanthanoid.
of Mn is of little importance.

38. Both Np and Pu show highest oxidation state of 46. Pe# = V«(« + 2) bM ; 5-93 = ^n(n + 2)
+7.
m n
or n(rt + 2) = 35-16BM
39. Prussian blue is KFe[Fe(CN)g] (Refer to or n = number of unpaired electrons = 5
page 8/33) .●. Mn^'*’ ion (3d^) has magnetic moment of
5-93 B.M.
40. A disproportionation reaction is one in which the
same substance undergoes oxidation as well as 47. Electronic configuration of Ti^'*’ is 3dK Thus Ti^"*"
reduction. In (c). contains an unpaired electron, so it will exhibit
colour in aqueous solution due to d-d transition.
0 -1
48. Pb'*"'', because it does not have vacant d-orbitals
CI2 > KCl
nor high nuclear charge and it does not belong to
0 +1
transition series.
and also CI2 ^ KOCl. 49. Ziegler-Natta catalyst [TiCl4 + A1(CH3)3] is used
for polymerization of olefin.
d- AND f-BLOCK ELEMENTS 8/59

50. Basic character of oxides decreases from left to 62. Lanthanoid contraction is the regular decrease in
right in a period of periodic table. atomic and ionic radii of lanthanides with
51. Cr03 + 2 NaOH ^ Na2Cr04 + H2O increase in atomic number. This is due to the
52. Na2S03 with dil. H2SO4 gives SO2 gas which imperfect shielding (or poor screening effect) of
reacts with K2Cr20-7 to give a green solution due /-orbitals due to their diffused shapes which is
to Cr2(S04)3. Since whole of the gas is unable to counterbalance the effect of the
consumed, the lime water will remain clear. increased nuclear charge. Hence, the net result is
K2Cr207 + 4 H.2SO4 a contraction in size of lanthanoids.
K2SO4 + Cr2(S04)3
+ 4 H2O + 3 I O I 63. (a) is not correct because Cd is not considered as

a transition element.
SO2 + I O I + H2O H2SO4
(c) is not correct because standard electrode
53. (NH4)2Cr207 N2 + 4 H2O + ^263 potential of Cu (E“Cu^+ZCu ) = + ve and that of

low
54. KMn04 titrations are carried out only in presence
of dilute H2SO4. HCl cannot be used as oxygen Cr (E!Cr2+/Cr ) =-ve.
given out by KMn04 "*■ partly used up to
oxidize HCl to CI2. If HNO3 used, it itself acts (d) is not correct becase 11 (3d*) is coloured
as an oxidizing agent. while Ti^'*' (3d®) is colourless.

ee
II. Assertion-Reason Type Questions
55. 2 Mn02 + 4 KOH + O2
Fusion
—^ ) 2K2Mn04

F
^

Fr
Air Dark green (X)
64. Correct explanation. Mercury has completely
+ 2H2O filled d- subshell.

for
ur
Dispropo 65. Correct A. Manganese shows a maximum
2K2Mn04 +4 HCl -itionation > 2KMn04
(X) Intense purple (Y) oxidation state of +7 (as it has 3d ^ 45 ^ config
uration and the energy difference between 3d and
ks
+ Mn02 + 4 KCl + 2 H2O
Yo
45 subshells is small).
oo

56. 2KMn04 + ^ ^2^04 + 10 KBr 66. Correct R. Ce'*^ has the tendency to change to
eB

2 moles
Ce^^ (as +3 oxidation state is more stable).
6 K2SO4 + 2 MnS04 + 5Bt2 + 8 H2O 67. Correct A. First ionization enthalpy of the 3d
r

3 moles transition series increases with increase in atomic

ou
ad

Thus, 2 moles of KMn04 when heated with number.

excess KBr in acidic medium produce 5 moles of 68. Correct R. The sum of the enthalpies of
Y

Br2. sublimation and ionization is not balanced by

57. In the balanced equation,
Re
nd

hydration enthalpy.
69. Correct A. The maximum oxidation state shown
2Mn04+I6H++C2O42 + 2Mn+2+2C02
Fi

by osminius + 8.
+ 8H2O 70. Correct explanation. Cu^'*' ion oxidize iodide to
x=l6 and y = 2
58. Tm > Thulium + Atomic number 69.
iodine and CUI2 itself is reduced to CU2I2
(2 Cu2+ + 41- > CU2I2 +12)
It is a lanthanoid because lanthanoids are present
71. R is the correct explanation of A,
in sixth period of periodic table from atomic
number 57 (lanthanum) to atomic number 71 72. R is the correct explanation of A.
(lutetium). 73. R is the correct explanation of A.
59. The electronic configuration of 74. Correct explanation. Due to lanthanoid
64Gd=[Xe] 4/”^ 5^*652 contraction Zr and Hf have nearly the same size.
60. Ce = [Xe] 4/2 5 d9 6s^\ oxidation states shown 75. Correct explanation. In lanthanoids, energy gap
are + 3 and + 4.
between 4/and 5d subshells is large while in case
61. La2+ due to lanthanoid contraction has the largest of actinoids, energy gap between 5/, 6d and
ionic radius. 7s subshell is very small.
8/60 'Pn/tdeefi.'a. New Course Chemistry (X11)CB19]

76. Correct explanation. Higher enthalpies of 80. Correct A. is colourless (because it has
atmosiation are due to stronger metallic bonding configuration, i.e., no unpaired 2d electron)
on account of greater number of unpaired electrons whereas V'*'*’ is coloured (because it has 2d^
present in them. configuration, i.e., one unpaired electron is present
in 3d).
77. Correct R. E° value for Cr^’^/Cr^'*’ is negative
81. Correct explanation. Lanthanoids show limited
(_ 0-41 V) whereas E° value for Mn^'*’/Mn^'^ is number of oxidation states because energy gap
positive (+ 1-57 V). Thus, Mn^'*' can undergo between 4/ and 5d subshells is large whereas
reduction to form Mn^"^ and hence is oxidizing. actinoids show large number of oxidation stales
78. Correct explanation. Mercury has no partly or because of very small energy gap between 5/, 6d
incompletely filled d-subshell in atomic state or and 7.V subshells.
common oxidation state of + 2. Hence, it is not 82. Correct explanation. The most stable oxidation

ow
considered as a transition element. state of lanthanoids is + 3. ions can lose
79. Correct explanation. Mn-'*’ has 3J-' config electron to form M"-’'*' and hence are reducing
uration which is half-filled and hence stable agents while ions can gain electron to form
whereas Fe-'*’ has the configuration 3d^ and hence and hence are oxidizing agents.
can lose one electron easily to form stable 3d^ 83. Correct A. Transition metals have high melting

e
Fl
re
configuration. points.

F
goNg-epxu/\i_ problems
ur
or
1. (ieneral introduction,
4. Why is there striking similarities (horizontal
sf
and vertical) in successive members of the
electronic configuration and
k
transition series ?
Yo
general characteristics and trends
oo

Ans. This is because of the fact that along a

in properties of Iraasition elements horizontal row, electrons enter an incomplete
B

1. Though copper, silver and gold have inner shell in building up an atom, while the
re

completely filled sets of J-orbitals yet they outer level remains almost unchanged. In
are considered as transition metals. Why ? vertical columns, similarities are due to similar
u

electronic configurations even in the d-

ad

Ans. These metals in their common oxidation states

Yo

siibshells.
have incompletely filled d-orbitals, e.g., Cu^"^
has 3d^ and Au^'*' has 5d^ configuration. 5. Why the properties of third transition series
are very similar to second transition series ?
d

2. Name the elements which are not really

Re
in

Or Why the second and third members in each

transition elements but are discussed with
group of the transition elements have very
F

them. Why is it so ? similar atomic radii ? (CBSE 2012)

Ans. Zinc, cadmium and mercury have completely Ans. In the third transition series after lanthanum,
filled sets of <7-orbitals, and therefore, are not there is lanthanoid contraction. Due to this
really transition metals. However, they are often contraction, the size of any atom of tlie third
discussed with them because of the similarities transition series is almost the same as that of
of some of their properties to those of transition the element lying just above in the second
metals. Moreover, they are the end members of transition series. This leads to similarity in their
these series.
properties.
3. Explain why Fe is a transition metal but Na 6. The melting and boiling points of Zn, Cd and
is not ?
Hg are low. Why ?
Ans. Fe contains incompletely filled 3d subshell (3d ^ Or Why Zn, Cd and Hg are soft and have
4.v^). Hence, it is a (/-block element, i.e., a low m.pt. ?
transition element. Na has no (/-subshell. Last
Ans. In Zn. Cd and Hg, all the electrons in ^/-subshel!
electron in it enters 3^ orbital (H^ 2.^ 2p^ 3^'). are paired. Hence, the metallic bonds present
Hence, it belongs to j-block.
d- AND f-BLOCK ELEMENTS 6/61

in them are weak. That is why they have iow The reaction proceeds through a new path with
melting and boiling points. lower activation energy and hence is faster.
7. K2ptClg is a well known compound whereas According to aclsorpiion theory, transition
correspondingNi compound is not known. metal provides a large surface area with free
State a reason for it. valencies on which the reactants are adsorbed.
An.s. This is because Pt'*'*' is more stable than Ni"^^ As a result, the concentration of the reactants
on the surface increases and hence the rate
as the sum of four ionization enthalpies of Pt increases.
is less than that of Ni.
12. Why generally there is an increase in density
8. Most of the transition metals do not displace
of elements from titanium (Z = 22) to copper
hydrogen from dilute acids. Why ? (Z = 29) in the first series of transition
Ans. This is because most of the transition metals elements. (CBSE 2010)
have negative oxidation potentials. Ans. Due to decrease in size and increa.se in mass.
9. The E® values in respect of the electrodes of 13. Giving reasons indicate which one of the
chromium (Z - 24), manganese (Z = 25) and following would be coloured ?
iron (Z = 26) are : Cr^VCr^+ = - 0-4 V,

w
Cu"^. VO-+, Sc^\ Ni~+ (At. Nos. Cu = 29,
Mn^^ Mn^+ =: + I S V, Fe3+/Fe2+ = + 0-8 V. V==23, Sc = 2I, Ni = 28)

F lo
On the ba.sis of the above information Ans. Ni-"*^ due to incompletely filled fr-orbitals.
compare the feasibilities of further oxidation 14. Scandium forms no coloured ions, yet it is
of their +2 oxidation states.
regarded as a transition element. Explain

e
Ans. -ve value of E® for Cr^'^/Cr-'*' shows that Cr^"^ is why ?

Fre
least stable. Greater +ve value for Mn^'^/Mir'^ Ans. Scandium in the ground .state has one electron in
than that for Fe^VEe-'*' shows that Mn^"^ i s more
stable than Fe^'*'. Hence, stability of +2
for
the >?cl- subshell. Hence, it is regarded as a
transition element.
oxidation state is in the order ; Cr-"^ < Fe-"^ < However, in its common oxidation stale +3, it
r
Mn^'^ or the oxidation of their +2 state to +3 has no electron in subshell (3d^). Hence, it
You
oks

state is in the order: Cr^"^ > Fe^"*" > Mir"^ does not form coloured ions.
eBo

10. Why do transition elements show variable 15. A substance is found to have a magnetic
oxidation states ? moment of 3.9 B.M. How many unpaired
electrons does it contain ?
(HP Board 2013, CBSE 2013)
ad
our

Ans. In the transition elements, the energies of Ans. Using the formula, ^ = (/h-2)B.M., n =3.
(n-1) i/-orbitals and orbitals are very close. 16. Decide giving reason which one of the
Hence, electrons from both can participate in following pairs exhibits the property
bonding. indicated :
Re
dY

3+
11. How would you account for the fact that the (0 Sc or
Cr^'*’ exhibits paramagnetism
transition metals and their compounds are (ii) V or Mn exhibits more number of
Fin

oxidation states
found to be good catalysts in many processes ?
(CBSE 2007) (Atomic numbers : Sc = 21, Cr = 24, V = 23,
Mn = 25)
Ans. According to activated complex theory
(intermediate compound formation theory), due Ans. (/■) 2iSc = [At] 3d * 4^“.
to variable osxidation slates, transition metal Sc^"*" = [Ar] —No unpaired electron
combines with the reactant/s to form an 24 Cr = [Ar] 3d^AsK
intermediate (an unstable species) called = [Ar] 3c/—Three unpaired electron
activated complex which finally decomposes Hence, Cr^'*' exhibits paramagneti.sm.
to form the products regenerating the catalyst (//) Mn exhibits more number of oxidation
A + C 4 [AC] 4
slates.
Reactant catalyst Activated complex
Reason. 23V = 3d ^ 4r^. Its ox. states can be
(Intermediate) +2, +3, +4. +5
B + C
2.*! Mn = 3d ^ As^ Its ox. states can be +2. +3,
Product Cataly.st +4, +5. +6, +7.
8/62 ‘P'tcuCeeft.‘‘A New Course Chemistry (XlI)BBSl

17. Why are the ionization energies of 5d Hence, their chemical properties are similar.
elements greater than 3d elements ? Moreover, their valence shell configuration
Ans. In the 5d series, after lanthanium (Z = 57), there remains the same because the electrons are
is lanthanoid contraction. In each group, the added into the inner 4/-subshell. Hence, they
size of 5d element is smaller while nuclear show similar chemical properties.
charge Is greater than 3d element. Hence, 21. Assign reason for each of the following
ionization energies of 5d elements are greater statements :
than 3d elements.
(i) The largest number of oxidation states are
18. Describe giving reason which one of the exhibited by the elements in the middle of
following pairs has the property indicated ? the first row transition elements.
(/) Fe or Cu has higher melting point. (h) The atomic radii decrease in size with the
(«) Co"'^ or Ni^"*" has lower magnetic moment. increasing atomic number in the lanthanoid
Ans. (/) Fe has higher melting point than Cu. This is series.
because Fe has four unpaired electrons in 3d- Ans. (/) The maximum number of oxidation states
subshell while Cu has only one electron in the are shown by that element in a transition series
4A-subshell. Hence, metallic bonds in Fe are

F low
which has maximum number of unpaired
much stronger than those in Cu. electrons. This is so in the middle of the series.
(i7) -yjCo = [Ar] 3d'^ 4^, (ii) Refer to page 8/36 (cause of lanthanoid
Co""^ = [At] 3d ^ (3 unpaired electron) contraction).
28 Ni = [Ar] 3d^4s^, 22. Assign reasons for each of the following ;
Ni^'^= [Ar] 3rf®(2 unpaired electrons)

re
(0 Manganese exhibits the highest oxidation
Hence, Ni has lower magnetic moment than Co.
19. The 4d and 5d series of transition metals
for F
state of + 7. (Raj. Board 2012)
(/£) Unlike Cr'% Mn^+, Fe-^+ and the
have more frequent metaUmetal bonding in subsequent ions of 3d series of elements,
their compounds than do the 3 d metals. the 4d and 5d series metals generally do not^
Your

Explain. form stable cationic species. (CBSE 2011)

s
eBook

Ans. In the same group of ^/-block elements, the 4d Ans. (/) Electronic configuration of 2sMn is 4a-.
and 5d transition elements have larger size than As all the 7 electrons of 3d and 4s can
that of 3d element. Thus, the valence electrons participate in bonding, it exhibits highest
ad

are less tightly held and hence can form metal-

our

oxidation state of + 7.
metal bond more frequently. (That is why (fV) In the 4d and 5d series higher oxidation
melting points of 4d and 5d series as well as states are more stable (because difference of
enthalpies of atomisation are higher than those
Re

of 3d series)
energy between 4d and 5a or 5d and 6a is
smaller than between 3d and 4a). In the higher
Y

20. Give reasons for each of the following :

Find

oxidation states, bonds formed are mostly

(/) Transition metal fluorides are ionic in covalent. Hence, they do not form cationic
nature whereas bromides and chlorides are
species.
usually covalent in nature.
23. Why is first ionization enthalpy of Cu higher
{ii) Chemistry of all the lanthanoids is quite than that of Na? (Assam Board 2013)
similar.
Ans. Cu = [At]** 3(/"’ 4a‘, Na = [Nej'° 3a I
Ans. (0 As electronegativity of halogens decreases
in the order F > Cl > Br, the ionic character of Though electron to be removed in both cases
transition metal halides decreases in the order is from outermost .v-orbilal but nuclear charge
M—F > M—Cl > M—Br. Hence, fluorides are of Cu is much higher than that of Na. Hence,
ionic whereas chlorides and bromides are attraction on 4a electron in Cu is much greater
covalent. than on 3a elecqon in Na. Moreover, removal

{ii) The change in the size of the lanthanoids of le~ from 3a in Na gives a stable noble gas
due to lanthanoid contraction is very small as configuration whereas in case of Cu, it gives
we proceed from La (Z = 57) to Lu (Z = 71). pseudo noble gas configuration.
d- AND f-BLOCK ELEMENTS 8/63

24. Give reasons : (/) Mn shows the highest As E" is the sum of enthalpy of atomisation,
oxidation state of + 7 with oxygen hut with ionization enthalpy and hydration enthalpy,
fluorine it shows the highest oxidationstate therefore, E° for Mn^^/Mn^'*' couple is more
of+4. positive than Fe^’^/Fe^'*'.
(j7) Transition metals show variable Note : The large positive E° for Mn^^/Mn-'*’
oxidation state.
means that Mn^'*' can be easily reduced to Mn^"^,
{Hi) Actinoids .show irregularities in their is less stable. E” value for Fe^'^/Fe-'*'
electronic configuration. (CBSE 2016) is positive but small, i.e.. Fe^'*' can also be re
Ans. (/) Manganese can form pK-dK bond with duced to Fe-'*’ but less easily. Thus, Fe^'*’ is more
oxygen by utilizing 2p orbital of oxygen and stable than Mn^'*'. It also explains why +3 state
3d orbital of Mn. Hence, it can show highest of Mn is of little importance.
oxidation state of + 7. With fluorine, Mn cannot
(b) 2^Fe = [ArJ 4.?' (5 unpaired electrons)

w
form such pn-dii bond. Hence, with fluorine it
can show a maximum oxidation stale of + 4. 29 Cu = [Ar] 4.y* (1 unpaired electron)
no Art. 8.4.7. (Hi) Art. 8.10.1. Greater the number of unpaired electrons,
stronger is the metallic bond and, therefore,

lo
25. Use the following data to answer the
questions given below and also Justify giving higher is the enthalpy of atomisation,

e
(c) 21 Sc = 3t/‘ Sc^'*' = 3(P

re
reasons.

Ti = 3d^ 4.?2, Ti^-" = 3d^

rF 22

F
Cr Mn Fe Co 3+
Thus, Sc has no unpaired electron in 3d
E -091 - M8 - 0*44 - 0-28 sub.shell. No d-d transition can occur. Hence,
it is colourless. Ti^'*’ has one unpaired electron

r
fo
u
E - 0-41 + 1*57 + 0*77 + 1-97 in 3d subshcli. Hence, it is coloured.
M^+/m2+
27. Based on the data, arrange Fe^'*', Mn^'*’ and
ks
(fl) Which is a stronger reducing agent in Cr^'*' in the increasing order of stability of +
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aqueous medium, Cr^'*’ or Fe^"^ and why ? 2 oxidation state
oo

(b) Which is the most stable ion in -1-2 E

Cr^/Cr 2+ =-0-4V,
= -t-l-5V,
B

oxidation state and why ? (CBSE 2019)

E.”Fc^^/Fe^* = -t-0-8V.
e

Ans. (a) The ion which is oxidized more easily is a

stronger reducing agent. Oxidationpotential of [CBSE Sample Paper 2018, CBSE 2018 (C)]
ur

Cr^"^ (Cr-"^ > Cr^"^ + e~) is -t- 0-41 V and Ans. Higher the standard reduction potential, more
ad

easily is M^'*’ reduced to M^"*", i.e., greater is

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that of Fe-+ (Fe--^ > Fe-'+ + e~) is - 0-77 V.

Thus, Cr2+ is a stronger reducing agent, the stability of -t- 2 oxidation state. Thus,
increasing order of stability of + 2 oxidation
d

(b) e: is maximum for cobalt which

state will be Mn^^ > Fe“-^ > Cr2+ or increasing
Re
in

order is Cr“'*‘ < Fe-'*' < Mn-'*’.

means Co^'*' is most easily reduced to Co"'*'.
28. On the basis of the standard electrode
F

Hence, Co-'*’ is most stable.

26. Give reasons :
potential values stated for acid solutions,
predict whether Ti'*'*’ species may be used to
{a) E“ value for Mn^'*'/Mn^''' couple is much oxidise Fe(ll) to Fe(III)
more positive than that of Fe^^/Fe^’*’. Ti‘*+-fe- ^ Ti-^+, E“ = ^■ 0 01 V
{b) Iron has higher enthalpy of atomisation Fe^-*- -1- e~ Fe^-*-, E" = + 0*77 V
than that of copper.
(CBSE Sample Paper 2018)
(c) Sc^'*' is colourless in aqueous solution Ans. We have to predict the spontaneity of the
whereas Ti^"*" is coloured. (CBSE 2018) reaction
Ans. (a) 25Mn^+ = [Ar] 3d ^ = fAr] 3d ; Ti'*'^ + Fe--*- ■) Ti3-*- pg3-h
26pe2-^ = [Ar] 3d ^ 26^®^^ = [Ar] 3d ^ e”cell = -I- 0 01 -I- (- 0-77 V) = - 0-76 V
Thus, Mn-'*’ has more stable configuration than
Mn^'*' while Fe^'*' has more stable configuration As E
cdl
is -ve, the above reaction is non-
than Fe-^. Consequently, large third ionisation spontaneous, i.e., Ti'*'*’ cannot oxidize Fe(II) to
enthalpy is required to change Mn-'*’ to Mn^'*'. Fe(in).
8/64 New Course Chemistry (XlI)CSslS]

29. Following are the transition metal ions of 3d 31. Following ions are given :
series :
Cr^-", Cu2+, Cu^ Fe2+, Fe^^, Mn^+
Ti-**, V2+, Mn^, Cr^ Identify the ion which is
(Atomic numbers : Ti = 22, V = 23, Mn = 25,
(/) a strong reducing agent
Cr = 24)
(/{) unstable in aqueous solution
Answer the following :
(i) Which ion is most stable in aqueous solu (m) a strong oxidizing agent
tion and why ? Give suitable reason in each case.

(/() Which ion is a strong oxidizing agent and (CBSE 2020)

why ? Ans. (0 Cr-'*’ is a strong reducing agent because ii
(Hi) Which ion is colourless and why ?
can be easily oxidized to Cr^’*' (E Cr^^/Cr^'*’ IS

(CBSE 2017)
Ans. (/) Their electronic configurations are : very low = - 0-41 V or E”Cr^’^/Cr^’*’ is very high

w
Ti'^’^ = \.s~ 2s^ Ip^ 3.V- 3p^
= + 04\ V).
\s^2x-2p^3s^3p^3cP
(ii) Cu’*’ in unstable in aqueous solution because
= 1^2 2s~ 2/ 3^2 3/ 3d^

F lo
it is easily reduced to Cu as E Cu'*/Cu is high
Mn2+ = I.s-2 2.y2 2p^ 3.y2 3/ 3d^
Thus, Ti'^ is the most stable because it has noble (= -I- 0-52 V).

ee
gas configuration. (Hi) Mn^’’’ is a strong oxidizing agent because

Fr
(//) Oxidizing agent is the substance which it it can be easily reduced to Mn2’'’( E° -)+
self is reduced most easily, i.e., can gain elec
tron easily.
for
is very high = + T57 V).
ur
V2+ and Cr^’*’ iire also stable because they have 32.

filled t2g level (i.e. ) discussed in unit 9. Thus. Cr Mn Fe Co Ni Cu Zn

s
E
Mn-^’*’ can gain electron easily. Moreover Mir"*"
ook
Yo
-0-91 -M8 -0-44 -»*28 -0*25 +0-34 -0-76
IS more stable than Mn-^'*’. Hence is the
strongest oxidizing agent. From given E° values of the first row
eB

(Hi) Ions are coloured if they have incompletely transition elements, answer the following
filled fr-orbilals, Those with fully-filled or empty questions :
r

rf-orbitals are colourless. As Ti'^ has empty d-

ad
ou

(0 Why E Mn2+/Mn value is highly negative

orbitals, hence it is colourless.
30. How is the variability of oxidation states of as compared to other elements ?
Y

the transition elements different from that of

(ii) What is the reason for irregularity in the
Re

the non-transition elements ? Illustrate with above E" values ?

nd

examples.
value exceptionally
Fi

Ans. The variability in oxidation states of transition (Hi) Why is E

Cu2+/Cu
metals is due to the incomplete filling of d-or- positive ? (CBSE 2022)
bitals in such a way that their oxidation states
differ from each other by unity, e.g.. Fe2’'’ and Ans. (0E° value is highly negative (=- M 8 V)
Mn2‘'’/Mn
Cu’*’ and Cu2+ etc.
because of greater stability of half-filled
In case of non-transition elements, the oxida
J-subshell of Mn2’'’ (d^).
tion states differ by units of two, e.g.. Pb2'*' and
Pb'^’*’, Sn2+ and Sn"^ etc. Moreover, in transi (ii) E** values depend upon ionization enthalpies
tion elements, the higher oxidation states are (1E| + IE9) and sublimation enthalpies which
more stable for heavier elements in a group. For do not show a regular trend.
example, in group 6, Mo (VI) and W (VI) are (Hi) E shows positive value because
Cu2+/Cu
more stable than Cr (VI). In />-block elements,
the lower oxidation stales are more stable for E° value is the sum of enthalpies of ionization
heavier membere due to inert pair effect, e.g., and sublimation which is not balanced by
in group 16, Pb (II) is more stable than Pb (IV). hydration enthalpy.
d- AND f-BLOCK ELEMENTS 8/65

II. Some important Ans. (0 4 FeCY204 + 16 NaOH + 7 O2

compounds of transition elements 8 Na2Cr04 + 2 F&>0-^ + 8 H-5O
(A) Sodium chromaio
33. Write the formulae of different oxides of
manganese. What is the oxidation state of
2 Na2Cr04 + H2SO4 — Na2Cr20y
(B) Sodium dichromaie
mangenesc in each of them ? Arrange them
in order of their decreasing acidic character. + Na2S04 + H20
+2 +8/3 +3 +4 +7 Na2Cr20y + 2 KCI K2Cr207
Ans. MnO Mn,0, Mn^O, MnO, Mn,0 7 (C) Potassium dichromaie
Basic - -
Acidic + 2NaCl
AmpluHclic
Thus, their acidic character decreases as : (ii) Potassium dichromaie is used as a powerful

ow
Mn^Oy > Mn02 > M112O3 > M113O4 > MnO oxidizing agent in industries and in the tanning
of leather.
34. What is the effect of pH on the solution of
K2Cr20y solution ? in. Study of inner-
Ans. In aqueous solution : transitton//>block elements
Cr,02- + H,0 V ^ 2Cr02- + 2H+

e
39. The +3 oxidation states of lanthanum (Z

re
Dichromaie ion Chronialeion

rFl
(orange red) (yellow) = 57), gadolinium (Z = 64) and lutetium (Z
= 71) are especially stable. Why ?

F
In acidic medium (i.e., decreasing pW),
equilibrium shifts backward and the colour is Ans. This is because in the +3 oxidation state, they
orange red. In basic medium (i.e.. increasing have empty, half-filled and completely filled

r
/?H), equilibrium will shift forward and the
ou
4/sub-shell respectively.
solution is yellow.
fo
40. The outer electronic configurations of two
ks
35. Give the relationship between the equivalent members of the lanthanoid series are as
weight and molecular weight of KMn04 in follows: 4/^ 5rf^ 6s^ and 4/^ 5d** 6^^
oo

ia) acidic (b) alkaline and (c) neutral media. What are their atomic numbers ? Predict the
Y

Ans. (a) E = M/5 (b) In strongly alkaline medium.

eB

oxidation states exhibited by these elements

E= M. In weakly alkaline medium E = M/3 in their compounds.
(c) E = M/3. Refer to page 8/29, 30.
Ans. Complete E.C. of 1st lanthanoid = [Xe]^'^ 4/ 1
r

36. Why is it not advisable to dissolve KMn04 5d * 6s^. Atomic No. = 58 (Ce)
ou

in cone. H2SO4 ?
ad
Y

Complete E.C. of 2nd lanthanoid = fXe]^"^ 4/^

Ans. With cone. H,S04, KMn04 reacts to form 5d^6s~. Atomic No. = 63 (Eu)
M112O7 which on warming decomposes to
Oxidation states of 1st lanthanoid = + 2 (4/’-),
d

Mn02 (see reaction on page 8/29).

+ 3(4/1),+ 4 (4/0)
Re
in

37. Why in permanganate ion, there is a

covalency between Mn and oxygen ? Oxidation states of 2nd lanthanoid = + 2 (4/^)
and +3 (4/^).
F

Ans. In MnO^ ion, Mn is in highest oxidation state, 41. Why Zr and Hf or Nb and Ta exhibit similar
viz., + 7. In high oxidation state, transition
properties ?
metals form covalent bonds (according to
Fajan’s rules as oxidation state increases, ionic Or Zirconium (atomic number 40) and hafnium
character decreases). (atomic number 72) occur together in
minerals and they exhibit similar properties.
38. When chromite ore FeCr204 is fused with Give reasons.
NaOH in presence of air, a yellow coloured
compound (A) is obtained which on Ans. Due to lanthanoid contraction. Hf (Z = 72) has
acidification with dilute sulphuric acid gives size similar to that of Zr (Z= 40). Hence, their
a compound (B). Compound (B) on reaction properties are similar and therefore, occur
with KCI forms an orange coloured together. For the same reason, Nb and Ta have
crystalline compound (C) (/) Write the similar size and hence similar properties.
formulae of the compounds (A), (B) and (C) 42. One among the lanthanides, Ce (III), can be
(//) Write one use of compound (C). easily oxidized to Ce (IV) (At. No. of Ce =
(CBSl- 2016 58). Explain why ?
8/66 1^ta.cicefi.'<x New Course Chemistry (XII)CZsl9]

Or Of the lanthanides, cerium (At. No. 58) forms Ans. La(0H)3 is more basic than Lu(OH)3. As the
a tetrapositive ion, in aqueous solution. size of the lanthanoid ions decreases from La^'*’
Why ? to the covalent character of the hydro
Ans. Ce (III) having the configuration 4/ * 5d^ 6s^ xides increases {Fajan’s rules). Hence, the basic
can easily lose an electron to acquire the strength decreases from La(OH)3 to Lu(OH)3.
configuration 4 / and form Ce(IV). In fact, IV. Miscellaneous
this is the only (+ IV) lanthanide which exists
in solution. 46. Account for the following :
43. The electronic configurations of actinide (/) Oxidizing power in the series :
elements are nut known with certainty. < Crp^- < Mn04
Explain.
(iV) Actinoid contraction is greater from
Ans. In actinides, 5/ and 6d subshells are close in element to element than lanthanoid
energy. The outermost 7s orbital remains filled

w
contraction.
with 2 electrons The.eleclron can easily (Hi) Oxoanions of a metal show higher
jump from 5/ to 6d or vice versa. Further, oxidation states.

F lo
irregularities in electronic configurations are Ans. (0 This is due to increasing stability of the lower
14
also related to the stabilities of and / species to which they are reduced.
occupancy of the 5/orbitals. Hence, they show

ee
(//) This is due to poor shielding effect of 5/
a large number of oxidation states (Moreover, electrons of actinoids than 4/ electrons of

Fr
they are radioactive with short half-lives. lanthanoids.
Hence, their properties cannot be studied easily.) (m) This is due to high electronegativity and

for
44. The actinide contraction differs from multiple bond formation with metal by oxygen.
ur
lanthanide contraction ? Explain. 47. Account for the following :
Ans. Lanthanide contraction refers to the gradual (a) Europium (II) is more stable than cerium
ks
decrease in the size of the lanthanoids and their (II).
Yo
trivalent ions. (Z = 58 to 71) whereas actinoid (b) Transition metals have high enthalpies
oo

of atomisation,
contraction refers to the gradual decrease in the
eB

size of the actinoids or their ions (M^'*' or M'^'*') (c) Actinoid ions are generally coloured.
(Z = 90 to 103). They differ in the fact that in Ans. (a) Europium has stable electronic
the actinoid series, the contraction is greater configuration, Le., [Xe] 4/^ 5cfi 6s^.
r
ou

from element to element due to poor shielding

ad

(b) Due to large number of unpaired electrons

by 5/electrons than by 4/electrons. in their atoms, there is larger and stronger
Y

45. Which out of the two, La(OH)3 and metallic bonding,

Lu(OH)3, is more basic and why ? (c) Unpaired electrons are present in their ions
nd
Re

(Manipur Board 2011) which undergo/-/ transition.

Fi

i| ● Very Short Answer

mu & II * Short Answer

● Long Answer

VERY SHORT ANSWER QUESTIONS Carrying 1 mark

1. General introduction, 2. Why are transition elements known as d-

electronic configuration and block elements ?
Ans. Transition elements are called cf-block element
general characteristics and trends
because the last electron enters (/j - 1) fr-orbital,
in properties of transition elements
i.e.,c/-orbital of the penultimate shell.
1. Why are transition elements so named ? 3. The electronic configuration of a transition
Ans. Transition elements are so named because their element in +3 oxidation state is [Ar] 3d^. Find
properties are inbetween those of r-block and out its atomic number.

p-block. (CBSE Sample Paper 2017)

 8/67
d- AND f-BLOCK ELEMENTS

Ans. = [Ar]'^ )2<f, i.e., it has 25 electrons. Ans. As we move from left to right along a transition
It has been formed by loss of 3 e~ from metal M. series, the nuclear charge increases which tends
to decrease the size but the addition of electrons
Hence, no. of electrons in neutral metal atom
in the ^/-subshell increases the screening effect
= 25 + 3 = 28 e".
which counterbalances the effect of increased
.-. No. of protons = Atomic no. = 28 nuclear charge.
4. How many elements are present in the 11. How does the ionic/covalent character of the
i/-block of the periodic table ?
compounds of a transition metal vary with
Ans. 40. its oxidation state ?
5. How many transition series of elements are Ans. As the oxidation state increases, more and more
there in the periodic table ? Name them. valence shell electrons are involved in bonding.
Ans. There are four main transition series of The atomic core becomes less shielded
elements corresponding to filling of 3d, Ad, 5d

ow
resulting in the increased attraction on the
and 6d sublevels in 4th, 5th, 6th and 7th electrons. Because of this, ionic character oi
periods. They are known as first transition bonds decreases with the increa.se in oxidation
series or 3d series, second transition or Ad state.
series, third transition or 5d series and fourth
transition or 6d series.
12. What is the common oxidation state of Cu,

e
Fl
Ag and Au ?

re
6. Why does a transition series contain 10
elements ?
Ans. All these three metals show a common

F
oxidation state of +1.
Ans. There are five r/-orbitals in an energy level and
each orbital can have two electrons. As we 13. Which metal in the first transition series (3rf
ur
or
move from one elcrient to the next, an electron series) exhibits + 1 oxidation state most
is added and for c.impletc filling of the five d frequently and why?
sf (CBSE 2013)
orbitals. 10 electrons are required. Ans. Copper in the first transition .series exhibits +
k
1 oxidation state most frequently. This is
Yo
7. In the transition scries, starting from
oo

lanthanum (57La), the next element hafnium because Cu is When one electron is
(72Hf) has an atomic number of 72. Why do lost, the configuration becomes most stable due
B

we observe this jump in atomic number ? to fully filled configuration.

Ans. This is because after jyLa, filling of 4/-orbital 14. Give an example of disproportionation
re

starts which is complete at atomic number 71. reaction.

These 14 elements, therefore, belong to /-block
u

Ans. Copper (I) salts decompose in water. Half the

ad

and placed separately at the bottom of the

Yo

Cu (I) changes to copper metal and the other

periodic table. ■> Cu + Cu^/
half to copper (II), i.e., 2Cu*
8. Which out of the following is/are transition This is known as disproportionation.
d

element/s and why ?

Re

15. Write all possible oxidation states of an

in

Zn, Ag, Cd, Au.

element having atomic number 25.
Ans. Ag and Au are transition elements because they
F

have incompletely filled ii-subshell. Ans. E.C. of 2sMn = [Ar]‘* 3d^ Ai~. It shows oxi
dation states of + 2, + 3, + 4, -I- 5, + 6 and + 7.
9. Both Cu and Zn have completely filled 3d
atomic orbitals. Cu is considered as 16. Why does vanadium pentoxide act as a
transition element but Zn is not. Explain. catalyst ?
(Assam Board 2012) Ans. Vanadium exists in multiple oxidation states.
3+ ions
s
Ans. This is because Cu in one of the oxidation states 17. Out of [Sc(H20)g]^^ and [TifHjO)^]
viz. + 2, has incompletely filled 3d subshell which is coloured and why ? Give reason.
(3c/^) while Zn has completely filled 3d^^ only. (HP Board 2011, Assam Hoard 2013)
3+
3+ is
:
10. The atomic sizes of Fe, Co and Ni are nearly Ans. [Ti(H20)6] coloured while [Sc(H20)g]'
same. Explain with reason. is colourless. This is because Ti^'*’ has one
(Raj. Board 2011) electron in the J-orbital (3d^) which can absorb
Or Atomic size of 3d series elements from energy corresponding to yellow wavelength and
chromium to copper is almost the same. Give undergo d-d transition (from t2g to e^). But Sc^"^
reason. (Karnataka Board 2012) is 3cP, i.e., it has no electron in 3d subshell.

i
8/68
'P'uuicefi.'A New Course Chemistry (XII)iasiHl
18. Which of the following cations are coloured Ans. Copper, silver and gold are called coinage metals
in aqueous solution and why? because in the early times, they were used in
Sc^\ Ti‘*^ Mn2+ milking coins.
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25) 26. Write the general electronic configuration
(CBSE 2013) of transition elements or (/-block elements.
Ans. 4.y® (No unpaired electron), V^'*' = (Karnataka Hoard 2012)
3(fi 4.9^^ (2 unpaired electrons), = 3iP 4s*^ Ans. (n- 1) d
1 - 10
ns
2

(No unpaired electron), = 3d^ 4i*^ 27. Zinc, cadmium and mercury are generally
(5 unpaired electrons). Thus, only and not considered as transition metals. Give
Mn-'*' are coloured becau.se they have unpaired reasons.
electrons.
(HPB i d 2011, Assam Board 2013)
19. Copper (I) compounds are white and Ans. These elements in their most common
diamagnetic while copper (II) compounds oxidation state of + 2 have completely filled
are coloured and paramagnetic and form (/-orbitals. Hence, d-d transition cannot occur.
coloured compounds. Explain.

w
28. Name the element in the 3d series that
(Hr. Board 2012, MP Board 2013) (/) shows maximum oxidation state

F lo
Ans. In copper (I) ion, all orbitals are completely (h) is diamagnetic (Karnalak' n.
-..-cl 2012)
filled (b^ 2s- 2p^ 3s^ 3p^ 35*®) and, therefore,
Ans. (/) Manganese (ii) Zinc.
it is diamagnetic and forms white compounds.
Copper (II) ion has electronic configuration 1 ,r 29. Why Zn^ salts are white while Ni^-^ salts or

ee
Cu^^ salts are blue?
2j-2p®3.y^3p^3(/^(ha .s one unpaired electron) (HP Boar.- 2011,

Fr
and, therefore, it is paramagnetic and forms .Assam Board 2012, HP Board 2013)
Ans. Zn^'*' has completely filled J-orbitals (3</'®)
coloured compounds.
20. Which divalent metal ion has maximum
for
while Ni^"^ has incompletely filled (/-orbitals
r
paramagnetic character among the first {3d ®) or Cu"'*’ has 3cfi. Hence, d-d transition
You
transition metals ? Why ? can take place in Ni^-^ salts or Cu^"^ salts but
s
ook

not in Zn‘-^ salts.

Ans. Mn^"^ because of maximum number of unpaired
electrons, viz., 5. 30. Calculate the magnetic moment of Ni^"^.
eB

21. Which of the two : V (IV) or V (V) is para (Raj. Board 2012)
magnetic ? Give reasons. At No of V = 23. Ans. ForNi^-^ =3c^ =|U|U|U|T |T .
our

Ans. 23V = [ArJ 3d^ As~, V (IV) = [Ar] 3d \

ad

Hence, n = 2.
0
V (V) = [Ar] 3d
p = ^n(n + 2) = 72(2-5-2) = ^/8 = 2-84BM .
Thus, V (IV) has one unpaired electron while
dY

V (V) has no unpaired electron. Hence, V (IV) 11. Some important

Re

is paramagnetic. compounds of transition elements

Fin

22. What do you mean by 18 carat gold ? 31. Write the formula of an oxoanion of
Ans. 18 carat gold contains 18 parts by weight of manganese (Mn) in which it shows the
pure gold in 24 parts by weight of the alloy, oxidation state equal to its group number.
i.e. 75% gold and 25% copper. (CBSE 2017)
23. Which elements are called ferrous metals ?
Ans. MnO^ (permanganate ion). Oxidation state of
Ans. The elements of the first triad of Groups 8, 9 Mn = + 7
and 10. i.e., iron, cobalt and nickel are called
Group number of Mn = 7
ferrous metals.
or Write the formula of an oxoanion of
24. Which elements are called platinum metals ? chromium (Cr) in which it shows the
Ans. The elements of the second and third triad of oxidation state equal to its group number,
Groups 8, 9 and 10, i.e., Ru, Rh, Pd ; Os. Ir ●●’.sr 2017'
and Pt are called platinum metals.
25. What are coinage metals ? Why are they so Ans. Cr20^“ (dichromate ion). Oxidation stale of
called ? Cr = + 6. Group number of Cr = 6.

4
d- AND f-BLOCK ELEMENTS 8/69

32. What happens when chromates are kept in III. Study of inner-
acidic solution and dichroinates in the
transition//-block elements
alkaline solution ?
Ans. Chromates in the acidic solution are converted 40. What is the basic difference between the
into dichromates and thus the colour becomes electronic configurations of transition and
inner transition elements ?
orange whereas dichromates in the alkaline
solution are converted into chromates and the Ans. General electronic configuration of transition
colour becomes yellow. elements = [Noble gas] (n - 1) d and
1-14
for inner transition elements = (/i 2)f
33. Complete the following reaction :
(/I- 1) Thus, in transition elements,
Heat
last electron enters r/-orbital of penultimate
(NH4)2Cr207 shell while in inner transition elements, it enters
Heat
Ans. (NH4)2Cr207 (iO > N2 (g) + 4 H2O (0 /-orbital of ante-penultimate shell.
+ Cr203 (5) 41. Write any two uses of pyrophoric alloys.

w
34. Name one ore each of manganese and Ans. Pyrophoric alloys emit sparks when struck.
chromium. (Assam Board 2013) Hence, they are used in making flints for

F lo
lighters and in making of bullets and shells.
Ans. Mn = Pyrolusite, Mn02;
42. Give the general electronic configuration of
Cr = Chromite/ferrochrome, Fe0.Cr203. Actinides.

ee
35. What happens when KMn04is heated with Ans.

Fr
cone. H2SO4 or acidified oxalic acid or
43. Write the electronic configuration of the
treated with acidified FeS04 ?
element with atomic number 102.
Ans. Refer to pages 8/29 to 8/31,
for
Ans. [Rn]'^^5/‘'’4^^7.r.
ur
36. Write one reaction using alkaline KMn04.
44. What is the maximum oxidation state shown
s
CH2 by actinides ?
ok

Alkaline KMn04
Yo
Ans. + H2O + (O) Ans. + 7.
(Baeyer’s reagent)
o

CH2 45. Write the electronic configuration of

eB

Ethylene Gadolinium (Z = 64) and its most stable

CH2OH
oxidation state. (AIPMT Mains 2008)
r

CH2OH Ans. Gd (Z = 64) = [Xe] 4/^ 5d ' 6s^

ad
ou

Ethylene glycol Most stable oxidation state = +3.

46. Which is the most common oxidation state
Y

37. Write ionic reaction for the reaction between

of lanthanides ?
Mn04 ions and oxalate ions at 333 K.
Re
nd

Ans. +3.
Heal
Ans. 2M11O4-hI6H+-h5C20j~ 47. What is meant by lanthanoid contraction ?
Fi

(CBSE 2011, Assam Board 2013)

2Mn2++8H2O + 10CO2, Ans. The regular decrease in the atomic and ionic
radii of lanthanoids with increasing atomic
38. Complete the following : number is called lanthanoid contraction.

2 MnOJ {aq) + 5 H2O2 (ag) 4

48. Why is the separation of lanthanide elements
2+ difficult ?
2Mn + 8 H2O + ....
Ans. Due to lanthanide contraction, the change in
(Manipur Board 2011) the atomic or ionic radii of these elements is
Ans. 2 Mn04 + 5 H2O2 (oq) + 6 {aq) ■>
very small. Hence, their chemical properties are
similar. This makes their separation difficult.
2 Mn--*- + 8 H2O + 5 O2
49. Explain why lanthanoids are paramagnetic
39. Why KMnO^ is used in cleaning surgical in nature ?
instruments in hospitals ?
Ans. All lanthanoids except La^'*' and Lu^'*' contain
Ans. This is because KMn04 has a germicidal action. unpaired electrons and hence are paramagnetic.
8/70 New Course Chemistry fXinPTSTwn

SHORT ANSWER QUESTIONS Carrying 2 or 3 marks

I. General introduction, 7. Why do transition elements exhibit the

electronic contiguratiun and tendency for complex formation ? Explain with
suitable examples. (Raj. Board 2011)
general characteristics and trends
[Art. 8.4.11]
in properties of transition elements
8. What are interstitial compounds ? Why are such
1. What are transition elements ? Write electronic compounds well known for transition metals ?
configurations of 24Cr and 29CU and also (Hr. Board 2011, 2012) [Art. 8.4.12]
mention their valencies and cations.
9. Explain the following about transition metals :
(Jharkhand Board 2011) (0 Magnetic behaviour (//) Metallic character
[Art 8.1, 8.2 (Cr = +2, +3, Cu = +1, +2)] (in) Oxidation states.

low
or What are transition elements ? Why are they (Uttarakhand Board 2012)
called i/-block elements ? Write the electronic [Art. 8.4.10, 8.4.2 & 8.4.7]
configuration of the first and the last member 10. Mention the direct consequence of the
of 3d transition series. (J & K Board 2012) following factors on the chemical behaviour
[Art. 8.1 & 8.2] of the transition elements:

ee
or What are transition elements ? Write two (/) They have incompletely filled tZ-orbitals in

F
characteristics of the transition elements. the ground state or in one of oxidized states of

Fr
(CBSE2015) [Page 8/3,4] their atoms.

2. The ionization energies of the elements of first (») They contribute more valence electrons per

for
atom in the formation of metallic bonds.
ur
transition series do not vary much with
increasing atomic number. Explain why. Hint. (/) form complexes (//) strong metallic
(Art. 8.4.5] bond, hard, high densities and high enthalpies
s
of atomisation.
ok
3. The sum of the first and second ionization
Yo
enthalpies and third and fourth ionization 11. Discuss briefly the following physico-chemical
Bo

enthalpies of nickel and platinum are : properties of transition metals

IE, + IE2 IE3 + IE4 (i) Ionisation potential (H) Ionic radii
re

(MJ mol-i) (MJ mol-‘) (Hi) Complex formation. [Art. 8.4]

Ni 2-49 8-80 12. Explain the following about the transition
ou
ad

elements.
Pt 2-66 6-70
(0 Oxidation states are variable
Y

Based on the above information, answer the

(//) Exhibit good catalytic properties
following :
(Hi) Form interstitial compounds
nd
Re

(a) Which is most common oxidation state for

(HP Board 2011) [Art. 8.4]
Ni and Pt ? Why ?
Fi

(6) Out of the two, name the metal which can 13. Explain ; (/) compounds of transition metals
are often coloured. (HP Board 2013)
easily form compounds in + 4 oxidation state
and why ? [Art. 8.4.6] (//) copper (I) is diamagnetic whereas copper
4. Transition metal compounds generally act as (II) is paramagnetic.
catalysts. Give reasons. (Hi) A transition metal forms alloys with other
(Hr. Board 2012, CBSE 2013) transition metals easily. (MP Board 2012)

[Art. 8.4.8] [Art 8.4.9,8.4.10 & 8.4.13]

5. Why do the transition elements form coloured 14. Account for the following ;
compounds ? Explain. (/■) Transition metals and their compounds show
(CBSE 2008, MP Board 2011, Hr. Board 2012, catalytic activities.
Uttarakhand Board 2012, UP Board 2012) (H) Zn, Cd and Hg are non-transition elements.
[Art. 8.4.9] (i'/i) Zr and Hf are of almost identical atomic
6. Explain why transition metal ions usually show radii. (CBSE 2022)
paramagnetic behaviour. [(/) Art 8.4.8 (ii) Page 8/3 (Hi) due to lanthanoid
(Hr. Board 2012, UP Board 2012) [Art 8.4.10] contraction]
d- AND f-BLOCK ELEMENTS 8/71

II. Some important 22. (fl) Write the balanced chemical equations
involved in the preparation of KM11O4 from
compounds of transition elements (CBSE 2020)
pyrolusite ore (Mn02)-
15. How is potassium dichromate prepared from (b) Using balanced chemical equations,
chromite ore ? Gi/e its three oxidizing illustrate the oxidising properties of potassium
properties. Write the chemical equations of the permanganate in acidic and basic media.
reactions involved.
[Art. 8.7.2]
(J & K Board 2011, 2012 Hr. Board 2012, 23. What happens when (t) KMn04 is healed
Kerala Board 2012)
(HP Board 2013)
[Art. 8.7.1]
(ji) K^Cr207 is heated. [Art 8.7.1 & 8.7.2]
16. Write the structure of potassium dichromate. 24. What happens when acidified KM11O4 reacts
Give its uses also. [Art. 8.7.1] with (0 H^S (n) Oxalic acid (tit) KNO2
17. What happens when potassium dichromatc Write ionic reactions. (Hr. Board 2012)

w
reacts with (i) acidified solution of ferrous [Art. 8.7.2]
sulphate (») acidified solution of potassium 25. Describe the oxidizing property of KMn04 in
iodide (Hi) sodium chloride in the presence of

F lo
neutral or faintly alkaline medium for its
cone. H^S04 (tv) when sulphur dioxide is reactionwith iodideions and thiosulphateions.
passed through its acidified solution [Art 8.7.2]

ee
(v) K,Cr207 is heated. (J & K Board 2012) 26. What chemical changes take place when

Fr
[Art. 8.7.1] (i) MnO^ is fused with KOH in air ?
or What happens when acidified K2Cr207 reacts (it) pH of a chromate solution is progressively
with
for
lowered ? [Art. 8.7.2]
ur
(t) H2S (it) FeS04 (m) K1 (Hr. Board 2012) 27. Write ionic equations for the conversion of
[Art. 8.7.1] (a) manganate to permanganate
s
(b) permanganate to manganese (U)
ook

18. Give equation for chromyl chloride lest.

Yo

[Art. 8.7.1] (c) chromate to dichromate. [Art. 8.7]

28. Write complete chemical equations for :
eB

19. Name the gas evolved when cone HCl is added

to powdered potassium dichromate. On passing (f) Oxidation of Fc^'*' by Cr207~ in acid
medium.
the evolved gas through acidified KBr solution,
r
ad
ou

the solution turns brown. Write the balanced (I'O Oxidation of S2O5 by Mn04 in neutral
equations for the reactions involved. aqueous medium. (CBSE 2008)
(West Bengal Board 2013) [(i) Page 8/25, (u) Page 8/31]
Y

[CI2 gas (page 8/26, property (vi)), 29. Write chemical equations for the following
Re
nd

reactions :
CI2 + 2 KBr > 2 KCI + Brj]
(a) Oxidation of nitrite ion by MnO^ in acidic
Fi

20. Write an equation in ionic form to represent

medium
the oxidizing action of Cr^O^" in acidic (b) Acidification of potassium chromate
medium. Also draw the structure of Cr207 solution

ion. [Art. 8.7.1] (c) Disproportionation of Mn (VI) in acidic

solution,
21. Assign reasons for the following observations :
(0 Mn-'*' compounds are more stable than Fe-'*' [(a) 5NO2 + 2Mn04 + 6 H+
compounds towards oxidation to their +3 stale. 2Mn^+ + 3H20+ SNOj
(ii) Interstitial compounds are well known for
transition elements.
{*) 2 K2Cr04 + 2 H+ > K2Cr207 + 2 K+
+ H2O
(Hi) An aqueous solution of potassium
chromate is yellow but changes its colour on (c) 3MnOj- +4H+ ■>

decreasing the pH of the solution. 2MnO" + Mn02 + 2H20]

[Art. 8.4 & 8.7.1] (Permanganate ion)
8/72
New Course Chemistry (XI!)BZS1S]

30. Complete the following chemical equations : 37. Identify the following ;
(/) Transition metal of 3r/ series that exhibits
(0 Cn,0“-+ 6Fe2++ I4H-" ■>
the maximum number of oxidation states.
07) 2Cr02- + 2H-" (/7) An alloy consi.sting of approximately 95%
lanlhiinoid metal used to produce bullet, shell and
077) 2Mn0- + 5C,0^- + 16H-^ ■> lighter flint. [CBSE 2018 (C)]
[Ans. 0) Mn 07) niLsch metal
(CBSE 2013)
(pyrophoric alloy) (Page 8/38, 39)]
[(i) and (i7), Art. 8.7.1 (i77) Art. 8.7.2J
38. Complete the following equations :
31. Complete the following chemical equations ;
(a) 2MnO-+5SO“-+6H+ ■>

(/) 8Mn0;;+3S20“- + H2O

(h) Cr,0^^ + 6Fe--^+14H-^
((7) Cr^O^^ +3Sn“+ +14H+ ■>
[CBSE 2018 (C)l

w
(CBSE 2016)
[Ans. (a) 2Mn0“ + 5S02- + 6H+ 4

[(/) Page 8/31, (i7) Cr^O^- + 3 Sn^+ 2Mn2+ + 3H20+5S02- (Page 8/31)

F lo
+ 14 2 +3 + 7 H2O]
32. What is meant by disproportionation of (b) Cr202- + 6Fe-^ + 14H+
oxidation state. Give one example. 2 + 6 Fe^+ + 7 H2O (Page 8/25)]

ee
(Hr. Board 2011) 39. Write the preparation of the following ;

Fr
[Art. 8.7.2/Ans. to Q. 8.9, page 8/84] ((■) KMn04 from K^Mn04
33. What happens when (/) H2S reacts with (/(') Na2Cr04 from FeCr-,04
acidified K2Cr20y solution ?” (ii) Ethanol is for
ur
oxidized with acidified KMn04 solution ? (i77) Cr^O^- from CrOf- [CBSE 2018 (C)]
(Bihar Board 2011) [Art. (0 8.7.1 (H) Art. 8.7.2] [Ans. (i) Page 8/28 (ii) Page 8/24 (i77) Page 8/24]
s
ook

34. What happens when

Yo

in. Study of inner

(/) (NH4)2Cr20y is heated ? transition/y-block elements
eB

(t7) H3PO3 is heated ? (CBSE 2017)

473 K
40. What are inner transition metals ? Why are they
[(0 Point 5, page 8/33 (ii) 4 H3PO3 so called ? Write their general electronic
our

configuration. Which of the following atomic

ad

3 H3PO4 + PH3 (Refer to Unit 7)]

numbers are the atomic numbers of inner
35. Complete the following equations : transition elements ?
Y

(0 2MnO“ + 16H+ + 5S- ■>

25, 59, 74, 95, 104.
Re

heat (Uttarakhand Board 2012)

nd

(i7) KMn04 (CBSE 2017)

[Art. 8.8]
Fi

[Ans. (i) 2MnO“ + 16 H-*- + 5 ■>

41. What is lanthanoid contraction ? Give two
2 + 8 H2O + 5 S (Page 8/30) examples of lanthanoid elements. What are
(ii) 2 KMn04 ^ K2Mn04 + MnOj + O2 oxidation states exhibited by lanthanoid
(Page 8/29)] elements ? (.lharkhand Board 2012)
36. Complete and balance the following chemical [Art. 8.9]
equations : 42. What is lanthanide contraction ? What is its
(a) Fe-^ + MnO" + H-" -> cause and what are its consequences ?
(HP Board 2011, 2013, Hr. Board 2011, 2012,
(b) MnO- + H^O + r ■ ^ (CBSE 2018) J & K Board 2012, Kerala Board 2012)
[Art. 8.9.3]
[Ans. (a) 10 Fe^"^ + 2MnO“ + 16 ●>
or Write a note on lanthanoid contraction.
10 Fe-^+ + 2 Mn^^ + 8 HjO (Page 8/31) (Maharashtra Board 2012,
Uttarakhand Board 2012)
(b) 2 MnO" + 16H++ 101- ●>
43. What is Misch metal ? Give its one use.
2 Mn^+ + 8 HjO + 5 (Page 8/31)] [Art. 8.9.5]
d- AND f-BLOCK ELEMENTS 8/73

44. Write general electronic configuration of ih) Complete the following equation
/-block elements. Write any two uses of (CBSE 2015)
lanthanides and any two uses of actinides. 2MnO-+6H-"-H5NO; 4

(MP Board 2011) [Art. 8.8, 8.9.5, 8.10.5] [(a) (/} Art. 8.10.3 (ii) Art. 8.4.9
45. How would you account for the following ib) 2Mn0"+6H+-^5N02
observations ? 2+

(i) The enthalpies of atomisation of the

2Mn
-nSNOj-i-aH^O]
transition metals are high. 49. Account for the following ;
{ii) Of the lanthanoids,only cerium (Z = 58) is (a) CuCl-, is more stable than CU2CI2.
known to exhibit quite stable +4 state in (b) Atomic radii of Ad and 5d series elements
solutions. [Art. 8.4.3 & 8.9.2] are neaily same,
46. How would you account for the following ? (c) Hydrochloric acid is not used in per
{/) Many of the transition elements and their manganate titrations. (CBSE Foreign 2017)
compounds can act as good catalysts [Ans. (a) Point (ix), page 8/13

w
{ii) The metallic radii of the third {5d) series of (b) Art. 8.4.1. (due to lanthanoid
transition elements are virtually the same as contraction (c) Page 8/32]

F lo
those of the corresponding members of the or Account for the following :
second series
{a) Eu"'*' is a strong reducing agent.
or Zr (Z = 40) and Hf (Z = 72) have almost
{b) Orange colour of dichromatc ion changes

ee
identical radii. (CBSE 2013)
to yellow in alkaline medium,

Fr
{Hi) There is a greater range of oxidation states
among the actinoids than among the (c) E° (M^'^/M) values for transition metals show
inegular variation. (CBSE Foreign 2017)
lanthanoids (CBSE 2009)
[(i) Art 8.4.8 (ii) Page 8/36 for
[Ans. (a) can lose electron to form
ur
(iii) Q. 12 (b) (i), Page 8/76] more stable Eu-^* Property (10), page 8/38
{b) Property 4, page 8/24 (c) Page 8/10]
oks

47. Compare lanthanides with actinides (write any

50. (a) Write two consequencies of lanthanoid
Yo

four points). (MP Board 2013) [Art. 8.11]

o

contraction.
eB

IV. Miscellaneous
{b) Name the element of 3d series which
48. {a) How would you account for the following : exhibits the largest number of oxidation states.
Give reason. (CBSE 2022)
our

(/) Actinoid contraction is greater than

ad

lanthanoid contraction. [(a) Page 8/36 {b) Mn (+2 to +7).

{ii) Transition metals form coloured compounds. Point (m)» Page 8/12]
Y
Re
nd

LONG ANSWER QUESTIONS Carrying 5 or more marks

Fi

1. What is a transition element ? Give reasons for involved. What is the effect of increasing pH
the following : of a solution of potassium dichromate ?
(i) Transition elements have high enthalpies of (Hr. Board 2011) [Art. 8.7.1]
atomisation. {b) Represent the reaction of acidified
(//) Most of the compounds formed by potassium dichromate with (0 K1 solution and
transition metals are coloured. {ii) FeS04 solution with the help of chemical
equations, [Art. 8.7.1]
{Hi) Transition metals and most of their
compounds show a paramagnetic behaviour, (c) Draw the structure of chromate ion.
(/v) Transition metals form complex [Art. 8.7.1]
compounds. [Art. 8.4] 3. (a) Describe how potassium permangamate is
2. (a) Describe the preparation of potassium made from pyrolusite. Write the chemical
dichromate from chromite (FeCr204) ore. Write equations for the reactions involved.
the chemical equations of the reactions [Art. 8.7.2]
8/74 ^%€ideef.a'4. New Course Chemistry (XIl)E!Zs29]
{!>) Whai happens when :
(ii) The E° for copper is positive (+ 0-34
(/■) Acidified solution of KMn04 is heated with M“^/M
oxalic acid. V). It is the only metal in the first series of
(ii) K2Cr207 is heated with sodium chloride and transition elements showing this type of
behaviour
cone. H2SO4 ?
[Art. (i) Art. 8.7.2, (ii) Art. 8.7.1] (Hi) The E° value for Mn^VMn^'^ couple is
4. Give reasons : (a) Transition metals have high much more positive than for Cr^'^/Cr^'*’ or
enthalpies of atomization. Fe^'^/Fe^'*’ couple (CBSE 2009)
[(a) (0 Page 8/25, (ii) Page 8/31

ow
(b) Among the lanthanoids, Ce(III) is easily
oxidised to Ce(IV). (b) (i) Art 8.4.1. (ii) Page 8/10 (iii) Page 8/10]
(c) Fe^'^ I Fe-~ redox couple has less positive or (fl) What is meant by the term ‘lanthanoid
contraction’ ? What is it due to and what
electrode potential than Mn^"*" I Mn*^ couple.
consequences does it have on the chemistry of
(d) Copper (I) has d configuration, while

e
elements following lanthanoids in the periodic
copper (II) had configuration, still copper

re
table ?
(II) is more stable in aqueous solution than
copper (I). (h) Explain the following observations :

Flr
F
(e) The second and third transition series (/) Cu'*' ion is unstable in aqueous solution
elements have almost similar atomic radii.ou (//) Although Co^'*’ ion appears to be stable, it
(/) Among transition metals, the highest is easily oxidized to Co^"^ ion in the presence
of a strong ligand

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oxidation state is exhibited in oxoanions of a
metal.

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(iii) The E?Mn2+/Mn value for manganese is

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[(a) Art 8.4.3 (b) Page 8/35 (c) Art 8.4.6
much more than expected from the trend for
(d) Page 8/14 (e) Page 8/5 (f) Page 8/14] other elements in the serie.s
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(CBSE 2009)
5. (n) Out of Ag2S04, Cup2, Mgp2 and CuCl, [(a) Art. 8.9.3 (ft) (i) Page 8/14
which compound will be coloured and why ?
Y
(ii) Unit 9, page 9/56 (iii) Page 8/10]
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(h) Explain : (/) CrO^“ is a strong oxidizing 7. (a) Complete the following chemical
equations:
agent while MnO^" is not. (//)Zr and Hf have
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identical sizes. (0 (a^) + H2S(g) + H'^(a^) ■>

(iii) The lowest oxidation state of manganese (i7) Cu^+ (aq) + 1“ (aq) ■>
ad
do

is basic while the highest is acidic. (ft) How would you account for the following ?
(iv) Mn (II) shows maximum paramagnetic (0 The oxidizing power of oxoanions is the
characteramongst the divalent ion.s of the first
in

transition series. order : VO^ < Cr20^“ < MnO^ .

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[Alls, (a) CUF2 (b) (/) Cr in CrO^“ is in the (ii) The third ionization enthalpy of manganese
F

(Z = 25) is exceptionally high.

highest oxidation state, i.e., +6 while Mn in
(iii) Cr-'*' is stronger reducing agent than Fe^'*'.
MnO^' is in oxidation state +6 and its most (CBSE 2010)
stable oxidation state is +7 (ii) due to
lanthanoid contraction (Hi) Refer to Art 8.6 [Ans. (a) (i) Cr^O^- (a?) + 3 H2S (g) + 8H-*- (ag)
(iv) Mn^"^ has 3d ^ configuration] > 2 Cr^ (ag) + 7 HjO (/) + 3 S (s).
6. (rt) Complete the following chemical reaction (ii) 2 Cu^ (ag) + 41" (aq) -> CU2I2 (s) + I2 (g).
equations :
(ft) (0 This is due to increasing stability of
(/) Cr20^“ (aq) + l~ (aq) + H+ (aq) — + the lower species to which they are reduced.
(ii) Because 3rd electron has to be removed from
(//) MnO;] (aq) + (aq) + (aq) ■ +
stable half-filled 3 ^-orbitals (25Mn = 3 4 s^).
(b) Explain the following observations : (iu) E” (Cr^/Cr2+) = -ve (- 0-41 V),
(/) In general the atomic radii of transition E" (Fe^/Fe2+) = +ve (+ 0-77 V)
elements decrease with atomic number in a Hence, Cr^ is easily oxidized to Cr^ but
given series Fe^ cannot be easily oxidized to Fe^]
d- AND f-BLOCK ELEMENTS 8/75

or (fl) Complete the following chemical (Hi) Out of Cr-^'^ and which is stronger
equations: oxidizing agent and why ?
(iv) Name a member of the lanthanoid series
(0 MnO-(aq) + Spl-(aq) + B^O(l) which is well known to exhibit +2 oxidation
state,

(ii) Cr^O'^-(aq) + Fe2+ (aq) + (aq) ■ (v) Complete the equation :

(b) Explain the following observations :
Mn04 + 8H-^+5e" ^ (CBSE 2014)
- (i) La^+ (Z = 57) and Lu^’*' (Z = 71) do not show
[Ans. (/) Mn, page 8/12 (h) Cu, page 8/10
any colour in the solutions.
(Hi) page 8/10
(//) Among the divalent cations in the first series
(E Mn^+/Mn2+ = +l-57V)
of transition elements, manganese exhibits the

ow
maximum paramagnetism. (iv) Eu^'*' due to exactly half filled/^
(Hi) Cu"^ ion is not known in aqueous solution. configuration and due to completely
filled / configuration (page 8/35)
(CBSE 2010)
(v) Mn04 + 8 H+ + 5 c' ^ Mn^+ + 4 H2O]

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[Ans. (a) (i) 8MnO“ (aq) + 3820^“(a?) + HjO 9. (rt) Account for the following :

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8Mn02