Vedantu’s

Vedantu’s
Tatva
Vedantu’s TatvaPractice
Tatva Practice
Practice Book| Biology
Book | Physics
Book- Vol.
- Vol.
2 !

Genetics and
Evolution
Class 11 CBSE
Class 11 JEE

SCAN CODE
to explore
to know how to
use this Book
Vedantu
 Academic Progress for every Student

Regular tests &

assignments

VEDANTU
IMPROVEMENT Report card with
PROMISE detailed analysis

VIP

Parent-teacher
meetings

Our Extraordinary Results 2022

5.2X
HIGHER

1414 869
CBSE 10 scored above
3.8X
Vedantu students
90% JEE Main
Vedantu students
scored above 99%ile
HIGHER
6.6X
HIGHER

1000+ 1500+
Vedantu students Vedantu students aced
NEET cleared NEET 2022 JEE Adv. JEE Advanced 2022

student scoring above 90%

SCAN CODE
to know more
about VIP
Published by
Vedantu Innovations Pvt. Ltd.
D. No. 1081, 3rd Floor, Vistar Arcade,
14th Main Rd, Sector 3, HSR Layout
Bangalore, Karnataka, India 560 102
www.vedantu.com

All rights reserved. No part of this book may be reproduced or utilized in any form or by any
means, electronic or mechanical, including photocopying, recording, or by any information
storage and retrieval system, without permission in writing from the publishers.
Notice: Vedantu is committed to serving students with the best resources and knowledge.
Bearing that in mind, we have obtained all the information in this book from sources regarded
as reliable, and taken utmost care in editing and printing this book. However, as authors and
publishers, we are not to be held responsible for unintentional mistakes that might have crept
in. Having stated that, errors (if any) brought to our notice shall be gratefully acknowledged

Printed by
Sanjay Printers, U.P.
sanjayprinters1156@gmail.com
Tatva Practice Book
Features

1
Solve all types of
exercise questions
based on the latest
CBSE exam pattern.

Scan the QR code in the

Answers Section to view
detailed solutions for all
exercise questions.
Founder’s Message
Dear Student,

I am delighted to present to you a comprehensive Ready Reckoner with amazing content that will guide
you through your exams—‘TATVA’. Tatva means ‘Core’ and the word is fully aligned with the culture,
mission, and vision of Vedantu, which is why it gives me immense pleasure and joy to share this book
with you. Vedantu has always aimed to revolutionize the teaching and learning process and we have
speedily progressed towards bringing a superior quality of educational content to your table. Tatva is a
step forward in this direction. This book is your guide, your practice guru, and your companion in

consistent effort, diligence, and research by our experienced team of subject experts and teachers.
This book has summarised points of each chapter in the form of “Chapter at a Glance” to suit the needs
of Grades 11-12 students.
Here are a few guiding points to optimally use Tatva with a planned approach:
•
believe that a daily dose of Tatva will keep all your exam blues at bay.
• Use the Tatva booklet to mark notes so that it always comes in handy for last-minute revisions
before your exams. Notes should include key points along with questions which you couldn't solve

• Exercise 1 of the book includes subjective board pattern questions and Exercise 2 includes SCQs,
assertion reasoning questions, and case-based questions.

Before wrapping up, remember the practice mantra: “Practice what you know, and it will help to make
clear what now you do not know.” – Rembrandt

bright. All the very best!

Anand Prakash
Founder and Academic Head, Vedantu

Anand Prakash Sir has been a pioneer in producing Top Ranks in JEE/NEET
and Olympiads. He has personally taught and mentored AIR 1, 6, 7 (JEE
Advanced), AIR-1, 7, 9(AIIMS), and thousands of more students who have
successfully cleared these competitive exams in the last few years.
Credits
“Happiness lies in the joy of achievement
and the thrill of creative effort.”
—Franklin D. Roosevelt

Tatva is the brainchild of a group of creative Vedans who have strived tirelessly to weave success stories for you.

starting with our leaders who have been guiding and encouraging us at every step of the way:
Vamsi Krishna Sir, Anand Prakash Sir and Pulkit Jain Sir

instrumental in making Tatva a reality:

Sahil Bhatia, Sudhanshu Jain, Shubam Gupta, Ajay Mittal, Arshad Shahid, Jaideep Sontakke

The managers who embodied every aspect of what Tatva aimed to accomplish and brought their ideas and
diligence to the table to execute this vision immaculately:
Harish Rao, Neha Surana, Charubak Chakrabarti, Prashant Palande

Chemistry Team

into reality. Our heartfelt gratitude to our creative content developers and the typesetting team, who have put
in their hard work, insight, and eagerness to nurture and execute Tatva into ‘your ready handbook’ and bring a
positive learning experience to you.
Teachers Subject Matter Experts
Dignesh Barot Bhoop Kumar Seema Agarwal (Team Lead)
Nikhil Sharma Malika Arora
Garima Bhutani Priyanka Kumari

Typesetting Team Graphic Designers

Anita Maurya Mahaveer Singh Elavarasan V Moorthy R
Ramesh Kumar Zakir Hussain

Nilanjan Chowdhury Rabin Jacob Jayamol Joseph Mohit Kamboj

The journey of bringing Tatva to life, from an idea to the book you are holding, would not have been possible
without the extensive support of our diligent Operations Team, our amazing Academic Team, our dedicated
team of Teachers, and our talented Tech Team.
6

TABLE OF CONTENTS

Some Basic Concepts of Chemistry

Chapter at a Glance..................................................................................................................................... 11

Solved Examples......................................................................................................................................... 14

Exercise - 1 : Basic Subjective Questions....................................................................................................... 18

Exercise - 2 : Basic Objective Questions......................................................................................................... 20

Answer Key .............................................................................................................................................. 210

Structure of Atom
Chapter at a Glance..................................................................................................................................... 25

Solved Examples......................................................................................................................................... 28

Exercise - 1 : Basic Subjective Questions....................................................................................................... 33

Exercise - 2 : Basic Objective Questions......................................................................................................... 35

Answer Key .............................................................................................................................................. 211

Classification of Elements and Periodicity in Properties

Chapter at a Glance..................................................................................................................................... 40

Solved Examples......................................................................................................................................... 43

Exercise - 1 : Basic Subjective Questions....................................................................................................... 49

Exercise - 2 : Basic Objective Questions......................................................................................................... 51

Answer Key .............................................................................................................................................. 212

 7

Chemical Bonding and Molecular Structure

Chapter at a Glance..................................................................................................................................... 55

Solved Examples....................................................................................................................................... 57

Exercise - 1 : Basic Subjective Questions....................................................................................................... 61

Exercise - 2 : Basic Objective Questions......................................................................................................... 63

Answer Key ............................................................................................................................................. 213

Redox Reactions
Chapter at a Glance..................................................................................................................................... 68

Solved Examples........................................................................................................................................ 70

Exercise - 1 : Basic Subjective Questions....................................................................................................... 76

Exercise - 2 : Basic Objective Questions......................................................................................................... 78

Answer Key ............................................................................................................................................. 214

Hydrogen
Chapter at a Glance..................................................................................................................................... 83

Solved Examples........................................................................................................................................ 85

Exercise - 1 : Basic Subjective Questions....................................................................................................... 89

Exercise - 2 : Basic Objective Questions......................................................................................................... 91

Answer Key ............................................................................................................................................. 215

8

Organic Chemistry-Some Basic Principles & Techniques

Chapter at a Glance..................................................................................................................................... 96

Solved Examples........................................................................................................................................ 98

Exercise - 1 : Basic Subjective Questions....................................................................................................... 104

Exercise - 2 : Basic Objective Questions......................................................................................................... 106

Answer Key ............................................................................................................................................. 216

Hydrocarbons
Chapter at a Glance..................................................................................................................................... 111

Solved Examples........................................................................................................................................ 114

Exercise - 1 : Basic Subjective Questions....................................................................................................... 118

Exercise - 2 : Basic Objective Questions......................................................................................................... 120

Answer Key ............................................................................................................................................. 217

States of Matter
Chapter at a Glance..................................................................................................................................... 125

Solved Examples........................................................................................................................................ 127

Exercise - 1 : Basic Subjective Questions....................................................................................................... 132

Exercise - 2 : Basic Objective Questions......................................................................................................... 134

Answer Key ............................................................................................................................................. 218

 9

Thermodynamics
Chapter at a Glance..................................................................................................................................... 139

Solved Examples........................................................................................................................................ 142

Exercise - 1 : Basic Subjective Questions....................................................................................................... 149

Exercise - 2 : Basic Objective Questions......................................................................................................... 151

Answer Key ............................................................................................................................................. 219

Equilibrium
Chapter at a Glance..................................................................................................................................... 156

Solved Examples........................................................................................................................................ 159

Exercise - 1 : Basic Subjective Questions....................................................................................................... 164

Exercise - 2 : Basic Objective Questions......................................................................................................... 166

Answer Key ............................................................................................................................................. 220

The s-Block Elements

Chapter at a Glance..................................................................................................................................... 171

Solved Examples........................................................................................................................................ 173

Exercise - 1 : Basic Subjective Questions....................................................................................................... 178

Exercise - 2 : Basic Objective Questions......................................................................................................... 180

Answer Key ............................................................................................................................................. 221

10

The p-Block Elements

Chapter at a Glance..................................................................................................................................... 185

Solved Examples........................................................................................................................................ 187

Exercise - 1 : Basic Subjective Questions....................................................................................................... 192

Exercise - 2 : Basic Objective Questions......................................................................................................... 194

Answer Key ............................................................................................................................................. 222

Environmental Chemistry
Chapter at a Glance..................................................................................................................................... 199

Solved Examples........................................................................................................................................ 201

Exercise - 1 : Basic Subjective Questions....................................................................................................... 205

Exercise - 2 : Basic Objective Questions......................................................................................................... 206

Answer Key ............................................................................................................................................. 223

SOME BASIC CONCEPTS OF CHEMISTRY
12 SOME BASIC CONCEPTS OF CHEMISTRY

Chapter at a Glance
• All substances contain matter, which can exist in three states • The atomic mass of an element is expressed relative to 12C
– solid, liquid or gas. The constituent particles are held in isotope of carbon, which has an exact value of 12u. Usually,
different ways in these states of matter and they exhibit their the atomic mass used for an element is the average atomic
characteristic properties. Matter can also be classified into mass obtained by taking into account the natural abundance
elements, compounds or mixtures. An element contains of different isotopes of that element. The molecular mass of
particles of only one type, which may be atoms or molecules. a molecule is obtained by taking sum of the atomic masses of
The compounds are formed where atoms of two or more different atoms present in a molecule. The molecular formula
elements combine in a fixed ratio to each other. Mixtures occur can be calculated by determining the mass per cent of different
widely and many of the substances present around us are elements present in a compound and its molecular mass.
mixtures.

• The number of atoms, molecules or any other particles present

• When the properties of a substance are studied, measurement in a given system are expressed in the terms of Avogadro
is inherent. The quantification of properties requires a system constant (6.022 × 1023). This is known as 1 mol of the respective
of measurement and units in which the quantities are to be particles or entities.
expressed. Many systems of measurement exist, of which the
English and the Metric Systems are widely used. The scientific
community, however, has agreed to have a uniform and • Chemical reactions represent the chemical changes undergone
common system throughout the world, which is abbreviated by different elements and compounds. The coefficients
as SI units (International System of Units). indicate the molar ratios and the respective number of particles
taking part in a particular reaction. The quantitative study of
the reactants required or the products formed is called
• The uncertainty is taken care of by specifying the number of stoichiometry. Using stoichiometric calculations, the amount
significant figures, in which the observations are reported. of one or more reactant(s) required to produce a particular
amount of product can be determined and vice-versa.

• The combination of different atoms is governed by basic laws

of chemical combination — these being the Law of • The amount of substance present in a given volume of a
Conservation of Mass, Law of Definite Proportions, Law of solution is expressed in number of ways, e.g., mass per cent,
Multiple Proportions, Law of Reciprocal Proportion, Gay mole fraction, molarity and molality
Lussac’s Law of Gaseous Volumes and Avogadro Law. All
these laws led to the Dalton’s atomic theory, which states
that atoms are building blocks of matter.
SOME BASIC CONCEPTS OF CHEMISTRY 13

Scheme for Calculation of Mole, Mass and Number of Particles

Various Relationships of Mole

14 SOME BASIC CONCEPTS OF CHEMISTRY

Solved Examples
Example-1 Example-7
t What do you mean by significant figure? Carbon and oxygen are known to form two compounds.
Sol. The digits in a properly recorded measurement are known The carbon content in one of these is 42.9% while in the
as significant figures. It is also defined as the total numbers other it is 27.3%. Show that this data is in agreement with
of figures in a number including the last digit whose value the law of multiple proportions.
is uncertain is called number of significant figures. Sol. Oxide 1 Carbon Oxygen
Example-2 42.9% 57.1%
How many significant figures are present in the following?  Amount of oxygen that combines with 1 g carbon
(i) 0.0025 (ii) 208 (iii) 5005 57.1
 = 1.33 g
(iv) 126,000 (v) 500.0 (vi) 2.0034 (NCERT) 42.9
Sol. (i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5. Oxide 2 Carbon Oxygen
Example-3 27.3% 72.7%
8 –1
If the speed of light is 3.0 × 10 m s , calculate the distance  Amount of oxygen that combines with 1 g carbon
covered by light in 2.00 ns. (NCERT)
72.7
Sol. Distance covered = Speed × Time = 3.0 × 108m s–1 × 2.00 ns  = 2.66 g
27.3
10 9 s Ratio of oxygen in oxide (1) and (2) = 1 : 2
= 3.0 × 108m s–1 × 2.00 ns × = 6.00 × 10–1m
1 ns Thus, Law of multiple proportion is verified.
= 0.600 m Example-8
Example-4 Classify the following substances into elements,
compounds and mixtures.
What is the S.I. unit of mass ?
(i) Air (ii) Diamond (iii) LPG (iv) Dry ice
Sol. S.I. unit of mass is kilogram (kg).
(v) Graphite (vi) Steel (vii) Marble (viii) Smoke
Example-5 (ix) Glucose (x) Laughing gas
Why Law of conservation of mass should better be called Sol. Elements : Diamond; Graphite
as Law of conservation of mass and energy ?
Compounds : Marble; Glucose; Laughing gas; Dry ice
Sol. In nuclear reactions, it is observed that the mass of the Mixtures : Air; LPG; Steel; Smoke
products is less than the mass of the reactants. The
Example-9
difference of mass, called the mass defect, is converted
into energy according to Einstein equation, E =  m c2. Calculate the molecular mass of the following:
Hence, we better call it as a law of conservation of mass (i) H2O (ii) CO2 (iii) CH4
and energy. Sol. (i) Molecular mass of H2O = 2(1.008 amu) + 16.00 amu
Example-6 = 18.016 amu
Is the law of constant composition true for all types of (ii) Molecular mass of CO2 = 12.01 amu + 2 ×16.00 amu
compounds ? Explain why or why not.
= 44.01 amu
Sol. No, law of constant composition is not true for all types of
compounds. It is true only for the compounds obtained (iii) Molecule mass of CH4 = 12.01 amu + 4 (1.008 amu)
from one isotope. For example, carbon exists in two common = 16.042 amu
isotopes, 12C and 14C.
SOME BASIC CONCEPTS OF CHEMISTRY 15
Example-10
3.01 10 22 x
Carbon occurs in nature as a mixture of carbon-12 and So, 
6.02  10 23
17
carbon-13. The average atomic mass of carbon is 12.011.
What is the percentage abundance of carbon-12 in nature ? 17  3.01  10 22
or x = 0.85 g
Sol. Let x be the percentage abundance of carbon-12; then 6.02  10 23
(100 – x) will be the percentage abundance of carbon-13.
Example-15
12x  13 (100  x) How many molecules and atoms of oxygen are present in
Therefore,  12.011
100 5.6 litres of oxygen (O2) at NTP ?
or 12x + 1300 – 13x = 1201.1 Sol. We know that 22.4 litres of oxygen at NTP contain
x = 98.9 6.02 × 1023 molecules of oxygen.
Abundance of carbon-12 is 98.9% So, 5.6 litres of oxygen at NTP contain
Example-11 5.6
 ×6.02 × 1023 molecules
What will be the mass of one C atom in g?
12
(NCERT) 22.4
Sol. 1 mol of 12C atoms = 6.022 × 1023 atoms = 12 g = 1.505 × 1023 molecules
Thus, 6.022 × 1023 atoms of 12C have mass = 12 g 1 molecule of oxygen contains
= 2 atoms of oxygen
12
1 Atom of 12C will have mass = g 23
So, 1.505 × 10 molecules of oxygen contain
6.022  10 23
= 1.9927 × 10–23 g = 2 × 1.505 × 1023 atoms
= 3.01 × 1023 atoms
Example-12
Example-16
Calculate the number of atoms in each of the following :
(i) 52 moles of He (ii) 52 u of He (NCERT) How many electrons are present in 1.6 g of methane ?
Sol. (i) 1 mol of He = 6.022 × 1023 atoms Sol. Gram-molecular mass of methane
 mole of He = 52 × 6.022 × 1023 atom (CH4) = 12 + 4 = 16 g
= 3.131 × 1025 atoms
(ii) 1 atom of He = 4 u of He 1.6
Number of moles in 1.6 g of methane =  0.1
4 u of He = 1 atom of He 16
Number of molecules of methane in 0.1 mole
1
 52 u of He = × 52 atoms = 13 atoms = 0.1 × 6.02 × 1023
4
= 6.02 × 1022
Example-13
One molecule of methane has = 6 + 4 = 10 electrons
Calculate the mass of 2.5 gram atoms of oxygen.
Sol. We know that So, 6.02 × 1022 molecules of methane have
= 10 × 6.02 × 1022 electrons
Mass of an element in grams
Number of gram atoms = Atomic mass of the element in grams = 6.02 × 1023 electrons
Example-17
So, Mass of oxygen = 2.5 × 16 = 40.0 g
Calculate the mass per cent of different elements present in
Example-14
sodium sulphate (Na2SO4). (NCERT)
What is the mass of 3.01 × 1022 molecules of ammonia ?
Sol. Mass % of an element
Sol. Gram-molecular mass of ammonia = 17 g
Number of molecules in 17 g (one mole) of NH3 = 6.02 × 1023 Mass of that element in the compound
  100
Molar mass of the compound
Let the mass of 3.01 × 1022 molecules of NH3 be = x g
16 SOME BASIC CONCEPTS OF CHEMISTRY

Now, molar mass of Na2SO4 = 2 (23.0) + 32.0 + 4 × 16.0 Example-20

Determine the empirical formula of an oxide of iron which
= 142 g mol–1
has 69.9 % iron and 30.1 % dioxygen by mass. (NCERT)
46 Sol.
Mass percent of sodium =  100 = 32.39 %
142 El eme nt Percen- Ato mi c Moles Mole Si mp l e s t
tage mass of ratio who le
32 element no.ratio
Mass per cent of sulphur =  100 = 22.54 %
142 69.9 1.25
Iron 69.9 55.85  1.25 1 2
55.85 1.25
64
Mass per cent of oxygen =  100 = 45.07% 30.1
 1.88
1.88
1.5
142 Oxygen 30.1 16.00 3
16.00 1.25
Example-18  Empirical formula = Fe2O3.
Calculate the empirical formula of a compound that contains Example-21
26.6% potassium, 35.4% chromium and 38.1% oxygen Hydrogen chloride (HCl) on oxidation gives water and
[Given Atomic weight of K = 39.1; Cr = 52; O = 16] chlorine. How many litres of chlorine at STP can be obtained
Sol. starting with 36.50 g HCl ?
El eme nt P ercen- Ato mic Moles Mole Simplest Sol. Oxidation of HCl takes place according to the following
tage mass of ratio whole equation :
element no.ratio

26.6 0.68 4HCl  O2 

 2Cl2  2H 2 O
Potassium 26.6 39.1  0.68 1 1×2=2 4 mol 2 mol
39.1 0.68

35.4 0.68 Mass 36.5

Chromium 35.4 52.0  0.68 1 1×2=2
52 0.68 Moles of HCl = Molecular mass  36.5  1 mole
38.1 2.38
Oxygen 38.1 16.0  2.38  3.5 3.5 × 2 = 7
16 0.68  4 moles HCl give 2 moles Cl2
Therefore, empirical formula is K2Cr2O7. 2
Example-19  1 mole will give moles Cl2 = 0.5 moles Cl2
4
An organic compound containing C,H and N gave the fol- Volume of Cl2 at STP = 22.4 × 0.5 = 11.2 litre
lowing analysis: C: 40% H:13.3%, N:46.67%. If its molecular
Example-22
formula weight is three times its empirical formula weight
How much copper can be obtained from 100 g of copper
then find out its empirical and molecular formula of the com-
sulphate (CuSO4)? (Atomic mass of Cu = 63.5 amu)
pound.
(NCERT)
Sol. Relative no. of atoms of C = 40/12 = 3.33
Sol. 1 mole of CuSO4 contains 1 mole (1 g atom) of Cu
Relative no. of atoms of H = 13.3/1 = 13.3 and that for N = Molar mass of CuSO4 = 63.5 + 32 + 4 × 16 = 159.5 g mol–1
46.67/14 = 3.33 Thus, Cu that can be obtained from 159.5 g of CuSO4
Thus, simplest atomic ratio C:H:N = 63.5 g
 Cu that can be obtained from 100 g of CuSO4
= 3.33:13.33:3.33 = 1:4:1
63.5
Therefore the empirical formula of the compound is CH4N =  100 g = 39.81 g.
159.5
M olecular F o rm u la M ass
Also, given: E m pirical F o rm u la M ass = 3 = n-factor

Therefore, molecular formula is (CH4N)3 i.e. C3H12N3

SOME BASIC CONCEPTS OF CHEMISTRY 17
Example-23
100 g
In the reaction, A + B2 AB2, identify the limiting reagent, Volume of 100 g nitric acid solution = 1.41g mL1
if any, in the following mixtures
= 70.92 mL = 0.07092 L
(i) 300 atoms of A + 200 molecules of B
1.095 mole
(ii) 2 mol A + 3 mol B Conc. of HNO3 in moles per litre = 0.07092 L =15.44 M
Sol. (i) According to the given reaction, 1 atom of A reacts
Example-25
with 1 molecule of B.
 200 molecules of B will react with 200 atoms of A and If the density of methanol is 0.793 kg L–1, What is its volume
100 atoms of A will be left unreacted. Hence, B is the needed for making 2.5 L of its 0.25 M solution? (NCERT)
limiting reagent while A is the excess reagent. Sol. Let us calculate moles of methanol present in 2.5L of 0.25
(ii) According to the given reaction, 1 mol of A reacts solution.
with 1 mol of B2.
Moles of CH 3OH
 2 mole of A will react with 2 mol of B. Hence, A is the Molarity 
Volume in litre
limiting reactant.
Example-24 Moles of CH 3OH
0.25 =
Calculate the concentration of nitric acid in moles per litre 2.5
in a sample which has a density 1.41 g mL–1 and the mass Moles of CH3OH = 0.25 × 2.5 = 0.625 moles
percent of nitric acid in it being 69%. (NCERT) Mass of CH3OH = 0.625 × 32 = 20 g
Sol. (Molecular mass of CH3OH = 32)
Mass percent of 69% means that 100 g of nitric acid solution Now 0.793 × 103g of CH3OH is present in 1000 mL
contain 69g of nitric acid by mass.
Molar mass of nitric acid HNO3 = 1 + 14 + 48 = 63 gmol–1 1000
20 g of CH3OH will be present in = × 20
69 g 0.793×10 3
oles in 69g HNO3 = 63 g mol  1  1.095 mole
= 25.2 mL
18 SOME BASIC CONCEPTS OF CHEMISTRY

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)
15. 23.72 g of a substance ‘X’ occupies 5.56 cm3. What
1. What is the SI unit of time? will be its density measured in correct significant
figures?
2. Define atomic mass.
16. What is difference between mass and weight?
3. How many atoms are present in 3 moles of carbon?
17. Volume of a gas at STP is 6.57 × 10–8 cm3. Find out
4. Define Density. the number of molecules in it ?

5. What is the relation between °C (degree celsius) and °F 18. If 8.7g of Na2CO3 is added to 18g of CH3COOH
(degree Fahrenheit) ? solution, the residue is found to weigh 22g. What is the
mass of CO2 released in the reaction?
6. What is a homogeneous mixture?
19. The mass of nitrogen per gram hydrogen in the
7. What is the molecular mass of ethane? compound hydrazine is exactly one and half times of
the mass of nitrogen in the compound ammonia. Which
8. Define the law of conservation of mass. law is illustrated by the given fact?

9. What is the SI unit of weight? 20. Zinc sulphate solution contains 22.65% zinc and 43.9%
of water of crystallization. If the law of constant
10. Give two examples of molecules having molecular proportions is true then what is the weight of zinc
formula same as empirical formula. required to produce 20 g of the zinc sulphate crystals?

Section – B (2 Marks Questions) Section – C (3 Marks Questions)

11. Calculate the molar mass of the following: 21. Carbon has the following three isotope with relative
(i) H2O (ii) NH3 abundances and masses (amu) shown in the table.
Isotopes Relative Atomic
12. One atom of an element weighs 4.2 × 10–22 g. What is
Abundance (%) mass(amu)
its atomic mass?
12
C 98.892 12
13. How many significant figures are present in the 13
C 1.108 13.00335
following? 14
C 2 × 10–10 14.00317
(i) 0.0657
(ii) 456 From the above data, Find out the average atomic mass
(iii) 878.0 of carbon.

14. Which set of figures will be obtained after rounding off 22. 45.4 L of dinitrogen reacted with 22.7 L of dioxygen
the following to three significant figures? and 45.4 L of nitrous oxide was formed. The reaction is
(i) 34.216 given below
(ii) 0.06597 2N2 (g) + O2 (g) → 2N2O (g)
(iii) 15.8107 Which law is being obeyed in this experiment?
SOME BASIC CONCEPTS OF CHEMISTRY 19

23. Calculate the molarity of pure water using the density Section – D (5 Marks Questions)
to be 1000 kg m-3.
29. In a compound C, H and N are present in the ratio of
9:1:3.5 by weight. If molecular weight of the compound
24. In five moles of ethane (C2H6), calculate the following: is 108, then what is the molecular formula of the
(i) Number of moles of carbon compound?
(ii) Number of moles of hydrogen atoms
(iii) Number of molecules of ethane 30. (a) Calculate the amount of carbon dioxide that could
be produced when
25. Calculate the weight of CO having the same number of (i) 1 mole of carbon is burnt in air.
oxygen atoms as present in 22g of CO2? (ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
26. How many number of aluminium ions are present in (b) What is the SI unit of mass? How is it defined?
0.051 g of aluminium oxide?

27. What is the percentage of carbon in ethanol?

28. An organic compound on analysis was found to contain

10.06 % carbon, 0.84% hydrogen and 89.10% chlorine.
What will be the empirical formula of the substance?
20 SOME BASIC CONCEPTS OF CHEMISTRY

Exercise – 2: Basic Objective Questions

Section – A (Single Choice Questions)

1. Chemistry is called the science of 5. Calculate the volume of the given cube in mL.
(a) Atoms (b) Molecules
(c) Both (a) and (b) (d) None of the above

2. Observe the pictures carefully and label A, B and C

appropriately

(a) A  atom, B  compound C  molecule

(b) A  compound, B  molecule C  atom (a) 100 mL (b) 1000 mL
(c) A  molecule, B  atom C  compound (c) 10mL (d) None of these
(d) A  compound, B  atom C  molecule
6. The S.I. unit of density is
3. Some examples of …..A…..arecolour, melting point, (a) gm−3 (b) kg/cm3
−3
odour, boiling point, density etc. The examples of (c) gcm (d) Kg/m3
…..B….. are characteristic reactions of different
substance. These include acidity or basicity, 7. What should be the volume of the milk (in m3) which
combustibility etc. measures 5 L?
Here, A and B refer to (a) 5 × 10–3 m3 (b) 5 × 103 m3
6 3
(a) A  physical properties, B  biochemical (c) 5 × 10 m (d) 5 × 10000 m3
properties
(b) A  physical properties, B  chemical properties 8. The metric unit for volume is …..A…..and the SI unit
(c) A  chemical properties, B  physical properties for volume is …..B….. Here, A and B refer to
(d) A  biophysical properties, B  biochemical (a) L, m3 (b) m3, L
3 3
properties (c) cm , m (d) mL, m3

4. Any quantitative observation or measurement is 9. What is the value of temperature 75oF on the Kelvin
represented by a …..A…..followed by …..B….. in scale?
which it is measured. Here, A and B refers to (a) 24 K (b) 215 K
(a) A  alphabet, B  units (c) 297 K (d) 348 K
(b) A  units, B  number
10. 0.0016 can be written as …..A…..in scientific notation.
(c) A  Roman numeral, B  number
Here A refers to
(d) A  number, B  units
(a) 1.6 × 10–3 (b) 1.6 × 10–2
–1
(c) 1.6 × 10 (d) 1.6 × 10–4
SOME BASIC CONCEPTS OF CHEMISTRY 21

11. Calculate (9.8 × 10–2) × (2.5 × 10–6). The result should 20. Arrange the following in the order of increasing mass
be (Atomic mass; O = 16, Cu = 63, N = 14).
(a) 24.50 109 (b) 2.450 107 I. One atom of oxygen
(c) 24.50 107 (d) 2.450 108 II. One atom of nitrogen
III. 1 × 10–10 Mole of oxygen atom
IV. 1 × 10–9 Mole of copper
12. …..A….. is the sum of atomic masses of the elements
(a) I < II < III < IV (b) II < I < III < IV
present in a molecule. Here, A refers to
(c) II < IV < I < III (d) IV < II < III < I
(a) Molar mass (b) Formula mass
(c) Atomic mass (d) Molecular mass
Section - B (Assertion & Reason Type Questions)
13. The mass of one mole of a substance in grams is called
(a) Avogadro mass (b) Formula mass 21. Assertion (A):Science is sub-divided into various
(c) Atomic mass (d) Molar mass disciplines like chemistry, physics, biology, geology
etc.
14. Which of the following represents the largest unit? Reason (R):For the sake of convenience.
(a) Kilolitre (b) Mililitre (a) Both A and R are correct; R is the correct
(c) Decilitre (d) Megalitre explanation of A
(b) Both A and R are correct; R is not the correct
A  100 explanation of A
15. Mass % of an element 
molar mass of compound (c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Here, A refers to
(a) % of that element in the compound
22. Assertion (A): Chemistry deals with the composition,
(b) Moles of that element in the compound
structure and properties of matter. These aspects can be
(c) Mass of that element in the compound
best described and understood in terms of basic
(d) Moles of compound
constituents of matter: atoms and molecules.
Reason (R):That is why chemistry is called the science
16. A solution is prepared by adding 4g of a substance A to
of atoms and molecules.
18 g of water. Calculate the mass percent of the solute.
(a) Both A and R are correct; R is the correct
(a) 12.5 % (b) 15.18 %
(c) 18.18% (d) 20% explanation of A
(b) Both A and R are correct; R is not the correct
17. What will be the molality of the solution made by explanation of A
dissolving 10 g of NaOH in 100 g of water? (c) A is correct; R is incorrect
(a) 10 m (b) 5 m (d) R is correct; A is incorrect
(c) 2.5 m (d) 1.25 m
23. Assertion (A):Everything around us, for example pen,
18. A solution is made by dissolving 49 g of H2SO4 in 250 pencil, fan, water, air, all living beings etc., are
mL of water. The molarity of the solution is composed of matter.
(a) 4M (b) 1M Reason (R):All of them, they possess mass and occupy
(c) 2M (d) 5M space.
(a) Both A and R are correct; R is the correct
19. The concentration of a solution or the amount of explanation of A
substance present in its given volume can be expressed
(b) Both A and R are correct; R is not the correct
in which of the following ways?
(a) Mass per cent or weight per cent (w/w%) explanation of A
(b) Molality (c) A is correct; R is incorrect
(c) Mole fraction or molarity (d) R is correct; A is incorrect
(d) All of the above
22 SOME BASIC CONCEPTS OF CHEMISTRY

24. Assertion (A): Zero at the end or right of a number are (a) Both A and R are correct; R is the correct
significant provided they are not on the right side of the explanation of A
decimal point. (b) Both A and R are correct; R is not the correct
Reason (R): Significant figures for 0.200 is 3 whereas explanation of A
for 200 it is 1. (c) A is correct; R is incorrect
(a) Both A and R are correct; R is the correct (d) R is correct; A is incorrect
explanation of A
(b) Both A and R are correct; R is not the correct 28. Assertion (A):The molality of the solution is
explanation of A independent of temperature.
(c) A is correct; R is incorrect Reason (R): The molality of the solution is expressed
(d) R is correct; A is incorrect in units of moles per unit volume of solvent.
(a) Both A and R are correct; R is the correct
25. Assertion (A):Dalton’s theory could explain the laws explanation of A
of chemical combination. (b) Both A and R are correct; R is not the correct
Reason (R): The origin of the idea that matter is explanation of A
composed of small indivisible particles called ‘a-tomio’ (c) A is correct; R is incorrect
(meaning indivisible); chemical reactions involve (d) R is correct; A is incorrect
recognition of atoms. These are neither created nor
destroyed in a chemical reaction. Section – C (Case Study Questions)
(a) Both A and R are correct; R is the correct
explanation of A Case Study – 1
(b) Both A and R are correct; R is not the correct
explanation of A Matter is defined as anything that occupies space and
(c) A is correct; R is incorrect possesses mass. Matter can exist in 3 physical states
(d) R is correct; A is incorrect viz. solid, liquid, gas. Solid - a substance is said to be
solid if it possesses a definite volume and a definite
26. Assertion (A): Molarity of a solution does not depends shape, e.g., sugar, iron, gold, wood etc. Liquid- A
upon temperature. substance is said to be liquid, if it possesses a definite
Reason (R): Volume of a solution is temperature volume but no definite shape. They take up the shape of
dependent entity. the vessel in which they are put, e.g., water, milk, oil,
(a) Both A and R are correct; R is the correct mercury, alcohol etc. Gas- a substance is said to be
explanation of A gaseous if it neither possesses definite volume nor a
(b) Both A and R are correct; R is not the correct definite shape. This is because they fill up the whole
explanation of A vessel in which they are put, e.g., hydrogen, oxygen etc.
(c) A is correct; R is incorrect The three states are inter convertible by changing the
(d) R is correct; A is incorrect conditions of temperature and pressure as follows:

27. Assertion (A): The temperatures in degree Celsius and

degree Fahrenheit are related to each other by the
following relationship.
9
F   C   32
5
Reason (R):The temperature on degree Celsius and
Kelvin scale are related to each other by the following
relationship.
C  K  273.15
SOME BASIC CONCEPTS OF CHEMISTRY 23

29. A substance having definite shape is called 34. One of the statements of Dalton’s atomic theory is
(a) Solid (b) liquid given below
(c) gas (d) gel “Compounds are formed when atoms of different
elements combine in a fixed ratio”
30. Which of the following substance is liquid? Which of the following laws is not related to this
(a) Sugar (b) Salt statement?
(c) Water (d) oxygen (a) Law of conservation of mass
(b) Law of definite proportions
31. Liquid can convert into a gas by
(c) Law of multiple proportions
(a) Freezing (b) Sublimation
(d) None of these
(c) Melting (d) Vaporization

32. Solid can convert into a liquid by 35. Dalton's atomic theory fails to explain
(a) Freezing (b) Sublimation (a) Binding forces between atoms in compounds.
(c) Melting (d) Vaporization (b) Existence of allotropes
(c) Structure of atom
Case Study – 2 (d) All of above

In 1808, Dalton published ‘A New System of Chemical 36. Compounds are formed due to
Philosophy’ in which he proposed the following: (a) combination of atoms of different elements in a
(i) Matter consists of indivisible atoms. fixed ratio
(ii) All the atoms of a given element have identical (b) combination of molecules
properties including identical mass. Atoms of (c) combination of same atoms
different elements differ in mass. (d) combination of atoms and molecules
(iii) Compounds are formed when atoms of different
Case Study – 3
elements combine in a fixed ratio.
(iv) Chemical reactions involve the reorganization of A mole is defined as the amount of substance that
atoms. These are neither created nor destroyed in a contains as many particles or entities (atoms,
chemical reaction. molecules, or ions) as there are in exactly 12g of carbon
- 12 isotope. One mole contains exactly 6.02214076 ×
Drawbacks of Dalton’s Theory:
1023 elementary entities. This number is the fixed
(i) Does not explain the structure of the atom. numerical value of the Avogadro constant, NA, when
(ii) Fails to explain binding forces between atoms in expressed in the unit mol–1 and is called the Avogadro
compounds. number.
(iii) Does not explain Gay Lussac's law. One mole of an element = 6.022 × 1023 atoms
Mass of element
Moles of an element =
Atomic mass
33. In …..A….. Dalton published …..B….. in which he Atomic mass
proposed the theory. Here, A and B refer to Mass of one atom =
6.022  1023 atoms
(a) A  1708, B  An old system of chemical
37. A mole contains as many particles as in
philosophy (a) 12g of carbon - 12 isotope
(b) A  1908, B  A new system of physical (b) 12g of carbon - 14 isotope
philosophy (c) 10g of carbon - 12 isotope
(c) A  1808, B  A new system of chemical (d) 24g of carbon - 12 isotope
philosophy
(d) A  1808, B  An old system of physical 38. Moles of an element defined as
philosophy (a) Mass of element / Atomic mass
(b) Atomic mass / Mass of element
(c) Mass of element × Atomic mass
(d) Atomic mass / Volume
24 SOME BASIC CONCEPTS OF CHEMISTRY

39. One mole of an element contains 40. Volume occupied by 1 mole of a gas at N.T.P is
(a) 6.022 × 10-23 atoms (b) 6.022 × 1021 atoms (a) 22.4 L (b) 22.4 mL
23
(c) 6.022 × 10 atoms (d) 6.022 × 1027 atoms (c) 22400 L (d) 2240 mL

Find Answer Key and Detailed Solutions at the end of this book

SOME BASIC CONCEPTS OF CHEMISTRY

STRUCTURE OF ATOM
26 STRUCTURE OF ATOM

Chapter at a Glance
 Constituents of atom: Atom is no longer considered as Direction of propagation of waves. All of them travel with
indivisible. It is made up of electrons, protons and neutrons the velocity of light.
called fundamental particles.  Relationship between velocity, frequency & wavelength:
 Electron: A fundamental particle which carries one unit c = 
negative charge and has a mass nearly equal to 1/1837th of where c : speed of light i.e. 3 × 108 m/s in vacuum
that of hydrogen atom.  : frequency;  : wavelength
 Proton: A fundamental particle which carries one unit
 Electromagnetic spectrum: When all the electromagnetic
positive charge and has a mass nearly equal to that of
hydrogen atom. radiations are arranged in increasing order of wavelength
 Neutron: A fundamental particle which carries no charge or decreasing frequency the band of radiations obtained is
but has a mass nearly equal to that of hydrogen atom. termed as electromagnetic spectrum.
 Thomson’s model of atom: An atom is a sphere of positive  Black body radiation: If the substance being heated is a
electricity in which sufficient number of electrons were black body (which is a perfect absorber and perfect radiator
embedded to neutralize the positive charge just as seeds in of energy) the radiation emitted is called black body radiation.
a melon or raisins in pudding. It could not explain results of  Photoelectric effect: When radiation of certain minimum
Rutherford’s  -rays scattering experiments. frequency () strike the surface of a metal, electrons are
 Rutherford’s model of atom: A thin foil of gold was ejected. This minimum energy (h0) is called wave function
bombarded with  -particles. Most of the  -particles (W0).
passed through the foil undeflected, a few were deflected
through small angle while very few were deflected back. It h  h 0  (1 / 2) m e v 2
was therefore, concluded that there was sufficient empty
space within the atom and small heavy positively charged  Planck’s quantum theory: This theory was put forward to
body at the center called nucleus. Thus, atom consists of a explain the limitations of electromagnetic wave theory. It
heavy positively charged nucleus in the centre containing suggests that radiant energy is emitted or absorbed
all protons and the electrons were revolving around the discontinuously in the form of small packets of energy called
nucleus so that the centrifugal force balances the force of quanta (called photons in case of light). Energy of each
attraction. quantum (E) = hv where ‘h’ is Planck’s constant
 Atomic number and mass number: The general notation (= 6.626 × 10-34 Js). Total energy emitted or absorbed = nhv
that is used to represent the mass number and atomic number where n is an interger.
of a given atoms is A
ZX
 Emission and Absorption Spectra: When light emitted from
Where, X – symbol of element any source is directly passed on to prism and resolved, the
spectrum obtained is called emission spectrum. In case of
A – Mass number white light, e.g., from sun, it is resolved into seven colours
Z – atomic number (VIBGYOR). The spectrum obtained is called contiuous
 Isotopes: Isotopes are the atoms of the same element having spectrum. If light emitted from a discharge tube is resolved,
some coloured lines are obtained. The spectrum obtained
identical atomic number but different mass number. The
is called line spectrum. It white light is first passed through
difference is due to the difference in number of neutrons.
the solution of a compound or vapour of a substance and
 Isobars: Atoms of different elements having different atomic then resolved, the spectrum obtained is called absorption
numbers but same mass numbers are called isobars. spectrum. It has dark lines in the continuous spectrum.
 Isotones: Atoms of different elements which contain the  Absorption spectrum of hydrogen: When H2 gas is taken in
same number of neutrons are called isotones. the dischange tube, series of lines obtained and the regions
 Isoelectronic species: The species (atoms or ions) in which they lie are as under:
containing the same number of electrons are called
isoelectronic. Series: Lyman Balmer
 Paschen Brackett Pfund
     
 Electromagnetic radiation: Energy is emitted continuously
from any source in the form of radiations travelling in the Region: UV Visible Infrared
form of waves and associated with electric and magnetic
fields, oscillating perpendicular to each other and to the
STRUCTURE OF ATOM 27

 Rydberg formula: This formula is used to calculate wave  Principal quantum number (n): It determines the size of the
number of different series of lines of the spectrum of orbital. Its values are 1, 2, 3, etc. or K, L, M, etc. It also
determines the energy of the main shell in which the elecron
hydrogen or hydrogen like particles as :
is present and maximum number of electrons present in the
nth shell (= 2 n2).
 1 1   Azimuthal quantum number (l): It determines the number
v  R  2  2  Z 2 (Z = 1 for hydrogen)
n  of subshells present in any main shell (n) and the shape of
 i nf  the subshell. For a given value of n, l = 0 to n - 1. Thus, for
n = 1, l = 0 (one subshell), for n = 2 , l = 0, 1, (2 subshell), for
where R = Rydberg constant = 109677 cm-1 or 1.097×107 m-1 n = 3, l = 0, 1, 2 (3 subshells), for n = 4, l = 0, 1, 2, 3 (4
 Bohr’s Model: subshells). For l = 0, 1, 2 and 3. designation are s, p, d and f
respectively. Thus, subshells present are : n = 1 (1s), n = 2
1312Z2 (2s, 2p), n = 3 (3s, 3p, 3d), n = 4 (4s, 4p, 4d, 4f).
En  kJ mol1
n2  Magnetic quantum number (m): It determines the number
of orbitals present in any subshell and the orientation of
2.178  10 18 Z 2 13.6Z2 each orbital. For a given value of l, m = - l to + l including ‘0’.
 J / atom  eV / atom
n2 n2 Spin quantum number (s): It tells about the spinning motion
of the electron, i.e., clockwise or anti-clockwise. For a given
2.165  106 Z
Velocity of electron, v n  m/s 1 1
n value of m, s   and  . It helps to explain magnetic
2 2
0.529 n 2 properties of the substances.
Radius of orbit  Å
Z  Shapes of atomic orbitals: The shape of an orbital is found
 Dual behaviour of particle: According to de Broglie, every by finding the probability   2  of the electron in that orbital
object in motion has a wave character. The wavelengths
associated with ordinary objects are so short (because of at different points around the nucleus and representing by
their large masses) that their wave properties cannot be the densiy of points. The shape of the electron cloud thus
detected. The wavelengths associated with electrons and obtained gives the shape of the orbital. Some orbitals are
other subatomic particles (with very small mass) can found to have a region of space within it where probability
however be detected experimentally. is zero. This is called a node. It may be spherical/radial or
planar/angular.
h h h  Rules for filling of electrons in orbitals:
  
mv p 2m(KE)  Aufbau principle: Orbitals are filled in order of their
increasing energy. The order of energy and hence that of
 Heisenberg’s Uncertainty Principle: It is impossible to
filling orbitals is found by (n + l) rule. It states “lower the (n
measure simultaneously the position and momentum of a
+ l) value, lower is the energy. If two orbitals have same (n
small particle with absolute accuracy. If an attempt is made
+ l) value, orbital with lower value of n has lower energy.”
to measure any of these two quantities with higher accuracy,
Thus, the order is:
the other becomes less accurate. The product of the
uncertainty in the position (x) and the uncertainty in 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d....
momentum (p) is always a constant and is equal to or (n + l) 1 2 3 3 4 4 5 5 5 6 6 6 7 7
greater than h/4.  Hund’s rule of maximum multiplicity: Pairing of electrons
(x). (v)  h/4m does not occur in orbitals of the same subshell (degenerate
 Quantum mechanical Model of atom: Quantum mechanics orbitals) until each of them is first singly occupied.
is a theoretical science tshat deals with the study of the  Pauli exclusion principle: No two electrons in an atom can
motion of microscopic objects which have both particle like have the same set of four quantum numbers or an orbital
and wave like properties. The fundamental equation of can have maximum two electrons and these must have
quantum mechanics was developed by Schrodinger. opposite spin.
 Quantum number: It is a set of four numbers which give  Electronic configuration of elements: Distribution of
complete information about any electron in an atom. These electrons of an atom into different shells, subshells and
are: orbitals is called its electronic configuration. Complete
electronic configuration is obtained by following the above
rules, e.g.,
17
Cl = 1s22s22p63s23p2x3p2y3p1z
28 STRUCTURE OF ATOM

Solved Examples
Example-1 Cosmic rays < X-rays < < amber light < radiation from
microwave ovens < radiation of FM radio.
The threshold frequency  0 for a metal is 7 × 1014s-1.
Example-5
Calculate the kinetic energy of an electron emitted when
radiation of frequency 1 × 1015s-1 hits the metal. Yellow light emitted from a sodium lamp has a wavelength

Sol. 0 = 7 × 1014 s–1 ;  = 1015 s–1 (  ) of 580 nm. Calculate the frequency (  ) and wave

According to photo electric effect number ( v ) of the yellow light. (NCERT)

h = h0 + K.E. Sol. As per the formula of frequency,
K.E. = h ( – 0) speed of light in vacuum
Frequency 
= 6.626 × 10–34 (1015 – 7 × 1014) wavelength
= 1.9878 × 10–19 J.
c
Example-2 v …….. (i)

Calculate the number of protons, neutrons and electrons in
3  108 m / s
80
Br . (NCERT)
35
v  5.17  1014 s –1
Sol. Number of protons = number of electron 580  10 m
–9

= 35 = atomic number Yellow light emitted from a sodium lamp has a frequency
Number of neutron = Mass number (A) –Number of proton  5.17 1014 s 1
 80  35  45
1
Example-3 Wave number of yellow lights, v =

Write the complete symbol for the atom with the given
atomic number (Z) and Atomic mass (A) 1
v=  1.72  106 m 1
(i) Z = 17, A = 35 580  10 9
(ii) Z = 92, A = 233 Example-6
(iii) Z = 4, A = 9 (NCERT) Calculate frequency of yellow radiation having wavelength
35
Sol. (i) 17 Cl 5800Å .
c 3 108 ms 1
(ii) 233
U Sol. v  
92
 5800 1010 m
(iii) 9 Be  5.172  1014 s1
4

Example-4 Example-7
Arrange the following type of radiations in increasing order The energy required to break one mole of Cl-C1 bonds in
of frequency and wavelength Cl2 is 242 kJ mol-1. Calculate the largest wavelength of light
capable of breaking a single Cl-Cl bond.
(a) radiation from microwave oven
(b) amber light from traffic signal 242  103
(c) radiation from FM radio Sol. Energy required for one Cl2 molecule  J
NA
(d) cosmic rays from outer space and
hc
(e) X-rays. (NCERT) E

Sol. The increasing order of frequency is as follows:
hc 6.626 10 –34  3  108  6.02  1023
Radiation from FM radio < < radiation from microwave oven Or   
< amber light from traffic signal < X- rays < cosmic rays E 242  103
The increasing order of wavelength is as follows:  494  109 m  494nm
STRUCTURE OF ATOM 29

Example-8 Kinetic energy of emission:

What is the number of photons of light with wavelength of =(hv – hv0)
4000 pm that provide 1 J of energy ? (NCERT) =(E – W)eV
Sol. As per the formula of energy (E) of a photon Energy = =3.01eV – 2.13eV
Planck’s constant × frequency of light =0.97 ev
 c (iii) Kinetic energy of the ejected electron = Energy of
E  nhv v  
   striking phton-threshold energy
nhc 1
E mv 2  (hv  hv 0 )
 2
E 1(hv hv0 )
n v
hc m
Where, h = Planck’s constant  6.626 10 –34 Js 2  0.97  1.60  10 19
v J
The formula of n: 9.10 1031 kg
 0.3415  1012 m 2 s 2
1J    4000  1012 m 
n  2.012  1016 v  5.84 105 ms 1
 6.626 1034 Js  3 108 m / s  Example-10
A gas absorbs photon of 355 nm and emits two wavelengths.
Therefore, the number of photons  2.012 1016 .
If one of the emission is at 680 nm, calculate the wavelength
Example-9 of other emission.
A photon of wavelength 4 × 10–7 m strikes on metal surface, Sol. Energy values are always additive
the work function of the metal being 2.13 eV. Calculate Etotal = E1 + E2
(i) the energy of the photon (eV),
hc hc hc
(ii) the kinetic energy of the emission, and  
 1 2
(iii) the velocity of the photoelectron
(1 eV= 1.6020 × 10–19 J).

hc
Sol. (i) Energy which emits photons E 

where,
h = Planck’s constant = 6.62× 10–34Js
c = speed of light = 3 × 108ms–1
 = wavelength of radiation = 4 × 10–7m 1 1 1
 
 1 2
E
 6.62 10  3 10  4.97 10
–34 8
19
J
4  107 1 1 1
 
355 680 2
Convert into eV : 1eV = 1.6 × 10–19J

4.97  1019 2  742.77nm

E(eV)  eV  3.1eV  743nm
1.6  1019
Therefore, the energy of the photon is 3.1eV.
(ii) Kinetic energy of the ejected electron = Energy of striking
photon- threshold energy
=(3.01 × 2.13)eV
= 0.97 eV
30 STRUCTURE OF ATOM

Example-11 Sol. For Lyman series

Calculate the frequency of the first line in the balmer series 1 1 1 
in the H-atom. R 2  2
 1 n 2 
Sol. Frequency of first line in Balmer series can be calculated as
15R 1 1 
R 2  2
1 1 16 1 n 2 
v  3.29  1015  2  2  s 1
n
 1 n 2  15R  n 22  1 
 
16  n 22 
 1 1  1 1 15 n 22  1
 3.29  1015  2    3.29  1015     2
  2   3 
2
4 9 16 n2
15n 22  16n 22  16
5 n 22  16, n 2  4
 3.29  1015   4.57  1014 s 1
36 Example-14
Example-12 Calculate the energy ratio of photon of wavelength 3000Å
What is the maximum number of emission lines when the and 6000Å.
excited electron of an H atom in n = 6 drops to the ground Sol. 1 =3000Å, 2 = 6000Å
state? (NCERT) hc hc
E1  
n  n  1 1 3000
Sol. Number of spectral lines 
2 hc hc
E2  
Where n is the level drops down to the ground state. 2 6000
n=6 hc
6  6  1 E1 3000

Number of spectral lines   15 E2 hc
2
6000
hc 6000 2
  
3000 hc 1
E1 : E2 = 2 : 1
Example-15
Find out the ratio of energy of 3rd and 4th orbit of H-atom.

13.6  Z 2
Sol. KE of 3rd energy level  eV
n2

13.6  12 13.6
 eV  eV
32 9

Through the given diagram also we can calculate the number 13.6eV 13.6
KE of 4th energy level   eV
of spectral lines in emission spectrum. 42 9
Total number of spectral lines are = (5 + 4 + 3 + 2 + 1) = 15
E3
lines Ratio of 3rd and 4th energy level  E
4
Example-13
The wavelength of a spectral line emitted by hydrogen atom 13.6eV
16 9 10
in the Lyman series is cm . Calculate the value of n .  
15R 2 13.6eV 9
(R = Rydberg constant). 10
STRUCTURE OF ATOM 31

Example-16 Example-19
An electron is moving in Bohr’s fourth orbit. Its de-Broglie Show that the circumference of the Bohr orbit for the
wavelength is  . Calculate the circumference of the fourth hydrogen atom is an integral multiple of the de Broglie
orbit. wavelength associated with the electron revolving around
Sol. According to Bohr’s concept, an electron always move in the orbit. (NCERT)
Sol. We all know that orbit angular momentum of an electron is
nh given by
the orbit with angular momentum (mvr) equal to .
2
h
mvr  n
nh n  h  2
 mvr  or r  . 
2 2  mv  Suppose there are ’n’ no. of wavelengths associated with
an electron in an orbit  .
n h
or r  (from de-Broglie equation,   ) Now, the circumference of the orbit will be
 mv
2 r  n (where, = 1, 2, 3, …)
For fourth orbit (n = 4)
But According to de Broglie’s equation:
2
r
 h

2 mv
circumference  2 r  2   4

h
Example-17 2 r  n 
mv
What will be the kinetic energy of radiation having
 h
wavelength ? mvr  n 
2 2
Sol. We know that It was concluded that the permitted orbits are those for
which the angular momentum of an electron is an integral
h multiple of h/2  . Hence, the circumference of the Bohr

2mE orbit ‘2  r’ represents the circumference of the Bohr orbit
of the hydrogen atom is an integral multiple of de Broglie’s
h2 wavelength.
2 
2mE Example-20
Calculate the de-Broglie wavelength of a tennis ball of mass
h2 4h 2 2h 2 60 g moving with a velocity of 10 m/s.
E  

2
2m 2
m 2
2m.   Sol. De Broglie gave an expression for the wavelength and
2 momentum of all material particles.
Example-18 h

Find the momentum of a photon of frequency 50 ×10 s 17 –1
mv
h Where,
Sol.   or mv = momentum
mv m = mass of the particle = 0.1 kg
h h = Planck’s constant = 6.63 × 10–34 Js
 Momentum  [c  v  ]

 = wavelength of radiation
hv
 Momentum  v = velocity of the particle = 4.37×105
c
6.62  1034  50 1017 h 6.63  1034
  
3  108 mv 60  103 10
= 1.1 × 10–23kg ms–1 = 1.105 × 10–33m
32 STRUCTURE OF ATOM
Example-21 v  uncertainty in veslocity
Uncertainty in position of a particle of 25g in space is According to question
10–5m. Find out the uncertainty in velocity (ms–1). (Planck’s
constant, h = 6.6 × 10–34 Js) h
x A  m  0.05 
4
h
Sol. x.v 
4 m h
x B  5m  0.02 
4
6.62  1034
x  Equation (i) and divided by equation (ii), then
4  3.14  25 103  105
= 2.10 × 10–28m x A  m  0.05 x A
 1 or 2
Example-22 x B  5m  0.02 x B
Find out the uncertainty in the position of an electron (mass Example-24
= 9.1 × 10–31 kg) moving with a velocity 300 ms-1, accurate Using s, p, d notations, describe the orbital with the
upon 0.001 %. (h = 6.63 ×10–34Js) following quantum numbers.
h (a) n = 1, l = 0;
Sol. x.v  (b) n = 3; l = 1
4 m
(c) n = 4; l = 2;
6.63  1034 (d) n = 4; l = 3. (NCERT)
x 
4  3.14  9.11031  300  0.00110 2 Sol. Principal quantum number = n value
= 0.01933 = 1.93 × 10–2m Azimuthal quantum number (l) = 0 to (n-1)
Example-23 (a) for n = 1, l = 0; orbital is 1s.
The uncertainties in the velocities of two particles A and B (b) For n = 3 and l = 1; orbital is 3p.
are 0.05 and 0.02 ms-1 respectively. The mass of B is five (c) For n = 4 and l = 2; orbital is 4d.
time to that of mass A. What is the ratio of uncertainties (d) For n = 4 and l = 3; orbital is 4f.
 x A 
  in their positions? Example-25
 x B 
Arrange 3d, 4s, 4p orbitals in the order of increasing energy
Sol. According to Heisenberg as per Aufbau principle.
h Sol. According to Aufbau principle, as electron enters the orbital
x  m  v  of lowest energy first and subsequent electron are fed in
4
the order of increasing energies. The relative energies of
Where, x = uncertainty in position various orbital in increasing order are
m = Mass of particle 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5s,6p,7s
STRUCTURE OF ATOM 33

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Which of the following rays/particles will not show 15. Calculate the frequency of the radiation emitted when
deflection from their path on passing through an an electron jumps from n = 3 to n = 2 in a hydrogen
electric field? atom.
Proton, cathode rays, electron, neutron
16. Calculate the wavelength and frequency of a light wave
2. An atom having mass number 13 has 7 neutrons. What whose time period is 2.0 × 10–10 s.
is the atomic number of the atom?
17. Energy of an electron in H atom in ground state is 13.6
3. What do you mean by valence electrons? eV. What is the value in first excited state?

4. What are the α - particles? 18. In the lowest energy level of hydrogen atom, what is
the angular momentum of electron?
5. Why some atoms shows radioactivity?
19. Why was a change in the Bohr Model of atom required?
6. What does nucleon contains ? Due to which important development (s), concept of
movement of an electron in an orbit was replaced by,
7. Why Bohr’s orbits are called stationary states? the concept of probability of finding electron in an
orbital? What is the name given to the changed model
8. Which of the following orbitals are degenerate of atom?
4d xy ,4d yz ,4dz2 , 3d xy ,3d yz ,3dz2 ?
20. Show the distribution of electrons in oxygen atom
9. Nickel atom can lose two electrons to form Ni2+ ion. (atomic number 8) using orbital diagram.
The atomic number of nickel is 28. From which orbital
will nickel lose two electrons? Section – C (3 Marks Questions)

10. Calculate the total number of angular nodes and radial 21. How many protons, electrons and neutrons are there in
nodes present in 3p orbital. the following nuclei?
(i) 17 25
8 O (ii) 12 Mg (iii) 80
35 Br
Section – B (2 Marks Questions)
22. The wavelength of first spectral line in the Balmer
11. What is the reason behind fractional atomic mass?
series is 6561 Å. Calculate the wavelength of the
second spectral line in Balmer series.
12. What is the difference between mass number and
atomic mass?
23. Threshold frequency, ν0 is the minimum frequency
which a photon must possess to eject an electron from a
13. What are the general features of isotope?.
metal. It is different for different metals. When a
14. Wavelengths of different radiations are given below : photon of frequency 1.0 × 1015 s–1 was allowed to hit a
λ(A) = 300 nm λ(B) = 300 μm metal surface, an electron having 1.988 × 10 -19 J of
λ(C) = 3 nm λ (D) = 30 A° kinetic energy was emitted. Calculate the threshold
Arrange these radiations in the increasing order of their frequency of this metal.
energies.
34 STRUCTURE OF ATOM

24. Radius of the fourth orbit in hydrogen atom is 0.85 nm. Section – D (5 Marks Questions)
Calculate the velocity of the electron in this orbit (mass
of electron = 9.1 × 10-31 kg).
29. When an electric discharge is passed through hydrogen
gas, the hydrogen molecules dissociate to produce
25. Mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × excited hydrogen atoms. These excited atoms emit
10-25 J, calculate its wavelength. electromagnetic radiation of discrete frequencies which
can be given by the general formula
26. The ionization energy of He+ is 8.72 × 10-18 J atom-1.
1 1
Calculate the energy of first stationary state of Li2+. v = 109677  2 − 2 
 ni nf 
27. Calculate the ratio of the radius of 2nd orbit of H atom What points of Bohr’s model of an atom can be used to
and that of 3rd orbit. arrive at this formula? Based on these points derive the
above formula giving description of each step and each
28. I. Based upon the above information, arrange the term.
following orbitals in the increasing order of energy.
(a) 1s, 2s, 3s, 2p (b) 4s, 3s, 3p, 4d 30. (a) What is the significance of the negative electronic
(c) 5p, 4d, 5d, 4f, 6s (d) 5f, 6d, 7s, 7p energy.
II. Based upon the above information, solve the (b) Calculate the longest wavelength of light that will
questions given below : be needed to remove an electron from the third orbit
(a) Which of the following orbitals has the lowest of He+ ion.
energy? 4d, 4f, 5s, 7p
(b) Which of the following orbitals has the highest
energy? 5p, 5d, 5f, 6s, 6p
STRUCTURE OF ATOM 35

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions)

1. Thomson suggested that the amount of deviation of the 6. Major development(s) responsible for the formulation
particles from their path in the presence of electrical or of Bohr’s model of atom were.
magnetic field depends upon (a) Dual character of the electromagnetic radiation
I. The magnitude of the negative charge on the which means that radiations possess both wave like
particle. and particle like properties
II. The mass of a particle. (b) Experimental results regarding atomic spectra which
III. The strength of the electrical or magnetic field. can be explained only by assuming quantized
Choose the appropriate option with the correct set of electronic energy level in atoms.
statements. (c) Both (a) and (b)
(a) I and II (b) II and III (d) None of the above
(c) I and III (d) I, II and III
7. The energy of an electron in first Bohr orbit of H-atom
2. Thomson determined charge to mass ratio of electron as is -13.6 eV. The possible energy value of electron in the
(a) 1.758820 1010 Ckg −1 (b) 17.58820 1010 Ckg −1 excited state of Li2+is
(a) -122.4 eV (b) 30.6 eV
(c) 1.758820 109 Ckg−1 (d) 17.58820 1011 Ckg −1 (c) -30.6 eV (d) 13.6 eV

3. Calculate the number of protons, neutrons and electrons 8. The angular momentum of an electron in a given
80
in 35 Br stationary state can be expressed as
h
(a) Protons = 80, electrons =80, neutrons =35 (a) me vr = n. ( n = 1, 2,3...)
(b) Protons = 35, electrons =55, neutrons =80 2π
mv h
(c) Protons = 35, electrons =35, neutrons =80 (b) e = n. ( n = 1, 2,3...)
(d) Protons = 35, electrons =35, neutrons =45 r 2π
nhr
4. The oscillating electric and magnetic fields produced by
(c) me v = ( n = 1, 2,3...)
2π
oscillating charged particles are h
I. Perpendicular to each other.
(d) me vr = ( n = 1, 2,3...)
2πn
II. Parallel to each other
III. Both are perpendicular to the direction of o

propagation of the wave. 9. Calculate the mass of a photon with wavelength 3.6 A
IV. Both are parallel to the direction of the waves. (a) 6.135 10−27 kg (b) 0.615 10−28 kg
Choose the correct option (c) 0.613 10−32 kg (d) 6.135 10−30 kg
(a) I and III (b) II and IV
(c) I and IV (d) II and III
10. An electron is moving in Bohr’s fourth orbit. Its de-
Broglie wavelength is λ Calculate the circumference of
5. The number of electron ejected, in the photoelectric
the fourth orbit.
experiment is proportional to the
(a) Intensity of light (b) Brightness of light (a) 2λ (b) 4λ
(c) Both (a) and (b) (d) Neither (a) nor (b) 4 2
(c) (d)
λ λ
36 STRUCTURE OF ATOM

11. The correct mathematical expression of Heisenberg’s 17. Identify the correct order of increase in the energy of
Uncertainty principle is the orbitals of hydrogen atom is
Δx h h (a) 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
(a)  (b) Δx  ΔVx 
ΔVx 4πm 4πm (b) 1s > 2s = 2p > 3s = 3p = 3d > 4s = 4p = 4d = 4f
(c) 1s = 2s = 3s = 4s > 2p = 3p = 4p > 3d = 4d > 4f
h h
(c) Δx  ΔVx = (d) Δx  Vx  (d) 1s = 2s = 3s = 4s < 2p = 3p = 4p < 3d = 4d < 4f
4πm 4πm

18. The orbital diagram(s) which follow the Aufbau

12. In atom, an electron is moving with a speed of 600 m/s
principle is/are
with an accuracy of 0.005 %. Certainty with which the
position of the electron can be located is (h = 6.6 × 10–34
kg m2 s–1, mass of electron, em = 9.110−31 kg )
(a) 1.52 10−4 m (b) 5.10 10−3 m
Choose the correct option
(c) 1.92 10−3 m (d) 3.84 10−3 m
(a) Only I and II (b) Only II and III
2
(c) Only II, III and IV (d) All I, II, III and IV
13. If probability density ψ is constant on a given
surface, 19. The correct valence electronic configuration of Cr and
I. ψ is constant over the surface Cu among the following electronic configurations are
II. the boundary surface for ψ
2
and ψ are different. [Ar] 3d5 4s1 – I [Ar] 3d10 4s1 – II
[Ar] 3d 4s – III
4 2
[Ar] 3d9 4s2 – IV
The correct statement is/are (a) Cr-I; Cu-IV (b) Cr-III; Cu-IV
(a) Only I (b) Only II (c) Cr-II; Cu-III (d) Cr-I; Cu-II
(c) Both I and II (d) None of these
20. The completely filled and completely half-filled
14. Which combinations of quantum number n, l, m and s
subshells are stable due to
for the electron in an atom does not provide a
(a) symmetrical distribution of electrons
permissible solution of the wave equation?
(b) exchange energy
1 1 (c) Both (a) and (b)
(a) 3, 2,1, (b) 3,1,1,
2 2 (d) None of the above
1 1
(c) 3, 3,1, (d) 3, 2, −2,
2 2 Section - B (Assertion & Reason Type Questions)
15. Which one of the following sets of quantum numbers 21. Assertion (A): Kinetic energy of the ejected electron is
represents the highest energy level in an atom? proportional to the frequency of the electromagnetic
1
(a) n = 4, l = 0, m = 0, s = + radiation.
2 Reason(R): A more intense beam of light consists of
1 larger number of photons (hv), consequently, the
(b) n = 3, l = 1, m = 1, s = +
2 number of electrons ejected is also larger as compared
1 to that from light of weaker intensity.
(c) n = 3, l = 2, m = 2, s = +
2 (a) Both Assertion and Reason are true, and the Reason
1 is the correct explanation of the Assertion.
(d) n = 3, l = 0, m = 0, s = + (b) Both Assertion and Reason are true, but the Reason
2
is not the correct explanation of the Assertion.
16. What is the correct order for the size of the s-orbital? (c) Assertion is true, but Reason is false.
(a) 4s > 2s > 3s > 1s (b) 4s > 3s > 2s > 1s (d) Both Assertion and Reason are false.
(c) 1s > 2s > 3s > 4s (d) 1s > 3s > 2s > 4s
STRUCTURE OF ATOM 37

22. Assertion (A): The spectrum of white light is known as 26. Assertion (A): Angular nodes are directly related to
continuous spectrum. magnetic quantum number.
Reason (R): It is continuous because violet merges into Reason (R): It is because, magnetic quantum number
blue, blue into green and so on. specifies the shape of orbitals.
(a) Both Assertion and Reason are true, and the Reason (a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion. is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason (b) Both Assertion and Reason are true, but the Reason
is not the correct explanation of the Assertion. is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false. (c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false. (d) Both Assertion and Reason are false.

23. Assertion (A) : The radius of first stationary state 27. Assertion (A): Ten different set of four quantum
called Bohr’s radius for hydrogen atom is 0.529 Å. numbers are possible for d-subshell.
Reason (R) : Radius for each circular orbit is Reason (R): The d-subshell splits has five degenerate
( rn ) = a 0 n 2 , where, n (energy shell) = 1,2,3,….. orbitals.
(a) Both Assertion and Reason are true, and the Reason
(a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion.
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason
(b) Both Assertion and Reason are true, but the Reason
is not the correct explanation of the Assertion.
is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
(d) Both Assertion and Reason are false.

28. Assertion (A): Electronic configuration of an element

24. Assertion (A): The path of an electron in an atom can
is 1s2 2s2 2p6 3s2 3p6 3d5 4s1.
never be determined or known accurately.
Reason (R): Last electron of element present in 4s-
Reason (R): Both the exact position and exact velocity
subshell.
of an electron in an atom cannot be determined
(a) Both Assertion and Reason are true, and the Reason
simultaneously (Heisenberg’s uncertainty principle).
is the correct explanation of the Assertion.
(a) Both Assertion and Reason are true, and the Reason
(b) Both Assertion and Reason are true, but the Reason
is the correct explanation of the Assertion.
is not the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason
(c) Assertion is true, but Reason is false.
is not the correct explanation of the Assertion.
(d) Both Assertion and Reason are false.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
Section – C (Case Study Questions)
25. Assertion (A) : Isotopes have same chemical properties
but different physical properties. Case Study - 1
Reason (R) : Chemical properties of an element depend
on its number of electrons. Physical properties depends The spectrum of radiation emitted by a substance that
on mass number. has absorbed energy is called an emission spectrum.
(a) Both Assertion and Reason are true, and the Reason Atoms, molecules or ions that have absorbed radiation
is the correct explanation of the Assertion. are said to be “excited”. To produce an emission
(b) Both Assertion and Reason are true, but the Reason spectrum, energy is supplied to a sample by heating it
or irradiating it and the wavelength (or frequency) of
is not the correct explanation of the Assertion.
the radiation emitted, as the sample gives up the
(c) Assertion is true, but Reason is false. absorbed energy, is recorded.
(d) Both Assertion and Reason are false. An absorption spectrum is like the photographic
negative of an emission spectrum. A continuum of
38 STRUCTURE OF ATOM

radiation is passed through a sample which absorbs Electron can make jump from one stationary orbit to
radiation of certain wavelengths. The missing another stationary orbit by absorbing or emitting a
wavelength which corresponds to the radiation photon of energy equal to difference in the energies of
absorbed by the matter, leaves dark spaces in the bright the stationary orbit i.e. energy change does not take
continuous spectrum. place in continuous manner.

29. The emission spectra of atoms in the gas phase do not 33. Bohr’s atomic model suggest that
show a continuous speed of wavelength from red to (a) Electrons have a particle as well wave character
violet, rather they emit light only at specific (b) Atomic spectrum of atom should contain only five
wavelengths with dark spaces between them. Such lines
spectra is/are called (c) Electron on H atom can have only certain values of
(a) line spectra angular momentum
(b) atomic spectra (d) All of the above
(c) Both (a) and (b) are true
(d) None of these 34. When electron jumps from 1s to 2s orbit in H-atom
(a) Energy is released
30. Which of the following statements is incorrect (b) Then atom becomes cation
regarding emission spectra? (c) Size of the atom decreases
(a) The emission spectrum is the spectrum of radiation (d) Energy is absorbed
absorbed by a substance.
(b) To produce an emission spectrum, energy is 35. I. The energy of radiation increase with decrease in
supplied to a sample by heating it or irradiating it . wavelength.
(c) It is also known as fingerprint of an element. II. The spectrum of H-atom is exactly same as that of
(d) All of the above He+ ion.
III. Energy of radiation increases with increase in v .
31. The only series of lines appear in the visible region of The correct statement(s) is/are
the electromagnetic spectrum of hydrogen is (a) I and II (b) II and III
(a) Lyman series (b) Balmer series (c) I and III (d) I, II and III
(c) Paschen series (d) Pfund series
36. According to the Bohr’s model of hydrogen atom
32. The wave number of the spectral line in the emission of (a) total energy of the electron is quantized
8 (b) angular momentum of the electron is quantized and
hydrogen will be equal to times the Rydberg’s
9 h
given as l ( l +1).
constant if the electron jumps from 2π
(a) n = 10 to n =1 (b) n = 3 to n = 1 (c) Neither (a) nor (b)
(c) n = 9 to n = 8 (d) n = 0 to n = 1 (d) Both (a) and (b)

Case Study – 2 Case Study - 3

Postulates of Bohr’s model: There are certain orbits Quantum Numbers defined as a set of four numbers
around the nucleus such that if electron will be with the help of which we can get complete information
revolving in these orbit, then it does not emit any about all the electrons in an atom i.e. location, energy,
electromagnetic radiation. These are called stationary type of orbital occupied, shape and orientation of that
orbit for the e–. The necessary electrostatic force of orbital etc.
attraction is produced by attraction forces of nucleus. The three quantum numbers called as Principal,
Angular momentum of the electron in these stationary Azimuthal and Magnetic quantum number are derived
orbit is always an integral multiple of h/2π from Schrodinger wave equation. The fourth quantum
STRUCTURE OF ATOM 39

number i.e. the Spin quantum number was proposed 39. In a multi-electron atom, which of the following
later on. orbitals described by the three quantum numbers will
have the same energy in the absence of magnetic and
37. The correct statement(s) is/are electric fields?
I. The shape of the orbitals is given by magnetic A. n = 1, l = 0, m = 0
quantum number. B. n = 2, l = 0, m = 0
II. If the value of l = 0, the electron distribution is C. n = 2, l = 1, m = 1
spherical.
D. n = 3, l = 2, m = 1
III. Angular momentum of 1s, 2s, 3s electrons are equal.
(a) II and III (b) I and III E. n = 3, l = 2, m = 0
(c) I and II (d) II only (a) D and E (b) C and D
(c) B and C (d) A and B
38. Which of the following statements are incorrect for an
electron that has n = 4 and m = -2 ? 40. Which of the following statements concerning quantum
I. The electron is in the fourth principle electronic numbers are correct?
shell. I. Angular quantum number determines the three
II. The electron may be in a d-orbital. dimensional shape of the orbital.
III. The electron must have the spin quantum number II. The principal quantum number determines the
1 orientation and energy of the orbital.
=−
2 III. Magnetic quantum number determines the size of
IV. The electron may be in a p-orbital. the orbital.
(a) I and II (b) III and IV IV. Spin quantum number of an electron determines the
(c) I and IV (d) I, II and III orientation of the spin of the electron relative to the
chosen axis.
(a) I and II (b) I and IV
(c) III and IV (d) II, III and IV

Find Answer Key and Detailed Solutions at the end of this book

STRUCTURE OF ATOM
CLASSIFICATION OF ELEMENTS AND
PERIODICITY IN PROPERTIES
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 41

Chapter at a Glance
 Mendeleev’s periodic table was based on atomic masses of (iii) Elements which lie in the border line between metals and
the elements. When Mendeleev presented the periodic table, non-metals are called semimetals or metalloids. B, Si, Ge, As,
only 63 elements were known. He left 29 places in the table Sb, Te, Po and At are regarded as metalloids.
for unknown elements.  IUPAC given a new scheme for assigning a temporary name
to the newly discovered elements. The name is derived directly
 Modern Mendeleev periodic table is based on atomic numbers
from the atomic number of the elements.
of the elements. The modern periodic law is : “The physical
and chemical properties of the elements are periodic function  The recurrence of similar properties of the elements after
of their atomic numbers”. certain definite intervals when the elements are arranged in
order of increasing atomic numbers in the periodic table is
 The horizontal row in the periodic table is called a period and
termed periodicity. The cause of periodicity is the repetition
vertical column is called group. There are seven periods and
of similar electronic configuration of the atom in the valence
eight groups in the modern Mendeleev periodic table.
shell after certain definite intervals. These definite intervals
 The long or extended form of periodic table consists of seven are 2, 8, 8, 18, 18 and 32. These are known as magic number.
periods and eighteen vertical columns (groups or families). Periodicity is observed in a number of properties which are
The elements in a period have same number of energy shells, directly or indirectly linked with electronic configuration.
i.e., principal quantum number (n). These are numbered 1 to 7.
 Effective nuclear charge increases across each period.
 At present 114 elements are known.
 Atomic radii generally decrease across the periods.
 In a vertical column (group), the elements have similar valence  Atomic radii generally increase on moving from top to bottom
shell electronic configuration and therefore exhibit similar in the groups.
chemical properties.
 Atomic radius is of three types :
 There are four blocks of elements: s-, p-, d- and f-block
(a) Covalent radius
depending on the orbital which gets the last electron. The
(b) Crystal or metallic radius
general electronic configuration of these blocks are :
(c) Van der Waals’ radius
s-block : [Noble gas] ns1 or 2.
 Cations are generally smaller than anions.
p-block : [Noble gas] ns2np1–6
 Cations are smaller and anions are larger than neutral atoms
d-block : [Noble gas] (n – 1)d1–10ns1 or 2
of the elements.
f-block : [Noble gas] (n – 2)f1–14(n – 1)d0 or 1ns2

 The elements are broadly divided into three types :

(i) Metals comprise more than 78% of the known elements.  The first element is each group of the representative elements
s-block, d-block and f-block elements are metals. The higher shows abnormal properties, i.e., differs from other elements of
members of p-block are also metals. the group because of much smaller size of the atom.
(ii) Non-metals are less than twenty. (C, N, P, O, S, Se, H, F, Cl,
Br, I, He, Ne, Ar, Kr, Xe and Rn are non-metals).
42 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

 The ions having same number of electrons but different nuclear  Electron gain enthalpy becomes more negative from left to right
charge are called isoelectronic ions. in a period and less negative from top to bottom in a group.

In isoelectronic ions, the size decreases if Z/e increases i.e.,  Successive electron gain enthalpies are always positive.
greater the nuclear charge, smaller is the size of the ion.
 The elements with higher ionisation enthalpy have higher
 The energy required to remove the most loosely held electron negative electron gain enthalpy.
from the gaseous isolated atom is termed ionisation enthalpy.
 Electronegativity is the tendency of an atom to attract the
 Ionisation enthalpy values generally increase across the periods. shared pair of electrons towards itself in a bond.

 Ionisation enthalpy values generally decrease down the group.  Electronegativity increases across the periods and decreases
down the groups.
 Removal of electron from filled and half filled shells requires
higher energy. For example, the ionisation enthalpy of nitrogen  Metals have low electronegativities and non-metals have
is higher than oxygen. Be, Mg and noble gases have high high electronegativities.
values.
 Metallic character decreases across the periods and
 Metals have low ionisation enthalpy values while non-metals increases down the group.
have high ionisation enthalpy values.
 Valence of an element belonging to s- and p- block (except
 Successive ionisation enthalpies of an atom have higher values. noble gases) is either equal to the number of valence
electrons or eight minus number of valence electrons.
IEI < IEII < IEIII ...
 The reducing nature of the elements decreases across the
 The enthalpy change taking place when an electron is added to
period while oxidising nature increases.
an isolated gaseous atom of the element is called electron gain
enthalpy. The first electron gain enthalpy of most of the elements  The basic character of the oxides decreases while the acidic
is negative as energy is released in the process but the values character increases in moving from left to right in a period.
are positive or near zero in case of the atoms having stable
configuration such as Be, Mg, N, noble gases, etc.
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 43

Solved Examples
Example-1 Example-4
Which important property did Mendeleev use to classify (a) What is modern periodic law ? Discuss the main features
the elements in his periodic table and did he stick to that? of long form of periodic table.
(NCERT) (b) Give the general electronic configuration of s, p, d & f-
block elements.
Sol. According to the Mendeleev’s Periodic Law, “the
properties of elements are a periodic function of their atomic Sol. (a) The physical and chemical properties of the elements are
mass”. Mendeleev arranged known 63 elements in periodic functions of their atomic numbers. The main features
horizontal rows and vertical columns of a table. Mendeleev of long form of periodic table are as follows :
violated his periodic law at certain places to give 1. The aufbau (build up) principle and the electronic
appropriate position to the few elements based on their configuration of atoms provide a theoretical foundation for
properties. Assume that atomic measurements might be the periodic classification.
incorrect, placed elements with similar properties together. 2. The long form of the periodic table consists of horizontal
For example: Iodine having lower atomic mass than rows called periods and vertical columns called groups.
tellurium was placed ahead of tellurium because iodine
3. There are altogether seven periods. The period number
showed similar properties with fluorine. The primary aim
corresponds to the highest principal quantum number (n) of
of Mendeleev was to arrange the elements with similar
the elements in the period.
properties in the same group
4. The first period contains 2 elements. The subsequent
Example-2 periods consists of 8, 8, 18, 18 and 32 elements, respectively.
What is the basic theme of organisation in the periodic The seventh period is incomplete and like the sixth period
table. (NCERT) would have a theoretical maximum (on the basis of quantum
numbers) of 32 elements.
Sol. The elementary idea of organization in the periodic table is
to classify the elements into groups and periods. This table 5. In this form of the Periodic Table, 14 elements of both sixth
is arranged according to the properties of the elements. and seventh periods (lanthanoids and actinoids,
Same properties elements belong to same group. It is very respectively) are placed in separate panels at the bottom.
easy and systematic method to learn the properties of 6. Elements having similar outer electronic configurations in their
elements. atoms are arranged in vertical columns referred to as groups or
families. There are in all 18 vertical column or groups.
Example-3
7. The elements of groups 1 (alkali metals), 2 (alkaline earth
What is the basic difference in approach between the
metals) and 13 to 17 are called the main group elements.
Mendeleev’s Periodic Law and the Modern Periodic Law?
These are also called typical or representative or normal
(NCERT) elements.
Sol. According to the Mendeleev’s Periodic Law, “the 8. The elements of group 3 to 12 are called transiation elements.
properties of elements are a periodic function of their atomic 9. lanthanoids & actinoids are together referred to as inner
mass”. Mendeleev arranged known 63 elements in transition elements.
horizontal rows and vertical columns of a table on the other
(b) (i) General outer electronic configuration of s-block elements
hand, the Modern periodic Law states that “the physical
is ns1–2 i.e., either ns1 or ns2.
and chemical properties of elements are periodic functions
of their atomic numbers”. This modern law was proposed (ii) General outer electronic configuration of p-block elements
is ns2np1–6.
by Henry Moseley.
(iii) General outer electronic configuration of d-block elements
is (n – 1) d1–10 ns0–2.
(iv) General outer electronic configuration of f-block elements is
(n – 2) f1–14 (n – 1) d0–1 ns2.
44 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Example-5 Example-9
On the basis of quantum numbers, justify that the sixth What is the group number, period and block of the element
period of the periodic table should have 32 elements. with atomic number 43 ?
(NCERT) Sol. The electronic configuration of the element with atomic
Sol. Sixth period corresponds to the filling of the sixth energy number 43 is
level i.e; n=6. Since in this period only sixteen orbitals (one 1s2, 2s22p6, 3s23p63d10, 4s24p64d5, 5s2
6s, seven 4f, five 5d and three 6p)are available, thereby
sixth period contains thirty two elements. It begins with Since, the last electron is accommodated in d-subshell, the
Caesium in which electrons enters the 6s orbital and ends element belongs to d-block. The principal quantum number
at radon in which the filling of 6p orbitals is complete. After of outermost shell is 5, the element belongs to 5th period.
filling of 6s orbital, the next electron enters the 5d orbitals Group number of the element = 5 + 2 = 7 i.e.,
against the Aufbau principle and thereafter the filling of
seven 4f orbital begins with cerium and end up with The element belongs to group 7.
lutetium. Maximum 2 electrons can be accommodated in Example-10
each orbital. Thus, 16 orbitals can accommodate a maximum
of 32 electrons. Therefore, the sixth period of the periodic How does atomic radius vary in a period and in a group?
table should have 32 elements. How do you explain the variation?

Example-6 Sol. Atomic Radius: It is defined as average distance from the

centre of nucleus upto the centre of outermost shell
Write the atomic number of the element present in the third
electrons. It is measured in angstrom or picometer. It is not
period and seventeenth group of the periodic table.
possible to measure the exact atomic radius because an
(NCERT)
atom is unstable, and it can’t be isolated to get its radius
Sol. Chlorine is belonging to the 3rd period and 17th group. Hence, and electron cloud around the atom do not have sharp
Chlorine atomic number is 17. boundary.
Example-7
Trend in the periodic table:
Which element do you think would have been named by
(a) Across a period: Atomic radii decreases across a period
(i) Lawrence Berkeley Laboratory from left to right. It is due to increase in atomic number; the
(ii) Seaborg’s group? (NCERT) number of electrons increases but number of shells remain
Sol. (i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with same due to which electrons fill in same shell which in turn
Z = 97 increase the nuclear force of attraction between nucleus
and electrons of outermost shell and thus atomic size
(ii) Seaborgium (Sg) with Z = 106 decreases.
Example-8 Down the group: Atomic radii increases down in a group
To which block (s, p, d or f) does the element with atomic from top to bottom. It is due to increase in atomic number;
number 50 belong ? the number of shells also increases in which electron in the
Sol. The electronic configuration of element with atomic number outer shell of each succeeding element lie farther and farther
50 is : away from nucleus. Due to increase in atomic number, the
nuclear charge should also increase but it is dominated by
1s2, 2s22p6, 3s23p63d10, 4s24p64d10, 5s25p2
screening effect on the valence electrons by the electron
The last electron enters into 5p-orbital. Hence, it is a present in the inner shells and hence atomic size increase
p-block element. down the group.
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 45

Example-11 Example-13
Consider the following species: N , O , F , Na , Mg and
3– 2– – + 2+
Define atomic radius. Explain various factors affecting it ?
Al 3+
(NCERT) Sol. Atomic radius is defined as the distance of valence shell of
(a) What is common in them? electrons from the centre of the nucleus of an atom.
(b) Arrange them in the order of increasing ionic radii. Factors affecting atomic radius :

Sol. (a) All species are isoelectronic. The species in which atoms (i) No. of shells : The atomic radius increases with the increase
in the no. of the shells.
or ions of different elements that have same number of
electrons but have different magnitude of nuclear atomic radius  no of shells
charge are known as isoelectronic species. They may (ii) Nuclear charge : Atomic radius decreases with the increase
be neutral or ionic species. All species have 10 electrons. in the Nuclear charge. Due to high nuclear charge, the
nucleus attracts the electrons towards itself thereby reducing
(b) The size of cation is always smaller than its parent atom its own size
and the size of anion is always greater than its parent
atom. Therefore, the arrangement of the given species 1
atomic radius 
in order of their increasing ionic radii is as follows: Nuclear charge

Al3+< Mg2+< Na+< F–< O2–< N3– (iii) Shielding or screening effect : Atomic radius increases with
the increase in the shielding effect. This is because the
Example-12
electrons presents between the Nucleus and the valence
The increasing order of reactivity among group 1 elements shell shields the valence electrons from the Nucleus i.e. it
is Li < Na < K <Rb<Cs whereas that among group 17 reduces the force of attraction between the Nucleus and the
elements is F > Cl > Br > I. Explain. (NCERT)
Valence electrons. atomic radius  shielding effect
Sol. The group 1 belongs to the metals. The reactivity of metals
Example-14
depends upon their tendency to lose electrons to acquire
the inert gas configuration. As we move down in a group of Among the elements Li, K, Ca, S and Kr, which one is expected
to have the lowest first ionization enthalpy and which the
metal elements, the atomic size of the elements goes on
highest first ionization enthalpy ?
increasing. Due to increase in size of atoms the valence
electrons of metal atom, which take part in chemical reactions Sol. K has the lowest first ionization energy. Kr has the highest
first ionization energy.
becomes more far away from the
Example-15
nucleus and hence can be removed easily. Consequently,
the increasing order of reactivity among group 1 elements Among the second period elements, the actual ionization
energies are in the order :
is as follows: Li < Na < K <Rb<Cs
Li < B < Be < C < O < N < F < Ne.
In the group 17th, The reactivity of non-metals depends
Explain why (i) Be has higher i H than B (ii) O has lower
upon their tendency to gain electrons to acquire the inert
gas configuration. As we move down in a group of non- i H than N and F ?
metal elements, the atomic size of the elements goes on Sol. (i) The ionization enthalpy, among other things, depends upon
the type of electron to be removed from the same principal
increasing. Due to increase in size of atoms the distance of
shell. In case of Be (1 s2 2 s2) the outermost electron is
nucleus of atom from the valence shell increases. The force
present in 2s-orbital while in B (1 s2 2 s2 2 p1) it is present in
of attraction exerted by the nucleus of the atom on the 2p-orbital. Since 2s-electrons are more strongly attracted
valence shell electrons decreases, as a result tendency to by the nucleus than 2p-electrons, therefore, lesser amount
gain electron also decreases. Consequently, the decreasing of energy is required to knock out a 2p-electron than a 2s-
order of reactivity among group 17 elements is as follows: electron. Consequently, i H of Be is higher than i H of B.

F > Cl > Br > I

46 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

(ii) The electronic configuration of N (1s2 2s2 2p1x 2p1y 2p1z) Example-18
in which 2p-orbitals are exactly half-filled is more stable The first ionisation energy of carbon atom is greater than
than the electronic configuration of O (1s2 2s2 2p2x 2p1y 2p1z) that of boron atom, whereas reverse is true for the second
in which the 2p-orbitals are neither exactly half-filled nor ionisation energy. Explain.
completely filled. Therefore, it is difficult to remove an
Sol. The electronic configurations of carbon and boron are as
electron from N than from O. As a result, iH of N is higher
follows :
than that of O. Further, the electronic configuration of F is
1s2 2s2 2p2x 2p2y 2p1z. Because of higher nuclear charge (+9), C : 1s2, 2s2 2p1x 2Py1
the first ionization enthalpy of F is higher than that of O.
Further, the effect of increased nuclear charge outweights B : 1s2, 2s2 2p1x
the effect of stability due to exactly half-filled orbitals,
therefore, the iH of N and O are lower than that of F. Due to higher nuclear charge in carbon, the force of
Example-16 attraction towards valency electron is more in carbon atom
and hence the first ionisation energy is greater than boron
Among the elements B, Al, C and Si atom. After loss of one electron, the monovalent cations
(i) Which element has the highest first ionisation enthalpy ? have the configurations as follows :
(ii) Which element has the most metallic character ? B+ : 1s2, 2s2
Justify your answer in each case.
C+ : 1s2, 2s2 2p1x
Sol. Arrange the elements B, Al, C and Si into different groups and
periods in order of their increasing atomic numbers, we have, The B+ configuration is stable one and hence the removal of
Group  13 14 electron is difficult in comparison to C+. Hence, second
Period 2 B C ionisation potential of boron is higher than carbon.

Group 3 Al Si Example-19

(i) Since ionization enthalpy increases along a period and Which of the following pairs of elements would have a
decreases down a group, therefore, C has the highest first more negative electron gain enthalpy?
ionization enthalpy.
(i) O or F (ii) F or Cl (NCERT)
(ii) Since metallic character increases down a group and decreases
along a period, therefore, Al, is the most metallic element. Sol. (i) The electron gain enthalpy of F is more negative than
Example-17 that of O because when we move from left to right the
atomic size decreases. Fluorine size is smaller than oxygen.
How does the metallic and non metallic character vary on
Due to small size the interelectronic repulsion increases
moving from left to right in a period ?
causing for the decreases in the electron gain enthalpy
Sol. On moving from left to right in a period, the number of values.
valence electrons increases by one at each succeeding
(ii) The electron gain enthalpy of fluorine is less negative
element but the number of shells remains the same. As a
than that of chlorine because when an electron is added to
result, the nuclear charge increases and the tendency of the
F, the added electron goes to the smaller quantum level
element to lose electron decreases and hence the metallic
(n=2). Due to small size the interelectronic repulsion
character decreases as we move from left to right in a period.
increases causing for the decreases in the electron gain
Conversely, as the nuclear charge increases, the tendency
enthalpy values. On the other hand, in chlorine the electron
of the element to gain electrons increases and hence the
enters the high quantum level(n=3) which occupies a larger
non-metallic increases from left to right in a period.
region of space where the electron repulsion is less.
Alternatively, metallic character decreases and non-metallic Therefore, the electron gain enthalpy of Cl is more negative
character increases as we move from left to right in a period. than that of F.
It is due to increase in ionization and electron gain enthalpy.
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 47

Example-20 Example-22
Would you expect the second electron gain enthalpy of O as Why are electron gain enthalpies of Be and Mg positive ?
positive, more negative or less negative than the first ? Justify Sol. They have fully filled s-orbitals and hence have no tendency
your answer. (NCERT) to accept an additional electron. Consequently, energy has
Sol. The second electron gain enthalpy of O is positive as to be supplied if an extra electron has to be added to the
explained below : much higher energy p-orbitals of the valence shell. That is
When an electron is added to O atom to form O– ion, energy why electron gain enthalpies of Be and Mg are positive.
is released. Thus, first electron gain enthalpy of O is negative. Example-23
Distinguish between electronegativity and electron affinity.
O g   e–g   O–g  ;  eg H  141kJ mol1
Sol. Electron affinity Electronegativity
But when another electron is added to O to form O ion,
– 2–
1. Electron gain enthalpy or 1.Electronegativity of an
energy is absorbed to overcome the strong electrostatic
electron affinity is defined atom in a molecule is
repulsion between the negatively charged O– ion and the
second electron being added. Thus, the second electron as the amount of energy defined as the tendency
gain enthalpy of oxygen is positive. released when neutral of an atom to attract
gaseous atom, accepts an towards itself the shared
O–g   eg   O2g ; eg H  780 kJ mol 1
electron to form an anion. pair of electrons.
Example-21 2. It is expressed in eV/atom 2.It does not have any unit
‘Electron affinity of fluorine is less than that of chlorine’. (electron volt per atom) or (it is a number)
Explain.
in kil number) joules per mol (kJ mol–1).
Sol. (i) Def : Electron gain enthalpy or electron affinity is defined 3. It is a kind of absolute 3.It is a relative term (atoms
as the amount of energy released, when neutral gaseous
property of the elements. are compared with fluorine,
atom, accepts an electron to form an anion.
whose assigned value of
(ii) The electronic cofiguration of fluorine is 1s22s22p5, while
electronegativity is 4.0
that of chlorine, it is 1s22s22p63s23p5. In both the elements
there are 7 electrons in their outermost shell. The size of F- 4. Electron affinity value is 4. It is measured when the
atom is smaller than Cl-atom. measured when the atoms atoms are in their
(iii) In fluorine, 2p-orbitals are compact and closer to the are in their gaseous state. combined state (in state
nucleus. Thus, the screening effect is very low. Hence there molecules).
is electron-electron repulsion in the valence shell. Thus,
Example-24
when an electron is added to the p-orbital of a fluorine.
Thus, when an electron is added to the p-orbital of a fluorine Use the periodic table to answer the following questions:
it experiences less attraction and hence less energy is (a) Identify an element with five electrons in the outer
liberated to form fluoride ion. subshell.

(iv) In chlorine, the orbital accepting an electron to form (b)Identify the element that would tend to lose two electrons.
chloride ion is 3p-orbital, which is away form the nucleus. (c) Identify the element that would tend to gain two electons.

(v) Therefore, the electron-electron repulsion is less and (d) Identify the group having metal, non-metal, liquid as
more energy is liberated, when an electron is added to a well as gas at room temperature. (NCERT)
chlorine atom forming a chloride anion. Thus, fluorine has Sol. (a) The genral electronic configuration of the elements
less electron affinity than chlorine. having five electrons in the outer subshell is ns2 np5. This
electronic configuration is characteristic of elements of group
17, i.e., halogens and their examples are F, Cl, Br, I, At, etc.
48 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

(b) The elements which have a tendency to lose two electrons Example-25
must have two electrons in the valence shell. Therefore, their Halogens except Flourine shows positive oxidation state of
general electronic configuration should be n s2. This electronic +1, +2, +3, +5, and +7.
configuration is characteristic of group 2 elements, i.e., alkaline
Sol. (i) The outer Electronic configuration of halogens are ns2,
earth metals and their examples are Mg, Ca, Sr, Ba, etc.
np5 they can gain one electron and show a common
(c) The elements which have a tendency to accept two electrons oxidation state of –1.
must have six electrons in the valence shell. Therefore, their
(ii) The other halogen exhibit higher oxidation state as, +1,
general electronic configuration is ns2 np4. This electronic
+2, +3, +5, and +7 due to vacant d-orbitals in their shell.
configuration is characteristic of group 16 elements and their
examples are O and S. (iii) Since Flourine does not have d-orbital, it only exhibits
only ‘–1’ oxidation state.
(d) A metal which is liquid at room temperature is mercury. It is
a transition metal and belongs to group 12. A non-metal Therefore Halogens except fluorine shows positive oxidation
which is a gas at room temperature is hydrogen (group 1), state +1, +2, +3, +5 and +7.
nitrogen (group 15), oxygen (group 16), fluorine, chlorine
(group 17) and inert gases (group 18).
A non-metal which is a liquid at room temperature is bromine
(group 17).
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 49

Exercise – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Who was the first scientist to classify elements 14. In short give the features of the seven periods.
according to their properties?
15. How does the reactivity of non–metals changes in a
2. Define and state Mendeleev’s periodic law.
period and group?
3. What is the basis of triad formation of elements?
16. Give the properties of the oxides in a particular period.
4. Define electronic configuration.
17. Between fluorine and chlorine which one of these has
5. How does atomic size change in a group? low electron gain enthalpy? Give reason.
6. Name the metal which floats on water.
18. What are metalloids? Give an example.
7. Predict the position of the element in the periodic table
satisfying the electronic configuration (n – 1) d1 ns2 for 19. Why is ionisation enthalpy of oxygen less than that of
n = 4. nitrogen?

8. What is the electronic configuration when elements are 20. Among alkali metals which element do you expect to
classified group wise?
be least electronegative and why?
9. Why Li and Mg show resemblance in chemical
behaviour? Section – C (3 Marks Questions)
10. The size of an atom can be expressed by three radii.
Name them. Which of these have the highest, and the 21. How does electronegativity and non – metallic
lowest value of the atomic radius of an element? character related to each other?

Section – B (2 Marks Questions) 22. How does electronegativity vary along a period and
vary down a group? Give reason.

11. Justify the given statement with suitable examples “the 23. (a) How many periods are there in the Modern Periodic
properties of the elements are a periodic function of Table of elements?
their atomic numbers”. (b) How do atomic radius, valency and metallic
character vary down a group?
12. Write the name and deduce the atomic number of the (c) How do the atomic size and metallic character of
following element elements vary as we move from left to right in a period?
i. The second alkali metal
ii. The fourth noble gas 24. Give reason for the anomalous properties of 2nd period
iii. The third halogen elements when compared to the elements in their
iv. The first transition element respective groups.

13. Explain classification of elements based on Newland’s 25. How does electron gain enthalpy vary across a period
law of Octaves. Give its limitations. and vary down the group? Give reason
50 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

26. What is the difference between an amphoteric oxide (b) Which of the following will have the most negative
and a neutral oxide? electron gain enthalpy and which the least negative?
P, S, Cl, F
27. Explain why atomic size decreases along a period? Explain your answer.

28. Which group of elements have maximum value for

30. (a) All transition elements are d-block elements, but all
ionisation enthalpy? Give reason.
d-block elements are not transition elements. Explain.
(b) What would be the atomic number of the next
Section – D (5 Marks Questions) (i) alkali metal
(ii) halogen and
29. (a) The amount of energy released when one million of (iii) noble gas
atoms of iodine in vapour state are converted to I – ions if discovered in future.
(c) Arrange the ions: Li2+, He+ and Be3+ in the
is 4.9  10–13 J according to the reaction:
increasing order of their ionic radii.
I ( g ) + e − ⎯⎯
→ I− ( g )
Express the electron gain enthalpy of iodine in terms of
kJ mol–1 and eV per atom.
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 51

Exercise – 2: Basic Objective Questions

Section-A (Single Choice Questions)

1. Atomic number to IUPAC nomenclature, the name of 10. Which of the following has the highest ionization
element having atomic number 101 is enthalpy?
(a) Unnilbium (b) Unnilunium (a) P (b) N
(c) Unnilquadium (d) None of these (c) As (d) Sb

2. Group I elements of modern periodic table are called 11. The correct order of first ionization potential is
(a) alkali metals (b) alkaline earth metals (a) K > Na > Li (b) Be > Mg >Ca
(c) halogens (d) Both (a) and (b) (c) B > C > N (d) Ge> Si > C

3. Metals are good conductors of heat and….A….Here, 12. An atom of an electronegative element becomes an ion
‘A’ refers to by
(a) energy (b) electricity
(a) Gain of electrons (b) Loss of electrons
(c) Both (a) and (b) (d) None of these
(c) Loss of its radius (d) Serving as a reductant
4. Which one is the correct order of the size of the iodine
species. 13. Second electron gain enthalpy
(a) I > I+> I– (b) I > I–> I+ (a) is always negative
–
+
(c) I > I > I (d) I–> I > I+ (b) is always positive
(c) can be positive or negative
5. Which of the following ions has the smallest radius? (d) is always zero
(a) Li+ (b) Na+
(c) Be 2+
(d) K+ 14. Electron gainenthalpy is maximum for
(a) Cl (b) F
6. Atomic radii of F and Ne, in Å, are given by (c) Br (d) I
(a) 0.72, 0.71 (b) 0.72, 1.6
(c) 1.6, 1.58 (d) 0.71, 0.72 15. Electron affinity of the inert gases is
(a) High (b) Low but positive
7. Arrange the following elements in the order of (c) Moderate (d) Almost zero
increasing atomic size Cl, S, P, Ar
(a) Ar, Cl, S, P (b) Cl, S, P, Ar 16. The most electropositive element is
(c) S, Cl, P, Ar (d) Ar, P, S, Cl (a) Cs (b) Ga
(c) Li (d) Pb
8. Lowest ionization potential in periods is shown by
(a) inert gases (b) halogens 17. With respect to chlorine, hydrogen will be
(c) alkali metals (d) alkaline earth metals (a) Electropositive (b) Electronegative
(c) Neutral (d) None of these
9. Which element has the highest ionization energy?
(a) Hydrogen (b) Lithium 18. Which is the property of non-metal?
(c) Boron (d) Sodium (a) Electronegative
(b) Basic nature of oxide
(c) Reducing property
(d) Low ionization potential
52 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

19. In Periodic Table, the basic character of oxides 23. Assertion (A): Nitrogen has higher ionisation enthalpy
(a) increases from left to right and decrease from top of than oxygen.
bottom Reason (R): In a period, ongoing from left to right,
(b)decreases from right to left and increases from top to
ionisation enthalpy increases.
bottom
(c) decreases from left to right and increases from top to (a) Both Assertion and Reason are true, and the Reason
bottom is the correct explanation of the Assertion.
(d) decreases from left to right and increases from (b) Both Assertion and Reason are true, but the Reason
bottom to top is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
20. Among halogens, the correct order of amount of energy (d) Both Assertion and Reason are false.
released in electron gain (electron gain enthalpy) is:
(a) F >Cl> Br > I
24. Assertion (A): Electron gain enthalpy of oxygen is
(b) F <Cl<Br < I
more negative than that of nitrogen, but less negative
(c) F<Cl> Br > I
than fluorine.
(d) F<Cl< Br < I
Reason (R): N, O and F belong to the second period of
the periodic table.
Section - B (Assertion & Reason Type Questions) (a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion.
21. Assertion (A): Ionisation enthalpy of K is numerically (b) Both Assertion and Reason are true, but the Reason
the same as electron gain enthalpy of K+. is not the correct explanation of the Assertion.
Reason (R): Ionisation enthalpy and electron gain (c) Assertion is true, but Reason is false.
enthalpy both depend on screening effect. (d) Both Assertion and Reason are false.
(a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion. 25. Assertion (A): Metallic character is the highest at the
(b) Both Assertion and Reason are true, but the Reason extreme left side of the periodic table.
is not the correct explanation of the Assertion. Reason (R): Ionisation enthalpy increases across a
(c) Assertion is true, but Reason is false. period.
(a) Both Assertion and Reason are true, and the Reason
(d) Both Assertion and Reason are false.
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason
22. Assertion (A): Atoms which have loosely bound
is not the correct explanation of the Assertion.
valence electrons have high value of ionisation
(c) Assertion is true, but Reason is false.
potential.
(d) Both Assertion and Reason are false.
Reason (R): The energy required to gain an electron
from an isolated gaseous atom, is known as ionisation
26. Assertion (A): Among isoelectronic species, the cation
potential.
with the greater positive charge will have a smaller
(a) Both Assertion and Reason are true, and the Reason
radius.
is the correct explanation of the Assertion.
Reason (R): Greater is the attraction of the electrons to
(b) Both Assertion and Reason are true, but the Reason
the nucleus, smaller is the size of atom/ion.
is not the correct explanation of the Assertion.
(a) Both Assertion and Reason are true, and the Reason
(c) Assertion is true, but Reason is false.
is the correct explanation of the Assertion.
(d) Both Assertion and Reason are false.
(b) Both Assertion and Reason are true, but the Reason
is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false.
CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 53

27. Assertion (A): Atomic number of the element 30. In any period the valency of an element with respect to
Ununtrium is 113. oxygen
Reason (R): According to IUPAC systematic (a) Increases one by one from IA to VIIA
nomenclature, the numerical roots for 1,1 and 3 are un, (b) Decreases one by one form IA to VIIA
un and tri, respectively. (c) Increases one by one from IA to IVA and then
(a) Both Assertion and Reason are true, and the Reason decreases from VA to VIIA one by one
is the correct explanation of the Assertion. (d) Decreases one by one from IA to IVA and then
(b) Both Assertion and Reason are true, but the Reason increases from VA to VIIA one by one
is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false. 31. The order of increasing sizes of atomic radii among the
(d) Both Assertion and Reason are false. elements O, S, Se and As is :
(a) As < S < O < Se (b) Se < S < As < O
28. Assertion (A): Fluorine is the most electronegative (c) O < S < As < Se (d) O < S < Se < As
element.
Reason (R): Fluorine is stable as it has extra stable 32. Which group of the periodic table contains coinage
half-filled electronic configuration. metal ?
(a) Both Assertion and Reason are true, and the Reason (a) IIA (b) IB
is the correct explanation of the Assertion. (c) IA (d) None of these
(b) Both Assertion and Reason are true, but the Reason
is not the correct explanation of the Assertion. Case Study – 2
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false. Metals comprise more than seventy eight per cent of the
known elements. Nonmetals, which are located at the
Section – C (Case Study Questions) right side of the periodic table, are less than twenty in
number. Elements which lie at the border line between
metals and non-metals (e.g., Si, Ge, As) are called
Case Study - 1
metalloids or semi-metals. Metallic character increases
Periodic trends are observed in atomic sizes, ionization with increasing atomic number in a group whereas
enthalpies, electron gain enthalpies, electronegativity decreases from left to right in a period. The physical
and valence. The atomic radii decrease while going and chemical properties of elements vary periodically
from left to right in a period and increase with atomic with their atomic numbers.
number in a group. Ionization enthalpies generally
increase across a period and decrease down a group. 33. The only non-metal which is liquid at ordinary
Electronegativity also shows a similar trend. Electron temperature is
gain enthalpies, in general, become more negative (a) Hg (b) Br2
across a period and less negative down a group. There (c) NH3 (d) None of these
is some periodicity in valency, for example, among
representative elements, the valency is either equal to 34. Which of the following is non-metallic ?
the number of electrons in the outermost orbitals or (a) B (b) Be
eight minus this number. (c) Mg (d) Al

29. Periodic classification of elements is used to examine 35. Alkali metals are powerful reducing agents because
the (a) these are metals
(a) periodic trends in physical properties of elements (b) their ionic radii are large
(b) periodic trends in chemical properties of elements (c) these are monovalent
(c) Both (a) and (b) (d) their ionisation potential is low
(d) None of the above
54 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

36. Which of the following metals requires the radiation of 38. All the members in a group in long form of periodic
highest frequency to cause the emission of electrons? table have the same
(a) Na (b) Mg (a) valence
(c) K (d) Ca (b) number of valence electrons
(c) chemical properties
Case Study - 3 (d) All of the above

Mendeleev’s Periodic Table was based on atomic 39. Which of the scientists given below discovered that
masses. Modern Periodic Table arranges the elements periodic table should be based on the atomic number?
in the order of their atomic numbers in seven horizontal (a) Mendeleev (b) Newlands
rows (periods) and eighteen vertical columns (groups or (c) Moseley (d) Lothar Meyer
families). Atomic numbers in a period are consecutive,
whereas in a group they increase in a pattern. Elements 40. How many elements are there in 6th period of periodic
of the same group have similar valence shell electronic table?
configuration and, therefore, exhibit similar chemical (a) 18 (b) 8
properties. (c) 30 (d) 32

37. Mendeleev classified elements in

(a) increasing order of atomic groups
(b) eight periods and eight groups
(c) seven periods and nine groups
(d) eight periods and seven groups

Find Answer Key and Detailed Solutions at the end of this book

CLASSIFICATION OF ELEMENTS AND

PERIODICITY IN PROPERTIES
CHEMICAL BONDING AND
MOLECULAR STRUCTURE
56 CHEMICAL BONDING AND MOLECULAR STRUCTURE

Chapter at a Glance
• Kossel’s first insight into the mechanism of formation of • The valence bond (VB) approach to covalent bonding is
electropositive and electronegative ions related the process basically concerned with the energetics of covalent bond
to the attainment of noble gas configurations by the respective formation about which the Lewis and VSEPR models are silent.
ions. Electrostatic attraction between ions is the cause for Basically the VB theory discusses bond formation in terms
their stability. This gives the concept of electrovalency. of overlap of orbitals.
• The first description of covalent bonding was provided by • For explaining the characteristic shapes of polyatomic
Lewis in terms of the sharing of electron pairs between atoms molecules Pauling introduced the concept of hybridisation
and he related the process to the attainment of noble gas of atomic orbitals. sp, sp2, sp3 hybridizations of atomic
configurations by reacting atoms as a result of sharing of orbitals of Be, B, C, N and O are used to explain the formation
electrons. The Lewis dot symbols show the number of valence and geometrical shapes of molecules like BeCl2, BCl3, CH4,
electrons of the atoms of a given element and Lewis dot NH3 and H2O. They also explain the formation of multiple
structures show pictorial representations of bonding in bonds in molecules like C2H2 and C2H4.
molecules. • To predict hybridisation following furmula may be used:
• An ionic compound is pictured as a three-dimensional
aggregation of positive and negative ions in an ordered 1
No. of hybrid orbital (X) = [Total no. of valence e– in the
arrangement called the crystal lattice. In a crystalline solid 2
there is a charge balance between the positive and negative central atom + total no. of monovalent atoms – charge on
ions. The crystal lattice is stabilized by the enthalpy of lattice cation + charge on anion]
formation. Value of X 2 3 4 5 6 7
• While a single covalent bond is formed by sharing of an Type of sp sp2 sp3 sp3d sp3d2 sp3d3
electron pair between two atoms, multiple bonds result from hybridisation
the sharing of two or three electron pairs. Some bonded atoms
• The molecular orbital (MO) theory describes bonding in
have additional pairs of electrons not involved in bonding.
terms of the combination and arrangement of atomic orbitals
These are called lone pairs of electrons. A Lewis dot structure
to form molecular orbitals that are associated with the molecule
shows the arrangement of bonded pairs and lone pairs around
as a whole. The number of molecular orbitals are always equal
each atom in a molecule. Important parameters, associated
to the number of atomic orbitals from which they are formed.
with chemical bonds, like: bond length, bond angle, bond
enthalpy, bond order and bond polarity have significant effect • The electronic configuration of the molecules is written by
on the properties of compounds. filling electrons in the molecular orbitals in the order of
increasing energy levels. As in the case of atoms, the Pauli
• There is a very important and extremely useful concept called
exclusion principle and Hund’s rule are applicable for
resonance. The contributing structures or canonical forms
the filling of molecular orbitals. Molecules are said to be stable
taken together constitute the resonance hybrid which
if the number of elctrons in bonding molecular orbitals is
represents the molecule or ion.
greater than that in antibonding molecular orbitals.
• The VSEPR model used for predicting the geometrical
shapes of molecules is based on the assumption that electron 1
• Bond order = [No. of e– in bonding orbitals –
pairs repel each other and, therefore, tend to remain as far 2
apart as possible. According to this model, molecular No. of e– in antibonding orbitals.]
geometry is determined by repulsions between lone pairs
and lone pairs; lone pairs and bonding pairs and bonding • Hydrogen bond is formed when a hydrogen atom finds itself
pairs and bonding pairs. The order of these repulsions being: between two highly electronegative atoms such as F, O and
lp-lp > lp-bp > bp-bp N. It may be intermolecular (existing between two or more
molecules of the same or different substances) or
intramolecular (present within the same molecule).
CHEMICAL BONDING AND MOLECULAR STRUCTURE 57

Solved Examples
Example-1 Example-4
t Explain the formation of a chemical bond. (NCERT) Write the favourable factors for the formation of ionic bond.

Sol. According to Kossel and Lewis, atoms combine together (NCERT)

in order to complete their respective octets as to acquire Sol. 1. Low ionisation enthalpy of metal atoms
the stable inert gas configuration. This can occur in two 2. High electron gain enthalpy of non-metal atoms
ways, by transfer of one or more electrons from one atom to 3. High lattice enthalpy of compound formed.
other by sharing of electrons between two or more atoms. Example-5
Example-2 What type of bond is formed when atoms have
Write Lewis dot symbols for atoms of the following (i) Zero difference of electronegativity.
elements: Mg, Na, B, O, N, Br. (NCERT) (ii) Little difference of electronegativity.
Sol. (iii) High difference of electronegativity.
Sol. (i) Non-polar covalent
(ii) Polar covalent
(iii) Electro-valent.
Example-6
How do you express the bond strength in terms of bond
order? (NCERT)
Sol. Bond strength is directly proportional to the bond order.
Greater the bond order, more is the bond strength.
Example-7

Example-3 Write the resonance structures for SO3, NO2 and NO3–.

Write Lewis dot structures of the following molecules/ions: Sol.

(i) CO (ii) HCN

Sol. V = number of valence electron of molecule
R = Number of electron required for octet in molecule
S = Number of shared electron ( S = R – V)
U = Number of unshared electron (U = V – S)
(i) CO
V = 4 + 6 = 10 electrons
R = 8 + 8 = 16 electrons
S = R – V = 6 electrons
U = 10 – 6 = 4 electrons
C O or C O
(ii) HCN
V = 1 + 4 + 5 = 10 electrons
R = 2 + 8 + 8 = 18 electrons Example-8
S = 8 electrons Which out of NH3 and NF3 has higher dipole moment and
why? (NCERT)
U = 2 electrons
Sol. In both molecules i.e., NH3 and NF3 the central atom (N) has
H C N or H – C N a lone pair electron and there are three bond pairs. Hence,
both molecules have a pyramidal shape. Since fluorine is
58 CHEMICAL BONDING AND MOLECULAR STRUCTURE

more electronegative than hydrogen, it is expected that the Example-12

net dipole moment of NF3 is greater than NH3. However, the Discuss the shape of the following molecules using the
net dipole moment of NH3(1.46 D) is greater than that of NF3 VSEPR model.
(0.24 D). BeCl2, BCl3, SiCl4, AsF5, H2S, PH3. (NCERT)
The resultant moment of N–H bonds add up to the bond
Sol. BeCl2 : A linear molecule. Be atom has 2 electrons in it
moment of the lone pair (the two being in the same direction),
outermost orbit. Each chlorine atom has seven valence
whereas that of the three N–F bonds partly cancels the
electrons. The Lewis structure of BeCl2 is
moment of the lone pair.
Hence, the net dipole moment of NF3 is less than that of
NH3. There are two electrons pairs and to minimise the repulsion,
Example-9 these electron pairs tend to keep themselves far away from
Which of the following compounds has the largest dipole each other, i.e., 180º apart. This gives BeCl2 a linear structure.
moment ? BCl3 : In BCl3 molecule, the three bond pairs of electrons
(i) CH3OH (ii) CH4 (iii) CF4 (iv) CO2 (v) CH3F. are located around B in a triangular arrangement. Thus, the
molecule BCl3 has a triangular planar geometry.
Sol. CH4 and CF4 have tetrahedral structure and are symmetrical,
hence their dipole moment is zero. CO2 is linear and hence
its dipole moment is also zero. Of the remaining CH3OH and
CH3F, since F is more electronegative than O, CH3F will
have largest dipole moment.
Example-10
Apart from tetrahedral geometry, another possible SiCl4 : A tetrahedral molecule. Si has 4 electrons in its
geometry for CH4 is square planar with the four H atoms at outermost shell. Due to mutual sharing of electrons with Cl
the corners of the square and the C atom at its centre. there are 4 electron pairs around Si. To keep the repulsion at
Explain why CH4 is not square planar? (NCERT) the minimum, these 4 electron pairs should be arranged in a
Sol. According to VSEPR theory, if CH4 which were square tetrahedral manner around Si. Thus, SiCl4 is a tetrahedral
planar, the bond angle would be 90º. For tetrahedral molecule.
structure, the bond angle is 109º28’. Therefore, in square AsF5 : Trigonal bipyramidal molecule : As has five electrons
planar structure, repulsion between bond pairs would be in its outermost orbit. Due to sharing of 5 electrons from 5 F-
more and thus the stability will be less. atoms, there are in all 5 electron pairs. These are distributed
Example-11 in space to form a trigonal bypyramid.
What do you understand by bond pairs and lone pairs of H2S : Bent (V-shaped) structure : S has 6 electrons its
electrons? Illustrate by giving one example of each type. outermost shell. 2H-atoms contribute 2 electrons during
Sol. The electron pair involved in sharing between two atoms bonding. Thus, there are 8 electrons or 4 electron pairs
during covalent bonding is called shared pair or bond pair. around S. This gives a tetrahedral distribution of electron
At the same time, the electron pair which is not involved in pairs around S. The two corners of the tetrahedron are
sharing is called lone pair of electrons. occupied by H-atoms and the other two by the lone-pairs of
electrons. Thus, H2S has a bent structure.

For example, In CH 4 there are only 4 bond

PH3 : Trigonal pyramidal : Phosphorus atom has 5 electrons

pairs, but in H 2 O there are two bond pairs and in its outermost orbit. H-atoms contribute one electron each
to make in all 8 electrons around P-atom. Thus, 4 pairs of
two lone pairs. electrons would be distributed in a tetrahedral fashion
CHEMICAL BONDING AND MOLECULAR STRUCTURE 59

around the central atom. Three pairs form three P-H bonds Example-17
while the fourth pair remains unused. Due to repulsion
Out of p-orbital and sp-hybrid orbital which has greater
between the bonding and lone pairs of electrons, the angle.
directional character and why?
Example-13
Sol. sp-orbital has greater directional character than p-orbital.
Interpret the non-linear shape of H2S molecule and non This is because p-orbital has equal sized lobes with equal
planar shape of PCl3 using valence shell electron pair electron density in both the lobes wheres sp-hybrid orbital
repulsion (VSEPR) theory. has greater electron density on one side.
Sol. In H2S, two bonded pairs and two lone pairs of electrons
Example-18
are present. Tetrahedral configuration comes into existence.
Two of the positions are occupied by lone pairs, hence the Indicate the number of  and -bonds in the following
actual structure is bent or V-shaped. molecules.
In PCl3, three bonded pairs and one lone pair of electrons (i) CH 3  CH  CH 2
are present. Tetrahedral configuration comes into existence.
(ii) CH 3  CH 2  CH 2  CH 3
One position is occupied by lone pair, hence the actual
structure is pyramidal. (iii) CH 3  C  C  CH 3
Example-14
(iv)
What is the total number of sigma and pi bonds in the
following molecules? Sol. For finding out the number of  and -bonds in a molecule,
the following points should be kept in mind:
(a) C2H2 (b) C2H4 (NCERT)
1. All single bonds are sigma bonds.
Sol. (a) H  C  C  H 2. All double bonds consist of one  and two -bonds.
 bond = 3, bond = 2 3. All triple bonds consist of one  and two -bonds.
(i) This molecule consists of seven single bonds and
one double bond. Thus, number of -bonds = 8 and
number of -bonds = 1.
(ii) This molecule consists of 13 single bonds.
Thus, number of -bonds = 13.
(iii) This molecule consists of 8 single bonds and one
Example-15 triple bond. Thus, number of -bonds = 9 and
Explain how V.B theory differs from the Lewis concept. number of -bonds = 2.
Sol. The Lewis concept of describes the formation of bond in (iv) The molecule of benzene consists of nine single
terms of sharing of one or more electron pairs and the octet bonds and three double bonds.
rule. It does not explain the energetics of the bond formation Thus, number of -bonds = 12 and number of -bonds
and shapes of the polyatomic molecules. = 3.
The VB theory describes the bond formation in terms of Example-19
hybridization and overlap of the orbitals. The overlap of Is there any change in the hybridization of B and N atoms as
orbitals along the intermolecular axis increases the electron-
a result of the following reaction?
density between the two nuclei resulting in a decrease in
the energy and formation of a bond. BF3 + NH3 F3B NH3. (NCERT)
Example-16 Sol. Here, B atom in BF3 is sp hybridized and one of its p orbital
2

is empty. N in NH3 is sp3 hybridized and are of its hybrid

Out of  and –bonds which one is stronger and why? orbitals is occupied by a lone-pair of electrons. During the
Sol.  –bond is stronger. This is because  –bond is formed by reaction a coordinate bond is formed due to one-side
head-on overlapping of atomic orbitals and, therefore, the sharing of electron pair.
overlapping is large. –bond is formed by sideway BF3 + NH3 [F3B  : NH3]
overlapping which is small. There is no change in the hybridization of any of the two
atoms in this reaction.
60 CHEMICAL BONDING AND MOLECULAR STRUCTURE

Example-20 than that on H2. This is because 2s orbital of Li (involved in

Which hybrid orbitals are used by carbon atoms in the the bonding) is much larger than 1s orbital of H.
following molecules? Example-23
(a) CH3 – CH3 (b) CH3 – CH = CH2 (c) CH3 – CH2 – OH Compare the relative stabilities of O2– and N2+ and comment
(NCERT) on their magnetic (paramagnetic or diamagnetic) behaviour.
Sol. Sol. M.O. Electronic configuration of
O2  KK 22s *2
2s  2pz 2p x  2p x 2p x 2p y
2 2 2 *2 *1

1 3
Bond order = (8  5)   1.5
2 2
M.O. Electronic configuration of
N 2  KK  2s
2
*2
2s  2p x 2p y  2pz
2 2 1

1 5
Bond order = (7  2)   2.5
2 2
As bond order of N 2  bond order of O 2 , therefore, N 2 is
more stable than O 2 .
Example-21
Each of them contains unpaired electron, hence both are
Using molecular orbital theory to explain why the molecule
paramagnetic.
Be2 does not exist. (NCERT)
Example-24
Sol. E.C. of Be = 1s2 2s 2
Which of the two, peroxide ion or superoxide ion, has a
M.O.E.C. of Be2  1s  2  *1s 2  2s 2  *2s 2 larger bond length ?
Sol. The bond length in a molecule depends on bond order. the
1
Bond order = (4 – 4) higher the bond order, smaller will be the bond length.
2
= 0 Peroxide ion, O 22

Hence, Be 2 does not exist Configuration KK (2s)2 (2s)2  (2pz)2 (2px)2

Example-22 (2py)2 (2px)2 (2py)2

Does Li2 exist ? If so, estimate its bond order and compare 86
Bond order = 1
its bond dissociation energy with that of H2. 2
Sol. Li has a configuration : 1s2, 2s1. Superoxide ion, O 2
There are two s orbitals (1s and 2s) on each atom. These Configuration KK (2s)2 (2s)2 (2pz)2 (2px)2
combine to give four MO’s. These are
(2py)2 (2px)2 (2py)1
(1s)  (1s)  1s   *1s
85
(2s)  (2s)  2s   * 2s. Bond order =  1.5
2
Thus all the six electrons are accommodated in these four Bond order of superoxide is higher than peroxide ion,
orbitals. The electronic configuration for Li 2 is hence bond length of peroxide ion is larger.
(1s) 2 ( *1s) 2 ( 2s) 2 . Example-25
Nb = 4, Na = 2. Explain why HF is less viscous than H2O.
Sol. There is greater intermolecular hydrogen bonding in H2O
42 than that in HF as each H2O molecule forms four H-bonds
Bond order in Li 2   1.
2 with other water molecules whereas HF forms only two H-
Therefore, Li 2 should be a stable species. Its bond bonds with other HF molecules. Greater the intermolecular
dissociation energy is 105 kJ mol –1 , as compared to H-bonding, greater is the viscosity. Hence, HF is less
431 kJ mol–1 for H2. Thus, the bond in Li2 is much weaker viscous than H2O.
CHEMICAL BONDING AND MOLECULAR STRUCTURE 61

EXERCISE-1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. The given statement is True or False: 13. A molecule of PCl5 exists while that of NCl5 does not.
The shape of SF6 molecule is octahedral whereas that of
IF7 is square pyramidal. 14. What is difference between bond order and bond length
and state relation between them?
2. Why is NaCl a bad conductor of electricity in the solid
state? 15. Identify the compound/compounds which does not obey
octet rule:
3. Out of HF, HCl, HBr and HI, the lowest boiling point LiCl, SF6, H2SO4.
is of ________ and highest boiling point is of
________. 16. Which of the following has maximum bond angle?
H2O, CO2, NH3, CH4
4. The bond in peroxide ion is _________ then in
superoxide ion.(Weaker /Stronger) 17. List two conditions for forming hydrogen bonds.

5. In N2 molecule energy of 2px and 2py are ________ 18. Write down the M.O. configuration of N2 and calculate
then 2pz. (Lower/Higher) its Bond order.

6. N2 molecule is __________. 19. Why does type of overlap given in the following figure
(Paramagnetic/Diamagnetic) not result in bond formation?

7. Give one example each for a compound with

(a) An ionic bond
(b) A covalent bond

8. What is the SI unit of dipole moment?

20. AlF3 is a high melting solid whereas SiF4 is a gas.

9. Give two examples of molecules containing coordinate
Explain why?
bonding.

10. Name one compound each involving sp3, sp2, sp Section – C (3 Marks Questions)
hybridization.
21. Define octet rule. Write its significance and limitations.
Section – B (2 Marks Questions)
22. What is an ionic bond? With two suitable examples and
11. Write the significance of plus and minus sign shown in explain the difference between an ionic and a covalent
representing the orbitals. bond?

12. Define electronegativity. How does it differ from 23. Although geometries of NH3 and H2O molecules are
electron gain enthalpy? distorted tetrahedral, bond angle in water is less than
that of ammonia. Discuss.
62 CHEMICAL BONDING AND MOLECULAR STRUCTURE

24. Use Lewis symbols to show electron transfer between Section – D (5 Marks Questions)
the following atoms to form cations and anions:
(a) K and S (b) Ca and O (c) Al and N. 29. (i) Which one has high boiling point and why? Ethyl
alcohol or dimethyl ether.
25. BeF2 and H2O are both tri-atomic molecules but have (ii) XeF2 molecule is sp3d hybridized, but is linear in
different shapes. Discuss. shape. Why?

26. Explain the structure of CO32– ion in terms of 30. (i)Describe the hybridisation in case of PCl5 . Why are
resonance. the axial bonds longer as compared to equatorial bonds?
(ii) s – orbital does not show any preference for
27. What is meant by hybridization of atomic orbitals? direction.
Describe the shapes of sp, sp2 hybrid orbitals.

28. Compare the relative stability of the following species

and indicate their magnetic properties: O2 ,O+2 ,O−2 .
CHEMICAL BONDING AND MOLECULAR STRUCTURE 63

EXERCISE-2: Basic Objective Questions

Section – A (Single Choice Questions)

1. Kossel and Lewis approach was based on the 8. Which of the following cannot be used to measure
(a) Reactivity of elements bond lengths?
(b) Inertness of noble gases (a) Spectroscopy
(c) Reactivity of metals (b) X-ray diffraction
(d) Inertness of non-metals (c) Electron diffraction
(d) Young’s Double-slit method
2. Which of the following compound is covalent?
(a) H2 (b) CaO 9. The amount of energy required to break one mole of
(c) KCl (d) Na2S bonds of a particular type between two atoms in a
gaseous state is called
3. Calculate the formal charge of C in CH4. (a) Bond enthalpy (b) Bond angle
(a) 4 (b) 1 (c) Bond order (d) None of these
(c) –4 (d) 0
10. If the bond dissociation enthalpy of O2, N2 and H2 are
4. Which of the above statement(s) play(s) an important 498 kJ mol-1, 946 kJ mol-1 and 435.8 kJ mol-1
role in the formation of ionic compounds as per kossel respectively. Choose the correct order of decreasing
& Lewis ? The formation of Ionic Compounds would bond strength.
primarily depend on: (a) H2> N2> O2 (b) N2> O2> H2
I. The ease of formation of the positive and negative (c) O2> H2> N2 (d) H2> O2> N2
ions from the respective neutral atoms
II. The arrangement of the positive and negative ions in 11.
the solid that is the lattice of the crystalline compound.
(a) Only I (b) Only II
(c) Both I and II (d) None of these

5. Ionic bonds will be formed more easily between

elements with comparatively.
(a) Low ionization enthalpy and high electron affinity
(b) High ionization enthalpy and high electron affinity
(c) Low ionization enthalpy and low electron affinity
(d) High ionization enthalpy and low electron affinity In the given resonating structures, which is/are called
resonance hybrid?
6. Lattice energy of an ionic compound depends upon (a) Structure I (b) Structure II
(a) Charge on the ion and size of the ion (c) Structure III (d) Structure I and II
(b) Packing of ions only
(c) Size of the ion only 12. Dipole moment is usually designated by a Greek letter
(d) Charge on the ion only ‘μ ‘
µ=Q×r
7. Among the following the maximum covalent character Here Q and r represents
is shown by the compound (a) Q = charge, r= distance of separation
(a) FeCl2 (b) SnCl2 (b) Q = heat, r= radius
(c) AlCl3 (d) MgCl2 (c) Q = charge, r= radius of cations
(d) Q = charge, r= radius of anions
64 CHEMICAL BONDING AND MOLECULAR STRUCTURE

13. Which of the following molecule has dipole moment 20. The correct order of the strength of H-bonds is:
zero? (a) H....F > H....O > H....N (b) H....N > H....O > H....F
(a) HF (b) H2O (c) H....O > H....N > H....F (b) H....F > H....N > H....O
(c) BF3 (d) CHCl3
Section–B (Assertion & Reason Type Questions)
14. Which of the following molecules have same molecular
shape?
21. Assertion: The bond order of helium is zero.
I. CH4, II BF3 III. NH4+, IV. SF4
Reason: The number of electrons in bonding
(a) I and II (b) III and IV
molecular orbital and antibonding molecular orbital are
(c) I and III (d) I, III and IV
equal.
15. In which of the following structures all the bonds are (a) Assertion is correct, reason is correct; reason is a
present at 90o? correct explanation for assertion.
(a) Trigonal planar (b) Tetrahedral (b) Assertion is correct, reason is correct; reason is not
(c) Trigonal bipyramidal (d) Octahedral a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
16. Which of the molecules has trigonal bipyramidal (d) Assertion is incorrect, reason is correct.
geometry with bond angles 120o and 90o?
(a) SF6 (b) PCl5 22. Assertion: The lesser the lattice enthalpy more stable
(c) CH4 (d) BF3 is the ionic compound.
Reason: The lattice enthalpy is greater, for ions of
17. Correct statement regarding molecules SF4, CF4 and highest charge and smaller radii.
XeF4 are: (a) Assertion is correct, reason is correct; reason is a
(a) 2, 0 and 1 lone pairs of central atom respectively correct explanation for assertion.
(b) 1, 0 and 1 lone pairs of central atom respectively (b) Assertion is correct, reason is correct; reason is not
(c) 0, 0 and 2 lone pairs of central atom respectively a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) 1, 0 and 2 lone pairs of central atom respectively
(d) Assertion is incorrect, reason is correct.

18. Which of the following does not represent positive

23. Assertion: CCl4 is a non-polar compound.
overlap?
Reason: The dipole moments are cancelled out.
(a) (b)
(a) Assertion is correct, reason is correct; reason is a
correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not
a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
(c) (d)
24. Assertion: In NH3, N is sp3 hybridised, but angle is
found to be 107°.
Reason: The decrease in bond angle is due to repulsion
between the lone pair & bond pair .
19. Which one of the following species is diamagnetic in (a) Assertion is correct, reason is correct; reason is a
nature ? correct explanation for assertion.
(a) H−2 (b) H+2 (b) Assertion is correct, reason is correct; reason is not
a correct explanation for assertion
(c) H 2 (d) He+2
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
CHEMICAL BONDING AND MOLECULAR STRUCTURE 65

25. Assertion: Bonding molecular orbital has greater Section – C (Case Study Questions)
stability than corresponding antibonding molecular
orbital. Case-Study-1
Reason: The bonding molecular orbital has higher
energy Lewis concept is unable to explain the shapes of
(a) Assertion is correct, reason is correct; reason is a molecules. This theory provides a simple procedure to
correct explanation for assertion. predict the shapes of covalent molecules. Sidgwick and
(b) Assertion is correct, reason is correct; reason is not Powell in 1940, proposed a simple theory based on the
a correct explanation for assertion repulsive interactions of the electron pairs in the
(c) Assertion is correct, reason is incorrect valence shell of the atoms. It was further developed
(d) Assertion is incorrect, reason is correct. and redefined by Nyholm and Gillespie (1957).The
repulsive interaction of electron pairs decrease in the
26. Assertion: π bonds are weaker than σ bonds. order: Lone pair (lp) – Lone pair (lp) > Lone pair (lp) –
Reason: Overlapping in σ bonds takes place to a larger Bond pair (bp) > Bond pair (bp) – Bond pair (bp) For
extent the prediction of geometrical shapes of molecules with
(a) Assertion is correct, reason is correct; reason is a the help of VSEPR theory, it is convenient to divide
correct explanation for assertion. molecules into two categories as (i) molecules in which
(b) Assertion is correct, reason is correct; reason is not the central atom has no lone pair and (ii) molecules in
a correct explanation for assertion which the central atom has one or more lone pairs
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct. 29. The decreasing order of the repulsive interaction of
electron pairs is (Here, lp = lone pair, bp = bond pair).
27. Assertion: Lone pair-lone pair repulsive interactions (a) lp-lp > lp-bp > bp-bp (b) lp-bp > lp-lp > bp-bp
are greater than lone pair-bond pair and bond pair-bond (c) lp-lp > bp-bp > lp-bp (d) bp-bp > lp-lp > lp-bp
pair interactions.
Reason: The space occupied by lone pair electrons is 30. For the prediction of geometrical shapes of molecules
more as compared to bond pair electrons. with the help of VSEPR theory molecules are divided
(a) Assertion is correct, reason is correct; reason is a into two categories. These categories are
correct explanation for assertion. (a) Molecules in which the central atom has no lone
(b) Assertion is correct, reason is correct; reason is not pair
a correct explanation for assertion (b) Molecules in which the central atom has one or
(c) Assertion is correct, reason is incorrect more lone pairs.
(d) Assertion is incorrect, reason is correct. (c) Both (a) and (b)
(d) None of the above
28. Assertion: BF3 molecule has zero dipole moment.
Reason: F is electronegative and B–F bonds are polar 31. The shape of the molecule depends on the _______
in nature. (a) adjacent atom (b) valence electrons
(a) Assertion is correct, reason is correct; reason is a (c) bond order (d) None of these
correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not 32. In BrF3 molecule, the lone pairs occupy an equatorial
a correct explanation for assertion position to minimize.
(c) Assertion is correct, reason is incorrect (a) lone pair-bond pair repulsion only
(d) Assertion is incorrect, reason is correct. (b) bond pair-bond pair repulsion only
(c) lone pair-lone pair repulsion and lone pair-bond
pair repulsion
(d) lone pair-lone pair repulsion only
66 CHEMICAL BONDING AND MOLECULAR STRUCTURE

Case-Study-2 Case-Study-3

Nitrogen, oxygen and fluorine are the highly The valence bond theory (VBT) approach to covalent
electronegative elements. When they are attached to a bonding is basically concerned with the energetics of
hydrogen atom to form covalent bond, the electrons of covalent bond formation about which the Lewis and
the covalent bond are shifted towards the more VSEPR models are silent. Basically the VB theory
electronegative atom. This partially positively charged discusses bond formation in terms of overlap of
hydrogen atom forms a bond with the other more orbitals. For example the formation of the H 2 molecule
electronegative atom. This bond is known as hydrogen from two hydrogen atoms involves the overlap of the
bond and is weaker than the covalent bond. The 1s orbitals of the two H atoms which are singly
magnitude of H-bonding depends on the physical state occupied. It is seen that the potential energy of the
of the compound. It is maximum in the solid state and system gets lowered as the two H atoms come near to
minimum in the gaseous state. Thus, the hydrogen each other. At the equilibrium inter-nuclear distance
bonds have strong influence on the structure and (bond distance) the energy touches a minimum. Any
properties of the compounds. There are two types of H- attempt to bring the nuclei still closer results in a
bonds (i) Intermolecular hydrogen bond (ii) sudden increase in energy and consequent
Intramolecular hydrogen bond destabilization of the molecule. Because of orbital
overlap the electron density between the nuclei
33. Which one of the following molecules will form a increases which helps in bringing them closer. It is
linear polymeric structure due to hydrogen bonding? however seen that the actual bond enthalpy and bond
(a) NH3 (b) H2O length values are not obtained by overlap alone and
(c) HCl (d) HF other variables have to be taken into account.

34. o-nitrophenol can be easily steam distilled whereas p- 37. VBT theory is based on the
nitrophenol cannot be. This is because of : (a) Knowledge of atomic orbitals and electronic
(a) strong intermolecular hydrogen bonding in o- configuration of elements
nitrophenol (b) Overlap criteria and the hybridization of atomic
(b) strong intramolecular hydrogen bonding in o- orbitals
nitrophenol (c) The principles of variation and superposition
(c) strong intramolecular hydrogen bonding in p- (d) All of the above
nitrophenol
(d) dipole moment of p-nitrophenol is larger than that 38. The valence bond theory explains the shape, the
of o-nitrophenol formation and directional properties of bonds in
polyatomic molecules like CH4, NH3 and H2O etc., in
35. The order of the boiling points of the given compounds terms of
is (a) Overlapping of atomic orbital
(a) HF > H2O > NH3 (b) H2O > HF > NH3 (b) Hybridisation of atomic orbitals
(c) NH3> HF > H2O (d) NH3> H2O > HF (c) Both (a) and (b)
(d) None of the above
36. The boiling point of a substance increases with increase
in 39. The strength of bonds formed by s-s, p-p and p-s
(a) Intermolecular hydrogen bonding overlap has the order:
(b) Intramolecular hydrogen bonding (a) s-s > p-p > p-s (b) s-s > p-s > p-p
(c) Non-polarity (c) p-p > p-s > s-s (d) p-p > s-s > p-s
(d) Both (a) and (c)
CHEMICAL BONDING AND MOLECULAR STRUCTURE 67

40. The strength of covalent bond ___________ extent of

overlapping of orbitals.
(a) may be or may not be related to
(b) is independent on
(c) is dependent on
(d) is not related to

Find Answer Key and Detailed Solutions at the end of this book

CHEMICAL BONDING AND

MOLECULAR STRUCTURE
 Table of Contents

REDOX REACTIONS
REDOX REACTIONS 69

Chapter at a Glance
 Oxidation is a process which involves loss of electrons or  Decomposition reaction
increase in oxidation number.
2KClO3  2KCl  3O 2
 Reduction is a process which involves gain of electrons or
decrease in oxidation number.  Displacement reaction

 Oxidation agent is a substance which accepts one or more CuSO 4  Zn  Cu  ZnSO 4

electrons or its oxidation number decreases.
Sn  2HCl  SnCl 2  H 2
 Reducing agent is a substance which loses one or more
electrons or its oxidation number increases. Br2  2I   2Br   I 2

 Oxidation and Reduction always occur side by side.

 Disproportionation Reactions
Oxidation Number
The reactions in which the oxidation number of an element
 Oxidation numnber is the charge which an atom appears to both increases and decreases.
have when all other atoms ar removed from it as ions.
e.g., 2H 2 O 2  2H 2 O  O 2
For elementary state, O.N is zero.
In this case the oxidation number of O decreases from -1
e.g., He(O.N. = 0), H2 (O.N. = 0), Sn (O.N. = 0)
to -2 (in H2O) and increases from -1 to 0 (in O2)
 Oxidation number of hydrogen is +1 except in hydrides NaH,
Cl 2  2OH   OCl   Cl   H 2 O
LiH, CaH2 (-1).
–
 Oxidation number of oxygen is always -2 except in peroxides, O.N. of Cl increases from 0 to +1 (in OCl ) and decreases from
–
H2O, N2O2 (-1) 0 to -1 (in Cl ).

in superoxides KO2 (-1/2) 3ClO   ClO 3  2Cl is also a disproportionation reaction

– –
in OF2 (+2) because O.N. of Cl increase from +1 (in ClO ) to +5 (in ClO3 )
–
and decrease from +1 to -1 (in Cl ).
O2F2 (+1)
Oxidation and reduction reactions can be balanced by
 Combination reaction
oxidation number method and half reaction method.
2Mg  O 2  2MgO
70 REDOX REACTIONS

Solved Examples
Example-1
While sulphur dioxide and hydrogen peroxide can act as (b) Oxidation state of Ag is reduced from +1 to 0. Oxidation
oxidising as well as reducing agents in their reactions, ozone state of C is increased from 0 to 2.
and nitric acid act only as oxidants. Why? (c) Oxidation state of Cu is reduced from +2 to +1. Oxidation
(NCERT) state of C is increased from 0 to +2.
Sol. Sulphur dioxide and hydrogen peroxide can act as oxidizing (d) Oxidation state of O is reduced from -1 to -2. Oxidation
as well as reducing agents in their reactions because of the state of N is increased from -2 to 0.
range of the oxidation states of the elements. The Sulphur (e) Oxidation state of Pb is increased from 0 to +2. Oxidation
range is +6 to -2. In case of SO2 the Sulphur oxidation state state of Pb in PbO2 is reduced from +4 to +2.
is +4. It has a chance to oxidized as well as reduced. That’s
Oxidised Reduced Oxidising Reducing
why SO2 can act as an oxidising as well as a reducing agent. Reaction
substance substance agent agent
The oxygen range is 0 to -2. In case of H2O2 the oxygen
(a) C6H6O2 AgBr AgBr C6H6O2
oxidation state is -1. It has a chance to oxidized as well as
 Ag(NH3 ) 2   Ag(NH3 ) 2 
 
reduced. That’s why H2O2 can act as an oxidising as well as (b) HCHO HCHO
a reducing agent. (c) HCHO Cu+2 Cu+2 HCHO
In case of ozone and nitric acid, the oxygen and nitrogen (d) N2H4 H2O2 H2O2 N2 H4
can only decrease the oxidation state only. Hence, ozone (e) Pb PbO2 PbO2 Pb
and HNO3 acts only as an oxidant.
Example-4
Example-2 Justify giving reactions that among halogens, fluorine is
The compound AgF2 is an unstable compound. However, if the best oxidant.
formed, the compound acts as a very strong oxidizing agent. Sol. Like metals, there is also an activity series for nonmetals.
Why? (NCERT) Their reactivity depends upon their oxidizing power.
Sol. The stable oxidation state of silver is +1. Compound AgF2 Oxidizing power of halogens decreases as we move down
is an unstable compound. However, if formed, the compound the group 17 from F to I. thus, Fluorine is the strongest
acts as a very strong oxidizing agent due to convert into its oxidizing agent. It can displace in their aqueous solutions
stable oxidation state. Cl2 from Cl- ions, Br2 from Br- ions and I2 from I- ions. Thus,
Fluorine can show displacement reaction.
Example-3
Identify the substance oxidised, reduced, oxidising agent 2X   aq   F2  aq   X 2  g   2F  aq 
and reducing agent for each of the following reactions: The increases order of oxidizing power of halogens is
(a) 2AgBr (s) + C6H6O2(aq)  2Ag(s) + 2HBr (aq) + I2<Br2<Cl2<F2 .
Example-5
C6H4O2(aq)
Out of aluminium and silver vessel, which one will be more
(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq)  2Ag(s) +
suitable to store 1 M HCl solution and why ?
HCOO–(aq) + 4NH3(aq) + 2H2O(l)
E oAl3 |Al  1.66V, E oAg  |Ag  0.80V.
(c) HCHO (l) + 2Cu2+(aq) + 5 OH–(aq)  Cu2O(s) + HCOO–
(aq) + 3H2O(l) Sol. Since reduction potential of silver is more than that of
(d) N2H4(l) + 2H2O2(l)  N2(g) + 4H2O(l)
 
hydrogen E H |H2 , Pt  0 , silver vessel will be suitable to
o

(e) Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l)

o
(NCERT) store 1M HCl. On the other hand, E Al3 |Al is less than that
Sol. (a) Oxidation of Ag decreases from +1 to zero. removed of
hydrogen means oxidation C6H6O2 to C6H4O2.
o
 
of hydrogen E H |H2 , Pt so that hydrogen will be liberated
if stored in aluminium vessel.
REDOX REACTIONS 71

Example-6 Therefore, the oxidation number of Mn is + 6.

Assign oxidation numbers to the underlined elements in (e) CaO 2
each of the following species: (NCERT)
2 x
(a) NaH2PO4 (b) NaHSO4 Ca O 2
(c) H4P2O7 (d) K2MnO4 Then, we have
(e) CaO2 (f) NaBH4
(g) H2S2O7 (h) KAl(SO4)2.12 H2O  2   2  x   0
Sol. NaH 2 PO 4 Let us assume the oxidation number of P be x.  2  2x  0
Oxidation number of Na = +1  x  1
Oxidation number of H = +1 Therefore, the oxidation number of O is – 1.
Oxidation number of O = –2
1 1 x 2
(f) NaBH 4
 Na H 2 P O 4 1 x 1
Then, we have Na B H 4

1 1  2  1  1 x   4  2   0 Then, we have

 1 2  x  8  0 1 1  1 x   4  1  0
 x  5  1 x  4  0
Therefore, the oxidation number of P is +5.  x  3
(b) NaHSO 4 Therefore, the oxidation number of B is + 3.
1 1 x 2
(g) H 2 S2 O7
Na HSO 4
Then, we have 1 x 2
H 2 S2 O 7
1 1  1 1  1 x   4  2   0 Then, we have
 11 x  8  0
2  1  2  x   7  2   0
 x  6
Therefore, the oxidation number of S is + 6.  2  2 x  14  0
 2 x  12
(c) H 4 P 2 O7
 x  6
1 x 2
H 4 P2 O 7 Then, we have
Therefore, the oxidation number of S is + 6.
4  1  2  x   7  2   0 (h) KAl  SO4 2 .12H 2 O
 4  2 x  14  0
1 3 
 x 2  1 2
 2 x  10 K Al  SO4  .12 H 2 O
 2
 x  5
Therefore, the oxidation number of P is + 5. Then, we have
(d) K 2 MnO 4 1 1  1 3  2  x   8  2   24  1  12  2   0
1 2
K 2 Mn O 4
x
 1  3  2 x  16  24  24  0
Then, we have  2 x  12
 x  6
2  1  x  4  2   0
Therefore, the oxidation number of S is + 6.
 2 x 8  0
 x  6
72 REDOX REACTIONS

Example-7 Example-10
What sorts of informations can you draw from the following Calculate individual oxidation number of each S-atom in
reaction? (NCERT) Na2S4O6 (sodium tetrathionate) with the help of its structure.
Sol.
 CN 2 g   2OH  aq   CN aq   CNO aq  H 2 Ol 
Sol. Let us write the oxidation number of the Carbon.

C3
 
2

 N   2OH  aq   C N  aq   H 2 O  l 
  2 g 
We can see the oxidation number changes in the L.H.S and
R.H.S. Carbon is oxidized as well as reduced so this reaction Example-11
is disproportionation reaction. Disproportionation are those Write the formulae for the following compounds:
reactions in which the atom of same element is
simultaneously oxidized as well as reduced. (a) Mercury(II) chloride
Example-8 (b) Nickel(II) sulphate
Consider the elements: (c) Tin(IV) oxide
Cs, Ne, I and F (d) Thallium(I) sulphate
(a) Identify the element that exhibits only negative oxidation (e) Iron(III) sulphate
state.
(f) Chromium(III) oxide (NCERT)
(b) Identify the element that exhibits only positive oxidation
state. Sol. (a) Mercury (II) chloride: HgCl2
(c) Identify the element that exhibits both positive and (b) Nickel (II) sulphate: NiSO4
negative oxidation states. (c) Tin (IV) oxide: SnO2
(d) Identify the element which exhibits neither the negative (d) Thallium (I) sulphate: Tl2SO4
nor does the positive oxidation state. (NCERT)
(e) Iron (III) sulphate: Fe2(SO4)3
Sol. (a) Fluorine is only element in the periodic table that shows
always only negative oxidation state of –1. (f) Chromium (III) oxide: Cr2O3
(b) Cesium is the metal and it shows positive oxidation state Example-12
of +1. Fluorine reacts with ice and results in the change:
(c) Iodine is nonmetal but it will show both positive and H2O(s) + F2(g)  HF(g) + HOF(g)
negative oxidation states. Justify that this reaction is a redox reaction (NCERT)
(d) Neon is noble gas and it oxidation sate is zero due to Sol. Let’s write the oxidation number of each atom involved in
stable noble gas configuration. the given reaction:
Example-9
Find the average and individual oxidation number of Fe &
Pb in Fe3O4 & Pb3O4, which are mixed oxides.
Sol. (i) Fe3O4 is mixture of FeO & Fe2O3 in 1 : 1 ratio
so, individual oxidation number of Fe = +2 & +3

1(2)  2 ( 3) Oxidation number of F increases from 0 to +1 in HOF and

& average oxidation number =  8/3 the oxidation number decreases from 0 to –1 in HF. F is
3
oxidized as well as reduced. So that the given reaction is a
(ii) Pb3O4 is a mixture of PbO & PbO2 in 2 : 1 molar ratio redox reaction.
so, individual oxidation number of Pb are +2 & +4

2(2)  1 ( 4)
& average oxidation number of Pb =  8/ 3
3
REDOX REACTIONS 73

Example-13 Example-15
Refer to the periodic table given in your book and now Balance the following equations :
answer the following questions: (a) H 2 O 2  MnO 4 
 Mn 2  O 2 (acidic medium)
(a) Select the possible non-metals that can show
disproportionation reaction. (b) Zn  HNO3 (dil) 
 Zn(NO 3 ) 2  H 2O  NH 4 NO 3
(b) Select three metals that can show disproportionation
(c) CrI3  KOH  Cl2 
 K 2CrO4  KIO4  KCl  H 2O
reaction. (NCERT)
Sol. In disproportionation reactions, one of the reacting (d) P2 H 4 
 PH 3  P4
substances always contains an element that can exist in at
least three oxidation states. (e) Ca 3 (PO 4 ) 2  SiO 2  C 
 CaSiO 3  P4  CO
(a) Phosphorous(P4), Chlorine (Cl), Iodine(I), and sulphur Sol. (a) 6H   5H 2 O 2  2MnO 4 
 2Mn 2  5O 2  8H 2 O
(S 8 ) can show disproportionation reactions.
Disproportionation are those reactions in which the (b) 4Zn  10HNO 3 (dil) 

atom of same element is simultaneously oxidized as well
as reduced. 4Zn (NO3 ) 2  3H 2 O  NH 4 NO3
(b) Manganese (Mn), Chromium(Cr) and Copper(Cu), can (c) 2CrI3  64KOH  27Cl 2 
 2K 2 CrO 4  6KIO 4
show disproportionation reactions. Disproportionation
are those reactions in which the atom of same element  54KCl  32H 2O
is simultaneously oxidized as well as reduced.
(d) 6P2 H 4 
 8PH 3  P4
Example-14
The Mn 3+ ion is unstable in solution and undergoes (e) 2Ca 3 (PO 4 ) 2  6SiO 2  10C 

disproportionation to give Mn2+, MnO2, and H+ ion. Write a
balanced ionic equation for the reaction. (NCERT) 6CaSiO3  P4  10CO
Sol. The Mn 3+ ion is unstable in solution and undergoes Example-16
disproportionation to give Mn2+, MnO2, and H+ ion. Let us The number of moles of ferrous oxalate oxidised by one
write the reaction: mole of KMnO4 in acidic medium is :
Mn 3aq   Mn 2aq   MnO 2s   H aq  5 2 3 5
(a) (b) (c) (d)
The oxidation half equation by balancing the electrons and 2 5 5 3
charge are: Sol. Eq. of FeC2O4 = Eq. of KMnO4
Mn 3 aq   MnO 2 aq   4H aq  e  moles of FeC2O4 × 3 = moles of KMnO4 × 5
The number of moles of ferrous oxalate oxidised by one
Now balance the O atoms and H+ ions by adding water mole of KMnO4 = 3/5
molecules,
Example-17
Mn 3aq   2H 2 O l   MnO 2s   4H aq   e  ...(i) How many moles of KMnO4 are needed to oxidise a mixture
of 1 mole of each FeSO4 & FeC2O4 in acidic medium ?
On the other hand, the reduction half equation by balancing
the electrons are: 4 5 3 5
(a) (b) (c) (d)
Mn 3 
 e  Mn 2 5 4 4 3
 aq   aq  ...(ii)
Sol. Eq. of KMnO4 = Eq. of FeSO4 + Eq. of FeC2O4
Combine the both equation (i) and (ii) by adding as:
moles of KMnO4 × 5
2Mn 3 aq   2H 2 O l   MnO 2 s   2Mn 2aq   4H aq  = moles of FeSO4 × 1 + moles of FeC2O4 × 3
 moles of KMnO4 = 4/5 Ans. (a)
74 REDOX REACTIONS

Example-18 Example-21
Arrange the following metals in the order in which they Calculate the normality of a solution containing 13.4 g of
displace each other from the solution of their salts. sodium oxalate in 100 mL Sol.
Al, Cu, Fe, Mg and Zn. (NCERT)
Sol. Lower the standard reduction potential value higher will be wt. in g / eq. wt
Sol. Normality =
the activity of metal as it will have higher tendency to lose vol of solution in litre
electrons. The order of the increasing reducing power of
Here, eq. wt. of Na2C2O4 = 134/2 = 67
the given metals is Cu < Fe < Zn < Al <Mg.
(Acidic solution) 13.4 / 67
so N  2N
Example-19 100/1000
Predict the products of electrolysis in each of the following:
Example-22
(NCERT)
40.05 mL of 1.0 M Ce+4 are required to titrate 20.0 mL of 1.0
(i) An aqueous solution of AgNO3 with silver electrodes
mL of 1.0 M Sn2+ to Sn4+. What is the oxidation state of
(ii) An aqueous solution AgNO3 with platinum electrodes
cerium in the reduction product ?
(iii) A dilute solution of H2SO4 with platinum electrodes
Sol. (i) An aqueous solution of AgNO3 ionizes and form Ag+ Sol. The reaction occurring are :

Ce 4   ne   Ce
4n 
and NO3 ions. On electrolysis, there are two positive ..............(i)
charge ions and two negative charge ions are in the solution. Sn 2   Sn 4   2e  ..............(ii)
It is aqueous solution so that water also dissociates and
To balance the equations, (the no. of electrons lost = no. of
form H+ and OH-. But in the given condition, it is active
electrons gained) multiply eq. (i) by 2 and eq. (ii) by n and
electrode. So Silver only oxidized and reduced.
add
Hence, Ag+ ions are reduced at the cathode and Ag metal
gets oxidized at the anode. 2Ce 4   nSn 2   Ce 4  n    Sn 4 
(ii) Pt is inert electrode; it will not take participate in the
Moles of Ce4  in 40.05 mL of 1.0 M solution,
chemical reaction. An aqueous solution of AgNO3 ionizes
and form Ag+ and ions. On electrolysis, there are two positive 1.0
charge ions and two negative charge ions are in the solution.   40.05  40.05  10 3 mol
1000
It is aqueous solution so that water also dissociates and
form H+ and OH-. At the anode, oxidation of water occurs to Now 2 mol of Ce4  will oxidise n mole of Sn 2
liberate O2. At the cathode, Ag+ ions are reduced and get
40.05  103 mol of Ce+4 will oxidise Sn2+
deposited.
(iii) Pt is inert electrode; it will not take participate in the n
 40.05 10 3 mol  20.02n  10 3 mol
chemical reaction. An aqueous solution of H2SO4 ionizes 2
and form H+ and SO4-2 ions. On electrolysis, there are two
But moles of Sn2+ in 20.0 mL of 1.0 M solution
positive charge ions and two negative charge ions are in
the solution. It is aqueous solution so that water also 1.0
  20.0  20.0  103 mol
dissociates and form H+ and OH-. Hence, at the cathode, H+ 1000
ions are reduced to liberate H2 gas. At the anode H2O is
oxidized to liberate O2 molecules.  20.02n  103 mol  20.0 10 3 mol
Example-20 n 1
Calculate the normality of a solution obtained by mixing Hence 1 mol of electrons are required in the reduction of
50 mL of 5M solution of K2Cr2O7 and 50 mL of 2 M K2Cr2O7
each mol of Ce4  ion.
in acidic medium.
Sol. v.f. of K2Cr2O7 = 6  Ce 4   e   Ce 3
N1V1  N 2 V2
so N f  Ce3 is the reduction product.
V1  V2

5  6  50  2  6  50
  21 N
50  50
REDOX REACTIONS 75

Example-23 The oxidation number of N decreases from +5(in HNO3) to

Which of the following is not an example of redox reaction? +3 (in NOCl)and therefore, HNO3 and therefore, HNO3 acts
as an oxidizing agent.
(a) CuO  H 2 
 Cu  H 2 O
Thus, reaction (i) is a redox reaction.
(b) Fe 2 O 3  3CO 
 2Fe  3CO 2 2 1 1 1 2 1 1 1
(ii) HgCl 2 (aq)  2 KI (aq) 
 Hg I 2 ( aq)  KCl(aq)
(c) 2K  F2 
 2KF
Here oxidation number of none of the atoms undergo a
(d) BaCl 2  H 2 SO 4 
 BaSO 4  2HCl chang and therefore, this is not a redox reaction.
Ans. (d) Example-25

2 2
Chlorine is used to purify drinking water. Excess of chlorine
Sol. Ba Cl 2  H 2SO 4  Ba SO 4  2HCl is harmful. The excess of chlorine is removed by treating
with sulphur dioxide. Present a balanced equation for this
There is no change in oxidation number. Hence, not a redox
redox change taking place in water. (NCERT)
reaction.
Sol. Chlorine is used to purify drinking water. Excess of chlorine
is harmful. The excess of chlorine is removed by treating
Example-24 with sulphur dioxide due to the formation of two acids like
Identify the redox reactions out of the following reactions H2SO4 and HCl.
and identify the oxidising and reducing agents in them. The balanced equation:
(i) 3HCl(aq)  HNO3 (aq) 
 Cl2  SO 2  SO 42  Cl1

Cl2 (g)  NOCl(g)  2H 2 O(l ) Reduction half reaction

Cl 2  2e   2Cl 
(ii) HgCl 2 (aq)  2KI(aq) 
 HgI 2 (s)  2KCl(aq)
Oxidation half reaction:
1 1 1  5  2
Sol. (i) 3HCl(aq)  H N O3 (aq) 
 SO 2  2H 2 O  SO 42   4H   2e 
Balanced, reaction:
0 3  2 1 1 2
Cl 2 (g)  NO Cl (g)  2H 2 O(l ) Cl 2  SO 2  2H 2 O  2Cl   SO 42   4H 
Here, Oxidation number of Cl increases from -1 (in HCl) to 0
(in Cl2). Therefore, Cl– is oxidized and hence HCl acts as a
reducing agent.
76 REDOX REACTIONS

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Define oxidation reaction in terms of classical concept. 15. Justify that the following reactions are redox reactions :
(a) CuO ( s ) + H 2 ( g ) → Cu ( s ) + H 2 O ( g )
2. Define reduction reaction in terms of classical concept.
(b) Fe2 O3 ( s ) + 3CO ( g ) → 2Fe ( s ) + 3CO2 ( g )
3. What is oxidation in terms of electron transfer?
16. The following reaction represents the process of
4. What is meant by reduction in terms of electrons bleaching. Identify and name the species that bleaches
transfer ? the substances due to its oxidizing action.
Cl2 ( g ) + 2OH − ( aq ) → ClO− ( aq ) + Cl− (aq) + H 2O ( l )
5. Write formula for Mercury(II) chloride.

17. PbO and PbO2 reacts with HCl according to following

6. Define equivalent mass.
chemical equations :
(i) 2PbO + 4HCl → 2PbCl 2 + 2H 2 O
7. At what concentration of Zn2+ (aq) will its electrode
potential become equal to its standard electrode (ii) PbO2 + 4HCl → PbCl2 + Cl2 + 2H 2 O
potential? Why do these compounds differ in their reactivity?

8. What is the Stock notation of chromium trioxide? 18. Define :

(i) Electrochemical series
9. What is the oxidation number of nitrogen in HN3? (ii) Redox titrations

10. CuSO 4 (aq) + Zn(s) → Cu(s) + ZnSO 4 (aq) 19. Calculate average oxidation state of S in Tetrathionate
This reaction is an example of ________. ion.

− −
Section – B (2 Marks Questions) 20. Which of the two ClO2 or ClO4 show disproportio-
nation reaction and why?
11. Name the different types of redox reaction.
Section – C (3 Marks Questions)
12. (a) Define disproportionation reaction.
(b) Identify the type of given reaction. 21. Classify the following redox reaction giving reasons :
2H 2 O 2 → 2H 2 O + O 2 (a) N 2 ( g ) + O2 ( g ) → 2NO ( g )
(b) 2Pb ( NO3 )2 ( s ) → 2PbO ( s ) + 4NO2 ( g ) + O2 ( g )
13. Calculate oxidation number of underlined element :
(i) K2CrO4 (ii) H2SO4 (c) NaH ( s ) + H 2 O ( l ) → NaOH ( aq ) + H 2 ( g )

14. Check the feasibility of the following redox reaction 22. Write the disproportionation reaction of the following
with the help of electrochemical series. species :
Ni ( s ) + 2Ag + ( aq ) → Ni 2 + ( aq ) + 2Ag ( s ) (a) ClO–
−
(b) ClO3
−
(c) ClO2
REDOX REACTIONS 77

23. Identify the oxidant and reductant in the following Section – D (5 Marks Questions)
reactions :
1
(a) Zn ( s ) + O 2 ( g ) → ZnO ( s )
29. Write balanced equation for the following reactions :
2 (i) Reaction of liquid hydrazine (N2H4) with chlorate
(b) CH 4 ( g ) + 4Cl2 ( g ) → CCl4 ( g ) + 4HCl ( g ) ion ( ClO3− ) in basic medium produces nitric oxide
(c) I2 ( aq ) + 2S2 O32 − ( aq ) → 2I − ( aq ) + S4 O62 − ( aq ) gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(ii) Balance the following equation by half reaction
24. MnO24− undergoes disproportionation reaction in acidic
method
−
medium but MnO4 does not. Give reason. Cr2 O72− + H+ + I− → Cr 3+ + I2 + H2O

25. Calculate the oxidation number of each sulphur atom in

30. Consider the following table of standard reduction
the following compounds :
potentials:
(a) Na 2S2 O3 (b) Na 2SO3 (c) Na 2SO 4
Reaction E0 (V)

26. Using stock notation, represent the following A3+ + 2e− → A+ 1.36
compounds : HAuCl4, Tl2O, FeO.
B2 + + 2e− → B 0.72
27. Copper reacts with nitric acid. A brown gas is formed C2 + + 2e− → C –0.28
and the solution turns blue. The equation may be
written as: D+ + e − → D –1.42
− 2+
Cu + NO ⎯⎯
→ NO2 + Cu
3

Balance the equation by oxidation number method. (a) Which substance is

(i) Strongest oxidizing agent ?
28. It is possible to store (ii) Strongest reducing agent ?
(i) Copper sulphate in a zinc vessel? (b) Which substances can be oxidized by B2+ ?
(ii) Copper sulphate in a silver vessel? (c) Which substance can be reduced by C ?
(iii) Copper sulphate in a nickel vessel? (d) Write a balanced chemical equation for the overall
cell reaction that gives the highest cell voltage and
calculate Eo for the reaction.
78 REDOX REACTIONS

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions)
8. Consider the following standard reduction potentials:
1. An oxidation process involves Ca 2+ + 2e− → Ca; Eo = −2.87 V
(a) Increase in oxidation number
Pb2+ + 2e− → Pb; Eo = −0.13 V
(b) Decrease in oxidation number
(c) Both decrease and increase in oxidation number Cu 2+ + 2e− → Cu; Eo = 0.34 V
2+
(d) No change in oxidation number Hg2 + 2e− → Hg; Eo = 0.92 V

2. In the reaction, SO 2 + 2H 2S → 3S + 2H 2 O the substance Pt 2+ + 2e− → Pt; Eo = 1.20V

oxidized is Which of the following metals is the strongest
(a) H2S (b) SO2 REDUCING AGENT?
(c) S (d) H2O (a) Ca (b) Pb
(c) Cu (d) Hg
3. The number of electrons lost in the following change is
Fe + H 2 O → Fe3O 4 + H 2 9. The correct order of reactivity of K, Mg, Zn and Cu
with water according to the electrochemical series is
(a) 2 (b) 4
(a) K > Mg > Zn > Cu (b) Mg > Zn > Cu > K
(c) 6 (d) 8
(c) K > Zn > Mg > Cu (d) Cu > Zn > Mg > K

4. In the reaction
10. The eq. wt. of K2CrO4 as an oxidizing agent in acid
3Br2 + 6CO32− + 3H2 O → 5Br − + BrO3− + 6HCO3− medium is
(a) Bromine is oxidized and carbonate is reduced mol. wt. 2  mol. wt.
(b) Bromine is reduced and water is reduced (a) (b)
2 3
(c) Bromine is neither reduced nor oxidised
mol. wt. mol. wt.
(d) Bromine is both reduced and oxidised (c) (d)
3 6

5. The oxidation state of S in H2SO4is

11. The mass of oxalic acid crystals (H2C2O4.2H2O)
(a) +2 (b) +4
required to prepare 50 mL of a 0.2 N solution is :
(c) +6 (d) +7
(a) 4.5 g (b) 6.3 g
(c) 0.63 g (d) 0.45 g
6. The reaction, 3ClO− (aq) ⎯⎯
→ ClO3− (aq) + 2Cl− (aq) is
an example of 12. In the conversion NH2OH → N2O, the equivalent
(a) oxidation reaction weight of NH2OH will be :
(b) reduction reaction
(a) M/4 (b) M/2
(c) disproportionation reaction
(c) M/5 (d) M/1
(d) decomposition reaction
(M = molecular weight of NH2OH)

7. For the redox reaction 13. Which of the following is the best description of the
MnO−4 + C2O42− + H+ → Mn 2+ + CO2 + H2O, behaviour of bromine in the reaction given below?
H 2 O + Br2 → HOBr + HBr
The correct stoichiometric coefficients of MnO−4 ,
(a) Proton acceptor only
C2 O42− and H+ are respectively: (b) Both oxidized and reduced
(a) 2,5,16 (b) 16,5,2 (c) Oxidised only
(c) 5,16,2 (d) 2,16,5 (d) Reduced only
REDOX REACTIONS 79

14. The oxidant which cannot act as a reducing agent is Section - B (Assertion & Reason Type Questions)
(a) SO2 (b) NO2
(c) CO2 (d) ClO2
21. Assertion (A): H2O2 can actas an oxidizing agent.
Reason (R): All the peroxides behave as oxidizing
15. An element that never has a positive oxidation state in
agents only.
any of its compound is
(a) If both (A) and (R) are correct and (R) is the correct
(a) Boron (b) Oxygen
explanation of (A)
(c) Chlorine (d) Fluorine
(b) If both (A) and (R) are correct but (R) is not the
correct explanation of (A)
16. Which one of the following reactions involves
(c) If (A) is correct but (R) is incorrect.
oxidation-reduction?
(d) If (A) is incorrect but (R) is correct
(a) H2 + Br2→ 2HBr
(b) NaBr + HCl → NaCl + HBr
22. Assertion (A): SO2 and Cl2 are both bleaching agents.
(c) HBr + AgNO3→AgBr + HNO3
Reason (R): Both are reducing agents.
(d) 2NaOH + H2SO4→ Na2SO4 + 2H2O
(a) If both (A) and (R) are correct and (R) is the correct
explanation of (A)
17. In which of the following compounds, the oxidation
(b) If both (A) and (R) are correct but (R) is not the
number of iodine is fractional?
correct explanation of (A)
(a) IF7 (b) I3–
(c) If (A) is correct but (R) is incorrect.
(c) IF5 (d) IF3
(d) If (A) is incorrect but (R) is correct

18. Which of the following statements about the following

23. Assertion (A): F2 undergoes disproportionation reaction.
reaction is correct?
Reason (R): Fluorine does not show positive oxidation
2Cu 2 O ( s ) + Cu 2S ( s ) → 6Cu ( s ) + SO2 ( g ) states.
(a) Both Cu2O and Cu2S are reduced. (a) If both (A) and (R) are correct and (R) is the correct
(b) Only Cu2S is reduced. explanation of (A)
(c) only Cu2S is the oxidant (b) If both (A) and (R) are correct but (R) is not the
(d) Only Cu2O is reduced correct explanation of (A)
(c) If (A) is correct but (R) is incorrect.
19. In the equation NO2− + H2 O ⎯⎯
→ NO3− + 2H + ne− (d) If (A) is incorrect but (R) is correct
n stands for
(a) 1 (b) 2 24. Assertion (A): HNO3 acts only as an oxidizing agent,
(c) 3 (d) 4 while HNO2 acts both as an oxidizing agent and a
reducing agent.
20. The oxidation number of an element in a compound is Reason (R): The oxidation number of N in HNO3 is
evaluated on the basis of certain rules, which of the maximum.
following rules is not correct in this respect ? (a) If both (A) and (R) are correct and (R) is the correct
(a) The oxidation number of hydrogen is always +1 explanation of (A)
(b) The algebraic sum of all the oxidation numbers in a (b) If both (A) and (R) are correct but (R) is not the
compound is zero correct explanation of (A)
(c) An element in the free or the uncombined states (c) If (A) is correct but (R) is incorrect.
bears oxidation number zero (d) If (A) is incorrect but (R) is correct
(d) In alkali compounds, the oxidation number of
fluorine is −1
80 REDOX REACTIONS

25. Assertion (A): In a redox reaction, the oxidation 28. Assertion (A): O3 can act as an oxidizing agent as well
number of oxidant decrease while that of reductant as a reducing agent, but SO2 can act only as an oxidant.
increases. Reason (R):The oxidation number of O in O3 is zero,
and the oxidation number of S in SO2 is +4.
Reason (R): Oxidant gains electron(s) and reductant
(a) If both (A) and (R) are correct and (R) is the correct
loses electron(s)
explanation of (A)
(a) If both (A) and (R) are correct and (R) is the correct (b) If both (A) and (R) are correct but (R) is not the
explanation of (A) correct explanation of (A)
(b) If both (A) and (R) are correct but (R) is not the (c) If (A) is correct but (R) is incorrect.
correct explanation of (A) (d) If (A) is incorrect but (R) is correct
(c) If (A) is correct but (R) is incorrect.
(d) If (A) is incorrect but (R) is correct Section – C (Case Study Questions)

Case Study – 1
26. Assertion (A): Equivalent mass of KMnO4 is equal to
one-fifth of its molecular mass when it acts as the The oxidation and reduction reactions or redox
oxidizing agent in mild basic medium. reactions are very common in our daily life. Oxidation
Reason (R): The oxidation number of Mn in KMnO4 is process involves loss of electrons whereas reduction
+7. process involves gain of electrons. During transfer of
(a) If both (A) and (R) are correct and (R) is the correct electrons in a reaction, the species which loses electron
explanation of (A) is said to be oxidised. In other words, oxidation &
(b) If both (A) and (R) are correct but (R) is not the reduction reactions takes place simultaneously. After
correct explanation of (A) reading above passage answer the following questions.
(c) If (A) is correct but (R) is incorrect.
29. In the following reaction
(d) If (A) is incorrect but (R) is correct
P4 + 3KOH + 3H 2 O → 3KH 2 PO 2 + PH 3

27. Assertion (A): The change in oxidation state of Mn in (a) Phosphorus is both oxidized and reduced
(b) Only phosphorus is reduced
KMnO4 (acidic medium) is less than the change in
(c) Phosphorus is oxidized
oxidation state of Mn in KMnO4 in basic medium.
(d) None of these
Reason (R): In acidic medium, oxidation state of Mn
changes from +7 to +2, while in basic medium, it
30. Gain of oxygen in a compound is an example of….
changes from +7 to +4. (a) oxidation (b) reduction
(a) If both (A) and (R) are correct and (R) is the correct (c) hydrogenation (d) None of these
explanation of (A)
(b) If both (A) and (R) are correct but (R) is not the 31. Zn 2 + ( aq.) + 2e − → Zn ( s ) . This is
correct explanation of (A) (a) oxidation (b) reduction
(c) If (A) is correct but (R) is incorrect. (c) redox reaction (d) None of the above
(d) Both (A) and (R) are incorrect.
32. Removal of oxygen from a compound is an example
of….
(a) oxidation (b) reduction
(c) oxygenation (d) dehydrogenation
REDOX REACTIONS 81

Case Study – 2 Case Study – 3

Equivalent weight Redox reactions form an important class of reactions in

Molecular weight / Atomic weight which oxidation and reduction occur simultaneously.
=
n − factor Oxidation numbers are assigned in accordance with a
n-factor is very important in redox as well as non- constant set of rules. Oxidation number denotes the
redox reactions. With the help of n-factor we can oxidation state of an element in a compound ascertained
predict the molar ratio of the reactant species taking according to a set of rules formulated on the basis that
part in reactions. In general n-factor of acid/base is electron pair in a covalent bond belongs entirely to
number of moles of H+/OH– furnished per mole of
more electronegative element. It is not always possible
acid/base. n-factor of a reactant is no. of moles of
electrons lost or gained per a mole of reactant. to remembereasily in a compound that which element is
Example 1: more electronegative than the other. Therefore, a set of
1. In acidic medium: KMnO4 (n = 5) → Mn 2+ rules has been formulated to determine the oxidation
number of an element in a compound. If two or more
2. In neutral medium: KMnO4 (n = 3) → MnO2
than two atoms of an element are present in the
3. In basic medium: KMnO4 (n = 1) → Mn 6+ molecule, the oxidation number of the atom of that
element will be the average of the oxidation number of
33. n-factor of Ba(MnO4)2 in acidic medium is all the atoms of that element.
(a) 2 (b) 6
(c) 10 (d) None of these 37. The oxidation number of carbon in CH 2 Cl2 is
34. For the reaction, (a) 0 (b) 2
H3 PO2 + NaOH → NaH 2 PO2 + H 2 O (c) 3 (d) 5
What is the equivalent weight of H3PO2?
38. Oxidation state of oxygen in F2O is
(mol. wt. is M)
(a) +1 (b) -1
M
(a) M (b) (c) +2 (d) -2
2
M
(c) (d) None of these 39. The oxidation number of S in Na2S4O6 is
3
(a) + 0.5 (b) +2.5
(c) + 4 (d) + 6
35. The equivalent mass of oxidizing agent in the
following reaction is
SO 2 + 2H 2S → 3S + 2H 2 O 40. In the conversion of Br2to BrO3− , the oxidation number

(a) 32 (b) 64 of Br changes from

(c) 16 (d) 8 (a) zero to +5 (b) +1 to +5
(c) zero to -3 (d) +2 to +5
36. M is the molecular weight of KMnO4. The equivalent
weight of KMnO4 when it is converted into K2MnO4 is :
(a) M (b) M/3
(c) M/5 (d) M/7
82 REDOX REACTIONS

Notes:

Find Answer Key and Detailed Solutions at the end of this book

REDOX REACTIONS
HYDROGEN
84 HYDROGEN

Chapter at a Glance
• Hydrogen is the lightest atom with only one electron. Loss • Water is the most common and abundantly available
of this electron results in an elementary particle, the proton. substance. It is of a great chemical and biological
Thus, it is unique in character. It has three isotopes, namely: significance. The ease with which water is transformed from
protium (11H), deuterium (D or 21H) and tritium (T or 31H). liquid to solid and to gaseous state allows it to play a vital
Amongst these three, only tritium is radioactive. role in the biosphere.

• Inspite of its resemblance both with alkali metals and • The water molecule is highly polar in nature due to its bent
halogens, it occupies a separate position in the periodic structure. This property leads to hydrogen bonding which
table because of its unique properties. is the maximum in ice and least in water vapour.

• Dihydrogen on the industrial scale is prepared by the water- • Presence of calcium and magnesium salts in the form of
gas shift reaction from petrochemicals. It is obtained as a hydrogencarbonate, chloride and sulphate in water makes
byproduct by the electrolysis of brine. water hard.Water free from soluble salts of calcium and
magnesium is called Soft water.
• Among the chemical reactions of dihydrogen, reducing
reactions leading to the formation hydrogen halides, water, • Both temporary and permanent hardness can be removed
ammonia, methanol, vanaspati ghee, etc. are of great by the use of zeolites, and synthetic ion-exchangers.
importance.
• Heavy water, D2O is another important compound which is
• Though dihydrogen is rather inactive at room temperature manufactured by the electrolytic enrichment of normal water.
because of very high negative dissociation enthalpy, it It is essentially used as a moderator in nuclear reactors.
combines with almost all the elements under appropriate
• Hydrogen peroxide, H 2O2 has an interesting non-polar
conditions to form hydrides.
structure and is widely used as an industrial bleach and in
• All the type of hydrides can be classified into three pharmaceutical and pollution control treatment of industrial
categories: ionic or saline hydrides, covalent or molecular and domestic effluents.
hydrides and metallic or non-stoichiometric hydrides.
• It acts as an oxidising as well as reducing agent in both
acidic and alkaline media.
HYDROGEN 85

Solved Examples
Example-1
FeCrO4
Justify the position of hydrogen in the periodic table on
CO  g   H 2 O  g 
 catalyst  CO
the basis of its electronic configuration.  2 g   H2 g 
673K
(NCERT)
With the removal of CO2 the reaction shifts in the forward
Sol. Hydrogen has been placed at the top of the alkali metal in
direction and thus, the production of dihydrogen will be
group, but it is not a member of the group.
increased.
Its position is not justified properly because of its electronic
Example-5
configuration as (1s1). It can be placed with alkali metals
because it also has similar configuration (ns1) as alkali metals. Describe the bulk preparation of hydrogen by electrolytic
method. What is the role of an electrolyte in this process?
However, it can also be placed along with halogen in group
17 since just like halogen it can acquire inert gas (NCERT)
configuration by accpepting one electron. Sol. The electrolyte (15-20% NaOH solution) increases
Example-2 conductivity of water.
Why does hydrogen occur in a diatomic form rather than in Cathode (iron) : Reduction of water occurs.
a monoatomic form under normal conditions ? 2H 2 O  2e   2H 2  2OH 
(NCERT)
Anode (nickel coated iron) : Oxidation of OH  occurs
Sol. Hydrogen atom has only one electron and thus, to achieve
stable inert gas configuration of helium, it shares its single 2OH   H 2 O  1 / 2O 2  2e 
electron with electron of other hydrogen atom to form a Example-6
stable diatomic molecule. The stability of H2 is further Can dihydrogen act as oxidising agent ? If so give chemical
confirmed by the fact, that formation of one mole of gaseous reactions to support the statement.
H2 molecules results in the release of 435.8 kJ of energy.
Sol. Dihydrogen can act as oxidising agent when it forms metal
H  g   H  g   H 2  g  ; H  435.8 KJ mol 1
hydrides.
Example-3 2Li  H 2  2LiH
Write the names of isotopes of hydrogen. What is the mass Example-7
ratio of these isotopes? Can conc. H2SO4 be used for drying H2 gas? Justify.
(NCERT) Sol. Conc. H2SO4 cannot be used for drying H2 gas because
Sol. The various, isotopes of hydrogen are : H2SO4 does absorb moisture from moist H2, but the process
is highly exothermic. The heat so produced causes
1
1 H  Protium  ; 12 H or D  Deuterium  ; 13 H or T  Tritium 
hydrogen to catch fire because of its inflammable nature.
The mass ratio of 11 H: 12 H: 13 H is 1: 2 : 3 Example-8
Example-4 Complete the following reactions:
How can the production of dihydrogen obtained from ‘Coal 
(i) H 2  g   MmOn  s   
gasification’ be increased?

(NCERT) (ii) CO  g   H 2  g  
Catalyst
Sol. The production of dihydrogen in coal gasification can be
increased by reacting CO(y) present in syngas with steam (NCERT)
in the presence of iron chromate catalysts. 
Sol. (i) nH 2  g   MmOn  s    mM  s   nH 2 O  l 
86 HYDROGEN

  CH OH l methanol B2 H 6  2NH 3   BH 2  NH 3  2   BH 4 

(ii) CO  g   2H 2  g  
Catalyst 3   
B2 H 6  2NaH  2Na   BH 4   Sodium boronhydride 

Example-9
Which gaseous compound on treatment with dihydrogen Example-14
produces methanol?
How do you expect the metallic hydrides to be useful hy-
Sol. Carbon Monoxide drogen storage ? Explain.
Example-10 (NCERT)
Give an example of an ionic hydride and a covalent hydride. Sol. In some of the transition metal hydrides, hydrogen is
Sol. Ionic Hydride: NaH absorbed as H atoms. Due to the inclusion of H-atoms, the
Covalent Hydride: NH3 metal lattice expands and thus becomes less stable.
Example-11 Therefore, when such metallic hydride is heated, it
Arrange the following: decomposes to release hydrogen gas and very finely divided
metal. The hydrogen evolved in this manner can be used as
(i) CaH2, BeH2 and TiH2 in order of increasing electrical
a fuel. Thus, transition metals or their alloys can act as
conductance.
sponge and can be used to store and transport hydrogen
(ii) LiH, NaH and CsH in order of increasing ionic character.
to be used as a fuel.
(iii) H-H, D–D and F–F in order of increasing bond
dissociation enthalpy. Example-15
(iv) NaH, MgH2 and H2O in order of increasing reducing What so you understand by the term ‘auto-protolysis’ of
properly. water? What is its significance?
(NCERT) (NCERT)
Sol. (i) BeH2 < TiH2 < CaH2 Sol. Auto-protolysis means self-ionisation of water. It may be
(ii) LiH < NaH < CsH represented as
(iii) F–F < H–H < D–D H 2 O  l  H 2 O  l   H 3 O   aq  OH   aq 
(iv) H2O < MgH2 < NaH Acid1 Base2 Acid 2 Base
Example-12 Due to auto-protalysis water is amphoteric in nature, i.e., it
Do you expect the carbon hydride of the type (CnH2n+2) to can act as an acid as well as base.
act as Lewis acid or base ? Justify your answer. Example-16
(NCERT) Consider the reaction of water with F2 and suggest, in terns
Sol. Carbon hydride of the type (CnH2n+2) are electron precise of oxidation and reduction, which species are oxidised/
hydrides. In other words, they have exact numbers of reduced?
electrons required to form covalent bonds. Therefore, they (NCERT)
do not have tendency to either gain or lose electrons and
hence, they do not act as Lewis acids or Lewis bases. Sol. 2F2  aq   2H 2 O  l   O 2  g   4H   aq   4F  aq 
In this reaction water acts as a reducing agent and itself
Example-13 gets oxidised to O2 while F2 acts as an oxidising agent and
What characteristics do you expect from electron deficient hence itself reduced to F  ions.
hydrides with respect to their structure and chemical Example-17
reactivity ? Explain the following :
(NCERT) (a) Water is excellent solvent for ionic compounds.
Sol. Electron deficient hydrides do not have sufficient number (b) Lakes freeze from top to bottom.
of electrons to form normal covalent bonds. They generally Sol. (a) Water has a high dielectric consstnt (78.39) due to the
exist in polymeric forms such as B2H6, B4H10, (AIH3)n, etc. polar character of its molecule. Water is an excellent solvent
Due to deficiency of electrons, these hydrides act as Lewis for many ionic as well as covalent compounds. Dissolution
acids and thus, form complex entities with Lewis bases such of ionic compounds takes place because of ion-dipole
as : NH3, H  ions, etc. interactions. Dissolution of molecular compounds such as
HYDROGEN 87
alcohols, amides, urea, sugar, glucose. honey, etc., in water Example-22
takes place because of the tendency of these substances What happens when ?
to form hydrogen bonds with water molecules. (i) Heavy water reacts with magnesium nitride.
(b) This is due to the fact that the frozen water does not (ii) Heavy water reacts with sodium.
sink to the bottom but keeps floating at the surface due to
Sol. (i)
its lesser density. This provides thermal insulation to the
water below it. The lesser density of ice can be attributed to Mg3 N 2  6D2 O  Mg  OD 2  2ND3  Deutrammonia 
open cage-like structure on account of hydrogen bonding.
Example-18 (ii) Na  2D 2 O  2NaOD  D 2 .

What is 'demineralised water' and how can it be obtained?

(NCERT) Example-23
Sol. Water which is free from all cations and anions is called Knowing the properties of H2O and D2O, do you think D2O
demineralised water. It is obtained by passing hard water can be used for drinking purpose.
first through cation exchange resin and then through anion (NCERT)
exchange resin. Sol. No, D2O is injurious to human beings, plants and animals.

Example-19 Example-24
What properties of water make it useful as a solvent? what Explain the following :
types of compound can it (i) dissolve (ii) hydrolyse? (i) Why hydrated barium peroxide is used in the preparation
(NCERT) of hydrogen peroxide instead of the anhydrous variety ?
Sol. Water is highly polar in nature thats why it has high dielectric (ii) Phosphoric acid is preferred to sulphuric acid in the
constant and high dipole moment. Because of these preparation of H2O2 from barium peroxide.
properties, water is a universal solvent. (iii) Statues coated with white lead on long exposure to
It can hydrolyse many oxides metallic or non-metallic, atmosphere turns black and the original colour can be
hydrides, carbides. nitrides etc. restored on treatment with H2O2.
Sol. (i) If anhydrous barium peroxide is used in the preparation,
Example-20 the barium sulphate, thus formed forms an insoluble
What causes the temporary and permanant hardness of protective coating on the surface of solid barium peroxide
water? hydrated. This prevents the further reaction of the acid.
i.e., causing the reaction to stop. If, however, hydrated
(NCERT)
barium peroxide (in the form of the paste) is used, the water
Sol. Temporary hardness of water is due to the presence of causes to dislodge the insoluble BaSO4 from the surface of
bicarbonantes of calcium and magnesium in water i.e.,
BaO2. Hence BaSO4 settles at the bottom of the reaction
Ca(HCO3)2 and Mg(HCO3). Permanent hardness of water is
vessel and the reaction continues without any difficulty.
due to the presence of soluble chlorides and sulphates of (ii) When phosphoric acid is used in the preparation of
calcium and magnesium i.e., CaCl2, CaSO4, MgCl2 and
H2O2 from BaO2 it plays the dual role. It liberates H2O2 and
MgSO4.
also acts as a preservator by retarding its decomposition.
(iii) White lead is used as a pigment. The statues coated
Example-21 with white lead get blackened due to the action of H2S
Hard water is softened before using in boilers. present in atmosphere in traces.
Sol. Hard water on boiling forms precipitates of MgCO3, and Pb(OH)2  white lead  .2PbCO3  3H 2S
CaSO4 which form scales in the boilers.As of result of these
scales in the boilers, the boiler gets deteriorated due to  3PbS  2CO 2  4H 2 O
over heating. Moreover, these scales are non-conducting
and therefore, more fuel is consumed. Therefore, in order to
prevent the formation of scales, hard water is softened
before using in boilers.
88 HYDROGEN
Example-25
Calculate the volume strength of 2N H2O2 solution.
Sol. Strength = Normality × Equivalent weight
Normality = 2N, Equivalent weight of H2O2 = 17
 Strength = 2 × 17 = 34g/L
Hydrogen peroxide decomposes as:
2H2O2  2H2O + O2
2 × 34 = 68g 22.4L at N.T.P.
68 g of H2O2 produce O2 at N.T.P = 22.4L

22.4  34
34 g of H2O2 will produce O2 at N.T.P =
68
= 11.2 L
So given solution of H2O2 produces 11.2L of O2 at N.T.P.
Volume strength = 11.2 volumes
HYDROGEN 89

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Which isotope of hydrogen

(i) does not contain neutron? 17. Show with reaction the amphoteric nature of water.
(ii) is radioactive?
18. Distinguish between:
2. Name the isotopes of hydrogen. Temporary hardness & permanent hardness of the
water.
3. Give the laboratory method of preparation of hydrogen.
19. Give the main characteristics of isotopes.
4. Give the commercial method of preparation of
dihydrogen. 20. Classify the following as covalent, ionic
polymer or interstitial hydrides:
5. Why is dihydrogen gas not preferred in balloons? (i) BeH2 (ii) CaH2 (iii) GeH4 (iv) TiH2

6. How is ammonia prepared using dihydrogen? Section – C (3 Marks Questions)

7. Give an example of each of an ionic hydride and a 21. (i)Why is the ionisation enthalpy of hydrogen higher
covalent hydride. than that of sodium?
(ii)Why does hydrogen occupy unique position in the
8. What happens when water is added to calcium hydride? periodic table?

9. What is Calgon? 22. What are the advantages in using hydrogen as a fuel?

10. How is heavy water obtained from ordinary water? 23. Compare the structure of H2O and H2O2.

Section – B (2 Marks Questions) 24. Complete the following equations.

(i) PbS + H2O2 →
11. Define hard water.What makes water hard ?Does hard (ii) MnO4– + H2O2+ H+ →
water give lather with soap ?
25. Comment upon the reactions of dihydrogen with
12. Why is ice less dense then water and what kind of (i) Chlorine
attractive forces must be overcome to melt ice? (ii) Sodium and
(iii) Copper (II) oxide
13. Why is water an excellent solvent for ionic or polar
substances? 26. Describe the usefulness of water in biosphere and
biological systems.
14. What do you understand by : (i)water – gas shift
reaction (ii) syn-gas? 27. Calculate the strength of 10 volume solution of
hydrogen peroxide.
15. Write two uses of interstitial hydrides.

16. What happens when sodium hydride reacts with water?

90 HYDROGEN

28. Dihydrogen reacts with dioxygen (O2) to form water. On the basis of this data explain in which of these
Write the name and formula of the product when the liquids intermolecular forces are stronger?
isotope of hydrogen which has one proton and one
neutron in its nucleus is treated with oxygen? Will the (ii) When the first element of the periodic table is
reactivity of both the isotopes be the same towards treated with dioxygen, it gives a compound whose
oxygen? Justify your answer. solidstate floats on its liquid state. This compound
has an ability to act as an acid as well as a base.
Section – D (5 Marks Questions) What products will be formed when this compound
undergoes auto ionization?
29. (i) Melting point, enthalpy of vapourisation and (iii) Complete the following reaction:
Boil
viscosity data of H2O and D2O is given below: Mg(HCO3 )2 ⎯⎯⎯
→ _______ + 2CO2 .
H2O D2O
Melting point/K 273.0 276.8 30. (i) Give a method for the manufacture of hydrogen
peroxide and explain the reactions involved
Enthalpy of vapourisation at(373 40.66 41.61 therein.
K) kJ mol–1 (ii) Illustrate oxidising and reducing properties of
Viscosity/centipoise 0.8903 1.107 hydrogen peroxide with equations.
HYDROGEN 91

EXERCISE – 2: Basic Objective Questions

Section-A (Single Choice Questions)
8. What is the trend of boiling points of hydrides of N, O
1. Which of the following is not an isotope of hydrogen? and F?
(a) Protium (b) Ortho-para hydrogen (a) Due to lower molecular masses, NH3, H2O and HF
(c) Deuterium (d) Tritium have lower boiling points than those of the
subsequent group member hydrides.
2. Hydrogen is prepared by the reaction of (b) Due to higher electronegativity of N, O and F; NH 3,
(a) Electrolysis of acidified water using Pt H2O and HF show hydrogen bonding and hence
(b) Granulated zinc with dilute HCl higher boiling points than the hydrides of their
(c) By water gas shift reaction subsequent group members.
(d) All of the above (c) There is no regular trend in the boiling points of
hydrides.
3. Which of the following metals does not liberate
(d) Due to higher oxidation states of N, O and F, the
hydrogen from acids?
(a) Fe (b) Cu boiling points of NH3, H2O and HF are higher than
(c) Mg (d) Zn the hydrides of their subsequent group members.

4. Which of the following statement is correct regarding 9. Which one of the following statements about water is
properties of hydrogen? FALSE?
(a) It has a strong odour (a) Water can act both as an acid and as a base.
(b) It is lighter than air (b) There is extensive intramolecular hydrogen bonding
(c) It can behave as a metalloid in the condensed phase.
(d) None of these (c) Ice formed by heavy water sinks in normal water.
(d) Water is oxidized to oxygen during photosynthesis.
5. Hydrogenation of vegetable oils using nickel as a
catalyst gives edible fats which is/are known as 10. Which of the following reactions shows reduction of
(a) Coconut oil water?
(b) Soyabean oil (a) 2H2O + 2Na → 2NaOH + H2
(c) Margarine and vanaspati ghee (b) 2F2 + 2H2O → 4H+ + 4F− + O2
(d) Vanaspati ghee
(c) P4O10 + 6H2O → 4H3PO4
6. Dihydrogen, under certain reaction condition, combines (d) None of these
with almost all elements except noble gases to form
binary compounds. The binary compounds are called 11. Which of the following compounds is used for water
(a) Oxides (b) Saline hydrides softening?
(c) Molecular hydrides (d) Hydrides (a) Ca 3 ( PO 4 )2 (b) Na 3 PO 4
(c) Na 6 P6 O18 (d) Na 2 HPO 4
7. Which of the following statement regarding hydrides is
not correct?
(a) Ionic hydrides are crystalline, non-volatile and 12. The temporary hardness of water due to calcium
non-conducting in solid state. bicarbonate can be removed by adding
(b) Electron-deficient hydrides act as Lewis acids (a) CaCO3 (b) CaCl2
or electron acceptors.
(c) HCl (d) Ca(OH)2
(c) Elements of group-13 form electron-deficient
hydrides.
(d) Elements of group 15-17 form electron-precise
hydrides.
92 HYDROGEN

13. Chemical A is used for water softening to remove 20. Like alkali metals, hydrogen forms
temporary hardness. A reacts with sodium carbonate to I. Oxides and oxyacids.
generate caustic soda. When CO2 is bubbled through A, II. Halides and interhalides.
it turns cloudy. What is A? III. Oxides, halides and sulphide.
(a) CaCO3 (b) CaO (a) I and II (b) III only
(c) Ca(OH)2 (d) Ca(HCO3)2 (c) I only (d) II and III

14. Heavy water is used as Section-B (Assertion & Reason Type Questions)
(a) drinking water (b) detergent
(c) washing water (d) a moderator 21. Assertion (A): Hydrogen is the first element in the
periodic table.
15. In which media H2O2 acts as an oxidising as well as Reason (R): It has electronic configuration 1s1.
reducing agent?
(a) Both A and R correct; R is the correct
(a) Acidic
explanation of A
(b) Alkaline
(b) Both A and R are correct; R is not the correct
(c) Acidic and alkaline both
explanation of A.
(d) Acidic, alkaline and neutral
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect.
16. Which of the following statements(s) is /are correct
about fuel cells?
22. Assertion (A): Soft water are free from the soluble
(a) They are highly efficient
salts of calcium and magnesium.
(b) They are free from pollution
Reason (R): It does not give lather with soap easily.
(c) They run till reactants are active
(a) Both A and R correct; R is the correct
(d) All of these
explanation of A
(b) Both A and R are correct; R is not the correct
17. Carbon hydride of the type, CnH2n+2 do not act as Lewis
explanation of A
acid or Lewis base. They behave as normal covalent
(c) A is correct; R is incorrect
hydrides because
(d) R is correct; A is incorrect
(a) carbon hydrides are electron-rich hydrides
(b) carbon hydrides are electron-deficient hydrides
23. Assertion (A): Hard water forms scum/precipitate with
(c) carbon hydrides are electron-precise hydrides
soap.
(d) carbon hydrides are non-stoichiometric hydrides
Reason (R): Formation of scum occurs as
2C17 H35COONa ( aq ) + M 2 + ( aq ) →
18. The H−O−H angle in water molecule is about
(a) 90° (b) 180° ( C17 H35COO )2 M  +2Na + ( aq ) ; M is Ca / Mg
(c) 102° (d) 104.5° (a) Both A and R correct; R is the correct
explanation of A
19. Hydrogen resembles halogens in many respects for (b) Both A and R are correct; R is not the correct
which several factors are responsible. Of the following explanation of A
factors, which one is the most important in this respect? (c) A is correct; R is incorrect
(a) Its tendency to lose an electron to form a cation (d) R is correct; A is incorrect
(b) Its tendency to gain a single electron in its
valence shell to attain stable electronic configuration
(c) Its low negative electron gain enthalpy value
(d) Its small size
HYDROGEN 93

24. Assertion (A):H2O2 is not stored in glass bottles. 28. Assertion (A):Hydrogen combines with other elements
Reason (R):Alkali oxides present in glass catalyse the by losing, gaining or sharing of electrons.
decomposition of H2O2. Reason (R):Hydrogen forms electrovalent and covalent
bonds with other elements.
(a) Both A and R correct; R is the correct
(a) Both A and R are correct; R is the correct
explanation of A
explanation of A
(b) Both A and R are correct; R is not correct
(b) Both A and R are correct; R is not the correct
explanation of A
explanation of A
(c) A is correct; R is incorrect
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
(d) R is correct; A is incorrect

25. Assertion (A): Increasing pressure on pure water

decreases its freezing point.
Section – C (Case Study Questions)
Reason (R): Density of water is maximum at 4° C.
(a) Both A and R are correct; R is the correct Case Study-1
explanation of A
(b) Both A and R are correct; R is not the correct Hydrogen has electronic configuration 1s1. It’s
electronic configuration is similar to the outer
explanation of A
electronic configuration (ns1) of alkali metals, which
(c) A is correct; R is incorrect belong to the first group of the periodic table. Also, like
(d) R is correct; A is incorrect halogens (with ns2 np5 configuration belonging to the
seventeenth group of the periodic table), it is short by
26. Assertion (A): In alkaline medium, H2O2 reacts with one electron to the corresponding noble gas
potassium ferricyanide. configuration, helium (1s2). Therefore, hydrogen has
Reason (R): H2O2 is a strong reducing agent. resemblance to alkali metals, which lose one electron to
form unipositive ions, as well as with halogens, which
(a) Both A and R are correct; R is the correct gain one electron to form uninegative ion. Hydrogen
explanation of A
has three isotopes: protium 11 H, deuterium 12 H or D
(b) Both A and R are correct; R is not the correct
and tritium, 13 H or T.
explanation of A
(c) A is correct; R is incorrect 29. Hydrogen has resemblance to alkali metals as well as
(d) R is correct; A is incorrect with halogens. The reason is that
(a) It loses one electron to form unipositive ion and
27. Assertion (A): Permanent hardness of water can be gains one electron to form uninegative ion
removed by using washing soda. (b) It loses one electron to form uninegative ion and
Reason (R): Washing soda reacts with soluble calcium gains one electron to form unipositive ion
and magnesium chlorides and sulphates in hard water to (c) It gains one electron like alkali metals and six
electrons like halogens
form insoluble carbonates.
(d) None of the above
(a) Both A and R are correct; R is the correct
explanation of A 30. Which of the following explanations justifies not
(b) Both A and R are correct; R is not the correct placing hydrogen in either the group of alkali metals or
explanation of A halogens?
(c) A is correct; R is incorrect (a) The ionisation energy of the hydrogen is too high
(d) R is correct; A is incorrect for group of alkali metals and too low for the
halogen group
(b) Hydrogen atom does not contain any neutron
(c) Hydrogen is much lighter than the alkali metals or
halogens
(d) Hydrogen can form compounds with almost all
other elements
94 HYDROGEN

31. In what respect electronic configuration of hydrogen 35. Non-stoichiometric hydrides are produced by
and halogens are similar? (a) palladium, vanadium (b) manganese, lithium
(a) Hydrogen and halogens have one electron in their (c) nitrogen, fluorine (d) carbon, nickel
outermost shell. 36. CH 4 , NH 3 , H 2 O and HF are the example of
(b) Hydrogen and halogens have one electron less than (a) Molecular hydrides
the noble gas configuration. (b) Metallic hydrides
(c) Hydrogen and halogens can lose one electron to (c) Ionic hydrides
form positive ions. (d) Electron deficient covalent hydrides
(d) Hydrogen and halogens show noble gas
configuration. Case Study-3

32. Which of the following statement regarding protium, In the gas phase water is a bent molecule with a bond
deuterium and tritium is not correct? angle of 104.5°, and O–H bond length of 95.7 pm.In
(a) They are isotopes of hydrogen the liquid phase water molecules are associated
(b) They have similar electronic configurations together by hydrogen bonds. The crystalline form of
(c) They exist in the nature in the ratio 1:2:3 water is ice. At atmospheric pressure ice crystallises in
(d) Their atomic masses are in the ratio 1:2:3 the hexagonal form, but at very low temperatures it
condenses to cubic form. Density of ice is less than that
of water. Therefore, an ice cube floats on water. In
Case Study-2 winter season ice formed on the surface of a lake
provides thermal insulation which ensures the survival
Dihydrogen is inactive at room temperature because of of the aquatic life. This fact is of great ecological
very high negative dissociation enthalpy, still it significance.
combines with almost all the elements under
appropriate conditions to form hydrides. All the type of 37. The maximum number of hydrogen bonds formed by a
hydrides can be classified into three types: ionic or water molecule in ice is
saline hydrides, covalent or molecular hydrides and (a) 4 (b) 1
metallic or non-stoichiometric hydrides. Alkali metal (c) 2 (d) 3
hydrides are good reagents for preparing other hydride
compounds. Molecular hydridesare of great importance 38. How many hydrogen bonded water molecules are
in our daily life. Metallic hydrides are useful for ultra- associated with CuSO4.5H2O?
purification of dihydrogen and as dihydrogen storage (a) Five (b) One
media. (c) Four (d) Three

33. Ionic hydrides react with water to give 39. The density of water is less in its solid state because
(a) Acidic solutions (b) Hydride ions (a) in solid state (ice), water molecules are arranged in
(c) Basic solutions (d) Electrons highly ordered open cage like structure.
(b) more extensive hydrogen bonding is present in solid
34. Earlier, it was thought that in the non-stoichiometric state ice
hydrides, hydrogen occupies interstices in the metal (c) the water molecules are closest in solid state of
lattice producing distortion without any change in its water.
type. Consequently, they were termed as (d) water in rigid crystalline, closely packed structure
………X…….. Here, X refers to in its solid state.
(a) Interstitial hydrides (b) Molecular hydrides
(c) Ionic hydrides (d) None of the above 40. Liquid water is denser than ice due to
(a) higher surface tension (b) hydrogen bonding
(c) van der Waals forces (d) covalent bonding
HYDROGEN 95

Notes:

Find Answer Key and Detailed Solutions at the end of this book

HYDROGEN
ORGANIC CHEMISTRY - SOME BASIC
PRINCIPLES AND TECHNIQUES
ORGANIC CHEMISTRY - SOME BASIC PRINCIPLES & TECHNIQUES 97

Chapter at a Glance
 The nature of the covalent bonding in organic compounds  A heterolytic cleavage yields carbocations or carbanions,
can be described in terms of orbitals hybridisation concept, while a homolytic cleavage gives free radicals as reactive
according to which carbon can have sp 3, sp 2 and sp intermediate.
hybridised orbitals.
 In organic chemistry, a number of organic compounds
 A sp3 hybrid orbital can overlap with 1s orbital of hydrogen
having different physical and chemical properties can be
to give a carbon - hydrogen (C–H) single bond (sigma, 
represented by the same molecular formula. The property
bond). by virtue of which organic compounds possessing different
 Overlap of two sp2 orbital carbon results in the formation of physical and chemical properties and having the same
a carbon–carbon  bond. molecular formula is known as isomerism and the different
compounds are known as isomers.
 The unhybridised p-orbitals on two adjacent carbons can
undergo side-by side overlap to give a pi (  ) bond.

 Organic compounds can be represented by various  It is of two types: Structural Isomerism & Stereo

structural formulas. Isomerism.

 Organic compounds can be classified on the basis of their  There are various methods of purification like sublimation,
structure or the functional groups they contain. distillation and differential extraction are based on the
difference in one or more physical properties.
 The naming of the organic compounds is carried out by a
set of rules laid down bythe International Union of Pure Chromatography is a useful technique of separation, identi-
and Applied Chemistry (IUPAC). fication and purification of compounds.

 In IUPAC nomenclature, the names are written with the  Nitrogen, sulphur, halogens and phosphorus are detected
structure in such a way that the reader can deduce the by Lassaigne’s test. Carbon and hydrogen are estimated by
determining the amounts of carbon dioxide and water
structure from the name.
produced.
 A covalent bond may be cleaved in heterolytic or homolytic
fashion.  Nitrogen is estimated with Duma’s or Kjeldahl’s method
and halogens by Carius method. Sulphur and phosphorus
 The inductive, resonance, electromeric and
are estimated by oxidising them to sulphuric and
hyperconjugation effects may help in the polarisation of a
bond making certain carbon atom or other atom positions phosphoric acids respectively.
as places of low or high electron densities.
98 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

Solved Examples
Example-1 Example-4
What are hybridisation states of each carbon atom in the Explain, how is the electronegativity of carbon atoms
following compounds? related to their state of hybridization in an organic
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6. compound?
(NCERT) Sol. Electronegativity of carbon atoms increases with increase
Sol. in s-character of the hybridized carbon.
sp3  sp 2  sp
25% s-character 33.3% s-character 50% s-character
Thus, sp hybridized carbon is most electronegative.
Example-5
Indicate the σ- and π-bonds in the following molecules.
C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3
(NCERT)
Sol.
Example-2
What is the state of hybridisation of C2, C3, C5 and C6 of
the hydrocarbon?

Sol. If number of σ bonds = 2; hybridisation is sp

If number of σ bonds = 3; hybridisation is sp2
If number of σ bonds = 4; hybridisation is sp3
Example-6
Write bond-line formulas for: Isopropyl alcohol, 2,3-
dimethylbutanal, heptan-4-one. (NCERT)
Sol.

Example-3
What is the hybridization of each carbon in
H2C = C = CH2?
Sol.
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 99

Example-7 Sol. 1. Electrophiles: The neutral or positively charged species

Give condensed and bond line structural formulas and with deficiency of electrons and capability of accepting
identify the functional group(s) present, if any, for a pair of electrons (Lewis acid) are called electrophiles.
(a) 2,2,4-trimethylpentane Some examples are,
(b) 2-hydroxy-1,2,3-propanetricarboxylic acid Neutral electrophiles: BF3, AlCl3, SO3, FeCl3, SiCl4, :CH2
(c) Hexanedial. (NCERT) (carbenes),
Sol.
(a)
Charged electrophiles H+, Cl+, Br+, NO+, NO 2 ,

2. Nucleophiles: The neutral or negatively charged species

which are capable of acting as donor of electron pair
(Lewis base) are called nucleophiles. Some examples
(b) are,
Neutral nucleophiles:
      
N H3 , R N H2 , R 2 N H, R 3 N , H 2 O:, R O H, R O R
 

Charged nucleophiles: CN , OH , X , OR , SH , RCOO–

– – – – –

(c) Example-9
Identify the reagents shown in bold in the following
OH C C H 2  C H 2  C H 2  C H 2  C HO
1 2 3 4 5 6 equations as nucleophiles or electrophiles.
Hexanedial
(a) CH 3 COOH  HO   CH 3 COO   H 2 O

Condensed Bond line Functional (b) CH3 COCH 3  C N   CH3 2 C  CN  OH 
formula formula groups 
(c) C6 H 6  CH 3 CO  C 6 H 5 COCH 3 (NCERT)
(a)
(CH3)3CCH2CH-(CH3) 2 Sol. (a) HO– is a nucleophile (b) –CN is a nucleophile (c) CH3C+O
is an electrophile
Example-10
Give the IUPAC name of the following compounds whose
bond line formulae are given below.
(b) HOOCCH2C(OH)

(COOH)CH2COOH (carboxyl) and

–OH(hydroxyl)

Sol.
(c) OHC(CH2)4CHO (i)

(aldehyde)

Example-8
What are electrophiles and nucleophiles? Explain with
examples. (NCERT)
100 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

(ii) Sol. IUPAC name of the given compound is

2,3-dimethyl pentanoyl chloride

Example-14
Example-11 Give the IUPAC names of the following compounds:
What is the IUPAC name of the compound having the (NCERT)
formula (a) (b)
CH  C – CH = CH2 is
Sol. In IUPAC nomenclature double bond is given more
preference than triple bond.
4 3 2 1
CH  C  C H  CH 2 (c) (d)
1-butene-3-yne
But-1-en-3-yne
Example-12
What is the IUPAC name of the compound?

(e)

Sol. (f) Cl2CHCH2OH

Sol.

(a)
Propylbenzene
CHO group gets higher priority over ketone group in
numbering of principle carbon chain. IUPAC name is 3-
keto-2-methylhex-4-enal
Example-13
What is the IUPAC name of
(b)
3-methylpentanenitrile

(c)

2,5-dimethylheptane
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 101

(d)
(d)
3-bromo-3-chloroheptane
Example-16
Which of the two: O2NCH2CH2O- or CH3CH2O- is expected
to be more stable and why? (NCERT)
(e)
Sol. O2N—CH2—CH2O- is expected to be more stable than
CH3—CH2O- because —NO2 group has -I effect, this leads
3-chloropropanal to the dispersal of negative charge. On the other hand
—CH3 group has +I effect, this leads to the intensification
of the negative charge. Dispersal of the charge leads to
(f)
the stability of ion while intensification of negative charge
2,2-dichloroethanol leads to the un stability of ion.

Example-15 More stable

For the following bond cleavages, use curved arrows to
show the electron flow and classify each as homolysis or
Less stable
heterolysis. Identify reaction intermediate produced as
Example-17
free radical, carbocation and carbanion. (NCERT)
Identify the most stable species in the following set of
 
(a) CH 3 O  OCH 3  CH 3 O  O CH 3 ions and give the reason:
   
(i) C H 3 , C H 2 Br, C HBr2 , C Br3
(b)
Θ Θ Θ Θ
(ii) C H 3 , C H 2 Cl, C HCl2 , C Cl3

(c) 
Sol. (i) C H 3 is the most stable species because the
replacement of H by Br increases positive charge on
carbon atom and destabilises the species.
(d) Θ
(ii) C Cl3 is the most stable species because on replacing

Sol. H by Cl, negative charge on carbon is reduced and

species is stabilized.
Example-18
Draw the possible resonance structures for
 
CH 3  O C H 2 and predict which of the structures is more


stable. Give reason for your answer.

Sol. The given carbocation has two resonance structures,

(c)

Structure (II) is more stable because both the carbon atoms

and the oxygen atom have an octet of electrons.
102 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

Example-19 Example-22
Explain why alkyl groups act as electron donors when 0.2475 g of an organic substance gave on combustion
attached to a π-system? 0.4950 g of carbon dioxide and 0.2025 g of water. Calculate
Sol. Due to hyperconjugation, alkyl groups act as electron the percentage of carbon and hydrogen in it.
donors when attached to a π-system as shown below.
Sol. Weight of the substance taken = 0.2475 g
Weight of CO2 formed = 0.4950 g
Weight of H2O formed = 0.2025 g
Now, we know
CO2 = C and H2O = 2H
44 12 18 2
Weight of C in 0.4950 g of CO2

0.4950  12
 g
44
Example-20
Weight of H in 0.2025 g of H2O
In Lassaigne’s test for nitrogen, the blue colour is due to
the formation of 0.2025  2
 g
(a) Ferriferrocyanide 18

(b) Potassium ferrocyanide Hence, percentage of C

(c) Sodium ferrocyanide 0.4950  12 100

   54.54
(d) Sodium cyanide 44 0.2475
Ans. (a) and percentage of H
Sol. In Lassaigne’s test substance is heated strongly with 0.2025  2 100
sodium metal then water extract is boiled with alkaline FeSO4    9.09
18 0.2475
solution and after cooling FeCl3 solution and excess of HCl
is added in it. If prussian blue or green ppt. is obtained then Example-23
Nitrogen is confirmed – 0.1877 g of an organic substance when analysed by the
Dumas method yield 31.7 ml of moist nitrogen measured at
Na  C  N 
 NaCN 14C and 758 mm mercury pressure. Determine the
percentage of nitrogen in the substance. (Aqueous tension
FeSO 4  2NaOH 
 Fe(OH) 2  Na 2SO 4
at 14C = 12 mm).
6NaCN  Fe(OH) 2 
 Na 4 [Fe(CN)6 ]  2NaOH v(P  a) 273
Sol. Volume of N2 at NTP  
3Na 4  Fe  CN 6   4FeCl3  t  273 760
= V ml
Fe 4 [Fe(CN)6 ]3  12NaCl
substituting the various values in the above equation,
Ferriferrocyanide (Prussian blue)
Example-21 31.7(758  12) 273
V   29.6 ml
14  273 760
In sodium fusion test of organic compounds, the nitrogen
of an organic compound is converted to 28
Weight of 29.6 ml of nitrogen =  29.6 g
(a) Sodamide (b) Sodium cyanide 22400
(c) Sodium nitrite (d) Sodium nitrate
28  29.6 100
Ans. (b)  Percentage of nitrogen    19.72
22400 0.1877
fusion
Sol. Na  C  N   NaCN
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 103
Example-24 Example-25
Explain, why an organic liquid vapourises at a temperature Name a suitable technique of separation of the components
below its boiling point in its steam distillation? (NCERT) from a mixture of calcium sulphate and camphor. (NCERT)
Sol. In steam distillation, the organic liquid starts to boil when Sol. Sublimation, because camphor can sublime wheras CaSO4
the sum of vapour pressure due to the organic liquid (p1) does not.
and the vapour pressure due to water (p2) becomes equal
to atmospheric pressure (p), i.e. p = p1 + p2. Since p1 < p2,
organic liquid will vapourise at a lower temperature than its
boiling point.
104 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Define homologous series. 14. Give the structure and hybridization of

I. CH3+
2. Define carbanion. II. CH3–

3. Give the reason for the fusion of an organic compound 15. What is the type of hybridization of each carbon in the
with sodium metal for testing nitrogen, sulphur and following compounds?
halogens. (a) CH3Cl
(b) HCONH2
4. Write the expanded form of the following condensed
formulas into their complete structural formulas. 16. Write the structural formula of:
(a) CH3CH2COCH2CH3. (a) o-Ethylanisole (b) p-Nitroaniline
(b) CH3CH = CH(CH2)3CH3.
17. Identify electrophilic centre in the following:
5. What is a functional group? CH3CH = O, CH3l.

6. A species having a carbon atom possessing ………….. 18. In which C–C bond of CH3CH2CH2Br, the inductive
of electrons and a positive charge is called a effect is expected to be the least?
carbocation.
19. What is the basic principle of Crystallization &
7. Free radicals are formed due to ………….. bond Chromatography?
fission.
20. Write formulas for the next four members of
8. The Prussian blue colour obtained during Lassaigne’s homologous series for the compound, HCOOH.
test for nitrogen is due to the formation of ………
Section – C (3 Marks Questions)
9. The different fractions of crude oil can be separated by

10. Inductive effect involves displacement of electrons 21. Giving justification, categorise the following
(True/False). molecules/ions as nucleophile or electrophile.
HS− , BF3 ,C2 H5 O− ,(CH3 )3 N :,
Section – B (2 Marks Questions) + + +
C l,CH3 − C = O, H 2 N − , N O2

11. What are reactive intermediates? How are they

22. Give reasons why the following two structures I and II
generated by bond fission?
cannot be the major contributors to the real structures
of CH3COOCH3.
12. Why does SO3 act as an electrophile?

13. Write structural formulae for compounds named as

(a) 1-bromoheptane
(b) 5-bromoheptanoic acid
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 105

23.
+
Explain why (CH3 )3 C is more stable than CH3 C H 2
+
Section – D (5 Marks Questions)
+
and CH3 is the least stable cation? 29. (a) What conclusions would you draw if the
Lassaigne’s extract gives a blood red colouration
24. In Dumas’ method for estimation of nitrogen, 0.3g of with FeCl3?
an organic compound gave 50mL of nitrogen collected (b) Suggest methods for the separation of the
at 300K temperature and 715mm pressure. Calculate following mixture:
the percentage composition of nitrogen in the a mixture of liquid A (b.p. 365 K) and liquid B
compound. (Aqueous tension at 300K=15 mm) (b.p. 355 K)
(c) On complete combustion, 0.246 g of an organic
25. During estimation of nitrogen present in an organic compound gave 0.198g of carbon dioxide and
compound by Kjeldahl’s method, the ammonia evolved 0.1014g of water. Determine the percentage
from 0.5 g of the compound in Kjeldahl’s estimation of composition of carbon and hydrogen in the
nitrogen, neutralized 10 mL of 1 M H2SO4. Find out compound.
the percentage of nitrogen in the compound
30. (a) What type of structural isomerism is shown by
26. What is the difference between distillation, distillation
under reduced pressure and steam distillation?

27. Discuss the principle of estimation of halogens and

sulphur present in an organic compound. (b) Write all structural isomers of molecular formula
C3H6O.
28. What is the relationship between the members of the (c) Draw the resonance structures of
following pairs of structures? Are they, structural,
geometrical or resonance contributors?
(a)

(b)

(c)
106 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

EXERCISE-2: Basic Objective Questions

Section – A (Single Choice Questions)

1. The hybridization of carbon atoms in C—C single bond 7. The IUPAC name of
of H — C  C — CH = CH2 is:
(a) sp3 – sp3 (b) sp2 – sp3
(c) sp – sp
3
(d) sp – sp2
is:
(a) 2,2-diethyl pentanoic acid
2. Nucleophile: ….?.... : : electrophile: electron seeking
(b) 2,4-dimethylhexanoic acid
(a) proton seeking (b) neutron seeking
(c) 2-methyl-4-ethyl pentanoic acid
(c) atom seeking (d) electron rich
(d) 4-ethyl-2-methyl pentanoic acid
3. Which of the following species is a nucleophile?
8. Consider the following carbocations
(a) H2O (b) H3O
(I) Cl3C+ (II) Cl2CH+
(c) both (a) and (b) (d) none of these +
(III) ClCH2 (IV) CH3+
4. Which of the following species is an electrophile? The stability sequence on the basis of inductive effect,

(a) H (b) H- follows the order
(c) both (a) and (b) (d) none of these (a) IV > I > II < III (b) I < II < III < IV
(c) IV < I < III < II (d) IV < II < I < III
5. The IUPAC name of the following compound is:
9.

(a) 2, 5, 6-trimethylheptane
(b) 1,3-isopropyl-3-methylpropane Which of the following orders is correct for the
(c) 2, 6, 3-trimethylheptane stability of these carbanions ?
(d) 2,3,6-trimethylheptane (a) I > II > III (b) III > II > I
(c) II > I > III (d) III > I > II
6. The name of
10. Consider the following carbocations,
+ +
(I) C6 H5 CH2 (II) C6 H5CH 2 C H 2
according to IUPAC nomenclature system is: + +

(a) 2,3-dibromo-1,4-dichlorobut-2-ene (III) C6 H5 C HCH3 (IV) C6 H 5 C ( CH 3 )2

(b) 1,4-dichloro-2,3-dibromobut-2-ene The correct sequence of the stability of these is
(c) dichlorobromobutene (a) II < I < III < IV (b) II < III < I < IV
(d) dichlorobromobutane (c) III < I < II < IV (d) IV < III < I < II

11. Which of the following has minimum-I-effect?

(a) –NO2 (b) – COOH
+
(c) –F (d) − N R 3
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 107

12. Electromeric effect is Section – B (Assertion & Reason Type Questions)

(a) permanent effect (b) temporary effect
(c) resonance effect (d) inductive effect 21. Assertion (A): The C  C distance is longer than
C = C distance
13. The kind of delocalisation involving sigma bond
orbitals is called Reason (R): Both C atoms in alkynes are sp2
(a) inductive effect hybridized.
(b) hyperconjugation effect (a) Both assertion and Reason are true and the Reason
(c) electromeric effect is the correct explanation of the Assertion.
(d) none of these (b) Both Assertion and Reason are true but Reason is
not the correct explanation of the Assertion.
14. The compounds CH3CH = CHCH3 and (c) Assertion is true but Reason is false.
CH3CH2CH = CH2 (d) Both Assertion and Reason are false.
(a) are tautomers
(b) are position isomers
22. Assertion (A): Ethylene is more reactive than ethane.
(c) are chain isomers Reason (R): Ethylene possess pi bond.
(d) exist together in dynamic equilibrium (a) Both assertion and Reason are true and the Reason
is the correct explanation of the Assertion.
15. The number of metamers possible for C4H10O is (b) Both Assertion and Reason are true but Reason is
(a) 1 (b) 2 not the correct explanation of the Assertion.
(c) 3 (d) 4 (c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
16. Which of the following compound will show
metamerism? 23. Assertion (A): Carbon is tetravalent and forms
(a) CH3 – CO – C2H5 (b) C2H5 – S – C2H5 covalent bonds.
(c) CH3 – O – CH3 (d) CH3 – O – C2H5 Reason (R): This can be explained in terms of its
electronic configuration and hybridization of s and p-
17. How will you purify a liquid having non-volatile orbitals
(a) Both assertion and Reason are true and the Reason
impurities?
is the correct explanation of the Assertion.
(a) Sublimation (b) Crystallization (b) Both Assertion and Reason are true but Reason is
(c) Simple Distillation (d) Chromatography not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
18. When 0.32g of a compound is heated with conc. HNO 3 (d) Both Assertion and Reason are false.
and BaCl2, 0.932g BaSO4 is obtained. The percentage
of sulphur in the compound is 24. Assertion (A): The two structures,
(a) 20 (b) 40
(c) 60 (d) 80

19. In Lassaigne’s test when both N and S are present,

blood red colour obtained is due to the formation of are chain isomers.
(a) Ferric ferrocyanide (b) Ferric thiocyanate Reason (R): The chain isomers differ in the chain of
carbon atoms.
(c) Ferric cyanide (d) None (a) Both assertion and Reason are true and the Reason
is the correct explanation of the Assertion.
20. Kjeldahl’s method is used in the estimation of (b) Both Assertion and Reason are true but Reason is
(a) Nitrogen (b) Halogens not the correct explanation of the Assertion.
(c) Sulphur (d) Oxygen (c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
108 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

25. Assertion (A): Methyl carbanions have pyramidal Section – C (Case Study Questions)
shape.
Reason (R): The carbon atom carrying negative charge Case – Study 1
has an octet of electrons.
(a) Both assertion and Reason are true and the Reason Organic compounds can be classified on the basis of
their structure or the functional groups they contain. A
is the correct explanation of the Assertion. functional group is an atom or group of atoms bonded
(b) Both Assertion and Reason are true but Reason is together in a unique fashion and which determines the
not the correct explanation of the Assertion. physical and chemical properties of the compounds.
(c) Assertion is true but Reason is false. The naming of the organic compounds is carried out by
following a set of rules laid down by the International
(d) Both Assertion and Reason are false. Union of Pure and Applied Chemistry (IUPAC). In
IUPAC nomenclature, names are correlated with the
26. Assertion (A): A mixture of o-nitrophenol and p- structure in such a way that the reader can deduce the
nitrophenol can be separated by steam distillation structure from the name.
Reason (R): p-Nitrophenol is steam volatile while o-
nitrophenol is not steam volatile. 29. Which of these contains the carbonyl group?
(a) ketones (b) aldehydes
(a) Both assertion and Reason are true and the Reason
(c) esters (d) all of these
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true but Reason is 30. What is the IUPAC name of t-butyl alcohol?
not the correct explanation of the Assertion. (a) Butan-2-ol (b) 2-Methylpropan-2-ol
(c) Butan-1-ol (d) Propan-2-ol
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false. 31. The suffixes for alcohols, aldehydes and ketones
according to IUPAC system are respectively :
27. Assertion (A): During test for nitrogen with (a) -alc, -ald, -ket (b) -ol, -al, -ket
Lassaigne’s extract on adding FeCl3 solution (c) -ol, -al, -one (d) -coh, -ald, -one
sometimes a red precipitate is obtained.
Reason (R): Sulphur is also present 32. Which of the following compounds has incorrect
(a) Both assertion and Reason are true and the Reason IUPAC nomenclature?
(a)
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true but Reason is
not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
(b)
(d) Both Assertion and Reason are false.

28. Assertion (A):

S
|| (c)
NH 2 − C − NH 2 gives red colour in Lassaigne’s test
Reason (R): Compounds having N along with C give
red colour in Lassaigne’s test.
(a) Both assertion and Reason are true and the Reason
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true but Reason is (d)
not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES 109

Case – Study 2 These organic reactions taking place involves the

breaking and making of covalent bonds. A covalent
Purification, qualitative and quantitative analysis of bond may be cleaved in heterolytic or homolytic way.
organic compounds are carried out for determining A heterolytic cleavage yields carbocations or
their structures. The methods of purification like carbanions, while a homolytic cleavage gives free
sublimation, distillation and differential extraction are radicals as reactive intermediate. Reactions proceeding
based on the difference in one or more physical through heterolytic cleavage involve the
properties. Chromatography is a useful technique of complimentary pairs of reactive species. These are
separation, identification and purification of electron pair donor known as nucleophile and an
compounds. It is classified into two categories: electron pair acceptor known as electrophile.
adsorption and partition chromatography.
Crystallization is one of the most commonly used 37. Due to the presence of an unpaired electron, free
techniques for the purification of solid organic radicals are
compounds. It is based on the difference in the (a) cations (b) anions
solubilities of the compound and the impurities in a (c) chemically inactive (d) chemically reactive
suitable solvent. After getting the compound in a pure
form, its qualitative analysis is carried out for detection
38. Which of the following species is a nucleophile?
of elements present in it. Nitrogen, sulphur, halogens
and phosphorus are detected by Lassaigne’s test. (a) CH3O- (b) CH3OH
Carbon and hydrogen are estimated quantitatively by (c) both (a) and (b) (d) none of these
determining the amounts of carbon dioxide and water
produced. Nitrogen is estimated quantitatively by 39. In heterolytic fission, _____ are formed. In homolytic
Dumas or Kjeldahl’s method and halogens by Carius fission, _____ are formed.
method. (a) one free radical and one cation; one free radical and
one anion
33. Which of the following will give Lassaigne’s test for
(b) free radicals; a cation and anion
nitrogen?
(c) a cation and anion; free radical
(a) NH4NO3 (b) NH2NH2
(d) one free radical and one anion; one free radical and
(c) KNO3 (d) Cyanogen
one cation

34. Dumas’ method involves the determination of nitrogen

40. Which of the following carbocations will be most
content in the organic compound in form of
stable?
(a) NH3 (b) N2 +
(c) NaCN (d) (NH4)2SO4 (a) Ph 3C+ (b) CH3 − CH 2
+
(c) (CH3 )2 CH (d) Both (a) & (b)
35. How will you separate iodine from sodium chloride?
(a) Sublimation (b) Distillation
(c) Both (a) and (b) (d) Crystallization

36. The best method for the separation of naphthalene and

benzoic acid from their mixture is
(a) sublimation (b) chromatography
(c) crystallization (d) distillation

Case – Study 3

Organic reaction mechanism concepts are based on the

fission of a covalent bond, structure of the substrate
molecule, the attacking reagents, the electron
displacement effects and the conditions of the reaction.
110 ORGANIC CHEMISTRY-SOME BASIC PRINCIPLES & TECHNIQUES

Notes:

Find Answer Key and Detailed Solutions at the end of this book

ORGANIC CHEMISTRY-SOME
BASIC PRINCIPLES & TECHNIQUES
HYDROCARBONS
112 HYDROCARBONS

Chapter at a Glance
HYDROCARBONS 113
114 HYDROCARBONS

Solved Examples
Example-1 From the above mechanism, it is evident that during
Write IUPAC names of the following compounds: propagation step, CH3 free radicals are produced which
may undergo three reactions, i.e., (i) — (iii). In the chain
(a) CH 3 CH  C  CH 3 2 termination step, the two CH3 free radicals combine together
to form ethane (CH3—CH3) molecule.
(b) CH 2  CH  C  C  CH 3 (NCERT)
Example-4
Isomeric pentanes have different value of boiling point for
example, n-pentane has highest boiling point among its
Sol. (a) three isomers. This is due to ______ ?
Sol. With increase in number of branched chains the molecule
attains the shape of a sphere. This result in smaller area of
(b) contact and therefore weak intermolecular forces between
spherical molecules, which are overcome at relatively lower
Example-2 temperatures.

In the alkane, CH3 CH 2  C  CH3  2  CH 2  CH  CH3 2 , Example-5

Higher alkanes on heating to higher temperature decompose
Identify 1o , 2o , 3o carbon atoms and give the number of H- into lower alkanes, alkenes etc, this process is known as
atoms bonded to each one of these. (NCERT) ______ ?
Sol. The expanded formula of the given compound is Sol. Decomposition of higher alkanes into smaller alkanes is
known as pyrolysis or cracking.

Pt/Pd/ Ni  C H  C H  Other products

C12 H 26  7 16 5 10
dodecane 973K heptane pentane

Example-6
Why is Wurtz reaction not preferred for preparation of
Example-3 alkanes containing odd number of carbon atoms? Illustrate
How do you account for the formation of ethane during your answer by taking one example. (NCERT)
chlorination of methane? (NCERT) Sol. For preparation of alkanes containing odd number of carbon
Sol. Chlorination of methane is a free radical reaction which atoms, a mixture of two alkyl halides has to be used. Since
occurs by the following mechanism; two alkyl halides can react in three different ways, therefore,
a mixture of three alkanes instead of the desired alkane
would be formed. For example, Wurtz reaction between ‘1-
bromopropane and 1-bromobutane gives a mixture of three
alkanes i.e., hexane, heptane and octane as shown below:
HYDROCARBONS 115

Sol. (i) C4 H10  g  13 / O2  g  


 4CO2  g   5H2 O  g 
Butane

(ii) C5 H10  g  15 / 2O 2  g  

 5CO2  g   5H 2 O  g 
Pentene

Example-7 (iii) C6 H10  g  17 / 2O 2  g  


 6CO 2  g   5H 2 O  g 
Hexyne
Spatial arrangement of atoms which can be converted into
one another by rotation around a C-C single bond in alkane (iv)
is termed as conformers. In the above context one must
remember that the rotation around the C-C single bond is
______ .
Sol. The rotation is hindered by a weak energy barrier of 1-
20KJ/mol due to weak repulsive interactions between
Example-11
adjacent bonds.
What effect does branching of an alkane chain has on its
Example-8
boiling point? (NCERT)
Give the representation of Eclipsed and the staggered
Sol. Branching of carbon atom chain decreases the boiling point
conformations.
of alkane.
Sol. Eclipsed and the staggered conformations can be
Example-12
represented by Sawhorse and Newman projection.
Which of the following has higher dipole moment; cis-2-
butene and trans-2-butene?
Sol. Cis from of alkene is found to be more polar than the trans
form. For example, dipole moment of cis-but-2-ene is 0.33
Dedye, whereas dipole moment of the trans form is almost
zero or it can be said that trans-but-2-ene is non-polar. This
can be understood by drawing 3 geometries of the two
forms as given below from which it is clear that in the trans-
but-2-ene, the two methyl groups are in opposite directions.
Therefore, dipole moments of C-CH2 bonds cancel, thus
making the transform non-polar.

Example-9
When 2-methyl propane is oxidized in the presence of
KMnO4, then product formed is
Example-13

Sol.  CH3 3 CH 

  CH3 3 COH
KMnO4
Oxidation
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and
 2  methypropane   2  methylpropan  2  ol  pentan-3-one. Write the structure and IUPAC name of ‘A’
Example-10 (NCERT)
Write chemical equations for the combustion reaction of Sol. Step-1. Write the structure of the products side by side
the following hydrocarbons, (i) Butane (ii) Pentane (iii) with their oxygen atoms pointing towards each other.
Hexyne (iv) Toluene. (NCERT)

Pentan-3-one
116 HYDROCARBONS

Step 2. Remove the oxygen atoms and join the two ends by Sol. (a) Isomers of C4H8 having one double bonds are:
a double bond, the structure of the alkene ‘A’ is

Example-14
An alkenen ‘A’ contains three C-C, eight C-H, a-bonds, and
one C-C -bond. ‘A’ on ozonolysis gives two moles of an
aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
(NCERT)
Sol. (i) An aldehyde with molar mass of 44 u is ethanal, CH3CH
=0
(ii) Write two moles of ethanal side by side with their oxygen
atoms pointing towards each other.

CH 3 CH  O O  CHCH 3
Ethanal Ethanal Example-17
(iii) Remove the oxygen atoms and join then by a double By treating calcium carbide with water ...A... prepared.
bond, the structure of alkene ‘A’ is Here, A refers to

Sol. CaC2  2H 2 O  Ca  OH  2  C 2 H 2
 ethyne 
A
As required, but-2-ene has three C  C , eight C  H a- Example-18
bonds and one C  C   bond . When proene reacts with one mole of HBr then product
obtained is
Example-15
Carbon-carbon double bond in alkene consists of ____ Br
|
bonds & _____ pi bonds. Sol. CH3  CH  CH2  HBr  CH2  CH CH3
Sol. Ethene is a class of alkene consists of one sigma bond and
one pi bond. Example-19
Explain why the following system is not aromatic?

(NCERT)
Sigma bond is formed by head on cverlapping and pi bond
is formed by sideways overlapping of orbitals.
Sol.
Example-16
For the following compounds, write structural formulas end
IUPAC names for all possible isomers having the number
of double or triple bond as indicated: (NCERT)
(a) C4H8 (one double bond) (b) C5H8 (one triple bond) Due to the presence of a sp3-hybridized carbon, the system
is not planar. It does contain six -electrons but the system
is not fully conjugated since all the six -electrons do not
form a single cyclic electron cloud which surrounds all the
atoms of the ring. Therefore, it is not an aromatic compound.
HYDROCARBONS 117

Example-20 Example-22
What are the necessary conditions for any system to be Why does benzene undergo electrophilic substitution
aromatic? (NCERT) reactions easily and nucleophilic substitutions with
Sol. The necessary conditions for a molecule to be aromatic are: difficulty? (NCERT)

It should have a cyclic cloud of delocalised Sol. Due to the presence of an electron cloud containing 6 -
-electrons above and below the plane of the molecule. electrons above and below the plane of the ring, benzene is
a rich source of electrons. Consequently, it attracts the
It should be planar. This is because complete delocalization
electrophiles (electron-deficient) reagents towards it and
of -electrons is possible only if the ring is planar to allow
repels nucleophiles (electron- rich) reagents. As a result,
cyclic overlap of p-orbitals.
benzene undergoes electrophilic substitution reactions
It should contain Huckel number of electrons, i.e., (4n + 2) easily and nucleophilic substitutions with difficulty.
n-electrons where n = 0, 1, 2, 3 etc. A molecule which does
Example-23
not satisfy any one or more of the above conditions is said
to be non-aromatic. Out of benzene, m-dinitrobenzene and toluene which will
undergo nitration most easily and why? (NCERT)
Example-21
Sol. CH 3 group is electron-donating while—NO 2 group is
How will you convert the following compounds into
electron-withdrawing. Therefore, maximum electron density
benzene? (NCERT)
will be in toluene, followed by benzene and least in m-
(i) Ethyne (ii) Ethene dinitrobenzene. Therefore, the ease of nitration decreases
in the order toluene > benzene > m-dinitrobenzene.
Example-24
Suggest the name of another Lewis acid instead of
Sol. (i) anhydrous aluminium chloride which can be used during
ethylation of benzene. (NCERT)
Sol. Anhydrous Ferric Chloride (FeCl3) another Lewis acid
(ii) Ethene is first concverted into ethyne and then to which can be used.
benzene as shown above. Example-25
(iii) Arrange the following set of compounds in order of their
decreassing relative reactivity with an electrophile, E+.
Chlorobenzene, 2, 4-dinitrochlorobenzene,
p-nitrochlorobenzene (NCERT)
Sol. The typical reactions of benzene are electrophilic
substitution reactions. Higher the electron-density in the
benzene ring, more reactive is the compound towards these
reactions. Since NO 2 is a more powerful electron-
withdrawing group than Cl, therefore, more the number of
nitro groups, less reactve is the compound. Thus, the
overall reactivity decreases in the order.
Chlorobenzene > p-nitrochlorobenzene >
2, 4-dinitrochlorobenzene,
118 HYDROCARBONS

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions) 15. Write the structural formulas for the following
compounds whose IUPAC names are given:
(i) 3,4,8-Trimethyldecane
1. The boiling point of hydrocarbons decreases with
(ii) 3,4,4,5-Tetramethylheptane
increase in branching. Give reason.
(iii) 2,5-Dimethylhexane
2. Unsaturated compounds undergo addition reactions.
16. Which of the following has the highest boiling point ?
Why?
(i) 2-Methylpentane
(ii) 2, 3-Dimethylbutane
3. State kharasch effect.
(iii) 2, 2-Dimethylbutane
4. How is alkyne prepared from calcium carbide?

5. How will you distinguish between ethylene and

methane? 17. Which among(a) and (b) are
aromatic?
6. Why are alkanes relatively unreactive?
18. Draw cis and trans isomers of hex-2-ene. Which
isomers will have higher boiling point and why?
7. What is the difference between isomers and
conformers?
19. Write the product and their IUPAC names obtained
when hex-1-ene reacts with HBr
8. Write the structural formulas for the following
(i) in the absence of peroxide
compounds whose IUPAC names are given:
(ii) in the presence of peroxide
3-Ethyl-2-methylpentane
20. Propanal and pentan-3-one are the ozonolysis products
9. Why do alkynes not show geometrical isomerism?
of an alkene. What is the structural formula and IUPAC
name of alkene.
10. Sodium salt of which acid will be need for the
preparation of propane? Write chemical equation for
Section – C (3 Marks Questions)
the reaction.
21. Predict the major product(s) of the following reactions
Section – B (2 Marks Questions) and explain their formation.
( Ph − CO − O)
11. How is aromaticity of a compound judged? H3C − CH = CH 2 ⎯⎯⎯⎯⎯
HBr
2
→
H3C − CH = CH2 ⎯⎯ ⎯
HBr
→
12. Ethyne is acidic in nature in comparison to ethene and
ethane. Why is it so? 22. A hydrocarbon containing two double bonds on
reductive ozonolysis produces glyoxal, ethanal and
13. Write the chemical equations of reactions involved in propanone. Predict the structure of the hydrocarbon and
ozonolysis of alkenes. also give its IUPAC name.

14. How polythene obtained by polymerization. 23. Alkynes on reduction with sodium in liquid ammonia
form trans-alkenes. Will the butene thus formed on
reduction of 2-butyne show geometrical isomerism?
HYDROCARBONS 119

24. Write an equation of the reaction of propyne with water Section – D (5 Marks Questions)
in the presence of H2SO4 and HgSO4. Show the
intermediate. 29. Write structures of different chain isomers of alkanes
corresponding to the molecular formula C6H14. Write
25. Write the reaction for the preparation of butane by their IUPAC names and classify each carbon atoms as
(i) Wurtz reaction (ii) Kolbe’s reaction 1o, 2o, 3o or 4o.

26. Write the IUPAC names of the following 30. How would you carry out the following conversions?
(i) (C2H5)4C (ii) (CH3)2CHCH(CH3)2 (a) H3 CCH 2 CH = CH 2 → H3 C − CH 2 CH 2 CH 2 OH
(b)
27. Give complete mechanism for the nitration of benzene. H3 C CH 2 − CH = CH 2 → H 3C − CH 2 − CH(OH)CH 2OH
(c) H3 CC  CH → H 3C COCH 3
28. How can the following conversion be carried out ?
(d) Br2 CHCHBr2 → HC  CH
(a) Propene to propyne
(b) Propene to methanal and ethanal
120 HYDROCARBONS

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions) 8. Eclipsed and the staggered conformations can be
represented by
1. Which of the following is correct IUPAC name of (a) Sawhorse (b) Newman projection
given hydrocarbon? (c) Both {a) and (b) (d) None of these

9. Number of possible structural isomers of C4H8 is

(a) 3 (b) 4
(c) 5 (d) 2

(a) 2-ethylpentane (b) 4-ethylpentane 10. Anti-Markownikoff’s rule addition reaction is not
(c) 3-methylpentane (d) 3-methyhéxane observed in
(a) pent-2-ene (b) but-2-ene
2. The IUPAC name of CH 2 = C ( CH3 )2 is (c) 1-butene (d) propene
(a) 2-methylbutene (b) 1-methylpropene
(c) 2-methylbut-2-ene (d) 2-methyl prop-1-ene 11. Reaction of HBr with propene in the presence of
peroxide gives
3. Below reaction is an example of: (a) isopropyl bromide (b) 3-bromo propane
AlCl (c) allyl bromide (d) n-propyl bromide
CH3 − CH 2 − CH 2 − CH3 ⎯⎯⎯
 ⎯
3
→ CH3 – CH– CH3
|
CH3 12. When one mole of an alkene on ozonolysis produces 2
(a) isomerization (b) polymerization moles of propanone, the alkene is
(c) cracking (d) de-hydrogenation (a) 3-methyl-1-butene (b) 2,3-dimethyl-1-butene
(c) 2,3-dimethyl-2-pentene (d) 2, 3-dimethyl-2-butene
4. The given structures,
13. An alkene, on ozonolysis gives formaldehyde and
acetaldehyde. The alkene is
(a) ethene (b) propene
are
(c) but-1-ene (d) but-2-ene
(a) conformational isomers (b) chain isomers
(c) position isomers (d) functional isomers
14. Which of the following will decolorise alkaline KMnO4
solution?
5. X is heated with soda lime and gives ethane. X is
(a) C3H8 (b) CH4
(a) ethanoic acid (b) methanoic acid
(c) CCl4 (d) C2H4
(c) propanoic acid (d) Either (a) and (c)
15. When 2-butyne is treated with H2 in presence of
6. Which of the following compounds can be best
prepared by Wurtz-reaction? Pd-BaSO4; the product formed will be
(a) Iso-butane (b) n-butane (a) cis-2-butene (b) trans-2-butene
(c) n-pentane (d) Iso-pentane (c) 1-butene (d) 2-hydroxybutane

7. The compound with highest boiling point is 16. Which of the following will give H2 gas with Na?
(a) n-hexane (b) n-pentane (a) CH4 (b) C2H6
(c) 2, 2-dimethylpropane (d) 2-methylbutane (c) C2H4 (d) C2H2
HYDROCARBONS 121

17. Which is most acidic of the following? 23. Assertion (A) Decolourisation of KMnO4 solution is
(a) Methane (b) Acetylene used as a test for unsaturation.
(c) 1-butene (d) Neo-pentane Reason (R) Alkenes on reaction with cold, dilute
aqueous solution of potassium permanganate produce
18. Strength of the double bond is ...A.,. than that of C—C vicinal glycols.
single bond. Here, A refers to (a) If both assertion and reason are true and reason is
(a) lower (b) constant the correct explanation of assertion.
(c) higher (d) variable (b) If both assertion and reason are true but reason is
not the correct explanation of assertion.
19. Bayer’s reagent is (c) If assertion is true but reason is false
(a) alkaline permanganate solution (d) If both assertion and reason are false.
(b) acidified permanganate solution
(c) neutral permanganate solution 24. Assertion (A) Sodium salt of butanoic acid on heating
(d) aqueous bromine solution with soda lime gives butane.
Reason (R) Decarboxylation reaction yields alkanes
20. The presence of the chlorine atom on benzene ring having same number of carbon atoms as the parent acid.
makes the second substituent enter at a position (a) If both assertion and reason are true and reason is
(a) ortho (b) meta the correct explanation of assertion.
(c) para (d) ortho/para (b) If both assertion and reason are true but reason is
not the correct explanation of assertion.
Section-B (Assertion & Reason Type Questions) (c) If assertion is true but reason is false
(d) If both assertion and reason are false.
21. Assertion (A) The addition of HBr on 2-butene gives
25. Assertion (A) Methane on chlorination gives a mixture
two isomeric products.
of products.
Reason (R) Addition of HBr on 2-butene follows
Reason (R) Chlorination proceeds through the
Markownikoff’s rule.
formation of chloronium ion.
(a) Both Assertion and Reason are true and Reason is
(a) Both Assertion and Reason are true, and the Reason
the correct explanation of Assertion.
is the correct explanation of the Assertion
(b) Both Assertion and Reason are true but Reason is
(b) Both Assertion and Reason are true, but the Reason
not the correct explanation of Assertion.
is not the correct explanation of the Assertion
(c) Assertion is true but Reason is false.
(c) Assertion is true; but Reason is false
(d) Both Assertion and Reason are false.
(d) Both Assertion and Reason are false
22. Assertion (A) Wurtz reaction is not preferred for the
26. Assertion (A) The bond length of C  C is longer than
preparation of alkanes containing odd number of carbon
atoms. C = C and C − C .
Reason (R) It is not possible to prepare alkanes with Reason (R) Each carbon has same hybridisation.
odd number of carbon atoms Wurtz reaction. (a) Both Assertion and Reason are true, and the Reason
(a) If both assertion and reason are true and reason is is the correct explanation of the Assertion
the correct explanation of assertion. (b) Both Assertion and Reason are true, but the Reason
(b) If both assertion and reason are true but reason is is not the correct explanation of the Assertion
not the correct explanation of assertion. (c) Assertion is true; but Reason is false
(c) If assertion is true but reason is false (d) Both Assertion and Reason are false
(d) If both assertion and reason are false.
122 HYDROCARBONS

27. Assertion (A) Alkenes are easily attacked by 29. Which of the following statement(s) is/are correct
electrophilic reagents. regarding structure of benzene?
Reason (R) The presence of pi bond makes alkenes I. All the six carbons in benzene are sp2 hybridised.
behave as sources of loosely held mobile electrons. II. sp2 hybridised orbital of carbon combines with
(a) Both Assertion and Reason are true, and the Reason another sp2 hybridised orbital of carbon to give C—
is the correct explanation of the Assertion C sigma bonds.
(b) Both Assertion and Reason are true, but the Reason III. C–H sigma bond is obtained by lateral overlapping
is not the correct explanation of the Assertion of sp2 hybridised orbital of C and 1s orbital of H-
(c) Assertion is true; but Reason is false atom.
(d) Both Assertion and Reason are false Choose the correct statement(s).
(a) I, II and III are correct (b) I and II are correct
(c) I and III are correct (d) II and III are correct

30. What is the ratio of  - and  -bonds in benzene ?

28. Assertion (A) is aromatic in nature. (a) 1 : 4 (b) 2 : 1
Reason (R) It does not contain Huckel number of (c) 1 : 1 (d) 2 : 2
electrons.
(a) Both Assertion and Reason are true, and the Reason 31. Benzene can be obtained by heating either benzoic acid
is the correct explanation of the Assertion with X or phenol with Y. X and Y are respectively.
(b) Both Assertion and Reason are true, but the Reason (a) zinc dust and soda lime
is not the correct explanation of the Assertion (b) soda lime and zinc dust
(c) Assertion is true; but Reason is false (c) zinc dust and sodium hydroxide
(d) Both Assertion and Reason are false (d) soda lime and copper

Section-C (Case Study Questions) 32. X-ray diffraction study of benzene reveals that
I. It is a planar molecule.
Case Study - 1
II. C–C bond lengths are of the same order which is
Benzene was isolated by Michael Faraday in 1825. The intermediate between C–C single bond and C–C
molecular formula of benzene. C6H6, indicates a high double bond,
degree of unsaturation. This molecular formula did not III. Benzene is reluctant towards the addition reaction
account for its relationship to corresponding alkanes, under normal conditions.
alkenes and alkynes. Due to its unique properties and IV. It explains the unusual behavior of benzene.
unusual stability, it took several years to assign its
structure. Benzene was found to be a stable molecule Which of the above statements are true?
and found to form a triozonide which indicates the (a) I, II and III
presence of three double bonds. Benzene was further. (b) II, III and IV
Found to produce one and only one monosubstituted (c) I, III and V
derivative which indicated that all the six carbon and (d) All the given statements are true
six hydrogen atoms of benzene are identical. On the
basis of this observation August Kekule in 1865
Case Study – 2
proposed structure for benzene having cyclic
arrangement of six carbon atoms with alternate single Hydrocarbons are the compounds of carbon and
and double bonds and one hydrogen atom attached to hydrogen only. Hydrocarbons are mainly obtained from
each carbon atom. coal and petroleum, which are the major sources of
Benzene is commercially isolated from coal tar. energy. Petrochemicals are the prominent starting
However, it may be prepared in the laboratory by the materials used for the manufacture of a large number of
following methods. commercially important products. LPG (liquefied
(i) Cyclic polymerisation of ethyne petroleum gas) and CNG (compressed natural gas), the
(ii) Decarboxylation of aromatic acids main sources of energy for domestic fuels and the
(iii) Reduction of phenol
HYDROCARBONS 123

automobile industry, are obtained from petroleum. Case Study - 3

Hydrocarbons are classified as open chain saturated
(alkanes) and unsaturated (alkenes and alkynes), cyclic Aromatic hydrocarbons are non- polar molecules and
(alicyclic) and aromatic, according to their structure. are usually colourless liquids or solids with a
The important reactions of alkanes are free radical characteristic aroma.. Aromatic hydrocarbons are
substitution, combustion, oxidation and aromatization. immiscible with water but are readily miscible with
Alkenes and alkynes undergo addition reactions, which organic solvents. They burn with sooty flame. Arenes
are mainly electrophilic additions. Aromatic are characterised by electrophilic substitution reactions.
hydrocarbons, despite having unsaturation, undergo However, under special conditions they can also
mainly electrophilic substitution reactions. These undergo addition and oxidation reactions.
undergo addition reactions only under special The common electrophilic substitution reactions of
conditions. arenes are nitration, halogenation, sulphonation, Friedel
Craft’s alkylation and acylation reactions in which
33. Complete combustion of CH4 gives: attacking reagent is an electrophile (E+)
(a) CO2 + H2O (b) CO2 + H2
(c) COCl2 (d) CO + CO2 + H2O 37. Which of the following represents Friedel-Craft’s
alkylation reaction?
34. On mixing alkane with chlorine and irradiating with (a) C6 H6 + C2 H5 Cl ⎯⎯⎯
AlCl3
→ C6 H5 C2 H5 + HCl
ultraviolet light, it forms only one monochloroalkane. (b) C6 H5 OH + HCl ⎯⎯⎯
ZnCl2
→ C6 H5 Cl + H 2 O
The alkane is
(a) propane (b) pentane (c) C6 H5 Cl + CH3COCl ⎯⎯⎯
AlCl3
→ C6 H5 COCH3 + Cl 2
(c) iso-pentane (d) neo-pentane (d) C2 H5 Br + Mg ⎯⎯⎯
ether
→ C2 H5 MgBr

35. Select the structure of the major product formed in the 38. Identify the correct order of reactivity in electrophilic
reaction substitution reactions of the following compounds

(I) (II)

(a) (b)
(III) (IV)
(a) I > II > III > IV (b) IV > III > II > I
(c) II > I > III > IV (d) II > III > I > IV

39. Identify the substitute group, that acts as ortho-para

(c) (d) directing, during electrophilic substitution in aromatic
36. Which of the following products is formed when n- compounds
heptane is passed over (Al2O3 + Cr2O3) catalyst at 773
(a) –NH2 (b) – NO2
K?
(a) Benzene (b) Toluene (c) –SO3H (d) N2
(c) Polyheptane (d) Cycloheptane
40. The reaction of toluene with Cl2 in presence of FeCl3
gives predominantly
(a) benzoyl chloride (b) benzyl chloride
(c) o - and p – chlorotoluene (d) m-chlorotoluene
124 HYDROCARBONS

Notes:

Find Answer Key and Detailed Solutions at the end of this book

HYDROCARBONS
STATES OF MATTER
126 STATES OF MATTER

Chapter at a Glance
 Boyle’s Law : PV = constant  P1V1 = P2V2  uMP : uAVG: uRMS = 1:1.128:1.224

V V1 V2 Vreal
 Charles’ Law : = constant  T  T  Compressibility Factor, Z  V
T 1 2 ideal

 Van der waals Equation :

P P1 P2
 Gay Lussac’s Law : = constant  T  T
T  an 2 
 P  2   V  nb   nRT
1 2

 V 
 Avogadro’s Law : V  n
 At very low pressure or at high temperature and moderate/
 Ideal Gas Equation : PV = nRT
low pressure: Z = 1 and PV = nRT
 Variation of ideal gas equation: PM = dRT
8a
 Dalton’s Law of Partial Pressure : Pressure due a gas in a  Critical Temperature : Tc 
27Rb
gas mixture PA = XAPTOTAL

3RT 2RT a
 u RMS  u MP   Critical Pressure : Pc 
M M 27b 2

8RT
u AVG   Critical Volume : VC = 3b
M
STATES OF MATTER 127

Solved Examples
Example-1
P1 V1
Which type of intermolecular forces exist among the follow- V2 
P2
ing molecules ?
(i) H2S molecules (ii) H2O molecules 1 bar  2.27 L

(iii) Cl2 and CCl4 molecules 0.2 bar
(iv) SiH4 molecules (v) Helium atoms = 11.35 L
(vi) He atoms and HCl molecules Since ballon bursts at 0.2 bar pressure, the volume of the
Sol. (i) Dipole-dipole interactions (because H2S is polar). ballon should be less than 11.35 L.
(ii) Hydrogen bonding. Example-4
(iii) London dispersion force (because both are non-polar). On a ship sailing in a pacific ocean where temperature is
(iv) London dispersion forces (because SiH4 is non-polar). 23.4ºC, a balloon is filled with 2L air. What will be the volume
of the balloon when the ship reaches Indian ocean, where
(v) London dispersion forces (because He atoms have
temperature is 26.1ºC ?
symmetrical electron clouds).
Sol. According to Charles’ law
(vi) Dipole-induced dipole forces (because HCl is polar while
He atom has symmetrical electron cloud). V1 V2

Example-2 T1 T2
A gas occupies 200 mL at a pressure of 0.820 bar at 20ºC.
V1 = 2L V2 = ?
How much volume will it occupy when it is subjected to
external pressure of 1.025 bar at the same temperature ? T1 = 273 + 23.4 = 296.4 K T2 = 273 + 26.1 = 299.1

Sol. P1 = 0.820 bar P2 = 1.025 bar V1T2

 V2 
V1 = 200 ml V2 = ? T1
Since temperature is constant, therefore, by applying Boyle’s
law, 2L  299.1K
  2.018 L
296.4 K
P1 V1
P1V1 = P2V2 or V2  Example-5
P2
What is the increase in volume when the temperature of 800
(0.820bar)  (200 mL) mL of air increases from 27ºC to 47ºC under constant pressure
V2   160 ml
1.025 bar of 1 bar ?
Sol. Since the amount of gas and the pressure remains constant,
Example-3
Charles’ law is applicable. i.e.
A balloon is filled with hydrogen at room temperature. It
will burst if pressure exceeds 0.2 bar. If at 1 bar pressure V1 V
 2
the gas occupies 2.27 L volume, upto what volume can T1 T2
the balloon be expanded ?
V1 = 800 mL V2 = ?
Sol. According to Boyle’s law, at constant temperature,
T1 = 273 + 27 = 300 K T2 = 273 + 47 = 320 K
P1 V1 = P2 V2
If P1 = 1 bar then V1 = 2.27 L 800 mL V2

300 K 320 K
if P2 = 0.2 bar, then
128 STATES OF MATTER

Example-9
(800 mL)
or V2   (320 K) The density of a gas at 27ºC and 1 bar presssure is
(300 K)
2.56 g L–1. Calculate the molar mass.
= 853.3 mL Sol. We are given
 Increase in volume of air = 853.3 – 800 = 53.3 mL P = 1.0 bar, T = 27 + 273 = 300 K,
Example-6 d = 2.56 g L–1 or 2.56 g dm–3
What will be the minimum pressure required to compress We know that
500 dm3 of air at 1 bar to 200 dm3 at 30oC? (NCERT)
dRT
Sol. P1 =1bar, P2 = ?, V1 =500dm 3 ,V2 =200dm 3 M
P
P1V1 = P2V2
where R = 0.083 bar dm3 mol–1 K–1
P1 V1 1 bar×500dm 3
P2    2.5bar 2.56g dm 3  0.083 bar dm3 mol1K 1  300 K
V2 200dm3 M
1 bar
Example-7
= 63.74 g mol–1
A vessel of 120 mL capacity contains a certain amount of
gas at 35oC and 1.2 bar pressure. The gas is transferred to Molar mass of gas = 63.74
another vessel of volume 180 mL at 35oC. What would be Example-10
its pressure? (NCERT) Using the equation of state PV = nRT ; show that at given
Sol. P1 = 1.2 bar, P2 = ?, V1 = 120 mL, V2 = 180 mL temperature, density of a gas is proportional to gas pressure
From Boyle’s law, P1V1 = P2V2 p. (NCERT)
Sol. PV = nRT
P V 1.2bar×120mL
P2  1 1 
V2 180mL m  m Mass of gas (g) 
PV  RT n   
M  M Molar mass of gas 
P2 = 0.8 bar.
Example-8 mRT dRT  m
A gas occupies volume of 250 mL at 745 mm Hg and 25ºC. or P or P =  Density, d = 
VM M  V
What additional pressure is required to reduce the gas
volume to 200 mL at the same temperature ? PM
or d ; If T = cosntant, d  P
Sol. P1 = 745 mm Hg P2 = ? RT
V1 = 250 mL V2 = 200 mL Example-11
Since temperature remains constant, therefore, by applying What will be the pressure exerted by a mixture of 3.2 g of
Boyle’s law, methane and 4.4 g of carbon dioxide contained in a 9 dm3
P1V1 = P2V2 flask at 27oC? (NCERT)
Sol. Moles of CH4,
P1V1 (745 mm Hg)  (250 mL)
or P2 = V  (200 mL)
2 Mass of CH 4
n CH 4 
Molar mass of CH 4
= 931.25 mm Hg
The additional pressure required = 931.25 – 745 [Molar mass of CH4 = 12 + 4 × 1 = 16]
= 186.25 mm. 3.2
  0.2 mol
16
Similarly, moles of CO2,
STATES OF MATTER 129

[mol. wt. of H2 = 1+1 = 2]

4.4
n CO2   0.1 mol
44 Total number of moles = 0.25 + 2.0 = 2.25 mol

[Molar mass of CO2 = 12 + 2 × 16 = 44] P = ? ; n = 2.25 mol ; V = 1 dm3 ;

Total moles = 0.2 + 0.1 = 0.3 mol R = 0.083 bar dm3 K-1 mol-1

PV = nRT T = 27oC = 300 K

PV = nRT
nRT
Pressure, P  Pressure,
V
nRT 2.25mol  0.083 bar dm 3 K 1mol 1  300K
P 
0.3mol  0.0821dm 3 atmK 1 mol1  300K V 1 dm3
  0.821atm
9 dm 3
P = 56.025 bar
In terms of SI units : Example-14

0.3mol  8.314 Pa m3 K 1mol 1  300 Calculate the volume occupied by 8.8 g of CO2 at 31.1oC
Pressure, P  and 1 bar pressure. R = 0.083 bar LK-1 mol-1. (NCERT)
9  103 m3
Sol. PV = nRT
P = 8.314 × 104 Pa
Example-12 m
PV  RT
M
Calculate the temperature of 4.0 mole of a gas occupying 5
dm3 at 3.32 bar. (R = 0.083 bar dm3 K-1 mol-1). (NCERT) P = 1 bar, V = ?, m = 8.8 g
Sol. Apply ideal gas equation, PV = nRT M = 44 g mol-1 (CO2)
Pressure, P = 3.32 bar R = 0.083 bar LK-1 mol-1 and T = 304.1 K
Volume, V = 5 dm3 Volume occupied by 8.8 g of CO2,
Number of moles, n = 4 mol
mRT 8.8g  0.083 bar LK 1mol 1  304.1
Gas constant, R = 0.083 bar dm3 K-1 mol-1 V 
PM 1bar  44g mol1
Temperature, T = ?
V = 5.048 L
PV 3.32bar  5 dm 3
Example-15
T 
Rn 0.083 bar dm3 K 1mol 1  4 mol
Density of a gas is found to be 5.46 g/dm3 at 27°C and 2 bar
T = 50 K pressure. What will be its density at STP? (NCERT)
Example-13 pM
Sol. Density, d 
Calculate the total pressure in a mixture of 8g of dioxygen RT
and 4g of dihydrogen confined in a vessel of 1 dm3 at 27oC. For same gas at different temperatures and pressures:
R = 0.083 bar dm3 K-1 mol-1. (NCERT) d 2 p 2 T1

Sol. To calculate the total pressure exerted by a mixture of d1 p1T2
gases first find total number of moles of gas and then
d1 = 5.46g/dm3, d2 = ?
apply the relation, PV = nRT.
T1 = 27°C = 300K, T2 = 0°C = 273K
Mass 8 p1 = 2 bar, p2 = 1 bar
Moles of O2, n O2    0.25mol
Mol.wt. 32
p 2 T1d1 1 300  5.46
[mol. wt. of O2 = 16 + 16 = 32] d2    3gdm –3
p1T2 2  273
4
Moles of H2, n H2   2.0 mol
2
130 STATES OF MATTER

Example-16 Example-19
A 10 dm flask at 298 K contains a gaseous mixture of CO
3
Calculate the root mean square speed of methane molecules
and CO2 at a total pressure of 2.0 bar. If 0.20 mole of CO is at 27ºC.
present, find its partial pressure and also that of CO2. Sol. Root mean square speed,
Sol. According to Dalton’s law of partial pressure
3RT
Ptotal = P(CO)  P(CO2 )  2.0 bar u r.m.s. 
M
n(CO) RT T = 27 + 273 = 300 K, M = 16, R = 8.314 × 107
Partial pressure of CO, P(CO) =
V
(0.2 mol)  (0.083 bar dm3K 1mol1 )  (298 K) 3  8.314 107  300
 u r.m.s. 
10 dm3 16

= 0.495 bar. = 683.9 × 102 cm s–1

= 683.9 m s–1
Partial pressure of CO2, P(CO2 )
Example-20
 Ptotal  P(CO)
At what temperature will the RMS speed of a gas be twice
= 2.0 – 0.495 = 1.505 bar. its MP speed at 100oC.
Example-17 Sol. uMP =  2RT1 /M and uRMS =  3RT2 /M ;
A mixture of dihydrogen and dioxygen at one bar pressure
contains 20% by weight of dihydrogen. Calculate the partial where T1 = 373K.
pressure of dihydrogen. (NCERT) Given, uRMS = 2uMP
Sol. Mass of H2 = 20g, Mass of O2 = 80 g thus, T2 = (8×373/3) = 994.67 K
20 80 Example-21
Moles of H2 = = 10 mol, Moles of O2 =
2 32 Out of N2 and NH3, which one will have greater value for
= 2.5 mol van der Waals constant ‘a’ and which one will have greater
value for van der Waals constant ‘b’ ?
10
Mole fraction of H2 =  0.8 Sol. (i) As NH3 is more easily liquefiable (due to hydrogen
10  2.5
Partial pressure = 0.8 × P = 0.8 × 1 bar = 0.8 bar bonding), intermolecular forces of attraction are stronger
than in N2. Hence, NH3 will have greater value for ‘a’.
Example-18
(ii) s N2 molecule is larger in size than NH3, hence N2 will
A neon-dioxane mixture contains 70.6g of dioxygen and 167.5 have greater value for ‘b’.
g of neon. If pressure of the mixture of gases in the cylinder
(For NH3, a = 4.17 L2 atm mol–2, b = 0.0371 L mol–1
is 25 bar, what is the partial pressure of dioxygen and neon
in the mixture. For N2, a = 1.39 L2 atm mol–2, b = 0.0391 L mol–1)
70.6 Example-22
Sol. No. of moles of dioxygen = = 2.21 mol
32 Explain the significance of van der Waals parameters ?
167.5 (NCERT)
No. of moles of neon = = 8.375 mol
20 Sol. ‘a’ is a measure of the magnitude of the intermolecular forces
2.21 of attraction while b is a measure of the effective size of the
Mole fraction of dioxygen, x O2 =
2.21  8.375 gas molecules.
8.375
Mole fraction of neon, xNe = = 0.79
2.29  8.375
STATES OF MATTER 131

Example-23
8 (27Pc b 2 ) 8Pc b
1 mole of sulphur dioxide occupies a volume of 350 ml at  Tc  
27Rb R
27ºC and 5 × 106 Pa pressure. Calculate the compressibility
factor of the gas. Is it less or more compressible than an Pc = 100 atm, b = 0.050 dm3 mol–1
ideal gas ?
8  (100 atm)  (0.050 dm3mol 1 )
Sol. Compressibility factor,  Tc 
0.082 dm3atm mol1 K 1
PV
Z
nRT = 487.2 K.
n = 1 mol, P = 5 × 106 Pa, V = 350 mL = 0.350 × 10–3 m3 Example-25
R = 8.314 Nm K mol , T = 27 + 273 = 300 K
–1 –1
Critical temperature for CO2 and CH4 are 31.1ºC and – 81.9ºC
respectively. Which of these has stronger intermolecular
5 106  0.350  103 forces and why? (NCERT)
Z  = 0.702
1.0  8.314  300
Sol. Higher the critical temperature, more easily the gas can be
SO2 is more compressible than an ideal gas (which has Z=1). liquefied, i.e., greater are the intermolecular forces of
Example-24 attraction. Hence, CO2 has stronger intermolecular forces
than CH4.
Calculate the critical temperature of a Van der Waals gas
for which Pc is 100 atm and b is 0.050 dm3 mol–1.

8a
Sol. Tc =
27Rb
2 2 2
But a = 3Pc Vc  3Pc  (3b)  27 Pc b
132 STATES OF MATTER

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions) 14. How is the pressure of a gas related to its density at a
particular temperature?
1. Define Boyle’s law.
15. At what temperature will both the Celsius and
2. How is the pressure of a given sample of gas related to Fahrenheit scale read the same value?
temperature at constant volume?
16. Calculate the density of ammonia (NH3) at 30oC and 5 bar
3. Define an ideal gas. pressure.

4. What property of molecules of real gases is indicated 17. The Van der Waals constant for two gases A and B are as
by Van der Waals constant ‘a’? follows:
Gas a (atm L2 mol–2) b (L mol–1)
5. What is Boyle temperature?
A 1.63 0.0326

6. Why does a liquid boil at a lower temperature at the top of B 3.72 0.0521
a mountain than at sea level?
Which of these
7. How is compressibility factor expressed in terms of molar (i) is more easily liquified?
volume of a real gas and that of the ideal gas? (ii) has larger molecular size?

8. Convert 40℃ into Kelvin 18. At what temperature will hydrogen molecules have the
same root mean square speed as nitrogen molecules at
9. Is dalton’s law of partial pressure valid for a mixture of SO2 27oC?
and O2?
19. One of the assumptions of kinetic theory of gases is that
10. Arrange solid, liquid and gas in order of increasing energy? there is no force of attraction between the molecules of a
gas. State and explain the evidence that shows that the
Section – B (2 Marks Questions) assumption is not applicable for real gases.

11. Name the intermolecular forces between pV

20. Compressibility factor, Z, of a gas is given as Z =
(i) SiH4 molecules nRT
(ii) HCl molecules in liquid HCl (i) What is the value of Z for an ideal gas?
(iii) He and a polar molecule (ii) For real gas what will be the effect on value of Z above
(iv) Water molecules Boyle’s temperature?

12. Name the energy which arises due to motion of atoms Section – C (3 Marks Questions)
or molecules in a body. How is this energy affected
when the temperature is increased? 21. A sample of nitrogen occupies a volume of 1.0 L at a
pressure of 0.5 bar at 40oC. Calculate the pressure if the
13. A gas that follows Boyle’s law, Charle’s law and gas is compressed to 0.225 mL at –6oC.
Avogadro’s law is called an ideal gas. Under what
conditions a real gas would behave ideally?
STATES OF MATTER 133

22. A gas cylinder containing cooking gas can withstand a 28. A gas occupies a volume of 250 mL at 745 mm Hg and
pressure of 14.9 atmosphere. The pressure gauge of the 25oC. What additional pressure is required to reduce the
cylinder indicates 12 atmosphere at 27oC. Due to a sudden gas volume to 200 mL at the same temperature?
fire in the building. The temperature starts rising. At what
temperature will the cylinder explode? Section – D (5 Marks Questions)

23. A 10 litre flask contains 0.2 mole of methane, 0.3 mole of 29. (i) State Avogadro’s Law.
hydrogen and 0.4 mole of nitrogen at 25oC. What is the (ii) A 2L vessel contains oxygen at a pressure of 380 mm
partial pressure of each component and what is the pressure Hg at 27oC. 1.40 g of N2 gas is introduced in the vessel.
inside the flask? Will the pressure of mixture increases or decreases and
to what extent?
24. Two moles of ammonia were found to occupy a volume of (iii) Density of a gas is found to be 5.46g dm–3 at 127oC and
5L at 27oC. Calculate the pressure using van der Waals 2 bar pressure. What will be its density at STP?
equation (a = 4.17 bar L2 mol–2, b = 0.0371 L mol–1).
30. Two flask A and B have equal volume. Flask A contains
25. Compare the temperature of 3 mol of SO2 at 15 bar H2 and is maintained at 600 K while flask B contains an
occupying a volume of 10L obtained by the ideal gas equal mass of CH4 gas and is maintained at 1200 K.
equation and van der Waals equation (a = 6.7 bar L2 mol–2, (i) Which flask contains a greater number of molecules?
b = 0.0564 L mol–1). How many times more?
(ii) In which flask is the pressure greater? How many times
26. Calculate the root mean square, average and most greater?
probable speed of oxygen at 27oC. (iii) In which flask will the molecules move faster?

27. The mass of 525 ml of a gaseous compound at 28oC

and 0.970 bar pressure was found to be 0.900g.
Calculate the molar mass of the compound.
134 STATES OF MATTER

EXERCISE – 2: Basic Subjective Questions

Section – A (Single Choice Questions)
5. The pressure exerted by the gas is due to
1. van der Waals forces include (a) the repulsion between gas molecules
I. London forces. (b) the attraction between gas molecules
II. chemical bonding forces. (c) collision of the particles with the wall of the
III. dipole-dipole forces. container
IV. dipole-induced dipole forces. (d) None of the above
Which of the following option is correct?
(a) Only I and III (b) Only III and IV 6. The unit of a in van der Waals equation,
(c) Only I and II (d) I, III and IV  an 2 
 P + 2  ( V − nb ) = nRT . is
 V 
2. Hydrogen bonding is
(a) a weak type of dipole-dipole interaction (a) atm L2 mol–2 (b) atm L mol–2
–3
(b) a strong type of dipole-dipole interaction (c) atm L mol (d) None of these
(c) a weak type of dipole-induced dipole interaction
7. For a real gas, the compressibility factor Z has different
(d) All of the above
values at different temperatures and pressures. Which
3. Which of the following graphs represents the correct of the following is not correct under the given
Boyle’s law? conditions?
(a) Z < 1 at very low pressure.
(b) Z > 1 at high pressure.
(c) Z = 1 under all conditions.
(d) Z = 1 at intermediate pressure

8. Which one of the following option is correct?

(a) vmp : vav : vrms :: 1 : 1.224 : 1.128
(b) vmp : vav : vrms :: 1 : 1.128 : 1.224
(c) vmp : vav : vrms :: 1.224 : 1.128
(d) vmp : vav : vrms :: 1.128 : 1.24 : 1

9. What are the most favourable conditions to liquefy a

gas?
(a) Low temperature and low pressure
(a) (i), (ii) and (iii) (b) (i) and (iv) (b) Low temperature and high pressure
(c) (ii) and (iii) (d) (i), (ii) and (iv) (c) High temperature and high pressure
(d) High temperature and low pressure
4. Which of the following relationship is correct for an
ideal gas? 10. In a flask of volume V litres, 0.2 mol of oxygen, 0.4
V pT MV p mol of nitrogen, 0.1 mol of ammonia and 0.3 mol of
(a) = (b) =
n R m RT helium are enclosed at 27oC. If the total pressure
d p exerted by these non-reacting gases is one atmosphere,
(c) = (d) Both (a) and (b) the partial pressure exerted by nitrogen is
M RT
(a) 0.1 atmosphere (b) 0.3 atmosphere
(c) 0.8 atmosphere (d) 0.4 atmosphere
STATES OF MATTER 135

11. At high altitudes, water boils at a lower temperature 18. Measurement of the amount of dry gas collected over
because water from volume of moist gas is based on the
(a) the atmospheric pressure is high at high altitudes (a) Charle’s law
(b) the viscosity of water is reduced at high altitudes (b) Dalton’s law of partial pressure
(c) the atmospheric pressure is low at high altitudes (c) Avogadro’s law
(d) None of these (d) Boyle’s law

12. The rms speed of CH4 is 3 times of the rms speed of O3. 19. 2.5 L of sample of a gas at 27oC and 1 bar pressure is
Which of the following option is correct? compressed to a volume of 500 mL keeping the
(a) TO3  TCH 4 (b) TO3 = TCH 4 temperature constant, the percentage increase in the
pressure is:
(c) TO3  TCH 4 (d) None of these
(a) 100% (b) 400%
(c) 500% (d) 80%
13. van der Waals equation of state is obeyed by real gases.
For n moles of a real gas, the expression will be 20. If a gas is heated at constant pressure, its density:
 p na  V  (a) will decrease
(a)  + 2   = nRT
 n V  n − b  (b) will increase
 a  (c) may increase or decrease
(b)  p + 2  ( V − b ) = nRT (d) will remain unchanged
 V 
 na 
(c)  p + 2  ( nV − b ) = nRT Section-B (Assertion & Reason Type Questions)
 V 
 n 2a 
(d)  p + 2  ( V − nb ) = nRT 21. Assertion (A) Three states of matter are the result of
 V  balance between intermolecular forces and thermal
energy of the molecules.
14. Which of the following properties of water can be used Reason (R) Intermolecular forces tend to keep the
to explain the spherical shape of rain droplets? molecules together but thermal energy of molecules
(a) Viscosity (b) Surface tension tends to keep them apart.
(c) Critical phenomena (d) Pressure (a) Both A and R are correct; R is the correct
explanation of A
(b) Both A and R are correct; R is not the correct
15. What is SI unit of viscosity coefficient ( ) ? explanation of A
(a) pascal (b) N s m–2 (c) A is correct; R is incorrect
(c) km–2 s (d) N m–2 (d) R is correct; A is incorrect

16. The total pressure of a mixture of two gases is: 22. Assertion (A) At constant temperature, pV versus p
(a) the sum of the partial pressures plot for Ideal gas is not a straight line.
(b) the difference between the partial pressures Reason (R) At high pressure all gases have z > 1 but at
(c) the product of the partial pressures intermediate pressure most gases have z < 1.
(d) the ratio of the partial pressures (a) Both A and R are correct; R is the correct
explanation of A
(b) Both A and R are correct; R is not the correct
17. The pressure exerted by 6.0g of methane gas in a 0.03 explanation of A
m3 vessel at 129oC is (Atomic masses: C = 12.01, H = (c) A is correct; R is incorrect
1.01 and R = 8.314 JK–1 mol–1) (d) R is correct; A is incorrect
(a) 31684 Pa (b) 215216 Pa
(c) 13409 Pa (d) 41648 Pa
136 STATES OF MATTER

23. Assertion (A): Molar volume of an ideal gas at 273.15 27. Assertion (A): Avogadro’s law is applicable only
K and 1 bar is 22.4 L. under similar conditions of temperature and pressure.
Reason (R): Volume of a gas is inversely proportional Reason (R): Real gases follow avogadro’s law
to temperature. (a) Both A and R are correct; R is the correct
(a) Both A and R are correct; R is the correct explanation of A
explanation of A (b) Both A and R are correct; R is not the correct
explanation of A
(b) Both A and R are correct; R is not the correct
(c) A is correct; R is incorrect
explanation of A (d) R is correct; A is incorrect
(c) A is correct; R is incorrect
(d) Both A and R are incorrect 28. Assertion (A): CH4 molecules have higher value of
most probable speed than Cl2 molecules at any given
24. Assertion (A): Compressibility factor for the helium temperature.
gas varies with pressure with positive slope (at all Reason (R): CH4 is polyatomic gas and Cl2 is diatomic
pressures). gas.
Reason (R): Repulsive forces dominate in helium gas, (a) Both A and R are correct; R is the correct
even at low pressure. explanation of A
(a) Both A and R are correct; R is the correct (b) Both A and R are correct; R is not the correct
explanation of A explanation of A
(b) Both A and R are correct; R is not the correct (c) A is correct; R is incorrect
explanation of A (d) R is correct; A is incorrect
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Section – C (Case Study Questions)
25. Assertion (A): Gases obey Boyle's law at high
Case Study-1
temperature and low pressure only.
Reason (R): At low pressure and high temperature,
Intermolecular forces are the forces of attraction and
gasses would behave like ideal gas.
repulsion between interacting particles (atoms and
(a) Both A and R are correct; R is the correct
molecules). This term does not Include the electrostatic
explanation of A
forces that exist between the two oppositely charged
(b) Both A and R are correct; R is not the correct
ions and the forces that hold atoms of a molecule
explanation of A
together i.e., covalent bonds.
(c) A is correct; R is incorrect
Attractive intermolecular forces are known as van der
(d) R is correct; A is incorrect
Waals forces. Van der Waals forces vary considerably
in magnitude and include dispersion forces or London
26. Assertion (A): P-V graph (at constant temperature) for
forces, dipole-dipole forces, and dipole induced dipole
ideal gas is rectangular hyperbola.
forces.
Reason (R): Ideal gas obeys Charle’s law
(a) Both A and R are correct; R is the correct
explanation of A 29. Attractive intermolecular forces are known as
(b) Both A and R are correct; R is not the correct (a) Magnetic forces (b) van der Waals forces
explanation of A (c) nuclear forces (d) mechanical forces
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect 30. Dispersion force is also known as
(a) Newton’s force (b) London force
(c) Centrifugal forces (d) gravitational force
STATES OF MATTER 137

31. London force is a/an 35. A sample of a gas occupies 600 mL at 27oC and 1 atm.
(a) attraction force that acts between two temporary What will be the volume at 127oC if the pressure is kept
dipoles constant?
(a) 800 mL (b) 300 mL
(b) attraction force that acts between two permanent
(c) 250 mL (d) 900 mL
dipoles
(c) repulsion force that acts between two permanent 36. The pressure in well inflated tyres of automobiles is
dipoles almost constant, but on a hot summer day this increases
considerably and tyre may burst. This phenomena is
(d) repulsion force that acts between two temporary
explained by
dipoles (a) Boyle’s law (b) Charles’ law
(c) Gay Lussac’s law (d) Avogadro law
32. In case of dipole-induced dipole forces, interaction
energy is proportional to Case Study-3
1 Dalton law of partial pressure states that the pressure
(a) 6 , where r = distance between two molecules
r exerted by the mixture of non-reactive gases is equal to
1 the sum of the partial pressure of individual gases i.e.,
(b) 2 , where r = distance between two molecules the pressure which these gases would exert if they were
r
enclosed separately in the same volume and under the
1 same conditions of temperature. In a mixture of gases,
(c) 3 , where r = distance between two molecules
r the pressure exerted by the individual gas is called
1 partial pressure. Mathematically,
(d) 8 , where r = distance between two molecules
r PTotal = p1 + p 2 + p3 + ..... ( at constant T,V )
where Ptotal is the total pressure exerted by the mixture
Case Study-2 of gases and p1, p2, p3 etc. are partial pressure of gases.

There are four laws which relate the state variables of a 37. In a mixture of gases, the pressure exerted by the
gas in two states. These four laws are : individual gas is called
Boyle’s Law It states that at constant temperature, the (a) total pressure (b) partial pressure
pressure of a fixed amount (i.e. number of moles n) of (c) individual pressure (d) osmotic pressure
gas varies inversely with its volume.
38. Which of the following relationships between partial
Charles’s Law It states that at constant pressure, the
pressure, volume and temperature is correct?
volume of a fixed mass of a gas is directly proportional
nRT
to its absolute temperature. (i) P =
Gay Lussac’s Law It states that at constant volume, V
pressure of a fixed amount of a gas varies directly with P =
(ii) total p1 + p 2 + p3
the temperature. RT
(iii) Ptotal = ( n1 + n 2 + n 3 )
V
33. Representing P, V and T as pressure, volume and (a) (i) and (ii) (b) (i) and (iii)
temperature, which of the following is the correct (c) (ii) and (iii) (d) (i), (ii) and (iii)
representation of Boyle’s law?
1 1 39. The pressure of a mixture of equal weights of two gases
(a) V  (P constant) (b) V  (T constant) X and Y with molecular weight 4 and 40 respectively is
T P
11 atm. The partial pressure of the gas X in the mixture
(c) PV = RT (d) PV = nRT
is
(a) 10 atm (b) 11 atm
34. A balloon is filled with Argon at room temperature. It (c) 12 atm (d) 13 atm
will burst, if pressure exceeds 0.2 bar. If at 1 bar
40. Equal masses of helium and oxygen are mixed in a
pressure the gas occupied 2.27 L volume, up to what
container at 60oC. The fraction of the total pressure
volume can the balloon be expanded? exerted by oxygen in the mixture of gases is
(a) 8L (b) 11.35 L (a) 5/3 (b) 1/3
(c) 8.35 L (d) 15 L (c) 1/9 (d) 6/7
138 STATES OF MATTER

Notes:

Find Answer Key and Detailed Solutions at the end of this book

STATES OF MATTER
THERMODYNAMICS
140 THERMODYNAMICS

Chapter at a Glance
 System: A part of universe which is under investigation.  Enthalpy (H): It is sum of internal energy and pressure-
volume energy of the system at a particular temperature and
 Surroundings: The rest of the universe which is not a part
pressure. It is also called heat content (H = U + PV).
of the system.
 Enthalpy change ( ΔH ): It is the measure of heat change
 State of the system: The conditions of existence of a system
when its macroscopic properties have definite values. taking place during the process at constant temperature and
constant pressure.
 State functions: The thermodynamic quantities which
depend only on the initial and final state of the system. q p  H
 Energy is exchanged between the system and the Enthalpy H and internal energy change (U) are related as
surroundings as heat if they are at different temperatures.
H  U  PV
 The properties of the system whose value is independent
of the amount of substance are called intensive properties, H  U  n g RT
e.g., temperature, pressure, viscosity, surface tension,
dielectric, specific heat capacity.
Where n g  Gaseous moles of products – Gaseous moles
 The properties of the system whose value depends upon of reactants
the amount of substance present in the system are called
extensive properties. e.g., mass, volume, surface area,
energy, enthalpy, entropy, free energy, heat capacity.

Θ
 
Enthalpy of reaction Δr H : The enthalpy change
accompanying a chemical reaction when the number of moles
 Work is also a mode of transference of energy between
of reactants react to give the products as given by the
system and the surroundings. Work done by the system on
balanced chemical equation.
the surroundings is given by PV.

 Internal energy (U): The energy associated with the system 

Θ
 
Standard enthalpy of formation Δf H : The enthalpy
at a particular conditions of temperature and pressure. change accompanying the formation of one mole of the
 Internal energy change ( ΔU ): It is a measure of heat change compound from its element at standard conditions and all
occurring during the process at constant temperature and the substances being in their standard states.
constant volume.
As a convention, the  f H of every element is assumed to
q v  U be zero.

 First law of thermodynamics: It states that the energy of

the universe always remains constant during chemical and

Θ
 
Enthalpy of combustion Δc H : The amount of heat
physical changes. Mathematically. change when 1 mole of the substance is completely burnt in

U  q  w excess of oxygen or air.  c H  is always negative.

 Internal energy of an ideal gas depends only on temperature.

Therefore, in an isothermal process involving ideal gas,
  Θ

Enthalpy of solution Δsol H : The enthalpy change when

U  0 . one mole of a substance is dissolved in large excess of

solvent so that further dilution does not give any further
 Work of expansion, w  –PV . enthalpy change.
 Work of expansion in vacuum.  Enthalpy of hydration: It is the enthalpy change during the
hydration of 1 mole of anhydrous salt to a specific hydrate.
pext  0, w  0
THERMODYNAMICS 141

 Hess law: The enthalpy change in a particular reaction is the  Gibbs’ energy: Gibbs’ energy is the energy in a system that
same whether the reaction takes place in one step or in a can be converted into useful work
number of steps. G   w useful
 Bond enthalpy: The average amount of energy required to G  H  T S
break one mole of the bonds of a particular type in gaseous
For a spontaneous process G  0
molecules.
For a non - spontaneous process G  0
q rev For a process at equilibrium G  0
 Entropy: S 
T  Relation of Gibbs’ energy and equilibrium constant:
 r G    2.3 03R T log K
  r Sº  Sº (products) – Sº (reactants)

i.e., Stotal  Ssys  Ssurr must be positive

142 THERMODYNAMICS

Solved Examples
Example-1 Now from first law of thermodynamics
Choose the correct answer. A thermodynamic state q = U – W
function is a quantity
800 = U + 405.3  U = 394.7 Joule
(i) used to determine heat changes Example-4
(ii) whose value is independent of path Calculate w, q and U when 0.75 mol of an ideal gas expands
(iii) used to determine pressure volume work isothermally and reversibly at 27oC from a volume of 15 L to
(iv) Whose value depends on temperature only. 25 L.

Sol. State function is quantity whose value is independent of Sol. For isothermal reversible expansion of an ideal gas,
path, So statement (ii) is correct. V2
w = – 2.303 nRT log V Putting n = 0.75 mol,
Example-2 1

Calculate the internal energy change in each of the V1 = 15L, V2 = 25 L, T = 27 + 273 = 300 K and
following cases :
R = 8.314 J K–1 mol–1, we get
(i) A system absorbs 15 kJ of heat and does 5 kJ of work.
25
(ii) 5 kJ of work is done on the system and 15 kJ of heat is w = – 2.303 × 0.75 × 8.314 × 300 log  955.5 J
15
given out by the system.
Sol. (i) Here, q = +15 kJ (–ve sign represents work of expansion)

w = – 5 kJ For isothermal expansion of an ideal gas, U = 0

 According to first law of thermodynamics, U = q + w = U = q + w gives q = – w = + 955.5 J.

15 + (–5) = 10 kJ Thus, internal energy of the system Example-5
increases by 10 kJ. A 5-litre cylinder contained 10 moles of oxygen gas at 27oC.
(ii) Here, w = + 5 kJ Due to sudden leakage through the hole, all the gas escaped
into the atmosphere and the cylinder got empty. If the
q = – 5 kJ
atmospheric pressure is 1.0 atmosphere, calculate the work
 According to first law of thermodynamics, done by the gas. (1 L atm = 101.3 J)
U = q + w = – 15 + (+ 5) = – 10 kJ
Sol. Vinitial = 5L, T = 27oC = 27 + 273 K = 300 K
Thus, the internal energy of the system decreases by 10 kJ.
Example-3 nRT 10  0.0821  300
Vfinal =   246.3 L
P 1.0
A gas present in a cylinder fitted with a frictionless piston
expands against a constant pressure of 1 atm from a V = final – Vinitial = 246.3 – 5 = 241.3 L
volume of 2 litre to a volume of 6 litre. In doing so, it wexp = – PV = – 1 × 241.3 L atm = – 241.3 × 101.3 J
absorbs 800 J heat from surroundings. Determine increase = – 24443.7 J.
in internal energy of process.
Sol. Since, work is done against constant pressure and thus,
irreversible.
Given, V = (6 – 2) = 4 litre; P = 1 atm
 W = – 1 × 4 litre-atm = – 4 × 1.01325 × 102 J
= 405.3 J
THERMODYNAMICS 143

Example-6 = – 1 (40 – 20) = – 20L atm

10g of argon gas is compressed isothermally and reversibly = – 20 × 1.01325 × 102 J
at a temperature of 27ºC from 10L to 5L. Calculate q, W, U = – 2026.5 J
and H for this process. R = 2.0 cal K–1 mol–1. log10 2 = 0.30. WBC = O (Isochoric)
VA
Atomic wt. of Ar = 40 WCA = – 2.303 nRT log10 V
C
V2 Example-8
Sol. W = – 2.303 nRT log10 V
1 5.6 dm3 of an unknown gas at S.T.P. required 52.25J of heat
to raise its temperature by 10oC at constant volume. Calculate
10 5 Cv, Cp and  of the gas
= – 2.303 × × 2 × 300 log10
40 10
Sol. The 22.4 dm3 of a gas at S.T.P. = 1 mol
W = 103.991 cal 1
 5.6 dm3 of the gas at S.T.P. =  5.6  0.25 mol
22.4
U = 0 ; H = 0 (Constant temperature)
Thus, for 10o rise, 0.25 mol of the gas at constant volume
 q = U – W q = – W = – 103.991 cal require heat = 52.25 J
Example-7  For 1o rise, 1 mol of the gas at constant volume will
Two mole of a perfect gas undergo the following processes:
52.25
require heat =  20.9
(a) a reversible isobaric expansion from (1.0 atm, 20.0L) to 10  0.25
(1.0 atm, 40.0L).
 Cv = 20.9 J K–1 mol–1
(b) a reversible isochoric change of state from (1.0 atm, 40.0L)
Now, Cp = Cv + R = 20.9 J K–1 mol–1 + 8.314 J K–1 mol–1
to (0.5 atm, 40.0L)
= 29.214 J K–1 mol–1
(c) a reversible isothermal compression from (0.5 atm, 40.0L)
Cp 29.214
to (1.0 atm, 20.0L).    = 1.4
Cv 20.9
(i) sketch with labels each of the processes on the same P-
Example-9
V diagram.
The reaction of cyanamide, NH2CN(s) with dioxygen was
(ii) Calculate the total work (w) and the total heat change
carried out in a bomb calorimeter and U was found to be -
(q) involved in the above processes.
742.7 KJ mol-1 at 298 K. Calculate the
(iii)What will be the value of U, H and S for the overall enthalpy change for the reaction at 298 K.
process?
3
Sol. The overall is cyclic one, i.e., initial state is regained, thus NH 2 CN (s) + O 2 (g)  N 2 g  +CO 2 g  +H 2 O l (NCERT)
2
U = 0; H = 0 and S = 0. Sol. Enthalpy change for given reaction:

3
NH 2 CN (s) + O 2 (g)  N 2 g  +CO 2 g  +H 2 O l
2
H  U   ng RT

Given value:
U = 742.7 kJ mol 1
n g = n g (products) n g (reactants)
= (2 1.5) moles = +0.5 moles
Now, total work W = WAB + WBC + WCA T = 298 K
WAB = – P (VB – VA)
R = 8.314 × 10-3 kJ mol-1 K -1
144 THERMODYNAMICS

H = ( 742.7 kJ mol 1) + (+0.5 mol) (298 K) Sol. The reaction is:

8.314 × 10-3 kJ mol-1 K -1 1

C6 H 6    7 O 2 g  
 6CO 2 g   3H 2 O  
2
= 742.7 + 1.2
In this reaction, O2 is the only gaseous reactant and CO2
H = 741.5kJ mol 1
is the only gaseous product.
Example-10
1 1 3
Given  n g  n p  n r  6  7  1  
2 2 2
N 2 g   3H 2 g  
 2NH 3 g ; Also, we are given U (or qv) = – 3263.9 kJ mol–1
 r H  = - 92.4 kJ mol-1 T = 25oC = 298K

What is the standard enthalpy of formation of NH3 gas? 8.314

R = 8.314 J K–1 mol– = kJ K 1 mol1
(NCERT) 1000
Sol. H (or qp) = U + ng RT = – 3263.9 kJ mol–1 +
Standard enthalpy of formation of a substance is defined as
 3   8.314 
the enthalpy change, i.e accompanied in the formation of   mol   kJ K 1 mol1   298 K 
 2  1000 
one mole of a compound from its elements in their most
stable state of aggregation. For 1 mole of ammonia, = – 3263.9 – 3.7 kJ mol–1 = – 3267.6 kJ mol–1.
Example-13
1 3
N 2 g   H 2 g  
 NH3 g  A heated copper block at 130oC loses 340 J of heat to the
2 2
surroundings which are at room temperature of 32 oC.
1 Calculate
 rH
2
(i) the entropy change of the system (copper block)
= ½ ( 92.4 kJ mol 1) (ii) the entropy change in the surroundings
= 46.2 kJ mol 1 (iii) the total entropy change in the universe due to
Example-11 this process
Enthalpies of formation of CO(g), CO2(g), N2O(g)and Assume that the temperature of the block and the
N2O4(g) are 110 , 393, 81 kJ and 9.7 kJ mol 1
surroundings remains constant.
respectively. Find the value of  r H for the reaction: Sol. Tsystem = 130o C = 130 + 273 K = 403 K,
N 2 O4(g) + 3CO(g) 
 N 2 O(g) + 3CO2(g) (NCERT) Tsurr = 32oC = 32 + 273 K = 305 K
qsystem = – 340 J, qsurr = + 340 J
Sol. Standard enthalpy of formation of a substance is defined
as the enthalpy change. rH for a reaction = fH value of q system 340 J
(i) Ssystem  T  =-0.84 JK -1a
products - fH value of reactants. According to the given system 403 K
reaction, Substituting the values, we can get the heat of q surr 340 J
(ii)  Ssurr  T  =+1.11 J K -1
formation: 305 K
surr
Therefore, the value of rH for the reaction is. (iii) Stotal or Suniverse = Ssystem + Ssurr
Example-12 = – 0.84 + (+1.11) J K–1 = 0.27 J K–1
The heat of combustion of benzene in a bomb calorimeter
(i.e., constant volume) was found to be 3263.9 kJ mol–1 at
25 oC. Calculate the heat of combustion of benzene at
constant pressure.
THERMODYNAMICS 145

Example-14  Tf = 300 K
Calculate the entropy change involved in conversion of
V2
one mole (18 g) of solid ice at 273 K to liquid water at the Ssys = R ln V = R ln 4 = 11.52 JK–1
1
same temperature (latent heat of fusion = 6025 J mol–1).
Sol. Entropy change for ice  water is given by Ssurr = 0  qsys = qsurr = 0

 H Suniv = 11.52 JK–1

f S  f
Tf G = – TSuniv = – 300 × 11.52 = – 3456 J/mol.
Here, f H = 6025 J mol–, Tf = 273 K Example-17
Calculate the standard free energy change for the reaction,
6025 J K  1 mol 1 4NH3 (g) + 5O2 (g) 
  f S  =22.1 J K -1 mol -1 .  4NO (g) + 6H2O (l)
273 K
Given that the standard free energies of formation (f Go) for
Example-15 NH3 (g), NO (g) and H2O (l) are – 16.8, + 86.7 and – 237.2 kJ
Calculate the entropy change in surroundings when 1.00 mol –1 respectively. Predict the feasibility of the above
mol of H 2O(l) is formed under standard conditions. reaction at the standard state.

 f H  286 kJ mol1 (NCERT) Sol. Here, we are given

f Go (NH3) = – 16.8 kJ mol–1
Sol. Entropy changes can be measured by the equation:
f Go (NO) = + 86.7 kJ mol–1
q
Ssurr  Surr ...........  i 
T f Go (H2O) = – 237.2 kJ mol–1
Amount of heat is evolved on the formation of 1 mol of  r Go =  f Go (Products) –  f Go (Reactants) =
H2O(l). [14 × f Go (NO) + 6 × f Go (H2O)] – [4 × f Go (NH3)
qsurr = +286 kJ mol 1 + 5 × f Go (O2)]
Substitute the value in the eq.(i) at 298K = [4 × (86.7) + 6 × (–237.2)] – [4 × (– 16.8) + 5 × 0] = – 1009.2 kJ
286 kJ mol1 -1 1 Since r Go is negative, the process is feasible.
 = 959.73 J mol K
298k Example-18
Hence, the value of entropy change in surroundings is
A + B 
 C + D ; H = – 10,000 J mol–1, S = – 33.3 J
959.73 J/mol/K.
mol–1 K–1
Example-16
(i) At what temperature the reaction will occur
1.0 mol of an ideal gas, initially present in a 2.00 L insulated spontaneously from left to right ?
cylinder at 300 K is allowed to expand against vacuum to
(ii) At what temperature, the reaction will reverse?
8.00 L. Determine W, E, H, Suniv and G .
Sol. G = H – TS
Sol.
At equilibriu, G = 0 so that H = TS or

H 10000 J mol 1
T  =300.03 K
S 33.3 JK 1 mol 1

(i) For spontaneity from left to right. G should be –ve

for the given reaction. This will be so if T < 300.3 K
W = – pext V = 0, q = 0, E = 0 = H
(ii) For reverse reaction to occur, G should be +ve forward
reaction. This will be so if T > 300.3 K.
146 THERMODYNAMICS

Example-19  Heat evolved when 0.04 mole of H+ ions combine with 0.04
The enthalpies of all elements in their mole of OH– ions
standard states are: = 57.1 × 0.04 = 2.284 kJ
(i) unity
(ii) zero
(ii) 200 cm3 of 0.2 M H2SO4  0.2  200 mole of H2SO4 =
qSurr 1000
(iii) < 0 Ssurr  ...........  i 
T 0.04 mole of H2SO4 = 0.08 mole of H+ ions 400 cm3 of 0.5 M
0.5
(iv) different for each element (NCERT) KOH   400 mole of KOH = 0.2 mole of KOH = 0.2
1000
Sol. (ii) is correct. The enthalpies of all elements in their mole of OH ions
–

standard states are zero. Thus, 0.08 mole of H+ ions will neutralize 0.08 mole of OH–
Example-20 ions. (out of 0.2 mole of OH– ions) to form 0.08 mole of H2O.
Hence, heat evolved = 57.1 × 0.08 = 4.568 kJ
U of combustion of methane is X kJ mol 1. The value of
In case (i), heat produced = 2.284 kJ = 2284 J
H is
(i) = H Total volume of the solution = 500 + 200 = 700 mL

(ii) > U Assuming density of solution = 1g/mL

(iii) < U So mass of solution = 700g

Specific heat = 4.18 J K–1 g–1
(iv) = 0 (NCERT)
Sol. Option (iii) is correct. Q 2284
Q  m  C   T  T  
m  C 700  4.18
H  U  n g RT.
= 0.78ºC In case (ii), heat produced = 4.568 kJ = 4568 J Total
H  X  n g RT  H  U
mass of the solution = 200 + 400 = 600 g
Example-21
Q 4568
Calculate the amount of heat evolved when  T   = 1.82ºC
m  C 600  4.18
(i) 500 cm3 of 0.1 M hydrochloric acid is mixed with 200 cm3
Example-22
of 0.2 M sodium hydroxide solution
Calculate the standard enthalpy of formation of CH3OH(l)
(ii) 200 cm3 of 0.2 M sulphuric acid is mixed with 400 cm3
from the following data:
of 0.5 M potassium hydroxide solution.
3
Assuming that the specific heat of water is 4.18 J K–1 g–1 CH 3 OH (l)  O 2 (g)  
2
and ignoring the heat absorbed by the container,
CO 2(g)  2H 2 O l  ;  r H   -726 kJ mol-1
thermometer, stirrer etc., what would be the rise in
temperature in each of the above cases ? C(g ) + O2(g) 


0.1 CO 2(g) ;  c H   -393 kJ mol-1

Sol. (i) moles of HCl =  500 = 0.05 = 0.05 mole of H+ ions
1000
0.2 1
200 cm3 of 0.2 M NaOH   200 mole of NaOH = 0.04 H 2(g)  O 2(g )  
1000 2
(NCERT)
mole = 0.04 mole of OH ions Thus, 0.04 mole of H+ ions will H 2 O(l) ;  f H   -286 kJ mol-1 .
combine with 0.04 mole of OH– ions to from 0.04 mole of H2O
Sol. Standard enthalpy of formation of a substance is defined
and 0.01 mole of H+ ions will remain unreacted.
as the enthalpy change. i.e accompanied in the formation
 Heat evolved when 1 mole of H+ ions combine with 1 mole
of one mole of a compound from its elements in their most
of OH– ions = 57.1 kJ.
stable state of aggregation.
THERMODYNAMICS 147

rH for a reaction = fH value of products - fH value 1

of reactants. According to the given reaction, T   H  G 
S
3
CH 3 OH (l)  O 2 (g)   At equilibrium change in free energy = 0. Substitut the free
2
energy value we get:
CO 2(g)  2H 2 O l  ;  r H   -726 kJ mol-1
H
C(g) + O2(g) 
 =
S
CO2(g) ;  c H   -393 kJ mol -1
1 400kJ mol1
H 2(g)  O 2(g )    = 200K
2 0.2kJ K 1mol1
H 2 O(l) ;  f H   -286 kJ mol-1 .
T should be more than 2000K then the given reaction to be
For the standard enthalpy= Eq (ii) + 2 × eq (iii) eq(i) spontaneous.
Example-24
fH CH 3OH  l    cH  2 f H 
 H 2 O  l    rH For the reaction 2A g   B g   2D g  U = 10.5 kJ and
S = 44.1 JK -1.
= ( 393 kJ mol 1) + 2( 286 kJ mol 1) ( 726 kJ mol 1)
Calculate G  for the reaction, and predict whether there
= ( 393 572 + 726) kJ mol 1
action may occur spontaneously.
  f H  CH 3 OH  l    239kJ mol1 Sol. To decide the spontaneity of a process we have to consider
the change in entropy of the system and surrounding which
Example-23
is not easy to be determined while dealing with chemical
For the reaction at 298 K,
reactions. The new thermodynamic function is the Gibbs
2A  B  C energy (G).

H  400kJ mol1 and S  0.2 kJ K 1 mol1 We know that:

At what temperature will the reaction become spontaneous G  H  T S

considering H and S to be constant over the Where,
temperature range? (NCERT) H = Enthalpy change
Sol. To decide the spontaneity of a process we have to consider S = Entropy change
the change in entropy of the system and surrounding which
First calculate the enthalpy change by using the given
is not easy to be determined while dealing with chemical
formula below:
reactions. The new thermodynamic function is the Gibbs
 H   U   n g RT
energy (G).
We know that: Given value:
G  H  T S U  10.5kJ mol1
Where,
n g = n g (products) n g (reactants)
H = Enthalpy change = (2 -3) moles = -1 moles
S = Entropy change T = 298 K
We can write the expression based on temperature:
R = 8.314  103 kJ mol1 K 1

Substitute the value in the expression of H .

148 THERMODYNAMICS

 H   U   n g RT Example-25

= ( 10.5 kJ) + (-1) (8.314 × 10 3 kJ K-1 mol-1) (298 K) The equilibrium constant for a reaction is 10. What will be

= 10.5 kJ +2.48 kJ the value of Go? (Given, R = 8.314 JK-1 mol-1, T = 300 K).
(NCERT)
H = 12.98 kJ
Substituting the values of “Ho and “So in the expression Sol. At chemical equilibrium, G  2.303 RT log K eq
of G  :
Substitute the value in the above equation.
G  H  TS
= (2.303) (8.314 JK-1 mol-1) (300 K) log10
= 12.98 kJ - (298 K) ( 44.1 J K-1)
= 5744.14 Jmol-1
= 12.98 kJ - 13.14 kJ = - 0.16 kJ
Hence, free energy value is negative so the reaction is Hence, the free energy value is -5744.14J/mol.
spontaneous.
THERMODYNAMICS 149

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. What do you mean by system? 17. Explain specific heat.

2. What do you mean by surroundings? 18. For solids or liquids, there is no significant difference
between H and U .
3. Explain first law of thermodynamics.
19. Define extensive and intensive properties?
4. ‘Coffee held in a cup’ is what type of system?
20. C(graphite) + O2(g) → CO2(g) ΔH= - 393.5 kJ mol–1 .
5. List one example of isolated system. This is combustion reaction of C and CO2 is forming.
Write the H° values of the two preocesses.
6. If work is done by the system, what will be the effect
on internal enery of the system. Section – C (3 Marks Questions)

7. In thermodynamics we discuss about open and close 21. What are different type of system. Explain
sytsem. So, animals and plants belongs to?
22. Define the relation between the enthalpy of reaction and
8. Explain state of thermodynamic system? bond enthalpy?

9. What do you mean by Enthalpy? 23. Given below is the reaction :

2Ag 2 O ( s ) → 4Ag ( s ) + O2 ( g ) . H and S are given
10. How do you define enthalpy mathematically? + 61.17 kJ mol–1 and + 132 Jk–1 respectively. At what
temperature reaction will be spontaneous?
Section – B (2 Marks Questions)
24. Since, H = U + pV then prove H = q p
11. Internal energy is state function while work done is not.
Why? 25. Prove Cp – Cv = R

12. Predict the sign of S for the following reaction 26. During the combustion of ethylene, carbon dioxide and

CaCO3(s) ⎯⎯
→ CaO(s) + CO2(g). water is formed. The heats of formation of CO2, H2O
and C2H4 are –393.7, –241.8, +52.3 kJ per mole
13. What do you mean by spontaneous process. respectively. Calculate heat of combustion of ethylene.

14. What do you mean by non-spontaneous proces? 27. Heat released in the both processes will be same or
different. Why or why not?
1
15. The standard heat of formation of Fe2O3(s) is 824.2kJ H 2 ( g ) + O2 ( g ) → H 2O ( g )
mol–1. Calculate heat change for the reaction. 2
4Fe(s) + 3O2(g) → 2Fe2O3(s) 1
H 2 ( g ) + O2 ( g ) → H 2O ( l )
2
16. Explain Heat capacity and what are the units of heat
capacity?
150 THERMODYNAMICS

28. The molar heat capacity under constant volume 30. 221.4 J is needed to heat 30g of ethanol from 15 oC to
conditions is equal to 3/2 R. Derive. 18oC. Calculate
(a) specific heat capacity, and
Section – D (5 Marks Questions) (b) molar heat capacity of ethanol.

29. a) Give the relationship between U and H for gases.

b) Standard enthalpy of formation. Explain
THERMODYNAMICS 151

EXERCISE– 2: Basic Objective Questions

Section – A (Single Choice Questions) 8. Which property is evaluated by the third law of
thermodynamics?
1. Hess law can be explained by which of the following? (a) Absolute Energy (b) Absolute enthalpy
(a) 1st law of Thermodynamics (c) Absolute Entropy (d) Absolute Free Energy
(b) 2nd law of Thermodynamics
(c) Entropy change 9. What is/are the reasons that water is a desirable heat
(d) H = U + PV sink for calorimeters?
I) Waters heat specific capacity is very precisely
2. The expansion of 5 moles of an ideal gas from a known.
pressure of 10 atm against a constant external pressure II) Water is readily available
of 1 atm at 300 K is done isothermally and irreversibly. III) Water has an unusually large specific heat capacity.
How much work is done at 300 K: (a) I only (b) I and II
(a) –15.921 kJ (b) –11.224 kJ (c) I, II and III (d) II only
(c) –110.83 kJ (d) None of these
10. Chemical reactions occur when the bond energy of the
3. In absolute zero, an ideal crystal has a zero entropy. reactants is greater than the bond energy of the
According to which law of thermodynamics is this true? products.
(a) Zeroth Law (b) First Law (a) Exothermic (b) Diathermic
(c) Second Law (d) Third Law (c) Endothermic (d) Endergonic

4. The one with the highest entropy is? 11. In a reversible process the system absorbs 400 kJ heat
(a) Mercury (b) Hydrogen and performs 200 kJ work on the surroundings. What is
(c) Water (d) Graphite the increase in the internal energy of the system?
(a) 850 kJ (b) 600 kJ
5. System in which there is no exchange of matter, work (c) 350 kJ (d) 200 kJ
or energy from surroundings is
(a) closed (b) adiabatic 12. Which of the following neutralization reactions is
(c) isolated (d) isothermal. maximum exothermic?
(a) HCl and NaOH
6. Which has the highest entropy per mole among the (b) HCN and NaOH
following? (c) HCl and NH4OH
(a) Liquid Nitrogen (b) Hydrogen Gas (d) CH3COOH and NH4OH
(c) Mercury (d) Diamond
13. A student runs a reaction in a closed system. In the
7. An ideal gas is taken around the cycle ABCA as shown course of the reaction, 54.7 kJ of heat is released to the
in P-V diagram. The next work done by the gas during surroundings and 16.3 kJ of work is done on the
the cycle is equal to: system. What is the change in internal energy ( U ) of
the reaction?
(a) –79.0 kJ (b) 50.4 kJ
(c) 79.0 kJ (d) –38.0 kJ

(a) 12P1V1 (b) 6P1V1

(c) 5P1V1 (d) P1V1
152 THERMODYNAMICS

14. Which of the following statement is correct. Section-B (Assertion & Reason Type Questions)
(a) The entropy always increases
(b) The change in entropy along with suitable change in 21. Assertion (A): Refractive index is an intensive
enthalpy decides the fate of a reaction property.
(c) The enthalpy always decreases Reason (R): Refractive index does not depend on the
(d) Both the enthalpy and the entropy remain constant quantity or size of matter present.
(a) Both Assertion and Reason are true, and the Reason
15. 3 mole of an ideal gas at 27oC expands isothermally and is the correct explanation of the Assertion.
reversibly from a volume of 4 litres to 40 litre. The (b) Both Assertion and Reason are true, but the Reason
work done (in kJ) is: is not the correct explanation of the Assertion.
(a) w = –28.72 kJ (b) w = –17.232 kJ (c) Assertion is true, but Reason is false.
(c) w = –5.736 kJ (d) w = –4.988 kJ (d) Both Assertion and Reason are false.

16. The latent heat of vapourization of a liquid at 500 K 22. Assertion (A): H2O(l) - > H2O(s), then internal energy
and 1 atm pressure is 10.0 Kcal/mol. What will change will be zero.
be the change in internal energy ( U ) of 2 moles of Reason (R): In this case there is negligible change in
liquid at the same temperature volume
(a) 13.0 kcal/mol (b) –13.0 kcal/mol pΔV = Δn g RT = 0
(c) 18.0 kcal (d) –7.0 kcal/mol ΔH = U
(a) Both Assertion and Reason are true, and the Reason
17. When CaC2 is formed from CaO(s) and C(s) from the is the correct explanation of the Assertion.
reaction: CaO(s)+ 3 C(s) → CaC2(s) + CO (g) (b) Both Assertion and Reason are true, but the Reason
given that  f H  (CaC2) = –14.2Kcal.  f H  (CO) = is not the correct explanation of the Assertion.
–26.4 Kcal. Calculate the heat required to make 6.4 Kg (c) Assertion is true, but Reason is false.
CaC2 (s). (d) Both Assertion and Reason are false.
(a) 5624 Kcal (b) 1.11 x 104Kcal
3
(c) 86.24 x 10 Kcal (d) 1100 Kcal 23. Assertion (A): Heat absorbed by an ideal gas during
isothermal expansion in vacuum is zero.
18. The entropy of universe is: Reason (R): As q = ΔU + pΔV
(a) Continuously Increasing In isothermal expansion, ΔT = 0, ΔU = 0, Pext = 0
(b) Continuously Decreasing internal energy and pressure is zero so heat is zero.
(c) Zero (a) Both Assertion and Reason are true, and the Reason
(d) Constant is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason
19. If ΔU and ΔW represent the increase in internal energy is not the correct explanation of the Assertion.
and work done by the system respectively in a (c) Assertion is true, but Reason is false.
thermodynamical process, which of the following is (d) Both Assertion and Reason are false.
true?
24. Assertion (A): qv = Δ U(heat absorbed at constant
(a) ΔU = -ΔW, in a adiabatic process
volume is equal to internal energy).
(b) ΔU = ΔW, in a isothermal process Reason (R): As ΔU = q + w  ΔU = q + PΔV
(c) ΔU = ΔW, in a adiabatic process
At constant volume ΔV = 0 , therefore, ΔU = q .
(d) ΔU = -ΔW, in a isothermal process
(a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion.
20. The bond energy (in Kcal mol–1) of a C-C single bond (b) Both Assertion and Reason are true, but the Reason
is approximately is not the correct explanation of the Assertion.
(a) 1 (b) 10 (c) Assertion is true, but Reason is false.
(c) 83-85 (d) 1000 (d) Both Assertion and Reason are false.
THERMODYNAMICS 153

25. Assertion (A): In a cyclic process, net work done is Section – C (Case Study Questions)
equal to negative amount of heat energy transferred.
Reason (R): In cyclic process, system returns to its Case Study – 1
initial state ΔU = q + w therefore, ΔU = 0  w = -q Hess’s Law of Constant Heat Summation. This law was
(a) Both Assertion and Reason are true, and the Reason presented by Hess in 1840. According to the law, if a
is the correct explanation of the Assertion. chemical reaction can be made to take place in a
(b) Both Assertion and Reason are true, but the Reason number of ways in one or in several steps, the total
is not the correct explanation of the Assertion. enthalpy change is always the same. Thus, the total
(c) Assertion is true, but Reason is false. enthalpy change of a chemical reaction depends on the
(d) Both Assertion and Reason are false. initial and final stages only.

26. Assertion (A): Changes takes place in adiabatic

process are irreversible.
Reason (R): Adiabatic changes occur suddenly.
(a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion. Total enthalpy change from A to D,
(b) Both Assertion and Reason are true, but the Reason ΔH total = ΔH1 + ΔH 2 + ΔH3
is not the correct explanation of the Assertion. ΔH total + ΔH direct = 0
(c) Assertion is true, but Reason is false.
i.e., ΔH total = −ΔH direct
(d) Both Assertion and Reason are false.
For example, formation of CO, from C in two different
manners involves a total heat change of –393.5 kJ/ mol
27. Assertion (A): Temperature of a gas can be raised
Single step process:
without heating it.
Reason (R): While compression in adiabatic process, C(s ) + O2( g ) → CO2( g ) ; H = −393.5 kJ/mol
work done in compressing the gas increases its internal Two-step process:
1
energy. (i) C ( s ) + O 2( g ) → CO( g ) ; H = −110.5 kJ/mol
(a) Both Assertion and Reason are true, and the Reason 2
1
(ii) CO ( g ) + O( g ) → CO 2( g ) ; H = −283.0 kJ/mol
is the correct explanation of the Assertion.
(b) Both Assertion and Reason are true, but the Reason 2
is not the correct explanation of the Assertion. HTotal = −110.5 + ( −283.0 ) = −393.5 kJ/mol
(c) Assertion is true, but Reason is false.
(d) Both Assertion and Reason are false. 29. For the given reactions,
H 2( g ) + F2( g ) → 2HF( g ) ; H = −124 kcal
28. Assertion (A): 2H2(g) + O2(g) → 2H2O(l); H = –
H 2( g ) → 2H ( g ) ; H = 104 kcal
571.6 kJ mol–1
Does not represent enthalpy of formation of water. F2( g ) → 2F( g ) ; H = 37.8 kcal
Reason (R): This is because H in the above reaction then the value of H for H( g ) + F( g ) → HF( g ) is
is for 2 moles of H2O.
(a) 142 kcal (b) – 132.9 kcal
Note: H f of pure liquid or solid is constant. (c) 132 keal (d) 134 keal.
(a) Both Assertion and Reason are true, and the Reason
is the correct explanation of the Assertion. 30. From the below given cases, Hess’s law is not
(b) Both Assertion and Reason are true, but the Reason applicable?
is not the correct explanation of the Assertion. (a) Determination of lattice energy
(c) Assertion is true, but Reason is false. (b) Determination of resonance energy
(d) Both Assertion and Reason are false. (c) Determination of enthalpy of transformation of one
allotropic form to another
(d) Determination of entropy
154 THERMODYNAMICS

31. Calculate the energy change for given reaction. 35. How Lattice enthalpies can be determined:
(a) Born-Haber cycle (b) Hess’s law
(c) lattice cycle (d) none of these.

36. In which of the following thermochemical changes

The bond energies of C — H and C — Cl are 413 and H is always negative?
328 kJ mol–1 respectively. (a) Enthalpy of solution
(a) –1465 kJ/mol (b) 1465 kJ/mol (b) Enthalpy of hydrogenation
(c) –1482 kJ/mol (d) 1482 kJ/mol (c) Enthalpy of reaction
(d) Enthalpy of transition
32. Steps are given for a reaction , A → 2B:
A ⎯⎯
→ C; ΔH = q1 Case Study – 3
C ⎯⎯
→ D; H = q 2 A thermodynamically irreversible process is always
1 accompanied by an increase in the entropy of the
D ⎯⎯ → B; H = q 3
2 system and its surroundings taken together while in a
The heat of reaction is thermodynamically reversible process, the entropy of
the system and its surrounding taken together remain
(a) q1 – q2 + 2q3 (b) q1 + q 2 − 2q3
unchanged i.e. for reversible process Ssys + Ssurr = 0
(c) q1 + q 2 + 2q 3 (d) q1 + 2q 2 − 2q 3 and for irreversible process Ssys + Ssurr  0
combining the two we have Ssys + Ssurr 0 where
Case Study – 2
‘equal’ to sign refers to a reversible process while the
The enthalpy of a system is defined as the sum of the ‘greater than’ sign refers to an irreversible process.
internal energy of the system and the energy that arises Change in entropy for an ideal gas under different
due to its pressure and volume. Mathematically, the conditions may be calculated as
enthalpy is defined by the equation, V P
H = U + PV S = 2.303 nR log10 2 = 2.303nR l og10 1
V1 P2
Enthalpy change ( H ) of a system is the heat absorbed
(for isothermal process)
or evolved by the system at constant pressure.
T V
H = qp , H = U + PV S = 2.303 nCP log10 2 = 2.303nCP log10 2
T1 V1
(for isobaric process)
33. Which statement is correct about enthalpy?
T2 P
(a) It is an extensive property. S = 2.303 nCV log10 = 2.303nCV log10 2
(b) It is not a state function. T1 P1
(c) Its absolute value cannot be determined.
(d) Enthalpy of a compound is equal to enthalpy of 37. Calculate entropy change when 3 moles of an ideal gas
formation of that compound. expands reversibly and isothermally from an initial
volumeof 5 litre to 50 litre at 27°C.
34. In which of the following reactions will U be equal to (a) 190.15 JK–1 (b) 57.44 JK–1
–1
H ? (c) 87.25 JK (d) 90.13 JK–1
1
(a) H 2( g ) + O 2( g ) ⎯⎯
→ H 2 O( l ) 38. The entropy of the universe
2
(a) increasing and tending towards maximum value
(b) H2(g) +I2(g ) ⎯⎯
→ 2HI(g )
(b) decreasing and tending to be zero
(c) N2 O4(g) ⎯⎯
→ 2NO2(g) (c) remains constant
(d) decreasing and increasing with a periodic rate.
(d) 2SO2(g) + O2(g) ⎯⎯
→ 2SO3(g)
THERMODYNAMICS 155

39. For which process, total entropy of system and 40. For which among the following given change S is
surrounding increases. positive:
(a) reversible (b) irreversible (a) mixing of two gases (b) boiling of liquid
(c) exothermic (d) endothermic. (c) melting of solid (d) all of these.

Find Answer Key and Detailed Solutions at the end of this book

THERMODYNAMICS
EQUILIBRIUM
EQUILIBRIUM 157

Chapter at a Glance
158 EQUILIBRIUM
EQUILIBRIUM 159

Solved Examples
Example-1 H 2 O(g)  CO(g)  H 2 (g)  CO 2 (g)
Define the term ‘amorphous’. Give a few examples of
Calculate the equilibrium constant for the reaction.
amorphous solids.
1  0.40
The following reaction has attained equilibrium Sol. At equilibrium  H 2 O   mol L1 = 0.06 mol L–1,
10
CO(g)  2H 2 (g)  CH 3 OH(g), H o  92.0kJ mol1
[CO]= 0.06 mol L–1.
What will happen if (i) volume of the reaction vessel is
0.4
suddenly reduced to half? (ii) the partial pressure of H2   mol L1  0.04 mol L1 ,
10
hydrogen is suddenly doubled? (iii) an inert gas is added
[CO2] = 0.04 mol L–1,
to the system?
K
 H 2  CO2   0.04  0.04  0.444
CH 3OH  , K  PCH OH  H 2 OCO 0.06  0.06
Sol. K c 
3

 CO  H 2 
p
2
PCO  p H2 2
Example-4
At 700 K, equilibrium constant for the reaction
(i) When volume of the vessel is reduced to half, the
concentration of each reactant or product becomes double. H 2 (g)  I2 (g)  2HI(g)is 54.8. If 0.5 mol L–1 of HI (g) is

2 CH OH
present at equilibrium at 700K, what are the concentrations
1
Thus, Qc  2[CO] (2[H ])2  4 Kc As Q c  K c , equilibrium will
3
of H2(g) and I2(g) assuming that we initially started with HI
2
(g) and allowed it to reach equilibrium at 700 K. (NCERT)
shift in the forward direction, producing more of CH 3 OH
1
Sol. 2HI(g)  H 2 (g)  I 2 (g), K 
to make Q c  K c . 54.8
At equilibrium [HI]= 0.5 mol L–1,
PCH3OH 1 1 x x 1
(ii) Q P  P 
(2PH 2 ) 2
 Kp
4  H2   [I2 ]  x mol L1  K  (0.5) 2 
CO 54.8
Again Qp < Kp, equilibrium will shift in the forward direction This gives x  0.068.i.e., [H2]= [I2] = 0.068 mol L–1
to make Qp= Kp. Example-5
(iii) As volume remains constant, molar concentration will What is the equilibrium concentration of each of substances
not change. Hence there is no effect on the state of in the equilibrium when the initial concentration of ICI was
equilibrium. 0.78 M?
Example-2 2I Cl(g)  I 2 (g)+Cl 2 (g),K c =0.14 (NCERT)
A sample of HI (g) is placed in a flask at a pressure of 0.2 atm.
At equilibrium, the partial pressure of HI (g) is 0.04 atm. Sol. Suppose at equilibrium,  I 2  = Cl 2  = x mol L1 . Then
What is Kp for the given equilibrium? 2ICl  I 2 (g) + Cl 2 (g)
Sol. 2HI(g)  H 2 (g)  I 2 (g) Initial conc. 0.78M 0 0
Initial pressure 0.2 atm 0 0 At eqm. 0.78-2x x x

At eqm. 0.04 atm

0.16
atm
0.16
atm Kc =
 I2  Cl2   0.14 
x x
2 2
 ICl  (0.78  2 x) 2
2
(Decrease in the pressure of HI = 0.2 – 0.04 = 0.16 atm.)
p H 2  p I2 0.08atm  0.08atm x
 Kp    4.0. or x 2  0.14(0.78  2x)2 or  0.14  0.374
p 2HI (0.04atm)2 0.78  2x
Example-3 or x  0.292  0.748x or 1.748x  0.292 or x  0.167
One mole of H2O and one mole of CO are taken in a 10 litre Hence, at equilibrium,
vessel and heated at 725 K. At equilibrium 40% of water (by
mass) reacts with CO according to the equation:  I2  = Cl2  =0.167M,  ICI =0.78-2×0.167M=0.446M
160 EQUILIBRIUM
Example-6
x2
Ethyl acetate is formed by the reaction between ethanol and  Kc  1
 8.3 103 (Given)
0.5  10
acetic acid and the equilibrium represented as:
CH 3COOH(l )  C2 H 5 OH(l )  CH3COOC 2 H 5 (l )  H 2 O(l ) or x 2  (8.3  10 3 )(0.5  101 )  4.15  104 or
(i) Write the concentration ratio (reaction quotient), Qc for
x  4.15  104  2.04 102 M  0.02M
this reaction (Note: water is not in excess and is not a solvent
in this reaction) Hence,  PCl3 eq   Cl 2 eq  0.02M
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18
mol of ethanol, there is 0.171 mol of ethyl acetate in the final Example-8
equilibrium mixture. Calculate the equilibrium constant. Equilibrium constant, Kc for the reaction,
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid N 2 (g)  3H 2 (g)  2NH 3 (g) at 500 K is 0.061. At a
and maintaining it at 293 K, 0.214 mol of ethyl acetate is particular time, the analysis shows that composition of the
found after some time. Has equilibrium been reached? reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol
(NCERT) L–1 NH3. Is the reaction at equilibrium? If not, in which
direction does the reaction tend to proceed to reach
CH3COOC2 H5 H 2O
Sol. (i) Qc  CH COOH C H OH equilibrium? (NCERT)
 3  2 5 
 NH3    0.5 2
2

(ii) CH3COOH  C2H5OH  CH3COOC2 H5  H2O Sol. Qc   0.0104

 N 2  H 2 3  3.0  2.0 3
Initial 1.00 mol 0.180 mol
As Q c  K c , reaction is not in equilibrium
At eqm.1–0.171 0.180–0.171 0.171 mol 0.171 mol
0.829 mol = 0.009 mol As Q c  K c , reaction will proceed in the forward direction.
Molar Example-9
concs 8.29/V 0.009/V 0.171/V 0.171/V The equilibrium constant for the reaction

Kc 
 CH3COOC2 H5  H 2 O   0.171/ V  0.171/ V   3.92 H 2 (g)  Br2 (g)  2HBr(g) at 1024 K is 1.6 ×10 . Find the 5

CH3COOH C2 H5OH   0.829 / V  0.009 / V  equilibrium pressure of all gases if 10.0 bar of HBr is
introduced into a sealed container at 1024 K.
(iii) CH3COOH  C2H5OH  CH3COOC2 H5  H2O
Initial 1.000 mol 0.500 mol 1
Sol. 2HBr(g)  H2 (g)  Br2 (g). K 
After time t 1–0.214 0.500–0.214 0.214 mol 0.214 mol (1.6 105 )
0.786 mol = 0.286 mol Initial 10 bar
 0.214 / V  (0.214 / V)  0.204 At eqm. 10–p p/2 p/2
Reaction quotient (Qc) =
(0.786 / V)(0.286 / V)
(p / 2)(p / 2) 1 p2 1
Kp    
As Q c  K c , equilibrium has not been attained (10  p) 2
1.6 105
4(10  p) 1.6 105
2

Example-7 Taking square root of both sides, we get

A sample of pure PCl5 was introduced into an evacuated p 1

2(10  p) 4  10 2 or 4  10 p  2(10  p) or
2
vessel at 473 K. After equilibrium was attained, concentration
of PCl5 was found to be 0.5×10–1 mol L–1. If value of Kc is 8.3
× 10–3, what are the concentrations of PCl3 and Cl2 at 20
equilibrium? 402p  20 or p   4.98 10 2 bar
402
Sol. PCl5 (g)  PCl3 (g)  Cl2(g)
At eqm 0.5× 10 mol L
–1 –1
x mol L –1
x mol L–1 (suppose)
EQUILIBRIUM 161
Example-13
Hence, at equilibrium p H2  p Br2
The reaction, CO(g)+3H2(g)  CH4(g) + H2O (g), is at
= p/2 = 2.5×10–2 bar, p HBr  10  p  10 bar equilibrium at 1300 K in a 1 L flask. It also contains 0.30 mol
Example-10 of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown
Dihydrogen gas used in Haber’s process is produced by amount of CH4 in the flask. Determine the concentration of
reacting methane from natural gas with high temperature CH4 in the mixture. The equilibrium constant, Kc, for the
steam. The first stage of two stage reaction involves the reaction at the given temperature is 3.90. (NCERT)
formation of CO and H2. In second stage, CO formed in first
 CH 4  H 2 O
state is reacted with more steam in water gas shift reaction, Sol. K c 
CO H 2 
3

CO(g)  H2 O(g)  CO2 (g)  H2 (g) . If a reaction vessel at

400ºC is charged with an equimolar mixture of CO and steam
 3.90 
 CH 4   0.02 
such that pCO  p H2O = 4.0 bar, what will be the partial  0.30  0.10 
3 (molar conc= No. of moles
pressure of H2 at equilibrium? Kp = 0.1 at 400ºC. (NCERT)
because volume of flask = 1L)
Sol. Suppose the partial pressure of H2 at equilibrium= p bar
or [CH4]= 0.0585 M=5.85×10–2 M
CO(g)  H 2 O(g)  CO2(g) + H2 (g) Example-14
Initial pressure 4.0 bar 4.0 bar At temperature T, a compound AB2(g) dissociates according
At eqm (4–p) (4–p) p p to the reaction

p2 p 2AB2  2AB  g   B2  g 
Kp   0.1(given)   0.1  0.316
(4  p)2
4p with a degree of dissociation, x, which is small compared
p = 1.264–0.316p or 1.316p=1.264 or p=0.96 bar. Hence, with unity. Deduce the expression for x in terms of the
equilibrium constant Kp and the total pressure, P.
(pH2 )eq = 0.96 bar
Sol. 2AB2  2AB(g)  B2 (g)
Example-11
Initial 1 mole 0 0
Predict which of the following reaction will have appreciable
concentration of reactants and products: At equilibrium 1 – x 2x x
Total number of moles at equilibrium = 1 – x + 2x + x = 1 + 2x
(a) Cl 2 (g)  2Cl(g), K c  5  10 39
1 x 2x x
(b) Cl2 (g)  2NO(g)  2NOCl(g),Kc  3.7 108 p AB2   P, p AB   P, p B2  P
1 2 x 1  2x 1  2x
(c) Cl2 (g)  2NO2 (g)  2NO2 Cl(g),K c  1.8 (NCERT)
p2AB  pB2 2
 2x  P   x  P   1  x 
2

Sol. For reaction (c), as Kc is neither high nor very low, reactants  Kp       P
 1  2x   1  2x   1  2x
2
p AB2 
and products will be present in comparable amounts.
Example-12
4x 3 P
The value of K c for reaction, 3O 2 (g)  2O 3 (g), is or K p 
1  x 1  2x 
2.0 × 10–50 at 25ºC. If the equilibrium concentration of O2 in air
at 25ºC is 1.6 × 10–2, what is the concentration of O3? Neglecting x in comparison to unity,
(NCERT) 1/ 3
 Kp 
K p  4x 3 P or x   
 O3   2.0 1050   O3 
2 2
 4P 
Kc 
Sol.  O2  1.6 102 
3 3

or [O3]2 = (2.0 × 10–50) (1.6×10–2)3 = 8.192×10–56 or

[O3]= 2.86×10–28 M
162 EQUILIBRIUM
Example-15 Example-16
Calculate the pH of the following solutions: The degree of ionization of a 0.1 M bromoacetic acid solution
(a) 2g of TlOH dissolved in water to give 2 litre of the solu is 0.132. Calculate the pH of the solution and the pKa
tion bromoacetic acid. (NCERT)
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the (a) OH– (b) F–
solution (c) H+ (d) BCl3.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the Sol. CH 2 (Br)COOH CH 2 (Br)COO   H 
solution Intial Conc C 0 0
(d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of Conc at eqm C – C C C
the solution (NCERT)
0.1  0.132
2
C.C C2
Ka     2 103
2g 1 C(1  ) 1   0.868
Sol. Molar conc. of TlOH = (204 16  1)gmol1  2L
pK a   log(2  10 3 )  3  0.301  2.699
= 4.52 ×10 M–3

 H    C   0.1 0.132  1.32  10 2 M

  OH     TlOH   4.52  10 3 M
pH   log(1.32  102 )  2  0.1206  1.88
 H    1014 /(4.52  10 3 )  2.21 10 12 M
Example-17
12
pH   log(2.21  10 )  12  (0.3424)  11.66 The pH of 0.005 M codeine (C18H21NO3) solution is 9.95.
Calculate the ionization constant and pKb.
0.3g 1 3
Molar conc. of Ca(OH)2 =  40  34 gmol1  0.5L  8.1110 M Sol. Cod  H 2 O CodH   OH 
pH  9.95  pOH  14  9.95  4.05i.e.
Ca(OH) 2  Ca 2   2 OH 
 log  OH    4.05
  OH    2  Ca(OH) 2   2  (8.11  10 3 ) M

or log  OH   4.05  5.95


= 16.22×10–3 M
pOH   log(16.22  10 3 )  3  1.2101  1.79
or  OH   8.913  10 M
 5

pH  14  1.79  12.21

8.91105 
2 2
(c) Molar conc. of NaOH CodH  OH  OH 
Kb    
0.3g 1 Cod Cod 5 103
 1
  3.75  10 2 M
40g mol 0.2L
= 1.588 × 10–6
 OH    3.75  10 2 M pK b   log 1.588  10 6   6  0.1987  5.8 .

pOH   log(3.75  10 2 )  2  0.0574  1.43 Example-18

Calculate the hydrogen ion concentration in the following
 pH  14  1.43  12.57
biological fluids whose pH are given:
(d) M1V1  M 2 V2 13.6M  1mL  M 2  1000mL (a) Human muscle–fluid, 6.83
 M2 = 1.36 × 10–2 M (b) Human stomach fluid, 1.2
(c) Human blood, 7.38
 H     HCl   1.36  10 2 M.
(d) Human saliva, 6.4 (NCERT)
2
pH   log(1.36  10 )  2  0.1335  1.87 Sol. (a) log[H  ]   pH  6.83  7.17

[H  ]  Antilog 7.17  1.479  10 7 M

EQUILIBRIUM 163
(b) log[H  ]   pH  1.2  2.8 (4.57 10 3 )(4.57 103 )
Ka   2.09  104
 [H  ]  Antilog 2.8  6.31 102 M 0.1

(c) log[H  ]   pH  7.38  8.62   K a / C  2.09 104 / 0.1  0.0457

Example-22
[H  ]  Antilog 8.62  4.169  10 8 M
The ionic product of water at 310 K is 2.7 ×10–14. What is the

(d) log[H ]   pH  6.4  7.60 pH of neutral water at this temperature?

[H  ]  Antilog 7.60  3.981  10 7 M Sol.  H    K w  2.7  1014  1.643  107 M

Example-19
If 0.561 g KOH is dissolved in water to give 200 mL of solution pH   log  H     log(1.643  107 )
at 298 K, calculate the concentration of potassium, hydrogen  7  0.2156  6.78
and hydroxyl ions. What is its pH?
Example-23
0.561 1000
Sol.  KOH    M  0.050M One the basis of the equation pH =– log [H+], the pH of
56 200 10–8 mol dm–3 solution of HCl should be 8. However, it is
As KOH  K   OH  ,  K    OH    0.05M observed to be less than 7.0. Explain the reason.
 H    K w /  OH    10 14 / 0.05  10 14 /(5  102 ) Sol. pH of 10–8 M HCl solution is not 8 because this concentration
= 2.0×10–13 M. is so low that H+ ion produced from H2O in the solution (viz.
pH = –log [H+] = – log (2.0×10–13) 10–7M) cannot be neglected. Total [H+] = 10–8 + 10–7 M.
= 13 – 0.3010 = 12.699 Solving and calculating pH, we get the value close to 7 but
Example-20 less than 7 as the solution is acidic.
The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Example-24
Calculate the concentration of strontium and hydroxyl ions pH of a solution of a strong acid is 5.0. What will be the pH
and the pH of the solution. (Atomic mass of Sr = 87.6) of the solution obtained after diluting the given solution
(NCERT) 100 times?
Sol. Molar mass of Sr(OH)2= 87.6+34=121.6g mol –1 Sol. pH = 5 means [H+]= 10 –5 M. On diluting 100 times,
Solubility of Sr(OH)2 in moles 105
 H     107 M
100
19.23gL1
L–1 =  0.1581M
121.6gmol1 This should give pH =7 but it cannot be so because solution
is acidic and pH should be less than 7. The reason is that
Assuming complete dissociation, [H+] from H2O cannot be neglected. Thus, total [H+] = 10–7
Sr(OH)2  Sr2+ + 2 OH–
M (from HCl) + 10–7 M (from H2O) = 2 × 10–7 M
 [Sr 2  ]  0.1581M
7
[OH  ]  2  0.1581  0.3162M  pH   log(2  10 )  7  0.3010  6.699.
Example-25
pOH   log 0.3162  0.5,  pH  14  0.5  13.5
pH of 0.08 mol dm–3 HOCl solution is 2.85. Calculate its
Example-21
Ionisation constant.–7
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34.
Sol. pH of HOCl=2.85 i.e., –log [ H+] = – 2.85
Calculate the ionization constant of the acid and its degree
of ionization in the solution. (NCERT) or log [H+] = – 2.85 = 3.15

Sol. HCNO H   CNO or [H+] = antilog 3.15 = 1.413 × 10–3 M

pH=2.34 mean –log[H+]= 2.34 or log [H+]=–2.34= 3.66 For weak monobasic acid HA H+ + A–

1.413  10 3 
2 2
or  H   Anti log 3.66  4.57  10 M
 3
 H  
Ka   = 2.4957 × 10–5.
 CNO     H    4.57  10 3 M
 HA  0.08
164 EQUILIBRIUM

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. What is dynamic equilibrium? 15. List factors on which equilibrium constant depends.

2. Define physical equilibrium with example. 16. Why the addition of inert gas does not change the
equilibrium?
3. If pH of solution is 7. Calculate the pOH value.
17. Justify the statement that water behaves like an acid and
4. How does diluting with water affect the pH of a buffer also like a base on the basis of protonic concept.
solution?
18. Ksp for HgSO4 is 6.4 x 10-6. What is the solubility of the
5. List one factor on which boiling point depends? salt.

6. Define Henry’s law. 19. Calculate the pH of the solution

0.002 M HBr
7. What happens at the high-altitude boiling point in the
water? 20. The concentration of H+ in a soft drink is 3.8 x 10–4 M.
What is its pH?
8. What is the concentration of H3O+ and OH- ions in
water at 25oC? Section – C (3 Marks Questions)

9. Draw the conclusion from the following equilibrium

21. Calculate the hydrolysis constant and degree of
Solid Liquid hydrolysis and pH of 0.10 M KCN solution at 25°C. Ka
H2O(s) H2O(l) for HCN = 6.2 × 10-10.

10. What happens to the pH if a few drops of acid are 22. Give the generalizations concerning the composition of
added to CH3COONH4 solution? equilibrium mixtures.

Section – B (2 Marks Questions) 23. The solubility product of AgCl in water is 1.5 × 10-10.
Calculate its solubility in 0.01 M NaCl solution.
11. List some characteristics of equilibrium which involve
physical processes. 24. (i) In the reaction equilibrium
A+B C + D,
12. Write the expression for the equilibrium constant for What will happen to the concentrations of A, B and
the given reaction: D if concentration of C is increased.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) (ii) What will happen if concentration of A is
increased?
13. Explain when total number of reactants and products
are equal there , K is unitless. 25. The degree of dissociation of N2O4,
N2O4(g) 2NO2(g), at temperature T and total
14. What is the unit of equilibrium for the reaction pressure is a. Find the expression for the equilibrium
N2(g) + 3H2(g) 2NH3(g). constant of this reaction at this temperature and
pressure?
EQUILIBRIUM 165

26. Calculate the pH of 0.01 M solution of NH4CN. Given Section – D (5 Marks Questions)
that the Ka for HCN = 6.2 × 10-10 and Kb for NH3 = 1.6
x 10-5. 29. From a stock solution of 0.300 M NaOAc and 0.300 M
HOAc with a given Ka value of 1.8 x 10–5. What is the
27. Will dissociation of HI and Pcl5 depends on pressure volume of components needed for the preparation of
applied . Explain 500. mL of 0.300 M acetate buffer with a pH of 5.0?
(Report as: volume of acetic acid, volume of sodium
28. A student needs to prepare a buffer solution with a pH acetate respectively)
of 4.65. Assuming a pKa of 4.57, how many mL of 0.1
M B- would need to be added to 16.0 mL of 0.1 M HB 30. Name and explain the factors which influence the
prepare this buffer? equilibrium state.
166 EQUILIBRIUM

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions)
5. Calculate the pOH of a solution at 25°C that contains
1. Name the processes in which chemical equilibria is 1 × 10–11 M of hydronium ions, i.e. H3O+.
important. (a) 3.000 (b) 9.000
(a) biological, chemical (c) 1.000 (d) 7.000
(b) chemical, biological
(c) biological, environmental 6. If, in the reaction N2O4 ⇌ 2NO2, x is that part of N2O4
(d) chemical, physical which dissociates, then the number of molecules at
equilibrium will be
2. Equilibria involving O2 molecules and the protein (a) 1 (b) 3
hemoglobin play a crucial role (c) (1 + x) (d) (1 + xy)2
(a) In the transport of O2 form our lungs to our
muscles 7. Which of the following is the weakest base?
(b) In the transport and delivery of O2 from our lungs (a) F– (b) Cl–
to our muscles (c) Br (d) I–
(c) In the transport of CO2 and delivery of O2 from our
lungs to our muscles 8. The Ksp for Cr (OH)3 is 1.6 x 10−30. The molar
(d) In the transport and delivery of CO2 and O2 from solubility of this compound in water is:
our lungs to our muscles (a)  (1.6 x 10–30)
(b) ( (1.6 x 10–30))(1/4)
3. H2O (l ) H 2 O ( vap.) (c) ( (1.6 x 10–30)/(27)))(1/4)
In this equation, the double half arrows indicate that (d) (1.6 x 10–30)/(27)
(a) the processes in which the forward direction is
more than the backward direction 9. The solubility product of a salt having general formula
(b) the processes in which the forward direction is less MX2. In water is: 4 x 10–12. The concentration of M2+
than the backward direction ions in the aqueous solution of the salt is
(c) the processes in both the direction are going on (a) 4.0 x 10–10 M (b) 1.6 x 10–4 M
–4
simultaneously (c) 1.0 x 10 M (d) 2.0 x 10–6 M
(d) the processes in both the direction are not going on
simultaneously 10. Why buffer solution have constant acidity and
alkalinity?
4. Look at the following reactions carried out in closed (a) They have large excess of H+ or OH– ions
vessels (b) They have fixed value of pH
I. PCl5 → PCl3 + Cl2 (c) These give unionised acid or base on reaction with
added acid or alkali
II. N2 + O2 → 2NO
(d) Acids and alkalies in these solutions are shielded
III. 2KClO3 → 2KCl + 3O2 from attack by other ions
IV. Fe3+ + SCN– → [Fe(SCN)]2+
Which one of them is irreversible reaction? 11. In a chemical reaction, A B the system will be in
(a) Only II (b) Only I equilibrium when
(c) Only III (d) Only IV (a) 90% of A changes to B
(b) A completely changes to B
(c) only 50% of A changes to B
(d) the rate of change of A to B and B to A on both the
sides are same
EQUILIBRIUM 167

12. Which of the following is correct for equilibrium state? 19. The solubility product of a sparingly soluble salt AB at
(a) The number of molecules leaving the liquid equals room temperature is 1.21 × 10-6, its molar solubility is
the number returning to liquid from the vapour (a) 1.21 × 10-6 M (b) 1.1 × 10-4 M
-3
(b) The number of molecules leaving the liquid is not (c) 1.1 × 10 M (d) None of these
equal the number returning to liquid from the
vapour 20. What is the pH of a 0.10 M solution of barium
(c) The number of molecules leaving the liquid is less hydroxide, Ba (OH)2?
than the number returning to liquid from the (a) 11.31 (b) 11.7
vapour (c) 13.30 (d) None of these
(d) The number of molecules leaving the liquid is
more than the number returning to liquid from the Section-B (Assertion & Reason Type Questions)
vapour
21. Assertion (A): There is only one temperature
13. Equilibrium can be established for (melting point) at 1 atm (1.013 bar) at which the two
(a) chemical reactions (b) physical processes phases can coexist. If there is no exchange of heat with
the surroundings, the mass of the two phases remains
(c) Both (a) and (b) (d) None of the above
constant.
Reason (R): It is the state of the solid liquid
14. The pH of solution on mixing equal volume of solution equilibrium.
having pH = 3 and pH = 4. [log 5.5 = 0.7404] (a) Both A and R are correct; R is the correct
(a) 4.38 (b) 3.54 explanation of A
(c) 3.26 (d) 4.12 (b) Both A and R are correct; R is not the correct
explanation of A
(c) A is correct; R is incorrect
15. A solution of an acid has pH = 4.70. Find out the
(d) R is correct; A is incorrect
concentration of OH- ions (pKw = 14).
(a) 5 × 10-10 (b) 6 × 10-10
22. Assertion (A): The vapour pressure of gas in liquid is
(c) 2 × 10-5 (d) 9 × 10-5
constant at a given temperature.
Reason (R): It is the state of liquid vapour
16. Which one of the following statement is incorrect?
equilibrium.
(a) The conjugate base of H2PO4- is HPO42-
(a) Both A and R are correct; R is the correct
(b) pH + pOH = 14 for all aqueous solutions
explanation of A
(c) The pH of 1 × 10-8 M HCl is 8
(b) Both A and R are correct; R is not the correct
(d) pH decreases with rise in temperature
explanation of A
(c) A is correct; R is incorrect
17. The solution which resist change in pH on dilution or
(d) R is correct; A is incorrect
with the addition of small amounts of acid or alkali are
called
23. Assertion (A): For dissolution of solids in liquids, the
(a) homogeneous solutions
solubility is constant at a given temperature.
(b) heterogeneous solutions
Reason (R): For dissolution of gases in liquids, the
(c) ionic solutions concentration of a gas in liquid is inversely proportional
(d) buffer solutions
to the pressure (concentration) of the gas over the
liquid.
18. Equimolar solutions of the following were prepared in
(a) Both A and R are correct; R is the correct
water separately. Which one of the solutions will record explanation of A
the highest pH? (b) Both A and R are correct; R is not the correct
(a) CaCl2 (b) SrCl2 explanation of A
(c) BaCl2 (d) MgCl2 (c) A is correct; R is incorrect
(d) R is correct; A is incorrect
168 EQUILIBRIUM

24. Assertion (A): An aqueous solution of ammonium 28. Assertion (A): For the equilibrium mixture
acetate can act as a buffer. CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g) if the volume is
Reason (R): Acetic acid is a weak acid and NH4OH is decreased, reaction proceeds in the forward direction.
a weak base. Reason (R): For the methanation reaction (above),
(a) Both A and R are correct; R is the correct decrease in volume causes Qc > Kc
explanation of A (a) Both A and R are correct; R is the correct
(b) Both A and R are correct: R is not the correct explanation of A
explanation of A (b) Both A and R are correct; R is not the correct
(c) A is correct; R is incorrect explanation of A
(d) R is correct; A is incorrect (c) A is correct; R is incorrect
(d) R is correct; A is incorrect
25. Assertion (A): pH of 10–7 M HCl is 6.69 at room
temperature Section – C (Case Study Questions)
Reason (R): pH of acidic solution is always below 7 at
25°C. Case Study – 1
(a) Both A and R are correct; R is the correct
explanation of A Le Chatelier's principle is also known as the
(b) Both A and R are correct: R is not the correct equilibrium law, used to predict the effect of change on
explanation of A a system at chemical equilibrium.
This principle states that equilibrium adjust the forward
(c) A is correct; R is incorrect
and backward reactions in such a way as to accept the
(d) R is correct; A is incorrect change affecting the equilibrium condition. When
factor like concentration, pressure, temperature, inert
26. Assertion (A): The value of Kw at 25°C is 1.0 x 10–14 gas that affect equilibrium are changed, the equilibrium
mol2 dm–6. will shift in that direction where the effects that caused
Reason (R): Kw of water changes with change in by these changes are nullified. This principle is also
temperature. used to manipulate reversible reaction in order to obtain
(a) Both A and R are correct; R is the correct suitable outcomes.
explanation of A 29. Under which condition maximum product will form?
(b) Both A and R are correct: R is not the correct A2(g) + B2(g) ⇌ X2(g), ∆rH = –X kJ ?
explanation of A (a) Low temperature and high pressure
(c) A is correct; R is incorrect (b) Low temperature and low pressure
(d) R is correct; A is incorrect (c) High temperature and high pressure
(d) High temperature and low pressure
27. Assertion (A): Kp can be equal to, less than or greater
than the value of Kc. 30. Pick out the reaction upon increase the pressure
Reason (R): Kp = Kc(RT)∆n where ∆n is the change in reaction move in backward direction?
the number of moles of gaseous reactants and products. (a) H2(g) + I2(g) ⇌ 2HI(g)
(a) Both A and R are correct; R is the correct (b) 2NH3(g) ⇌ N2(g) + 3H2(g)
explanation of A (c) C(s) + O2(g) ⇌ CO2(g)
(b) Both A and R are correct; R is not the correct (d) 2H2(g) + O2(g) ⇌ 2H2O(g)
explanation of A
(c) A is correct; R is incorrect 31. For the reversible reaction,
(d) R is correct; A is incorrect N2(g) + 3H2(g) ⇌ 2NH3(g) + heat
The equilibrium shifts in backward direction
(a) by decreasing the concentration of NH3(g)
(b) by decreasing the pressure
(c) by increasing the concentrations of N2(g) and H2(g)
(d) by increasing pressure and decreasing temperature
EQUILIBRIUM 169

32. Given below is a reaction, list suitable conditions by Case Study – 3

which ammonia formation is there.
N2 + 3H2 ⇌ 2NH3; ∆H = –21.9 kcal are For the following reaction,
(a) low temperature, low pressure and catalyst N2(g) + 3H2(g) ⇌ 2NH3(g); ∆H = —22.4 kcal; the
(b) low temperature, high pressure and catalyst favourable conditions are
(c) high temperature, low pressure and catalyst (i) high concentration of N2 and H2.
(d) high temperature, high pressure and catalyst (ii) high pressure (as ∆ng = —2).
(iii) low temperature (as the reaction is exothermic).
Case Study – 2 Although the low temperature favour the formation of
NH3, yet an optimum temperature is set, as the rate of
Reactants and products coexist at equilibrium, so that reaction is very slow at low temperature. A temperature
the conversion of reactant to products is always less of 500°C is selected and a pressure of 200 atm is
than 100%. Equilibrium reaction may involve the applied in practice.
decomposition of a covalent (non-polar) reactant or
ionization of ionic compound into their ions in polar
solvents. Ostwald dilution law is the application of the 37. What is the expression of equilibrium constant for
law of mass action to the weak electrolytes in solution. synthesis of NH3.
A binary electrolyte AB which dissociates into A+ and
 NH3   NH3 
2 2

B– ions i.e. (a) (b)

AB ⇌ A+ + B–  N 2  H 2 
3
 N2 
for every weak electrolyte, Since  << 1 (1 – ) = 1
 N2  H2 
2 2

K (c) (d)
C ⇒= KV H2   NH3 
K = C2 ⇒  =

33. A monobasic weak acid solution has a molarity of 38. What happens to the yield of ammonia produced when
0.005M and pH of 5. What is its percentage ionization the pressure is increased?
in this solution? (a) Increases
(a) 2.0 (b) 0.2 (b) Decreases
(c) 0.5 (d) 0.25 (c) Remain constant
(d) Increases and then decreases
34. Calculate ionisation constant for pyridinium hydrogen
chloride. (Given that H+ ion concentration is 3.6 × 10–3 39. Which substance is used as a catalyst promotes in the
M and its concentration is 0.02 M.) synthesis of ammonia?
(a) 6.03× 10–12 (b) 6 × 10–18 (a) Mn (b) Fe
(c) 1.5 × 10 –12
(d) 12 × 10–8 (c) Cr (d) Cu

35. The hydrogen ion concentration of a 10–8 M HCl 40. In the reaction, N2(g) + 3H2(g) ⇌ 2NH3(g)
aqueous solution at 298 K(Kw = 10–14) is Which side equilibrium is shifted when we increase the
(a) 9.525 x 10–8 M (b) 1.0 x 10–8 Al pressure of the system?
(c) 1.0 x 10–6 M (d) 1.0525 x 10–7 M (a) Right
(b) Left
36. Ostwald dilution law is applicable to (c) No change
(a) weak electrolytes (b) non-electrolyte (d) First toward right and then left
(c) strong electrolyte (d) all type of electrolyte
170 EQUILIBRIUM

Notes:

Find Answer Key and Detailed Solutions at the end of this book

EQUILIBRIUM
THE s-BLOCK
ELEMENTS
172 THE s-BLOCK ELEMENTS

Chapter at a Glance
Atomic and Physical Properties of s block Elements

Property Alkali Metals Trend Alkaline earth metals Trend

Electronic configuration ns1
ns2
Atomic radius Increases down the group Increases down the group
Ionisation energy Decreases down the the group Decreases down group
Hydration energy Decreases down the group Decreases down the group
Ionic Mobility Increases down the group Increases down the group
Melting Point Decreases down the group Irregular
Boiling point Decreases down the group Irregular
Density Increases down the group (exception: K) Increases down the group
Flame colouration Li - Crimson Red Be, Mg - Doesn’t give
Na - Yellow K - Violet, Rb - Red violet Ca - Brick red, Sr - Crimson,
Cs - Blue Ba - Apple green

Chemical Properties of s block Elements

Reaction with Alkali Metals Alkaline earth metals

Excess of Oxygen Li forms monoxide, All form oxides,
Na peroxide and rest superoxide Ba also forms peroxide.
Water Forms hydroxide Forms hydroxide
Air Dry air - corresponding Forms oxide as well as nitride
oxide/peroxide/superoxide (Be and Mg react in powder form)
Moist air - Hydroxide
Li forms nitride also
Hydrogen and halogens Forms ionic hydride Forms ionic hydride
(except Li) and halides respectively and halides
Reducing nature Li - Most powerful Ba - Most powerful
Na - Least powerful Be - Least powerful
Liquid ammonia Forms deep blue conducting solution of Forms deep blue black conducting
ammoniated metal ion and ammoniated solution of ammoniated metal ion
electron and ammoniated electron
THE s-BLOCK ELEMENTS 173

Solved Examples
Example-1
 Atomic and ionic radii increase with increase in
What are the common physical and chemical features of atomic number.
alkali metals? (NCERT)  The alkaline earth metals owing to their large size of
Sol. The alkali metals show a regular trend in their physical and atoms have fairly low values of Ionization energy.
chemical properties with the increasing atomic number.  They are harder than alkali metals.
Physical properties of alkali metals:  Electropositive character increases on going down the
 Alkali metals have low melting and boiling point. group.
 Alkali metals (except Li) exhibit photoelectric effect. Example-3
 Alkali metals impart characteristic colours to the flame. Why are alkali metals not found in nature? (NCERT)
 Alkali metals are silvery white in colour and are generally Sol. Alkali metals are very reactive in nature. They will react
soft and light metals. very fast with air and water. Therefore, they always exist in
 Alkali metals have low ionization enthalpies. combine state not in free state in nature.
 Alkali metals are highly electropositive in nature Example-4
 Alkali metals show +1 oxidation states.
Find out the oxidation state of sodium in Na2O2. (NCERT)
 Alkali metals have the largest size in their respective
Sol. Oxidation state of sodium in Na2O2:
periods.
Let x be the oxidation state of Na
Chemical properties of alkali metals: Alkali metals are
2x + 2 (-1) = 0
highly reactive elements. The cause for their
(here, oxygen is in peroxide, so we must use the peroxide
high chemical reactivity is:
oxidation state of oxygen)
 Alkali metals have low value of first ionization enthalpy
2x – 2 = 0
 Alkali metals have low heat of atomization.
x = +1
 Alkali metals are highly reactive in nature. Example-5
 Alkali metals have strong tendency to get oxidized and
Explain why is sodium less reactive than potassium.
acts as strong reducing agents.
(NCERT)
 Alkali metals hydroxides are highly basic in nature.
Sol. Sodium less reactive than potassium due to ionization
 Alkali metals dissolve in liquid ammonia and give deep
blue colour solution which are conducting in nature. enthalpy.
Example-2 Ionization enthalpy of sodium = 496 kJ/mol.
Discuss the general characteristics and gradation in Ionization enthalpy of potassium = 419 kJ/mol
properties of alkaline earth metals. (NCERT) Ionization enthalpy of potassium is less than that of sodium
Sol. The alkaline earth metals have 2 electrons in the s- orbital due to this potassium is more reactive than sodium.
Example-6
of the valence shell.
Compare the alkali metals and alkaline earth metals with
 The alkaline earth metals are silvery white, lustrous and
respect to
relatively soft but harder than the alkali metals.
(i) ionization enthalpy,
 The hydration enthalpies of alkaline earth metal ions
(ii) basicity of oxides,
are larger than those of alkali metal ion.
(iii) solubility of hydroxides. (NCERT)
174 THE s-BLOCK ELEMENTS

Sol. Alkaline earth metals are very much like that of the alkali This property makes Cs and K useful as electrodes in
metals. photoelectric cells.
(i) Ionization enthalpy: Alkaline earth metals is higher than Example-10
alkali metals due to high nuclear charge. the second When alkali metal dissolves in liquid ammonia, the solution
ionization enthalpy of alkaline earth metals is smaller can acquire different colours.Explain the reason for this
than those of the corresponding alkali metals. type of colour change. (NCERT)
(ii) Basicity of oxides: Basicity of oxides of alkali earth metals Sol. All alkali metals dissolve in liquid ammonia and give deep
blue colour solution which are conducting in nature. These
are lower than that of alkaline metals.
solutions contain ammoniated cations and ammoniated
Solubility of hydroxides: Alkali metals are higher than electrons. Ammoniated electrons absorb energy
that of alkaline earth metals. they are soluble in water. corresponding to red region of visible light for their
excitation to higher energy levels
Example-7
 –
M   x  y  NH3   M  NH3  x   e  NH 3  y 
In what ways lithium shows similarities to magnesium in its  
Ammoniated
chemical behaviour? (NCERT) electrons
Sol. The similarity between lithium and magnesium is because Example-11
of their similar sizes; comparable atomic radii and Beryllium and magnesium do not give colour to flame
electronegativities. whereas other alkaline earth metals do so. Why?
The major similarities are: (NCERT)
Sol. Beryllium and magnesium do not give colour to flame
 Both are harder and lighter than other elements of
whereas other alkaline earth metals due to the small size,
the respective groups. the ionization enthalpies of Be and Mg are much higher.
 Carbonates of both decompose easily on heating Hence, a large amount of energy is required to excite their
and form oxides and CO2 and both do not form solid outermost electron. That’s why they do not impart colour
to the flame.
hydrogen carbonate.
Example-12
 Both LiCl and MgCl2 are soluble in ethanol.
Discuss the various reactions that occur in the Solvay
 Both react slowly with water.
process. (NCERT)
 Both form nitride.
 Both can form complex compounds. Sol. The reaction involved in Solvay process are:

Example-8 NH3 + H2O + CO2 ——— NH4HCO3

Explain why can alkali and alkaline earth metals not be NaCl+ NH4HCO3 ——— NaHCO3 + NH4Cl

obtained by chemical reduction method. (NCERT) 2NaHCO3——— Na2CO3 +H2O + CO2

Sol. Alkali and alkaline earth metals are strong reducing agents. Example-13
That’s why alkali and alkaline earth metals are not obtained Potassium carbonate cannot be prepared by Solvay
by chemical reduction methods. process. Why? (NCERT)
Example-9
Sol. This is due to the reason that potassium bicarbonate formed
Why are potassium and caesium, rather than lithium used
as an intermediate (when CO2 is passed through ammoniated
in photoelectric cells? (NCERT)
solution of potassium chloride) is highly soluble in water
Sol. The ionization enthalpy of lithium is 520kJ/mol but and cannot be separated by filtration.
potassium and caesium having 419 and 376 kJ/mol. These
elements when irradiated with light, the light energy
absorbed may be enough to make an atom lose electron.
THE s-BLOCK ELEMENTS 175

Example-14 Example-17
Why is Li2CO3 decomposed at a lower temperature whereas Beryllium has …A… Property from rest of alkaline earth
Na2CO3 at higher temperature? (NCERT) metals and show diagonal relationship with …B… Here, A
Sol. The carbonates of alkali metals are quite stable towards and B refer to
heat. Li2CO3 is considerably less stable and decompose (a) Same aluminium
readily on heating due to small polarizes a large carbonate
(b) Different, aluminium
ion leading to the formation of stable Li2O and CO2 on
(c) Sami silicon
heating. As the electropositive character increase down
the group, the stability of carbonates increases. Lattice (d) Different boron

energy of Na2CO3 is higher than that of Li2CO3. Na2CO3 is Ans. (b)

ionic compound and Li2CO3 is a covalent compound. Sol. Beryllium has different property from rest of alkaline earth
Example-15 metals and show diagonal relationship with aluminium..
Give the solubility and thermal stability of the following Example-18
compound of the alkali metals with those of the alkaline The correct reason of expanded coordination number of
earth metals. alkaline earth metal other than beryllium is due to
(a) Nitrates
(a) Presence of vacant p-orbital
Sol. (a) Hydration energy is more than lattice energy due to this
(b) Presence of vacant d-orbital
nitrates of both group 1 and group 2 elements are soluble in
water. (c) Presence of vacant f-orbital in some cases

Nitrates of both group 1 and group 2 elements are thermally (d) All of the above
unstable. Group 1 and 2 nitrates decomposed as follows: Ans. (b)

heat  2NaNO  O Sol. Metal other than Be form more than 4 coordinates complex
2NaNO3  2 2
due to presence of vacant d-orbital.

2KNO  2KNO  O
3 2 2 Example-19

  2LiO  4NO  O Magnesium burns in air to produce MgO and …X… with
4LiNO3  2 2 2
…Y… flame. Here, X and Y refer to

 
2Mg NO  2MgO  4NO  O
3 2 2 2
(a) Mg2N3, Oxidizing
Example-16
(b) Mg3N2, reducing
…X and …Y…Shown diagonal similarity.
(c) Mg3N2, dazzling
Here, X and Y refer to
(d) Mg2N3, dazzling
(a) Lithium and magnesium
Ans. (c)
(b) Lithium and beryllium
Sol. Mg is more electropositive metal and buns withdazzling
(c) Sodium and magnesium flame in air to give MgO and Mg3N2.
(d) None of these
Ans. (a)
Sol. Lithium and magnesium shows similarity in many of their
properties hence, they are diagonally related to each other.
176 THE s-BLOCK ELEMENTS

Example-20 IV. Biological function of s-block elements is due to

Which of the following is correctly represented according maintenance of ion balance and nerveimpulse condition.
to their existence? (a) Only I
I. MgCl2 .6H2O II. NaCl. 6H2O (b) II and III
III. KCl. 6H2O IV. CaCl2. 6H2O (c) II and III
(a) I and II (b) II and III (d) Only III
(c) II and IV (d) I and IV Ans. (c)
Ans. (d) Sol. Statement II and III are incorrect, because Oxides and
Sol. MgCl2 and CaCl2 exist as hydrates while KCl and NaCl hydroxides of alkaline earth metals and alkali metal are
does not exist in the hydrate form. alkaline in nature.
Example-21
Apart from Na and K, Ca and Mg are also found in biological
Which of the following is/are correct statement(s) regarding fluid.
carbonates of alkaline earth metals? Example-23
I. They are thermally unstable Match the column I with Column II and choose the correct
II. They are insoluble in water option from the codes given below.
III. They cannot be precipitated by adding sodium or
Column I Column II
ammonium carbonate.
A.Sodium 1. Present in biological fluid
IV. Beryllium carbonate is thermally unstable.
B.Beryllium 2. Radioactive element
(a) I, II and III are correct
C.Francium 3. Lower abundance
(b) II and IV are correct
D. Calcium 4. Alkali metals
(c) I, II and IV are correct
5. Alkaline earth metal
(d) All are correct
A B C D
Ans. (c)
Sol. Statement III is incorrect because carbonates can be (a) 1 4 2 5

precipitated by addition of sodium or ammonium carbonate (b) 1 2 3 4

due to common ion effect. (c) 3 1 2 5
Example-22
(d) 4 5 2 1
Which of the following statements(s) is/are incorrect
Ans. (d)
regarding the s-block elements?
Sol. Correct matching is represented as
I. Francium is highly radioactive element.
A. Sodium- Alkali metal
II. Oxides and hydroxides of alkali metals and alkaline earth
metals are not alkaline in nature. B. Beryllium- Alkaline earth metal

III. Sodium and potassium are the only two s- block C. Francium- radioactive element
elements are found in large proportion in biological D. Calcium - Present in biological fluid
fluids.
THE s-BLOCK ELEMENTS 177

Example-24 Example-25
Milk of lime reacts with chlorine to form …A…a constituent Which of the following is/are correct statement (s)?
of …B… Here, A and B refer to (a) Ca3(PO4)2 is part of bones
(a) Hypochlorite, cement (b) 3Ca3(PO4).CaF2 is part of enamel on teeth
(b) Hypochlorite, bleaching powder (c) Ca2+ ions are important in blood clotting
(c) Hypochlorite, bleaching powder (d) All of the above are correct
(d) Hypochlorous cement Ans. (d)
Ans. (b) Sol. All the given statement are correct.
Sol. Milk of lime i.e., Ca(OH)2 reacts with chlorine to form Ca3(PO4)2 is part of bones
hypochlorite Ca(OCl2) which is a constituent of bleaching
3Ca3(PO4).CaF2 is part of enamel on teeth
powder.
Ca2+ ions are important in blood clotting
Ca  OH 2  2Cl 2  CaCl 2  CaOCl 2  2H 2 O
178 THE s-BLOCK ELEMENTS

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Melting and boiling points of alkali metals is low. 15. S-block elements never occur in free state. Explain.
Explain? What are their usual modes of occurrence?

2. Explain the diagonal relationship in periodic table. 16. Why second ionization enthalpy of Calcium is more
than the first? How is that calcium forms CaCl2 and not
3. Why are lithium compounds soluble in organic CaCl . Explain
solvents?
17. Solubility of alkaline metal hydroxides increases down
4. By which process we can prepapre sodium carbonate? the group . Explain.

5. What is the formula of soda ash? 18. Solution of Na2CO3 is alkaline. Give reason.

6. Why do alkaline earth metals have low ionization 19. Discuss the various reactions that occur in the solvay
enthalpy? process

7. What is the nature of oxide formed by Be? 20. Give reason:

Lithium halides are covalent in nature.
8. Beryllium show similarities with Al. Why?
Section – C (3 Marks Questions)
9. What happens when gypsum is heated to 390K?
21. Explain each of the following:
10. Why anhydrous calcium sulphate cannot be used as (i) The mobilities of the alkali metal ions in aqueous
plaster of Paris? solution are Li+ < Na+ < Rb+ < Cs+
(ii) Lithium is the only alkali metal to form directly a
Section – B (2 Marks Questions) nitride.
(iii) Eo for M2+(aq) + 2e– → M(s) (where M = Ca, Sr
11. Why metals like potassium and sodium cannot be or Ba) is nearly constant.
extracted by reduction of their oxides by carbon?
22. Explain :
12. Complete the reactions and balance them. (i) Why Lithium on being heated in air mainly forms
(a) Na2O2 and water the monoxide and not peroxide.
(b) KO2 and water (ii) K, Rb and Cs on being heated in the presence of
excess supply of air form superoxides in preference to
13. Potassium carbonate can not be prepared by the oxides and peroxides.
SOLVAY process. Give reason. (iii) An aqueous solution of sodium carbonate is
alkaline in nature.
14. Explain what happens when
(i) Sodium hydrogen carbonate is heated. 23. The hydroxides and carbonates of sodium and
(ii) Sodium with mercury reacts with water. potassium are easily soluble in water while the
corresponding salts of magnesium and calcium are
sparingly soluble in water. Explain.
THE s- BLOCK ELEMENTS 179

24. Action of heat on the following: Section – D (5 Marks Questions)

(i) Na2CO3 and CaCO3
(ii) MgCl2.6H2O and CaCl2.6H2O 29. Arrange the following in the decreasing order of the
(iii) Ca(NO3)2 and NaNO3 property mentioned:
25. What are the main uses of calcium and magnesium? (i) Li+, Na+, K+, Rb+ (ionic mobility)
(ii) Be, Mg, Ca, Sr (melting point)
26. Explain diagonal relationship. What is it due to? (iii) BeO, MgO, CaO (Enthalpy of formation)
(iv) Be, Mg, Ca (metallic radius)
27. The hydroxides and carbonates of sodium and
potassium are easily soluble in water while the 30. How do the following properties change on moving
corresponding salts of magnesium and calcium are from group 1 to group 2 in the periodic table.
sparingly soluble in water . Explain. (i) Atomic size (ii) Ionization enthalpy
(iii) Density (iv) Melting points.
28. Why Beryllium shows anomalous behaviour. Write
three such properties of beryllium which makes it
anomalous in the group.
180 THE s-BLOCK ELEMENTS

EXERCISE-2: Basic Objective Questions

Section–A (Single Choice Questions)

1. General electronic configuration for alkaline earth 9. Alkali metal ions are colored due to?
metals is (a) presence of single valence electron
(a) [Noble gas] ns1 (b) [Noble gas] ns2 (b) excitation of valence electron
(c) Both (a) and (b) (d) None of the above (c) emission of radiation in visible region during come
back of electron to the ground state energy level
2. Pick put the one which does not belongs to alkaline (d) none of the above
earth metal.
10. How does the color of alkali metal ions in any sample
(a) Beryllium (b) Strontium
(c) Francium (d) Calcium can be detected?
(a) flame photometry
3. Diagonally related elements is / are: (b) atomic adsorption method
(c) both (a) and (b)
(a) Beryllium and aluminium
(d) None of these
(b) Lithium and magnesium
(c) Sodium and aluminium
11. In solid state, hydrogen carbonate of lithium doesn’t
(d) Both (a) and (b)
exist because:
(a) low polarization value of Li+
4. Which of the following ion(s) is/are responsible for
(b) low hydration energy of Li+
biological function such as maintenance of ion balance
(c) high hydration energy of Li+
and nerve impulse conduction? (d) high polarization value of Li+
(a) Sodium (b) Potassium
(c) Magnesium (d) All of these 12. On moving top to bottom the negative value of
formation ….A… for fluoride of alkali metals. Here, A
5. Along the group in alkali metals the density refers to
(a) decrease top to bottom (a) becomes less negative
(b) increase top to bottom (b) becomes more negative
(c) remains same (c) remains constant
(d) follow an irregular trend (d) none of these

6. The low melting point of alkali metals is due to 13. In which part of human body, sodium ions are found?
(a) weak metallic bonding (a) Outside of cells (b) In blood plasma
(b) presence of only one valence electron (c) In interstitial fluid (d) All of these
(c) both (a) and (b)
(d) none of the above 14. Which ion is most abundant within cell fluid?
(a) Sodium ion
7. Which of the following metal has stable carbonates? (b) Potassium ion
(a) Na (b) Mg (c) Both are equally abundant
(c) Al (d) Si (d) None of these

8. The reaction of Cl2 with X gives bleaching powder x is 15. Which of the following is correctly order regarding
(a) CaO (b) Ca(OH)2 hydration enthalpies of alkaline earth metals?
(c) Ca(OCl)2 (d) Ca(O3Cl)2 (a) Be2+ < Mg2+ < Ca2+ < Sr2+ < Ba2+
(b) Be2+ > Mg2+ > Ca2+ > Sr2+ > Ba2+
(c) Be2+ < Mg2+ > Ca2+ > Sr2+ > Ba2+
(d) Mg2+ > Be2+ > Ca2+ > Sr2+ > Ba2+
THE s- BLOCK ELEMENTS 181

16. Why Beryllium and magnesium do not impart any 22. Assertion (A): In general, alkali and alkaline earth
color to the flame test? metals form superoxides.
(a) lower value of ionization value Reason (R): There is a single bond between O and O
(b) higher value of ionization energy in superoxides.
(c) larger size of alkaline earth metal (a) Both A and R are correct; R is the correct
(d) Both (b) and (c) are correct explanation of A
(b) Both A and R are correct; R is not the correct
17. Which salt is used in radiotherapy? explanation of A
(a) Barium salt (b) Calcium salt (c) A is correct; R is incorrect
(c) Radium salt (d) Strontium salt (d) R is correct; A is incorrect

18. Antacid used in medicine is. 23. Assertion (A): BeSO4 is soluble in water while BaSO4
(a) Milk of magnesia is not.
(b) Suspension of hydroxide of magnesium in water Reason (R): Hydration energy decreases down the
(c) Suspension of hydroxide of calcium in water group from Be to Ba and lattice energy remains almost
(d) Both (a) and (b) constant.
(a) Both A and R are correct; R is the correct
19. What are Oxo-Acids? explanation of A
(a) Acids containing Oxygen (b) Both A and R are correct; R is not the correct
(b) Acid containing Sulphur explanation of A
(c) Acid containing Carbon (c) A is correct; R is incorrect
(d) None of these (d) R is correct; A is incorrect

20. The hydroxides having maximum basic strength among 24. Assertion (A): Lithium salts are hydrated.
the following is : Reason (R): Lithium has higher polarizing power than
(a) LiOH (b) NaOH other alkali metal group members.
(c) Ca(OH)2 (d) KOH (a) Both A and R are correct; R is the correct
explanation of A
(b) Both A and R are correct; R is not the correct
Section-B (Assertion & Reason Type Questions)
explanation of A
(c) A is correct; R is incorrect
21. Assertion (A): The carbonate of lithium decomposes (d) R is correct; A is incorrect
easily on heating to form lithium oxide and CO 2.
Reason (R): Lithium being very small in size polarizes 25. Assertion (A): The compounds of alkaline earth metals
large carbonate ion according to Fajan’s rule leading to are more extensively hydrated than those of alkali
the formation of more stable Li2O and CO2. metal.
(a) Both A and R are correct; R is the correct
Reason (R): This is due to low hydration energy of
explanation of A
alkaline earth metal as compared to alkali metal.
(b) Both A and R are correct; R is not the correct
(a) Both A and R are correct; R is the correct
explanation of A
explanation of A
(c) A is correct; R is incorrect
(b) Both A and R are correct; R is not the correct
(d) R is correct; A is incorrect
explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
182 THE s-BLOCK ELEMENTS

26. Assertion (A): Ba, Ca and Sr produces individual M ( g ) ⎯⎯⎯ → M+ + e–

st
1 I.E.
 noble gas
characteristic color to flame test.  noble gas ns1
Reason (R): This is due to excitation of loosely held Metal Ionization Energy (kJ mol-1)
electrons from outermost orbit to excited state of IE1 IE2
orbital.
Li 520.1 7296
(a) Both A and R are correct; R is the correct
Na 495.7 4563
explanation of A
(b) Both A and R are correct; R is not the correct K 418.6 3051
explanation of A Rb 402.9 2633
(c) A is correct; R is incorrect Cs 375.6 2230
(d) R is correct; A is incorrect
29. The characteristic of Alkali metals are
27. Assertion (A): Compounds of alkaline earth metal are (a) good conductors of heat and electricity
less ionic than the compounds of alkali metals. (b) high melting points
Reason (R): This is due to small size and increased (c) low oxidation potentials
nuclear charge of alkaline earth metals. (d) high ionization potentials
(a) Both A and R are correct; R is the correct
explanation of A 30. Metals dissolve in liquid ammonia giving coloured
(b) Both A and R are correct; R is not the correct solutions which are conducting in nature. The colour of
explanation of A the solution and reason of its conductance is
(c) A is correct; R is incorrect (a) yellow, NH4+
(d) R is correct; A is incorrect (b) blue, ammoniated metals
(c) orange, [M(NH3)x]+
28. Assertion (A): Beryllium and aluminium show (d) blue, ammoniated electron.
diagonal relationship.
Reason (R): This is due to exactly same charge/radius 31. Alkali metals displace hydrogen from water forming
ratio. bases due to the reason that
(a) Both A and R are correct; R is the correct (a) they are far above the hydrogen in electrochemical
explanation of A series bases on oxidation potential
(b) Both A and R are correct; R is not the correct (b) they are far below the hydrogen in electrochemical
explanation of A series based on oxidation potential
(c) A is correct; R is incorrect (c) their ionization potential is less than that of other
(d) R is correct; A is incorrect elements.
(d) they contain only one electron in their outermost
Section – C (Case Study Questions) shell.

Case Study–1 32. Alkali metals have high second ionization energy due
to
Case I: Read the passage given below and answer the (a) Removal of electrons from monovalent cation is
following questions from 29 to 32. difficult due to electrostatic repulsion
Alkali metals have the lowest ionization energy in their (b) Removal of electrons is easy from the noble gas
configuration.
corresponding period in periodic table because they
(c) Removal of electrons from the monovalent cation
have large size which results in a large distance
is difficult due to formation of stable noble
between the nucleus and the outermost electron.
configuration
Ionization energy of alkali metals decreases from Li to (d) None of the above
Cs due to increases in atomic size. First ionization
energy of alkali metals is very low but they have very
high value of second ionization energy.
THE s- BLOCK ELEMENTS 183

Case Study–2 Case Study–3

Case-II: Read the passage given below and answer the Case-III: Read the passage given below and answer
following question from 33 to 36. Alkaline earth the following question from 37 to 40. All alkali metals
metals are less reactive with water as compared to dissolve and form blue solution in liquid ammonia.
alkali metals. Their reactivity with water increases When alkali metals are dissolved in liquid ammonia,
down the group. Be does not react with water increases there is a considerable expansion in total volume hence
down the group. Be does not react with water at all, such solution of an alkali metal in ammonia shows
magnesium reacts only with hot water while other certain characteristic properties which are explained on
metals Ca, Sr and Ba react with cold water. the basis of formation of ammoniated (solvated) metal
Order of the reactivity with water: cations and solution in the following way:
Ba > Sr > Ca > Mg M → M + + e–
Be(OH)2 is amphoteric, but the hydroxides of Mg, Ca, e– + yNH3 → [e(NH3)y]–
Sr and Ba are basic. The basic strength increases from Thus, M + (x + y)NH3 → [M(NH3)x]+ + [e(NH3)y]–
Mg to Ba. Ammoniated Ammoniated
Be ( OH2 ) Mg ( OH )2 Ca ( OH )2 Sr ( OH )2 Ba ( OH )2 Metal cation electron
Amphoteric Weakly basic Basic Strongly basic Strongly basic The blue solution is paramagnetic and has high
electrical conductivity due to the presence of unpaired
33. Which of the following statements is false? electron in the cavities in ammoniacal solution.
(a) Strontium decomposes water readily than
beryllium 37. Nitride of a metal has chemical formula ‘A’ (M3N).
(b) Barium carbonate melts at a higher temperature On heating ‘A’, it produces metal metal M again. It
than calcium carbonate. reacts with H2O to produce gas ‘B’. When ‘B’ passes
(c) Barium hydroxide is more soluble in water than through CuSO4 solution it increases color intensity of
magnesium hydroxide solution. M and B can be
(d) Beryllium hydroxide is more basic than barium (a) Mg and NH3 (b) Na and NH3
hydroxide. (c) Li and NH3 (d) Al and NH3

34. Be(OH)2 is amphoteric in nature because: 38. When Sodium dissolves in liquid NH3 , it give a deep
(a) It can act as both acid and base blue solution due to
(b) It can act as acid only (a) ammoniated Na+
(c) It can act as base (b) ammoniated Na–
(d) It can behave as neutral compound (c) formation of Na+/Na– pair
(d) ammoniated electrons
35. CaCl2 + 2H2O → X + 2HCl
Here X will be 39. The increasing order of the density of alkali metals is
(a) Ca(OH)2 (b) CaO (a) Li < K < Na < Rb < Cs (b) Li < Na < K < Rb < Cs
(c) Ca (d) Ca(OH)Cl (c) Cs < Rb < Na < K < Li (d) Cs < Rb < K < Na < Li

36. Solubility order of hydroxides of alkaline earth metal. 40. The reaction between sodium and water can be made
(a) Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 less vigorous by
< Ba(OH)2 (a) lowering the temperature
(b) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 (b) adding a little alcohol
< Be(OH)2 (c) amalgamating sodium
(c) Be(OH)2 > Mg(OH)2 > Ca(OH)2 >Sr(OH)2 (d) adding a little acetic acid
> Ba(OH)2
(d) Be(OH)2 < Mg(OH)2 > Ca(OH)2 < Sr(OH)2
> Ba(OH)2
184 THE s-BLOCK ELEMENTS

Notes:

Find Answer Key and Detailed Solutions at the end of this book

The s-BLOCK ELEMENTS

THE p-BLOCK
ELEMENTS
186 THE p-BLOCK ELEMENTS

Chapter at a Glance
Group 13 Group 14

Physical Property Remarks Physical Property Remarks

Atomic Radii  B < Al > Ga < In < Tl Atomic Radii  C < Si < Ge < Sn < Pb

Ionisation Energy  B > Tl > Ga > Al > In Ionisation Energy  C > Si > Ge > Sn < Pb

Electronegativity  B > Tl > In > Ga > Al Electronegativity  C > Si = Ge = Sn < Pb

Oxidation state  General oxidation states: +1, +3 Oxidation state  General oxidation states: +2, +4

Metallic character  Increases down the group Metallic character  Increases down the group

Melting point  Decreases up to Ga then increases Melting point  Decreases up to Sn then increases

Boiling point  Decreases down the group Boiling point  Decreases down the group

Density  Increases down the group Density  Increases down the group

Chemical Property Remarks Chemical Property Remarks

Reactivity with  Reacts with air to form oxide (at mod Reactivity with air  Reacts with air to form monoxide and
air erate temp.) and nitride (at high temp.)
dioxides
Crystalline boron unreactive, amor
Reactivity with  Only Sn reacts with steam. Rest all do
phous boron reacts. Aluminium forms
oxide layer and does not react further. water not react with water in any form

 Boron does not react. Reactivity with  Reacts with halogens to form
Reactivity with
halogens halogens dihalides and tetrahalides
acids and alkalis Aluminium reacts with both acids and
alkalis to liberate hydrogen gas.

Aluminium passive with conc.

nitric acid.

Reactivity with  Reacts with halogens to form

halogens trihalides (except tri-iodide of

Thallium)

Nature of  Boron - acidic, Al and

trioxides Ga - amphoteric, In and Tl - Basic

THE p-BLOCK ELEMENTS 187

Solved Examples
Example-1 state is more stable than the Tl. In Tl, the +1 state is more
Discuss the pattern of variation in the oxidation states of stable.
(i) B to Tl and Example-3
(ii) C to Pb. (NCERT) Why does boron trifluoride behave as a Lewis acid?
Sol. (i) B to Tl: This elements are belongs to group 13. The Sol. Boron trifluoride behave as a Lewis acid. Boron trifluroides
electronic configuration of this group are ns2 np1. Most are electron deficient species. boron halides have three
common oxidation state should be +3 but it is only boron covalent bonds hence, require two electrons to complete
and aluminium. Ga, In, Tl, show both the +1 and +3 oxidation octet.
states. On moving down the group, the +1 state becomes
more stable due to transition contraction. But for Tl (+1) is
more stable than Tl (+3) because of the inert pair effect. The
stability of the +3 oxidation state decreases on moving
down the group.
Group 13 element Oxidation state
B +3
Al +3
Example-4
Ga, In, Tl +1, +3
Consider the compounds, BCl3 and CCl4. How will they
(ii) C to Pb
behave with water? Justify. (NCERT)
These elements are belonging to group 14. The electronic
configuration of this group are ns2 np 2. Most common Sol. BCl3 is electron deficient species. When it reacts with
oxidation state should be +4. The elements of group 14 electron rich species like water, it reacts fastly and formed
show tetra valency by sharing four of its valence electrons.
boric acid.
But Ge, Sn, Pb also show +2 oxidation state due to inert pair
effect. the stability of the lower oxidation state increases BCl3  3H 2 O  3HCl  B  OH 3
and that of the higher oxidation state decreases on moving
down the group. In case of CCl4, it cannot increase the coordination number
Group 14 element Oxidation state due to unavailability of vacant d orbital.
C +4 Example-5
Si +4 Is boric acid a protic acid? Explain.
Ge, Sn, Pb +2, +4
Sol. Boric acid only partially reacts with water to form H3O+ and
Example-2
[B(OH)4]-, it behaves as a weak acid.
How can you explain higher stability of BCl3 as compared

to TlCl3? (NCERT) B  OH 3  2HOH   B  OH 4   H 3 O 
Sol. Elements of 13 group reacts with halogens to form trihalides.
It does not donate protons like other acids.
BCl3 is more stable than TlCl3 because the boron +3 oxidation
188 THE p-BLOCK ELEMENTS
Example-6 (ii)
Explain what happens when boric acid is heated. 
2Al(s) 2NaOH (aq) 6H 2 O(l) 2Na   Al  OH  4   3H 2(g )
(aq)
(NCERT)
Example-9
 
H3 BO3   HBO2  B2 O3 What are electron deficient compounds? Are BCl3 and SiCl4
370 K red hot
Sol.
Metaboric acid Boric oxide electron deficient species?
Example-7 Explain.
Describe the shapes of BF 3 and BH 4 – . Assign the Sol. Electron deficient species are those species in which the
hybridisation of boron in these species. octet of the central metal atom is not complete.
Sol. (i) BCl3
(i) BF3 : Broon trifluoride are planar molecule in which central Boron trichloride behave as a Lewis acid. It is electron
atom is sp hybridized. An sp hybridized boron atom has
2 2 deficient species. boron tri-chloride have three covalent
an empty p-orbital which can accept a pair of electron. BF3 bonds hence, require two electrons to complete octet.
is shorter and stronger due to p  -p  back bonding. (ii) SiCl4
Silicon is belonging to group 14. The electronic
configuration of this group are ns2 np2. Most common
oxidation state should be +4. The elements of group 14
show tetra valency by sharing four of its valence electrons.
it forms four covalent bonds with four chlorine atoms.
Therefore, SiCl4 is not an electron-deficient compound.
Example-10
Write the resonance structures of CO32  and HCO 3 .
(ii) BH 4 : Boron-hydride ion hybridization is sp 3. it is (NCERT)
2–
tetrahedral in structure. Sol. (a) CO 3

Example-8
Write reactions to justify amphoteric nature of aluminium.
(NCERT)
Sol. Amphoteric: When the substance reacts with acid as well
as base then substance known as amphoteric in nature. There are three resonating structures for the carbonate
ion.
Aluminum is amphoteric because it reacts with acid as well
(b) HCO 3–
as base.

(i) 2Al(s)  6HCl(aq)  2Al3 (aq)  6Cl  (aq)  3H 2(g)

THE p-BLOCK ELEMENTS 189

(a) Lead belongs to group 14. The electronic configuration

of this group are ns2 np2. Most common oxidation state
should be +4. The elements of group 14 show tetra valency
by sharing four of its valence electrons. But Pb show +2
There are only two resonating structures for the bicarbonate oxidation state due to inert pair effect. the stability of the
ion.
lower oxidation state increases and that of the higher
Example-11 oxidation state decreases on moving down the group.
What is the state of hybridisation of carbon in Hence, PbCl4 is much less stable than PbCl2. However, the
(a) CO 32  formation of PbCl4 takes place when chlorine gas is bubbled
through a saturated solution of PbCl2.
(b) Diamond
(c) Graphite? (NCERT) PbCl2(s)  Cl 2(g)  PbCl4(l)

Sol. The state of hybridisation of carbon in the given substance (b) The stability of the lower oxidation state increases and that
are: of the higher oxidation state decreases on moving down
Substance Hybridization of carbon the group. Pb(IV) is highly unstable and when heated, it
reduces to Pb(II).
CO32– sp 2

Diamond sp 3 PbCl4(l)   PbCl 2(s)  Cl 2(g)
Graphite sp 2 (c) Lead is known not to form PbI4 due to I- is strong reducing
Example-12 agent which reduces
Explain the difference in properties of diamond and graphite
Pb+4 to Pb+2 and also stability of +4 oxidation state of lead is
on the basis of their structures. (NCERT)
lesser than +2 state.
Sol.
Example-14
Diamond Graphite
Suggest reasons why the B–F bond lengths in BF3 (130
It is hardest substance. It is soft substance.
Hybridisation of each Hybridization of each pm) and BF4 (143 pm) differ.. (NCERT)
carbon is sp3 carbon atom is sp2 Sol. Boron trifluoride are planar molecule in which central atom
Geometry is tetrahedral Geometry is planar is sp2 hybridized. An sp2 hybridized boron atom has an
Bad conductor of electricity Good conductor of electricity empty p-orbital which can accept a pair of electron. BF3 is
due to free electron.
shorter and stronger due to p  -p  back bonding. This
Not layered structure Layered structure
imparts a double bond character to the B–F bond.
Used for making cutters Used as lubricant
Example-13
Rationalise the given statements and give chemical
reactions:
• Lead (II) chloride reacts with Cl2 to give PbCl4.
• Lead (IV) chloride is highly unstable towards heat.
• Lead is known not to form an iodide, PbI4. (NCERT)
Sol. Bond length to shorten in BF3 (130 pm) due to double-bond
character.
190 THE p-BLOCK ELEMENTS

But in case of BF4- a change in hybridisation from sp2 to hydrolysed back to give boric acid and NaOH. In order to
prevent the reversibility of reaction add some polyhydroxy
sp3. There is no double bond character because vacant
compound such as glycerol, mannitol or 1, 2 cis- diols.
orbital filled by other fluoride ion and B–F bond length of These polyhydroxy compounds can form chelate complex
143 pm in BF4 ion. with sodium metaborate and prevent it to get hydrolysed.
Example-17
B (OH)3 + NaOH  NaBO2 + Na [B (OH)4] + H2O
How can this reaction is made to proceed in forward
direction?
(a) Addition of cis 1, 2 diol
(b) Addition of borax
Example-15 (c) Addition of trans 1, 2 diol
If B–Cl bond has a dipole moment, explain why BCl3 (d) Addition of Na2HPO4
molecule has zero dipole moment. (NCERT) Ans. (a)

Sol. Boron trichloride are planar molecule in which central atom Sol. B  OH 3  NaOH  NaBO 2  Na  B  OH 4   H 2 O
is sp2 hybridized. It is a symmetrical molecule. Hereafter, in this reaction, if you added cis 1,2-diol, then the product
the dipole-moments of the B–Cl bond cancel each other,
Na  B  OH 4  reacts with cis 1,2 -diol and produces a
thereby causing results zero-dipole moment
cyclic product and free water molecules. Due to the
formation of cyclic products and four water molecules the
entropy of the reaction increases, so the reaction becomes
more feasible. Therefore, this reaction is made to proceed
in the forward direction by the addition of cis 1, 2 diols. The
reaction is shown below,
OH O OH
HC HC
+ B(OH)—4 B + 2H2O
HC OH HC O OH
Example-16
1 : 1 polyol-boron complex
 Na  B(OH) 4 
B(OH)3  NaOH 
How can this reaction be made to proceed in forward direc- OH O O CH
HC HC
2 + B(OH)—4 B + 4H2O
tion?
HC OH HC O O CH
(a) Addition of cis-1,2-diol
bis-1, 2-diol 2 : 1 polyol-boron complex
(b) Addition of borax
Example-18
(c) Addition of trans-1, 2-diol
Which of the following minerals does not contain aluminium?
(d) Addition of Na2HPO4
(a) Cryolite (b) Mica
Ans. (a)
(c) Feldspar (d) Fluorspar
Sol. Boric acid on reaction with NaOH gives sodium
metaborate as product. Because of aqueous conditions, Ans. (d)
this reaction is reversible in nature. Sodium metaborate gets Sol. Cryolite, mica, feldspar are ores of aluminium with
chemical formula given below:
THE p-BLOCK ELEMENTS 191

Ans. (c)
Mica  KAl2  AlSi3 O10   OH 2
Sol. CO 2 have low solubility in water..
Cryolite  Na 3 AlF6
Fluorspar  CaF2 Example-23
H 2O Heat NaOH
Feldspars are the aluminosilicates with general formula SiCl 4   X  Y  Z
X, Y and Z in the above reaction are
AT4 O8 .
X Y Z
Example-19
(a) SiO2 Si NaSi
Aluminium chloride exists as dimer, Al2Cl6 in solid state as
(b) Si(OH)4 SiO2 Na2SiO3
well as in solution of non-polar solvents such as benzene.
When dissolved in water, it gives (c) Si(OH)4 Si SiO2
(a) [Al(OH)6]3– + 3HCl (b) [Al(H2O)6]3+ + 3Cl– (d) SiO2 SiCl4 Na2SiO3
(c) Al3+ + 3Cl– (d) Al2O3 + 6HCl Ans. (b)
Ans. (b) Sol. Silicon tetrachloride on reaction with water gives o-silicilic
acid and HCl. This acid on heating gives silica which is
Sol. Aluminium chloride can form a dimer like Al 2 Cl 6 . In acidic in nature. Therefore, silica on reaction with base like
solid-state as well as in the solution of non-polar solvents NaOH gives sodium salt of silicate and water as shown in
such as benzene aluminum chloride exists as a dimer. When the reactions:
it dissolves in water, it gives aluminium hexa hydroxide
SiCl 4  4H 2 O  Si  OH 4 (X)  4HCl
anion with hydrochloric acid, as follows.
Si  OH 4 

 SiO 2 (Y)  H 2 O
3 
Al2 Cl6  12H 2 O  2[ Al(H 2 O) 6 ]  6Cl
SiO 2  2NaOH  Na 2SiO 3 (Z)  H 2 O
Example-20 Example-24
Which of the following hydrides is least stable to hydrolysis? An oxide X in its normal form is almost non-reactive due to
(a) CH4 (b) SiH4 very high X – O bond enthalpy. It resists the attack by
(c) SnH4 (d) PbH4 halogens, hydrogen and most of acids and metals even at
Ans. (d) elevated temperatures. It is only attacked by HF and NaOH.
The oxide X is
Sol. PbH 4 is the least stable due to inert pair effect.
(a) SiO2 (b) CO2
Example-21
(c) SnO2 (d) PbO2
Which of the following is not a use of graphite ?
Ans. (a)
(a) For electrodes in batteries.
Sol. SiO2 have giant covalent structure and is not attacked by
(b)Crucibles made from graphite are used for its
inertness to dilute acids and alkalies. halogen, metal at high temp. But react with HF and
NaOH.
(c) For adsorbing poisonous gases.
Example-25
(d) Lubricant at high temperature.
Silicon has a strong tendency to form polymers like silicones.
Ans. (c)
The chain length of silicone polymer can be controlled by
Sol. Graphite is not used for absorbing poisonous gas.
adding
Example-22
(a) MeSiCl3 (b) Me2SiCl2
Which property of CO 2 makes it of biological and
(c) Me3SiCl (d) Me4Si
geochemical importance ?
Ans. (c)
(a) Its acidic nature.
Sol. Chain length of silicones polymer can be controlled by add-
(b) Its colourless and odourless nature.
ing Me3SiCl act as inhibitor of reaction.
(c) Its low solubility in water.
(d) Its high compressibility.
192 THE p-BLOCK ELEMENTS

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. How does metallic and non-metallic character vary in a 17. Covalence of carbon not expand beyond four. Explain
group? briefly.

2. Third-period elements can expand their covalence 18. Boron forms no compounds in a unipositive state but
above four. Briefly explain. thallium in a unipositive state is quite unstable. Why?

3. Give two examples of electron deficient molecules. 19. Why Borazine is more reactive than benzene?

4. Arrange the following halides of boron in the increasing 20. Why does CO2 have a linear shape with no dipole
order of acidic character: BF3, BCl3, BBr3, BI3. moment?

5. Why CCl4 behaves as an electron precise molecule? Section – C (3 Marks Questions)

6. Lead unaffected by water. Why? 21. Carbon does not form ionic compounds. Why?

7. Why Diamond is the hardest substance known? 22. Heavier elements do not form
p − p multiple bond
as carbon do. Explain
8. What is water gas composed of ?
23. Why is carbon monoxide considered poisonous?
9. What is the maximum covalence shown by N?
24. Why there is difference in C-C bond length of graphite
10. Bi(V) is a stronger oxidizing agent than Bi(III). Why? and diamond as C–C bond length is shorter in graphite
than the C–C bond length of diamond.
Section – B (2 Marks Questions)
25. Briefly explain three different methods of obtaining
11. SiCl4 forms [SiCl6]2– while CCl4 does not form [CCl6]2– elemental boron.
Explain.
26. Give reasons:
12. Why does carbon form covalent compounds whereas (i) A mix of dil. NaOH & aluminium pieces are used to
lead forms ionic compounds? open drains.
(ii) Diamond is used as an abrasive.
13. What are halides of carbon? Give few examples. (iii) Aluminium wires are used to make transmission
cables.
14. Why are boron halides & diborane referred to as
“Electron deficient compounds”? 27. Write some properties of carbon monoxide.

15. All the elements of group 13 except thallium show a +3 28. Write some important uses of borax.
oxidation state while it, shows a +1 oxidation state.
Give reasons.

16. Carbon different from other member of the group.

Why?
THE p-BLOCK ELEMENTS 193

Section – D (5 Marks Questions)

29. (a) [SiF6]2– is known whereas [SiCl6]2– is not known. 30. (a) Why carbon shows catenation but silicon does not?
Give reasons (b) What are the Similarities between boron and
(b) Select the member (s) of group 14 that silicon?
(i) forms the most acidic oxide
(ii) is commonly found in the +2 oxidation state
(iii) used as a semi-conductor.
(c) Why diamond has high melting point ?
194 THE p-BLOCK ELEMENTS

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions)

1. The property which does not support anomalous 7. Graphite is soft solid lubricant extremely difficult to
behaviour of boron is melt. The reason for this anomalous behaviour is that
(a) formation of trihalides graphite.
(b) high ionization energy (a) Has carbon atoms arranged in large plates of rings
(c) small size of strongly bound carbon atoms with weak
(d) high electronegativity interpolate bonds.
(b) Is a non-crystalline substance
2. Which among the following is dissimilarity between (c) Is an allotropic form of carbon
silicon and boron? (d) Has molecules of variable molecular masses like
(a) both form acidic oxides polymers.
(b) both form halides which are Lewis acids
(c) their hydrides are stable 8. Which is strongest Lewis acid?
(d) their chlorides hydrolyse to their respective acids (a) BF3 (b) BCl3
(c) BBr3 (d) BI3
3. Elements of oxygen family are ?
9. Generally the Boron Trihaides act as
(a) Non metals (b) Metalloids
(a) Strong reducing agent
(c) Radioactive (d) Polymorphic
(b) Lewis Acids
(c) Lewis Bases
4. Borax is used as a cleansing agent because on
(d) Dehydrating Agents
dissolving in water, it gives
(a) Alkaline solution
10. Phosphorus which is used as a rat poison?
(b) Acidic solution
(a) White (b) Violet
(c) Bleaching solution
(c) Red (d) Black
(d) Amphoteric solution.

11. How Nitrogen (I) oxide is produced ?

5. Boron fibres are used in making
(a) Thermal decomposition of ammonium nitrate
(a) bullet-proof jacket
(b) Disproportionation of N2O4
(b) light composite material for aircraft
(c) Thermal decomposition of ammonium nitrite
(c) Both (a) and (b)
(d) None of the above
(d) None of the above

12. Why Red phosphorus is chemically less reactive ?

6. When boron and caustic potash is heated, what will be
(a) It does not contain P – P bonds
the products.
(b) It does not contain tetrahedral P4 molecules
(a) Potassium Borate + Dihydrogen
(c) It does not catch fire in air even upto 400 oC
(b) Potassium Borate + Water
(d) It has a polymeric structure
(c) Potassium Borate + H2
(d) Borax + Dihydrogen
13. What is the order of the stability of +1 oxidation state
among Al, Ga, ln and Tl increases in the sequence?
(a) Al < Ga < ln < Tl
(b) Tl < ln < Ga < Al
(c) ln < Tl < Ga < Al
(d) Ga < ln < Al < Tl
THE p-BLOCK ELEMENTS 195

14. Why aluminium (III) chloride forms a dimer because Section-B (Assertion & Reason Type Questions)
aluminium
(a) belongs to 3rd group 21. Assertion (A): +2 oxidation state of Pb is more stable
(b) can have higher coordination number than +4 state.
(c) cannot form a trimer Reason (R): PbI4 is an unstable compound.
(d) has high ionization energy. (a) Both A and R are correct and R is the correct
explanation of A
15. The correct order of atomic radii in group 13 elements (b) Both A and R are correct but R is not the correct
is explanation of A
(a) B < Al < ln < Ga < Tl (c) A is correct; R is incorrect
(b) B < Al < Ga < ln < Tl (d) R is correct; A is incorrect
(c) B < Ga < Al < Tl < ln
(d) B < Ga < Al < ln < Tl 22. Assertion (A): The volatility and thermal stability of
halides of carbon shows the order CF4 < CCl4 > CBr4 >
16. The type of hybridisation of boron in diborane is CI4.
(a) sp3 hybridisation (b) sp2 hybridisation Reason (R): The bond energy increases in the order
(c) sp hybridisation (d) sp3d2 hybridisation C – I < C – Br < C – Cl < C – F
(a) Both A and R are correct and R is the correct
17. Which of the following statements regarding ozone is explanation of A
not correct? (b) Both A and R are correct but R is not the correct
(a) The oxygen-oxygen bond length in ozone is explanation of A
identical with that of molecular oxygen (c) A is correct; R is incorrect
(b) The ozone is response hybrid of two structures. (d) R is correct; A is incorrect
(c) The ozone molecule is angular in shape
(d) Ozone is used as a germicide and disinfectant for 23. Assertion (A): The heaviest element in each p-block
the purification of air. group is the most metallic in nature.
Reason (R): The non-metallic character decreases
18. Among the C–X bond (where, X = Cl, Br, I) the correct down the group.
decreasing order of bond energy is (a) Both A and R are correct and R is the correct
(a) C – I > C – Cl > C – Br explanation of A
(b) C – I > C – Br > C – Cl (b) Both A and R are correct but R is not the correct
(c) C – Cl > C – Br > C – I explanation of A
(d) C – Br > C – Cl > C – I (c) A is correct; R is incorrect
(d) R is correct; A is incorrect
19. Which of the following oxidation states are most
characteristic for lead and tin respectively? 24. Assertion (A): On mixing AuCl3 and SnCl2, purple
(a) 2, 2 (b) 4, 2 colour solution is formed.
(c) 2, 4 (d) 4, 4 Reason (R): SnCl2 reduces AuCl3 to colloidal gold.
(a) Both A and R are correct and R is the correct
20. The hybridisation of the central atom in SiF62–, explanation of A
[GeCl6]2–, [Sn(OH)6]2– is (b) Both A and R are correct but R is not the correct
(a) sp3d (b) sp3d2 explanation of A
3
(c) sp (d) sp3d3 (c) A is correct; R is incorrect
(d) R is correct; A is incorrect
196 THE p-BLOCK ELEMENTS

25. Assertion (A): Carborundum is used as an abrasive. Section – C (Case Study Questions)
Reason (R): Its structure is same as that of diamond.
(a) Both A and R are correct and R is the correct Case Study – 1
explanation of A
(b) Both A and R are correct but R is not the correct The heavier members of 13 and 14 groups besides the
explanation of A group oxidation state also show another oxidation state
(c) A is correct; R is incorrect which is two units less than the group oxidation state.
(d) R is correct; A is incorrect Down the group (  ), the stability of higher oxidation
state decreases and that of lower oxidation state
26. Assertion (A): Al forms [AlF6]3– but B does not form increases. This concept which is commonly called inert
[BF6]3– pair effect has been used to explain many physical and
Reason (R): B does not react with F2. chemical properties of the element of these groups.
(a) Both A and R are correct and R is the correct
explanation of A 29. More stable oxidation state exhibit by heavier members
(b) Both A and R are correct but R is not the correct of groups 13?
explanation of A (a) +3 only (b) +1 only
(c) A is correct; R is incorrect (c) +1 and +3 both (d) +1, +2, +3
(d) R is correct; A is incorrect
30. Which among the following will undergo reduction
27. Assertion (A): The tendency for catenation decreases easily.
in the order C > Si > Ge > Sn. (a) SiO2 (b) GeO2
Reason (R): The catenation depends on the strength of (c) SnO2 (d) PbO2
the element-element bond.
(a) Both A and R are correct and R is the correct 31. Which among the following will undergo oxidation
explanation of A easily?
(b) Both A and R are correct but R is not the correct (a) GaCl (b) InCl
explanation of A (c) BCl3 (d) TlCl
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect 32. The strongest reducing agent among the following is.
(a) SnCl2 (b) SnCl4
28. Assertion (A): Boron differs from aluminium and other (c) PbCl2 (d) GeCl2
members of group 13 in a number of properties.
Reason (R): Boron shows anomalous behavior. Case Study – 2
(a) Both A and R are correct and R is the correct
explanation of A Allotropy: The phenomenon of existence of the same
substance (element or compound) in two or more
(b) Both A and R are correct but R is not the correct
forms, in the same physical state, having different
explanation of A properties. Different forms are called allotropes or
(c) A is correct; R is incorrect allotropic modifications.
(d) R is correct; A is incorrect Except lead, all other elements of group 14 show
allotropy.
Element Allotropic form
C Crystalline: Graphite and diamond
Amorphous: coal, coke and charcoal
Si Crystalline and amorphous
Ge Two crystalline forms
Sn Three forms: grey tin, white tin, rhombic
tin
THE p-BLOCK ELEMENTS 197

33. Why wood charcoal is used in gas masks? 37. Which of the following statements about anhydrous
(a) is poisonous aluminium chloride is correct?
(b) liquefies gases (a) It exists as Al2Cl6 dimer in vapour form.
(c) is porous (b) It is not easily hydrolysed.
(d) adsorbs poisonous gases (c) It sublimes at 100oC under vacuum.
(d) It is a strong Lewis base.
34. Which of the following is not sp2 hybridised?
(a) Graphite (b) Graphene 38. Which one of the following statements is correct?
(c) Fullerene (d) Dry ice (a) All boron trihalides form back bonding.
(b)Anhydrous aluminium chloride is an ionic
35. Fullerene with formula C60 has a structure where every compound.
carbon atom is (c)Aluminium bromide make up the electron deficiency
(a) sp-hybridized (b) sp2-hybridized by bridging with other aluminium bromide.
3
(c) sp -hybridized (d) not hybridized (d) None of these.

36. Thermodynamically the most stable form of carbon is 39. The dimeric structure of aluminium chloride disappear
(a) diamond (b) graphite when
(c) fullerene (d) coal (a) it dissolves in water
(b) it reacts with donor molecules like R3N
Case Study – 3 (c) it dissolves in benzene
(d) both (a) and (b)
The high charge and small size of Al3+ ion gives it a
high charge density which is responsible for its 40. Which of the following reaction is incorrect?
tendency to show
(a) BF3(g) + F−(aq) → BF4−
(a) covalency in its compounds in the gaseous state
(b) BF3( g ) + 2H 2 O →  BF3OH  + H 3O +
−
(b) high hydration enthalpy which stabilizes its
compounds in solution, and
(c) BCl3( g ) + 3C2 H5OH( l) → B ( OC2 H5 )3( l) + 3HCl
(c) high lattice enthalpy of its compounds in the solid
state. (d) BCl3( g ) + 2C2 H5 N( l) → Cl3 B ( C5 H5 N )2(s )
Hence, aluminium can form both covalent and ionic
bond. Like halides of boron, halides of aluminium do
not show backbonding because of increase in size of
aluminium. In fact, aluminium atoms complete their
octets by forming dimers. Thus, chloride and bromide
of aluminium exist as dimers. Thus, chloride and
bromide of aluminium exist as dimers, both in the
vapor state and in polar solvents like benzene, while
the corresponding boron halides exist as monomer. In
boron trihalides, the extent of back bonding decreases
with increase in size of halogens and thus, Lewis acid
character increases. All BX3 are hydrolysed by water
but BF3 shows as different behavior.
198 THE p-BLOCK ELEMENTS

Notes:

Find Answer Key and Detailed Solutions at the end of this book

THE p-BLOCK ELEMENTS

ENVIRONMENTAL CHEMISTRY
200 ENVIRONMENTAL CHEMISTRY

Chapter at a Glance
 Environmental pollution: It is the effect of undesirable  Biochemical Oxygen Demand (BOD): The amount of
changes in our surroundings that have harmful effects on oxygen required by bacteria to break down the organic matter
plants, animals and human beings. present in a certain volume of a sample of water.

 Major Environmental Pollutions are :  Eutrophication: The process in which nutrient enriched
water bodies support a dense plant population, which kills
(i) Air pollution
animal life by depriving it of oxygen and results in
(ii) Water pollution subsequent loss of biodiversity.
(iii) Soil pollution  Pesticides: These are organic compounds which are used
 Green house effect: About 75% of the solar energy reaching to protect plants from pests.
Earth is absorbed by the Earth’s surface, which increases  Herbicides: They are used to kill weeds or undesirable
its temperature. The rest of the heat radiates back to the vegetation. Examples: sodium chlorate (NaClO3) and sodium
atmosphere. Some of the heat is trapped by gases such as arsinite (Na3AsO3).
carbon dioxide, methane, ozone, chlorofluorocarbon
 Green chemistry: Green chemistry is a strategy to design
compounds (CFCs) and water vapour in the atmosphere.
chemical processes and products which reduces or
Thus, they add to heating of the atmosphere. This causes
eliminates the use and generation of hazardous substances.
global warming.
The chemical reactions should be such that the reactants
 Global warming: An increase in the average temperature of are fully converted into useful environmental friendly
the Earth’s atmosphere (especially a sustained increase that products by using an environment friendly medium so that
causes climatic changes) which may be caused by additional no chemical pollutants introduced in the environment.
heat being trapped by the greenhouse gases.

 Smog: Smoke is a mixture of smoke, dust particles and small

drops of fog.
ENVIRONMENTAL CHEMISTRY 201

Solved Examples
Example-1
Ans. (c)
Which of the following factors have harmful effect on our Sol. Particulate pollutants -These are dust mist, fumes smog,
environment ? smoke etc.,
(i) Environmental pollution Example-5
(ii) Natural environmental gases Sulphur dioxide causes:
(iii) Deforestation
I. respiratory diseases in human being
(iv) Uncontrolled birth rate II. Red haze in the traffic
(a) (i), (ii) and (iii) (b) (ii) and (iii) III. Irritation of the eyes.
(c) (i) and (iv) (d) (i), (iii) and (iv)
(a) I and II (b) II and III
Ans. (d)
(c) I and III (d) I, II and III
Sol. Factors causing harmful effect of environment are (i)
Ans. (c)
pollution (ii) increasing population, i.e., uncontrolled birth
Sol. Sulphur dioxide causes respiratory diseases in human
rate (iii) deforestation etc.,
being and irritation of the eyes.
Example-2
Example-6
Pollutants which are slowly degraded by natural process
Carbon monoxide gas is more dangerous than carbon
rapidly known as
dioxide gas. Why? (NCERT)
(a) Biodegradable pollutants
Sol. Carbon monoxide combines with haemoglobin to form a
(b) Non-Biodegradable pollutants
very stable compound known as carboxyhaemoglobin
(c) Both (a) and (b)
when its concentration in blood reaches 3-4%, the oxygen
(d) None of the above
carrying capacity of the blood is greatly reduced. This
Ans. (a)
results into headache, nervousness and sometimes death
Sol. Pollutants which are degraded by natural process rapidly
of the person. On the other had CO2 does not combine
are known as biodegradable pollutants.
with haemoglobin and hence is less harmful than CO.
Example-3
Example-7
The lowest region of atmosphere in which the human
Which gases are responsible for greenhouse effect? List
beings along with other organisms live is called…A… Here
some of them.
A refers to.
Sol. CO2 is mainly responsible for greenhouse effect. Other
(a) Stratosphere (b) Troposphere
greenhouse gases are methane, nitrous oxide, water
(c) Hydrosphere (d) Mesosphere
vapours, CFCs and ozone.
Ans. (d)
Example-8
Sol. The lowest region of atmosphere in which the human
beings along with other organisms live is called What is smog? How is classical smog different from
Troposphere photochemical smog? (NCERT)
Example-4 Sol. The word smog is a combination of smoke and fog. It is a
Which of the following are major particulate pollutants? type of air pollution that occurs in many cities throughout
I Dust II. Mist the world. Classical smog occurs in cool humid climate. It
III. Smoke IV. Smog is also called as reducing smog. Whereas photochemical
smog occurs in warm and dry sunny climate. It has high
(a) I and IV (b) II and IV
concentration of oxidising agents and therefore, it is also
(c) I ,II,III and IV (d) I, II and III
called as oxidising smog.
202 ENVIRONMENTAL CHEMISTRY

Example-9 (d) A  Photochemical smog, B  Normal smog

What do you mean by ozone hole? What are its Ans. (c)
consequences? (NCERT) Sol. Classical smog occurs in cold humid climate. It is a mixture
of smoke fog and SO2 Chemically it is a reducing mixture
Sol. Depletion of ozone layer creates some sort of holes in the
and so it is also called as reducing smog. Photochemical
blanket of ozone which surround us, this is known as smog occurs in warm dry and sunny climate the main
ozone hole. components of the photochemical smog result from the
1. With the depletion of the ozone layer, UV radiation action of sunlight on unsaturated, hydrocarbons and
filters into the troposphere which leads to aging of nitrogen oxides produced by automobile and factories.
skin, cataract, sunburn, skin cancer etc. Example-14
2. By killing many of the phytoplanktons, it can damage Catalyst converters are fitted automobiles to reduce
the fish productivity. emission of harmful gases. Catalytic converters change
3. Evaporation rate increases through the surface and unburnt hydrocarbons into
stomata of leaves which can decrease the moisture (a) carbon dioxide and water
content of the soil. (b) carbon monoxide

Example-10 (c) methane

(d) carbon dioxide and methane
What would have happened if the greenhouse gases were
totally missing in the earth’s atmosphere? Discuss. Ans. (a)
Sol. Catalytic converters are fitted into automobiles to reduce
(NCERT)
emission of harmful gases. Catalytic converts change
Sol. The solar energy radiated back from the earth surface is unburnt hydrocarbons into CO2 and H2O.
absorbed by the green house gases (CO2, CH4, O3, CFCs)
Example-15
are present near the earth’s surface.
When rain is accompanied by a thunderstorm, the
They heat up the atmosphere near the earth’s surface and
collected rain water will have a pH value
keep it warm. As a result of these, there is growth of
(a) slightly lower than that of rain water without
vegetation which supports the life. In the absence of this
thunderstorm
effect, there will be no life of both plant and animal on the
(b) slightly higher than that when the thunderstorm is not
surface of the earth.
there
Example-11 (c) uninfluenced by occurrence of thunderstorm
What is chlorosis? (d) Which depends on the amount of dust in air
Sol. Slowdown of process of formation of chlorophyll in plants Ans. (a)
with the presence of SO2 is called chlorosis. Sol. Thunderstorm creates high temperature condition. At this
Example-12 temperature O2 and N2 react to each other and form oxides
Which zone is known as ozonosphere? of nitrogen. Now, oxides of nitrogen react with rain water
and form HNO3. So that collected rain in this condition is
Sol. Stratosphere
slightly acidic.
Example-13
Example-16
…A…smog occurs in warm dry and sunny climate while What do you mean by Biochemical Oxygen Demand
… B… smog occurs in cool humid climate. Here, A and B (BOD)? (NCERT)
refer to Sol. The amount of oxygen required by bacteria to breakdown
(a) ) A  Classical smog, B  Photochemical smog the organic matter present in a certain volume of a sample
(b) A  Normal smog, B  Photochemical smog of water is called Biochemical Oxygen Demand (BOD).
(c) A  Photochemical smog, B  Classical smog
ENVIRONMENTAL CHEMISTRY 203

Example-17 Example-21
The value of BOD of clean water is : Which of the following pesticides is introduced during
(a) < 5 ppm World War II to control malaria :
(b) > more than 5 ppm (a) Nicotine
(c) 1 ppm (b) DDT
(d) 5 ppm (c) Aldrin
Ans. (a) (d) Dieldrin
Sol. Clean water would have BOD value of less than 5 ppm. Ans. (b)
Example-18 Sol. During World War II DDT was found to be of great use in
Pollution from domestic sewage and animal excreta when the control of malaria and other insect-borne diseases.

dissolved in river water then it. Example-22

(a) increases BOD of water Soil is polluted by
(b) decreases BOD of water I. Pesticied
(c) does not affect BOD of water II. Synthetic fertilizers
(d) increases DO of water III. Green manure
Ans. (a) (a) I and II
Sol. Pollution from domestic sewage and animal excreta when (b) I and II
dissolved in river water then it increases biochemical (c) II and III
oxygen demand of H2O (d) I, II and
Example-19 Ans. (b)
What are pesticides and herbicides? Explain giving Sol. Pesticides and synthetic fertilizers pollute the soil.
examples. (NCERT) Example-23
Sol. Pesticides are the chemical compounds used in agriculture Waste management
to control the damages caused by insects, rodents, weeds (a) The proper disposal of wastes
and various crop diseases.
(b) The improper disposal of wastes
Example: Aldrin, Dilldrin, B.H.C. etc.
(c) Both (a) and (b)
Herbicides: These are the chemicals used to control weeds.
(d) The oxidation of wastes in open air
Example: Triazines
Ans. (a)
Example-20
Sol. Waste management includes the proper disposal of
A large number of fish are suddenly found floating dead wastes.
on a lake. There is no evidence of toxic dumping but you
find an abundance of phytoplankton. Suggest a reason
for the fish kill. (NCERT)
Sol. Excessive phytoplankton (organic pollutants such as
leaves, grass trash etc.) present in water are biodegradable.
Bacteria decomposes these organic matters in water.
During this process when large number of bacteria
decomposes these organic matters, they consume the
dissolved oxygen in water. When the level of dissolved
oxygen falls below 6 ppm the fish cannot survive.
204 ENVIRONMENTAL CHEMISTRY

Example-24 Example-25
Why the management wastes is of utmost importance? Which of the following chemical has more toxic effect when
(a) The improper disposal of wastes is one of the major used for the purpose dry cleaning of clothes?
causes of environmental degradation (a) Tetrachloroethene (b) H2O2
(b) The poor management causes health problems leading (c) Liquified CO2 (d) None of the above
to epidemic Ans. (a)
(c) Both (a) and (b) Sol. Tetra chloroethene has more toxic effect when used for
(d) It is directly linked with our economy the purpose of dry cleaning of clothes.
Ans. (c)
Sol. As a normal practice, all domestic wastes should be
properly collected and disposed. The poor management
of wastes causes heath problems.
ENVIRONMENTAL CHEMISTRY 205

EXERCISE – 1: Basic Subjective Questions

Section – A (1 Mark Questions)

1. Define Troposphere? 18. Discuss ‘greenhouse effect’? And its effects on the
global climate?
2. Give some name of gaseous air pollutants.
19. Describe pneumoconiosis?
3. Explain the diseases caused by sulphur dioxide?
20. List some preventive measures for photochemical
4. Greenhouse effect is caused by which gases? List some smog?
names.
Section – C (3 Marks Questions)
5. What happens when CFCs come in contact with ozone
layer?
21. List some harmful effects of oxides of nitrogen in
atmosphere?
6. Describe greenhouse effect?
22. What are the reactions involved for ozone layer
7. Disease caused due to ozone layer depletion? depletion in the atmosphere?

8. Define smog? 23. pH of rain is about 5.6? When does it become acid rain.

9. List two names of gases which cause acid rain. 24. How industrial wastes cause water pollution?

10. In which season London smog is caused and time of the

25. Explain the reason behind ozone depletion.
day?
26. (i) When does rain water become acid rain?
Section – B (2 Marks Questions) (ii) What is BOD?
(iii) What is PAN?
11. Photochemical smog is composed of what?
27. How the Taj Mahal in India is affected by pollutant?
12. The maximum and minimum BOD value for pure water
and polluted water respectively. 28. What is the root cause for damaging occurs at Tajmahal
and other monuments near Agra?
13. Role of ozone layer in the stratosphere?
Section – D (5 Marks Questions)
14. Define Stratosphere pollutants ? Give examples.
29. What is the difference between primary and secondary
15. Carbon dioxide is highly poisonous, State reason. air pollutants?

16. Hydrocarbons is not good for human beings as well as 30. Define biodegradable and non-biodegradable waste?
plants?

17. Which zone is called ozonosphere?

206 ENVIRONMENTAL CHEMISTRY

EXERCISE – 2: Basic Objective Questions

Section – A (Single Choice Questions)

1. Substance which causes pollution is termed as 9. Aerosol particles of oxides or ammonium salts in rain
(a) catalyst (b) pollutant drops results in?
(c) activator (d) inhibitor (a) dry-deposition (b) wet-deposition
(c) Both (a) and (b) (d) normal depositon
2. An average human being requires nearly how much
times more air than the food. 10. Ozone is thermodynamically
(a) 2-4 (b) 200-400 (a) stable
(c) 12-15 (d) None of these (b) unstable
(c) sometimes stable and sometimes unstable
3. Which pollutants are degraded by natural process (d) None of the above
rapidly
(a) biodegradable pollutants 11. Chlorofluorocarbons also known as
(b) non-biodegradable pollutants (a) Pyrenes (b) Freons
(c) Both (a) and (b) (c) Radons (d) Both (a) and (b)
(d) None of the above
12. CFCs are used as refrigerators, air conditioners because
4. Distance of stratosphere above sea level lies between (a) non-reactive in nature (b) non-toxic in nature
(a) 10-50 km (b) 0-10 km (c) non-flammable in nature (d) All of these
(c) 50-100 km (d) None of these
13. Wastes of organic water are
5. Atomospheric layer contains clouds, water vapour and (a) non-biodegradable (b) biodegradable
air is known as (c) Both (a) and (b) (d) None of these
(a) Stratosphere (b) Troposphere
(c) Mesosphere (d) All of these 14. What is the concentration of dissolve oxygen in clear
water is
6. Which one of the following is major source for Cabon (a) 10 ppm (b) 100 ppm
monoxide in air? (c) 5 ppm (d) 1 ppm
(a) automobile exhaust
(b) incomplete combustion of coal 15. Pollutants which cause water pollution?
(c) incomplete combustion of firewood (a) Oil spills in ocean (b) Pesticides
(d) All of the above (c) Detergents (d) All of these

7. Why carbon monoxide is poisonous 16. Chemical nicotine is obtained from

(a) it binds to hemoglobin to form stable compound (a) power plants (b) cotton
(b) it is carcinogenic in nature (c) paddy (d) tobacco plants
(c) Both (a) and (b)
(d) it induces the chance of heart attack 17. Non-biodegradable industrial wastes are generated by
(a) thermal power plants (b) steel plant
8. What are the major componenets of Acid rain? (c) fertilizer industries (d) All of these
(a) H2SO4 (b) HNO3
(c) Both (a) and (b) (d) HCl
ENVIRONMENTAL CHEMISTRY 207

18. “Reducing potentially hazardous waste through smarter 23. Assertion (A): In upper stratosphere, Ozone is
production”. destroyed by solar radiation
This represents a great step forward for Reason (R): Thinning of the ozone layer allows
(a) green chemistry excessive UV radiations to reach the surface of earth.
(b) green revolution (a) Both A and R are true and R is the correct
(c) industrial revolution explanation of A
(d) green biotechnology (b) Both A and R true but R is not the correct
explanation of A
19. What happens if there were no green house gases in the (c) Both A & R are not correct
atmosphere, then the temperature of the earth (d) A is not correct but R is correct
(a) increases (b) decreases
(c) remain same (d) Either (a) or (b) or (c) 24. Assertion (A): Soil and water pollution is caused by
excessive use of chlorinated synthetic pesticides .
20. What are the source(s) of dissolved oxygen in water Reason (R): Such pesticides are non-biodegradable.
is/are (a) Both A and R are true and R is the correct
(a) photosynthesis (b) natural aeration explanation of A
(c) mechanical aeration (d) All of these (b) Both A and R true but R is not the correct
explanation of A
Section-B (Assertion & Reason Type Questions) (c) Both A & R are not correct
(d) A is not correct but R is correct
21. Assertion (A): What if BOD level of water in a pond is
more than 20 ppm, it is said to be highly polluted. 25. Assertion (A): Photochemical smog is oxidizing in
Reason (R): High BOD means low activity of bacteria nature.
in water. Reason (R): During the sequence of reactions, NO2 and
(a) Both A and R are correct; R is the correct O3, are formed which are constituents of photochemical
explanation of A smog.
(b) Both A and R correct; R is not the correct (a) Both A and R are true and R is the correct
explanation of A explanation of A
(c) A is correct; R is incorrect (b) Both A and R true but R is not the correct
(d) R is correct; A is incorrect explanation of A
(c) Both A & R are not correct
22. Assertion (A): Excess nitrate in drinking water can (d) A is not correct but R is correct
cause disease such as methemoglobinemia.
Reason (R): In drinking water, the maximum limit of 26. Assertion (A): Carbon dioxide is one of the important
nitrate is 50 ppm. greenhouse gases.
(a) Both A and R are correct; R is the correct Reason (R): Carbon dioxide is largely produced by
explanation of A respiratory function of animals and plants.
(b) Both A and R correct; R is not the correct (a) Both A and R are true and R is the correct
explanation of A explanation of A
(c) A is correct; R is incorrect (b) Both A and R true but R is not the correct
(d) R is correct; A is incorrect explanation of A
(c) Both A & R are not correct
(d) A is not correct but R is correct
208 ENVIRONMENTAL CHEMISTRY

27. Assertion (A): Greenhouse effect was observed in 31. Name a lung disease caused due to the inhalation of
houses used to grow plants and these are made of green coal dust from coal mining industry?
glass. (a) Pneumoconiosis (b) Tuberclosis
Reason (R): Greenhouse name has been given because (c) Diarhea (d) Anoxia
glass houses are made of green glass.
(a) Both A and R are true and R is the correct 32. Write the full form of PAN.
explanation of A (a) Peroxyacryl nitrate (b) Peroxyacetyl nitrate
(b) Both A and R true but R is not the correct (c) Peroxyacryl nitride (d) Peroxyacetyl nitride
explanation of A
(c) Both A & R are not correct Case Study – 2
(d) A is not correct but R is correct
Ozone gas is continuously formed by the action of UV
rays on molecular oxygen, and also degraded into
28. Assertion (A): The pH of acid rain is less than 5.6 molecular oxygen in the stratosphere. There should be
Reason (R): Carbonic acid is formed when Carbon a balance between production and degradation of ozone
dioxide present in the atmosphere dissolves in rain in the stratosphere. Of late, the balance has been
water. disrupted due to enhancement of ozone degradation by
(a) Both A and R are true and R is the correct chlorofluorocarbons (CFCs). CFCs find wide use as
explanation of A refrigerants. CFCs discharged in the lower part of
atmosphere move upward and reach stratosphere. In
(b) Both A and R true but R is not the correct
stratosphere, UV rays act on them releasing Cl atoms.
explanation of A Cl degrades ozone releasing molecular oxygen, with
(c) Both A & R are not correct these atoms acting merely as catalysts. Cl atoms are not
(d) A is not correct but R is correct consumed in the reaction. Hence, whatever CFCs are
added to the stratosphere, they have permanent and
Section – C (Case Study Questions) continuing effects on Ozone levels. Although ozone
depletion is occurring widely in the stratosphere, the
depletion is particularly marked over the Antarctic
Case Study – 1 region. This has resulted in formation of a large area of
thinned ozone layer, commonly called as the ozone
The effect of undesirable changes in our surroundings hole
which have adverse effects on plants, animals and
human beings is termed as environmental pollution. 33. Ozone can be used as?
The changes may originate from either a natural source (a) Strong oxidizing agent (b) weak oxidizing agent
or human activity. The living or biotic (i.e. plants and (c) Strong reducing agent (d) weak reducing agent
animals) and non-living or abiotic components that are
affected adversely are called receptors or targets. 34. What is the full form of CFC?
(a) Chlorylfluorocarbon (b) Chlorofluorylcarbon
29. Pollutant which do not occur in nature and exist into the (c) Chlorofluoidocarbon (d) Chlorofluorocarbon
environment by human activity is
(a) Plastic (b) CO2 35. What problem arises in the absence of ozone layer.
(c) Sulpur dioxide (d) Methane (a) non-melanoma skin cancer
(b) lung cancer
30. The gaseous air pollutant which is a lung irritant and (c) melanoma skin cancer
can lead to acute respiratory diseases in children is (d) Pneumoconiosis
(a) NO2 (b) CO2
(c) SO2 (d) CH4 36. Which free radical is responsible for causing break
down of ozone into oxygen due to use of CFCs?
(a)Cl (b) F
(c) H (d) None of the above
ENVIRONMENTAL CHEMISTRY 209

Case Study – 3 37. Which of the following groups contain only

biodegradable items?
We produce a lot of wastes on a daily routine and (a) Grass, flowers and leather
throw them away or discard them. These substances (b) Grass, wood and plastic
include kitchen waste like vegetables and fruit peels, (c) Fruit peels, cade and lime juice
empty cartoons, used tea leaves, and so many (d) Cake, wood and grass
expendable items like juices, plastic bags, paper, old
clothes, old footwear etc. Many of these materials like 38. Which of the following is the best way for disposal of
paper, vegetable and fruit peels can be easily broken vegetable and fruit peels?
down by the action of bacteria or other decomposers. (a) Landfill (b) Recycling
Such substances which can be easily broken down by (c) Composting (d) Burning
the action of bacteria are named biodegradable
substances. Other substances or materials like plastics, 39. We should reduce the use of the plastic bags, bottles
metallic cans and pesticides which cannot be broken etc. because:
down easily by biological processes are named non- (a) They are not durable
biodegradable substances. (b) They are non-biodegradable
Since non-biodegradable does not not broken down (c) They are made of toxic materials
with the help of bacteria and hence, they cause various (d) They react with the atmospheric gases
types of pollution: air pollution, water pollution and
land pollution etc. 40. What pollution type is caused by plastics?
(a) Noise pollution
(b) water pollution
(c) Soil pollution
(d) Both marine and soil pollution

Find Answer Key and Detailed Solutions at the end of this book

ENVIRONMENTAL CHEMISTRY
210 ANSWER KEY

Answer Key
CHAPTER - 1: SOME BASIC CONCEPTS OF CHEMISTRY

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. Second 3. 1.8066 × 1024 1. (c) 2. (c) 3. (b) 4. (d)

7. 30.07u 9. Newton 5. (b) 6. (d) 7. (a) 8. (a)
10. H2O, CO2 11. 18g mol–1, 17g mol–1 9. (c) 10. (a) 11. (b) 12. (d)
12. 252.96g 13. (i) 4 (ii) 3 (iii) 4 13. (d) 14. (d) 15. (c) 16. (c)
14. (i) 34.2 (ii) 0.066 (iii) 15.8 15. 4.27g cm –3
17. (c) 18. (c) 19. (d) 20. (b)
17. 1.765 × 10 12
18. 4.7g 21. (a) 22. (a) 23. (a) 24. (d)
20. 4.53g 21. 12.011u 25. (a) 26. (d) 27. (c) 28. (c)
23. 55.55M 29. (a) 30. (c) 31. (d) 32. (c)
24. (i) 10 (ii) 30 (iii) 3.011×1024 33. (c) 34. (a) 35. (d) 36. (a)
25. 28g 26. 6.023 × 10 20 37. (a) 38. (a) 39. (c) 40. (a)
27. 52.17 % 28. CHCl3
29. C6H8N2 30. 44g, 22g, 22g
ANSWER KEY 211

Answer Key
CHAPTER - 2: STRUCTURE OF ATOM

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. Neutron 2. 6 1. (d) 2. (b) 3. (d) 4. (a)

9. 4s 5. (c) 6. (c) 7. (c) 8. (a)
10. angular nodes = 1, radial nodes = 1 9. (c) 10. (b) 11. (d) 12. (c)
15. 4.57 × 10 s 14 –1
13. (a) 14. (c) 15. (c) 16. (b)
16. Wavelength = 6.02 ×10–2 m, Frequency = 5.0 × 109s–1 17. (a) 18. (b) 19. (d) 20. (c)
17. –3.4 eV 21. (a) 22. (a) 23. (a) 24. (a)
21. (i) 8, 8, 9 (ii) 12, 12, 13 (iii) 35, 35, 45 25. (a) 26. (d) 27. (a) 28. (c)
22. 4860 Å 23. 7 × 10 s 14 –1
29. (c) 30. (a) 31. (b) 32. (b)
24. 5.45 × 10 m s
5 –1
25. 8.9625 × 10 m –7
33. (c) 34. (d) 35. (c) 36. (d)
26. –19.62 × 10 J atom
–18 –1
27. 4:9 37. (a) 38. (b) 39. (a) 40. (b)
30. 205.5nm
212 ANSWER KEY

Answer Key
CHAPTER - 3: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. Johann Dobereiner 1. (b) 2. (a) 3. (b) 4. (d)

6. Li, Na and K 5. (c) 6. (b) 7. (b) 8. (c)
7. Fourth period and III A group 9. (a) 10. (b) 11. (b) 12. (a)
12. Na, Kr, Br, Sc 13. (b) 14. (a) 15. (d) 16. (a)
17. F 17. (a) 18. (a) 19. (c) 20. (c)
20. Cs 21. (b) 22. (d) 23. (b) 24. (b)
28. Halogens 25. (a) 26. (a) 27. (a) 28. (c)
29. (c) 30. (c) 31. (d) 32. (b)
33. (b) 34. (a) 35. (d) 36. (b)
37. (c) 38. (d) 39. (c) 40. (d)
ANSWER KEY 213

Answer Key
CHAPTER - 4: CHEMICAL BONDING AND MOLECULAR STRUCTURE

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. False 1. (b) 2. (a) 3. (d) 4. (c)

4. Weaker 5. (a) 6. (a) 7. (c) 8. (d)
5. Lower 9. (a) 10. (b) 11. (c) 12. (a)
6. Diamagnetic 13. (c) 14. (c) 15. (d) 16. (b)
7. Sodium Chloride, Methane 17. (d) 18. (b) 19. (c) 20. (a)
8. Coulomb meter 21. (a) 22. (d) 23. (a) 24. (a)
9. H3O+, O3 25. (c) 26. (a) 27. (a) 28. (b)
10. CH4, C2H4, C2H2 29. (a) 30. (c) 31. (b) 32. (c)
33. (d) 34. (b) 35. (b) 36. (a)
37. (d) 38. (c) 39. (c) 40. (c)
214 ANSWER KEY

Answer Key
CHAPTER - 5: REDOX REACTIONS

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. (a) 2. (a) 3. (d) 4. (d)

1
5. HgCl2 8. Cr2(III)O3 9.  5. (c) 6. (c) 7. (a) 8. (a)
3
9. (a) 10. (c) 11. (c) 12. (b)
10. Metal displacement reaction 13. (i) +6 (ii) +6
19. 2.5 25. (a) +2 (b) +4 (c) +6 13. (b) 14. (c) 15. (d) 16. (a)
28. (i) No (ii) Yes (iii) No 17. (b) 18. (a) 19. (b) 20. (a)
30. (a) (i) A 3+
(ii) D (b) C, D 21. (c) 22. (c) 23. (d) 24. (a)
(c) A , B
3+ 2+
(d) 2.78V 25. (a) 26. (d) 27. (d) 28. (d)
29. (a) 30. (a) 31. (b) 32. (b)
33. (c) 34. (a) 35. (c) 36. (a)
37. (a) 38. (c) 39. (b) 40. (a)
ANSWER KEY 215

Answer Key
CHAPTER - 6: HYDROGEN
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

1. Protium, Tritium 1. (b) 2. (d) 3. (b) 4. (b)

2. Protium, Deuterium, Tritium 5. (c) 6. (d) 7. (d) 8. (b)
8. Ca(OH)2 is formed 9. (b) 10. (a) 11. (c) 12. (d)
9. Na6P6O18 13. (c) 14. (d) 15. (c) 16. (d)
17. (c) 18. (d) 19. (b) 20. (b)
21. (a) 22. (c) 23. (a) 24. (a)
25. (b) 26. (a) 27. (a) 28. (a)
29. (a) 30. (a) 31. (b) 32. (c)
33. (c) 34. (a) 35. (a) 36. (a)
37. (a) 38. (b) 39. (a) 40. (b)
216 ANSWER KEY

Answer Key
CHAPTER - 7: ORGANIC CHEMISTRY -SOME BASIC PRINCIPLES & TECHNIQUES

EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

6. Sextext 1. (d) 2. (d) 3. (a) 4. (a)

7. Homolytic 5. (d) 6. (a) 7. (b) 8. (b)
8. Ferric Ferrocyanide 9. (b) 10. (a) 11. (c) 12. (b)
9. Fractional Distillation 13. (b) 14. (b) 15. (c) 16. (b)
10. True 17. (c) 18. (b) 19. (b) 20. (a)
24. 17.46% 21. (d) 22. (a) 23. (a) 24. (a)
25. 56.0% 25. (b) 26. (c) 27. (a) 28. (c)
28. Structural Isomers, Resonance contributors, Geometrical 29. (d) 30. (b) 31. (c) 32. (d)
Isomers 33. (d) 34. (b) 35. (a) 36. (c)
29. 21.95%, 4.58% 37. (d) 38. (c) 39. (c) 40. (a)
30. Metamerism
ANSWER KEY 217

Answer Key
CHAPTER - 8: HYDROCARBONS
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

8. 3-Ethyl-2-methylpentane 1. (d) 2. (d) 3. (a) 4. (b)

10. Butanoic acid 5. (c) 6. (b) 7. (a) 8. (c)
16. 2-Methylpentane 9. (a) 10. (b) 11. (d) 12. (d)
17. Compound (b)-Aromatic 13. (b) 14. (d) 15. (a) 16. (d)
18. cis form 17. (b) 18. (c) 19. (a) 20. (d)
19. 2-Bromohexane, 1-Bromohexane 21. (d) 22. (c) 23. (a) 24. (d)
20. 3-Ethylhex-3-ene 25. (c) 26. (d) 27. (a) 28. (a)
21. 1-Bromopropane, 2-Bromopropane 29. (b) 30. (a) 31. (b) 32. (d)
22. 2-Methyl-2, 4-Hexadiene. 33. (a) 34. (d) 35. (b) 36. (b)
26. 3,3-Diethylpentane, 2-3-Dimethylbutane 37. (a) 38. (c) 39. (a) 40. (c)
218 ANSWER KEY

Answer Key
CHAPTER - 9: STATES OF MATTER
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

2. P  T 1. (d) 2. (b) 3. (a) 4. (c)

4. Intermolecular attraction 5. (c) 6. (a) 7. (c) 8. (b)
Vreal 9. (b) 10. (d) 11. (c) 12. (c)
7. Z  V
ideal 13. (d) 14. (b) 15. (b) 16. (a)
8. 313 K 17. (d) 18. (a) 19. (b) 20. (b)
9. No
21. (a) 22. (d) 23. (d) 24. (a)
10. Solid < liquid < gas
11. Dispersion force, Dipole-dipole interactions, Dipole-induced 25. (a) 26. (b) 27. (c) 28. (b)
dipole, Hydrogen bonding 29. (b) 30. (b) 31. (a) 32. (a)
12. Thermal energy 33. (b) 34. (b) 35. (a) 36. (c)
13. Low pressure and High temperature 37. (b) 38. (d) 39. (a) 40. (c)
dRT
14. p 
M
15. –40
16. 3.37 g dm–3
17. B, B
18. 21.43 K
20. Z = 1, Z >1
21. 1895.6 bar
22. 99.5°C
23. 0.489 atm, 0.734 atm, 0.979 atm, 2.202 atm
24. 9.32 bar
25. 609.8 K, 623.54 K
26. 48356 cm/sec , 44542 cm/sec, 39483 cm/sec
27. 44.1 g mol-1
28. 186.25 mm Hg
29. Increase to1.12 atm, 4 g dm-3
30. A, A, A
ANSWER KEY 219

Answer Key
CHAPTER - 10: THERMODYNAMICS
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

7. Open system 1. (a) 2. (b) 3. (d) 4. (b)

12. Positive 5. (c) 6. (b) 7. (c) 8. (c)
15. 1648.4 kJ 9. (c) 10. (c) 11. (d) 12. (a)

23. above 463.4K 13. (d) 14. (b) 15. (b) 16. (c)

26. -1323.3KJ 17. (b) 18. (a) 19. (a) 20. (c)

30. 2.46 J/g/oC , 113.2J/mol/oC 21. (a) 22. (a) 23. (a) 24. (a)
25. (a) 26. (a) 27. (a) 28. (a)
29. (b) 30. (d) 31. (c) 32. (c)
33. (b) 34. (b) 35. (a) 36. (b)
37. (b) 38. (a) 39. (b) 40. (d)
220 ANSWER KEY

Answer Key
CHAPTER - 11: EQUILIBRIUM
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

3. 7 1. (c) 2. (b) 3. (c) 4. (c)

5. (a) 6. (a) 7. (d) 8. (c)
5. Atmospheric pressure
9. (c) 10. (c) 11. (d) 12. (a)
7. Decreases
13. (c) 14. (c) 15. (a) 16. (c)
8. 10-7 M 17. (d) 18. (c) 19. (c) 20. (c)
10. Unchanged 21. (a) 22. (a) 23. (c) 24. (d)

18. 8 x10-3 M 25. (a) 26. (b) 27. (a) 28. (c)
29. (a) 30. (b) 31. (b) 32. (b)
19. 2.7
33. (b) 34. (a) 35. (d) 36. (a)
20. 3.42 37. (a) 38. (a) 39. (b) 40. (a)
21. 11.1

23. 1.5 x 10-8 M

26. 9.31

28. 1.923

29. 178.6 ml , 321.4 ml

ANSWER KEY 221

Answer Key
CHAPTER - 12: THE s-BLOCK ELEMENTS
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

4. Solvay’s process 1. (b) 2. (c) 3. (d) 4. (d)

5. Na2CO3 5. (b) 6. (c) 7. (a) 8. (c)
9. (c) 10. (c) 11. (d) 12. (a)
13. (d) 14. (b) 15. (b) 16. (b)
17. (c) 18. (d) 19. (a) 20. (d)
21. (a) 22. (c) 23. (a) 24. (c)
25. (c) 26. (a) 27. (a) 28. (c)
29. (a) 30. (d) 31. (b) 32. (c)
33. (d) 34. (a) 35. (a) 36. (a)
37. (c) 38. (d) 39. (a) 40. (c)
222 ANSWER KEY

Answer Key
CHAPTER - 13: THE p-BLOCK ELEMENTS
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

3. BF3, Al2Cl6 1. (a) 2. (b) 3. (d) 4. (a)

4. BF3 < BCl3 < BBr3 < BI3 5. (c) 6. (a) 7. (a) 8. (d)
9. (b) 10. (a) 11. (c) 12. (d)
13. (a) 14. (b) 15. (d) 16. (a)
17. (a) 18. (c) 19. (c) 20. (b)
21. (b) 22. (a) 23. (a) 24. (a)
25. (b) 26. (c) 27. (a) 28. (b)
29. (c) 30. (d) 31. (a) 32. (d)
33.(d) 34. (d) 35. (b) 36. (b)
37. (a) 38. (c) 39. (a) 40. (b)
ANSWER KEY 223

Answer Key
CHAPTER - 14: ENVIRONMENTAL CHEMISTRY
EXERCISE - 1: Basic Subjective Questions EXERCISE - 2: Basic Objective Questions

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

Scan the QR code and check detailed solutions. Scan the QR code and check detailed solutions.

9. NO 2 ,SO 2 1. (b) 2. (c) 3. (a) 4. (a)

5. (b) 6. (d) 7. (a) 8. (c)
17. Stratosphere zone.
9. (b) 10. (b) 11. (b) 12. (d)
26. 2 to 3.5, Biochemical Oxygen Demand, Peroxyacetyl nitrate
13. (c) 14. (a) 15. (d) 16. (d)
17. (d) 18. (a) 19. (b) 20. (d)
21. (c) 22. (a) 23. (d) 24. (a)
25. (a) 26. (b) 27. (c) 28. (b)
29. (a) 30. (a) 31. (a) 32. (b)
33. (a) 34. (d) 35. (a) 36. (a)
37. (a) 38. (c) 39. (b) 40. (d)
 Monthly
Youtube Views

Monthly Where Students

Web + App Visitors Take Live Classes

Hours of
Users Live Learning

43
Parents
Happy
Trust
Students
Vedantu
MILLION+ MILLION+
 Our Results Speak for Us

JEE Main 2023

170
Vedantu Students
JEE Scored Above 99%ile

1000+ Vedantu Students Scored

Above 90%ile

JEE Advanced 2022 JEE Advanced 2021

2700+ Vedantu Students Cleared JEE Main in 2022

1500+ Vedantu Students Cleared JEE Adv. 2022 1500+ Vedantu Students Cleared JEE Adv. 2021

NEET (UG) 2022 NEET (UG) 2021

1000+ Vedantu Students Qualified NEET 2022 1172+ Vedantu Students Qualified NEET 2021

CBSE Class 12 ISC Class 12 CBSE Class 10

Annmary
Aatman Upreti Shreya Roshan Aastha N Raj Gitanjali Rajulal Shreya Nigam Khushi Arora Anshika Singha
Santhosh
98.4% 98.4% 98.0% 99.8% 98.4% 99.8% 99.6% 99.4%

ICSE Class 10

M.D.Sriya Varshil J Patel Mohammad Y Devika Sajeev Sakshi Semwal Aloki Upadhyay Ishita Surana Saumya Gupta

99.4% 97.4% 97.2% 97.2% 97.2% 99.4% 99.2% 99.2%

#HereForRealAchievers