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 How to use your
Tatva Practice Book

1
2. Hexagonal Packing
Hexagonal packing is more
efficient. Its coordination
number is 6 and voids in the
packing are smaller than square
packing. If we place another Scan the QR Code in each
layer on square packing then chapter’s theory section to
there are the following
view micro concept videos
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Solid State

Exercise - 2:

2
Solve all types of
1. CsBr crystallizes in a body
centered cubic lattice. The edge
length of unit cell is 436.6 pm.
Given that the atomic mass of
Cs = 133u and Br = 80u, the
exercise questions density of CsBr is:
based on the latest (NEET 2019)
NEET pattern.
(a) 42.5 g/cm3 (b) 0.425 g/cm3
(c) 4.25 g/cm3 (d) 8.5 g/cm3

Answer Key

3
CHAPTER-1: SOLID STATE
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Exercise-1: Basic Objective preparation content,
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Founder’s Message
Dear Student,
I am delighted to present to you an amazing book, a ready reckoner to guide you through your NEET
exams-‘TATVA’. Tatva- which means the ‘Core’ is fully aligned with the culture, the mission, and the
vision of Vedantu and therefore it gives me immense pleasure and joy to share this book with you. We
at Vedantu have always aimed to revolutionize the teaching and learning process and have speedily
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one such book series that will help you keep up with the pace and competitive nature of NEET. This
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companion in moving towards your dreams. Tatva is a result of the consistent effort, diligence, and
research by our experienced team of subject experts and teachers.
This book has been curated to suit the needs of NEET aspirants like you, to strengthen your
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A few guiding points to optimally use Tatva with a planned approach:
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• We suggest a revision of the theory followed by practice of solved examples.
• Practice relevant questions daily after finishing Vedantu lectures and session assignments. We
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• Use the Tatva booklet to mark notes so that it always comes in handy for last-minute revision
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questions which you couldn't solve in the first attempt.
• Exercise 1 contains easy to moderate questions which are relevant for the exam (level-wise),
Exercise 2 has past year questions and Exercise 3 is the Achiever’s Section that will throw you with
challenges that will prepare you for the challenging questions during the entrance and make things
seamless for you.
• Before wrapping up, here is your practice mantra: “Practice does not make perfect. Only perfect
practice makes perfect” - Vince Lombardi
We strongly believe in you and wish to make the journey to your success
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Credits
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We thank our leaders for their insight and mentorship. They steered the project in the right direction and were
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6

TABLE OF CONTENTS

Structure of an Atom

Theory........................................................................................................................................................ 8

Solved Examples......................................................................................................................................... 30

Exercise - 1 : Basic Objective Questions....................................................................................................... 35

Exercise - 2 : Previous Year Questions......................................................................................................... 45

Exercise - 3 : Achiever’s Section ................................................................................................................. 49

Answer Key .............................................................................................................................................. 172

Periodic Properties

Theory........................................................................................................................................................ 53

Solved Examples......................................................................................................................................... 71

Exercise - 1 : Basic Objective Questions....................................................................................................... 77

Exercise - 2 : Previous Year Questions......................................................................................................... 86

Exercise - 3 : Achiever’s Section ................................................................................................................. 90

Answer Key .............................................................................................................................................. 174

 7

Chemical Bonding

Theory...................................................................................................................................................... 94

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

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

Exercise - 2 : Previous Year Questions....................................................................................................... 130

Exercise - 3 : Achiever’s Section................................................................................................................ 134

Answer Key ............................................................................................................................................. 176

Mole & Equivalent concept

Theory....................................................................................................................................................... 138

Solved Examples........................................................................................................................................ 151

Exercise - 1 : Basic Objective Questions...................................................................................................... 155

Exercise - 2 : Previous Year Questions........................................................................................................ 165

Exercise - 3 : Achiever’s Section ................................................................................................................ 168

Answer Key ............................................................................................................................................. 178

 01
STRUCTURE OF AN ATOM
Chapter 01

Structure Of An Atom

1. Introduction to Atom
The atomic existence has been proposed since the time of early
Indian and Greek philosophers who were of the view that atoms
are the fundamental building blocks of matter. According to the
philosphers, the continued subdivisions of matter would yield
atoms which would be indivisible. This word ‘atom’ has been
derived from the Greek word ‘a-tomio’ which means non-divisible.
These earlier ideas were mere speculations and there was no way
to test them experimentally. These ideas remained untouched for
a very long time and were revived again by scientists in the
nineteenth century. In this chapter, we will explore the inside world
of atoms which is full of mystery and surprises. Whole chemistry
is based on atoms and their structures. We will also study the
Fig. 1.1: Production of Cathode Rays
behaviour exhibited by the electrons and their consequences.
2.1.1 Properties of Cathode Rays
2. Subatomic Particles 1. They produce sharp shadow of the solid object in their path
suggesting that they travel in straight line.
Dalton’s atomic theory was able to explain the law of conservation
of mass, law of constant composition and law of multiple 2. They are deflected towards the positive plate in an electric
proportion very successfully but it failed to explain the results of field suggesting that they are negatively charged. They were
many experiments like it was known that substances like glass or named as electrons by Stoney.
ebonite when rubbed with silk or fur generate electricity. 3. Similarly, when a magnetic field is applied, they are deflected
in direction which shows that they carry negative charge.
4. They can make a light paddle wheel placed in their path to
2.1 Discovery of Electron rotate. This means they possess kinetic energy and are
William Crookes in 1879 studied the electrical discharge in material particles.
partially evacuated tubes known as cathode ray discharge tubes. 5. They ionise gases through which they travel.
A discharge tube is made of glass, about 60cm long containing 6. They produce X-rays when they strike on a metallic target.
two thin pieces of metals called electrodes, sealed in it. This is
7. The characterstics of cathode rays (electrons) do not depend
known as Crookes Tube. The negative electrode is also known
on the material of electrodes and nature of the gas present in
as cathode and positive electrode is known as anode.
the cathode ray tube.
When a gas enclosed at low pressure (  10-4 atm) in discharge
tube is subjected to a high voltage (  10,000V), invisible rays
The negatively charged material particles constituting the cathode
originating from the cathode and producing a greenish glow
rays are called electrons.
behind the perforated anode on the glass wall coated with
Thus, we can conclude that electrons are basic constituents of
phosphorescent material ZnS is observed. These rays were called
all matter.
cathode rays.

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Structure of an Atom
10 STRUCTURE OF AN ATOM

2.1.2 Charge to Mass Ratio of Electron R.A Millikan devised a method known as oil drop experiment to
determine the charge on the electrons.
In this method, oil droplets produced by the atomiser, were allowed
to enter in the form of mist through a tiny hole in the upper plate
of electrical condenser. The downward motion of the oil droplets
was viewed through the telescope, equipped with a micrometer
eye piece. By measuring the rate of fall of these droplets, Millikan
was able to measure the mass of oil droplets. The air inside the
chamber was ionized by passing a beam of X-rays through it.
The charge on these oil droplets was acquired by collisions with
the gaseous ions. The fall of these charged oil droplets can be
accelerated, retarded or made stationary depending upon the
charge on the droplets and the polarity and strength of the voltage
Fig. 1.2: The Apparatus to determine the charge to applied to the plate. By carefully measuring the effects of electrical
the mass ratio of electron field strength on the motion of oil droplets, Millikan concluded
that the magnitude of electrical charge, q, on the droplets is always
In 1897, J.J. Thomson measured e/m ratio of electron by using an integral multiple of the electrical charge e, that is, q = n e,
cathode ray tube and applying electric and magnetic field where n = 1, 2, 3... .
perpendicular to each other as well as to the path of electrons. Charge on the electron is found to be –1.6022 × 10–19C.
The extent of deviation of electrons from their path in the presence
The mass of electron was thus calculated as
of electric and magnetic field depends on:
(a) Charge on the electron e 1.6022 10 19 C
m   9.1094  1031 kg.
(b) Mass of the particle e / m 1.758820  1011 C / kg
(c) The strength of electric or magnetic field
When only electric field is applied, the electrons are deflected to
the point A. When only magnetic field is applied, the electrons
2.1.4 Origin of Cathode Rays
are deflected to the point C. By balancing the strengths of electric The cathode rays were first produced from the material of the
or magnetic fields, the electrons are allowed to hit the screen at cathode and then from the gas inside the discharge tube due to
point B i.e. the point where electrons hit in the absence of electric bombardment of the gas molecules by the high speed electrons
and magnetic field. By measuring the amount of deflections, emitted first from the cathode.
Thomson was able to calculate the value of e/m as
1.758820 × 1011C/kg. 2.2 Discovery of Proton
Since the atom as a whole is electrically neutral and the presence
2.1.3 Charge on The Electron of negatively charged particles in it was established, therefore it
was thought that some positively charged particles must also be
present in the atom. So, during the experiments with cathode
rays, the scientist Goldstein designed a special type of discharge
tube. He discovered new rays called Canal rays. The name canal
rays is derived from the fact that the rays travelled in straight line
through a vacuum tube in the opposite direction to cathode rays,
pass through and emerge from a canal or hole in the cathode.
They are also known as anode rays.

Fig. 1.3: The Millikan oil drop apparatus to

determine charge on electron

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Structure of an Atom
STRUCTURE OF AN ATOM 11

positive particle.

e 1.602  10 19
Mass of the proton = 
e/m 9.579  10 4
= 1.672 × 10–24 g
or = 1.672 × 10–27kg
Mass of a proton = 1.672 × 10–27kg

Fig. 1.4: Production of Anode rays 2.3 Discovery of Neutron

The theoretical requirement for the existence of a neutron particle
2.2.1 Properties of Anode Rays in the atomic nucleus was put forward by Rutherford in 1920.It
1. They travel in straight lines. was proposed to be a particle with no charge and having mass
almost equal to that of a proton. He named it as neutron. In 1932,
2. They carry a positive charge.
Chadwick proved its existence. He observed that, when a beam
3. They are made up of material particles.
4. The mass of the particles constituting the anode rays is of  particles  4
2 He 2  is incident on Beryllium (Be), a new type
found to depend on the nature of gas taken.
of particle was ejected. It had mass almost equal to that of a
5. The charge to mass ratio(e/m) of the particles is also found
proton ( 1.672 ×10–27kg) and carried no charge.
to depend on the gas taken.
9
6. Their behaviour in electric and magnetic field is opposite to 4 Be  42 He  12 1
6 C  0n
that observed for electron.

2.2.2 Origin of Anode Rays 2.4 Definitions

These rays are believed to be produced as a result of the knock Electron
out of the electrons from the gaseous atoms by the bombardment A fundamental particle which carries one unit negative charge
of high speed electrons of the cathode rays on them. These and has a mass nearly equal to 1/1837th of that of hydrogen atom.
anode rays are not emitted from the anode but are produced in Proton
the space between the anode and the cathode. A fundamental particle which carries one unit positive charge
The lightest positively charged particles were obtained when and has a mass nearly equal to that of hydrogen atom.
the gas taken in the discharge tube was hydrogen. The e/m value Neutron
of these particles were maximum. They had minimum mass and A fundamental particle which carries no charge but has a mass
unit positive charge. The particle was called a proton. nearly equal to that of hydrogen atom.
Charge on a proton = + 1.6022 × 10–19C
These particles were found to have the e/m value as + 9.579×104
coulomb/g. This was the maximum value of e/m observed for any

Table 1.1: Properties of Fundamental Particles

Name Discoverer Symbol Charge Relative Mass

Charge

Electron J.J. Thomson e –1.6022 × 10–19 C –1 9.1094×10–31 Kg

Proton Goldstein p +1.6022 × 10–19C +1 1.6726 × 10–27 kg

Neutron Chadwick n 0 0 1.6749 × 10–27 kg

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Structure of an Atom
12 STRUCTURE OF AN ATOM

3. Thomson’s Model of an Atom

Sir J. J. Thomson, who discovered the electron, was the first to
suggest a model of atomic structure.
(i) All atoms contain electrons.
(ii) The atom as a whole is neutral. The total positive charge
and total negative charge must be equal.
He visualised all the positive charge of the atom as being sprea
out uniformly throughout a sphere of atomic dimensions (i.e.
approx. 10–10 m in radius). The electrons were smaller particles
together carrying a negative charge, equal to the positive charge
in the atom. They were studded in the atom like plums in a pudding.
The charge distribution was such, that it gave the most stable
arrangement. This model of the atom was often called the plum –
pudding model. Also the raisin pudding model or watermelon
model.

Fig. 1.6: Rutherford's Scattering Experiment

4.1.1 Observations
Rutherford carried out a number of experiments, involving the
scattering of   particles by a very thin foil of gold. Observations
Fig. 1.5: Thomson's proposed model of atom were:
(i) Most of the   particles (99%) passes through it, without
any deviation or deflection.
3.1 Drawbacks
(ii) Some of the   particles were deflected through small angles.
Though the model was able to explain the overall neutrality of
(iii) Very few   particles were deflected by large angles and
the atom, it could not satisfactorily explain the results of scattering
occasionally an   particle got deflected by 180o
experiments carried out by Rutherford.

4. Rutherford’s Nuclear Model of 4.1.2 Conclusions

an Atom (i) An atom must be extremely small and must consist of mostly
empty space because most of the particles passed through
it without any deflection.
4.1 Rutherford’s - Scattering experiment (ii) Very few particles were deflected to a large extent. This
Rutherford conducted - particles scattering experiments in 1909. indicates that:
In this experiment, a very thin foil of gold (0.004nm) is bombarded (a) Electrons because of their negative charge and very low
by a fine stream of alpha particles. A phosphorescent screen mass cannot deflect heavy and positively charged 
(ZnS) is placed behind the gold foil, where points were recorded particles
which were emerging from -particles. Polonium was used as the (b) There must be a very heavy and positively charged body in
source of -particles. the atom i.e. nucleus which does not permit the passage of
positively charged  -particles.
(c) Because the number of  -particles which undergo deflection
of 180º, is very small, therefore the volume of positively
charged body must be extremely small fraction of the total
volume of the atom. This positively charged body must be
at the centre of the atom which is called nucleus.

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Structure of an Atom
STRUCTURE OF AN ATOM 13

of nucleus and the planets that of revolving electrons and

the model is known as planetary model.
(iv) Electrons and nucleus are held together by the electrostatic
force of attraction.
(v) Forces of attraction operating on the electron are exactly
balanced by centrifugal forces.

4.2.1 Drawbacks
(i) According to classical mechanics, any charged body in
motion under the influence of attractive forces should
continuously radiate energy and if this is so, the electron
will follow a spiral path and finally fall into the nucleus and
the structure would collapse. But this behaviour is never
observed.
Fig. 1.7: Scattering of -particles by thin gold foil
(ii) It says nothing about the electronic structure of atoms i.e.
how the electrons are distributed around the nucleus and
It has been found that radius of atom is of the order of
what are the energies of these electrons.
10 –10 m while the radius of the nucleus is of the order of
10–15 m. Thus if a cricket ball represents a nucleus, then the
radius of atom would be about 5 km. 5. Special Terms
4.2 Rutherford’s Nuclear Atomic Model 5.1 Atomic Number (Z)
Atomic number of an element is equal to the number of unit
positive charges or number of protons present in the nucleus of
the atom of the element. It also represents the number of electrons
in the neutral atom. Eg. Number of protons in
Na = 11 , thus atomic number of Na = 11

5.2 Mass Number (A)

The elementary particles (protons and neutrons) present in the
nucleus of an atom are collectively known as nucleons.
The mass number (A) of an atom is equal to the sum of protons
and neutrons. It is always a whole number. Thus,
Fig. 1.8: Rutherford’s Nuclear atomic model Mass number (A) = Number of protons(Z) + Number of
neutrons(n)
Therefore, number of neutrons (n) = Mass Number (A) – Number
(i) An atom consists of tiny positively charged nucleus at the
of protons (Z)
centre and it is surrounded by hollow portion called extra
nuclear part. n =A– Z
(ii) The atom is electrically neutral, as the number of electrons
is equal to number of protons in it. Thus, total positive NOTE :
charge of the nucleus is balanced by the total negative The general notation that is used to represent the mass
charge of electrons. A
number and atomic number of a given atoms is ZX
(iii) The electrons in the extra nuclear part are revolving around
Where, X – symbol of element
the nucleus in circular paths called orbits. Thus, an atom A – Mass number
resembles the solar system in which the sun plays the role Z – atomic number

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Structure of an Atom
14 STRUCTURE OF AN ATOM

5.3 Isotopes, Isobars, isotones and Table 1.5: Example of Isobars

Isobar Atomic Mass No. of No. of No. of
Isoelectronic species number number electrons protons neutrons
40
18 Ar 18 40 18 18 22
5.3.1 Isotopes
Isotopes are the atoms of the same element having identical 40
K 19 40 19 19 21
19
atomic number but different mass number. The difference is due
to the difference in number of neutrons. 40
Ca 20 40 20 20 20
20
The chemical properties of atoms are controlled by the number
of electrons. Thus, isotopes of an element show same chemical
behaviour. 5.3.3 Isotones
Table 1.2: Isotopes of Hydrogen Atoms of different elements which contain the same number of
Isotope Formula Mass No. of No. of neutrons are called isotones. Eg
number protons neutrons Table 1.6: Example of Isotones
1 Isotones Atomic number Mass number Neutrons
Protium 1 H (H) 1 1 0
36
16 S 16 36 20
2
Deuterium 1 H (D) 2 1 1
37
17 Cl 17 37 20
3
Tritium 1 H (T) 3 1 2
38
18 Ar 18 38 20

Table 1.3: Isotopes of Oxygen 39

19 K 19 39 20
Isotope Mass number No. of No. of
40
protons neutrons 20 Ca 20 40 20
16
8 O 16 8 8

17
5.3.4 Isoelectronic species
8 O 17 8 9
The species (atoms or ions) containing the same number of
18 electrons are called isoelectronic. Eg. O2–, F–, Na+, Mg+2, Al+3, Ne
8 O 18 8 10
etc.
To go further into the atomic mysteries, we will have to
Table 1.4: Isotopes of some common elements understand the nature of electromagnetic radiations and study
Element Isotopes Maxwell’s Electromagnetic Wave theory”.
James Maxwell was the first to give a comprehensive explanation
12
Carbon (C) 6 C, 13 14
6 C, 6 C about the interaction between the charged bodies and the
behaviour of electric and magnetic fields.
14
Nitrogen (N) 7 N, 15
7 N

234 235 238

Uranium 92 U, 92 U, 92 U
32 33 34 36
Sulphur 16 S, 16 S, 16 S, 16 S

5.3.2 Isobars
Atoms of different elements having different atomic numbers
but same mass numbers are called isobars. Eg

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Structure of an Atom
STRUCTURE OF AN ATOM 15

6. Wave Nature of In vacuum, all types of electromagnetic radiations travel at

the same speed i.e. 3 × 108 m/s. This is called speed of light.
Electromagnetic Radiations 5) Wavelength (  ) : The distance between two adjacent crests
Electromagnetic Radiations are waves which are formed as a result or troughs is called wavelength. Units : Angstrom(Å)
of oscillating magnetic and electric fields which are perpendicular [1 Å=10–10 m]
to each other and both are perpendicular to direction of 6) Wave Number (  ): It is the number of wavelengths per
propagation of wave.
unit length. Units : m-1
  1/ 

7.1 Relationship Between Speed (c),

Frequency (  ) & Wavelength (  )
c  
where c : speed of light i.e. 3 × 108 m/s in vaccum
 : frequency; : wavelength
Fig. 1.9: The electric and magnetic field components of an
electromagnetic wave
8. Electromagnetic Spectrum
They do not require any medium and can move in vacuum unlike When all the electromagnetic radiations are arranged in increasing
sound waves. order of wavelength or decreasing frequency the band of
radiations obtained is termed as electromagnetic spectrum.

7. Characteristics of Wave
Light is a form of radiation and has wave characterstics. The
various characterstics of a wave are:

Fig. 1.11: The spectrum of electromagnetic radiation and

Visible spectrum
Fig. 1.10: Propagation of a Wave
The visible spectrum is a subset of this spectrum (VIBGYOR)
1) Amplitude (A) : It is height of the crest or depth of trough of whose range of wavelength is 380-760nm.
a wave. Units : metre (m)
The wavelengths increase in the order:
2) Frequency (  ) : The number of waves passing through a
Gamma Rays < X-rays < Ultra-violet rays < Visible< Infrared <
point in one second. Units : Hertz (Hz) or s–1 Micro-waves <Radio waves.
3) Time Period (T) : The time taken by a wave to complete one
vibration is called time period. Units : second (s).
4) Velocity (v) : The distance travelled by a wave in one second
is called velocity. Units : m/s

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Structure of an Atom
16 STRUCTURE OF AN ATOM

8.1 Electromagnetic Wave Theory (2) Photoelectric Effect

When radiations with certain minimum frequency (0) strike the
The main points of this theory are:
surface of a metal, the electrons are ejected from the surface of
(1) A source (like the heated rod) emits energy continuously in
the metal. This phenomena is called photoelectric effect. The
the form of radiations (i.e. no change in wavelength or
electrons emitted are called photoelectrons.
frequency of the emitted radiations even on increasing the
energy radiated).
(2) These radiations are Electromagnetic in nature.

8.2 Failure of EM Wave Theory

The theory failed because of 2 experiments:
(1) Black Body Radiation
According to Maxwell’s theory on heating a body the intensity
should increase, that is, energy radiated per unit area should
increase without having any effect on the wavelength or
frequency.
But we observe that when we heat an iron rod, it first turns to red
then white and then becomes blue at very high temperatures. Fig. 1.13: Photoelectric effect
This means that frequency of emitted radiations is changing.
According to Maxwell’s Theory, the ejection of electrons should
An ideal body, which emits and absorbs radiations of all
depend on intensity of radiation that is if electrons are not being
frequencies is called black body and radiation emitted by a black
ejected, then on increasing the intensity they can be ejected.
body is called black body radiation
The variation of intensity with wavelength at different
temperatures for a black body is shown below:

Fig. 1.14: Apparatus for studying the photoelectric effect

The following observations are made:

Fig. 1.12: Wavelength-intensity relationship (i) The electrons are ejected from the metal surface as soon as
the beam of light strikes the surface, i.e., there is no time lag
So it is observed that with increasing temperature, the dominant between the striking of light beam and the ejection of
wavelength in the emitted radiations decreases and the frequency electrons from the metal surface.
increases. (ii) The number of electrons ejected is proportional to the
That is at higher temperatures, though the intensity rises as intensity or brightness of light.
predicted by Maxwell’s theory but the wavelength decreases. If
T1>T2>T3 then 1< 2< 3.

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(iii) For each metal, there is a characteristic minimum frequency, 9.3 Explanation of Photoelectric Effect
0 (also known as threshold frequency) below which (i) When light of some particular frequency falls on the surface
photoelectric effect is not observed. At a frequency of metal, the photon gives its entire energy to the electron
 > 0, the ejected electrons come out with certain kinetic of the metal atom. The electron will be ejected from the
energy. The kinetic energies of these electrons increase with metal only if the energy of the photon is sufficient to
the increase of frequency of the light used. overcome the force of attraction of the electron by the
Thus, these findings were contradictory to the Maxwell’s theory. nucleus. So, photoelectrons are ejected only when the
The number of electrons ejected and kinetic energy associated incident light has a certain minimum frequency (threshold
with them should depend on the intensity of light. It has been frequency 0). The Threshold energy required for emission
observed that though the number of electrons ejected does is called “Work Function” that is “h0”.
depend upon the brightness of light, the kinetic energy of the (ii) If the frequency of the incident light () is more than the
ejected electrons does not.
threshold frequency (0), the excess energy is imparted to
To justify these findings Max Planck gave his Quantum theory. the electron as kinetic energy. Hence,
Energy of one quantum = Threshold Energy + Kinetic
9. Particle Nature of Energy
Electromagnetic Radiations h  h 0  (1 / 2) m e v 2

9.1 Planck’s Quantum Theory (iii) When  > 0, then on increasing the intensity, the number
of quanta incident increases thereby increasing the number
The main points of this theory are:
of photoelectrons ejected.
(i) The energy is emitted or absorbed not continuously but
discontinuously in the form of small discrete packets of (iv) When  > 0, then on further increasing the frequency, the
energy. Each such packet of energy is called a ‘quantum’. energy of each photon increases and thus kinetic energy of
In case of light this quantum of energy is called a photon. each ejected electron increases.
(ii) At a time only one photon can be supplied to one electron
or any other particle.
(iii) One quantum cannot be divided or distributed.
(iv) The energy of each quantum is directly proportional to the
frequency of radiation.

hc
E   or E  h 

Fig. 1.15: Plot of Kinetic energy of photoelectrons emitted
h = Planck’s constant = 6.626 × 10-34Js versus frequency of absorbed photons
(v) The total energy emitted or absorbed by a body will be in
whole number quanta. NOTE :
nhc Energy can also be expressed in Electron Volt(eV).
Hence E = nh  The energy acquired by an electron when it is accelerated

through a potential difference of one Volt.
This is also called “Quantisation of energy”.
1eV = 1.602 × 10–19J

9.2 Explanation of Black Body Radiation Conclusion:

As the temperature is increased the energy emitted increases Light has both the Wave nature (shows the phenomena of
thereby increasing the frequency of the emitted radiations. As diffraction and interference) and Particle nature (could explain
the frequency increases the wavelength shifts to lower values. the black body radiation and photoelectric effect) . Thus, light
has dual nature.

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Bohr’s Model is based on “Atomic Spectra”, therefore before

moving further we will study:

10. Hydrogen Spectra

10.1 Spectrum
A spectrum is a group or band of wavelengths/colours and the
study of emission or absorption spectra is known as
spectroscopy.

10.2 Types of Spectrum

There are two types of spectrum:
1) Emission Spectrum
2) Absorption Spectrum

Fig. 1.17: Discontinuous Emission Spectra

10.3 Emission Spectrum
When radiations emitted from a source are incident on a prism
and are separated into different wavelengths and obtained on a
10.4 Absorption Spectra
photographic plate. When light from any source is first passed through the solution
of a chemical substance and then analysed, it is observed that
sthere are some dark lines in the continuous spectra.
(a) Continuous Emission Spectra:
There are no gaps between various wavelengths, one
wavelength merges into another.

Fig. 1.16: Continuous Emission Spectra

(b) Discontinuous Emission Spectra:

It is also known as Line Spectra or atomic spectra.
In this, certain wavelengths go missing from a group and
that leaves dark spaces in between giving discontinuity to
the spectrum. It is also known as fingerprint of an element. Fig. 1.18: Absorption Spectra

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11. Bohr’s Model

Bohr studied the atomic spectra of hydrogen and based on that
he proposed his model.

NOTE :
This model is applicable to H-atom or H-like species like
He+, Li2+ and Be3+ .

11.1 Postulates
1) An atom consists of a small, heavy, positively charged
nucleus in the centre and the electrons revolve around it in
circular orbits.
2) Electrons revolve only in those orbits which have a fixed
value of energy. Hence, these orbits are called energy levels
Fig. 1.19: Atomic Excitation and de-excitation
or stationary states.
They are numbered as 1,2,3,...... These numbers are known (a) Radii of the stationary states:
as Principal Quantum Numbers.
3) Since the electrons revolve only in those orbits which have 52.9n 2
rn  pm
fixed values of energy, hence electrons in an atom can have Z
only certain definite values of energy and not any of their
For H-atom (Z = 1), the radius of first stationary state is
own. Thus, energy of an electron is quantised.
4) Like energy, the angular momentum of an electron in an atom called Bohr orbit (52.9 pm)
can have certain definite values and not any value of their (b) Energy of an electron is given by:
own.
En= –RH (Z2/n2) n = 1,2,3.......
nh where RH is Rydberg’s constant and its value is
mvr 
2 2.18 × 10–18 J.
Where m is the mass of the electron, v is the velocity of the
Z = atomic number
electron, r is the radius of the orbit, h is Planck’s constant
n=1,2,3...... and so on. Z2
E n  2.18 1018 J / atom
5) An electron does not lose or gain energy when it is present n2
in the same shell. Z2
E n  13.6 eV / atom
6) When an electron gains energy, it gets excited to higher n2
energy levels and when it de-excites, it loses energy in the Z2
form of electromagnetic radiations and comes to lower E n  1312 kJ / mol
n2
energy values.
Thus, energies of various levels are in the order:
K < L < M < N...... and so on.

Energy of the lowest state(n=1) is called ground state.

(c) Velocities of the electron in different orbits:

2.188 106 Z
vn  m/s
n

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11.2 What Does Negative Energy For

Hydrogen Atom Means?
The negative sign means that the energy of the electron in the
atom is lower than the energy of a free electron at rest. A free
electron at rest is an electron that is infinitely far away from the
nucleus (n  ) and is assigned the energy value of zero. As
the electron gets closer to the nucleus (as n decreases), En
becomes more and more negative. The most negative energy
value is given by n=1 which corresponds to the most stable
orbit.

11.3 Transition of Electron

We know that energy is absorbed or emitted when electron excites
or de-excites respectively. The energy gap between the two orbits
is Fig. 1.20: Electronic Transitions in the hydrogen spectrum

E  E f  E i
Lyman Series
 R   R  When an electron jumps from any of the higher states to the
E    2H     2H 
 nf   ni  ground state or first state (n = 1) ,the series of spectral lines
emitted lies in the ultra violet region and are called as Lyman
 1 1   1 1  series.
E  R H  2  2   2.18 10 18  2  2  J / atom
 ni nf   ni nf  Therefore , in Rydberg’s formula n1= 1, n2 = 2,3,4,5...
The wavelength associated with the absorption or emission of Balmer Series
the photon is:
When an electron jumps from any of the higher states to the
state with n = 2, the series of spectral lines emitted lies in the
1 E R H  1 1  7  1 1  1
      1.09677 10  2  2m visible region and are called as Balmer series.
 hc hc  n i2 n f2   ni nf  Therefore, in Rydberg’s formula n1 = 2, n2 = 3,4,5,6....
This is known as Rydberg’s formula. Paschen Series
When an electron jumps from any of the higher states to the
state with n = 3, the series of spectral lines emitted lies in the
NOTE :
infrared region and are called as Paschen series.
1.09677 × 107 m-1 is known as Rydberg’s constant.
Therefore, in Rydberg’s formula n1 = 3, n2 = 4,5,6...
Brackett Series
11.4 Line Spectrum of Hydrogen When an electron jumps from any of the higher states to the
When an electric discharge is passed through gaseous hydrogen, state with n = 4, the series of spectral lines emitted lies in the
the H 2 molecules dissociate and the energetically excited infrared region and are called as Brackett series.
hydrogen atoms produced emit electromagnetic radiations of Therefore, in Rydberg’s formula n1 = 4, n2 = 5,6,7...
discrete frequency. The hydrogen spectra consists of several Pfund Series
lines named after their discoverer.
When an electron jumps from any of the higher states to the
We get discrete lines and not a continuous spectra because the state with n = 5, the series of spectral lines emitted lies in the
energy of an electron cannot change continuously but can have infrared region and are called as Pfund series.
only definite values. Thus we can say that energy of an electron
Therefore, in Rydberg’s formula n1 = 5, n2 = 6,7...
is quantised.

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11.5 Ionisation Energy mvr  n

h
It is the energy required to remove the electron completely from 2
the atom so as to convert it into a positive ion. Suppose there are ’n’ no. of wavelengths associated with an
electron in an orbit  .
Thus, n1 = 1 and n2 = 
Now, the circumference of the orbit will be
21.8  1019 1312
En   J atom 1 or  2 kJ mol1 2 r  n (where, = 1, 2, 3, …)
n2 n
But According to de Broglie’s equation:
h
11.6 Limitations of Bohr’s Model 
mv
1) Inability to explain line spectra of multi-electron atoms. h
2 r  n 
2) It fails to account for the finer details (doublet-two closely mv
spaced lines) of the hydrogen spectra. h
3) Inability to explain splitting of lines in the magnetic field mvr  n 
2
(Zeeman Effect) and in the electric field (Stark Effect)- If the
source emitting the radiation is placed in magnetic or electric
field, it is observed that each spectral line splits up into a
12.2 Heisenberg’s Uncertainty Principle
number of lines. Splitting of spectral lines in magnetic field
It is impossible to measure simultaneously the position and
is known as Zeeman Effect while splitting of spectral lines
momentum of a small particle with absolute accuracy. If an attempt
in electric field is known as Stark Effect.
is made to measure any of these two quantities with higher
4) It could not explain the ability of atoms to form molecules
accuracy, the other becomes less accurate. The product of the
by covalent bonds.
uncertainty in the position (x) and the uncertainty in momentum
5) It could not explain dual behaviour of matter and also (p) is always a constant and is equal to or greater than h/4.
contradicts Heisenberg uncertainty principle.
(x). (p)  h/4

12. Quantum Mechanical Model

or (x). (mv)  h/4
or (x). (v)  h/4m
This model was based on two concepts:
(1) de Broglie Concept of dual nature of matter 12.2.1 Explanation
(2) Heisenberg uncertainty Principle Suppose we attempt to measure both the position and momentum
of an electron. To pin point the position of the electron we have
to use light so that the photon of light strikes the electron and
12.1 Dual Nature of Particle the reflected photon is seen in the microscope. As a result of the
de Broglie proposed that matter should also exhibit dual behaviour hitting, the position as well as the velocity of the electron are
i.e. both particle and wave like properties. disturbed.
h h h
  
mv p 2m(KE) 12.2.2 Significance of Uncertainity Principle
It rules out the existence of definite paths or trajectories of
Where p is linear momentum of a particle.
electrons as stated in Bohr’s Model.
According to de Broglie, every object in motion has a wave
character. The wavelengths associated with ordinary objects are
so short (because of their large masses) that their wave properties NOTE :
cannot be detected. The wavelengths associated with electrons The effect of Heisenberg Uncertainty Principle is significant
and other subatomic particles (with very small mass) can however only for motion of microscopic objects, and is negligible for
be detected experimentally. that of macroscopic objects.
We all know that orbit angular momentum of an electron is given
by

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13. Schrodinger’s Wave 1) Principal Quantum Number (n)

It tells about the energy shell to which an electron belongs.
Equation n = 1,2,3,4,5..... and so on.
Quantum mechanics is a theoretical science that deals with the This number helps to explain the main lines of the spectrum
study of the motion of microscopic objects which have both on the basis of electronic jumps between these shells.
particle like and wave like properties. The fundamental equation (a) It gives the average distance of the electron from the
of quantum mechanics was developed by Schrodinger. nucleus. Larger the value of n, larger is the distance from
For a system (such as an atom or a molecule whose energy does the nucleus.
not change with time) the Schrödinger equation is written as (b) It completely determines the energy in hydrogen atom or
H  = E  where H is a mathematical operator called Hamiltonian. hydrogen like species.
Schrödinger gave a recipe of constructing this operator from the The energy of H-atom or H-like species depends only on
expression for the total energy of the system. An atomic orbital is the value of n.
the wave function  for an electron in an atom. The probability Order of energy : 1 < 2 < 3 < 4 < 5....... and so on.
of finding an electron at a point within an atom is proportional to For multi-electron species, energy depends on both principal
the square of the orbital wave function i.e., |  | 2 at that point. and azimuthal quantum number.
The maximum number of electrons present in any shell = 2n2
|  | 2 is known as probability density and is always positive.
2) Azimuthal Quantum Number (l)
From the value of |  | 2 at different points within an atom, it is
It is the quantum number that determines orbital angular
possible to predict the region around the nucleus where electron momentum. It is also known as subsidiary quantum number.
will most probably be found. Within the same shell, there are number of sub-shells, so
This wave function stores all the information about an electron number of electronic jumps increases and this explains the
like energy, position, orbital etc. As such it does not have any presence of fine lines in the spectrum. This quantum number
physical significance. The information stored in  about an tells about :
electron can be extracted in terms of Quantum Numbers. (a) The number of subshells present in an energy shell.
(b) Angular momentum of an electron present in subshell.
13.1 Probability Density (c) Shapes of various subshells present within the same shell.
(d) Relative energies of various subshells.
|  |2 is the probability of finding the electron at a point within an
Value of l varies from 0 to n – 1
atom.
For 1st shell (n = 1): l = 0
For 2nd shell (n = 2): l = 0,1
13.2 Concept of Orbital For 3rd shell (n = 3): l = 0,1,2
It is a three dimensional space around the nucleus within which For 4th shell (n = 4): l = 0,1,2,3
the probability of finding an electron of given energy is maximum
(say upto 90%).
Table 1.7: Notation for Subshell with respect to
Azimuthal Quantum Number
14. Quantum Numbers
They may be defined as a set of four numbers with the help of Value of l Designation of subshell
which we can get complete information about all the electrons in 0 s
an atom i.e. location, energy, type of orbital occupied, shape and 1 p
orientation of that orbital etc. 2 d
The three quantum numbers called as Principal, Azimuthal and 3 f
Magnetic quantum number are derived from Schrodinger wave 4 g
equation. The fourth quantum number i.e. the Spin quantum 5 h
number was proposed later on.

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The notations s,p,d,f represent the initial letters of the word sharp,
These orbitals of the same subshell having equal energy are
principal, diffused and fundamental. In continuation l = 4 is called
g subshell and l = 5 is called h subshell and so on. .
called degenerate orbitals Eg.
The three p-orbitals of a particular principal shell have the
Table 1.8: Notation for Subshell with respect to same energy in the absence of magnetic field.
Principal Quantum Number Similarly, all five orbitals of d-subshell of a particular shell have
the same energy.
Principal shell Subshells Thus, for H-atom order of energy is:
1st shell l = 0 (s-subshell) 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < ..........
2nd shell l = 0,1 (s & p subshell)
3rd shell l = 0,1,2 (s,p & d subshell) The total possible values of m in a given subshell = 2l + 1
4th shell l = 0,1,2,3 (s,p,d & f subshell)
Total no. of orbitals in a given shell = n2
NOTE : 4) Spin Quantum Number (s)
The number of subshells present in any principal shell is equal The electron in an atom not only moves around the nucleus
to the number of the principal shell. but also spins about its own axis. Since the electron in an
Energies of various subshell present within the same shell is: orbital can spin either in clockwise or anti-clockwise
direction. Thus s can have only two values
s< p < d < f
1 1
Angular momentum of an electron in orbital :  or 
2 2
h This quantum number helps to explain the magnetic
l (l  1)   l (l  1)
2 properties of substances.

(3) Magnetic Quantum Number (m)

NOTE :
This quantum number is required to explain the fact that An orbital cannot have more than two electrons and these
when the source producing the line spectrum is placed in a electrons should be of opposite spin.
magnetic field, each spectral line splits up into a number of
Thus, maximum number of electrons in s-subshell = 2
line (Zeeman effect).
Maximum number of electrons in p-subshell = 6
Under the influence of external magnetic field, electrons of
a subshell can orient themselves in a certain preferred Maximum number of electrons in d-subshell = 10
regions of space around the nucleus called orbitals. Maximum number of electrons in f-subshell = 14
The magnetic quantum number determines the number of
preferred orientations of the electrons present in a subshell. 14.1 Shapes of Atomic Orbitals
Since each orientation corresponds to an orbital, thus
(1) Shape of s-orbitals
magnetic quantum number determines the number of
orbitals present in any subshell. (a) They are non-directional and spherically symmetric i.e.
probability of finding the electron at a given distance is
Value of m ranges from – l to +l including zero.
equal in all directions.
(b) 1s orbital and 2s orbital have same shape but size of 2s is
Table 1.9: Number of orbitals in a subshell larger.
Subshell Orbitals (m) Number of orbitals (c) There is a spherical shell within 2s orbital where electron
s-subshell (l=0) m = 0 1 density is zero and is called a node.
p-subshell (l=1) m = -1, 0, 1 3 (d) The value of azimuthal quantum number(l) is zero (l=0)
d-subshell (l=2) m = –2, –1, 0, 1 ,2 5 and magnetic quantum number can have only one value
f-subshell (l=3) m = –3, –2, –1, 0, 1, 2, 3 7 i.e. m = 0

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(3) Shapes of d-orbitals

(a) They are designated as dxy, dyz, dzx and d x2 y2 . They
have a shape like a four leaf clover. The fifth d orbital
designated as d z2 looks like a dumb-bell with doughnut
shaped ring.
(b) All five d orbitals have same energy in the absence of
magnetic field.
(c) The d orbitals have azimuthal quantum number l = 2 and
magnetic quantum number values –2, –1,0,+ 1,+ 2.
Fig. 1.21: Boundary surface diagrams of the s orbital
(d) For principal shell number 1 and 2, there are no d orbitals.

(2) Shape of p-orbitals

(a) It consists of two lobes present on either side of the
plane that passes through the nucleus. The p-orbital is
dumb-bell shaped.
(b) There are three possible orientations of electron cloud in
p-orbitals. Therefore, the lobes of p-orbital may be
considered to be along x,y and z axis. Hence they are
designated as px, py, pz. The three p-orbitals are oriented
at right angles to one another.
(c) First main energy level( Principal quantum number n = 1)
does not contain any p-orbital.
(d) The three p-orbitals of a particular energy level have same
energy in absence of an external electric and magnetic
field and are called degenerate orbitals.
(e) Like s orbitals, p-orbitals increase in size with increase in
the energy of main shell of an atom. Thus, value of Fig. 1.23: Boundary surface diagrams of five 3d orbitals
azimuthal quantum number is one (l=1) and magnetic
quantum number has three values (m= –1, 0, +1) We have drawn boundary surface diagrams i.e the surface is drawn
in space for an orbital on which probability density ()2 is
constant. It encloses the region where probability of finding the
electron is very high. We do not draw a boundary surface diagram
that encloses 100% probability of finding the electron because
probability density has some value, howsoever small it may be, at
any finite distance from the nucleus. Thus it is not possible to
draw a boundary surface diagram of a rigid size in which the
probability of finding the electron is 100%.

Fig. 1.22: Boundary surface diagrams of three 2p orbitals

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(2) Planar or Angular Nodes: They are the planes cutting
through the nucleus on which probability of finding the
electron is zero.
Number of Planar/Angular Nodes: l
Total Number of nodes: n - 1

Fig. 1.26: Nodal planes in p-orbitalN

15. Electronic Configuration

15.1 Filling of Orbitals in an Atom
(1) Aufbau Principle: In the ground state of the atoms, the
orbitals are filled in order of their increasing energies. In
other words, electrons first occupy the lowest energy orbital
available to them and enter into higher energy orbitals only
when the lower energy orbitals are filled. Unlike H-atom
where energy of orbitals depend only on n, energy of orbitals
of multi-electron orbitals depend on both n and l. Their
order of energy can be determined using (n+l) rule.
Fig. 1.24: The plots of the orbital wave function, According to this rule, lower the value of (n+l) for an orbital,
probability density and Radial Probability lower is its energy. If two different types of orbitals have
density of 1s and 2s orbitals the same value of (n+l), the orbital with lower value of n has
lower energy.

14.2 Nodes and nodal planes

Node:
It is a region of zero probability.
There are two types of nodes:
(1) Radial or Spherical nodes: Three dimensional regions in
an orbital where probability of finding the electron
becomes zero.
Number of radial/ spherical nodes = n – l – 1

Fig. 1.27: Order of filling of orbitals

Fig. 1.25: Nodes in s-orbital
1s22s22p63s23p64s23d104p65s24d105p66s2 ......

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(2) Pauli Exclusion Principle: An orbital can have maximum (3) Hund’s rule of maximum multiplicity: Electron pairing in
p,d and f orbitals cannot occur until each orbital of a given
two electrons and these must have opposite spin. subshell contains one electron each. Also all the singly
occupied orbitals will have parallel spin.

Table 1.10: Electronic configurations of elements in the ground state

Atomic No. Element Electronic Configuration

1 H 1s1
2 He 1s2
1
3 Li [He] 2s
2
4 Be [He] 2s
2 1
5 B [He] 2s 2p
2 2
6 C [He] 2s 2p
2 3
7 N [He] 2s 2p
2 4
8 O [He] 2s 2p
2 5
9 F [He] 2s 2p
2 6
10 Ne [He] 2s 2p
1
11 Na [Ne] 3s
2
12 Mg [Ne] 3s
2 1
13 Al [Ne] 3s 3p
2 2
14 Si [Ne] 3s 3p
2 3
15 P [Ne] 3s 3p
2 4
16 S [Ne] 3s 3p
2 5
17 Cl [Ne] 3s 3p
2 6
18 Ar [Ne] 3s 3p
1
19 K [Ar] 4s
2
20 Ca [Ar] 4s
1 2
21 Sc [Ar] 3d 4s
2 2
22 Ti [Ar] 3d 4s
3 2
23 V [Ar] 3d 4s
5 1
24 Cr [Ar] 3d 4s
5 2
25 Mn [Ar] 3d 4s
6 2
26 Fe [Ar] 3d 4s
7 2
27 Co [Ar] 3d 4s
8 2
28 Ni [Ar] 3d 4s
10 1
29 Cu [Ar] 3d 4s
10 2
30 Zn [Ar] 3d 4s

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15.2 Exceptional Configuration of Cr &

Cu
The completely filled and completely half filled sub-shells are
stable due to the following reasons:
1. Symmetrical distribution of electrons: It is well known that
symmetry leads to stability. The completely filled or half
filled subshells have symmetrical distribution of electrons
in them and are therefore more stable. Electrons in the same
subshell (here 3d) have equal energy but different spatial
distribution. Consequently, their shielding of one another
is relatively small and the electrons are more strongly
attracted by the nucleus.
2. Exchange Energy : The stabilizing effect arises whenever
two or more electrons with the same spin are present in the
degenerate orbitals of a subshell. These electrons tend to
exchange their positions and the energy released due to
this exchange is called exchange energy. The number of
exchanges that can take place is maximum when the subshell
is either half filled or completely filled. As a result the
exchange energy is maximum and so is the stability.
eg. for Cr : [Ar] 4s1 3d5 Fig. 1.28: Possible exchange for a d5 configuration

Thus, total number of exchanges = 10

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Summary
 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.
A  Emission and Absorption Spectra: When light emitted from
of a given atoms is ZX
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

SCAN CODE
Structure of an Atom
STRUCTURE OF AN ATOM 29

 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’.
 2
J / atom  eV / atom
n 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

SCAN CODE
Structure of an Atom
30 STRUCTURE OF AN 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 (  ) of
14 –1 15 –1
Sol. 0 = 7 × 10 s ;  = 10 s 580 nm. Calculate the frequency (  ) and wave number ( v )
According to photo electric effect of the yellow light.
h = h0 + K.E. Sol. As per the formula of frequency,
K.E. = h ( – 0) speed of light in vacuum
–34 15 14
Frequency 
= 6.626 × 10 (10 – 7 × 10 ) wavelength
–19
= 1.9878 × 10 J.
c
Example-2 v …….. (i)

Calculate the number of protons, neutrons and electrons in
80
Br . 3  108 m / s
35
v  5.17  1014 s –1
Sol. Number of protons = number of electron 580  10 –9 m
= 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 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. The largest wavelength of light capable
of breaking a single Cl-C1 bond is
(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. 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 AN ATOM 31

Example-8 Kinetic energy of emission:

What is the number of photons of light with wave =(hv – hv0)
length of 4000 pm that provide 1 J of energy ? =(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 mv2  (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  16
n  2.012  10 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, the other is at
Example-9 Sol. Energy values are always additive
A photon of wavelength 4 × 10–7 m strikes on metal surface, Etotal = E1 + E2
the work function of the metal being 2.13 eV. Calculate
hc hc hc
(i) the energy of the photon (eV),  
 1 2
(ii) the kinetic energy of the emission, and
(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
1 1 1
 = wavelength of radiation = 4 × 10–7m  
 1 2

E
 6.62 10  3 10  4.97 10
–34 8
19
J 1 1 1
4  107  
355 680 2
Convert into eV : 1eV = 1.6 × 10–19J
2  742.77nm
4.97  1019  743nm
E(eV)  eV  3.1eV
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
32 STRUCTURE OF AN ATOM

Example-11 Sol. For Lyman series

The first line in the balmer series in the H-atom will have the 1 1 1 
frequency 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
 n1 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
  2   3  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? 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
Ratio of energy of 3rd and 4th orbit of H-atom is

13.6  Z2
Sol. KE of 3rd energy level  eV
n2

13.6  12 13.6
 2
eV  eV
3 9

13.6eV 13.6
Through the given diagram also we can calculate the number KE of 4th energy level   eV
42 9
of spectral lines in emission spectrum.
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
13.6eV
The wavelength of a spectral line emitted by hydrogen atom
9 10
16  
cm . Calculate the value of n . 13.6eV 9
in the Lyman series is
15R 2
10
(R = Rydberg constant).
STRUCTURE OF AN ATOM 33

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.
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
The de-Broglie wavelength of a tennis ball of mass 60 g
h2 4h 2 2h 2 moving with a velocity of 10 m/s is approximately
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 ×1017s–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 kg ms–1
–23
= 1.105 × 10–33m
34 STRUCTURE OF AN ATOM
Example-21 v  uncertainty in veslocity
–5
Uncertainty is positon of a particle of 25g in space is 10 m. According to question
Hence, uncertainty in velocity (ms–1) is (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
Uncertainty in the position of an electron (mass = 9.1 × 10– Example-24
31
kg) moving with a velocity 300 ms-1, accurate upon 0.001 Using s, p, d notations, describe the orbital with the
% will be (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.
x 
4  3.14  9.1 1031  300  0.001102 Sol. Principal quantum number = n value
= 0.01933 = 1.93 × 10 m–2 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 
According to Aufbau principle, the correct order of energy
Sol. According to Heisenberg of 3d, 4s and 4 p-orbitals is
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 AN ATOM 35

Exercise – 1: Basic Objective Questions

Subatomic Particles 10. Neutrons are …A… particles.
Identify A from the given option in order to complete
1. Early Indian and Greek philosophers were of the view the above statement.
that …A…are the fundamental building blocks of (a) Positively charged (b) Negatively charged
matter. Here, ‘A’ refers to (c) Neutral (d) None of these
(a) Atoms (b) Molecules 11. Which of the following statements about the electron is
(c) Protons (d) Electron incorrect?
2. Atoms can be further divided into sub-particles like (a) It is a negatively charged particle
(a) Electrons (b) Protons (b) The mass of electron is equal to the mass of
(c) Neutrons (d) All of these neutron
3. The e/m for anode rays in comparison to cathode rays (c) It is basic constituent of all atoms
is: (d) It is a constituent of cathode rays
(a) very low (b) high 12. Cathode rays travel
(c) same (d) none (a) From negative electrode (cathode) to positive
4. Mass of neutron is about ..............times the mass of the electrode (anode)
electron. (b) From negative electrode (anode) to positive
(a) 1840 (b) 1480 electrode (cathode)
(c) 1000 (d) None of these (c) In any direction
5. Positive rays or canal rays are: (d) None of the above
(a) electromagnetic waves 13. Mass of electron is equal to
(b) a stream of positively charged gaseous ions (a) 1.6  10 19 kg (b) 1.6  10 18 kg
(c) a stream of electrons (c) 9.1094  10 31 kg (d) 9.1094  10 30 kg
(d) neutrons
6. The electrical discharge through the gases could be
observed only at Thomson's model and Rutherford’s model of an
(a) Very low pressure and very low voltage atom
(b) Very high pressure and very high voltage
(c) Very high pressure and very low voltage 14. Thomson model of atom is also known as
(d) Very low pressure and very high voltage (a) Plum pudding model (b) Raisin pudding model
7. Oil drop experiment was developed by (c) Watermelon model (d) All (a), (b) and (c)
(a) R. A Millikan (b) J.J Thomson 15. According to the Thomson model of an atom, mass of
(c) Michael Faraday (d) None of these the atom is assumed to be
8. The smallest and lightest positive ion, obtained from (a) Uniformly distributed over the atom
hydrogen was called …A.. Here, ‘A’ refers to (b) Randomly distributed over the atom
(a) Proton (c) In some cases (a) and in some cases (b)
(b) Neutron (d) None of the above
(c) Electron 16. Which of the following properties of an atom could be
(d) Both proton and neutron explained correctly by Thomson model of atom?
9. Electrical discharge carried out in the modified cathode (a) Overall neutrality of atom
ray tube led to the discovery of particles carrying (b) Spectra of hydrogen atom
positive charge are also known as ..A… rays. Here, ‘A’ (c) Position of electrons, protons and neutrons in atom
refers to (d) Stability of atom
(a) X (b) Canal
(c) Beta (d) Gamma
36 STRUCTURE OF AN ATOM

17. When a gold sheet is bombarded by a beam of 25. Atomic number, (Z) =
α–particles, only a few of them get deflected whereas (a) Number of protons in the nucleus of an atom
most go straight, undeflected. This is because (b) Number of electrons in a neutral atom
(a) The force of attraction exerted on the α–particles by
(c) Both (a) and (b)
the oppositely charged electrons is not sufficient.
(d) None of the above
(b) A nucleus has a much smaller volume than that of
26. Nucleons are
an atom.
(a) Only neutrons
(c) The force of repulsion acting on the fast-moving α–
(b) Neutrons + protons
particles are very small.
(c) Neutrons + protons + electrons
(d) The neutrons in the nucleus do not have any effect
(d) Neutron + electrons
on the α–particles.
27. Atoms with same mass number but different atomic
18. Rutherford’s scattering experiment is related to the size
numbers are called
of the
(a) Isotopes (b) Isobars
(a) Nucleus (b) Atom
(c) Isotones (d) None of these
(c) Electron (d) Neutron
28. An isoelectronic pair is
19. An  -particle scattering experiment of Rutherford, (a) Ca and K
bombards very thin …A… foil with  -particles. (b) Ar and Ca2+
Fill the blank with an appropriate option. (c) K and Ca2+
(a) Silver (b) Gold (d) Ar and K
(c) Aluminium (d) Copper 12
20. According to Rutherford, the nucleus surrounded by
29. 6 C , 13
6 C,
14
6 C and 35
17 Cl , 37
17 Cl are the examples of
electrons that move around the nucleus with a very (a) Isotopes
high speed in …A… paths called orbits. Here, A refers (b) Isobars
to (c) Isotopes and isobars respectively
(a) Elliptical (b) Parabolic (d) Isobars and Isotopes respectively
(c) Circular (d) None of these 30. The number of electrons and protons in an atom of
21. Rutherford model could not explain the third alkaline earth metal is
(a) Electronic structure of an atom (a) e  20, p  20
(b) Stability of an atom (b) e  18, p  20
(c) Both (a) and (b) (c) e  18, p  18
(d) None of the above
(d) e  19, p  20
31. The triad of nuclei that is isotonic is :
Special terms 14 15 17 12 14 19
(a) 6 C, 7 N, 9 F (b) 6 C, 7 N, 9 F
22. The ion that is isoelectronic with CO is: (c) 14 14 17
(d) 14 14 19
6 C, 7 N, 9 F 6 C, 7 N, 9 F
(a) CN  (b) O2
(c) O2 (d) N 2
76 Electromagnetic spectrum
23. An isotone of 32Ge is
77 77
(i) 32Ge (ii) 33As 32. Maxwell suggested that when electrically charged
77 78
(iii) 34Se (iv) 34Se particles move under acceleration, alternating electrical
(a) Only (i) and (ii) (b) Only (ii) and (iii) and magnetic fields are produced and transmitted.
(c) Only (ii) and (iv) (d) (ii), (iii), and (iv) These fields are transmitted in the form of waves called
24. Which of the following atoms and ions are (a) Electromagnetic waves
isoelectronic i.e. have the same number of electrons (b) Electromagnetic radiations
with the neon atom (c) Both (a) and (b)
–
(a) F (b) Oxygen atom (d) None of the above
–
(c) Mg (d) N
STRUCTURE OF AN ATOM 37

33. The oscillating electric and magnetic fields produced 12 1

(c) 5.17210 s
by oscillating charged particles are 14 1
I. Perpendicular to each other. (d) 1.74 10 s
II. Parallel to each other 41. Rank the following types of radiations from the highest
III. Both are perpendicular to the direction of energy to the lowest.
propagation of the wave. ultraviolet/visible/X-ray/microwave/infrared
IV. Both are parallel to the direction of the waves. (a) X-ray, ultraviolet, microwave, infrared, visible
Choose the correct option (b) ultraviolet, X-ray, visible, infrared, microwave
(a) I and III (b) II and IV (c) infrared, microwave, ultraviolet, visible, X-ray
(c) I and IV (d) II and III (d) X-ray, ultraviolet, visible, infrared, microwave
34. According to the electromagnetic theory of Maxwell,
(a) Charged particles when accelerated should emit Planck's quantum theory
electromagnetic radiation
42. The relation between energy of a radiation and its
(b) Charged particles when accelerated should absorb
frequency was given by :
electromagnetic radiation (a) de Broglie (b) Einstein
(c) In some cases (a), in some cases (b) (c) Planck (d) Bohr
(d) None of the above 43. Which is not characteristics of Planck’s quantum
14 –1
35. The frequency of a wave of light is 12 × 10 s . The
theory of radiation?
wave number associated with this light is
–7 –8 –1 (a) Radiation is associated with energy.
(a) 5 × 10 m (b) 4 × 10 cm
–7 –1 4 –1 (b) Energy is not absorbed or emitted in whole number
(c) 2 × 10 m (d) 4 × 10 cm
14 or multiples of quantum.
36. The frequency of a green light is 6 × 10 Hz. Its
(c) The magnitude of energy associated with a quantum
wavelength is :
is proportional to the frequency.
(a) 500 nm (b) 5 nm
(d) Radiation energy is neither emitted nor absorbed
(c) 50,000 nm (d) None of these
continuously but in small packets called quanta.
37. Value of speed of light is
44. According to Planck, the smallest quantity of energy
(a) 3.0  10 8 m s  1
that can be emitted or absorbed in the form of
(b) 3.0  10 7 m s  1
electromagnetic radiation is called
(c) 3.0  1010 ms  1 (a) Photon (b) Quantum
(d) 3.0  10 9 m s  1 (c) Both (a) and (b) (d) None of these
38. Relationship between frequency (v), wavelength    45. The number of photons of light of  = 2.5 × 10 m
6 –1

and velocity of light (c) is necessary to provide 1 J of energy are

v (a) 2 × 1018 (b) 2 × 1017
(a) c  (b) c  v  (c) 2 × 1020 (d) 2 × 1019

46. The number of photons emitted in 10 hours by a 60 W
(c) c  v   (d) c  v  
sodium lamp (λ of photon = 6000 Å)
39. The number of wavelengths per unit length is called 24 23
(a) Wavelength (v) (a) 6.50 × 10 (b) 6.40 × 10
23 23
(c) 8.40 × 10 (d) 3.40 × 10

(b) Wavelength v o
47. The energy ratio of photon of wavelength 3000 A and
(c) Wave number  v  o
6000 A is
(d) Wave number (v)
(a) 2:1 (b) 1:2
40. Calculate frequency of yellow radiation having
o
(c) 1:1 (d) 1:4
wavelength 5800 A . 48. The energy required to break one mole of Cl-C1 bonds
(a) 5.17210 s
14 1 in Cl2 is 242 kJ mol-1. The largest wavelength of light
1 capable of breaking a single Cl-C1 bond is
(b) 174s (a) 700 nm (b) 494 nm
(c) 596 nm (d) 640 nm
38 STRUCTURE OF AN ATOM

(d) None of these

Particle nature of electromagnetic radiations 57. Ultraviolet light of 6.2 eV falls on aluminium surface
(work function = 4.2 eV). The kinetic energy (in joule)
49. Einstein’s theory of photoelectric effect is based on : of the fastest electron emitted is approximately:
(a) Newton's corpuscular theory of light –21 –19
(a) 3 × 10 (b) 3 × 10
(b) Huygen’s wave theory of light (c) 3 × 10
–17
(d) 3 × 10
–15

(c) Maxwell’s electromagnetic theory of light 58. The threshold wavelength for the photoelectric effect
(d) Planck’s quantum theory of light on sodium is 5000 Å. Its work function is:
50. In photoelectric effect the number of photo-electrons (a) 4 × 10 J
–19
(b) 1J
emitted is proportional to : (c) 2 × 10 J
–19 –10
(d) 3 × 10 J
(a) intensity of incident beam 59. If the threshold frequency of metal for the photoelectric
(b) frequency of incident beam effect is ν0, then which of the following will not
(c) velocity of incident beam happen?
(d) work function
(a) If the frequency of the incident radiation is ν0, the
51. Increase in the frequency of the incident radiations kinetic energy of the electrons ejected is zero.
increase the : (b) If the frequency of the incident radiation is ν, the
(a) rate of emission of photo-electrons kinetic energy of the electrons ejected will be hν –
(b) work function hν0.
(c) kinetic energy of photo-electrons
(c) If the frequency is kept the same at  but the
(d) threshold frequency
intensity is increased, the number of electrons
52. Photoelectric effect shows
ejected will increase.
(a) particle-like behaviour of light
(d) If the frequency of incident radiation is further
(b) wave-like behaviour of light
increased, the number of photoelectrons ejected
(c) both wave-like and particle-like behaviour of light
will increase.
(d) neither wave-like nor particle-like behaviour of
light
Bohr's model of an atom
53. The kinetic energy of the photoelectrons does not
depend upon 60. The energy of electron in 3rd orbit of hydrogen atom is
–1 –1
(a) Intensity of incident radiation (a) –1311.8 kJ mol (b) –82.0 kJ mol
–1 –1
(b) Frequency of incident radiation (c) –145.7 kJ mol (d) –327.9 kJ mol
(c) Wavelength of incident radiation 61. The ionization energy of H atom is 13.6 eV. The
2+
(d) Wave number of incident radiation. ionization energy of Li ion will be
54. The ideal body which emits and absorbs radiations of (a) 54.4 eV (b) 40.8 eV
all frequencies, is called a black body and the radiation (c) 27.2 eV (d) 122.4 eV
emitted by such a body is called
(a) White body radiation (b) Black body radiation 62. Energy of electron of hydrogen atom in second Bohr
(c) Black body emission (d) None of these orbit is
–19 –19
55. The exact frequency distribution of the emitted (a) –5.44 × 10 J (b) –5.44 × 10 kJ
–19 –19
radiation (i.e., intensity versus frequency curve of the (c) –5.44 × 10 cal (d) –5.44 × 10 eV
radiation) from a black body depends on
(a) Temperature (b) Pressure 63. The Bohr orbit radius for the hydrogen atom (n = 1) is
(c) Both (a) and (b) (d) Neither (a) nor (b) approximately 0.530 Å. The radius for the first excited
56. Minimum frequency (v0), below which photoelectric state (n = 2) orbit is
effect is not observed is called (a) 0.13 Å (b) 1.06 Å
(a) Threshold frequency (c) 4.77 Å (d) 2.12 Å
(b) Minimum frequency
(c) Lowest frequency
STRUCTURE OF AN ATOM 39

64. According to Bohr model, angular momentum of an 71. Major development(s) responsible for the formulation
electron in the 3rd orbit is : of Bohr’s model of atom were
3h 1.5h (a) Dual character of the electromagnetic radiation
(a) (b) which means that radiations possess both wave like
 
and particle like properties
3 9h
(c) (d) (b) Experimental results regarding atomic spectra
h 
which can be explained only by assuming
65. Electronic energy is a negative energy because
quantized electronic energy level in atoms.
(a) Electron carries a negative charge.
(c) Both (a) and (b)
(b) Energy is zero near the nucleus and decreases as
(d) None of the above
the distance from the nucleus increases.
72. The ratio of the difference in energy between the first
(c) Energy is zero at an infinite distance from the
and the second Bohr orbit to that between second and
nucleus and decreases as the electron comes closer
third Bohr orbit is
to the nucleus.
1 1
(d) There are interelectronic repulsions. (a) (b)
66. The energy of an electron in the orbit 2 3
(a) Changes with time 27 4
(c) (d)
(b) Does not change with time 5 9
(c) Sometime change and sometime does not change 73. The energy of second Bohr orbit in the hydrogen atom
+
(d) None of the above is –3.4 eV. The energy of fourth orbit of He ion would
67. The value of a0 in radius of nth state of hydrogen atom, be
(a) –3.4 eV (b) –0.85 eV
rn  n 2 a0 is (c) –13.64 eV (d) +3.4 eV
(a) 5.27 pm (b) 52.9 pm 74. The energy of an electron in the first Bohr orbit of H
(c) 529 pm (d) 0.529 pm atom is –13.6 eV. The possible energy value(s) of the
68. Bohr’s atomic model suggest that excited state(s) for electrons in Bohr orbits to hydrogen
(a) Electrons have a particle as well wave character is (are)
(b) Atomic spectrum of atom should contain only five (a) –3.4 eV (b) –4.2 eV
lines (c) –6.8 eV (d) +6.8 eV
(c) Electron on H atom can have only certain values of 75. The ionization energy of the hydrogen atom is
angular momentum 13.6 eV. The energy required to excite the electron in a
(d) All of the above hydrogen atom from the ground state to the first
69. Which of the following statements does not form a part excited state is
–18 –23
of Bohr’s model of hydrogen atom? (a) 1.69 × 10 J (b) 1.69 × 10 J
23 25
(a) Energy of the electrons in the orbit is quantized (c) 1.69 × 10 J (d) 1.69 × 10 J
(b) The electron in the orbit nearest to the nucleus has 76. The energy required to dislodge electron from excited
the lowest energy isolated H-atom, IE1 = 13.6 eV is
(c) Electrons revolve in different orbits around the (a) = 13.6 eV (b) > 13.6 eV
nucleus (c) < 13.6 and > 3.4 eV (d) ≤ 3.4 eV
(d) The position and velocity of the electrons in the 77. The energy of an electron in first Bohr orbit of H-atom
orbit cannot be determined simultaneously is -13.6 eV. The possible energy value of electron in
70. For a Bohr atom, angular momentum of the electron is the excited state of Li2+ is
(n = 0, 1, 2 , 3…) (a) -122.4 eV (b) 30.6 eV
n2h2 nh (c) -30.6 eV (d) 13.6 eV
(a) (b)
4 2
h 2 2
(c) (d)
4 nh
40 STRUCTURE OF AN ATOM

Hydrogen spectra 87. The third line in Balmer series corresponds to an

electronic transition between which Bohr’s orbits in
78. The line spectrum observed when electron jumps from hydrogen atom
higher level to M level is known as (a) 5 → 3 (b) 5 → 2
(a) Balmer series (b) Lyman series
(c) 4 → 3 (d) 4 → 2
(c) Paschen series (d) Brackett series
88. Which transition in the hydrogen atomic spectrum will
79. In hydrogen spectrum, the series of lines appearing in
have the same wavelength as the transition, n = 4 to n
ultraviolet region of electromagnetic spectrum are
= 2 of He+ spectrum?
called
(a) n = 4 to n = 3 (b) n = 3 to n = 2
(a) Balmer lines (b) Lyman lines
(c) n = 3 to n = 1 (d) n = 2 to n = 1
(c) Pfund lines (d) Brackett lines
89. The wave number of the spectral line in the emission of
80. A ray of white light is spread out into a series of
8
coloured band called …. of visible light. hydrogen will be equal to times the Rydberg’s
9
(a) Visible band (b) Spectrum
(c) Both (a) and (b) (d) None of these constant if the electron jumps from
81. In the line spectrum of hydrogen, the lines described by (a) n = 10 to n =1
the formula (b) n = 3 to n = 1
(c) n = 9 to n = 8
 1 1  (d) n = 0 to n = 1
v  109, 677  2  2  cm 1
2 n  90. The wavelength of a spectral line emitted by hydrogen
Where, n  integer, n  3 constitutes 16
atom in the Lyman series is cm . Calculate the
(a) Balmer series (b) Lyman series 15 R
(c) Pfund series (d) Paschen series value of n2 . (R = Rydberg constant).
82. In the line spectrum of hydrogen, the first five series of
(a) 3 (b) 4
lines that correspond to n1=1,2,3,4,5 are known as
(c) 2 (d) 1
(a) Lyman, Paschen, Pfund, Bracket, Balmer
respectively
(b) Paschen, Pfund, Bracket, Lyman, Balmer Dual nature of particle
respectively
91. An electron is moving in Bohr’s fourth orbit. Its de-
(c) Lyman, Balmer, Pfund, Paschen, Bracket
Broglie wavelength is . Calculate the circumference
respectively
of the fourth orbit.
(d) Lyman, Balmer, Paschen, Bracket, Pfund
(a) 2  (b) 4 
respectively
4 2
83. Value of Rydberg constant (RH) is (c) (d)
18 18
 
(a) 21.8 10 J (b) 21.8 10 J 92. The de-Broglie wavelength of a tennis ball of mass 66
18 18
(c) 2.18 10 J (d) 2.18 10 J g moving with the velocity of 10 metres per second is
84. What transition in He+ ion shall have the same wave approximately
–35 –33
number as the first line in Balmer series of H atom? (a) 10 metres (b) 10 metres
–31 –36
(a) 7 → 5 (b) 5 → 3 (c) 10 metres (d) 10 metres
(c) 6 → 4 (d) 4 → 2 93. The wavelength of a cricket ball weighing 100 g and
85. An electron jumps from 6th energy level to 3rd energy travelling with a velocity of 50 m/s is
–28 –37
level in H-atom, how many lines belong to the visible (a) 1.3 × 10 m (b) 1.3 × 10 m
region? –34
(c) 1.3 × 10 m
–30
(d) 1.3 × 10 m
(a) 1 (b) 2 –23
94. An electron has kinetic energy 2.8 × 10 J de–Broglie
(c) 3 (d) Zero –31
86. The wave number for the shortest wavelength wavelength will be nearly(me = 9.1 × 10 kg)
–4 –7
transition in the Balmer series of atomic hydrogen is (a) 9.24 × 10 m (b) 9.24 × 10 m
–8
(a) 27420 cm
–1
(b) 28420 cm
–1 (c) 9.28 × 10 m (d) 9.24 × 10–10 m
–1 –1
(c) 29420 cm (d) 12186 cm
STRUCTURE OF AN ATOM 41

95. Light possesses Heisenberg's uncertainty principle

(a) Particle like behaviour (b) Wave like behaviour
(c) Both (a) and (b) (d) None of these
103. If uncertainty in the position of an electron is zero, the
96. Relation between wavelength,    and momentum, uncertainty in its momentum would be
(p) of a material particle is h
(a) zero (b) 
h 4
(a)   hp (b)  
p h
(c)  (d) infinite
(c)   h  p (d)   h  p 4
o 104. For an electron, if the uncertainty in velocity is (v) ,
97. Calculate the mass of a photon with wavelength 3.6 A .
27 28
the uncertainty in its position (x) is given by :
(a) 6.13510 kg (b) 6.13510 kg
h 
33
(c) 6.13510 kg
30
(d) 6.13510 kg (a) m (b)
2 hm
98. The de-Broglie equation applies h 2m
(c) (d)
(a) to electrons only 4m h
(b) to protons only 105. “It is impossible to determine simultaneously the exact
(c) to neutron only position and exact momentum (or velocity) of an
(d) All the object in motion electron”. This principle is
99. An electron with velocity ‘v’ is found to have a certain (a) Heisenberg Uncertainty principle
value of de-Broglie wavelength. The velocity that the (b) dual behaviour of matter
neutron should possess to have the same de Broglie (c) de-Broglie’s Principle
wavelength is (d) None of the above
(a) v (b) v/1840 106. Which of the following is responsible to rule out the
(c) 1840v (d) 1840/v existence of definite paths or trajectories of electrons?
100. If EA, EB and EC represents kinetic energies of an (a) Pauli’s exclusion principle
electron, alpha particle and proton respectively and (b) Heisenberg’s uncertainty principle
each moving with same de-Broglie wavelength then (c) Hund’s rule of maximum multiplicity
choose the correct among the following. (d) Aufbau principle
(a) E A  EB  EC (b) EA  EB  EC 107. The uncertainty principle and the concept of wave
(c) EB  EC  EA (d) EA  EC  EB nature of matter was proposed by …A… and …B…
respectively. Here, A and B refer to
101. The momentum of a photon of frequency 50 1017 s1
(a) A  de-Broglie; B  Heisenberg
nearly
40 1
(b) A  Heisenberg; B  de-Broglie
(a) 2.27 10 kg ms (c) A  Heisenberg; B  Planck
40 1
(b) 3.3310 kg ms (d) A  Planck; B  de-Broglie
23 1 108. The uncertainty in the momentum of an electron is
(c) 1.110 kg ms
23 1
1.0 105 kg ms1 . The uncertainty in its position will
(d) 4.2 10 kg ms
be
102. What will be the kinetic energy of radiation having
(a) 1.50 1028 m (b) 1.05  1026 m

wavelength ? (c) 5.27  10 30 m (d) 5.25  1028 m
2
109. A ball of mass 200g is moving with a velocity of 10m
m 2 4h –1
sec . If the error in measurement of velocity is 0.1%,
(a) (b)
4h m 2 the uncertainty in its position is :
2h 2h 2 –31
(a) 3.3 × 10 m
–27
(b) 3.3 × 10 m
(c) (d)
m m 2 –25
(c) 5.3 × 10 m (d) 2.64 × 10 m
–32
42 STRUCTURE OF AN ATOM

110. The uncertainties in the velocities of two particles A 119. How many spherical nodes are present in 4s orbital in a
and B are 0.05 and 0.02 ms-1 respectively. The mass of hydrogen atom?
B is five times to that of mass A. what is the ratio of (a) 0 (b) 2
 x A  (c) 3 (d) 4
uncertainties   in their positions? 120. The number of nodes possible in radial wave function
 x B  of 3d orbital is
(a) 2 (b) 0.25
(a) 1 (b) 2
(c) 4 (d) 1
(c) 0 (d) 3
Quantum numbers 121. The d-orbital with the orientation along X and Y axes
is called:
111. In the Schrodinger’s wave equation,  represents (a) d z2 (b) d zy
(a) Orbit (b) Wave function
(c) d yz (d) dx2  y2
(c) Wave (d) Radial probability
112. The quantum number not obtained from Schrodinger 122. 3py orbital has .......... nodal plane
equation is (a) XY (b) YZ
(a) n (b) l (c) ZX (d) Any
(c) m (d) s 123. The number of angular nodes in a 3s atomic orbital is
113. Probability of finding an electron at a point within an (a) 0 (b) 1
atom is (c) 2 (d) 3
(a) equal to the  2 at that point 124. The number of radial nodes in a 3s atomic orbital is
(a) 0 (b) 1
(b) proportional to the  2 at that point (c) 2 (d) 3
(c) inversely proportional to the  2 at that point 125. ….‘X’… is a type of quantum number. It is a positive
integer with value of n = 1,2,3… It determines the size
(d) None of the above
and to large extent the energy of the orbital and also
114. A subshell with n = 6, l = 2 can accommodate a
identifies the shell. Identify ‘X’.
maximum of
(a) Azimuthal quantum number
(a) 10 electrons (b) 12 electrons
(b) Principal quantum number
(c) 36 electrons (d) 72 electrons
(c) Magnetic orbital quantum number
115. Which of the following sets of quantum number is
(d) Spin orbital quantum number
correct for an electron in 4f orbital?
126. The region of an orbital where probability density
(a) n = 3, l = 2, m = –2, s = +1/2
function reduces to zero is called
(b) n = 4, l = 4, m = –4, s = –1/2
(a) nodal surface (b) node
(c) n = 4, l = 3, m = +1, s = +1/2
(c) both (a) and (b) (d) None of these
(d) n = 4, l = 3, m = +4, s = +1/2
127. The total number of nodes are given by
116. The correct designation of an electron with n = 4, l = 3,
m = 2, and s = 1/2 is: (a)  n1 (b)  n  l 1
(a) 3d (b) 4f (c)  n1 (d)  n  l  1
(c) 5p (d) 6s
128. The maximum number of electrons in the shell with
117. A 3d-electron having s = +1/2 can have a magnetic
principal quantum number ‘n’ is equal to
quantum no:
(a) 2n (b) 2n2
(a) +2 (b) +3
(c) 2n3 (d) 2n4
(c) –3 (d) +4
118. The maximum number of 4d-electrons having spin
quantum number, s = +1/2
(a) 10 (b) 7
(c) 1 (d) 5
STRUCTURE OF AN ATOM 43

1 135. In manganese atom, Mn (Z = 25), the total number of

129. An electron with values 4, 3, -2, and  for the set of
2 orbitals populated by one or more electrons (in ground
four quantum number n, l, ml and ms respectively state) is
belongs to (a) 15 (b) 14
(a) 4s orbital (b) 4p orbital (c) 12 (d) 10
(c) 4d orbital (d) 4f orbital 136. Presence of three unpaired electrons in phosphorus
130. The nodes present in 3p-orbitals are atom can be explained by
(a) Two spherical (a) Pauli’s rule
(b) Two planar (b) Uncertainty principle
(c) One planar (c) Aufbau’s rule
(d) One planar, one spherical (d) Hund’s rule
131. The stationary states for electron are numbered n = 137. Which electronic configuration does not follow Pauli’s
1,2,3… These integral numbers are known as exclusion principle?
(a) Principal quantum number (a) 1s2, 2s2 2p4
2 2 4 2
(b) Azimuthal quantum number (b) 1s , 2s 2p , 3s
2 4
(c) Spin quantum number (c) 1s , 2p
2 2 6 3
(d) None of the above (d) 1s , 2s 2p , 3s
132. The electrons identified by n and l 138. Magnetic quantum number for the last electron in
(i) n = 4, l = 1 (ii) n = 4, l = 0 sodium is:
(iii) n = 3, l = 2 (iv) n = 3, l = 1 (a) 3 (b) 1
can be placed in order of increasing energy, from (c) 2 (d) zero
lowest to highest 139.
2 2 1
Nitrogen has the electronic configuration 1s ,2s 2px
(a) (iv) < (ii) < (iii) < (i)
(b) (ii) < (iv) < (i) < (iii) 2p1y 2p1z and not 1s 2 , 2s 2 2p 2x 2p1y 2p 0z . It was proposed
(c) (i) < (iii) < (ii) < (iv) by:
(d) (iii) < (i) < (iv) < (ii) (a) Aufbau principle
133. Which of the following statements concerning the (b) Pauli’s exclusion principle
quantum numbers are correct? (c) Hund’s rule
I. Angular quantum number determines the three (d) Uncertainty principle
dimensional shape of the orbital. 140. What is the correct order for the size of the s-orbital?
II. The principal quantum number determines the (a) 4s  2s  3s  1s (b) 4s  3s  2s  1s
orientation and energy of the orbital. (c) 1s  2s  3s  4s (d) 1s  3s  2s  4s
III. Magnetic quantum number determines the size of 141. In the ground state of the atoms, the orbitals are filled
the orbital. in order of their
IV. Spin quantum number of an electron determines (a) decreasing energies
the orientation of the spin of electron relative to the (b) increasing energies
chosen axis. (c) no role of energy in filling of orbitals
The correct set of option is (d) None of the above
(a) I and II (b) I and IV 142. Pauli exclusion principle state that
(c) III and IV (d) II, III and IV (a) no two electrons in an atom can have the same set
of four quantum numbers
Electronic configuration (b) only two electrons may exist in the same orbital and
these electrons must have opposite spin
134. The correct ground state electronic configuration of
chromium atom (Z = 24) is (c) Both (a) and (b) statement are true
5
(a) [Ar] 3d 4s
1 4
(b) [Ar] 3d 4s
2 (d) None of the above
6 0 1 5
(c) [Ar] 3d 4s (d) [Ar] 4s 4p
44 STRUCTURE OF AN ATOM

143. Pairing of electrons in the orbitals belonging to the 147. For which one of the following sets of four quantum
same subshell (p, d or f ) does not take place until each numbers an electron will have the highest energy
orbital belonging to that subshell has got one electron n  m s
each, i.e., it is singly occupied. This is called (a) 3 2 1 1/2
(a) Hund’s rule of maximum multiplicity (b) 4 1 0 –1/2
(b) Pauli’s exclusion principle (c) 4 2 –1 1/2
(c) Aufbau principle (d) 5 0 0 –1/2
(d) None of the above 148. Consider the ground state of Cr atom (Z = 24). The
144. Aufbau principle does not give the correct arrangement number of electrons with azimuthal quantum numbers,
of filling up of the atomic orbitals in l = 1 and 2 are respectively
(a) Cu and Zn (b) Co and Zn (a) 12 and 4 (b) 12 and 5
(c) Mn and Cr (d) Cu and Cr (c) 16 and 6 (d) 16 and 5
145. Ground state electronic configuration of nitrogen atom 149. Which of the following statements in relation to the
can be represented as hydrogen atom is correct?
(a) 3s, 3p and 3d-orbitals all have the almost same
I.     
energy
     (b) 3s and 3p-orbitals are of lower energy than 3d-
II.
orbital
III.      (c) 3p-orbitals is lower in energy than 3d-orbital
(d) 3s-orbitals is lower in energy than 3p-orbital
IV.      150. Which of the following option does not represent
Choose the correct option ground state electronic configuration of an atom?
(a) I and II (b) III and IV 2 2 6 2 6 8 2
(a) 1s 2s 2p 3s 3p 3d 4s
(c) I and IV (d) II and III
2 2 6 2 6 9 2
146. Identify the correct order of increase in the energy of (b) 1s 2s 2p 3s 3p 3d 4s
the hydrogen atom 2 2 6 2 6 10 1
(c) 1s 2s 2p 3s 3p 3d 4s
(a) 1s  2s  2p  3s  3p  3d  4s  4p  4d  4f
2 2 6 2 6 5 1
(d) 1s 2s 2p 3s 3p 3d 4s
(b) 1s  2s  2p  3s  3p  3d  4s  4p  4d  4f
(c) 1s  2s  3s  4s  2p  3p  4p  3d  4d  4f
(d) 1s  2s  3s  4s  2p  3p  4p  3d  4d  4f
STRUCTURE OF AN ATOM 45

Exercise – 2: Previous Year Questions

1. The nucleus of an atom can be assumed to be spherical. 8. For f-orbital, the values of m are
The radius of the nucleus of mass number A is given (Manipal 2009)
by 1.25 × 10–13 × A1/3 cm. The radius of an atom is (a) –2, –1, 0, +1, +2
1.0 Å. If the mass number is 64 then the fraction of the (b) –3, –2, –1, 0, +1, +2, +3
atomic volume that is occupied by the nucleus is (c) –1, 0, +1
(Manipal 2008) (d) 0, +1, +2, +3
–3 9. The uncertainty in the position of an electron
(a) 1.0 × 10 (b) 5.0 × 10–5
(mass = 9.1 × 10–28 g) moving with a velocity of
(c) 2.5 × 10–2 (d) 1.25 × 10–13
2. Deflection back of a few particles on hitting thin foil of 3.0 × 104 cm s–1 accurate upto 0.011% will be,
gold show that [where h = 6.626×10–27 erg sec]
(Manipal 2008) (Manipal 2009)
(a) Nucleus is heavy (a) 1.92 cm (b) 7.66 cm
(b) Nucleus is small (c) 0.176 cm (d) 3.84 cm
(c) Both (a) and (b) 10. Which of the following is not a permissible
(d) electrons create hindrance in the movement of arrangement of electrons in an atom?
  particles. (CBSE AIPMT 2009)
3. If uncertainty in position and momentum are equal then (a) n = 3, l = 2, m = –3, s = – 1
2
uncertainty in velocity is (b) n = 5, l = 3, m = 0, s = + 1
(CBSE AIPMT 2008) 2
(c) n = 4, l = 0, m = 0, s = – 1
(a) 1 h
(b) h
2
2m  2
(d) n = 3, l = 2, m = –2, s = – 1
2
(c) 1 h
(d) h
2 2 1
m   11. Nitrogen has the electronic configuration 1s ,2s 2px
4. Probability of finding an electron at the nodal surface
2p1y 2p1z and not 1s2 2s2 2p2x 2p1y 2p0z which is determined
is (AMU 2009)
(a) unity (b) low by (Haryana PMT 2010)
(c) high (d) zero (a) Pauli’s exclusion principle
5. The number of radial nodes of 3s and 2p-orbitals are (b) Aufbau principle
(Haryana PMT 2009) (c) Hund’s rule
(a) 2, 0 (b) 0, 2 (d) Uncertainty principle
(c) 1, 2 (d) 2, 1 12. Principal, azimuthal and magnetic quantum number are
6. The number of nodal planes in px orbital is respectively (RPMT 2010)
(Haryana PMT 2009) (a) size, orientation and shape
(a) 1 (b) 2 (b) size, shape and orientation
(c) 3 (d) 0 (c) shape, size and orientation
7. The radius of the Bohr orbit of the H-atom is r. Then (d) None of the above
the radius of the orbit of Li2+ will be 13. A particle having a mass of 1.0 mg has a velocity of
(AMU 2009) the 3600 km/h. Calculate the wavelength of the
(a) r/9 (b) r/3 particle.  h  6.626  10 27 erg  s 
(c) 3r (d) 9r (Guj CET 2010)
(a) 6.626  10 28 cm (b) 6.626  1029 cm
(c) 6.626  10 30 cm (d) 6.626  10 31 cm
46 STRUCTURE OF AN ATOM

14. When the electron of a hydrogen atom jumps from n = 22. The energies, E1, and E2 of two radiations are 25 eV
4 to n = 1 state, the number of spectral lines emitted is and 50 eV respectively. The relation between their
(AMU 2010) wavelengths i.e., 1 and 2 will be
(a) 15 (b) 9
(CBSE AIPMT 2011)
(c) 6 (d) 3
15. The relation between wave motion of photons and (a)   1  (b)   
1 2 1 2
stream of particle is (MP PMT 2010) 2
h (c)   2 (d)   4
2 1 2 1 2
(a) x. p  (b) E  mc
4 23. Which one of the following statements in relation to
h the hydrogen atom is correct ? (Manipal 2010)
(c)   (d) E  h
 (a) 3s, 3p and 3d-orbitals all have the same energy
16. The wave number of the spectral line in the emission (b) 3s and 3p-orbitals are of lower energy than 3d-
spectrum of hydrogen will be equal to  times the orbital
9
(c) 3p-orbital is lower in energy than 3d-orbital
Rydberg’s constant if the electron jumps from
(d) 3s-orbital is lower in energy than 3p-orbital
(KCET 2010)
24. When the electrons of hydrogen atom return to L shell
(a) n = 3 to n = 1 (b) n = 10 to n = 1
from shells of higher energy, we get a series of lines in
(c) n = 9 to n = 1 (d) n = 2 to n = 1
the spectrum. This series is called
17. The wavelength of which series lie towards the
(KCET 2012)
ultraviolet region? (RPMT 2010)
(a) Balmer series (b) Lyman series
(a) Lyman (b) Balmer
(c) Brackett series (d) Paschen series
(c) Paschen (d) None of these
25. The nitride ion in lithium nitride is composed of
18. Which of the following pairs is isoelectronic?
(KCET 2012)
(VMMC 2010)
(a) 7 protons + 10 electrons
(a) Ne, Cl 2+
(b) Ca , F –
(b) 10 protons + 10 electrons
(c) Mg2+, K+ (d) N3–, Na+ (c) 7 protons + 7 electrons
19. If n = 6, the correct sequence of filling of electrons will (d) 10 protons + 7 electrons
be (CBSE AIPMT 2011) 26. Maximum number of electrons in a subshell with
(a) ns  np  (n – 1) d  (n – 2) f l = 3 and n = 4 is (AIPMT 2012)
(b) ns  (n – 2) f  (n – 1) d  np (a) 14 (b) 16
(c) ns  (n – 1) d  (n – 2) f  np (c) 10 (d) 12
(d) ns  (n – 2) f  np  (n – 1) d 27. The value of azimuthal quantum number (l) is 2 then
20. The total number of atomic orbitals in fourth energy the value of principal quantum number (n) is
level of an atom is (CBSE AIPMT 2011) (Manipal 2013)
(a) 4 (b) 8 (a) 2 (b) 3
(c) 16 (d) 32 (c) 4 (d) 5
34
21. Calculate the velocity of an electron having a 28. The value of Planck’s constant is 6.6310 Js. The
wavelength of 0.15 nm. The mass of an electron is 17 1
speed of light is 310 nms . Which value of closest
9.109×10-28g. (Guj.CET 2011) to the wavelength in nanometer of a quantum of light
(h = 6.626 × 10–27 erg-s) 15 1
with frequency of 6 10 s ?
(a) 4.85 × 108 cm s–1 (NEET 2013)
(a) 10 (b) 25
(b) 2.062 × 10–15 cm s–1
(c) 50 (d) 75
(c) 2.068 × 10–10 cm s–1
(d) 4.85 × 10–8 cm s–1
STRUCTURE OF AN ATOM 47

29. What is the maximum numbers of electrons that can be 35. Be2 is isoelectronic with which of the following ions?
associated with the following set of quantum numbers? (AIPMT 2014)
n  3, l  1 and , m  1 (NEET 2013)  
(a) H (b) Li
(a) 4 (b) 2  2
(d) Na (d) Mg
(c) 10 (d) 6
 2
36. The orbital angular momentum of electrons in d orbital
30. Based on equation E = – 2.178 × 10–18 J  Z2  certain is equal to (CBSE AIPMT 2015)
n 
conclusions are written. Which of them is not correct? (a) 6h (b) 2h
(CBSE AIPMT 2013) (c) 2 3 h (d) 0h
(a) For n = 1, the electron has more negative energy 37.
2
The number of d-electrons in Fe  Z  26 is not equal
than it does for n = 6 which means that the electron
to the number of electrons in which one of the
is more loosely bound in the smallest allowed orbit.
following ? (CBSE AIPMT 2015)
(b) The negative sign in equation simply means that the
energy of electron bound to the nucleus is lower (a) s  electrons in Mg  Z  12
than it would be if the electrons were at the infinite (b) p  electrons in Cl  Z  17
distance from the nucleus
(c) Larger the value of n, the larger is the orbit radius (c) d  electrons in Fe  Z  26
(d) Equation can be used to calculate the change in (d) p  electrons in Ne  Z  10
energy when the electron changes orbit.
38. Which is the correct order of increasing energy of the
31. The maximum number of electrons which can be held
listed orbitals in the atom of titanium ?
by subshell with azimuthal quantum number ‘l’ in an
(At. No., Z = 22) (NEET 2015)
atom is given by (Kerala CEE 2014)
(a) 4s3s3p3d (b) 3s3p3d4s
(a)  2l  1 (b)  2l  2
(c) 3s3p4s3d (d) 3s4s3p3d
(c) 2  2l  1 (d) 2 2l  2 39. How many electrons can fit in the orbital for which
32. The ratio of de-Broglie wavelengths of a deuterium n = 3 and l = 1 ? (NEET 2016)
atom to that of an -particle, when the velocity of the (a) 6 (b) 10
former is five times greater than that of the latter is (c) 14 (d) 2
(Kerala CEE 2014) 40. Two electrons occupying the same orbital are
(a) 4 (b) 0.2 distinguished by (NEET 2016)
(c) 2 (d) 0.4 (a) principal quantum number
33. The uncertainty in the velocity of particle of mass (b) magnetic quantum number
6.6261028 kg is 1106 ms1 . What is the uncertainty (c) azimuthal quantum number
(d) spin quantum number.
in its position (in nm) ?  h  6.626  10 34 Js 
41. Number of photons emitted by a 100 W (Js1) yellow
(Kerala CEE 2014) lamp in 1.0 s is ( of yellow light is 560 nm)

(a)
1/ 2 (b)
 2.5 /  (JIPMER 2017)
18 18
(a) 1.6 10 (b) 1.4 10
(c)  4 /  (d) 1/ 4
(c) 2.8 1020 (d) 2.11020
34. Calculate the energy in joule corresponding to light of
wavelength 45 nm.
(Planck’s constant, h = 6.63  1034 Js, speed of light c
= 3  108 ms1) (NEET 2014)
15 11
(a) 6.67 10 (b) 6.67 10
15 18
(c) 4.42 10 (d) 4.42 10
48 STRUCTURE OF AN ATOM

42. Which one is a wrong statement? (NEET 2018) (c) Balmer series (d) Paschen series
(a) An orbital is designated by three quantum numbers 48. The number of protons, neutrons and electrons in
while an electron in an atom is designated by four 175
Lu , respectively, are (NEET 2020)
71
quantum numbers.
(a) 71, 104 and 71 (b) 104, 71 and 71
(b) The electronic configuration of N atom is
(c) 71, 71 and 104 (d) 175, 104 and 71
49. From the following pairs of ions which one is not an
iso-electronic pair ? (NEET 2021)
(c) Total orbital angular momentum of electron in ‘s’ (a) Mn 2  , Fe3 (b) Fe2  , Mn 2 
 2
orbital is equal to zero (c) O2  , F (d) Na ,Mg
(d) The value of m for d 2 is zero. 50. A particular station of All India Radio, Net Delhi,
z
43. If two atoms have equal number of electrons, it is broadcasts on a frequency of 1,368 KHz (kilohertz).
called (JIPMER 2019) The wavelength of the electromagnetic radiation
(a) isoelectronic (b) isotone emitted by the transmitter is : [Speed of light, c = 3 
(c) isobar (d) None of these 108 ms1 ] (NEET 2021)
44. Orbital having 3 angular nodes and 3 total nodes is (a) 2192 m (b) 21.92 cm
(NEET 2019) (c) 219.3 m (d) 219.2 nm
(a) 5p (b) 3d 51. Identify the incorrect statement from the following.
(c) 4f (d) 6d (NEET 2022)
45. 4d, 5p, 5f and 6p orbitals are arranged in the order of (a) All the five 5d orbitals are different in size when
decreasing energy. The correct option is compared to the respective 4d orbitals.
(NEET 2019) (b) All the five 4d orbitals have shapes similar to the
(a) 5f  6p  4d  5p (b) 5f  6p  5p  4d respective 3d orbitals.
(c) In an atom, all the five 3d orbitals are equal in
(c) 6p  5f  5p  4d (d) 6p  5f  4d  5p
energy in free state.
46. In hydrogen atom, the de-Broglie wavelength of an (d) The shapes of dxy, dyz, and dzx orbitals are similar
electron in the second Bohr orbit is [Given that Bohr to each other; and d x2  y2 and d x2 are similar to each
radius, a 0  52.9 pm ] (NEET 2019)
other.
(a) 211.6 pm (b) 211.6  pm 52. If radius of second Bohr orbit of the He+ ion is
(c) 52.9  pm (d) 105.8 pm 105.8pm, What is the radius of third Bohr Orbit of Li2+
47. Which of the following series of transitions in the ion ? (NEET 2022)
spectrum of hydrogen atoms falls in visible region? (a) 158.7 pm (b) 15.87 pm
(NEET 2019) (c) 1.587 pm (d) 158.7 Å
(a) Brackett series (b) Lyman series
STRUCTURE OF AN ATOM 49

Exercise – 3: Achiever’s Section

7. When sodium compounds are heated in a Bunsen
Single Choice Questions burner flame, they emit light at a wavelength of 5890
–4
Å. If 1.0 × 10 moles of sodium atoms each emit a
1. Let mp be the mass of proton, mn that of a neutron,
20 40
photon of this wavelength, how many kilojoules of
M1 that of a 10 Ne nucleus and M2 that of a 20 Ca energy are emitted?
nucleus. Then –2
(a) 2.03 × 10 kJ
–3
(b) 8.08 × 10 kJ
(a) M 2  2M1 –2 –3
(c) 6.20 × 10 kJ (d) 2.03 × 10 kJ
(b) M1  10  m p  m n  8. Find the frequency of light that corresponds to
5
(c) M 2  2M1 photons of energy 5.0 10 erg
21 1 21
(d) M1  M 2 (a) 7.510 sec (b) 7.510 sec
21 1 21
How long would it take a radiowave of frequency, 6 (c) 7.510 sec (d) 7.5 10 sec
2.
× 103 sec–1 to travel from Mars to Earth, a distance of 9. If a certain metal was irradiated by using two
8 × 107km? different light radiations of frequency ‘x’ and ‘2x’,
(a) 266 sec (b) 246 sec the maximum kinetic energy of the ejected electrons
(c) 280 sec (d) None of these are ‘y’ and ‘3y’ respectively. The threshold
3. Suppose 10–17J of light energy is needed by the frequency of the metal will be:
interior of the human eye to see an object. The (a) x/3 (b) x/2
photons of green light (= 550 nm) needed to see the (c) 3x/2 (d) 2x/3
10. As electron moves away from the nucleus, its
object are :
potential energy
(a) 27 (b) 28
(a) Increases (b) Decreases
(c) 29 (d) 30
(c) Remains constant (d) None of these
4. 4000 Å photon is used to break the iodine molecule,
11. According to Bohr’s theory, the angular momentum
then the % of energy converted to the K.E. of iodine
of an electron in 5th orbit is
atoms if bond dissociation energy of I2 molecule is
h h
246.5 kJ/mol (a) 25 (b) 1.0
 
(a) 8% (b) 12%
(c) 18% (d) 25% h h
(c) 10 (d) 2.5
5. A photon of 300 nm is absorbed by gas and then re  
emits two photons. One re-emitted photon has 12. Bohr’s theory is not applicable to :
wavelength 496 nm, the wavelength of second re- (a) H (b) He+
2+
emitted photon is: (c) Li (d) H+
(a) 757 (b) 857 13. The velocity of electron in second shell of hydrogen
(c) 957 (d) 657 atom is :
6 1 6 1
6. Consider a 20 W light source that emits (a) 10.94 10 ms (b) 18.88 10 ms
monochromatic light of wavelength 600 nm. The 6 1 6 1
(c) 1.888 10 ms (d) 1.094 10 ms
number of photons ejected per second in the form of
Avogadro’s constant NAV is approximately : 14. The most probable radius (in pm) for finding the

(a) NAV (b) 10–2NAV electron in He is
(a) 0.0 (b) 52.9
(c) 10–4 NAV (d) 10–6 NAV
(c) 26.5 (d) 105.8
50 STRUCTURE OF AN ATOM

15. Which of the following electronic transition in a 22. The correct set of four quantum numbers for the
hydrogen atom will require the largest amount of valence electrons of rubidium atom (Z=37) is :
energy? 1 1
(a) 5,1, 0  (b) 5,1,1 
(a) From n = 1 to n = 2 (b) From n = 2 to n = 3 2 2
(c) From n =  to n = 1 (d) From n = 3 to n = 5 1 1
16. If the total energy of an electron in the 1st shell of H (c) 5, 0,1  (d) 5, 0,0 
2 2
atom = 0.0 eV then its potential energy in the 1st 23. The number of unpaired d-electrons retained in
excited state would be
Fe2+(At. no. of Fe = 26) ion is
(a) +6.8 eV (b) +20.4 eV
(a) 6 (b) 3
(c) –6.8 eV (d) +3.4 eV
(c) 4 (d) 5
17. The emission spectra are observed by the
consequence of the transition of electrons from a
higher energy state to the ground state of He+ ion. Six Assertion-Reason Type Questions
different photons are observed during the emission
spectra, then what will be the minimum wavelength While answering these questions, you are required to choose
during the transition? any one of the following four responses.
4 4 (A) If both Assertion and Reason are correct and the Reason is
(a) (b)
27R H 15R H a correct explanation of the Assertion.
15 16 (B) If both Assertion and Reason are correct but Reason is not
(c) (d)
16R H 15R H a correct explanation of the Assertion.
18. The frequency of light emitted for the transition n = 4 (C) If the Assertion is correct but Reason is incorrect.
(D) If the Reason is correct but Assertion is incorrect.
to n = 2 of He+ is equal to the transition in H atom
corresponding to which of the following?
24. Choose the correct answer out of the following
(a) n = 3 to n = 1 (b) n = 2 to n = 1
choices.
(c) n = 3 to n = 2 (d) n = 4 to n = 3
Assertion (A): Atoms are electrically neutral.
19. The ionization enthalpy of the hydrogen atom is
Reason (R): Number of protons and electrons are
1.312 × 106 J mol–1. The energy required to excite different.
the electron in the atom from n = 1 to n = 2 is (a) A (b) B
(a) 8.51 × 105 J mol–1 (b) 6.56 × 105 J mol–1 (c) C (d) D
(c) 7.56 × 105 J mol–1 (d) 9.84 × 105 J mol–1 25. Assertion (A): The radius of the first orbit of
20. A stream of electrons from a heated filament was hydrogen atom is 0.529Å.
passed between two charged plates kept at a potential Reason (R): Radius of each circular orbit
difference ‘V’ esu. If e and m are charge and mass of (rn)=0.529Å (n2/Z), where n = 1, 2, 3 and Z = atomic
an electron, respectively, then the value of h/  number.
(a) A (b) B
(where  is wavelength associated with electron
(c) C (d) D
wave) is given by :
26. Assertion (A): For Balmer series of hydrogen
(a) 2meV (b) meV
spectrum, the value n1  2 and n 2  3, 4,5 .
(c) 2meV (d) meV
Reason (R): The value of n for a line in Balmer
21. In any subshell, the maximum number of electrons
series of hydrogen spectrum having the highest
having same value of spin quantum number is
wavelength is 4 and 6.
(a) l  l  1 (b) l + 2 (a) A (b) B
(c) 2l + 1 (d) 4l + 2 (c) C (d) D
STRUCTURE OF AN ATOM 51

27. Assertion (A): It is impossible to determine the exact 29. The wavelength of light emitted in the visible region by
position and exact momentum of an electron He+ ions after collisions with H atoms is :
simultaneously. (a) 6.5 × 10–7 m (b) 5.6 × 10–7 m
–7
Reason (R): The path of an electron in an atom is (c) 4.8 × 10 m (d) 4.0 × 10–7 m
clearly defined. 30. The ratio of the potential energy of the n = 2 electron
(a) A (b) B for the H atom to that of He+ ion is :
(c) C (d) D (a) 1/4 (b) ½
(c) 1 (d) 2
Comprehension based questions

In a mixture of H – He+ gas (He+ is singly ionized He

atom), H atoms and He+ ions are excited to their
respective first excited states. Subsequently, H atoms
transfer their total excitation energy to He+ ions (by
collisions). Assuming that the Bohr model of the atom
is applicable, answer the following questions.
28. The quantum number, n of the state finally populated in
He+ ions is:
(a) 2 (b) 3
(c) 4 (d) 5
52 STRUCTURE OF AN ATOM

Notes:

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

STRUCTURE OF AN ATOM
 02
PERIODIC PROPERTIES
Chapter 02

Periodic Properties
1. Introduction Table 2.2: Newland’s Octaves

Periodic table may be defined as the table which classifies all the
known elements in accordance with their properties in such a way
that elements with similar properties are grouped together in the
same vertical column and dissimilar elements are separated from
one another.

2. Historical Development of the Drawback or Limitation of Newland Octave Rule:

Periodic Table
(a) This rule is valid only up to Ca and fails in the case of heavier
elements.
All earlier attempts of the classification of the elements were based (b) When noble gas elements were discovered at a later stage,
upon their atomic weights. their inclusion in these octaves disturbed the entire
arrangement.

2.1 Dobereiner’s Triads 2.3 Lother Meyer’s Curve

In 1829, Dobereiner classified certain elements in the groups of
Meyer presented the classification of elements in the form of a
three called triads. The three elements in a triad had similar curve between atomic volume and atomic masses and stated that
chemical properties. When the elements in a triad were arranged the properties of the elements are the periodic functions of their
in the order of increasing atomic weights, the atomic weight of the atomic volumes.
middle element was found to be approximately equal to the He concluded that “Physical properties of elements are periodic
arithmetic mean of the other two elements. functions of their atomic masses.”

Table 2.1: Dobereiner’s Triads

Drawback or Limitation of Dobereiner’s Triads:

Dobereiner could not arrange all the elements known at that time
into triads. He could identify only three such triads that have Fig. 2.1: Lother Meyer’s curve
been mentioned.

2.4 Mendeleev’s Periodic Law

2.2 Newland’s Law of Octaves Mendeleev arranged elements in horizontal rows and vertical
In 1865, an English chemist, John Alexander Newlands observed columns of a table in order of their increasing atomic weights.
that When the lighter elements were arranged in order of their
increasing atomic weights, the properties of every eighth element “The physical and chemical properties of the elements are a periodic
were similar to those of the first one like the eighth note of a musical function of their atomic masses.”
scale. This generalization was named as Newlands’s law of octaves.

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Characteristics of Mendeleev’s Periodic Table - Corrections were done in atomic wt. of elements - U, Be, In,
(a) It is based on atomic weight. Au, Pt.

(b) 63 elements were known, noble gases were not discovered. Demerits of Mendeleev’s periodic table -

(c) He was the first scientist to classify the elements in (a) Position of hydrogen - Hydrogen resembles both, the alkali
systematic manner i.e. in horizontal rows and in vertical metals (IA) and the halogens (VIIA) in properties so
columns. Mendeleev could not decide where to place it.

(d) Horizontal rows were called periods and there were 7 periods (b) Position of isotopes- As atomic wt. of isotopes differs, they
in mendeleev’s Periodic table. have been placed in different position in Mendeleev’s
periodic table. But there were no such places for isotopes in
(e) Vertical columns were called groups and there were 8 groups Mendeleev’s table.
in mendeleev’s Periodic table.
(c) Anomalous pairs of elements - There were some pair of
(f) Each group upto VII is divided into A & B subgroups ‘A’ sub elements which did not follow the increasing order of atomic
groups element are called nomal elements and ‘B’ sub groups weight.
elements are called transition elements.
For example, Ar (atomic weight = 39.9) precedes K (atomic
(g) The VIII group consisted of 9 elements in rows (Transitional weight = 39.1) and similarly Co (atomic weight = 58.9) has
metals group). been placed ahead of Ni (atomic weight = 58.7).
(h) The elements belonging to same group exhibit similar (d) Unlike elements were placed in same group -
properties.

Merits or advantages of Mendeleev’s periodic table -

(a) Study of elements - First time all known elements were classified
in groups according to their similar properties. So study of
the properties become easier of elements.
(b) Prediction of new elements - It gave encouragement to the
discovery of new elements as there were some gaps in it. Sc
(Scandium), Ga (Gallium), Ge (Germanium), Tc (Technetium)
were the elements for whom position and properties were
defined by Mendeleev even before their discoveries and he More reactive Less reactive
left the blank spaces for them in table.
Alkali metal Coin metal
e.g.-Blank space at atomic wt. 72 in silicon group was called
Eka silicon (means properties like silicon ) and element Normal elements Transition element
discovered later was named Germanium. • Cu, Ag and Au placed in I st group along with Na, K etc.
Similarly other elements discovered after mendeleev periodic While they differ in their properties.
table were: (Only similar in having ns1 electronic configuration)
Eka aluminium - Gallium (Ga), Eka Boron - Scandium (Sc)
Eka Silicon - Germanium (Ge), Eka Manganese - Technetium Table 2.3: Mendeleev’s Predictions for the
(Tc) Elements
(c) Correction of doubtful atomic weights - Correction were done
in atomic wt. of some elements.
Atomic Wt.=Valency  Equivalent weight.
Initially, it was found that equivalent wt. of Be is 4.5 and it is
trivalent (V=3), so the weight of Be was 13.5 and there is no
space in Mendeleev’s table for this element. But, after
correction, it was found that Be is actually divalent (V = 2).
So, the weight of Be became 2  4.5=9 and there was a space
between Li and B for this element in Mendeleev’s table.

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56 PERIODIC PROPERTIES

2.5 Modern Periodic Law (a) Position of isotopes - Atomic No. of isotopes are similar, so,
different isotopes can be placed at same place in periodic
In 1913, the English physicist, Henry Moseley observed table.
regularities in the characteristic X-ray spectra of the elements. A
(b) (Ar-K) (Co-Ni) (Te-I) are now in increasing order of atomic
plot of v (where v is frequency of X-rays emitted) against atomic number.
number (Z) gave a straight line and not the plot of v vs atomic (c) Lanthanides and actinides are in IIIB group.
mass. (d) In modern periodic table diagonal line separates out metals,
Mendeleev’s Periodic Law was, therefore, accordingly modified. metalloids and non metals.
This is known as the Modern Periodic Law and can be stated as: (e) Elements of same group have same general formula of
electronic configuration of outer most shell.
The physical and chemical properties of the elements are periodic
functions of their atomic numbers. Demerits of long form of periodic table -
Characteristics of Modern Periodic Table - (a) Position of hydrogen is still controversial.
(a) 18 vertical columns called groups. (b) ‘He’ is a inert gas but it has different electronic configuration
than other inert gas elements.
st
(b) I to VIII group +0 group of inert gases. (c) Lanthanides and actinides are still not placed in main frame.
(c) Inert gases were introduced in periodic table by Ramsay.
(d) Isotopes have different physical properties but have same
(d) 7 horizontal series called periods. place in periodic table.
Merits of long form of periodic gtable -

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PERIODIC PROPERTIES 57

Fig. 2.2: Periodic Table of Elements

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58 PERIODIC PROPERTIES

Table 2.4: Nomenclature of Elements with Atomic Number Above 100

Atomic Name according to Symbol IUPAC IUPAC

Number IUPAC nomenclature Official Name Symbol
101 Unnilunium Unu Mendelevium Md
102 Unnilbium Unb Nobelium No
103 Unniltrium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordium Rf
105 Unnilpentium Unp Dubnium Db
106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
110 Ununnillium Uun Darmstadtium Ds
111 Unununnium Uuu Rontgenium Rg
112 Ununbium Uub Copernicium Cn
113 Ununtrium Uut Nihonium Nn
114 Ununquadium Uuq Flerovium Fl
115 Ununpentium Uup Moscovium Mc
116 Ununhexium Uuh Livermorium Lv
117 Ununseptium Uus Tennessine Ts
118 Ununoctium Uuo Oganesson Og

Nomenclature of elements with Atomic Numbers > 100 3. Prediction of Block, Period &
The naming of the new elements had been traditionally the privilege Group
of the discoverer and the suggested name was ratified by the
IUPAC. 1. Block - last electron enters into which orbital
2. Period - Max value of principal quantum number
Table 2.5: Notation for IUPAC Nomenclature of
Elements 3. Group - s block - No. of valence electron
p block - No. of valence electron + 10
Digit Name Abbreviation
0 nil n d block – ns + No. of (n – 1) d electron
1 un u f block – III B
2 bi b
3 tri t
4 quad q
4. Properties of an Element
5 pent p 4.1 Atomic Radius
6 hex h
We cannot measure the exact size of an isolated atom because its
7 sept s
outermost electron have a remote chance of being found quite far
8 oct o from the nucleus. So different types of atomic radius can be used
based on the environment of atoms i.e; covalent radius, van der Waals’
radius, metallic radius.

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PERIODIC PROPERTIES 59

4.1.1 Covalent Radius 4.1.3 Metallic Radius (Crystal radius)

The half of the distance between the nuclei of two identical atoms It is one-half of the distance between the nuclei of two adjacent
joined by single covalent bond in a molecule is known as covalent metal atoms in the metallic crystal lattice.
radius.

Fig. 2.3: Representation of Covalent Radius

So covalent radius for A-A Fig. 2.5: Representation of Metallic Radius

So metallic radius for A-A
d
rA  A A d = rA + rA
2
If covalent bond is formed between two different elements then d
rA 
2
d A B  rA  rB  0.09 ( A   B )
* rcovalent < rmetallic < rvanderwaals
where A and B are electronegativity of A and B

4.2 Variation of Atomic Radii in the

4.1.2 Vander Waal’s Radius
It is half of the internuclear distance between adjacent atoms of Periodic Table
the two neighbouring molecules in the solid state. (a) Variation along a period
In general, the covalent and van der Waals radii decrease with
increase in atomic number as we move from left to right in a period.
It is because with in the period the outer electrons are in the same
valence shell & the effective nuclear charge increases as the
atomic number increases resulting in the increased attraction of
electrons to the nucleus.

rcovalent rvander Waal’s Table 2.6: Atomic Radii across the period
Fig. 2.4: Representation of Vander Waal’s Radius

d A A
rvander waal 
2

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60 PERIODIC PROPERTIES

(c) Ionic Radius

The removal of an electron from an atom results in the formation of a
cation, whereas gain of an electron leads to an anion.
In general, the ionic radii of elements exhibit the same trend as the
atomic radii. A cation is smaller than its parent atom because it
has fewer electrons while its nuclear charge remains the same.
The size of an anion will be larger than that of the parent atom
because the addition of one or more electrons would result in
increased repulsion among the electrons and a decrease in
effective nuclear charge. For example, the ionic radius of fluoride
ion (F–) is 136 pm whereas the atomic radius of fluorine is only 64
pm. On the other hand ,the atomic radius of sodium is 186 pm
compared to the ionic radius of 95 pm for Na+.
(d) Isoelectronic Species

Fig. 2.6: Variation of atomic radius with atomic number Isoelectronic species are those which have same number of
electrons. For example, O2–, F–, Na+ and Mg2+ have the same number
across the second period
of electrons (10). Their radii would be different because of their
(b) Variation along a group different nuclear charges. The cation with the greater positive
Atomic radius in a group increase as the atomic number increases. charge will have a smaller radius because of the greater attraction
It is because with in the group, the principal quantum number (n) of the electrons to the nucleus. Anion with the greater negative
increases and the valence electrons are farther from the nucleus. charge will have the larger radius. In this case, the net repulsion of
the electrons will outweigh the nuclear charge and the ion will
expand in size.
Table 2.7: Atomic Radii down the group
Order of atomic radii is
Mg2+ < Na+ < F– < O2–
General Trend :

4.3 Ionization Energy

The minimum amount of energy required to remove the electron
from the outermost orbit of an isolated atom in the gaseous state
is known as ionization energy.
IE (First Ionization Energy)
2+ IE IE
1  M + 
M  2
–e–  M 
3
–e– 
IE4
M3+ 
–e–  M
4+

IE1, IE2, IE3 and IE4 are successive ionization energies.

Fig. 2.7: Variation of atomic radius with atomic number
for alkali metals and halogens

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PERIODIC PROPERTIES 61

= 96.49 kJ mol–1
IE 4  IE3  IE 2  IE1
or  i H 4  i H3  i H 2   i H1  1 electron volt (eV) per atom = 23.06 kcal mol–1
= 96.49 kJ mol–1
Variation of Ionisation Energy in Periodic Table Important Points
(a) Variation along a period  Ionization energy increases with decreasing the size of an
In a period, the value of ionisation enthalpy increases from left to atom or an ion
right which breaks where the atoms have some-what stable
 Ionization energy increases with decreasing screening effect.
configurations. The observed trends can be easily explained on
the basis of increased nuclear charge and decrease in atomic radii.  Ionization energy increases with increasing nuclear charge
Both the factors increase the force of attraction towards nucleus  Ionization energy increases if atom having half filled and fully
and consequently, more and more energy is required to remove filled orbitals
the electrons and hence, ionisation enthalpies increase.
 The penetrating power of orbitals is in the order
(b) Variation along a group s>p>d>f
On moving down the group, the atomic size increases gradually Increases
due to an addition of one new principal energy shell at each
Period
succeeding element. On account of this, the force of attraction
towards the valence electrons decreases and hence the ionisation

Decreases
Io
enthalpy value decreases. n
en izat
erg ion
y

Group

Application of ionisation potential:

(a) Metallic and non metallic character:
Metallic  I.P. Low (Na, K, Rb etc.)
Non metallic  I.P.High (F, Cl, Br etc.)
(b) Reactivity:

1
Fig. 2.8: Variation of first ionization enthalpies (iH) with atomic Reducing character 
Ionisation potential
number for elements with Z = 1 to 60

(a) IA group has minimum I.P. so they are strong reducing agents
in gaseous state (Li < Na < K < Rb < Cs )
4.4 Units of I.E./I.P. (b) VIIA group has maximum I.P. so they are strong oxidising
It is measured in units of electron volts (eV) per atom or kilo agents (F < Cl > Br > I )
calories per mole (kcal mol–1) or kilo Joules per mole (kJ mol–1). One
electron volt is the energy acquired by an electron while moving (c) Stability of oxidation states:
under a potential difference of one volt. (a) If the difference between two successive I.P.  16eV then
lower oxidation state is stable.
1 electron volt (eV) per atom = 3.83 × 10–20 cal per atom
= 1.602 × 10–19 J per atom (1 cal = 4.184 J) Na  Na  IP1 
= 3.83 × 10–20 23
× 6.023 × 10 cal mol –1 e.g.  2  42.7 eV
Na  Na IP2 
–1
= 23.06 kcal mol
So Na+ is stable.
–19 23 –1
= 1.602 × 10 × 6.023 × 10 J mol

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62 PERIODIC PROPERTIES
(b) If the difference between two successive I.P.  11eV, higher (b) Variation along a group
oxidation state is stable. The electron gain enthalpies, in general, become less negative in
going down from top to bottom in a group. This is due to increase
Mg  Mg  IP1  in size on moving down a group. This factor is predominant in
e.g  2  7.4 eV
Mg  Mg IP2  comparison to other factor, i.e., increase in nuclear charge.

So Mg2 is stable. Table 2.8: Electron gain enthalpies (in kJ/mol)

Al  Al  
 2 
12.8eVSo Al  is stable
Al  Al 

Al2+  Al3+} 6.0eV So Al3+ is stable

 Al  is stable only in gaseous state

 3 Increases
 Al is stable in liquid and solid state
Period

Electron Gain Enthalpy: El

Decreases
e
af ctro
fin n
When an electron is added to a neutral gaseous atom (X) to convert ity
it into a negative ion, the enthalpy change accompanying the
process is defined as the Electron Gain Enthalpy (egH). Electron Group
gain enthalpy provides a measure of the ease with which an atom
adds an electron to form anion as represented by
Factors Affecting Electron Affinity (E.A.):
X(g) + e–  X–(g)
1
Depending on the element, the process of adding an electron to (a) Atomic size  EA  Atomic size
the atom can be either endothermic or exothermic. For many
elements energy is released when an electron is added to the atom
and the electron gain enthalpy is negative. For example, group 17 1
(b) Screening effect  EA  Screening effect
elements (the halogens) have very high negative electron gain
enthalpies because they can attain stable noble gas electronic
configurations by picking up an electron. On the other hand, noble (c) Effective nuclear charge (Zeff )  EA  Zeff
gases have large positive electron gain enthalpies because the (d) Stability of completely filled or half filled orbitals 
electron has to enter the next higher principal quantum level leading
to a very unstable electronic configuration. Electron affinity of filled or half filled orbital is very less or zero as
energy is given to introduce any electron. It is because of its
Variation of Electron Gain Enthalpy: stability.
(a) Variation along a period Applications of Electron Affinity:
Electron gain enthalpy becomes more and more negative from left (a) Electron affinity  Oxidising nature
to right in a period. This is due to decrease in size and increase in
But F has more oxidising power than Cl because F has more
nuclear charge as the atomic number increases in a period. Both
standard reduction potential.
these factors favour the addition of an extra electron due to higher
force of attraction by the nucleus for the incoming electron. (b) Electron affinity  Reactivity

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 They form anions by gaining electron (c) Hybridisation state of an atom -

 Their bond nature is ionic Electronegativity  % of s-character in hybridised atom

(c) Electron affinity electronegativity sp > sp2 > sp3

(d) Elements of high electron affinity form oxide and hydroxides, s-character 50% 33% 25%
which are acidic in nature
Electronegativity 3.25 2.75 2.5
4.5 Electronegativity Because s-orbital is near so by increasing s-character in
The tendency of an atom to attract the shared pair of electrons hybridisation state, EN also increases.
towards itself is known as its electronegativity. (d) Positive oxidation state -
According to Pauling, the electronegativity of F is 4.0 and Electronegativity  Positive oxidation state
electronegativity of other elements can be calculated as
Mn 2  Mn 4  Mn 7
1/2 1/2
(A – B) = 0.208 [EA–B – (EA–A × EB–B) ]
O  O  O 
According to Mulliken
Fe  Fe2  Fe3
IP  EA • As atomic radius decreases by increasing oxidation state of
Electronegativity 
2 cation species, EN increases.

(where IP = Ionization potential, EA = Electron affinity) • In anionic species, the order of electronegativity is
If IP and EA are taken in electron volt O2  O  O
 Percentage ionic character = 16 (A – B) + 3.5 (A – B)2 where (e) Electronegativity does not depends on filled or half filled
A and B are electronegativities of A and B. orbitals, because it is a tendency to attract bonded electron,
not to gain electron from outside.
• If the difference in the electronegativities of combining atoms
is 1.7, the bond is 50% covalent and 50% ionic.
Table 2.9: Electronegativity values across the period
• If the difference in electronegativities of oxygen and element
is very high, the oxide shows a basic character.
Increases

Period
El
ec
tro
ne
Decreases

ga
tiv
ity
Table 2.10: Electronegativity values down the group

Group

The periodic trends of elements in the periodic table

Factors Affecting electronegativity :
(a) Atomic size -
1
Electronegativity 
Atomic size
(b) Effective nuclear charge (Zeff ) 

Electronegativity  Zeff

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Application of electronegativity : X B  Electronegativity of B
(A) Metallic and non metallic nature -
If X A  X B  2.1 Ionic %  50% i.e. Ionic bond
Low electronegativity  Metals
If X A  X B  2.1 Ionic %  50% i.e. Covalent bond
High electronegativity  Non metals
(b) Gallis experimental values are-
Metallic character increases down the group but decreases along
a period. X A  X B  1.7 Ionic

(B) Bond length - X A  X B  1.7 Covalent

1  If X A  X B ; then A  B will be non polar, e.g. H  H, F  F

EN 
Bond length
If X A  X B and difference of EN is small then
Here EN  difference in electronegativities of bonded atoms A   B  bond will be polar covalent
HF < HCl < HBr < HI
e.g. H 2 O (H   O   H   )
 HF has minimum bond length because of much difference in
the electronegativities of H and F.  If X A  X B and X A  X B EN is high then,

(C) Bond energy - By increasing EN , bond length decreases A  Β bond will be polar or ionic
and hence bond energy increases.
e.g. Na  Cl
Bond energy  Electronegativity difference
 In HF, X A  X B  1.9, which is more than 1.7, even then it is
HF > HCl > HBr > HI covalent compound.
(D) Reactivity - (F) Nature of oxides - Consider an oxide AO

1 If X A  X O  2.3 Basic oxide

Bond energy  Stability 
Reactivity
If X A  X O  2.3 Amphoteric oxide
As bond energy  difference of electronegativities
If X A  X O  2.3 Acidic oxide
1 (a) Along the period, acidic nature increases.
So,  Electronegativity  Stability  Reactivity
(b) Down the group, basic nature increases.
HF HCl HBr HI
 
Reactivity

 HI is most reactive hydrohalides or strongest acid among all

hydrohalides.
(E) Nature of bonds -
(a) It can be determined by Hannay & Smith formula- i.e. when in periodic table the distance bewteen the element and
oxygen increases, basic character increases.
Ionic %  16 (X A  X B )  3.5 (X A  X B ) 2
NO 2  ZnO  K 2 O
Here X A  Electronegativity of A

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
acidic character decreases
 BeO, Al 2 O3 , ZnO,SnO, PbO,SnO 2 , PbO 2 ,Sb 2 O3
etc. are amphoteric oxides.

 CO, H 2 O, NO, N 2 O are neutral oxides.

PCl3  3HOH  3ΗCl  H 3 PO3
Acidic strength of oxide and oxyacid
B2O3 CO2 N 2O 5 PCl3  Cl2  H 2 O  2ΗCl  POCl3

EN increase, acidic nature increase

POCl3  3HOH  3ΗCl  H 3 PO 4
H 3 BO3  H 2 CO3  HNO3 (b) If electronegativity of X < electronegativity of A then on
hydrolysis product will be HOX
(hypohalous acid)

e.g. Cl 2 O
H 3 PO4  H 3 AsO4  H 3SbO 4

H 2 SO 4  H 2 SeO 4  H 2 TeO 4

HOF  HOCl  HOBr  HOI

HClO4  HBrO 4  HIO 4 So On hydrolysis

N 2 O5  P2 O5  As 2 O5

N 2 O3  P2 O3  As 2 O3  Sb 2 O3

Acidic nature  oxidation state

Acidic properties increases with increasing oxidation state of an
element

HClO 4  HClO3  HClO2  HClO

4.6 Periodic Trends in Chemical Properties

HNO3  HNO 2

H 2SO 4  H 2SO3

SO3  N 2 O 3

Sb 2 O3  Sb 2 O5  N 2 O  NO  NO 2  N 2 O5

(I) Hydrolysis of AX - Where A  Other element and

X  Halogen
(a) If electronegativity of X > Electronegativity of A then on
hydrolysis product will be HX.

 For example (BCl3 ), EN of Cl  EN of B

Fig. 2.9: Periodic trends in various chemical properties

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66 PERIODIC PROPERTIES

4.6.1 Periodicity of Valence or Oxidation States may be defined as the charge acquired by its atom on the basis of
The electrons present in the outermost shell of an atom are called electronegativity of the other atoms in the molecule.
valence electrons and the number of these electrons determine (b) Variation within a group
the valence or the valency of the atom. It is because of this reason
When we move down the group, the number of valence electrons
that the outermost shell is also called the valence shell of the
remains the same, therefore, all the elements in a group exhibit the
atom and the orbitals present in the valence shell are called valence
same valence. For example, all the elements of group 1 (alkali metals)
orbitals.
have valence one while all the elements of group 2 (alkaline earth
In case of representative elements, the valence of an atom is metals) exhibit a valence of two.
generally equal to either the number of valence electrons (s- and
Noble gases present in group 18 are zerovalent, i.e., their valence
p-block elements) or equal to eight minus the number of valence
is zero since these elements are chemically inert.
electrons.
4.6.2 Anomalous Properties of Second Period
Elements
Table 2.11: Valence of atoms of various groups
It has been observed that some elements of the second period
show similarities with the elements of the third period placed
diagonally to each other, though belonging to different groups.
For example, lithium (of group 1) resembles magnesium (of group
2) and beryllium (of group 2) resembles aluminium (of group 13)
and so on. This similarity in properties of elements placed
In contrast, transition and inner transition elements, exhibit variable diagonally to each other is called diagonal relationship.
valence due to involvement of not only the valence electrons but
d- or f-electrons as well. However, their most common valence are Table 2.13: Elements showing Diagonal relationship
2 and 3.

Table 2.12: Periodic Trends in Valence of Elements

The anomalous behaviour is due to their small size, large charge/

radius ratio and high electronegativity of the elements. In addition,
the first member of group has only four valence orbitals (2s and
2p) available for bonding, whereas the second member of the
groups have nine valence orbitals (3s, 3p, 3d). As a consequence
of this, the maximum covalency of the first member of each group
is 4 (e.g., boron can only form [BF4]–), whereas the other members
Let us now discuss periodicity of valence along a period and of the groups can expand their valence shell to accommodate
within a group. more than four pairs of electrons e.g., aluminium forms [AlF6]3–.
(a) Variation along a period Furthermore, the first member of p-block elements displays greater
ability to form p-p multiple bonds to itself (e.g., C=C, CC, NN)
As we move across a period from left to right, the number of
and to other second period elements (e.g., C=O, CN, N=O)
valence electrons increases from 1 to 8. But the valence of elements,
compared to subsequent members of the same groups.
w.r.t. H or O first increases from 1 to 4 and then decreases to zero.
In the formation of Na 2O molecule, oxygen being more
electronegative accepts two electrons, one from each of the two
sodium atoms and thus shows an oxidation state of –2. On the
other hand, sodium with valence shell electronic configuration as
3s1 loses one electron to oxygen and is given an oxidation state of
+1. Thus, the oxidation state of an element in a given compound

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PERIODIC PROPERTIES 67
4.6.3 Periodic Trends and Chemical Reactivity The reactivity of non-metals increases from left to right in a period.
During reaction, non-metals tend to form anions.
Reactivity of Metals For example, in the second period, the reactivity of non- metals
increases in the order.
The reactivity of metals is measured in terms of their tendency to
lose electrons from their outermost shell.
C <N <O <F
In a period Less reactive More reactive
The tendency of an element to lose electrons decreases in going –— Reactivity increases 
from left to right in a period. So, the reactivity of metals decreases In a group
in a period from left to right. For example, the reactivity of third
The reactivity of non-metals in a group decreases as we go down
period elements follows the order.
the group. This is because the tendency to accept electrons
Na > Mg > Al decreases down the group. The reactivity of halogens follows
More reactive Reactive the order
In a group
F < Cl < Br I
The tendency to lose electrons increases as we go down a group. Most reactive Least reactive
So, the reactivity of metals increases down the group. Thus, in –— Reactivity decreases  
group 1, the reactivity follows the order.
The normal oxide formed by the element on extreme left is the
Li < Na < K < Rb < Cs most basic (e.g., Na2O) whereas that formed by the element on
Least reactive Most reactive extreme right is the most acidic (e.g., Cl2O7). Oxides of elements in
–— Reactivity increases 
 the centre are amphoteric (e.g., Al2O3, As2O3) or neutral (e.g., CO,
NO, N2O). Amphoteric oxides behave as acidic with bases and as
Reactivity of Non-Metals
basic with acids, whereas neutral oxides have no acidic or basic
The reactivity of a non-metal is measured in terms of its properties.
tendency to gain electrons to form an anion.
In a period

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Some Important Facts about
Elements:
 Bromine is a non-metal which is liquid at room temperature.  Iodine is the element which sublimes.
 Mercury is the only metal that is liquid at room temperature.  Hydrogen is the most abundant element in the universe.
 Gallium (m.pt. 29.8ºC), caesium (m.pt. 28.5ºC) and francium  Only ozone is the coloured gas with garlic smell.
(m.pt. 27ºC) are metals having low melting points.  Metalloids have electronegativity values closer to 2.0.
 Tungsten (W) has the highest melting point (3380ºC) among  First synthetic (i.e., man-made) element is technetium
metals. (At. No. 43).
 Carbon has the highest melting point (4100ºC) among non-  Most poisonous metal-Plutonium.
metals.  Rarest element in earth’s crust-Astatine.
 Oxygen is the most abundant element on the earth.  The elements coming after uranium are called transuranic
 Aluminium is the most abundant metal. elements. The elements with Z = 104 – 112, 114 and 116 are
called trans-actinides or super heavy elements. All these
 Iron is the most abundant transition metal.
elements are synthetic, i.e., man-made elements. These are
 Highest density is shown by osmium (22.57 g cm–3) or iridium radioactive elements and not found in nature.
(22.61 g cm–3).  The elements ruthenium (Ru), germanium (Ge), polonium (Po)
 Lithium is the lightest metal. Its density is 0.54 g cm–3. and americium (Am) were named in honour of the countries
 Silver is the best conductor of electricity. named Ruthenia (Russia), Germany, Poland and America,
respectively.
 Diamond (carbon) is the hardest natural substance.
 The members of the actinide series are radioactive and majority
 Francium has the highest atomic volume. of them are not found in nature.
 Boron has the lowest atomic volume.
 The element rutherfordium (Rf, 104) is also called
 The most abundant gas in atmosphere is nitrogen. Kurchatovium (Ku) and element dubnium (Db, 105), is also
 Fluorine is the most electronegative element. called hahnium.
 Chlorine has the maximum negative electron gain enthalpy.  Promethium (Pm, 61) a member of lanthanide series is not
 Helium has the maximum ionisation enthalpy. found in nature. It is a synthetic element.
 Cesium or francium has the lowest ionisation enthalpy.  Special names are given to the members of these groups in
 Helium and francium are smallest and largest atoms respectively. periodic table.
 H– and I– ions are the smallest and largest anions respectively. Group 1 or IA Alkali metals
 H+ and Cs+ ions are the smallest and largest cations respectively. Group 2 or IIA Alkaline earth metals
 Cesium is the most electropositive element. Group 15 or VA Pnicogens
 Element kept in water is phosphorus, P4 (white or yellow). Group 16 or VIA Chalcogens
 Element kept in kerosene are Na, K, Rb, Cs, etc. Group 17 or VIIA Halogens
Group 18 or VIIIA Inert or noble gases
(zero)

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Summary

 Mendeleev’s periodic table was based on atomic masses of of similar electronic configuration of the atom in the valence
the elements. When Mendeleev presented the periodic table, shell after certain definite intervals. These definite intervals
only 63 elements were known. He left 29 places in the table are 2, 8, 8, 18, 18 and 32. These are known as magic number.
for unknown elements. Periodicity is observed in a number of properties which are
directly or indirectly linked with electronic configuration.
 Modern Mendeleev periodic table is based on atomic numbers
of the elements. The modern periodic law is : “The physical  Effective nuclear charge increases across each period.
and chemical properties of the elements are periodic function  Atomic radii generally decrease across the periods.
of their atomic numbers”.  Atomic radii generally increase on moving from top to bottom
 The horizontal row in the periodic table is called a period and in the groups.
vertical column is called group. There are seven periods and  Atomic radius is of three types :
eight groups in the modern Mendeleev periodic table. (a) Covalent radius
 The long or extended form of periodic table consists of seven (b) Crystal or metallic radius
periods and eighteen vertical columns (groups or families).
(c) Van der Waals’ radius
The elements in a period have same number of energy shells,
i.e., principal quantum number (n). These are numbered 1 to 7.  Cations are generally smaller than anions.
 Cations are smaller and anions are larger than neutral atoms
 At present 114 elements are known.
of the elements.
 In a vertical column (group), the elements have similar valence
shell electronic configuration and therefore exhibit similar
chemical properties.
 The first element is each group of the representative elements
 There are four blocks of elements: s-, p-, d- and f-block shows abnormal properties, i.e., differs from other elements of
depending on the orbital which gets the last electron. The the group because of much smaller size of the atom.
general electronic configuration of these blocks are :
 The ions having same number of electrons but different
s-block : [Noble gas] ns1 or 2. nuclear charge are called isoelectronic ions.
p-block : [Noble gas] ns2np1–6 In isoelectronic ions, the size decreases if Z/e increases i.e.,
d-block : [Noble gas] (n – 1)d 1–10 1 or 2
ns greater the nuclear charge, smaller is the size of the ion.
f-block : [Noble gas] (n – 2)f 1–14
(n – 1)d0 or 1ns2  The energy required to remove the most loosely held electron
from the gaseous isolated atom is termed ionisation enthalpy.
 The elements are broadly divided into three types :
(i) Metals comprise more than 78% of the known elements.  Ionisation enthalpy values generally increase across the periods.
s-block, d-block and f-block elements are metals. The higher  Ionisation enthalpy values generally decrease down the group.
members of p-block are also metals.  Removal of electron from filled and half filled shells requires
(ii) Non-metals are less than twenty. (C, N, P, O, S, Se, H, F, Cl, higher energy. For example, the ionisation enthalpy of nitrogen
Br, I, He, Ne, Ar, Kr, Xe and Rn are non-metals). is higher than oxygen. Be, Mg and noble gases have high
(iii) Elements which lie in the border line between metals and values.
non-metals are called semimetals or metalloids. B, Si, Ge, As,  Metals have low ionisation enthalpy values while non-metals
Sb, Te, Po and At are regarded as metalloids. have high ionisation enthalpy values.
 IUPAC given a new scheme for assigning a temporary name
 Successive ionisation enthalpies of an atom have higher values.
to the newly discovered elements. The name is derived directly
from the atomic number of the elements. IEI < IEII < IEIII ...
 The recurrence of similar properties of the elements after  The enthalpy change taking place when an electron is added
certain definite intervals when the elements are arranged in to an isolated gaseous atom of the element is called electron
order of increasing atomic numbers in the periodic table is gain enthalpy. The first electron gain enthalpy of most of the
termed periodicity. The cause of periodicity is the repetition elements is negative as energy is released in the process but

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the values are positive or near zero in case of the atoms having  Metals have low electronegativities and non-metals have
stable configuration such as Be, Mg, N, noble gases, etc. high electronegativities.
 Electron gain enthalpy becomes more negative from left to right in  Metallic character decreases across the periods and
a period and less negative from top to bottom in a group. increases down the group.
 Successive electron gain enthalpies are always positive.  Valence of an element belonging to s- and p- block (except
 The elements with higher ionisation enthalpy have higher noble gases) is either equal to the number of valence
negative electron gain enthalpy. electrons or eight minus number of valence electrons.

 Electronegativity is the tendency of an atom to attract the  The reducing nature of the elements decreases across the
shared pair of electrons towards itself in a bond. period while oxidising nature increases.

 Electronegativity increases across the periods and decreases  The basic character of the oxides decreases while the acidic
down the groups. character increases in moving from left to right in a period.

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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.
Sol. According to the Mendeleev’s Periodic Law, “the (b) Give the general electronic configuration of s, p, d & f-
properties of elements are a periodic function of their atomic block elements.
mass”. Mendeleev arranged known 63 elements in Sol. (a) The physical and chemical properties of the elements are
horizontal rows and vertical columns of a table. Mendeleev periodic functions of their atomic numbers. The main features
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
corresponds to the highest principal quantum number (n) of
showed similar properties with fluorine. The primary aim
the elements in the period.
of Mendeleev was to arrange the elements with similar
4. The first period contains 2 elements. The subsequent
properties in the same group periods consists of 8, 8, 18, 18 and 32 elements, respectively.
The seventh period is incomplete and like the sixth period
Example-2 would have a theoretical maximum (on the basis of quantum
What is the basic theme of organisation in the periodic numbers) of 32 elements.
table. 5. In this form of the Periodic Table, 14 elements of both sixth
and seventh periods (lanthanoids and actinoids,
Sol. The elementary idea of organization in the periodic table is respectively) are placed in separate panels at the bottom.
to classify the elements into groups and periods. This table
is arranged according to the properties of the elements. 6. Elements having similar outer electronic configurations in their
Same properties elements belong to same group. It is very atoms are arranged in vertical columns referred to as groups or
easy and systematic method to learn the properties of families. There are in all 18 vertical column or groups.
elements. 7. The elements of groups 1 (alkali metals), 2 (alkaline earth
metals) and 13 to 17 are called the main group elements.
Example-3 These are also called typical or representative or normal
What is the basic difference in approach between the elements.
Mendeleev’s Periodic Law and the Modern Periodic Law? 8. The elements of group 3 to 12 are called transiation elements.
Sol. According to the Mendeleev’s Periodic Law, “the 9. lanthanoids & actinoids are together referred to as inner
properties of elements are a periodic function of their atomic transition elements.
mass”. Mendeleev arranged known 63 elements in (b) (i) General outer electronic configuration of s-block elements
horizontal rows and vertical columns of a table on the other is ns1–2 i.e., either ns1 or ns2.

hand, the Modern periodic Law states that “the physical (ii) General outer electronic configuration of p-block elements
is ns2np1–6.
and chemical properties of elements are periodic functions
(iii) General outer electronic configuration of d-block elements
of their atomic numbers”. This modern law was proposed
is (n – 1) d1–10 ns0–2.
by Henry Moseley.
(iv) General outer electronic configuration of f-block elements is
(n – 2) f1–14 (n – 1) d0–1 ns2.
72 PERIODIC 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 ?
Sol. Sixth period corresponds to the filling of the sixth energy Sol. The electronic configuration of the element with atomic
level i.e; n=6. Since in this period only sixteen orbitals (one number 43 is
6s, seven 4f, five 5d and three 6p)are available, thereby 1s2, 2s22p6, 3s23p63d10, 4s24p64d5, 5s2
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
The element belongs to group 7.
seven 4f orbital begins with cerium and end up with
Example-10
lutetium. Maximum 2 electrons can be accommodated in
How does atomic radius vary in a period and in a group?
each orbital. Thus, 16 orbitals can accommodate a maximum
How do you explain the variation?
of 32 electrons. Therefore, the sixth period of the periodic
table should have 32 elements. Sol. Atomic Radius: It is defined as average distance from the
centre of nucleus upto the centre of outermost shell
Example-6
Write the atomic number of the element present in the electrons. It is measured in angstrom or picometer. It is not
third period and seventeenth group of the periodic possible to measure the exact atomic radius because an
table. atom is unstable, and it can’t be isolated to get its radius
Sol. In the periodic table, each period starts with alkali metal and electron cloud around the atom do not have sharp
and ends with noble gases. Element in the 3rd period and boundary.
seventeenth group belongs to halogen family. Chlorine is
Trend in the periodic table:
belonging to the 3rd period and 17th group. Hence, Chlorine
(a) Across a period: Atomic radii decreases across a period
atomic number is 17.
from left to right. It is due to increase in atomic number; the
Example-7
number of electrons increases but number of shells remain
Which element do you think would have been named by same due to which electrons fill in same shell which in turn
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group? increase the nuclear force of attraction between nucleus
and electrons of outermost shell and thus atomic size
Sol. (i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with Z =
97 decreases.
(ii) Seaborgium (Sg) with Z = 106 Down the group: Atomic radii increases down in a group
Example-8 from top to bottom. It is due to increase in atomic number;
To which block (s, p, d or f) does the element with atomic the number of shells also increases in which electron in the
number 50 belong ? outer shell of each succeeding element lie farther and farther
Sol. The electronic configuration of element with atomic number away from nucleus. Due to increase in atomic number, the
50 is :
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.
PERIODIC PROPERTIES 73

Example-11 Example-13
Consider the following species: N3–, O2–, F–, Na+, Mg2+ and Define atomic radius. Explain various factors affecting it ?
Al3+
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.
Sol. (a) All species are isoelectronic. The species in which atoms Factors affecting atomic radius :
(i) No. of shells : The atomic radius increases with the increase
or ions of different elements that have same number of
in the no. of the shells.
electrons but have different magnitude of nuclear charge
atomic radius  no of shells
are known as isoelectronic species. They may be neutral or
(ii) Nuclear charge : Atomic radius decreases with the increase
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.
1
Therefore, the arrangement of the given species in order of atomic radius 
Nuclear charge
their increasing ionic radii is as follows: Al3+< Mg2+< Na+<
(iii) Shielding or screening effect : Atomic radius increases with
F–< O2–< N3–
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
reduces the force of attraction between the Nucleus and the
is Li < Na < K <Rb<Cs whereas that among group 17
Valence electrons. atomic radius  shielding effect
elements is F > Cl > Br > I. Explain.
Example-14
Sol. The group 1 belongs to the metals. The reactivity of metals
depends upon their tendency to lose electrons to acquire Among the elements Li, K, Ca, S and Kr, which one is expected
to have the lowest first ionization enthalpy and which the
the inert gas configuration. As we move down in a group of highest first ionization enthalpy ?
metal elements, the atomic size of the elements goes on Sol. K has the lowest first ionization energy. Kr has the highest
increasing. Due to increase in size of atoms the valence first ionization energy.
electrons of metal atom, which take part in chemical reactions Example-15
becomes more far away from the Among the second period elements, the actual ionization
energies are in the order :
nucleus and hence can be removed easily. Consequently,
Li < B < Be < C < O < N < F < Ne.
the increasing order of reactivity among group 1 elements
Explain why (i) Be has higher i H than B (ii) O has lower i
is as follows: Li < Na < K <Rb<Cs H than N and F ?
In the group 17th, The reactivity of non-metals depends Sol. (i) The ionization enthalpy, among other things, depends upon
upon their tendency to gain electrons to acquire the inert the type of electron to be removed from the same principal
shell. In case of Be (1 s2 2 s2) the outermost electron is present
gas configuration. As we move down in a group of non-
in 2s-orbital while in B (1 s2 2 s2 2 p1) it is present in 2p-orbital.
metal elements, the atomic size of the elements goes on Since 2s-electrons are more strongly attracted by the nucleus
increasing. Due to increase in size of atoms the distance of than 2p-electrons, therefore, lesser amount of energy is
required to knock out a 2p-electron than a 2s-electron.
nucleus of atom from the valence shell increases. The force Consequently, i H of Be is higher than that i H of B.
of attraction exerted by the nucleus of the atom on the
valence shell electrons decreases, as a result tendency to
gain electron also decreases. Consequently, the decreasing
order of reactivity among group 17 elements is as follows: F
> Cl > Br > I
74 PERIODIC PROPERTIES

(ii)The electronic configuration of N (1s2 2s2 2p1x 2p1y 2p1z) in Example-18

which 2p-orbitals are exactly half-filled is more stable than
the electronic configuration of O (1s2 2s2 2p2x 2p1y 2p1z) in The first ionisation energy of carbon atom is greater than
which the 2p-orbitals are neither exactly half-filled nor that of boron atom, whereas reverse is true for the second
completely filled. Therefore, it is difficult to remove an ionisation energy. Explain.
electron from N than from O. As a result, iH of N is higher Sol. The electronic configurations of carbon and boron are as
than that of O. Further, the electronic configuration of F is follows :
1s2 2s2 2p2x 2p2y 2p1z. Because of higher nuclear charge (+9),
the first ionization enthalpy of F is higher than that of O. C : 1s2, 2s2 2p1x 2Py1
Further, the effect of increased nuclear charge outweights
the effect of stability due to exactly half-filled orbitals,
B : 1s2, 2s2 2p1x
therefore, the iH of N and O are lower than that of F.
Example-16 Due to higher nuclear charge in carbon, the force of
Among the elements B, Al, C and Si attraction towards valency electron is more in carbon atom
(i) Which element has the highest first ionisation and hence the first ionisation energy is greater than boron
enthalpy ? atom. After loss of one electron, the monovalent cations
(ii) Which element has the most metallic character ? have the configurations as follows :
Justify your answer in each case. B+ : 1s2, 2s2
Sol. Arrange the elements B, Al, C and Si into different groups and
C+ : 1s2, 2s2 2p1x
periods in order of their increasing atomic numbers, we have,
Group  13 14 The B+ configuration is stable one and hence the removal of
Period 2 B C electron is difficult in comparison to C+. Hence, second
ionisation potential of boron is higher than carbon.
Group 3 Al Si
(i) Since ionization enthalpy increases along a period and Example-19
decreases down a group, therefore, C has the highest first Which of the following pairs of elements would have a
ionization enthalpy.
more negative electron gain enthalpy?
(ii)Since metallic character increases down a group and decreases
along a period, therefore, Al, is the most metallic element. (i) O or F (ii) F or Cl
Example-17 Sol. (i) The electron gain enthalpy of F is more negative than
How does the metallic and non metallic character vary on that of O because when we move from left to right the
moving from left to right in a period ? atomic size decreases. Fluorine size is smaller than oxygen.
Sol. On moving from left to right in a period, the number of Due to small size the interelectronic repulsion increases
valence electrons increases by one at each succeeding
causing for the decreases in the electron gain enthalpy
element but the number of shells remains the same. As a
result, the nuclear charge increases and the tendency of the values.
element to lose electron decreases and hence the metallic
(ii) The electron gain enthalpy of fluorine is less negative
character decreases as we move from left to right in a period.
Conversely, as the nuclear charge increases, the tendency than that of chlorine because when an electron is added to
of the element to gain electrons increases and hence the F, the added electron goes to the smaller quantum level
non-metallic increases from left to right in a period. (n=2). Due to small size the interelectronic repulsion
Alternatively, metallic character decreases and non-metallic increases causing for the decreases in the electron gain
character increases as we move from left to right in a period.
enthalpy values. On the other hand, in chlorine the electron
It is due to increase in ionization and electron gain enthalpy.
enters the high quantum level(n=3) which occupies a larger
region of space where the electron repulsion is less.
Therefore, the electron gain enthalpy of Cl is more negative
than that of F.
PERIODIC PROPERTIES 75

Example-20 Example-23
Would you expect the second electron gain enthalpy of O as Distinguish between electronegativity and electron affinity.
positive, more negative or less negative than the first ? Justify Sol. Electron affinity Electronegativity
your answer.
Sol. The second electron gain enthalpy of O is positive as 1. Electron gain enthalpy or 1.Electronegativity of an
explained below : electron affinity is defined atom in a molecule is
When an electron is added to O atom to form O– ion, energy as the amount of energy defined as the tendency
is released. Thus, first electron gain enthalpy of O is negative. released when neutral of an atom to attract

O g   e–g   O–g  ;  eg H  141kJ mol1 gaseous atom, accepts an towards itself the shared
electron to form an anion. pair of electrons.
But when another electron is added to O– to form O2– ion,
2. It is expressed in eV/atom 2.It does not have any unit
energy is absorbed to overcome the strong electrostatic
repulsion between the negatively charged O– ion and the (electron volt per atom) or (it is a number)
second electron being added. Thus, the second electron in kil number) joules per mol (kJ mol–1).
gain enthalpy of oxygen is positive.
3. It is a kind of absolute 3.It is a relative term (atoms
–  2 1
O g e
g  O ; eg H  780 kJ mol
g property of the elements. are compared with fluorine,
Example-21 whose assigned value of
electronegativity is 4.0
‘Electron affinity of fluorine is less than that of chlorine’.
Explain. 4. Electron affinity value is 4. It is measured when the
Sol. (i) Def : Electron gain enthalpy or electron affinity is defined measured when the atoms atoms are in their
as the amount of energy released, when neutral gaseous are in their gaseous state. combined state (in state
atom, accepts an electron to form an anion.
molecules).
(ii) The electronic cofiguration of fluorine is 1s22s22p5, while that Example-24
of chlorine, it is 1s22s22p63s23p5. In both the elements there
are 7 electrons in their outermost shell. The size of F-atom is Use the periodic table to answer the following questions:
smaller than Cl-atom. (a) Identify an element with five electrons in the outer
(iii) In fluorine, 2p-orbitals are compact and closer to the nucleus. subshell.
Thus, the screening effect is very low. Hence there is (b) Identify the element that would tend to lose two electrons.
electron-electron repulsion in the valence shell. Thus, when
(c) Identify the element that would tend to gain two electons.
an electron is added to the p-orbital of a fluorine. Thus,
when an electron is added to the p-orbital of a fluorine it (d) Identify the group having metal, non-metal, liquid as
experiences less attraction and hence less energy is liberated well as gas at room temperature.
to form fluoride ion. Sol. (a) The genral electronic configuration of the elements
(iv) In chlorine, the orbital accepting an electron to form chloride having five electrons in the outer subshell is ns2 np5. This
ion is 3p-orbital, which is away form the nucleus. electronic configuration is characteristic of elements of group
17, i.e., halogens and their examples are F, Cl, Br, I, At, etc.
(v) Therefore, the electron-electron repulsion is less and more
energy is liberated, when an electron is added to a chlorine (b) The elements which have a tendency to lose two electrons
atom forming a chloride anion. Thus, fluorine has less must have two electrons in the valence shell. Therefore, their
electron affinity than chlorine. general electronic configuration should be n s2. This electronic
configuration is characteristic of group 2 elements, i.e., alkaline
Example-22 earth metals and their examples are Mg, Ca, Sr, Ba, etc.
Why are electron gain enthalpies of Be and Mg positive ? (c) The elements which have a tendency to accept two electrons
Sol. They have fully filled s-orbitals and hence have no tendency must have six electrons in the valence shell. Therefore, their
to accept an additional electron. Consequently, energy has general electronic configuration is ns2 np4. This electronic
to be supplied if an extra electron has to be added to the configuration is characteristic of group 16 elements and their
much higher energy p-orbitals of the valence shell. That is examples are O and S.
why electron gain enthalpies of Be and Mg are positive.
76 PERIODIC PROPERTIES

(d) A metal which is liquid at room temperature is mercury. It is

a transition metal and belongs to group 12. A non-metal
which is a gas at room temperature is hydrogen (group 1),
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).
Example-25
Halogens except Flourine shows positive oxidation state of
+1, +2, +3, +5, and +7.
Sol. (i) The outer Electronic configuration of halogens are ns2,
np5 they can gain one electron and show a common oxidation
state of –1.

(ii) The other halogen exhibit higher oxidation state as, +1, +2,
+3, +5, and +7 due to vacant d-orbitals in their shell.
(iii) Since Flourine does not have d-orbital, it only exhibits only
‘–1’ oxidation state.
Therefore Halogens except fluorine shows positive oxidation
state +1, +2, +3, +5 and +7.
PERIODIC PROPERTIES 77

Exercise – 1: Basic Objective Questions

Dobereiner's Triads, Newlands' Law of Octaves, (c) Mendeleev’s system of classifying elements was
more elaborate than that of Lothar
Lothar Meyer's Curve and Mendeleev's Periodic
(d) Both (a) and (b)
Table
Modern Periodic Table
1. Which of the following is not a Dobereiner’s triad?
(a) Cl, Br, I (b) Ca, Sr, Ba 9. The basis of modern periodic table is
(c) Li, Na, K (d) Fe, Co, Ni (a) atomic volume (b) atomic number
2. John Alexander Newlands relationship referred to as (c) atomic weights (d) atomic size
the …….A…… Here, A refers to 10. Physical and chemical properties of elements are
(a) Law of triads (b) Law of Octaves determined by
(c) Law of Duet (d) Law of Nonanes (a) electronic configuration
3. According to Alexander Newlands, elements are (b) valency
arranged in…..A….order of their atomic weights. Here (c) atomic size
A refers to (d) all of these
(a) Increasing (b) Decreasing 11. In 1913, Henry Moseley observed regularity in the
(c) Similar (d) None of these characteries…A…of the elements. Here, ‘A’ refers to
4. Lothar Meyer plotted the physical properties such as (a) ultraviolet spectra (b) X-ray spectra
atomic volume, melting point and ….A…..against (c) infrared spectra (d) visible spectra
atomic weight. Here, A refers to 12. Atomic number is equal to the…. A…. Here, ‘A’ refers
(a) mass (b) boiling point to
(c) Surface tension (d) molecules (a) number of neutrons
5. Lothar Meyer drew a graph showing the relation (b) number of protons
between (c) number of electrons
(a) Atomic number and atomic weight (d) All of these
(b) Atomic number and atomic size 13. The vertical columns in the Periodic Table are termed
(c) Atomic weight and atomic volume as
(d) Atomic weight and atomic size (a) periods (b) groups
6. According to Mendeleev, elements are arranged in (c) series (d) None of these
order of their increasing ……A….Here, A refers to 14. The periods in the Periodic Table are numbered from
(a) Atomic radius (b) Atomic number (a) 1 to 16 (b) 1 to 18
(c) Electron number (d) Atomic weight (c) 1 to 7 (d) None of the above
7. Mendeleev left the gap under aluminum and a gap 15. Elements whose outer electronic configuration vary
under silicon and called these elements Eka-Aluminium from ns2np1 to ns2np6 constitute
and……A……Here, A refers to (a) s-Block of elements
(a) Eka-Gallium (b) Eka-Germanium (b) p-Block of elements
(c) Eka-Hydrogen (d) Eka-Silicon (c) d-Block of elements
8. Which of the following is incorrect explanation about (d) f-Block of elements
Mendeleev’s periodic law ? 16. In the fourth period of the periodic table, how many
(a) Mendeleev arranged elements in horizontal rows elements have one or more 4d electrons?
only (a) 2 (b) 18
(b) Mendeleev arranged elements with increasing (c) 0 (d) 6
atomic weight
78 PERIODIC PROPERTIES

17. Which of the following pairs has both members from 26. Which pair has both members from the same period of
the same group of periodic table Periodic Table?
(a) Mg, Ba (b) Mg, Na (a) Cl, Br (b) Ca, Cl
(c) Mg, Cu (d) Mg, Cl (c) Na, Ca (d) Na, Cl
18. Without looking at the periodic table, select the 27. The elements with atomic numbers 9, 17, 35, 53, 85 are
elements of IIIA group of the periodic table all
(atomic numbers are given) : (a) noble gases (b) halogens
(a) 3, 11, 19, 37 (b) 5, 13, 21, 39 (c) heavy metals (d) light metals
(c) 7, 15, 31, 49 (d) 5, 13, 31, 49 28. Identify the electronic configuration of transition
19. The transition elements have a characteristic electronic element.
configuration which can be represented as (a) 1s2, 2s22p6, 3s23p6, 4s2
(a) (n – 2) s2p6d1–10 (n – 1) s2p6ns2 (b) 1s2, 2s22p6, 3s23p63d2, 4s2
(b) (n – 2) s2p6d1–10 (n – 1) s2p6d1 or 2 ns1 (c) 1s2, 2s22p6, 3s23p63d10, 4s24p2
(c) (n – 1)s2p6d10 ns2np6nd1–10 (d) 1s2, 2s22p6, 3s23p63d10, 4s24p1
(d) (n – 1) s2p6d1–10ns0–2 29. Molybdenum metal is related to which block?
(a) s-block (b) p-block
20. The element whose electronic configuration is 1s2, 2s2,
(c) d-block (d) None of these
2p6, 3s2 is 30. Manganese is related to which block of the Periodic
(a) metal (b) metalloid Table?
(c) inert gas (d) non – metal (a) s-block (b) p-block
21. The electronic configuration of an element is 1s2, (c) d-block (d) f-block
2s22p6, 3s23p3. What is the atomic number of the 31. Which one of the following pair of elements has similar
element which is just below the above element in the properties?
periodic table? (a) 10, 12 (b) 11, 20
(a) 34 (b) 49 (c) 21, 33 (d) 13, 31
(c) 33 (d) 31 32. Which one of the following is not a chalcogen?
(a) S (b) Se
22. The element having electronic configuration [Kr] 4d10,
(c) O (d) Na
4f14, 5s2, 5p6, 5d4, 6s2 belongs to 33. Atomic to IUPAC nomenclature, the name of element
(a) s – block (b) p – block having atomic number 101 is
(c) d – block (d) f – block (a) Unnilbium (b) Unnilunium
23. Which of the following represents an excited state of an (c) Unnilquadium (d) None of these
atom? 34. According to IUPAC nomenclature, the name of
(a) [Ne] 3s2 3p6 4s2 3d8 element having atomic number 116 is
(b) [Ne] 3s2 3p6 4s1 3d5 (a) Unnilunium (b) Ununoctium
(c) Ununhexium (d) Unnilhexium
(c) [Ne] 3s2 3p6 4s2 3d1
35. Group I elements of modern periodic table are called
(d) 1s2 2s2 2p5 3s1 (a) alkali metals (b) alkaline earth metals
24. An element with atomic number 21 is a (c) Both (a) and (b) (d) None of these
(a) halogen
36. ns1 and ns2 outermost electronic configuration belongs
(d) representative element
to the
(c) transition element
(a) s-block elements (b) p-block elements
(d) alkali metal
(c) d-block elements (d) f-block elements
25. An atom has electronic configuration 1s2, 2s2, 2p6,
3s2, 3p6, 3d3, 4s2, you will place it in
(a) fifth group (b) fifteenth group
(c) second group (d) third group
PERIODIC PROPERTIES 79
37. Metals are good conductors of heat and….A….Here, 46. An element ‘X’ belongs to the fourth period and
‘A’ refers to fifteenth group of the Periodic Table. Which one of the
(a) energy (b) electricity following is true regarding the outer electronic
(c) Both (a) and (b) (d) None of these configuration of ‘X’. It has
38. The elements having characteristics of both metals and (a) Partially filled ‘d’orbitals and completely filled ‘s’
non-metals is/are termed as orbitals.
(a) semi-metals (b) metalloids (b) Completely filled ‘s’orbitals and completely filled
(c) Either (a) and (b) (d) None of these ‘p’ orbitals.
39. The element with atomic number 57 belongs to (c) Completely filled ‘s’orbitals and half-filled ‘p’
(a) s-block (b) p-block orbitals.
(c) d-block (d) f-block (d) Half-filled ‘d’orbitals and completely filled ‘s’
40. The last element of the p-block in the 6th period is orbitals.
represented by the outermost electronic configuration. (e) Completely filled‘d’, ‘s’ and ‘p’ orbitals.
47. The number of elements present in fifth period of
(a) [Rn] 7 s2 7p6
Periodic Table is
(b) [Kr] 5f 14 6d107s2 7p0 (a) 10 (b) 8
(c) [Xe] 4f 14 5d10 6s2 6p6 (c) 32 (d) 18
48. The element with atomic number 36 belongs to
(d) [Xe] 4f 14 5d106s2 6p4
........block in the Periodic Table.
41. Which of the elements whose atomic numbers are given (a) p (b) s
below, cannot be accommodated in the present set up of (c) f (d) d
the long form of the periodic table?
(a) 107 (b) 118
(c) 126 (d) 102
Atomic and Ionic Radii
42. The electronic configuration of the element which is
just above the element with atomic number 43 in the 49. Which of the following atom has largest size
same group is ............ (a) Cs (b) K
(c) Kr (d) Xe
(a) 1s2 2s2 2p6 3s2 3p6 3d5 4s2
50. In comparison to the parent atom, the size of the
(b) 1s2 2s2 2p6 3s2 3p6 3d5 4s3 4p6 (a) Cation is smaller but anion is larger
(c) 1s2 2s2 2p6 3s2 3p6 3d6 4s2 (b) Cation is larger but anion is smaller
(c) Cation and anion are equal in size
(d) 1s2 2s2 2p6 3s2 3p6 3d7 4s2
(d) All the three are correct depending upon the atom
43. The elements with atomic numbers 35, 53, and 85 are
51. Which one is the correct order of the size of the iodine
all ................
species.
(a) Noble gases (b) Halogens
(c) Heavy metals (d) Light metals (a) I > I+> I– (b) I > I–> I+
44. The element with the lowest atomic number that has a (c) I+ > I–> I (d) I–> I > I+
ground state electronic configuration of (n-1) d6ns2 is 52. Which of the following ions has the smallest radius?
located in the (a) Li+ (b) Na+
(a) fifth period (b) sixth period
(c) Be2+ (d) K+
(c) fourth period (d) third period
53. When a chlorine atom becomes chloride ion, its size
45. Element having electronic configuration 1s2, 2s2, 2p6, (a) remains unaltered (b) increases
3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2, 5p3 belongs to the (c) decreases (d) none of these
group in the periodic table 54. Which of the following pairs has almost the same
(a) III group (b) II group atomic radii?
(c) V group (d) VII group (a) Al, Ga (b) Be, Mg
(c) Mg, Al (d) B, Be
80 PERIODIC PROPERTIES

55. Identify the correct order of the size of the following (a) Ar, Cl, S, P (b) Cl, S, P, Ar
(a) Ca2+< K+<Ar< S2–<Cl– (c) S, Cl, P, Ar (d) Ar, P, S, Cl
65. The single covalent radius of P is 0.11 nm The single
(b) Ca2+< K+<Ar<Cl–< S2–
covalent radius of Cl will be :
(c) Ar<Ca2+< K+<Cl–< S2– (a) smaller than P (b) greater than P
(d) Ca2+<Ar< K+<Cl–< S2– (c) same as P (d) twice of P
56. Which one of the following ions has the highest value 66. In isoelectronic species of Mg2+, N3–, Al3+, the order
of ionic radius? of decreasing ionic radii will be
(a) O2– (b) B3+
(a) N3–> Mg2+> Al3+ (b) Mg2+> Al3+> N3–
(c) Li+ (d) F–
(c) Al3+> N3–> Mg2+ (d) Al3+ = Mg2+< N3–
57. Atomic radii of F and Ne, in Å, are given by
67. Identify the correct order in which the ionic radius of
(a) 0.72, 0.71 (b) 0.72, 1.6
the following ions increases
(c) 1.6, 1.58 (d) 0.71, 0.72
58. Among 3rd-row elements atomic size is maximum for (I) F– (II) Na+
(a) sodium (b) aluminium (III) N3–
(c) magnesium (d) chlorine (a) III, I, II (b) I, II, III
59. Which of the following order is correct for the size of (c) II, III, I (d) II, I, III
Fe3+, Fe, and Fe2+? 68. Correct order of radii is
(b) F < O2– < N3–

(a) Fe < Fe2+< Fe3+ (b) Fe2+< Fe3+< Fe (a) N < Be < B

(c) Fe < Fe3+< Fe2+ (d) Fe3+< Fe2+< Fe (c) Na < Li < K (d) Fe3+< Fe2+< Fe4+

60. The ions O2–, F–, Na+, Mg2+and Al3+are

isoelectronic. Their ionic radii show Ionization Enthalpy
(a) an increase from O2– to F– and then decrease from
69. Shielding of the nuclear charge by the inner core
Na+ to Al3+
of……A…….does not increase. Here, A refers to
(b) a decrease from O2– to F– and then increase from
(a) electron (b) proton
Na+ to Al3+
(c) neutron (d) Both (a) and (c)
(c) a significant increase from O2– to Al3+ 70. Lowest ionization potential in periods is shown by
(d) a significant decrease from O2– to Al3+ (a) inert gases (b) halogens
61. The correct order of the sizes of C, N, P, S, is (c) alkali metals (d) alkaline earth metals
(a) N < C < P < S (b) C < N < S < P
71. The electronic configurations of some atoms are given
(c) C < N < P < S (d) N < C < S < P
below, which of these should have the highest second
Atomic radius ionization energy?
(a) 1s2, 2s2, 2p6, 3s2
62. In which of the following pair, both the species are
isoelectronic but first one is large in size than the (b) 1s2, 2s2, 2p6, 3s1
second? (c) 1s2, 2s2, 2p6
(a) S2–, O2– (b) Cl–, S2–
(d) 1s2, 2s2, 2p5
(c) F–, Na+ (d) N3–, P3– 72. The correct arrangement of the elements in the order of
63. The correct order of ionic size of N3–, Na+, F–, Mg2+ decreasing ionization energies is
and O2–is: (a) Na > Mg > Al (b) Mg > Na > Al
(a) Mg2+> Na+> F–> O2–< N3– (c) Al > Mg > Na (d) Mg > Al > Na
(b) N3–< F–> O2–> Na+> Mg2+
(c) Mg2+< Na+< F–< O2–< N3–
(d) N3–> O2–> F–> Na+< Mg2+
64. Arrange the following elements in the order of
increasing atomic size Cl, S, P, Ar
PERIODIC PROPERTIES 81
73. The maximum tendency to form uni-positive ion is for 82. Alkali metals in each period have
the element which has the following electronic (a) Smallest size
configuration (b) Lowest ionization enthalpy
(a) 1s2, 2s2, 2p6, 3s2 (c) Highest ionization enthalpy
(d) Highest electronegativity
(b) 1s2, 2s2, 2p6, 3s2, 3p1
83. Amongst the elements with the following electronic
(c) 1s2, 2s2, 2p6 configurations which one of them may have the highest
(d) 1s2, 2s2, 2p6, 3s2, 3p3 ionization energy?
74. Which element has the highest ionization energy? (a) [Ne]3s23p3 (b) [Ne]3s23p2
(a) Hydrogen (b) Lithium (c) [Ar]3d104s24p3 (d) [Ne]3s23p1
(c) Boron (d) Sodium 84. The bond enthalpy is the highest for
75. An element will have lowest ionisation potential when (a) F2 (b) Cl2
its electronic configuration is (c) Br2 (d) H2
(a) 1s1 (b) 1s2, 2s2, 2p2 85. Which of the following has the highest ionization
(c) 1s2, 2s2, 2p5 (d) 1s2, 2s2, 2p6, 3s1 enthalpy?
(a) P (b) N
76. The order of ionization potential between He+ ion and (c) As (d) Sb
H-atom (both species are in gaseous state) is 86. The correct order of ionization energy of C, N, O, F is
(a) I.P. (He+) = I.P. (H) (b) I.P. (He+) < I.P. (H) (a) F < N < C < O (b) C < N < O < F
(c) C < O < N < F (d) F < O < N < C
(c) I.P. (He+) > I.P. (H) (d) cannot be compared
87. Which one of the following order is correct for the first
77. The correct order of second I.E. of C, N, O and F are in
ionization energies of the elements?
the order
(a) B < Be < N < O (b) Be < B < N < O
(a) F > O > N > C (b) C > N > O > F
(c) B < Be < O < N (d) B < O < Be < N
(c) O > N > F > C (d) O > F > N > C
88. The correct order of first ionization potential is
78. Arrange the following atoms in order of increasing first
(a) K > Na > Li (b) Be > Mg >Ca
ionization energy K, Cs, Rb, Ca
(a) Cs, Rb, K, Ca (b) Rb, Cs, K, Ca (c) B > C > N (d) Ge > Si > C
89. Which of the following has the highest ionization
(c) Cs, Rb, Ca, K (d) Rb, Cs, Ca, K
energy?
79. The element which has highest first ionization energy in
(a) K+ (b) Cl–
the periodic table is
(a) H (b) Rn (c) Ar (d) Cs+
90. The correct order in which the first ionization potential
(c) F (d) He
increases is
80. Correct order of first ionization potential among the
(a) Na, K, Be (b) K, Na, Be
following elements Be, B, C, N, O is
(c) K, Be, Na (d) Be, Na, K
(a) B < Be < C < O < N
91. The element with the lowest ionization potential is
(b) B < Be < C < N < O
(a) Na (b) K
(c) Be < B < C < N < O
(c) Rb (d) Cs
(d) Be < B < C < O < N
92. Which of the following has the lowest ionization
81. In a periodic table, on moving from left to right along a
energy?
period, the metallic character of an element
(a) Oxygen (b) Nitrogen
(a) decreases
(c) Fluorine (d) Sulphur
(b) increases
93. Ionisation energy decreases down the group due to
(c) remains the same
(a) increase in charge
(d) first increases and then decreases
(b) increase in atomic size
(c) decrease in atomic size
(d) decrease in shielding effect
82 PERIODIC PROPERTIES

94. Ionization potential for a noble gas is M+ (g) → M2+ (g) + e– (4)
(a) maximum in a period
M(g) →M2+ (g) + 2e– (5)
(b) minimum in a period
The second ionization energy of M(g) could be
(c) either minimum or maximum
calculated from the energy values associated with:
(d) constant
(a) 1 + 3 + 4 (b) 2 – 1 + 3
95. The highest first ionization potential is of
(c) 1 + 5 (d) 5 – 3
(a) carbon (b) boron
105. Which of the following statements is incorrect?
(c) oxygen (d) nitrogen
(a) The second ionization energy of sulphur is greater
96. A neutral atom will have the lowest ionization potential
than that of chlorine
when electronic configuration is
(b) The third ionization energy of phosphorus is greater
(a) 1s2 (b) 1s2, 2s2 2p2 than that of aluminium
(c) 1s2, 2s22p6 (d) 1s2, 2s22p6, 3s1 (c) The first ionization energy of aluminium is
97. The ionization energy will be maximum for approximately the same as that of gallium
(a) H (b) Li (d) The second ionization energy of boron is greater
(c) Be (d) B than that of the carbon
98. The correct order of second ionization potential of C, 106. Which of the following order is wrong character
N, O and F is (a) NH3< PH3<AsH3 – Acidic
(a) C > N > O > F (b) O > N > F > C (b) Li < Be < B < C – IE
(c) O > F > N > C (d) F > O > N > C (c) Al2O3<MgO< Na2O< K2O – Basic
99. Which of the following isoelectronic ions has the
lowest ionization energy? (d) Li+< Na+< K+< Cs+ – Ionic radius
107. Which of the following orders regarding ionization
(a) K+ (b) Ca2+
energy is correct?
(c) Cl–1 (d) S2– (a) N > O > F (b) N < O < F
100. The ionization potential of X– ion is equal to (c) N > O < F (d) N < O > F
(a) The electron affinity of X atom 108. The correct order of decreasing second ionization
(b) The electronegativity of X atom enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is
(c) The ionization potential of X atom (a) Cr > Mn > V > Ti (b) V > Mn > Cr > Ti
(d) None of these (c) Mn > Cr > Ti > V (d) Ti > V > Cr >Mn
101. Second ionization potential of Li, Be and B is in the 109. The atomic numbers of vanadium (V), chromium (Cr),
order manganese (Mn), and iron (Fe) are respectively 23, 24,
(a) Li > Be > B (b) Li > B > Be 25, and 26. Which one of these may be expected to have
(c) Be > Li > B (d) B > Be > Li the highest second ionization enthalpy?
102. The first four I.E. values of an element are 284, 412, (a) V (b) Cr
656 and 3210 kJ mol–1. The number of valence (c) Mn (d) Fe
electrons in the element are 110. The first ionization potentials of four consecutive
(a) one (b) two elements, present in the second period of the Periodic
(c) three (d) four Table, are 8.3, 11.3, 14.5, and 13.6 eV respectively.
103. Which of the following electronic configuration is Which one of the following is the first ionization
associated with the biggest jump between the second potential (in eV) of nitrogen?
and third ionization energies? (a) 13.6 (b) 11.2
(c) 8.3 (d) 14.5
(a) 1s2 2s2 2p2 (b) 1s2 2s2 2p6 3s1
(c) 1s2 2s2 2p6 3s2 (d) 1s2 2s2 2p1
104. Consider the following changes
M(s) →M(g) (1)
2+
M(s) → M (g) + 2e – (2)
M(g) → M+ (g) + e– (3)
PERIODIC PROPERTIES 83

Electron Gain Enthalpy or Electron Affinity 121. Electron affinity of the inert gases is
(a) High (b) Low but positive
111. Arrange, N, O and S in order of decreasing electron (c) Moderate (d) Almost zero
affinity. 122. First electron gain enthalpy of oxygen is –141 kJ mol–1
(a) S > O > N (b) O > S > N second electron gain enthalpy will be
(c) N > O > S (d) S > N > O (a) +141 kJ mol–1 (b) –141 kJ mol–1
112. The correct order for electron affinities is a
(c) +780 kJ mol–1 (d) –780 kJ mol–1
(a) F > Br > I (b) F < Br < I
123. In which of the following processes, energy is liberated
(c) F < I > Br (d) Br < I < F
113. Which of the following element is expected to have (a) ClCl+ + e– (b) HClH+ + Cl–
highest electron affinity? (c) O– + e–O2– (d) F + e–F–
(a) 1 s22 s22 p6 3 s2 3 p5 124. The correct order of increasing electron affinity of the
following elements is
(b) 1 s22 s22 p3
(a) O < S < F <Cl (b) O < S <Cl< F
(c) 1 s22 s22 p4 (c) S < O < F <Cl (d) S < O <Cl< F
(d) 1 s22 s22 p5 125. Which of the following represents the correct order of
114. An atom of an electronegative element becomes an ion increasing electron gain enthalpy without sign for the
by elements O, S, F, and Cl?
(a) Gain of electrons (a) S < O <Cl< F (b) Cl< F < O < S
(b) Loss of electrons (c) O < S < F <Cl (d) F < S < O <Cl
(c) Loss of its radius 126. The electronic configurations of four elements are
(d) Serving as a reductant given below. Arrange these elements in the correct
115. The value of electron affinity for noble gases is likely to order of the magnitude (without sign) of their electron
be affinity.
(a) high (b) low (i) 2s2 2p5 (ii) 3s2 3p5
(c) zero (d) positive
(iii) 2s2 2p4 (iv) 3s2 3p4
116. Second electron gain enthalpy
Select the correct answer using the codes given below
(a) is always negative
(a) (i) < (ii) < (iv) < (iii)
(b) is always positive
(b) (ii) < (i) < (iv) < (iii)
(c) can be positive or negative
(c) (i) < (iii) < (iv) < (ii)
(d) is always zero
(d) (iii) < (iv) < (i) < (ii)
117. The element having very high ionization enthalpy but
127. Among halogens, the correct order of amount of energy
zero electron gain enthalpy is?
released in electron gain (electron gain enthalpy) is:
(a) H (b) F
(a) F >Cl> Br > I (b) F <Cl<Br < I
(c) He (d) Br
(c) F<Cl> Br > I (d) F<Cl< Br < I
118. Electron affinity is maximum for
(a) Cl (b) F
Electronegativity and Its Applications
(c) Br (d) I
119. The correct order of electron affinity of halogens is 128. The most electropositive element is
(a) F > Cl > Br > I (b) I > Br > Cl > F (a) Cs (b) Ga
(c) Cl > F > Br > I (d) Cl > F < Br < I (c) Li (d) Pb
120. Which of the following species has the highest electron 129. Which of the following represent highly electropositive
affinity? as well as highly electronegative element in its period?
(a) F (b) O (a) Nitrogen (b) Fluorine
(c) O – (d) Na+ (c) Hydrogen (d) None
84 PERIODIC PROPERTIES

130. What will be the electropositive character when we (a) acidic, basic (b) acidic, acidic
move from left to right in a period (c) basic, acidic (d) basic, basic
(a) Increases (b) Decreases 141. The electronegativity of the following elements
(c) No change (d) First increases then increases in the order
decreases (a) C, N, Si, P (b) N, Si, C, P
131. Outermost electronic configuration of least (c) Si, P, C, N (d) P, Si, N, C
electronegative element in the periodic table is 142. The electronegativity of Cl, F, O, S increases in the
(a) 2s2 2p5 (b) 3s2 3p5 order of
(a) S, O, Cl, F (b) S, Cl, O, F
(c) 2s2 2p4 (d) 6s2 6p6 7s1
(c) Cl, S, O, F (d) S, O, F, Cl
132. In any period of the periodic table the electronegativity
143. Which is the property of non-metal?
of elements as we move from left to right.
(a) Electronegative
(a) Increases (b) Decreases
(b) Basic nature of oxide
(c) Remains constant (d) None
(c) Reducing property
133. The element with highest electronegativity value is
(d) Low ionization potential
(a) F (b) Cl
(c) P (d) N
134. Which of the following element has the highest Valency and Oxidation State
electronegativity?
(a) As (b) Sb 144. In Periodic Table, the basic character of oxides
(c) P (d) S (a) increases from left to right and decrease from top of
135. With respect to chlorine, hydrogen will be bottom
(a) Electropositive (b) Electronegative (b) decreases from right to left and increases from top to
(c) Neutral (d) None of these bottom
136. The outermost electronic configuration of the most (c) decreases from left to right and increases from top to
electronegative element is bottom
(d) decreases from left to right and increases from
(a) ns2, np3 (b) ns2, np4
bottom to top
(c) ns2, np5 (d) ns2, np6 145. Which one of these is most basic
137. Which of the following is the second most (a) CO2 (b) SnO2
electronegative element?
(c) NO2 (d) SO2
(a) Chlorine (b) Oxygen
(c) Sulfur (d) Fluorine 146. Which of the following order is wrong?
138. Which of these have no unit? (a) NH3< PH3<AsH3 – acidic
(a) Electronegativity (b) Electron affinity (b) Li < Be < B < C – first IP
(c) Ionization energy (d) Excitation potential (c) Al2O3<MgO< Na2O < K2O – basic
139. Which one of the following statements is incorrect? (d) Li+< Na+< K+< Cs+ – ionic radius
(a) Greater is the nuclear charge, greater is the electron 147. A metal ion with +3 charge has five electrons in the 3d-
gain enthalpy subshell, the metal is
(b) Nitrogen has almost zero electron gain enthalpy (a) Fe (b) Cr
(c) Electron gain enthalpy decreases from fluorine to (c) Mn (d) Zn
iodine in the group
(d) Chlorine has highest electron gain enthalpy
140. Aqueous solutions of two compounds M1 – O – H and
M2 – O – H are prepared in two different beakers. If,
the electronegativity of M1 = 3.4,
M2 = 1.2, O = 3.5 and H = 2.1, then the nature of two
solutions will be respectively
PERIODIC PROPERTIES 85
148. In any period the valency of an element with respect to 149. Which of the following pairs do not show a diagonal
oxygen relationship?
(a) increases one by one from IA to VIIA (a) Li and Mg (b) Be and Al
(b) decreases one by one from IA to VIIA (c) B and Si (d) C and S
(c) increases one by one from IA to IVA and then 150. The diagonal partner of element B is
decreases from VA to VIIA one by one (a) Li (b) Al
(d) decreases one by one from IA to IVA and then (c) Si (d) Mg
increases from VA to VIIA one by one
(e) changes randomly
86 PERIODIC PROPERTIES

Exercise – 2: Previous Year Questions

1. Which one of the following ions has the highest value rather than 8; what would be the formula of the stable
of ionic radius? (DUMET 2008) fluoride ion? (DUMET 2009)
(a) O
2
(b) B3 (a) F – (b) F+
(c) Li  (d) F (c) F 2+ (d) F3+
2. Increase in atomic size down the group is due to 10. The oxide of an element whose electronic configuration
(JCECE 2008) is 1s2, 2s2 2p6, 3s1 is (KCET 2009)
(a) increase in number of electrons (a) amphoteric (b) basic
(b) increase in number of protons and neutrons (c) acidic (d) neutral
(c) increase in number of protons 11. In which of the following arrangements, the order is not
(d) increase in number of protons, neutrons and according to the property indicated after it in the
electrons bracket? (Manipal 2009)
3. Diagonal relationship is not shown by (BHU 2008) 3+ 2+
(a) Al < Mg < Na < F + 

(a) Li and Mg (b) Be and Al (Increasing ionic size)

(c) B and Si (d) C and P (b) B < C < N < O
4. Which two elements in the Periodic Table would you (Increasing first ionization energy)
expect to combine in the most violent fashion? (c) I < Br < F <Cl
(Kerala CEE 2008) (Increasing electron gain enthalpy)
(a) H and O (b) Cl and F (d) Li < Na < K <Rb
(c) Mg and N (d) Cs and F (Increasing metallic radius)
5. Element with atomic number 38, belongs to 12. Increasing order of ionic size for the ions, F–, O2–, Na+,
(MP PMT 2008) Al3+ is (AMU 2009)
(a) IIA group and 5th period 2– – + 3+
(a) O < F < Na < Al
(b) IIA group and 2nd period
(c) VA group and 2nd period (b) Al3+< Na+< F–< O2–
(d) IIIA group and 5th period (c) O2–< Na+< F–< Al3+
6. An element with atomic number 112 has been made (d) Al3+< F–< Na+< O2–
recently. It should be (Haryana PMT 2008) 13. Which represents the correct order of the first ionization
(a) an actinide (b) a transition metal potential of third-period elements? (AFMC 2009)
(c) a noble gas (d) a lanthanide (a) Na > Ma > Al > Si
7. The discovery of which of the following group of (b) Na < Mg < Al < Si
elements gave a death blow to Newland’s law of (c) Na < Si < Al < Mg
octaves? (Manipal 2008) (d) Na < Al < Mg < Si
(a) Inert gases (b) Alkaline earths 14. The element with atomic number 117 has not been
(c) Rare earths (d) Actinides discovered yet. In which family would you place this
8. Amongst the elements with following electronic element if discovered? (Kerala CEE 2009)
configurations, which one of them may have the highest (a) Alkali metals
ionisation energy? (AIPMT 2009) (b) Alkaline earth metals
(a) Ne 3s 2 3p 2  (b) Ar 3d10 4s 2 4p 3  (c) Halogens
(d) Noble gases
(c) Ne 3s 2 3p1  (d) Ne 3s 2 3p3 
(e) Coinage metals
9. It is believed that atoms combine with each other such 15. Which one of the following is correct order of the size?
that the outermost shell acquires a stable configuration (Haryana PMT 2010)
of 8 electrons. If stability were attained with 6 electrons (a) I  I  I (b) I  I  I
(c) I   I  I (d) I  I  I
PERIODIC PROPERTIES 87
16. Ease of formation of the cation is favoured by 25. The increasing order of the density of alkali metal is
(RPMT 2010) (Kerala CEE 2011)
(a) lower value of ionisation potential (a) Li < K < Na < Rb < Cs
(b) lower value of electron affinity (b) Li < Na < K < Rb < Cs
(c) higher value of electron affinity (c) Cs < Rb < Na < K < Li
(d) lower value of electronegativity (d) Cs < Rb < K < Na < Li
17. Mg and Li are similar in their properties due to (e) Li < Na < Rb < K < Cs
(VMMC 2010) 26. The electronegativity of the following elements
(a) same e/m ratio increases in the order (AFMC 2011)
(b) same electron affinity (a) C, N, Si, P (b) N, Si, C, P
(c) same group (c) Si, P, C, N (d) P, Si, N, C
(d) same ionisation potential 27. Which of the following has the highest electron
18. Electronic configuration of an element X having atomic affinity? (AFMC 2011)
number 24 is (UP CPMT 2010) (a) F– (b) O–
(a)  Ar  3d 5 4s1 (b)  Ar  3d 4 4s 2 (c) O (d) Na
(c)  Ne  2p 3s
5 1
(d)  Ar  3d 4s
6 2 28. The correct order of decreasing electronegativity values
among the elements I-beryllium, II-oxygen, III-nitrogen
19. Which one of the following ions has electronic
and IV-magnesium, is (Kerala CEE 2011)
configuration  Ar  3d 6 ? (CBSE AIPMT 2010) (a) (II) > (III) > (I) > (IV)
(Atomic numbers: Mn = 25, Fe = 26, Co = 27, Ni = 28) (b) (III) > (IV) > (II) > (I)
(a) Co
3
(b) Ni
3 (c) (I) > (II) > (III) > (IV)
3 (d) (I) > (II) > (IV) > (III)
(c) Mn3 (d) Fe
29. The electronic configuration of an element is
20. The property of attracting electrons by the halogen
atoms in a molecule is called (JCECE 2010) 1s 2 , 2s 2 2p6 ,3s 2 3p6 3d10 , 4s 2 4p6 4d10 ,5s 2 5p3 .
(a) ionization potential (b) electron affinity From which group of the period table, it belongs?
(c) electronegativity (d) electronic attraction (KCET 2012)
21. Generally, the first ionization energy increases along a (a) 2nd (b) 15th
period. But there are some exceptions. One which is not (c) 3rd (d) 7th
an exception is (KCET 2010) 30. Identify the wrong statement in the following
(a) N and O (b) Na and Mg (AIPMT 2012)
(c) Mg and Al (d) Be and B (a) Amongst isoelectronic species, smaller the positive
22. Among the following the third ionization energy is charge on the cation, smaller is the ionic radius
highest for (AMU 2010) (b) Amongst isoelectronic species, greater the negative
(a) magnesium (b) boron charge on the anion, larger is the ionic radius
(c) beryllium (d) aluminium (c) Atomic radius of the elements increases as one
23. The element with the electronic configuration as moves down the first group of the periodic table
[Ar]3d10 4s2 4p3 represents a (JCECE 2010) (d) Atomic radius of the elements decreases as one
(a) metal (b) non-metal moves across from left to right in the 2nd period of
the periodic table
(c) metalloid (d) transition element
31. The elements ‘X’, ‘Y’, and ‘Z’ form oxides which are
24. The correct order of the decreasing ionic radii among acidic, basic, and amphoteric respectively. The correct
the following isoelectronic species is order of their electronegativity is (EAMCET 2012)
(CBSE AIPMT 2010) (a) X > Y > Z (b) Z > Y > X
+ 2+
(a) K >Ca >Cl > S– 2– (c) X > Z > Y (d) Y > X > Z
(b) Ca2+> K+> S2–>Cl– 32. The correct order of decreasing first ionisation potential
is (UP CPMT 2013)
(c) Cl–> S2–>Ca2+> K+
(a) Ca  K  Rb  Cs (b) Cs  Rb  K  Ca
(d) S2–>Cl–> K+>Ca2+
(c) Ca  Cs  Rb  K (d) K  Rb  Cs  Ca
88 PERIODIC PROPERTIES

33. The electron affinity of Be is similar to (c) addition of electron in oxygen results in larger size
(UP CPMT 2013) of the ion
(a) He (b) B (d) electron repulsion outweights the stability gained by
(c) Li (d) Na achieving noble gas configuration
34. Which one of the following arrangements represents the 41. The stability of +1 oxidation state among Al, Ga, In and
correct order of least negative to most negative electron TI increases in the sequence (AIPMT 2015)
gain enthalpy for C, Ca, Al, F and O? (NEET 2013) (a) Tl < ln < Ga < Al (b) ln < Tl < Ga < Al
(a) Al < Ca < O < C < F (c) Ga < ln < Al < Tl (d) Al < Ga < In < TI
(b) Al < O < C < Ca < F 42.
2
The species Ar, K  and Ca contain the same number
(c) C < F < O < Al < Ca
of electrons. In which order do their radii increase?
(d) Ca < Al < C < O < F
(AIPMT 2015)
35. Which one of the following is the correct order of the 2  2 
(a) Ca  Ar  K (b) Ca  K  Ar
size of the ions? (EAMCET 2014)  2  2
(c) K  Ar  Ca (d) Ar  K  Ca
(a) Na   Mg 2  F  O2 43. In which of the following options, the order of
(b) O2  F  Na   Mg 2 arrangements does not agree with the variation of the
2   2 property indicated against it? (NEET 2016)
(c) Mg  Na  F  O
2 
(d) O  F  Mg  Na
2  (a) Al3  Mg 2  Na   F (increasing ionic size)
36. The correct arrangement for the ions in the increasing (b) B < C < N < O (increasing first ionization enthalpy)
order of their radius is (KCET 2014) (c) Li < Na < K <Rb (increasing metallic radius)
(a) Na  ,Cl ,Ca 2 (b) Ca 2  , K ,S2  (d) None of these
44. Which of the following is the correct order of stability
(c) Na  , Al3 ,Be2 (d) Cl , F ,S2  for the given superoxide’s? (JIPMER 2017)
37. The compound not acting as a reducing agent is (a) KO 2  RbO2  CsO 2 (b) CsO 2  RbO 2  KO 2
(Manipal 2014)
(c) RbO 2  CsO 2  KO 2 (d) KO 2  CsO 2  RbO 2
(a) SO2 (b) SeO2
45. Which of the following pairs contain metalloid
(c) TeO2 (d) All of these
elements in the periodic table? (Kerala CEE 2017)
38. Amongst the following, select the element having
(a) Na and K (b) F and Cl
highest ionisation enthalpy. (MHT CET 2014)
(c) Ca and Mg (d) As and Si
(a) Sodium (b) Potassium
46. The plot of square root of frequency of X-ray emitted
(c) Beryllium (d) Magnesium
against atomic number led to the suggestion of which
39. Which of the following orders of ionic radii is correctly
law/rule? (MHT CET 2017)
represented? (AIPMT 2014)
  2
(a) Periodic law (b) Modern periodic law
(a) H  H  H (b) Na  F  O (c) Hund’s rule (d) Newland’s law
(c) F  O2  Na  (d) None of these 47. The element Z = 114 has been discovered recently. It
40. The formation of the oxide ion, O 2g from oxygen atom will belong to which of the following family/group and
requires first an exothermic and then an endothermic electronic configuration? (NEET 2017)
step as shown below: (NEET 2015) (a) Carbon family,  Rn  5f 14 6d10 7s 2 7p 2
O  g   e   O g  ;  f H o  141kJ mol 1 (b) Oxygen family,  Rn  5f 14 6d10 7s 2 7p 4
O g   e   O 2g ;  f H o  780 kJ mol 1 (c) Nitrogen family,  Rn  5f 14 6d10 7s 2 7p 6
2
Thus, process of formation of O in gas phase is (d) Halogen family,  Rn  5f 14 6d10 7s 2 7p 5
2
unfavorable even though O is isoelectronic with neon.
It is due to the fact that,
(a) O ion has comparatively smaller size than oxygen
atom
(b) oxygen is more electronegative
PERIODIC PROPERTIES 89
48. Which of the following oxides is most acidic in nature? 50. Identify the incorrect match. (NEET 2020)
(NEET 2018) Name IUPAC Official Name
(a) MgO (b) BeO (A) Unnilunium (i) Mendelevium
(c) BaO (d) CaO (B) Unniltrium (ii) Lawrencium
49. For the second period elements the correct increasing (C) Unnilhexium (iii) Seaborgium
order of first ionization enthalpy is (NEET 2019) (D) Unununnium (iv) Darmstadium
(a) Li < Be < B < C < O < N < F < Ne (a) (A), (i) (b) (B), (ii)
(b) Li < Be < B < C < N < O < F < Ne (c) (C), (iii) (d) (D), (iv)
(c) Li < B < Be < C < O < N < F < Ne 51. The lUPAC name of an element with atomic number
(d) Li < B < Be < C < N < O < F < Ne 119 is (NEET 2022)
Li < B < Be < C < O < N < F < Ne (a) ununennium (b) unnilennium
(c) unununnium (d) ununoctium
90 PERIODIC PROPERTIES

Exercise – 3: Achiever’s Section

Single choice questions
8. Of the following orders of relative sizes, choose the
incorrect order
1. The period number in the long form of the periodic
table is equal to (a) Li < Na < K (b) Al3+< Mg2+ < Na+
(a) Magnetic quantum number of any element of the (c) C < Si < Al (d) Mg < Al < Na
period. 9. Consider the isoelectronic species, Na+, Mg2+,F –and
(b) The Atomic number of any element of the period. O2–. The correct order of increasing length of their radii
(c) Maximum Principal quantum number of any is ...........
element of the period.
(a) F–< O2– < Mg2+< Na+
(d) Maximum Azimuthal quantum number of any
element of the period. (b) Mg2+< Na+ < F– < O2–
2. The statement that is not correct for periodic (c) O2–<F–< Na+< Mg2+
classification of elements is: (d) O2–< F– < Mg2+< Na+
(a) The properties of elements are the periodic function 10. The order of screening effect of electrons of s, p, d and f
of their atomic numbers. orbitals of a given shell of an atom on its outer shell
(b) Non-metallic elements are less in number then electrons is:-
metallic elements. (a) s> p > d >f (b) f> d > p > s
(c) For transition elements, the 3d-orbitals are filled (c) p< d < s <f (d) f> p > s > d
with electrons after 3p-orbitals and before 4s- 11. Which of the following isoelectronic ions has the
orbitals. lowest ionization energy?
(d) The first ionization enthalpies of elements generally
(a) K+ (b) Ca2+
increase with an increase in atomic number as we go
along a period. (c) Cl– (d) S2–
3. La (Lanthanum) having atomic number 57 is a member 12. IE1 and IE2 of Mg(g) are 178 and 348 kcal mol–1
of: respectively. The energy required for the reaction Mg
(a) s-block elements (b) p-block elements
(g)  Mg2+ (g) + 2e– is
(c) d-block elements (d) f-block elements
(a) + 170 kcal (b) + 526 kcal
4. If the Aufbau principle had not been followed, Ca
(c) – 170 kcal (d) – 525 kcal
(Z = 20) would have been placed in the:
13. For a given value of n (principal quantum number),
(a) s-block (b) p-block
ionization energy is highest for
(c) d-block (d) f-block
(a) d-Electrons
5. Which of the following represents an excited state of an
(b) f-Electrons
atom?
(c) p-Electrons
(a) [Ne] 3s2 3p6 4s2 3d8 (b) [Ne] 3s2 3p6 4s1 3d5 (d) s-Electrons
(c) [Ne] 3s2 3p6 4s2 3d1 (d) 1s2 2s2 2p5 3s1 14. From which of the following species it is easiest to
6. Of the following which will have the minimum atomic remove one electron?
radius? (a) O (g) (b) O2– (g)
(a) N (b) Na
(c) O+ (g) (d) O– (g)
(c) K (d) F
15. The electronic configurations of the elements X, Y, Z,
7. In which of the following sets of elements, they have
and J are given below. Which element has the highest
nearly the same atomic size
metallic character?
(a) Li, Be, B, C (b) Mg, Ca, Sr, Ba
(a) X = 2, 8, 4 (b) Y = 2, 8, 8
(c) O, S, Se, Te (d) Fe, Co, Ni, Cu
(c) Z = 2, 8, 8, 1 (d) J = 2, 8, 8, 7
PERIODIC PROPERTIES 91
16. The ionization energy of boron is less than that of (c) I and II (d) II and III
beryllium because: 22. Which of the following properties shows a gradual
(a) beryllium has a higher nuclear charge than boron decrease with an increase in atomic number across a
(b) beryllium has a lower nuclear charge than boron period in the periodic table?
(c) the outermost electron in boron occupies a (a) Electron Affinity
2p-orbital (b) Ionization potential
(d) the 2s and 2p-orbitals of boron are degenerate (c) Electronegativity
17. An element with lowest ionization energy in their (d) Size of atom.
respective period is
(a) chalcogen (b) halogen Assertion-Reason
(c) alkali metals (d) inert gas
18. Second electron affinity While answering these questions, you are required to choose
(a) Is always negative (energy is absorbed) any one of the following four responses.
(b) Is always positive (energy is released) (A) If both Assertion and Reason are correct and the Reason is
(c) Can be positive or negative a correct explanation of the Assertion.
(d) Is always zero (B) If both Assertion and Reason are correct but Reason is not
19. The lower electron affinity of fluorine than that of a correct explanation of the Assertion.
chlorine is due to (C) If the Assertion is correct but Reason is incorrect.
(a) Smaller size (D) If the Reason is correct but Assertion is incorrect.
(b) Smaller nuclear charge
(c) Difference in their electronic configurations 23. Assertion: Third ionization energy of sulphur is larger
(d) Its highest reactivity than phosphorous.
20. A, B, and C are hydroxy-compounds of the elements X, Reason: There is a larger amount of stability associated
Y, and Z respectively. X, Y, and Z are in the same with filled s- and p- sub-shells (a noble gas electron
period of the periodic table. A gives an aqueous configuration) which corresponds to having eight
solution of pH less than seven. B reacts with both electrons in the valence shell of an atom or ion.
strong acids and strong alkalis. C gives a strongly (a) A (b) B
alkaline aqueous solution. (c) C (d) D
Which of the following statements is/are true? 24. Assertion: Generally, ionization enthalpy increases
I: The three elements are metals. from left to right in a period.
II: The electronegativities decrease from X to Y to Z. Reason: When successive electrons are added to the
III: The atomic radius decreases in the order X, Y, and orbitals in the same principal quantum level, the
Z. shielding effect of the inner core of electrons does not
IV:X, Y, and Z could be phosphorus, aluminum, and increase very much to compensate for the increased
sodium respectively. attraction of the electron to the nucleus.
(a) I, II, III only correct (a) A (b) B
(b) I, III only correct (c) C (d) D
(c) II, IV only correct 25. Assertion: Boron has a smaller first ionization enthalpy
(d) II, III, IV only correct than beryllium.
21. Consider the following statements: Reason: The penetration of a 2s electron to the nucleus
(I) The radius of an anion is larger than that of the is more than the 2p electron hence 2p electron is more
parent atom. shielded by the inner core of electrons than the 2s
(II) The ionization energy generally increases with the electrons.
increasing atomic number in a period. (a) A (b) B
(III) The electronegativity of an element is the tendency (c) C (d) D
of an isolated atom to repel an electron.
Which of the above statements is/are correct?
(a) I alone (b) II alone
92 PERIODIC PROPERTIES

26. Assertion: Manganese (atomic number 25) has a less Comprehension based questions
favorable electron affinity than its neighbors on either Paragraph for Questions 28 to 30
side.
Reason: The magnitude of an element’s electron Elements with their electronic configurations are given below:
Answer the following questions:
affinity depends on the element’s valence shell
I: 1s2 2s2
electrons configuration.
II: 1s2 2s2 2p6
(a) A (b) B
III: 1s2 2s2 2p6 3s2
(c) C (d) D
27. Assertion: Electron gain enthalpy becomes less IV: 1s2 2s2 2p3
negative as we go down a group. V: 1s2 2s2 2p5
Reason: The size of the atom increases on going down
28. The element with highest I.E. is:
the group and the added electron would be farther from (a) I (b) III
the nucleus. (c) II (d) V
29. The element with lowest electron gain enthalpy is:
(a) A
(a) I (b) II
(b) B
(c) III (d) IV
(c) C
30. The most ionic compound will be formed between:
(d) D
(a) I and IV (b) I and V
(c) III and IV (d) III and V
PERIODIC PROPERTIES 93

Notes:

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

PERIODIC PROPERTIES
 03
CHEMICAL BONDING
Chapter 03

Chemical Bonding

1. Introduction (ii) Expansion of Octet: The compounds having more than 8

electrons in the outermost shell of central atom.
Atoms are usually not capable of free existence but groups of For example: PCl5, SF6
atoms of the same or different elements exist as one species, e.g.,
(iii) Odd Electron Bonds: In molecules with an odd number of
H2, O2, P4, S8, H2O.
electrons like nitric oxide, NO and nitrogen dioxide, NO2, the
A group of atoms existing together as one species and having octet rule is not satisfied for all the atoms.
characteristic properties is called a molecule.
Obviously, there must be some force which holds these atoms
together within the molecules. 3. Lewis Dot Structure

2. Kossel-Lewis Approach (Octet 3.1 Valency

Rule) The valence or valency of an element is the measure of its
combining capacity with other atoms when it forms chemical
The atoms of different elements combine with each other in order
compounds or molecules.
to complete their respective octets (i.e., 8 electrons in their
outermost shell) or duplet (i.e., outermost shell having 2
electrons) in case of H, Li and Be to attain stable nearest noble 3.2 Lewis Symbols
gas configuration.
In the formation of a molecule, only the outer shell electrons are
involved and they are known as valence electrons. The inner
2.1 Modes of Chemical Compositions shell electrons are well protected and are generally not involved
This can occur in two ways : in the combination process. It is, therefore, quite reasonable to
consider the outer shell electrons, i.e., valence shell electrons
1. By complete transference of one or more electrons from
while discussing chemical bonds.
one atom to another: This process is referred to as
electrovalency and the chemical bond formed is termed as G.N. Lewis introduced simple symbols to denote the valence shell
electrovalent bond or ionic bond. electrons in an atom. The outer shell electrons are shown as dots
surrounding the symbol of the atom. These symbols are known
2. By sharing of electrons: This can occur in two ways as
as Lewis symbols or electron dot symbols. These symbols ignore
follows:
the inner shell electrons. A few examples are given below:
(i) When the shared electrons are contributed by the two
combining atoms equally, the bond formed is called covalent
bond.
(ii) When these electrons are contributed entirely by one of the Fig.3.1 : Lewis Symbols
atoms but shared by both, the bond formed is known as a
coordinate bond, also called dative bond.
4. Covalent Bond
The bond formed between the two atoms by mutual sharing of
2.2 Exceptions of Octet Rule electrons between them so as to complete their octets or duplets
(i) Electron Deficient Compounds: Compounds having less than in case of elements having only one shell is called covalent bond
8 electrons in the outermost shell of the central atom. or covalent linkage and the number of electrons contributed by
For example: BF3, BeCl2, AlCl3 each atom is known as covalency.

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For example:
6. Formal Charge
The formal charge on an atom in a molecule or ion is defined as the
difference between the number of valence electrons of that atom
in the free state and the number of electrons assigned to that atom
in the Lewis structure, assuming that in each shared pair of
electrons, the atom has one electron of its own and the lone pair
on it belongs to it completely. Thus, it can be calculated as follows:
 Formal charge  F.C.  on   Total no. of valence electrons 
  –
 an atom in a molecule/ion   in the free atom 
Fig.3.2 : Formation of Covalent Bond
 Total no. of electrons of  1  Total no. of shared electrons 
For Example:  lone pairs non  bonding electrons  –  
    2  bonding electrons  
Draw the Lewis dot structure of HCN molecule.
Solution 1
i.e., FC  V  L  S
2
Step-1 : Total number of valence electrons in HCN = 1 + 4 + 5 = 10
(1H = 1, 6C = 2 ,4, 7N = 2, 5) For Example:
Step-2 : Write the skeletal structure that is HCN. Calculate formal charge on each O-atom of O3 molecule.
Sol. Lewis structure of O3 is:
Step-3 : Putting one shared pair of electrons between H and C and
one between C and N, and the remaining as lone pairs, we have

In this structure, duplet of H is complete but octets of C and N are

not complete. Hence, multiple bonding is required between C and The atoms have been numbered as 1, 2 and 3.
N. Octets of C and N will be complete if there is triple bond between Formal charge on end O-atom numbered 1
C and N. Thus,
1
 64  4  0
2
Formal charge on central O-atom numbered 2
5.Lewis Representation of Some 1
 6   1
Molecules and Ions
 62
2
Table : 3.1 : The Lewis Representation of Some Molecules Formal charge on end O-atom numbered 3

1
 66  2   1
2
Hence, we represent O3 along with formal charges as :

6.1 Significance of Formal Charge

(Perchloric
acid) The main advantage of the calculation of formal charges is that it
helps to select the most stable structure, i.e., the one with least
(Nitrous acid)
energy out of the different possible Lewis structures. The most
stable is the one which has the smallest formal charges on the
(Nitric acid) atoms.

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7. Ionic Bond 7.3 Lattice Energy of an Ionic Compound

When a bond is formed by complete transference of electrons Depends upon Following Factors
from one atom to another so as to complete their outermost orbits (i) Size of the ions: Smaller the size, greater will be the lattice
by acquiring 8 electrons (i.e., octet) or 2 electrons (i.e., duplet) in energy.
case of hydrogen, lithium etc. and hence acquire the stable nearest (ii) Charge on the ions: Greater the magnitude of charge, greater
noble gas configuration, the bond formed is called ionic bond or the interionic attraction and hence higher the lattice energy.
electrovalent bond.
For example:
7.4 Hydration Energy of Ionic Bond
Whenever any ionic or polar covalent compound dissolve in polar
solvent (water), compound will get separated in gaseous cation
and anion.
For example:

LiF  s   Li   g   F  g 
(i) Cation and anion in the gaseous state reacts with a solvent
and will get surrounded by polar solvent molecule.
(ii) This process is known as solvation or hydration ( if water is
Fig.3.3 : Formation of Ionic Bond
solvent).
(iii) Energy released in this process is known as solvation energy
or hydration energy.
7.1 Electrovalency
The number of electrons lost or gained during the formation of an
electrovalent linkage is termed as the electrovalency of the element. NOTE
For example, sodium and calcium lost 1 and 2 electrons respectively Hydration energy of an ion is directly proportional to the charge
and so their valencies are 1 and 2. Similarly, chlorine and oxygen of the ion and inversly proportional to its ionic radii. Thus,
gain 1 and 2 electrons respectively, so they possess an higher the charge to size ratio i.e charge density, higher will be
electrovalency of 1 and 2. In other words, valency is equal to the the hydration energy.
charge on the ion.
7.4.1 Application of Hydration Energy
7.2 Factors Governing the Formation of (i) Size of the hydrated ions - Greater the hydration of the ion ,
greater will be its hydrated radii.
Ionic Bonds
For example:
There are three main factors:
Li + (aq) > Na+ (aq)
(i) Ionisation energy (IE) : The lower the value of the ionisation
(ii) Mobility of the ion - More is the hydration, smaller will be the
energy of an atom greater will be the ease of formation of
mobility of the ions in the electric field.
cation from it. The size of the cation is always smaller than
For example:
the atom from which it is derived.
(ii) Electron affinity (EA) : The higher the electron affinity of an Li + < Na + < K + < Rb+ < Cs+
atom the greater will be the ease of formation of anion from it. (iii) Electrical conductance - It is related to mobility of ions.
The size of the anion is always larger than the atom from As Mobility increases, electrical conductance increases.
which it is derived.
For example:
(iii) Lattice energy (U) : The amount of heat evolved when one
mole of ionic compound is formed from positive and negative Li + < Na + < K + < Rb+ < Cs+
ions in the gaseous form is known as the lattice energy. Lattice
energy must be very high.

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7.5 Characteristics of Ionic Compounds 8. Fajan’s Rule

(i) Physical State This rule is for covalent character of an ionic bond.Covalent
These compounds usually exist in the solid state. character of an ionic bond is favoured by :
(ii) Crystal Structure (i) Small positive ion
The crystalline ionic compounds have well defined crystal (ii) Large negative ion
structure or crystal lattice. (iii) Large charge on ions.
For example: In NaCl each Na+ ion is surrounded by six Cl– ions  Thus for a fixed cation, the larger the size of anion, the more
and vice versa. the magnitude of the charge, the more is covalent character
For eg. covalent character of sodium halides follows the order:
NaI > NaBr > NaCl > NaF
 For fixed anion, the smaller the size of cation, the more the
magnitude of the charge, the more is the covalent character
For eg. BeCl2 > MgCl2 > CaCl2 > SrCl2 > BaCl2.
 It has been observed that a cation having 18 (s2p6d10) electrons
in outermost shell (pseudo noble gas configuration) can
polarise the anion more than cation having 8 (s2p6 ) noble gas
electronic configuration.
Hence CuCl is more covalent than NaCl. Similarly AgCl is
more covalent than KCl.

Fig.3.4 : The Crystal Structure of Sodium Chloride, NaCl 8.1 Polarising Power of Cation
(iii) High melting and boiling points It is generally represented by φ . The ability of a cation to polarize
Ionic compounds possess high melting and boiling points. the anion is referred to as polarizing power. It is directly
This is because ions are tightly held together by strong proportional to the charge density, which in turn is directly related
electrostatic forces of attraction and hence a huge amount of to the charge on cation, while inversely related to the size of
energy is required to break the crystal lattice. anion.
(iv) Solubility The polarizing power increases with decrease in the size of cation
Electrovalent compounds are soluble in solvents like water i.e. smaller cations are very effective in the polarization of anion.
which are polar in nature and have high dielectric constant. However, the polarizing power increases with increase in the
Non-polar solvents like carbon tetrachloride, benzene etc. charge on cation.
having low dielectric constants are not capable of dissolving charge on cation
ionic solids. Hence, ionic solids are soluble in polar solvents φ 
radius of cation
and insoluble in non-polar solvents.
(v) Electrical conductivity
Ionic compounds are good conductors of electricity in 8.2 Polarisibility of Anion
solution or in the molten state. In solution or molten state,
It is the tendency of an anion to undergo polarization. It indicates
their ions are free to move. As the ions are charged, they are
attracted towards electrodes and thus act as carriers of electric the easiness with which an anion undergoes distortion in presence
current. of a cation.
(vi) Ionic Reactions It is directly proportional to the size as well as the negative charge
The reactions of the ionic compounds are, in fact, the reactions on the anion.
between the ions produced in solution. As the oppositely The larger anions can undergo distortion very easily than the
charged ions combine quickly, these reactions are, therefore, smaller ones.
quite fast. It is also important to note that the anions with greater negative
For example: charge also undergo polarization easily.
Na+ Cl– (aq) + Ag+ NO3– (aq) AgCl (s) + NaNO3 (aq)

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9. Bond Parameters

9.1 Bond Length

The equilibrium distance between the centres of the nuclei of the
two bonded atoms is called its bond length. Fig.3.5 : Bond Angle in Different Molecules

9.1.1 Factors Affecting Bond Length 9.4 Bond Order

(i) Size of the atoms : The bond length increases with increase
In the Lewis representation of a molecule or ion, the number of
in the size of the atoms. For example, bond lengths of H–X
bonds present between two atoms is called bond order. For
are in the order :
example, the bond orders of a few molecules are given below :
HI > HBr > HCl > HF
Molecules : H—H O=O NN CO
(ii) Multiplicity of bond: The bond length decreases with the
multiplicity of the bond. Thus, bond length of carbon-carbon Bond order : 1 2 3 3
bonds are in the order : For odd electron molecules, as the three electron bond is
CC<C=C<C–C considered as equivalent to half covalent bond, bond order can
be fractional also. For example, Lewis structure of NO is

9.2 Bond Energy .

The amount of energy required to break one mole of bonds of a

particular type so as to separate them into gaseous atoms is called
bond dissociation enthalpy or simply bond enthalpy.
10. Resonance
It is often observed that a single Lewis structure is inadequate for
the representation of a molecule in conformity with its
9.2.1 Factors Affecting Bond Energy
experimentally determined parameters. For example, the ozone, O3
(i) Size of the atoms: Greater the size of the atoms, greater is the molecule can be equally represented by the structures I and II
bond length and less is the bond dissociation enthalpy, i.e., shown below:
less is the bond strength.
(ii) Multiplicity of bonds : For the bond between the same two
atoms, greater is the multiplicity of the bond, greater is the
bond dissociation enthalpy. This is firstly because atoms
come closer and secondly, the number of bonds to be broken
is more. For example, bond dissociation enthalpies of H2, O2
and N2 are in the order :
H–H < O = O < N  N
(iii) Number of lone pairs of electrons present: Greater the
number of lone pairs of electrons present on the bonded Fig. 3.6 : Resonance in the O3 molecule (structures I and II
atoms greater is the repulsion between the atoms and hence represent the two canonical forms while the structure III is the
less is the bond dissociation enthalpy. resonance hybrid)
For example, for a few single bonds, we have In both structures we have a O–O single bond and a O=O double
Bond enthalpy in the order bond. The normal O–O and O=O bond lengths are 148 pm and 121
C-C(348kJ/mol) > N-N(163kJ/mol) > O-O(146kJ/mol) > pm respectively. Experimentally determined oxygen-oxygen bond
F-F(139kJ/mol) lengths in the O3 molecule are same (128 pm). Thus the oxygen-
oxygen bonds in the O3 molecule are intermediate between a double
and a single bond. Obviously, this cannot be represented by either
9.3 Bond Angle of the two Lewis structures shown above.
The angle between the lines representing the directions of the The concept of resonance was introduced to deal with the type of
bonds, i.e., the orbitals containing the bonding electrons is called difficulty experienced in the depiction of accurate structures of
the bond angle. molecules like O3 .

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Thus, according to the concept of resonance, whenever a single

Lewis structure cannot explain all the properties of the molecule,
11. Dipole Moment
the molecule is then supposed to have many structures with similar It can be defined as the product of the magnitude of the charge
energy. Positions of nuclei, bonding and nonbonding pairs of and the distance between the centres of positive and negative
electrons are taken as the canonical structure of the hybrid which charge. It is usually designated by a Greek letter ‘µ’. Mathematically,
describes the molecule accurately. For O3, the two structures shown it is expressed as follows :
above are canonical structures and the III structure represents the Dipole moment (µ) = charge (Q) × distance of separation (r)
structure of O3 more accurately. This is also called resonance hybrid. Dipole moment is usually expressed in Debye units (D). The
–30
conversion factor is 1 D = 3.33564 × 10 C m where C is coulomb
and m is meter.
10.1 Rules to Write Canonical Forms
For Example:
(i) The relative position of all the atoms in each of the cononical
(i) In BF3, the dipole moment is zero although the B – F bonds
forms must be the same. They should differ only in the position o
are oriented at an angle of 120 to one another, the three bond
of electrons.
moments give a net sum of zero as the resultant of any two is
(ii) The number of unpaired and paired electrons in each of the equal and opposite to the third.
canonical forms must me same.
(iii) The contributing structures should not differ much in energy.
(iv) The contributing structures should be such that negative
charge resides on more electronegative and positive charge
on the electropositive. Like charges should not reside on
atoms close together in the canonical forms.
Formal charges on the atoms in the molecule help us in choosing Fig.3.7 : Representation of Bond Dipoles
the most appropriate resonance structure. For example, nitrous (ii) In NH3, N is more electronegative than H. So, N pulls the
oxide molecule is represented by two resonance electron dot electrons from H towards itself and therefore, the direction of
structures, one of which has a negative formal charge on the moment due to the N-H bonds is in the same direction as that
oxygen atom and the other of which has a negative charge on the of the lone pair of electrons on Nitrogen.
terminal nitrogen atom.
In NF3, F is more electronegative than N. So, all F atoms pull
the electrons towards themselves. And, thus resultant
moment is opposite to the directions of that of the lone pair
Since oxygen is a more electronegative element than nitrogen, the of electrons on the N atom, thereby the net dipole decreases.
structure that places a negative formal charge on oxygen is And, so dipole moment of NH3 is much higher than that of
probably lower in energy than the structure that has a negative NF3.
formal charge on nitrogen.
Thus, the actual structure of N2O is

10.2 Stability of Resonating Structure

(i) The more covalent bonds a structure has, the more stable it is.
(ii) Structure in which all of the atoms have a complete valence Fig.3.8 : Representation of Bond Dipoles in Compounds with
shell of electrons (i.e., the noble gas structure) are especially Lone Pair
stable and make large contributions to the hybrid.
(iii) Charge separation decreases stability.
(iv) Resonance contributors with negative charge on highly
electronegative atoms are more stable than ones with negative
charge on less or nonelectronegative atoms.

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11.1 Applications of Dipole Moment  Pairs of electrons in the valence shell repel one another since
their electron clouds are negatively charged.
(i) The molecules having zero dipole moment are non-polar
 These pairs of electrons tend to occupy such positions in
molecules and those having net  0 are polar in nature.
space that minimise repulsion and thus maximise distance
(ii) The value of dipole moment can be used for determining the
between them.
amount of ionic character in a bond.
 The valence shell is taken as a sphere with the electron pairs
Percentage of ionic character
localising on the spherical surface at maximum distance from
Experimental value of dipole moment one another.
=  100
Theoretical value of dipole moment  A multiple bond is treated as if it is a single electron pair and
the two or three electron pairs of a multiple bond are treated

12.Valence Shell Electron Pair 

as a single super pair.
Where two or more resonance structures can represent a
Repulsion (VSEPR) Theory molecule, the VSEPR model is applicable to any such
The first simple theory that was put forward to explain the shapes structure.
of molecules is known as Valence Shell Electron Pair Repulsion The repulsive interaction of electron pairs decrease in the order:
(VSEPR) theory. This theory was given by Sidgwick and Powell in Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) >
1940 and was further improved by Nyholm and Gillespie in 1957. Bond pair (bp) – Bond pair (bp)
For the prediction of geometrical shapes of molecules with the
12.1 The Main Postulates of VSEPR help of VSEPR theory, it is convenient to divide molecules into
two categories as
Theory (i) molecules in which the central atom has no lone pair and
 The shape of a molecule depends upon the number of valence (ii) molecules in which the central atom has one or more lone
shell electron pairs (bonded or nonbonded) around the central pairs
atom.

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Table : 3.2 : Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons

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Table : 3.3 (a) : Shape (geometry) of Some Simple Molecules with Central Ions having One or More Lone Pairs of Electrons(E).

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Table : 3.3 (b) : Shape (geometry) of Some Simple Molecules with Central Ions having One or More Lone Pairs of Electrons(E).

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12.2 Predicting the Shapes of the Molecules 13. Valence Bond Theory
and Ions.

(i) Total no. of electron pairs around the central atom 

1
(No.
13.1 Postulates of Valence Bond Theory
2 • Covalent bonds are formed when two valence orbitals (half-
of valence electrons of central atom + No. of atoms linked to filled) belonging to two different atoms overlap on each other.
central atom by single bonds) The electron density in the area between the two bonding
For negative ions, add number of electrons equal to the units atoms increases as a result of this overlapping, thereby
of negative charge on the ion to the valence electrons of the increasing the stability of the resulting molecule.
central atom. • The presence of many unpaired electrons in the valence shell
of an atom enables it to form multiple bonds with other atoms.
For positive ions, subtract number of electrons equal to the
The paired electrons present in the valence shell do not
units of positive charge on the ion from the valence electrons
participate in the formation of chemical bonds as per the
of the central atom.
valence bond theory.
(ii) No. of bond pairs (shared pairs) = No. of atoms linked to
• Covalent chemical bonds are directional and are also parallel
central atom by single bonds. to the region corresponding to the atomic orbitals that are
(iii) No. of lone pairs = Total no. of electron pairs – No. of shared overlapping.
pairs. • Sigma bonds and pi bonds differ in the pattern that the atomic
For Example: orbitals overlap in, i.e. pi bonds are formed from sidewise
On the basis of VSEPR theory, predict the shapes of the following: overlapping whereas the overlapping along the axis
containing the nuclei of the two atoms leads to the formation
(i) ClF3 (ii) BrF5
of sigma bonds.
Solution.
(i) Shape of ClF3
No. of valence electrons of the central Cl atom = 7
13.2 Overlapping of Atomic Orbital
No. of atoms linked to it by single bonds = 3. When two atoms come in contact with each other to form a bond,
their overlap can be positive, negative or even zero depending
73 upon the sign (phase) and direction of orientation of amplitude of
 Total no. of electron pairs around Cl  5
orbital wave function in space.
2
No. of bond pairs = No. of atoms linked to Cl = 3. (i) Positive Overlapping of Atomic Orbital: When the phase of
two interacting orbitals is same, then the overlap is positive
 No. of lone pairs = 5 – 3 = 2. and in this case, the bond is formed. The phase of the two
Thus, the molecule is of the type AB3L2. interacting orbital (+ or -) comes from the sign of orbital wave
Hence, it is T-shaped. function and is not related to the charge in any sense.
(ii) Shape of BrF5
No. of valence electrons of central Br atom = 7
No. of atoms linked to it by single bonds = 5.

75
 Total no. of electron pairs around Br  6
2
No. of bond pairs = No. of atoms linked to Br = 5.
 No. of lone pairs = 6 – 5 = 1.
Thus, the molecule is of the type AB5L. Hence, it has square
pyramidal shape.

Fig.3.9 : Positive Overlapping of Atomic Orbital

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(ii) Negative Overlapping of Atomic Orbital: When the phase of (i) s-s overlapping
two interacting atomic orbital is opposite, then the overlap is
negative and in this case, the bond is not formed.

Fig.3.12 : s-s overlapping forming a sigma bond

(ii) s-p overlapping

Fig.3.13 : s-p overlapping

(iii) p-p overlapping

Fig.3.14 : p-p overlapping

Fig.3.10 : Negative Overlapping of Atomic Orbital 2. Pi () Bond
(iii) Zero Overlapping of Atomic Orbital: When the orientation Pi-bond is formed by lateral (sideways) overlapping of p-orbitals,
of two interacting atomic orbital is such that there is no i.e., by overlapping of p-orbitals in a direction at right angles to
overlapping of the orbital, that is known as zero overlapping. the internuclear axis.

Fig.3.11 : Zero Overlapping of Atomic Orbital

13.3 Types of Overlapping and Nature of Fig.3.15 : p-p overlapping forming a pi bond
Covalent Bonds For eg.
Depending upon the type of overlapping, the covalent bonds are In case of oxygen molecule (each oxygen atom having electronic
mainly of two types: configuration, 1s2 2s2 2p2x 2p1y 2p1z), the two atoms are held
together by one -bond and one -bond as shown in figure.
1. Sigma () bond
When a bond is formed between two atoms by the overlap of
their atomic orbitals along the internuclear axis (end to end or
head on overlap), the bond formed is called sigma () bond.

Fig.3.16 : Formation of oxygen molecule

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13.4 Comparison of Sigma and Pi bonds X

1   No. of valence electrons  
  
Sigma () Bond Pi () Bond 2  of the central atom 
(i) This bond is formed by (i) This is formed by sideway Charge on the cation 
overlap of orbitals along overlapping of orbitals  No. of monovalent atoms / groups   
   if the given species is 
their internuclear axis (lateral overlapping). surrounding the central atom   
a polyatomic cation 
(end to end overlap).
Charge on the anion 
(ii) This is formed by (ii) This is formed by the  
 if the given species is 
overlapping between s-s, overlap of p-p orbitals a polyatomic anion 
 
s-p or p-p orbitals. only.
(iii) Overlapping is quite large (iii) Overlapping is to a small 1
i.e., X   VE  MA  c  a 
2
and hence sigma bond extent. Hence, -bond is
Note that only monovalent atoms (MA) or groups are to be
is a strong bond. a weak bond.
considered. For divalent ions, MA = 0.
(iv) Electron cloud in this case (iv) Electron cloud of -bond
If X = 2, it means two hybrid orbitals are to be formed. Hence,
is symmetrical. is unsymmetrical. hybridization is sp. If X = 3, it means three hybrid orbitals are to be
(v) Sigma bond consists of (v) Pi () bond consists of formed. Hence, hybridisation is sp2 and so on, as given in the
only one electron cloud, electron clouds, one following table :
symmetrical about the above the plane of atomic
internuclear axis. nuclei and the other Value of X 2 3 4 5 6 7
below it. Type of
sp sp2 sp3 sp3d sp3d2 sp3d3
(vi) Free rotation about a (vi) Free rotation about a hybridisation
-bond is possible. -bond is not possible

15. Molecular Orbital Theory

because on rotation,
overlapping vanishes and
so the bond breaks. When the atoms to be bonded come close together, the orbitals of
the bonded atoms lose their individual character and fuse (overlap)
to form larger orbitals called molecular orbitals. Like atomic orbitals,
14. Hybridisation there are molecular orbitals in a molecule. The only difference is
Hybridisation is defined as the mixing of the atomic orbitals that in atomic orbitals, electrons move under the influence of only
belonging to the same atom but having slightly different energies one nucleus (i.e. Atomic orbital are monocentric), while in molecular
so that a redistribution of energy takes place between them resulting orbitals, electrons move under the influence of many nuclei, they
in the formation of new orbitals of equal energies and identical are polycentric.
shapes. The new orbitals thus formed are known as hybrid orbitals.
15.1 Important Features of M.O.T.
14.1 Some Important Points About (i) Like an Atomic orbital which is around the nucleus of an
atom there are molecular orbital which are around the nuclei
Hybridisation of a molecule.
(i) Only those orbitals which have approximately equal energies
(ii) The molecular orbitals are entirely different from the atomic
and belong to the same atom or ion can undergo hybridisation.
orbitals from which they are formed.
(ii) Number of hybrid orbitals produced is equal to the number of
(iii) The valence electrons of the constituent atoms are considered
atomic orbitals mixed.
to be moving under the influence of nuclei of participating
atoms in the molecular orbital.
(iv) The molecular orbitals possess different energy levels like
14. 2 Predicting Hybridisation
atomic orbitals in an isolated atom.
Calculate the number of hybrid orbitals (X) to be formed by the
(v) The shape of molecular orbitals are dependent upon the
central atom as follows :
shapes of atomic orbitals from which they are formed.

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(vi) Molecular orbitals are arranged in order of increasing energy

just like atomic orbitals.
(vii) The number of molecular orbitals formed is equal to the
number of atomic orbitals combining in bond formation.
(viii) Like atomic orbitals, the filling of electrons in molecular
orbitals is governed by the three principles such as Aufbau
principle, Hund’s rule and Pauli’s exclusion principle.

15.2 Conditions for Atomic Orbitals to

Form Molecular Orbital (M.O.)
(i) The combining atomic orbitals should be of a comparable energy.
(ii) The combining atomic orbitals must overlap to a large extent.
Greater the overlap, stable is the molecule formed.

Fig.3.17 : Energy level diagram for 1s atomic orbitals

15.3 Linear Combination of Atomic
Orbitals (LCAO)
(i) The most convenient way of working out the wave functions
for molecular orbitals is to adopt the method of linear
combination of atomic orbitals (LCAO) i.e. by addition or
subtraction of wave funtions of individual orbitals thus
 MO   A   B
b  A  B
a  A – B
 b : Bonding molecular orbital,  a : Antibonding molecular
orbital
(ii) Quantum mechanics shows that linear combination of two
wave functions gives, two linear combination and hence two
Fig.3.18 : Energy level diagram for 2pz atomic orbitals
molecular orbitals:
(a) Bonding orbital : Molecular orbital of lower energy.
(b) Antibonding orbital : Molecular orbital of higher energy.

15.3.1 Energy Level Diagram for Molecular Orbitals

(i) Two 1s atomic orbitals combine to form two molecular orbitals:
(a) Bonding MO : 1s
(b) Antibonding MO : 1s
(ii) 2s and 2p atomic orbitals give rise to the following eight
molecular orbitals:
(a) Bonding MO : 2s, 2pz, 2px, 2py
(b) Antibonding MO : 2s, 2pz, 2px, 2py

Fig.3.19 : Energy level diagram for 2px atomic orbitals

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15.4 Relative Energies of M.O. and Filling 15.5 Electronic Configuration/Bond Order
of Electron of Simple Diatomic Molecules
The electronic configuration and the bond order in case of simple
diatomic molecules can be obtained by filling the molecular orbitals
by applying Aufbau principle and Hund's rule.

1
Bond order = [Number of bonding electrons – number of
2
antibonding electrons]

15.6 Bonding in Some Diatomic Molecules

and Ions
(i) Hydrogen molecule ( H 2 ): Total number of electrons = 2,
filling in molecular orbitals we have  1s 2

Nb  Na 2  0 2
Bond order =   1
2 2 2
Hence there is a single bond between two hydrogen atoms
(H - H). Since there is no unpaired electron, H 2 is
diamagnetic
(ii) Helium molecule ( He2 ): The total number of electrons = 4
Fig. 3.20 : M.O Energy level diagram for O2, F2 and Ne.
and filling in molecular orbitals we have  1s 2   *1s 2

Nb  Na 2  2
Bond order =  0
2 2
Hence He2 molecule cannot exist
(iii) Lithium molecule (Li2): There are six electrons in lithium
molecule. The electronic configuration of Li molecule may
be written as: Li2 : 1s 2  1s *2  2s 2
Nb – Na 4 – 2
Bond order   1
2 2
Thus, there is one Li–Li sigma bond.
It is diamagnetic.
(iv) Beryllium molecule (Be2): Total number of electrons = 8.
The molecular orbital electronic configuration is:
Be 2 : 1s 2   *1s 2   2s 2   * 2s 2

Nb – Na 1
Bond order    4 – 4  0
2 2
The zero bond order suggests that Be2 molecule does not
exist.
Fig. 3.21 : M.O energy diagram for Li2, Be2, B2, C2 and N2
molecule

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(v) Boron molecule (B2): Total number of electrons= 10. The It is paramagnetic
molecular orbital electronic configuration is: (xi) O -2 (Super oxide ion): Total number of electrons
2 2 2 2 1 1
B2 : 1s   *1s  2s   * 2s  2p  2p x y (16 + 1) = 17 .
Nb – Na 1 Electronic configuration
Bond order    6 – 4  1
2 2
It is paramagnetic.
 1s2   *1s2   2s2   *2s2   2p2z
(vi) Carbon molecule (C2): Total number of electrons= 12. The
molecular orbital electronic configuration is:   2p2x   2p2y   *2p2x   *2p1y

C2 : 1s 2   *1s 2   2s 2   * 2s 2  2p x 2  2p y 2 Bond order

Nb – Na 1 N b  N a 10  7 3 1
Bond order   8 – 4  2    1
2 2 2 2 2 2
Thus the C2 molecule has two bonds. It is paramagnetic.
It is diamagnetic.
2– (xii) Peroxide ion O 22  - Total number of electrons(16 + 2) = 18
(vii) C2 : Total number of electrons= 14. The molecular orbital
. The electronic configuration is
electronic configuration is:
C2– 2 2 2 2 2 2 2  1s 2   *1s 2   2s 2   * 2s2   2p z2   2p 2x
2 : 1s   *1s   2s   * 2s   2p x  2p y  2p z

Nb – Na 1   2p 2y   * 2p 2x   * 2p 2y
Bond order   10 – 4   3
2 2
It is diamagnetic. N b  N a 10  8 2
Bond order =   1
2 2 2
(viii) Nitrogen molecule ( N 2 ) - The total number of electrons =
It is diamagnetic
14 and filling in molecular orbitals we have
(xiii) Fluorine molecule (F2)- Total number of electrons = 18.
 1s 2   *1s 2   2s 2   * 2s 2   2p 2x   2p 2y   2p 2z The electronic configuration is

N b  N a 10  4 6 .. ..  1s 2   *1s 2   2s 2   * 2s 2   2p 2z   2p 2x
Bond order     3 N  N
2 2 2   2p 2y   * 2p 2x   * 2p 2y
It is diamagnetic. 10 – 8
Bond order  1
(ix) Oxygen molecule ( O 2 )- Total number of electrons =16 and 2
electronic configuration is (xiv) Carbon monoxide (CO) - Total number of electrons
= (6 + 8) = 14
 1s 2   *1s 2   2s 2   * 2s 2   2p z2   2p 2x The electronic configuration is
  2p 2y   * 2p1x   * 2p1y
 1s2   *1s 2   2s2  pz2   2p x2   2p 2y   * 2s2
N b – N a 10 – 6 4
Bond order =   2 N b  N a 10  4 6
2 2 2 Bond order =   3
2 2 2
As shown by electronic configuration the O2 molecule
contains two unpaired electrons, hence it is paramagnetic It is diamagnetic
+
in nature. (xv) CO - Total number of electrons
(x) O +2 - Total number of electrons (16 – 1) = 15. = (6 + 8 – 1) = 13
Electronic configuration The electronic configuration is

 1s 2   *1s 2   2s 2   * 2s 2  p z2  1s2   *1s 2   2s2   2p x2   2p2y   2p z 2   * 2s1

  2p 2x   2p 2y   * 2p1x Nb – Na 1
Bond order   10 – 3  3.5
2 2
N b  N a 10  5 5 1 It is paramagnetic.
Bond order   2
2 2 2 2

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NOTE: As a result of hydrogen bonding, a H–atom links the two

+ electronegative atoms simultaneously, one by a covalent bond
It may be noted that in CO and CO , *2s MO is higher in energy
and the other by a hydrogen bond. Hence it is said to form a
2px, 2py and 2pz
hydrogen bridge. It is merely a strong electrostatic attractive forces
and not a normal chemical bond. It is very weak (strength about 2-
(xvi)Nitric Oxide (NO)- Total number of electrons 10 K cal/mol).
= (7 + 8) = 15
 1s 2   *1s 2   s 2   * 2s 2   2p 2z 17.1.1 Conditions for Hydrogen Bonding
(i) The molecule must contain a highly electronegative atom linked
  p 2x   2p 2y   * 2p1x
to H-atom. The higher the electronegativity, more is the
N b  N a 10  5 5 1 polarisation of the molecule.
Bond order    2
2 2 2 2 (ii) The size of the electronegative atom should be small. The
It is paramagnetic smaller the size the greater is the electrostatic attraction.

17.1.2 Types of Hydrogen Bonding

16. Metallic Bond
 The constituent particles of metallic solids are metal atoms 17.1.2.1 Intermolecular Hydrogen Bonding
which are held together by metallic bond. A metal atom is When hydrogen bonding takes place between different molecules
supposed to consist of two parts, valence electrons and the of the same or different compounds, it is called intermolecular
remaining part (the nucleus and the inner shells) called hydrogen bonding e.g. HF, H2O, ROH (same compound) water-
kernel. alcohol, water-ammonia (different compound) etc.
 The kernels of metal atoms occupy the lattice sites while the
space between the kernel is occupied by valence electrons.
 Due to small ionisation energy the valence electrons of metal
atoms are not held by the nucleus firmly. Therefore, the
electrons leave the field of influence of one kernel and come Fig. 3.22 : Intermolecular Hydrogen Bonding
under the influence of another kernel.
 Thus the electrons are not localised but are mobile. The 17.1.2.1 Intramolecular Hydrogen Bonding
simultaneous attraction between the kernels and the mobile The hydrogen bonding which takes place within a molecule itself.
electrons which hold the kernel together is known as metallic It takes place in compounds containing two groups such that one
bond. group contain a H-atom linked to an electronegative atom and the
 This model is known electron sea model. other group contains a highly electronegative atom linked to a
lesser electronegative atom. The bond is formed between the H-
atom of one group with the more electronegative atom of the
17. Intermolecular Forces other group.
Examples are as shown below:

17.1 Hydrogen Bonding

When a hydrogen atom linked to a highly electronegative atom
(like F,O, or N) comes under the influence of another strongly
electronegative atom, then a weak bond is developed between
them which is called as hydrogen bond. It is represented by dotted
line as follows Fig. 3.23 : Intramolecular Hydrogen Bonding

 –  –  –
H– X     H– X     H– X

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17.2 Ion Dipole Attraction

+
This force is between an ion such as Na and a polar molecule 17.3.1.2 Dipole-Induced Dipole Attraction
such as HCl A dipole-induced dipole attraction is a weak attraction that results
when a polar molecule induces a dipole in an atom or in a non
polar molecule by disturbing the arrangement of electrons in the
non polar species.

Na+ HCl
17.3.1.3 Induced Dipole (London Dispersion Force)
Fig. 3.24 : Ion Dipole Attraction
The London dispersion force is the weakest intermolecular force.
The London dispersion force is a temporary attractive force that
17.3 Vander Waals Forces results when the electrons in two adjacent atoms occupy positions
that make the atoms form temporary dipoles. This force is
(i) This type of attractive forces occurs in case of non polar
sometimes called an induced dipole induced dipole attraction.
molecules such as H2, O2, Cl2, CH4, CO2, etc.
(ii) The existence of weak attractive forces among the nonpolar
molecule was first proposed by Dutch scientist J.D. Vander NOTE:
Waal. The relative strength of various bonds is as follows :
(iii) Vander waal force  molecular weight  Boiling point. Ionic bond > Covalent bond > Metallic bond > H-bond > Vander
waal bond.
17.3.1 Types of Vander Waals Force

17.3.1.1 Dipole-Dipole Attraction

A dipole-dipole force is when the positive side of a polar molecule
(a molecule with a slightly positive side and a slightly negative
side) attracts the negative side of another polar molecule.

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Summary

• Kossel’s first insight into the mechanism of formation of Basically the VB theory discusses bond formation in terms
of overlap of orbitals.
electropositive and electronegative ions related the process
to the attainment of noble gas configurations by the respective • For explaining the characteristic shapes of polyatomic
ions. Electrostatic attraction between ions is the cause for molecules Pauling introduced the concept of hybridisation
their stability. This gives the concept of electrovalency. of atomic orbitals. sp, sp2, sp3 hybridizations of atomic
• The first description of covalent bonding was provided by orbitals of Be, B, C, N and O are used to explain the formation
and geometrical shapes of molecules like BeCl2, BCl3, CH4,
Lewis in terms of the sharing of electron pairs between atoms
NH3 and H2O. They also explain the formation of multiple
and he related the process to the attainment of noble gas
bonds in molecules like C2H2 and C2H4.
configurations by reacting atoms as a result of sharing of
electrons. The Lewis dot symbols show the number of valence • To predict hybridisation following furmula may be used:
electrons of the atoms of a given element and Lewis dot
1
structures show pictorial representations of bonding in No. of hybrid orbital (X) = [Total no. of valence e– in the
molecules. 2
central atom + total no. of monovalent atoms – charge on
• An ionic compound is pictured as a three-dimensional
cation + charge on anion]
aggregation of positive and negative ions in an ordered
Value of X 2 3 4 5 6 7
arrangement called the crystal lattice. In a crystalline solid
there is a charge balance between the positive and negative Type of sp sp 2
sp 3 3
sp d 3
sp d 2
sp d3
3

ions. The crystal lattice is stabilized by the enthalpy of lattice hybridisation

formation. • The molecular orbital (MO) theory describes bonding in
• While a single covalent bond is formed by sharing of an terms of the combination and arrangement of atomic orbitals
electron pair between two atoms, multiple bonds result from to form molecular orbitals that are associated with the molecule
the sharing of two or three electron pairs. Some bonded atoms as a whole. The number of molecular orbitals are always equal
have additional pairs of electrons not involved in bonding. to the number of atomic orbitals from which they are formed.
These are called lone pairs of electrons. A Lewis dot structure • The electronic configuration of the molecules is written by
shows the arrangement of bonded pairs and lone pairs around
filling electrons in the molecular orbitals in the order of
each atom in a molecule. Important parameters, associated
increasing energy levels. As in the case of atoms, the Pauli
with chemical bonds, like: bond length, bond angle, bond
exclusion principle and Hund’s rule are applicable for
enthalpy, bond order and bond polarity have significant effect
the filling of molecular orbitals. Molecules are said to be stable
on the properties of compounds.
if the number of elctrons in bonding molecular orbitals is
• There is a very important and extremely useful concept called greater than that in antibonding molecular orbitals.
resonance. The contributing structures or canonical forms
taken together constitute the resonance hybrid which 1
• Bond order = [No. of e– in bonding orbitals –
represents the molecule or ion. 2
• The VSEPR model used for predicting the geometrical No. of e– in antibonding orbitals.]
shapes of molecules is based on the assumption that electron • Hydrogen bond is formed when a hydrogen atom finds itself
pairs repel each other and, therefore, tend to remain as far between two highly electronegative atoms such as F, O and
apart as possible. According to this model, molecular N. It may be intermolecular (existing between two or more
geometry is determined by repulsions between lone pairs molecules of the same or different substances) or
and lone pairs; lone pairs and bonding pairs and bonding intramolecular (present within the same molecule).
pairs and bonding pairs. The order of these repulsions being:
lp-lp > lp-bp > bp-bp
• The valence bond (VB) approach to covalent bonding is
basically concerned with the energetics of covalent bond
formation about which the Lewis and VSEPR models are silent.

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Solved Examples
Example-1 Example-4
t Explain the formation of a chemical bond. Write the favourable factors for the formation of ionic bond.
Sol. According to Kossel and Lewis, atoms combine together Sol. 1. Low ionisation enthalpy of metal atoms
in order to complete their respective octets as to acquire 2. High electron gain enthalpy of non-metal atoms
the stable inert gas configuration. This can occur in two 3. High lattice enthalpy of compound formed.
ways, by transfer of one or more electrons from one atom to
Example-5
other by sharing of electrons between two or more atoms.
What type of bond is formed when atoms have
Example-2
(i) Zero difference of electronegativity.
Write Lewis dot symbols for atoms of the following (ii) Little difference of electronegativity.
elements: Mg, Na, B, O, N, Br.
(iii) High difference of electronegativity.
Sol. 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?
Sol. Bond strength is directly proportional to the bond order.
Greater the bond order, more is the bond strength.
Example-7
Write the resonance structures for SO3, NO2 and NO3–.
Example-3
Sol.
Write Lewis dot structures of the following molecules/ions:
(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 Example-8
R = 2 + 8 + 8 = 18 electrons Which out of and has higher dipole moment and why?
S = 8 electrons Sol. In and , the difference in electronegativity is nearly same but
U = 2 electrons the dipole moment of = (1.46D) is greater than, NF3 = (0.24D)
H C N or H – C N In , the dipole moments of the three N–H bonds are in the
same direction as the lone pair of electron.
CHEMICAL BONDING 115

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

pairs, but in H 2 O there are two bond pairs and

two lone pairs. PH3 : Trigonal pyramidal : Phosphorus atom has 5 electrons
Example-12 in its outermost orbit. H-atoms contribute one electron each
to make in all 8 electrons around P-atom. Thus, 4 pairs of
Discuss the shape of the following molecules using the
electrons would be distributed in a tetrahedral fashion
VSEPR model.
around the central atom. Three pairs form three P-H bonds
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3. while the fourth pair remains unused. Due to repulsion
between the bonding and lone pairs of electrons, the angle.
116 CHEMICAL BONDING

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 PCl 3 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. Indicate the number of  and -bonds in the following
Two of the positions are occupied by lone pairs, hence the molecules.
actual structure is bent or V-shaped.
(i) CH 3  CH  CH 2
In PCl3, three bonded pairs and one lone pair of electrons
are present. Tetrahedral configuration comes into existence. (ii) CH 3  CH 2  CH 2  CH3
One position is occupied by lone pair, hence the actual
(iii) CH 3  C  C  CH 3
structure is pyramidal.
Example-14 (iv)
What is the total number of sigma and pi bonds in the
Sol. For finding out the number of  and -bonds in a molecule,
following molecules?
the following points should be kept in mind:
(a) C2H2 (b) C2H4 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
orbitals along the intermolecular axis increases the electron- Is there any change in the hybridization of B and N atoms as
density between the two nuclei resulting in a decrease in a result of the following reaction?
the energy and formation of a bond. BF3 + NH3 F3B NH3.
Example-16 Sol. Here, B atom in BF3 is sp2 hybridized and one of its p orbital
Out of  and –bonds which one is stronger and why? is empty. N in NH3 is sp3 hybridized and are of its hybrid
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
Example-17 atoms in this reaction.
Out of p-orbital and sp-hybrid orbital which has greater
directional character and why?
CHEMICAL BONDING 117

Example-20 431 kJ mol–1 for H2. Thus, the bond in Li2 is much weaker
Which hybrid orbitals are used by carbon atoms in the than that on H2. This is because 2s orbital of Li (involved in
following molecules? the bonding) is much larger than 1s orbital of H.
(a) CH3 – CH3 (b) CH3 – CH = CH2 (c) CH3 – CH2 – OH Example-23
Sol. Compare the relative stabilities of O2– and N2+ and comment
on their magnetic (paramagnetic or diamagnetic) behaviour.
Sol. M.O. Electronic configuration of
O2  KK 2s
2
*2 2 2 2 *2 *1
2s  2pz 2p x 2p x 2p x 2p y

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

1 5
Bond order = (7  2)   2.5
2 2
Example-21 As bond order of N 2  bond order of O 2 , therefore, N 2 is
Use molecular orbital theory to explain why the molecule
more stable than O 2 .
does not exist.
Each of them contains unpaired electron, hence both are
Sol. E.C. of Be = 1s2 2s 2 paramagnetic.
M.O.E.C. of Be2  1s 2  *1s 2  2s 2  *2s 2 Example-24

1 Which of the two, peroxide ion or superoxide ion, has a

Bond order = (4 – 4) larger bond length ?
2
Sol. The bond length in a molecule depends on bond order. the
= 0
higher the bond order, smaller will be the bond length.
Hence, Be2 does not exist
Peroxide ion, O 22
Example-22
Configuration KK (2s)2 (2s)2  (2pz)2 (2px)2
Does Li2 exist ? If so, estimate its bond order and compare
its bond dissociation energy with that of H2. (2py)2 (2px)2 (2py)2
Sol. Li has a configuration : 1s2, 2s1. 86
Bond order = 1
There are two s orbitals (1s and 2s) on each atom. These 2
combine to give four MO’s. These are Superoxide ion, O 2
(1s)  (1s)  1s   *1s Configuration KK (2s)2 (2s)2 (2pz)2 (2px)2
(2s)  (2s)  2s   * 2s. (2py)2 (2px)2 (2py)1
Thus all the six electrons are accommodated in these four 85
orbitals. The electronic configuration for Li 2 is Bond order =  1.5
2
(1s) 2 ( *1s) 2 ( 2s) 2 . Bond order of superoxide is higher than peroxide ion,
hence bond length of peroxide ion is larger.
Nb = 4, Na = 2.
Example-25
42
Bond order in Li 2   1. Explain why HF is less viscous than H2O.
2
Sol. There is greater intermolecular hydrogen bonding in H2O
Therefore, Li 2 should be a stable species. Its bond than that in HF as each H2O molecule forms four H-bonds
dissociation energy is 105 kJ mol –1 , as compared to with other water molecules whereas HF forms only two H-
bonds with other HF molecules. Greater the intermolecular
H-bonding, greater is the viscosity. Hence, HF is less
viscous than H2O.
118 CHEMICAL BONDING

Exercise – 1: Basic Objective Questions

Kossel and Lewis Approach

1. The attractive force which holds various constituents Codes :

(atoms, ions etc.) together in different chemical species A B C
is called a (a) 2 3 1
(a) Chemical bond (b) 2 1 3
(b) Chemical compound (c) 1 2 3
(c) Ionic bond (d) 1 3 2
(d) Covalent bond 7. According to Lewis and Kossel approach, which of the
2. Kossel and Lewis approach was based on the following molecules has complete octet of the central
(a) Reactivity of elements atom?
(b) Inertness of noble gases (a) LiCl
(c) Reactivity of metals (b) BeH2
(d) Inertness of non-metals (c) BCl3
3. According to Kossel and Lewis approach, the outer (d) CO2
shell can accommodate a maximum of …X… electrons.
Here, X refers to Lewis Dot Structures
(a) 1 (b) 8
(c) 18 (d) 32 8. The Lewis dot structures can be written by adopting the
4. Lewis postulated that atoms achieve the stable octet following steps:
when they are linked by I. The total number of electrons required for writing the
(a) Only ionic bonds structure are obtained by adding the valence electrons
(b) Only covalent bonds of the combining atoms.
(c) Only coordinate bonds II. Knowing the chemical symbols of the combining
(d) Chemical bonds atoms and having knowledge of the skeletal structure of
5. Atoms can combine either by transfer of valence the compound, it is easy to distribute the total number
electrons from one atom to another or by sharing of of electrons as bonding shared pairs between the atoms
valence electrons in order to have an octet in their in proportions to the total bonds.
valence shells. This is called. III. In general the least electropositive atom occupies
(a) Octet rule the central position in the molecule/ion.
(b) Duplet rule IV. After accounting for the shared pairs of electrons
(c) quartet rule for single bonds, the remaining electron pairs are either
(d) None of these utilized for multiple bonding or remain as the lone
6. Match the following species of Column I with pairs.
information given in Column II choose the correct Which of the above step(s) is/are incorrect? Choose the
option from the codes given below. correct option.
Column I Column II (a) Only I
(b) Only III
A. BCl3 1. Odd electron molecules (c) Both I and III
(d) Both II and IV
B. SF6 2. Incomplete octet
C. NO2 3. The expanded octet of central atom
CHEMICAL BONDING 119

9. The important condition(s) for Lewis dot structure of 16. How many double bonds are present in carbon dioxide
covalent compounds is/are molecule?
(a) Each bond is formed as a result of sharing of an (a) One (b) Two
electron pair between the atoms (c) Three (d) Four
(b) Each combining atom contributes at least one 17. Which of the following Lewis representation of the
electron to shared pair molecules NF3, O3 and HNO3 is correct?
(c) The combining atom attain the outer shell noble gas
configurations as a result of the sharing of electrons
(d) All of the above
10. The number of dots around the Lewis symbol
represents: (a) Only I (b) Only II
(a) The number of valence electrons (c) Only III (d) I, II and III
(b) The number of protons 18. Lewis dot structure of CO,NO2- and CO32- are I, II and
(c) The number of inner electrons III respectively.
(d) All of the above
11. Lewis introduced simple notations to represent valence
electrons in an atom called.
(a) Lewis symbols
(b) Lewis element
(c) Lewis molecule Which of these structure(s) is/are wrong?
(d) Lewis structure (a) I (b) II
12. The group valence of the element is generally equal to (c) III (d) I, II and III
the 19. Calculate the formal charge of C in CH4.
(a) Number of dots in Lewis symbol (a) 4 (b) 1
(b) Eight minus the number of dots (c) –4 (d) 0
(c) Valence electrons 20. In the given structure, the formal charge on the oxygen
(d) Any of the above may be possible atom of P ─ O bond is
13. Find out the correct Lewis symbol for the atom carbon
among the following options.
(a) C (b) : C 
(c) : C : (d) C 

Lewis Representation of Some Molecules & Ions, (a) +1

Formal Charge (b) –1
(c) –0.75
14. Which of the following compound is covalent? (d) +0.75
(a) H2 21. Assertion (A) Formal charges do not indicate real
(b) CaO charge separation within the molecule.
(c) KCl Reason (R) Formal charges help in the selection of the
(d) Na2S lowest energy structure from a number of possible
15. When two atoms share two electron pairs they are said Lewis structure for a given species.
to be joined by a (a) Both A and R are correct, R is the correct
(a) Single bond explanation of A
(b) Double bond (b) Both A and R are correct, R is not the correct
(c) Triple bond explanation of A
(d) None of the above (c) A is correct, R is incorrect
(d) R is correct, A is incorrect
120 CHEMICAL BONDING

22. Which of the following is/are correct? (c) Anions, anions (d) None of these
I. Generally the lowest energy structure is the one with 27. Lattice energy of a solid increases if.
the smallest formal charges on the atoms. (a) Size of ions is small
II. The formal charge is a factor based on a pure (b) Charges of ions are small
covalent view of bonding in which electron pairs are (c) Ions are neutral
shared equally by neighboring atoms: (d) None of these
(a) Both A and R are correct, R is the correct 28. Lattice energy of an ionic compound depends upon
explanation of A (a) Charge on the ion and size of the ion
(b) Both A and R are correct, R is not the correct (b) Packing of ions only
explanation of A (c) Size of the ion only
(c) A is correct, R is incorrect (d) Charge on the ion only
(d) R is correct, A is incorrect 29. Consider the following statements;
I. In ionic solids, the sum of the electron gain enthalpy
Ionic Bond and the ionisation enthalpy may be positive but still the
crystal structure gets stabilised due to the energy
23. Consider the following statement, released in the formation of crystal lattice.
I. The formation of the positive and negative ions from II. A qualitative measure of the stability of an ionic
the neutral atoms by loosing and gaining electrons compound is provided simply by achieving octet of
respectively. electrons around the ionic species in gaseous state.
II. The arrangement or position of the positive and (a) I is correct, II is wrong
negative ions in the lattice of the crystalline compound. (b) I is correct, II is also correct
Which of the above statement(s) play(s) an important (c) I is wrong, II is also wrong
role in the formation of ionic compounds (d) I is wrong, II is correct
(a) Only I 30. The most stable compound is:
(b) Only II (a) LiF (b) LiCl
(c) Both I and II (c) LiBr (d) LiI
(d) None of these 31. Which combination will give the strongest ionic bond?
–
24. The formation of positive and negative ions involve (a) K+ and Cl (b) K+ and O2
–
I. Removal and addition of electrons from neutral gas (c) Ca2+ and Cl (d) Ca2+ and O2
atoms respectively 32.  
An ionic compound A B is most likely to be formed
II. Ionisation enthalpy and electron gain enthalpy when:
respectively. The correction option is. (a) the ionization energy of A is high and electron
(a) I is correct; II is incorrect affinity of B is low
(b) I is incorrect; II is correct (b) the ionization energy of A is low and electron
(c) Both I and II are incorrect affinity of B is high
(d) Both I and II are correct (c) both, the ionization energy of A and electron affinity
25. Ionic bonds will be formed more easily between of B are high
elements with comparatively. (d) both, the ionization energy of A and electron
(a) Low ionization enthalpy and high electron affinity affinity of B are low
(b) High ionization enthalpy and high electron affinity 33. Lattice energy of an ionic compound depends upon
(c) Low ionization enthalpy and low electron affinity (a) Charge on the ion and size of the ion
(d) High ionization enthalpy and low electron affinity (b) Packing of ions only
26. Ionic compounds in the crystalline state consist of (c) Size of the ion only
orderly three dimensional arrangements of …X… and (d) Charge on the ion only
…Y… held together by coulombic interaction energies.
In the above statement X and Y refer to
(a) Cations, anions (b) Cations, Cations
CHEMICAL BONDING 121

34. Which of the following compounds contain(s) no 40. Among the following the maximum covalent character
covalent bond(s)? is shown by the compound
KCl, PH , O , B H , H SO4 (a) FeCl2 (b) SnCl2
3 2 2 6 2
(a) KCl, B2H6 (c) AlCl3 (d) MgCl2
(b) KCl, B2H6, PH3 41. Consider the following statements :
(c) KCl, H2SO4 I. The smaller the size of cation and larger the size of
(d) KCl anion, the greater the covalent character of an ionic
35. In which of the following species the bonds are non- bond.
directional? II. The greater the charge on the cation the greater will
(a) NCl3 be the covalent character of the ionic bond.
(b) RbCl Which of the above statements belongs to Fajan’s rule?
(c) BeCl2 Choose the correct option.
(d) BCl3 (a) Only I (b) Only II
36. The melting point of RbBr is 682ºC while that of NaF is (c) Neither I nor II (d) Both I, II
988ºC. The principal reason for this fact is : 42. In an ionic compound A X  the degree of covalent
(a) the molar mass of NaF is smaller than that of RbBr bonding is greatest when :
(b) the bond in RbBr has more covalent character than (a) A and X ion are small
the bond in NaF (b) A is small and X is large
(c) the difference in electronegativity between Rb and
(c) A and X ions are approximately of the same size
Br is smaller than the difference between Na and F
(d) X is small and A is large
(d) the internuclear distance, rc + ra is greater for RbBr
43. The compound with the highest degree of covalency is :
than for NaF
(a) NaCl (b) MgCl2
37. The amount of energy released when one mole of ionic
(c) AgCl (d) CsCl
solid is formed by packing of gaseous ion is called :
(a) Ionisation energy
(b) Solvation energy Bond Parameters
(c) Lattice energy
(d) Hydration energy 44. The lowest O – O bond length in the following
38. I. When a gas phase atom in its ground state gains an molecule is :
electron, the enthalpy change is called electron gain (a) O2F2 (b) O2
enthalpy. (c) H2O2 (d) O3
II. The electron gain process is always exothermic. 45. Bond length if defined as the equilibrium distance
III. The ionization enthalpy process is always between the nuclei of two bonded atoms in a molecule.
endothermic Bond length can be calculated in the case of
(a) I and II are correct while III is incorrect (a) Covalent compounds only
(b) All I, II and III are correct (b) Ionic compounds only
(c) I and III are correct while II is incorrect (c) Both covalent as well as ionic compounds
(d) All I, II and III are incorrect (d) None of the above.
46. Which of the following cannot be used to measure bond
Fajan’s Rule lengths?
(a) Spectroscopy
39. The lattice energy of KF, KCl, KBr and KI follow the
(b) X-ray diffraction
order:
(c) Electron diffraction
(a) KF > KCl > KBr > KI
(d) Young’s Double-slit method
(b) KI > KBr > KCl > KF
(c) KF > KCl > KI > KBr
(d) KI > KBr > KF > KCl
122 CHEMICAL BONDING

47. The amount of energy required to break one mole of Resonance

bonds of a particular type between two atoms in a
gaseous state is called
52.
(a) Bond enthalpy (b) Bond angle
(c) Bond order (d) None of these
48. If the bond enthalpy of O2, N2 and H2 are 498 kJ mol-1 ,
946 kJ mol-1 and 435.8 kJ mol-1 respectively. Choose
the correct order of decreasing bond strength.
(a) H2 > N2 > O2 (b) N2 > O2 > H2
(c) O2 > H2 > N2 (d) H2 > O2 > N2
49. The bond order in H2, O2 and N2 is respectively
(a) 1, 2 and 3 (b) 3, 2 and 1 In the given resonating structures, which one is called
(c) 2, 1 and 3 (d) 3, 1 and 2 resonance hybrid?
50. Which of the following is correct: (a) Structure I (b) Structure II
1 1 (c) Structure III (d) Structure I and II
(a) Bond order    stability
bond length bond enthalpy 53. The O─O bond length in ozone molecule is:
1 1 1 (a) 121 pm (b) 148 pm
(b) Stability   
bond order bond length bond enthalpy (c) 128 pm (d) None of the above
(c) Stability  bond order  bond length  bond 54. Whenever a single Lewis structure cannot describe a
enthalpy molecule accurately, a number of structures with almost
(d) Stability  bond order  bond enthalpy similar energy, position of nuclei, bonding and non-
1 bonding pairs of electrons are taken as the canonical
 structures of the hybrid which describes the molecule
bond length
accurately. This concept is known as
51. Match the bond type given in Column I with the bond
(a) Resonance (b) Hybridisation
length in column II and choose the correct option from
(c) Inductive effect (d) None of the above
the codes given below
55. The structure which represents the molecule’s structure
Column I Column II
more accurately is called
A. O-H 1. 154 pm
(a) Resonance hybrid (b) Canonical structure
B. C-C 2. 133 pm (c) Resonating structure (d) None of the above
3. 96 pm 56. Resonating structure are represented through which
C. C-O
type of arrows?
D. C=C 4. 143 pm (a)  (b) 
Codes : (c)  (d) 
A B C D 57. Deviation of O─O bond length in ozone molecule from
(a) 3 1 4 2 the normal bond length seems as
(b) 1 2 3 4 (a) Single bond length increases while double bond
(c) 4 3 2 1 length decreases
(d) 2 4 1 3 (b) Single bond length decreases while double bond
length increases
(c) Single bond length increases while double bond
length remains same
(d) Single bond length remains same while double bond
length increases.
CHEMICAL BONDING 123

58. Resonance structures can be written for: 63. Dipole moment is usually expressed in Debye units (D)
(a) O3 from the formula the unit of dipole moment is coulomb
(b) NH3 metre (Cm). The relation between both units of dipole
(c) CH4 moment is.
(d) H2O (a) 1 D = 1 Cm
59. Which of the following is correct representation of (b) 1 D = 3.33564 × 10-30 Cm
resonating structures? (c) 1 D = 2.22564 × 10-30 Cm
(d) 1 D = 1.11564 × 10-30 Cm
64. The dipole moment of HF molecule may be represented
as

(a) (b)

(c) (d)
65. Correct statement for water would be
(a) It is a non-polar molecule
(b) It is a polar molecule
(a) Only A (c) It has linear structure
(b) Only B (d) It has bond angle of 109° 28'
(c) Both A & B 66. Which of the following molecule has dipole moment
(d) None of the above zero?
60. Which of the following molecules represents (a) HF (b) H2O
resonance? (c) BF3 (d) CHCl3
(a) O3 67. The decreasing order of dipole moments of the
(b) CO 32  molecules HF, H2O, BeF2, NF3 is
(c) CO2 (a) HF > H2O > BeF2 > NF3
(d) All of these (b) H2O > HF > NF3 > BeF2
(c) BeF2 > NF3 > HF > H2O
Dipole Moment of a Bond (d) NF3 > BeF2 > H2O > HF
68. NH3 and NF3, both the molecules have pyramidal shape
61. The product of the magnitude of the charge and the with a lone pair of electrons on nitrogen atom. Why
distance between the centres of positive and negative does the dipole moment of NH3 is greater than that of
charge is called. NF3? Choose the correct option.
(a) Charge ratio (a) F is more electronegative than N atom
(b) Dipole moment (b) In NF3 the orbital dipole is in the direction opposite
(c) Current flow to the resultant dipole moment of the three N-F bonds.
(d) Magnetic moment (c) In NH3 the orbital dipole is in the direction same to
62. Dipole moment is usually designated by a Greek letter ‘ the resultant dipole moment of the N-H bonds.
μ‘ (d) All of the above
69. Polarity in a molecule and hence the dipole moment
µ=Q×r
depends primarily on electronegativity of the
Here Q and r represents
constituent atoms and shape of a molecule. Which of
(a) Q = charge, r = distance of separation
the following has the highest dipole moment?
(b) Q = heat, r = radius
(a) CO2 (b) HI
(c) Q = charge, r = radius of cations
(c) H2O (d) SO2
(d) Q = charge, r = radius of anions
124 CHEMICAL BONDING

70. The highest dipole moment is of: (b) The bond pair electrons in a molecule occupy more
(a) CF4 (b) CH3OH space as compare to the lone pairs of electrons.
(c) CO2 (d) CH3F (c) Both (a) and (b) are incorrect
71. Which of the following molecules does not have a (d) None of the above
dipole moment? 76. For the prediction of geometrical shapes of molecules
(a) IBr (b) CHCl3 with the help of VSEPR theory molecules are divided
(c) CH2Cl2 (d) BF3 into two categories. These categories are
72. Among the following, the molecule with highest dipole (a) Molecules in which the central atom has no lone
moment is: pair
(a) CH3Cl (b) CH2Cl2 (b) Molecules in which the central atom has one or
(c) CHCl3 (d) CCl4 more lone pairs.
(c) Both (a) and (b)
VSEPR Theory (d) None of the above
77. Which of the following molecules have same molecule
73. Consider the following statements, geometry?
1. The shape of a molecule depends upon the number I. CH4, II BF3 III. NH4+, IV. SF4
of valence shell electron pairs (bonded or non- (a) I and II (b) III and IV
bonded) around the central atom. (c) I and III (d) I, III and IV
2. Pairs of electrons in the valence shell repel one 78. In which of the following structures all the bonds are
another since their electron clouds are negatively present at 90o?
charged. (a) Trigonal planar (b) Tetrahedral
3. These pairs of electrons tend to occupy such (c) Trigonal bipyramidal (d) Octahedral
positions in space that minimize repulsion and thus 79. Which of the molecules has trigonal bipyramidal
maximise distance between them. geometry with bond angles 120o and 90o?
4. The valence shell is taken as a sphere with the (a) SF6 (b) PCl5
electron pairs localizing on the spherical surface at (c) CH4 (d) BF3
maximum distance from one another. 80. Correct statement regarding molecules SF , CF and
4 4
5. A multiple bond is treated as if it is a single XeF are:
4
electron pair and the two or three electron pairs of a (a) 2, 0 and 1 lone pairs of central atom respectively
multiple bond are treated as a single super pair. (b) 1, 0 and 1 lone pairs of central atom respectively
6. Where two or more resonance structures can (c) 0, 0 and 2 lone pairs of central atom respectively
represent a molecule, the VSEPR model is (d) 1, 0 and 2 lone pairs of central atom respectively
applicable to any such structure. 81. The pair of species with similar shape is :
Which of the above statements are the postulates of (a) PCl3, NH3 (b) CF4, SF4
VSEPR theory? Choose the correct option. (c) PbCl2, CO2 (d) PF5, IF5
(a) 1, 2 and 3 (b) 4, 5 and 6 82. The geometrical arrangement of orbitals and shape of
(c) 2, 3 and 5 (d) All of these
I3 are respectively:
74. The decreasing order of the repulsive interaction of
electron pairs is (Here, lp=lone pair, bp=bond pair). (a) trigonal bipyramidal geometry, linear shape
(a) lp-lp > lp-bp > bp-bp (b) hexagonal geometry, T-shape
(b) lp-bp > lp-lp > bp-bp (c) triangular planar geometry, triangular shape
(c) lp-lp > bp-bp > lp-bp (d) tetrahedral geometry, pyramidal shape
(d) bp-bp > lp-lp > lp-bp 83. The shape of the noble gas compound XeF4 is:
75. Why does the repulsion is greater in lp-lp as compared (a) square planar (b) distorted tetrahedral
to lp-bp and bp-bp ? (c) tetrahedral (d) octahedral
(a) The lone pair electrons in a molecule occupy more
space as compare to the bonding pair of electrons
CHEMICAL BONDING 125

84. BrF3 molecule, the lone pairs occupy an equatorial (d) XeF2 – Linear shape
position to minimize. 93. Which of the following are isoelectronic and
(a) lone pair-bond pair repulsion only isostructural?
(b) bond pair-bond pair repulsion only NO3- ,CO32- ,ClO3- ,SO3
(c) lone pair-lone pair repulsion and lone pair-bond pair
repulsion (a) NO3- ,CO32- (b) SO3 ,NO3-
(d) lone pair-lone pair repulsion only (c) ClO3- ,CO32- (d) CO32- ,ClO3-
85. The shape of the molecule depends on the _______
94. FAsF bond angle in AsF3 Cl2 molecule is:
(a) adjacent atom (b) valence electrons
(c) surroundings (d) atmosphere (a) 90º and 180º (b) 120º
86. The species having pyramidal shape is (c) 90º (d) 180º
(a) SO3 (b) BrF3
(c) SiO32- (d) OSF2 Valence Bond Theory
87. Shape of XeF4 is
(a) T-shaped (b) Tetrahedral 95. VBT theory is based on the
(c) Octahedral (d) Square planar (a) Knowledge of atomic orbitals and electronic
88. Match the following Column and choose the correct configuration of elements
option from the codes given below. (b) Overlap criteria and the hybridization of atomic
Column I Column II orbitals
A. BeCl2 1. Linear (c) The principles of variation and superposition
(d) All of the above
B. BF3 2. Trigonal planar 96. The valence bond theory explains the shape, the
C. CO2 formation and directional properties of bonds in
polyatomic molecules like CH4, NH3 and H2O etc., in
Codes :
terms of
A B C
(a) Overlapping of atomic orbital
(a) 1 1 2
(b) Hybridisation of atomic orbitals
(b) 2 2 1
(c) Both (a) and (b)
(c) 1 2 1
(d) None of the above
(d) 2 1 2
97. Which of the following does not represent positive
89. The molecule exhibiting maximum number of non-
overlap?
bonding electron pairs (l.p.) around the central atom is:
(a) XeOF4 (b) XeO2F2
(c) XeF2 (d) XeO3
90. Which of the following pairs of species have identical
shapes?
(a) NO 2 and NO 2 (b) PCl and BrF (a) (b)
5 5
(c) XeF4 and ICl 4 (d) TeCl4 and XeO4
91. The shapes of XeF4, XeF5Θ and SnCl2 are: (c) (d)
(a) octahedral, trigonal bipyramidal and bent 98. The strength of covalent bond ___________ extent of
(b) square pyramidal, pentagonal planar and linear overlapping of orbitals.
(c) square planar, pentagonal planar and angular (a) may be or may not be related to
(d) see-saw, T-shaped and linear (b) is independent on
92. Which is not correctly matched? (c) is dependent on
(a) XeO3 – Trigonal bipyramidal (d) is not related to
(b) ClF3 – bent T-shape
(c) XeOF4 – Square pyramidal
126 CHEMICAL BONDING

99. The strength of bonds formed by s-s, p-p and p-s 107. In sp2 orbital, character of pz orbital will be
overlap has the order: (a) always 33.33%
(a) s-s > p-p > p-s (b) always 0%
(b) s-s > p-s > p-p (c) always 66.66%
(c) p-p > p-s > s-s (d) either 33.33% or 0%
(d) p-p > s-s > p-s 108. Which of the following orders is correct for
100. In a compound electronegativity?
NC M(CO)3
C=C (a) sp3 > sp2 > sp (b) sp > sp2 > sp3
NC C2H5 (c) sp2 > sp > sp3 (d) sp3 > sp > sp2
the number of sigma and pi bonds respectively are : 109. The orbitals used by C in forming C–H and C–Cl bonds
(a) 19, 11 (b) 19, 5 of CH2Cl2 are
(c) 13, 11 (d) 7, 3 (a) four sp orbitals
101. Assuming the bond direction to the z-axis, which of the (b) four sp2 orbitals
overlapping of atomic orbitals of two-atom (A) and (B) (c) four sp3 orbitals
will result in bonding? (d) None of these
(I) s-orbital of A and p orbital of B 110. CH 2  CH  CN
x
3 2 1
(II) s-orbital of A and p orbital of B
z Cl–C2 bond of this molecule is formed by
(III) p -orbital of A and p orbital of B (a) sp3-sp2 overlap
y z
(IV) s-orbital of both (A) and (B) (b) sp2 -sp3 overlap
(a) I and IV (b) I and II (c) sp-sp2 overlap
(c) III and IV (d) II and IV (d) sp2-sp2 overlap
111. CH2 = C = CH2 In this molecule (allene)
Hybridisation (a) all three C-atoms are sp2 hybridized
102. Diagonal hybridisation is the another name of (b) both terminal C-atoms are sp2 hybridized while
(a) sp3 hybridisation (b) sp2 hybridisation central C-atom is sp-hybridized
(c) sp hybridization (d) All of the above (c) both terminal C-atoms are sp-hybridized while
103. Which of the following angle corresponds to sp2 central C-atom is sp2 hybridized
hybridisation? (d) none of these
(a) 90o (b) 120o 112. What type of hybridization does a BCl3 molecule
o
(c) 180 (d) 109o undergo?

104. The hybridization of the central atom in ICl2 is: (a) sp (b) sp2
3
(a) dsp2 (b) sp (c) sp (d) sp3d
(c) sp 2
(d) sp3 113. Isostructural species are those which have the same
105. The state of hybridization of the central atom is not the shape and hybridisation. Among the given species
same as in the others: identify the isostructural pairs.
(a) B in BF3 (a) [NF3 and BF3] (b) [BF4- and NH4+]
(b) O in H3O+ (c) [BCl3 and BrCl3] (d) [NH3 and NO3-]
(c) N in NH3 114. In which of the following pairs, both the species have
(d) P in PCl3 the same hybridization?

106. CH3 – CH2 – CH = CH2 has hybridisation: (I) SF4 , XeF4 (II) I3 ,XeF2
(a) sp, sp, sp2, sp2   3
(III) ICl4 ,SiCl4 (IV) ClO3 ,PO4
(b) sp3, sp3, sp2, sp
(c) sp3, sp3, sp2, sp2 (a) I, II (b) II, III
(d) sp3, sp2, sp2, sp (c) II, IV (d) I, II, III
CHEMICAL BONDING 127

115. Which one of the following is the correct set with correct for B2, C2 and N2.
respect to molecule, hybridization and shape ? (d) Both order I and II are correct for all molecules.
(a) BeCl , sp2, linear 122. The condition to form a molecular orbital from atomic
2
2
(b) BeCl , sp , triangular planar orbital is
2
(a) Atomic orbitals must have comparable energies and
(c) BCl , sp2, triangular planar
3 of proper symmetry
(d) BCl , sp3, tetrahedral (b) Atomic orbitals must be in proper symmetry only
3
116. Which of the following statements is incorrect for PCl5? (c) Atomic orbitals must be of comparable energies
(a) Its three P–Cl bond lengths are equal only
(b) It involves sp3d hybridization (d) None of the above
(c) All bond angles are 90° 123. The number of molecular orbitals formed is equal to the
(d) Its shape is trigonal bipyramidal number of combining atomic orbitals. When two atomic
117. Select pair of compounds in which both have different orbitals combine, two molecular orbitals are formed one
hybridization but have same molecular geometry: is known as bonding molecular orbital while the other
is called anti-bonding molecular orbitals. Which M.O.
(a) BF3, BrF3 (b) ICl-2 ,BeCl 2
is more stable?
(c) BCl3, PCl3 (d) PCl3, NCl3
(a) Bonding molecular orbital
118. The hybridization of atomic orbitals of the central atom
(b) Anti-bonding molecular orbital
“Xe” in XeO4, XeO2F2, and XeOF4 respectively.
(c) Both orbitals have same stability
(a) sp3, sp3d2, sp3d2 (d) None of the above
(b) sp3d, sp3d, sp3d2 124. The molecular orbitals like atomic orbitals are filled in
(c) sp3, sp3d2, sp3d accordance with the
(d) sp3, sp3d, sp3d2 (a) Aufbau principle
  (b) Pauli’s exclusion principle
119. The hybridization of atomic orbitals of N in NO2 , NO3
(c) Hund’s rule

and NH4 are respectively: (d) All of the above
(a) sp, sp2, sp3 (b) sp, sp3, sp2 125. Which of the following have identical bond order?
–
(c) sp2, sp, sp3 (d) sp2, sp3, sp (I) CN (II) O
2
+
120. The structure and hybridization of Si(CH ) is : (III) NO (IV) N2
3 4
(a) bent, sp (b) trigonal, sp2 (a) I, III, IV (b) II, IV
(c) octahedral, sp3d (d) tetrahedral, sp3 (c) I, II, III (d) I, IV
126. Which one of the following species is diamagnetic in
Molecular Orbital Theory nature ?
 
(a) H2 (b) H2
121. The increasing order of energies of various molecular (c) H 2 (d) He2

orbitals are
I.  ls <   ls <  2s <   2s   2 pz  127. Take NA as the number of Anti-bonding molecular
  
orbitals and NB as the number of Bonding molecular
( 2 px   2 p y )  ( 2 px   2 p y )   2 pz
orbitals. The molecule is stable when NA
II.  ls <   ls <  2s <   2 s  ( 2 px   2 p y ) ____________ NB.
  2 p z  (  2 p x    2 p y )    2 pz (a) is greater than
(a) Order I is correct for all molecules and II is (b) is equal to
(c) is less than
incorrect
(b) Order II is correct for all molecules and I is (d) is greater than or equal to
incorrect
(c) Order I is correct for O2 and F2 while order II is
128 CHEMICAL BONDING

128. Which of the following is paramagnetic? +

(c) O2 <O2 <O2 <O2
- 2-
2
(a) O 2 (b) NO + - 2-
– (d) O2 <O2 <O2 <O2
(C) CO (d) CN 
129. The ground state electronic configuration of valence 137. When N is formed from N2 bond-order .......... and
2
shell electrons in nitrogen molecule (N ) is written 
2 when O is formed from O2 bond–order .......... :
2
 2s2 , * 2s2 , 2 px2 , 2py 2   2pz2 .Hence, the bond (a) increases (b) decreases
(c) increases, decreases (d) decreases, increases
order in nitrogen molecule is :
(a) 0 (b) 1
(c) 2 (d) 3 Intermolecular Forces
130. The ion that is isoelectronic with CO is:
– +
(a) O2 (b) N 138. Metallic bond is.
2
+ –
(a) a type of covalent bond.
(c) O2 (d) CN (b) a type of ionic bond
 (c) an attraction between positive and negative ions.
131. The number of anti-bonding electron pairs in O2
molecular ion on the basis of molecular orbital theory is (d) an attraction between positive ions and electrons.
(atomic no. of O is 8): 139. The boiling point of a substance increases with increase
(a) 5 (b) 2 in
(c) 3 (d) 6 (a) Intermolecular hydrogen bonding
132. Which one of the following is not paramagnetic? (b) Intramolecular hydrogen bonding
+
(c) Non-polarity
(a) NO (b) N 2 (d) Both (a) and (c)
(c) CO (d) O 140. Which of the following molecules do show the
2
133. Using MO theory, predict which of the following hydrogen bonding?
species has the shortest bond length? I. CCl3CH(OH)2 II. o-hydroxy benzoic acid
+ - III. o-nitro phenol
(a) O2 (b) O2
(a) II and III (b) Only I
2- 2+
(c) O2 (d) O2 (c) I and III (d) I, II and III
134. According to MO theory 141. Which of the following atom can form a hydrogen
 bond?
(a) O2 is paramagnetic and bond order is greater than
(a) Sodium (b) Oxygen
O2. (c) Aluminum (d) Rubidium

(b) O2 is paramagnetic and bond order is less than O2. 142. What type of intermolecular forces are due to the
 attraction between temporary dipoles and their induced
(c) O2 is diamagnetic and bond order is less than O2.
temporary dipoles?

(d) O2 is diamagnetic and bond order is more than O2. (a) metallic bond (b) London dispersion
135. Which of the following molecules have same bond (c) hydrogen bond (d) ionic bond
order? 143. Which one of the following molecules will form a
H2, He2, C2 and O2 linear polymeric structure due to hydrogen bonding?
(a) H2 and C2 (b) C2 and O2 (a) NH3 (b) H2O
(c) He2 and C2 (d) H2 and He2 (c) HCl (d) HF\
+ – 2–
136. Bond order of O , O , O and O is in order: 144. The correct order of the strength of H-bonds is:
2 2 2 2
(a) H....F > H....O > H....N
- 2- +
(a) O <O <O2 <O
2 2 2 (b) H....N > H....O > H....F
2- - + (c) H....O > H....N > H....F
(b) O2 < O2 < O2 <O2
(d) H....F > H....N > H....O
CHEMICAL BONDING 129

145. o-nitrophenol can be easily steam distilled whereas 148. An ether is more volatile than alcohol having the same
p-nitrophenol cannot be. This is because of : molecular formula. This is due to
(a) strong intermolecular hydrogen bonding in (a) alcohols having resonance structures
o-nitrophenol (b) inter-molecular hydrogen bonding in ethers
(b) strong intramolecular hydrogen bonding in (c) inter-molecular hydrogen bonding in alcohols
o- nitrophenol (d) dipolar character of ethers
(c) strong intramolecular hydrogen bonding in 149. The order of the boiling points of the given compounds
p-nitrophenol is
(d) dipole moment of p-nitrophenol is larger than that (a) HF > H2O > NH3
of o-nitrophenol (b) H2O > HF > NH3
146. Which of the following hydrogen bonds is the (c) NH3 > HF > H2O
strongest? (d) NH3 > H2O > HF
(a) O—H...N (b) F—H...F 150. The boiling points at atmospheric pressure of HF, H S,
2
(c) O—H...O (d) O—H...F NH can be arranged in the following order:
3
147. What happens to the London forces, if the number of
(a) HF > NH > H S (b) HF > H S > NH
electrons increases? 3 2 2 3

(a) Increases (b) Decreases (c) HF < H S < NH (d) HF < NH < H S
2 3 3 2
(c) No effect (d) Unpredictable
130 CHEMICAL BONDING

Exercise – 2: Previous Year Questions

1. Electron deficient molecule is 9. The geometry of sulphate ion is
(RPMT 2007) (AFMC 2009)
(a) CCl4 (b) PCl5 (a) square planar
(c) BF3 (d) SF6 (b) tetrahedral
2. Least ionized salt is (c) square pyramidal
(RPMT 2007) (d) octahedral
(a) KCl (b) AgCl 10. HCl molecule contains
(c) MgCl2 (d) CaCl2 (AFMC 2009)
3. Which of the following is weakest? (a) ionic bond
(BHU 2008) (b) covalent bond
(a) Ionic (c) hydrogen bond
(b) Covalent (d) coordinate bond
(c) Van der Waals’ forces 11. Some of the properties of the species, NO3– and H3O+ are
(d) Metallic bond described below. Which one of them is correct?
4. The angular shape of ozone molecule (O3) consists of (AIPMT 2010)
(CBSE AIPMT 2008) (a) Dissimilar in hybridization for the central atom with
(a) 1 sigma and 2 pi bonds different structures
(b) 2 sigma and 2 pi bonds (b) Isostructural with same hybridization for the central
(c) 1 sigma and 1 pi bond atom.
(d) 2 sigma and 1 pi bond (c) Isostructural with different hybridization for the
5. Which of the following has the lowest boiling point? central atom.
(Punjab PMET 2008) (d) Similar in hybridization for the central atom with
(a) NaCl (b) CuCl different structures.
(c) CuCl2 (d) CsCl 12. Which of the following has electrovalent linkage?
6. What is the dominant intermolecular force or bond that (MP PMT 2010)
must be overcome in converting liquid CH3OH to a gas? (a) CaCl2 (b) AlCl3
(AIPMT 2009) (c) SiCl4 (d) PCl3
(a) Dipole-dipole interaction 13. Which of the following substance has the highest
(b) Covalent bonds melting point?
(c) London dispersion force (VMMC 2010)
(d) Hydrogen bonding (a) BaO (b) MgO
7. A covalent molecule AB3 has a pyramidal structure. The (c) KCl (d) NaCl
number of lone pair and bond pair of electrons in the 14. The higher lattice energy corresponds to
molecule are respectively. (AIPMT 2010)
(KCET 2009)
(a) MgO (b) CaO
(a) 0 and 4 (b) 3 and 1
(c) SrO (d) BaO
(c) 1 and 3 (d) 2 and 2
15. In which of the following molecules the central atom
8. If I2 is dissolved in aqueous KI, the intense yellow
does not have sp3 hybridization?
 
species I3 is formed. The structure of I3 ions is (AIPMT 2010)
(DUMET 2009) (a) CH4 (b) SF4
(a) square pyramidal (b) trigonal bipyramidal (c) BF4– (d) NH4+
(c) octahedral (d) pentagonal bipyramidal
CHEMICAL BONDING 131

16. In which one of the following species the central atom 24. The pair of species that has the same bond order in the
has the type of hybridization which is not the same as following is
that present in the other three? (NEET 2013)
(AIPMT 2010) (a) CO, NO+ (b) NO–, CN–
(a) SF4 (b) I3– (c) O2, N2 (d) O2, B2
(c) SbCl52– (d) PCl5 25. Which of the following is paramagnetic?
17. Which of the two ions from the list given below that (NEET 2013)
have the geometry that is explained by the same (a) CN– (b) NO+
hybridization of orbitals, NO2–, NO3–, NH2–, NH4+, (c) CO (d) O2–
SCN–? 26. The outer orbitals C in ethene molecule can be
(AIPMT 2011) considered to be hybridized to give three equivalent sp2
– –
(a) NO2 and NO3 (b) NH4 and NO3–
+
orbitals. The total number of sigma   and pi  
(c) SCN– and NH2– (d) NO2– and NH2–
bonds in ethene molecule is.
18. The correct order of increasing bond length of
(NEET 2013)
C–H, C–O, C–C and C = C is
(AIPMT 2011) (a) 3 sigma   and 2 pi   bonds
(a) C  H  C  C  C  O  C  C (b) 4 sigma   and 1 pi   bonds
(b) C  C  C  C  C  O  C  H
(c) 5 sigma   and 1 pi   bonds
(c) C  O  C  H  C  C  C  C
(d) C  H  C  O  C  C  C  C (d) 1 sigma   and 2 pi   bonds
19. Bond order of 1.5 is shown by 27. Which of the following is polar molecule?
(AIPMT 2012) (NEET 2013)
  (a) SiF4 (b) XeF4
(a) O 2 (b) O 2
2
(c) BF3 (d) SF4
(c) O 2 (d) O2 28. XeF2 is isostructural with
20. Which of the following species contains three bond pairs (NEET 2013)
and one lone pair around the central atom? (a) SbCl3 (b) BaCl2
(AIPMT 2012) (c) TeF2 (d) ICl2–
(a) H2O (b) BF3 29. Dipole induced dipole interactions are present in which
(c) NH2– (d) PCl3 of the following pairs:-
21. During change of O2 to O2– ion, the electron adds on (CBSE AIPMT 2013)
which one of the following orbitals? (a) SiF and He atoms
4
(AIPMT 2012)
(b) H O and alcohol
2
(a)  * Orbital (b)  Orbital
(c) Cl2 and CCl4
(c)  *Orbital (d)  Orbital
(d) HCl and He atoms
22. Shape and hybridization of SO2 are
30. Which one of the following molecules contains no
(CPMT 2012)
bond? (CBSE AIPMT 2013)
(a) V shape, sp
(a) NO 2 (b) CO 2
(b) triangular planar, sp2
(c) V shape, sp2 (c) H 2 O (d) SO2
(d) tetrahedral, sp2 31. Which one of the following species has a planar
23. In which of the following pairs both the species have triangular shape? (NEET 2014)
sp3 hybridisation? (a) N3

(b) NO3
(NEET 2013)
(c) NO 2 (d) CO 2
(a) SiF4, BeH2 (b) NF3, H2O
(c) NF3, BF3 (d) H2S, BF3
132 CHEMICAL BONDING

32. In which of the following pairs, both the species are not 39. Which of the following pairs of species have the same
isostructural? bond order?
(NEET 2015) (NEET 2017)
(a) Diamond, Silicon carbide (a) O2 , NO
 
(b) CN , CO
(b) NH3, PH3 
(c) XeF4, XeO4 (c) N2 , O2 (d) CO, NO
(d) SiCl4, PCl4+ 40. The species, having bond angles of 120º is
33. Maximum bond angle at nitrogen is present in which of (NEET 2017)
the following? (a) ClF3 (b) NCl3
(NEET 2015) (c) BCl3 (d) PH3
(a) NO2+ (b) NO3– 41. Consider the following species:
(c) NO2 (d) NO2– (NEET 2018)
34. Which of the following species contains equal number of  
CN , CN , NO and CN
 and  - bonds? Which one of these will have the highest bond order?
(NEET 2015)  
(a) CN (b) CN
(a) (CN)2 (b) CH2(CN)2
(c) NO (d) CN
(c) HCO3– (d) XeO4
35. Which of the following pairs of ions are isoelectronic 42. In the structure of CIF3 , the number of lone pairs of
and isostructural? electrons on central atom ‘Cl’ is
(NEET 2015) (NEET 2018)
2   2 (a) two (b) four
(a) SO , NO
3 3 (b) ClO ,SO
3 3
2 2  2
(c) one (d) three
(c) CO3 ,SO3 (d) ClO3 ,CO3 43. Which of the following is paramagnetic?
36. Which one of the following compounds shows the (NEET 2019)
presence of intramolecular hydrogen bond? (a) N2 (b) H2
(NEET 2016) (c) Li2 (d) O2
(a) H2O2 44. Which of the following diatomic molecular species has
(b) HCN only  bonds according to Molecular Orbital Theory?
(c) Cellulose (NEET 2019)
(d) Concentrated acetic acid (a) Be2 (b) O2
37. The correct geometry and hybridization for XeF4 are (c) N2 (d) C2
(NEET 2016) 45. Which of the following is the correct order of dipole
(a) Octahedral, sp3d2 moment?
(b) Trigonal bipyramidal, sp3d (NEET 2019)
(c) Planar triangle, sp3d3 (a) NH3 < BF3 < NF3 < H2O
(d) Square planar, sp3d2 (b) BF3 < NF3 < NH3 < H2O
38. Consider the molecules CH4, NH3 and H2O. Which of (c) BF3 < NH3 < NF3 < H2O
the given statements is false? (d) H2O < NF3 < NH3 < BF3
(NEET 2016) 46. Identify a molecule which does not exist.
(a) The H–O–H bond angle in H2O is smaller than the (NEET 2020)
H–N–H bond angle in NH3
(a) He2 (b) Li2
(b) The H–C–H bond angle in CH4 is larger than the H–
N–H bond angle in NH3. (c) C2 (d) O2
(c) The H–C–H bond angle in CH4, the H–N–H bond
angle in NH3, and the H–O–H bond angle in H2O
are all greater than 90º.
(d) The H–O–H bond angle in H2O is larger than the H–
C–H bond angle in CH4.
CHEMICAL BONDING 133

47. Which of the following set of molecules will have zero 51. Which amongst the following is incorrect statement?
dipole moment? (NEET 2022)
(NEET 2020) (a) The bond orders of O2+, O2, O2– and O22– are 2.5, 2,
(a) Ammonia, beryllium difluoride, water, 1,4- 1.5 and 1, respectively.
dichlorobenzene. (b) C2 molecule has four electrons in its two
(b) Boron trifluoride, hydrogen fluoride, carbon dioxide, degenerate molecular orbitals.
1,3-dichlorobenzene (c) H2+ ion has one electron.
(d) O2+ ion is diamagnetic.
(c) Nitrogen trifluoride, beryllium difluoride, water, 1,3-
52. Amongst the following which one will have maximum
dichlorobenzene.
'lone pair- lone pair' electron repulsions?
(d) Boron trifluoride, beryllium difluoride, carbon
(NEET 2022)
dioxide, 1,4-dichlorobenzene (a) ClF3 (b) IF5
48. Which of the following molecules is non-polar in nature? (c) SF4 (d) XeF2
(NEET 2021) 53. Match List - I with List- II.
(a) SbCl5 (b) NO2 Choose the correct answer from the options given
(c) POCl3 (d) CH2O below : (NEET 2022)
49. Match List-I with List-II List-I(Hydrides) List-II(Nature)
(NEET 2021) (a) MgH2 (i) Electron precise
Column I Column II (b) GeH4 (ii) Electron deficient
1. Square pyramidal (c) B2H6 (iii) Electron rich
A. PCl5
(d) HF (iv) Ionic
B. SF6 2. Trigonal planar (a) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
C. BrF5 3. Octahedral (b) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
(c) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
(a) A – iii, B – i, C – iv, D - ii (d) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
(b) A – iv, B – iii, C – ii, d - i
(c) A – iv, B – iii, C – i, D – ii
(d) A – ii, B – iii, C – iv, D – i
50. BF3 is planar and electron deficient compound.
Hybridization and number of electrons around the
central atom, respectively are
(NEET 2021)
(a) sp2 and 6 (b) sp2 and 8
(c) sp3 and 4 (d) sp3 and 6
134 CHEMICAL BONDING

Exercise – 3: Achiever’s Section

Single Choice Questions

1. Which of the following substances has the least covalent

character? 6. Given the correct order of initials T or F for the
(a) Cl2O (b) NCl3 following statements. Use T is statement is true and F if
(c) PbCl2 (d) BaCl2 it is false
2. Which has a maximum dipole moment? (I) The order of repulsion between different pair of
electrons is LP – LP > LP –BP > BP – BP
(II) In general, as the number of lone pairs of electrons
(a) (b) on the central atom increases, the value of bond angle
from normal bond angle also increases.
(III) The number of lone pair on O in H 2 O is 2 while on
N in NH 3 is 1
(c) (d) (IV) The structures of xenon fluorides and xenon
oxyfluorides could not be explained on the basis of
VSEPR theory
3. The dipole moment of o, p and m-dichlorobenzene will (a) TTTF (b) TFTF
be in the order : (c) TFTT (d) TFFF
(a) o> p > m 7. Which molecule has regular geometry?
(b) p > o > m (I) SnCl 2 (II) NH 3
(c) m> o > p (III) PCl5 (IV) SF6
(d) o > m > p (a) I, III & IV
4. Given the correct order of initials T or F for the (b) II, III & IV
following statements. Use T if statements is true and F if (c) III & IV
it is false: (d) All
(I)  CH3 2 P  CF3 3 is non-polar and  CH3 3 P  CF3 2 8. Which of the following compounds have the same no. of
is polar molecule lone pairs with their central atom?

(II)  CH3  P  CH3  bond angles are equal in both above- (I) XeF5 (II) BrF3

mentioned molecules (III) XeF2 (IV) H 3S

(III) PF3 will be more soluble in polar solvent than (V) Triple Methylene
SiF4 (a) IV and V (b) I and III
(c) I and II (d) II, IV and V
(a) TTF
(b) FFT 9. The compound MX 4 is tetrahedral. The number of
(c) FFF XMX angles in the compound is:
(d) FTT (a) three (b) four
5. Which molecular geometry is least likely to result from a (c) five (d) six
trigonal bipyramidal electron geometry? 10. Which of the following are isoelectronic and
(a) Trigonal planar isostructural? NO 3 , CO 32  , ClO 3 , SO 3
(b) See-saw (a) NO 3 , CO 23  (b) SO 3 , NO 3
(c) Linear
(c) ClO 3 , CO 32  (d) CO 23  , SO 3
(d) T-shaped
CHEMICAL BONDING 135

11. Match List I and II and pick out correct matching codes 19. Which one of the following pairs of species has the same
from the given choices: bond order?
List I List II (a) CN  and NO (b) CN  and CN 
Compound Structure
(c) O2 and CN  (d) NO and CN 
A. ClF3 1. See-Saw
B. PCl5 2. Tetrahedral
C. IF5 3. Trigonal bipyramidal Assertion-Reason Type Questions
D. CCl4 4. Square pyramidal
E. SeF4 5. T-shaped While answering these questions, you are required to
Codes choose any one of the following four responses.
(a) A-5, B-4, C-3, D-2, E-1 (A) If both Assertion and Reason are correct and the
(b) A-5, B-3, C-4, D-2, E-1 Reason is a correct explanation of the Assertion.
(c) A-5, B-3, C-4, D-1, E-2 (B) If both Assertion and Reason are correct but Reason
(d) A-4, B-3, C-5, D-2, E-1 is not a correct explanation of the Assertion.
12. The hybridization of the central atom will change when: (C) If the Assertion is correct but Reason is incorrect.
(a) NH3 combines with H+ (D) If the Reason is correct but Assertion is incorrect.
(b) H3BO3 combines with OH- (E) Both Assertion & Reason are incorrect.
(c) NH3 forms NH2- 20. Assertion: The atoms in a covalent molecule are said to
(d) H2O combines with H+ share electrons, yet some covalent molecules are polar.
13. The species in which the N atom is in a state of sp Reason: In polar covalent molecules, the shared
hybridization is: electrons spend more time on average near one of the
atoms.
(a) NO2 (b) NO3
(a) A (b) B
(c) NO 2 (d) NO2 (c) C (d) D (e) E
21. Assertion: Water is a good solvent for ionic compounds
14. The hybridization of orbitals of N atom in NO3 , NO2
but poor for covalent compounds.
and NH 4 are respectively Reason: Hydration energy of ions releases sufficient
(a) sp,sp2 ,sp3 (b) sp2 ,sp,sp3 energy to overcome lattice energy and break hydrogen
bonds in the water while covalent compounds interact so
(c) sp,sp3 ,sp2 (d) sp2 ,sp3 ,sp weakly that even van der Waal’s forces between
15. The species having pyramidal shape is molecules of covalent compounds cannot be broken.
(a) SO3 (b) BrF3 (a) A (b) B
(c) SiO32  (d) OSF2 (c) C (d) D (e) E
22. Assertion: CCl4 is a non-polar molecule.
16. Which of the following species is not paramagnetic?
(a) NO (b) CO Reason: CCl4 has polar bonds.
(c) O 2 (d) B 2 (a) A (b) B
17.  
Stability of the species Li 2 , Li and Li increases in the (c) C (d) D (e) E
2 2
23. Assertion: C3O2 has linear structures.
order of
Reason: Each C atom in C3O2 is sp-hybridized.
(a) Li 2  Li 2  Li 2 (b) Li 2  Li 2  Li 2 (a) A (b) B
(c) Li 2  Li 2  Li 2 (d) Li 2  Li 2  Li 2 (c) C (d) D (e) E
18. In which of the following ionization processes. the bond 24. Assertion: B2 molecule is diamagnetic.
order has increased and the magnetic behavior has Reason: The highest occupied molecular orbital is of σ
changed? type.
(a) N 2  N 2 (b) C2  C2 (a) A (b) B
(c) C (d) D (e) E
(c) NO  NO  (d) O2  O2
136 CHEMICAL BONDING

25. Assertion: Fluorine molecule has bond order one. 29. Which of the following has correct placement of lone
Reason: The number of electrons in the anti bonding pairs and bond pairs?
molecular orbitals is two less than that in bonding
molecular orbitals.
(a) A (b) B
(c) C (d) D (e) E
26. Assertion: The HF2- ion exists in the solid state & also (a)
in liquid state but not in aqueous state.
Reason: The magnitude of hydrogen bonds among HF
molecules is weaker than that in between HF and H2O.
(a) A (b) B (b)
(c) C (d) D (e) E

Comprehension Based Question

Comprehension
(c)
In order to explain the characteristic geometrical shapes
of polyatomic molecules, Pauling introduced the concept
of hybridisation. The orbitals undergoing hybridisation
should have nearly same energy. There are various types
of hybridisations involving s, p and d - type of orbitals. (d)
The type of hybridisation gives the characteristic shape
30. Which molecule does not have the same type of
of the molecule or ion.
hybridisation as P has in PF5?
27. Which of the following statement is not correct?
(a) CIF3 (b) SF4
(a) The hybridized orbitals are always equivalent in
(c) XeF4 (d) XeF2
energy and shape
(b) sp hybridized orbitals has more s-character than sp2
hybridised orbital.
(c) Promotion of electron is essential condition prior to
hybridisation.
(d) The hybridized orbitals are directed in space in some
preferred directions to have minimum repulsion
between electron pairs.
28. Which other molecule or ion has same shape as NO2+
ion?
(a) SF2 (b) H3O+
(c) XeF2 (d) CO32–
CHEMICAL BONDING 137

Notes:

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

CHEMICAL BONDING
 04
MOLE & EQUIVALENT CONCEPT
Chapter 04

Mole & Equivalent Concept

1. Introduction to Chemistry Liquid: A substance is said to be liquid, if it possesses a definite

volume but no definite shape. They take up the shape of the vessel
Chemistry is defined as the study of the composition, properties
in which they are kept, e.g., water, milk, oil, mercury, alcohol etc.
and interaction of matter. Chemistry is often called the central
science because of its role in connecting the physical sciences, Gas: A substance is said to be gaseous if it neither possesses
which include chemistry, with the life sciences and applied definite volume nor a definite shape. This is because they fill up
sciences such as medicine and engineering. the whole vessel in which they are kept, e.g., hydrogen, oxygen etc.

Various branches of chemistry are:

1.1 Physical Chemistry

The branch of chemistry concerned with the way in which the
physical properties of substances depend on and influence their
chemical structure, properties, and reactions.

Fig. 4.1 :Arrangement of particles in solid, liquid and gaseous

1.2 Inorganic Chemistry state
The branch of chemistry which deals with the structure,
The three states are interconvertible by changing the conditions
composition and behavior of inorganic compounds. All the
substances other than the carbon-hydrogen compounds are of temperature and pressure as follows:
classified under the group of inorganic substances.

1.3 Organic Chemistry

The discipline which deals with the study of the structure,
composition and the chemical properties of organic compounds
is known as organic chemistry.

1.4 Biochemistry
The discipline which deals with the structure and behavior of the
components of cells and the chemical processes in living beings
is known as biochemistry.

1.5 Analytical Chemistry Fig. 4.2 : Interconversion of the three states of matter
The branch of chemistry dealing with separation, identification
and quantitative determination of the compositions of different
substances. 3. Classification of Matter at
Macroscopic Level
2. Matter At the macroscopic or bulk level, matter can be classified as
Matter is defined as any thing that occupies space and possesses
mass. Matter can exist in 3 physical states viz. solid, liquid, gas. (a) mixtures
Solid: A substance is said to be solid if it possesses a definite (b) pure substances.
volume and a definite shape, e.g., sugar, iron, gold, wood etc.

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140 MOLE & EQUIVALENT CONCEPT

These can be further sub-divided as shown below:

Oxygen Hydrogen

Fig. : 4.4: Atoms of different elements

Fig. : 4.3 Classification of matter

(a) Mixtures : A mixture contains particles of two or more pure

substances present in it (in any ratio) which are called its
components. A mixture may be homogeneous or
heterogeneous. Fig. : 4.5 : A representation of atoms and molecules
Homogeneous mixture: In homogeneous mixture the Compound: A compound is a pure substance containing two
components completely mix with each other and its or more than two elements combined together in a fixed
composition is uniform throughout i.e it consist of only one proportion by mass. Further, the properties of a compound
phase. Sugar solution and air are thus, the examples of are completely different from those of its constituent elements.
homogeneous mixtures. Moreover, the constituents of a compound cannot be
Heterogeneous mixtures: In heterogeneous mixture the separated into simpler substances by physical methods. They
composition is not uniform throughout and sometimes the can be separated by chemical methods.
different phases can be observed. For example, grains and
For example, hydrogen and oxygen are gases, whereas, the
pulses along with some dirt (often stone) pieces, are
compound formed by their combination i.e., water is a liquid.
heterogeneous mixtures.

NOTE:
Any distinct portion of matter that is uniform throughout in
composition and properties is called a Phase.

(b) Pure substances: Pure substances are substances that are

made up of only one kind of particles and the constituent
particles of pure substances have fixed composition.
H2O Water
Pure substances can be further classified into elements and
compounds. Fig. 4.6 : A depiction of molecule of water
Element: Particles of an element consist of only one type of
atoms. These particles may exist as atoms or molecules. 4. Properties of Matter
Depending upon the physical and chemical properties, the Every substance has unique or characteristic properties. These
elements are further subdivided into three classes, namely properties can be classified into two categories – physical
(1) Metals (2) Non-metals and (3) Metalloids. properties and chemical properties.
Sodium, copper, silver, hydrogen, oxygen, etc., are some
examples of elements. Their all atoms are of one type.
Whereas, in some others, the constituent particles are 4.1 Physical Properties
molecules which are formed by two or more atoms. For
Physical properties are those properties which can be measured
example, hydrogen, nitrogen and oxygen gases consist of
or observed without changing the identity or the composition of
molecules, in which two atoms combine to give their
the substance. Some examples of physical properties are colour,
respective molecules.
odour, melting point, boiling point, density etc.

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MOLE & EQUIVALENT CONCEPT 141

4.2 Chemical Properties given below.

These units pertain to the seven fundamental scientific quantities.
Chemical properties are those in which a chemical change in the
The other physical quantities such as speed, volume, density
substance occurs. The examples of chemical properties are
etc. can be derived from these quantities.
characteristic reactions of different substances; these include
acidity or basicity, combustibility etc.
6. Some Important Definitions
5. Measurement
6.1 Mass and Weight
5.1 Physical Properties Mass of a substance is the amount of matter present in it while
Many properties of matter, such as length, area, volume, etc., are weight is the force exerted by gravity on an object. The mass of
quantitative in nature. Quantitative measurement of properties is a substance is constant whereas its weight may vary from one
required for scientific investigation. Evidently, the measurement place to another due to change in gravity. The SI unit of mass is
of any physical quantity consists of two parts: the kilogram (kg). The SI derived unit (unit derived from SI base
(1) The number, and (2) The unit (standard of reference chosen units) of weight is newton.
to measure any physical quantity.)
For example, length of a room can be represented as 6 m; here, 6
is the number and m denotes metre, the unit in which the length
6.2 Volume
is measured. Volume is the amount of space occupied by a substance. It has
the units of (length)3. So in SI system, volume has units of m3.
But again, in chemistry laboratories, smaller volumes are used.
5.2 S.I. Units Hence, volume is often denoted in cm3 or dm3 units.
The International System of Units (in French Le Systeme A common unit, litre (L) which is not an SI unit, is used for
International d’Unités – abbreviated as SI) was established by measurement of volume of liquids.
the 11th General Conference on Weights and Measures 1 L = 1000 mL , 1000 cm3 = 1 dm3
The SI system has seven base units and they are listed in table

6.3 Density
The density of a material is defined as its mass per unit volume.
The symbol most often used for density is  (the lower case
Greek letter rho). SI unit of density is kg m–3.

Table : 4.1 : Base Physical Quantities and their Units

Base Physical Quantity Symbol for Quantity Name of SI unit Symbol for SI unit
Length l metre m
Mass m kilogram kg
Time t second s
Electric current I ampere A
Thermodynamic temperature T kelvin K
Amount of substance n mole mol
Luminous intensity Ir candela cd

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6.4 Temperature (iv) Zeros at the end or right of a number are significant, provided
they are on the right side of the decimal point. For example,
Temperature is a physical property of matter that quantitatively
0.200 g has three significant figures. But, if otherwise, the
expresses the common notions of hot and cold. There are three
terminal zeros are not significant if there is no decimal point.
common scales to measure temperature — °C (degree celsius), °F
For example, 100 has only one significant figure, but 100. has
(degree fahrenheit) and K (kelvin). The temperature on two scales
three significant figures and 100.0 has four significant figures.
is related to each other by the following relationship:
Such numbers are better represented in scientific notation.
°F = 9/5 (°C) + 32 We can express the number 100 as 1×102 for one significant
K = °C + 273.15 figure, 1.0×102 for two significant figures and 1.00×102 for
three significant figures.

7. Uncertainity in Measurement (v) Counting the numbers of object, for example, 2 balls or 20
eggs, have infinite significant figures as these are exact
numbers and can be represented by writing infinite number
7.1 Scientific Notation of zeros after placing a decimal i.e., 2 = 2.000000 or 20 =
20.000000.
In Chemistry, we come across very large and very small numbers.
It is very tedious to write down such numbers in the ordinary
way. These numbers are usually expressed in a simple way known 7.3 Precision and Accuracy
as exponential form or scientific notation. Precision refers to the closeness of various measurements for
For example, the number 246.38 may be expressed as: the same quantity. However, accuracy is the agreement of a
246.38 = 2.4638 × 10 × 10 = 2.4638 × 102 particular value to the true value of the result. For example, if the
Thus, in general, number is written in scientific notation as: true value for a result is 2.00g and 3 students takes two
measurements each as
number with a single number of places
non-zero digit decimal point was moved
+ moved left Table : 4.2 : Data to Illustrate Precision and Accuracy
– moved right
N × 10n Measurements/g
1 2 Precision and Accuracy
7.2 Significant Figures Student A 1.95 1.93 Precise but not accurate
Significant figures are meaningful digits which are known with Student B 1.94 2.05 Neither precise nor accurate
certainty plus one which is estimated or uncertain. The uncertainty Student C 2.01 1.99 Both precise and accurate
is indicated by writing the certain digits and the last uncertain
digit. Thus, if we write a result as 11.2 mL, we say the 11 is certain
and 2 is uncertain and the uncertainty would be +1 in the last 7.4 Addition and Subtraction of Significant
digit. Unless otherwise stated, an uncertainty of +1 in the last Figures
digit is always understood.
The result cannot have more digits to the right of the decimal
There are certain rules for determining the number of significant point than either of the original numbers.
figures. These are stated below:
For example, 12.11 + 18.0 + 1.012 = 31.1
(i) All non-zero digits are significant. For example in 285 cm,
there are three significant figures and in 0.25 mL, there are
two significant figures. 7.5 Multiplication and Division of Significant
(ii) Zeros preceding to first non-zero digit are not significant. Figures
Such zero indicates the position of decimal point. Thus, 0.03
In these operations, the result must be reported with no more
has one significant figure and 0.0052 has two significant
significant figures as in the measurement with the few significant
figures.
figures.
(iii) Zeros between two non-zero digits are significant. Thus, 2.005
For example, 2.5 × 1.25 = 3.1
has four significant figures.

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8. Laws of Chemical Combination 8.4 Law of Reciprocal Proportions

If two different elements combine separately with a fixed mass of
a third element, the ratio of the masses in which they do so are
8.1 Law of Conservation of Mass either the same as or a simple multiple of the ratio of the masses
“In a chemical reaction the mass of reactants consumed and mass in which they combine with each other. This Law was given by
of the products formed is same, that is mass is conserved.” This Richter in 1792.
is a direct consequence of law of conservation of atoms. This law For example, Consider 3 g of C reacting with 1 g of H to form
was put forth by Antoine Lavoisier in 1789. methane. Also, 8 g of O reacting with 1 g of H to form water. Here,
For example,12 g of carbon combines with 32 g of oxygen to form the mass ratio of carbon and oxygen is 3:8. Similarly, 12 g of C
44 g of CO2. reacts with 32 g of O to form CO2. The mass ratio of carbon and
oxygen is 12:32 = 3:8.
8.2 Law of Constant/Definite Proportions The mass ratio in which C and O combine with each other is the
same as the mass ratio in which they separately combine with a
This law states that a given compound always contains exactly
fixed mass of H.
the same proportion of elements by weight. This law was given
by, a French chemist, Joseph Proust.
For example, carbon dioxide can be obtained by a number of 8.5 Gay Lussac’s Law of Gaseous Volumes
methods, such as This law was given by Gay Lussac in 1808. He observed that
(i) By burning coal or candle when gases combine or are produced in a chemical reaction they
C + O2  CO2 do so in a simple ratio by volume provided all gases are at same
(ii) By heating limestone (CaCO3) temperature and pressure.
For example, Two volumes of hydrogen combine with one volume
C aC O 3  Heat
  C aO  CO 2 of oxygen to give two volumes of water vapour.
(iii) By the action of dilute hydrochloric acid on marble pieces.
CaCO3 + 2HCl  CaCl2 + CO2 + H2O
It has been observed that each sample of carbon dioxide
contains carbon and oxygem elements in the ratio of 3 : 8 by
weight.

Hydrogen gas Oxygen gas Steam

8.3 Law of Multiple Proportions
According to this law, if two elements can combine to form more
than one compound, the masses of one element that combine Fig. 4.7 : Two volumes of hydrogen react with one volume of
with a fixed mass of the other element, are in the ratio of small oxygen to give two volumes of water vapour(steam)
whole numbers. This law was proposed by Dalton in 1803.
For example,
Hydrogen combines with oxygen to form two compounds namely
8.6 Avogadro Law
water and hydrogen peroxide. In 1811, Avogadro proposed that equal volumes of gases at the
same temperature and pressure should contain equal number of
Hydrogen + Oxygen  Water
molecules.
2g 16g 18g
For example, The deflation of automobile tyres. When the air
Hydrogen + Oxygen  Hydrogen Peroxide
trapped inside the tyre escapes, the number of moles of air present
2g 32g 34g in the tyre decreases. This results in a decrease in the volume
Therefore, the masses of oxygen (16g or 32g) which combine occupied by the gas, causing the tyre to lose its shape and deflate.
with fixed mass of hydrogen (2 parts) bear a simple ratio i.e.,
16 : 32 or 1 : 2.

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9. Dalton’s Atomic Theory For example,

Isotope Relative abundance (%) Atomic mass (amu)
12
C 98.892 12
9.1 Postulates of Dalton’s Atomic Theory 13
C 1.108 13.00335
In 1808, Dalton published ‘A New System of Chemical Philosophy’ 14 –10
C 2 × 10 14.00317
in which he proposed the following:
(i) Matter consists of indivisible atoms.
(ii) All the atoms of a given element have identical properties From the above data, the average atomic mass of carbon will
including identical mass. Atoms of different elements differ come out to be:
in mass. (0.98892) (12 u) + (0.01108) (13.00335 u) + (2 × 10–12) (14.00317 u) =
(iii) Compounds are formed when atoms of different elements 12.011 u
combine in a fixed ratio.
(iv) Chemical reactions involve reorganization of atoms. These 10.3 Molecular Mass
are neither created nor destroyed in a chemical reaction.
Molecular mass is the sum of atomic masses of the elements
present in a molecule. It is obtained by multiplying the atomic
9.2 Drawbacks of Dalton’s Theory mass of each element by the number of its atoms and adding
them together. For example,
(i) Does not explain structure of atom.
Molecular mass of methane,
(ii) Fails to explain binding forces between atoms in compounds.
(CH4) = (12.011 u) + 4 (1.008 u) = 16.043 u
(iii) Does not explain Gay Lussac's law.

10. Atomic and Molecular Masses 10.4 Formula Mass

Some substances, such as sodium chloride, do not contain
discrete molecules as their constituent units.
10.1 Atomic Mass The formula, such as NaCl, is used to calculate the formula mass
Atomic mass of an element is defined as the average relative instead of molecular mass.
mass of an atom of an element as compared to the mass of an Thus, the formula mass of sodium chloride is atomic mass of
atom of carbon (12C) taken as 12 . sodium + atomic mass of chlorine
Mass of an atom = 23.0 u + 35.5 u = 58.5 u
Atomic mass =
1
mass of a carbon atom  12 C 
12
This scale of relative masses of atoms is called atomic mass unit
11. Mole Concept
scale and is abbreviated as a.m.u (new symbol ). A mole is defined as amount of substance that contains as many
One atomic mass unit is defined as a mass exactly equal to one- particles or entities (atoms, molecules or ions) as there are in
twelfth of the mass of one carbon - 12 atom. exactly 12g of carbon - 12 isotope.
And 1 amu = 1.66056 × 10–24 g One mole contains exactly 6.02214076 × 1023 elementary entities.
This number is the fixed numerical value of the Avogadro constant,
NA , when expressed in the unit mol–1 and is called the Avogadro
10.2 Average Atomic Mass number.
Many naturally occurring elements exist as more than one isotope. One mole of atoms = 6.022 × 1023 atoms
The atomic mass of such an element is the average of the atomic
Mass of element
masses of the naturally occurring isotopes of the element taking Moles of an element 
into account their relative abundance. This is called average Atomic mass
atomic mass.

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Mass percentage of an element

Mass of one atom  Atomic mass
6.022  10 23 mass of that element in the compound × 100

23
One mole of molecules = 6.022 × 10 molecules molar mass of the compound

Mass of compound Molar mass of water = 18.02 g

Moles of a compound 
Molecular mass 2  1.008
Mass percentage of hydrogen =  100
18.02
Molecular mass
Mass of one molecule  = 11.18
6.022  10 23
Volume occupied by 1 mole of a gas at N.T.P. = 22.4 L. 16.00
Mass percentage of oxygen =  100
18.02
= 88.79

13. Empirical Formula and Molecular

Formula
An empirical formula represents the simplest whole number ratio
of various atoms present in a compound, whereas, the molecular
formula shows the exact number of different types of atoms
present in a molecule of a compound.
 Molecular formula and empirical formula are related as
Molecular formula = n (Empirical formula)
where n is a simple whole number and may have values 1,2,3...
It is equal to
Fig. 4.8: Various relationships of mole
Molecular mass
n
Empirical formula mass

13.1 Determination of the Empirical

Formula af a Compound
The empirical formula of a compound can be determined from the
percentage composition of different elements and atomic masses
Fig 4.9: Scheme for calculation of mole, mass and number of of the elements.
particles The various steps involved in determining the empirical formula
are:
12. Percentage Composition Step I: Divide the percentage of each element by its atomic mass.
This gives the moles of atoms of various elements in the molecule
Information regarding the percentage of a particular element
of the compound.
present in a compound is given by percentage composition.
Percentage of an element
For example, water contains hydrogen and oxygen, the Moles of atoms 
percentage composition of both these elements can be calculated Atomic mass of the element
as follows:

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Step II: Divide the result obtained in the above step by the smallest (ii) It should be balanced i.e., the total number of atoms on both
value among them to get the simplest ratio of varoius atoms. sides of the equation must be equal.
Step III: Make the values obtained above to the nearest whole (iii) It should be molecular i.e., the elementary gases like
number and multiply, if necessary, by a suitable integer to make hydrogen, oxygen, nitrogen, etc. must be represented in
the values whole numbers. This gives the simplest whole number molecular form as H2, O2, N2 etc.
ratio.
Step IV: Write the symbols of the various elements side by side
14.2. Stoichiometry
and insert the numerical value at the right hand lower corner of
each symbol. The formula thus obtained represents the empirical It is calculation of masses or volumes of reactants and products
formula of the compound. involved in a chemically balanced reaction. Consider the formation
of ammonia.
N2(g) + 3H2(g)  2NH3(g)
13.2 Steps of Determination of the
All are gases indicated by letter (g) and coefficients 3 for H2 and
Molecular Formula of a Compound 2 for NH3 are called stoichiometric coefficients. The formation of
Step I: Determine the empirical formula as described above. ammonia can be interpreted in many ways.
Step II: Calculate the empirical formula mass by adding the atomic • One mole of N2(g) reacts with three moles of H2(g) to give
masses of the atoms in the empirical formula. two moles of NH3(g).
Step III: Determine the molecular mass by a suitable method. • 28g of N2(g) reacts with 6g at H2(g) to give 34g of NH3(g).
Step IV: Determine the value of n as • 22.4L of N2(g) reacts with 67.2L of H2(g) to give 44.8L of
Molecular mass NH3 (g).
n
Empirical formula mass
Change n to the nearest whole number. 15. Limiting Reagent (LR) and
Step V: Multiply empirical formula by n to get the molecular
formula.
Excess Reagent (ER)
Molecular formula = n × Empirical formula If the reactants are not taken in the stoichiometric ratios then the
reactant which is less than the required amount determines how
much product will be formed and is known as the Limiting

14. Chemical Equations and Reagent and the reactant present in excess is called the Excess
Reagent.
Stoichiometry For example, in the reaction
2H2 + O2  2H2O
If the reaction mixture contains 2 mol of H2 and 2 mol of O2, then
14.1 Chemical Equation only 1 mol of O2 will be used up and 1 mol of O2 will be left over.
The chemical equation may be defined as: Theerefore, in this case, hydrogen is the limiting reactant. Oxygen
A brief representation of a chemical change in terms of symbols is excess reactant.
and formulae of substances involved in it.
For example, the reaction of silver nitrate with sodium chloride to 16. Percentage Yield
give silver chloride and sodium nitrate may be represented as
Due to practical reasons the amount of product formed by a
AgNO 3  NaCl  AgCl  NaNO 3 chemical reaction is less than the amount predicted by theoretical
 
Reactants Products calculations. The ratio of the amount of product formed to the
Essentials of Chemical Equation amount predicted when multiplied by 100 gives the percentage
yield.
(i) It must be consistent with the experimental facts i.e., a chemical
equation must represents a true chemical change. If a chemical Actual Yield
Percentage Yield = ×100
reaction is not possible between certain substances it cannot Theoretical Yield
be represented by a chemical equation.

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MOLE & EQUIVALENT CONCEPT 147

17. Reaction in Aqueous Media ppm :

mass of solute
× 106
and Strength of solution mass of solution
 Important relations
There are many reactions which take place in solutions. In solution
generally one component is present in lesser amount and is called (i) Relation between molality (m), Molarity (M), density (d) of
solute while the other present in excess is called the solvent. The solution and molar mass of solute (M0)
amount of solute present in a given quantity of solvent or solution d : density in g/mL
is expressed in terms of concentration. M0: molar mass in g mol–1
The concentration of the solution is usually expressed in the M  1000
following ways: Molality, m 
1000d  MM 0
 Mass percentage: Mass of solute (in g) present in 100g of
(ii) Relationship between molality (m) and mole fraction (B) of
solution.
the solute
Mass of solute
Mass percentage   100 B 1000 1   A 1000
Mass of solution m  m 
 Mass by Volume percentage: Mass of solute (in g) present in 1  B M A ; A MA
100mL of solution Where MA = molecular mass of solvent, A = mole fraction of
Mass of solute solvent.
Mass by volume percentage  100
Volume of solution
POINTS TO REMEMBER:
 Volume by Volume percentage: Volume of solute present per
100 parts by volume of solution {only for liq-liq solutions}  Molarity is the most common unit of measuring strength of
solution.
Volume of solute  The product of Molarity and Volume of the solution gives
Volume by volume percentage  100
Volume of solution the number of moles of the solute, n = M × V
 Molarity (M): The number of moles of solute dissolved per
litre of solution. 18. Dilution Law
moles of solute When a solution is diluted, more solvent is added, the moles
M
volume of solution  L  of solute remains unchanged. If the volume of a solution
 Molality (m): The number of moles of solute dissolved per having a Molarity of M1 is changed from V1 to V2 we can write
kg of solvent. that:
moles of solute M1V1 = M1V2
m=
mass of solvent  kg 
 Mole fraction (): The ratio of number of moles of one
component to the total number of moles (solute and solvent)
19. Effect of Temperature
present in solution. Let nA represents moles of solute and nB Volume of the solvent increases on increasing the temperature.
represents moles of solvent. Then, But it shows no effect on the mass of solute in the solution
nA assuming the system to be closed i.e. no loss of mass.
Mole fraction of solute   A   The formulae of strength of solutions which do not involve volume
nA  nB
of solution are unaffected by changes in temperature. e.g. molality
nB
Mole fraction of solvent   B   remains unchanged with temperature. Formulae involving volume
nA  nB
are altered by temperature e.g. Molarity.
 Parts per milion: Whe a solute is present in trace quantities,
it is convenient to express concentration in parts per million
(ppm) and is defined as:

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148 MOLE & EQUIVALENT CONCEPT

20.Introduction to Equivalent 20.4 Methods of Determining Equivalent

Concept Masses
(i) Hydrogen displacement method: It is for metals which
Equivalent concept is a way of understanding reactions and
can displace H2 from acids.
processes in chemistry which are often made simple by the use
of Equivalent concept. Equivalent mass of metal
Weight of metal
 1.008
Weight of displaced hydrogen
20.1 Valency Factor (Z)
Weight of metal in gram
Equivalent weight is the ratio of molar mass and a factor which is   11.2litre
Volume of H 2 in litre
called n-factor or valency factor
(ii) Oxide Method:
Molecular Mass
Equivalent mass, E 
Z Weight of metal
Equivalent mass of metal   8.0
Weight of oxygen

20.2 Equivalent Mass (iii) Chloride method:

It is the number of parts by weight of the substance that combines Weight of metal
Equivalent mass of metal   35.5
or displaces, directly or indirectly, 1.008 parts by mass of hydrogen Weight of chlorine
or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine. It
Weight of metal in gram
can be calculated as:   11.2 litre
Volume of Cl2 in litre
Atomic mass
(i) Equivalent mass for elements  Valency
21. Law of Chemical Equivalence
Molecular mass According to it, during a chemical reaction, equivalent of each
(ii) Equivalent mass for acids  reactant reacted will be equal to equivalent of each product
Basicity of acids
produced. Irrespective of stoichiometric coefficients of reactants
 Basicity: It is the number of H+ ions that can be displaced and products.
from one molecule of a substance. aA + bB  cC + dD
Molecular mass Equivalent of A = Equivalent of B = Equivalent of C
(iii) Equivalent mass for bases  Reacted Reacted Produced
Acidity of Base
= Equivalent of D
 Acidity: It is the number of OH– ions that can be displaced
Produced
from one molecule of a substance.
22. Normality
20.3 Gram Equivalent Mass (GEM) The normality of a solution is the number of equivalents of solute
It is the mass of a substance expressed in grams or equivalently present in 1L of the solution.
the quantity of substance whose mass in grams is equal to its equivalents of solute
N
equivalent mass is called one gram equivalent or gram equivalent volume of solution (L)
mass. The number of equivalents of solute present in a solution is
given by Normality × Volume (L).
No. of gm equivalents  Mass in gm
GEM On dilution of the solution the number of equivalents of the
solute is conserved and thus, we can apply the formula :
N1V1 = N2V2
Relationship between Normality and Molarity
N = M × Z ; where ‘Z’ is the Valency factor

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MOLE & EQUIVALENT CONCEPT 149

Summary
• All substances contain matter, which can exist in three states of different isotopes of that element. The molecular mass of
– solid, liquid or gas. The constituent particles are held in a molecule is obtained by taking sum of the atomic masses of
different ways in these states of matter and they exhibit their different atoms present in a molecule. The molecular formula
characteristic properties. Matter can also be classified into can be calculated by determining the mass per cent of different
elements, compounds or mixtures. An element contains elements present in a compound and its molecular mass.
particles of only one type, which may be atoms or molecules.
The compounds are formed where atoms of two or more
• The number of atoms, molecules or any other particles present
elements combine in a fixed ratio to each other. Mixtures occur
in a given system are expressed in the terms of Avogadro
widely and many of the substances present around us are
constant (6.022 × 1023). This is known as 1 mol of the respective
mixtures.
particles or entities.

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

• The uncertainty is taken care of by specifying the number of • The amount of substance present in a given volume of a
significant figures, in which the observations are reported. solution is expressed in number of ways, e.g., mass per cent,
mole fraction, molarity and molality
• The combination of different atoms is governed by basic laws
of chemical combination — these being the Law of • Equivalent mass is the number of parts by weight of the
Conservation of Mass, Law of Definite Proportions, Law of substance that combines or displaces, directly or indirectly,
Multiple Proportions, Law of Reciprocal Proportion, Gay 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen
Lussac’s Law of Gaseous Volumes and Avogadro Law. All or 35.5 parts by mass of chlorine. It can be calculated as:
these laws led to the Dalton’s atomic theory, which states
that atoms are building blocks of matter. Molecular mass
Equivalent mass, E  Z valency factor
 
• The atomic mass of an element is expressed relative to 12C
isotope of carbon, which has an exact value of 12u. Usually,
• The normality of a solution is the number of equivalents of
the atomic mass used for an element is the average atomic
solute present in 1L of the solution.
mass obtained by taking into account the natural abundance

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150 MOLE & EQUIVALENT CONCEPT

Scheme for calculation of mole, mass and number of particles

Various relationships of mole

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MOLE & EQUIVALENT CONCEPT 151

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 (iv) 126,000 (v) 500.0 57.1
(vi) 2.0034  = 1.33 g
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%
If the speed of light is 3.0 × 108 m s–1, calculate the distance
 Amount of oxygen that combines with 1 g carbon
covered by light in 2.00 ns.
Sol. Distance covered = Speed × Time = 3.0 × 108m s–1 × 2.00 ns 72.7
 = 2.66 g
27.3
10 9 s
8 –1
= 3.0 × 10 m s × 2.00 ns × = 6.00 × 10–1m Ratio of oxygen in oxide (1) and (2) = 1 : 2
1 ns
Thus, Law of multiple proportion is verified.
= 0.600 m Example-8
Example-4 Classify the following substances into elements,
What is the S.I. unit of mass ? compounds and mixtures.
Sol. S.I. unit of mass is kilogram (kg). (i) Air (ii) Diamond (iii) LPG (iv) Dry ice (v) Graphite
(vi) Steel (vii) Marble (viii) Smoke (ix) Glucose
Example-5
(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
products is less than the mass of the reactants. The Mixtures : Air; LPG; Steel; Smoke
difference of mass, called the mass defect, is converted Example-9
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 (iii) Molecule mass of CH4 = 12.01 amu + 4 (1.008 amu) =
compounds. It is true only for the compounds obtained
from one isotope. For example, carbon exists in two common 16.042 amu
isotopes, 12C and 14C.
152 MOLE & EQUIVALENT CONCEPT
Example-10
3.01 1022 x
Carbon occurs in nature as a mixture of carbon-12 and So, 23

6.02  10 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  1022
Sol. Let x be the percentage abundance of carbon-12; then or x = 0.85 g
6.02  10 23
(100 – x) will be the percentage abundance of carbon-13.
Example-15
12x  13 (100  x)
Therefore,  12.011 How many molecules and atoms of oxygen are present in
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 6.02 ×
x = 98.9 1023 molecules of oxygen.
Abundance of carbon-12 is 98.9% So, 5.6 litres of oxygen at NTP contain
Example-11 5.6
12
 ×6.02 × 1023 molecules
What will be the mass of one C atom in g? 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
12 = 2 atoms of oxygen
1 Atom of 12C will have mass = g
6.022  1023 23
So, 1.505 × 10 molecules of oxygen contain
= 1.9927 × 10–23 g = 2 × 1.505 × 1023 atoms
Example-12 = 3.01 × 1023 atoms
Calculate the number of atoms in each of the following : Example-16
(i) 52 moles of He (ii) 52 u of He
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
= 3.131 × 1025 atoms (CH4) = 12 + 4 = 16 g
(ii) 1 atom of He = 4 u of He
1.6
4 u of He = 1 atom of He Number of moles in 1.6 g of methane =  0.1
16
1 Number of molecules of methane in 0.1 mole
 52 u of He = × 52 atoms = 13 atoms
4 = 0.1 × 6.02 × 1023
Example-13 = 6.02 × 1022
Calculate the mass of 2.5 gram atoms of oxygen. One molecule of methane has = 6 + 4 = 10 electrons
Sol. We know that
So, 6.02 × 1022 molecules of methane have
Mass of an element in grams = 10 × 6.02 × 1022 electrons
Number of gram atoms = Atomic mass of the element in grams
= 6.02 × 1023 electrons
So, Mass of oxygen = 2.5 × 16 = 40.0 g Example-17
Example-14 Calculate the mass per cent of different elements present in
What is the mass of 3.01 × 1022 molecules of ammonia ? sodium sulphate (Na2SO4).
Sol. Gram-molecular mass of ammonia = 17 g Sol. Mass % of an element
Number of molecules in 17 g (one mole) of NH3 = 6.02 × 1023 Mass of that element in the compound
22
  100
Let the mass of 3.01 × 10 molecules of NH3 be = x g Molar mass of the compound
MOLE & EQUIVALENT CONCEPT 153
Now, molar mass of Na2SO4 = 2 (23.0) + 32.0 + 4 × 16.0 Molecular Formula Mass
–1
Also, given: Empirical Formula Mass = 3 = n-factor
= 142 g mol

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

Mass percent of sodium =  100 = 32.39 %
142 Example-20
Determine the empirical formula of an oxide of iron which
32 has 69.9 % iron and 30.1 % dioxygen by mass.
Mass per cent of sulphur =  100 = 22.54 %
142
Sol.
64 El eme nt Percen- Ato mic Moles Mole S i mp l e s t
Mass per cent of oxygen =  100 = 45.07% tage mass of ratio who le
142 element no.ratio

Example-18 69.9
 1.25
1.25
1
Iron 69.9 55.85 2
55.85 1.25
Calculate the empirical formula of a compound that contains
26.6% potassium, 35.4% chromium and 38.1% oxygen 30.1 1.88
Oxygen 30.4 16.00  1.88 1.5 3
[Given K = 39.1; Cr = 52; O = 16] 16.00 1.25

Sol.  Empirical formula = Fe2O3.

El eme nt Percen- Ato mic Moles Mole Simplest Example-21
tage mass of ratio whole
element no.ratio Hydrogen chloride (HCl) on oxidation gives water and
chlorine. How many litres of chlorine at STP can be obtained
26.6 0.68
Potassium 26.6 39.1  0.68 1 1×2=2 starting with 36.50 g HCl ?
39.1 0.68
Sol. Oxidation of HCl takes place according to the following
35.4 0.68 equation :
Chromium 35.4 52.0  0.68 1 1×2=2
52 0.68

38.1 2.38
4HCl  O2 
 2Cl2  2H2 O
Oxygen 38.1 16.0  2.38  3.5 3.5 × 2 = 7 4 mol 2 mol
16 0.68
Therefore, empirical formula is K2Cr2O7. Mass 36.5
Example-19 Moles of HCl = Molecular mass  36.5  1 mole
An organic compound containing C,H and N gave the fol-
lowing analysis: C: 40% H:13.3%, N:46.67%. If its molecular  4 moles HCl give 2 moles Cl2
formula weight is three times its empirical formula weight 2
 1 mole will give moles Cl2 = 0.5 moles Cl2
then find out its empirical and molecular formula of the com- 4
pound. Volume of Cl2 at STP = 22.4 × 0.5 = 11.2 litre
Sol. Relative no. of atoms of C = 40/12 = 3.33 Example-22
Relative no. of atoms of H = 13.3/1 = 13.3 and that for N = How much copper can be obtained from 100 g of copper
46.67/14 = 3.33 sulphate (CuSO4)? (Atomic mass of Cu = 63.5 amu)
Thus, simplest atomic ratio C:H:N Sol. 1 mole of CuSO4 contains 1 mole (1 g atom) of Cu
Molar mass of CuSO4 = 63.5 + 32 + 4 × 16 = 159.5 g mol–1
= 3.33:13.33:3.33 = 1:4:1
Thus, Cu that can be obtained from 159.5 g of CuSO4
Therefore the empirical formula of the compound is CH4N = 63.5 g
 Cu that can be obtained from 100 g of CuSO4
63.5
=  100 g = 39.81 g.
159.5
154 MOLE & EQUIVALENT CONCEPT
Example-23
100g
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?
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 3 OH
 2 mole of A will react with 2 mol of B. Hence, A is the Molarity 
Volume in L
limiting reactant.
Example-24 Moles of CH 3 OH
Calculate the concentration of nitric acid in moles per litre
0.25 =
2.5
in a sample which has a density 1.41 g mL–1 and the mass
Moles of CH3OH = 0.25 × 2.5 = 0.025 moles
percent of nitric acid in it being 69%.
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
contain 69g of nitric acid by mass. Now 0.793 × 103g of CH3OH is present in 1000 mL
–1
Molar mass of nitric acid HNO3 = 1 + 14 + 48 = 63 gmol
1000
69 g 20 g of CH3OH will be present in = × 20
oles in 69g HNO3 = 63 g mol 1  1.095 mole 0.793×10 3
= 25.2 mL
MOLE & EQUIVALENT CONCEPT 155

Exercise – 1: Basic Objective Questions

Physical Quantities and Their Measurements in 7. Which of the following is the most accurate
Chemistry measurement?
(a) 9 m (b) 9.0 m
(c) 9.00 m (d) 9.000 m
1. Which of the following statements(s) is false about the
8. How many significant figures are present in 0.10100 ×
arrangement of particles, in different states of matter?
103?
(a) Solids have definite volume and definite shape
(a) 7 (b) 5
(b) Liquids have definite volume and definite shape
(c) 3 (d) 10
however they take the shape of the container in which
9. _________is referred to as the closeness of different
they are placed
measurements for the same quantity.
(c) Gases have neither definite volume nor definite
(a) Accuracy
shape they completely occupy the shape of the
(b) Precision
container in which they are placed
(c) Analysis
(d) All the above
(d) Dimension
2. Statement I: The temperatures in degree Celsius and
10. How many significant figures does 63180 have?
degree Fahrenheit are related to each other by the
(a) 5 (b) 4
following relationship.
(c) 1 (d) 2
9
ºF   º C   32 11. Which of the following has the largest volume ?
5 (a) 445g of water at 4oC (density = 1g/cm3)
Statement II: The temperature on degree Celsius and (b) 600g of chloroform at 20ºC(density=1.5g/cm3)
Kelvin scale are related to each other by the following (c) 155 cm3 of steel
relationship. (d) 0.50L milk
º C  K  273.15 12. Assertion : Density is expressed as g mL–1 whereas
(a) Statement I is correct specific gravity is dimension less.
(b) Statement II is correct Reason : Specific gravity is ratio of the masses of
(c) Both the statement are correct solution and solvent.
(d) Neither the statement I not II is correct (a) If both assertion and reason are correct and reason is
3. Which of the following represents the largest unit? the correct explanation of assertion.
(a) Decilitre (b) Decalitre (b) If both assertion and reason are true but reason is
(c) Kilolitre (d) Megalitre not the correct explanation of assertion.
4. How can we write 232.508 in scientific notation? (c) If assertion is true but reason is false.
(a) 23.2508 × 102 (b) 2.32508 × 102 (d) If reason is true but assertion is false.
1
(c) 232.528×10 (d) 2325.08 × 100
5. 2.005 has …..A….. significant figures. Here, A refers
to
Laws of Chemical Combination & Dalton's
(a) Four (b) Three Atomic Theory
(c) Two (d) One
6. The result of the operation 2.5 × 1.5 should be which of 13. …A… are formed when atoms of different elements
the following on the basis of significant figures? combine in a …B… ratio. Here, A and B refer to
(a) 3.125 (b) 3.13 (a) A  Compound; B  Fixed
(c) 3.1 (d) 31.25 (b) A  Molecules; B  Any
(c) A  Compound; B  Any
(d) A  Molecule; B  Fixed
156 MOLE & EQUIVALENT CONCEPT

14. Which of the following statement is false? 20. One of the statements of Dalton’s atomic theory in
(I) Matter consists of indivisible atoms. given below
(II) All the atoms of a given element have identical “Compounds are formed when atoms of different
properties including identical mass. elements combine in a fixed ratio”
(III) Compounds are formed when atoms of different Which of the following laws is not related to this
elements combine in any ratio. statement?
(IV) Chemical reactions involve reorganization of (a) Law of conservation of mass
atoms. These are neither created nor Destroyed in a (b) Law of definite proportions
chemical reaction. (c) Law of multiple proportions
(a) Only I (d) None of these
(b) Only II 21. If 6.3g of NaHCO3 are added 15.0g CH3COOH
(c) Only III solution, the residue is found to weigh 18.0 g. What is
(d) Only IV the mass of CO2 released in the reaction?
15. What did Dalton’s Theory couldn’t explain? (a) 4.5g (b) 3.3 g
(a) gaseous volumes (c) 2.6 g (d) 2.8 g
(b) conservation of mass 22. Who proposed Law of Conservation of Mass?
(c) chemical philosophy (a) Antoine Lavoisier
(d) none of these (b) Joseph Proust
16. Which law states that matter can neither be created nor (c) Lorenzo Romano
destroyed? (d) Joseph Louis
(a) Law of definite proportions 23. Which of the following reaction is not according to the
(b) Law of conservation of mass law of conservation of mass
(c) Law of multiple proportions (a) 2Mg(s) + O2(g)  2MgO(s)
(d) Gay Lussac’s law of gaseous volumes (b) C3H8(g) + O2(g)  CO2(g) + H2O(g)
17. Which law states that if two elements can combine to (c) P4(s) + 5O2(g)  P4O10(s)
form more than one compound, the masses of one (d) CH4(g) + 2O2(g)  CO2(g) + 2H2O
element that combine with a fixed mass of other 24. Carbon and oxygen combine to form two oxides,
element, are in the ratio of small whole number? carbon monoxide and carbon dioxide in which the ratio
(a) Avagadro law of the weights of carbon and oxygen is respectively
(b) Law of definite composition 12:16 an 12:32. These figures illustrate the:
(c) Law of conservation of mass (a) Law of multiple proportions
(d) Gay Lussac’s law of gaseous volumes (b) Law of reciprocal proportions
18. The Gay Lussac’s law was explained properly by the (c) Law of conservation of mass
work of …A… in 1811. Here, A refers to (d) Law of constant proportions
(a) Dalton 25. The law of multiple proportion is illustrated by
(b) Avogadro (a) Carbon monoxide and carbon dioxide
(c) Gay Lussac (b) Potassium bromide and potassium chloride
(d) Antoin Lavoisier (c) Water and heavy water
19. In 1811, Avogadro proposed that equal volumes of (d) Calcium hydroxide and barium hydroxide
gases at the same …A… and …B… should contain 26. The percentage of copper and oxygen in samples of
equal number of molecules. Here, A and B refer to CuO obtained by different methods were found to be
(a) Temperature and pressure the same. This illustrates the law of
(b) density and mass (a) constant proportions
(c) molarity and normality (b) conservation of mass
(d) Avogadro constant and Planck's constant (c) multiple proportions
(d) reciprocal proportions
MOLE & EQUIVALENT CONCEPT 157

27. Hydrogen and oxygen combine to form H2O2 and H2O 34. A sample of calcium carbonate (CaCO3) has the
containing 5.93% and 11.2% hydrogen respectively. following percentage composition: Ca = 40%, C =
The data illustrates: 12%, O = 48%. If the law of constant proportions is
(a) law of conservation of mass true, then the weight of calcium in 4g of a sample of
(b) law of constant proportion calcium carbonate obtained from another source will
(c) law of reciprocal proportion be:
(d) law of multiple proportion (a) 0.016 g (b) 0.16 g
28. n g of substance X reacts with m g of substance Y to (c) 1.6 g (d) 16 g
form p g of substance R and q g of substance S. This 35. Assertion : Pure water obtained from different states of
reaction can be represented as follows : India always contains hydrogen and oxygen in the ratio
X+Y=R+S of 1 : 8 by mass.
The relation which can be established in the amounts of Reason: Total mass of reactants and products during
the reactants and the products will be chemical change is always the same.
(a) n – m = p – q (a) If both assertion and reason are correct and reason is
(b) n + m = p + q the correct explanation of assertion.
(c) n = m (b) If both assertion and reason are true but reason is
(d) p = q not the correct explanation of assertion.
29. A sample of CaCO3 has Ca = 40%, C = 12% and O = (c) If assertion is true but reason is false.
48%. If the law of constant proportions is true, then the (d) If reason is true but assertion is false.
mass of Ca in 5 g of CaCO3 from another source will
be : Atomic and Molecular Masses
(a) 2.0g (b) 0.2g
(c) 0.02g (d) 20.0g
36. Consider the following statements,
30. If water sample are taken from sea, rivers or lake, they
(I) Mixture may contain the components in any ratio
will be found to contain hydrogen and oxygen in the
and their composition is variable.
approximate ratio of 1:8. This indicates the law of:
(II) Pure substances have fixed composition.
(a) Multiple proportion
(III) The constituents of pure substances cannot be
(b) Definite proportion
separated by simple physical methods.
(c) Reciprocal proportions
(IV) Copper, glucose, silver, gold, water are the
(d) None of these
examples of pure substances.
31. Zinc sulphate contains 22.65% Zn and 43.9% H2O. If
(V) Glucose contains carbon, hydrogen and oxygen
the law of constant proportions is true, then the mass of
The true statements are.
zinc required to give 40g crystals will be :
(a) Only I, II, III (b) Only III, IV, V
(a) 90.6 g (b) 9.06 g
(c) Only II, IV, V (d) All of these
(c) 0.906 g (d) 906 g
37. Statement I: The constituents of a compound can be
32. 3 g of a hydrocarbon on combustion in excess of
separated into simpler substances by physical method.
oxygen produces 8.8g of CO2 and 5.4 g of H2O. The
Statement II: The constituents of compound can be
data illustrates the law of:
separated by chemical method.
(a) conservation of mass
Choose the correct option.
(b) multiple proportions
(a) Statement I is correct
(c) constant proportions
(b) Statement II is correct
(d) reciprocal proportions
(c) Both the statements are correct
33. The law of conservation of mass holds good for all of (d) None of the above statement are correct
the following except:
38. ‘amu’ has been replaced by ‘u’ which is known as
(a) All chemical reactions (b) Nuclear reactions
…A… Here, A refers to
(c) Endothermic reactions (d) Exothermic reactions (a) Unified mass (b) Uni mass
(c) Unitech mass (d) Unit mass
158 MOLE & EQUIVALENT CONCEPT

39. What’s the formula mass of NaCl? 49. A gaseous mixture contains oxygen and nitrogen in the
(a) 23 u (b) 35.5 u ratio of 1 : 4 by weight. Therefore, the ratio of their
(c) 58 u (d) 58.5 u number of molecules is
40. What is the mass of hydrogen in terms of amu? (a) 1 : 4 (b) 1 : 8
(a) 1.0020 amu (b) 1.0180 amu (c) 7 : 32 (d) 3 : 16
(c) 1.0070 amu (d) 1.0080 amu 50. How many H2O molecules are in a 0.10g sample of
41. Calculate the molecular mass of sucrose (C12 H22O11) CuSO4.5H2O (M = 249.7) ?
molecule? (a) 1.2 × 1021 (b) 4.1 × 1021
22
(a) 342 amu (b) 343 amu (c) 2.4 × 10 (d) 1.2 × 1023
(c) 341 amu (d) 340 amu 51. The number of molecules in 4.25 g of ammonia is about
42. Given that the abundances of isotopes 54Fe, 56Fe and (a) 1.0 × 1023 (b) 1.5 × 1023
57 23
Fe are 5%, 90% and 5%, respectively, the atomic (c) 2.0 × 10 (d) 2.5 × 1023
mass of Fe is 52. The weight of one molecule of the compound C60H122 is
(a) 55.85 (b) 55.95 (a) 1.4 × 10–21 g (b) 1.09 × 10–21 g
23
(c) 55.75 (d) 56.05 (c) 5.025 × 10 g (d) 16.023 × 1023 g
53. Choose the wrong statement:
(a) 1 mole means 6.02 × 1023 particles
Mole Concept
(b) Molar mass is mass of one molecule
(c) Molar mass is mass of one mole of a substance
43. The mass of one mole of a substance in grams is called (d) Molar mass is molecular mass expressed in grams
its …A… Here, A refers to 54. The number of moles of SO2Cl2 in 13.5 g is :
(a) Avogadro mass (b) Molar mass (a) 0.1 (b) 0.2
(c) Atomic mass (d) Formula mass (c) 0.3 (d) 0.4
44. Which of the following gases will have the least 55. M g of a substance when vaporised occupy a volume of
volume if 10 g of each gas is taken at same temperature 5.6 litre at NTP. The molecular mass of the substance
and pressure? will be :
(a) CO2 (b) N2 (a) M (b) 2M
(c) CH4 (d) HCl (c) 3M (d) 4M
45. According to S.I. system, ______ was used to measure 56. Number of molecules in 1 litre of oxygen at STP is :
the amount of substance. 6.02×1023 6.02×1023
(a) mole (b) weight machine (a) (b)
32 22.4
(c) weight (d) mass
46. If one mole of ammonia contains “y” number of 32
(c) 32 × 22.4 (d)
particles, then how many particles do 1 mole of glucose 22.4
contain? 57. The number of molecules in 89.6 litre of a gas at NTP
(a) 2y (b) 0.5y are :
(c) 3y (d) y (a) 6.02×1023 (b) 2×6.02×1023
23
47. What’s the number of particles in 10 moles of (c) 3×6.02×10 (d) 4×6.02×1023
3
hydrochloric acid? 58. The mass of 112 cm of CH4 gas at STP is
(a) 6.022 × 1024 particles (a) 0.16 g (b) 0.8 g
(b) 6.022 × 1023 particles (c) 0.08 g (d) 1.6 g
(c) 6.22 × 1023 particles 59. Which of the following represents 180 g of water
(d) 3.22 × 1022 particles correctly?
48. How many moles of magnesium phosphate, (I) 5 moles of water.
Mg3 (PO4)2 will contain 0.25 mole of oxygen atoms ? (II) 10 moles of water.
(a) 0.02 (b) 3.125×10–2 (III) 6.023 × 1023 molecules of water.
(c) 1.25×10–2 (d) 2.5×10–2 (IV) 6.023 × 1024 molecules of water.
MOLE & EQUIVALENT CONCEPT 159

Choose the correct option. 68. The number of water molecules present in a drop of
(a) I and II (b) I and IV water (volume = 0.0018 ml) at room temperature is
(c) II and IV (d) II and III (density of H2O = 1 g/mL)
60. Which of the following contains atoms equal to those in (a) 6.023 × 1019 (b) 1.084 × 1018
17
12 g Mg ? (At. wt. Mg = 24) (c) 4.84 × 10 (d) 6.023 × 1023
(a) 12 gm C (b) 7 gm N2 69. The ratio of masses of oxygen and nitrogen in a
(c) 32 gm O2 (d) None of These particular gaseous mixture is 1: 4. The ratio of number
1 of their molecule is :
61. If 1 moles of oxygen combine with Al to form (a) 7 : 32 (b) 1 : 8
2
Al2O3, the weight of Al used in the reaction is (Al = 27) (c) 3 : 16 (d) 1 : 4
(a) 27 g (b) 54 g 70. Which has maximum number of atoms?
(c) 40.5 g (d) 81 g (a) 24g of C (12)
62. Which has the highest mass ? (b) 56g of Fe (56)
(a) 50 g of iron (b) 5 moles of N2 (c) 27g of Al (27)
(c) 0.1 mol atom of Ag (d) 1023 atoms of carbon (d) 108g of Ag (108)
63. Which of the following weighs the least ? 71. The weight of 1 × 1022 molecules of CuSO4. 5H2O is
(a) 2 g atom of N (at. wt. of N = 14) (a) 41.59 g (b) 415.9g
(b) 3 × 1023 atoms of C (at. wt. of C = 12) (c) 4.159 g (d) none of the three
(c) 1 mole of S (at. wt. of S = 32) 72. If 1021 molecules are removed from 200mg of CO2,
(d) 7 g silver (at. wt. of Ag = 108) then the number of moles of CO2 left are
64. If NA is Avogadro’s number then number of valence (a) 2.85 × 10–3 (b) 28.8 × 10–3
electrons in 4.2 g of nitride ions (N3–) is (c) 0.288 × 10–3 (d) 1.68 × 10–2
(a) 2.4 NA (b) 4.2 NA
(c) 1.6 NA (d) 3.2 NA
Percentage Composition ; Empirical &
65. Which among the following is the heaviest?
(a) One mole of oxygen Molecular Formula
(b) One molecule of sulphur trioxide
(c) 100 amu of uranium 73. Mass % of an element
(d) 44g of carbon dioxide A  100
66. Rearrange the following I to IV in order of increasing 
Molar mass of compound
masses and choose the correct answer [At. wt. of N =
Here, A refers to
14 u, O = 16 u, Cu = 63 u]
(a) Volume of that elements in the compound
I 1 molecule of oxygen
(b) Mol of that element in the compound
II 1 atom of nitrogen
(c) Mass of that element in the compound
III 1 × 10–10mol molecule of oxygen
(d) % of that element in the compound
IV 1 × 10–10mol atom of copper
74. What is the percentage of carbon, in ethanol?
(a) II < I < III < IV
(a) Mass percent of carbon = 50.00%
(b) IV < III < II < I
(b) Mass percent of carbon = 52.14 %
(c) II > I > III > IV
(c) Mass percent of carbon = 55.00%
(d) I < II < IV < III
(d) Mass percent of carbon = 51.04%
75. If two compounds have the same empirical formula but
67. The largest number of molecules is in
different molecular formula, they must have
(a) 36 g of water
(a) Different percentage composition
(b) 28 g of carbon monoxide
(b) Different molecular weights
(c) 46 g of ethyl alcohol
(c) Same viscosity
(d) 54 g of nitrogen pentoxide.
(d) Same vapour density
160 MOLE & EQUIVALENT CONCEPT

76. A Compound contains 69.5% oxygen and 30.5% 85. Determine the empirical formula of Kelvar, used in
nitrogen and its molecular weight is 92. The molecular making bullet proof vests, is 70.6% C, 4.2% H, 11.8%
formula of compound is N and 13.4% O
(a) N2O (b) NO2 (a) C7H5NO2 (b) C7H5N2O
(c) N2O4 (d) N2O5 (c) C7H9NO (d) C7H5NO
77. An organic compound containing C and H has 92.3% of
carbon, its empirical formula is
Chemical Equations and Stoichiometry
(a) CH (b) CH3
(c) CH2 (d) CH4
78. A gas has molecular formula (CH)n. If vapour density 86. According to the law of conservation of mass …A…
of the gas is 39, what should be the formula of the has the same number of atoms of each element on both
compound? sides of the equation. Here, A refers to
(a) CH4 (b) C3H8 (a) Unbalanced chemical equation
(c) C2H6 (d) C6H6 (b) Balanced chemical equation
79. An oxide of metal (M) has 40% by mass of oxygen. (c) Stoichiometric chemical equation
Metal M has atomic mass of 24. The empirical formula (d) Non-stoichiometric chemical equation
of the oxide is 87. Which of the given reactions are counted as balanced
(a) M2O (b) M2O3 reactions?
(c) MO (d) M3O4 (a) H 2 +O 2  2H 2 O
80. What is the empirical formula of a compound (b) 4Al+3O2  2Al2 O3
composed of O and Mn in equal weight ratio ?
(c) Mg  OH 2 +2HNO3  2Mg  NO3 2 +2H2O
(a) MnO (b) MnO2
(c) Mn2O3 (d) Mn2O7 (d) N 2 +3H 2  NH 3
81. A compound was found in contain 21.67% Mg, 21.4% 88. What is the weight of oxygen required for the complete
C and 57.0% O by mass of oxygen. What is the combustion of 2.8 kg of ethylene ?
simplest formula of this compound ? (a) 2.8 kg (b) 6.4 kg
(a) MgCO3 (c) 9.6 kg (d) 96 kg
(b) MgC2O4 89. NaOH is formed according to the reaction
(c) Mg2CO3 1
(d) Mg(CO)4 2Na  O2  Na 2O
2
82. The simplest formula of a compound containing 50% of
Na 2 O  H 2 O  2 NaOH
an element X (atomic weight 10) and 50% of element Y
(atomic weight 20) is : To make 4g of NaOH, Na required is
(a) XY (b) X2Y (a) 4.6g (b) 4.0g
(c) XY2 (d) X2Y3 (c) 2.3g (d) 0.23g
83. The empirical formula of a compound is CH2. One 90. How many moles of ferric alum
mole of this compound has a mass of 42 g. Its (NH 4 ) 2 SO 4 Fe 2 (SO 4 )3 . 24 H2O can be made from the
molecular formula is sample of Fe containing 0.0056 g of it ?
(a) C3H6 (b) C3H8 (a) 10–4mol
(c) CH2 (d) C2H2 (b) 0.5 × 10–4mol
84. An organic compound has an empirical formula CH 2 O (c) 0.33 × 10–4mol
. It’s vapour density is 45. The molecular formula of the (d) 2 × 10–4mol
compound is 91. 1 mole of oxalic acid is treated with conc. H2SO4. The
(a) CH2O (b) C2H5O resultant gaseous mixture is passed through a solution
(c) C2H2O (d) C3H6O3 of KOH. The mass of KOH consumed will be (where
KOH absorbs CO2.)
MOLE & EQUIVALENT CONCEPT 161

2H SO
4  CO  CO  H O (c) Molality
 COOH 2  2 2
(d) All of the above
2 KOH + CO2 K2CO3 + H2O 98. A solution is prepared by adding 2g of a substance A to
(a) 28 g (b) 56 g 18 g of water. Calculate the mass percent of the solute.
(c) 84 g (d) 112 g (a) 8% (b) 9%
92. If 80.00 g of X combines with 1.5 × 1023 atoms of Y to (c) 10% (d) 11%
form X2Y without any of either element remaining, 99. A solution is made by dissolving 49 g of H2SO4 in 250
what is the atomic weight of X ? mL of water. The molarity of the solution prepared is
(a) 8.0  10 (b) 2.0  10 (a) 2M (b) 1M
2
(c) 1.6  10 (d) 1.2  10 2 (c) 4M (d) 5M
93. Calculate the number of oxygen atoms required to 100. What is the molality of a solution of 10 g NaOH in 500
combine with 7.0g of N2 to form N2O3 if 80% of N2 is g water?
converted into products. (a) 0.5 mol/kg (b) 2.1 mol/kg
3 (c) 1.5 mol/kg (d) 3.4 mol/kg
N2  O2 
 N2O3
2 101. A mixture contains 9 gram each of H 2 O and NaCl,
(a) 3.24 × 1023 what is the mole fraction of NaCl?
(b) 3.6 × 1023 (a) 0.15 (b) 0.23
(c) 18 × 1023 (c) 0.45 (d) 0.64
(d) 6.02 × 1023 102. If 100 ml of H2SO4 (A) and 100 ml of H2O (B) are
94. If 0.5 mole of BaCl2 is mixed with 0.20 mole of mixed. Then the mass per cent of H2SO4 would be
Na3PO4, the maximum number of moles of Ba3(PO4)2 (Given density of H2SO4 = 0.9 g/ml; density of H2O =
then can be formed is 1.0 g/ml)
(a) 0.1 (b) 0.2 (a) 60 (b) 50
(c) 0.5 (d) 0.7 (c) 47.36 (d) 90
103. The density of 1 M solution of NaCl is 1.0585g mL–1.
The molality of the solution is
Percent Yield; Reactions in Aqueous Media and (a) 1.0585 (b) 1.00
Strength of Solution (c) 0.10 (d) 0.0585
104. Density of a 2.05 M solution of acetic acid in water is
95. NH3 is produced according to the following reaction: 1.02 g/mL. The molality of the solution is
N2(g) + 3H2(g)  2NH3(g) (a) 0.44 mol Kg–1 (b) 1.14 mol kg–1
In an experiment 0.25 mol of NH3 is formed when 0.5 (c) 3.28 mol kg–1 (d) 2.28 mol kg–1
mol of N2 is reacted with 0.5 mol of H2. What is % 105. What is the mole fraction of glucose in 10% w/w
yield? glucose solution?
(a) 75% (b) 50% (a) 0.01 (b) 0.02
(c) 33% (d) 25% (c) 0.03 (d) 0.04
96. The amount of solute in moles per unit volume 106. Dissolving 120g of urea (mol. wt. 60) in 1000g of water
is_______. gave a solution of density 1.15 g/mL. The molarity of
(a) Mole the solution is
(b) Molarity (a) 1.78 M (b) 2.00 M
(c) Density (c) 2.05 M (d) 2.22 M
(d) None of the mentioned 107. A molal solution is one that contains one mole of a
97. The concentration of a solution or the amount of solute in
substance present in its given volume can be expressed (a) 1000 g of the solvent
in any of the following ways. (b) one litre of the solvent
(a) Mass percent or weight percent (w/w%) (c) one litre of the solution
(b) Mole fraction or molarity (d) 22.4 litres of the solution
162 MOLE & EQUIVALENT CONCEPT

108. An aqueous solution of ethanol has density 1.025 g/mL density of this solution is 1.05 g/mL, what is the molar
and it is 2 M. What is the molality of this solution ? concentration?
(a) 1.79 (b) 2.143 (a) 6.77 M (b) 6.45 M
(c) 1.951 (d) None of these (c) 0.0017 M (d) 16.9 M
109. The mole fraction of a given sample of I2 in C6 H6 is 119. What is the molarity of SO24 ion in aqueous solution
0.2. The molality of I2 in C6H6 is that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete
(a) 0.32 (b) 3.2 dissociation and density of solution 1 g/mL)
(c) 0.032 (d) 0.48 (a) 3 × 10–4 M (b) 2 × 10–4 M
110. In which mode of expression, the concentration of a –4
(c) 10 M (d) None of these
solution remains independent of temperature ? 120. What is the weight % sulphuric acid in an aqueous
(a) Molarity (b) Normality solution which is 0.502 M in sulphuric acid? The
(c) Formality (d) Molality specify gravity of the solution is 1.07
111. With increase of temperature, which of these changes? (a) 4.6 % (b) 5.67 %
(a) molality (c) 9.53 % (d) 22.0 %
(b) weight fraction of solute 121. Mole fraction of ethanol in ethanol - water mixture is
(c) fraction of solute present in unit volume of water 0.25. Hence, percentage concentration of ethanol
(d) mole fraction. (C2H6O) by weight of mixture is
112. Molarity and Normality changes with temperature (a) 25 (b) 75
because they involve : (c) 46 (d) 54
(a) Moles 122. The molarity of a solution obtained by mixing 750 mL
(b) equivalents of 0.5 M HCl with 250 ml of 2 M HCl will be
(c) weights (a) 0.875 M (b) 1.00 M
(d) volumes (c) 1.75 M (d) 0.0975 M
113. Normality of 0.74 g Ca(OH)2 in 5 mL solution is 123. The density of a solution prepared by dissolving 120g
(a) 8 N (b) 4 N of urea (mol. mass = 60 u) in 1000 g of water is
(c) 0.4 N (d) 2 N 1.15g/mL. The molarity of this solution is
114. Normality of a 2 M sulphuric acid is (a) 0.50 M (b) 1.78 M
(a) 2 N (b) 4 N (c) 1.02 M (d) 2.05 M
(c) N / 2 (d) N / 4 124. Assertion (A) : Molarity of a solution depends upon
115. What volume of 0.232 N solution contains 3.17 temperature.
milliequivalent of solute ? Reason (R) : Volume of a solution is temperature
(a) 137 mL (b) 13.7 mL dependent entity.
(c) 27.3 mL (d) 12.7 mL (a) Both A and R are correct; R is the correct
116. What volume of 0.4 M FeCl3 . 6H2O will contain 600 explanation of A
mg of Fe3+ ? (b) Both A and R are correct; R is not the correct
(a) 49.85 mL explanation of A
(b) 26.78 mL (c) A is correct; R is incorrect
(c) 147.55 mL (d) R is correct; A is incorrect
(d) 87.65 mL
117. The density (in g mL–1) of a 3.60 M sulphuric acid
Applications of Strength of Solutions
solution that is 29% H2SO4 (molar mass = 98 g mol-1)
by mass will be
125. Molarity equation of a mixture of solutions of same
(a) 1.45 (b) 1.64
substance is given by
(c) 1.88 (d) 1.22
M1  V1  M 2  V2  M3  V3  ...
118. An antifreeze mixture contains 40% ethylene glycol (a)
(C2H6O2) by weight in the aqueous solution. If the  M1  M 2  M 3
MOLE & EQUIVALENT CONCEPT 163

(b) M1V1  M 2 V2  M3 V3  ...  M  V1  V2  V3  Equivalent Weight & Equivalent Concept

M1 M 2 M 3  1 1 1 
(c)    ...  M     132. 10 mL of N/2 HCl, 20 mL of N/2 H SO and 30 mL
2 4
V1 V2 V3  V1 V2 V3  N/3 HNO are mixed together and solution made to one
3
M M M  1 1 1  litre. The normality of the resulting solution is
(d) 1  2  3  ...  M1    
V1 V2 V3  V1 V2 V3  (a) 0.20 N (b) 0.10 N
126. The final molarity of a solution made by mixing 50 mL (c) 0.50 N (d) 0.025 N
of 0.5 M HCl, 150mL of 0.25 M HCl and water to 133. 1 L of a normal solution is diluted to 2000 ml. The
make the volume 250 mL is resulting normality is :
(a) 0.5 M (b) 1 M (a) N/2 (b) N/4
(c) 0.75 M (d) 0.25 M (c) N (d) 2 N
127. A standard solution of 0.165 M HCl is being used to 134. 1L solution of NaOH contains 4.0 g of it. What shall be
determine the concentration of an unknown NaOH the difference between molarity and the normality?
solution. If 25.5 mL of an acid solution are required to (a) 0.10 (b) zero
neutralize 15.0 mL of the base, what is the molarity of (c) 0.05 (d) 0.20
the NaOH solution ? 135. 100 ml of 0.3 N HCl is mixed with 200 ml of 0.6 N
(a) (0.165)/ (25.5 + 15.0) M H2SO4. The final normality of the resulting solution
(b) (15.0) / (0.165) (25.5) M will be
(c) (0.165) (15.0/25.5) M (a) 0.1 N (b) 0.2 N
(d) (0.165) (25.5/15.0) M (c) 0.3 N (d) 0.5 N
128. Assertion : When a solution is diluted from volume 136. Normality of a mixture of 30 mL of 1N H2SO4 and 20
V1 to V2 by adding solvents, its molarity before mL of 4N H2SO4 is
dilution M1 and after dilution M2 are related as : (a) 1.0 N (b) 1.1 N
M1V1 = M2V2 (c) 2.0 N (d) 2.2 N
Reason: During dilution, moles of the solute remains 137. Normality of solution obtained by mixing 10 mL of 1N
conserved. HCl, 20 mL of 2N H2SO4 and 30 mL of 3N HNO3 is
(a) A (b) B (a) 1.11 N (b) 2.22 N
(c) C (d) D (c) 2.33 N (d) 3.33 N
129. 25 mL of a solution of Ba(OH)2 on titration with a 138. 0.115 g of pure sodium metal was dissolved in 400 ml
0.1 M solution of HCl gave a titre value of 35 mL. distilled water. The normality of the above solution
The molarity of barium hydroxide solution was (a) 0.010 N (b) 0.0115 N
(a) 0.07 (b) 0.14 (c) 0.0125 N (d) 0.046 N
(c) 0.28 (d) 0.35 139. 50 ml of 10 N H2SO4, 25 ml of 12 N HCl and 40 ml of
20 5N HNO3 were mixed together and the volume of the
130. 6.023 × 10 molecules of urea are present in 100 mL of
its solution. The concentration of urea solution is mixture was made 1000 ml by adding water. The
(a) 0.001 M normality of the resulting solution will be
(b) 0.1 M (a) 1 N (b) 2 N
(c) 0.02 M (c) 3 N (d) 4 N
(d) 0.01 M 140. The normality of 0.3 M phosphorous acid (H3PO3) is
131. To neutralize completely 20 mL of 0.1 M aqueous (a) 0.1 (b) 0.9
solution of phosphorus (H3PO3) acid, the volume of 0.1 (c) 0.3 (d) 0.6
M aqueous KOH solution required is 141. 2H3PO4 + 3Ca(OH)2Ca3(PO4)2 + 6H2O
(a) 60 mL (b) 20 mL Equivalent weight of H3PO4 in this reaction is
(c) 40 mL (d) 10 mL (a) 98 (b) 49
(c) 32.66 (d) 24.5
164 MOLE & EQUIVALENT CONCEPT

142. The Ew of H3PO4 in the reaction is (c) 112 g (d) 168 g

Ca(OH)2 + H3PO4  CaHPO4 + 2H2O 147. 10 mL of 0.2 N HCl and 30 mL of 0.1 N HCl together
(Ca = 40, P = 31, O = 16) exactly neutralises 40 mL of solution of NaOH, which
(a) 49 (b) 98 is also exactly neutralised by a solution in water of 0.61
(c) 32.66 (d) 147 g of an organic acid. What is the equivalent weight of
143. 0.1 g of metal combines with 46.6 mL of oxygen at the organic acid ?
STP. The equivalent weight of metal is (a) 61 (b) 91.5
(a) 12 (b) 24 (c) 122 (d) 183
(c) 6 (d) 36 148. When 100 ml of 1 M NaOH solution and 10 ml of 10 N
144. The vapour density of a chloride of an element is 39.5. H2SO4 solution are mixed together, the resulting
The Ew of the elements is 3.82. The atomic weight of solution will be :
the element is (a) alkaline (b) neutral
(a) 15.28 (b) 7.64 (c) acidic (d) none of these
(c) 3.82 (d) 11.46 149. Which of the following 1 g L–1 solution has the highest
145. What weight of a metal of equivalent weight 12 will normality ?
give 0.475 g of its chloride? (a) NaOH (b) H2SO4
(a) 0.12 g (b) 0.24 g (c) HCl (d) HNO3
(c) 0.36 g (d) 0.48 g 150. 5.6 g of a metal forms 12.7 g of metal chloride. Hence
146. How many grams of phosphoric acid would be needed equivalent weight of the metal is
to neutralise 100 g of magnesium hydroxide ? (The (a) 127 (b) 254
molecular weights are : H3PO4 = 98 and Mg (OH)2 = (c) 56 (d) 28
58.3)
(a) 66.7 g (b) 252 g
MOLE & EQUIVALENT CONCEPT 165

Exercise – 2: Previous Year Questions

1. The weight of one molecule of a compound C60H122 is 10 How many moles of lead (II) chloride will be formed
(BHU 2007) from a reaction between 6.5 g of PbO and 3.2 g of
(a) 1.3×10–20 g (b) 5.01× 10–21 g HCl? (CBSE AIPMT 2008)
(c) 3.72×1012 g (d) 1.4×10–21 g (a) 0.044 (b) 0.333
2. 25 g of MCl4 contains 0.5 mole chlorine then its (c) 0.011 (d) 0.029
molecular weight is (Punjab PMET 2007) 11. Of two oxides of iron, the first contained 22% and the
(a) 100 g mol–1 (b) 200 g mol–1 second contained 30% of oxygen by weight. The ratio
(c) 150 g mol–1 (d) 400 g mol–1 of weights of iron in the two oxides that combine with
3. The number of gram molecules of chlorine in 6.02×1025 the same weight of oxygen is (J&K CET 2008)
hydrogen chloride molecule is (KCET 2007) (a) 3 : 2 (b) 2 : 1
(a) 10 (b) 100 (c) 1 : 2 (d) 1 : 1
(c) 50 (d) 5 12. The equivalent weight of a metal is 9 and vapour
4. Gram molecular volume of oxygen at STP is density of its chloride is 59.25. The atomic weight of
(KCET 2007) metal is
3
(a) 3200 cm (b) 5600 cm3 (DUMET 2008)
(c) 22400 cm3 (d) 11200 cm3 (a) 23.9 (b) 27.3
5. The crystalline salt Na2SO4.xH2O on heating loses (c) 36.3 (d) 48.3
55.9% of its weight. The formula of crystalline salt is 13. A metal M of equivalent mass E forms an oxide of
(Kerala CEE 2007) molecular formula MxOy. The atomic mass of the
(a) Na2SO4.5H2O (b) Na2SO4.7H2O metals is given by the correct equation (BCECE 2008)
(c) Na2SO4.2H2O (d) Na2SO4.10H2O (a) 2E(y/x) (b) xyE
6. An element, X has the following isotopic composition: (c) E/y (d) y/E
200 199 202
X : 90% X : 8.0% X : 2.0% 14. 1.520 g of hydroxide of a metal on ignition gave 0.995
The weighted average atomic mass of the naturally g of oxide. The equivalent weight of metal is
occurring element X is closest to (AIPMT 2007) (JCECE 2008)
(a) 201 amu (b) 202 amu (a) 1.52 (b) 0.995
(c) 199 amu (d) 200 amu (c) 190 (d) 9
7. Mass of 0.1 mole of methane is (KCET 2008) 15. A gas mixture contains O2 and N2 in ratio of 1:4 by
(a) 1 g (b) 16 g weight. The ratio of their number of molecules is
(c) 1.6 g (d) 0.1 g (JIPMER 2008)
8. The maximum number of molecules are present in (a) 1:8 (b) 1:4
(BCECE 2008) (c) 3:16 (d) 7:32
(a) 15 L of H2 gas at STP 16. An organic compound contains carbon, hydrogen and
(b) 5 L of N2 gas at STP oxygen. Its elemental analysis gave C, 38.71% and H,
(c) 0.5 g of H2 gas 9.67%. The empirical formula of the compound would
(d) 10 g of O2 gas be (AIPMT 2008)
9. Number of atoms of He in 100 u of He (atomic weight (a) CHO (b) CH4O
of He is 4) are (BCECE 2008) (c) CH3O (d) CH2O
(a) 25 17. One atom of an element weighs 1.8 × 10–22 g. Its
(b) 100 atomic mass is (Manipal 2009)
(c) 50 (a) 29.9 (b) 154
(d) 100 × 6 × 10–23 (c) 108.36 (d) 18
166 MOLE & EQUIVALENT CONCEPT

18. The number of molecules in 18 mg of water in terms of 27. Volume of a gas at NTP is 1.12×10–7 cm3. The number
Avogadro number NA is (J&K CET 2009) of molecules in it is (Manipal 2010)
–3 –2 12
(a) 10 NA (b) 10 NA (a) 3.01×10 (b) 3.01×1024
(c) 10–1 NA (d) 10 NA (c) 3.01×1023 (d) 3.01×1020
19. A bivalent metal has an equivalent mass of 32. The 28. A mixture of CaCl2 and NaCl weighing 4.44 g is
molecular mass of the metal nitrate is (KCET 2009) treated with sodium carbonate to precipitate all the
(a) 168 (b) 192 Ca2+ ions as calcium carbonate. The calcium carbonate
(c) 188 (d) 182 so obtained is heated strongly to get 0.56 g of CaO.
20. The volume of 2 N H2SO4 solution is 0.1 dm3. The The percentage of NaCl in the mixture (atomic mass of
volume of its decinormal solution (in dm3) will be Ca = 40) is (KCET 2010)
(MHT CET 2009) (a) 75 (b) 30.6
(a) 0.1 (b) 0.2 (c) 25 (d) 69.4
(c) 2 (d) 1.7 29. The number of atoms in 0.1 mol of a triatomic gas is
21. Excess of carbon dioxide is passed through 50 mL ( N A =6.022×10 23 ) (CBSE AIPMT 2010)
of 0.5 M calcium hydroxide solution. After the (a) 6.026 × 1022 (b) 1.806 × 1023
completion of the reaction, the solution was (c) 1.800 × 10 22
(d) 3.600× 1023
evaporated to dryness. The solid calcium carbonate 30. 60g of a compound on analysis produced 24g of
was completely neutralized with 0.1 N hydrochloric carbon, 4g hydrogen and 32g oxygen. The empirical
acid. The volume of hydrochloric acid required is (At. formula of the compound is. (BVP 2010)
mass of calcium=40 (KCET 2009) (a) CH2O2 (b) CH2O
3 3
(a) 200cm (b) 500cm (c) CH4O (d) C2H4O2
(c) 400cm3 (d) 300 cm3 31. 20.0 kg of N2(g) and 3.0 kg of H2(g) are mixed to
22. 10 g of hydrogen and 64 g of oxygen were filled in a produce NH3 (g). The amount of NH3(g) formed is
steel vessel and exploded. Amount of water produced (Kerala CEE 2011)
in this reaction will be (AIPMT 2009) (a) 17 kg (b) 34 kg
(a) 3 mol (b) 4 mol (c) 20 (d) 3 kg
(c) 1 mol (d) 2 mol 32. Which has the maximum number of molecules among
23. In an experiment, 4 g of M2Ox oxide was reduced to 2.8 the following? (AIPMT 2011)
g of the metal. If the atomic mass of the metal is (a) 44 g CO2 (b) 48 g O3
56 g mol–1, the number of O atoms in the oxide is (c) 8 g H2 (d) 64 g SO2
(AFMC 2010) 33. The vapour density of a gas is 11.2. The volume
(a) 1 (b) 2 occupied by one gram of the gas at STP is
(c) 3 (d) 4 (Manipal 2013)
24. To dissolve 0.9 g metal, 100 mL of 1N HCl is used. (a) 1.0 L (b) 11.2 L
What is the equivalent weight of metal? (c) 22.4 L (d) None of these
(MP PMT 2010) 34. 20
6.02 × 10 molecules of urea are present in 100 mL of
(a) 7 (b) 9 its solution. The concentration of solution is
(c) 10 (d) 6 (NEET 2013)
25. 74.5 g of a metallic chloride contain 35.5 g of chlorine. (a) 0.001 M (b) 0.1 M
The equivalent weight of the metal is (Manipal 2010) (c) 0.02 M (d) 0.01 M
(a) 19.5 (b) 35.5 35. Equal masses of H2, O2 and methane have been taken
(c) 39.0 (d) 78.0 in a container of volume V at temperature 27°C in
26. If 1 mL of water contains 20 drops the number of identical conditions. The ratio of the volumes of gases
molecules in a drop of water is (AFMC 2010) H2 : O2 : methane would be (NEET 2014)
(a) 1.67×1021 molecules (b) 1.376×1026 molecules (a) 8 : 16 : 1 (b) 16 : 8 : 1
(c) 1.344×1018 molecules (d) 4.346×1020 molecules (c) 16 : 1 : 2 (d) 8 : 1 : 2
MOLE & EQUIVALENT CONCEPT 167

36. When 22.4 litres of H2(g) is mixed with 11.2 litres of weighs 10g and 0.05 mol of X3Y2 weighs 9g, the
Cl2(g), each at STP, the moles of HCl(g) formed is atomic weights of X and Y are (NEET 2016)
equal to (NEET 2014) (a) 60, 40 (b) 20, 30
(a) 1 mol of HCl(g) (c) 30, 20 (d) 40, 30
(b) 2 mol of HCl(g) 45. Which of the following is dependent on temperature?
(c) 0.5 mol of HCl(g) (NEET 2017)
(d) 1.5 mol of HCl(g) (a) Molarity (b) Mole fraction
37. 1.0 g of magnesium is burnt with 0.56 g O2 in a closed (c) Weight percentage (d) Molality
vessel. Which reactant is left in excess and how much? 46. In which case is number of molecules of water
(At. wt. Mg = 24, O = 16) (NEET 2014) maximum? (NEET 2018)
(a) Mg, 0.16 g (a) 18 mL of water
(b) O2, 0.16 g (b) 0.18 g of water
(c) Mg, 0.44 g (c) 0.00224 L of water vapours at 1 atm and 273 K
(d) O2, 0.28 g (d) 10–3 mol of water
38. Which one of the following data has only four 47. The number of moles of hydrogen molecules required
significant figures? (Kerala CEE 2014) to produce 20 moles of ammonia through Haber’s
(a) 6.023 × 1023 (b) 285 cm process is (NEET 2019)
(c) 0.0025 L (d) 0.200 g (a) 40 (b) 10
39. A mixture of gases contains H 2 and O 2 gases in the (c) 20 (d) 30
ratio of 1 : 4 (w/w). What is the molar ratio of the two 48. The density of 2 M aqueous solution of NaOH is 1.28
gases in the mixture ? (NEET 2015) g/cm3. The molality of the solution is [Given that
(a) 4 : 1 (b) 16 : 1 molecular mass of NaOH = 40 g mol–1]
(c) 2 : 1 (d) 1 : 4 (Odisha NEET 2019)
40. 20.0 g of a magnesium carbonate sample decomposes (a) 1.20 m (b) 1.56 m
on heating to give carbon dioxide and 8.0g magnesium (c) 1.67 m (d) 1.32 m
oxide. What will be the percentage purity of 49. Which one of the following has maximum number of
magnesium carbonate in the sample ? (NEET 2015) atoms? (NEET 2020)
(a) 60% (b) 84% (a) 1 g of Ag(s) [Atomic mass of Ag = 108]
(c) 75% (d) 96% (b) 1 g of Mg(s) [Atomic mass of Mg = 24]
41. If Avogadro number NA, is changed from 6.022 × 1023 (c) 1 g of O2(g) [Atomic mass of O = 16]
mol–1 to 6.022 × 1020 mol–1, this would change (d) 1 g of Li(s) [Atomic mass of Li = 7]
(NEET 2015) 50. An organic compound contains 78% (by wt.) carbon
(a) The mass of one mole of carbon and remaining percentage of hydrogen. The right
(b) The ratio of chemical species to each other in a option for the empirical formula of this compound is:
balanced equation [Atomic wt. of C is 12, H is 1] (NEET 2021)
(c) The ratio of elements to each other in a compound (a) CH3 (b) CH4
(d) The definition of mass in units of grams. (c) CH (d) CH2
42. The number of water molecules is maximum in 51. What mass of 95% pure CaCO3 will be required to
(NEET 2015) neutralise 50 mL of 0.5 M HCl solution according to
(a) 1.8 gram of water the following reaction?
(b) 18 gram of water CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) +
(c) 18 moles of water 2H2O(l)
(d) 18 molecules of water. [Calculate upto second place of decimal point]
43. What is the mass of the precipitate formed when 50 mL (NEET 2022)
of 16.9% solution of AgNO3 is mixed with (a) 1.25g (b) 1.32g
50 mL of 5.8% NaCl solution? (c) 3.65g (d) 9.50g
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5) 52. In one molal solution that contains 0.5 mole of a
(NEET 2015) solute, there is (NEET 2022)
(a) 3.5 g (b) 7 g (a) 500 mL of solvent
(c) 14 g (d) 28 g (b) 500 g of solvent
44. Suppose elements X and Y combine to form two (c) 100 mL of solvent
compounds XY2 and X3Y2. When 0.1 mol of XY2 (d) 1000 g of solvent
168 MOLE & EQUIVALENT CONCEPT

Exercise – 3: Achiever’s Section

Single Choice Questions

1. 100 mL of mixture of NaOH and Na2SO4 is 8. In a compound C, H and N atoms are present as 9%,
neutralised by 10 mL of 0.5 M H2SO4. Hence, NaOH 1%, 3.5% respectively. Molecular weight of
in 100 mL solution is compound is 108. Molecular formula of compound
(a) 0.2 g (b) 0.4 g is
(c) 0.6 g (d) None (a) C 2 H 6 N 2 (b) C3 H 4 N
2. P and Q are two elements which forms P2Q3 and (c) C 6 H 8 N 2 (d) C9 H12 N3
PQ2. If 0.15 mole of P2Q3 weights 15.9g and 0.15
9. 1 mole of oxalic acid is treated with conc. H2SO4.
mole of PQ2 weights 9.3g, the atomic weight of P
The resultant gaseous mixture is passed through a
and Q is (respectively) :
solution of KOH. The mass of KOH consumed will
(a) 18, 26 (b) 26,18
be (where KOH absorbs CO2.)
(c) 13, 9 (d) None of these H SO
2 4  CO  CO  H O
3. If we consider that 1/6, in place of 1/12, mass of  COOH 2  2 2
carbon atom is taken to be the relative atomic mass 2 KOH + CO2 K2CO3 + H2O
unit, the mass of one mole of a substance will (a) 28 g (b) 56 g
(a) be a function of the molecular mass of the (c) 84 g (d) 112 g
substance 10. A solution of NaOH is prepared by dissolving 4.0 g
(b) remain unchanged of NaOH in 1 L of water. Calculate the volume of
(c) increase two-fold the HCl gas at stp that will neutralize 50 mL of this
(d) decrease twice solution.
4. Number of atoms in 560g of Fe (atomic mass (a) 224 mL (b) 56 mL
56 g mol–1) is (c) 112 mL (d) 448 mL
(a) twice that of 70 g N2 11. A solution containing 12.0% NaOH by mass has a
(b) half that of 20 g H2 density of 1.131 g/mL. What volume of this solution
(c) Both (a) and (b) contains 5.00 mol of NaOH ?
(d) None of the above (a) 0.0240 L (b) 1.67 L
5. The total number of electrons present in 18 ml of (c) 1.47 L (d) 1.00 L
water (density of water is 1 g ml–1) is 12. Given the reaction :
(a) 6.02 × 1023 (b) 6.02 × 1023 Na 2 O(s)  H 2 O(l ) 
 2NaOH(aq)
(c) 6.02 × 1024 (d) 6.02 × 1025
What is the molarity of the solution formed if 1.35g
6. A 27.0 g sample of an unknown hydrocarbon was
of Na2O is mixed with H2O such that the final
burned in excess O2 to form 88g of CO2 and 27g of
volume is 100 mL?
H2O. What is possible molecular formula of
(a) 0.0435 M (b) 0.108 M
hydrocarbon ?
(c) 0.217 M (d) 0.435 M
(a) CH4 (b) C2H2
(c) C4H3 (d) C4H6 13. N2 + 3H2  2NH3
7. When a hydrate of Na2CO3 is heated until all the Molecular weight of NH3 and N2 are x1 and x2,
water is removed, it loses 54.3 per cent of its respectively. Their equivalent weights are y1 and y2,
mass. The formula of the hydrate is respectively. Then (y1 – y2) is
(a) Na2CO3.10H2O  2 x1  x 2 
 6 
(b) Na2CO3.7 H2O (a)   (b) (x1 – x2)
(c) Na2CO3.5 H2O (c) (3x1 – x2) (d) (x1 – 3x2)
(d) Na2CO3. 3H2O 14. 5 mL of N-HCl, 20 mL of N/2 H2SO4 milli
equivalent and 30 mL of N/3 HNO3 are mixed
together and the volume is made to 1L. The
normality of the resulting solution is
(a) N/5 (b) N/10
(c) N/20 (d) None of these
MOLE & EQUIVALENT CONCEPT 169

Assertion-Reason Type Questions 21. Assertion : 1mole of H2SO4 is neutralised by 2

moles of NaOH but 1 equivalent of H2SO4 is
(A) If both assertion and reason are correct and neutralised by 1 equivalent of NaOH.
reason Reason : Equivalent weight of H2SO4 is half of its
is the correct explanation of assertion. molecular weight while equivalent weight of NaOH
(B) If both assertion and reason are true but reason is is 40.
not the correct explanation of assertion.
(C) If assertion is true but reason is false. (a) A (b) B
(D) If reason is true but assertion is false. (c) C (d) D
22. Assertion:Equivalent weight of a base
15. Assertion :Pure water obtained from different states molecular weight
=
of India always contains hydrogen and oxygen in the acidity
ratio of 1 : 8 by mass. Reason:Acidity is the number of replaceable
Reason: Total mass of reactants and products during hydrogen atom in one molecule of the base.
chemical change is always the same. (a) A (b) B
(a) A (b) B (c) C (d) D
(c) C (d) D
16. Assertion : Both 138 g of K2CO3 and 12 g of carbon Comprehension Type questions
have same number of carbon atoms.
Comprehension
Reason : Both contains 1 g atom of carbon which
contains 6.022 × 1023 carbon atoms.
(a) A (b) B Often more than one reaction is required to change starting
(c) C (d) D materials into the desired product. This is true for many
17. Assertion :Both 12g of carbon and 27g of reaction that we carry out in the laboratory and for many
aluminium contain 6.023 × 1023 atoms. industrial process. These are called sequential reactions. The
Reason :Molar mass of an element contains amount of desired product from each reaction is taken as the
Avogadro number of atoms. starting material for the next reaction.
(a) A (b) B I: 2KClO 3  2KCl+3O 2
(c) C (d) D II. 4Al+3O 2  2Al2 O3
18. Assertion :Number of molecules present in SO2 is KClO3 decomposes in step I to give O2, which in turn, is used
twice the number of molecules present in O2. by Al to form Al2O3 in step II. First we determine O2 formed
Reason : Molecular mass of SO2 is double to that of in step I and then Al used by this O2 in step II. Both reactions
O2. can be added to determine amount of KClO3 that can give
(a) A (b) B required amount of O2 needed for Al.
(c) C (d) D Net : 2KClO3 + 4Al  2KCl + 2Al2O3
19. Assertion : If 30 mL of H2 and 20 mL of O2 react to Thus, 2KClO3  4Al
form water, 5 mL of H2 is left at the end of the
Or KClO3  2Al
reaction
Reason :H2 is the limiting reagent. I :CaO + 3C CaC2 + CO
(a) A (b) B II :CaC2 + 2H2O Ca(OH)2 + C2H2
(c) C (d) D CaC2 (calcium carbide) is prepared in step I. It is used to
20. Assertion :For a binary solution of two liquids, A prepare acetylene (C2H2) in step II. Suppose we want to
and B, with the knowledge of density of solution, determine amount of CaO that can give enough CaC2 to
molarity can be converted into molality. converted required amount of C2H2. Amount of CaO is
Reason: Molarity is defined in terms of volume and determined in step I and then amount of C2H2 in step II. We
molality in terms of mass, and mass and volume are can relate CaO and C2H2 stoichiometrically by writing net
related by density. reaction which is
(a) A (b) B CaO + 3C + 2H2 O  Ca(OH)2 + C2H2 + CO
(c) C (d) D
Thus, CaO  C2H2
170 MOLE & EQUIVALENT CONCEPT

23. NX is produced by the following step of reactions equilibrium molarity of NaOH equal to zero because HCl is
M + X2  M X2 completely dissociated.
MX2 + X2 M3X8
25. Calculate the analytical molarity of Cl– ion in
M3X8 + N2CO3 N X + CO2 + M3O4 solution which is prepared by mixing 100 ml of 0.1
How much M (metal) is consumed to produce 206 M NaCl and 400 ml of 0.01 M BaCl2.
gm of NX. (Take At. wt of M = 56, N=23, X = 80) (a) 0.018 M (b) 0.036 M
(a) 42 gm (b) 56 gm (c) 0.084 M (d) 0.046 M
14 7 26. The molarity of 68 % of H2SO4 whose density is
(c) gm (d) gm 1.84 g/cc is
3 4
(a) 12.76 M (b) 6.84 M
24. The following process has been used to obtain iodine (c) 18.4 M (d) 6.8 M
from oil-field brines in California.
27. HCl is 80% ionised in 0.01 M aqueous solution. The
NaI + AgNO3 AgI + NaNO3 equilibrium molarity of HCl in the solution is
AgI + Fe FeI2 + Ag (a) 0.002 (b) 0.06
FeI2 + Cl2 FeCl3 + I2 (c) 0.02 (d) 0.008
If 381 kg of iodine is produced per hour then mass of
AgNO3 required per hour will be Comprehension
[atomic mass Ag– 108, I– 127, Fe–56, N–14, Cl–
35.5] HNO3 used as a reagent has specific gravity of 1.42g
(a) 170 kg (b) 340 kg mL–1 and contains 70% by strength HNO3.
(c) 255 kg (d) 510 kg 28. Normality of acid is.
(a) 16.78 (b) 15.78
Comprehension (c) 14.78 (d) 17.78
29. Volume of acid that contains 63g pure acid is.
(a) 100 mL (b) 40.24 mL
The analytical molarity of a solution gives the total number of
(c) 63.38 mL (d) 70.68 mL
moles of a solute in one litre of the solution. The equilibrium
30. Volume of water required to make 1N solution from
molarity represents the molar concentration of particular 2 mL conc. HNO3.
species in a solution at equilibrium. In order to specify the (a) 29.56 mL (b) 30.56 mL
equilibrium molarity of a particular species it is necessary to (c) 28.56 mL (d) 31.56 mL
know how the solute behaves when it is dissolved in a
solvent. e.g., if analytical molarity of HCl is 0.1 M then
MOLE & EQUIVALENT CONCEPT 171

Notes:

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

MOLE & EQUIVALENT CONCEPT

172 ANSWER KEY

Answer Key
CHAPTER - 1: Structure of An Atom
Exercise - 1: Basic Objective Questions

DIRECTIONS FOR USE-

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1. (a) 2.(d) 3. (a) 4. (a) 77. (c) 78. (c) 79. (b) 80. (b)
5. (b) 6. (d) 7. (a) 8. (a) 81. (a) 82. (d) 83. (d) 84. (c)
9. (b) 10. (c) 11. (b) 12. (a) 85. (d) 86. (a) 87. (b) 88. (d)
89. (b) 90. (b) 91. (b) 92. (b)
13. (c) 14. (d) 15. (a) 16. (a)
93. (c) 94. (c) 95. (c) 96. (b)
17. (b) 18. (a) 19. (b) 20. (c)
97. (c) 98. (d) 99. (b) 100. (d)
21. (c) 22. (a) 23. (c) 24. (a)
101. (c) 102. (d) 103. (d) 104. (c)
25. (c) 26. (b) 27. (b) 28. (b)
105. (a) 106. (b) 107. (b) 108. (c)
29. (a) 30. (a) 31. (a) 32. (c)
109. (d) 110. (a) 111. (b) 112. (d)
33. (a) 34. (a) 35. (d) 36. (a)
113. (b) 114. (a) 115. (c) 116. (b)
37. (a) 38. (b) 39. (c) 40. (a)
117. (a) 118. (d) 119. (c) 120. (c)
41. (d) 42. (c) 43. (b) 44. (b)
121. (d) 122. (c) 123. (a) 124. (c)
45. (a) 46. (a) 47. (a) 48. (b)
125. (b) 126. (c) 127. (c) 128. (b)
49. (d) 50. (a) 51. (c) 52. (a)
129. (d) 130. (d) 131. (a) 132. (a)
53. (a) 54. (b) 55. (a) 56. (a)
133. (b) 134. (a) 135. (a) 136. (d)
57. (b) 58. (a) 59. (d) 60. (c)
61. (d) 62. (a) 63. (d) 64. (b) 137. (d) 138. (d) 139. (c) 140. (b)
65. (c) 66. (b) 67. (b) 68. (c) 141. (b) 142. (c) 143. (a) 144. (d)
69. (d) 70. (b) 71. (c) 72. (c) 145. (b) 146. (a) 147. (c) 148. (b)
73. (a) 74. (a) 75. (a) 76. (d) 149. (a) 150. (b)
ANSWER KEY 173
Exercise - 2: Previous Year Questions Exercise - 3: Achiever’s section

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

5. (a) 6. (a) 7. (b) 8. (b) 5. (a) 6. (c) 7. (a) 8. (c)
9. (c) 10. (a) 11. (c) 12. (b) 9. (b) 10. (a) 11. (d) 12. (d)
13. (b) 14. (c) 15. (d) 16. (a) 13. (d) 14. (c) 15. (a) 16. (a)
17. (a) 18. (d) 19. (b) 20. (c) 17. (b) 18. (b) 19. (d) 20. (c)
21. (a) 22. (c) 23. (a) 24. (a) 21. (c) 22. (d) 23. (c) 24. (c)
25. (a) 26. (a) 27. (b) 28. (c)
25. (a) 26. (c) 27. (c) 28. (c)
29. (b) 30. (a) 31. (c) 32. (d)
29. (c) 30. (a)
33. (d) 34. (d) 35. (b) 36. (a)
37. (b) 38. (c) 39. (a) 40. (d)
41. (c) 42. (b) 43. (a) 44. (c)
45. (b) 46. (b) 47. (c) 48. (a)
49. (b) 50. (c) 51. (d) 52. (a)
174 ANSWER KEY
CHAPTER - 2: PERIODIC PROPERTIES

Exercise - 1: Basic Objective Questions

DIRECTIONS FOR USE-

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1. (d) 2.(b) 3. (a) 4. (b) 77. (d) 78. (a) 79. (d) 80. (a)
5. (c) 6. (d) 7. (d) 8. (a) 81. (a) 82. (b) 83. (a) 84. (d)
9. (b) 10. (a) 11. (b) 12. (b) 85. (b) 86. (c) 87. (c) 88. (b)
13. (b) 14. (c) 15. (b) 16. (c) 89. (a) 90. (b) 91. (d) 92. (d)
17. (a) 18. (d) 19. (d) 20. (a) 93. (b) 94. (a) 95. (d) 96. (d)
21. (c) 22. (c) 23. (d) 24. (c) 97. (a) 98. (c) 99. (d) 100. (a)
25. (a) 26. (d) 27. (b) 28. (b) 101. (b) 102. (c) 103. (c) 104. (d)
29. (c) 30. (c) 31. (d) 32. (d) 105. (a) 106. (b) 107. (c) 108. (a)
33. (b) 34. (c) 35. (a) 36. (a) 109. (b) 110. (d) 111. (a) 112. (a)
37. (b) 38. (c) 39. (c) 40. (c) 113. (a) 114. (a) 115. (c) 116. (b)
41. (c) 42. (a) 43. (b) 44. (c) 117. (c) 118. (a) 119. (c) 120. (a)
45. (c) 46. (c) 47. (d) 48. (a) 121. (d) 122. (c) 123. (d) 124. (a)
49. (a) 50. (a) 51. (d) 52. (c) 125. (c) 126. (d) 127. (c) 128. (a)
53. (b) 54. (a) 55. (b) 56. (a) 129. (c) 130. (b) 131. (d) 132. (a)
57. (b) 58. (a) 59. (d) 60. (d) 133. (a) 134. (d) 135. (a) 136. (c)
61.(d) 62. (c) 63. (c) 64. (b) 137. (b) 138. (a) 139. (c) 140. (a)
65. (a) 66. (a) 67. (d) 68. (b) 141. (c) 142. (b) 143. (a) 144. (c)
69. (a) 70. (c) 71. (b) 72. (d) 145. (b) 146. (b) 147. (a) 148. (a)
73. (b) 74. (a) 75. (d) 76. (c) 149. (d) 150. (c)
ANSWER KEY 175

Exercise - 2: Previous Year Questions Exercise - 3: Achiever’s section

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

5. (a) 6. (b) 7. (a) 8. (d) 5. (d) 6. (d) 7. (d) 8. (d)
9. (b) 10. (b) 11. (b) 12. (b) 9. (b) 10. (a) 11. (d) 12. (b)
13. (d) 14. (c) 15. (d) 16. (a) 13. (d) 14. (b) 15. (c) 16. (c)
17. (a) 18. (a) 19. (a) 20. (c) 17. (c) 18. (a) 19. (a) 20. (c)
21. (b) 22. (c) 23. (c) 24. (d) 21. (c) 22. (d) 23. (a) 24. (a)
25. (a) 26. (c) 27. (c) 28. (a)
25. (a) 26. (b) 27. (a) 28. (c)
29. (b) 30. (a) 31. (c) 32. (a)
29. (b) 30. (d)
33. (a) 34. (d) 35. (b) 36. (b)
37. (c) 38. (c) 39. (d) 40. (d)
41. (d) 42. (b) 43. (b) 44. (a)
45. (d) 46. (b) 47. (a) 48. (b)
49. (c) 50. (d) 51. (a)
176 ANSWER KEY

CHAPTER -3 : CHEMICAL BONDING

Exercise - 1: Basic Objective Questions

DIRECTIONS FOR USE-

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1. (a) 2.(b) 3. (b) 4. (d) 77. (c) 78. (d) 79. (b) 80. (d)
5. (a) 6. (a) 7. (d) 8. (b) 81. (a) 82. (a) 83. (a) 84. (c)
9. (d) 10. (a) 11. (a) 12. (d) 85. (b) 86. (d) 87. (d) 88. (c)
13. (c) 14. (a) 15. (b) 16. (b) 89. (c) 90. (c) 91. (c) 92. (a)
17. (d) 18. (a) 19. (d) 20. (b) 93. (a) 94. (a) 95. (d) 96. (c)
21. (b) 22. (b) 23. (c) 24. (d) 97. (b) 98. (c) 99. (c) 100. (a)
25. (a) 26. (a) 27. (a) 28. (a) 101. (d) 102. (c) 103. (b) 104. (d)
29. (a) 30. (a) 31. (d) 32. (b) 105. (a) 106. (c) 107. (d) 108. (b)
33. (a) 34. (d) 35. (b) 36. (d) 109. (c) 110. (c) 111. (b) 112. (b)
37. (c) 38. (c) 39. (a) 40. (c) 113. (b) 114. (c) 115. (c) 116. (c)
41. (d) 42. (b) 43. (c) 44. (b) 117. (b) 118. (d) 119. (a) 120. (d)
45. (a) 46. (d) 47. (a) 48. (b) 121. (c) 122. (a) 123. (a) 124. (d)
49. (a) 50. (d) 51. (a) 52. (c) 125. (a) 126. (c) 127. (c) 128. (b)
53. (c) 54. (a) 55. (a) 56. (b) 129. (d) 130. (d) 131. (c) 132. (c)
57. (b) 58. (a) 59. (c) 60. (d) 133. (d) 134. (a) 135. (b) 136. (b)
61. (b) 62. (a) 63. (b) 64. (b) 137. (c) 138. (d) 139. (a) 140. (d)
65. (b) 66. (c) 67. (b) 68. (d) 141. (b) 142. (b) 143. (d) 144. (a)
69. (c) 70. (d) 71. (d) 72. (a) 145. (b) 146. (b) 147. (a) 148. (c)
73. (d) 74. (a) 75. (a) 76. (c) 149. (b) 150. (a)
ANSWER KEY 177

Exercise - 2: Previous Year Questions Exercise - 3: Achiever’s section

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

5. (c) 6. (d) 7. (c) 8. (b) 5. (a) 6. (b) 7. (c) 8. (c)
9. (b) 10. (b) 11. (a) 12. (a) 9. (d) 10. (a) 11. (b) 12. (b)
13. (b) 14. (a) 15. (b) 16. (c) 13. (d) 14. (b) 15. (d) 16. (b)
17. (a) 18. (a) 19. (b) 20. (d) 17. (b) 18. (c) 19. (a) 20. (a)
21. (a) 22. (c) 23. (b) 24. (a) 21. (a) 22. (b) 23. (a) 24. (e)
25. (d) 26. (c) 27. (d) 28. (d)
25. (a) 26. (c) 27. (c) 28. (c)
29. (d) 30. (c) 31. (b) 32. (c)
29. (b) 30. (c)
33. (a) 34. (d) 35. (b) 36. (c)
37. (a) 38. (d) 39. (b) 40. (c)
41. (a) 42. (a) 43. (d) 44. (d)
45. (b) 46. (a) 47. (d) 48. (a)
49. (c) 50. (a) 51. (d) 52. (d)
53. (a)
178 ANSWER KEY

CHAPTER - 4: MOLE & EQUIVALENT CONCEPT

Exercise - 1: Basic Objective Questions

DIRECTIONS FOR USE-

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1. (b) 2.(a) 3. (d) 4. (b) 77. (a) 78. (d) 79. (c) 80. (d)
5. (a) 6. (c) 7. (d) 8. (b) 81. (b) 82. (b) 83. (a) 84. (d)
9. (b) 10. (b) 11. (d) 12. (c) 85. (d) 86. (b) 87. (b) 88. (c)
13. (a) 14. (c) 15. (a) 16. (b) 89. (c) 90. (b) 91. (d) 92. (c)
17. (c) 18. (b) 19. (a) 20. (a) 93. (b) 94. (a) 95. (a) 96. (b)
21. (b) 22. (a) 23. (b) 24. (a) 97. (d) 98. (c) 99. (a) 100. (a)
25. (a) 26. (a) 27. (d) 28. (b) 101. (b) 102. (c) 103. (b) 104. (d)
29. (a) 30. (b) 31. (b) 32. (a) 105. (a) 106. (c) 107. (a) 108. (b)
33. (b) 34. (c) 35. (b) 36. (d) 109. (b) 110. (d) 111. (c) 112. (d)
37. (b) 38. (a) 39. (d) 40. (d) 113. (b) 114. (b) 115. (b) 116. (b)
41. (a) 42. (b) 43. (b) 44. (a) 117. (d) 118. (a) 119. (a) 120. (a)
45. (a) 46. (d) 47. (a) 48. (b) 121. (c) 122. (a) 123. (d) 124. (a)
49. (c) 50. (a) 51. (b) 52. (a) 125. (b) 126. (d) 127. (d) 128. (a)
53. (b) 54. (a) 55. (d) 56. (b) 129. (a) 130. (d) 131. (c) 132. (d)
57. (d) 58. (c) 59. (c) 60. (b) 133. (a) 134. (b) 135. (d) 136. (d)
61.(b) 62. (b) 63. (b) 64. (a) 137. (c) 138. (c) 139. (a) 140. (d)
65. (d) 66. (a) 67. (a) 68. (a) 141. (c) 142. (a) 143. (a) 144. (b)
69. (a) 70. (a) 71. (c) 72. (a) 145. (a) 146. (c) 147. (c) 148. (b)
73. (c) 74. (b) 75. (b) 76. (c) 149. (c) 150. (d)
ANSWER KEY 179

Exercise - 2: Previous Year Questions Exercise - 3: Achiever’s section

DIRECTIONS FOR USE- DIRECTIONS FOR USE-

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

5. (d) 6. (d) 7. (c) 8. (a) 5. (c) 6. (d) 7. (b) 8. (c)
9. (a) 10. (d) 11. (a) 12. (a) 9. (d) 10. (c) 11. (c) 12. (d)
13. (a) 14. (d) 15. (d) 16. (c) 13. (a) 14. (d) 15. (b) 16. (a)
17. (c) 18. (a) 19. (c) 20. (c) 17. (a) 18. (d) 19. (d) 20. (a)
21. (b) 22. (b) 23. (c) 24. (b) 21. (b) 22. (c) 23. (a) 24. (d)
25. (c) 26. (a) 27. (a) 28. (a)
25. (b) 26. (a) 27. (a) 28. (b)
29. (b) 30. (b) 31. (a) 32. (c)
29. (c) 30. (a)
33. (a) 34. (d) 35. (c) 36. (a)
37. (a) 38. (a) 39. (a) 40. (b)
41. (a) 42. (c) 43. (b) 44. (d)
45. (a) 46. (a) 47. (d) 48. (c)
49. (d) 50. (a) 51. (b) 52. (b)
MASTER INDEX
VOLUME 1:
Structure of an Atom
Periodic Properties
Chemical Bonding
Mole & Equivalent concept

VOLUME 2:
Redox Reactions
Gaseous State
General Organic Chemistry
Hydrocarbons
Hydrogen

VOLUME 3:
Thermodynamics and Thermochemistry
Chemical Equilibrium
Ionic Equilibrium

VOLUME 4:
s- Block
p-Block (group 13 and 14)
Environmental Chemistry
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