UNIT

2 ATOMIC STRUCTURE
2| XI Std. Chemistry

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure |3

John Dalton coined the term atom. The atom is the

fundamental particle of matter and considered to be
indivisible and indestructible.
In fact, the atom as the whole is electrically neutral as
number of protons in it is equal to number of electrons.

How small is an atom?

Atoms are very small – they are about
0.00000001 cm wide. Think about the thickness of a
crisp. The number of atoms you would need to stack up to
make the thickness of a crisp, is approximately the same number of crisps you would
need to stack up to make the height of Mount Everest!
That’s roughly 7 million crisps!
Electron, proton, neutron are the main fundamental particles of an atom.

Dalton’s Atomic Theory

ACTIVE SITE EDUTECH-9844532971

4| XI Std. Chemistry

Limitations of Dalton’s Theory

• It fails to explain why atoms of different kinds should differ in mass and valency

etc.

• The discovery of isotopes and isobars showed that atoms of same elements may

have different atomic masses (isotopes) and atoms of different kinds may have

same atomic masses (isobars).

• The discovery of various sub-aomic particles like X-rays, electrons, protons etc.

during late 19th century lead to the idea that the atom was no longer an indivisible

and smallest particle of the matter.

DISCOVERY OF FUNDAMENTAL PARTICLES

The atom consists of several sub-atomic particles like electron, proton, neutron,
positron, neutrino etc. Out of these particles, the electron, proton and the neutron are
called fundamental subatomic particles.

Discovery of electron – study of Cathode rays:

J.J. Thomson observed that, when a high voltage is applied between the electrodes
fitted in discharge tube, at a very low pressure, some invisible radiations are emitted
from the cathode. At this stage wall of the discharge tube near cathode starts
glowing.

Glowing is due to the bombardment of glass wall by the cathode rays. It may be noted
that when the gas pressure in the tube is 1 atm, no electric current flows through the
tube. This is because the gases are poor conductor of electricity.
Origin of Cathode rays:
Cathode rays are first produced in cathode due to bombardment of the gas molecules
by the high-speed electrons emitted first from the cathode.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure |5

Properties of Cathode rays

i. They travel in straight lines with high speed. Because they cast the shadow of an
object on the ZnS screen placed in its path.

ii. They are made up of material particles having both mass and velocity. Because,
they rotate a light paddle wheel placed in its path.

iii. They carry negative charge, the negatively charged material particles constituting
the cathode rays are called electrons.
Cathode rays get deflected when they placed in an electric field & magnetic field.
Direction of deflection shows that they are negatively charged.

iv. They produce heating effect.

v. They cause ionization of the gas through which they pass.
vi. They produce X-rays when they strike against the surface of hard metals like
tungsten, molybdenum etc.

ACTIVE SITE EDUTECH-9844532971

6| XI Std. Chemistry

vii. They produce green fluorescence on the glass walls of the discharge tube exp:
ZnS.

viii. They affect the photographic plates.

ix. They possess penetrating effect (i.e., they can easily pass-through thin foils of
metals).
x. The nature of the cathode rays does not depend upon the nature of the gas,
taken in the discharge tube and the nature of cathode material.
xi. For each cathode rays, the ratio of charge (e) to mass (m) is constant

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure |7

Discovery of proton – study of Anode rays:

Goldstein discovered the presence of positive rays. He performed discharge tube
experiment in which he took perforated cathode and a gas at low pressure was kept
inside a discharge tube.
On applying high voltage between electrodes, new rays were coming from the side of
anode and passing through the hole in the cathode gives fluorescence on the opposite
glass wall coated with zinc sulphide.
These rays were called anode rays or canal rays or positive rays.

Origin of anode or positive rays:

In the discharge tube the atoms of gas lose negatively charged electrons. These
atoms, thus, acquire a positive charge. The positively charged particle produced from
hydrogen gas was called the proton.
H → H+(proton) + e-

Properties of Anode rays:

i) They travel in straight lines. However, their speed is much less than that of the
cathode rays.
ii) They are made up of material particles.
iii) They are positively charged, hence they called as canal rays or anode rays.
iv) The nature of anode rays depends on the gas taken in the discharge tube.
v) For different gases taken in discharge tube the charge to mass ratio (e/m) of the
positive particles constituting the positive rays is different.

ACTIVE SITE EDUTECH-9844532971

8| XI Std. Chemistry

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure |9

Fundamental particles:
1) Electron: Electron is a universal constituent discovered by the J.J. Thomson.
 Charge: It was determined by Mullikan by oil drop experiment as -1.602x10-
19
coulombs
or 4.803x10-10 e.s.u.
 Mass:9.11x10-28g (nearly equal to 1/1837th of mass of hydrogen atom).
 Specific charge:e/m ratio is called specific charge & is equal to 1.76x108
coulombs/gm.
 Mass of one mole of electrons: It is 0.55 mg.
 Charge on one mole of electron is 96500 coulombs or 1 faraday.
 Density: 2.17x1017 g/cc.

2. Proton: (+1p0 or 1H1)

 It was discovered by Goldstein.
 Charge:It carries positive charge i.e.1.602 x 10-19coulombs or 4.803x10-10 esu.
 Mass:1.672x10-24g or 1.672x10-27kg.It is 1837 times heavier than an electron.
 Specific charge (e/m):9.58x104coulomb/gm.

3. Neutron (0n1)
* It was discovered by Chadwick by bombarding Be atom with high speed -particles.
𝟒𝐁𝐞𝟗 + 𝟐 𝐇𝐞𝟒 → 𝟔𝐂
𝟏𝟐
+𝟎 𝐧𝟏
* Charge: Charge less or neutral particle.
* Mass:1.675x10-24 g or 1.675x10-27 kg.
* Density:1.5x1014 g/cm3 and is heavier than proton by 0.18%.
* Specific charge: It is zero.
* Among all the elementary particles neutron is the heaviest and least stable.

Summary 
Electron Proton Neutron

Discover J.J. Thomson Goldstein James Chadwick

Symbol e or e−1 P+ n0

Absolute charge −1.6022 × 10−19 +1.6022 × 10−19 0

Relative charge −1 +1 No charge

Mass in kg 9.109 × 10−31 1.673 × 10−27 1.675 × 10−27

Mass in amu 0.0005486 1.007 1.008

Relative mass (app) 1/1836 1 1

ACTIVE SITE EDUTECH-9844532971

10 | XI Std. Chemistry

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 11

Classical Models of Atom:

1) Thomson’s Atomic Model

According to Thomson, an atom is a sphere of positive charge having a number of

embedded electrons in it and sufficient enough to neutralize the positive charge.

This model is compared with a water melon in which seeds are embedded or pudding in

ACTIVE SITE EDUTECH-9844532971

12 | XI Std. Chemistry

which raisins are embedded. Therefore, this model, sometime called watermelon model

or raisin or plum pudding model.

Important Features

• It shows the electrical neutrality of the atom.

• The mass of the atom is assumed to be uniformly distributed over the atom.

The model was rejected as it was not consistent with the results of the

alpha scattering experiment.

Limitation:

• It is failed explain the results of scattering experiment of Rutherford & the

stability of atom.

• It is a static model. It does not reflect the movement of electrons.

2) Rutherford’s Atomic Model:

Rutherford, performed -ray scattering experiment in which he bombarded thin foils
of metals like gold, silver, platinum or copper with a beam of fast-moving radioactive
particles originated from a lead block. The presence of 𝛼 particles at any point around
the thin foil of gold after striking it was detected with the help of a circular zinc

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 13

sulphide screen. The point at which a𝛼 particle strikes this screen; a flash of light is
given out.

Observations and Conclusions

i. Most of -particles passed through the gold foil without any deflection from their
original path.
Because atom has largely empty space as most of the -particles passed through
the foil undeflected.
ii. A few of alpha particles are deflected fairly at large angles while some are
deflected through small angles.
Because there is heavy positive charge at the center of the atom which causes
repulsions. The entire mass of the atom is concentrated in
the nucleus.
iii. A very few -particles are deflected back along their path.
According to Rutherford,
1. Atom is spherical & mostly hollow with lot of empty space in it.
2. It has a small +ly charged part at its center known as nucleus.

ACTIVE SITE EDUTECH-9844532971

14 | XI Std. Chemistry

3. The nucleus is surrounded by electrons. Electrons revolve round the nucleus with
very high speeds in circular paths called orbits.
4. The number of extra nuclear electrons is equal to the number of units of positive
charge in the nucleus. Therefore, the atom is electrically neutral. Electrons and
the nucleus are held together by electrostatic forces of attraction.
5. Rutherford’s model has resemblances with solar system. Hence, it’s also known as
planetary model of the atom.
6. There is an empty space around the nucleus called extra nuclear part. In this part
electrons are present. As the nucleus of the atom is responsible for the mass of
the atom, the extra nuclear part is responsible for its volume.

Drawbacks:
1. According to the electromagnetic theory of
Maxwell, when a charged particle moves under
the influence of attractive force it loses energy
continuously in the form of electromagnetic
radiation. Therefore, an electron in an orbit will
emit radiation.
As a result of this, the electron should lose energy at every turn and move closer
and closer to the nucleus following a spiral path. Ultimate result is that it will fall
into the nucleus thereby making the atom unstable.
i.e., Rutherford’s model cannot explain the stability of the atom.

2. If the electrons lose energy continuously, the

spectrum is expected to be continuous but the
actual observed spectrum consists of well-defined
lines of definite frequencies. Here the loss of
energy by the electrons is not continuous in an
atom.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 15

Atomic number(Z): Atomic number denotes the number of protons or the number

of electrons in the neutral atom.

Atomic number (Z) = Number of protons in the nucleus of an atom or ion

= Number of electrons in a neutral atom.
Each element has a different atomic number.

A The atomic number of sulfur (S) is 16.

B The atomic number of iron (Fe) is 26.

C The atomic number of silver (Ag) is 47.

Each element has a definite and fixed number of protons.

If the number of protons changes, then the atom becomes a

different element.

Changes in the number of particles in the nucleus (protons or neutrons)

are very rare. They only take place in nuclear processes such as:

⚫ radioactive decay

⚫ nuclear bombs

⚫ nuclear reactors.

Remember!
In an atom… APE!
A= P= E
Atomic number = number of protons = number of electrons

Mass number (A): The mass number is the total number of protons and neutrons

present in the nucleus of an atom of an element and indicated as A.

Protons and neutrons present in the nucleus of an atom are collectively known as

nucleons. Therefore, the mass number is also known as nucleon number.

Mass number (A) = Number of protons (Z) + Number of neutrons (n)

ACTIVE SITE EDUTECH-9844532971

16 | XI Std. Chemistry

The number of neutrons (n) in an atom is equal to the difference between the
mass number and the atomic number. n = A – Z

Mass Number

Atomic Number
A
Z X OR
Z XA

Symbol of Element

Atomic number= Protons= Electrons

Mass number= Atomic number + number of Neutrons

Isotopes, Isobars and Isotones:

Isotopes: The atoms of the same element which have the same atomic number but
different mass numbers are called isotopes.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 17

Exp- 6 C12 , 6 C13 , 6 C14 , 8 O16 , 8 O17 , 8 O18 , 17 Cl35 , 17 Cl37

deuterium
hydrogen tritium
1 proton
1 proton 1 proton
1 neutrons
0 neutrons 2 neutrons
1 electron 1 electron
1 electron

Isotopes of an element differ in the number of neutrons present in the nucleus. But
they have the same number of protons and electrons.
Because of same number of electrons, they show same chemical properties.
They, have different number of neutrons, so they will have different masses and
hence different physical properties.

Isobars: The atoms of different elements which have the same mass number but
different atomic numbers are called isobars.
Exp: 18 Ar 40 , 19 K 40 , 20 Ca40

They have same number of nucleons. But they are differed chemically because the
chemical characteristics depend upon the number of electrons which is determined by
the atomic number.

ACTIVE SITE EDUTECH-9844532971

18 | XI Std. Chemistry

Isotones: Isotones are the atoms of different elements which have the same number
of neutrons.
Eg: 6 C14 , 7 N15 , 8 O16 (n = 8)

14 Si30 , 15 P31, 16 S32 (n = 16)

Isotones show different physical and chemical properties.

CLASS EXERCISE
1. The number of neutrons present in 19K39 is:
a) 39 b) 19 c) 20 d) None of these
2. The nucleus of the atom (Z > 1) consists of:
a) Proton and neutron b) Proton and electron
c) Neutron and electron d) Proton, neutron and electrons
3. The number of electrons in a neutral atom of an element is equal to it’s:
a) Atomic weight b) Atomic number c) Equivalent weight d)
Electron affinity
4. The specific charge of the canal rays:
a) Is not constant but changes with gas filled in discharge tube
b) Remains constant irrespective of the nature of gas in discharge tube
c) Is maximum when gas present in discharge tube is hydrogen
d) Is 9.58 x 104 coulombs/g
5. Proton is:
a) Nucleus of deuterium b) Ionized hydrogen molecule

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 19

c) Ionized hydrogen atom d) An α-particle

6. According to the Rutherford which statement is correct?
a) Electron revolves in fixed circular path around the nucleus
b) Electron revolves around the nucleus
c) Electron does not decrease its energy at the time of revolution
d) Electron obeys law of conservation of momentum at the time of revolution.
7. Rutherford's scattering experiments led to the discovery of
a) Nucleus
b) Presence of neutrons in the nucleus
c) Both a and b
d) Revolving nature of electrons around the nucleus
8. Deflection back of a few particles on hitting thin foil of gold shows that:
a) Nucleus is heavy b) Nucleus is small c) Both a and b
d) Electrons create hindrances in the movement of α –particles
9. α-particles are represented by
a) Lithium atoms b) Helium nuclei c) Hydrogen nucleus d) None of the above

HOME EXERCISE:
1. The species in which one of the fundamental particles is missing is
a) Helium b) Protium c) Deuterium d) Tritium
2. The discovery of neutron is late because neutron has
a) +ve charge b) –ve charge c) neutral charge d) lightest particle
3. Which of the following statements are correct?
a) Isotopes have same number of protons
b) Isobars have same nucleon number.
c) Isobars have same number of protons
d) Both a and b
4. The charge on electron is calculated by
a) Mullikan b) J J Thomson c) Ruther ford d) Newton
5. J J Thomson Model could able to explain the following?
a) Stability of Atom b) electrical neutrality of atom
c) Stability of nucleus d) all of these
6. The thickness of the gold foil used in Ruther Ford α ray scattering experiment
a) 0.0004 cm b) 0.0004 m c) 0.0004 mm d) 0.004 cm
7. What is the size of atom predicted by Ruther Ford?
a) 10-13cm b) 10-14cm c) 10-12cm d) 10-8cm
8. If Thomson Model is correct what should be the observation in α-ray scattering
experiment
a) All the α-rays should pass through the gold foil
b) Only few α-rays should pass through the gold foil
c) 98% of α-rays should get reflected back
d) Both b & c
9. Which part of atom is responsible for volume of atom?
a) Nucleus b) extra nuclear part c) protons d)unknown particle

ACTIVE SITE EDUTECH-9844532971

20 | XI Std. Chemistry

CLASS EXERCISE:
1)c 2)d 3)b 4) a&c 5)c 6)b 7)c 8)b 9)b

HOME EXERCISE:
1)b 2)c 3)d 4)a 5)b 6)a 7)d 8)d 9)b

Nature of Light (Electromagnetic Radiation):

Electromagnetic radiation does not need any medium for propagation e.g visible, ultra
violet, infrared, x-rays, -rays, radio waves, radiant energy etc.
Two theories were proposed to explain the nature and the propagation of light
i. Corpuscular theory: This theory was proposed by Newton. According to this
theory light is propagated in the form of invisible small particles. i.e. light has
particle nature.
The particle nature of light explained some of the experimental facts such as
reflection and refraction of light but it failed to explain the phenomenon of
interference and diffraction. Therefore, was discarded and ignored.
ii. Wave theory of light (electromagnetic wave theory): was explained by James
Clark Maxwell in 1864 to explain & understand the nature of
electromagnetic radiation.

Features of this theory are:

a) The light is a form of electromagnetic radiations.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 21

b) Light radiations consist of electric & magnetic fields oscillating perpendicular to

each other.
c) Vertical component of wave, ‘E’ indicates the change in the strength of the electric
field and the horizontal component of the wave ‘H’ indicates the change in the
strength of the magnetic field.
d) These radiations do not require any medium for propagation.
e) The radiations possess wave character and travel with the velocity of light
i.e. 3x108 m/sec because of the above characteristics, the radiation is called
electromagnetic radiations or waves.

Electromagnetic radiation is explained by following characteristics:

1. Wave length:
The distance between two successive crests, troughs or between any two
consecutive identical points in the same phase of a wave is called wave length. It is
denoted by the letter (lambda).
The wave length is measured in terms of meters (m), centimeters (cm), angstrom
units (A0) nanometers (nm), picometers (pm) and also in millimicrons (m).
The S.I. unit of wavelength is meter, m
0 –10 –8
1A = 10 m or 10 cm
–9 –7
1nm = 10 m or 10 cm = 10A0
–12 –10
1pm = 10 m or 10 cm =10−2 A0

2.Frequency:
The number of waves that pass-through a given point in one second is known as
frequency of radiation. It is denoted by the ‘v’ (nue).

SI unit of frequency is per second(s–1) or Hertz (Hz). A cycle is said to be completed

when a wave consisting of a crest and a trough passes through a point.

3.Velocity:
Distance travelled by the wave in one second is called velocity or speed of the wave
(C).
SI unit is meters per second (ms-1).

ACTIVE SITE EDUTECH-9844532971

22 | XI Std. Chemistry

C of electromagnetic radiation in vaccum is a constant commonly called the speed of

light and is denoted by ‘c’. It is equal to 3 × 108ms–1.
4.Wave number:
Number of waves that can be present at any time in unit length is called wave
number.
It is denoted by  (nue bar).
It is the reciprocal of wave length.
1
Wave number =  =

It is expressed in per centimeter (cm–1) or per meter (m–1).
The SI unit of wave number is m–1.
Wave length, wave number𝝂̅ , frequency 𝝂 and velocity c are related as follows
.
⇒ 𝑪 = 𝜐𝜆
5.Amplitude:
The height of the crest or the depth of the trough of the wave is called amplitude
of the wave. It is denoted by A.
The amplitude determines the strength or intensity or brightness of radiation.

6.Time period:
It is the time taken by the wave for one complete cycle or vibrations. It is
denoted by T. It is expressed in second per cycle.
1 1
T= (Where  = frequency)
𝑉 

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 23

ACTIVE SITE EDUTECH-9844532971

24 | XI Std. Chemistry

Electromagnetic spectrum:
The arrangement of different types of electromagnetic radiations in the order of
increasing wavelengths or decreasing frequencies is known as electromagnetic
spectrum.

S.No. Name Wavelength(A0) Frequency(Hz) Source

1. Radio wave 3  10 – 3  10
14 7
1  10 – 1  10
5 9
Alternating current of high
frequency
2. Microwave 3  107 – 6  106 1  109 – 5  1011 Klystron tube
3. Infrared (IR) 6  106 – 7600 5  1011 – 3.95  1016 Incandescent objects
4. Visible 7600–3800 3.95  10 – 7.9  10
16 14
Electric bulbs, sun rays
5. Ultraviolet(UV) 3800–150 7.9  1014 – 2  1016 Sun rays, arc lamps with
mercury vapours
6. X-Rays 150–0.1 2  1016 – 3  1019 Cathode rays striking metal
plate
7.  -Rays 0.1–0.01 3  1019 – 3  1020 Secondary effect of radioactive
decay
8. Cosmic Rays 0.01–zero 3  1020 –Infinity Outer space

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 25

Limitations of Electromagnetic Wave Theory:

This theory was successful in explaining the properties of light such as interference,
diffraction etc.
But it could not explain the following:
(i) The phenomenon of black body radiation.
(ii) The photoelectric effect.
(iii) The change heat capacity of solids as a function of T.
(iv) The line spectra of atoms with special reference to hydrogen.
These phenomena could be explained only if electromagnetic waves are supposed to
have particle nature.

Black body radiation:

When a radiant energy falls on the surface of a body, a part of it is absorbed, a part of
it is reflected and the remaining energy is transmitted.
An ideal body is expected to absorb completely the radiant energy falling on it is known
as a black body. A black body is not only a perfect absorber but also a perfect
emitter of radiant energy.
A hollow sphere coated inside with a platinum black, which has a small hole in its wall
can act as a near black body.

ACTIVE SITE EDUTECH-9844532971

26 | XI Std. Chemistry

The radiation emitted by a black body kept

at high temperature is called black body
radiation. A black body radiation is the
visible glow that the solid object gives off
when heated.
A graph is obtained by plotting the
intensity of radiation against wave
length gives the following details.
1. The nature of radiation depends upon the T of the black body.
2. If the energy emitted is continuous the curve should be as shown by the dotted
lines.
3. At a given temperature the intensity of radiation increases with the wave length,
reaches maximum and then decreases.
4. The intensity of radiation is greatest at the medium wave lengths and least at
highest and lowest wave lengths.
5. As the temperature increases the peak of maximum intensity shifts towards the
shorter wave lengths.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 27

2.5 MODERN APPROACH TO THE STRUCTURE OF ATOM

Planck’s quantum theory:

In order to explain black body radiation, Max Planck proposed quantum theory of
radiation.
Postulates
1. Emission of radiation from a body is due to vibrations of the charged particles in the
body.
2. Energy is emitted or absorbed by a body discontinuously in the form of small
packets of energy called quanta.

3. Energy of each quantum of light is directly proportional to the frequency of the

radiation.
E   or E = h 
Where ‘h’ is known as Planck’s constant.
The value of ‘h’, 6.6256 × 10–34 Jsec- or 6.6256 × 10–27ergs sec-
4. In case of light, the quantum of energy is called a photon.
Total amount of energy emitted or absorbed by a body is some whole number multiple
of quantum,
E = nh  , where n is an integer such as 1,2,3 . . . . .
This means that a body can emit or absorb energy equal to hv, 2hv, 3hv . . . . . Or any
other integral multiple of h. This is called quantization of energy.
5. The emitted radiant energy is propagated in the form of waves.

ACTIVE SITE EDUTECH-9844532971

28 | XI Std. Chemistry

Photoelectric Effect:
When radiations with certain minimum frequency (ν0 ) strike the surface of a metal, the
electrons are ejected from the surface of the metal. It is called photoelectric effect,
electrons emitted are called photoelectron.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 29

For each metal a certain minimum frequency is needed to eject the electrons called
as threshold frequency ( o ) which differs from metal to metal.

K.E. of photoelectron
K.E. constant
K.E. of photoelectron

o
Intensity of Incident radiation
Frequency of absorbed photon

K.E. as a function of frequency K.E. as a function of intensity

It was explained by Einstein. When light of suitable frequency falls on a metal surface,
the light photon gives its energy to the electron of metal atom and the electron is
ejected from metal surface by absorbing this energy. The minimum energy of a
photon required to eject an electron from a metal is called work function () of the
metal. The remaining part of the energy (h  - ) of photon is used to increase the
kinetic energy of the ejected electron. If o is the threshold frequency and  , the
frequency of incident light then
Work function,  = h o .
 According to Einstein, E = h 

 Kinetic energy of photo electron Ek = E -  = h − ho

CLASS EXERCISE
1. The frequency of a radiation whose wave length is 600 nm is
a) 3 x 1014 sec-1 b) 4 x 1014 sec-1 c) 5 x 1014 sec-1 d) 3 x 1015 sec-1
2. The wavelength of light having wave number 4000 cm-1 is
a) 2.5  m b) 250  m c) 25  m d) 25 nm  m

ACTIVE SITE EDUTECH-9844532971

30 | XI Std. Chemistry

3. What is energy of photons that corresponds to a wave number of 2.5 × 10-5 cm-1?
a) 2.5 × 10-20 erg b) 5.1 × 10-23 erg c) 5.0 × 10-22 erg d) 8.5 ×
10-22 erg
4. The frequency of radiation having wave number 10m-1 is:
a) 10s-1 b) 3×107s-1c) 3×1010s-1 d) 3×109s-1
5. The wavelengths of two photons are 2000Å and 4000Å respectively. What is the
ratio of their energies?
a) 1/4 b) 4 c) 1/2 d) 2
6. In photo electric effect the number of photo electrons emitted is proportional to
a) Intensity of incident beam b) Frequency of incident beam
c) Velocity of incident beam d) Work function of photo cathode
7. The kinetic energy of the photo electrons does not depend upon
a) Intensity of incident radiation b) frequency of incident
radiation
c) Wavelength of incident radiation d) wave number of incident radiation
8. The work function of a metal is 3.1 x 10-19 J. Which frequency of photons will not
cause the ejection of electrons?
a) 5 x 1014 s-1 b) 5 x 1015 kHz c) 6 x 1014 s-1 d) 5 x 1012 Hz
9. The work function of a metal is 4.2 eV. If radiation of 2000 falls on the metal, then
the kinetic energy of the fastest photo electrons is
a) 1.6 × 10-19 J b) 16 × 1010 J c) 3.2 × 10-19 J d) 6.4 × 10-10 J
10. A photo electric emitter has a threshold frequency v0. When light of frequency 2v0
is incident, the speed of photo electrons is V. When light of frequency 5v0 is
incident, the speed of photo electrons will be
a) 4V b) 2V c) 2.5V d) 2.5V

HOME EXERCISE
1. Wave theory failed to explain the following properties
a) diffraction b) interference c) black body radiation d) all the above
2. Plank’s quantum theory is explained which of the following properties
a) quantization b) black body radiation c) diffraction d) both a &b
3. The electromagnetic radiation with high energy
a) radio waves b) X-rays c) Infra-red radiation d) visible light
4
4. The atomic transition gives rise to radiation of frequency 10 Hz. The change in
energy per mole of atoms taking place would be:
a) 3.99 × 10–6J b) 3.99J c) 6.62×10––24J d) 6.62× 10–30J
5. Two electromagnetic radiations having energy ratio 3:2 is falling on metal surface
and producing metallic luster what is the ratio of wave numbers of those radiation?
a) 1:2 b) 2:3 c) 3:2 d) 9:4
6. The energy of the photons which corresponds to light of frequency 3 1015 sec-1 is
a) 1.9876×10-15 ergs b) 2.9876×10-8 ergs
c) 1.9876×10-10 ergs d) 1.9876×10-11 ergs
7. Find the frequency of light that correspond to photons energy 5.0 x 10-5 erg

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 31

a) 2.2 x 1011 sec-1 b) 7.5 x 1021 sec-1c) 4.0 x 10-5 sec-1 d) 4.0 x 104 sec-1
8. Photoelectric effect shows:
a) Particle-like behavior of light b) Wave like behavior of light
c) Both wave like and particle-like behavior of light behavior of light d) none
9. When frequency of light incident on a metallic plate is doubled, the KE of the
emitted photoelectrons will be:
a) Doubled b) Halved
c) Increased but more than doubles of previous KE d) Unchanged

CLASS EXERCISE:
1)c 2)a 3)c4)d 5)d6)a 7)a 8)d9)c 10)b

HOME EXERCISE:
1)c 2)d 3)b 4)b 5)c 6)d 7)b 8)a 9)c

ATOMIC SPECTRA
Spectrum is the impression produced on a screen when radiations of a particular
wavelengths are analyzed through a prism or diffraction grating.

ACTIVE SITE EDUTECH-9844532971

32 | XI Std. Chemistry

Spectra are broadly classified into two types.

1. Emission Spectrum: When the radiation emitted from some source, e.g., from the
sun or by-passing electric discharge through a gas at low pressure or by heating some
substance to high temperature etc. is passed directly through the prism and then
received on the photographic plate, the spectrum obtained is called ‘Emission
spectrum’.
Spectrum of radiation emitted by a substance in its excited state is an emission
spectrum.

Emission Spectrum is of two types:

a. Continuous Spectrum: When white light from any source such as sun, a bulb or
any hot glowing body is analyzed by passing through a prism, it is observed that
it splits up into seven different colors from violet to red, (like rainbow), as
shown in fig.

ACTIVE SITE EDUTECH- 9844532971

 Atomic Structure | 33

These colors are so continuous that each of them merges into the next. Hence, the
spectrum is called continuous spectrum.
It may be noted that on passing through the prism, red colour with the longest
wavelength is dedicated least while violet colour with shortest wavelength is deviated
the most.

b. Discontinuous Spectrum: When gases or vapours of a chemical substance are

heated in an electric Arc or in a Bunsen flame, light is emitted. If the ray of
this light is passed through a prism, a line spectrum is produced.
• A discontinuous spectrum consisting of distinct and well-defined lines with dark
areas in between is called line spectrum. It is also called atomic spectrum.
• The emission spectrum consisting of a series of very closely spaced lines is called
band spectrum.

Band spectrum is the characteristic of molecules. Hence it is also known as molecular

spectrum. The band spectrum is due to vibrations and rotations of atoms present in a
molecule.
Differences between line and band spectrum
Line spectrum Band spectrum
The line spectrum has sharp, distinct well- 1. The band spectrum has many closed lines.
1.
defined lines.
2. The line spectrum is the characteristic of 2. The band spectrum is characteristic of
atoms and is also called atomic spectrum. molecules and is also called molecular
spectrum.
3. The line spectrum is due to transition of 3. The band spectrum is due to vibrations and
electrons in an atom. rotations of atoms in a molecule

4. The line spectrum is given by inert gases, 4. The band spectrum is given by hot metals and
metal vapors and atomized nonmetals. molecular nonmetals.

2. Absorption spectra: When white light from any source is first passed through the
solution or vapours of a chemical substance and then analyzed by the spectroscope, it is
observed that some dark lines are obtained. Further, it is observed that the dark lines
are at the same place where coloured lines are obtained in the emission spectra for the
same substance.

ACTIVE SITE EDUTECH-9844532971

 34 | XI Std. Chemistry

Difference between emission spectra and absorption spectra

EMISSION SPECTRA ABSORPTION SPECTRA
1. Emission spectrum is obtained when the 1. Absorption spectrum is obtained when the
radiation from the source are directly white light is first passed through the
analyses in the spectroscope. substance and the transmitted light is
analyzed in the spectroscope.
2. It consists of bright coloured lines 2. It consists of dark lines in the otherwise
separated by dark spaces. continuous spectrum.
3. Emission spectrum can be continuous 3. Absorption spectrum is always discontinuous
spectrum (if source emits white light) or spectrum of dark lines.
discontinuous, i.e., line spectrum if source
emits some coloured radiation.

ACTIVE SITE EDUTECH- 9844532971

 Atomic Structure | 35

Emission Spectrum of Hydrogen:

When hydrogen gas at low pressure is taken in the discharge tube and the light emitted
on passing electric discharge is examined with a spectroscope, the spectrum obtained is
called the emission spectrum of hydrogen which contain large number of lines which
are grouped into different 5 different series,
• Lyman series,
• Balmer series
• Paschen series
• Brackett series
• Pfund series.
• Humphrey series

Wave numbers of all the lines in all the series can be calculated by the Rydberg
equation.
1 1 1
ν̅ = = RZ2 ( 2 − 2 )
λ n1 n2
Where n1 and n2 are whole numbers, n2> n1.

For one electron species like He+, Li2+ and Be3+, the value of R is 109677cm–1× Z2,
where Z is the atomic number of the species.

n=7 Humphrey series

n=6 n1=5, n2=6,7,8----
n=5
ELECTRONIC
Pfund series
TRANSITIONS n1=4, n2=5,6,7,8---- Infra-red region
n=4 Brakett series
n1=3, n2=4,5,6----
n=3 Near
Paschen series
Infra-red region
n1=2, n2=3,4,5,6----
υ
n=2
Balmer series Visible region

n1=1, n2=2,3,4,5----
n=1 Lyman series U.V region

ACTIVE SITE EDUTECH-9844532971

36 | XI Std. Chemistry

Different series of spectral lines in hydrogen emission spectrum

The wave number for any single electron species like He+, Li2+ and Be3+ can be
1 1
calculated from the equation ν̅ = Z2 R H ( − )
n21 n22

CLASS EXERCISE
1. Number of spectral lines possible when an electron falls from fifth orbit to ground
state in hydrogen atom is
a) 4 b) 15 c) 10 d) 21
2. Which of the following electronic transitions require the largest amount of energy?
a) n = 1 to n = 2 b) n = 2 to n = 3 c) n = 3 to n = 4 d) n = 4 to n = 5
3. Which of the following spectral line is associated with a minimum wavelength?
a) n = 5 to n = 1 b) n = 4 to n = 1 c) n = 3 to n = 1 d) n = 2 to n = 1
4. Of the following transitions in hydrogen atom the one which gives an absorption line
is lowest frequency is
a) n = 1 to n= 2 b) n = 3 to n = 5 c) n = 2 to n = 1 d) n = 5 to n = 3
5. The first emission line of Balmer series in H spectrum has wave number equal to
9𝑅𝐻 7𝑅 3𝑅𝐻 5𝑅
a) 400
𝑐𝑚−1 b) 144𝐻 𝑐𝑚−1 c) 4
𝑐𝑚−1 d) 36𝐻 𝑐𝑚−1
6. If the series limit of wave length of the Lyman series for hydrogen atoms is 912
0
A .then the series limit of wave length for the Balmer series of hydrogen atom
is
912
a) 912𝐴𝑜 b) 2 × 912𝐴° c) 4×912A° d) 2
𝐴°

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 37

HOMEEXERCISE
1. There are three energy levels in an atom. How many spectral lines are possible in
its emission spectra?
a) One b) Two c) Three d) Four
2. The wave length of second line in the Balmer series of hydrogen spectrum is equal
to (R=Rydberg constant)
a) 36/5R b) 5R/36 c) 3R/16 d) 16/3R
3. When an electron falls from higher orbit to third orbit in hydrogen atom, the
spectral time observed
a) Balmer series b) Lyman series c) Brackett series d) Paschen series
4. Which of the following electronic transitions require the largest amount of energy?
a) n = 1 → n =2 b) n = 2 → n = 3 c) n = 3 → n = 4 d) n = 4 → n = 5
5. The wave number of the series limiting line for the Lyman series for hydrogen atom
is
(R = 109678 cm-1).
a) 82259 cm-1 b) 109678 cm-1 c) 1.2157 x 10-5 cm d) 9.1176 x 10-6 cm

CLASS EXERCISE:
1) c 2)a 3)a 4)b 5)d 6)c

HOME EXERCISE:
1)c 2)d 3)d 4)a 5)b

ACTIVE SITE EDUTECH-9844532971

38 | XI Std. Chemistry

Bohr’s and Sommerfeld’s Atomic models

To overcome the objections of Rutherford model and to explain the hydrogen
spectrum, Bohr proposed a quantum mechanical model.

POSTULATES OF BOHR’S THEORY

• The electrons revolve round the nucleus with definite velocity in
certain fixed closed circular paths called orbits (or) shells (or)
stationary state. These shells are numbered as 1, 2, 3, 4 or
termed as K, L, M, N from the nucleus.
• Each orbit is associated with a definite amount of energy. As
long as an electron is revolving in an orbit it neither loses nor
gains energy. Hence these orbits are called stationary states or stable
orbits.

• The centrifugal force of the revolving electron in a stationary orbit is balanced by

the electrostatic attraction between the electron and the nucleus.
• Electron can revolve only in orbits whose angular momentum are an integral multiple
of the factor h/2 π.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 39

• The energy of an electron changes only when it moves from one orbit to
another. Outer orbits have higher energies while inner orbits have lower energies.
The energy is absorbed when an electron moves from inner orbit to outer
orbit. The energy is emitted when the electron jumps from outer orbit to
inner orbit.

n=2 n=2
+energy −energy

energy is absorbed energy is released

+ +

Absorption of energy during excitation Release of energy during de-excitatior

• The energy emitted or absorbed in a transition is equal to the difference

between the energies of the two orbits (E2 – E1). Energy emitted or absorbed is
in the form of quanta.
E = E2 – E1 = hv
Here E1 and E2 are the lower and higher allowed energy states.

• Expressions for radius of orbit: Hydrogen atom contains

one proton in the nucleus and one electron revolving around
the nucleus in a circular orbit of radius r.
The electron maintains the same circular motion in given orbit as
centripetal and centrifugal forces are equal in magnitude and
opposite in direction.
Centripetal force = centrifugal force (columbic forces of
attraction provides necessary centripetal force)
− e2 mv 2 e2
i.e. 2
=− ; = mv 2
r r r
nh nh
According to Bohr's quantum condition; mvr = ; v= ;
2 2mr

ACTIVE SITE EDUTECH-9844532971

40 | XI Std. Chemistry

n 2h 2 e2 m  n 2h 2 n 2h 2 n 2h 2
v2 = ; = ; e2 = ; r=
4 2 m 2 r 2 r 4 2 m 2 r 2 4 2 mr 4 2 me 2
 h2 
The radius of the nth orbit is given by, rn = n 2   = 0.529 x 10–8 n2cm
 4 2me 2 
 

Where h = Planck's constant; m = mass of electron; e = charge of electron; r n =

radius of nth orbit
The radius of the first orbit of hydrogen atom is called Bohr's radius which is
denoted by r0.
r0 = 0.529 x 10–8 cm = 0.529 A0
• Expression for Energy of electron: The total energy of the electron in a
stationary orbit is equal to sum of its kinetic and potential energies.
Total energy of electron E = K.E + P.E.
K.E. is always positive and P.E is always negative.
1 e 1 e2 e2 e2 1 e2
K.E. is half to that of P.E. in magnitude. mv 2 − = − ( mv 2 = )= −
2 r 2 r r r 2 r
2 2 e 4 m z2
Energy of electron for single electron species is En = – x .
h2 n2

1 e 2 4 2 me 2 2 2 e 4 m
By substituting the value of r. En = – ; En = – ;
2 n 2h 2 n 2h 2

k 2 2 e 4 m
En = − (k is constant ; k = );
n2 h2

13.6 2.18 x10 −11 2.18x10 −18 313.6

En = − eV/atom (or) − ergs/atom (or) − J/atom (or) −
2 2 2
n n n n2
1312
kcal/mole or − kJ/mole
n2
The energy of electron is negative in the atom. As the value of n increases energy
increases. When n is infinity the value of E is zero. When n value decreases the energy
of electron also decreases.
• Derivation of Rydberg equation:
When an electron jumps from outer energy level (n2) to inner energy level (n1), energy is
released.
i.e. E2 – E1 = E = h
E2 = energy of electron in higher orbit (n2)
E1 = energy of electron in lower orbit (n1)
− 2 2 e 4 m 2 2 e 4 m
E2 – E1 = + ;
n 22 h 2 n12 h 2

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 41

2 2 e 4 m  1 1 
E = − ;
h2 n2 n2 
 1 2

2 2 e 4 m  1 1 
h = − ;
h2 n2 n2 
 1 2

2 2 e 4 m  1 1 
= − ;
h3 n2 n2 
 1 2

1 v
Wave number  = =
 c

2 2 e 4 m  1 1  2 2 e 4 m  1 1 
 = c ; c = − ; = − ;
h3 n2 n2  ch 3  n12 n 22 
 1 2

1  1 1 
= = R − 
 n2 n2 
 1 2

2 2 e 4 m
R= = 1,09,681 cm–1.
3
ch
This value Rydberg constant (R) calculated by Bohr as above is in good agreement
with experimental value.
• Expression for velocity of electron:
h nh
• As per Bohr’s quantum condition, mvr = n ; Vn = ;
2 2 mr
2e 2
Substituting ‘r’; Vn = (for ‘H’ atoms for any other single electron species; Vn =
nh
2e 2 z
x
h n
• Sub situating the values of constants,
2.188 z
Vn = x cm/sec.
10 −8 n
• Number of revolutions per second, made by electron in circular orbit is =
velocity v
=
circumference 2r

Explanation of Hydrogen Spectrum by Bohr’s Theory:

Bohr’s theory successfully explains the origin of lines in hydrogen emission spectrum.
Hydrogen atom has only one electron. It is present in K shell of the atom (n = 1). When
hydrogen gas is subjected to electric discharge, energy is supplied. The molecules
absorb energy and split into atoms. The electrons in different atoms absorb different
amounts of energies. By the absorption of energy, the electrons are excited to
different higher energy levels.

ACTIVE SITE EDUTECH-9844532971

42 | XI Std. Chemistry

Atoms in the excited state are unstable. Therefore, the electrons jump back into
different lower energy states in one or several steps. In each step the energy is
emitted in the form of radiation and is indicated by a line.
Each line has a definite frequency and thus the emission spectrum of hydrogen has
many spectral lines.
• Lyman series are obtained in UV region, when electron returns to the ground state
from higher energy levels 2, 3, 4, 5 ......... and so on.
• Balmer series are obtained in visible region when electron returns to second energy
level from higher energy levels 3, 4, 5, 6 and so on.
• Paschen series are obtained in near infrared region, when electron returns to third
energy level from higher energy levels 4, 5, 6.... And so on.
• Brackett series are obtained in mid infrared region when electron returns to fourth
energy level from higher energy levels 5, 6, 7 . . . and so on.
• Pfund series are obtained in far infrared region when electron returns to the fifth
energy level from higher energy levels 6, 7…….
n(n + 1)
• No.of possible spectral lines = ; n = n2 – n1
2
Where, n2 = higher energy level.

n1 = lower energy level.

n = difference in the two energy levels.

• As the value of 'n' increases

i) the total energy of electron increases
ii) the energy difference between the successive orbits decreases
iii) P.E increases and K.E. decreases
iv) radius of orbits increases
v) velocity of electron decreases

Merits and demerits of Bohr’s Atomic model:

1. Bohr’s model explains the stability of the atom. The electron revolving in a
stationary orbit does not lose energy and hence it remains in the orbit forever.
2. Bohr’s theory successfully explains the atomic spectrum of hydrogen.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 43

3. This theory not only explains hydrogen spectrum but also explains the spectra of
2+ 3+
one electron species such as He+, Li and Be etc.
4. The experimentally determined frequencies of spectral lines are in close
agreement with those calculated by Bohr’s theory.
5. The value of Rydberg constant for hydrogen calculated from Bohr’s equation tallies
with the value determined experimentally.

Limitations of Bohr’s model:

1. Bohr’s theory applicable for single electron species and fails to explain the spectra
of multielectron atoms.

2. It could not explain the fine structure of atomic spectrum.

3. It does not explain the splitting of spectral lines into a group of finer lines under
the influence of magnetic field (Zeeman Effect) and electric field (Stark effect).

4. Bohr’s theory predicts definite orbits for revolving electron. It is against the wave
nature of electron.
5. Bohr’s theory is not in agreement with Heisenberg’s uncertainty principle.
6. He couldn't give a theoretical reason for most of his assumption, for example,
he couldn't justify why angular momentum of the electron should be quantized.
7. Spectrum of isotopes of hydrogen were expected to be same according to the
Bohr's model but found different experimentally.

ACTIVE SITE EDUTECH-9844532971

44 | XI Std. Chemistry

Sommerfeld’s Atomic Model:

It is an extension of Bohr’s model. In this model, the electrons in an atom Radial Velocity
revolve around the nuclei in elliptical orbit. The circular path is a special Tar velocity
•
case of ellipse. Association of elliptical orbits with circular orbits explains Avg Velocity
the fine line spectrum of atoms. • major axis
focus

The main postulates are:

Minor axis
i) The motion of electron in closed circular orbits is influenced by its
n=4,k=4
own nucleus and is set up into closed elliptical paths of definite energy levels. n=4,k=3
n=4,k=2
ii) The nucleus is one of the foci for all these orbits.
iii) The angular momentum of electron in closed elliptical paths is also quantized i.e.
k (h/2), where k is another integer except zero. •
n length of major axis
iv) The ratio k
= length of min or axis
suggests for the possible
number of subshells in a shell. Possible values of k for n = 4 are 1,
2, 3, 4 respectively. For any given value of n, k cannot be zero as in
that case, the ellipse would degenerate into a straight line passing
through the nucleus. When n = k, path becomes circular.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 45

CLASS EXERCISE
1. The ratio of radius of 2nd and 3rd Bohr orbit is
a) 3 : 2 b) 9 : 4 c) 2 : 3 d) 4 : 9
2. According to Bohr’s model, the angular momentum of an electron in 4th orbit is
a) h/3 b)h/2 c) 2h/ d) 3h/2
3. The radius of Bohr’s first orbit in hydrogen atom is 0.053 nm. The radius of second
orbit
of He+ would be
a) 0.0265 nm b) 0.053 nm c) 0.116 nm d) 0.212 nm
4. The minimum energy required to excite a hydrogen atom from its ground state
a) 13.6 eV b) -13.eV c) 3.4 eV d) 10.2 eV
5. The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a
hydrogen atom is
a) 1:2 b) -1:2 c) 1:1 d) -1:1
6. The ratio of potential energy and total energy of an electron in a Bohr orbit of a
hydrogen atom is
a) 2:1 b)-1:2 c) 1:1 d) -1:2
7. The ratio of kinetic energy and total energy of an electron in a Bohr orbit of a
hydrogen atom is
a) 1:-1 b) -2:1 c) 1:1 d) -1:2
8. In a certain Bohr orbit the total energy is - 4.9 eV for this orbit, the kinetic
energy and potential energy are respectively.
a) 9.8 eV, - 4.9 eVb) 4.9 eV, - 98 eV c) 4.9 eV, - 4.9 eV d) 9.8 eV, - 9.8 eV
9. If speed of electron in first Bohr orbit of hydrogen be ‘x’, then speed of the
electron in second orbit of He+ is:
a) x/2 b) 2x c) x d) 4x
10. The ratio of the difference in energy between the first and second Bohr orbits to
that between the second and third Bohr orbit is
a) 1/2 b) 1/3 c) 4/9 d) 27/5
HOME EXERCISE
1. Calculate the ratio of the radius of in 3rd energy level of Li+2ion of 2nd energy
level of He+ ion
a) 3:2 b)1:2 c)2:3 d)1:1
2. Of the following, which of the statement(s) regarding Bohr’s theory is wrong?
a) Kinetic energy of an electron is half of the magnitude of its potential energy
b) Kinetic energy of an electron is negative of total energy of electron
c) Energy of electron decreases with increase in the value of the principal quantum
number
d) The ionization energy of H-atom in the first excited state is negative of one
fourth of the energy of an electron in the ground state.
3. If first ionization energy of hydrogen is E, then the ionization energy of He+ would
be:
a) E b) 2E c) 0.5E d) 4E
+ 2
4. The ratio of radii of first orbits of H, He and Li is:
a) 1:2:3 b) 6:3:2 c) 1:4:9 d) 9:4:1

ACTIVE SITE EDUTECH-9844532971

46 | XI Std. Chemistry

5. The angular momentum of an electron in the M shell of hydrogen atom is

a) 3h /2  b) h/ 2  c) h /  d) 2h / 
6. If ionization potential of H-atom is 13.6 eV, the ionization potential of He+ is
a) 54.4 eV b) 6.8eV c)13.6eV d) 27.2eV
7. The ionization energy of H-atom is its ground state is 2.17 x 10-18J. The ionization
energy of Li +2 in the ground state will be
a) 1.953 x 10-15 J b) 1.953 x 10-16 J c) 1.953 x 10-17J d) 1.953 x 10-18 J
8. If the value of E = - 78.5 K.cal /mole. The order of the orbit in hydrogen atom is
a) 4 b) 3 c) 2 d) 1
9. The ionization potential of hydrogen atom is 13.6 eV. The energy required to
remove an electron in the n = 2 state of the hydrogen atom is
a) 3.4 eV b) 6.8 eV c) 13.6 eV d) 27.2 eV
10. The minimum energy (numerical value) required to be supplied to H-atom to push its
electron from 2nd orbit to the 3rd orbit
a) 1.9 eV b) 2.2 eV c) 2.7 eV d) 7.0 eV

CLASS EXERCISE:
1)d 2)c 3)c 4) d 5) b 6) a 7) a 8) b 9)c 10)d

HOME EXERCISE:
1)a 2)c 3)d 4)b 5)a 6)a 7)c 8)c 9)a 10)a

DUAL NATURE OF MATTER (de-BROGLIE’S WAVE THEORY)

Light exhibits different properties such as diffraction,
interference, photoelectric effect, Compton effect, reflection
and refraction. The phenomenon of diffraction and interference
can be explained by the wave nature of the light. But the
phenomenon of photoelectric effect and Compton Effect can be
explained by the particle nature of the light. Thus, light has dual
nature. De-Broglie proposed that matter like radiation, should also
exhibit dual behavior.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 47

hc
Einstein’s generalization of Planck’s theory is given as, E = hν = λ
2
Einstein’s mass energy relationship is E = mc
hc 2 h h
Equating above two equations, we get, λ
= mc or λ
= mc or λ = mc
h
Where ‘c’ is the velocity of light. If the velocity of micro particle is ‘v’ then, λ = mV
This is de Broglie’s equation,
Where ‘λ’ is the de Broglie’s wave length, ‘m’ is the mass of the
moving particle and ‘h’ is Planck’s constant.
h
P = mv or λ = P
Here 𝜆 signifies wave nature and P signifies particle nature.
It is applicable to microparticles like electron, proton, etc.,
and not applicable for macro-bodies like cricket ball, bullet
etc.
The electron moving with high speed possesses both the particle nature and the wave
nature. The waves associated with material particles are known as matter waves or
particle waves.

ACTIVE SITE EDUTECH-9844532971

48 | XI Std. Chemistry

The Heisenberg’s uncertainty principle:

“It is impossible to determine simultaneously and accurately the exact position and
momentum or velocity of a sub-atomic particle like electron in an atom”.
One can determine the position of a particle very accurately, and then the
determination of its velocity becomes less accurate. Similarly, one can determine the
velocity of a particle very accurately, and then the determination of its position
becomes less accurate. The certainty in one factor introduces the uncertainty in
another factor.
If the uncertainty in the determination of the position of a small particle is given by Δx
and uncertainty in its momentum is Δp, then
ℎ
(Δx) (Δp) ≥ 𝑛𝜋, Where n = 1,2,3,4.........
For an electron revolving around the nucleus in an atom the value of n is nearly 4.
Thus Heisenberg’s principle can also be stated as the product of uncertainty in position
and momentum of an electron like micro particle moving with high speed cannot be less
than h/4.

Where m is the mass of the particle and Δv is uncertainty in velocity.

Where,

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 49

If the position of the particle is known exactly (Δx = 0), Δv becomes infinity (∞) and
vice versa. Heisenberg's uncertainty principle is not applicable to those objects which
cannot change their position by themselves when a light falls on them. It is applicable
for micro particles like electrons.

EXPLANATION OF HEISENBERG’S 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 the reflected
photon is seen in the microscope. As a result of the hitting, the position
as well as the velocity of the electron is distributed. The accuracy with
which the position of the particle can be measured depends upon the
wavelength of the light used. The uncertainty in position is +  . The
shorter the wavelength, the greater is the accuracy. But shorter
wavelength means higher frequency and hence higher energy. This high
energy photon on striking the electron changes its speed as well as
direction. But this is not true for macroscopic moving particle. Hence
Heisenberg’s uncertainty principle is not applicable to macroscopic
particles.

Significance of Heisenberg’s uncertainty principle:

Like de Broglie equation, although Heisenberg’s uncertainty principle holds good for all
objects but it is significance only for microscopic particles. The reason for this is quite
obvious. The energy of the photon is insufficient to change the position and velocity of
bigger bodies when it collides with them. For example, the light from a torch falling on
a running rat in a dark room, neither change the speed of the rat nor its direction, i.e.,
position.
This may be further illustrated with the following examples:
For a particle of mass 1 mg, we have
ℎ 6.625×10−34 𝑘𝑔𝑚2 𝑠 −1
Δx.Δ𝜐 = = = 10−28 𝑚2 𝑠 −1
4𝜋𝑚 4×3.1416×(10−6 𝑘𝑔)
Thus, the product of Δx and Δ𝜐 is extremely small. For particles of mass greater than 1
mg, the product will still smaller. Hence, these values are negligible.
For a microscopic particle like an electron, we have
ℎ 6.625×10−34 𝑘𝑔𝑚2 𝑠−1
Δx.Δ𝜐 = 4𝜋𝑚 = 4×3.1416×(9×10−31 𝑘𝑔) ≈ 10−4 𝑚2 𝑠 −1

ACTIVE SITE EDUTECH-9844532971

50 | XI Std. Chemistry

POINTS OF DISTINCTION BETWEEN A PARTICLE AND A WAVE

PARTICLE WAVE
1. A particle occupies a well-defined position 1. A wave is spread out in space, e.g., on
in space, i.e., a particle is localized in throwing a stone in a pond of water, the
space, e.g., a grain of sand, a cricket ball, waves start moving out in the form of
etc. concentric circles. Similarly, the sound of
the speaker reaches everybody in the
audience. Thus, a wave is delocalized in
space.
2. When a particular space is occupied by 2. Two or more waves can coexist in the
one particle, the same space cannot be same region of space and hence interfere.
occupied simultaneously by any other
particle. In other words, particles do not
interfere.
3. When a number of particles are present in 3. When, a number of waves are present in a
a given region of space, their total value is given region of space, due to interference,
equal to their sum, i.e., it is neither less the resultant wave can be larger or smaller
nor more. than the individual waves, i.e.,
interference may be constructive or
destructifve.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 51

DIFFERENCE BETWEEN ELECTROMAGNETIC WAVES AND MATTER WAVES

ELECTROMAGNETIC WAVES MATTER WAVES
1. The electromagnetic waves are associated 1. Matter waves are not associated with
with electric and magnetic fields, electric and magnetic fields.
perpendicular to each other and to the
direction of propagation.
2. They do not require any medium for 2. They require medium for their
propagation, i.e., they can pass through propagation, i.e., they cannot pass
vacuum. through vacuum.
3. They travel with the same speed as that of 3.
They travel with lower speeds.
light. Moreover, it is not constant for all matter
waves.
4. They leave the source, i.e., they are 4. They do not leave the moving particle,
emitted by the source. i.e., they are not emitted by the particle.
c h
5. Their wavelength is given by  = 5. Their wavelength is given by  =
v mv

CLASS EXERCISE
1. A ball of 100 g mass is thrown with a velocity of 100 ms–1. The wavelength of the
de Broglie wave associated with the ball is about

a) 6.63 × 10–35 m b) 6.63 × 10–30 m c) 6.63 × 10–35 cm d) 6.63 × 10–33 m

2. If kinetic energy of a proton is increased nine times the wavelength of the de-
Broglie wave associated with it would become
a) 3 times b) 9 times c) 1/3 times d) 1/9 times
3. Number of waves made by a Bohr electron in one complete revolution in the 3rd
orbit
a) 1 b) 2 c) 3 d) 4
4. The uncertainty in position and velocity of a particle are 10-10m and 5.27 x 10-24ms-1
respectively. Calculate the mass of mass of the particle.(h=6.625 10-34 J-s)
5. Calculate the uncertainty in velocity a cricket ball of mass 150g. if the uncertainty
0
in its positionis the order of 1 A (h=6.6 x 10-34kg m2 s-1)
6. In an atom, an electron is moving with a speed of 600 m sec-1 with an accuracy of
0.005% certainty with the position of the electron can be
located is:(h=6.6 x 10-34kg m2 s-1, mass of electron=9.1 x 10-31kg)
a) 1.52 x 10-4 m b) 5.1 x 10-3 m c) 1.92 x 10-3 m d) 3.84 x 10-3
HOME EXERCISE
1. The de Broglie wavelength of 1 mg grain of sand blown by a 20ms-1 wind is:
a) 3.3 x 10-29 b) 3.3 x 10-21 m c) 3.3  10-49 M d) 3.3  10-42 m

ACTIVE SITE EDUTECH-9844532971

52 | XI Std. Chemistry

2. If the kinetic energy of an electron is increased 4 times, the wavelength of the

Broglie wave associated with it would become:
1 1
a) 4times b) 2times c) times d) times
2 4
3. The momentum of the particle having the wave length of 1Å is
a) 6.6 x 10-19 gram cm/sec b) 6.6 x 1019 gram cm/sec
c) 6.6 x 1034 gram cm/sec d) 6.6 x 10-34 gram cm/sec
4. If the uncertainty in the position of an electron is 10-8cm, the uncertainty in its
velocity is
a) 3×108 cm/sec b) 5.8×107 cm/sec
c) 6.625×109 cm/sec d) 7.35 × 10-8 cm/sec
5. The uncertainty in momentum of an electron is 1 x 10-5 kg-m/s. The uncertainty in
its position will be (h = 6.6 x 10-34 Joule-sec)
a) 1.05 x 10-28m b) 1.05 x 10-26 m c) 5.27 x 10-30 m d) 5.25 x 10-28 m
6. Uncertainty in the momentum of an electron is 10-5kg.m/sec. The uncertainty in its
position will be
a) 1.05 x 10-28 m b) 1.05 x 10-26m c) 5.27 x 10-30m d) 5.25 x 10-25m
CLASS EXERCISE:
1)a 2)c 3)c 4)0.099kg 5)3.499×10–24 m/s–1 6)c

HOME EXERCISE:
1)a 2)c 3)a 4)b 5)c 6)c

Classical mechanics, based on Newton’s laws of motion, was successful in

explaining the motion of macroscopic bodies like falling stones or motion of planets
around the sun etc. But it failed when applied to microscopic particles like electrons,
atoms, molecules etc. Hence, new branch introduced called as ‘Quantum mechanics.

Schrodinger Wave Equation:

Quantum mechanics, as developed by Erwin Schrodinger is based on the wave motion
associated with the particles. The Schrodinger differential wave equation is given by
∂2 ψ ∂2 ψ ∂2 ψ 8π2 m
+ + ∂y2 + (E − V)ψ
∂x2 ∂z2 h2
Here x, y, z are Cartesian coordinates of the electron
m = mass of electron
h = Planck’s constant
E = total energy of the electron (KE + PE)
V = potential energy of the electron (PE)
ψ= wave function of the electron.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 53

Significance of ψ: ψ is the wave function. It gives the amplitude of the electron wave.

The intensity of light is proportional to the square of amplitude (ψ2). Just as 𝛙2

indicates the density of photons in space, 𝛙2 in case of electron wave denotes the
probability of finding an electron in the space or probability of finding the electron is
also maximum.\

ACTIVE SITE EDUTECH-9844532971

54 | XI Std. Chemistry

QUANTUM NUMBERS
The behavior of an electron in an atom is described mathematically by a wave function
or orbital. They are principal quantum number, azimuthal quantum number, magnetic
quantum number and spin quantum number.
‘Set of numbers used to describe energy, size, shape of orbitals in an atom’ called
as quantum numbers.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 55

1.Principal quantum number(n):

• ‘n’ can be any whole number value such as 1,2,3,4, etc. The energy shells
corresponding to these numbers are K, L, M, N, etc.
• Principal Quantum no. indicates the main energy level to which the
electron belongs. It also indicates the average distance of an
electron from nucleus and also the speed of the atomic electron.

• As the ‘n’ value increases the distance of electron from the nucleus increases and its
energy also increases.
• The maximum no. of electrons that can be present in an orbit is given by 2𝑛2 . The
maximum no. of electron in K, L, M, and N shells are 2,8,18 and 32 respectively.
0.529×𝑛2 o
• The radius of the orbit is given by the expression: rn = 𝑍
A.
−13.6×𝑍2
• The energy of the electron/orbit is given by the expression. En = 𝑛
cm/sec
8
2.18×10 ×𝑍
• The velocity of the electron is given by the expression. Vn = 𝑛
cm /sec.

2. Azimuthal Quantum Number:

• Azimuthal Quantum number was introduced by Sommerfeld’s to explain the fine
spectrum.
• It is also called as secondary quantum no. or orbital angular momentum quantum
number or subsidiary quantum number.
• It is denoted by l.
• ‘l’ can have the values from 0 to (n-1), a total of ‘n’ values. ‘l’ values 0,1,2,3 indicates
s,p,d,f. s,p,d and f are spectroscope terms which indicates sharp. Principle, diffuse
and fundamental respectively.

ACTIVE SITE EDUTECH-9844532971

56 | XI Std. Chemistry

• Azimuthal Quantum number indicates the sub-shell to which the electron belongs. It
also determines the shapes of the orbital in which the electron
n l is present.
• Each main energy shell can have ‘n’ number of sub-shells. 1 0 (1s)

• The orbital angular momentum (L) of an electron is given by

2 0 (2s), 1 (2p)
ℎ
the expression: L = √𝑙(𝑙 + 1) 3 0 (3s), 1 (3p), 2(3d)
2𝜋
4 0 (4s), 1(4p), 2(4d), 3(4f)
• As the value of 'n' increases
i) the total energy of electron increases
ii) the energy difference between the successive orbits decreases
iii) P.E increases and K.E. decreases
iv) radius of orbits increases
v) velocity of electron decreases
Orbital Value of n Value of l Value of (n + l)
1s 1 0 1+0=1
2s 2 0 2+0=2
2p (n = 2)
2p 2 1 2+1=3
Has lower energy than 3s (n = 3)
3s 3 0 3+0=3 3p (n = 3)
Has lower energy than 4s (n = 4)
3p 3 1 3+1=4
3d (n = 3)
4s 4 0 4+0=4 Has lower energy than 4p (n = 4)
3d 3 2 3+2=5
4p 4 1 4+1=5

Name of the
Principal Azimuthal
sub-
Shell Quantum Quantum Notation
stationary
number (n) number (l)
state
K 1 0 S 1s

L 2 0 S 2s
1 P 2p

M 3 0 s 3s
1 p 3p
2 d 3d

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 57

N 4 0 s 4s
1 p 4p
2 d 4d
3 f 4f

3. Magnetic Quantum number:

• Magnetic quantum number was introduced by Lande to explain Zeeman Effect.
• It is denoted by m or ml.

• This quantum number refers to different orientations of electron could in a

particular subshell. These orientations are called the orbitals.
• An electron due to its orbital motion around the nucleus generates an electric .This
electric field in turn produces a magnetic field which can interact with the external
magnetic field. Thus, under the influence of the external magnetic field, the
electrons of a subshell can orient themselves in certain preferred regions of space
around the nucleus called orbitals. The magnetic quantum number determines the
number of preferred orientations of the electron present in a subshell.
Since each orientation corresponds to an orbital, therefore, the magnetic
quantum number determines the number of orbitals present in any subshell.
• ‘m’ can have values from – 𝑙 to +𝑙 including zero, a total (2 𝑙+1) values.
Subshell 𝒍 m values No. of orientations (Orbitals)
s 0 0 1
p 1 -1, 0, +1 3
d 2 -2, -1, 0, +1, +2 5
F 3 -3, -2, -1, 0, +1, +2, +3 7
• When l = 0, m has only one value, m = 0. The sub-level‘s’ has one orbital called s
orbital.
• When l =1, m can have 3 values m = –1, 0, +1. The sub-level ‘p’ has three space
orientations or three orbitals. The three orbitals are designated as px, py and pz.

ACTIVE SITE EDUTECH-9844532971

58 | XI Std. Chemistry

• When l = 2, m can have 5 values m = –2,–1, 0, +1, +2. The sub-level ‘d’ has five space
orientations or five orbitals. The five orbitals are designated as dxy, dyz, dzx,
dx2 −y2 and dz2 .
• When l = 3, m can have 7 values m = –3,–2,–1,0,+1,+2,+3. The sub-level ‘f’ has seven
space orientations or seven orbitals.
The magnetic quantum number gives orientation of orbitals in space. All the orbitals
present in a sublevel have same energy and shape. They are called ‘degenerate orbitals’,
which differ in their spatial orientation.
• Each value of ‘m’ constitutes an orbital in the sublevel.
• Maximum no. of electrons in subshell: 2(2𝑙+1) or (4 𝑙+2).

4. Spin Quantum Number:

• Spin Quantum number was proposed by Uhlenbeck and Goudsmith.
• It is denoted by ‘s’ or ‘ms’.
• It indicates the direction of spinning of electron present in any orbital.
• Since the electron in an orbital can spin either in the clockwise direction or in anti-
clockwise direction, hence for a given value of m, s can have only two values, i.e.,
+1/2 and -1/2 or these are very often represented by two arrows pointing in the
opposite direction,
i.e.,↑and ↓.
If an orbital contains 2 electrons, the two magnetic moments oppose and cancel each
other.
Thus, in an atom, if all the orbitals are fully filled, net magnetic moment is zero and the
substance is diamagnetic (i.e., repelled by the external magnetic field). However, if
some half-filled orbitals are present, the substance has a net magnetic moment and is
paramagnetic (i.e., attracted by the external magnetic field).
• The spin angular momentum (𝜇s) of an electron is given by
h
μs = √s(s + 1) 2π

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 59

What will be the values of other quantum numbers of an electron for which the principal quantum
number is 2?

n l m Srms
2 0(s) 0 +1 / 2, −1 / 2

1(p) +1 +1 / 2, −1 / 2

0 +1 / 2, −1 / 2

+1 / 2, −1 / 2

ACTIVE SITE EDUTECH-9844532971

60 | XI Std. Chemistry

Summary 
Principal
Type of Azimuthual Magnetic Quantum Spin Quantum
Quantum
information Quantum No.(l) No.(m or ml) No.(s or ms)
No.(n)
To explain the
To explain the To explain the To explain the splitting
1. Why is it magnetic
main lines of a fine structure of of lines in a magnetic
required ? properties of
spectrum the line spectrum field
substances
No. of orbitals present Direction of
(i) Main shell in (i) No. of
2. What does in any subshell or the electron spin, i.e.,
which the subshells present
it tell ? number of orientations clockwise or anti-
electron resides in any main shell
of each subshell clockwise
(ii) Approx.
(ii) Relative
distance of the
energies of the
electron from
subshells
the nucleus
(iii) Energy of
the shell
(iv) Max. no. of (iii) Shapes of
electrons orbitals
present in the
shell (2n2)
Principal
Type of Azimuthual Magnetic Quantum Spin Quantum
Quantum
information Quantum No.(l) No.(m or ml) No.(s or ms)
No.(n)
3. What are
N l m or ml s or ms
the symbols?
For a particular For a particular value For a particular
4. What are 1, 2, 3, 4 etc., value of m,
value of n, l = 0 of l, m = – l to + l
the values? i.e., any integer
to n – 1 including zero S = + ½, –½
l = 0,s – subshell; Two arrows
l = 1,p – subshell; For p-subshell m = – 1, pointing in
5. Other
K, L, M, N etc. 0 + 1 dsignated as px, py opposite
designations ? l = 2,d – subshell; and pz directions, i.e., 
l = 3,f – subshell; and 

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 61

Atomic Orbital:
The three-dimensional space around the nucleus where the probability of finding
the electron is maximum is called an atomic orbital.

Differences between orbit and orbital:

Orbit Orbital
1. n orbit is a well-defined circular path 1. An 1. An orbital is the region of space around the
orbital is the region of space around the around the nucleus where the probability
around the nucleus in which the electron of finding the electron is maximum
revolves. (95%)
2. An orbit represents the movement of 2. An orbital represents the movement of
electron in one plane. electron in three dimensional spaces.
3. An orbit means the position as well as the 3. In an orbital it is not possible to find the
velocity of the electron can be known with position as well as velocity of the electron
Certainty. can be known with certainty.
4. Orbits are circular or elliptical shaped. 4. They have different shapes like spherical,
dumbbell etc
Orbitals have different shapes. s-orbital is
Spherical and p orbital is dumb bell shaped.
5. Orbits do not have directional 5. Except ‘s’ orbitals, all other orbitals have
characteristics. directional characteristics
6. An orbit can have a maximum number of2n2 6. An orbital can accommodate a maximum of
electrons. only two electrons.

Node- The three-dimensional space around the nucleus where the probability of
finding the electron is minimum or zero.

Types of Nodes:
Nodes are of two types: a) Radial Node b) Angular Node

ACTIVE SITE EDUTECH-9844532971

62 | XI Std. Chemistry

A radial node is the spherical region around then nucleus, where the probability if
finding the electron is zero (Ψ 2 = 0).

Similarly, nodal plane (angular plane) has zero probability of finding electron.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 63

Calculation of no. of nodes:

Exp: In a 3p -orbital
No. of Radial nodes = 3-1-1 = 1
No. of angular nodes = 1
Total no. of nodes = 2.

ACTIVE SITE EDUTECH-9844532971

64 | XI Std. Chemistry

NUMBER OF NODAL PLANES

Orbitals Number of nodal planes Nodal

s 0 Nil
px 1 YZ
py 1 ZX
pz 1 XY
dxy 2 YZ, ZX
dyz 2 ZX, XY
dzx 2 XY, YZ
d x2 - y2 2 YZ, ZX

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 65

Shapes of Orbitals:

• s –Orbitals: s- Orbital can accommodate electrons with l = 0 and these orbitals are
present in every orbit starting from 1st orbit.

Orbital in which e-s with

n=1 , l = 0 are present is called 1s - orbital.
All s-orbitals are spherical in shape and the size of sphere increases with ‘n’ value. s -
Orbitals are spherically symmetrical because the probability of finding the electron
around the nucleus is same in all directions.

• p – Orbitals:
p- Sublevel begins from 2nd orbit. For p -
sublevel l = 1, indicates that each p - sub level
contains three orbitals with ‘m’ values –1, 0, +1.
These are designated as px, py and pz, depending
on the axis in which electron density is present.
In px-orbital, electron density is concentrated along the x-axis.
p-Orbitals have dumb-bell shape. Each p -orbital has two lobes separated by one nodal
plane. The probability density function is zero on the plane where the two lobes touch
each other. The nodal planes for px, py and pz - orbitals are YZ, ZX and XY - planes
respectively.
The three orbitals present in a given p - sublevel will have same shape, size and energy
but different orientations (differ in m value). These three orbitals are perpendicular to
each other and the angle between any two p - orbitals is 90o.
For 2p orbital, number of nodes calculated as below

ACTIVE SITE EDUTECH-9844532971

66 | XI Std. Chemistry

For 3p orbital, number of nodes calculated as below

• d - Orbitals: begins from 3rd orbit (n = 3). For d- sub level l= 2, indicates that each
d - sublevel contains five orbitals with ‘m’ values –2, –1, 0, +1, +2. These are designated
as dxy,dyz,dzx, 𝑑𝑥 2 −𝑦2 and d𝑧2 .
All d-orbitals (except d𝑧2 ) have double dumb-bell shape. Each d-orbital has 4 lobes
separated by two nodal planes.
In case of dxy, dyz and dzxorbitals, lobes are present in between the corresponding
axes. i.e., between x and y axis in case of dxy orbital. Where as in d𝑥 2 −𝑦2 and d𝑧2
orbitals lobes are present along the axes. dxy Orbital contains yz and zx as nodal
planes. dyz and dzx contain (xy, zx) and (xy, yz) planes respectively. d𝑥 2 −𝑦2 orbital

contains two nodal planes perpendicular to each other and which make an angle of 45o
with respect to x and y axes. 𝑑𝑧2 orbital does not contain nodal planes.

For 3d orbital, number of nodes calculated as below

5d orbitals present in a given d- sublevel will have same energy in the ground state.

CLASS EXERCISE
1. If the above radial probability curve indicates ‘2s’ orbital, the distance between the
peak points X.Y is:

a) 2.07Å b) 1.59Å c) 0.53Å d) 1.1Å

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 67

2. The wave function curve which crosses ‘x’ axis maximum number of times in the
graph drawn between distance from nucleus r(on x axis) and radial wave function R(ψr).
a)4d b)4p c)4s d)4f

3. The number of nodal planes is greatest for the orbital:

a) 4s b) 2p c) 3d d) 2s

4. The radial distribution curve of the orbital with double dumbbell shape in the 4th
principle shell consists of ‘n’ nodes, n is
a) 2 b) 0 c)1 d) 3
5. Which one of the following sets of quantum numbers represents as impossible
arrangements?
n 1 m s
a) 3 2 –2 ½
b) 4 0 0 ½
c) 3 2 –3 ½
d) 5 3 0 –1/2.
6. Correct set of four quantum numbers for the valence (outermost) electron of rubidium
(Z = 37) is
1 1 1 1
a) 5, 0, 0, + b) 5, 1, 0, + c) 5, 1, 1, + d) 6, 0, 0, +
2 2 2 2
7. The maximum number of electrons in an orbital having same spin quantum number
will be:
a) l + 2 b) 2l + 1 c) l(l + 1) d) l (l + 1)
1
8. The four quantum number of last electron of an atom are 4, 0, 0, + then atomic
2
number of that element could be
a) 19 b) 55 c) 36 d) 37
9. The number of atomic orbitals with quantum numbers n = 3, l = 1, m = 0
a) 1 b) 6 c) 3 d) 5
10. The number of electrons that can have n = 4 andl = 3 is
a) 10 b) 14 c) 6 d) 5

HOME EXERCISE
1. Which of the following can be negative?
a) 4πr2ψ2 b) 4πr2ψ2dr c) ψ d) ψ2
2. The quantum number not obtained from the Schrodinger’s wave equation is
a) n b) l c) m d) s
3. Maxima’s in Radial probability distribution curve of 2s is
a) One b) Two c) Three d) Four
4. In which of following case would the probability of finding an electron in dxyorbital be
zero?
a) Xy and yz plane b) xy and planes
c) xz and yz planes d) z-direction, yz and xz planes

ACTIVE SITE EDUTECH-9844532971

68 | XI Std. Chemistry

5. The principal quantum number of an atom is related to the

a) Size of the orbital b) spin angular momentum
c) Orbital angular momentum d) orientation of the orbital in space
6. The orbital angular momentum of an electron in 2s orbital is:
1 h h h
a) . b) zero c) d) 2.
2 2 2 2
7. What will be all 4-Sets of Quantum Number for last electron of sodium?
a) n = 3 1=0 m=0 s = +1/2
b) n = 3 1=1 m=1 s = +1/2
c) n = 2 1=0 m=0 s = +1/2
d) n = 2 1=1 m=1 s = +1/2
8. p-orbitals of an atom in presence of magnetic field are:
a) Three fold degenerate b) Two fold degenerate c) Non-degenerate d)
none of these
9. The quantum number that is no way related to an orbital
a) principalb) azimuthal c) magnetic d) spin
10. Which one of the following set of quantum number is not possible for a 4pelectron?
1 1
a) n = 4 l = 1, m = +1 s = + b) n = 4 l = 1 m = 0 s = +
2 2
1 1
c) n= 4, l = 0, m = 2, s = + d) n = 4, l = 1, m = 1, ms = -
2 2

CLASS EXERCISE:
1)a 2)c 3)c 4)c 5)c 6)a 7)b 8)a 9)a 10)b

HOME EXERCISE:
1)c 2)d 3)b 4)c 5)a 6)b 7)a 8)a 9)d 10)c

ENERGY OF ORBITALS
The energy of an electron in a hydrogen atom is determined only by the principal
quantum number. Within a shell, all hydrogen orbitals have the same energy,
independent of the other quantum numbers.
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f
Although the shapes of 2s and 2p orbitals are different, an electron has the same
energy when it is in 2s orbital or 2p orbital. The energy of an electron in a
multielectron atom depends, not only on its principal quantum number, but also on its
azimuthal quantum number. The s, p, d and f orbitals within a given shell have slightly
different energies in a multi electron atom.

Electronic configuration of multi electron atoms:

The distribution and arrangement of electrons in the main shells, subshells and
orbitals of an atom is called electronic configuration of the element.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 69

• 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 after the lower energy orbitals are filled.
The relative energy of an orbital is given by
(n + l ) rule or Bohr-Burry’s rule. As (n + l) value increases, the energy of orbital
increases.
• The orbital with the lowest (n + l) value is filled first.

• When two or more orbitals have the same (n +l) value, the one with the lowest
‘n’ value (or) highest l value is preferred in filling.

Exp- Consider two orbitals 3d and 4s.

n+l value of 3d = 3 + 2 = 5 and of 4s = 4 + 0 = 4. Since
4s has lowest (n +l) value, it is filled first before filling
taking place in 3d.
Consider the orbitals 3d, 4p and 5s
The (n + l) value of 3d = 3 + 2 = 5
The (n +l) value of 4p = 4 + 1 = 5
The (n +l) value of 5s = 5 + 0 = 5

ACTIVE SITE EDUTECH-9844532971

70 | XI Std. Chemistry

These three values are same. Since the ‘n’ value is lower to 3d orbitals, the electrons
prefer to enter in 3d, then 4p and 5s.

The order of increasing energy of atomic orbitals is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s
and so on.
The sequence in which the electrons occupy various orbitals can be easily remembered
with the help of Moeller’s diagram as shown in above Fig.
Sub-shell n l (n+l)

1s 1 0 1

2s 2 0 2

2p 2 1 3 Lowest value of n

3s 3 0 3

3p 3 1 4 Lowest value of n

4s 4 0 4

3d 3 2 5 Lowest value of n

4p 4 1 5

5s 5 0 5

4d 4 2 6 Lowest value of n

5p 5 1 6

6s 6 0 6

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 71

4f 4 3 7 Lowest value of n

5d 5 2 7

6p 6 1 7

7s 7 0 7

5f 5 3 8 Lowest value of n

6d 6 2 8

7p 7 1 8

• Pauli’s Exclusion principle: stated as “No two electrons in an atom can have the
same set of values for all the four quantum numbers”. This means that two
electrons in an orbital may have the same n, same l and same m but differ in spin
quantum number. In an orbital if one electron has clockwise spin, the other has
anticlockwise spin. It follows that an orbital can hold a maximum of two electrons
with opposite spins.
Exp- helium atom has two electrons in its 1s orbital. Quantum numbers for first
electron are n =1, l = 0, m = 0 and s = +1/2. Quantum numbers for second electron are: n
=1, l = 0, m =0, s = –1/2.
The two electrons have the same value for n, same value for l and same value for m but
differ in s.

ACTIVE SITE EDUTECH-9844532971

72 | XI Std. Chemistry

• Hund’s rule of maximum multiplicity:

According to this rule, when electrons are filled in degenerate orbitals of a
subshell, pairing of an electron takes place only when each orbital of the subshell
is filled with one electron each. It can be also stated that, in ground state of an
atom, the configuration which has more number of unpaired electrons is most
stable.

Thus in s, p, d and f subshells, pairing starts from 2nd, 4th, 6th and 8th electrons
respectively.
Ex: Electronic configuration of N (7) is 1s2 2s2 2p3.
The electrons in 2p subshell are occupied sing ally. i.e., 1s2 2s2 2𝑝𝑥1 2𝑝𝑦1 2𝑝𝑧1

Electronic configuration of elements from 1 to 30 (up to Zinc)

Electronic configuration
Atomic Valence Electronic
Element In terms of inert
number In terms of Orbitals configuration
gases
H 1 1s1 1s1 1s1
He 2 2s2 1s2 1s2
Li 3 1s22s1 [He]2s1 2s1
Be 4 1s22s2 [He]2s2 2s2
B 5 1s22s23p1 [He]2s22p1 2s22p1

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 73

C 6 1s22s22p2 [He]2s22p2 2s22p2

N 7 1s22s22p3 [He]2s22p3 2s22p3
O 8 1s22s22p4 [He]2s22p4 2s22p4
F 9 1s22s22p5 [He]2s22p5 2s22p5
Ne 10 1s22s22p6 [He]2s22p6 2s22p6
Na 11 1s22s22p63s1 [Ne]3s1 3s1
Mg 12 1s22s22p63s2 [Ne]3s2 3s2
Al 13 1s22s22p63s23p1 [Ne]3s23p1 3s23p1
Si 14 1s22s22p63s23p2 [Ne]3s23p2 3s23p2
P 15 1s22s22p63s23p3 [Ne]3s23p3 3s23p3
S 16 1s22s22p63s23p4 [Ne]3s23p4 3s23p4
Cl 17 1s22s22p63s23p5 [Ne]3s23p5 3s23p5
Ar 18 1s22s22p63s23p6 [Ne]3s23p6 3s23p6
K 19 1s22s22p63s23p64s1 [Ar]4s1 4s1
Ca 20 1s22s22p63s23p64s2 [Ar]4s2 4s2
Sc 21 1s22s22p63s23p64s23d1 [Ar]4s23d1 4s23d1
Ti 22 1s22s22p63s23p64s23d2 [Ar]4s23d2 4s23d2
V 23 1s22s22p63s23p64s23d3 [Ar]4s23d3 4s23d3
Cr 24 1s22s22p63s23p64s13d5 [Ar]4s13d5 4s13d5
Mn 25 1s22s22p63s23p64s23d5 [Ar]4s23d5 4s23d5
Fe 26 1s22s22p63s23p64s23d6 [Ar]4s23d6 4s23d6
Co 27 1s22s22p63s23p64s23d7 [Ar]4s23d7 4s23d7
Ni 28 1s22s22p63s23p64s23d8 [Ar]4s23d8 4s23d8
Cu 29 1s22s22p63s23p64s13d10 [Ar]4s13d10 4s13d10
Zn 30 1s22s22p63s23p64s23d10 [Ar]4s23d10 4s23d10

ACTIVE SITE EDUTECH-9844532971

74 | XI Std. Chemistry

Stability of atoms
Extra stability is associated with atoms in which degenerate orbitals are either half-
filled or completely filled due to

(1) Symmetrical distribution of electrons

(2) Exchange energy. Greater the exchange energy greater is the stability.
The presence of half-filled and completely filled degenerate orbitals gives greater
stability to atoms.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 75

ACTIVE SITE EDUTECH-9844532971

76 | XI Std. Chemistry

It is for this reason the electronic configurations of Cr and Cu are represented as [Ar]
1 5 1 10
4s 3d and [Ar] 4s 3d respectively.

CLASS EXERCISE
1. When 3d-orbital is complete, the newly entering electron goes into:
a) 4f b) 4s c) 4p d) 4d
2. An electron will have the highest energy in the set:
a) 3, 2, 1, 1/2 b) 4, 2, –1, 1/2 c) 4, 1, 0, –1/2 d) 5, 0, 0, ½

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 77

3. Which has minimum number of unpaired d -electrons?

3+
a) Fe 3+ b) Co 3+ c) Co 2+ d) Mn
4. Which of the following is violation of Pauli’s exclusion principle?
2s 2p 2s 2p 2s 2p 2s 2p

a) b) c) d)
5. The number of electrons in the M shell of an element with atomic number 24 is
a) 24 b) 12 c) 13 d) 8

HOME EXERCISE
1. The maximum number of unpaired electrons present in 4f -energy level is:
a) 5 b) 7 c) 10 d) 6
2. The number of unpaired electrons in fluorine atom is:
a) 7 b) 5 c) 1 d) 2
3. Which set has the same number of unpaired electrons in their ground
state?
a) N, P, V b) Na, P, Cl c) Na + , Mg 2+ , Al d) Cl – , Fe 3+ , Cr 3+
4. In which of the following electron distributions in ground state, only the Hund’s rule
is violated
2s 2p 2s 2p 2s 2p 2s 2p

a) b) c) d)
5. Electronic configuration of Ni is [Ar] 3d 8 , 4s 2 . The electronic
configuration of next element is:
a) [Ar] 3d 10 , 4s 1 b) [Ar] 3d 9 , 4s 2
c) [Ar] 3d 8 , 4s 2 , 4p 1 d) none of these

CLASS EXERCISE:
1)c 2)b 3)a 4)b 5)c

HOME EXERCISE:
1)b 2)c 3)a 4)a 5)a

Exercise 1
1. The heaviest particle among the following is
(A) meson (B) proton (C) neutron (D) electron

Ans (C)

2. The mass of an electron is

(A) 9.1  10–10 g (B) 9.1  10–18 g (C) 9.1  10–25 g (D) 9.1  10–28 g

Ans (D)

3. The mass of a proton is in the order of

ACTIVE SITE EDUTECH-9844532971

78 | XI Std. Chemistry

(A) 10–23 kg (B) 10–24 kg (C) 10–26 kg (D) 10–27 kg

Ans (D)

4. Compared to mass of lightest nucleus, the mass of an electron is only

1 1 1 1
(A) (B) (C) (D)
80 360 1840 1000

Ans (C)

5. The size of nucleus is measured in

(A) amu (B) Angstrom (C) Fermi (D) cm

Ans (B)

6. The charge on one mole of electrons is the same as

(A) Faraday (B) Coulomb (C) Ampere (D) unit charge

Ans (A)

7. Nucleons consist of
(A) protons + electrons (B) neutrons + electrons

(C) protons + nucleus (D) protons + neutrons

Ans (D)

8. An element X with atomic number 13 has a mass number 27, then the number of electrons,
protons and neutrons present in the atom are respectively
(A) 13, 13, 27 (B) 14, 13, 13 (C) 13, 14, 13 (D) 13, 13, 14

Ans (D) Atomic number = number of protons = number of electrons = 13

9. An atom has 26 electrons, its atomic mass is 56, the number of neutrons in the nucleus of the
atom will be
(A) 26 (B) 30 (C) 35 (D) 56

Ans (B)

10. The charge of the atom containing 17 protons, 18 neutrons and 18 electrons is
(A) +1 (B) –1 (C) –2 (D) zero

Ans (B)

11. The atomic number of an element is equal to

(A) sum of the protons and neutrons (B) sum of the protons and
electrons

(C) number of neutrons (D) number of protons

Ans (D)

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 79

12. A species X contains 9 protons, 10 electrons, 11 neutrons. Then it is

(A) a neutral atom (B) an anion (C) an isotope (D) a cation

Ans (B) Number of electrons is more than number of protons. Hence it is an anion.

13. The number of neutrons in dipositive zinc ion with mass number 70 (Z = 30) is
(A) 34 (B) 36 (C) 40 (D) 38
2+
Ans (C) For Zn (Z = 30)  number of neutrons = A − Z = 70 − 30 = 40

14. The mass number of an element is 23 and atomic number is 11. The number of protons,
electrons and neutrons, respectively are
(A) 11, 11, 12 (B) 12, 12, 11 (C) 11, 12, 11 (D) 12, 11, 12

Ans (A)

15. Electromagnetic radiation with minimum wavelength is

(A) ultra violet (B) radio wave (C) X-ray (D) infra red

Ans (C)

16. Electromagnetic radiation with maximum wavelength is

(A) ultra violet (B) radio wave (C) X-ray (D) infra red

Ans (B)

17. Band spectrum is given by

(A) gaseous ions (B) gaseous atoms (C) gaseous molecules (D) radicals

Ans (C)

18. The spectrum obtained from visible light is

(A) continuous spectrum (B) band spectrum

(C) non-continuous spectrum (D) line spectrum

Ans (A)

19. The spectral lines of Lyman series lie in

(A) IR region (B) far IR region (C) UV region (D) visible region

Ans (C)

20. Lines in the Balmer series of hydrogen atom lie in

(A) I R region (B) visible region (C) UV region (D) far infrared
region

Ans (B)

21. The series of lines in the visible region of the hydrogen spectrum is called
(A) Lyman series (B) Balmer series (C) Pfund series (D) Paschen series

ACTIVE SITE EDUTECH-9844532971

80 | XI Std. Chemistry

Ans (B)

22. When an electron jumps from L shell to M shell,

(A) energy is emitted (B) energy is absorbed (C) X-rays are emitted (D) -rays are
emitted

Ans (B)

23. The fourth line in the Balmer series of hydrogen spectrum corresponds to which one of the
following electronic transition?
(A) 3 → 1 (B) 5 → 1 (C) 5 → 2 (D) 6 → 2

Ans (D)

24. The wave number of the light emitted by a certain source is 2  106 m−1. The wavelength of this
light is
(A) 500 nm (B) 500 m (C) 200 nm (D) 5  107 m

Ans (A)

1 1 1
 = ,= = = 0.5  10−6 m
  2  10 6
 = 5  10−7 m = 500  10−9 m = 500 nm

25. With increasing quantum number, the energy difference between adjacent orbits of hydrogen
atom
(A) increases (B) decreases

(C) remains constant (D) increases initially and then decreases

Ans (B)

26. Value of n1 in Balmer series is

(A) 2 (B) 1 (C) 3 (D) 0
Ans (A)
27. The transition of electron in the hydrogen atom that emits radiation of higher energy is from
(A) n = 2 to n = 1 (B) n = 3 to n = 2 (C) n = 2 to n = 3 (D) n = 1 to n = 2

Ans (A) The energy difference between n = 2 to n = 1 is more.

28. Which electronic level would allow the hydrogen atom to absorb energy but not emit energy?
(A) 3s (B) 2p (C) 2s (D) 1s

Ans (D) Electron in 1s level can only absorb energy. It cannot emit energy. It is the ground state.
29. For the second line of the Balmer series, the value of n1 and n2 are respectively
(A) 2 and 3 (B) 1 and 2 (C) 2 and 4 (D) 1 and 3

Ans (C) For Balmer series value of n1 = 2, n2 = 3, 4, 5, 6

For the second line value of n2 = 4

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 81

30. The quantum number which specifies the size of the atomic orbital and also the energy of the
electron is
(A) principal quantum number (B) azimuthal quantum number

(C) magnetic quantum number (D) spin quantum number

Ans (A)
31. How many orbitals are possible in 3rd main energy level?
(A) 4 (B) 3 (C) 9 (D) 14

Ans (C)
n = 3; l = 0, 1, 2
l Number of orbital (2l + 1)
0 1
1 3
2 5
Total number of orbitals = 1 + 3 + 5 = 9

Or

Number of orbitals in any energy level is given by n2

 Number of orbitals in 3rd energy level = 32 = 9.

32. The angular momentum of an electron in 2s orbital is

1 h h h
(A) + (B) zero (C) (D) 2
2 2 2 2

Ans (B)
h
For 2s orbital l = 0, angular momentum = l (l + 1)  =0
2

33. The difference in angular momentum associated with an electron in two successive orbits of
hydrogen atom is
nh h h h
(A) (B) (C) (D) ( n − 1)
2 2 2 2
Ans (B)
nh
mvr =
2
2h 1h 1h h
now ( mvr ) 2 − (mvr )1 = − = or
2 2 2 2
34. If uncertainty in position and velocity are equal, then uncertainty in momentum will be
1 mh 1 h h mh
(A) (B) (C) (D)
2  2 m 4m 4
Ans (A)
h h
x . p = or x . v =
4 4m

ACTIVE SITE EDUTECH-9844532971

82 | XI Std. Chemistry

h h
if x = v , then v 2 =  v =
4m 4m
h 1 mh
p = m . v = m =
4m 2 
35. The shortest wavelength transition in Balmer series of atomic hydrogen is
o o o o
(A) 4215 A (B) 3942 A (C) 1437 A (D) 3647 A
Ans (D)
1 1 1
 = = RZ 2  2 − 2 
  n1 n 2 
For the shortest wavelength in Balmer series; n 2 = 
1 1 1 
= 109678  1 2 − 2 
 shortest 2  
1 109678
= cm −1
 4
4 4 4  10 10 o
= = m or = = 3647 A
109678 109678  100 109678  100
36. What is the wavelength of radiation emitted producing a line in the Lyman series when an
electron falls from fourth stationary state in a hydrogen atom? [RH = 1.1  107 m−1]
(A) 96.97 nm (B) 969.7 nm (C) 9.697 nm (D) none of 1, 2, 3
Ans (A)
Use Rydberg’s formula.
37. Wavelength of a photon with energy 1 electron volt is
o o o o
(A) 12420 A (B) 1242 A (C) 124.2 A (D) 1.24 A
Ans (A)
Energy = 1 eV = 1.602  10−19 J
hc 6.626  10 −34 Js  3  10 8 m s −1 o
= = −19
= 12420 A
E 1.602  10 J
38. Number of lines observed in the visible region of the hydrogen spectrum when an electron
jumps from 5th orbit to ground state is / are
(A) 1 (B) 3 (C) 5 (D) 6
Ans (B)

5
4
3
2

1
39. An electron of velocity ‘v’ is found to have a certain value of de Broglie wavelength. The velocity
possessed by neutron to have the same de Broglie wavelength is
v 1840
(A) 1820 v (B) (C) 3 v (D)
1840 v
Ans (B)
Since the mass of the proton is 1840 times the mass of an electron.

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 83

40. The wavelength of the radiation emitted, when an electron falls from infinity to energy level 1 in
an hydrogen atom is (R = 1.097  107 m−1)
(A) 406 nm (B) 192 nm (C) 91 nm (D) 9.1  10−8 nm

Ans (C)
1 1 
 = R  2 − 2  = R = 1.097  107 m−1
1  

1 1
= = = 91 nm
 1.097  10 7

41. If the Planck’s constant, h = 6.6  10−34 Js, then the de-Broglie wavelength of a particle having
momentum of 3.3  10−24 kg m s−1 will be
(A) 0.02 Å (B) 0.5 Å (C) 2 Å (D) 500 Å

Ans (C)
h 6.6  10 −34
= = = 2Å
p 3.3  10 − 24

42. The de-Broglie wavelength of a golf ball weighing 0.2 kg and moving with a speed of 1.4  10−3 m
s−1 is about
(A) 2.35  10−30 m (B) 1.8  10−30 m (C) 4.6  10−28 m (D) 3.31  10−20 m

h 6.6  10 −34
Ans (A)  = = −3
= 2.35  10 −30 m
mv 0.2  1.4  10
43. The maximum number of electrons in the nth main energy level is
(A) n2 (B) 2n2 (C) 3n2 (D) 4n2

Ans (B)

44. The de-Broglie equation treats an electron as

(A) a particle (B) wave (C) both (A) and (B) (D) neither (A) or (B)

Ans (C)

45. It is impossible to know simultaneously the exact position and momentum of a moving particle
at any instant. This is known as
(A) Aufbau principle (B) Hund’s rule

(C) Heisenberg’s principle (D) Pauli’s exclusion principle

Ans (C)

46. Which of the following statements is false with respect to an electron?

(A) It is a particle (B) It has wave property

(C) Its motion is affected in the magnetic field (D) It loses energy while moving in orbitals

Ans (D)

ACTIVE SITE EDUTECH-9844532971

84 | XI Std. Chemistry

47. Electrons prefer to occupy a subshell which has lower (n + l) value. This is called
(A) Hund’s rule (B) Pauli’s principle (C) uncertainty principle (D) aufbau principle

Ans (D)

48. The valence shell of calcium contains

(A) 2 electrons (B) 4 electrons (C) 6 electrons (D) 3 electrons

Ans (A) Calcium is an alkaline earth metal, hence valence shell contains 2 electrons, (ns 2).
49. In potassium, the order of energy level is
(A) 3s, 3d (B) 3p, 4s (C) 4s, 4p (D) 4s, 3d

Ans (B) K (Z = 19) 1s2 2s2 2p6 3s2 3p6 4s1

50. An ion which has 18 electrons in the outermost shell is
(A) Cu+ (B) Fe3+ (C) Cs+ (D) K+

Ans (A) Cu, Z = 29 has the electronic configuration, 1s2 2s2 2p6 3s2 3p2 4s1 3d10

Cu + : 1s2 2s2 2p6 3s2 3p6 3d10

Cu+ has 18 electrons in the outermost shell.

51. Two electrons of any orbital are distinguished by

(A) spin quantum number (B) principal quantum number

(C) azimuthal quantum number (D) magnetic quantum number

Ans (A)

52. In the following electronic configuration, some rules have been violated
(I) Hund (II) Pauli’s exclusion (III) aufbau

4p4

4s1

3d5
(A) I and II (B) I and III (C) II and III (D) I, II and III

Ans (D)

53. Which of the following arrangements of electrons is most likely to be stable?

3d 4s 3d 4s
(A) (B)

(C) (D)

Ans (A)

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 85

Interelectronic repulsions are minimum.

54. Number of electrons that F (Z = 9) has in p-orbital is equal to,

(A) number of electrons in s-orbitals in Na (11 e−)

(B) number of electrons in d-orbitals in Fe3+ (23 e−)

(C) number of electrons in d-orbitals in Mn (25 e−)

(D) as in 1, 2 and 3

Ans (D)

55. If aufbau rule is not followed, the 19th electron in Sc (Z = 21) will have
(A) n = 3, l = 0 (B) n = 3, l = 1 (C) n = 3, l = 2 (D) n = 4, l = 0

Ans (D)

Sc (Z = 21)  [Ar]18 4s2 3d1 ; 19th electron enters the 4s orbital

For the 4s orbital, n = 4; l = 0.

56. The shape of atomic orbitals is indicated by

(A) magnetic quantum number (B) spin quantum number

(C) principal quantum number (D) subsidiary quantum number

Ans (D) Azimuthal quantum number is also known as subsidiary quantum number.

57. The d-orbital that does not have four lobes is

(A) d x 2 − y 2 (B) dxy (C) dyz (D) d z 2

Ans (D)

58. For a dumb-bell shaped orbital, the value of ‘l’ is

(A) 0 (B) 1 (C) 2 (D) 3

Ans (B)

59. The maximum number of electrons present in any orbital is

(A) 2 (B) 8 (C) 18 (D) 32

Ans (A)

60. The maximum number of electrons that can be accommodated in a p-orbital is

(A) 2 (B) 6 (C) 3 (D) 4

Ans (A)
The maximum number of electrons in any orbital is 2.

ACTIVE SITE EDUTECH-9844532971

86 | XI Std. Chemistry

Exercise 2
1. For how many orbitals, the quantum numbers n = 3, l = 2, m = +2 are possible?
(A) 1 (B) 2 (C) 3 (D) 4

Ans (A) n = 3, l = 2 represent 3d subshell. M = +2 represents one particular orbital of this

subshell.

2. An element M has an atomic mass 19 and atomic number 9. Its ion is represented by
(A) M+ (B) M2+ (C) M– (D) M2–

Ans (C) The element is F. Its ion is F–, i.e., M–

(E.C. of M = ns2 np5. It will gain one electron to acquire. Stable configuration)

3. The correct order of increasing energy of atomic orbital’s is?

(A) 5p < 4f < 6s < 5d (B) 5p < 6s < 4f < 5d

(C) 4f < 5p < 5d < 6s (D) 5p < 5d < 4f < 6s

Ans (B) Energies are : 5p < 6s < 4f < 5d

4. Chloride ion and potassium ion are isoelectronic then?

(A) their sizes are same

(B) Chloride ion is bigger than potassium ion

(C) potassium ion is relatively bigger

(D) depends upon the other cation or anion

Ans (B) Cl– ion has 17 protons in the nucleus while K+ has 19 protons. Thus, nuclear charge of K+
is greater and bence its size is smaller.

5. In hydrogen atom, energy of first excited state is –3.4 eV. Then find out the KE of the same orbit
of hydrogen atom
(A) +3.4eV (B) +6.8eV (C) –13.6eV (D) +13.6eV

Ans (A) Energy in the excited state is nothing but kinetic energy.
6. In the Bohr’s orbit, what is the ratio of total Kinetic energy and total energy of the electron?
(A) –1 (B) –2 (C) +1 (D) +2
Ans (A)
7. The value of planck’s constant is 6.63  10–34J.s. The velocity of light is 3.0 108 ms–1. Which
value is closest to the Wavelength in nanometers of a quantum of light with frequency of 8 
1015 s–1?
(A) 2  10–25 (B) 5  10 –18 (C) 4  101 (D) 3  107
Ans (C)
c 3  108 ms−1
= = = 0.375  10−7 m
 8  1015 s−1
= 37.5 10−9 m = 37.5nm

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 87

 4.0  10nm

8. A proton is about 1840 times heavier than an electron. When it is accelerated by a potential
difference of 1 kV, its kinetic energy will be
(A) 1840 ke V (B) 1/1840 ke V (C) 1 ke V (D) 920 ke V

Ans (C) Energy enquired by the electron (as K.E.) on being accelerated by a potential difference
of 1 kV = 1Ke V

As proton has the same charge and energy = potential difference  charge, hence proton will
have same kinetic energy.

9. For Principal quantum number n = 4, the total number of orbital’s having l = 3 is

(A) 3 (B) 5 (C) 7 (D) 9
Ans (C)

For n = 4, E.C. = 1s2 2s2 2p6 3s2,

3p6 3d10 4s2 4p6 4d10 4f14

L = 3 means f subshell which has 7 orbitals.

10. The frequency of the radiation emitted when the electron falls from n 4 to n = 1 in a hydrogen
atom will be (given ionization energy of H = 2.18  10–18 J atom–1 and h = 6.625  10–34 Js)
(A) 1.54  10–15s–1 (B) 1.03  1015 Js–1 (C) 3.08  1015 Js–1 (D) 2.0  1015 s–1
Ans (C)

I.E. = E − E1 = 0 − E1z

= 2.18  10–18 J atom–1

2.18  10−18
Thus, E n = − j atom −1
n2

 1 1
E = E4 − E1 = −2.18  10−18  2 − 2 
4 1 

= 2.044 10−18 j atom−1

E 2.044  10−18 J
E = hv or v = =
h 6.625  10−34 Js

= 3.085  1015s–1.

11. Among the following transition metal ions, the one where all the metal ions have 3d2 electronic
configuration is
(A) Ti3+ , V 2 +,Cr 3+ , Mn 4+ (B) Ti 4 +, V 4 +,Cr 6 +, Mn 7+

(C) Ti 4 +, V 3 +,cr 2 +, Mn 3+ (D) Ti 2 +, V3+ ,Cr 4 + , Mn 5+

ACTIVE SITE EDUTECH-9844532971

88 | XI Std. Chemistry

Ans (D)

Ti 2 +, V 3+ ,Cr 4 + and Mn5+ all have 3d2 electronic configuration.

12. The radius of the area covered by second orbital to the first orbital is
(A) 1 : 1 (B) 1 : 16 (C) 8 : 1 (D) 16 : 1

Ans (D)

rn  n 2

 Area covered , (An  n4) ( Area = r2)

A 2 24 16
 = =
A1 14 1

13. Correct order of radii is

(A) N  Be  B (B) F−  O2−  N3− (C) Na  Li  K (D)
3+ 4+
Fe  Fe  Fe 2

Ans F–, O2−, N3– are isoelectronic. Greater the nuclear charge, greater is the attraction on the
electrons, smaller is the size.

14. The energy of the second bohr orbit of the hydrogen atom is – 238 kJ mol–1 ; hence the energy of
fourth bohr orbit would be
(A) –1312 kJ mol–1 (B) –82 kJ mol–1 (C) – 41 kJ mol–1(D) –164 kJ mol–1

Ans (B)

k
E2 = − = −328 kJ mol −1
22

k = 1312 kJ mol −1

k 1312
E4 = − 2
=− = −82 kJ mol −1
4 16

15. The isoelectronic pair is

(A) Cl2O,ICl2− (B) ICl2− ,ClO2 (C) IF2+ ,I3− (D) ClO2− ,CIF2+

Ans (D)

Electrons in CIO2− = 17 + 2  8 + 1 = 34

Electrons in ClF2 = 17 + 9  2 −1 = 34

16. The most probable radius (in pm) for finding the electron in He+ is
(A) 100.2 (B) 52.9 (C) 26.5 (D) 105.8

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 89

Ans (C)

52.9  12
rn = pm
2

52.9  12
 For He + , r1 = = 26.5 Pm.
2

17. Given : The mass of electron is 9.11  10-31 kg, planck constant is 6.626  10–34 Js, the uncertainty
involved in the measurement of velocity within a distance of 0.1 Å is
(A) 5.70  10−8 ms–1 (B) 5.79  10−5 ms–1

(C) 5.79  10−6 ms–1 (D) 5.79  10−7ms–1

Ans (C)

x.m  v = h / 4

 ( 0.1  10−10 m )( 9.11  10−31 kg ) ( v )

0.626  10−34 kg m2 s−1

=
4  3.14

18. The de-Broglie wavelength associated with a ball of mass 1 kg having kinetic energy 0.5 J is
(A) 6.626  10–34 m (B) 13.20  10–34m

(C) 10.38  10–21m (D) 6.626  10–34Å

Ans (A)

1
mv 2 = 0.5J
2

1
 1kg  v 2 = 0.5J
2

or v 2 = 1 or v = 1m s−1

h 6.626 10−34 kg m2 s −1
= =
mv 1kg  1ms−1

= 6.626  10-34 m.

19. If the uncertainty in position and momentum are equal, then uncertainty in velocity is
1 h h 1 h h
(A) (B) (C) (D)
2m  2 m  

Ans (A)

ACTIVE SITE EDUTECH-9844532971

90 | XI Std. Chemistry

h h
.if x = p, then ( p ) =
2
x.p =
4 4x

h h
Or p = ,id., m  =
4 4

1 h 1 h
Or Gn = =
m 4 2m 

20. The measurement of the electron position is associated with an uncertainty in momentum which
is equal to 1  10–18 g cm s–1. The uncertainty in electron velocity is, (mass if an electron is 9  10–
28
g)
(A) 1  109 cm s–1 (B) 1  106 cm s–1 (C) 1  1011 cm s–1 (D) 1  1012 cm s–1

Ans (B)

p = m  

 1  10–18 g cms–1 = 9  10–28 g  

Or v = 1  10 9 cms–1

21. Among the following, the metal that cannot be used for photoemission is
(A) potassium (B) rubidium (C) cesium (D) magnesium

Ans (D)

22. Which of the following graphs explains photoelectric effect?

KE KE KE KE

Frequency (ν) Frequency ν Frequency ν Frequency ν

(A) (B) (C) (D)

Ans (C)

KE = hν – hν0 (h is the slope of the line)

23. The radiation with highest frequency among the following is

(A) X-rays (B) γ- rays (C) UV rays (D) IR rays

Ans (B)

γ- rays. Frequency is in the order γ- rays > X-rays > UV rays > IR rays

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 91

24. Black body radiation supports

(A) particle nature of electromagnetic radiation.

(B) wave nature of electromagnetic radiation

(C) dual nature of electromagnetic radiation

(D) ionizing character of electromagnetic radiation

Ans (A)

25. If the specific charge on a proton is x , then the specific charge on alpha-particle is
x x
(A) x (B) (C) 2x (D)
4 2

Ans (D)

specific charge is e/m

26. Diffraction of electrons supports

(A) particle character of electrons.

(B) wave character of moving electrons.

(C) dual nature of electrons.

(D) none

Ans (B)

27. A univalent negative ion has 36 electrons. It has 31.7% less protons as compared to neutrons.
The number of neutrons is
(A) 36 (B) 35 (C) 46 (D) 40

31.7  35
Ans (C) Let the number of neutrons be x, x − = 35
100

100x – 31.7  35 = 35  100

x = 35  131.7 10−2 = 46 neutrons

28. Match the following

Column − I Column − II

(i) Dual nature of electrons (P) J. J. Thomson

(ii) Charge of electrons (Q) James Chadwick

(iii) Charge to mass ratio of electrons (R) R.A Millikan

(iv) Neutrons (S) Einstein

ACTIVE SITE EDUTECH-9844532971

92 | XI Std. Chemistry

(A) (i) → (R) (ii) → (P) (iii) → (Q) (iv) → (S)

(B) (i) → (S) (ii) → (R) (iii) → (P) (iv) → Q)

(C) (i) → (P) (ii) → (Q) (iii) → (R) (iv) → (S)

(D) (i) → (R) (ii) → (S) (iii) → (Q) (iv) → (P)

Ans (B)

29. The frequency of the first line in the Lyman series in the hydrogen atom is 0. The frequency of
corresponding line emitted in He+ is
0 0
(A) 20 (B) 40 (C) (D)
2 4

Ans (B)

30. In the Balmer series if the first line has a wave length 6500 Å, the wave length of the third
Balmer line is
(A) 5014 Å (B) 5200 Å (C) 4814 Å (D) 4414 Å

Ans (D)

31. The energy of an electron in a hydrogen like species is proportional to

n2 Z Z2 Z2
(A) 2 (B) 2 (C) (D)
Z n n n2

Ans (D)

mZ 2 e 4 Z2
E= − = a constant 
8 02 n 2 h 2 n2

32. Which of the following is not a statement of Bohr’s theory?

(A) Energy of an electron in an atom is quantized.

(B) Angular moment of the electron in an orbit is quantized.

(C) Electrons revolve around the nucleus in circular orbits of definite energy.

(D) At a certain distance from the nucleus the probability of finding the electron is maximum.

Ans (D)

Probability does not feature in Bohr’s model

33. The ratio of velocity of the electron in its ground state of He+ and Li2+ ions is
(A) 3 : 2 (B) 2 : 3

(C) 2: 2 (D) cannot be determined

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 93

Ans (B)

Z v Z n 2 1 2
v  1 = 1  2 = =
n v 2 n1 Z 2 1 3 3

34. What transition in He+ would have the same wavelength as that of the second line in Balmer
series of hydrogen atom?
(A) n1 = 4, n2 = 8 (B) n1 = 5, n2 = 8 (C) n1 = 3, n2 = 4 (D) n1 = 2, n2 = 4

Ans (A)

1  1 1 
= RHZ2  2 − 2 
 n 1 n 2 

1 1  1 1 
RH  2 − 2  = R H  4  2 − 2 
2 4  n 1 n 2 

1 1 1  1 1 
RH   − 2  = RH  2 − 2 
4 2 2
4  n 1 n 2 

1 1  1 1 
 4 2 − 8 2  =  n 21 − n 2 2  , n1 = 4 and n2 = 8
   

35. If the ionization energy of He+ is xJ , then the energy of the electron in the first excited state is
x x
(A) − J (B) − J (C) –x J (D) −4x J
4 2

Ans (A)

IE = E  − E1 = 0 – E1 = −E1 = x J

E1 x
En = 2
=− J
n 4

36. The lowest angular momentum that a Bohr electron in hydrogen atom can have is
h h h h
(A) (B) (C) 2 (D)
 2  4

Ans (B)

nh
Angular momentum = , for the lowest value, n = 1
2

37. The velocity of an electron in the nth orbit of a hydrogen like species is given by
e2 Z Ze 2 Ze2 Ze 2
(A) (B) (C) (D)
mr n mr n 4m 0 r mrn

Ans (C) Centripetal force = Electrostatic force of attraction.

ACTIVE SITE EDUTECH-9844532971

94 | XI Std. Chemistry

mv 2 1 Ze 2 Ze 2 Ze 2
 =  2  v2 = v=
r 4 0 r 4m 0 r 4m0 r

38. De-Broglie equation from the following is

p hc 2r c
(A)  = (B)  = (C)  = (D)  =
h E n 2

Ans (C)

2r Circumference
= =
n No. of waves

 h h 2r 
 mvr = n or 2r = n or 2r = nl  
2 mv n 

39. The wavelength of a beam of -particles moving with a velocity of 6.6  103 m/s is
(A) 1.5  10–10 m (B) 1.5  10–11 m (C) 1.1  10–15 m (D) 1.1  10–12 m

Ans (B)

h
= =
mv
6.6  10 −34  6.02  10 23  4  10 −3 
−3
= 1.5  10 −11   mass of an alpha particle = 
4  10  6.6  10 3
 6.02  10 23 

40. Radius of second Bohr orbit of hydrogen atom in terms of its wavelength will be
h   2
(A) (B) (C) (D)
   

Ans (B)

2r
λ= , substitute n = 2
n

41. The principal, azimuthal and magnetic quantum numbers are respectively related to the
(A) size, shape and orientation (B) shape, size and orientation

(C) size, orientation and shape (D) none of the above

Ans (A)

42. An atom has 2 electrons in the K shell, 8 electrons in the L shell and 6electrons in the M shell.
The total number of electrons in the s-orbital s is
(A) 6 (B) 5 (C) 7 (D) 10

Ans (A)

43. 36Kr has an electronic configuration [Ar] 4s23d104p6. For the next element, 37th electron will go
into which of the sub shells?

ACTIVE SITE EDUTECH- 9844532971

Atomic Structure | 95

(A) 4f (B) 4d (C) 3p (D) 5s

Ans (D)

44. For n = 3, how many times does the wave function cross zero for l = 1?
(A) Once (B) Twice (C) Thrice (D) Does not cross

Ans (A)

Radial wave function = (n − l − 1) = 3 −1 − 1 = 1

1
45. What is the orbital degeneracy of the level that has energy, − hcRH in the ground state of
9
hydrogen atom?
(A) 1 (B) 4 (C) 9 (D) 16

Ans (C)

= RH  1 − 1  , E =
1 hc 1
= hcRH
  32  2  
  9

This energy corresponds to energy level n = 3 which has 9 degenerate orbitals.

46. According to Bohr’s theory, the angular momentum of an electron in 5th orbital is
(A) 25 h/ (B) 1.0 h/ (C) 10 h/ (D) 2.5 h/

Ans (D)

h h h
Angular momentum = n = 5 = 2 .5
2 2 

47. Which one of the following pairs of ions have the same electronic configuration?
(A) Cr3+, Fe3+ (B) Fe3+, Mn2+ (C) Fe3+, Co3+ (D) Sc3+, Cr3+

Ans (B)

Fe3+ and Mn2+ : [Ar] 3d5.

h
48. The orbital angular momentum for an electron revolving in an orbit is given by l (l + 1) . .
2
This momentum for an s-electron will be given by
h 1 h h
(A) 2 . (B) . (C) zero (D)
2 2 2 2

Ans (C)

For s-electron, l = 0.

ACTIVE SITE EDUTECH-9844532971

96 | XI Std. Chemistry

In the following questions two statements are given. One labelled as statement I and the other
labelled as statement II. Examine both the statements and mark the correct choice according to
the instructions given below.
(A) Both the statement I and statement II are true and the statement II is the correct explanation
of the statement I

(B) Both are true, but the statement II is not the correct explanation of the statement I

(C) The statement I is true, but the statement II is false

(D) Both are false or statement I is false and the statement II is true on its own

49. Statement-I : Charge to mass ratio of cathode ray particles is same irrespective of the nature of
cathode used or of the gas taken in the discharge tube.
Statement-II : Electrons are universal constituents of all matter.

Ans (A)

50. Statement-I: The spectra of He+ is similar to that of Li2+.

Statement-II: Both are positively charged.

Ans (B)

ACTIVE SITE EDUTECH- 9844532971

02.
Structure of Atom
4. A dinegative ion of the element X consists of 10
1. Atomic Number, Mass Number, electrons and 8 neutrons. A dipositive ion of the
element Y consists of 12 protons. The number
Atomic Species of neutrons in Y is 1.5 times the number of
electrons in atom X. Then the mass numbers of
1. The pair, in which ions are isoelectronic with X and Y would be in the ratio
Al3+ is: (a) 1 : 2 (b) 2 : 3
(a) Br– and Be2+ (b) Cl– and Li+ (c) 3 : 2 (d) 2 : 5
2– +
(c) S and K (d) O2– and Mg2+ (e) 1 : 3
JEE Main-25.06.2022, Shift-I Kerala CEE -03.07.2022
Ans. (d) : O2–, Mg2+ and Al3+ are isoelectronic. Because Ans. (b) : Proton Neutron Electrons
all have 10 electrons. X2– 8 8 10
2. Given below are two statements: one is labelled Y2+ 12 1.5 × 8 = 12 12
as Mass of X 2−
=
(8 + 8) = 16
Assertion (A) and the other is labelled as Mass of Y 2+ 12 +12 24
Reason (R). 2– 2+ 2– 2+
X : Y = 16 : 24 Or X : Y = 2 : 3
Reason (A): The ionic radii of O2– and Mg2+ are
same. 5. The oxide which contains an odd electron at the
Reason (R): Both O2– and Mg2+ are nitrogen atom is
isoelectronic species. (a) N2O (b) NO2
In the light of the above statements, choose the (c) N2O3 (d) N2O5
correct answer form the options given below JEE Main-26.06.2022, Shift-II
(a) Both (A) and (R) and true and (R) is the Ans. (b) : Species Total electron
correct explanation of (A) N 2O 22
(b) Both (A) and (R) are true but (R) is not the NO2 23
correct explanation of (A) N 2O 3 38
(c) (A) is true but (R) is false N 2O 5 54
(d) (A) is false but (R) is true Hence, NO2 contain an odd electron at the nitrogen
JEE Main-27.06.2022, Shift-I atom is 23.
Ans. (d) : For ionic radii– 6. Which of the following is isoelectronic?
Ion O2– Mg2+ (a) CO2, NO2 (b) NO 2− ,CO 2
Electrons (8 + 2) = 10 12 – 2 = 10 (c) CN − ,CO (d) SO 2 ,CO 2
Atomic number (Z) 8 12 NEET-2002
O2– is greater than Mg2+. O2– and Mg2+ are isoelectronic
species both have 10 electrons. Ans. (c) : CN − and CO are isoelectronic because they
3. Which one of the following molecules contains have equal number of electrons. Therefore, in CN − the
an incomplete octet of the central atom? number of electrons ⇒6+7+1=14
(a) SF6 (b) AlCl3 And, in CO the number of electrons ⇒ 6+8=14
(c) CH4 (d) PF5 7. The cause of instability of nucleus is:
(e) H2O (a) High proton to electron ratio
Kerala CEE -03.07.2022 (b) High proton to neutron ratio
Ans. (b) : As we know that octet means 8 electrons in (c) Low proton to electron ratio
valence shell. Here, AlCl3 contains an incomplete octet (d) Low proton to neutron ratio
atom. MPPET - 2012
K L M N Ans. (d) : The cause of instability of a nucleus is due to
AlCl3 ⇒ 13Al = 2 8 3 low proton to neutron ratio. when neutron to proton
ratio is less than 1, alpha emission occurs. When it is
17Cl = 2 8 7
greater than 1.5, beta emission occurs.
8. The difference between number of Neutrons
and Protons is positive for :
(a) Hydrogen atom (b) Deuterium atom
(c) Tritium atom (d) None of these
MPPET-2013
Objective Chemistry Volume-I 142
Ans. (c) : Isotopes n PDifference Codes :
(n- P) (A) (B) (C) (D)
Hydrogen (H) 0 1 –1 (a) 4 1 2 3
Deuterium 1 1 0 (b) 5 1 2 3
Tritium 2 1 1 (c) 4 1 3 2
Hence, the difference between number of Neutrons and (d) 1 4 2 3
Protons is positive for tritium atom. AP-EAMCET-2008
∆
9. One atom of 19 39
K contains : Ans. (a) : (A) CaCO3 
Dcomposition
→ CaO + CO 2
100gm 22.4L
(a) 19p; 20n and 19e– (b) 19p; 20n and 20e–
(c) 20p; 19n and 20e– (d) 20p; 19n and 19e– 22.4 ×10
Q 10 g CaCO3 decomposition =
AP-EAMCET/1991 100
Ans. (a) : In potassium (K), no. of proton (p) = 19 = 2.24 L CO2
Number of neutron (n) = 39 – 19 = 20 (B) Na 2 CO3 
Excess HCl
→ 2NaCl + H 2 O + CO 2
Number of electron = 19 22.4L

10. Which one of the following is not an 22.4 ×1.06

∴ 1.06 g Na2CO3 = L CO2
isolectronic pair ? 106
(a) Mg2+, C4– (b) N3–, O2– = 0.224 L CO2
2– 2–
(c) N , O (d) F–, Al3+ (C) C 
Excess O2
→ CO 2
AP-EAMCET-2002 12g 22.4L

Ans. (c) : Isoelectronic species have same number of ∴ 2.4 g carbon on combustion will give,
electrons. 22.4 × 2.4
N2– = 7 + 2 = 9 = = 4.48L CO 2
12
O2– = 8 + 2 = 10
(D) 2CO  Excess O2
→ 2CO 2
N2– and O2– is not an isoelectronic species. 56gCombustion
2×22.4L
11. Assertion (A) : Equal moles of different 2 × 22.4 × 0.56
substances contain same number ∴ 0.56 gm Carbon monoxide =
of constituent particles. 56
Reason (R) : Equal weights of different = 0.448 L CO2
substances contain the same 13. Among the following, the isoelectronic species
number of constituent particles. is/are
The correct answer is : (i) O2–, F–, Na+, Mg2+ (ii) Na+, Mg+, Al3+, F–
(a) Both (A) and (R) are true and (R) is the (iii) N3–, O2–, F–, Ne
correct explanation of (A) (a) (i) and (ii) (b) (i), (ii) and (iii)
(b) Both (A) and (R) are true, but (R) is not the (c) (ii) and (iii) (d) (i) and (iii)
correct explanation of (A) TS-EAMCET-2016
(c) (A) is true, but (R) is false Ans. (d) : The isoelectronic species are elements that
(d) (A) is false, but (R) is true have the same number of electrons. O2–, F–, Na+, Mg2+,
AP-EAMCET-2007 have 10 electron and N3–, O2–, F– and Ne have also 10
Ans. (c) : Equal moles of different substances contain electrons.
same number of constituent particles, but equal weights 14. There are six electrons, six protons and six
of different substance do not contain the same number neutrons in an atom of an element. What is the
of constituent particles. atomic number of the element?
Hence, assertion (A) is true but reason (R) is not true. (a) 6 (b) 12
12. Match of following columns : (c) 18 (d) 24
Column-I Column-II NDA (II)-2016
(At STP) Ans. (a) Given, No. of electron = 6
(A) 10 g CaCO3 (1) 0.224 L CO2 No. of proton = 6
 ∆
→ No. of neutron = 6
Decomposition
We know that,
(B) 1.06 g Na2CO3 (2) 4.48 L CO2 Atomic No. = No. of Proton

Excess HCl
→ Atomic No. = 6
(C) 2.4 g C (3) 0.448 L CO2 Hence, the atomic number of element = 6.
→Excess O2
combustion
15. What is the atomic number of the element with
symbol Uus?
(D) 0.56 g CO (4) 2.24 L CO2
(a) 117 (b) 116
→
Excess O2
combustion (c) 115 (d) 114
(5) 22.4 L CO2 TS-EAMCET-2016
Objective Chemistry Volume-I 143
Ans. (a) : The atomic number of the element with Ans. (c) : Given, 35 Cl
17
symbol Uus is 117. Ununseptium (Uus) is the second 35
heaviest known element and its electronic configuration No. of electron in 17 Cl =17
14 10 2 6
is 5f 6d 7s 7p . 32
No. of electron in 16 S = 16
16. The sum of the total number of neutrons +
present in protium, deuterium and tritium is No. of electron in 34
16 S = 16–1=15
(a) 5 (b) 3 +
No. of electron in 40
18 Ar = 18–1=17
(c) 4 (d) 6 35 2 −
TS-EAMCET (Engg.), 05.08.2021 Shift-II No. of electron in 16 S = 16 + 2 = 18
+
Ans. (b) : No. of neutrons –
40
Hence, 18 Ar contain same no. of electron as 17 35
Cl .
Protium ( 1 H ) = 0
1
20. S + Conc. H2SO4  → X+Y
Deuterium ( 1 H ) = 1
2 Here X is a gas and Y is a liquid and both are
triatomic molecules. The number of electron
Tritium ( 13 H ) = 2 lone pars present on the central atoms of X and
The sum of the total number of neutrons is Y are respectively.
⇒ 0 +1+ 2 (a) 2, 1 (b) 1, 0
(c) 1, 2 (d) 2, 2
⇒3
AP EAMCET-2017
17. The number of protons, electrons and neutrons
in a species are equal to 17, 18 and 18 Ans. (c) : S + 2H2SO4  → 3SO2 + 2H2O
respectively. Which of the following will be the X and Y are SO2 and H2O
proper symbol of this species ? The structure of SO2 and H2O
35 35
(a) 17 Cl (b) 17 Cl —
36 36
(c) 17 Cl (d) 17 Cl —
AP EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (b) : p = 17, e = 18, n = 18 means, Z = 17 ⇒ Hence the option (c) is correct
element is chlorine, which contains one electron more
21. The characteristics of elements X, Y and Z with
(17 + 1), so it is chloride ion, Cl–. atomic numbers, respectively, 33, 53 and 83
⇒ Symbol: 17 35
Cl − ; are:
Mass number, A = n + p = 18 + 17 = 35 (a) X, Y and Z are metals.
Atomic number, Z = 17 (b) X and Z are non-metals and Y is a metalloid.
18. Which of the following has magic number of (c) X is a metalloid, Y is a non-metal and Z is a
protons and neutrons? metal.
(a) 8 O17 (b) 13 Al27 (d) X and Y are metalloids and Z is
metal.
(c) 9 F17 (d) 20 Ca 40 JEE Main 16.03.2021, Shift-II
AP EAMCET (Medical) -1998 Ans. (c) : The characteristics of elements X, Y and Z
Ans. (d): Magic numbers are 2, 8, 20, 28, 50, 82 with atomic numbers respectively 33, 53 and 83 are the
protons and 2, 8, 20, 28, 50, 82, 126 neutrons in the arsenic, iodine and bismuth. As we know the properties
nucleus. These numbers impart stability to the nucleus. of that iodine, arsenic is metalloid in nature whereas
O 17
 → proton = 8 Bismuth is metal.
(a) 8 22. Which of the following is true about sodium
neutron = 9
chloride ?
13 Al
27
 → proton = 13 (a) Molecular mass = 58.5 amu
(b)
neutron = 14 (b) Formula mass = 58.5 amu
F17  → proton = 9 (c) Molecular mass = 5.85 amu
(c) 9 (d) Formula mass = 5.85 amu
neutron = 8 AP EAMCET (Engg.) 18.9.2020 Shift-I
Ca 
40
→ proton = 20 Ans. (b) : Sodium chloride is an ionic crystal of rock
(d) 20
neutron = 20 salt structure in which 4 NaCl units are present in one
Thus, the calcium nucleic has the both proton and unit cell of coordination number, Na+ : Cl = 6 : 6. So,
neutrons number is 20. actual molecule of sodium chloride is not NaCl, NaCl is
19. The species that has the same number of the molecular formula of formula mass = (23 + 35.5) =
35 58.5 amu.
electrons as 17 Cl is
+
23. An alloy of metals X and Y weighs 12g &
(a) 32 16 S (b) 3416 S contains atoms X and Y is the ratio of 2: 5 The
+ 35 2 − percentage of Metal X in the alloy is 20 by
(c) 40 18 Ar (d) 16 S
mass. If the atomic mass of X is 40. What is the
NDA (II)-2017 atomic mass of metal Y.
Objective Chemistry Volume-I 144
 (a) 64 amu (b) 32 amu No. of electrons in the neutral atom = x+3
(c) 60 amu (d) 50 amu ∴No.of protons = x + 3 = 23 + 3 = 26
AP EAPCET 20.08.2021 Shift-I 3+
Hence, the ion is 5626 Fe
12 27. Which of the following clement represents is
Ans. (a): Mass of metal X in alloy = × 20 = 2.40g
100 isoelectronic sequence?
Mass of metal Y in alloy =12–2.4=9.6g (a) N, O, F, Ne (b) Na+, Mn2+, Al3+, Si4+
2.4 (c) Cl , Ar, Ca , Ti (d) Be, Mg2+, Ca, Si2+
– 2+ 4+

No. of atoms of X = × 6.02 ×1023 AP-EAMCET (Medical), 2006

40
Ratio of atoms of X and Y = 2:5 Ans. (c) : The species which have same number of
electrons are called the isoelectronic
2.4 × 6.02 × 1023 5
No of atoms of Y = × Cl– = atomic no. ± no. of charge = 17+1 =
40 2 Similarly,
22
= 9.03×10 atom
Ar = 18 + 0 =
Now, 9,03×1022 atoms of Y are present in 9.6 g
∴ 6.02×1023 atoms of Y area present in Ca 2 + = 20-2 =
9.6 Ti4+ = 22-4 =
⇒ × 6.02 ×1023 =64 g ∴ All these are isoelectronic. Hence, the correct
9.03 ×1022
∴ atomic mass of Y = 64 amu. option is (c).
28. Elements X and Y belong to the same group.
24. The masses of an electron, a proton and a Which one of the following set of atomic
neutron respectively will be in the ratio –– numbers represent X and Y?
(a) 1836.15 : 1838.68 (b) 1856.15 : 1858.68 (a) 15, 30 (b) 20, 31
(c) 1834.15 : 1836.68 (d) 1846.15 : 1848.68 (c) 19, 55 (d) 33, 16
AP EAPCET 20.08.2021 Shift-I AP-EAMCET (Medical), 2002
Ans. (a) : Mass of proton = 1.673 × 10–27 kg Ans. (c) : The atomic number 19, 55 belongs to the
Mass of neutron = 1.675 × 10–27 kg alkali metal. The name of the atomic number 19, 55
Mass of electron = 9.109 × 10–31 elements is potassium, cesium respectively.
mp 29. The number of protons in a negatively charged
∴ Mass ratio of proton and electron = atom (anion) is
me (a) more than the atomic number of the element
1.673 × 10 –27 (b) less than the atomic number of the element
= (c) more than the number of electrons in the
9.1×10 –31 atom
= 1835.16 (d) less than the number of electrons in the atom
mn NDA (II)-2011
And Mass ratio of neutron and electron =
me Ans : (d) The number of protons in a negatively
charged atom. It is less than the number of electron in
1.675 × 10–27 the atom.
=
9.1×10 –31 • The atoms of a component are impartial in nature
= 1838.68 since it contains equivalent number of electrons and
25. In the following sets of ions, which one is not protons.
isoelectronic with the rest of the species? 30. Which one among the following most correctly
O2–,F–,Na,Mg2+,Al3+, Ne determines the atomic number of an element ?
(a) Mg2+ (b) Al3+ (a) Number of protons
2–
(c) O (d) Na (b) Number of protons and electrons
COMEDK-2014 (c) Number of ions
(d) Number of nucleons
Ans. (d) : O 2– (10e – ),F – (10e – ), Na(11e – ), Mg 2+ (10e – ),
NDA (II)-2011
Al3+ (10e – ) and Ne(10e – ) Ans : (a) Number of protons are most correctly
Thus, Na is not isoelectronic with the rest of the species. determines the atomic number of an element.
26. An ion with mass number 56 contains three • Atomic number the quantity of a chemical
units of positive charge and 30.4% more component in the periodic system by which the
neutrons than electrons. The ion is component are organized arranged by expanding
3+ 3+ number of protons in the nucleus.
(a) 56 28 Ni (b) 56
26 Fe
31. A stable nucleus (light with A < 10) has
(c) 56
27 Co 3+ (d) Cr 3+
56
24 (a) exactly the same number of neutrons and
COMEDK-2016 protons
(b) more neutrons than protons
Ans. (b) : Let no. of electrons in the ion M3+ = x
(c) no neutrons
30.4x (d) no protons
∴No. of neutrons = x + = 1.304x
100 NDA (II)-2011
Objective Chemistry Volume-I 145
Ans : (a) A stable nucleus has exactly the same number For 23 Na
11
of neutrons and protons.
atomic number = 11
32. Which of the following ratio will give stability atomic mass = 23
to daughter element, when radioactive parent neutrons = 23 –11
element has less number of protons compared
to number of neutrons? = 12
Hence, (I) and (II) statements are correct.
N +1 N −1
(a) (b) 35. The radius of La3+ (Atomic number of La = 57)
Z +1 Z +1 is 1.06 Å. Which one of the following given
N −1 N +1 values will be closest to the radius of Lu3+
(c) (d)
Z −1 Z −1 (Atomic number of Lu = 71) ?
GUJCET-2011 (a) 1.40 Å (b) 1.06 Å
Ans. (b): 0 n →1 H
1 1 (c) 0.85 Å (d) 1.60 Å
VITEEE-2018
or 11 p + −0 1β (β - particle) Ans. (c) : As we move left to right in lanthanide series,
i.e. emitting β-particles the newly formed nuclide has its there is increase in atomic number but the size of cation
n/p value lower than its parent nucleide and hence get decreases.
approaches closer to the zone of stability and become So, possible radius for Lu3+ is 0.85 Å
stable.
36. Assertion: The chemical properties of different
N −1 isotopes are same.
Hence, stability ratio of daughter element .
Z +1 Reason: Isotopes have same number of neutron.
33. Isotope used in brain scan is (a) If both Assertion and Reason are correct and
(a) 1H3 (b) 6C11 Reason is the correct explanation of
(c) 6C14 (d) 80Hg197 Assertion.
SRMJEEE-2010 (b) If both Assertion and Reason are correct, but
Ans. (b) :The isotope of 6C11 is (Choline) used in the Reason is not the correct explanation of
scanning of the brain in medical sector. Assertion.
(c) If Assertion is correct but Reason is incorrect.
34. Observe the following statements regarding
isotones: (d) If both the Assertion and Reason are incorrect.
39 40
I. K and Ca are isotones. AIIMS 25 May 2019 (Morning)
II. Nuclides having different atomic (Z) and mass Ans. (c) : The chemical proporties of different isotopes
numbers (A) but same number of neutrons (n) are same. But isotopes does'nt have same number of
are called isotones. neutron.
III. 19F and 23Na are isotones. 37. The number of electrons and neutrons of an
The correct answer is element is 18 and 20 respectively. Its mass
(a) 1, 2 and 3 are correct number is
(b) Only 1 and 2 are correct (a) 37 (b) 17
(c) Only 1 and 3 are correct (c) 38 (d) 22
(d) Only 2 and 3 are correct AIIMS-1994
AP-EAMCET (Engg.)-2005 Ans. (c) : Mass no = No. of protons + No. of neutrons
No. of protons = No of electrons
Ans. (b) : We know that when nuclides have same
number of neutrons (N) but have different mass number ⇒ No of Protons ⇒ 18
(A) and proton number (Z) are called isotones. ⇒ Mass No. = 20+18= 38
38. The number of electrons in [19K40]-1 is
(i)
(a) 28 (b) 19
atomic number (Z) = 19 (c) 40 (d) 20
AIIMS-1994
atomic mass (A) = 39
Ans. (d): The electronic configuration of the given
neutrons (N) = A – Z = 39 – 19 = 20 K(19): 1s2, 2s2, 2p6, 3s2, 3P6, 4P1
The number of electrons present originally = 19
∴ Total number of electrons = 19+1= 20 (from negative
atomic number = 20 charge).
atomic mass = 40
neutrons = 40 – 20 = 20 39. Which of the following is the correct sequence
39 40
of atomic weights of the given elements?
Hence, K and Ca are isotones. (a) Ni > Co > Fe (b) Fe > Co > Ni
(iii) For 9 F 19 (c) Co > Fe > Ni (d) Co > Ni > Fe
AIIMS-1996
atomic number = 9
atomic mass = 19 Ans. (d) : Atomic weight of Fe (z=26)=55.85 amu.
neutrons = 19–9 = 10 Atomic weight of Co(z=27)⇒58.90 amu
Objective Chemistry Volume-I 146
 Atomic weight of Ni (z=28) ⇒58.69 amu (a) M2=2M1 (b) M1<10(mp+mn)
Hence, the correct order of atomic weight is follow (c) M2>2M1 (d) M1=M2
CO>Ni>Fe. AIIMS-2015
40. Assertion: Nuclear binding energy per nucleon Ans. (a): 20
10 Ne contains 10 protons and 10 neutrons
in the order 94 Be > 73 Li > 42 He ∴ M1= 10mp+10 Mn
Reason: Binding energy per nucleon increases 40
20 Ca contains 20 protons and 20 neutrons
linearly with difference in number of neutrons
and protons. ∴ M2= 20Mp+20 Mn
(a) If both Assertion and Reason are correct and ∴ M2 = 2M1
the Reason is a correct explanation of the 43. Deuterium nucleus contains:
Assertion. (a) 1 proton, 1 electron (b) 1 proton, 1 neutron
(b) If both Assertion and Reason are correct but (c) 2 protons, 1 electron (d) 1 proton, 2 electrons
Reason is not a correct explanation of the AIIMS-1998
Assertion. Ans. (b): Deuterium nucleus contains 1 proton and 1
(c) If the Assertion is correct but Reason is neutron because it is isotope of hydrogen.
incorrect.
(d) If both the Assertion and Reason, are e 11
44. If e = 1.60206 ×10-19 C, = 1.75875 ×10 Ckg -1
incorrect. m
(e) If the Assertion is incorrect but the Reason is then the mass of electron is
correct. (a) 7.5678 × 10 −31 kg (b) 9.1091 ×10 −31 kg
AIIMS-2004
(c) 11.2531 × 10 −31 kg (d) 13.0513 × 10 −31 kg
Ans. (d): Binding energy per nucleon of 3Li7 (5.38
Mev) is less than 2He4 (7.08 Mev) as helium is found to AIIMS-1999
be more stable than Li. As the atomic mass number 1.60206 ×10−19
increases the binding energy per nucleon decreases. As Ans. (b): Mass of electron =
the atomic number and the atomic mass number 1.75875 ×1011
increase the repulsive electrostatic force within the = 9.1091×10–31
nucleus increase due to the greater number of protons in OR
the heavy elements. To overcome this increased e
repulsion, the proportion of neutrons in the nucleus = 1.75875 × 1011 C kg −1
m
must increase to maintain stability. This increase in the e = 1.60206×10–19 C
neutron to proton ratio only partially compensates for
the growing proton-proton repulsive force in the heavier e / e 1.60206 × 10−19
m= = Kg
naturally occurring element. m 1.75875 × 1011
41. Assertion: Atoms are not electrically neutral. = 9.1091×10–31 Kg
Reason: Number of protons and electrons are 45. Positron is:
different (a) Electron with positive charge
(a) If both Assertion and Reason are correct and (b) A nucleus with one neutron and one proton
the Reason is a correct explanation of the (c) A nucleus with two protons
Assertion. (d) A helium nucleus
(b) If both Assertion and Reason are correct but AIIMS-1998
Reason is not a correct explanation of the
Ans. (a): The positron is the antiparticle or the
Assertion.
antimattery counter part of the electron. The positron
(c) If the Assertion is correct but Reason is has an electric charge of +1, an spin of , and has the
incorrect. same mass as an electron. When a low–energy positron
(d) If both the Assertion and Reason, are collides with a low energy electron, annihilation occurs,
incorrect. resulting in the production of two or more gamma ray
(e) If the Assertion is incorrect but the Reason is photons Positrons may be generated by positron
correct. emission radiative energetic photon which is in
AIIMS-1999 teracting with an atom is a material.
Ans. (d): Atoms are electrically neutral because they 46. The nitride ion in lithium nitride is composed of
have an equal number of protons and electrons, and as (a) 7 protons + 7 electrons
we know that the protons and electrons contain opposite (b) 10 protons + 7 electrons
types of charges, hence their negatively charged (c) 7 protons + 10 electrons
electrons are completely balanced with their positively (d) 10 protons + 10 electrons
charged protons.
Hence, given both the assertion and reason are false. CG PET -2018
42. Let mp be the mass of a proton, mn that of a Ans. (c) : Nitride ion can be represented as N3–,
Number of protons = 7
neutron, M1 that of a 20 10 Ne nucleus and M 2 that Number of electrons = 7+3= 10
40
of a 20 Ca nucleus. Then Therefore, nitride ion contains 7 protons +10 electrons.
Objective Chemistry Volume-I 147
47. Increasing order (lowest first) for the values of 52. Which of the following is not isoelectronic?
e/m for electron (e), proton (p), neutron (n) and (a) Na+ (b) Mg 2+
α-particles is (c) O 2−
(d) Cl−
(a) e, p, n, α (b) n, α, p, e CG PET -2007
(c) n, p, e, α (d) n, p, α, e
Ans. (d) : Na , Mg , O contain 10 electrons each Cl −
+ 2+ –

CG PET -2009 has 18 electrons.

0 53. The correct symbol of the species with number
Ans. (b) : (e / m) n =
1.675 × 10−27 of electrons, protons and neutrons as 18, 16 and
16 respectively is
2 ×1.6602 × 10−19 C
(e / m)α = 32
(a) 16 S 32
(b) 18 S
4 × 1.675 × 10−27 kg
(c) 32
S2− (d) 32
S2−
1.602 × 10−19 C
16 18

(e / m) p = AMU-2014
1.675 × 10−27 kg Ans. (c) :
1.602 ×10−19 C Species No of No of No of
(e / m)e =
9.108 ×10−31 kg Electron Protons neutrons
48. Isoelectronic pair among the following is
32
16 S
16 16 16
2+
(a) Ca and K (b) Ar and Ca 32 18 18 14
2+ 18 S
(c) K and Ca (d) Ar and K –
32 2 − 16+2e 18 16 16
UPTU/UPSEE-2008 16 S
Ans. (b) : Ar and Ca2+ are isoelectronic species as they 32 2
18 S
18+2e– 20 18 14
have same number of electrons i.e. 18.
54. The magnitude of the charge on the electron is
49. Which of the following is isoelectronic pair? 4.8 × 10–10 esu. What is the magnitude of the
–
(a) CN , O3 (b) ClO2, BrF2 charge on the proton on the nucleus of helium
(c) BrO−2 ,BrF2+ (d) ICl2, ClO3 atom?
JCECE - 2013 (a) 4.8 × 10–10 esu (b) 9.6 × 10–10 esu
–10
(c) 6.4 × 10 esu (d) 14.4 × 10–10 esu
Ans. (c) : Both BrO−2 (35 + 2 × 8 + 1 = 52)
AMU-2013
and BrF2+ (35 + 2 × 9 − 1 = 52) have 52 electrons. Ans. (b) : Since, the charge on the proton is equal in
50. N2 and CO are magnitude to that on the electron, the charge on the
(a) Isomers (b) Isoelectronic proton is also 4.8×10–10 esu. Since, a helium nucleus
contains 2 protons, its charge is 9.6×10–10 esu.
(c) Isotopes (d) Isobars
J & K CET-(2002) 55. Atomic number equal to the
(a) Atomic mass of the element
Ans. (b) : N2 have 14×2=28 electrons
(b) Sum of protons and neutrons
CO have 12+16 = 28 electrons
(c) Number of the protons in the nucleus
Thus both are iso electronic (d) Number of the neutrons in the nucleus.
OR AMU–2001
Both N2 and CO contain 14 electrons each thus N2 and
CO are isoelectronic. Ans. (c) : Atomic number is equal to the number of
protons in the nucleus.
51. Pick out the isoelectronic structures from the
56. The ratio of electron, proton and neutron in
following
tritium is
CH 3+ H 3O + NH 3 CH 3− (a) 1 : 1 : 1 (b) 1 : 1 : 2
I II III IV (c) 1 : 1 : 3 (d) 1 : 2 : 3
(a) I and II (b) I and VI Assam CEE-2014
(c) I and III (d) II,III and IV Ans. (b) : The atomic number and the mass number of
CG PET- 2010 tritium are 1 and 3 respectively. Tritium contains 1
electron, 1 proton and 2 neutrons.
Ans. (d) : CH 3+ has 6+3–1= 8 Hence, the ratio of the number of electrons. Protons and
Electrons H3O + has 3+8–1=10 electrons NH3 neutrons in tritium is 1:1:2
has 7+3=10 electrons CH 3− 57. The atomic numbers of elements X, Y and Z
has 6+3+1=10 electrons so II, III, IV are isoelectronic are 19, 21 and 25 respectively. The number of
structures as they have the same number of electrons. electrons present in the M shells of these
Hence, the correct option (d). elements follow the order.
OR (a) Z > X > Y (b) X > Y > Z
(c) Z > Y > X (d) Y > Z > X
H3O+, NH3 and CH 3− have same number of electrons.
Assam CEE-2014
Objective Chemistry Volume-I 148
Ans. (c) : The electronic configuration of X with atomic (a) (1) and (2) (b) (2) and (3)
number 19 is 1s2 2s2 2p6 3p6 4s1. Now the M shell is the (c) (1) and (3) (d) (1), (2) and (3)
3rd energy level and hence, it contains (2+6=8) BCECE-2017
electrons. Ans. (c) : B is a noble gas so the next element will be in
The electronic configuration of with atomic number 21 the next period and will be alkali metals. Next element
is 1s2 2s2 2p6 3x2 3p6 4s2 3d1. Now the M shell is the 3rd will be than alkaline earth metal.
energy level and hence, it contains (2+6+1=9) electron. 1 ‘A’ has higher electron affinity. It belongs to halogen
The electronic configuration of Z with atomic number family.
25 is 1s2 2s2 2p6 3s2 4s2 3d5. Now the M shell is the 3rd 2 ‘C’ exists in +1 oxidation state
energy level and hence it contains (2+6+5=13 3 ‘D’ is an alkaline earth metal
electrons). Hence, the Statements (1) and (3) are correct.
Since Z contains more electrons in the M shell than Y
and Y contains more electrons in the M shell than X. So 63. n/p ratio during positron decay
the number of electrons present in the M–shell of these (a) Increases (b) Decreases
elements follows the order Z>Y>X Hence, the correct (c) Remains constant (d) All of these
option is c. CG PET- 2015
58. The number of electrons, protons and neutrons Ans. (a) 0 n1 → 1 p1 + –1 p0 + v i.e. n decreases, p
in phosphide ion (P3-) is- increases.
(a) 15, 15, 16 (b) 15, 16, 15
64. The electronic configuration of an element is
(c) 18, 15, 16 (d) 15, 16, 18
1s2, 2s2 2p6, 3s2 3p6, 3d10, 4s2 4p3.
Assam CEE-2021
To which of the following elements it is similar
Ans. (c) : P3– (PHOSPHIDE) in properties?
Electron = 15 + 3 = 18 (a) Boron (b) Oxygen
Proton = 15 (c) Nitrogen (d) Chlorine
Neutron = 31 – 15 = 16
CG PET -2008
59. Nuclides :
(a) Have specific atomic numbers Ans. (c) : The valance shell electronic configuration is
4s2 4p3 similar to nitrogen 3s23p2.
(b) Have same number of protons
(c) Have specific atomic number and mass 65. If the de-Broglie wavelength of the electron in
numbers nth Bohr orbit in a hydrogenic atom is equal to
(d) Are isotopes 1.5πa0 (a0 is Bohr radius), then the value of n/Z
is
BCECE-2006
(a) 1.0 (b) 0.75
Ans. (c): Nuclides have a definite number of protons (c) 0.40 (d) 1.50
and neutrons and consequently definite atomic number
and mass number. Such as oxygen nuclei contain 8 [JEE Main 2019, 12 Jan Shift-II]
protons and 8 neutrons (8O16). n2
Ans. (b) : 2πrn= nλ; 2πao × = nλ
60. The energy released in an atom bomb explosion z
is mainly due to n n 1.5πa o
(a) Release of neutrons 2πao= =1.5πao = = =0.75
(b) Release of electrons z z 2πa o
(c) Greater mass of products than initial material 66. The introduction of a neutron into the nucleus
(d) Lesser mass of products than initial material of an atom would lead to a change in
BCECE-2006 (a) Atomic number
Ans. (d) : The source of large energy, produced during (b) Atomic mass
atom bomb explosion, is the mass defect occurring (c) Chemical nature of the atom
during the fission reaction, which is converted into (d) Number of electron
energy equivalent to mass defect. CG PET -2019
61. Number of neutron in C12 is : Ans. (b) : The introduction of a neutron into the nuclear
(a) 6 (b) 7 composition, of an atom would lead to a change in its
(c) 8 (d) 9 atomic mass. However, its atomic number, the chemical
BCECE-2005 nature of the atom and the number of the electron s will
Ans. (a) : Number of neutron = atomic mass– atomic remain unchanged as they are related to the number of
number. For C12 number of neutron = 12–6=6 protons and they are independent of the number of
neutron.
62. The atomic number of elements A, B, C, and D
are Z–1, Z+1, Z+2 respectively. If B is a noble 67. The element with atomic number 55 belongs to
gas, choose the correct option. which block of the periodic table?
(1) (A) has higher electron affinity. (a) s-block (b) p- block
(2) (C) exists in +2 oxidation state. (c) d- block (d) f- block
(3) (D) is an alkaline earth metal. CG PET -2004
Objective Chemistry Volume-I 149
Ans. (a) : The electronic configuration of the above Ans. (b) : It will gain two electron to form a compound
element is [Xe]6s1. Since the valence electrons are in with sodium (Na).
the S subshell, the element belongs to S–block. Given, atomic no. of element = 8 i.e. oxygen
68. Neutrons are found in atoms of all elements Electronic configuration of oxygen= 1s2, 2s2, 2p4.
except in Therefore, valency = 2, 6
(a) Chlorine (b) Oxygen Electronic configuration of Na= 2, 8, 1
(c) Argon (d) Hydrogen Valency = 1 (As it is easier to donate one electron)
CG PET -2004 Hence, the given element will gain 2 electron form two
Ans. (d) : Neutrons can be found in all atomic nuclear sodium atoms to completes its octet.
except hydrogen. Hydrogen is that atomic nucleus and it Reaction involved –
consists of no charge and a mass of slightly over 1 amu. 4Na+ + O2→ 2Na2O
69. The atomic weight of an element is 39. The 72. The triad of the nuclei that is isotonic :
number of neutrons in its nucleus is one more (a) 6C14, 7N14, 9F19 (b) 6C14, 7N15, 9F17
14 14 17
than the number of protons. The number of (c) 6C , 7N , 9F (d) 6C12, 7N14, 9F19
protons, neutrons and electrons-respectively in HP CET-2018
its atoms would be
Ans. (b) : Isotones are nuclides with same number of
(a) 19, 20, 19 (b) 19, 19, 20 neutrons
(c) 20, 19, 19 (d) 20, 19, 20
∴ 6C14, 7N15, 9F17 are isotones with 8 neutrons in each
CG PET -2004 case.
Ans. (a) : Atomic weight of an element = 39 unit 73. Three elements X, Y and Z are in the 3rd
∴ Atomic weight = no of proton + no of neutron period of the periodic table. The oxides of X, Y
Hence According to question. and Z, respectively, are basic, amphoteric and
no. of neutron = no. of proton +1 acidic. The correct order of the atomic number
So, of X, Y and Z is
39= no. of neutron + no. of proton (a) Z < Y < X (b) X < Y < Z
39= no. of proton +1 +no. of proton (c) X < Z < Y (d) Y < X < Z
2×no. of proton = 38 (JEE Main 2020, 2 Sep Shift-II)
no. of proton = 19 Ans. (b) : X<Y<Z MgO<Al2O3<SiO2
Also, no. of electron = no. of proton 74. The group having isoelectronic species is
They, no. of electron = 19 (a) O2–, F–, Na+, Mg2+ (b) O–, F–, Na, Mg+
And no. of neutron = 19+1 = 20 (c) O2–, F–, Na, Mg2+ (d) O–, F–, Na+, Mg2+
Hence, no. of proton = 19 (JEE Main-2017)
no. of neutron = 20 Ans. (a) : Isoelectronic species are those which
no. of electron=19 contains same number of electrons
70. In the following reaction Species Atomic Number of
3 Li + ? → 2 He + 1H
6 4 3
number electrons
2–
the missing particle is O 8 10
(a) Electron (b) Neutron F– 9 10
(c) Proton (d) Deuteron Na+ 11 10
CG PET- 2010
Mg2+ 12 10
Ans. (b) : In the given reaction
O– 8 9
3 Li + ? → 2 He + 1H
6 4 3
Na 11 11
Mass number and atomic number on LHS and RHS will
be equal. Mg+ 12 11
Let us assume the missing particle be xAy Option (a) is correct which contains isoelectronic
Atomic number species O2–, F–, Na+, Mg2+
3+x=2+1 75. Which one of the following sets of ions
x=0 represents a collection of isoelectronic species ?
Mass number, (a) K+, Cl–, Ca2+, Sc3+ (b) Ba2+, Sr2+, K+, S2–
6+y=4+3 (c) N3–,O2–, F–, S2– (d) Li+, Na+, Mg2+,
y=1 Ca2+
It represents neutron, Assam CEE-2020
(AIEEE 2006)
3 Li + 0n → 2 He + 1H
6 1 4 3

Ans. (a) : The ions which have the same number of

71. The atomic number of an element is 8. How electrons are called isoelectronic
many electrons will it gain to form a compound K+ Cl– Ca2+ Sc3+
with sodium?
(a) One (b) Two (2,8,8) (2,8,8) (2,8,8) (2,8,8)
(c) Three (d) Four All these ions contains 18 electrons each. So all these
NDA (II)-2018 ions are isoelectronic.
Objective Chemistry Volume-I 150
76. According to the periodic law of elements, the Ans. (d) : Isoelectronic species contain same number of
variation in properties of elements is related to shells so their size is effected by the nuclear charge.
their 80. The atomic number of unnilunium is ......
(a) Atomic masses (JEE Main 2020, 6 Sep Shift-II)
(b) Nuclear masses Ans. (101) : Synthetic radioactive element, symbol=md
(c) Atomic numbers with atomic number 101 made by bombarding lighter
(d) Nuclear neutron-proton number ratios elements with light nuclei accelerated in cyclotrons.
(AIEEE 2003) 81. Which one of the following constitutes a group
Ans. (c) : The modern form of periodic table is the one of the isoelectronic species?
in which the chemical elements are arranged in order of (a) C 22− , O −2 , CO, NO
increasing atomic number.
Elements with similar properties are arranged in the (b) NO + , C 22− , CN − , N 2
same column (called a group) and elements with the (c) CN − , N 2 , O 22− , CO 22−
same number of electron shells are arranged in the same
row (called a period). (d) N 2 , O 2− , NO + , CO
77. The group number, number of valence JEE Main-09.10.2018
electrons and valency of an element with Ans. (b) : Number of electrons in each species are
atomic number 15, respectively, are given below
(a) 16, 5 and 2 (b) 15, 5 and 3 N2=14 CN − =14
(c) 16, 6 and 3 (d) 15, 6 and 2 −
O 2 = 17 C 22− = 14
(JEE Main 2019, 12 April Shift-I)
Ans. (b) : The group number, number of valence NO+ =14 O 2− = 18
electrons and valency of an element with atomic CO = 14 No = 15
number is are 15, 5 and 3 respectively. Modern periodic It is quite evident from the above that NO+, C 22− , CN − ,
table is based on the atomic number. Modern periodic
table is based on the atomic number. Number of valence N2 and Co are isoelectronic in natures.
electrons present in an atom decides the group number. 82. Hydrogen has three isotopes (A), (B) and (C). If
Electronic configuration of element have atomic the number of neutron(s) in (A), (B) and (C)
number 15–1s2 2s2 2p6 3s2 3p3 respectively, are (x), (y) and (z), the sum of (x),
As five electrons are present in valence shell. Its group (y) and (z) is
number is 15. Valency of element having atomic (a) 4 (b) 3
number 15 is +3 (8–5=3). (c) 2 (d) 1
78. The isoelectronic set of ions is [JEE Main 2020, 8 Jan Shift-II]
(a) F–, Li+, Na+ and Mg2+ Ans. (b) : Number of neutrons in protium ( H)
1
1 is zero
(b) N3–, Li+, Mg2+ and O2–
(c) Li+, Na+, O2– and F– (x), number of neutron in deuterium ( 2
1 H or D ) is 1(y)
2
1

(d) N3–, O2–, F– and Na+ and number of neutrons in tritium ( H or ,3 T ) is 2(z) the
3
1
(JEE Main 2019, 10 April Shift-I) sum of x, y and z is x+y+z = 3.
Ans. (d) : Key idea = Isoelectronic species contains 83. Atoms with identical atomic number but
same number of electron. different atomic mass number are known as
The species with its atomic number and number of (a) Polymers (b) Isobars
electrons are as follows: (c) Isotopes (d) Isomers.
Species Atomic Number of J & K CET-(2014)
(ions) number (z) electrons Ans. (c) : Isotope, one of two or more species of atoms
N3– 7 7+3=10 of a chemical element with the same atomic number and
2– position in the periodic table and nearly identical
O 8 8+2=10
chemical behavior but with different atomic masses and
F− 9 9+1=10
physical properties every chemical element has one or
Na+ 11 11–1=10 more isotopes.
Li+ 3 3–1=2 Example:– H11 , H12 , H13 are the isotopes of hydrogen.
Mg2+ 12 12–2=10 84. Negatively charged particles are called
Isoelectronic set of ions are N3–, O2–, F–, Na+ and Mg2+. (a) Electrons (b) Protons
79. The size of the iso-electronic species Cl–, Ar (c) Neutrons (d) None of the above
and Ca2+ is affected by J & K CET-(2014)
(a) azimuthal quantum number of valence shell Ans. (a) :
(b) electron-electron interaction in the outer
Particles Symbol Relative charge
orbitals
(c) principal quantum number of valence shell Electrons e –1
(d) nuclear charge Protons p +1
(JEE Main 2019, 8 April Shift-I) Neutrons n 0

Objective Chemistry Volume-I 151

85. Mass number of an atom is the sum of 90. Which of the following species is isotonic with
86
(a) Number of protons + number of neutrons + 37Rb ?
84 85
number of electrons (a) 36Kr (b) 37Mg
87 89
(b) Number of protons + number of neutrons (c) 38Sr (d) 39Y
(c) Number of protons + number of electrons J & K CET-(1997)
(d) Number of electrons + number of neutrons. Ans. (c) : The number of neutrons in Rb= (86–37)= 49
J & K CET-(2014) and in Sr = (87–38) = 49.
Ans. (b) : The total number of nucleous is termed as Therefore, both of them are isotonic species.
mass number (A) of the atom. 91. Atoms with same atomic number and different
Hence, mass number (A)= No. of protons (p)+No. of mass numbers are called
neutrons(n). (a) isobars (b) isomers
(c) isotones (d) isotopes
86. Mass of a proton is
JCECE - 2009
(a) 0.00727 amu (b) 1.0087 amu
Ans. (d) : Isobars, are the atom having the same mass
(c) 0.00054 amu (d) 1.00727 amu. number but different atomic number. For example, the
J & K CET-(2014) atomic number of carbon and nitrogen is 6 and 7
Ans. (d) : A mass of proton is 1.00727 amu. A proton is respectively. Isotopes are the atoms with same atomic
14 13
a subatomic particle, with a positive electrical charge of no and different atomic mass Ex. 6C , 6C .
+1e elementary charge and a mass slightly less than that 92. The number of electrons, neutrons and protons
of a neutrons. Protons and neutrons, each with masses in a species are equal to 10,8 and 8 respectively.
of approximately one atomic mass unit, are collectively The proper symbol of the species is
referred to as “nucleons” (particles present in atomic (a) 16 (b) 188 O
8 O
nuclei).
39 (c) 18
Ne (d) 16
O 2−
87 19 K and 40
20 C are
10 8
JIPMER-2011
(a) Isomers (b) Isobars
Ans. (d) : From the given data, it is clear that the atomic
(c) Isotones (d) Isotopes number z, of the species is 8 (number of protons).
J & K CET-(2001) Since the number of electron are two more than the
Ans. (c) : No. of neutron in K= 39–19= 20 number of protons, hence, it is a binegative species.
No. of neutron in Ca= 40–20= 20 Thus, the species is 168 O 2− .
So, No, of neutron are equal
93. In long form of periodic Table the properties of
∴ They are isotones. the elements are a periodic function of their
88. Neutron is a fundamental particle carrying (a) Atomic size (b) Ionization energy
(a) A charge of +1 unit and a mass of one unit (c) Atomic mass (d) Atomic number
(b) No charge and a mass of one unit JIPMER-2010
(c) No charge and no mass Ans. (d): In modern periodic table, the physical and
(d) A charge of–1 unit and no mass chemical properties of the element are a periodic
J & K CET-(2000) function their atomic number, i.e. if the element are
Ans. (b) : Proton and e– which are electrically charged, arrange in order of their atomic number, the element
neutron have no charge they are electrically neutral. The with similar properties are repeated after certain regular
intervals.
mass of a neutron is slightly greater than the mass of a
proton, which is 1 atomic mass unit (amu.) 94. An element with mass number 81 contains
31.7% more neutrons as compared to protons.
89. The specific heat of a metal is 0.11 and its It symbol is
equivalent weight is 18.61. Its exact atomic (a) 40X81 (b) 35X81
weight is (c) 81X 317
(d) 40X81
(a) 58.2 (b) 29.1 JIPMER-2019
(c) 55.83 (d) 27.91
Ans. (b) : Let, the number of protons= x
J & K CET-(1998)
31.7x 131.7x
Ans. (c) : Approximate atomic wt. of the metal Number of neutrons= x + =
100 100
6.4 6.4
= = = 58.18 Mass number= number of protons + number of neutrons
specific heat 0.11 131.7x
81= x+
Approx at.wt 100
Valency of the metal=
Eq.wt. ∴ x= 35
58.18 Thus, the symbol of the element is 35X81.
= = 3(whole no.) 95. Which of the following is not iso electronic with
18.61 H2S?
So, exact at. wt. of the element = eq.wt.×Valency (a) F2 gas (b) Oxide ion
= 18.61×3 (c) Ca+2 (d) Sc+3
= 55.83 JIPMER-2019
Objective Chemistry Volume-I 152
Ans. (b) : Iso electronic species are those which have 100. 1 u (amu) is equal to
same number of electron. The total number of electron (a) 1.492×10–10 J (b) 1.492×10–7 J
in the given species are as follows. (c) 1.492×10–13 J (d) 6.023×1023 J
H2S= 2+16 = 18 electron MHT CET-2010
F2= 9+9 = 18 electron Ans. (a) : 1u (unified mass) is equal to exactly 1/12th of
O2–= 8+2=10 electron the mass of C12 atom.
Ca2+ = 20–2 18 electron 1 12g
Sc3+ = 21–3 = 18 electron Thus, 1u = ×
2–
Thus O (Oxide ion) is not iso–electronic with H2S. 12 6.022 ×10 23
1u =1.66×10–24 = –1.66×10–27 Kg
96. If two atoms have equal number of electron it Since, E = mc2, thus
is called
= 1.66×10–27×(3×108) Kg m2s2
(a) isoelectronic (b) isotone
(c) isobar (d) None of these = 1.492×10–10 J
JIPMER-2019 101. Chlorine (Cl17) free radical contains how many
electrons around the nucleus?
Ans. (a) : The species having same number of electrons
are called isoelectronic. For example Na+ and Mg2+ (a) 16 (b) 17
have 10 electron thus are isoelectronic. (c) 18 (d) 19
97. An odd electron molecule among the following MHT CET-2010
is Ans. (b) : In free radicals, the number of electrons are
(a) CO (b) SO2 same as that in isolated neutral atom.
(c) CO2 (d) NO Q In Cl atom, the number of electrons= 17
(e) OF2 ∴ In Cl free radicals, the number of electrons, = 17.
Kerala-CEE-2015 102. From the following pairs of ions which one is
Ans. (c) : N(z=7) and O (z=8) not an iso-electronic pair?
have 7 and 8 electrons respectively (a) Fe 2+ , Mn 2+ (b) O 2 , F
∴ number of electrons in NO= 7+8= 15 (c) Na + , Mg 2+ (d) Mn 2+ , Fe3+
Hence, no is an odd electron molecule.
NEET-2021
98. The effective nuclear charge of an elements
with three valence electrons is 2.60. What is the Ans. (a) : Total no. of e–
6 2
atomic number of the element? 26Fe → 3d 4s , Fe2+→3d6 24
5 2 2+ 5
(a) 1 (b) 2 25Mn → 3d 4s , Mn →3d 23
(c) 3 (d) 4 103. The number of protons, neutrons and electrons
(e) 5 in 175
71 Lu , respectively, are
Kerala-CEE-2018 (a) 71, 107 and 71 (b) 104, 71 and 71
Ans. (e) : Effective nuclear charge = 2.60 and valence (c) 71, 104and 71 (d) 175, 104 and 71
shell contain 3 electrons. NEET-2020
Thus, the minimum number of main shells for the given
element are two i.e. (x=2) and its configuration will be Ans. (c) : 104
71 Lu
1s2, 2s2, 2p1 Number of protons = number of element = 71
So, that the given element has 5 electrons in all also, for Number of neutrons = 175 = 71 104
a neutral atom. 104. Be2+ is isoelectronic with which of the following
Therefore, no. of electrons= Atomic number ions?
There, the atomic number of the element is 5. (a) H+ (b) Li+
+
99. The number of electrons, protons and neutrons (c) Na (d) Mg+
in a species are equal to 10, 11 and 12 NEET-2014
respectively. The proper symbol of the species is 2+
Ans. (b) : Be = (4–2) = 2 is isoelectronic with Li+ (3–
(a) 1122
Na + 23
(b) 11 Na 1=2) since, both have the same number of electron in
(c) 23
10 Ne (d) 23
11 Na + their outermost shell.
105. Isoelectronic speicies are
(e) 23
Na 2 +
(a) CO, CN–, NO+, C 22− (b) CO–, CN, NO, C −2
11
Kerala-CEE-2020
Ans. (d) : No. of electron present = no. of atomic no. (c) CO+, CN+, NO–, C 2 (d) CO, CN–, NO+, C 22−
No. of protons = No. of electron = atomic no. AMU-2011
+
22
11 Na ⇒ No. of electron = 10 NEET-2000
⇒ No. of proton = 11 Ans. (a) : Species having same no. of electrons are
⇒ No. of neutron = A – Z called isoelectronics.
= 23 – 11 The no of electrons in CO = CN − = NO + = C 22− = 14
= 12 So, these are isoelectronic species.
Objective Chemistry Volume-I 153
106. Which one of the following is not isoelectronic 112. The atomic number of an element is 17. The
with O2–? number of orbitals containing electron pairs in
(a) T1+ (b) Na+ its valence shell is
3–
(c) N (d) F– (a) 3 (b) 4
NEET-1994 (c) 6 (d) 8
Ans. (a) : O2–, N3– F– and Na+ have 10 electrons. UP CPMT-2001
107. The mass number and atomic number of Z in Ans. (a) : Write the electronic configuration and then
the following sequence are respectively count number of paired electrons in valence shell.
B X 
A −2 α −1β
→ Y → Z Atomic number 17 = 1s2, 2s2, 2p6, 3s2, 3p5
(a) A – 4, B + 1 (b) A, B –1 Valence shell has 3s2 and 3p5 orbitals
(c) A – 4, B – 2 (d) A – 4, B – 1
UP CPMT-2010
Ans. (d) : AB X −2 α
→ AB −− 42 Y →−1β A− 4
B− 1 Z
∴ Valence shell has 3 paired electrons.
Thus, the mass umber and atomic number of Z are 113. The binding energy of an atom is 128 MeV. The
A – 4 and B – 1 respectively. binding energy per nucleon is 8, the number of
nucleon is
108. The atomic number of an element ‘M’ is 26.
How many electrons are present in the M-shell (a) 4 (b) 8
of the element in its M3+ state? (c) 16 (d) 32
(a) 11 (b) 15 UP CPMT-2001
(c) 14 (d) 13 Ans. (c) : Number of nucleons in an atom
UP CPMT-2008 binding energy of atom
Ans. (d): K L M N =
binding energy per nucleon
M(26) 1s2, 2s2 2p6, 3s23p63d6 4s2
3+ 2 2 6 2 6 5
M 1s , 2s 2p , 3s 3p 3d 4s0 =
128
= 16
Hence, 13 electrons are present in the M-shell of M3+ 8
state.
∴ Atom has 16 nucleons.
109. Atomic number of an element is equal to the
number of 114. Which of the following is correct increasing
(a) protons (b) neutrons e
order for the value of ?
(c) protons+electrons (d) protons+neutrons m
UP CPMT-2005 (a) e < p < n < α (b) n < p < e < α
Ans. (a) : Atomic number= number of protons (c) n < p < α < e (d) n < α < p < e
=number of electrons UPTU/UPSEE-2015
Mass number = number of protons Ans. (d) : n < α < p < e
+ number of neutrons n α p e
110. How many neutrons are present in tritium e 0 +2 +1 1
nucleus?
m 1 4 1 1/1836
(a) 2 (b) 3
(c) 1 (d) 0 e 1
0 1 1836
UP CPMT-2003 m 2
3
Ans. (a) : Tritium ( 1 H) is an isotope of hydrogen 115. For which element the inertness of the electron
pair will not be observe?
∴ atomic number = 1 (a) Sn (b) Fe
∴ protons = electrons = 1
(c) Pb (d) In
No. of neutrons = Atomic mass–No. of protons
= 3–1 WB-JEE-2010
=2 Ans. (b) : Inert pair effect is exhibited only by heavy
metals of p–block elements. Fe belongs to d–block
111. Which one of the following has unit positive
elements.
charge and 1 amu mass?
(a) Electron (b) Neutron 116. If species has 16 protons, 18 electrons 16
(c) Proton (d) None of these neutrons, find the species and its charge
UP CPMT-2003 (a) S1– (b) Si2+
3–
Ans. (c) : Mass of neutron = 1.675 × 10–24 (c) P (d) S2–
g = 1.008 amu, AP EAMCET (Engg.) 21.09.2020, Shift-II
Charge = 0 COMEDK-2017
Mass of electron = 9.11 × 10–28 g, WB-JEE-2010
Charge unit negative = –1.602 × 10–19C Ans. (d) : 16p means z=16
Proton has mass of 1.677×10–24 g = 1.007 amu and 18 e means 2 unit negative charge is present.
proton has one unit positive charge = 1.602×10–19C Hence, species is S2–

Objective Chemistry Volume-I 154

 122. According to Aufbau principle, the sub-shell
2. Atomic Models which is occupied by the electron first has :
(a) higher energy (b) lower stability
117. Which one of the following statements is (c) lower energy (d) can't b predicted
correct? AP-EAMCET-1993
(a) Rutherford's alpha-particle scattering Ans. (c) : The Aufbau Principle states that in the ground
experiment lead to the discovery of electron state of atom or ion, electron fill atomic orbital of the
(b) J. J. Thomson suggested that the nucleus of lowest available energy level's before occupying higher
an atom contains protons levels (e.g. is before 2s). In this way, the electron of an
(c) The atomic number of an element is the same atom or ion from the most stable electron configuration
as the number of protons in the nucleus of its possible.
atom e
(d) The mass number of an atom is equal to the 123. The constancy of ratio for electrons inspite
number of electrons in its shells m
NDA (II)-2015 of variations of gas present in the discharge
tube or of the material of the cathode shows
Ans : (c) The atomic number of an element is the same that :
as the number of proton in the nucleus of its atom.
(a) electrons are negatively charged
118. Rutherford's alpha-particle scattering
(b) electrons are universal constituents of matter
experiment was responsible for the discovery of
(a) Electron (b) Proton (c) electrons are the lightest of all particles
(c) Nucleus (d) Helium 1
NDA (I)-2017 (d) mass of the electron is of the mass of
1838
Ans : (c) Rutherford's alpha–particle scattering was H-atom
responsible for the discovery of atomic nucleus. AP-EAMCET-1994
Nucleus contains most of the mass. Most of atom has
empty space. Ans. (b) : The constancy of e/m ratio for electrons
inspite of variation of gas present in the discharge tube
119. The number of periods present in the long form
of the periodic table is : or the material of the cathode shows that electrons are
(a) 6 (b) 7 universal constituents of matters.
(c) 8 (d) 18 124. Rutherford's experiment on scattering of α-
AP-EAMCET (Med.)-1999 particles showed for the first time that the atom
Ans. (b) : The number of groups in the long form of has :
periodic table is 18. The number of periods in the long (a) nucleus (b) electron
form of periodic table is 7. (c) proton (d) neutron
120. Who discovered that cathode rays are made up AP-EAMCET-1995
of electrons Ans. (a) : Rutherford's show that for first time atom has
(a) William Crookes (b) G. J. Stoney positively charged nucleus through his -particles
(c) R. A. Millikan (d) J. J. Thomson scattering experiment.
MPPET- 2009 Number of scattered particles–
Ans. (d) :
• J.J. Thomson discovered that cathode rays are made
up of electrons.
• William Crookes investigated cathode rays in 1879
and found that they were bent by a magnetic field.
• Robert Millikan was a physicist who discovered the
elementary charge of electron using the oil drop 125. In Rutherford's α-ray scattering experiment,
experiment. the alpha particles are detected using a screen
• G.J. Stoney – He is introducing the term electron as coated with :
the "Fundamental unit quantity of electricity". (a) carbon black
121. When 4p orbital in any atom is filled (b) platinum black
completely, the next electron goes in : (c) zinc sulphide
(a) 5s (b) 3d
(d) polytetrafluoro ethylene
(c) 4d (d) 4f
AP-EAMCET-1999
AP-EAMCET-1991
Ans. (a) : According to the Aufbau principle and (n + Ans. (c) : Rutherford's Alpha Scattering Experiment
1) rule, the orbitals with increasing order their energies directed high energy streams of α-particles from a
is as follows, radioactive source at a thin sheet (100 nm thickness) of
gold.
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
• In order to study the deflection caused to the α-
Therefore, after 4p orbital is completely filled, the next particles, he placed a fluorescent zinc sulphide screen
electron goes in 5s orbital. around the thin gold foil.
Objective Chemistry Volume-I 155
126. Which of the following statements is correct There, n = 2 and z = 3 for Li
about the cathode rays ? Putting these values, we get-
1. These rays start from anode and move 2.18 × 106 × 3
towards cathode. ⇒ V= = 3.27 × 106 ms −1
2. They are visible with human eye. 2
3. In presence of electric and magnetic fields, So, option (b) is correct.
they behave like positively charged 130. The number of protons. Neutrons and electrons
particles. in 136 C respectively are ––
4. Their characteristic does not depend on the (a) 6, 7, 6 (b) 13, 6, 6
nature of material electrode. (c) 6, 7, 13 (d) 6, 6, 13
(a) 1 (b) 2
(c) 3 (d) 4 AP EAPCET 20.08.2021 Shift-I
AP EAMCET (Engg.) 21.09.2020, Shift-I Ans. (a) : For 13C6
Ans. (d) : Characteristics of cathode rays are: Atomic number = Number of protons = 6
(1) The cathode rays start from cathode and move Number of neutrons = Atomic mass number of protons
towards the anode. Thus, statement (1) is incorrect. = 13 – 6 = 7
(2) These rays themselves are not visible but their Number of electrons = number of protons = 6
behaviour can be observed with the help of 131. The wavelength of a spectral line emitted by
phosphor materials such as ZnS. Thus, statement (2)
is incorrect. 16
hydrogen atom in the Lyman series is cm.
(3) In the absence of electrical or magnetic field, these 15R
rays travel in straight lines. In the presence of What is the value of n2? (R = Rydberg
electrical or magnetic field, the behaviour of constant)
cathode rays are similar to that expected from (a) 2 (b) 3
negatively charged particles, suggesting that the (c) 4 (d) 1
cathode rays consist of negatively charged particles, AIIMS-2011
called electrons. Thus statement (3) is incorrect. Ans. (c) : For Lyman series.
(4) The characteristics of cathode rays (electrons) do
not depend upon the material of electrodes and the 1 1 1
nature of the gas present in the cathode ray tube. =R 2 − 2
λ  n1 n 2 
Hence, statement (4) is correct.
127. The nucleus of an atom contains 15R 1 1 
=R 2 − 2
(a) Proton and electron 16 1 n 2 
(b) Neutron and electron
(c) Proton and neutron 15R  n 22 − 1 
= 
(d) Proton, neutron and electron 16R  n 22 
MPPET-2008
15  n 22 − 1 
Ans. (c) : The nucleus of an atom contain of proton and = 
neutron. 16  n 22 
128. "The properties of elements are periodic 16n 22 − 15n 22 − 16 = 0
functions of their atomic weights." This
periodic law was given by n 22 − 16 = 0
(a) Dobereiner (b) Lother Meyer n2 = 4
(c) Mendeleev (d) Alexander 132. In which of the following options, the law of
AP EAMCET (Engg.) 17.09.2020 Shift-I triad is applicable?
Ans. (c) : "The properties of elements are periodic (a) Na, K, Rb (b) Cl, Br, I
functions of their atomic weights." (c) C, N, O (d) Ca, Sr
It is Mendeleev's periodic ion. TS-EAMCET 09.08.2021, Shift-I
129. If the velocity of the electron in the hydrogen Ans. (b) : For Dobereiner’s triads, the atomic weight of
atom in its first orbit is 2.18 × 106 ms–1, then a middle element is almost average of the other two.
will be the velocity of the electron in the second The law of triad is applicable to following species.
orbit of Li2+ ? 35.5 + 126.9
(a) 2.18 × 106 ms −1 (b) 3.27 × 106 ms −1 1) For Cl Br and I = = 81.2
35.5 80 126.9 2
−1
(c) 6.54 × 10 ms
6
(d) 1.45 × 106 ms −1
SCRA - 2009 7 + 39
2) For Li Na and K = = 23
Ans. (b) : Given: The velocity of electron in the first 7 23 39 2
orbit = 2.18 × 106 ms −1
40 + 137
z 3) For Ca Sr and Ba = = 88.5
We know that, V = 2.18 × 106 ms −1 40 88 137 2
n
Objective Chemistry Volume-I 156
133. Match the following 136. Television picture tube is basically
List-I List-II (a) cathode ray tube
A. Chadwick I. Cathode rays (b) anode ray tube
B. Rutherford II. X-rays spectra (c) hybrid of cathode and anode tube
C. Mosley III. Discovery of neutrons (d) none of the above
D. J.J. Thomson IV. Nuclear atom model AMU-2014
The correct match is Ans. (a) : The CRT (cathode ray tube) is basically the
A B C D picture tube of a television set. A television screen is the
(a) IV I II III front surface of a large picture tube, properly called a
(b) III II IV I cathode ray tube.
(c) III II I IV 137. The charge on an electron was discovered by :
(d) III IV II I (a) J.J. Thomson (b) Neil Bohr
TS-EAMCET (Engg.), 06.08.2021 (c) James Chadwick (d) Mullikan
Ans. (d): List-I List-II BCECE-2004
Chadwick → Discovery of neutrons Ans. (d) : The charge of the electron was measured by
Rutherford → Nuclear atom model Mullikan in 1909 and it was found to be 1.602×10–19
Mosley → X-rays spectra coulombs.
J.J. Thomson → Cathode rays 138. Rutherford's famous experiment with
134. According to Moseley, a straight line group is α-particles used this metal
obtained on plotting (a) Ni (b) Au
(a) the frequencies of characteristics X-rays of (c) Fe (d) Zn
elements against their atomic numbers BCECE-2009
(b) the square of the frequencies of Ans. (b) : Rutherford conducted an experiment by
characteristics X-rays of elements against bombarding a thin sheet of gold (Au) with α-particles
their atomic numbers and then studied the trajectory of these particles after
(c) the square root of the frequencies of their interaction with gold foil.
characteristics X-rays of elements against the 139. Which of the following transition of an electron
atomic numbers in H-atom will emit maximum energy?
(d) the reciprocal of the frequencies of (a) n 6  → n5 (b) n1  → n2
characteristics X-rays of elements against the
atomic numbers. (c) n 3 
→ n2 (d) n 4 
→ n3
SRMJEEE – 2008 BCECE-2017
Ans. (c) : According to Moseley, the square root of the Ans. (c) : We know that energy can we calculated by
frequencies of characteristic X-rays of elements against hc
the atomic numbers. ν = a(Z − b) using formula (E) = .
λ
where, a = Proportionality constant 1 1 1
b = Another constant = RZ2  2 − 2 
Z = Atomic number λ  n1 n 2 
135. The position of both an electron and helium 1 1
atom is known within 1.0 nm. The momentum E = RZ2 hc  2 − 2 
n
 1 n 2
of the electron is known within 5.0 × 10-26 kg
ms-1. The minimum uncertainty in the When, transition is from n3 – n2, it produce highest
measurement of the momentum of the helium energy.
atom is 140. The energy of an electron in nth orbit of
(a) 7.0 × 10-26 kg ms-1 (b) 5.0 × 10-26 kg ms-1 hydrogen atom is
(c) 8.0 × 10-26 kg ms-1 (d) 6.0 × 10-26 kg ms-1 13.6 13.6
(a) eV (b) eV
AIIMS-1994 n 4
n3
Ans. (b) : The Heisenberg uncertainty principle 13.6 13.6
h (c) − 2 eV (d) eV
∆x × ∆p ≥ n n
4π CG PET -2007
Where ∆x = Uncertainty in position,
Ans. (c) : Energy of an electron in Bohr’s orbit is given
∆p = Uncertainty in momentum by the relationship.
h 13.6
= constant E n = − 2 eV
4π n
h 141. Which of the following is incorrect regarding
As ∆x is same for electron and helium and is a
4π Rutherford’s atomic model?
constant, therefore minimum uncertainty in the (a) Atom contains nucleus
measurement of the momentum of the helium atom will (b) Size of nucleus is very small in comparison to
be same as that of an electron which is 5.0×10–26 kg ms–1. that of atom
Objective Chemistry Volume-I 157
 (c) Nucleus contains about 90% mass of the atom Ans. (a) : Neutron was discovered by J. Chadwick in
(d) Electrons revolve around the nucleus with may 1932. According to Chadwick it is another sub-
uniform speed. atomic particle and part of an atom. It has no charge i.e.
HP CET-2018 it is electrically neutral and it is represented by n.
Ans. (c) : According to Rutherford's majority of the 146. Which of the following is not true in
mass was concentrated in a minute positively changed Rutherford's nuclear model of atom?
region (the nucleus surrounded by electrons). (a) Protons and neutrons are present inside
142. Given below are two statements: nucleus
Statement-I Rutherford's gold foil experiment (b) Volume of nucleus is very small as compared
cannot explain the line spectrum of hydrogen to volume of atom
atom. (c) The number of protons and neutrons are
Statement-II Bohr's model of hydrogen atom always equal
contradicts Heisenberg's uncertainty principle. (d) The number of electrons and protons are
In the light of the above statements, choose the always equal
most appropriate answer from the options Manipal-2019
given below:
Ans. (c) : The sum of protons and neutrons give the
(a) Statement I is false but statement II is true. mass number, the number of neutrons can be very
(b) Statement I is true but statement II is false. different to the number of protons.
(c) Both statement I and statement II are false.
147. Rutherford's experiment on the scattering of
(d) Both statement I and statement II are true.
α-particles showed for the first time that the
[JEE Main 2021, 27 July Shift-I] atom has:
Ans. (d) : Rutherford’s gold foil experiment only (a) electrons (b) Protons
proved that electrons are held towards nucleus by (c) Nucleus (d) neutrons
electrostatic forces of attraction and move in circular
orbit with very high speeds. UPTU/UPSEE-2005
Bohr’s model gave exact formula for simultaneous Ans. (c) : The conclusion of Rutherford from his α-ray
calculation of speed and distance of electrons from the experiment are:
nucleus, something which was deemed impossible (i) Most of the space inside the atom is empty
according to Heisenberg. because most of the alpha–particles passed
143. If the Thomson model of the atom was correct, through the gold foil without getting deflected.
then the result of Rutherford's gold foil (ii) Very few particles were deflected from their path,
experiment would have been indicating that the positive charge of the atom
(a) All of the α-particles pass through the gold occupies very little space.
foil without decrease in speed. (iii) A very small fraction of alpha–particles were
(b) α-particles are deflected over a wide range of deflected by 180o.
angles Indicating that all the positive charges and mass of the
(c) all α-particles get bounced back by 1800. gold atom were concentrated in very small volume
(d) α-particles pass through the gold foil within the atom. From the data he also calculated that
deflected by small angles and with reduced the radius of the nucleus is about 105 times less the
speed. radius of the atom.
[JEE Main 2021, 27 July Shift-II]
Ans. (d) : As in Thomson model, protons are diffused 3. Dual Nature of Electron
(charge is not centered) α–
Particles deviate by small angles and due to repulsion148. The energy of mole of photons of radiation of
from protons this speed decreases. wavelength 300 nm is (given h = 6.63 × 10–34Js,
144. For d-electron, the orbital angular momentum NA= 6.02 × 1023 mol–1 C = 3 × 108 ms–1)
is (a) 235 kJ mol–1 (b) 325 kJ mol–1
–1
6h 2h (c) 399 kJ mol (d) 435 kJ mol–1
(a) (b) JEE Main-24.06.2022, Shift-II
2π 2π
(c) h / 2π (d) 2h / π Ans. (c) : Given,
J & K CET-(2004) Wavelength of photon = 300 nm
Planck’s constant = 6.63 × 10–34 Js
h
Ans. (a) : The angular momentum L = l(l + 1) ⋅ NA = 6.02 × 1023mol–1
2π C = 3 × 108 ms–1
For d orbital, l = 2 As we know that–
h h hc
∴L= 2(2 + 1) ⋅ = 6. Energy =
2π 2π λ
145. Neutron is discovered by
(a) Chandwick (b) Rutherford 6.63 × 10 –34 × 3 × 108
Energy =
(c) Yukawa (d) Dalton 300 × 10 –9
JIPMER-2008 = 6.63 × 10–19 J

Objective Chemistry Volume-I 158

Energy of one mole of photon We know that –
= 6.63 × 10–19 × 6.02 × 1023 mol–1 h
λ=
= 399 kJ/mol. mv
149. If wavelength of photon is 2.2 × 10 −11 m and 6.6 × 10−34
−34 λ=
h = 6.6 × 10 J s, then momentum of photon 6.6 × 10−31 × 107
−44 −1 −1
(a) 1.452 × 10 kg m s (b) 6.89 × 10 kg m s 43
λ = 10–34 × 1024
−23
(c) 3 × 10 kg ms −1 −22
(d) 3.33 × 10 kg m s −1 λ = 10–10
o
λ = 1A
Karnataka CET-17.06.2022, Shift-II
∴ 1A = 10
o –10
Ans. (c) : Given that –
Wavelength of proton (λ) = 2.2 × 10–11 m 152. What is the approximate wavelength in Å of a
photon having energy 2eV?
Planck's constant = 6.6 × 10–34 Js.
[Planck’s constant = 6.63 ×10–34J.s]
Momentum of photon = ?
(a) 6200 (b) 5100
According to de – Broglie equation
(c) 4600 (d) 3900
h
λ= SCRA-2010
p
Ans. (a) : We know that, E = nhν
h Where, h = Planck’s constant (6.67×10–34 Js)
p=
λ ν = Frequency of radiation
6.6 × 10−34 c = Velocity of light (3×108 m/s)
p= −11
2.2 × 10 c
∴ E = nh
p = 3 × 10–34 × 1011 λ
p = 3 × 10–23 kg. ms–1 nhc
or λ=
150. The minimum energy that must be possessed E
by photons in order to produce the
6.67 × 10−34 × 3 ×108
photoelectric effect with platinum metal is or λ= (1eV) = 1.6×10–19 J
[Given: The threshold frequency of platinum is 1.6 × 10−19
1.3× 1015 s–1 and h = 6.6 × 10–34 Js.] 20.01× 10−7
(a) 3.21 × 10–14 J (b) 6.24 × 10–16 J λ=
2 × 1.6
(c) 8.58 × 10–19 J (d) 9.76 × 10–20 J
λ = 6.2531 ×10−7
JEE Main-25.06.2022, Shift-II
Ans. (c) : Given, λ = 6253 × 10 −10
Threshold frequency of platinum = 1.3 × 1015 s–1 or λ = 6253Å
Planck’s constant (h) = 6.6 × 10–34 Js 153. The longest wavelength present in Balmer
As we know that– series lines is
Energy =
hc [Given Rydberg constant = 1.097 x 107 m-1]
λ (a) 640 nm (b) 656 nm
c (c) 662 nm (d) 670 nm
And frequency (v) = SCRA-2014
λ
Then, Ans. (b) : In the Balmer series, the energy gap is very
Energy = hv less between n = 2 to n = 3.
= 6.6 × 10–34 Js × 1.3 × 1015 s–1 So, the wavelength high in nature.
= 8.58 × 10–19 J Given that, Rydberg constant = 1.097 × 107 m-1
151. A particle of mass 6.6 × 10–31 kg is moving with Using the following Formula–
a velocity of 1 × 107 ms–1. The de Broglie  1 1 
wavelength (in Å) associated with the particle, v = 1.097 × 107  2 − 2 
is (h = 6.6 × 10-34 Js)  n1 n 2 
(a) 1 (b) 10 1 1
or v = 1.097 ×107  − 
(c) 5 (d) 2 4 9
(e) 4 5
Kerala CEE -03.07.2022 or v = 1.097 ×107 ×
36
Ans. (a) : Given that – 7 −1
Mass particle (m) = 6.6 × 10 kg –31 or v = 0.1523 × 10 m
Velocity of particle (v) = 1 × 107 m/s We know that,
Wave length (λ) = ? 1
λ=
h = 6.6 × 10 Js
–34 v

Objective Chemistry Volume-I 159

 1 Ans. (a) : Given that :
λ= velocity of A = 0.05 m/sec.
0.1523 × 107 m −1
velocity of B = 0.02 m/sec.
λ = 6.56 × 10−7 m Let, the mass of particle A = x
or λ = 656nm ∴ mass of particle B = 5x
154. The energy of an electromagnetic radiation is According to de-Broglie equation
19.875×10–13 erg. What is its wave number in
h
cm–1 ? (h=6.625×10–27 erg-s; c=3×1010 cm s–1) λ=
(a) 1000 (b) 106 mv
(c) 100 (d) 10,000 h
For particle A : λ A = (i)
AP-EAMCET-2002 x × 0.05
Ans. (d) : Given that : h
E = 19.875 × 10–13 erg For particle B : λ B = (ii)
5x × 0.02
h = 6.625 × 10–27 erg sec. Divide equation (i) and (ii) we get
C = 3 × 1010 cm.sec–1
λ A 5x × 0.02 2
∴ E = hv = =
λB x × 0.05 1
E
v= λA : λB = 2 : 1
h
158. Two particles of masses m & 2m have equal
19.875 ×10 –13 kinetic energies. The de-Broglie wavelength are
v=
6.625 × 10–27 × 3 × 1010 in the ratio of_____
v = 104 = 10,000 (a) 1 : 1 (b) 1 : 2
155. If the wavelength of an electromagnetic (c) 1 : 2 (d) 2 :1
radiation is 2000 Å, what is its energy in erg? AP EAPCET 23-08-2021 Shift-I
(a) 9.94 × 10–12 (b) 9.94 × 10–19 Ans. (d) : de-Broglie wavelength equation is-
–12
(c) 4.97 × 10 (d) 4.97 × 10–19
h h
AP-EAMCET-2003 λ= =
Ans. (a) : Given,  = 2000 Å = 2000 × 10–10m, E = ? p mv
where,
hc 6.626×10 –27 erg sec×3×108 m/sec λ = Wavelength
∴ E= =
λ 2000×10 –10 m h = Planck's constant
–12
= 9.94 × 10 erg P = Momentum
156. The energy of a photon is 3×10–12 erg. What is m = Mass of the particle
its wavelength in nm? v = Velocity
(h = 6.62 × 10–27 erg-s, c = 3 × 1010 cm s–1) ∴ P = mv
(a) 662 (b) 1324 Squaring both side and divide 1/2 we get-
(c) 66.2 (d) 6.62 1 2 1 2 2
AP-EAMCET-2006 P = mv
2 2
Ans. (a) : Given that :
E = 3 × 10–12 erg 1 2  1 2
P = (K.E.).m ∴ K.E. = mv 
h = 6.62 × 10–27 erg – s 2  2 
C = 3 × 1010 cm/sec P = 2(K.E.)m
∴ we know that :
h
hc So, λ =
E= 2(K.E.)m
λ
Given, two particles of mass m and 2m and kinetic
6.62×10 –27 ×3×1010 energy are same –
λ=
3×10 –12 h
λ = 6.62 × 10–27+22 λ1 = .......(i)
2(K.E.).m
λ = 6.62 × 10–5 cm
h
λ = 662 × 10–7 cm λ2 = .......(ii)
λ = 662 × 10–9 m 2(K.E.)2m
λ = 662 nm dividing (i) & (ii) we get
157. The velocities of two particles A and B are 0.05 λ1 h 2(KE)2m
and 0.02 ms–1 respectively. The mass of B is five = ×
times the mass of A. The ratio of their de- λ2 2(KE)m h
Broglie's wavelength is : λ1 2
(a) 2 : 1 (b) 1 : 4 =
(c) 1 : 1 (d) 4 : 1 λ2 1
AP-EAMCET-2008 ∴ Wavelength ratio = 2 :1
Objective Chemistry Volume-I 160
159. With what velocity must an electron travel so
hν φ
that its momentum is equal to that of a photon V0 = −
of wavelength 663 nm? e e
(a) 1098 m/s (b) 109.8 m/s Where, V0 = Stopping potential
(c) 10.98 m/s (d) 1.098 m/s ν = Frequency
AP EAPCET 23-08-2021 Shift-I φ = Work function
Ans. (a) : Given, λ = 663 nm = 663×10–9 m 162. The energy required to overcome the attractive
me = 9.1×10–31 kg forces on the electrons, w, of some metals is
V=? listed below. The number of metals showing
photoelectric effect when light of 300 nm
h h wavelength falls on it is
∴ λ= =
p me v (1eV = 1.6 × 10–19 J)
Where, h = Planck's constant Metal Li Na K Mg Cu Ag Fe Pt W
p = Momentum w(eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75
h (a) 6 (b) 8
V= (c) 5 (d) 4
meλ AP EAMCET (Medical) - 2013
6.626 × 10−34 Ans. (d) : Given:
V= −31
9.1 × 10 × 663 × 10 −9 wavelength (λ) = 300 nm = 300×10–9 m
6.626 l eV = 1.6 × 10–19 J
V= × 10 6
∴Energy of a photon of radiation of wavelength 300nm
9.1 × 663 will be -
V = 1098 m/sec.
E = hv
160. The ratio of de-Broglie wavelength of two
particles A and B is 2 : 1. If the velocities of A or hc
E=
–1 –1
and B are 0.05 ms and 0.02 ms respectively λ
then the ratio of their masses MA : MB must be
______. or E=
( )
6.6 × 10−34 ( JS ) × 3 × 108 ms −1
(a) 5 : 1 (b) 10 : 1 300 × 10−9 m
(c) 1 : 5 (d) 1 : 8 or E = 6.6×10–19J
AP-EAMCET 25-08-2021 Shift - I 6.6 × 10−19
Ans. (c) : From de-Broglie equation– E = eV
1.6 × 10−19
h or E = 4.14 eV
λ=
mv There are 4 metals for which the energy required to
Where, λ = Wavelength overcome the attractive forces on the electrons, w, is
h = Planck's constant less than 4.14 eV, the energy of light of wavelength 300
nm. Thus the number of metals showing a photoelectric
λ A m B VB
∴ = effect is 4.
λ B m A VA 163. If, the binding energy of electrons in a metal is
m A λ B VB 250 kJ/mol, what should be the threshold
= frequency of the striking photons in order to
m B λ A VA
free an electron from the metal surface ?
1 × 0.02 (a) 6.26 × 1014 s–1 (b) 12.4 × 1014 s–1
= 12 –1
2 × 0.05 (c) 6.26 × 10 s (d) 12.4 × 1012 s–1
0.01 AP EAMCET (Engg.) 21.09.2020, Shift-I
= Ans. (a) Binding energy of 1 mole of electrons = 250 kJ
0.05
Binding energy of 1 electron = (250)/(6.022 × 1023) kJ
m A : m B = 1: 5
= 4.15 × 10–22 kJ
161. The relation between the stopping potential = 4.15 × 10–19 J
(V0) and frequency ( ν ) is correctly represented Threshold energy (hv0) = Binding energy
in [φ = Work function] ∴ hv0 = 4.15 × 10–19 J
or v0 = (4.15 × 10–19 J)/ h
φ hν 2 he φ
(a) V0 = − (b) V0 = + = (4.15 × 10–19 J)/(6.626 × 10–34 J s)
e e ν e = 6.26 × 1014 s–1
hν φ hν Hence, the correct option is (a).
(c) V0 = − (d) V0 = 2
e e e 164. de Broglie was awarded the Nobel Prize in the
TS-EAMCET (Engg.), 05.08.2021 Shift-II year
Ans. (c) : The relation between the stopping potential (a) 1921 (b) 1922
(V0) and frequency ( ν ) is correctly represented by the (c) 1929 (d) 1928
following equation. TS EAMCET 10.08.2021, Shift-I
Objective Chemistry Volume-I 161
Ans. (c) : de Broglie was awarded by Nobel Prize in Na = 2.30 × 1.6022 × 10–19 = 3.68 × 10–19 Joule
1929 for his discovery of the wave nature of electrons. Thus Mg, K and Na metal will eject electron only when
165. Upon irradiation with radiation of a suitable W ≤ E(300) .
wavelength on the cathode, the photocurrent Hence the option (b) is correct
produced was reduced to zero by applying a 167. [Ti(H2O)6]3+ absorbs light of wavelength 498
stopping potential of 2.63 V. If the work nm during a d - d transition. The octahedral
function of the cathode is 4.3 eV. Find the splitting energy for the above complex is --------
approximate wavelength of the radiation (in -×10–19 J. (Round off to the Nearest Integer). h
nm)How many among the given species have = 6.626×10–34 Js; c = 3×108 ms–1
the highest bond order? CN–, CO, NO+, O2+,
O 2, N 2
(a) 224 (b) 179
JEE Main 16.03.2021, Shift-II
(c) 190 (d) 165
TS EAMCET 10.08.2021, Shift-I Ans. (4) : Given- λ = 498nm
The octahedral splitting energy is:
Ans. (b) : By Einstein photoelectric equation
hc
KEmax = hν – , KE max = ev ∆0 or E =
λ
hc
ev = −φ φ = wort function 6.66 × 10−34 × 3 × 108
λ E=
hc 498 ×10−9
= ev + φ c = 3 × 108 m / s or E = 4.00×10–19J (rounding off)
λ 168. Calculate the wave number and frequency of
hc
λ= h = 6.63 ×10−34 / s orange radiation having wavelength 6300 Å.
ev + φ (a) 1.587 × 108 m −1 , 4.761× 1016 s −1
6.63 × 10−34 × 3 ×108 (b) 1.587 × 10 4 m −1 , 4.761× 1014 s −1
λ=
1.6 × 10−19 × 2.63 + 1.6 ×10−19 × 4.3 (c) 1.587 × 106 m −1 ,4.761 × 1014 s −1
19.89 × 10−26 (d) 1.587 × 106 m −1 ,4.761 × 1016 s −1
λ=
1.6 × 10−19 (2.63 + 4.3) AP EAMCET (Engg.) 17.09.2020 Shift-I
19.89 ×10−26 × 1019 Ans. (c) : For orange radiation (EMR) of λ = 6300Å
λ=
1.6 × 6.93 (i) Wave number,
λ = 1.792 × 10−7 v= =
1 1
=
1
λ = 179.2 ×10−9 λ 6300Å 6300 × 10−10 m
= 1.587 × 106 m-1
λ = 179nm
c 3 × 108 m s −1
166. The work functions of Ag, Mg, K and Na (ii) Frequency, =
respectively in eV are 4.3, 3.7, 2.25, 2.30, When λ 6300 × 10−10 m
an electromagnetic radiation of wavelength of = 4.761 × 1014s-1
300 nm is allowed to fall on these metal surface, 169. Which of the following statements is correct
the number of metals from which the electrons about photoelectric effect ?
are ejected is (1ev = 1.6022 × 10–19J) (A) The number of electrons ejected from metal
(a) 4 (b) 3 surface is inversely proportional to intensity
(c) 2 (d) 5 of light.
AP EAMCET-2017 (B) Below threshold frequency, photoelectric
effect can be observed.
Ans. (b) : Given that, (C) At higher frequency than threshold frequency,
λ = 300 nm = 300 × 10–9 m the ejected electrons have certain kinetic
hc energy.
E=
λ (D) At higher frequency than threshold frequency,
the electron is still on the metal surface.
6.6 × 10 –34 × 3 × 108
E= (a) 1 (b) 2
300 × 10 –9 (c) 3 (d) 4
19.8 ×10 –26 AP EAMCET (Engg.) 18.09.2020, Shift-I
E=
300 × 10 –9 Ans. (c) : Kinetic energy of photoelectrons
E = 6.6 × 10–19 Joule KE = hv – hv0
And work function for Ag, Mg, K and Na in Joules are – ⇒ When > V0, KE > 0
Ag = 4.3 × 1.6022 × 10–19 = 6.89 ×10–19 Joule So, statement C is correct.
Mg = 3.7 × 1.6022 × 10–19 = 5.93 × 10–19 joule Where, v = Frequency of incident light
K = 2.25 × 1.6022 × 10–19 = 3.60 × 10–19 joule V0 = Threshold frequency

Objective Chemistry Volume-I 162

Now, other options will be correct when 173. If two particles A and B are moving with the
A. Number of photoelectrons ejected from metal surface same velocity. But wavelength of A is found to
intensity of light. be double than that of B. Which of the
B. When, v < V0, KE < 0 phtoelectric effect cannot be following statements are correct?
observed. (a) Both A and B have same mass
D. When, v = v0, KE = 0 the electrons will be still on (b) Mass of A is half that of B
the metal surface. (c) Mass of B is half that of A
170. Calculate energy of half mole of photons of a (d) Mass of B is one-fourth that of A
radiation with frequency 3 × 1012 Hz. AP EAPCET 19-08-2021 Shift-I
(a) 598.2 kJ mol–1 (b) 0.598 kJ mol–1 Ans. (b) : From the equation of De-Broglie
–1
(c) 1.196 kJ mol (d) 119.6 kJ mol–1
h
AP EAMCET (Engg.) 21.09.2020, Shift-II λ = Where λ = wavelength
Ans. (b) : Energy of a photon, E = hv, where, p
v = Frequency = 3 × 1012 Hz Or λ =
h
h = Plank’s constant
Therefore, energy m.v
= 6.6 × 10–34 × 3 × 1012 Hz = 1.9878 × 10–21 J P = Momentum
1 mole = 6.02 × 1023 photons, For particle A –
1 h
Therefore, mole = 3.01 × 1023 photons 2λ = ……. (i)
2 mA v
1 For particle B –
Therefore, energy of mole of photons
2 h
= 1.9878 × 10 × 3.01 × 1023
–21 λ= ……. (ii)
mBv
= 0.5982 kJ/mol
Hence, the correct option is (b). From equation (i) and (ii)
171. One mole of alkene X on ozonolysis gave one h m v
− 2= × B
mole of acetaldehyde and one mole of acetone. mA v h
The IUPAC name of X is mB
− Or 2=
(a) 2-methyl-2butene (b) 2-methyl-1-butene mA
(c) 2-butene (d) 1-butene
mB
AP EAMCET (Engg.)-2009 mA =
2
Ans. (a): X O3
→ CH 3CHO + CH 3COCH 3 174. Identify the correct statements regarding
The molecule X must be a compound Balmer series: The wave number of each line in
containing 5 carbon atoms, and thus, X cannot be 1– the spectrum of atomic hydrogen is given the
butene or 2–butene as they contain 4 carbon atoms only.
 1 1 
equation v = R H  2 − 2  where RH is a
n
 1 n 2 
constant and n1 and n2 are integers.
2–methyl–2–butene (i) As wave length decrease, the lines in the
series constant
(ii) The integer n1 is equal to 2
(iii) The ionization energy of hydrogen can be
calculated from the wave number of these
2-methyl – 1 – butene lines
2–methybut –2– ene gives desired products. (iv) The line of longest wavelength corresponds to
172. The de-Broglie wavelength of a tennis ball of n2 = 3
mass 60 g moving with a velocity of 10 ms–1 is Options:
approximately ....... (Planck's constant h = 6.63 (a) (i), (ii) & (iii) only (b) (i), (iii) & (iv) only
× 10–34 J. s) (c) (ii) & (iv) only (d) (ii), (iii) & (iv) only
(a) 1.1 × 10–31 m (b) 1.1 × 10–33 m AP EAPCET-6 Sep. 2021, Shift-II
–34
(c) 1.1× 10 m (d) 1.1 × 10–32 m Ans. (c) : Explanation:– For Balmer series in the
AP EAMCET (Engg.) 18.9.2020 Shift-I spectrum of hydrogen,
Ans. (b) : de-Broglie's wave equation is
1 1
h 6.63 × 10−34 v = RH  2 − 2 
λ= = = 1.10 × 10−33 m  n1 n 2 
mv ( 60 × 10−3 ) × 10
Where, RH = Rydberg’s constant Incase of
[m = Mass of the ball = 60g = 60 × 10–3kg, Balmer series, transition takes place from various
v = velocity of the ball = 10 ms–1] energy levels such as n2 =3,4,5…..to n1=2.

Objective Chemistry Volume-I 163

For longest wavelength, minimum energy transition or λ = 6.625 ×10–5 cm
takes place n2 = 3 to n1 = 2. Since, energy is inversely or λ = 6.625 × 10–7 m
proportional to wavelength. or λ = 662.5 × 10–9 m
1 or λ = 662.5 nm
i.e, E ∝ As, the wavelength decreases, the lines in
λ 178. Calculate the velocity of an electron having
the series converge. wavelength of 0.15 nm. Mass of an electron is
–28
175. It the work function for the photoelectron 9.109 × 10 grams. (h = 6.626 × 10–27 erg-
emission of a mental is 3.75 eV, then the second).
threshold wave length of the radiation needed (a) 4.85 × 108 cm. sec.–1
for the ejection of the electron is (b) 2.062 × 1015 cm. sec.–1
approximately_____. (c) 2.062 × 1010 cm. sec.–1
(a) 315 nm (b) 280 nm (d) 4.85 × 109 cm. sec.–1
(c) 330 nm (d) 290 nm GUJCET-2011
AP EAPCET 20.08.2021 Shift-II Ans. (a): Given that,
ο 12.375 ο ο
Ans. (c): λ(∈ A) = A ≈ 3300 A λ = 0.15 nm = 0.15 × 10–9 m
3.75 M = 9.109 × 10–28 gm = 9.109 × 10–31 kg
ο
Or λ ≈ 330 nm [Q 1 A = 0.1 nm] h = 6.626 × 10–27 erg-second = 6.626 × 10–34 J sec.
According to de Broglie equation–
176. A 150 watt bulb emits light of wavelength 6600
Å and only 8% of the energy is emitted as light. λ = h = h
How many photons are emitted by the bulb per p mv
second?
h 6.626 ×10−34
(a) 4×1019 (b) 3.24×1019 v= =
(c) 4.23×1020 (d) 3×1020 mλ 9.109 × 10−31 × 0.15 × 10−9
6 8
COMEDK-2014 v = 4.8494 × 10 m/sec = 4.849 × 10 cm/sec.
Ans. (a) : Power of the bulb= 150 watt= 150 J s–1 179. Which of the following relations is correct, if
As only 8% of the energy is emitted as light so, the total the wavelength (λ) is equal to the distance
energy emitted per second travelled by the electron in one second?
150J × 8 h is the Planck's constant and m is the mass of
= = 12J electron
100
(a) λ = h/p (b) λ = h/m
hc
Energy of one photon, E = hυ = (c) λ = h/p (d) λ = h/m
λ
(6.626 × 10 Js) × (3 × 10 m s )
–34 8 –1 TS EAMCET 04.08.2021, Shift-I
= = 3.0118×10 –19
J Ans. (d) : We know that–
6600 × 10 –10 m
∴ Number of photons emitted λ=
h
.....(i)
12J mv
= = 3.98 × 10 ≈ 4.0 × 10
19 19
Here, m = Mass of electron
3.0118 × 10 –19 J
177. The energy of an electromagnetic radiation is v = Velocity of electron
3 × 10–12 ergs. What is the wavelength in λ = Wavelength of electron
nanometers? If wavelength (λ) is equal to distance travelled by the
h = 6.625 × 10–27 erg sec., electron in one second i.e.
c = 3 × 1010 cm sec.–1 λ=v
(a) 400 (b) 228.3 Put in equation (i), we get–
(c) 3000 (d) 662.5
h
AP-EAMCET (Medical), 2002 λ=
Ans. (d) : Given that – mλ
–12
E = 3 × 10 ergs h
⇒ λ =
2

λ = ? (nanometer) m
h = 6.625 × 10–27 erg. sec
C = 3 × 1010 cm sec–1 h
λ=
hc m
Now, E =
λ 180. When light of wavelength 248nm falls on a
metal of threshold energy 3.0 eV, the de-
hc
or λ= Broglie wavelength of emitted electrons
E is______Å. (Round off to the Nearest Integer).
6.625 ×10−27 (erg.sec) × 3 × 1010 (cm sec −1 ) [me = 9.1×10−31 kg;c = 3.0 ×108 ms−1;1eV = 1.6 ×10−19 Js]
or λ= −12
3 ×10 (ergs) JEE Main 16.03.2021, Shift-I
Objective Chemistry Volume-I 164
Ans. (9Å) : Given that, Wavelength ( λ ) = 248nm Dividing equation (i) by (ii) we get–
λ1 hc 100eV
Threshold energy ( φ ) = 3.0eV = ×
λ 2 25 eV hc
de–Broglie wavelength ( λ ) = ? λ1 : λ2 = 4 : 1
hc 182. de Broglie relationship has no significance for
Now, Energy incident = (a) An electron (b) A proton
λ
Where, h = Planck's constant (c) A neutron (d) An iron ball.
c = Speed of light SRMJEEE – 2009
λ = Wavelength Ans. (d) : According to the de-Broglie, every object in
motion has a wave character. The wavelength
6.6×10-34 × 3 ×108
∴ E= eV associated with ordinary objects are so sort because of
248 × 10−19 × 1.6 × 10−19 their large masses that their wave properties cannot be
E = 5eV detected. The light mass of the particle can be detected
From the Einstein photoelectric equation– experimentally but not in the case of large mass.
E = φ + K.E. 183. The wavelength associated with a particle of
or K.E. = 5 eV – 3 eV mass 3.313 × 10–31 kg moving with velocity 103
or K.E. = 2 eV m/s is
or K.E. = 2×1.6×10–19=3.2×10–19 J (a) 2 × 10–6 m (b) 2 × 10–6 cm
h (c) 2 × 10 m–7
(d) 2 × 10–7 cm
For de-Broglie wavelength ( λ ) = SRMJEEE – 2010
mv
1 Ans. (a) : Given that, Mass of particle = 3.313 × 10–
31
K.E. = mv 2
kg
2 v = 103 m/s, λ = ?
2 ( K.E.) Now, from de–Broglie equation –
So, v=
m h
λ=
h p
Hence, λ =
2 ( K.E.) × m Where, λ = Wavelength
h = Planck's constant
6.6 ×10−34 p = Momentum
or λ=
2 × 3.2 × 10−19 × 9.1× 10−31 6.626 ×10−34 kgm 2s −1
∴ λ=
6.6 ×10−34 3.313 × 10−31 kg ×103 ms −1
or λ=
58.24 ×10−50 = 2 × 10–6 m
or λ=9Å 184. Which one of the following frequencies of
181. If the energies of two light radiations E1 and E2 radiation (in Hz) has a wavelength of 600 nm?
are 25 eV and 100 eV respectively, then their (a) 2.0×1013 (b) 5.0×1016
14
respective wavelength λ1 and λ2 would be in the (c) 2.0×10 (d) 5.0×1014
ratio λ1 : λ2 = ______ AP-EAMCET- (Engg.)-2011
(a) 2:1 (b) 4:1 Ans. (d) : Given that,
(c) 1:4 (d) 1:2 Wavelength (λ) = 600 nm = 600 × 10–9 m
AP EAPCET 19-08-2021, Shift-II We know that–
Ans. (b) : Given that, E1 = 25 eV, E2 = 100 eV c 3 ×108
λ1 : λ2 = ? ν = =
λ 600 × 10−9
From the equation of energy - ν = 5 × 1014 Hz.
hc
E= 185. If the kinetic energy of a particle is reduced to
λ half, de-Broglie wavelength becomes :
Where, E = Energy 1
h = Planck's constant (a) 2 times (b) times
λ = Wavelength 2
c = Speed of light (c) 4 times (d) 2 times
For first radiation– AP-EAMCET (Engg.) 2015
hc hc
E1 = λ ⇒ = λ1 ...( i ) Ans. (d) : Given that, (K.E.) 2 = 1/2 (K.E.)1
1 25 eV According to de-Broglie equation
For second radiation– h h h
λ= = =
hc hc mv p 2m K.E.
E2 = ⇒ = λ 2 ...( ii )
λ2 100 eV Where, K.E. = Kinetic energy

Objective Chemistry Volume-I 165

 th
So, the de-Broglie wavelength is inversely proportional 189. What is the wave number of 4 line in Balmer
to square root of kinetic energy (K.E.) series of hydrogen spectrum?
(R = 1,09,677 cm–1)
λ1 (K.E.)2 (a) 24,630 cm–1 (b) 24,360 cm–1
Therefore, =
λ2 (K.E.)1 (c) 24,730 cm –1
(d) 24,372 cm–1
λ1 1 (K.E.)1 AP - EAMCET (Medical) - 2007
Then, = Ans. (d) : We know that,
λ2 2 (K.E.)1
 1 1 
or λ 2 = 2λ1 v = R 2 − 2 
n
 1 n 2 
Thus, de-Broglie wavelength becomes 2 times to the Where, v = Wave number
original wavelength. R = Rydberg constant
186. The frequency of radiation emitted, when an 4th line in Balmer series that means–
electron falls from n = 3 to n = 1. in a hydrogen n1 = 2, n2 = 6
atom would be _____
 1 1 
(a) 2.92 ×103hour–1 (b) 2.92 ×10-15s–1 ∴ v = 1,09,677  2 − 2  cm −1
(c) 2.92 ×10 s 15 –1
(d) 2.92 min –1 2 6 
AP- EAPCET- 07-09-2021, Shift-I 1 1 
v = 1,09,677  −  cm −1
Ans. (c) : Given that, n1= 1, n2 = 3  4 36 
According to the Rydberg's formula– or v = 24372cm −1
1  1 1  190. What is the wavelength (in m) of a particle of
∴ = RH  2 − 2 
λ  n1 n 2  mass 6.62 × 10–29 g moving with a velocity of
103 ms–1?
1 1 1  (a) 6.62×10–4 (b) 6.62×10–3
= 1.096 × 105  − 2 
λ 1 3  (c) 10 –5
(d) 105
1 8 AP-EAMCET (Engg.)-2005
= 1.096 × 105 × = 0.968 × 105 cm −1
λ 9 Ans. (c) : Given that, m = 6.62×10–29g = 6.62×10–32 kg
We know that, v = 103 ms–1, λ (in m) = ?
c Now, from the de-Broglie equation–
∴ v=
λ h
Q λ=
= 3 × 10 × 0.968 × 10
10 5
mv
= 2.92 ×1015 sec –1 6.6 × 10 −34 J s
∴ λ =
187. Which one of the following transitions of an 6.62 × 10 −32 (kg) × 103 (ms −1 )
electron in hydrogen atom emits radiation of
the lowest wavelength? or λ = 1×10–5 m
(a) n 2 = ∞ to n1 = 2 (b) n 2 = 4 to n1 = 3 191. The wavelengths of two photons are 2000Å and
4000Å respectively. What is the ratio of their
(c) n 2 = 2 to n1 = 1 (d) n 2 = 5 to n1 = 3 energies?
AP-EAMCET- (Engg.) - 2010 (a) 1/4 (b) 4
hc (c) 1/2 (d) 2
Ans. (c) : ∆E = VITEEE 2019
λ
When the value of 'n' increases the difference of energy Ans. (d) : Given that, λ 1 = 2000 Å , λ 2 = 4000 Å
between successive orbit decreases. We know that,
i.e, E 2 − E1 > E 3 − E 2 > E 4 − E 3 so on--- hc
∴ E=
∴ λ is lowest for n2=2 to n1 =1 transitions. λ
188. The basis of quantum mechanical model of an So, E hc λ
1
= × 2
atom is E 2 λ1 hc
(a) Angular momentum of electron
E1 4000 Å
(b) Quantum numbers =
(c) Dual nature of electron E 2 2000 Å
(d) Black body radiation E1
=2
AP-EAMCET (Engg.) 2013 E2
Ans. (c) : The basis of quantum mechanical model of an 192. The values of Planck’s constant is 6.63 × 10–34
atom is dual nature of electron i.e. electron possesses Js. The velocity of light is 3.0 ×108 m s–1. Which
particle nature as well as wave nature. It means when value is closest to the wavelength in nanometers
the matter is moving its shows the wave property are of a quantum of light with frequency of 8 × 1015
associated with it and when it is in the state. s–1?
Objective Chemistry Volume-I 166
 (a) 5 × 10–18 (b) 4 × 101 (a) 1060 nm (b) 496 nm
(c) 3 × 107 (d) 2 × 10–25 (c) 300 nm (d) 215 nm
AIPMT 2003 VITEEE 2013
−34
Ans. (b) : Given that, h = 6.626 × 10 J sec. Ans. (b) : E = E1 + E2
c = 3 × 108 m / sec. , v = 8 × 1015 sec −1 hc hc hc
= +
hc λ λ1 λ 2
Q E = hv =
λ 1 1 1
= +
3 ×108 300 760 λ 2
8 × 1015 =
λ 1 1 1
= −
3 × 108 λ 2 300 760
λ= = 0.375 × 10−7 ≈ 4 × 101 nm
8 × 1015 λ2 = 496 nm
193. The ratio of slopes of Kmax vs v and v 0 vs v 196. Electrons with a kinetic energy of 6.023 × 104 J/
curves in the photoelectric effects gives (v = mol are evolved from the surface of a metal, when
frequency, Kmax = maximum kinetic energy, v0 it is exposed to radiation of wavelength of 600 nm.
= stopping potential) The minimum amount of energy required to
(a) The ratio of Planck’s constant of electronic remove an electron from the metal atom is
charge
(b) Work function (a) 2.3125 × 10−19 J (b) 3 × 10−19 J
(c) Planck’s constant (c) 6.02 × 10−19 J (d) 6.62 × 10−34 J
(d) Charge of electron VITEEE- 2009
VITEEE 2015
Ans. (a) : Given that, KE of 1 mol = 6.023 × 104 J
Ans. (d) : We know that, h v = h v 0 + e v 0
or KE of 6.023 × 1023 atoms = 6.023 × 104 J
h h
ev0 = hv − hv0 , v0 = v − v0 6.023 × 104
e e ∴ KE of 1 atom =
On comparing this equation with the straight line 6.023 × 1023
equation i.e y = mx + c we know that, = 1.0 × 10–19 J
The slope of v 0 vs v is hc 6.626 × 10−34 × 3 × 108
h Q hν energy = =
(slope)1 = λ 600 × 10−9
e = 3.313 × 10–19 J
h v = h v 0 + KE Now Threshold energy = hν − KE
KE = h v – h v 0 (constant) = 3.313 × 10–19 –1.0 × 10–19
y = mx + (–h v 0) =2.313 × 10–19J
Thus, slope of Kmax vs v is Hence minimum amount of energy required to remove
( slope )2 = h an electron from the metal ion will be 2.313 × 10–19 J
197. The wavelengths of electron waves in two
( slope )2 h
∴ = = e (charge of electron) orbits is 3 : 5. The ratio of kinetic energy of
( slope )1 h / e electrons will be
194. A certain metal when irradiated by light (r = (a) 25 :9 (b) 5 : 3
3.2 × 1016Hz) emits photoelectrons with twice (c) 9 : 25 (d) 3 : 5
kinetic energy as did photoelectrons when the VITEEE- 2009
same metal is irradiated by light (r = 2. 0 × Ans. (a) : According to de-Broglie's equation -
1016Hz). The v0 of metal is
(a) 1.2 × 1014 Hz (b) 8 × 1015 Hz h
λ=
16
(c) 1.2 × 10 Hz (d) 4 × 1012 Hz mν
Ans. (b) : According to question squaring both side, we get −
( KE )1 = 2 × ( KE )2 h2
λ2 =
hv1 − hv0 = 2 ( hv2 − hv0 ) m2ν 2
hv0 = 2hv2 − hv1 h2
or mν 2 =
v0 = 2v2 − v1 mλ 2
v0 = 2 × ( 2 × 1016 ) − ( 3.2 × 1016 ) 1
Q KE = mν 2
2
v0 = 8 × 10 Hz 15
2 2
195. A near UV photon of 300 nm is absorbed by a Therefore, K1  λ 2   5 
=  = 
gas and then re-emitted as two photons. One K 2  λ1   3 
photon is red with wavelength of the second
photon is ∴ K1 : K2 = 25 : 9
Objective Chemistry Volume-I 167
198. A bulb emitted electromagnetic radiation of Ans. (d): Energy of a photon = hν
660 nm wavelength. The total energy of
c
radiation is 3 × 10–18 J. The number of emitted Qv=
photon will be: λ
(h = 6.6 × 10–34 Js, c = 3 × 108 m/s) hc
Hence, E =
(a) 1 (b) 10 λ
(c) 100 (d) 1000 So, energy depends on wavelength.
AIIMS 26 May 2019 (Morning) 201. Wavelength of a particular transition for H
Ans. (b): Given that, total energy of radiation = 3×10– atom is 400 nm. What can be wavelength of
18
J He+ for same transition?
h = 6.6 × 10 –34 Js, c = 3 × 108 , λ = 660 × 10 –9 (a) 400 nm (b) 100 nm
Number of photons emitted = n (c) 1600 nm (d) 200 nm
We know that, AIIMS-26 May, 2018
nhc Ans. (b): According to the Rhyberg's equation–
E=
λ 1  1 1 
Where, h = the plank's constant = R H  2 − 2  × Z2
λ  n1 n 2 
c = the speed of light
and λ = wavelength 1  1 1 
= R H  2 − 2  (1) 2 ..…(i)
Now, by substituting the value of these, we get– 400 n
 1 n 2 
6.6 × 10−34 × 3 ×108 × n
3 × 10−18 = 1  1 1 
660 × 10−9 = R H  2 − 2  (2)2 …..(ii)
λ He+  n1 n 2 
30
n= =10 On dividing equation (i) by (ii), we get
3
199. Assertion: Threshold frequency is the 400 400
λ He+ = 2 = = 100 nm
maximum frequency required for the ejection 2 4
of electron from the metal surface. 202. Find the frequency of light that corresponds to
Reason: Threshold frequency is characteristic photons of energy 5.0×10−5 erg?
of a metal. (a) 7.5 × 10−21 sec −1 (b) 7.5 × 10 −21 sec
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of (c) 7.5 × 1021 sec −1 (d) 7.5 × 1021 sec
Assertion AIIMS-2010
(b) If both Assertion and Reason are correct, but Ans. (c): We know that, E= hv.
Reason is not the correct explanation of
E 5.0 × 10−5 erg
Assertion. v= =
(c) If Assertion is correct but Reason is incorrect. h 6.63 × 10−34 Js
(d) If both the Assertion and Reason are 5.0 ×10−5 erg
incorrect. v=
6.63 × 10−34 × 107 erg sec
AIIMS-26 May, 2018
∴[1J= 107 erg]
Ans. (d): Threshold frequency is a minimum frequency
required for the emission of electron from the metal v = 7.54×1021 sec–1
surface. 203. Ratio of energy of photon of wavelength 3000Å
200. Assertion: All photons possess the same and 6000Å is
amount of energy. (a) 3 : 1 (b) 2 : 1
Reason: Energy of photon does not depend (c) 1 : 2 (d) 1 : 3
upon wavelength of light used. AIIMS-2012
(a) If both Assertion and Reason are correct and Ans. (b): Given that, ratio of energy of photon of
the Reason is a correct explanation of the wavelength 3000Å and 6000Å
Assertion.
hc
(b) If both Assertion and Reason are correct but E=
Reason is not a correct explanation of the λ
Assertion. E1 λ 2 6000
(c) If the Assertion is correct but Reason is = = = 2 :1
E 2 λ1 3000
incorrect.
(d) If both the Assertion and Reason, are 204. The de Broglie wavelength associated with a
incorrect. ball of mass 1 kg having kinetic energy 0.5 J is :
(e) If the Assertion is incorrect but the Reason is (a) 6.626 × 10 −34 m (b) 13.20 × 10 −34 m
−21
correct. (c) 10.38 × 10 m (d) 6.626 × 10 −34 Å
AIIMS-1998 AIIMS-2006
Objective Chemistry Volume-I 168
Ans. (a): Given data, mass of ball = 1 kg 207. The wavelength of visible light is:
Kinetic energy (K.E) = 0.5 J (a) 2000Å−3700Å (b) 7800Å−8900Å
de-Broglie wavelength (λ) = ? (c) 3800Å−7600Å (d) None of these
Now, as we know that, AIIMS-1998
1 Ans. (c): The visible spectrum is the portion of the
K.E. = mv 2 electromagnetic spectrum that is visible to the human
2 eye.
1
0.5 = × 1× v 2 Electromagnetic radiation in this range of wavelengths
2 is called visible light or simple light. A typical human
v2 = 1 eye will respond to wavelengths from about.
v = 1 ms–1 380 to 700 nm or 3800Å – 7600Å
h The visible spectrum is VIBGYOR. Violet has the
Therefore, de-Broglie wavelength λ = shortest wavelength, at around 380 nanometers, and red
mv has the longest wavelength, at around 700 nanometers
6.63 × 10−34 in this spectrum.
=
1× 1 208. The wavelength of a 150 g rubber ball moving
=6.63×10–34 m with a velocity of 50 ms−1 is :
205. The de- Broglie wavelength of an electron in (a) 3.43 × 10−33 cm (b) 5.86 × 10 −33 cm
the ground state of hydrogen atoms is: (c) 7.77 × 10−33 cm (d) 8.83 × 10 −33 cm
(K.E.= 13.6eV; 1ev=1.602×10−19 J) AIIMS-1998
(b) 33.28nm (b) 3.328nm Ans. (d): Given that, v = 50 m/sec, m = 150 gm =
(c) 0.3328nm (d) 0.0332 nm 150×10–3 kg
AIIMS-2000 According to de-Broglie –
Ans. (c): For electron in the ground state,
h 6.626 × 10−34
h λ= =
Q mvr = mv (150 × 10−3 ) × 50
2π
⇒ 8.83×10–35 m
h
Or mv = ⇒ 8.83×10–33 cm
2πr 209. The de-Broglie wavelength of a particle with
h mass 1 g and velocity 100 m/s is
Now, mv=
λ (a) 6.63 × 10–35 m (b) 6.63 × 10–34 m
h h (c) 6.63 × 10 m
–33
(d) 6.65 × 10–35 m
So, = NEET-1999
λ 2πr
Ans. (c) : Given that, m=1g, v=100 m/s
λ=2πr h= 6.63×10–34 J-s
According to de-Broglie wave equation –
λ=2×3.14×0.53 Å =3.328 Å
h 6.63 ×10−34 J-s
= 3.328×10–10 m Q 1 Å = 10-10m λ= =
= 0.3328×10–9 m mv 1× 10−3 Kg × 100m / s
= 0.3328 nm
206. The de- Broglie wavelength associated with a = 6.63×10–33 m
–34
particle of mass 10−6 Kg moving with a velocity 210. If the Planck's constant h = 6.6 × 10 Js, the
−1
of 10 ms is : de Broglie wavelength of a particle having
−7 −16 momentum of 3.3 × 10–24 kg ms–1 will be
(a) 6.63 × 10 m (b) 6.63 × 10 m
(a) 0.002 Å (b) 0.5 Å
(c) 6.63 × 10 −21 m (d) 6.63 × 10 −29 m
AIIMS-2001 (c) 2 Å (d) 500 Å
Ans. (d): According to the de Broglie equation, BITSAT 2018
h Ans. (c) : Given that,h = 6.6 × 10 , p = 3.3 × 10-24
-34
λ=
mv h
We know that, λ = (Q p = mv)
Where, h = Planck’s constant mv
v = velocity h
m = mass of the particle Or λ=
p
λ = wavelength of the particle.
6.63 ×10−34 Js h 6.6 × 10 –34
λ = −6 λ = = –24
= 2 × 10 –10 m = 2Å
10 Kg × 10ms −1 p 3.3 × 10
211. The energy required to remove an electron
6.63 × 10−34 from metal X is E = 3.31 × 10-20 J. Calculate the
= −5
10 maximum wavelength of light that can photo
= 6.63×10–29 m eject an electron from metal X.
Objective Chemistry Volume-I 169
 (a) 6.01 × 10–6 m (b) 3.01 × 10–3 m hc
(c) 5.01 × 10–6 m (d) None of these Ans. (a) : We know that, E=hv=
λ
AMU-2014 Again, E=mc2
Ans. (a) : Given that, E = 3.31 × 10-20 J hc
h = 6.626 × 10-34 J.sec ,c = 3 × 108 ∴ mc2=
m/s λ
hc h 6.626 × 10−37
Energy (E) = or m= = kg
λ cλ 3 × 108 × 3.6 × 10−10
=0.6135×10-32 kg =61.35×10-34 kg
hc 6.626 × 10−34 × 3 × 108
Or λ = = 215. The de-Broglie wavelength (λ) associated with
E 3.31× 10−20 a photoelectron varies with the frequency (ν) of
⇒ 6.01×10 m –6
the incident radiation as, [ν0 is threshold
212. The energy of one mole of photons of radiation frequency]
whose frequency is 5 × 1014 Hz will be
1 1
(a) 19.951 KJ mol–1 (b) 199.51 KJ mol–1 (a) λ ∝ 1
(b) λ ∝ 3
–1
(c) 39.90 KJ mol (d) 399.0 KJ mol–1 (ν − ν 0 ) 4 (ν − ν 0 ) 2
AMU-2015, 2007
1 1
Ans. (b) : Energy of quantum (or photon), E=hv (c) λ ∝ (d) λ ∝
Where, Planck’s constant, h = 6.62 × 10–34 J-s (ν − ν 0 ) 1
(ν − ν 0 ) 2
1 mol of photon = 6.023 × 1023 molecules of photon so [JEE Main 2019, 11 Jan Shift-II]
the Energy of 1 mole of photon will be,
Ans. (d) : For electron
E= N0hv
λ
=6.023×1023×6.62×10–34×5×1014 J/mole λ DB = (de − broglie wavelength)
=199.36 × 103 J/mole 2mK.E
or 199.36KJ mol–1 Where λ = Wavelength of particles
213. In hydrogen atom, the de Broglie wavelength of h = Planck’s constant
an electron in the second Bohr orbit is [Given K.E = Kinetic energy
that Bohr radius, a0 =52.9 pm] According to photoelectric effect
(a) 211.6 pm (b) 211.6 π pm hv = hv0 +KE
(c) 52.9 π pm (d) 105.8 pm KE= hv – hv0
NEET-Odisha 2019
h 1
Ans. (b) : According to Bohr, λ DB = , λ DB ∝ 1
nh 2m × (hv − hv0 )
mvr = (v − v0 ) 2
2π 216. The energy of a photon is given as, ∆E/atom =
nh 3.03×10–19 J atom–1 then, the wavelength (λ) of
2πr= =nλ ………..(i)
mv the photon is
Where, r= Radius (a) 0.656 nm (b) 6.56 nm
λ= Wavelength (c) 65.6 nm (d) 656 nm
n= Number of orbit AMU–2001
a n2 Ans. (d) : Given that, ∆E = 3.03×10-19J atom–1
Also, r= o ………..(ii) According to formula –
z
Where, ao = Bohr radius= 52.9 pm hc 6.626 × 10−34 × 3 ×108
Z= Atomic number ∆E = =
λ λ
On substituting the value of ‘r’ from equation (ii) to
hc
equation (i) we get λ= = 656 nm
λ
2πn a o
2
nλ = 217. The wavelength of a ball of mass 100 g moving
z with a velocity of 100 ms-1 be
2πna o (a) 6.626 × 10–30 m (b) 6.626 × 10–35 m
λ=
z (c) 6.626 × 10 m –32
(d) 6.626 × 10–34 m
λ = 2π× 2 × 52.9 [Qn = 2, z = 1] Assam CEE-2020
λ = 211.6π pm Ans. (b) : Given that, m = 100g , v = 100 m/s
214. The mass of a photon with wavelength 3.6Å According to the de-Broglie equation –
shall be h 6.626 ×10−34 kg − m 2 / sec
-34
(a) 61.35×10 kg -29
(b) 6.135×10 kg λ = =
-29 -29
mv 0.1kg × 100 m/sec
(c) 7.185×10 kg (d) 71.85×10 kg
AMU-2006 = 6.626 ×10−35 m

Objective Chemistry Volume-I 170

218. The energy ratio of a photon of wavelength Ans. (a) : Given that, λ1 = 2000 Å and λ2 = 8000 Å
3000 Å and 6000 Å is Using the equation of energy–
(a) 1 : 1 (b) 2 : 1 hc
(c) 1 : 2 (d) 1 : 4 E=
BCECE-2007 λ
E1 λ 2  1
Ans. (b) : Given that, λ1=3000 Å, λ2=6000 Å = Q E ∝ 
hc hc hc hc E 2 λ1  λ
E1 = = and E 2 = = E 8000
λ1 3000 λ 2 6000 Thus, 1
= =4
E 2 2000
hc
E1 3000 hc 6000 2 ∴ E1= 4E2
= = × = 222. Which of the following equations represent de-
E2 hc 3000 hc 1
Broglie relation?
6000
E1:E2=2:1 h v
(a) =p (b) λm =
219. A photon having a wavelength of 845 Å, causes mv p
the ionisation of N atom. What is the ionisation h h
energy of N? (c) λ = (d) λ =
(a) 1.4 kJ (b) 1.4 × 10 kJ
4 mp mv
(c) 1.4 × 102 kJ (d) 1.4× 103 kJ CG PET -2008
BCECE-2003 WB-JEE-2008
Ans. (d) : The ionisation energy is the amount of J & K CET-(1999)
energy required to take out most loosely bonded AIIMS-1994
electron from isolated gaseous atom. Ans. (d) : De-Broglie proposed that an electron like
∴ Ionisation energy of nitrogen light behaves both as a material particle and as a wave.
= energy of photon This proposal gave birth to a new theory known as wave
c mechanical theory of matter. De-Broglie equation is
= Nh given as–
λ
Where N = 6.02 × 1023 h
λ=
c = 3 × 108 mv
0 h
λ = 854 A = 854 × 10−10 M λ=
p
6.02 ×1023 × 6.6 × 10−34 × 3 ×108 Where, λ = Wavelength of light
=
854 × 10−10 h = Planck's constant
= 1.4 × 106 J mol-1 = 1.4 × 103 kJ mol-1 m = Mass of particle
220. The increasing order of wavelength for He+ ion, v = Velocity of particle
neutron (n) and electron (e) particles, moving 223. If radius of first Bohr's orbit of hydrogen atom
with the same velocity is– is 'X', then the de- Broglie wavelength of
(a) λ He+ < λ e < λ n (b) λ He+ = λ n = λ e electron in 3rd orbit is nearly
(a) 2πx (b) 6πx
(c) λ He+ < λ n < λ e (d) λ e < λ n < λ He+
x
BCECE-2016 (c) 9x (d)
3
h CG PET- 2016
Ans. (c) : λ =
m.v Ans. (b) : For a particular element, radius of an orbit–
(As per de–Broglie relation) rn ∝ n 2
1
∴ λ∝ (h and ν are constant) Where, n is the principal quantum number–
m r1 1
Also; Q m He+ > mn > me ∴ =
r2 32
∴ λHe+ < λn < λe If, r1 = X then r3= 9X
For third shell,
221. The relationship between energy (E) of 2πr3 = 3λ
wavelengths 2000 Å and 8000 Å, respectively 2π.9X
λ= ⇒ 6πX
is– 3
(a) E1 = 4E 2 (b) E1 = 2E 2 224. A particle 'A' moving with a certain velocity
E2 E2 has the de- Broglie wavelength 1 Å. For a
(c) E1 = (d) E1 = particle 'B' with mass 25%of 'A' and velocity
2 4 75% of 'A'. The de- Broglie wave length of 'B'
BCECE-2016 will be
Objective Chemistry Volume-I 171
 (a) 3 Å (b) 5.33 Å 226. What is the work function of the metal, if the
(c) 6.88 Å (d) 0.68 Å light of wavelength 4000Å generates
CG PET- 2013 photoelectron of velocity 6×105 ms–1 from it?
(Mass of electron = 9×10–31 kg Velocity of light
Ans. (b) : Given, = 3×108 ms–1 Planck's constant = 6.626×10–34 Js
For, particle 'A'– Charge of electron = 1.6×10–19 JeV–1
λA = 1 Å (a) 4.0 eV (b) 2.1 eV
Let, the mass of particle A(mA) = x (c) 0.9 eV (d) 3.1 eV
Velocity of particle A(VA) = y [JEE Main 2019, 12 Jan Shift-I]
For, particle 'B'– °
x × 25 x Ans. (b) : Given that, λ= 4000 A
Mass of particle (B) = = v = 6 × 10 ms 5 –1
100 4 From the Einstein's photoelectric equation–
y × 75 3y
Velocity of particle (B) = = hv = wo+
1
mv2
100 4 2
de-Broglie equation for the particle 'A', we get–
1
h wo= hv – mv2
λA = 2
m A VA
6.626×10-34 ×3×108 1
h = − × 9 × 10−31 × (6 × 105 ) 2
1× 10−10 = .....(i) 4000 × 10−10–19 2
x×y = (4.9695–1.62)10
de-Broglie equation for the particle 'B', we get– = 3.3495×10–19 J
h 3.3495 ×10−19
λB = wo = eV = 2.0934 = 2.1eV
m B VB 1.6 ×10−19
h 227. If p is the momentum of the fastest electron
λB = ejected from a metal surface after the
 x  3y  irradiation of light having wave length λ then
  
 4  4  for 1.5 p momentum of the photoelectron, the
16 h wavelength of the light should be (Assume
or λB = .....(ii) kinetic energy of ejected photoelectron to be
3 xy very high in comparison to work function)
From equation (i) and (ii), we get– 4 3
16 (a) λ (b) λ
−10
λ B = × 1× 10 9 4
3 2 1
λB = 5.33 × 10–10 (c) λ (d) λ
3 2
or λB = 5.33 Å [JEE Main 2019, 8 April Shift-II]
225. The wavelength of associated wave of a Ans. (a) : hν –φ= KE
particle moving with a speed of one-tenth that
of light is 7Å. The particle must be  hc 
  = KE + φ
(a) Electron (b) Proton  λ incident
(c) Nanoparticle (d) Photo 1 m 1 P2
CG PET -2017 K.E= mv 2
× =
2 m 2m
°
Ans. (a) : Given that, λ = 7 A = 7 × 10−10 m P2 hc hc
KE = = = ………..(i)
th 2m λincident λ
Particle moves with a speed of   of speed of light–
1
 10  P 2 × (1.5) 2 hc
= ………..(ii)
1 2m λ'
∴ Velocity of particle = × 3 ×108 = 3 × 107 ms −1 On dividing equation (ii) by (i), we get
10 4λ
From de-Broglie equation we get– λ'=
h 9
Q λ= 228. The work function of sodium metal is 4.41×10–
mv 19
J. If photons of wavelength 300 nm are
h 6.6 × 10−34 Js incident on the metal, the kinetic energy of the
m= = ejected electrons will be
λv 7 ×10−10 × 3 × 107
(h = 6.63×10–34 J s; c = 3×108 m/s)...........×10–21J.
m = 0.3142 × 10 −31 kg [JEE Main 2020, 2 Sep Shift-II]
m = 0.31× 10−31 ≈ 10 −31 kg Ans. (2.23×10-19J) : Given that:–
–19
Which is in accordance with mass of electron Thus φ work function of sodium = 4.41×10 J
among the given particles, electron is the correct choice. Wavelength (λ) = 300 nm

Objective Chemistry Volume-I 172

 We have to find kinetic energy, Ek=? Ans. (9) : λ = 248 ×10−9 m; w 0 = 3 × 1.6 × 10−9 J
hc
We know the equation, = Ek + φ hc
λ = w 0 + K.E
λ
hc
Ek= −φ 6.63 × 10−34 × 3 × 108
λ K.E = − 3 × 1.6 ×10−19
246 × 10−9
6.64 × 10−34 × 3 × 108 K.E = 3.2 × 10–19J
= − 4.41× 10−19 J
300 × 10−9
p = 2m K.E
= 6.64 × 10 −34+8+9−2 − 4.41 × 10 −19 J
E k = 2.23 ×10 −19 J = 2 × 9.1×10−31 × 3.2 × 10−19
229. A source of monochromatic radiation of = 7.63 × 10–25
wavelength 400 nm provides 1000J of energy in h 6.63 × 10−34
10 seconds. When this radiation falls on the λ= =
surface of sodium, x×1020 electrons are ejected p 7.63 × 10−25
per second. Assume that wavelength 400 nm is λ = 8.7 × 10–10
sufficient for ejection of electron from the λ = 8.7Å Q1Å = 10 −10 m
surface of sodium metal. The value of x is..........
(Nearest integer) λ≈9
(h = 6.626×10-34 Js) 232. A 50 watt bulb emits monochromatic red light
[JEE Main 2021, 25 July Shift-I] of wavelength of 795 nm. The number of
Ans. (2) : According to the question, work function of photons emitted per second by the bulb is x ×
metal is equal to energy of photon 1020. The value of x is .............
λphoton = 400nm [Given, h = 6.63×10–34 Js and c = 3.0×108 ms–1
hc 6.626 ×10−34 × 3 × 108 [JEE Main 2021, 1 Sep Shift-II]
E photon = = Ans. (2) : Total energy per sec = 50 J
λ 400 × 10−9
= 4.95 × 10–19J nhc
E=
In 10 second, the amount of energy provided is 1000J. λ
In 1 second, the amount of energy provided is 100J.
n × 6.63 × 10−34 × 3 × 108
Number of photons emitted per second 50 =
Total energy 795 × 10−9
= n = 1998.49 × 1017 [n = no. of photons per second]
Energy of one photon
⇒ 1.998×1020
100
Number of photons = ⇒ 2×1020
4.95 × 10−19 ⇒ x×1020
= 2.02 × 1020 photons
So, 2.02 × 1020 photons eject 2.02 × 1020 electrons from x=2
the sodium metal. 233. The number of photons emitted by a
230. The wavelength of electrons accelerated from monochromatic (single frequency) infrared
rest through a potential difference of 40 kV is range finder of power 1 mW and wavelength of
x×10–12m. The value of x is ............(Nearest 1000 nm, in 0.1 second is x × 1013. The value of
integer) Given : Mass of electron = 9.1×10–31 kg x is ............(Nearest integer)
Charge on an electron = 1.6×10–19C, Plank's (h = 6.63×10–34 Js, c = 3.00×108 ms–1)
constant = 6.63×10–34 Js [JEE Main 2021, 27 Aug Shift-II]
[JEE Main 2021, 20 July Shift-II] Ans. (50) : Given data, P = 1mW
h h 10–3 J → 1 sec.
Ans. (6) : λ = =
p 2meV 10–4 → 0.1 sec.
6.6 × 10−34 nhc
= m ∴E =
λ
2 × 9.1× 10−31 ×1.6 × 10−19 × 40 × 103
= 0.614 × 10-11m n × 6.63 × 10−34 × 3 × 108
10−4 =
= 6.14 × 10-12m 1000 × 10−9
13
So, n=50.2×10
x=6 ∴ x = 50
231. When light of wavelength 248 nm falls on a 234. A metal surface is exposed to 500 nm radiation.
metal of threshold energy 3.0 eV, the de– The threshold frequency of the metal for
Broglie wavelength of emitted electrons photoelectric current is 4.3×1014 Hz. The
is.........Å. [Round off to the nearest integer] velocity of ejected electron is ..........×105 ms–1
[Use : 3 =1.73, h = 6.63×10–34 Js, me = 9.1×10– (Nearest integer) (Use h = 6.63×10–34 Js, me =
31
kg, c = 3.0×108 ms–1, 1eV = 1.6×10–19J] 9.0×10–31 kg]
[JEE Main 2021, 16 March Shift-I] [JEE Main 2021, 26 Aug Shift-II]
Objective Chemistry Volume-I 173
Ans. (5) : Given data, 1 RH
v = RH × −
n 2 82
1 R
v = RH × 2 − H
n 64
Compounding it with y = mx+c
m=RH
Let v be speed of electron having max. K.E. Linear with slope RH.
We know that, E = φ + K.Emax 237. The de Broglie wavelength of particle is
hc 1 (a) Proportional to its mass
= hvo + mv 2 (b) Proportional to its velocity
λ 2
(c) Inversely proportional to its momentum
6.63 × 10−34 × 3 × 108 −34 1 (d) Proportional to its total energy
= 6.63 × 10 × 4.3 × 10 + mv
14 2

500 × 10−9 2 J & K CET-(2012)

6.63 × 30 × 10−20 1 Ans. (c) : De-Broglie wavelength (λ) is given by–
= 6.63 × 4.3 ×10−20 + mv 2
5 2 h h k
λ= = =
−20 1 −31 mv p p
11.271× 10 J = × 9 ×10 × v 2

2 1
v = 5 × 105 m / sec. ∴ λ∝
p
235. A gas absorbs photon of 355 nm and emits at 238. Calculate the wavelength associated with an
two wavelengths. If one of the emission is at 680 electron moving with a velocity of 106 m/s.
nm, the other is at (mass of e- =9.1 × 10-31 kg, h = 6.6 × 10-34 kg
(a) 1035 nm (b) 325 nm m2s–1)
(c) 743 nm (d) 518 nm (a) 6.2 × 10–8m (b) 7.25 × 10–8m
[AIEEE-2011] 0
Ans. (c) : We know that – (c) 6.25 Α (d) none of these
λ = 355 mm J & K CET-(2013)
λ1 = 680 nm Ans. (d) : From de– Broglie equation –
hc h
E = hv = λ=
λ mv
hc hc hc 6.6 × 10−34
E = E1 + E 2 or = + =
λ λ1 λ 2 9.1× 10−31 ×106
1 1 1 1 1 1 = 7.325 ×10–10 m.
= + ⇒ = +
λ λ1 λ 2 355 680 λ 2 239. The de Broglie wavelength of a ball of mass 10
1 1 1 g moving with a velocity of 10 ms–1 is
= − [h = 6.626 × 10–34 Js]
λ 2 355 680 –33
(a) 6.626 × 10 m (b) 6.626 × 10–29m
355 × 680 (c) 6.626 × 10 m–31
(d) 6.626 × 10–36m
= 742.769 nm ≈ 743nm.
680 − 355 J & K CET-(2011)
236. For emission line of atomic hydrogen from ni = Ans. (a) : We know that –
8 to nf = n, the plot of wave number (ν) against h
 1  λ=
 2  will be (The Rydberge constant, RH is in mv
n  10
wave number unit) h= 6.626×10–34Js, m = =0.01 kg,
1000
(a) non linear v=10ms–1
(b) linear with slope –RH Putting the values,
(c) linear with slope RH
6.626 ×10−34
(d) linear with intercept – RH λ= = 6.626 ×10−33 m
[JEE Main 2019, 9 Jan Shift-I] 0.01× 10
240. Two oxides of a metal contain 36.4% and
1  1 1
Ans. (c) : = v = R H z 2  2 − 2  53.4% of oxygen by mass respectively. If the
λ  nf 8  formula of the first oxide is M2O then that of
Let nf = n and z = 1 for H, the second is
(a) M2O3 (b) MO
 1 1
v = RH × 2 − 2  (c) MO2 (d) M2O5
n 8  J & K CET-(2011)
Objective Chemistry Volume-I 174
Ans. (b) : Consider atomic mass of metal = x (a) A positron (b) A photon
We know that, (c) An α-particle (d) A neutron
16 JCECE - 2014
% of oxygen in M2O = × 100 = 36.4
2x + 16 Ans. (c) : The de-Broglie equation is λ =
h
=
h
.
1600 p mv
⇒ = 36.4
2x + 16 1
Here, h and v are constant. So, λ ∝ . Since, the α-
⇒ 1600 = 72.8x + 582.4 m
x = 13.978 particle has the highest mass among the given entities, it
Second oxide oxygen % = 53.4% has the smallest de-Broglie wavelength.
Second oxide metal % = 46.6% 244. A 600 W mercury lamp emits monochromatic
Atomic Ratio M : O radiation of wavelength 331.3 nm. How many
46.6 53.4 photons are emitted form the lamp per
: second?(h=6.626×10-34J-s; velocity of light
13.978 16
=3×108 ms-1)
3⋅3 : 3⋅3
(a) 1×1019 (b) 1×1020
1 : 1 21
Hence, the formula of metal oxide = MO. (c) 1×10 (d) 1×1023
JIPMER-2011
241. If the de-Broglie wavelength of a particle of
mass m is 100 times its velocity, then its value Ans. (c): We know that–
in terms of its mass (m) and planck’s constant nhc
E=
(h) is λ
1 m h Eλ
(a) (b) 10 n=
10 h m hc
1 h m 600 × 331 ⋅ 3 ×10−9
(c) (d) 10 n=
10 m h 6.626 × 10−34 × 3 × 108
J & K CET-(2009) So, n = 1 × 1021 photons/second
Ans. (b) : Let wavelength of particle be x 245. What is the wavelength (in m) of a particle of
mass 6.62 × 10–29 g moving with a velocity of
x 103 ms–1?
Velocity, v =
100 (a) 6.62 × 10–4 (b) 6.62 × 10–3
We know that – (c) 10 –5
(d) 105
h JIPMER-2009
λ=
mv Ans. (c) : According to de–Broglie equation
h × 100 h
x= λ=
m×x mv
h 6.62 × 10−34
x = 100
2
λ=
m 6.62 × 10−29 × 10−3 × 103
h λ = 10–5 m.
⇒ x = 100
m 246. Dual nature of particle was given by
(a) Bohr theory
h
x = 10 (b) Thomson model
m (c) Heisenberg principle
242. The de-Broglie wavelength of helium atom at (d) de-Broglie equation
room temperature is J&K CET (2010)
(a) 6.6 × 10–34 m (b) 4.39 × 10–10 m JIPMER-2005
–11
(c) 7.34 × 10 m (d) 2.335 × 10–20 m Ans. (d): Dual nature of particle was given by de–
JCECE - 2013 Broglie in 1923.
3RT 3 × 8.314 × 298 De-Broglie equation relate the particle character with
Ans. (c) : Vrms = = = 1363ms−1 the wave character of matter.
M 4 × 10−3
247. Number of photons emitted by a 100 W (Js–1)
h 6.626 ×10−34 × 6.023 × 1023
λ= = yellow lamp in 1.0 s. Take the wavelength of
mv 4 × 10−3 × 1363 yellow light as 560 nm, and assume 100 percent
–11
= 7.32 × 10 m efficiency.
243. Which particle among the following will have (a) 1.6 × 1018 (b) 1.4 × 1018
20
the smallest de-Broglie wavelength, assuming (c) 2.8 × 10 (d) 2.1 × 1020
that they have the same velocity? JIPMER-2018
Objective Chemistry Volume-I 175
Ans. (c): As per the Plank's quantum theory, From equation (i) and (ii), we get
hc λA 1
E= Nhν = N = ⇒ λ B = 33.33 × 10−9 × 3 = 99.99 × 10−9 m
λ λB 3
Eλ = 1.0×10–7 m.
N= …………(i) 250. A body of mass x kg is moving with a velocity
hc
E= 100 w (Js–1)×1s= 100 J of 100 ms–1. Its de-Broglie wavelength is 6.62 ×
10–35 m. Hence, x is (h = 6.62 × 10–34 Js)
Putting the value in equation (i), we get (a) 0.1 kg (b) 0.25 kg
100J × 560 × 10−9 m (c) 0.15 kg (d) 0.2 kg
N=
6.626 × 10−34 Js × 3 ×108 ms −1 Karnataka-CET, 2009
= 2.82×1020. Ans. (a): Given,
h = 6.62×10–34 Js
248. With regard to photoelectric effect, identify the v = 100 ms–1
correct statement among the following. λ = 6.62×10–35
(a) Energy of electron ejected increases with the According to de-Broglie equation-
increase in the intensity on incident light.
h
(b) Number of electron ejected increases with the λ=
increase in the frequency of incident light. mv
(c) Number of electron ejected increases with the 6.62 ×10−34
increase in work function x=
6.62 × 10−35 × 100
(d) Number of electron ejected increases with the 10−34 1
increase in the intensity of incident light. x = −34 −1
=
Karnataka-CET-2020 10 × 10 ×10 10 2

= 0.1 kg
Ans. (d) : The number of electron ejected increases 251. A body of mass 10 mg is moving with a velocity
with the increase in the intensity of incident light, an of 100 ms–1. The wavelength of de-Broglie wave
increase in the intensity of incident light means that the associated with it would be (h = 6.6. × 10–34 Js)
number of photons incident per unit surface area of the (a) 6.63 × 10–35 m (b) 6.63 × 10–34 m
metal increase. (Provided the incident photons has its (c) 6.63 × 10 m –31
(d) 6.63 × 10–37 m
frequency more than threshold frequency). Karnataka-CET-2007
The equation for the photoelectric effect is Ans. (c) : Given that,
hv = hv0 + KE m = 10 mg = 10 × 10–6 kg
Where, v = frequency of the incident radiation v = 100 ms–1
hv0 = work function From de-Broglie equation,
It is clear that the number of electron elected does not h
depend of work function. λ=
mv
249. Two particle A and B are in motion. If the h 6.63 ×10−34
wavelength associated with 'A' is 33.33 nm, the λ= =
wavelength associated with 'B' whose mv 10 × 10–6 ×100
1 6.63 ×10−34
momentum is rd of 'A' is =
3 10−3
(a) 1.0 × 10–8 m (b) 2.5 × 10–8 m m = 6.63 × 10–31 m
–7
(c) 1.25 × 10 m (d) 1.0 × 10–7 m 252. The number of photons emitted per second by
a 60 W source of monochromatic light of
Karnataka-CET-2019 wavelength 663 nm is (h = 6.63 × 10-34 Js)
Ans. (d) : Given that, (a) 4× 10–20 (b) 1.5× 1020
–20
λA=33.3 nm or 33.33×10–9 m (c) 3× 10 (d) 2× 1020
–20
1 (e) 1× 10
PB = PA Kerala-CEE-2009
3
According to de-Broglie– Ans. (d) : Given that,
h = 6.63 × 10–34 J sec.
h
λ= λ = 663 nm = 663 × 10–9 m
p c = 3 × 108 m/sec
h n=?
λA= ……… (i)
PA E = 60 W = 60 J sec–1
Q We know that,
h 3h
λB = = ……… (ii) nhc
PA PA E=
λ
3
Objective Chemistry Volume-I 176
∴ n=
E×λ ∴ n×3×10–19= 200 (where n= no. of photons)
hc 200
n= = 6.66 ×1020
60 × 663 × 10 −9 3 ×10−19
=
6.63 × 10 −34 × 3 × 108 256. The de Broglie wavelength of the matter wave
6 × 663 × 10 −8 associated with an object dropped from a
= height x, when it reaches the ground is
3 × 663 × 10 −28
proportional to
= 2 × 10–8 + 28
= 2 × 1020 1
(a) x 2 (b)
253. The relationship between the energy E1 of the x
o (c) x (d) x 3 / 2
radiation with a wavelength 8000 A and the
energy E2 of the radiation with a wavelength (e) x
o Kerala-CEE-2020
16000 A is: Ans. (b) : According to de-Broglie wavelength–
(a) E1 = 6E2 (b) E1 = 2E2 h
1 λ=
(c) E1 = 4E2 (d) E1 = E2 mv
2 Since, an object dropped from a height x then,
(e) E1 = E2 v2 = u2 – 2gx
Kerala-CEE-2005
v = 2gx
Ans. (b) : We know that,
h
1 So, λ=
E∝ m 2gx
λ
E1 λ 2 16000 1
= = ⇒ E1 = 2E 2 or λ∝
E 2 λ1 8000 x
254. The work function of a metal is 5 eV. What is 257. Calculate the energy of photons having
the kinetic energy of the photoelectron ejected wavelength, 5 × 10–7 m falls on a metal surface
from the metal surface if the energy of the of work function, 3.4 × 10–19 J :
-19
incident radiation is 6.2 eV? (1 eV = 1.6 × 10 ) (a) 3.97 × 10–19 J (b) 3.55 × 10–19 J
-19 -19
(a) 6.626 × 10 J (b) 8.10 × 10 J (c) 2.97 × 10 J –19
(d) 2.57 × 10–19 J
-18 -18
(c) 1.92 × 10 J (d) 8.01 × 10 J Manipal-2016
(e) 1.92 × 10-19 J hc
Kerala-CEE-2014 Ans. (a) : Energy of photons, E= hv =
λ
Ans. (e): Given that, Given that-
Work function (hv0) = 5eV
h = 6.626 × 10−34 Js
Energy of incident radition (hv) = 6.2 eV
The kinetic energy of photoelectron: c = 3 ×108 m / s
K.E = hν − hν 0 λ = 5 ×10−7 m
= 6.2–5
(6.626 × 10−34 Js) × (3 ×108 ms −1 )
= 1.2 eV So, E=
= 12×1.6×10–19 J= 1.92×10–19 J 5 × 10−7
–19
= 3.975 ×10 J
255. A 250 W electric bulb of 80% efficiency emits a
o 258. Which one is the wrong statement?
light of 6626 A wavelength. The number of h
photons emitted per second by the lamp is (h = (a) The uncertainty principle is ∆E × ∆t ≥
6.636 × 10-34 Js) 2π
17 16 (b) Half filled and fully filled orbital have greater
(a) 1.42 × 10 (b) 2.18 × 10
stability due to greater exchange energy,
(c) 6.66 × 1020 (d) 2.83 × 1016 greater symmetry and more balanced
16
(e) 4.25 × 10 arrangement.
Kerala-CEE-2014 (c) The energy of 2s-orbital is less than the
hc energy of 2p-orbital in case of hydrogen like
Ans. (c) : Energy of one photon = atoms.
λ
h
6.626 × 10−34 × 3 × 108 (d) de-Broglie's wavelength is given by λ = ,
−10
= 3 ×10−19 J mv
6626 × 10 where m = mass of the particle, v = group
250 × 80 velocity of the particle
Energy emitted by bulb= = 200J
100 NEET-2017

Objective Chemistry Volume-I 177

Ans. (c): (a) According to heisenberg uncertainity On applying ∆E = hv
principle, the uncertainties of position (∆x) and –2.04 × 10–18 = 6.626 × 10–34 × v
momentum (p=m∆v) are related as −2.04 × 10−18
h h v=
∆ x ⋅ ∆p ≥ or ∆x ⋅ m∆v ≥ 6.626 ×10−34
4π 4π v = 3.08 × 1015 s–1
h  ∆v   Z2 
∆x ⋅ m ⋅ ∆a ⋅ ∆t ≥  ∆t = ∆a,a = acceleration 
hπ 261. Based on equation E = -2178 × 10–15 J  2 
h n 
Or, ∆x ⋅ F ⋅ ∆t ≥ [Q F = m ⋅ ∆a ] certain conclusions are written. What of them
hπ is not correct?
h (a) Equation can be used to calculate the change
∆E.∆t ≥ [∵∆E = F.∆x, E = energy] in energy when the electron changes orbit.
hπ
Thus, statement (a) is correct. (b) For n=1, the electron has a more negative
(b) The half filled and fully filled orbitals have greater energy than it does for n = 6 which means
stability due to greater exchange energy, greater that the electron is more loosely bound in the
symmetry and more balanced arrangement. smallest allowed orbit.
Thus, statement (b) is correct. (c) The negative sign in equation simply means
(c) For a single electronic species like H, energy that the energy of electron bound to the
depends on value of x and does not depend on l. Hence nucleus is lower than it would be if the
energy of 2s–orbital and 2p–orbital is equal in case of electrons were at the infinite distance from
hydrogen like species. the nucleus.
Therefore, statement (c) is incorrect. (d) Larger the value of n, the larger is the orbit
(d) According to de-Broglie equation, radius.
h NEET-2013
Wavelength (λ) = Ans. (b): For n=1 the electron has more negative
mv energy than it does for n=6 which means that the
Where, h= Planck’s constant electron is less loosely bound in the smallest allowed
Thus, statement (d) is correct. orbit.
259. A 0.66 kg ball is moving with a speed of 100 We know that,
m/s. The associated wavelength will be
 Z2 
(h = 6.6 ×10–34) Js) E = −R H  2 
(a) 6.6 × 10–32 m (b) 6.6 × 10–34 m n 
(c) 1.0 × 10 m –35
(d) 1.0 × 10–32 m If n = 1
NEET-Main 2010  Z2 
E = −R H   = −2.178 × 10−18 J
Ans. (c): According to de-broglie equation-  1 
h If n = 6
λ=
mv  Z2  −2.178
E = −R H  2  = × 10−18 J
6.6 × 10−34 6  36
= = 1× 10−35 m
0.66 × 100 = −6.05 × 10 −20 J
260. The frequency of radiation emitted when the E n =1 > E n =6
electron falls from n = 4 to n = 1 in a hydrogen
atom will be (Given ionization energy of H = 262. Calculate the energy in joule corresponding to
2.18 ×10–18 J atom–1 and h = 6.626 × 10–34J s) light of wavelenth 45 nm.
(a) 1.54 × 1015 s–1 (b) 1.03 × 1015 s–1 Planck'e constant, h = 6.63m × 10–34 J s, speed
(c) 3.08 × 10 s 15 –1
(d) 2.00 × 1015 s–1 of light, c = 3 × 108 m s–1)
NEET-2004 (a) 6.67 × 1015 (b) 6.67 × 1011
Ans. (c): Hydrogen like atom– (c) 4.42 × 10 –15
(d) 4.42 × 10–18
NEET-2014
E −2.18 × 10−18
E n = − 21 = J / atom Ans. (d) : Using planck’s quantum theory –
n n2
hc 6.63 ×10−34 × 3 × 108
−2.18 ×10−18 E= =
E1 = J / atom λ 45 × 10−9
(1)2 = 4.42×10 J–18

2.18 ×10−18 263. The value of Planck's contant is 6.63 × 10–34 Js.
E4 = − J / atom Speed of light is 3 × 1017 nm s–1. Which a value
( 4 )2
On applying ∆E = E1 – E4 is closed to the wavelength in nanomter of a
quantum of light with frequency of 6×1015 s–1?
 −2.18 × 10−18  (a) 50 (b) 75
= −2.18 ×10−18 −  
 16  (c) 10 (d) 25
–18
= –2.04 × 10 J/atom NEET-2013
Objective Chemistry Volume-I 178
Ans. (a) : We know that – We know that,
c = νλ c
λ=
c υ
λ=
ν 3 × 108
λ= = 0.375 × 10−7 m
3 × 1017 8 × 1015
λ= = 50nm = 0.375×10–7×109 nm
6 ×1015
264. According to law of photochemical equivalence = 0.375×102 nm
the energy absorbed (in ergs/mole) is given as = 37.5 nm
(h = 6.62 × 10–27 ergs, c = 3 × 1010 cm s–1, ≈ 4×101 nm
NA = 6.02 × 1023 mol–1) 267. For given energy, E = 3.03 × 10–19 Joules
corresponding wavelength is
1.196 ×108 2.859 ×105
(a) (b) (h = 6.626 ×10–34) J sec, c = 3 × 108 m/sec
λ λ (a) 65.6 nm (b) 6.36 nm
2.859 ×1016 1.196 × 1016 (c) 3.4 nm (d) 656 nm
(c) (d) NEET-2000
λ λ
Karnataka NEET-2013 Ans. (d) : We know that –
Ans. (a) : Given data, hc
E=
h = 6.62 × 10–27 ergs λ
c = 3 × 1010 cm s–1 Given that –
NA = 6.02 × 1023 mol–1 E = 3.03 × 10–19J
hcN A h = 6.626 × 10–34 Js
E= c = 3 × 108 m/s
λ
6.6 × 10−34 × 3 × 108
6.62 × 10−27 × 3 × 1010 × 6.02 ×1023 λ=
= 3.03 × 10−19
λ
= 6.56 ×10 m = 656 ×10 −9 m = 656 nm
−7

1.196 ×10 8
= ergs mol−1. 268. A particular station of All India Radio, New
λ Delhi, broadcasts ona frequency of 1,368 kHz
265. The energies E1 and E2 of two radiations are 25 (kilohertz). The wavelength of the electro-
eV and 50 eV respectively. The relation magnetic radiation emitted by the transmitter
between their wavelengths i.e. λ1 and λ2 will be is [speed of light, c = 3.0 × 108 m s–1]
(a) λ1 = λ2 (b) λ1 = 2λ2 (a) 21.92 cm (b) 219.3 m
1 (c) 219.2 m (d) 2192 m
(c) λ1 = 4λ2 (d) λ1 = λ2 NEET-2021
2
NEET-2011 Ans. (b) : We know that –
Ans. (b) : Given, E1= 25ev and E2= 50 ev c
Wave length (λ) =
hc hc v
E1 = , E 2 =
λ1 λ2 3 × 108
λ= = 219.298m = 219.3m
By dividing E1 and E2 1368 ×103
25 λ 2 269. What will be the wavelength of a photon having
⇒ = energy 1eV (1eV=1.601×10–19J)?
50 λ1
(a) 1.241×10–7m (b) 12.41×10–9m
λ 1 –7
(d) 1.241×10–9m
⇒ 2 = (c) 12.41×10 m
λ1 2 Tripura JEE-2019
⇒ λ1=2λ2 Ans. (c): According to formula
266. The value of Planck's constant is 6.63 ×10–34 Js. hc
E=
The velocity of light is 3.0 ×108 ms–1. Which λ
value is closest to the wavelength in nanometers
hc 6.626 ×10−34 × 3 × 108
of a quantum of light with frequency of λ= =
8×1015s–1? E 1.6 × 10−19
(a) 2 × 10–25 (b) 5 × 10–18 λ = 12.41×10 m.–7

(c) 4 × 10 1
(d) 3 × 107 270. What is the wavelength (in m) of a particle of
mass 6.62 × 10–29 g moving with a velocity of
NEET-2003 103ms–1?
Ans. (c) : Given that – (a) 6.62 × 10–4 (b) 6.62 × 10–3
c = 3 × 108 m/s (c) 10–5 (d) 105
λ = 8 × 1015 s–1 UP CPMT-2008
Objective Chemistry Volume-I 179
Ans. (c): According to de-Broglie equation Ans. (c) : According to de–Broglie wavelength,
h h
λ= λ=
mν mv
Given that – Where, h = Planck's constant
m = 6.62 × 10–29g m = Mass of particle
v = 103 m/s v = Velocity
h = 6.62 × 10–34Js Given, m= 100 g, v= 100 cm s–1
putting these value in the equation. h=6.6×10–34 Js= 6.6 × 10–27 ergs
6.62 ×10−34 −5 6.6 × 10−27
λ= = 10 m λ =
6.62 × 10−29 × 103 × 10−3 100 × 100
271. If the energy difference between the ground λ= 6.6×10 –31
cm
state and its excited state of an atom is 275. Which one of the following corresponds to a
4.4 ×10-14 J. The wavelength of photon required photon of highest energy?
to produce the transition: (a) λ = 300 mm (b) ν = 3 × 108 s −1
(c) −
–12 –12
(a) 2.26 × 10 m (b) 1.13 × 10 m ν = 30 cm −1 (d) ε = 6.626 × 10 −27 J
(c) 4.52 × 10–16 m (d) 4.52 × 10–12 m WB-JEE-2017
UPTU/UPSEE-2006
hc  1 
Ans. (d) : We know that – Ans. (a) : ∴ E= hv = = hc.v Q = v 
hc λ  λ 
∆E = hv = Where E= energy of photon
λ
c = velocity of photon (=light)
hc 6.62 × 10−34 × 3 × 108 λ = wavelength of photon
λ= =
∆E 4.4 × 10−14 h= plank’s constant.
For (a)
λ = 4.52×10–12m. E= 6.63×10–34 ×3×108/300×10–9
272. What is the wavelength associated with a tennis E= 1.98×10–25 J/300×10–9
3
ball of mass 10 g and travelling at a velocity of 1.98 ×10−25 J
6.626 ms–1? E=
[h=6.626×10–34Js–1] 300 × 10−9 m
(a) 10–34 m (b) 10–31 m (a) → E = 6.6 × 10–19J
(c) 6.626m (d) 6.626×10–31m For (b)
UPTU/UPSEE-2011 E= hν =6.63×10–34 ×3×108
Ans. (a) : According to de-Broglie equation – (b) → E= 1.98×10–25 J
h  1 
λ= For (c) E = hc.v Q = v 
mv  λ 
–34 8 –2
−34
6.626 × 10 Js E= 6.63×10 ×3×10 ×30×10
−34
λ= = 10 m (c) → E= 5.96×10 –26
1kg × 6.626 ms −1
For (d) → E= 6.62 ×10–27 J
273. Time period of a wave is 5 × 10–3 sec what is the Hence highest energy for photon is in (a).
frequency.
(a) 5×10-3 s-1 (b) 2×102 s-1 4. Uncertainty Principle
(c) 23×103 s-1 (d) 5×102s-1
UPTU/UPSEE-2008 276. If the work function of a metal is 6.63 × 10–19 J,
1 the maximum wavelength of the photon
Ans. (b) : Frequency (n) = required to remove a photoelectron from the
time period (T) metal is ––––– nm. (Nearest integer)
Here, T= 5×10–3 s [Given : h = 6.63 × 10–34 Js, and c = 3 × 108 m s–
1
1 ]
n= −3
= 0.2 × 103
5 × 10 JEE Main-28.06.2022, Shift-I
n = 2×102 s–1 Ans. (300) : Given,
274. As per de-Broglie's formula a macroscopic Work function of metal = 6.63 × 10–19 J
particle of mass 100 g and moving at a velocity
–1
Planck’s constant = 6.63 × 10–34Js
of 100 cm s will have a wavelength of c(velocity) = 3 × 108 m/s
(a) 6.6 × 10 cm–29
(b) 6.6 × 10 cm
–30
As we know that–
(c) 6.6 × 10–31 cm (d) 6.6 × 10–32 cm hc
Work function =
WB-JEE-2014 λ
Objective Chemistry Volume-I 180
 hc So,
Then λ=
work function 6.626 ×10−34
m=
6.63 × 10 –34 × 3 × 108 4 × 3. 14 × 10−7 × 2.4 × 10−26
λ=
6.63 ×10 –19 6.626 ×10−34 ×1033
λ = 3000 nm m=
4 × 3. 14 × 2.4
277. A fast moving particle of mass 6.63 × 10–28 g m = 21.89 × 10–3 g
can be located with an accuracy of 1Å. The
uncertainty in its velocity (in ms–1) is about (h = 279. The uncertainties in the velocities of two
6.63 × 10–34 Js) particles A and B are 0.05 and 0.02 ms–1
(a) 8 × 103 (b) 8 × 104 respectively. The mass of B is five times to that
(c) 8 × 105 (d) 8 × 106  ∆x 
(e) 8 × 10 7 of A. What is the ratio of uncertainties  A 
Kerala CEE -03.07.2022  ∆x B 
their positions in :
Ans. (c) : Given that –
(a) 2 (b) 0.25
Mass of particle (m) = 6.63 × 10–28g
= 6.63 × 10–31 kg (c) 4 (d) 1
Planck's constant (h) = 6.63 × 10−34 Js AP-EAMCET-2006
o
∆x = 1A = 10 m
–10 Ans. (a) : Given, ∆vA = 0.05 ms and mA = m
–1

According to Heisenberg’s uncertainty principle ∆vB = 0.02 ms–1 mB = 5 m

According to Heisenberg uncertainty principle–
∆x. ∆ p = h
4π h
m∆v.∆x ≥
∆x. m∆v = h 4π
4π Where, h = Planck's constant
h ∆x = Uncertainity in position
∆v =
4 π × m × ∆x ∆v = Uncertainity in velocity
6.63 × 10−34 m = Mass
∆v = For particle A :
4 × 3.14 × 6.63 × 10−31 ×10−10
h
10−34 × 1041 m × 0.05 × ∆ xA = .... (i)
∆v = 4π
12.56
For particle B :
107 h
∆v = 5 m × 0.02 × ∆ xB = .... (ii)
12.56 4π
100 × 105 Dividing equation (i) and (ii) we get :
∆v =
12.56 m × 0.05 × ∆x A
=1
∆v = 8 × 105 5m × 0.02 × ∆x B
278. If the uncertainty in velocity and position of a ∆x A 0.02 × 5
minute particle in space are, 2.4 × 10–26 (m s–1) =
–7
and 10 (m) respectively. The mass of the ∆x B 0.05
particle in g is _____. (Nearest integer) ∆x A
(given : h = 6.626 × 10–34 Js) =2
JEE Main-27.06.2022, Shift-I ∆ x B

Ans. (29.89×10–3) : 280. If the uncertainty in velocity of a moving object

We know the uncertainty principle– is 1.0×10–6 ms–1 and the uncertainty in its
h position is 58 m, The mass of this object is
∆x. ∆p = approximately equal to that of
4π
(h = 6.626×10–34 Js)
h
∆x. m∆v ≥ Q ∆p = m∆v (a) helium (b) deuterium
4π (c) lithium (d) electron
h
m= AP EAMCET (Medical) - 2013
4π . ∆x. ∆v Ans. (d) : Given-
Given that –
Uncertainty in velocity ( ∆v ) = 1.0×10–6 ms–1
∆v = 2.4 × 10–26
∆x = 10–7 Uncertainty in position (∆x) = 58 m
n = 6.626 × 10–34 Mass (m) = ?

Objective Chemistry Volume-I 181

Now, from the Heisenberg principle- From equation of Heisenberg uncertainity principle –
h h
∆x.∆p ≥ ∆x.∆v ≥
4π 4πm
where - ∆x = uncertainty in position Where, h = Planck's constant
∆p = uncertainty in momentum ∆x = Uncertainity in position
∆v = Uncertainity in velocity
h
∴ ∆x.∆v = From equation (1) we get
4πm h
6.6 × 10−34 m∆v2 ≥
or m= 4πm
4 × 3.14 × 58 × 1.0 × 10−6 h
∆v ≥2
6.6 4πm 2
or m= × 10−34+ 6
728.48 h
or m = 9.05×10–28–3 ∆v ≥
4πm 2
or m = 9.05×10–31
Thus, the mass of moving object is approximately equal 1 h
∆v ≥
to that of electron. 2m π
281. If a cricket ball of weight 0.1 kg has an 283. Find the uncertainlity in the position of an
uncertainty of 0.15 ms-1 in its velocity then the electron which is moving with a velocity of 2.99
uncertainty in the position of the cricket balls is × 104 cm.s–1. accurate up to 0.0016%. (Given,
________ me = 9.1 × 10–28 g. h = 6.626 × 10–27 erg.s)
(a) 3.5×10–33 m (b) 35×10–33m (a) 1.211 mm (b) 2.99 × 10–10 mm
–34
(c) 1.7×10 m (d) 1.7×10–33 m (c) 0.121 mm (d) 12.11 mm
AP EAPCET-6 Sep. 2021, Shift-II AP EAPCET 24.08.2021, Shift-I
Ans. (a) : According to Heisenberg’s Uncertainty Ans. (d) : Given –
principle– ∆v = 2.99 × 104 cm s–1, me = 9.1 × 10–28 gm,
⇒ ∆p.∆x ≥
h h = 6.626 × 10–27 g cm2 s–2
4π Hence,
Where, h = Planck’s constant (6.67×10–34 J-s) 0.0016
Q ∆p = m∆v ∆v = 2.99 × 104 cm s–1 ×
100
h ∆v =0.478 cm s–1
⇒ m∆v∆x ≥ From equation of Heisenberg uncertainity principle –
4π
Given, m = 0.1kg, ∆x=0.15m/sec h
∆x.∆v ≥
h 4πm
∆x= Where, h = Planck's constant
4πm∆v ∆x = Uncertainity in position
6.67 × 10−34 ∆v = Uncertainity in velocity
=
4 × 3.14 × 0.1× 0.15 m = Mass
6.67 π = 3.14
× 10−34
=
0.1884
or ∆x =
( )
6.626 × 10 –27 g cm 2 s –2 ⋅ s
⇒ 35.40 × 10 −34 4 × 3.14 × 0.478(cm s ) × 9.1× 10 –28 (gm)
–1

⇒ 3 ⋅ 5 × 10−33 m ∆x = 1.21 cm = 12.1 mm

282. If the uncertainty in momentum and 284. If a proton is accelerated to a velocity of 3 × 107
uncertainty in the position of a particle are m.s–1 which is accurate up to ± 0.5%, then the
equal. then the uncertaintiy in its velocity uncertainty in its position will be ____ [mass of
would be given by _______. proton = 1.66 × 10–27 kg, h =6.6 × 10–34 J.s]
h 1 h (a) 1.55 × 10–12 m (b) 3.24 × 10–13 m
(a) ∆v ≥ (b) ∆v ≥ (c) 1.58 × 10 m
–13
(d) 2.11 × 10–13 m
2π 2m π
AP EAPCET 24.08.2021, Shift-I
h 1 h Ans. (d): Given –
(c) ∆v ≥ (d) ∆v ≥
π m π ∆v = 3 × 107 m s–1
AP EAPCET 24.08.2021, Shift-I ∆x = ?
CG PET -2019 m = 1.66 × 10–27 kg
Ans. (b) : Given – h = 6.626 × 10–34 J-s
Uncertainity in momentum = Uncertainity in position It is given that velocity have the ± 0.5% So, the velocity
i.e. m∆v = ∆x ... (1) will be –

Objective Chemistry Volume-I 182

 m = Mass
∆v = 3 × 107 ms–1 × 0.5 π = 3.14
100
∆v = 3 × 5 × 104 ms–1 6.63 × 10–34
10 –2 × 4.5 × ∆x ≥
From the Heisenberg uncertainity principle – 4 × 3.14
h 6.63 ×10 –34
∆x.∆v ≥ ∆x ≥
4πm 4 × 3.14 ×10 –2 × 4.5
h
or ∆x = ∆x ≥ 1.17 × 10 –33
4πm ⋅ ∆v So, x= 1.17 ≈ 1.
∆x =
(
6.6 × 10–34 kgm 2s –1 ) 287. Both the position and exact velocity of an
(
4 × 3.14 × 1.66 × 10 –27 ( kg ) × 1.5 ×105 ms –1 ) electron in an atom cannot be determined
simultaneously and accurately. This is known
−12
6.6 × 10 as
= (a) De Broglie principle
31.2744
∆x = 2.11 × 10–13 m (b) Hamiltonian law
(c) Heisenberg uncertainty principle
285. In an atom, an electron is moving with a speed
of 600 m/s with an accuracy of 0.005%. (d) Bohr theory of hydrogen atom
Certainty with which position of the electron TS-EAMCET 09.08.2021, Shift-I
can be located is Ans. (c) : Heisenberg uncertainty principle– It is
(h= 6.6 ×10–34 kg m2 s–1, mass of electron em=9.1 impossible to determine the exact position and the exact
×10–31 kg) momentum of the particle simultaneously.
(a) 1.52× 10–4 m (b) 5.10× 10–3 m According to Heisenberg uncertainty principle–
–3
(c) 1.92× 10 m (d) 3.84× 10–3 m h
COMEDK-2017 ∆x. ∆p ≥
4π
Ans. (c) : Given: velocity of electron, v = 600 ms –1
Thus, lesser the error in the momentum more will be the
Accuracy of velocity= 0.005% error in the position of the particle.
600 × 0.005 288. A microscope using appropriate photons is
∴ ∆v = = 0.03m / s
100 engaged to track an electron in an atom within
According to Heisenberg’s uncertainty principle, distance of 0.001 nm. What will be the
h uncertainty involved in measuring its velocity?
∆x.m∆v ≥ (a) 5.79 × 107 m/s (b) 5.79 × 106 m/s
4π 7
(c) 4.79 × 10 m/s (d) 3.7 × 106 m/s
h
∆x = TS EAMCET 05.08.2021, Shift-I
4πm∆v Ans. (a):
Where, h = Planck's constant
According to heisenberg’s uncertainty principle-
∆x = Uncertainity in position
h
∆v = Uncertainity in velocity ∆v ≥
m = Mass 4π m∆x
6.6 × 10 –34 Where, h = Planck's constant
⇒ ∆x = = 1.92 ×10 –3 m ∆x = Uncertainity in position
4 × 3.14 × 9.1×10 –31 × 0.03
∆v = Uncertainity in velocity
286. A ball weighing 10 g is moving with a velocity m = Mass
of 90 ms–1. If the uncertainty in its velocity is
5%, then the uncertainty in its position is 6.626 ×10−34
______ ×10–33 m. (Rounded off to the nearest ∆v ≥
4 × 3.14 × 9.11× 10−31 × 0.01× 10−10
integer) [Given: h=6.63×10–34 Js] ∆v ≥ 0.0579 × 109 m/s
JEE Main 26.02.2021,Shift-II ∆v ≥ 5.79 × 107 m/s
Ans. (1) : Given that,
289. Heisenberg's uncertainty principle is in general
m= 10gm =10×10–3=10–2 kg significant to
v = 90 m/sec. (a) Planets
5 (b) Cricket bal of 500 g
∆v = 90 × = 4.5m / sec
100 (c) Cars
According to Heisenberg’s principle (d) Micro particles having a very high speed
h TS EAMCET 04.08.2021, Shift-I
m. ∆v.∆x ≥
4π Ans. (d) : According to Heisenberg's uncertainty
Where, h = Planck's constant principle we cannot simultaneously determine both the
∆x = Uncertainity in position precise velocity and position of a particle. It is general
∆v = Uncertainity in velocity significant to micro particles having a very high speed.

Objective Chemistry Volume-I 183

The mathematical expression of law is given below : 292. The uncertainties in the velocities of two
h particles, A and B are 0.05 and 0.02 ms−1
∆x × ∆y ≥ respectively. The mass of B is five times to that
4π of the mass of A. What is the ratio of
∆x = Change in position of the particle. ∆x A
∆y = Change in momentum of particle. uncertainties in their positions?
h = Planck's constant. ∆x B
290. A proton and a Li3+ nucleus are accelerated by (a) 2 (b) 0.25
the same potential. If λLi and λP denote the de (c) 4 (d) 1
Broglie wavelengths of Li3+ and proton AIIMS-2008
λ Ans. (a): According to Heisenberg’s uncertainty
respectively, then the value of Li is x ×10–1. principle,
λP
h
The value of x is _____(Rounded off to the ∆x × m∆v=
nearest integer) 4π
[Mass of Li3+ = 8.3 mass of proton] Where, h = Planck's constant
JEE Main 24.02.2021, Shift-I ∆x = Uncertainity in position
Ans. (2) : Given – ∆v = Uncertainity in velocity
m = Mass
Potential of proton = e p– V
π = 3.14
Potential of Li3+ = 3e p– V For particle A, ∆x = ∆xA
λ Li M=m, ∆v=0.05
and = x × 10–1 h
λp So, ∆xA × m × 0.05= ………. (i)
According to de-Broglie equation 4π
For particle B, ∆x = ∆xB m = 5m, ∆v = 0.02
h
Now, λ = So, ∆xB × 5m ×0.02 =
h
2M ( K.E ) 4π
………..(ii)
Where, h = Planck's constant ∆x A 5 × 0.02
m = Mass So, =
∆x B 0.05
K.E. = Kinetic energy
=2
λ Li h 2M p ( e p− V ) 293. The product of uncertainty in velocity and
∴ = × uncertainty in position of a micro particle of
λp 2M Li3+ ( 3e p– V ) h mass 'm' is not less than in which of the
following?
M p ( e p− V ) 3π h
∴ x ×10 –1 = (a) h × (b) ×m
M Li3+ (3e –p V) m 3π
h 1 h
 Mp 1  (c) × (d) ×m
 M Li3+ = 8.3M p ⇒ =  4π m 4π
 M Li3+ 8.3  AP-EAMCET – 2016
Ans. (c) : According to uncertainty principle
1
x × 10–1 = ⇒ x=2 h
8.3 × 3 ∆p.∆x ≥ (i)
4π
291. Assertion: It is impossible to determine the Where, ∆p = Uncertainty in momentum
exact position and exact momentum of an
electron simultaneously. ∆x = Uncertainty in position
Reason: The path of an electron in an atom is h = Planck's constant
clearly defined. ∴ ∆p = m ⋅∆v
(a) If both Assertion and Reason are correct and Putting the value of ∆p in equation (i) we get -
the Reason is the correct explanation of
Assertion h
m ⋅ ∆v ⋅ ∆x ≥
(b) If both Assertion and Reason are correct, but 4π
Reason is not the correct explanation of
h 1
Assertion. ∆x ⋅ ∆v ≥ ×
(c) If Assertion is correct but Reason is incorrect. 4π m
(d) If both the Assertion and Reason are incorrect. 294. The exact path of electron in any orbital cannot
AIIMS-2016 be determined. The above statement is based
Ans. (c): According to Heisenberg’s uncertainty on ______
principle, we cannot determine the exact position and (a) Hund's Rule (b) Bohr's Rule
exact momentum of an electron simultaneously. So (c) Uncertainty Principal (d) Aufbau Principal
assertion is correct but reason is not correct. AP- EAPCET- 07-09-2021, Shift-I
Objective Chemistry Volume-I 184
Ans. (c) : According to Heisenberg uncertainty ∆x = Uncertainty in position
principle, the position and the velocity of an object ∆ν = Uncertainty in velocity
cannot both be measured exactly at the same time. m = Mass of particle
Hund’s Rule- Every orbitals in a subshell is singly h = Planck’s constant
occupied with one electron before any one orbital is
doubly occupied, 298. The measurement of the electron position if
Aufbau Principle- Electrons fill lower energy atomic associated with an uncertainty in momentum,
orbitals before filling higher energy ones. which is equal to 1 × 10–18g cm s–1. The
uncertainty in electron velocity is, (mass of an
295. The uncertainty in position and velocity of a electron is 9 × 10–28 g)
particle are 10–10m and 5.27 × 10–24 ms–1
(a) 1 × 109 cm s–1 (b) 1 × 106 cm s–1
respectively. Calculate the mass of the particle 5 –1
is (h = 6.625 × 10–34 Js) – (c) 1 × 10 cm s (d) 1 × 1011 cm s–1
(a) 0.099 kg (b) 0.99 kg BITSAT 2007
(c) 0.92 kg (d) None Ans. (a) : Momentum, ∆p = m∆v
VITEEE 2019 Substituting the given values of ∆x and m, we get
Ans. (a): Given, 1×10−18 g cms −1 = 9 ×10−28 g × ∆v
∆n = 10–10 m
∆v = 5.27×10–24 m.s–1 1× 10−18
or ∆v =
h = 6.625×10–34 J-s 9 × 10−28
According to Heisenberg’s uncertainty principle– = 1.1× 109 cms −1 = 1×109 cms −1
h
∆x.m∆v ≥ [∴p = m∆v] 299. A stream of electrons from a heated filament
4π was passed between two charged plates kept at
h a potential difference V esu. If e and m are
m=
4π.∆x × ∆v charge and mass of an electron, respectively,
6.625 × 10−34 J sec then the value of h/λ (where, λ is wavelength
m= associated with electron wave) is given by
4 × 3.14 × 10−10 × 5.27 × 10−24
m = 0.099 kg. (a) 2 meV (b) meV
296. Which of the following pairs is not correctly (c) 2 meV (d) meV
matched?
[JEE Main-2016]
(a) Hund’s rule In orbitals of equivalent
energy electron spins remain h h
Ans. (c): Wavelength, λ = =
unpaired if possible. mv 2m(KE)
(b) Pauli’s No two electrons can have all
h
exclusion the four quantum number ∴ = 2meV
principle identical. λ
(c) Zeeman effect the effect of magnetic field on 300. The de Broglie wavelength of an electron in the
the atomic spectra. 4th Bohr orbit is
(d) Uncertainty It is impossible to determine (a) 6πa0 (b) 2πa0
principle the position of an electron. (c) 8πa0 (d) 4πa0
J & K CET-(2013) [JEE Main 2020, 9 Jan Shift-I]
Ans. (d) : According to uncertainty principle, it is Ans. (c) : De–Broglie wavelength of an electron
impossible to measure simultaneously the position and 2πr= nλ
momentum of an electron with absolute accuracy.
Where, r = Radius of orbit
297. The Heisenberg uncertainty principle may be n = no. of orbit
stated as
n2
(a) ∆x.∆v = h (b) ∆x.∆v = h/4πm 2π × a0 = n ⋅ λ
z
(c) ∆x.∆v ≥ h/4πm (d) ∆x.∆v = h/4π
n
WB-JEE-2012, AMU-2004 2π× a 0 = λ
z
Ans. (c) : According to Heisenberg's uncertainty
4
principle : "It is impossible to determine simultaneously 2π× a 0 = λ ⇒ 8πa0
both the position and momentum (or velocity) of a 1
microscopic particle with absolute accuracy. 301. A ball weighing 10 g is moving with a velocity
h of 90 ms–1. If the uncertainty in its position is
Mathematically. ∆x × ∆ν ≥ 5%, then the uncertainty in its position is
4πm ..........×10–33 m(Rounded off to the nearest
Or ∆x.m∆v ≥
h integer). [Given, h = 6.63×10–34 J-s]
4π [JEE Main 2021, 26 Feb Shift-II]

Objective Chemistry Volume-I 185

Ans. : According to Heisenberg uncertainty principle hc
Ans. (a): ∴Energy, E =
∆ x ∆p ≥
h λ
4π 6.62 × 10−27 × 3 ×1010
h 3 × 10−12 =
∆x = (∆p = m∆v) λ
4πm∆v 6.62 × 10−27 × 3 ×1010
λ=
6.63 × 10−34 Js 3 × 10−12
⇒
4 × 3.14 × 10 ×10−3 kg × 90 ms −1 × 0.05 ⇒ 6.62 × 10 −5 cm ⇒ 662×10–9 m
⇒ 1.173×10–33 m ⇒ 662×10–7 cm ⇒ 662 nm
⇒ 1×10–33 m 305. A dust particle has mass equal to 10–11 g,
302. A proton and a Li3+ nucleus are accelerated by diameter 10–4 cm and velocity 10–4 cm/s. The
the same potential. If λLi and λP denote the de- error in measurement of velocity is 0.1%. What
Broglie wavelengths of Li3+ and proton will be the uncertainty in its position?
(a) 0.527 × 1010 cm (b) 5.27 × 109 cm
λ –15
respectively, then the value of Li is x×10–1. (c) 0.527 × 10 cm (d) 0.527 × 10–9 cm
λP JCECE - 2014
The value of x is..........(Rounded off to the Ans. (d) : v = 10–4 cm s–1
nearest integer)
0.1×10−4
Mass of Li3+ = 8.3 mass of proton) ∴ ∆v = ( due to 1% error )
[JEE Main 2021, 24 Feb Shift-I] 100
= 1 ×10 cm s–1
–7

h According to Heisenberg’s uncertainty principle –

Ans. λ =
2mqV h
Now, ∆v.∆x ≥
4πm
λ Li m p (e)(V)
= , m Li = 8.3 Where, h = Planck's constant
λp m Li (3e)(V) m = Mass of particle
π = 3.14
λ Li 1 1
mp= = = = 2 × 10−1 6.626 × 10−27
λp 8.3 × 3 5 ∆x =
4 × 3.14 × 10−11 × 10−7
x=2 = 0.527 × 10–9 cm
303. An accelerated electron has a speed of 5×106 306. In an atom, an electron is moving with a speed
ms–1 with an uncertainty of 0.02%. The of 600 m/s with an accuracy of 0.005%.
uncertainty in finding its location while in Certainty with which the position of the
motion is x×10–9m. The value of x is electron can be located is
..............(Nearest integer) (Given, h= 6.6×10–34 kg m2s–1, mass of electron
[Use mass of electron = 9.1×10–31 kg, h = em=9.1×10–31 kg)
6.63×10–34 Js, π = 3.14] (a) 2.15× 10 −3 m (b) 2.78× 10 −3 m
−3
[JEE Main 2021, 25 July Shift-II] (c) 1.92× 10 m (d) 3.24× 10 −3 m
0.02 JIPMER-2015
Ans. Given, ∆v= × 5 ×106 = 103 m / sec Ans. (c) : Given that velocity of electron, v= 600 m/s
100
According to Heisenberg’s uncertainty principle Accuracy of velocity = 0.005 %
600 × 0.005
∆x. ∆v ≥
h ∆v = ⇒ 0.03
4πm 100
According to Heisenberg’s uncertainty principle -
6.63 × 10−34
x ×10−9 ×103 = ∆x.m∆v ≥
h
4 × 3.14 × 9.1× 10−31 4π
x × 10 × 10 = 0.058 × 10−3
−9 3
6.6 × 10−34
∆x =
0.058 × 10−6 4 × 3.14 × 9.1× 10−31 × 0.03
x= = 58
10−9 = 1.92 ×10–3 m
x = 58 307. The de-Broglie wavelength of a particle with
304. The energy of a photon is 3×10–12 erg. What is mass 1 kg and velocity 100 m/s is
(a) 6.6 × 10–33 m (b) 6.6 × 10–36 m
its wavelength in nm? +33
(c) 3.3 × 10 m (d) 3.3 × 10–36 m
(h = 6.62 × 10–27 erg-s; c = 3 × 1010 cm/s) JIPMER-2008, JCECE - 2007
(a) 662 (b) 1324 AP-EAMCET (Engg.) 1997, 1996
(c) 66.2 (d) 6.62 AP – EAMCET - (Medical)-1997
JCECE - 2009 NEET-1999
Objective Chemistry Volume-I 186
Ans. (b) : de-Broglie equation– 6.626 × 10−34
h m=
λ= 4 × 3.14 × 10−10 × 10−10
mv ⇒ 5.270×10–15 kg
Where, h = Planck's constant 311. Uncertainty principle is valid for
m = Mass (a) Proton
v = Velocity (b) Methane
6.62 × 10−34 kg m 2s −1 (c) Both (a) and (b)
=
1kg ×100ms −1 (d) 1 µ m sized platinum particles
⇒ 6.62 × 10 −36 m (e) 1 µ m sized NaCl particles
308. Heisenberg uncertainty principle can be Kerala-CEE-2017
explained as Ans. (a) : Uncertainty Principle states that the position
∆P × h h and momentum for a subatomic particle is uncertain to
(a) ∆x ≥ (b) ∆x × ∆P ≥ find simultaneously. Thus, the uncertainty principle is
4π 4π
valid for protons (subatomic particle).
h πh
(c) ∆x × ∆P ≥ (d) ∆P ≥ 312. Maximum number of photons emitted by a
π ∆x bulb capable of producing monochromatic
JIPMER-2008 light of wavelength 550 nm is ––––––, if 100 V
JCECE - 2007 and 1A is supplied for one hour.
BCECE-2005 (a) 1 × 1024 (b) 5 × 1024
23
Ans. (b) :According to Heisenberg’s uncertainty (c) 1 × 10 (d) 5 × 1023
principle. (e) 5 × 10 22

h h Kerala-CEE-2019
∆x.∆p ≥ or ∆x.(m.∆v) ≥
4π 4π Ans. (a) : Given that,
309. A golf-ball weigh 40.0g. If it is moving with a λ = 550 nm = 550 × 10–9 m
velocity of 20.0 ms-1, it’s de-Broglie wave length i = 1A, V = 100V
is
(a) 1.66×10-34 nm (b) 8.28×10-32nm
hc 6.626 × 10−34 × 3.0 × 108
∴E= = =0.036×10–17
(c) 8.28×10 nm–25
(d) 1.66×10–24nm λ 550 ×10−9
JIPMER-2004 So, each proton carries energy, E = 0.036 × 10–17
Ans. (c) : From de-Broglie equation– total energy output in one second
No. of photons =
h energy carried per photon
λ=
mv 100 × 3600
Where, h = Planck's constant n=
0.036 × 10−17
m = Mass of Golf-ball
n = 1 × 1024
v = Velocity
313. The ratio of de-Broglie wavelengths for
6.626 × 10−34 J.s electrons accelerated through 200 V and 50 V is
λ=
40 ×10−3 kg × 200 ms −1 :
⇒ 8.28×10–34 m (a) 1 : 2 (b) 2 : 1
⇒ 8.28×10–34 ×109nm (c) 3 : 10 (d) 10 : 3
⇒ 8.28×10–25 nm Manipal-2020
310. The minimum values of uncertainties involved Ans. (a) : As we know,
in the determination of both the position and 1
eV = mv 2
velocity of a particle respectively are 1× 10-10 m 2
and 1 × 10-10ms-1. Then, the mass (in kg) of the
particle is 2eV
So, v=
(a) 5.270×10-15 (b) 5.270×10-20 m
-16
(c) 5.270×10 (d) 5.270×10-10 Ans, mv = 2meV
-14
(e) 5.270×10 According to de Broglie equation–
Kerala-CEE-2015 h
Ans. (a) : According to Heisenberg uncertainty λ=
principle - mv
h
∆x.∆p=
h or λ=
4π 2meV
h h When it is accelerated through 200 volt
⇒ ∆x.m∆v = ⇒ m= h
4π 4π∆x.∆v λ1 = …….(i)
Given ∴ ∆x=1×10–10 m, ∆v=1×10–10ms–1 2meV. 200

Objective Chemistry Volume-I 187

When it is accelerated through 50 volt
h
λ2 = …….(ii)
2meV. 50
On solving equation (i) and (ii), we get
(d) The value of m for d 2z is zero.
λ1 50 1
= = NEET-2018
λ2 200 2
Ans. (c) : While filling electron to the orbitals, direction
314. If uncertainty in position and velocity are of spin cannot change unless the orbital is half–filled.
equal, then uncertainty in momentum will be : 317. If uncertainty in position and momentum are
1 mh 1 h equal, then uncertainty in velocity is
(a) (b)
2 π 2 πm 1 h h
(a) (b)
h mh m π π
(c) (d)
4πm 4π 1 h h
Manipal-2018 (c) (d)
2m π 2π
Ans. (a) Given, ∆x=∆v NEET-2008
According to Heisenberg Uncertainity Principle– Ans. (c) : According to Heisenberg,
h h
∆x × ∆p ≥ Uncertainty Principle - ∆x.∆p =
4π 4π
Where, ∆x = Change in position h
∆p = Change in momentum (∆p) =2
[∴∆x = ∆p]
4π
h
∆x ×m∆v= (Q ∆p = m∆v) h
4π (m.∆v) 2 =
4π
h
(∆v2)= [Q ∆x=∆v] h
4πm m 2 .∆v 2 =
4π
h mh
∆p = m∆v = m ⇒ h 1 h
4πm 4π ∆v = =
4πm 2 2m π
1 mh 318. The measurement of the electron position is
∆p =
2 π associated with uncertainty in momentum,
315. Uncertainty in the position of an electron which is equal to 1 × 10–18 g cm s–1. The
(mass = 9.1×10–31 kg) moving with a velocity uncertainty in election velocity is (mass of an
–1
300 ms accurate upon 0.001% will be electron is 9 × 10–28g)
(h = 6.63×10–34 J-s) (a) 1 × 109 cm s–1 (b) 1 × 106 cm s–1
(a) 19.2×10–2 m (b) 5.76×10–2 m (c) 1 × 105 cm s–1 (d) 1 × 1011 cm s–1
(c) 1.92×10 m –2
(d) 3.84×10 m–2 NEET-2008
MHT CET-2011 Ans. (a) : Momentum, ∆p= m∆v
0.001 1×10–18 g cm s–1= 9×10–28g ×∆v
Ans. (c) : ∆v = × 300 ms −1 = 3 × 10−3 ms −1 ∴ ∆v = 1×109 cms–1
100
h 319. Given : The mass of electron is 9.11 × 10–31 kg.
∆x.m∆v = Planck constant is 6.626×10–34Js, the
4π uncertainty involved in the measurement of
h velocity within a distance of 0.1 Ǻ is
∆x =
4πm∆v (a) 5.79 × 105 m s–1 (b) 5.79 × 106 m s–1
−34 2 −1
6.63 ×10 kg m s (c) 5.79 × 10 m s
7 –1
(d) 5.79× 108 m s–1
⇒ −31 −3 −1 NEET-2006
4 × 3.14 × 9.1× 10 kg × 3 ×10 ms
= 0.019336 Ans. (b) : According to Heisenberg’s uncertainty
h
∆x ≃ 1.933 ×10–2 m. principle, ∆x∆v = ,
316. Which one is a wrong statement? 4 πm
(a) Total orbital angular momentum of electron or h
∆v =
in s-orbital is equal to zero. 4πm∆x
(b) An orbital is designated by three quantum Where ∆x is uncertainty involved in the measurement of
numbers while an electron in an atom is position ∆v is uncertainty involved in the measurement
designed by four quantum numbers. of velocity h is planck’s constant m is mass of the
(c) The electronic configuration of N atom is electron

Objective Chemistry Volume-I 188

 ° Where, ∆x = Uncertainty in position or change in
Q Given ∆x = 0.1 A
= (0.1×10–10) position
∆v = uncertainty in velocity
= 10–11 m
m = 9.11 ×10–31 kg h = Planck 's constant ( 6.626 × 10−27 Js )
h= 6.626 ×10–34 Js
m = mass of electron ( 9.1× 10−28 kg )
On substituting the values–
6.626 ×10−34 Here, ∆v = 0.001% of 3 × 10 4
∆v = 0.001
4 × 3.14 × 9.11× 10−31 ×10−11 = × 3 × 104 = 0.3cm / s
= 0.0579 × 108 ms–1 = 5.79 × 106 ms–1 100
320. The uncertainty in momentum of an electron is h
∴∆x =
1 × 10–5 kg m/s. The uncertainty in its position 4πm∆v
will be (h = 6.62 × 10–34 kg m2/s) 6.626 ×10−27
(a) 5.27 × 10–30 m (b) 1.05 × 10–26 m = = 1.93cm
4 × 3.14 × 9.1× 10−28 × 0.3
(c) 1.05 × 10 m –28
(d) 5.25 × 10–28 m
323. If Ee, Eα and Ep represent the kinetic energies
NEET-1998 of an electron, α-particle and a proton
Ans. (a) : The uncertainty in the position of an electron respectively each moving with same de-Broglie
h wavelength then
is : ∆x=
4π∆p (a) Ee = Eα = Ep (b) Ee > Eα > Ep
Where, ∆x = The position of the particle (c) Eα > Ep > Ee (d) Ee > Ep > Eα
∆p = The momentum of the particle UP CPMT-2011
h = Planck’s constant Ans. (d) : De-Broglie wavelength,
6.626 × 10−34 h
∆x = λ=
4 × 3.146 × 10−5 mν
= 5.27×10–30 m h
ν= ..... (i)
321. The position of both, an electron and a helium m. λ
atom is known within 1.0 nm. Further the 1
momentum of the electron is known within 5.0 K.E = mν 2 ..... (ii)
2
× 10–26 kg m s–1. The minimum uncertainty in Now, put the value of v in Eq. (ii)
the measurement of the momentum of the 2
helium stone is 1  h 
K.E = m  
(a) 8.0 × 10–26 kg m s–1 (b) 80 kg m s–1 2  m.λ 
(c) 50 kg m s–1 (d) 5.0 × 10–26 kg m s–1
1  h2 
NEET-1998 K.E =  
2  m × λ2 
Ans. (d) : Uncertainty principle state that the product of
uncertainty position and uncertainty in momentum is 1
Hence, K.E ∝ (If λ and h-constant)
constant for a particle m
h and the order of K.E is as–
∆x, ∆p =
4π Ee > Ep > Eα
Here, given ∆x = 1.0nm for both electron and helium
atom, so ∆p charge of momentum is also same for 5. Schrodinger Wave Equation
both the particle
Therefore uncertainty in momentum of the helium stone 324. When an electron in an excited hydrogen atom
is also 5.0 × 10−26 kg ms −1 jumps from an energy level for which n = 5 to
level for which n = 2, the spectral line is
322. Uncertainty in position of an electron (mass = observed in the which series of the hydrogen
9.1 × 10–28 g) moving with a velocity of 3×104 spectrum?
cm/s accurate upto 0.001% will be (a) Lyman (b) Balmer
(use h/(4π) in uncertainity expression where h (c) Paschen (d) pfund
= 6.626 × 10–27 erg second) AP-EAMCET-1991
(a) 5.76 cm (b) 7.68 cm
Ans. (b): In Balmer series, hydrogen spectrum n1 = 2
(c) 1.93 cm (d) 3.84 cm
and n2 = 3, 4 .....
NEET-1995
• An electron in an excited hydrogen atom (n = 2)
Ans. (c) : According to Heisenberg’s uncertainty jump to n = 5 energy level, then the spectral line is
h observed is Balmer series.
principle ∆x × ∆v =
4πm Electron transitions for the Hydrogen atom

Objective Chemistry Volume-I 189

 (a) 2 (b) 3
(c) 4 (d) 1
AP-EAMCET-2007
16
Ans. (c) : Given, λ = cm, n 2 = ?
15R
For Lyman series–
1 1 1
= R 2 − 2
λ  n1 n 2 
15R 1 1 
=R 2 − 2
16 1 n 2 
15R 1
or = 1− 2
16R n2
325. Which one of the following series of lines is 1 15 1
found in the UV region of atomic spectrum of or = 1− =
hydrogen? n 22 16 16
(a) Balmer (b) Paschen 1 1
or =
(c) Brackett (d) Lyman n2 4
AP-EAMCET-1991
or n2 = 4
Ans. (d) : Lyman series of hydrogen spectrum lies in
UV region. 329. An electronic transition in hydrogen atom
• Balmer series lies in visible region. results in the formation of Hα line of hydrogen
• Paschen series lies in infrared region. in Lyman series, the energies associated with
• Brackett series lies in infrared region. the electron in each of the orbits involved in the
Hence, Lyman series lies in UV region and Paschen transition (in kcal mol–1) are :
series lies in IR region of spectrum. (a) –313.6, –34.84 (b) –313.6, –78.4
326. Among the first lines of Lyman, Balmer, (c) –78.4, –34.84 (d) –78.4, –19.6
Paschen and Brackett series in hydrogen AP-EAMCET-2008
atomic spectra, which has the highest energy? Ans. (b) : Energy of electron in nth orbit–
(a) Lyman (b) Balmer
2π 2 Z 2 e 4
(c) Paschen (d) Brackett En = a
AP-EAMCET-1999 n 2h 2
Putting the value of h, e and π, we get
 1 1 
Ans. (a) : Energy, ∆E = 13.6  2
− 2  eV −1311.8Z2
n
 1 n 2  En = kJ/mole
For the first line of Lyman series, n2
n1 = 1, n2 = 2 –313.52Z2
En = kcal/mole
1 1  n2
∆E = 13.6  2 − 2  eV = 10.2eV
1 2  (Q1 kcal = 4.184 kJ)
and energy decreases as we move on to the next series. For H atom, Z = 1 and Lyman series have the value of
327. What are the values of n1 and n2 for the 2nd line n1 = 1, n2 = 2.
in the Lyman series of hydrogen atomic n1=1, n2=2
spectrum ? Since, Energy of electron in n1 orbit
(a) 3 and 5 (b) 2 and 3
–313.52 × (1) 2
(c) 1 and 3 (d) 2 and 4 =
AP-EAMCET-2000 (1) 2
Ans. (c) : 2nd line (Hβ line) is formed when electron = – 313.6 kcal/mole
rd st
jumps from 3 orbit to 1 orbit in Lyman series. and energy of electron in n2 orbit
∴ n1 = 1, n2 = 3 –313.52 × (1) 2
=
(2) 2
= – 78.38 kcal/mole
330. The ratio of the highest to the lowest
328. The wavelength of spectral line emitted by wavelength of Lyman series is
16 (a) 4 : 3 (b) 9 : 8
hydrogen atom in the Lyman series is cm.
15R (c) 27 : 5 (d) 16 : 5
What is the value of n2? (R=Rydberg constant) TS-EAMCET (Engg.), 05.08.2021 Shift-II
Objective Chemistry Volume-I 190
Ans. (a) : The equation of Lyman series wavelength for Ans. (a) : The spectrum of any element depends upon
hydrogen atom is the no. of electron in outer shell.
1  1  Electronic configuration of He = 1s2
= R H 1 − 2  Reason:- Li+ has the same electronic configuration like
λ  n  as He atom i.e.
Where, λ = Lyman series wavelength for hydrogen
Li+ = 1s2, 2s1 (Removing of one electron form outer shell)
atom
Li+ = 1s2
RH = Rydberg Constant
i.e Li+ = 2
For lowest wavelength, n = ∞
334. Which of the following electron transitions in
1  1  the H-atom will release the largest amount of
∴ = R H 1 − 2 
λL  ∞  energy?
1 (a) n=3 to n=2 (b) n=2 to n=1
or = RH ..... (1) (c) n=5 to n=2 (d) n=6 to n=2
λL
COMEDK-2012
For highest wavelength, n = 2
Ans. (b) : As we know,
1  1  3R H
∴ = R H 1 −  = ..... (2) hc 1 1 1
λH  4 4 E= and = R  2 − 2 
λ λ n
 1 n 2
From equation (1) and (2)
So, we can say–
λH 4
= ⇒4:3 1 1
λL 3 E∝ 2 − 2
331. Which one of the following is extracted through n
 1 n 2

alloy formation ? Energy released when electron jumps from

(a) Manganese (b) Silver (a) For, n= 3 to n=2
(c) Nickel (d) Lead  1 1  1 1 5
SCRA - 2009  2 − 2 = − =
 2 3   4 9  36
Ans. (b) : Silver is the metal extracted through alloy (b) For, n = 2 to n=1
formation.
Note-Alloys are made by mixing two or more elements,  1 1  1 1  3
 2 − 2 = − =
at least one of which is a metal. 1 2  1 4  4
332. Which among the following represents (c) For, n = 5 to n=2
Schrodinger wave equation ?
 1 1   1 1  21
d 2 ψ d 2 ψ d 2 ψ 4πm  2 − 2 = − 5 =
(a) + + + (E − v) ψ = 0  2 5   4 2  100
dx 2 dy 2 dz 2 h (d) For, n = 6 to n = 2
h  d2 d2 d2   1 1  1 1  8
(b) Ĥ =  + + +V  2 − 2 = − =
4π2 m  dx 2 dy 2 dz 2   2 6   4 36  36
−h 2  d 2 d2 d2  3 21 8 5
(c) Ĥ = > > >
 2 + 2 + 2 +P 4 100 36 36
8π m  dx
2
dy dz 
Hence, largest amount of energy is released when
d ψ d ψ d ψ 8π2 m
2 2 2
electron jumps from n=2 to n=1.
(d) + + + 2 (E − v)ψ = 0
dx 2 dy 2 dz 2 h 335. A body of mass 10 mg is moving with a velocity
AP EAMCET (Engg.) 18.9.2020 Shift-I of 100 m s–1. The wavelength of de Broglie wave
GUJCET-2008 associated with it would be (h=6.63×10–34 J s)
Ans. (d) : The Schrondinger wave equation is (a) 6.63 ×10–35 m (b) 6.63 ×10–34 m
–31
(c) 6.63 × 10 m (d) 6.63 × 10–37 m
d 2 ψ d 2 ψ d 2 ψ 8π2 m
+ + + 2 (E − v)ψ = 0 COMEDK-2019
dx 2 dy 2 dz 2 h Ans. (c) Given, Mass, m = 10 mg = 10×10–3 g
Where, ψ = Wave function = 10×10–6 kg= 10–5 kg
m = Mass of electron, v = 100 ms –1 ,de Broglie, λ = ?
h = Planck's constant
h = 6.63×10–34 J s
E = Total energy of electron
V = Potential energy of electron h
de Broglie relation, λ =
333. The spectrum of Helium is expected to be mv
similar to that of–––– 6.63 × 10 –34 6.63 ×10 –34
(a) Li+ (b) H λ= = = 6.63 × 10–31 m
10 –5 × 100 10 –3
(c) Na (d) He+ 336. The first emission line on hydrogen atomic
AP EAPCET 19-08-2021 Shift-I spectrum in the Balmer series appears at (R =
NEET-1998 Rydberg constant):
Objective Chemistry Volume-I 191
 5R 3R 339. Which one of the following conditions is
(a) cm −1 (b) cm −1 incorrect for a well behaved wave function (ψ)?
36 4
(a) ψ must be finite
7R 9R (b) ψ must be single valued
(c) cm −1 (d) cm −1
144 400 (c) ψ must be infinite
AP-EAMCET (Medical), 2006 (d) ψ must be continuous
AP EAMCET (Medical) -1998 AP-EAMCET- (Engg.) - 2010
AP-EAMCET (Engg.) -1998 Ans. (c) : For a well behaved wave function (ψ) is
Ans. (a) : The first emission line on hydrogen atomic defined as the wave function (ψ) must be finite, single
spectrum in Balmer series contains the following valued and continuous.
values- 340. Assertion (A) : The probability of finding an
n1 = 2 and n2 = 3 electron in a small volume around a point (x, y,
z) at a distance 'r' from the nucleus is
 1 1  proportional to ψ2.
now, ν = R  2 − 2  Cm −1
 n1 n 2  Reason (R) : Subatomic particles have both
where – ν = wave number wave and particle nature.
The correct answer is
R = Rydberg constant
(a) Both (A) and (R) are true and (R) is the
1 1 correct explanation of (A)
∴ ν= R −  (b) Both (A) and (R) are true but (R) is not the
4 9
correct explanation of (A)
9−4 −1 (c) Only (A) is true but (R) is not true
ν =R   Cm
 36  (d) (A) is not true but (R) is true
5R AP - EAMCET(MEDICAL) - 2009
or ν = Cm −1 Ans. (b) : The square of the wave function, ψ2 is called
36 probability density and is always positive. The
337. What are the values of n1 and n2 respectively probability of finding an electron at a distance 'r' from
for Hβ line in the Lyman series of hydrogen the nucleus is expressed by following relation–
atomic spectrum? Probability = 4πr2dr⋅ψ2
(a) 3 and 5 (b) 2 and 3 ∴ Probability ∝ ψ2
(c) 1 and 3 (d) 2 and 4 The de-Broglie explains any particle shows the both
AP-EAMCET (Medical), 2006 nature i.e. particle and wave nature. Both (A) and (R)
are true but (R) is not the correct explanation of (A).
Ans. (c) : The Lyman series of lines are the lines in the
hydrogen spectrum which appear in the ultraviolet 341. Wave number of spectral line for a given
region. The value of n1 and n2 is given as follows– transition is x cm–1 for He+, then its value for
n1 = 1 Be3+ (isoelectronic of He+) for same transition
n2 = 2,3…. is–
x −1
338. The radial probability distribution curve (a) cm (b) x cm–1
obtained for an orbital wave function (ψ) has 3 4
peaks and 2 radial nodes. The valence electron (c) 4x cm–1 (d) 16x cm–1
of which one of the following metals does this BCECE-2013
wave function (ψ) correspond to ? Ans. (c) : The spectral lines of the transitions are
(a) Cu (b) Li directly proportional to square of atomic number i.e. z2.
(c) K (d) Na v ∝ z2
AP- EAMCET(Medical) -2010 Given, v1 = x
Ans. (d): The radial probability distribution curve
v1 z12
obtained for an orbital wave function has 3 peaks and 2 Now,=
radial nodes i.e. n =3 and l= 0 ( for radial node =2) v2 z 22
He+, z1 = 2
Hence, the valence electron of the metal must present in For,
Be3+, z2 = 4
3s-orbital. The electronic configuration of the given For,
metal is – x (4) 2
=
29 Cu = [ Ar ] 3d , 4s
10 1
v2 (2) 2
3 Li = [ He ] 2s
1
v2 = 4x cm–1

19 K = [ Ar ] 4s
1 342. What is effective nuclear charge and the
periphery of nitrogen atom when an extra
11 Na = [ Ne ] 3s 1 electron is added in the formation of an anion?
(a) 1.20 (b) 2.45
Thus, the given wave function corresponds to the (c) 3.55 (d) 5.95
valence electrons of the sodium metal. BCECE-2013
Objective Chemistry Volume-I 192
Ans. (c) : As an extra electron is added in nitrogen atom
then added electron will be screened by five electrons in
second orbit (2s22p3) and two electrons in first orbit
(1s2)
Now screening constant–
σ = 5×0.35 in nth orbit + 2×0.85 in (n–1)th orbit
⇒ 1.75 + 1.70 ⇒ 3.45
∴ Effective nuclear charge
The graph between wave function ψ and distance (r)
⇒ 7 – 3.45 = 3.55
from the nucleus helps in determining the shape of
343. The graph between |ψ|2 and r (radial distance) orbital.
is shown below. This represents. 345. The frequency of light emitted for the
transition n = 4 to n = 2 of He+ is equal to the
transition in H atom corresponding to which of
the following?
(a) n = 3 to n = 1 (b) n = 2 to n = 1
(c) n = 3 to n = 2 (d) n = 4 to n = 3
AIEEE-2011
Ans. (b) : For He+
1  1 1 
(a) 1s-orbital (b) 2p-orbital v = = R H 22  2 − 2 
λ 2 4 
(c) 3s-orbital (d) 2s-orbital For H
[JEE Main 2019, 10 April Shift-I]  1
1 1 
Ans. (d) : The given probability density curve is for 2s v = = R H × 12  2 − 2 
λ n
 1 n 2 
orbital because it has only one radial node. Among
other given orbitals, 1s and 2p do not have any radial For same frequency–
node and 3s has two radial nodes.  1 1   1 1 
22  2 − 2  =  2 − 2 
 2 4   n1 n 2 
1 1 1 1
⇒ 2
− 2 = 2− 2
n1 n 2 1 2
⇒ n1 =1and n 2 = 2
346. Energy of an electron is given by E = –2.178 ×
 Z2 
10–18 J  2  . Wavelength of light required to
344. The electrons are more likely to be found n 
excite an electron in an hydrogen atom from
level n = 1 to n = 2 will be (h = 6.62 × 10–34) Js
and c = 3.0 ×108 ms–1)
(a) 1.214×10–7 m (b) 2.816×10–7 m
–7
(c) 6.500 × 10 m (d) 8.500×10–7 m
[JEE Main 2013]
Ans. (a) : Energy change–
 1 1  hc
∆E= 2.178 ×10–18  2 − 2  =
1 2  λ
−34
 1 1  6.62 × 10 × 3.0 × 10
8
= 2.178×10–18  2 − 2  =
(a) in the region a and c 1 2  λ
(b) in the region a and b ∴ λ=1.214×10–7 m
(c) only in the region a 347. For any given series of spectral lines of atomic
(d) only in the region c hydrogen, let ∆ν = ν max − ν min be the difference
[JEE Main 2019, 12 April Shift-I] in maximum and minimum frequencies in cm-1.
Ans. (a) : The electrons are more likely to be found in The ratio ∆ν Lyman / ∆ν Balmer is
the region a and c as shown in figure. At b, wave (a) 27 : 5 (b) 5 : 4
function becomes zero and is called radial nodal surface (c) 9 : 4 (d) 4 : 1
or simple node. [JEE Main 2019, 9 April Shift-I]
Objective Chemistry Volume-I 193
Ans. (c) : For Lyman series–
1 1  1 1 
∆ V = V max − V min = R  2 − 2  − R  2 − 2 
1 ∞  1 2 
3R R
⇒R– =
4 4
and the shape of d x 2 − y2 and d z 2 are different
For Balmer series
 1 1   1 1
∆ V = V max − V min = R  2 − 2  − R  2 − 2 
2 ∞  2 3 
R 5R 4R
⇒ − =
4 36 36
R
∆ VLyman 36
∴ = 4 = = 9:4
∆ VBalmer 4R 16 350. The electronic configuration of Pt (atomic
36 number 78) is:
348. Which of the following orbitals will have zero (a) [Xe] 4f14 5d9 6s1 (b) [Kr] 4f14 5d10
14 10
probability of finding the electron in the yz (c) [Xe] 4f 5d (d) [Xe] 4f14 5d8 6s2
plane? JEE Main-29.06.2022, Shift-I
(a) px (b) py Ans. (a) : Sub-electronic configuration of Pt (atomic
(c) pz (d) dyz number 78) is–
14 9 1
78Pt = 54[Xe] 4f 5d 6s
WB-JEE-2010
351. Which of the following statements are correct?
Ans. (a) : px orbital lies along x–axis. Hence, the (a) The electronic configuration of Cr is (Ar) 3d5
probability of finding an electron is zero in the yz plane. 4s1
(b) The magnetic quantum number may have a
negative value
(c) In the ground state of an atom, the orbitals are
filled in order of their increasing energies
(d) The total number of nodes are given by n–2
Choose the most appropriate answer from the
options given below:
(a) (a), (c) and (d) only (b) (a) and (b) only
(c) (a) and (c) only (d) (a), (b) and (c) only
JEE Main-29.06.2022, Shift-I
Ans. (c) : Statements –
6. Electronic Configuration and (a) Electronic configuration of Cr
5 1
Shape of Orbital's. 24Cr = [Ar] 3d 4s
(d) Total number of nodes are (n–1)
349. Identity the incorrect statement from the So, option (c) is correct
following 352. Element "E" belongs to the period 4 and group
16 of the periodic table. The valence shell
(a) All the five 4d orbitals have shapes similar to electron configuration of the element, which is
the respective 3d orbitals. just above "E" in the group is
(b) In an atom, all the five 3d orbitals are equal in (a) 3s2 , 3p4 (b) 3d10, 4s2, 4p4
10 2 4
energy in free state. (c) 4d , 5s , 5p (d) 2s2, 2p4
(c) The shapes of dxy , dyz and dxz orbitals are JEE Main-28.06.2022, Shift-I
similar to each other; and d x 2 − y2 and d z 2 are Ans. (a) : Element “E” belong to the period 4 and group
similar to each other. 16 of the periodic table.
(d) All the five 5d orbitals are different in size Then, element is 34Se = [Ar] 3d10 4s2 4p4. [4 period, 16
group]
when compared to the respective 4d orbitals.
So, “E” is sulphur because “E” is above the ‘Se’.
NEET-17.07.2022 Therefore, electronic configuration is 3s2, 3p4.
Ans. (c) : The size of orbital depends on the principle 353. The number of nodal planes present in *s
quantum number and the shape of orbital's depends on antibonding orbitals is :
the azimuthal quantum number (l). In all the five 3d (a) 1 (b) 2
orbital are equal in energy in free state and the shape of (c) 0 (d) 3
dxy, dyz and dzx are similar. Karnataka-CET, 2008
Objective Chemistry Volume-I 194
Ans. (a) : In an antibonding molecular orbital, most of Ans. (b): According to Aufbau principle electron are
the electron density is located away from the space first occupy those orbitals whose energy is lowest.
between the nuclei, as a result of which there is a nodal Hence the electronic configuration is–
plane between the nuclei. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s
Note- A plane at which the electron density is zero. 5f, 6d, 7p.
359. The correct electronic configuration of
potassium is
(a) ls2,2s2,2p6,3s1
(b) ls2,2s2,2p6,3s2,3p6,4s2
(c) ls2,2s2,2p6,3s2,3p5,4s2
354. Mg2+ is isoelectronic with (d) ls2,2s2,2p6,3s2,3p6,4s1
(a) Cu2+ (b) Zn2+ AP EAMCET- 1992
+
(c) Na (d) Ca2+ Ans. (d) : The electronic configuration of Potassium is-
Karnataka-CET-2007 2 2 6 2 6 1
19K= ls 2s 2p 3s 3p 4s
Ans. (c) : Isoelectronic species have same number of 360. Chlorine atom, in its third excited state, reacts
electron. Mg2+ and Na+ both have 10 electrons hence, with fluorine to form a compound X. The
they are isolectronic species. formula and shape of X are
355. Which of the following atomic orbitals is not (a) ClF5, pentagonal
directed along the axis? (b) ClF4, tetrahedral
(a) px (b) d x 2 − y2 (c) ClF4, pentagonal bipyramidal
(c) dxy (d) d z 2 (d) ClF7, pentagonal bipyramidal
AP EAMCET- 2003
AP-EAMCET (Med.)-1999
Ans. (d) : The electronic configuration of 17 Cl is given
Ans. (C) : The lobe of the of three orbital i.e. dxy, d y2 ,
below-
d x 2 exists in between the axis and the lobe of d x 2 − y2 2 2 6 2 5 0
17 Cl = 1s 2s 2p 3s 3p 3d

and d z 2 exist in along the axis.

356. How many electrons are present in the M shell

of the atom of an element with atomic number
24?
(a) 5 (b) 6 Hence, it forms the ClF7 compound and have the
(c) 12 (d) 13 pentagonal bipyramidal geometry.
AP-EAMCET (Med.)-1999 361. The electronic configuration of curium (Z = 96)
Ans. (d) : The electronic configuration of atomic is
number 24 is 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5. The name (a) [Rn] 5f8 6d0 7s2 (b) [Rn] 5f7 6d1 7s2
3 5 2
of the element is Chromium. (c) [Rn] 5f 6d 7s (d) [Rn] 5f6 6d2 7s2
24Cr = 2, 8, 13, 1
JHARKHAND – 2019
K L M N
Ans. (b) : The electronic configuration of Curium (Z =
So, M shell contains 13 electrons.
96)
357. The symbol of the element 'Tungeston' is: 7 1 2
96Cm = [Rn] 5f 6d 7s
(a) Ta (b) W Curium (Cm) is a synthetic chemical element of the
(c) Tl (d) Tc actinoid series in the periodic table.
NDA (II)-2015 362. The number of unpaired electrons in carbon
Ans : (b) The symbol of the element 'Tungsten' is W atom is
and atomic number is 74. It is the rarest and toughest (a) one (b) Two
metal found in nature. (c) Three (d) Four
• It have 5 stable and 21 unstable isotopes. MPPET- 2009
• It have highest melting point and tensile strength. Ans. (b) : The electronic configuration of carbon is–
2 2 2
358. Which one of the following orbitals has the 6C = 1s , 2s , 2p
highest energy?
(a) 5d (b) 5f
(c) 6s (d) 6p Hence, the number of unpaired electron in carbon atom
SCRA-2015 is 2.
Objective Chemistry Volume-I 195
363. In the change of NO+ to NO, the electron is 368. In the ground state, an element has 13 electrons
added to a in M shell. The element is :
(a) σ orbital (b) π orbital (a) Copper (b) Chromium
(c) σ* orbital (d) π* orbital (c) Nickel (d) Iron
SCRA 2010 AP-EAMCET-2001
Ans. (d) : According to molecular orbital diagram– Ans. (b) : Given that :
electronic configuration of. In the ground state an element has 13 electron in M
NO + (14e) − σ1s 2 , σ*1s 2 , σ2s 2 , σ* 2s 2 , π2p 2x = π2p 2y , σ 2p 2z , shell is Chromium.
K L M N
π* 2p x = π* 2p y 24Cr = 1s2 2s2 2p6 3s2 3p6 4s1 3d5 = 2 8 13 1
NO(15e− ) − σ1s 2 , σ*1s 2 , σ2s 2 , σ* 2s 2 σ 2p z2 , π2p x2 = π2p2y , π* 2p1x = π* 2p y No. of electron in M shell = 2 + 6 + 5 = 13.
* *
The last electron into π 2px orbital i.e. π orbital. 369. X litre of carbon monoxide is present at STP. It
364. For the iso-electronic series, which among the is completely oxidised to CO2. The volume of
following species requires the least energy to CO2 formed is 11.207 L. What is the value of X
remove an outer electron? in litres ?
(a) K+ (b) Ag (a) 22.414 (b) 11.207
(c) Cl– (d) S2– (c) 5.6035 (d) 44.828
SCRA-2014 AP-EAMCET-2002
Ans. (d) : The isoeletronic species are K , Cl– and S2–.
+
1
The electronic configuration are– Ans. (b) : CO + O 2 → CO 2
22.4L 2
+ 2 2 6 2 6 0 22.4L
19 K = 1s 2s 2p 3s 3p 4s
xL 11.207 L

− For STP,
17 Cl = 1s 2 2s 2 2p6 3s 2 3p6 4s 0
22.4 L of CO, oxidize to CO2 = 22.4 L
2−
= 1s 2 2s 2 2p6 3s 2 3p6 4s 0
16 S 22.4
The S ion has more in size than K+ and Cl– because 16
2– 1 L of CO, oxidize to CO2 = L =1
22.4
proton of the nuclear pull on 18 electron. So, the energy
required to removing the electron from outer orbital is 22.4
11.207 L of CO, oxidize to CO 2 = ×11.207
less in quantity. 22.4
365. The maximum number of electrons that can be = 11.207 L
accommodated in all the orbitals for which l = 370. If the electron of a hydrogen atom is present in
3, is : the first orbit, the total energy of the electron
(a) 15 (b) 14 is:
(c) 10 (d) 6
AP-EAMCET-1991 −e 2 −e 2
(a) (b) 2
Ans. (b) : For l = 3, orbital is f. r r
−e 2 −e 2
(c) (d)
2r 2r 2
Hence, maximum number of electron is 14. AP-EAMCET-2003
366. The rule that explains the reason for chromium Ans. (c) : We know that :
to have [Ar] 3d5, 4s1 configuration instead of
[Ar] 3d4, 4s2, is : 1
Kinetic energy, K.E. = mv 2
(a) Pauli's exclusion principle 2
(b) Aufbau principle −e 2
(c) Hund's rule Potential energy =
(d) Heisenberg principle r
AP-EAMCET-1996 1 e2  e2 
or K.E. = Q mv 2
= 
Ans. (c) : Hund's rule, state that electrons do not pair up 2 r  r 
in the orbitals unless all the degenerated orbital have got
one electron each to get stability. ∴ Total energy = K.E. + PE
Hence, to get more stability chromium has [Ar] 3d5, 4s1 1 e2 e2
configuration instead of [Ar] 3d4 4s2. = −
2 r r
367. The electronic configuration of sodium is :
e2  1 
(a) [Ne] 3s2 (b) [Ne] 3s1 =  − 1
(c) [Ar] 4s 1
(d) [Ar] 4s2 r 2 
AP-EAMCET-1999 e 2  −1 
Ans. (b) : Electronic configuration of sodium is =  
2 2 6 1 r  2 
11Na = 1s 2s 2p 3s
or 11Na = [Ne] 3s . 1 −e 2
=
Hence, the correct option is (b). 2r

Objective Chemistry Volume-I 196

371. Which of the following elements have least
number of electrons in its M shell ?
(a) K (b) Mn
(c) Ni (d) Sc
AP-EAMCET-2004
Ans. (a) : Potassium (K) elements has least number of
electrons in it's M shell.
19 K = 1s
2 2
 2s 2p 6 3s 2 3p 6 4s

1

K
     N
L M
Number of electrons in M shell (n = 3) = 8 So, the option (b) is correct.
Similarly, 375. In which of the following pair are the ions
isoelectronic ?
K L M N (a) Mg2+, Ar (b) Na–, O2–
25 Mn = 2 8 13 2 3+
(c) Al , Cl –
(d) K+, Ne
Ni = 2 8 16 2
28 NDA (I)-2019
21 Sc = 2 8 9 2
Ans : (b) Isoelectronic species have the same number
Hence, the correct option is (a). of electrons but different in atomic number. In the given
372. The atomic numbers of elements X, Y and Z option, Na+ and O2– are isoelectronic species having 10
are 19, 21 and 25 respectively. The number of electron.
electrons present in the M shell of these Electronic configuration of Na+ and O2–;
elements follow the order : Na+= 1s2 2s2 2p6
(a) Z > X > Y (b) X > Y > Z O2–= 1s2 2s2 2p6
(c) Z > Y > X (d) Y > Z > X
376. Assertion (A) : Atoms with completely filled and
AP-EAMCET-2005 half-filled subshells are stable.
Ans. (c) : Given atomic number of element : Reason (R) : Completely filled and half-filled
Configuration, 19 X = 1s 2
 2s
2
2p 6 3s 2 3p 6 4s

1
subshells have symmetrical distribution of
K
     N electrons and have maximum exchange energy.
L M

Y = 1s 2 2
2p 3s 3p 4s2 3d
6 2 16 The correct answer is
21  2s
      (a) (A) and (R) are correct, (R) is the correct
K L N MM
explanation of (A)
25 Z = 1s
2 2
 2s 2p 6 3s 2 3p 6 4s2 3d 5

K
     N

M
(b) (A) and (R) are correct, (R) is not the correct
L M
explanation of (A)
For the order of number of electrons in M shell is : (c) (A) is correct, but (R) is not correct
Z>Y>X (d) (A) is not correct, but (R) is correct
373. The maximum number of sub-levels, orbitals TS-EAMCET-2016
and electrons in N shell of an atom are
Ans. (a) : Atoms with completely filled and half filled
respectively :
subshells have symmetrical distribution of electrons and
(a) 4, 12, 32 (b) 4, 16, 30 have maximum exchange energy and the electrons are
(c) 4, 16, 32 (d) 4, 32, 64 more stable in the atom.
AP-EAMCET-2007
377. The element with the electronic configuration
Ans. (c) : For N shell : 1s22s22p63s23p63d104s1 is
(i) Number of shell (n) = 4 (a) Cu (b) Ca
(ii) The number of sub-shell (l) = 4 (c) Cr (d) Co
(iii) Number of orbital = n2 = 42 = 16 TS-EAMCET-2016
(iv) Number of electron = 2n2 Ans. (a) : The element with electronic configuration
= 2 × 42 1s22s22p63s23p63d104s1 is Copper and Copper has
= 32 atomic number is 29.
374. Which orbital among the following has zero 378. The number of unpaired electrons in Co2+, is
radial nodes and 2 angular nodes? (a) 2 (b) 3
(a) 4s (b) 3d (c) 4 (d) 5
(c) 2p (d) 5s TS-EAMCET (Engg.), 07.08.2021 Shift-II
AP EAPCET 23-08-2021 Shift-I Ans. (b) : The electronic configuration of Co2+ is 1s2
Ans. (b) : Radial node = n – l – 1 2s2 2p6 3p6 3d7.
Where, n = Principle quantum number or Co2+ = [Ar]3d7
l = Azimuthal quantum number
Angular node = l
l value for s = 0 , p = 1, d = 2, f = 3

Objective Chemistry Volume-I 197

379. [Ar]3d104s1 electronic configuration belongs to h
(a) Ti (b) Tl Ans. (a) Angular momentum, µ = l ( l +1)
2π
(c) Cu (d) V Where l = Azimuthal quantum number for orbital
MPPET-2008 Azimuthal quantum no. (l)
Ans. (c) : The electronic configuration of [Ar] 3d10 451 s 0
is belong to Cu. p 1
It is in 4th period and IB group. d 2
380. The electronic configuration of Cs is f 3
(a) [ Kr ] 5s1 (b) [ Xe] 6s1 So, for 3s orbital an electron having zero angular
momentum.
(c) [ Rn ] 7s1 (d) [ Ar ] 4s1 384. To which group of the periodic table does an
AP EAMCET (Engg.) 21.09.2020, Shift-II element having electronic configuration [Ar]
Ans. (b) : Caesium atomic number is 55. 3d54s2 belong?
Thus, electron configuration of Caesium is 1s22s22p63s2 (a) Second (b) Fourth
3p64s23d104p65s24d105p66s1 or [Xe] 6s1. (c) Seventh (d) Third
Hence, correct option is (b). AP EAPCET 19-08-2021 Shift-I
381. How many unpaired electrons will be present Ans. (c) : The given electronic configuration–
5 2
in the ground state of an atom which has 18[Ar], 3d , 4s
valence electronic configuration 3d6 in its +3 Which is the electronic configuration of Mn and Mn
oxidation state ? element belongs to the 7th group of periodic table.
(a) 1 (b) 3 385. The element with atomic number 12 belongs to
(c) 4 (d) 7 ........ group and ......... period
AP EAMCET (Engg.) 18.9.2020 Shift-I (a) IA, third (b) III A, third
Ans. (b) : Ground state electronic configuration of the (c) II A, third (d) II A, second
element (Co, z = 27) will be [Ar] 3d7 4s2. AP EAMCET (Engg.) 2001
Ans. (c): The element with atomic number 12 is
magnesium.
The electronic configuration for Mg = 1s2, 2s2, 2p6 3s2.
There are 2 electron in the outermost shell so it belongs
to the II A group. Mg has an outermost shell, n=3
therefore, the third period.
382. Which among the following represents zero 386. The maximum number of sub-level, orbitals
overlap ? and electrons in N shell of an atom are
Z respectively
(a) + + -
(a) 4, 12, 32 (b) 4, 16, 30
(b) - + - +
Z (c) 4, 16, 32 (d) 4, 32, 64
AP EAMCET (Engg.) -2007
Ans. (c) : For ‘N’ shell
+

Z Q The number of shell (n) = 4

(c) +
∴The number of sublevels or subshell (l) = 4
The number of orbitals = n2 = 42 = 16
and the number of electrons = 2n2=2×42=32
Z 387. Electronic configuration of X is 1s2 2s2 2p6 3s2
(d) 3p1. It belongs to
(a) third group and third period
(b) thirteenth group and third period
AP EAPCET 24.08.2021 Shift-II (c) first group and third period
Ans. (c): Both orbitals p and s are aligned to different (d) third group and first period.
axis so their total overlap become zero. COMEDK-2017
Ans. (b) : 1s2 2s2 2p6 3s2 3p1
i.e., n=3, third period
And group number for p-block elements =
10+number of electrons in s and p orbitals =
383. To which orbital among the following may an 10+2+1=13
electron having zero angular momentum 388. Elements A, B, C, D and E have the following
belong? electronic configuration:
(a) 3s (b) 3p (a) 1s2 2s2 2p2 (b) 1s2 2s2 2p6 3s2
2 2 5
(c) 3d (d) 4f (c) 1s 2s 2p (d) 1s2 2s2 2p6 3s2 3p2
2 2 6 2 6
AP EAPCET 24.08.2021 Shift-II (e) 1s 2s 2p 3s 3p

Objective Chemistry Volume-I 198

 Which of these will belong to the same group in 394. Which of the following is correct for number of
the periodic table? electron, number of orbitals and type of
(a) A and B (b) A and C orbitals respectively in N orbit?
(c) A and D (d) A and E (a) 4, 4 and 8 (b) 4, 8 and 16
COMEDK-2018 (c) 32, 16 and 4 (d) 4, 16 and 32
Ans. (c) : A and D belongs to same group because in GUJCET-2011
same group. Number of valence electrons is same. Ans. (c): Orbits K L M N
A: 1s2 2s2 2p2 No of electrons 2 8 18 32
D: 1s2 2s2 2p6 3s2 3p2 No. of orbitals s s,p s,p,d, s,p,d,f
2 4 9 16
389. An orbital with n=3, l=1 is designated as
∴ For N orbit, No. of e– = 32, No. of orbital = 16 and
(a) 1s (b) 3s type of orbital = 4
(c) 3p (d) 3d
395. In which of the following pairs, the outer most
COMEDK-2014 electronic configuration will be the same?
Ans. (c) : For 3p-orbital, n=3 and l =1. (a) Fe2+ and Co+ (b) Cr+ and Mn2+
2+ +
390. The total number of orbitals in the fifth energy (c) Ni and Cu (d) V2+ and Cr+
…….is: JEE Main 25-02-2021, Shift-I
(a) 5 (b) 10
Ans. (b) :
(c) 18 (d) 25 (a) Fe2+ : [Ar] 3d6 Co+ : [Ar] 3d7 4s1
AP-EAMCET (Medical), 2006 +
(b) Cr : [Ar] 3d 5
Mn2+ : [Ar] 3d5
Ans. (d) : Given:- n = 5 2+
(c) Ni : [Ar]3d 8
Cu2+ : [Ar] 3d9
2+ 3
Total number of orbital's in the fifth energy level (d) V : [Ar] 3d Cr+ : [Ar] 3d5
= n2 So, (b) option is correct.
= 52 396. An orbital with one angular node shows three
= 25. maxima in its radial probability distribution
391. The atomic number of an element is 35. What curve, the orbital?
is the total number of electrons present in all (a) 3s (b) 4p
the p-orbitals of the ground state atom of that (c) 5d (d) 3p
element? TS EAMCET 05.08.2021, Shift-I
(a) 6 (b) 11 Ans. (b) : The atomic orbital with one angular node
(c) 17 (d) 23 shows 3 maxima, the number of radial nodes must be 2.
AP-EAMCET (Medical), 2003 The 3s, 5d and 4p orbital's have two radial nodes.
However, only the 4p orbital's have one angular node.
Ans. (c) : The atomic number of element is 35. The
name of the element is bromine. The electronic
configuration of Br is 1s2,2s2,2p6,3s2,3p6 ,4s2,3d10,4p5.
Electrons present in p- orbital is 17.
392. The total number of electrons present in all the
‘s’ orbitals, all the ‘p’ orbitals and all the ‘d’
orbitals of cesium ion are respectively:
(a) 8, 26, 10 (b) 10, 24, 20
(c) 8, 22, 24 (d) 12, 20, 22 397. Assertion: Yb+2 is more stable in comparison to
AP-EAMCET (Medical), 2003 Gd+2
Ans. (b) : According to the Aufbau Principle-the Reason: The electronic configuration of Gd is
electronic configuration of Cs+ is 1s2, 2s2, 2p6, 3s2,3p6, [Xe] 4f7 5d2 6s2.
4s2, 3d10, 4p6, 5s2, 4d10, 5p6 (a) If both Assertion and Reason are correct and
Total electron in s-orbital = 10 Reason is the correct explanation of
Total electron in p-orbital = 24 Assertion.
(b) If both Assertion and Reason are correct, but
Total electron in d- orbital = 20 Reason is not the correct explanation of
393. The electronic configuration of four different Assertion.
elements is given below, identify the group XIV (c) If Assertion is correct but Reason is incorrect.
element among these: (d) If both the Assertion and Reason are
(a) [He]2s1 (b) [Ne]3s2 incorrect.
2 2
(c) [Ne]3s 3p (d) [Ne]3s23p5 AIIMS 25 May 2019 (Morning)
AP-EAMCET (Medical), 2001 Ans. (c): The electronic configuration of Yb2+ and Gd2+
Ans. (c) : The element of group XIV belongs to the p- are:
block elements. As we know, outer configuration of p- Yb2+: [Xe] 4f14 5d0 6s0
block goes to the ns2 np2. The XIV group elements Gd2+: [Xe] 4f7 5d1 6s0
contains the four electron in outer shell. Hence, the Thus, the given electronic configuration of
correct option is C. Gd2+ is not correct.

Objective Chemistry Volume-I 199

398. Assertion: Helium and beryllium having Ans. (a): The outermost electronic configuration of the
similar outer electronic configuration of types most electronegative element is ns2 np5. It represents a
ns2. Reason: Both are chemically inert. halogen atom. In a period, halogens have the highest
(a) If both Assertion and Reason are true and the electro negativity due to the small size and high
Reason is a correct explanation of the Assertion. effective number charge.
(b) If both Assertion and Reason are true but 403. The configuration 1s 2 , 2s 2 2p 5 , 3s1 shows:
Reason is not a correct explanation of the
Assertion. (a) excited state of O −2
(c) If Assertion is true but the Reason is false (b) excited state of neon atom
(d) If both Assertion ad Reason are false (c) excited state of fluorine atom
AIIMS-1994 (d) ground state of fluorine atom
Ans. (c): Although He and Be have similar outer AIIMS-1997
electronic configuration of the type ns2, only He (s2) has Ans. (b): The ground state of neon is 1s2 2s2 2p6 on
inert gas configuration and hence is chemically inert but excitation an electron from 2p jumps to 3s orbital. The
Be (1s2, 2s2) close not have inert gas configuration and excited neon configuration is 1s2 2s2 2p5 3s1.
hence is not chemically inert. 404. The element with outer electronic
399. Assertion: d5 configuration is more stable than d4 configuration (n – 1)d2ns2. where n = 4. would
Reason: d5 has more exchange energy as belong to ___.
compared to d4 because 10 & 6 exchanges are (a) 2nd period. 2nd group (b) 4th period. 4th group
possible in d5 & d4 respectively. (c) 4th period. 2nd group (d) 2nd period. 4th group
(a) If both Assertion and Reason are correct and AP EAPCET 19-08-2021, Shift-II
Reason is the correct explanation of Ans. (b) : Given,
Assertion. Outer electronic configuration = (n - 1)d2 ns2 where n =
(b) If both Assertion and Reason are correct, but 4
Reason is not the correct explanation of ∴ = (4 – 1)d2, 4s2
Assertion. = 3d2, 4s2
(c) If Assertion is correct but Reason is incorrect. This is the outer electronic configuration of titanium
(d) If both the Assertion and Reason are which is belongs to the 4th period and 4th group.
incorrect. 405. Transition metal elements exhibit general
AIIMS 26 May 2019 (Morning) electronic configuration _______
Ans.(a): d5 configuration has exactly haif–filled (a) ns1–2 nd1–10 (b) ns1–2 (n – 1) d1–10
configuration so has more exchange energy. While d4 2
(c) ns (n –1)d 10
(d) ns2 (n – 1)d0
has only four such (unpaired) electron these has less AP- EAPCET- 07-09-2021, Shift-I
exchange energy. Maximum exchange energy leads to Ans. (b) : Transition metal element exibit general
the stabilization of the atoms. electronic configuration is ns1–2(n – 1) d1–10.
Therefore, d5 configuration is more stable them that of 406. If an element with atomic number Z = 115 has
d4 configuration. been discovered today in which of the following
400. Spectrum of Li2+ is similar to that of family would it had been placed and identify its
(a) H (b) Be electronic configuration.
(c) He (d) Ne (a) Boron Family : [Ar] 4f145d106s26p3
AIIMS-2002 (b) Carbon Family : [Kr] 5f146d107s27p3
Ans. (a): Electronic configuration of Li is 1s2 2s1 and of (c) Nitrogen Family : [Rn] 5f14 6d10 7s27p3
Li2+ is 1s1. That is similar to the electronic configuration (d) Oxygen Family : [Xe] 4f145d106s26p3
of H (1s1) Which has only one electronic in its valance AP- EAPCET- 07-09-2021, Shift-I
shell, thus it has spectrum similar to that of H. Ans. (c) : Atomic number Z = 115 is Moscovium (Mc)
401. Which of the following element is represented is a synthetic chemical element and an extremely
by electronic configuration radioactive element. It belongs to nitrogen family and
1s 2 2s 2 2p1x 2p1y 2p1z electronic configuration is
115Mc =[Rn] 5f14 6d10 752 7p3
(a) Nitrogen (b) oxygen
407. The atomic number of an element 'M' is 26.
(c) fluorine (d) sulphur How many electrons are present in the M-shell
AIIMS-2001 of the element in its M3+ state?
Ans. (a): 7N– 1s2, 2s2 2px1 2py1 2pz1 (a) 11 (b) 15
This element contains 7 electrons (c) 14 (d) 13
402. The outermost configuration of most AP - EAMCET (Medical) - 2007
electronegative element is Ans. (d) : The atomic number of element (M) is 26.
(a) ns2 np5 (b) ns2 np6 According to Aufbau principle, the electronic
2 4
(c) ns np (d) ns2 np3 configuration of 26M will be 1s2, 2s2, 2p6, 3s2, 3p6, 3d6,
AIIMS-2000 4s2.

Objective Chemistry Volume-I 200

 3+
For 26M = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5 Ans. (c) : Give that,
3+
or 26M = 2, 8, 13 Angular node (l) = 3
KL M Total node = 3
Hence, the electrons are present in M shell is 13. Total node = Radial node + Angular node
408. Which of the following graph represents 3=n–l–1+l
variation of 2p-orbital wave function with 3=n–1
distance from the nucleus? n=4
Subshell nl = 4f
412. The number of radial nodes of 3s and 2p
orbitals are respectively
(a) 2, 0 (b) 0, 2
(c) 1, 2 (d) 2, 2
BITSAT-2017
Ans. (a) : For a given orbital with principal quantum
number (n) and azimuthal quantum number (l)
number of radial nodes = (n – l – 1)
for 3s orbital: n = 3 and l = 0
therefore, number of radial nodes
=3–0–1=2
for 2p orbital; n = 2 and l = 1
therefore, number of radial nodes
VITEEE-2016 =2–1–1=0
Ans. (a) : In option (a) show the variation of 2p– 413. According to Bohr’s theory, the angular
orbitals wave function with distance from the nucleus. It momentum for an electron in 5th orbit is
has one maxima and passes through origin. (a) 2.5 h / π (b) 5 h / π
409. The orbital angular momentum of an electron (c) 25 h / π (d) 5 π / 2h
in 2p orbital is UPTU/UPSEE-2010
(a) h/4 π (b) zero
Ans. (a) : The angular momentum (l) of an electron in a
(c) h/2 π (d) 2 h/2 π nh
Assam CEE-2019 Bohr’s orbit is given as L=
2π
Ans. (d) : Angular momentum in an orbital h
h It is an integral multiple of
= l ( l + 1) , ( l = 1) 2π
2π In the fifth Bohr orbit, the angular momentum of
  5h
=  2h  electron is (L) =
 2π  2π
410. The orbital angular momentum of a p-electron 2.5h
L=
given as π
h h = 2.5 h\π
(a) (b) 3
2π 2π 414. Which one of the following decides the shapes
of orbitals in an energy shell?
3h h (a) Magnetic quantum number
(c) (d) 6
2π 2π (b) Principle quantum number
NEET-Mains 2012 (c) Azimuthal quantum number
Ans. (a) : Orbital angular momentum (m) (d) Spin quantum number
h AMU-2013
= l(l + 1) Ans. (c) : Principal quantum number tells the principal
2π
energy level or shell to which the electron belongs or
For P electrons; l=1 the average distance of the electron from the nucleus.
h 2h h Azimuthal quantum number tells, the shape of the
Thus, m = l(l + 1)
= = various subshells present within the same principal
2π 2π 2π
shell.
411. Orbital having 3 angular nodes and 3 total Magnetic quantum number determines the number of
nodes to preferred orientations of the electrons present in
(a) 5p (b) 3d subshell. Spin quantum number helps to explain state of
(c) 4f (d) 6d each electron in an orbital.
Odisha NEET-2019

Objective Chemistry Volume-I 201

415. The electrons indentified by quantum numbers ls 2 , 2s 2 2p 6 ,3s 2 3p 6 3d10 , 4s 2 4p3
n and l (i) n = 4, l = 1; (ii) n = 4, l =0; (iii) n = 3,
l = 2; (iv) n = 3, l = 1 can be placed in order of Its properties would be similar to which of the
increasing energy from the lowest to highest as following elements?
(a) (iv) < (ii) < (iii) < (i) (a) Boron (b) Oxygen
(b) (ii) < (iv) < (i) < (iii) (c) Nitrogen (d) Chlorine
(c) (i) < (iii) < (ii) < (iv) CG PET -2004
(d) (iii) < (i) < (iv) < (ii) Ans. (c) : The valence shell electronic configuration is
AMU-2011 4s2 4p3 similar to nitrogen 2s2, 2p3.
Ans. (a) : The greater is the value of (n+1), the greater 421. Out of the following electronic configurations
si the energy of orbitals. the one of a transition element is
(i) n=4, l=1 = 4p orbital (a) ls 2 , 2s 2 2p 2 ,3s 2 3p 6 3d10 ,4s 2 4p 6
(ii) n=4, l=0 = 4s orbital
(iii) n=3, l=2 = 3d orbital (b) ls 2 , 2s 2 2p 6 ,3s 2 3p 6 3d 2 ,4s 2
(iv) n=3, l=1 = 3p orbital (c) 1s 2 , 2s 2 2p 6 ,3s 2 p 6 , 4s 2
According to the Aufbau principle, energies of above–
mentioned orbitals are in the order of – (d) ls 2 , 2s 2 2p 6 ,3s 2 3p 6 3d10 , 4s 2 4p1
The increasing order of energy CG PET -2004
(iv) 3p < (ii) 4s< (iii) 3s< (i) 4p Ans. (b) : 1s2 2s2 2p6 3s2 3p6 3d2 4s2 is a transition
416. The element whose electronic configuration is element the last electron here enters the 3d sub shell.
1s2 2s2 2p6 3s2 is a 422. Aufbau principle is not satisfied by
(a) metoalloid (b) metal (a) Cr and Cl (b) Cu and Ag
(c) noble gas (d) non-metal (c) Cr and Mg (d) Cu and Na
AMU-2004
CG PET -2004
Ans. (b) : 1s2 2s2 2p6 3s2 is electronic configuration of
alkaline earth metal. Ans. (b) : The correct option is Cu and Ag. Cu as we
know should have the configuration 3d9 4s2 But it does
417. The electronic configuration of P in H3 PO4 not obey the Aufbau Principal and has a configuration
(a) 1s 2 2s 2 , 2p 6 ,3s 2 3p 6 (b) 1s 2 , 2s 2 , 2p 6 ,3s 2 of 3d10 4s1.
(c) 1s 2 2s 2 2p 6 (d) 1s 2 , 2s 2 , 2p 6 ,3s 2 3p 6 , 4s1 Similarly Ag lies below Cu in the periodic table [same
group] and thus disobeys.
CG PET- 2011 aufbau principal.
Ans. (a) : The electronic configuration of pin H3PO4 is
1s2 2s2 2p6 3s2 3p6 423. Which of the following can be represented by
the configuration [Kr]5s2?
P atom has 5 valance electron it shares 3 with 3–OH
groups to form 3P–OH bonds it shares 2 with one O (a) Ca (b) Sr
atom to form p=O bond. (c) Ba (d) Ra
418. The pair having the similar shape is CG PET -2004
(a) BF3 and NF3 (b) BF4− and NH +4 Ans. (b) : Electronic configuration of strontium is
[Kr]5s2
(c) SiC14 and SC14 (d) CH 3+ and N 3O +
424. Which of the following describes the shape of
CG PET- 2011 orbital?
− + (a) Principal quantum number
Ans. (b) : Both BF4 and NH 4 ion are tetrahedral.
(b) Azimuthal quantum number
419. Quantum numbers l = 2and m = 0 represent (c) Magnetic quantum number
the orbital
(d) Spin quantum number
(a) dxy (b) d x 2 − y2
CG PET -2006
(c) d z 2 (d) dxz Ans. (b) : Principal quantum number represents the
CG PET- 2016 principal energy shell where the electron is revolving
Ans. (c) : Quantum number l= 2 and m=0 Azimuthal quantum describe the slope of an orbital
Respresent dz2 orbital. Magnetic quantum number describes the orientation of
Note– For s, p, d and f orbitals, the value of the the orbital in space.
azmithual quantum number ‘l’ is 0, 1, 2, 3. 425. Electronic configuration of H+ is
When l=2, m can have values (a) 1s0 (b) 1s1
2
–2, –1, 0, +1 +2. (c) 1s (d) 1s1,2s1
A d–subshell can be have five different CG PET- 2010
orientations and orbitals Ans. (a) : Electronic configuration
Corresponding to these orientations are H+ = 1s0
dxy, dxz, dyz, d x 2 − y2 ,d z2 . H– = 1s2
He+ = 1s1
420. The electronic configuration of an element is

Objective Chemistry Volume-I 202

426. The ion which does not have configuration of 429. The correct electronic configuration and spin-
argon (Ar) is: only magnetic moment (BM) of Gd3+ (Z=64),
(a) I- (b) K+ respectively, are
-
(c) Cl (d) Ca2+ (a) [Xe]4f7 and 7.9 (b) [Xe]5f7 and 7.9
7
HP CET-2018 (c) [Xe]5f and 8.9 (d) [Xe]4f7 and 8.9
Ans. (a) : Argon has 18 electrons Cl–, K+, Ca2+ have 18 (JEE Main 2020, 5 Sep Shift-I)
electrons I– have 54 electron so I– does not have Ans. (a) : Given, [Gd]3+
electronic configuration with Ar. To find electronic configuration and magnetic moment
427. Match List-I with List-II. of [Gd]3+ -
0 0
List-I List-II Electronic configuration of [Gd]3+:[Xe]4f7 5d 6s
(Electronic (∆i in kJ mol–1) Hence number of unpaired electron n=7
configuration of
elements) Hence magnetic moment n ( n + 2) = 63 = 7.9 BM
A. 1s22s2 (i) 801 Hence electronic configuration is [Xe]4f7
2 2 4 Magnetic moment is 7.9 BM
B. 1s 2s 2p (ii) 899
C. 1s22s22p3 (iii) 1314 430. In the sixth period, the orbitals that are filled
D. 1s22s22p1 (iv) 1402 are
(a) 6s, 5f, 6d, 6p (b) 6s, 4f, 5d, 6p
Choose the most appropriate answer from
(c) 6s, 5d, 5f, 6p (d) 6s, 6p, 6d, 6f
the options given below.
Codes (JEE Main 2020, 5 Sep Shift-I)
A B C D Ans. (b) : As per (n+l) rule in 6th period, order of orbital
(a) (ii) (iii) (iv) (i) filling is 6s, 4f, 5d, 6p.
(b) (i) (iv) (iii) (ii) 431. Outermost electronic configuration of a group-
(c) (i) (iii) (iv) (ii) 13 element E is 4s2 4p1. The electronic
(d) (iv) (i) (ii) (iii) configuration of an element of p-block period-
(JEE Main 2021, 26 Feb Shift-I) five placed diagonally to element, E is
Ans. (a) : The correct option is A (a) -(ii), (b) -(iii), (a) [Kr]3d104s24p2 (b) [Ar]3d104s24p2
10 2 2
C(iv), d-(i) (c) [Xe]5d 6s 6p (d) [Kr]4d105s25p2
1s2 2s2→Be (JEE Main 2021, 20 July Shift-II)
1s2 2s22p4→O Ans. (d) : [Kr]4d 5s2 5p2
10

1s2 2s22p3→N The element E is Ga and the digonal element of 5th

1s2 2s22p1→B period is 50Sn having outer electronic configuration will
The ionization enthalpy order is B < Be < O < N. Be be [Kr]4d10 5s2 5p2.
has more I.E compared to B due to extra state fully 432. Among the following, the energy of 2s-orbital is
filled 2s orbital N has more I.E Compared to O due to lowest in
extra stable half filled 2p orbital. (a) K (b) H
Hence, (c) Li (d) Na
N→1402 kJ /mol [JEE Main 2019, 12 April Shift-II]
O→1314 kJ/mol Ans. (a) : The energy of 2s-orbital is lowest in K
B→801 kJ/mol (potassium). An orbital gets larger as the principle
Be→899 kJ /mol quantum number n increases.
428. The electronic configuration of bivalent Correspondingly the energy of the electron in such an
europium and trivalent cerium are orbital become less negative, meaning that the electron
(atomic number : Xe = 54, Ce = 58, Eu = 63) is less strongly bound and has less energy. The graph of
(a) [Xe]4f7 and [Xe]4f1 principle quantum number with atomic number is
(b) [Xe]4f7 6s2 and [Xe]4f26s2
(c) [Xe]4f2 and [Xe]4f7
(d) [Xe]4f4 and [Xe]4f9
(JEE Main 2020, 9 Jan Shift-I)
Ans. (a) : The electronic configuration of europium and
cerium in the ground state is–
Eu(63) ⇒ [Xe]4f76s2
Ce(58) ⇒ [Xe]4f15d16s2
Bivalent means it lost two electrons and trivalent means 433. Ge (Z = 32) in its ground state electronic
it lost three electrons configuration has x completely filled orbitals
Eu2+ ⇒ [Xe]4f7 with ml = 0. The value of x is
Ce3+ ⇒ [Xe]4f1 [JEE Main 2021, 31 Aug Shift-I]

Objective Chemistry Volume-I 203

Ans. (7) : 32Ge : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2
7 orbitals are fully filled with ml = 0.
The value of x is 7. 438. Which electronic configuration will show the
434. A certain orbital has n = 4 and ml = –3. The highest first ionization potential?
number of radial nodes in this orbital (a) 1s22s22p1 (b) 1s22s22p5
is........(Round off to the nearest integer). 2 2
(c) 1s 2s 2p 3
(d) 1s22s2
[JEE Main 2021, 17 March Shift-I] J & K CET-(2018)
Ans. (0) : n = 4, ml = –3 so l = 3 Ans. (b) : 1s22s22p1=B
Number of radial nodes = n – l – 1 1s22s22p5=F
=4–3–1 1s22s22p3=N
=0 1s22s2=Be
435. The number of orbitals with n = 5, m1 = +2 is The order of first ionization potential is F > N > Be > B.
...............(Round off to the nearest integer). 439. The orbital diagram in which Aufbau principle
[JEE Main 2021, 16 March Shift-II] is violated is
Ans. (3) : For, n = 5
(a) (b)
l = 0, 1, 2, 3, 4
For, l = 2, m = –2, –1, 0, 1, 2
For, l = 3, m = –3, –2, –1, 0, 1, 2, 3 (c) (d)
For, l = 4, m = –4, –3, –2, –1, 0, 1, 2, 3, 4 J & K CET-(2013)
Hence, the number of orbitals with n = 5, ml = +2 is 3. Ans. (c) : According to Aufbau principle in the ground
436. Which of the following atoms in its ground state of the atoms the electron fill subshells of the
state has the highest number of unpaired lowest available energy, then they fill subshells of
electrons? higher energy. So, in option (c) 2s subshell should filled
(a) Chromium (24) (b) Iron (26) before the 2p subshell is occupied while in (a) Hund’s
rule is violated.
(c) Manganese (25) (d) Vanadium (23)
J & K CET-(2016) 440. What is the total number of electrons that can
have the values n = 2, l = 1, s = ½ in the
Ans. (a) : The electronic configuration of the give electronic configuration 1s22s22p3?
element is– (a) 1 (b) 3
Chromium (24): (3d54s1) – 6 unpaired electron. (c) 5 (d) 7
Iron (26) : (3d64s2) – 4 unpaired electron. J & K CET-(2013)
Manganese (25) : (3d54s2) – 5 unpaired electron. Ans. (b) The given electronic configuration is 1s22s22p3
Vanadium (23) : (3d33s2) – 3 unpaired electron.
So, 24Cr has the highest number of unpaired electrons. 2p
437. The sulphur compound in which the sulphur Then, n=2, l=1 i.e. 2p
atom has octet configuration in its valence shell Hence, the total number of electron is 3.
among the following is
441. According to Aufbau principle, the correct
(a) Sulphur trioxide order of energy of 3d, 4s and 4p orbitals is
(b) Sulphur hexafluoride (a) 4p < 3d < 4s (b) 4s < 4p < 3d
(c) Slphur dichloride (c) 4s < 3d < 4p (d) 3d < 4s < 4p
(d) Sulphur dioxide J & K CET-(2006)
J & K CET-(2016) Ans. (c) : According to Aufbau principle the increasing
Ans. (c) : Octet rule dictates that atoms are most stable order of the energies of the orbital's is
when their valence shells are filled with eight electrons. 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s…..
The moleculer of the halogens, oxygen, nitrogen and 442. Electronic configuration of deuterium atom is
carbon are known to obey the octet rule.
(a) 1sl (b) 2s2
1
(c) 2 s (d) 1s2
J & K CET-(2005)
Ans. (a) : Deuterium is an isotope of hydrogen its
atomic number is one
Hence, its electronic configuration is as 1D2: 1s1
443. Two nodal planes are present in
(a) π * 2p x (b) σ 2p z
(c) π 2p x (d) π 2p y
SCl 2 : Cl − S − Cl (complete octet) J & K CET-(2004)

Objective Chemistry Volume-I 204

Ans. (a) : π*2px has two nodal planes perpendicular to
each other

444. Ground state electronic configuration of

nitrogen atom can be represented as
(a) From the above equation option (a) i.e. 1s2, 2s2 2p2, 3s2
does not follow Aufbau Rule.
447. The element with the electronic configuration
as [Ar] 3d104s24p3 represents a
(b) (a) metal (b) non-metal
(c) metalloid (d) transition element
JCECE - 2010
Ans. (c) : Electronic configuration shows that the p-
(c) orbital of the element is not complete. That is why, it is
a p-block element. Moreover, the atomic number of the
element is 33(As). Therefore, it is a metalloid.
448. The electronic configuration of element with
atomic number 24 is
(d) (a) 1s2, 2s22p6, 3s23p63d4, 4s2
(b) 1s2, 2s22p6, 3s23p63d10
(c) 1s2, 2s22p6, 3s23p63d6
(d) 1s2, 2s22p6, 3s23p63d5, 4s1
J & K CET-(2003) JCECE - 2010
Ans. (a) : Nitrogen (N) atomic number = 7 Ans. (d) : The electronic configuration of element with
Electronic configuration (7N) = 1s2, 2s2, 2p3 atomic number 24 is
Hence, 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
(Q Exactly half-filled orbitals are more stable than
nearly half-filled orbitals.)
449. Which of the following is correct for number of
electrons, number of orbitals and type of
445. A 3p-orbital has orbitals respectively in N-orbit?
(a) two non-spherical nodes (a) 4, 4 and 8 (b) 4, 8 and 16
(b) two spherical nodes (c) 32, 16 and 4 (d) 4, 16 and 32
(c) one spherical and one non-spherical node JCECE - 2013
(d) one spherical and two non-spherical nodes Ans. (c) : For N shell, types of orbitals = 4s, 4p, 4d, 4f
J & K CET-(1998) Number of orbitals = 1+ 3 + 5 + 7 = 16
number of electrons = 32
Ans. (c) : Given 3p orbital
450. The outermost electronic configuration of the
In 3p orbital, n = 3 and l =1
most electronegative element is
We know, (a) ns2np3 (b) ns2np4
Spherical Node = n − l − 1 = 3 – 1 – 1 = 1 2
(c) ns np 5
(d) ns2np6
Non spherical Node = l = 1 JCECE - 2014
Hence, one spherical Node and one non spherical Node. Ans. (c) : ns2np5 configuration shows the most
446. Impossible configuration is: electronegative element as after gaining one electron it
becomes more stable (inert gas configuration)
(a) 1s2, 2s2, 2p2, 3 s2 (b) 1s2, 2s2 [electronegativity is the tendency of attracting electron].
2 2 6
(c) 1s , 2s , 2p (d) none of these 451. If n=6, the correct sequence for filling of
JCECE - 2003 electrons will be.
Ans. (a): Aufbau Principal: it state that electron are (a) ns →(n-1) d→(n-2) f→np
filled in the atomic orbitals in the increasing order of (b) ns →(n-2) f→np→(n-1)d
orbital energy level It means the available atomic (c) ns → np→(n-1) d→(n-2)f
orbitals with the lowest energy levels are occupied (d) ns →(n-2) f→(n-1)d→np
before those with higher energy levels. JIPMER-2014

Objective Chemistry Volume-I 205

Ans. (d) : If n=6, then the correct sequence for filling of 455. Among the elements form atomic number 1 to
electrons will be 36, the number of elements which have an
ns→(n–2)f→ (n–1)d→np. unpaired electron in their s subshell is
It will be (a) 2 (b) 7
6s→4f→ 5d→6p. (c) 6 (d) 9
The energy of 4f and 5d orbitals are in between the Karnataka-CET-2014
energies of 4s and 6p orbitals. Ans. (c) : Key elements among 1 to 36 atomic number
Note:– Higher is the (n+ l ), higher will be the energy of which have one-unpaired electron in their s-subshell-
the orbital. Q The alkali metals have 1 electron in s-subshell-
For n=6 H = 1 → 1s1
6s= 6+0=6 Li = 3 → 1s2 2s2
6p=6+1=7 Na = 11 → 1s2 2s2 2p6 3s1
5d=5+2=7 K = 19 → [Ar] 4s1
4f=4+3=7 Cr = 24 → [Ar] 4s1 3d5
Hence, the correct order is:
Cu = 29 → [Ar] 4s1 3d10
ns→(n–2)f→(n–1)d→np Therefore, there are total 6 elements between atomic
452. The total number of orbitals in the fifth energy number 1 to 36 which have one electron in their outer
level is most s-subshell.
(a) 5 (b) 10 456. Impossible orbital among the following is
(c) 18 (d) 25 (a) 3f (b) 2p
JIPMER-2009 (c) 4d (d) 2s
UP CPMT-2008 Karnataka-CET-2012
Ans. (d) : For fifth energy level Ans. (a) : According to Bohr Bury's scheme maximum
The value of n will be 5 number of e- in 3rd orbital 2n2=2(3)2=18
Therefore, the total number of orbitals is given by n2 = Maximum number of e- in s-subshell =2, p-subshell =6,
52 = 25 d-subshell =10 and f-subshell =14 therefore, 18 e-
453. The last element of the p-block in 6th period is present in 3rd orbital enter in s, p and d subshell.
represented by the outer most electronic There after there is no e- available for f-subshell that is
configuration why 3f subshell is not possible.
(a) 7s2, 7p6 (b) 5f14, 6d10, 7s2, 7p5 Thus, the value of magnetic quantum number is-l to +l,
14 10 2 4
(c) 4f , 5d , 6s , 6p (d) 4f14, 5d10, 6s2, 6p6 then –2 to +2, i.e –2–1, 0,+1, +2
Karnataka-CET-2020 457. The number of naturally occurring p-block
Ans. (d) : As we know form periodic table, the last elements that are diamagnetic is
element of the p-block in 6th period is Radon (86Rn). (a) 18 (b) 6
The electronic configuration of 86Rn (c) 5 (d) 7
2 2 6 2 6 10 2 6 10 2
86Rn = 1s , 2s , 2p , 3s , 3p , 3d , 4s , 4p , 4d , 5s , Karnataka-CET-2011
14 10 2 6
4f , 5d , 6s , 6p Ans. (a) : The group 18 elements have completely filled
Or it can be also written as [Xe] 4f14, 5d10,6s2, 6p6 subshells, therefore, they are diamagnetic. They are He,
454. The orbital nearest to the nucleus is Ne, Ar, Kr and Xe.
(a) 4f (b) 5d 458. Which one of the following is not correct in
(c) 4s (d) 7p respect of hybridization of orbitals?
Karnataka-CET-2018 (a) The orbitals present in the valence shell only
Ans. (c) : In general orbitals with lower energy are are hybridized
present near to the nucleus. This can be calculate by (b) The orbitals undergoing hybridization have
(n+1) rule. almost equal energy
(i) Lower be the sum of (n+1) value, nearest the orbital (c) Promotion of electron is not essential
to the nucleus. condition for hybridization
(ii) And for the same sum, if the value of n will be (d) It is not always that only partially filled
lower, then its energy will also be low and it will be orbitals participate in hybridization in some
nearest to the nucleus. cases even filled orbitals in valence shell take
Therefore, from the given options- part
(e) Pure atomic orbitals are more effective in
(a) 4f→ ( l =3) ∴ n+ l = 4+3=7 forming stable bonds than hybrid orbitals
(b) 5d→ ( l =2) ∴ n+ l = 5+2=7 Kerala-CEE-2013
(c) 4s→ ( l =0) ∴ n+ l = 4+0=4 Ans. (e) : Pure atomic orbitals are less effective in
(d) 7p→ ( l =1) ∴ n+ l = 7+1=8 forming stable bonds than hybrid orbitals.

Objective Chemistry Volume-I 206

459. Which of the following are isoelectronic Ans. (b) : According to question.
species? n=4
(i) NH3 (ii) CH 3+ Now,
Number of atomic orbital in particular energy level is
(iii) NH 2– (iv) NH +4 given by formula =n2
Choose the correct answer from the codes Number of orbital's in fourth energy level ⇒42⇒ 16
given below. orbital's.
(a) (i), (ii), (iii) (b) (ii), (iii), (iv) 464. Which of the following pairs of d-orbital's will
(c) (i), (ii), (iv) (d) (i), (iii), (iv) have electron density along the axes?
(e) (ii), (iii) (a) d z 2 , dxy (b) dxy, dyz
Kerala-CEE-2014
(c) d z2 ,d x 2 − y2 (d) d xy ,d x 2 − y2
Ans. (d) : Species Number of electrons
(1) NH3 7+1×3=10 NEET-II 2016
+ Ans. (c) : The shapes of various d-orbitals are as follow
(2) CH 3 6+1×3–1=8

(3) NH −2 7+1×2+1=10
+
(4) NH 7+1×4–1=10
4
Since, xy xz yz
(i) (iii) and (iv) have the equal number of electrons
therefore these are called as isoelectronic species.
460. Electronic configuration of only one p-block
element is exceptional. One molecule of that
element consists of how many atoms of it?
(a) One (b) Two
(c) Three (d) Four
MHT CET-2015
Ans. (a) : Helium is the element with exceptional So, the pairs of d-orbitals having electron density along
configuration in p-block with only 2 electrons in the axis are dz2 and dx2–y2.
outermost shell Helium is monoatomic hence it consists 465. 4d, 5p, 5f and 6p orbital's are arranged in the
of only 1 atom. order of decreasing energy. The correct option
Electronic configuration of helium (He) = 1s2 is
(a) 5f > 6p> 4d > 5p (b) 5f > 6p > 5p > 4d
461. The number of spherical nodes in 3p orbitals
(c) 6p > 5f > 5P > 4d (d) 6p > 5f > 4d > 5p
are/is
NEET-2009
(a) one (b) three
(c) none (d) two Ans. (b) : As we know, Greater the (n+1) values greater
is the energy (n+1) values for,
NEET-1988
4d=4+2=6
Ans. (a) : Given, 3p orbital 5p=5+1=6
In 3p orbital, n = 3 and l = 1 5f=5+3=8
We know, 6p=6+1=7
Spherical Node = n–l – 1 = 1 Order of energy would be 5f > 6p > 5p > 4d
462. The orientation of an atomic orbital is 466. The stable electronic configuration of
governed by chromium is
(a) principle quantum number (a) 3d6 4Sl (b) 3d5 4S2
5 l
(b) azimuthal quantum number (c) 3d 4S (d) 3d6 4So
(c) spin quantum number J & K CET-(2007)
(d) magnetic quantum number Ans. (c) : The electronic configuration of 24Cr is 1s2 2s2
NEET-2006 2p6 3s2 3p6 3d5 4s1
Ans. (d) : Magnetic Quantum number represents the 467. Which one of the following electronic
orientation of an orbital around the nucleus. It is arrangements is absurd?
represented by ml. (a) n=3,l=1,m= –1 (b) n=3, l =0, m= 0
463. The total number of atomic orbitals in fourth (c) n=2, l =0, m= –1 (d) n=2, l =1, m= 0
energy level of an atom is WB-JEE-2019
(a) 8 (b) 16 Ans. (c) : The value of magnetic quantum number m
(c) 32 (d) 4 depends upon aziamuthal quantum number l m varies
NEET-2011 form –l to +l
UPTU/UPSEE-2004 So, if the value of l. is O then value of m will also be O.

Objective Chemistry Volume-I 207

468. The ground state term symbol for an electronic 473. An element belongs to group 15 and third
state is governed by: period of the periodic table, Its electronic
(a) Heisenberg’s principle configuration will be
(b) Hund’s rule (a) 1s2 2s2 2p3 (b) 1s2 2s2 2p4
2 2 6 2 3
(c) Aufbau principle (c) 1s 2s 2p 3s 3p (d) 1s2 2s2 2p6 3s2 3P2
(d) Pauli exclusion principle WB-JEE-2011
UPTU/UPSEE-2004 Ans. (c) : The element belongs to group 15 and third
Ans. (b) : The ground state term symbol for an period is phosphorus (15P).
electronic state is governed by Hund's Rule. Electronic configuration of phosphorus.
2 2 6 2 3
469. The compound in which cation is isoelectronic 15P = 1s , 2s , 2p , 3s , 3p
with anion is: 474. The representation of the ground state
(a) NaCl (b) CsF electronic configuration of He by box-diagram
(c) NaI (d) K2S
UPTU/UPSEE-2004 as is wrong because it violates
Ans. (d) : NaCl: No of e– in Na+ = 11–1=10 (a) Heisenberg's uncertainty principle
No of e- in Cl– = 17+1=18 (b) Bohr's quantization theory of angular
CsF No of e- in Cs+ = 55–1=54 momenta
No of e- in F- = 9+1=10 (c) Pauli's exclusion principle
(d) Hund's rule
NaI No of e- in Na+ ⇒11–1=10
WB-JEE-2011
No of e- in I- ⇒53+1=54
K2S No of e- in K+ =19–1=18 Ans. (c) : According to Pauli Exclusion principle, in
any orbital, maximum two electrons can exist, having
No of e- in S2- =16+2=18 opposite spin.
So, K2S anion has isoelectronic with cation
475. Which of the following electronic configuration
470. Which of the following will violate Pauli’s is not possible?
exclusion principle?
(a) n=3, l = 0, m=0 (b) n=3, l = 1, m= –1
(c) n=2, l = 0, m= –1 (d) n=2, l = 1, m= 0
(a) WB-JEE-2018
Ans. (c) : When l=0, m will also be zero.
Hence electronic configuration is not possible.
(b)
476. The number of unpaired electrons in Ni
(c) Both (a) and (b) (atomic number = 28) are
(d) None of the above (a) 0 (b) 2
UPTU/UPSEE-2014 (c) 4 (d) 8
Ans. (b) : Pauli’s exclusion principle states that in a WB-JEE-2018
8 2
single atom no two electrons will have an identical set Ans. (b) 28Ni has electronic configuration = [Ar]3d ,4s
or the same quantum numbers. It means orbital must
have opposite spins or it should be anti parallel.
471. The electronic configuration of the oxide ion is
much most similar to the electron configuration ∴ No of unpaired electron =2
of the
(a) sulphide ion (b) nitride ion
(c) oxygen atom (d) nitrogen atom 7. Quantum Number
UPTU/UPSEE-2009
Ans. (b) : Sulphide ion (S2-)=No of electrons =16+2=18 477. Consider the following pairs of electrons
Nitride ion (N3-)=No of electrons =7+3=10 1
(A) (a) n=3, l=1, m1=1, ms= +
Oxide ion (O2-)=No of electrons =8+2=10 2
No of electros in oxygen atom =8 1
(b) n=3, l =2, m1=1, ms= +
No of electrons in nitrogen atom=7 2
Electronic configuration of O2- or N3- (10) 1
1s2 2s2 2p6 (B) (a) n=3, l =2, m1= –2, ms= –
2
472. The electronic configuration 1s22s22p63s23p63d9 1
represents a (b) n=3, l =2, m1= –1, ms= –
(a) Metal atom (b) Non metal atom 2
(c) Non metallic anion (d) Metallic cation 1
(C) (a) n=4, l =2, m1= 2, ms= +
WB-JEE-2008 2
Ans. (d) : The given configuration represents the 1
(b) n=3, l =2, m1= 2, ms= +
electron configuration of Cu2+ i.e a metallic cation. 2
Objective Chemistry Volume-I 208
 The pairs of electrons present in degenerate (a) (A), (B) and (C)
orbitals is/are: (b) (A), (C), (D) and (E)
(a) Only (A) (b) Only (B) (c) (A), (C) and (D)
(c) Only (C) (d) (B) and (C) (d) (A), (B), (C) and (D)
JEE Main-24.06.2022, Shift-I
JEE Main-28.06.2022, Shift-II
Ans. (b) :
(A) (a) n = 3, l = 1 Ans. (c) : Here, statements (A), (C) and (D) are correct.
⇒ 3p And statements (B) and (E) are incorrect.
(b) n = 3, l = 2 The correct form of statements (B) and (E)–
⇒ 3d (B) The azimuthal quantum number ‘l’ for a given ‘n’
(B) (a) n = 3, l = 2 (principle quantum number) can have values as ‘l’ = 0,
1, 2 ……..(n – 1)
⇒ 3d
(b) n = 3, l = 2 (E) for l = 5, the total number of orbital
⇒ 3d = (2l + 1) =11
(C) (a) n = 4, l = 2 481. For n=3 energy level the number of possible
⇒ 4d orbitals (all kinds) are
(b) n = 3, l = 2 (a) 1 (b) 3
⇒ 3d (c) 4 (d) 9
478. The number of radial and angular nodes in 4d CG PET -2005
orbital are, respectively Ans. (d) : Given, n = 3
(a) 1 and 2 (b) 3 and 2 and we know,
(c) 1 and 0 (d) 2 and 1 Number of orbitals =n2
JEE Main-26.06.2022, Shift-II =(3)2 =9
Ans. (a) : We know that, 482. The number of radial nodes for 4p is
Radial nodes = n–l–1
(a) 4 (b) 3
Angular node = l
(c) 2 (d) 1
Where, n = Principle quantum number
l = Azimuthal quantum number SCRA-2015
For 4d, n = 4, l =2 Ans. (c): Radial node = n – l –1
∴ Radial node = 4–2–1 = 1 Where, n = principle quantum number
And angular node = 2 l = Azimuthal quantum number
479. Consider the following set of quantum For 4p orbital
numbers. n = 4, l = 1
The number of correct sets of quantum ∴ Radial node = 4–1–1 = 2
numbers is _____. 483. Which of the following sets of quantum
n l ml numbers is restricted:
(a) 3 3 –3 (b) 3 2 –2 (a) n = 3, 1 = 1, m = + 2
(c) 2 1 +1 (d) 2 2 +2 (b) n = 3, 1 = 2, m = –2
JEE Main-27.06.2022, Shift-II (c) n = 3, 1 = 1, m = + 1
Ans. (2) : The value of azimuthal quantum number (d) n = 3, l = 1, m = –1
s = 0, p = 1, d = 2, f = 3. MPPET - 2012
For n = 2, l = 0, 1, ml = +1, 0, –1
For n = 3, l = 0, 1, 2, ml = +2, +1, 0, –1, –2 Ans. (a) : If n = 3
Therefore, the correct set of quantum number in option Where, n = Principle Quantum No.
(b) and (c). Then, l (Azimuthal Quantum No.)
Hence, The number of correct sets of quantum number l = 0, 1, 2
is 2. ↓ ↓ ↓
480. Consider the following statement :
(A) The principle quantum number ‘n’ is a positive and m = 0. ,
integer with values of ‘n’ = 1, 2, 3,….
(B) The azimuthal quantum number ‘I’ For a 484. The magnetic quantum number m for the
given ‘n’ (principle quantum number) can have outermost electron in the Na atom, is :
values as ‘l’ = 0, 1, 2, …n (a) 1 (b) 2
(C) Magnetic orbital quantum number ‘mI’ for a (c) 3 (d) 0
particular ‘l’ (azimuthal quantum number) has AP-EAMCET-1991
(2I+1) values. Ans. (d) : The electronic configuration of sodium atom
(D) ± 1/2 are the two possible orientations of
electron spin. is :
11 Na = 1s
2 2
(E) For l =5, there will be a total of 9 orbital Which  2s 2p6 3s

1
 
of the above statement are correct? K L M

Objective Chemistry Volume-I 209

 For 3s1 Assertiion : Energy of the orbital decreases
Principle quantum number (n) = 3 with increase of 'n'.
Azimuthal quantum number (l) = 0 Reason : Energy is required in shifting away
Magnetic quantum number (m) = 0 the negatively charged electron from positively
485. Which one of the following sets of the quantum charged nucleus
numbers is not possible for a 4p electron ? (a) Assertion and Reasoning are correct
statements and Reason is the correct
1 explanation for Assertion
(a) n = 4, l = 1, m = +1, s = +
2 (b) Assertion and Reasoning are correct
1 statements and Reason is not the explanation
(b) n = 4, l = 1, m = 0, s = + for Assertion
2
(c) Assertion is correct, Reason is wrong
1
(c) n = 4, l = 1, m = 2, s = + (d) Assertion is wrong, Reason is correct
2 AP EAMCET (Engg.) 21.09.2020, Shift-I
1 Ans. (d) : As we move away from the nucleus, in atom
(d) n = 4, l = 1, m = –1, s = −
2 the potential energy of electron increases, as the
AP-EAMCET-1998 distance for nucleus increases, the value of n increases.
Ans. (c) : For 4p electron, Consequently the kinetic energy of the electron
decreases, but total energy as a whole increases. Thus,
1
n = 4, l = 1, m = +1, 0, –1, S = ± the Assertion is incorrect but Reason is correct. Hence,
2 the correct option is (d).
Hence the option (c) is not correct. 489. The four quantum numbers of the valence
486. Based on the Bohr’s theory of hydrogen atom, electron of potassium are:
the speed of the electron, energy of the electron,
and the radius of its orbit varies with the
principal quantum number, respectively as
1 1
(a) n, n2, n2 (b) 2 , , n
n n
1 1 1 1
(c) , , n 2 (d) , 2 , n 2
n n n n 1 1
TS-EAMCET (Engg.), 07.08.2021 Shift-II (a) 4,0,1, (b) 4,1,0,
2 2
Ans. (d) : 1 1
Z (c) 4,0,0, (d) 4,1,1,
Speed of electron (V) = × 2.188 × 108 cm / sec 2 2
n AP EAMCET (Medical) -1998
1 Ans. (c): The electronic configuration of potassium is
or V ∝ [Where n is energy level]
n 1s 2 , 2s 2 , 2p 6 ,3s 2 ,3p 6 , 4s1. We find the four quantum of
Z2 4s– shell–
Energy of e– (E) = –13.6 2 eV [Z is atomic number]
n Principal quantum number (n) = 4
1 Azimuthal quantum number ( l ) = 0
or E∝ 2 Magnetic quantum number (m) = 0
n
1
n2 Spin quantum numbers (s) = +
Radius of orbital (rn)= 0.529 × Å 2
Z
or rn ∝ n2 490. "No two electrons in an atom can have the
same set of all four quantum numbers". This
1 1
Hence, V : E : rn = : 2 : n 2 principle of called
n n (a) Zeeman effect
487. Find the exact orbital from the following for (b) Pauli's exclusion principle
which n = 4 and l = 1. (c) Stark effect
(a) 4s (b) 3d (d) Heisenberg principle
(c) 4d (d) 4p AP EAMCET (Engg.) 18.09.2020, Shift-I
AP EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (b) : Pauli's exclusion principle is, "No two
Ans. (d) : For, n (principal quantum number) = 4 an l electrons in an atom can have the same set of all four
(azimuthal quantum number) = 1 quantum numbers".
The orbital is 4p. e.g. for 4s2 configuration, its
For, 4s → n = 4, l = 0 n l m s
3d → n = 3, l = 2 1st e– = 4 0 0 +1/2
4d → n = 4, l = 2 2nd e– = 4 0 0 –1/2
488. Choose the correct option regarding:

Objective Chemistry Volume-I 210

491. How many emission spectral lines are possible 1
when hydrogen atom is excited to nth energy (b) n=3, l=1, m=1, s = +
level ? 2
n(n + 1) (n + 1) 1
(a) (b) (c) n=3, l=2, m= –2, s = +
2 2 2
1
(n − 1)n n2 (d) n=3, l=0, m=0, s = +
(c) (d) 2
2 4 COMEDK 2018, Assam CEE-2018
TS EAMCET-2017
JIPMER-2012
n(n − 1) Karnataka-CET-2011
Ans. (c) : Number of spectral lines =
2 NEET-1994
Where, n is the orbital of electron. Ans. (c) : The orbital with highest (n+l) value will have
492. A subshell n = 3, l = 2 can accommodate the highest energy. In the given sets n=3, l=2, m =–2, s
maximum of______ =+1/2 have n+l =5 i.e., 3d-orbital has the highest
(a) 10 electrons (b) 6 electrons energy.
(c) 18 electrons (d) 16 electrons 498. Four different sets of quantum numbers for
AP EAPCET 20.08.2021 Shift-II four electrons are given below:
Ans. (a): Given that subshell n = 3, l = 2 1 1
e1 =4, 0, 0, – ;e 2 = 3,1,1, –
So, 2 2
l = 0 (s = subshell) 1 1
l = 1 (p = subshell) e3 =3, 2, 2, + ;e 4 = 3, 0, 0, +
l = 2 (d = subshell) 2 2
The order of energy of e1, e2, e3 and e4 is
l = 3 (f = subshell)
(a) e1> e2>e3>e4 (b) e4>e3>e2>e1
In n = 3 and l = 2 maximum no. of electron is 10.
(c) e3>e1>e2>e4 (d) e2>e3>e4>e1
493. The quantum number which explains the line COMEDK-2018
spectra observed as doublets in case hydrogen
and alkali metals and doublets are triplets in Ans. (c) : Higher (n+l) value, higher is the energy and
case of alkaline earth metals is for same (n+l) value higher n, higher will be the energy.
(a) spin (b) azimuthal Thus, e3>e1>e2>e4
(c) magnetic (d) principal 499. Which one of the following expressions
AP EAMCET (Engg.) 2012 represent the electron probability function (D)?
Ans. (a) : Spin quantum number explains the line (a) 4 πr dr ψ 2 (b) 4 πr 2 dr ψ
spectra observed as doublets in case of hydrogen and
(c) 4 πr 2 dr ψ 2 (d) 4 πr dr ψ
alkali metals and doublets and triplets in case of
alkaline earth metal. AP-EAMCET (Medical), 2003
494. The number of angular and radial nodes of 4d Ans. (c) : According to the quantum mechanics, the plot
orbital respectively are of ψ2 versus distance from the nucleus (r) indicates the
(a) 3, 1 (b) 1, 2 probability of an electron in a certain volume of space is
(c) 3, 0 (d) 2, 1 the plot of the probability density. It explains a method
AP EAMCET (Engg.) -2014 for estimating an electrons probability at a certain
location. This amputation generates a quantity called an
Ans. (d) : For 4d subshell, n=4, l = 2 electron may be found in a certain region at a certain
No. of angular nodes =l = 2 place. Hence, the electron probability function is
Number of radial nodes = n–l – 1⇒ 4–2–1=1 4 πr 2 dr ψ 2 .
496. An electron having spin quantum number, s =
–1/2 and magnetic quantum number, m=+3 can 500. The number of radial nodes in 3s and 2p
be present in orbitals, respectively are
(a) both s-orbital and p-orbital (a) 2 ; 2 (b) 2 ; 0
(b) p-orbital only (c) 0 ; 0 (d) 3 ; 2
(c) f-orbital only TS-EAMCET (Engg.), 06.08.2021
(d) both d-orbital and f-orbital Ans. (b): No. of radial node = n – l – 1
COMEDK-2012 n = Principle quantum number
Ans. (c) : For f-orbital, l=3; m= –l to +l l = Azimuthal quantum number
i.e., –3 to +3 The value of l are–
497. Which one of the following sets of quantum l =012 3
s p d f
numbers represents the highest energy level in For 3s, No. of radial node = 3 – 0 – 1
an atom? =2
1 For 2p, No. of radial node = 2 – 1 – 1
(a) n=4, l=0, m=0, s = +
2 =0
Objective Chemistry Volume-I 211
501. The orbital with 4 radial and 1 angular nodes Ans. (c): For 4p electron n=4, l=1, m=1, 0+1
is
1
(a) 5py (b) 6pz ms =−+ As l = 1, m cannot be equal to 2.
(c) 4dxy (d) 5dyz 2
TS EAMCET 05.08.2021, Shift-I Therefore set of quantum number is not possible.
Ans. (b) : Radial nodes = n –l– 1 505. In the electronic configuration of which of the
Angular node = l given elements, ‘Aufbau principle’ or ( n+l )
where, n = Principle quantum number. rule is violated ?
l = Azimuthal quantum number. (a) Mn (b) Ga
For 5py, n = 5, l = 1 (c) La (d) Pu
Radial node = 5 – 1 – 1 = 3 AP EAPCET 24.08.2021 Shift-II
Angular node = 1 Ans. (c) : Aufbau principle states that the electrons are
For 6pz, n = 6, l = 1 filled into an atomic orbital in the increasing order of
Radial node = 6 – 1 – 1 = 4 orbital energy level.
Angular node = 1 Therefore La electronic configuration of the elements
violated the Aufbau principle rule because La is
For 4dxy, n = 4, l = 2 lanthanoid element.
Radial node = 4 – 2 – 1 = 1
506. With increasing Principal Quantum number,
Angular node = 2
the energy difference between adjacent energy
For 5dyz, n = 5, l = 2 levels in H-atom______
Radial node = 5 – 2 – 1 = 2 (a) Decreases
Angular node = 2 (b) Increases
Hence, the orbital node 4 radial and 1 angular nodes is (c) remain constant
6pz. (d) decreases at low level of 'n' & increases for
502. The orbital having two radial as well as two higher value of 'n'
angular nodes is: AP EAPCET 20.08.2021 Shift-II
(a) 3p (b) 4d Ans. (a): With increasing principle Quantum number,
(c) 4f (d) 5d the energy difference between adjacent energy levels in
JEE Main 26.02.2021, Shift-I H-atom is decreases.
Ans. (d) : Angular node = l 507. The maximum number of electrons, present in
Where, l = azimuthal quantum number an orbit that is represented by azimuthal
quantum number (l) = 3, will be
Angular Radial (a) 8 (b) 2
Orbital
node(l ) node(n - l -1) (c) 14 (d) 6
5d 2 2 AIIMS-1996
Ans. (c): l=3 corresponds to f-orbitals. Since there are
4f 3 0 seven f-orbitals and each orbital accommodates 2
3f 1 1 electrons, So maximum number of electrons is 14.
4d 2 1 508. What is maximum wavelength of line of
Balmer series of hydrogen spectrum?
503. Which of the following elements outermost (R=1.09×107m−1)
orbit's last electron has magnetic quantum
(a) 400 nm (b) 654 nm
number m = 0?
(c) 486 nm (d) 434 nm
(a) Na (b) O
AIIMS-26 May, 2018
(c) Cl (d) N
AIIMS 25 May 2019 (Evening) Ans. (b): For maximum wavelength in the Balmer
2 2 6 1 series energy difference is lowest.
Ans. (a) : Na 11Na=1s 2s 2p 3s
For last electron, l=0, m=0  1 1 
Thus, v = R × Z2  2 − 2  , n 2 = 3and n1 = 2
504. Which one of the following set of quantum  n n 2 
numbers is not possible for 4p electron? 1  1 1
1 ∴ = 1.09 × 107 × 12  2 − 2 
(a) n = 4, l =1, m = –1, ms = + λ 2 3 
2
1  1 1
1 ⇒ = 1.09 ×107 × 1 − 
(b) n = 4, l =1, m = 0, ms = + λ 4 9
2
1  5
(c) n = 4, l =1, m = 2, ms = + ⇒ 1.09 × 107 ×1 
2  36 
1 36
(d) n = 4, l =1, m = –1, ms = – ⇒λ=
2 5 × 1.09 ×107m
AIIMS 25 May 2019 (Evening) ⇒ 6.60×10 -7
m⇒660 nm
Objective Chemistry Volume-I 212
509. Which transition in the hydrogen atomic Ans. (a): According to Aufbau's principle, the filling
spectrum will have the same wavelength as the of electrons in various subshells of an atom takes
transition, n=4 to n=2 of He+ spectrum? place in the increasing order of energy, starting with
(a) n=4 to n= 3 (b) n=3 to n= 2 the lowermost.
(c) n=4 to n= 2 (d) n=2 to n= 1 According to the Bohr–Bury rule i.e. (n+l) sum rule,
AIIMS-2016 the sub shell with the lower value of (n+l) is filled
first if the values for (n+l) are equal, the one with the
1  1 1  smaller value of n is filled first.
Ans. (d): Q = z 2 .R  2 − 2 
λ  n1 n2  n l (n+l)
∴ For He ion, z=2
+
(i) 4 1 5
1 1 1 (ii) 4 0 4
= 22 R  2 − 2 
λ 2 4  (iii) 3 2 5
1 3 3 (iv) 3 1 4
= 4× R × = R So, the correct order is iv < ii < iii < i.
λ 16 4
The same value for H-atom is possible when electron 513. Azimuthal quantum number defines
jumps from n=2 to n=1 i.e. (a) e/m ratio of electron
1 1 1  3 (b) angular momentum of electron
= 1× R  −  ⇒ R (c) spin of electron
λ 1 4  4
(d) magnetic momentum of electron
510. In Bohr series of lines of hydrogen spectrum,
AIIMS-2002
the third line from the red end corresponds to
which one of the following inter- orbit jumps of Ans. (b): Generally azimuthal quantum number defines
the electron for Bohr orbits in an atom of angular momentum.
hydrogen 514. The quantum number 'm' of a free gaseous
(a) 5→2 (b) 4 →1 atom is associated with:
(c) 2→5 (d) 3→2 (a) the effective volume of the orbital
AIIMS-2017 (b) the shape of the orbital
Ans. (a): In hydrogen spectrum coloured radiaton (c) the spatial orientation of the orbital
means visible radiation corresponds to Balmer series. (d) the energy of the orbital in the absence of a
(n1=2, n2=3, 4…….) magnetic field
AIIMS-1998
Ans. (c): Magnetic quantum number m is associated
with spatial orientation of the orbital. It is also called
orientation quantum number because it gives the
orientation or distribution of the electron clouds.
515. For principle quantum number n=4, the total
number of orbitals having l=3 is:
nd (a) 3 (b) 7
2 line from the red end, it means 5→2
(c) 5 (d) 9
511. Which of the following combinations of AIIMS-2004
quantum numbers is allowed?
Ans. (b): For n=4, l =3 (f–subshell), number of values
ms 1
(a) n l m (b) 2 0 0 − of m=2l+1=7 values
3 2 1 0 2 ⇒ Number of orbitals in f-subshell is 7
1 1 516. Quantum numbers of an atom can be defined
(c) 3 −3 −2 + (d) 1 0 1 +
2 2 on the basis of
AIIMS-2013 (a) Hund's rule
Ans. (b): (a) is allowed as s cannot be zero (b) Pauli's exclusion principle
(c) is not allowed as l cannot be –3 because l can be any (c) Aufbau's principle
whole number from 0 to n–1 and it cannot be n. (d) Heisenberg's uncertainty principle
(d) is not allowed as for l=o and m cannot be =1 AIIMS-2002
512. The electrons identified by quantum numbers n Ans. (b): Each electron in an atom is designated by a
and l for (i) n=4, l=1(ii) n=4, l=0(iii) set of four quantum numbers. According to pauli's
n=3,l=2(iv)n=3,l=1 can be placed in order of exclusion principle, no two electron in an atom have
increasing energy, from the lowest to highest, same values of all the four quantum numbers. Therefore
as consequently, an orbital accommodates two electron
(a) (iv)<(ii)<(iii)<(i) (b) (ii)<(iv)<(i)<(iii) with opposite spins, these two electron have the same
(c) (i)<(iii)<(ii)<(iv) (d) (iii)<(i)<(iv)<(ii) value of quantum number 'n' , 'l' , and 'm' but value of 's'
AIIMS-2014 will be different.

Objective Chemistry Volume-I 213

517. The total number of orbitals in a shell with (a) K (b) Ti
principal quantum number 'n' is: (c) Na (d) Sc
(a) n2 (b) n+1
AP-EAMCET (Engg.)-2004
(c) 2n (d) 2n2
AIIMS-1997 Ans. (a) : The four quantum number are given as –
Ans. (a): No. of orbital in a shell =n2. n=4
518. In an atom the order of increasing energy of l=0
electrons with quantum numbers m=0
(i) n = 4, l = 1 (ii) n = 4, l = 0 1
and s = +
(iii) n = 3, l = 2 and (iv) n = 3, l = 1 is 2
(a) iii < (i) < iv < ii (b) ii < iv < i < iii The value of principle quantum number is 4, it means
(c) i < iii < ii < iv (d) iv < ii < iii < i that elements belongs to the fourth period. The value of
AP-EAMCET (Engg.) - 2014 azimuthal quantum number is 0 that means electron
Ans. (d) : The order of increase of energy can be present at s-orbital. m = 0 indicates that the valence
calculated from (n + l) rule. If two orbitals have same electron is present in orbital of s-subshell. The value of
value of (n + l), the orbital with lover value of n will be s denote the orientation of electron. Hence, from the
filled first. above data, it is clear that atom belongs to s-block and
(i) For n = 4, l = 1, (n + l) = 4+1 = 5 the name of metal is K.
(ii) For n = 4, l = 0, (n + l) = 4+0 = 4 522. The number of radial nodes of 3s and 2p
(iii) For n = 3, l = 2, (n + l) = 3+2 = 5 orbitals respectively are
(iv) For n = 3, l = 1, (n + l) = 3+1 = 4
(a) 0, 2 (b) 2, 0
So, the order is (iv) < (ii) < (iii) < (i)
(c) 1, 2 (d) 2, 1
519. The number of angular and radial nodes of 4d
AP-EAMCET (Engg.) 2013
orbital respectively are
(a) 3, 1 (b) 1, 2 Ans. (b) Number of radial nodes = n − l − 1
(c) 3, 0 (d) 2, 1 Where n = no. of principle quantum number.
AP-EAMCET (Engg.) - 2014 l = Azimuthal quantum number.
Ans. (d) : Number of radial node = n – l – 1 For 3s For 2p
Where, n = Principle quantum number n = 3, l = 0 n = 2, l =1
l = Azimuthal quantum number Radial nodes = 3–0–1 Radial nodes =2–1–1
∴ For 4d =2 =0
n = 4, l = 2 The number of radial nodes for 3s and 2p are 2, 0
Radial node = 4 – 2 – 1 = 1 respectively.
Angular node = l = 2
523. With increase in principal quantum number n,
Thus, the number of angular and radial node of 4d the energy difference between adjacent energy
orbital is 2, 1 respectively.
levels in hydrogen atom
520. Which of the following sets of quantum (a) increases
numbers is correct for an electron in 3d-
orbital? (b) decreases
(c) remains constant
1
(a) n = 3, l = 2, m = – 3, s = + (d) decreases for lower values of n and increases
2 for higher values of n
1
(b) n = 3, l = 3, m = + 3, s = – AP - EAMCET(MEDICAL) - 2009
2
Ans. (b) : As we know, the energy difference between
1
(c) n = 3, l = 2, m = – 2, s = + adjacent energy levels is inversely proportional to the
2 principal quantum number, n. Therefore, with increase
1 in n, the energy difference will decrease.
(d) n = 3, l = 2, m = – 3, s = –
2 1 1
∆E ∝  2 − 2 
AP-EAMCET (Engg.) 2015  n1 n 2 
Ans. (c) : An electron for 3d-orbitals–
524. The correct set of quantum numbers for Rb
n = 3, l = 2, m = –2, s = + ½
(atomic no. 37) is
Note : Azimuthal quantum number of s, p, d and f
orbital is 0, 1, 2 and 3 respectively. 1 1
(a) 5,0,0, − (b) 5,1,0,
521. The values of four quantum numbers of 2 2
valence electron of an element are n = 4, l = 0, 1 1
(c) 6,0,1, (d) 5,1,1,
1 2 2
m = 0 and s = + . The element is :
2 JEE Main-2013

Objective Chemistry Volume-I 214

Ans. (a) : The electronic configuration for Rb (37) is (a) 4,3,2,+1/2 (b) 4,2,1,0
Rb (37) (c) 4,3,-2,+1/2 (d) 4,2,1,–1/2
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 AMU-2003
1 Ans. (d) : For 4d electron.
For 5s1, n = 5. l= 0, m = 0, s = n=4, l=2
2
525. How many electrons in 19K have n = 3; l = 0 ? 1
m=–2, –1, 0+1, +2; s= ±
(a) 1 (b) 2 2
(c) 4 (d) 3 532. The nineteenth electron of chromium has
JCECE - 2011 which of the following set of quantum number?
Ans. (b) : n = 3 and l = 0 means 3s-orbitals and s- n l m s
orbital occupy 2 electrons. (a) 3 0 0 ±1/2 (b) 3 2 –2 ±1/2
526. The Balmer series in atomic hydrogen is (c) 4 0 0 ±1/2 (d) 4 1 –1 ±1/2
observed in the following spectral region AMU EXPLORER-2002
(a) infrared (b) ultraviolet Ans. (c) : Atomic number of chromium is 24. The
(c) visible (d) far IR electronic configuration of chromium is 1s2, 2s2, 2p6,
AMU-2014 3s2, 3p6, 4s4, 3d5. The nineteenth electron goes to the 4s
Ans. (c) : Balmer series in atomic hydrogen is observed orbital. The quantum number is illustrated below for 4s
in visible region. orbital.
Note:– Lyman series lies in UV region and paschen ↑
series lies in IR region and bracket and other lies in for
IR region. 4s
Principle quantum number (n) = 4
527. For the 19th electron of K the values of Azimuthal quantum number (l) = 0
quantum number will be
Magnetic quantum number (m) = 0
1 1
(a) 4, 1, 0, + (b) 4, 0, 0, + 1
2 2 Spin quantum number (s) = ±
2
1 1 Hence, the correct option of set of quantum number is 0.
(c) 3, 2, 0, + (d) 3, 0, 0, +
2 2 533. The ion that is isoelectronic with CO is
AMU-2010
(a) O −2 (b) N +2
Ans. (b) : 19K → 1S2 2s2 2p6 3s2 3p6 4s1
Values of quantum number - (c) CN– (d) O +2
1 Assam CEE-2018
n=4, I = 0, m = 0 and s= + UP CPMT-2012
2
528. An electron with values 4,1,0 and +1/2 for the JCECE - 2006
set of four quantum numbers n, l, ml and ms J & K CET-(1998)
respectively, belongs to NEET-1997
(a) 4s - orbital (b) 4p - orbital Ans. (c) : Number of electrons in
(c) 4d- orbital (d) 4f- orbital CO = 6+8=14
–
AMU-2006 CN = 6+7+1=14
+
Ans. (b) : n = 4, l = 1 indicates 4p orbital. N 2 = 7+7–1=13
529. An electron with values 4, 2,–2 and +1/2 for the N 22− = 7+7+2=16
set of four quantum numbers n, l, ml and ms,
respectively, belongs to NO = 7+8+1=16
–
(a) 4s-orbital (b) 4p-orbital CO is isoelectronic with CN because both species have
(c) 4d-orbital (d) 4f-orbital same number of electrons.
AMU-2005 Isoelcetronic species refers to the species which have
same number of electrons.
Ans. (c) : The value of ‘n’ and ‘ l ’ equal to 4 and 3
CN–1 is isoelectronic with CO, both loose 14 electron.
respectively, corresponds to 4f–orbital, hence the
electron will belong to 4f–orbital. 534. Which of the following is not possible?
530. For l = 1, the correct set of values of m are (a) n = 2, l = 1, m = 0 (b) n = 2, l = 0, m = −1
(a) 1, 2, 3 (b) 0, 1, 2 (c) n = 3, l = 0, m = 0 (d) n = 3, l = 1, m = −1
(c) 0 (d) -1, 0, +1 BCECE-2007
AMU-2004 Ans. (b) : (a) n=2, l =1, m=0, it is possible
Ans. (d) : For l =1 (b) n=2, l =0, m=–1, it is not possible because, if l =0,
m = – 1,0,+ 1[ m = – l,.......0,.....+ l ] m must be 0. The value of m totally depends on upon
531. Which of the following represents correct set of the value of l (m= – l to l )
the four quantum numbers ( n, l, m and s, (c) n=3, l =0, m= –0, it is possible
respectively) of a 4d electron? (d) n=3, l =1, m= –1, it is possible
Objective Chemistry Volume-I 215
535. In n = 3, l = 0 and m = 0, then atomic number 540. An e– has magnetic quantum number as –3,
is : what is its principal quantum number?
(a) 12 or 13 (b) 13 or 14 (a) 1 (b) 2
(c) 10 or 11 (d) 11 or 12 (c) 3 (d) 4
BCECE-2004 BITSAT 2016
Ans. (d) : When m = –3, l = 3, ∴ n = 4.
Ans. (d) : n=3, l =0, m=0 indicates last shell is 3s so,
electronic configuration will be 1s22s2 2p6 3s1-2 541. Which of the following sequences of the energy
So atomic number is 11 or 12. levels of the subshells related to principal
536. The correct set of four quantum number for quantum number four (n=4)?
the valence electron of rubidium (Z = 37) is : (a) s < p < d < f (b) s < d < p < f
(a) n = 5, l = 0, m = 0, s = + 1/2 (c) s < f < p < d (d) p < s < d < f
(b) n = 5, l = 1, m = 1, s = + 1/2 CG PET -2008
(c) n = 5, l = 1, m = 1, s = + 1/2 Ans. (a): The sequence of energies of the sub shells is
(d) n = 6, l = 0, m = 0, s = + 1/2 compassed of using m + l rule where 'n' is principle
quantum and 'l' is azimuthal quantum no. for same
BCECE-2003
value of 'n' higher value of l have higher energy.
Ans. (a) : Electronic configuration of rubidium (z=37)
is [Kr]36 5s1 So,
s<p<d<f
∴ for valence electron (5s1) 542. Four different sets of quantum numbers for 4
1 electrons are given below
n=5, l =0, m=0, s =
2 1 1
e1 = 4, 0, 0, − e 2 = 3,1,1, −
537. Azimuthal quantum number determines the 2 2
(a) size 1 1
(b) spin e3 = 3, 2, 2, + e 4 = 3, 0, 0, +
2 2
(c) orientation
The order of energy of e1 ,e 2 ,e 3 ,e4 is
(d) angular momentum of orbitals
BCECE-2011 (a) e1 > e 2 > e3 > e 4 (b) e 4 > e3 > e 2 > e1
h (c) e3 > e1 > e 2 > e 4 (d) e 2 > e3 > e 4 > e1
Ans. (d) : Angular momentum of orbital = l ( l +1) CG PET- 2013
2π
538. What is the correct orbital designation of an Ans. (c) : Higher (n+l) value higher is the energy and
electron with the quantum number, for same (n+l) value higher n, higher will the energy.
Thus, e3 > e1 > e2 > e4
1
n = 4, l = 3,m = −2,s = ? 543. 2p orbitals have
2 (a) n = 1, l = 2 (b) n = 1, l = 0
(a) 3s (b) 4f
(c) 5p (d) 6s (c) n = 2, l = 1 (d) n = 2, l = 0
BCECE-2010 CG PET -2004
th Ans. (c) : In 2 p-orbital , 2 denotes principle quantum
Ans. (b) : n=4 represent 4 orbit
l =3 represent f subshell number (n=2) and p denotes azimuthal quantum number
m= –2 represent orientation of f–orbital (l=1).
544. Which of the ions having following electronic
1
s= represents direction of spin of electron structure would have maximum magnetic
2 moment?
∴ the orbital is 4f. (a) 1s 2 , 2s 2 2p 6 ,3s 2 3p 6 3d 3
539. The correct set of quantum numbers for an
(b) 1s 2 , 2s 2 2p 6 ,3s 2 3p 6 3d 5
element (Z = 17) for the unpaired electron will
be – (c) 1s 2 , 2s 2 2p 6 ,3s 2 3p 6 3d 7
(a) 3,1,1, ±1/ 2 (b) 2, 0, 0, ±1/ 2 (d) 1s 2 , 2s 2 2p 6 ,3s 2 3p 6 3d 9
(c) 3, 0, 0, ±1/ 2 (d) 2,1,1,0
CG PET -2005
BCECE-2017
Ans. (a): Electronic configuration of (Z=17) 1s2, 2s2, Ans. (b) : Magnetic moment = n ( n + 2 ) , where n is
2p6, 3s2, 3p5 unpaired electron is present in 3p–orbital no. of unpaired electrons.
that can be shown as Higher the number of unpaired electrons, higher is the
magnetic moment.
Number of unpaired electrons are maximum in option b
(5 unpaired electrons).
1 Option a and c :3 unpaired electrons
Thus, for 3p, n=3, l=1, m= ± 1, s= ±
2 Option d : 1 unpaired electrons.
Objective Chemistry Volume-I 216
545. A particle of mass nearly equal to proton is Ans. (a):
moving with a velocity nearly equal to the Electronic configuration of Rubidium is (Z = 37)
velocity of light. The wavelength of wave = 1s22s22p63s23p64s23d104p65s1
associated with it is
The valence electron is in 5s orbital
(a) directly proportional to its velocity
(b) inversely proportional to its energy 1
Its 4 quantum number are- n = 5, l = 0, m = 0, s = +
(c) (a) is wrong while (b) is correct 2
(d) Both (a) and (b) are correct 549. The number of orbitals associated with
CG PET -2018 1
quantum number n = 5, m s = + is
Ans. (a) : 2
1 (a) 25 (b) 50
m v2 (c) 15 (d) 11
λ1 m 2 v 2 2 2 2 v1 K.E.1 v1
= = = = [JEE Main 2020, 7 Jan Shift-I]
λ 2 m1v1 1 m v 2 v 2 K.E 2 v 2
2
1 1 Ans. (a) : Total no. of electrons in n=5 are 2n2 i.e 2×52
As K.E1 =K.E2 =50
λ v +1 50
∴ 1 = 1 or λ × v Thus electrons having, ms = = = 25 which is
λ2 v2 2 2
equal to no. of orbitals.
546. A particle moving with a velocity 106 m/s will
550. The quantum number of four electrons are
have de-Broglie wavelength nearly
given below:
[Given, m = 6.62 × 10–27 kg, h = 6.62 × 10–34 J-s]
1
(a) 10–9 m (b) 10–13 m I. n = 4, l = 2, m1 = –2, ms = −
(c) 10 m–19
(d) 1 Å 2
1
CG PET -2009 II. n = 3, l = 2, m1 = 1, ms = +
Ans. (b): Given, 2
Velocity of particle =106 m/s 1
III. n = 4, l = 1, m1 = 0, ms = +
mass =6.62×10-27 kg 2
Plancks constant (h)=6.62×10-34 JS–1 1
IV. n = 3, l = 1, m1 = 1, ms = −
h 2
According to de-broglie wavelength λ =
p The correct order of their increasing energies
h will be
or = [∴ p = mv ] (a) IV < III < II < 1 (b) I < II < III < IV
mv (c) IV < II < III < I (d) I < III < II < IV
6.62 × 10−34 [JEE Main 2019, 8 April Shift-I]
λ= = 10−13 m
6.62 × 10−27 ×106 Ans. (c) : Smaller the value of (n+l), smaller the energy
547. From the given sets of quantum numbers, the if two or more sub-orbits have same values of (n+ l)
one that is inconsistent with the theory is suborbits with lower values of n has lower energy. The
1 (n+ l) value of the given option are as follows:
(a) n = 3, l = 2, m = −3,s = + (I) n=4, l =2; n + l =6
2
1 (II) n=3, l =2; n + l =5
(b) n = 4, l = 3, m = 3,s = + (III) n=4, l =1; n + l =5
2
(IV) n=3, l =1; n + l =4
1
(c) n = 2, l = 1, m = 0,s = − Among (II) and (III) n=3 has lower energy. Thus the
2 correct order of their increasing energies will be IV < II
1 < III < I
(d) n = 4, l = 3, m = 2,s = +
2 551. The number of subshells associated with n = 4
CG PET- 2010 and m = –2 quantum number is
Ans. (a) : When l=2, m≠–3 (a) 8 (b) 2
548. The correct set of four quantum number for (c) 16 (d) 4
the valence electrons of rubidium atom (Z = 37) [JEE Main 2020, 2 Sep Shift-II]
is Ans. (b) : For m=-2, it represent 2 subs hells.
1 1 552. The value of magnetic quantum number of the
(a) 5,0,0, + (b) 5,1,0, +
2 2 outermost electron of Zn+ ion is
1 1 [JEE Main 2021, 31 Aug Shift-II]
(c) 5,1,1, + (d) 5,0,1, +
2 2 Ans. (0) : Zn+ →1s2 2s2 2p6 3s2 3p6 3d10 4s1
[JEE Main 2014] outermost electron is in 4s subshell m=0.

Objective Chemistry Volume-I 217

553. The figure that is not a direct manifestation of Ans. (a) : The electronic configuration of Cu in ground
the quantum nature of atoms is state [Ar] 3d10 4s1
Hence, for outermost subshell, n=4 , l=0
1
(a) ml=0 and ms =+
2
556. The total number of orbitals associated with
the principal quantum number n = 3, is
(a) 9 (b) 8
(c) 5 (d) 7
J & K CET-(2014)
Ans. (a) : by using the relation, no of orbital's
(b) =n2=(3)2=9.
557. How many quantum numbers are required to
define the electron in an atom?
(a) 2 (b) 3
(c) 1 (d) 4
J & K CET-(2019)
Ans. (d) : Four quantum numbers (principal quantum
(c) number, azimuthal quantum number, magnetic quantum
number and spin quantum number) are required to
define or electron in an atom.
558. The electrons identified by quantum numbers n
and l, (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l =
2 and (iv) n = 3, l = 1 can be placed in order of
increasing energy as
(a) (i) < (ii) < (iii) < (iv)
(b) (iv) < (iii) < (ii) < (i)
(d) (c) (iv) < (ii) < (iii) < (i)
(d) (iv) < (i) < (ii) < (iii)
J & K CET-(2011)
Ans. (c) : j
[JEE Main 2020, 2 Sep Shift-I]
n l nl sub orbit n+l
Ans. (c) : The internal energy of 'Ar' or any gas has
nothing to do with quantum nature of atom hence, 4 1 4p 5
4 0 4s 4
3 2 3d 5
3 1 3p 4
Higher the value of (n + l), higher the energy if (n + l)
are same, sub orbit with lower value of n has lower
energy.
554. Maximum number of electrons in a shell Thus, 3p < 4s < 3d < 4p
principle quantum number n is given by Hence, correct order is (iv) < (ii) < (iii) < (i)
(a) n (b) 2n
(c) n2 (d) 2n2 559. Out of the following which is the correct set of
J & K CET-(2012) quantum numbers for outermost electron of
potassium (Z= 19)?
Ans. (d) : The maximum number of electrons that can
n l m s
be present in a shell is 2n2. Where n corresponds to a
shell this was stated by Pauli exclusion principle. (a) 4 3 2 –1/2
(b) 4 2 0 –1/2
555. The correct set of four quantum numbers for
the electron in the outermost subshell of Cu (c) 4 1 0 +1/2
atom (atomic number 29) in its ground state, is (d) 4 0 0 –1/2
n l ml ms J & K CET-(2010)
(a) 4 0 0 +1/2 Ans. (d) : Ans. (d) : K (z=19):[Ar]18 4s1
(b) 3 0 0 +1/2 Outermost electron :4s1
(c) 4 1 0 +1/2 1 −1
(d) 3 1 +1 +1/2 n=4, l =0, m=0, s=+ or .
J & K CET-(2016) 2 2
Objective Chemistry Volume-I 218
560. The number electrons accommodated in an Ans. (d) : n = 4, l = 3
orbit with principal quantum number 2 is l=0 for s sub-shell
(a) 2 (b) 6 l=1 for p sub- shell
(c) 10 (d) 8 l=2 for d sub- shell
J & K CET-(2007) l=3 for f sub- shell
Ans. (d) : The number of electrons =2n2 ∴ 4f
Where n = principal quantum number. For n =2 565. Which of the following quantum numbers
Number of electrons = 2(2)2 =8 distinguishes the two electrons present in an
561. The correct set of the four quantum numbers orbital from each other?
of a 4d electron is (a) principal quantum number
(a) 4, 2, 1, –1/2 (b) 4, 2, 1, 0 (b) azimuthal quantum number
(c) 4, 3, 2, +1/2 (d) 4, 3, –2, +1/2 (c) spin quantum number
J & K CET-(2005) (d) none of these
Ans. (a) : For 4d arbitral J & K CET-(1999)
Principal quantum Number (n) = 4 Ans. (c) : Spin quantum of any two electrons in an
Azimuthal quantum number (l) –2 (For d-subshell l=2) orbital cannot be same. If one has a value +1/2 other
Magnetic quantum number (m) can have values ranging have –1/2.
from –l to +l i.e –2 to +2 +1 1
,
Spin quantum number (s) has only two values 2 2
1 1 e.g. for 1s2
+ and −
2 2 566. The correct set of quantum numbers for the
562. The total number of orbital's possible for unpaired electron of chlorine atom is
principal quantum number n is (Atomic No. of Cl = 17 i.e. 3s2, 3p5)
(a) n (b) n2 N l m
(c) 2n (d) 2n2 (a) 3 1 0
J & K CET-(2004) (b) 3 0 1
Ans. (b) : The total number of orbital is a shell is n2 the (c) 3 1 1
principal quantum number (symbolized) is one of four (d) 3 0 0
quantum number assigned to each electron in an atom to J & K CET-(1999)
describe that electron is state. The total number of Ans. (c) : The correct set of quantum number for the
orbital possible for principal quantum n is n2. unpaired electron of chlorine atom is
563. The electrons identified by quantum numbers n = 3, l = 1 and m = 0.1
(i) n = 4, l = 1 (ii) n = 4, l = 0 For chlorine atom (Z = 17) the last
(iii) n = 3, l = 2 (iv) n = 2, l = 1 Electron enters into 3p orbital for which
Can be placed in order of increasing energy from 1
the lowest to highest as n = 3, l = 1, m = –1, 0 or + 1 and s = ±
2
(a) (iv) < (ii) < (iii) < (i)
567. The maximum number of electrons in all those
(b) (ii) < (iv) < (i) < (iii) orbitals for which principal quantum number
(c) (i) < (iii) < (ii) < (iv) is 3 and azimuthal quantum number 2, is
(d) (iii) < (i) < (iv) < (ii) (a) 2 (b) 8
J & K CET-(2003) (c) 10 (d) 18
Ans. (a) : According to Aufbau's principle the filling of J & K CET-(1997)
electrons in Various subshells of an atom takes place in Ans. (c) : Principal quantum number, n = 3, azimuthal
the increasing under of energy starting with lowest most. quantum ; l = 2,
According to the Bohr Bury rule i.e (n + l)Sum rule, the Hence, for l = 2, it is d – subshell.
sub-shell with the lowest value of (n + l) if filled first. If d – Subshell has 5 degenrate orbitals and 1 orbital
the values for (n + l) are equal, the one with the smaller contain maximum 2 electron
value of n is filled first.
Hence, 5 orbital contain maximum 2 × 5 = 10 electron.
N l (n + l)
568. Which of the following has highest energy?
i. 4 1 5 (a) n = 2, l = 1 (b) n = 3, l = 2
ii. 4 0 4 (c) n = 3, l = 1 (d) n = 2, l = 0
iii. 3 2 5 JCECE - 2006
iv. 3 1 4 Ans. (b) : Energy is highest in the outermost shell
The correct order is iv < ii < iii < i n = 2, l = 1 = 2p
564. n = 4 and l = 3 is designated as n = 3, l = 2 = 3d
(a) 4s (b) 4p n = 3, l = 1 = 3p
(c) 4d (d) 4f n = 2, l = 0 =2s
J & K CET-(1999) According to Dufbaus principle 3d has highest energy.

Objective Chemistry Volume-I 219

569. The shape of the orbital with the value of l =2 1
and m = 0 is (c) 5 3 0 –
2
(a) spherical (b) dumb-bell
1
(c) trigonal planar (d) square-planar (d) 3 2 –2
JCECE - 2009 2
JIPMER-2017
Ans. (b) : s p d f
l = 0, 1 2 3d – Orbital is Ans. (a) : For each value of n, l = n –1 (max) and for
double dumb-bell shaped. each value of 1.
m = 1 to + 1
570. The electron, identified by quantum numbers n
Thus, for 1 = 2, m = –2, –1, 0, 1, 2 and m = –3
and l, (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l =
2 (iv) n = 3, l = 1 can be placed in order of 574. Maximum number of electrons in a subshell of
increasing energy, from the lowest to highest, an atom determined by the following?
as (a) 4l + 2 (b) 2n2
(a) (iv) < (ii) < (iii) < (i) (c) 4l – 2 (d) 2l +1
(b) (ii) < (iv) < (i) < (iii) JIPMER-2013
(c) (i) < (iii) < (ii) < (iii) Ans. (a) : For a given shell, l the number of subshells,
m1 = (2l + 1) since each subshell can accommodate 2
(d) (iii) < (i) < (iv) < (ii) electrons of opposite spin,
JCECE - 2014 So, maximum number of electrons in a subshell = 2 (2l
Ans. (a) : (i) 4p (ii) 4s (iii) 3d (iv) 3p. + 1) = 4 l + 2.
According to Aufbau rule, order of increasing energy is 575. If magnetic quantum number of a given atom
3p < 4s < 3d < 4p. represented by 3, then what will be its principal
571. The orbital angular momentum of an electron quantum number?
is 2s orbital is (a) 2 (b) 3
1 h (c) 4 (d) 5
(a) + ⋅ (b) zero
2 2π JIPMER-2006
h h Ans. (c) : If the magnetic quantum number of a given
(c) (d) 2 ⋅ atom is represented by –3, then its principal quantum
2π 2π
JCECE - 2015 number will be 4.
For a given value of l,
Ans. (b) : Orbital angular momentum
m = –l, ……..0 …… + l when
h
= l ( l + 1) = 0 (Q for 2s-electron, l = 0) m = –3, l = 3
2π For given value of n , l
572. The electrons identified by quantum numbers n Can have values from 0 to n –1 when
and l, are as follows l = 3, n = 4, So that
I. n = 4, l = 1 II. n = 4, l = 0 n = –1, –4 –1 = 3
III. n = 3, l = 2 IV. n = 3, l = 1 576. Number of orbitals in L energy level
If we arrange them in order of increasing (a) 1 (b) 2
energy, i.e. from lowest to highest, the correct (c) 3 (d) 4
order is JIPMER-2005
(a) IV < II < III < I (b) II < IV < I < III
Ans. (d) : It is second energy level. Second energy level
(c) I < III < II < IV (d) III < I < IV < II
contains one s-orbital and three P orbitals. Therefore
JIPMER-2017 total number of orbital in L level is four.
Ans. (a) : (i) More be the sum of n + 1 → more be the 577. The shape of an orbital is determined by
energy
(a) n (b) l
(ii) Same sum, the more value of n have higher energy.
(c) m (d) s
∴ n + 1 for → JIPMER-2004
(i) 4 + 1 = 5 (ii) 4+0=4
Ans. (b) : The azimuthal quantum number (l)
(iii) 3 + 2 = 5 (iv) 3+1=4
determines the shape of an orbital.
Order is (iv) < (ii) < (iii) < (i).
For l = 0 (spherical, s)
573. Among the following set of quantum numbers, l = l (dumb-shape)
the impossible set is
578. Azimuthal quantum number (l) defined
n l m s
(a) shape of orbitals
1
(a) 3 2 –3 – (b) orientation of orbitals
2 (c) energy of orbitals
1 (d) size of orbitals
(b) 4 0 0
2 JIPMER-2019
Objective Chemistry Volume-I 220
Ans.(a) : The azimuthal quantum number (l) determines Ans. (c) : As the electronic configuration of Na is :
the shape of an orbital Is2 2s2 2p6 3s1
l = 0 (sperical, s) So, the outermost electron is 3s1 therefore quantum
l = 1 (dumb – bell, p) etc. 1
numbers for 3s’:3,0, 0, .
579. The correct set of quantum number for the 2
unpaired electrons of chlorine atom is 583. The set of quantum numbers for the outermost
1 1 electron for copper in its ground state is
(a) 2, 1, – 1, + (b) 2, 0, 0, +
2 2 1 1
(a) 4, 1, 1, + (b) 3, 2, 2, +
1 1 2 2
(c) 3,1,1, ± (d) 3,0, 0, ±
2 2 1 1
Karnataka-CET-2017 (c) 4, 0, 0, + (d) 4, 2, 2, +
2 2
Ans. (c) : The correct set of quantum number of the Karnataka-CET, 2010
unpaired electron of chlorine atom is Ans. (c) : The electronic configuration of the Cu atom
1 is
3, 1, 1, ± 10 1
2 29Cu = [Ar] 3d 4s
For chlorine atom ( Z = 17 ) the last electron enters into Since, the outermost shell is 4s, thus outermost electron
3p orbital for is in it.
Which n = 3, l = 1, m = –1, 0 For 4s1,
+1 1
or + 1 and. s = . n = 4, l = 0, m = 0, s = +
2 2
580. Consider the following sets of quantum 584. The correct set of four quantum number for
numbers. Which of the following setting in not outermost electron of potassium (Z = 19) is
permissible arrangement of electrons in an 1 1
atom? (a) 4, 1, 0, (b) 3, 1, 0,
n l m s 2 2
1 1 1
(a) 4 0 0 − (c) 4, 0, 0, (d) 3, 0, 0,
2 2 2
Karnataka-CET, 2009
1
(b) 5 3 0 + Ans. (c) : K (19) : 1s2, 2s22p6, 3s23p6, 4s1
2 4s1 is the valence electron is potassium, hence the correct set
1 1
(c) 3 2 –2 − of four quantum numbers for outermost electron is 4, 0, 0, .
2 2
1 585. When the azimuthal quantum number has the
(d) 3 2 –3 +
2 value of 2, the number of orbitals possible are
Karnataka-CET-2016 (a) 7 (b) 5
Ans. (d) : For a given value of ‘l’ the permission (c) 3 (d) 0
permissible value of ‘m’ are –l, – (l –1) …. 0….. (l–1) l Karnataka-CET, 2008
Thus for l = 2, m cannot have a value of –3. Ans. (b) : Given, azimuthal quantum number (l) = 2
581. The two electrons have the following set of Number of orbitals = (2l + 1)
quantum number, P = 3, 2, –2 + 1/2, Q = 3, 0, 0 = (2 × 2 + 1) = 4 + 1 = 5
+ 1/2 586. The number of angular and radial nodes in 3p
Which of the following statement is true? orbital respectively are
(a) P and Q have same energy (a) 3, 1 (b) 1, 1
(b) P has greater energy than Q (c) 2, 1 (d) 2, 3
(c) P has lesser energy than Q Kerala-CEE-29.08.2021
(d) P and Q represent same electron Karnataka-CET-2021
Karnataka-CET-2015
Ans. (b) : For angular node = l
Ans. (b) : P has greater energy than Q as the l value P is
greater than Q and the energy is give by ( n + l ) value. and Radial node = n – l – 1
582. The correct set of four quantum numbers for Where, n = Principle quantum no.
the outermost electron of sodium (Z = 11) is l = Azimuthal quantum no.
1 1 For 3p orbitals the value of l for p-orbital is 1
(a) 3, 1, 1, (b) 3, 2, 1,
2 2 ∴ No. of angular node (l) = 1
1 1 and No. of radial nodes = n – l –1 = 3 – 1 – 1 = 1
(c) 3, 0, 0, (d) 3, 1, 0,
2 2 587. The electrons, identified by quantum number n
Karnataka-CET-2012 and l,

Objective Chemistry Volume-I 221

 (I) n = 3; l = 2 n =5; l = 0 (III) n = 4; l =1 (IV) n n =1 l=0 m=0 s
= 4; l = 2 (V) n = 4; l = 0
can be placed in order of increasing energy, as n=2 l = 0,1 m = 0, ±1 s and p
(a) I < V < III < IV < II n=3 l = 0.1, 2 m = 0, ±1, ±2 s, pand d
(b) I < V < III < II < IV
592. Which of the following sets of quantum
(c) V < I < III < II < IV
numbers is impossible arrangement?
(d) V < I < II < IIII < IV
(e) V < I < IV < III < II 1
(a) n = 3, m = –2, s = +
Kerala-CEE-2009 2
Ans. (d) (i) n = 3, l = 2 ⇒ 3d 1
(ii) n = 5, l = 0 ⇒ 5s (b) n = 4, m = 3, s = +
2
(iii) n = 4, l = 1 ⇒ 4p
1
(iv) n = 4, l = 2 ⇒ 4d (c) n = 5, m = 2, s = –
(v) n = 4, l = 0 ⇒ 4s 2
According to Aufbau principle, An electron enters the 1
(d) n = 3, m = – 3 s = –
orbital with lowest energy in order to increasing 2
energies. Manipal-2018
4s < 3d < 4p < 5s < ud
Ans. (d) :
588. Which one of the following set of quantum
numbers is not possible for electron in the ground n = 1, 2, 3, …..
state of an atom with atomic number 19? l = 0, 1, 2, …. (n – 1) ex if
(a) n = 2, l = 0, m = 0 (b) n = 2, l = 1, m = 0 n = 3, l = 0, 1, 2
(c) n = 3, l = 1, m = -1 (d) n = 3, l = 2, m = +2 m = -l to l
(e) n = 4, l = 0, m = 0 −1
Kerala-CEE-2006 s= +1 ,
2 2
Ans. (d) : 19K = 1s2 2s2 2p6 3s2 3p64s1, l = 2 means
d – subshell which is not present in K. n ≠ m so the following configuration is not possible.
589. The correct set of quantum numbers (n, l and n = 3, m = -3, s = −1
2
m respectively) for the unpaired electron of
chlorine atom is: 593. For f-orbital the values of m are :
(a) 2, 1, 0 (b) 2, 1, 1 (a) –2, –1, 0, +1, +2
(c) 3, 1, 1 (d) 3, 2, 1 (b) –3, –2, –1, 0, +1, +2, +3
(e) 3, 2, –1 (c) –1, 0, +1
Kerala-CEE-2004 (d) 0, +1, +2, +3
Ans. (c) Cl (Z = 17) : 1s2 2s2 2p6 3s2 3p2x 3py2 3pz1 Manipal-2017
1s2 2s2 2p 2x 2p1y 2p1z Ans. (b) : For f-orbital , l = 3 thus m lies from -3 to +3
For the unpaired electron Hence m, = -3, -2, -1, 0, +1, +2 , +3
n = 3, l = 1, m = –1, 0 or + 1
594. What is the orbital angular momentum of an
590. The number of electrons with azimuthal electron in 'f' orbital?
quantum number l = 1 and l = 2 for Cr in
ground state respectively are 1.5h 6h
(a) (b)
(a) 16, 5 (b) 16, 4 π π
(c) 12, 4 (d) 16, 3
(e) 12, 5 3h 3h
(c) (d)
Kerala-CEE-2015 π 2π
Ans. (e) : Electrons configuration of MHT CET-2014
2 2 6 2 6 1 5
24Cr = 1s 2s sp 3s 3p 4s 3d Ans. (c) : Orbital angular momentum
Given,
h
azimuthal quantum number (l) = 1= p orbital mvr = l (l + 1) for f-orbital
azimuthal quantum number (l) = 2 = d orbital 2π
So, number of electrons in p– orbital l=3
⇒ 12 and number of electrons in d–orbital = 5. ∴Orbital angular momentum
591. The magnetic quantum number for d-orbital is
h 3h
given by : mvr = 3(3 + 1) =
(a) 2 (b) 0, ±1, ±2 2π π
(c) 0, 1, 2 (d) 5 3h
Manipal-2019 =
π
Ans. (b) :

Objective Chemistry Volume-I 222

595. The maximum number of electrons in a 600. Maximum number of electrons in a subshell of
subshell is given by the expression an atom is determined by the following
(a) 4l – 2 (b) 4l + 2 (a) 2l+1 (b) 4l –2
(c) 2l + 2 (d) 2n2 (c) 2n2 (d) 4l +2
NEET-1989 NEET-2009
Ans. (b) : For an azimuthal quantum number l there are Ans. (d) : For a given shell l,
2l + 1 orbitals in a subshell since, each orbital can the number of subshells. m1 = (2l + 1)
accommodate two electrons of opposite spin, the Since each subshell can accommodate 2 electrons of
maximum number of electrons in a subshell is given by opposite spin, so maximum number of electrons in a
2 × (2l + 1) = 4l + 2 subshell
596. The total number of electrons that can be ⇒ 2(2l + 1) = 4l + 2.
accommodated in all the orbitals having 601. Which of the following is not permissible
principal quantum number 2 and azimuthal arrangement of electrons in an atom?
quantum number 1 are (a) n = 5, l = 2, m = 0, s + 1/2
(a) 2 (b) 4 (b) n = 5, l = 3, m = 0, s + 1/2
(c) 6 (d) 8 (c) n = 3, l = 3, m = 0, s + 1/2
NEET-1990 (d) n = 4, l = 0, m = 0, s = -1/2
Ans. (c) : When n = 2 and l = 1 then subshell is 2p the NEET-2009
number of orbital’s in p-subshell. Ans. (c) : If n = 3
⇒ (2l + 1 ) = (2 × 1 + 1) l = 0 to (3 – 1) = 0, 1, 2
=3 m = - l to + l = -2, -1, 0 +1, +2
Total (maximum) number of electrons 1
2 × number of orbital’s S=±
2
⇒2×3 Is not a permissible set of quantum number
⇒6 602. The correct set of four quantum numbers for
(as each orbital contains 2 electrons) the valence electron of rubidium atom (Z= 37)
597. For azimuthal quantum number l = 3, the is
maximum number of electrons will be (a) 5, 1, 1, + 1/2 (b) 6, 0, 0 +1/2
(a) 2 (b) 6 1 1
(c) 0 (d) 14 (c) 5, 0, 0, + (d) 5, 1, 0, +
2 2
NEET-1991 NEET-2012
Ans. (d) : As we known azimuthal quantum number l Ans. (c) : Electronic configuration of rubidium
=3 denotes f – subshell and f-subshell contain seven
( z = 37 ) is [ Kr ] 5S1
36
orbitals. Each orbital can contain maximum of two
electrons with opposite spin. Thus f-subshell contain For valence electron (5s1)
total 14 electrons. 1
598. The following quantum numbers are possible n = 5, l = 0, m = 0, s =
2
for how many orbitals? 603. The angular momentum of electron in 'd'
n = 3, l = 2, m = +2 orbital is equal to
(a) 1 (b) 2
h
(c) 3 (d) 4 (a) 2 3 (b) 0 h
NEET-2001 2π
Ans. (a) : The three quantum number n, l, and m1 h h
(c) 6 (d) 2
enable us to label completely an orbital. It is one of the 2π 2π
d-orbital’s present in 3d subshell. NEET-2015, cancelled
599. If n = 6, the correct sequence for filling of
h
electrons will be Ans. (c) : Angular momentum = l ( l + 1)
(a) ns → (n-2)f → (n-1)d → np 2π
(b) ns → (n-1)d → (n-2)f → np For d orbital, l = 2
(c) ns → (n-2)f → np → (n–1)d h
Angular momentum = 2 ( 2 + 1)
(d) ns → np → (n–1)d → (n–2)f 2π
NEET-2011 h
= 6
Ans. (a) : For n = 6 electron are filled as 2π
6s → 4f → 5d → 6p 604. What is the maximum number of orbital's that
This is because electron first enters in that orbital’s for can be identified with the following quantum
which (n + 1) is lower. In case, if (n + 1) is same it goes numbers?
first in that orbital for which n is lowest. n= 3, l =1, m1 = 0
Objective Chemistry Volume-I 223
 (a) 1 (b) 2 Ans. (a) : Cr (Z = 24) : 1s2, 2s2, 2p6, 3s2, 3p2 4s1, 3d5
(c) 3 (d) 4
NEET-2014 ↑
For 19th electron
Ans.(a): Only one orbital 3pz has following set of 4s '
quantum number, n = 3, l = 1 and m1 = 0
1
605. What is the maximum numbers of electrons n = 4, l = 0, m = 0, s = +
that can be associated with the following set of 2
quantum numbers? 611. No two electron can have the same values of
n = 3, l = 1 and m = –1 …..quantum numbers.
(a) 4 (b) 2 (a) One (b) Two
(c) 10 (d) 8 (c) Three (d) Four
NEET-2013 UPTU/UPSEE-2004
Ans. (b) : The orbital associated with n = 3, l = 1 is 3p. Ans. (d) : According to pauli principle 2 elelctron does
One orbital (with m = –1 of Subshell Can accommodate not have the same value of all four quantum number.
maximum 2 electrons. They have maximum same value are 3.
606. Two electrons occupying the same orbital are 612. Which of the following electron has minimum
distinguished by
energy?
(a) azimuthal quantum number
(b) spin quantum number 1
(a) n=4, l=0. m = 0.s= +
(c) principal quantum number 2
(d) magnetic quantum number 1
NEET-I 2016 (b) n=4, l=1. m = + − 1.s= + 2
Ans. (b) : If two electrons in an atom are in the same
1
atomic orbital, then they must have the same n, l and m (c) n=5, l=0. m = 0.s= +
values. Thus in order to satisfy pauli’s exclusion 2
principal, they must have opposite spins. 1
607. How many electrons can fit in the orbital for (d) n=3, l=2. m = -2.s= +
2
which n = 3 and l = 1?
UPTU/UPSEE-2017
(a) 2 (b) 6
(c) 10 (d) 14 Ans. (a) : (i) Energy of electron depends on sum of n +
NEET-II 2016 1 values.
Ans. (a) : An orbital can always have a maximum no. (ii) For the same sum of n + 1, the orbital with lower
of two electrons only. value of n is filled first. Lower be the sum, lower the
energy of electron.
608. Which set of quantum numbers is not possible?
(iii) Electrons are filled in increasing order of energy.
1 1 Thus among the given options, n + 1 for :
(a) 3,2,–2, (b) 3,2,–3,
2 2 (a) 4 + 0 = 4 (b) 5 + 0 = 5
1 1 (c) 4 + 1 = 5 (d) 3 + 2 = 5
(c) 4,0,0 (d) 5,3,0,
2 2 4 + 0 = 4 is the
Tripura JEE-2022 613. Which of the following set of quantum
Ans. (b) : When l = n, m, cannot be -3 value of m1 numbers represents the highest energy of an
cannot numerically greater than the value of l. atom?
609. Maximum number of electrons in a subshell 1
with l = 3 and n = 4 is (a) n = 3, l = 2, m = 1, s = +
(a) 14 (b) 16 2
(c) 10 (d) 12 1
(b) n =4, l = 0, m=0, s = −
NEET-2012 2
Ans. (a) : l = 3 and n = 4 represents 4f so total number 1
of electrons in a subshell (c) n = 3, l = 0, m = 4, s = +
2 (2l + 1) 2
⇒2(2 × 3 + 1) = 14 electrons. Hence 1
(d) n = 3, l = 0, m =1, s = +
f-subshell can contain maximum 14 electrons. 2
610. The set of quantum number for 19th electron of UPTU/UPSEE-2016
chromium (Z = 24) is Ans. (a): n = 3 , l = 0 represents 3s – orbital
1 1 n = 3, l = 01 represents 3p – orbital
(a) 4, 0, 0, + (b) 4, 1, –1, +
2 2 n = 3, l = 2 represents 3d – orbital
1 1 n = 4, l = 0 represents 4s – orbital
(c) 3, 2, 2, + (d) 3, 2, –2, + The order of increasing energy of the orbitals is
2 2
UP CPMT-2011 3s < 3p < 4s < 3d

Objective Chemistry Volume-I 224

614. Consider the ground state of Cr atom (Z = 24).
The number of electrons with the azimuthal Ans. (a): n = 4, l =1 4P orbital
quantum number, l = 1 and 2 are respectively n = 4, l =0 4s orbital
(a) 12 and 4 (b) 12 and 5 n = 3, l = 2 3d orbital
(c) 16 and 4 (d) 16 and 5
UPTU/UPSEE-2012 n = 3, l = 1 3p orbital
Ans. (b) : Electronic configuration of 24Cr According to (n + l) rule,
1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s1
No. of electrons in l = 1(p-orbital) is increasing order of energy
6 + 6 = 12.
No of electrons in l = 2 (d-orbital) is 5 the number of
electrons with the azimuthal quantum number l = 1 and (iv) < (ii) < (iii) < (i)
2 respectively are 12 and 5
615. 'No two electrons in an atom can have the same
set of quantum numbers.' This principle is 618. Which of the following sets of quantum
known by which one of the following? number represents the 19th electron of Cr (Z =
(a) Zeeman's exclusion principle 24)?
(b) Stark's exclusion principle  1  1
(c) Pauli's exclusion principle (a)  4,1, −1, +  (b)  4,0,0, + 
(d) Hersbach's exclusion principle  2  2
UPTU/UPSEE-2011  1  1
(c)  3, 2,0, −  (d)  3, 2, −2, + 
Ans. (c) : the Pauli’s exclusion principle states that in  2  2
an atom or molecule, no two electrons can have the WB-JEE-2017
same set of all four electronic quantum numbers.
Cr = [ Ar ] 4s1 3d 5
18
As an orbital can contain a maximum no. of only two Ans. (b) : 24
electrons, the two electrons must have opposing spins,
this means if one electron is assigned as a spin up 19 th electron enters in 4s1 subshell
( )
+1 electron, the other electron must be spin-down
2
For 4s1
+1
( )
−1 electron.
2
n = 4, l = 0, m = 0, s =
2
When n = 5, l = 0, 1, 2, 3, or 4 and m = -4 to +4 n = 5, l 619. The difference between orbital angular
= 4, m = 0, s = +1 is a correct set of quantum number. momentum of an electron in a 4f-orbital and
2 another electron in a 4s-orbital is
616. Identity the correct statement. (a) 2 3 (b) 3 2
(a) Quantum numbers (n, l, m, s) are arbitratily
(b) All the quantum numbers (n, l, m, s) for any (c) 3 (d) 2
pair of electron in an atom can be identical WB-JEE-2020
under special circumstance Ans. (a) : Orbital angular momentum is given by the
(c) All the quantum numbers (n, l, m, s)may not formula,
be required to describe an electron of an atom
completely h
(l + 1) ×
(d) All the quantum numbers (n, l, m, s) are 2π
required to describe an electron of an atom ∴ For 4f, l = 3
completely)
h
WB-JEE-2013 → orbital angular momentum = 12 ×
Ans. (d) : All the quantum numbers (n, l, m, s) are 2π
required to describe an electron of an atom completely. For 4s, l = 0,
n describes the position and energy of the electron in an orbital angular momentum = 0
orbit or shell. l is used to describe subshell and the ∴ Difference = 12 − 0 = 12 = 2 3 .
space of the orbital occupied by the electron.
m describes the preferred orientation of orbital’s in
space. s describes the spinning of an electron on its axis. 8. Bohr's Model for Hydrogen
617. If the given four electronic configurations.
(i) n = 4, l = 1 (ii) n = 4, l = 0 Atom
(iii) n = 3, l = 2 (iv) n = 3, l = 1
are arranged in order of increasing energy, 620. If radius of second Bohr orbit of the He+ ion is
then the order will be 105.8 pm, what is the radius of third Bohr orbit
(a) (iv) < (ii) < (iii) < (i) of Li2+ ion?
(b) (ii) < (iv) < (i) < (iii) 0
(c) (i) < (iii) < (ii) < (iv) (a) 158.7 A (b) 158.7 pm
(d) (iii) < (i) < (iv) < (ii) (c) 15.87 pm (d) 1.587 pm
WB-JEE-2017 NEET-17.07.2022
Objective Chemistry Volume-I 225
 n2 Ans. (d) : Given that,
Ans. (b) : r = × a0
Z 13.6 z 2
+ En = − 2
eV atom–1
For He , n = 2 n
Z=2 For 3rd Oribit
rHe+ = 105.8 pm −13.6(z)2
E3 = –E = .....(i)
(2) 2 32
rHe+ = × a0
2 −13.6(z)2
E1 = .....(ii)
4 × a0 (1)2
105.8 =
2 Divided equation (i) and (ii) we get
105.8 × 2 −E −13.6 12
a0 = = 52.9 = ×
4 E1 32 −13.6
For Li2+, n = 3, Z = 3
E1 = –9E
32
rLi2+ = × a 0 624. If the radius of 1s electron orbit of a hydrogen
3 atom is 53 pm, then the radius of the 3p
= 3 × 52.9 = 158.7 pm electron orbit would be
621. If the radius of the 3rd Bohr’s orbit of hydrogen (a) 0.477 nm (b) 477 nm
atom is r3 and the radius of 4th Bohr’s orbit is (c) 159 pm (d) 17.66 nm
r4. Then: SCRA-2014
9 16 Ans. (a) : Given that,
(a) r4 = r3 (b) r4 = r3
16 9 n = 1, r1 = 53 pm, n = 3, r3 = ?
3 4 Now, from the radii expression-
(c) r4 = r3 (d) r4 = r3
4 3 52.9n 2
JEE Main-26.06.2022, Shift-I rn = pm
z
2
n Where, n = no. of orbit.
Ans. (b) : r =
z z = atomic number
For H– atom, we know = 1 52.9 × 9
∴ r3 = pm
So, r ∝ n2 1
r3 ∝ 9 r3 = 476.1 pm
r4 ∝ 16 or r3 = 0.476 nm
r3 9 625. The velocity of an electron in the first Bohr
From ratio = = orbit is v1. What is its velocity in the third
r4 16
Bohr's orbit?
16 (a) v1/9 (b) v1/3
r4 = r3
9 (c) v1 (d) 3v1
622. The hydrogen line spectrum provides evidence SCRA-2010
for the
(a) Heisenberg Uncertainty principle Ans. (b) : The velocity of an electron in the nth
(b) Wave like properties of light stationary orbit for H- like species is expressed as,
(c) Diatomic nature of hydrogen z
Vn = 2.18×106 m sec–1
(d) Quantized nature of atomic energy states n
SCRA-2012 Here, z = 1 and n= 3
–
Ans. (d) : The hydrogen like spectrum is an important So, Velocity in the third { Q velocity of e in first
piece of evidence to show the quantized electronic Bohr orbit = V1}
structure of an atom. v
Atomic orbits have fixed energies. Bohr’s orbit = 1
3
623. The energy of an electrons in the 3rd orbit of an
atom is –E. 626. The energy of the electron in the hydrogen
The energy of an electron in the first orbit will atom is given by the expression :
be
E −e 2 −n 2 h 2
(a) –3E (b) − (a) (b)
3 r2 2π2Z2 e 4 m
E −2 π 2 Z 2 e 4 nh
(c) − (d) –9E (c) 2 2
(d)
9 n h 2π
MPPET- 2009 AP-EAMCET-1991
Objective Chemistry Volume-I 226
Ans. (c) : The energy of the electron in the hydrogen hc
atom is given as : Also, E =
λ
–2π2 Z2 e 4 Emitting light of longest wavelength will have low
En = energy and if the principle quantum number increases,
n 2h 2 the difference between the energy of two orbits
627. The basic assumption of Bohr's model of decreases. Therefore, n = 4 to n=3 refers the lower
hydrogen atom is that : energy.
(a) the energy of the electron is quantised Thus that should be the increasing order of wavelength.
(b) the angular momentum of the electron is 630. The energy of an electron present in Bohr's
quantised second orbit of hydrogen atom is :
(c) the radial distance of the electron is quantised (a) – 1312 J atom–1 (b) – 328 kJ mol–1
(d) the orbital velocity of the electron is (c) – 328 J mol –1
(d) – 164 kJ mol–1
quantized
AP EAMCET (Engg.) 2001
AP-EAMCET-1994
Ans. (b) : For Bohr's second orbit
Ans. (b) : Bohr's Model–
(1) Electrons are revolving around nucleus K, L, M, N –1312
EA = kJ. mol –1
in specific energy level with centripetal force given by n2
nucleus. –1312
E2 = = –328 kJ / mol.
mu 2 ZKe 2 22
= 2
r r 631. Which one of the following statements is not
(2) Only those orbits are allowed whose angular correct?
h (a) Rydberg's constant and wave number have
momentum (p) is integral multiple of . same units
2π
(b) Lyman series of hydrogen spectrum occurs in
nh the ultraviolet region
p = mvr =
2π (c) The angular momentum of the electron in the
628. The radius of the second Bohr's orbit is : h
ground state of hydrogen atom is equal to
(a) 0.053 nm (b) 0.106 nm 2π
(c) 0.212 nm (d) 0.0265 nm (d) The radius of first Bohr orbit of hydrogen
AP-EAMCET-1995 atom is 2.116 × 10–8 cm
Ans. (c) : Bohr radius (rn) = ∈0n2h2 AP-EAMCET-2002
n 2h 2 Ans. (d) : The radius of first Bohr orbit of hydrogen
rn = atom is 0.53 Å or 0.53×10–8 cm. Hence, statement is
4π me 2 kZ
2
false.
1
k= • The unit of Rydberg constant as well as wave number
4π∈0 (v ) is cm–1. Hence, statement is true.
n 2 h 2 ∈0 a • The Lyman series of hydrogen spectrum occurs at the
∴ rn = = n2 0
πme 2 Z Z ultra violet region. Hence, the given statement is true.
Where, m = mass of electron • According to Bohr's theory, Angular momentum,
e = charge of electron nh
h = Planck's constant mvr = .
2π
k = Coulomb constant
h
n 2 × 0.53 So, in ground state, angular momentum = .
rn = Å 2π
Z
632. An electron is moving in Bohr's fourth orbit.
Radius of nth Bohr orbit for H-atom
Its de-Broglie wavelength is λ. What is the
= 0.53 n2 Å [Z = 1 for H-atom] circumference of the fourth orbit?
∴ Radius of 2 Bohr orbit for H-atom
nd
2
= 0.53 × (2)2 = 2.12 Å (a) (b) 2λ
= 0.212 nm [1Å = 0.1 nm] λ
629. In the Bohr hydrogen atom, the electronic 4
(c) 4λ (d)
transition emiting light of longest wavelength is: λ
(a) n = 2 to n = 3 (b) n = 4 to n = 3 VITEEE-2014
(c) n = 3 to n = 2 (d) n = 2 to n = 1 Ans. (c) : According to Bohr's concept, an electron
AP-EAMCET-1997 always move in the orbit with angular momentum (mvr)
 1 1  nh
Ans. (b) : We know that ∆E = R H  2
− 2 equal to .
 n1 n 2  2π
Thus the decreasing order of energy transition is– nh
∴ mvr =
n=2 to n=3 > n=2 to n=1 > n=3 to n=2 > n=4 to n=3 2π

Objective Chemistry Volume-I 227

 n  h  putting the vale in eqn (1) we get
r= ⋅  h
2π  mv  λ = 4π.∆n
h
 h 
Q λ = de - Broglie equation  λ = 4π∆n
 mv  o o

nλ λ = 4π × 0.529 A (∴ given ∆n = 0.529 A )

r= o
2π λ = 2.116 A
For 4th orbit– 636. If the wavelength of the first line of Balmer
4λ 2λ series is 656 nm, then the wavelengths of its
r= = second line and limiting line respectively are ––
2π π –––––
2λ (a) 485.9 nm & 434 nm
∴ Circumference = 2πr = 2π×
π (b) 485.9 nm & 364.4 nm
(c) 715 nm & 434 nm
= 4λ
(d) 608 nm & 415.2 nm
633. The energy (in ev) associated with the electron AP EAPCET 25.08.2021, Shift-II
in the 1st orbit of Li2+ is
Ans. (b) : Given that λ1 = 656 mm and n1 = n2 and n2 =
(a) – 122.4 (b) – 61.15 3.
(c) – 30.5 (d) – 244.6 We know that or fist wavelength
TS-EAMCET (Engg.), 07.08.2021 Shift-II  1
1 1 
Ans. (a) : Given that, v = = RH  2 − 2 
λ1  n1 n 2 
n = 1, Z = 3 (for Li3+)
Z2 1  1 1 
Q Energy of electron ( E n ) = −13.6 2 ev = RH  2 − 2 
n 656 2 3 
st
For 1 orbit– 1 1 1
= RH  − 
(3) 2 656 nm 4 9
E1 = −13.6 2 ev 1 5
(1) = RH ––––––– (1)
or E1 = –122.4 ev 656 nm 36
634. Which energy level transition among the For second line, n1 = 2, n2 = 4
following will have the least wavelength? 1  1 1 
= RH  2 − 2 
(a) n4 → n3 (b) n4 → n2 λ2 2 4 
(c) n4 → n1 (d) n2 → n1 1 1 1 
AP EAPCET 25.08.2021, Shift-II = RH  − 
λ2  4 16 
Ans. (c) : n4 → n1 transition will have the least 1 3
wavelength. The option (c) is correct. Energy is = RH –––––– (2)
inversely proportional to wavelength. For an electron to λ 2 16
undergo electronic transition from higher to lower Dividing equation (1) and (2) we get
energy level wavelength is least. λ2 5 16
= ×
635. Calculate the de Broglie's wavelength of an 656 36 3
electron residing in the 2nd Bohr's orbit of a λ 2 = 485.9 nm
Hydrogen atom. (Bohr's radius, a0 = 0.529Å)
Wavelength of limiting line –
(a) 0.2116 nm (b) 2.116 πÅ n1 = 2, n2 = ∞
(c) 21.16 m (d) 20116 µm So,
AP EAPCET 25.08.2021, Shift-II 1  1 1  1
= RH  2 − 2  = RH ×
Ans. (b) : λ 2 ∞  4
h h Dividing (1) and (2) we get
λ= = ––––––– (1)
p mv λ 5
= ×4
h 656 36
∴ ∆V ∆n = 20
4πm λ= × 656 nm = 364.4 nm
36
h
m∆V = –––––––– (2) λ nm = 364.4 nm
4π∆n

Objective Chemistry Volume-I 228

637. Which of the following corresponds to the n = no of orbit
energy of the possible excited state of 2 = atomic number
hydrogen? Given- n = 3
(a) –13.6 eV (b) 13.6 eV
52.9 ( 3)
2
(c) –3.4 eV (d) 3.4 eV ∴ R3 = pm
TS EAMCET-2017 2
BCECE-2012 9 × 52.9
r3 = pm …….(i)
NEET-2002 2
st
Ans. (c) : Energy level of an atoms are For the 1 orbit i.e. n = 1
2
E n = −13.6Z / n eV 2
52.9 × (1) 2
r1 = pm
Where Z = atomic number 2
Energy level (n) = 1, 2, 3, ….. 52.9
For hydrogen atom, Z = 1 r1 = pm …….(ii)
2
And first excited state, n = 2 From (i) and (ii)
−13.6 ×12 r3 = 9r1
∴E = Hence, the radius of the 3rd orbit is 9 times the radius of
22 st
1 orbit
E = –3.4 eV
640. Energy associated with the first orbit of He+ is
638. The electron in the hydrogen jump on
(a) 8.72×10–18 joules (b) 0.872 ×10–18 joules
absorbing 12.75 eV of energy would jump to
____orbit (c) –0.872 ×10 joules (d) –8.72×10–18 joules
–18

(a) 3 (b) 2 COMEDK-2015

(c) 5 (d) 4 AMU-2015
AP EAPCET 24.08.2021 Shift-II Ans. (d) : For H-like particles,
Ans. (d) : We know that –21.78 × 10–19 2
En = ZJ
−19 n2
−21.8 × 10 +
En = Joule For He , Z = 2, n = 1
2
n –21.78 × 10 –19
−19 = × (2) 2 = –8.712 × 10 –18 J
−21.8 × 10 1 2

E1 = 2 641. Assuming Rydberg constants are equal, the

1 ground state energy of the electron in hydrogen
E1 = – 21.8×10-19J (n = 1 for ground state at H-atom) atom is equal to
After absorbing 12.75eV of energy the energy of (a) the ground state energy of the electron in He+
electron will be -
(b) the first excited state energy of the electronic
– 21.8×10-19 + 12.75×1.6×10-19 He+
– 21.8×10-19 + 20.4×10-19 (c) the first excited state energy of the electron in
– 1.4×10-19 J Li2+
−19 (d) the ground state energy of the electron in Be3+
−21.8 × 10 -19
Thus 2
= – 1.4 × 10 COMEDK-2020
n 2
Z
21.8 × 10
−19 Ans. (b) : E n = –R H . 2
2 n
n = –19
1.4 × 10 For ground state hydrogen atom : Z=1, n=1
n2 = 15.57 1
∴E1 = –13.6 × = –13.6eV
n=4 1
Thus the e- will jump to the 4th orbit. For the ground state in He+; Z=2, n=1
639. On the basis of Bohr’s model. The radius of the E1= –13.6 × 4 = –54.4eV
3rd orbit is––– 1
(a) Equal to the radius of 1st orbit For the first excited state in He+; Z=2, n=2;
(b) 3times the radius of 1st orbit 4
(c) 5 times the radius of 1st orbit E 2 = –13.6 × = –13.6eV
4
(d) 9 times the radius of 1st orbit For the first excited state in Li2+; Z=3, n=2;
AP EAPCET 19-08-2021 Shift-I 9
E1 = –13.6 × = –30.6eV
Ans. (d) : From the equation of Bohr’s model- 4
52.9(n 2 ) For the ground state energy of the electron in
rn = pm Be3+; Z=4, n=1;
2
th
Where - rn = radius of n orbital = –13.6 × 16 = –217.6eV

Objective Chemistry Volume-I 229

642. Match the following 644. The ratio of potential energy (PE) and total
List-I List-II energy of an electron in a Bohr orbit of the
hydrogen atom is
nh (i) Paschen series (a) 1 (b) 2
(A) mvr =
2π 1
(B) Infra-red (ii) Electron total energy (c) –1 (d)
2
h (iii) de-Broglie equation TS-EAMCET 09.08.2021, Shift-I
(C) λ =
p Ke2
Ans. (b) : Potential energy (P.E.) = − .....(i)
–e 2 (iv) Schrodinger equation rn
(D)
2π 1Ke 2
And, Total energy (T.E.) = − .....(ii)
(v) Bohr’s equation 2rn
The correct answer is P.E. −Ke 2 / rn
∴ =
A B C D T.E. −1Ke 2 / 2rn
(a) (v) (ii) (iii) (i)
P.E.
(b) (iii) (ii) (v) (iv) or =2
(c) (v) (i) (iii) (ii) T.E.
(d) (iv) (i) (ii) (iii) ∴ P.E. : T.E. = 2 : 1
AP-EAMCET (Medical), 2008 645. Electromagnetic radiation of wavelength 663
nm is just sufficient to ionise the atom of metal
Ans. (c) : A. The ionization energy of metal A in kJ mol–1
List-I List-II is . (Rounded-off to the nearest integer)
nh Bohr's equation [h = 6.63×10–34 Js, c = 3.00×108 ms–1, NA =
(A) mvr = 6.02×1023 mol–1]
2π
JEE Main 25.02.2021, Shift-II
(B) Infra-red Paschen series
Ans. : Given, wavelength (λ) = 663 nm {1nm = 10–9m}
de-Broglie equation = 663×10–9m
h
(C) λ = hc
p We know that, E =
λ
–e 2 Electron total energy For 1 mole of atom, total energy is given by,
(D)
2π hc
⇒ E = NA ×
643. The wavelength (in Å) of an emission line λ
obtained for Li2+ during an electronic 6.023 ×1023 × 6.63 × 10−34 × 3 ×108
transition from n2 = 2 to n1 = 1 is (R = Rydberg ⇒E=
constant) 663 × 10−9 × 1000
–19
= 3×10 J
3R 3R Ionisation energy per mol
(a) (b)
4 4 = 3 × 10–19 ×10–3 × 6.02 × 1023
4 4 = 180.6 KJ mol–1
(c) (d)
3R 27R = 181 kJ mol–1
AP-EAMCET (Medical), 2008 646. Assertion: A spectral line will be seen for a 2Px
→2py transition.
Ans. (d) : Given data: Reason: Energy is released in the form of waves
n1 = 1 of light when the electron drops from 2px to 2py
n2 = 2 orbital.
Z = 3 (for Li2+) (a) If both Assertion and Reason are true and the
λ =? Reason is a correct explanation of the
Now, from Rydberg equation- Assertion
(b) If both Assertion and Reason are true but
 1 1  Reason is not a correct explanation of the
υ = RZ2  2 – 2 
 n1 n2  Assertion
(c) If Assertion is true but the Reason is false
1 2 1 1
Or = R ( 3)  –  (d) If both Assertion and Reason are false
λ 1 4  AIIMS-1996
1 3 Ans. (d): In this case both assertion and reason are
= 9R × false. Both 2px and 2py orbitals have equal energy (2p
λ 4
orbitals are degenerate), there is no possibility of
4 electron transition and hence, no energy is released and
Or λ=
27R thus, no spectral line will be observed.
Objective Chemistry Volume-I 230
647. The maximum energy is possessed by an 651. In hydrogen atomic spectrum, a series limit is
electron, when it is present found at 12186.3 cm−1. Then, it belongs to
(a) in first excited state (a) Lyman Series (b) Balmer series
(b) in nucleus (c) Paschen series (d) Brackett series
(c) at infinite distance from the nucleus AIIMS-2014
(d) in ground energy state Ans. (c): Series limit is the last line of the series
AIIMS-1996 i.e. n2 = ∞
Ans. (c): When electron is placed at an infinite distance
from the nucleus the energy increases sharply. 1  1 1   1 1 
∴ v= =R 2 − 2  =R 2 − 2
The potential energy of the electron is minimum at λ  n1 n 2   n1 ∞ 
equilibrium distance from the nucleus. R
648. Assertion: Angular momentum of an electron v= 2
n1
in any orbit is given by angular momentum
nh 109677.76
= , where n is the principal quantum 12186.3 =
2π n12
number 109677.76
Reason: The principal quantum number n can n12 = = 9, n1 = 3
12186.3
have any integral values. The line belongs to Paschen series.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of  Z2 
Assertion 652. Based on equation, E = -2.178 ×10-18 J  2  ,
n 
(b) If both Assertion and Reason are correct, but certain conclusions are written. Which of them
Reason is not the correct explanation of
is not correct?
Assertion.
(a) Larger the value of n, the larger is the orbit
(c) If Assertion is correct but Reason is incorrect. radius.
(d) If both the Assertion and Reason are
(b) Equation can be used to calculate the change
incorrect.
in energy when the electron changes orbit.
AIIMS-2012-13
(c) For n=1, the electron has a more negative
Ans. (b): Both assertion and reason are correct. Reason energy than it does for n=6 which means that
is not the correct explanation of assertion. the electron is more loosely bound in the
649. What is the energy (kJ/mol) associated with the smallest allowed orbit.
de-excitation of an electron from n=6 to n=2 in (d) The negative sign in equation simply means
He+ ion? that the energy of electron bound to the
(a) 1.36×106 (b) 1.36×103 nucleus is lower than it would be if the
3
(c) 1.16×10 (d) 1.78×103 electrons were at the infinite distance from
AIIMS-27 May, 2018 the nucleus.
Ans. (c): Given that, AIIMS-2015
n = 6 to n = 2 Ans. (c): For n=1, the electron has a more negative
energy than it does for n=6 which means that the
E = 13.6 × Z12  − 
1 1 electron is more loosely bound in the smallest allowed
 4 36  orbit.
 9 −1  653. The energy of electron in first energy level is
= 13.6 × 4  
 36  -21.79 ×10-12 erg per atom. The energy of
8 electron in second energy level is:
= 13.6 × 4 × (a) −54.47 × 10 −12 erg atom −1
36
= 12.08 eV × 96 kJ/mol (b) −5.447 × 10 −12 erg atom −1
= 1.16 × 103 KJ/mol (c) −0.447 × 10 −12 erg atom −1
650. In Second orbit of H atom the velocity of e− is:
(d) −0.05447 × 10−12 erg atom −1
(a) 2.18×106 m/sec (b) 3.27×106 m/sec
5
(c) 10.9×10 m/sec (d) 21.8×106 m/sec AIIMS-2000
AIIMS-27 May, 2018 Ans. (b): Assume that atom to be hydrogen like,
AIIMS-2001 Energy of nth energy level
z −E
Ans. (c): V = 2.18 × 106 × E n = 2 1 , where E1 is energy of first energy level
n x
1 −E −E −21.79 × 10−12
V = 2.18 × 106 × = 10.9 × 105 m/sec. E2 = 21 = 1 =
2 2 4 4
Hence, V = 10.9 × 105 m / sec = −5.447 × 10 −12 erg per atom

Objective Chemistry Volume-I 231

654. If velocity of an electron in the first Bohr orbit Therefore, the angular momentum is an integral
of H is v1 then velocity in second orbit will be h
(a) v1 (b) 2v1 multiple of .
2π
v1 v Hence option (a) is not permitted for angular
(c) (d) 1
2 4 momentum of hydrogen atom.
SRMJEEE – 2007
657. In a hydrogen atom, the electron is at a
Ans. (c) : The velocity of an electron of the nth distance of 4.768 Å from the nucleus. The
stationary state for hydrogen like species is expressed
as– angular momentum of the electron is
z 3h h
Vn = 2.18 × 106 × m / s (a) (b)
n 2π 2π
Where z = atomic number and n = orbit h 3h
(c) (d)
Now, V1 = 2.18 × 106 ×
(1) m / s π π
(1) AP- EAMCET(Medical) -2010
The velocity in second orbit is– Ans. (a): Given data:-
(1) rn = 4.768Å
V2 = 2.18 × 106 ×
(2) As we know the radius of first Bohr's orbit is –
V1 n2
or V2 = rn = 0.529 Å
2 Z
Z 4.768 = 0.529 × n 2
V∝ since, Z and n both have become twice,
n 4.768
velocity of electron in second orbit of H+ will be v. n2 =
0.529
655. If the energy of an electron in the second Bohr
orbit of H-atom is –E, what is the energy of the or n2 = 9
electron in the Bohr's first orbit? n=3
(a) 2E (b) –4E ∴ Angular momentum of the electron is
(c) –2E (d) 4E
nh 3h
SRMJEEE – 2010 (mvr) = =
Ans. (b) : The expression of energy of hydrogen atom 2π 2 π
is : 658. In the Bohr hydrogen atom, the electronic
transition emitting light of longest wave length
 z2 
En = –2.18 × 10–18  2  J is:
n  (a) n = 2 to n =3 (b) n = 4 to n = 3
Given– E2 = – E (c) n = 3 to n =2 (d) n = 2 to n = 1
E1 = ? AP – EAMCET - (Medical)-1997
(1) 2 Ans. (b) : The spectrum of hydrogen atom obey the
∴ E2 = – 2.18 × 10–18 × J=−E ...(i)
(2) 2 following formula–
(1) 2  1 1 
and E1 = – 2.18 × 10–18 × J (from equation i) ν = RH  2 − 2 
(1) 2  n1 n 2 
or E1 = – 4E
Where RH = Rydberg constant
656. According to Bohr's theory which one of the
following values of angular momentum of
hydrogen atom is not permitted? In the Paschen series of hydrogen atom spectrum, the
1.25h h minimum energy difference found in n =4 to n = 3 due
(a) (b) to which they have longest wavelength.
π π
659. What is the lowest energy of the spectral line
1.5h 0.5h
(c) (d) emitted by the hydrogen atom in the Lyman
π π series?
AP-EAMCET- (Engg.)-2011 (h = Planck's constant, c = Velocity of light,
Ans. (a): According to Bohr's theory. R=Rydberg's constant).
nh 5hcR 4hcR
mvr = (a) (b)
2π 36 3
Where, m = mass of atom 3hcR 7hcR
(c) (d)
v = velocity 4 144
r = radius of nth orbit AP-EAMCET (Engg.)-2005
Objective Chemistry Volume-I 232
Ans. (c): Given data:- For the Lyman series- 1 1 4 −1 3 3
n1 = 1, n2 = 2  2 − 2  = 4× = 4× =
From the Rydberg equation-  n1 n 2  16 16 4
for first line n2 = 2, n1 = 1
 
ν = R 1 − 1   1 1  1 1  3
 n2 n2  12 − 22  = 1 − 4  = 4
 1 2     
where- ν = wave number Hence n1 = 1, n2 = 2
R = Rydberg constant So, transition n2 = 2 to n1 = 1 will give spectrum of the
same wavelength as that of Balmer transition n2 = 4 to
1 1  n1 = 2 in He+.
ν = R − 
1 4  662. What is the degeneracy of the level of H-atom
3R  R 
or ν= that has energy  − H  ?
4  9 
1 (a) 16 (b) 9
Q λ=
υ (c) 4 (d) 1
4 VITEEE 2013
∴ λ= Ans. (b): Energy of electron form nth terms is
3R
Now, R Z2
En = − H2
hc n
E=
λ For H–atom,
hc × 3R −R H −R H × 12
E= = , n2 = 9
4 9 n2
3hcR 663. The ground state energy of hydrogen atom is –
or E= 13.6 eV. The energy of second excited state of
4
For lowest energy of the spectral line in Lyman series n1 He+ ion in eV is
= 1, n2 = 2. (a) –54.4 (b) –3.4
(c) –6.04 (d) –27.2
660. According to Bohr’s theory, the angular
[JEE Main 2019, 10 Jan Shift-II]
momentum for an electron of 3rd orbit is
(a) 3 h (b) 1.5 h Ans. (c) : Given that,
energy of hydrogen atom = 13.6 eV.
(c) 9 h (d) 2
h Second excited state is 3rd orbit–
π z2 4
VITEEE 2014 E = –13.6 2 eV = −13.6 × eV
n 9
Ans. (a) : According to Bohr's theory, the angular = –6.04 eV.
momentum of e– in nth orbit is equal to
664. The energy of electron in the first Bohr orbit of
nh H atom is - 13.6 eV. The possible energy value
mvr =
2π of the excited state (s) for electrons in Bohr
For 3rd orbit, orbits of hydrogen is
3h (a) -3.4 eV (b) -4.2 eV
mvr = (c) -6.8 eV (d) +6.8 eV
2π AMU-2011
 h 
So, mvr = 3.h ⇒ Q h =  Ans. (a): The energy of electron on Bohr orbits of
 2 π hydrogen atoms is given by the expression
661. What transition in the hydrogen spectrum −13.6 × Z2
would have the same wavelength as the Balmer En = eV
n2
transition, n = 4 to n = 2 of He+ spectrum? For first excited state, n = 2
(a) n = 4 to n = 2 (b) n = 3 to n = 2
(c) n = 2 to n = 1 (d) n = 4 to n = 3 −13.6 × (1) 2 −13.6
∴ E2 = = = −3.4eV
VITEEE-2013 22 4
Ans. (c): According to the question 665. The energy of electron in nth orbit of hydrogen
atom is
λ H = λ He+
13.6 13.6
(a) eV (b) eV
1 1 1 1 n n2
R H ZH2  2 − 2  = R H Z2He+  2 − 2 
n
 1 n 2 2 4  13.6 13.6
(c) 3
eV (d) eV
n n4
1 1 1 1 
R H ×1  2 − 2  = R H × 4  −  AMU EXPLORER-2002
 n1 n 2   4 16  Karnataka-CET-2016

Objective Chemistry Volume-I 233

Ans. (b) : The energy of electron in nth orbit of −13.6z 2
hydrogen atom is given below- Ans. (d) : For H atom , E X = eV
n2
13.6 For second orbit, n = 2
or En = 2 eV
n z = 1 (for hydrogen)
666. The spectrum of H+ is expected to be similar to −13.6 × (1) 2 13.6
that of ∴ E2 = =− eV
(a) Be2+ (b) Li+ (2) 2 4
(c) Na (d) He+ −13.6 ×1.6 ×10−19
AMU EXPLORER-2002 = J
4
Ans. (d) : Helium ion (He+) and hydrogen has one −19
= −5.44 × 10 J
electron in their outermost shell so both show the same
671. The wave number of the spectral line in the
spectrum having similar spectral lines of transitions.
emission spectrum of hydrogen will be equal to
667. For which of the following species, Bohr’s 8/9 times the Rydberg's constant if the electron
theory is not applicable? jumps from–
(a) 07+ (b) Be3+ (a) n = 3 to n = 1 (b) n = 10 to n = 1
2+
(c) Li (d) He2+ (c) n = 9 to n = 1 (d) n = 2 to n = 1
Assam CEE-2014 BCECE-2014
Ans. (d) : Bohr's theory is applicable to the species
1  1 1 
containing only one electron. He2+ contain no electron. Ans. (a) : wave number, v = = RH  2 − 2  ,
668. The ratio of energy of the electron in ground λ n
 1 n 2 
state of hydrogen to the electron in first excited 8  1 1 
state of Be3+ is where RH is Rydberg’s constant R H = R H  2 − 2 
(a) 1 : 4 (b) 1 : 8 9 n
 1 n 2 

(c) 1 : 16 (d) 16 : 1 When, transition is from n2 = 3 to n1 = 1 then,

Assam CEE-2014 1 1 1 8
v = = R H  2 − 2  = . RH
 z2  λ 1 3  9
Ans. (a) : E x = −13.6  2  eV
n  672. In hydrogen atom, if energy of an electron in
ground state is 13.6 eV, then that in the 2nd
E1 (H) = -13.6 eV in ground state n = 1, for H atom z = excited state is –
1 (a) –1.51 eV (b) –34 eV
−13.6 × ( 4 ) 
2 (c) –6.04 eV (d) +13.6 eV
3+
E 2 (Be ) =  in first excited state BCECE-2018
( 2)
2
 Ans. (a) : 2nd excited state means third energy level
3+
n = 2 for Be , z = 4 E 13.6
E 3 = 21 = − = −1.51eV
∴ E1 : E 2 = 1: 4 . 3 9
669. The Bohr's orbit radius for the hydrogen atom 673. The wave number of the limiting line in Lyman
(n = 1) is approximately 0.53 Å. The radius for series of hydrogen is 109678 cm–1. The wave
the first excited state (n = 2) orbit is : number of the limiting line in Balmer series of
(a) 0.27 Å (b) 1.27 Å He+ would be :
(a) 54839 cm–1 (b) 109678 cm–1
(c) 2.12 Å (d) 3.12 Å –1
(c) 219356 cm (d) 438712 cm–1
BITSAT-2013
BITSAT-2014
BCECE-2004
th Ans. (b) : Given that,
Ans. (c) : Radius of n orbit hydrogen atom = 0.530Å
RH = 109678 cm–1
Number of excited state = (n) = 2
Wave number of the limiting line in Balmer series of
Atomic number of hydrogen atom (z) = 1 He+
We know that the Bohr radius
1 1
n2 (2)2 v = R H .Z2  2 − 2 
r = × Radiusof atom = × 0.530
z 1  n1 n 2 
= 4 × 0.530  1 1
= 2.12 Å v = 109678 × (2)2  2 − 
 (2) ∞
670. The energy of an electron in second Bohr orbit
of hydrogen atom is– v = 109678 cm–1
(a) –5.44 × 10–19 eV (b) –5.44 × 10–19 cal 674. If the radius of H is 0.53 Å, then what will be
(c) –5.44 × 10 kJ
–19
(d) –5.44 × 10–19 J the radius of 3 Li 2+ ?
AIIMS 26 May 2019 (Evening) (a) 0.17 Å (b) 0.36 Å
BITSAT 2017 (c) 0.53 Å (d) 0.59 Å
BCECE-2010 BITSAT-2012
Objective Chemistry Volume-I 234
Ans. (a) : We know that, n2
r (H - atom) × n 2 Ans. (d) : Bohr radius for nth orbit = 0.53A ×
(H-like) = n z
Z Where, z = atomic number
For ground state, n = 1
∴ Bohr radius of 2nd orbit of Be3+ =
0.53 × 1
∴ rn ( 3 Li 2+ ) = 0.53 × (2) 2 0
3 = 0.53A
= 0.17Å 4
675. The first emission line in the atomic spectrum 0.53 × (1) 2 0

of hydrogen in the Balmer series appears at (d) Bohr radius of 1st orbit of H = = 0.53A
1
9R 7R orbit of Be3+ is equal to
(a) cm −1 (b) cm −1 Hence, Bohr’s radius of 2nd
400 144 that of first orbit of hydrogen.
3R −1 5R −1
(c) cm (d) cm 679. The IE of hydrogen atom is 13.6 eV. The
4 36 energy required to remove a electron in the
BITSAT-2016 n=2 state of the hydrogen atom is
Ans. (d) : For Balmer n1 = 2 and n2 = 3; (a) 27.2eV (b) 13.26eV
 1 1  5R −1 (c) 6.8eV (d) 3.4eV
v = R 2 − 2  = cm
 2 3  36 J & K CET-2015
The first emission line in the atomic spectrum of CG PET-2010
5R −1 Ans. (d) : The ionization energy of hydrogen atom is
hydrogen in the Balmer series appears at cm .
36 13.6 eV energy required to move to second state is,
676. The wavelength of an electron having kinetic The change in energy involve in removing the electron
energy equal to 4.55×10−25J is (h=6.6 from n = 2
×10−34kgm2s−1mass of electron =9.1×10−31kg) ∆E = E∞ – E(2)
(a) 7.25×10−7nm (b) 725m
(c) 7.25×10−7m (d) 7.25×107m  −13.6 
∆E = 0 –  2 
CG PET-2011  2 
Ans. (c) : Applying de Broglie wavelength +13.6
∆E =
h 4
(λ ) =
2me ∆E = 3.4 eV
Where, m = mass of electron = 9.1 × 10-31 kg, h = 6.6 × 680. Which one of the following about an electron
10–34 kg m2 s–1 occupying the 1s-orbital in a hydrogen atoms is
E = K.E. =4.55 × 10-25 J incorrect? (The Bohr radius is represented by
6.6 × 10−34 a 0)
λ=
2 × 9.1× 10−31 × 4.55 ×10−25 (a) The electron can be found at a distance 2a0
from the nucleus.
6.6 × 10−34
= = 7.25 × 10−7 m. (b) The magnitude of the potential energy is
9.1×10−28 double that of its kinetic energy on an
677. For which of the following species, Bohr theory average.
does not apply?
(a) H (b) He+ (c) The probability density of finding the electron
(c) H +
(d) Li2+ is maximum at the nucleus.
(d) The total energy of the electron is maximum
CG PET -2005, AIIMS-2000
when it is at a distance a0 from the nucleus.
Ans. (c) : One of the limitations of Bohr’s atomic
model is that it does not explain the spectra of multi- [JEE Main 2019, 9 April Shift-II]
electron atoms all these species like H, He+ and Li2+ are Ans. (d) : Explanation of point a :
iso electronic and have only one electron. Their
electronic configurations are same and so their spectra
is explained by Bohr’s atomic model But H- has 2
electron.
678. Bohr's radius of 2nd orbit of Be3+ is equal to
that of
(a) 4th orbit of hydrogen
(b) 2nd orbit of He+
(c) 3rd orbit of Li2+
(d) first orbit of hydrogen
CG PET -2009
Objective Chemistry Volume-I 235
 683. The radius of the second Bohr orbit in terms of
the Bohr radius, a0, in Li2+ is
2a 0 4a 0
(a) (b)
3 3
4a 0 2a 0
(c) (d)
9 9
[JEE Main 2020, 8 Jan Shift-II]
Ans. (b) : For 2nd Bohr orbit of Li2+,
z = 3, n = 2
The radial propability graph shows that electron in is a 0 .n 2
can be formed at 2a0 r=
z
Explanation of point b, c, d
En = –13.6 a 0 .(2)2
=
z2 v 3
E n = –13.6 × e ............(i)
n2 4a 0
=
2
z v 3
K. E. = 13.6 × e ............(ii) 684. According to Bohr's atomic theory,
n2
z2 Z2
P. E. = – 27.2 × e v ............(iii) I. kinetic energy of electron is ∝
n2
n2
From (ii) and (iii) equalion, |PEl = l K. E.| and P. E. = II. the product of velocity (v) of electron and
2En principal quantum number (n), 'vn' ∝Z2.
From eq. (i), if n increases; en becomes less negative III. frequency of revolution of electron in an orbit
and vice versa. Z3
So, if n = 1 is ∝ 3
n
i.e. when e– is at distance a0 from nucleolus En has
maximum value. IV. coulombic force of attraction on the electron is
681. The radius of the second Bohr orbit for Z3
∝ 4
hydrogen atom is (Planck's constant (h) = n
6.6262×10–34 Js; mass of electron = 9.1091×10–31 Choose the most appropriate answer from the
kg; charge of electron (e) = 1.60210×10–19 C; options given below.
permittivity of vacuum (∈0) = 8.854185×10–12
kg–1m–3A2) (a) Only III (b) Only I
(a) 1.65Å (b) 4.76Å (c) I, III and IV (d) I and IV
(c) 0.529Å (d) 2.12Å [JEE Main 2021, 24 Feb Shift-II]
[JEE Main 2017] Ans. (d) : According to Bohr’s theory :
Ans. (d) : Radius of nth orbit is– z 2 eV z2
0.529n 2 (A) K.E. = 13.6 ⇒ K.E. ∝
rn = Å n 2 atom n2
Z
z
n = 2, Z = 1 (B) speed of e – ∝
r2 = 0.53 × 4 Å = 2.12 Å n
∴ v× n ∝ z
682. Which of the following is the energy of a
possible excited state of hydrogen? v
(C) Frequency of revolution of e– =
(a) +13.6 eV (b) –6.8 eV 2πr
(c) –3.4 eV (d) +6.8 eV
[JEE Main-2015] z2
∴ Frequency ∝
Ans. (c) : E1 for H - atom = 13.6 eV n3
13.6 eV z
E2 for H - atom = – = – 3.4 eV (D) F ∝
22  n 2 2
 
13.6  z 
E3 for H - atom = – 2 eV = –1.5 eV  
3
Hence, energy of possible excited state of hydrogen is z3
F∝
–3.4 eV. n4

Objective Chemistry Volume-I 236

685. The kinetic energy of an electron in the second Ans. (c) : According to Bohr's model of an atom,
Bohr orbit of a hydrogen atom is equal to Z
h2 V∝
. The value of 10× x is ..........(a0 is radius n
xma02 Z = atomic number of atom, corresponds to the +ve
of Bohr's orbit) charge so as Z increases velocity increases so statement
I is worng.
(Nearest integer) (Given, π = 3.14)
and as ‘n’ decreases velocity increases so, statement-II
[JEE Main 2021, 27 Aug Shift-I] is correct.
We know that, 687. Given below are two statements.
1 Statement-I Bohr's theory accounts for the
K.E. = mv 2 .....(i)
2 stability and line spectrum of Li+ ion.
nh Statement-II Bohr's theory was unable to
mvr = → (Bohr 's model) explain the splitting of spectral lines in the
2π
presence of a magnetic field. In the light of the
n 2h 2 above statements, choose the most appropriate
(mv) 2 =
4π 2 r 2 answer from the options given below:
1 n 2h 2 (a) Both statement I and II are true.
mv 2 = × 2 2 .....(ii) (b) Statement I is false but statement II is true.
m 4π r
(c) Both statements I and II are false.
Put (ii) in (i)–
(d) Statement is true but statement II is false.
1 1 n2h 2 [JEE Main 2021, 18 March Shift-II]
K.E. = × × 2 2
2 m 4π r Ans. (b) : Statement -1 is false since, Bohr’s theory
Now, n0 = 2, r1 = a0, r2 = a0 × (2)2 = 4a0 accounts for the stability and spectrum of single
electronic species (e.g. He+, Li2+ etc)
1 22 h 2
K.E. = × 2 statement – II is true since, Bohr's theory was unable to
2 4π (4a 0 ) 2 m explain the splitting of spectral lines in the presence of a
h2 magnetic field.
⇒ 688. Bohr model of hydrogen atom was unable to
32π2 a 20 m
explain
from question– (a) Rydberg's formula of atomic spectra
h2 h2 (b) Heisenberg's uncertainty principle
=
xma 0 32π 2 ma 02
2 (c) Planck's law of energy quantization
(d) Rutherford' model of atomic structure
1 1 J & K CET-(2012)
=
x 32π 2 Ans. (b) : Bohr model of hydrogen atom could not
x = 32π 2 explain :
x = 315.5 (i) de Broglie concept of dual nature of matter and
10x = 315.5 × 10 Heisenberg’s uncertainty principle.
= 3155 (ii) the splitting of lines in magnetic and electric field.
686. Given below are two statements. 689. Energy of one mole of photons of radiation
whose frequency is 5 × 1014 Hz is
Statement-I According to Bohr's model of an
(a) 199.51 kJ mol–1 (b) 189.51 kJ mol–1
atom, qualitatively the magnitude of velocity of –1
(c) 198.51 kJ mol (d) 188.51 kJ mol–1
electron increases with decrease in positive
J & K CET-(2014)
charges on the nucleus as there is no strong
hold on the electron by the nucleus. Ans. (a) : Given, Frequency v = 5×1014 ∵ Energy of 1
Statement-II According to Bohr's model of an photon, E = hυ
atoms, qualitatively the magnitude of velocity ∴ Energy of 1 mole of photon, E’ = N A hυ
of electron increases with decrease in principal = 6.022×1023×6.626×10–34×5×1014
quantum number. In the light of the above = 199.508×103J mol–J = 199.51 kJmol–1
statements, choose the most appropriate
690. Let the energy of an nth orbit of H atom be
answer from the options given below.
–21.76 × 10-19 / n2J. What will be the longest
(a) Both statement I and statement II are false. wavelength of energy required to remove an
(b) Both statement I and statement II are true. electron from the third orbit?
(c) Statement I is false but statement II are true. (a) 0.628 nm (b) 1.326 × 10-7 m
(d) Statement I is true but statement II is false. (c) 0.798 pm (d) 0.821 µm
[JEE Main 2021, 26 Aug Shift-I] J & K CET-(2017)
Objective Chemistry Volume-I 237
Ans. (d) : Given, energy of an nth orbit of H atom 1 1
λ= = cm −1 =1.216 ×10−5 cm
−21.76 ×10−19 v 82258
E= J
n2 o

We know that, =1216 × 10−8 = 1216 A

hc hc 695. What is the energy (in eV) required to excite
E = ⇒λ = the electron from n = 1 to n = 2 state in
λ E hydrogen atom? (n = principal quantum
Where, c = speed of light = 3×108m/s number).
h = Plank's constant = 6.6 ×10–34 (a) 13.6 (b) 3.4
6.6 × 10−34 × 108 × n 2 (c) 17.0 (d) 10.2
λ= VITEEE-2012
−21.76 × 10−19
For n = 3 Ans. (d) : Energy required to excite the electron from n
= 1 to n = 2
6.6 × 10−34 × 3 × 108 × (3)2
λ=  1 1  1 1 
−21.76 × 10−19 ∆E = 13.6z 2  2 − 2  = 13.6 × (1)2  2 − 2 
λ = 8.2×10 = 0.82 × 10–6 m = 0.82µm
–7 n
 1 n 2  1 2 
691. With respect to atomic spectrum, each line is  1 3
the Lyman series is due to electrons returning 13.6  1 −  = 13.6 ×
 4 4
(a) from a particular higher energy level to n=3 = 10.2 eV
(b) from a particular higher energy level to n=2
696. The first emission line in the electronic
(c) from a particular higher energy level to n=1 spectrum of hydrogen in the Balmer series
(d) from a particular higher energy level to n=4 appears at cm–1
J & K CET-(2018) (a) 9R/400 cm–1 (b) 7R/144 cm–1
–1
Ans. (c): Lyman series : n = 2, 3, 4, 5 to n = 1 (c) 3R/4 cm (d) 5R/36 cm–1
692. What is the maximum no of emission lines J & K CET-(2003)
obtained when the excited electrons of a Ans. (d) :
hydrogen atom in n=5 drop to ground state? For Balmer series n1= 2 and n2 = 3
(a) 10 (b) 5 For the first line
(c) 12 (d) 15
 1 1  1 1  5R −1
J & K CET-(2010) v = R 2 − 2  = R −  = cm
 2 3   4 9  36
Ans. (a) : Number of emission lines in the spectrum be
equal to (n2-n1) (n2–n1+1) where n2=5 and n1=1 697. Which of the following atomic models involve
the concept of stationary orbitals?
(5 –1)(5 –1+1) 20 (a) Bohr’s model (b) Thomson model
∴ = = 10 lines
2 2 (c) Rutherford model (d) wave model
693. The value of Rydberg constant is J & K CET-(2002)
(a) 109678 cm–1 (b) 109876 cm–1 Ans. (a) : According to Bohr’s concept electrons
–1
(c) 108769 cm (d) 108976 cm–1 revolve only is those orbits which have a fixed value of
J & K CET-(2007) energy these orbital are called stationary orbitals.
Ans. (a) : Rydberg constant is a physical constant 698. Which of the following value of nl in the
relating to atomic spectra. Denoted by R ∞ for heavy 1 1 1
atoms and RH for hydrogen. value of Rydberg constant relationship is = R H  2 - 2  is correct when
λ  n l n 2 
is 109678 cm–1.
n2 > n1 corresponds to Paschen lines in the
694. The wavelength of a spectral line in Lyman Hydrogen spectrum?
series, when electron jumping back to 2nd orbit, (a) 1 (b) 2
is (c) 3 (d) 4
(a) 1162 (b) 1216 J & K CET-(2001)
(c) 1362 (d) 1176
Ans. (c): The paschen line all lie in the infrared band.
J & K CET-(2007) This series overlaps with the next (Brackett) series, i.e.
1 1 1 the shortest line in the bracket series has a wavelength
Ans. (b) : = RH  2 - 2  that falls among the paschen series. For paschen lines in
λ  n1 n 2  the hydrogen spectrum the value of n is 3.
For Lyman Series n1=1, n2=2 699. Transition from n=4, 5, 6 to n = 3 in hydrogen
1 1  3 spectrum gives
v = 109 677 cm –1  2 – 2  = 109677× cm –1 (a) Lyman series (b) Balmer series
1 2  4
(c) Paschen series (d) Pfund series
= 82258 cm –1 J & K CET-(2000)
Objective Chemistry Volume-I 238
Ans. (c) : For Paschen series, 704. The wavelength of the radiation emitted, when
n1= 3, n2 > n1, i.e. n2 = 4,5, 6, ………∞ in a hydrogen atom electron falls from infinity
to stationary state one, would be (Rydberg
700. According to Bohr’s theory of hydrogen atom, constant = 1.097 × 107 m–1)
which of the following is quantized for an
electron? (a) 91 nm (b) 192 nm
(a) acceleration (b) velocity (c) 406 nm (d) 9.1 × 10–8 nm
(c) angular momentum (d) angular acceleration JCECE - 2008
J & K CET-(1997) 1  1 1 
Ans. (c) : According to Bohr’s theory, electron can Ans. (a) : = R H z2  2 − 2 
λ  n1 n 2 
move in that circular path around the nucleous, in which
h 1 1 1
the angular momentum of an electron is either or = 1.097 ×107 × (1) 2 m −1  2 − 
2π λ 1 ∞ 
nh Therefore, λ = 91×10–9 m
, n=1 ,2 ,3 ……...
2π λ = 91 nm
So, angular momentum is quantized for the electron. 705. The energy of second Bohr orbit of the
701. The ionization energy of hydrogen atom in its hydrogen atom is –328 kJ mol–1; hence the
ground state is–13.12 × 105J mol–1. The energy energy of fourth Bohr orbit would be
of the electron corresponding to the first (a) – 41 kJ mol–1 (b) – 1312 kJ mol–1
–1
excited state will be (c) – 164 kJ mol (d) – 82 kJ mol–1
(a) –42.48 × 105 J mol–1 JCECE - 2008, 2012
(b) –26.24 × 105 J mol–1 Ans. (d) : The energy of second Bohr orbit of hydrogen
(c) –6.56 × 105 J mol–1 atom (E2) is –328 kJ mol–1 therefore
(d) –3.28 × 105 J mol–1 E
J & K CET-(1997) −328 = 2 ⇒ En = −328 × 4 = −1312kJ mol−1
2
Ans. (d) : First excited state means, n= 2 −1312
−13.12 × 105 ∴ En = kJ mol−1
E2 = J mol–1 n2
2
2 If n = 4,
= – 3.28 ×105 J mol–1 1312
702. The expression of angular momentum of an ∴ E4 = − 2 kJ mol−1 = −82 kJ mol−1
4
electron in a Bohr's orbit is: 706. When an electron in hydrogen spectrum jumps
nh nh from n = 7 to n = 2, the total number of spectral
(a) (b)
3π 2π lines possible are
h h (a) 21 (b) 15
(c) (d) l(l + 1). (c) 5 (d) 9
4π 2π
JCECE - 2003 JCECE - 2016
Ans. (b): The expression of angular momentum of an Ans. (b) : Total number of spectral lines for the
nh transition from n (Higher) → n(lower) will be =
electron in a Bohr's orbit is . ∆n.(∆n + 1)
2π
703. Ratio of kinetic energy of hydrogen and helium 2
gas at 300 K is : Where,
(a) 2 : 1 (b) 4 : 5 ∆n = Difference of nH – nL = 7 – 2 = 5
(c) 1 : 1 (d) 1 : 2 5× 6
∴ Total number of lines = = 15
JCECE - 2006 2
Ans. (c) : We know that, 707. An electron is moving in Bohr's fourth orbit.
3 Its de-Broglie wave length is λ. What is the
K.E. = nRT circumference of the fourth orbits?
2
Therefore, 2
(a) (b) 2λ
K.E. ∝ n λ
w 4
or K.E. ∝ (c) 4λ (d)
M λ
K.E.H2 M He JIPMER-2014
= Ans. (c) : According to Bohr’s concept an electron
K.E.He M H2 always move in the orbit with angular momentum (mvr)
2 1 equal to nh/2π
= =
2 1 nh
mvr =
K.E.H2: K.E.He = 1:1 2π
Objective Chemistry Volume-I 239
 n  h  711. If the energies of the two photons in the ratio of
r=   3 : 2, their wavelength will be in the ratio of
2π  mv  (a) 2 : 3 (b) 9 : 4
nλ (c) 3 : 2 (d) 1 : 2
r=
2π Karnataka-CET-2011
h hc
(From de–Broglie equation, λ= ) Ans. (a) : E =
mv λ
For fourth orbit (n=4) 1
2λ E∝
r= λ
π E1 λ 2
2λ =
∴ circumference= 2πr = 2π × = 4λ E 2 λ1
π
3 λ2
708. The ratio of the difference in energy between =
the first and the second Bohr orbit to that 2 λ1
between the second and the third Bohr orbit is ∴ λ1 : λ2 = 2 : 3
(a) 1/2 (b) 1/3 712. The radius of the first Bohr orbit of hydrogen
(c) 4/9 (d) 27/5 atom is 0.529Å. The radius of the third orbit of
JIPMER-2012 H+ will be
(a) 8.46Å (b) 0.705Å
1312 1312 3
Ans. (d) :∆E=E2–E1 = 2 − = 1312   (c) 1.59Å (d) 4.29Å
1 22 4 (e) 2.98Å
1312 1312  5  Kerala-CEE-2007
= E3 – E2 = 2
− 2 = 1312   Ans. (d) : According to Bohr model,
2 3  36 
0.529 × n 2
Radius of hydrogen atom (rn)= Å
3 5 z
∴ E2– E1: E3–E2 = : = 27:5
4 36 0.529 × (3) 2
r3 =
709. Ratio of radii of second and first Bohr orbits of 1
H atom is =4.79Å
(a) 2 (b) 4 713. Which diagram best represents the appearance
(c) 3 (d) 5 of the line spectrum of atomic hydrogen in the
JIPMER-2005 visible region?
Ans. (b) : According to Bohr's rule, r ∝ n2
Where r is radii and n is the number of orbit
r1 (2)2
So, = =4
r2 (1) 2
Hence, ratio of radii is 4.
710. What will be the number of waves formed by a
Bohr electron in one complete revolution in its
second orbit?
(a) Three (b) Two
(c) One (d) Zero
JIPMER-2016
Ans. (b) : Number of waves = circumference of orbit Kerala-CEE-2007
de − Broglie wavelength of electron Ans. (c) : Line spectrum of atomic hydrogen in the
h visible region
λ=
mv
2πr 2π
µ= = × mvr
h h
mv 714. Which transition in the hydrogen atomic
nh 2π nh spectrum will have the same wavelength as the
mvr = , µ = × transition, n = 4 to n = 2 of He+ spectrum?
2π h 2π
(a) n = 4 to n = 3 (b) n = 3 to n = 2
µ= n
(c) n = 4 to n = 2 (d) n = 3 to n = 1
For second orbit, n =2 (e) n = 2 to n = 1
∴ number of waves,µ = 2 Kerala-CEE-2011
Objective Chemistry Volume-I 240
 717. In the hydrogen atom spectrum, the emission of
1 1 1
Ans. (e) : = z2R H  2 − 2  the least energetic photon taken place during
λ  n1 n 2  the transition from n= 6 energy level to n =
........energy level.
1 1 1 3 3
For He2, = z 2 R H  2 − 2  = 4 × = (a) 1 (b) 3
λ 2 4  16 4 (c) 5 (d) 4
1 2 1 1  3 (e) 2
For H, =1 RH  2 − 2  = Kerala-CEE-2016
λ 1 2  4
Ans. (c) : As we move towards outer shell the energy
Hence, for hydrogen n=2, n=1 increases but the difference between the successive
715. The ratio of frequency corresponding to the shalls decreases
third line in Lyman series of hydrogen atomic E2–E1>E3–E2>E4–E3---------
spectrum to that of the first line in Balmer Hence, the emission of least energetic photon
series of Li2+ spectrum is takes place during the transition from n=6 energy level
4 5 to n=5 energy level.
(a) (b) 718. The energy of an electron is the 3S orbital
5 4
(excited state) of H – atom is
4 3 (a) – 1.5eV (b) – 13.6eV
(c) (d)
3 4 (c) – 3.4eV (d) 4.53eV
3 (e) 4.53eV
(e) Kerala-CEE-2017
8
−13.6
Kerala-CEE-2012 Ans. (a) : E n = eV
Ans. (d) : Lyman series n1=1 n2
13.6 ×1
For third line of Lyman series, n2 = 4 E1 = − eV
For hydrogen, z = 1 n2
13.6 13.6
c  1 1  E n = − 2 eV = − = −1.5eV
vH = = c.R H Z2  2 − 2  (3) 9
λ  n1 n 2  = – 1.51ev (n=3, for 3s–orbital)
1 1  15 719. The shortest wavelength of Paschen series in
= cRH (1)2  − 2  = R H ⋅ c hydrogen spectrum is 8182 Å. The first
 1 (4)  16 member of the Paschen series is nearly.
For lithium, Z = 3 (a) 15400 Å (b) 12200 Å
For first line of Balmer series, n1 = 2, n2=3 (c) 13400 Å (d) 18700 Å
 1 1  5 (e) 16700 Å
vLi2+ = c.RH (3)2  2 − 2  = c.RH×9× Kerala-CEE-2020
 (2) (3)  36
Ans. (d) : For the calculation of wavelength of spectral
5 lines Rydberg gave a formula,
= c.R H
4 1 1 1
 15  = R H Z2  2 − 2 
  cR H 15 4 3 λ n
 1 n 2

= 
vH 16
= × = Where, λ = Wavelength
v Li 2+ 5 16 5 4 RH = Rydberg’s constant
  cR H
4 Z = Atomic number of atom
716. The shortest wavelength of the line in hydrogen n1 and n2 = integers
atomic spectrum of Lyman series when RH = RH = 1.0973 × 107 m–1
109678 cm-1 is 1  1 1 
= 1.0973 × 107  2 − 2 
(a) 1002.7Å (b) 1215.67 Å λ  (3) (4) 
(c) 1127.30 Å (d) 911.7 Å 1 7
(e) 1234.7 Å = 1.0973 × 107 ×
λ 16 × 9
Kerala-CEE-2014 1 16 × 9
Ans. (d) : For Lyman series, n1=1 and for shortest =
λ 7 × 1.0973 ×107
wavelength, n2=∞ 144
1 1 1  λ= ×10−7
∴ v = = R H  2 − 2  = 109678 cm −1 7.6811
λ 1 ∞  λ = 18.747 × 10–7
1 λ = 18747 ×10–10 ( Q 1m = 10–10 Å)
λ= cm = 911.7 × 10−8 cm = 911.7Å
109678 λ ≈ 18,800 Å

Objective Chemistry Volume-I 241

720. In the atomic spectrum of hydrogen, the Ans. (d) : Radius of hydrogen atom = 0.53Å
spectral lines pertaining to electronic transition Number of excited state (n) = 2
of n = 4 to n = 2 refers to : Atomic number of hydrogen atom (Z)=1
(a) Lyman series (b) Balmer series We know that the Bohr radius.
(c) Paschen series (d) Brackett series
n2 (2) 2
Manipal-2018 r= × rn = × 0.530
Ans. (b) : In Balmer series, electrons transfer from Z 1
= 4×0.530
higher orbits to 2nd orbit.
= 2.12Å
721. Which of the following statement do not form a
726. In a Bohr's model of an atom, when an electron
part of Bohr's model of hydrogen atom? jumps from n = 1 to n = 3, how much energy
(a) Energy of the electrons in the orbits are will be emitted or absorbed?
quantized. (a) 2.389 × 10–12 ergs (b) 0.239 × 10–10 ergs
(b) The electron in the orbit nearest the nucleus (c) 2.15 × 10–11 ergs (d) 0.1936 × 10–12 ergs
has the lower energy. NEET-1996
(c) Electrons revolve in different orbits around Ans. (d) : According to Bohr’s model of atom
the nucleus
(d) The position and velocity of the electrons in  1 1 
∆E = 2.18×10–18  2 − 2  J
the orbit cannot be determined simultaneously n
 1 n 2 
NEET-1989
 1 1 
Ans. (d) : According is the Heisenberg uncertainty =2.18×10–18  −  J
principle, the position and velocity of the electrons in 2 9
the orbit cannot be determined simultaneously. This =1.9×10–18 J
statement is not correct according to Bohr’s model = 1.9×10–18×107 erg
because according to Bohr’s model it was possible to = 0.19×10–10 erg
determine both position and velocity simultaneously. 727. According to the Bohr theory, which of the
following transitions in the hydrogen stom will
722. If r is the radius of the first orbit, the radius of
give the ratio the least energic photon?
nth orbit of H-atom is given by
(a) n = 6 to n = 1 (b) n = 5 to n = 4
(a) rn2 (b) rn (c) n = 6 to n = 5 (d) n = 5 to n = 3
(c) r/n (d) r2n2 NEET-Mains 2011
NEET-1988 Ans. (c) : We know that
Ans. (a) : Radius of nth orbit= r1n2 (for H–atom).
1 1
723. The energy of an electron in the nth Bohr orbit ∆E ∝  2 − 2  , where n2>n1
of hydrogen atom is n
 1 n 2

13.6 13.6 ∴ n=6 to n=5 will give least energetic photon

(a) − 4 eV (b) – 3 eV
n n 728. The energy of second Bohr orbit of the
13.6 13.6 hydrogen atom is –328 kJ mol–1. hence the
(c) – 2 eV (d) – eV energy of fourth Bohr orbit would be
n n (a) –41 kj mol–1 (b) –82 kj mol–1
NEET-1992 (c) –164 kj mol –1
(d) –1312 kj mol–1
th
Ans.(c): Energy of an electron in n Bohr orbit of NEET-2005
13.6 Ans. (b) : Energy of electron in nth orbit of hydrogen
hydrogen atom (E) = − 2 eV (Z = 1)
n E
En = − 2
724. Who modified Bohr's theory by introductin n
elliptical orbits for electrons path? Where E is a constant
(a) Rutherford (b) Thomson E
(c) Hund (d) Sommerfeld E 2 = 2 = −328kJ mol−1
2
NEET-1999 E= 4×328 kJ mol–1
Ans. (d) : Sommerfeld modified Bohr’s theory. −E −4 × 328
According to him electrons move in elliptical orbits in E4 = 2 = kJ mol −1
4 16
addition to circular orbits. = –82 kJ mol–1
725. The Bohr orbit radius for the hydrogen atom 729. The radius of which of the following orbit is
(n =1) is approximately 0.530Å. The radius for same as that of the first Bohr's orbit of
the first excited state (n=2) orbit is (in Å) hydrogen atom
(a) 4.77 (b) 1.06 (a) He+ (n = 2) (b) Li2+ (n = 2)
2+
(c) 0.13 (d) 2.12 (c) Li (n = 3) (d) Be4+ (n = 2)
NEET-1998 UP CPMT-2013

Objective Chemistry Volume-I 242

Ans. (d) : For H-like particles, the radii of the first Ans. (c): Hβ line is formed when electron jumps from
stationary states are given by the expression 3rd orbit in Lyman series.
a0n2 n1 = 1, n2 = 3
rn = 733. The energy of second Bohr orbit of the
Z
For H-atom, n = 1 and Z = 1 hydrogen atom is – 328 kJ mol–1; hence the
∴ rn = a0 = Bohr energy of fourth Bohr orbit would be
radius = 52.9 pm. (a) –41kJ mol–1 (b) –1312 kJ mol–1
–1
(a) For He+ ion, n = 2 and Z = 2 (c) –164kJ mol (d) –82 kJ mol–1
UPTU/UPSEE-2007
a 0 (2) 2 Ans. (d) : Energy of 2nd orbit= –328 kJ/mol
∴ rn = = 2a 0
2 328
(b) For Li2+ ion, n = 2 and Z = 3 Now, energy of 4th orbit will be = − = − 82 kJ mol−1
4
a 0 (2) 2 4a 0 734. The wavelength of the radiation emitted, when
∴ rn = =
3 3 in a hydrogen atom electron falls from infinity
(c) For Li2+ ion, n = 3 and Z = 3 to stationary state, would be (Rydberg constant
a 0 (3) 2 = 1.097 × 107 m–1)
∴ rn = = 3a 0 (a) 91 nm (b) 192 nm
3 (c) 406 nm (d) 9.1 × 10–8 nm
(d) For Be4+ ion, n = 2 and Z = 4
UPTU/UPSEE-2007
a 0 (2) 2
∴ rn = = a 0 = Bohr radius = 52.9 pm 1  1 1 
4 Ans. (a) : = νH = R H  2 − 2 
730. Number of spectral lines of Lyman series of λ  n1 n 2 
electron when it jumps from 6 to first level (in 1 1 
Lyman series), is = 1.097 ×107  2 − 2 
(a) 9 (b) 12 1 ∞ 
(c) 15 (d) 18 1
= m = 9.11× 10−8 m
UP CPMT-2009 1.097 × 107
n ( n − 1) = 91.1×10–9 m
Ans. (c): Number of spectral lines = = 91.1 nm
2
735. The radius of hydrogen atom in the ground
6 ( 6 − 1) state is 0.53Å.The radius of Li2+ ion (atomic
=
2 number = 3) in a similar state is:
= 15 (a) 0.176Å (b) 0.30Å
731. What is the wave number of 4th line in Balmer (c) 0.53Å (d) 1.23Å
series of hydrogen spectrum? UPTU/UPSEE-2014, 2005
(R = 1,09,677 cm–1) Ans. (a) :
(a) 24,630 cm–1 (b) 24,360 cm–1
n2 12
(c) 24,730 cm –1
(d) 24,372 cm–1 Radius of Li 2+ ion = r1 × = 0.53 × = 0.17 Ǻ
UP CPMT-2008 z 3
736. For a Bohr atom angular momentum M of the
 1 1  electron is : (n=0,1,2,....)
Ans. (d): ν = R  2 − 2 
n
 1 n 2  nh 2 n 2h2
th (a) (b)
For Balmer series n1 = 2 and for 4 line in Balmer 4π 4π
series n2 = 6 2
R = 109677 cm–1 nh nh
(c) (d)
 1 1 4π 2π
ν =109677  2 − 2  UPTU/UPSEE-2005
2 6 
Ans. (d) : For a Bohr atom, angular momentum m of
1 1 
= 109677  −  the electron (mvr) =
nh
 4 36  2 π
ν = 24,372 cm-1 –
737. If an e is revolving in the first bohr orbit of a
732. What are the values of n1 and n2 respectively H-atom with a velocity of 2.19 × 108 cms–1,
for Hβ line in the Lyman series of hydrogen what will be the velocity of the e– in the third
atomic spectrum? orbit of H-atom ?
(a) 3 and 5 (b) 2 and 3 (a) 2.19 × 108 cms–1 (b) 7.3 × 107 cms–1
8 –1
(c) 1 and 3 (d) 2 and 3 (c) 6.57 × 10 cms (d) 1.09 × 108 cms–1
UP CPMT-2008 UPTU/UPSEE-2018
Objective Chemistry Volume-I 243
Ans. (b) : The velocity of electron in different orbits is 740. Which one of the following represents the
given by the following expression, correct ratio of the energy of electron in
z ground state of H atom to that of the electron
Vn = Vo × in the first excited state of Li+?
n
Given, for first Bohr orbital of H–atom (a) 4:9 (b) 9:4
Vo = 2.19×108 cm s–1 (c) 1:4 (d) 4:1
∴ the velocity of e– in third orbit of H–atom is (Q n=3) UPTU/UPSEE-2011
2
1 Z
Vn= 2.19×108× Ans. (a) : E n ∝
3 n2
= 0.73×108 cm s–1 1
2

= 7.3×107 cm s–1 For H atom, E1 ∝ 2 [QZ = 1 for hydrogen]

1
738. The radius of which of the following orbit is E1 = 1
same as that of the first Bohr's orbit of 2
hydrogen atom? 3
For Li+ ion, E 2 ∝ 2 [QZ = 3for Li + ion]
(a) Li2+ (n = 2) (b) Li2+ (n = 3) 2
3+
(c) Be (n = 2) (d) He+ (n = 2) 9
UPTU/UPSEE-2015 E2 =
4
n2 9
Ans. (c) : rn = rH × Thus, E1/E2 = 1 / or 4 : 9
z 4
For Li2+ (n=2) 741. What is the energy required to move the
(2)2 rH × 4 electron from ground state of H atom to the
rLi2+ = rH × = first excited state? Given that the ground state
3 9
For Li2+ (n=3) energy of H atom is 13.6 eV and that the energy
En of an electron in nth orbital of an atom or
(3) 2 ion of atomic number Z is given by the
rLi2+ = rH × = 3rH
3 equation
For Be3+ (n=2) En= – (13.6Z2/n2).
(2)2 (a) 13.6 eV (b) 3.4 eV
rBe3+ = rH × = rH (c) 10.2 eV (d) –10.2 eV
4
For He+ (n=2) UPTU/UPSEE-2011
rH × (2) 2 −13.6
rHe+ = = 2rH Ans. (c) : E n = × Z2
4 n2
3+
Thus, Be (n=2) has same radius as that of the first The energy required to move from ground state (n = 1)
Bohr's orbit of H–atom. of H to first excited state (n = 2) is
739. The ratio of the difference in energy between  1 1 3
∆E = E2 – E1 = –13.6  2 − 2  = 13.6 × = 10.2 eV
the first and second Bohr orbit to that between  2 1  4
the second and third Bohr orbit is
742. An electron from one Bohr stationary orbit
1 1 can go to next higher orbit
(a) (b)
2 27 (a) by emission of electromagnetic radiation
4 27 (b) by absorption of any electromagnetic
(c) (d)
9 5 radiation
UPTU/UPSEE-2013 (c) by absorption of electromagnetic radiation of
1 1 particular frequency
Ans. (d) : : E1 – E2 = –13.6 × Z2  2 − 2  (d) without emission or absorption of
1 2  electromagnetic radiation
3 UPTU/UPSEE-2008
E1 – E2 = –13.6 × Z2   ––––– (i)
4 Ans. (c) : According to Bohr's atomic model, if energy
 1 1 is supplied to an electron it may jump from a lower
E2 – E3 = –13.6 × Z2  2 − 2  energy level to higher energy level. Energy is absorbed
2 3  in the form of quanta (or photon)
 5  ∆E = h ν
E2 – E3 = –13.6 × Z2   ––––– (ii)
 36  Where, ν is the frequency.
From eqn. (i) and (ii) According to above postulate an electron from one Bohr
 E1 − E 2  3 × 36 27 stationary orbit can go to next higher orbit by the
 = = absorption of electromagnetic radiation of particular
 E 2 − E3  4 × 5 5 frequency.
Objective Chemistry Volume-I 244
743. For the Paschen series the values of n1 and n2 in 747. The emission spectrum of hydrogen discovered
the expression first and the region of the electromagnetic
spectrum in which it belongs, respectively are
 1 1  (a) Lyman, ultraviolet (b) Lyman, visible
∆E = R H c  2 − 2 
 n1 n 2  (c) Balmer, ultraviolet (d) Balmer, visible
(a) n1=1,n2=2,3,4,… (b) n1=2,n2=3,4,5,… WB-JEE-2014
(c) n1=3,n2=4,5,6,… (d) n1=4,n2=5,6,7,… Ans. (d) : Balmar spectrum of hydrogen was discovered
first and it lies in the visible region.
WB-JEE-2009
748. The time taken for an electrin to complete one
Ans. (c) : Paschen series is the third series of hydrogen revolution in Bohr orbit of hydrogen atom is
spectrum and lies in the infrared region, therefore for
4m 2π r 2 n2h2
this series, n1 = 3 and n2 > n1, i.e. 4, 5, 6, ...... (a) 2 2
(b)
n h 4mr 2
744. In Sommerfeld's modification of Bohr's theory,
4π 2 mr 2 nh
the trajectory of an electron in a hydrogen (c) (d)
atom is nh 4π 2
mr 2
(a) a perfect ellipse WB-JEE-2016
(b) a closed ellipse like curve, narrower at the Ans. (c) : We know according to Bohr's theory
perihelion position and flatter at the aphelion nh
mvr =
position 2π
(c) a closed loop on a spherical surface nh
V=
(d) a rosette. 2πmr
WB-JEE-2010 ∴ Time required for one complete revolution
Ans. (a) : According to Sommerfield's modification, the 2πr 2πr 4π2 mr 2
T= = × 2πmr =
trajectory of an electron in a hydrogen atom is a closed V nh nh
ellipse like curve in addition to circular orbits. 749. The energy required to break one mole of
745. The electronic transitions from n = 2 to n = 1 hydrogen-hydrogen bonds in H2 is 436 KJ.
will produce shortest wavelength in (Where n = What is the longest wavelength of light
principal quantum state) required to break a single hydrogen- hydrogen
2+ + bond?
(a) Li (b) He (a) 68.5 nm (b) 137 nm
(c) H (d) H+ (c) 274 nm (d) 548 nm
WB-JEE-2011 WB-JEE-2016
1  1 1  Ans. (c) : Amount of energy required to break one H-H
Ans. (a) : = Z2 R H  2 − 2 
λ  n1 n 2  436 ×10 3
436
bond is = J mol−1
n1 = 1, n2 = 2 6.023 × 10 23
6.023 × 10 20

Now,
1 1 1  1 3 hc
= Z2 R H  −  = = R H Z2 E=
λ 1 4  λ 4 λ
∴λ∝ 2
1 6.626 ×10−34 × 3 × 108
λ= × 6.023 ×1020
Z 436
Hence, elements having high atomic number will 6.626 × 6.023 × 3
produce shortest wavelength. = ×10−6 = 274 nm
436
746. The energy of an electron in first Bohr orbit of 750. The radius of the first Bohr orbit of a hydrogen
H-atoms is 13.6 eV. the possible energy value of atom is 0.53×10-8 cm. The velocity of the
2+
electron in the excited state of Li is electron in the first Bohr orbit is
(a) -122.4 eV (b) 30.6 eV (a) 2.188×108cm s–1 (b) 4.376×108cm s–1
8 –1
(c) -30.6 eV (d) 13.6 eV (c) 1.094×10 cm s (d) 2.188×109cm s–1
WB-JEE-2011 WB-JEE-2020
Ans. (a) : The formula used for the velocity of electron
Z2 in Bohr orbit of a hydrogen atom is:
Ans. (c) : En = E1× 2
n nh
E1 = energy of hydrogen in first orbit V=
2πmr
n = 2, Z = 3
1× 6.6 × 10−34
(3) 2 V=
En = – 13.6 × 2
= – 30.6 eV 2 × 3.14 × 9.1× 10−31 × 0.53 × 10−10
(2) = 2.18 × 106 ms–1

Objective Chemistry Volume-I 245

 9. Isobars and Isotopes 10. Line Spectrum of Hydrogen
751. ( Ge 76 , 34Se76 ) and(14 Si 30 , 16S 32 ) are examples of
32
757. The ratio of the shortest wavelength of two
spectral series of hydrogen spectrum is found
(a) isotopes and isobars to be about 9. The spectral series are
(b) isobars and isotones (a) Lyman and Paschen
(c) isotones and isotopes (b) Brackett and Pfund
(d) isobars and isotopes
(c) Paschen and Pfund
WB-JEE-2014
(d) Balmer and Brackett
Ans. (b) : 32Ge76 and 34Se76 have the same mass number [JEE Main 2019, 10 April Shift-II]
but different atomic number, so they are isobars.
Number of neutrons in 14Si30 = 30 – 14 = 16 1  1 1 
Number of neutrons in 16S32 = 32 – 16 = 16 = R H  2 − 2  Z2
λ  n1 n 2 
Because of the presence of some number of neutrons, Ans. (a) : 2
30 32 1  1 1 
14Si and 16S are called isotones. = R H  2 − 2  Z2
752. 19K40 and 20Ca40 are known as λ1  m1 m 2 
(a) isotopes (b) isobars As for shortest wavelengths both n2 and m2 are ∞ and z
(c) isotones (d) isodiaphers for H is 1
UP CPMT-2002 λ
40 40 ∴ 1 = 9 ( given )
Ans.(b): 19K and 20Ca both have same mass no but λ2
different atomic number, are called isobars. 2
753. Isotopic pair is m 
∴ 9= 1
(a) 20X40, 21Y40 (b) 20X40, 20Y41  n1 
20 20
(c) 40X , 41Y (d) None of these 2
2
UP CPMT-2005 3  m 
or   =  1 
Ans. (b) : Key Idea: Isotopes are the atoms of same  1   n1 
element which have different atomic masses (due to
different number of neutrons). They had same atomic m1 =3 (lyman)
numbers. n1 = 1 (paschen)
∴ choice (b) is the correct answer because both (20X40, 758. The shortest wavelength of H atom in the
41
20X ) have same atomic number but different atomic Lyman series is λ1. The longest wavelength in
mass. the Balmer series of He+ is
754. O2 and O3 are 36λ1 5λ1
(a) allotropes (b) isotopes (a) (b)
5 9
(c) isomorphs (d) polymorphs
9λ1 27λ1
UP CPMT-2010 (c) (d)
Ans. (a) : Since, in O2 and O3 different numbers of 5 5
same element i.e. oxygen are present, these are [JEE Main 2020, Sep Shift-II]
allotropes. Ans. (c) : As we know
Note. Different crystalline structures, different number hc
of atoms and different nuclear spins all result in ∆E =
allotropy. λ
755. An isobar of 20Ca40 is hc
So, λ = for λ minimum i.e. shortest; ∆E =
(a) 18Ar40 (b) 20Ca38 ∆E
(c) 20Ca 42
(d) 18Ar38 maximum For Lyman series n = 1 & for ∆E max.
MHT CET-2008 Transition must be form n = ∝ to n = 1
Ans. (a) : Isobars have same atomic mass but different So,
atomic numbers. Therefore, the isobars of 20Ca40 is  1
40 1 1 
18Ar . = RH Z2  2 − 2 
756. Isotones have λ  n1 n 2 
(a) same number of protons 1
(b) same number of electrons = RH Z2 (1 − 0 )
λ
(c) same number of neutrons
(d) same isotopic mass 1 1
= R × (1)2 ⇒ λ1 =
UP CPMT-2010 λ R
Ans. (c): Species having the same number of neutron For longest wavelength ∆E = minimum for Balmer
but different atomic number as well as atomic mass are Series n = 3 to n = 2 will have ∆E = minimum.
39 40
called isotones. E.g. 18 Ar, 19 K. for He+ (Z) = 2

Objective Chemistry Volume-I 246

So, 762. Which is the shortest wavelength line in the
Lyman series of the hydrogen spectrum?
1  1 1  (R=1.097×10–2nm–1)
= RH × Z2  2 − 2 
λ2  n1 n 2  (a) 94.21 nm (b) 91.16 nm
(c) 911.6 nm (d) 933.6 nm
1 1 1
= RH × 4  −  JIPMER-2004
λ2 4 9
1  1 1 
1 5 Ans. (b) : : For Lyman Series = R 2 − 2 
= RH × λ  (1) n 
λ2 9
Hence for λ to be smallest n should be greatest
λ 2 = λ1 × 5 / 9 1  1 
= 1.097 ×10−2 1 − 
759. Match the type of series given in Column I with λ min  ∞2 
the wavelength range given in Column II and ⇒ 1.097 × 10–2
choose the correct option. or
Column I Column II 1
A. Lyman 1. Ultraviolet λ min = = 91.16nm
1.097 × 10−2
B. Paschen 2. Near infrared 763. The wave number of the spectral line in the
C. Balmer 3. Far infrared emission spectrum of hydrogen will be equal to
D. Pfund 4. Visible 8
times the Radberg’s constant if the electron
Codes 9
(a) 1 2 4 3 (b) 4 3 1 2 jumps from
(c) 3 1 2 4 (d) 4 3 2 1 (a) n = 3 to n = 1 (b) n = 10 to n = 1
JIPMER-2017 (c) n = 9 to n = 1 (d) n = 2 to n = 1
Ans. (a) : The correct match is Karnataka-CET, 2010
(A) → Lyman → (1) Ultraviolet Ans. (a) : Wave number of spectral line in emission
spectrum of hydrogen,
(B) → Paschen → (2) Near infrared
(C) → Balmer → (4) Visible  1 1 
ν = RH  2 − 2  … (i)
(D) → Pfund → (3) Far infrared n
 1 n 2 
760. What is the wave number of 4th line in Balmer 8
series of hydrogen spectrum? (R = 1, 09, 677 Given ν = R H
cm–1) 9
On putting the value of ν in Eq. (i), we get
(a) 24,630 cm–1 (b) 24,360cm −1
8  1 1 
(c) 24,730 cm–1 (d) 24,372 cm–1 RH = RH  2 − 2 
JIPMER-2009 9  n1 n 2 
−  1 1  8 1 1
Ans. (d) : V = R  2 − 2  = 2− 2
9 (1) n
 n1 n 2  2
th
For Balmer series n1= 2 and for 4 line in Balmer series 8 1
−1 = − 2
n2 = 6 9 n2
–1
R = 109677 cm 1 1
−  1 1 =
V = 109677  2 − 2  3 n2
2 6  ∴ n2 = 3
1 1  Hence, electron jumps from n2 = 3 to n1 = 1
= 109677  − 
 4 36  764. X-rays are electromagnetic radiation whose
− wavelengths are of the order of:
–1
V = 24, 372 cm (a) 1 metre (b) 10–1 metre
–5
761. What are the values of n1 and n2 respectively (c) 10 metre (d) 10–10 metre
for Hβ line in the Lyman series of hydrogen NDA (II)-2015
atomic spectrum 44? Ans : (d) X-ray are electromagnetic radiation whose
(a) 3 and 5 (b) 2 and 3 wavelength are the order of 10–10 meter.
(c) 1 and 3 (d) 2 and 4 • X-ray are produces when high velocity electron
JIPMER-2009 collide with metal plates
Ans. (c) : HB line is formed when electron jumps from • It have shorter wavelength of the electromagnetic
3rd orbit to 1st orbit in Lyman Series spectrum.
∴n1 = 1, n2 =3 • It have capability to travel in vaccum.
Objective Chemistry Volume-I 247
765. The shortest wavelength in hydrogen spectrum Or
of Lyman series when RH = 109678 cm-1, is n2 = 2
(a) 1002.7Å (b) 1215.67Å ∴ The electron jumps from second orbit (n2) to ground
(c) 1127.30Å (d) 911.7Å state (n1)
(e) 1234.7Å 768. Splitting of spectrum lines in magnetic field is
Kerala-CEE-2010 (a) Stark effect (b) Raman effect
Ans. (d) : For Lyman Series, n1 = 1 (c) Zeeman effect (d) Rutherford effect
− 1  1 1  1 1  UPTU/UPSEE-2008
V = = RH  2 − 2  = RH  2 − 2  Ans. (c) : Stark effect ⇒ The splitting of spectral lines
λ  n1 n 2  1 ∞ 
under the influences of electric field is called stark
Or effect.
1 1 effect ⇒ The splitting of spectral lines under
λ= = Zeeman
R H 109678cm −1 the influences of magnetic field is called Zeeman effect.
= 911.7 × 10–8 769. Heat treatment of muscular pain involves
= 911.7Å radiation of wavelength of about 900 nm.
766. What will be the longest wavelength line in Which spectral line of H-atoms is suitable for
Balmer series of spectrum? this purpose? [RH = 1×105cm–1, h = 6.6 × 10–34
(a) 546 nm (b) 656 nm Js,c = 3 × 108 ms–1]
(c) 566 nm (d) 356 nm (a) Paschen, 5→3 (b) Paschen, ∞→3
NEET-1996 (c) Lyman, ∞→1 (d) Balmer, ∞→2
Ans. (b) : The longest wavelength means that lowest [JEE Main 2019, 11 Jan Shift-I]
energy. We known that relation for wavelength
 1 1  1 1  1 1
1 1  Ans. (b) : = R H  2 − 2  = 107  2 − 
= RH  2 − 2  λ n n 3 ∞
λ  n1 n 2   1 2 

Here n1 = n2 n2 = 3 λ = 9 × 10–7 m
RH is Rydberg constant = 109677 cm–1 ∴ λ = 900 nm
1  1 1  770. In hydrogen spectrum, the series of lines
= 109677  2 − 2  = 15233 appearing in ultra violet region of
λ  (2) (3)  electromagnetic spectrum are called :
Or (a) Balmer lines (b) Lyman lines
1 (c) Pfund lines (d) Brackett lines
λ= = 6.56 × 10–5 cm
15233 Manipal-2017
⇒ 6.56 × 10–7 λ ⇒ 656 nm Ans. (b) : Lyman series was found out in the ultraviolet
767. The wave number of hydrogen atom in Lymen region by Lyman.
series is 82200 cm−1. The electron goes from Balmer series was found out in the visible region by
Balmer.
(a) n3 → n2 (b) n2 → n1
Bracket series was found out in the infrared region by
(c) n4 → n3 (d) None of these Brackett.
UPTU/UPSEE-2013 P fund series was found out in the infrared region by P
Ans. (b) : According to Rydberg formula fund.
−  1 1 
V = R 2 − 2 
 n1 n 2  11. Filling of Orbital’s in atom
Here, R = 109677 = 109678 cm–1
− 771. In a given atom no two electrons can have the
V = 82200 cm–1 same values for all the four quantum numbers.
82200  1 1  This is called
Thus, = −  (a) Hund's Rule
109678  12 n 22 
(b) Aufbau principle
Or (c) Uncertainty principle
3  1  (d) Pauli's Exclusion principle
= 1 − 
4  n 22  NEET-1991
Ans. (d) : In a given atom no two electrons can have the
1 3 1
= 1− = same values for all the four quantum numbers. This is
n 22 4 4 called Pauli's Exclusion principle.

Objective Chemistry Volume-I 248

772. Wave nature of electrons was demonstrated by Ans. (d) : Hund's rule of maximum multiplicity states
(a) Schrodinger that, the pairing of electrons in the orbital's of a
(b) de-Broglie particular subshell (p, d or f) does not take place until
(c) Davisson and Germer all the orbital of as subshell are singly occupied. Thus
all he given statements are not related to Hund's rule.
(d) Heisenberg
775. The scientist who proposed the atomic model
UP CPMT-2004 based one the quantization of energy for the
J & K CET-(1999) first time was
Ans. (c) : (a) Schrodinger He put forward new model (a) Max Planck (b) Neils Bohr
of atom by taking into account the de-Broglie concept (c) de-Broglie (d) Heisenberg
of dual nature matter and Heisenberg's uncertainly J & K CET-(2008)
principle. He described the motion of electron in threo Ans. (b) : Neils Bohr utilized the concept of
dimensional space in the form of mathematical equation quantization of energy (Proposed by max Planck) first
known as Schrödinger wave equation. time to give a new model of atom.
(b) de-Broglie According to him, all the material 776. Who used the quantum theory for the first time
particles possess wave character as well as particle to explain the structure of atom?
character. (a) de Broglie (b) Bohr
h (c) Heisenberg (d) Einstein
λ= (de - Broglie equation) J & K CET-(2005)
mv
Ans. (b) : To overcome the objections of Rutherford's
Water, λ = Wavelength model and to explain the hydrogen spectrum, Bohr
h = Planck's constant proposed a quantum mechanical model of the atom,
m = mass of particle which is based on the quantum theory of radiation and
v = Velocity of particle the classical laws of physics.
(c) Davisson and Germer They carried out experiment 777. Which is not in accordance to aufbau
to show the wave character of electrons. They observed principle?
when a beam of electrons is allowed to fall on the
surface of nickel crystal are received on photographic
plate, a diffraction pattern similar to that of X-rays is
obtained. Since, X-rays are electromagnetic rays, it
means electrons have wave character also.
(d) Heisenberg's uncertainty principle It states that BCECE-2008
position and momentum of an electron cannot be Ans. (c) : Aufball principle states that in the ground
measured accurately and simultaneously. state of an atom, the orbital with lower energy is filled
∴ Correct answer is (c) because wave nature of electron up first before the filling of the orbital's with a higher
is demonstrated by Davission and Germer. energy commences, increasing order of energy of
various orbital's is
773. The observation that the ground state of 1s, 2s, 2p, 3s, 4s, 3d, 3p, 5s etc.
nitrogen atom has 3 unpaired electrons in its
electronic configuration and not otherwise is 778. The Pauli exclusion principle applies to the
following option.
associated with:
(a) H (b) H+
(a) Pauli’s exclusion principle –
(c) H (d) He+
(b) Hund’s rule of maximum multiplicity Manipal-2016
(c) Heisenberg’s uncertainty principle
Ans. (c) : Since H has one electron and H ⊕ has no
(d) Ritz combination principle electron the Pauli exclusion principle does not apply to
JCECE - 2005
them however H has two electrons; hence this principle
Ans. (b) : According to Hund's rule while filling applies on it.
orbital's of a Subshell pairing of electrons does not take 779. Select the correct options according to Hund's
place until all the orbitals of the subshell are singly rule and Pauli exclusion principle.
occupied.
774. Hund’s rule state that
(a) Number of two e– can be in two separate
orbitals Codes :
(b) Number of two e– can be present with similar (a) Both I and II (b) Both II and III
spin in a orbital (c) Both I and III (d) Both III and IV
(c) No e– can exist in d orbital Manipal-2016
(d) None of the above Ans. (d) : Figure given in option (d) follow the Hund's
JIPMER-2019 rule and Pauli exclusion principle.
Objective Chemistry Volume-I 249