2 Atomic Structure

Introduction
(a) The word atom was first introduced by Ostwald (1803 - 1807) in scientific world.
(b) According to him matter is ultimately made up of extremely small indivisible particles called atoms.
(c) It takes part in chemical reactions.
(d) Atom is neither created nor destroyed

Dalton’s Atomic Theory:

Dalton proposed the atomic theory on the basis of the law of conservation of mass and law of definite
proportions. The salient features of this theory are
(a) Each element is composed by extremely small particles called atoms.
(b) Atoms of a particular element are all alike but differ with the atoms of other elements.
(c) Atom of each element is an ultimate particle, and has a characteristic mass but is structureless.
(d) Atom is indestructible i.e. it can neither be destroyed nor created.
(e) Atom of an element takes part in chemical reaction to form molecule.

Discovery of Fundamental particles:

Atom consist of several sub-atomic particles like neutron, proton, electron, neutrino, positron etc. Out
of these particles, electron, proton and the neutron are called fundamental subatomic particles.
(1) Electron
Electron discovered by J.J. Thomson (1897) and it is negatively charged particle.

Cathode Ray Experiment

• William Crookes in 1879 studied the electrical discharge in partially evacuated tubes called as
cathode ray discharge tube.
• Discharge tube is a hollow tube that is made of glass. It has vaccum inside and two electrode.
Negative electrode is called as cathode and positive electrode is called anode.
• When a gas enclosed at low pressure (~10–4 atm) in discharge tube is subjected to a high voltage
(~10,000 V), invisible rays originating from the cathode and producing a greenish glow behind the
perforated anode on the glass wall coated with phosphorescent material ZnS is observed. These
rays

Atomic Structure 33
 Properties:
• Cathode rays travel in straight line.
• It has been observed that these rays can create mechanical effect. During experiment it succeds
to rotate the wheel.
• Cathode rays consist negatively charged particles called as electron.
• Cathode rays travel with high speed.
• Cathode rays can cause fluorescence.
• Cathoderays heat the object on which they fall due to transfer of kinetic energy to the object.
• When cathode rays fall on heavy metals, X-rays produced.
• Cathode rays possess ionizing power that is they ionize the gas through which they pass.
• The cathode rays produce scintillation on the photographic plates.
• They can penetrate through thin metallic sheets

(2) Proton:
• Proton was discovered by Goldstein and it is positively charged particle.

Anode Ray Experiment

A German scientist named Eugen Goldstein performed canal ray experiment hence discovers the
proton. The discovery of proton followed by the discovery of the electron strengthened the structure
of the atom. In the experiment, Goldstein applied high voltage across a discharge tube which had
a perforated cathode. A faint luminous ray was seen extending from the holes in the back of the
cathode

Properties:
• Anode rays travel in straight line.
• Anode rays are material particle.
• Anode rays are positively charged.
• Anode rays may get swing by external magnetic field.
• Anode rays also affect the photographic plate.
• e/m ratio of anode rays is lesser than electrons.
• These anode rays produce flashes of light on ZnS screen.

34 Atomic Structure
(3) Neutron:
• In 1932, Chadwick bombarded Be with a stream of -particles (He+2). He noticed that penetrating
radiations were produced which not affected by magnetic field and electric field. These
radiations consisted of neutral particles, which were called neutrons.
4 Be9 + 2He4 → 6C12 + 0n1

Fundamental Particles Table

Atoms are made up of three fundamental particles. The charge and mass of these fundamental
particles are as follows:

Electron Proton Neutron

Symbol e or –1e or e¯
0
P or 1p 1
n or 0n1
kg 9.109534 × 10–31 1.6726485 × 10–27 1.6749543 × 10–27
amu 5.4858026 × 10–4 1.007276471 1.008665012
Mass

1
Relative 1 1
1837
Actual (in C) 1.6021892 × 10–19 1.6021892 × 10–19 0
Charge
relative –1 +1 0

One unit charge = 4.80298 × 10–10 esu

= 1.60210 × 10–19 coulombs
1
one amu = × mass of 6C12 atom
12
= 1.67 × 10–24 g

Thomson’s Atomic Model

• Thomson was the first to propose a detailed model of the atom.
• Thomson proposed that an atom consists of a uniform sphere of positive charge in which the
electrons are distributed more or less uniformly.
• This model of atom is known as “Plum-Pudding model” or ‘’Raisin Pudding Model” or “Water Melon
Model”

e–

Sphere of
+ve charge

Drawbacks:
• An important drawback of this model is that the mass of the atoms is considered to be evenly
spread that atom.
• It is a static model. It does not reflect the movement of electron.
• It couldn't explain the stability of an atom.

Atomic Structure 35
 Rutherford’s Atomic Model
Rutheford Gold Foil Experiment:
Rutherford carried out experiment on the bombardment of atoms by high speed positively charged -
particles emitted from radium and gave the following observations, which was based on his experiment.
Very few Thin gold Few
Radioactive
substance foil

Most
Lead block
Lead plate
with hole Few

Observation:
(a) Most of the  - particles (nearly 99%) continued with their straight path.
(b) Some of the  - particles passed very close to the centre of the atom and deflected by small
angles.
(c) Very few particles thrown back (180°) .
Atom of metal foil
Few
Beam of –particles Majority
 of –rays

Very Few
Majority
of –rays
Few
Atom of metal foil
Atomic Model:
(a) Most of the  - particles were continued their straight path that means most of the space of the
atom is empty.
(b) The centre of an atom has a positively charged body called nucleus which repel positively charged
 - particles and thus explained the scattering phenomenon.
(c) Whole mass of an atom is concentrated in its nucleus and very few throw back means the size of
the nucleus is very small 10–13 cm. It showed that the nucleus is 10–5 times small in size as compared
to the total size of atom.

Applications of Rutherford Model:

On the basis of scattering experiments, Rutherford proposed model of the atom, which is known as
nuclear atomic model. According to this model -
(i) An atom consists of a heavy positively charged nucleus where all the protons and neutrons are
present. Protons & neutrons are collectively reffered to as nucleons. Almost whole of the mass of
the atom is contributed by these nucleon. The magnitude of the +ve charge on the nucleus is
different for different atoms.
(ii) The volume of the nucleus is very small and is only a minute fraction of the total volume of the
atom. Nucleus has a diameter of the order of 10–12 to 10–13 cm and the atom has a diameter of the
order of 10–8 cm

36 Atomic Structure
 DA Diameter of theatom 10−8
= = = 105 , DA = 105 DN
DN Diameter of thenucleus 10−13
Thus diameter (size) of the atom is 105 times the diameter of the nucleus.
• The radius of a nucleus is proportional to the cube root of the number of nucleons within it.
R  A1/3  R = R0A1/3 cm
Where R0 = 1.33  10-13(a constant) and A = mass number (p + n) and R = radius of the nucleus.

(iii) There is an empty space around the nucleus called extra nuclear part. In this part electrons are
present. The number of electrons in an atom is always equal to number of protons present in the
nucleus. As the nuclear part of atom is responsible for the mass of the atom, the extra nuclear
part is responsible for its volume. The volume of the atom is about 1015 times the volume of the
nucleus.
volumeof theatom (10–8 )3
= = 1015
volumeof thenucleus (10–13 )3

(iv) Electrons revolve around the nucleus in closed orbits with high speeds. The centrifugal force acting
on the revolving e- is being counter balanced by the force of attraction between the electrons and
the nucleus.
• This model was similar to the solar system, the nucleus representing the sun and revolving
electrons as planets.

Drawbacks of Rutherford Model:

(1) This theory could not explain stability an atom. According to Maxwell electron loses it energy
continuously in the form of electromagnetic radiations. As a result of this, the e– should loss energy
at every turn and move closer and closer to the nucleus following a spiral path. The ultimate result
will be that it will fall into the nucleus, thereby making the atom unstable.
Nucleus
+

–
e
(2) If the electrons loses energy continuously, the observed spectrum should be continuous but the
actual observed spectrum consists of well defined lines of definite frequencies. Hence, the loss of
energy by electron is not continuous in an atom.

Some Important Definitions

(A) Mass Number: It is represented by capital A. The sum of number of Neutrons and protons is called
the mass Number of the element. i.e. A = number of protons + number of Neutrons
(B) Atomic Number: It is represented by Z. The number of protons present in the Nucleus is called
atomic number of an element. It is also known as nuclear charge.
• For neutral atom: Number of proton = Number of electron
• For charged atom: Number of e– = Z – (charge on atom), Z= number of protons only

Atomic Structure 37
 Example 17 Cl35 → n = 18, p = 17, e = 17
Two different elements can not have the same Atomic Number
Number of Neutrons = Mass number – Atomic number
= A – Z = (p + n) – p = n
Representation of element → ZXA (where X→ symbol of element)

(C) Isotopes: Given by Soddy, are the atoms of a given element which have the same atomic number
but different mass number i.e. They have same Nuclear charge but different number of Neutrons.

Example 1. 17 Cl37 17 Cl37 Example 2. C12

6 C13
6 C14
6

n = 18 n = 20 e=6 e=6 e=6

e = 17 e = 17 p=6 p=6 p=6
p = 17 p = 17 n=6 n=7 n=8

Example 3. (Proteium Dueterium Tritium)

1 H 1
1 H 2
1 H3
e=1 e=1 e=1
p=1 p=1 p=1
n=0 n=1 n=2

1 H1 is the only normal hydrogen which have n = 0 i.e. no nuetrons

Duetrium is also called as heavy hydrogen. It represent by D
• Isotopes have same chemical property but different physical property.
• Isotopes do not have the same value of e/m.

(D) Atomic Weight: The atomic weight of an element is the average of mass of all the isotopes of that
element.
• An element have three isotopes y1, y2 and y3 and their isotopic weights are w1, w2, w3 and their
percentage/possibility/probability/ratio of occurrence in nature are x1, x2, x3 respectively, then
the average atomic weight of element is
w 1x 1 + w2x2 + w 3x3
Average atomic weight =
x 1 + x2 + x 3
Eg. Cl35 Cl37
Probability 75% 25%
Ratio 3 : 1

Average atomic weight =

(35  3) + (37  1) 142
= = 35.5
3+1 4
(E) Isobars: Given by Aston, isobars are the atoms of different element which have the same mass
number but different Atomic number i.e They have different number of Electron, Protons &
Neutrons But sum of number of neutrons & Protons remains same.

38 Atomic Structure
 H3 
1 2
He3  19
K 40  20
Ca 40 
   
p = 1 p = 2 p = 19 p = 20 
Example 1. Example 2.
e = 1 e = 2 n = 21  n = 20 
   
n = 2 n = 1  e = 19  e = 20 

p+n=3 p+n=3 n+p = 40 n + p = 40

• Isobars do not have the same chemical & physical property.

(F) Isodiaphers: They are the atoms of different element which have the same difference of the number
of Neutrons & protons.
Example 1. 5 B 11
6 C13 Example 2. 7 N15 9 F 19

p=5 p=6 p=7 p=9

n=6 n=7 n=8 n = 10
e=5 e=6 e=7 e=9
n – p =1 n – p =1 n – p =1 n – p =1

(G) Isotones/ Isonuetronic Species / Isotonic: They are the atoms of different element which have the
same number of neutrons.
Example 1. 1 H3 2He4 Example 2. 19 K 39
20 Ca40
p=1 p=2 e = 19 e = 20
n=2 n=2 p = 19 p = 20
e=1 e=2 n = 20 n = 20
(H) Isosters: They are the molecules which have the same number of atoms & electrons.
Ex.1 CO2 N2O Ex.2 CaO KF
Atoms = 1 + 2 Atoms = 2 + 1 Atoms = 2 Atoms = 2
=3 =3
Electrons Electrons Electrons = Electrons =
= 6 +8 × 2 =7×2+8 20 + 8 = 28 e¯ 19 + 9 = 28 e¯
= 22 e¯ = 22 e¯

(I) Isoelectronic Species: They are the atoms, molecules or ions which have the same number of
electrons.
Ex.1 Cl¯ Ar Ex.2 H2O NH3 Ex.3 BF3 SO2
Electron 18 e¯ 18 e¯ Electron 2+8 7+3 Electron 5+9×3 16 + 8 × 2
= 10 e¯ = 10 e¯ = 5 + 27 16 + 16
= 32 e¯ = 32 e¯

Concept Builder (1)

1. If S1 be the specific charge (e/m) of cathode rays and S2 be that of positive rays, then which is
true ?
(1) S1 = S2 (2) S1 < S2 (3) S1 > S2 (4) Either of these

Atomic Structure 39
 2. An increasing order (lowest first) for the values of e/m for electron (e), proton (p), neutron (n)
and alpha () particle is:
(1) e, p, n,  (2) n, , p, e (3) n, p, e,  (4) n, p, , e

3. Select iso electronic set:-

(a) Na+, H3O+, NH4+ (b) CO3–2, NO3–, HCO3–
(c) P–3, HCl, C2H6, PH3 (d) N–3, O–2, F
(1) a, b, d (2) b, c, d (3) a, b, c, d (4) a, b, c

4. The ratio of specific charge of a proton and an –particle is:–

(1) 2: 1 (2) 1: 2 (3) 1: 4 (4) 1: 1

5. An isotone of Ge76 is:-

32

(i) 32 Ge77 (ii) As77

33 (iii) 34 Se77 (iv) 34 Se78
(1) (ii) & (iii) (2) (i) & (ii) (3) (ii) & (iv) (4) (ii) & (iii) & (iv)

6. 13
C6 and 12C6 and differ from each in respect of number of
(1) electrons (2) protons (3) neutrons (4) none of these

7. Atomic weight of Ne is 20.2. Ne is mixture of Ne 20 and Ne22, Relative abundance of heavier

isotope is:-
(1) 90 (2) 20 (3) 40 (4) 10

Wave
• A wave motion is a mean of transfer of energy from one point to another point without any
conveying of matter between the points.
• When we throw the piece of stone particles on water surface in a pond, we observe circles of ever
increasing radius, till these strike on the wall of the pond.
• When we put piece of cork on the surface of this water, we observe that the cork moves up and
down as the wave passes, but the piece does not travel along with the waves.

A Wave is Characterized by Six Characterstics

(1) Wavelength ()
It is defined as distance between two nearest crest or nearest through. It is measured in term of a
Å (Angstrom), pm (Picometer), nm (nanometer), cm (centimeter), m (meter)
1Å = 10–10 m, 1Pm = 10–12 m
1nm = 10–9m, 1cm = 10–2 m

(2) Frequency (n)

Frequency of a wave is a number of waves which pass through a point in 1 sec. it is measured in
term of Hertz (Hz), sec–1, or cycle per second (cps)
1 Heartz = 1 sec–1 = 1 cps

40 Atomic Structure
(3) Time period (T)
Time taken for one complete oscillation of wave is known as period (T). Time taken by wave to

travel a distance equal to one wavelength. If C is the speed of wave, then C =
T

(4) Wave Number ( n )

Number of wavelength per unit length.
1
v=


(5) Amplitude (A)

It is the height of depth or crest of a through of a wave.

(6) Velocity (C)

It is defined as distance covered by a wave in 1 sec.
C
=


Electromagnetic Waves (EMW)

• It contains electric and magnetic field.
• Energy is always transferred in the form of waves with the speed of light (3 × 108 m/s).
• It’s a pure energy waves.
• It does not contain mass no medium is required for transmission.
• Direction of propagation is perpendicular from both electric field and magnetic field.
• There are various types of electromagnetic waves (radiation) which differs from one another in
wavelengths.
Example : Cosmic Rays, g-rays , X-rays, U.V, visible, I.R, Micro, Radio.

Electromagnetic Spectrum
It is defined as the array of various types of electromagnetic radiations in order of their increasing (or
decreasing) wavelengths or frequencies.

Atomic Structure 41
 Maxwell Theory of Electromagnetic Wave
• All the radiations have wave nature which explains interference (linear superposition) and
diffraction.
• They are composed of oscillating electric and magnetic field which are perpendicular to each other
and to the direction of propagation.
• All the radiations (radio waves, micro waves, infra red waves, visible, UV, X-rays, g-rays) travel at
the speed of light in vaccum.
• Energy of electromagnetic wave is proportional to amplitude and not linked with frequence of
waves.

Limitations of Maxwell Theory of Electromagnetic

Wave
• Phenomenon of black body radiations.
• Photoelectric effect.
• Line spectra of atoms

The Electromagnetic Spectrum

Planck's Quantum Theory

According to Planck's Quantum Theory:
1. The radiant energy emitted or absorbed by a body not continuously but discontinuously in the form
of small discrete packets of energy and these packets are called quantum.
2. In case of light, the smallest packet of energy is called as 'photon' but in general case the smallest
packet of energy called as quantum.

42 Atomic Structure
3. The energy of each quantum is directly proportional to frequency of the radiation i.e.

hc  c
E   E = h or E=  = 
  
h is proportionality constant or Planck's constant
h = 6.626 × 10–37 kJ s or 6.626 × 10–34 J s or 6.626 × 10–27 erg s

4. Total amount of energy transmitted from one body to another will be some integral multiple of
energy of a quantum. E = nh Where n is an integer and n = number of quantum

nhc
E = nh = = nhc

Bohr’s Atomic Model
Bohr developed a model for hydrogen and hydrogen like atoms one-electron species (hydrogenic
species). He applied quantum theory in considering the energy of an electron bond to the nucleus.

Important Postulates of Bohr’s Model:

(1) Electrons revolve around the nucleus along certain circular paths known as “Orbit” or “Shells”

K, L, M, ................. are shells

K → 1st orbit
L → 2nd orbit
M → 3rd orbit
(2) Electron is associated with a fixed energy in a particular orbit. The change in electronic energy is
possible only when; electron changes its orbit number.
(a) When an electron goes to higher orbit from lower orbit, then energy is absorbed.

E2 – E1 = E (Absorbed energy)

(b) If an electron jumps to lower orbit from higher orbit, then energy is radiated.

E2 – E1 = E (Emission of energy)

Thus E = Ehigher – Elower = En – En = Energy of a quantum = h = Bohr's frequency condition

2 1

Atomic Structure 43
 (3) The electrostatic force of attraction acting between the electron and the nucleus is counterbalance
by the centrifugal force
mV 2 Ze2
=K 2
r r

Ze2
or mv2 = K
r

h
(4) Electrons can revolve only in those orbits in which angular momentum is integral multiple of
2

h
mvr = n .
2
where, h = Planck’s constant = 6.62 × 10–34 J-s
m = mass of electron
v = Electronic velocity
n =orbit number
Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or absorb only those
quanta which can take it to a certain higher energy level i.e., all those photons having energy less than
or more than a particular energy level will not be absorbed by hydrogen atom. But if energy supplied
to hydrogen atom is more than 13.6 eV then all photons are absorbed and excess energy appear as
kinetic energy of emitted photo electron.

Applications of Bohr's Model

1. Radius of Various Orbits (Shell):
n2h2
r=
4 mKZe2
2

Derivation:
Kq1q2 K.Ze.e KZe2
Columbic force = = =
r2 r2 r2
(Tangential
velocity)

r
+ Ze e–
Coulabic or
Nucleus Electrostatic
force Kq1q2
( )
r2

(Where K is Constant) K = 9 × 109 Nm2 /coulumb2

44 Atomic Structure
 As we know that Coulombic force = Centrifugal force
KZe2 mv 2 KZe2
= or v 2 = ....(1)
r2 r mr
nh nh
As we know that mvr = or v = ....(2)
2 2mr
Now putting the value of v from eqn.(2) to eqn.(1)
2
 nh  KZe2 n2h2 KZe2
  = or =
 2mr  mr 42m2r2 mr

n2h2
r= ....(3)
42mKZe2

Putting the value of , h, m, K, & e (Constants) in the above eqn. (3)

n2
r = 0.529 × 10–8 × cm { 1Å = 10–10m = 10–8 cm}
Z

n2
rn = 0.529  Å
Z
This formula is only applicable for hydrogen and hydrogen like species i.e. species contains single
electron.

2. Velocity of an Electron:
KZe2 mv 2
Since Coulombic force = Centrifugal force or = or KZe2 = (mvr)(v)
r 2
r
nh nh
now putting the value of Angular momentum m.v.r. =  KZe2 = (v)
2 2

2KZe2
v=
nh

Z
now putting the value of  k, e & h v = 2.188  106  ms−1
n
3. Energy of an Electron:
Let the total energy of an electron be E. It is the sum of kinetic and potential Energy.
i.e. E = K.E.+ P.E.
1   Kq q   Kze2 
E =  mv 2  +  1 2  P.E. = − 
2   r   r 

1 K.Ze.( −e) 1 KZe2

E= mv 2 + = mv 2 −
2 r 2 r
now putting the value of mv2 from eq. (1)
KZe2 KZe2 KZe2
E= − =–
2r r 2r
now putting the value of r from eq. (3)

Atomic Structure 45
 KZe2  42mKZe2 22m  K2Z2e4
En = − or En = −
2n2h2 n2h2
now putting the value of , K, e, m, h, we get:

21.8  10−19  Z2 Z2
En = − J / atom or En = −13.6  eV / atom
n2 n2
This formula is applicable for hydrogen atom & hydrogen like species i.e. single electron species.
Since n can have only integral values, it follows that total energy of the e – is quantised.
The –ve sign indicats that the e¯ is under attraction towards nucleus.
Some Extra Points:
KZe2 1
(i) K.E = i.e., K.E. 
2r r
On increasing radius, K.E. decreases.
KZe2 1
(ii) P. E. = – i.e. P.E.  –
r r
On increasing radius, P.E. increases.
KZe2 1
(iii) E = – i.e. E.  –
2r r
On increasing radius, total energy increases.

P.E =– 2KE KE =– E P.E = 2E

Energy difference between two energy levels:

 1 1
= – 13.6 × Z2  2 − 2 
 n2 n1 
Energy level for H atom can be represented as follows:
Shell O E5
Shell N
n = 6 or P E6 = – 0.38 eV Shell M
E4
Shell L E3
n = 5 or O E5 = –0.54 eV
Shell K E2
n = 4 or N E4 = – 0.85 eV E5 – E4 = 0.36 eV E1
Nucleus +
n = 3 or M E3 = – 1.51 eV E4 – E3 = 0.66 eV
Shell 1
n = 2 or L E2 = – 3.4 eV E3 – E2 = 1.89 eV
Shell 2
n = 1 or K E1 = – 13.6 eV E2 – E1 = 10.2 eV Shell 3
Shell 4
i.e. (E2 – E1) > (E3 – E2) > (E4 – E3) > (E5 – E4)..... Shell 5

Energy Definitions
(a) Ground State (G. S.):
The lowest energy state of an atom or ion or molecule
G. S. for H-atom n=1
G. S. for He - ion
+
n=1

(b) Excited State (E. S.):

The energy states above the ground state are referred to as an excited state
n=2 1st E. S.

46 Atomic Structure
 n=3 2nd E. S.
n=4 3rd E. S.
Total E. S. = (n – 1)

(c) Excitation Potential (E. P.):

The amount of energy needed to promote an electron from ground state to any excited state
1st E. P. = E2 – E1 = –3.4 – (13.6) = 10.2 eV
2nd E. P. = E3 – E1 = –1.51 – (–13.6) = 12.09

(d) Ionization Potential (I. P..)/Energy/Enthalpy:

Corresponding energy required to remove the electron from G.S. to the infinite excited state.
I. P. = E – E 1
= 0 – (–13.6) = 13.6 eV

(e) Separation Energy (S. E.):

The amount of energy needed to remove the electron to infinity from the excited state.
Ist S. E. = E – E2 = 0 – (–3.4) = 3.4 eV
Also 2nd S. E. = E – E3 = 0 – (–1.51) = 1.51 eV

Concept Builder (2)

1. If the radii of first orbits of H, He+, Li+2 and Be+3 are r1, r2, r3 and r4 respectively, then their correct
decreasing order will be:
(1) r1 > r2 > r3 > r4 (2) r3 < r2 > r4 < r1
(3) r1 < r2 < r3 > r4 (4) Radius of all are equal

x
2. An electron in an atom jumps in such a way that its kinetic energy changes from x to . The
4
change in potential energy will be:
3 3 3 3
(1) + x (2) − x (3) + x (4) − x
2 8 4 4

3. The distance between 4th and 3rd Bohr orbits of He+ is:
(1) 2.645 × 10–10m (2) 1.322 × 10–10m (3) 1.851 × 10–10 m (4) None

4. What atomic number of an element “X” would have to become so that the 4 th orbit around X
would fit inside the 1st Bohr orbit of H atom?
(1) 3 (2) 4 (3) 16 (4) 25

5. The angular momentum of electron of H-atom is proportional to:

1 1
(1) r2 (2) (3) r (4)
r r

Atomic Structure 47
 Spectrum
When a radiation is passed through a spectroscope (prism) for the dispersion of the radiation, the
pattern (photograph) obtained on the screen (photographic plate) is called as spectrum of the given
radiation
Classification of Spectrum

(1) Emission (1) Absorption

Hydrogen Line Spectrum or Hydrogen Spectrum

When an electric excitation is applied on hydrogen atomic gas at Low pressure,a bluish light is emitted.
when a ray of this light is passed through a prism, a spectrum of several isolated sharp line is obtained.
The wavelength of various lines show that spectrum lines lie in visible, Ultraviolet and Infra red region.
These lines are grouped into different series.

–0.28 7 Q
0.10
–0.38 6 Far I.R. region P
0.16 Humphery series
–0.54 5 O
I.R. region
0.31
P-fund series
–0.85 4 N
I.R. region
0.66 Bracket series
–1.51 3 M
Infra Red region
1.89 or

12.09
12.75

13.06
Paschen series
–3.4 2 L
Visible region
10.2 or

10.2
Balmer series
–13.6 1 K
Ultra violet region
or
Lyman series

Series Discovered by regions n2 → n1

lyman lyman U.V. region n2 =2, 3, 4, ... / n1=1
Balmer Balmer Visible region n2 = 3,4,5, ... / n1 =2
Paschen Paschen Infra red (I. R.) n2 =4, 5, 6, ... / n1 =3
Bracket Bracket I.R. region n2 =5, 6, 7, ... / n1 =4

Pfund Pfund I.R. region n2 =6, 7, ... / n1 = 5

Humphery Humphery far I.R. region n2 = 7, ... / n1 = 6

48 Atomic Structure
Similar Words:
• First line / Starting line / Initial line (max. and min)
• Last line / limiting line / marginal line (min and max.)
• First line of any series =  line
Second line of any series =  line
Third line of any series =  line

Rydberg Formula:
In 1890, Rydberg gave a very simplest theoretical Equation for the calculation of the wavelength of
various lines of hydrogen like spectrum and the equation is
1  1 1
= = RZ2  2 − 2  where R = Rydberg constant = 109678 cm–1 109700 cm–1
  n1 n2 
1
= 9.12 × 10–6 cm = 912 Å
R
n1 and n2 are orbits and for a particular series n1 is constant and n2 varies.
for Lyman n1 = 1, n2 = 2, 3, 4,.....
for Balmer n1 = 2, n2 = 3, 4, 5,....
for Paschen n1 = 3, n2 = 4, 5, 6,....
for Bracket n1 = 4, n2 = 5, 6, 7,...
for Pfund n1 = 5, n2 = 6, 7, 8,...
for Humphery n1 = 6, n2 = 7, 8, 9,...
Derivations of Rydberg Formula:
 E = En − En
2 1

−22mK2Z2e4  −22mK2Z2e4 
E= − 
n22h2  n21 h2 

22mK2Z2e4 22mK2Z2e4  hc 
= −  E = h = 
2 2
nh
1
n22h2   

hc 22mK2Z2e4  1 1 1 22mK2e4Z2  1 1
=  2 − 2 or =  2 − 2
 h2  n1 n2   ch3  n1 n2 

22mK2e4
where is a constant which is equal to rydberg constant (R).
ch3
1  1 1
= RZ2  2 − 2 
  n1 n2 

Calculation of Number of Spectral Lines:

(n2 –n1 )(n2 − n1 + 1)
(a) Total number of spectral lines = 1 + 2 + ..... (n2 – n1) =
2
if n1 = 1(ground state)

Total number of spectral lines =

(n 2
− 1) n2
=
n ( n − 1)
2 2
(b) Number of spectral lines which falls in a particular series (n2 – n1)
where n2 = higher energy series, n1 = lower energy series

Atomic Structure 49
 Limitation of The Bohr's Model:
 Bohr's theory does not explain the spectrum of multi electron atom.
nh
2. Why the Angular momentum of the revolving electron is equal to , has not been explained by
2
Bohr's theory.
3. Bohr interrelate quantum theory of radiation and classical law of physics without any theoretical
explanation.
4. Bohr's theory does not explain the fine structure of the spectral lines. Fine structure of the spectral
line is obtained when spectrum is viewed by spectroscope of more resolution power.
5. Bohr theory does not explain the spiliting of spectral lines in the presence of magnetic field
(Zemman's effect) or electric field (Stark's effect)

Wave Mechanical Model of an Atom

Two experiments:
1. de-Brogle concept (Dual nature of Matter)
2. Heisenberg's Uncertainty principle.

1. The Dual Nature of Matter (The Wave Nature of Electron):

(i) In 1924, a French physicist, Louis de-Broglie suggested that if the nature of light is both that of
a particle and of a wave, then this dual behavior should be true for the matter also.
(ii) The wave nature of light rays and X-rays is proved on the basis of their interference and
diffraction and many facts related to radiations can only be explained when the beam of light
rays is regarded as composed of energy corpuscles or photons whose velocity is 3 × 10 10 cm/s.
(iii) According to de-Broglie, the wavelength  of an electron is inversely proportional to its
momentum p.
1 h
 or = (Here h = Planck's constant, p = momentum of electron)
p p

h
Momentum (p) = Mass (m) × Velocity (v)   =
mv

(iv) The above relation can be confined as follows by using Einstein's equation, Planck's quantum
theory and wave theory of light.
Einstein's equation, E = mc2 where E is energy, m is mass of a body and c is its velocity.
 E = h (According to Planck's quantum theory)

and c =  (According to wave theory of light)

c c
 =  E=h×
 
But according to Einstein's equation E = mc2
c h h h
E = mc2 = h × or mc = or p = or  =
   p
(v) It is clear from the above equation that the value of  decreases on increasing either m or v or
both. The wavelength of many fast-moving objects like an aeroplane or a cricket ball, is very
low because of their high mass.

50 Atomic Structure
 Bohr's Theory and de-Broglie Concept:
(i) According to de-Broglie, the nature of an electron moving around the nucleus is like a wave
that flows in circular orbits around the nucleus.
(ii) If an electron is regarded as a wave, the quantum condition as given by Bohr in his theory is
readily fulfilled.
(iii) If the radius of a circular orbit is r, then its circumference will be 2r.
nh
(iv) We know that according to Bohr theory, mvr =
2
nh nh  h 
or 2r = (mv = p momentum) or 2r =  =  de-Broglie equation 
mv p  p 

 2r=n (where n = total number of waves 1, 2, 3, 4, 5, ..... and  = Wavelength

nh nh
(v)  2r = or mvr = mvr = Angular momentum
mv 2

h
Thus mvr = Angular momentum, which is a integral multiple of .
2
(vi) It is clear from the above description that according to de-Broglie there is similarity between
wave theory and Bohr theory.

n= 5
n= 6

Figure : Similarity between de-Broglie Waves and Bohr’s orbit

2r
number of waves made by electron in one complete revolution = = n (orbit number)


2. Heisenberg Uncertainty Principle:

In 1927, Werner Heisenberg presented a principle known as Heisenberg Uncertainty principle which
states as: "It is impossible to measure simultaneously the exact position and exact momentum of
a body as small as an electron."
The Uncertainty of measurement of position (x) and the Uncertainty of momentum (p) or (mv),
are related by Heisenberg's relationship as:
h
x. p 
4
h
or x .  (mv) >
4

h
or x. v 
4m

Atomic Structure 51
 p h
or t × x × 
t 4
h
or F × t × x 
4
h
or E. t  (for energy and time)
4
(where h is Planck's constant.)
For an electron of mass m(9.10 × 10–28 g), the product of Uncertainty is quite large.
When x = 0, v =  and vice-versa.
In the case of bigger particles (having considerable mass), the value of Uncertainty product is
negligible. If the position is known quite accurately, i.e., x is very small, v becomes large and
vice-versa.
On the basis of this principle, therefore, Bohr's picture of an electron in an atom, which gives a
fixed position in a fixed orbit best we can think of in terms of probability of locating an electron
with a probable velocity in a given region of space at a given time. The space or a three-dimensional
region round the nucleus where there is maximum probability of finding an electron of a specific
energy is called an atomic orbital.

Golden Key Points

• de-Broglie wavelength in terms of kinetic energy.

1 1 2 2
Kinetic Energy (K.E.) = mv2 or m × K.E. = m v or m2v2 = 2m K.E. or mv = 2m K.E.
2 2

But  =
h
mv
 =
h
2m K.E.
( mv = 2m K.E. )
• When a charged particle carrying Q coulomb is accelerated by applying potential difference V then
K.E. = Q×V Joule

h h  150  12.25
But =  = For electron   = Å = Å
2m K.E. 2m QV  V  V

• The wave nature of electron was verified experimentally by Davisson and Germer.
• de-Broglie hypothesis is applicable to macroscopic as well as microscopic objects but it has no
physical significance for macroscopic objects.

h
• Remember = 0.527  10–34 J sec
4

52 Atomic Structure
 Concept Builder (3)

1. What electronic transition in Li2+ produces the radiation of the same wavelength as the first line
in the Lyman series of hydrogen?
(1) n = 4 to n = 2 (2) n = 9 to n = 6
(3) n= 9 to n = 3 (4) n = 6 to n = 3
2. The first Lyman transition in the hydrogen spectrum has E = 10.2 eV. The same energy change
is observed in the second Balmer transition of:
(1) Li2+ (2) Li+ (3) He+ (4) Be3+

3. A hydrogen atom in the ground state is excited by monochromatic radiation of wavelength  Å.

The resulting spectrum consists of maximum 15 different lines. What is the wavelength ?
(RH = 109737 cm–1)
(1) 937.3 Å (2) 1025 Å
(3) 1236 Å (4) None of these

4. Which of the following series of lines in the atomic emission spectrum of hydrogen is in the
visible region?
(1) Lyman (2) Paschen
(3) Balmer (4) Brackett

Quantum Mechanical Model of Atom

• In 1926 Erwin Schrodinger was developed this model.
• This atomic model is based on wave and particle nature of electron.
• This model describes that in electronic field of (+ve)ely charged nucleus the electron has 3D wave.
• Wave motion of an electron was explained by the equation given by Schrodinger. The differential
equation is:

d2  d2  d2  82m
+ + + (E – V)  = 0
dx2 dy 2 dz2 h2

where x, y and z are cartesian co-ordinates of the electron

m = mass of the electron
E = total energy of electron
V = potential energy of electro
h = planck's constant
 = wave function of the electron

Significance of 
Coordinations x, y and z expresses amplitude function which is also termed as wave function.
Depending upon the coordinates values, wave function can have positive or negative values.

Atomic Structure 53
 Significance of 2
2 is a probability factor. It discribes the probability for finding an electron within a small space. The
space where there is a maximum probability for finding an electron is termed as orbital.
Variations of Radial Wave Function (R)
(i) Plots of radial wave function R against the distance r:
The variation of the radial part of the orbital wave funcitons for 1s, 2s and 2p orbitals. The radial
funcitn value changes form positive to zero then to negative. The region where this function reduces
to zero is called nodal surfaces or simply nodes.

(ii) Radial probability density (R2):

Square of radial wave function, R2, is the measure of the probability of finding the electron in a unit
volume around a particular point and is called probability density.

(iii) Radial distribution function, (4r2R2):

Probability density of finding electron at a point at a distance r form the nucleus. It is more useful
for discussing the probability for finding the electron in a spherical shell between the spheres of
radii r + dr and r as the atoms have spherical symmetry.

Quantum Numbers:
As we know to search a particular person in this world 4 things are needed:-
• Country to which the person belongs.
• The city in that country where the person is residing
• The area in that city
• House number

54 Atomic Structure
 Similarly to locate the position of an electron in the atom 4 identification number are required and
these identification number are called as quantum number.
1. Principal quantum number (n) → Shell (Orbit)
2. Azimuthal quantum number (l) → Sub shell
3. Magnetic quantum number (m) → Orbital
4. Spin quantum number (S) → Spin of e–

1. Principal Quantum Number (n): Given By → Bohr's

It represents the name, size and energy of the shell to which e – belongs
• The value of n lies between 1 to 
i.e n = 1,2,3,4– – – – –  corresponding name of shells are K, L, M, N, O, – – – – –
• Greater the value of n, greater is the distance from the nucleus.
n2
r = 0.529 × Å
z
r 1 < r2 < r 3 < r 4 < r 5 – – – – – – – –
• Greater the value of n, greater is the energy of shell
z2
E = – 13.6 × eV/atom
n2
E 1 < E2 < E 3 < E 4 – – – – – – – –
nh
• The angular momentum of a revolving electron is mvr =
2
Where n = Principal quantum number.
• The number of electrons in a particular shell is equal to 2n2

2. Azimuthal Quantum Number / Angular Quantum Number / Secondary Quantum Number /Subsidiary
Quantum Number (l):
Represented by '' (Given by – Sommerfeld)

• It represents the shape of the subshell and orbital and orbital angular momentum
• Value of  between 0 to (n – 1)

i.e.  = 0,1,2 – – – – – – – – – (n–1)

 = 0(s Subshell)

  = 1(p Subshell)

 = 2(d Subshell)

 = 3(f Subshell)

Example: If n = 1 then  = 0  1s i.e. in n =1 shell, only one subshell 's' is present.

If n = 2 then  = 0,1  2s,2p i.e. in n =2 shell, two subshell 's' & 'p' are present.

If n = 3 then  = 0,1,2 3s, 3p, 3d i.e. in n =3 shell, three subshell 's' , 'p' & 'd' are present.

If n = 4 then  = 0,1,2,3  4s,4p, 4d, 4f i.e. in n =4 shell, four subshell 's' , 'p' , 'd' & 'f' are present.

Atomic Structure 55
 • If the value of n is same then the order of energy of the various subshell will be
s<p<d<f
Example: 4s < 4p < 4d < 4 f, 3s < 3p< 3d, 2s < 2p
• If Value of is same but value of n is different then the order of energy will be.

Example: 1s < 2s < 3s < 4s < 5s < 6s

3d < 4d < 5d < 6d
4p < 5p <6p

h  h
• The orbital angular momentum = ( + 1)
2
or ( + 1) ħ  =  { is called as 'hash' }
2 

Orbital angular momentum: For s subshell = 0
h
For p Subshell = 2 or 2
2
• The number of electron in a particular subshell is equal to 2(2+1)

for s subshell number of electrons = 2 e—

for p subshell number of electrons = 6 e—
for d subshell number of electrons = 10 e—
for f subshell number of electrons = 14 e—
• Shape of the subshell: s → spherical
p → dumb bell shape
d → double dumb bell shape
f → complex shape

3. Magnetic Quantum Number /Orientation Quantum Number (m): given by → linde

It represents the shape of different orbitals and the orientation of electron cloud (orbital)
• Under the influence of magnetic field each subshell is further subdivided into orbitals ( The
electron cloud is known as orbital)
Magnetic quantum number describe these different distribution of electron cloud.
• Value of m = all integral value from –  to + including zero.

i.e. Value of m = –→ 0 → + 

Case-I
If = 0 then m = 0 it implies that s subshall has only one orbital called as s orbital.

Shapes of s-Orbitals:
The s-orbitals are spherically symmetrical about the nucleus, i.e., the probability of finding s electron
is same in all directions from the nucleus. The size of the orbital depends on the value of principal
quantum number, there is one spherically symmetrical orbital. The 1s orbital is smaller than 2s-orbital
and 2s-orbital is smaller than 3s, but all are spherical in shape as shown in figure.

56 Atomic Structure
 2S 3S
1S Node Node
Although the s-orbitals belonging to different shells are spherically symmetrical, yet they differ in
certain respects as explained below:
• The probability of 1s electron is found to be maximum near the nucleus and decreases as the
distance from the nucleus increases. In case of 2s electrons, the probability is again maximum near
the nucleus and then decreases to zero and increases again and then decreases as the distance
from the nucleus from the nucleus increases. The intermediate region (a spherical shell) where the
probability is zero is called a nodal surface of simply node. Thus, 2s-orbital differs from 1s-orbital
in having one node within it. Similarly, 3s has two nodes. in general, any ns orbital has (n -1) nodes.
• The size and energy of the s-orbital increases as the principal quantum number inreases, i.e., the
size and energy of s-orbital increases in the order
1s < 2s < 3s ....
The s orbital of higher energy levels are also symmetrically spherical and can be represented as
follows:
y
Nodal
suface

1s
z 2s
3s
Case-II
+1 0 –1
If  = 1 ( p - subshell) then m =
px pz py

It implies that, p subshell have three orbitals called as px, py and pz.

Shape of p-Orbitals:
There are three p-orbitals, commonly
referred to as px, py and pz. These three p-orbitals, posses equivalent energy and therefore, have same
relation with the nucleus. They, however, differ in their direction & distribution of the charge.

y y y
z z z

x x x

px py pz

Atomic Structure 57
 These three p-orbitals are situated at right angle to another and are directed along x, y and z axes
(figure)
• Each p orbital has dumb bell shape ( 2 lobes which are separated from each other by a point of
zero probability called nodal point or node or nucleus).
• The two lobes of each orbital are separated by a plane of zero electron density called nodal plane.
• Each p orbital of higher energy level are also dumb bell shape but they have nodal surface.

Nodal Surface: Orbital Nodal surface Nodal Plane: Orbital Nodal plane 2
px 0 px yz plane
3 px 1 py xz plane
4 px 2 pz xy plane
npx (n – 2)

(Nodal plane)
y yz plane y

xy plane
Nodal y y
surface (Nodal plane)
xz plane
x (Nodal plane)
2Px x x x
z 3Px Px
4Px z z z
Py Pz
Nodal point

Case-III
When  = 2, 'm' has five values – 2, –1, 0, +1, +2. It implies that d subshell of any energy shell has five

orbitals. All the five orbitals are not identical in shape. Four of the d orbitals d xy, dyz, dxz, contain four
lobes while fifth orbital consists of only two lobes. The lobes dxy orbital lie between x and y axes.
Similar is the case for dyz and dxz. Four lobes of orbital are lying along x and y axes while the two lobes
of orbital are lying along z axes and contain a ring of negative charge surrounding the nucleus in xy
plane. Geometry of d orbital is Double Dumb bell
−2 −1 0 +1 +2
m =
dx2 − y2 dzx dz2 dyz dxy

Shape of d–Orbitals:
• It implies that d subshell has 5 orbitals i.e. five electron cloud and can be represented as follows.
y z z y z
x

y x x y
x

x
dx –y
2 2

dxy dyz dxz dz 2

In d-Orbital:
Nodal y
surface

x
3P
3dxy
x
4P
4dxy
z 5P
x
5dxy
x

58 Atomic Structure
 Number of nodal surface = n––1

number of nodal plane = 

     total node = n –– 1 + 

= (n - 1)


• Nodal plane: dxy → xz & yz nodal plane: 2

dxz → xy & zy nodal plane: 2
dzy → dzx & yx nodal plane: 2

dx2 –y2 → 2, nodal plane

dz2 → 0, nodal plane

Note: Orbitals of d subshell are Equivalent in energy.
Representation of The Orbitals:
s
s
s subshell →
p
px py pz
p subshell →
d
dxy dyz dz dxz dx – y
d subshell →
2 2 2

4. Spin Quantam Number (s): (Given by → Gold schmidt)

• It represents the direction of electron Spin around its own axis

1
• for clock wise spin/spin up() electron → 
2

1
• for anticlock wise spin/spin down() electron →
2
h
Spin angular momentum of an e –
= s ( s + 1) . or s ( s + 1)
2
• Each orbital can accomodate 2 electrons with opposite spin or spin paired.
Correct  Spin paired e— Wrong  Spin parallel e—

Electronic Configuration
Filling of electron in different energy subshell is electronic configuration.
Rules for Filling Subshell:
1. (n +) rule

2. Aufbau Principle
3. Hund's Rule of Maximum Multiplicity
4. Pauli's Exclusion Principle

Atomic Structure 59
 1. (n+) rule: According to it the sequence in which various subshell are filled up can also be

determined with the help of ( n +) value for a given subshell.

Principle of (n+) rule: The subshell with lowest (n+) value is filled up first, When two or more

subshell have same (n+) value then the subshell with lowest value of n is filled up first.
In case of H-atom: Energy only depends on principle quantum number
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < ...................
Sub Shell n  n +

1s 1 0 1
2s 2 0 2
2p 2 1 3 (1)
3s 3 0 3 (2)
3p 3 1 4 (1)
4s 4 0 4 (2)
3d 3 2 5 (1)
4p 4 1 5 (2)
5s 5 0 5 (3)
4d 4 2 6 (1)
5p 5 1 6 (2)
6s 6 0 6 (3)
Order: 1s , 2s , 2p , 3s , 3p , 4s , 3d , 4p , 5s , 4d , 5p , 6s , 4f14, 5d10, 6p6,7s2, 5f14, 6d10 , ....
2 2 6 2 6 2 10 6 2 10 6 2

2. Aufbau Principle:
Aufbau is a German word and its meaning ' Building up'
• Aufbau principle gives a sequence in which various subshell are filled up depending on the
relative order of the Energies of various subshell.
• Principle: The subshell with minimum energy is filled up first when this subshell obtained
maximum quota of electrons then the next subshell of higher energy starts filling.
• The sequence in which various subshell are filled are as follows.

1s 2s 3s 4s 5s 6s 7s
Starting
point 2p 3p 4p 5p 6p 7p

3d 4d 5d 6d

4f 5f

1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6,7s2, 5f14, 6d10 , ....

For Example:
1 H → 1s1
2 He → 1s2

60 Atomic Structure
 3 Li → 1s2 , 2s1
4 Be → 1s2, 2s2
5 B → 1s2, 2s2, 2p1
6 C → 1s2 , 2s 2,2p2
7N → 1s2 , 2s2 , 2p3
8 O → 1s2 , 2s2 , 2p4
9 F → 1s2 , 2s2 , 2p5
10 Ne → 1s2 , 2s2 , 2p6
11 Na → 1s2 , 2s2 , 2p6 , 3s1
12 Mg → 1s2 , 2s2 , 2p6 , 3s2
13 A → 1s2 , 2s2 , 2p6 , 3s2 , 3p1

14 Si → 1s2 , 2s2 , 2p6 , 3s2 , 3p2

15 p → 1s2 , 2s2 , 2p6 , 3s2 , 3p3
16 S → 1s2 , 2s2 , 2p6 , 3s2 , 3p4
17 Cl → 1s2 , 2s2 , 2p6 , 3s2 , 3p5
18 Ar → 1s2 , 2s2 , 2p6 , 3s2 , 3p6
19 K → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1
20 Ca → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2
21 Sc → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d1
22 Ti → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d2
23 V → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d3
24 Cr → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1, 3d5 [Exception]
25 Mn → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d5
26 Fe → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d6
27 Co → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d7
28 Ni → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d8
29 Cu → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s1, 3d10 [Exception]
30 Zn → 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d10
Electronic Configuration Can be Written by Following Different Methods:
• Fe
26 → (1) 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 4s2, 3d6
(2) 1s2 , 2s2 , 2p6 , 3s2 , 3p6, 3d6 , 4s2
(3) 1s2 , 2s2p6 , 3s2p6d6 , 4s2
(4) [Ar] 4s2 3d6
2
• Fe → 1s2 2s2 2p6
26 3s2 3p6 3d6 4s
(n–2) (n−1) n
n → Outer most Shell or Ultimate Shell or Valence Shell
In this Shell e– are Called as Valance electron or this is called core charge
(n–1) → Penultimate Shell or core or pre valence Shell
(n–2) → Pre Penultimate Shell
• If we remove the last n Shell (ultimate Shell) then the remaining shell collectivly be called as
Kernal.

Atomic Structure 61
 3. Hund's Rule of Maximum Multiplicity: (Multiplicity: Many of the same kind)
• According to Hund's rule electrons are distributed among the Orbitals of subshell in such a
way as to give maximum number of unpaired electron with parallel spin.
• Thus the Orbital available in the subshell are first filled singly with parallel spin electron before
they begin to pair this means that pairing of electrons occurs with the introduction of second
electron in 's' subshell, fourth electron in 'p' subshell, 6th electron in 'd' Subshell & 8th e– in 'f'
subshell.
Example 5 B→ 1s2 2s2 2p1
  

6 C→ 1s2 2s2 2p2

   

7N→ 1s2 2s2 2p3

    

8 O→ 1s 2
2s 2
2p4
    

9 F→ 1s 2
2s 2
2p5
    

10 Ne → 1s2 2s2 2p6

    

4. Pauli's Exclusion Principle: In 1925 Pauli stated that no two electron in an atom can have same
values of all four quantum numbers.i.e. An orbital can accomodates maximum 2 electrons with
opposite spin.
Example 1. C12 →
6 1s2 2s2 2p2
   

px pz py
n 1 2 2
   0 0 1

m 0 0 – 1, 0 +1
s +½, –½ +½, –½ +½, +½,
Example 2. Cl →
17 1s2 2s2 2p6 3s2 3P5
        

n = 1 2 2 3 3
  = 0 0 1 0 1

m= 0 0 –1, 0, +1, 0 – 1, 0, +1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
+ ,− + ,− + ,− + ,− + ,− + ,− + ,− + ,− +
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

62 Atomic Structure
Exception of Aufbau Principle: In some cases it is seen that the electronic configuration is slightly
different from the arrangement given by Aufbau principle. A simple region behind this is that half filled
& full filled subshell have got extra stability.

Example 1. 24 Cr → 1s2 2s2 2p6 3s2 3p6 4s2 3d4 (wrong configuration)
    

24 Cr → 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (right configuration)

     

Example 2. 29 Cu → 1s2 2s2 2p6 3s2 3p6 4s2 3d9 (wrong configuration)
     

29 Cu → 1s2 2s2 2p6 3s2 3p6 4s1 3d10 (right configuration)

     

Stability Order of Completely Filled and Half Filled Sub-Shells

For an atom the electronic configuration of its ground state implies to the lowest energy state and
higher stability. The electronic configuration of most of the atoms follows the basic rules.
Certain elements such as Cr or Cu do not follow the rules because the two sub-shells 4s and 3d slightly
differ in energy, i.e., 4s is slightly lower in energy than 3d orbital. So the valence electronic configuration
are 3d54s1 and 3d104s1 respectively, and not 3d44s2 and 3d94s2. In terms of exchange energy and
symmetry, the extra stability of fully and half-filled electronic configuration are explained.
Symmetrical distribution of electrons
There is relatively large symmetrical distribution of electrons for either completely filled or half-filled
orbitals of the same sub-shell. Therefore, the electrons have large attraction towards nucleus because
of small shielding effect between them.
Example: Cr

3d 4s
(less symmetrical)

3d 4s
(more symmetrical)

Exchange Energy
Exchange means shifing of electrons from one orbital to another within same sub-shell. Energy gets
released when electrons exchange their positions and the energy is called exchange energy. For
maximum number of exchanges, the maximum the energy released and the maximum the stabilisation.
Half-filled and fully-filled degenerate orbitals have more number of elctron exchanges, and
consequently, they have larger exchange energy of stabilisation.

Atomic Structure 63
 Ex. Cr (For 3d44s2)

Total Exchanges = 3 + 2 + 1 = 6

Ex. Cr (For 3d54s1)

Total Exchanges = 4 + 3 + 2 + 1 = 10

Concept Builder (4)

1. Which of the following set of quantum number is impossible for an electron?

1 1
(1) n = 1, l = 0, ml = 0, ms = + (2) n = 9, l = 7, ml = –6, ms = –
2 2
1 1
(3) n = 2, l = 1, ml = 0, ms = + (4) n = 3, l = 2, ml = –3, ms = +
2 2

2. Which of the following statements is correct for an electron having azimuthal quantum number
l = 2?
(1) The electron may be in the lowest energy shell.
(2) The electron is in a spherical orbital.
1
(3) The electron must have spin ms = +
2
(4) The electron may have a magnetic quantum number = – 1

64 Atomic Structure
3. Four electrons in an atom have the sets of quantum numbers as given below. Which electron
in at the highest energy level?
(1) n = 4, l = 0, ml = 0, ms = +1/2 (2) n = 3, l = 1, ml = 0, ms = –1/2
(3) n = 3, l = 2, ml = 0, ms = +1/2 (4) n = 4, l = 1, ml = –1, ms = –1/2

4. Which set of quantum numbers is possible for the last electron of Mg + ion ?
(1) n = 3, l = 2, m = 0, s = +1/2 (2) n = 2, l = 3, m = 0, s = +1/2
(3) n = 1, l = 0, m = 0, s = +1/2 (4) n = 3, l = 0, m = 0, s = +1/2

5. In any subshell, the maximum number of electrons having same value of spin quantum number
is:
(1) ( + 1) (2) l + 2 (3) 2l + 1 (4) 4l + 2

6. Which of the following statement is (are) correct ?

(1) The electronic configuration of Cr is [Ar]3d5,4s1 (Atomic no. of Cr = 24)
(2) The magnetic quantum number may have a negative value
(3) In chlorine atom, 9 electrons have a spin of one type and 8 of the opposite type, (Atomic
no. of Cl = 17)
(4) All of the above

7. A subshell n = 5, l = 3 can accommodate:

(1) 10 electrons (2) 14 electrons (3) 18 electrons (4) None of these

ANSWER KEY

Concept Builder (1)

1. (3) 2. (2) 3. (4) 4. (1) 5. (3) 6. (3) 7. (4)

Concept Builder (2)

1. (1) 2. (1) 3. (3) 4. (3) 5. (3)

Concept Builder (3)

1. (4) 2. (3) 3. (1) 4. (3)

Concept Builder (4)

1. (4) 2. (4) 3. (4) 4. (4) 5. (3) 6. (4) 7. (2)

Atomic Structure 65
 Exercise – I

Atomic Number, Mass Number and 5. The relative abundance of two rubidium
Important Definitions isotopes of atomic weights 85 and 87 are
75% and 25% respectively. The average
1. Which one of the following constitutes a
atomic wt. of rubidium is:–
group of the isoelectronic species?
(1) 75.5 (2) 85.5 (3) 86.5 (4) 87.5
(1) C2- -
2 ,O2 ,CO,NO
(2) NO+ ,C2- -
2 ,CN ,N2

(3) CN- ,N2 ,O22- ,C22- (4) N2 ,O-2 ,NO+ ,CO 6. Atomic weight of Ne is 20.2. Ne is mixture
of Ne20 and Ne22, Relative abundance of
2. List I
heavier isotope is:-
(i) 54, 56, 57, Fe28
26Fe 26Fe 26Fe 26
(1) 90 (2) 20 (3) 40 (4) 10
(ii) 1H3, 2He3
(iii) 76, 77
32Ge 33As 7. Naturally occurring boron has two
(iv) 92 U235, 90Th
231
isotopes whose atomic weights are 10.00
(v) 1 H1 , 1 D2, 1 T3 (I) and 11.00 (II). Atomic weight of natural
List II boron is 10.80. The percentage of
(a) Isotopes isotopes (I) and (II) respectively is:-
(b) Isotones (1) 20 and 80 (2) 10 and 20
(c) Isodiaphers (3) 15 and 75 (4) 30 and 70
(d) Isobars
Match the above correct terms:-
8. Let mass of electron is half, mass of
(1) (i) → a, (ii) → d, (iii) → b, (iv) → c, (v) → a
proton is two times and mass of neutron
(2) (i) → a, (ii) → d, (iii) → d, (iv) → c, v → a
is three fourth of original. The find out
(3) (v) →a, (iv) → c. (iii) → d, (ii)→b, (i) → a
new atomic wt. of O16 atom:-
(4) None of them
(1) increases by 37.5%
(2) Remain constant
3. The atom A, B, C have the configuration
(3) increases by 12.5%
A → [Z(90) + n(146)], B → [Z(92) + n(146)],
(4) decreases by 25%
C → [Z(90) + n(148)] So that:-
(a) A and C – Isotones
(b) A and C - Isotopes 9. In 7N14 if mass attributed to electron were
(c) A and B – Isobars doubled & the mass attributed to protons
(d) B and C – Isobars were halved, the atomic mass would
(e) B and C - Isotopes become approximately:-
The wrong statement's are:- (1) Halved (2) Doubled
(1) a, b only (2) c, d, e only (3) Reduced by 25% (4) Remain same
(3) a, c, d only (4) a, c, e only
10. A certain negative ion X–2 has in its
4. In an atom 13Al27. number of protons is (a) nucleus 18 neutrons and 18 electrons in
electron is (b) and neutron is (c). Hence its extra nuclear structure. What is the
ratio will be [in order c: b: a] mass number of the most abundant
(1) 13: 14: 13 (2) 13: 13: 14 isotope of 'X':
(3) 14: 13: 13 (4) 14: 13: 14 (1) 35.46 (2) 32 (3) 36 (4) 39

66 Atomic Structure
11. Isotopes of an element have 18. The value of Planck’s constant is
(1) similar chemical properties but 6.63 × 10–34 Js. The velocity of light is
different physical properties 3.0 × 108 ms–1, Which value is closest to the
(2) similar chemical and physical wavelength in nanometers of a quantum of
properties light with frequency of 8 × 1015 s–1?
(1) 5 × 10–18 (2) 4 × 101
(3) similar physical properties but different
(3) 3 × 107 (4) 2 × 10–25
chemical properties.
(4) different chemical and physical
19. The ratio of the energy of a photon of 2000
properties. Å wavelength radiation to that of 4000 Å
radiation is
12. The charge to mass ratio of protons is (1) 1 / 4 (2) 4 (3) 1 / 2 (4) 2
(1) 9.55 × 10–4 C/g (2) 9.55 × 104 C/g
(3) 1.76 × 108 C/g (4) 1.76 × 10–8 C/g Bohr’s Atomic Model

20. If the velocities of first, second, third and

13. The charge to mass ratio of -particles is fourth orbits of hydrogen atom are v1, v2,
approximately ..... the charge to mass v3 and v4 respectively, then which of the
ratio of protons following should be their increasing
(1) twice (2) half order?
(3) four times (4) Six times (1) v1 > v2 > v3 > v4 (2) v4 < v3 < v1 < v2
(3) v1 > v2 < v3 > v4 (4) Equal for all
Electromagnetic Waves and
Planck’s Quantum Theory 21. If v1, v2, v3 and v4 are velocities of the
electron present in the first orbit of H,
14. The energy of a photon of radiation having He+, Li+2 and Be+3, then which of the
wavelength 3000Å is nearly following should be their increasing
(1) 6.63 × 10–19 J (2) 6.63 × 10–18 J order?
(3) 6.63 × 10–16 J (4) 6.63 × 10–49 J (1) v1 < v2 < v3 < v4 (2) v4 = v3 = v2 = v1
(3) v1 < v2 < v3 > v4 (4) v1 > v2 < v3 > v4
15. Energy of a photon having wave number
22. Which orbits of H, He+ and Li+2 have
1.00 cm–1 is identical energies?
(1) 6.62 × 10–34 J (2) 1.99 × 10–23 J (1) Second orbits of all the three
(3) 6.62 × 10–32 J (4) 6.62 × 10–36 J (2) First orbit of H, second orbit of He+
and third orbit of Li+2
16. The frequency of wave is 6 × 1015 s–1. Its (3) Third orbit of all the three
wave number would be (4) Fourth orbit of H, First orbit of He+
and Fifth orbit of Li+2
(1) 105 cm–1 (2) 2 × 10–5 cm–1
(3) 2 × 10–7 cm (4) 2 ×105 cm–1 23. What should be the correct order of
energies of the first orbits of H, He+ and
17. The number of photons of light having Li+2, if these are represented as E1, E2 and
wavelength 100 nm which can provide
E3 respectively?
1.00 J energy is nearly
(1) E1 < E2 < E3 (2) E3 < E2 < E1
(1) 107 photons (2) 5 × 1018 photons
(3) E1 < E2 > E3 (4) E1 = E2 = E3
(3) 5 × 1017 photons (4) 5 × 107 photons

Atomic Structure 67
 24. What should be the ratio of energies of 31. What should be the ratio of energies of
the fifth orbits of Li +2
and He ? + the third and fifth orbits of He+?
(1) 4: 9 (2) 9: 4 (3) 12: 16 (4) 7: 2 (1) 25: 9 (2) 5: 3
(3) 16: 9 (4) None of these
25. The ratio of radii of the third and fifth
orbits of Li+2 will be 32. The ratio of the radii of first orbits of H,
(1) 9: 25 (2) 25: 9 He+ and Li+2 is:
(3) 2: 3 (4) 4: 5 (1) 1: 2: 3 (2) 6: 3: 1
(3) 9: 4: 1 (4) 6: 3: 2

26. What should be the circumference of the

33. The ratio of the radius of the atom to the
second orbit of hydrogen atom?
radius of the nucleus is of the order of
(1) 13.3 Å (2) 3.7 Å
(1) 105 (2) 106 (3) 10–5 (4) 10–8
(3) 26.6 Å (4) 3.4 Å

34. The energy of an electron in the first Bohr

27. The diameter of the second orbit of
orbit of H atom is–13.6 eV. The possible
hydrogen atom should be:
energy value(s) of the excited state for
(1) 2.12 Å (2) 4.23 Å
electrons in Bohr orbits of Li2+ is
(3) 2.10 Å (4) 4.01 Å
(1) –61.2 eV (2) –13.6 eV
(3) –122.4 eV (4) +6.8 eV
28. What should be the ratio of the radii of
the orbits of electron in Na+10 and 35. In a hydrogen atom, if the energy of
hydrogen atom? electron in the ground state is –13.6 eV,
(1) 11:1 (2) 1: 11 then that in the 2nd excited state is:
(3) 1: 1 (4) 1: 2 (1) –1.51 eV (2) – 3.4 eV
(3) –6.04 eV (4) –13.6 eV
29. What should be the velocity of the
electron present in the fourth orbit of 36. In hydrogen atom energy of first excited
hydrogen atom, if the velocity of the state is –3.4 eV. Then find out the K.E. of
electron present in its first orbit is 2.188 the electron in the same orbit of
× 10+8 cm per second? hydrogen atom:
(1) 1.094 × 108 cm per second (1) 3.4 eV (2) + 6.8 eV
(2) 5.47 × 107 cm per second (3) –13.6 eV (4) +13.6 eV
(3) 4.376 × 108 cm per second
(4) 2.188 × 106 cm per second 37. According to Bohr’s theory, the angular
momentum of an electron in 5th orbit is:
30. What should be the velocity of the (1) 25 h/ (2) 1.0 h/
electrons present in the first, second and (3) 10 h/ (4) 2.5 h/
third orbits of H, He+ and Li+2,
respectively? 38. Angular momentum for P-shell electron
(1) 2.188 × 108 cm per second is:
(2) 5.47 × 107 cm per second 3h 2h
(1) (2) Zero (3) (4) None
(3) 4.376 × 108 cm per second  2
(4) 2.188 × 106 cm per second

68 Atomic Structure
39. The radius of a shell for H-atom is 4.761Å. 45. The graphical representation of energy of
The value of n is: e– and atomic number is:
(1) 3 (2) 9 (3) 5 (4) 4

(1) (2)
40. The ratio of the radii of two Bohr orbits
of H-atoms is 4: 1, what would be their
nomenclature:
(1) K & L (2) L & K
(3) N & L (4) 2 & 3 both (3) (4)

41. The velocity of electron in third excited

state of Be3+ ion will be: 46. Maximum frequency of emission is
3 obtained for the transition:
(1) (2.188 × 108)ms–1
4 (1) n = 2 to n = 1 (2) n = 6 to n = 2
3 (3) n = 1 to n = 2 (4) n = 2 to n = 6
(2) (2.188 × 106)ms–1
4
(3) (2.188 × 106) Kms–1 47. If the ionization energy of hydrogen is
(4) (2.188 × 103) Kms–1 313.8 K cal per mole, then the energy of
the electron in 2nd excited state will be:
(1) – 113.2 Kcal/mole
42. The energy of H-atom in nth orbit is En
(2) – 78.45 Kcal/mole
then energy in nth orbit of singly ionised (3) – 313.8 Kcal/mole
helium atom will be: (4) – 35 Kcal/mole
(1) 4En (2) En/4
(3) 2En (4) En/2 48. Which of the following electron transition
will require the largest amount of energy
in a hydrogen atom:
43. The energy of second Bohr orbit of the (1) From n = 1 to n = 2
hydrogen atom is – 328 kJ/mol. Hence (2) From n = 2 to n = 3
the energy of fourth Bohr orbit should (3) From n =  to n = 1
be: (4) From n = 3 to n = 5
(1) – 41 KJ/mol
(2) – 1312 KJ/mol 49. A single electron orbits a stationary
(3) – 164 KJ/mol nucleus (z = 5). The energy required to
(4) – 82 KJ/mol excite the electron from third to fourth
Bohr orbit will be:
(1) 4.5 eV (2) 8.53 eV
44. Potential energy is – 27.2 eV in second
(3) 25 eV (4) 16.53 eV
orbit of He+ then calculate, double of
total energy in first excited state of
50. En = – 313.6/n2. If the value of En = – 34.84
hydrogen atom:
then to which of the following values
(1) – 13.6 eV (2) – 54.4 eV
does ‘n’ correspond:
(3) – 6.8 eV (4) – 27.2 eV
(1) 1 (2) 2
(3) 3 (4) 4

Atomic Structure 69
 51. Which of the following is a correct 56. The energy of an electron in the first Bohr
relationship: orbit of H atom is − 13.6 eV. The possible
(1) E1 of H = 1/2 E2 of He = 1/3 E3 of Li =
+ +2
energy value(s) of the excited state(s) for
1/4 E4 of Be+3 electrons in Bohr orbits of hydrogen
(2) E1(H) = E2(He ) = E3(Li ) = E4(Be )
+ +2 +3
is/are:
(3) E1(H) = 2E2(He ) = 3E3(LI ) = 4E4(Be )
+ +2 +3
(1) − 3.4 eV (2) − 4.2 eV
(4) No relation (3) − 6.8 eV (4) + 6.8 eV

52. Energy required to remove an e– from M 57. The radius of which of the following orbit
shell of H-atom is 1.51 eV, then energy of is same as that of the first Bohr’s orbit of
first excited state will be: hydrogen atom?
(1) – 1.51 eV (2) +1.51 eV (1) He+ (n = 2) (2) Li2+ (n = 2)
(3) – 3.4 eV (4) – 13.6 eV (3) Li2+ (n = 3) (4) Be3+ (n = 2)

53. The ionisation energy for excited

hydrogen atom in eV will be: Spectrum and Spectrum Line
(1) 13.6
58. The spectrum of He+ is expected to be
(2) Less than 13.6
similar to that of
(3) Greater than 13.6
(1) Li+ (2) H
(4) 3.4 or less
(3) Na (4) He

54. The energy required to excite an electron 59. Which of the following should be the
of H-atom from first orbit to second orbit expression for the last line of Paschen
is: series?
3 1 1 1  1  1 1
(1) of its ionisation energy (1) = R − (2) = R − 
4  2  
9   4 9
1
(2) of its ionisation energy 1 1 1  1  1 1
2 (3) = R −  (4) = R − 
  9 16    16  
1
(3) of its ionisation energy
4
60. Which of the following should be the
(4) None
correct value of the wave number of first
line in Balmer series of hydrogen atom?
55. Which of the following is a correct graph:
(1) 5R/36 (2) –5R/36
(3) R/9 (4) –R/9
KE KE
(1) (2) 61. What should be the frequency of
n n radiation of the emission spectrum when
the electron present in hydrogen atom
KE KE undergoes transition from n = 3 energy
(3) (4) level to the ground state?

(1) 3 × 1015 second–1 (2) 3 × 105 second–1
n Z2
(3) 3 × 1010 second–1 (4) 3 × 108 second–1

70 Atomic Structure
62. The ratio of minimum frequency of 69. If the shortest wavelength of Lyman
Lyman & Balmer series will be: series of H atom is x, then the wave
(1) 1.25 (2) 0.25 (3) 5.4 (4) 10 length of first line of Balmer series of H
atom will be:
63. The wavelength of the radiation emitted, 9x 36x 5x 5x
(1) (2) (3) (4)
when in a hydrogen atom electron falls 5 5 9 36
from infinity to stationary state 1 would
be 70. What transition in He+ will have the same
(Rydberg constant = 1.097 × 107 m–1)  as the I line in Lyman series of H - atom:
(1) 91 nm (2) 192 nm (1) 5 → 3 (2) 3 → 2
(3) 406 nm (4) 9.1 × 10–8 nm (3) 6 → 4 (4) 4 → 2

64. The ratio of minimum wavelengths of

71. In H-atom, electron transits from 6th orbit
Lyman & Balmer series will be:
to 2nd orbit in multi step. Then total
(1) 1.25 (2) 0.25 (3) 5 (4) 10
spectral lines (without Balmer series) will
be:
65. Which transition emits photon of
(1) 6 (2) 10
maximum frequency:
(3) 4 (4) 0
(1) second spectral line of Balmer series
(2) second spectral line of Paschen series
72. The figure indicates the energy level
(3) fifth spectral line of Humphery series
diagram for the origin of six spectral lines
(4) first spectral line of Lyman series
in emission spectrum (e.g line no. 5 arises

66. The wavelength of photon obtained by from the transition from level B to X)

electron transition between two levels in which of the following spectral lines will
not occur in the absorption spectrum:
H-atom and singly ionised He are 1 and
2 respectively, then:
(1) 2 = 1 (2) 2 = 21
(3) 2 = 1/2 (4) 2 = 1/4
(1) 1, 2, 3 (2) 3, 2
67. Find out ratio of following for photon
(3) 4, 5, 6 (4) 3, 2, 1
(max.)Lyman: (max)Bracket
(1) 1: 16 (2) 16: 1
73. A certain electronic transition from an
(3) 4: 1 (4) 1: 4
excited state to ground state of the H-
68. If H-atom is supplied with 12.1 eV energy atom in one or more step gives rise to
and electron returns to the ground state three lines in the ultra violet region of the
after excitation the number of spectral spectrum. How many lines does this
lines in Balmer series would be: transition produce in the infrared region
(use energy of ground state of H-atom = of the spectrum:
– 13.6 eV) (1) 1 (2) 2
(1) 1 (2) 2 (3) 3 (4) 4
(3) 3 (4) 4

Atomic Structure 71
 74. The energy photon emitted corresponding 81. A ball of 100 g mass is thrown with a
to transition n = 3 to n = 1 is: velocity of 100 ms–1. The wavelength of
[h = 6 ×10–34 J-sec.] the de Broglie wave associated with the
(1) 1.76 ×10–18 J (2) 1.76 ×10–16 J
ball is about
–17
(3) 1.76 ×10 J (4) None of these
(1) 6.63 × 10–35 m (2) 6.63 × 10–30 m
(3) 6.63 × 10–35 cm (4) 6.63 × 10–33 m
75. The difference in the wavelength of the
1st line of Lyman series and 2nd line of
Balmer series in a hydrogen atom is 82. Which of the following has least de-
9 4 Broglie ?
(1) (2)
2R R (1) electron (2) proton
5 (3) CO2 (4) SO2
(3) (4) None
R
83. If the de-Broglie wavelength of the fourth
de-Broglie Concept Bohr orbit of hydrogen atom is 4Å, the
76. No. of wave in fourth orbit:– circumference of the orbit will be:
(1) 4 (2) 5 (1) 4Å (2) 4 nm
(3) 0 (4) 1
(3) 16 Å (4) 16 nm

77. What is the ratio of the De-Broglie wave

84. What is the de-Broglie wavelength
lengths for electrons accelerated through
200 volts and 50 volts:- associated with the hydrogen electron in
(1) 1: 2 (2) 2: 1 its third orbit:
(3) 3: 10 (4) 10: 3 (1) 9.96 × 10–10 cm (2) 9.96 × 10–8 cm
(3) 9.96 × 104 cm (4) 9.96 × 108 cm
78. A particle X moving with a certain velocity
has a de broglie wavelength of 1Å. If
85. The correct order of wavelength of
particle Y has a mass of 25% that of X and
Hydrogen (1H1), Deuterium ( 1 H2 ) and
velocity 75% that of X, de broglie's
wavelength of Y will be:- Tritium (1H3) moving with same kinetic
(1) 3Å (2) 5.33 Å energy is
(3) 6.88 Å (4) 48Å
(1) H > D > T (2) H = D = T
79. The wavelength associated with a golf (3) H < D < T (4) H < D > T
ball weighing 200 g and moving at a speed
of 5 m/h is of the order
Heisenberg Uncertainty Principle
(1) 10–10m (2) 10–20m
(3) 10 m
–30
(4) 10–40 m 86. For the electron if the uncertainty in
velocity is v, the uncertainty in its
80. The de-Broglie wavelength of a tennis
position (x) is given below:
ball of mass 60 g moving with a velocity
of 10 metres per second is approximately h 2
(1) mv (2)
[Planck’s constant, h = 6.63 × 10–34 Js] 2 hmv
(1) 10–25 metres (2) 10–33 metres h 2m
(3) (4)
(3) 10–31 metres (4) 10–16 metre 4mv hv

72 Atomic Structure
87. The uncertainty in momentum of moving 94. An electron has magnetic quantum
particle is 1.0 × 10 –15
kg m s , the minimum
–1
number as -3, what is its principal
uncertainty in its position would be quantum number:-
(1) 5.28 × 10 –20
m (2) 5.28 × 10 –49
m (1) 1 (2) 2 (3) 3 (4) 4
(3) 6.63 × 10 –49
m (4) 6.63 × 10 –22
m
95. The total spin resulting from a d7
88. The uncertainty in the position of an configuration is:-
electron (mass 9.1 × 10–28gm) moving with 1 3
(1) (2) 2 (3) 1 (4)
a velocity of 3 × 104 cm sec–1, Uncertainty 2 2
in velocity is 0.011% will be:-
(1) 1.92 cm (2) 7.68 cm 96. No. of all subshells of n +  = 7 is:-
(3) 0.175 cm (4) 3.84 cm
(1) 4 (2) 5 (3) 6 (4) 7
89. The Uncertainty in position of an electron
& helium atom are same. If the 97. The quantum number of 20th electron of
Uncertainty in momentum for the Fe(Z = 26) ion would be:–
electron is 32 × 105 , then the Uncertainty (1) 3, 2, – 2, – ½ (2) 3, 2, 0, ½
in momentum of helium atom will be (3) 4, 0, 0, + ½ (4) 4, 1, – 1, + ½
(1) 32 × 105 (2) 16 × 105
(3) 8 × 105 (4) None 98. Which orbitlal has two angular nodal
planes:-

Quantum Numbers (1) s (2) p (3) d (4) f

90. In an atom, for how many electrons, the

99. In an atom having 2K, 8L, 8M and 2N
quantum numbers will be , n = 3, l = 2, m
electrons, the number of electrons with
1
= + 2, s = + :- 1
2 m = 0; S = + are
2
(1) 18 (2) 6
(1) 6 (2) 2 (3) 8 (4) 16
(3) 24 (4) 1

100. Orbital angular momentum of a 3d

91. Spin angular momentum for electron:-
electron is:-
h h
(1) s(s + 1) (2) 2s(s + 1) h h h h
2 2 (1) 2 (2) 6 (3) (4)
2 2 2 4
(3) s(s + 2) (4) None

101. An orbital with = 0 is symmetrical about

92. If l = 3 then type and number of orbital
the:-
is:–
(1) x-axis only (2) y-axis only
(1) 3p, 3 (2) 4f, 14
(3) z-axis only (4) The nucleus
(3) 5f, 7 (4) 3d, 5

102. For azimuthal quantum number l = 3, the

93. The total value of m for the electrons
maximum number of electrons will be:
(n = 4) is -
(1) 2 (2) 6 (3) zero (4) 14
(1) 4 (2) 8 (3) 16 (4) 32

Atomic Structure 73
 103. Which d - orbital has different shape from 112. Which of the following sets of quantum
rest of all d - orbitals ? number is not possible?
(1) d (2) d (3) d xy (4) dxz 1
x2 − y 2 z2
(1) n = 3; l = 0 , ml = 0, ms = +
2
104. The total number of orbitals in a shell 1
(2) n = 3; l = 0 , ml = 0, ms = –
with principal quantum number ‘n’ is: 2
(1) 2 n (2) 2 n2
1
(3) n2 (4) n + 1 (3) n = 3; l = 0, ml = –1, ms = +
2

105. What is the correct orbital designation for 1

(4) n = 3; l = 1, ml = 0, ms = –
the electron with the quantum numbers , 2
n = 4 , l = 3 , m = – 2 , s = 1/2
(1) 3 s (2) 4 f (3) 5 p (4) 6 s 113. Which quantum number is sufficient to
determine the energy of the electron in
106. The total number of electrons that can be hydrogen atom?
accommodated in all the orbitals having
(1) l (2) n (3) ms (4) ml
principal quantum number 2 and
azimuthal quantum number 1 is:
114. The number of radial nodes in 3s and 2p
(1) 2 (2) 4 (3) 6 (4) 8
respectively are:
107. Which atom has as many as s - electron (1) 2 and 0 (2) 1 and 2
as p - electron ?
(3) 0 and 2 (4) 2 and 1
(1) H (2) Mg (3) N (4) Na

115. The following quantum number are

108. The probability of finding an electron
residing in a px orbital is not zero: possible for how many orbitals?

(1) In the yz plane n = 3, l = 2, m = +2

(2) In the xy plane (1) 3 (2) 2 (3) 1 (4) 4

(3) In the y direction
(4) In the z direction 116. The 19th electron of chromium has which
of the following sets of quantum
109. The number of nodal planes in a px
numbers?
orbital is:
n l m s
(1) 1 (2) 2 (3) 3 (4) zero
1
(1) 3 0 0
110. The orbital angular momentum of an 2
electron in p-orbital is: 1
(2) 3 2 –2
h h 1 h 2
(1) zero (2) (3) (4)
2 2 2 2 1
(3) 4 0 0
2

111. The value of Azimuthal quantum number 1

(4) 4 1 –1
for all electrons present in 5p orbitals is 2
(1) 4 (2) 5 (3) 2 (4) 1

74 Atomic Structure
117. The orbital angular momentum for an Electronic Configuration
electron revolving in an orbit is given by
122. A transition metal 'X' has a configuration
h
l(l + 1) This momentum for an s- [Ar] 3d5 in its + 3 oxidation state. Its
2 atomic number is:–
electron will be given by
(1) 22 (2) 26 (3) 28 (4) 19
h 1 h
(1) 2. (2) .
2 2 2 123. 4s2 is the configuration of the outermost
h orbit of an element. Its atomic number
(3) zero (4)
2 would be:–
(1) 29 (2) 24 (3) 30 (4) 19
118. Which of the following sets of quantum
numbers is correct for an electron in 4f- 124. A neutral atom of an element has 2K, 8L,
orbital?
11 M and 2N electrons. The number of s-
1
(1) n = 4, l = 3, m = 4, s = + electron in the atom are
2
(1) 2 (2) 8 (3) 10 (4) 6
1
(2) n = 4, l = 4, m = –4, s = –
2
125. The explaination for the presence of
1
(3) n = 4, l = 3, m = +1, s = + three unpaired electrons in the nitrogen
2
atom can be given by:-
1
(4) n = 3, l = 2, m = –2, s = + (1) Pauli's exclusion principle
2
(2) Hund's rule
119. For principal quantum number n = 4, the (3) Aufbau's principle
total number of orbitals having l = 3 is (4) Uncertainty principle
(1) 3 (2) 7 (3) 5 (4) 9

126. n and  values of an orbital 'A' are 3 and

120. Which of the following sets of quantum
numbers represents the highest energy of 2, of another orbital 'B' are 5 and 0. The
an atom? energy of
1 (1) B is more than A
(1) n = 3, l = 1, m = 1, s = +
2 (2) A is more than B
1 (3) A and B are of same energy
(2) n = 3, l = 2 m = 1 s = +
2 (4) None
1
(3) n = 4, l = 0, m = 0, s = +
2
127. Sum of the paired electrons present in
1
(4) n = 3, l = 0, m = 0, s = + the orbital with  = 2 in all the species
2
Fe2+, Co2+ and Ni+2 are:–
121. The orbital angular momentum of an
(1) 9 (2) 12 (3) 6 (4) 15
electron in 2s orbital is:
1 h
(1) + . (2) Zero 128. The number of electrons in the M-shell of
2 2
the element with atomic number 24 is:-
h h
(3) (4) 2 (1) 24 (2) 12 (3) 8 (4) 13
2 2

Atomic Structure 75
 129. Which among the following is correct of 135. The atomic number of an element is 17,
5
B in normal state? the number of orbitals containing
2s 2p electron pairs in the valency shell is:-

(1) : Against Hund’s (1) 8 (2) 2 (3) 3 (4) 6

rule
136. Which of the following transition neither
(2) : Against Aufbau shows absorption nor emission of energy
in case of Hydrogen atom:–
principle as well as Hunds rule
(1) 3px → 3s (2) 3dxy → 3dyz
(3) :Violation of
(3) 3s → 3dxy (4) All the above
Pauli’s exclusion principle and not
Hund’s rule 137. In potassium the order of energy level for

(4) :Against Aufbau 19th electron is:

(1) 3 s > 3 d (2) 4 s < 3 d
principle
(3) 4 s > 3 p (4) 4 s = 3 d

130. The atomic orbitals are progressively

138. Remaining part of atom except outer
filled in order of increasing energy. This
orbit is called:-
principle is called.
(1) Kernel (2) Core
(1) Hund’s rule
(3) Empty space (4) None of these
(2) Aufbau principle
(3) Exclusion principle
139. For H atom, the energy required for the
(4) de- Broglie rule.
removal of electron from various sub-
shells is given as under:-
131. For a given principal level n = 4, the
energy of its subshells is in the order
(1) s < p < d < f (2) s > p > d > f
(3) s < p < f < d (4) f < p < d < s.
The order of the energies would be:-
(1) E1 > E2 > E3 (2) E3 > E2 > E1
132. The maximum number of unpaired
(3) E1 = E2 = E3 (4) None of these
electrons are in
(1) Fe+2 (2) Fe+3
(3) Fe4+ (4) Fe 140. The maximum number of electrons in a
sub-shell is given by the expression
133. The number of d-electrons retained in (1) 4l – 2 (2) 4l + 2
Fe2+ (At. no. of Fe = 26) ion is: (3) 2l + 1 (4) 2n2
(1) 6 (2) 3 (3) 4 (4) 5
141. Total number of nodal planes, angular
134. Which of the following ions has the and spherical nodes in 3s-subshell are
maximum value of magnetic moment? respectively:
(1) Cu+ (2) Cu2+ (1) zero, zero, 2 (2) 2, 2, 2
(3) Fe2+ (4) Fe3+ (3) zero, zero, zero (4) zero, 2, 2

76 Atomic Structure
142. The maximum number of electrons with 146. How many electrons in argon have m = 0?
clockwise spin that can be (1) 6 (2) 8 (3) 10 (4) 12.
accommodated in a f-sub-shell is
(1) 14 (2) 7 (3) 5 (4) 10 147. Consider the ground state of Cr atom
(z = 24). The numbers of electrons with
143. In magnesium atom, in ground state, the the azimuthal quantum numbers, l = 1
number of electrons with m = 0 is: and 2 are, respectively.
(1) 4 (2) 6 (3) 2 (4) 8 (1) 12 and 4 (2) 12 and 5
(3) 16 and 4 (4) 16 and 5
144. In chromium atom, in ground state, the
number of occupied orbital is 148. Which of the following statements in
(1) 14 (2) 15 (3) 7 (4) 12 relation to the hydrogen atom is correct?
(1) 3s, 3p and 3d orbitals all have the
145. A sub-shell with n = 6, l = 2 can same energy.
accommodate a maximum of (2) 3s and 3p orbitals are of lower energy
(1) 12 electrons than 3d orbital.
(2) 36 electrons (3) 3p orbital is lower in energy than 3d
(3) 10 electrons orbital.
(4) 72 electrons (4) 3s orbital is lower is energy than 3p
orbital.

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

Atomic Structure 77
 Exercise – II

1. What should be the wavelength and energy 7. The maximum probability of finding
respectively of the emitted light when the electron in the dxy orbital is:
electron of hydrogen atom undergoes
(1) along the x-axis
transition from first excited state to
(2) along the y-axis
ground state?
(3) at an angle of 45º from the x & y-axis
(1) 1215 Å and 21.8 × 10–12 erg
(4) at an angle of 90º from the x & y-axis
(2) 6.560 Å and 16.35 × 10–12 erg
(3) 1215 Å and 16.35 × 10–12 erg
(4) 6560 Å and 21.8 × 10–12 erg 8. The ratio of (E2– E1) to (E4 – E3) for the
hydrogen atom is approximately equal to:
2. What should be the ratio of energies of the (1) 10 (2) 15 (3) 17 (4) 12
electrons of the first orbits of Na+10 and H?
(1) 11 :1 (2) 121 : 1
9. The number of waves made by a Bohr
(3) 1 : 121 (4) 1 : 11
electron in an orbit of maximum magnetic

3. What should be the frequency (in cycles quantum number 3 is:

per second) of the emitted radiation when (1) 3 (2) 4 (3) 2 (4) 1
an electron undergoes transition from M
energy level to K energy level, if the value 10. The wave number of the limiting line in
of R is 10 cm .
5 –1
Lyman series of hydrogen is 109678 cm–1.
(1) 3/2 × 10 15
(2) 8/3 × 10 15
The wave number of the limiting line in
(3) 8/5 × 10 15
(4) 9/4 × 1015
Balmer series of He+ would be:
(1) 54839 cm–1 (2) 219356 cm–1
4. The ionization energy of the ground state
(3) 109678 cm–1 (4) 438712 cm–1
hydrogen atom is 2.18 × 10-18 J. The energy
of an electron in second orbit of He+ will
be 11. The ratio of specific charge of an electron
(1) –1.09 × 10-18 J (2) –4.36 × 10-18 J to that of a proton is
(3) –3.18 × 10-18 J (4) –2.18 × 10-18 J (1) 1 : 1 (2) 1837 : 1
(3) 1 : 1837 (4) 2 : 1
5. If kinetic energy of a proton is increased
nine times the wavelength of the de-
12. The electrons, identified by quantum
Broglie wave associated with it would
numbers n and l, (i) n = 4, l = 1 (ii) n= 4, l =
become
0 (iii) n = 3, l = 2 (iv) n = 3, l = 1 can be
(1) 3 times (2) 9 times
(3) 1/3 times (4) 1/9 times placed in order of increasing energy, from
the lowest to highest, as
6. An ion Mna+ has the magnetic moment (1) (iv) < (ii) < (iii) < (i)
equal to 4.9 B.M. The value of ‘a’ is: (2) (ii) < (iv) < (i) < (iii)
(1) 3 (2) 4 (3) (i) < (iii) < (ii) < (iv)
(3) 2 (4) 5 (4) (iii) < (i) < (iv) < (ii)

78 Atomic Structure
13. The ionization enthalpy of hydrogen atom 19. The frequency of light emitted for the
is 1.312 × 10 J mol . The energy required to
6 –1 transition n = 4 to n = 2 of He+ is equal to
excite the electron in the atom from n = 1 the transition in H atom corresponding to
which of the following:
to n = 2 is
(1) n = 3 to n = 1 (2) n = 2 to n = 1
(1) 8.51 × 105 J mol–1 (2) 6.56 × 105 J mol–1
(3) n = 3 to n = 2 (4) n = 4 to n = 3
(3) 7.56 × 105 J mol–1 (4) 9.84 × 105 J mol–1
20. Energy of an electron is given by
14. The frequency of radiation emitted when  Z2 
the electron falls from n = 4 to n = 1 in a E= – 2.178 ×10–18J   . Wavelength of light
 n2 
 
hydrogen atom will be (Given ionization
required to excite an electron in an
energy of H = 2.18 × 10–18 J atom–1)
hydrogen atom from level n = 1 to n = 2 will
(1) 1.03 × 1015 s–1 (2) 3.08 × 1015 s–1
be :
(3) 2.00 × 1015 s–1 (4) 1.54 × 1015 s–1 (h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1)
(1) 1.214 × 10–7 m (2) 2.816 × 10–7 m
15. What is the maximum number of electrons (3) 6.500 × 10 m–7
(4) 8.500 × 10–7 m
which can be accommodated in an atom in
which the highest principal quantum 21. Supposing the I.P. of hydrogen atom is 960
eV. Find out the value of principal quantum
number value is 4?
number having the energy equal to
(1) 10 (2) 18 (3) 36 (4) 54
– 60 eV:
(1) n = 2 (2) n = 3 (3) n = 4 (4) n = 5
16. The wavelength of radiation emitted when
22. If the ionisation potential of an atom is
an electron in a hydrogen atom makes a
20V, its first excitation potential will be:
transition from an energy level with n = 3
(1) 5 V (2) 10 V (3) 15 V (4) 20 V
to a level with n = 2 is:
−1312 23. A single electron orbits a stationary
[Given that En = kJmol–1]
n 2 nucleus of charge +Ze, where Z is a
(1) 6.56 × 10–7 m (2) 65.6 nm constant, It requires 47.2 eV to excite
electron from second Bohr orbit to third
(3) 65.6 × 10–7 m (4) any of the above
Bohr orbit, find the value of Z:
(1) 1 (2) 3 (3) 5 (4) 4
17. The energy required to escape the electron
from ground state of H is 13.6 eV then the 24. A photon of energy 12.75 eV is completely
same for Ist excited state of H atom: absorbed by a hydrogen atom initially in
(1) 3.4 ground state. The principle quantum
(2) 13.6 number of the excited state is:
(3) 27.2 (1) 1 (2) 3 (3) 4 (4) 
(4) can’t say anything
25. An electron in hydrogen atom (ionisation
energy 13.6eV) jumps from third excited
18. A gas absorbs a photon of 355 nm and
state to first excited state. The energy of
emits at two wavelengths. If one of the photon emitted in the process is :
emissions is at 680 nm, the other is at: (1) 1.89 eV (2) 2.55 eV
(1) 743 nm (2) 518 nm (3) 12.09 eV (4) 12.75 eV
(3) 1035 nm (4) 325 nm

Atomic Structure 79
 26. If a photon of energy 14 eV. is incident on 29. In the following transition which statement
an H-atom, what is true: is correct:
(1) Atom will be ionised and electron will E3
have a kinetic energy of 14 eV E2
(2) Atom will be ionised and electron will E1
have a kinetic energy of 0.4 eV (1) E3 – 1 = E3-2 – E2-1 (2) 3 = 1 + 2
(3) Photon passes through atom without (3) 3 = 2 + 1 (4) All of these
interacting with it
(4) More than one electrons will make 30. In which transition, one quantum of energy
transitions is emmited:

27. The ratio of the difference in energy (1) n = 4 → n = 2 (2) n = 3 → n = 1

between the first and second Bohr orbit to (3) n = 4 → n = 1 (4) All of them
that between second and third Bohr orbit
in H-atom is : 31. When a hydrogen sample in ground state is
(1) 4/9 (2) 1/3 (3) 27/5 (4) ½ bombarded then what potential is required
to accelerate electron so that first Paschen
28. Match the following: line is emitted:
(A) Energy of ground (i) + 6.04 eV (1) 2.55 V (2) 0.65 V
State of He+ (3) 12.09 V (4) 12.75 V
(B) Potential energy (ii) –27.2 eV
of I orbit of H- 32. Given that in the H-atom the transition
atom energy for n = 1 to n = 2, is 10.2 eV. The
(C) Kinetic energy of (iii) 8.72 × 10–18 J energy for the same transition in Be3+ is :
II excited state (1) 20.4 eV (2) 163.2 eV
of He +
(3) 30.6 eV (4) 40.8 eV
(D) Ionisation (iv) –54.4 eV
potential of He+ 33. The wavelength of first line of the Lyman
(1) A-(i), B-(ii), C-(iii), D-(iv) series for hydrogen is 1216Å. The
(2) A-(iv), B-(iii), C-(ii), D-(i) wavelength for the first line of this series
(3) A-(iv), B-(ii), C-(i), D-(iii) for a 10 time ionised sodium atom (z = 11)
(4) A-(ii), B-(iii), C-(i), D-(iv) will be:
(1) 1000 Å (2) 100 Å
(3) 10 Å (4) 1 Å

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 3 2 2 3 3 1 3 2 2 3 2 1 4 2 3 1 1 1 2 1 3 3 3 3 2
Que. 26 27 28 29 30 31 32 33
Ans. 2 3 3 3 4 4 2 3

80 Atomic Structure
 Exercise – III (Previous Year Questions)
1. If uncertainty in position and momentum 7. If n = 6, the correct sequence for filling of
are equal, then uncertainty in velocity is? electrons will be: [AIPMT PRE 2011]
[AIPMT 2008] (1) ns → (n–2)f → (n–1)d → np
h 1 h (2) ns → (n–1)d → (n–2)f → np
(1) (2)
 2m  (3) ns → (n–2)f → np → (n–1)d
h 1 h (4) ns → np → (n–1)d → (n–2)f
(3) (4)
2 m 
8. According to the Bohr Theory, which of
2. The measurement of the electron the following transitions in the hydrogen
position is associated with an uncertainty atom will give rise to the least energetic
in momentum, which is equal to 1× 10–18 g
photon? [AIPMT MAINS 2011]
cm s–1. The uncertainty in electron
(1) n = 5 to n = 3 (2) n = 6 to n = 1
velocity is: (mass of electron = 9×10–28g)
(3) n = 5 to n = 4 (4) n = 6 to n = 5
[AIPMT 2008]
(1) 1 × 10 cm s
11 –1
(2) 1 × 109 cm s–1
(3) 1 × 10 cm s
6 –1
(4) 1 × 105 cm s–1 9. Maximum number of electrons in a
subshell with  = 3 and n = 4 is:
3. Maximum number of electrons in a
subshell of an atom is determined by the [AIPMT PRE 2012]
following:- [AIPMT 2009] (1) 10 (2) 12 (3) 14 (4) 16
2
(1) 2n (2) 4 + 2
10. The correct set of four quantum numbers
(3) 2 + 1 (4) 4 – 2
for the valence electron of rubidium atom
4. Which of the following is not permissible (Z = 37) is:- [AIPMT MAINS 2012]
arrangement of electrons in an atom ? (1) 5, 0, 0, + ½ (2) 5, 1, 0, + ½
[AIPMT 2009] (3) 5, 1, 1, + ½ (4) 6, 0, 0 + ½
(1) n = 3,  = 2, m = –2, s = –1/2
11. The orbital angular momentum of a p-
(2) n = 4,  = 0, m = 0, s = –1/2
electron is given as:
(3) n = 5,  = 3, m = 0, s = +1/2 [AIPMT MAINS 2012]
(4) n = 3,  = 2, m = –3, s = –1/2 3 h h
(1) (2) 6.
2  2
5. A 0.66 kg ball is moving with a speed of
h h
100 m/s. The associated wavelength will (3) (4) 3
2 2
be
–34
(h = 6.6 × 10 Js):- [AIPMT 2010] nh
–34 –35 12. In Bohr's orbit indicates:-
(1) 6.6 × 10 m (2) 1.0 × 10 m 2
–32 –32
(3) 1.0 × 10 m (4) 6.6 × 10 m [AIIMS 2012]
(1) Momentum
6. The total number of atomic orbitals in (2) Kinetic energy
fourth energy level of an atom is:-
(3) Potential energy
[AIPMT PRE 2011]
(4) Angular momentum
(1) 8 (2) 16 (3) 32 (4) 4

Atomic Structure 81
 13. The value of Planck’s constant is 18. Calculate the energy in joule corresponding
6.63 × 10–34 Js. The speed of light is 3 × to light of wavelength 45 nm:
1017 nms–1. Which value is closest to the (Planck’s constant h = 6.63 × 10–34 Js;
wavelength in nanometer of a quantum of speed of light c = 3 × 108 ms–1)
light with frequency of 6 × 1015 s–1?
[AIPMT 2014]
[NEET UG 2013]
(1) 6.67 × 1015 (2) 6.67 × 1011
(1) 75 (2) 10 (3) 25 (4) 50
(3) 4.42 × 10 –15
(4) 4.42 × 10–18
14. Based on equation E = –2.178 × 10–1Js
 Z2  19. The energy of an electron of 2py orbital is
 2  certain conclusions are written.
n  [AIIMS 2014]
 
(1) greater than 2p, orbital
Which of them is not correct?
[NEET UG 2013] (2) Less than 2pz orbital
(1) For n = 1, the electron has a more (3) same as that of 2px and 2pz orbital
negative energy than it does for n = 6 (4) Equal to 2s orbital
which means that the electron is
more lossely bound int he smallest
20. The number of d-electrons in Fe2+(Z = 26)
allowed orbit.
(2) The negative sign in equation simply is not equal to the number of electrons in
means that the energy of electron which one of the following?
bound to the nucleus is lower than it [AIPMT 2015]
would be if the electrons were at the (1) p-electrons in Cl (Z=17)
infintite distance form the nucleus. (2) d-electrons in Fe (Z=26)
(3) Larger the value of n, the larger is the (3) p-electrons in Ne (Z=10)
orbit radius
(4) s-electrons in Mg (Z=12)
(4) Equation can be used to calculate the
change in energy when the electron
change orbit. 21. The angular momentum of electron in ‘d’
orbital is equal to:_ [AIPMT 2015]
15. What is the maximum number of (1) 2 (2) 2 3
electrons that can be associated with the
following set of quantum numbers? n = 3; (3) 0 (4) 6
l = 1 and m = –1 [NEET UG 2013]
(1) 2 (2) 10 (3) 6 (4) 4 22. Which is the correct order of increasing
energy of the listed orbitals in the atom
16. A particle is moving with 3 times faster
of titanium? [Re-AIPMT 2015]
than speed of e–. Ratio of wavelength of
(At. no. Z = 22)
particle & electron is 1.8 × 10–4 then
particle L is:- [AIIMS 2013] (1) 3s 3p 3d 4s (2) 3s 3p 4s 3d
(1) Neutron (2) - particle (3) 3s 4s 3p 3d (4) 4s 3s 3p 3d
(3) Deutron (4) Tritium
23. In which tranistion of hydrogen atom
have same wavlength as in Balmer series
17. What is the maximum number of orbitals
that can be identified with the following transition of He+ ion(n =4 to n = 2)
[AIIMS 2015]
quantum numbers? n = 3, = 1, m = 0 ,
(1) 4 to 2 (2) 3 to 2
[AIPMT 2014]
(3) 2 to 1 (4) 4 to 1
(1) 1 (2) 2 (3) 3 (4) 4

82 Atomic Structure
24. Two electrons occupying the same orbital 29. Match the metal ions given in Column I
are distinguished by [NEET(I) 2016] with the spin magnetic moments of the
(1) Principal quantum number ions given in column II and assign the
(2) Magnetic quantum number correct code: [NEET 2018]
(3) Azimuthal quantum number Column I Column II
(4) Spin quantum number
a. Co3+ i. 8 B.M.
25. Which of the following pairs of d-orbitals b. Cr3+ ii. 35 B.M.
will have electron density along the axis?
[NEET(II) 2016] c. Fe3+ iii. 3 B.M.
(1) d 2 ,d (2) dxy ,d d. Ni2+ iv. 24 B.M.
z x2 − y2 x2 − y2

(3) d 2 ,dxz (4) dxz ,dyz v. 15 B.M.

z
a b c d
26. How many electrons can fit in the orbital (1) iv i ii iii
for which n = 3 and l = 1? [NEET(II) 2016] (2) i ii iii iv
(1) 10 (2) 14 (3) 2 (4) 6 (3) iv v ii i
(4) iii v i ii
27. Which one is the wrong statement?
[NEET 2017] 30. Wave length of particular transition for H
(1) de–Broglie's wavelength is given by atom is 400 nm. What can be wavelength
h of He+ for same transition: [AIIMS 2018]
= ; where m = mass of the
mv (1) 400 nm (2) 100 nm
particle, v = group velocity of the (3) 1600 nm (4) 200 nm
particle.
(2) The uncertainty principle is 31. A gas metal in bivalent state have
h approximately 23e– what is spin
E  t  .
4 magnetic moment in elemental state :
(3) Half filled and fully filled orbitals have [AIIMS 2018]
greater stability due to greater (1) 2.87 (2) 5.5
exchange energy, greater symmetry (3) 5.9 (4) 4.9
and more balanced arrangement.
(4) The energy of 2s orbital is less than 32. What is maximum wavelength of line of
the energy of 2p orbital in case of Balmer series of Hydrogen spectrum (R =
Hydrogen like atoms. 1.09 × 107 m–1): [AIIMS 2018]
(1) 400 nm (2) 654 nm
28. Which one is a wrong statement ? (3) 486 nm (4) 434 nm
[NEET 2018]
(1) The electronic configuration of N 33. In second orbit of H atom what is velocity
of e– [AIIMS 2018]
atom is (1) 2.18 × 106m/sec (2) 3.27 × 106m/sec
(3) 10.9 × 105m/sec (4) 21.8 × 106m/sec
(2) An orbital is designated by three
quantum numbers while an electron 34. 4d, 5p, 5f and 6p orbitals are arranged in
in an atom is designated by four the order of decreasing energy. The
quantum numbers. correct option is: [NEET 2019]
(3) Total orbital angular momentum of (1) 6p > 5f > 4d > 5p
electron in 's' orbital is equal to zero. (2) 5f > 6p > 4d > 5p
(4) The value of m for d 2 is zero. (3) 5f > 6p > 5p > 4d
z
(4) 6p > 5f > 5p > 4d

Atomic Structure 83
 35. Which of the following series of 40. A particular station of All India Radio,
transitions in the spectrum of hydrogen New Delhi. Broadcasts on a frequency of
atom falls in visible region ? 1,368 kHz (kilohertz) The wavelength of
[NEET 2019]
the electromagnetic radiation emitted by
(1) Paschen series (2) Brackett series
the transmitter is:
(3) Lyman series (4) Balmer series
[Speed of light, c = 3.0 × 108 ms–1]
36. The number of angular nodes and radial [NEET-2021]
nodes in 3s orbital are [NEET-2020] (1) 219.3 m (2) 219.2 m
(1) 0 and 2, respectively (3) 2192 m (4) 21.92 m
(2) 1 and 0, respectively
(3) 3 and 0, respectively
41. Identify the incorrect statement from the
(4) 0 and 1, respectively
following. [NEET-2022]
(1) All the five 5d orbitals are different in
37. The number of protons, neutrons and
size when compared to the respective
175
electrons in 71
L , respectively are: 4d orbitals.
[NEET-2020] (2) All the five 4d orbitals have shapes
(1) 71, 104 and 71 (2) 104, 71 and 71 similar to the respective 3d orbitals.
(3) 71, 71 and 104 (4) 175, 104 and 71 (3) In an atom, all the five 3d orbitals are
equal in energy in free state.
38. For which one of the following, Bohr (4) The shapes of dxy, dyz, and dzx orbitals
model is not valid? [NEET-2020]
are similar to each other; and dx2 − y2
(1) Hydrogen atom
(2) Singly ionized helium atom (He+) and dz2 are similar to each other.
(3) Deuteron atom 42. If radius of second Bohr orbit of the He+
(4) Singly ionized neon atom (Ne+)
ion is 105.8 pm, what is the radius of third
Bohr orbit of Li2+ ion? [NEET-2022]
39. From the following pairs of ions which
(1) 158.7 pm (2) 15.87 pm
one is not an iso-electronic pair?
(3) 1.587 pm (4) 158.7 Å
[NEET-2021]
(1) O2–, F– (2) Na+, Mg2+
(3) Mn2+, Fe3+ (4) Fe2+, Mn2+

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 2 2 2 4 2 2 1 4 3 1 3 4 4 1 1 1 1 4 3 1 4 2 3 4 1
Que. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42
Ans. 3 4 1 3 2 3 2 3 3 4 1 1 4 4 1 4 1

84 Atomic Structure