KAPIL DHIMAN CHEMISTRY CLASSES

A Path From Darkness To Light Of Knowledge

TOPIC: ATOMIC STRUCTURE

EXERCIS-1 (CONCEPTUAL QUESTIONS)
DPT-1
INTRODUCTION (2) Isotones 30 31
Si,35 32
P,16 S
α -particle
14
1. Rutherford’s scattering
experiment proved that atom has :- (3) Isobars 16
8 O,17 18
8 O,8 O

(1) Electrons (2) Neutrons (4) Isoelectronic N −3 , O −2 , Cr +3

(3) Nucleus (4) Orbitals 7. The atom A, B, C have the configuration
2. A and B are two elements which have A → [ Z (90) + n(146)], B → [ Z (92) + n(146)] ,
same atomic weight and are having
C → [ Z (90) + n(148)] so that :-
atomic number 27 and 30 respectively.
(a) A and C - Isotones
If the atomic weight of A is 57 then
(b) A and C - Isotopes
number of neutron in B is :-
(c) A and B - Isobars
(1) 27 (2) 33
(d) B and C - Isobars
(3) 30 (4) 40
(e) B and C - Isotopes
3. Find out the nucleus which are
The wrong statements are:
isoneutronic :-
(1) a, b only (2) c, d, e only
(1) 14
C , 15 17
7 N, 9 F (2) 12
C , 14 19
7 N, 9 F
6 6
(3) a, c, d only (4) a, c, e only
(3) 14
6
14
C, N , F
7
17
9 (4) 14
6
14
C, N , F
7
19
9 8.(i) Fe54 ,26 Fe56 ,26 Fe57 ,26 Fe58 (a) Isotopes
26

4. Species which are isoelectronic to one

(ii) 1 H 3 ,2 He3 (b) Isotones
another are
(iii) Ge76 ,33 As 77 (c) Isodiaphers
(a) CN − (b) OH − 32

(iv) U 235 , 90Th 231 (d) Isobars

(c) CH 3+ (d) N2 92

(e) CO (v) 1 H 1 ,1 D 2 ,1 T 3
Correct answer is :- Match the above correct terms:
(1) a, b, c (2) a, c, d (1) [(i),-a], [(ii)-d], [(iii)-b], [(iv)-c], [(v)-a]
(3) a, d, e (4) b, c, d (2) [(i),-a], [(ii)-d], [(iii)-d], [(iv)-c], [(v)-a]

5. For any anion X −3 the mass number is (1) [(v),-a], [(iv)-c], [(iii)-d], [(ii)-b], [(i)-a]

14. If anion has 10 electrons, then (4) None of them

number of neutrons in X 2 nucleus :— 9. Choose the false statement about

deuterium
(1) 10 (2) 14
(1) It is an isotope of hydrogen
(3) 7 (4) 5
(2) It contains [(1e − ) + (1 p + ) + (1(n)]
6. Which of the following pairs is correct
matched (3) It contains only [(1( p + ) + (1(n)]

(1) Isotopes 40
20
40
Ca,19 K (4) D2O is called as heavy water

1
10. If the table of atomic masses is 18. Let mass of electron is half, mass of
established with the oxygen atom and proton is two times and mass of neutron
assigned value of 200, then the mass of is three fourth of original masses, then
carbon atom would be, approximately: new atomic weight of O16 atom:-
(1) 24 (2) 150 (1) increases by 37.5%
(3) 50 (4) 112 (2) remain constant
11. The relative abundance of two rubidium (3) increases by 12.5%
isotopes of atomic weights 85 and 87 (4) decreases by 25 %
are 75% and 25% respectively. The
19. An isotone of 32 Ge76 is :-
average atomic weight of rubidium is :-
(i) Ge77 (ii) As 77
(1) 75.5 (2) 85.5 32 33

(3) 86.5 (4) 87.5 (iii) 34 Se77 (iv) 34 Se78

12. The ratio of specific charge of a proton (1) (ii) & (iii) (2) (i) & (ii)
and an α —particle is (3) (ii) & (iv) (4) (ii)&(iii)&(iv)
(1) 2:1 (2) 1:2 20. In N 14 if mass attributed to electrons
7
(3) 1:4 (4) 1:1
were doubled & the mass attributed to
13. In an atom 13 Al
27
number of proton is protons were halved, the atomic mass
(a) electron is (b) and neutron is (c). would become approximately :-
Hence ratio will be [ order c: b: a] (1) Halved
(1) 13:14:13 (2) 13:13:14 (2) Doubled
(3) 14 : 13: 13 (4) 14 : 13 : 14 (3) Reduced by 25%
14. Atomic weight of Ne is 20.2. Ne is (4) Remain same
mixture of Ne and Ne relative 21. The value of planck’s constant is
abundance of heavier isotope is: 6.63 ×10−34 Js . The velocity of light is
(1) 90 (2) 20
3.0 ×108 ms −1 . Which value is closest to
(3) 40 (4) 10
the wavelength in metres of a quantum
15. Number of protons, neutrons &
of light with frequency of 8 × 1015 s −1
electrons in the element 89 γ 231
is :-
(1) 3 ×107 (2) 2 ×10−25
(1) 89, 231, 89 (2) 89, 89, 242
(3) 5 ×10−18 (4) 3.75 ×10−8
(3) 89, 142,89 (4) 89,71,89
22. If change in energy (∆E ) =3 ×10−8 J ,
16. Atoms 6 C 13 and 8 O17 are related to each other

as :- h=6.64 ×10−34 J − s and c= 3 ×108 m / s ,

(1) Isotones (2) Isoelectronic then wavelength of the lights is

0 0
(3) Isodiaphers (4) Isosters (1) 6.64 ×103 A (2) 6.64 × 105 A
17. The e/m ratio is maximum for :- 0 0
(3) 6.64 ×10−8 A (4) 6.64 ×1018 A
(1) D+ (2) He +
BOHR’S ATOMIC MODEL
(3) H+ (4) He 2+ 23. Angular momentum in second Bohr
orbit of H-atom is x. Then find out

2
 angular momentum in 1st excited state 30. The ratio of the radii of two Bohr orbits

of Li +2 ion: of H-atoms is 4: 1, what would be their

(1) 3x (2) 9x nomenclature :-

x (1) K&L (2) L&K

(3) (4) x
2 (3) N&L (4) 2 & 3 both
24. Angular momentum for P—shell 31. The velocity of electron in third excited
electron is :- state of Be3+ ion will be :-
3h 3
(1) (2) zero (1) (2.188 ×108 )ms −1
π 4
2h 3
(3) (4) None (2) (2.188 ×106 )ms −1
2π 4
25. Multiplication of electron velocity and (3) (2.188 ×106 )Kms −1
radius for a orbit in an atom is
(4) (2.188 ×103 )Kms −1
(1) Proportional to mass of electron
32. The Bohr orbit radius for the hydrogen
(2) Proportional to square of mass of
0
electron atom (n = 1)is approximately 0.530 A .
(3) Inversely proportional to mass of The radius for the first excited state (n =
electron 2) will be
(4) Does not depend upon mass of 0 0
(1) 0.13 A (2) 1.0 A
electron 0 0

26. The radius of a shell for H-atom is (3) 4.77 A (4) 2.12 A
0 33. According to Bohr theory, the radius (r)
4.761 A . The value of n is :-
and velocity (v) of an electron vary with
(1) 3 (2) 9
the increasing principal quantum
(3) 5 (4) 4
number ‘n’ as :-
27. In Bohr’s atomic model radius of 1st
(1) r increases, v decreases
orbit of Hydrogen is 0.053 nm then
(2) r and v both increases
+2
radius of 3rd orbit of Li is : (3) r & v both decreases
(1) 0.159 (2) 0.053 (4) r decreases, v increases
(3) 0.023 (4) 0.026 34. The ratio of radius of first orbit in
28. The first three radius ratio of Bohr hydrogen to the radius of first orbit in
orbits deuterium will be :-
(1) 1 : 05 : 0.5 (2) 1:2:3 (1) 1:1 (2) 1:2
(3) 1: 4 : 9 (4) 1 : 8 : 27 (3) 2:1 (4) 4:1
+2
29. For Li ion, r2 : r5 will be 35. For any H like system, the ratio of
(1) 9:25 (2) 4:25 velocities of I, II and III orbit i.e.
(3) 25:4 (4) 25:9 V1 : V2 : V3 will be
(1) 1:2:3 (2) 1 : ½ : 1/3
(3) 3:2:1 (4) 1:1: 1

3
36. The energy of H—atom in nth orbit is En

then energy in nth orbit of singly ionised

helium atom will be
(1) 4 En (2) En / 4

(3) 2 En (4) En / 2
37. The energy of second Bohr orbit of the
hydrogen atom is —328 KJ/mol. Hence 43. Going from K-shell to N-shell in case of
the energy of fourth Bohr orbit should H-atom :-
be (1) Kinetic energy decreases
(1) -41 KJ/mol (2) -1312 KJ/mol (2) Total energy decreases
(3) -164 KJ/mol (4) -82 KJ/mol (3) Potential energy decreases
38. In a hydrogen atom, if energy of an (4) None of these
ground state is —13.6 eV, then energy 44. Maximum frequency of emission is
excited state is : obtained for the transition
(1) -1.51 eV (2) -3.4 eV (1) n = 2 to n = 1
(3) -6.04 eV (4) -13.6 eV (2) n = 6 to n = 2
39. The ratio between kinetic energy and (3) n = 1 to n = 2
the total energy of the electrons of (4) n = 2 to n = 6
hydrogen atom according to Bohr’s 45. If the ionization energy of hydrogen is
model is 313.8 K cal per mole, then the energy of
(1) 2:1 (2) 1:1 the electron in 2nd excited state will be
(3) 1 : -1 (4) 1:2 (1) - 113.2 Kcal/mole
40. Potential energy is — 27.2 eV in second (2) - 78.45 Kcal/mole
orbit of He + then calculate, double of (3) - 313.8 Kcal/mole
total energy in first excited state of (4) -35 Kcal/ mole
hydrogen atom 46. Which of the following electron
(1) — 13.6eV (2) -54.4 eV transition will require the largest
(3) —6.8 eV (4) -25.2 eV amount of energy in a hydrogen atom :-

41. The energy levels for A( + Z −1) can be (1) From n= 1 to n=2
Z
(2) From n = ∞ to n=3
given by:
(3) From n = ∞ to n= 1
(1) En for A( + Z −=
1)
Z 2 × En for H
(4) From n=3 to n=5
( + Z −1)
(2) En for A = Z × En for H 47. If the potential energy (PE) of hydrogen
1 electron is —3.02 eV then in which of
(3) En for A( + Z −=
1)
× En for H
Z2 the following excited level is electron
1 present
(4) En for A( + Z −1)= × En for H
Z (1) 1st (2) 2nd
42. The graphical representation of energy (3) 3rd (4) 4th
of e- and atomic number is:-

4
48. The radiation of low frequency will be (4) No relation
emitted in which transition of hydrogen 55. Which is correct for any H like species:-
atom (1) ( E2 − E1 ) > ( E3 − E2 ) > ( E4 − E3 )
(1) n= 1 to n=4 (2) n=2 to n=5
(2) ( E2 − E1 ) < ( E3 − E2 ) < ( E4 − E3 )
(3) n =3 to n=1 (4) n=5 to n=2
(3) ( E2 − E1 ) = ( E3 − E2 ) = ( E4 − E3 )
49. A single electron orbits a stationary
nucleus (z = 5). The energy required to (4) ( E2 − E=
1) 1/ 4( E3 − E=
2) 1/ 9( E4 − E3 )
excite the electron from third to fourth 56. Which of the following is a correct
Bohr orbit will be graph:-
(1) 4.5 eV (2) 8.53 eV
(3) 25 eV (4) 16.53eV
50. The ratio of energies of hydrogen atom
for first and second excited state is
(1) 4/1 (2) 1/4
(3) 4/9 (4) 9/4
57. First excitation potential of H atom is
51. En = −313.6 / n 2 . If the value of
(1) 10.2 eV (2) 3.4 eV
En = −34.84 then to which of the
(3) 0 (4) - 3.4 eV
following values does ‘n’ correspond :- 58. Energy required to remove an e- from M
(1) 1 (2) 2 shell of H-atom is 1.51 eV, then energy
(3) 3 (4 )4 of first excited state will be :-
52. The ratio of potential energy and total (1) —1.51 eV (2) +1.51 eV
energy of an electron in a Bohr orbit of (3) —3.4 eV (4) —13.6 eV
hydrogen like species is 59. The ionisation potential of the hydrogen
(1) 2 (2) —2 atom is 13.6 eV. The energy needed to
(3) 1 (4) —1 ionise a hydrogen atom which is in its
53. Which is not a cored order of energy for second excited state is about
1, 2 & 3 orbit :- (1) 13.6 eV (2) 10.2 eV
(1) E1 > E2 > E3 (3) 3.4 eV (4) 1.5 eV

(2) ( PE )1 < ( PE ) 2 < ( PE )3 60. The ionisation energy for excited

hydrogen atom in eV will be :-
(3) ( KE )1 > ( KE ) 2 > ( KE )3
(1) 13.6
(4) 1 & 3 both
(2) Less than 13.6
54. Which of the following is a correct
(3) Greater than 13.6
relationship :-
(4) 3.4 or less
(1) E1 of H = 1/2 E2 of He + = 1/ 3E3 of 61. The energy required to excite an
Li +2 = 1/ 4 E4 of Be +3 electron of H-atom from first orbit to
+ +2 second orbit is
(2)=
E1 ( H ) E=
2 ( He ) E=
3 ( Li ) E4 ( Be +3 )
3
=
(3) E2 ( He + ) 3=
E1 ( H ) 2= E3 ( Li +2 ) 4 E4 ( Be +3 ) (1) of its ionisation energy
4

5
 1 (1) 1.25 (2) 0.25
(2) of its ionisation energy
2 (3) 5.4 (4) 10
1 69. Which transition emits photon of
(3) of its ionisation energy
4 maximum frequency :—
(4) None (1) second spectral line of Balmer series
62. The ionisation potential of a singly (2) second spectral line of Paschen
ionised helium ion is equivalent to :- series
(1) Kinetic Energy of first orbit (3) fifth spectral line of Humphery series
(2) Energy of last orbit (4) first spectral line of Lyman series
(3) Average energy in orbits 70. Which one of the following species will
(4) Maximum energy in orbits give a series of spectral lines similar to
63. The ionisation energy for the H— atom that of Mg
is 13.6 eV, then the required energy to (1) Al 3+ (2) Na
excite it from the ground state to next
(3) Mg + (4) F
higher state will be (in eV) :-
71. The ratio of minimum wavelengths of
(1) 3.4 (2) 10.2
Lyman & Balmer series will be :-
(3) 12.1 (4) 1.5
(1) 1.25 (2) 0.25
SPECTRUM AND SPECTRAL LINES
(3) 5 (4) 10
64. The spectrum of He is expected to be
72. The wavelength of photon obtained by
similar to that of:-
electron transition between two levels in
(1) H (2) Na
H—atom and singly ionised He are λ1
(3) He + (4) Li +
65. Third line of Balmer series is produced and λ2 . respectively, then :-

by which transition in spectrum of H- (1) λ2 = λ1 (2) λ2 = 2λ1

atom
(3) λ2 = λ1 / 2 (4) λ2 = λ1 / 4
(1) 5 to 2 (2) 5 to 1
73. Find out ratio of following for photon
(3) 4 to 2 (4) 4 to 1
(1) 1:16 (2) 16:1
66. Which one of the following electron
(3) 4:1 (4) 1:4
transitions between energy levels
74. The ratio of wavelengths of first line of
produces the line of shortest wavelength
Lyman series in Li +2 and first line of
in hydrogen spectrum?
(1) n2 → n1 (2) n3 → n1 Lyman series in deuterium (1 H 2 ) is :-

(1) 1:9 (2) 9:1

(3) n4 → n1 (4) n4 → n3
(3) 1:4 (4) 4:1
67. Which series have highest energy in
75. In an electronic transition atom cannot
hydrogen spectrum :-
emit :-
(1) Balmer (2) Bracket
(1) Visible light (2) γ -rays
(3) Pfund (4) Lyman
(3) Infra red light (4) Ultra violet light
68. The ratio of minimum frequency of
Lyman & Balmer series will :—

6
76. The first Lyman transition in the (3) 1 + 2 + 3 ……(x- 1)
hydrogen spectrum has ∆E =
10.2eV . (4) (x + 1) (x + 2) (x + 4)
The same energy change is observed in 83. The figure indicates the energy level
the second Balmer transition of :- diagram for the origin of six spectral

(1) Li 2+ (2) Li + lines in emission spectrum(e.g. line no.

5 arises from the transition from level B
(3) He + (4) Be3+
to X) which of the following spectral
77. The limiting line in Balmer series will
lines will not occur in the absorption
have a frequency of :-
spectrum
(1) 3.65 ×1014 sec −1 (2) 3.29 ×1015 sec −1
(3) 8.22 ×1014 sec −1 (4) −8.22 ×1014 sec −1
78. If the shortest wavelength of Lyman
series of H atom is x, then the wave
length Balmer series of H atom will be :-
(1) 1, 2, 3 (2) 3, 2
9x 36 x (3) 4,5,6 (4) 3, 2, 1
(1) (2)
5 5
84. A certain electronic transition from an
5x 5x excited state to ground state of the 112
(3) (4)
9 36
atom in one or more step gives rise to
79. The first emission line in the H-atom
three lines in the ultra violet region of
spectrum in the Balmer series will have
the spectrum. How many lines does this
wave number
transition produce in the infrared region
5 R −1 3R −1
(1) cm (2) cm of the spectrum
36 4
(1) 1 (2) 2
7R 9 R −1
(3) cm −1 (4) m (3) 3 (4) 4
144 400
85. Four lowest energy levels of H— atoms
80. What transition in He + will have the
are shown in the figure. The number of
same λ as the I line in Lyman series of
emission lines could be
H - atom
(1) 5→3 (2) 3→2
(3) 6→4 (4) 4→2
81. In H—atom, electron transits from 6th (1) 3 (2) 4
orbit to 2nd orbit in multi step. Then (3) 5 (4) 6
total spectral lines (without Balmer 86. In the above problem, the number of
series) will be absorption lines could be
(1) 6 (2) 10 (1) 3 (2) 4
(3) 4 (4) 0 (3) 5 (4) 6
82. An atom has x energy level, then total 87. If 9.9 eV energy s supplied to H atom,
number of lines in its spectrum are: the no. of spectral lines emitted is equal
(1) 1+2+3…… (x+1) to
(2) 1+2+3 …….(x2)

7
 (1) 0 (2) 1 0 0
(1) 3A (2) 5.33 A
(3) 2 (4) 3
0 0
DE-BROGLIE CONCEPT AND HEISENBERG (3) 6.88 A (4) 48 A
PRINCIPLE 95. The number of waves made by a Bohr
88. An electron has kinetic energy
electron in an orbit of maximum
2.8 ×10−23 J de- Broglie wavelength will
magnetic quantum number +2:-
be nearly (m=
e 9.1×10−31 kg ) (1) 3 (2) 4
−24 −7
(1) 9.28 ×10 m (2) 9.28 ×10 m (3) 2 (4) 1

(3) 9.28 ×10−8 m (4) 9.28 ×10−10 m 96. The uncertainity in position of an

89. What is the de-Broglie wavelength electron & helium atom are same. If the

associated with the hydrogen electron in uncertainity in momentum for the

its third orbit electron is 32 ×105 , then the

(1) 9.96 ×10−10 cm (2) 9.96 ×10−8 cm uncertainity in momentum of helium

atom will be
(3) 9.96 ×104 cm (4) 9.96 ×108 cm
90. If the de-Broglie wavelength of the (1) 32 ×105 (2) 16 ×105

fourth Bohr orbit of hydrogen atom is (3) 8 ×105 (4) None

0 97. The uncertainty in the position of an
4 A , the circumference of the orbit will
electron (mass 9.1×10−28 gm) moving
be
0 with a velocity of 3 ×104 cm sec −1 ,
(1) 4A (2) 4nm
uncertainity in velocity is 0.011% will
0
(3) 16 A (4) 16nm be:
91. No. of wave in fourth orbit :- (1) 1.92 cm (2) 7.68cm
(1) 4 (2) 5 (3) 0.175 cm (4) 3.84 cm
(3) 0 (4) 1 98. Heisenberg Uncertainity principle is not
92. What is the ratio of the De-Broglie wave valid for
lengths for electrons accelerated (1) Moving electron
through 200 volts and 50 volts :- (2) Motor car
(1) 1: 2 (2) 2:1 (3) Stationary particles
(3) 3: 10 (4) 10 : 3 (4) 2 & 3 both
93. For a valid Bohr orbit, its circumference 99. What should be the momentum (in
should be gram centimetre per second) of a
(1) = nλ (2) = (n − 1)λ particle if its de-Broglie wavelength is

(3) > nλ (4) < nλ 0

1 A and the value of h is 6.6252 x 10
94. A particle X moving with a certain second ?
velocity has a debroglie wavelength of
(1) 6.6252 ×10−19 gcm / s
0
1 A . If particle Y has a mass of 25% that (2) 6.6252 ×10−2 gcm / s
of X and velocity 75% that of X,
(3) 6.6252 ×10−24 gcm / s
debroglies wavelength of Y will be

8
 (4) 6.6252 ×10−27 gcm / s (1) 1 (2) 2

100. What should be the mass of the photon (3) 3 (4) 4

0 109. Which of the following is correct for a

of sodium if its wavelength is 5894 A , 44—electron
the velocity of light is 3 ×10 8
1
(1) =
n 4,=
 2,=
s +
metre/second and the value of h is 2
6.6252 ×10−34 kg m 2 / s ? (2) =
n 4,=
 2,=
s 0

(1) 3.746 ×10−26 kg (2) 3.746 ×10−30 kg (3) =

n 4,=
 3,=
s 0

3.746 ×10−34 kg (4) 3.746 × 10−36 kg 1

(4) n = 4,  = 3, s = +
2
101. Which of the following has least de-
110. If n = 3, then which value of  is correct
Broglie A?
(1) 0 (2) 1
(1) e− (2) p
(3) 2 (4) All of them
(3) CO2 (4) SO2 111. Energy of atomic orbitals in a particular
QUANTUM NUMBERS shell is in order:-
102. The following quantum no. are possible (1) s < p <d < f (2) s> p > d > f
for how many orbitals n = 3,  = 2 , (3) p<d<f<s (4) f >d>s>p
m=+2 112. Which statement is not correct for n = 5,
(1) 1 (2) 2 m = 2 :-
(3) 3 (4) 4 (1) = 4
103. Number of possible orbitals (all types) in (2)  = 0, 1,2,3 ; s = + 1/2
n = 3 energy level is :- (3)  =3
(1) 1 (2) 3 (4)  =2, 3, 4
(3) 4 (4) 9 113. Spin angular momentum for electron:-
104. Which sub—shell is not permissible :- h h
(1) s ( s + 1) (2) 2 s ( s + 1)
(1) 2d (2) 4f 2π 2π
(3) 6p (4) 3s h
(3) s ( s + 2) (4) None
105. Nodal plane is found in which orbital : 2π
(1) n = 2,  = 0 (2) n = 3,  =0 114. The maximum number of electrons in a
(3) n = 2,  = 1 (4) n = 1,  =0 p—orbital with n=6 and m=0 can be :-
106. No. of nodal surface in 2s orbital :- (1) 14 (2) 6
(1) 0 (2) 1 (3) 2 (4) 10
(3) 2 (4) 3 15. The total value of m for the electrons
107. Number of orbitals in h sub-shell is (n=4) is -
(1) 11 (2) 15 (1) 4 (2) 8
(3) 17 (4) 19 (3) 16 (4) 32
108. How many quantum numbers are
required to specify the position of
electron

9
116 In an atom, for how many electrons, the go into which one of the following sub-
quantum numbers will be, levels
1 (1) 4f (2) 4d
n=
3,  =
2, m =
+2, s =
+ :-
2 (3) 3p (4) 5s
(1) 18 (2) 6 124. The maximum probability of finding an
(3) 24 (4) 1 electron in the d xy orbital is
117. Which orbital is represented by the
(1) Along the x-axis
complete Wave function ψ 420 :- (2) Along the y-axis
(1) 4d (2) 3d (3) At an angle of 45° from the x and
(3) 4p (4) 4s y axis
118. An electron is in one of 44 orbital. (4) At an angle of 90° from the x and
Which of the following orbital quantum y axis
number value is not possible :- 125. Which orbitlal has two angular nodal
(1) n=4 (2)  =1 planes :-
(3) m= 1 (4) m=2 (1) s (2) p
119. A neutral atom of an element has 2K, (3) d (4) f
8L, 1 M 2N electrons. The number of s- 126. An orbital with  = 0 is symmetrical
electron in the atom are about the :-
(1) 2 (2) 8 (1) x-axis only (2) y-axis only
(3) 10 (4) 6 (3) z-axis only (4) The nucleus
120. If  = 3 then type and number of orbital 127. If n &  are principal and azimuthal
is quantum no. respectively then the
(1) 3p, 3 (2) 4f, 14 expression for calculating the total no.
(3) 5f, 7 (4) 3d, 5 of electron in any energy level is :-
121. Any nf—orbital can accomodate upto :—
(1) 14 electron
(2) Six electrons
(3) Two electrons with parallel spin
(4) Two electrons with opposite spin
RULES FOR FILLING OF ORBITALS
122. n,  and m values of an electron in 3 p y
128. Which configuration does not obey
orbital are :-
pauli’s exclusion principle:-
(1) =
n 3;=
 1 and m=1
(2) n = 3;  = 1 and m = —1
(3) Both 1 and 2 are correct
(4) None of these 129. Which of the following configuration
123. 36 Kr has the electronic configuration follows the Hund’s rule :-

(18 Ar ) As 2 3d 10 4 p 6 . The 39th electron will

10
 137. In ground state of Cr24 , number of

orbitals with paired and unpaired

130. The basis of three unpaired electrons
electron
present in the configuration of nitrogen
(1) 10 (2) 12
is :—
(3) 15 (4) 18
(1) Aufbau principle
138. For Na (Z = 11) set of quantum numbers
(2) Pauli’s principle
for last electron is:—
(3) Hund’s principle
1
(4) Uncertainty principle (1) n = 3,  = 1, m = 1, s = +
2
131. The orbital with maximum energy is
1
(1) 3d (2) 5p (2) n = 3,  = 0, m = 0, s = +
2
(3) 4s (4) 6d
1
132. n and  values of an orbital ‘A’ are 3 (3) n = 3,  = 0, m = 1, s = +
2
and 2, of another orbital ‘B’ are 5 and 0.
1
The energy of (4) n = 3,  = 1, m = 1, s = −
2
(1) B is more than A 139. Which of the following set of quantum
(2) A is more than B numbers is correct for the 19th electron
(3) A and B are of same energy of Chromium
(4) None n  m s
133. No. of all subshells of n +  = 7 is:- (1) 3 0 0 1/2
(1) 4 (2) 5 (2) 3 2 —2 1/2
(3) 6 (4) 7 (3) 4 0 1 1/2
134. Electronic configuration (4) 4 1 —1 1/2
has violated :— 140. Which set of quantum number is correct
(1) Hund’s rule for an electron in 3p orbital
(2) Pauli’s principle 1
(1) n = 3,  = 2, m = 0, s = +
(3) Aufbau principle 2
(4) (n +  ) rule (2) n=
3,  =
0, m =
+1, s =
+
1
2
135. The total spin resulting from a d9
1
configuration is:- (3) n=
3,  =
−2, m =
−1, s =
+
2
1
(1) (2) 2 1
2 (4) n = 3,  = 1, m = 0, s = +
2
3
(3) 1 (4) 141. An atom of Cr [Z= 24] loses 2 electrons.
2
How many unpaired electrons shall be
136. Which of the following transition neither
shows absorption nor emission of there in Cr +2 :

energy in case of Hydrogen atom (1) 4 (2) 3

(3) 2 (4) 1
(1) 3 p x → 3s (2) 3d xy → 3d yz
142. The atomic weight of an element is
(3) 3s → 3d xy (4) All the above
double its atomic number. If there are
11
 three electrons in 2p sub-shell, the (3) 4,0,0,+1/2 (4) 4,1,-1,+1/2
element is 149. The atomic number of the element
(1) C (2) N having maximum number of unpaired
(3) O (4) Ca 3p electrons is (in ground state):-
143. The atomic number of an element is 17, (1) 15 (2) 10
the number of orbitals containing (3) 12 (4) 8
electron pairs in the valance shell IS: 150. Which one represent is in ground state
(1) 8 (2) 2 configuration
(3) 3 (4) 6
144. A transition metal ‘X’ has a

configuration [ Ar ]3d 5 in its + 3

oxidation state, Its atomic number is:—
(1) 22 (2) 26
(3) 28 (4) 19

145. 4s 2 is the configuration of the

outermost orbit of an element. Its
atomic number would be :-
151. The electronic configuration of a
(1) 29 (2) 24
(3) 30 (4) 19 dipositive metal ion M 2+ is 2, 8, 14 and

146. Sum of the paired electrons present in its ionic weight is 58 a.m.u. The number

the orbital with  = 2 in all the species of neutrons in its nucleus would be :-
(1) 30 (2) 32
Fe 2+ , Co 2+ and Ni +2 are:—
(3) 34 (4) 42
(1) 9 (2) 12
152. In an atom having 2K, 8L, 8M and 2N
(3) 6 (4) 15
electrons, the number of electrons with
147. What is the electronic configuration of
1
an element in its first excited state m = 0; S = + are
2
which is isoelectronic with O2
(1) 6 (2) 2
(1) 2 3
[ Ne]3s 3 p 3d 1
(2) 2
[ Ne]3s 3 p 4
(3) 8 (4) 16
153. The number of electrons in the M-shell
(3) [ Ne]3s1 3 p 3 3d 2 (4) [ Ne]3s1 3 p 5
of the element with atomic number 24 is
148. The quantum number of 20th electron
(1) 24 (2) 12
of Fe(Z = 26) ion would be
(3) 8 (4) 13
(1) 3,2,—2,—½ (2) 3,2,0,½

12
KAPIL DHIMAN CHEMISTRY CLASSES
A Path From Darkness To Light Of Knowledge

TOPIC: ATOMIC STRUCTURE

ANSWER KEY
EXERCIS-I (CONCEPTUAL QUESTIONS)

Q. ANSWER Q. NO. ANSWER Q. NO. ANSWER Q. NO. ANSWER

NO.
1 3 41 1 81 1 121 4
2 1 42 4 82 3 122 3
3 1 43 1 83 3 123 2
4 3 44 1 84 1 124 3
5 2 45 4 85 4 125 3
6 2 46 1 86 1 126 4
7 4 47 2 87 1 127 4
8 1 48 4 88 3 128 2
9 3 49 3 89 2 129 1
10 2 50 4 90 3 130 3
11 2 51 3 91 1 131 4
12 1 52 1 92 1 132 1
13 3 53 1 93 1 133 1
14 4 54 2 94 2 134 1
15 3 55 1 95 1 135 1
16 3 56 3 96 1 136 4
17 3 57 1 97 3 137 3
18 1 58 3 98 4 138 2
19 3 59 4 99 1 139 3
20 3 60 4 100 4 140 4
21 4 61 1 101 4 141 1
22 3 62 1 102 1 142 2
23 4 63 2 103 4 143 3
24 1 64 4 104 1 144 2
25 3 65 1 105 3 145 3
26 1 66 3 106 2 146 2
27 1 67 4 107 1 147 1
28 3 68 3 108 3 148 3
29 2 69 4 109 1 149 1
30 4 70 1 110 4 150 3
31 4 71 2 111 1 151 2
32 4 72 4 112 2 152 1
33 1 73 2 113 1 153 4
34 1 74 1 114 3
35 2 75 2 115 3
36 1 76 3 116 4
37 4 77 3 117 1
38 1 78 2 118 2
39 3 79 1 119 2
40 3 80 4 120 3

13
 KAPIL DHIMAN CHEMISTRY CLASSES
A Path From Darkness To Light Of Knowledge

TOPIC: ATOMIC STRUCTURE

EXERCIS-II (PREVIOUS QUESTIONS)
DPT-2
1. If uncertainty in position and 6. The total number of atomic orbitals in
momentum are equal, then uncertainty fourth energy level of an atom is
in velocity is? (1) 8 (2) 16

h 1 h (3) 32 (4) 4
(1) (2)
π 2m π 7. The energies E1 and E2 of two

h 1 h radiations are 25 eV and 50eV

(3) (4)
2π m π respectively. The relation between their
2. The measurement of the electron wavelengths i.e. λ1 and λ2 will be :
position is associated with an
(1) λ1 = λ2 (2) λ1 = 2λ2
uncertainty in momentum, which is
1
equal to 1× 10−18 gcm s −1 the uncertainty (3) λ1 = 4λ2 (4) λ1 = λ2
2
in electron velocity is (mass of electron = 8. If n =6, the correct sequence for filling of
−28
9 ×10 g) electrons will be :

(1) 1×1011 cm s −1 (2) 1×109 cm s −1 (1) ns → (n − 2) f → (n − 1)d → np

(2) ns → (n − 1)d → (n − 2) f → np
(3) 1×106 cms −1 (4) 1×105 cm s −1
3. Maximum number of electrons in a (3) ns → (n − 2) f → np → (n − 1)d

subshell of an atom is determined by (4) ns → np (n − 1)d → (n − 2) f

the following :- 9. According to the Bohr Theory, which of
(1) 2n 2 (2) 4 + 2 the following transitions in the hydrogen

(3) 2 + 1 (4) 4 − 2 atom will give rise to the least energetic

4. Which of the following is not permissible photon?

arrangement of electrons in an atom? (1) n = 5 to n = 3

(1) n = 3,  = 2, m = —2, s = —1/2 (2) n = 6 to n = 1

(2) n = 4,  = 0, m = 0, s = —1/2 (3) n = 5 to n = 4

(3) n = 5,  = 3, m = 0, s = +1/2 (4) n = 6 to n = 5

(4) n = 3,  = 2, m = —3, s = —1/2 10. Smallest wavelength occurs for

5. A 0.66 kg ball is moving with a speed of (1) Lyman series

100 m/s. The associated wavelength (2) Balmar series

(3) Paschen series
will be (h = 6.6 × 10−34 Js )
(4) Brackett series
(1) 6.6 ×10−34 m (2) 1.0 ×10−35 m
11. Which of the following is wrong for Bohr
−32 −32
(3) 1.0 ×10 m (4) 6.6 ×10 m model

14
 (1) It establishes stability of atom quantum of light with frequency of
(2) It is contradicted with Heisenberg 6 ×1015 s −1 ?
uncertainity principle (1) 75 (2) 10
(3) It explain the concept of spectral (3) 25 (4) 50
lines 18. Based on equation
(4) e − behaves as particle & wave  Z2 
−2.178 ×10−18 J  2 
E= certain
12. Maximum number of electrons in a n 
subshell with  = 3 and n = 4 is: conclusions are written. Which of them
(1) 10 (2) 12 is not correct?
(3) 14 (4) 16 (1) For n = 1, the electron has a more
13. The correct set of four quantum negative energy than it does for n =6
numbers for the valence electron of which means that the electron is more
rubidium atom (Z = 37) is:- lossely bound in the smallest allowed
(1) 5, 0, 0,+½ (2) 5, 1, 0,+½ orbit.
(3) 5, 1, 1, +½ (4) 6, 0, 0 + ½ (2) The negative sign in equation simply
14. The orbital angular momentum of a p- means that the energy of electron bound
electron given as :- to the nucleus is lower than it would be
3 h h if the electrons were at the infinite
(1) (2) 6.
2π 2π distance from the nucleus
h h (3) Larger the value of n, the larger is
(3) (4) 3
2π 2π the orbit radius
15. Threshold frequency of a metal is (4) Equation can be used to calculate
5 ×1013 sec −1 which 1×10+14 sec −1 the change in energy when the electron
frequency light is focused then change orbit
maximum kinetic energy of emitted 19. What is the maximum numbers of
electron :- electrons that can be associated with

(1) 3.3 ×10−21 (2) 3.3 ×10−20 the following set of quantum numbers?
n = 3; l= l and m = —1
(3) 6.6 ×10−21 (4) 6.6 ×10−20
(1) 2 (2) 10
nh
16. In Bohr’s orbit indicates (3) 6 (4) 4
2π
20. A particle is moving with 3 times faster
(1) Momentum
(2) Kinetic energy than speed of e − . Ratio of wavelength of

(3) Potential energy particle & electron is 1.8 × 10−4 then

(4) Angular momentum particle is :-
17. The value of Planck’s constant is (1) Neutron (2) α -particle
6.63 ×10−34 Js . The speed of light is (3) Deutron (4) Tritium
21. What is the maximum number of
3 ×1017 nm s −1 . Which value is closest to
orbitals that can be identified with the
the wavelength in nanometer of a
following quantum numbers?

15
 n = 3,  = 1, m  = 0 (1) Ti 3+ (2) Cr 2+
(1) 1 (2) 2 (3) Co 2+ (4) Ni 2+
(3) 3 (4) 4 28. The angular momentum of electron in
22. Calculate the energy in joule ‘d’ orbital is equal to
corresponding to light of wavelength 45 (1) 2 (2) 2 3
−34
nm (Planck’s constant=h 6.63 ×10 Js ;
(3) 0 (4) 6
−1
speed of light c = 3 × 10 ms )
8
29. Which is the coned order of increasing
(1) 6.67 ×10 15
(2) 6.67 ×10 11
energy of the listed orbitals in the atom
(3) 4.42 ×10−15 (4) 4.42 ×10−18 of titanium? (At. no. Z = 22)
23. Magnetic moment 2.83 BM the following (1) 3s 3p 3d 4s (2) 3s 3p 4s 3d
ions? (At. nos. Ti = 22, Cr = 24, Mn=25, (3) 3s 4s Sp 3d (4) 4s 3s 3p 3d
Ni=28) :- 30. In which transition of hydrogen atom

(1) Ti 3+ (2) Ni 2+ have same wavelength as in Balmer

(3) Cr 3+ (4) Mn 2+ series transition of He + ion (n=4 to n=2)

(1) 4 to 2 (2) 3 to 2
24. The energy of an electron of 2 p y , orbital
(3) 2 to 1 (4) 4 to 1
is
****************
(1) greater than 2 px orbital

(2) Less than 2 pz orbital

(3) same as that of 2 px and 2 pz ANSWER KEY

EXERCIS-II (PREVIOUS YEAR
orbital QUESTIONS)
(4) Equal to 2s orbital
Q. NO. ANSWER Q. NO. ANSWER
25. Which of the following pairs of ions are 1 2 16 4
isoelectronic and isostructural? 2 2 17 4
(1) ClO3− , CO32− (2) SO32− , NO3− 3 2 18 1
4 4 19 2
(3) ClO3− , SO32− (4) CO32− , SO32− 5 2 20 1
6 2 21 1
26. The number of d-electrons in Fe 2+
7 2 22 4
(Z=26) is not equal to the number of
8 1 23 2
electrons in which one of the following? 9 4 24 3
(1) p-electrons in Cl (Z = 17) 10 1 25 3
(2) d-electrons in Fe (Z = 26) 11 4 26 1
(3) p-electrons in Ne (Z = 10) 12 3 27 4
13 1 28 4
(4) s-electrons in Mg (Z = 12)
14 3 29 2
27. Magnetic moment 2.84 B.M. is given 15 2 30 3
by:-
(At. no.), Ni = 28, Ti = 22, Cr =
24,Co=27)

16