Choose the best answer
1.
Electronic configuration of species M^{2+} is 1s^{2} 2s^{2}
2p^{6} 3s^{2} 3p^{6} 3d^{6} and its atomic
weight is 56. The number of neutrons in the nucleus of species M is
a) 26
b) 22
c) 30
d) 24
Answer: (c) 30
Solution:
M^{2+} : 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}
M : 1s^{2} 2s^{2}
2p^{6} 3s^{2} 3p^{6} 3d^{8}
Atomic number = 26
Mass number =56
No. of neutrons = 56 -26 = 30
2.
The energy of light of wavelength 45 nm is
a) 6.67 × 10^{15}J
b) 6.67 × 10^{11}J
c) 4.42 × 10^{–18}J
d) 4.42 × 10_{–15}J
Answer: c) 4.42 × 10^{–18}J
Solution:
E = hν = hc/λ
=6.626 × 10^{-}^{34} J s × 3 ×10^{8} m s^{-}^{1 } /_{ }45 ×10^{-}^{9} m
= 4 .42 ×10^{-}^{18} J_{}
3.
The energies E_{1} and E_{2} of two radiations are 25 eV and 50
eV respectively. The relation between their wavelengths ie λ_{1} and λ_{2}
will be
a)
λ_{1}/λ_{2} = 1
b)
λ_{1} = 2λ_{2}
c)
λ_{1} = √(25 × 50 λ_{2})
d)
2λ_{1} = λ_{2}
Answer: b) λ_{1} = 2λ_{2}
Solution:
4.
Splitting of spectral lines in an electric field is called
a) Zeeman effect
b) Shielding effect
c) Compton effect
d) Stark effect
Answer: d) Stark effect
Solution:
Spliting of spectral lines in
magnetic field is called Zeeman effect and spliting of spectral lines in
electric field, is called Stark effect.
5.
Based on equation E = – 2.178 × 10^{–18} J (z^{2}/n^{2})
certain conclusions are written. Which of them is not correct ?
a) Equation can be used to calculate the change in energy
when the electron changes orbit
b) For n = 1, the electron has a more negative energy than
it does for n = 6 which means that the electron is more loosely bound in the
smallest allowed orbit
c) The negative sign in equation simply means that the
energy of electron bound to the nucleus is lower than it would be if the
electrons were at the infinite distance from the nucleus.
d)
Larger the value of n, the larger is the orbit radius.
Answer: b) For n = 1, the electron
has a more negative energy than it does for n = 6 which means that the electron
is more loosely bound in the smallest allowed orbit
Solution:
Correct
statement: For n=1, the
electron has more negative energy than it does for n=6 which means that the
electron is strongly bound in the
smallest allowed orbit.
6.
According to the Bohr Theory, which of the following transitions in the
hydrogen atom will give rise to the least energetic photon ?
a)
n = 6 to n = 1
b)
n = 5 to n = 4
c)
n = 5 to n = 3
d)
n = 6 to n = 5
Answer: d) n = 6 to n = 5
Solution:
n = 6 to n=5
E_{6} = -13.6 / 6^{2} ;
E_{5} = - 13.6 / 5^{2}
E_{6} - E_{5} = (-13.6 /6^{2})
- (-13.6/5^{2})
= 0.166 eV atom^{-1}
E_{5} - E_{4} = (-13.6 /5^{2})
- (-13.6/4^{2})
= 0.306 eV atom^{-1}
7. Assertion : The spectrum of He^{+} is expected to
be similar to that of hydrogen
Reason
: He^{+} is also one electron system.
(a) If both assertion and reason are true and reason is the
correct explanation of assertion.
(b) If both assertion and reason are true but reason is not
the correct explanation of assertion.
(c) If assertion is true but reason is false
(d) If both assertion and reason are false
Answer: (a) If both assertion and
reason are true and reason is the correct explanation of assertion.
8. Which of the following pairs of d-orbitals will have
electron density along the axes ?
a)
d_{z}2, d_{xz}
b)
d_{xz}, d_{yz}
c)
d_{z}2, d_{x}2_{-y}2
d)
d_{xy}, d_{x}2_{-y}2
Answer: c) d_{z}2, d_{x}2_{-y}2
9.
Two electrons occupying the same orbital are distinguished by
a)
azimuthal quantum number
b)
spin quantum number
c)
magnetic quantum number
d)
orbital quantum number
Answer: b) spin quantum number
Solution:
Spin quantum number
For the first electron m_{s} = +½
For the second electron m_{s} = -½
10.
The electronic configuration of Eu (Atomic no. 63) Gd (Atomic no. 64) and Tb
(Atomic no. 65) are
a) [Xe] 4f^{6} 5d^{1} 6s^{2}, [Xe]
4f^{7} 5d^{1} 6s^{2} and [Xe] 4f^{8} 5d^{1}
6s^{2}
b) [Xe] 4f^{7} , 6s^{2}, [Xe] 4f^{7}
5d^{1} 6s^{2} and [Xe] 4f^{9} 6s^{2}
c) [Xe] 4f^{7} , 6s^{2}, [Xe] 4f^{8}
6s^{2} and [Xe] 4f^{8} 5d^{1} 6s^{2}
d) [Xe] 4f^{6} 5d^{1} 6s^{2}, [Xe] 4f^{7} 5d^{1} 6s^{2} and [Xe] 4f^{9} 6s^{2}
Answer: b) [Xe] 4f^{7} , 6s^{2},
[Xe] 4f^{7} 5d^{1} 6s^{2} and [Xe] 4f^{9} 6s^{2}
Solution:
Eu : [Xe] 4f^{7}, 5d^{0}, 6s^{2}
Gd : [Xe] 4f^{7}, 5d^{1}, 6s^{2}
Tb: [Xe] 4f^{9}, 5d^{0}, 6s^{2}
11.
The maximum number of electrons in a sub shell is given by the expression
a)
2n^{2}
b)
2l + 1
c)
4l + 2
d)
none of these
Answer: c) 4l + 2
Solution:
2(2l+1) = 4l+2
12.
For d-electron, the orbital angular momentum is
a)
√(2h)/2π
b)
√(2)h/2π
c)
√(2x4) h/2π
d)
√(6)h/2π
Answer: d) √(6)h/2π
Solution:
Orbital angular momentum = √(l(l+1) h/2π
For d orbital = √(2 x 3) h/2π = √6 h/2π
13) What is the maximum numbers of electrons that can be
associated with the following set of quantum numbers ? n = 3, l = 1 and m = –1
a) 4
b) 6
c) 2
d) = 10
Answer: c)
2
Solution:
n = 3; l=1; m = -1
either 3p_{x} or 3p_{y}
i.e., Maximum two electrons can be accomodated either in 3p_{x} or in 3p_{y}
14. Assertion : Number of radial and angular nodes for 3p
orbital are 1, 1 respectively.
Reason
: Number of radial and angular nodes depends only on principal quantum num-ber.
a) both assertion and reason are true and reason is the
correct explanation of as-sertion.
b) both assertion and reason are true but reason is not the
correct explanation of assertion.
c) assertion is true but reason is false
d) both assertion and reason are false
Answer: c) assertion is true but
reason is false
Solution:
No. of radial node = n-l-1
No. of angular node = l
for 3p orbital
No. of angular node = l = 1
No. of radial node = n-l-1 = 3-1-1 = 1
15.
The total number of orbitals associated with the principal quantum number n = 3
is
a)
9
b)
8
c)
5
d)
7
Answer: a) 9
Solution:
n = 3; l = 0; m_{l} = 0 - one s orbital
n = 3; l = 1; m_{l} = -1, 0, 1 - three p
orbitals
n = 3; l = 2; m_{l} = -2, -1, 0, 1, 2 -
five d Orbitals
Overall nine orbitals are possible.
16.
If n = 6, the correct sequence for filling of electrons will be,
a)
ns → (n – 2) f → (n – 1)d → np
b)
ns → (n – 1) d → (n – 2) f → np
c) ns → (n – 2) f → np → (n – 1) d
d) none of these are correct
Answer: a) ns → (n – 2) f → (n – 1)d
→ np
Solution:
n = 6
According Aufbau principle,
6s →4f →5d →6p
ns → (n-1)f →(n-2)d → np
17.
Consider the following sets of quantum numbers :
Which
of the following sets of quantum number is not possible ?
a)
(i), (ii), (iii) and (iv)
b)
(ii), (iv) and (v)
c)
and (iii)
d)
(ii), (iii) and (iv)
Answer: b) (ii), (iv) and (v)
Solution:
(ii) l can have the values from 0 to n-1
n=2; possible 'l' values are 0, 1 hence l = 2 is not possible.
(iv)for l=0; m=-1 not possible
(v)for n=3 l = 4 and m = 3 not possible
18.
How many electrons in an atom with atomic number 105 can have (n + l) = 8 ?
a)
30
b)
17
c)
15
d)
unpredictable
Answer: b) 17
Solution:
n+l = 8
Electronic configuration of atom with atomic number 105 is [Rn] 5f^{14} 6d^{3} 7s^{2}
19.
Electron density in the yz plane of 3dx 2 −y2
orbital is
a)
zero
b)
0.50
c)
0.75
d)
0.90
Answer: a) zero
Solution:
Option (a) - Zero (Refer to Figure 2.9)
20.
If uncertainty in position and momentum are equal, then minimum uncertainty in
velocity is
Solution:
Answer: (c) 1/2m √h/π
21. A macroscopic particle of mass 100 g and moving at a
velocity of 100 cm s^{–1} will have a de Broglie wavelength of
a)
6.6 × 10^{–29} cm
b)
6.6 × 10^{–30} cm
c)
6.6 × 10^{–31} cm
d)
6.6 × 10^{–32} cm
Answer: c) 6.6 × 10^{–31} cm
Solution:
m = 100 g = 100 x 10^{-3} kg
v= 100 cm s^{-1} = 100 x
10^{-2} m s^{-1}
λ =h/mv
22)
The ratio of de Broglie wavelengths of a deuterium atom to that of an α -
particle, when the velocity of the former is five times greater than that of
later, is
a)
4
b)
0.2
c)
2.5
d)
0.4
Answer: d) 0.4
23)
The energy of an electron in the 3rd orbit of hydrogen atom is –E. The energy
of an electron in the first orbit will be
a)
–3E
b)
−E/3
c)
−E/9
d)
–9E
Answer: d) –9E
Solution:
24)
Time independent Schnodinger wave equation is
Answer: (a) H^{∧} ψ
= Eψ
25.
Which of the following does not represent the mathematical expression for the
Heisenberg uncertainty principle ?
a)
∆x.∆p ≥h/4π
b)
∆x.∆v ≥h/4πm
c)
∆E.∆t ≥h/4π
d)
∆E.∆x ≥h/4π
Answer: d) ∆E.∆x ≥h/4π