atomic structure - multiple choice questions for iit-jee, neet, sat,kvpy

ATOMIC STRUCTURE

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Multiple Choice Questions for IIT – JEE/NEET/KVPY/SAT

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(a) 325 nm

(b) 743 nm

(c) 518 nm

(d) 1035 nm

1. A gas absorbs a photon of 355 nm and emits at two wavelengths.

If one of the emissions is at 680 nm, the other is at:

The wavelength of absorbed radiation is related to those of emitted radiation as

Therefore,

Solving, we get l2 = 743 nm.

absorbed 1 2

1 1 1λ λ λ

2

1 1 1355 680 λ


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(a)8.82 x 10-17 J atom-1

(b) 4.41 x 10-16 J atom-1

(c) -4.41 x 10-17 J atom-1

(d) 2.2 x 10-15 J atom-1

2. Ionization enthalpy of He+ is 19.6 x 10-18 J atom-1. The energy of the first stationary state (n = 1) of Li2+ is

Solution in next Slide


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The ionization enthalpy (IE) is expressed as

Given that

So,

(1)Now, for Li2+, we have

Multiplying and dividing by , we get

From Eq. (1), we have

+ + + +2 2

2 2He He He He

1 1IE 13.6Z 13.6 Z (where Z 2)1

+18

HeIE 19.6 10 J

+2 18 1He

13.6 Z 19.6 10 J atom

2+ 2+ 2+2 2

1 2 2 2Li Li Li

1 1 1( ) 13.6Z 13.6Z1 1

E

+2He

Z

2+

2+ + +

+

22 2Li

1 2Li He HeHe

Z 9( ) 13.6Z 13.6ZZ 4

E

2+18 17 1

1 Li

9( ) 19.6 10 4.41 10 J atom4

E

(a)1.52 x 10-4 m

(b) 5.10 x 10-3 m

(c) 1.92 x 10-3 m

(d) 3.84 x 10-3 m

3. In an atom, an electron is moving with a speed of 600 m s-

1 with an accuracy of 0.005%. Certainty with which the position of the electron can be located is (h = 6.6 x 10-34 kg m2 s-1, mass of electron me = 9.1 x 10-31 kg)

Using Heisenberg’s uncertainty principle,

or

Given that = 0.005% of 600 ms-1, so

Hence,

.4πhx m v

4πhxm v

v

10.005600 0.03 ms100

v

343

31

6.6 10 1.92 10 m4 3.14 9.1 10 0.03

x


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(a)0.032 nm

(b) 0.40 nm

(c) 2.5 nm(d) 14.0 nm

4. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 x 103 m s-1 (mass of proton = 1.67 x 10-27 kg and h = 6.63 x 10-34 J s).

Using de Broglie relationship

34

27 3

6.63 10λ 0.40 nm1.67 10 10

hmv


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(a)8.51 x 105 J mol-1

(b) 6.56 x 105 J mol-1

(c) 7.56 x 105 J mol-1

(d) 9.84 x 105 J mol-1

5. The ionization enthalpy of hydrogen atom is 1.312 x 106 J mol-1. The energy required to excite the electron in the atom from n = 1 to n = 2 is

The energy for stationary states n = 1 and n = 2 are, respectively,

Therefore,

6 6

1 22 2

1.312 10 1.312 10and E(1) (2)

E

6 65 1

2 1 2

1.312 10 1.312 10 9.84 10 J mol2 1

E E E


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(a)n = 3, l = 0, ml = 0, ms = +1/2

(b) n = 3, l = 1, ml = 1, ms = +1/2

(c) n = 3, l = 2, ml = 1, ms = +1/2(d) n = 4, l = 0, ml = 0, ms = +1/2

6. Which of the following sets of quantum numbers represents the highest energy of an atom?

According to (n + l) rule, more is the value of (n + l) more is the energy. Hence, when n = 3 and l = 2, then (n + l) = (3 + 2) = 5.


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(a)4

(b) 8

(c) 16

(d) 32

7. The total number of atomic orbitals in fourth energy level of an atom is

When n = 4, the possible values of l are

Therefore, the total number of atomic orbitals = 1 + 3 + 5 + 7 = 16.


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(a)

(b)

(c)

(d)

8. The energies E1 and E2 of two radiations are 25 eV and 50 eV, respectively. The relation between their wavelengths, that is, and will be

1λ 2λ

1 21λ λ2

1 2λ λ

1 2λ 2λ

1 2λ 4λ


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(a)

(b)

(c)

(d)

9. If n = 6, the correct sequence of filling of electrons will be

( 1) ( 2)ns np n d n f

( 2) ( 1)ns n n f n d np

( 1) ( 2)ns n n d n f np

( 2) ( 1)ns n f np n d

Filling is done in accordance with Aufbau’s (n + l) rule. Lower value of (n + l) is preferred. If there is the same value of (n + l), then lower value of n is preferred. Hence, the correct sequence will be ( 2) ( 1)ns n n f n d np


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(a)Mn3+

(b) Fe3+

(c) Co3+

(d) Ni3+

10. Which one of the following ions has electronic

configuration

[Ar] 3d 6 ?

The atomic number of Co is 27. Now, for Co3+ (24), the electronic configuration is

1s2 2s2 2p6 3s2 3p6 4s0 3d6 = [Ar]3d6


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(a)2n2

(b) 4l + 2(c) 2l + 1

(d) 4l - 2

11. Maximum number of electrons in a subshell of an atom is determined by the following:

Each shell has n subshells and each orbital has 2 electrons. Therefore, the maximum number of electrons is 2n2.


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(a)n = 3, l = 2, ml = -2, ms = -

1/2

(b) n = 4, l = 0, ml = 0, ms = -

1/2

(c) n = 5, l = 3, ml = 0, ms =

+1/2

(d) n = 3, l = 2, ml = -3, ms = -

1/2.

12. Which of the following is not permissible arrangement of electrons in an atom ?

Possible value of ml = -l to +l. Hence, for l = 2, the possible values of ml = -2 to +2 and not -3.


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(a)1 x 1011 cm s-1

(b) 1 x 109 cm s-1

(c) 1 x 106 cm s-1

(d) 1 x 105 cm s-1

13. The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 x 10-18 g cm s-1. The uncertainty in electron velocity is (mass of an electron is 9 x 10-28 g).

Uncertainty in momentum ( p) = m v (uncertainty in velocity). Therefore,

1810 9 1

28

10 0.1 10 1 10 cm s9 10

v


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(a)470.5 nm

(b) 587.5 nm

(c) 387.5 nm

(d) 487.5 nm

14. Applying Bohr’s model when electron in hydrogen atoms comes from n = 4 to n = 2, calculate the wavelength of the line. In this process, write whether energy is released or absorbed.

Also write the range of radiation. (Given that RH = 2.18 x

10-18 J, h = 6.63 x 10-34 J s.) Solution in next Slide


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Applying Rydberg’s formula,

2H 2 2

1 2

1 1λhcE R Z

n n

For hydrogen atom, Z = 1. The transition from n = 4 to n = 2 means n2 = 4, n1 = 2. Therefore,

Therefore,

or

In this process, energy is released. The transition lies in the visible region.

18 18 192 2

1 1 32.18 10 2.18 10 4.08 10 J2 4 16

E

194.08 10 Jλhc

34 8

19

6.63 10 3 10λ 487.5 nm4.08 10

Solution in next Slide

15. Consider the following sets of quantum numbers:

n l ml ms

(i) 3 0 0 +1/2(ii) 2 2 1 +1/2(iii) 4 3 -2 -1/2(iv) 1 0 -1 -1/2(v) 3 2 3 +1/2

Which of the following sets of quantum number is not possible?

(a)(i), (ii), (iii) and (iv)

(b) (ii), (iv) and (v)(c) (i) and (iii)

(d) (ii), (iii) and (iv)


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(ii) is not possible for any value of n because l varies from 0 to (n - 1) thus for n = 2, l can be only 0, 1, 2.

(iv) is not possible because for l = 0, ml = 0.

(v) is not possible because for l = 2, ml varies from -2 to +2.


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16. The maximum number of electrons that can have principal quantum number, n = 3, a spin quantum

number, ms = -1/2, is ______.

For n = 3, total number of orbitals = n2

= 9 which means that 18 electrons can

be accommodated in 9 orbitals in

which 9 will have clockwise spin (ms =

+1/2) and 9 will have anticlockwise

spin (ms = -1/2).

9


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17. The work function (W0) of some metals is listed below.

The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is __________.

M eta l

W 0 (eV )

L i

2 .4

N a

2 .3

K

2 .2

M g

3 .7

C u

4 .8

A g

4 .3

F e

4 .7

P t

6 .3

W

4.7 5

Given that wavelength is l = 300 nm = 300 x 10-9 m = 3 x 10-7 m. Therefore, energy is

For a metal to show photoelectric effect, its work function has to be less than or equal to 4.1 eV. So, the number of metals having work function less than 4.1 eV are Li, Na, K and Mg.

4

4

34 8 1919

7 19

6.6 10 3 10 6.6 106.6 10 J 4.1 eVλ 3 10 1.6 10hcE hv


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(a)1 and diamagnetic.

(b) 0 and diamagnetic.

(c) 1 and paramagnetic.

(d) 0 and paramagnetic.

18. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is

The electronic configuration of B2 molecule is

Therefore, bond order is

So, the nature is diamagnetic as there are no unpaired electrons.

2 2 2 2 22B (10) σ1 σ*1 σ2 σ*2 π2 xs s s s p

6 4 1.2


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19. Read the following paragraph and answer the questions that follow:

Paragraph for Questions 19(i) - (iii): The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light, the ion undergoes transition to a state S2. The state S2 has radial node and its energy is equal to the ground state energy of the hydrogen atom.

(i) The state S1 is (a) 1s (b) 2s (c) 2p (d) 3s

(ii) Energy of the S1 in units of the hydrogen atom ground state energy is (a) 0.75 (b) 1.50 (c) 2.25

(d) 4.50

(iii) The orbital angular momentum quantum number of the state S2 is (a) 0 (b) 1 (c) 2 (d) 3


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(i) (b) For S1 (spherically symmetrical) state, number of nodes = 1 so n - 1 = 1 and n = 2 For S2 state, radial node = 1. Therefore, energy of S2 state is

Now,

So, state in S1 is 2s and S2 is 3p.

(ii) (c) The energy of the S1 in units of the hydrogen atom ground

state energy is

(iii) (b)Azimuthal quantum number for S2 is l = 1.

2

2

2

13.6 in ground state 13.6S HZE E

n

2 2

13.6 9 3SE nn

1 13.6 9 2.254 ( 13.6)

S

H

EE

(A)Azimuthal quantum number determines the orbital angular momentum of electron.(B)Pauli’s exclusion principle states that an orbital can contain maximum of two electrons.(C) Principal quantum number determines the size of the orbital, azimuthal quantum number determines the shape of the orbital and magnetic quantum number determines the orientation of the orbital.(D) Three of the quantum numbers (principal, azimuthal, magnetic) together represent the probability density of electron.

A (q); B (s); C (p, q, r); D (p, q, r).


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