PSEB Class 11 Chemistry Chapter 2 Structure of Atom Multiple Choice Questions

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Question 1. How many electrons can fit in the orbital for which n = 3 and l = 1?

  1. 2
  2. 6
  3. 10
  4. 14

Answer: 1. 2

For n=3 and l = 1, the subshell is 3p and a particular 3p orbital can accommodate only 2 electrons.

Question 2. Which of the following pairs of d-orbitals will have electron density along the axes?

  1. \(d_{z^2}, d_{x z}\)
  2. \(d_{x z}, d_{y z}\)
  3. \(d_{z^2}, d_{x^2-y^2}\)
  4. \(d_{x y} d_{x^2-y^2}\)

Answer: 3. \(d_{z^2}, d_{x^2-y^2}\)

∴ \(d_{x^2-y^2}\) and \(d_z^2\) orbitals have electron density along the axes while dxy, dyz, and dxz orbitals have electron density in between the axes.

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Question 3. Two electrons occupying the same orbital are distinguished by

  1. Azimuthal quantum number
  2. Spin quantum number
  3. Principal quantum number
  4. Magnetic quantum number.

Answer: 2. Spin quantum number

PSEB Class 11 Chemistry Chapter 2 Structure of Atom Multiple Choice Questions

For the two electrons occupying the same orbital values of n, I and ml are the same but ms is different, -31, i.e., \(\frac{1}{2}\) and –\(\frac{1}{2}\)

Question 4. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium? (Atomic number Z = 22)

  1. 4s 3s 3p 3d
  2. 3s 3p 3d 4s
  3. 3s 3p 4s 3d
  4. 3s 4s 3p 3d

Answer; 3. 3s 3p 4s 3d

Ti(22) : ls² 2s² 2p6 3s² 3p6 4s² 3d²

∴ The order of increasing energy is 3s,3p,4s,3d.

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PSEB Class 11 Chemistry Chapter 2 Structure of Atom MCQs

Question 5. The number of d-electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which one of the following?

  1. d-electrons in Fe (Z = 26)
  2. p-electrons in Ne (Z = 10)
  3. s-electrons in Mg (Z = 12)
  4. p-electrons in Cl (Z= 17)

Answer: 4. p-electrons in Cl (Z= 17)

Number of d-electrons in Fe2+ = 6

Number of p-electrons in Cl = 11

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Question 6. The angular momentum of an electron in the ‘d’ orbital is equal to

  1. 2√3 h
  2. h
  3. √6 h
  4. √2 h

Answer: 3. √6 h

Angular momentum = \(\sqrt{l(l+1)} \hbar\)

For d-orbital, l = 2

Angular momentum = \(\sqrt{2(2+1)} \hbar=\sqrt{6} \hbar\)

Question 7. What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l= 1, ml = 0

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 1. 1

Only one orbital, 3pz has the following set of quantum numbers, n = 3, l = 1, and ml = 0.

Question 8. What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = -1

  1. 4
  2. 2
  3. 10
  4. 6

Answer: 2. 2

The orbital associated with n = 3, l = 1 is 3p. One orbital (with m= -1) of 3p-subshell can accommodate a maximum of 2 electrons.

Question 9. The outer electronic configuration of Gd (Atomic number 64) is

  1. \(4 f^5 5 d^4 6 s^1\)
  2. \(4 f^7 5 d^1 6 s^2\)
  3. \(4 f^3 5 d^5 6 s^2\)
  4. \(4 f^4 5 d^5 6 s^1\)

Answer: 2. \(4 f^7 5 d^1 6 s^2\)

The electronic configuration of 64Gd is [Xe]4f7 5d1 6s2

Question 10. The maximum number of electrons in a subshell with l = 3 and n = 4 is

  1. 14
  2. 16
  3. 10
  4. 12

Answer: 1. 14

l = 3 and n= 4 represents 4f. So, total number of electrons in a subshell = 2(2l + 1) = 2(2 x 3 + 1) = 14 eiectrons.

Hence, the f-subshell can contain a maximum of 14 electrons.

Structure of Atom Multiple Choice Questions PSEB Class 11

Question 11. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is

  1. 5, 1,1,+1/2
  2. 6, 0, 0,+1/2
  3. 5, 0, 0, +1/2
  4. 5,1, 0, +1/2

Answer: 3. 5, 0, 0, +1/2

Rb(37): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1

For 5s, n =5,l=0,m=0, s = +1/2 or -1/2

Question 12. The orbital angular momentum of a p-electron is given as

  1. \(\frac{h}{\sqrt{2} \pi}\)
  2. \(\sqrt{3} \frac{h}{2 \pi}\)
  3. \(\sqrt{\frac{3}{2}} \frac{h}{\pi}\)
  4. \(\sqrt{6} \frac{h}{2 \pi}\)

Answer: 1. \(\frac{h}{\sqrt{2} \pi}\)

Orbital angular momentum (m) = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\)

For p-electrons; l = 1

Thus, m = \(\sqrt{1(1+1)} \frac{h}{2 \pi}=\frac{\sqrt{2} h}{2 \pi}=\frac{h}{\sqrt{2} \pi}\)

Question 13. The total number of atomic orbitals in the fourth energy level of an atom is

  1. 8
  2. 16
  3. 32
  4. 4

Answer: 2. 16

The total number of atomic orbitals in any energy level is given by n².

Question 14. If n = 6, the correct sequence for the filling of electrons will be

  1. ns → (n – 2)f → (n – 1)d → np
  2. ns → (n – 1 )d  → (n – 2)f → np
  3. ns → (n – 2)f → np → (n -1)d
  4. ns → np → (n – 1)d → (n – 2)f

Answer: 1. ns → (n – 2)f →(n – 1)d → np

Question 15. The maximum number of electrons in a subshell of an atom is determined by the following

  1. 2l +1
  2. 4l – 2
  3. 2n²
  4. 4l + 2

Answer: 4. 4l+ 2

For a given shell, n the number of subshells = (2l + 1)

Since each orbital can accommodate 2 electrons of opposite spin, so maximum number of electrons in a subshell = 2(2l+1) =4l +2.

Question 16. Which of the following is not a permissible arrangement of electrons in an atom?

  1. n = 5, l = 3, m = 0, s = +1/2
  2. n = 3, l = 2, m = -3, s = -1/2
  3. n – 3, l = 2, m = -2, s = -1/2
  4. n = 4, l = 0, m = 0, s = -1/2

Answer: 2. n = 3, l = 2, m = -3, s = -1/2

In an atom, for any value of ru, the values of l = 0 to (n-1).

For a given value of l, the values of ml = -l to 0 to +l and the value of s = +1/2 or -1/2.

In option (2), l = 2 and ml = -3

This is not possible, as values of m7 which are possible for l = 2 are -2, -1,0, +1 and +2 only

Structure of Atom Multiple Choice Questions PSEB Class 11

Question 17. The orientation of an atomic orbital is governed by

  1. Principal quantum number
  2. Azimuthal quantum number
  3. Spin quantum number
  4. Magnetic quantum number.

Answer: 4. Magnetic quantum number.

The principal quantum number represents the name, size, and energy of the shell to which the electron belongs.

The azimuthal quantum number describes the spatial distribution of electron cloud and angular momentum. The magnetic quantum number describes the orientation or distribution of the electron cloud.

The spin quantum number represents the direction of electron spin around its own axis.

Question 18. In a given atom no two electrons can have the same values for all four quantum numbers. This is called

  1. Hunds Rule
  2. Aufbau principle
  3. Uncertainty principle
  4. Pauli’s Exclusion Principle.

Answer: 4. Pauli’s Exclusion principle.

This is a Pauli’s exclusion principle

Question 19. The order of filling of electrons in the orbitals of an atom will be

  1. 3d, 4s, 4p, 4d, 5s
  2. 4s, 3d, 4p, 5s, 4d, 4s
  3. 5s, 4p, 3d, 4d, 5s
  4. 3d,4p,4s,4d,5s

Answer: 2. 4s, 3d, 4p, 5s, 4d, 4s

The higher the value of n +l) for an orbital, the higher is its energy. However, if two different types of orbitals have the same value of (n +l), the orbital with the lower value of n has lower energy

Question 20. The electronic configuration of Cu (atomic number 29) is

  1. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^9\)
  2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)
  3. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 4 p^6 5 s^2 5 p^1\)
  4. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 4 p^6 3 d^3\)

Answer: 2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)

The electronic configuration of Cu+ is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)

Question 21. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are

  1. 2
  2. 4
  3. 6
  4. 8

Answer: 3. 6

n=2, l = 1

It means 2p-orbitals.

Total no. of electrons that can be accommodated in all the 2p orbitals = 6

PSEB Class 11 Chemistry Structure of Atom MCQs with Answers

Question 22. An ion has 18 electrons in the outermost shell, it is

  1. Cu+
  2. Th4+
  3. Cs+
  4. K+

Answer: 1. Cu+

Cu+ ion has 18 electrons in its outermost shell. The electronic configuration of Cu+ is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10}\)

Question 23. The number of unpaired electrons in N2+ is/are

  1. 2
  2. 0
  3. 1
  4. 3

Answer: 3. 1

N2+ = 1s²2s³2p¹

∴ No. of unpaired electrons = 1

Question 24. The maximum number of electrons in a subshell is given by the expression

  1. 4l – 2
  2. 4l + 2
  3. 2l + 2
  4. 2n²

Answer: 2. 4l + 2

No. of orbitals in a subshell = 2l + l

No. of electrons = 2(2l + 1) = 4l + 2

Question 25. The number of spherical nodes in 3p orbitals is/are

  1. One
  2. Three
  3. None
  4. Two

Answer: 1. One

No. of radial nodes in 3p-orbita = n – l – 1 = 3 – 1 -1 = 1

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