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JEE Chemistry 2023 Question with Solution

What is the number of unpaired electrons(s) in the highest occupied molecular orbital of the following species: N2,N2+,O2,O2−N_2, N_2^+, O_2, O_2^-?

  • A

    0,1,2,10, 1, 2, 1

  • B

    2,1,2,12, 1, 2, 1

  • C

    1,0,1,01, 0, 1, 0

  • D

    2,1,0,12, 1, 0, 1

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The species are N2,N2+,O2N_2, N_2^+, O_2 and O2−O_2^-.

Find: The number of unpaired electrons in the highest occupied molecular orbital (HOMO) of each species.

Using molecular orbital theory:

For N2N_2,

σ(1s)2 σ∗(1s)2 σ(2s)2 σ∗(2s)2 π(2px)2 π(2py)2\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \pi(2p_x)^2\, \pi(2p_y)^2

The HOMO has all electrons paired, so unpaired electrons =0= 0.

For N2+N_2^+,

σ(1s)2 σ∗(1s)2 σ(2s)2 σ∗(2s)2 π(2px)2 π(2py)1\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \pi(2p_x)^2\, \pi(2p_y)^1

The HOMO contains one unpaired electron, so unpaired electrons =1= 1.

For O2O_2,

σ(1s)2 σ∗(1s)2 σ(2s)2 σ∗(2s)2 π(2px)2 π(2py)2 π∗(2px)1 π∗(2py)1\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \pi(2p_x)^2\, \pi(2p_y)^2\, \pi^*(2p_x)^1\, \pi^*(2p_y)^1

The HOMO has two unpaired electrons, so unpaired electrons =2= 2.

For O2−O_2^-,

σ(1s)2 σ∗(1s)2 σ(2s)2 σ∗(2s)2 π(2px)2 π(2py)2 π∗(2px)2 π∗(2py)1\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \pi(2p_x)^2\, \pi(2p_y)^2\, \pi^*(2p_x)^2\, \pi^*(2p_y)^1

The HOMO contains one unpaired electron, so unpaired electrons =1= 1.

Therefore, the numbers are 0,1,2,10, 1, 2, 1, so the correct option is A.

Common mistakes

  • Confusing the HOMO with the entire molecular orbital configuration. The question asks for unpaired electrons only in the highest occupied molecular orbital, not the total number in the whole species. Identify the topmost occupied orbital first, then count unpaired electrons there.

  • Using the wrong molecular orbital order for N2N_2 and O2O_2 species. The ordering of 2p2p orbitals differs across lighter and heavier diatomic molecules, so write the correct MO configuration before counting electrons.

  • Pairing electrons prematurely in degenerate π\pi or π∗\pi^* orbitals. According to Hund's rule, electrons occupy degenerate orbitals singly before pairing. Fill degenerate orbitals correctly before deciding the number of unpaired electrons.

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