Chemistry · Chemical Bonding

NEET (UG) 2023 — Question 57

The correct order of energies of molecular orbitals of $N_2$ molecule, is:

Option A:
$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$
Option B:
$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \sigma^*_{2p_z} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^*_{2p_x} = \pi^*_{2p_y})$
Option C:
$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma_{2p_z} < \sigma^*_{2p_z}$
Option D:
$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$
Correct

Answer: D

Step-by-step solution

For homonuclear diatomic molecules of second-period elements with atomic number $\leq 7$ (Li₂ to N₂), the order of molecular orbital energies differs from that for O₂ and F₂. The molecular orbital configuration of N₂ (14 electrons) follows the sequence: $\sigma_{1s}$ , $\sigma^*_{1s}$ , $\sigma_{2s}$ , $\sigma^*_{2s}$ , $(\pi_{2p_x} = \pi_{2p_y})$ , $\sigma_{2p_z}$ , $(\pi^*_{2p_x} = \pi^*_{2p_y})$ , $\sigma^*_{2p_z}$ . Thus the increasing order of energy is: $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$ . Hence option D is correct.

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Chemistry
Chapter: Chemical Bonding
Topic: Molecular Orbital Theory