Question

In the molecular orbital diagram for the molecular ion, \(\mathrm{N_2^+}\), the number of electrons in the \(\sigma_{2p_z}\) molecular orbital is :

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

N2 has s-p mixing, so \(\sigma_{2p_z}\) lies above \(\pi_{2p}\); the electron removed to make N2+ comes from \(\sigma_{2p_z}\).

Answer 1

The Correct Option is B

\(\mathrm{N_2^+}\) has \(7+7-1=13\) electrons.
For \(\mathrm{N_2}\) and its ions (because of s-p mixing) the order is \(\sigma_{1s}\lt\sigma^*_{1s}\lt\sigma_{2s}\lt\sigma^*_{2s}\lt\pi_{2p_x}=\pi_{2p_y}\lt\sigma_{2p_z}\lt\pi^*_{2p_x}=\pi^*_{2p_y}\lt\sigma^*_{2p_z}\).
Filling 13 electrons: \(\sigma_{1s}^2\sigma^{*2}_{1s}\sigma_{2s}^2\sigma^{*2}_{2s}\pi_{2p_x}^2\pi_{2p_y}^2\sigma_{2p_z}^1\).
So \(\sigma_{2p_z}\) holds 1 electron.
Answer: (2) 1