Question

Electron will be lost from which molecular orbital when $N_2$ is converted to $N_2^{+}$ ?

• A. ${\sigma }^{\star }2p_x$
• B. $\sigma 2p_z$
• C. $\pi 2p_x$
• D. ${\pi }^{\star }2p_x$

Answer 1

Answer: (B)

This can be done by the molecular orbital theory.
First, write the molecular orbital electronic configuration of $N_2$ molecule.
The total number of electrons present in the dinitrogen molecules is $=14$ .
All the $14$ electrons distributed in the bonding and antibonding molecular orbitals given below.
$N_2\to K{K}_0\sigma 2s^2\sigma \star 2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2$
$N_2^{+}\to KK\sigma 2s^2\sigma \star 2s^2\pi 2p_x^2=\pi 2p_y^{2}2p_z^1$
As per the molecular orbital electronic configuration, we can understand that the last electron is removing from the $\sigma 2p_z$ bonding molecular orbital.