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.
$N_{2} \rightarrow KK_{0}\sigma 2s^{2}\sigma \star2s^{2}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}\sigma 2p_{z}^{2}$
$N_{2}^{+} \rightarrow KKσ2s^{2}\sigma \star2s^{2}\pi 2p_{x}^{2}=\pi 2p_{y}^{2}2p_{z}^{1}$
As per the molecular orbital electronic configuration, we can understand that the last electron is removing from the $\sigma 2p_{z}$ bonding molecular orbital.