Question

In the formation of $N_2^{+}$ from $N_2$, the electron is removed from:

• A. $\sigma \text{orbital}$
• B. $\pi \text{orbital}$
• C. ${\sigma }^{\ast }\text{orbital}$
• D. ${\pi }^{\ast }\text{orbital}$

Answer 1

The correct option is A $\sigma \text{orbital}$

The molecular electronic configuration of $N_2$ is:
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2P_y}{)}^2\right({\pi }_{2P_z}{)}^2({\sigma }_{2P_z}{)}^2$
So the molecular electronic configuration of $N_2^{+}$ is:
$\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^2\left({\pi }_{2P_y}{)}^2\right({\pi }_{2P_x}{)}^2({\sigma }_{2P_z}{)}^1$
Hence in the formation of $N_2^{+}$ from $N_2$, the electron is lost from a $\sigma$ - orbital.