Question

In the conversion of $N_2$ into $N_2^{+}$, the electron will be lost from which of the following molecular orbital?

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

Answer 1

The correct option is B $\sigma 2p_z$

$N_2\to 14$ electrous

$MOT$ configuration:

$\sigma {15}^2\sigma {15}^2\sigma {25}^2\sigma {25}^2\left[\begin{array}{c}\pi 2{px}^2\\ \pi 2{py}^2\end{array}\right]\sigma 2p{z}^2$

$N_2^{+}\to 13$ electrons

$MOT$ configuration:

$\sigma {15}^2{\sigma }^x{15}^2\sigma {25}^2{\sigma }^x{25}^2\left[\begin{array}{c}\pi 2{px}^2\\ \pi 2{py}^2\end{array}\right]\sigma 2p{z}^1$

$\Rightarrow$ Electron is lost from $\sigma 2pz$ orbital.

Therefore, the answer is option (B).