Question

How many number of electrons will be present in the ${\sigma }_{2p_z}$ molecular orbital of $N_2^{+}$ ion ?

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

The molecular orbital diagram of $N_2^{+}$

$N_2^{+}\to \left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2,({\sigma }_{2s}^{\ast }{)}^2[{\pi }_{2p_x}^2={\pi }_{2p_y}^2]{\sigma }_{2p_z}^1$
Therefore, number of electrons in $\sigma 2p_z$ is 1.