Question:
Two pi and half sigma bonds are present in:
Correct Option: , 4
Solution:
$\mathrm{N}_{2}^{+}=13 e^{-}$
$=\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}$
$=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{1}$
B.O. $=\frac{\begin{array}{c}\text { Bonding } \\ \text { electrons }\end{array}-\begin{array}{c}\text { Antibonding } \\ \text { electrons }\end{array}}{2}$
B.O. $=\frac{9-4}{2}=2.5=2 \pi$ bond $+0.5 \sigma$ bond