What is the bond order in $\mathrm{N}_2^+$ ?

Question

Aicharya · Accepted Answer

Electronic configuration of $\mathrm{N}_2^{+}:(\sigma 1 \mathrm{~s})^2\left(\sigma^* 1 \mathrm{~s}\right)^2$ $(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{x}}\right)^2\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)^2\left(\sigma 2 \mathrm{p}_{\mathrm{z}}\right)^1$ Bond order $=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{1}{2}(9-4)=2.5$

What is the bond order in $\mathrm{N}_2^+$ ?

Solution