What is the bond order of the nitrogen molecule, ${\text{N}}_2$?

Solution

1. The electronic configuration of ${\text{N}}_2$ is $(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2(\sigma 2p_z{)}^2$.
2. The number of bonding electrons is $N_b=10$.
3. The number of antibonding electrons is $N_a=4$.
4. Bond order is calculated as $$\begin{array}{rl}\frac{N_b-N_a}{2}& =\frac{10-4}{2}\\ & =3\end{array}$$.
5. Hence the answer is (C).