Question

Consider the following species:
$C{N}^{+},C{N}^{-},NO\text{and}CN$
Which one of these will have the highest bond order?

• A. $NO$
• B. $C{N}^{-}$
• C. $CN$
• D. $C{N}^{+}$

Answer 1

The correct option is B $C{N}^{-}$

MO electronic configuration of $NO$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2=\pi 2p_y^2,{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0$
$\text{Bond order}=\frac{10-5}{2}=2.5$

MO electronic configuration of $C{N}^{-}$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^2$
$\text{Bond order}=\frac{10-4}{2}=3$

MO electronic configuration of $CN$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^1$
$\text{Bond order}=\frac{9-4}{2}=2.5$

MO electronic configuration of $C{N}^{+}$ is $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2=\pi 2p_y^2$
$\text{Bond order}=\frac{8-4}{2}=2$

Hence, option (b) is the correct answer.