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Consider the following species: $CN^+$, $CN^-$, NO and CN. Which one of these will have the highest bond order?
Explanation
$CN^-$ is isoelectronic with $N_2$ (14 electrons), bond order 3.
Detailed Solution
NO: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\sigma2p_z)^2(\pi2p_x)^2 = (\pi2p_y)^2(\pi^*2p_x)^1$; BO $= \frac{10-5}{2} = 2.5$
$CN^-$: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2(\sigma2p_z)^2$; BO $= \frac{10-4}{2} = 3$
CN: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2(\sigma2p_z)^1$; BO $= \frac{9-4}{2} = 2.5$
$CN^+$: $(\sigma1s)^2(\sigma^*1s)^2(\sigma2s)^2(\sigma^*2s)^2(\pi2p_x)^2 = (\pi2p_y)^2$; BO $= \frac{8-4}{2} = 2$
So $CN^-$ has the highest bond order (3).