Explanation

(a)
$CO(14) = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2},\pi 2p_{y}^{2} = \ \pi 2p_{z}^{2},\ \sigma 2p_{x}^{2}$
BO = $\frac{N_{b - \ }N_{a\ }}{2} = \frac{10 - 4}{2} = 3$
$NO^{-}(16) = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2},\sigma 2p_{x}^{2},(\pi 2p_{y}^{2} = \pi 2p_{z}^{2})$,
$\ \ _{\pi}^{*}2p_{y}^{1} = \ _{\pi}^{*}2p_{z}^{1}$
BO = $\frac{10 - 6}{2} = 2$
$NO^{+}(13);\$BO$= 3$
$CN^{-}(14);\$BO$= 3$
$N_{2}(14);BO = 3$
Hence, bond order of $NO^{-}$ is different from that of CO.