Explanation

(a)
Bond length $\propto \frac{1}{Bond\ order}$
$NO^{-} = 16e^{-} = \ \sigma 1s^{2},\ _{\sigma}^{*}1s^{2}\ ,\sigma 2s^{2},\ _{\sigma}^{*}2s^{2}\ ,\ \sigma 2p_{x}^{2},\pi 2p_{y}^{2} = \pi 2p_{z}^{2},\ _{\pi}^{*}2p_{y}^{1} = \ _{\pi}^{*}2p_{z}^{1}$
BO = $\frac{N_{b - \ }N_{a\ }}{2}$
$\ \ \ \ \ \ \ = \frac{10 - 6}{2} = 2$
Similarly BO of $NO^{+}$ will be calculated as
$NO^{+} = 14e^{-}$
BO $= \frac{10 - 4}{2} = 3$
$CN^{-} = 14e^{-},\ BO = 3$
$CN^{+} = 13e^{-},\ BO = \frac{9 - 4}{2} = 2.5$
Bond order is least for $NO^{-}$ . So, its bond length is highest.