Increasing- order of bond length in $\mathrm{NO},{\mathrm{NO}}^{+}$ and ${\mathrm{NO}}^{-}$

Electronic configuration of No:

$\begin{array}{l}\sigma 1s^2\sigma 1s^2\sigma 2s^2\sigma 2s^2\pi 2P_x^2=\pi 2P_y^2\sigma 2P_z^2.\\ \pi 2P_x=\pi 2P_y\end{array}$

Bond order $=\frac{N_b-Na}{2}=\frac{10-5}{2}=2.5$

Electronic configuration of  ${\mathrm{NO}}^{-}$:

$\begin{array}{l}\sigma 1s^2\sigma 1s^2\sigma 2s^2\sigma 2s^2\pi 2P_x^2=\pi 2P_y^2\sigma 2P_z^2\\ \pi 2P_x=\pi 2P_y\end{array}$

Bond order $=\frac{N_b-Na}{2}=\frac{10-6}{2}=2$

Here the

Bond order  $\propto \frac{1}{\text{ Bond length }}$

So order of bond length in increasing order will be

${\mathrm{NO}}^{+}<{\mathrm{NO}}^2<{\mathrm{NO}}^{-}$

Hence correct option is B.