${\mathrm{N}}_2$and ${\mathrm{O}}_2$ are converted to monocations ${\mathrm{N}}_2^{+}$ and ${\mathrm{O}}_2^{+}$ respectively, which is wrong statement?

Bond order $=1/2\times \left(N_B-N_{AB}\right)$

Order of filling MO is:

1   ${\mathrm{N}}_2:$ 

valence electrons  $=5+5=10$
$\begin{array}{l}\left(\sigma 2\mathrm{s}{)}^2\right(\sigma \ast 2\mathrm{s}{)}^2\left(\sigma 2\mathrm{p}{)}^2\right(\pi 2\mathrm{p}{)}^4\\ \mathrm{BO}=1/2\times (6-0)=3\end{array}$

2   ${{\mathrm{N}}^{+}}_2:$
valence electrons  $=5+5-1=9$

$\begin{array}{l}\mathrm{BO}=1/2\times (5-0)=2.5\\ 3\cdot {\mathrm{O}}_2:\end{array}$

valence electrons  $=6+6=12$

$\begin{array}{l}\left(\sigma 2\mathrm{s}{)}^2\right(\sigma \ast 2\mathrm{s}{)}^2\left(\sigma 2\mathrm{p}{)}^2\right(\pi 2\mathrm{p}{)}^4{\left({\pi }^{\ast }2\mathrm{p}\right)}^2\\ \mathrm{BO}=1/2\times (6-2)=2\end{array}$

3    ${\mathrm{O}}_2^{+}:$

valence electrons $=6+6-1=11$

$\begin{array}{l}\left(\sigma 2\mathrm{s}{)}^2\right(\sigma \ast 2\mathrm{s}{)}^2\left(\sigma 2\mathrm{p}{)}^2\right(\pi 2\mathrm{p}{)}^4{\left({\pi }^{\ast }2\mathrm{p}\right)}^1\\ \mathrm{BO}=1/2\times (6-1)=2.5\end{array}$

Bond strength  $\propto$ bond order

bond strength/order ${\mathrm{N}}_2>{\mathrm{N}}_2^{+}$ and ${\mathrm{O}}_2^{+}>{\mathrm{O}}_2$ 

${\mathrm{N}}_2^{+}$ is paramagnetic with one unpaired $e^{-}$

Hence correct option is D.