Bond stability $\propto$ bond order

Bond order $=1 / 2 \times\left(\mathrm{N}_{\mathrm{B}}-\mathrm{N}_{\mathrm{AB}}\right)$

Order of filling MO is:

1. $\mathrm{N}_{2}$ :

valence electrons $=5+5=10$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{4}$

$\mathrm{BO}=1 / 2 \times(6-0)=3$

2. $\mathrm{N}_{2}^{+}$

valence electrons $=5+5-1=9$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{3}$

$\mathrm{BO}=1 / 2 \times(5-0)=2.5$

$3 . \mathrm{N}_{2}^{-}$

valence electrons $=5+5+1=11$

$(\sigma 2 s)^{2}(\sigma * 2 s)^{2}(\sigma 2 p)^{2}(\pi 2 p)^{4}\left(\pi^{*} 2 p\right)^{1}$

$\mathrm{BO}=1 / 2 \times(6-1)=2.5$

4. $\mathrm{N}_{2}^{2-}$

valence electrons $=5+5+2=12$

$(\sigma 2 \mathrm{s})^{2}(\sigma * 2 \mathrm{s})^{2}(\sigma 2 \mathrm{p})^{2}(\pi 2 \mathrm{p})^{4}\left(\pi^{*} 2 \mathrm{p}\right)^{2}$

$\mathrm{BO}=1 / 2 \times(6-2)=2$

bond strength or stability: $\mathrm{N}_{2}>\mathrm{N}_{2}^{+}>\mathrm{N}_{2}^{-}>\mathrm{N}_{2}^{2-}$

$\mathrm{N}_{2}^{+}$

here $\mathrm{N}_{2}^{+}$ and $\mathrm{N}_{2}^{-}$ have same bond order but later has one electron in antibonding orbital which reduces its stability as compared to $\mathrm{N}_{2}^{+}$

Therefore, A is correct.