The correct statement with regard to $H_2^{+}$ & $H_2^{-}$ is

The correct option is B

$H_2^{+}$ is more stable than $H_2^{-}$

Bond order $=\frac{1}{2}(Electronsinbondingorbitals-Electronsinanti-bondingorbitals)$

The electronic configuration of $H_2^{+}={\sigma }_{{1s}^1}$

B.O $=\frac{1}{2}(1-0)=\frac{1}{2}$

The electronic configuration of $H_2^{-}={\sigma }_{{1s}^2}{\sigma }_{{1s}^1}^{\ast }$

B.O $=\frac{1}{2}(2-1)=\frac{1}{2}$

Therefore, the bond order of both ions is the same. But the $H_2^{+}$ is more stable than $H_2^{-}$ even though the bond order is the same in both cases. This is because the fact that presence of an antibonding electron in $H_2^{-}$ lowers its stability as compared to $H_2^{+}$.