Answer:
Option B
Explanation:
Li2 (3+3=6) $\sigma 1s^{2},\sigma^{*}1s^{2},\sigma 2s^{2}$
Bond order $=\frac{N_{b}-N_{a}}{2}=\frac{4-2}{2}=\frac{2}{2}=1$
Li2+ (3+3-1=5) = $\sigma 1s^{2},\sigma^{*}1s^{2},\sigma 2s^{1}$
Bond order= $\frac{3-2}{2}=\frac{1}{2}=0.5$
Li2- (3+3+1=7) = $\sigma 1s^{2},\sigma^{*}1s^{2},\sigma 2s^{2},\sigma^{*}2s^{1}$
Bond order $=\frac{4-3}{2}=\frac{1}{2}=0.5$
Stability order is $Li_{2}^{}$ < $Li_{2}^{+}$ < $Li_{2}^{-}$
(because Li2- has more number of electrons in antibonding orbitals which destabilites the species)