Discussion Forum

Stability of the spcies Li₂, Li₂^-, and Li₂⁺  increases in the order of

Answer:

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)