Discussion Forum



1)

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


A) $Li_{2}^{}$ < $Li_{2}^{+}$ < $Li_{2}^{-}$

B) $Li_{2}^{-}$ < $Li_{2}^{+}$ <$Li_{2}^{}$

C) $Li_{2}^{}$ < $Li_{2}^{-}$ < $Li_{2}^{+}$

D) $Li_{2}^{- }$ < $Li_{2}^{}$ < $Li_{2}^{+}$

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)