Stability of the species $$L{i_2},Li_2^ - $$  and $$Li_2^ + $$  increases in the order of:

Solution :
$$L{i_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2},$$     ∴ Bond order $$\, = \frac{1}{2}\left( {4 - 2} \right) = 1$$
$$Li_2^ + = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^1},$$     $$B.O.$$  $$ = \frac{1}{2}\left( {3 - 2} \right) = 0.5$$
$$Li_2^ - = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^1},$$      $$B.O.$$  $$ = \frac{1}{2}\left( {4 - 3} \right) = 0.5$$
The bond order of $$Li_2^ + $$ and $$Li_2^ - $$ is same but $$Li_2^ + $$  is more stable than $$Li_2^ - $$  because $$Li_2^ + $$  is smaller in size and has $$2$$  electrons in Anti bonding orbital whereas $$Li_2^ - $$ has $$3$$ electrons in Anti bonding orbital. hence $$Li_2^ + $$  is more stable than $$Li_2^ - $$