Question

Stability of the species $L{i}_2,L{i}_2^{-}$ and $L{i}_2^{+}$ increases in the order of

• A. $L{i}_2^{-}<L{i}_2<L{i}_2^{+}$
• B. $L{i}_2^{-}<L{i}_2^{+}<L{i}_2$
• C. $L{i}_2<L{i}_2^{-}<L{i}_2^{+}$
• D. $L{i}_2<L{i}_2^{+}<L{i}_2^{-}$

Answer 1

Answer: (B)

$L{i}_2:{\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2$ Bond order $=1$
$L{i}_2^{+}:{\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^1$ Bond order $=0.5$
$L{i}_2^{-}:{\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 1}$ Bond order $=0.5$
$L{i}_2^{-}$ is less stable than $L{i}_2^{+}$ because the incoming electron goes to anti-bonding molecular orbital. So, the stability order is $L{i}_2>L{i}_2^{+}>L{i}_2^{-}$ .