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

The correct option is B $L{i}_2^{-}<L{i}_2^{+}<L{i}_2$

The molecular electronic configuration of,
$L{i}_2=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^2$
$\therefore$ Bond order = $\frac{1}{2}(4-2)=1$
The molecular electronic configuration of,
$L{i}_2^{+}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2({\sigma }_{2s}{)}^1$
$\therefore$ Bond order = $\frac{1}{2}(3-2)=0.5$
The molecular electronic configuration of,
$L{i}_2^{-}=\left({\sigma }_{1s}{)}^2\right({\sigma }_{1s}^{\ast }{)}^2\left({\sigma }_{2s}{)}^2\right({\sigma }_{2s}^{\ast }{)}^1$
​​​​​​​$\therefore$ Bond order = $\frac{1}{2}(4-3)=0.5$
Since $L{i}_2$ has highest bond order hence $L{i}_2$ is the most stable species. The bond order of $L{i}_2^{+}$ and $L{i}_2^{-}$ is same but $L{i}_2^{+}$ is more stable than $L{i}_2^{-}$ because $L{i}_2^{+}$ has less number of electrons in anti bonding orbital compared to that of $L{i}_2^{-}$