Question

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

• 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 $L{i}_2^{-}<L{i}_2^{+}<L{i}_2$

$L{i}_2\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2$
$\therefore B.O=\frac{N_b-N_A}{2}=\frac{4-2}{2}=1$

$L{i}_2^{+}\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^1$ $\therefore BO=\frac{3-2}{2}=\frac{1}{2}=0.5$

$L{i}_2^{-}\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^1$ $\therefore B.O=\frac{4-3}{2}=\frac{1}{2}=0.5$

Stability order is $L{i}_2^{-}<L{i}_2^{+}<L{i}_2$ as $L{i}_2^{-}$ has more electrons in the antibonding orbitals which destabilises the species.