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

${L{i}_2}^{-}$ < ${L{i}_2}^{+}$ < $L{i}_2$

${L{i}_2}^{-}<{L{i}_2}^{+}<L{i}_2$

$L{i}_2$ has 6 electrons. The configuration will be $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2$

$B.O.=\frac{4-2}{2}=1$

${L{i}_2}^{+}$ has 5 electrons. The configuration will be $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^1$

$B.O.=\frac{3-2}{2}=0.5$

${L{i}_2}^{-}$ has 7 electrons. The configuration will be $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^1$

$B.O.=\frac{4-3}{2}=0.5$

We see that both ${L{i}_2}^{-}$ and ${L{i}_2}^{+}$ have the same bond order. But ${L{i}_2}^{-}$ has a higher number of antibonding electrons. So, ${L{i}_2}^{+}$ will be more stable.