Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\mathrm{L}{\mathrm{i}}_2:{\mathrm{\sigma }}_{1\mathrm{s}}^2\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2}\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^2$Bond order$=1$

$\mathrm{L}{\mathrm{i}}_2^{+}:{\mathrm{\sigma }}_{1\mathrm{s}}^2\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2}\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^1$Bond order$=0.5$

$\mathrm{L}{\mathrm{i}}_2^{-}:{\mathrm{\sigma }}_{1\mathrm{s}}^2\mathrm{ }{\mathrm{\sigma }}_{1\mathrm{s}}^{\mathrm{*}2}\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^2\mathrm{ }{\mathrm{\sigma }}_{2\mathrm{s}}^{\mathrm{*}1}$Bond order$=0.5$

$\mathrm{L}{\mathrm{i}}_2^{-}$is less stable than$\mathrm{L}{\mathrm{i}}_2^{+}$because the incoming electron goes to anti bonding molecular orbital. So, the stability order is$\mathrm{L}{\mathrm{i}}_2>\mathrm{L}{\mathrm{i}}_2^{+}>\mathrm{L}{\mathrm{i}}_2^{-}$.