JEE Main 2019 — Chemical Bonding and Molecular Structure Question with Solution

According to molecular orbital theory $L{i}_2^{+}$ has 5 electrons $-\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^1$

$L{i}_2^{-}$ has 7 electrons $-\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^1$

$\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}=\frac{\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{l}-\mathrm{E}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{A}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{b}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{l}}{2}$

Bond order of $L{i}_2^{+}=\frac{3-2}{2}=\frac{1}{2}$

Bond order of $L{i}_2^{-}=\frac{4-3}{2}=\frac{1}{2}$

As both have similar bond order, both $L{i}_2^{+}$ and $L{i}_2^{-}$ will be stable.

Note: $L{i}_2^{-}$ will be relatively unstable to $L{i}_2^{+}$ as it has more electrons in anti bonding orbitals, so have more energy and having more repulsion so $L{i}_2^{-}$ is less stable as $L{i}_2^{+}$