The formation of the oxide ion $\mathrm{O}^{2-}(\mathrm{g})$, from oxygen atom requires first an exothermic…

The formation of the oxide ion $\mathrm{O}^{2-}(\mathrm{g})$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} ightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} ightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Thus process of formation of $\mathrm{O}^{2-}$ in gas phase is unfavourable even though $\mathrm{O}^{2-}$ is isoelectronic with neon. It is due to the fact that
  1. Electron repulsion outweighs the stability gained by achieving noble gas configuration
  2. $\mathrm{O}^{-}$ ion has comparatively smaller size than oxygen atom
  3. Oxygen is more electronegative
  4. Addition of electron in oxygen results in larger size of the ion.

Solution

Incoming electrons occupies the smaller $\mathrm{n}=2$ shell, also negative charge on oxygen $\left(\mathrm{O}^{-}ight)$ is another factor due to which incoming electron feel repulsion. Hence electron repulsion outweigh the stability gained by achieving noble gas configuration.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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