The formation of the oxide ion, \( \mathrm{O}^{2-}(\mathrm{g}) \), ...
0The formation of the oxide ion, \( \mathrm{O}^{2-}(\mathrm{g}) \), from oxygen
P atom requires first an exothermic and then an
W endothermic step as shown below:
\( \mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
\( \mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1} \) 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) Oxygen is more electronegative
(2) Addition of electron in oxygen results in larger size of the ion.
(3) Electron repulsion outweighs the stability gained by achieving noble gas configuration.
(4) \( \mathrm{O}^{-} \)ion has comparatively smaller size than oxygen atom
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