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The formation of the oxide ion, $O^{2-}(g)$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O(g) + e^- \rightarrow O^-(g)$; $\Delta_fH^\circ = -141$ kJ $mol^{-1}$
$O^-(g) + e^- \rightarrow O^{2-}(g)$; $\Delta_fH^\circ = +780$ kJ $mol^{-1}$
Thus process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,
$O(g) + e^- \rightarrow O^-(g)$; $\Delta_fH^\circ = -141$ kJ $mol^{-1}$
$O^-(g) + e^- \rightarrow O^{2-}(g)$; $\Delta_fH^\circ = +780$ kJ $mol^{-1}$
Thus process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,
A
Oxygen is more electronegative
B
Addition of electron in oxygen results in larger size of the ion
C
Electron repulsion outweighs the stability gained by achieving noble gas configuration
D
$O^-$ ion has comparatively smaller size than oxygen atom
Detailed Solution
Adding the second electron to the negatively charged $O^-$ ion is opposed by strong electron–electron repulsion.
This repulsion outweighs the stability gained by reaching the neon configuration, so the second step is strongly endothermic.
This repulsion outweighs the stability gained by reaching the neon configuration, so the second step is strongly endothermic.