The formation of the oxide ion, $O^{2–}$ (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below : <br> $O(g) + e^- \to O^- (g), \Delta_f H^{\ominus} = -141 kJ\;mol^{-1}$ <br> $O^-(g) + e^- \to O^{2-} (g), \Delta_f H^{\ominus} = +780 kJ\;mol^{-1}$ <br> 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 :

Question

( A ) addition of electron in oxygen results in larger size of the ion
( B ) $O^–$ ion has comparatively smaller size than oxygen atom
( C ) oxygen is more electronegative
( D ) electron repulsion outweighs the stability gained by achieving noble gas configuration