The formation of the oxide ion, ${\mathrm{O}}^{2-}\left(\mathrm{g}\right)$ , from oxygen atom requires first an exothermic and then an endothermic step as shown in below:

$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}⟶{\mathrm{O}}^{-}\left(\mathrm{g}\right)\mathrm{ }\mathrm{ }\mathrm{ };\mathrm{ }\mathrm{ }{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\ominus }=-141\mathrm{ }\mathrm{k}\mathrm{J}\mathrm{ }\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$

${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}⟶{\mathrm{O}}^{2-}\mathrm{ }\left(\mathrm{g}\right)\mathrm{ };\mathrm{ }\mathrm{ }{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\ominus }=+780\mathrm{ }\mathrm{k}\mathrm{J}\mathrm{ }\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$

Thus process of formation of ${\mathrm{O}}^{2-}$ in gas phase is unfavourable though ${\mathrm{O}}^{2-}$ is isoelectronic with neon. It is due to the fact that,

• Correct answer
  A

  Electron repulsion outweighs the stability gained by achieving noble gas configuration.
• B

  ${\mathrm{O}}^{-}$ Ion has comparatively smaller size than oxygen atom.
• C

  Oxygen is more electronegative.
• D

  Addition of electron in oxygen results in larger size of the ion.

Hence, the correct answer is option: 'A'.