What is the value of electron gain enthalpy of ${\mathrm{Na}}^{+}$ if ${\mathrm{IE}}_1$ of $\mathrm{Na}=5.1\mathrm{eV}$?

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 below:

$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-}\left(\mathrm{g}\right); {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\ominus }=-141 {\mathrm{kJmol}}^{-1}$

${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{2-}\left(\mathrm{g}\right); {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^{\ominus }=+780 {\mathrm{kJmol}}^{-1}$

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