The formation of oxide ion $\mathrm{O}^{2-}$ from oxygen atom requires an exothermic reaction, followed by…
The formation of oxide ion $\mathrm{O}^{2-}$ from oxygen atom requires an exothermic reaction, followed by an endothermic step as shown below. The process of formation of $\mathrm{O}^{2-}$ in gas phase is unfavorable $\left(\Delta H^{\ominus}=+\right.$ ve $)$, even though it has stable configuration of the nearest noble gas neon. This is because
$\begin{aligned} & \mathrm{O}(g)+e^{-} \longrightarrow \mathrm{O}^{-}(g) ; \Delta H^{\ominus}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{O}^{-}(g)+e^{-} \longrightarrow \mathrm{O}^{2-}(g) ; \Delta H^{\ominus}=+760 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
$\mathrm{O}^{2-}$ has larger size than that of neon
oxygen is more electronegative than neon
$\mathrm{O}^{2-}$ has larger size than oxygen atom
electron repulsion in oxide ion is more which over comes the stability achieved by noble gas configuration.
Solution
The process of formation of $\mathrm{O}^{2-}$ in the gas phase is unfavorable even through $\mathrm{O}^{2-}$ is isoelectronic with neon. It is due to that electronic repulsion outweighs the stability gained by achieving noble gas configuration. The formation of oxide ion, $\mathrm{O}^{2-}(g)$ from oxygen atom require first an exothermic and then endothermic step.
$
\begin{aligned}
& \mathrm{O}(g)+e^{-} \longrightarrow \mathrm{O}^{-}(g) ; \Delta H^{\ominus}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{O}^{-}(g)+e^{-} \longrightarrow \mathrm{O}^{2-}(g) ; \Delta H^{\ominus}=+760 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$
When an electron is added to $\mathrm{O}^{-}$anion, there is strong electrostatic repulsion between two negative charge due to this, the second electron gain enthalpy of oxygen is positive