Question

The formation of oxide ion, $O^{2-}\left(g\right)$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O\left(g\right)+e^{-}\to O_{\left(g\right)}^{-};{}_{\Delta f}{\text{H}}^{\ominus }=-141{\text{kJ mol}}^{-1}$
$O^{-}\left(g\right)+e^{-}\to O_{\left(g\right)}^{2-};{}_{\Delta f}{\text{H}}^{\ominus }=+780{\text{kJ mol}}^{-1}$
Thus the process of formation of $O^{2-}$ in gas phase is unfavourable even thought $O^{2-}$ is isoelectric with neon. It is due to the fact that,

• A. Oxygen is more electropositive
• 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

Answer 1

The correct option is C Electron repulsion outweighs the stability gained by achieving noble gas configuration

Second electron gain enthalpy is positive (energy absorbed) because the incoming electron causes more repulsion when it enters the uninegative anion.