Q. The formation of the oxide ion, $O^{2-}(g)$, from oxygen atom requires first an exothermic and then an endothermic step as shown below : $O(g) + e^- \rightarrow O^-(g)$ ; $\Delta_fH^\circ$ = -141 kJ $mol^{-1}$ $O^-(g) + e^- \rightarrow O^{2-}(g)$ ; $\Delta_fH^\circ$ = +780 kJ $mol^{-1}$ 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,
A) Oxygen is more electronegative
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
Explanation
[Solution] Adding an electron to the negatively charged $O^-$ ion faces strong repulsion, which outweighs the stability of the noble gas configuration. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.