• # question_answer 61)   The formation of the oxide ion, ${{O}^{2-}}$, from oxygen atom requires first an exothermic and then an endothermic step as shown below : $O(g)+e\to {{O}^{-}}(g);\,\,\,\,\,\,\,\,\,\Delta {{H}^{O-}}=-141\,kJ\,mo{{l}^{-1}}$ ${{O}^{-}}(g)+e\to {{O}^{2-}}(g);\,\,\,\,\,\,\,\,\,\Delta {{H}^{O-}}=+780\,kJ\,mo{{l}^{-1}}$ 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 an 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

(c) There is a lot of repulsion when similar charges approach each other as $O_{(gas)}^{-}$ and electron are both negatively charged. To add an electron under such situation the force of repulsion is to be overcome by applying external energy.