The formation of the oxide ion, {\O}^{2-}{ }_{( {\g})} from oxygen atom requires first an exthermic and then an endothermic step as shown below: {\O}_{( {\g})}+ {\e}^{-} \to {\O}^{-}{ }_{( {\g})} ; ∆_{ {\f}} {\H}^{\circ}=-141 {\kJ} {\mol}^{-1} {\O}^{-}{ }_{( {\g})}+ {\e}^{-} \to {\O}^{2-}{ }_{( {\g})} ; ∆_{ {\f}} {\H}^{\circ}=+780 {\kJ} {\mol}^{-1} Thus, process of formation of {\O}^{2-} is isoelectronic with neon. It is due to the fact that, [NEET 2015]

Correct Answer is : electron repulsion outweighs the stability gained by achieving noble gas configuration.

{\O}^{-}ion has comparatively smaller size than oxygen atom

addition of electron in oxygen results in larger size of the ion

electron repulsion outweighs the stability gained by achieving noble gas configuration.

Class 12 NEET Chemistry P Block Elements Questions

Question : 29 of 101

Marks: +1, -0

The formation of the oxide ion, O2−‌(g) from oxygen atom requires first an exthermic and then an endothermic step as shown below:
O(g)+e−→O−‌(g);∆fH∘=−141‌kJ‌mol−1
O−‌(g)+e−→O2−‌(g);∆fH∘=+780‌kJ‌mol−1
Thus, process of formation of O2− is isoelectronic with neon. It is due to the fact that,

[NEET 2015]

O−ion has comparatively smaller size than oxygen atom
oxygen is more electronegative
addition of electron in oxygen results in larger size of the ion
electron repulsion outweighs the stability gained by achieving noble gas configuration.

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