Ncert Exercises & Solutions

The formation of oxide ion O^2-(g), from oxygen atom, requires first an exothermic and then an endothermic step as shown below

$O (g) + e^{-} \to O^{-} (g); ∆ H^{⊝} = - 141 kJ {mol}^{- 1}$
$O^{-} (g) + e^{-} \to O^{2} (g); ∆ H^{⊝} = + 780 kJ {mol}^{- 1}$

Thus, the process of formation of O^2- in the gas phase is unfavourable even though O^2- is isoelectronic with neon. It is due to the fact that:

1.	Oxygen is more electronegative
2.	Addition of electron in oxygen results in a larger size of the ion
3.	Electron repulsion outweighs the stability gained by achieving noble gas configuration
4.	O^- ion has comparatively smaller size than oxygen atom

Hint: Use the concept of electronic repulsion

Although O^2- has noble gas configuration isoelectronic with neon but its formation is unfavourable due to the strong electronic repulsion (interelectronic repulsion) between the negatively charged O^- ion and the second electron being added.

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

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