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  • The formation of the oxide ion, from oxygen atom requires first an exothermic and then

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

\begin{aligned} &\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta_{f} H^{\ominus}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta_{f} H^{\ominus}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

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

Option: 1

Electron repulsion outweighs the stability gained by achieving noble gas configuration
 


Option: 2

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


Option: 3

Oxygen is more electronegative
 


Option: 4

Addition of electron in oxygen results in larger size of the ion.


Answers (1)

best_answer

Factual.

Due to small size of Oxygen atom, there are large interelectronic repulsions when it is added with two additional electrons. 

These interelectronic repulsions outweigh the noble gas configuration achieved by the oxide ion and hence the overall process is endothermic.

Posted by

Irshad Anwar

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