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  • The formation of the oxide ion O2-(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:  <img alt="O(g)+e^{-}\rightarrow O^{-}(g)

The formation of the oxide ion O2-(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below: 

O(g)+e^{-}\rightarrow O^{-}(g); \Delta_{f}H^{\ominus}=-141kJ\:\:mol^{-1}

O^{-}(g)+e^{-}\rightarrow O^{2-}(g); \Delta_{f}H^{\ominus}=+780kJ\:\: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

Option: 1

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


Option: 2

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

The first electron gain enthalpy of the Oxygen atom is negative, as when the electron is added, the attractive forces between the incoming electron and nucleus outweigh the repulsions in between electrons.  

But on adding the second electron, the repulsions take over attractive forces, hence it's an endothermic process.

The correct answer is option 1.

Posted by

Gunjita

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