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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: 

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

As learnt in

Variation of electron gain enthalpy with Zeff -

The higher the Zeff the greater is the tendency of the atom to attract the incoming electron towards itself and hence higher the amount of energy released.

- wherein

\bigtriangleup _{eg}H^{-}\:\:or\ \:EA\propto Z_{eff}

 

 We know that  E_{a}\ \alpha \ Z_{eff}

After gaining 1 electron, the value of Zeff falls and thus gaining another electron becomes unfavourable to such an extent that the process becomes endothermic. Electron repulsion at this point outweighs the stability gained by achieving noble gas configuration.


Option 1)

Electron repulsion outweighs the stability gained by achieving noble gas configuration

This option is correct

Option 2)

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

This option is incorrect

Option 3)

Oxygen is more electronegative.

This option is incorrect

Option 4)

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

This option is incorrect

Posted by

Plabita

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