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

O_{(g)}+e^{-}=O^{-}\, _{(g)};\; \Delta H^{\circ}=-142\; kJmol^{-1}

O^{-}\, _{(g)}+e^{-}=O^{2-}\, _{(g)};\; \Delta H^{\circ}=844\; kJmol^{-1}

This is because

  • Option 1)

    oxygen is more electronegative

  • Option 2)

    oxygen has high electron affinity

  • Option 3)

    O^{-} ion will tend to resist the addition of another electron

  • Option 4)

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

 

Answers (1)

As we learnt in 

 

Successive Ionization enthalpies -

The ionisation enthalpies to remove first, second, third etc.Electrons from an isolated gaseous atom are called successive ionisation enthalpies.

- wherein

\bigtriangleup _{i}H_{2}^{-}>\bigtriangleup _{i}H_{1}^{-}

\bigtriangleup _{i}H_{3}^{-}>\bigtriangleup _{i}H_{2}^{-}

and\:\:so\:\:on.

 

 As we observed before, successive ionization energies are higher. Similarly, successive electron gain enthalpies are higher.

 

That is because it is harder to gain another electron in O^{\Theta}which is already negatively charged. 

 


Option 1)

oxygen is more electronegative

This option is incorrect.

Option 2)

oxygen has high electron affinity

This option is incorrect.

Option 3)

O^{-} ion will tend to resist the addition of another electron

This option is correct.

Option 4)

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

This option is incorrect.

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