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The formation of oxide ion O^(2-)(g) fro...

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^(-) rarr O^(-)(g), DeltaH^(-) = - 141 kj mol^(-1)`
`O^(-)(g) +e^(-) rarr O^(2-) (g), DeltaH^(-) =+ 780 kj mol^(-1)`
Thus, process of formation of `O^(2-)` in gas phase is unfavourable even through `O^(2-)` is isoelectronic with neon. It is due to the fact that A) oxygen is more electronegative B) addition of electron in oxygen results in larget size of the ion C) electron repulsion outweights the stability gained by achieving noble gas configuration D) `O^(-)` ion has comparatively smaller size than oxygen atom

A

oxygen is more electronegative

B

addition of electron in oxygen results in larger size of the ion

C

electron repulsin outweighs the stability gained by achieving noble gas configuration

D

`O^(-)` ion has comparitively smaller size of the oxygen atom

Text Solution

Verified by Experts

The correct Answer is:
C

Although `O^(2-)` has noble gas configuration isolectric with neon but its formation is unfavourable due to the strong electronic 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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The formation of the oixde ion, O^(2+)(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g)+e^(-) rarr O^(-)(g), Delta H^(O)=-141kJ mol^(-) O(g)+e^(-) rarr O^(-)(g), H^(O)=-kJ mol^(-) Thus the process of formation of O^(2-) in gas phase is neon. It is due to the fact that