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

The formation of the oxide ion, `O^(2-)`(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:
`O(g) + e^(-) to O^(-) (g), DeltaH^(@) = -141 "kJ mol"^(-1)`
`O^(-)(g) + e^(-) to O^(2-) (g), DeltaH^(@) = +780"kJ 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,

A

oxygen is more electronegative.

B

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

C

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

D

Lattice energy of oxide formation compensate energy gained during anion formation.

Text Solution

Verified by Experts

The correct Answer is:
D
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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

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