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

The formation of the oxide ion `O_(g)^(2-)` requires first an exothermic and then an endothermic step as shown below:
`O_(g)+e^(-) rarr O_(g)^(-)`, `DeltaH=-142 kJ mol^(-1)`
`O(g)+e rarr O_(g)^(2-)`, `DeltaH=844kJ mol^(-1)`
This is because:

A

`O^(-)` ion has comparatively larger size than oxygen atom

B

oxygen has high electron affinity

C

`O^(-)` ion will tend to resist the addition of another electron

D

oxygen is more electronegative

Text Solution

Verified by Experts

The correct Answer is:
C
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The formation of the oxide ion O_((g))^(2-) requires first an exothermic and then an endothermic step as shown below. O_((g)) +e^(-) = O_((g))^(-) DeltaH^(@) =- 142 kJ mol^(-1) O_((g))^(-) + e^(-) = O_((g))^(2-) DeltaH^(@) = 844 kJ mol^(-1) This is because of :

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

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

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