The formation of oxide ion {\O}^{2-}(g) require first an exothermic and then an endothermic step as shown below: {O}(g)+e^{-} \to{O}^{-}(g); \\Delta \H^\circ=-142 {\kJ} {\mol}^{-1} {\O}^{-}(q)+e^{-} \to{\O}^{2-}(q) ; \\Delta \H^\circ=844 {\kJ} {\mol}^{-1} This is because

Correct Answer is : O− ion will tend to resist the addition of another electron

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

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

Oxygen has high electron affinity

Manipal Entrance Test 2018 Chemistry Paper

Question : 23 of 50

Marks: +1, -0

The formation of oxide ion O2−(g) require first an exothermic and then an endothermic step as shown below:
O(g)+e−→O−(g); ΔH°=−142‌kJ‌mol−1
O−(q)+e−→O2−(q) ; ΔH°=844‌kJ‌mol−1
This is because

Oxygen is more electronegative
O− ion has comparatively larger size than oxygen atom
O− ion will tend to resist the addition of another electron
Oxygen has high electron affinity

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