The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic step as shown below O(g) + e- = $O_{(g)}^{-}$ $\Delta $Ho = -142 kJmol-1 $O_{(g)}^{-}$ + e- = $O_{(g)}^{2-}$ $\Delta $Ho = 844 kJmol-1 This because

Question

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^o = -142 kJmol^-1

$O_{(g)}^{-}$ + e^- = $O_{(g)}^{2-}$ $\Delta $H^o = 844 kJmol^-1

This because

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

Oxygen has high electron affinity

Oxygen is more electronegative

O^- ion has comparatively larger size than oxygen atom

Solution

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