The formation of the oxide ion, O2$-$(g) from oxygen atom requires first an exthermic and then an endothermic step as shown below: O(g) + e$-$ $ \to $ O$-$(g); $\Delta $fHo = $-$141 kJ mol$-$1 O$-$(g) + e$-$ $ \to $ O2$-$(g); $\Delta $fHo = +780 kJ mol$-$1 Thus, process of formation of O2$-$ is isoelectronic with neon. It is due to the fact that,

Question

The formation of the oxide ion, O^2$-$_(g) from oxygen atom requires first an exthermic and then an endothermic step as shown below:
O_(g) + e^$-$ $ \to $ O^$-$_(g); $\Delta $_fH^o = $-$141 kJ mol^$-$1

O^$-$_(g) + e^$-$ $ \to $ O^2$-$_(g); $\Delta $_fH^o = +780 kJ mol^$-$1

Thus, process of formation of O^2$-$ is isoelectronic with neon. It is due to the fact that,

O^$-$ ion has comparatively smaller size than oxygen atom

oxygen is more electronegative

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

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

Solution

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