The gas-phase conversion of an oxygen atom into oxide ion, O²⁻(g), occurs in two steps: first an exothermic electron gain and then an endothermic one, as shown below: O(g) + e⁻ → O⁻(g); ΔH° = -141 kJ mol⁻¹ O⁻(g) + e⁻ → O²⁻(g); ΔH° = +780 kJ mol⁻¹ Even though O²⁻ is isoelectronic with neon,

Question

The gas-phase conversion of an oxygen atom into oxide ion, O²⁻(g), occurs in two steps: first an exothermic electron gain and then an endothermic one, as shown below: O(g) + e⁻ → O⁻(g); ΔH° = -141 kJ mol⁻¹ O⁻(g) + e⁻ → O²⁻(g); ΔH° = +780 kJ mol⁻¹ Even though O²⁻ is isoelectronic with neon, its formation in the gaseous state is not favourable. This is because

Accepted Answer

oxygen has a higher electronegativity. The correct option highlights that oxygen's high electronegativity means it has a strong tendency to attract electrons, which contributes to the instability of the O²⁻ ion due to increased electron-electron repulsion, outweighing the stability gained from achieving a noble gas configuration.

Solution

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