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

the repulsion between added electrons is greater than the gain in stability from attaining a noble gas arrangement. Adding a second electron to the already-negative O- ion is opposed by electron-electron repulsion that outweighs the stability gained from the noble-gas configuration (+780 kJ/mol step). That repulsion, not electronegativity, is why gaseous O2- formation is unfavourable.

Solution

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