The formation of the oxide ion O²⁻(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g) + e⁻ → O⁻(g); \Delta H° = -141 kJ mol⁻¹ O⁻(g) + e⁻ → O²⁻(g); \Delta H° = +780 kJ mol⁻¹ Thus, the process of formation of O²⁻ in gas phase is unfavourable even though O²⁻ is isoelectronic with neon. It is due to the fact that:

The formation of the divalent oxide ion ($O^{2-}$) involves two steps:

1. First Electron Gain: $O + e^- \rightarrow O^-$. This is exothermic ($\Delta H = -141$ kJ/mol) because the neutral oxygen atom attracts the electron.
2. Second Electron Gain: $O^- + e^- \rightarrow O^{2-}$. This step is strongly endothermic ($\Delta H = +780$ kJ/mol). Even though $O^{2-}$ achieves the stable noble gas configuration of Neon ($2s^2 2p^6$), energy is required to overcome the strong electrostatic repulsion between the negatively charged $O^-$ ion and the incoming negatively charged electron.

This repulsive force dominates the energy landscape, making the overall process in the gas phase energetically unfavourable (endothermic), despite the stability of the octet.