Eduprobe
Question:
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); ΔfH° = -kJmol⁻¹
O⁻(g) + e⁻ → O²⁻(g); ΔfH° = +780kJmol⁻¹
Thus, the process of formation of O²⁻ in the gas phase is unfavorable even though O²⁻ is isoelectronic with neon. It is due to the fact that:
electron repulsion outweighs the stability gained by achieving noble gas configuration
addition of electron in oxygen results in larger size of the ion
oxygen is more electronegative
O⁻ ion has comparatively smaller size than oxygen atom
Solution:
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); ΔfH° = -kJmol⁻¹
O⁻(g) + e⁻ → O²⁻(g); ΔfH° = +780kJmol⁻¹
The overall process is endothermic because the second step requires a significant amount of energy (+780 kJmol⁻¹) to overcome the electron-electron repulsion in the already negatively charged O⁻ ion. Adding another electron to the O⁻ ion increases electron-electron repulsion, which outweighs the stability gained by achieving a noble gas configuration (isoelectronic with Neon). Therefore, the correct answer is that electron repulsion outweighs the stability gained by achieving a noble gas configuration.