Knowing the electron gain enthalpy values for `O rarr O^(Ө)` and `O rarr O^(2-)` as `- 141 kJ mol^-1` and `+ 702 kJ mol^-1` respectively, how can you account for the formation of a large number of oxides having `O^(2-)` species and not `O^(Ө)` ?

Explanation

Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be.
Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving `O^(2−)` ion is much more than the oxide involving `O^(−)` ion. Hence, the oxide having `O^(2−)` ions are more stable than oxides having `O^(−)`. Hence, we can say that formation of `O^(2−)` is energetically more favourable than formation of `O^(−)`.