Knowing the electron gain enthalpy values for ` O to O^(-) and O to O^(2-) ` as `-141 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^(–)`?
(Hint: Consider lattice energy factor in the formation of compounds).

Explanation

Althrough the formation of `O^(2-)` anion requires more energy in comparison to the formation of `O^(Ө)` anion (actually energy is released), yet in large number of oxides, oxygen is divalent in nature. This is due to the fact that lattice energies of the oxides having `O^(2-)` anions are very high on account of greater magnitude of electrostatic forces of attraction.