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^(-)` ?

Explanation

The values of electron gain enthalpy suggest that formation of `O^(2-)` is endothermic. However, `O^(2-)` has most stable configuration in `s_2 np_6`. In solid state, the lattice energies of the oxides having `O^(2-)` anions are very high which compensates the positive electron gain enthalpy of oxide (`O^(2-)`) formation. Thus, more oxides with `O^(2-)` ion are formed.