A large number of oxides can be formed with `O^(2-)` species but not with `O^(-)` species-explain.
[Given, `Delta_(eg)H_(OtoO-)=-141kJcdotmol^(-1) ,Delta_(eg)H_(OtoO^(2-))=702kJcdotmol^(-1)`]

Explanation

Electron gain enthalpy of the process `OtoO^(-)` is negative but that of `OtoO^(2-)` is positive. There are a large number of oxides containing `O^(2-)` species and not `O^(-)`, because 1. `O^(2-)` has a stable noble gas electronic configuration and 2. due to higher charge on `O^(2-)` than on `O^(-)` ions, lattice energy released during the formation of oxides containing `O^(2-)` species in the solid state is much higher than that during the formation of oxides containing `O^(-)` species. It more than compensates the higher electron gain enthalpy required during the process `O^(-)toO^(2-)`. Therefore, the formation of oxides containing `O^(2-)` species is energetically more favourable than the formation of oxides containing `O^(-)` species. Hence, oxygen forms a large number of oxides having `O^(2-)` species instead of `O^(-)` .