First and second electron gain enthalpies of oxygen are `-141 and + 702 kJ mol^(-1)` How is large number of oxides accounted for ? 

Explanation

`O(g)+e^(-)rarrO^(-)(g),DeltaH=-141kJmol^(-1),O^(-)(g)+e^(-)rarrO^(2-)(g),DeltaH=+702kJ mol^(-1)`.
Formation of `O^(-)` is exothermic and `O^(2-)` is endothermic. However, oxygen is divalent and forms oxide with stable `ns^(2)np^(6)` configuration. Further, the lattice energies of oxides are very high on account of greater electrostatic forces of attraction.