Knowing the electron gain enthalpy values for $O \rightarrow O^{-}$ and $O \rightarrow O^{2-}$ 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 $O^{-}$?
(Hint: Consider lattice energy factor in the formation of compounds).

(N/A) The stability of an ionic compound is primarily determined by its lattice energy. $A$ higher lattice energy leads to greater stability of the compound.
Lattice energy is directly proportional to the product of the charges on the ions. When a metal reacts with oxygen,the lattice energy of an oxide containing $O^{2-}$ ions is significantly higher than that of an oxide containing $O^{-}$ ions.
Although the second electron gain enthalpy for oxygen is endothermic $(702 \ kJ \ mol^{-1})$,this energy is more than compensated for by the large amount of lattice energy released during the formation of the crystal lattice with $O^{2-}$ ions. Therefore,oxides containing $O^{2-}$ are more stable than those containing $O^{-}$.