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

Formation of `MO_2` and `MO` involves the following steps:
`M_(g)overset(triangle_iH_l)toM^(o+)overset(triangle_iH_2)toM^(2+)`
`Ooverset(triangle_(eg)H_l)toO^(ɵ)underset("Energy")overset("Lattice")toM^(2+)(O^(ɵ))_(2(s))`.
`M^(2+)+O^(2-)underset("energy")overset("lattice")toM^(2+)O^(2-)`
(i). `O^(2-)` has a stable noble gas configuration.
(ii). Due to higher charge on `O^(2-)` than on `O^(ɵ)` ions, tha lattice energy released during the formation of oxides containing `O^(2-)` ion is much larger than during the formation of oxides containing `O^(ɵ)` ion. Hence formation of oxides containing `O^(2-)` species is energetically more favourable than oxides containing `O^(ɵ)` species.