Knowing the electron gain enthalpy values for `O to O^(-)` and `OtoO^(2-)` as -141 and `702 kJcdotmol^(-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

In the reaction between a bivalent metal and oxygen, the formation of `MO_(2)` and MO involves the following steps:
`M(g)overset(Delta_(i)H_(1))toM^(+)(g)overset(Delta_(i)H_(2))toM^(2+)(g)`
`O(g)overset(Delta_(eg)H_(1))toO^(-)(g)overset(Delta_(eg)H_(2))toO^(2-)(g)`
`M^(2+)(g)+2O^(-)(g)overset("Lattice energy")toM^(2+)(O^(-))_(2)(s)`
`M^(2+)(g)+O^(2-)(g)overset("Lattice energy")toM^(2+)O^(2-)(s)`
If the formation of `O^(2-)` ions was dependent on only one factor, electron-gain enthalpy, then O would have formed `O^(-)` in presence to `O^(2-)`. However, the `O^(2-)` state in most of the oxides is due to: (i) the stable, noble gas configuration of `O^(2-)` ions (ii) higher charge on `O^(2-)` than on `O^(-)` ions resulting in higher lattice energy of `O^(2-)` than `O^(-)` ions. These two factors more than compensates the higher electron gain enthalpy `(Delta_(eg)H_(2))` needed for the formation of `O^(2-)` from `O^(-)` ions. Thus formation of oxides with `O^(2-)` species is energetically more favourable than those of `O^(-)` species. Therefore in most of the oxides, oxygen contains `O^(2-)` ions.