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-) speciesand not O^(-) ?

Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2

1.	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-) speciesand not O^(-) ?
Answer» Solution :Consider the reaction of a divalent metal ( M) with oxygen . The formation of `M_2O and MO` involves the following steps : `M(g) overset(Delta_i, H_1)to M^(+) (g) overset(Delta_i/ H_2) to M^(2+) (g)` `Deltai H_1 and Delta_i H_2` are first and second ionisation enthalpies of the metal M . `O(g) overset(Delta_(eg) H_1) to O^(-) ( g) overset(Delta_(eg)H_2)to O^(2-)(g)` `Delta_(eg) H_1 and Delta_(eg)H_2` are first and second ELECTRON gain enthalpies `2M^(+)(g) + O^(-)(g) overset("Lattice energy")to M_2O (s)` `M^(2+)(g) + O_(2)^(-)(g) overset("Lattice energy ")to MO(s)` Although `Delta_i H_2` is much more than `Delta_i H_1 and Delta_(eg) H_2` is much higher than `Delta_(eg)H_1`, yet the lattice nenergy of formation of MO(s) due to higher charges is muchmore than that of `M_2O(s)`. In other words, formation MO is energetically more favourable than `M_2O`. It is due to this reason that oxygen forms PREFERABLY oxides having the `O^(2-)` species and not `O^(-)`.