Knowing the electron gain enthalpy values of O → O– and O → O2– as – 141 and 702 kJ mol–1 respectively, how can you account for the formation of a large number of oxides having O2– species and not O– ?
Consider the reaction of a divalent metal (M) with oxygen. The formation of MO2 and MO involves the following steps :
If electron gain-enthalpy were the only factor involved in the formation of O2– ions, we would expect that O will prefer to form O– rather than O2– ions. Actually a large number of oxides have O2– species and not O–. This is due to the following two reasons :
(i) O2– has the stable noble gas configuration
(ii) Due to higher charge on O2– than on O– ions, the lattice energy released during the formation of oxides contanining O2– species in the solid state in much higher than lattice energy released during the formation of oxides having O– species. In other words, it is the higher lattice energy of formation of oxides containing O2– species which more than compensates the higher electron gain enthalpy (DegH2) needed during formation of O2– from O– ions. Thus, formation of oxides containing O2– species is energetically more favourable than oxides containing O– species.
It is because of this reason that oxygen forms a large number of oxides having O2– species and not O–.