  Question

# Knowing the electron gain enthalpy values for O→O− and O→O−2  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− ?

Solution

## Given,  O+e−→O−;ΔegH=−141kJmol−1O+2e−→O2−;ΔegH=+702kJmol−1 As it is obvious from the given data, the formation of divalent anion [O2−] needs more energy as compared to monovalent anion [O−] where energy is actually released. Yet the number of oxides with -2 state of oxygen (e.g., Na2O,K2Oetc) is more. Stability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2-ion is much more than the oxide involving O-ion. Hence, the oxide having O2-ions are more stable than oxides having O-. Hence, we can say that the formation of O2-is energetically more favourable than the formation of O-  Suggest corrections   