Knowing the electron gain enthalpy values forO 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^(-) ?

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

1.	Knowing the electron gain enthalpy values forO 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^(-) ?
Answer» SOLUTION :The second electron enthalpy of oxygen for the formation of `O^(2-)`is positive while first electron gain enthalpy of oxygen for the formation of `O^-`is negative. This means that if electron gain enthalpy is the only factor involved for the formation of divalent ions, we would expect that oxygen would prefer to form `O^-`ions rather than `O^(2-)`ions. ACTUALLY, a large number of oxides have `O^(2-)`species and not `O^-`. This is because (i) Divalent `O^(2-`) has the stable noble gas CONFIGURATION (II) In the solid state, large amount of energy KNOWN as lattice enthalpy is released to form divalent `O^(2-)`ions than monovalent `O^-` ions. It is the greater lattice enthalpy of `O^(2-)`ions which compensates for the high energy required to remove the second electron. This is responsible for greater stability of `O^(2-)`ions as compared to `O^-`ions.