Question

Knowing the electron gain enthalpy values for $O\to O^{-}$ and $O\to O^{-2}$ as -141 and 702 kJ $mo{l}^{-1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$ ?

Answer 1

Given,

$O+e^{-}\to O^{-};\Delta e_{g}H=-141kJmo{l}^{-1}O+2e^{-}\to O^{2-};\Delta e_{g}H=+702kJmo{l}^{-1}$

As it is obvious from the given data, the formation of divalent anion $\left[O^{2-}\right]$ needs more energy as compared to monovalent anion $\left[O^{-}\right]$ where energy is actually released. Yet the number of oxides with -2 state of oxygen (e.g., $N{a}_{2}O,K_{2}Oetc$) 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 O^2-ion is much more than the oxide involving O^-ion. Hence, the oxide having O^2-ions are more stable than oxides having O^-. Hence, we can say that the formation of O^2-is energetically more favourable than the formation of O^-