Question

Knowing the electron gain enthalpy values for 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⁻? (Hint: Consider lattice energy factor in the formation of compounds).

Answer 1

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 formation of O²⁻ is energetically more favourable than formation of O⁻.

Answer 2

Let us consider the reaction of oxygen with monopositve metal, we can have two compounds. MO(O in -1 state) and M₂O (O in -2 state). The energy required for formation of O^-2 is compensated by increased coulombic attraction between M⁺and O^-2. Coulombic force of attraction, F_A is proportional to product of charges on ions i.e.

`F_A prop (q_1q_2)/r^2`

where q₁ and q₂ are charges on ions and r is distance between ions. Same logic can be applied if metal is dispositive.