Question

Knowing the electron gain enthalpy values for \[\ce{O -> O-}\] and \[\ce{O -> O^{2-}}\] 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⁻?

(Hint: Consider lattice energy factor in the formation of compounds).

Answer 1

Let us consider the reaction of oxygen with monopositive metal; we can have two compounds. MO(O in −1 state) and M₂O (O in −2 state). The energy required for the 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 the distance between ions. The same logic can be applied if metal is dispositive.

Answer 2

Because of the larger charge on O²⁻ and M²⁺ species, the lattice energy of the synthesis of oxide-containing O²⁻ species (i.e., oxide of the type MO) is significantly higher than that of oxide-containing O-species (i.e., oxide of the type M₂O).

Therefore, inspite of the fact that \[\ce{\Delta_{eg}H_{(O \leftarrow O^{2-})} >> \Delta_{eg}H_{(O \leftarrow O-)}}\] and \[\ce{\Delta_iH_{(M^+ \leftarrow M^{2+})} >> \Delta_iH_{(M \leftarrow M^{+})}}\], In terms of energy, MO formation is more favourable than M₂O formation. This explains why the number of oxides of type MO is significantly higher than that of oxides of type M₂O.