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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).
Solution 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−.
Solution 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, FA 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.
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