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Question

Knowing the electron gain enthalpy values for OO and OO2  as -141 and 702 kJ mol1 respectively, how can you account for the formation of a large number of oxides having O2 species and not O ?


Solution

Given, 

O+eO;ΔegH=141kJmol1O+2eO2;ΔegH=+702kJmol1

As it is obvious from the given data, the formation of divalent anion [O2] needs more energy as compared to monovalent anion [O] where energy is actually released. Yet the number of oxides with -2 state of oxygen (e.g., Na2O,K2Oetc) 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 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 the formation of O2-is energetically more favourable than the formation of O-

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