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First and second electron gain enthalpies of oxygen are -141 and + 702 kJ mol^(-1) How is large number of oxides accounted for ? |
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Answer» SOLUTION :`O(g)+e^(-)rarrO^(-)(g),DeltaH=-141kJmol^(-1),O^(-)(g)+e^(-)rarrO^(2-)(g),DeltaH=+702kJ mol^(-1)`. Formation of `O^(-)` is EXOTHERMIC and `O^(2-)` is endothermic. However, oxygen is divalent and forms OXIDE with stable `ns^(2)np^(6)` configuration. Further, the lattice ENERGIES of oxides are very HIGH on account of greater electrostatic forces of attraction. |
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