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InterviewSolution
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Knowing the electron gain enthalpy values for ` O to O^(-) " and " O to O^(2-) " as " -141 " kJ mol"^(-1) " and " 702 " kJ mol"^(-1)` respectively, how can you account for the formation of a large number of oxides having `O^(2-)` species and not `O^(-)` ? |
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Answer» According to available data : `O +e^(-) to O^(-) , Delta_(eg) (H)=-141 " kJ mol"^(-1)` ` O=2e^(-) to O^(2-) , Delta_(eg) H=+ 702 " kJ mol"^(-1)` Although the formation of divalent anion `(O^(2-))` needs more energy as compared to monovalent anion `(O^(-))` where energy is actually released, still in large number of oxides (e.g., `Na_(2)O, K_(2)O, CaO` etc.) oygen is divalent in nature. This is on account of a more stable crystal lattice because of greater magnitude of electrostatic forces of attraction involving divalent oxygen than the oxides in which oxygen is monovalent in nature. |
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