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, ho

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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, ho

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^(-)` ?

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

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.