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| 1. |
Knowing the electron gain enthalpy values for O to O^(-) and O to O^(2-)as -141 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^(-)? |
| Answer» Solution :The VALUES of ELECTRON gain ENTHALPY suggest that formation of `O^(2-)`is endothermic. However, `O^(2-)`has most stable CONFIGURATION in `s_2 np_6`. In SOLID state, the lattice energies of the oxides having `O^(2-)`anions are very high which compensates the positive electron gain enthalpy of oxide (`O^(2-)`) formation. Thus, more oxides with `O^(2-)`ion are formed. | |