Saved Bookmarks
| 1. |
Knowing the electron gain enthalpy values of O rarr O^(-) and O rarr 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 :Consider the reaction of a divalent metal (M) with oxygen. The formation of `MO_(2)` and MO involves the following steps : `{:(M(g) overset(Delta_(i)H_(1))RARR M^(+)(g) overset(Delta_(i)H_(2))rarr M^(2+)(g) ","O(g) overset(Delta_(eg)H_(1))rarr O^(-)(g)overset(Delta_(eg)H_(2))rarr O^(2-)(g)),(M^(2+)(g)+2O^(-)(g) overset("Lattice energy")rarr M^(2+)(O^(-))_(2)(s)","M^(2+)(g) overset("Lattice energy")rarr M^(2+) O^(2-)(s)):}` If electron gain enthalpy were the only factor involved in the formation of `O^(2-)` ions, we would expect that O will prefer to form `O^(-)` rather than `O^(2-)` ions. ACTUALLY a large number of oxides have `O^(2-)` species and not `O^(-)`. This is due to the following two reasons : (i) `O^(2-)` has the stable noble GAS configuration (ii) Due to higher charge on `O^(2-)` than on `O^(-)` ions, the lattice energy released during the formation of oxides containing `O^(2-)` species in the solid state is MUCH higher than lattice energy released during the formation of oxides having `O^(-)` species. In other words, it is the higher lattice energy of formation of oxides containing `O^(2-)` species which more than compensates the higher electron gain enthalpy `(Delta_(eg)H_(2))` needed during formation of `O^(2-)` from `O^(-)` ions. Thus, formation of oxides containing `O^(2-)` species is energyetically more favourable than oxides containing `O^(-)` species. It is because of this reason that oxygen forms a larger number of oxides having `O^(2-)` species and not `O^(-)`. |
|