How would you account for the following :
(i) `H_(2)S` is more acidic than `H_(2)O` ?
(ii) The `N - O` bond in `NO_(2)^(-)` is shorter than the `N - O` bond in `NO_(3)^(-)`.
(iii) Both `O_(2) and F_(2)` stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine.

Explanation

(i) Because bond dissociation enthalpy of `H - S` bond is lower that of `H - O` bond. Oxygen is more electronegative than S.
(ii) In the resonance structure of these two species, in `NO_(2)^(-),2` Bonds are sharing a double bond while in `NO_(3)^(-),3` bonds are sharing a double bond which means that bond in `NO_(2)^(-)` will be shorter than in `NO_(3)^(-)`.
OR
In `NO_(2)^(-)` bond order is 1.5 while in `NO_(3)^(-)`, bond order is 1.33 because of the tendency of form multiple bonds with metal.
(iii) `O_(2) and F_(2)` both stabilize higher oxidation states of metals but `O_(2)` exceeds `F_(2)` in doing so because fluorine is highly electronegative and smaller in size in comparison to oxygen.