How would you account for the following :
(i) `H_(2)S` is 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 florine.

Explanation

(i) It is because bond dissociation energy of H-S bond than H-O bond due to longer bond length.
(ii)
Since in `NO_(3)^(-)` there are more resonating structures then `NO_(2)^(-)`. Thus the `N-O` bond in `NO_(2)^(-)` is shorter than `N-O` bond in `NO_(3)^(-)`.
(iii) It is because oxygen can gain two electrons, therefore, lattice energy can overcome ionisation energy of the metal to show higher oxidation state.