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 inNO_(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.

How would you account for the following : (i) H_(2)S is more acidic t

1.	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 inNO_(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.
Answer» SOLUTION :(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.