Draw the structures of the following :
(i) `H_(2)S_(2)O_(8)`
(ii) `HClO_(4)`
(b) How would you account for the following :
(i) `NH_(3)` is a stronger base than `PH_(3)`.
(ii) Sulphur has a greater tendency for catenation than oxygen.
(iii) `F_(2)` is stronger oxidising agent than `Cl_(2)`.

Explanation

(a) `H_(2)S_(2)O_(8)` (Peroxodisulphurc acid )

(ii) `HClO_(4)` (Perchloric acid)

(ii) The property of catenation depends upon the element - element bond. Since sulphur S-S bond strength is much more than O-O bond strength . So sulphur has greater tendency for catenation than oxygen.
(iii) Since the oxidising power of halogens decreases from `F_(2)` to `I_(2)`. Therefore `F_(2)` is the stronger oxidising agent than `Cl_(2)` due smaller size and absent of d- orbital.