Account for the following :
(i) Sulphur in vapour from exhibits paramagnetic behavious.
(ii) `SnCl_(4)` is more covalent the `SnCl_(2)`.
(iii) `H_(3)PO_(2)` is a stronger reducing agent then `H_(3)PO_(3)`.

Explanation

(i) Sulphur partly cxists as S2 molecular in vapour state, which has two unpaired electrons
in the antibondig `P^(ast)` molecular orbitals and hence, it exhibits paramagnetism.
(ii) According to the Fajan's rule, the central metal with more positive oxidation number,
Shows more polarising power. This in ture shows more cavalent charater.
In `SnCl_(4), Sn` exists in +4 oxidation state, while in `SnCl_(2)`, Sn exists in +2 oxidation state. So
`SnCl_(4)` is more covalent then `SnCl_(2)`.
(iii) Greater the number of element hydrogen (E -H) bonds present ina compound, greater is
the reducing power of the compound. `H_(3)PO_(2)` is a stronger reducing agent as it contains
P-H bonds, while `H_(3)PO_(2)` has only one P-H bond.