(a) Give reasons :
(i) `H_(3)PO_(3)` undergoes disproportionation reaction but `H_(3)PO_(4)` does not .
(ii) When `Cl_(2)` reacts with excess of `F_(2), ClF_(3)` is formed and not `FCl_(3)`.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature .
(b) Draw the structure of the following :
(i) `XeF_(4) " " (ii) HClO_(3)`

Explanation

(a) (i) It is because in `H_(3)PO_(3)` , 'P' is in `+3` , intermediate oxidation state which can increase to `+5` and decrease to `-3` where as in `H_(3)PO_(4)` , 'P' is in highest oxidation state `+5` which can only gain electrons , i.e., undergoes reduction only , acts as oxidising agent and cannot disproportionate .
(ii) As chlorine has vacant d-orbitals and hence show an oxidation state of +3 but F has no d-orbitals , therefore , it cannot show positive oxidation states . Further , since F can show only-1 oxidation state . Hence state . Hence , it can form only `ClF_(3)` and `FCl_(3)`. Moreover , because of bigger size , Cl can accommodate 3 small F atoms around it while F being smaller cannot accommodate three bigger sized Cl atoms around it .
(iii) Due to small size , high electronegativity , oxygen forms `Ppi - P pi` multiple bond . As a result , oxygen exists as `O_(2)` molecules , which are held together by weak van der waals forces of attraction which can be easily overcome at room temperature . Hence `O_(2)` is a gas at room temperature . Sulphur on the other hand , bigger size and lower electronegativity , does not form `Ppi - Ppi` multiple bond .
Hence it forms S-S single bonds which is stronger bond and has tendency of catenation and exist as in `S_(8)` form . Because of bigger size , the forces of attraction holding the `S_(8)` molecules together are much stronger which cannot be overcome at room temperature . Consequently , sulphur is a solid at room temperature.
(b) (i) `XeF_(4)`