(a) (i) Write the disproportionation reaction of `H_(3)PO_(3)`
(ii) Draw the structure of `XeF_(4)`
(b) Account for the following:
(i) Although Fluorine has less negative electron gain enthalpy yet `F_(2)` is strong oxidizing agent.
(ii) Acidic character decreases from `N_(2)O_(3)` to `Bi_(2)O_(3)`, in group 15.
(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved.

Explanation

(i) See Q. 14 (ii), Paper-2013, Outside Delhi, Set-I, IPage 202].
(ii) See Q 11 (i), Paper-2009, Delhi Board, Set-I, [Page 28]
(b) (i) F has although low electron affinity but low dissociation energy and have high hydration energy of its ion therefore, fluorine is strongest oxidizing agent.
(ii) On moving down the group, the atomic size increases, electronegativity decrease5 and metallic character increases and hence, acidic character decreases.
(c) Aqueous acidified solution of potassium dichromate is orange in colour. When sulphur dioxide gas is bubbled through the solution the colour changes to green. This is because sulphur dioxide reduces the dichromate `CrO_(7)^(2-)` which contains chromium in +6 oxidation state to `Cr^(3+)`.
`K_(2)Cr_(2)O_(7)(aq)+3SO_(2)(g)+H_(2)SO_(4)(aq)toCr_(2)(SO_(4))_(3)+K_(2)SO_(4)(aq)+H_(2)O(l)`