Give appropriate reasons for each of the following :
(a)Addition of `Cl_2` to Kl solution gives a brown colour but excess of `Cl_2` turns it colourless
(b)Perchloric acid is a stronger acid than sulphuric acid.
( c)HI can not be prepared by heating Nal with concentrated `H_2SO_4`

Explanation

(a)`Cl_2` being a stronger oxidising agent than `I_2`, first oxidises KI to `l_2` which imparts brown colour to the solution.But when `Cl_2` is passed in excess , the `I_2` so formed gets further oxidised to `HIO_3` (colourless)
`2KI(aq)+Cl_2(g)to2KCl(aq)+I_2(s), 5Cl_2+I_2+6H_2Oto10HCl+2HIO_3`
(b)Oxidation state of Cl in `HCIO_4` is `+7` and that of S in `H_2SO_4` is +6.(Cl is more electronegative than S).As a result, `ClO_(3)` part of `HCIO_4` can break the O-H bond more easily to liberate a proton than `SO_2` part in `H_2SO_4`Thus `HCIO_4` is a stronger acid than `H_2SO_4`
( c)HI is a stronger reducing agent It, therefore, reduces `H_2SO_4` to `SO_2` and itself gets oxidised `l_2`
`Nal+H_(2)SO_(4)toNaHSO_(4)+HI]xx2`
`H_(2)SO_(4)+2HItoSO_(2)+l_(2)+2H_(2)O`
`overline(2Nal+3H_2SO_4toSO_2+l_2+ 2H_2O+SO_2`