When `Cl_(2)` is passed through KI solution, the solution becomes brown in colour. However, when excess of `Cl_(2)` is used, the solution becomes colourless. Explain.

Explanation

Chlorine, a stronger oxidising agent than iodine, first oxidises KI to produce `I_(2)` for which the solution becomes brown in colour (due to formation of the complex `KI_(3)`).
`2KI(aq)+Cl_(2)(g)to2KCl(aq)+I_(2)(s)`
`I_(2)+KIhArrKI_(3)`(brown)
When excess `Cl_(2)` is passed through the solution, the `I_(2)` thus formed is further oxidised to colourless iodic acid `(HIO_(3))`.
`5Cl_(2)+I_(2)+6H_(2)O-10HCl+ 2HIO_(3)` (colourless)