A solution of `KBr` is treated with each of the following. Which one would liberate bromine ?

Options

A. `Cl_2`
B. HI
C. `I_2`
D. `SO_3`

Answer

B,C

Explanation

`Cl_2 and SO_3` being stronger oxidising agent , oxidise `Br^-` to `Br_2` gas
`2Br^(-)+Cl_(2)toBr_2+2Cl^(-)`
`SO_3 to SO_2 + [O]`
`2Br^(-)+2H^(+)+[O]to Br_2+H_2O`
`2Br^(-)+SO_3 + 2H^(+)to Br_2 + H_2O +SO_2`
HI being reducing agent and `I_2` being weak oxidising agent (comes after bromine) don't liberate bromine from `Br^-`.