Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Explanation

The oxidising power of halogens decreases in the order `F_(2)gtCl_(2)gtBr_(2)gtl_(2).F_(2)` being the strongest oxidant can oxidise `Cl^(-),Br^(-)` and `l^(-)` ions. `Cl_(2)` oxidises `Br^(-)` and `l^(-)` ions, whereas `Br_(2)` can oxidise only `l^(-)` ions. `l_(2)` is unable to oxidise none of these. The reactions are as follows:
Oxidising reactions of `F_(2)` :
`{:(F_(2)(g)+2Cl^(-)(aq)to2F^(-)(aq)+Cl_(2)(g)),(F_(2)(g)+2Br^(-)(aq)to2F^(-)(aq)+Br_(2)(l)),(F_(2)(g)+2l^(-)(aq)to2F^(-)(aq)+l_(2)(s)):}}`
Oxidising reactions of `Cl_(2)` :
`{:(Cl_(2)(g)+2Br^(-)(aq)to2Cl^(-)(aq)+Br_(2)(l)),(Cl_(2)(g)+2l^(-)(aq)to2Cl^(-)(aq)+l_(2)(g)):}}`
Oxidising reaction of `l_(2)` :
`Br_(2)(l)+2l^(-)(aq)to2Br^(-)(aq)+l_(2)(s)}`
Thus, fluorine is the best oxidant.
The reducing power of hydrohalic acids decreases in the order `HlgtHBrgtHClgtHF.HI` and HBr reduce `H_(2)SO_(4)` to `SO_(2)` while HCl and HF are unable to do so.
`2HBr+H_(2)SO_(4)toSO_(2)+2H_(2)O+Br_(2)`
`2HI+H_(2)SO_(4)toSO_(2)+2H_(2)O+l_(2)`
HCl reduces `MnO_(2)` to `Mn^(2+)` but HF is unable to do so. This indicates that HCl is a stronger reducing agentd than HF.
`MnO_(2)+4HCl toMnCl_(2)+Cl_(2)+2H_(2)O`
`MnO_(2)+4HFto` No reaction
Thus, HI is the best reductant among hydrohalic compounds.