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

Explanation

The decreasing oxidising power of halogens is, `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`
`F_(2)` being the strongest oxidant can oxidise `Cl^(-) , Br^(-), I^(-)` ions. While `Cl_(2)` oxidises `Br^(-)`, `I^(-)` ions, whereas `Br_(2)` oxidises only `I^(-)` ions. `I_(2)` cannot oxidise all the three halide ions. Oxidising reactions of Fluorine.
`F_(2)(g) + 2Cl^(-) (aq) rarr 2F^(-) (aq) + Cl_(2)(g)`
`F_(2)(g) + 2Br^(-) (aq) rarr 2F^(-) (aq) + Br_(2)(g)`
`F_(2)(g) + 2I^(-) (aq) rarr 2F^(-) (aq) + I_(2)(g)`
Oxidising reactions of Chlorine
`Cl_(2)(g) + 2Br^(-) (aq) rarr 2Cl^(-) (aq) +Br_(2)(g)`
`Cl_(2)(g) + 2I^(-) (aq) rarr 2Cl^(-) (aq) + I_(2)(g)`
Oxidising reactions of Bromine
`Br_(2)(g) + 2I^(-) (aq) rarr 2Br^(-) (aq) + I_(2)(g)`
Thus, fluorine is the best oxidant. The decreasesing order of reducing power of hydrohalic acids is,
`HI gt HBr gt HCl gt HF`
HI and HBr reduce `H_(2)SO_(4) " to " SO_(2)`, while HCl and HF do not undergo such reduction reactions.
`2HBr + H_(2)SO_(4) rarr SO_(2) + 2H_(2)O + Br_(2)`
`2HI + H_(2)SO_(4) rarr SO_(2) + 2H_(2)O + I_(2)`
HCl reduces `MnO_(2) " to " Mn^(2+)` but HF is unable to do so, showing HCl stronger reducing agent than HF.
`MnO_(2) + 4HCl rarr MnCl_(2) + Cl_(2) + 2H_(2)O`
`MnO_(2) + 4HF rarr` No reaction
Thus, HI is the best reductant among hydrohalic compounds.