Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds,

$\mathrm{F}_{2}$ can oxidize $\mathrm{Cl}^{-}$to $\mathrm{Cl}_{2}$, Br to $\mathrm{Br}_{2}$, and $\mathrm{I}^{-}$to $\mathrm{I}_{2}$ as:

$\mathrm{F}_{2(\text { av })}+2 \mathrm{Cl}_{(s)}^{-} \longrightarrow 2 \mathrm{~F}_{(\text {aq })}^{-}+\mathrm{Cl}_{(g)}$

$\mathrm{F}_{2(a q)}+2 \mathrm{Br}_{(a q)}^{-} \longrightarrow 2 \mathrm{~F}_{(a q)}^{-}+\mathrm{Br}_{2(n)}$

$\mathrm{F}_{2(\text { ax })}+2 \mathrm{I}_{\text {(ap) }}^{-} \longrightarrow 2 \mathrm{~F}_{(\text {aqu })}^{-}+\mathrm{I}_{2(s)}$

On the other hand, $\mathrm{Cl}_{2}, \mathrm{Br}_{2}$, and $\mathrm{I}_{2}$ cannot oxidize $\mathrm{F}^{-}$to $\mathrm{F}_{2}$. The oxidizing power of halogens increases in the order of $\mathrm{I}_{2}<\mathrm{Br}_{2}<\mathrm{Cl}_{2}<\mathrm{F}_{2}$. Hence, fluorine is the best oxidant among halogens.

$\mathrm{HI}$ and $\mathrm{HBr}$ can reduce $\mathrm{H}_{2} \mathrm{SO}_{4}$ to $\mathrm{SO}_{2}$, but $\mathrm{HCl}$ and $\mathrm{HF}$ cannot. Therefore, $\mathrm{Hl}$ and $\mathrm{HBr}$ are stronger reductants than $\mathrm{HCl}$ and $\mathrm{HF}$.

$2 \mathrm{HI}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{I}_{2}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

$2 \mathrm{HBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Br}_{2}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

Again, $\mathrm{I}^{-}$can reduce $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}^{+}$, but $\mathrm{Br}^{-}$cannot.

$4 \mathrm{I}_{(\text {ap })}^{-}+2 \mathrm{Cu}_{\text {(ial) }}^{2+} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2(s)}+\mathrm{I}_{2(\text { aq })}$

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$.