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

F₂ can oxidise Cl^- to Cl₂, Br^- to Br₂and I^- to I₂ as:

F_2(aq) + 2Cl^-_(s)→ 2F^-_(aq) + Cl_2(g)

F_2(aq) + 2Br^-_(aq)→ 2F^-_(aq) + Br_2(l)

F_2(aq) + 2I^-_(aq)→ 2F^-_(aq) + I_2(s)

On the other hand, Cl_2, Br₂and I₂ cannot oxidise F^- to F₂. The oxidising power of halogens increases in the order of I₂˂Br₂˂Cl₂˂F₂. Hence, fluorine is the best oxidant among halogens. HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.

2HI + H₂SO₄→ I₂ + SO₂ + 2H₂O

2HBr + H₂SO₄→ Br₂ + SO₂ + 2H₂O

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

4I^-_(aq) + 2Cu²⁺_(aq) → Cu₂I_2(s) + I_2(aq)

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

Thus, the reducing power of hydrohalic acids increases in the order of HF ˂ HCl ˂ HBr ˂ HI.