Question

$HBr$ and $HI$ reduce $H_{2}S{O}_4$, $HCl$ can reduce $KMn{O}_4$ and $HF$ reduces:

• A. $H_{2}S{O}_4$
• B. $KMn{O}_4$
• C. $K_2{Cr}_2{O}_7$
• D. none of these

Answer 1

Answer: (D)

none of these

$HF$ is not a reducing agent. That happens because the fluoride ion, $F^{-}$, is a very small ion, which implies that its outermost electrons are actually very close to the nucleus.

$HI$ is the strongest reducing agent among hydrogen halides.

The reducing property increases in the order $HCl<HBr<HI$.

$HF$ reduces neither $H_{2}S{O}_4$ nor $KMn{O}_4$ nor $K_2{Cr}_2{O}_7$.

Hence, option $D$ is correct.