Select ALL correct statements.

A $IO X_{3} X^{-} + 5 I X^{-} + 6 H X^{+} 3 I X_{2} + 3 H X_{2} O$ is a comproportionation reaction
BIn $2 H X_{2} S + SO X_{2} 3 S + 2 H X_{2} O$ , each $SO X_{2}$ molecule gains 4 electrons
C0.10 mol of $IO X_{3} X^{-}$ with excess $I X^{-}$ and acid gives 0.30 mol of $I X_{2}$
DIn $2 H X_{2} S + SO X_{2} 3 S + 2 H X_{2} O$ , the sulphur of $H X_{2} S$ is reduced

Answer: A, B, C

In the first reaction I (+5) and I (-1) converge on I (0), which is comproportionation, and the mole ratio

IO X_{3} X^{-}

I X_{2}

is 1 : 3, so 0.10 mol gives 0.30 mol. In the second, S goes from +4 to 0, a gain of 4 electrons per

SO X_{2}

. The S of

H X_{2} S

goes from -2 to 0, so it is oxidised, not reduced.