Acidic $K M n O_{4}$ can be decolourised be:

{S O}_{2}

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H_{2} O_{2}

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F e {S O}_{4}

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{F e}_{2} {({S O}_{4})}_{3}

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Solution

The correct option is D ${S O}_{2}$
Acidified $K M n O_{4}$ can be decolorized by $S O_{2}$ .Potassium permanganate( $K M n O_{4}$ ) has a purple colour. Sulphur dioxide is a reducing agent.
Potassium permanganate is an oxidizing agent. When sulphur dioxide reacts with potassium permanganate the solution decolourizes.

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1.5

{K M n O}_{4}

H_{2} O_{2}

:

2 F e S O_{4} + H_{2} S O_{4} + H_{2} O_{2} \to {F e}_{2} {(S O_{4})}_{3} + 2 H_{2} O

{K M n O}_{4}

{F e}^{2 +} (o r C_{2} O_{4}^{2 -}, {S O}_{2} e t c .)

{F e}_{2} S_{3}

{F e}_{2} S_{3}

{F e}_{3} S_{3} + O_{2} ⟶ {F e S O}_{4} + {S O}_{2}

K M n O_{4}

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