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For the given aqueous reactions, which of the statement(s) is / are true?
Excess
$KI + K_{3} [Fe (CN)_{6}] \overset{dil. H_{2} {SO}_{4}}{\to} Brownish - yellow solution ↓ {ZnSO}_{4} White ppt. + Brownish-yellow filtrate ↓ {Na}_{2} s_{2} O_{3} Colourless solution$

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The first reaction is a redox reaction..

White precipitate is ${Zn}_{3} {[Fe (CN)_{6}]}_{2}$ .

Addition of filtrate to starch solution gives blue colour.

White precipitate is soluble in NaOH solution.

answer is A, C, D.

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Detailed Solution

The solutions are,

Brownish yellow solution is $K_{4} [Fe (CN)_{6}]$ .

White ppt is $K_{2} Zn {[Fe (CN)_{6}]}_{or K_{2}} {Zn}_{3} {[Fe (CN)_{6}]}_{2}$

Brownish yellow filterate is $I_{2}$ .

Thus, option a, c and d are correct.

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