Prussian blue is formed when :

Ferrous sulphate reacts with

F e C l_{3}

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Ferric sulphate reacts with

K_{4} [F e {(C N)}_{6}]

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Ferrous ammonium sulphate reacts with

F e C l_{3}

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Ammonium stilphate reacts with

F e C l_{3}

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Solution

The correct option is B Ferric sulphate reacts with $K_{4} [F e {(C N)}_{6}]$ .
Prussian blue is formed when ferric sulphate reacts with $K_{4} [F e {(C N)}_{6}]$ .
$\begin{matrix} 2 F e_{2} {(S O_{4})}_{3} + 3 K_{4} [F e {(C N)}_{6}] ⟶ & F e_{4} {[F e {(C N)}_{6}]}_{3} & + 6 K_{2} S O_{4} P r u s s i a n b l u e \end{matrix}$ .

Option (B) is correct.

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F e C l_{3}

K_{4} [F e (C N)_{6}]

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