The common formula of Turn bull's blue and Prussian blue is:

F e_{3} [F e (C N)_{6}]_{2}

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F e_{3} [F e (C N)_{6}]_{3}

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

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K_{4} F e_{2} [F e (C N)_{6}]_{2}

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Solution

The correct option is C $K F e [F e (C N)_{6}]$
$4 F e C l_{3} + 3 K_{4} [F e (C N)_{6}] \to F e_{4} [F e (C N)_{6}]_{3} + 12 K C l$
or
$F e C l_{3} + K_{4} [F e (C N)_{6}] \to K F e [F e (C N)_{6}] + 3 K C l$
$F e S O_{4} + K_{3} [F e (C N)_{6}] \to K F e [F e (C N)_{6}] + K_{2} S O_{4}$
Turn bull's blue

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Similar questions

F e_{3} {[F e {(C N)}_{6}]}_{2}

In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]₄

{F e}^{2 +}

K_{3} [{F e (C N)}_{6}]

{F e}^{3 +}

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

F e^{3} + {[F e {(C N)}_{6}]}_{6}^{3 -} \to \times H_{2} O_{2} S o l u t i o n - ------- \to Z

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