$K_{3} [{F e (C N)}_{6}]$ is used in the detection of ${F e}^{2 +}$ ion which it gives a deep blue colour. This colour is due to the formation of:

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

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{F e}_{4} {[{F e (C N)}_{6}]}_{3}

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

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

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Solution

The correct option is B ${F e}_{3} {[{F e (C N)}_{6}]}_{2}$
The reaction of ${F e}^{2 +}$ with potassium ferricyanide forms blue precipitate of Turnbull's blue which is used as pigment in ink and paint.
$3 {F e}^{2 +} + 2 K_{3} [{F e (C N)}_{6}] \to {F e}_{3} {[{F e (C N)}_{6}]}_{2}$

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Similar questions

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_{4} [F e (C N)_{6}]_{3}

{N H}_{2} - {N H}_{2}

{F e}_{4} {[F e ({C N}_{6})]}_{3}

F e {S o}_{4}, F e {C I}_{3}

C u^{2 +}

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

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

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

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