Ferric ion forms a Prussian blue precipitate due to the formation of:

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

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

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F e (C N S)_{3}

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

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Solution

The correct option is D $F e_{4} [F e (C N_{6})]_{3}$
Ferric ion forms a Prussian blue precipitate due to formation of $F e_{4} [F e (C N)_{6}]_{3}$ .

$F e^{+ 3} + [F e (C N)_{6}]^{- 4} \to F e_{4} [F e (C N)_{6}]_{3} P r u s s i a n b l u e$ .

Hence, option D is correct.

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Similar questions

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

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 (O H)_{3}

{F e}^{2 +}

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

{F e}^{3 +}

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

F e C l_{3} (a q)

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

F e C l_{3}

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

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

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