Turnbull’s blue is

Ferricyanide

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copper ferricyanide

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ferrous cyanide

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

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Solution

The correct option is D $F e_{4} [F e (C N)_{6}]_{3}$
Both Prussian blue and turn bull’s blue have identical structure

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Similar questions

F e^{3 +}

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

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_{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}

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