Colour of $F e_{4} [F e (C N)_{6}]_{3}$ is

Purple

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Blue

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White

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None of the above

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Solution

The correct option is B Blue
$F e_{4} [F e (C N)_{6}]_{3}$ is Prussian Blue which is blue in colour. This is due to intervalence charge transfer where the excitation and subsequent transfer of electrons from a lower oxidation state cation to neighbouring cation of higher oxidation state.
Hence, Option "b" is the correct answer.

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Similar questions

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

{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}

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

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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