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Central Atom and Ligands

What is the c...

Question

What is the common formula of Turnbull's blue and Prussian blue?

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

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

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

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

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Solution

The correct option is B $K F e [F e (C N)_{6}]$
The common formula of Turnbull's blue and Prussian blue is $K F e [F e (C N)_{6}]$
In this complex, one iron atom is in +2 oxidation state and other iron atom is in +3 oxidation state.

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Similar questions

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

{F e}^{2 +}

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

{F e}^{3 +}

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

{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 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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