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Question

The formula of prussian blue is:
$N a_{4} [F e (C N)_{6}]$
$F e (C N)_{2}$
$F e_{4} [F e (C N)_{6}]_{3}$
$F e (C N S)_{3}$

A

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

B

F e (C N)_{2}

C

F e (C N S)_{3}

D

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

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Solution

Verified by Toppr

Ferric ferrocyanide /Potassium ferrocyanide/ Sodium ferrocyanide is prussian blue.
$F e_{4} [F e (C N)_{6}]_{3}$
Option D is correct,

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Similar Questions

Q1

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

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Q2

Assertion :

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

gives Prussian blue colour during Lassaigne test (element detection test). Reason: NaCN gives Prussian blue colour of

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

with

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

.

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Q3

Colour of

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

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Q4

In the complex

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

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Q5

Two solutions (A) containing

F e C l_{3} (a q)

and (B) containing

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are separated by semipermeable membrane as shown below. If

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