The blue compound is formed in the positive test for nitrogen with Lassaigne's solution of an organic compound is:

{N a}_{4} [F e ({C N)}_{5} (N O S)]

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

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[F e ({C N)}_{3}

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

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Solution

The correct option is D $F e_{4} [F e (C N)_{6}]_{3}$
If Nitrogen is present in the organic compound, then sodium extract contains ${N a}_{4} [F e {(C N)}_{6}]$ .
$N a + C + N F u s e - -- \to N a C N$
$F e S O_{4} + 6 N a C N ⟶ {N a}_{4} [F e {(C N)}_{6}] + {N a}_{2} S O_{4}$
This ${F e {(C N)}_{6}}^{- 4}$ turns to ${F e}_{4} {[F e {(C N)}_{6}]}_{3}$ (Prussian blue) on reaction with $F e {C l}_{3}$
$3 {N a}_{4} [F e {(C N)}_{6}] + 4 F e {C l}_{3} ⟶ {F e}_{4} {[F e {(C N)}_{6}]}_{3} Prussian Blue + 12 N a C l$
Hence, the blue compound is ${F e}_{4} {[F e {(C N)}_{6}]}_{3}$ .

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Similar questions

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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