The Lassaignes solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of

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

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

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

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

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Solution

The correct option is D $F e_{4} [F e (C N)_{6}]_{3}$
$F e S O_{4} + N a O H ⟶ F e (O H)_{2} + N a_{2} S O_{4}$
$6 N a C N + F e (O H)_{2} ⟶ N a_{4} [F e (C N)_{6}]$
sodium ferrocyanide
$3 N a_{4} [F e (C N)_{6}] + 4 F e C l_{3} ⟶ F e_{4} [F e (C N)_{6}]_{3} + 12 N a C l$
Ferric ferro cyanide
(Prussian blue)
To a portion of SFE, freshy prepared ferrous sulphate, $F e S O_{4}$ solution is added and warmed. Then about 2 to 3 drops of $F e C l_{3}$ solution are added and acidified with conc. $H C l$ . The appearance of a Prussian blue color indicates the presence of nitrogen.
Detection of nitrogen in Lassaigne's test

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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

{F e}^{2 +}

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

{F e}^{3 +}

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

{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}]

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

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