Question

In Lassaigne’s test for detection of nitrogen in an organic compound, the blue color appears due to the formation of
A)$N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
B) $F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3}$
C) $F{e_2}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$
D) $F{e_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$

Answer 1

Hint: We know that the Lassaigne test is that the chemical analysis that's finished the identification of nitrogen, halogen, and sulfur in a compound. It’s also called a sodium fusion test.

Complete step by step answer:First, we discuss the principle of the Lassaigne test:
The principle behind the sodium fusion test is that the conversion of the presented element into its ionic form by the sodium metal.
-The nitrogen element is converted to its ionic form by sodium is, ${\text{Na + C + N}}\xrightarrow{{}}{\text{NaCN}}$
The Sulfur element is converted to its ionic form by sodium is, ${\text{2Na + S}}\xrightarrow{{}}{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}$
-The halogen is converted to its ionic form by sodium is, ${\text{Na + X}}\xrightarrow{{}}{\text{NaX}}$
-Now we see the procedure to hold out the Lassaigne test:
-Gently heat the tiny piece of sodium metal during a fusion tube and melt the metal till it forms a shining globule.
-After melting, add a given substance and warmth strongly and crush the hotdog tube in water taken during a china dish.
-Boil the contents then cool and filter. The filtrate is understood as Lassaigne’s extract.
-Divide the extract and perform the respective test for the element.
-In the Lassaigne's test for the detection of nitrogen in a compound, the looks of a blue colored compound is because of ferric ferrocyanide.
-In this test, sodium fusion extract is boiled with ferrous sulphate then acidified with conc. acid. Prussian blue color complex ferric ferrocyanide confirms the presence of nitrogen.
-It is ferric ferrocyanide.

Therefore, the right option is C.

Note:Lassaigne’s test additionally wont to detect the presence of sulfur is that the sodium nitroprusside test.
If the sample contains sodium sulfide within the extraction when the extraction is acidified with and on the addition of sugar of lead it forms sugar of lead (black precipitate) which confirms the presence of sulfur.
${\text{N}}{{\text{a}}_{\text{2}}}{\text{S + Pb}}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}} \right)_{\text{2}}}\xrightarrow{{}}{\text{PbS}} \downarrow {\text{ + 2C}}{{\text{H}}_{\text{3}}}{\text{COONa}}$