Question

Three students, Manish Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne's extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid $FeS{O_4}$ and dilute sulphuric acid to a part of Lassaigne's extract. Manish and Rajni obtained Prussian blue colour but Ramesh got a red colour. Ramesh repeated the test with the same Lassaigne's extract, but again got a red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this. observation. Also, write the chemical equations to explain the formation of compounds of different colours.

Answer 1

Hint: Lassaigne’s extract of Lassaigne’s test is used to detect nitrogen, sulphur and halogens present in the organic compound. When a small piece of sodium metal is heated in a fusion tube with organic compound sodium converts all the elements present into its ionic form.
$Na + C + N \to {\text{NaCN}}$
$2Na + S \to {\text{N}}{{\text{a}}_2}{\text{S}}$
$Na + X \to {\text{NaX}}$ where ($X = Cl,\;Br\;or\;I$)

Complete answer:
We know that depending upon the amount of sodium metal we use during fusion, it may form either sodium thiocyanate $\left( {NaSCN} \right)$ or a mixture of sodium cyanide $\left( {NaCN} \right)$ and sodium sulphide $\left( {N{a_2}S} \right)$ if the organic compound contains both $N$ and $S$ . When sodium metal is used very less, only $NaSCN$ is formed.
We know that during the preparation of Lassaigne’s extract due to oxidation of $F{e^{2 + }}$ some $F{e^{3 + }}$ ions are formed. This $F{e^{3 + }}$ ions form then react with $NaSCN$ to form ferric thiocyanate. This ferric thiocyanate has a red colouration.
$F{e^{2 + }}\xrightarrow{{Aerial\;oxidation}}F{e^{3 + }}$
$F{e^{3 + }} + 3NaSCN \to Fe{\left( {SCN} \right)_3} + 3N{a^ + }$
Now let’s consider the situation when we use excess sodium metal. When excess sodium metal is used the sodium thiocyanate formed will decompose to form a mixture of sodium cyanide and sodium sulphate.
${\text{NaSCN}} + 2{\text{Na}}\mathop \to \limits^\vartriangle {\text{NaCN}} + {\text{N}}{{\text{a}}_2}{\text{S}}$
Now we have ${\text{NaCN}}$ formed. Which when reacting with $FeS{O_4}$. We know that Prussian blue colour is due to ferric ferrocyanide. When more ${\text{NaCN}}$ and $F{e^{3 + }}$ are added we get ferric ferrocyanide.
$2NaCN + FeS{O_4} \to N{a_2}S{O_4} + Fe{\left( {CN} \right)_2}$
To this, we are adding more ${\text{NaCN}}$
${\text{Fe}}{\left( {{\text{CN}}} \right)_2} + 4{\text{NaCN}} \to {\text{N}}{{\text{a}}_4}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]$
We got sodium hexacyanoferrate, to this we are adding again $F{e^{3 + }}$
$3N{a_4}[Fe\left( {CN{)_6}} \right] + 4F{e^{3 + }} \to F{e_4}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_3} + 12N{a^ + }$
Thus we got the ferric ferrocyanide and the blue Prussian colour.

Thus Manish and Rajni used excess sodium and Ramesh used less sodium that is why he got the red colouration.

Note:
When a small piece of sodium metal is heated in a fusion till it melts into a shining globule. At this stage we are adding the substance and heated strongly, it is then plunged into distilled water containing china dish. These contents are boiled, cooled and then filtered. This filter is known as Lassaigne’s extract.