Question

Why is the Fresh solution of $FeS{O_4}$ used in a test for nitrogen?

Answer 1

Hint: This question is based on the knowledge of conceptual details of the reaction which is used to test the presence of nitrogen in a solution. Ferrous sulphate reacts with nitrogen to form colored ions after reduction of ferrous sulphate. This test is widely used to test the presence of nitrogen.

Complete answer:
Lassaigne's test can identify nitrogen, sulphur, and halogens in organic molecules. A little bit of Na metal is heated with the organic compound in a fusion tube. The idea is that Na changes all of the elements present into ionic form in this way.
By boiling the fused mixture with distilled water, the produced ionic salts are removed. Sodium fusion extract is what it's called.
The extract is acidified with concentrated $H_2SO_4$ and heated with $FeSO_4$. The presence of nitrogen is indicated by the appearance of Prussian blue.
The following reactions take place:
\[ F{e^{2 + }}\; + {\text{ }}6C{N^-}\; \to {[Fe{\left( {CN} \right)_6}]^{4 - }} \\
  F{e^{2 + }}\; + {\text{ }}{H^ + } \to F{e^{3 + }}\; + {\text{ }}{e^-} \\
  {[Fe{\left( {CN} \right)_6}]^{4 - }}\; + {\text{ }}4F{e^{3 + }}\; \to F{e_4}[Fe{\left( {CN} \right)_6}].{H_2}O\; \\
\]
Ferrous ions are oxidised to ferric ions with the help of the acid. The presence of nitrogen is indicated by the production of ferriferous cyanide.
The compounds with N but no C atoms do not pass this test.$N{H_2}N{H_2}$, for example, despite having a N atom, fails this test. This is because the $C{N^ - }$ ion requires both C and N to form. Diazonium salts cannot be used for this test because when heated, they breakdown into nitrogen gas.

Note:
$HN{O_3}$ is used to acidify the extract, which is subsequently treated with $AgN{O_3}$. The presence of $Cl$ is indicated by a white precipitate soluble in $N{H_4}OH$, the presence of $Br$ is shown by a yellowish precipitate sparingly soluble in $N{H_4}OH$and the presence of I is indicated by a yellow precipitate insoluble in $N{H_4}OH$.