Discuss the chemistry of Lassaigne's test.

(N/A) Lassaigne's test is employed to detect the presence of nitrogen,sulphur,halogens,and phosphorus in an organic compound. These elements are present in the covalent form in an organic compound and are converted into the ionic form by fusing the compound with sodium metal.
$Na + C + N \xrightarrow{\Delta} NaCN$
$2Na + S \xrightarrow{\Delta} Na_2S$
$Na + X \xrightarrow{\Delta} NaX$ (where $X = Cl, Br, I$)
The cyanide,sulphide,and halide of sodium formed are extracted by boiling the fused mass in distilled water. This is called Lassaigne's extract.
$(a)$ Test for nitrogen:
Lassaigne's extract is boiled with iron $(II)$ sulphate and then acidified with concentrated sulphuric acid. Sodium cyanide reacts with iron $(II)$ sulphate to form sodium hexacyanoferrate $(II)$. Upon heating with sulphuric acid,some iron $(II)$ is oxidised to iron $(III)$,which reacts to form iron $(III)$ hexacyanoferrate $(II)$,known as Prussian blue.
$6 CN^{-} + Fe^{2+} \longrightarrow [Fe(CN)_6]^{4-}$
$3[Fe(CN)_6]^{4-} + 4Fe^{3+} \xrightarrow{xH_2O} Fe_4[Fe(CN)_6]_3 \cdot xH_2O$ (Prussian blue)
$(b)$ Test for sulphur:
$(i)$ Lassaigne's extract is acidified with acetic acid and treated with lead acetate. $A$ black precipitate of lead sulphide indicates sulphur.
$S^{2-} + Pb^{2+} \to PbS$ (Black)
$(ii)$ Treatment with sodium nitroprusside gives a violet colour.
$S^{2-} + [Fe(CN)_5NO]^{2-} \to [Fe(CN)_5NOS]^{4-}$ (Violet)
If both nitrogen and sulphur are present,$NaSCN$ is formed,which gives a blood-red colour with $Fe^{3+}$.
$Fe^{3+} + SCN^{-} \to [Fe(SCN)]^{2+}$ (Blood red)
$(c)$ Test for halogens:
The extract is acidified with nitric acid and treated with silver nitrate. The resulting precipitate ($AgCl$ white,$AgBr$ pale yellow,$AgI$ yellow) confirms the halogen.