The nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide ($\mathrm{NaCN}$) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate.

$6\mathrm{NaCN} + {\mathrm{FeSO}}_4 \to \underset{\text{Sodium ferrocyanide}}{{\mathrm{Na}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]} + {\mathrm{Na}}_2{\mathrm{SO}}_4$Ferric ions generated during the process react with ferrocyanide to form Prussian blue precipitate of ferric ferrocyanide.

${\mathrm{Na}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6] + {\mathrm{Fe}}^{3+} \to \underset{\text{prussian blue}}{\underset{\text{Ferric ferrocyanide}}{{\mathrm{Fe}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}_3}}$

 

Sulphur present in organic compounds are detected by Lassaigne’s test. Here, a small piece of $\mathrm{Na}$ metal is heated in a fusion tube with the organic compound. The principle is that, in doing so, $\mathrm{Na}$ converts all the elements present into ionic form.

$2\mathrm{Na}+\mathrm{S} \to {\mathrm{Na}}_2\mathrm{S}$

Test of sulphur :       

 The extract is treated with sodium nitroprusside. The appearance of violet colour indicates the presence of sulphur.The following reaction occurs:

 ${\mathrm{S}}^{2-}+ \left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\mathrm{NO}{]}^{2-} \to \left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\mathrm{NOS}{]}^{4-}$

The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

The halogen present in the compound converted into sodium halide, and it will be precipitated with silver nitrate.