(N/A) Nitrogen $(N)$,Sulphur $(S)$,Halogens $(Cl, Br, I)$,and Phosphorus $(P)$ present in an organic compound are detected by "Lassaigne's test".
$(a)$ The elements $(N, X, S)$ present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. The following reactions take place:
$(i)$ $Na + C + N \xrightarrow{\Delta} NaCN$
$(ii)$ $2Na + S \xrightarrow{\Delta} Na_{2}S$
$(iii)$ $Na + X \xrightarrow{\Delta} NaX$ (where $X = Cl, Br, \text{ or } I$)
Cyanide,sulphide,and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
$(b)$ Test for Nitrogen:
Procedure: The sodium fusion extract is boiled with iron $(II)$ sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron $(II)$ sulphate and forms sodium hexacyanidoferrate $(II)$. On heating with concentrated sulphuric acid,some iron $(II)$ ions are oxidised to iron $(III)$ ions which react with sodium hexacyanidoferrate $(II)$ to produce iron $(III)$ hexacyanidoferrate $(II)$ (ferriferrocyanide),which is Prussian blue in colour.
$6CN_{(aq)}^{-} + Fe_{(aq)}^{2+} \rightarrow [Fe(CN)_{6}]_{(aq)}^{4-}$
$3[Fe(CN)_{6}]_{(aq)}^{4-} + 4Fe^{3+} \xrightarrow{xH_{2}O} Fe_{4}[Fe(CN)_{6}]_{3} \cdot xH_{2}O$ (Prussian blue)
$(c)$ Test for Sulphur: Following two tests occur for detection of sulphur:
$(i)$ The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. $A$ black precipitate of lead sulphide indicates the presence of sulphur.
$S_{(aq)}^{2-} + Pb_{(aq)}^{2+} \rightarrow PbS_{(s)}$ (Black precipitate)
$(ii)$ On treating sodium fusion extract with sodium nitroprusside,the appearance of a violet colour further indicates the presence of sulphur.