Nitrogen Detection (Lassaigne's Test)

Nitrogen in organic compounds is detected by converting it into sodium cyanide, which further reacts with iron(II) sulfate to form sodium hexacyanoferrate(II). By reacting this with iron(III) ions, Prussian blue is formed:

$ \begin{aligned} &\mathrm{Na}+\mathrm{C}+\mathrm{N} \rightarrow \mathrm{NaCN} \\\\ &6 \mathrm{NaCN}+\mathrm{FeSO}_4 \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Na}_2 \mathrm{SO}_4 \\\\ &\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \text{ (Prussian blue)} \end{aligned} $

Sulphur Detection

Sulphur can be detected using Sodium nitroprusside:

$ \mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{Na}_2 \mathrm{S} \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right] \text{ [Purple]} $

Phosphorous Detection

Phosphorus detection involves reaction with ammonium molybdate to form a canary yellow compound:

$ \begin{aligned} &\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3 \\\\ &\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \rightarrow\\\\& \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O} \text{ (canary yellow)} \end{aligned} $

Halogen Detection

Halogens react with silver nitrate to form specific colored precipitates:

$ \begin{aligned} &\mathrm{NaCl}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_3 \text{ (White)} \\\\ &\mathrm{NaBr}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgBr}+\mathrm{NaNO}_3 \text{ (Pale yellow)} \\\\ &\mathrm{NaI}+\mathrm{AgNO}_3(\mathrm{aq}) \rightarrow \mathrm{AgI}+\mathrm{NaNO}_3 \text{ (Yellow)} \end{aligned} $

Based on the above reactions, the correct matching is:

Option A

A-III (Nitrogen), B-I (Sulphur), C-IV (Phosphorous), D-II (Halogen).