In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of:

Step 1: Recall the Carius Method for Halogens

In the Carius method, the organic compound containing halogen is placed in a thick-walled sealed glass tube (Carius tube) along with:

1. Fuming nitric acid ($\ce{HNO3}$, conc.) — strong oxidising agent
2. Silver nitrate ($\ce{AgNO3}$) — to precipitate the halide as silver halide

Step 2: The Reaction Sequence

$\ce{Organic-X + HNO3(fuming) -> CO2 + H2O + HX}$ $\ce{HX + AgNO3 -> AgX (v) + HNO3}$

The halogen (X) is released as HX, which immediately reacts with $\ce{AgNO3}$ to form the insoluble silver halide precipitate:

• $\ce{Cl}$ → $\ce{AgCl}$ (white)
• $\ce{Br}$ → $\ce{AgBr}$ (pale yellow)
• $\ce{I}$ → $\ce{AgI}$ (yellow)

Step 3: Evaluate Each Option

| Reagent | Role in Carius Method? | |---|---| | (a) $\ce{AgNO3}$ ✅ | Precipitates halide as $\ce{AgX}$ — essential reagent | | (b) $\ce{HNO3}$ | This is fuming HNO₃ — already implied; but the question asks what's added alongside it | | (c) $\ce{BaSO4}$ | Used for sulphur estimation — not halogens | | (d) $\ce{CuSO4}$ | Catalyst in Kjeldahl's method — not Carius |

Answer: (a) $\ce{AgNO3}$

Key Points to Remember:

• Carius method (halogens): fuming $\ce{HNO3}$ + $\ce{AgNO3}$ → $\ce{AgX}$ precipitate
• Carius method (sulphur): fuming $\ce{HNO3}$ + $\ce{BaCl2}$ → $\ce{BaSO4}$ precipitate
• $\ce{BaSO4}$ is the product of sulphur estimation, not the reagent
• The tube must be sealed (high pressure) — fuming HNO₃ is required for complete oxidation