Which of the following reactions gives yellow precipitate for the following sequence
1.`NaOH`
2.dil `HNO_3`
3. `AgNO_3`

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the sequence of reactions step by step and identify which compound will yield a yellow precipitate when treated with the given reagents: NaOH, dilute HNO3, and AgNO3.

### Step 1: Identify the Halogenated Compounds
The question involves compounds that contain halogens (Cl, Br, I) attached to a benzene ring. The key here is to identify which halogen can form a yellow precipitate with silver nitrate (AgNO3).

### Step 2: Reaction with NaOH
When NaOH is added to the halogenated compound, it will lead to the substitution of the halogen with a hydroxyl group (OH) or the formation of a corresponding alkali metal halide.
- For example, if we have a compound with Iodine (I), the reaction would lead to the formation of sodium iodide (NaI) and the corresponding alcohol.

### Step 3: Addition of Dilute HNO3
The next step involves adding dilute HNO3. This step will acidify the medium but does not directly affect the halide ions. The iodide ion (I-) remains in the solution.

### Step 4: Addition of AgNO3
Finally, when AgNO3 is added to the solution, it will react with the iodide ions present. The reaction is as follows:
\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \]
This reaction produces silver iodide (AgI), which is known to form a yellow precipitate.

### Conclusion
From the above steps, we conclude that the presence of iodine (I) in the halogenated compound is necessary to produce a yellow precipitate of silver iodide (AgI) when AgNO3 is added. Therefore, the correct option that gives a yellow precipitate is the one containing iodine.

### Final Answer
The compound that gives a yellow precipitate when treated with the sequence of reagents is the one containing iodine (I).

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