`Br_(2)+NaOH overset(R.T.) to Y+Z`
If Y gives precipitate with `AgNO_(3)`, then Z does not undergo reaction with:

Options

A. `Cr^(3+)(aq.)`
B. `Fe^(2+)(aq.)`
C. `Al^(3+)(aq.)`
D. `Sn^(2+)(aq.)`

Explanation

To solve the question, we need to analyze the reaction between bromine (Br₂) and sodium hydroxide (NaOH) and identify the products Y and Z. We also need to determine which cation does not react with Z.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between bromine and sodium hydroxide can be written as:
\[ \text{Br}_2 + \text{NaOH} \rightarrow \text{Y} + \text{Z} \]
At room temperature, this reaction produces sodium bromide (NaBr) and sodium hypobromite (NaOBr).

2. **Determine Y and Z**:
- From the reaction, we can identify:
- \( Y = \text{NaBr} \)
- \( Z = \text{NaOBr} \)

3. **Analyze the Properties of Y**:
- Y (NaBr) gives a precipitate with silver nitrate (AgNO₃). The reaction is:
\[ \text{NaBr} + \text{AgNO}_3 \rightarrow \text{AgBr (s)} + \text{NaNO}_3 \]
- AgBr is a pale yellow precipitate, confirming that Y is indeed NaBr.

4. **Understand the Nature of Z**:
- Z (NaOBr) acts as an oxidizing agent. We need to check its reactivity with different cations.

5. **Check Reactivity of Z with Given Cations**:
- **Cr³⁺**:
\[ \text{NaOBr} + \text{Cr}^{3+} \rightarrow \text{Cr}^{6+} \quad (\text{oxidation occurs}) \]
- **Fe²⁺**:
\[ \text{NaOBr} + \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \quad (\text{oxidation occurs}) \]
- **Al³⁺**:
- Al³⁺ is already in its maximum oxidation state, so no reaction occurs:
\[ \text{NaOBr} + \text{Al}^{3+} \quad (\text{no reaction}) \]
- **Sn²⁺**:
\[ \text{NaOBr} + \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} \quad (\text{oxidation occurs}) \]

6. **Conclusion**:
- The only cation that does not undergo a reaction with Z (NaOBr) is **Al³⁺**. Therefore, the answer to the question is:
\[ \text{Al}^{3+} \text{ does not undergo reaction with } Z. \]

### Final Answer:
**Al³⁺ does not undergo reaction with Z (NaOBr).**