`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 given and determine the products Y and Z, as well as the reactivity of Z with the provided options.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ \text{Br}_2 + \text{NaOH} \xrightarrow{\text{R.T.}} Y + Z \]
Here, bromine (Br2) reacts with sodium hydroxide (NaOH).

2. **Determine the Products**: The reaction of bromine with sodium hydroxide at room temperature leads to a disproportionation reaction, where bromine is converted into sodium bromide (NaBr) and sodium hypobromite (NaBrO). Therefore, we can identify:
- \( Y = \text{NaBr} \)
- \( Z = \text{NaBrO} \)

3. **Analyze the Given Condition**: It is stated that Y gives a precipitate with AgNO3. Sodium bromide (NaBr) will react with silver nitrate (AgNO3) to form a white precipitate of silver bromide (AgBr):
\[ \text{NaBr} + \text{AgNO}_3 \rightarrow \text{AgBr (s)} + \text{NaNO}_3 \]
This confirms that \( Y \) is indeed NaBr.

4. **Determine the Properties of Z**: Now, we need to analyze the properties of \( Z \) (NaBrO). Sodium hypobromite is known to be an oxidizing agent.

5. **Evaluate Reactivity of Z**: We need to check which of the given options Z (NaBrO) does not react with:
- **Option A: Cr3+**: Can be oxidized to Cr6+.
- **Option B: Fe2+**: Can be oxidized to Fe3+.
- **Option C: Al3+**: Aluminum is already in its highest oxidation state (+3) and cannot be oxidized further.
- **Option D: Sn2+**: Can be oxidized to Sn4+.

6. **Conclusion**: Since NaBrO is an oxidizing agent, it will not react with aluminum ions (Al3+) because aluminum is already in its highest oxidation state. Therefore, the answer is:
\[ \text{Z does not undergo reaction with Al}^{3+}. \]

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

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