Assertion: `NaBO_(3) //OH^(-)` can be used for oxidation of `Cr^(3+)` to `Cr^(6+)`
Reason: IN alkaline medium `NaBO_(3)` produces `H_(2)O_(2)`

Options

A. If both (A) and (R) are true and (R) is the correct explanation of (A)
B. if both (A) and (R) are true but (R) is not correct explanation of (A)
C. If (A) is true but (R) is false
D. If (A) is false but (R) is false

Answer

To solve the question regarding the assertion and reason provided, we will analyze both statements step by step.

### Step 1: Understanding the Assertion
The assertion states that `NaBO3 // OH^(-)` can be used for the oxidation of `Cr^(3+)` to `Cr^(6+)`.

**Explanation:**
- `NaBO3` is sodium perborate, which is known for its oxidizing properties.
- In the presence of hydroxide ions (OH^-), sodium perborate can facilitate the oxidation of chromium ions.

### Step 2: Understanding the Reason
The reason given is that in alkaline medium, `NaBO3` produces `H2O2` (hydrogen peroxide).

**Explanation:**
- In alkaline conditions, sodium perborate decomposes to produce hydrogen peroxide, which is a strong oxidizing agent.
- The reaction can be summarized as:
\[ \text{NaBO}_3 + \text{OH}^- \rightarrow \text{H}_2\text{O}_2 + \text{other products} \]

### Step 3: The Oxidation Reaction
Now, we need to understand how `H2O2` can oxidize `Cr^(3+)` to `Cr^(6+)`.

**Explanation:**
- The oxidation of `Cr^(3+)` to `Cr^(6+)` can be represented as:
\[ \text{Cr}^{3+} + \text{H}_2\text{O}_2 \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O} \]
- Here, `CrO4^(2-)` indicates that chromium is in the +6 oxidation state.

### Step 4: Conclusion
Both the assertion and the reason are correct. The assertion is true because `NaBO3` can indeed oxidize `Cr^(3+)` to `Cr^(6+)`, and the reason is also true as `NaBO3` produces `H2O2` in alkaline medium, which is responsible for the oxidation.

### Final Answer
- **Assertion:** True
- **Reason:** True
- **Conclusion:** The reason is the correct explanation of the assertion.