A : `Sn^(2+)" and "Fe^(3+)` can't remain together in a solution.
R : `Sn^(2+)" and "Fe^(3+)` will react mutually to form `Sn^(4+)" and "Fe^(2+)`.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).
C. If Assertion is true statement but Reason is false, then mark (3).
D. If both Assertion and Reason are false statements, then mark (4).

Answer

To analyze the statement regarding the inability of `Sn^(2+)` and `Fe^(3+)` to coexist in a solution, we can break down the reasoning into a step-by-step solution.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- `Sn^(2+)` is in the +2 oxidation state.
- `Fe^(3+)` is in the +3 oxidation state.

2. **Understand the Reactivity**:
- `Fe^(3+)` can be reduced to `Fe^(2+)`.
- `Sn^(2+)` can be oxidized to `Sn^(4+)`.

3. **Determine the Reaction**:
- The presence of `Sn^(2+)` provides a reducing environment for `Fe^(3+)`. Thus, `Sn^(2+)` can reduce `Fe^(3+)` to `Fe^(2+)`.
- Simultaneously, `Fe^(3+)` being reduced means that `Sn^(2+)` is oxidized to `Sn^(4+)`.

4. **Write the Balanced Reaction**:
- The overall reaction can be represented as:
\[ \text{Sn}^{2+} + \text{Fe}^{3+} \rightarrow \text{Sn}^{4+} + \text{Fe}^{2+} \]

5. **Conclusion**:
- Since `Sn^(2+)` reduces `Fe^(3+)` to `Fe^(2+)`, it cannot coexist with `Fe^(3+)` in a stable manner, as they will react with each other.

### Final Statement:
- Therefore, the statement A is true: `Sn^(2+)` and `Fe^(3+)` cannot remain together in a solution because they will react mutually to form `Sn^(4+)` and `Fe^(2+)`, making statement R also true.