The pair of compounds that can exist together is:

Options

A. `FeCl_(3), SnCl_(2)`
B. `HgCl_(2), SnCl_(2)`
C. `FeCl_(2), SnCl_(2)`
D. `FeCl_(3), Kl`

Explanation

To determine which pair of compounds can coexist without undergoing oxidation or reduction, we need to analyze the oxidation states of the elements in each pair. Here’s a step-by-step solution:

### Step 1: Identify the oxidation states of each compound in the pairs.
- **FeCl3**: Iron (Fe) is in the +3 oxidation state.
- **SnCl2**: Tin (Sn) is in the +2 oxidation state.
- **HgCl2**: Mercury (Hg) is in the +2 oxidation state.
- **FeCl2**: Iron (Fe) is in the +2 oxidation state.
- **KI**: Potassium (K) is in the +1 oxidation state.

### Step 2: Analyze each pair for potential oxidation or reduction.
1. **FeCl3 and SnCl2**:
- Fe (+3) can be reduced to Fe (+2).
- Sn (+2) cannot be oxidized further.
- **Conclusion**: This pair can undergo reduction; hence, they cannot coexist.

2. **HgCl2 and SnCl2**:
- Hg (+2) can be reduced to Hg (+1).
- Sn (+2) cannot be oxidized further.
- **Conclusion**: This pair can undergo reduction; hence, they cannot coexist.

3. **FeCl2 and SnCl2**:
- Fe (+2) is in its lowest oxidation state and cannot be reduced further.
- Sn (+2) is also in its lowest oxidation state and cannot be oxidized further.
- **Conclusion**: This pair cannot undergo oxidation or reduction; hence, they can coexist.

4. **FeCl3 and KI**:
- Fe (+3) can be reduced to Fe (+2).
- K (+1) cannot be oxidized further.
- **Conclusion**: This pair can undergo reduction; hence, they cannot coexist.

### Step 3: Finalize the answer.
The pair of compounds that can coexist without undergoing oxidation or reduction is **FeCl2 and SnCl2**.

### Summary of the Solution:
- The answer is **FeCl2 and SnCl2** as both are in their lowest oxidation states and cannot be further reduced or oxidized.