`AIF_(3)` is soluble in `HF` only in presence of `KF`. It is due to the formation of

Options

A. `K_(3)[AlF_(3)H_(3)]`
B. `K_(3)[AlF_(6)]`
C. `AlH_(3)`
D. `K[AlF_(3)H]`

Explanation

To solve the question regarding the solubility of \( \text{AlF}_3 \) in \( \text{HF} \) only in the presence of \( \text{KF} \), we need to analyze the formation of a specific complex. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Components
- **Aluminum Fluoride (\( \text{AlF}_3 \))**: In this compound, aluminum is in the +3 oxidation state. It has a tendency to form complexes due to its ability to coordinate with ligands.
- **Hydrofluoric Acid (\( \text{HF} \))**: This is a source of fluoride ions and can help solubilize aluminum fluoride.
- **Potassium Fluoride (\( \text{KF} \))**: This provides additional fluoride ions which are crucial for the complex formation.

**Hint**: Identify the oxidation states and the roles of each component in the reaction.

### Step 2: Coordination Chemistry of Aluminum
- Aluminum can form coordination complexes with a coordination number of 4 or 6. The most common complexes are \( \text{AlX}_4 \) (tetrahedral or square planar) and \( \text{AlX}_6 \) (octahedral).
- In the case of \( \text{AlF}_3 \), when it reacts with \( \text{HF} \) and \( \text{KF} \), it can form a complex with a higher coordination number.

**Hint**: Consider the geometry and coordination number of aluminum complexes.

### Step 3: Formation of the Complex
- When \( \text{AlF}_3 \) reacts with \( \text{HF} \) in the presence of \( \text{KF} \), it leads to the formation of the complex \( \text{K}_3[\text{AlF}_6] \).
- This complex has an octahedral geometry, where aluminum is surrounded by six fluoride ions.

**Hint**: Look for the specific complex that forms from the reaction of aluminum fluoride with fluoride ions.

### Step 4: Conclusion
- The presence of \( \text{KF} \) is essential because it provides additional fluoride ions that coordinate with aluminum, allowing the formation of the stable octahedral complex \( \text{K}_3[\text{AlF}_6] \).
- Therefore, \( \text{AlF}_3 \) is soluble in \( \text{HF} \) only when \( \text{KF} \) is present due to the formation of this complex.

**Final Answer**: The solubility of \( \text{AlF}_3 \) in \( \text{HF} \) in the presence of \( \text{KF} \) is due to the formation of the complex \( \text{K}_3[\text{AlF}_6] \).