`F_2` is the formed by reacting `K_2MnF_6` with

Options

A. `SbF_5`
B. `MnF_3`
C. `KSbF_6`
D. `MnF_4`

Explanation

To determine the compound that forms fluorine gas (F₂) when potassium hexafluoromanganate (K₂MnF₆) reacts with another substance, we will analyze the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant we have is potassium hexafluoromanganate, K₂MnF₆. We are looking for a Lewis acid that can react with it.

2. **Choose a Suitable Lewis Acid**:
A strong Lewis acid that can react with K₂MnF₆ is sulfur tetrafluoride (SF₄) or sulfur pentafluoride (SF₅). In this case, we will consider SF₅.

3. **Write the Reaction**:
When K₂MnF₆ reacts with SF₅, the following reaction occurs:
\[ K_2MnF_6 + SF_5 \rightarrow KSbF_6 + MnF_3 + \frac{1}{2} F_2 \]
Here, SF₅ displaces the weaker Lewis acid MnF₄ from the salt.

4. **Analyze the Products**:
The products of this reaction include:
- KSbF₆ (potassium hexafluoroantimonate)
- MnF₃ (manganese trifluoride)
- F₂ (fluorine gas)

5. **Conclusion**:
From the reaction, we can conclude that fluorine gas (F₂) is produced when K₂MnF₆ reacts with SF₅.

### Final Answer:
Fluorine gas (F₂) is formed by reacting K₂MnF₆ with sulfur pentafluoride (SF₅).

---