`XeF_(2)` reacts with `PF_(5)` to give :

Options

A. `XeF_6`
B. `[XeF]^(+) [PF_6]^(-)`
C. `XeF_4`
D. `[PF_4]^(+)[XeF_3]^(-)`

Answer

To solve the question regarding the reaction between \( \text{XeF}_2 \) and \( \text{PF}_5 \), we can follow these steps:

### Step 1: Understand the Reactants
- We have \( \text{XeF}_2 \) (Xenon difluoride) and \( \text{PF}_5 \) (Phosphorus pentafluoride).
- \( \text{PF}_5 \) is known to act as a fluoride ion acceptor due to the presence of phosphorus in a high oxidation state.

### Step 2: Identify the Reaction Type
- The reaction involves \( \text{XeF}_2 \) donating a fluoride ion to \( \text{PF}_5 \).
- This results in the formation of a cationic species from \( \text{XeF}_2 \) and an anionic species from \( \text{PF}_5 \).

### Step 3: Write the Reaction Equation
- The reaction can be represented as:
\[ \text{XeF}_2 + \text{PF}_5 \rightarrow \text{XeF}^+ + \text{PF}_6^- \]
- Here, \( \text{XeF}^+ \) is the cation formed from \( \text{XeF}_2 \) and \( \text{PF}_6^- \) is the anion formed from \( \text{PF}_5 \).

### Step 4: Analyze the Products
- The products of the reaction are \( \text{XeF}^+ \) and \( \text{PF}_6^- \).
- We need to check the provided options to identify which one corresponds to these products.

### Step 5: Check the Options
- Based on the reaction, we can conclude that:
- Option A: Incorrect (does not match the products)
- Option B: Correct (matches \( \text{XeF}^+ \) and \( \text{PF}_6^- \))
- Option C: Incorrect (does not match the products)
- Option D: Incorrect (does not match the products)

### Final Answer
- The correct answer is **Option B**.

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