Complete the following reaction : `Xe+PtF_(6)`

Options

A. `XeF_(4)+O_(2)F_(2)rarrXeF_(6)+O_(2)`
B. `Xe+PtF_(6)rarrXeF_(6)+Pt`
C. `Xe+PtF_(6)rarrXe^(+)[PtF_(6)]^(-)`
D. `Xe+PtF_(6)rarrXeO_(2)F_(4)+Pt`

Explanation

To complete the reaction between xenon (Xe) and platinum hexafluoride (PtF6), we need to consider the nature of the reactants and the type of reaction that occurs.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are xenon (Xe), a noble gas, and platinum hexafluoride (PtF6), a strong fluorinating agent.

2. **Understand the Reaction Type**:
- Noble gases like xenon can form compounds under certain conditions, especially with strong fluorinating agents. In this case, PtF6 can react with Xe to form a complex.

3. **Write the Reaction**:
- When xenon reacts with platinum hexafluoride, it forms a complex ion. The reaction can be represented as:
\[ \text{Xe} + \text{PtF}_6 \rightarrow \text{[XePtF}_6]^+ \]
- In this reaction, xenon donates a pair of electrons to the platinum, forming a positively charged complex ion \([XePtF_6]^+\) and a negatively charged fluoride ion \([PtF_6]^-\).

4. **Final Equation**:
- The complete balanced reaction can be written as:
\[ \text{Xe} + \text{PtF}_6 \rightarrow \text{[XePtF}_6]^+ + \text{F}^- \]

### Final Answer:
The complete reaction is:
\[ \text{Xe} + \text{PtF}_6 \rightarrow \text{[XePtF}_6]^+ + \text{F}^- \]