In the preparation of compounds of Xe, Bartlett has taken `O_(2)^(+)Pt F_(6)^(-)` as a base compound. This is becase

Options

A. both `O_(2)` and Xe have same size.
B. both `O_(2)` and Xe have same electron gain enthalpy.
C. both `O_(2)` and Xe have almost same ionisation enthalphy.
D. both Xe and `O_(2)` are gases.

Explanation

To understand why Bartlett chose \( O_2^{+}PtF_6^{-} \) as a base compound in the preparation of xenon compounds, we can break down the reasoning into several steps:

### Step 1: Understanding Ionization Enthalpy
- Ionization enthalpy is the energy required to remove an electron from an atom or molecule in the gas phase.
- The ionization enthalpy values for xenon (Xe) and oxygen (O) are quite similar, with xenon having an ionization enthalpy of approximately 1176 kJ/mol and oxygen around 1180 kJ/mol.

### Step 2: Similarity in Chemical Behavior
- Because the ionization enthalpies of xenon and oxygen are close in value, it suggests that they may exhibit similar chemical behavior.
- This similarity allows for the possibility of forming stable compounds when xenon interacts with other elements or compounds.

### Step 3: Selection of the Base Compound
- Neil Bartlett chose \( O_2^{+}PtF_6^{-} \) as a base compound because it provides a suitable environment for xenon to react and form compounds.
- The positive charge on \( O_2^{+} \) indicates that it can accept an electron, which is important for reactions involving noble gases like xenon.

### Step 4: Formation of Xenon Compounds
- By using \( O_2^{+}PtF_6^{-} \), Bartlett was able to successfully synthesize various xenon compounds, demonstrating the reactivity of xenon under specific conditions.
- The choice of this base compound was crucial in facilitating the formation of xenon-fluorine compounds.

### Conclusion
- In summary, Bartlett's choice of \( O_2^{+}PtF_6^{-} \) as a base compound was due to the similar ionization enthalpy of xenon and oxygen, which allowed for the successful synthesis of xenon compounds.