A) \[Xe{{F}_{2}}\]
B) \[Xe{{F}_{4}}\]
C) \[Xe{{O}_{3}}\]
D) \[XeFe\]
Correct Answer: C
Solution :
\[Xe{{F}_{2}},Xe{{F}_{4}}\]and\[Xe{{F}_{6}}\]can be directly prepared. \[Xe+{{F}_{2}}\xrightarrow[673\,K]{Ni\,tube}\text{ }Xe{{F}_{2}}\] \[Xe+2{{F}_{2}}\xrightarrow[6\,atm]{673\,K}Xe{{F}_{4}}\] \[Xe+3{{F}_{2}}\xrightarrow[50-60\,atm]{523-573K}Xe{{F}_{6}}\] \[Xe{{O}_{3}}\]is obtained by the hydrolysis of\[Xe{{F}_{6}}\] \[Xe{{F}_{6}}+3{{H}_{2}}O\xrightarrow{{}}Xe{{O}_{3}}+6HF\]
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