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