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 \[\operatorname{Xe} + {{F}_{2}}~~\xrightarrow{Ni\,\,tube} Xe{{F}_{2}};\, Xe+ 2{{F}_{2}}\] \[\xrightarrow[6\,\,atm]{673\,\,K}\,\,\,\,Xe{{F}_{4}}\] \[Xe+3{{F}_{2}}\,\,\xrightarrow[50\,-\,60\,\,atm]{523\,-\,573\,\,K}\,Xe{{F}_{6}}\] \[Xe{{O}_{3}}\] 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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