A) \[XeO{{F}_{2}}\]
B) \[XeO{{F}_{4}}\]
C) \[Xe{{O}_{3}}\]
D) \[Xe{{O}_{2}}{{F}_{2}}\]
Correct Answer: A
Solution :
\[Xe{{F}_{4}}\]reacts with water at\[-80{}^\circ C\]to give xenon oxyfluoride. \[Xe{{F}_{4}}+{{H}_{2}}O\xrightarrow[{}]{-80{}^\circ C}\underset{\begin{smallmatrix} \,\,\,\,\,\,xlnon \\ oxyfluoride \end{smallmatrix}}{\mathop{XeO{{F}_{2}}}}\,+2HF\]
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