A) \[Xe{{O}_{3}}\]
B) \[Xe{{F}_{4}}\]
C) \[Xe{{F}_{2}}\]
D) \[XeO{{F}_{4}}\]
Correct Answer: D
Solution :
\[Xe{{F}_{6}}+{{H}_{2}}O\to \underset{(X)}{\mathop{XeO{{F}_{4}}}}\,+2HF\] \[2Xe{{F}_{6}}+Si{{O}_{2}}\to \underset{(X)}{\mathop{2XeO{{F}_{4}}}}\,+Si{{F}_{4}}\]You need to login to perform this action.
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