A) \[XeSi{{O}_{4}}+HF\]
B) \[Xe{{F}_{2}}+Si{{F}_{4}}\]
C) \[XeO{{F}_{4}}+Si{{F}_{4}}\]
D) \[Xe{{O}_{3}}+Si{{F}_{2}}\]
Correct Answer: C
Solution :
[c] \[2Xe{{F}_{6}}+Si{{O}_{2}}\to Si{{F}_{4}}+2XeO{{F}_{4}}\]
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