A) \[Xe{{F}_{4}}+{{O}_{2}}{{F}_{2}}\to Xe{{F}_{6}}+{{O}_{2}}\]
B) \[Xe{{F}_{2}}+P{{F}_{5}}\to {{[XeF]}^{+}}P{{F}_{6}}^{-}\]
C) \[Xe{{F}_{6}}+{{H}_{2}}O\to XeO{{F}_{4}}+2HF\]
D) \[Xe{{F}_{6}}+2{{H}_{2}}O\to Xe{{O}_{2}}{{F}_{2}}+4HF\]
Correct Answer: A
Solution :
In the reaction \[\overset{+4}{\mathop{X}}\,e{{F}_{4}}+{{\overset{+1}{\mathop{O}}\,}_{2}}{{F}_{2}}\to \overset{+6}{\mathop{X}}\,e{{F}_{6}}+O_{2}^{0}\] Xenon undergoes oxidation while oxygen undergoes reduction.
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