A) + 2
B) + 4
C) + 6
D) 0
Correct Answer: C
Solution :
Xenon hexafluoride reacts with silica to form \[\text{XeO}{{\text{F}}_{\text{4}}}\]as: \[2Xe{{F}_{6}}+Si{{O}_{2}}\xrightarrow{{}}\text{ }XeO{{F}_{4}}+Si{{F}_{4}}\] The oxidation state of xenon in \[\text{XeO}{{\text{F}}_{\text{4}}}\] is calculated as: \[\overset{x}{\mathop{X}}\,e\overset{-2}{\mathop{O}}\,{{\overset{-1}{\mathop{F}}\,}_{4}}\] \[x+(-2)+4\times (-1)=0\] \[x-2-4=0\] \[x=+\,6\]
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