Question

Xenon hexaflouride reacts with silica to form a xenon compound X. The oxidation state of xenon in X is:

• A. 2
• B. 4
• C. 6
• D. 0

Answer 1

Answer: (C)

6

Answer 2

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: $\overset{x}{\mathop{X}}\,e\overset{-2}{\mathop{O}}\,{{\overset{-1}{\mathop{F}}\,}_{4}}$ $x+(-2)+4\times (-1)=0$ $x-2-4=0$ $x=+\,6$