This reaction was asked in my college entrance exam. I searched through the entire Internet, but I couldn't find any papers/sources.

I suspect since O₂F₂ is a strong fluorinating agent, one of the xenon fluorides (XeF₂, XeF₄, or XeF6) will form, but I am not sure. Reaction conditions were not given in the question.

  • 1
    $\begingroup$ This was asked in JEE advance this year (right ?). I gave it too btw. $\endgroup$
    – Ankit
    Sep 3, 2022 at 17:11
  • $\begingroup$ $$\ce{XeF4 + O2F2 ->[143K] XeF6 + O2}$$ source:- ncert.nic.in/ncerts/l/lech107.pdf page 40 in the pdf $\endgroup$
    – Harshil
    Sep 28, 2022 at 18:27

1 Answer 1


From my research, it really depends on the experimental conditions and procedure. One paper in 1965 found that primarily $\ce{XeF2}$ was formed when xenon reacted with $\ce{O2F2}$ at low temperatures$^1$. In contrast, another paper found a mixture of $\ce{XeF2}$ and $\ce{XeF4}$ was formed, and prolonged exposure to $\ce{O2F2}$ generated mostly $\ce{XeF4}$ but no $\ce{XeF6}$ was observed$^2$.


  1. https://pubs.acs.org/doi/10.1021/ic50027a038
  2. https://pubs.acs.org/doi/10.1021/ic00334a037
  • $\begingroup$ How does a Xenon fluorine bond work exactly? I thought Xenon is a noble gas so it wouldn't react $\endgroup$ Sep 2, 2022 at 17:23
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    $\begingroup$ @SidharthGhoshal, Xenon is the least-noble of the non-radioactive noble gasses, and fluorine is extremely reactive. This lets the fluorine bond to the outer-shell electrons of xenon. $\endgroup$
    – Mark
    Sep 2, 2022 at 23:50
  • $\begingroup$ @SidharthGhoshal This should answer your question: chemistry.stackexchange.com/questions/31845/… $\endgroup$ Sep 3, 2022 at 6:07

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