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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.

Mithoron
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Ayush Kumar
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1 Answers1

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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$^2$.

Sources:

  1. https://pubs.acs.org/doi/10.1021/ic50027a038
  2. https://pubs.acs.org/doi/10.1021/ic00334a037
Mithoron
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M.L
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  • How does a Xenon fluorine bond work exactly? I thought Xenon is a noble gas so it wouldn't react – Sidharth Ghoshal Sep 02 '22 at 17:23
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    @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. – Mark Sep 02 '22 at 23:50
  • @SidharthGhoshal This should answer your question: https://chemistry.stackexchange.com/questions/31845/why-are-there-more-fluoride-compounds-formed-with-xenon – Nilay Ghosh Sep 03 '22 at 06:07