# What is the geometric configuration of the four fluorine atoms during the synthesis of xenon tetrafluoride?

Neil Bartlett (1932–2008) first synthesized $\ce{XeF4}$ (and $\ce{XeF6}$) in 1962. In the synthesis, a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF4}$, which Wikipedia claims is not part of the reaction.

Due to the nature of the noble gas, xenon, one might imagine that the fluorines would need to could be "forced" onto the $\ce{Xe}$ atom at the same time, otherwise with each addition, the molecule would be even less energetically likely. I was trying to ascertain in what geometric form the four fluorine atoms would configure themselves to make this addition possible.

I had first assumed that the nickel was being used as a catalyst, and the fluorines could be plucked from its surface accordingly, but this does not jive with the description of the synthesis above. According to the response below, one synthesis of the molecule does rely on forming fluoride radicals on a catalyst, but it's mysterious that the synthesis above does not mention this.

Assuming there is no catalyst, how are the fluorines at 400°C oriented to allow the addition of 2 molecules at presumably the same time?

F - F
.   .
.   .
F - F


(and is that even possible?)

• You are correct about that, I have edited the question. Of what form does the $F{_2}$ take, then? – jonsca May 4 '12 at 12:43
• $\ce{F_{2}}$ is a diatomic gas. Bear in mind that the $\ce{Xe}$ is in the +4 or +6 oxidation state in the corresponding fluorides. Housecroft and Sharpe pp. 496 describe a synthesis of xenon fluorides involving UV irradiation or electric discharge. This would suggest to me homolytic cleavage of $\ce{F_{2}}$ followed by $\ce{F^{\cdot}}$ radical attack on $\ce{Xe}$. – Richard Terrett May 4 '12 at 12:55