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

Neil Bartlett (1932–2008) first synthesized $\ce{XeF{_4}}$ (and $\ce{XeF{_6}}$) in 1962. In the synthesis, a nickel chamber is used, and heated to 400°C, causing the formation of $\ce{NiF{_4}}$, 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?)

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The answer appears to be right there in the linked synthesis details - the nickel is non-catalytic and is only there to generate a passivated surface that won't degrade in the high-temperature fluorine atmosphere.

The plot thickens! this paper suggests that xenon may in fact abstract fluorine off the passivated nickel fluoride surface due to observed zeroth order kinetics, which lends some serious credence to jonsca's original inference of Ni being catalytic. This will take some more investigation...

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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
Ah, okay, I was assuming that the fluorines all came as a package. I know nil about inorganic, so thanks for pointing that out. If you edit your answer to reflect that, I'll accept it. – jonsca May 4 '12 at 12:59
@jonsca - I'd have to confirm that that's how the reaction actually goes but it makes sense to me. – Richard Terrett May 4 '12 at 13:04
@jonsca - I did a bit of poking around about the mechanism and found an interesting paper (I can only access the abstract from here). I've added an addendum to my answer. Note that nickel is not required to make the reaction go. – Richard Terrett May 4 '12 at 14:47
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