# Xe–F bond length in xenon fluorides

The order of bond length of $\ce{Xe-F}$ in $\ce{XeF2}$, $\ce{XeF4}$, $\ce{XeF6}$ is?

My thoughts:

As hybridization of $\ce{Xe}$ in $\ce{XeF2}$ is $\ce{sp^3d}$, $\ce{XeF4}$ is $\ce{sp^3d^2}$, $\ce{XeF6}$ is $\ce{sp^3d^3}$.

%$\ce{s}$ decreases from $\ce{XeF2}$, $\ce{XeF4}$ to $\ce{XeF6}$, so bond length should increase, but my book says the opposite, i.e. $\ce{XeF2} > \ce{XeF4} > \ce{XeF6}$.

• Downvoting for incorrect use of d orbital participation for main-group compounds.
– Jan
Nov 3 '17 at 13:33
• @Jan I don't think a misconception in a question (and a misconception based on something which continues to be taught in general chemistry courses in the US) warrants a downvote. Nov 3 '17 at 14:34
• Something seems fishy... Multiple sources say that the Xe-F bond is strongest in XeF2 and weakest in XeF6... Nov 30 '17 at 9:09
• The explanation for the trend in bond strength of Xe-F in the xenon fluorides is the steric crowding about xenon, resulting in weaker bonds when there are more fluorine atoms packed around the central xenon atom. Nov 30 '17 at 9:11

Yes, your book is correct. To rationalize it, throw away hydridization and electron exitation and work with concept of three-center four-electron bond. The idea is that in $\ce{XeF2}$ the real structure is a mix of $\ce{F-Xe+...F^-}$ and $\ce{F^-...Xe+-F}$, so the bonding is half-covalent and half-ionic. By packing more fluorine around the xenone atom, we force xenone into state with higher effective atomic charge, so the ionic part of the bonding becomes stronger as $\ce{Xe^{3+}}$ pulls fluorine anions much stronger.

same logic applies to many other cases, in particular sulfur oxides and sulfur fluorides.

For context, I quote experimental bond lengths from paper. Note, the paper itself quotes the values from other sources with slightly different values from different sources.

XeF2 ... 1.977

XeF4 ... 1.94

XeF6 ... 1.94 & 1.850 (two bond types)

• Your resonance structure doesn't seem correct. In your structure, there is a negative formal charge on fluorine and the Xe-F portion is neutral, giving an overall negative charge to the molecule. However, the XeF2 molecule is neutral. Nov 24 '17 at 3:38

This happens because higher number of fluorine atoms induce a higher positive oxidation state on Xe which causes the Xe atom to shrink in size resulting in a smaller bond length.

XeF2 has Xe in +2 oxidation state while XeF6 has it in +6 oxidation state