# The real structure of difluoro(dioxo)xenon (XeO2F2)

I have a problem regarding structure of difluoro(dioxo)xenon ($$\ce{XeO2F2}$$). I have posted a image above. my teacher is saying that the structure B is correct as there is very less repulsion in structure B that is what I too think but many of the sources and books have written the structure A to be correct and me as well as my teacher is in a doubt that why structure A is correct not structure B as there is an angle of 180° between oxygen atoms. If I am wrong please guide me and also please tell me what are factors should I look at for judging a correct shape for a compound. Any hint will be very very helpful to me .

Xenon is a noble gas and thus does not need to form any bonds to complete its octet. When such a bond is formed, this is known as a hypercoordinate structure: it contains more bonding partners than the most simple of simple Lewis structure theories would allow. Thanks to Linus Pauling, for many years chemists have been trying to invoke d orbitals to explain this bonding situation when the actual case is far more trivial and can be explained with simple octet considerations.

The two oxygen atoms which are depicted as double-bonded in the d orbital model are actually connected by a single bond. If you want, this bond can be formed by oxygen keeping its six valence electrons in three orbitals and using the forth, empty orbital to bond with a single orbital of xenon. Both electrons can be described as originally donated by xenon and a good way to draw it in the Lewis formalism is to use charge separation:

$$\ce{\overset{-}{O}-\overset{2+}{Xe}-\overset{-}{O}}$$

This leaves the two fluorine atoms. They actually form a slightly more complex bond known as a 4-electron-3-centre bond (thus fluorine only bonds to xenon in neutral compounds in multiples of 2). For a deeper explanation of the 4-electron-3-centre bond, see this previous answer of mine. In a Lewis formalism, this can only described with resonance structures as shown below:

$$\ce{F-\overset{+}{Xe}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Xe}-F}$$

This bonding situation means that the fluoride atoms must occupy opposite positions with $$\angle(\ce{F-Xe-F}) \approx 180°$$. Thus, A is the correct depiction of $$\ce{XeO2F2}$$ which was confirmed in a neutron diffraction experiment by Peterson, Willett and Huston.[1]

The paper determines the $$\ce{O-Xe-O}$$ angle as approximately $$105°$$. This is somewhat in-between a pure p orbital and an $$\mathrm{sp^2}$$ description of xenon, $$15°$$ from either. At first approximation it would, however, be fine to assume unhybridised xenon and derive the bonding situation from there with each oxygen bonded to one of the p orbitals and the two fluorines bonding to the third.

Reference:

[1]: S. W. Peterson, R. D. Willett, J. L. Huston, J. Chem. Phys. 1973, 59, 453–459. DOI: 10.1063/1.1679827.

Think for yourself. The sole objective of the VSEPR Theory is to determine the nearest shape of a molecule such that there is minimum repulsion. In this example as we can see, we have to take into account lp-lp and lp-bp repulsions. Since Fluorine is more electronegative than Oxygen, the electronic cloud density is more in the former. Hence, it is more favorable to put the F-atoms in the equatorial plane with the O-atoms occupying the axial plane. This is to reduce the bp repulsions with the lp : Option B) correct.