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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 .

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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.

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  • $\begingroup$ Then in $\ce{XeOF5-}$ the axial position should be occupied by two fluorine atoms, since then at least two fluorine atoms out of the 5 will be in a 4-electron-3-centre bond. But that is not the case; the oxygen and the lone pair occupy the axial positions (see 3.10 (i)). So is this something which cannot be generalised? $\endgroup$ Oct 22 at 8:06
  • $\begingroup$ @AshishAhuja Instead of linking to a problem set, you could also have linked directly to the corresponding research article which the problem set actually references. But that's minor. As for the structure of $\ce{XeOF5-}$, this is complicated both to rationalise and to identify. The authors of the paper note that it has not been possible to crystallise the salt as it violently decomposes. Therefore, the structural analysis is based on inferrals from other analytic data. In addition, the entire picture is more complicated, as we are not dealing with ... $\endgroup$
    – Jan
    Oct 22 at 12:18
  • $\begingroup$ a neutral molecule but a molecular anion. For corresponding halogen-centred anions (e.g. $\ce{IF6-}$), the rule of thumb is to start off with the central atom carrying the necessary overall charge (i.e. $\ce{I-}$) and then adding ligand pairs using 4e3c bonds as needed. This rule already fails for xenon as we would start off with $\ce{[Xe-]^\dagger}$ which requires an electron occupying a higher shell. This already means that $\ce{XeOF5-}$ defies the standard rules of thumb and would require a different, more general set of assumptions. $\endgroup$
    – Jan
    Oct 22 at 12:21
  • $\begingroup$ On the other hand, $\ce{XeOF5-}$ is synthesised from $\ce{XeOF4}$ -- whose structure is easily predictable -- by addition of fluoride and the literature states that it is subject to rapid fluoride exchange and fluctuation (to the point where no fluorine-xenon coupling is observed in the NMR -- but a small side peak corresponding to $\ce{XeOF4}$ is clearly visible in the $\ce{^19F}$ NMR) so I'm not even sure if we can give a good answer to the question on its electronic structure. All things considered, I think it would be a great standalone question! $\endgroup$
    – Jan
    Oct 22 at 12:23
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    $\begingroup$ Ok, thank you. I think I will avoid asking a new question for now since my understanding of the chemistry which is involved here is very limited and I would not be able to phrase one properly with the relevant details, like those which you have mentioned in your third comment. And I probably wouldn't be able to understand the answers either; I'm in high school and I don't even understand NMR properly. Nevertheless, I'm satisfied for now. $\endgroup$ Oct 23 at 7:14
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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.

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