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I have recently being studying structures of different types of compounds, when I came across the structure of $\ce{[XeF5]+}$ which have a structure distorted square pyramidal. I could not understand what "distorted" square pyramidal is and how it is different from square pyramidal.

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Going by a standard argumentation, the $\ce{[XeF5]+}$ cation would be expected to have two different bond types: a classical 2-electron-2-centre bond to the axial fluorine and a pair of two 4-electron-3-centre bonds making up the base of the pyramid with the equatorial fluorines. This structure is shown in figure 1.

Structure of the pentafluoridoxenon(VI) cation
Figure 1: Hypothetical perfect structure of $\ce{[XeF5]+}$. Image taken from Wikipedia, where a full list of authors is available.

The actual structure of $\ce{[XeF5]+}$, reported by Leary et al.,[1] is not perfect, though. The axial fluorine atom does not sit at the centre of the pyramid, destroying the perfect $C_\mathrm{4v}$ point group. This is probably an effect of core orbitals on the central xenon atom (I cannot access the paper where I am right now). Because the pyramid no longer has four equal faces, it is distorted.

Reference:
K. Leary, D. H. Templeton, A. Zalkin, N. Bartlett, Inorg. Chem. 1973, 12, 1726. DOI: 10.1021/ic50126a004.

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    $\begingroup$ Jan, I think it is the other way around. When all of the angles are 90° (e.g. your Figure 2) you have a regular square pyramid. When the angles deviate from 90° you have a distorted square pyramid. $\endgroup$ – ron Aug 20 '16 at 19:41
  • $\begingroup$ 1) Chemistry borrowed the term from mathematics were it was originally defined, so how common a chemical structure is doesn't matter. 2) At least in the solid phase $\ce{XeF5^{+}}$ is distorted, the $\ce{F_{ax}-Xe-F_{eq}}$ angle is 79°, not 90° $\endgroup$ – ron Aug 20 '16 at 20:10
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    $\begingroup$ @ron Sorry, I misinterpreted what you (and Mith, and probably OP’s textbook, too) meant by distorted, not knowing about the crystal structure (and not reading the other answer in good enough detail). I know get your point and see that your reasoning is correct. Will edit the answer in a sec. $\endgroup$ – Jan Aug 20 '16 at 20:22
  • $\begingroup$ Jan, just read your revised answer. I think the article says the axial fluorine does sit at the top of the pyramid, in the center, maintaining approximate $\ce{C_{4v}}$ symmetry. It is just that the 4 equatorial fluorines are pushed up towards the axial fluorine ($\ce{F_{ax}-Xe-F_{eq}}$ ~ 79°) and away from the axial lone pair on the other side of the pyramid. There are still 4 equivalent faces. $\endgroup$ – ron Aug 20 '16 at 21:01
  • $\begingroup$ @ron Atm I just feel like deleting for simplicity :D well, whatever, I’ll check out the article once I get a chance … $\endgroup$ – Jan Aug 20 '16 at 21:05
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The structure of $\ce{XeF5}$ is approximately described by the $\ce{C_{4v}}$ point group. This means that the four equatorial F atoms may lie in the same plane as the Xe atom and that the fifth, axial, F atom lies above the Xe atom at an angle of 90 degrees to the plane of the other F atoms (and so in the principal axis). If the point group is $\ce{C_{4v}}$ the adjacent angle $\ce{F_{eq} - Xe - F _{eq}}$ has to be close to 90 degrees. The lone pair of electrons occupies the sixth position, opposite to the axial F atom. In the xray - diffraction data of $\ce{XeF5+ PdF6-}$ crystals the $\ce{F_{eq} - Xe - F _{ax}}$ angle is ~80. In the crystal this may be due to packing or interaction of adjacent ions, as there seems to be no reason, based on symmetry, why the $\ce{F_{eq} - Xe - F _{ax}}$ angle is not very close to 90 degrees. (The structure has been reported by Leary et. al. Inorg. Chem. 1973, vol 12, p 1726)
Edit. Considering the comments below if the Xe atom is not in the plane of the four equatorial F atoms then the axial F atom can still be on the principal axis, but the angle $\ce{F_{eq} - Xe - F_{ax}}$ will differ from 90. The structure will appear distorted from square pyramidal in the sense that the Xe is not in the plane.

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  • $\begingroup$ Nope, this is wrong. Even regular square pyramids where the central atom is not in the plane of the pyramid’s bottom for atoms have $C_\mathrm{4v}$ (try it!). $\endgroup$ – Jan Aug 20 '16 at 18:52
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    $\begingroup$ "there seems to be no reason, based on symmetry..." Since the $\ce{F_{eq}-Xe-F_{ax}}$ interaction is different from the $\ce{F_{eq}-Xe-LP_{ax}}$ interaction (LP=lone pair), symmetry tells us that the $\ce{F_{eq}-Xe-F_{ax}}$ angle cannot be 90°. It may be 89.999° or ~80°, but it can't be 90°. Consequently the structure must be a distorted square pyramid. $\endgroup$ – ron Aug 20 '16 at 20:34
  • $\begingroup$ @jan Yes, of course, in $\ce{ C_{4V}}$ they can be in or out of the plane but only as long as they are on the principal axis , but in the crystal structure they are in the plane as far as in can see, and we are discussing this molecule. So my statement is not "wrong", just incomplete. The down vote is misleading to the OP, if not others, as this answer does describe the actual experimental structure not any theorising as to what it might be. Also the equatorial and axial bond lengths are 0.181 and 0.184 nm so almost identical, which suggests very little difference in bonding. $\endgroup$ – porphyrin Aug 20 '16 at 20:35
  • $\begingroup$ @ron, I cannot see why without any external forces on the molecule why the angle should not be arbitrarily close to 90 $^o$. I can see how the molecule might distort to make the square containing the four F atoms into a diamond shape but that would not tip the axial F off axis as it appears to be. Any displacement of the axis F to one side would cause more interaction only to be countered by a lessening to another and so restore the atom to be on axis. With an external force in one direction a non-right angle is possible, and in fact expected. $\endgroup$ – porphyrin Aug 20 '16 at 20:46
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    $\begingroup$ I am not saying that the axial fluorine tips off center. I am just saying that since the $\ce{F_{ax}-Xe-F_{eq}}$ interaction must be different from the $\ce{LP_{ax}-Xe-F_{eq}}$ interaction, that the $\ce{F_{ax}-Xe-F_{eq}}$ angle cannot be exactly 90° and, in fact, it turns out to be 79°. $\endgroup$ – ron Aug 20 '16 at 21:07

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