# How many equal Xe–O bond length are present in (XeO_6)-3? [closed]

I tried to solve this problem by drawing a structure but the structure did not give me any answer and I was not able to draw the proper structure please help me out.

• what level of theory are you looking at? VSEPR? MO? I don't want to delve into anything MO (those stuffs that abuses hydrogen atom solution. meh. not a believer.). With VSEPR, if you show what you have done so far, I can take a look. – Argyll Sep 5 '18 at 3:34
• @Argyll are you serious about MO theory? You don’t have to believe in it, but that would put you at odds with the majority of chemical literature published in the last few decades. If you’re joking, it’s hard to tell. – orthocresol Sep 5 '18 at 4:57
• On topic, I have some suspicion that this odd-electron ion does not actually exist. $\ce{XeO6^4-}$ does, though. – orthocresol Sep 5 '18 at 4:58
• This compound exists. – user584880 Sep 5 '18 at 6:57
• Probably a misprint because I got it from a book – user584880 Sep 5 '18 at 16:43

Most likely, the structure is a distorted octahedron. This means, that the electron structure of the compound can be derived from the orbital diagram of $\ce{XeF6}$ octahedron (see here What is the molecular structure of xenon hexafluoride?)
Electron count in $\ce{XeF6}$ and $\ce{[XeO6]^{3-}}$ differs by 3 electrons, so we have to remove three electrons from the diagram found in the reference. Thus, the HOMO are two orbitals of $e_\mathrm{g}$ symmetry populated by 3 electrons. When two orbitals of the same energy have different electron population, a Jahn-Teller distortion occurs. The orbitals of similiar symmetry with similar occupation can be found, for example, in $\ce{Cu^{2+}}$ compounds. They typically show a geometry of distorted octahedrons with 2 opposite ligands significantly further from the central atom, than the 4 other ligands.
So, with quick and dirty treatment suggests that there are 2 different bond lengths in the ion $\ce{[XeO6]^{3-}}$.