I already referred to this similar question but it yielded no explanation for the reason. Cations and anions of similar sizes stabilize each other through lattice energy effects in the solid state. This explains the stability of the alkali metal carbonates and sulphates as one moves down the group. Applying a similar logic here, the $\ce{[XeF2-]}$ anion is larger than the $\ce{F-}$ anion, yet $\ce{XeF6}$ exists with the $\ce{F-}$ anion.

this paper seems to contain exactly what I need but I am unable to access it. Could someone kindly shed light on the structure of $\ce{XeF6}$ in the solid state?

  • 7
    $\begingroup$ XeF2− is not a thing, and besides, would give a wrong overall composition. $\endgroup$ – Ivan Neretin Feb 21 at 5:46
  • 5
    $\begingroup$ As has already been pointed out, $\ce{[XeF5+][XeF2-]}$ is a rather problematic formulation; but $\ce{[XeF5+][XeF7-]}$ is more sensible and would be a valid question. $\endgroup$ – orthocresol Feb 21 at 10:37

First and most obviously, if something in solid state forms ions by shuffling atoms around, the end result must yield the same ratio of atoms. Take for example $\ce{PF5}$: this exists as $\ce{[PF4+][PF6-]}$. If you add the atoms together, you get $\ce{P2F10}$ which is $2\times\ce{PF5}$. The same is true for the proposed structure $\ce{[XeF5+][F-]}$ which adds up to $\ce{XeF6}$ as it should. Your proposed version would give $\ce{Xe2F7}\ne\ce{XeF6}$.

Second, the proposed anion does not actually work out. $\ce{XeF2}$ exists itself as a neutral compound obeying the octet rule and featuring an all-paired Lewis structure (and I believe also all-paired MO scheme). If you were to create an anion, you would need an additional electron making it $\ce{XeF2^.-}$. This species surely exists but not for long as it will react and degrade to something else.

Without having to go into detail why a different solution is more viable, it is elementary to show how impossible an option your proposed solution is.

| improve this answer | |
  • $\begingroup$ You showed conclusively(atleast for me) that the OP's proposition won't work out. However, is there any way to make an educated guess a priori for the formulation of $\ce{[XeF5+][F-]}$ ,given $\ce{[XeF6]}$ ? $\endgroup$ – Yusuf Hasan Feb 21 at 10:49
  • $\begingroup$ The formulation $[XeF_5]^+[F]^-$ is at best a simplification. In reality the various solid state structures consist of a number of oligomers with bridging fluorine see e.g. the abstract for sciencedirect.com/science/article/abs/pii/S0022113906001266 - halides with bridging halogen are very common, with $Al_2Cl_6$ possibly being the best known example $\endgroup$ – Ian Bush Feb 21 at 11:04
  • $\begingroup$ @IanBush Ohh, I see. But still, is there anyway to guess before hand how many bridging fluorines to expect(like the one $\ce{[F]-}$ acting as bridge in the above formulation)? $\endgroup$ – Yusuf Hasan Feb 21 at 11:15
  • $\begingroup$ @YusufHasan I don’t know if there’s an easy way to guess. In the $\ce{PF5}$ case, a fluorine atom is essentially transferred from one phosphorus to another. In the xenon case, fluorine atoms bridge between xenon centres. If anything, I would want to assume that octahedral and tetrahedral structures around central atoms might be preferred due to the higher symmetry but take that with a grain of salt (pun intended). $\endgroup$ – Jan Feb 21 at 12:10
  • $\begingroup$ @Jan Just a theory: in $\ce{PF5}$ , the axial bonds are a bit longer than the equatorial ones in the solid state. So, would it be probable to think that one of the axial bonds might break off from one of the monomers, and get transferred to another $\ce{PF5}$, thus allowing $\ce{[PF4]+}$ to adopt a stucture with less bp-bp repulsions? $\endgroup$ – Yusuf Hasan Feb 21 at 12:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.