Since Neil Bartlett's 1962 discovery that xenon was capable of forming chemical compounds, a large number of xenon compounds have been discovered and described. Almost all known xenon compounds contain the electronegative atoms fluorine or oxygen. The oxidation state of xenon in its compounds is generally +2, +4, +6, or +8.

Xenon is known to form three fluorides: $\ce{XeF_{$n$}} \;(n = 2, 4, 6)$ where the oxidation states of xenon are +2, +4, and +6. $\ce{XeF8}$ is not known to exist even though the oxidation state would be +8. Why is this so? Could it possibly be due to the fact that 8 fluorine atoms can't fit around a xenon atom?

Xenon is known to form three oxides: $\ce{XeO_{$n$}} \;(n = 2, 3, 4)$ where the oxidation states of xenon is +4, +6, and +8. $\ce{XeO2}$ was not known until 2011. Why did it take so long to be discovered? Moreover, the lowest oxide of xenon $\ce{XeO}$ is not known though the oxidation state of xenon would be +2. Why?

  • $\begingroup$ en.wikipedia.org/wiki/Xenon_dioxide $\endgroup$
    – Mithoron
    Apr 23 '16 at 20:34
  • $\begingroup$ I think this has something to do with XeO4 having sp3 hybridization while XeF6 has sp3d2 hybridization. I've often heard the generalization that despite fluorine's higher electronegativity, oxygen can generally bring out higher oxidation states in its compounds because of its ability to form double bonds. I don't know why you couldn't just keep adding more fluorines though $\endgroup$
    – gannex
    Apr 23 '16 at 22:53
  • $\begingroup$ @Mithoron The link does not describe why it took so long time to discovered while the other two oxides were well known. $\endgroup$ Apr 24 '16 at 3:23
  • $\begingroup$ This pdf says that: "Our findings imply that xenon monoxide, XeO, has no region of stability up to a pressure of at least P = 200 GPa." And then later state that: "we do not find XeO to be stable at any pressure" $\endgroup$
    – manshu
    Apr 24 '16 at 16:24

$\ce{XeF8}$ is not known to exist though O.N is +8. Why is this so?

At least 2 compounds have been reported that contain the $\ce{XeF8^{2-}}$ unit. See, for example:

  • $\ce{(NO^+)2[XeF8]^{2-}}$ (reference)
  • Metal salts of the form $\ce{(M^{+})_2[XeF8]^{2-}}$ where M is a metal salt such as $\ce{Cs, Rb}$ (see the above reference) or $\ce{Na}$ (see p. 62 in Advances in Inorganic Chemistry, Volume 46, A. G. Sykes editor; link

The $\ce{XeF8}$ portion of the molecules approximates a square-antiprism geometry.

enter image description here

(image source)

the lowest oxide of xenon $\ce{XeO}$ is not known though O.N is +2. Why?

Perhaps the molecule is unstable, favoring disproportionation to other xenon oxides plus oxygen. For example, Andreas Hermann and Peter Schwerdtfeger suggest the following pathway at high pressure:

$\ce{3XeO → Xe3O2 + 1/2O2}$

The authors go on to note "we do not find $\ce{XeO}$ to be stable at any pressure" (link to full paper).

Edit: See orthocresol's comment below. He argues that at ambient pressure the following decomposition pathway might be more likely:

$\ce{XeO -> Xe + 1/2 O2}$

  • 2
    $\begingroup$ 1) The paper writes that XeO is "metastable toward the decomposition reaction", which implies a thermodynamic instability. 2) It's worth a mention that the calculations were ran at a whopping pressure of 75 GPa (~740,000 atm), so under more normal conditions the instability of XeO cannot be attributed to this pathway. If I have read the paper correctly, Xe3O2 is unstable wrt the elements under 75 GPa, so I would hazard a guess that the likely decomposition pathway of XeO at 1 atm is XeO -> Xe + 1/2 O2. $\endgroup$
    – orthocresol
    Apr 24 '16 at 18:39
  • $\begingroup$ Thanks orthocresol, I'll edit my answer to include your thinking. $\endgroup$
    – ron
    Apr 24 '16 at 19:02
  • $\begingroup$ One more thing: The higher oxides: XeO3 and XeO4 were discovered earlier but the oxide XeO2 was discovered recently! What took it so long to discover? $\endgroup$ Apr 25 '16 at 4:42
  • $\begingroup$ Wikipedia says it is prepared at 0 C and quickly decomposes to $\ce{XeO3}$ with a half-life around 2 minutes. Sounds like it may have been difficult to tell that something other than $\ce{XeO3}$ was being produced. $\endgroup$
    – ron
    Apr 25 '16 at 13:26
  • 6
    $\begingroup$ @NilayGhosh No, any chemical that decomposes or transforms can be said to have a half-life, not just radioactive compounds. $\endgroup$
    – ron
    Apr 25 '16 at 15:07

According to Heats of Formation of XeF3+, XeF3-, XeF5+, XeF7+, XeF7-, and XeF8 from High Level Electronic Structure Calculations Inorganic Chemistry 2010, vol. 49, pages 261–270:

Unlike the previously studied XeF2, XeF4, and XeF6,
$\ce{XeF8}$ is predicted to be thermodynamically unstable with respect to loss of $\ce{F2}$, and the reaction is calculated to be exothermic by 22.3 kcal/mol at 0K.

Actually, $\ce{XeO}$ is known, but as an eximer.

The first observation was the work published in 1946 as New Band System in the Green Excited in a Mixture of Xenon and Oxygen and the Energy of Dissociation of CO Phys. Rev. vol. 69, pages 36– 37.

A more-recent article about XeO is Optical and Electron Spin Resonance Studies of Xenon–Nitrogen–Helium Condensates Containing Nitrogen and Oxygen Atoms J. Phys. Chem. A, 2015, 119, pp 2438–2448

Emissions from excimer XeO* molecules have been observed and extensively studied in both gaseous(1-4) and condensed(5-11) phases. Up to now, experiments with XeO* in condensed phases were limited to solid and liquid rare gas (RG) matrices. Beams of electrons, α-particles, or protons, as well as ultraviolet (UV) irradiation (with wavelengths of 260 nm and shorter) were used to form excited xenon–oxygen complexes in neon, argon, or krypton matrices doped with xenon atoms and O2, N2O, or CO2 molecules as precursors of oxygen atoms. It is worth noting that excitation of a xenon matrix doped with O atoms causes an emission with a much simpler spectrum. It consists of two intense bands with maxima at 370 and 740 nm,(8, 12) on the edges of the visible range.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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