I can't find the answer to this question on this SE website, and I apologize for the repetition if it has been answered before. It is my understanding that compound formation has only been observed for the noble gases argon, xenon, and krypton, while no compounds have ever been observed for neon and helium, and that this might have something to do with both the decreased ionization potential of larger noble gas atoms and the high value of electronegativity of the halogen (usually fluorine) that bonds with the noble gas atoms to form compounds. So, my question is why aren't compounds formed from argon, xenon, and krypton bonding with chlorine, bromine, and iodine as prevalent as those formed from those noble gases and fluorine? Could the smaller size of the fluorine atom be a reason for its apparently higher reactivity? -John

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    $\begingroup$ 1) Molecular ions, containing Ne, He, Ar, were observed. 2) The question is more about relative stability. Many molecules, that are stable in vacuum, can't be isolated as pure compounds. 3) Noble gases have very low-energy, small outer orbitals, so they need similar partner to interact. Fluorine and to some extent oxygen have lowest-energy valence orbitals, so they give relatively stable compounds with noble gases. $\endgroup$
    – permeakra
    Jul 22, 2014 at 21:01
  • $\begingroup$ The bond between fluorines in F2 is weak while bonds between fluorine and other atoms tend to be strong. $\endgroup$ Jul 23, 2014 at 1:20

1 Answer 1


You know that the bigger the noble gas atom, the easier it is to ionize it. By the same token it is also easier for a Xe atom than for a He atom to share some electron density in a covalent bond. That is, Xe is more reactive (a very little bit) than He (not at all).

For halogens, the tendency is in some sense the opposite: The smaller the halogen atom the more eager it is to gain electron density -- that is, the more electronegative it is. Fluorine is so electronegative that it can persuade Xe, Kr and even Ar (with difficulty) to give up some electron density to form a covalent bond. Cl is not quite so electronegative, and Br and I even less. Basically you need the least "noble" of the noble gases and the greediest halogen of all to even have a chance at a bond.

One can justify this handwaving with thermodynamic calculations and molecular orbital arguments (though the latter are often handwavy too).

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    $\begingroup$ >The smaller the halogen atom the more eager it is to gain electron density || The funny thing is that electron affinity of fluorine, chlorine and bromine is 328, 349 and 325 kJ/mol with highest for chlorine. So, assuming your logic is correct, noble gases' chlorides should be more widespread, than fluorides. $\endgroup$
    – permeakra
    Jul 23, 2014 at 2:12
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    $\begingroup$ But electron affinity isn't only determinant of electronegativity, as you know. In the present case, we must look at the energy change between (say) $\ce{X_2{+}Xe}$ and $\ce{XeX_2}$. The lower dissociation energy of $\ce{F_2}$ compared to $\ce{Cl_2}$ (alluded to in brinb's comment) is enough to insure the combination is thermodynamically possible for F but not for Cl. $\endgroup$ Jul 23, 2014 at 4:08
  • $\begingroup$ Yeah, cuz orbital energies are so very well defined. Go to cccbdb.nist.gov/gap2.asp and type H2O -- what is the HOMO-LUMO gap for water? $\endgroup$ Jul 23, 2014 at 7:54
  • $\begingroup$ ACtually, yes, their energies may be well defined in terms of peaks of spectra of electrons, knocked out by, say, x-rays. $\endgroup$
    – permeakra
    Jul 23, 2014 at 8:27
  • $\begingroup$ @permeakra Perhaps you can flesh out your comment into an answer, using actual values of the energies. Table 3-3 in Harry Gray's "Chemical Bonds" gives the valence-orbital ionization energies as 115,000 cm${}^{-1}$ for Kr (4p), 111 for Cl (3p) and 151 for F (2p). Are those the relevant numbers for your argument? $\endgroup$ Jul 23, 2014 at 8:49

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