Is bonding specifically for and between electrons? Why cant two atoms share muons, which are different particles of same charge, spin and different mass?

Why there aren't muon-electron bonds?

Why is the octet (or eighteen valence electron) rule only for electrons, and not for all particles with similar charge and spin compared to the electron?

  • $\begingroup$ I wrote 32 valence electron rule for f-block elements but then I remembered f orbitals dont participate in bonding.:) $\endgroup$ Dec 16, 2019 at 0:22
  • $\begingroup$ for lanthanides at least . $\endgroup$ Dec 16, 2019 at 0:22
  • 2
    $\begingroup$ Remember that the Pauli exclusion principle applies to identical fermions. A muon is not identical to an electron, so if you introduce a muon into an atom it will decay into the lowest possible orbital. This orbital is 1s like of course, so in principle you could make a molecule out of protons and muons; however muons decay very quickly and are captured by the nucleus on a similar time scale. $\endgroup$
    – PJ R
    Dec 16, 2019 at 1:29
  • 2
    $\begingroup$ Also, wouldn't the fact that muons "orbit" much closer to the nucleus mean that it would be very hard to get two nuclei close enough together to have enough interaction of both nuclei with the muon that there is a net benefit? I think the cost of bringing the nuclei together would likely be greater than the benefit of bonding. $\endgroup$
    – Andrew
    Dec 16, 2019 at 13:23
  • 1
    $\begingroup$ I've tried adjusting the question title and content, let me know if it's not what you meant. $\endgroup$ Dec 16, 2019 at 23:48

1 Answer 1


You are correct; electrons and muons are fermions with different quantum numbers (specifically, they differ in the electron number and the muon number), so the Pauli exclusion principle does not apply between them (though it of course applies among electrons and muons separately). A somewhat similar case happens with protons and neutrons (also fermions) in the nuclear shell model, which attempts to describe nuclei as containing proton and neutron shells, analogous to electron shells. The proton and neutron shells are filled independently.

Because (negative) muons are the second generation Standard Model equivalent of the electron, whatever electrons do, muons can copy - since there are "electronic" orbitals, there are also "muonic" orbitals. However, there are two main differences.

First, for electron orbitals, it's a good approximation to assume the nucleus is stationary with respect to the electrons (the Born-Oppenheimer approximation), due to the great difference in their masses (the lightest nucleus, a proton, has approximately 1836 times the mass of an electron). Because the muon is approximately 207 times heavier than an electron, now a muon has an appreciable mass relative to a proton (approximately one-ninth), and therefore the BO approximation is considerably worse. This is a scenario in-between a regular atom and positronium, where an electron "orbits" a positron "nucleus", which has the same mass. See Phys. Rep. 1982, 86 (4), 169-216 for a quantitative analysis of BO approximation errors in a simple muonic molecule.

The second and more striking difference is that, again due to the muons being 207 times heavier, muonic orbitals are accordingly around 207 times smaller meaning a muonic orbital has a typical radius of around 0.5-1 pm compared to 100-200 pm for an electronic orbital. The energies are also 207 times larger in magnitude - the 1s electronic orbital in hydrogen has an energy of -13.6 eV, whereas for "muonic hydrogen", the 1s muonic orbital has an energy of -2815 eV. These facts can be determined simply by solving the Schrödinger equation, except inputting a mass 207 times greater for the negatively-charged particle.

As you can see, this leads to a severe mismatch between the realms of electronics and muonics. There is no kind of shared electron-muon bond - at best, there would be a separate electron bond and a muon bond. However, because the energetics involved in the muon bond are so much higher, the system in the ground state is essentially equivalent to just having the muon bond, plus a small correction due to a grossly warped electron bond. The muon bond is formed normally, and the electron bond has to deal with whatever geometry is forced by the muons, however crazy.

As an example, imagine a neutral atom of regular hydrogen and a neutral atom of muonic hydrogen interacting. The muonic hydrogen atom basically pierces into the depth of the electron cloud of the regular hydrogen atom (the electron can hardly repel the muon efficiently, since it is so tightly bound to its nucleus), until both nuclei get quite close. Then the muon latches onto the other proton. The system is stabilised when the two hydrogen nuclei are approximately 0.5 pm apart, and the lone muon forms half of a muonic sigma bond.

From the "point of view" of the muon, the system looks like a singly-ionised muonic dihydrogen molecule ($\ce{\mu-H_2^+}$), with slight corrections due to the electron buzzing around, most of the time far away.

However, from the "point of view" of the electron, it sees a bizarrely elongated nucleus (from the typical ~1 fm sphere to a ~500 fm spindle) with a total charge of +1e (since the muon almost perfectly screens out a full positive charge). This spindly nucleus is still quite small relative to the electron cloud, and so the electron likely behaves similarly to a normal isolated hydrogen atom, with some corrections due to the non-spherical distribution of charge at its "nucleus" (the two separate protons bound by a muon). The electron still provides a slight amount of bonding between the protons, but it is much less than normal due to the odd geometry forced by the muon.

Muonic chemistry in its full glory would be a fascinating (and extremely dangerous!) copy of the electronic chemistry we know, but the two would operate almost completely independently. The muon is actually one of the most stable subatomic particles, with a lifetime of 2.2 µs. That sounds like almost nothing, but it's many orders of magnitude more than what is necessary to observe "chemistry". Unfortunately, it's just too difficult to produce for how ephemeral it is...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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