According to this Wikipedia page, Mendeleev originally believed that the inert gases belonged in Group 0 (to the left of the alkali metals). Thus, helium would be placed in the second period to the left of lithium and argon would be placed in the third period to the left of sodium. Apparently this made sense to him at the time, since atomic theory was in its infancy (and Mendeleev didn't like it) and quantum mechanics was only a twinkle in the eyes of Planck, Bohr, and others.

Based on this arrangement, there is an open spot above helium and to the left of hydrogen for an ultralight inert gas he called "proto-helium". We now know that hydrogen must be the lightest known element, meaning Mendeleev was wrong about the placement of the inert gases. However, host of the other holes in Mendeleev's table were filled with real elements having properties nearly identical to what Mendeleev predicted.

Mendeleev predicted that proto-helium would have a miniscule mass (ca. $ 5.3\times10^{-11}$ amu) and high (near relativistic) velocity. Mendeleev believed that this near massless particle would be able to permeate all matter, rarely interacting chemically. He thought that they might be responsible for radioactive decay.

A neutrino is massless (or nearly so), and travels at the speed of light (if massless) or just under (if it has mass). Otherwise, it ignores the electromagnetic force, and so has no chemical interaction. Neutrinos are often produced during radioactive decay.

The relevant references point back to Mendeleev's original writings (in Russian), which I cannot access, so I cannot verify if he made a prediction about "proto-helium". Did he predict this particle? Has anyone previously made the connection between Mendeleev's proto-helium and neutrinos?

  • $\begingroup$ The free neutron is the element with atomic number 0. It's unstable, of course, but it is still reasonably denoted at $^1_0\mathrm{n}$. It also doesn't really engage in chemistry on account of the lack of electrons, but it captures quite handily on—for instance—Hydrogen-1 at room temperature. $\endgroup$ May 21, 2013 at 3:47
  • $\begingroup$ Conceptually, I agree with you, but a neutron is far from massless. It's mass (1.008664 u) is almost identical to hydrogen (1.007825 u). Free neutrons are also not stable. They undergo beta decay with a half-life of 881.5 seconds. $\endgroup$
    – Ben Norris
    May 22, 2013 at 18:30

2 Answers 2


While it's an interesting thought, I don't think it's very useful to suggest Mendeleev predicted neutrinos. It's far more sensible for it to be a curious coincidence. His suggestions of ghost-like properties similar to neutrinos (very hard to detect, tiny mass, little interaction with matter, etc) are more easily explained as an attempt to justify (properly) why it was so hard to detect: if it had a much larger mass, someone should've weighed it by then; if it interacted more, someone should've managed to capture it physically or chemically, and so on. It's not uncommon for theoretical predictions to temporarily "take refuge" just beyond the limit of observability (for example, proton decay), with varying degrees of justification. In this case, Mendeleev's prediction was wrong in its original terms, and happened to almost match something else entirely because neutrinos really are hard to detect! Also, the connection with radioactivity could very likely have been made merely due to good timing with its discovery. Since no one had a good explanation, why not forward your own!

And as a sidenote, neutrinos definitely have mass, otherwise they could not suffer the observed flavour oscillation.

  • $\begingroup$ The presumption that flavor oscillation implies mass is not obviously true (although likely true in the context of the standard model). $\endgroup$
    – Lighthart
    Apr 19, 2013 at 5:32
  • $\begingroup$ @Lighthart how do you argue that? I haven't heard of any theoretical framework, especially not one neutrino physicists take seriously, that produces flavor oscillation from massless neutrinos. $\endgroup$
    – David Z
    May 18, 2013 at 7:52
  • $\begingroup$ I would love to see any direct physical evidence that refutes my claim. Otherwise, 'not one neutrino physicists take seriously' is just politics, not science. $\endgroup$
    – Lighthart
    May 18, 2013 at 17:47
  • 3
    $\begingroup$ @Lighthart Massless particles have velocity $c$ in every frame of reference and experience no proper time in their travels. Oscillations are the result of a time evolution operator. It follows that oscillating particles are non-luminal which implies mass. QED. Only special relativity and QM required. $\endgroup$ May 19, 2013 at 2:58
  • $\begingroup$ And the experimental evidence? $\endgroup$
    – Lighthart
    May 19, 2013 at 4:01

Neutrinos are chargeless, which is a concept that does not fit nicely in Mendeleev's periodic table; it seems improbable he would predict this particle. The closest bet would be positronium.

  • 1
    $\begingroup$ Positronium could not be Mendeleev's "proto-helium", since it would definitely interact chemically. It does meet the apparent mass requirement. $\endgroup$
    – Ben Norris
    Apr 18, 2013 at 1:02
  • $\begingroup$ Its not so obvious it would interact. It has a full 1s shell. $\endgroup$
    – Lighthart
    Apr 18, 2013 at 1:04
  • 1
    $\begingroup$ Good point. Though it is my understanding that the Pauli principle does not apply for electron-positron interactions, I don't know why! Atkins' Molecular Quantum Mechanics suggests that the need for antisymmetry with respect to the interchange of identical fermions can be rationalized using relativistic arguments, so maybe a relativistic formulation of quantum mechanics is required to explain this? $\endgroup$ Apr 30, 2013 at 12:40
  • 1
    $\begingroup$ Well I think so... You also might be right about the Pauli principle being necessary here... It just not so obvious to me. I don't exactly have a bottle of positronium to experiment with ;)... which means my speculation epsitomologically violates the firest law of quantum mechanics (integral of wavefunction times its conjugate normalized is non-zero) $\endgroup$
    – Lighthart
    Apr 30, 2013 at 14:36
  • 1
    $\begingroup$ Note that positronium exists in both singlet and triplet states. That is to say that the positron and the electron are decidedly not identical fermions, which is no surprise because the have different intrinsic quantum numbers. The deciding factor is the symmetry of the combined wavefunction under particle exchange (it must be anti-symmetric). $\endgroup$ May 21, 2013 at 3:45

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