by Brian Dodson

Created on: November 20, 2009   Last Updated: November 22, 2009

Helium is a monatomic inert gas that is the second least reactive element in nature. (Neon is still less reactive.) It is also one of the least complex elements, having only 2 protons in its nucleus. Helium's 2 valence electrons reside in a closed orbital, which explains the extremely low chemical activity.

The chemistry of helium is extremely limited. There are no known stable compounds of helium, although the theoretical compounds HHeF, CsFHeO, and N(CH3)4FHeO have been predicted to exist. An endohedral fullerene complex, HeC60, has been formed by diffusing helium into the hollow interior of the fullerene. Although stable, this is not a true chemical compound.

Helium has 9 known isotopes, ranging from helium-2, which has only the 2 protons, to helium-10, consisting of 2 protons and 8 neutrons. Only helium-3 and helium-4 are stable, with 99.996% of the helium found on Earth (mostly found along with natural gas) consisting of helium-4.

Helium-5 is an exotic helium isotope having 2 protons and 3 neutrons. It appears as an intermediate product in the process of deuterium-tritium fusion, and decays within 7x10^-22 seconds into helium-4, a free neutron, and about 17 million electron volts of kinetic energy. Helium-5 is difficult to characterize owing to the extremely short lifetime, but it is probably a halo nucleus, in which the extra neutron is only weakly bound to the remainder of the nucleons. This is suggested by the fact that the mass of the helium-5 nucleus is almost exactly that of its decay products, a helium-4 nucleus and a neutron.

The most important fact for the discussion of its chemistry is the half-life of the nucleus. This may seem an odd statement, since chemistry takes place through exchange of the valence electrons. The details of the nuclear structure have very little influence on chemical reactions.

Let's take a look at making helium-5. It results from the fusion of a deuterium nucleus and a tritium nucleus. The reaction can be written as:

1H2 + 1H3 + 10 keV 2He5 + 17 MeV

The 10 kiloelectronvolts of energy on the left side of the reaction is the energy which is required to force the deuterium and tritium nuclei close enough that they can fuse together. As this is well above the ionization threshold for deuterium and tritium, both will have lost their electrons by the time they fuse into the helium-5 nucleus.

A bare nucleus does not have any chemistry. It would seem, however, that under conditions where electrons are being stripped off atoms, there must be excess electrons floating around. The helium-5 nucleus has a double positive charge, so it will attract free electrons in an attempt to capture them.

The problem is that it doesn't have enough time. The helium-5 nucleus splits apart after 7x10^-22 sec. In that length of time, light will travel only .002 angstroms - anything else will only travel shorter distances. A neutral helium atom is about 0.6 angstroms in diameter, or about 300 times larger. There is nowhere near enough time for electrons to be captured before the helium-5 nucleus splits into helium-4 and a free neutron. In the absence of valence electrons, helium-5 does not undergo any chemical reactions. Despite its intrinsic interest, and again owing to the extremely short half-life, it seems extraordinarily unlikely that it will have any practical applications.

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