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Why does a collision between an excited helium molecule ($\ce{He^*}$) and a ground-state helium atom create $\ce{He2^*}$?

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The electronic configurations are $\mathrm{(1s)^2}$ for $\ce{He}$ and $\mathrm{(1s)^1(2s)^1}$ for $\ce{He^*}$.

Put this information in an energy diagram and see what the molecular orbitals look like, especially how they are filled. (Further hint: Filled-unfilled orbital interactions are generally favorable.)

I have attached a graphic:

enter image description here

This leads to a bond order of $\frac{1}{2}(2-1+1) = 1$, and therefore $\ce{He2^*}$ is stable with respect to $\ce{He + He^*}$. The excited dimer is known as an excimer.

As an aside: The relative energy levels of the molecular orbitals are very hard to get right out of the blue, so take them with a grain of salt. As was kindly pointed out by Martin, the excitation of helium will initially form a singlet excited configuration, depicted above. This should be able to relax to a triplet excited state.

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  • $\begingroup$ Just a little nitpick. I think the excited Helium should not be described as a triplet, as excitation usually does not flip the spin, keeping it a singlet. $\endgroup$ – Martin - マーチン Nov 20 '14 at 3:24
  • $\begingroup$ @Martin I mostly see these diagrams in org chem, where they don't care all that much about spin-correctness. I've amended my answer, hopefully to your satisfaction. Thanks for the nitpick :) $\endgroup$ – tschoppi Nov 20 '14 at 9:58
  • $\begingroup$ Lovely, I appreciate the appreciation :D. Most of the times it might not even matter and a triplet Helium is of course also an excited state, probably just not the first. $\endgroup$ – Martin - マーチン Nov 20 '14 at 10:49
  • $\begingroup$ @Martin-マーチン the triplet state would probably be the "first excited state" in that it is the lowest-energy excited state, but yeah, probably not the initially formed excited state ;) Anyway, I will take the liberty to edit the diagram. $\endgroup$ – orthocresol Nov 29 '16 at 10:25
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In any chemical collision process, one of the most important physical parameters to consider is the angular momentum. Throughout a simple chemical reaction, the total angular momentum of the system is preserved. In the case of a simple diatomic reaction, we need to conserve both the spin and the orbital angular momentum.

Consider the ground state of orthoheium He (1s)$^2$. This atom has two spin paired s-electrons. Therefore the total spin is 0 and the total orbital angular momentum is also 0.

Consider now the ground state of parahelium, He(1s)(2s). This atom should have (in the simple case) have two same spin electrons in each of the 1s and 2s orbitals. Therefore the total spin is 1, and the orbital angular momentum is 0.

So, we need to create a final molecular orbital picture with orbital angular momentum 0 (easy: all $\sigma$ orbitals) and a total spin of 1. This can only be achieved if our final molecule also has two unpaired, aligned spins. The lowest energy configuration possible is therefore (1$\sigma_g$)(1$\sigma_g$)(1$\sigma_u^*$)(2$\sigma_g$), which is indeed the excited state helium dimer He$_2^*$.

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  • $\begingroup$ Why should the excited Helium be a triplet? Can you back that up? $\endgroup$ – Martin - マーチン Nov 20 '14 at 3:25
  • $\begingroup$ Ah, what I should have said was that the "excited state" helium atom was actually the ground state of parahelium. The question does, after all, ask how two group state helium atoms can create He* (ie ground state of orthohelium and parahelium). I will update. $\endgroup$ – DrHarps Nov 21 '14 at 8:31
  • $\begingroup$ Orthohelium is the triplet with S=1 and Parahelium is a singlet with S=0. And I still do not understand why the excited Helium has to be a triplet, the most simplest case would be still singlet. $\endgroup$ – Martin - マーチン Nov 21 '14 at 8:55

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