How can an excited helium dimer be formed?

Question

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

Answer 1

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:

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.

Answer 2

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$_2^*$.