# Dysprosium electron configuration

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 5.14.

The rare-earth element dysprosium has atomic number 66, and its ground state is listed as ${^{5}I_{8}}$. This means that ${S=2}$, ${L=6}$ (since after ${F=3}$, the labels go in alphabetical order, so ${G=4}$, ${H=5}$ and ${I=6}$), and ${J=8}$. This isn’t enough information on its own to determine the electron configuration, since the outer shells don’t fill in strict order of ${n}$ and ${L}$ due to shielding effects. For these shells, the ${s}$ shell of level ${n+1}$ is filled before the ${p}$ shell of level ${n}$, and the ${d}$ shell of ${n+1}$ before the ${p}$ of ${n}$ and so on. Given the maximum populations of the various shells (${s}$ has a maximum of 2, ${p}$ of 6, ${d}$ of 10 and ${f}$ of 14), a possible configuration for dysprosium is

$\displaystyle \left(1s\right)^{2}\left(2s\right)^{2}\left(2p\right)^{6}\left(3s\right)^{2}\left(3p\right)^{6}\left(4s\right)^{2}\left(3d\right)^{10}\left(4p\right)^{6}\left(5s\right)^{2}\left(4d\right)^{10}\left(5p\right)^{6}\left(6s\right)^{2}\left(4f\right)^{10}$

We can check this against the given values. The last shell (${4f}$) contains 10 out of a possible 14 electrons. According to Hund’s first rule, these should be arranged to give the maximum possible spin, which would mean 3 pairs and 4 unpaired electrons with parallel spin. This gives ${S=4\times\frac{1}{2}=2}$ which matches the listing above. The value of ${L}$ is difficult to check, since it depends on symmetry requirements which would be difficult (though possible, if you’re persistent) to calculate for the 4 unpaired electrons. However, the 4 unpaired electrons in the ${4f}$ shell have a maximum possible ${L}$ of ${L=12}$, so ${L=6}$ is certainly possible. Having ${L}$ and ${S}$, we can apply Hund’s third rule which in this case says that ${J=L+S}$ since the last shell is more than half full.