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.

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