I know what the electron configurations of the lanthanides are, but I was asking myself, why they are so irregular.

The configutation of Lanthanum is $\mathrm{5d^1\ 6s^2}$, but according to the Klechkowski rule, the $\mathrm{4f}$ orbital has a lower energy level and should be filled up before the $\mathrm{5d}$ orbital. Why isn't that happening?

The irregulatity of Gadolinium is more comprehensive. The electron pairing energy is higher than the energy difference of fitting that electron in the $\mathrm{5d}$ shell.

Then we come to Ytterbium which has every shell filled up. So why does it's most stable oxidation number is still $\mathrm{+III}$? I would understand why it can leave 2 electrons (it's outermost shell is the $\mathrm{6s}$-shell) but after that, we have to remove a single electron of the filled $\mathrm{4f}$ shell. Isn't that extremely unfavorable?

I hope that somebody can explain why all this happens, I would really appreciate it.

  • $\begingroup$ related: chemistry.stackexchange.com/questions/151/… $\endgroup$ – getafix Aug 28 '16 at 19:01
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    $\begingroup$ The relative energy of the orbitals is not static. It depends on the charge of the nuclei and electron configuration. Klechkowski rule is just a rationalization and just as most rationalizations it does fail. In general, the less nodes an orbital has, the less is its energy (which is why $n+1 s/p $ orbitals are filled before $n d$ orbital) However, the precise relative energy is in a very delicate balance between 'node' factor, charge factor and repulsion with core electrons. It cannot be fit into universal simple rules, even if ad-hock explanations may be invented. $\endgroup$ – permeakra Aug 28 '16 at 22:44

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