Why is the 5d orbital instead of the 4f orbital filled in lanthanum?

Question

For orbitals with the same value of $n + l$ (e.g. the $\mathrm{3d}$ and $\mathrm{4p}$ orbitals), we usually expect the orbital with the lower value of $n$ to be filled earlier. Hence, for example, the $\mathrm{3d}$ orbitals are filled before the $\mathrm{4p}$ orbitals in the transition metals.

However, the electronic configuration of lanthanum is $[\ce{Xe}](\mathrm{5d})^1(\mathrm{6s})^2$, and not $[\ce{Xe}](\mathrm{4f})^1(\mathrm{6s})^2$.

I know that the $\mathrm{5d}$ and $\mathrm{4f}$ orbitals are of similar energies, but is there any better explanation for this?

Answer 1

According to Ground-state configurations of ionic species I through XVI for Z = 57-74 and the interpretation of 4d-4f emission resonances in laser-produced plasmas Phys. Rev. A 25, 275 (where "I" means the neutral atom) :

We begin therefore by showing in the left-hand columns of Table I the configurations given by Martin et al. for lanthanum through lutetium. Here it is seen that at the fourth spectrum the 4f level has in all cases fallen below 5d and 6s. This is due to the well-known phenomenon of 4f wavefunction collapse [reference 5]. The effective potential in which the 4f electron moves is made up of an attractive Coulomb term and a repulsive centrifugal term, which, in general, combine to give a double well potential, the height of the intervening barrier depending on the nuclear charge Z. For neutral atoms the field in the region of the outer potential is essentially hydrogenic. The occurrence of the neutral lanthanides coincides with the appearance of bound 4f levels in the inner well, which, in turn, is due to the deepening of the well and the lowering of the potential barrier as a result of the increasing nuclear charge.

Here, "reference 5" is Spectral Distribution of Atomic Oscillator Strengths Rev. Mod. Phys. 40, 441 which goes into quantitative equations for the attractive Coulomb term and a repulsive centrifugal term.