# How do you find the effective nuclear charge of lanthanides?

Jun 27, 2017

You do not have to understand this throughout your chemistry curriculum (or at least, I never had to), because it is the result of multiple complicated competing trends. The effective nuclear charge in general is not predictable for the Lanthanides.

Due to scalar relativistic effects, the $6 s$ orbital contracts and the $4 f$ and $5 d$ orbitals (slightly) expand, which we call the "lanthanide contraction". What that says for the effective nuclear charge is that it is not a good approximation to try to use Slater's rules to predict ${Z}_{e f f}$ for the Lanthanides.

Slater's rules incorrectly predict something like the following for ${Z}_{e f f}$, i.e. a monotonous increase:

However, the atomic radii are all over the place:

This is likely due to spin-orbit relativistic effects, where the $4 f$, $5 d$, and $6 s$ orbital energies vary in a certain way as a result of the coupling of the electron spin angular momentum and the orbitals' angular momenta. This alters the shielding extents in certain ways...

Normally, we use ${Z}_{e f f}$ to judge how large the radii might be. Thus, there is no clear trend for ${Z}_{e f f}$ here...