So I understand than the lanthanide contraction is due to poor shielding of the 4f electrons which decreases the radius. However, if Im not mistaken the relativistic effects lead to a contraction of the core orbitals (s,p) while the f and d orbitals expand. So what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right? They usually say relativistic effects are one of the reasons for the lanthanide contraction Thanks


The lanthanide contraction is caused by two effects

  • The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals
  • relativistic effects

This Wikipedia article presents a nice discussion of the lanthanide contraction. The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.

what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right?

That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see this SE Chem post for a more detailed description). In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius. A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.

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  • $\begingroup$ Both are relevant, but IMHO, the poor screening is an important part. The effective nuclear charge (Zeff) increases across the lanthanides, but the added 4f electrons don't do a good job screening that larger nuclear charge. Thus, the atomic radius contracts. A similar trend is seen across other rows. $\endgroup$ – Geoff Hutchison Nov 24 '14 at 15:48
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    $\begingroup$ @GeoffHutchison Yes, like I said the relativistic effect is estimated to account for only 10% of the contraction, the poor screening by the 4f electrons being the major factor. $\endgroup$ – ron Nov 24 '14 at 15:52

Only orbitals with l=0 (s orbitals) have radial components that are non-zero at the nucleus:

enter image description here

s-shells, including and inparticular the 6s shell, are much more sensitive to the relativistic effects (compared to l>0 shells) due to s-orbitals allowing electronc density AT the nucleus. If s-shell electron density is non-zero at the nucleus, it's reasonable to suspect that s-shell electron density is contracted more by relativistic effects.

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