8
$\begingroup$

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

$\endgroup$
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.