6
$\begingroup$

I was looking at this question:

How can f-f transitions happen?

And got some answers that answered some of the questions I had, but not all. I am still trying to conceptually grasp why f-f transitions happen at all because a) there should be no crystal field splitting because of the lanthanide contraction (orbitals are closer to the nucleus, are not perturbed by the ligand sphere as there is next to no interaction) and b) then all f-electrons should be degenerate.

This is in contrast to d-elements, where d-d interactions can be explained by crystal field splitting and, more precisely, by looking at Tanabe-Sugano-diagrams (because not every d-d transition is the same due to the symmetry of the orbitals involved and spin-orbit-coupling).

So how to best explain this?

Edit:

I am asking because I was confused by this diagram:

enter image description here

In this diagram, one excited Eu(III) state of 5D(0) can decay to 7F(6) state. Is this a d-f transition or a f-f transition?

Sorry if this is obvious, but even though the term symbol is a D, this not always coincides with the orbital; for example, Sm(III) has the ground state term symbol 6H(5/2), but obviously the outer electrons don't reside in a "h-orbital".

$\endgroup$

1 Answer 1

6
+100
$\begingroup$

To say that there is no crystal-fiekd splitting is to overstate the fact. There is a small amount of splitting and $f-f$ transitions can be seen, but they are much weaker than $d-d$ transitions in transition-metal complexes.

Both $d-d$ and $f-f$ transitions are Laporte-forbidden if we have pure $f$ orpure $d$ orbitals. In the real case, molecular vibrations and the formation of molecular orbitals between metal and ligands break the symmetry that causes the Laporte-forbidden condition, allowing these transitions to take place. However, the orbital mixing process is less effective for $f$ orbitals than for $d$ orbitals because the former are relatively isolated from the ligands and undergo only weak metal-ligand bonding interactions. Therefore $f-f$ transitions are much weaker than $d-d$ transitions.

In contrast transitions heyween $f$ and $d$ orbitals ($f-d$ transitions) are not Laporte-forbidden because of the different inhetent symmetry of $f$ and $d$ orbitals. These $f-d$ transitions are therefore quite stron. It is $f-d$ transitions, not $f-f$ transitions, that predominate in the UV and visible spectra of lanthanide complexes.

Experimental spectra are given fir some lanthanide(III) complexes by Purohit and Bhojak[1].

Reference

  1. Suresh Purohit and N. Bhojak (2013). "The Absorption Spectra of Some Lanthanide (III) Ions". Research & Reviews: Journal of Chemistry. 2(2), e-ISSN: 2319-9849.
$\endgroup$
1
  • $\begingroup$ Could you maybe take a look at my edit? Seems to incorporate the f-f or d-f transition question $\endgroup$
    – Mäßige
    Jul 3, 2023 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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