Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Interesting behavior of strong correlation between electrons occur in metals with partially filled d or f orbitals (transition metals). Why these strong correlations do not appear with elements with incomplete p or s orbital for example ?

share|cite|improve this question
probably because s and p orbitals are more tightly and symmetrically held to nucleus... – Vineet Menon Apr 18 '12 at 12:23
Can you give some examples of the phenomena you're interested in? At the moment it's not clear what you're asking. – John Rennie Apr 18 '12 at 13:07
@JohnRennie See: – Tarek Apr 18 '12 at 13:53

At a fundamental level, many of the interesting properties of transition metals and transition metal compounds can be understood qualitatively from the simple but elegant Hubbard model.

The basic idea is that electronic structure is a competition between kinetic energy and electron-electron repulsion. For solids that have s and p orbitals only, there is a lot of overlap between the valence orbitals of neighboring atoms. Having a lot of overlap means that the system resembles an electron gas, in which case kinetic energy is in some sense more important than e-e interactions. This is why simple metals such as sodium behave a lot like Fermi gasses.

On the other hand, d and f orbitals are more tightly bound to the nucleus because the electrons in the higher s and p shells do not screen the nucleus very well. In this case there is not so much overlap between the d and f orbitals and the neighboring sites, and so e-e interactions become more important relative to kinetic energy. These interactions result in strong correlations between electrons and interesting behavior such as metal-insulator transitions.

One small aside: you can make correlation important in s and p systems if you stretch them to reduce the overlap between orbitals. For example, the stretched H2+ ion is a pathological case for which many electron structure methods such as DFT will actually have serious problems. See Phys. Rev. B 56, 16021 (1997).

share|cite|improve this answer

It isn't transition metals that display the strongly correlated behaviour, it's materials like transition metal oxides. The link you provided in the comments makes this clear.

The d and f orbitals are not populated in the order you expect. For example the 4s atomic orbital is populated before the 3d, and the 5p and 6s are populated before the 4f. That means the electrons in d and f orbitals tend to be screened by electrons in higher shells e.g. the 3d electron density lies nearer the nucleus than the 4s electron density and is partially screened by it.

This screening means, and there is considerable hand waving here, that in the sort of materials described in the Wikipedia article the d and f orbitals don't give you the free electron bands you get in metals. Instead you find these materials tend to be on the cusp between metals and insulators. It's because the electrons are neither free nor tightly bound that you get the strong electron correlations.

Later: if anyone is still interested, the f orbitals are even more deeply buried behind s and p orbitals than the d orbitals are. That's why the lanthanides are chemically so similar (and hard to separate). Since the f orbitals are almost completely screened they do not affect the chemical properties of the atoms very much. The d orbitals are screened to a lesser extent, so there is more variation in chemistry along a row of the transition metals than along the lanthanides.

share|cite|improve this answer
I think you are using the term screening here in a different sense than the usual one, where electrons closer to the nucleus screen the positive nuclear charge from electrons at a larger distance, and hence those later electrons feels a smaller nuclear charge. – Tarek Apr 20 '12 at 17:05
Have a look at This explains in a bit more detail. – John Rennie Apr 20 '12 at 17:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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