I've learnt in Chemistry that for example in Chromium, the 4s subshell only has one electron in favor of making the 3d subshell half-full. Also, for Copper the same happens.

What other circumstances does this happen under? Do ions obey the same pattern, e.g. Mn+?

  • 1
    $\begingroup$ It's a bit of a complicated matter. Though it can be explained qualitatively for the lighter elements with some finer arguments than merely comparing orbital energy levels via the aufbau principle, things start to get messy when $f$ (and eventually $g$) sublevels and strong relativistic effects are present. If I find the time, I may write up an answer, but for now I shall link this NIST webpage, which has very recent data on ground state electronic configurations of atoms and ions. Just type in an element and play around! Lots to see. $\endgroup$ May 25 '13 at 15:31
  • $\begingroup$ Do note however that the listed data uses a slightly different nomenclature than usual in chemistry for atoms/ions. For example, neutral iron ($\ce{Fe^0}$) is listed as "Fe I", while the $\ce{Fe^3+}$ ion, often called iron(III), is listed as "Fe IV". $\endgroup$ May 25 '13 at 15:35

If we take the definition of "3d subshell takes precedence over the 4s subshell" as being that the configuration has zero 4s electrons, yet at least one 3d electron, then all such cases are ions.


V: 3d^4
Cr: 3d^5
Co: 3d^8
Ni: 3d^9
Cu: 3d^10


All elements from Sc to Zn


All elements from Ti to Ga



Rules for filling up atomic orbitals can be understood by considering the governing principles:

Aufbau principle: electrons fill from lowest energy level to highest (attraction of positive nucleus on electrons: the farther they are from nucleus, higher is their energy or shallower potential well)

Pauli exclusion principle: no two electrons can have all quantum numbers the same; in particular, only two electrons with opposite spins can occupy an orbital. Pairing up of electrons with opposite spin raises its energy. Electrons with parallel spins have a repulsive exchange interaction between themselves.

Hund's rule: electrons are stabilized if they have parallel spins spatially distributed among different orbitals of a subshell. They'll get paired up only after each subshell has one electron, all having parallel spins.

Formation of half filled or fully filled orbitals thus has a stabilizing effect. Energy difference between $n$s and $(n-1)$d is not so high. In case of some special elements, transition of an outer electron from $n$s to $(n-1)$d is energetically favorable if it effects net stabilization due to maximization of electrons with parallel spins in d-shell, thereby over-riding the aufbau principle.

Nb, Mo, Pd, and Ag are some examples of special configurations. These principles should hold true for ions as well if we consider the remaining valence electrons in the system.


I agree with @Nicolau Saker Neto that it's complicated. You might want to look at these two papers, if you're really interested in more of the details:

The links might be paywalled. Sorry. :-/

  • 2
    $\begingroup$ Welcome to Chemistry.SE! Unfortunatlely, we usually don't allow answers that are little more than a link to another resource. Please quote relevant portions of the papers and/or explain them in brief. Thank you :) $\endgroup$ Jul 8 '13 at 21:18
  • 1
    $\begingroup$ So would it have been better as a comment? It was more of a "here's where to learn more" than a full-out explanation. It is, in fact, complicated, and a little post here, even attempting to summarized the papers, would hardly do it justice. $\endgroup$
    – BLHaas
    Jul 17 '13 at 17:25
  • $\begingroup$ No one had "answered" the question, so I was attempting to note that @Nicolau Saker Neto basically had answered it, and provide further resources to back up the explanation. How should I approach this in the future? $\endgroup$
    – BLHaas
    Jul 17 '13 at 17:28
  • 1
    $\begingroup$ Well, in the future try to leave a comment OR add some meat to the answer. Even a partial summary is good. :) Unfortunately, you need 50 reputation to comment, which you don't have. $\endgroup$ Jul 17 '13 at 17:30

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.