To be explicit, $\text{Cr}^{3+}$ and $\text{Sc}$ have the same number of electrons, the only difference is the nuclear charge.

This is not an isolated anomaly, it seems: the book I'm using claim that with all fourth period transition metal cations, the $4 \text{s}$ electrons are removed first. This indicates that the $4 \text{s}$ electrons are at a higher energy than $3\text{d}$.

My question: why is this the case, given that the $4\text{s}$ electrons are 'added' before the $3\text{d}$ electrons as one moves rightwards across the fourth period, indicating that they are of a lower energy? I cannot reconcile this apparent contradiction.


1 Answer 1


The electron configuration of chromium is : $\ce{[Ar] 4s^1 3d^5 }$ (chromium is one of those examples of special electron configurations : the electron configuration is not $\ce{[Ar] 4s^2 3d^4 }$ as many people might suspect, an electron is moved to the 3d orbital because this configuration is more stable --> that is the reason why chromium has this special electron configuration)

So the $\ce{Cr^3+}$ ion loses an electron from $\ce{4s}$ and loses 2 electrons from $\ce{3d}$ to give it an electron configuration of $\ce{[Ar] 3d^3 }$

Scandium has an electron configuration of $\ce{[Ar] 4s^2 3d^1 }$ and isn't one of those exceptions.

These are the reasons why the chromium(+III) ion doesn't have the same electron configuration as scandium, despite the fact that they have equal amounts of electrons.

As for why the 4s orbital is emptied first:

In all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only seems to apply to building the atom up in the first place. (Using the Aufbau principle) In all other respects, the 4s electrons are always the electrons you need to think about first.

(this link is interesting for you in case you wanted a more detailed explanation about the 4s / 3d problem :http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html )

EDIT : Chromium, just like all other elements, tries to obtain more stable configurations by losing certain amounts of electrons. Now, in the case of $\ce{Cr^3+}$ half/fully filled subshells can't be achieved ( you get $\ce{[Ar] 3d^3}$ ) but that doesn't necessarily mean that it is not stable. But chromium does have other ions in which it does have more stable electron configurations. Take a look at the $\ce{Cr^6+}$ ion : here the ion has an electron configuration that is equal to the one of argon, which does have completely filled valence subshells. ($\ce{[Ne] 3s^2 3p^6}$)

So ions can't always achieve these more stable half/fully filled subshells, but that doesn't necessarily mean that they are not stable.

  • $\begingroup$ I was under the impression that the cause of the stability that gives rise to the electron movingg to the 3d orbital in Chromium was that all the 3d subshells would become half-filled if an electron moved from 4s to 3d. However, this is not the case in $\text{Cr}^{3+}$ so I don't see how this logic applies (I'm assuming my dilemma arises from the argument in my first sentence being flawed). $\endgroup$
    – Meow
    Commented Mar 8, 2014 at 21:14
  • $\begingroup$ I've edited my question - I hope that is the answer that you were looking for :) $\endgroup$
    – user2117
    Commented Mar 9, 2014 at 10:13

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.