What is the ground state electron configuration of chromium?

Is it $\ce{[Ar]}4s^23d^4$ or

Is it $\ce{[Ar]}4s^13d^5$

  • 3
    $\begingroup$ At worst, your chances are 50:50. If you remember the rules concerning half-filled and filled d orbitals, they are much better ;) $\endgroup$ Commented Mar 12, 2014 at 16:52
  • $\begingroup$ @Philipp my textbook says the answer is [Ar]4s2 3d4 But when i googled it, the answer was opposite. So i am confused.. $\endgroup$
    – Pradeep
    Commented Mar 12, 2014 at 17:06
  • 6
    $\begingroup$ @Googleuser Hmm, sorry then. What textbook are you using? Usually it is a rather well known thing that chromium has $\ce{[Ar] 4s^1 3d^5}$. Maybe you should consider using a different textbook. $\endgroup$
    – Philipp
    Commented Mar 12, 2014 at 17:12
  • $\begingroup$ See chemistry.stackexchange.com/questions/151/… $\endgroup$
    – Ben Norris
    Commented Jun 30, 2015 at 16:19

2 Answers 2


WebElements page about chromium (and a number of resources) agree with the comment by @Philipp:

The ground state electron configuration of ground state gaseous neutral chromium is $\ce{[Ar]}3d^54s^1$

Which in some resources is written as $\ce{[Ar]}4s^13d^5$

enter image description here

Based on the Royal Society of Chemistry article The trouble with the aufbau principle:

it appears that the most stable configuration for atoms of chromium, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum and gold involves only moving one electron into an $s$ orbital.

Chromium is one of a handful of Transition elements that share this electron configuration.


Chromium and copper are examples of elements with "anomalous" electron configurations, meaning that they don't follow the normal rules we use for populating the configurations of other elements.

The commonly given reason for this is that the energy of a shell is minimized when the number of electrons with the same spin is maximized (Hund's rule). As a result, when the energy levels of two successively filled sub-shells are already close together (as they are with 4s and 3d sub-shells), the slightly favored half-filled configuration can "win" over the energy increase needed to move an electron to an even-more-slightly higher energy level. In the case of chromium, this means that one of the 4s electrons will go to the 3d orbital, resulting in two half-filled sub-shells where all electrons within each sub-shell have the same spin.

In the case of copper, a similar thing happens. The difference is that the 4s electron moves into an almost-filled 3d shell in order to completely fill it. You get a slight energy decrease when all electrons are paired within a sub-shell. This, in combination with the decrease obtained from achieving a half-filled s orbital, ends up being enough to overcome the increase in energy required to move that electron to the 3d orbital in the first place.

It would be nice if these empirical rules were consistent across the entire table, but unfortunately they are not. If you look up the actual electron configuration for other d- and f-block elements, you will see there are some patterns, and similar things happen for other elements, but because they are so dependent on the delicate balance between energy levels, it is not possible to reliably predict them with simple rules for all elements. In "real life" we use spectroscopy and quantum mechanical calculations to find the actual electron configurations.

However, since chromium and copper are common enough and reliably predictable with simple rules, we tend to use those as classroom examples to demonstrate that the reality of electron configurations is more complex than the simple rules we give you in school.


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