Cu has an anomalous electron configuration. $\ce{Cu ~=~ 1s^2~2s^2~2p^6~3s^2~3p^6~4s^1~3d^{10}}$, it does not follow the usual pattern. In this case, the 3d subshell is filled before the 4s, which usually happens in the reverse order ($\ce{1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^{9}})$.

My question is, can you tell by an element's position on the periodic table (Group #, Row #, Block, etc.) that it will have an anomalous electron configuration? Do I have to memorize which elements have this property?

I am noting that Cu is in the 1B Group of Transition metals which implies that it has one valence electron, this is consistent with its configuration. will this be true for all cases?

Also: $\ce{Fe ~=~ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$ and is a transition metal as well (Group 8). This implies it must have 8 valence electrons. But following the definition of valence electrons (The number of electrons in outermost shell), $n = 4$ is the outermost shell and I get a conundrum. Obviously if I add $\ce{4s + 3d}$ electrons everything is good. Why is this?


2 Answers 2


Many related questions have been asked on this topic before. I suggest you read these previous posts:

Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?

How can one explain Niobium’s weird electronic configuration?

The Aufbau Principle is a general rule of thumb to find the electronic configuration of an element but usually many exceptions arise in the d and f block elements due to electron-electron interactions.

The concept of valence electrons does get a little fuzzy when you consider $d$ block elements since the energies of the $\ce{3d}$ and $\ce{4s}$ subshells are quite similar so I think its better to avoid it. $\ce{d}$ block elements can show a staggeringly large number of oxidation states (for example manganese which shows almost all oxidation states from 0 to +7). This can't be explained using the concept of valence electrons and the octet rule.

$\ce{Fe}$ itself can lose 2 electrons to form $\ce{Fe^{2+}}$ or 3 electrons to form $\ce{Fe^{3+}}$ so now again you can see that you can't explain this by saying that $\ce{Fe}$ has 8 valence electrons.

Other exceptional configurations: (there are quite a few): niobium, platinum, gold, cerium, gadolinium, etc.

  • $\begingroup$ Fe can have even more oxidation states than magnesium. All the way from -2 to +6. Example of iron in +6 oxidation state: Potassium ferrate, Example of Iron in -2 oxidation state: Disodium tetracarbonylferrate $\endgroup$
    – Caters
    Commented Oct 21, 2014 at 15:14

First of all the configuration of $\ce{Cu}$ you see has anomalous electron configuration because completely filled or half filled sub shells are more stable than any other configuration and you can see the configuration at last it is $\ce{4s^13d^{10}}$ and the basic configuration is $\ce{4s^23d^{10}}$ so you can see that $\ce{4s}$ is half filled and $\ce{3d}$ is completely filled. It is said this configuration is more stable due to symmetry. And it has no connection to their position in periodic table, and you only have to remember the concept and be careful while writing the configuration to notice this type of conditions.

As much I know they both are to be taken as outer most shell because $\ce{4s}$ is a lower energy cell as compared to $\ce{3d}$ and it filled first. I don't know the details about this, I hope some one will give a better answer to the second part as I also wish to know about that.


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