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Copper has an electron configuration of $\ce{[Ar] 3d^10 4s^1}$. Now sometimes the noble state is written as $\ce{[Ar] 3d^10 4s^1}$ or as $\ce{[Ar] 4s^2 3d^9}$.

Now the first noble state seems to be the same as his normal configuration and the latter seems to have equal electrons but divided in another way.

Now copper has two oxidation numbers $+1$ and $+2$ and as copper is d-element of the $4$ period it follows the $18$-electrons rule. But probably the $18$ electrons aren't created to get the krypton noble state. So how does the noble state configuration of copper work?

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  • $\begingroup$ "nobel state" is a very unusual notion. Cu looses electrons to be +1 or +2, the electronic state of Krypton would be -7, that makes no sense. Cu +1 has a full 3d and an empty 4a orbital. Check wikipedia von "Hund's rules' and "Aufbau principle". $\endgroup$ – Karl Dec 12 '16 at 13:49
  • $\begingroup$ See "the order of filling of $3\mathrm{d}$ and $4\mathrm{s}$ orbitals" by chemguide for some perspectives. $\endgroup$ – DHMO Jan 17 '17 at 9:10
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$\ce{[Ar] 4s^1 3d^10}$ is the correct configuration for copper. Why?

In simplest terms, some elements do not follow the Aufbau principle, so there are some alternate ways that electrons can arrange themselves that give these elements better stability.

Copper is a transition element and transition elements like either half filled orbitals or fully-filled orbitals. This makes them more stable.

For example:

Using the Aufbau principle, you would write the following electron configurations:

  • $\ce{Cr = [Ar] 4s^2 3d^4}$
  • $\ce{Cu = [Ar] 4s^2 3d^9}$

However, the actual electron configurations are:

  • $\ce{Cr = [Ar] 4s^1 3d^5}$
  • $\ce{Cu = [Ar] 4s^1 3d^10}$
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  • $\begingroup$ Copper still follows the Aufbau principle (electrons go to orbitals with lowest energies), it is just that the energies change when you are filling the electrons. The assumption that the $4\mathrm{s}$ orbital has a lower energy than the $3\mathrm{d}$ orbital is the Madelung rule instead of the Aufbau principle. $\endgroup$ – DHMO Jan 17 '17 at 9:07

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