What is the noble gas configuration of copper? 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?
$\ce{[Ar] 4s^1 3d^10}$  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:


$\ce{Cr = [Ar] 4s^1 3d^5}$
$\ce{Cu = [Ar] 4s^1 3d^10}$