Copper has an electron construction of $\ce 3d^10 4s^1$. Currently sometimes the noble state is composed as $\ce 3d^10 4s^1$ or as $\ce 4s^2 3d^9$.

You are watching: Give the ground-state electron configuration for copper (cu) using noble-gas shorthand.

Now the very first noble state seems to be the very same as his normal configuration and also the last seems to have equal electrons yet divided in an additional way.

Now copper has two oxidation number $+1$ and also $+2$ and also as copper is d-element the the $4$ duration it follows the $18$-electrons rule. However probably the $18$ electrons aren"t developed to acquire the krypton noble state. So exactly how does the noble state construction of copper work?

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edited january 17 "17 at 9:11

DHMO
inquiry Dec 12 "16 in ~ 11:16

Marijn Marijn
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$\ce 4s^1 3d^10$ is the correct construction for copper. Why?

In most basic terms, some elements do no follow the Aufbau principle, so there room some alternate ways that electrons can arrange themselves that give these elements far better stability.

Copper is a change element and transition elements prefer either half filled orbitals or fully-filled orbitals. This renders them much more stable.

For example:

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

$\ceCr = 4s^2 3d^4$$\ceCu = 4s^2 3d^9 However, the actual electron configurations are: \ceCr = 4s^1 3d^5$$\ceCu = 4s^1 3d^10$
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edited jan 17 "17 in ~ 9:08

DHMO
answer Dec 17 "16 in ~ 3:35

AniketAniket
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