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I understand that copper has the electronic configuration $\ce{[Ar] 4s^1 3d^10}$ because a fully filled $\ce{4d}$ orbital provide a more stable configuration.

But I don't understand why the two copper ions have the following electronic configurations: $\ce{4s^0 3d^10}$ and $\ce{4s^0 3d^9}$.

Specifically, why is the electron from the $\ce{4s}$ subshell is removed before the electron from the $\ce{3d}$ subshell?

It does have a higher value of $n$ placing it farther than the latter, but it also has lower energy. And how does the extra stability provided due to the fully filled orbital accounted for?

My knowledge of chemistry is very basic at this point, so I would appreciate if any mistakes in the logic of my question or otherwise were brought to my attention.

sushant_padha
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  • While adding electrons, see the energy level (generally, and add some $d^5$ and $d^{10}$ considerations too). While removing electrons, the OUTERMOST electron gets removed first. – insipidintegrator Sep 09 '22 at 17:21
  • @insipidintegrator thank you for your comment. This might be a dumb question but does "outermost" simply mean the highest subshell of the valence shell? Ex: 3p in Cl, and 4s in Na? – sushant_padha Sep 09 '22 at 17:25
  • Yes. That’s it. – insipidintegrator Sep 09 '22 at 17:26
  • highest $n+l$ value – Safdar Faisal Sep 09 '22 at 17:33
  • Don't waste your time in hair splitting of electron configurations. Most chemists remember electron configurations like a parrot and when you ask the details-nobody knows! I have yet to encounter one who knows the experimental details of electron configuration determination. – AChem Sep 09 '22 at 23:54

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The electrons will be removed first from the 4s orbital and not the 3d orbital. It is because the 3d orbital is less in energy than the 4s orbital. The 3d electrons screen the outer most electrons, so electrons can be easily removed.

ClearlyConfused
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