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Copper has an electron structure of $\ce{[Ar] 3d^10 4s^1}$. In salts it may form two stable ions, the $\ce{Cu^+}$ and $\ce{Cu^2+}$. (It's actually not very clear to me why the 2+ ion is common, why does it readily lose one of its d orbital electrons, but I'm willing to wave my hands and say "something quantum").

In pure copper metal, in the usual high school picture of metallic bonding, the conduction band electrons delocalise leaving a structure with ions "floating in a sea of delocalised elections". In copper, how many of its electrons delocalise? Does it make sense to describe copper metal as $\ce{Cu^+}$ or $\ce{Cu^2+}$ ions surrounded by an electron ocean?

metal ions surrounded by delocalised electrons

Image of metallic bonding from BBC bitesize

The linked page says that in group 2 metals, both valence electrons are delocalised, but I've also read that in copper, the whole $\ce{3d^10 4s^1}$ electrons are all valence electrons.

Proscionexium
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James K
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  • https://chemistry.stackexchange.com/questions/24873/why-do-copper-ii-complexes-contain-so-many-valence-electrons – Mithoron Jun 22 '23 at 13:29
  • Use the word "delocalized" carefully, this sea has a bonding contribution, $11$ $e^-$ will not have a collective motion when $\Delta V$ is applied, however only the states near to the Fermi surface will have such a behavior, from the density of states about $n_e = 1.4 $ in agreement with the result of the linked answer. – M06-2x Jun 22 '23 at 19:35
  • Can you exapand on that comment, if necessary in an answer to the duplicate. – James K Jun 22 '23 at 21:20
  • I guess he means there's plenty of covalent component to the bonding - electrons that don't delocalize. – Mithoron Jun 23 '23 at 00:15

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