# Why does copper have a more common oxidation state of +2, but only one valence electron?

Jun 14, 2016

The $3 d$ electrons can act as valence electrons along with the $4 s$ one. But in most cases only one of the $3 d$ electrons can be donated at a time, so we get a maximum oxidation state of $+ 2$.
It's an energy balance. Donating additional valence electrons to make higher oxidation states costs energy. This has to be made up with the electron affinity of the nonnetal and greater electrostatic attraction to negative ions and dipoles. In the case of copper it usually works for donating one $3 d$ electrons plus the $4 s$ electron for $+ 2$, but getting to $+ 3$ is too much.
There are exceptions. With highly electronegative oxygen and fluorine atoms there may be enough electron affinity and electrostatic attraction to draw off additional $3 d$ valence electrons so the oxidation state can reach $+ 3$ or possibly even $+ 4$ in some fluoride and oxide compounds (https://en.wikipedia.org/wiki/Copper). The ceramic superconductor ${\text{YBa"_2"Cu"_3"O}}_{7}$ (https://en.wikipedia.org/wiki/Yttrium_barium_copper_oxide) has both $\text{Cu (II)}$ and $\text{Cu(III)}$.
On the other hand, low-electronegativity "soft base" nonmetals often have enough electron withdrawing power to bind only the energetically "easy" $4 s$ electron. We often see copper in the $+ 1$ oxidation state when it's bounded to such elements like phosphorous, sulfur, or iodine.