# Why does copper assume a 3d^10 4s^1 configuration if both "Cu"^(+) and "Cu"^(2+) are possible?

Aug 30, 2017

Well, the $3 d$ orbitals drop in energy as ${Z}_{e f f}$ increases across the 1st-row transition metals:

By the time we get to copper ($Z = 29$), the $3 d$ is apparently low enough in energy that it is more energetically favorable to pair a $3 d$ electron than a $4 s$ electron. So, $3 {d}^{10} 4 {s}^{1}$ is more stable for $\text{Cu}$ than $3 {d}^{9} 4 {s}^{2}$...

As for why ${\text{Cu}}^{2 +}$ is more prevalent than ${\text{Cu}}^{+}$, that's a mystery to me. Apparently, ${\text{Cu}}^{+}$ is less stable than ${\text{Cu}}^{2 +}$ in aqueous solution.

$2 {\text{Cu"^(+)(aq) -> "Cu"(s) + "Cu}}^{2 +} \left(a q\right)$

And it turns out that during the hydration of ${\text{Cu}}^{+}$, forming ${\text{Cu}}^{2 +}$ by releasing one more electron to reduce another ${\text{Cu}}^{+}$ allows the release of excess hydration energy.