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

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

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Answer 1 · 2017-08-30T19:34:51

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

Graphed from Appendix B.9

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

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

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

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

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

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