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

1 Answer
Truong-Son N.
Aug 30, 2017

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

Graphed from Appendix B.9Graphed from Appendix B.9

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

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

2"Cu"^(+)(aq) -> "Cu"(s) + "Cu"^(2+)(aq)

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

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