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

1 Answer
Aug 30, 2017

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

Graphed 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.