Chemistry Electronic Configuration

[chriswwt]

Homework Statement

Explain why copper(II) is the more common oxidation state than copper (I) by giving the electronic configurations of the ions involved.

The Attempt at a Solution

i've considered about the configurations of Cu+ and Cu2+ but the result seems to contradict the fact.
Cu+:[Ar]3d^10
Cu2+:[ar]3d^9
Base on the configuration, Cu2+ should be readily reduced to Cu+,so Cu+ should be the more common oxidation state in this regard.
Please help correct mistakes I've make, or did i think in a wrong way?
thanks.

 

[GCT]

You should reassess the electronic configuration

http://www.webelements.com/webelements/elements/text/Cu/econ.html

 

[chriswwt]

so now the new configuration is
Cu+:3d^8 4s^2
Cu2+:3d^7 4s^2
cu2+ has 3 unpaired 3d electrons whereas Cu+ has 2 unpaired electrons, in theory the unpaired electrons in Cu2+ will repelled from the inner electrons and thus more unstable?

 

[chriswwt]

well i find that the original configuration i proposed should be correct since the atom tends to retain the extra stability of fully filled 3d subshell...