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.
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?