Why is Copper(II) More Common Than Copper(I)?

AI Thread Summary
Copper(II) is more common than Copper(I) due to its electronic configurations, with Cu+ having a configuration of [Ar]3d^10 and Cu2+ as [Ar]3d^9. Initially, it was thought that Cu2+ would easily reduce to Cu+, suggesting Cu+ should be more prevalent. However, the corrected configurations reveal Cu+ as [Ar]3d^8 4s^2 and Cu2+ as [Ar]3d^7 4s^2, indicating that Cu2+ has three unpaired electrons, making it less stable. The stability of fully filled 3d subshells contributes to the prevalence of Cu2+. Ultimately, the electronic structure and stability considerations explain why copper(II) is the more common oxidation state.
chriswwt
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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.
 
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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?
 
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...
 
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