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

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SUMMARY

Copper(II) is the more common oxidation state than Copper(I) due to its electronic configurations. The correct configurations are Cu+: 3d10 and Cu2+: 3d9. The stability of Cu2+ arises from its ability to form stronger interactions due to the presence of unpaired electrons, despite the initial assumption that Cu2+ should be readily reduced to Cu+. The presence of unpaired electrons in Cu2+ contributes to its prevalence in various chemical environments.

PREREQUISITES
  • Understanding of electronic configurations in transition metals
  • Knowledge of oxidation states and their stability
  • Familiarity with concepts of electron pairing and unpaired electrons
  • Basic principles of chemical bonding and interactions
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  • Research the role of unpaired electrons in transition metal chemistry
  • Study the stability of different oxidation states in transition metals
  • Explore the implications of electronic configurations on chemical reactivity
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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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