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

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Discussion Overview

The discussion revolves around the relative stability and commonality of the oxidation states of copper, specifically comparing Copper(I) and Copper(II). Participants explore the electronic configurations of the ions and their implications for stability, addressing the theoretical aspects of oxidation states.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the electronic configurations of Cu+ and Cu2+ indicate that Cu2+ should be readily reduced to Cu+, implying that Cu+ should be more common.
  • Another participant prompts a reassessment of the electronic configuration of copper ions, providing a link to an external resource.
  • A subsequent reply presents revised configurations, stating that Cu+ has 3d^8 4s^2 and Cu2+ has 3d^7 4s^2, and raises the idea that the unpaired electrons in Cu2+ may lead to greater instability.
  • Another participant defends their original configuration, arguing that the stability of a fully filled 3d subshell supports the correctness of their initial claim.

Areas of Agreement / Disagreement

Participants express differing views on the electronic configurations and stability of the copper oxidation states, with no consensus reached on which configuration is correct or which oxidation state is more common.

Contextual Notes

There are unresolved assumptions regarding the stability of the oxidation states based on electronic configurations, and participants have not fully agreed on the correct configurations or their implications.

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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