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Homework Help: Chemistry Electronic Configuration

  1. Dec 29, 2006 #1
    1. The problem statement, all variables and given/known data
    Explain why copper(II) is the more common oxidation state than copper (I) by giving the electronic configurations of the ions involved.


    3. 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.
     
  2. jcsd
  3. Dec 29, 2006 #2

    GCT

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  4. Dec 29, 2006 #3
    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?
     
  5. Dec 30, 2006 #4
    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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