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The complex [Co(ox)3]4- has a high-spin electron configuration. How many unpaired electrons are there in the complex?

I know because it's a high-spin, it's tetrahydral shaped. But I don't know what (ox) is so, I'm stuck in finding the oxidation state of Co which helps me find the number of electrons.

  • Chemistry -

    Oxalic acid is H2C2O4 so oxalate is the C2O4^2- part. So that makes Co 2+. right?
    (Co usually is +2 or +3)

  • Chemistry -

    Okay, cool. So, the oxidation state of Co is 2+.

    Co2+ = [Ar]4s2 3d7

    Metal cations get rid of the s orbital before the d, so can I say then the number of unpaired electrons is 7?

  • Chemistry -

    For high spin, yes.

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