Oxalic acid is H2C2O4 so oxalate is the C2O4^2- part. So that makes Co 2+. right?
(Co usually is +2 or +3)
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.
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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?
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?
For high spin, yes.
The structure is octahedral because [ox] is a bidenatate molecule. One way to tell this is by looking at the overall charge of the complex ion. since it is -4 and we know that cobalt can either be +2 or +3, the only two ways to get an over charge of -4 is if the ox is either -6 or -7. Because we have 3 [ox] we know that the charges on them are identical and since your almost never hear about fractioned charges, you can safely assume that the ox has a charge of -2 with a total of -6. Keeping this in mind, let us go back to the Co. Since we have verified that the cobalt has a charge of -2, figure out the d-electron configuration on the periodic table to determine how many electrons are in the d-orbital then fill in the electrons according to the rules of high spin. You should remain with 3 unpaired electrons.
i think the answer is 3.
1 answer