Question

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.

Answers

DrBob222
Oxalic acid is H2C2O4 so oxalate is the C2O4^2- part. So that makes Co 2+. right?
(Co usually is +2 or +3)
Sara
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?
DrBob222
For high spin, yes.
Nani
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.
midoriya izuku
i think the answer is 3.

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