posted by Sara on .
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.
Oxalic acid is H2C2O4 so oxalate is the C2O4^2- part. So that makes Co 2+. right?
(Co usually is +2 or +3)
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?
For high spin, yes.