How does one determine the oxidation state of an organometallic complex?
There are a few ways to approach this problem (and, of course, every chemist thinks the way they use is the most correct), but let’s look at what is hopefully the simplest way to think about this. We’re going to only talk about complexes with metal-carbon bonds and no other types of ligands. If you’re curious enough to want to know how nitroxide binds a transition metal, chances are you have an inorganic chemistry textbook at home anyway.
Okay, let’s look at tetra(methyl) zirconium, which has no net charge. In this counting technique, all of the electrons in the metal-carbon bonds are placed on the carbon atoms. So we get four methyl anions and one zirconium ion. Since this complex has no net charge, the zirconium center must balance out the net four negative charges from the methyl groups. Zirconium must be in the +4 oxidation state. Easy, right?
One more example: K3[Fe(CN)6] or potassium hexacyanoferrate. We again move the electrons in the metal-carbon bonds to the carbon groups, making six cyanide ions (CN−), an iron ion, and three potassium ions. We have six negative charges (6 CN−) and three positive charges (3 K+ ions), so to balance out the charges, the iron center must be in the +3 oxidation state.
It’s important to remember that when we count electrons in this way, we’re doing only that—counting. Don’t get the idea that all of these metal-carbon bonds are the same, or that they’re all ionic bonds—this is just for electron bookkeeping!