[CrimpJiggler]

Heres the Monsanto process:

I can't figure out how to determine the electron count for these rhodium complexes. Starting with the one at the top. Heres how I count its electrons: rhodium is a group 9 element so it has 2s electrons and 7d electrons which means it has 9 out of 18 valence electrons to begin with. The 2 CO ligands add an extra 4 electrons, bringing the total to 13. I'm a bit confused about how the iodide ligands work, do they also donate 2 electrons each? If so, that brings the electron count to 17. The ion has a charge of -1, bringing the total electron count to 18. How can this be a coordinatively unsaturated complex? If you add 2 more ligands to it then you end up with an electron count of 22.

[Dan]

For the first one, the oxidation state of Rh is +1, Rh⁺ and the ligands to be 2 CO and 2 I^- (each donate 2 electrons).

Alternatively, you can take Rh(0), add 1 electron for the overall negative charge, then add 2 electrons from each CO and consider the I ligands neutral - in this case they are I^. radicals and donate 1 electron each.

Either way you should get 16.

[CrimpJiggler]

Thanks, that answers a few of the questions I had but I'm still confused. How do you know rhodiums oxidation state is +1?

[Yakimikku]

Thanks, that answers a few of the questions I had but I'm still confused. How do you know rhodiums oxidation state is +1?

The CO ligands are neutral and the iodides have -1 charge each. Thus, the rhodium's oxidation state (Rh^x+) must be such that the overall charge is -1. So, it comes down to x - 2 = -1 and solve. The same analysis works in all cases except when the charge on a ligand is ambiguous (say, NO for example).

[CrimpJiggler]

Thats one thing thats been confusing me. If its a I radical that adds, I can see how it raises the oxidation state of the metal but if its a I^- that adds to the metal then it shouldn't alter the oxidation state at all because both of the electrons belonged to the I atom to begin with.

EDIT: Ah yeah, I think I'm getting it now. If I do an oxidative addition of HI to a metal, the I^- anion will donate 2 electrons but the H⁺ ion will also add to the metal so theres a net increase in oxidation state.

[Yakimikku]

Thats one thing thats been confusing me. If its a I radical that adds, I can see how it raises the oxidation state of the metal but if its a I^- that adds to the metal then it shouldn't alter the oxidation state at all because both of the electrons belonged to the I atom to begin with.

EDIT: Ah yeah, I think I'm getting it now. If I do an oxidative addition of HI to a metal, the I^- anion will donate 2 electrons but the H⁺ ion will also add to the metal so theres a net increase in oxidation state.

I'm confused to whether you are still asking about electron counting/determining the oxidation state of a metal or redox chemistry of a metal complex. In any case, oxidative addition is an oxidation of the metal and reduction of what is being added. In your comment about oxidative addition of HI, you get both an iodide and hydride ligand (H^-)--a 2 electron reduction. The electrons come from the metal and so the metal is oxidized by two electrons, increasing the oxidation state by two. There is an oxidative addition step in the Monsanto process you included. To bundle all of this discussion together, identify the oxidative addition step, state the oxidation states of the added species and metal centers before and after the oxidative addition, and then count the electrons for the metal complex before and after the addition.