In metal complex electron counting, is the neutral or ionic formalism more correct?

Question

I have been practicing finding total valence electron counts by both methods, and a few points have arisen that have confused me.

1) Which method gives the true d-electron count?

From what I understand, the ionic method assumes that both electrons from a bond go to the more electronegative atom, and the bond is 100% ionic. This lines up with formal oxidation state which uses the same premise. As a result, d-electrons are usually calculated as completely lost from the metal, giving a positive formal oxidation state.

In the neutral method, the d-electron count doesn't change as all calculations assume that M-L bonds are cleaved such that both species are neutral, and metal oxidation state stays at 0.

This results in 2 different values for a d-electron count, and I was wondering if the value actually means anything. I guess it doesn't matter in the end as the same number of electrons are put into the overall MO scheme, but I was wondering if being asked the d-electron count has any point to it. This can be seen in the below example from here:

2) If the neutral formalism is used, where does the electron deficiency in positively charged complexes come from?

In the ionic method, we assume everything is charged and just take electrons out of the metal to account for the overall charge (even this seems strange, why do we take electrons from the metal when we could take them from anywhere else to give the same MO scheme?). But in the neutral formalism, if everything is assumed neutral, where are we supposed take the electrons from?

3) If the electrons don't have to be taken from the metal, what is the point of assigning metal oxidation state?

I always hear statements like "d8 metals such as Pt(II) commonly form square planar complexes", but both the assignment of d8 and Pt(II) differ when using a different formalism. So is the use of the ionic method and oxidation of the metal specifically simply a convention that allows us to compare complexes quickly by their total electron count?

It seems to me like the correct answer is somewhere in between the 2 methods (between 100% ionic and 100% covalent), and the use of one or the other is arbitrary, but any corrections would be appreciated.

Also, if oxidation state is arbitrary, is the argument that "metals with lower ionisation energies reach higher oxidation states so can form more bonds" the same as saying "metals that have lower ionisation energies generally have larger radii so can fit more covalent bonds around them"?

Answer 1

I think, the answer to number 1 will remove the need to answer the subsequent questions:

1) Which method gives the true d-electron count?

Neither!

You cannot, and I want to stress that, derive the oxidation state of a metal in a complex solely by looking at its ligands. You can make educated guesses and in some cases only one oxidation state is plausible but there are many cases—most notably those denoted $\ce{\{FeNO\}^n}$ where such an assingment is meaningless to impossible. Without consulting spectroscopic data, it is absolutely impossible to tell whether a $\ce{\{FeNO\}^7}$ compound is $\ce{[\overset{+II}{Fe}(NO^.)]}$ or a $\ce{[\overset{+III}{Fe}(NO^-)]}$. In fact, entire papers have been published arguing one way or the other.

This essentially renders your follow-up questions entirely moot, but I do wish to take the time to point out that the metal’s oxidation state along with the corresponding oxidation state of the ligands is theoretically important information that can predict the electrochemical behaviour of a complex. The only problem is that it cannot always be neatly predicted as shown above. It is worth noting, though, that certain complexes have been assigned metal oxidation states beyond reasonable doubts (e.g. $\ce{[Cu(NH3)4(H2O)2]}$), and in such cases the predictions of reactivity are absolutely adequate.