Mathematical equivalence of ionic-bond resonance structures and d-orbital participation in hypervalent molecules

Question

It is now routine in the scientific community that "main-group hypervalent molecules, under certain conditions, tend to be better represented by resonance structures involving ionic bonds, rather than the utilisation of d-orbitals". For example, PF5 is better represented by a total of 5 "[PF4]+ [F]-" resonance structures, than sp3d hybritals on the phosphorus atom. However, after reading the Wikipedia article for "molecular orbital"s carefully, I began to think that they are, strictly speaking, mathematically the same-

The article said that "When the energy difference between the atomic orbitals of two atoms is quite large, one atom's orbitals contribute almost entirely to the bonding orbitals, and the other atom's orbitals contribute almost entirely to the antibonding orbitals. Thus, the situation is effectively that one or more electrons have been transferred from one atom to the other. This is called an (mostly) ionic bond.": maybe the ionic nature of a molecular orbital formed by a 3d orbital of P and a 2p orbital on F can be considered a special case of this statement, as P's 3d orbitals >>>>> F's 2p orbitals in terms of energy levels.

The question is- is my guess true?

Answer 1

You are essentially correct, but the distinction is the degree of bonding.

As you described, in a polarized or partially ionic bond, the atomic orbital of the more electronegative atom is lower in energy. Thus, the bonding molecular orbital is closer in energy to that atomic orbital, and the antibonding molecular orbital is closer in energy to the atomic orbital of the less electronegative atom. Because of this, the relative contribution of the lower energy AO to the bonding orbital is greater, and the contribution from the higher energy AO correspondingly lesser.

The distinction between "covalent" and "ionic" is therefore blurry. There's a continuum of degree of ionic bonding. In the extreme case of a completely ionic bond, there is no stabilization of the "bonding" MO relative to the lower energy AO. Furthermore, the higher energy AO no longer contributes at all. At that point, we no longer have bonding and antibonding MO's, but rather both are nonbonding orbitals comprised entirely of a single AO each.

But there are plenty of cases where there is just a little bit of stabilization, and we still describe the bond as "ionic". That's partly because it's something of a judgement call at what point we conclude that there is no bonding interaction. We can't solve the Schrodinger equation exactly, so we use computational approximations. Our computational methods aren't precise enough to tell the difference between a tiny bit and none at all. For the $d$ orbitals in the compounds in question, most methods predict at least a small participation of the $d$ orbitals in most compounds. But how much above 0 does it have to be before it's no longer correct to say that $d$ orbitals are not involved? We can say with certainty that the participation of the $d$ orbitals is nowhere near the $sp^3d^2$ view that used to be taught.

In your question, you described the compounds as "better represented" by ionic bonding "rather than utilization of $d$ orbitals."

You're correct that this distinction isn't either/or, but the key words are "better represented". Even if the $d$ orbitals contributed as much as 5% to the bonding MO, isn't the ionic representation "better"?