Why can only orbitals of similar energy interact significantly?

Question

In molecular orbital theory, it is often said that only orbitals with a) the same symmetry and b) similar energy can interact to a significant degree. I understand the necessity for the same symmetry, but the only explanation for b) I've been able to find is this one: https://www.quora.com/Why-is-interaction-strongest-between-orbitals-of-similar-energies

"It's fairly common to think of orbitals as static, but their phase changes in time. The frequency of this change is proportional to the energy of the orbital. So even if two orbital with identical shape and initial phase but different energy could exist, their overlap would only be constructive some of the time because they would not stay in phase. They can only be in phase all of the time if they have the same frequency, which is only possible if they have the same energy."

However, if this were the case, then every molecular orbital made out of AOs of different energy would be bonding exactly half of the time, while being anti-bonding the other half. Wouldn't this make it, on average, non-bonding?

Is there a better explanation for why the interaction is the strongest between orbitals of similar energy?

Answer 1

The closest thing to a correct and intuitive explanation is the following:

We can treat a molecular orbital as approximately a linear combination of two atomic orbitals, and solve the time-independent Schrodinger equation (i.e. we assume that some linear combinations of atomic orbitals are stationary states).

When we do this, we find that the energies of possible molecular orbitals depend on four quantities: the energies of the atomic orbitals $\alpha_A$ and $\alpha_B$, the "resonance integral" $\beta$ which is the transition matrix element $\langle \psi_A|H|\psi_B\rangle$ between the two orbitals, and the "overlap integral" $S$, which is just the usual overlap $\langle \psi_A|\psi_B\rangle$ of the two atomic orbitals. Solving the time-independent Schrodinger equation yields the following polynomial in energy:

$$(\alpha_A-E)(\alpha_B-E)-(\beta-ES)^2=0$$

This polynomial has the following approximate solutions $E_1$ and $E_2$, which correspond to the energy of the two molecular orbitals. Important note: the particular approximation taken means that the following equations are not valid for $\alpha_A=\alpha_B$; in fact, they are only valid when $|\alpha_A-\alpha_B|>|\beta-\alpha_A S|$.

$$E_1\approx \alpha_A+\frac{(\beta-\alpha_A S)^2}{\alpha_A-\alpha_B}$$

$$E_2\approx \alpha_B+\frac{(\beta-\alpha_B S)^2}{\alpha_B-\alpha_A}$$

If we chose our ordering of atoms such that $\alpha_A<\alpha_B$, then we have that $E_1<\alpha_A<\alpha_B$, so $E_1$ is the bonding orbital energy. Meanwhile, $E_2>\alpha_B>\alpha_A$, so $E_2$ is the antibonding orbital energy. (If the opposite relationship is chosen, then $E_1$ and $E_2$ are switched.)

So we have that the bonding orbital energy $E_1$ is close to the energy of the lower-energy atomic orbital $\alpha_A$, with a negative "correction". The bigger the negative correction, the more stable the bonding orbital is relative to the atomic orbitals. As you can see, the negative correction grows smaller when the energy difference $\alpha_A-\alpha_B$ is larger. So, the bigger the energy difference is between the atomic orbitals, the less stable the bonding orbital will be.

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The above was summarized from passages in Nguyên Trong Anh's Frontier Orbitals; the relevant passage is available here: https://dasher.wustl.edu/chem478/reading/frontier-orbitals-anh.pdf