From here, the Frost Circles for both benzene and pyridine:

enter image description here

The $\pi$-MOs formed by 6 $\pi$-AOs of an aromatic ring are:

enter image description here

Listed with respect to the ordering in the above image, these MOs in benzene's $D_{6h}$ symmetry correspond to the group representations:


$\pi_4 / \pi_5=E_{2u}$

$\pi_2 / \pi_3=E_{1g}$


The $D_{6h}$ symmetry of benzene is reduced to $C_{2v}$ in pyridine. I believe the pyridine MOs transform in $C_{2v}$ symmetry as:







Onto my question:

Because the doubly degenerate $E$ representation does not exist in the $C_{2v}$ symmetry of pyridine, the corresponding $\pi$ orbitals are no longer degenerate. I could just stop there and accept that symmetry dictates degeneracy and there's nothing more for me to understand, but is there a way to dig a level deeper and connect the degeneracy to a physical explanation?

For example, I can imagine nodes breaking the extended conjugation of the $\pi$-atomic orbitals, which would explain why the $\pi$-MO with zero nodes is the lowest energy bonding orbital, and the $\pi$-MO with the most nodes the highest energy antibonding orbital. In this way, I can understand why the pyridine MOs climb in energy with increasing nodes. However, if it were solely about nodes and conjugation, then the benzene $\pi_2 / \pi_3$ and $\pi_4 / \pi_5$ MOs should not be degenerate; the one with more nodes would be higher lying in energy.

Or maybe the $\pi_3$ MO in pyridine is higher in energy because it has no $\pi$-atomic orbital situated on the nitrogen atom—the node is located on the nitrogen. Nitrogen being more electronegative than carbon, there should be a lower potential with some electron density localized in an atomic orbital on nitrogen. Therefore, the $\pi_2$ MO would lie lower in energy than its $\pi_3$ counterpart because nitrogen possesses an atomic orbital in the $\pi_2$ MO.

Is there a physical explanation why these orbitals in benzene are degenerate, and vice versa why the corresponding orbitals in pyridine are not?


2 Answers 2


My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs:

$$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$

where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$.

In Hückel theory the main difference between benzene and pyridine is that one atomic orbital is shifted in energy, so one of the $\alpha$ terms turns into, say, $\gamma$. Without loss of generality we will label the nitrogen with the index $n = 1$. In theory, the resonance integrals $\beta$ involving the nitrogen should change as well, but to a first approximation we assume they are the same.

In this case therefore the Hamiltonian matrix, when cast in the AO basis, changes from $\mathbf{H}_0$ to $\mathbf{H}$:

$$\mathbf{H}_0 = \begin{pmatrix} \alpha & \beta & 0 & 0 & 0 & \beta \\ \beta & \alpha & \beta & 0 & 0 & 0 \\ 0 & \beta & \alpha & \beta & 0 & 0 \\ 0 & 0 & \beta & \alpha & \beta & 0 \\ 0 & 0 & 0 & \beta & \alpha & \beta \\ \beta & 0 & 0 & 0 & \beta & \alpha \end{pmatrix} \qquad \longrightarrow \qquad \mathbf{H} = \begin{pmatrix} \gamma & \beta & 0 & 0 & 0 & \beta \\ \beta & \alpha & \beta & 0 & 0 & 0 \\ 0 & \beta & \alpha & \beta & 0 & 0 \\ 0 & 0 & \beta & \alpha & \beta & 0 \\ 0 & 0 & 0 & \beta & \alpha & \beta \\ \beta & 0 & 0 & 0 & \beta & \alpha \end{pmatrix}$$

Most of the Hamiltonian is unchanged, so we could consider this within the framework of perturbation theory, where we have a perturbed Hamiltonian $\hat{H} = \hat{H}_0 + \hat{V}$. The relevant perturbation is therefore

$$\mathbf{V} = \begin{pmatrix} \gamma - \alpha & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$

or to condense this back into operator form, $\hat{V} = (\gamma - \alpha)|1\rangle\langle 1|$.

The proper mathematical analysis is slightly annoying because the (unperturbed) eigenstates that we're interested in looking at are degenerate, which then necessitates that you cast the matrix $\hat{V}$ in the basis of the degenerate states of interest. Here, the two states we're interested in are $|\psi_2\rangle$ and $|\psi_3\rangle$. The application of Hückel theory to benzene gives the formulae for these unperturbed states as

$$\begin{align} |\psi_2\rangle &= \frac{\sqrt 3}{3}|1\rangle + \frac{\sqrt 3}{6}|2\rangle - \frac{\sqrt 3}{6}|3\rangle - \frac{\sqrt 3}{3}|4\rangle - \frac{\sqrt 3}{6}|5\rangle + \frac{\sqrt 3}{6}|6\rangle \\ |\psi_3\rangle &= \frac{1}{2}|2\rangle + \frac{1}{2}|3\rangle - \frac{1}{2}|5\rangle - \frac{1}{2}|6\rangle \end{align}$$

Atom labels and graphical scheme of π bonding MOs in benzene/pyridine

Note that this choice of coefficients is not unique, as the two states are degenerate: any linear combination of these two MOs is also a permissible MO with the same energy. In any case, the relevant matrix elements of $\hat{V}$ in the new basis $\{|\psi_2\rangle,|\psi_3\rangle\}$ are given by

$$V_{i-1,j-1} = \langle \psi_i | \hat{V} | \psi_j \rangle$$

and if you assume that $\langle n | n' \rangle = \delta_{nn'}$ (i.e. the AOs have no overlap with each other, which is a key assumption of simple Hückel theory anyway), then it becomes very simple:

$$\mathbf{V} = \begin{pmatrix} \frac{\gamma-\alpha}{3} & 0 \\ 0 & 0\end{pmatrix}$$

Perturbation theory tells us that the eigenvectors of $\hat{V}$ are the "states which are stable towards the perturbation". What this means is that, since the degeneracy is lifted by the perturbation, you also remove the free choice of having any linear combinations be a permissible MO: the coefficients of AOs in the new MOs must be uniquely determined (ignoring the phase factor). Since $\hat{V}$ is diagonal in the basis $\{|\psi_2\rangle,|\psi_3\rangle\}$, it conveniently means that both $|\psi_2\rangle$ and $|\psi_3\rangle$ are the eigenvectors of $\hat{V}$.

More interesting here is the first-order corrections to the eigenvalues of $\hat{H}_0$, which are simply the eigenvalues of $\hat{V}$. In this case, since $\hat{V}$ is diagonal, the eigenvalues are just $(\gamma-\alpha)/3$ and $0$: that is to say, the MO $\psi_2$ is shifted in energy by $(\gamma-\alpha)/3$ and the MO $\psi_3$ is unshifted in energy.

  • 1
    $\begingroup$ On the bonus lesson of applying perturbation theory: Is there a name for what you did to derive $V_{i-1,j-1}=\langle \psi_i | \hat{V} | \psi_j$ ? I'd like to read more on how you knew to do that then maybe make a new question on constructing the full perturbation matrix for all states. I noticed the off-diagonal elements in the 2x2 matrix fell away because $\psi_3$ has no $n=1$ term. If the perturbative matrix included e.g. $\psi_1$, I'd expect a non-zero off-diagonal element for $\psi_2$ and $\psi_1$. Then the perturbative correction to $\psi_2$ EV wouldnt actually be $(\gamma-\alpha)/3$, no? $\endgroup$
    – Blaise
    Aug 11, 2018 at 16:32
  • 2
    $\begingroup$ @Blaise well, the usual matrix representation of an operator in the basis $\{|i\rangle\}$ has $A_{ij} = \langle i | \hat{A} | j \rangle$ - that is standard QM. The only issue here is that our basis set begins its numbering at 2, so I had to minus one for consistency's sake. $\endgroup$ Aug 11, 2018 at 16:50
  • 1
    $\begingroup$ Regarding adding in $|\psi_1\rangle$ into the basis set, after looking up my old notes, I'm fairly sure that the derivation of the results I cited requires that all of the basis functions have the same (unperturbed) energy. Hence, you're only supposed to deal with the wavefunctions within the degenerate set. You're correct that adding it in would lead to (unphysical) off-diagonal elements $V_{12}$. It's unphysical because if you then went on to diagonalise $V$ and find its eigenstates, you'd find that there's mixing between $|\psi_1\rangle$ and $|\psi_2\rangle$, which shouldn't happen... $\endgroup$ Aug 11, 2018 at 16:58
  • 1
    $\begingroup$ ... because mixing between states of different energy only happens (1) when the states are of the same symmetry which I don't think is the case here; (2) when you go beyond the first-order energy correction, to find the first-order correction to the wavefunctions themselves (ugly maths!). Sorry for the somewhat excessive rambling, but I like it! $\endgroup$ Aug 11, 2018 at 17:06

In a benzene molecule draw the mirror line from one carbon atom, called C-1, through the opposite carbon. For each degenerate energy level you can construct one orbital that's symmetric under reflection through this mirror, and a second orthogonal orbital that's anti-symmetric under reflection. Note that the anti-symmetric orbital cannot have any electron density on the mirror itself; then, to keep the total electron density balanced when the energy level is occupied, the symmetric orbital must have maximum electron density along the mirror.

Now replace C-1, which lies on the mirror plane, with a nitrogen atom. The molecule is still symmetric about this mirror plane, so molecular orbitals must still be symmetric or anti-symmetric under reflection about this mirror. But whereas the anti-symmetric orbital has no electron density on the mirror and so is unaffected by the carbon/nitrogen trade, the symmetric orbital has maximum electron density right where we have replaced the carbon with nitrogen. So the symmetric orbital is singled out as being most "vulnerable" when benzene is changed to pyridine even as the anti-symmetric orbital is not "vulnerable" at all. This selectivity breaks the degeneracy.


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