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The Frost Circle is a useful mnemonic to determine the relative energies for the molecular orbitals of an aromatic ring. One vertex of the polygon of the cycle is drawn pointing downwards and the entire polygon is inscribed in a circle. At this point, the location of each vertex corresponds to the MO relative energies. What is the physical explanation as to why a Frost Circle can accurately be used to predict the relative energies of molecular orbitals in aromatic compounds? What quantum mechanical phenomena lead to this emergent property? I'm looking for a connection between an equation and the circle I suppose.

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  • $\begingroup$ There is no property and no connection; it is just a mere coincidence. $\endgroup$ – Ivan Neretin May 8 '18 at 19:19
  • $\begingroup$ @IvanNeretin in order to develop the visual representation in the first place, thought had to be taken in its development. I doubt anyone would intuitively happen upon the vertices of a polygon corresponding to MOs... surface level they don’t seem connected. From the below answer you can see that the original explanation did in fact stem from an equation and that equation was central to developing a more approachable visual model. $\endgroup$ – Joe May 8 '18 at 22:45
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The MO energies are obtained from Hückel theory: see Pi molecular orbitals of polyenes (and the Internet in general, there are many excellent expositions) for more details.

Regarding the Frost circles, they do not actually have any meaning: they are simply visual representations of the energies obtained from Hückel theory for cyclic polyenes. The last line of porphyrin's answer is relevant here:

In the case of circular polyenes the solutions are $E_j=\alpha + 2\beta \cos(2\pi j/n)$.

where $n$ is the number of p-orbitals (e.g. for benzene $n = 6$) and $j$ is an integer satisfying $1 \leq j \leq n$. (The convention $0 \leq j \leq n-1$ is also sometimes seen, but it is mathematically equivalent.)

I am not a mathematician so some of the notation may be less than ideal, but the idea is as follows. Note that if you draw a circle with radius $r$, then the coordinates of any point on the circle $P$ may be described as $(r\sin\theta, -r\cos\theta)$ where $\theta$ is the (clockwise) angle that $\overrightarrow{OP}$ makes with the negative $y$-axis $\overrightarrow{OY}$. For example, here is a circle with $r = 2$. The angle $\theta$ here is equal to $\pi/4$, so the coordinates of the point $P$ are given by $(2\sin(\pi/4), -2\cos(\pi/4)) = (\sqrt{2}, -\sqrt{2})$.

Circle

Now we simply need to choose a circle with radius $r = -2\beta$ (note that $\beta < 0$, hence we need the negative sign) and select the six points on the circle where $\theta = 2\pi j/n$. The $y$-coordinates of these six points will then be the six MO energies. I have illustrated this for $n = 6$ and $\beta = 1$. We need to pick the points with

$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$

and these are $A$, $B$, $C$, $D$, $E$, and $F$ in order. Note that $\angle YOA = \pi/3$, $\angle YOB = 2\pi/3$, and so on. And note that whatever the value of $n$, there will always be the case where $j = n$ and hence $\theta = 2\pi$: that corresponds to the vertex at the bottom of the circle.

Frost circle

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