Chemistry - How does a Frost diagram reproduce the solutions to the wave equation?

Solution 1:

Algebraically, the levels of cyclic polyenes may be derived using simple Hückel theory (see also: Pi molecular orbitals of polyenes). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is:

$$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$

where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ orbital before interaction (Coulomb integral), $\beta$ is the interaction energy between two adjacent $\mathrm p_{\pi}$ orbitals (the resonance integral) and $j= 0, \pm 1, \pm 2,\ldots, \pm \frac{N - 1}{2}, +\frac{N}{2}$ for even $N$, and $j= 0, \pm 1, \pm 2, \ldots, \pm \frac{N - 1}{2}$ for odd $N$. The very simple form of this equation leads to a useful mnemonic for remembering the energy levels of these molecules. Draw a circle of radius $2\beta$ and inscribe an $N$-vertex polygon such that one vertex lies at the bottom position. The points at which the two figures touch define the Hückel energy levels. And that is what is called a Frost diagram.

Solution 2:

Frost developed this mnemonic patterning as an extension of the Hückel ($4n+2$) rule. A Frost diagram is usually applied to all-carbon, monocyclic, π systems. It allows one to find the number of molecular orbitals in the molecule's π system and their energetic positions. To construct a Frost diagram, proceed as follows:

• Draw a circle and inscribe a regular polygon with a vertex located at the bottom of the circle. The polygon has the same shape as the ring you are interested in. For example, if you are interested in benzene, draw a hexagon; for the tropylium ion, draw a heptagon.
• Energy-wise, the top and bottom of the circle are defined as $\alpha+2\beta$ and $\alpha-2\beta$ respectively (so the circle has radius $2\beta$); the center of the circle is located at $\alpha$; other points can be interpolated accordingly; the bottom of the circle is at lower energy than the top of the circle.
• Wherever a vertex of the polygon touches the circle, that is the energetic location of a molecular orbital.

Using benzene as an example, the lowest MO has energy $\alpha-2\beta$; the HOMO is degenerate (2 MO's) and located at $\alpha-\beta$; the LUMO is also degenerate and located at $\alpha+\beta$. Any orbital below the center of the circle is bonding, any orbital at the center is non-bonding and any orbital in the top-half of the circle is antibonding.