Particularly for longer alkyl chains. Or is predicting these molecular orbitals from quantum mechanics too complex to be able to be translated easily as a geometric pattern? If Frost Circles exist, I just figured something could for this as well. The only pattern I've been able to identify is that each molecular orbital is symmetrical, starts with no nodes and goes up by one with each higher energy orbital but for systems with many carbon atoms, this isn't a specific enough set of rules. An example of what I'm talking about in case it's unclear

  • $\begingroup$ How many carbon atoms are you intending to look at? From looking at this figure there does seem to be a pretty clear pattern in the same vein as a frost circle (at least for the even numbered case). @Elmer $\endgroup$
    – Tyberius
    Commented Apr 18, 2018 at 18:34
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    $\begingroup$ It all boils down to eigenvalues of adjacency matrices. $\endgroup$ Commented Apr 18, 2018 at 18:42
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    $\begingroup$ @IvanNeretin That isn't really useful. Even if the questioner knows what those words are, they wouldn't know that they should look for the Huckel method. $\endgroup$ Commented Apr 18, 2018 at 18:48
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    $\begingroup$ @pentavalentcarbon That's why this was a comment and not an answer. Anyway, thank you for making it more useful. $\endgroup$ Commented Apr 18, 2018 at 18:53
  • $\begingroup$ @Tyberius Let's say we're looking at 10 carbons. And what I'm looking for are the positions of the node(s) at each orbital. All I know now is that they go up by one each time starting at zero and ending between each pi orbital and that the positions of the nodes are arranged symmetrically, but this doesn't help me predict the orbitals for molecules beyond a certain length. $\endgroup$
    – Elmer
    Commented Apr 18, 2018 at 20:00

1 Answer 1


This should work for any even number of conjugated carbons (and with slight modification, for cations/radicals/anions of uneven numbers of carbons). You can follow a simple procedure to generate all the $\pi$ orbitals.

  1. Generate the lowest orbital with no nodes
  2. Generate next orbital by creating a node through the middle bond
  3. For all subsequent orbitals, generate by taking the previous orbital, placing nodes to the left and right of the original node, and then removing the original node.
  4. End when all bonds are separated by nodes

We can write these as binary strings where a 1 is one orientation and 0 is the other. So for example, you could write the lowest orbital of N=10 chain as 0000000000, the next as 0000011111, the next as 0000110000, and so on.

The Huckel method mentioned in the comments is a more formal way of obtaining the orbitals by solving for the eigenvalues and eigenvectors of an adjacency matrix. The adjacency matrix is formed by numbering the carbon chain and forming an NxN matrix with the rows and columns numbered like the carbons. In each row, put a 1 if the carbon of the row and column are adjacent and put an x if they are the same. A simple online calculator with a few examples and instructions for this can found here (this only gives the eigenvalues, you would probably have to download or code your own version to give the eigenvectors). You only really need these eigenvectors to get an equation for the orbitals, but for your case you can do that just as well with the simple procedure by simply writing the MOs as linear combinations of the AOs.


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