# Why are these molecular orbitals invalid for hexatriene?

Here are the MOs (more accurately, the breakdown of the MOs) for conjugated hexatriene:

Now, when first asked to draw the MOs myself, I drew this one for $\psi_3$:

It still has two nodes, and is still symmetric. I see nothing wrong with it, except that it doesn't match the "correct" MOs.

Similarly, for $\psi^*_4$, I can draw

Now, as I understand it, MOs are formed from various combinations of atomic orbitals, subject to a few rules regarding symmetry. From what I know, the two MOs I've given above should be valid MOs for hexatriene, or at least components of an MO of hexatriene.

Why aren't these two MOs valid? Also, is there a methodical way to obtain the sequence of MOs in the general case(of a conjugated polyene)?

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The general method to find molecular orbitals of π electrons in conjugated hydrocarbons in first approximation is called Hückel’s method. It is an undergrad-level topic, and you should find lots of explanations on the internet or in undergrad physical chemistry.

The wave function you propose is actually equal to $(-\psi_1 +\psi_3 + \psi_5^*)$   (with normalization factor not indicated). Thus, it is not an eigenstate of the system, and not a molecular orbital. But… unless you know Hückel's method, I am not sure there is an intuitive reason to discard this possibility: it has the right number of nodes (i.e. +/– inversions) for $\psi_3$.

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Ah, makes more sense now :) I'll read up on Huckel's method then. Thanks! –  ManishEarth Feb 12 '13 at 13:49

I read up a bit about Huckel theory. There's a pdf that explains this particularly well.

At one point, the following formula is derived at one point:

$$c_s\propto\sin\left(\frac{\pi k s}{n+1}\right)$$

which gives the coefficient (sans a positive normalisation constant) of the atomic orbital of the $s$th atom in the $k$th MO of a conjugated linear or monocyclic system on $n$ atoms.

A quick and dirty method has also been provided.

Using this, one can get a general idea of how the MO should look, and then one can get the exact +/- sequence by considering the number of nodes. Note that this is really the same thing as applying the $c_s$ formula given above, just in a graphical manner.

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