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The textbooks I have read introduce LCAO by considering the H2 molecule. In this example, there are only two possible combinations of the two 1s orbitals (in phase and out of phase).

When there are more than two valence orbitals, however, a greater amount of combinations arise. A common description of the pi-MOs in butadiene is shown below: Butadiene MOs There are two (and I think only these two, assuming all combining p-orbitals are identical) possible combinations of the p-orbitals which (apparently) do not contribute to the bonding in butadiene. These are shown at the bottom of the illustration above.

Why is it that n AOs only form n MOs, even when more than n LCAO are possible? You would think that all possible combinations would contribute a possible electron energy state to the molecule.

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marked as duplicate by Mithoron, Tyberius, Todd Minehardt, Jon Custer, airhuff Sep 19 '17 at 22:01

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    $\begingroup$ I believe that is more about the number of nodes than the look of MOs. Moreover, symmetry plays not the last part in such distribution. $\endgroup$ – MEL Science Sep 19 '17 at 8:45
  • $\begingroup$ Yes, symmetry is very important (all MOs must transform according to the system’s irreps). Also, this is a conservation-type thing. But the MO guys will answer soon enough. $\endgroup$ – Jan Sep 19 '17 at 9:09
  • $\begingroup$ While symmetry is important, the question will be perfectly valid in a system without any symmetry at all. $\endgroup$ – Ivan Neretin Sep 19 '17 at 9:34
  • $\begingroup$ related chemistry.stackexchange.com/questions/23578/… $\endgroup$ – Mithoron Sep 19 '17 at 19:04
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If you have $n$ functions (e.g. AOs) you can make a maximum of $n$ new linearly independent functions (e.g. MOs). If you try to make $n+1$ MOs, then any one of them can be expressed as a linear combination of the other $n$ MOs.

The most usual way to enforce linear independence is to enforce orthogonality, i.e. all your MOs have to have zero net overlap with each other. All the "correct" MOs you drew are orthogonal to each other, but the "incorrect" MOs are not orthogonal to any of the "correct" MOs.

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    $\begingroup$ Upvoted! MOs are required to be orthogonal, orthogonality implies linear independence, and one can not build more than $n$ linear independent functions (MOs) by making linear combinations of $n$ basis functions (AOs). The only problem is that I have no idea how to explain all this at the introductory MOT level. :| $\endgroup$ – Wildcat Sep 19 '17 at 13:16
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You're correct in saying that p-orbitals are identical, and it follows that the following configurations are the same:

enter image description here

You must only consider their relative alignments (i.e. their symmetry).

To consider your question, approach it using the Hückel method. The wavefunction of a molecular orbital is given as a linear combination of the atomic orbitals, which mathematically looks like:

\begin{equation} \tag{1} \Psi = c_1\phi_1 + c_2\phi_2 + c_3\phi_3 + c_4\phi_4 \end{equation}

where $c_i$ is the coefficient of $\phi_i$ in the LCAO-MO picture. Using the Hückel method we build the secular determinant for the molecular orbitals (see here if you're unfamiliar with this):

\begin{equation} \tag{2} \begin{vmatrix} x & 1 & 0 & 0 \\ 1 & x & 1 & 0 \\ 0 & 1 & x & 1 \\ 0 & 0 & 1 & x \end{vmatrix} = 0 \end{equation}

where $x$ is defined as

\begin{equation} \tag{3} x = \frac{\alpha - E}{\beta} \end{equation}

The expansion of the secular determinant is trivial and gives the following roots:

\begin{equation} \tag{4} x = \pm 1.62 \hspace{0.5cm} \text{or} \hspace{0.5cm} x = \pm 0.62 \end{equation}

Since the 4x4 secular determinant is made from 4 atomic orbitals, it is mathematically restricted to have 4 solutions. Using Equation 3 one may rearrange these solutions to find the orbital energies in terms of $\alpha$ and $\beta$. These roots are actually the eigenvalues of the secular matrix $ \ \textbf{H} - E\textbf{S} \ $ used in the Hückel method. The corresponding eigenfunctions are the coefficients of the atomic orbitals in Equation 1. The molecular orbitals for butadiene are found to be:

\begin{align} \tag{5} \Psi_a &= \phantom{-}0.372\phi_1 + 0.602\phi_2 + 0.602\phi_3 + 0.372\phi_4 & E &= \alpha + 1.618 \beta \\ \Psi_b &= -0.602\phi_1 - 0.372\phi_2 + 0.372\phi_3 + 0.602\phi_4 & E &= \alpha + 0.618 \beta \\ \Psi_c &= -0.602\phi_1 + 0.372\phi_2 + 0.372\phi_3 - 0.602\phi_4 & E &= \alpha - 0.618 \beta \\ \Psi_d &= \phantom{-}0.372\phi_1 - 0.602\phi_2 + 0.602\phi_3 - 0.372\phi_4 & E &= \alpha - 1.618 \beta \end{align}

If you look closely at the signs of each coefficient, you can see that they correspond to the phases in the allowed molecular orbital configurations. The origin of this behaviour follows from the discrete nature of quantum mechanics. If we call the coefficients amplitudes of a sine wave fitted to the length of the molecule this is more clear:

Ref: https://chem.libretexts.org/

There is no solution to our secular determinant which allows for orbital symmetry in the way which you queried. This is explained by the picture formed by the sine waves, along with the particle in a box model.

References:

  1. P. Atkins & R. Friedman, Molecular Quantum Mechanics, Oxford University Press, Oxford, 5th edn., 2011.

  2. https://chem.libretexts.org/

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