# How do we obtain the π molecular orbitals for allene via Hückel theory?

For a conjugated system, for instance 1,3-butadiene, we can try to obtain the π molecular orbital energies and wavefunctions via Hückel theory. In our given molecule, there are 4 interacting Pz orbitals. By creating a 4x4 Hückel matrix and solving for the energy eigenvalues, we can arrive at 4 possible linear combinations (of p-orbitals) to get 4 molecular orbitals.

However, how can we use Hückel theory to deal with a molecule like 1,2-propadiene (allene)?

My attempt: I thought that since this molecule has essentially 2 separate π-systems: we may be able to apply Hückel theory to each one individually. Then, we can take all the possible combinations of the individual energies to arrive at the possible total energies. But I am not sure if this is correct. What would be the correct way of approaching this problem?

For p-orbitals $$i$$ and $$j$$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero:

$$\langle i | H | j \rangle = 0$$

If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form:

$$\mathbf{Hc} = E\mathbf{c},$$

or in explicit form, using the fact that the matrix elements of $$\mathbf{H}$$ are $$\mathbf{H}_{ij} = \langle i | H | j \rangle$$,

$$\begin{pmatrix} \alpha & \beta & 0 & 0 \\ \beta & \alpha & 0 & 0 \\ 0 & 0 & \alpha & \beta \\ 0 & 0 & \beta & \alpha \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{pmatrix} = E \begin{pmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{pmatrix}$$

This matrix has block-diagonal form, so solving for the eigenvalues of the entire matrix simply reduces to solving for the eigenvalues of each individual $$2\times 2$$ block (lots of discussion can be found on Google). For a physical interpretation, this means that both π systems have no interaction with each other, so you can indeed solve them individually.

Each block gives rise to one MO with energy $$\alpha + \beta$$ ("bonding") and one with energy $$\alpha - \beta$$ ("antibonding"), so all in all you will have two degenerate MOs with energy $$\alpha + \beta$$ and two degenerate MOs with energy $$\alpha - \beta$$.

* If you don't ignore the overlap matrix $$\mathbf{S}$$, whose matrix elements are defined by $$\mathbf{S}_{ij} = \langle i | j \rangle$$, this becomes a "generalised eigenvalue equation"

$$\mathbf{Hc} = E\mathbf{Sc}$$ However, because the overlap matrix also has the same block-diagonal form, ignoring the overlap matrix doesn't change the fact that the problem is separable into two $$2\times 2$$ blocks. It merely changes the exact eigenvalues and eigenstates by a small amount.

• Does this mean that our OVERALL molecule has 4 MOs? I am having trouble understanding how each MO could represent the entire molecule, since each come from a single 2x2 (representing one of the π-systems) . Apr 22 '20 at 18:52
• @JamesBond yes; well, four π-type MOs, at least. And yes, MOs are typically more delocalised than the conventional Lewis bonds, but the keyword is "typically". There is no requirement that they must stretch over the entire molecule. In fact, even localised lone pairs on an atom can be perfectly valid MOs sometimes. (Think e.g. the 2px and 2py lone pairs in HF, which cannot mix with the other orbitals due to symmetry mismatch.) Apr 22 '20 at 19:52
• That makes sense. I completely forgot that there isn't a strict requirement for MOs to always be delocalized throughout the entire molecule. Like some sigma MOs: formed via overlap of orbitals between 2 atoms (so mainly localized), but nonetheless considered a MO. Apr 22 '20 at 20:08
• Indeed. In this case, if you were to go beyond simple Huckel theory, there is probably going to be some overlap between the π-type orbitals and the C–H σ-type orbitals on the opposite side of the molecule. The overlap will not be amazing anyway, though, because they are relatively far apart. Huckel theory assumes complete separation between the π and σ frameworks, so that overlap is set to zero, which allows you to treat the π system separately (this is completely true in systems like ethene or benzene, where the π and σ orbitals are always orthogonal because of symmetry). Apr 23 '20 at 5:56