When drawing molecular orbital diagrams for mixing two orbitals we always end up with one orbital which is lower in energy than the initial orbitals and another which is higher in energy. The principle extends to mixing of more than two orbitals with the total number of orbitals always being conserved.

Often we will use arguments involving mixing of molecular orbitals to form MOs of a slightly lower energy to explain phenomena such as hyperconjugation with diagrams such as this.

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In this situation, when you have one filled and one empty orbital, the electrons always end up in a lower energy than they started.

What stops you from mixing these orbitals again to produce even lower energy orbitals, even if the energy difference is very small? I understand that symmetry might have something to do with it but I am not well versed in the more mathematical elements of MO theory such as symmetry groups. Or is my initial assumption that mixing two orbitals always results in a lower energy wrong?

  • $\begingroup$ You start with two functions $f_1$ and $f_2$, and you are trying to mix them to get new functions $g_1 = c_{11}f_1 + c_{12}f_2$ and $g_2 = c_{21}f_1 + c_{22}f_2$, in order to minimise the energy of $g_1$. The problem is that there are constraints on the constants $c_{ij}$, namely $c_{i1}^2 + c_{i2}^2 = 1$ and $c_{1i}^2 + c_{2i}^2 = 1$ ($i = 1,2)$. These mean that there is a fixed value for which the energy is minimised and this is the closest approximation to the ground state energy that you can get (by the variational principle) using the basis set that you have $\endgroup$ May 23, 2016 at 17:29
  • $\begingroup$ In less mathematical terms, one could look at it as the strength of mixing decreasing with the size of the energy gap between the two orbitals. The larger the energy gap, the weaker the mixing. This means that the total mixing you can get is asymptotic and will eventually reach an end (the most stable configuration). Also, symmetry has nothing to do here, symmetry only dictates which wavefunctions can mix with which but not how much they mix. $\endgroup$ May 23, 2016 at 17:32
  • $\begingroup$ I object. Symmetry has very much to do with this. Being different eigenfunctions of the same problem, our resulting functions are orthogonal. They can't interact (that is, "mix"). $\endgroup$ May 23, 2016 at 17:52
  • $\begingroup$ @Ivan That is true but I don't think that was what was meant by symmetry, symmetry was referring to the irreducible representations under which the orbitals transform. $\endgroup$ May 23, 2016 at 18:02
  • $\begingroup$ @orthocresol Well, take two H atoms so they produce $\sigma$ and $\sigma^*$ orbitals, and voila, these two belong to different irreducible representations, so (in a way) they are orthogonal "because of symmetry". $\endgroup$ May 23, 2016 at 18:05

1 Answer 1


The orbital $\psi$ is obtained as a mixture of the orbitals {$\phi_i$}

$$\psi= c_1 \phi_1 + c_2 \phi_2$$

choosing the {$c_i$} through energy minimization.

Once obtained $\psi$, if you try to mix it with the initial {$\phi_i$} to obtain a new function, lets say $\mu$, you'll have:

$$\mu = c_3 \psi + c_4 \phi_1 + c_5 \phi_2$$

(notice that the coefficients don't need to be the same)

or equally

$$\mu = c_3 (c_1 \phi_1 + c_2 \phi_2) + c_4 \phi_1 + c_5 \phi_2$$ $$\mu = c_3 c_1 \phi_1 + c_3c_2 \phi_2 + c_4 \phi_1 + c_5 \phi_2$$

$$\mu = (c_3 c_1 + c_4) \phi_1 + (c_3c_2+c_5) \phi_2$$

that has the same functional form that $$\psi= c_1 \phi_1 + c_2 \phi_2$$

so as the criteria is the energy minimization it should be no surprising that we can not get better energy that making $(c_3 c_1 + c_4) = c_1$ and $(c_3c_2+c_5)=c_2$.

The idea behind this is that you want to obtain the "better" function $\psi$, but you are restricted to a set of functions of the form: $c_1 \phi_1 + c_2 \phi_2$. You impose this restriction because it only requires to obtain two coefficients, that is far simpler than obtaining an arbitrary function (defined with infinite images). One restriction is that the equation $c_i \phi_i = c_k \phi_k$ can only be true if $c_i = c_k = 0$.

You can expand your search to a larger set of functions, for example: $c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3$, but it requires that the set still be linearly independent, that is, you can not get any of the functions with a linear combination of the others, (i.e. it is not possible to get $c_1 \phi_1 = c_2 \phi_2 + c_3 \phi_3$ if at least one $c_i$ is different from 0). If not, as in the discussion above, the space of function will contain the same functions so the inclusion of the new function is in some sense redundant or useful.

You can add infinite linearly independent functions, but, it does not mean that the energy will be always get lower.


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