The energy of molecule orbitals increases with more nodal planes. W1 (in the attached picture) has no nodal plane. I'd like to know how to draw the nodal planes in cyclopropane molecule orbitals but do not know how. I found no picture in the web with the nodal planes of cyclopropane.

How many nodal planes have W3, W4 and W5?

expected nodal planes in cyclopropane in the picture from walsh orbitals, picture modificated from www.enu.kz/repository/repository2014/orbital-interaction.pdf page 85

  • 1
    $\begingroup$ Not all nodes are planar. Some may be points, radial, angular, etc. $\endgroup$
    – Ben Norris
    Jul 22, 2014 at 1:54
  • $\begingroup$ Could you please provide the source of this picture? I would like to see the context, as I think they assigned W4 as the LUMO, which is not true at all. $\endgroup$ Jul 22, 2014 at 4:40
  • $\begingroup$ @Martin the source appears to be Rauk's book: books.google.com/books?id=Kn1YXlMlbPMC&pg=PA85&lpg=PA85 $\endgroup$ Jul 22, 2014 at 6:54
  • $\begingroup$ @SilvioLevy Thank you. I was not mistaken, he assigned W4 as the LUMO. $\endgroup$ Jul 22, 2014 at 7:13
  • $\begingroup$ It seems he's not the only one to think so. See for example page 2 of arkat-usa.org/get-file/19033 (and if true, this just goes to show what I wrote in my answer: the number of nodal surfaces is not an absolute predictor of the energetic order). $\endgroup$ Jul 22, 2014 at 7:32

2 Answers 2


The assessment of nodal planes is not as trivial as it might seem at first, because they are usually not planes at all.

Orbitals can be explained in terms of symmetry operations. There are three different operations and each of them has a unique element (in cartesian coordinates):

  • Rotation: Axis
  • Mirror: Plane
  • Inversion: Point

When we talk about hydrogen type atomic orbitals, we usually refer to their cartesian form. It is already fairly obvious for the $\ce{2s}$, that there is no nodal plane, while there are certainly nodes.

The representations of $\ce{s}$ orbitals are always trivial with respect to the total symmetry of the external field (= Point group). That means any symmetry operation, which is given through the external field will be matched by the $\ce{s}$ orbital. Another term for this is totally symmetric.

The representation of $\ce{p}$ orbitals are always antisymmetric with respect to inversion in any given external field.

The representation of $\ce{d}$ orbitals is always antisymmetric to two mirror planes. That excludes of course the $\ce{d_{z^2}}$ orbital, which is not a cartesian, but a spherical representation. (Note: In cartesian space there are 6 $\ce{d}$ orbitals. As quantum numbers only allow 5 spherical orbitals, they have to be transformed. See for example at chemissian.com or Schlegel and Frisch.) Early basis sets for quantum chemical calculations used the cartesian representation, because it wa easier to code and calculate.

Due to the geometric arrangement of nuclei in a molecule, the external field is given through the point group of the molecule. Every molecular orbital can be expressed through symmetry operations of that point group.

Cyclopropane has $D_\mathrm{3h}$ symmetry. And every molecular orbital has to respect that. I have already written about orbitals these orbitals in Why does cyclopropane give bromine water test? However, I would like to shed some more light on this, because I believe the picture you found is not quite correct.
The following picture shows the depicted orbitals from your source, as well as the LUMO.
orbs of cyclopropane
Where W1 corresponds to HOMO-2, W2 and W3 correspond to both HOMO, W4 corresponds to LUMO+3, finally W5 and W6 correspond to LUMO+2.
The ordering of these orbitals is taken from a BP86/cc-pVTZ calculation of Gaussian09. A methodology that I believe gives fairly accurate results for molecular orbitals. (It is however noteworthy, that unoccupied orbitals are constructed from occupied orbitals. That means, that their physical meaning is limited.)

Let's have a closer look at the orbitals. The HOMO-2 is of $A_1'$ symmetry and therefore behaves like an $\ce{s}$ orbital. There is no antibonding nodal surface in this orbital. That means at any bonding axis of the ring structure the sign of the wavefunction does not change. There is however a nodal surface that does not affect bonding, as was already pointed out in some comments.

The HOMOs are of $E'$ symmetry and therefore behave like $\ce{p_x}$ and $\ce{p_y}$ orbitals. These orbitals have one antibonding nodal surface. In a result of symmetry restrictions these nodal surfaces are perpendicular to each other.

This is also true for the LUMO+2s. In addition to this there is another antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO again is of $A_1'$ symmetry and behaves like a $\ce{s}$ orbital. There is again no antibonding nodal surface with respect to the $\ce{C-C}$ bonds, but the main electron density would be outside of any bond and there is an antibonding nodal surfaces with respect to the $\ce{C-H}$ bonds.

The LUMO+3 is of $A_2'$ symmetry and does therefore not behave like any other atomic orbital. Here we find three antibonding nodal surfaces with respect to the $\ce{C-C}$ bonds.

The bonding situation in cyclopropane is very complicated and can be explained in many different ways. It is not as obvious, that the molecule is a $\sigma$ aromate, having in plane $\pi$ orbitals (HOMO). The explanation of its molecular orbitals were first composed in terms of a Walsh like behaviour on the basis of an extended Hückel calculation with a minimum basis.

When you disregard the $\ce{C-H}$ bonds and only allow three carbon $\ce{sp^2}$ and three carbon $\ce{p}$ orbitals, then you can only arrive at the boning picture that was depicted in your source (and many other publications), as you can only have six molecular orbitals formed from these. There are, however, many more atomic orbitals that mix into the final wavefunction. First of all it is necessary to state, that the used $\ce{sp^2}$ orbitals are not of the ideal $\frac13\ce{s}+\frac23\ce{p}$ composition (see more about that in the linked question and the answer of ron therein).

Now finally for the discussion about the correlation of nodal planes and increasing energy. It is a very common statement, that is missing it's most important limitation. It is only completely true for orbitals of the same symmetry. (This is also a necessary requirement.)

For example, the energy of the $\ce{s}$ orbital series increase with the main quantum number, as you add one more nodal surface per one increment in $n$: $\ce{1s->0; 2s->1; 3s->2;...}$. The same applies to the $\ce{p}$ orbital series: $\ce{2p->1; 3p->2; ...}$.

Now this statement became popular, when talking about aromatic systems. In the framework of Hückel's molecular orbital method, this must be true, since all regarded orbitals are of the same symmetry.

However, it is usually true, that an orbital with fewer nodal planes is more stable than another with more, but this is more a gut feeling, than actual, factual science.

  • 1
    $\begingroup$ +1 for an excellent answer overall, but especially for your point about the necessity of like symmetry when correlating nodes to energy. It's a subtle and important point that gets glossed over routinely in most qualitative MO texts (as I know to my own detriment). $\endgroup$
    – Greg E.
    Jul 22, 2014 at 9:43
  • $\begingroup$ Nice answer. There is nothing better for checking the validity of a qualitative MO scheme than a real calculation. $\endgroup$
    – Philipp
    Jul 22, 2014 at 10:32
  • $\begingroup$ @Martin Thanks for taking the time to write a long answer and providing the results of your calculations. It's nice that the rule of thumb about energetic ordering is not being broken in this example. But I'm still uneasy about its status as a rigorous rule, even with the symmetry caveat. Take a molecule lacking all symmetry, which is the commonest case once you leave the realm of textbook examples. Here all molecular orbitals belong the the same (trivial) symmetry class. Does it follow that more nodal planes implies higher energy? $\endgroup$ Jul 22, 2014 at 18:09
  • $\begingroup$ @SilvioLevy Yes it still is valid. Every nodal surface adds a sign change to the orbital wave function and therefore the energy of this orbital has to go up as it gets larger (less dense). For the asymmetric case it is not trivial to see the sign changes, as orbitals mix with a lot of functions and it is not easy see it as clearly. Sometimes then it helps to break it down to local symmetry and look there. However, there are more factors at play than only the sign change, overlap integral, etc. This is why a computation will always be superior to the rule of thumb. $\endgroup$ Jul 23, 2014 at 2:37
  • 1
    $\begingroup$ @Martin: Thanks for your very helpful explanation! $\endgroup$
    – laminin
    Jul 24, 2014 at 10:23

Instead of nodal planes think of nodal surfaces. For W1 there is one nodal surface, a cylinder that looks like a circle seen from above (the view in the book).

Continuing with the hints. All the combinations that come from orbitals that have radial symmetry (the ones where the bulges point toward the center and away from the center) have the same cylinder as a nodal surface. If you draw the circle representing this cylinder seen from above, you'll basically have completed the nodal surface diagram for W1, W5, W6.

The use of nodal surfaces helps build intuition, but it only goes so far. For example, it doesn't have to be true that more nodal surfaces equals higher energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.