From the attached picture (MO diagram of cyclopentadienyl anion) I can figure out where the nodal planes are, but why are they exactly at these positions?

enter image description here

Why is this nodal plane invalid?


  • $\begingroup$ The picture is not quite right. As these are all pi orbitals, you have to add one nodal surface (in this case a plane) to what it says there. This plane is identical for all orbitals. $\endgroup$ – Martin - マーチン Aug 15 '14 at 19:23
  • $\begingroup$ I believe that the figure you added has a higher energy than any of the others. So if it is possible, it would be an anti-bonding orbital. $\endgroup$ – LDC3 Aug 16 '14 at 0:39
  • $\begingroup$ How do you know that? $\endgroup$ – RBW Aug 16 '14 at 8:47

First a descriptive explanation:

The Molecule is symmetric. Thus the resulting mocular orbitals are symmetric too. If you are constructing different MOs where the nodal planes are different than in the picture above they wouldnt be symmetric.

Mathematical Explanation:

These nodal planes result from a LCAO-MO consideration with hückel approximation. You have to form the secular determinant for such a problem at which the coulomb integrals are Alpha and the resonance is only dissimilar zero when the two considered atoms are neighbors. After solving this hückel-matrix you can determinate the coefficents of the several molecular orbitals. Doing that you will see that the nodal planes exist as you drew them above.
(More Information about the Hückel-Method: here)

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  • $\begingroup$ I see more ways to put the planes .For example, a molecular orbital with one nodal plane can be constructed so that the nodal plane passes through two atomic orbitals. I don't understand how MOs in the image are symmetric and the one I proposed here isn't. $\endgroup$ – RBW Aug 15 '14 at 13:54
  • $\begingroup$ @Marko the molecule is of $D_{5h}$ symmetry, with a $C_5$ symmetry axis passing through the centre of the ring perpendicular to the carbon plane. This axis is accompanied by five vertical mirror planes. It can be shown, that these mirror planes are also the nodal planes for high lying orbitals. In the case you explain it would break this symmetry. $\endgroup$ – Martin - マーチン Aug 15 '14 at 19:30
  • $\begingroup$ I attached the nodal plane I proposed. I still don't see how it isn't allowed while the others are. $\endgroup$ – RBW Aug 15 '14 at 20:29

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