I am a little confused about how point groups give information about MO of polyatomic molecules.

Is it true that all molecules of a given point group such as, $\mathrm{C_{2v}}$, will have the same MO diagram, the only thing that will change is the number of electrons?

  • 4
    $\begingroup$ You also have to consider the atoms being used to form the MO's and the orbitals available. For instance if one atom is highly electronegative the diagram will change (different orbitals may interact than in the case where both atoms are close in energy). Equally an atom further down the periodic table has more orbitals available for bonding, also changing the diagram. $\endgroup$
    – NotEvans.
    Nov 14 '15 at 23:54
  • $\begingroup$ Not in the slightest. Point groups provide a simple tool to divide orbitals into isolated interacting sets. That's all. $\endgroup$
    – permeakra
    Nov 15 '15 at 16:26

Is it true that all molecules of a given point group such as, $C_{2v}$, will have the same MO diagram, the only thing that will change is the number of electrons?

No. There are so many different possibilities of how molecules with $C_{2v}$ may look, that they cannot possibly all have the same energy level diagram. Take water and acetone for mild examples.

What you can do, however, is estimate how different molecular orbitals will look based on the atomic orbitals you put into the estimation. Let’s take formaldehyde as it is a nice and simple example.

We have two hydrogen atoms, one carbon and one oxygen. Carbon and oxygen both lie on the $C_2$ axis so we can consider them separately, but we should use group orbitals for the hydrogens (I am defining the $yz$ plane to be the plane of the molecule). Luckily, they only have s orbitals, so it doesn’t get too complicated. Check out how the s-orbitals transform with the different symmetry operations to find their irreducible representations: $\mathrm{A_1 + B_2}$. Carbon and oxygen each have an s orbital and three p orbitals which will transform in the same way: the s orbital as $\mathrm{A_1}$ (naturally), the p orbitals transform as $\mathrm{A_1 + B_1 + B_2}$ (for $\mathrm{p}_z, \mathrm{p}_x$ and $\mathrm{p}_y$, respectively).

Thus, we have five atomic (or group) orbitals of the $\mathrm{A_1}$ type which will combine to form most of the σ bonds, three of the $\mathrm{B_2}$ which will also combine to σ bonds and two of the $\mathrm{B_1}$ type which will combine to form π bonds. Had we chosen water, we would only have had six orbitals altogether transforming as $\mathrm{3 A_1 + B_1 + 2 B_2}$ — a clear distinction showing that the diagrams cannot always be the same.

We now have to estimate the energy levels the orbitals initially have. It is a good guess to put the highest-energy atomic orbital at the respective atom’s ionisation energy; then you would just have to esimate a lower level for the s orbitals of carbon and oxygen. With these levels in hand, you can educatedly guess the different energy levels of the resulting molecular orbitals and educatedly guess what the orbital would look like: The resulting MO must transform in the same way as the AOs used to build it. So any $\mathrm{A_1}$ orbital will be symmetric to any symmetry operation while $\mathrm{B_2}$ will always have a nodal plane in the $\ce{C=O}$ bond axis, etc. I’m sure you can now come up with a well-educated guess of the formaldehyde orbitals; you can check out what they actually look like on the homepage of Professor Zipse. (You can ignore the accompanying German text. The higher-level antibonding MOs LUMO+2 and LUMO+3 that are required by theory are not shown; one must be $\mathrm{B_2}$, the other $\mathrm{A_1}$.)


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