Structure & Reactivity in Chemistry

Molecular Orbitals

MO11. Geometry and Orbital Contribution to Bonding

What if a molecule had three or more atoms in it? We can no longer draw a simple molecular orbital interaction diagram, in which we show on atom on each side and the molecular results in the middle, because we don't have two atoms coming together to share electrons.

Instead, we look at slightly simplified MO diagrams in which we keep track of the result all those atomic orbitals coming together to make new orbitals.  At a very simple level, the qualitative result is always the same and can be roughly predicted based on the Lewis structure.

Problem MO11.1.

a) This is the MO diagram for ethane.
     Draw a Lewis structure of ethane.
     Populate the electrons into the MO diagram.

b) This is the MO diagram for propane.
     Draw a Lewis structure of propane.
     Populate the electrons into the MO diagram.
     Add in the missing MO cartoons for the different levels.
     Label the HOMO and LUMO.

c) This is the MO diagram for butane.
    Draw a Lewis structure of butane.
    Populate the electrons into the MO diagram.
    What happpens to the MO diagram as the alkane chain length increases?

Problem MO11.2.

 a)  This is the MO diagram for ethene.
     Draw a Lewis structure of ethene.
     Populate the electrons into the MO diagram.

b)  This is the MO diagram for propene.
     Draw a Lewis structure of propene.
     Populate the electrons into the MO diagram.
     Add in the missing MO cartoons for the different levels.
     Label the HOMO and LUMO.

c)  This is the MO diagram for 1-butene.
     Draw a Lewis structure of 1-butene.
     Populate the electrons into the MO diagram.
    What happpens to the MO diagram as the alkane chain length increases?

d)  This is the MO diagram for 2-butene.
     Draw a Lewis structure of 2-butene.
     Populate the electrons into the MO diagram.
     Does the location of the pi bond change the MO diagram?

e)  This is the MO diagram for 1-butyne.
     Draw a Lewis structure of 1-butyne.
     Populate the electrons into the MO diagram.
     How does an alkyne change the MO diagram?

Methane is a moderately small molecule. Its Lewis structure is pretty easy to draw, with a carbon in the middle bonded to four hydrogens. Trying to find the interactions of the hydrogen 1s orbitals with the orbitals on carbon could be challenging. However, molecular orbital calculations generally start with a good guess at the geometry of the structure.

In quantitative calculations, that starting geometry is often modified after the energy of the molecule has been calculated. A long series of iterative calculations are done, changing bond lengths or bond angles each time, until the geometry with the lowest possible energy has been found. If this geometry has the lowest energy, then it is probably the shape that the molecule would really adopt in nature. In fact, it is assumed in quantum chemistry that if a lower energy structure can be found, it is always a more accurate picture of the real molecule (assuming no mistakes have been made in the calculation).

Since the carbon in methane has four groups attached to it, we can be pretty sure that the carbon is tetrahedral.

We will assume the four hydrogens are arranged at the corners of a tetrahedron around the central carbon.  Overlap with the carbon 2s orbital is easy to see, because the s orbital can overlap with neighbours in all directions.  Of course, there could be an in-phase combination and an out-of-phase combination.

The complicated part arises when we want to introduce the carbon electrons and have them interact with the hydrogens. There is no universal orientation of p orbitals. Which way should they point?

In principle, it does not matter too much what direction we choose for the p orbitals.  We should eventually end up with the same conclusions in the end.  However, things may get extremely complicated and mistakes are likely to occur if we don't choose an orientation that makes the interactions simple to see.  Since it doesn't matter which way the p orbitals are pointing, we should choose a direction that makes the calculation easiest.

Orienting the p orbitals this way simplifies things a lot. It means each lobe of each p orbital is directed at a space (on the face of a cube) in between two hydrogen atoms (on the opposite corners of the face). The other lobe of the same p orbital is directed at a space (on the opposite face of the cube) in between the other two hydrogen atoms (on the opposite corners of that face). The same is true for each p orbital, but they point at different faces and they interact with different combinations of the same four hydrogens.

As a result of this choice, there are really only a few interactions to consider in methane.

Note that here the hydrogen atoms are often treated as being in phase with each other, or at least pairs of them are. This assumption simplifies the calculation and cuts down on the number of non-allowed combinations (i.e. combinations that produce the total number of orbitals over eight).

Organizing these combinations in terms of energy:

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