Wednesday, May 7, 2014

Molecular Orbital Theory Notes III

The first two posts in the series gave brief information about the d-orbitals and the metal-ligand orbital interactions. Now it is time to construct a molecular orbital diagram for a metal complex in the form of ML6. 

A few things you should never forget:

- Ligands are mostly more electronegative than the transition metal. Therefore, ligand orbitals should be drawn lower than metal orbitals. This also tells us that the bonding orbitals are mainly on ligand orbitals! 

- Nonbonding orbitals will be drawn at the same level as the atomic orbitals. So, they are mainly on the metals!

- Antibonding orbitals are closer to the metal. Actually, if you learn more about the MO theory and transition metal reactions, you can see the importance of this fact.

There is a really nice method to draw the diagram. 

1. Find the point group (I have many posts and a simple flowchart to determine point groups here.) 

2. Go to the character table for the point group and assign the symmetry properties (those t2g, eg, B1u etc. things) to each orbital. Sounds hard, but the character table gives you all. Very easy!

3. Because, there has to be symmetry and there is to have interactions, do the same thing for the ligand orbitals.

4. Following the rule that Sigma interactions > Pi > Delta, match the ligand and metal atomic orbitals. The ones that are the same (bonding) will go down, the ones that has no corresponding orbitals will be the nonbonding d-orbitals and the rest will be the antibonding ones.

Here is the first and the most important example:

For a perfect octahedral metal complex like ML6, the point group is Oh and the character table for Oh looks like this:

Now we are in the second step. The shortcut is that the first column with x, y, z is for the P orbitals.  So, our p-orbitals for this character group have the T1u "label." The first row (A1g) is always the s orbital. The second column with xy, xz etc. is for d-orbitals. So, in an octahedral field; dxz, dxy, dyz orbitals will be degenerate! And dz^2 and dx^2-y^2 are degenerate. If you are careful enough, you will notice that this looks like the "famous" d-orbital splitting in an octahedral field. I hope now you can see the relation.

Finally, let's draw the diagram. Once the orbitals are written for the metals on left, and the ligand orbitals on right (lower than the metal atomic orbitals), look for the symmetry and just connect them. A very important thing not to be forgotten is that for a transition metal, the d-orbital is always a (n-1) orbital. For example, Cr is in the first row and the electron configuration for the free atom is "3d5, 4s1" That's why we draw the d-orbitals lower than the s and p orbitals. I guess it is clear. The diagram below is from wikipedia. The antibonding t1u and a1g orbitals can change places. It is not a big deal. In the end, you should always count the number of MO's. The number should be equal to the some of the ALL atomic orbitals.  In this case; 

9 (from the metal) + 6 (ligands) = 15 MO's

Then you can place the electrons obeying the Aufbau Principle, Hund's Rule and obviously Pauli Exclusion Principle.  I hope this was helpful. 

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