General steps

1. First, determine the point group of the molecule.

2. Draw out the orbitals of the central atoms.

3. Determine the irreducible representation of the orbitals of the central atoms

4. Draw out the possible bonding interaction of the orbitals from the peripheral atoms

5. Determine the reducible representation of the orbitals of the peripheral atoms

6. From the reducible representation, determine the irreducible representation of the orbitals of the peripheral atoms

7. Determine the orbital energies from a table.

8. Place the orbitals on the energy scale, qualitatively.

9. Start forming bonding, non-bonding, and antibonding interaction . Remember ONLY ORBITALS THAT HAVE THE SAME IRREDUCIBLE MULLIKEN SYMBOL AND POINT AT THE RIGHT DIRECTION CAN THEY FORM A BONDING INTERACTION.

10. Fill in the electrons

Things to remember:

The left side is the metal center. The right side of the diagram is for ligands.

Non bonding pairs do not bond. Therefore they move right into the center of the diagram, not changing height. This is for non bonding electrons and such.

Count all electrons donated by ligands and metal center before filling in the center bonding electrons.

Remember that for every non bonding orbital there will be a bonding orbital.

Symmetry adapted linear combination models will illuminate how the molecule looks while bonding. What electron densities are charged etc.

Molecular orbital diagrams are done to help us gain a better understanding of the molecule in terms of binding and interactions. There is Bond Order that is determined in a Molecular orbital diagram. The bond order describes number of bonds between a distinct pair of atoms. Lewis dot structures are used to calculate a bond order, or there is also a method using a molecular orbital diagram and it's orbitals.

Bond order = [bonding electrons - non bonding electrons] / 2

Molecular orbital diagrams also tell if a specific molecule is magnetic or not. A molecule is either paramagnetic or diamagnetic. Paramagnetic means the molecule has an electron without the opposite spin pair. This ultimately leads to the molecule displaying magnetic characteristics. Diamagnetic means all of the electrons in the molecular orbital diagram is paired. Therefore having no magnetic qualities. This is important because it shows how we can use the molecular orbital diagram as a tool to tell us real characteristics of a molecule. The molecular orbital diagrams is a tool that was created to help us understand and think of molecules binding in a new light. Ultimately it does not show the overall complexity involved in bonding, but helps provide foundations of understanding to students. ^[1]

↑ molecular orbital diagrams, November 14th, 2012.

Example Water

1. Water belongs to C2v point group. Refer to point group. Below is the character table of C2v point group

2. The orbitals of central atom, oxygen, is as followed,

3. In order to determine the irreducible representations of these s and p orbitals, the symmetry operations must be done individually on the orbitals. If the orbital does not change sign, the character returns a +1 for the symmetry operation. If the orbital changes sign, the character returns a -1 for the symmetry operation. a. S orbital has A1 for the Mulliken symbol

b. Px orbital has B1 for the Mulliken symbol

c. Py orbital has B2 for the Mulliken symbol

d. Pz orbital has A1 for the Mulliken symbol

These Mulliken symbols for the P orbitals are also given by the second to last column by the character table.

4. Since hydrogen atom belongs to the first period, there is only s orbital involved in bonding. S orbitals exhibit solely sigma bonding.

5. Now, we need to determine the reducible representation of the orbitals above. Remember, there are TWO orbitals in the figure above, so there a total of TWO SALCS. If the symmetry operation does not change the orbital position = +1 If the symmetry operation changes the orbital position = 0 If the symmetry operation does not change the position, but changes sign = -1 Remember, changing position takes precedence than changing sign

6. To determine the irreducible representation from the reducible representation, some educated guess can be done. First of all, the central S orbital must almost always involve in sigma bonding, so one of the irreducible representation must be A1. If A1 is determined, then B1 can be determined by simple arithmetic.

7. The S orbitals from hydrogen atom have -13.61eV of energy. The S orbital from oxygen has -32.38 eV of energy, and the P orbitals have -15.85 eV.

8. The scale is in electron volts (eV)

9. Now we are able to connect the lines to form the complete molecular orbital diagram

I. The A1 SALC together with A1 on the central atom form the lowest 1a1 bonding orbital and 3a1 antibonding orbital. The bonding orbital must have lower energy than both A1 from the central atom and A1 from the SALC. The antibonding orbital must have higher energy than both A1 from the central atom and A1 from the SALC.

II. The B1 SALC together with B1 on the central atom form 1b1 bonding orbital and 2b1 antiboding orbital.

III. Because 2S and 2Pz on the central atom have the same Mulliken symbol, they mix to form 2a1. The mixing lower the energy of Pz orbital.

IV. Because there is no corresponding B2 orbital of the SALC, B2 on the central becomes nonbonding.

On the right, are the corresponding figures of the orbitals. Note that 1a1 has more central atom contribution than the SALC because the bonding molecular orbital is closer in energy than the SALC. On the contrary, 2b1 has more SALC contribution than the central atom because the antibonding molecular orbital is closer in energy to the SALC.

10. On the left side, the central atom, oxygen has 6 valance electrons. Fill the electrons from bottom to top according to Hund’s Rule. On the other hand, the SALCs only have 2 electrons because each H atom donates 1 electron. As the result, there are 8 electrons in the system. Then, they can be filled from bottom to top in the molecular diagram.

Now we have a complete molecular orbital diagram of the water molecule.

Note that there are two lone pairs in water molecule in Lewis dot structure.

So, how do we locate these two lone pairs in the molecular orbital diagram? The obvious one is the one in non-bonding orbital, which is the b2 orbital. The other one would be the bonding orbital, 1a1. Even though 1a1 is a bonding orbital, it also has some non-bonding character because the S orbital on the central atom is used way less than 50% in forming the 1a1 bonding. The S orbital only contributes less than 50% of its orbital in forming 1a1 bonding orbital because it is too low in energy.

Because 1a1 and b2 are really close in energy to the S orbital and the P orbitals in oxygen, these two molecular orbitals are more oxygen in character. As the result, the electrons located in these two orbitals are located on the oxygen, not on the hydrogen. This theory corresponds really well with Lewis’ theory.

References

Figueroa, Joshua. "Character Tables, Irreducible Representations of Central Atom." Lecture 9. University of California, San Diego, La Jolla. Lecture. Figueroa, Joshua. "SALCs, Molecular Orbital Diagrams, and High Symmetry Point Groups." Lecture 10. University of California, San Diego, La Jolla. Lecture. Figueroa, Joshua. "More SALC’s Molecular Orbital Diagrams and High Symmetry Point Groups." Lecture 11. University of California, San Diego, La Jolla. Lecture. Figueroa, Joshua. "Molecular Orbital Diagrams Featuring and Pi-Bonding and High Symmetry Point Groups." Lecture 13. University of California, San Diego, La Jolla. Lecture. Figueroa, Joshua. "Degenerate Orbitals, Methods for Determining Irreducible Representations and SALCs." Lecture 14. University of California, San Diego, La Jolla. Lecture.

[Molecular_orbital_diagrams-1] ular orbital diagrams, November 14th, 2012.

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