Molecular Orbital Theory
Thomas M. Moffett Jr., SUNY Plattsburgh, 2007
During the past century there have been many advancements in explaining molecular structure.
Today there are two dominant theories, valence bond (VB) theory and molecular orbital (MO)
theory. In VB theory the principle thought is that pairs of electrons repel each other, and
therefore orient themselves in a way that minimizes these repulsions. Covalent bonds are formed
by overlapping of atomic orbitals. Thus the electrons that form a bond between two atoms are
localized between the atoms.
In MO theory atomic orbitals are combined into new orbitals, called molecular orbitals (MOs).
While MOs are generated from atomic orbitals (1s, s2, 2p, etc.), they need not be localized
between two atoms like covalent bonds are in VB theory. The number of MOs formed is equal to
the sum of atomic orbitals (AOs) in all the atoms comprising the molecule. Thus if a hydrogen
atom is described by three AOs, the H2 molecule would be described by six MOs.
Molecular orbitals can be classified as bonding or antibonding. Depending on how the AOs are
combined the MO will either have an increase in electron density between two atoms (bonding),
or a decrease in electron density (antibonding). More specifically antibonding orbitals contain a
node between atoms. A node is an area with zero electron density. MOs are also classified
according to their orientation to the atoms making them up. MOs that have their densities
centered on the line connecting two atoms together are sigma orbitals. Denoted as (bonding)
and * (antibonding)
Figure 1: Sigma bonding (left) and antibonding (right) MOs in F2
Molecular orbitals that have their densities above and below the line connecting the atoms are pi
orbitals, (bonding) and * (antibonding).
Figure 2: Pi bonding (left) and antibonding (right) MOs in F2
When atomic orbitals combine to form bonding MOs, the resulting MO is more stable than the
AOs from which it is formed, thus the MO is lower in energy. Antibonding MOs are less stable
than the AOs, and therefore have a higher energy. Figure 3 shows how MOs may be formed
from AOs for a diatomic molecule.
When electrons fill MOs they obey the Aufbau principle,
the Pauli Exclusion Principle, and Hunds Rule – just like
filling atomic orbitals.
While figure 3 shows each MO forming from the
combination of 2 AOs, reality is of course a little more
complicated. MOs are a combination of many AOs, the
closer in energy they are the more they mix. With some
diatomic molecules the 2s and 2p sublevels are close
enough in energy to allow for mixing when forming
MOs. When this happens the ordering of the 2p and 2p
MOs reverses ( 2p is lower in energy).
Figure 3: AOs combining to form MOs in a
diatomic molecule
Bond order is the number of pairs of electrons shared
between two atoms. To calculate bond order from a MO
electronic configuration, it is necessary to determine the
number of electrons in bonding MOs and the number of
electrons in antibonding MOs.
(numberof bondinge- ) - (numberof antibonding e- )
BondOrder=
2
(1)
A bond order of one is equivalent to a single bond, a bond order of two is a double bond etc.
When a molecule contains an equal number of bonding and antibonding electrons it has a bond
order of zero, thus indicating that the molecule is not stable and won’t form.
HOMO and LUMO
The most important MOs are the highest (energy) occupied molecular orbital (HOMO) and the
lowest (energy) unoccupied molecular orbital (LUMO). These are the orbitals that are most
likely to be involved in any chemical reactions. For this reason, MOs are generally referenced
from the HOMO or LUMO orbital. That is, the occupied orbital with the second highest energy
is referred to as HOMO-1. LUMO+1 would represent the second lowest unoccupied orbital.
One clear advantage of MO theory over VB theory is that electrons can be delocalized. That
means that the electrons in an MO are not confined to the space between two atoms. One way of
thinking about delocalization is that it is similar to resonance energy from VB theory. Figure 4
shows the HOMO and LUMO for trinitrotoluene (TNT), notice how the orbitals are spread
throughout the whole molecule. In general the more delocalized the orbitals are, the more stable
the molecule is (lower energy).
Figure 4: HOMO (left) and LUMO (right) for trinitrotoluene
Multiplicity and Magnetism
Electrons can have one of two spins alpha (ms = +1/2) or beta (ms = -1/2). Spinning electrons will
interact with a magnetic field (in opposite ways). When two electrons are paired in an orbital the
magnetic effects are cancelled. Atoms or molecules that contain unpaired electrons are weakly
attracted to magnetic fields, and are said to be paramagnetic. Atoms or molecules in which all the
electrons are paired are not attracted to magnetic fields, and are said to be diamagnetic.
Multiplicity is a property that describes an atom’s or molecule’s electronic structure. The total
number of alpha and beta electrons in an atom/molecule are needed to calculate multiplicity.
Before calculating multiplicity you first calculate total spin (S).
S = (# of alpha electrons)(+1/2) + (# of beta electrons)(-1/2)
(2)
Multiplicity = |2S| + 1
(3)
Molecules that contain only paired electrons would have a total spin of 0 (all the alpha electrons
cancel all of the beta electrons), and a multiplicity of one. A multiplicity of 1 is referred to as a
singlet, two is a doublet, etc. Since a pair of electrons in an orbital will cancel out, it is only
necessary to determine the number of unpaired electrons. Thus any molecule with 1 unpaired
electron would be a doublet, any with two would be a triplet, etc.
Energy Profiles and Stearic Hindrance
In general the atoms in molecules are arranged in a way that minimizes interactions between
atoms. In a molecule such as n-butane that contains four carbon atoms in a chain (CH3-CH2-CH2CH3), the two end carbons are most stable when they are as far away from each other as possible.
This orientation minimizes stearic hindrance. Stearic hindrance is the repulsion that occurs when
two electron clouds are trying to occupy the same space.
The most stable conformation of n-butane is called the staggered conformation and has a
dihedral angle of 180 o. A dihedral angle is the angle formed by four atoms (actually the angle
formed between two planes formed by the atoms). In this case the four carbon atoms. When
looking down the bond between the second and the third carbon, the two end carbons are
opposite of each other. The least stable conformation is referred to as eclipsed and has 0 o
dihedral angle.
Figure 5: Eclipsed (left) and staggered (right) conformations
of n-butane
Figure 6: Energy profile for n-butane
By incrementally rotating the C2-C3 bond in n-butane, and calculating the energy for each
increment it is possible to create an energy profile. The energy profile (figure 6) clearly shows
that the staggered conformation is the most stable (lowest in energy). Since the C2-C3 bond is a
single bond it is free to rotate, although free might not be the correct word. In fact it takes a small
amount of energy for the bond to rotate, this is referred to as the rotation barrier.
Rotation Barrier = Eleast stable conformation – Emost stable conformation
(4)
ab initio Calculations
The calculations that are used in this lab are so called ab initio calculations. The term ab initio
refers to “first principles”, meaning that the computer will attempt to approximate a solution to
the Schrödinger equation using only the equations of quantum mechanics and a few physical
constants (Planck’s constant, speed of light, etc.) In a calculation the computer will attempt to
optimize a set of MOs for a given geometry. Every change in geometry will result in a slightly
different set of MOs.
In this lab you will utilize several types of calculations, geometry optimizations and single point
energy calculations. In a geometry optimization, the computer will attempt to calculate the most
stable geometric arrangement of the atoms, and the most stable set of MOs for that geometry. In
a single point calculation the computer will optimize the MOs for the given geometry.
Saving Images
Pictures speak louder than words, so you may want to save pictures of your molecules (MOs)
and include them in your lab report. To do so, first display the molecules on the screen as you
want them to appear in your report. Then click on File Save As, in the box for “Save as type:”,
select either “Bitmap file” or “JPEG file”. You can then transfer the files to your flash drive or
email them to yourself. In general bitmaps are better quality and will look better when inserted
into a word document. Of course bitmaps are also larger files.
Procedure:
A. VB Failure for O2
1.) Draw a Lewis structure for the oxygen molecule. From your structure predict whether
oxygen is diamagnetic or paramagnetic. In a Lewis structure each pair of electrons
(bonding and lone pairs) contains one alpha and one beta electron.
2.) Build O2.
3.) Select Setup
Calculations.
4.) The first line of the window should read “Calculate: Equilibrium Geometry at Ground
State with Hartree-Fock 3-21G”. We will refer to this as an HF/3-21G geometry
optimization, HF represents the method, and 3-21G indicates the number of mathematical
functions describing each atom. In the box for multiplicity, singlet should be selected.
Click on Submit (you will now be prompted to save your file).
5.) When the calculation is finished measure the bond length.
6.) Click on Display
Properties, a new window will appear. Record the Energy.
7.) Setup another HF/3-21G geometry optimization, this time select triplet for multiplicity.
Submit the calculation and repeat steps 5 and 6.
8.) Close the molecule.
B. Molecular Orbitals of N2
1.) Build the nitrogen molecule.
2.) Setup an HF/3-21G geometry optimization.
3.) In the “Print” section select the box for “Orbitals & Energies”.
4.) Click Submit, the program will prompt you to save your molecule.
5.) When the calculation finishes click on Display Surfaces. Click on Add, select HOMO
from the surface list, click on OK. The HOMO surface should now be listed in the
Surfaces window, click on the yellow box to display the surface. Determine if the MO is
bonding or antibonding, and if it is a sigma or pi MO.
6.) Repeat step 5 for all occupied MOs, (HOMO-1, HOMO-2, … HOMO-6) and the three
lowest unoccupied MOs (LUMO, LUMO+1, LUMO+2). When examining the MOs
make sure that you are looking at only one at a time (make sure you have only one yellow
box selected).
7.) In order to draw an accurate MO diagram we need to determine the energies of the MOs.
Click on Display Output, scroll down until you come to the section titled “Closed
Shell Molecular Orbital Coefficients”. See the supplemental handout on Angel for
information on reading the output file. Record the energies of all of the MOs examined in
steps 5 and 6.
8.) Construct an MO diagram for nitrogen, every occupied MO contains two electrons (one
alpha, one beta). Compare this diagram to the one in your textbook, are they in
agreement?
9.) Close the molecule.
C. Energy Profiles of n-Butane and Hydrazine
Hydrazine (H2N-NH2) is made up of two sp3 nitrogen atoms connected by a single bond. The
energy of hydrazine depends on the orientation the two NH2 groups with respect to one another.
In this part of the lab you will create an energy profile for hydrazine by rotating the N-N bond
and recording energies.
1.) Build hydrazine. Select the N-N bond (a red marker should now appear on the bond),
rotate the bond by holding down the alt key and the left mouse button. Put the molecule
into an eclipsed conformation (figure 1). (do not minimize). Select View.
Figure 7: eclipsed conformation of hydrazine
2.) Constrain the dihedral angle by selecting
Geometry Constrain Dihedral , be sure to click the
atoms in the following order, H-N-N-H. Click on the lock
and set the dihedral angle to 0.0 o, hit enter.
3.) Open the properties window (Display Properties), click
on the constraint marker. In the properties window, check
the box for dynamic. Set the values from 0.0 o to 180 o.
Set the number of steps to 19. Be sure to hit the enter
button after making each change.
4.) Setup an HF/6-31G* calculation. Instead of Geometry
optimization, select “energy profile”. Make sure that the
box for global calculations is selected. Submit the
calculation.
5.) When the job is finished, close your hydrazine molecule. A new file named
filename.M001.Spartan has been created, open this file. In the lower left hand corner of
the screen are controls that will allow you to animate your molecule, or step through one
molecule at a time.
6.) Open the properties window, then click on the dihedral constraint. Click on the red P
( ), this will post the calculated dihedral angles to a spread sheet.
7.) Open the spreadsheet by selecting Display Spreadsheet, a spreadsheet window should
appear with a column listing the molecules and a second column listing dihedral angles
(constraint 1). The header labeled constraint 1 can be renamed by clicking on the cell and
typing in dihedral angle.
8.) Add a column for relative energies by selecting Add, and then choosing “Rel E” from the
list of selections. Save your spreadsheet data.
9.) An energy surface can be created by selecting Display Plots. Select dihedral angle for
the X-axis and Rel E for the Y-axis. A plot of energy vs. dihedral angle should now
appear on the screen.
10.) You can print out the image on the screen by selecting File Print. Make sure that the
graph is clearly visible before printing.
Build n-butane (CH3CH2CH2CH3) and set up an HF/3-21G energy profile calculation. Constrain
the C-C-C-C dihedral angle from 0.0 o to 180.0 o, set the calculation to 19 steps. When the
calculation finishes repeat steps 5-10 from the hydrazine procedure.
To examine why the most stable structures of hydrazine and n-butane are so different it is
necessary to look at the HOMO surface on hydrazine.
1.) Open your hydrazine molecule (the energy profile file). Add A HOMO surface, be sure to
make sure the global calculations box is checked. Set up a single point HF/3-21G
calculation. To run a single point calculation, select “Energy" instead of “Equilibrium
Geometry” when setting up the calculation. Again make sure the global calculations box
is selected.
2.) When the calculation is completed you’ll be able to display the HOMO surface and then
cycle through the structures. Pay particular attention to the surface at the minimum.
D. Resonance Effects
The carbonate ion (CO32-) can be drawn with the following resonance structures:
Figure 8: Resonance structures of the carbonate ion.
In reality the carbonate ion is a combination of all three structures. All of the carbon-oxygen
bonds are equivalent, and the charge is equally distributed among all the atoms.
1.) Build the carbonate ion, use the delete ( ) function to remove any unused valence
spaces. If you don’t delete them a hydrogen atom will automatically be added.
2.) Setup an HF/3-21G geometry optimization calculation. In the box for total charge select
dianion. Submit the calculation.
3.) When the calculation finishes measure the bond lengths, and record the energy.
4.) Determine the charges on all the atoms. Click on an atom and record the electrostatic
charge in the properties window. Repeat this for all four atoms.
5.) Display the HOMO surface, the HOMO represents the MO in which the extra two
electrons have been added.
Now you will build another carbonate ion, in which we will remove the resonance effects.
6.) Build the carbonate ion as before. Set the single bonds to 1.44 by selecting Geometry
Measure Distance, click on a single bond. In the text box at the bottom right corner
of the screen enter the 1.44 and press enter. Set the other single bond to 1.44 and the
double bond to 1.21 .
7.) Set up a single point HF/3-21G calculation. When the calculation finishes record all of
the same data that you did in steps 3-5.
Data Analysis and Questions
Part A
• Draw the Lewis structure for O2. Determine if O2 is paramagnetic or diamagnetic from
your structure.
• Record the energy of the singlet and triplet forms of O2.
• Draw the MO diagram for O2. Determine if O2 is paramagnetic or diamagnetic from your
diagram.
• Calculate the bond order.
Part B
• Classify the first 10 MOs of the N2 molecule (bonding or anti-bonding, and sigma or pi).
• Record the energies of the first 10 MOs.
• Draw the MO diagram for N2. Does this diagram agree with the diagram in your
textbook?
• Calculate the bond order.
Part C
• Print out the energy profiles for hydrazine and n-butane.
• Calculate the barrier to rotation in both molecules.
Part D
• Record all bond lengths, and the energy for both molecules.
• Calculate the delocalization energy (the difference in energy between the two structures).
Q1) Explain what is wrong with the Lewis Structure for O2. Does the MO diagram correctly
predict oxygen’s magnetic properties?
Q2) According to VB theory, the most stable conformation of hydrazine would be staggered.
Does MO theory agree with this, if not explain why?
Q3) Comment on the differences between the HOMO surfaces of the two carbonate ions. How
does the surface correlate with the observed charges?
Q4) VB theory can quickly and easily be used to draw structures of very complicated molecules.
While it’s possible to draw MOs for only the simplest molecules, for anything more
complicated computers are needed to calculate the MOs. At the same time MO theory
correctly predicts many properties that VB theory can’t. Discuss whether you would
consider one theory or the other to be more useful, or perhaps they are equally useful.
(There is not a correct answer to this question, we just want you to evaluate the benefits and
limits to both theories.)
Q5) If a hydrogen atom is described with 5 atomic orbitals and an oxygen atom is described by
9, how many MOs would be formed for the calculation of a water molecule? How many
would be occupied MOs (water is a singlet).
Pre-Lab Questions
1.) Define the following terms:
HOMO –
LUMO –
2.) Explain Hund’s rule, the Aufbau principle, and the Pauli exclusion principle.
3.) Two possible MO diagrams for B2 are posted below.
a.) Given that B2 is a triplet, which diagram is correct?
b.) Predict whether B2 would be paramagnetic or diamagnetic.
c.) Calculate the bond order.