thomas m. moffett jr- molecular orbital theory

8/3/2019 Thomas M. Moffett Jr- Molecular Orbital Theory

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Molecular Orbital TheoryThomas 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 formedby 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 hydrogenatom 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 arecombined 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 classifiedaccording 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


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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 stablethan 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 morecomplicated. 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 formingMOs. When this happens the ordering of the 2p and 2p

MOs reverses (

2p is lower in energy).

Bond order is the number of pairs of electrons shared

between two atoms. To calculate bond order from a MO

Figure 3: AOs combining to form MOs ina electronic configuration, it is necessary to determine the

diatomic molecule number of electrons in bonding MOs and the number of

electrons in antibonding MOs.

2

)egantibondinof(number-)ebondingof(numberOrderBond

--

= (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 wont form.

HOMO and LUMO

The most important MOs are the highest (energy) occupied molecular orbital (HOMO) and thelowest (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 energyis 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. Thatmeans 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 stablethe molecule is (lower energy).


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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 willinteract 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 weaklyattracted 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 atoms or molecules 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 asinglet, 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 betweenatoms. In a molecule such as n-butane that contains four carbon atoms in a chain (CH 3-CH2-CH2-

CH3), 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 whentwo electron clouds are trying to occupy the same space.


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The most stable conformation of n-butane is called the staggered conformation and has a

dihedral angle of 180o. 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 0o

dihedral angle.

Figure 5: Eclipsed (left) and staggered (right) conformations Figure 6: Energy profile for n-butaneof 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 asingle 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 Schrdinger equation using only the equations of quantum mechanics and a few physicalconstants (Plancks 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 moststable geometric arrangement of the atoms, and the most stable set of MOs for that geometry. Ina 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


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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 oremail 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 whetheroxygen 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 SetupCalculations.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 DisplayProperties, 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.) ClickSubmit, the program will prompt you to save your molecule.5.) When the calculation finishes click on DisplaySurfaces. Click on Add, select HOMO

from the surface list, click on OK. The HOMO surface should now be listed in theSurfaces 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.


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6.) Repeat step 5 for all occupied MOs, (HOMO-1, HOMO-2, HOMO-6) and the threelowest 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 DisplayOutput, scroll down until you come to the section titled ClosedShell 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 (onealpha, 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. Theenergy 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.

2.) Constrain the dihedral angle by selectingGeometryConstrain Dihedral , be sure to click theatoms 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 (DisplayProperties), clickon the constraint marker. In the properties window, checkthe 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 theFigure 7: eclipsed conformation of hydrazine calculation.

5.) When the job is finished, close your hydrazine molecule. A new file namedfilename.M001.Spartan has been created, open this file. In the lower left hand corner of


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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 thelist of selections. Save your spreadsheet data.

9.) An energy surface can be created by selecting DisplayPlots. Select dihedral angle forthe X-axis and Rel E for the Y-axis. A plot of energy vs. dihedral angle should nowappear on the screen.

10.)You can print out the image on the screen by selecting FilePrint. Make sure that thegraph is clearly visible before printing.

Build n-butane (CH3CH2CH2CH3) and set up an HF/3-21G energy profile calculation. Constrainthe C-C-C-C dihedral angle from 0.0

oto 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 tomake sure the global calculations box is checked. Set up a single point HF/3-21Gcalculation. 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 youll be able to display the HOMO surface and thencycle 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.


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1.) Build the carbonate ion, use the delete ( ) function to remove any unused valencespaces. If you dont delete them a hydrogen atom will automatically be added.

2.) Setup an HF/3-21G geometry optimization calculation. In the box for total charge selectdianion. 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 twoelectrons 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 GeometryMeasure 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 thedouble bond to 1.21 .

7.) Set up a single point HF/3-21G calculation. When the calculation finishes record all ofthe 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 fromyour 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).


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Q1) Explain what is wrong with the Lewis Structure for O2. Does the MO diagram correctly

predict oxygens 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 its possible to draw MOs for only the simplest molecules, for anything morecomplicated computers are needed to calculate the MOs. At the same time MO theory

correctly predicts many properties that VB theory cant. 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 andlimits to both theories.)

Q5) If a hydrogen atom is described with 5 atomic orbitals and an oxygen atom is described by9, how many MOs would be formed for the calculation of a water molecule? How many

would be occupied MOs (water is a singlet).


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Pre-Lab Questions

1.) Define the following terms:

HOMO

LUMO

2.) Explain Hunds 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.

thomas m. moffett jr- molecular orbital theory

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