CHEMISTRY 2000
Topic #1: Bonding – What Holds Atoms Together?
Spring 2010
Dr. Susan Lait
Taking MO Theory Past the Diatomics
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Molecular orbital theory is valid for all molecules. Thus far, we’ve
focused on the diatomics because it’s only convenient to manually
derive complete MO diagrams for molecules that are either very
small or highly symmetric.
It was easier for us to predict MOs for symmetric molecules (like
O2) than asymmetric molecules (like CO). For this reason, our
studies of larger molecules will be confined to highly symmetric
ones – like BeH2, CO2 and C6H6. Computers are used to predict
MOs for less symmetric molecules (“computational chemistry”).
The principles used to predict MOs and MO diagrams for the
diatomics are the same as those used for larger molecules.
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Continue to look for valence atomic orbitals with:
 COMPATIBLE SYMMETRY ( vs. )
 COMPATIBLE ENERGY (as close as possible; no farther apart
than ~ 1 Ry)
Make sure you make the same number of MOs as you had AOs!
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Keep electron distribution in MOs symmetric if molecule is symmetric.
Molecular Orbitals for BeH2
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BeH2 is the simplest triatomic molecule. In the gas phase, it is
linear (as we’d expect from the Lewis structure):
The relative energies for the atomic orbitals of Be and H are:
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1s (Be) = -9.38 Ry
2s (Be) = -0.61 Ry
2p (Be) = +0.14 Ry
▪
1s (H) = -0.99 Ry
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Which atomic orbitals will combine to make  MOs?
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Which will combine to make  MOs?
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Which will not combine at all (leaving them as either  or 
nonbonding MOs)?
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Molecular Orbitals for BeH2
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We are combining four -symmetry AOs to make four -MOs.
To predict what these four MOs will look like, we use symmetry
and an increasing number of nodes:
To predict their energies, we expect the lowest energy -MO to
be lower in energy than all AOs from which it’s made, and we
expect the highest energy -MO to be higher in energy than all
the AOs from which it’s made. We then use these two MOs to
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estimate the energies of the middle MOs.
Molecular Orbitals for BeH2
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As in the MO diagrams for diatomics, the MOs are tracked in the
middle of the diagram with the AOs at the sides. Equivalent
atoms (such as the Hs in BeH2) are grouped at one side, making
sure to include all their valence AOs (one 1s AO for each H):
2p
2s
1s
Energy
Be
H (x2)
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MO Diagram for BeH2
4
3
1
2p
2s
1s
2
Energy
1
Be
BeH2
H (x2)
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Molecular Orbitals for CO2
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CO2 may not appear much more complicated than BeH2 – both
of them being linear triatomic molecules – but it has rather
more valence orbitals to consider. As usual, we will split them
into -symmetric AOs and -symmetric AOs:
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Valence Molecular Orbitals for CO2
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The six -symmetric AOs can be subdivided into two sets:
the 2px AOs and the 2py AOs. Each of those sets can be
combined to make three MOs. We use symmetry and nodes to
predict the shapes of these MOs and their relative energies:
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Valence Molecular Orbitals for CO2
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The six -symmetric AOs must be treated as a group. They are
combined to make six -symmetric MOs. Again, we use
symmetry and nodes to predict the shapes of these MOs and
their relative energies:
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Valence MO Diagram for CO2
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A computer helps us find the relative energies of the MOs
(particularly  vs. ):


3
2p
2
Energy
2p
2s
4
3
1
2
1
C
CO2
2s
O (x2)
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Valence MO Diagram for CO2
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Use the MO diagrams on the previous pages to:
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Write the valence orbital occupancy for CO2
Sort the MOs into bonding, antibonding and nonbonding MOs. Is the
number of each type of MO consistent with CO2’s Lewis structure?
What is the main difference between the picture of CO2 provided by
MO theory and the picture of CO2 provided by the Lewis structure?
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MO Theory and Delocalized  Electrons
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Once we move beyond the linear triatomics, the -MOs quickly
become relatively complicated and difficult to predict.
e.g. the valence -MOs of BH3:
The -MOs, on the other hand, are still relatively straightforward
to predict. When we can draw multiple resonance structures for
a molecule, we are often interested in the electrons that “move”
between resonance structures. MO theory treats these
electrons as being in delocalized -MOs. Since -MOs and MOs are always treated separately (a phenomenon called
sigma-pi separation), we often ignore the -MOs and discuss
just the -MOs for large molecules with multiple resonance
structures. This is similar to our practice of ignoring core MOs
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and discussing bonding using only the valence MOs.
MO Theory and Delocalized  Electrons
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Important points to remember:
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We are approximating that, for every bond (single, double or triple),
there are two electrons in  bonding MOs. This does not mean that
there is a -MO focussed on the bonding region. It may, for
example, mean that there are three bonding -MOs holding together
three different bonds (as shown for BH3 on the previous page – the
three leftmost MOs are -bonding MOs, each containing 2 electrons)
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When discussing bond order, only  bond order can be determined
from a -MO diagram. Total bond order will typically be 1 greater.
Total bond order =  bond order +  bond order
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Recall that:
 A  orbital with lower energy than p orbitals is net bonding.
 A  orbital with the same energy as p orbitals is net nonbonding.
 A  orbital with higher energy than p orbitals is net antibonding.
Looking at the nodes tells us which atoms are pulled together and
which atoms are pulled apart by electrons in a particular MO.
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MO Theory and Delocalized  Electrons
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Before we look at delocalized  electrons, let’s construct a
couple of -MO diagrams for localized  electrons:
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Ethyne (CHCH) has two perpendicular  systems:
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Ethene (CH2CH2) has only one  system:
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Resonance and MO Theory: Benzene (C6H6)
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The classic example of delocalized  electrons and MO theory is
benzene. The Lewis structure of benzene is:
H
H
H
C
C
H
C
C
H
H
C
C
H
H
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C
C
H
H
C
C
C
C
H
H
If we assign molecular geometries to each carbon atom, we see
that benzene is a planar molecule:
=
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As usual, we divide the valence atomic orbitals into -symmetric
and -symmetric. Since -symmetric orbitals are perpendicular
to the plane/line of the molecule, only one set of 2p orbitals is
-symmetric in benzene (usually designated 2pz).
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Resonance and MO Theory: Benzene (C6H6)
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This gives us ____ -MOs and ____ -MOs.
To divide the electrons between these MOs, we recall that:
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There are 2 electrons in -MOs for every bonded pair of electrons
in the molecule. We also count 2 electrons for every lone pair of
electrons that doesn’t move * between resonance structures.
Benzene contains ____ C-C bonds, ____ C-H bonds and no lone
pairs, so it has ____ electrons in its -MOs.
Any electrons which move between resonance structures are in MOs. In benzene, ____ electrons are in different places in the two
resonance structures, so benzene has ____ electrons in its -MOs.
H
H
H
C
C
C
C
H
H
H
C
C
C
C
H
*
H
C
C
H
H
C
C
H
H
There are rare cases where lone pairs that remain in place for all reasonable resonance structures are, in fact,
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in -MOs. These exceptions can be explained by MO theory, but will not be addressed in CHEM 2000.
Resonance and MO Theory: Benzene (C6H6)
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So, we know that benzene has ____ -MOs containing ____
electrons. We also know that these -MOs were made by
combining six 2pz orbitals on the six carbon atoms.
Cyclic molecules tend to have degenerate  energy levels.
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A “quick and dirty” trick to predict the relative energy levels for this
kind of molecule is to draw a circle. Inside the circle, draw a
polygon the same shape as your molecule. Make sure the tip of
the polygon is at the bottom of the circle. The corners of the
polygon are at the same height as the relative energy levels. If
you draw a horizontal line across the middle of the circle
(representing the energy of the p atomic orbitals), all orbitals below
it are net bonding and all orbitals below it are net antibonding.
This method is a huge oversimplification but it works.
H
H
C
C
C
H
4
3
H
C
C
C
H
H
E
2
1
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Resonance and MO Theory: Benzene (C6H6)
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We’ve worked out the energy levels, but what do the MOs look
like? To answer that, we go back to our old standby method –
symmetry and nodes!
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The lowest energy -MO (1) has one nodal plane – the one cutting
through the plane of the molecule. It has no nodes perpendicular to
the plane of the molecule.
Every time you go up to the next energy level, add a node
perpendicular to the plane of the molecule, keeping them as evenly
spaced as possible. If there are two degenerate -MOs, keep their
nodes as askew as possible (perpendicular if possible, at a 45º angle
as next choice, etc.)
The highest energy -MO should be “as antibonding as possible” –
i.e. nodes between every pair of neighbouring atoms
Once you’ve determined the energy levels and the shapes of the
-MOs, the last step is to fill in the electrons. Recall that we have
____  electrons to add to the -MO diagram.
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Resonance and MO Theory: Benzene (C6H6)
H
H
C
C
C
H
H
C
E
4
C
C
H
3
H
Energy of
2p AOs
2
1
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This MO diagram confirms that benzene can’t be considered to
have “three double bonds and three single bonds”. It really does
have three bonds with bond order _____. This has been proven
experimentally: all six C-C bonds in benzene are 140 pm (whereas
pure C-C bonds are 154 pm and pure C=C bonds are 134 pm). 19
Resonance and MO Theory: Ozone (O3)
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Another molecule whose resonance structures we looked at in
CHEM 1000 was ozone. The resonance structures for ozone are:
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The molecular geometry of ozone is _____________________.
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Ozone has:
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____ -symmetric valence AOs therefore ____ valence -MOs.
____ -symmetric valence AOs therefore ____ valence -MOs.
Ozone has:
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____ electrons in valence -MOs.
____ electrons in valence  -MOs.
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Resonance and MO Theory: Ozone (O3)
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Use symmetry and nodes to work out what the ____ -MOs of
ozone look like, ranking them from lowest to highest energy.
Then, fill in the electrons to give a complete -MO diagram.
How does this MO diagram relate to the Lewis structures you
drew? Can it justify the formal charges of the atoms?
E
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Resonance and MO Theory: Nitrate (NO3-)
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The resonance structures for nitrate are:
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The molecular geometry of nitrate is _____________________.
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Nitrate has:
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____ -symmetric valence AOs therefore ____ valence -MOs.
____ -symmetric valence AOs therefore ____ valence -MOs.
Nitrate has:
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____ electrons in valence -MOs.
____ electrons in valence  -MOs.
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Resonance and MO Theory: Nitrate (NO3-)
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Use symmetry and nodes to work out what the ____ -MOs of
nitrate look like, ranking them from lowest to highest energy.
Then, fill in the electrons to give a complete -MO diagram.
How does this MO diagram relate to the Lewis structures you
drew? Can it justify the formal charges of the atoms?
E
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