Recap: The MO diagram for CO

The diagram below is consistent with an AM1 calculation on CO
CHEMISTRY 2000
Topic #1: Bonding – What Holds Atoms Together?
Spring 2009
Prof. René Boeré
Preparing for the 6th Class
For the third lecture, please (re)read pages 447-452
and pages 462-463 (Focus on Photoelectron Spectroscopy)
The topic is: delocalized electrons
Molecular Orbitals for CO


Summary of the approximate LCAO method
Write the orbital occupancy for CO.

Draw a Lewis structure for CO, and compare the bond order
obtained by the Lewis method to the bond order obtained by MO.




Comment on the MO diagram on the previous page in relation to
the formal charges for CO.


Brain teaser: how does the MO diagram of CO harmonize with the
diagram and the
negative formal charge on C in the Lewis diagram,
coordination of the CO to heme via the carbon atom?
it is remarkably reliable - although we understand MO theory in terms of
qualitative arguments, we will wherever it is critical use MO diagrams
substantiated by high-level calculations or experimental evidence like PES
atomic orbitals can and will interact with orbitals on other nuclei which have
the same symmetry
the strongest interactions occur between those orbitals which are closest in
energy
for heteronuclear bonds, the relative order of the atomic orbitals can be
deduced by the electronegativities of the bonded elements
for heteronuclear bonds, the resulting MO's are polarized such that the
more bonding orbital is dominated by contribution from the more
electronegative element, whereas the more antibonding orbital is
dominated by contribution from the less electronegative element
Contrary to what Petrucci states, the MO diagram for CO has the orbitals in the same
order as N2 does. This has been verified experimentally. Why does this make sense?
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1
Bond properties
Summary of the approximate LCAO method (2)


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

to be useful, an MO treatment for a molecule should have both an energy
level diagram and topologically correct sketches of the resulting MO's
ultra-violet PES is an experimental "snapshot" of the inverse of a MO
diagram; sharp peaks emanate from orbitals which are essentially nonbonding; peaks with lots of fine-structure emanate from strongly bonding or
anti-bonding
g MO's
PES of molecules with degenerate orbitals containing unpaired electrons
often show two peaks for ionizations from those orbitals, corresponding to
ionizations with and without spin-pairing
the bond order for a molecule is defined as:
½  (number of electrons in bonding orbitals - number of electrons in
antibonding orbitals)
ignore non-bonding orbitals in this count! Bond order harmonizes MO
theory to the popular Lewis dot structures; partial bond-orders are
permissible in MO theory


Taking MO Theory Past the Diatomics
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

Bond enthalpy for a given pair of bonds increases are bond order increases.
Bond length decreases as bond order increases.
For a given pair of elements
elements, the bond enthalpy increases as bond length
decreases.

Molecular Orbitals for BeH2
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
y
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.

Continue to look for 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!

Keep electron distribution in MOs symmetric if molecule is symmetric.


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:



▪
1s (H) = -0.99
0 99 Ry

Which atomic orbitals will combine to make  MOs?

Which will combine to make  MOs?

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1s (Be) = -9.38
9 38 Ry
2s (Be) = -0.61 Ry
2p (Be) = +0.14 Ry
Which will not combine at all (leaving them as either  or 
nonbonding MOs)?
8
2
Molecular Orbitals for BeH2

Molecular Orbitals for BeH2
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:

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 AOs (one 1s AO for each H):
2p
2s
1s

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
MO to be higher in energy than all the AOs from
which it’s made. We then use these two MOs to estimate the energies
of the middle MOs.
Energy
1s
Be
BeH2
H (x2)
9
MO Diagram for BeH2
10
Molecular Orbitals for CO2
5

4
2p
CO2 may not appear much more complicated than BeH2 – both of them
being linear triatomic molecules – but it has rather more orbitals to
consider. As usual, we will split them into core AOs, -symmetric AOs
and -symmetric AOs:
1
2s
1s
3
Energy

2
The three core AOs can be readily dealt with (and will then be ignored
as we construct a valence MO diagram for CO2):
LARGE
energy gap
1s
1
Be
BeH2
H (x2)
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12
3
Valence Molecular Orbitals for CO2

Valence Molecular Orbitals for CO2
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:

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:
13
Valence MO Diagram for CO2
Valence MO Diagram for CO2
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A computer helps us rank the energies of the -MOs vs. -MOs:

Without the lines connecting AOs and MOs, we get:




3
3
2p
Energy
2p
2
2p
Energy
2s
1
7
C
2
1
7
2s
6
6
5
4
CO2
5
4
CO2
O (x2)
C
15
2p
2s
2s
O (x2)
16
4
Valence MO Diagram for CO2

MO Theory and Delocalized  Electrons
Use the MO diagrams on the previous pages to:
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Write the 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?
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What is the main difference between the picture of CO2 provided by
MO theory and the picture of CO2 provided by the Lewis structure?
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
l (a
( phenomenon
h
called
ll d sigma-pi
i
i separation),
i ) we often
f
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 and discussing bonding using only the valence MOs.
17
MO Theory and Delocalized  Electrons
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Resonance vs. MO Theory: Benzene (C6H6)
Important points to remember:


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
H
H
H
H
C
H

When discussing bond order, only  bond order can be determined
from a -MO diagram. Total bond order will typically be 1 greater.
C
H
C
C
C
C
C
C
C
C
C
H
H
H
C
H
H
If we assign molecular geometries to each carbon atom, we see
that benzene is a planar molecule:
=
Total bond order =  bond order +  bond order

The classic example of delocalized  electrons and MO theory is
benzene. The Lewis structure of benzene is:
H
three leftmost MOs are -bonding MOs, each containing 2 electrons)
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As a general rule:
 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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As usual, we divide the atomic orbitals into core, -symmetric
and -symmetric. Since -symmetric orbitals are perpendicular
t th
to
the plane/line
l
/li off the
th molecule,
l
l only
l one sett off 2p orbitals
bit l is
i
-symmetric in benzene (usually designated 2pz).
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5
Resonance vs. MO Theory: Benzene (C6H6)
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Resonance vs. MO Theory: Benzene (C6H6)
This gives us _____ core MOs, ____ -MOs and ____ -MOs.
To divide the electrons between these MOs, we recall that:

Core MOs are always full. If we have ____ core MOs, there are
____ electrons in those core MOs.

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
H
H
C
C
C
C
C
C
C
C
H

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.

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
H
C
H
H
C
C
H
H
* There
are cases where lone pairs that remain in place for all reasonable resonance structures are, in fact,
in -MOs. These exceptions can be explained by MO theory, but will not be addressed in CHEM 2000.
4
3
H
C
C
H
E
C
C
H
H
2
1
21
Resonance vs. MO Theory: Benzene (C6H6)
Resonance vs. MO Theory: Benzene (C6H6)
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22
We’ve worked out the energy levels, but what do the MOs look like? To
answer that, we go back to our old standbies – symmetry and nodes!

The lowest energy -MO (1) has one nodal plane – the one cutting
through 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
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
H
H
C
C
H
H
C
C
E
4
C
C
H
H
3
Energy of
2p AOs
2
1
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.

This MO diagram confirms that benzene can’t be considered to have
“h
“three
d
double
bl bonds
b d and
d three
h
single
l bonds”.
b d ” It really
ll does
d
have
h
three
h
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).
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6
Resonance vs. MO Theory: Ozone (O3)
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Resonance vs. MO Theory: Ozone (O3)
Another molecule whose resonance structures we looked at in CHEM
1000 was ozone. The resonance structures for ozone are:
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?

The molecular g
geometryy of ozone is _____________________.
Ozone has:

____ core AOs and therefore ____ core MOs.

____ -symmetric AOs and therefore ____ -MOs.

____ -symmetric AOs and therefore ____ -MOs.
E
Ozone has:

____ core electrons

____ electrons in -MOs.

____ electrons in  -MOs.
25
Resonance vs. MO Theory: Nitrate (NO3)

The resonance structures for nitrate are:

The molecular geometry of nitrate is _____________________.

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Resonance vs. MO Theory: Nitrate (NO3)
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?

Nitrate has:

____ core AOs and therefore ____ core MOs.

____ -symmetric AOs and therefore ____ -MOs.

____ -symmetric AOs and therefore ____ -MOs.
E
Nitrate has:

____ core electrons

____ electrons in -MOs.

____ electrons in  -MOs.
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