Physical 5.61 Chemistry
Lecture #31
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HCKEL MOLECULAR ORBITAL THEORY
In general, the vast majority polyatomic molecules can be thought of as
consisting of a collection of twoelectron bonds between pairs of atoms. So
the qualitative picture of and bonding and antibonding orbitals that we
developed for a diatomic like CO can be carried over give a qualitative
starting point for describing the C=O bond in acetone, for example. One
place where this qualitative picture is extremely useful is in dealing with
conjugated systems that is, molecules that contain a series of alternating
double/single bonds in their Lewis structure like 1,3,5hexatriene:
Now, youmay have been taught in previous courses that because there are
other resonance structures you can draw for this molecule, such as:
that it is better to think of the molecule as having a series of bonds of
order 1 rather than 2/1/2/1/ MO theory actually predicts this
behavior, and this prediction is one of the great successes of MO
theory as a descriptor of chemistry. In this lecture, we show how even a
very simple MO approximation describes conjugated systems.
Conjugated molecules of tend to be planar, so that we can place all the atoms
in the xy plane. Thus, the molecule will have reflection symmetry about the
zaxis:
z
Now, for diatomics, we had reflection symmetry about x and y and this gave
rise to x and y orbitals that were odd with respect to reflection and
orbitals that were even. In the same way, for planar conjugated systems the
orbitals will separate into orbitals that are even with respect to reflection
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and z orbitals that are odd with respect to reflection about z. These z
orbitals will be linear combinations of the pz orbitals on each carbon atom:
z
In trying to understand the chemistry of these compounds, it makes sense
to focus our attention on these z orbitals and ignore the orbitals. The z
orbitals turn out to be the highest occupied orbitals, with the orbitals
being more strongly bound. Thus, the forming and breaking of bonds as
implied by our resonance structures will be easier if we talk about making
and breaking bonds rather than . Thus, at a basic level, we can ignore the
existence of the orbitals and deal only with the orbitals in a qualitative
MO theory of conjugated systems. This is the basic approximation of
Hckel theory, which can be outlined in the standard 5 steps of MO theory:
1) Define a basis of atomic orbitals. Here, since we are only interested
in the z orbitals, we will be able to write out MOs as linear
combinations of the pz orbitals. If we assume there are N carbon
atoms, each contributes a pz orbital and we can write the th MOs as:
N
i
= ci pz
i =1
2) Compute the relevant matrix representations. Hckel makes some
radical approximations at this step that make the algebra much
simpler without changing the qualitative answer. We have to compute
two matrices, H and S which will involve integrals between pz orbitals
on different carbon atoms:
i
i
H ij = p z H pzj d
Sij = pz p zj d
The first approximation we make is that the pz orbitals are
orthonormal. This means that:
1 i = j
Sij =
0 i j
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Equivalently, this means S is the identity matrix, which reduces our
generalized eigenvalue problem to a normal eigenvalue problem
Hi c = E c
Hic = E Sic
The second approximation we make is to assume that any Hamiltonian
integrals vanish if they involve atoms i,j that are not nearest
neighbors. This makes some sense, because when the pz orbitals are
far apart they will have very little spatial overlap, leading to an
integrand that is nearly zero everywhere. We note also that the
diagonal (i=j) terms must all be the same because they involve the
average energy of an electron in a carbon pz orbital:
i
i
H ii = p z H p z d
Because it describes the energy of an electron on a single carbon, is
often called the onsite energy. Meanwhile, for any two nearest
neighbors, the matrix element will also be assumed to be constant:
i
H ij = pz H pzj d
i,j neigbors
This last approximation is good as long as the CC bond lengths in the
molecule are all nearly equal. If there is significant bond length
alternation (e.g. single/double/single) then this approximation can be
relaxed to allow to depend on the CC bond distance. As we will see,
allows us to describe the electron delocalization that comes from
multiple resonance structures and hence it is often called a resonance
integral. There is some debate about what the right values for the
, parameters are, but one good choice is =11.2 eV and =.7 eV.
3) Solve the generalized eigenvalue problem. Here, we almost always
need to use a computer. But because the matrices are so simple, we
can usually find the eigenvalues and eigenvectors very quickly.
4) Occupy the orbitals according to a stickdiagram. At this stage, we
note that from our N pz orbitals we will obtain N orbitals. Further,
each carbon atom has one free valence electron to contribute, for a
total of N electrons that will need to be accounted for (assuming the
molecule is neutral). Accounting for spin, then, there will be N/2
occupied molecular orbitals and N/2 unoccupied ones. For the ground
state, we of course occupy the lowest energy orbitals.
5) Compute the energy. Being a very approximate form of MO theory,
Hckel uses the noninteracting electron energy expression:
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N
Etot = Ei
i =1
where Ei are the MO eigenvalues determined in the step.
To third illustrate how we apply Hckel in practice, lets work out the energy of
benzene as an example.
1
6
2
5
3
4
1) Each of the MOs is a linear combination of 6 pz orbitals
c1
c2
6
c
i
c = 3
= ci pz
i =1
c4
c
5
c
6
2) It is relatively easy to workout the Hamiltonian. It is a 6by6 matrix.
The first rule implies that every diagonal element is :
H=
The only other nonzero terms will be between neighbors: 12, 23, 34, 45,
56 and 61. All these elements are equal to :
H=
All the rest of the elements involve nonnearest neighbors and so are zero:
5.61 Physical Chemistry
0
H=
0
0
3) Finding the eigenvalues of H is
energies:
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0 0 0
0 0 0
0 0
0 0
0 0
0 0 0
easy with a computer. We find 4 distinct
E6=2
E4=E5=
E2=E3=+
E1=+2
The lowest and highest energies are nondegenerate. The second/third and
fourth/fifth energies are degenerate with one another. With a little more
work we can get the eigenvectors. They are:
+1
+1
+1
+1
+1
+1
1
2
0
0
+2
+1
1 +1 5
1 +1 4
1 1 3
1 1 2
1 +1 1
1 +1
c6 =
c=
c=
c=
c=
c=
6 1
12 +1
4 +1
4 1
12 1
6 +1
+1
2
0
0
2
+1
1
1
+1
+1
1
+1
The pictures at the bottom illustrate the MOs by denting positive (negative)
lobes by circles whose size corresponds to the weight of that particular pz
orbital in the MO. The resulting phase pattern is very reminiscent of a
particle on a ring, where we saw that the ground state had no nodes, the
first and second excited states were degenerate (sine and cosine) and had
one node, the third and fourth were degenerate with two nodes. The one
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difference is that, in benzene the fifth excited state is the only one with
three nodes, and it is nondegenerate.
4) There are 6 electrons in benzene, so we doubly occupy the first 3 MOs:
E6=2
E4=E5=
E2=E3=+
E1=+2
5) The Hckel energy of benzene is then:
E = 2 E1 + 2 E2 + 2 E3 = 6 + 8
Now, we get to the interesting part. What does this tell us about the
bonding in benzene? Well, first we note that benzene is somewhat more
stable than a typical system with three double bonds would be. If we do
Hckel theory for ethylene, we find that a single ethylene double bond has
an energy
EC =C = 2 + 2
Thus, if benzene simply had three double bonds, we would expect it to have a
total energy of
E = 3EC =C = 6 + 6
which is off by 2. We recall that is negative, so that the electrons in
benzene are more stable than a collection of three double bonds. We call
this aromatic stabilization, and Hckel theory predicts a similar stabilization
of other cyclic conjugated systems with 4N+2 electrons. This energetic
stabilization explains in part why benzene is so unreactive as compared to
other unsaturated hydrocarbons.
We can go one step further in our analysis and look at the bond order. In
Hckel theory the bond order can be defined as:
occ
Oij ci c
j
=1
This definition incorporates the idea that, if molecular orbital has a bond
between the ith and jth carbons, then the coefficients of the MO on those
carbons should both have the same sign (e.g. we have pzi + pzj). If the orbital
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is antibonding between i and j, the coefficients should have opposite
signs(e.g. we have pzi pzj). The summand above reflects this because
ci c > 0
if ci , c have same sign
j
j
ci c < 0
j
if ci , c have opposite sign
j
Thus the formula gives a positive contribution for bonding orbitals and a
negative contribution for antibonding. The summation over the occupied
orbitals just sums up the bonding or antibonding contributions from all the
occupied MOs for the particular ijpair of carbons to get the total bond
order. Note that, in this summation, a doubly occupied orbital will appear
twice. Applying this formula to the 12 bond in benzene, we find that:
O12 2c1 =1c2 =1 + 2c1 =2 c2 =2 + 2c1 =3 c2
=3
+1 +1
+1 +2
+1 0
= 2
+ 2
+ 2
12 12
4 4
6 6
1
22
=2 +2 =
6
12 3
Thus, the C1 and C2 formally appear to share 2/3 of a bond [Recall that we
are omitting the orbitals, so the total bond order would be 1 2/3 including
the bonds]. We can repeat the same procedure for each CC bond in
benzene and we will find the same result: there are 6 equivalent bonds,
each of order 2/3. This gives us great confidence in drawing the Lewis
structure we all learned in freshman chemistry:
Youmight have expected this to give a bond order of 1/2 for each CC
bond rather than 2/3. The extra 1/6 of a bond per carbon comes directly
from the aromatic stabilization: because the molecule is more stable than
three isolated bonds by 2, this effectively adds another bond to the
system, which gets distributed equally among all six carbons, resulting in an
increased bond order. This effect can be confirmed experimentally, as
benzene has slightly shorter CC bonds than nonaromatic conjugated
systems, indicating a higher bond order between the carbons.
Just as we can use simple MO theory to describe resonance structures and
aromatic stabilization, we can also use it to describe crystal field and ligand
field states in transition metal compounds and the sp, sp2 and sp3 hybrid
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orbitals that arise in directional bonding. These results not only mean MO
theory is a useful tool in practice these discoveries have led to MO theory
becoming part of the way chemists think about molecules.