CHEM 2060 Lecture 30: Cyclic MOs L30-1
Cyclic Molecules …let’s start with simple “molecules” using only 1s orbitals
Rule 1: For a system with n atoms, position the atoms evenly in a “circular”
arrangement.
For 3 atoms, this is an equilateral triangle.
For 4 atoms, this is a square.
For 5 atoms, this is a pentagon
For 6 atoms, this is a hexagon
etc...
Rule 2: For a system with n atoms, there are n “like” atomic orbitals (e.g., n 1s
orbitals), and therefore there will be n molecular orbitals. These will increase in
energy as the number of nodes increases.
Rule 3: The lowest energy MO has no nodes (completely bonding). The next
energy level is doubly degenerate (two MOs with identical energy, each with
one node). Every successive energy level is also doubly degenerate unless n is
an even number, in which case the highest MO (most antibonding) is unique.
Rule 4: Nodes must be evenly spaced and pass through the center!
CHEM 2060 Lecture 30: Cyclic MOs L30-2
Note that the cyclic MOs look like the linear MOs “wrapped around”.
CHEM 2060 Lecture 30: Cyclic MOs L30-3
This may seem a little esoteric (putting 3, 4 and 5 H atoms in a circle).
However it is actually incredibly useful.
As before, the circles can be turned into p orbitals very easily. (pz in this case)
These form the π- molecular orbitals of planar aromatic (or resonance stabilized)
molecules (like benzene) and control reactivity.
In benzene, each C atom contributes
one p electron to the π system.
CHEM 2060 Lecture 30: Cyclic MOs L30-4
Summary:
We can draw the MOs of some largish organics.
That’s not all – we can order them in the correct energy sequence from # of
nodes alone.
We can identify the “Frontier Orbitals” of the molecules – those that are
involved in reactivity.
But first, a sidestep: we are going to determine the relative energies of the
orbitals. This will require some math – but then I will show you how to do it
with the “magic circles”
ENERGIES
Let’s go back to good old molecular hydrogen.
CHEM 2060 Lecture 30: Cyclic MOs L30-5
Recall:
Bonding MO: ψ = C1 1s1 + C2 1s2
2
∫ ψ 2∂τ = ∫ (C11s1 + C2 1s2 ) ∂τ = 1
C12 ∫ 1s 2∂τ + C2 2 ∫ 1s 2∂τ + 2C1C2 ∫ 1s11s2∂τ = 1




=1
€
=1
S=0*
overlap integral
* The approximation that S = 0 involves a fairly substantial error in the case of
H2+. The overlap in H2+ is actually 0.585. In most cases however S = 0 is an
O.K. approximation.
CHEM 2060 Lecture 30: Cyclic MOs L30-6
2
2
So, C1 + C2 = 1
Therefore
€
therefore
€
and
and
2C12 = 1
C1 = C2 by symmetry
1
and C1 =
2
1
(1s1 +1s2 )
2
1
ψ− =
(1s1 −1s2 )
2
ψ+ =
Bonding MO wavefunction
Antibonding MO wavefunction
So bonding and antibonding MOs are
formed by addition and subtraction of AOs.
Formed by adding/subtracting generic
orbital wavefunctions φa and φb
CHEM 2060 Lecture 30: Cyclic MOs L30-7
Use 2 AOs and form 2 MOs, one bonding and one antibonding.
Bonding MO:
ψ = C aφ a + C bφ b
Let’s assume Ca = Cb, then (Ca = Cb = C)
Essentially, we are stating that there is no dipole moment and that there is an
equal contribution to the MO from both AOs.
This is reasonable for a homonuclear diatomic.
∫ ψ 2∂τ = C 2 ∫ φa 2∂τ + C 2 ∫ φb 2∂τ + 2C ∫ φaφb∂τ





=1
=1
solve
set to zero
So:
Bonding MO:
ψ+ =
Antibonding MO:
1
(φa + φb )
2
ψ− =
1
(φa − φb )
2
€
C=
1
2
CHEM 2060 Lecture 30: Cyclic MOs L30-8
How do you get the energy of a system that is described by a wavefunction?
Schrodinger Equation: Ĥ is the Hamiltonian Operator and E is the energy.
CHEM 2060 Lecture 30: Cyclic MOs L30-9
In order to get energies, we use the Schrödinger equation.
Hˆ ψ = Eψ
Multiply both sides of the equation by ψ*
(RECALL: If the wavefunctionψ includes an imaginary number, we need to
multiply by the complex conjugate ψ* to get the square of the function. If ψ
does not contain an imaginary number, then ψ simply equals ψ*.)
So:
ψ * Hˆ ψ = ψ * Eψ
Because E is a constant, we can rearrange the right hand side of the equation.
We cannot do the same for the left hand side because Ĥ is an operator.
€
So:
ψ * Hˆ ψ = E ψ * ψ

i.e., ψ 2
QUESTION: If ψ is a normalized wavefunction, we can easily solve for the
energy by doing what?
€
CHEM 2060 Lecture 30: Cyclic MOs L30-10
Taking the integral over all space, we get
ψ * ψ∂τ
∫ (ψ * Hˆ ψ )∂τ = E 
∫


=1
if normalized
Therefore:
E = ∫ (ψ * Hˆ ψ )∂τ






€
α
−13.6 eV for H 1s
The integral is an energy (-13.6 eV for H 1s) and is given the symbol α.
Okay, now here’s potential source of confusion…
€
Textbooks use ψ* to mean the complex conjugate of ψ.
But they also use ψ* to mean the antibonding MO.
To avoid this confusion, we are using:
ψ+ =bonding MO
ψ = antibonding MO
CHEM 2060 Lecture 30: Cyclic MOs L30-11
EXAMPLE: H2
For the hydrogen molecule we add the 1s orbitals (φ) on atoms Ha and Hb.
The normalized bonding MO is then:
ψ+ =
So:
Hˆ
1 φ
(
2 a
1 φ
(
2 a
+ φb )
+ φb ) = E
1 φ
(
2 a
+ φb )
€
Because
ψ+ has no imaginary numbers, we can just multiply both sides by ψ+
€
1 φ
(
2 a
+ φb ) Hˆ
1 φ
(
2 a
+ φb ) =
1 φ
(
2 a
+ φb ) E
1 φ
(
2 a
+ φb )
Remember that E is a real number and that ψ+ and φ are normalized.
Integrate both sides over all space:
€
1 ∫φ H
ˆ φa∂τ + 1 ∫ φb Hˆ φb∂τ + 1 ∫ φa Hˆ φb∂τ + 1 ∫ φb Hˆ φa∂τ
a
2 
 2  2  2 
=α
=α
=β
=β
=E
CHEM 2060 Lecture 30: Cyclic MOs L30-12
So: 12 α + 12 α + 12 β + 12 β = E
Or more succinctly:
€
E (ψ + ) = α + β
α = ∫ φa Hˆ φa∂τ
This is called the Coulomb Integral
This is numerically equivalent to the Valence Orbital Ionization Potential
€
(VOIP) of an electron in orbital φa.
β = ∫ φa Hˆ φb∂τ
This is called the Resonance Integral
This is directly related to the amount of overlap between the atomic orbitals.
NOTE:
Remember that when we normalized the wavefunction ψ
+
we made the approximation that the overlap integral S = 0
=
1 φ
(
2 a
+ φb )
A better treatment of this problem would be to optimized the overlap integral in
order to stabilize the wavefunction and get a€better (lower) energy value.
CHEM 2060 Lecture 30: Cyclic MOs L30-13
If we do the same thing for the antibonding MO:
ψ− =
1 φ
(
2 a
− φb )
We get:
€
1 ∫φ H
ˆ φa∂τ + 1 ∫ φb Hˆ φb∂τ − 1 ∫ φa Hˆ φb∂τ − 1 ∫ φb Hˆ φa∂τ
a
2 
 2  2  2 
=α
=α
=β
=β
1α + 1α − 1 β − 1 β
2
2
2
2
=E
E =α −β
HOMEWORK: Prove this to yourself.
Summary: Bonding MO energy:
Antibonding MO energy:
€
E ψ+ = α + β
( )
E ψ− = α − β
( )
=E
CHEM 2060 Lecture 30: Cyclic MOs L30-14
MO energy level diagram of H2
So we can now plot the energy levels of the bonding and antibonding MOs and
the atomic H 1s orbitals as a function of α and β.
Important points:
For this to make sense, the value of β
must be a negative number!
The value of α gives the energy of the
atomic orbitals that are being used (H 1s
orbitals in this case).
A larger absolute value of β generates a
larger gap between the bonding and the antibonding MO energies.
This makes sense because β is related to the amount of overlap between the
atomic orbitals. (Remember, we looked at this pictorially!)
CHEM 2060 Lecture 30: Cyclic MOs L30-15
Important consequences
We get more strongly bonding and more strongly antibonding orbitals when:
Δα is small … little difference in energy between atomic orbitals
β is large … more overlap between atomic orbitals.
Energy separation between b. and a.b. MOs increases with increasing overlap.
Recall – as the internuclear separation (between s orbitals) decreases, the
overlap integral S increases.
φ1
φ2
Only orbitals of like symmetry give a non-zero overlap integral, S.
Orthogonal
S=0
Non-orthogonal
S>0