Chemistry 5350
Advanced Physical Chemistry
Fall Semester 2013
Problem Assignment
October 22, 2013
Quantum Theory - Molecular Electronic Structure
1. The H2 Molecule
(a) Evaluate the energy for the two molecular orbitals (MOs)
1
1
(φH1s1 + φH1s2 )
and σ2 = √
(φH1s1 − φH1s2 )
σ1 = √
2 + 2S12
2 − 2S12
generated by combining two H1s atomic orbitals (AOs), using the equations
H11 + H12
H22 − H12
ε1 =
and ε2 =
1 + S12
1 − S12
and carry out the calculation for S12 = 0.1, 0.2, 0.6 to mimic the effect of atomic
√ separation in
the molecule. Use the parameters H11 = H22 = -13.6 eV with H12 = −1.75 S12 H11 H22 .
Explain the trend that you observe in the results.
Energy, eV
Energy of Bonding and Antibonding MOs
as a Function of Overlap
4
2
0
-2
-4
-6
-8
-10
-12
-14
-16
-18
-20
ε1
ε2
0.1
0.2
0.3
0.4
0.5
0.6
Overlap, S
(b) Calculate the value for the coefficients of the AOs for S12 = 0.6 using the two molecular orbitals
(σ1 and σ2 ) given above.
How do they differ from the values calculated for S12 = 0.3? Can you offer an explanation for
the changes?
MO Coefficient
1
2+2S12
√ 1
2−2S12
√
S12 = 0.6
0.56
1.12
S12 = 0.3
0.62
0.85
Notice, that the antibonding MO raises higher in energy than the bonding MO. In addition, the
coefficient for the antibonding MO is more sensitive to the value of the overlap, S12 .
2. Calculate the bond order in each of the following species.
Predict which of the two species in the above pairs has the higher vibrational frequency.
Bond Order =
Total Bonding Electrons - Total Antibonding Electrons
2
For the sequence H2 → N2 , the orbital energies increase in the sequence
1σg < 1σu∗ < 2σg < 2σu∗ < 1πu < 3σg < 1πg∗ < 3σu∗
Moving across the periodic table to O2 and F2 , the order of the 1πu and the 3σg changes.
1σg < 1σu∗ < 2σg < 2σu∗ < 3σg < 1πu < 1πg∗ < 3σu∗
Species Pair
(a) Li2 , Li+
2
(b) C2 , C+
2
(c) O2 , O+
2
(d) F2 , F–2
Total Bonding
Electrons
4, 3
8, 7
10, 10
10, 10
Total Antibonding
Electrons
2, 2
4, 4
6, 5
8, 9
Bond Order
1.0,
2.0,
2.0,
1.0,
0.5
1.5
2.5
0.5
Higher Vibrational
Frequency
Li2
C2
O+
2
F2
3. The Allyl Radical
The allyl radical, (C1 −C2 −C3 )• , has a delocalized π network that can be described by the Hückel
method.
(a) Derive the molecular orbital (MO) energy levels of this species using the Hückel method.
(b) Sketch what you would expect the MOs to look like, place the electrons in the energy levels
appropriate for the ground state, and classify the MOs as bonding, nonbonding, or antibonding.
φ1 is a bonding molecular orbital, φ2 is a nonbondng molecular orbital, and φ3 is an antibonding
molecular orbital.
(c) Calculate the delocalization energy for the allyl radical.
(d) Using calculated bond orders, justify whether or not the C-C bonds in the allyl radical are
−−− C2 −
−−− C3 )• .
localized like this, (C1 −C2 −C3 )• ←−→ (C1 −C2 −C3 )• , or delocalized like this, (C1 −
The π Electron Charge (qµ ) on atom µ is given by
Occupied
MOs
qµ =
X
ni c2iµ
i=1
where ni is occupation number (0, 1, 2) of MO i, and ciµ is the coefficient of atom µ in MO i.
The π Bond Order (pµν ) between two atoms µ and ν is given as
Occupied
MOs
pµν =
X
ni ciµ ciν
i=1
where ni is the occupation number (0, 1, 2) of MO i, ciµ is the coefficient of atom µ in MO i,
and ciν is the coefficient of atom ν in MO i.
Please note that in the allyl radical there is one C1 −C2 σ bond and one C2 −C3 σ bond, and
so the total bond order for the two C−C bonds is 1.71. Thus the allyl radical behaves as a
−−− C2 −
−−− C3 )• with two partial π bonds.
delocalized system, (C1 −
4. The Cycloheptatrienyl Cation.
(a) Derive the energies of the cycloheptatrienyl cation using the Hückel method, and place the
electrons in the energy levels appropriate for the ground state.
(b) How many unpaired electrons will this species have?
The cycloheptatrienyl cation, or tropylium ion, will have no unpaired electrons.
(c) Would you expect the cation, the neutral species, or the anion to be aromatic? Justify your
answer.
Use the 4n + 2 rule to determine aromaticity where n must be an integer.
For the cycloheptatrienyl cation, also called the tropylium ion, there are 6 electrons, and so
n = 2.
The neutral radical and the anion have 7 and 8 electrons, respectively. For these species there
is no value of n which will obey the 4n + 2 rule.
Thus, the cycloheptatrienyl cation is aromatic, but the corresponding neutral radical and anion
are not.