Chemistry 2000 (Spring 2010)
Problem Set #1: Molecular Orbital Theory and Diatomic Molecules
Solutions
Answers to Questions in Petrucci (only those w/out answers at the back of the book)
11.28. If the σ2p MO was lower energy than the two π2p MOs, B2 would be diamagnetic since
the final two valence electrons would both be placed in the lower energy σ2p MO. Since
B2 is paramagnetic, the final two valence electrons must be placed in the two degenerate
π2p MOs which means that these two MOs must be lower in energy than all of the other
MOs made from the 2p AOs.
11.30. No, it is not correct to say that the bond energy always decreases when a diatomic
molecule loses an electron. F2 and O2 are counterexamples to this point. When a
molecule loses an electron, it will come from the highest occupied molecular orbital. In
both O2 and F2, this MO is an antibonding MO. Removing an electron from an
antibonding MO *increases* the bond energy.
11.32.
(a)
The C2+ MO diagram should resemble the diagram for C2 on p.444 of the text, but with
one less electron. (The format used in the lecture notes or on p.443 is ideal.)
The O2- MO diagram should resemble the diagram for O2 but with one more electron.
The F2+ MO diagram should resemble the diagram for F2 but with one less electron.
The NO+ MO diagram should resemble the diagram for N2 (except that the AOs for O are
lower energy than the AOs for N).
O2- has a bond order of 1.5
(b)
C2+ has a bond order of 1.5
F2+ has a bond order of 1.5
NO+ has a bond order of 3
All four have positive bond orders and can therefore be stable.
(c)
NO+ is diamagnetic. The other three species are paramagnetic, and they each have one
unpaired electron.
11.34. NO+, CO and CN- are isoelectronic since they each have 10 valence electrons.
CN+ and BN are isoelectronic since they each have 8 valence electrons.
11.36.
CN- has a bond order of 3.
(a)
CO+ has a bond order of 2.5.
(b)
CN is paramagnetic.
CO+ is diamagnetic.
(c)
CO+ should have a longer bond length as it has a smaller bond order.
11.55. NeF would have a bond order of 0.5. NeF+ would have a bond order of 1. NeF- would
have a bond order of 0. It’s probably that the spectroscopists observed either NeF+ or
NeF since those two species have positive bond orders. (NeF+ is more likely.)
11.67. Pi bonding MOs have nodal planes; however, these nodes lie along the internuclear axis
NOT perpendicular to it. Pi bonding MOs still have more electron density between the
two nuclei than the atomic orbitals from which they’re made.
Antibonding MOs have nodal planes PERPENDICULAR to the internuclear axis and less
electron density between the two nuclei (compared to the atomic orbitals).
Additional Practice Problems
1.
Draw the potential energy curve for a diatomic molecule. Clearly label the bond
dissociation energy and equilibrium bond length on your drawing.
Note that the bond dissociation energy (a positive value since energy is required to break
a bond) is equal to the difference between E = 0 and the lowest energy vibrational energy
level (close to the bottom of the energy curve but not exactly at the bottom).
2.
(a)
Two 3d orbitals can overlap in either a σ fashion or in a π fashion.
Show how two 3d orbitals can have σ overlap. Draw the resulting molecular orbitals.
(b)
3dz2 orbitals used because bond is, by definition, along the z axis
Show how two 3d orbitals can have π overlap. Draw the resulting molecular orbitals.
These are the 3dxz or 3dyz orbitals (since the bond is along the z-axis).
3.
(a)
For a diatomic molecule of oxygen (O2), we define the bond as lying along the z axis.
Draw the valence MO diagram for O2. Include labels for the molecular orbitals.
4σ ∗
2π ∗
2p
2p
1π
Energy
3σ
2σ ∗
2s
2s
1σ
O
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
O2
O
Which valence atomic orbitals combine to form σ MOs in O2? Be specific.
2s(OA) and 2s(OB) combine to make 1σ and 2σ*
2pz(OA) and 2pz(OB) combine to make 3σ and 4σ*
Which valence atomic orbitals combine to form π MOs in O2? Be specific.
2px(OA) and 2px(OB) combine to make one pair of 1π and 2π*
2py(OA) and 2py(OB) combine to make one pair of 1π and 2π*
2
* 2
2
4
* 2
Write the valence orbital occupancy for O2. (1σ) (2σ ) (3σ) (1π) (2π )
Is O2 diamagnetic or paramagnetic? paramagnetic
What is the net σ bond order for this molecule?
1
What is the net π bond order for this molecule?
1
What is the overall bond order for this molecule? 2
How many electrons must be added to O2 to reduce the bond order to zero? If this
number of electrons is added, what product(s) will be formed?
If four electrons are added to O2, the bond order will be reduced to zero. This will
generate two oxide anions: O2 + 4 e- → 2 O2-
For questions 4 and 5, you will want to
sketch the MO diagram for N2 in order to
work out the orbital occupancies. There is
no need to draw out the same diagram twice
(once for each question).
Remember that N2+ has one less electron
than N2 and N2- has one more electron than
N2.
4σ ∗
2π ∗
2p
2p
Energy
3σ
1π
2σ ∗
2s
2s
1σ
N
4.
N2
N
The bond dissociation enthalpies for N2 and N2- are 945 kJ/mol and 765 kJ/mol
respectively. (There is only a small difference between enthalpies and energies.) Using
an argument based on MO theory, explain why N2- has a smaller bond dissociation
energy than N2.
Ground-state valence orbital occupancy of N2: (1σ)2(2σ∗)2(1π)4(3σ)2
Bond order: 3
Ground-state valence orbital occupancy of N2-: (1σ)2(2σ∗)2(1π)4(3σ)2(2π∗)1
Bond order: 5
2
Bond dissociation enthalpy increases with bond order. Since N2 has a higher bond order,
we expect it to have a higher bond dissociation enthalpy, as observed.
5.
(a)
Draw Lewis diagrams for N2+ and N2-. What bond orders would you predict from the
Lewis diagrams?
I would probably draw the Lewis diagram of N2+ as the following pair of resonance
structures:
The bond order predicted from the Lewis diagram would be 3. On the other hand, the
valence MO orbital occupancy is (1σ)2(2σ∗)2(1π)4(3σ)1, with a bond order of 5 . In this
2
case, the Lewis diagram and MO theory disagree. Note that we could draw a Lewis
diagram with a bond order of 5 as follows:
2
(The formal charge on each N atom would be + 1 2 .) However, people seldom draw
Lewis diagrams in which a single electron is shared.
(b)
Determine the bond orders for these two ions using MO theory. Do they agree with the
values obtained for your Lewis diagrams?
For N2-, we would draw the following resonance structures:
The bond order predicted from these resonance structures would be 2. The valence MO
orbital occupancy is (1σ)2(2σ∗)2(1π)4(3σ)2(2π∗)1, again with a bond order of 5 . In this
2
case too, the Lewis diagram and MO theory disagree.
Note that Lewis diagrams are not always good at describing radicals (molecules in which
there are unpaired electrons).
6.
When we draw Lewis diagrams, we ignore the core electrons and focus only on the
valence electrons. Discuss how MO theory provides support for this practice.
In MO theory, the core electrons form molecular orbitals which are either essentially
identical to the original atomic orbitals (e.g. HF), or which are non-overlapping
superpositions of core orbitals (e.g. in the homonuclear diatomics). They are in either
case not involved in bonding. This justifies focusing on the valence electrons, both in MO
theory and in Lewis diagrams.
7.
Use an MO diagram to show that Be2 should not exist.
4σ ∗
2π ∗
2p
2p
Energy
3σ
1π
2σ ∗
2s
2s
1σ
Be
2
Be2
* 2
Valence orbital occupancy for Be2 is (1σ) (2σ )
Bond order for Be2 is zero therefore Be-Be bond should not form.
Be
8.
(a)
(b)
In the gas phase, it is possible to make exotic diatomic molecules like LiF (which
normally exists in a lattice rather than as a diatomic molecule).
Atomic Orbital Energies
Li
F
1s
-4.77 Ry
1s
-51.2 Ry
2s
-0.40 Ry
2s
-2.95 Ry
2p
-1.37 Ry
*
-18
1 Ry = RH = 2.179×10 J
The above energies were determined experimentally. Name and briefly describe the
technique used to make these measurements.
These energies would be obtained by photoelectron spectroscopy.
High-energy photons ionize a sample. Part of the energy of the photon goes into the
ionization process, and the rest shows up as kinetic energy of the ejected electron. The
ionization energy is −εI, where εI is the orbital energy. By measuring the energies of the
ejected electrons, we can therefore calculate the corresponding orbital energies.
Develop the MO diagram for LiF. Clearly indicate which atomic orbitals mix to make
which molecular orbitals. Show the orbital occupancy in LiF and estimate the bond
order.
The core 1s orbitals from both fluorine and lithium are much lower in energy than any of
the orbitals shown in the following diagram. They become the 1σ and 2σ MOs.
The 1σ and 1π orbitals are essentially nonbonding (consist primarily/only of atomic
orbitals from fluorine). Thus, there are two electrons in the only bonding orbital, the 2σ,
implying a bond order of 1.
Valence orbital occupancy is (1σ)2(2σ)2(1π)4
(c)
(d)
We would expect LiF to have substantial ionic character. Does your MO work agree
with this expectation? Explain.
There is still a substantial difference in energy between the fluorine 2p and lithium 2s
orbitals (almost 1 Ry). Thus, the linear combination which makes up the 2σ orbital will
have a large coefficient for the fluorine 2pz orbital and a smaller coefficient for the
lithium 2s orbital. In other words, there is more electron density on the fluorine atom in
the 2σ MO. This corresponds to a partial negative charge on F and a partial positive
charge on Li.
Does your MO work agree with a Lewis diagram for LiF?
The Lewis diagram is as follows:
There is a single bond and three non-bonding pairs on the fluorine atom, exactly as in the
MO diagram.