Chemistry 3830
1.
Answers to Assignment #3
Draw a qualitative correlation diagram between the MOs for the linear and cyclic polyhydrogen systems, H3 up to H6. Use
the symmetry-adapted orbitals from resource section 5 of your textbook. For a qualitative study of the orbitals and
electron occupation of the linear and cyclic polyhydrogen systems, H3 up to H6 use HyperChem calculations at the AM1
level. Correlate your results for the two versions of each molecule. Predict what the charge of the most stable (most
strongly bonded) version of each structure is (consider only neutral, 1+ and 1– charges as possibilities; higher charges
would be unrealistic in the extreme!)
My answers all have used the model builder to create the molecules. A few cases can be minimized, but others cannot, so
to be consistent I have used all model-built structures, with single-point AM1 calculations corrected only for charge and
multiplicity. H3 has an odd number of electrons, so it is most logical to consider both the monocation and the monoanion.
In the monocation, the lowest energy structure is clearly the cyclic form, since the a’ orbital is overall more bonding than
is the 1σg orbital:
cyclic
linear
H3+
e'
-6.5 eV
a'
-29.8 eV
+4.5 eV
2σg
-8.4 eV
1σu
-27.0 eV
1σg
In the monoanion, the situation is reversed, because occupancy of e’ is worse than the less antibonding 1σu. The slight
loss in bond energy for 1σg is compensated for by the bigger gains in the upper orbital; we would predict a linear structure
for the H3– molecule. NOTE: the cyclic form must be specified as a triplet ground state to get this (correct) picture. The
singlet form is a higher-energy species.
H3-
cyclic
linear
2σg
+1.5 eV
1σu
-7.7 eV
1σg
e'
+10.3 eV
-7.7 eV
+15.4 eV
a'
H4 is even-electron. Hence, we need only consider one electron count for this system. However, note again that the cyclic
form must be specified as a triplet ground state because of the degeneracy of the eu orbitals. The energetic preference is
for the linear system, because of the great stabilization of the non-bonding eu orbital as it becomes 1σu. The loss of
bonding in the a1g level on going to 1σg is small compared to this big gain; the system is more stable as linear.
H4
cyclic
+5.7 eV
linear
b2g
eu
-1.9 eV
-21.8 eV
a1g
1-10
2σu
+5.7 eV
+3.4 eV
2σg
-10.5 eV
1σu
-19.5 eV
1σg
The H5 system is again an odd-electron system. I will only present the monocation (you should try both!); interestingly
both the cation and the anion are more stable as the linear form. The preference is again caused by the conversion of the
non-bonding (weakly anti-bonding) e1’ orbitals to bonding 1σu.
cyclic
H5+
linear
e2'
-4.5 eV
e1'
-13.7 eV
-33.9 eV
a1'
+2.7 eV
+3.7 eV
+6.9 eV
3σg
2σu
2σg
-20.7 eV
1σu
-29.2 eV
1σg
H6 is again even, so there is only one possibility to consider. Here the more stable form is the cyclic form.
H6
cyclic
6.5 eV
linear
b2u
+4.6 eV
e2g
eu
-10.0 eV
-25.6 eV
a1g
+5.9 eV
+5.1 eV
3σg
3σg
+2.8 eV
2σu
-8.8 eV
2σg
-14.9 eV
1σu
-21.7 eV
1σg
So there we have the cylic and linear polyhydrogens. These AM1 results differ from simplistic one-electron LCAO theory
– the main difference is that in one-electron theory, electrons are added and removed with no change in orbital energy,
which is overly simplistic. The basic topologies are important ones that we will see repeatedly in this course.
Note also that the symmetry labels of the orbitals change from one point group to another, but the relationship between
related orbitals is easily recognized from the topologies. We will use these topological patterns at several times during this
course, and they form the basis of the s-type symmetry-adapted orbitals developed in Appendix 4.
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2.
a) Use HyperChem to build a model of H2S, and minimize it in the AM1 semi-empirical model. What are the bond
lengths and angle? Sketch the atomic energy levels and all the valence molecular orbitals in H2S using HyperChem as
a tool. Your final picture should include the energy levels of all the AO and MO’s, and topological sketches of the
MO’s. What is the bond-order expected to be in your molecule (overall bond order, and bond order per bond)?
We start with a full diagram for the AM1 minimized H2S geometry, which has quite a small bond angle. We put the
AO’s in reasonable position based on relative electronegativity:
We need to contrast the bonding in
H2S at the AM1 level of theory
H2S with that in water (in your notes). The
H SAO's
S
primary difference is that the S atomic
orbitals are closer in energy to that of H
1.7 eV
2b2
than the O atomic orbitals in water. First,
+0.9 eV
we expect the overlap to be better (more
3a1
covalence). This is actually countermanded
somewhat by the greater disparity in orbital
size, so that in fact the overall strength of
b1
-9.5 eV
-9.5 eV
the bonds is weaker in H2S, although they
2a1
-12.2 eV
are definitely less polar.
-15 eV
-15.4 eV
The lower polarity reduces the charge
1b2
disparity; sulfur is less negative than
oxygen is in water, so hydrogen sulfide is a
weaker Brønsted base than water.
What is not so easy to explain is the
greater bond angle in H2S – it is in fact due
-19.7 eV
1a1
to greater mixing between 3a1 and 2a1.
Evidence for this is that the gap between b1
Bond length = 1.32 Å; bond order = 2.5; 1.25 per S-H bond
and 2a1 is larger in H2S than in H2O (both
in AM1 and in the PES). Since lowering 2a1 only occurs by bending the molecule, the driving force for greater
bonding is the stabilization of 2a1.
b) Now force H2S to be linear (Select the bond angle; then use Build Constrain to Linear, and Model Build). Do a
single point AM1 calculation of the linear version of the molecule. Now construct a graph linking the MO’s of linear
and angular H2S (this kind of graph is known as a “Walsh” diagram). Use your diagram to explain why hydrogen
sulfide adopts the geometry it does.
Walsh diagram for H2S at the AM1 level of theory
bent
3.3 eV
1.7 eV
2b2
+0.9 eV
3a1
-9.5 eV
-12.2 eV
-15.4 eV
-1.69 eV
b1
2σu
2σg
πu
-9.7 eV
2a1
1b2
-17.6 eV
-18.6 eV
-19.7 eV
1σu
1σg
Bond length = 1.34 Å;
bond order = 2; 1 per S-H bond
1a1
3.
linear
We see that in the bent form, the orbital 2a1 is stabilized from a completely n.b. orbital of π-symmetry, lowering the
total energy of the molecule. Now the bonds are distributed between the 2a1 and 1b2 orbitals, which have partial bond
character. The PES bears this out with (A) a sharp peak at –10.6 for the completely orthogonal b1 orbital, ionization
from which does not change the bond length or angle. And (B) peaks at –13.5 and –15.5 eV for the 2a1 and 1b2
orbitals, which have vibrational fine structure, and therefore result in changes in either bond length or angle. Again, as
in H2O, ionization of a primarily non-bonding orbital such as 2a1 will still change the bond angle significantly (imagine
removing one electron from an approximately sp2 hybridized orbital containing a “lone pair”.
a) Use HyperChem to build a model of BH3, and minimize it in the AM1 semi-empirical model. What are the bond
lengths and angle? Sketch the atomic energy levels and all the valence molecular orbitals in BH3 using HyperChem as
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a tool. Your final picture should include the energy levels of all the AO and MO’s, and topological sketches of the
MO’s. What is the bond-order expected to be in your molecule (overall bond order, and bond order per bond)?
My calculations for neutral BH3 give the following energies (AM1 method): 1a1’ –21.8; 1e’ –11.9; 1a2” +1.6; 2 a1’
+2.7; 2e’ +5.8 eV. The orbitals are all double occupied up to 1e’ (6 electrons in total). The bond order is ½ [6 – 0] = 3
or a BO of 1 per B–H bond. Note that the LUMO is the unhybridized pz orbital on the Boron atom. This is the
acceptor orbital when BH3 acts as a Lewis Acid. This is the most common reaction that BX3 compounds undergo. The
bond length is 1.19 Å, and the bond angles are all 120°.
b) Now force BH3 to be pyramidal (Select all three bond angles; then use Build Constrain to tetrahedral, and Model
Build). Do a single point AM1 calculation of the pyramidal version of the molecule. Now construct a graph linking
the MO’s of planar and pyramidal BH3 (this kind of graph is known as a “Walsh” diagram). Use your diagram to
explain why boron hydride adopts the geometry it does.
In the Walsh diagram, the planar BH3 is shown at the right-hand side, and the pyramidal form is at the left. Compare
the schematic orbitals for pyramidal BH3 with your HyperChem calculations (with “ideal” tetrahedral angles H–B–H.)
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The orbitals change their label on traversing the change in bond angle, so it is important to look at the shapes to make
sure that you track the same species. The planar orbitals shown are 1a1’, degenerate 1e’ (HOMO) and empty 1a2”
(LUMO). These correspond to the following C3v orbitals: 1a1, degenerate 1e and 2a1. In the pyramidal form, the
energies are 1a1 –22.38; 1e –11.62; 2a1 +0.2 eV. This corresponds quite clearly to a spot along the Walsh diagram
shown above where the grey vertical line is drawn.
We can now say quite definitively: an AH3 molecule will be most stable in the planar form if it has six electrons,
because that will fill the 1a1’ and 1e’ orbitals and leaves the high-energy 1a2” orbital empty. However, filling this latter
orbital with either one or two electrons is expected to drive the shape towards pyramidal because of the great
stabilization that occurs for the 2a1 orbital on bending.
4.
5.
Use HyperChem to build a model of CH4, and minimize it in the AM1 semi-empirical model. What are the bond
lengths and angle? Sketch the atomic energy levels and all the valence molecular orbitals in CH4 using HyperChem as
a tool. Your final picture should include the energy levels of all the AO and MO’s, and topological sketches of the
MO’s. What is the bond-order expected to be in your molecule (overall bond order, and bond order per bond)?
Correlate your results to the PES provided (next page).
The geometry has C–H of 1.11 Å and ideal tetrahedral angles. The calculated energies (AM1 level) are: 1a1 –28.9; 1t2 –
13.3; 2t2 +4.7; 2a1 5.18 eV (note that AM1 places the 2a1 above 2t2; since the energies of empty orbitals are extremely
susceptible to calculational method, this discrepancy is immaterial.) The orbitals are double occupied to the 1t2 level (8
electrons in total). Note the “t” type symmetry label, indicating triply degenerate sets of wavefunctions. Such symmetry is
only seen for cubic and higher point groups. The bond order is ½
[8 – 0] = 4 or BO=1 per C–H bond. However, contrary to simple
VB theory, the four C–H bonds in methane are not identical. 6
electrons are in the HOMO, and two in the much-lower lying 1a1
orbital. For verification of this surprising result, we turn to the
UV-PES spectrum shown at right. Indeed, there is a large peak
(indicating lots of emitted electrons) corresponding to ionization
from a level at about –13 eV, and a small peak (fewer electrons)
corresponding to a –23 eV level. This is an amazing fit to the
AM1 calculational result, and clearly shows the correctness of the
delocalized MO description of the bonding in methane. In fact,
the VB orbitals we commonly use are non-orthogonal, and thus
not strictly valid solutions to the Schrödinger equation. They can
be obtained by taking linear combinations of the four orbitals
shown in the diagram above.
Use HyperChem to build a model of PH3, and minimize it in the AM1 semi-empirical model. What are the bond
lengths and angles? Sketch the atomic energy levels and all the valence molecular orbitals in PH3 using HyperChem
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as a tool. Your final picture should include the energy levels of all the AO and MO’s, and topological sketches of the
MO’s. What is the bond-order expected to be in your molecule (overall bond order, and bond order per bond)?
PH3 at the AM1 level of theory
H SAO's
N
2e
2e
3a1
e
3a1
a1
2a1
2a1
-10 eV
-10.4 eV
1e
-18 eV
-13.0 eV
1e
-21.3 eV
1a1
1a1
Bond length = 1.36 Å; bond angle 96.4 deg; bond order = 3; 1 per P-H bond
In PH3, the bond angles are much sharper than they are in NH3. In this way, it closely resembles H2S compared to
H2O. This results in the 2a1 orbital becoming a very clearly non-bonding orbital with a high degree of phosphorus s
character.
Again, just as in the H2S case, this pyramidalization in which the 2a1 orbital becomes stabilized (see the Walsh diagram
for question 3) is actually driven by mixing with the 3a1 level. As the 3a1 becomes more strongly antibonding (it is
indeed extremely antibonding in PH3, 2a1 becomes more an more a lone-pair, non-bonding type of orbital, despite the
fact that it is becoming stabilized. The strongly localized lone-pair character of the LUMO accounts for the excellent
Lewis basicity of phosphine. Its unusually low Brønsted basicity is easily explained by considering the
electronegativity of P and H, which are almost identical. Thus the charge calculated on P is only –0.04 by AM1, versus
–0.4 units on the nitrogen atom of ammonia.
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6.
Multiple bonds. Use HyperChem to build a model of CO2, and minimize it in the AM1 semi-empirical model. What
are the bond lengths and angle? Sketch the atomic energy levels and all the valence molecular orbitals in CO2 using
HyperChem as a tool. Your final picture should include the energy levels of all the AO and MO’s, and topological
sketches of the MO’s. What is the bond-order expected to be in your molecule (overall bond order, and bond order per
bond)? Correlate your results to the PES provided (next page).
A schematic orbital
energy level diagram
is provided on this
page, along with the
PES for reference.
The shapes of the
orbitals are given on
the diagram on the
next page.
Both AM1 and the
PES tell us that the
first order interaction
diagram shown here
must be modified.
Thus 1σg and 1σu both
gain bond character
and hence 2σg and 2σu
lose bond character to
become primarily nonbonding (lone pair)
orbitals.
In
this
linear
molecule, sharp PES
bands mean nonbonding MO’s, and
broad or split bands
mean bonding or antibonding
orbitals.
There is excellent
agreement
between
the
AM1
orbital
energies and the PES
data. Thus the 1πg
orbitals are completely
non-bonding,
and
correspond to p-type lone pair orbitals
localized on oxygen.
1πu is a doubly
degenerate set of π orbitals spreading over the
whole OCO linkage.
The bond order is ½ [8 – 0] = 4, or two per
C=O bond.
However, unlike the VB
treatment, there are two π bonds extending
over the whole length of the molecule!
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8-10
7.
More multiple bonds. Use HyperChem to build a model of ethene, and minimize it in the AM1 semi-empirical model.
What are the bond lengths and angle? Now construct an energy level diagram for ethene in the following manner: (1)
use AM1 to model a “carbene” fragment, i.e. CH2. Specify a triplet ground state. Answer “ignore” to the warning
about minimizing a triplet – it works well enough in this case. Then construct a diagram that correlates the orbitals in
two carbene fragments with the orbitals in ethene. Note that the symmetry labels will not be the same, since the point
groups of carbene and ethene are different. Therefore use your own judgment to decide which orbitals are symmetry
related to one another. In this way, sketch the atomic energy levels and all the valence molecular orbitals in ethene
using HyperChem as a tool. Your final picture should include the energy levels of all the AO and MO’s, and
topological sketches of the MO’s. What is the bond-order expected to be in your molecule (overall bond order, and
bond order per bond)? Correlate your results to the PES provided (next page).
The two CH2 fragments. In the correct triplet ground state, each has doubly-filled 1a1 and 1b2 orbitals, and singly filled 2a1 and
1b1 orbitals. Now arrange these on either side of an interaction diagram, and combine them based on the shape of the fragment
orbitals. The labels will change because we are going to move from C2v to D2h point groups!
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Ethene has 12 electrons in 6 doubly-filled orbitals. The bonding between C–H and C-C is mixed together in the
delocalized MO approach. The bond order is ½ [12 – 0] = 6, which correspond to the five σ and one π of the VB
description of ethene.
There is remarkable agreement between AM1 and the PES spectrum. All bands are broad due to bonding character,
but the first band which is ionization from the π-HOMO is sharper than the others, in accordance with the idea that the
π-bond is less-strongly bonding than the σ-bonds.
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