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Fall 2014
CHEM 104a
Advanced Inorganic Chemistry
Solution Set #4
Problem 1:
(a) Prepare the MO diagram for the cyanide ion, 𝐂𝐂𝐍𝐍− . Use sketches to show clearly how the
atomic orbitals interact to form MOs.
The MO diagram can be prepared nearly identically to that of CO, however with different atomic
orbital energies. As a result of the closer energy matching between C-N versus C-O, we expect
the mixing to be stronger, giving a larger bonding/antibonding stabilization/destabilization. Note
that as with CO, a significant s-p mixing occurs, resulting in a downward shift of 1σ orbitals and
an upward shift of the 2σ orbitals.
(b) What is the bond order, and how many unpaired electrons does cyanide have?
The bond order of CN- is 3. The 1σg and 1πu contribute a total of 6 bonding electrons, while the
slightly antibonding 1σu* and slightly bonding 2σg cancel each other out.
(c) Which molecular orbital of 𝐂𝐂𝐍𝐍 − would you predict to interact most strongly with a
hydrogen 1s orbital to form an H-C bond in the reaction: 𝐂𝐂𝐍𝐍 − + 𝐇𝐇 + → 𝐇𝐇𝐇𝐇𝐇𝐇? Explain.
The energy of the H+ 1s orbital is closest in energy matching to the 2σg orbital of CN, and
therefore would give the strongest mixing. Additionally, we know that H+ can only contribute an
unfilled orbital (LUMO), meaning that CN- would have to contribute an occupied MO (HOMO)
for this reaction.
(d) Many inorganic compounds contain 𝐂𝐂𝐍𝐍− as a ligand bonded to a transition metal center.
Based on the shape of the orbital used to form the H-CN bond, would 𝐂𝐂𝐍𝐍 − be more likely
to bind through an M-CN bond, M-NC bond, or both?
Similar to our analysis for CO, in which we showed that an M-CO bond would always proceed
through bonding with C, we can make similar assessments here. As noted for CO, the most
reactive orbitals of CN would be the HOMO (2σg) and the LUMO (1πg*). For both of these
orbitals the larges lobe of electron density is centered on the C side of the molecule, in part due
to the s-p mixing in forming the 2σg orbital (consider mixing 2 atomic orbitals of carbon with 1
atomic orbital of N), and the fact that the 1πg* orbital is closer in energy to the C 2p orbitals
(closer energy, more C 2p character). Though C and N are better energy matched than C and O,
metal-cyanide bonds still tend to form through a M-C bond.
Problem 2:
The first ionization energies of BF, CO, and N2 are 11.06 eV, 14.01 eV, and 15.57 eV,
respectively. Explain the increase in ionization energy for this isoelectronic series on the
basis of atomic orbital composition of the highest occupied molecular orbital (HOMO).
We can see this increase in ionization energies either through directly considering the MO
diagrams, or through intuition based on the energy matching of these diatomic molecules. In the
above MO diagrams, we plot the s and triply degenerate p orbitals for all the molecules. For N2,
the strong energy matching between the N 2p orbitals gives a very low bonding sigma orbital
(lowest HOMO). The energy is slightly pushed up due to sp mixing. For CO, the increased
energy mismatch between the C 2s and 2p orbitals with the O 2p orbitals gives a slightly higher
MO (stronger sp mixing compared to N2 pushes this orbital up higher). Finally, for BF, the
HOMO is the s-p antibonding sigma MO. This yields the lowest MO due to the antibonding
character and increased energy mismatch.
Problem 3:
(a) For the molecule 𝐍𝐍𝐇𝐇𝟐𝟐 − , construct the MO diagram for the bent conformation.
(b) Draw the Walsh Diagram for a bent and a linear 𝐍𝐍𝐇𝐇𝟐𝟐 − molecule. Should the molecule
be linear or bent? Why?
The Walsh diagram can be constructed by considering the MO energy associated with each N
atomic orbital. The orthogonality of the 2pz orbital is broken upon going to the bent geometry
(originally non-bonding in the linear form). Additionally, the s-p mixing in the bent
configuration serves to stabilize the 1σg to the 1a1 (better overlap due to s-p hybridization) and
destabilize the 3a1 compared to the 2σg*. Finally, the 2px overlap gets worse as you go from the
linear to bent configuration. All taken together, the 2a1 slight bonding orbital gives a net lower
energy for the preferred bent configuration.
(c) Draw the photoelectron spectrum of 𝐍𝐍𝐇𝐇𝟐𝟐 − .
See the solution for part (a). The 2a1 peak will be relatively narrow due to its slight bonding
(fairly non-bonding) nature.
Problem 4:
Fluorinated compounds often exhibit unique, extreme chemistries due to the high
electronegativity of fluorine. Of particular interest is a class of compounds dubbed
“superacids”. The strongest superacid is fluoroantimonic acid, 𝐒𝐒𝐒𝐒𝐅𝐅𝟔𝟔 − , also called “magic
acid” as it is 106 times stronger than 100% sulfuric acid, and is obtained from the reaction:
𝐒𝐒𝐒𝐒𝐅𝐅𝟓𝟓 + 𝟐𝟐𝟐𝟐𝟐𝟐 → 𝐒𝐒𝐒𝐒𝐅𝐅𝟔𝟔 − + 𝐇𝐇𝟐𝟐 𝐅𝐅 +
The high basicity of 𝐒𝐒𝐒𝐒𝐅𝐅𝟔𝟔 − is sufficient to produce the extremely reactive fluoronium cation
𝐇𝐇𝟐𝟐 𝐅𝐅 + , which is even able to protonate alkanes. To account for the high acidity of 𝐇𝐇𝟐𝟐 𝐅𝐅 + :
(a) Draw the MO diagram for 𝐇𝐇𝟐𝟐 𝐅𝐅 + , which adopts a bent geometry.
The fluoronium ion MO can be constructed by taking the LGOs of the H- -H fragment, the same
geometry as water, in which you get a symmetric LGO and an antisymmetric LGO, both of
similar energies due to the lack of strong mixing between the hydrogens (too far apart). In
contrast to water, the fluorine 2s is too low to contribute any appreciable mixing. As a result, two
sets of mixing occur, between the 2pz and symmetric LGO, and the 2px and antisymmetric LGO.
The overlap of the 2pz-LGO set should be slightly higher due to the more σ-type interaction (the
2px has more of a π-type interaction), giving a lower bonding, and higher antibonding MO. The
2py does not share the correct symmetry and is therefore orthogonal to the hydrogen LGOs, and
therefore remains non-bonding.
(b) Construct the MO diagram for the hydronium ion, 𝐇𝐇𝟑𝟑 𝐎𝐎+ . What does the relative
position of the lowest unoccupied molecular orbital (LUMO) tell you about the relative
acidity/reactivity of 𝐇𝐇𝟐𝟐 𝐅𝐅 + versus 𝐇𝐇𝟑𝟑 𝐎𝐎+ ?
For the hydronium ion MO, take the H3 molecular fragment, which is isolobal to NH3. The
resulting LGOs of the H3 cluster are depicted, for which you should refer to the lecture slides or
either text for a more rigorous derivation. The O 2s orbital is nearly too far to participate strongly
in bonding, and can be treated as essentially non-bonding (though experiment tells us that there
is some s-p type mixing). This means the predominant orbital mixing is between the 2px and 2py
on O with the two degenerate LGOs (they both have the doubly degenerate E irreducible
representation, therefore are also degenerate in energy), as well as the 2pz with the symmetric
LGO (A1 symmetry).
The acidity of a molecule is dictated by the position of its LUMO, as acids are electron acceptors,
with lower LUMOs being more reactive than higher ones. For H2F+, the energy match between
the H 1s and the F 2p is only fair, giving a relatively low antibonding LUMO. In comparison, the
LUMO of H3O+ is formed from O 2p interactions with much better energy matching, giving a
higher LUMO, which is less reactive and less acidic than H2F+. Additionally, the O 2s orbital is
much higher than the F 2s orbital, giving some modest s-p mixing to H3O+, which serves to raise
the LUMO even higher. As a sanity check, we can also recognize that the weaker acid is the
stronger base, which are characterized by having a high HOMO. The HOMO of H2F+ is the nonbonding F 2p orbital. The HOMO of H3O+ in contrast is the essentially non-bonding O 2p orbital,
which is higher in energy due to the starting energy of O 2p (to understand the non-bonding
character, consider the modest s-p mixing in H3O+, or make reference to the discussion of the
MO diagram of NH3 in Miessler and Tarr). Therefore, H2F+ is the stronger acid, and H3O+ is the
stronger base.
Another very strong acid is HF, which is capable of dissolving glass, SiO2. Concentrated
HF will undergo a homoassociation reaction to also produce 𝐇𝐇𝟐𝟐 𝐅𝐅 + , as well as a very stable
counter anion 𝐇𝐇𝐅𝐅𝟐𝟐 − :
𝟐𝟐𝟐𝟐𝟐𝟐 → 𝐇𝐇𝟐𝟐 𝐅𝐅 + + 𝐇𝐇𝐅𝐅𝟐𝟐 −
(c) Construct the MO diagram for the linear F-H-F- anion.
The MO diagram and its derivation is covered quite nicely in Miessler and Tarr. Some of the
highlights are: 1) While the F- -F 2s LGOs are too low to exhibit strong mixing with H, there still
exists some s-p mixing due to the favorable symmetry and overlap with the 2p LGOs (this results
in the slight downward shift of 1ag and 1b1u and the slight upward shift of 2ag, 3ag and 2b1u). 2)
One could alternatively look at the slight upshift of 2b1u as resulting from a slight anti-bonding
interaction between the two fluorines. While we typically ignore the energy changes upon LGO
mixing, the short bond distance (they’re only separated by a proton) gives slight anti-bonding
character.
(d) 𝐇𝐇𝐅𝐅𝟐𝟐 − exhibits one of the strongest known systems of “hydrogen bonding”. Describe the
feature(s) of the F-H-F bonds that account for this high stability.
The hydrogen bond here is the two H-F bonds that form the molecule. From our MO, these two
bonds are essentially composed of a single orbital, 2ag. The 2ag MO consists of a large bonding
lobe that spans over both H-F bonds. This results in extra stability, and this unusually strong
hydrogen bond. One way to understand this is to consider what it would take to break this bond.
Because the MO spans the entire molecule, in order to break either H-F bond, you would have to
simultaneously break both of them at once. This is similar in effect to the stability afforded by
delocalized π-bonds in aromatic systems. Again, reference Miessler and Tarr for additional
discussion.
(e) Predict the bond order of the H-F bond.
HF2- is essentially composed of one bonding MO for both H-F bonds, in effect giving a bond
order of ½.
(f) Based on your MO diagram, would 𝐇𝐇𝐅𝐅𝟐𝟐 − be paramagnetic or diamagnetic?
HF2- contains no unpaired electrons, and therefore would be diamagnetic.
(g) Would you be able to distinguish HF, 𝐇𝐇𝟐𝟐 𝐅𝐅 + , and 𝐇𝐇𝐅𝐅𝟐𝟐 − in a photoelectron spectrum? If
so, how?
With reference to the MO diagram for HF, the PE spectra are expected to be distinguishable
from one another (given perfect resolution). While all spectra will display a narrow non-bonding
peak from the F 2s, the HF2- spectrum will appear as two close non-bonding peaks. The PE
spectrum for H2F+ will display two adjacent, wide bonding peaks. Other differences can be
pointed out, serving as a basis for distinguishing the different hydrogen-fluorine combinations.
Problem 5:
Consider the hypothetical square-planar molecule XeH4 in the following coordinate system:
(a) Write the appropriate SALCs of the hydrogen 1s atomic orbitals which will overlap
with each of the following xenon orbitals that can form molecular orbitals: 5s, 5px, 5py, and
5dx2-y2.
(b) Draw atomic-orbital overlaps with the appropriate Xe orbital for each of these SALCs.
There are several ways to generate the SALCs for XeH4: 1) an intuitive approach by inspection,
2) mathematically solving for the wavefunction of the H4 cluster that overlaps with each atomic
orbital and 3) the projection operator method. We’ll mostly focus on method 1) and 3).
To determine them by inspection, we can recognize that the only contributions that give the
greatest net overlap with the 5s, 5px, 5py, and 5dx2-y2 orbitals are the ones depicted above.
For a more rigorous approach, such as if we don’t know which atomic orbitals we want to bond
with, we apply the projection operator method.
For the planar configuration, the D4h character table gives a reducible representation of:
D4ℎ
Γ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝐸𝐸
4
2𝐶𝐶4
0
𝐶𝐶2
0
2𝐶𝐶2 ′
2
2𝐶𝐶2 ′′
0
𝑖𝑖
0
2𝑆𝑆4
0
𝜎𝜎ℎ
4
2𝜎𝜎𝑣𝑣
2
2𝜎𝜎𝑑𝑑
0
This gives a set of irreducible representations:
Γ𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐴𝐴1𝑔𝑔 + 𝐵𝐵1𝑔𝑔 + 𝐸𝐸𝑢𝑢
Applying the projection operator to each gives four MOs:
𝑃𝑃� 𝐴𝐴1𝑔𝑔 (𝐻𝐻𝑎𝑎 ) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. (𝐻𝐻𝑎𝑎 + 𝐻𝐻𝑏𝑏 + 𝐻𝐻𝑐𝑐 + 𝐻𝐻𝑑𝑑 )
𝑃𝑃�𝐵𝐵1𝑔𝑔 (𝐻𝐻𝑎𝑎 ) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. (𝐻𝐻𝑎𝑎 − 𝐻𝐻𝑏𝑏 + 𝐻𝐻𝑐𝑐 − 𝐻𝐻𝑑𝑑 )
𝑃𝑃�𝐸𝐸𝑢𝑢 (𝐻𝐻𝑎𝑎 ) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. (𝐻𝐻𝑎𝑎 − 𝐻𝐻𝑐𝑐 )
𝑃𝑃�𝐸𝐸𝑢𝑢 (𝐻𝐻𝑏𝑏 ) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. (𝐻𝐻𝑏𝑏 − 𝐻𝐻𝑑𝑑 )
From inspection of the character table, we can match up the 5s and 5dz2 orbitals with the A1g
LGO/SALC, the 5px and 5py with the doubly degenerate Eu LGO/SALC, and the 5dx2-y2 with the
B1g LGO/SALC.
(c) Sketch the MO diagram for XeH4.
As before, the s-p mixing from the 5s, 5dz2 and a1g LGO give three MOs, one bonding (5s + 5dz2),
one slightly anti-bonding (resulting from a hybrid mix of 5s - 5dz2) , and one antibonding (-5s 5dz2).
(d) Indicate which molecular orbitals are occupied, and calculate the net Xe-H bond order,
based on the molecular orbital model.
The MO diagram is filled up to the 5p nonbonding MO. The bond order results from four filled
bonding MOs (1a1g, 1eu, 1b1g) and one filled antibonding (2a1g). Per Xe-H bond, this gives a
bond order of:
Problem 6:
𝐵𝐵. 𝑂𝑂. =
3
1 1
� � (8 − 2) =
4
4 2
Consider the molecule silicon tetrachloride, SiCl4, useful for the synthesis of silicon
nanowires and other silicon based solid state materials.
(a) Draw the MO diagram for the molecule in a planar configuration.
Due to the isolobality between XeH4 and SiCl4, we can adopt the LGOs for the H4 cluster to the
Cl4 cluster, assuming each Cl has one half filled 3p orbital to bond with:
Thus,
(b) Draw a Walsh diagram and predict the stabilization/destabilization of the molecule
upon distortion to a tetrahedral configuration.
Upon distortion to the tetrahedral configuration, the Eu and B1g LGOs form a degenerate set of 3
Td LGOs. This can be readily seen by applying the isolobal analogy of CH4 from the texts and
lecture slides. As such,
Since the planar configuration only has 3 bonding MOs (1a1g, 1eu), the tetrahedral configuration
with 4 bonding MOs (1a1, 1td) is preferred.
Another silicon precursor commonly used in the semiconductor community is silane, SiH4,
and silicon tetrafluoride, SiF4.
(c) Modify your MO diagram for SiH4 and SiF4. How would you expect the bond strength
of Si-H and Si-F to differ from that of Si-Cl?
We can simply change the energy position of LGOs for H4 and F4, and re-plot the MOs taking
into account the new energy matches and overlap. In SiH4, the similar energy of H 1s and Cl 3p
means that no dramatic changes to the MO diagram are made. Since s orbitals generally have
poorer overlap compared to p orbitals, the mixing of SiH4 is slightly weaker, though the bond
strength is still a fairly strong polar covalent bond. However, as we move to SiF4, the large
energy mismatch between Si and F mean that the orbital mixing is poorer, leading to a more
ionic, weaker bond. Note that although the absolute energy of the 1td and 1a1 orbitals are lower
compared to SiCl4 and SiH4, this does not mean that it has a stronger bond. The bond strength is
additionally related to the energy required to go from the bonded molecule, to the unbonded
molecule (i.e. the energy difference between 1td and F 2p).