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Fall 2014
CHEM 104a
Advanced Inorganic Chemistry
Problem Set #5
(Due 11/13)
Problem 1:
While we have previously talked about an MO theory approach to bond formation, we
have not quite touched upon bond breaking.
One of the first high power lamps of the early 20th century was the carbide lamp: a metal
housing containing calcium carbide upon which water slowly dripped, yielding flammable
acetylene by the following reaction:
CaC2 + 2H2O → C2H2 + Ca(OH)2
Starting with a C22- MO diagram, draw the MO diagram describing the above reaction,
and sketch the shapes of the relevant MOs to illustrate the bond formation and bond
breaking in the above reaction. How favorable would this reaction be if instead CaC2 were
replaced by CuC2?
In order to form the desired C-H bonds in acetylene, the only appropriate orbitals with the
correct symmetry are the HOMO of C22- and the LUMO of H2O. While there are a few possible
ways to draw the interacting MOs (above it is depicted as a single H2O giving up both hydrogens,
though you could also envision a mechanism where two water molecules are involved), it is
essential that one utilizes the H2O LUMO, as it is an anti-bonding MO, containing a nodal plane
(denoted by the dashed lines) across the O-H bond. By mixing with this orbital, and populating it
with electrons (denoted by the center mixed MO), these nodal planes effectively break the O-H
bond. One could alternatively take the LUMO+1 (the second lowest unoccupied MO) as it also
contains an O-H anti-bonding node.
There is however, the issue of breaking both O-H bonds at once, leaving an O2- anion, which is
not one of our final products. One way to justify breaking only one O-H bond at a time is to take
linear combinations of the LUMO and LUMO+1, which would look something like this:
Again, we are utilizing an MO with an anti-bonding node along the bond we would like to break.
However, by taking this linear combination, one of the O-H bonds becomes non-interacting (i.e.
no bonding or anti-bonding e- density), meaning we can break one O-H bond without affecting
the other. This approach is superficially akin to the orbital predictions derived from VBT.
If we were to replace CaC2 with CuC2, the products formed would be Ca(OH)2 and Cu(OH)2,
respectively. Based on our understanding of Hard-Soft Acid-Based (HSAB) Theory, we would
predict that the hard base, OH-, would prefer to bind to the hard acid, Ca2+. Upon switching to
the softer acid, Cu2+, the complex becomes much less favorable to form, resulting in a small
thermodynamic driving force for this reaction, perhaps slowing down the evolution of acetylene.
Problem 2:
Predict whether the reactants (left) or the products (right) will be favored in each of the
following equilibria, and briefly explain why.
(a) As2S3 + 3HgO ↔ As2O3 + 3HgS
Products favored. Hg2+ is the prototypical soft acid, more favorable to bind to the soft
2base, S . As3+, conversely, is a harder acid (higher in periodic table, more electronegative), and
O2- is a fairly hard base.
(b) H3O+ + HSO4- ↔ H2O + H2SO4
Reactants favored. H2O is a hard base, and H+ is a hard acid.
(c) MgSO4 + CuI2 ↔ MgI2 + CuSO4
Reactants favored. Cu2+ is softer than Mg2+, I- is softer than SO42-.
(d) NaCN + RbOH ↔ NaOH + RbCN
Products favored. Na+ and OH- are hard acids and bases, respectively.
Problem 3:
In some cases, CO can act as a bridging ligand between main-group and transition-metal
atoms. When it forms a bridge between Al and W in the compound having the formula:
(C6H5)3Al-[bridging CO]-W(CO)2(C5H5)
is the order of the atoms Al-CO-W, or Al-OC-W? Explain, providing a simple MO diagram
to illustrate your point.
There are several levels of complexity to this answer. At the most superficial level, Al is a harder
acid/base than W, and O is the harder/more electronegative side of CN. Therefore, you would
expect the bridging CO to adopt the structure, Al-OC-W.
However, a more detailed analysis of the MO diagrams would reveal a somewhat interesting
bonding scheme. Since W is a softer acid/base, it will bind with the CO end, through the usual σ
bonding and π back-bonding that we have discussed previously for CO ligands.
What about Al? Since Al is harder, we would indeed expect it to bind to the oxygen side. A first
order answer would be that it would σ bond with the CO HOMO (even though there is less edensity on this end, there is still some available). However, if we expand our frontier MO search
to the HOMO-1 (second highest occupied MO, still relatively close in energy to the HOMO), we
would see that we can gain much better overlap due to the higher e- density on O in this MO. In
order to have positive overlap, the Al UMO would have to only overlap with one of the lobes of
CO’s π HOMO-1, which would give a bent bond. Experimentally, it is usually observed that
such bridging carbonyls have a bent C-O-Al bond.
Problem 4:
One of the contributions J. H. Hildebrand (namesake of Hildebrand Hall) made to modern
chemistry is the concept of adduct formations, for which he demonstrated the various
colors of iodine, I2, when dissolved in various aromatic solvents.
Similarly, when bromine, Br2, is dissolved in hexane, a single absorption band is seen
centered on 500nm. However, when Br2 is dissolved in methanol, this absorption band
shifts, and a new band is formed. Using the relevant frontier MOs, draw the MO diagrams
for Br2 in hexane and Br2 in methanol. Draw the two UV-Vis spectra, and denote on the
MO diagram the corresponding electronic transitions.
A very similar problem is discussed in MT 6.4.2 (5th ed.). Here, methanol is a relatively soft base,
which interacts with the relatively soft acid, Br2. From this HOMO-LUMO mixing, as depicted
below, new spectral features can be observed.
The originally observed π*g→σ*u transition is shifted to a shorter wavelength (higher energy) as
the σ*u is shifted up as a result of mixing with the methanol HOMO. A new, higher energy
absorption band appears as a result of the newly formed Br2-methanol bonding orbital,
representing a charge transfer from methanol to Br2. In comparison, hexane is not a very soft
base, so the interaction with Br2 is negligible. With reference to Fig. 6.10 and 6.11 in MT, I leave
it to you to determine what the corresponding UV-Vis spectra should look like.
--------------------------------------------------------------------------------------------------------------------Extra Credit
Problem 5:
Zintl phases are a form of somewhat exotic chemistry, possessing a range of unique
properties arising from their cluster structures. Among these properties, the unusual ionic
charges of many such clusters remain intriguing, particularly for such applications as
lithium ion batteries, where a high anionic charge means more Li+ can be stored for higher
capacity batteries . For the following simple Zintl ions, utilize Wade’s Rules to predict and
draw the shape of the cluster:
(a) [Sb4]2Each Sb contributes 3 valence p electrons to bonding, plus the extra 2 e- gives a total of 14 eavailable for bonding. Two e- allocated towards delocalized cluster bonding (essentially the
perfectly symmetric A1 MO), leaving 6 e- pairs for the framework. This means, according to
Wade’s Rules, that we start with a 6 vertex polyhedral (octahedral), occupying only 4 vertices
with the 4 Sb atoms (n+2, arachno, square planar, possibly distorted):
2Sb
Sb
Sb
Sb
Sb
Sb
Sb Sb
You might guess it is:
2-
Sb
Sb
Sb
Sb
Sb
Sb
Sb
Sb
Though experimentally we know it is the first structure.
(b) [Te6]2+
Each Te contributes 4 valence p e-, minus 2 e- for the cationic charge, giving 22 bonding e-.
Subtracting out the 2 e- for delocalized bonding, this leaves 10 pairs to form the framework. Thus,
we start with a 10 vertex polyhedral, in which we only occupy the neighboring 6 positions (klado,
trigonal prismatic).
Te
Te
Te
Te
Te
Te
Te
Te
Te
2+
Te
Te
Te
(c) [Ge9]4Each Ge contributes 2 valence e-, plus the 4 e- for the cationic charge, giving 22 e- available for
bonding. As above, this yields a 10 vertex polyhedral, in which we occupy 9 of the sites (n+1,
nido)
4Ge
Ge
Ge
Ge
Ge
Ge
Ge Ge
Ge
(d) [Sb5]5Sb each contributes 3 valence e-, plus the 5 e- for the anionic charge, gives a total of 20 bonding
e-. This would call for a 9 vertex polyhedral, where we only occupy 5 of the positions (klado).
5Sb
Sb
Sb
Sb
Sb
Sb
Sb
Sb Sb
Sb
Interestingly enough, people have also claimed to have isolated Zintl anions of the formula
[Sb8]8-.
For each of the above, is there a BxHy structural analogue? If so, what?
(a) [B4H4]6- or B4H10
(b) [B6H6]10- or B6H16
(c) [B9H9]2- or B9H11
(d) [B5H5]10- or B5H15
No comment on whether these boranes actually exist. Due to the high ionic character, these
compounds are often unstable.
Problem 6:
Let’s take the Zintl anion [Sb4]2- from above. Prepare a Walsh type diagram to justify why
it adopts such a structure, as opposed to the structure of [Ge4]2-. Consider only valence p
orbitals, oriented radially (pointing towards the center of the cluster), or tangentially.
For each of these clusters (square planar for [Sb4]2- and tetrahedral for [Ge4]2-), we’ll take the
three p orbitals for each atom, and group them into the following basis sets:
X
Y
1) radially oriented p-orbitals, 2) tangential px/py orbitals, 3) pz orbitals.
Since these sets are largely orthogonal to each other, we can treat them separately.
For each basis set, we get 4 irreducible representations which derived previously.
Eu
B1g
A1g
In which A1g represents a bonding MO, Eu are relatively non-bonding MOs, and B1g is an antibonding MO.
For the px/py tangential set:
B2g
Eu
A2g
One might predict that the Eu MOs from this tangential set will mix with the Eu set from the
radial set.
And for the pz orbitals:
A2u
Eg
B2u
From these interactions, we can expect an MO diagram that looks like:
For the tetrahedral shape of [Ge4]2-, we’ll take a similar approach in splitting up the MOs.
1) radially oriented p orbitals
2) 3x2 of the tangential p orbitals
3) which we will then combine with the remaining tangential p orbital
*note, while it is possible to treat all 4 tangential p orbitals at once, I think this approach is a bit
visually more accessible.
For the radial p orbital set, we’ll get:
T2
A1
x3
Where A1 is bonding, and the triply degenerate T2 is antibonding.
For the 3x2 tangential p orbitals, we get:
'
A2
E
'
"
A2
E
"
Combining these with the remaining p orbital gives (left), and the full diagram for the cluster
(right):
A Walsh-type diagram for these two isoforms would look like:
From this, we clearly see the underlying mechanism behind Wade’s Rule’s predictions. With
only 10 valence electrons, [Ge4]2- only fills up to the slightly bonding e orbitals. However, [Sb4]2has 14 valence electrons, which would fill all the way up to the high energy t2 antibonding MOs
for the tetrahedral form. Instead, it opts for the square planar geometry, in which it only fills up
to non-bonding MOs.