Chem. 312, Au13
Problem Set 6
Due in class, Friday, November 8
1. Enthalpies of precipitation are known for the following halides: LiF, -5 kJ/mol; NaCl, -4
kJ/mol; AgCl, -65 kJ/mol; CaCl2, +83 kJ/mol.
(a) Use the data on hydration enthalpies in Tables 2.1 and 2.5 in the text to calculate the lattice
energies of these compounds.
ΔHprecipitation = U (lattice energy) - [ΔHhydration,cation + ΔHhydration,anion]
For LiF (all in kJ/mol), -5 = U(LiF) - (-515 + -497), so U(LiF) = -1017 kJ/mol.
For NaCl, -4 = U(NaCl) - (-405 + -355), so U(NaCl) = -764 kJ/mol.
For AgCl, -65 = U(AgCl) - (-475 + -355), so U(AgCl) = -895 kJ/mol.
For CaCl2, + 83 = U(CaCl2) - [-1592 + 2(-355)], so U(CaCl2) = -2219 kJ/mol.
(b) All but one of these halides have the NaCl structure -- which one? [Don’t overthink this
question – it’s trivial.] Use the radius ratio to choose the lattice type for this halide, from
Table 3.4.
CaCl2 cannot have the NaCl structure because there are two chlorides to one calcium.
Since rCa/rCl = 0.68, six coordinate cations are predicted. The typical 1:2 structure with 6coordinate cations is rutile (TiO2); the actual structure is like rutile, but not identical.
For the 1:1 salts: rLi/rF = 0.756; rNa/rCl = 0.695; rAg/rCl = 0.772. The strict radius rules would
predict that LiF and AgCl should have the CsCl structure, but they both have the NaCl
structure. Evidently the radius ratio has to be significantly above the critical value before
the structure shifts.
(c) Calculate the lattice energies of the compounds with the NaCl structure, using their ionic
radii, the Madelung constant, etc. For which compound is there the greatest discrepancy
with your answer to (a), the experimental value? Any idea why?
Using U = 138,900 MZ+Z–[1-1/n]/[rc+ ra), with M = 1.748 for the NaCl structure:
U(LiF) is calculated to be -968 kJ/mol, using n = 6. This is 5% different from
experimental value of -1017 kJ/mol.
U(NaCl) is calculated to be -751 kJ/mol, using n = 8. This is 2% different from
experimental value of -764 kJ/mol.
U(AgCl) is calculated to be -734 kJ/mol, using n = 9.5. This is 18% different from the
experimental value of -895 kJ/mol.
AgCl has the largest deviation, probably because the bonding in AgCl is partially covalent
(the difference in electronegativity is only 1.2). These formulas assume only ionic bonding.
Another way of saying the same thing is that Ag+ and Cl- are both fairly soft and
polarizable so their electron clouds will deform in the presence of the other, which gives
overlap and covalent bonding.
2. Cesium and gold form an ionic compound, Cs+Au-, with a cesium-gold distance of 369 pm.
(a) What type of lattice will CsAu adopt? Show your work.
The radius of Cs+ is 181 pm (from the text), so the radius of Au- must be 188 pm. This is a
1:1 salt with a radius ratio rCs/rAu of 0.96, so the CsCl structure is predicted.
(b) Calculate the predicted lattice energy of CsAu.
Using U = 138,900 MZ+Z–[1-1/n]/[rc+ ra), with Z+ = Z- = 1, M = 1.76267, r+ + r- = 369 pm,
and n = 12, gives U = 608 kJ/mol.
Problem Set 6, p. 2
3. Calculate the Madelung constant
for 2D structure at right. First, find
a unit cell and determine the stoichiometry. Then, just as we did in
class, pick a single ion and look at
all the neighbors at a distance r,
those at a distance 2r, etc. Give the
first five terms of the infinite
series. What value for M do you
get if you use only the first term? If
you use the first two terms? Three
terms? Four terms? Five terms?
Do you think the expansion
converges to a positive number,
indicating a stable structure?
Assume that the ion charges are
±1, ±2, and/or ±3. Are the gray
ions likely the cations?
Chem. 312, Au13
r
This is a structure based on 90° angles; I would call it a square net. There are a
number of possible unit cells; I like the one that is a square with a grey ion at the center,
with the edges through the middle of the four surrounding white ions. Then the cell has
one grey ion and two white ions, or AB2 stoichiometry. Most likely the gray ion is the cation
because the cations are smaller. Then the ion charges are [A2+ (gray)][B– (white)]2.
As shown in the diagram below, there are 4 A2+–B– contacts at a distance r. These
give a favorable (negative) interaction of 4 × ZAZB/r or -8/r since ZA = +2, ZB = -1). [Since
all we’re calculating for the moment is the Madelung constant, we don’t need to worry
about the e2N/4πεο(1 - 1/n)].
Then there are four A2+-A2+ contacts at a distance 2r, for an unfavorable interaction of
4 × (2×2)/2r or +8r. Then there are eight favorable A-B interactions at a distance of r√5: 8
× (2×-1)/r√5 = -16/r√5. Next farther away are four unfavorable A-A interactions at a
distance of 2r√2, or 4 × (2×2)/2r√2 = +4√2/r. The fifth closest interaction is four A–B
interactions at a distance of 3r, for an energy of –8/3r. (The last interaction is not indicated
in the diagram, but you can see it if you take the 1r horizontal vector and extend it two
more units to the next B site.)
Problem Set 6, p. 3
Chem. 312, Au13
2r
2r√2
r√5
r
r
The first five terms are therefore:
M = 8 – 8 + 16/√5 – 4√2 + –8/3
The value for the Madelung constant M after one term is 8. After two terms, M = zero;
after three terms, 7.16; after four terms, 1.50; and after five terms, 4.17.
The value for M looks like it is converging to a positive number, so the structure looks
stable. This is reinforced by the next term not being a positive (unfavorable) interaction, as
one might have expected, but rather eight attractive A-B interactions at r√13. To see
these, look three steps to the right and two steps up from the cation in question -- √13
comes from √(32 + 22).
4. We have discussed in class the "holes" in the close packed lattices, hcp and
ccp (fcc). This discussion assumed that the stacking of the close packed
layers is perfect, so that the distance of any atom to all twelve of its
neighbors is the same. But in hcp structures, the spacing between the layers
is often not the ideal spacing. Consider the case where the spacing between
the layers is larger than the ideal, in other words when the dimension “c”
indicated on the drawing at left is larger than the ideal value.
c
(a) What happens to the tetrahedral holes in this case? Are they still regular tetrahedra? If
not, describe the shape of the new polyhedron, in other words the distortion that occurs.
(b) What happens to the octahedral holes in this case? Are they still regular octahedra? If not,
describe the distortion that occurs.
(a) The tetrahedral holes distort to an elongated trigonal prism:
a triangle with the fourth anion pulled away. The drawing at
right is exaggerated.
(b) The octahedral holes also distort. Think of an octahedron
as a trigonal antiprism: two triangles facing each other but
pointed in opposite directions (rotated 60˚ vs. one another).
The distortion in this case is that the triangles are pulled apart,
as indicated at right.
(a)
(b)
Problem Set 6, p. 4
Chem. 312, Au13
5. Read sections 4.6, 4.7, and 4.8 in our text (Wulfsberg). The important minerals spinel and
olivine are mentioned and their typical compositions described. [I’m told the Twin Sisters
mountains, southwest of Mount Baker, contain a great deal of olivine (a variety called dunite).]
Could these two classes of minerals adopt the same structure? Explain which cations are in
which types of holes, and what fraction of octahedral and tetrahedral holes are occupied in
each structure.
The Twin Sisters mountains are shown at right
(the olivine/dunite hiding under the trees and
the snow). Spinels have the formula AB2O4,
with the oxide ions in a cubic close packed
lattice. Olivine is the orthosilicate (Mg,Fe)2SiO4,
which can also be described as an AB2O4
structure (A = Si4+, B = Mg2+ or Fe2+), so it is
possible that they could have the same
structure.
In olivine, the small silicon ion clearly has tetrahedral coordination and the larger 2+ ions
likely are octahedral. In spinels, the book says that the A2+ ions normally occupy tetrahedral
holes while the B3+ ions have octahedral coordination. This seems somewhat unusual, since
the +3 ions would often be expected to be smaller than the +2 ions, and the smaller ions
usually occupy the smaller tetrahedral holes.
With these conclusions, both structures have AB2O4 stoichiometry, and in both the A ion is in
a tetrahedral hole and the B ions are in octahedral holes. So yes, they could have the same
structure. This is the correct answer to my question. Looking up the structures in Wells
Structural Inorganic Chemistry, the difference between the two structures is in part that
olivine has hexagonal close packed oxide ions, while spinels have cubic close packed oxides
ions. This book also says that the cation ordering in spinels is complex, and the B ions often
find themselves in the tetrahedral holes.
There are two tetrahedral holes and one octahedral hole per close packed atom or ion. In
AB2O4, therefore, the A ions occupy one-eighth of the tetrahedral holes and the B ions onehalf of the octahedral holes.
6. The graph below shows the lattice energies (U) of MCl2 salts for +2 cations from Ca2+ to
Zn2+ (purple points). For this problem, assume that all of the solids have the same structure
(the same Madelung constant) with octahedral coordination of the metal ions. The graph
plots the negative of the lattice energy, so the more strongly bound lattices are at the right.
Problem Set 6, p. 5
Chem. 312, Au13
(a) Based on the equation for the lattice energy discussed in class and given in the book, and
the general trend for ionic radii moving left to right, would you expect lattice energies to
get more negative, less negative, or stay the same on moving from Ca2+ to Zn2+? Explain
your reasoning.
The overall trend is that the lattice energies (U) of MCl2 salts for +2 cations become more
favorable from Ca2+ to Zn2+, because the ionic radii get smaller across this series (moving
left to right).
(b) Does the actual data fit your prediction?
Yes, sort of. There is a general overall trend to more negative U values toward Zn2+,
which is the dotted line trend in the figure, but it’s not very accurate.
(c) Look up the “actual” ionic radii for these ions using the table inside the back cover of the
textbook. (I put “actual” in quotes because these are not real measured numbers like
ionization energies, they are the values Shannon and Pruitt chose that fit a set of structural
data best.) Do the actual radii perfectly fit the general trend? Do the exceptions from the
radius trend explain some or all of the lattice energy variation?
No, the Shannon-Pruitt radii do not perfectly fit the trend. The values (in pm) are:
Ca2+, 114; Ti2+, 100; V2+, 93; Cr2+, 94; Mn2+, 97; Fe2+, 92; Co2+, 88; Ni2+, 83 pm; Cu2+, 91;
Zn2+, 88;
Some of the exceptions to the left-right trend do fit the lattice energy deviations. For
instance, the larger radius for Mn2+ than the elements on either side is consistent with its
smaller |U|. However, the larger radius for Cu2+ vs. Zn2+ is opposite to the trend in |U|.
(d) Why is there no value for d1 (scandium)?
There is no value for scandium because “ScCl2” is not a known compound (to my
knowledge). Scandium is like aluminum, it’s compounds almost entirely contain Sc3+ ions.
8. Transition metal chemists explain the data in question 7 (and other data) using a concept
called Ligand Field Stabilization Energy (LFSE). We discussed this briefly right before the
first midterm. The trend you predicted in question 7 ignored the splitting of the d orbitals,
unless this splitting affected the Shannon-Pruitt radii. But the d orbitals are not at the same
energy in these ionic solids, they split two above three. Relative to the average d-orbital
energy, the t2g orbitals are stabilized by –2/5 Δo, and the eg orbitals are raised by 3/5 Δo. The
LFSE is defined as the amount of stabilization versus this average:
LFSE for an ion with configuration t2gn egm = (n × –2/5Δo) + (m × +3/5Δo)
Problem Set 6, p. 6
Chem. 312, Au13
For more information on LFSE, you could look at ocw.nctu.edu.tw/upload/classbfs120912062973271.pdf
or other resources on the web.
When all the d orbitals are filled, in a d10 ion (t2g6 eg4), the LFSE is zero. In this case, the
energy from populating the more stable t2g orbitals [6 × (–2/5Δo)] is exactly equal to the
amount of destabilization from populating the destabilized (raised) e2g orbitals [4 × (+3/5Δo)].
There is no net stabilization versus the case where the d orbitals didn’t split.
For a d1 ion, (t2g1 eg0), the LFSE is –2/5Δo. Now there are more electrons (1) below the
imaginary average energy than there are above it (zero). So there is a net stabilization.
(a) For each of the points in the graph in question 7, give the electronic configuration (t2gn egm),
draw a two-above-three orbital pattern showing the electrons, and calculate the LFSE. Plot
the LFSE vs. the number of d orbitals, like the graph below. Does this explain the doublehumped curve? What do you have to assume about the high-spin vs. low-spin state of the
ions to make this work out?
You have to assume that all the ions are high spin to reproduce the trend. For instance,
the d5 ion falls pretty much on the line between the d0 and d10 ions that both have LFSE =
0. When the d5 ion is high spin it has LFSE = 0 and when it is low spin it has a very
negative LFSE of –10/5Δo.
d1
d5
d9
eg
d2
t2g
eg
t2g
eg
t2g
d6
d10
eg
t2g
eg
t2g
d3
d7
eg
t2g
eg
t2g
d4
d8
eg
t2g
eg
t2g
eg
t2g
d0, t2g0 eg0, LFSE = (0 × –2/5Δo) + (0 × +3/5Δo) = 0.
d2, t2g2 eg0, LFSE = (2 × –2/5Δo) + (0 × +3/5Δo) = –4/5Δo.
d3, t2g3 eg0, LFSE = (3 × –2/5Δo) + (0 × +3/5Δo) = –6/5Δo.
d4, t2g3 eg1, LFSE = (3 × –2/5Δo) + (1 × +3/5Δo) = –3/5Δo.
d5, t2g3 eg3, LFSE = (3 × –2/5Δo) + (2 × +3/5Δo) = 0.
d6, t2g4 eg2, LFSE = (4 × –2/5Δo) + (2 × +3/5Δo) = –2/5Δo.
d7, t2g5 eg2, LFSE = (5 × –2/5Δo) + (2 × +3/5Δo) = –4/5Δo.
d8, t2g6 eg2, LFSE = (6 × –2/5Δo) + (2 × +3/5Δo) = –6/5Δo.
d9, t2g6 eg3, LFSE = (6 × –2/5Δo) + (3 × +3/5Δo) = –3/5Δo.
d10, t2g6 eg4, LFSE = (6 × –2/5Δo) + (4 × +3/5Δo) = 0.
Yes, the LFSEs predict a double hump just like the measurements. This explains the data
very well.
(b) In your view, is the LFSE important?
Yes, particularly in a relative sense: one metal ion vs. another. However, in the overall
lattice energy, the LSFE is a pretty small effect, on the order of 10-15% looking at the
difference between the dotted line and the experimental data.
Problem Set 6, p. 7
Chem. 312, Au13
9. In class we are discussing mostly the thermodynamics of solids, mostly focused on perfectly
ordered (crystalline) solids. In this context, “thermodynamics” means the energy of the solid,
∆H° (or perhaps ∆G° including T∆S°). What we don’t teach – because it’s much more complicated and a lot of isn’t known – is the kinetics (rate) and mechanism by which solids form or
dissolve or change structure. This problem and the next one, like the De Yoreo paper on the
last problem set, involve recent papers that discuss very different aspects of these questions.
Look at the paper “Abnormal Grain Growth Induced by Cyclic Heat Treatment” T. Omori, T.
Kusama, S. Kawata, I. Ohnuma, Y. Sutou, Y. Araki, K. Ishida, R. Kainuma Science 2013,
341, 1500. I also found very use the commentary that accompanies this article: “A New
Route for Growing Large Grains in Metals” E. M. Taleff and N. A. Pedrazas Science 2013,
341, 1461.
(a) What is a “polycrystalline” material? How is it different than a “single crystal” material?
How is this related to “grains”?
A single crystal has all of the atoms lined up pretty perfectly in the arrangement you would
expect from the unit cell, all the way across the crystal. A polycrystalline material is
composed of lots of grains, each one of which is a small single crystal.
(b) In one sentence, what is the question that this paper addresses? Is this a topic of purely
academic interest, or are there potential commercially-valuable applications of this?
This paper presents a new way to make metal alloys with unusually large grains. From
their introduction and final paragraph, it sounds like this could be very valuable in a
number of applications.
(c) The unusual alloy that they are working with contains three different metals and has both
an α and a β phase. [A phase is a form of the material with a different structure, what
organic chemists would call an isomer.] What are the three metals, and what are the
structures of the α and a β phases? [See the middle column of page 1 of the paper.]
The alloy contains Cu, Al, and Mn. The β phase is body-centered cubic and the α phase is
face-centered cubic.
(d) What is the scale of the images in this paper? Do they show individual atoms? How do
they compare with the images in the De Yoreo paper we looked at last week?
These images in this paper are at micron or mm scale, and each feature encompasses
many many atoms (a 1 mm particle might have something like 1010 atoms. The De Yoreo
images had scale bars as small as 1 nm and showed individual atoms.
(e) Figures 3 and 4 in the paper are brightly colored. Are these the real color, or are these
‘false-color” images where the color shows some other property?
These are false-color images. The images highlight the different orientations of the grains.
I assume that this material looks like any other silvery metal. I bet they look a bit like the
images in Figure 2.
10. Look at the paper “Mg2+ Tunes the Wettability of Liquid Precursors of CaCO3: Toward
Controlling Mineralization Sites in Hybrid Materials” J. K. Berg, T. Jordan, Y. Binder, H.
G. Börner, and D. Gebauer J. Am. Chem. Soc. 2013, 135, 12512-12515.
(a) Why is calcium carbonate important? What do you think is are “off-equilibrium
morphologies”?
Calcium carbonate is probably the most important mineral made by organisms
(biomineral). Most shells are primarily CaCO3. There is a most stable structure of CaCO3,
the one present “at equilibrium.” But many biominerals are under active control of proteins
and do not adopt the most stable morphology. See:
Problem Set 6, p. 8
Chem. 312, Au13
http://en.wikipedia.org/wiki/Calcium_carbonate and
http://www.nature.com/srep/2013/130402/srep01587/full/srep01587.html
(b) Do you think that the formation of different morphologies or polymorphs is related to the
“wettability”? For a definition of that term, see: http://en.wikipedia.org/wiki/Wetting
I think that the wettability is probably critical to the formation of different polymorphs.
(c) What is the scale of the images in this paper? Do these look like crystals, with sharp edges
and clear faces? Why do you think they look this way?
They don’t look like crystals to me. That’s because they grow in the presence of organic
polymers, which are not crystalline.
(d) Is this a paper about solid-state chemistry, that would fit into this middle section of Chem
312, or a paper about hydration of cations that would fit into the first section of the class?
Or maybe an bio-organic paper because it uses peptides and organic polymers? Or is it all
of the above, and the way we teach Chemistry an arbitrary way to divide the subject?
It’s all of the above. There are good reasons why it makes sense to teach this subject in
pieces. But the growth of these materials, by the organism or in the flask, doesn’t care
how we learn the Chemistry.