P.S. 9 Answer Key, p. 1
Chem 165 Spring14
Chem 165 Homework #9 not to be turned in.
1. For ionic solids, the principle of isomorphous substitution states that one cation can substitute
for another in a lattice if they have the same charge and differ in radii by not more than 1020%. In light of this, answer the following questions. You can find the most commonly used
ionic radii (Shannon or Shannon-Prewitt radii) at http://en.wikipedia.org/wiki/Ionic_radius or
doi:10.1107/S0567739476001551
(a) Explain why the olivine minerals (Mg,Fe)2SiO4 contain varying amounts of Mg2+ and Fe2+,
but not, for instance, Ca2+ and Fe2+ or Li+ and Fe2+.
2+
2+
Mg (86 pm) and Fe (92 pm) have similar radii so they can substitute for each other in minerals
2+
+
such as olivine. Ca (114 pm) is too big. Li (90 pm) has the right size but the wrong charge.
(b) Most table salt (NaCl) in the U.S. is “iodized,” that is it contains iodide (a necessary
nutrient). Do you think that the iodide ions in table salt are substituted into the NaCl lattice, or
present in separate crystals of NaI or KI? Why or why not?
–
–
I (206 pm) is 23% larger than Cl (167 pm) and is therefore unlikely to substitute for it. I believe
that table salt contains some KI. [Side note: the potassium salt is used, I think, because it’s
cheaper. It’s cheaper because KI is less soluble than NaI, making it easier to isolate from solution
(Li, Rb, Cs are all much less abundant and more expensive).]
(c) A few years ago, a young environmental chemist interviewed for a faculty position here at
UW. One of his ideas for new research projects involved using the trace lead and bismuth ions
in coral as a way to date when the coral was formed (using radioactive isotopes of Pb and Bi).
I asked him why he thought there would be trace lead or bismuth to study. He responded that
there was a lot of cadmium in the coral, which is mostly CaCO3, so there was likely to be lead
and bismuth as well. What do you think of his answer?
2+
2+
Cd (95 pm) readily substitutes for Ca (100 pm) because they are of similar size and the same
2+
2+
charge. Pb (119 pm) seems pretty big to substitute for Ca (19% larger) but it’s possible.
3+
5+
Bismuth, as an odd numbered element, forms Bi and Bi ions, but not a 2+ ion, so it’s unlikely to
2+
be found to any great extent in place of Ca .
2. The alloy CuAu has the ccp/fcc structure. Under certain conditions, this is an ordered structure,
with alternating layers of gold and layers of copper. Under other conditions, the copper and
gold atoms are randomly distributed, so that the odds that any given site in the solid is
occupied by a copper or a gold atom is 50/50. In solid state chemistry terms, these are two
different phases of CuAu (an organic chemist would call them isomers).
(a) Which phase is favored at higher temperatures, and which at lower temperatures? Recall that
equilibria in chemistry are determined by free energies, ΔG. Which phase -- the high
temperature or the low temperature -- has the more favorable enthalpy? Which phase has the
more favorable entropy? Does this make sense based on our intuitive concepts of bonding
and entropy?
CuAu has the ordered structure below 380 ˚C, and the disordered (random) structure above 420
˚C.
The equilibrium between the two phases is governed by the free energy,
ΔG = ΔH - TΔS. The entropy term becomes more important at higher temperatures. Since the
disordered phase is favored at higher temperature, this phase must be favored by entropy. The
ordered phase must be more favored by enthalpy.
The same answer is (of course) obtained from focusing on the equilibrium constant, Keq, which is
directly related to the free energy ∆G = -RT ln(Keq):
Keq = e-ΔG/RT = e-(ΔH-TΔS)/RT = e-ΔH/RT × eΔS/R
As the temperature goes up, ΔH becomes less important. This makes sense, because there’s
more average thermal energy at higher temperatures.
P.S. 9 Answer Key, p. 2
Chem 165 Spring14
Entropy is associated with more random or more disordered structures. The best bonding – the most
favorable enthalpy – usually occurs when the structure is ordered, because a copper atom likely has a
preference for which of its neighbors are copper and which are gold.
(b) Do you think that this is a general principle, that more disordered structures should be
favored at higher temperatures?
Yes, it is a general principle that more disordered structures are favored at higher temperatures.
More disorder is entropically more favorable and entropy becomes more important at higher
temperatures. Typically atoms or ions have a preferred ordered structure, in which the bonding
(the enthalpy) is most favorable.
3. Consider an intrinsic defect in an ionic solid such as LiF or MgO, such as one anion and one
cation missing from their normal lattice positions.
(a) Does this defect make the lattice energy (an enthalpy) more favorable or less favorable?
An intrinsic defect -- atoms/ions missing or in the wrong places -- will make the lattice energy less
favorable. The ions are less tightly held together when there is a defect in the structure. This
effect of a defect is an enthalpy term because the lattice energy is an enthalpy.
(b) Is the enthalpic effect of a specific defect larger in LiF or MgO, if all other things are equal?
Explain, and be quantitative if possible.
The enthalpic effect of an intrinsic defect such as one anion and one cation missing from their
normal lattice positions is more unfavorable in MgO than in LiF. This is because the charges are
twice as large in magnesium oxide while the radii are comparable. Thus MgO should have roughly
four times the lattice energy of LiF, and the enthalpy of a specific defect should be four times less
favorable.
4. (a) Oxtoby 22.31. By “the nature of electrical conduction”, Oxtoby is asking whether electrons or
holes are the primary charge carriers.
(a) Si doped with P is n-doped, because P has one more valence electron than Si. An n-doped
material has excess electrons and therefore the electrical conductivity is primarily by electron
movement.
(b) InSb doped with Zn is p-doped, because Zn has one less valence electron than In. A p-doped
material has excess holes and therefore the electrical conductivity is primarily by hole movement.
(b) The primary charge carriers have compensating charges in the solid. In a simple Lewis-dot
picture, what are the compensating charges?
+
When the “extra” electron leaves the vicinity of the phosphorus atom, it leaves behind an P ion.
This is the compensating charge.
5. Oxtoby 22.32.
Following the problem 3 answers: (a) Ge doped with In is p-doped; (b) CdS doped with As is ndoped.
6. (a) Questions 22.33 and 22.34.
(a) 680 nm, red. (b) 580 nm, yellow. See http://en.wikipedia.org/wiki/Color
(b) GaAs1-xPx materials can be viewed as a solid solutions of GaAs and GaP. One way to predict the
relative band gaps of materials like this is to think of the band gap transition as a metal-to-ligand
charge transfer transition. In NaCl, a highly ionic compound, the valence band is almost entirely
Cl– orbitals and electrons, and the conduction band is almost entirely empty Na+ orbitals. The
transition involves moving an electron from Cl– to Na+. Using this model, NaI should have a
smaller bandgap than NaCl (and it does). This is because it is easier [requires less energy] to
remove an electron from I– than from Cl–, based on their electronegativities. If it’s easier to
remove an electron in general, then it’s easier to move an electron from I– to Na+.
P.S. 9 Answer Key, p. 3
Chem 165 Spring14
This argument is less convincing for materials such as GaAs and GaP, which have polar covalent
bonding, not ionic bonding. Still, this argument gives the correct answer, so please use it to
predict the relative band gaps for GaAs vs. GaP.
3–
P is more electronegative than As, so removing an electron from “P ” is harder than removing e
3–
from “As ”. So the band gap of GaP is larger (higher in energy).
–
(c) Typically the band bap of a solid solution varies linearly with composition of the material, so the
band gap of GaAs0.5P0.5 is halfway between the band gaps of GaAs and GaP. Are the bandgaps
given in questions 23.13 and 23.14 consistent with this rule about solid solutions and the relative
bandgaps from (b)? Explain.
Yes, this is consistent. The bandgap is higher for GaP than for GaAs, so the bandgap for the
material with the higher % P (GaAs0.14P0.86) has the higher bandgap.
7. Oxtoby 22.35.
The ZnO band gap decreases upon heating. ZnO is white so its bandgap is in the UV part of the
spectrum. The compound turning yellow means that some of the absorption occurs in the visible
region, so the bandgap must have moved towards the visible region, toward lower energy.
8. Based on your first-hand experience with these materials, which has a larger band gap, diamond or
graphite? Explain. (Both are pure, elemental C; graphite is the major component of pencil “lead.”)
Graphite is black and diamond is clear (no absorption of visible light). Thus diamond must have a
large band gap (it is an insulator). Graphite is actually a metal, with essentially no band gap, and
that's why it is black. (Yes, graphite conducts electricity.)
9. (a) The UV/visible absorption spectrum of a molecular compound usually has a peak. But the
absorption spectrum of a semiconductor or insulator usually has a sharp “edge”, absorbing all
photons above this energy. Explain this difference.
Optical absorption in a molecular compound involves the promotion of an electron from one orbital
to another (typically from a bonding or nonbonding orbital to an antibonding orbital). In principle
this could be a sharp line, but in practice the presence of vibrations and different solvation states
etc. lead to a broad line, with a peak. A semiconductor or insulator, however, has a band gap. A
photon with energy below the band gap (wavelength above a certain value) cannot be absorbed.
But any photon with energy above the band gap can be absorbed, because the nature of a band is
that there are lots of very closely spaced states.
(b) In the transition metal chemistry portion of the class, we discussed the optical spectrum of a
metal complex has having d-d or charge transfer bands (we focused on LMCT transitions).
Using this perspective, how would you describe the band-gap absorption of NaCl or Al2O3?
Both NaCl and Al2O3 are colorless because their band gaps are larger than the energy of visible
light. Since these are pretty well described as ionic compounds, essentially all of the valence
electrons are on the anions. Therefore the filled valence band is primarily chloride (for NaCl) or
+
3+
oxide (for Al2O3). The empty valence band is primarily Na or Al . The band-edge absorption
–
+
moves an electron from the valence band to the conduction band, in other words from Cl to Na ,
2–
3+
or from O to Al , so this resembles a ‘ligand-to-metal charge transfer’ transition (LMCT).
(c) Using your answer to (b), can you make a rough connection between the activity series for
metal redox potentials and the relative energies of the band edge absorptions for MX
compounds?
The LMCT transition involves adding an electron to the cation. In a rough way, the ability to reduce
a cation is related to its position in the activity series. A more active metal has a cation that is more
difficult to reduce. Therefore this transition should be higher in energy for NaCl than for CuCl,
+
+
because it’s easier to put an electron onto Cu than it is onto Na , from the activity series.
10. In class we talked about why metals are malleable and ductile, and did a demonstration of
“work hardening” of a metal. A nice summary of this can be found at the end of:
http://courses.chem.psu.edu/chem112/materials/metals.html
P.S. 9 Answer Key, p. 4
Chem 165 Spring14
(this whole page is a nice review of many of the concepts we’ve been talking about).
At the very end of this page there’s a discussion of why the addition of carbon makes iron
harder (into “carbon steel”) and why different treatments of iron (fast vs. slow cooling, for
instance) lead to quite different properties. Can the formation of Fe3C be viewed as a
precipitation? In light of why metals are malleable and ductile, why does a very small amount
of Fe3C cause such a large change in the strength of the steel?
Yes, in a way, the formation of Fe3C be viewed as a precipitation, occurring when the iron changes
phase and the carbon is no longer as soluble. The small grains of Fe3C get in the way of the
migration of defects that allow metals to be ductile – they “pin” the dislocations. Thus a small
amount of Fe3C can have a very large change in the materials properties.
11. Take a quick look at M. Behrens et al. Science 2012, 336, 893 “The Active Site of Methanol
Synthesis over Cu/ZnO/Al2O3 Industrial Catalysts” DOI: 10.1126/science.1219831
http://www.sciencemag.org/content/336/6083/893.full?sid=8e555794-f8b8-43ae-9ed6-d4f9fbd3d5ee
(a) I find the pictures in Figure 3 remarkable, for their resolution of individual atoms. What do
the A, B, etc. letters in the little squares in Figure 3D signify?
These indicate the A, B, C layers of the copper nanocrystals.
(b) Look briefly at http://en.wikipedia.org/wiki/Crystal_twinning to see what a crystal “twin”
means. How many twin boundaries are there in Figure A? Is the large circle of atoms in this
figure one close-packed hexagonal sheet?
Most of the circle seems to be one sheet. The left side could be considered a bent part of that
sheet, or part of a separate sheet.
(c) Does the left-most panel of Figure 2A represent a hexagonal close-packed layer of copper
atoms? What does the middle panel of Figure 2A show?
Yes, the left panel is a-close packed sheet. The middle panel shows a close-packed crystal of Cu
that has been cleaved at an angle to the layers, so you can see the edges of about 4 different
layers.
(d) This paper argues that copper atoms on step edges, such as those in Figure 2A middle, are
the active catalytic sites for the production of methanol from CO + 2H2. Is it reasonable to you
that atoms at step edges would be more chemically reactive?
Yes, this is reasonable. Atoms in metals typically have lots of nearest neighbors, so those at an
edge that have only a few neighbors are likely to be high energy, reactive species, easily making
bonds with added reagents.