Chem 165 Problem Set #8 due in class Monday June Oxtoby Problem 21.2.

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Answer Key, p. 1 Chem 165 Problem Set #8 due in class Monday June 2. 1. Oxtoby Problem 21.2. 2. Derive the Bragg equation (equation 21.

1 Answer Key, p. 1 Chem 165 Problem Set #8 due in class Monday June Oxtoby Problem Derive the Bragg equation (equation 21.1) from the requirement that the difference in scattering path length is an integral multiple of the wavelength (Figures and 21.11). That the difference in scattering path length is an integral multiple of the wavelength gives the equation directly above Figure 21.11: EF + FG = nλ. Figure is constructed such that line segments BE and BG are perpendicular to the x-ray beams (DF and FH, respectively). Then BEF and BGF are right triangles. From the construction of the Figure, the angle between line segment AB and the horizontal lattice plane is θ. Then angle ABF ( ABF) is 90 + θ and EBF is θ. From trigonometry, sinθ = opposite / hypotenuse = EF/BF. Similarly, sinθ = FG/BF. BF = the lattice spacing d. So EF + FG = 2d sinθ. So nλ = 2d sinθ, QED. 3. Problem Problem Problem Show your work.

2 Answer Key, p Problem Show your work.

3 Answer Key, p Problem (We did this in class.) 8. Consider the infinite two-dimensional structure which is illustrated below. (a) What is the repeating unit, the unit cell, of the structure? What is the stoichiometry of the solid? [In other words, if we call the large circles A and the small shaded circles B, what are x and y in the formula for the solid A x B y?]. (b) Does the structure have 4-fold, 5-fold, 6-fold, or some other symmetry? (c) Are the shaded circles more likely the anions or the cations? Explain. drawing for question 3: (a) There are a variety of ways to describe the repeat unit or unit cell of this 2D structure. One way is shown at right. The unit cell contains one shaded ion, and 2 1 /6 plus 2 1 /3, or one, unshaded ion. So the formula is AB. (b) The structure has three-fold symmetry. One drawback of the unit cell choice at right is that it does not reveal this symmetry. (c) B + A - is more likely because the B ions are smaller than the A ions. Cations are typically smaller than anions. 9. Cesium and gold form an ionic compound, Cs + Au, with a cesium-gold distance of 369 pm. Using the cation radius of 181 pm for Cs +, what type of lattice will CsAu adopt? Show your work. Since the radius of Cs + is 181 pm, the radius of Au - must be 188 pm. This is a 1:1 salt with a radius ratio r Cs /r Au of 0.96, so the CsCl structure is predicted.

4 Answer Key, p Consider a group of two lithium ions and an oxide ion, fixed in space in two different arrangements, (i) and (ii): In both cases the three particles lie in a line. Assume that the energy of interaction of any two particles of charge ±1 is a /r, where r is the distance between them. Calculate the total energy (the sum of all the interactions) in both arrangements, in units of a /r. Which is more stable? (i) Li + O 2- Li + (ii) O 2- Li + Li + r [Note: because this is just three ions, not an infinite lattice, you have to sum all of the interactions. For the infinite lattice, the lattice energy calculation is the same for every ion in a 1:1 salt, so we do the computation for one ion and then multiply by Avogadro s number to get the energy per mole.] The attraction of one O 2- for one Li + at a distance r is -2a/r, because O 2- has charge -2. (i) E = 2-2a/r + a/2r = -3.5a/r. (ii) E = -2a/r - 2a/2r + a/r = -2a/r. The more stable arrangement has the oxide in the middle. 11. (a) Draw a unit cell for the twodimensional structure at right and determine its stoichiometry. Call the smaller shaded circles A and the larger unshaded circles B. (b) Calculate a type of Madelung constant for this structure as we did in class. Pick one of the A ions 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 do you get if you use only the first term? If you use the first two terms? Three terms? Four terms? Five r 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. (c) The Madelung constant is conceptually more complicated in this case; why? Hint: would you get the same constant if part (b) had stated that you should pick a B ion? Don t do the calculation, just think about it. This is tricky. (a) I like the unit cell that is a square with the center of the A ions as the corners (and a hole in the middle). This unit cell contains four quarter A ions, and four half B ions. So the structure has AB 2 stoichiometry. (b) As shown in the diagram below, there are 4 A 2+ B contacts at a distance r. These give a favorable (negative) interaction of 4 Z A Z B /r or -8/r since Z A = +2, ZB = -1). [Since all we re calculating for the moment is the Madelung constant, we don t need to worry about the e 2 N/4πεο(1-1/n)].

5 Answer Key, p. 5 Then there are four A 2+ -A 2+ 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 4/3r. (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.) The first five terms are therefore: M = / /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, The value for U 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 comes from ( ). 2r 2r 2 r 5 r r 12. 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. (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. c

6 Answer Key, p. 6 (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. 13. Describe the trend in the measured lattice energies below, and explain it. salt CaSO 4 SrSO 4 BaSO 4 lattice energy kj/mol kj/mol kj/mol The lattice energies become less favorable (smaller absolute values of negative numbers) upon changing Ca 2+ to Sr 2+ to Ba 2+. This is because the cations get larger, leading to a larger structure, and the lattice energy depends inversely on the distance between the ions. The sulfate is common to all three structures so you don t have to worry about it. 14. Oxtoby Problem (c) Which do you think is more accurate, the value from the approach in part (a) or the approach in part (b)? Why? How many significant figures do you think should be given in the answers to parts (a) and (b)? (a) (b) (c) The Born-Haber cycle of part (a) should be much more accurate, since the values that go into it can be accurately measured (the heats of hydration being a little tricky to measure). The calculation

7 Answer Key, p. 7 in part (b) has a number of approximations in it, such as the neglect of repulsive forces mentioned in Oxtoby s answer, that all ions have a repulsive interaction once their electron clouds start to touch (which is why the ions don t simply get closer and closer together). Normally this is modeled with a Lennard-Jones potential, but Oxtoby chooses to do that analysis only for molecular crystals (Section 21.5), and not extend it to ionic crystals. 15. Look over the paper Synthesis, Computed Stability, and Crystal Structure of a New Family of Inorganic Compounds: Carbonophosphates by H. Chen, G. Hautier, and G. Ceder J. Am. Chem. Soc. 2012, 134, (dx.doi.org/ /ja ). This is a long article and you just need to get the flavor of it or look at specific sections for this question. (a) Where are the authors working? Are they in a chemistry department? Does the work feel like chemistry to you? Can you detect a perspective in the Introduction that indicates the background of the authors? The paper comes from the MIT department of Materials Science and Engineering (MS&E). To me, this feels like pretty basic chemistry but this only serves to indicate how the divisions between fields are blurring. The engineering focus is evident (at least to me) in the very first sentence with the word devices. They are not simply aimed at discovering new science, they want to build devices. (b) The paper reports both experiments and ab initio calculations. What does ab initio mean in this context? Are these calculations like the Coulomb s Law/Madelung constant calculations that we are doing? Ab initio is from the Latin and translates as from the beginning. In the context of chemical calculations, it means that the calculations were done directly from the theory (the Schrödinger equation or the equivalent) without any parameters that might come from experiments. Our Madelung constant computations are not even close to ab initio because they use parameters such as the ionic radii and the parameter describing e -e repulsion. (c) Analyze the composition of the sidorenkite structure shown in Figure 2. What are the charges on the cations and oxyanions? You can predict the oxidation state of the transition metal M based on the requirement for charge balance. The structure is (Na + ) 3 (M 2+ )(PO 4 3 )(CO 3 2 ). (d) All the simple binary carbonates and phosphates (M n+ ) x (CO 3 ) y and (M n+ ) x (PO 4 ) y are known. What is the algebraic relationship between n, x and y that gives stable solids in each case? For (M n+ ) x (CO 3 ) y, xn must equal 2y for charge balance; for (M n+ ) x (PO 4 ) y, xn = 3y. (e) The Figure 2 caption says M = Mn, Fe, etc. Where are the transition metal ions in the drawings of Figure 2? The official definition of sidorenkite is as the pure manganese mineral. The natural mineral is formed in some complex geochemical environment. Do you think that the natural mineral has 100% manganese, or do you think that there are typically a mixture of 2+ cations at the transition metal sites? The transition metal ions are at the center of the purple octahedra in the drawings of Figure 2. Almost all natural minerals have mixtures of cations, because the minerals form in complex environments, not in a laboratory beaker containing purified materials. (f) Explain why one of the compounds in Figure 3 is white while the others are colored. The compounds containing transition metal ions are colored (due to transitions within the d shell). The Mg 2+ compound is white because Mg 2+ is not a transition metal ion and has no d electrons.

8 Answer Key, p. 8 (g) Are you surprised at the color of the Cu derivative? Explain, using the coordination environment of the Cu and a demonstration done in class. The Cu 2+ ions are surrounded by an octahedron of oxide ligands. This is not so different than the Cu(H 2 O) 6 2+ ion demonstrated in class to be light blue due to an absorption at the low-energy edge of the visible range. This absorption (at hν = o ) is not so different in the solid shown here, so it is also blue (although a different shade). (h) The X-ray diffraction (XRD) data shown in Figures 4 and 5 show that the materials are crystalline. Yet the scanning electron microscope (SEM) images in Figure 6 don t show nice crystals with shiny flat faces. How can these both be correct? [Hint: What roughly are the unit cell dimension (Table 3) and what are the scale bars in Figure 6?] The two experiments probe very different length scales. The materials are crystalline on the chemical scale, with hundreds of unit cells nicely aligned next to each other in a crystalline fashion. The unit cells all have dimensions of less than 10 Å or 1,000 pm = 1 nm. So 100 unit cells is only 100 nm, less then 10% of the scale bar in the images. If you look closely, particularly at Figure 6c, you can see little crystallites of about this size. (i) Are you surprised that these materials have not been discovered before? I m not surprised, because mixed anion solids are rare. The anions set the lattice so making a lattice with two different sized anions is more complicated. (j) From the conclusions, what potential technological application do the authors have in mind for these new materials? Batteries. 16. Look at Direction-Specific Interactions Control Crystal Growth by Oriented Attachment D. Li, M. H. Nielsen, J R. I. Lee, C. Frandsen, J. F. Banfield, J. J. De Yoreo Science 2012, 336, What that we ve done in class has to do with the thermodynamics of forming solids, such as lattice energies. We have not discussed the kinetics (the rates) of forming or dissolving solids. This is much more complicated and not yet fully understood. This article presents some neat recent work in this area. (a) Who is the person to whom correspondence should be addressed? What institution is he associated with? Is this a University? (Actually, he has moving to Pacific Northwest National Laboratory in Richland, WA and now has an affiliation with the UW Chemistry Department.) The corresponding author is Jim De Yoreo, who was at Lawrence Berkeley National Laboratory (LBNL). This is a national lab not a university, although it is associated with the University of California (Berkeley). (b) What is a biomineral? Isn t that kind of an odd term, since minerals are not alive? It s not an odd term at all. Many organisms create minerals, often as their homes or part of their internal or external structure (skeletons, teeth, etc.). A biomineral is a mineral made by an organism: (c) How long are the scale bars in Figure 1 of this paper. Note that the scale bars in parts H and I are different from those in A-G. Roughly to one significant figure how big is a typical atom or ion? Give your answer in nm, pm and Å. The images in this paper are quite high resolution, with scale bars of 2 nm (H and I) and 5 nm for the others. A typical atom or ion has a radius of about 2 Å = 200 pm = 0.2 nm.

9 Answer Key, p. 9 (d) What is the chemical composition of the material being imaged in Figure 1? What are the white dots in Figures 1H and 1I? The paper refers to them as iron oxyhydroxide [Fe III (O)(OH)] nanoparticles closely related to six-line ferrihydrite (5Fe 2 O 3 9H 2 O). (e) Watch one of the movies that are available online as Supplementary Materials [Link under Article Views at the top of the left margin.] Is this cool? I think this is very neat, to be able to watch this solid grow in real time. Amazing.

Experiment 7: Understanding Crystal Structures

Experiment 7: Understanding Crystal Structures To do well in this laboratory experiment you need to be familiar with the concepts of lattice, crystal structure, unit cell, coordination number, the different