FIRST MIDTERM EXAM Chemistry March 2011 Professor Buhro

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1 FIRST MIDTERM EXAM Chemistry March 2011 Professor Buhro Signature Print Name Clearly ID Number: Information. This is a closed-book exam; no books, notes, other students, other student exams, or any other resource materials may be consulted or examined during the exam period. Calculators are permitted. Partial credit will be given for partially correct reasoning in support of incorrect or correct final answers. Additional space for answers is provided at the end of this exam; please clearly label any answers you place there. Please find Potentially Useful Information attached as the last pages of this exam. 1. (15 pts) 2. (15 pts) 3. (15 pts) 4. (15 pts) 5. (15 pts) 6. (10 pts) 7. (15 pts) Total (100 pts) 1

2 1. 15 total pts. An incomplete Mooser-Pearson plot for MX 2 compounds is given below. A version of this plot was given on the practice exam and key, although the answer given was partially incorrect. There are four structure fields (regions) on this plot, as shown below. One field (region) contains compounds having the fluorite (CaF 2 ) structure, one contains compounds having the rutile structure, and one contains compounds having either CdI 2 or CdCl 2 structures. The remaining field contains the compound SiO 2 having the -quartz or related structures in which Si atoms have a coordination number of 4 and the O atoms have a coordination number of 2. Please assign each of these four fields by adding the labels CaF 2, rutile, CdI 2, and SiO 2. Then, in the space below, please write a brief justification for the assignments you have made. n 2

3 2. 15 total pts. The XRD pattern below was obtained from a powdered metallic specimen having a conventional crystal structure for a metal. Note that the sine-squared-theta ratios are also given. Please assign Miller indices to all the reflections in the pattern, identify the crystal structure of the metal by name (type), and calculate the lattice parameter or parameters. Show your work, using this and the following page. Line No. 2 sin 2 ( ) m hkl /m 100 *m hkl /m 100 m hkl /m m hkl = (h 2 + k 2 +l 2 ) *Assuming first reflection is 100 Assuming first reflection is 110 Assuming first reflection is 111 3

4 2. (cont.) 4

5 3. 15 total pts. The XRD pattern below was obtained from a powdered metallic specimen having a conventional crystal structure for a metal. Note that the sine-squared-theta ratios are also given. Please assign Miller indices to all the reflections in the pattern, identify the crystal structure of the metal by name (type), and calculate the lattice parameter or parameters. Show your work, using this and the following page. Line No. 2 sin 2 ( ) m hkl /m 100 *m hkl /m 100 m hkl /m m hkl = (h 2 + k 2 +l 2 ) *Assuming first reflection is 100 Assuming first reflection is 110 Assuming first reflection is 111 5

6 3. (cont.) 6

7 4. 15 total pts. The XRD pattern below was obtained from a powdered specimen having the CsCl structure. Note that the sine-squared-theta ratios are also given. Please assign Miller indices to all the reflections in the pattern, and calculate the lattice parameter or parameters. Show your work, using this and the following page. Line No. 2 sin 2 ( ) m hkl /m 100 *m hkl /m 100 m hkl /m m hkl = (h 2 + k 2 +l 2 ) *Assuming first reflection is 100 Assuming first reflection is 110 Assuming first reflection is 111 7

8 4. (cont.) 8

9 5. 15 total pts. (a) 5 pts. The unit cell for the rock-salt (NaCl) structure is shown below. The smaller cations are black and the larger anions are gray. Note that an anion is positioned at the origin of the unit cell. Please resketch the cell after moving a cation to the origin, using the frame on the lower right. (b) 5 pts. The unit cell for the zinc blende (ZnS) structure is shown below. The smaller cations are black and the larger anions are gray. Note that an anion is positioned at the origin of the unit cell. Please resketch the cell after moving a cation to the origin, using the frame on the lower right. 9

10 5. (c) 5 pts. Although many crystal structures have anti relatives, such as fluorite and antifluorite, other crystal structures do not. Please explain why there are no anti-rock-salt, anti-zinc-blende, or anti-cscl structures total points. (a) 5 pts. The cubic-close-packed crystal structure can be described by a cubic unit cell or a hexagonal unit cell. Please explain briefly why the cubic unit cell is conventionally used. (b) 5 pts. Please briefly explain the purpose of refining a fitted pattern to experimental XRD data. What primary goals are achieved by refinement? 10

11 7. 15 total pts. Unit cells from a cubic lattice are depicted below. Please assign Miller indices to the crystallographic directions depicted by each of the arrows placed within the unit cells. Write your answers in the brackets provided. c a b [ ] [ ] [ ] [ ] [ ] 11

12 Extra work space for any prior problem Please clearly label any work here that is to be graded. 12

13 Potentially Useful Information Engel-Brewer rules: s, p electrons structure bcc hcp fcc 4 diamond (cubic) 1 Å = 10 8 cm = m = 0.1 nm = 100 pm Cu K radiation ( = Å) n = 2dsin d hkl = a(h 2 + k 2 + l 2 ) -1/2 for cubic systems (a = b = c, = = = 90 o ) 1/ l d hkl ( ) 2 h k hk 2 for hexagonal systems (a = b c, = = 90 o, = 120 o ) 3a c d hkl = (h 2 /a 2 + k 2 /b 2 + l 2 /c 2 ) -1/2 for orthonormal (including tetragonal and orthorhombic) crystal systems ( = = = 90 o ); for tetragonal systems, a = b c for orthorhombic systems, a b c sin 2 ( 1 )/sin 2 ( 2 ) = m 1 /m 2, where m = (h 2 + k 2 + l 2 ) In an ideal hcp metal, in which all 12 nearest neighbors are equidistant, c/a = The area of a parallelogram is: b = base h = height area = b h h parallelepiped The volume of a parallelepiped is: b base area = parallelogram area h = height parallelepiped volume = (base area) h h The volume of a sphere is: V sphere = r 3 13

14 Definitions of sine and cosine functions: sin = opposite side/hypotenuse (of a right triangle) cos = adjacent side/hypotenuse (of a right triangle) Lowest-angle reflections in XRD powder patterns: Primitive-cubic lattice 100 Primitive but non-cubic lattice 100, or 010, or 001 bcc lattice 110 fcc lattice 111 simple hcp-based structure 100 or 002 Reflections present (allowed by symmetry): primitive (P) all hkl may be present* body-centered (I) h + k + l = 2n (even)* face-centered (F) hkl are all odd or all even* A-centered (A) k + l = 2n* B-centered (B) h + l = 2n* C-centered (C) h + k = 2n* Systematic absences (extinctions): primitive (P) no absences required by lattice type* body-centered (I) h + k + l = odd* face-centered (F) 100, 110, 210, 211, (300, 221), 310, 320, 321, etc.* diamond (F) 100, 110, 200, 210, 211, (300, 221), 310, 222, 320, 321, etc. 14