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. [2 pts. for each correct label] CdI 2 CaF 2 n rutile SiO 2 The fields for highest cation/anion coordination numbers and ionicity are at the upper right, and those for lowest cation/anion coordination numbers and highest covalency are at the lower left. Thus CaF 2 (CN Ca = 8; CN F = 4) is at the upper right [2 pts]. The bonding in SiO 2 is highly covalent, and the coordination numbers for this compound are the smallest (CN Si = 4; CN O = 2), so it is at the lower left [2 pts]. Rutile and CdI 2 have the same intermediate coordination numbers (CN M = 6; CN X = 3), but compounds with the rutile structure are more ionic (larger ), and compounds with the CdI 2 structure are less ionic (smaller ). Thus CdI 2 is at the middle left and rutile is at the middle right [3 pts]. 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. [9 pts.] or Line No. 2 sin 2 () m hkl /m 100 *m hkl /m 100 m hkl /m /330 m hkl = (h 2 + k 2 +l 2 ) *Assuming first reflection is 100 Assuming first reflection is Assuming first reflection is 111 3

4 2. (cont.) The crystal structure of this metal is bcc. [3 pts.] Calculate the lattice parameter a: d Å Å 2sin( 2 ) 2sin( ) 2 2 rearrange the d-spacing formula: a d (1 1 0 ) 2( Å) Å [3 pts.] 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. 101 [5 pts.] 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 Assuming first reflection is 111 The structure is not cubic; it must be hcp [1 pt.]. If so, line 3 must be the 101 reflection. Assume that line 1 is 100, line 2 is 002, and calculate the lattice parameters a and c. Then confirm that the 101 reflection appears at the correct 2 value. See the work on the next page. 5

6 3. (cont.) Use the first two reflections to solve for the lattice parameters a and c: d 2sin d100 2 a 3a 2 combine : 3 a 2sin Å a Å [2 pts.] 3sin sin 2 similarly : 12 2 c d002 2 c c 2sin002 2(1.542 Å) c Å [2 pts.] 2sin002 2sin Now check the position of the 101 reflection: d Å 2 2 3(2.653 Å) (4.334 Å) Å sin 2sin 44.64,which matches line 3 [2 pts.] 2 d101 2(2.030 Å) Now guess that line 4 or 5 is the reflection: d (1 1 11) Å 2 3(2.653 Å) Å 2 2sin 71.07,which matches line 5 [2 pts.] 2(1.327 Å) Line 4 must be the 111, 102, or 201 reflection (smallest available indices). Given the values of a and c, the 102 will be at lowest 2 : d Å 2 2 3(2.653 Å) (4.334 Å) Å sin 58.58,which matches line 4 [1 pt.] 2(1.576 Å) 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. [11 pts.] or 221 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 Assuming first reflection is 111 7

8 4. (cont.) Calculate the lattice parameter a: d Å Å 2sin( 2 ) 2sin( ) 2 2 rearrange the d-spacing formula: a d (1 0 0 ) d Å [4 pts.] 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. -1 pt. each missing ion, up to a max. of -2 (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. -1 pt. each missing ion, up to a max. of -2 Anions may occupy either T + or T sites (alternating octants must be occupied). 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. The anti structures in these cases do not exist because the structures generated by exchanging the cation and anion positions are identical to the original 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. The cubic cell is conventional because it has higher symmetry [5 pts.], or because it reveals the full symmetry of the crystal structure [5 pts.]. (b) 5 pts. Please briefly explain the purpose of refining a fitted pattern to experimental XRD data. What primary goals are achieved by refinement? The two primary goals are to confirm the crystal structure [2.5 pts.] and to obtain precise lattice parameters [2.5 pts.]. 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. [3 pts. for each correct set of indices] c a b [ ] [ ] [ ] [ ] [ ] 11

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