Homework 1 (not graded) X-ray Diffractometry CHE 380.45 Multiple Choice 1. One of the methods of reducing exposure to radiation is to minimize. a) distance b) humidity c) time d) speed e) shielding 2. X-rays are different from some other types of electromagnetic radiation because x-rays are. a) electrons b) photons of very long wavelength c) nonionizing d) atoms e) ionizing 3. Which of the following is not a method used to grow single crystals: a) solvent diffusion b) vapor diffusion c) reactant dilution d) sublimation e) slow solvent evaporation 4. What is the most significant problem in solving a crystal structure data set? a) the unit cell problem b) the reduction problem c) the structure factor problem d) the phase problem e) the refinement problem 1
5. A Level A alert in checkcif indicates. a) a general alert to is mostly FYI b) something that should be checked and explained, but is unlikely to be serious c) potentially a serious problem d) generally a serious problem that must be corrected before publication e) the structure has no hope of being refined suitably for publication 6. Which of the following programs is typically used for structure refinement? a) SIR-97 b) ORTEP3 c) XCAD4 d) Mercury e) SHELXL 7. Which command, when placed in the.ins file, instructs the computer to write a.cif file which contains final refinement and bond parameter information? a) ACTA b) PLAN c) HFIX d) MERG e) ANIS Short Answer 8. Unit Cells Identify the most likely crystal system for each of the following: a) Crystal Structure Determination of C 91.5 H 160.5 I 0.5 KO 14.5 Sm 6. Data were collected at 173 K on a Enraf-Nonius CAD4 diffractometer with graphite monochromated Mo Kα radiation (λ= 0. 71073 Å). Crystal data: M = 2497.35,????? space group??? (No.???), a = 26.6451(9) Å, b = 26.6451(9) Å, c = 26.6451(9) Å, α = 90, β = 90, γ = 90, V = 18917.0(11) Å 3, Z = 8, D calcd = 1.75 Mg/m 3, abs coeff = 3.93 mm -1, R (F; F 2 > 2σ) = 0.059, Rw (F 2 ; all data) = 0.105, theta range for data collection 3.42-26.02, largest diff peak and hole 1.24 and -0.83 e Å-3 (near I), reflections collected 11862, independent reflections 3095 [R(int) = 0.053], reflections with I >2σ(I) 1938. Data collection KappaCCD, program package WinGX, abs correction MULTISCAN, refinement using SHELXL- 97, drawing using ORTEP-3 for Windows. cubic b) Compound 6a CH 3 OH in the table below monoclinic c) Compound 1b in the table below triclinic 2
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9. Once a quality single crystal is obtained, list, in chronological order, the major steps involved in determining an X-ray structure and publishing the results, and briefly describe what takes place at each step. Prepare Crystals see notes & text Collect diffraction data see notes & text Solve crystal structure see notes & text Refine crystal structure see notes & text Validation and checking see notes & text Repeat last two as needed see notes & text Prepare figures, tables, etc see notes & text Archive data and results see notes & text Interpret and communicate results see notes & text 10. Unit Cells a) List the seven crystal systems: (See table 3.3.) triclinic, monoclinic, orthorhombic, trigonal, hexagonal, tetragonal, cubic b) Provide the common metric dimensions (a, b, c, α, β, γ ) tabulated for 3 of the above crystal systems (3 pts): (See table 3.3.) triclinic given given given given given given monoclinic given given given given 90 given orthorhombic given given given 90 90 90 trigonal given = a given 90 90 120 hexagonal given = a given 90 90 120 tetragonal given = a given 90 90 90 cubic given = a = a 90 90 90 c) Can a triclinic crystal system, have metric dimensions α = β = γ = 90º? Explain why or why not. Yes, there is no restriction as to the cell lengths and angles for triclinic; usually when α = β = γ = 90º, the crystal system is of higher symmetry, but if the molecules pack in lower (triclinic) symmetry but happen to have α = β = γ = 90º, then the crystal system is triclinic. 4
d) Draw in the smallest unit cell on the following diagram : 11. a) The ISU diffractometer uses Mo Kα radiation (0.71 Å). We rarely collect data above θ = 29º because the diffraction becomes too weak to observe. From the Bragg equation, what is the minimum d-spacing that can be observed to this resolution in θ? 2dsin(theta) = n lambda 2 d sin29 = 1 (0.71) d = 0.732 angstroms b) Cr Kα (λ = 2.29 Å) radiation is rarely used compared to Mo Kα radiation (0.71 Å) or Cu Kα radiation (1.54 Å). Keeping your answer to part a. in mind, propose an explanation. longer wavelength means a larger measurable d spacing at a given theta; measuring to d = 0.732 angstroms with Cr is impossible, theta = 90º is the maximum theoretical resolution and for Cr, that is d = 1.145 Å. The real resolution will be significantly worse since diffraction intensity fades with increasing theta. 12. Galactose (C 6 H 12 O 6 ) is a monosaccharide which may be formed as the direct product of the photosynthetic combination of carbon dioxide and water. The β-d cyclic form of galactose crystallizes in an orthorhombic unit cell with a = 7.699 Å, b = 7.773 Å, c = 12.641 Å. With Z = 4, crystallographically the density, ρ, (to 3 sig. figs.) is: g cm -3. ρ = mz(n A V) V = l x w x h when all angles are 90º; ρ = (179.94 g mol -1 )(4)/[(6.02x10 23 )(7.699 x 10-8 cm)(7.773 x 10-8 cm)(12.641 x 10-8 cm) ρ = 1.58 g cm -3 5
13. (An orthorhombic crystal (diffraction group mmm) has an intense diffraction peak at (8 6 2). By definition (and excluding anomalous dispersion effects), list any other peaks in the complete diffraction pattern which must have the same intensity. (-8-6 -2), (-8 6-2), (8-6 2), (8-6 -2), (-8 6 2), (-8-6 2), (8 6-2) 14. Shown below is the observed diffraction pattern when k = 0, + h is plotted to the right on the horizontal axis and + l is plotted downward on the vertical axis. Any intensity less than 17 (dark blue +) may be considered noise. (Assume that all h, k, or l greater than 1 were measure and any systematically absent would show as a dark blue +.) a) Place an X at (h k l) = (0 0 0). (Red X in center) b) What is the significance of (h k l) = (0 0 0). It is the direct undiffracted X-ray beam. c) Circle the (-3 0 7) reflection. d) Which reflection: (-1 0 2), (0 0-14), (3 0 3), or (4 0 6) is at the largest value of θ? e) What information does the intensity of each reflection provide? Gives atomic arrangement f) What information do the locations of the reflections provide? Gives crystal lattice (unit cell) information g) The space group of this structure is P2 1 2 1 2 1 so it must contain several 2 1 screw axes. Place a box around the reflections below that are consistent with one of the 2 1 screw axes. There are two choices, both shown. 6
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15. Consider the.ins file shown on the next two pages to completed the following: a) What is the formula of the asymmetric unit in the structure? C 7 H 15 N 3 O 5 ClNi 0.5 b) What do the occupancy and/or the coordinates of the Ni atom in this structure indicate about the structure? The 10.5000 indicates only half of a Ni is in the asymmetric unit the positions 0.5 0 0.5 tell you the Ni atom is centered on a face of the unit cell c) Three of the atoms have labels that are inconsistent with their structure factor. Which are they, and what type of atom would they be refined as? O2 would be refined as a H; O3 would be refined as a N; O5 would be refined as a Cl; d) Which non-hydrogen atoms have been refined as isotropic? O2, O3, O4, O5 e) Which non-hydrogen atoms have been refined as anisotropic? (-0.5 for each extra or missing) Cl1, NI1, N1, N2, N3, O1, C1, C3, C5, C9, C11, C12, C13, C12A, C13A (3 pt) f) What is the purpose of the CONF command? To have shelx calculate torsion angles and list them in the <name>.cif g) Which hydrogen atoms have been refined using a riding model (have been fixed based on the location of parent atoms)? H1A/B, H3A/B, H9A/B h) Which hydrogen atoms have been freely refined (likely assigned based on Q peaks)? H1, H5 (1 pt) i) Which carbon atom(s) will have (a) hydrogen atom(s) fixed to it after the next refinement sequence? C11, C12, C12A and C13A j) Why are the errors on the unit cell parameters α and γ zero? The crystal system is monoclinic, so those angles are defined to be exactly 90 degrees k) How big was the measured crystal? 0.10 0.40 0.50 mm (1 pt) l) How many Q peaks will appear after the next refinement sequence? 4 (1 pt) 8
TITL import01 in P 21/n CELL 0.71073 7.9150 17.2430 9.1270 90.000 106.442 90.000 ZERR 4.00 0.0030 0.0020 0.0030 0.000 0.001 0.000 LATT 1 SYMM 1/2 - X, 1/2 + Y, 1/2 - Z SFAC C H N O CL NI UNIT 28 60 12 20 4 2 MERG 2 EADP C13 C13A EADP C12 C12A HFIX 33-1.50 C11 HFIX 23-1.20 C12 C13 C12A C13A FMAP 2 PLAN 4 SIZE 0.10 0.40 0.50 ACTA BOND $H CONF WGHT 0.14670 2.81140 L.S. 4 TEMP -100.00 FVAR 0.98638 MOLE 1 CL1 5 0.620755 0.328449 0.582968 11.00000 0.03898 0.05259 = 0.04139-0.00464 0.00903-0.00274 O2 2 0.507692 0.367964 0.650053 11.00000 0.08703 O3 3 0.734453 0.382211 0.540188 11.00000 0.10463 O4 4 0.527757 0.286656 0.452478 11.00000 0.08894 O5 5 0.731124 0.277339 0.693504 11.00000 0.11069 MOLE 2 NI1 6 0.500000 0.000000 0.500000 10.50000 0.02327 0.02689 = 0.03809 0.00305 0.00806-0.00230 N1 3 0.720296 0.017294 0.695696 11.00000 0.03159 0.04588 = 0.04654-0.00157 0.00381 0.00161 N2 3 0.358340 0.057074 0.641134 11.00000 0.02934 0.04003 = 0.05508-0.00607 0.01603-0.00187 N3 3 0.581650 0.102199 0.402772 11.00000 0.04284 0.03009 = 0.05942 0.00704 0.02260-0.00220 O1 4-0.000727 0.131524 0.653259 11.00000 0.03990 0.05265 = 0.09181-0.00697 0.03602-0.00815 H1 2-0.095796 0.153943 0.609332 11.00000 0.05281 C1 1 0.248146-0.002878 0.686507 11.00000 0.04209 0.05385 = 0.06293 0.01421 0.02925 0.00833 AFIX 23 H1A 2 0.135420-0.007521 0.605006 11.00000-1.20000 H1B 2 0.220660 0.014462 0.780606 11.00000-1.20000 AFIX 0 C3 1 0.243533 0.119883 0.548912 11.00000 0.05620 0.04301 = 0.06219 0.00793 0.03031 0.00627 AFIX 23 H3A 2 0.319504 0.152748 0.504968 11.00000-1.20000 H3B 2 0.157213 0.094410 0.462188 11.00000-1.20000 AFIX 0 C5 1 0.142953 0.172715 0.623313 11.00000 0.04308 0.05710 = 0.07353-0.00423 0.02737-0.00530 H5 2 0.223409 0.191001 0.722641 11.00000 0.05103 C9 1 0.666248 0.081474 0.285247 11.00000 0.06214 0.04941 = 9
0.06720 0.01313 0.03386 0.00182 AFIX 23 H9A 2 0.756453 0.120896 0.283111 11.00000-1.20000 H9B 2 0.577102 0.081618 0.184196 11.00000-1.20000 AFIX 0 C11 1 0.079146 0.241683 0.523519 11.00000 0.06709 0.05175 = 0.10221 0.00938 0.04489 0.01500 PART 1 C12 1 0.665362 0.052934 0.823846 10.60000 0.05277 0.05730 = 0.02709-0.00069-0.00389 0.01210 C13 1 0.493688 0.100517 0.754282 10.60000 0.04630 0.05864 = 0.03221-0.00956 0.00723-0.00049 PART 2 C12A 1 0.669878 0.082846 0.781288 10.40000 0.05277 0.05730 = 0.02709-0.00069-0.00389 0.01210 C13A 1 0.493643 0.069258 0.803373 10.40000 0.04630 0.05864 = 0.03221-0.00956 0.00723-0.00049 AFIX 0 MOLE 0 HKLF 4 END 16. In the textbook, problem 2.1 See page 353 of text. 17. In the textbook, problem 4.1 See page 355 of text. 18. In the textbook, problem 5.1 See page 356 of text. 19. In the textbook, problem 5.2 (Skip part c) See page 356 of text. Hint: on 16 thru 19, it is quite reasonable for some that you may need to look at the answer to figure out how to work this type of question, but once you have worked one, it should be relatively easy to work analogous questions (which I can dream up.) 10
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