edlled by FRANK DeHAAN Occidental College LOSAngeles, California 90041
Interpretation of a Patterson Map-A Dry-Lab Experiment in X-Ray Crystallography R i c h a r d E. M a r s h A r l h u r Amos N a ~ c s Lahorolor) of Chrmical Physics C'olrfornia l n s r i t u l r of Twhnolnx, 125 ~ a s a d e n a California91 , a n d C h r i s t e r E. N o r d m a n Uniuersity of Michigan A n n Arbor, Michigan 48104 I n this paper we describe another "dry lab" problem,l whose purpose i s to enable students without access t o some of t h e more modem, sophisticated instruments of chemical research t o become familiar with t h e techniques of interpreting d a t a obtained with thnse instruments. T h e problem we describe here is concerned with t h e interpretation of single-crystal X-ray diffraction data. More specifically, i t involves t h e interpretation of a Patterson m a p , calculated from t h e diffraction d a t a , s o as t o obtain detailed information concernine t h e arraneement of atoms in t h e diffracting crystal. Thin prohlem, o r o n e similar t o it, h a s been resented for several w a r s in undereraduate classes at b o t h t h e University of ~ i c h i ~ aannd a t t h e California Institute of Technology, Experience h a s shown t h a t a s t u d e n t having completed a one-term (ten weeks) course in X-ray diffraction will require a b o u t 10-15 hr t o complete t h e problem.
Figure 1. A typical X-ray diffraction photograph, obtained by the "precession" method (see second reference in Reading Assignment. PP. 122-1331, The darkness of the spot9 measures the relative intensitf I(hk0 due to ditlraction horn the various sets of lanice planer (h.k.0.
is diffracted, in turn, by each set of accessible lattice planes hkl, where h , k, and i are the Miller indices describing each set of planes. The intensity [(hkl) of each beam is corrected for geometric factors and absorptjon to yield a set of values JF(hk1)J" These quantities are the squared magnitudes of the "structure factors," F(hk1). A Fourier transformation of the com~leteset of F(hk1)'s will result in the elecBackground tron density function p ( x i r )throughout the unit cell: since regions of high rlcctnm density must represent the positions of the various The ibllowing is i~ h r i d ncrounl 01' tho exprriments and measureatnms, this function yields a reliableand accurate picture of the entire ments Ihding up 1 0 the cmnputittion ofthe I'nt.terson function. ntrurturp ofthe crystal. However, F(hk1) is a complex quantity, conAsinglr cryr;t;tld : ~ s u l W ~ n cinr .thiscaw l.*l-rlithi;inr,,ismountrd sisting ol'hnth an amplitudeand a phase;since the phase information on :in S-my diffr;whm~tcrur phutngraphic S-my diil'raetian camera. isnot contained in themeasuredvalue of IF(hkllJ', it is not possible and tlw diffwc.lion pallern is rwonled. Figurr I shows part. 01'a to calculate the electron density function from the primary data alone. photographically rvn,rdcd pattern: the entire difl'ractim pattern 01' Unravelling the phases of the structure factors F(hk1) is the notorious the crystal cc,nsisls. 241. the tro peaks 5 an0 6 represent S C vectors Inat have a component y, - yo close to %.
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Volume 54, Number 5, May 1977 1 319
Zy,, %) will lie on the Harker line u = %, w = K, and Peak No. 2 suggests y. = 0.16. The vector (%., 2y., 22.) should then lie a t u = 0.438, u = 0.32, w = 0.078, and Peak No. 8 has these approximate coordinates. Note that this latter peak represents but a single S-S vector, while each of the Harker peaks represents two identical S-S vectors-position (1)-position (3) and (2)-(4), for example. With the sulfur atoms located, we next w i d e r peaks representing the S-C vectors. We might, for example, try Peak No. 10, which lies at about 1.8 A from the origin of the Patterson map and could correspond to a C-S covalent bond. By adding, in turn, the coordinates of the four eouivalent oositions (a)-(d) . . . . of this oeak to the coordinates assigned td the sulf& atom, we obtain four G i b l e sets of coordinates for a carbon atom;of these, theset (0.188, -0.14, -0.041) showsadditional S-C vectors all of which are represented by additional Patterson eaks (Peaks No. 4,5, and 9). Similarly, fm; "2ak No. 7, also a t 1.8 from the origin (taking u = -0.228, which is equivalent to 0.772), we can obtain the set of coordinates (-0.009,0.16,0.150) for the second carbon atom, and these cwrdinates give rise to other S-C vedors at the approximate positionsof Peak No. 6,11, and 12. Slightly improved coordinates for the atoms can be obtained by averaging the ewrdinates of the various Patterson vectors, Leading to a final set of coordinates approximately as follows
8:
Figure 3. A drawing of the crystal stwcture of 1,MiIhianeas viewed along baxis. The decimal numbers are the approximate Y cwldinates Of the various atoms. Vectors 3.5, and 6 are those represented by Prominent maxima in the Haner section (Fig. 2). The symmetry of the Patterson map will generate equivalent peaks a t the four positions
u, u, w 1- u), -0, -w (4 u,-0, w (d) -u, u, -w (a)
(h)
-U(OI
thus, Peak No. 12 will also occur a t (0.272, -0.17, -0.388), (0.728, -0.17,0.388) and (0.272,0.17, -0.388). Peak No. 1, at the origin of the Patterson map, represents the overlap of vectors from each atom to itself-a total of 4 S-S vectors and 8 C-C vectors. Since an S-S vector should be about nine times as high as a C-C vector, we can estimate that an S-S vector should be about 350 in height, a C-S vector about 115,and a C-C about 40. The eleven remaining peaks should represent S-S and C-S vectors, with the largest peaks being S-S. Referring now to the coordinates y., z,of the sulfur atoms, we note that there should he three distinct S-S vectors, with coordinates (Zx,, 2y,, 223, (5, K Zy,, #), and (5 - 2x8, #, % - 22.). This last vector, a t (% - Zx,, %, # - Zz.), must lie on the section u = %of the Patterson map--the "Harker" section; this section is shown in Figure 2. Peak No. 3 is the logical candidate for the Harker vector, suggesting the coordinates I, = 0.219,~.= 0.039. Similarly, the vector (K,'h -
I.,
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Finally, we can explain the relative heights of the various Patterson peaks by taking account of the number of interatomic vectors that contribute. Thus. Peaks 2 and 3 each reoresent two identical S-S vectors. Peaks 4.5.6. and 7 each mores& four aoormimatelv . . eaual . S.C ( o h ) vec&&l'cak R is asingles-s vector, and Pesks9.12 are double S-(.'vectors. Some C-Cvectorseontribute to Peak No. 2; the remaining C-C vectom are repwsented hy additional, small Patterson peaks. 3) From the above coordinateg and the dimensions of the unit cell, we obtain the following dimensions of the 1,4-dithiane molecule
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Bond lengths
S- C1 S- Cs CI- Cz'
1.82 A 1.82 A 1.45 A
Bond angles CI-S- Cs S- CI- CZ' S- C2- C1'
96' 112O 113'
where the primed atoms have coordinates -x, -y, -r. Since the molecule has a center of symmetry, the dimensions of theother half are identical. These dimensions are in good agreement with the more accurate values obtained by least-squares refinement methods.'
