Crystal data Formula sum K 0.5 N 0.5 O 1.5 Formula weight 50.55

1. Crystal data Formula sum K0.5 N0.5 O1.5 Formula weight 50.55 Crystal system orthorhombic Space group P n m a (no. 62) Unit cell dimensions a = 6.4360 b = 5.4300 c = 9.1920 Cell volume 321.24 Å3 Z Crystal data Formula sum Cs3 Cl O Formula weight 450.17 Crystal system orthorhombic Space group P n m a (no. 62) Unit cell dimensions a = 9.4430 b = 4.4520 c = 16.3200 Cell volume 686.10Å3 Z

2. Atomic coordinates Atom Wyck. Occ. x y z K1 4c 0.5 0.25510 1/4 0.41640 N1 4c 0.5 0.41560 1/4 0.75510 O1 4c 0.5 0.40980 1/4 0.89070 O2 8d 0.41290 0.45010 0.68640 TITL *Niter-K(NO3)-[Pnma]-Holden J R, Dickinson C W CELL 1.54180 6.436 5.430 9.192 90.0 90.0 90.0 SYMM P n m a (62) UNIT 8 60 24 4 SFAC K N O K1 1 0.25510 0.25000 0.41640 10.5000 = 0.03110 0.23800 0.02480 0.00000 0.00110 0.00000 N1 2 0.41560 0.25000 0.75510 10.5000 = 0.01920 0.02500 0.02970 0.00000 0.00160 0.00000 O1 3 0.40980 0.25000 0.89070 10.5000 = 0.05000 0.04070 0.02720 0.00000 -0.0080 0.00000 O2 3 0.41290 0.45010 0.68640 11.0000 = 0.04930 0.02680 0.03920 -0.0038 0.00520 0.00640 END

4. p. 58

5. Cell unchanged but with lower crystal class Cell changed with the same symmetry Cell changed with a different Bravais Lattice

6. Space groups (and enantiomorphous pairs) that are uniquely determinable from the symmetry of the diffraction pattern and from systematic absences are shown in bold-type. Point groups w/o inversion centers or mirror planes are emphasized by boxes.

7. Space groups (and enantiomorphous pairs) that are uniquely determinable from the symmetry of the diffraction pattern and from systematic absences are shown in bold-type. Point groups w/o inversion centers or mirror planes are emphasized by boxes.

9. (hk) (hk) (hk) (hk) (hk) (hk) (hk) (hk) Equivalent planes (hk) {hk} [hk] <hk> - - - - - - - - - - - - * - - - * - - - - - - * * * * * * * * a* =60o a* A complete data set covers all 8 octants of r.l. points. p.61

10. Hexagonal Axes vs. Rhombohedral Axes Two ways to relate rhombohedral indices to hexagonal indices, the obverse and reverse relationship. The hexagonal cell : a1, a2, c Rhombohedral cell: r1, r2 , r3 Obversehexagonal axes: a1 = r2 – r3 a2 = r3 – r1 c = r1 + r2+ r3 Reverse hexagonal axes: a1 = r3 – r2 a2 = r1 – r3 c = r1 + r2+ r3

11. Hexagonal Axes vs. Rhombohedral Axes The hexagonal cell : a1, a2, c and indices (h k .) Rhombohedral cell: r1, r2 , r3and indices (m n p) Obverse H vs. R: Obversehexagonal axes: a1 = r2 – r3 a2 = r3 – r1 c = r1 + r2+ r3 h = n – p k = -m + p  = m + n + p -h + k +  = 3p Reverse hexagonal axes: a1 = r3 – r2 a2 = r1 – r3 c = r1 + r2+ r3 h - k +  = 3p

12. 5. Parameters in intensity data collection Choice of wavelength Parameters in intensity data collection Choice of wavelength Resolution: 2/  |hmax| = dhk-1 dhk  ½  Resolution: 2/  |hmax| = dhk-1 dhk  ½ 

14. Data Processing 1. Data Reduction Preliminary manipulation of intensities—their conversion to a correct, more usable form Decay correction Cause 1: due to unstable crystal – decomposing or slipping Cause 2: due to unstable X-ray source – instrument misalignment or instability in tube voltage Point-detector case: Corrections can be made based on a certain standard reflections Typical behavior of the relative intensities of threereflections of a crystal monitored with a diffractometer at different times after exposure begins

15. Lp correction “L” stands for Lorenz factor: Cause : the r.l. point have a non-negligible volume so that it will have different angular speed when passing through the Ewald sphere (i.e. a higher intensity when in diffracting position for a longer time). angular velocity linear velociy at point p • Ltime = / (|r*|cos) • |r*| = 2sin /  • L = 1/sin2 (the simplest possible form) Different forms of the L factor may be given for different experimental arrangements.

16. unpolarized rays E+ E//cos2θ E+ E// Io E2+ E// 2 Diffraction circle “p“ stands for polarization factor The diffracted X-ray beam is polarized relative to the incident beam. s1 s0 IE2+ (E// cos2θ)2 p = I/Io Case 1: no monochromator : p = (1+ cos22θ)/2 Case 2: with monochromator :The beam is further polarized when monochrator(s) is (are) used. (a) when s0, s1,ands2 are co-planar, p = (1+ cos22θcos22θM)/ (1+ cos22θM) (b) when s0, s1,and s2 are not co-planar, p = (cos22θ + cos22θM)/ (1+ cos22θM) Lp factor: (1+ cos22θ)/2sin2 andIrel= Iobs/Lp ; Irel= I/Lp

17. Absorption correction (1) Relative transmission factor plotted as a function of angle  for a reflection chosen with a  value close to 90o. The  rotation curve can be used to make an absorption correction. (2) Empirical absorption correction may also be applied based on the intensityvariations in symmetry-equivalent reflections.  rotation Transmission factor ABSORPTION Correction (i) applied before refinement (in the data reduction stage) (ii) applied during refinement by an input of the precise description of the crystal shape I = Ioe-t The absorption effect depends on the crystal’s shape, size, and density. This effect is much more severe at low 2θ angles. The path lengths of the beams reflected from the two small elements of the crystal, A and B, are different for different reflections

18. The intensity data are averaged over all symmetry-equivalent reflections. Friedel’s Law indicating that Ihk Ihk (Ihk Ihk ) Eleven Laue Symmetry Groups 1 2/m mmm 4/m 4/mmm 3 3m 6/m 6/mmm m3 m3m – – – – – 2. Data Averaging Iiave = N Ii /N ( i = 1 to n, n = no. symmetry equivalents ) The value of I is used to decide which datamis a real signal or just a noise. Rint = (Ii -Iiave)/Ii HW: Listthe intensities and their esd’s for all symmetry equivalents of the reflections (3 2 6), (0 2 6), (9 0 6) and (3 3 0) in Xtal01. Calculate their average I and sigma I. What is the Rint just for this group of reflections?

19. Pattern Decompostion Extract Bragg-peak intensity from powder pattern

21.  (x,y,z) {Fhkl} FT FT Electron-density function

22. (x,y,z) Fhklexp(ihkl) FT FT Chapter 8 Structure Solution (1)The phase problem To solve a crystal structure is to solve the phase problem. Why? Simply because the “phase” of the diffracted wave is missing in diffraction intensity measurements, i. e., only the amplitudes of the diffracted waves are measured in experiments: Ihkl FhklF*hkl→Fhkl phase angle hkl = tan-1(B/A) Complex form Fhkl= Ahkl + iBhkl = |Fhkl|exp(ihkl)

25. How to Solve the Phase Problem? 1. Patterson manipulation methods {Fhkl} FT cal Ihkl→ P(u,v,w)→ (xH,yH,zH) → {hkl} cal Bragg intensity Patterson Function some located atoms initially derived phases Heavy-Atom methods; Superposition methods

29. Heavy-Atom methods; To find the position of a heavy atom, one must utilize the “Harker vectors”, which correspond to vectors formed between t symmetry-related atoms. For example, in the space group P21/c, there are three kinds of Harker vectors, namely, (u,v,w), (u,½,w), and (0,v, ½).

30. The two chlorine atoms are at (0.113, 0.912, 0.080) and (0.295, 0.731, 0.383).The first 23 strongest Patterson peaks are shown to the right: Harker lines of (0,v, ½) type are: peaks #2, #8, #16 Harker planes of (u,½,w) type are: #3, #10, #11, #15 It is clear to see that from peaks #2 and #3, the atomic coordinate of the first chlorine atom, Cl1, could be derived; and from peaks #8 and #10, the coordinates of the second chlorine atom , Cl2, could be obtained.

31. (x,y,z) Fhklexp(ihkl) FT obs cal FT 2. Direct methods

33. Crystal Structure Determination and Refinement Using the Bruker AXS SMART APEX System

34. Flowchart for Method Adapted from William Clegg “Crystal Structure Determination” Oxford 1998.

35. Select and Mount the Crystal • Use microscope • Size: ~0.4 (±0.2) mm • Transparent, faces, looks single • Epoxy, caulk, oil, grease to affix • Glass fiber, nylon loop, capillary

36. Goniometer Head

37. Goniometer

38. Goniometer Assembly

39. SMART ASTRO setup data collection strategy sample screening data collection SAINTPLUS new project change parameters SAINT: integrate SADABS: scale & empirical absorption correction SHELXTL new project XPREP: space group determination XS: structure solution XL: least squares refinement XCIF: tables, reports project database default settings detector calibration

40. George M. Sheldrick Professor, Director of Institute and part-time programming technician1960-1966: student at Jesus College and Cambridge University, PhD (1966)    with Prof. E.A.V. Ebsworth entitled "NMR Studies of Inorganic Hydrides"1966-1978: University Demonstrator and then Lecturer at Cambridge University; Fellow of Jesus College, CambridgeMeldola Medal (1970),  Corday-Morgan Medal (1978)1978-now: Professor of Structural Chemistry at the University of GoettingenRoyal Society of Chemistry Award for Structural Chemistry (1981)Leibniz Prize of the Deutsche Forschungsgemeinschaft  (1989)Member of the Akademie der Wissenschaften zu Goettingen (1989)Patterson Prize of the American Crystallographic Association (1993) Author of more than 700 scientific papers and of a program called SHELX Interested in methods of solving and refining crystal structures (both small molecules and proteins) and in structural chemistryemail:  gsheldr@shelx.uni-ac.gwdg.defax:  +49-551-392582

41. (1) Concept of the least-squares refinements Mathematical basis of Least Squares method • A series of unknowns: X1, X2, …., Xm • A series of observations: f1, f2, …., fn a11X1 + a12X2 + …+ a1m Xm = f1 • the coefficients a’s are known and (i) more equations than unknowns, i. e. n > m, (ii) the observations are not perfect (iii) these n equations are not fully consistent Need “Least squares” method !

42. i. e. min (S) = min (i wi ei2 ) weighting factor A: Linear case: n x m m x1 a11X1 + a12X2 + …+ a1m Xm = f1 a21X1 + a22X2 + …+ a2m Xm = f2 … an1X1 + an2X2 + …+ anm Xm = fn {aij} are knownand n > m The error: e1 =a11X1 + a12X2 + …+ a1m Xm– f1 e2 = a21X1 + a22X2 + …+ a2m Xm – f2 … en = an1X1 + an2X2 + …+ anm Xm – fn A X = F n x1 n equations for n observations and to solve m unknowns We want to get Xi’s when S = e12 + e22 + … + en2 is minimum S is the sum of squares of what you calculate minus what you observed

43. Minimizing “S” (a) substitute equations for S (b) find the minimum For all j = 1 , 2 , 3 , · · · · · · , m (c) then we obtain the normal equation i = 1 , 2 , · · · · · · · , m the normal equation (d) Solve the m simultaneous eqn for x, i.e. the estimate of

44. or * must calculate matrix B mm B-1 B Normal eqns the soln: common variance Question: How good are X’s? *must estimate the “precision of the derived unknowns ( parameters )” Define : variance - covariance matrix Correlation coefficient: (ij= ji)

45. Common variance Example: x1 = 2 x2 = 4 m = 3 x3 = 6 n = 5 2x1+3x2+x3 = 21 x1+2x2+x3 = 17

46. If accept x1,x2,x3 = 2 ,4 ,6 at first The L.S yields And the correlation function is B: Non-linear case let

47. = if Single-crystal case: the Structure Factor Unknowns : (xj,yj,zj), Bj,·······etc

48. The function to be minimized is ( ) å 2 = - o c R w F F r h r h ¶ c m F å - = o c F F ¶ P = 1 j j p. 115 p.117

49. The normal equation L. S. Procedure for non-linear case : (i) guess Xjo (ii) form F: fi = fi - fio (fiobs - fical) (iii) calculate or approximate aij, i.e. fi/ Xi (iv) set up normal equations and solve for Xj (v) Xj‘ = Xjo +  Xj (vi) go back to (ii) unless  Xj<< (Xj), i. e. convergence obtained when { Xj/ (Xj)} << 0.05

50. When a=b , =  =  = 90° Find a* c* for series of hkl powder lines at position q We need Powder case Powder indexing: having series of powder lines knowning their Bragg angles at which the lines occur Normal equation: where