Basic Crystallography Part 1. Theory and Practice of X-ray Crystal Structure Determination

Transcription

1 Basic Crystallography Part 1 Theory and Practice of X-ray Crystal Structure Determination

2 Course Overview Basic Crystallography Part 1 n Introduction: Crystals and Crystallography n Crystal Lattices and Unit Cells n Generation and Properties of X-rays n Bragg's Law and Reciprocal Space n X-ray Diffraction Patterns from Crystals Basic Crystallography Part 2 n Review of Part 1 n Selection and Mounting of Samples n Unit Cell Determination n Intensity Data Collection n Data Reduction n Structure Solution and Refinement n Analysis and Interpretation of Results

3 Introduction to Crystallography

4 What are Crystals? 4

5 Examples of Crystals

6 Examples of Protein Crystals

7 Growing Crystals Kirsten Böttcher and Thomas Pape

8 We have a crystal How do we get there? we want a structure!

9 Crystal Systems and Crystal Lattices

10 Foundations of Crystallography n Crystallography is the study of crystals. n Scientists who specialize in the study of crystals are called crystallographers. n Early studies of crystals were carried out by mineralogists who studied the symmetries and shapes (morphology) of naturally-occurring mineral specimens. n This led to the correct idea that crystals are regular threedimensional arrays (Bravais lattices) of atoms and molecules; a single unit cell is repeated indefinitely along three principal directions that are not necessarily perpendicular.

11 What are Crystals? n A crystal or crystalline solid is a solid material whose constituent atoms, molecules, or ions are arranged in an orderly, repeating pattern extending in all three spatial dimensions. 11

12 The Unit Cell Concept Ralph Krätzner

13 Unit Cell Description in terms of Lattice Parameters n a,b, and c define the edge lengths and are referred to as the crystallographic axes. a β c b γ α n α, β, and γ give the angles between these axes. n Lattice parameters à dimensions of the unit cell.

14 A crystal is a homogenous solid formed by a repeating, three-dimensional pattern of atoms. A A A A A A A A A A A A A A A A The unit cell is the repeating unit with dimensions of a, b, c and angles α, β and γ. A crystal can be described completely by translations of the unit cell along the unit cell axes. + b A A + a A A + a A b a A + a

15 Choice of the Unit Cell (sometimes confusing)

16 Choice of the Unit Cell A B A B C D C No symmetry - many possible unit cells. A primitive cell with angles close to 90º (C or D) is preferable. The conventional C-centered cell (C) has 90º angles, but one of the primitive cells (B) has two equal sides.

17 The shortest possible introduction to crystallography There are seven types of unit cells (crystal systems). Combined with centering, we obtain the 14 Bravais lattices. α = β = γ = 90 α = γ = 90 Crystal systems: triclinic monoclinic orhorhombic tetragonal trigonal/hexagonal cubic Bravais lattices: ap mp, mc op, oa, oi, of tp, ti hp, hr cp, ci, cf α = β = 90 γ = 120 a = b α = β = γ = 90 a = b α = β = γ = 90 a = b = c P : primitive, A,B,C : face centered I : body centered F : (all-)face centered R : rhombohedral centered

18 The shortest possible introduction into crystallography The space group is the combination of Bravais lattice + symmetry of the crystal. Point group symmetry of a molecule does not necessarily imply that this symmetry is also present in the crystal. A A A A A A A A A A A A A A A A Pm The unit cell contains: two molecules one molecule A A A A A A A A A A A A A A A A Independent values for both distances Both distances are identical due to symmetry

19 The shortest possible introduction into crystallography The asymmetric unit is the part of the unit cell, from which the rest of the unit cell is generated using symmetry operations. To build the complete crystal we need only the space group and the atom positions in the asymmetric unit. A A A A A A A A A A A A A A A A Pm A A A A A A A A A A A A A A A A Independent values for both distances Both distances are identical due to symmetry

20 7 Crystal Systems - Metric Constraints n Triclinic - none n Monoclinic - α = γ = 90, β 90 n Orthorhombic - α = β = γ = 90 n Tetragonal - α = β = γ = 90, a = b n Cubic - α = β = γ = 90, a = b = c n Trigonal - α = β = 90, γ = 120, a = b (hexagonal setting) or α = β = γ, a = b = c (rhombohedral setting) n Hexagonal - α = β = 90, γ = 120, a = b

21 Bravais Lattices n Within each crystal system, different types of centering produce a total of 14 different lattices. n P Simple n I Body-centered n F Face-centered n B Base-centered (A, B, or C-centered) n All crystalline materials can have their crystal structure described by one of these Bravais lattices.

22 Bravais Lattices

23 Bravais Lattices

24 Crystal Families, Crystal Systems, and Lattice Systems

25 Example: n The monoclinic space groups: n P2 P2 1 C2 n Pm Pc Cm Cc n P2/m P2 1 /m C2/m C2/c 2: two fold axis 2 1 : screw axis or improper axis m: mirror plane c: sliding mirror or improper mirror

26 Crystal Families, Crystal Systems, and Lattice Systems

27 Generation of and Properties of X-rays

28 A New Kind of Rays n Wilhelm Conrad Röntgen n German physicist who produced and detected Röntgen rays, or X-rays, in n He determined that these rays were invisible, traveled in a straight line, and affected photographic film like visible light, but they were much more penetrating.

29 Properties of X-Rays n Electromagnetic radiation (λ = 0.01 nm 10 nm) n Wavelengths typical for XRD applications: 0.05 nm to 0.25 nm or 0.5 to 2.5 Å 1 nm = 10-9 meters = 10 Å n E = ħc / λ

30 Generation of Bremsstrahlung Radiation Electron (slowed down and changed direction) nucleus Fast incident electron electrons Atom of the anode material X-ray n Braking radiation. n Electron deceleration releases radiation across a spectrum of wavelengths.

31 Generation of Characteristic Radiation Photoelectron M L Emission Kα n Incoming electron knocks out an electron from the inner shell of an atom. Electron K Lα n Designation K,L,M correspond to shells with a different principal quantum number. Kβ

32 Generation of Characteristic Radiation n Not every electron in each of these shells has the same energy. The shells must be further divided. n K-shell vacancy can be filled by electrons from 2 orbitals in the L shell, for example. Bohr`s model n The electron transmission and the characteristic radiation emitted is given a further numerical subscript.

33 Generation of Characteristic Radiation Energy levels (schematic) of the electrons M L Intensity ratios Kα 1 : Kα 2 : Kβ = 100 : 50 : 20 K Kα 1 Kα 2 Kβ 1 Kβ 2

34 Emission Spectrum of an X-Ray Tube

35 Emission Spectrum of an X-Ray Tube: Close-up of Kα Cullity, B.D. and Stock, S.R., 2001, Elements of X-Ray Diffraction, 3 rd Ed., Addison-Wesley

36 Sealed X-ray Tube Cross Section Cullity, B.D. and Stock, S.R., 2001, Elements of X-Ray Diffraction, 3 rd Ed., Addison-Wesley n Sealed tube n Cathode / Anode n Beryllium windows n Water cooled

37 Characteristic Radiation for Common X-ray Tube Anodes Anode Kα 1 (100%) Kα 2 (50%) Kβ (20%) Cu Å Å Å Mo Å Å Å

38 Modern Sealed X-ray Tube n Tube made from ceramic n Beryllium window is visible. n Anode type and focus type are labeled.

39 Sealed X-ray Tube Focus Types: Line and Point Target Take-off angle n The X-ray beam s cross section at a small take-off angle can be a line shape or a spot, depending on the tube s orientation. Filament Target Spot n The take-off angle is the targetto-beam angle, and the best choice in terms of shape and intensity is usually ~6. Line n A focal spot size of mm: mm (line) mm (spot)

40 Interaction of X-rays with Matter

41 Interactions with Matter d incoherent scattering λco (Compton-Scattering) wavelength λpr intensity I o coherent scattering λpr (Bragg-scattering) absorption Beer s law I = I0*e-µd fluorescence λ> λpr photoelectrons

42 Coherent Scattering n Incoming X-rays are electromagnetic waves that exert a force on atomic electrons. e - n The electrons will begin to oscillate at the same frequency and emit radiation in all directions.

43 Interaction of X-rays with matter - Thomson-Diffusion - hν e - hν The interaction with an electromagnetic field induces the oscillation of an electron Being an accelerated charged particle, the electron emits another electromagnetic wave.

44 Constructive and Deconstructive Interference = =

45 Interaction of X-rays with matter - Thomson-Diffusion - The intensity of the diffracted X-ray beam depends on the diffusion angle. y e - 2θ x For polarisation in y- direction: The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. I Th = I i m e r c 4 cos 2 2θ

46 Thomson diffusion is coherent: ϕ Th = ϕ i + α (α = 180 pour e- ) Thomson diffusion Extension to non-polarised light yields: e cos 2 2 θ I Th = I i m 2 r 2 c 4 2 Polarisation factor P (later) The maximum intesity of the diffracted beam is less than 2% I diminishes with distance I Th of neutrons is zero I Th (protons) = 10-7 I Th (electrons) Thomson diffusion is elastic: ω Th = ω i

47 X-ray experiment Why do we need a single crystal? Primary X-ray beam Amorphous sample diffuse reflection

48 X-ray experiment Why do we need a single crystal? Crystalline sample Amorphous sample localised reflections diffuse reflection

49 X-ray experiment Why do we need a single crystal? Polycrystalline sample overlapping reflections Crystalline sample localised reflections Amorphous sample diffuse reflection

50 Coherent Scattering by an Atom n Coherent scattering by an atom is the sum of this scattering by all of the electrons. 2θ n Electrons are at different positions in space, so coherent scattering from each generally has different phase relationships. n At higher scattering angles, the sum of the coherent scattering is less.

51 Localised reflections - Laue construction Interactions of an X-ray beam with several diffracting centers µ υ a a cos µ a cos ν The intensity is only different from zero, when: Δ = a cos ν + a cos µ = n λ

52 Why n λ? Δ = ¼ λ I=0 Δ = 1/3 λ I=0 Δ = λ I=I 0

53 The Laue construction in 3 dimensions µ υ a a cos µ a cos ν a cos µ a + a cos ν a = h λ b cos µ b + b cos ν b = k λ c cos µ c + c cos ν c = l λ 3 equations, 6 angles, 3 distances too complicated

54 Bragg construction Thomson diffusion is coherent: ϕ Th = ϕ i θ θ d sinθ θ d hkl Glancing reflections at the lattice planes hkl of the crystal, which obey the Laue condition. Bragg law: 2d hkl sinθ = n λ (n = 1, 2, 3 )

55 X-ray Diffraction by Crystals

56 Diffraction of X-rays by Crystals n The science of X-ray crystallography originated in 1912 with the discovery by Max von Laue that crystals diffract X-rays. n Von Laue was a German physicist who won the Nobel Prize in Physics in 1914 for his discovery of the diffraction of X-rays by crystals. Max Theodor Felix von Laue ( )

57 X-ray Diffraction Pattern from a Single-crystal Sample Rotation Photograph

58 Diffraction of X-rays by Crystals After Von Laue's pioneering research, the field developed rapidly, most notably by physicists William Lawrence Bragg and his father William Henry Bragg. William Henry Bragg In , the younger Bragg developed Bragg's law, which connects the observed scattering with reflections from evenly-spaced planes within the crystal. William Lawrence Bragg

59 Bragg s Law n X-rays scattering coherently from 2 of the parallel planes separated by a distance d. n Incident angle and reflected (diffracted angle) are given by θ.

60 Bragg s Law n n n The condition for constructive interference is that the path difference leads to an integer number of wavelengths. Bragg condition à concerted constructive interference from periodically-arranged scatterers. This occurs ONLY for a very specific geometric condition. λ = d sinθ n λ = 2d sinθ 2

61 Bragg s Law nλ = 2d sin(θ) θ θ d We can think of diffraction as reflection at sets of planes running through the crystal. Only at certain angles 2θ are the waves diffracted from different planes a whole number of wavelengths apart (i.e., in phase). At other angles, the waves reflected from different planes are out of phase and cancel one another out.

62 Reflection Indices n These planes must intersect the cell edges rationally, otherwise the diffraction from the different unit cells would interfere destructively. n We can index them by the number of times h, k and l that they cut each edge. n The same h, k and l values are used to index the X-ray reflections from the planes. z y x Planes (or )

63 Examples of Diffracting Planes and their Miller Indices n Method for identifying diffracting planes in a crystal system. c n A plane is identified by indices (hkl) called Miller indices, that are the reciprocals of the fractional intercepts that the plane makes with the crystallographic axes (abc). b a

64 Miller indices of lattice planes 110 b = 1/k = 1 ð k = 1 b b a a = 1/l = 1 ð l = 1 a The plane nearest to the origin (not the plane through the origin), intersects the axes a, b and c at 1/h, 1/k and 1/l. An index of 0 indicates a plane parallel to an axis. hkl are the Miller indices of the lattice planes The higher the indices, the smaller the lattice spacing d hkl.

65 Miller indices of lattice planes b = 1/k = 1 ð k = b a c c=1/4 210 b a a = 1/l = 1 ð l = 1 b=1/3 b a b=1/1 x b x a a=1/2 a=1/0 = a=1/3 b=1/2 b b=1/1 b a=1/-1 a a 324

66 What is 030? 030 b a There are no atoms in these planes. Why do we see reflections with them?

67 What is 030? θ θ θ θ d 100 d 200 2d 100 sinθ 1 = λ 2d 100 sinθ 2 = 2λ 2d 100 sinθ 3 = 3λ 2d 200 sinθ 2 = λ, d 200 =d 100 /2 2d 300 sinθ 3 = λ, d 300 =d 100 /3 2d 100 sinθ n = n λ 1 reflection for each plane, but additional planes n reflections for each plane 2d n00 sinθ = λ, d n00 =d 100 /n Thus, we have only first order reflections, but we have to add additional virtual hkl-planes.

68 Diffraction Patterns Two successive CCD detector images with a crystal rotation of one degree per image: For each X-ray reflection (black dot), indices h,k,l can be assigned and an intensity I = F 2 measured

69 Reciprocal Space n The immediate result of the X-ray diffraction experiment is a list of X-ray reflections hkl and their intensities I. n We can arrange the reflections on a 3D grid based on their h, k and l values. The smallest repeat unit of this reciprocal lattice is known as the reciprocal unit cell; the lengths of the edges of this cell are inversely related to the dimensions of the real-space unit cell. n This concept is known as reciprocal space; it emphasizes the inverse relationship between the diffracted intensities and real space.

70 X-ray experiment θ θ θ d hkl From the position of primary and diffracted beam: Orientation of the lattice planes in the crystal (perpendicular to the bisecting of the two beams) 180 2θ Reflection angle θ : Distance between lattice planes Knowing the distances between lattice planes (d hkl ) and their orientations, we obtain the unit cell.

71 Very fast: the reciprocal lattice The distance d (dhkl) between lattice planes can be calculated from the unit cell parameters: 1 d 2 = h a k b l c 2 2 (orthorhombic system) With the reciprocal values d* = 1/d, a* = 1/a, b* = 1/b, c* = 1/c we obtain: *2 2 *2 2 *2 d = h a + k b + Each reflection hkl can thus be described as a vector d* = (h k l) in the reciprocal space formed by the basis vectors a*, b* and c*. From the orientation of the primary and reflected beams, we obtain the direction of d* for each reflection, from the reflection angle theta the lattice spacing d and thus d* = 1/d. Indexing is the art to find a set of basis vectors a*, b*, c* which allow the description of each reflection with integer values of h, k and l. l 2 c * Finding the longest vectors which can describe all reflections A certain error must be allowed If necessary, move to shorter basis vectors From a*, b* and c*, the unit cell parameters and the Miller indices are known.

72 Structure determination Crystallisation Single crystal selection Molecular structure: Atomic positions Solution and Refinement Crystalline structure: Unit cell and space group Crystal: Macroscopic dimensions Dataset collection H K L I σ Space group determination Détermination of the unit cell Unit cell + space group: Dimensions and symmetry of the crystalline structure Intensity of the reflections: Atomic positions Unit cell H K L I σ Raw data

73 The seven crystal systems Which possibilities exist for a unit cell? Dimension Angle conditions conditions Triclinic - - Monoclinic - α = γ = 90 o Orthorhombic - α = β = γ = 90 o Tetragonal a = b α = β = γ = 90 o Trigonal, hexagonal a = b α = β = 90 o, γ = 120 o Cubic a = b = c α = β = γ = 90 o

74 The 14 Bravais lattices We obtain 14 Bravais lattices, when we combine the crystal systems with the centering. Centering describes that more than one unit /molecule is present in the unit cell (additional translational symmetry). c a b Primitive: P Face centered: A, B, C All-faces centered: F Body centered: I

75 Why do we need centered lattices? A A A A A A A A A In the absence of symmetry, we can choose every possible unit cell. If a C 2 rotation axis or a mirror plane is present, only a monoclinc unit cell (or higher) is compatible with these symmetry elements. Without that symmetry it wouldn t even be a monoclinic cell: it would be a triclinic cell with angles very, very close to 90 o! To correctly describe our structure, we have always to choose the highest possible symmetry.

76 Why do we need centered lattices? A A A A A A A A A A A A A A A A A A A A A A A A C 2 or σ How can we describe a crystal which contains a certain symmetry, for example a C 2 axis or a mirror plane, but the smallest cells are incompatible with these symmetry elements. We choose a cell of higher volume, containing more than one lattice point, a so called centered cell. In this example, we have a centered monoclinic cell. In crystallographic language, a face centering C adds to the existing translations (x+1,y,z; x,y+1,z; x,y,z+1) another one with x+0.5, y+0.5, z.

77 The 14 Bravais lattices 7 crystal systems and 6 centerings: Why do we not have 42 Bravais lattices? Triclinic: Only P (primitive). Every centered lattice can be transformed into a primitive one.

78 The 14 Bravais lattices 7 crystal systems and 6 centerings: Why do we not have 42 Bravais lattices? Triclinic: Only ap Monoclinic: Only mp and mc A can be transformed into C by exchanging the axes a and c. B can be transformed into P. mi mc I can be transformed into C. c b c b mb mp a a

79 The 14 Bravais lattices 7 crystal systems and 6 centerings: Why do we not have 42 Bravais lattices? Triclinic: Only ap Monoclinic: Only mp and mc Orthorhombic: op, oa, oi, of ob and oc can be transformed into oa by simple axis exchange op oa oi of

80 The 14 Bravais lattices tp ti cp ci cf 7 crystal systems and 6 centerings: Why do we not have 42 Bravais lattices? Triclinic: Only ap Monoclinic: Only mp and mc Orthorhombic: op, oa, oi, of Tetragonal: tp and ti (Because of the C 4 -symmetry, ta becomes tf, which can be transformed into ti. tc can be transformed into tp.) Trigonal, hexagonal Cubique: cp, ci et cf (C 3 symmetry: ca/cb/cc become automatically cf)

81 Les 14 réseaux de Bravais («Bravais lattice») Triclinic: ap Monoclinic: mp et mc Orthorhombic: op, oa, oi, of Tetragonal: tp et ti Trigonal, hexagonal: hp, hr (obverse et reverse) Cubic: cp, ci et cf

82 Centered cells have a higher volume than the corresponding primitive cell. What happens if we increase the size of our unit cell? θ θ θ d sinθ d hkl Equiangular reflection at the lattice plane hkl of the crystal, which obeys the Laue conditions. Bragg law: 2d hkl sinθ = λ (n = 1, 2, 3 )

83 What happens if we increase the size of our unit cell? Bragg Law: Tetragonal unit cell: 2d hkl sinθ = λ sinθ = ½ λ / d hkl a=b=5 Å, c=10 Å a=b=5 Å, c=20 Å h k l d hkl sinθ θ h k l d hkl sinθ θ Å Å Å Å Å Å Å Å Å Å Å Å Å Å Å Å

84 What happens if we increase the size of our unit cell? A unit cell with two times the volume has twice the number of reflections in the same θ region. Can we thus increase the number of reflections by increasing the size of our unit cell? On doubling the axis length a reflection {0 0 1} becomes { 0 0 2}. What about the new reflection { 0 0 1} of our increased unit cell?

85 What happens if we increase the size of our unit cell? What about the new reflection { 0 0 1} of our increased unit cell? We find systematically another atom at d hkl /2. (In other words, we introduced by doubling of the unit cell a new translation operation x, y, z+0.5). The path length difference to this atom is ½ λ. 0 0 ½ Due to the additional translational symmetry, only reflections with {h k l} with l = 2n are present (path length difference = 2nλ). For reflections {h k l} with odd h, the reflections are systematically absent, since the atom z+0.5 causes destructive interference.

86 What happens if we increase the size of our unit cell? Tetragonal unit cell: a=b=5 Å, c=10 Å a=b=5 Å, c=20 Å h k l d hkl sinθ θ h k l d hkl sinθ θ Å Å Å Å Å Å Å Å Å Å Å Å Å Å Å Å An artificial increase of an axis length adds new reflections, which are, however, systematically absent (I = 0).

87 Systematic absences and centering The introduction of a translational symmetry (x+0.5, y, z) causes the systematic absence of all reflections {h k l} with h 2n. All translational symmetries introduces systematic absences. Since all centerings introduce additional translations, centerings are associated with the presence of systematic absences, which we can use to DETERMINE the centering from the reflection list.: translation Z restrictions P A x, y+0.5, z+0,5 2 k+l = 2n B x+0.5, y, z h+l = 2n C x+0.5, y+0.5, z 2 h+k = 2n F A + B + C 4 h+k = 2n, h+l = 2n, k+l = 2n I x+0.5, y+0.5, z h+k+l = 2n Thus from investigating systematic absences in the reflection list, we can determine the Bravais lattice.

88 Systematic absences In the same way, symmetry elements which include translations also cause systematic absences: Screw axes (axes hélicoïdal): translation restrictions 2 1, 4 2, 6 3 a x+0.5 h00 with h = 2n b y+0.5 0k0 with k = 2n c z l with l = 2n 3 1, 3 2, 6 2, 6 4 c z+1/3 00l with l = 3n 4 1, 4 3 c z l with l = 4n 6 1, 6 5 c z+1/6 00l with l = 6n

89 Systematic absences Glide planes: translation zonal restrictions b a y+0.5 0kl avec k = 2n c a z+0.5 0kl avec l = 2n n a y+0.5, z+0.5 0kl avec k+l = 2n d a y+0.25, z kl avec k+l = 4n (F) c b z+0.5 h0l avec l = 2n a c x+0.5 hk0 avec h = 2n c [110] z+0.5 hhl avec l = 2n (t, c) c [120] z+0.5 hhl avec l = 2n (trigonal) d [110] x+0.5, z+0.25 hhl avec 2h+l = 4n (t, ci)

90 Small Molecule Example YLID Space Group Determination hk0 layer 0kl layer

91 Symmetry and Space Groups

92 Small Molecule Example YLID Unit Cell Contents and Z Value n Chemical formula is C 11 H 10 O 2 S n Z value is determined to be 4.0 n Density is calculated to be n The average non-h volume is calculated to be 17.7

93 Small Molecule Example YLID Set Up File for Structure Solution

94 Space groups The combination of the translations, defined by the Bravais lattice, and the elements of symmetry possible in an infinite crystal result in the 230 possible space groups. (Group theory: a space group must be closed, i. e. the action of a new element cannot generate an element which is not already in the group. Example: The most common space group: P2 1 /c x+1, y, z x, y+1, z a P1 x, y, z+1 c 0.25 Z=1 P1 Z=2 P2 1 /c Z=4 -x, -y, -z i -y -y 0.5-x, 0.5-y, 0.5-z i 0.5-x, y, z i x, 0.5-y, z i x, y, 0.5-z i x, 0.5-y, 0.5-z i 0.5-x, y, 0.5-z i 0.5-x, 0.5-y, z i -y 0.5+y.... y 0.5-y -y 0.5+y.... y 0.5-y x, 0.5-y, z+0.5 c -x, 0.5+y, 0.5-z y Asymmetric unit.. y

95 Space group determination The determination of the correct space group is essential since without a correct description of the symmetry neither the solution nor the refinement of a structure is possible.. To determine the space group we can use the following information: The Laue group (Symmetry of the reflections) The presence of screw axes and glide planes (systematic absences) The presence of an inversion center (value E(E-1), chirality of the molecule)

96 Crystal systems and Laue group The combination of the crystal systems with allowed symmetry elements yields 32 crystal classes (or crystallographic point groups): Crystal Crystal system class Triclinic C 1 1 C i -1 Monoclinic C 2 2 C s m C 2h 2/m Orthorhombic D C 2v mm2 D 2h mmm Tetragonal C 4 4 S 4-4 C 4h 4/m D C 4v 4mm D 2d -42m D 4h 4/mmm Crystal Crystal system class Trigonal C 3 3 C 3i -3 D 3 32 C 3v 3m D 3d -3m Hexagonal C 6 6 C 3h -6 C 6h 6/m D C 6v 6mm D 3h -6m2 D 6h 6/mmm Cubic T 23 T h m3 O 432 T d -43m O h m3m

97 32 crystal classes, but 11 Laue groups Adding the inversion symmetry caused by Friedel s law, we obtain 11 Laue groups. The Laue group describes the symmetry of our observed reflections! Crystal Crystal Laue system class group Triclinic C C i -1 Monoclinic C 2 2 2/m C s m C 2h 2/m Orthorhombic D mmm C 2v mm2 D 2h mmm Tetragonal C 4 4 4/m S 4-4 C 4h 4/m D C 4v 4mm 4/mmm D 2d -42m D 4h 4/mmm Crystal Crystal Laue system class group Trigonal C C 3i -3 D m1 C 3v 3m -31m D 3d -3m Hexagonal C 6 6 6/m C 3h -6 C 6h 6/m D /mmm C 6v 6mm D 3h -6m2 D 6h 6/mmm Cubic T 23 m3 T h m3 O 432 m3m T d -43m O h m3m

98 The structure factor F The intensity of an X-ray beam diffracted at an hkl-plane depends on the structure factor F hkl for this reflection. The structure factor is the sum of all the formfactors (atomic scattering factors) in the unit cell. F F F F H F H F H F H F H F H F H F H F H F 100 = f F + f H H d 100 d 010 H H F F F H H H F 010 = f F - f H The structure factor F hkl thus contains information on the spatial distribution of atoms.

99 The structure factor F Δ F = f F = f + f F = f - f Δ for general lattice plane F = = f f F 1 + cos 2 1 e 2πi Δ Δ 0 1 πδ + f f 2 e πi Δ 2 F hkl Structure factor = N j= 1 f j e 2π i( h x j + k y j + l z The structure factor F hkl depends on the spatial distribution of the atoms and, more specifically, on their distance to the reflection plane. j )

100 The Crystallographic Phase Problem n In order to calculate an electron density map, we require both the intensities I = F 2 and the phases α of the reflections hkl. n The information content of the phases is appreciably greater than that of the intensities. n Unfortunately, it is almost impossible to measure the phases experimentally! This is known as the crystallographic phase problem and would appear to be unsolvable!

101 Structure determination Crystallisation Single crystal selection Molecular structure: Atomic positions Crystalline structure: Unit cell and space group Crystal: Macroscopic dimensions How to find F hkl? The intensities of the reflections are proportional to the square of the amplitude of the structure factor: 021 Détermination of the 123 elemental cell Dataset collection I ~ F hkl 2 Unit cell H K L I σ Raw data

102 The phase problem The value of each structure factor F hkl depends on the distribution of the atoms in the unit cell (i. e. the electronic density). We can thus obtain this electronic density, from the combination of all structural factors using a Fourier transformation ρ( x, y, z) = 1 V hkl F hkl e 2πi( hx + ky+ lz) The factor F hkl takes the form of a cosinus function with an amplitude F hkl and a phase α hkl. These two, F hkl and α hkl, vary for each reflection hkl and depend on the atomic positions relative to the hkl plane. α hkl F hkl F hkl N = j= 1 f j e 2π i( h x j + k y j + l z j ) = F hkl e iα hkl The only formula you have to know by heart! ρ( = 1 iαhkl 2πi( hx + ky+ lz) x, y, z) Fhkl e e V hkl

103 The phase problem ρ ( = 1 x, y, z) F e iαhkl e 2πi( hx + ky+ lz) V hkl hkl One small problem: Since we can determine F hkl = I hkl, we do not know the phases α hkl! FT ρ( x, y, z) F hkl = F hkl e iα hkl FT known unknown

104 A (very) short introduction to phases h,k,l = 2,3,0; Centers on planes, α 230 = 0º, strong reflection h,k,l = 2,1,0; centers between planes, α 210 = 180º, strong reflection

105 A (very) short introduction to phases The phase φ of a reflection where the atoms are situated on the hkl planes has a phase of approximately 0º; if the atoms are found between the planes the phase is approximately 180º. For randomly distributed atoms, we cannot predict the phase and the reflection is weak due to strong destructive interference. h,k,l = 0,3,0; weak reflection, α 030 =? Do we really need the phases?

106 First estimation of phases (Patterson, direct methodes): Experiment α hkl Structure solution F hkl = F o α hkl FT ρ 1 (x,y,z) = Atomic coordinates F o = I ρ 2 (x,y,z) We want to optimise: ρ(xyz) FT F c = F c α hkl Optimisation criterium: M = w( F o 2 - F c 2 ) 2 F hkl = TF F o α hkl ρ c (x,y,z) TF α hkl Manual confirmation Refinement The phases improve with each cycle ρ c (x,y,z) Manual confirmation Δρ=1/V (F o -F c )e -2π Difference Fourier map

107 Crystal Structure Solution by Direct Methods n Early crystal structures were limited to small, centrosymmetric structures with heavy atoms. These were solved by a vector (Patterson) method. n The development of direct methods of phase determination made it possible to solve non-centrosymmetric structures on light atom compounds

108 Rapid Growth in Number of Structures in Cambridge Structural Database

109 Initial Solution XP / XShell Graphical programs for the visualisation and modification of your structure

110 Initial solution S B F 5 C 6 C6 F 5 C 6 F 5 The initial solution proposes a electron density map in which the highest points of electron densities are identified (Q-peaks).

111 Initial solution S B F 5 C 6 C6 F 5 C 6 F 5 For some centers of electron density an element is already proposed. You have to verify if they are correctly assigned.

112 Initial solution Several centers of electron densities are not related to your structure. Do not hesitate to delete them! A deleted electron density will show up again in later cycles, if it is correct. If a structure is difficult to identify, look for regular structures, such as aromatic rings etc.

113 Initial solution S B F 5 C 6 C6 F 5 C 6 F 5 Q-peaks are sorted after their electronic density with Q1 being the highest electron density. The peaks with the smallest numbers have the highest electron density and are thus fluorine atoms.

114 Initial solution Now you have to assign correct elements to the remaining electron densities: Carbon

115 Initial solution Now you have to assign correct elements to the remaining electron densities: Carbon Fluor

116 Initial solution Now you have to assign correct elements to the remaining electron densities: Carbon Fluor Boron At the ends no Q- peaks must be left!

117 Refinement organigram Initial solution XS (SHELXS).ins,.hkl.res,.lst Sort the atoms XP / XSHELL.res.ins Identify the atoms Delete the rest XP / XSHELL.res.ins Refinement XL (SHELXL).ins,.hkl.res,.lst Anisotropic refinement XL (SHELXL).ins,.hkl.res,.lst Addition of hydrogen atoms XP, XSHELL, XL (SHELXL).res,.hkl.ins,.lst Refinement of the weighing scheme XP, XSHELL, XL (SHELXL).ins,.hkl.res,.lst,.cif yes Are atoms missing? no Final refinement XP, XSHELL, XL (SHELXL).ins,.hkl.res,.lst,.cif

118 Which parameters are refined? Ins-file: CELL Sl Atom position The atomic positions are given as fractions of the basis vectors. The values given are thus not in Å and Si1 is found at x = u a = Å = Å y = v b = Å = Å z = w c = Å = Å The fractional atomic positions are normally symbolized by the letters u, v and w. Since each atom is present in the unit cell u, v and w have values between 0 and 1. For practical reasons, we find sometimes values <0 or >1, but never <-1 or >2.

119 Which parameters are refined? Sl Occupation factor The occupation factor indicates how many atoms are occupying this position. The maximal occupation factor is 1, but in special cases, smaller numbers are possible. Disorder Ins-file: O H1A H2A

120 Which parameters are refined? Sl Occupation factor The occupation factor indicates how many atoms are occupying this position. The maximal occupation factor is 1, but in special cases, smaller numbers are possible. An atom on a symmetry element Disorder is duplicated on its position. To Special positions avoid this, the program would have to recognize and exclude atoms on special positions from some symmetry operations. Sl By applying an occupation factor of ½, the atom can be treated in the same way as all other atoms.

121 Thermal motion of atoms / Temperature factor Atom position Occupation factor U iso Sl U iso? The distribution of the electronic density depends on the thermal motion of the atoms Thermal motion is not identical for all atoms An X-ray experiment takes significantly more time than thermal motion we obtain an averaged distribution of the electron density Isotropic motion: The vibration of an atom is identical in all directions It is described by a Gaussian function ρ r 2 r' = 2πUe ( ) U ' 2 ρ : distribution of the electron density r : distance from the equilibrium position

122 Fourier transformation: q Thermal motion of atoms 2 sin θ sin θ 2 8π U B ( r*) = e 2π Ur* = e 2 2 λ = e λ 2 2 U = r' 2 [ ] 2 Å Average squared displacement B = 2 8π U [ ] 2 Å Atomic temperature factor (Debye-Waller factor) x% probability that the atom is found inside the radius of the indicated sphere F F F F F

123 It is described in our model by an ellipsoid (different Gaussian functions in three dimensions) or, mathematically speaking, by a symmetric tensor: = ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' z z y z x z y y y x z x y x x U ') ' 2 ' ' 2 ' ' 2 ' ' ' ( *) ( z y U z x U y x U z U y U x U e r q = π Anisotropic motion: The atomic motion is not isotropic, i. e. identical in all directions. Instead of 1 parameter (U iso ) we require 6 parameters for the ellipsoid (3 radii and the direction of the principal axis). Thermal motion of atoms

124 Difference Fourier map The refinement result does not show any new atoms. There are no missing atoms The remaining peaks can be attributed to: Thermal motion of atoms Hydrogen atoms Errors

125 Difference Fourier map After anisotropic refinement: Les pics restants peut être attribués aux: Thermal motion of atoms Hydrogen atoms Errors The 8 highest Q- peaks (Q1-Q8) correspond to hydrogen atoms!

126 Using constraints and restraints: Hydrogen treatment The position of hydrogen atoms is a challenge in X-ray diffraction studies due to their low electron density. In addition, the C-H (and OH, NH etc.) distances are determined systematically to short for two reasons: 1. «Libration» The effect is especially strong for hydrogen atoms (very light atoms) 1.04 Å (neutron)

127 Using constraints and restraints: Hydrogen treatment The position of hydrogen atoms is a challenge in X-ray diffraction studies due to their low electron density. In addition, the C-H (and OH, NH etc.) distances are determined systematically to short for two reasons: 1. «Libration» The effect is especially strong for hydrogen atoms (very light atoms) 1.04 Å (neutron) 0.96 Å (rayons X)

128 Using constraints and restraints: Hydrogen treatment The position of hydrogen atoms is a challenge in X-ray diffraction studies due to their low electron density. In addition, the C-H (and OH, NH etc.) distances are determined systematically to short for two reasons: 1. «Libration» The effect is specificially strong for hydrogen atoms (very light atoms) 1.04 Å (neutron) 2. The maximum of the electron density is located in the bond, not on the nucleus Å (rayons X) X H

129 Hydrogen treatment With data of moderate quality, you can find the hydrogen atoms in the difference Fourier map.

130 Hydrogen treatment : 1. Free refinement 1. Free refinement The anisotropic refinement of hydrogen atoms requires 9 parameters per hydrogen atom. We encounter problems with the data/parameter ratio The electron density of hydrogen is very low Isotropic refinement Conditions for free refinement of hydrogen atoms: Sufficient data quality The obtained distances, angles and U iso must be reasonable: CH 0.96 Å ± 10% = HCX ideal ± 10% U iso < 0.2 Å 2 and not too varied

131 Summary of Part 1 n Introduction to crystals and crystallography n Definitions of unit cells, crystal systems, Bravais lattices n Introduction to concepts of reciprocal space, Fourier transforms, d-spacing, Miller indices n Bragg s Law geometric conditions for relating the concerted coherent scattering of monochromatic X-rays by diffracting planes of a crystal to its d-spacing. n Part 2 We will use these concepts and apply them in a practical way to demonstrate how an X-ray crystal structure is carried out.