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

Size: px

Start display at page:

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

Johnathan Barnaby Taylor
5 years ago
Views:

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

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

3 The Unit Cell Concept Ralph Krätzner

4 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.

5 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

6 Choice of the Unit Cell (sometimes confusing)

7 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.

8 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

9 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

10 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.

11 Bravais Lattices

12 Space groups (monoclinic): 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

13 Crystal Families, Crystal Systems, and Lattice Systems

14 Generation of and Properties of X-rays

15 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

16 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.

17 Emission Spectrum of an X-Ray Tube

18 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.

19 Constructive and Deconstructive Interference = =

20 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

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

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

23 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.

24 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

25 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 )

26 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.

27 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 )

$Examples of Diffracting Planes and their Miller Indices n Method for identifying diffracting planes in a crystal system.$

28 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

29 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.

30 Miller indices of lattice planes 210 b = 1/k = 1 ð k = 1 b b=1/1 b a c c=1/4 a a = 1/l = 1 ð l = 1 b=1/3 b a 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

31 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.

32 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.

33 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

34 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?

35 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.

36 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).

37 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.

38 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

39 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)

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

41 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.

42 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 )

43 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

44 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

45 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

46 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?

47 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

48 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.

49 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.

50 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.

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

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

53 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!

54 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.

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

Basic Crystallography Part 1. Theory and Practice of X-ray Crystal Structure Determination Basic Crystallography Part 1 Theory and Practice of X-ray Crystal Structure Determination Course Overview Basic Crystallography Part 1 n Introduction: Crystals and Crystallography n Crystal Lattices and