Definition of the crystalline state: Crystals are solids (but not all solids are crystals!)‏ Crystals are the most ordered form of the matter Crystal are 3-D (2-D) regular arrays of ions, atoms, molecules; they have triple (double) periodicity Crystals have long range order. Each repeating unit (whatever it is) within a crystal has an identical environment

X-ray Diffraction is the essence of the X-ray crystal structure analysis (XRA)‏ The main aim of XRA is the determination of 3-D structure of the chemical entity = structural motive which is forming the crystal that is being repeated periodically in the whole volume of the crystal Crystal forming chemical entities = motives: metals, ions, atoms (e.g. diamond), organic compounds, peptides, proteins, lipids, oligosaccharides, DNA, RNA etc.

X-ray analysis is the most accurate method to determine: - structure of the crystal hence structure of the crystal motif e.g. molecule: - bond distances and angles (C-C 1.542(2) Å, C-C-N 123.72(12)o) - conformation of the compound - absolute configuration of the compound/atom X-ray crystal analysis is the best source of the above data: they are the key components of structural databases for further chemical (e.g. quantum) and physical calculations Co(III)(Ph4porphyrin) (Cl) Mn(H2O)62+ Rh(en)2Cl2+ trans

 phase What are the components of waves? A B  |F| wavelength | F | = amplitude phase

How to obtain the structure of the motif (compound)‏ from its crystal structure? The foundation of recognition/”visibility” of all structures: Scattering and Diffraction All objects – irrelevant of their size – scatter radiation which is shine on them. http://www.acoustics.salford.ac.uk/feschools/waves/super.htm#phase http://www.numathics.com/arens/scattering/Scattering.html

Difference between scattering and diffraction: - scattering: spherical - diffraction: more directional as it results from interference of scattering from many centres, - (or it results from interference of incoming = = source wave with new, scattered waves.)

How we can retrieve the information about the scattering objects? Because we can focus back the scattered/diffracted waves again. The image of the object Lenses: focussing Scattering/Diffraction The object

Why can we focus back the image of the scattering/diffracting object? Because light travels with different media with different speed: - different media have different refractive index n n is a measure how much is the speed of light (or other waves such as sound waves) is reduced inside the medium > nG > nA Air nAc nG Glass nG = 1.5 Here

What should be the relationship between effective scattering and the size of the object ? Web examples http://www.acoustics.salford.ac.uk/feschools/waves/super.htm#phase The power of scattering/diffraction by an object is: directly proportional to the similarity between the wavelength of the incident radiation and the size of the scattering object Larger object – larger waves needed for an effective scattering Smaller objects – smaller wave needed for an effective scattering

Hence: X-ray radiation in X-ray analysis!  of radiation 3.7 pm 400 – 700 nm Resolution 2 nm 10 nm 200nm Typical C-C bond: Å, so The most useful radiation to study crystal Structure has to have  in the range Å Hence: X-ray radiation in X-ray analysis! 1.54 ~1-2 1 meter = 102 cm = 103 mm = 106 mm = 109 nm =1010 Å

Intensity of the wave: I ~ F 2 I = √F 2 = | F | wave = | F | x  = F Intensity of the wave: I ~ F 2 If we know I then: I = √F 2 = | F | Lenses are focussing back all information that is contained in the scattered or diffracted waves: - amplitudes | F | (intensities I)‏ - phases  so they can produce back the image of the scattering/diffracting object

What?!! NO LENSES!!!! Usually Monochromatic: one, well defined  In X-ray Diffraction we do not have lenses which could focus diffracted rays back to the crystal structure, n=1!. We can only register: - directions of the diffracted X-rays and their - intensities I hence | F | amplitudes only of the diffracted rays: !their phases are missing! = phase problem

1. Intensities I of diffracted X-rays therefore | F | - amplitudes of Information directly available from an X-ray single crystal diffraction experiment: 1. Intensities I of diffracted X-rays therefore | F | - amplitudes of diffracted X-rays 2. Directions of the diffracted X-rays The phases  must be reconstructed in rather complex/difficult experimental and computing methods: phase problem = phase solution methods

The key-feature of XRD and XRA is the interaction between the crystal and the incoming X-ray radiation (l in range off 0.8 – 2 Å). X-rays in the crystal are: scattered by the electrons: - Thomson scattering: the electron oscillates in the electric field of the incoming X-ray beam and an oscillating electric charge radiates electromagnetic waves - this is elastic and coherent scattering: frequencies and wavelengths of the incoming X-rays and scattered-diffracted X-rays are the same/unchanged this scattering is becoming very discreet in terms of directions some scattered X-ray waves are reinforced, some weakened as we are dealing here with the diffraction - REFLECTIONS which is amplified by millions copies of the same atoms (electrons!)‏ in the same positions in the crystal space due to crystal periodic, repetitive (in 3-D) unique character But why not to measure scattering from one molecule and determine its structure this way?…..

We use crystals as 3-D amplifiers of scattering We cannot measure (yet) the X-ray scattering produced by single chemical entity (organic molecule): it is too weak. We use crystals as 3-D amplifiers of scattering coming from single crystal motif. X-ray Diffraction is well welcomed “side effect” of this process due to amplifying or cancelling effect of scattered radiation emitted by electrons There are also other types of interactions of X-rays with electrons: e.g. excitations. These type of high energy phenomena would damage the single molecule almost immediately. In crystal there are thousands of molecules – some of them survive long enough to give a measurable radiation. .

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Crystal structure = Crystal Lattice + motif. web: http://marie. epfl The 3-D periodicity of the crystal can be simplified and represented by an abstract crystal lattice.

 X ? b a Crystal lattice is described by three translations: a, b, c They can not be just any translations: they have to reproduce all crystal motives (lattice points) if applied to any single lattice point It is NOT the unit cell It is the right unit cell a b ? Two atom compound: X 

They determine the unit cell, which has to be: a lattice ‘building block’, which edges correspond to a, b, c it should give the whole crystal lattice if moved by a, b, c it has to be of the right handed system it has to have the smallest possible volume it has posses the highest possible symmetry characteristic for the lattice (this is why some unit cells are not primitive)‏

a z c b   y  x The unit cell in three dimensions. The unit cell is defined by three vectors a, b, and c, and three angles , , .    b c a y x z Unit cells are defined in terms of the lengths of the three vectors and the three angles between them. For example, a = 94.2 Å, b = 72.6 Å, c = 30.1 Å,  = 90.0°,  = 102.1°,  = 90.0° a = 8.32 Å, b = 15.23 Å, c = 9.28 Å, a = 90.0°, b = 90.0°, g = 90.0°

Content of the Unit cell c Size and the arrangement Crystal structure (symmetry) of the unit cell Crystal structure Motif = molecule, atoms To get the structure of the motive we have to: get the information about the unit cell size and its arrangement

In the crystal lattice we can distinguish: - lattice points - lattice directions - lattice planes Co-ordinates of the lattice points are given in the fractions u,v,w of the a,b,c lattice translations (u,v,w) a b c x y z

Crystal Lattice directions symbol: examples: [uvw] [100] [010] [001] : z : c b a c y x b a

Crystal planes The planes are “imaginary” = inter-plane spacing measured at 90o to the planes The planes are “imaginary” All planes in a set of planes are identical - equivalent The perpendicular distance between pairs of adjacent planes is called d: interplanar spacing Need to label planes to be able to identify them…………

(1 3 0) plane plane (2 1 0) (h k l) (h k l) Miller index (hkl) Find intercepts on a, b, c: 1/2, 1, 0 (1 1/3 0)‏ Take reciprocals 2, 1, 0 (1 3 0)‏ plane (2 1 0) (1 3 0) plane Miller index (hkl) (h k l) (h k l) General label is (h k l) which intersects at a/h, b/k, c/l (hkl) is the MILLER INDEX of that plane (round brackets, no commas).

x y z 1a 1b 1c O (111) Standard triangle x y z 1a 1b 1c O (222)

(0 1 0) plane (0 0 1) plane Plane perpendicular to x cuts at 1,  ,  b c Plane perpendicular to x cuts at 1,  ,   (1 0 0) plane a b c a b c (0 1 0) plane (0 0 1) plane NB an index 0 means that the plane is parallel to that axis

Planes - conclusions Miller indices define the orientation of the plane within the unit cell The Miller Index defines a set of planes parallel to one another (remember the unit cell is a subset of the “infinite” crystal All possible sets of planes in a particular lattice may be described by (hkl) values Any of these sets of planes may contain scattering electrons (atoms) (or be close to): this is crucial for scattering and diffraction. Distance between planes is given by dhkl Reciprocal dependence between (hkl) and dhkl : Larger (hkl) values (finely spaced planes) then smaller dhkl.

Diffraction – on the optical grating Web example! Path difference XY between diffracted beams 1 and 2: sin = XY/a  XY = a sin  For 1 and 2 to be in phase and give constructive interference, XY = , 2, 3, 4…..n so a sin  = n where n is the order of diffraction It is so-called grating relationship where a = is the distance between scattering centres

a sin  = n a

Non-diffracted X-rays (97%)‏ Principles of BRAGG X-ray diffraction experiment: diffracted X-rays Crystal Incoming X-ray Non-diffracted X-rays (97%)‏ detector

Diffraction – on the crystal lattice “grating” Incident X-ray radiation “Reflected” radiation 1 (ca. 3%)‏ 2   dhkl    Set of crystal planes (h k l) X Z Y  dhkl dhkl XY = YZ = dhkl sin  Transmitted radiation (ca. 97%!)‏ Beam 2 lags beam 1 by XY + YZ = 2d sin  So 2dhkl sin  = n  Bragg Law as inter-atomic distances are in the range of 0.5 - 2 Å so  must be in the range of 0.5 - 2 Å X-rays, electrons, neutrons suitable

Difference between light and X-ray reflections + + + + + + + + Light: - + - + - + - + X-ray: Reflection of the Light: is not coherent (multi )‏ does not depend on  is happening only on the surface can be focused back almost 100% of the incident light is reflected X-ray Diffraction/Reflection: coherent (usually single )‏ strictly depends on  is happening in 3-D volume of the crystal can not be focused back only about 3% of the incoming X-rays is diffracted; ~97% goes through the crystal unchanged