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Mi 10:15-12:00, Hörsaal II George Sheldrick gsheldr@shelx.uni-ac.gwdg.de

Methods in Chemistry III – Part 1 Modul M.Che.1101 W S 201 0/11 – 3 Modern Methods of Inorganic Chemistry. Mi 10:15-12:00, Hörsaal II George Sheldrick gsheldr@shelx.uni-ac.gwdg.de. The lattice.

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Mi 10:15-12:00, Hörsaal II George Sheldrick gsheldr@shelx.uni-ac.gwdg.de

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  1. Methods in Chemistry III – Part 1Modul M.Che.1101 WS 2010/11 – 3Modern Methods of Inorganic Chemistry Mi 10:15-12:00, Hörsaal II George Sheldrick gsheldr@shelx.uni-ac.gwdg.de

  2. The lattice Crystals are made up of identical ‘bricks’ (unit-cells) that constitute a three-dimensional translation lattice. The cell is defined by the 3 vectors a, b and c; the 3 angles between them are ,  and  as shown in the diagram. V = Volume of the unit-cell = abc [ 1 – cos2 – cos2 – cos2 + 2cos.cos.cos ]½ c   b a 

  3. The unit-cell The unit-cell is the smallest unit that can generate the entire structure by translation operations alone. Within the cell there can be several symmetry related ‘asymmetric units’ with identical contents, but in general in different orientations. Although in the NaCl structure there is an atom at each corner of the unit-cell, this is often not the case. If the structure is centro-symmetric, the origin of the cell is always placed on an inversion center. The cell is always chosen so that the symmetry elements are positioned in accord with volume A of the International Tables for Crystallography.

  4. The choice of the unit-cell When the symmetry is low, there can be a wide choice of possible unit-cells. In certain cases it is better to choose a non-primitive, centered cell in order to show the full symmetry of the structure more clearly: A A B B C D C When there is no symmetry, the position and shape but not volume of the cell may be chosen freely. Here a primitive cell with angles close to 90º (C or D) is preferable. Here the conventional C-centered cell C has 90º angles, but one of the primitive cells (B) has two equal sides.

  5. The 14 Bravais lattices (part 1) P P C triclinic: a = b = c  =  =  monoclinic: a = b = c;  =  = 90º =  I F P C orthorhombic: a = b = c;  =  =  = 90º

  6. The 14 Bravais lattices (part 2) P I P R tetragonal: a = b = c;  =  =  = 90º hexagonal: a = b = c;  =  = 90º,  = 120º and rhombohedral: a = b = c,  =  =  P I F cubic: a = b = c,  =  =  = 90º

  7. Unconventional lattices Monoclinic B is not one of the 14 Bravais lattices, because it can be transformed to monoclinic P with half the volume. Monoclinic I can be transformed to monoclinic C with the same volume. Despite this, the I cell is sometimes used when the C cell would have a -angle far from 90º.

  8. Screw axes 43-axis 41-axis

  9. Selected tetragonal space groups a = b = c,  =  =  = 90º ½+ .. .. + P41 P4 .. .. + + ¾+ ¾+ .. .. .. .. + + ¼+ .. .. .. + .. + ¼+ ½+ .. .. + .. .. + + .. .. + + P42 P4 .. .. + + ½+ ½+ .. .. – – ½+ – .. – .. ½+ .. .. + + .. .. + +

  10. Atom co-ordinates The positions of atoms within a unit-cell are described by co-ordinates 0  x< 1, 0  y< 1 and 0  z< 1. The vector distance between two atoms x1y1z1 und x2y2z2 is given by: d = (x2-x1)a + (y2-y1)b +(z2-z1)c oder d = ax + by + cz from which follows: d2 = (ax)2+(by)2+(cz)2+2bc(yz)cos+2ac(xz)cos+2ab(xy)cos The x, y and z co-ordinates are also used to describe the action of the symmetry operations, for example: x+1, y, z is the equivalent atom in the next cell in the direction a; –x, –y, –z is an atom generated by an inversion center at 0, 0, 0; x, y, z –y, x, z+¼ –x, –y, z+½ y, –x, z+¾ describes a 41-axis along c

  11. Example: PbO (red crystal modification) The red modification of PbO is tetragonal, a = b = 3.98 Å, c = 5.02 Å with two formula units in the unit-cell. Pb lies on 0, 0.5, 0.237 and 0.5, 0, 0.763 and O on 0, 0, 0 and 0.5, 0.5, 0. The lead atom is co-ordinated by four oxygen atoms with a lone pair of at the apex of a square pyramid; oxygen is tetrahedrally surrounded by four lead atoms. The structure possesses 2-, 21-, 4- and 4-axes, inversion centers and mirror planes. The shortest Pb–O-distance corresponds to x = 0.5, y = 0.0, z = 0.237, so: d = [(0.53.98)2+(0.2375.02)2]½ = 2.32 Å +0.237 +0.237 –0.237 –0.237 –0.237 +0.237 +0.237 –0.237 –0.237 –0.237 +0.237 +0.237 Projection of 4 unit-cells viewed down c; O Pb

  12. Exercises • Why do the 14 Bravais lattices not include tetragonal C and tetragonal F? PtS (mineral name Cooperite) ist tetragonal, a = b = 3.47 Å, c = 6.10 Å. The following atoms are present in the unit-cell: Pt: 0 ½ 0 and ½ 0 ½; S: 0 0 ¼ and 0 0 ¾. 2. Draw four unit-cells in projection viewed down b. 3. Draw four unit-cells in projection viewed down c. Describe the co-ordination geometry of the Pt and S atoms. Does it make chemical sense? 4. What is the shortest Pt–S-distance? 5. Which type of lattice (P, A, B, C, I or F) is present? 6. Which of the following symmetry elements are present: 1, 4, 41, 42, 43, 4?

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