Chemical Crystallography
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1 Chemical Crystallography Prof Andrew Goodwin Michaelmas 2014 Recap: Lecture 1
2 Why does diffraction give a Fourier transform? k i = k s = 2π/λ k i k s k i k s r l 1 = (λ/2π) k i r l 2 = (λ/2π) k s r
3 Total additional pathlength is (λ/2π) (k i k s ) r Total phase shift is (k i k s ) r k i k s Q = k i k s k s Total phase shift is Q r
4 F (Q) = 1 + exp(iq r) = exp(iq 0) + exp(iq r) F (Q) = exp(iq r 1 ) + exp(iq r 2 ) + exp(iq r 3 )+... = exp(iq r j ). j F (Q) = (r) exp(iq r)dr
5 ki Q Q 2 = Q 2 = k i k s 2 = ki 2 + ks 2 = ki 2 + ks 2 2k i k s cos = 2 2 cos 2 ks = sin 2, 2θ Q = 4 sin. Q = 4 sin to the Bragg d = /2 sin Q = 2 d
6 1 d 2 = h2 a 2 + k2 b 2 + l2 c 2. = F crystal (Q) = [Q (ha + kb + lc )] F unitcell (Q) hkl F crystal (Q = hkl) = F unitcell (hkl) F crystal (Q = hkl) = 0
7 = j = j (Q) ) Q) B j =8 2 u 2 j F (hkl) = j F (hkl) = j f(q hkl ) exp[2 i(hx j + ky j + lz j )] exp( B j sin 2 / 2 ) b j exp[2 i(hx j + ky j + lz j )] exp( B j sin 2 / 2 ) 4-fold axis z (x,y,z) ( y,x,z) )
8 mirror plane x (x,y,z) ( x,y,z)
9 m 2 / m 3m centre of symmetry (x,y,z) ( x, y, z)
10 4-fold axis z (x,y,z) [ ( y,x,z) ] (x,y,z) [ (y, x, z)
11 Crystal system Symmetry requirements Point groups Triclinic None 1 1 Monoclinic One 2-fold axis 2 m 2/m Orthorhombic Three 2-fold axes 222 mm2 mmm Trigonal One 3-fold or 3 axis m 3m Tetragonal One 4-fold or 4 axis 4 4 4/m 422 4mm 42m 4/mmm Hexagonal One 6-fold or 6 axis 6 6 6/m 622 6mm 6m2 6/mmm Cubic Four 3-fold or 3 axes 23 m m m 3m Schönflies International (Hermann-Mauguin) C1 S2 1 1 C2 C1h C2h 2 m 2/m D2 C2v D2h 222 mm2 mmm C3 S6 D3 C3v D3v m 3m C4 S4 C4h D4 C4v 4 4 4/m 422 4mm D2v D4h 42m 4/mmm C6 C3h C6h D6 C6v 6 6 6/m 622 6mm D3h D6h 6m2 6/mmm T Th O Td Oh 23 m m m 3m
12 2-fold axis z (x,y,z) ( x, y,z) Group 1: (0,0,z) Group 2: (x,y,z) Group 3: ( x, y,z) F (hkl) = X fj exp[2 i(hxj + kyj + lzj)] j X = fj exp[2 ilzj]+ fj exp[2 i(hxj + kyj + lzj)] + fj exp[2 i(hxj + kyj + lzj)] j2group1 X j2group2 j2group3 = fj exp[2 ilzj]+ fj exp[2 i(hxj + kyj + lzj)] + fj exp[2 i( hxj kyj + lzj)] j2group1 X j2group2 j2group2 = fj exp[2 ilzj]+ fj exp[2 i({ h}xj + { k}yj + lzj)] + X fj exp[2 i( { h}xj { k}yj + lzj)] j2group1 j2group2 j2group2 = F ( h kl)
13 X X X X F ( h k l) = X j = X j f j exp[2 i({ h}x j + { k}y j + { l}z j )] f j exp[ 2 i(hx j + ky j + lz y )] = F (hkl) I( h k l) = F ( h k l) 2 = F (hkl) 2 = F (hkl) 2 = I(hkl) Reciprocal space is centrosymmetric Symmetry of the diffraction pattern The point group of the diffraction pattern is known as the Laue class. The point group of the diffraction pattern is the point group of the crystal plus a centre of symmetry. The Laue classes are the centrosymmetric crystallographic point groups
14 Crystal system Laue class Point groups of the Laue group Triclinic Monoclinic 2/m 2 m 2/m Orthorhombic mmm 222 mm2 mmm Trigonal m 32 3m 3m Tetragonal 4/m 4 4 4/m 4/mmm 422 4mm 42m 4/mmm Hexagonal 6/m 6 6 6/m 6/mmm 622 6mm 6m2 6/mmm Cubic m3 23 m3 m 3m m m 3m I 4 1 cd 4mm 4/mmm P /m
15 Where are we heading? Translational symmetry does not affect directly the symmetry of a diffraction pattern, but does affect which reflections are observed.
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