Structure Factors F HKL. Fobs = k I HKL magnitude of F only from expt

Size: px

Start display at page:

Download "Structure Factors F HKL. Fobs = k I HKL magnitude of F only from expt"

Reynard Gray
6 years ago
Views:

1 Structure Factors F HKL Fobs = k I HKL magnitude of F only from expt F calc = Σ f i {cos(2π(hx i + ky i + lz i )) + i sin(2π(hx i + ky i + lz i ))} sum over all atom locations (x,y,z) where f i is the scattering power of atom i. At low angles f i = number of electrons. f c = 6 f N = 7 etc At higher angles f i decreases as explained later. For centrosymmetric structures the sin terms drop out since for every atom at (xyz) another is at (-x,-y,-z). AND sin(θ) = -sin(-θ)

2 Non-centrosymmetric cases. i F(hkl) F = A + B i is a vector in the complex plane A = Σ cos terms B = Σ sin terms real α magnitude ( vector length) = (A 2 + B 2 ) F(-h,-k,-l) phase angle, α = tan -1 (B/A) angle between F vector and real axis centro case B = 0 if A + phase angle α = 0 if A negative α = π we will not concern ourselves with this here. Friedel pairs : / F(hkl)/ = /F(-h,-k,-l)/ and α HKL = -α -H,-K,-L except in chiral systems and cases of anomalous dispersion where this is not true the effects of anomalous dispersion are used to determine absolute configuration.

3 What is a structure factor anyhow? it s the fraction of electrons scattering in phase at the angle θ HKL it s the vector sum of scattering vectors, f i from each atom in the unit cell. it has units of electrons they are the coefficients you need to write out the Fourier series for the electron density F(000) would = total electrons in cell. The Bragg angle θ 000 = 0 for this unobservable reflection - X-rays go straight thru with no phase lag. At any other angle objects will be more or less in or out of phase reducing the F HKL below that of F 000.

4 Note that the path difference increases with the angle and hkl. f drops off with increasing angle because scattering from top and bottom electrons are increasingly slightly out of phase.

5 Sample calculation for CsCl Z =1 Pm3m centrosymmetric Cs at (0,0,0) Cl at (1/2,1/2,1/2) or reversed in figure below. summing over the two atom locations (neglect f i vsθ) F hkl = f CS cos(2π(hx cs + ky cs + lz cs )) + f Cl cos(2π(hx cl + ky cl + lz cl )) F hkl = 55 cos(0) + 17 cos(2π(h/2 + k/2 + l/2)) F 100 = cos(π) = = 38 F 200 = cos(2π) = = 72 F 111 = cos(3π) = = 38 etc. Note that the alternate choice of origin with Cl in the centre gives negative F s. the phase depends on the choice of cell origin.

6 No absences in Pm3m CsCl uses Wy a and b sites.

7 Sample calculation for body centredα-fe Fe at (0,0,0) Fe at (1/2,1/2,1/2). Z =2 Im3m summing over the two atom locations (neglect f i vsθ) F hkl = f Fe {cos(2π(0h + 0k + 0l)) + cos(2π(1/2h + 1/2k + 1/2l)) F hkl = 26 {cos(0) + cos(2π(h/2 + k/2 + l/2)) F 100 = 26{cos(0) + cos(π)} = 0 F 200 = 26{cos(0) + cos(2π) = 52 F 111 = 26{cos(0)+ cos(3π) = 0 F 211 = 52 F 221 = 0 etc absences occur for hkl when h+k+l = 2n+1 (odd) (condition for observing reflection h+k+l = 2n (even)

8 NaCl Fm3m z=4 α-fe Im3m Z =2 CsCl Pm3m Z =1

9 NaCl Fm3m Z =4 a =b=c = 5.6 Å Counting atoms Na 12 on each edge X ¼ + 1 in centre = 4 Cl 8 on corners X 1/8 + 6 on faces X ½ = 4 density = mass/volume = Z x MW / N o a 3 = gm/cm 3 for a in angstroms density = Z x MW / a 3 = 4 x 58.5/ (0.6 x ) Exercise. Compute the structure factors for NaCl. Obtain the NaCl bond length.

10 48 = 6x4x2 what you get if you take all permutations of xyz + or - each. see Pm3m Then add face centring to get 192. NaCl uses Wy a and b Z =4 CaF 2 Z =4 Ca in Wy a F in Wy c F centre gives absences m3m has 3 mirrors

11 Prove that a 2 1 along b requires 0k0 k =2n F 0k0 = Σ cos(2π(0x + ky+0z) For every atom at xyz another identical one is at -x, y+1/2, -z. summing over half the atoms and their screw equivalents separately F 0k0 = Σ f i cos(2πky) + Σf i cos(2πk(y+1/2)) = Σ f i cos(2πky) + Σ f i cos(2πky+ πk) for k even + + = F 0k0 for k odd + - = 0 or sketch cos(q) and cos(q+π) and add the two.

12 Prove that a C centred lattice requires hkl h+k = 2n F hkl = Σ cos(2π(hx + ky+lz) For every atom at xyz another identical one is at x+½,y+1/2, z. summing over half the atoms and C centred equivalents separately {A = 2π(hx + ky+lz) } F hkl = Σ f i cos(a) + Σf i cos(a +π(h+k)) using cos(a+b) = cos(a)cos(b)-sin(a)sin(b) F hkl = Σ f i cos(a) + Σf i cos(a)cos(π(h+k)-sin(a)(sin(π(h+k)) noting (sin π or 2π etc = 0 and cosπ(h+k) = 1 or -1 for even or odd h+k for h+k even F = 2 Σ f i cos(a) for h+ k odd F = Σ f i cos(a) -Σf i cos(a) = 0

13 Solving a Structure from HKL file. 1. Examine symm of unit cell, reflection intensities, and systematic absences to determine space group. 2. Use Direct or Patterson Methods to obtain trial phases. 3. Calculate e-density map and pick out peaks in it. 4. Assign peaks to C, N,O, etc. 5. Using tables of f i and atom locations compute F calc and its sign (phases). 6. Use I HKL to obtain F obs. Use phases from (5) and F obs to compute e-density map A. 7. Compute another e-density map using F calc s from (5). 8. Examine the difference = Difference Map for peaks and holes. 9. Add or delete atoms. Repeat steps 7 and Run least squares to minimize R = Σ {F o -F c }/ Σ F o

SOLID STATE 9. Determination of Crystal Structures

SOLID STATE 9. Determination of Crystal Structures SOLID STATE 9 Determination of Crystal Structures In the diffraction experiment, we measure intensities as a function of d hkl. Intensities are the sum of the x-rays scattered by all the atoms in a crystal.