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2.1 Identifying crystal systems and lattice types.
2.1.1 Explain what is meant by the term lattice. What are the lattice types of the following structure types.
(a) CsCl
(b) NaCl
(c) cubic ZnS
A lattice is an array of equivalent lattice points, in one, two, or three dimensions that shows the full translational symmetry.
(a) Primitive (cubic)
(b) Face-centred (cubic)
(c) Face-centred (cubic)
2.2 Drawing lattice planes and Miller indices.
2.2.1 Draw a diagram to represent an orthorhombic unit cell of dimensions 3 Å x 5 Å x 4 Å. Mark on it (or copies of it) lattice planes with Miller indices
(1, 2, 4) (1, 0, −1) (0, 0, 1)
Mark the distance corresponding to d124 on the appropriate diagram.
2.3 d-Spacing calculations and Miller indices
2.3.1 For an orthorhombic unit cell of dimensions a = 3, b = 5 and c= 8 Å, draw on a diagram (or diagrams) of the unit cell, planes with the Miller indices
(2, 1, 2) and (4, 2, 4)
Use the Eqn 2.4 to calculate d-spacing values for the (2, 1, 2) and (4, 2, 4) planes. How are these related?
Using 
For the (2,1,2) reflection
For the (4,2,4) reflection
Thus as expected the d-spacing of the (4,2,4) plane is half that of the (2,1,2). The figure also shows the position of the (4,2,4) half the distance from the origin of the (2,1,2) plane. Note also that alternate planes with Miller indices (4,2,4) lie on top of planes with (2,1,2) indices.
2.4 Bragg equation calculations
2.4.1 A powder diffractometer uses the refection from the (1,1,1) plane of a cubic germanium crystal (a = 5.658 Å) as a monochromator. Calculate the experimental diffraction angle, 2θ, from this plane for an X-ray of wavelength 2.23 Å.
First of all we need to calculate the d-spacing of the (1,1,1) plane in germanium using the equation (Eqn 2.6)
So d = a/√3 = 3.267 Å. Then using the Bragg equation 2dsinθ = nλ 2 ×3.267 × sinθ = 2.23 gives θ=19.956° and 2θ = 39.91°
2.5 Systematic absences
2.5.1 Li2O adopts the antifluorite structure with an F-type cubic lattice. The lattice parameter of Li2O is 4.61 Å, what will be the 2θ value of the first reflection observed in its powder X-ray diffraction pattern?
As the lattice type is F the first reflection will have the Miler indices (1,1,1). The d-spacing corresponding to this will be given (as in problem 2.4) as a/√3 = 2.662 Å. Using λ=1.54 Å sand the Bragg equation 2 × 2.662 sinθ = 1.54 so 2θ= 33.63°.
2.5.2 Diamond has an F-type cubic lattice. What will be the Miller indices of the first five allowed reflections observed in its X-ray diffraction pattern?
For an F –centred lattice h, k , l have to be all odd or all even so the first five expected reflections would have Miller indices (1,1,1), (2,0,0), (2,2,0) (3,1,1) and (2,2,2). (In fact additional crystallographic symmetry beyond that of the F-centred lattice means that the (2,0,0) reflection is not observed experimentally).
2.5.3 The first reflection in the X-ray diffraction pattern of CsCl is found at 2θ = 21.55°, calculate its lattice parameter.
CsCl has a primitive cubic unit cell (dimensions a) so the first reflections will have the Miller indices (1, 0 ,0). For this reflection d = a, so in the Bragg equation, 2dsinθ = nλ, 2 × a × sin(21.11/2) = 1.5418. Therefore, a=4.208 Å.
2.5.4 Draw a unit cell of iron metal, which adopts a face-centred, cubic arrangement of metal atoms. Mark on this diagram two adjacent planes with the Miller indices (1, 1, 0). Hence explain why the (1, 1, 0) reflection is absent in the powder X-ray diffraction pattern of iron metal.
The (1, 1, 0) reflection is absent in iron metal as X-rays scattering from the atoms in the interleaving planes are exactly out of phase from those scattering from those scattered by atoms in the (1, 1, 0) planes. There is an equivalent number in each so the scattering cancels out exactly.
2.6 Indexing of PXD data
2.6.1 The powder X-ray diffraction pattern (λ=1.5418 Å) collected from a sample of radium (Ra) metal showed reflections at the following values of 2θ (degrees)
24.45 34.85 43.04 50.12 56.53 62.50 68.15 73.60 78.89
Index the data, determine the lattice type and the lattice parameter.
Heating radium in fluorine gas afforded the product A containing 14.39 % F whose powder diffraction pattern could be indexed using a cubic lattice parameter of 6.381 Å with systematic absences indicative of an F-centred lattice. Describe a likely structure for A and calculate the shortest Ra - F distance.
The data from radium metal can be indexed as follows
2θ |
sin2θ x 1000 |
ratio |
hkl |
24.45 |
44.84 |
2 |
110 |
34.85 |
89.67 |
4 |
200 |
43.04 |
134.56 |
6 |
211 |
50.12 |
179.41 |
8 |
220 |
56.53 |
224.25 |
10 |
310 |
62.50 |
269.13 |
12 |
222 |
68.15 |
313.91 |
14 |
321 |
73.60 |
358.83 |
16 |
400 |
78.89 |
403.65 |
18 |
411/330 |
From the systematic absences the lattice type is body centred, I, and the lattice parameter can be determined as 5.148 Å.
The composition of the product from radium –fluorine reaction is RaF2 and with an F-centred lattice type and the AB2 stoichiometry the most common structure type found is the fluorite (CaF2) structure which has close packed Ra2+ ions with F− ions in all the tetrahedral holes. The shortest Ra-F distance is from an Ra2+ ion at the unit cell origin (0,0,0) to an F− ion at (¼,¼,¼) which is a quarter of the distance along the unit cell body diagonal of length √3 × a . So the shortest Ra-F distance is (√3a)/4 = 2.229 Å.
2.6.2
(a) Cadmium selenide and cadmium sulphide form a complete series of compounds with the general composition CdS1-xSex .This material is used as a pigment whose colour varies from yellow at x = 0 to red at x = 1. Both CdS and CdSe adopt the cubic ZnS structure which can also be described as Cd2+ ions occupying half the tetrahedral holes in a cubic close packed arrangement of sulfide ions.
(i) Draw the structure of CdS clearly labelling and marking the atom positions.
(ii) Given the following ionic radii
S2– 164 pm Se2– 182pm Cd2+ 67 pm
Calculate the expected lattice parameters for CdS and CdSe
(iii) Sketch a diagram showing how the lattice parameter of CdS1-xSex would be expected to vary as a function of x.
(iv) A sample of orange paint from a crime scene was analysed by powder X-ray diffraction (λ=1.5418Å) and showed reflections at the following values of 2θ (degrees)
28.47 32.99 47.35 56.18 58.92 69.21
Index the data, determine the lattice type and the lattice parameter.
(v) Hence determine the composition of the pigment in the organic paint assuming it to be from the CdS1-xSex solid solution.
(i) 
(ii) The shortest Cd-S(Se) distance ( similar to problem 2.5) is (√3a)/4. Setting this equal to the sum of the ionic radii of Cd2+ and S2− gives
164 pm + 67 pm = (√3a)/4
Giving a = 533.5 pm for the lattice parameter of CdS. For CdSe the equivalent calculation give a = 575.0 pm.
(iii) CdS and CdSe have the same structure type and substituting one ion with another should lead to a smooth and gradual change in the lattice parameter. (When this is a linear variation of lattice parameter with composition this is known as Vegard’s law; sometimes small positive or negative curved deviations from linear behaviour occur.)
(iv)
2θ |
sin2θ |
ratio |
hkl |
28.47 |
0.0605 |
3 |
111 |
32.99 |
0.0806 |
4 |
200 |
47.35 |
0.1612 |
8 |
220 |
56.18 |
0.2217 |
11 |
311 |
58.92 |
0.2419 |
12 |
222 |
69.21 |
0.3225 |
16 |
400 |
The data can be indexed using a common ratio of 0.0201 and in agreement with the structure shown in the Figure for CdS andCdSe the lattice type is face-centred.
The lattice parameter using a wavelength of 154.18 pm is 543.0 pm.
(v) Using the graph or the linear expression that gives this graph for x in CdS1-x Sex.
543.0 = 533.5 (1-x) +575.1(x)
Gives x = 0.23 and the composition CdS0.77Se0.23 for the pigment.
2.6.3 (a) The ionic radii (Å) relevant to the caesium halides (CsF, CsBr and CsI) may be taken as
Cs+ 1.88 F– 1.29 Br– 1.96 I– 2.20
(i) Use radius ratio rules to predict the structure type adopted by each of these three caesium halides.
The radius ratio, γ, is the radius of the smaller ion divided by that of the larger. For CsF, CsBr and CsI the values of γ are respectively 0.686, 0.959 and 0.854. Radius ratio rules predict that CsF would have the rocksalt structure (with γ between 0.414 and 0.732) and CsBr and CsI the CsCl structure type ( γ > 0.732)
(ii) Use the ionic radii to calculate values for the lattice parameter of each structure.
For the rocksalt structure the lattice parameter (from the repeat Cs-F-Cs along the unit cell edge) will be 2(r+ + r−) = 6.34 Å. For the Cs-Cl structure type the lattice parameter, a, can be derived from the fact that the unit cell body diagonal is √3 x a which is equal to 2(r+ + r−). So for CsBr a = 4.43 Å and CsI a = 4.78 Å
(iii) Describe the three lattice types known for the cubic crystal system and state the systematic absences expected for each of in powder X-ray diffraction data.
|
Lattice type and condition for reflection to be seen |
||
Miller indices |
Primitive, P All possible values of h, k and l |
Body centred, I h + k + l = 2n |
Face centred, F h, k, l all odd or all even |
(iv) Predict the position of the first reflection observed in the powder X-ray diffraction patterns from CsF, CsBr and CsI. Take λ = 1.54 Å.
For CsF with the F-centred lattice of rocksalt, the first reflection would be expected to have Miller indices (1,1,1). Using
2θ of the first reflection is 24.29°.
For the primitive lattice types of CsBr and CsI the first expected refection has Miller indices (1,0,0). So the first reflection is predicted at 20.02° (CsBr) and 18.54°(CsI)
(v) The first reflection observed in the powder X-ray diffraction pattern of CsI is measured at 26.73°. Explain any difference between this value and the value calculated in (iv).
Cs+ and I− are isoelectronic and will scatter X=rays to an almost identical degree. Considering the CsI unit cell as a set of atoms that are all “equivalent” changes the lattice type to body centred – as “seen” by X-rays. Therefore the first observed reflection with measureable intensity will have Miller indices (1,1,0) which corresponds to the 26.73° peak.
2.7 Structure factor calculations
2.7.1 Silver metal adopts a simple F-centred cubic structure with atoms at the coordinate positions (0,0,0), (½,½,0), (½,0,½), and (0,½,½). The structure is centrosymmetric so the structure factor expression simplifies to
Calculate a value of the structure factor, in terms of fAg, for the (1,0,0), (1,1,0) and (1,1,1) reflections. Comment on your answer in respect of the systematic absences seen for F-centred lattices.
Similarly F100 = 0
This calculation demonstrates the mathematical and theoretical origin of the face centred lattice systematic absence conditions. In that for the (1,0,0) plane scattering from two atoms at (½,½,0) and (½,0,½) is exactly out of phase with that from the two other atoms at (0,0,0) and (0,½,½) giving zero intensity. The same is true for the (1,1,0) reflection but for the (1,1,1) reflection the scattering is in phase for all the four atoms in the unit cell giving intensity to this reflection.
2.8 Structure solution methods
2.8.1 Decide whether Patterson or Direct Methods would be the most appropriate starting point for the structure solution of the following compounds.
(a) NaSiAlO4 (b) PbCO3 (c) AgSnI3
(a) Na+, Al3+ Si4+ and O2− are all, isoelectronic eight electron species. For such materials direct methods would be most appropriate
(b) PbCO3 contains the heavy ion Pb2+ and two light atom species. So Patterson methods would be used to locate the lead atom position. Once the lead site(s) had been located Fourier difference calculations and map techniques would yield the carbonate anion atom positions.
(c) Ag+, Sn2+ and I− have similar electron counts of 46, 48, 54 and again Direct Methods would be the preferred structure solution method.