Chapter 2 Answers to Self-test Questions

2.1 Identify the crystal system of β-Hg shown in Figure 2.4(c).

The unit cell has all length equal and a single angle between them not equal to 90°, so the unit cell is rhombohedral from Table 2.1.

2.2 Identify the lattice types of(a) α-Hg and (b) β-Hg shown in Figures 2.4(b) and (c).

1. The unit cell has mercury atoms on the corners and at the cell centre. Thus the additional translational symmetry (+½ , +½, +½) is present, making this a body centred I lattice type.
2. This rhombohedral unit cell only has mercury atoms at the corners so the translational symmetry of the structure is just that of the unit cell making it a primitive, P, lattice type  

2.3 For the sameorthorhombic cell of dimensions 5 × 3 ×4 Å draw planes with Miller indices (a)  (2, −1, −2)  and   (b)   (4, 0, 0).

 

2.4 By consideration of the expressions for d-spacing values in Table 2.2 determine for which crystal systems the d-spacing values of the (1, 1, 1) and (−1, −1, −1) planes will be equal.

Any crystal system in which pairs of Miller indices are squared or multiplied together h² (as −1 × −1  = 1 ×1),  or (h²+hk+l²)  That is all crystal systems.

 

 

2.5 Calculate diffraction angles using λ= 1.5406 Å for the (1, 0, 0) and (0, −1, 0) planes of a cubic crystal structure of lattice parameter 6.31 Å.

Substituting values into the expression

and then calculating 2θ as the measured diffraction angle give 2θ = 14.02°  for (1, 0, 0)  As the h²+k²+l² values for the two reflections (1, 0, 0) and (0, −1, 0) are identical the reflections occur at the same 2θ value

 

2.6 Calculate the monochromator angle with the same germanium crystal needed to select MoK_α radiation.

MoK_α radiation has a wavelength of 0.71073 Å so substituting into 2dsinθ = nλ with n=1 and d = 3.266 Å gives 2 × 3.266 sinθ = 0.71073. Solving to give 2θ = 12.49°

2.7 Index the following PXD data collected from polonium (λ = 1.5418Å) and calculate the lattice parameter.

Using a Table similar to that of the examples in the text         

 

The lattice parameter can be determined from any reflection but using the highest angle one

 

So a = 3.362 Å

 

2.8 Show that for a face centred cubic lattice the(1, 1, 1) plane has no interleaving planes and will, therefore, be present in a PXD pattern.

2.9 Determine the diffraction angle of the next highest angle diffraction peak, above the (4,2,0) reflection, expected to be seen for crystalline C₆₀ in its PXD data.

The next reflection for an F centred lattice with hkl all odd or all even after the (4,2,0) (h²+k²+l² = 20) is the (4,2,2) ( h²+k²+l² = 24). Substituting appropriate values, a = 14.11 Å, into  

gives 2θ = 31.0°.

 

2.10 Which of the Group II sulfides, MgS, CaS, SrS, or BaS, which all adopt the rock salt structure, will display a PXD pattern that can be indexed using a primitive lattice type?

The material will appear to be primitive to X-rays when the cation and anion are isoelectronic. With the sulfide ion, an 18 electron species, it is Ca²⁺ that is isoelectronic so the pattern from CaS will present in its PXD pattern as being primitive cubic with half the true unit cell parameter 

    

2.11. The structure of solid C₆₀ is face-centred cubic arrangement C₆₀ molecules. Use the lattice parameter for solid C₆₀ derived in Example 2.9 to obtain an estimate for the size of a single C₆₀ molecule.

The face centred cubic arrangement of spheres is the same as cubic close packing. It these close packed spheres are in contact then two such touching spheres are those at the corner of the  unit cell and one in the centre of a face. The distance from the corner of a cube of side a to the centre of a face can be calculated using Pythagoras’ theorem to be (√2 × a)/2. For C₆₀ this distance is 14.11/1.414 = 9.98 Å. This distance would be equivalent to twice the radius of a C₆₀ molecule for two spheres in contact so an estimate for the diameter of a C₆₀ molecule is ~10 Å. This calculation assumes the C₆₀ molecules touch but they will be separated by the distance equivalent to twice the van der Waals radius of carbon so the size of a C₆₀ icosahedron is smaller than this calculation suggests.        

2.12. The transition between the two polymorphs of TlBr occurs at 232 K. Sketch a figure, similar in form to that of Figure 2.29, but over the angular range 2θ =10-60 ° showing the expected evolution of the diffraction pattern (λ = 1.5418 Å) of TlBr between 200 and 300 K. 

The key features of plot should be

1. a shift in peak positions to larger 2θ values at the sample is cooled below 300 K. This represents a contraction in the unit cell, smaller d-spacings and therefore larger 2θ values. The shift will be larger in 2θ terms towards higher angles as d α 1/sinθ. The reflections seen will be the same as those at 300 K.
2. When the material goes through the phase change there is an abrupt change in pattern with the new unit cell and systematic absences
3. On further cooling the peaks with move to higher 2θ values.    

2.13 (a) Calculate |F_hkl| for the (2, 0, 0) reflection for CsCl using   f_Cs = 45 and f_Cl = 15. (b)  Repeat the calculation of |F_hkl| for the (2,1,0) reflection but for CsI instead of  CsCl and using f_I = 41. Comment on your answer.

 

(a) Using again for the (2, 0, 0) reflection gives 

(b)

Thus zero intensity will be observed for the (2,1,1) reflection. In fact all reflections with h+k+l =2n+1 will have zero intensity. This is equivalent the systematic absence condition for a body-centred unit cell. Because Cs⁺ and I⁻ are isoelectronic they are observed as equivalent in the X-ray diffraction experiment. Making the unit cell corner and centre sites equivalent in the CsCl structure generates a unit cell with body-centring.

This calculation shows the origin of the body-centring systematic absence in that for a general reflection with Miller Indices (h, k, l)  |F_hkl|  is given by      

And if, as in body centring, the atoms, with f_j = f_A always occur in pairs at (x, y, z) and (x+½,y+½,z+½) then

When h+k+l is odd then for all values of h, k and l the second term in this expression is equal and opposite to the first and they cancel out to give F_hkl =0 

2.16 Calculate the expected position of the (1,1,1) reflection from chromium metal, a reflection not observed in the experimental data.

Substituting appropriate values into

With h = k = l = 1 and A = 0.1069 and C = 0.0302, sin²θ = 3A +C = 0.3509, giving 2θ = 72.65°.    

2.17 Calculate the size of a gold nanoparticle that gives rise to a (1,1,1) reflection of width 0.5°.

As in the example using the Scherrer formula and converting 0.5° to 0.5/57.296= 0.0087 radians

 

2.18 The structure of the alloy FeCo is shown in Figure 2.52. Predict which reflections would be observed in (i) the powder X-ray diffraction pattern from FeCo and (ii) the powder neutron diffraction pattern from FeCo. Take the X-ray scattering powers of the neighbouring elements Fe and Co to be identical while the relative neutron scattering powers are 9.45 (Fe) and 2.49 (Co).

Figure 2.52 The structure of FeCo with ordered Fe and Co positions.

1. In X-ray diffraction Fe and Co atoms scatter to an almost equivalent degree being very close to isoelectronic. Making all the atoms equivalent in Fig 2.52 produces a body centred cubic structure type so the systematic absence h+k+l = 2n+1 would be observed.
2. In neutron diffraction the significant differences in scattering length mean that the unit cell would be observed as primitive and no systematic absence observed.