1. CHE-20031 (Structural Inorganic Chemistry) X-ray Diffraction & Crystallography lecture 2
Dr Rob Jackson
LJ1.16,
Twitter: #che20031

2. X-ray Diffraction & Crystallography: lecture 2 plan
Introduction to X-ray diffraction
Waves and diffraction
X-rays
What happens in an XRD experiment?
The Bragg equation
Introduction to XRD patterns
Indexing patterns
che-20031: XRD & Crystallography lecture 2

3. che-20031: XRD & Crystallography lecture 2
Waves & Diffraction
Interference of Waves:
Constructive (add displacement)
Destructive (subtract displacement)
Diffraction:
Interference of waves caused by an object
Pattern of enhanced and diminished intensities (dark and light spots)
Occurs when dimensions of objects are comparable to the wavelength of the radiation
che-20031: XRD & Crystallography lecture 2

4. che-20031: XRD & Crystallography lecture 2
X-rays
Electromagnetic radiation
Interacts with electrons in matter
Electron clouds scatter X-rays
Wavelength of X-rays are comparable with the separation between atoms in compounds
Interference of scattered x-rays occurs
λ = 60–190 pm (1 pm = m)
These are comparable to the bond lengths in molecules and spacing of atoms in crystals.
che-20031: XRD & Crystallography lecture 2

5. What happens in an X-ray diffraction experiment?
Generate a collimated X-ray beam of single wavelength (60–190 pm, monochromatic) radiation
Use crystal monochromator and exit slits
Beam interacts with sample
Detect scattered X-rays
Measure intensity
che-20031: XRD & Crystallography lecture 2

6. But what is actually happening?
We are dealing with crystalline solids.
X-rays interact with different lattice planes
Scattering of x-rays from crystalline solids is best described by consideration of diffraction from a set of planes defined by Miller indices (hkl)
This is shown on the next slide which also introduces the ideas behind the Bragg equation.
che-20031: XRD & Crystallography lecture 2

7. The Bragg equation – (i)θ A
lattice planes B θ
dhkl C D
Consider scattering from lattice points A and D
diffracted X-rays
They will be in phase (constructive interference) if the extra distance travelled by an X-ray scattered by D is a whole number of wavelengths, λ.
che-20031: XRD & Crystallography lecture 2

8. The Bragg equation – (ii)
Extra distance = BD + DC
Depends on lattice spacing, dhkl, and angle of incidence of x-ray beam, θ
BD + DC = 2 x dhkl x sin θ
lattice planes
dhkl
che-20031: XRD & Crystallography lecture 2

9. The Bragg equation – (iii)
nλ = 2 dhkl sin θ
λ is the wavelength, and θ is the angle of incidence.
Each set of planes, (hkl), has a particular separation, dhkl
For each set of planes there will be a diffraction maximum (peak) at a particular angle, θ.
che-20031: XRD & Crystallography lecture 2

10. How the Bragg equation is used
The relationship between d-spacing and lattice parameters of crystalline solid can be determined by geometry
Combine equation relating d-spacing to lattice parameters for particular crystal system, with the Bragg equation
This gives a direct relationship between diffraction angle and lattice parameters
X-rays are scattered by a crystalline solid, giving a diffraction pattern (series of peaks). The location of the peaks are related to the lattice parameters of the materials.
Gives structural information about crystalline materials.
che-20031: XRD & Crystallography lecture 2

11. Relating the diffraction angle to the lattice parameters
For a cubic material
From lecture 1 slide 26:
The Bragg equation is: λ = 2 dhkl sin θ (n=1)
Combine these equations! (This will be shown).
Gives
che-20031: XRD & Crystallography lecture 2

12. An XRD pattern: peak intensity vs 2
che-20031: XRD & Crystallography lecture 2

13. Understanding the pattern
Each peak represents diffraction of X-rays from a lattice plane in a crystalline solid.
Indexing is the process of assigning Miller Indices to each peak in an X-ray diffraction pattern.
We will now look at how to index a pattern – i.e. how to get the Miller indices for each peak.
che-20031: XRD & Crystallography lecture 2

14. che-20031: XRD & Crystallography lecture 2
Indexing
Starting with the equation from slide 11:
We can assume that (2/4a2) is a constant, and write it as A.
Next we have to read the  values from an XRD pattern, and convert them to sin2.
che-20031: XRD & Crystallography lecture 2

15. che-20031: XRD & Crystallography lecture 2
Indexing procedure
Obtain θ from 2θ in your diffraction pattern for the first 7 or 8 reflections (fewer if specified). The accuracy of your answer will depend on how well you can read the scale on the powder pattern and the accuracy of your calculations.
Calculate sin2 θ [which is the same as (sinθ)2] for all reflections.
Divide the value of sin2θ for each reflection by the value for the lowest reflection. This eliminates A in the equation above and gives the ratio.
The ratio is a number corresponding to h2+ k2 + l2 so you must determine the three integers (0,1,2…) that will give the ratio when squared and added together.
che-20031: XRD & Crystallography lecture 2

16. che-20031: XRD & Crystallography lecture 2
Example
The data below comes from a diffraction pattern. The procedure on slide 15 has been followed, and is explained for the first 2 points on the right.
If 2 = ,  = 9.607
sin  =
sin2 =
Calculate for all values
Divide each value of sin2 by the lowest reflection value (0.0279)
Determine the Miller indices – e.g. for point 1:
1 = 2
sin2
Ratio
Miller indices
19.213
0.0279 1
100
27.302
0.0557 2
110
33.602
0.0836 3
111
38.995
0.1114 4
200
43.830
0.1393 5
210
48.266
0.1672 6
211
56.331
0.2228 8
220
60.093
0.2507 9
221, 300
che-20031: XRD & Crystallography lecture 2

17. Missing ratios and systematic absences
Notice that you never* get a ratio of 7, because there are no three integers that can be squared and added together to give 7 (or 15).
Also, depending on the lattice type, certain reflections may be missing, called systematic absences.
*But see table for body-centred I cell.
The table below applies to cubic systems. Depending on the lattice type, the following rules apply:
Lattice type
Rules for observed reflections
Primitive P
All reflections observed
Body-centred I
h + k + l = 2n (2,4,6,8,10,12,14,16)
Face-centred F
h,k,l are either all odd or all even
Side-centred C
h + k = 2n
This can help determine the lattice type.
che-20031: XRD & Crystallography lecture 2

18. Tetragonal and orthorhombic systems
The equation sin2θ = A (h2 + k2 + l2) is only true for cubic systems
Tetragonal (a=b≠c): 1/d2 = (h2 + k2) /a2 + (l2 /c2)
Orthorhombic (a≠b≠c): 1/d2 = h2/a2 + k2/b2 + l2 /c2
Determining the lattice parameters for tetragonal and orthorhombic systems is more complicated. Use the positions of the (100), (010) and (001) reflections to determine a, b and c.
che-20031: XRD & Crystallography lecture 2

19. Summary and learning objectives
Having attended this lecture, and read and understood the notes, you should be able to:
Understand what is happening in an XRD experiment.
Write down and explain the Bragg equation.
Calculate Miller indices from reflections for diffraction patterns from cubic materials.
Understand the effect and meaning of systematic absences for cubic lattices.
che-20031: XRD & Crystallography lecture 2