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CHE-20031 ( S tructural I norganic C hemistry) X-ray Diffraction & Crystallography lecture 2

CHE-20031 ( S tructural I norganic C hemistry) X-ray Diffraction & Crystallography lecture 2. Dr Rob Jackson LJ1.16, 01782 733042 r.a.jackson@keele.ac.uk www.facebook.com/robjteaching Twitter: #che20031. X-ray Diffraction & Crystallography: lecture 2 plan.

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CHE-20031 ( S tructural I norganic C hemistry) X-ray Diffraction & Crystallography lecture 2

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  1. CHE-20031 (Structural Inorganic Chemistry)X-ray Diffraction & Crystallography lecture 2 Dr Rob Jackson LJ1.16, 01782 733042 r.a.jackson@keele.ac.uk www.facebook.com/robjteaching 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. 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. 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 = 10-12 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) A θ 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) lattice planes dhkl 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 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. 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. 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. 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 = 19.213,  = 9.607 • sin  = 0.1669 • sin2 = 0.0279 • 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 = 12 + 0 + 0 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: 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

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