Chap23 solutions

Physical Chemistry

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23 The solid state Solutions to exercises Discussion questions E23.1(b) We can use the Debye–Scherrer powder diffraction method, follow the procedure of Example 23.3, and in particular look for systematic absences in the diffraction patterns. We can proceed through the following sequence 1. Measure distances of the lines in the diffraction pattern from the centre. 2. From the known radius of the camera, convert the distances to angles. 3. Calculate sin 2 θ . 4. Find the common factor A = λ 2 / 4a 2 in sin 2 θ = 2 / 4 a 2 )(h 2 + k 2 + l 2 ) . 5. Index the lines using sin 2 θ/A = h 2 + k 2 + l 2 . 6. Look for the systematic absences in ( hkl ). See Fig. 23.22 of the text. For body-centred cubic, diffraction lines corresponding to h + k + l that are odd will be absent. For face-centred cubic, only lines for which h , k , and l are either all even or all odd will be present, other will be absent. 7. Solve A = λ 2 / 4 a 2 for a . E23.2(b) The phase problem arises with the analysis of data in X-ray diffraction when seeking to perform a Fourier synthesis of the electron density. In order to carry out the sum it is necessary to know the signs of the structure factors; however, because diffraction intensities are proportional to the square of the structure factors, the intensities do not provide information on the sign. For non-centrosymmetric crystals, the structure factors may be complex, and the phase α in the expression F hkl =| F hkl | e i α is indeterminate. The phase problem may be evaded by the use of a Patterson synthesis or tackled directly by using the so-called direct methods of phase allocation. The Patterson synthesis is a technique of data analysis in X-ray diffraction which helps to circum- vent the phase problem. In it, a function P is formed by calculating the Fourier transform of the squares of the structure factors (which are proportional to the intensities): P(r) = 1 V X hkl | F hkl | 2 e 2 π i (hx + ky + lz) The outcome is a map of the separations of the atoms in the unit cell of the crystal. If some atoms are heavy (perhaps because they have been introduced by isomorphous replacement), they dominate the Patterson function, and their locations can be deduced quite simply. Their locations can then be used in the determination of the locations of lighter atoms. E23.3(b) In a face-centred cubic close-packed lattice, there is an octahedral hole in the centre. The rock-salt structure can be thought of as being derived from an fcc structure of Cl ions in which Na + ions have ±lled the octahedral holes. Thecaesium-chloridestructurecanbeconsideredtobederivedfromtheccpstructurebyhavingCl ions occupy all the primitive lattice points and octahedral sites, with all tetrahedral sites occupied by Cs + ions. This is exceedingly dif±cult to visualize and describe without carefully constructed ±gures or models. Refer to S.-M. Ho and B. E. Douglas, J. Chem. Educ . 46 , 208, 1969, for the appropriate diagrams.
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Chap23 solutions - 23 The solid state Solutions to...

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