1. Molecular Crystals

2. Molecular Crystals: Consist of repeating arrays of molecules and/or ions.

5. C 17 H 24 NO 2 + Cl -. 3 H 2 O

8. Although Z = 2, the unit cell contains portions of a number of molecules.

9. Cl -

11. H2OH2O

12. H2OH2O Hydrogen bonds Cl OH 2

14. Hydrogen bond

16. Model with atoms having VDW radii.

19. C 17 H 24 NO 2 + Cl -. 3 H 2 O Although this material is ionic, the + and - charges are not close enough to contribute to the formation of the crystal.

20. Molecular crystals tend to be held together by forces weaker than chemical bonds. van der Waal’s forces are always a factor. Hydrogen bonding is often present.

22. A layer in an ionic solid with ions of similar radii.

23. Metallic crystal – single layer of like sized atoms forms hexagonal array.

24. Second layer can start at a point designated b or c.

25. At this point, the third layer can repeat the first and start at a or it can start at c.

26. Third layer repeats first layer.

28. Unit Cell Unit cell volume = V

29. Unit Cell Unit cell volume = V V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  Note: text page 807 may not be correct.

30. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos 

32. cos 90 o = 0

33. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0

34. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0 V = abc 1 - cos 2 

35. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0 V = abc 1 - cos 2  sin x = 1 - cos 2 x

36. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0 V = abc 1 - cos 2  sin x = 1 - cos 2 x V = abc sin 

37. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0

38. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0

39. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  cos 90 o = 0 V = abc

40. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos 

41. a = b

42. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  a = b V = a 2 c

43. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos 

45. V = a 3

46. V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos 

50. V = a 2 c 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos 

52. V = a 2 c 1- cos 2 

53. V = a 2 c sin 

54. V = a 2 c 1- cos 2  V = a 2 c sin  = a 2 c sin 120 o

55. Cell volume and cell contents:

56. A unit cell will usually contain an integral number of formula units.

57. Cell volume and cell contents: A unit cell will usually contain an integral number of formula units. The number of formula units in the cell is often related to the symmetry of the cell.

58. The number of formula units in the unit cell is designated by Z.

59. Space group General Positions Z

60. P1 x, y, z 1

61. Space group General Positions Z P1 x, y, z 1 P1 x, y, z 2 -x, -y, -z

62. Unit Cell Unit cell volume = V V = abc 1- cos 2  - cos 2  - cos 2  + 2 cos  cos  cos  Note: text page 807 may not be correct. Triclinic P1 Z = 2

63. If Z = 2 then the total mass in the unit cell is the formula weight x 2.

64. If Z = 2 then the total mass in the unit cell is the formula weight x 2. If the volume is V then the density of the crystal is formula wt. X 2 V x N o

65. Triclinic cell: a = 6.8613 Å  = 74.746 o b = 9.1535 Å  = 81.573 o c = 16.8637 Å  = 73.339 o V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol

66. V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol Z = 2 Density = 363.87 g (2) 974.45 Å 3 x 6.02 x 10 23

67. V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol Z = 2 Density = 363.87 g (2) 974.45 Å 3 x 6.02 x 10 23 1Å = 1 x 10 -8 cm

68. V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol Z = 2 1Å = 1 x 10 -8 cm Density = 363.87 g (2) 974.45 x 10 -24 x 6.02 x 10 23 Density = 363.87 g (2) 974.45 Å 3 x 6.02 x 10 23

69. V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol Z = 2 Density = 727.74 g 5866.19 x 10 -1 Density = 363.87 g (2) 974.45 x 10 -24 x 6.02 x 10 23

70. V = 974.45 Å 3 C 17 H 24 NO 2 + Cl -. 3 H 2 O FW = 363.87 g/mol Z = 2 Density = 727.74 g 5866.19 x 10 -1 cm 3 Density = 363.87 g (2) 974.45 x 10 -24 x 6.02 x 10 23 = 1.241 g/cm 3

71. Infinitely repeating lattices

75. Three possible unit cells; one lattice.

76. Crystal lattices include a large number of repeating sets of planes.

84. These sets of planes can act as a diffraction grating for waves of the proper wavelength.

85. These sets of planes can act as a diffraction grating for waves of the proper wavelength. d

86. These sets of planes can act as a diffraction grating for waves of the proper wavelength. d = < 1 to 250 Å

87. When radiation on the order of 1 ångstöm wavelength interacts with a crystal lattice having interplanar spacings on the order of ångstöms, diffraction occurs.

88. Where do we find 1 ångstöm Wavelength radiation?

90. 1 Å

91. What is the source of 1 Å radiation?

92. 1 Å Emission spectrum for hydrogen in visible range

93. Electron transitions for H atom.

94. Electron transitions for H atom. Transitions in visible region.

95. It is possible to cause certain metals to emit X-rays by temporarily removing a core electron.

97. X-ray emission

98. e-e- +- HV

99. e-e- +- If the potential difference is large enough, core electrons will be ejected from the metal. source of electrons Metal target

100. hot filament – e - source

101. metal target

102. hot filament – e - source metal target +- Accelerating potential

104. 1 x 10 mm

105. KV = 50 mA = 40

106. 1 x 10 mm KV = 50 mA = 40 = 2000 watts

107. hot filament – e - source metal target +- Accelerating potential

109. X-ray scattering is due to the interaction of X-rays and the electron density around atoms.

110. d = < 1 to 250 Å

111. B’ E E’

112. d = < 1 to 250 Å B’ E E’ If (B B’)-(E E’) = an integral # of wavelengths, 100% reinforcement.

113. B’ E E’ If (B B’)-(E E’) = an integral # of wavelengths, 100% reinforcement. 2dsin  = n

114. B’ E E’ If (B B’)-(E E’) = an integral # of wavelengths, 100% reinforcement. 2dsin  = n Bragg’s Law

115. B’ E E’ If (B B’)-(E E’) = an integral # of wavelengths, 100% reinforcement. 2dsin  = n Bragg’s Law = wavelength n = integer (order of diffraction)

116. 2dsin  = n Bragg’s Law = wavelength n = integer (order of diffraction)

117. 2dsin  = n Bragg’s Law = wavelength n = integer (order of diffraction) If d becomes larger,  must decrease.

118. 2dsin  = n Bragg’s Law = wavelength n = integer (order of diffraction) If d becomes larger,  must decrease. There is a reciprocal relationship between The crystal lattice and the diffraction pattern.

122. 2dsin  = n d*d* n = 0 1 2 3 4

123. 2dsin  = n d*d* n = 0 1 2 3 4 is set by the X-ray target

124. 2dsin  = n d*d* n = 0 1 2 3 4 is set by the X-ray target  can be measured by determining the angle between the direct and diffracted beam.

125. 2dsin  = n d*d* n = 0 1 2 3 4 is set by the X-ray target  can be measured by determining the angle between the direct and diffracted beam. Unit cell can be determined from this data.

126. Note that intensities of the diffraction spots vary.

127. Note that intensities of the diffraction spots vary. Diffraction intensities tend to decease as  increases.

128. Note that intensities of the diffraction spots vary. The derivation of Bragg’s Law is correct but the conditions are more complicated. Each diffraction spot is the sum of the waves from all atoms in the unit cell.

129. Note that intensities of the diffraction spots vary. The derivation of Bragg’s Law is correct but the conditions are more complicated. Each diffraction spot is the sum of the waves from all atoms in the unit cell. This includes a significant amount of destructive interference.

130. d d Each diffraction maximum includes information on the electron density in the repeat distance.

131. Conversion of X-ray intensities to electron densities is a very complicated process.

132. Conversion of X-ray intensities to electron densities is a very complicated process. A major step is determining the atomic coordinates for a model.

133. Once the coordinates for a model are determined, it is possible to calculate what the intensity data for that model would look like.

134. Once the coordinates for a model are determined, it is possible to calculate what the intensity data for that model would look like. The observed intensities and the model intensities are compared.

135. A least-squares refinement of model intensities against observed intensities allows the model structure to become the actual crystal structure.