1. Disorder in crystals

2. All lattice points are not always the same.

3. Apatite Ca 3 (PO 4 ) 2

4. Ca 2+

5. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å

6. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å Sr 2+ 1.12Å

7. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å Sr 2+ 1.12Å

8. Group II Be Mg Ca Sr Ba Ra

9. Group II Be Mg Ca Sr Ba Ra 2+ in ionic compounds

10. Group II Be Mg Ca Sr Ba Ra 2+ in ionic compounds 88 Sr – 86% of naturally occuring 38

11. Group II Be Mg Ca Sr Ba Ra 2+ in ionic compounds 88 Sr – 86% of naturally occuring 38 90 Sr – radioactive isotope product of nuclear weapons testing 38

12. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å Sr 2+ 1.12Å If Sr 2+ replaces Ca 2+ consistently, the structure changes.

13. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å Sr 2+ 1.12Å If Sr 2+ replaces Ca 2+ consistently, the structure changes. This is not disorder.

14. Apatite Ca 3 (PO 4 ) 2 Ca 2+ 0.98Å Sr 2+ 1.12Å If Sr 2+ replaces Some Ca 2+ randomly, the structure is disordered.

16. If a crystal contains 90% Ca and 10% Sr, each M 2+ site will appear to be Ca/Sr 90/10% based on diffraction data.

21. Defects in Crystals

22. Disorder implies that all positions are occupied, but the occupation of some sites may not be consistent.

23. Defects in Crystals A defect is a break in the infinite lattice.

24. Defects in Crystals A defect is a break in the infinite lattice. Some sites that would normally be occupied in a perfect lattice, are open.

25. Color center defect - h + Cl - Cl + e -

26. Color center defect - h + Cl - Cl + e - Cl 0.99 Å Cl - 1.81 Å

27. The uncharged Cl is not affected by the + charges and is considerably smaller than the Cl -.

28. The uncharged Cl is not affected by the + charges and is considerably smaller than the Cl -. The Cl can move through, and leave, the lattice.

29. The uncharged Cl is not affected by the + charges and is considerably smaller than the Cl -. The Cl can move through, and leave, the lattice. The electron can be trapped in the octahedral vacancy left by the Cl -.

30. Anion missing; replaced by e -.

31. The overall lattice is not disturbed.

32. Anion missing; replaced by e -. This does not have to be the same site vacated By the Cl -.

33. Color center defect Anion missing; replaced by e -.

34. Color center defect The presence of e - in a void leads to an electronic transition in the visible range.

35. In a real (as opposed to a ‘perfect’) Crystal, a small portion of the sites will be unoccupied.

36. In a real (as opposed to a ‘perfect’) Crystal, a small portion of the sites will be unoccupied. This is called a Shottky defect.

37. + - Perfect

38. + - Real

39. + - Perfect Real In ionic crystals, charges still must balance.

40. Shottky Defect

41. Shottky Defect: a void that does not disturb the structure.

42. Shottky Defect in metal.

43. Other defects may alter the lattice.

44. + - Interstitial site:

45. + - Interstitial site: position between ions or atoms which can be occupied by another ion or atom.

46. + - Interstitial site: position between ions or atoms which can be occupied by another ion or atom.

47. + - Move ion from normal site to interstitial site.

48. Frenkel defect: lattice is distorted when an ion is moved to an interstitial site.

49. Defects tend to be dynamic.

51. Nonstoichiometric Compounds

53. Wüstite

54. FeO

55. Wüstite FeO = O = Fe

56. Wüstite FeO = O = Fe +2 -2

57. Wüstite FeO = O = Fe +2 -2 Fe 0.85-0.95 O If there is less than 1 Fe per O, Fe must be in more than 1 ox. State.

58. Wüstite FeO = O = Fe 2+, Fe 3+ +2 -2 Fe 0.85-0.95 O

59. Fe 0.85 O

60. Fe 0.85-0.95 O Fe 0.85 O Fe 2+ x ; Fe 3+ 0.85-x

61. Fe 0.85-0.95 O Fe 0.85 O 2x + 3(0.85-x) = 2 Fe 2+ x ; Fe 3+ 0.85-x

62. Fe 0.85-0.95 O Fe 0.85 O 2x + 3(0.85-x) = 2 Fe 2+ x ; Fe 3+ 0.85-x 2x + 2.55 –3x = 2

63. Fe 0.85-0.95 O Fe 0.85 O 2x + 3(0.85-x) = 2 Fe 2+ x ; Fe 3+ 0.85-x 2x + 2.55 –3x = 2 -x = -0.55

64. Fe 0.85-0.95 O Fe 0.85 O 2x + 3(0.85-x) = 2 Fe 2+ x ; Fe 3+ 0.85-x 2x + 2.55 –3x = 2 x = 0.55

65. Fe 0.85-0.95 O Fe 0.85 O 2x + 3(0.85-x) = 2 Fe 2+ x ; Fe 3+ 0.85-x 2x + 2.55 –3x = 2 x = 0.55 (Fe 2+ 0.55, Fe 3+ 0.30 )O

66. Fe 0.85 O

67. (Fe 2+ 0.85, Fe 3+ 0.10 )O Fe 0.95 O

68. Thermodynamics of Crystals

71. Na + Cl - ionic bond

72. Na +

75. Account for ionic attractions and repulsions based on the distance of the ions and their charges.

76. - + r

77. - + r The energy of this pair depends on coulombic attraction and repulsion. E p = - z 1 z 2 e 2 b + r rnrn

78. - + r The energy of this pair depends on coulombic attraction and repulsion. E p = - z 1 z 2 e 2 b + r rnrn attraction term (decreases energy)

79. - + r The energy of this pair depends on coulombic attraction and repulsion. E p = - z 1 z 2 e 2 b + r rnrn attraction term (decreases energy) Repulsive term (increases energy)

80. - + r E p = - z 1 z 2 e 2 b + r rnrn z = charge number

81. - + r E p = - z 1 z 2 e 2 b + r rnrn z = charge number e = electron charge

82. - + r E p = - z 1 z 2 e 2 b + r rnrn z = charge number e = electron charge r = internuclear separation

83. - + r E p = - z 1 z 2 e 2 b + r rnrn z = charge number e = electron charge r = internuclear separation b, n are repulsion constants

84. - + r E p = - z 1 z 2 e 2 b + r rnrn z = charge number e = electron charge r = internuclear separation b, n are repulsion * constants * repulsion due to physical contact, not coulombic repulsion

85. E p = - z 1 z 2 e 2 b + r rnrn The lattice energy for a mole of NaCl can be evaluated by multiplying the energy by N o and including a factor that accounts for all ion-ion interactions.

86. E p = - U = N o Az 1 z 2 e 2 B + r rnrn z 1 z 2 e 2 b + r rnrn -

87. E p = - U = N o Az 1 z 2 e 2 B + r rnrn z 1 z 2 e 2 b + r rnrn Lattice energy -

88. E p = - U = N o Az 1 z 2 e 2 B + r rnrn z 1 z 2 e 2 b + r rnrn Lattice energy Avagadro’s number -

89. E p = - U = N o Az 1 z 2 e 2 B + r rnrn z 1 z 2 e 2 b + r rnrn Lattice energy Avagadro’s number Madelung constant -

90. Repeat S&P pg 80

92. When an individual ion is considered in a cubic lattice, there is a group of oppositely charged ions at a given distance followed by a group of like charged ions at a longer distance.

93. If r = a in NaCl then there are 6 Cl - at distance a from Na +.

94. If r = a in NaCl then there are 6 Cl - at distance a from Na +. There are 12 Na + at a distance of 2 a from the initial Na +.

95. a

96. a 2 a

97. Madelung constant for NaCl Potential energy for nearest neighbors = -6e 2 a Potential energy for next-nearest = 12e 2 2 a

98. Madelung constant for NaCl Potential energy for nearest neighbors = -6e 2 a Potential energy for next-nearest = 12e 2 2 a e2e2 a - 6 12 8 + -+......... 13 2

99. Madelung constant for NaCl Potential energy for nearest neighbors = -6e 2 a Potential energy for next-nearest = 12e 2 2 a e2e2 a - 6 12 8 + -+......... 13 2  1.75

100. U = N o Az 1 z 2 e 2 B + a anan B = Az1z2e2Az1z2e2 n a n-1 -

101. B = U = N o Az1z2e2Az1z2e2 + a anan - Az 1 z 2 e 2 B + a anan - Az1z2e2Az1z2e2 n a n-1 Az1z2e2Az1z2e2 n

102. B = U = N o Az 1 z 2 e 2 B + a anan - Az1z2e2Az1z2e2 n a n-1 U = -N o Az1z2e2Az1z2e2 a U = N o Az1z2e2Az1z2e2 + a anan - Az1z2e2Az1z2e2 n a n-1 1- 1 n

103. U = -N o Az1z2e2Az1z2e2 a 1- 1 n n varies from 9 to 12; it is determined from the compressibility of the material

104. U = -N o Az1z2e2Az1z2e2 a 1- 1 n U calc U exp kJ/mol NaCl 770 770 KF 808 803 NaH 845 812

105. Where do experimental values for U come from?

106. EE

107. EE

108. K (g) K + (g) + e - 419 kJ/mol

109. First ionization energy

110. K (g) K + (g) + e - 419 kJ/mol First ionization energy Cl (g) + e - Cl - (g) 349 kJ/mol

111. K (g) K + (g) + e - 419 kJ/mol First ionization energy Cl (g) + e - Cl - (g) 349 kJ/mol Electron affinity

112. K (s) K (g)  H sublimation

113. ½ Cl 2(g) Cl (g)  H dissociation

114. K (s) K (g)  H sublimation ½ Cl 2(g) Cl (g)  H dissociation K (g) K + (g) + e - 419 kJ/mol Cl (g) + e - Cl - (g) 349 kJ/mol

115. K (s) K (g)  H sublimation ½ Cl 2(g) Cl (g)  H dissociation K (g) K + (g) + e - 419 kJ/mol Cl (g) + e - Cl - (g) 349 kJ/mol K + (g) + Cl - (g) KCl (s) U

116. K + (g) + Cl - (g) KCl (s) -U-U K (g) + Cl (g) K (s) + ½ Cl 2(g)  H sub + ½ D I -A-A -e-e+e+e -  H f

117. K + (g) + Cl - (g) KCl (s) -U-U K (g) + Cl (g) K (s) + ½ Cl 2(g)  H sub + ½ D I -A-A -e-e+e+e -  H f Born-Haber Cycle

118. K + (g) + Cl - (g) KCl (s) -U-U K (g) + Cl (g) K (s) + ½ Cl 2(g)  H sub + ½ D I -A-A -e-e+e+e -  H f Born-Haber Cycle Only term not from experiment

119. K + (g) + Cl - (g) KCl (s) -U-U K (g) + Cl (g) K (s) + ½ Cl 2(g)  H sub + ½ D I -A-A -e-e+e+e -  H f Born-Haber Cycle Only term not from experiment U = -  H f +  H sub + ½ D + I - A

120. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A NaCl -414 109 113 490 347 kJ/mol

121. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A NaCl 779 -414 109 113 490 347 kJ/mol

122. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A NaCl 779 -414 109 113 490 347 NaBr -377 109 96 490 318 kJ/mol

123. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A NaCl 779 -414 109 113 490 347 NaBr 754 -377 109 96 490 318 NaI -322 109 71 490 297 kJ/mol

124. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A NaCl 779 -414 109 113 490 347 NaBr 754 -377 109 96 490 318 NaI 695 -322 109 71 490 297 kJ/mol

125. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A U = -N o Az1z2e2Az1z2e2 a 1- 1 n

126. Born-Haber Cycle U = -  H f +  H sub + ½ D + I - A U = -N o Az1z2e2Az1z2e2 a 1- 1 n Thermo(B-H) Theory NaCl 779 795 NaBr 754 757 NaI 695 715

127. 1. Construct a diagram for the Born-Haber cycle for the various thermodynamic properties associated with the formation of magnesium chloride. Homework problems for 10/3 continued

128. The important values are:  H sub Mg 147.7 kJ/mol IE 1 Mg 737.7 kJ/mol IE 2 Mg 1450.7 kJ/mol D Cl 2 243 kJ/mol A Cl 348.6 kJ/mol  H f MgCl 2 -642 kJ/mol continued

129. CsCl 2. In the CsCl structure, how many ions would be included in the first attractive term for the Madelung constant. continued

130. CsCl 2. In the CsCl structure, how many ions would be included in the first repulsive term for the Madelung constant.