1. Chemistry 481(01) Spring 2015 Instructor: Dr. Upali Siriwardane
Office: CTH 311 Phone
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 9: :30 a.m.
April 7 , 2015: Test 1 (Chapters 1,  2, 3)
April 30, 2015: Test 2 (Chapters  5, 6 & 7)
May 19, 2015: Test 3 (Chapters. 19 & 20)
May 19, Make Up: Comprehensive covering all Chapters

2. Chapter 3. Structures of simple solids
Crystalline solids: The atoms, molecules or ions pack together in an ordered arrangement Amorphous solids: No ordered structure to the particles of the solid. No well defined faces, angles or shapes Polymeric Solids: Mostly amorphous but some have local crystiallnity. Examples would include glass and rubber.

3. The Fundamental types of Crystals
Metallic: metal cations held together by a sea of electrons Ionic: cations and anions held together by predominantly electrostatic attractions Network: atoms bonded together covalently throughout the solid (also known as covalent crystal or covalent network). Covalent or Molecular: collections of individual molecules; each lattice point in the crystal is a molecule

4. Metallic Structures Metallic Bonding in the Solid State:
Metals the atoms have low electronegativities; therefore the electrons are delocalized over all the atoms.
We can think of the structure of a metal as an arrangement of positive atom cores in a sea of electrons. For a more detailed picture see "Conductivity of Solids".
Metallic: Metal cations held together by a sea of valanece electrons

5. Packing and Geometry
Close packing ABC.ABC... cubic close-packed CCP gives face centered cubic or FCC(74.05% packed) AB.AB... or AC.AC... (these are equivalent). This is called hexagonal close-packing HCP
HCP
CCP

6. Packing and Geometry Loose packing Simple cube SC
Body-centered cubic BCC

7. The Unit Cell
The basic repeat unit that build up the whole solid

8. Unit Cell Dimensions The unit cell angles are defined as:
a, the angle formed by the b and c cell edges
b, the angle formed by the a and c cell edges g, the angle formed by the a and b cell edges
a,b,c is x,y,z in right handed cartesian coordinates
a g b a c b a

9. Bravais Lattices & Seven Crystals Systems
In the 1840’s Bravais showed that there are only fourteen different space lattices. Taking into account the geometrical properties of the basis there are 230 different repetitive patterns in which atomic elements can be arranged to form crystal structures.

10. Fourteen Bravias Unit Cells

11. Seven Crystal Systems

12. Number of Atoms in the Cubic Unit Cell
Coner- 1/8
Edge- 1/4
Body- 1
Face-1/2
FCC = 4 ( 8 coners, 6 faces)
SC = 1 (8 coners)
BCC = 2 (8 coners, 1 body)
Face-1/2
Edge - 1/4
Body- 1
Coner- 1/8

13. Close Pack Unit Cells
CCP
HCP
FCC = 4 ( 8 coners, 6 faces)

14. Unit Cells from Loose Packing
Simple cube SC Body-centered cubic BCC
BCC = 2 (8 coners, 1 body)
SC = 1 (8 coners)

15. Coordination Number
The number of nearest particles surrounding a particle in the crystal structure. Simple Cube: a particle in the crystal has a coordination number of 6 Body Centerd Cube: a particle in the crystal has a coordination number of 8 Hexagonal Close Pack &Cubic Close Pack: a particle in the crystal has a coordination number of 12

16. Holes in FCC Unit Cells
Tetrahedral Hole (8 holes) Eight holes are inside a face centered cube. Octahedral Hole (4 holes) One hole in the middle and 12 holes along the edges ( contributing 1/4) of the face centered cube

17. Holes in SC Unit Cells
Cubic Hole

18. Octahedral Hole in FCC
Octahedral Hole

19. Tetrahedral Hole in FCC

20. Structure of Metals
Crystal Lattices A crystal is a repeating array made out of metals. In describing this structure we must distinguish between the pattern of repetition (the lattice type) and what is repeated (the unit cell) described above.

21. Uranium is a good example of a metal that exhibits polymorphism.
Metals are capable of existing in more than one form at a time Polymorphism is the property or ability of a metal to exist in two or more crystalline forms depending upon temperature and composition. Most metals and metal alloys exhibit this property.
Uranium  is  a  good example of    a    metal    that exhibits polymorphism.

22. Alloys
Substitutional Second metal replaces the metal atoms in the lattice Interstitial Second metal occupies interstitial space (holes) in the lattice

23. Properties of Alloys
Alloying substances are usually metals or metalloids. The properties of an alloy differ from the properties of the pure metals or metalloids that make up the alloy and this difference is what creates the usefulness of alloys. By combining metals and metalloids, manufacturers can develop alloys that have the particular properties required for a given use.

24. Metallic Bonding Models
The difference in chemical properties between metals and non-metals lie mainly in the fact those atoms of metals fewer valence electrons and they are shared among all the atoms in the substance: metallic bonding.

25. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Metallic solids
Repeating units are made up of metal atoms, Valence electrons are free to jump from one atom to another + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

26. Electron-sea model of bonding
The metallic bond consists of a series of metals atoms that have all donated their valence electrons to an electron cloud, referred to as an electron sea which permeates the entire solid. It is like a box (solid) of marbles (positively charged metal cores: known as Kernels) that are surrounded by water (valence electrons).

27. Electron-sea model Explanation
Metallic bond together is the attraction between the positive kernels and the delocalized negative electron cloud. Fluid electrons that can carry a charge and kinetic energy flow easily through the solid making metals good electrical and thermal conductor. The kernels can be pushed anywhere within the solid and the electrons will follow them, giving metals flexibility: malleability and ductility.

28. Delocalized Metallic Bonding
Metals are held together by delocalized bonds formed from the atomic orbitals of all the atoms in the lattice. The idea that the molecular orbitals of the band of energy levels are spread or delocalized over the atoms of the piece of metal accounts for bonding in metallic solids.

29. Molecular orbital theory
Molecular Orbital Theory applied to metallic bonding is known as Band Theory. Band theory uses the LCAO of all valence atomic orbitals of metals in the solid to form bands of s, p, d, f bands (molecular orbitals) just like simple molecular orbital theory is applied to a diatomic molecule, hydrogen(H2).

30. 13. Describe metallic bonding and properties in terms of:
Electron-sea model of bonding:
Band Theory:

31. Types of conducting materials
a) Conductor (which is usually a metal) is a solid with a partially full band. b) Insulator is a solid with a full band and a large band gap. c) Semiconductor is a solid with a full band and a small band gap.

32. Linear Combination of Atomic Orbitals

33. Linear Combination of Atomic Orbitals

34. 14. Draw the s band (molecular orbitals) for ten Na on a line (one dimensional) and show bonding and anti-bonding molecular orbitals and fill electrons.

35. 15. Describe the metallic properties of sodium in terms of band theory

36. Conduction Bands in Metals

37. 16. Using a band diagram, explain how magnesium can exhibit metallic behavior even though its 3s band is completely full.

38. Types of Materials
A conductor (which is usually a metal) is a solid with a partially full band An insulator is a solid with a full band and a large band gap A semiconductor is a solid with a full band and a small band gap Element Band Gap C 5.47 eV Si 1.12 eV Ge 0.66 eV Sn 0 eV

39. Band Gaps

40. 17. Draw a Band diagram for carbon/silicon/germanium/tin, and label valence band, conduction band and band gap?

41. 18. Draw a band diagrams to show the difference between(Band gaps: C =  5.47, Si  = 1.12, Ge =  0.66,   Sn =  0)
Conductor (Sn):
Insulator (C):
Semiconductor (Ge):

42. Band Theory of Metals

43. Band Theory
Insulators – valence electrons are tightly bound to (or shared with) the individual atoms – strongest ionic (partially covalent) bonding. Semiconductors - mostly covalent bonding somewhat weaker bonding. Metals – valence electrons form an “electron gas” that are not bound to any particular ion

44. Bonding Models for Metals
Band Theory of Bonding in Solids Bonding in solids such as metals, insulators and semiconductors may be understood most effectively by an expansion of simple MO theory to assemblages of scores of atoms

45. Band Gaps

46. Doping Semiconductors

47. 19. Draw a band diagram for thermal/photo (Intrinsic) and doped (Extrinsic) semiconductors and explain the origin of semicondictivity?
d)    Thermal/photo (Intrinsic) (Ge):
e)    Doped (Extrinsic) (Si/As):

48. 20. Draw a band diagram for a p-type (Si/Ga) and n-type (Si/As) semiconductors and show holes and electrons that is responsible for semiconductivity. f) p-type(Si/Ga): g) n-type(Si/As):

49. 21. What is a transistor with emitter (E), collector(C) and base (B), and how it works?

50. 21. What is a transistor with emitter (E), collector(C) and base (B), and how it works?

51. Vacuum tubes and Transisters

52. 22. What the difference between a transistor (semiconductor device) and vacuum tube?

53. 23. Using the diagram explain how a diode works.

54. 24. What is an integrated circuit?

55. Structure of Ionic Solids
Crystal Lattices A crystal is a repeating array made out of ions. In describing this structure we must distinguish between the pattern of repetition (the lattice type) and what is repeated (the unit cell) described above.
Cations fit into the holes in the anionic lattice since anions are lager than cations.
In cases where cations are bigger than anions lattice is considered to be made up of cationic lattice with smaller anions filling the holes

56. Basic Ionic Crystal Unit Cells

57. 1) Give coordination number for both anion and cation
of the following ionic lattices.
a) CsCl Structure:
b) Rock Salt Structure:
c) Fluorite Structure:
d) Sphalrite Structure:
e) Wurtzite:
f) Rutile:

58. 1) Calculate the number of formula units in the unit cell of the following metallic and ionic compounds.
NaCl(fcc)
CsCl(bcc)
ZnS(fcc)
CaF2(bcc)

59. Radius Ratio Rules
r+/r- Coordination Holes in Which Ratio Number Positive Ions Pack tetrahedral holes FCC octahedral holes FCC cubic holes BCC

60. Cesium Chloride Structure (CsCl)

61. Reproduced with permission from Soli-State Resources.
Rock Salt (NaCl)
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Soli-State Resources.

62. Sodium Chloride Lattice (NaCl)

63. NaCl Lattice Calculations

64. CaF2

65. Reproduced with permission from Solid-State Resources.
Calcium Fluoride
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.

66. Zinc Blende Structure (ZnS)

67. Reproduced with permission from Solid-State Resources.
Lead Sulfide
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.

68. Wurtzite Structure (ZnS)

69. Summary of Unit Cells Volume of sphere in SC = 4/3p(½)3 = 0.52
Volume of a sphere = 4/3pr3
Volume of sphere in SC = 4/3p(½) = 0.52
Volume of sphere in BCC = 4/3p((3)½/4)3 = 0.34
Volume of sphere in FCC = 4/3p( 1/(2(2)½))3 = 0.185

70. Density Calculations
Aluminum has a ccp (fcc) arrangement of atoms. The radius of Al = 1.423Å ( = 143.2pm). Calculate the lattice parameter of the unit cell and the density of solid Al (atomic weight = 26.98).
Solution:
4 atoms/cell [8 at corners (each 1/8), 6 in faces (each 1/2)]
Lattice parameter: a/r(Al) = 2(2)1/2
a = 2(2)1/2 (1.432Å) = 4.050Å= x 10-8 cm
Density = g/cm3

71. Miller Indices
Miller indices are used to specify directions and planes
• These directions and planes could be in lattices or in
crystals
• The number of indices will match with the dimension of the
Lattice or the crystal
• (h, k, l) represents a point on a plane
• To obtain h, k, l of a plane Identify the intercepts on the a- , b- and c- axes of the unit cell.

72. Miller Indices
Eg. intercept on the x-axis is at a, b and c ( at the point (a,0,0) ), but the surface is parallel to the y- and z-axes - strictly therefore there is no intercept on these two axes but we shall consider the intercept to be at infinity ( ∞ ) for the special case where the plane is parallel to an axis.
The intercepts on the a- , b- and c-axes are thus
Intercepts :    1 , ∞ , ∞
Take the reciprocals of the fractional intercepts: 1/1 , 1/ ∞, 1/ ∞
• (h, k, l) for this plane becomes 1,0,0

73. Reproduced with permission from Soli-State Resources.
Rock Salt (NaCl)
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Soli-State Resources.

74. Sodium Chloride Lattice (NaCl)
0,0,1
0,0,2
2,2,2
1,1,1

75. CaF2
0,0,1
0,0,4
0,0,2
0,0,4
0,0,2
0,0,2
0,0,4

76. Reproduced with permission from Solid-State Resources.
Calcium Fluoride
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.

77. Zinc Blende Structure (ZnS)
0,0,1
0,0,4
0,0,2
0,0,4

78. Reproduced with permission from Solid-State Resources.
Lead Sulfide
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.

79. Wurtzite Structure (ZnS)

80. Antifluorite Structure

81. ρ = r+ Radius ratio rule Radius ratio rule states As
the size (ionic radius, r+) of a cation increases,
more anions of a
particular size can pack around it.
Thus, knowing the size of the ions, we should be able to predict
a priori
which type of crystal packing
will be observed.
We can account for the relative size of both ions by using the RATIO of
the ionic radii:
ρ = r+ r−

82. Radius Ratio Rules
r+/r- Coordination Holes in Which Ratio Number Positive Ions Pack tetrahedral holes FCC octahedral holes FCC cubic holes BCC

83. Radius Ratio Appplications
Suggest the probable crystal structure of (a) barium fluoride; (b) potassium bromide; (c) magnesium sulfide. You can use tables to obtain ionic radii.
a) barium fluoride; Ba2+= 142 pm F- = 131 pm
b) potassium bromide; K+= 138 pm Br- = 196 pm
c) magnesium sulfide; Mg2+= 103 pm S2- = 184 pm
Radius ratio(barium fluoride): 142/131 =1.08
Radius ratio(potassium bromide): 138/196=0.704
Radius ratio(magnesium sulfide): 103/184= 0.559

84. Radius Ratio Appplications
Radius ratio(barium fluoride): 142/131 =1.08
Radius ratio(potassium bromide): 138/196=0.704
Radius ratio(magnesium sulfide): 103/184= 0.559
Barium fluoride: 142/131 =1.08 ( ) CN 8 FCC fluorite
Potassium bromide: 138/196=0.704 ( ) CN 6 FCC K+ in octahedral holes
Magnesium sulfide: 103/184= ( ) CN 6 FCC
r+/r Coordination Holes in Which
Ratio Number Positive Ions Pack
tetrahedral holes FCC
octahedral holes FCC
cubic holes BCC

85. Radius Ratio Applications
Barium fluoride: 142/131 =1.08 ( ) CN 8 FCC
Potassium bromide: 138/196=0.704 ( ) CN 6 FCC K+ in octahedral holes
Magnesium sulfide: 103/184= ( ) CN 6 FCC

86. Unit Cells dimensions and radius
a = 2r or r = a/2

87. Summary of Unit Cells Volume of sphere in SC = 4/3p(½)3 = 0.52
Volume of a sphere = 4/3pr3
Volume of sphere in SC = 4/3p(½) = 0.52
Volume of sphere in BCC = 4/3p((3)½/4)3 = 0.34
Volume of sphere in FCC = 4/3p( 1/(2(2)½))3 = 0.185

88. Density Calculations
Aluminum has a ccp (fcc) arrangement of atoms. The radius of Al = 1.423Å ( = 143.2pm). Calculate the lattice parameter of the unit cell and the density of solid Al (atomic weight = 26.98).
Solution:
4 atoms/cell [8 at corners (each 1/8), 6 in faces (each 1/2)]
Lattice parameter: a/r(Al) = 2(2)1/2
a = 2(2)1/2 (1.432Å) = 4.050Å= x 10-8 cm
Density = g/cm3

89. 3. What is Coulombs law how it applies to ionic bond?

90. Coulomb’s Law
k = constant q+ = cation charge q- = anion charge r = distance between two ions

91. Coulomb’s Model where e = charge on an electron = 1.602 x 10-19 C
e0 = permittivity of vacuum = x C2J-1m-1
ZA = charge on ion A
ZB = charge on ion B
d = separation of ion centers

92. Ions with charges Q1 and Q2: The potential energy is given by:
Ionic Bonds
An ionic bond is simply the electrostatic attraction between opposite charges.
Ions with charges Q1 and Q2: d d Q E 2 1
The potential energy is given by:

93. Estimating Lattice Energy
Arrange with increasing lattice energy:
KCl
NaF
MgO
KBr
NaCl
701 kJ d Q E 2 1
910 kJ K+
Cl
3795 kJ
671 kJ d
788 kJ K+
Br d

94. Lattice Energy
The Lattice energy, U, is the amount of energy required to separate a mole of the solid (s) into a gaseous atoms (g) of its ions. Lattice Enthalpy:

95. 4. What is lattice energy? Take NaCl as an example.

96. Lattice energy Lattice energy Compound kJ/mol LiCl 834
NaCl 769
KCl 701
NaBr 732
Na2O
Na2S
MgCl
MgO
The higher the lattice energy, the stronger the attraction between ions.

97. Lattice Energy

98. 5. Place the following compounds in order of increasing lattice energy: a) magnesium oxide b) lithium fluoride c) sodium chloride. Give the reasoning for this order.

99. Properties of Ionic Compounds
Crystals of Ionic Compounds are hard and brittle Have high melting points When heated to molten state they conduct electricity When dissolved in water conducts electricity

100. Trends in Melting Points
Compound Lattice Energy (Enthalpy) (kcal/mol) NaF -201 NaCl -182 NaBr -173 NaI -159

101. Trends in Melting Points
Compound Lattice Energy (kcal/mol) NaF -201 NaCl -182 NaBr -173 NaI -159

102. Trends in Properties LiCl 0.68 1.81 605 834 NaCl 0.98 1.81 801 769
Compound q+ radius q- radius M.P (oC) L.E. (kJ/mol)
LiCl
NaCl
KCl
LiF
NaF KF
MgCl
CaCl
MgO
CaO

103. 6. Explain the lattice energy and melting point trends:
Compound
Cation radius (Angstroms)
Anion radius (Angstroms)
Melting Point (Centigrade)
Lattice Energy (kcal/mol)
MgCl2
0.65
1.81
714
2326
CaCl2
0.94
782
2223
MgO
1.45
2852
3938
CaO
2614
3414

104. Madelung Constant
Madelung constant is geometric factor that depends on the lattice structure.

105. Madelung Constant Calculation

106. 7. Why is Madalung constant for NaCl is significantly different from CaF2 value and why is it different for different ionic lattice types?
Ionic Solid
Madelung Constant
Coor. #
A : C
Lattice Type
NaCl
6 : 6
Rock salt
CaF2
8 : 4
Fluorite

107. 8. Calculate the first two terms of the series for the Madelung constant for the cesium chloride lattice. How does this compare with the limiting value?

108. Degree of Covalent Character
Fajan's Rules (Polarization)Polarization will be increased by:
1. High charge and small size of the cation
2. High charge and large size of the anion
3. An incomplete valence shell electron configuration

109. Trends in Melting Points Silver Halides
Compound M.P. oC AgF 435 AgCl 455 AgBr 430 AgI 553

110. 9. In calculating lattice energy, why should Coulombs law equation is multiplied by the Avogadro’s number, N and Madelung constant, A?
N z2e N A z2e2
Lattice Energy = x N x A =
4peor peor

111. Born-Lande Model:
This modes include repulsions due to overlap of electron electron clouds of ions. eo = permitivity of free space A = Madelung Constant ro = sum of the ionic radii n = average Born exponent depend on the electron configuration n = Born exponent, typically a number between 5 and 12, determined experimentally by measuring the compressibility of the solid

112. 10. In the correction to lattice energy what is factor n in Born-Lande equation accounted for and how it relate to electronic configuration?

113. 11. Using the Born-Lande equation, calculate the lattice energy of cesium chloride. N A z2e2 1 Lattice Energy = ( ) 4peor n

114. 12. What are Born-Mayer and Kapustinskii equations
12. What are Born-Mayer and Kapustinskii equations? How are they different from the Born-Lande equation?

115. Born_Haber Cycle
Energy Considerations in Ionic Structures

116. Born-Haber Cycle?
Relates lattice energy ( L.E) to: Sublimation (vaporization) energy (S.E) Ionization energy metal (I.E) Bond Dissociation of nonmetal (B.E) DHf formation of NaCl(s) L.E. = E.A.+ 1/2 B.E. + I.E. + S.E. - DHf

117. 13. What is a Born-Haber cycle?

118. Ionic bond formation

119. Energy and ionic bond formation
Example - formation of sodium chloride.
Steps DHo, kJ
Vaporization of Na(s) Na(g)
sodium
Decomposition of 1/2 Cl2 (g) Cl(g)
chlorine molecules
Ionization of sodium Na(g) Na+(g)
Addition of electron Cl(g) + e Cl-(g)
to chlorine
( electron affinity)
Formation of NaCl Na+(g)+Cl-(g) NaCl

120. Energy and ionic bond formation
Na(s) + 1/2 Cl2(g)
Na(g) + 1/2 Cl2(g)
Na(g) + Cl(g)
Na+(s) + Cl(g)
Na+(s) + Cl-(g)
NaCl(s)
+496 kJ(I.E.)
+121 kJ(1/2 B.D.E.)
+92 kJ(S.E.)
-349 kJ (E.A.)
-771 kJ (L.E.)
-411 kJ(DHf)

121. Calculation of DHf from lattice Energy

122. 14. Calculate the Lattice energy of NaCl from following thermodynamic data:
Steps
DHo, kJ 1.
Vaporization of sodium: Na(s)  Na(g)
+92 2.
Decomposition of Cl2: 1/2 Cl2 (g)  Cl(g)
+121 3.
Ionization of sodium: Na(g)  Na+(g)
+496 4.
Electron affinity to chlorine: Cl(g) + e-  Cl-(g)
-349 5.
Formation of NaCl(s): Na(g)+1/2Cl2 (g)  NaCl(s)
-411

123. 15. Construct a Born-Haber cycle for the formation of aluminum fluoride. Do not perform any calculation.

124. Hydration of Cations

125. Solubility: Lattice Energy and Hydration Energy
Solubility depends on the difference between lattice energy and hydration energy holds ions and water. For dissolution to occur the lattice energy must be overcome by hydration energy.

126. Solubility: Lattice Energy and Hydration Energy
For strong electrolytes lattice energy increases with increase in ionic charge and decrease in ionic size H hydration energies are greatest for small, highly charged ions Difficult to predict solubility from size and charge of ions. we use solubility rules.

127. Thermodynamics of the Solution Process of Ionic Compounds
Heat of solution, DHsolution : Enthalpy of hydration, DHhyd, Lattice Energy, Ulatt

128. Solution Process of Ionic Compounds

129. 16. Define following terms:
Enthalpy of solution, DHsolution:
Enthalpy of hydration, DHhydration:
Solvent-solvent intermolecular attractions,
DH°solvent-solvent:

130. 17 . How is Enthalpy of solution, DHsolution, Enthalpy of hydration: DHhydration, and Lattice energy:Ulatt are related?

131. Enthalpy from dipole – dipole Interactions
The last term, DH L-L, indicates the loss of enthalpy from dipole - dipole interactions between solvent molecules (L) when they become solvating ligands (L') for the ions.

132. Hydration Process

133. Different types of Interactions for Dissolution

134. Hydration Energy of Ions

135. Hydration Process

136. Calculation of DHsolution

137. Heat of Solution and Solubility

138. 18.Predict the solubility of following ionic compounds: a) LiF:
b) LiI:
c) CsI:
d) MgF2:
Lattice Energy(U)
DHhyd, M+
DHhyd, M-
LiF
1030
-950
-60
LiI
720
-80
CsI
585
-700
MgF2
3100
-2800
-120

139. 19. Give rational explanation to the solubility rules in terms of ion sizes, lattice energy(U), DHhyd, and DHsolution.
a) All compounds containing alkali metal cations and the ammonium ion are soluble.   
b) All compounds containing NO3-, ClO4-, ClO3-, and C2H3O2- anions are soluble.

140. 20. Calculate the enthalpy of formation of calcium oxide using a Born-Haber cycle. Obtain all necessary information from the data tables in the Appendices. Compare the value that you obtain with the actual entropy measured value of DHf(CaO(s)).

141. 21. Although the hydration energy of the calcium ion, Ca2+, is much greater than that of the potassium ion, K+, the molar solubility of calcium chloride is much less than that of potassium chloride. Suggest an explanation.