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Transcript of Chapter-3-1 Chemistry 481, Spring 2015, LA Tech Instructor: Dr. Upali Siriwardane e-mail:...
Chapter-3-1Chemistry 481, Spring 2015, LA Tech
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311 Phone 257-4941
Office Hours:
M,W 8:00-9:00 & 11:00-12:00 am;
Tu,Th, F 9:30 - 11:30 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
Chemistry 481(01) Spring 2015
Chapter-3-2Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-3Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-4Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-5Chemistry 481, Spring 2015, LA Tech
Packing and GeometryClose 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
CCPHCP
Chapter-3-6Chemistry 481, Spring 2015, LA Tech
Loose packing
Simple cube SC
Body-centered cubic BCC
Packing and Geometry
Chapter-3-7Chemistry 481, Spring 2015, LA Tech
The Unit CellThe basic repeat unit that build up the whole solid
Chapter-3-8Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-9Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-10Chemistry 481, Spring 2015, LA Tech
Fourteen Bravias Unit Cells
Chapter-3-11Chemistry 481, Spring 2015, LA Tech
Seven Crystal Systems
Chapter-3-12Chemistry 481, Spring 2015, LA Tech
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
Coner- 1/8Edge - 1/4Body- 1
Chapter-3-13Chemistry 481, Spring 2015, LA Tech
Close Pack Unit Cells
CCP HCP
FCC = 4 ( 8 coners, 6 faces)
Chapter-3-14Chemistry 481, Spring 2015, LA Tech
Simple cube SC Body-centered cubic BCC
Unit Cells from Loose Packing
SC = 1 (8 coners) BCC = 2 (8 coners, 1 body)
Chapter-3-15Chemistry 481, Spring 2015, LA Tech
Coordination NumberThe 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
Chapter-3-16Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-17Chemistry 481, Spring 2015, LA Tech
Holes in SC Unit Cells
Cubic Hole
Chapter-3-18Chemistry 481, Spring 2015, LA Tech
Octahedral Hole in FCC
Octahedral Hole
Chapter-3-19Chemistry 481, Spring 2015, LA Tech
Tetrahedral Hole in FCC
Tetrahedral Hole
Chapter-3-20Chemistry 481, Spring 2015, LA Tech
Structure of MetalsCrystal 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.
Chapter-3-21Chemistry 481, Spring 2015, LA Tech
PolymorphismMetals 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.
Chapter-3-22Chemistry 481, Spring 2015, LA Tech
AlloysSubstitutional
Second metal replaces the metal atoms in the lattice
Interstitial
Second metal occupies interstitial space (holes) in the lattice
Chapter-3-23Chemistry 481, Spring 2015, LA Tech
Properties of AlloysAlloying 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.
Chapter-3-24Chemistry 481, Spring 2015, LA Tech
Metallic Bonding ModelsThe 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.
Chapter-3-25Chemistry 481, Spring 2015, LA Tech
Metallic solidsRepeating units are made up of metal atoms,
Valence electrons are free to jump from one atom to another
++ + +
++ + +
++ + +
++ + +
++
++
++ +
+
++ ++
++ + +
++ + +
Chapter-3-26Chemistry 481, Spring 2015, LA Tech
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).
Chapter-3-27Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-28Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-29Chemistry 481, Spring 2015, LA Tech
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).
Chapter-3-30Chemistry 481, Spring 2015, LA Tech
13. Describe metallic bonding and properties in terms of:
Electron-sea model of bonding:
Band Theory:
Chapter-3-31Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-32Chemistry 481, Spring 2015, LA Tech
Linear Combination of Atomic Orbitals
Chapter-3-33Chemistry 481, Spring 2015, LA Tech
Linear Combination of Atomic Orbitals
Chapter-3-34Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-35Chemistry 481, Spring 2015, LA Tech
15. Describe the metallic properties of sodium in terms of band theory
Chapter-3-36Chemistry 481, Spring 2015, LA Tech
Conduction Bands in Metals
Chapter-3-37Chemistry 481, Spring 2015, LA Tech
16. Using a band diagram, explain how magnesium can exhibit metallic behavior even though its 3s band is completely full.
Chapter-3-38Chemistry 481, Spring 2015, LA Tech
Types of MaterialsA 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 eVSi 1.12 eVGe 0.66 eVSn 0 eV
Chapter-3-39Chemistry 481, Spring 2015, LA Tech
Band Gaps
Chapter-3-40Chemistry 481, Spring 2015, LA Tech
17. Draw a Band diagram for carbon/silicon/germanium/tin, and label valence band, conduction band and band gap?
Chapter-3-41Chemistry 481, Spring 2015, LA Tech
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):
Chapter-3-42Chemistry 481, Spring 2015, LA Tech
Band Theory of Metals
Chapter-3-43Chemistry 481, Spring 2015, LA Tech
Band TheoryInsulators – 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
Chapter-3-44Chemistry 481, Spring 2015, LA Tech
Bonding Models for MetalsBand 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
Chapter-3-45Chemistry 481, Spring 2015, LA Tech
Band Gaps
Chapter-3-46Chemistry 481, Spring 2015, LA Tech
Doping Semiconductors
Chapter-3-47Chemistry 481, Spring 2015, LA Tech
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):
Chapter-3-48Chemistry 481, Spring 2015, LA Tech
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):
Chapter-3-49Chemistry 481, Spring 2015, LA Tech
21. What is a transistor with emitter (E), collector(C) and base (B), and how it works?
Chapter-3-50Chemistry 481, Spring 2015, LA Tech
21. What is a transistor with emitter (E), collector(C) and base (B), and how it works?
Chapter-3-51Chemistry 481, Spring 2015, LA Tech
Vacuum tubes and Transisters
Chapter-3-52Chemistry 481, Spring 2015, LA Tech
22. What the difference between a transistor (semiconductor device) and vacuum tube?
Chapter-3-53Chemistry 481, Spring 2015, LA Tech
• 23. Using the diagram explain how a diode works.
Chapter-3-54Chemistry 481, Spring 2015, LA Tech
24. What is an integrated circuit?
Chapter-3-55Chemistry 481, Spring 2015, LA Tech
Structure of Ionic SolidsCrystal 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
Chapter-3-56Chemistry 481, Spring 2015, LA Tech
Basic Ionic Crystal Unit Cells
Chapter-3-57Chemistry 481, Spring 2015, LA Tech
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:
Chapter-3-58Chemistry 481, Spring 2015, LA Tech
1) Calculate the number of formula units in the unit cell of the following metallic and ionic compounds.
a) NaCl(fcc)
b) CsCl(bcc)
c) ZnS(fcc)
d) CaF2(bcc)
Chapter-3-59Chemistry 481, Spring 2015, LA Tech
Radius Ratio Rules
r+/r- Coordination Holes in Which
Ratio Number Positive Ions Pack
0.225 - 0.414 4 tetrahedral holes FCC
0.414 - 0.732 6 octahedral holes FCC
0.732 - 1 8 cubic holes BCC
Chapter-3-60Chemistry 481, Spring 2015, LA Tech
Cesium Chloride Structure (CsCl)
Chapter-3-61Chemistry 481, Spring 2015, LA Tech
Rock Salt (NaCl)
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Soli-State Resources.
Chapter-3-62Chemistry 481, Spring 2015, LA Tech
Sodium Chloride Lattice (NaCl)
Chapter-3-63Chemistry 481, Spring 2015, LA Tech
NaCl Lattice Calculations
Chapter-3-64Chemistry 481, Spring 2015, LA Tech
CaF2
Chapter-3-65Chemistry 481, Spring 2015, LA Tech
Calcium Fluoride
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.
Chapter-3-66Chemistry 481, Spring 2015, LA Tech
Zinc Blende Structure (ZnS)
Chapter-3-67Chemistry 481, Spring 2015, LA Tech
Lead Sulfide
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.
Chapter-3-68Chemistry 481, Spring 2015, LA Tech
Wurtzite Structure (ZnS)
Chapter-3-69Chemistry 481, Spring 2015, LA Tech
Summary of Unit Cells
Volume of a sphere = 4/3pr3
Volume of sphere in SC = 4/3p(½)
3 = 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
Chapter-3-70Chemistry 481, Spring 2015, LA Tech
Density CalculationsAluminum 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= 4.050 x 10-8 cm
Density = 2.698 g/cm3
Chapter-3-71Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-72Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-73Chemistry 481, Spring 2015, LA Tech
Rock Salt (NaCl)
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Soli-State Resources.
Chapter-3-74Chemistry 481, Spring 2015, LA Tech
Sodium Chloride Lattice (NaCl)
0,0,1 0,0,2
1,1,12,2,2
Chapter-3-75Chemistry 481, Spring 2015, LA Tech
CaF2
0,0,1 0,0,4 0,0,2 0,0,4
0,0,20,0,2 0,0,4
Chapter-3-76Chemistry 481, Spring 2015, LA Tech
Calcium Fluoride
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.
Chapter-3-77Chemistry 481, Spring 2015, LA Tech
Zinc Blende Structure (ZnS)
0,0,1 0,0,4 0,0,40,0,2
Chapter-3-78Chemistry 481, Spring 2015, LA Tech
Lead Sulfide
© 1995 by the Division of Chemical Education, Inc., American Chemical Society.
Reproduced with permission from Solid-State Resources.
Chapter-3-79Chemistry 481, Spring 2015, LA Tech
Wurtzite Structure (ZnS)
Chapter-3-80Chemistry 481, Spring 2015, LA Tech
Antifluorite Structure
Chapter-3-81Chemistry 481, Spring 2015, LA Tech
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−
Radius ratio rule
Chapter-3-82Chemistry 481, Spring 2015, LA Tech
Radius Ratio Rules
r+/r- Coordination Holes in Which
Ratio Number Positive Ions Pack
0.225 - 0.414 4 tetrahedral holes FCC
0.414 - 0.732 6 octahedral holes FCC
0.732 - 1 8 cubic holes BCC
Chapter-3-83Chemistry 481, Spring 2015, LA Tech
Radius Ratio AppplicationsSuggest 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
a) Radius ratio(barium fluoride): 142/131 =1.08
b) Radius ratio(potassium bromide): 138/196=0.704
c) Radius ratio(magnesium sulfide): 103/184= 0.559
Chapter-3-84Chemistry 481, Spring 2015, LA Tech
Radius Ratio Appplicationsa) Radius ratio(barium fluoride): 142/131 =1.08
b) Radius ratio(potassium bromide): 138/196=0.704
c) Radius ratio(magnesium sulfide): 103/184= 0.559
• Barium fluoride: 142/131 =1.08 (0.732-1) CN 8 FCC fluorite• Potassium bromide: 138/196=0.704 (0.414-0.732) CN 6 FCC K+ in
octahedral holes• Magnesium sulfide: 103/184= 0.559 (0.414-0.732) CN 6 FCC
r+/r- Coordination Holes in Which
Ratio Number Positive Ions Pack
0.225 - 0.414 4 tetrahedral holes FCC
0.414 - 0.732 6 octahedral holes FCC
0.732 - 1 8 cubic holes BCC
Chapter-3-85Chemistry 481, Spring 2015, LA Tech
Radius Ratio Applications• Barium fluoride: 142/131 =1.08 (0.732-1) CN 8 FCC
• Potassium bromide: 138/196=0.704 (0.414-0.732) CN 6 FCC K+ in octahedral holes
• Magnesium sulfide: 103/184= 0.559 (0.414-0.732) CN 6 FCC
Chapter-3-86Chemistry 481, Spring 2015, LA Tech
Unit Cells dimensions and radius
a = 2r or r = a/2
Chapter-3-87Chemistry 481, Spring 2015, LA Tech
Summary of Unit Cells
Volume of a sphere = 4/3pr3
Volume of sphere in SC = 4/3p(½)
3 = 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
Chapter-3-88Chemistry 481, Spring 2015, LA Tech
Density CalculationsAluminum 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= 4.050 x 10-8 cm
Density = 2.698 g/cm3
Chapter-3-89Chemistry 481, Spring 2015, LA Tech
3. What is Coulombs law how it applies to ionic bond?
Chapter-3-90Chemistry 481, Spring 2015, LA Tech
Coulomb’s Law
k = constant
q+ = cation charge
q- = anion charge
r = distance between two ions
Chapter-3-91Chemistry 481, Spring 2015, LA Tech
Coulomb’s Model
where e = charge on an electron = 1.602 x 10-19
C
e0
= permittivity of vacuum = 8.854 x 10-12
C2
J-1
m-1
ZA = charge on ion A
ZB = charge on ion B
d = separation of ion centers
Chapter-3-92Chemistry 481, Spring 2015, LA Tech
An ionic bond is simply the electrostatic attraction between opposite charges.
Ions with charges Q1 and
Q2:
The potential energy is given by:
d
· ·
d
QQE
21µ
Ionic Bonds
Chapter-3-93Chemistry 481, Spring 2015, LA Tech
Arrange with increasing lattice energy:
KCl
NaF
MgO
KBr
NaCl788 kJ
671 kJ
3795 kJ
910 kJ
701 kJ
d
· ·K
+Cl
· ·K
+Br
d
d
QQE
21µ
Estimating Lattice Energy
Chapter-3-94Chemistry 481, Spring 2015, LA Tech
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:
Chapter-3-95Chemistry 481, Spring 2015, LA Tech
4. What is lattice energy? Take NaCl as an example.
Chapter-3-96Chemistry 481, Spring 2015, LA Tech
Lattice energy
The higher the lattice energy, the stronger the attraction between ions.
Lattice energy
Compound kJ/mol
LiCl 834
NaCl 769
KCl 701
NaBr 732
Na2O 2481
Na2S 2192
MgCl2 2326
MgO 3795
Lattice energy
Compound kJ/mol
LiCl 834
NaCl 769
KCl 701
NaBr 732
Na2O 2481
Na2S 2192
MgCl2 2326
MgO 3795
Chapter-3-97Chemistry 481, Spring 2015, LA Tech
Lattice Energy
Chapter-3-98Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-99Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-100Chemistry 481, Spring 2015, LA Tech
Trends in Melting Points
Compound Lattice Energy (Enthalpy)
(kcal/mol)
NaF -201
NaCl -182
NaBr -173
NaI -159
Chapter-3-101Chemistry 481, Spring 2015, LA Tech
Trends in Melting Points
Compound Lattice Energy
(kcal/mol)
NaF -201
NaCl -182
NaBr -173
NaI -159
Chapter-3-102Chemistry 481, Spring 2015, LA Tech
Compound q+ radius q- radius M.P (oC) L.E. (kJ/mol)
LiCl 0.68 1.81 605 834
NaCl 0.98 1.81 801 769
KCl 1.33 1.81 770 701
LiF 0.68 1.33 845 1024
NaF 0.98 1.33 993 911
KF 1.33 1.33 858 815
MgCl2 0.65 1.81 714 2326
CaCl2 0.94 1.81 782 2223
MgO 0.65 1.45 2852 3938
CaO 0.94 1.45 2614 3414
Trends in Properties
Chapter-3-103Chemistry 481, Spring 2015, LA Tech
6. Explain the lattice energy and melting point trends:
CompoundCation radius
(Angstroms)
Anion radius
(Angstroms)
Melting Point
(Centigrade)
Lattice Energy (kcal/mol)
MgCl2 0.65 1.81 714 2326CaCl2 0.94 1.81 782 2223
MgO 0.65 1.45 2852 3938
CaO 0.94 1.45 2614 3414
Chapter-3-104Chemistry 481, Spring 2015, LA Tech
Madelung ConstantMadelung constant is geometric factor that
depends on the lattice structure.
Chapter-3-105Chemistry 481, Spring 2015, LA Tech
Madelung Constant Calculation
Chapter-3-106Chemistry 481, Spring 2015, LA Tech
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 1.747558 6 : 6 Rock salt
CaF2 2.51939 8 : 4 Fluorite
Chapter-3-107Chemistry 481, Spring 2015, LA Tech
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?
Chapter-3-108Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-109Chemistry 481, Spring 2015, LA Tech
Trends in Melting Points Silver Halides
Compound M.P. oC
AgF 435
AgCl 455
AgBr 430
AgI 553
Chapter-3-110Chemistry 481, Spring 2015, LA Tech
9. In calculating lattice energy, why should Coulombs law equation is multiplied by the Avogadro’s number, N and Madelung constant, A?
N z2e2 N A z2e2
Lattice Energy = - ------ x N x A = - --------
4peor 4peor
Chapter-3-111Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-112Chemistry 481, Spring 2015, LA Tech
10. In the correction to lattice energy what is factor n in Born-Lande equation accounted for and how it relate to electronic configuration?
Chapter-3-113Chemistry 481, Spring 2015, LA Tech
11. Using the Born-Lande equation, calculate the lattice energy of cesium chloride.
N A z2e2 1Lattice Energy = - -------- ( 1 - ---) 4peor n
Chapter-3-114Chemistry 481, Spring 2015, LA Tech
• 12. What are Born-Mayer and Kapustinskii equations? How are they different from the Born-Lande equation?
Chapter-3-115Chemistry 481, Spring 2015, LA Tech
Born_Haber CycleEnergy Considerations in Ionic Structures
Chapter-3-116Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-117Chemistry 481, Spring 2015, LA Tech
13. What is a Born-Haber cycle?
Chapter-3-118Chemistry 481, Spring 2015, LA Tech
Ionic bond formation
Chapter-3-119Chemistry 481, Spring 2015, LA Tech
Energy and ionic bond formationExample - formation of sodium chloride.
Steps DHo, kJVaporization of Na(s) Na(g) +92sodium
Decomposition of 1/2 Cl2 (g) Cl(g) +121chlorine molecules
Ionization of sodium Na(g) Na+(g) +496
Addition of electron Cl(g) + e- Cl-(g) -349to chlorine
( electron affinity)Formation of NaCl Na+(g)+Cl-(g) NaCl -771
Chapter-3-120Chemistry 481, Spring 2015, LA Tech
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)
Chapter-3-121Chemistry 481, Spring 2015, LA Tech
Calculation of DHf from lattice Energy
Chapter-3-122Chemistry 481, Spring 2015, LA Tech
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
Chapter-3-123Chemistry 481, Spring 2015, LA Tech
15. Construct a Born-Haber cycle for the formation of aluminum fluoride. Do not perform any calculation.
Chapter-3-124Chemistry 481, Spring 2015, LA Tech
Hydration of Cations
Chapter-3-125Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-126Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-127Chemistry 481, Spring 2015, LA Tech
Thermodynamics of the Solution Process of Ionic Compounds
Heat of solution, DHsolution :
Enthalpy of hydration, DHhyd,
Lattice Energy, Ulatt
Chapter-3-128Chemistry 481, Spring 2015, LA Tech
Solution Process of Ionic Compounds
Chapter-3-129Chemistry 481, Spring 2015, LA Tech
16. Define following terms:Enthalpy of solution, DHsolution:
Enthalpy of hydration, DHhydration:
Solvent-solvent intermolecular attractions,
DH°solvent-solvent:
Chapter-3-130Chemistry 481, Spring 2015, LA Tech
17 . How is Enthalpy of solution, DHsolution, Enthalpy of hydration: DHhydration, and Lattice energy:Ulatt are related?
Chapter-3-131Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-132Chemistry 481, Spring 2015, LA Tech
Hydration Process
Chapter-3-133Chemistry 481, Spring 2015, LA Tech
Different types of Interactions for Dissolution
Chapter-3-134Chemistry 481, Spring 2015, LA Tech
Hydration Energy of Ions
Chapter-3-135Chemistry 481, Spring 2015, LA Tech
Hydration Process
Chapter-3-136Chemistry 481, Spring 2015, LA Tech
Calculation of DHsolution
Chapter-3-137Chemistry 481, Spring 2015, LA Tech
Heat of Solution and Solubility
Chapter-3-138Chemistry 481, Spring 2015, LA Tech
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 -950 -80
CsI 585 -700 -80
MgF2 3100 -2800 -120
Chapter-3-139Chemistry 481, Spring 2015, LA Tech
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.
Chapter-3-140Chemistry 481, Spring 2015, LA Tech
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)).
Chapter-3-141Chemistry 481, Spring 2015, LA Tech
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.