Chapter-3-1 Chemistry 481, Spring 2015, LA Tech Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office: CTH 311 Phone 257-4941 Office Hours:

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. 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.


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


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


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


Loose packing

Simple cube SC

Body-centered cubic BCC

Packing and Geometry


The Unit CellThe basic repeat unit that build up the whole solid


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


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.


Fourteen Bravias Unit Cells


Seven Crystal Systems


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


Close Pack Unit Cells

CCP HCP

FCC = 4 ( 8 coners, 6 faces)


Simple cube SC Body-centered cubic BCC

Unit Cells from Loose Packing

SC = 1 (8 coners) BCC = 2 (8 coners, 1 body)


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


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


Holes in SC Unit Cells

Cubic Hole


Octahedral Hole in FCC

Octahedral Hole


Tetrahedral Hole in FCC

Tetrahedral Hole


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.


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.


AlloysSubstitutional

Second metal replaces the metal atoms in the lattice

Interstitial

Second metal occupies interstitial space (holes) in the lattice


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.


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.


Metallic solidsRepeating units are made up of metal atoms,

Valence electrons are free to jump from one atom to another

++ + +

++ + +

++ + +

++ + +

++

++

++ +

+

++ ++

++ + +

++ + +


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).


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.


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.


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).


13. Describe metallic bonding and properties in terms of:

Electron-sea model of bonding:

Band Theory:


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.


Linear Combination of Atomic Orbitals


Linear Combination of Atomic Orbitals


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.


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


Conduction Bands in Metals


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


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


Band Gaps


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


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):


Band Theory of Metals


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


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


Band Gaps


Doping Semiconductors


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):


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):


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


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


Vacuum tubes and Transisters


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


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


24. What is an integrated circuit?


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


Basic Ionic Crystal Unit Cells


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:


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)


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


Cesium Chloride Structure (CsCl)


Rock Salt (NaCl)

© 1995 by the Division of Chemical Education, Inc., American Chemical Society.

Reproduced with permission from Soli-State Resources.


Sodium Chloride Lattice (NaCl)


NaCl Lattice Calculations


CaF2


Calcium Fluoride


Reproduced with permission from Solid-State Resources.


Zinc Blende Structure (ZnS)


Lead Sulfide




Wurtzite Structure (ZnS)


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


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


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.


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


Rock Salt (NaCl)


Reproduced with permission from Soli-State Resources.


Sodium Chloride Lattice (NaCl)

0,0,1 0,0,2

1,1,12,2,2


CaF2

0,0,1 0,0,4 0,0,2 0,0,4

0,0,20,0,2 0,0,4


Calcium Fluoride




Zinc Blende Structure (ZnS)

0,0,1 0,0,4 0,0,40,0,2


Lead Sulfide




Wurtzite Structure (ZnS)


Antifluorite Structure


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


Radius Ratio Rules







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


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







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


Unit Cells dimensions and radius

a = 2r or r = a/2


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


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


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


Coulomb’s Law

k = constant

q+ = cation charge

q- = anion charge

r = distance between two ions


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


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


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


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:


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


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


Lattice Energy


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.


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


Trends in Melting Points

Compound Lattice Energy (Enthalpy)

(kcal/mol)

NaF -201

NaCl -182

NaBr -173

NaI -159


Trends in Melting Points

Compound Lattice Energy

(kcal/mol)

NaF -201

NaCl -182

NaBr -173

NaI -159


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


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


Madelung ConstantMadelung constant is geometric factor that

depends on the lattice structure.


Madelung Constant Calculation


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


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?


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


Trends in Melting Points Silver Halides

Compound M.P. oC

AgF 435

AgCl 455

AgBr 430

AgI 553


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


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

http://en.wikipedia.org/wiki/Compressibility


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


11. Using the Born-Lande equation, calculate the lattice energy of cesium chloride.

N A z2e2 1Lattice Energy = - -------- ( 1 - ---) 4peor n


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


Born_Haber CycleEnergy Considerations in Ionic Structures


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


13. What is a Born-Haber cycle?


Ionic bond formation


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


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)


Calculation of DHf from lattice Energy


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


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


Hydration of Cations


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.


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.


Thermodynamics of the Solution Process of Ionic Compounds

Heat of solution, DHsolution :

Enthalpy of hydration, DHhyd,

Lattice Energy, Ulatt


Solution Process of Ionic Compounds


16. Define following terms:Enthalpy of solution, DHsolution:

Enthalpy of hydration, DHhydration:

Solvent-solvent intermolecular attractions,

DH°solvent-solvent:


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


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.


Hydration Process


Different types of Interactions for Dissolution


Hydration Energy of Ions


Hydration Process


Calculation of DHsolution


Heat of Solution and Solubility


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


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.


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)).


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.

Documents

Chapter-3-1 Chemistry 481, Spring 2015, LA Tech Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office: CTH 311 Phone 257-4941 Office Hours: