Chapter3 Solid States

8/3/2019 Chapter3 Solid States

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Chapter 3. The Structures of Simple Solids

A.Fundamental Aspects of Solids & Sphere Packing

B. Common unit cells

C. The crystal structures of metals (or other elements)

D. The Crystal Structures of Ionic Solids

E. The Energetic Aspect of Ionic Solids

Outline

References:1.Inorganic Chemistry, Catherine E. Housecroft and Alan G. Sharpe, 3rd Ed.,

Pearson Education Ltd, 2008, Chapter 6.2.Inorganic Chemistry, D.F.Shriver, P.W.Atkins et al, 3rd Ed., Oxford University

Press, 2006, Chapter 3.3. Basic Inorganic Chemistry, F. A. Cotton, G. Wilkinson and P. L. Gaus, 3rd

Ed., John Wiley & Sons, Inc. 1995, Chapter 4.4. Concepts and Models of Inorganic Chemistry, B. Douglas, D. McDaniel andJ. Alexander, 3rd Ed., John Wiley Sons, Inc. 1994, Chapter 5.

5. Introduction to Coordination, Solid State, and Descriptive Inorganic

Chemistry, G. E. Rodgers, McGraw- Hill, 1994, Chapters 7 and 8.6. Inorganic Chemistry, G. L. Miessler, D. A. Tar, Prentice Hall: Pearson

Education Inc., 2004, Chapters 7.


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A. Fundamental Aspects of Solids & Sphere Packing.

Many inorganic materials are crystalline solids. e.g.

Metals Ionic solids (e.g. NaCl) Covalent solid (e.g. diamond) Atomic or Molecular solid (e.g. Ar, CO2)

In term of crystal structures, they can be thought as formed by packing of

atoms, ions or molecules to extended crystallattices.


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Concept of unit cell How to describe the structures of solids?A crystal structure is composed ofa motif, a set of atoms arranged in a

particular way, and a lattice (,). Motifs are located upon the points

of a lattice, which is an array of points repeating periodically in three

dimensions

Examples of two-dimensional Lattices


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Acrystal structure can be

represented by a three-

dimensional lattice

Lattice are usually described in terms of unit cell.


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A unit cellis a subdivision (or smallest component) of a crystal

that, when stacked together without rotation or reflection,

reproduces the crystal.

Start with a

lattice, one

can define

unit cells

a lattice point

a lattice point

For a given

periodical

arrangement, the

number ofchoices in unit

cell can be more

than one.

Preferred one

are those of

smaller size

and higher


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3-D Lattices and unit cells: The seven crystal systems or essential unit cells

or rhombohedral


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primitive hexagonalunit cell

conventional hexagonalunit cell

A note on Hexagonal unit cell

120o 60o


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Each essential system can have different lattice types,

e.g. for cubic system:

Primitive (P) Body-centered (I) Face-centered (F)


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The 14 possible BRAVAIS LATTICES


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Location of motif in a unit cell:

Fractional atomic coordinates and projections

x

y

z

x

y

(0,1)

(0,1)

(0,1)

(0,1)

(1/2)x

y


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Another example

x

y

(1/2)

(0,1)

(0,1)

(1/2)

(0,1)

(1/2)


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Exercise. Give the fractional coordinates of the followingunit cell.


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Counting Atoms in 3D Cells

Atoms in different positions in a cell are shared by differing numbers of unitcells

Vertex atom shared by 8 cells => 1/8 atom per cell

Edge atom shared by 4 cells => 1/4 atom per cell

Face atom shared by 2 cells => 1/2 atom per cell

Body unique to 1 cell => 1 atom per cell


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B. Common unit cells1. Unit cells from close packing of spheres

A single layerof spheres is closest-packed with a HEXAGONALcoordination of each sphere


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A second layerof spheres is placed in the indentations left by the first layer.When a third layerof spheres is placed in the indentations of the second layer

there are TWO choices.

Two common types of close packing


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Two common types of close packing: another view


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Is the packing shown below also a

close packing?

A

B

A

CA


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Cubic close packing (also called face centered closepacking)

abbreviation:ccp orfcc

Pattern: ABCABCABC. Contains a cuboctahedron

Coordination number of each sphere =


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Choosing unit cell for cubic close packing (also calledface centered close packing)

Choice 1: corners and face-centered

Number of balls in a unit cell:


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Examples


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Choice 2: edge and body centered



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Hexagonal close packing: hcpPattern: ABABAB

Number of

balls in a unit

cell = ?

unit cell of hcp: choice 1

Contains an anti-cuboctahedron

Coordination number of each sphere:


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Choice 2


Choice 1, another view


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Structural characteristics of

fcc and hcp

Holes in close-packed structures


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Oh d Td H l


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Oh and Td Holes

View the structures of cuboctahedron or anti-cuboctahedron toobtain the following properties:

No. of balls : no. of octahedral holes : no. of tetrahedral holes =

1. Number of holes surrounding a ball:#Td =, #Oh =

An octahedral hole is shared by __ balls.

A tetrahedral hole is shared by __ balls.

# balls : # Oh holes=

# balls : # Td holes=


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Consider the unit cell for fcc (or ccp) again:

(a) Coordination number for each

sphere (number of nearest balls):

(b) Number of spheres in the unit cell:

(c) Number of tetrahedron holes in

the unit cell:

(d) Number of tetrahedron holes inthe unit cell:

(e) Relationship between cell size (l)and radius of ball (r):


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Where are the holes?

Tetrahedral holes:

under each corner

Octahedral holes:center + middle of the edges


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Consider the unit cell for hcp again:

(a) Coordination number foreach sphere (number ofnearest balls):

(b) Number of spheres in theunit cell:

(c) Number of tetrahedron holesin the unit cell:

(d) Number of tetrahedron holesin the unit cell:


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Relationship between cell size and radius of ball

(r):

If the sphere radius is r,

a = ____

c = ___


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Relationship between cell size and radius of ball (r):

If the sphere radius is r,

a = 2r

c = 2 h

c

h

c = 2 h

h =8r

3

24r

3

6

3

2

=

8r

3

6 r

3

42 2r

3

4 4r(2/3)1/2


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2. Body-centered cubic packing (bcc)

(a) Coordination number for eachsphere:

(b) irregular holes only

(c) number of spheres in the unit cell:

(d) Relationship between cell size (l)

and radius of ball (r):


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3. Primitive cubic packing (simple cubic)

(a) Coordination number for each sphere:

(b) Type of holes:

(c) Number of spheres in the unit cell:

(d) Number of spheres in the unit cell:

(e) Relationship between cell size (l) and radius of ball (r):


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4. Fraction of space occupied by spheres

in different unit cells.

Fraction of space occupied by spheres.

For FCC,

Vsphere

Vunit cell= =

4 (4/3) r3

l3

l = 2(2)1/2

r

= 74.05%

The fraction of space occupied by spheres for hcp is________. (Both hcp and fcc are most close-packed

structures.)


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5. CALCULATION OF DENSITY OF UNIT CELL

Mass of unit cell = Number of atoms in unit cell x Mass of each atom

Mass of unit cell =

Mass of each atom =

Z = No of atoms in unit cellNis Avogadros Number

M is the Molar mass of the Crystal

a3 is the volume of the unit cell

Where a is the edge length of a unit cell

a is the edge length of a cube usually expressed in pico metre or nano metre1pm =10-12 m and 1nm = 10-9m, Mis the molar mass of the element.

Density of Unit CellMass of unit Cell

Volume of unit Cell

=

Z m

m

M

N=

=

Z M

Na3


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Summary on fraction of space occupation

Name coord. no. sphere fraction of touching occupation

simple 6 cell edge 0.52

cubic

body body 0.68

center 8 diagonal

ccp 12 face 0.74

or fcc diagonal

hcp 12 0.74


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C. The crystal structures of metals (or other elements)

They can be best described in terms of close-packed and non-close-packedstructures. E.g.

simple cubic (sc)primitive cubic (cubic-P)

body-centered cubic (bcc)fcc


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Table: The most stable crystal structures assumed by the elements in their solid

phase

He

hcp

Li Be B C N O F Ne

bcc hcp rh d hcp sc - ccp

Na Mg Al Si P S Cl Arebcc hcp ccp d sc or tet ccp

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

bcc ccp hcp hcp bcc bcc bcc bcc hcp ccp ccp hcp sc d rh hcp or ccp

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

bcc ccp hcp hcp bcc bcc hcp hcp ccp ccp ccp hcp tet tet rh hcp or ccp

Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

bcc bcc hcp hcp bcc bcc hcp hcp ccp ccp ccp rh hcp ccp rh mono - ccp

Note: sc = simple cubic, tet = tetragonal, rh = rhombohedral, d = diamond

or = orthorhombic, mono = monoclinic

The crystal structures of metals


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Polymorphism

Many metals adoptdifferent structures atdifferent temperatureor pressure.

Polymorphism: Asolid adopts differentcrystal forms

In elemental solids,this is called

Allotropy.

-iron

-iron

-iron

e.g. Structure of Fe changes with Temperature


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Structures of alloys a blend of metallic elements

Maybe formed, if-radii of the elements are similar.

-structure of pure metals are the

same.

-electropositive character are

similar.

(a)Substitutional solid solutions:Replacement of one type of metal

atoms in a structure by another.

(b) Interstitial solid solutions

of nonmetals:

Additional smallatoms (e.g. C, B,

N) occupy holes within the latticeof original metal structure


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Structures of alloys

(c) Intermetallic compounds.

Structure is different from the

structures of either component metals

e.g. Cu, ccp, Zn, hcp, but alloy CuZn has the following structure

f


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Bronze is a substitutional alloy

Examples of alloys


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Carbon Steel is an interstitial alloy

Th C t l St t f I i lid


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The Crystal Structures ofIonic solids

Many ionic solid structures can be regarded as derived from arrays in which theanions (sometimes the cations) stack together in fcc or hcp (or others) patterns

and the counter ions occupy the octahedral and tetrahedral holes in the close

packing.

e.g. Nail


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(a) MX structures

Rock salt structure

Close packing type of anion:Type of holes occupied by cation:

Hole occupancy:

Coordination no.of Na+ :

Coordination no.of Cl- :


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Zinc blende (ZnS) structure

Close packing type of anion:Type of holes occupied by cation:

Hole occupancy:

Coordination no.of Zn2+ :

Coordination no.of S2- :

Cesium chloride structure


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Cesium chloride structure

Close packing type of anion:

Type of holes occupied by cation:

Hole occupancy:Coordination no.of NH4

+ :

Coordination no.of Cl- :

NH4ClCsCl

Nickel arsenide (NiAs) str ct re


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Nickel-arsenide (NiAs) structure

Close packing type of AsType of holes occupied by Ni

Hole occupancy:Coordination no.of As :

Coordination no.of NI :


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Wurtzite (ZnS) structure

Close packing type of anion:

Type of holes occupied by cation:

Hole occupancy:

Coordination no.of Zn2+ :

Coordination no.of S2-

:


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(b) MX2 structures Fluoride (CaF2)

Close packing type of Ca2+

:Type of holes occupied by F-:

Hole occupancy:

Coordination no.of Ca2+ :

Coordination no.of F- :


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Rutile (TiO2)

Coordination numberfor Ti:

Coordination number

for O:


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(c) More complicated structures:

Perovskite structure ABO3 (CaTiO3)

Ca and O form close packing.

Ti's fill % of the holes.

Coordination number of Ti:

Coordination number of Ca:Coordination number of O:

Ca

O

Ti

A[B ]O


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The Spinel structure( ) MgAl2O4

A[B2]O4

O: fcc close-packingA: occupy 1/8 of T holes

B: occupy of O holes

Fe3O4 = Fe2+[Fe3+]2O4

A[B2]O4


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Relationship between some important structures

4 R ti li i St t B d


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4. Rationalizing Structures Based on

Radius Ratio (r+/r-)

Why NaCl and CsCl adopt different structures?

Related to size of the cavity or r+/r-?

Here, r+ and r- are ionic radii of cation and anion, respectively.

Filli i i h i l


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Filling cavity with a particle

Too bigToo small

Just fitting

Perfect packing occurs

when anions touch andcations fit perfectly in thepocket. Then, anions

touch anions and cationstouch anions.

If cations are too large(largerrM/rX), anions

cannot touch (not ideal).

If anions are too large

(smallerrM

/rX

), cationsrattle in pocket (worse).

Conclusion:The listed rM/rX ratios

are

the minima, not maxima.


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For fcc structure MX

Forfcc structure MX, X- isclose-packed.

If X- has its radius r-, howbig in size is the octahedralhole?

r+/r-=?

(r++r-)/(r-+r-) = sin(45) r+/r- = 0.414

r-

+ r-

r-+ r

+

r+

+ r+

45o


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Forsimple cubic structure:

L

L

L2

+L2

= X2

X

Y 2 L

2

= X

2

X = 2

1/2

L

X2

+ L2

= Y2

2 L2

+ L2

= Y2

Y = 31/2

L

r-+r

-

2r-+2r

+

Y = (3)1/2 (L)Y = 2r-+2r+ = (3)1/2 (2r-)

r+/r- = 0.732

M+

M-

Summary:


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y

r+/r- 0.155 to 0.225 to 0.414 to 0.732 to 1

maximum

coordination 3 4 6 8 12

numbertype of holes trigonal tetrahedral octahedral cubic Close-packing

The listed rM/rX ratios are the minima, not maxima.


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Some examples:

Anions CationsCrystal r+/r-

Structure C.N. Structure C.N.

NaCl 1.16/1.67=0.69 fcc 6 all oct. holes 6

ZnS 0.88/1.70=0.52 fcc 4 half tet. holes 4(Zinc Blende)

ZnS 0.88/1.70=0.52 hcp 4 hall tet. holes 4

(Wurtzite)

CsCl 1.81/1.67=1.08 simple 8 all cubic holes 8

cubic

One can see that not all examples fit the table in the summary. This imply

that one should be careful in using the pure ionic model.


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Radius ratios are only correct ca. 50% of the time, not very good for afamily of archetypal ionic solids -random choice might be just as

successful as radius ratio rules and saying that all adopt the NaClstructure more so!

Ionic radii change withcoordination number r8 > r6

> r4

It is hard to determine Ionic radii


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D.The Energetic Aspect of Ionic Solids

An indication of stability for a solid: Lattice energy:

Definition. The standard enthalpy change in the formation ofgaseous ions from a solid.

MX(s) M+(g) + X-(g) HL > 0

(note: some books refer the Lattice energy to the reversedprocess!).

Lattice disruption endothermic

HL , Larger

HL, stability

How to obtain HL? (two methods)


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1. Thermodynamic Model: An Experimental Approach

IE-EAM(g) + X(g) M+(g) + X-(g)

S D HL

Hf0

M(s) + X2(g) MX(s)

Hf

0

+

HL = S +

D + IE - EA

IE: Ionization Energy; EA: Electron Affinity

S: Sublimation Energy; D: Dissociation Energy

Some thermo-chemical data (kJ/mol) at 298K for the alkali metal halides:

Make comparison of data obtained by the two approaches and comment on their differences.

-Hf0 S D IE EA HL(exp) HL(theo)

LiF 616.9 160.7 78.9 520.5 328.0 1049.0 966

LiI 270.1 160.7 106.8 520.5 295.4 762.7 723

Lattice enthalpy and theBorn-Haber cycle

2. Theoretical models for calculating lattice enthalpies (purely ionic model)


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g ( y )

Consider electrostatic and repulsive interactions

between the ions.

Electrostatic

+ -

+

attractive

repulsive

Repulsive forces (Born forces)

- electron-electron- nucleus-nucleus

electrostatic


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Equations proposed to estimate HLThe Born-Lande equation

=NA|Z+Z-|e

2

40doAHL (1 - )

d

d0

The Born-Meyer equation

The Kapustinski equation

=NA|Z+Z-|e

2

40doHL (1 - )

1

n

=n|Z+Z-|

doKHL (1 - )

d

d0

+ -

d0

HLZ

+ Z-

r+

+ r-

Z+ Z

-

d0

=NA|Z+Z-|e

2

40doHL (1 - )

1

n

electrostatic Born

forces

The Born-Lande equation 0


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The Born Lande equation

=

NA|Z+Z-|e2

40doA

HL (1 - )

1

n

Z+Z-e2

40do=Ec

d0

U = EC + ER

=ER

B

d0n

NA = Avogadro constantA = Madelung constant, relating to the

geometry of the crystal.z+ = charge number of cation

z = charge number of anione = elementary charge, 1.6022 1019 C

0 = permittivity of free space

d0 = distance to closest ion

n = Born exponent, a number between 5and 12, determined experimentally by

measuring the compressibility of thesolid, or derived theoretically

+ -

0


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How is Born Lande Equation derived? 0


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How is Born-Lande Equation derived?

=

NA|Z+Z-|e2

40doA

HL (1 - )

1

n

Z+Z-e2

40do

=Ec

d0

U = EC + ER

=ER

B

d0n

+ -

0

Consider

Mz+(g) + Xz-(g) MX (s) U =- HL = ??

U = EC + ER

EC: from electrostatic interactionER: from Born forces

(1) Contribution from Electrostatic interactions between ions


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(1) Contribution from Electrostatic interactions between ions.

+ -

d0

Ec =

(Z+)(Z-)e2

40doTwo ions:

In cryatals,e.g. 1 mol of NaCl, EC = ?

e g NaCl type crystals


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e.g. NaCl type crystals

A = Madelung ConstantEc =

NA(Z+)(Z-)e2

40do

=

NA(Z+)(Z-)e2

40do ANaCl

ANaCl= = 1.74756

NA(Z+)(Z-)e2

40doAEc =

(2) C t ib ti f B f


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(2) Contribution from Born forces.

=ER

NAB

d0n

+ -

d0

n: born component, related

to electronicconfiguration

Z+Z-e2

40do=Ec

d0

=ER

B

d0

n

(3) T t l


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(3) Total energy

B =NA(Z+)(Z-)e

2

40don

Ad0n-1

NA(Z+)(Z-)e2

40doAU =

NAB

d0n

+

=NAZ+Z-e

2

40do(1 - )

1

nU

U = EC + ER

=NA|Z+Z-|e

2

40doAHL (1 - )

1

n

Z+Z-e2

40do=Ec

d0

=ER

B

d0n

It can be shown that

B M E tid0


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Born-Mayer Equation

Similar to Born-Lande equation.

The term related to repusive force ismodified.

=NA|Z+Z-|e

2

40(A(1 - )

d

= NA|Z+Z-|e2

40doA(1 - )d

d0

r+ + r-) r+ + r-

HL

d is a parameter that is approximately 34.5pm if d0 is in pm.

=ER Be-d0/d

Z+Z-e2

40do

=Ec

d0

U = EC + ER

+ -


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Why different?


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Consequences of lattice enthalpies

(a) Thermal stabilities of ionic solids

MCO3 MO + CO2

greater lattice energy


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Lattice energy contribution favors the decomposition process because

r(O-2) is less than r(CO32-) so that HL(MO) is larger than HL(MCO3).

= HL(MO) - HL(MCO3) lattice energy change

MCO3 MO + CO2

M CO3

M O

small %changein lattice scale

M CO3

M O

large %changein lattice scale

islargerin%

issmallerin%

A greatly exaggerated representation of the change in lattice parameterfor cations of different sizes

disfavors decompositionfavors decomposition

Conclusion: Large cations stabilize large anionsThe lattice energy change is also proportional to the charge of cations.

Therefore, MCO3 (M2+) are less stable than M2CO3 (M

+).

greater lattice energy

smaller lattice energy


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(b) Solubility ofionic solids in water

Observations:(i) Salts of one small ion and

one large counter-ionhave higher solubility.

(ii)Salts of anion and cationof similar size have low

solubility.

H

Explanation?


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HLMX(s) M+(g) + X-(g)

H(M+) H(X-)

M+(solv) + X-(solv)

Hs

Hs = HL + [H(M+) + H(X-)]

HL: positive, proportional to

H(M+) and H(X-) are all negative, proportional to

HL: disfavors the solution process

[H(M+) + H(X-)]: favors the solution process

1(rM+ + rX-)

1rM+

1rX-

+


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Size of M+ and X- HL [Hsol(M+) + Hsol(X-)] solubility

Both are

Large

Both are

Small

One large

One small

Conclusion: salts having one small ion and one large counter-ionhave higher solubility.

Examples: CsF and LiI have the highest solubility when compared other

alkali halides.


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End of Chapter 3

Chapter3 Solid States

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