Transcript
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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
Number of balls in a unit cell:
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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
Number of balls in a unit cell:
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
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