Chapter 2 Slide 1 of 85
Chapter Two
Structures of Solids
Chapter 2 Slide 2 of 85
States of Matter Compared
Chapter 2 Slide 3 of 85
Chapter 2 Slide 4 of 85
Some Characteristics of Crystalline Solids
Chapter 2 Slide 5 of 85
Network Covalent Solids• These substances contain a network of covalent bonds that extend
throughout a crystalline solid, holding it firmly together.• In material science, polymorphism is the ability of a solid material to
exist in more than one form or crystal structure. Diamond, graphite and the Buckyball are examples of polymorphs of carbon. α-ferrite, austenite, and δ-ferrite are polymorphs of iron. When found in elemental solids the condition is also called allotropy.
• The allotropes of carbon provide a good example1. Diamond has each carbon bonded to four other carbons in a
tetrahedral arrangement using sp3 hybridization.
2. Graphite has each carbon bonded to three other carbons in the same plane using sp2 hybridization.
3. Fullerenes and nanotubes are roughly spherical and cylindrical collections of carbon atoms using sp2 hybridization.
Chapter 2 Slide 6 of 85
Crystal Structure of Diamond
Covalent bond
Crystal Structure of Graphite
Covalent bond
van der Waalsforce
Structure of a Buckyball
sp2 hybrid Carbon
Chapter 2 Slide 9 of 85
sp2 hybrid Carbon
Carbon Nano-tube
A nanotube (also known as a buckytube) is a member of the fullerene structural family, which also includes buckyballs. Whereas buckyballs are spherical in shape, a nanotube is cylindrical, with at least one end typically capped with a hemisphere of the buckyball structure. Their name is derived from their size, since the diameter of a nanotube is on the order of a few nanometers , while they can be up to several centimeters in length. There are two main types of nanotubes: single-walled nanotubes (SWNTs) and multi-walled nanotubes (MWNTs).
Chapter 2 Slide 10 of 85
Experimental Determinationof Crystal Structures
Bragg’s Law2 d sinθ = n λ
Chapter 2 Slide 11 of 85
Chapter 2 Slide 12 of 85
X-Ray Diffraction Image & Pattern
Single crystal Powder
Chapter 2 Slide 13 of 85
Crystal Lattices
• To describe crystals, three-dimensional views must be used.• The repeating unit of the lattice is called the unit cell.• The simple cubic cell (primitive cubic) is the simplest unit
cell and has structural particles centered only at its corners.• The body-centered cubic (bcc) structure has an additional
structural particle at the center of the cube.• The face-centered cubic (fcc) structure has an additional
structural particle at the center of each face.
Chapter 2 Slide 14 of 85
Unit Cells InCubic Crystal Structures
simple cubic(primitive cubic)
bcc fcc
Chapter 2 Slide 15 of 85
Face-centered cubicBody-centered cubicPrimitive cubic
Chapter 2 Slide 16 of 85
Occupancies per Unit CellsPrimitive cubic: a = 2r
1 atom/unit celloccupancy = [4/3(πr3)]/a3 = [4/3(πr3)]/(2r)3
= 0.52 = 52%Body-centered cubic: a = 4r/(3)1/2
2 atom/unit celloccupancy = 2 x [4/3(πr3)]/a3 = 2 x [4/3(πr3)]/[4r/(3)1/2]3
= 0.68 = 68%Face-centered cubic: a = (8)1/2 r
4 atom/unit celloccupancy = 4 x [4/3(πr3)]/a3 = 4 x [4/3(πr3)]/[(8)1/2 r]3
= 0.74 = 74%Closest packed
Chapter 2 Slide 17 of 85
a
ab
aba abcunoccupied holes
Closest packed
Chapter 2 Slide 18 of 85
abab
abcabc
Cubic close-packed structure
= Face-centered cubic
Hexagonal close-packed structure
polytypes
Chapter 2 Slide 19 of 85
Crystal Structures of Metals
Interionic Forces of Attraction
E = (Z+Z-e2)/4πεr e = 1.6 x 10-19 C
In vacuum, ε0 = 8.85 x 10-12 C2m-1J-1
In water, εH2O = 7.25 x 10-10 C2m-1J-1= 82 ε0
In liquid ammonia, εNH3 = 2.2 x 10-10 C2m-1J-1= 25 ε0
Chapter 2 Slide 21 of 85
Unit Cell of Rock-Salt(Sodium Chloride)
Cl- at fcc
Na+ at Oh holesCoord. #: Na+: 6; Cl-: 6atom/ unit cellNa: Cl= 4: 4 = 1: 1 NaCl
Chapter 2 Slide 22 of 85
Unit Cell of Cesium Chloride
Cl- at primitive cubicCs+ at Cubic holes
Coord. #: Cs+: 8; Cl-: 8atom/ unit cellCs: Cl= 1: 1 CsCl
Chapter 2 Slide 23 of 85
Unit Cell of Cubic Zinc Sulfide(Sphalerite or Zinc blende)
Coord. #: Zn2+: 4; S2-: 4atom/ unit cellZn: S = 4: 4 = 1: 1 ZnS
S2- at fccZn2+ at ½ Td holes
Chapter 2 Slide 24 of 85
A Born-Haber Cycle to CalculateLattice Energy
1/2D(Cl-Cl)
∆Ηsublimation (Na)
IE(Na) -EA(Cl)
lattice energy U0 = - ∆Ηf0(NaCl) + ∆Ηsublimation (Na) + 1/2D(Cl-Cl)
+ IE(Na) - EA(Cl)
= (+411 +107 +122 + 496 –349) kJ/mol
= +787 kJ/mol
∆Ηf0(NaCl)
U0NaCl(s) Na+(g) + Cl-(g)
Na(g) + Cl(g)Na(s) + 1/2Cl2(g)
Chapter 2 Slide 25 of 85
Lattice Energy & Madelung Constant
( ) ( ) ( ) 0 U- H 0 <=∆→+ −+sgg MXXM
U0: lattice energy
Factors contributed to Lattice energy•electrostatic energy ~90%•repulsion of close shells ~8%•dispersion forces ~1% •zero-point energy (lattice vibration at 0K)•correction for heat capacity ~1%
Chapter 2 Slide 26 of 85
Electrostatic energy in a crystal lattice, between a pair of ions
For NaCl crystal, Z+ =Z- =1r = d
( )reA
0
2
c 4ZZ
Eπε
−+
= A : Madelung Constant
Chapter 2 Slide 27 of 85
The value of Madelung constant is determined only by the “geometry of the lattice”, and independent of the “ionic radius” and “charge”.
Chapter 2 Slide 28 of 85
electrostatic energy repulsion of close shells
B: constantn: Born exponent
Born equation
12Xe, Au+
10Kr, Ag+
9Ar, Cu+
7Ne
5He
nIon configuration
Chapter 2 Slide 29 of 85
Let U0 = -(PE)0 N N: Avogadro’s number
Born-Lande equation
At minimum PE, attractive and repulsive forces are balanced, and d= d0.
0)(
=∂
∂d
PE
Chapter 2 Slide 30 of 85
For NaCl
A = 1.74756Z1 = 1, Z2 = -1d0 = rNa+ + rCl- = 2.81 x10-10 mn = (nNa+ + nCl-) /2 = (7+9)/2 = 8
⇒ U0= 755.2 kJ/mol
experimental U0= 770 kJ/mol
Chapter 2 Slide 31 of 85
Modification of Born-Lande equation
)345.0
1(4
ZZ
000
2
0 ddeNA
U −=−+
πε
Unit in ? , 10-10 m
2. Kapustinskii equation
Improving the repulsion
)345.0
1(ZZ
00
0 ddn
U −−
=−+κ
κ = 1.21 MJ. ? mol-1
n= ions per formulae.g. n= 2 for NaCl
n= 5 for Al2O3
A ? crystal lattice ? r+/r- ? r0A/n ~constant
1. Born-Mayer equation
Chapter 2 Slide 32 of 85
+218.1+183.8+130.4+48.3∆G0 (kJ/mol)
13001100840300T (?C)+172.1+171.0+160.6+175.0∆S0 (J/K mol)
+269.3+234.6+178.3+100.6∆H0 (kJ/mol)
BaSrCaMgData at 298K
( ) ( ) ( )g2ss3 CO MO MCO +→
A small cation increases the lattice enthalpy of the oxide more than that of a carbonate.
0
0
000
Tion decomposit
T-
SH
SHG
∆∆
=
∆∆=∆
Chapter 2 Slide 33 of 85
< 0 > 0
Solubility of ionic compounds
0000 T- T- SLSHG ∆=∆∆=∆
Enthalpy of solution Hydration enthalpy
Free energy of solution
Chapter 2 Slide 34 of 85
Hydration enthalpy
−+
−+
+∞=∆+∆=∆
+∞
−+
rrHHH
rrU
XMhydration11
1 0
−−=∆
ε1
12
2
rZ
Hhyd
•In general, difference in ionic size favors solubility in water.
•Ionic compound MX tends to be most soluble when rX-rM > 0.8?
Chapter 2 Slide 35 of 85
Correlation between ∆Hsolution and the differences between the hydration enthalpy of the ions
Chapter 2 Slide 36 of 85
Fajan's rules
Fajan's Rule: the degree of covalent character of ionicbond
Polarization effects: (a) idealized ion pair with no polarization; (b) mutually polarized ion pair; (c) polarization sufficient to formcovalent bond. Dashed lines represent hypothetical unpolarized ions.
Chapter 2 Slide 37 of 85
In 1923, Fajan suggested rules to predict the degree of covalent character in ionic compounds:
The polarization of an ionic bond and thus the degree of covalency is high if :1) the charges on the ions are high.eg., Al3+ ; Ti4+ ----- favors covalent character.
Na+ ; K+ ----- favors ionic character.
2) the cation is small.e.g., Na+ ion is larger than that of Al3+
Thus, Al3+ favors covalent character.
3) the anion is large.e.g. F- ionic radius : 0.136 nm favors ionic character.
I- ionic radius : 0.216 nm favors covalent character.
Chapter 2 Slide 38 of 85
4) An incomplete valence shell electron configurationNoble gas configuration of the cation better shielding and less polarizing power e.g. Hg2+ (r = 102 pm) is more polarizing than Ca 2+ (r = 100 pm)
HgO decomposed at 500 ?CCaO m.p. 2613 ?C
783
742
775
1418
m.p. (?C)
hexagonal
rhom.
cubic
cubic
Crys str
259
236
276
645 (dec)
m.p. (?C)
tetragonal
rhombohedral
orthorhombic
cubic
Crys str
CaI2
CaBr2
CaCl2
CaF2
Halides
HgI2
HgBr2
HgCl2
HgF2
Halides
Chapter 2 Slide 39 of 85
a) Charge factori) Cations
Na+ Mg2+ Al3+
Examples:For ions with noble gas (ns2 np6 ) structure, the only factor that have to be considered are size and charge factor.
ii) AnionsN3- O2- F-
Increasing charge : increase in polarizing power
Increase in ionic charge, more ready to be polarized
Chapter 2 Slide 40 of 85
1.5 x 10-5183sublimationAlCl3
57-70SiCl4
14101440
boiling point /°C
715800
melting point /°C
Conductance in molten state
Chlorides
29MgCl2
133NaCl
AlCl3 has covalent character. In fact, AlCl3 exists as dimer Al2Cl6at room conditions.
covalency
Chapter 2 Slide 41 of 85
b) Size factor
i) CationsBe2+ Mg2+ Ca2+ Sr2+ Ba2+
The smaller the cation, the higher is its polarizing power
52774CaCl2
56870SrCl2
Conductance in molten state
melting point /°CChlorides
955BaCl2
29715MgCl2
0.056404BeCl2
covalency
Chapter 2 Slide 42 of 85
ii) Anions
F- Cl- Br- I-
The larger the anion, the more polarizable is the anion.
eg.,NaF NaCl NaBr NaI
melting point /°C 990 800 755 651
covalency
Chapter 2 Slide 43 of 85
( ) ( ) ( )g2ss3 CO MO MCO +→
1360BaCO3
1289SrCO3
900CaCO3
540MgCO3
250BeCO3
Decomp. Temp. (?C)Carbonates
1923BaO
2430SrO2613CaO
2826MgO
2530BeO
m.p. (?C)Oxides
Ionic compounds
covalency
wurtzitestructure
NaClstructure
Some covalent character
Chapter 2 Slide 44 of 85
Close-packing of Spheresin Three Dimensions
Chapter 2 Slide 45 of 85
Octahedral holesTetrahedral holes Cubic holes
Close packed structure
Chapter 2 Slide 46 of 85
rh/r = 0.156
rh/r = 0.225
rh/r = 0.414
Chapter 2 Slide 47 of 85
Chapter 2 Slide 48 of 85
Chapter 2 Slide 49 of 85
Considering anion-anion
repulsion
Chapter 2 Slide 50 of 85
Unit Cell of Rock-Salt(Sodium Chloride)
Cl- at fcc
Na+ at Oh holesCoord. #: Na+: 6; Cl-: 6atom/ unit cellNa: Cl= 4: 4 = 1: 1 NaCl
Chapter 2 Slide 51 of 85
Rock-salt structure
Chapter 2 Slide 52 of 85
Unit Cell of NiAS
As3- at hcp
Ni3+ at Oh holesPolariable cation& anion
Chapter 2 Slide 53 of 85
Unit Cell of Cesium Chloride
Cl- at primitive cubicCs+ at Cubic holes
Coord. #: Cs+: 8; Cl-: 8atom/ unit cellCs: Cl= 1: 1 CsCl
Chapter 2 Slide 54 of 85
Unit Cell of Cubic Zinc Sulfide(Sphalerite or Zinc blende)
Coord. #: Zn2+: 4; S2-: 4atom/ unit cellZn: S = 4: 4 = 1: 1 ZnS
S2- at fccZn2+ at ½ Td holes
Chapter 2 Slide 55 of 85
Unit Cell of Hexagonal Zinc Sulfide (Wurtzite)
S2- at hcp
Zn2+ at ½ Td holesPolymorph of ZnS
Chapter 2 Slide 56 of 85
Pt2+ at fccS2- at ½ Td holesPtS4 unit is planar Pt-S more covalent
PtS
Chapter 2 Slide 57 of 85
Unit Cell of Fluorite Structure(Calcium Fluoride)
Ca2+ at fccF- at Td holesCoord. #: Ca2+: 8; F-: 4
atom/ unit cellCa: F= 4: 8 = 1: 2 CaF2
Chapter 2 Slide 58 of 85
•Fluorite Structure- MX2 Large M
•Antifluorite structure- M2XLarge X, e.g. Na2O
Chapter 2 Slide 59 of 85
Unit Cell of Rutile TiO2
O2- at hcpTi4+ at ½ Oh holes
Coord. #: Ti: 6; O: 3atom/ unit cellTi: O= 2: 4 = 1: 2
Chapter 2 Slide 60 of 85
Chapter 2 Slide 61 of 85
Chapter 2 Slide 62 of 85
A Structural Map for compounds of MX
Increasing covalency
Chapter 2 Slide 63 of 85
Chapter 2 Slide 64 of 85
Chapter 2 Slide 65 of 85
Chapter 2 Slide 66 of 85
Chapter 2 Slide 67 of 85
A Structural Map for compounds of MX2
Increasing covalency
Chapter 2 Slide 68 of 85
Salts of highly polarizing cations and easily polarizable anions have layered structures.
Chapter 2 Slide 69 of 85
Salts of highly polarizing cations and easily polarizable anions have layered structures.
CdCl2Cl- at ccpCd 2+ at ½ Oh sites
(alternate O layers)
van der Waals force
Chapter 2 Slide 70 of 85
CdI2I- at hcpCd 2+ at ½ Oh sites
(alternate O layer)
Chapter 2 Slide 71 of 85
MoS2Mo4+ at hcpS2- at Td sites
298pm
366pm
Chapter 2 Slide 72 of 85
Unit Cell of PerovskiteCaTiO3
AIIBIVO3
AIIIBIIIO3
Coord. #: A: 12; B: 6atom/ unit cellA: B: O= 1: 1: 3
A and O together at ccpB at 1/4 Oh holes
Chapter 2 Slide 73 of 85
Unit Cell of ReO3
Perovskite structure CaTiO3 without Ca
Chapter 2 Slide 74 of 85
Unit Cell of SpinelMgAl2O4
Normal SpinelAII[BIII]2O4, AIV[BII]2O4 , AVI[BI]2O4
e.g. NiCr2O4, Co3O4 , Mn3O4
Inverse SpinelB[AB]O4e.g. Fe3O4
O 2- at fccA at 1/8 Td holesB at 1/2 Oh holes
Chapter 2 Slide 75 of 85
Chapter 2 Slide 76 of 85
YBa2Cu3O7
Chapter 2 Slide 77 of 85
Quartz SiO2
Chapter 2 Slide 78 of 85
Chapter 2 Slide 79 of 85
Chapter 2 Slide 80 of 85
Chapter 2 Slide 81 of 85
Chapter 2 Slide 82 of 85
Mineral Ideal formulab CECc (meq/100 g)
Dioctahedral minerals Pyrophyllite Al2(Si4O10)(OH)2 0 Montmorillonite Nax(Al2-xMgx)(Si4O10)
(OH)2.zH2O 60 - 120
Beidellite Mx(Al2)(AlxSi4-xO10) (OH)2.zH2O
60 - 120
Nontronite Mx(Fe3+,Al)2(AlxSi4-xO10) (OH)2.zH2O
60 - 120
Trioctahedral minerals
Talc Mg3(Si4O10)(OH)2 0 Hectorite (Na2Ca)x/2(LixMg3-x)
(Si4O10)(OH)2.zH2O 60 - 120
Saponite Cax/2Mg3(AlxSi4-xO10) .zH2O
60 - 120
Sauconite Mx(Zn,Mg)3(AlxSi4-xO10) .zH2O
a: Only major cations are shown. b: x depends on the origin of the mineral; montmorillonites can show a degree of substitution x in the octahedral sheet in the range 0.05- 0.52. Natural samples generally show substitutions in both octahedral and tetrahedral sheets, which renders the real situations more complex. c: Cation-exchange capacity.
Chapter 2 Slide 83 of 85
Chapter 2 Slide 84 of 85
Chapter 2 Slide 85 of 85