1

STRUCTURES

OF

SOLIDS

2

SOLIDS

can be divided into two categories.

Crystalline Amorphous

Crystalline has long range order

Amorphous materials have short range order

3

4

Crystal Type

Particles Interparticle Forces

Physical Behaviour Examples

Atomic

Molecular

Metallic

Ionic

Network

Atoms

Molecules

Atoms

Positive and negative ions

Atoms

Dispersion

Dispersion

Dipole-dipole

H-bonds

Metallic bond

Ion-ion attraction

Covalent

• Soft

• Very low mp

• Poor thermal and electrical conductors Fairly soft Low to moderate mp Poor thermal and electrical conductors Soft to hard Low to very high mp Mellable and ductile Excellent thermal and electrical conductors Hard and brittle High mp Good thermal and electrical conductors in molten condition

• Very hard

• Very high mp

• Poor thermal and electrical conductors

Group 8A

Ne to Rn

O2, P4, H2O, Sucrose

Na, Cu, Fe

NaCl, CaF2, MgO

SiO2(Quartz)

C (Diamond)

TYPES OF CRYSTALLINE SOLIDS

5

Molecular Solids Covalent Solids Ionic solids

Metallic solids

Na+

Cl-

STRUCTURES OF CRYSTALLINE SOLID TYPES

6

DIAMOND QUARTZ

GRAPHITE

7

• The unit cell and, consequently, the entire lattice, is uniquely determined by the six lattice constants: a, b, c, α, β and γ.

• Only 1/8 of each lattice point in a unit cell can actually be assigned to that cell.

• Each unit cell in the figure can be associated with 8 x 1/8 = 1 lattice point.

Unit CellUnit Cell

Lattice the underlying periodicity of the crystal

Basis Entity associated with each lattice pointsLattice how to repeat

Motif what to repeat

Crystal = Lattice + MotifMotif or Basis: typically an atom or a group of atoms associated with each lattice point

Translationally periodic arrangement of motifs

Crystal

Translationally periodic arrangement of points

Lattice

9

ONE DIMENTIONAL LATTICE

ONE DIMENTIONAL UNIT CELL

a

a

UNIT CELL : Building block, repeats in a regular way

a

10

TWO DIMENTIONAL LATTICE

11

a

ba b, 90°

a b, = 90°

a

b

a = b, = 90°

a

a

a b, = 90°

a

b

a = b, =120°

a

a

TWO DIMENTIONAL UNIT CELL TYPES

12

TWO DIMENTIONAL UNIT CELL POSSIBILITIES OF NaCl

Na+

Cl-

Five 2D lattices

a=b =90ab =90

a=b =120ab =90

ab 90, 120

unit cell

Centered

Primitive

There are literally thousands of crystalline materials, there are only 5 distinct planar lattices

14

Primitive ( P ) Body Centered ( I )

Face Centered ( F ) C-Centered (C )

LATTICE TYPES

Crystalline Solids

There are several types of basic arrangements in crystals, such as the ones shown above.

# of Atoms/Unit Cell

simple cubic unit cell = 1 atom

body centered cubic = 2 atoms

face-centered cubic = 4 atoms

Crystalline Solids

We can determine the empirical formula of an ionic solid by determining how many ions of each element fall within the unit cell.

18

BRAVAIS LATTICES

7 UNIT CELL TYPES + 4

LATTICE TYPES = 14

BRAVAIS LATTICES

19

Isometric or Cubic

20

Tetragonal

21

Hexagonal

22

Rhombohedral

23

Orthorhombic

24

Monoclinic

25

Triclinic

Summary of Crystal Structures

ab

c

The Choice of a Unit Cell: Have the highest symmetry and minimal size

a b

c

a b

c

• Rare due to low packing density (only Po – Polonium -- has this structure)• Close-packed directions are cube edges.

• Coordination No. = 6 (# nearest neighbors) for

each atom as seen

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Figure 3.15 Illustration of coordinations in (a) SC and (b) BCC unit cells. Six atoms touch each atom in SC, while the eight atoms touch each atom in the BCC unit cell.

Calculations Using Unit Cells

2) Finding the radius of the atom given the type of unit cell and the length of one side of the lattice.

(This can be a little more complicated to derive a nice formula.)

By looking at the unit cells, we can determine how the length of one side, (a), is related to the radius, (r), of one atom…

•Simple Cubic: a = 2r or r = ½a…easy to see!

•Face Centered: a2 +a2 = (4r)2 … Pythagorean’s Theorem

Simplifying… 2a2 =16r2… a2=8r2…a = r (√8)

•Body Centered: We need a better 3-D view in order to derive a formula!

a

a

a

Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point.

Example 3.2 Determining the Relationship between Atomic Radius and Lattice Parameters

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Figure 3.14 The relationships between the atomic radius and the Lattice parameter in cubic systems (for Example 3.2).

Example 3.2 SOLUTION

Referring to Figure 3.14, we find that atoms touch along the edge of the cube in SC structures.

3

40

ra

In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this length—two radii from the face-centered atom and one radius from each corner, so:

2

40

ra

ra 20

In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so

• APF for a simple cubic structure = 0.52

APF = a3

4

3(0.5a) 31

atoms

unit cellatom

volume

unit cell

volume

Atomic Packing Factor (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

Adapted from Fig. 3.23, Callister 7e.

close-packed directions

a

R=0.5a

contains (8 x 1/8) = 1 atom/unit cell Here: a = Rat*2

Where Rat is the ‘handbook’ atomic radius

• Coordination # = 8

Adapted from Fig. 3.2, Callister 7e.

(Courtesy P.M. Anderson)

• Atoms touch each other along cube diagonals.--Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

Body Centered Cubic Structure (BCC)

ex: Cr, W, Fe (), Tantalum, Molybdenum

2 atoms/unit cell: (1 center) + (8 corners x 1/8)

Calculations Using Unit Cells

There are 2 basic calculations involving unit cells:

1) Finding the density of an element given the type of unit cell and the length of one side of the lattice.

Density = mass/volume

mass =

volume = LxWxH = (length of one side)3… we will use “a” as the length of one side, so… V= a3

Putting them together…Density=

(# of atoms in the unit cell) x (1 mole)_____ (6.02 x 1023 atoms)

x (formula mass) (1 mole)

(# of atoms in the unit cell) x _(formula mass)_

(6.02 x 1023 atoms)(a3)

Remember: simple cubic = 1 atom; body centered = 2 atoms; face-centered = 4 atoms

Example 3.3 Calculating the Packing Factor

74.018)2/4(

)34

(4)( Factor Packing

24r/ cells,unit FCCfor Since,

)34

)(atoms/cell (4 Factor Packing

3

3

0

30

3

r

r

r

aa

Calculate the packing factor for the FCC cell.

Example 3.3 SOLUTION

In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is .

30a

Example: Cubic unit cell of CsCl, a=b=c ===90Cs:(0,0,0)Cl: (1/2,1/2,1/2)

Single Crystal: Composed of only one particular type of space lattice.

Polycrystalline matter: Clusters of multiple crystals.

Face Centered Cubic (FCC)

ra 42 0

a0

a0

r

r

2r

Calculations Using Unit Cells

2) Finding the radius of the atom given the type of unit cell and the length of one side of the lattice.

(This can be a little more complicated to derive a nice formula.)

By looking at the unit cells, we can determine how the length of one side, (a), is related to the radius, (r), of one atom…

•Simple Cubic: a = 2r or r = ½a…easy to see!

•Face Centered: a2 +a2 = (4r)2 … Pythagorean’s Theorem

Simplifying… 2a2 =16r2… a2=8r2…a = r (√8)

•Body Centered: We need a better 3-D view in order to derive a formula!

a

a

a

Calculations Using Unit Cells

• A body-centered lattice is slightly trickier than the face-centered lattice because our diagonal doesn’t lie on the face of the cube. Instead, it lies within the body of the cube. We will also assume that the particles come in contact with each other unlike this drawing.

a

• Solving for the triangle in blue…

c2 = a2 + b2 … (4r)2 = a2 + b2

• We don’t have a value for “b”, but we can recognize that it is also a hypotenuse of a right triangle…

b2 = a2 +a2

•Substituting… (4r)2 = a2 + a2 + a2

•Simplifying…16r2 = 3a2

•Solving for “a”… a = r(√5⅓ )

a

a

Now we get to do practice problems!

(body-centered lattice)

Body Centered Cubic (BCC)

02a

03aa0

ra 43 0

Atomic Packing Factor: BCC

a

APF =

4

3 ( 3a/4 )32

atoms

unit cell atom

volume

a3unit cell

volume

length = 4R =Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

aR

Adapted from Fig. 3.2(a), Callister 7e.

a 2

a 3

• Coordination # = 12

Adapted from Fig. 3.1, Callister 7e.

(Courtesy P.M. Anderson)

• Atoms touch each other along face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

Face Centered Cubic Structure (FCC)

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

4 atoms/unit cell: (6 face x ½) + (8 corners x 1/8)

• APF for a face-centered cubic structure = 0.74

Atomic Packing Factor: FCC

The maximum achievable APF!

APF =

4

3( 2a/4 )34

atoms

unit cell atom

volume

a3unit cell

volume

Close-packed directions: length = 4R = 2 a

Unit cell contains: 6 x 1/2 + 8 x 1/8

= 4 atoms/unit cella

2 a

Adapted fromFig. 3.1(a),Callister 7e.

(a = 22*R)

• Ex: Cr (BCC)

A = 52.00 g/mol

R = 0.125 nm

n = 2

a = 4R/3 = 0.2887 nm

aR

= a3

52.002

atoms

unit cellmol

g

unit cell

volume atoms

mol

6.023 x 1023

Theoretical Density,

theoretical

actual

= 7.18 g/cm3

= 7.19 g/cm3

THEORETICAL DENSITY, Density = mass/volume

mass = number of atoms per unit cell * mass of each atom

mass of each atom = atomic weight/avogadro’s number

n AVcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)Avogadro's number (6.023 x 1023 atoms/mol)

Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen

Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H

At. Weight (amu) 26.98 39.95 137.33 9.012 10.81 79.90 112.41 40.08 12.011 132.91 35.45 52.00 58.93 63.55 19.00 69.72 72.59 196.97 4.003 1.008

Atomic radius (nm) 0.143 ------ 0.217 0.114 ------ ------ 0.149 0.197 0.071 0.265 ------ 0.125 0.125 0.128 ------ 0.122 0.122 0.144 ------ ------

Density (g/cm3) 2.71 ------ 3.5 1.85 2.34 ------ 8.65 1.55 2.25 1.87 ------ 7.19 8.9 8.94 ------ 5.90 5.32 19.32 ------ ------

Crystal Structure FCC ------ BCC HCP Rhomb ------ HCP FCC Hex BCC ------ BCC HCP FCC ------ Ortho. Dia. cubic FCC ------ ------

Adapted fromTable, "Charac-teristics ofSelectedElements",inside frontcover,Callister 6e.

Characteristics of Selected Elements at 20C

n AVcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm3/unit cell)Avogadro's number (6.023 x 1023 atoms/mol)

Example: CopperData from Table inside front cover of Callister (see previous slide):

• crystal structure = FCC: 4 atoms/unit cell• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)• atomic radius R = 0.128 nm (1 nm = 10 cm)-7

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3

Compare to actual: Cu = 8.94 g/cm3Result: theoretical Cu = 8.89 g/cm3

THEORETICAL DENSITY,

THEORETICAL DENSITY

n AVcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell (cm3/unit cell)

Avogadro's number (6.023 x 1023 atoms/mol)

Example problem on Density Computation

Problem: Compute the density of CopperGiven: Atomic radius of Cu = 0.128 nm (1.28 x 10-8 cm)

Atomic Weight of Cu = 63.5 g/mol Crystal structure of Cu is FCC

Solution: = n A / Vc NA

n = 4

Vc= a3 = (2R√2)3 = 16 R3 √2

NA = 6.023 x 1023 atoms/mol

= 4 x 63.5 g/mol / 16 √2(1.28 x 10-8 cm)3 x 6.023 x 1023 atoms/mol

Ans = 8.98 g/cm3

Experimentally determined value of density of Cu = 8.94 g/cm3

Example 3.4 Determining the Density of BCC Iron

Determine the density of BCC iron, which has a lattice parameter of 0.2866 nm.

Example 3.4 SOLUTION

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell

Avogadro’s number NA = 6.02 1023 atoms/mol

30a

3

2324 /882.7)1002.6)(1054.23(

)847.55)(2(

number) sadro'cell)(Avogunit of (volume

iron) of mass )(atomicatoms/cell of(number Density

cmg

53

Calculate Unit-Cell Dimension from Unit-Cell Type and Density: Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm3 and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. Compare this with the value of 392.4 pm obtained from X-ray diffraction. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of

a unit cell and then the edge length of a unit cell. Mass of one Pt atom:195.08 g Pt x 1 mol Pt = 3.239 x 10-22g Pt 1 mol Pt 6.022 x 1023Pt atoms 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass per unit cell = 4 x 3.239 x 10-22 = 1.296 x 10-21 g Pt 1 unit cell Since d = m/V; the Volume of the unit cell is: V = m/d V = (1.296 x 10-21 g Pt/cell) /(21.45 g/cm3) V = 6.042 x 10-23 cm3 /cell But the Volume of the cell = a3; therefore, a = V1/3 a = ( 6.042 x 10-23 cm3)1/3 = 3.924 x 10-8 cm a = 3.924 x 10-10 m = 3.924 Angstrom = 392.4 pm This is exactly the number obtained by the x-ray diffraction experiment.

54

Calculate Atomic Mass from Unit-Cell Dimension and Density: The unit cell length for Ag was determined to be 408.6 pm. Ag crystallizes in a fcc lattice with all atoms at the lattice points. Ag has a density of 10.50 g/cm3. Calculate the mass of a Ag atom, and, using the known atomic weight (107.87 amu), calculate Avogadro's #. STRATEGY: We can calculate the unit cell volume from a. Then, from the density, we can find the mass of the unit cell and then the mass of a single Ag atom. From this, we can find Avogadro's number. The Volume of the cell = a3: V = (408.6 x 10-12 m)3 V = 6.822 x 10-29 m3 The density of silver (in g/m3) is: d =10.50 g/cm3 x (1 cm/10-2m)3 = 1.050 x 107 g/m3 The mass of one unit cell is: m = d*V m = 1.050 x 107 g/m3 x 6.822 x 10-29 m3

m = 7.163 x 10-22 g Ag/cell Because there are 4 atoms in a fcc cell, the mass of a single Ag atom in the cell (7.163 x 10-22 g Ag/cell) / (4 atoms/cell) = 1.791 x 10-22 g Ag/atom The molar mass (Avogadro's number) = N 107.87 g Ag /mol Ag = 6.023 x 1023 atoms/mol 1.791 x 10-22 g Ag/atom OR: rho = Z*M / V*N N = Z*M/rho*V = (4 atoms)(107.87 g/mol) = 10.50 g/cm3(4.086 Angstroms)3(1x10-8cm/Angstrom)3 6.023 x 1023/mol

3.6 Polymorphism and Allotropy• Polymorphism The phenomenon in some metals, as well as

nonmetals, having more than one crystal structures.

• When found in elemental solids, the condition is often called allotropy.

• Examples:

– Graphite is the stable polymorph at ambient conditions, whereas diamond is formed at extremely high pressures.

– Pure iron is BCC crystal structure at room temperature, which changes to FCC iron at 912oC.

Allotropy - The characteristic of an element being able to exist in more than one crystal structure, depending on temperature and pressure.

Polymorphism - Compounds exhibiting more than one type of crystal structure.

Allotropic or Polymorphic Transformations

POLYMORPHISM AND ALLOTROPY

BCC (From room temperature to 912 oC)

Fe

FCC (at Temperature above 912 oC)

912 oC

Fe (BCC) Fe (FCC)

d=n/2sinc

x-ray intensity (from detector)

c

• Incoming X-rays diffract from crystal planes, followingBraggs law: n = 2dsin()

• Measurement of: Critical angles, c, for X-rays provide atomic spacing, d.

Adapted from Fig. 3.2W, Callister 6e.

X-RAYS TO CONFIRM CRYSTAL STRUCTURE

reflections must be in phase to detect signal

spacing between planes

d

incoming

X-rays

outg

oing

X-ra

ys

detector

extra distance travelled by wave “2”

“1”

“2”

“1”

“2”

7.3.2 Laue equation and Bragg’s Law

1. Laue equations

Laue first mathematically described diffraction from crystals

• consider X-rays scattered from every atom in every unit cell in the crystal and how they interfere with each other

• to get a diffraction spot you must have constructive interference

Max Von Laue

Interference condition:the difference in path lengths of adjacent lattice points must be a multiple integral of the wavelength.AD-CB = h AD = a·S = acos CB = a·S0 = acos0

a(cos-cos0) = h, h=0, 1, 2 ….

Where, a— lattice parameter0—angle between a and s0

— angle between a and s

Laue equation (Based on diffraction by 1D lattice)

0

For each h value, the diffraction rays from a 1D lattice make a cone.

Expanded to 3D lattice casea·(S-S0) = a(cos-cos0) = hb·(S-S0) = b(cos-cos0) = kc·(S-S0) = c(cos-cos0) = l

where, a,b,c—lattice parameter0,0,0—angle between a,b,c and s0

,, —angle between a,b,c and sh,k,l — indices of diffraction, integers

This is the Laue Equation!

which may not be prime to each other and are different from Miller indices for crystal plane!

In the diffraction direction, the difference between the incident and the diffracted beam through any two lattice points must be an integral number of wavelengths.The vector from (000) to (mnp):Tmnp = ma + nb +pcThe differences in wavelengths: =Tmnp · (S-S0) =(ma + nb +pc) ·(S-S0)= ma ·(S-S0)+nb ·(S-S0)+pc ·(S-S0)=mh+ nk+pl=(mh+nk+pl)

2. The Bragg’s Law

Bragg discovered that you could consider the diffraction arising from reflection from lattice planes

s0 sO

dhkl

s0 sO

dhkl

Conditions to obtain constructive inferences, a. the scattered x-ray must be coplanar with the

incident x-ray and the normal of the lattice plane. b. S=S0

P

=AD+DB = 2d(hkl)sinn

Condition for diffraction:

2d(hkl) sinn = n (n=1, 2, 3, … )

n: the angle of reflection

n: the order of the reflection

2dnhnknlsinnh,nk,nl=

However, the reflection order n is not measurable!

Reformulated Bragg equations:

2dhkl sin =

s0 s

O

Adhkl

D

B

(dnhnknl = dhkl/n)

reflection indices

n=2

2d(hkl) sinn = n (n=1, 2, 3, … )

2dhkl sin = (dnhnknl = dhkl/n)

Virtual reflection plane

The Bragg equation defines the direction of diffraction beams from a given set of lattice planes!

Diffraction indices or virtual reflection plane indices.

Lattice plane indices

66

X-Ray Diffraction & Bragg Equation

• English physicists Sir W.H. Bragg and his son Sir W.L. Bragg developed a relationship in 1913 to explain why the cleavage faces of crystals appear to reflect X-ray beams at certain angles of incidence (theta, θ).This observation is an example of X-ray wave interference. Sir William Henry  Bragg (1862-1942),

William Lawrence  Bragg (1890-1971)

o 1915, the father and son were awarded the Nobel prize for physics "for their services in the analysis of crystal structure by means of Xrays".

67

Bragg Equation• Bragg law identifies the angles of the incident radiation

relative to the lattice planes for which diffraction peaks occurs.

• Bragg derived the condition for constructive interference of the X-rays scattered from a set of parallel lattice planes.

68

BRAGG EQUATION• W.L. Bragg considered crystals to be made up of parallel planes of

atoms. Incident waves are reflected specularly from parallel planes of atoms in the crystal, with each plane is reflecting only a very small fraction of the radiation, like a lightly silvered mirror.

• In mirrorlike reflection the angle of incidence is equal to the angle of reflection.

өө

69

Diffraction Condition

• The diffracted beams are found to occur when the reflections from planes of atoms interfere constructively.

• We treat elastic scattering, in which the energy of X-ray is not changed on reflection.

70

Bragg Equation

• When the X-rays strike a layer of a crystal, some of them will be reflected. We are interested in X-rays that are in-phase with one another. X-rays that add together constructively in x-ray diffraction analysis in-phase before they are reflected and after they reflected.

Incident angleReflected angleWavelength of X-ray

Total DiffractedAngle

2

2

71

The line CE is equivalent to the distance between the two layers (d)

Bragg Equation

• These two x-ray beams travel slightly different distances. The difference in the distances traveled is related to the distance between the adjacent layers.

• Connecting the two beams with perpendicular lines shows the difference between the top and the bottom beams.

sinDE d

72

Bragg Law

• The length DE is the same as EF, so the total distance traveled by the bottom wave is expressed by:

• Constructive interference of the radiation from successive planes occurs when the path difference is an integral number of wavelenghts. This is the Bragg Law.

sinEF d

sinDE d

2 sinDE EF d

2 sinn d

73

Bragg Equation

where, d is the spacing of the planes and n is the order of diffraction.

• Bragg reflection can only occur for wavelength

• This is why we cannot use visible light. No diffraction occurs when the above condition is not satisfied.

• The diffracted beams (reflections) from any set of lattice planes can only occur at particular angles pradicted by the Bragg law.

nd sin2

dn 2

74

Scattering of X-rays from adjacent lattice points A and B

X-rays are incident at an angle on one of the planes of the set.

There will be constructive interference of the waves scattered from the two successive lattice points A and B in the plane if the distances AC and DB are equal.

A B

CD

2

75

Constructive interference of waves scattered from the

same planeIf the scattered wave makes the same angle to the plane as the

incident wave

The diffracted wave looks as if it has been reflected from the plane

We consider the scattering from lattice points rather than atoms because it is the basis of atoms associated with each lattice point that is the true repeat unit of the crystal; The lattice point is analoque of the line on optical diffraction grating and the basis represents the structure of the line.

76

Diffraction maximum Coherent scattering from a single plane is not

sufficient to obtain a diffraction maximum. It is also necessary that successive planes should scatter in phase

• This will be the case if the path difference for scattering off two adjacent planes is an integral number of

wavelengths

nd sin2

77

Labelling the reflection planes

• To label the reflections, Miller indices of the planes can be used.

• A beam corresponding to a value of n>1 could be identified by a statement such as ‘the nth-order reflections from the (hkl) planes’.

• (nh nk nl) reflection

Third-order reflection from (111) plane

(333) reflection

78

n-th order diffraction off (hkl) planes

• Rewriting the Bragg law

which makes n-th order diffraction off (hkl) planes of spacing ‘d’ look like first-order diffraction off planes of spacing d/n.

• Planes of this reduced spacing would have Miller indices (nh nk nl).

sin2

n

d

79

X-ray structure analysis of NaCl and KCl

The GENERAL PRINCIBLES of X-RAY STRUCTURE ANALYSIS to DEDUCE the STRUCTURE of NaCl and KCl

Bragg used an ordinary spectrometer and measured the intensity of specular reflection from a cleaved face of a crystal

found six values of for which a sharp peak in intensity occurred, corresponding to three characteristics wavelengths (K,L and M x-rays) in

first and second order (n=1 and n=2 in Bragg law)

• By repeating the experiment with a different crystal face he could use his eqn. to find for example the ratio of (100) and (111) plane spacings, information that confirmed the cubic symmetry of the atomic arrangement.

80

Details of structure Details of structure were than deduced from the differences between the

diffraction patterns for NaCl and KCl.

• Major difference; absence of (111) reflection in KCl compared to a weak but detectable (111) reflection in NaCl.

This arises because the K and Cl ions both have the argon electron shell structure and hence scatter x-rays almost equally whereas Na and Cl ions have different scattering strengths. (111) reflection in NaCl corresponds to one wavelength of path difference between neighbouring (111) planes.

81

Experimental arrangements for x-ray

diffraction• Since the pioneering work of Bragg, x-ray

diffraction has become into a routine technique for the determination of crsytal structure.

82

Bragg Equation

Since Bragg's Law applies to all sets of crystal planes, the lattice can be deduced from the diffraction pattern, making use of general expressions for the spacing of the planes in terms of their Miller indices. For cubic structures

Note that the smaller the spacing the higher the angle of diffraction, i.e. the spacing of peaks in the diffraction pattern is inversely proportional to the spacing of the planes in the lattice. The diffraction pattern will reflect the symmetry properties of the lattice.

2 sind n

2 2 2

ad

h k l

83

Bragg Equation A simple example is the difference between the

series of (n00) reflections for a simple cubic and a body centred cubic lattice. For the simple cubic lattice, all values of n will give Bragg peaks.

However, for the body centred cubic lattice the (100) planes are interleaved by an equivalent set at the halfway position. At the angle where Bragg's Law would give the (100) reflection the interleaved planes will give a reflection exactly out of phase with that from the primary planes, which will exactly cancel the signal. There is no signal from (n00) planes with odd values of n. This kind of argument leads to rules for identifying the lattice symmetry from "missing" reflections, which are often quite simple.

84

Types of X-ray camera

There are many types of X-ray camera to sort out reflections from different crystal planes. We will study only three types of X-ray photograph that are widely used for the simple structures.

1.Laue photograph

2.Rotating crystal method

3.Powder photograph

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X-RAY DIFFRACTION METHODS

X-Ray Diffraction Method

Laue Rotating Crystal Powder

OrientationSingle Crystal

Polychromatic BeamFixed Angle

Lattice constantSingle Crystal

Monochromatic BeamVariable Angle

Lattice ParametersPolycrystal (powdered)Monochromatic Beam

Variable Angle

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CLOSE-PACKING OF SPHERES

87

Example Problem:

What is the packing efficiency in the simple cubic cell of CsCl? What is the percentage of empty space in the unit cell? The chloride ions are at the corners with the cesium in the middle of the unit cell.rCl- = 181 pm; rCs+ = 169 pm

efficiencypacking100%xV

V

cell)unitin(total

cell)unitin(particles

Example Problem

What is the packing efficiency of NaCl? What is the percentage of empty space in the NaCl unit cell? )

rCl- = 181 pm; rNa+ = 98 pm; edge dist.NaCl = 562.8 pm

efficiencypacking100%xV

V

cell)unitin(total

cell)unitin(particles

90

Hard spheres touch along cube diagonal cube edge length, a= 4R/3

The coordination number, CN = 8

Number of atoms per unit cell, n = 2Center atom not shared: 1 x 1 = 18 corner atoms shared by eight cells: 8 x 1/8 = 1

Atomic packing factor, APF = 0.68

Corner and center atoms are equivalent

Body-Centered Cubic Crystal Structure (II)

a

91

Close-packing-HEXAGONAL coordination of each sphere

SINGLE LAYER PACKING

SQUARE PACKING CLOSE PACKING

92

TWO LAYERS PACKING

93

THREE LAYERS PACKING

94

95

Hexagonal close packing Cubic close packing

96

Cubic close packing

4 atoms in the unit cell (0, 0, 0) (0, 1 /2,

1 /2) (1 /2, 0, 1 /2) (

1

/2, 1 /2, 0)

Hexagonal close packing

2 atoms in the unit cell (0, 0, 0) (2/3,

1 /3, 1 /2)

74% Space is occupied

Coordination number = 12

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NON-CLOSE-PACKED STRUCTURES

68% of space is occupied

Coordination number = 8

a) Body centered cubic ( BCC )

b) Primitive cubic ( P)

52% of space is occupiedCoordination number = 6

98

6

ABCABC…12Cubic close packed

ABABAB…12Hexagonal close packed

ABABAB…8Body-centered Cubic

AAAAA…Primitive Cubic

Stacking pattern

Coordination number

Structure

Non-close packing

Close packing

99

8 12Coordination

number 6

Primitive cubic Body centered cubic Face centered cubic

100

101

ALLOTROPES

Existence of same element in different crystal structures.

eg. Carbon

Diamond Graphite Buckminsterfullerene

102

TETRAHEDRAL HOLES

OCTAHEDRAL HOLES

TYPE OF HOLES IN CLOSE PACKING

103

LOCATION OF OCTAHEDRAL HOLES IN CLOSE PACKING

104

LOCATION OF TETRAHEDRAL HOLES IN CLOSE PACKING

105

Ionic structures may be derived from the

occupation of holes by oppositely charged

ions (interstitial sites) in the close-packed

arrangements of ions.

IONIC CRYSTAL STRUCTURES

106

Radius ratio Coordinate number

Holes in which positive ions pack

0.225 – 0.414 4 Tetrahedral holes

0.414 – 0.732 6 Octahedral holes

0.732 – 1 8 Cubic holes

Hole Occupation - RADIUS RATIO RULE

Radius of the positive ion

Radius ratio =

Radius of the negative ion

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IONIC CRYSTAL TYPES

Ionic crystal type

Co-ordination number

A X

Structure type

AX

AX2

AX3

6 6

8 8

6 3

8 4

6 2

NaCl

CsCl

Rutile(TiO2)

Fluorite (CaF2)

ReO3

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a) ROCK SALT STRUCTURE (NaCl)

• CCP Cl- with Na+ in all Octahedral holes

• Lattice: FCC

• Motif: Cl at (0,0,0); Na at (1/2,0,0)

• 4 NaCl in one unit cell

• Coordination: 6:6 (octahedral)

• Cation and anion sites are topologically identical

STRUCTURE TYPE - AX

CLOSE PACKED STRUCTURES

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• CCP S2- with Zn2+ in half Tetrahedral holes ( T+ {or T-}

filled)

• Lattice: FCC

• 4 ZnS in one unit cell

• Motif: S at (0,0,0); Zn at (1/4,1/4,

1/4)

• Coordination: 4:4 (tetrahedral)

• Cation and anion sites are topologically identical

b) SPHALERITE OR ZINC BLEND (ZnS) STRUCTURE

110

• HCP with Ni in all Octahedral holes

• Lattice: Hexagonal - P

• Motif: 2Ni at (0,0,0) & (0,0,1/2) 2As at (2/3,1/3,

1/4)

& (1/3,2/3,

3/4)

• 2 NiAs in unit cell

• Coordination: Ni 6 (octahedral) : As 6 (trigonal

prismatic)

c) NICKEL ARSENIDE (NiAs)

111

• HCP S2- with Zn2+ in half Tetrahedral holes ( T+ {or T-}

filled )

• Lattice: Hexagonal - P

• Motif: 2 S at (0,0,0) & (2/3,1/3,

1/2); 2 Zn at (2/3,1/3,

1/8) &

(0,0,5/8)

• 2 ZnS in unit cell

• Coordination: 4:4 (tetrahedral)

d) WURTZITE ( ZnS )

112

COMPARISON OF WURTZITE AND ZINC BLENDE

113

STRUCTURE TYPE - AX

NON – CLOSE PACKED STRUCTURES

CUBIC-P (PRIMITIVE) ( eg. Cesium Chloride ( CsCl ) )

• Motif: Cl at (0,0,0); Cs at (1/2,1/2,

1/2) • 1 CsCl in one unit cell • Coordination: 8:8 (cubic) • Adoption by chlorides, bromides and iodides of larger cations, • e.g. Cs+, Tl+, NH4

+

114

• CCP Ca2+ with F- in all Tetrahedral holes

• Lattice: fcc

• Motif: Ca2+ at (0,0,0); 2F- at (1/4,1/4,

1/4) & (3/4,3/4,

3/4)

• 4 CaF2 in one unit cell

• Coordination: Ca2+ 8 (cubic) : F- 4 (tetrahedral)

• In the related Anti-Fluorite structure Cation and

Anion positions are reversed

STRUCTURE TYPE - AX2

CLOSE PACKED STRUCTURE eg. FLUORITE (CaF2)

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• CCP Ca2+ with F- in all Tetrahedral holes

• Lattice: fcc

• Motif: Ca2+ at (0,0,0); 2F- at (1/4,1/4,

1/4) & (3/4,3/4,

3/4)

• 4 CaF2 in one unit cell

• Coordination: Ca2+ 8 (cubic) : F- 4 (tetrahedral)

• In the related Anti-Fluorite structure Cation and

Anion positions are reversed

STRUCTURE TYPE - AX2

CLOSE PACKED STRUCTURE eg. FLUORITE (CaF2)

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ALTERNATE REPRESENTATION OF FLUORITE STRUCTURE

Anti–Flourite structure (or Na2O structure) – positions of

cations and anions are reversed related to Fluorite structure

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RUTILE STRUCTURE, TiO2

• HCP of O2- ( distorted hcp or Tetragonal)

• Ti4+ in half of octahedral holes

118

• HCP of Iodide with Cd in Octahedral holes of alternate layers

• CCP analogue of CdI2 is CdCl2

STRUCTURE TYPE - AX2

NON-CLOSE PACKED STRUCTURE

LAYER STRUCTURE ( eg. Cadmium iodide ( CdI2 ))

119

COMPARISON OF CdI2 AND NiAs

120

HCP ANALOGUE OF FLOURITE

(CaF2) ?

• No structures of HCP are known with all Tetrahedral sites (T+ and T-) filled. (i.e. there is no HCP analogue of the Fluorite/Anti-Fluorite Structure).

• The T+ and T- interstitial sites above and below a layer of close-packed spheres in HCP are too close to each other to tolerate the coulombic repulsion generated by filling with like-charged species.

Unknown HCP analogue of Fluorite

Fluorite

121

HOLE FILLING IN CCP

122

Formula Type and fraction of sites occupied

CCP HCP

AX All octahedral

Half tetrahedral (T+ or T-)

Rock salt (NaCl)

Zinc Blend (ZnS)

Nickel Arsenide (NiAs)

Wurtzite (ZnS)

AX2 All Tetrahedral

Half octahedral (ordered framework)

Half octahedral (Alternate layers full/ empty)

Fluorite (CaF2),

Anti-Fluorite (Na2O)

Anatase (TiO2)

Cadmium Chloride (CdCl2)

Not known

Rutile (TiO2)

Cadmium iodide (CdI2)

A3X All octahedral & All Tetrahedral

Li3Bi Not known

AX3 One third octahedral YCl3 BiI3

SUMMARY OF IONIC CRYSTAL STRUCTURE TYPES

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Rock salt(NaCl) – occupation of all octahedral holes

• Very common (in ionics, covalents & intermetallics )

• Most alkali halides (CsCl, CsBr, CsI excepted)

• Most oxides / chalcogenides of alkaline earths

• Many nitrides, carbides, hydrides (e.g. ZrN, TiC, NaH)

Fluorite (CaF2) – occupation of all tetrahedral holes

• Fluorides of large divalent cations, chlorides of Sr, Ba

• Oxides of large quadrivalent cations (Zr, Hf, Ce, Th, U)

Anti-Fluorite (Na2O) – occupation of all tetrahedral holes

• Oxides /chalcogenides of alkali metals

Zinc Blende/Sphalerite ( ZnS ) – occupation of half tetrahedral holes

• Formed from Polarizing Cations (Cu+, Ag+, Cd2+, Ga3+...) and Polarizable Anions (I-, S2-, P3-, ...)

e.g. Cu(F,Cl,Br,I), AgI, Zn(S,Se,Te), Ga(P,As), Hg(S,Se,Te)

Examples of CCP Structure Adoption

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Examples of HCP Structure Adoption

Nickel Arsenide ( NiAs ) – occupation of all octahedral holes

• Transition metals with chalcogens, As, Sb, Bi e.g. Ti(S,Se,Te);

Cr(S,Se,Te,Sb); Ni(S,Se,Te,As,Sb,Sn)

Cadmium Iodide ( CdI2 ) – occupation half octahedral (alternate) holes

• Iodides of moderately polarising cations; bromides and chlorides of strongly polarising cations. e.g. PbI2, FeBr2, VCl2

• Hydroxides of many divalent cations. e.g. (Mg,Ni)(OH)2

• Di-chalcogenides of many quadrivalent cations . e.g. TiS2, ZrSe2, CoTe2

Cadmium Chloride CdCl2 (CCP equivalent of CdI2) – half octahedral holes

• Chlorides of moderately polarising cations e.g. MgCl2, MnCl2

• Di-sulfides of quadrivalent cations e.g. TaS2, NbS2 (CdI2 form as well)

• Cs2O has the anti-cadmium chloride structure

125

PEROVSKITE STRUCTURE

Formula unit – ABO3

CCP of A atoms(bigger) at the corners

O atoms at the face centers

B atoms(smaller) at the body-center

126

• Lattice: Primitive Cubic (idealised structure)

• 1 CaTiO3 per unit cell

• A-Cell Motif: Ti at (0, 0, 0); Ca at (1/2, 1/2,

1/2); 3O at (1/2, 0, 0), (0, 1/2, 0), (0, 0, 1/2)

• Ca 12-coordinate by O (cuboctahedral)

• Ti 6-coordinate by O (octahedral) • O distorted octahedral (4xCa + 2xTi)

PEROVSKITE

• Examples: NaNbO3 , BaTiO3 ,

CaZrO3 , YAlO3 , KMgF3

• Many undergo small distortions: e.g. BaTiO3 is ferroelectric

Problem (text)

Cobalt(II) oxide is used as a pigment in pottery. It has the same type of crystal structure as NaCl. When exposed to X-rays (=153 pm) reflections were observed at 42.38°, 65.68°, and 92.60°. Determine the values of n to which these reflections correspond, and calculate the spacing between the crystal layers.

nλ = 2dsin θ

128

SPINEL STRUCTURE

Formula unit AB2O4 (combination of Rock Salt and Zinc Blend Structure)

Oxygen atoms form FCC

A2+ occupy tetrahedral holes

B3+ occupy octahedral holes

INVERSE SPINEL

A2+ ions and half of B3+ ions occupy octahedral holes

Other half of B3+ ions occupy tetrahedral holes

Formula unit is B(AB)O4