CRYSTAL STRUCTURE

Crystal Structure

Objectives

Relationships between structures-bonding-properties of engineering materials. Arrangements in crystalline solids Give examples of each: Lattice, Crystal Structure, Unit Cell and Coordination Numbers Describe hard-sphere packing and identify cell symmetry. Define directions and planes (Miller indices) for crystals

Outline

Structure of the Atom and Atomic Bonding Electronic Structure of the Atom Lattice, Unit Cells, Basis, and Crystal Structures Points, Directions, and Planes in the Unit Cell Crystal Structures of Ionic Materials Covalent Structures

Crystal Structure

Crystal Structure

Lattice- A collection of points that divide space into smaller equally sized segments.

Unit cell - A subdivision of the lattice that still retains the overall characteristics of the entire lattice.

Atomic radius - The apparent radius of an atom, typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number).

Packing factor - The fraction of space in a unit cell occupied by atoms.

Types of Crystal Structure

Body centered cubic (BCC) Face centered cubic (FCC) Hexagonal close packed (HCP)

Crystal Structure

A number of metals are shown below with their room temperature crystal structure indicated. There are substances without crystalline structure at room temperature; for example, glass and silicone. All metals and alloys

are crystalline solids, and most metals assume one of three different lattice, or crystalline, structures as they form: body-centered cubic (BCC),

face-centered cubic (FCC), or hexagonal close-packed (HCP).

Aluminum (FCC)Chromium

(BCC)Copper (FCC) Iron (alpha)

(FCC)

Iron (gamma) (BCC) Iron (delta) (BCC)

Lead (FCC) Nickel (FCC)

Silver (FCC) Titanium (HCP)

Tungsten (BCC)

Zinc (HCP)

Number of Lattice Points in Cubic Crystal Systems

In the SC unit cell: point / unit cell = (8 corners)1/8 = 1

In BCC unit cells: point / unit cell = (8 corners)1/8 + (1 center)(1) = 2

In FCC unit cells: point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4

Crystal Structure

In SC, BCC, and FCC structures when one atom is located at each lattice point.

Relationship between Atomic Radius and Lattice Parameters

Packing Factor

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 a0 3

Crystal Structure

Density

Density of BCC iron, which has a lattice parameter of 0.2866 nm.

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

32324

/882.7)1002.6)(1054.23(

)847.55)(2(

number) sadro'cell)(Avogunit of (volume

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

cmg

Crystal Structure

Points, Directions, and Planes in the Unit Cell

Atomic configuration in Face-Centered-Cubic

Geometry

RR

R

R

a

Unit Cell: The basic structural unit of a crystal structure. A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition.

Atomic configuration in Simple Cubic

Crystal Structure

Bravais Lattices

Crystal Structure

Unit Cells Types

PrimitiveFace-Centered

Body-Centered End-Centered

A unit cell is the smallest component of the crystal that reproduces the whole crystal when stacked together with purely translational repetition.

• Primitive (P) unit cells contain only a single lattice point.• Internal (I) unit cell contains an atom in the body center.• Face (F) unit cell contains atoms in the all faces of the planes composing the cell.• Centered (C) unit cell contains atoms centered on the sides of the unit cell.

Crystal Classes (cubic, tetragonal, orthorhombic, hexagonal, monclinic, triclinic, trigonal)  with 4 unit cell types (P, I, F, C) symmetry allows for only 14  types of 3-D lattice.

Crystal Structure

Counting Number of Atoms Per Unit Cell

Counting Atoms in 3D CellsAtoms in different positions are shared by differing numbers of unit cells.

• 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.

Simple Cubic

8 atoms but shared by 8 unit cells. So, 8 atoms/8 cells = 1 atom/unit cell

How many atoms/cell forBody-Centered Cubic?And, Face-Centered Cubic?

Crystal Structure

Atomic Packing Fraction for FCC

Face-Centered-CubicArrangement

APF = vol. of atomic spheres in unit cell total unit cell vol.

No. of atoms per unit cell = Volume of one atom= Volume of cubic cell = “R” related to “a” by

4/cell

4R3/3 a3

= 0.74APF =

a3

4

3( 2a/4)34

atoms

unit cell atomvolume

unit cell

volume

2a4RUnit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

Crystal Structure

APF for BCC

Again, For BCC

APF VatomsVcell

243

a 34

3

a3

38

0.68

• BCC: a = b = c = a and angles a = b =g= 90°.• 2 atoms in the cubic cell: (0, 0, 0) and (1/2, 1/2, 1/2).

Crystal Structure

AB

C

FCC Stacking

Highlighting the faces

Highlighting the stacking

Crystal Structure

ABCABC.... repeat along <111> direction gives Cubic Close-Packing (CCP)• Face-Centered-Cubic (FCC) is the most efficient packing of hard-spheres of any lattice.• Unit cell showing the full symmetry of the FCC arrangement : a = b =c, angles all 90°• 4 atoms in the unit cell: (0, 0, 0) (0, 1/2, 1/2) (1/2, 0, 1/2) (1/2, 1/2, 0)

FCC Stacking

Crystal Structure

A

B

HCP Stacking

Highlighting the cell

Highlighting the stacking

A

Layer A

Layer A

Layer B

Crystal Structure

ABABAB.... repeat along <111> direction gives Hexagonal Close-Packing (HCP)• Unit cell showing the full symmetry of the HCP arrangement is hexagonal • Hexagonal: a = b, c = 1.633a and angles a = b = 90°, g = 120°• 2 atoms in the smallest cell: (0, 0, 0) and (2/3, 1/3, 1/2).

HCP Stacking

Crystal Structure

To define a point within a unit cell….Express the coordinates uvw as fractions of unit cell vectors a, b, and c (so that the axes x, y, and z do not have to be orthogonal).

a

b

c

origin

pt. coord.

x (a) y (b) z (c)

0 0 0

1 0 0

1 1 1

1/2 0 1/2

pt.

Crystallographic Points, Directions, and Planes.

Crystal Structure

2]21[ .4

Crystallographic directions and coordinates.

Direction B1. Two points are 1, 1, 1 and 0, 0, 02. 1, 1, 1, -0, 0, 0 = 1, 1, 13. No fractions to clear or integers to reduce4. [111]

Direction A1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 03. No fractions to clear or integers to

reduce4. [100]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

Crystallographic Points, Directions, and Planes.Crystal Structure

Procedure:1. Any line (or vector direction) is specified by 2 points.

• The first point is, typically, at the origin (000).

2. Determine length of vector projection in each of 3 axes in units (or fractions) of a, b, and c.• X (a), Y(b), Z(c) 1 1 0

3. Multiply or divide by a common factor to reduce the lengths to the smallest integer values, u v w.

4. Enclose in square brackets: [u v w]: [110] direction.

a b

c

DIRECTIONS will help define PLANES (Miller Indices or plane normal).

[1 1 0]5. Designate negative numbers by a bar • Pronounced “bar 1”, “bar 1”, “zero” direction.

6. “Family” of [110] directions is designated as <110>.

Crystallographic Points, Directions, and Planes.

Crystal Structure

Crystallographic Points, Directions, and Planes.

Figure 2.9 Crystallographic planes and intercepts

Plane B1. The plane never intercepts the z

axis, so x = 1, y = 2, and z = 2.1/x = 1, 1/y =1/2, 1/z = 0

3. Clear fractions:1/x = 2, 1/y = 1, 1/z = 0

4. (210)

Plane A1. x = 1, y = 1, z = 12.1/x = 1, 1/y = 1,1 /z = 13. No fractions to clear4. (111)

Plane C1. We must move the origin, since

the plane passes through 0, 0, 0. Let’s move the origin one lattice

parameter in the y-direction. Then, x = ∞ , y = -1, and z = ∞

2.1/x = 0, 1/y = 1, 1/z = 03. No fractions to clear.

4 (o1-o)

Crystal Structure

Miller Indices for HCP Planes

As soon as you see [1100], you will know that it is HCP, and not [110] cubic!

4-index notation is more important for planes in HCP, in order to distinguish similar planes rotated by 120o.

1. Find the intercepts, r and s, of the plane with any two of the basal plane axes (a1, a2, or a3), as well as the intercept, t, with the c axes.

2. Get reciprocals 1/r, 1/s, and 1/t.3. Convert reciprocals to smallest integers in same ratios.4. Get h, k, i , l via relation i = - (h+k), where h is associated

with a1, k with a3, i with a2, and l with c.5. Enclose 4-indices in parenthesis: (h k i l) .

Find Miller Indices for HCP:

r s

t

Crystal Structure

Miller Indices for HCP Planes

What is the Miller Index of the pink plane?

1. The plane’s intercept a1, a3 and c at r=1, s=1 and t= , respectively.

2. The reciprocals are 1/r = 1, 1/s = 1, and 1/t = 0.

3. They are already smallest integers.

4. We can write (h k i l) = (1 ? 1 0).

5. Using i = - (h+k) relation, k=–2.

6. Miller Index is

(12 11)

Crystal Structure

Linear Density in FCC

LD =Number of atoms centered on a direction vector

Length of the direction vector

Example: Calculate the linear density of an FCC crystal along [1 1 0].

ANSWERa. 2 atoms along [1 1 0]

in the cube.b. Length = 4R

ASKa. How many spheres along blue line? b. What is length of blue line?

LD110 2atoms

4R

12R

XZ = 1i + 1j + 0k = [110]

Crystal Structure

Planar Packing Density in FCC

Ra 22

R4

PD =Area of atoms centered on a given plane

Area of the plane

Example: Calculate the PPD on (1 1 0) plane of an FCC crystal.

• Find area filled by atoms in plane: 2R2

• Find Area of Plane: 8√2 R2

PPD 2R2

8 2R2

4 20.555

Hence,

Always independent of R!

Crystal Structure

n AVcNA

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

Volume/unit cell

(cm3/unit cell)Avogadro's number

(6.023 x 1023 atoms/mol)

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

Compare to actual: Cu = 8.94 g/cm3

Result: theoretical Cu = 8.89 g/cm3

Theoretical Density,

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

Crystal Structure

Crystal Structures of Ionic Materials

Factors need to be considered in order to understand crystal structures of ionically bonded solids are:

• Ionic Radii• Electrical Neutrality• Connection between Anion Polyhedra• Visualization of Crystal Structures Using Computers

The cesium chloride structure, a SC unit cell with two ions (Cs+ and CI-) per lattice point. (b) The sodium chloride structure, a FCC unit cell with two ions (Na+ + CI-) per lattice point

Crystal Structure

Crystal Structures of Ionic Materials

Fluorite unit cell, (b) plan view.The zinc blende unit cell, (b) plan view.

Crystal Structure

Crystal Structures of Ionic Materials

The perovskite unit cell showing the A and B site cations and oxygen ions occupying the face-center positions of the unit cell.

Crystal Structure

Covalent Structures

The silicon-oxygen tetrahedron and the resultant β-cristobalite form of silica

Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding

Crystal Structure

Covalent Structures

Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding

Crystal Structure