00 Crystallography Basics

Brief introduction to Crystallography • • • • •

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Crystalline vs. Amorphous Materials

Crystalline material material – atoms atoms are are situated situated in a periodic array over large distances. Amorphous (or non-crystalline non-crystalline)) mate materia riall – wher where e long range order is absent. Therefore no real symmetry. takenfrombdhuey

Crystalline vs. Amorphous Materials

Crystalline material material – atoms atoms are are situated situated in a periodic array over large distances. Amorphous (or non-crystalline non-crystalline)) mate materia riall – wher where e long range order is absent. Therefore no real symmetry. takenfrombdhuey

SIMPLE CUBIC STRUCTURE (SCC) • Close packed packed directions directions are cube edges. edges. • Coordination # = 6

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BODY CENTERED CUBIC STRUCTURE (BCC) • Close packed directions are cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing.

• Coordination # = 8

Adapted from Fig. 3.2, Callister 6e. (Courtesy P.M. Anderson)

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FACE CENTERED CUBIC STRUCTURE (FCC) • Close packed directions are face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

• Coordination Number = 12 • Atomic packing factor (APF) = 0.74 (ideal)

Adapted from Fig. 3.1(a), Callister 6e. (Courtesy P.M. Anderson)

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FCC STACKING SEQUENCE A B C

• ABCABC... Stacking Sequence along cube diagonal • 2D Projection: A B B C A B B B A sites C C B sites B B C sites takenfrombdhuey

• Coordination #= 12

HEXAGONAL CLOSE-PACKED STRUCTURE (HCP) • ABAB... Stacking Sequence • 3D Projection

Adapted from Fig. 3.3, Callister 6e.

• 2D Projection A sites

Top layer

B sites

Middle layer

A sites

Bottom layer

• Coordination # = 12

• Only difference from FCC is in the stacking (and the orientation we generally consider when describing/drawing the stacking): - ABAB along z-axis, instead of ABCABC along cube diagonal takenfrombdhuey

Lattices and Unit Cells

•

What if we want to consider more complicated crystals?

•

Take advantage of symmetry and well known structures  –  A “  ” is a repeating array of special points in space that have identical surroundings (nearest neighbors, etc).  –  A

defines a repeating unit within the lattice 

• strictly the unit cell is the smallest repeating cell (though this is sometimes totally impractical and thus more convenient, but larger, unit cells are employed).

 –  A lattice does not define the crystal, only its symmetry.  –  Atoms (or ions, molecules, etc.) will be placed at any number of points at given distances/directions (i.e. vectors) from the lattice points . ONLY THEN do we have a crystal. takenfrombdhuey

2-d Lattices • A lattice repeats in all directions, and defines the symmetry. Square unit cell

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• The unit cell is the simplest repeating structure.

“Basis” • BUT, a lattice only identifies repeating points and their symmetry.  – What’s the lattice for the wall or floor in the classroom?

• Must be combined with a “basis” to know where atoms are in the crystal  – What kind of bricks are hung at each lattice point?  – Are there smaller stones patterned regularly in between the bricks? • Only by knowing a lattice AND a basis can a crystal structure (or anything else symmetric) truly be built. takenfrombdhuey

2-d Lattice square

Lattice:

Basis:

• • •

Anions at lattice positions (0,0) Cations at (½,0) Cations at (0,½)

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tetragonal

• Lattice:  – A minimized pattern that will be followed • Basis:  – a) A vector that defines the position of hooks on which something (atoms, molecules, bricks) will be hung with relation to the underlying pattern (lattice)   – and b) What exactly will be hung on each hook (atom, molecule) • Rules:  – At least one hook (basis) per lattice point  • (but can be more than 1)

 – Basis vectors must put atoms within the unit cell (not in an adjacent cell)  – Each basis vector applies to every single lattice point  takenfrombdhuey

Seven Crystal Systems

In 3 dimensions, there are only 7 basic crystal systems. Easily defined by: 3 edge lengths (a, b and c) and 3 interaxial angles (  and ).

This class

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Cubic

a=b=c

===90°

Tetragonal

a=b≠c

===90°

Orthorhombic

a≠ b≠c

===90°

Hexagonal

a=b≠c

==90° =120°

Monoclinic

a≠ b≠c

==90°≠

Triclinic

a≠ b≠c

≠≠

Rhombohedral

a=b=c

==≠90°

Cubic Bravais Lattices

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Why no base centered cubic?

Tetragonal Bravais Lattices

Tetragonal / a=b≠c °

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/

Why no base centered tetragonal?

Orthorhombic Bravais Lattices

/ / / /

b

/ / / /

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/ /

Other Bravais Lattices

a=b°c

  90° =120°

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Fourteen Bravais Lattices (13-14)

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Defining Basis Positions Using Fractional Co-ordinates

To locate atom positions in cubic unit cells we use rectangular x, y, and z axes.

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The FCC Structure

Each position might be provided, but probably the fewest number of positions will be given—the rest fill in as long as you know the Bravais lattice (ie the basic symmetry).

Lattice=Face Centered Cubic Basis=red atoms at (0,0,0)

Note: All atoms are not always shown because technically there are only 4/unit cell (for FCC)

-of course it looks like there are more atoms, but these simply result from the repeating lattice and symmetry; ie they are not distinct takenfrombdhuey

Simple unit cells Based on Simple Cubic Lattice with a basis of: 1) anions at the SC points 2) cations at the center [offset by (½, ½, ½)]

Based on FCC Lattice with a basis of: 1) anions at the FCC points 2) cations at the center [offset by (½, 0, 0)]

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More common unit cells

Based on FCC Lattice with a basis of: 1) anions at the FCC points 2) cations at (¼,¼,¼)

Based on SC Lattice with a basis of: 1) Big cation at the SC points 2) Small cation at (½,½,½) 3) Anion at (½,½,0), (½,0,½), (0,½,½) takenfrombdhuey

AmXp STRUCTURES • Consider CaF2 : based on the CsCl structure. • BUT: there are half as many Ca2+ as F- ions—how can we accommodate this? CsCl structure w/only half of cation sites occupied. The initial ‘unit cell’ is no longer valid since every lattice site doesn’t have a basis attached to it (some are void). Instead, we use a ‘supercell.’ 2 possibilities here:

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AmXp STRUCTURES

m is not equal to p

• Consider CaF2 : Based on the CsCl structure, BUT: there are half as many Ca2+ as F- ions  – This is accommodate with ‘empty cubes’.  – i.e. CsCl based structure w/only half of cation sites occupied. • Thus, the proposed CsCl ‘unit cell’ is no longer valid since every Simple Cubic lattice site doesn’t have a basis attached to it (half are void!).  – Instead, use a ‘supercell.’  – Other possibilities:

takenfrombdhuey Adapted from Fig. 12.5, Callister 6e.

Crystallographic Directions and Planes • Terminology  – Positions

(,,,)

 – Directions

[]

note no commas

 – Planes

()

note no commas

 – Families of directions commas  – Families of planes

{}

note no

note no commas

and NOT “one bar zero zero”

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Directions in Cubic Unit Cells • For cubic crystals, the crystallographic direction indices are the vector components of the direction resolved along each of the co-ordinate axes (x,y,z) • and reduced to the

.

Position co-ordinates of vector OM = 1, 0.5, 0 (but these aren’t integers) Position co-ordinates must be multiplied by 2 to obtain integers. Direction indices of OM = 2 (1, 0.5, 0) = [2, 1, 0] A negative index direction (e.g. see vector ON) is written with a bar over the index takenfrombdhuey

[Directions]

Group exercise z

• [100] • [110] • [111] • [021]

x

y

1. Draw cell + origin 2. Draw vector 

• [200]

3. “ r educe” to smallest integer values

• [210]

4. [xyz]

• [011]

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more [Directions]

z x

y

For negative directions : a. Add more unit cells.

• [100] • [011] • [011]

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OR b. Shift the origin.

z

Directions in Cubic Unit CellsGroup exercise z

z

[00 1 ]

[0 1 1 ] [1 1 1] y

y

x

y

x

z

x

z

z

[22 1 ]

[112]

[2 1 2] y

x

y x

y x

The origin may be shifted. Thus: a) If all positive, origin is at(0,0,0)

b) If z is negative, origin will be at (x,y,1)

c) If y is negative, origin will be at (x,1,z)

d) If x is negative, origin will be at (1,y,z)

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=family of directions

z x

y

• [100] • [010] • [001] • [100] • [010] • [001]

 A includes all possible directions with the same basic coordinates. takenfrombdhuey

z

• (xyz) • (100) • (110) • (111) • (100) • (020) • (040) • (04/30) • (½½½)

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x

(Planes) y

1. Draw the origin , cell, and normal vector. 2. Draw the plane at a distance from the origin of 1/sqrt(a2+b 2+c 2).

{Planes} =family of planes

z x

• {xyz} multiplici ty • {100} • {110}

3

• {111}

6

• {100}

? ?

How many equivalent planes are there in each family (“multiplicity”)? takenfrombdhuey

y

For a simple cubic lattice and (0,0,0) basis: • Along (xyz) plane • (100) • (110) ±

z x

y

±

Should be able to draw atoms in, and above/below, a given plane in a crystal.

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For CsCl (simple cubic lattice, basis of Cl-(0,0,0) and Cs(½½,½)) • Along (xyz) plane

Group exercise

• (100)

±

• (110) ± ±

±

±

z x

y

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Crystallographic Planes –  Miller Indices

Intercept;

1,

Reciprocal;

1/1, 1/ , 1/

Simplify;

∞

,

∞

1, 0, 0

Miller indices; (1 0 0)

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1,1,

∞

∞

1,1,1

∞

1/1,1/1,1/

1/1, 1/1, 1/1

1,1,0

1,1,1

(1 1 0)

(1 1 1 )

∞

Crystallographic Planes – Miller Indices When planes are more complicated, follow these rules: 1. Choose a plane that does not pass through the origin at (0,0,0) 2. Determine the intercepts of the plane in terms of the x, y, and z axes of the cube. 1/3,2/3,1 3. Form the reciprocals of these intercepts 3, 3/2, 1 4. Bring to smallest integers by multiplying or dividing through by a common factor. 6, 3, 2 5. Enclose integers in parentheses (6 3 2) takenfrombdhuey