UNIT -I

Crystallography

Crystal Structures

&

Matter

SOLIDS LIQUIDS GASES

CRYSTALLINE

AMORPHOUS

(Solidification occurs gradually – crystal)

(Solidification occurs suddenly – Amorphous)

Crystal is a substance in which the atoms or molecules are arranged in a definite, repeating pattern in three dimension.

Crystalline solid

Single crystal has an atomic structure that repeats periodically across its whole volume. Even at infinite length scales, each atom is related to every other equivalent atom in the structure by translational symmetry.

Single Crystal

Single Pyrite Crystal

AmorphousSolid

Polycrystal is a material made up of an aggregate of many small single crystals.(also called crystallites or grains)

The grains are usually 100 nm - 100 microns in diameter.

Polycrystals with grains that are < 10 nm in diameter are called nano-crystalline

Polycrystal PolycrystallinePyrite form

(Grain)

Amorphous (OR) isotropic solid is composed of randomly oriented atoms, ions (OR) molecules that do not form defined patterns (OR) lattice structures.

Crystal?Just a solid

A three dimensional translational periodic arrangement of atoms in space is called a crystal.

Crystal means…

The periodic array of atoms, ions, or molecules that form the solids is called Crystal Structure.

(OR)

For better understanding of crystal structure, we must know the following technical terms of crystals like….. Lattice Space Lattice Basis Unit cell Lattice parameters Lattice constant Atomic packing factor ( APF ) Co-ordination number…….

Space latticeA three dimensional translational periodic arrangement of points in space is called a lattice.

(OR)An infinite array of points in three dimensional in which every point has an identical environment to all others is called space lattice.

Basis (or) Motif

An atom or a group of atoms associated with each lattice point is called as basis or motif.

Crystal = Space Lattice + Motif (basis)

What is the relation between the lattice, basis & the crystal ?

Lattice

+

+ Basis

=

Crystal structure=

Lattice

Basis

The underlying periodicity of the crystal

How to repeat

Atom (OR) group of atoms associated with each lattice points

What to repeat

Facts about crystal cells

A cell is a finite representation of the infinite lattice.

A cell is a parallelogram (2D) or a parallelopiped (3D) with lattice points at their corners.

1-D Lattice

∞ ∞

2-D Lattice

∞

∞

∞

∞

3-D Lattice∞

∞

∞

∞

∞

∞

The smallest volume from which the entire crystal is build up by

translational repetition in three dimensions, called as unit cell.

It is the fundamental elementary pattern.

UNIT CELL

i.e.

X

Y

Z

Primitive cellIf the lattice points are only at the corners, the

cell is a primitive cell.

Non-primitive cell If there are lattice points in the cell other than

the corners, the cell is non-primitive cell.

Example : Unit cell of a simple cubic structure.

Lattice parameters

X

Y

Z

αβ

γ a

c

b

Since, a, b, & c --- the primitives of the unit cell. AND α, β, & γ --- the interfacial angles.

The primitives( a,b,c) and the interfacial angles(α,β,γ) are called the lattice parameters.

Lattice parameters decide the actual size and shape of

the unit cell.

Relation between latti ce constant(a)And crystal density(ρ)

Volume of the unit cell = a3

Therefore, mass of the unit cell = volume x density m = a3ρ -------------(1)

Let say ‘n’ molecules / unit cell , M is the molecular weight and N is the Avogadro number.

Then mass of the each molecule = M/N

Mass of the each unit cell = n x M/N -----------(2)

Therefore, Eq.(1) = Eq.(2)

a3ρ = n x M/N

Lattice constant(a) = (nM/Nρ)1/3 ---------(3)

Eq.(3) gives the relation between lattice constant & crystal density.

Stacking sequenceA

BC

The arrangement of atoms in the crystal.

Square packing

CONCLUSION: Not most space efficient

Hexagonal packing

CONCLUSION: Most space efficient

The simplest repeating motif

Unit Cells

The simplest repeating motif can be different shapes and sizes.

BRAVAIS LATTICELattices are classified on the basis of

their symmetry

In 1850, M. A. Bravais showed that identical

points can be arranged spatially to produce

14 types of regular pattern. These 14 space

lattices are known as ‘Bravais lattices’.

M. A. Bravais

14 types of space lattices in the 7 systems of crystal, which are called Bravais lattices.

Cubic 3Tetragonal 2Orthorhombic 4Monoclinic 2Triclinic 1Rhombohedral 1Hexagonal 1

07

14

1. Cubic crystals(3)Three lengths of the unit cell are same and they are at right angles.

i.e. a = b = c &

α = β = γ = 900

2. Tetragonal crystals(2)Two lengths of the unit cell are equal third one is longer and all are at right.

i.e. a = b ≠ c &

α = β = γ = 900

Y

Z

αβ

γ a

c

b

X

X

Z

b

c

aαβγ

Y

Ex: NaCl,Ag,Au,CaF,Pb,Cu,NaClo2 etc

Ex: Tio2 , Sno2 , KH2Po4

3. Orthorhombic crystals(4)Lengths of the unit cell are different, but they are at right angles.

i.e. a ≠ b ≠ c &

α = β = γ = 900

4. Monoclinic crystals(2)Lengths of the unit cell are different, two axes are at right angles and third one is obliquely inclined.

i.e. a ≠ b ≠ c &

α = β = 900 ≠ γ

XZ

Y

a

c

b

αβγ

Ex: C15H20O2, PbCo3, BaSo4, KNo3, K2So4, α-S

Ex: CaSo4 2H20(Gypsum), K2MgSo4 .6H2o

5. Triclinic crystal(1)Lengths of the unit cell are different, and they are obliquely inclined to each other.

i.e. a ≠ b ≠ c &

α ≠ β ≠ γ ≠ 900

6. Rhombohedral(Trigonal)crystal(1)Lengths of the unit cell are equal, and axes are equally inclined to eachother at an angle other than 900.

i.e. a = b = c &

α = β = γ ≠ 900

Ex: K2Cr2O7, CuSo4 5H2o

Ex: Calcite, As, Sb, Bi etc

7. Hexagonal crystal(1)Two axes of the unit cell are equal in length in one plane at 1200 with each other and third axis is perpendicular to this plane.

i.e. a = b ≠ c &

α = β = 900 , γ = 1200

CONCLUSIONBravais lattices are explained on the basis of lattice parameters only.

Ex: Quartz, Zn, Cd etc

a=2r b=2r

c = 4r(2/3)1/2

a = 90o

b=90o

g=120o

PRIMITIVES(a, b, c)

INTERFACIAL(α,β,γ) NAME OF THE CRYSTAL No. OF LATTICES

/ TYPES EXAMPLE

a = b = c α = β = γ = 900 Cubic 3 P,I,F

CsCl, NaCl

a = b ≠ c α = β = γ = 900Tetragonal

2P,I

TiO2 (Rutile), SnO2 (Cassiterite)

a ≠ b ≠ c α = β = γ = 900Orthorhombic

4P,I,F,C

KNO3,BaSO4 (Baryte)

a = b = c α = β = γ <1200 ≠900

Rhombohedral (Trigonal)

1P Ice, Al2O3

a = b ≠ c α = β = 900 ; γ= 1200 Hexagonal 1

P Mg, Zn, PbI2

a ≠ b ≠ c α =γ = 900 ; β ≠ 900 Monoclinic

2P,C

Monoclinic Sulphur,

Na2SO4.10H2O

a ≠ b ≠ c α ≠ β ≠ γ ≠ 900 Triclinic 1P CuSO4.5H2O

P: simple cubic I: body centered F: face centered C: base centered

ATOMIC RADIUS

Half the distance between the nearest neighbouring atoms in a crystal is called atomic radius.

A B

AB = X

Therefore, r = X/2

Since, r – atomic radius

Nearest neighbour distance(2r)

The distance between the centers of two nearest neighbouring atoms is called nearest neighbour distance. It will be 2r if r is the radius of the atom.

Co-ordination number

The number of equidistant nearest neighbours that an atom has in the given structure is known as co-ordination number.

NOTE:

By the help of co-ordination number we may say that given structure is either closely packed (or) loosely packed structure.

Atomic packing fractionThe ratio between the volume occupied by the atomsin the unit cell (v) to the total volume of the same unit cell (V) is called the atomic packing factor.

i.e. APF = v/VNOTE:

From APF value also we must say that the given Structure is either closed (OR) loosely packed structure.

SIMPLE CUBIC STRUCTURE(SC)1. All atoms are kept at corners of the unit cell only, consequently these atoms touch along cube edges.

a

Each atom has only 6 nearest neighbours . Hence, co-ordination number is six.

Because each corner atom has four neighbours in the same plane one vertical above and one immediately below.

2. Co-ordination number

+X

+Z

+Y

-X

-Y

-Z

R/A

a

3. Each corner atom is shared by 8 surrounding unit cells, share of each corner of unit cell comes to one-eight of an atom. Hence, each unit cell contains eight corner atoms. So the unit cell of simple cube contains only 1 atom.

i.e. 8 x 1/8 = 1 atom

How many atoms…….

So in this case, the unit cell is equal to the primitive cell.

4. Therefore, the total number of atoms in a unit cell is one.

5. It is a 1-layer structure. Hence, it’s stacking sequence is A A A A A …….

6. Atomic packing factor

APF = v / V

= 1x 4/3 π R3

a3

1x 4/3 π R3

(2R)3 =

since, a =2R

= π / 6 = 0.523 OR 52%Therefore, APF

It’s APF value is only 52%.hence, it is a loosely packed structure.

Ex: Only polonium (Po) exhibits in a certain temperature region.

BODY CENTERED CUBIC STRUCTURE(BCC)1. Atoms are kept at corners of the unit cell including one atom exactly centre of the unit cell.

Consequently these atoms touch along the body diagonals, hence the nearest concept is along the body diagonals.

In a crystal it appears as

Hence ,each atom has only 8 nearest neighbours.Hence, its co-ordination number is 8.

2. Co-ordination number

The nearest neighbouring atoms of any case are the body centered atoms, not the other corner atoms.

3. Each corner has one atom, but shares only one-eight part of an atom like simple cubic structure & one more atom exactly occupies center of the body.

Hence, it is known as Body centered cubic structure.

4. Total number of atoms in a unit cell of BCC is 2.

i.e. 8 x 1 /8 + 1 = 1 + 1 = 2 atoms

5. It is a 2-layer structure. Hence, it’s stacking sequence is AB AB AB AB AB AB …………….

6. Atomic packing factor

A

D

C B

a

Figure 1

r

2rr

D

A

Figure 2

From figure 1,

∆ ABC AC2 = AB2 + BC2 = a2 + a2 = 2a2 --------- (1)

∆ ACD AD2 = AC2 + CD2 = 2a2 + a2 = 3a2 AD2 = 3a2 (OR) AD = √3 a ------------- (2)

If ‘r‘ is the atomic radius, from figure 2 we can write AD = 4r --------(3)From eq.2 & 3, we get,

√3 a = 4r

r = √3 a / 4 ---------(4)

Therefore, APF = v / V 2 x 4/3 π r3

a3=

= 8π / 3a3 ( √3 a/4)3

= √3π /8 = 0.68 = 68%

Therefore, it’s APF is 68% .Hence we can say that it is also a loosely packed structure.

Since, r = √3 a/4

EX: Tungsten, Sodium, Iron, Chromium etc

FACE CENTERED CUBIC STRUCTURE(FCC)

AB

C

1. It is a 3-layer structure. Hence, it’s stacking sequence is ABC,ABC,ABC…..

AB

C

In a unit cell a 3-layer structure follows as…

2. Eight atoms are occupied at eight corners of a unit cell including six atoms are situated exactly center of it’s face.

3. Atoms are in contact with face diagonals, hence the nearest neighbours of any corner atom are the face centered atoms of the surrounding unit cells.

4. Each atom has 12 nearest neighbours hence it’s co-ordination number is 12.

Any corner atoms has 4 such atoms in it’s own plane, 4 in a plane above it and 4 in a plane below.

5. Total number of atoms in a unit cell are 4.

8 x 1/8 + 6 x ½ = 1 + 3 = 4 atoms

A

B

C

6. Atomic packing factor

From figure,

∆ ABC AC2 = AB2 + BC2 = a2 + a2 = 2a2 ---- (1)

But, AC = r + 2r + r = 4r --- (2)

Substituting the eq.2 in eq.1, we get,

(4r)2 = 2a2 r2 = 2a2 /16 = a2/8 Therefore, r = a / √8 --- (3)

Therefore, APF = v / V 4 x 4/3 π r3

a3=

4 x 4 x π x a3

(√8)3 x 3 x a3=

16 x π x a3

(√2)16 x 3 x a3=

π 3(√2)

=

0.74 = 74% =Therefore, APF

Therefore, it’s APF value is 74%. Hence, we can say that it is a closely packed structure.

EX: Most of the metals like Copper, Lead, Aluminium, Silver etc

DIAMOND CUBIC STRUCTURE1. The interpenetration of two FCC sub-lattices give the diamond cubic structure.

2. One sub-lattice say X has it origin at the point (0,0,0) and the other sub-lattice say Y has its origin quarter of the way along the body diagonal(i.e. a/4, a/4, a/4).

X

Y

ZPa/4

a/4

a/4

3. There are 8 corner atoms, 6 face centered atoms & 4 an interstitial atoms in a unit cell of a diamond cubic structure.

Interstitial atom (OR) Void

An interstitial void An interstitial atom

4. Its co-ordination number is 4

5. Eight corner atoms share one-eight part, six face centered atoms share half part of an atom, extra four interstitial atoms are there inside a unit cell of a diamond cubic structure.

i.e. 1/8 x 8 + 6 x 1/2 + 4 = 8 atoms6. Atomic packing factor

Y

X

Z

a/4

a/4

a/4

P

From figure,

∆ XPZ XZ2 = XP2 + PZ2 = (a/4)2 + (a/4)2 = a2 /8 --------- (1)

From ∆ XYZ XY2 = XZ2 + YZ2 = (a2/8) + a2/16 = 3a2/16 --------- (2)

But, XY = 2r ----------------(3)

Substituting eq(3) in eq (2), we get,

(2r) 2 = 3a2/16 (OR) 2r = √3 a/4 Therefore, a = 8r/√3 ----------- (4)

Therefore, APF = v / V 8 x 4/3 π r3

a3=

32 π r3

3(8r /√3)3 =

π √3

16 = = 0.34 (OR) 34%Therefore, APF

From the value of APF we can say that this structure is a loosely packed structure.

EX: Germanium, Silicon, Carbon etc

Ionic crystals consist of the negative and positive ions, attracted to each other.

IONIC CRYSTALS

There are no free electrons, ionic crystals are insulators.

The binding between the ions is mostly electrostatic and rather strong (binding energies around 1000 kJ/mol); it has no directionality

Many salts and oxides have this structure, e.g. KCl, AgBr, KBr, PbS, Or MgO, FeO, ...

1. This structure consists of two FCC sub-lattices, with two atoms in the base: one at (0, 0, 0), the other one at (a/2, 0, 0).

NaCl-structure

NaCl

2. The structure follows as..

Cl Na+

3. In 3-dimension the NaCl structure is …

4. Its co-ordination number is 6.

NaCI

CI

CI

CINa

Na

Na

Na

Cl

Its co-ordination number is 6.

5. Each unit cell in NaCl has 4 sodium ions and 4 associated chloride ions. Thus there are 4 molecules in a unit cell.

6. In NaCl crystal the distance between adjacent atoms are given by d = (M/2Nρ)1/3 cm

M ----- mass of the moleculeN ----- No. of moleculesΡ ----- density of NaCl

CsCl-structure

Cl

Cs(Almost all same in size)

1. Cesium ion is the body centre of the cube and the chlorine ion is present at each corner of the cube as shown in figure.

2. It is similar to BCC structure . Hence, its co-ordination number is 8 because each Cs ion have 8 nearest Cl ions and vice-versa.

3. Cs and Cl ions have approximately the same size.

The lattice points of CsCl are two interpenetrating simple cubic lattices.(i.e. One sub-lattice is at corner occupied by Cl and

second is at body centre by Cs.)

Ex: CaBr, Cal, NH4Cl

ZnS-structure(Zinc blende)1. ZnS structure is the interpenetration of two FCC sub-lattices, similar to the diamond cubic structure.

Cl

2. In diamond cubic structure, if the corner atoms and face centered atoms are replaced by sulphur (S) atoms and the four atoms present inside the unit cell are replaced by zinc (Zn) atoms OR vice-versa we get zinc sulphide structure as shown in figure.

Ex: Insb, Ga, As etc

Hexagonal closely packed(HCP) structure

Top layer

Middle layer

Bottom layer

A sites

B sites

A sites

1. In a HCP structure a unit cell contains three types of atoms,

(i) 12 corner atoms each of the hexagon corner.

(ii) 2 base centered atoms one at top & one at bottom face.

(iii) 3 atoms are situated in between the top and bottom face of hexagon in alternate vertical faces (i.e. inside the unit cell they can’t shared by other unit cells).

Therefore, each unit cell shares 14 atoms(12 corner & 2 base centered) with other unit cells and contains 3 atoms with in the unit cell.

An actual STM image of a Ni surface. Note the hexagonal arrangement of atoms. This image is the property of IBM Corporation.

2. Its co-ordination number is 12.

C

3. The number of atoms per unit cell are 6

(i) 12 corner atoms are sharing by six surrounding unit cells hence 12 x 1/6 = 2 atoms

(ii) Each base atoms are shared by two unit cells hence,

2 x ½ = 1 atom (iii) Three atoms are inside the unit cell. These atoms are never shared by other unit cells.

Therefore, 12 x 1/6 + 2 x ½ + 3 = 2 + 1 + 3 = 6 Atoms

a

rr

3. Atomic radius(r)

From figure, we can write a = 2r r = a/2Where, r = atomic radius a = edge of the hexagon

A B A B A B A

4. Stacking sequence

Its stacking sequence is AB AB AB AB ………..

A sites

B sites

A sites

5. Relation between C and a (or) c/a ratio

Let height of the unit cell of hexagon is C, the distance between two neighbouring

atoms is a say.

I,J,K,L,M,N are the corner atoms and O is the base atom in the bottom layer. P,Q,T are the middle layer atoms as shown in figure.

Consider the arrangement of atoms at I, N, O(bottom layer atoms) and P(middle layer atom).

Draw the normal line OR from O to the line IN, which bisects the line IN at a point R.

If S is the ortho centre for triangle INO then it is found that the length of OS is equal to 2/3 of OR i.e. OS = 2/3 OR ---------(2)

Consider a triangle IRO, cos 300 = OR/OI OR = OI cos 300 OR = a √3/2 -----------(1)

Since, OI = a cos 300 = √3/2

Sub. Eq(1) in eq(2), we get

OS = 2/3 a √3/2 OS = a/√3 ------ (3)

In triangle SOP, OP2 = OS2 + SP2 - (4) a2 = (a/√3)2 + (c/2)2

Simplifying the above equation, we get,

c/a = (8/3)1/2 = 1.6329 ---- (5)

Eq.(5) gives the c/a ratio as 1.6329

Since, OP = a OS = a/√3 SP = c/2

6. Atomic packing factor (APF)

Volume occupied by the number of atoms in a unit cell(v) = 6 x 4/3 π r3

= 24/3 π (a/2)3

= π a3 ---(1)

The volume of the unit cell(V) = Area of the base x height ------ (2)

Area of the base = area of the one triangle(ONI) x 6

Therefore, area of the one triangle(ONI) = ½ base x height = ½ (IN) (OR) -----(3)

since, r = a/2

Therefore, ONI = ½ x a x a √3 /2 = a2 √3 /4

Hence, area of the base = 6 x a2 √3/4 = 3 √3 a2/2 -- (4)

But, we know that height of the unit cell = C --- (5) Sub. eq(4) & (5) in eq(2), we get

Total volume of the unit cell (V) = 3√3a2 x c/2 ----(6)

Therefore, the APF = v/V ---------- (7)

Sub. eq(1) & (6) in eq(7), we get,

APF = πa3 / 3 √3a2 c/2 = 2π /3√3 (a/c)

= π / 3√2 = 0.74 = 74% ---- (8)

Therefore, its APF is 74%. Hence, it is a closely packed structure.

Ex : Magnesium, Zinc, Zirconium, Cadmium etc

MILLER INDICESWilliam Hallowes Miller

The set of three integers used to describe the orientation of a crystal planes are called Miller Indices.

(OR)

The directional cosine values of the line perpendicularto the plane are called Miller Indices.

General form for Miller Indices of a plane is (h k l)

Procedure for finding Miller IndicesSTEP:01 Find the intercepts of the plane along the axes a, b & c .

STEP:02 Take reciprocals of the intercepts.

STEP:03 Convert into smallest integers in the same ratio. (Taking its LCM)STEP:04 Enclose in parentheses .

NOTE: The reciprocal procedure avoids the intercept of infinity for a plane parallel to an axis by making it to zero.

Example:2

1

2STEP:01 Intercepts are 2,1,2

STEP:02 Take the reciprocals of the intercepts 2,1,2 as 1/2,1,1/2 STEP:03 Convert into smallest integers as 1,2,1 STEP:04 Enclose in the parentheses as (1,2,1).

Gives the Miller Indices for the above plane.

NOTE:

(1,2,1) gives only about three integers

1,2,1

OR

1,2,1Gives the all possible integers (family of integers)

Some Important Miller planes

011 Plane

001 Plane 110 Planes

111 Planes

Some Important Miller directions

INTER PLANAR DISTANCE

X

Z

YO

M

Nγ’

C

B

A

α’β’

From figure in a cubic lattice of cube edge ‘a’ the intercepts of theplane of the three axes are given by

OA = a/hOB = a/kOC = a/l

---- (1)

Let ON = d1 = perpendicular distance from the origin to the plane and the direction cosines of ON be Cos α’, Cosβ’ & Cosγ’ .

Cosα’ = ON/OA = d1/a/h = d1h/aCosβ’ = ON/OB = d1/a/k = d1k/aCosγ’ = ON/OC = d1/a/l = d1l /a

--- (2)

But,

Cos2α’ + cos2β’ +cos2γ’ = 1 --- (3)

Substituting eq(3) in eq(2), we get

(d1h/a)2 + (d1 k/a)2 + (d1 l/a)2 = 1(OR)

d1 = ON = a/√h2 + k2 + l2 -- (4)

Let ‘OM’ be the perpendicular distance of the next parallel plane from the origin.

Hence, its intercepts are

OM = d1 + d1 = 2d1 = 2a/√ h2 + k2 + l2 --- (5)

Therefore, the interfacing between two adjacent planes is equal to OM – ON = NM

Hence, NM = 2d1 – d1 = d1

d1= a/ √h2 + k2 + l2 -----(6)Eq(6) gives the interplanar spacing between two adjacent parallel planes of Miller Indices.