Permanent Dipoles

Secondary Bonding

• Secondary bonds are due to attractions of electric dipoles in atoms or molecules.

• Dipoles are created when positive and negative charge centers exist.

Dipole moment=

μ

q= Electric charge

d = separation distance +q -q

d

Fluctuating Dipoles

• Weak secondary bonds in noble gasses.

• Dipoles are created due to asymmetrical distribution of electron charges.

• Electron cloud charge changes with time.

Symmetrical distribution of electron charge

Asymmetrical Distribution (Changes with time)

Permanent Dipoles

• Dipoles that do not fluctuate with time are called Permanent dipoles.

¾ Examples:- Symmetrical Arrangement Of 4 C-H bonds

CH

Hydrogen Bonds

• Hydrogen bonds are Dipole-Dipole interaction

between polar bonds containing hydrogen atom.

(Hydrogen is covalently bonded to F, O or N atoms)

¾ Example :-

™ In water, dipole is created due to asymmetrical arrangement of hydrogen atoms.

™ Attraction between positive oxygen pole and negative hydrogen pole.

105 ⁰

O

H

H Hydrogen

Bond

van der Waals bonds

• Fluctuation-induced dipole bonds

– Charge on atom in motion being re- distributed

– Electric dipole

– Induced electric dipole

• Polar molecule induced dipole bonds – Neutral atoms form dipole

• Permanent dipole bonds: HCl

Van der Waal bonds

6 _n 12

N

A B

U

Architecture of solids

• Single crystals

– The arrangement of atom is regular and repeated in 3D pattern throughout the entirety of the crystalline solid without interruption

• Poly-crystals

– Solids consisting of many crystals or grains, divided by grain boundaries, where two grains meet

• Amorphous (Non-crystalline)

– Solids lack a systematic, repeated and regular arrangement of atoms

– Heat treatment usually would transform a random structure to an ordered state

The Space Lattice and Unit Cells

• Atoms or ions are arranged in repetitive 3-

Dimensional pattern, in long range order give rise to crystal structure.

• Properties of solids depends upon crystal structure and bonding force.

• An imaginary network of lines, with atoms at

intersection of lines, representing the arrangement of atoms is called space lattice.

Space Lattice

Atomic hard sphere model

– Lattice: a three-dimensional array of points coinciding with atomic sphere centers

– Unit cell: a small repeat entity representing the symmetry of the crystal structure

– Coordination number (Z): the number of nearest- neighbor atoms surrounding each atom in a

crystal structure

– Atomic packing factor (APF): the volume fraction of atomic spheres in a unit cell

– d-spacing: distance between lattice planes

Crystal Systems and Bravais Lattice

Principal Metallic Crystal Structures

• 90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.

BCC Structure FCC Structure HCP Structure

3-7

Body Centered Cubic (BCC) Crystal Structure

• Represented as one atom at each corner of cube and one at the center of cube.

• Each atom has 8 nearest neighbors.

• Therefore, coordination number is 8.

BCC Crystal Structure (Cont..)

• Each unit cell has eight 1/8 atom at corners and 1

full atom at the center.

• Therefore each unit cell has

• Atoms contact each other at cube diagonal

(8x1/8 ) + 1 = 2 atoms

3 Therefore, lattice 4 R

constant a =

3-9

Atomic Packing Factor of BCC Structure

Atomic Packing Factor = Volume of atoms in unit cell Volume of unit cell

V_atoms = = 8.373R³

3

3 4

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ R = 12.32 R³

Therefore APF = 8.723 R³

12.32 R³ = 0.68 V _{unit cell} = a³ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ Π 3 . 4 2

R3

Structure

• FCC structure is represented as one atom each at the corner of cube and at the center of each cube face.

• Coordination number for FCC structure is 12

• Atomic Packing Factor is 0.74

• Examples :-

¾ Aluminum (a = 0.405)

¾ Gold (a = 0.408)

FCC Crystal Structure (Cont..)

• Each unit cell has eight 1/8 atom at corners and six ½ atoms at the center of six faces.

• Therefore each unit cell has

• Atoms contact each other across cubic face diagonal

(8 x 1/8)+ (6 x ½) = 4 atoms

Atomic Packing Factor of FCC Structure

Atomic Packing Factor = Volume of atoms in unit cell Volume of unit cell

V_atoms = = 16.755R³

4 3

2

⎛ R ⎞

⎜ ⎟

⎝ ⎠ = 22.63 R³

Therefore APF = 16.755 R³

22.63 R³ = 0.74 V _{unit cell} = a³ =

4 3

4 3

⎛ ΠR ⎞

⋅⎜ ⎟

⎝ ⎠

3-10

Hexagonal Close-Packed Structure

• The HCP structure is represented as an atom at each of 12 corners of a hexagonal prism, 2 atoms at top and bottom face and 3 atoms in between top and bottom face.

• Atoms attain higher APF by attaining HCP structure than simple hexagonal structure.

• The coordination number is 12, APF = 0.74.

HCP Crystal Structure (Cont..)

• Each atom has six 1/6 atoms at each of top and

bottom layer, two half atoms at top and bottom layer and 3 full atoms at the middle layer.

• Therefore each HCP unit cell has

• Examples:-

¾ Zinc (a = 0.2665 nm, c/a = 1.85)

¾ Cobalt (a = 0.2507 nm, c.a = 1.62)

• Ideal c/a ratio is 1.633.

(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms

3-14

Atomic Packing Factor of HCP Structure

Atomic Packing Factor = Volume of atoms in unit cell Volume of unit cell

V_atoms =

Therefore APF = = 0.74

V _{unit cell} = 6(^{1 2} sin 60 ) ^{3 3 2}

2 2

a o ⋅ =c a c

=??

c=1.633a R=??a

4 3

6( )

3π R

Atom Positions in Cubic Unit Cells

• Cartesian coordinate system is use to locate atoms.

• In a cubic unit cell

¾ y axis is the direction to the right.

¾ x axis is the direction coming out of the paper.

¾ z axis is the direction towards top.

¾ Negative directions are to the opposite of positive directions.

• Atom positions are located using unit distances along the axes.

3-15

Directions in Cubic Unit Cells

• In cubic crystals, Direction Indices are vector

components of directions resolved along each axes, reduced to smallest integers.

• Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, converted to integers.

Direction Indices - Example

• Determine direction indices of the given vector.

Origin coordinates are (3/4 , 0 , 1/4).

Emergence coordinates are (1/4, 1/2, 1/2).

(3/4 , 0 , 1/4) - (1/4, 1/2, 1/2)

= (-1/2, 1/2, 1/4) reduced to smallest integers 4 x (-1/2, 1/2, 1/4) = (-2, 2, 1)

Therefore, the direction indices are 3-18

2 2 1

⎡ ⎤

⎣ ⎦

Miller Indices in Cubic Unit Cells

• Miller Indices are are used to refer to specific lattice planes of atoms.

• They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the

crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell.

z

Miller Indices =(111)

Miller Indices - Examples

• Intercepts of the plane at x,y & z axes are 1, ∞ and

∞

• Taking reciprocals we get (1,0,0).

• Miller indices are (100).

*******************

• Intercepts are 1/3, 2/3 & 1.

• taking reciprocals we get (3, 3/2, 1).

• Multiplying by 2 to clear fractions, we get (6,3,2).

• Miller indices are (632).

x x

y z

(100)

3-21

Miller Indices - Examples

• Plot the plane (101)

Taking reciprocals of the indices we get (1 ∞ 1).

The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.

******************************

• Plot the plane (2 2 1)

Taking reciprocals of the indices we get (1/2 1/2 1).

The intercepts of the plane are

Miller Indices - Example

• Plot the plane (110)

The reciprocals are (1,-1, ∞)

The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis)

To show this plane in a single unit cell, the

origin is moved along the positive direction of y axis by 1 unit.

x

y z

(110)

3-23

Miller Indices – Important Relationship

• Direction indices of a direction perpendicular to a crystal plane are same as miller indices of the plane.

• Example:-

• Interplanar spacing between parallel closest planes with same miller indices {h k l} is given by

{110} ⁽¹¹⁰⁾

x

y z

Crystallographic directions and planes

• A crystallographic direction is defined as a vector between two lattice sites. Indices are vector

components of direction reduced to smallest integer.

• Miller indices for cubic crystals – (1,0,0) plane

– {1,0,0} plane family – [1,0,0] direction

– <1,0,0> direction family

• Planes and directions of cubic crystals having the same indices are perpendicular to one another.

Ref: Ashby and Jones, 1980.

Structural Difference between HCP and FCC

Consider a layer of atoms (Plane ‘A’)

Another layer (plane ‘B’) of atoms is placed in ‘a’

Void of plane ‘A’

Third layer of Atoms placed in ‘b’ Voids of plane ‘B’. (Identical

to plane ‘A’.) HCP crystal.

Third layer of Atoms placed in ‘a’ voids of plane ‘B’. Resulting

In 3^rd Plane C. FCC crystal.

Plane A

‘a’ void

‘b’ void

Plane A Plane B

‘a’ void

‘b’ void

Volume Density

• Volume density of metal =

• Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm.

ρ

=

a=

2

4 R = ^{4 0.1278}

2

× nm

= 0.361 nm Volume of unit cell = V= a³ = (0.361nm)³ = 4.7 x 10^-23 cm³

ρ

FCC unit cell has 4 atoms.

Mass of unit cell = ⁽⁴ ₂₃^{)(63.54 / )}

6 10 /

atoms g mol atmos mol

× = 4.22 x 10^-22 g

22

23 3 3

4.22 10 4.7 10 8.98

m g g

V cm cm

−

−

= = × =

3-30 ×

Volume Density, cont’d

• Volumetric (material) density (ρ):

ρ

n: number of atoms associated with each unit cell A: atomic weight

V_c: volume of the unit cell

Planar Atomic Density

• Planar atomic density=

• Example:- In Iron (BCC, a=0.287), The (110) plane intersects center of 5 atoms (Four ¼ and 1 full atom).

¾ Equivalent number of atoms = (4 x ¼ ) + 1 = 2 atoms Area of 110 plane =

ρ

Selected area

ρ

_{( )}

287 .

0 2

2

= 2

17.2atoms

= nm

3-31

Figure 3.22 a&b

=

2

2 a × a = a

ρ

^{( )}

^{( )}

2 2

0.833

2 4

2 3

R R

a R

π π

= = =

⎛ ⎞

⎜ ⎟

⎝ ⎠

or

Linear Atomic Density

• Linear atomic density =

• Example:- For a FCC copper crystal (a=0.361), the [110] direction intersects 2 half diameters and 1 full diameter.

¾ Therefore, it intersects ½ + ½ + 1 = 2 atomic diameters.

ρ

Number of atomic diameters intersected by selected length of line in direction of interest

Selected length of line

mm

atoms nm

atoms nm

atoms 3.92 3.92 10⁶ 361

. 0 2

2 ×

=

=

×

ρ

nm 361 .

0 2 ×

Length of line =

Planer Density and Linear Density

• Linear atomic density (LD): the fraction of line length in a particular crystallographic direction that passes through atom centers.

EX: BCC along [110]

• Planer atomic density (PD): the fraction of total

crystallographic plane area that is occupied by atoms. The plane must pass through an atom’s center for the

particular atom to be included.

EX: BCC on (100)

LD = 2R 2a

PD =

1

4 × 4

⎛⎝⎜ ⎞

⎠⎟ πR² a × a

Density, cont’d

• Density of materials depend on (1) packing and (2) mass and size of atoms

• Metals

– Heavy atoms and closely packed

• Ceramics

– Light atoms and closely packed