chemistry. solid state-ii session objectives voids packing fraction

Chemistry

Solid State-II

Session Objectives

Voids

Packing fraction

Packing fraction for simple cubic unit cell

Volume of total number of atoms in a unit cellSo packing fraction =

Total volume of unit cell

3 3

3 3

4 41× r r

3 3= = = 0.52 or 52%a 2r

1×8 =1

8Effective number of atoms =

The fraction of total volume of a cube occupied by constituent particles.

Packing fraction for fcc unit cellThe number of effective atoms/anions/cations = 4

a

r

By the definition of packing fraction,

Taking the value of ‘a’ , we get

3

3

44 r 2 2

1 3.143PF 0.7432(4r)

For fcc unit cell, 4r

a2

3

3

44 r

3PFa

Packing fraction for bcc unit cell

The number of effective atoms/anions/cations = 2

4ra

3For bcc unit cell,

33

3

8 r × 3 3PF = = ×3.14=0.679

83× 4r

3

3

44× r

3PF =a

By the definition of packing fraction,

Packing fraction of hcp

Volume of the unit cell=Base area x height

Base area of regular hexagon =Area of six equilateral triangles each with side 2r and altitude 2rsin600

0 216 (2r)(2r) sin60 6 3r

2

First we will calculate the distance between base atomsurrounded by 6 other atoms and the centre of equilateral triangle formed by three atoms just abovebase atoms.

346 r

3PFBase area Height

r

h

2r c

or 3cos30

h 22r

h3

2

2 2 2 2

22 2 2

2r2r h c c

3

4r 2c 4r 4r

3 3

2c 2r

3

2Height of unit cell 2c 4r

3

Packing fraction of hcp

Packing fraction of hcp

2 3

Volume of unit cell

26 3r 4r 24 2 r

3

3

3

3

4Volume of six spheres 6 r

3

8 rpacking fraction

24 2 r

0.74

% volume occupied 74%

% volume empty space 26%

Interstitial sites or Voids

Surrounded by four spheres which lie at the vertices of a regular tetrahedron.

The number of tetrahedral voids is 2 × number of octahedral voids.

Surrounded by six spheres which lie at the vertices of a regular octahedron.

The number of octahedral voids is the number of atoms present in close packed arrangement.

Interstitial sites in ccp

Interstitial sites in fcc

1 at the center12 middle of the edge sites (each shared by 4 unit cells)

Net 4 Oh sites/unit cell

Octahedral (Oh) sites Tetrahedral (Td) sites

Net 8 Td sites/unit cell

Locating Tetrahedral and Octahedral Voids: fcc

• Number of octahedral voids are equal to number of ions present in the unit cell.

• Number of tetrahedral voids are double of octahedral voids.

Locating Tetrahedral and Octahedral Voids : bcc

• Number of octahedral voids are equal to number of ions present in the unit cell.

• Number of tetrahedral voids are double of octahedral voids.

Interstitial sites in hcp

3 Oh sites on top half of unit cell (by symmetry, 3 more on bottom half)

Total 6 Oh sites

6 Td sites on top half of unit cell (by symmetry, 6 more on bottom half)

Total 12 Td sites

Locating Tetrahedral and Octahedral Voids : hcp• Each body diagonal has two tetrahedral voids.

• Center of body and each edge center has octahedral void.

• Dividing cube into 8 minicubes, centre of each minicube has tetrahedral void.

Thank you