chemistry. solid state-ii session objectives voids packing fraction
TRANSCRIPT
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