Transcript Enthalpy - Career Launcher

Chemistry
Solid State-II
Session Objectives

Voids

Packing fraction
Packing fraction for simple cubic unit cell
The fraction of total volume of a cube occupied by
constituent particles.
Effective number of atoms =
So packing fraction =
1
×8 = 1
8
Volume of total number of atoms in a unit cell
Total volume of unit cell
4
4 3
1× r3
r
3
=
=3
= 0.52 or 52%
3
3
a
2r 
Packing fraction for fcc unit cell
The number of effective atoms/anions/cations = 4
For fcc unit cell,  a 
4r
2
By the definition of packing fraction,
4
PF 
4 3
r
3
a3
Taking the value of ‘a’ , we get
4
PF 
4 3
r  2 2
1 3.14
3


 0.74
3
3
2
(4r)
r
a
Packing fraction for bcc unit cell
The number of effective atoms/anions/cations = 2
For bcc unit cell, a 
4r
3
By the definition of packing fraction,
PF =
4 3
r
3
a3
4×
3
PF =
8r ×
 3
3
3
3×  4r 
=
3
×3.14 = 0.679
8
Packing fraction of hcp
4 3
r
3
PF 
Base area  Height
6
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
1

 6   (2r)(2r)sin600   6 3 r2
2

First we will calculate the distance between base atom
surrounded by 6 other atoms and the centre of
equilateral triangle formed by three atoms just above
base atoms.
Packing fraction of hcp
r
3
o
 cos 30 
h
2
2r
h
3
2
2r 
h
2
 2r 
2
 h2  c2  
 c
 3
r
2r
4r2
2
2
2
c  4r 
 4r2 
3
3
c  2r
2
3
Height of unit cell  2c  4r
2
3
c
Packing fraction of hcp
 Volume of unit cell
 6 3 r2  4r
2
 24 2 r3
3
Volume of six spheres  6 
 packing fraction 
8r3
24 2 r
3
4 3
r
3
 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
Octahedral (Oh) sites
12 middle of the
1 at the center edge sites (each
shared by 4 unit
cells)
Net 4 Oh sites/unit cell
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)
6 Td sites on top half of unit cell (by symmetry, 6 more on bottom half)
Total 6 Oh sites
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