(2096)lecture_notes_solid_state_e.pdf

DATE : 03-09-12 (Monday)
NOTE : Respected Faculty Member of R-Phase complete the Solid State according to
the following lecture plan.
[Collect Lecture Notes (No change) From SMDD] .
1-Lecture in Solid State
LECTURE = 1
Basic of solid state.
Total Lecture in this week (3-Physical)
LECTURE = 2
Simple cubic structure, BCC and Closest packing.
LECTURE = 3
Hcp structure
LECTURE = 4
Ccp structure
Total Lecture in this week (3-Physical)
LECTURE = 5
Radius ratio rule and Type of ionic structures
LECTURE = 6
Crystal defects and Properties of solids
LECTURE = 7
Solid State complete with discussion.

CHEMISTRY
SOLID STATE
LECTURE # 1 TO 7
Course : VIJAY (JR)
LECTURE NOTES

Lecture # 1
Section (A) Basics of Solid State
(1) Definition type of solid (Ionic/Covalent/Metallic/Molecular) + Properties. [10 Mins]
(2) Definition of lattice + Unit cell + Coordination Number + density. [15 Mins]
(3) 1–D & 2–D lattice  Square and Hexagonal packing. [15 Mins]
(4) Classification of unit cells in 3D ; 7 Crystal system + 14 Bravis lattice. [20 Mins]
(1) Definition type of solid (Ionic/Covalent/Metallic/Molecular) + Properties.
Matter can exist in three physical states namely: solid, liquid and gas.Lets us, very briefly, go through
them
COMPARISON OF STATES OF MATTER
Nature Gases Liquids Solids
Motion of
particles
Randomly moving
and no fixed
location of particles
Random motion is
possible but less than
that in gases
Particles have fixed
locations they can
only vibrate.
Forces b/w
particles
Forces of attraction
is zero
Intermediate value Strong inter-
particle attraction
Interparticle
Separation
Large interparticle
separation.
Small Interparticle
separation
Separation is close
to zero.
Volume/Shape No fixed volume, no
definite shape.
No fixed shape; but
fixed volume
Fixed volume and
shape (rigid)
Response to
external
changes
Compression and
expansion can be
done to large
extent.
Heat capacities
depend on process
Small extent of
compression and
expansion
Heat capacity approx.
constant
Compression /
expansion least
possible.
Constant heat
capacity
Whether a given system would exist as solid, liquid or gas depends upon two universal opposing tendencies
associated with the particles:
(i) Tendency of mutual attraction. (ii)Tendency to escape from each other.
Gases: Escaping tendency much higher than Attraction
Liquids: Escaping tendency higher than attraction but limited.
Solids: Attraction much higher that escaping tendency
Classification on the basis of forces among constituting particles:
Solid Constituent
particles
Bonding/
Attractive
forces
B.E.
(kJmol-1
)
Examples Physical
nature
mp
(K)
Electrical
Nature
Molecular Molecules i)Dispersion
ii)Dipole-dipole
interaction
iii)H-bonding
0.05-40
5.25
10-40
Argon
HCl
H2O(ice)
Soft
Soft
Hard
84
158
273
Insulator
Insulator
Insulator
Metallic Atoms
(kernels in
–e sea)
Positive ions
and electrons
(delocalized)
70-1000 Ag. Cu,
Mg
Hard ~800
To
1000
Conductor
Network
Or
Covalent
Atoms Covalent(short
range)
150-500 SiO2,SiC,
diamond,
graphite
Hard ~4000 Insulator
Ionic Ions Coulombic(infinit
e range)
400-4000 NaCl,
MgO
KCl,BsCl2
Hard but
Brittle
High
~1500
Insulator

Types of SOLIDS on the basis of arrangement of particles
Crystalline solids
(True solids)
Amorphous solids
(Super-cooled liquids)
1. Constituent particles
(ions, molecules, atoms)
will be arranged in a particular
fashion having long range order.
Ag. Fe,Cu,NaCl, HO(s)
2
Particles will be randomly arranged or
can have short range order.
Glass, plastics, anorphous silica.
2. Crystals can be generated by
cooling liquid slowly and in
controlled conditions. The
structure of the crystals is
dependent on these conditions.
So, same substance can have different.
crystalline structures in
conditions known as
and these structures are known as
polymorphs
POLYMORPHISM
different
When a liquid is suddenly cooled
amosphorus solid will result.
Amorphous glass
Sunlight
Slightly crystalline
results
milky appearance
3. Sharp / fixed melting point. These will be having a range of
temperature in which it melts.
# Cooling curve of a crystalline solid. # Cooling curve for amorphous solids.
t1 t2
liquid
only
solid
only
time
5ºC
Temp
(mp)
Crystalline solid
t1 t2
liquid
solid
only
time
T1
Temp T2
liquid + solid
4. Anisotropic
Physical property (mechanical strength,
Refractive index, electrical conductance)
will have diff.
value in different directions.
Isotropic
Same value in all directions.
Reason : Random arrangement of
molecules.
(diff. prop. in
diff. direction)
(same property
in all direction)
transition
(from liq. to solid)
(transition)

Characteristic of solids : Solids are characterized by their high density and low compressibility compared
with those of the gas phase. The values of these properties for solids indicate that the molecules (or ions)
in them are relatively close together. Solids can very easily be distinguished from liquids by their definite
shape, considerabla mechanical strength and rigidity. These properties are due to the existence of very
strong froces of attraction amongst the molecule (or ions) of the solids. It is because of these strong forces
that the structural units (atoms, ions, etc.) of the solid do not possess any translatory motion but can have
only vibrational motion about their mean positions.
Crystals and amorphous solids : Solids can generally be classified into two broad categories : crystals
and amorphous substances. The outsanding characteristics of a crystals are its sharp melting point and
its flast faces and sharp edges which, in a well developed form, are usually arranged symmetrical. These
properties are the result of a high degree of internal order which extends throughout the crystal (a definite
pattern constantly repeating in space), i.e. there exists what is known as the long range order. The pattern
is such that having observed it in some small region of the crystal, it is possible to predict accurately the
positions of particles in any region of the crystal, however far it may be from the region under observation.
But amorphous solids, such as glass, do not have this ordered arrangement. In many ways they are more
closely related to liquids than to crystalline solids and are, therefore, regarded as supercooled liquids with
high viscosity. In this chapter, we shall discuss the subject of crystalline structure.
In-Text Questions (NCERT)
1.1 Why are solids rigid?
Ans. In a solid, the constituent particles are very closely packed have fixed positions and can only oscillate
about thier mean positions.The forces of attraction among these particles are very strong. Hence,they are
rigid.
1.2 Why do solids have a definite volume?
Ans. The costiuent particles of a solid have fixed postions and are not free to move about, i.e., they posses
rigidity. That is why they have a definite volume.
1.3 Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon,
potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Ans. Amorphous : Polyurethane , teflon , cellophane , polyvinyl chloride , fibre glass
Crystalline solids : Benzoic acid , potassium nitrate , copper.
1.4 Why is glass considered a super cooled liquid?
Ans. Glass is an amorphous solid. Like liquids, it has a tendency to flow, through very slowly. The proof of this
fact is that the glass panes in the windows or doors of old building are invariably found to be slightly
thickers at the bottom than at the top. This is evidently due to the reason that glass flows down very slowly
and makes the bottom portion slightly thicker.
1.5 Refractive index of a solid is observed to have the same value along all directions. Comment on the nature
of this solid. Would it show cleavage property?
Ans. As the solid has same value of refractive index along all directions, this means that it is isotropic and hence
amorphous. Being an amorphous sloid, it would not show a clean cleavage when cut with a knife. Instead,
it would break into pieces with irregular surfaces.
1.6 Classify the following solids in different categories based on the nature of intermolecular forces operating in
them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon,
silicon carbide.
Ans. Potassium sulphate = Ionic , Tin = Metallic ,
Benzene = Molecular (non–polar) , Urea = Molecular(Polar),
Ammonia = Molecular (Hydrogen bonded), Water = Molecular (Hydrogen bonded) ,
Zinc sulphide = Ionic , Graphits = Covalent or Network,
Rubidium = Metallic , Argon = Molecular (Non–polar) ,
Silicon carbide = Covalent or Network.
1.7 Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high
temperature. What type of solid is it?
Ans. Covalent Network Solid like SiO2
(quartz) or SiC or (diamond)

1.8 Ionic solids conduct electricity in molten state but not in solid state. Explain.
Ans. In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However, in
the solid state, as the ions are not free but remain held together by strong electrostatic forces of attraction,
they cannot conduct electricity in the solid state.
1.9 What type of solids are electrical conductors, malleable and ductile?
Ans. Metallic solids.
(2) Definition of lattice + Unit cell + Coordination Number & Density & Packing fraction
STUDY OF STRUCTURE OF CRYSTALLINE SOLIDS To study crystalline solids we need some mathematical
and geometrical tools.Lets get familier with them.
1. Space lattice : It is the regular arrangement of points in space and forms the basis of classification of
all structures.This regular arrangement spread over whole space is the long range order.
2. Unit cell: Unit cell is the smallest arrangement of points which on repetition generates the whole space
lattice. It is the building block of the whole structure.
(a)Unit cell contains the minimum information in terms of edge lengths and angles sufficient to
determine the structure of the lattice.
(b)Generally unit cell is the smallest element of the space lattice. But sometimes a slightly bigger
element can be taken as unit cell to make calculations easy which will be representing the symmetry of
the lattice i.e., the choice of unit cell depends entirely on our convenience.
Various types of unit cells can be chosen within a particular lattice:
(i) Primitive:When lattice points are found only at the corners of the unit cell.
(ii) Centered/Non-primitive:When lattice points are found at the corners and as well as inside the unit
cell.
Space Lattice
Unit Cell
3. How to repete a unit cell to form whole lattice:
Repetition is done by translating the whole unit cell parallel to its edges and by the distances equal to
edge lengths of the unit cell.
4. How to view a crystal from lattice:
Constituent particles(in the form of spheres) are placed on lattice points to form the actual crystal
structure.
Space Lattice : The definiotn of a space lattice is strictlya geometrical concept and represents a three-dimensional
tranlational repettion of the centres of gravity of the units of pattern in the crystal.
These units of pattern can be anything, for example :
(i) In metals or inert gases, each lattice point may represent the position of each atom.
(ii) In a crystal of methane, each lattice point many represent the centre of a methane molecule.
(iii) In ionic crystals, the lattice points may represent ions or ion-pairs.
Coordination number : It is defined as the number of nearest neighbours or touching particles with other
particle present in a crystal is called its co-ordination number. It depends upon structure of the crystal.

(3) 1–D & 2–D lattice  Square and Hexagonal packing.
5. 1–DIMENSIONAL SPACE LATTICE
Uniformly separated lattice points in 1–dimension will represent 1–D lattice.Only one parameter is needed
to describe it.
d
6. 2 DIMENSIONAL SPACE LATTICE
It is the regular arrangement of points in a plane.Only three parameters are needed to describe the
lattice.Two sides a and b and angle between them().
(i) Square unit cell: a = b,  = 90º (ii) Rectangle: a  b, = 90º
a
90º
a
(Primitive Type)
b
90º
a
(Primitive Type)
(iii)Hexagonal unit cell: a = b,  = 60º
Unit cell can be chosen in more than one way.
a
60º
a
(Primitive Type)
(iv)Rhombic unit cell: a = b,  90º, 60º
Unit cell can be chosen in more than one way(since the diagonals of rhombus are perpendicular)
a
a
(Primitive Type)
Centered Unit Cell
(v)Parallelogram: a  b,   90º
This is the most unsymmetric case.
b
a
(Primitive Type)
STRUCTURE OF SOLIDS
(a) Solids having only one type of particles : e.g. metallic, molecular, covalent network solids.
The particle of the solid will be represented by a sphere of radius r. We assume that there is only attraction
force existing between those spheres.Hence in order to to get maximum stable arrangement of these
particles the seperation between the particles should be minimum.
E.g. Water kept in a container ,on cooling,results in freezing with initially at the circumference of surface
(1–D), then on the surface (2–D) and then throughout the volume (3–D).
ARRANGEMENT IN 1-D :
In one dimension it is possible to arrange the spheres in only one stable way.This is the position with
minimum separation and minimum PE of the system(most stable)
(1-D close packing stable arrangement)
This is the predominant way of packing in one dimensions and as such most of the space lattices will
show such an arrangement in one dimensions along the planes of close packing.

ARRANGEMENT IN TWO DIMENSION :
In two dimensions also there are two ways of packing the identical spheres(in moving from one dimension
to two dimensions it can be imagined that the two dimensional array will be made up of 1-D closed pack
arrays / lines which are stacked one on top of each other.
1. Square arangement(poor arrangement) : If the one dimensional arrays are placed on top of one
another, we get the square packing in twio dimensions.
One sphere will be in constant contact with 4 other spheres.
area of square = a2
= 4r2
; area of atoms in the square =
2
2
r
4
r
4
1





Packing fraction, fraction of area occupied by spheres = 4
r
4
r
2
2



= 0.78( 78% )
2. Hexagonal arrangement(better arrangement) : If each one dimensional array is placed in the cavity
of the just preceding array, we get the hexagonal packing in two dimensions.
area of hexagon = 6 ×
4
3
a2
= 6 ×
4
3
× 4r2
; area of atoms = r2
+
3
1
× 6r2
= 3r2
fraction of area occupied =
6
3
r
3
6
r
3
2
2



–
~ 0.91 ( 91% )
Better packing in 2d: Clearly more no. of spheres are in contact in the hexagonal arrangement, hence the
hexagonal arrangement is considered to be a better way of packing as compared to the square packing.
(4) Classification of unit cells in 3D ; 7 Crystal system + 14 Bravis lattice.
7. 3 DIMENSIONAL SPACE LATTICE
It is the regular arrangement of points in space.
Six parameters are required to completely describe the unit cell: 3 edge lengths and 3 angles.
a, b, c : edge lengths of unit cell (also known as the crystallographic axes.)
are known as the crystallographic angles.
Seven Primitive unit cells in crystals
Cubic
c
a
b
g
a
b ; Tetragonal
a
b
c ; Orthorhombic
a
b
c
;

Monoclinic
a
b
c
; Hexagonal
a
c
b
;
Rhombohedral
a
b
c
; Triclinic
a
b
c
Seven Primitive Unit Cells
Crystal
System
Axial
distances
Axial angles Examples Type of Unit
cells
Cubic a = b = c á = â = ã = 90° Cu, ZnS, KCl S,BC,FC
Tetragonal a = b  c á = â = ã = 90° Sn(white tin), SnO2,TiO2 S,BC
Orthorhombic a  b  c á = â = ã = 90° Rhombic sulphur, CaCO3 S,BC,FC,EC
Monoclinic a  b  c á = ã = 90°;â  90° Monoclinic sulphur, PbCrO2 S,BC
Hexagonal a = b  c á = â = 90°;ã = 120° Graphite, ZnO S
Rhombohedral a = b = c á = â = ã  90° CaCO3(Calcite), HgS(Cinnabar) S
Triclinic a  b  c á â ã  90° K2Cr2O7, CuSO4.5H2O S
Ways of choosing unit cells in 3-dimensions: Apart from unit cells with lattice points at the corners,
there are three other types of choosing unit cells called Non-Primitive/Centered unit cells.
Principle of choosing a particular type of unit cell:
"The surroundings of each and every lattice point in a lattice should be exactly identical."----(Very Imp.)
Hence all the lattice points are indistinguishible.
1. Primitive /Simple: lattice points only at the corners only.
Found in all crystal systems.
2. Body centered: Lattice points at corners & at the body center.
Found only in Cubic,Tetragonal,Orthorhombic and Monoclinic systems.
3. Face centered : Lattice points at corners & at each of the face centre.
Found only in Cubic and Orthorhombic systems.

4. End centered: Lattice point at corners & only at two oppsosite faces.
Found only in Orthorhombic system.
Observation : End centered unit cell are not found in Cubic system. It is found that each crystal class do
not have all the four different type of unit cells. For e.g. in cubic crystal class or system, only 3 types of unit
cells are found end centred is not found.
Total number of lattice arrangements in three dimensions = 14.These are called 14 BRAVAIS LATTICES
Ex.1 Identify molecular solid, covalent solid, ionic solid : P4
(s), S8
(s), SiC (s), Al2
O3
(s), He (s), Al2
Cl6
(s).
Sol. Molecular solid  P4
(s), S8
(s), He (s), Al2
Cl6
(s)
Covalent solid  SiC
Ionic solid  Al2
O3
(s).
Ex.2 The lattice parameters of a given crystal are a = 5.62 Å, b = 7.41 Å and c = 9.48 Å. The three
coordinate axes are mutually perpendicular to each other. The crystal is :
(A) tetragonal (B) orthorhombic (C) monoclinic (D) trigonal.
Ans. (B)
Sol. a  b  c &  =  =  = 90° the crystal system is orthorhombic.
Ex.3 Tetragonal crystal system has the following unit cell dimensions:
(A) a = b = c and  =  =  = 90° (B) a = b  c and  =  =  = 90°
(C) a  b  c and  =  =  = 90° (D) a = b  c and  =  = 90°,  = 120°
Ans. (B)
NCERT Point 1.4 (ONLY FOR FACULTY)
1.4 Crystals Lattice and unit cells : The main characteristic of crystalline
solids is a regular and repeating pattern of constituent particles. If the
three dimensional arrangement of constituent particles in a crystal is
represented diagrammatically, in which each particle is depicted as a
point, the arrangement is called crystal lattice. Thus, a regular three
dimensional arrangement is called crystal lattice. Thus, a regular three
dimensional arrangement of points in space is called a crystal lattice.
A portion of a crystal lattice is shown in fig :
There are only 14 possible three dimensional lattices. These are called Bravais Lattices (after the French
mathematician who first described them). The following are the characteristics of a crystal lattice:
(a) Each point in a lattice is called lattice point or lattice site.
(b) Each point in a crystal lattice represents one constituent particle which may be an atom, a molecule
(group of atoms) or an ion.
(c) Lattice points are joined by straight lines to bring out the geometry of the lattice.
Unit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates
the entire lattice.
A unit cell is characterised by :
(i) its dimensions along the three edges, a, b and c. These edges may or
may not be mutually perpendicular.
(ii) angles between the edges, á (between b and c) â (between a and c)
and ã (between a and b). Thus, a unit cell is characterised by six
parameters, a, b c, á, â and ã. These parameters of a typical unit cell are
shown in Fig.

Lecture # 2
Section (B) Simple cubic structure, BCC and CsCl structure Point Time
(1) Definitions of SC, BCC & CsCl unit cell. (1 + 2) [15 Mins]
(2) Contribution of an atom to a lattice point (formula of unit cell or compound for SC & BCC).
(3) Relation between edge length and radius & volume of unit cell. (3 + 4) [20 Mins]
(4) Coordination number in each case (SC, BCC), CsCl-Type.
(5) Structure of section nearest neighbours atom. (5 + 6) [25 Mins]
(6) Packing of fraction and density of SC & BCC & CsCl.
(1) Definitions of SC, BCC & CsCl unit cell .
Contribution of an atom to a lattice point.(formula of unit cell or compound for SC &
BCC)
Contribution of different lattice points in one unit cell (CUBIC SYSTEM)
Corner atom: Since eight cubes meet at one corner, each atom at the corner gives 1/8th contribution
to each cube.This means i/8th of the electron density, 1/8th of the nuclear density(in case of atoms), 1/
8th of the molecule are found completely inside one unit cell due to corner.
Face Center:Since each face is shared by two cubes, each atom/molecule is equally divided b/w them.
Body Centre:Whole of the atom/molecule is found inside.
Edge Centre: Each edge is shared by four cubes, each atom/molecule at the edge contributes 1/4th to
each cube.
Half
of
the
atom
Intext Questions of NCERT :
1.10 Give the significance of a ‘lattice point’.
Ans. Each lattice point represents one constituent particle of the solid. This constituent particle may be an
atom, a molecule(group of atoms) or an ion.
1.11 Name the parameters that characterise a unit cell.
Ans. A unit cell is characterized by the following six parameters :
(i) The dimensions ofthe unit cell along three edges. These are represented
by a, b and c. The edges may or may not be mutually perpendicular.
(ii) The angles between the edges. These are represented by  ,  . The
angle  is between b and c,  is between a and c and  is between a and b.
c
a
b
1.12 Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans. (i) For hexagonal unit cell, a = b  c,  =  = 90° ,  = 120°
For monoclinic unit cell, a  b  c,  =  = 90° ,  = 90°
(ii)Aface centered unit cell has one constituent particle present at the centre of each face in addition to the
particles present at the corners.
An end–face centered has one costituent particle each at the centre of any two opposite faces in addition
to the particles present at the corners.

1.13 Explain how much portion of an atom located at (i) corner and (ii) bodycentre of a cubic unit cell is part of
its neighbouring unit cell.
Ans. (i)An atom at the corner is shared by eight adjacent unit cells. Hence, portion of the atom at the corner that
belongs to one unit cell =
8
1
.
(ii) The atom at the body centre of a cubic unit cell is not shared by any other unit cell. Hence, it belongs
fully to the unit cell.
Ex.4 In a face centered cubic arrangement of A and B atoms whose A atoms are at the corner of the unit
cell and B atoms at the face centres. One of the A atom Is missing from one corner in unit cell.
The simplest formula of the compound is
(A) A7
B3
(B) AB3
(C) A7
B24
(D) A2
B3
Ans. (C)
Sol. A = 7 
8
1
=
8
7
B = 6 
2
1
= 3
Formula = 8
/
7
A B3
or A
A7
B24
Ex.5 A compound has cubical unit cell in which X atom are present at 6 corner, Y atom are at remaining corner
& only at those face centers which are not opposite to each other & Z atoms are present at remaining face
center & body center then find.
(i) Formula of compound (ii) Density if edge length = 2 Å.
Given : Atomic mass of X = 40 amu, Y = 60 amu, Z = 80 amu.
Sol. (i) X =
8
1
× 6 =
4
3
,
Y =
8
1
× 2 +
2
1
× 3 =
4
7
Z =
2
1
× 3 + 1 + 1 =
2
5
=
4
10
For formula :
4
10
4
7
4
3 Z
Y
X
= X3
Y7
Z10
(ii) 1 amu = 1.67 × 10–24
gram
1 amu = 23
10
02
.
6
1

gram.
Density =
Volume
Mass
=
3
8
)
10
2
(
80
4
10
60
4
7
40
4
3







amu/cc = 24
24
10
8
10
67
.
1
335





= 69.8 gram/cc.
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
NCERT Point 1.5, 1.5.1 to 1.5.3 (ONLY FOR FACULTY )
1.5 Number of Atoms in a unit Cell
We know that any crystal lattice is made up of a very large number of unit cells and every lattice point is
occupied by one constituent particle (atom, molecule or ion). Let us now work out what portion of each
particle belongs to a particular unit cell. We shall consider three types of cubic unit cells and for simplicity
assume that the constituent particle is an atom.

1.5.1 Primitive Cubic Unit cell : Primitive cubic unit cell has atoms only at its corner. Each atom at a corner
is shared between eight adjacent unit cells as shown in Fig. 1.8, four unit cells in the same layer and
four unit cells of the upper (or lower) layer. Therefore, only 1/8th of an atom (or molecule or ion) actually
belongs to a particular unit cell. In Fig. 1.9, a primitive cubic unit cell has been depicted in three different
ways. Each small sphere in Fig. 1.9 (a) represents only the centre of the particle occupying that
position and not its actual size. Such structures are called open structures. The arrangement of
particles is easier to follow in open structures. Fig. 1.9 (b) depicts space-filling representation of the unit
cell with actual particle size and Fig. 1.9 (c) shows the actual portions of different atoms present
different atoms present in a cubic unit cell.
In all, since each cubic unit cell has 8 atoms on its corners, the total number of atoms in one unit cell is
8 ×1/8 = 1 atom.
1.5.2 Body-Centered Cubic Unit Cell : A body-centred cubic (bcc) unit cell has an atom at each of its corners
and also one atom at its body centre. Fig. 1.10 depicts (a) open structure (b) space filling model and (c) the
unit cell with portions of atoms actually belonging to it. It can be seen that the atom at the body centre
wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell :
(i) 8 corners 1/8 per corner atom = 8 × 1/8=1 atom
(ii)1 body centre atom = 1 1= 1 atom  Total number of atoms per unit cell = 2 atom.
1.5.3 face-Centered Cubic Unit Cell : A face-centred cubic (fcc) unit cell contains atoms at all the corners
and at the centre of all the faces of the cube. It can be seen in Fig. 1.11 that each atom located at the
face-centre is shared between two adjacent unit cells and only 1/2 of each atom belongs to a unit cell.
Fig. 1.12 depicts (a) open structure (b) space-filling model and (c) the unit cell with portions of atoms
actually belonging to it. Thus, in a face-centred cubic (fcc) unit cell :
(i) 8 corners atoms 1/8 atom per unit cell= 8 × 1/8 = 1 atom
(ii) 6 face – centered atoms 1/2 atom per unit cell = 6 × 1/2 = 3 atoms
 Total number of atoms per unit cell = 4 atoms.
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

ARRANGEMENT IN THREE DIMENSIONS:
Density of the unit cell () : It is defined as the ratio of mass per unit cell to the total volume of unit cell.
 =
0
3
N
a
M
Z


Where Z = Number of particles per unit cell
M = Atomic mass or molecular mass
N0
= Avogadro number (6.023 × 1023
mol–1
)
a = Edge length of the unit cell= a × 10–10
cm
a3
= volume of the unit cell
i.e.  =
3
30
0
3
cm
/
g
10
N
a
M
Z




The density of the substance is same as the density of the unit cell.
Packing fraction (P.F.) : It is defined as ratio of the volume of the unit cell that is occupied by spheres of the unit
cell to the total volume of the unit cell.
Let radius of the atom in the packing = r
Edge length of the cube = a Volume of the cube V = a3
Volume of the atom (spherical)  =
3
4
r3
Packing density =
3
3
a
Z
r
3
4
V
Z



1. Simple cubical arrangement in three dimensions :
(will be made up of 2-D sheets arranged one over other)
The simple cubicle packing is obtained by arranging the square pack sheets of two dimension one over the
other such that spheres of the second sheet are exactly (vertically) above the spheres of first sheet.
(Note that , hence crystal thus formed will belong to the cubic crystal class, and as the
lattice points are only at the corners, hence the unit cell will be simple, therefore what we get is the simple
cubic )
(i) Relation between ‘a’ and ‘r’
a = 2r (because atoms along the edge are touching each other)
(ii) effective no. of atoms per unit cell Z:
Z 1
8
8
1



(iii) packing fraction :
= 3
3
3
3
3
3
r
8
3
r
4
r
)
r
2
(
3
4
a
r
3
4






= 0.52( 52% )
(Note:This is not a very efficient way of packing as the packing fraction is very low)

(iv) Coordination Number, CN:
It is defined as the No. of atoms touching any one particular atom = 6 for simple cubic
Type of neighbour Distance no.of neighbours
nearest a 6 (shared by 4 cubes)
(next)1
2
a 12 (shared by 2 cubes)
(next)2
3
a 8 (unshared)
(v)Density of unit cell :
Volume
cell
unit
of
Mass
 3
a
.
NA
M
Z 

BODY CENTRED CUBIC :
The body centred cubic is a unique way of packing, as the 2D - arrays that can be imagined to constitute
the space lattice are themselves formed in a unique way. The lattice points in the 2D array do not touch
each other. The spheres start touching each other only upon moving from 2D to 3D, i.e when the 2D arrays
are placed on top of each other in such a fashion that the spheres of the next plane are into the cavities of
the first plane of spheres.The third plane of spheres is then exactly identical to the first plane of spheres.
(i)Relation b/w a and r: a  2r (as atoms along the edge are not touching each other)
they touch along the body diagonal, hence 3 a = 4r.
.
(ii) Effective no-of atoms (f) = 1 +
8
1
× 8 = 2.
(iii) Packing fraction = 3
3
r
4
4
4
r
3
4
2




3
3
3 
 =
8
3

–
~ 0.68 ( 68% )
(iv)Coordination No. CN: 8 (the sphere at the body centre will be touching the spheres at the eight corners)
(v) Density = 3
a
3
a a
N
M
2
a
N
ZM


BCC :
Type of neighbour Distance no.of neighbours
nearest 2r =
2
3
a 8
(next)1
= a 6
(next)2
= 2
a 12
(next)3
=
2
11
a 24
(next)4
= 3
a 8
NCERT Ex.1.3
1.3 An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the
element is 7.2 g/cm3. How many atoms are present in 208 g of the element?
Sol. Volume of the unit cell = (288 pm)3
= (288×10–12
m) = (288×10–10
cm)3
= 2.39×10–23
cm3
Volume of 208 g of the element
density
mass
= 3
cm
g
2
.
7
g
208
 = 28.88 cm3
Number of unit cells in this volume
=
cell
unit
/
cm
10
39
.
2
cm
88
.
28
3
29
3


= 12.08×1023
unit cells
Since each bcc cubic unit cell contains 2 atoms, therefore, the total number of atoms in 208 g = 2 (atoms/
unit cell) × 12.08 × 1023
unit cells
= 24.16×1023
atoms.
Ex.6 How many 'nearest' and 'next nearest' neighbours respectively does potassium have in b.c.c. lattice
(A) 8, 8 (B) 8, 6 (C) 6, 8 (D) 8, 2
Ans. (B)
Ex.7 If a metal has a bcc crystal structure, the coordination number is 8, because :
(A) each atom touches four atoms in the layer above it, four in the layer below it and none i n
its own layer.
(B) each atom touches four atoms in the layer above it, four in the layer below it and one i n
its own layer.
(C) two atoms touch four atoms in the layer above them, four in the layer below them, and none
in their own layer.
(D) each atom touches eight atoms in the layer above it, eight in the layer below it and
none in its own layer.
Ans. (A)
Ex.8 Potassium crystallizes in body centered cubic lattice with a unit cell length a = 5.2 Å
(A) What is the distance between nearest neighbours
(B) What is the distance between next nearest neighbours
(C) How many nearest neighbours does each K atom have
(D) How many next nearest neighbours does each K have
(E) What is calculated density of crystalline K.
Sol. (A) 4.5 Å, (B) 5.2, (C) 8 , (D) 6 (E) 0.92 g/mL
(A) 2r =
2
a
3
=
2
2
.
5
3 
= 4.5 Å (B) distance = a = 5.2 Å
(C) 8 (D) 6 (E) d = 23
3
8
10
02
.
6
)
10
2
.
5
(
39
2




 = 0.92 g/ml

Lecture # 3
Section (C) Closest packing.
(1) Voids and their geometry and coordination number (2D). [10 Mins]
(2) Voids and their geometry and coordination number (3D) closest packing. [10 Mins]
(3) Radius of biggest and smallest atom that may fit in these voids only. [10 Mins]
(4) ABAB.........type and ABC.........ABC type closest packing. [20 Mins]
(5) Locating tetra and octahedral void in ccp and hcp, Number of voids per atom
in closest packing. [10 Mins]
(1) Voids and their geometry and coordination number (2D).
TYPES OF VOIDS FOUND IN CLOSE PACKINGS :
Although the close packed structures have the maximum packing efficiency, there are indeed empty
spaces left in the arrangements. Let us analyse the types of such voids and the maximum radius of a
particle that can be placed in such voids.
(VOID IN TWO DIMENSIONS)
(1) Triangular void (2-Dimensional 3-coordinate void) The triangular voids are found in the planes of
the close packed structures, whenever three speres are in contact in such a fashion.
2
R
30°
R
2
R
cos30° =
r
R
R


2
3
=
r
R
R


3
2
=
R
r
R 
 r = 0.155 R
where, R= Radius of the sphere, r = rmaximum radius of a sphere that can be placed inside the void.
(2) Square void (2-Dimensional 4-coordinate void) The square voids are found in the planes of the
close packed structures, whenever three speres are in contact in such a fashion.
AC2
= AB2
+ BC2
 [2(R + r)]2
= (2R)2
+ (2R)2
2(R + r) = 2
2 R r = )
1
2
(  R = 0.414 R
(2) Voids and their geometry and coordination number (3D) closest packing.
(3) Radius of biggest and smallest atom that may fit in these voids only.
(VOIDS IN THREE DIMENSIONS)
(2) Tetrahedral void (3-Dimensional 4-coordinate) :
The tetrahedral void is formed whenever a sphere is placed on top of the triangular arrangement.
a
b
c d
/2
A B
O
R+r
R
R
A
B
C
D
 = 109.28°
O

sin
2

=
r
R
R


r
R
R

= sin 54° 44  r = R
















1
2
sin
1  r = 0.225 R
(3) Octahedral void (3-Dimensional 6 coordinate void) The octahedral void is formed whenever two spheres
are placed, one on top and the other below a square arrangement of spheres.This can also be obtained by
placing two THREE-BALL arrangements on the top of each other such that a hexagonal view is obtained.
2
3
4 1
5
6
R
r
R
r
x
x = 2R; 2(R + r) = 2 x = 2
2 R r = )
1
2
(  R = 0.414 R
(4) Cubical void (3-Dimensional 8-coordinate void)
Along body diagonal : a
3 = 2 (R + r)
2 R
3 = 2 (R + r)
r
r
R
R
r = R
)
1
3
( 
r = R
)
1
3
(  r = 0.732 R
The cubical void is generally not found in closed packed structures, but is generated as a result of distortions
arising from the occupancy of voids by larger particles.
Ex.9(a) In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of
tetrahedral and octahedral voids?
(A) 0.543 (B) 0.732 (C) 0.414 (D) 0.637
Ans. (A)
Sol.
octahedral
l
tetrahedra
r
r
= R
414
.
0
R
225
.
0
= 0.543.
(b) The numbers of tetrahedral and octahedral holes in a ccp array of 100 atoms are respectively
(A) 200 and 100 (B) 100 and 200 (C) 200 and 200 (D) 100 and 100
Ans. (A)
Ex.10 Copper has a face-centred cubic structure with a unit-cell edge length of 3.61Å. What is the size of the
largest atom which could fit into the interstices of the copper lattice without distorting it?
Figure
(Hint. : Calculate the radius of the smallest circle in the figure)
Ans. 0.53Å
Sol. roctahedral = 0.414 R
for FCC 4R = a
2 ; R =
4
a
2
; r =
4
a
2
414
.
0
=
4
61
.
3
2
414
.
0 
= 0.53 Å

Lecture # 4
Section (D) hcp structure Point Time
(1) Derivation of unit cell from ABAB............. (1 ) [10 Mins]
(2) Calculation of a, b and c (height) (ABAB) ; volume of unit cell. (2, 3, 4) [30 Mins]
(3) Occupancy of unit cell.
(4) Packing fraction, density and coordination number.
(5) Next nearest neighbours.
(6) Location of tetrahedral and octahedral voids in hcp structure
(formula of unit cell or compound). (5 + 6) [15 Mins]
(4) ABAB.........type and ABC.........ABC type closest packing.
CLOSE PACKING IN THREE DIMENSIONS: This is the arrangements in which metallic crystals are generally
found. To generate close packing we must start with best possible arrangements in two dimensions i.e.,
the hexagonal arranged sheets and spheres of IInd sheet must rest depressions/voids of 1st sheet hexagonal
close pached
 All the voids of 1st sheet canot be occupied by shperes of 2nd sheet (because of same size of spheres)
Only 50% of the viods can be occupied. Out of six voids around anly sphere only three atternate voids can
be occupied by 2nd sheet of spheres. This arrangements is shown below.
a a
a
b
b b
b
b
In second layer we have two kinds of voids.
(i) Voids of second layer below which there are spheres of first layer (all voids of type ‘a’).
(ii) Voids of second layer below which there are voids of first layer (all voids of type ‘b’).
for third layer we have two possibilities.
(A) HEXAGONAL CLOSE PACKING( HCP )
If spheres of IIIrd
layer are placed in voids of IInd
layer which are lying exactly above the spheres of Ist layer
i.e., voids of type ’a’ then Ist
layer and IIIrd layer will become exactly identical and this generates
ABABAB pattern or the hexagonal close packing (HCP)
Unit cell : a = 2r = b;  = 120º
CALCULATIONS
1. For ‘c’: For the estimation of ‘c’, consider the spheres marked 1,2,3,4 in the unit cell as shown.These
four spheres form a regular tetrahedron. The length of the perpendicular from ‘4’ to the equilateral triangle 1-
2-3 will be equal to c/2.

cos 30º =
x
2
a
x =
30
cos
2
a
=
3
a
3
2
a2

apply pythagoras theorem.
x2
+ (C/2)2
= a2
 C =
3
2
4r
Volume of the hexagon = Area of base x Height.
= 6 .
4
3
a2
× c = 6 .
4
3
3
2
r
4 2
 4r = 24 2 . r3
2. Effective no. of atoms/unit cell, Z :
Z= 3 + 2 ×
2
1
+ 12 ×
6
1
= 3 + 1 + 2 = 6 [z = 6]
The three atoms at the centre of the hexagonal structure is not completely inside the unit cell. A part of
it lies lies the unit cell, but a part of other atoms in the same plane whose centre is not in the unit cell is
included inside the unit cell.
Hence eff. no. = 3.
3. Packing fraction : = = 2
3
r
2
24
r
3
4
6 

2
3

 = 0.74 = 74%
4. Coordination No. CN= 12( each spheres touches 6 spheres in its layer,3 above and 3 below)
5. Density =
Volume
Mass
= )
volume
(
Na
M
6
)
volume
(
N
ZM
a

NOTE: In closest packings, whenever two consecutive layers are of different kind then packing efficiency
will always be 74%.
Illustration 10. Show by simple calculation that the percentage of space
ccupied by spheres in hexagonal cubic packing (hcp) is 74%

Solution: Let the edge of hexagonal base =a And the height of hexagon = h And radius of sphere = r
The centre sphere of the first layer lies exactly over the void of 2nd
layer B.
The centre sphere and the spheres of 2nd
layer B are in touch
So, In ÄPQR (an equilateral triangle)
PR = 2r, Draw QS tangent at points
 In QRS QRS = 30°, SR = r
Cos30° = SR/QR
QR = r / 2
/
3 = 2r / 3
 PQ = 2
PR – QR2
= 2
r
4 – 4r2
/ 3
h1
= 2
r
8 / 3 = 2 2 /3r
 h = 2h1
= 4 2 /3r
Now, volume of hexagon = area of base x height
= 6 × 3 / 4 a2
× h  6 × 3 /4 (2r)2
× 4 2 /3r
[ Area of hexagonal can be divided into six equilateral triangle with side 2r)
No. of sphere in hcp = 12 × 1/6 + 1/2 × 2 + 3 = 2+1+3 = 6
 Volume of spheres = 6 × 4/3r3
 Percentage of space occupied by sphere = 6 × 4/3r3
/ 6 × 3 /4 × 4r2
× 4 2 /3 r × 100 = 74%.
LOCATION OF VOIDS HCP unit cell : (with the help of model)
(A) Tetrahedral voids:
Tetrahedral voids are found between any two spheres of alternate sheets that lie perpendicular to the
plane of the sheets.
8 are completely inside and 12 are shared
Effective number of tetrahedral voids in hcp unit cell :
effectively  8 + 12 ×
3
1
= 12
(B) Octahedral Voids: Total 6 octahedral voids completely inside the unit cell.
In HCP : Z = 6, tetrahedral voids = 12
In HCP: Z = 6, octahedral voids = 6
Lecture # 5
Section (E) ccp structure Point Time
(1) Derivation of unit cell from ABCABC..........ABC.......... (1) [10 Mins]
(2) Relation between edge length (a) and radius (R) for (ABCABC);
volume of unit cell. (2, 3, 4) [30 Mins]
(3) Occupancy of unit cell.
(4) Packing fraction, density and coordination number.
(5) Next nearest neighbours.
(6) Location of tetrahedral and octahedral voids in ccp structure. (5, 6, 7) [20 Mins]
(7) Formula of unit cell or compound
(B) ABC-ABC ARRANGEMENT/CUBIC CLOSE PACKING( CCP ):
If the third layer spheres are placed in those voids of IInd layer under which there are voids of the first layer
of spheres (voids of type ‘b’), then the first and the third layer of spheres will not be identical.Such an
arrangement will lead to anABC-ABC type of arrangement.It is also known as the cubical close packing or
also as the Face Centred Cubic structure(FCC).

a a
a
b
b b
b
b
In the ABC – ABC Pattem the spheres of 4th layer are vertically above the spheres of Ist layer then these
consecutive layers are different from each other, fourth layer will be idential to first layer so it will be called
ABC – ABC repeat pattern.It is also called the ccp (cubical close packing) because a cubical type of unit
cell is used for the study of this arrangement.
1. Relation between ‘a’ and ‘r’ :
a  2r
r
4
a
2  (as the spheres touch along the face diagonal)
2. Effective no. of atoms per unit cell, Z : Z= 4
6
2
1
8
8
1




3. Packing fraction : p.f. 2
2
r
4
4
4
r
3
4
4
3
3






 =
2
3

= .74( 74% )
4. Coordination number, CN = 12 ( each corner sphere touches three spheres per cube(each of them
being shared by two of the eight cubes surrounding that sphere) )
5. Density = 3
A a
.
N
M
Z 

Type of neighbour Distance no. of neighbours
nearest
2
a
12 = 




 
2
8
3
(next)1
a 6 = 




 
4
8
3
(next)2
2
3
a 24
(next)3
2
a 12
(next)4
2
5
a 24
LOCATION OF VOIDS : FCC/CCP unit cell: (with the help of model)
(A) Tetrahedral voids : The FCC unit cell has eight tetrahedral voids per unit cell. Just below every
corner of the unit cell, there is one Tetrahedral void.As there are eight corners, there are eight tetrahedral
voids.
1
2
3
4
The spheres 1, 2, 3, 4 form a tetrahedral void.
(B) Octahedral voids :In an FCC unit cell, there are four octahedral voids. They are present at all the
edgecentres and at the body centre. The contribution of the edge centre is
4
1
Hence, total number of octahedral voids = 






4
1
12 + (1) = 4
edge centres body centre
In CCP/FCC :Z = 4, tetrahedral voids = 8
number of tetrahedral voids = 2 × Z
no. of tetrahedral voids in close packings = 2 × eff. no. of spheres.
HENCE, there are two Tertahedra Voids per sphere in closed packing arrangements.
In CCP/FCC: Z = 4, octahedral voids = 4
no. of octahedral voids = Z
There is exactly one OV per sphere in close packing.
NCERT Intext Questions : 1.15 to 1.18
Ex.11 A metal crystallizes in two cubic phases i.e., FCC and BCC whose unit cell lengths are 3.5Å and 3.0
Å respectively. The ratio of their densities is :
(A) 3.12 (B) 2.04 (C) 1.26 (D) 0.72
Ans. (C)
Sol. d =
A
3
N
a
ZM
2
1
d
d
= 3
)
5
.
3
(
4
 2
)
3
( 3
= 1.26.
Ex.12 In a ccp structure, the :
(A) first and third layers are repeated (B) first and fourth layers are repeated
(C) second and fourth layers are repeated (D) first, third and sixth layers are repeated.
Ans. (B)

Ex.13 Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit
cell dimensions are a = 6.8 Å, b = 4.4 Å and c=7.2 Å. If the molar mass is 21.76, then the density of
crystals is :
(A) 0.6708 g cm–2
(B) 1.6708 g cm–3
(C) 2.6708 g cm–3
(D) None of these.
Ans. (A)
Sol. d =
A
3
N
a
ZM
= 23
8
8
8
10
023
.
6
10
2
.
7
10
4
.
4
10
8
.
6
76
.
21
4










 = 0.6708 g cm2–
.
Ex.14 An fcc lattice has lattice parameter a = 400 pm. Calculate the molar volume of the lattice including all
the empty space:
(A) 10.8 mL (B) 96 mL (C) 8.6 mL (D) 9.6 mL
Ans. (D)
Sol. Volume of 4 atoms = a3
= (4  10–8
)3
cm3
volume of NA
atoms =
4
)
10
4
( 8


 6.023  1023
= 9.6 ml.
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Lecture # 6
Section (F) Radius ratio rule & type of Ionic structures : Point Time
(1) Radius ratio rule (1) [5 Mins]
(2) Rock salt structure (2, 3, 4) [20 Mins]
(3) Anti fluorite structure
(4) ZnS structure
(5) Anion forms HCP & Cation in all octahedral void. (5, 6, 7) [10 Mins]
(6) Anion forms HCP & Cation in all tetrahedral void.
(7) Anion forms HCP & Cation in half tetrahedral void.
(8) Fluorite structure (8, 9, 10) [15 Mins]
(9) Perovskite structure
(10) Spinal structure (11, 12) [10 Mins]
(11) Diamond structure.
(12) Temperature & pressure effect on coordination number of compound.
Radius ratio rule: Radius ratio = 









r
r
,this gives the idea about the type of void occupied.
Radius ratio Type of void occupied Coordination No.


r
r
< 0.155 linear void 2
0.155 


r
r
< 0.225 triangular void 3
0.225 


r
r
< 0.414 tetrahedral void 4
0.414 


r
r
< 0.732 octahedral void 6
0.732 


r
r
< 1 cubical void 8
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
POSSIBILITIES OF THE RADIUS RATIOS(deal along with the table given)
Case I :


r
r
< 0.155 +
 It is unstable situation because cation is not touching any of anion and anions are in contact.

Case II :


r
r
= 0.155
+
a
a = 2r
This is the perfect arramgement for triangular void.Here cation will be touching three anions hence C.N.
= 3 and anions will be in limiting contact.
Case III : 0.155 <


r
r
< 0.225
+
a
 Here cation will be touching three anions hence C.N. = 3
but anions will not be in contact
3
a
= r+
+ r–
; a = 3 (r+
+ r–
)
Case IV :


r
r
= 0.225
Cation is perfectly in Tetrahedral void, C.N = 4 fcc  anion-anion contact will still be therefore
2 afcc
= 4r–
Case V : 0.225 <


r
r
< 0.414
Here fcc lattice will expand now anion will no longer be in contact therefore 2 afcc
> 4r–
Considering a minicube
Mini
Cube
a/2
Cation will lie at the centre of the body diagonal of the mini cube of side a/2.
r–
+ r+
=
2
1
× (body diagrol of minicube) r+
+ r–
=
2
1
×
2
afcc
3 a =
fcc
)
r
r
(
3
4
–


Case VI :


r
r
= 0.414
Cation is perfectly in Octahedral void, C.N. = 6
For fcc lattice 2 afcc
= 4r–
is valid and anions will be in limiting contact
+
a

Case VII : 0.414 <


r
r
< 0.732
FCC lattice undergoes symmetrical exapnsion. Anions loses contact 2 afcc
> 4r–
But the relative orientation of Cation and anions would remain same as in te case of perfect OV
situation.
afcc
= r–
+ 2r+
+ r–
Case VIII :


r
r
= 0.732 Cation is perfectly in cubical void C.N = 8 asc
= 2r–
Case IX : 0.732 <


r
r
< 1.00
Expansion undergoes and anions are not in contact.But relative orientation will remain same as that is
perfect cubic void.
3 asc
= 2(r+
+ r–
)
Structure of ionic compounds
Strcuture of compounds containing two different types of atoms.
The bigger atom or ion will form the lattice & smaller atom / ion will occupy the voids. Generally, bigger ion
is anion and smaller ion is cation.The type of void occupies by the cation is decided by radius ratio.
 In ionic compounds for maximum stability.
(1)ACation must be surrounded by maximum number of anions and vice versa.
(2)Anion-anion and cation–cation contact should be avoided
 There will 2 different kind of atoms, cation & anion, so there will be two different coordination number
(C.N.). C.N. of cation = no. of anions surroundings any cation.
C.N. of anion = no. of cations surrounding any anion.
STRUCTURE OF SOME REPRESENTATIVE IONIC COMPOUNDS
Ionic solids form rigid lattice structure.Large amount of energy is released when ions form lattice
arrangement form their gaseous states.This is expressed in terms of the Lattice energy(kJ/mol).This
much amount of energy is required to separate one mole of ionic-solid into its gaseous ions.
(1) NaCl type( AB-type ) of structure or rock salt structure or 6 : 6 C.N. compound
Experimental: –
Cl
Na
r
r 
= 0.51
Cl–
ions:fcc lattice Effective no. of Cl–
ions in a unit cell = 4
Eff. No. of Na+ ions in a unit cell must be 4 ( formula  Nacl)
Na+
ions must occupy all the octahedral voids.
formula of one unit cell = Na4
Cl4
= NaCl
effective no. of formula units , Z = 4 mass of unit cell =
A
N
ZM
M = 58.5 gm/mole
Edge length of unit cell :- a(fcc) = 2(rNa
+ + rCl
–) density  = 3
fcc
A a
.
N
ZM
Anion Cation

C.N.Na+ = 6 ; C.N.Cl– = 6
NaCl = 6 (cation) : 6 (anion) CN compound
For Cl–
:
Type of neighbour Distance No. of neighbours
Nearest Na+
,
2
a
6
(Next)1
Cl–
,
2
a
12
(Next)2
Na+
(B.C.),
2
3
a 8
(Next)3
Cl–
, a 6
Example : Halide of Li, Na, K and Rb, AgCl, AgBr, NH4
Cl.
Note : Lattice of NaCl is FCC of Cl–
in which all octahedral void are occupied by Na+
.
or  FCC of Na+
in which octahedral void are occupied by Cl–
.
(2) ZnS type( AB-type ) of unit cell or 4 : 4 CN compound
Zns – fcc (zinc blende) ; Zns – hcp (wurtzite)
experimental: 


2
2
s
zn
r
r
0.3
S2–
will form fcc lattice. effective no. of S2–
ions in one unit cell = 4.0
Zn2+
ions will occupy alternate (non-adjacent) four out of eight Tetahedral voids.
none of the face of the minicube should be common.
 effective no. of Zn2+
ions = 4 formula of unit cell = Zn4
S4
effective no. of formula units, Z = 4
mass =
A
N
ZM
M = 97 gm/mole








 

2
a
3
2
1
r
r fcc
s
zn 2
2  afcc
=
3
4
)
r
r
( 2
2
s
Zn 
  density  = 3
fcc
Aa
N
ZM
C.N. of Zn2+
ions = 4 C.N. of s2–
ions = 4
Ex : ZnS, CuCl, CuBr, CuI,AgI.
For S2–
Type of neighbour Distance No. of neighbours Ions
Nearest
4
3
aFCC
4 Zn2+
(Next)1
2
a
12 S2–
(Next)2
4
11
a 12 Zn2+
(Next)3
a 6 S2–
(Next)4
11
19
aFCC
12 Zn2+
3. CaF2
(Flourite structure)( AB2
-type) or 8 : 4 coordination no. compound
–
2
F
ca
r
r 
 Here Cation is bigger than Anion experimental: 0.225 >



ca
F
r
r
< 0.414
fcc lattice position are occupied by Ca2+
ion Effective no. of ca2+
ions = 4
F¯ ions occupy all the Tetrahedral voids Effective no. of F¯ ions = 8
formula = Ca4
F8
= CaF2
Effective no. of formula units, Z = 4
afcc
= )
r
r
(
3
4
–
2
F
Ca

 C.N. of Ca2+
= 8 C.N of F¯ = 4
Other examples : CaF2
, SrF2
, BaF2
, BaCl2
.

For Ca2+
, Type of neighbour Distance No. of neighbours Ions
Nearest
4
3
8 F–
(Next)1
2
a
12 Ca2+
(Next)2
16
11
24 F–
(Next)3
a 6 Ca2+
For F—
Type of neighbour Distance No. of neighbours Ions
Nearest
4
3
4 Ca2+
(Next)1
2
a
6 F–
(Next)2
2
a
12 F–
(Next)3
16
11
– Ca2+
Ex.15 A mineral having formula AB2
crystallize in the cubic close packed lattice, with the A atoms occupying
the lattice points. What is the co-ordination no. of A atoms? of the B atoms? what fraction of tetrahedral
sites is occupied by B atoms.
Ans. 8, 4, 100%.
It has fluorite (CaF2
) structure.
4. Na2
O (sodium oxide) (A2
B-type )or Anti - Fluorite structure or 4 : 8 coordination compound
O2–
– fcc lattice
Na+
– all the tetrahedral voids
5. CsCl type of structure or 8 : 8 CN compound : Experimantal : 

–
Cl
Cs
r
r
0.93
Cl–
forms Simple cubic lattice Effective Cl¯ ions = 1
Cs+
will occupy cubic void Effective Cs+
ion = 1
Formula = CsCl Z = 1 [mistake in NCERT not BCC lattice]
3
sc
a
Na
ZM

 asc
= )
r
r
(
3
2
–

 C.N of Cs+
= 8 C.N of Cl– = 8
Other example : CsCl, CsBr, CsI.
Ex.16 CsBr has b.c.c. structure with edge length 4.3 A. The shortest inter ionic distance in between Cs+
and Br–
is:
(A) 3.72 (B) 1.86 (C) 7.44 (D) 4.3
Ans. (A)
Sol. r+
+ r–
=
2
a
3
=
2
3
.
4
3 
= 3.72 Å .
Note : (1) Distance between two plane in FCC or HCP arrangement is
3
2
a or a = 2R = 2
3
2
R.
(2) If number of unit cell along one edge are ‘x’ then total number of unit cell in cube = x3
.

e.g.
Figure
Number of unit cell along one edge = 2, then total number of unit cell in cube = 23
= 8.
Other important structure :
(1) Spinel structure [AB2
O4
] :
Spinel is an oxide consisting of two type of metal ions with the oxides ion arranged in CCP layers in
normal spinel one eight of the tetrahedral holes occupied by one type of metal ion and one half of the
octahedral holes occupied by another type of metal ion.
Ex : A spinel is formed by Zn2+
, Al3+
and O2–
with Zn2+
ion in the tetrahedral holes. The formula of the
spinel is ZnAl2
O4
.
(2) Perovskite structure [ABO3
] :
This structure may be described as a cubic lattice, with barium ions occupying the corners of the unit
cell, oxide ions occupying the face centers and titanium ions occupying the centres of the unit cells.
Ex : BaTiO3
or MgTiO3
.
(3) Lattice of diamond :
 ZnS types in which all S2–
location and all Zn2+
location are occupied by C atoms.
So, Z = 8 atom per unit cell
aFCC
=
3
4
[dC – C
]
dC – C
= 2rC – C
where rC – C
is radius of C-atom.
 Packing efficiency = 34%.
 In crystal lattice of diamond, carbon atoms adopt FCC arrangement with occupancy of 50%
tetrahedral voids.
EFFECT OF PRESSURE AND TEMPERATURE COORDINATION NUMBER OF COMPOUND :
On increasing pressure  C.N. tends to increase
On increasing temperature  C.N. tends to decrease
e.g. 4 : 4
P 
6 : 6 ; 8 : 8
T 
6 : 6
–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
(ONLY FOR FACULTY)
Diamond Structure :
Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using
only covalent bonds, e.g., diamond structure. The latter is based on a face-centred cubic lattice where four
out of the eight tetrahedral, holes are occupied by carbon atoms. Every atom in this structure is surrounded,
tetrahedrally by four others. No discrete molecule can be discerned in diamond.
The entire crystal is a giant molecule, a unit cell of which is shown in Fig. 3.19.19. Germanium, silicon and
grey tin also crystallize in the same way as diamond.

Void volume in Diamond Crystal :
From Fig. we observe that
C – C bond = 2rc
, where rc
is the radius of sphere representing carbon
C-C-C = 109º28'
Diagonal face of the cube = 2 a
This distance will be equal to four times the distance dC–C
sin (54º44'), that is
4dC–C
sin(54º44) = 2 a or a =
2
)
44
º
54
sin(
d
4 C
–
C 
Area of the cube = a3
=
3
C
–
C
2
)
44
º
54
sin(
d
4





 
Total number of carbon atoms/unit cell = 8
Volume occupied by 8 carbons = 8 





 3
C
r
3
4
=
3
4
(dC – C
)3
Fraction of volume occupied = 3
C
–
C
3
C
–
C
2
)
44
º
54
sin(
d
4
)
d
(
3
4





 

= 0.34.
Graphite Structure : Graphite is another allotropic form of carbon and
is more stable than diamond. This has a layertype structure as shown in
Fig.
In each layer, we have carbons attached to each other through the
overlapping of sp2
hybrid orbitals, a stronger bond. Different layers are
held byweaker joinings, i.e. the 7t-bondings. The great difference between
graphite and diamond can be understood in terms of the above crystal
lattice. The distance between atoms in the plane is 143 pm, but the
distance between the two atomic lay planes is 335 pm. In two directions
the carbon atoms are tightly held as in the diamond, but in the third direction, the forces of attraction are
much less. As a result of this, one layer can slip over another. The crystals are flaky, and yet the material
is not wholly disintegrated by shearing action. This planar structureu a part of the explanation of the
lubricating action of graphite. Atoms with a valency of 2 cannot form isotropic three-dimensional
structures. Consequently, we find structures which consist of endless chains of atoms extending through
the crystal, the individual chains being held together by much weaker forces. Another example is that of
rhombic sulphur. Herethereareeight memberedringsof sulphur atoms. The bivalence of sulphur is maintained
and different molecules are held together by the van der Waals attractions. Generally, the covalent solids
have comparatively low densities as a result of the low coordination numbers. This effect is intensified in
those crystals in which covalently bound structural units are held in the crystal by van der Waals forces.
The distance between two units held by van der Waals forces is significantly greater than that between
units held by covalent, ionic or metallic bonds; these large distances result in a low density of the solid.
Ex. The edge length of the unit cube of diamond is 356.7 pm and this cube contains 8 carbon atoms.
Calculate: (a) the distance dC – C
between carbon atoms, assuming them to spheres in contact ; (b) the
fraction of the total volume that is occupied by carbon atom
Sol. (a) Edge length of the cube = 356.7 pm ; Face diagonal of the cube = 2 (356.7 pm) = 504.3 pm
This distance is also given by d = 4(dC – C
sin 54°44') = 4dC – C
(0.8164)
Thus, dC – C
=
8164
.
0
4
pm
3
.
504

= 154.4 pm Hence, Radius of carbon =
2
d C
–
C
=
2
pm
4
.
154
= 77.2 pm
Volume of the unit cell = (356.7 pm)3
= 4.538 x 10–29
m3
Volume occupied by 8 spheres = 8 





 3
)
pm
2
.
77
(
3
4
=
3
4
 (154.4 X 10–12
m)3
= 1.542 × 10–29
m3
Fraction of the volume occupied by carbon atoms = 3
29
3
29
m
10
538
.
4
m
10
542
.
1




= 0.34.

Lecture # 7
Section (G) Crystal defects & Properties of solid : Point Time
(1) Definition (1, 2, 3) [20 Mins]
(2) Stoichiometric defects
(3) Point defects – Schottky defect, Frenkel defect, Impurities defect
(4) Non-stoichiometric defect (4) [20 Mins]
Metal excess defects (F-center)
Non metal excess defects (metal deficient defect)
(5) Electrical Properties : (5, 6) [10 Mins]
Conductors, Insulators, Semiconductors.
(6) Doping : n-type semiconductors, p-type semiconductors, Superconductivity.
(7) Magnetic Properties (7,8,9) [10 Mins]
(i) Paramagnetism (ii) Diamagnetism (iii) Ferromagnetism
(iv)Anti-Ferromagnetism (v) Ferrimagnetism
(8) Application of semiconductor
(9) Effect of temperature on conductors, semiconductors, insulators.
CRYSTAL IMPERFECTIONS / DEFECTS
Imperfections can be because of :-
- Conditions under which crystals have been developed,
- Impurities,
- Temp (because of thermal conductivity some atoms/ions can get displaced)
These imperfections can be
(a) Point defects - defects will be only at certain lattice positions.
(b) Line defects - If atoms/ions are misplaced/missing/replaced by some other ions along a line.
(c) plane (screw) defects - If atoms/ions are misplaced/missing/replaced by some other ions along a line
in a plane.
POINT DEFECTS
1. Stoichiometric -The formula of compound remains same even after presence of these defects.
2.Non-stoichiometric -The formula of compound will get modified because of the presece of these
defects.
STOICHIOMETRIC DEFECTS
(a) Schottkey (b) Frenkel (c) Interstitial
Cation Anion
Cation Anion
Missing ions pairs
Schottky defect
Cation Anion
Frenkel defect
dislocated cation Cation Anion Interstitial entity
Interstitial defect
SCHOTTKEY : when atoms/point are totally missing from the lattice.
net density of crystal willl get decressed
exp
< theoritical
% missing units = %
100
–
th
exp
th












FRENKEL : When atoms/ions displaced from normal lattice positions and are present in some
interstitial voids.Density remains same

INTERSTITIAL :
When some small foreign atoms are trapped in interstitial voids of the lattice without any chemical reaction.
Formula remains the same exp
> theortical
NON-STOICHIOMETRIC DEFECTS :
(a) metal excess or cation excess defect :-
- If no. of missing anione is more than no. of missing cations.
- To maintain electrical neutrality some electron are trapped at anionic vacancies.
Because of these extra e–
the electrical and optical (colour) properties of the compound get modified.
So these locations of e–
s are also known as color centres (F - centres) .
ZnO - white in color at room temperature.
- on heating some O2–
ione get released in the form of O2
and e–
are trapped at their locations. because
of this it because yellow in color.
(b) metal deficiency (cation deficiency) defects
- no. of missing cations should be more than no. of missing anions.
- oxidation no. of cation will get modified to maintain electrical mentrality.
Cation Anion
-e
F-centre
Metal excess
Na+ Cl- Na+ Cl- Na+ Cl-
Na+ Cl- Na+ Cl- Sr+2 Cl-
Na+ Cl-
Cation Vacancy
Sr+2 replacing Na+
Na+ Cl- Na+ Cl- Sr+2 Cl-
Ex.17 Which of the following is incorrect
(A) The defect is known as schottky defect
(B) Density of compound in the defect decreases
(C) NaCl(s) is example which generally shows this defect
(D) Stoichiometry of compound will change slightly.
Ans. (D)
Ex-18 Ferrous oxide (FeO) is experimentally found to have the formula Fe0.93
O.Find the %age of Fe ions in +3
state.
Sol:
Fe O
0.93
Some Fe atom
are in +2 state
Some Fe atom
are in +3 state
Let there is xFe atom in +3 state
3x + 2(93 – x) = 200 x = 14
% Fe3+
=
93
14
× 100% =
93
1400
%  15.54%

PROPERTIES OF SOLIDS :
(i) Electrical Properties (ii) Magnetic Properties
(i) Electrical Properties : Solids exhibit an amazing range of electrical conductivities, the range of electrical
conductivities from 10–20
to 107
ohm–1
m–1
. Solids can be classified into three types on the basis of their
conductivities.
(1) Conductors : Metals are goods condcutors and have conductiveites in the order 107
(m)–1
.
(2) Insulators : Those solids which have very low conductivities ranging from 10–20
to 10–10
(m)–1
are
electrical insulators e.g.; MnO, CoO; NiO, CuO, Fe2
O3
, TiO2
.
(3) Semicondutors : Those solids which have intermideidate conductivities generally from 10–6
to 104
(m)–1
are termed as semicondcutors.
Intrinsic Semicondcutors : The conduction by pure subtances such as silicon and germanium is called
intrisic conduction and these pure substances exhibiting electrical condcutivity are called intrinsic
semicodcutors.
Causes of Conductance in Solids :
1. In most of the solids, conduction is through electron movement under an electric field.
2. In ionic solids conduction is by movement of ions in molten state.
3. The magnitude of electrical conductivity strongly depends upon the number of electrons available to take
part in conduction process.
4. In metals, conductivity strongly depends upon the number of electrons available per atom. The atomic
orbitals form moleuclar orbitals which are too close in energy to each other so as to form a band.
5. If conduciton band is not completely filled or it lies very close to a higher unoccupied band, then
electrons can flow easily under an electric field thereby showing conductivity.
6. In case of insulators, the gap between valence band conduction band is too large, so electrons cannot
jump from valence band to conduction band and very small conductivity is observed.
7. In case of semiconductors, the gap between valence band and conduction band is small and therefore
some of the electrons may jump from valence band to conduction and some conductivity is observed.
8. Electrical conductivity of semiconductors increases, with increase in temperature. This is due to the fact
that with increase in temperature, large number of valence electrons from the valence band can jump to
conduction band. Pure substances like silicon and germanium that exhibit this type of conducting behaviour
are called interinsic semiconductors.
9. For partical purpose, the conductivity of pure silicon and germanium is too low at room temperature,
therefore, there is need to increase the conductance by doping.
Figure
Doping : The conductivity of silicon and germanium can be increases by adding apporpirate amount of
suitable impurity. The process is called doping.
Type of Semiconductors :
(1) n-Type Semiconductors : Metal excess compounds conduct electricity through normal electron
conduction mechanism and are therefore n-type semiconductors.

Creation n-Type Semicondcutor :
1. When siliconk is doped with small amont of group -15 elements such as P, As or Sb, its electrical
condcutivity increass sharply.
2. In pure silicon each silocn atom uses its four valence electrons for the formation of four covalent bonds
with the neighbouring silicon atoms.
3. When silcon is doped with some group-15 element, the some of the positions in the lattice are substituted
by atoms fo groups -15 elements have five valence electrons. After forming the four covalent bonds with
silicon (or anyother group-14 element such as germanium). one excess electron is left on them.
4. Since this electron is not involed in bonding it becomes delocalized and contributre to electrical
conduction. Silicon dped with group 15 element behaves as a n-type semconductor.
(2) p-Type Semiconductors : Metal deficient compounds conduct electricity through positive hole conduction
mechanism and are therefore p-type semiconductors.
Creation p-Type Semiconductor :
1. Electrical conductivity of silicon or germinum can alos be increases by doping with some group-13
element such as B, Al or Ga.
2. Goup-13 elements have only three valence electrons. The combine with group-14 elements to form an
electron deficient bond or electron vacancy or a hole. These holes can move sthough the crystal like a
positve charge giving rise to electrical condcutivity.
3. Gorup-14 elements doped with group-13 elements behave as p-type semicondcutors. In the presence of
electricl field the holes move in direction opposite to that of electrons.
Application of n-type and p-type semicondcutiors.
1. Diode is a combination of n-type and p-type semicondcutors used as rectifier.
2. They are used for making transitors which contains n-p-n and p-n-p junctions to amplify radio and audio
singnals.
3. The solar cell is photo-diode used to convert light energy into electrical energy.
13-15 Compouds : The solid state materials are porudced by combination of elements of goups 13 and 15
the compounds thus obtianed are called 13-15 compouds e.g. InSb, AlP GaAs.
12-16 Compounds : The solid state compounds are obtianed by combination of elements fof groups 12
and 16 the compounds are called 12-16 compouds e.g. ZnS, CdS, CdSe and HgTe.
Super Conductivity : The electrical resistance of metals is found to depend on temperature. Electrical
resistance decreases with decrease in temperature and becomes almost zero near the absolute zero.
Materials in this state are said to possess. superconductivity. The phenonmenon of superconductivity was
first discovered by kammerlingh Onners in 1913 when he found that mercury becomes superconducting at
4.0 K temperature.
.
Transition Temperature : The temperature at which a substance starts behaving as super-condcutor is
called transition temperature.
.
(ii) Magnetic Properties : Every substance has some magnetic properties associated with it. The origin
of these properties lies in the electrons. Each electron in an atom behaves like a tiny magnet.
Its magnetic moment originates from two types of motions.
(i) its orbital motion around the nucleus.
(ii) its spin around its own axis.

Figure : Demonstration of the magnetic moment associated with
(a) an orbiting electron and (b) a spinning electron.
On the basis of their magnetic properties, substances can be classified into five categories:
.
(i) paramagnetic (ii) diamagnetic (iii) ferromagnetic (iv) antiferromagnetic and (v) ferrimagnetic.
.
(1) Paramagnetism : When substances which are attracted by the external magnetic field are called
paramagnetic substances and the phenomenon is called as paramagnetism. Atoms ion or molecules
containing unpaired electron show this property, eg. O2
Cu2+
, Fe3+
etc. these substances lost their magnetism
in the absence of magnetic field.
(2) Diamagnetic materials : Those materials which are repelled by magnetic field are called diamagnetic
materials e.g. Cu+
, TiO2
, NaCl and C6
H6
. They do not have unpaired electrons.
(3) Ferromagnetism : When substances show permanent magnetism even in the absence of the magnetic
field this phenomenon is called as Ferromagnetism and such substances as called as Ferromagnetic
substances e.g. Fe Ni Co and CrO2
.
This type of magnetism arise due to the spontaneous alignment of magnetic moments due to unpaired
electron in the same direction.
(4) Anti Ferro Magnetism : Substances which are expected to possess paramagnetism or ferro magnetism
on the basis of unpaired electrons but actually they possess zero net magnetic moment are called Anti-
ferromatic substances and the phenomenon is called asAnti-ferromagnetism. eg. MnO,Anti-ferromagnetism
is due to the presence equal number of magnetic moments in the opposite direction.
(5) Ferrimagnetism : Substances which are expected to possess large magnetism on the basis of unpaired
electrons, but actually have small magnetic momentum are called ferrimagnetic substances eg. Fe3
O4
,
ferrites of the formula M2+
, Fe2
O4
where M = Mg, Cu, Zn etc. Ferrimagnetism arises due to the unequal
moments in opposite direction resulting in same net magnetic moment. On heating these substance loss
their magnetism and convert in to paramagentic substance
Curie Temperature : The temperature at which a ferromagnetic substance loses its ferromagnetism and
becomes only paramagnetic. For iron the curie temperature is 1033 K and for nickel 629 K, for Fe3
O4
850
K. Below this temperature paramagnetic solid becomes ferromagnetic.
Domain : In solid state the metal ions of ferromagnetic substances are grouped together into small
regions called domains.
.
Effect of Temperature on Metal (Conductor) Semiconductor or Insulator :
1. The conductivity of semiconductors and insulators increases with increase in temperature
2. The conductivity of metal (conductors) decreases with increase in temperature.
NCERT Intext Questions : 1.19 to 1.24

Ex.19 What is a semiconductor? Name the two main types of semiconductors.
Ans. Substances whose conductance lies in between that of metals (conductors) and insulators are called
semiconductors. Two main types of semiconductors are n-type and p-type.
Ex.20 Explain the following with suitable examples:
(i) Ferromagnetism (ii) Paramagnetism (iii) 12-16 and 13-15 group compounds.
Ans. (i) Ferromagnetism : When substances show permanent magnetism even in the absence of the magnetic
field this phenomenon is called as Ferromagnetism and such substances as called as Ferromagnetic
substances e.g. Fe Ni Co and CrO2
.
This type of magnetism arise due to the spontaneous alignment of magnetic moments due to unpaired
electron in the same direction.
(ii) Paramagnetism : When substances which are attracted by the external magnetic field are called
paramagnetic substances and the phenomenon is called as paramagnetism. Atoms ion or molecules
containing unpaired electron show this property, eg. O2
Cu2+
, Fe3+
etc. these substances lost their magnetism
in the absence of magnetic field.
(iii) 13-15 Compouds : The solid state materials are porudced by combination of elements of goups 13
and 15 the compounds thus obtianed are called 13-15 compouds e.g. InSb, AlP GaAs.
.
12-16 Compounds : The solid state compounds are obtianed by combination of elements fof groups 12
and 16 the compounds are called 12-16 compouds e.g. ZnS, CdS, CdSe and HgTe.

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