Class XII notes

1. THE SOLID STATE
Introduction
Matter can exist in gaseous, liquid or solid state. Of these, solid have a definite volume and shape.
Discovering new solid materials is one of the most active areas of research in modern science and
technology. Materials like high temperature superconductors, biocompatible solids (for surgical
implants) and many other solid materials play vital role in evolution of technology. In gases and
liquids, atoms (or ions or molecules) are continually moving and this explains their ability to flow.
On the other hand atoms (or ions or molecules) in solid are held together by stronger
intermolecular/interionic forces.
Types of Solid
Amorphous Solid
Atoms are irregularly placed so these solids do not have characteristic shapes. e.g. Glass, rubber,
plastic.
(1) They do not have sharp m.pt.
(2) They are isotropic (show similar optical & electrical properties like refractive index, thermal
expansion etc. in all direction).
(3) These are super cooled liquid or pseudo solids because like liquids, amorphous solids have
tendency to flow though very slowly. Glass panes fixed to windows or doors of old buildings
are invariably found to be slightly thicker at the bottom than at the top. This is because the
glass flow down very slowly and makes the bottom portion slightly thicker. Some glass
objects from ancient civilizations are found to become milky in appearance because of some
crystalisation.
(4) They do not have well defined planes
(5) They do not have symmetry.
Crystalline Solid
Atoms are regularly or orderly arranged.
(1) Have sharp m. pt (2) Anisotropic (3) Crystal are bound by plane
faces
(4) They have plane of symmetry, centre of symmetry (always one) and axis of symmetry. eg. A
cubic crystal have total 23 elements of symmetry. Plane of symmetry = 9; Axes of
symmetry = 13; Centre of symmetry = 1.
Isotropy and Anisotropy
Amorphous substance are said to be isotropic because they exhibit the same optical & electrical
property in all directions. In amorphous solids; refractive index, thermal and electrical
conductivities, coefficient of thermal expansion, are independent of the direction along which they
are measured.
Crystalline substances, on the other hand, are anisotropic and the magnitude of a physical property
varies with directions. This arises due to different arrangement of particles in different directions.
For example, in a crystal of silver iodide, the coefficient of thermal expansion is positive in one
direction and negative in other.
Explanation of Isotropy and Anisotropy

2. In amorphous substance, as in liquids, the arrangement of particles
is random and disordered. Therefore, all directions are equivalent
and properties are independent of direction. On the other hand, the
particles in a crystal are arranged and well ordered. Thus, the
arrangement of particles may be different directions.
Distinction between Crystalline and Amorphous Solids
Property Crystalline Solids Amorphous Solids
Shape Definite characteristic geometrical
shape
Irregular shape
Melting point Melt at a sharp and characteristic
temperature
Gradually soften over a range of
temperature
Cleavage property When cut with a sharp edged tool,
they split into two pieces and the
newly generated surfaces are plain
and smooth
When cut with a sharp edged tool,
they cut into two pieces with
irregular surfaces
Heat of fusion They have a definite and
characteristic heat of fusion
They do not have a definite and
characteristic heat of fusion.
Anisotropy Anisotropic in nature. Mechanical
strength, electrical & properties
changes with direction
Isotropic in nature. Mechanical
strength, electrical & optical
properties are same in all direction.
Nature True solids Pseudo solids or super cooled
liquids
Order in arrangement
of constituent particles
Long range order Only short range order
Types of Solids:
Depending on binding forces, solids can be classified as
(1) Ionic Solid:
In ionic crystal the lattice is made of positive and negative ions. These are held together by
ionic bonds – the strong electrostatic attractions between oppositely charged ions.
Consequently, the cations and anions attract one another and pack together in an
arrangement so that the attractive forces maximize. Each ion is surrounded by neighbours of
opposite charge and there are no separate molecules. Since the ions are fixed in their lattice
sites, typical ionic solids are hard and rigid with high melting points. All ionic solids are
crystalline solids.
In spite of their hardness, ionic solids are brittle. They shatter easily by hammering. By
hammering, a layer of ions slips away from their oppositely charged neighbours and brings
them closer to ions of like charge. The increase of electrostatic repulsions along the displaced
plane causes the crystal to break. Ionic solids are non– conducting because the ions are in
fixed positions. However, in fused state the ions are allowed freedom of movement so that it
becomes possible for them to conduct electricity.
(2) Covalent Solid:

3. In covalent solids lattice sites are occupied by atoms which are bonded to one another by
covalent bonds. The atoms interlocked by a network of covalent bonds produce a crystal
which is considered to be a single giant molecule. These crystals are very hard and have very
high melting points. e.g. Diamond ,Graphite
(3) Molecular Solid:
In molecular solids molecules are the structural units. These are held together by weak
vander– waal’s forces. When this type of crystal melts, it is only the weak vanderwaal’s
forces that must be overcome. Therefore, molecular solids have low melting points. Most
organic substances are molecular solids. These are of three categories;
(i) Non–polar solid e.g. l2 (2) Polar solid (3) H–bonded solid e.g. ice
(4) Metallic Solid
The crystal of metals consists of atoms present at the lattice sites. The atoms are arranged in
different pattern, often in layers placed one above the other. The atoms in a metal crystal are
held together by a metallic bond. The valence electrons of the metal atoms are considered to
be delocalized leaving positive metal ions. The free electrons move throughout the vacant
spaces between the ions. The electrostatic attractions between the metal ions and the
electron cloud constitute the metallic bond. The electron sea model explains well the
properties of metals.
The mobile electrons in the crystal structure make metals excellent conductors of heat and
electricity. On application of force, say with a hammer, metal can be deformed. The metal
ions in the crystal change positions without making material difference in the environments.
The attractive force between ions and the electron cloud remain same. The crystal, therefore,
does not break.
Metallic solids are opaque as free electrons absorb all the incident light energy and do not
allow light to pass through.
Metallic bond results from the simultaneous attraction between mobile valence
electrons and positively charged kernels present in the electron sea.
Explanation for the Metallic Properties
The different physical properties of the metals mentioned earlier can be explained with the help of
electron sea model for the metallic bonds. These are disused as follows:
1. Electrical Conductivity
According to electron sea model, the electrons present in the sea are mobile but their
movement is not systematic. When a potential difference is applied by placing electrodes on
both sides of a metal, the movement of the electrons becomes systematic. They flow from
cathode to the anode. As a result, the electrical current flows across a metal or it becomes
conducting.
The electrical conductivity of metals decreases with the rise in temperature. Actually when
temperature increases, the average kinetic energy of the positively charged kernels also
increases. They start moving in the electron sea and obstruct the movement of the electrons
leading decreased electrical conductivity.
2. Thermal conductivity
When we heat a certain portion of a metal the kinetic energy of the electrons present
increases. They move towards the colder regions and collide with the electrons with less
kinetic energy. As a result, kinetic energy is transferred. Thus, heat is conducted through the
metal .
3. Malleability and ductility

4. We know that the metals are malleable (can be beaten into sheets) and ductile (can be drawn
into wires). Both these characteristics can be explained with the help of electron sea model.
When stress is applied on the metal surface, the kernels present in a particular layer slip
over the other layer. The crystal lattice or the shape of the crystal gets deformed as a result
of it. However, the environment of the kernels remains the same i.e. mobile electrons keep
on attracting the kernels. Only there is a shift in the position of the kernels from the one
lattice site to the other.
4. Metallic Lustre
Metals have generally a shining surface known as lustre. This lustre is also due to the mobile
electrons. When light falls on the surface of a metal, the electrons absorb photons and start
vibrating at a frequency equal to that of the incident light. These vibrating electrons emit
electromagnetic radiations in the form of light. As a result, the metal surface acquires a
shining appearance known as lustre.
5. High tensile strength
This is the property of the metal as a result of which they can be stretched without breaking.
The tensile strength is because of strong electrostatic force of attraction between the kernels
and mobile electrons.
6. Hardness of metals
The hardness of metals is related to the strength of the the metallic bonds between the
kernels and the mobile electrons. Greater the number of these electrons, stronger will be the
bond. Similarly, smaller the size of the kernels, greater will be their attraction for the
electrons. For example, sodium is very soft because the size oif the kernel is large and there
is only one valence electron.
Classification of Crystalline Solid:
Type of
Solid
Constituent
Particles
Nature of Forces Examples App. Binding
Energy
(kJmol–1)
Characteristics Properties
1. Ionic Positive
and
negative
ions
Strong
electrostatic forces
of attraction
NaCl, KNO3,
LiF, MgO
400–4000 Hard, brittle, high melting
points (~ 1500 K), soluble
in polar solvents, insulator
in the solid state, good
conductor in the molten
state or aqueous solution.
2.
Molecular
Molecules 1. vanderWaal’s
forces
2.Hydrogen bond
l2, solid CO2
(dry ice), and
Ar, CH4,
Napthalene,
glucose
sulphur, ice
< 40 Soft, low melting points,
low density, volatile and
insulators of heat and
electricity.

5. 3. Network
or Covalent
Atoms Covalent bonds C (Diamond
and graphite),
Si, SiO2, SiC
Boron, Black
150 – 500 Very hard with very high
melting point (~ 4000 K)
and insulators (except
graphite)
4. Metallic Atoms Metallic bond i.e.
Attraction
between positive
metal ions
(kernels) and
mobile electrons
All metals and
some alloys
70 – 1000 Range from soft to very
hard, high melting points
(~ 800 – 1000 K) malleable
and ductile, possess lustre,
good conductors of heat
and electricity.
The properties of the solids depend upon their structure and the nature of bonding involved. The
physical properties like molar enthalpy of fusion, electrical and thermal conductivity help us to
predict the type of which a given crystalline solid belongs.
Space Lattice and Unit Cell
Space Lattice:
A crystal is a homogeneous portion of a solid substance made of regular pattern of structural
units bonded by plane surfaces making definite angles with each other. The geometrical form
consisting only of a regular array of points (atoms, molecules, ion) in space is called space
lattice or it can be defined as an array of points showing how molecules, atoms or ions are
arranged in different sites in three dimensional space. The sites which are occupied by
constituent units (molecules ions or atoms) are called lattice points.
Unit Cell:
It is defined as the smallest building block of a solid which
when repeated over and again results into the space lattice.
We must specify the unit cell by length of its edges and the
angles between them.
Parameters of a unit cell:
A unit cell is characterized by
(i) Its dimensions (lengths) along the three edges as ‘a’, ‘b’ and ‘c’. These edges may or may not
be mutually perpendicular.
(ii) Angles ,  and  between the pair of edges. The angle  is between the edges b and c, angle
 is between the edges c and a and angle  is between the edges a and b. Thus, a unit cell is
characterized by six parameters, a, b, c, ,  and .
According to Bravais (1848), taking geometrical and symmetrical consideration into account
only seven shape for the unit cell are possible and within each crystal system there are only
four possible way arrangement of atoms or ions in a unit cell. These are:
i) Primitive or simple unit cell
When a unit cell contain constituent particles only at corners ,it is called primitive unit cell
i) Non –Primitive unit cells
When a unit cell contain one or more constituent particles at corners & some other
positions. It is called centered unit cell. There are three types non–primitive unit cell e.g.

6. 1. Cubic 2. Tetragonal
Centred Cubic Face Centred
Cubic
Simple Tetragonal
(Primitive)
Tetragonal
a = b ≠ c
 =  =  = 90o
i) Body centred : – When lattice points are at corners &centre of unit cell.
ii) Face centred: – When lattice points are at corners &centre of each face of unit cell
iii) End centred: – When lattice points are at corners &centre of two opposite faces
Since each arrangement can be considered of seven different unit cell
 Total number of unit cells possible are 7 × 4 = 28
Seven Primitive Unit Cells are
Sr.
No.
Crystal
Systems
Bravais
Lattices
Axes Angles Examples
1 Cubic cell Primitive,
Face centered,
Body centered
a = b = c  =  =  = 90o KCl, Cu, NaCl, CaF2, ZnS Diamond
2 Tetragonal cell Primitive,
Body centered
a = b  c  =  =  = 90o TiO2, SnO2, white Tin, , CaSO4
3 Orthorhombic Primitive,
Face centered,
Body
centered, End
centered
a  b  c  =  =  = 90o KNO3, BaSO4, K2SO4, Rhombic, S,
PbCO3
4 Monoclinic Primitive, End
centered
a  b  c  =  = 90o
 90o
Monoclinic S, Na2SO4.10H2O
5 Triclinic Primitive a  b  c  90o CuSO4.5H2O, H3BO3, K2Cr2O7
6 Rhombohedral
or Trigonal
Primitive a = b = c =  =  90o lCl, Quartz, Calcite (CaCO3),
NaNO3; HgS
7 Hexagonal Primitive a = b  c  =  = 90o,
 =120o
SiC, BN, graphite, CdS, ZnO

7. These 14 arrangements are found in various crystals and are
knownas BravaisLattice (outof possible 28 arrangements)
Effective Number of Atoms Per Unit Cell (Rank of Unit Cell)
1. Atom at the corner of unit cell is shared by 8 unit cells. Thus, contribution of each corner
atom (per unit cell) is 1/8.
2. Atom present at the face of a unit cell is shared by 2 unit cells. Contribution of face atom (per
unit cell) is 1/2.
3. For atom present at the centre of unit cell, contribution per unit cell is equal to 1.
4. For atom present at the edge of unit cell is shared by 4 unit cells. Contribution of edge atom
(per unit cell) is 1/4.
(1) Rank for Cubic Unit Cell
It is the effective number of atoms present in cell. It is represented by Z.
nc = no. of corner atoms,
nf = no. of face atoms
nb = no. of body atoms (inside the cube)
ne = no. of edge atoms
3. Orthorhombic
4. Monoclinic 5. Triclinic 6. Hexagonal 7. Rhombohedral

8. Rank = b
efC
n
4
n
2
n
8
n
 for cubic unit cell
(2) Rank for Hexagonal Unit Cell
4
n
&
3
n
;
2
n
;
6
n eefC
Assignment – I
Q.1 A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms
are at the corners of the cube and X atoms are at the alternate faces. What is the formula of
the compound?
Q.2 If three elements P, Q and R crystallize in a cubic solid lattice with P atoms at the corners, Q
atoms at the cube centre and R atoms at the centre of the faces of the cube, then write the
formula of the compound.
Q.3 Potassium crystallizes in a body–centred cubic lattice. Calculate the approximate number of
unit cells in 2g potassium. Atomic mass of potassium  39u.
Q.4 In a face centred lattice of X and Y, X atoms are present at the corners while Y atoms are at
face centres. Then the formula of the compound if one of the X atoms is missing from a
corner in each unit cell would be:
(a) X7Y24 (b) X24Y7 (c) XY24 (d) X24Y
Q.5 In a face centred lattice of X and Y, X atoms are present at the corners while Y atoms are at
face centres. Then the formula of the compound would be if one of the X atoms from a
corner is replaced by Z atoms (also monovalent)?
(a) X7Y24Z2 (b) X7Y24Z (c) X24Y7Z (d) XY24Z
Q.6 A solid has a structure in which W atoms are located at the corners of a cubic lattice. O atom
at the centre of the edges and Na atom at centre of the cubic. The formula for the compound
is
(a) NaWO2 (b) NaWO3 (c) Na2WO3 (d) NaWO4
Q.7 Which of the following are the correct axial distance and axial angles for rhombohedral
system?
(a) a = b = c,  =  =  90o (b) a = b  c,  =  =  = 90o
(c) a  b  c,  =  =  = 90o (d) a  b  c, 90o
Answers
1. XY 2. PQR3 3. 1.54 × 1022 4. a

9. 5. b 6. b 7. a
Packing of Spheres in Solids
In the formation of crystals, the constituent particles (atoms, ions or molecules) get closely package
together. The closely packed arrangement is that in which maximum available space is occupied
leaving minimum vacant space. This corresponds to a state of maximum possible density. The
closer the packing, the greater in the stability of the packed system.
As the constituent particle of crystals may be of varying shapes and, therefore, the mode of closest
packing of particles will vary according to their shapes and sizes. However, for understanding we
can use identical hard spheres of equal size to represent atoms in a metal in terms of closest
packing of identical spheres.
(A) Close Packing in One Dimension
There is only way of arranging spheres in one dimensional close packet structure in which
the spheres are placed in a horizontal row touching each other. This is shown in figure.
Close packing of spheres in one dimension.
As can be seen, in the arrangement each sphere is in contact with two of its negihbours. The
number of spheres which are touching a given sphere is called its co–ordination
number. Thus, in one dimensional close packed arrangement, the coordination number is
2.
(B) Close Packing in Two Dimensions
Two dimensional close packet structure can be generated by placing the rows of close
packed spheres. The rows can be combined in the following two ways with respect to the
first row to build a crystal plane:
i) The spheres are packed in such a way that the rows have a horizontal as well as
vertical alignment. In this arrangement, the spheres of second row are exactly above
those of the first row. The second row is exactly same as the first one. This type of
packing is also called square close packing in two dimensions.
Square close packing occupies 52.4% of available space [Co. No. 4]
ii) The spheres are packed in such a way that the spheres in the second row are placed
in the depression between the spheres of the first row. Similarly, the spheres in the
third row and placed in the depressions between the spheres of the second row and
so on. In this arrangement, the second row is different from the first row. But the
spheres in third row are aligned with those of the first row. Similarly, the spheres of
fourth row are aligned with those of second row. If the arrangement of spheres in
first row is called ‘A’ type, the one in the second row is different and may be called
‘B’ type. Now, the arrangement of spheres of third row is same as that of first row,
and therefore, it is also called ‘A’ type. Similarly, fourth row is called ‘B’ type. Hence
the arrangement is called ABAB …… type. This type of arrangement is called
hexagonal close packing of spheres in two dimensions and is shown in figure.

10. (C) Close Packing in Three Dimensions
We can now build other layers over the first layer to extent the packing in three dimensions.
This can be done by building layers on square packed and hexagonal close packed
arrangements of first layer.
(I) Placing Second Layer Over the First Layer
(a) Three dimensional close packing from two dimensional square
packed layers.
The second layer is placed over the first layer such that these
spheres of the second layers are exactly above those of first
layer. In this arrangement spheres of both layers are perfectly
aligned horizontally as well as vertically. Similarly, we may
place more layers one above the other. This type of packing
will have simple cubic unit cell. This type of packing will have
simple cubic unit cell.
(b) Threedimensional closepackingfromtwo dimensional hexagonal closepackedlayers
In the first layer there are some empty spaces or hollows called voids. These are triangular
in shape. These triangular voids are of two types marked as a and b. All the hollows are
equivalent but the spheres of second layer may be placed either on hollows which are
marked a or on other set of hollows marked b. It may be noted that it is not possible to place
spheres on both types of hollows. Let us place the spheres on hollows market b to make the
second layer which may be labeled as B layer. Obviously the holes marked ‘a’ remain
unoccupied while building the second layer.
Whenever a sphere of second layer is placed above the void of first layer, a tetrahedral void
is formed. These voids are called tetrahedral voids because a tetrahedral is formed when
the centres of these four spheres are joined. There are marked at ‘T’ voids.
The voids ‘a’ are double triangular voids. The triangle void in the second layer are above the
triangular voids in the first layer and the triangular shapes of these voids do not overlap. One
of them has the apex of the triangle pointing upward & other downward. Such voids are
surrounded by six spheres and are called octahedral voids.
Now the third layer can be build up by placing spheres above tetrahedral voids or octahedral
voids.

11. (II) Placing Third Layer over the Second Layer:
When third layer is placed over the second, there are two possibilities.
(a) Covering tetrahedral voids
When a third layer is to be added, again there are two types ofvoid available. One type of void
marked ‘a’ are unoccupied hollows of the first layer. The other type of void are void in the
second layer (marked c). Thus, there are two alternative to build the third layer.
SolidCircle RepresentLayerA andDotted Actual viewof hexagonalclosepacking
Tetrahhedral Void
Octahedral Void

12. Circle RepresentLayerB
The third layer of sphere may be placed on the tetrahedral voids marked (c) of the second
layer. In this arrangement, the spheres of the third layer lie directly above those in the first
layer. In other words, third layer becomes exactly identical to the first layer (labelled A).
As shown in figure.
Simplified view of Hexagonal close packing (ABAB …… pattern)
This type of packing is referred to as ABABA …. arrangement. This type of packing is also
known as hexagonal close packing. This type of arrangement is found in many metals like
Mg and Zn.
(b) Covering octahedral voids
The second way to pack spheres in the third layer is to place them over octahedral voids
marked ‘a’ (unoccupied void of first layer). This gives rise to a new layer labeled as C. Sphere
of third layer are not alligned with first layer or second layer. However, it can be shown that
the spheres in the fourth layer will correspond to those in the first layer.
As shown in figure.
This type packing is called as the ABCABCA ….. type of arrangement. It is also known as cubic
close packing(ccp) or face centred cubic arrangement (fcc).
It may be noted that both types of packing are equally economical though these have
different forms. In both cases, 74% of the available volume is occupies by the spheres and
co–ordination no equal to12 . Metals like Cu & Ag crystallies in FCC pattern.
(c) Body centred cubic close packing (ABAB …. Type) Co–ordination No. = 8

13. In this packing atom of each layer will not touch any other atom of the same layer but four
atoms of one layer will touch to same atom of second layer. In this packing unit cell will be
BCC and arrangement of first and third layer will be same
Important facts about Unit cells and the Packing of Spheres in Crystalline Solids
(i). Particles pack together in crystals so that they can be as close together as possible to
maximize intermolecular attractions.
(ii). Simple cubic packing – the spheres in one layer sit directly on top of those in the previous
layer.
(AAA……)
1. All layers are identical having both vertical and horizontal alignment of atoms.
2. Primitive – cubic unit cell.
3. Coordination number = 6. Sphere touches four neighbours in the same layer, one
above and one below.
4. Uses only 52% of available volume.
(iii). Body – centered cubic packing – spheres are in alternate layers in an ABAB….. arrangements
where the spheres in the B layers fit into the small depressions between spheres in the
neighbouringlayers. Atoms of each layer will not touch to each other but four atoms of same
layer will touch to same atom of 2nd layer.
1. Body – centered cubic cell.
2. Coordination number = 8; four neighbours above and four neighbours below
3. Occupies 68% of the available volume.
4. S–Block elements forms BCC unit cell
(iv). Hexagonal close packing– noncubic unit cell with two alternating layer (ABAB……)
1. Hexagonal arrangement of touching spheres.
2. Spheres in B layer fit into the small triangular depressions between spheres of A
layer.
3. Coordination number = 12
4. Six neighbours in the same layer, three above and three below.
(v). Cubic close packing– face centered cubic unit cell with three alternating layers,
(ABCABC …...)
1. A – B layers identical to hexagonal close packing
2. Third layer is offset from both A and B.

14. 3. Co–ordination number 12
4. Transition metal forms FCC unit cell
(vi) Cubic is the most symmetrical while triclinic is the most unsymmetrical system.
(vii).
Sr.
No.
Structure Type of packing Stacking pattern Coordinatio
n number
Space
used
(%)
Unit cell
(a) Simple
cubic
Square close
packing
AAA … 6 52 Primitive cubic
(b) Body
centred
cubic
Square close
packing
ABAB….. 8 68 Body centred
cubic
(c) Hexagonal
close
packed
Hexagonal ABAB… 12 74 (Non–cubic)
(d) Cubic close
packed
Hexagonal ABCABC… 12 74 Face centred cubic
(viii) In FCC or CCP there are three different layers which are constantly repeated in three
dimensions. In HCP every first & third layers are same and in ccp every first & fourth layers
are same.
Between A & B (in close packing) the Octahedral & tetrahedral voids are located.
If we place atoms of 3rd layer over the tetrahedral void, the layer A will be repeated, hence
ABAB….. structure or HCP will be formed.
If we place atoms of third layer over Octahedral void, then FCC will be generated as third
layer will be different layer & it is ABCABC ….. type.
The number of nearest neighbours for any atom, molecule or ion in a space lattice is called
the coordination number.
Type of Cubic Unit Cell & Packing Fraction
1. Simple or Primitive Cubic Unit Cell:
A cubic unit cell is said to be primitive if
(a) All the eight corners of the cube are occupied by the same
atoms & will touch each other.
(b) There is no such atoms present any where in the cube except
corners.
(c) Contribution of each atom or sphere for one unit cell is 1/8 of the volume of each sphere.
(d) Therefore, number of effective atoms in a unit cell. [Rank of unit cell] is one
Z = no. of corner atoms (nc) × contribution of each atoms = 1
8
1
8 
(e) Each face made by four corner atoms are in contact with each other.
(f) Relation between edge length (a) of cube and radius of a sphere (r).
a = 2r

15.  Atomic radius (r) = a/2
Hence, volume of unit cell = a3 = (2r)3 = 8r3 and
volume occupied by the these spheres in a unit cell
= .1isZwherer
3
4
Z 3

(g) Packing fraction (PF)
It is defined as the ratio of the volume occupied by the spheres in a unit cell to the volume of
the unit cell.
3
3
a
r
3
4
Z
cellunitofVolume
CellunitainspheressamethebyoccupiedVolume
PF


where, Z = no. of effective atoms or spheres in a unit cell;
For simple cubic unit cell z = 1 a = 2r
PF 3
3
r8
r
3
4
1 

6

 = 0.5236
In other words 52.36% space is occupied & 47.64% is vacant.
Void fraction: 1 – PF = 0.48
2. Body–Centred Cubic Unit Cell (BCC Unit Cell)
(i)Characteristics of BCC unit Cell
(a) All the eight corners of the cube are occupied by the same
atoms.
(b) One similar atom or different atom is also present exactly at
centre of the cubic lattice.
(c) Contribution of each atom present at eight corners for a unit
cell is 1/8 of the volume of a sphere and contribution of one
body centred atom is one.
(d) Therefore, number of effective atoms in a unit cell.
= no. of corner atoms (nc)×
8
1
+no of body centred atom (nb) × 1 = 211
8
1
8 
(e) The four spheres in bottom layer are not in contact with adjacent spheres as well as
the four spheres in top layer, but all the corners atoms are in contact with body
centred atom, So ‘a’ will be more than 2r
(ii) Relation between edge length (a) and radius of sphere (r)
Case I: When all the atoms are same & making contact along body diagonal.
In right angled ABC
AC2 = AB2 + BC2
AC2 = a2 + a2 or AC = a2
Now in right angled in ADC
AD2 = AC2 + DC2

16. AD2 =   222
a3aa2 
AD = a.3 r = radius of atom and a = edge length
 Body diagonal of cube = a3r4 
d = 2r

2
a3
dor
4
a3
r 
Packing fraction (PF) =
cellunitofVolume
cellunittheinspheresthebyoccupiedVolume
Volume of unit cell = 












3
r4
a
3
r4
a
3
3
And volume occupied by the spheres in the unit cell
= Z × volume of a sphere
= ]bccfor2Z[r
3
4
2 3

= %02.686802.0
3
r4
r
3
4
2
3
3








Void fraction = 0.32 = 32%
Case – II
When the body centred atom and the corner atoms are different
)r2r2(a3AB 
 …… (1)
= 3
3
a
3
c
a
r
3
4
r
3
4













rc = radius of cation, ra = radius of anion
Limiting condition or Ideal Condition:
2ra = a …… (ii)
The limiting condition corresponds to the maximum radius of the anion. Corner atoms are making a
contact along the edge of a cube. So, 2ra max = a
3 a = 2rc + 2ra Substitute volume of ‘a’ from eq. (ii)
3 (2ra) = 2rc + 2r a
2rc = 2ra  13 
 ac r13r 
732.0
r
r
a
c
 It is the minimum possible radius ratio in BCC

17. (r+) Cation smaller
If radius ratio is 0.732 or more than this the arrangement will be body centred& coordination
number will be 8.
Packing Fraction (in limiting condition)
There are two ions in BCC one cation and one anion.
P.F. =
 
 
  729.01)732.0(
6
1
r
r
6r
rr
6r2
rr
3
4
a
r
3
4
r
3
4
3
3
a
c
3
a
3
a
3
c
3
a
3
a
3
c
3
3
a
3
c






















 
















3. Face centred Cubic Unit Cell (FCC Unit Cell)
A face centred cubic unit cell is said to be an ideal face centred cubic unit cell if:
(a) All the eight corners of the cube are occupied by the same atoms.
(b) All the six face centred atoms are same (c) corner atoms or the face atoms may be same or
different
(d) Therefore, number of effective atoms in a unit cell
= no. of corner atoms (nc) × 1/8 + no. of face centred atoms (nf) × 1/2
= 431
2
1
6
8
1
8 
(e) Each corners atom are not in contact with each other but the face–centred atoms are in a
contact with the four spheres present at the four corners of a face.
(f) Also it is noted that each face centred atoms is in contact with adjacent face centred atom.
Case 1: When all atom are a2aar4 22

 face diagonal of the cube = a2r4 
 r = a
4
2
 r22aor,a
22
1
r 
Therefore, volume of unit cell = a3 =  3
r22 and volume occupied by the spheres in the unit cell
(If corner & face centred atoms are same)
= Z × volume of a sphere = 3
r
3
4
4 

18. Next nearest
neighbour [12]
Packing fraction (PF) =
cellunitofVolume
cellunittheinspheresthebyoccupiedVolume
=
 
7406.0
r22
r
3
4
4
3
3


= 74.06% space is occupied & 26% space is unoccupied
So FCC is also known as cubic closed packing (CCP)
Case II: The face centred atoms and the corner atoms are different
There are 3 face centred atoms (cation) and 1 corner (anion) atom in 1 unit
Packing Fraction = 3
3
a
3
c
a
r
3
4
1r
3
4
3 












Limiting or Ideal Conditions: If corner atoms are bigger & are touching
ac r2r2a2AB  …. (i)
a = edge length
Corner atoms are making contact along the edge of the cubes
2ra = a …. (ii)
On substituting maximum radius value of anion (r–) in (i)
2 (2ra) = 2rc + 2ra 2rc = 2 (2ra) – 2ra 2rc = 2ra  12 
  414.012
r
r
mina
c






It refers to the minimum limiting ratio: 414.0
r
r
a
c

Radius ratio range is equal to 0.414 – 0.732 ( value equal to 0.414 but less than 0.732)
So, packing fraction = 635.0]1)414.0(3[
6
1
r
r
3
6r2
rr3
3
4 3
3
a
c
3
a
3
c
3
c
























 

Co–ordination Number or Number of Nearest Neighbours
The nearest of closest equidistant atoms with respect to a given atom is called its nearest
neighbour. The number of nearest neighbor is called co–ordination number. Next nearest
neighbor represents the next close equidistant atom w.r.t a given atom.
(a) Simple cubic unit cell:
Number of Nearest neighbouris six& next nearest neighbour is 12.
Distance between nearest neighbour = a
Distance between next nearest neighbor = a 2

19. One corner is shared by 12 faces. The atoms meets the next nearest neighbor along the face
diagonal of each face. So the no. of next nearest neighbor will be 12(Number of atom in
each xy; yz; zx plane will be 4]
(b) Body centred Cubic unit cell
(i) The nearest neighbor for Body centred atom will be corner atom.
So the nearest neighbor = 8
distance between neighbouring atom (d) = .a
2
3
(i.e. half of the body diagonal)
(ii) The next nearest neighbour = 6 [i.e. atom of Body
centred of six cube]
Distance between next neighbour = a
(iii) For corner atom, the next nearest neighbour are going to be corner atoms i.e., next nearest
neighbour is 6.
3. Face centred cubic unit cell
No. of Nearest neighbour = 12; No. of next neighbour = 6
With respect to the corner atom, the nearest neighbour is the face centred atom. Number of
nearest neighbours is 12.
Number of Nearest neighbours = 12
Distance between nearest neighbours =
2
a2
(i.e. half of face diagonal)
(i) For corner atom, the next nearest atom is a corner atom i.e. next nearest neighbour = 6
(ii) For face centred atom (A) nearest neighbour atom will be (B) & (C)
(8 faces + 4 corners). So number of nearest neighbour is 12.

20. 8 faces 4 corner atom will be nearest neighbour Co–ordination Number = 12
distance AC = 1/2 face diagonal
22
2
a
2
a
AB 












distance AC = a
2
2
AB =
2
a
= 0.707 a = 0.7071 a AB = AC
For next nearest neighbour
Next nearest neighbour is the opposite face atom so that number of next nearest neighbour is six.
(all face atom). & distance of next nearest neighbour = a
Hexagonal Primitive Unit Cell (HCP)
In this geometry of hexagonal unit cell, only primitive cell is found
to be symmetrical. It is closest packing like FCC where 74% space
is occupied. A hexagonal primitive unit cell is said to be an ideal
hexagonal primitive unit cell is
(a) All the six corner of hexagonal contained same type of
atoms. i.e, bottom and top hexagon, both have same atom
at their corners
(b) Each hexagonal face contains same type of atom as that
corners atom. The corner atoms are touching each other
and they are making a contact with face centre as well.
(c) Upper and lower hexagonal planes are same type and it is
said to be A layer.
(d) Another layer is B layer, it is completely inside the unit cell.
There are three atoms, present at this layer which is
completely inside the unit cell and are in contact to each
other.
(e) Contribution of each atom present at the twelve corners
(six corner at bottom + six corner at top) of hexagonal
primitive unit cell is 1/6 of the volume of an atom &
contribution of each atom present inside hexagonal
primitive unit cell is 1.
(f) Therefore, number of effective atoms in a unit cell

21. = no. of corner atoms × 1/6 + no. of face centred atom × 1/2
+ no. of atoms inside the hexagonal primitive unit cell × 1
= 13
2
1
2
6
1
12  = 2 + 1 + 3 = 6
(g) The face centred atom is in contact with six spheres present at the corners of hexagon and
two corner sphere are also in contact with each other.
(h) If let us consider a is the length of hexagonal unit cell and r be the radius of sphere then a =
2r
(i) Packing fraction (PF) =
celltheofVolume
cellunittheinspheresthebyoccupiedVolume
=
heightareabase
r
3
4
6 3


Area of equilateral triangle =
4
a3 2
=
3
2
r4)r2(
4
3
6
r
3
4
6
3
2
r4a
4
3
6
r
3
4
6
2
3
2
3





















= 0.74 = 74%
CALCULATION OF HEIGHT:
YZ = r; XZ = XP = QZ = 2r
(XZ)2 = (YZ)2 + (XY)2
(2r)2 = (r)2 + (XY)2
(XY)2 = 3r2 XY = r3
Controid divide median in 2 : 1 ratio
3
2
r3XR 
(RP)2=(XP)2–(XR)2
;r
3
4
r4
2
h
;r
3
2
3r4
2
h 22
22
2
2


















3
2
r4h;
3
r22
2
h
;r
3
8
2
h 2
2






P.F.= 74.0
r224
r
3
4
6
3
2
r4)r2(
4
3
6
r
3
4
6
3
3
2
3





74% space is occupied.
Important Fact
ZnS has CCP lattice in zinc blende & HCP wurtzite structure.

22. In zinc blende; Zn+2 ions are placed in alternate Tetrahedral void & S2– form FCC lattice. One unit cell
is made up of four ZnS units.
In wurtzite S–2 form HCP unit cell i.e. 6 S–2 ions and Zn+2 ions goes in alternate Tetrahedral void i.e. 6
Zn+2 so one unit cell is made up of 6 ZnS units. The radius ratio of guest to host atom remains same
irrespective of the geometry of the unit cell.
P.F.(wurtzite) =
 








225.0
r
r
voidltetrahedrainas
r224
)rr
3
4
6
a
c
3
a
3
a
3
c
= ]1)225.0[(
23
3


= 0.74 (1.01) = 74.86%
So, the packing fraction of zinc blende &wurtzite will be same, but their formula weight will be
different.
Voids or Interstitial Site
Vacant space in the crystal lattice is called void. It is made up of host atoms (bigger) atoms which is
usually an anion. Small sized atoms are occupied in these voids. Sometimes an impurity atom can be
present in a void, which is called guest atom or foreign atom. The guest atom must make a contact
with host atom, but host atom may or may not make contact with each other. e.g. In II case, we get
the limiting radius ratio, which corresponds to the minimum value
There are four types of void existed:
(a) Triangular (b) Octahedral Voids (c) Tetrahedral Voids (d) Cubic Voids
Triangular Voids
(a) Triangular Voids

23. Triangular voids will be present in two dimensional layer where the Co. No. of guest atom
will be 3 as it is made up of three host atoms. Radius ratio will be
JK = KL = JL = 2ra
KM = rc + ra
MKN = 300
cos 300 = 155.0
r
r
rr
r
a
c
ac
a


(b) Octahedral Voids
The position in a close–packed unit cell where a foreign atom
is in contact with six host atoms in the form of an octahedral
is called an Octahedral Void.
(i) In FCC Unit Cell:
(a) Octahedral void is formed at the centre of each edge and hence there are 12 such
types of octahedral void.
(b) One octahedral void is located at the centre of unit cell.
(c) Contribution of each octahedral void existed at the
centre of each of the 12 edge is 1/4 of the volume of a
void and contribution of octahedral void existed at
the centre of unit cell is exactly volume of a void i.e.,
1.
(d) Hence, no. of effective octahedral void
= 
4
1
no. of octahedral void existed at each edge
+ 1 × no. of octahedral void at the centre of unit cell
=
4
1
× 12 + 1 × = 3 + 1 = 4
Hence, no. of effective octahedral void in fcc unit cell = 4
But we have also, no. of effective atoms (as host atoms) in fcc unit cell = 4
Number of octahedral void = Number of effective foreign atoms = No. of effective host atoms
(ii) In HCP Unit Cell:
Since, no. of effective host atoms = 6
No. of effective octahedral void = 6
Hence, no. of effective octahedral void in a face centred cubic unit cell is 4 and in a Hexagonal
close–packed unit cell is 6.
Radius Ratio for the octahedral void under ideal condition.
When anions are bigger & are in contact. Cross section through Octahedral site:
Cos FAE = )Base(
)hypotenus(
AF
AE
Foreign
atom
12 Octahedral Voids at the
edge centres

24. Cos 45o =
AF
AE
AFB = 90o
FAB = 45o
In  FAE: Cos 45o =
rR
R

rR
R
2
1


R
rR
2


R
r
12 
414.021
R
r

For an atom to occupy an octahedral void its radius must be 0.414 times the radius of the sphere.
Radius of guest atoms should be atleast 41.4% of radius of host atom. It can be greater than this but
not less than this. If ratio is more than this host atom will not be able to touch each other.
(c) Tetrahedral Void
(i) In FCC Unit Cell:
The vacant position in a close–packed unit cell where a foreign atom is in contact with four
host atoms in the form of a tetrahedral is called tetrahedral void.
If atoms are placed on the opposite faces at the alternate corners of a cube, the body centre
of a cube becomes the position of tetrahedral void. The centre of the smaller cube is the
position of tetrahedral void.
Lerger cube is the original FCC and the smaller cube is an imaginary cube inside the larger
cube.
There are 8 smaller cubes in a bigger cube & each having one void. In FCC there are four
atoms so the number of effective tetrahedral voids in a unit cell is double the number of
effective atoms in that unit cell. i.e., No. of effective tetrahedral voids = 2 × no. of effective
atoms
Hence in fcc unit cell, No. of effective tetrahedral voids = 2 × no. of effective atoms
= 2 × 4 = 8
(ii) In HCP Unit Cell:
No. of effective tetrahedral voids = 2 × no. of effective atoms = 2 × 6 = 12
Location of tetrahedral voids

25. Tetrahedral voids are present on the body diagonals, and one body diagonal contains two
tetrahedral voids, around each corner. Since there are four body diagonal, so, there are eight
tetrahedral void, each will be present at 1/4 of the body diagonal from each corners of the cube.
Radius Ratio of Tetrahedral Void
Body diagonal of smaller cube & larger cube are collinear & body diagonal of smaller cube is exactly
half the larger cube. Body diagonal of smaller cube =
2
1
× Body diagonal of larger cube
cubeerarglofdiagonalBody
4
1
rr ac 
a3
4
1
rr ac  ….. (1)
In FCC unit cell 4ra = a2 …… (2)
from (1) & (2)
aac r
2
3
rr 
1
2
3
r
r
a
c

= 0.225
Thus, for an atom to occupy a tetrahedral void, its radius must be 0.225 times the radius of the
sphere
Other Methods for Radius Ratio of Tetrahedral Void
(a) sin ABD = 0.8164 = AD/AB
or 816.0
rr
r
AB
AD
ac
a



or
816.0
1
r
rr
a
ac


 225.0
r
r
a
c

(b) A tetrahedral void may be represented by placing four spheres at the alternate corners of a
cube as shown in figure. It may be noted that a stable tetrahedral arrangement has four
spheres at the corners touching each other. However, for simplicity, the spheres are shown
by distant circles. Actually all the spheres are touching one another. Let us assume that the
length of each side of the cube is ‘a’ and radius of tetrahedral void is ‘r’ and the radius of
sphere is a face diagonal.
From right angled triangle ABC, Face diagonal
22
BCABAC 

26. 22
aaAC 
AC = a2 = 2R … (i)
AD = )rR(2a3  …. (ii)
Therefore from (i) & (ii)
2
3
R
rR


2
23
R
r 

225.0
R
r

(d) Cubic Void:
In simple cubic structure the space remaining at centre is termed as a cubic void. Single void
is present per cube. When another atom occupies this void, bcc structure is formed.
In the limiting cubical (body centered cubic) arrangement, a cation of radius, rc is touching all
the eight anions of radius, ra and the anions are also in contact with each other.
From the figure, it is obvious that
AB = BC = CD = 2ra = a
and AD = 2(rc + ra) = a3

a
a3
r2
)rr(2
a
ac


or 31
r
r
a
c

732.013
r
r
a
c

Important facts:
a) Number of octahedral voids = Number of particles in the closed packed arrangement.
b) Number of tetrahedral voids = 2 × Number of particles = 2 × Number of octahedral voids.
c) Radius (r) of the tetrahedral voids = 0.225 R
d) Radius (r) of the octahedral void = 0.414 R
In a CCP structure, there is 1 octahedral void at the centre of the body and 12 octahedral voids on
the 12 edges of the cube. Each void on the edge is common to four other unit cells. There are 8
tetrahedral voids. Each spheres of the corner is in contact with three others giving eight tetrahedral
void.
Sr. No. Voids Co. No. Radius Ratio
1. Triangular 3 0.155 to less than 0.225
2. Tetrahedral 4 0.225 to less than 0.414

27. 3. Octahedral 6 0.414 to less than 0.732
4. Cube 8 0.732 to less than 1
5. Closest packing of atoms of equal size 12 1
The number of nearest neighbours for only atom, molecule or ion in a space lattice is called the co–
ordination number. If an atom or ion is occupied in tetrahedral hole its Co. No. will be 4 and if goes
to octahedral holes its Co. No. will be 6.
Radius Ratio Rules
For the stability of an compound, each cation should be surrounded by maximum number of anions
and vice versa (for maximum electrostatic forces of attraction). The number of oppositely charged
ions surroundings each ions is called its co–ordination number.
Since ionic bond is non–directional, the arrangement of ions within the ionic crystal is determined
by the sizes of the ions.
The ratio of the radius of the cation to that of the anion is called radius ratio, i.e.
atomhostofRadius
atomquestofRadius
)r(aniontheofRadius
)r(cationtheofRadius
RatioRadius
a
c

The guest atom must make a contact with host atom but host atom may or may not make contact.
Evidently, greater is the radius ratio, the larger is the size of the cation and hence greater is its co–
ordination number. The relationships between the radius ratio and the co–ordination number and
the structural arrangement are called radius ratio rules, and are given in Table.
Radius ratio rules for AB type structures
Radius Ratio Co–ordination
no
Structural
arrangement
Structure type Examples
=0.155 but < 0.225 3 Planar triangular – B2O3
=0.225 but <0.414 4 Tetrahedral Sphalerite, ZnS CuCl, CuBr, Cul, BaS
=0.414 but < 0.732 6 Octahedral Sodium
chloride (Rock
salt)
HgS, NaBr, KBr, MgO,
MnO, CaO, CaS
=0.732 but <1.0 8 Body–centred
cubic
Caseium
chloride
Csl, CsBr, TlBr,
NH4Br
The increase in coordination number with increase in radius ratio can be easily understood by
considering a particular cation (so that its size i.e. r+ remains constant) and arranging anions of
decreasing size around it (so that r– decreases and hence r+/r– increases). We observe that as the
radius ratio increases, the number of anions touching the cation increases i.e. coordination number
increases
Illustrating effect of radius ratio on co–ordination number. As rc/ra increases, coordination number
increases
Assignment – II
Q.1 The co–ordination number of ccp structure for metals is 12, since
(a) each atom touches 4 others in same layer, 3 in layer above and 3 in layer below

28. (b) each atom touches 4 others in same layer, 4 in layer above and 4 in layer below
(c) each atom touches 6 others in same layer, 3 in layer above and 3 in layer below
(d) each atom touches 3 others in same layer, 6 in layer above and 6 in layer below
Q.2 Atoms of element B from hcp lattice and those of the element A occupy 2/3rd of tetrahedral
voids. What is the formula of the compound formed by these elements A and B?
Q.3 In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally
distributed between octahedral and tetrahedral voids. If all the octahedral voids are
occupied, what is the formula of the solid?
Q.4 In the mineral, spinel, having the formula MgAl2O4, oxide ions are arranged in the cubic
close packing, Mg2+ ions occupy the tetrahedral voids while Al3+ ions occupy the octahedral
voids.
(i) What percentage of tetrahedral voids is occupied by Mg2+ ions?
(ii) What percentage of octahedral voids is occupied by Al3+ ions?
Q.5 In a solid oxide ions are arranged in cubic close packing. One-sixth of the tetrahedral voids
are occupied by cations A while one-third of the octahedral void are occupied by the cations
B. What is the formula of the solid?
Q.6 B– ions form a close packed structure. If the radius of B– ion is 195 pm, calculate the radius
of the cation that just fits into the tetrahedral hole. Can a cation A+ having a radius of 82 pm
be slipped into the octahedral hole of the crystal A+ B–?
Q.7 If the close packed cations in an NaCl type solid have a radius of 75 pm, what would be the
maximum and minimum sizes of the anions forming the voids?
Q.8 A solid A+ B– has NaCl type close packed structure. If the anion has a radius of 250 pm, what
should be the ideal radius of the cation? Can a cation C+ having a radius of 180 pm be
slipped into the tetrahedral site of the crystal A+B–? Give reason for your answer.
Q.9 The mineral haematite, Fe2O3 consists of a cubic close packed array of oxide ions with Fe3+
ions occupying interstitial positions. Predict whether the iron ions are in the octahedral or
tetrahedral holes. Radius of Fe3+ = 0.65Å and that of O2– = 1.45Å.
Q.10 Xenon crystallizes in the face centred cubic lattice and the edge of the unit cell is 620 pm.
What is the nearest neighbour distance and what is the radius of xenon atom?
Q.11 AB has arrangement in which B is at the corner and A is at the body center and its unit cell
edge length is 400 pm. Calculate the inter-atomic distance in AB.
Answers
1. c 2 A4B3
3. A2B 4. (a) 12.5%, (b) 50% 5. ABO3
6. 43.875 pm, No 7. 102.5 pm, 181.2 pm 8. 103.4 pm, No
9. octahedral holes 10. 438.5 pm, 219.25 pm 11. 346.4 pm
Types of Ionic Solid
(l). AB type – Co. No. of both cation & anions are same
(1) CsCl Structure:
Cesium chloride structure is a simple cubicstructure (appears like bcc). Cs+ ions are large
enough to have eight Cl– ions around them. The Cl– ions are at the corners of a primitive cubic
array. There is a big central empty region in each Cl– array and Cs+ ion fits into it. The
structure may be considered to be built of two interpenetrating primitive cubic lattices (one
of Cs+ and another of
Cl–). Each Cs+ ion is surrounded by eight Cl– ion, and vice versa and so the structure has 8 : 8
coordination.

29. (2) Sodium Chloride Structure (Rock Salt):
NaCl has face centred cubic (fcc) structure. Imagine a fcc lattice formed by Na+ ions and
another fcc lattice formed by Cl– ions. When two lattices are interpenetrated half–way
through, the resulting structure represents the structure of NaCl. In complete structure, each
Na+ is surrounded by 6 Cl– ions and vice versa. In this octahedral arrangement, coordination
number of both Na+ and Cl– is 6 e.g. KCl, Nal, RbF, Rbl, alkali metal halides (except Cs),
alkaline earth metal oxide, (except BeO) sulphide, AgX (except Agl), NH4X, FeO (wustite)
pressure
Heating
NaCl type CsCl type
Formula weight = Na4Cl4
D =
.NoAvogadroa
wt.M4
3


D =
o
303
N10)a(
wt.M4



If a is in picometer& Density in g/ml
(3) Zinc blende (Sphalerite) Structure:
The main features of this structure are:
(i) Sulphide ions have CCP arrangement and zinc ions occupy tetrahedral voids.
(ii) There are zinc ions at one fourth of the distance along each body diagonal.

30. (iii) Only half of the alternate tetrahedral voids are occupied by Zn2+ ions. If S–2 ions are at lattice
point, there are Zn+2 ion at ¼ of the distance along each body diagonal.
(iv) Coordination no. of Zn2+ ions as well as S2– ions is 4. Thus, this structure has 4 : 4
coordination number.
Here electronegativity difference between Zn and S is
very small i.e., only 0.9 (electronegativity for Zn = 1.6
and S = 2.5), therefore, the bond between Zn and S
has significant covalent character.
Packing fraction of ZnS (Zinc blende):
= 3
3
a
3
c
a
4r
3
4
3
4







=
 
.]a2r4as[
2
r4
rr
3
16
a3
a
3
a
3
c








012.0)225.0(
392.0)732.0(
071.0)414.0(
3
3
3



=














1
r
r
23
3
a
c
Limiting value (tetrahedral void) = 225.0
r
r
a
c

=   1225.0
23
3


= 0.749 or 74.9%
(4) HCP structure of ZnS (Wurtzite)
It has hexagonal unit cell made up of six S–2 ion at corners, face centre& 3 atom in body centre i.e.
one unit cell has six sulphide ion. Zn+2 ions are placed in alternate tetrahedral void.
(5) Diamond:
Space lattice as ZnS, it has two interpenetrating FCC lattice, every carbon is surrounded by 4 other
carbon atom place at the corner of regular tetrahedron.
Carbon atom occupy corner, face centre as well as in 50% of tetrahedral void
 cr2
4
a3


31.  Packing fraction = 3
c
3
c
3
r8
r
3
4
8







= 34.
16
3


 In diamond only 34% of space is occupied & 66% is empty.
Density of Cubic Unit Cell
Density of cube unit cell = 3
a
atomof.noTotalatomoneofMass
cubeofVolume
cellunitofMass 

= 
oN
wt.At
Total no. of atom × 3
303
o
3
cm/gm
10)pm(aN
ZM
a
1




No = Avogadro’s number; a = edge length; Z = No. of atoms per unit cell;
d = distance between two neighbouring atoms.
In NaCl type unit cell, d = rc + ra = a/2
In CsCl type d = rc + ra = a
2
3
[2d = Body diagonal]
In ZnS type 4ra = a2 In CaF2 Type 4rc = a2
Theoretical density of solid is that which we are supposed to calculate, assuming crystal to be
perfect or ideal. Where as experimental density is defined for imperfect crystal where defects are
present.
(i) If theoretical density is more it will be a vacancy defect.
(ii) If theoretical density is less than observed density it will be metal excess defect without
anion vacancy.
Assignment – III
Q.1 CsCl has cubic structure. Its density is 4.0g cm–3. What is the distance between Cs+ and Cl–
ions? (At. mass of Cs = 133)
Q.2 An element has a body–centred cubic (bcc) structure with edge of 288 pm. The density of
the element is 7.2g/cm3. How many atoms are present in 208g of the element?
Q.3 KF has NaCl structure. If the distance between K+ and F– is 269 pm, find the density of
KF(NA = 6.02  1023mol–1, atomic masses K = 39, F = 19 amu).
Q.4 Calculate the density of silver which crystallizes in a face–centred cubic structure. The
distance between the nearest silver atoms in this structure is 287 pm. (Molar mass of Ag =
107.87 g
mol–1, NA = 6.02  1023mol–1)
Q.5 The density of a face centred cubic element (atomic mass = 60.2 amu) is 6.25g cm–3.
Calculate the length of the edge of the unit cell.
Q.6 What is the distance between Na+ and Cl– in a NaCl crystal if its density is 2.165g cm–3? NaCl
crystallizes in the fcc lattice.
Q.7 Iron (II) oxide has a cubic structure and each unit cell has side 5 Å. If the density of the oxide
is 4g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell (Molar mass of
FeO = 72g mol–1, NA = 6.02  1023mol–1)
Q.8 (a) An element crystallizes in BCC structure. The edge length of its unit cell is 288 pm. If the
density of the crystals is 7.2g cm–3, what is the atomic mass of the element?
(b) How many atoms of this element are present in 100gm ?

32. Q.9 An element (density 6.8g cm–3) occurs in the BCC structure with cell edge of 290 pm.
Calculate the number of atoms present in 200g of the element.
Q.10 Tungten has a density of 19.35g cm–3 and the length of the side of the unit cell is 316 pm.
The unit cell in the most important crystalline form of tungsten in the body centred cubic
unit cell. How many atoms of the element does 50g of the element contain?
Q.11 A compound AB crystallizes in lattice similar to bcc with unit cell edge length of 380 pm.
Calculate
(i) the distance between oppositely charged ions in the lattice
(ii) radius of B– if the radius of A+ is 190 pm
Q.12 An element A crystallizes in fcc structure. 200g of this element has 4.12  1024 atoms. The
density of A is 7.2g cm–3. Calculate the edge length of the unit cell.
Q.13 Sodium crystallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculate the
number of atoms in a unit cell. [Atomic mass of Na = 23.0 amu, Density of sodium = 0.9623g
cm–3, NA = 6.023  1023mol–1]
Q.14 Calculate the approximate number of unit cells present in 1g of ideal NaCl crystals.
Answers
1. 357 pm 2. 24.17 × 1023 atoms 3. 2.48 gm/cm3
4. 10.71 gm/cm3 5. 4 × 10–8 cm 6. 281 pm
7. 4 8. (a) 51.8 (b) 11.62 × 1023 9. 24.12 × 1023
10. 1.638 × 1023 11. (a) 329 pm (b) 139 pm 12. 299.9 pm
13. 2 14. 2.57 × 1021