UNIT-I

1. Bonding in solids

Ionic bond, covalent bond, Metallic Bond, Hydrogen Bond, Vander-Waal’s Bond, Calculation of Cohesive energy.

2. Crystallography and Crystal Structure

1. Space Lattice, 2. Unit cell, 3. Lattice parameters,4. Crystal systems, 5. Bravais Lattices, 6. Miller Indices, 7. Crystal Planes and Directions, 8. Spacing of orthogonal Crystal Systems, 9. Atomic Radius, Coordination number and packing factor of SC, BCC, FCC., 10. Diamond and hcp structures, 11. Structures of NaCl, ZnS, CsCl.

Bonding in solids

AIM AND OBJECTIVE

This lesson deals with the bonding in solids, which talks about Ionic bond, Covalent bond, Metallic Bond, Hydrogen Bond, Vander-Waal’s Bond, Calculation of Cohesive energy.

CONTENTS

IntroductionIonic bond, covalent bond, Metallic Bond, Hydrogen Bond, Vander-Waal’s Bond, Calculation of Cohesive energy.

INTRODUCTION

Electrical forces are responsible in binding the atoms and the molecules giving different physical properties of the solid. A solid is composed of billions of atoms packed together closely. According to the strength and directionality, the bonds are categorized into two main groups namely primary and secondary.  Primary bonds are formed by virtue of their nature and interatomic bonds. The secondary bonds are intermolecular bonds.

Primary bonds are classified into three types. 

(1)   Ionic bonds(2)   Covalent bonds(3)   Metallic bonds   

IONIC BOND 

                         In ionic crystals one or more electrons of one type of atoms are transferred to another atom and the result being a crystal that is composed of positive and negative ions .  The electronic configuration of the ion is essentially an inert gas configuration. The electrostatic interaction between the ions is the source of cohesive energy that binds the crystal together. 

                      In spite of electrostatic attraction there is also repulsive force between the negatively charge electron clouds which becomes operative when two ionic clouds try to overlap. But the electrostatic attractive force is greater than this coulombs repulsive force. 

 

     bonding in NaCl

 

                      Consider the case of NaCl molecule. The atomic number of Na is =  11 

The electronic configuration of Na  =  1s22s22p63s1

 

It has   One    electron in the outermost orbit.`This electron is called as   valence electron.The Na atom will try to lose this valence electron, in order to attain                           Inert gas configuration   

         The minimum energy required to detach an electron from Na atom is called as its ionization energy.    

          sodium       +  Ionisation Energy   ----------  Na+  +   electron                                     Ionisation energy required  =   5.1  eV 

The atomic number of chlorine is 17. Its electronic configuration is 1s22s22p63s23p5

It has seven electrons in its outermost orbit. It needs one electron to attain inertgas configuration. It readily an electron and releases some energy called as electron affinity. 

          Cl   +  electron  -----------   Cl _     +    Electron Affinity    

                  Electron affinity   =   3.6  eV

In order to produce two ions, NaCl needs energy equal to 5.1-3.6  =  1.5 eV                 Na   +   Cl  +  1.5 eV    --------    Na+    +     Cl_

              Na + and Cl- attract each other and form a bond. This bond is called as ionic bond. It is the bond between two oppositely charged io 

CHARACTERISTIC OF   IONIC  BOND MATERIALS  

Ionic bond is fairly strong. They have strong binding energy They have high melting and boiling points. They are soluble in water They are transparent to visible light Ionic crystals have close packed structure. Conductivity is less At high temperatures conductivity arises due to the mobility of ions. Non-directional because the charge distribution is spherical in nature

COVALENT BOND                    In this bonding the valence electrons are not transferred from one atom ato the other atom but the neighbouring atoms share their valence electrons under the formation of a strong covalent bond.                            

 

 

 

The simplest case of a single covalent bond occurs in the hydrogen molecule

 

BONDING IN HYDROGEN                        In the molecular orbital of hydrogen, the two electrons are equally shared between the nuclei and cannot be specifically identified with either nucleus.The two electrons are more likely to be found between the two nuclei and thus the electron density is relatively high.                    Consider the case of chlorine atom which has seven electrons in its outermost orbit. The spins of six electrons are paired, whereas the spin of the seventh electron is unpaired. The unpaired electron is always looking for another unpaired electron which comes closer to it. In such case when another chlorine atom comes near the first chlorine atom, the two unpaired electrons of the two chlorine atoms get paired such that their spins are anti parallel.  

CHARACTERISTICS OF COVALENT BOND MATERIALS 

o Covalent bond crystals are usually hard and brittle.o Binding energy is high so that their melting and boiling points are high but low

compared to ionic crystals.o Covalent bonds are highly directional in character.o These bonds have saturation property.o Covalent substances are insoluble in charactero These materials are soluble in non-polar solvents like benzene.o The conductivity of covalent crystals varies over a wide range. Some are excellent

insulators like diamond. Others are medium conductors like Silicon and Germanium. Some behave as poor metals like grey tin.

o The conductivity increases with increase in temperature.o These are transparent to longer wavelengths but opaque to shorter wavelengths.

Diamond is transparent to radiation longer than short ultraviolet wavelengths. Ge and Si are transparent for wavelengths longer than the infrared radiation.

o Carbon in the diamond structure is the hardest substance and has melting points of 3280

METALLIC BOND                             In metallic bonding, the valence electrons which hold the atoms together are not bound to individual atoms or pair of atoms but move freely throughout the whole metal. Metallic elements have low ionization energies and the atoms give their valence electrons to form and electron cloud or electron gas throughout the space occupied by the atoms. After losing their valence electrons, the atoms are in reality positive ions. These ions are held together by the forces that are similar to those of ionic bond and in that they are primarily electrostatic, but are between the ions and the electrons. Most of the metals have one or two valence electrons. These electrons are loosely held by their atoms and therefore can be easily released to form an electron cloud.        

 

        

electrons in metals -  electrons clouds moving in between positively charged ions 

                   The electrostatic interaction between the positive ions and the surrounding electron cloud hold the metal together. It is called as metallic bond. The metallic crystals have high electrical and thermal conductivities. Metals may be deformed without fracture because the electron gas permits atoms to slide fast on one another by acting as lubricant. Carbon material also available in the form of graphite. In graphite covalency is not fully achieved, hence they may easily reform into metallic bond and there by graphite can have good electrical conductivity. If any potential difference is applied between the two ends of a metallic rod, electron gas flows easily and constitutes an electric current. CHARACTERISTIC OF METALLIC BOND MATERIALS: 

They have high electrical and thermal conductivities. The metals are opaque to all electromagnetic radiation from very low frequency to

the middle ultraviolet, where they become transparent. The metals have high optical reflection and absorption coefficients. The bonding may be weak (mercury) or strong (tungsten). Therefore their melting

points are -39 degrees to 3410 degree Celsius. Due to the symmetrical arrangement of the positive ions in space lattice, metals

are crystalline. The metallic bond is inherent in typical metals and in many inters metallic

compounds.

HYDROGEN BONDING

As the name "hydrogen bond" implies, one part of the bond involves a hydrogen atom. The hydrogen must be attached to a strongly electronegative heteroatom, such as oxygen, nitrogen or fluorine, which is called the hydrogen-bond donor. This electronegative element attracts the electron cloud from around the hydrogen nucleus and, by decentralizing the cloud, leaves the atom with a positive partial charge. Because of the small size of hydrogen relative to other atoms and molecules, the resulting charge, though only partial, nevertheless represents a large charge density. A hydrogen bond results when this strong positive charge density attracts a lone pair of electrons on another heteroatom, which becomes the hydrogen-bond acceptor.

The hydrogen bond is not like a simple attraction between point charges, however. It possesses some degree of orientational preference, and can be shown to have some of the characteristics of a covalent bond. This covalency tends to be more extreme when acceptors bind hydrogens from more electronegative donors.

Strong covalency in a hydrogen bond raises the questions: "To which molecule or atom does the hydrogen nucleus belong?" and "Which should be labelled 'donor' and which 'acceptor'? According to chemical convention, the donor generally is that atom to which, on separation of donor and acceptor, the retention of the hydrogen nucleus (or proton) would cause no increase in the atom's positive charge. The acceptor meanwhile is the atom or molecule that would become more positive by retaining the positively charged proton. Liquids that display hydrogen bonding are called associated liquids.

Hydrogen bond in water

The most ubiquitous, and perhaps simplest, example of a hydrogen bond is found between water molecules. In a discrete water molecule, water has two hydrogen atoms and one oxygen atom. Two molecules of water can form a hydrogen bond between them. The oxygen of one water molecule has two lone pairs of electrons, each of which can form a hydrogen bond with hydrogens on two other water molecules. This can repeat so that every water molecule is H-bonded with four other molecules (two through its two lone pairs, and two through its two hydrogen atoms.)

H-O-H...O-H2

Liquid water's high boiling point is due to the high number of hydrogen bonds each molecule can have relative to its low molecular mass. Water is unique because its oxygen atom has two lone pairs and two hydrogen atoms, meaning that the total number of bonds of a water molecule is four.

In ice, the crystalline lattice is dominated by a regular array of hydrogen bonds which space the water molecules farther apart than they are in liquid water. This accounts for water's decrease in density upon freezing. In other words, the presence of hydrogen bonds enables ice to float, because this spacing causes ice to be less dense than liquid water.

Were the bond strengths more equivalent, one might instead find the atoms of two interacting water molecules partitioned into two polyatomic ions of opposite charge, specifically hydroxide and hydronium.(Hydronium ions are also known as 'hydroxonium' ions).

H-O- H3O+

Indeed, in pure water under conditions of standard temperature and pressure, this latter formulation is applicable only rarely; on average about one in every 107 molecules gives up a proton to another water molecule, in accordance with the value of the dissociation constant for water under such conditions.

Vander-Waal’s Bond

A weak force of attraction between electrically neutral molecules that collide with or pass very close to each other. The van der Waals force is caused by the attraction between electron-rich regions of one molecule and electron-poor regions of another (the attraction between the molecules seen as electric dipoles). The attraction is much weaker than a chemical bond. Van der Waals forces are the intermolecular forces that cause molecules to cohere in liquid and solid states of matter, and are responsible for surface tension and capillary action.

Calculation of Cohesive Energy in Nacl:

NaCl is the best example for ionic crystals . First consider Na atom with electronic configuration 1s2 2s2 2p6 3s1  .

There is a single electron in its outermost orbital . It can be detached easily . It takes minimum energy to detach an electron from Na. This energy is called Ionisation energy .

  Na  +  I.E     Na+ e-

Hence  I.E  is  5.1 eV .

Consider chlorine atom with electronic configuration 1s2 2s2 2p6 3s1 3s5 .

It needs one electron to attain inert gas configuration . It accepts an electron and releases some energy is called electron affinity .

        Cl  +  e-     Cl+  +  3.6 eV electron affinity .

The next energy required to create Nal+ and Cl- pair is equal to 5.1 eV - 3.6 eV  =  1.5 eV .

Using the below expression we get the potential energy .

       = 

        =      joule

    1 eV  =  1.6 x  10-19  joule

The energy released in the formation of NaCl  is  (5.1 - 3.6  )  =  15 eV

The bond energy or cohesive energy of NaCl is 1.5eV

2. CRYSTALLOGRAPHY AND CRYSTAL STRUCTURE

AIM AND OBJECTIVE

This lesson deals with the crystal structure of solids, which talks about (a) space lattice and (b) crystal structure, number of lattice points in the unit cell, Bravais lattices, various crystal systems, Miller indices, lattice constants and density, atomic packing, atomic radius, lattice constant and the crystal structure of sodium chloride.

CONTENTS:

2.1 Introduction

2.2 Space Lattice

2.3 Unit Cell

2.4 Lattice Parameters

2.5 Crystal Systems

2.6 Bravais Lattices

2.7 Miller Indices

2.8 Crystal Planes and Directions

2.9 Inter Planar Spacing of Orthogonal Crystal Systems

2.10 Atomic Radius

2.11 Co-ordination Number and Packing Factor of SC, BCC, FCC

2.12 Diamond and hcp Structures

2.13 Structures of NaCl, ZnS, CsCl.

2.1 INTRODUCTION

One has to understand the crystal structure (arrangement of atoms) thoroughly because solids have specific properties which depend on the structure. The materials are generally classified into (i) solids, (ii) liquids and (iii) gases. Solids are further classified into Crystalline and Non-Crystalline (Amorphous) solids depending on the arrangement of atoms. If the atoms are arranged periodically throughout the solid then it is said to be crystalline otherwise it is said to be non crystalline solid. The atoms / molecules are electrically neutral. But when atoms or molecules are brought closer together, a repulsive force operates between the similar charges in the atoms or molecules. An attractive force operates between the dissimilar charges. The ultimate force, holding the particles together in solids, is the resultant of attractive and repulsive forces (figure 1)

There appears a least distance, at which the particle cluster is the most stable. This minimum distance between the particles is the equilibrium distance (r0). The arrangement of particles in crystals is decided by the nature of bond between the particles and the value of the equilibrium distance. The atoms / molecules in the solids are held together either by (1) Ionic bonds, (2) Covalent bonds, (3) Metallic bonds or (4) Molecular bonds.

2.2 SPACE LATTICE

A space lattice can be generated by putting infinite number of points in space in such a way that the arrangement of points about a given point is same as at any other point. Each lattice point represents the location of an atom or particular group of atoms of the crystal. Intersection of any two lines in the figure 2 is a lattice point.

A set of an identical atom/s which is correlated to lattice points is called basis.

2.3UNIT CELL

The crystal structure of a material or the arrangement of atoms within a given type of crystal structure can be described in terms of its unit cell. The unit cell is a small box containing one or more atoms, a spatial arrangement of atoms. The unit cells stacked in three-dimensional space describes the bulk arrangement of atoms of the crystal. The crystal structure has a three dimensional shape. The unit cell is given by its lattice parameter which are the length of the cell edges and the angles between them, while the positions of the atoms inside the unit cell are described by the set of atomic positions (xi  , yi  , zi) measured from a lattice point.

2.3.1Number of lattice points per unit cell: Knowledge of the volume, density and molecular weight of the constituent atoms of the cell leads us to find the lattice points per unit cell.

Let us consider a unit cell with volume V(incm3), which can be calculated from the unit cell dimensions. Let ρ(gm/ cm3) be the density of the crystal. Then the weight of the matter in the unit cell will be V x ρ.

If n be the number of atoms or molecules per unit cell and M be the atomic (molecular) weight of one atom or molecule. The weight of the matter in the unit cell will be given by the expression-

n * M * 1.66*10-24 gms.

Where 1.66X 10-24 is the weight of hydrogen atom in gms. used in converting molecular weight in gms. Thus we have,

V*ρ = n * M * 1.66* 10-24

n = (V*ρ)/(M * 1.66* 10-24)

n = (V*ρ)/(M)* 6.023* 10-23

n = (V*ρ*N)/(M)

where N is the Avagadro number equal to 6.023* 10-23 The volume of the different lattice is as given here:

Crystal system Volume(V)

1. Cubic a3

2. Hexagonal abc sin(60)

3. orthorhombic abc

4. Rhombohedral ½ a3 sin(α/2)cos(α/2)

5. Trigonal a2c

6. Monoclinic abc sin(β)

2.4 CRYSTAL STRUCTURE

A crystal structure is formed when the basis is substituted in the space lattice i.e Lattice + Basis = Crystal structure (figure3).

2.5 LATTICE PARAMETERS

The lattice constant [or lattice parameter] refers to the constant distance between unit cells in a crystal lattice. Lattices in three dimensions generally have three lattice constants, referred to as a, b, and c. However, in the special case of cubic crystal structures, all of the constants are equal and we only refer to a. Similarly, in hexagonal crystal structures, the a and b constants are equal, and we only refer to the a and c constants. A group of lattice constants could be referred to as lattice parameters. However, the full set of lattice parameters consist of the three lattice constants and the three angles between them.

For example the lattice constant for a common carbon diamond is a = 3.57Å at 300 K. The structure is equilateral although its actual shape can not be determined from only the lattice constant. Furthermore, in real applications, typically the average lattice constant is given. As lattice constants have the dimension of length, their SI unit is the meter. Lattice constants are typically on the order of several angstroms (i.e. tenths of a nanometre). Lattice constants can be determined using techniques such as X-ray diffraction or with an atomic force microscope.

In epitaxial growth, the lattice constant is a measure of the structural compatibility between different materials. Lattice constant matching is important for the growth of thin layers of materials on other materials; when the constants differ, strains are introduced into the layer, which prevents epitaxial growth of thicker layers without defects.

2.6 CRYSTAL SYSTEMSTable 1  

Seven Crystal Systems

Crystal System No. Unit Cell Coordinate Description

1 Triclinic 1 Primitivea ≠ b ≠ cα ≠ β ≠ γ

2 Monoclinic2 Primitive

a ≠ b ≠ cα = β = 90° ≠ γ

3 Body   Centered

3 Orthorhombic 4 Primitive a ≠ b ≠ cα = β = γ = 90°

5 Base   Centered

6 Body   Centered

7 Face   Centered

4 Tetragonal8 Primitive

a = b ≠ cα = β = γ = 90°

9 Body   Centered

5 Trigonal 10 Primitivea = b = c

α = β = γ < 120°, ≠ 90°

6 Hexagonal 11 Primitivea = b ≠ c

α = β = 90°, γ = 120°

7 Cubic

12 Primitive

a = b ≠ cα = β = γ = 90°

13 Body   Centered

14 Face   Centered

2.7 BRAVAIS LATTICE

In geometry and crystallography, a Bravais lattice, studied by Auguste Bravais (1850),[1] is an infinite set of points generated by a set of discrete translation operations described by:

where ni are any integers and ai are known as the primitive vectors which lie in different planes and span the lattice. For any choice of position vector R, the lattice looks exactly the same.

A crystal is made up of a periodic arrangement of one or more atoms (the basis) repeated at each lattice point. Consequently, the crystal looks the same when viewed from any of the lattice points.

Two Bravais lattices are often considered equivalent if they have isomorphic symmetry groups. In this sense, there are 14 possible Bravais lattices in three-dimensional space. The 14 possible symmetry groups of Bravais lattices are 14 of the 230 space groups.

The Bravais Lattice has infinite number of lattice points in it. If a set of identical atoms / molecules are substituted in the space lattice then the lattice is said to be Bravais lattice. The surroundings of any atom/molecule is same as any other atom/molecule in the lattice. Otherwise it is said to be non-Bravais lattice. Below are the some figures 4(a) to 4(e) representing both Bravais lattice and non- Bravais lattices.

+ =

Fig 4(a)

Bravais Lattice

    / + =

Fig. 4(b)

Non-Bravais Lattice

2.8 MILLER INDICES

Figure-6

It is possible to define a system of parallel and equidistant planes which can be imagined to pass through the crystal structure are called as “Crystal Planes”. The position of a crystal plane can be expressed in terms of three integers namely “Miller indices”.

If x, y and Z are the starting co-ordinates for a plane then the Miller indices of the plane are obtained by the following procedure:

Consider the x-m y- and z co-ordinates of the lattice points of the plane lying on the x -, y- and the z-

directions of a reference frameTake the reciprocals of these x-, y- and z- co-ordinates values.Convert these fractions into whole numbers by multiplying all the numbers by a common multiplier.Then, if possible simplify these resulting numbers.

These simplified numbers, derived from the x-, y- and z- co-ordinates of the lattice points on the plane are named as the h, k and l values of the plane and is called the Miller index of the plane. All the planes parallel to this plane will have the same indices. So the hkl values for a plane also represent a family of all the parallel planes. The miller index a set of parallel planes is written as <hkl).

Example: Given that,

X= Y= Z=

Taking the ratio of intercepts with the basis vectors, we obtain

Taking reciprocals of the three fractions

Multiplying throughout by least common multiple ‘4’ for the denominator, we have the Miller indices

(5 8 4)

Which is read as “five eight four”

If a plane is oriented parallel to a coordinate axis, its intercept with the coordinate is taken as infinity, since the reciprocal of infinity is zero, the corresponding Miller indices value will also be zero.

Thus the Miller indices is a set of 3 lowest possible integers whose ratio taken in order is the same as that of the reciprocals of the Miller integers of the planes on the corresponding axes in the same order. Similar to the case of representation of directions in the space lattice, any given Miller indices set represents all parallel equidistant crystal planes for a given space lattice. Owing the rotational symmetry, certain planes which are not parallel to each other become in distinguishable from the crystallographic point of view. In such cases, Miller indices are enclosed in braces {} instead of parenthesis or brackets, which represents all the equivalent planes. For example in the case of cubic lattice, the 6 planes referring to 6 faces of a unit cell are represented by the Miller indices as (1 0 0) (0 0 1) (010) (0 0) (0 0 1) collectively designated as {1 0 0 }

2.8.1 EXPRESSION FOR INTERPLANAR SPACING IN TERMS OF MILLER INDICES:

Figure2

To get the interplanar distance, consider a plane ABC with Miller indices (hkl). In the reference frame, draw a normal to the plane from the origin. Let OP is the normal to the plane ABC. Let angle POA = ’, and POB = β’ and POC =’ be the angles made by the normal to the plane with the x, y and the z directions. OP = d is the interplanar distance

Then, from the figure, Cos = ON/OA = d/x

Cos β = ON/OB = d/y

Cos = ON/OC = d/z

But from the definition of Millers incise derived in the earlier section,

h = a/x therefore, x = a/h

k = b/y therefore, y=b/k

l = c/z therefore z= c/l

Writing the values of x, y and z in the above trigonometric equations we get,

Cos

Cos

Cos

From solid geometry, Cos2 + cos2 β + Cos2 =1

Substituting the values of the trigonometric relations we get

d2hkl =

Then, the interplanar distance‘d’ is given by

dhkl =

For cubic lattice a=b=c then,

2.10 CRYSTAL PLANES AND DIRECTIONS

Knowledge of the volume, density and molecular weight of the constituent atoms of the cell leads us to find the lattice points per unit cell.

Let us consider a unit cell with volume V(incm3), which can be calculated from the unit cell dimensions. Let ρ(gm/ cm3) be the density of the crystal. Then the weight of the matter in the unit cell will be V x ρ.

If n be the number of atoms or molecules per unit cell and M be the atomic (molecular) weight of one atom or molecule. The weight of the matter in the unit cell will be given by the expression-

n * M * 1.66*10-24 gms.

Where 1.66X 10-24 is the weight of hydrogen atom in gms. used in converting molecular weight in gms. Thus we have,

V*ρ = n * M * 1.66* 10-24

n = (V*ρ)/(M * 1.66* 10-24)

n = (V*ρ)/(M)* 6.023* 10-23

n = (V*ρ*N)/(M)

where N is the Avagadro number equal to 6.023* 10-23 The volume of the different lattice is as given here:

Crystal system Volume(V)

1. Cubic a3

2. Hexagonal abc sin(60)

3. orthorhombic abc

4. Rhombohedral ½ a3 sin(α/2)cos(α/2)

5. Trigonal a2c

6. Monoclinic abc sin(β)

2.5.1 Bravais Lattices and Crystal Systems: Bravais demonstrated mathematically that, in 3 dimensions there are only 14 different types of arrangements possible theoretically for Bravais lattices in seven crystal systems.

The 14 Crystal lattices are represented in table1 and the crystal systems with unit cells in table2

Table 1  Seven Crystal Systems and 14 Bravais lattices

Crystal System No. Unit Cell Coordinate Description

1 Triclinic 1 Primitivea ≠ b ≠ cα ≠ β ≠ γ

2 Monoclinic2 Primitive

a ≠ b ≠ cα = β = 90° ≠ γ

3 Body   Centered

3 Orthorhombic

4 Primitive

a ≠ b ≠ cα = β = γ = 90°

5 Base   Centered

6 Body   Centered

7 Face   Centered

4 Tetragonal8 Primitive

a = b ≠ cα = β = γ = 90°

9 Body   Centered

5 Trigonal 10 Primitivea = b = c

α = β = γ < 120°, ≠ 90°

6 Hexagonal 11 Primitivea = b ≠ c

α = β = 90°, γ = 120°

7 Cubic

12 Primitive

a = b ≠ cα = β = γ = 90°

13 Body   Centered

14 Face   Centered

2.4.2 Crystal Systems

Symbols used P - Primitive - simple unit cell

F - Face-centered - additional point in the center of each face

I - Body-centered - additional point in the center of the cell

C - Centered - additional point in the center of each end

R - Rhombohedral - Hexagonal class only

Table2

2.6 LATTICE CONSTANT AND DENSITY

Density is a macroscopic property. Basically it is the mass per unit volume. In case of crystals, mass of atoms packed in a conventional unit cell per unit volume of the cell gives the density of the crystal.

Density of a crystal =

Here m is the mass of atoms packed in the conventional unit cell of the crystal. V is the volume of the unit cell. Mass of atoms packed in the conventional unit cell of the crystal. V is the volume of the unit cell. Mass of an atom in the structure is given by the ratio of the atomic weight or Molecular weight ( M ) to the Avogadro number (NA). Mass of atoms contained in the conventional unit cell is then the number of atoms in the unit cell times the mass of an atom. If there are n atoms in the conventional unit cell, then, the mass of atoms in the unit cell is given by

----------(ii)

From equation ( i ), mass of atoms in the unit cell in terms of the density of the crystal is given by,

--------------(iii)

From equations ( ii ) and ( iii ),

From this equation, the density of the crystal is

In case of a cubic lattice the volume of the unit cell V = a3

Therefore

The lattice constant ‘a’ =

2.7 Atomic packing, Atomic radius, Lattice constant and crystal structure:

The number of atoms at equal and least distance from a given atom in the structure is the coordination number can be taken as the first nearest neighbors of an atom in the structure.

Atomic packing factor is the ratio of the volume of the unit cell occupied by atoms to the net volume of the unit cell

The following are the possible structures found in a cubic system of crystals, resulting due to the kinds of packing of atoms,

(a) Simple Cubic (SC),(b) Body Centered Cubic ( BCC ) and(c) Face Centered Cubic (FCC)

(a)Simple Cubic Structure (SC )In a simple cubic structure, the space lattice is cubic. Atoms are placed at all lattice locations. Therefore, an atom at a lattice location will be in contact with all its six nearest neighbors (figure12). Thus the co-ordination number for a simple cubic structure is N=6. In a simple cubic structure, the lattice constant is the cube edge of the conventional unit cell. Thus the lattice constant is the distance between the centers of two neighboring atoms. That is the lattice constant a = 2r, where r is the atomic radius. The atoms at the lattice locations are shared by eight unit cells. Each atom at the lattice location of the unit cell contributes (1/8) to the unit cell.

The number of atoms contained in a unit cell in the structure is given by the number of lattice locations in the unit cell multiplied by the contribution of the atom at each location. That is the number of atoms in a unit cell of the simple cubic structure is given by

n = (no. lattice locations X contribution of atoms at these location)

n = [8 X (1 /8) ] =1

Figure3

The parameters describing the structure of a simple cubic structure are:

1 Co-ordination Number N = 6

2 Number of atoms in a unit cell n = 1

3 Atomic radius R

4 Lattice constant r = 2r

5 Volume of a unit cell V = a3

6 Volume of an atom (4/3) r3

Volume of the unit cell occupied by the atom is given by

v = (number of atoms in a unit cell) X [volume of an atom]

v = (n) X [4/3 r3]

Atomic packing fraction in a simple cubic crystal structure is

APF =

(b) Body Centered Cubic Structure (BCC )

In body centered cubic structure, the space lattice is cubic. Atoms are placed at all lattice locations and there will be an additional atom at the body center of the cubic unit cell. In this structure, the atoms stacked along the body diagonal of the cubic unit cell are in contact. That is, the

atom at the body center of the unit cell will be in contact with all other atoms at the cubic lattice locations of the unit cell. Each atom in the structure will have eight nearest neighbors (figure 13). Thus, the co-ordination number for a body centered cubic structure is N = 8. In a body centered cubic structure, the lattice constant (a), is the cube edge of the conventional unit cell.

Figure4

Calculation of lattice constant in terms of atomic radius

Let d be diagonal distance between the atoms in a plane of the unit cell. Then,

If D is the diagonal of the cubic unit cell, then

But, the atoms along the body diagonal, are in contact, Therefore D = 4r

And D2 = (4r)2 - - - - - - - - (iii)

From equations (i) and ( iii) ,

D2 = (4r)2 = 3a2,

Thus, the relation between the lattice constant (a ) and atomic radius is,

The atoms at the lattice locations are shared by eight unit cell. Therefore, each atom at the lattice location of the unit cell contributes ( 1/8 ) to the unit cell. The number of atoms contained in the unit cell of the structure is given by the number of lattice locations in the unit cell multiplied by the contribution of the atom at each location plus the atom at the body center of the unit cell. That is the number of atoms in a unit cell the simple cubic structure is given by

n = (no. of lattice locations X contribution of atoms at these location ) + ( atom at the body center)

n = [ 8 X ( 1 X 8 ) ] + 1 =2

The parameters describing the structure of a body centered cubic structure are :

1 Co-ordination Number N = 8

2 Number of atoms in a unit cell n = 2

3 Atomic radius r

4 Lattice constant a =(4 / ) r

5 Volume of a unit cell V = a3

6 Volume of an atom a = (4/3) r3

Volume of the unit cell occupied by the atom is given by

v = (number of atoms in a unit cell ) X [ volume of an atom ]

v = (n ) X [ 4/3 r3]

Atomic packing fraction in a simple cubic crystal structure is

APF =

(C) Face Centered Cubic Structure (FCC)

In a face centered cubic structure, the space lattice is cubic. The conventional unit cell consists of eight small cubelets. Atoms are placed at alternate lattice locations. The atoms stacked along the face diagonal of the cubic unit cell are in contact. The atom at any lattice location of the unit cell will have twelve nearest neighbors (figure14)

Fig-5

Calculation of lattice constant in terms of atomic radius

Let d be the diagonal distance between the atoms in a plane of the conventional unit cell. Then,

d2= (a2 + a2 ) = 2a2 - - - - - - - ( i )

But, the atoms along the face diagonal of the conventional unit cell are in contact. Therefore,

d = 4r - - - - - - - ( ii )

And d2 = ( a2 + a2 ) = 2a2 - - - - ( iii )

From equations ( i ) and ( iii ) ,

d2 = ( 4r)2 = 2a2 ,

Thus, the relation between the lattice constant ( a ) and atomic radius is,

A =

The atoms at the lattice locations of the conventional unit cell are shared by eight such cells. Therefore each atom at the lattice location of the unit cell contributes ( 1 /8 ) to the unit cell. The number of atoms belonging to the unit cell in the structure is given by the number of lattice locations in the unit cell multiplied by the contribution of the atom at each location plus the contributions of the atoms at the face centered locations. The number of atoms in a unit cell of the face centered cubic structure is given by

n = ( No. of lattice locations X contribution of atoms at these location )

( No. of atoms at the face centers X their contribution to the unit cell )

n = [ 8 X (1 /8 )] + [ 6 X ( 1 / 2)] = ( 1 +3 ) = 4

The parameters describing the structure of a face centered cubic structure are :

1 Co-ordination Number N =12

2 Number of atoms in a unit cell n = 4

3 Atomic radius r

4 Lattice constant a =(4 / ) r

5 Volume of a unit cell V = a3

6 Volume of an atom a = (4/3) r3

Atomic packing factor is the ratio of the volume of the unit cell occupied by atoms to the net volume of the unit cell

Volume of the unit cell occupied by the atom is given by

v = (number of atoms in a unit cell ) X [volume of an atom]

v = (n ) X [ 4 / 3 ]

Atomic packing fraction in a face centered cubic crystal structure is

APF =

Hexagonal Close Packed Unit Cell

This is a hexagonal prism with:• one atom at each of the 12 corners,• one atom at the centre of each of the 2 hexagonal faces, and • a triangle of atoms in between the hexagons, which rest in the shaded valleys.

The co-ordination number is 12.

Note: This is closely related to fcc, because if the bottom hexagon is displaced so that the apex atom of the triangle rests in the non-shaded valley, then the structure is fcc!

Diamond Structure Unit Cell

This has tetrahedral units linked to each other.

It can also be described as two interpenetrating fcc structures, one at co-ordinates (0,0,0) and the other at (1/4,1/4,1/4).

The co-ordination number is 4.

2.8 CRYSTAL STRUCTURE OF SODIUM CHLORIDE:

Sodium chloride is an ionic crystal. It is a compound of the alkali halide family. Due to the proximity of the sodium atom with a chlorine atom, the valence electron from the sodium atom is transferred to the chlorine atom. With this transfer of an electron, the sodium atom becomes a cation and the chlorine is converted to an anion. Then, these ions are held by the ionic force and the bond that holds the atoms together is ionic.

A conventional unit cell of the structure consists of eight cubic primitive cells (figure 18). In the conventional unit cell of the crystal, sodium atoms occupy the FCC positions of a cubic structure. Chlorine atoms are placed at the intermediate positions between the sodium atoms. Chlorine atoms independently from an FCC structure similarly sodium atoms also from an independent FCC structure. The lattice constant of both the FCC structures is a = 5.63Au. A unit cell of a sodium chloride crystal can be regarded as the structure formed by the inter-penetration of sodium FCC lattice with the Chlorine FCC lattice through one half of the lattice constant. This can be treated as the graphical representation of the crystal structure.

The crystal structure can also be illustrated in terms of the co-ordinates of the atoms in the structure. For this, an orthogonal co-ordinate system is considered. The origins of the co-ordinate system can either be at the location of a chlorine atom or at the location of a sodium atom. With the origin at the location of the sodium atom, the co-ordinates of the sodium and chlorine atoms in the crystal structure are shown below. Sodium chloride is an ionic crystal formed by the transfer of the valence electron from sodium atom to the chlorine atom. Due to this transfer the sodium atom is converted to a cation and the chlorine is converted to an anion. These ions are held by the ionic force and the bond that holds the atoms together is ionic bond. In the unit cell of the compound, chlorine atoms occupy the FCC positions of a cubic structure. Sodium atoms occupy edge center positions.

Figure6

Atoms Co-ordinates

Sodium

000. 100, 101 , 001 , ½ 0 ½

½ ½ 0 , 1 ½ ½ , ½ ½ 1 , 0 ½ ½ ,

110 , 111 , 011 , 010 , ½ 1 ½

Chlorine

½ 0, 10 ½, ½ 01 , 00 ½ ,

1 ½ 0, 1 ½ 1, 0 ½ 1, 0 ½ 0, ½ ½ ½

½ 1 0, 1 ½ ½ , ½ 1 1, 0 1 ½

Density of sodium Chloride

Density of a crystal in terms of its lattice parameter is given by

Here n is the number of sodium chloride molecules in a conventional unit cell. In place of A the molecular weight of sodium chloride molecule is to be considered. V is the volume of the conventional unit cell with a lattice constant a = 5.63 Au. This is the distance between any two neighboring sodium atoms or the neighboring chlorine atoms. The number of molecules in a unit cell is equal to the sum of sodium atoms [n (Na)] contained in the cell and the chlorine atoms [n(CI)]. The number of sodium chloride molecules in a unit cell are :

N = [ n (Na) ] + [ n ( CI ) ]

The number of sodium atoms in the cell are :

[ n (Na)] = 8 ( 1/8 ) + 6 (1/2) = 1 + 3 = 4

The number of chlorine atoms in the cell are :

[ n ( CI ) ] = 4 ( 1/4 )+ 4 ( 1/4 )+ 1 = 4

Therefore, there will be four sodium chloride molecules in a conventional cell

Atomic weight of sodium is

A (Na) = 22.98 X 10-3 Kg

And that of Chlorine is A ( CI ) = 35.45 X 10-3 Kg

Therefore, the molecular weight of sodium chloride molecule is

A = A (Na ) + A ( CI ) = ( 23 X 10-3 ) + ( 33.45 X 10-3 ) kg

A = ( 58.43 X 10-3 ) Kg

Volume of the unit cell is V = a3 = ( 5.63 X 10-10 ) 3 m

And the Avogadro number is NA = ( 6.023 X 1023 )

The density of sodium chloride crystal is

AX-Type Crystal Structures

Some of the common ceramic materials are those in which there are equal numbers of cations and anions. These are often referred to as AX compounds, where A denotes the cation and X the anion. There are several different crystal structures for AX compounds; each is normally named after a common material that assumes

the particular structure.

Rock Salt Structure

Perhaps the most common AX crystal structure is the sodium chloride (NaCl), or rock salt, type. The coordination number for both cations and anions is 6, and therefore the cation–anion radius ratio is between approximately 0.414 and 0.732. A unit cell for this crystal structure (Figure 12.2) is generated from an FCC arrangement of anions with one cation situated at the cube center and one at the center of each of the 12 cube edges. An equivalent crystal structure results from a face-centered arrangement of cations. Thus, the rock salt crystal structure may be thought of as two interpenetrating FCC lattices, one composed of the cations, the other of anions. Some of the common ceramic materials that form with this crystal structure are NaCl, MgO, MnS, LiF, and FeO.

Cesium Chloride Structure

Figure 12.3 shows a unit cell for the cesium chloride (CsCl) crystal structure; the coordination number is 8 for both ion types. The anions are located at each of the corners of a cube, whereas the cube center is a single cation. Interchange of anions with cations, and vice versa, produces the same crystal structure. This is not a BCC crystal structure because ions of two different kinds are involved.

Zinc Blende Structure

A third AX structure is one in which the coordination number is 4; that is, all ions are tetrahedrally coordinated. This is called the zinc blende, or sphalerite, structure, after the mineralogical term for zinc sulfide (ZnS). A unit cell is presented in Figure 12.4; all corner and face positions of the cubic cell are occupied by S atoms, while the Zn atoms fill interior tetrahedral positions. An equivalent structure results

if Zn and S atom positions are reversed. Thus, each Zn atom is bonded to four S atoms, and vice versa. Most often the atomic bonding is highly covalent in compounds exhibiting this crystal structure (Table 12.1), which include ZnS, ZnTe, and SiC.