THE SOLID STATEIntroduction 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 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:

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

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 LustreMetals 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.

3. Network Atoms Covalent bonds C (Diamond 150 – 500 Very hard with very high

or Covalent

and graphite), Si, SiO2, SiC Boron, Black

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.

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

These 14 arrangements are found in various crystals and areknown as Bravais Lattice (out of 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.

3. Orthorhombic

4. Monoclinic 5. Triclinic 6. Hexagonal 7. Rhombohedral

nc = no. of corner atoms, nf = no. of face atoms nb = no. of body atoms (inside the cube)ne = no. of edge atoms

Rank =

nC8

+n f2

+ne4

+ nb for cubic unit cell

(2) Rank for Hexagonal Unit Cell nC6;n f2;ne3

∧ne4

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

Answers1. XY 2. PQR3 3. 1.54 × 1022 4. a5. 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.

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.

(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) Three dimensional close packing from two dimensional hexagonal close packed layers 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.

(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.

Solid Circle Represent Layer A and Dotted Actual view of hexagonal close packing

Tetrahhedral Void

Octahedral Void

Circle Represent Layer B 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

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 below3. 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.

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 Coordination 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 = 8 × 1

8= 1

(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 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

=

Z × 43πr3where Z is 1.

(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.

PF = Volume occupied by the same spheres in a unit CellVolume of unit cell

=Z × 4

3πr3

a3

where, Z = no. of effective atoms or spheres in a unit cell; For simple cubic unit cell z = 1 a = 2r

∴PF =1 × 4

3πr3

8 r3= π6 = 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)×

18 +no of body centred atom (nb) × 1 =

8 × 18

+ 1× 1= 2

(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 = √2 aNow in right angled in ADC

AD2 = AC2 + DC2

AD2 = (√2a )2 + a2 = 3a2

AD = √3 . a r = radius of atom and a = edge length

Body diagonal of cube = 4 r = √3 ad = 2r

r = √3a

4or d = √3 a

2

Packing fraction (PF) =

Volume occupied by the spheres in the unit cellVolume of unit cell

Volume of unit cell =

a3 = ( 4 r√3 )3

(a= 4 r√3 )

And volume occupied by the spheres in the unit cell = Z × volume of a sphere

= 2 × 4

3πr3 [Z = 2 for bcc ]

=

2× 43πr3

( 4 r√3 )3

0.6802 = 68 .02%

Void fraction = 0.32 = 32%Case – IIWhen the body centred atom and the corner atoms are different

AB= √3a = (2r+ + 2r− ) …… (1)

=

( 43 πrc3) + ( 43 πr a3 )a3

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

√3a = 2rc + 2ra Substitute volume of ‘a’ from eq. (ii)

√3 (2ra) = 2rc + 2r a

2rc = 2ra(√3 − 1 )

rc = (√3 − 1 ) r arcra

= 0 .732It is the minimum possible radius ratio in BCC

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. =

( 43 πrc3) + ( 43 πra3)a3

=

43π (r c3 + r a

3)

(2 r a)3= π6 [ rc3 + ra

3

ra3 ] = π

6 [( r cr a )3

+ 1] = π6 [( 0.732)3 + 1 ] = 0 .729

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

= 8 × 1

8+ 6 × 1

2= 1 + 3 = 4

(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 4 r = √a2 + a2 = √2a

face diagonal of the cube =4 r= √2a

(r+) Cation smaller

r =

√24a

r = 1

2√2a , or a = 2 √2r

Therefore, volume of unit cell = a3 = (2√2r )3

and volume occupied by the spheres in the unit cell (If corner & face centred atoms are same)

= Z × volume of a sphere = 4 × 4

3πr3

Packing fraction (PF) =

Volume occupied by the spheres in the unit cellVolume of unit cell

=

4 × 43πr3

(2√2 r )3= 0 .7406

= 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( 43 πrc3) + 1 ( 43 πr a3 )a3

Limiting or Ideal Conditions: If corner atoms are bigger & are touching

AB= √2a = 2 rc + 2 ra …. (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(√2 − 1 )

( rcra )min = (√2 − 1 ) = 0 .414

It refers to the minimum limiting ratio:

rcra

≥ 0 .414

Radius ratio range is equal to 0.414 – 0.732 ( value equal to 0.414 but less than 0.732)

So, packing fraction =

43π ( 3 rc

3 + rc3

2 ra3

) = π6 [3 ( rcra )

3

+ 1] = π6

[3 (0. 414)3 + 1 ] = 0.635

Co–ordination Number or Number of Nearest Neighbours

Next nearest neighbour [12]

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

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) =

√32

a .

(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 =

√2a2 (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.

8 faces 4 corner atom will be nearest neighbour Co–ordination Number = 12

distance AC = 1/2 face diagonal

AB= √( a2 )2+ ( a2 )

2

distance AC =

√22a

AB =

a√2

= 0.707 a = 0.7071 a AB = AC

For next nearest neighbourNext nearest neighbour is the opposite face atom so that number of next nearest neighbour is six. (all face atom). & distance of next nearest neighbour = aHexagonal 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 = no. of corner atoms × 1/6 + no. of face centred atom × 1/2 + no. of atoms inside the hexagonal primitive unit cell × 1

= 12 × 1

6+ 2× 1

2+ 3 × 1

= 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) =

Volume occupied by the spheres in the unit cellVolume of the cell

=

6 × 43πr3

base area × height

Area of equilateral triangle =

√3 a24

=

6 × 43πr 3

(6 × √34a2) × (4 r √ 23 )

=6 × 4

3πr3

6 × √34

(2 r )2 × 4 r √ 23= 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 = √3 rControid divide median in 2 : 1 ratio

XR = √3 r × 23

(RP)2=(XP)2–(XR)2

( h2 )2= 4 r2 − (√3 × 2

3× r )

2; ( h2 )

2= 4 r2 − 4

3r 2 ;

( h2 )2

=83r2 ; h

2=2√2 r√3

; h = 4 r √ 23

P.F.=

6 × 43πr 3

6 × √34

(2 r )2× 4 r √ 23=6 × 4

3πr3

24√2 r 3= 0 .74

74% space is occupied.

Important FactZnS has CCP lattice in zinc blende & HCP wurtzite structure. 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) =

6 × 43 [rc3 + r a

3 )]24√2 ra3 [as in tetrahedral void =

rcra

= 0 .225]=

π3√2

[ (0.225 )3 + 1 ]

= 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

Foreign atom

There are four types of void existed: (a) Triangular (b) Octahedral Voids (c) Tetrahedral Voids (d) Cubic Voids

Triangular Voids

(a) Triangular Voids 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 =

r arc + r a

∴r cra

= 0.155

(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.

12 Octahedral Voids at the edge centres

(d) Hence, no. of effective octahedral void

=

14

׿ ¿no. of octahedral void existed at each edge

+ 1 × no. of octahedral void at the centre of unit cell

=

14 × 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 =

AEAF (hypotenus)

(Base )

Cos 45o =

AEAF

AFB = 90o

FAB = 45o

In FAE: Cos 45o =

RR + r

1√2

= RR + r

√2= R + rR

√2= 1 + rR

rR

= 1− √2= 0.414

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 voidsTetrahedral 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 =

12 × Body diagonal of larger cube

rc + ra = 14

× Body diagonal of l arg er cube

rc + ra = 14

× √3a….. (1)

In FCC unit cell 4ra = √2 a …… (2) from (1) & (2)

rc + ra = √ 32 rarcra

= √ 32 − 1

= 0.225Thus, 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

ADAB

=ra

rc + ra= 0 .816

or

rc + rara

= 10 .816

rcra

= 0 .225

(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

AC = √AB2 + BC 2

AC = √a2 + a2

AC = √2 a= 2R … (i)

AD = √3 a = 2 (R + r ) …. (ii)

Therefore from (i) & (ii)

R + rR

= √3√2

rR

= √3 − √2√2

rR

= 0 .225

(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, r c 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) = √3a

2(rc + ra )2 ra

= √3 aa

or

rcra

+ 1= √3

rcra

= √3− 1= 0 .732

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 0.2252. Tetrahedral 4 0.225 to less than

0.4143. Octahedral 6 0.414 to less than 0.7324. Cube 8 0.732 to less than 15. 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.

Radius Ratio=Radius of the cation(rc )Radius of the anion (ra )

= Radius of quest atomRadius of host atom

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 r c/ra increases, coordination number increases

Assignment – IIQ.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 (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.

Answers1. 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, No9. 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.

(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)

Formula weight = Na4Cl4

D =

4 × M . wta3 × Avogadro No .

D =

4 × M . wt(a )3 × 10−30× No

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. (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):

=

[ 43 π c3 + 43π ra

3 ] × 4a3

=

163

π (rc3 + ra3 )

( 4 ra√2 )3

[as 4 ra = √2 a. ] (0. 414 )3= 0 .071(0 .732)3 = 0 .392(0 .225)3 = 0 .012

=

π3√2 [( rcra )

3

+ 1]Limiting value (tetrahedral void) =

rcra

= 0 . 225

=

π3√2

[ (0 .225 )3 + 1 ] = 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

√3 a4

= 2 rc

Packing fraction =

8× 43πr c3

( 8 r c√3 )3

=

√3 π16

= .34

In diamond only 34% of space is occupied & 66% is empty.

Density of Cubic Unit Cell

Density of cube unit cell =

Mass of unit cellVolume of cube

= Mass of one atom × Total no . of atoma3

=

At . wtNo

׿ ¿ Total no. of atom ×

1a3

= M × ZNo × a3 ( pm) × 10−30

gm /cm3

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 =

√32a

[2d = Body diagonal]

In ZnS type 4ra = √2 a In CaF2 Type 4rc = √2 aTheoretical 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 ?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