Lecture- 1 - Weebly · Web viewAnalysis shows that nickel oxide has the formula Ni0.98O1.00. What...

$Page 1: Lecture- 1 - Weebly · Web viewAnalysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions? Ans :- The formula of nickel oxide$
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Chemistrian Academy


Table of ContentsLecture- 1...................................................................................................................................................................................

General Characteristics of states:-.............................................................................................................................................

On the basis of different properties:-.........................................................................................................................................

Depending upon the nature of intermolecular forces operating in them:-................................................................................

Problems based on this Lecture.................................................................................................................................................

Lecture-2....................................................................................................................................................................................

Characteristics of a crystal lattice...............................................................................................................................................

Calculation of number of particles in a unit cell:-.....................................................................................................................

Problems based on this Lecture...............................................................................................................................................

Lecture-3..................................................................................................................................................................................

Relation between radius of atom, ion and edge length of SCC, BCC & FCC..............................................................................

Density of unit cell:-..................................................................................................................................................................


Lecture-4..................................................................................................................................................................................

Radius Ratio of octahedral And Tetrahedral void.....................................................................................................................


Assignment...............................................................................................................................................................................

Lecture:- 5................................................................................................................................................................................

IMPERFECTION OR DEFECTS IN SOLIDS....................................................................................................................................


Lecture-6..................................................................................................................................................................................

Electrical Properties..................................................................................................................................................................

Magnetic Properties.................................................................................................................................................................



Solid State

General Characteristics of states:-(1) Gaseous State(2) Liquid State(3) Solid State

Gaseous State:- (i) In this state no definite shape and volume.(ii) Very less attractive force between the gaseous particles.

Liquid State:-

(i) In this state definite volume but indefinite shape is there. Shape is according to the vessel size and shape.

(ii) Force of attraction between the particles is more as compare to gas particles.

Solid State:-(i) In this state shape and volume both are definite.(ii) Force of attraction between the particles is maximum.

Solids are having following properties:-

(i) The intermolecular distances in solids are short and intermolecular forces are strong.

(ii) The constituent particles (atoms, ions or molecules) of solids have fixed positions and can only oscillate about their mean positions.

(iii) Solids are almost incompressible.

Lecture- 1


(iv) The density of solids is greater than that of liquids and gases.(v) Solids diffuse very slowly as compared to liquids and gases.

On the basis of different properties:- We are having following types of solids:

(a) Crystalline solids (b) Amorphous solids

Distinction between crystalline & amorphous solids:-

Properties Crystalline solids Amorphous solidsShape The crystalline solids have

definite shapes.The amorphous solids have characteristics irregular shapes.

Order in arrangement of constituent particles

They have regular arrangement of constituent particles. They are said to exhibit long range order.

They do not have any regular arrangement of the constituent particles. They may have short range order.

Melting point They have sharp and characteristics melting point.

They do not have sharp m.p and they gradually soften over a range of temperature.

Cleavage property

When cut with a sharp edge tool, they split into two pieces and newly generated surfaces are plain and smooth.

When cut with a sharp edged tool, they cut into two pieces with irregular surfaces.

Enthalpy They have definite and characteristic enthalpy of fusion.

They do not have definite enthalpy of fusion.

Anisotropy They are anisotropic and have different physical

They are isotropic and have same physical properties in all


properties in different directions.

directions.

Nature They are true solids. They are isopseudo solids and super cooled liquids.

Common examples

Copper, silver, iron, common salt, zinc sulphide etc.

Glass, rubber, plastics etc.

Depending upon the nature of intermolecular forces operating in them:-The crystalline solids are four types:-

(i) Ionic solids(ii) Covalent or network solids(iii) Molecular solids(iv) Metallic solids

The characteristics of these solids summarized in the following:-

Types of solids characteristics

Ionic Covalent or Network

Molecular Metallic

1.

Constituent particles

+ve and –ve ions

Neutral atoms

Molecules(polar or non-polar)

+ve ions and electrons

2.

Bonding attractive forces

Electrostatic or columbic

Shared pairs of electrons

Weak polar forces or van der waals

Electrostatics attraction b/w +ve ions


forces or Hydrogen bonding.

and electrons.

3.

Binding energy(kj mol-1)

400-4000 150-500 0.05-4 or

5-25 or

10-40

70-1000

4.

Hardness Hard and brittle

Very hard Very soft Hard and soft

5.

Melting point(k) High-1500 Very high-4000

Low High or low

6.

Electrical conductivity

Bad conductors

(insulators) Good conductors in fused state or in solution.

Bad conductors

(insulators)

With few exceptions like graphite.

Bad conductors

(insulators)

Good conductors

Questions:-

Problems based on this Lecture Que.1:- Why are solids rigid? Ans:- The intermolecular forces of attraction that are present in solids are very strong. The

constituent particles of solids cannot move from their positions i.e., they have fixed


positions. However, they can oscillate about their mean positions. This is the reason solids are rigid.

Que.2:- Why do solids have a definite volume?Ans:- The intermolecular forces of attraction that are present in solids are very strong. The

constituent particles of solids have fixed positions i.e., they are rigid. Hence, solids have a definite volume.

Que.3:- Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene,benzoic acid, Teflon, potassium nitrate,cellophane,polyvinyl chloride, fiber glass, copper.

Ans:- Amorphous solids:-Polyurethane, Teflon, cellophane, polyvinyl chloride, fiber glass Crystalline solids:-Naphthalene, benzoic acid, potassium nitrate, copper.Que.4:- Why is glass considered a super cooled liquid?Ans:- Similar to liquids, glass has a tendency to flow, though very slowly. Therefore, glass is

considered as a super cooled liquid. This is the reason that glass windows and doors are slightly thicker at the bottom than at the top.

Que.5:- Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

Ans:- An isotropic solid has the same value of physical properties when measured along Different directions. Therefore, the given solid, having the same value of refractive

index along all directions, is isotropic in nature. Hence, the solid is an amorphous solid. When an amorphous solid is cut with a sharp edged tool, it cuts into two pieces with irregular surfaces.

Que.6:- Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.

Ans:- Potassium sulphate → Ionic solidTin → Metallic solid

Benzene → Molecular (non-polar) solid Urea → Polar molecular solid Ammonia → Polar molecular solid Water → Hydrogen bonded molecular solid Zinc sulphide → Ionic solid Graphite → Covalent or network solid Rubidium → Metallic solid Argon → Non-polar molecular solid Silicon carbide → Covalent or network solidQue.7: Solid A is a very hard electrical insulator in solid as well as in molten state and melts at

extremely high temperature. What type of solid is it?Ans:- The given properties are the properties of a covalent or network solid. Therefore, the

given solid is a covalent or network solid. Examples of such solids include diamond (C) and quartz (SiO2).

Que.8: Ionic solids conduct electricity in molten state but not in solid state. Explain.Answer:- In ionic compounds, electricity is conducted by ions. In solid state, ions are held

together by strong electrostatic forces and are not free to move about within the


solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity.

Que.9: What type of solids are electrical conductors, malleable and ductile?Ans:- Metallic solids are electrical conductors, malleable, and ductile.Que.10: What makes a glass different from a solid such as quartz? Under what conditions could

quartz be converted into glass?Ans:- The arrangement of the constituent particles makes glass different from quartz. In glass,

the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders. Quartz can be converted into glass by heating and then cooling it rapidly.

Que.11: Define the term 'amorphous'. Give a few examples of amorphous solids.Ans:- Amorphous solids are the solids whose constituent particles are of irregular shapes and

have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.

Que.12: Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

(i) Tetra phosphorus decoxide (P4O10) (vi) Graphite (ii) Ammonium phosphate (NH4)3PO4 (vii) Brass (iii) SiC (viii) Rb (iv) I2 (ix) LiBr (v) P4 (x) SiAns:- Ionic → (ii) Ammonium phosphate (NH4)3PO4, (ix) LiBr Metallic → (vii) Brass, (viii) Rb Molecular → (i) Tetra phosphorus decoxide (P4O10), (iv) I2, (v) P4. Covalent (network) → (iii) SiC, (vi) Graphite, (x) SiQue.13:Why does the window glass of the old building look milky?Ans:- It is heating during the day and cooling at night. Glass acquires some crystalline

character and these character look milky. Que.14:How can a material be made amorphous?Ans:- First melting the material and then cooling it rapidly from the liquid state.Que.15:How does amorphous silica differ from quartz?Ans:- In amorphous silica SiO4

2- tetrahedral are randomly joined to each other while in quartz they are joined in a regular manner.

Que.16:Ionic solids conduct electricity in the molten state but not in the solid state. Explain. Ans:- In molten state ionic solids dissociate to give free ions and hence can conduct electricity.

However in the solid state, the ions are not free to move but remain held together by strong electrostatic forces of attraction. Therefore they are electrically insulator.

Que.17:Refractive index of a solid is observed to have the same value along the directions. Comment on the nature of the solid. Would It show cleavage property?


Ans:- Since the solid has the same value of refractive index along all directions, it is isotropic and hence, amorphous. Being amorphous solid it would not show a clean cleavage when cut with a knife. Instead, it would break into pieces with irregular surfaces.

Que.18: What type of solids are electrical conductors, malleable and ductile?Ans:- Metallic solids.Que.19: Solids A is a very hard electrical insulator in solid as well as molten state and melt at an

extremely high temperature. What type of solid is it?Ans:- Covalent. Que.20: Urea has a sharp melting point but glass does not. Explain. Ans:- Urea is crystalline solid, so it has sharp melting point. On the other hand, glass is

molecular solid, so it melting point is not sharp. Que.21: Why is glass considered a super-cooled liquid?Ans:- Glass has characteristics similar to liquids. It can flow extremely slowly, so it is

considered to be super-cooled liquid. Que.22: Solid A is very hard electrical insulator in solid as well as in molten state and melts at

extremely high temperature. What type of solid is it?Ans:- It is covalent or network solid. Que.23: Write a distinguishing feature of metallic solids?Ans:- Electrical conduction in solid state. Que.24: How do metallic and ionic substance differ in conducting electricity? Ans:-

Metallic solids Ionic solids

1. Conduct electricity both in solid as well as in molten state.

Conduct electricity only in molten state or when dissolved in some polar solvent like water.

2. Conductivity is due to the migration of electrons.

Conductivity is due to the migration of cations and anions.

Que.25: (a) Define liquid crystals. (b) What are Sematic and Nematic liquids crystals?

Ans:- (a) In a temperature range just above the melting point, some substances are able to exist in a define pattern as in a solid but can flow as a liquid. These substance are known as liquid crystals.

(b) Sematic liquid crystals are having soap like phase and Nematic crystals are those which are having needle or thread-like phase.

Lecture-2

Characteristics of a crystal latticeLet us sum up the characteristics of a crystal lattice. The following are the characteristics of a crystal lattices.


(i) Each point in a crystal lattice is called lattice points or lattice site.(ii) Each point in a crystal lattices represents one constituent particles which

may be an atom, a molecule or an ion.(iii) Lattice points are joined by straight lines to bring out the geometry of

the lattices.

Type of crystal systems:- There are basically two types of unit cells constituting different crystal systems. There are:

(i) Primitive unit cells(ii) Centered unit cells Primitive unit cells:- These are unit cells which have points(or particles)

only at the corners. These are also called simple unit cells. Centered unit cells:- These are unit cells which have points(or particles) at

the corners as well as at some other positions.(a) Body centered unit cells (B.c.c):- The unit cell in which particles are

present at the corners as well as the body centre of the unit cell.(b) End centered unit cells: - The unit cell in which particles are present at

all the corners and at the centre of any two opposite faces.(c) Face centered unit cells (F.c.c):- The unit cell in which particles are

present at the corners as well as the centre of each face.


Space lattice or lattice points:-Space lattice:- It is an array of points showing how ions, atoms or molecules are situated(arranged) in different site in three dimensional space is called space lattice.Lattice point:- It is the point where atom, ion or molecule is placed.Unit cell:- It is the smallest unit which on repetition gives the required crystal lattice and gives almost all the information about the solid. For Ex. In 2-D

In three dimensional:-


Parameter of unit cell:-A unit cell is characterized by:-

(i) Its dimensions along the three edges as ‘a’, ‘b’ and ‘c’. these edges may or may not be mutually perpendicular.

(ii) Angles α, β and γ between the angle pair of edges. These are known as interstial angle or axial angles.

Crystal system:- on the basis of boundaries of unit cell. We are having seven types of crystal systems.


Bravais Lattice:- on these crystal system, arrangement of the atom or ions which gives space lattices known as Bravais lattice. We are having 14 types of bravais lattice.

Calculation of number of particles in a unit cell:-

Lattice point in the arrangement (cubic)Sc-8 Bcc- 9 Fcc- 14 Hcp- 17

Cubic systemAt corner, sharing of atom = 1/8At face, sharing of atom = 1/2At centre, sharing of atom = 1At edge, sharing of atom =1/4

In case of hexagonalAt corner =1/6 At face = 1/2 At centre = 3 NUMBER OF ATOMS PER UNIT CELL IN CUBIC CLOSE PACKED STRUCTURE OF ATOMSUnit cell Number of atoms Number of atoms per

unit cellVolume occupied by particles (%)

Corners Centers

Faces

Simple cube 1 0 0 1 52.4

Body centered (BCC) 1 1 0 2 68

Face centered (FCC) 1 0 3 4 74


Symmetry in the solidSolids are having three types of symmetries.

(1) Plane of Symmetry(2) Axis of symmetry(3) Centre of symmetry

Example:- In cubic→ Plan of symmetry = 9 → Axis of symmetry =13 → Centre of symmetry = 1

Coordination number (CN)It is the number of atoms or ions surrounded by the given atom or ion.Example : In three dimension coordination number is as;Hcp = 12 Bcc = 8 Sc = 6Fcc (Salt like) = 6 Ccp = 12

Questions :-

Problems based on this Lecture Que.1. In NaCl crystal, Cl- ions are in fcc arrangement. Calculate the number of Cl- ions in its unit

cell ? Ans:- Cl- ion per unit cell 8x1/8 + 6x1/2 = 4Que.2. Explain how much portion of an atom located at(i) corner and (ii) body- centre of a cubic

unit cell is part of its neighboring unit cell. Ans:- (i) A portion lying at the corner of a unit cell is shared equally by eight unit cells and

therefore, only one-eight (1/8) of each such point belongs to the given unit cell. (ii) A body centered point belongs entirely to one unit cell since it is not shared by any

other unit cell. Que.3. Define face-centered cubic structure. Ans:- There is a structural particle at the centre of each face as well as each corner.Que.4. What is body-centred cubic ? Ans :- Body centered cubic is an arrangement in which in addition to the particles at the

corners there is one particle present within the body of the unit cell.Que.5. How many lattice points are there in one unit cell of each of the following lattices ? (a) Face centered cubic (b)Face centered tetragonal (c) Body centered cubicAns :- (a) Lattice points located at the corners of the cube = 8 Lattice points located at the centered of each face = 6 Total no. of lattice point = 8 +6 = 14 (b) In face centered tetragonal, the number of lattice point is also the same, 8+6 = 14

(c)In body centered cubic arrangement, Lattice point located in the centre of the body = 1


Total no. of lattice points 8 + 1 = 9Que.6. A solid has structure in which Na, W and O atoms are arranged as follows: Na at the

centre, W atoms at the corners and O atoms at the edge of a cube. What type of lattice is this ? What is the simple formula of this compound ?

Ans:- The lattice is of simple cubic type : Na atoms per unit cell = 1x1 = 1 W atoms per unit cell = 8x1/8 = 1 O atoms per unit cell = 12 x 1/4 = 3 Simple formula is NaWO3.Que.7. What is the 2-dimensional coordination number of a molecules in square close packed

layer ?Ans :- Four Que.8. (a) What is meant by the term ‘coordination number’ ?

(b)What is the coordination number of atoms(i) In a cubic-close packed structure ?(ii) In a body- centred cubic structure ?

Ans:-(a) Coordination number defined as the number of nearst neighbours in a close packing. In ionic crystal, coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.

(b)(i) 12 (ii) 8Que.9. In a body centered cubic (bcc) unit cell, a metal atom at the center of the cell is

surrounded by how many other metal atoms ?Ans:- A metal atom at the center of the cell is surrounded by 8 atoms. Que.10. Give the significance of a ‘lattice point.Ans:- The significance of a lattice point is that each lattice point represents one

constituentparticle of a solid which may be an atom, a molecule (group of atom), or an ion.

Que.11. Name the parameters that characterize a unit cell.Ans:- The six parameters that characterise a unit cell are as follows. (i) Its dimensions along the three edges, a, b, and c These edges may or may not be equal. (ii) Angles between the edges These are the angle ∝ (between edges b and c), β

(between edges a and c), and γ (between edges a and b).Que.12. Distinguish between (i) Hexagonal and monoclinic unit cells (ii) Face-centred and end-centred unit cells.Ans:- (i) Hexagonal unit cell For a hexagonal unit cell,

Monoclinic unit cell For a monoclinic cell, a≠b≠c and α=γ=90, β≠90 (ii) Face-centred unit cell In a face-centred unit cell, the constituent particles are present at the corners

and one at the centre of each face. End-centred unit cell


An end-centred unit cell contains particles at the corners and one at the centre of any two opposite faces.

Que.13. Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

Ans:- (i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore,1/8 th portion of the atom is shared by one unit cell.

(ii) An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e., its contribution to the unit cell is 1.

Que.14. What is the two dimensional coordination number of a molecule in square close packed layer?

Ans:- In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two- dimensional coordination number of a molecule in square close packed layer is 4.

Relation between radius of atom, ion and edge length of SCC, BCC & FCC.

(SCC) (BCC) (FCC)

Packing fraction of unit cell:- For simple cubic unit cell (SCC)

In a simple cubic lattice the atoms are located only on the corners of the cube.

Lecture-3


Let take edge length or side of the cube = a, Let take radius of each particles = r The relation between radius and edge a a = 2r The volume of the cubic unit cell = side3

= a3

= (2r)3

= 8r3

Number of atoms in unit cell = 8 x 1 /8 = 1 The volume of the occupied space = (4/3)πr3

For Body centered cubic (BCC)

Let take edge length or side of the cube = a,Let take radius of each particles = rThe diagonal of a cube is always a√3 The relation between radius and edge a will a√3 = 4r divide by root 3 we get a = 4r/√3total number of atoms in body centered cubic number of atoms at the corner = 8 x 1/8 = 1


number of atoms at the center = 1total number of atoms = 2The volume of the cubic unit cell = side3

= a3

= (4r/√3)3

The volume of the occupied space = (4/3)πr3

For face centered cubic (FCC) Let take edge length or side of the cube = a,Let take radius of each particles = r

The diagonal of a square is always a√2 The relation between radius and edge a will a√2 = 4r divide by root 3 we get a = 4r/√2total number of atoms in body centered cubicnumber of atoms at the corner = 8 x 1/8 = 1number of atoms at the face = 6 x1/2 = 3total number of atoms = 4The volume of the cubic unit cell = side3


= a3 = (4r/√2)3

= (2√2 r)3

The volume of the occupied space = (4/3)πr3

Packing efficiency =Volumeoccupied by four sphere∈the unit cell X100Total volume of unit cell%

= 4 X ( 43 )πr 3 X 100

(2√2 r )3%

= 74 %

Density of unit cell:-

d = Z X M / a3 X NA

Where d is the density of unit cell Z is the rank of unit cell , M is the molar mass, a3 is the volume of cube, NA is Avogadro’s number

NA= 6.022 X 1023

Questions:-

Problems based on this LectureQue.1. Which of the following lattices has the highest packing efficiency ?


(i) Simple cubic (ii) Body centered cubic (iii) Hexagonal close-packed

Ans :- Hexagonal close-packed lattice has the highest packing efficiency (74%)

Que.2. A metal crystallizes into two cubic phases, face-centered cubic (FCC) and body-centered cubic (BCC) whose unit cell lengths are 3.5AO and 3.0 AO respectively. Calculate the ratio of the densities of FCC and BCC.

Ans :- Density of unit cell (d) = Z X M / a3 X NA

Density of fcc unit cell = 4 XM

N a X (3.5 )3

Density of bcc unit cell = 2 X M

Na X (3.0)3

density of f . c . cdensity of b . c . c= 4 X MNa X (3.5)3

X Na X (3.0)32 X M

= 10885.75=1.259

Que.3. How much space is occupied in (i) hcp (ii) ccp and (iii) bcc structure?

Ans:- (i) hcp :-74% (ii) ccp :- 74% (iii) bcc :- 68%.

Que.4. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?

Ans:- Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centred cubic lattices are 52.4% and 68% respectively.Que.5. An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge

length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?

Ans:- It is given that density of the element, d = 2.7 × 103 kg m−3

Molar mass, M = 2.7 × 10−2 kg mol−1

Edge length, a = 405 pm = 405 × 10−12 m = 4.05 × 10−10 m It is known that, Avogadro’s number, NA = 6.022 × 1023 mol−1

Applying the relation, d= ZxM

a3 xNa , Z= dxa3xNaM


Z=2.7 x 103 kgm−3 x ( 4.5x 10−10m ) 3x 6.022x 1023mol−12.7 x10−2kgmol−1

= 4.004 , = 4 This implies that four atoms of the element are present per unit cell. Hence, the unit

cell is face-centred cubic (fcc) or cubic close-packed (ccp).Que.6. How can you determine the atomic mass of an unknown metal if you know its density

and the dimension of its unit cell? Explain.Ans:- By knowing the density of an unknown metal and the dimension of its unit cell, the

atomic mass of the metal can be determined. Let ‘a’ be the edge length of a unit cell of a crystal, ‘d ’ be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z’ be the number of atoms in the unit cell.

Now, density of the unit cell

d= Z xMa3 ……..(i) [Since mass of the unit cell = Number of atoms in the unit cell × Atomic mass]

[Volume of the unit cell = (Edge length of the cubic unit cell)3] From equation (i),

we have:

Now, mass of the metal (m)

Therefore, If the edge lengths are different (say a, b and c), then equation (ii) becomes:

m=d (abc ) x Na

Z …….. (iv) From equations (iii) and (iv), we can determine the atomic mass of the unknown metal.Que.7. How many lattice points are there in one unit cell of each of the following lattice? (i) Face-centred cubic (ii) Face-centred tetragonal (iii) Body-centred.Ans:- (i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic. (ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal. (iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.Que.8. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and

density is 10.5 g cm−3, calculate the atomic mass of silver.Ans:- It is given that the edge length, a = 4.077 × 10−8 cm Density, d = 10.5 g cm−3 As the lattice is fcc type, the number of atoms per unit cell,

z = 4 We also know that, NA = 6.022 × 1023 mol−1 Using the relation:


= 107.13 gmol−1 Therefore, atomic mass of silver = 107.13 uQue.9. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the

cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Ans:- It is given that the atoms of Q are present at the corners of the cube. Therefore, number of atoms of Q in one unit cell

¿ 18x 8

It is also given that the atoms of P are present at the body-centre. Therefore, number of atoms of P in one unit cell = 1,This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1 Hence, the formula of the compound is PQ. The coordination number of both P and Q is 8.

Que.10. If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Ans:- A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that MPOQ is right-angled ∠POQ = 900

Now, applying Pythagoras theorem, we can write:

Que.11. Copper crystallises into a fcc lattice with edge length 3.61 × 10−8 cm. Show that

the calculated density is in agreement with its measured value of 8.92 g cm−3.Ans :- Edge length, a = 3.61 × 10−8 cm


As the lattice is fcc type, the number of atoms per unit cell, z = 4 Atomic mass, M = 63.5 g mol−1

We also know that, NA = 6.022 × 1023 mol−1

Applying the relation:

= 8.97 g cm−3

The measured value of density is given as 8.92 g cm−3. Hence, the calculated density 8.97 g cm−3 is in agreement with its measured value.

Que.12. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Ans :- The formula of nickel oxide is Ni0.98O1.00. Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100 Now, total charge on 100 O2− ions = 100 × (−2) = −200 Let the number of Ni2+ ions be x. So, the number of Ni3+ ions is 98 − x. Now, total charge on Ni2+ ions = x(+2) = +2x And, total charge on Ni3+ ions = (98 − x)(+3) = 294 − 3x Since, the compound is neutral, we can write: 2x + (294 − 3x) + (−200) = 0 ⇒ −x + 94 = 0 ⇒ x = 94 Therefore, number of Ni2+ ions = 94 And, number of Ni3+ ions = 98 − 94 = 4 Hence, fraction of nickel that exists as Ni2+=9498 = 0.959 And, fraction of nickel that exists as Ni3+ = 4

98 = 0.041 Alternatively, fraction of nickel that exists as Ni3+ = 1 − 0.959 = 0.041 Que.13. Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the

length of a side of the cell?Ans :- For a face-centred unit cell: a=2√2 It is given that the atomic radius, r = 0.144 nm So, a=2√2 X 0.144 = 0.407 nm Hence, length of a side of the cell = 0.407 nmQue.14. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm3 of aluminium?


Ans :- (i) For cubic close-packed structure: a= 2√2r = 2√2 X 125 = 353.55 pm = 354 pm (approximately) (ii) Volume of one unit cell = (354 pm)3

= 4.4 × 107 pm3

= 4.4 × 107 × 10−30 cm3

= 4.4 × 10−23 cm3

Therefore, number of unit cells in 1.00 cm3 = 14.4 X10−23

= 2.27 × 1022

Lecture-4Voids :- It is the three dimensional empty space present in the solids.

In two dimensions some of the voids are as,

Square voids Trigonal void

In three dimensions we are having two types of voids.

(1) Tetrahedral void (2) Octahedral void Tetrahedral void :- The void which are surrounded by four spheres which

lie at the vertices of a regular tetrahedron are called tetrahedral voids.


It is having following properties : Coordination number = 4 If radius of voids is rvoid and radius of the atom is ratom then,

r voidratom= 0.225

In the void generally cations are placed so ,

r+ ¿r−¿=0.225 ¿

¿

W.r.t the given atom number of tetrahedral voids are two. Octahedral void :- The void which are surrounded by six sphere located

on a regular octahedron are called octahedral void.

It is having following points. Coordination number is six for octahedral void. If radius of voids is rvoid and radius of the atom is ratom then,

r+ ¿r−¿=0.415¿

¿

If in the void cation is placed then

r+ ¿r−¿=0.415¿

¿

W.r.t given atom number of octahedral voids is one.

Radius Ratio of octahedral And Tetrahedral void Derivation of the relationship between the radius (r) of the

octahedral void and the radius (R) of the atoms in close packing.


A sphere into the octahedral void is shown in the diagram. A sphere above and a sphere below this small sphere have not been shown in the figure. ABC is a right angled triangle. The centre of void is A.

Applying Pythagoras theorem.

BC2 = AB2 + AC2

(2R)2 + (R + r)2 + (R + r)2 = 2(R + r)2 4R2/2 = (R + r)2

√2R = R + r (√2) = (R + r)2

r = √2R – R = (1.414 –1)R

r = 0.414 R

Derivation of the relationship between radius (r) of the tetrahedral void and the radius (R) of the atoms in close packing:

To simplify calculations, a tetrahedral void may be represented in a cube as shown in the figure. In

which there spheres form the triangular base, the fourth lies at the top and the sphere occupies the tetrahedral void.

Let the length of the side of the cube = a From right angled triangle ABC, face diagonal


AB = √AC2 + BC2 = √a2 + a2 = √2a As spheres A and B are actually touching each other, face diagonal AB = 2R ∴ 2R = √2a or R = 1/√2 a .... (i) Again from the right angled triangle ABD AD √AB2 + BD2 = √ (√2a)2 + a2 = √3aBut as small sphere (void) touches other spheres, evidently body diagonal AD = 2(R + r). ∴ 2(R + r) = √3a

R + r = √3/2 a ... (ii)

Dividing equation (ii) by equation (i)

R + r/R = √3 / 2 × a / a / √2 = √3 / √2

1 + r/R = √3 / √2 = 1.225 => r/R = 1.225 – 1 = 0.225

r = 0.225 R

Radius ratio in 1:1 or AB type structure

Questions:-

Problems based on this LectureQue.1.Define void.

Ans :- The empty spaces present between the metal atoms or the ions when they are packed within the crystal are called voids.

Que.2. What is number of tetrahedral voids in an unit cell of a cubic close-packed structure ?

Ans :- There are 8 tetrahedral voids in a unit cell.

Que.3. What is the coordination number of an octahedral void?

Ans:- six.

Assignment


Q.1. A compound forms hexagonal closed-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedaral voids?

Q.2. What is the relation of number of voids in terms of number of atoms?

Or

Lecture:- 5

IMPERFECTION OR DEFECTS IN SOLIDSDefect:- Any departure from perfectly ordered arrangement of constituent

particles in crystal is called imperfection or defect.

We are having two type of defects one is point defect another is line defect we

are discussing point defect.

Point defect:- The defects in which deviations or irregularities exist from the ideal arrangement around a point in a crystalline substance.

Stoichiometric defect:- In this types of defect, the crystal ratio between the cations and anions remains the same as represented by the molecular formula. Or the stoichiometry or formula of the solid is not disturbed the defects are called stoichiometric defects or intrinsic or thermodynamic defects.Basically these are of two types, vacancy defects and interstitial defects.


(i) Vacancy Defects:- When some of lattice sites are vacant in crystal are called vacancy defect. This defect can also develop when a substance is heated.

Effect on density:- this defect decrease the density of the substance.

(ii) Interstitial Defect:- When some constituent particles occupy an interstitial site is called interstitial defects.

Effect on density:- This defect increases the density of the substance.

Both the above defects shown by the non-ionic solids.for ionic solid they show they two type of defects which are following

(i) Schottky defects (ii) Frenkel defects Schottky defects:- In this type of defect equal number of cations and

anions are missing from their lattice sites so that electrical neutrality is maintained is called schottky defect. For example NaCl, KCl, KBr, CsCl etc.


Conditions causing schottky defects:-This defect is usually produced in strongly ionic compounds having High coordination number. Small difference in the size of cation and anion.

Effect on density:- The number of ions in an ionic crystal having schottky defect decrease, the density of the crystal also decreases.

Frenkel defects:- In this type of defect, when an ion is missing from its lattice site and occupies the vacant interstitial site existing between the cations and anions are called frenkel defect. For example Zns, AgCl, AgBr and AgI etc.

Conditions causing Frenkel defects:-This defect is generally produced in compounds having Low coordination number Large difference in the size of cations and anions.

Effect on density:- The number of ions in an ionic crystal having frenkel defect remains the same, there is no change in density.

NOTE: - AgBr is having both type of defects.


Consequences of schottky and frenkel defects: - Solids having these defects conducts electricity to a small extent. Due to the holes, the stability of the crystal decreases. In a crystal having schottky defect, the similar ions come closer to each

other and hence dielectric constant of crystal increases. Non-stochiometric Defects :- The defects which disturb the

stoichiometry of the compound are called non-stochiometric defects.These are two types :- (i) Metal excess defect (ii) Metal deficiency defect

Metal excess defect:- When one or more anion are missing from their positions and anion vacancies are produced.in the crystal metal ions are in excess amount that is why it is called metal excess defect.(a) Metal execess defect due to anions vacancies :

Alkali metal halides show this type of defect such as KCl and NaCl if crystals of NaCl are heated in Na vapours. Some sodium atoms

are deposited on the surface of the NaCl crystal. The Cl- ions diffuse to the surface of the crystal and combine with Na atoms to form NaCl. Na atoms lose electrons which diffuse into the crystal and occupy anionic vacancies so Na+ becomes more in crystal they impart yellow colour to the NaCl crystal because the unpaired electron absorbs energy from visible spectrum. In the same way LiCl crystals are sometimes pink and KCl becomes violet.

F-Centre:- The electron trapped in the anion vacancies are called F-Centre. They are responsible for colour to crystals.

This defect is found in those crystals which have schottky defects. The presence of F-Centres in an ionic crystal modifies the

properties of the crystals. These are following.(i) Colour:- Greater is the number of F- Centres greater is the

intensity of Colouration.(ii) Crystals having F-Centres are paramagnetic.(iii)When the crystals having F-center are irradiated with light, they

became photoconductor.


(b) Metal excess defect due to cation vacancies :

Sometimes if anions vacancies become less, metal ions

become extra which create metal excess defect. For example, on heating Zno, it gives oxygen.

Zno ∆ Zn2+ + 1/2O2 + 2e-

Zn2+ ions becomes extra and create metal excess defect.

Metal deficiency Defect:- This defect arise when metal show variable valency.

The defect arises due to the missing of cation from its lattice site and the presence of the cation having higher charge in the adjacent lattice site.

For example in the crystal of Feo, some Fe2+ ions are missing and loss of positive charge is balanced by the presence of required number of Fe3+ ions.


Questions:-

Problems based on this LectureQue.1. What is the meaning of the term imperfection in solids?

Ans:- Imperfection mean the departure from the perfect periodic arrangement of atoms, ions or molecules in the structure of crystalline substances.

Que.2. What is the type of the term imperfection in solids?

Ans:- (a) Stoichiometric defect (schottky defect and Frenkel defect) and (b) Non-stoichiometric defect (Metal excess defect, metal deficiency defect and impurity defects).

Que.3. What type of defect can arise when a solid heated? Which physical property is affected by it and in what way ?

Ans:- On heating a solid vacancy defect is produced in the crystal. This is because on heating. Some lattice sites become vacant. As a result of this defect the density of the substance decreases because some atoms or ions leave the crystal completely.

Que.4. CaCl2 will introduce schottky defect if added to AgCl crystal. Explain.


Ans:- Two Ag+ ions will be replaced by one Ca2+ ions to preserve electrical neutrality. Thus a hole is created at the lattice site for every Ca2+ ion introduced.

Que.5. Why is Frenkel defect not found in pure alkali metal halids?

Ans:- This is because of the fact that alkali metal ions have large size which can not fit into interstitial sites.

Que.6. Name the non-stoichiometric point defect responsible for colour in alkali halides.

Ans:- Excess of metal ions and formation of F-centres as a result of trapped electrons.

Que.7. What are F-centres?

Ans:- The free electrons trapped in the anions vacancies are called as F-centres.

Que.8. Name one solid in which both Frenkel and Schottky defects occur.

Ans:- AgBr

Que.9. Why does Frenkel defect not change the density of AgCl crystals?

Ans:- Because of the Frenkel defect no ions are missing from the crystal as a whole therefore there is no change in density.

Que.10. What is Schottky defect ?

Ans:- If equal number of cations and anions are missing from their lattice sites the defect is known as Schottky defect.

Que.11. What are non-stoichiometric compounds?

Ans:- If the actual ratio of the cations and anions is different as represented by the ideal chemical formula of the compound it is called a non-stoichiometric compound. Ex Fe0.9O, Cu1.5O

Que.12. Why is Frenkel defect found in AgCl?


Ans:- It is because of the reason that cation and anion differ in their size to more extent therefore cations occupy voids due to smaller size.

Que.13. What is the non-stoichiometric defect in crystals?

Ans:- A result of the imperfection in the crystal the ratio of the cation to the anions become different from that indicated by the ideal chemical formula then the defects are termed non-stoichiometric defects.

Que.14. Which point defect lower the density of crystal?

Ans:- Schottky defect.

Qus.15. What is Frenkel defect?

Ans:- When some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometry is maintained it is called Frenkel defect.

Que.16. What type of stoichiometric defect is shown by : (i) ZnS (ii) AgBr?

Ans:- (i) ZnS shows frenkel defects.

(ii) AgBr shows both frenkel and Schottky defects.

Que.17. Why does frenkel defect not change the density of AgCl crystal?

Ans:- This is because in frenkel defect no ion is missing from the crystal as a whole so the density remains unchanged.

Que.18. Which point defect of its crystal decreases the density of a solids?

Ans:- Schottky defct.

Que.19. Why are the solids containing F-centres paramagnetic?

Ans:- This is because the electrons occupying the vacant sites are ubnpaired.

Que.20. What are (a) vacancy defect (b) Interstitial defect? What type of compounds show these defect?


Ans:- (a) Vacancy defect :- When some lattice sites in a crystalline solid are vacant then the crystals suffer from vacancy defect. This defect can be created upon heating. As a result the density of the solid decrease.

(b) Interstitial defect:- When some constituent particles in a particular solid occupy interstitial sites this defect is created . as a result the density of the solid increases. Both these defect are shown by non- ionic solids.

Que.21. Why does white zinc oxide (ZnO) become yellow upon heating?

Ans:- ZnO on heating loses oxygen as the ions are trapped in interstitial site and electrons in the other. As a result the crystal appears yellow.

Lecture-6

Electrical Properties On the bases of electrical properties the solids divide into three parts.

(1) Conductors :- The solid which having conductivities range from 104 to 107

ohm-1 m-1.(2) Insulators :- The solid which having conductivities range from 10-20 to 10-10

ohm-1 m-1. (3) Semiconductors :- The solid which having conductivities range in between

10-6 to 104 ohm-1 m-1.


Causes of conductance in solids:-

Most of the solids conduction takes place due to migration of electrons under the influence of electric field.

In ionic compounds the ions are responsible for the conduction. In metals, conducting strongly depends upon the number of valence

electrons available in atom.

Explanation of behavior of conduction in the solids on the basis of band theory

Band :- When molecular orbitals closed to other molecular orbitals to formed band.

Impurity defect :- These defect arise when foreign atoms or ions are present in the lattice site.


Doping :- The addition of a foreign substance as impurity in semiconductors to increases the conductivity of the substance is called doping.

(1) Doping with electron rich impurities :- When the silicon or germanium(14- group) crystals doped with electron rich(15 group) impurities are called n-type semiconductors.

(2) Doping with electron deficit impurities :- When the silicon or germanium(14- group) crystals doped with electron deficit(13 group) impurities are called p-type semiconductors.

Group-13 Group-14 Group-15 B p-type C n-type N(gas) Al Si P Ga Ge As In Sn Sb Th Pb Bi

Application of n-type and p-type :-

(1) For making electronic components(2) For making npn and pnp type transistors.


Magnetic Properties The magnetic properties of the substance depend on electrons. Every electron revolves around the nucleus and also rotates around its own axis and thus every electrons behaviours like a magnet. On the basis of magnetic properties the substance are classified into following types.

(1) Diamagnetic :- The substance which are weakly repelled by magnetic field are called diamagnetic substances. They are weakly magnetized in opposite direction.Diamagnetism is shown by those substances which do not have unpaired electrons. For example NaCl , C6H6, H2O etc.

(2) Para-magnetism :- The substance which are weakly attracted by magnetic field are called paramagnetic substances. They are magnetized in a magnetic field in the same direction.Para-magnetism is shown by those substance which have one or more unpaired electrons. For example O2 , Cu2+ , Fe3+ .

(3) Ferromagnetic substance :- The substance which show magnetism even in the absence of the magnetic field are called ferromagnetic substances. For example Fe , Ni , Co etc.

(4) Anti-ferromagnetic substance :- The substance in which half of the magnetic moment vectors are aligned in one direction while the remaining half magnetic moment vectors are aligned in the opposite direction are called anti-ferromagnetic substance. For example V2O3 , Cr2O3 , MnO.

(5) Ferrimagnetic substance :- The substance in which the no of magnetic moment vectors aligned in parallel and antiparallel directions is unequal and hence such substance has some value of its resultant magnetic moment are called ferromagnetic substance. For example Mg2+ , Cu2+ , Zn2+


Questions:-

Problems based on this LectureQue.1. What is piezoelectricity? What type of crystals exhibit piezoelectricity ?

Ans :- The crystals, which on applying a mechanical stress produce electricity, are called piezoelectric crystals and the phenomenon is called piezoelectricity. Polar crystals in which the dipoles align themselves in an ordered manner under the influence of an electric field exhibit piezoelectricity.

Que.2. Define the ‘forbidden zone’ of an insulator.

Ans :- The energy gap between valence band and the conduction band in an insulator is called forbidden zone.

Que.3. Define superconductivity of a substance. Who discovered it ?

Ans :- A substance when it offers no resistance to flow of electricity is called superconductor and this property is called superconductivity. Most metals become superconducting at very low temperature(2-5K). Kammerling Onnes discovered it.

Que.4. Define the photovoltaic compound.

Ans :- A compound which generates current when exposed to light is called photovoltaic compound.


Que.5. A group 14 elements is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belongs?

Ans:- n-type semiconductor means conduction due to presence of excess of electrons therefore to convrt group 14 elements into n-type semiconductor it should be doped with group 15 elements.

Que.6. Classify each of the following as either a p-type or n-type semiconductor:

(a) Ge doped with In (b) B doped with Si

Ans:- (a) Ge is Group 14 element and In is Group 13 element. Thus an electron deficit hole is created and therefore it is p-type.

(b) B is Group 13 elements and Si is Group 14 element there will be a free electron. Thus it is n-type.

Que.7. What is doping? Why is it done?

Ans:- It is a process of adding impurities in a crystal lattice. Doping is done by adding calculated amount of impurities. It increases the conductivity.

Que.8. What is the difference in the semi-conductors obtained by doping silicon with Al or with P ?

Ans:- Silicon doped with Al forms p-type semiconductors, i.e. flow of current is because of creation of positive holes while silicon doped with P produces n-type semiconductors i.e. flow of current is due to extra electrons having negative charge.

Que.9. What type of crystal defect is produced when sodium chloride is doped with MgCl2?

Ans:- It is called impurity defect. A cation vacancy is produced. A substitutional solids solution is formed( because 2 Na+ ions are replaced by one Sr2+ ion the lattice site).


Que.10. How does the electrical conductivity of metallic conductors vary with temperature?

Ans:- Electrical conductivity decreases with rise in temperature because kernels began to vibrate and create hindrance in the flow of electrons.

Que.11. What is meant by ‘doping’ in a semiconductor?

Ans:- Doping means incorporating small amount of foreign impurity in crystal impurity in crystals. Doping of group 14 elements with 15 elements give rise to excess electron (n-type semiconductors) whereas with group 13 elements give rise to holes (p-type semiconductors).

Que.12. Account for the following :

(i) Phosphorus doped with silicon is a semiconductor.

(ii) Some of the glass objects recovered from ancient monuments look milky instead of being transparent.

(iii) Schottky defect lower the density of a solid.

Ans:- (i)Phosphorus doped with silicon has holes and hence acts as n-type semiconductor.

(ii) Glass objects recover from ancient monuments look milky because of some crystallization take place in them.

(iii) This is because some of the cations and anions are missing from the crystal lattice.

Que.13. How does the electrical resistivity of (i) a semiconductor (ii) metallic conductors vary with temperature ?

Ans:- Electrical resistivity of metallic conductors inceases with temperature while that of semiconductors and super conductors decreases with temperature.


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