Solid State
Solid State
Lecture-1
General Characteristics of Solids Crystalline and Amorphous Solids Classification of Crystalline Solids Based on Interparticle Binding Forces
Lecture-2
Crystal Lattices and Unit Cells Calculation of Number of Particles in a Unit Cell
Lecture-3
Close Packed Structures, Packing Efficiency Calculations Involving Unit Cell Dimensions
Lecture-4
Interstitial Voids
Lecture-5
Imperfections in Solids
Lecture-6
Electrical Properties of Solids Magnetic Properties of Solids
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100 Hours 100Marks
Believe where others doubt . Work where others refuse. Save where others waste. Stay where others quit and you will win where others lose.
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Solid State Polymers
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Solid State Lecture-1
1.
General characteristics of states (1) Gas State (2) Liquid State (3) Solid State Gas 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 postions and can only oscillate about their mean positions. (iii) Solids are almost incompressible. 3
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(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 solids
1. Shape
The crystalline solids have definite shape.
2. Order in arrangement of constituent particles 3. Melting Point
They have regular arrangement of constituent particles. They are said to exhibt long range order. They have sharp and characteristic melting point.
4. Cleavage Property
7. Nature
When cut with a sharp edged tool,they split into two pieces and the newly generated surfaces are plain and smooth. They have a definite and characteristic enthalpy of fusion They are anisotropic and have different physical properties in different directions. They are true solids.
The amorphous solids have characteristic shape irregular shapes. They do not have any regular arrangement of the constituent particles. They may have short range order. They do not have sharp m.p They gradually soften over a range of temperature. When cut with a sharp edged tool they cut into two pieces with irregular surfaces.
8. Common
Copper, silver, iron, common
5. Enthalphy 6. Anisotropy
examples
4
salt, zincsulphide potassium nitrate etc.
They do not have definite enthalpy of fusion. They are isotropic and have same physical properties in all directions They are isopseudo solids and supercooled liquids. Glass, rubber, plastics, etc.
Solid State
Depending upon the nature of intermolecular forces operating in them Note Isotropic in nature means their electrical and mechnical properties are same in all the directions. Anisotropic in nature means, their electrical and mechanical properties are different in different directions. Now we come to classification of crystalline solids. 1. √ √ √ √
Type of solid : Ionic: NaCl, MgO, KCl, CsCl, ZnS, Na2SO4, KNO3, CuSO4.5H2O. Covalent or network: Diamond(C), Graphite(C), Carborundum(SiC), Quartz(SiO2), Boron-nitride(BN). Molecular: Ice, Solid CO2, Sulphur (S8).Glucose, Napthalene etc. Metallic: s-block and d-block metals and some p-block elements, Na, CU, Ag, Au, Ni, Pt, Cr, etc.
Structure of quartz (crystalline) and quartz glass (amorphous) Quartz is a form of SiO2 (silica). It has tetrahedral SiO4– – units which are orderly arranged in crystalline quartz as shown in figure. When SiO2 is melted and the melt is cooled, it forms quartz glass which is amorphous. In this state, the SiO4 units are randomly joinded.
(a) crystal Quartz
(b) Amorphous Glass 5
6
6. Electrical Conductivity
5. Melting point (K)
4. Hardness
3.
2.
1. Constituent particles
Type of solid Characteristics
High -1500
400 - 4000
Positive and negative ions
Ionic Molecules (polar or non-polar)
Neutral atoms
Very high 4000
150 - 500
Molecular
Covalent or Network Metallic
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The characteristics of these solids summarised in the following
Solid State
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Questions Problems based on this Lecture Q.1 Why do solids have a definite volume ? Ans. The constituent particles in solids are bound to their mean positions by strong cohesive forces of attraction. The interparticle distances remain uncharged at a given temperature and thus solids have a definite volume. Q.2 Why are solids rigid ? Ans. In solids, the constituent atoms or molecules or ions are not free to move but can only oscillate about their mean positions due to strong interatomic or intermolecular or interionic forces. Q.3 What makes a glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ? Ans. Glass is an amorphous solid in which the constituent particles (SiO 4– – tetrahedral) have only a short range order and there is no long range order. In quartz the constituent particles (SiO4– – tetrahedral) have both short range as well as long range orders. On melting quartz and then cooling it rapidly is converted into glass. 7
100 Hours 100Marks Q.4 Why does the window glass of the old building look milky ? Ans. It is due to heating during the day and cooling at night, i.e., annealing a slow cooling process due to this after some years. Glass acquires some crystalline character and these character look milky. Q.5 How can a material be made amorphous ? Ans. First melting the material and then cooling it rapidly from the liquid state. Q.6 How does amorphous silica differ from quartz ? Ans. In amorphous silica SiO4– – tetrahedral are randomly joined to each other while in quartz they are joined in a regular manner. Q.7 Ionic solids conduct electricity in the molten state but not in the solid state. Explain. Ans. In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However, in the solids state, since the ions are not free to move but remain held together by strong electrostatic forces of attraction. Therefore, they are electrically insulator. Q.8 (a) ‘Stability of a crystal is reflected in the magnitude of its melting points’.Comments. (b) The melting points of some compounds are given below : Water = 273K, Ethyl alcohol = 155.7K, Diethyl ether = 156.8K, Methane = 90.5K. What can you say about the intermolecular forces between these molecules ? Ans. (a) Higher the melting point, stronger are the forces holding the constituent particles together and hence greater is the stability. (b) The intermolecular forces in water and ethyl alcohol are mainly the hydrogen bonding. Higher melting point of water as compared to alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are dipole-dipole attraction.Methane is a non polar molecule. The only forces present in them are the weak van der Waals’ forces. Q.9 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 an amorphous solid, it would not show a clean cleavege when cut with a knife. Instead, it would break into pieces with irregular surfaces. Q.10 Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic, acid, teflon, potassium nitrate, cellophone, polyvinyl chloride, fibre glass, copper. Ans. A m o r p h o u s s o l i d s : Po l y u re t h a n e , t e f l o n , c e l l o p h o n e , p o l y v i n y l chloride and fiber glass 8
Solid State Q.11 Ans. Q.12
Ans.
Q.13 Ans. Q.14 Ans. Q.16 Ans. Q.17 Ans. Q.18 Ans. Q.19 Ans.
Crystalline solids : Naphthalene, benzoic acid, potassium nitrate and copper. What type of solids are electrical conductors, malleable and ductile ? Metallic solids 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, rubidium, argon, silicon carbide. Ionic solids : Potassium sulphate, zinc sulphate. Covalent solids : Graphite, silicon carbide Molecular solids : Benzene, urea, ammonia, water, argon Metallic solids : Rubidium, tin. Solids A is a very hard electrical insulator in solid as well as molten state and melts at an extremely high temperature. What type of solid is it ? Covalent. Urea has a sharp melting point but glass does not. Explain. Urea is a crystalline solid, so it has a sharp melting point. On the other hand, glass is a molecular solid, so its melting point is not sharp. Why is glass considered a super-cooled liquid ? Glass has characteristics similar to liquids. It can flow extremely slowly, so it is considered to be a super-cooled liquid. 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 ? It is a covalent or network solid. Write a distinguishing feature of metallic solids ? Electrical conduction in solid state. How do metallic and ionic substance differ in conducting electricity ? Metallic solids 1. Conduct electricity both in solid as well as in molten state. 2. Conductivity is due to the migration of electrons.
1.
2.
Ionic solids Conduct electricity only in molten state or when dissolved in some polar solvent like water. Conductivity is due to the migration of cations and anions.
Q.20 (a) Define liquid crystals. (b) What are Semectic and Nematic liquid 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 substances are known as liquid crystals. (b) Semectic liquid crystals are having soap like phase and Nematic crystals are those which are having needle or thread-like phase.
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Lecture-2 Characteristics of a Crystal Lattice Let us sum up the charateristics of a cystal lattice. The following are the characteristics of a crystal lattices. (i) Each point in a crystal lattice is called lattices points or lattices site. (ii) Each point in a crystal lattices represents one constituent particles which may be an atom, a molecule or an ion. (iii) Lattics 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. These are : (i) Primitive unit cells (ii) Centred unit cells (i) Primitive unit cells: These are unit cells which have points (or particles) only at the corners. These are also called simple unit cells. (ii) Centred unit cells: These are unit cells which have points (or particles) at the corners as well as at some other positions. (a) Face centred unit cells in which the points are present at the corners as well as the centre of each face. (b) Body centred unit cells in which the points are present at all the corners as well as the body centre of the unit cell. (c) End centred unit cells in which the points are present at all the corners and at the centre of any two opposite faces. 10
Solid State
Space lattice or Lattice Points:— Now we will find about details of crystal solid. 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. Example – In simple cubic total numbers of lattice point are 8. Total no. of atoms are required For simple cubic = 8 For triangle arrangement no. of Lattice points three.
Space lattice = Triangle Lattice points = 3
Lattice points = 8 Space lattice = cubic
Unit cell : It is the smallest unit which on repeatation gives the required crystal lattice and gives almost all the information about the solid. Example – In two dimensional:
Square closed structure Unit cell is square
Hexagonal closed structure unit cell is hexagon
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In three dimensional: If cubic unit cell is there then we have as , , are axial angles. 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. and between the angle pair of edges,the (ii) Angles , angle is between the edges ‘b’ and ‘c’, angle is between the edges ‘c’ and ‘a’ and angle is between the edges ‘d’ and ‘b’. 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. Example– Cubic, Tetragonal, orthorhombic, hexagonal rhombohedral or trigonal, monoclinic, tirclinic. Seven Primitive Unit Cells and their Possible arrangement.
Variation as Centred Unit Cells Crystal system
Possible variations
Axial distances
Axial angles
Examples
or edge le ngths lengths Cubic
Primitive, Body
a=b=c
centred, Face-centred Tetragonal
Primitive,
blende, Cu a=b c
Body-centred, Orthorhombic
Primitive,Body-
NaCl, Zinc White tin, SnO2, TiO2, CaSo4
a b c
centred, Face-cent
Rhombic sulphur, KNO3, BaSO4
-red, End-centred Hexagonal
Primitive
a=b c
Graphite, ZnO, CdS
Rhombohedral
Primitive
a=b=c
HgS (Cinnabar)
or Trigonal Monoclinic
Calcite (CaCO3),
Primitive,
a b c
End-cented
Monoclinic sulphur, Na2SO 4, 10H 2O
Triclinic
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Primitive
a b c
K2Cr 2O 7 ,CuSO 4
Solid State
All crystals do not have simple lattices. There can be 14 different ways in which similar points can be arranged in a three dimensional space. Type of cell
No. of atoms at corners ingle or primitive cubic 8×(1/8) = 1 Single Body-centred cubic (bcc) 8×(1/8) = 1
Number of atoms in faces 0 0
Face-centred cubic (fcc) 8×(1/8) = 1
6×(1/2) = 3
No. of faces in the body of cube 0 1
0
Total 1
2 4
Bravais Lattices : On these crystal system, arrangement of the atom or ions which gives space lattices known as Bravais lattices. We are having 14 types of bravais lattice Example– In case of cubic system we are having three types of arrangements. (1) Simple Cubic (SC)
Total lattice point = 8 (2) Body Centred Cubic (BCC)
Total lattice points = 9 Eight at corners and one at centre. (3) Face centred Cubic (FCC)
Total lattice points = 14 13
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Eight at corners and six at face. So, in this we are having cube as crystal system and those arrangement of atoms as Bravais lattice. So, it is one crystal system is cubic and there bravais lattices s.c., b.c.c. and f.c.c. Calculation of number of particles in a Unit cell: Lattice point in the arrangement (Cubic) SC — 8 BCC — 9 FCC — 14 HCP — 17 In this we will calculate sharing of atoms, ions, molecules with the unit cell and accrording to the positions of the given atom, ion, molecule.
CUbic System
At corner, sharing of atom =
1 8
1 2 At centre, sharing of atom = 1 1 At edge, sharing of atom = 4
At face, sharing of atom =
14
Solid State
In case of hexagonal 1 1 At corner = At face = 6 2
At centre = 3
Symmetry in the solid Solids are having three types of symmetries. (1) Plane of symmetry (2) Axis of symmetry (3) Centre of symmetry Example :—
Plan of symmetry = 9 In cubic Axis of symmetry = 13 Centre of symmetry = 1
Formula of unit cell : When we combine the shared atoms, ions in the unit cell then formula is developped. Ex.-1 (i) If in the cubic system A at corners B at all the faces. Then, find out formula of compound. (ii) If one A is remove from one corner means, one corner is empty only seven corners are occupied by atoms. (iii) If one B is remove from face 1 Sol. (i) Each corner is having sharing= 8 1 Each face is having sharing= 2 A ‘A’ at a corner so each corner is having ‘A’ as = 8 B ‘B’ at a face so each face is having ‘B’ as = 6 15
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A 8 Only one ‘A’ is used in the unit cell Total face = 6 B =3B So total ‘B’ used = 6× 2 So in this case in the unit cell 1A and 3B Therefore, the formula is AB3 . (ii) If one A is remove from one corner means, one corner is empty only seven corners are occupied by atoms A 7A = So total ‘A’ used = 7× 8 8 A ( at each corner) 8 B =3B total ‘B’ used = 6× 2 B ( at each face) 2 7A : 3B or 7 A : 24B So formula 8 Therefore, formula is A7B24 . (iii) If one B is removed then from ‘six’ face one is empty and ‘five’ are occupied. B 5B 5 So, total ‘B’ used = 2 2 A =A total ‘A’ used = 8× 8 5B or 2A : 5 B So formula A : 2 Therefore, formula is A2B5.
So total ‘A’ used=
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Solid State
Ex.2 If in the above case A at corner B at face C at centre and D at edge, then find out formula of unit cell if following conditions are given. (a) If one A from one corner is removed. (b) If one B from one face is removed. (c) If one D from the edge is removed. (d) A, B, C are each are removed. 1 Sol. At corner sharing of atom = 8 1 At face sharing of atom= 2 1 At edge sharing of atom = 4 A B D 8, 6, C 1, 12 AB 3 CD 3 If no condition : 8 2 4 (a) If one A is removed from one corner then total occupied corners are 7 which are having ‘A’ Total ‘A’ = No. of corner occupied × Sharing of one atom at corner A Total ‘A’ used = 7 × 8 B B 3B B at face= 6 (6 faces with at each face) 2 2 C at centre = 1 × C = C(one centre is there) D D 3D (12-edge with at each edge) D at edge= 12 4 4 7A , 3B, C, 3D Formula is 8 So simple formula is 7A 24B 8C 24D A7B24C8D24
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(b) If one B is removed then total face occupied are = 5 A Total A used =8 A 8 Total B used =5
B 2
5B 2
Total C used = 1 × C = C D =12 3D Total D used 4 5B , C, 3D Therefore, A, and Simple formula is A2B5C2D6 . 2 (c) If one D is removed A Total A =8 A 8 B 3B Total B= 6 2 Total C = 1 × C = C D 11D Total D 11 4 4 11D Therefore, A, 3B, C, 4 and Simple formula is A4B12C4D11 (d) If one each removed A 7A Total A 7 8 8 B 5B Total B 5 2 2 Total C=0 × C=0 (only one C at centre which is removed) D 11D Total D 11 4 4 7A 5B 11D Therefore, 8 2 4 and Simple formula is A7B20D22
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Solid State
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Questions Problems based on this Lecture Q.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 8 Q.2
1 1 (from corners) + 6 (from face corners) = 4. 2 8
Explain how much portion of an atom located at (i) corner and (ii) bodycentreof a cubic unit cell is part of its neighbouring unit cell. Ans. (i) A point 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 centred point belongs entirely to one unit cell since it is not shared by any other unit cell. Q.3 Define face-centred cubic structure. Ans. There is a structural particle at the centre of each face as well as each corner.
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100 Hours 100Marks Q.4 What is body-centred cubic ? Ans. Body centred 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. Q.5
Calculate the number of atoms in a cubic based unit cell having one atom on each corner and two atoms on each body diagonal. Ans. There are four body diagonals passing through the center and connecting opposite corners. The atoms present on these diagonals are not shared by other unit cells. Contribution by atoms present on diagonals= 2×4 = 8 1 1 Contribution by atoms present at corners 8(at corner) 8 Total number of atom in the unit cell = 1 + 8 = 9 Q.6
How many lattice points are there in one unit cell of each of the following lattices? (a) face centred cubic (b) face centred tetragonal (c) body centred cubic Ans. (a) In face centred cubic arrangement, Lattice points located at the corners of the cube = 8 Lattice point locked at the centred of each face = 6 Total no. of lattice point = 8 + 6 = 14 (b) In face centred tetragonal, the number of lattice point is also the same, 8+6 = 14 (c) In body centred cubic arrangement, Lattice point located in the center of the body = 1 Total no. of lattice points 8 + 1 = 9 Q.7
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 = 1×1 = 1 1 1 8 1 3 O atoms per unit cell 12 4 Simple formula is NaWO3.
W atoms per unit cell
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Solid State
COORDINATION NUMBER (CN)
Coordination number is defined as number of atoms or ions surrounded by the given atom or ion.
Example :
X
for X CN is 4.
in three dimension coordination number is as; HCP = 12 for BCC, CN = 8 FCC (salt like) = 6 CCP = 12 , SC = 6
Questions Problems based on this Lecture Q.1
What is the 2-dimensional coordination number of a molecule in square close packed layer ? Ans. Four Q.2 (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) 8. Q.3 In a body-centred 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.
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Lecture-3 Relation between radius of atom, ion and edge length. Density, Packing fraction, Formula of unit cell. (i) For Simple Cubic Unit Cell(SCC) :
In this figure we are having as. r
a
If edge length of unit cell is ‘a’ and the radius of the atom is r. Then from the figure. a r= (a) Radius r + r = a 2 (b) Total no. of atoms in the unit cell at each corner 1 sharing. one atom is there with 8 So, total atoms = No. of corner × Part of sharing 1 8× = 1 8 One atom is shared in the unit cell. (c) If A and B atoms are palced at lattice points in the unit cell alternativily
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Solid State A B
B A A
B A
B
1
Sharing of A = 8 Total lattice points of A = 4 A A So, 8 ×4= Total sharing of A = 2 1 Sharing of B = 8 Total lattice points of B = 4 B ×4= Total sharing of B = So, B 8 2 A B Formula, 2 : 2 AB
(d) Density : Mass of unit cell Mass Volume a3 where a3 volume of unit cell Mass of unit cell is due to atoms in the unit cell. Total atoms in the unit cell=1 6.023×1023 atoms having mass = atomic weight d
Mass of one atom
Atomic weight
6.023 10 23 Atomic weight Atomic weight
So, density
6.023 10 23 a3
No a3
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(e) Packing Fraction : Volume of 1 atom
Packing fraction
Total vol. of cube
r3
4 3
a
3
52%
a from Ist 2
Where r
(ii) For face centred Cubic (FCC) : In this unit cell one atom is on the face of the cubic unit. This face centred atom touches each corner atom on one face. Analysis of face (a) Radius
So,
PQ 2
PR 2
PQ
a2
r
RQ 2 a2
2a
2a
r 2r r
a 2 4 P
S
R
Q
(b) Total number of atoms shared 1 1 at each corner so total sharing ×8=1 8 8 1 1 at each face so total sharing ×6=3 2 2 So, total atoms involved in the unit cell=1+3=4 24
Solid State
(c) If A at corner and B at face then formula total A used in unit cell total B used in unit cell A 3B, So AB3 is formula (d) Density of the unit cell Mass of unit cell d Volume of unit cell Mass is due to four atom
Atomic weight
4
No
d
a3 (e) Packing fraction of the unit cell Volume packed d Total Vol. of unit cell Volume is packed is due to four atoms. Which are involved in the unit cell. Volume of one atom Total volume packed by the atom Total volume of unit cell = a3 So, Packing fraction
4
4 3
a 2 2
a3 So, empty space is 26%.
3
74%
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(iii) In case of Body Centred Cubic (BCC) : In BCC one atom is placed at the centre of the unit cell, touching all the corner atoms. So, in this case digonal atoms are having touch with each other
a 3
P (a) Radius
Body centre
a Q
a 2
PQ
r 2r r
a 3
r
a 3 4
(b) Total number of atoms 1 at centre So, Total atoms = 1 + 1 = 2 Formula ‘A’ at corner and ‘B’ at centre A B ×8 : ×1 8 A:B or AB (c) Density
At. Wt. d
Mass Volume
(d) Packing fraction
26
Mass of 2 atoms
No V a3 Volume packed due to atoms Total volume of unit cell
2
Solid State
2
Volume due to 2 atoms a3
a 3 4
where r 2 Hence,
4 3
a 3 4
4 3 r 3 a3
68%
a3
(iv) For hexagonal close packing (hcp) (a) Suppose radius of each sphere = r From the figure , it is clear, Volume of unit cell = Base area × Height (h) Bass area of regular hexagon 6 Height of unit cell 4r.
3 (2r)2 4
6 3r 2
2 3
(b) Volume, Volume of unit cell
2 24 2r 3 3 Volume of atoms in hcp unit cell 6 3r 2
4r
1 (face center) 3(body) 6 2 4 3 Volume of 6 sphere 6 r 8 r3 3
12
1 (corner) 2 6
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Volume occupied in hcp arrangement = 74% Hexagonal Closed Packing(h.c.p.) Main points in this are : (a) It AB AB AB types (b) Packing fraction is 74%. So empty space is 26%. (c) Coordination number is 12. (d) Total atoms used in this = 6 3 (e) Volume of h.c.p. = 24 2r
Cubic Closed Packing(c.c.p.) Main points in this are : (a) ABC ABC ABC type. (b) Packing fraction is 74%. So empty space is 26%. (c) Coordination number is 12. (d) CCP is same as that of FCC for the calculation (e) Volume of c.c.p. = 16 2r 3 Note : Volume of h.c.p. is more than CCP. Note: In case of f.c.c. salt like Four formula in unit cell 28
Solid State
4 d
M No
a3 Ex. NaCl, Cl at corner and Faces, Na+ at edge centre and at centre. Edge length a = r – + 2r+ + r – = 2(r++r –) where r – = for Cl– and r + = for Na+ Cl—
Cl— Na+
In case of b.c.c. One formula in unit cell M 1 No d a3 Ex. CsCl, Cs+ at centre, Cl– at corner, analysis is as, r 2(r
2r
r
a 3
r ) a 3
Cl— Cs+ —
Cl
a a 2
Note:— c.c.p. is equivalent to f.c.c. for calculation. 29
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Questions Problems based on this Lecture Q.1
Which of the following lattices has the highest packing efficiency ? (i) Simple cubic (ii) Body-centred cubic (iii) Hexagonal close-packed lattice Ans. Hexagonal close-packed lattice has the highest packing efficiency (74%) Q.2 A metal crystallizes into two cubic phases, face-centred cubic (FCC) and body-centred cubic (BCC) whose unit cell lengths are 3.5Å and 3.0Å respectively. Calculate the ratio of the densities of FCC and BCC. Ans.
30
z M NA a3 4 M Density of f.c.c. unit cell N A (3.5 )3 Density of unit cell(d)
Solid State
2 M N A (3.0Å)3
Density of b.c.c. unit cell
Density of f.c.c. 2 M Density of b.c.c. N A (3.5Å)3 108 1.259 85.75
NA
(3.0Å)3 2 M
Q.3. An element has body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2. How many atoms are present in 208 g of the element? Ans. For the bcc structure, No. of atoms per unit cell, z = 2 Edge length of the unit cell, a = 288 pm = 288×10–12 m = 288×10–10 cm Density of the element, d = 7.2 g cm–3
We know that or
z M NA a3
d M
a3 NA z 7.2g cm
d
Substituting the values M or
M
(288
cm) 2
6.02 10
mol
51.8g mol
Now, 51.8g of the element contains
6.02 1023 atoms
6.02 1023 208 atoms 208 g of the element would be contain 51.8 23 24.16 10 atoms Q.4 Silver forms ccp lattice. X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (At. mass = 107.9- u). Ans. Given that, a = 408.6 pm = 408.6 ×10–10 cm In fcc, no. of atoms in unit cell, z = 4
We know that
d
z M N A a3
4 107.9 g mol (408 10 10 cm)3(6.022 1023 mol 1) 10.50 g cm Q.5
3
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. Desitys of the unit cell
Mass of the cell Volume of the unit cell 31
100 Hours 100Marks Edge of the cubic crystal in the unit cell = a Volume of the unit cell = a3 Mass of the unit cell = No. of atoms in the unit cell (z) × mass of each atom
Atomic mass Avogardro number z M Density of unit cell (d) a 3 NA Mass of each atom
Now, M
d
M NA
a3 NA z
Q.6
Aluminium crystallises in a cubic close-paked structure. Its metallic radius is 125 pm. (a) What is the length of the side of the unit cell ? (b) How many unit cells are there in 1.00 of aluminium ? Ans. (a) r = 125 pm a 2 =r + 2r + r = 4r a For cubic close paked structure, r 2 2 (b)
a r 2 2 125nm 2 1.414 354pm Volume of unit cell, a3 = (354×10–12cm) = 44.36×10–30 cm3 Now, 44.36×10–30 cm3 contains =1 unit cell
1 44.36 10 30 2.26 1022 unit Q.7 Silver crystallises in face centred unit cell. Each side of a unit cell has a length of 409 pm. What is the radius of an atom of silver ? (Assume that each face centred atom is touching the four corner atoms). a Ans. For fcc unit cell, r 2 2 (r+ 2r + r = a 2 ) a 409pm r 144.6 pm a 2 1.414 1 cm3 contains
Q.8
The well known mineral florite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are ions and ions an that ions are arranged in a lattice. The edge length is 5.4Å and the density of the solid is 3.2 g/cm3. Use this information to calculate Avogadro’s number (Molar mass of ) Ans. Since in a unit cell, there are ions and ions so there are formula units per unit cell. z = 4, a = 5.4 Å = 5.4×10–10 m = 5.4×10–8 cm d = 3.2 g cm–3 , M = 78.8 g mol–1 32
Solid State d
z M a3 NA
NA
z M a3 d
4 78.8 g mol 1 (5.4 10 8 cm)3 (3. 2 g cm 3 ) 6.02 1023 mol Q.9
1
Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 128 pm, calculate the density of copper metal. (Atomic mass of Cu=63.55 u and Avogadro’s number=6.02×1023)
Ans. For fcc lattice of copper, a 2 2 r (r+ 2r + r = a 2 ) = 2×1.414×127.8×10–8 cm = 3.61×10–8 cm
Now,
d
a
z M a3 NA
a
4 63.55 (3.61 10 8 cm)3 (6.02 1023 )
8.94 g cm
3
Q.10 Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316 pm, what is the radius of tungsten atom ? Ans. For body centred cubic unit cell, a = 316 pm 3 3 a or r 316.5pm 137.04 pm 4 4 Q.11 How much space is occupied in (i) hcp (ii) ccp and (iii) bcc structure ? Ans. (i) hcp : 74% (ii) ccp : 74% (iii) bcc : 68%. Q.12 Gold (atomic radius = 0.144 nm) crystallizes in a face centred unit cell. What is the lenght of the side of the cell ? Ans. For fcc, a 2 2 r 2 1.414 0.414 nm 0.407 nm. Q.13 In CaF2 crystal Ca2+ ions are present in FCC arrangement. Calculate the number of F— ions in the unit cell. r
Ans. Number of Ca2+ ions per unit cell = 8
1 1 + 6 = 4 2 8
Hence number of F— ions unit per cell = 2 × 4 = 8
33
100 Hours 100Marks
Lecture-4 VOIDS
Voids : It is three dimension space in the solid which is empty. In two dimension s0me of the voids are as,
Square void
Triangular void
In three dimension we are having two types of voids. (1) Tetrahedral Void (2) Octahedral Void Tetrahedral Void : In case of tetrahedral void four atoms or ions form tetrahedral. In which three atoms/ions are on the base and one is on the top. It is having following properties : (a) Coordination number = 4 (b) If radius of void is rvoid and radius of the atom is ratom then, rvoid ratom 0.225 (c) In the void generally cations are placed so, r+ 0.225 r— (d) w.r.t. the given atom number of tetrahedral voids are two. Octahedral Void : In case of octahedral void we are having four atoms on the base and two atom on the top and bottom of the four atoms. It is having following points. (a) Coordination number is six for octahedral void. (b) If radius of the void is rvoid and radius of atom is ratom then, 34
Solid State
rvoid ratom 0.415 (c) If in the void cation is placed then r+ r — 0.415 (d) W.r.t. given atom number of octahedral voids is one. Example : If in an oxide A in tetrahedral void B in octahedral void then the formula of the oxide is. Solution : Let number of oxygen atom is one then number of tetrahedral voids two and number of octahedral void one. So, formula will be, in tetrahedral void : 2A O B or 2A : B : O Therefore, formula, A2B O
Q.1 If w.r.t. oxygen A is having 60% of tetrahedral void and B is having 80% of octahedral void. Then find out, (a) Formula of oxide (b) Formula of nitride Solution: (a)With respect to oxygen number of octahedral void one. With respect to oxygen number of tetrahedral void two. But 60% of tetrahedral void and 80% of octahedral void are packed.
Tetrahedral void Octahedral void 60 80 2A O B 100 100 A 120 B 80 O 100 or A6 B4 O5 or A 12 B 8O 10 It is the formula of oxide. (b) If nitride is used then charged is balanced. Which is equal to charge of 5 atoms of oxygen, i.e., = –10 —10 charge is balanced by 5 oxygen atom —10 charge is balanced by 10/3 nitrogen atom 35
100 Hours 100Marks
So, in place of A6B4O5 we are having A6B4N10/3
or
A18B12N10
Means, A6B4O5×3 or A18 B12 O15 is replaced by A18B12N10 So, anionic vacancy, in this case is five, because five oxygen atoms are more than nitrogen atoms.
Radius Ratio of Octahedral Void Octahedral void is shown in figure. Though an octahedral void is surrounded by six spheres. Only four are shown. The spheres present above and below the void are not shown. Let us assume that the length of the unit cell is a cm. and radius of octahedral void is r and the radius of sphere is R. If the length of the unit cell is a cm, then In right angled DABC, AB = BC = a cm The diagonal AC is : AC
AB 2
Also, Now
AC
or
36
R
BC 2
a2 a 2
AC AB AB
2a a 2R
2r
R
2R
2 2R 2r 2R 1 r 2 1 R 1
2 1
2a D
R
c
2r
R A
B
2r Octahedral Void with Radius r
Solid State
r 2 1 0.414 R or r 0.414 R Thus, for an atom to occupy an octahedral void, its radius must be 0.414 the radius of the sphere. or
Radius Ratio of Tetrahedral Void A tetrahedral void may be represented by placing four sphere 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 sphere are shown by distance circles. Actually all the spheres are touching one another. Let us assume that the length of each side of the cube is a cm and radius of tetrahedral void is r and the radius of sphere is R. In the figure, AC is a face diagonal. In right angled ABC, AC2 = AB2 + BC2 or
AC
AB 2
BC 2
a2 a 2 2a As spheres A and C at the face diagonal (through shown by distant circles) are actually touching each other so that AC = R + R = 2R
2.a .....(i) 2 Now in the right angled triangle, ACD, AD is body diagonal and AD2 = AC2 + CD2 2R
2.a or R
37
100 Hours 100Marks
AD
AC 2
CD 2
2a2 a2 3a The tetrahderal void is present at the centre of the body diagonal AD show that half the length of this diagonal is equal to the sum of the radii of R and r. Thus,
AD 3.a .............(ii) 2 2 Dividing equation (ii) by (i), we get, R r
3.a 2 2 2a 3 2 r 3 3 or 1 R 2 1.732 1.414 1.414 r 0.225 R
3 2
Face diagonal C A
a B
a 2r
2 2
Body diagonal
R r R r 1 R
D
0.225
Thus, for an atom to occupy a tetrahedral void, its radius must be 0.225 times the radius of the sphere. For a given atom number of octahedral void = No. of atoms Number of tetrahedral void = 2×Number of atoms
38
Solid State
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Questions Problems based on this Lecture Q.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. Q.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. Q.3 What is the coordination number of an octahedral void ? Ans. Six. Q.4 A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedaral voids ? Ans. No. of atoms in the close packing 0.5 mol = 0.5×6.022×1023 = 3.011×1023 39
100 Hours 100Marks
Q.5
Ans. Q.6 Ans. Q.7 Ans. Q.8
Ans.
Q.9
Ans.
No. of octahedral voids = 1×No. of atoms in the packing = 3.011×1023 No. of tetrahedral voids = 2×No. of atoms in the packing = 2×3.011×1023 = 6.022×1023 Total No. of voids = 3.011×10 + 6.022×1023 = 9.33×1023. What is the relation of number of voids in terms of number of atoms ? Or In a close paked arrangement of N-spheres, how many (i) tetrahedral and (ii) octahedral sites are present? Number of octahedral voids = No. of atoms in the close paked arrangement. Number of tetrahedral voids = 2 × No. of atoms If the radius of the tetrahedral void is ‘r’ and radius of the atom close packing is R. What is the relation between ‘r’ and R ? ‘r’ = 0.225R If the radius of octahedral void is ‘r’ and radius of the atom in close packing is R. what is the relation between ‘r’ and R ? ‘r’ = 0.414 R. In a solid XY, ‘X’ atoms are in ccp arrangement and ‘Y’ atoms occupy all the octahedral sites. If all the face centred atoms along one of the axes are removed, then what will be the resultant stoichiometry of the compound? In a ccp types structure, 8 X atoms are at the corners and 6 X atoms are at the face centres of the cube. Removal of all the face centred atoms along one of axes means the removal of 2 X atoms. Thus, only 4 X atoms will be left on the centred of the faces. 1 1 (face centre) 3 No. of X atoms per unit cell 8(corner) 8 2 Y atoms are present in all the octahedral sites i.e. 12 at edge centres and 1 at body centre. 1 No. of Y atoms per unit cell 12 1 4 4 Therefore, Stoichiometry of the compound = X3Y4 . 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 the element A and B ? Suppose, the number of B atoms in the packing = n Number of tetrahedral voids = 2n As only 2/3rd of tetrahedral voids are occupied by atoms A,
2 4 2n n 3 3 4 Ratio of A and B n n 4 3 3 The formula of the compound is A4B3. No. of atoms A
40
Solid State
Lecture-5 IMPERFECT OR DEFECTS IN SOLIDS Any departure from perfectly ordered arrangment of constituent particles in crystal is called imperfection or defect.We are having two types of defects one is point defect another is line defect we are disscussing point defect. Type of point defects : (A) Stoichiometric defect (B) Non-stoichiometric defect (C) Impurity defects (A) Stoichiometric defect : In this type of defect, the crystal ratio between the cations and anions remains the same as represented by the molecular formula. means stoichiometry or formula of the solid is not distrubed the defects are called stoichiometric defects. we are having two stoichiometry defects mainly (i) Schottky defect (ii) Frenkel defect Schottky defect : In this type of defect equal number of cations and equal number of anions are missing from their lattice sites so that the electrical neutrality is maintained. It is called Schottky defect. In this type of defect we are having a pair of holes developed due to missing of cation and anion. ystal as , Ex .— L cry Ex.— Lee t A+B– is a cr A+
B—
A+
B—
B—
A+
B—
A+
A+
B—
A+
B—
In case of Schottky defect the given crystal will be like, 41
100 Hours 100Marks B— B—
A+
A+
B—
A+
B— A+
A+
B—
Due to this type of defect following properteics are there. (1) It become good conductor. (2) Mass is decreased because missing cations and anions goes away from the solid. So, density is decreased. (3) It is observed generally in that case where coordination is high and having small difference in the size of cations and anions. Ex.—NaCl, KCl, KBr, AgBr, and CsCl Frenkel defects : If an ion is missing from it’s lattice site to develop a hole and it occupies the interstial site. Electrically neutral and the stoichiometry of the compound are maintained. This type of defect is called Frenkel defect. Let in the crystal of the solid A+B–. The change is as, A+
B—
A+
B—
B—
A+
B—
A+
A+
B—
A+
B—
After defect,
42
Solid State
B— A+
B—
A+
B—
A+
B—
A+
B—
A+
B—
A+
Due to this defect,
(1) Conductivity is increased. (2) Density remains same because mass is constant. (3) It is observe in case of Low coordination number compound. (4) Large difference in the size of cations and anions. Ex.– AgCl, AgBr, AgI, ZnS. Note— AgBr is having both type of defects
(B)Non-stochiometric Defects : The defects which distrub the stochiometry of the compound are called non-stochiometric defects. These defects are either due to the presence of excess metal ions or excess non-metal ions. Metal excess defects: (a) Metal excess 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 dif-
fuse 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 studies. So Na+ becomes more in crystal 43
100 Hours 100Marks
The anion sites occupied by unpaired electrons are called F-centers. 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. (b) Metal excess defect due to extra cations : Sometimes if anions vacancies become less, metal ions become extra which create metal excess defect. For example, on heating ZnO, it gives oxygen. 1 ZnO Zn 2 O 2 2e 2 Zn2+ ions becomes extra and create metal excess defect.
Metal deficiency Defect : Many compounds contain less amount of metals or compared to the stochiometric proportion. This type of defect is generally shown by transition metals generally. 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. A+ +
A
B
-
+
A
+
A
+
A
+
A
B
-
+
A
B
-
A
+
B-
A+
B-
A+
B - B-
A A+
B-
A+
B-
A+
B-
A+
B-
A+ A+
B-
A+
B-
A+
-
+
-
+
-
+
-
+
+
B
A
Ex. Ans.
44
B
A
+
B
B
A
B
A
B-
B-
A+
+
B
-
B-
A+
B-
A+
B-
A+
B-
B-
A+
B-
A+
B
Find out % of Fe+2 and Fe+3 in Fe0.9O. In Fe0.9O or Fe9O10 Let Fe+2 are x and Fe+3 are (9–x) so, balancing charge in Fe9O10 +2(x) + 3(9–x) + 10×–2 = 0
Solid State
2x + 27 – 3x – 20 = 0 x=7
7 So, % of Fe = 9 100 +2
Ex. Ans.
77.7%
Fe+3 = 100 – 77.7 = 22.3% Find out % of Cu+ and Cu++ in Cu1.5O. In Cu1.5O or Cu15O10 Let Cu++ is x and Cu+ is (15–x) in Cu15O10 then Total charge (+2)x+(15–x)×(+1) + 10×–2 2x + 15 – x – 20 = 0 or x = 5 % of Cu++ =
5 15
100
33.33 %
% of Cu+ = 100 – 33.33 = 66.67
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Questions
100 Hours 100Marks
Problems based on this Lecture Q.1 What is the meaning of the term imperfection in solids ? Ans. Inperfection refers to the departure from the perfect periodic arrangement of atoms, ions or molecules in the structure of crystalline substances. Q.2 What is the type of lattice imperfections found in crystals ? Ans. (a) Stoichiometric defects, viz., Schottky defect and Frenkel defect and (b) Non-stoichiometric defects, viz., metal excess, metal deficiency and impurity defects. Q.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. Q.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. Q.5 Why is Frenkel defect not found in pure alkali metal halides ? Ans. This is because of the fact that alkali metal ions have large size which cannot fit into the interstitial sites. Q.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. Q.7 What are F-centres ? Ans. The free electrons trapped in the anion vacancies are called as F-centres. Q.8 Name one solid in which both Frenkel and Schoottky defects occur. Ans. AgBr Q.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. Q.10 What is Schottky defect ? Ans. If equal number of cations and anions are missing from their lattice sites, the defect is known as defect. Q.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 nonstoichiometric compound. Ex. Fe0.9O, Cu1.5O 46
Solid State
Q.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. Q.13 What is the non-stoichiometric defect in crystals ? Ans. A result of the imperfections in the crystal, the ratio of the cations to the anions become different from that indicated by the ideal chemical formula, then the defects are termed non-stoichiometric defects. Q.14 Which point defect bowers the density of a crystal ? Ans. Schottky defect Q.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. Q.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. Q.17 If NaCI is doped with 10–2 mol % of CaCl2 what is the concetration of cation vacancies ? Ans. Ions are replaced by 1 ion. Each ion causes one cation vacancy. Hence, concentration of cation vacancies on being doped with 10–2 mol % vacancies.
Q.18 Ans. Q.19 Ans. Q.20 Ans. Q.21 Ans.
10 2 6.023 1023 6.023 10 19 100 Why does frenkel defect not change the density of AgCl crystal ? This is because in Frenkel defects, no ion is missing from the crystal as a whole so the density remains unchanged. Which point defect of its crystals decreases the density of a solid ? Schottky defect. Why are the solids containing F-centres paramagnetic ? This is because the electrons occupying the vacant sites are unpaired. If 10–5 mole % ofAlCl3 is dopped in NaCl then find out vacancies developed. 1 AlCl3 means, 1Al+++ will replace 3 Na+ so, two vancancies are develop.
10 5 6.023 1023 will develop 100 = 2×6.023×1016 = 12.046×1016 vacancies. Q.22 What are (i) vacancy defect (ii) Interstitial defect ? What type of compounds show these defect ? Ans. (i) 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 decreases. (ii) 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. 47 So, 10–5 mole % or
100 Hours 100Marks Particles in the interstitial sits
Vacant lattice sites
Interstitial defect
Vacancy defect
Q.23 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 (a) Conductors : Solids having conductivity range from 104 to 107 ohm–1 m–1. (b) Insulators : Solids having conductivity range from 10–20 to 10–10 ohm–1 m–1. (c) Semi-conductors : Solids having conductivity range from 10–6 to 104 ohm–1 m–1. Conduction of electricity in metals Metals conduct electricity through electrons. Ionic conductor and electrolytes conduct electricity through ions. Conductivity of metals depend upon number of valence electrons. Atomic orbitals form molecular orbitals which are comparable in energy to form a bond. If this bond is partially filled or it overlaps with a higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field 48
Solid State
and metals conducts electricity. If this energy gap is large, the electrons cannot jump and conduction does not occur. Conduction in semi-conductors In semi-conductors, the energy gap between valence band and conduction band is small and some electrons may jump to conduction band and show same conductivity and this conductivity increases with temperature. Few substances like silicon and germanium are called intrinsic semi-conductors. This law of conductivity is increased by adding a foreign substances as impurity. This is called Doping.
(C) Impurites defect : (a) Electron-rich impurities :— When group-14 elements are doped with group 15 element like Phosphorus or Arsenic. This type of impurity occurs. Group 14 have 4 valence electrons which are used for the formation of 4 covalent bonds with 4 electrons of group -15 elements and thus one electron of group-15 remains free which conducts electricity. As the conduction is due to negatively charged electron So these type of semi-conductors are called n-typs semiconductor. (b) Electron - Deficit Impurities :— When group-14 elements are doped with group-13 elements. Out of four valence electrons of group-14, 3 are bonded with 3 valence electrons of group-13. The place where the fourth valence electrons is missing is called electron-hole. this electron hole can be filled by an electron from the neighbouring atom. But 49
100 Hours 100Marks
this electron leaves an electron hole at its original position. If it happens, it would appear as if the electron hole has moved in the direction opposite to that of the electron that filled it. On applying the electric fields, electrons moves towards the positively charged plate through electronic holes. But it seems that electrons holes are positively charged and are moving towards negatively charged plate. This is called p-type semi-conductor. Group-13 B Al Ga In Th
Group-14 C Si Ge Sn Pb
Group-15 N(gas) P As Sb 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 electrons revolves around the nucleus and also rotates around its own axis and thus every electron behaviours like a magnet. On the basis of magnetic properties the substance may be ; (i) Paramagnetism : The substances which are weakly attracted by magnetic field are called paramagnetic substances. They are magnetised in a magnetic field in the same direction This is due to presence of one or more unpaired electrons. These electrons attracted by magnetic field. 50
Solid State
(ii) Diamagnetic : The substances which are weakly repelled by magnetic field are called diamagnetic substances. They are weakly magnetised in opposite direction. Diamagnetism is shown by those substances which do not have unpaired electrons e.g., NaCl, H2O, C6H6 (iii) Ferromagnetism: many substances like Fe, Co, Ni, CrO2 , etc. are attracted strongly by magnetic field, such substances are called ferromagnetic substances. In solid state the metal ions of such substances are grouped together into small regions called domains. In the absence of magnetic field, these domains are randomly oriented and magnetic moments get cancelled. But when such substances are placed in
magnetic field, all the domains get oriented in the direction of the magnetic field and thus a strong magnetic field is produced. This ordering persist on removing the electrical field and thus it becomes a permanent magnet. (iv) Antiferromagnetism: If this alignment of domains is in such a way that they cancel magnetic moment of each other, then no-net magnetic moment is produced e.g., MnO. (v) Ferrimagnetism : If this alignment results a net magnetic moment, this is called ferrimagnetism. e.g., Fe3O4 .
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100 Hours 100Marks
Mechanism of Electrical Conduction
The conduction in most of the solids is through electron movement under an electrical field. However, in some ionic solids, the conduction is by ions. Therefore in the solids where the conduction is by the movement of electrons, the electrical conductivity depends on the number of electrons availiable to participate in the conduction process. The spaces between valence band and conduction band represent energies forbidden to electrons and we called energy gap or forbidden zone. (i) In metals, the conduction band is close to valence band and therefore, the electrons can easily go into the conduction band. Therefore, metals are good conductros. (ii) In insulators, the energy gap between valence band and conduction and therefore, electrons from valence band cannot move into the conduction band. (iii) Several solids have properties intermediate between metals and insulators. These are called semi-metals or semiconductors. 52
Solid State
Conduction of Electricity in Semiconductors Ge and Si are the most important commercial examples of semiconductors. The crystal structures of Ge and Si are similar to that of diamond. Atoms of both Ge and Si have four electrons in the outermost shell. Therefore each atoms is covalently bonded with neighbouring atoms through sp3 hybrid.
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53
Questions
100 Hours 100Marks
Problems based on this concept Q.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. Q.2 Define the ‘forbidden zone’ of an insulator. Ans. The large energy gap between valence band and the conduction band in an insulator called forbidden zone. Q.3 What is superconductivity ? Who discoverd it ? Or Define superconductivity of a substance. Ans. A substance is said to be superconductor when it offers no resistance to flow of electricity. This property is called superconductivity. Most metals become superconducting at very low temperature (2-5K). Kammerling Onnes discovered it. Q.4 Define photovoltaic compound. Ans. A compound which generates current when exposed to light is called photovoltaic compound. Q.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 belong ? Ans. n-type semiconductor means conduction due to presence of excess of electrons. Therefore, to convert group 14 elements into n-type semiconductor, it should be doped with group 15 elements. Q.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 element and Si is Group 14 element, there will be a free electron. Thus, it is n-type. Q.7 What is doping ? Why is it done ? Ans. It is process of adding impurities in a crystal lattice. Dopping is done by adding calculated amount of impurities. It increases the conductivity. Q.8 What is the difference in the semi-conductors obtained by doping silicon with Al or with P ? 54
Solid State 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. Q.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 2Na+ ions are replaced by one Sr2+ ion the lattice site). Q.10 How does the elecrical 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. Q.11 What is meant by ‘doping’ in a semiconductor ? Ans. Doping means incorporating small amount of forgien impurity in crystals. Doping of group 14 elements with group 15 elements give rise to excess electron (n-type semiconductors) whereas with group 13 elements give rise to holes (p-type semiconductors). Q.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 taken place in them. (iii) This because some of the cations and anions are missing from the crystal lattice. Q.13 How does the electrical resistivity of (i) a semiconductor (ii) metallic conductors vary with temperature ? Ans. Electrical resistivity of metallic conductors increases with temperature while that of semiconductors and super conductors decreases with temperature.
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100 Hours 100Marks
Extra Shoot Q.1. Explain why amorphous solids are isotropic. Q.2. How many different types of unit cells are possible for two dimensional lattices ? Q.3. If three elements X, Y and crystallize in a cubic solid lattice with X atoms at the corners, Y atoms at cube center and atoms at edges then what is the formula of the solid ? Q.4. What is Frankel defect ? Q.5. What is the coordination number of an atom present in an octahedral void ? Q.6. What is F-center ? Q.7.
Compare the size of energy gap in conductors, semiconductors and insulators?
Q.8. Define the term amorphous. Give four examples of amorphous solids. Q.9. A compound is formed by two elements M and N, The element N forms hcp and atoms M occupy 2/3rd of octahedral voids. What is the formula of the compound? Q.10. A compound contains two types of atoms- X and Y. It crystallises in a cubic lattice with atoms X at the corners of the unit cell and atoms Y at the body centres. What is the simplest possible formula of this compound? What is the coordination of Y ? Q.11. 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.12. Describe the following : (i) Hexagonal close packing (ii) Cubic close packing (iii)Body centred cubic packing Draw unit cell of each type.
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