Solid State Notes PDF

Chapter – 1  SOLID STATE 1 DEFINITION  Solid state represents the physical state of matter in which constituents (atoms/ions/molecules) have no translator moti...

Chapter – 1  SOLID STATE 1 DEFINITION  Solid state represents the physical state of matter in which constituents (atoms/ions/molecules) have no translator motion although vibratory or rotational motions are possible about their position in solid lattice or solid state is characterised by its compressibility, rigidity & mechanical strength. 2 CHARACTERISTICS OF SOLIDS a) Solids are rigid, almost incompressible. They have high density & mechanical strength. b) The constituent atoms/ ions/ molecules contain fixed positions & are not able to show translator motion owing to strong interparticle forces. However oscillatory (vibrational) or rotational motion about own axis, are possible. c) On heating, solids melt into liquids & the temperature at which melting is noticed at 1 atm pressure is known as melting point of solids. 3 AMORPHOUS AND CRYSTALLINE SOLIDS Crystalline Solids Amorphous Solids a) These have long range order of constituent a) These have only short range order of particles. constituent particles. b) These have definite characteristic geometrical b) These have irregular shape. shape. c) These melt at a sharp & characteristic c) These gradually soften over a range of temperature. temperature. d) When cut with a sharp edged tool, the newly d) When cut with a sharp edged tool, newly generated surfaces are plain & smooth. generated surfaces are irregular. e) These are true solids. e) These are pseudo solids or supercooled liquids. f) These have definite & characteristic heat of f) These do not have definite heat of fusion. fusion. g) These are anisotropic in nature. g) These are isotropic in nature. 1. Crystalline solids are also called true solids & principles given in this section holds good for crystalline solids. 2. Amorphous solids are also called as supercooled liquids or pseudo solids, as they have tendency to flow under own weight. E.g., glasses fitted in windows become slightly thicker at the bottom & thin at the top. DOYOU DO YOU 3. Crystalline nature of polymers: polymers may have crystalline as well as amorphous properties. E.g., high density polythene, fibres, stretched KNOW KNOW rubber, etc. on the other hand, amorphous polymers are low density polythene elastomers (synthetic rubber), etc. 4. Annealing: an amorphous substance, on heating at a certain temperature, may attain crystalline nature while on cooling regains its amorphous nature. This phenomenon is called annealing. E.g., glass on keeping in semi- solid state for a long time or on cooling, if allowed to melt very slowly, it becomes opaque due to formation of crystals in it.  Solid State 4 ISOTROPY AND ANISOTROPY a) Isotropy/anisotropy represents the directional dependence of properties of solids. b) A solid is said to be an isotropic, if its properties (like conductivity, strength, RI, etc) have same value in all directions. c) Amorphous solids are isotropic in nature. d) Crystalline solids, on the other hand, are anisotropic, i.e., their physical properties are different in different directions. For example, the speed of light passing through a crystal varies with the direction in which it is measured. 5 SPACE LATTICE AND UNIT CELL Space lattice: It is the three-dimensional arrangement of identical points in the space which represents how the constituent particles (atoms, ions, molecules) are arranged in a crystal. Each particle is depicted as a point. Unit cell: A unit cell is the smallest portion of a space lattice which, when repeated in different directions, generates the entire lattice. Note: A unit cell is characterised by six parameters, i.e., axial angles 𝛼, 𝛽 & 𝛾 and axial lengths a, b and c. Thus unit cell of a crystal possesses all the structural properties of a given crystal. c 𝛽 𝛼 b a 𝛾 𝛾 A portion of 3-D cubic lattice & its unit cell Crystal System: On the basis of the axial distances & the axial angles between the edges, the various crystals can be divided into 7 systems. These are listed in the below table. Crystal Systems Axial distance or edge lengths Axial angles Examples Cubic (most symmetrical) a=b=c 𝛼 = 𝛽 = 𝛾 = 900 Cu, KCl, NaCl, Zinc blende Tetragonal a=b≠c 𝛼 = 𝛽 = 𝛾 = 900 Sn(White tin), SnO2, TiO2, CaSO4 Orthorhombic or a≠b≠c 𝛼 = 𝛽 = 𝛾 = 900 Rhombic sulphur, Rhombic KNO3 , BaSO4 Monoclinic a≠b≠c 𝛼 = 𝛾 = 900 ; 𝛽 ≠ 900 Monoclinic sulphur, Na2SO4.10H2O Hexagonal a=b≠c 𝛼 = 𝛽 = 900 Graphite, ZnO, CdS ; 𝛾 ≠ 1200 Rhombohedral or Trigonal a=b=c 𝛼 = 𝛽 = 𝛾 ≠ 900 CaCO3 (Calcite), HgS (Cinnabar) Triclinic a≠b≠c 𝛼 ≠ 𝛽 ≠ 𝛾 ≠ 900 K2Cr2O7, H3BO3, (most unsymmetrical) CuSO4.5H2O  Solid State 6 TYPES OF CUBIC UNIT CELLS & NUMBER OF ATOMS PER UNIT CELL Types of Unit Calculation of No. of Atoms per Unit Cell No. of Atoms Cell per Unit Cell 1 Primitive 8(corner atoms) × 8 (atom per unit cell) 1 1 Body centred 8(corner atoms) × 8 (atom per unit cell) + 1(Body centered atom) 2 1 Face centred 8(corner atoms) × 8 (atom per unit cell) + 6 (Face centered atoms) 1 × (atom per unit cell) 4 2 7 PACKING EFFICIENCY  Packing efficiency is the percentage of total space filled by the particles. Volume occupied by atoms in unit cell (v)  Packing efficiency = × 100. Total volume of the unit cell (V) a) Packing efficiency in simple cubic structures: The spheres are in contact with each Let ‘a’ be the cube edge & ‘r’ the atomic radius. other along the edge of the cube. As the particles touch each other along the edge, Therefore a = 2r. G B Volume of the unit cell = 𝑎3 Since one atom is present in a unit cell, H A 4 4 𝑎 3 𝜋𝑎3 Its volume 𝑣 = 3 𝜋𝑟 3 = 3 𝜋 (2) = 6 F 𝜋𝑎3 𝑣 C ∴ Packing efficiency = × 100 = 6 × 100 𝑉 𝑎3 E D 𝜋 3.14 = 6 × 100 = 6 × 100 = 52%. Simple cubic unit cell. Therefore, 52% of unit cell is occupied by atoms & the rest 48% is empty space. b) Packing efficiency in ccp & hcp structures: The efficiencies of both type of packing, ccp & hcp, are equally good since in both, atom spheres occupy equal fraction (74%) of the available volume. We shall now calculate the efficiency of packing in ccp structure. Let ‘a’ be the unit cell length & ‘b’ the face diagonal (represented as AC in below figure). In this figure other sides are not shown for the sake of clarity. In triangle ABC, ∠𝐴𝐵𝐶 is 900, Therefore, AC2 = b2 = BC2 + AB2. = a2 + a2 = 2a2 G B ∴ b = √2 𝑎 A If r is the radius of the sphere, we find, b = 4r = √2 𝑎 H 4𝑟 𝑎 Or, a = = 2√2 𝑟 [Or, r = 2√2] b √2 As ccp structure has 4 atoms per unit cell, 4 F Therefore the total volume of 4 spheres (𝑣) = 4 × 3 𝜋𝑟 3. C 4 𝑣 4 ×( )× 𝜋𝑎3 E D 3 ∴ Packing efficiency = 𝑉 × 100 = 3 × 100 (2√2 𝑟) Cubic close packing 16 ( )× 𝜋𝑎3 𝜋 3 = × 100 = 3√2 × 100 16 ×√2 𝑟 3 = 74%. Therefore, 74% of unit cell is occupied by atoms & the rest 26% is empty space.  Solid State c) Packing efficiency in bcc structures: In this case the atom at the centre is in touch with other two atoms which are diagonally arranged. The spheres along the body diagonal are shown with solid boundaries in the given figure. In ∆EFD, b2 = a2 + a2 = 2a2 ∴ b = √2 𝑎 In ∆AFD, c = a + b2 = a2 + 2a2 =3a2 2 2 G ∴ c = √3 𝑎 B The length of the body diagonal ‘c’ is equal to 4r. H A ‘r’ being the radius of the sphere (atom). As all three spheres along the diagonal touch c a each other. Then, 𝑐 = 4𝑟. Therefore, 𝑐 = 4𝑟 = √3 𝑎 F a b C 4𝑟 √3 𝑎= [or, 𝑟 = 𝑎] E a D √3 4 As already established, the total number of atoms Body-centred cubic unit cell associated with a bcc unit cell is 2, 4 8 The volume (v) is, therefore, 2 × 3 𝜋𝑟 3 = 𝜋𝑟 3. 3 4𝑟 3 64𝑟 3 Volume of the unit cell (𝑉) = 𝑎3 = ( ) = 3√3. √3 𝑣 (8/3)𝜋𝑟 3 √3 Packing efficiency = 𝑉 × 100 = × 100 = 𝜋 × 100 = 68%. (64/3√3) × 𝑟 3 8 Therefore, 68% of unit cell is occupied by atoms & the rest 32% is empty space. Type of Cubic Unit Cell Number of Relation Coordination Packing Packing Empty Space Atoms per between Number Fraction Efficiency (100 – P.E) Unit Cell a&r (P.E) Simple Cubic 1 𝑎 = 2𝑟 6 0.524 52.4% 47.6% Face Centred Cubic 4 √2𝑎 = 4𝑟 12 0.74 74% 26% Body Centred Cubic 2 √3𝑎 = 4𝑟 8 0.68 68% 32% 8 DENSITY OF A UNIT CELL Mass of unit cell 𝑧.𝑚 Density of unit cell = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙 = 𝑎3 𝑧.𝑚 Or, 𝑑= 𝑎3.𝑁𝐴 Where, z = number of atoms per unit cell m = mass of one atom M = molar mass a = edge length 𝑁𝐴 = Avogadro number (i) For edge length ‘a’ in pm. Edge length = ‘a’ pm = 𝑎 × 10−12 𝑚 = 𝑎 × 10−10 𝑐𝑚 Volume of unit cell = (𝑎 × 10−10 𝑐𝑚 )3 = 𝑎3 × 10−30 𝑐𝑚3 𝑧 ×𝑚 Density = 𝑎3 × 10−30 × 𝑁 𝑔 𝑐𝑚−3 𝐴 (ii) In S.I units: For ‘a’ in meters and ‘m’ in kg 𝑚𝑜𝑙 −1 , 𝑧.𝑀 Density = 𝑎3. 𝑁 𝑘𝑔 𝑚−3 𝐴  Solid State Questions related to density: Q. 8. (i) Chromium metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be 287 pm. Calculation the atomic radius. What would be the density of chromium in g/cm3 ? √3 𝑎 √3 Solution: For bcc lattice, 𝑟 = = × 287 = 124.27 𝑝𝑚. 4 4 𝑍×𝑀 𝑛×𝑀 𝜌 = 𝑉×𝑁 = 𝑁 × 𝑎3 𝐴 𝐴 For bcc; 𝑍 = 2, 𝑎 = 287 × 10−10 𝑐𝑚 2 ×51.99 𝜌= = 7.30 𝑔/𝑐𝑚3. (287 ×10−10 )3 × 6.023 ×1023 Q. 8. (ii) The edge length of unit cell of a metal having molecular weight 75 g mol-1 is 5Å which crystallizes in cubic lattice. If the density is 2 g/cm3. Calculate the radius of metal atom. (NA = 6 × 1023) 𝑍×𝑀 Solution: 𝑟 = 𝑎 3 × 𝑁𝐴 Putting the given values; 𝑍 ×75 2 = (5 ×10−8 )3 × 6 ×1023 ⇒ Z = 2 (i.e., bcc lattice). √3 √3 Radius, 𝑟 = 4 ×𝑎 = 4 × 5 = 2.165Å = 2.165 × 10−8 𝑐𝑚 = 216.5 𝑝𝑚. Q. 8. (iii) An organic compound crystallizes in an orthorhombic system with two molecules per unit cell. The unit cell dimensions are 12.05Å, 15.05Å & 2.69Å. If the density of the crystal is 1.419 g cm-3. Calculate the molecular mass of the organic compound. 𝜌𝑁𝐴 𝑉 1.419 ×6.022 × 1023 ×(12.05 ×15.05 ×2.69 × 10−24 ) Solution: 𝑀 = = 𝑍 2 −1 = 209 𝑔𝑚𝑜𝑙 (∵ 𝑉 = 𝑎 × 𝑏 × 𝑐) Q. 8. (iv) A metallic element has a cubic lattice. Each edge of the unit cell is 2.88Å. The density of the metals is 7.20 g 𝑐𝑚−3. How many unit cells will be there in 100 g of the metal? Solution: Volume of the unit cell = (2.88Å)3 = (2.88 × 10−18 )3 = 23.9 × 10−24 𝑐𝑚3. Given that, mass = 100 g & density = 7.20 g/cm3. 𝑚 100 ∴ Volume of 100 g of the metal = 𝜌 = 7.20 = 13.9 𝑐𝑚3. 13.9 𝑐𝑚3 Or, number of unit cells in this volume = 23.9 × 10−24 𝑐𝑚3 = 5.82 × 1023. Q. 8. (v) A metal crystallizes into two cubic phases, face-centred cubic (fcc) & body-centred cubic (bcc) whose unit cell lengths are 3.5Å & 3.0Å, respectively. Calculate the ratio of densities of fcc & bcc. Solution: Unit cell length when metal crystallizes as fcc = 3.5 Å Unit cell length when metal crystallizes as bcc = 3.0 Å 𝑍 ×𝑀 ρ in 𝑓𝑐𝑐 = 𝑉 1× 𝑁 … … … … … (𝑖) 1 𝐴 𝑍2 ×𝑀 ρ in 𝑓𝑐𝑐 = 𝑉 × 𝑁𝐴 … … … … … (𝑖𝑖) 2 𝜌 𝑍 ×𝑉 Dividing eq. (i) by eq. (ii) & putting the data in equation (iii), 𝜌𝑓𝑐𝑐 = 𝑍1 × 𝑉2 𝑏𝑐𝑐 2 1 Now, For fcc, Z1 = 4; V1 = a3 = (3.5 × 10−8 )3 For bcc, Z2 = 2; V2 = a3 = (3.0 × 10−8 )3 𝜌𝑓𝑐𝑐 4 × (3.0×10−8 )3 = = 1.259. 𝜌𝑏𝑐𝑐 2 × (3.5×10−8 )3  Solid State 9 PACKING IN METALLIC CRYSTAL I. Packing in 2D a) Square close packing: In square close packing co-ordination number is four & it is less efficient as only 52.4% of the space is filled & 47.6% of the space is empty. Square close packing b) Hexagonal close packing: In hexagonal close packing, co-ordination number is six & it is more closely packed arrangement. In it 60.4% of the available space is filled. Hexagonal close packing II. Packing in 3D a) When extended to 3-D, square close packing forms a lattice having simple cubic unit cell in which co-ordination number is six. b) If the central empty space is filled by another sphere in a unit cell, then it becomes body-centred unit cell & co-ordination number becomes eight. c) When hexagonal arrangement is extended in 3-D by placing other closest packed layers on the top of voids (hollows) of the first layer (A), following arrangements may be obtained. According to the figure there are two types of voids in first layer, marked as ‘o’ & ‘t’. All the voids (o & t) are equivalent but the sphere of second layer can be placed either on ‘o’ voids or ‘t’ voids, all ‘o’ & ‘t’ voids cannot be covered simultaneously. According to the arrangement, second layer (B), covers only half of the voids of layer (A) Schematic representation of the first & second layers of hard spheres, ‘o’ & ‘t’ are octahedral & tetrahedral voids respectively.  Solid State When we want to arrange third layer, there are two possible arrangements. One is to put the spheres on voids ‘t’ of second layer. When spheres are placed in this way, it is observed that each sphere of third layer lies directly above those in the first layer. If this arrangement is continued indefinitely in the same sequence, it is represented as ABABAB...... a) ABAB type arrangement: Lattice of this arrangement (ABABAB...) is found to have hexagonal unit cell & thus it is also called as hexagonal close packing & abbreviated as hcp. E.g., Mg, Zn, etc. Building of second layer (B, shown shaded) covering hollows marked by dots (.). The hollows marked by crosses (x) remain unoccupied. Six-fold axis A B A B (i) (ii) (iii) ABABAB... system of close packing or hexagonal close packing (hcp) of spheres. b) ABCABC type arrangement: The second alternative to place the third layer over second layer is to place sphere over the unoccupied voids of the first layer. This arrangement gives a different lattice in which spheres of fourth layer will correspond with those of the spheres of first layer. If continued indefinitely the same sequence; it is represented as ABCABCABC.... Six-fold axis C B A C B A ABABAB... system of close packing or hexagonal close packing (hcp) of spheres. (i) (ii) (iii)  Solid State Lattice of this arrangement (ABCABCABC...) has a cubic symmetry & thus is called as cubic close packing & is abbreviated as ccp. Lattice of this arrangement is found to have face-centred unit cells & thus is also known as face-centred cubic structure. E.g., Cu, Ag, etc. 10 CRYSTAL DEFECTS  Crystal defects: The defects are basically irregularities in the arrangement of constituent particles.  Broadly, crystal defects are of two types: (a) Point defects & (b) Line defects.  Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance. Line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points. These irregularities are collectively called as crystal defects.  Point defects:  Interstitials: Atoms or ions which occupy normally voids in a crystal are called interstitials.  Vacancy: When one of the constituent particles is missing from the crystal lattice, this unoccupied position is called vacancy.  Point defects can be classified into three types: A. Stoichiometric defects: The point defects that do not disturb the stoichiometry of the solid are called Stoichiometric defects.  They are also called intrinsic or thermodynamic defects.  These are of two types, vacancy defects & interstitial defects. i. Vacancy defect: When some of the lattice sites are vacant, the crystal is said to have vacancy defect. It results in decreases in density of the substance. This defect can arise when a substance is heated. ii. Interstitial defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have interstitial defect. Due to this defect the density of the substance increases. Vacancy & interstitial defects are generally shown by non-ionic solids because ionic solids must always maintain electrical neutrality. Ionic solids show these defects as Schottky & Frankel defects as explained below: a) Schottky defect: This defect arises when equal number of cations & anions are missing from the lattice. It is a common defect in ionic compounds of high co-ordination number where both cations & anions are of the same size; e.g., KCl, NaCl, KBr, etc.  Due to this defect, density of crystal decreases & it begins to conduct electricity to a smaller extent. b) Frankel defect: This defect arises when some of the ions of the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation; e.g., AgBr, ZnS, etc.  Due to this defect, density does not change, electrical conductivity increases to a small extent & there is no change in overall chemical composition of the crystal. A+ 𝐵− A+ A+ A+ 𝐵− A+ 𝐵− A+ 𝐵− 𝐵− A+ 𝐵− 𝐵− 𝐵− A+ 𝐵− A+ 𝐵− A+ 𝐵− A+ A+ 𝐵− A+ 𝐵− A+ 𝐵− A+ 𝐵− A+ 𝐵− 𝐵− A+ 𝐵− A+ 𝐵− Schottky defect Frankel defect  Solid State B. Impurity defects: These defects arise when foreign atoms or ions are present in the lattice site (substitutional solid solutions) or in the interstitial sites (interstitial solid solutions).  For example, when molten NaCl containing a little amount of SrCl2 is crystallised, some of the sites of Na+ ions are occupied by Sr2+.  Each Sr2+ replaces two Na+ ions. It occupies the site of one ion & the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions. Na+ 𝐶𝑙 − Na+ 𝐶𝑙 − 𝐶𝑙 − Na+ 𝐶𝑙 − Na+ Na+ 𝐶𝑙 − 𝐶𝑙 − 𝐶𝑙 − Na+ 𝐶𝑙 − Na+ Induction of cation vacancy in NaCl by substitution of Na+ by Sr2+. C. Non-stoichiometry defects: These defects arises when stoichiometry of a substance is disturbed. These are of two types: i. Metal excess defect: This may occur in either of the following two ways: a) Metal excess defect due to anion vacancies: In this defect a negative ion from the crystal lattice may be missing from its lattice site leaving a hole or vacancy when is occupied by the electron originally associated with the anion. In this way crystal remains neutral. Alkali halides like NaCl & KCl are examples of this type of defect. b) Metal excess defect due to anion vacancies: In this defect an extra positive ion occupies interstitial position in the lattice & the free electron is trapped in the vicinity of this interstitial cation. In this way crystal remains neutral. For example, zinc oxide on heating loses oxygen & turns yellow. ZnO Heating Zn2+ + ½ O2 + 2e– 2+ The excess of Zn ions move to interstitial sites & the electrons to neighbouring interstitial sites. ii. Metal deficiency defect: This type of defect generally occurs when metal shows variable valency. The defect arises due to the missing of cation from its lattice site & the presence of the cation having higher charge in the adjacent lattice site. For example, FexO, where x = 0.93 to 0.96. A+ 𝐵− A+ 𝐵− A+ 𝐵− A2+ 𝐵− 𝐵− 𝐵− 𝐵− A2+ 𝐵− A+ 𝐵− A+ A+ 𝐵− A+ 𝐵− A+ 𝐵− A+ 𝐵− 𝐵− A+ 𝐵− A+ 𝐵− A+ 𝐵− A+ Metal deficiency defect due to missing cations Metal deficiency defect due to anions  Solid State F-Centres: These are the anionic sites occupied by unpaired electrons.  F-Centres impart colour to crystals.  They impart yellow colour to NaCl crystals, violet colour to KCl crystals & pink colour to LiCl crystals.  The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal. 11 ELECTRICAL PROPERTIES Solids can also be classified into three types based on their electrical conductivities: A. Conductors: The solids with conductivities ranging between 104 to 107 𝑜ℎ𝑚−1 𝑚−1are called conductors. B. Insulators: The solids with very low conductivities ranging between 10−20 to 10−10 𝑜ℎ𝑚−1 𝑚−1. C. Semi-conductors: The solids with conductivities in the intermediate ranging between 10−6 to 104 𝑜ℎ𝑚 −1 𝑚−1are called semi-conductor. CONDUCTION BAND THEORY  A metal is characterized by a band structure.  The highest filled band is called valence band  The lowest unoccupied band is called conduction band.  The gap between the two bands is called forbidden energy gap. Conductors Insulators Semi-conductors In case of conductors, either the In case of insulators, the forbidden In case of semi-conductors, valence band & conduction band energy gap is very large & the forbidden energy gap is small. overlap or the valence band is electrons are unable to excite to Therefore, some electrons may partially filled. the conduction band. jump to conduction band & show Conduction band some conductivity. Empty band Empty band Forbidden Zone (Large Energy Gap) Energy (Small Energy Gap) Filled band Partially filled Overlapping Band Bands Metal Insulator Semi-conductor DO YOU DO YOU Electrical conductivity of semi-conductors increases with rise in temperature, since more electrons can jump to the KNOW KNOW conduction band. Types Of Semi-Conductors 1. Intrinsic: These are those semi-conductors in which the forbidden energy gap is small. Only some electrons may jump to conduction band & show some conductivity. They have very low electrical conductivity. Example, pure silicon, pure germanium. 2. Extrinsic: When an appropriate impurity is added to an intrinsic semiconductor, their electrical conductivity increases & extrinsic semi-conductors are formed. Doping: The process of adding an appropriate amount of suitable impurity to increases the conductivity of semi-conductors is called doping. Normally, trivalent and pentavalent elements are used to dope silicon and germanium.  Solid State Types Of Extrinsic Semi-conductors 1. n-type semiconductors: They are formed when silicon is doped with electron rich impurity like group 15 elements. The increase in conductivity is due to the negatively charged free electrons. 2. p-type semiconductors: They are formed when silicon is doped with electron deficient impurity like group 13 elements. The increase in conductivity is due to the positively charged holes. Silicon atom Mobile electron Positive hole (no electron) As B Perfect Crystal n-type p-type Creation of n-type & p-type semiconductor Diode: When n-type & p-type semiconductors are combined, what results is known as diode. These diodes are used as rectifiers. DO YOU DO YOU Transistors: These are used to detect or amplify radio or audio signals. They consists of pnp & npn type of sandwich semiconductors. KNOW KNOW Photodiode: These are diodes which are capable of converting light energy into electrical energy & are used in solar cells. Uses Of Semi-conductors  These are used for making diodes, transistors, integrated circuits (IC) & microprocessors, etc.  Transistors are made by sandwich a layer of one type of semi-conductor between two layers of the other type of semi-conductor.  Npn & pnp type of transistors are used to detect or amplify radio & audio signals.  12-16 compounds: These compounds are formed by the combination of group 12 & group 16 elements. They possess an average valency of 4 & act as semi-conductors. Example: ZnS, CdS, CdSe & HgTe.  13-15 compounds: These compounds are formed by the combination of group 13 & group 15 elements. They possess an average valency of 4 & act as semi-conductors. Example: InSb, AlP & GaAs. Applications Of n-type & p-type semiconductors  Instead as diode, transistors, photo-diode there is another application of the semiconductors. Combination of elements such as Group – 13 – Group – 15 (example: InSb, AlP, GaAs) elements or Group – 12 – Group – 16 (example: ZnS, CdS, HgTe, CdSe) elements give rise to a large variety of solid state materials which stimulate average valence of four as in Ge or Si. Bonding of these compounds is not perfectly covalent and the extent of ionic character depends on the difference in electronegativities of the two elements.  Solid State 12 MAGNETIC PROPERTIES OF SOLIDS a) The magnetic properties of solids are due to the magnetic movements of electrons. b) The magnetic moments of electrons originates from two types of motions: i. Magnetic moment due to spin of electron around its own axis. This is known as spin magnetic moment. ii. Magnetic moment due to orbital motion of electron around the nucleus. This is known as orbital magnetic moment. c) Thus, each electron has permanent spin & orbital magnetic moments & acts as a tiny magnet. d) The magnitude of this magnetic moment is very small. It is measured in the unit Bohr magneton (𝜇𝐵 ) & is equal to 9.27× 10−24 𝐴 𝑚2. Curie temperature: The temperature at which a ferromagnetic substance loses its ferromagnetism & attains paramagnetism only is DO YOU YOU called Curie Temperature. DO Example: For iron, the curie temperature is 1033 K, for Nickel it is KNOW KNOW 629 K & for Fe3O4 it is 850 K. Below this temperature paramagnetic substances behaves as ferromagnetic substances. Born: September 22, 1791 London or Newington, England Died: August 25, 1867 (aged 75) Richmond upon Thames, England Awards & Honours: Copley Medal (1838) Copley Medal (1832) Subjects Of Study: Electromagnetism, Electric – motor, Electric generator, Chlorine, Benzene, etc.  Faraday invented the rubber balloon while experimenting with gases.  Elected a member of the Royal Society in 1824, he twice refused to become President. Memories  In June 1832, the University of Oxford granted Faraday a Doctor of Civil Law degree.  He became the first Fullerian Professor of Chemistry at the Royal Institution in 1833.  In 1845, Faraday became the first to classify substances as either diamagnetic or paramagnetic. He based this classification on his observation of the force exerted on substances in an inhomogeneous magnetic field.   Solid State Substance Characteristics Examples Application Diamagnetic  Repelled weakly by a magnetic field. Benzene, Insulators materials  All electrons are paired. NaCl, H2O. ↿⇂ ↿⇂ ↿⇂ ↿⇂  Weakly magnetised in a magnetic field in opposite direction. Paramagnetic  Weakly attracted by a magnetic field. O2, Cu2+, Fe3+, Electronic materials  Unpaired electrons are present. Cr3+. devices ↿⇂ ↑ ↑ ↑ ↑  Magnetised in a magnetic field in the same direction.  Permanent magnetisation is not possible. They lose their magnetism in the absence of magnetic field. Ferromagnetic  Strongly attracted by a magnetic field. Fe, Ni, Co, Gd, CrO2 is used in materials.  Unpaired electrons are present. CrO2. audio, video,  In solid state, the metal ions are grouped together into small regions called domains, which act as tiny tapes. magnets. In an un-magnetised ferromagnetic material, the domains are randomly oriented such that their magnetic moments get cancelled. However, in presence of a magnetic field, the domains spontaneously orient in the direction of the magnetic field producing a strong magnetic effect. ↑ ↑ ↑ ↑ ↑  Permanent magnetisation is possible, i.e., they do not lose their magnetism even in the absence of magnetic field.  Change to paramagnetic solid when heated.  Note: The characteristic temperature above which no ferromagnetism is observed is known as Curie Temperature. Anti-  The domains align in such a way that resultant MnO Used in ferromagnetic magnetic moment is zero. instruments of materials  The number of domains that are aligned in parallel magnetic direction is equal to the number of domains that are susceptibility aligned in anti-parallel direction. measurement. ↑ ↓ ↑ ↓ ↑ ↓ ↑  Change to paramagnetic solid when heated. Ferromagnetic  The domains align in such a way that there is a small materials net magnetic moment. Fe3O4 -------  The number of domains that are aligned in parallel (Magnetite), direction is not equal to the number of domains that Ferrites like are aligned in anti-parallel direction. MgFe2O4, ZnFe2O4, etc. ↑ ↓ ↓ ↑ ↓ ↓ Or, ↑ ↓ ↓ ↓ ↑ ↓  Weakly attracted by a magnetic field.  Change to paramagnetic solid when heated.  Solid State SOLIDS Isotropic Crystal Solids Amorphous Solids Ordered arrangement of Irregular arrangement of particles particles Sharp melting point Melt over a range of temperature Long range order Short range order Anisotropic Isotropic Crystal Lattice: The regular arrangement of an infinite set of points which describes the three- dimensional arrangement of constituent particles (atoms, ions or molecules) in space is called a crystal lattice or space lattice. Unit Cell: The smallest repeating unit of space lattice which when repeated over & over again in three-dimension, results into the whole of the space lattice of the crystal is called the unit cell. How to calculate the number of particles per unit cell ? Contribution of each atom present at the corner = 1/8 Contribution of each atom present on the face = 1/2 Contribution of each atom present on the edge centre = 1/4 Contribution of each atom present at the body centre = 1 Characteristics of Different Types of Unit Cells: Characteristics sc bcc fcc hcp Number of atoms per unit cell 1 2 4 6 Coordination number 6 8 12 12 Packing efficiency 52% 68% 74% 74% Radius 𝑎 𝑎 𝑎 √3 2 𝑎 2√2 2 4  Alloys: A substance that contains a mixture of elements and has metallic properties.  There are two types of alloys: (a) Substitutional alloys and (b) Interstitial alloys. (a) Substitutional alloys: In which atoms of host metal are replaced by other atoms of almost similar size. Examples: Pewter [Sn, Cu, Bi, Sb], solder [Pb, Sn]. (b) Interstitial alloys: In which holes (intersticies) of packed structures are occupied by small atoms. Examples: steel.  Solid State  Greenhouse glass Carbonia: CO2 is a greenhouse gas. If kept at high temperatures (> 700 K) and under extremely high pressure (≈ 6.4 × 105 atm), it forms a solid like silica (SiO2) which is about tem times harder than quartz but softer than diamond. It is called Carbonia and is hardest amorphous substance known, it disappears immediately when pressure is reduced.  Molecular bonds: This term is sometimes used for the weak van der Waal’s attraction forces. They are formed by polarizability originated at atomic level, as in noble gases, where orbitals are deformed due to mutual attraction between nuclei and electrons of neighbouring atoms.  Isotropy: A solid is said to be isotropic, if its properties (like conductivity, strength, RI, etc.) have same value in all directions. [Amorphous solids are isotropic in nature]  Anisotropic: Crystalline solids are anisotropic. i.e., their physical properties are different in different directions. Example: the speed of light passing through a crystal varies with direction in which it is measured. Seven Types Of Crystal Systems (In Vibgyor Colour) Crystal Systems Axial distance or edge lengths Axial angles Examples Cubic (most symmetrical) a=b=c 𝛼 = 𝛽 = 𝛾 = 900 Cu, KCl, NaCl, Zinc blende Tetragonal a=b≠c 𝛼 = 𝛽 = 𝛾 = 900 Sn(White tin), SnO2, TiO2, CaSO4 Orthorhombic or a≠b≠c 𝛼 = 𝛽 = 𝛾 = 900 Rhombic sulphur, Rhombic KNO3 , BaSO4 Monoclinic a≠b≠c 𝛼 = 𝛾 = 900 ; 𝛽 ≠ 900 Monoclinic sulphur, Na2SO4.10H2O Hexagonal a=b≠c 𝛼 = 𝛽 = 900 Graphite, ZnO, CdS ; 𝛾 ≠ 1200 Rhombohedral or Trigonal a=b=c 𝛼 = 𝛽 = 𝛾 ≠ 900 CaCO3 (Calcite), HgS (Cinnabar) Triclinic a≠b≠c 𝛼 ≠ 𝛽 ≠ 𝛾 ≠ 900 K2Cr2O7, H3BO3, (most unsymmetrical) CuSO4.5H2O Interstitial Sites In Closed Packed Structures Trigonal Void Tetrahedral Void Octahedral Void Cubic Void Octahedral Trigonal Void Void Trigonal void is formed Tetrahedral void Octahedral void is at the centre of three is formed by covering formed at the centre of spheres trigonal voids six spheres Cubic void  Solid State 𝑟 = 0.155𝑅 𝑟 = 0.225𝑅 𝑟 = 0.414𝑅 𝑟 = 0.732𝑅 𝑊ℎ𝑒𝑟𝑒, 𝑟 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓𝑣𝑜𝑖𝑑, 𝑅 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑙𝑜𝑠𝑒𝑙𝑦 𝑝𝑎𝑐𝑘𝑒𝑑 𝑠𝑝ℎ𝑒𝑟𝑒𝑠 Calculation Involving Unit Cell Dimensions Radius Ratio: Radius of cation 𝑟 Radius ratio = Radius of anion = 𝑟+ − Packing Fraction: 4 Volume of one sphere in the unit cell 𝑍 × 𝜋𝑟 3 Number of spheres/unit cell(𝑍) × 𝑓 = Volume of the unit cell (V) = 3 𝑎3 Density: 𝑍 ×𝑀 ρ = 𝑎3 × 𝑁 𝐴 Where, Z = no. of atoms/unit cell, M = atomic mass of the element, a = edge length of the unit cell, 𝑁𝐴 = Avogadro’s number Limiting Radius Ratio; Coordination Number & Geometry 𝑟+ ⁄𝑟− Coordination Number Geometry < 0.155 2 Linear 0.155 – 0.225 3 Plane triangular 0.225 – 0.414 4 Tetrahedral 0.414 – 0.732 6 Octahedral 0.732 – 1.000 8 Cubic (body centred) EXTERNAL FEATURES OF A CRYSTAL STRUCTURE Faces: Crystals are bounded by planar surfaces arranged Face A Face B in definite pattern called ‘faces’. Faces are of two types 900 900 named as like and unlike faces, with similar and different Interfacial angle faces respectively.  Forms: All faces corresponding to a crystal constitute a ‘form’. The form may be simple (in like faces) or Normal to Normal to combination (in unlike faces) type. Face B Face A  Edge and Solid Angle: The intersection of two faces Representation of Interfacial Angle The faces, edges and interfacial angles are constitute an edge and when three or more edges intersect related as, 𝑓 + 𝑐 = 𝑒 + 2 they form a solid angle. Where, 𝑓 = number of faces  Interfacial angle: The angle between the normals of the 𝑒 = number of edges two intersecting faces is called interfacial angle. 𝑐 = number of interfacial angles  Zone and Zone-axis: The faces of a crystal occur in sets. These sets are called zones. Each zone forms a complete belt around a crystal. A line drawn through the centre of crystal in a direction parallel to the edges of a zone is known as zone-axis. TYPES OF SYMMETRY IN CRYSTALS  All crystals of same substance possess the same element of symmetry.  A crystalline substance has the following three types of symmetry: (a) Centre of symmetry (b) Plane of symmetry and (c) Axis of symmetry  Solid State (a) Centre of symmetry: An imaginary point within a crystal through which if we draw any line, intersects the surface of crystal at equal distances in all directions. (b) Plane of symmetry: It is n imaginary plane passing through the crystal that divides the crystal structure into two equal half. In a crystal structure, two different types of plane of symmetry are present as shown in diagram. (c) Axis of symmetry: It is an imaginary line passing through the crystal, along which if a crystal is rotated, the same appearance is repeated more than once in a complete revolution. 3600  The number of times the appearance is repeated is given by 𝑛 = , 𝜃0 0 where, 𝜃 is the angle of rotation after which the 𝜃 crystal structure is repeated and the symmetry is called 𝑛 − 𝑓𝑜𝑙𝑑 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦.  Elements of Symmetry: The sum of the numbers of planes, axis and centre of symmetries possessed by a crystal is known as elements of symmetry of a crystal e.g., For cubic crystals, plane of symmetry = 3 + 6 = 9 Axis of symmetry = 3 + 4 + 6 = 13 Centre of symmetry = 1 Total elements symmetry = 9 + 13 + 1 = 23 Imperfections/Defects in Solids Point defects Line defects Deviations from ideal arrangement Deviations from ideal arrangement in around a point or an atom. entire rows of lattice points. Stoichiometric Impurity defects Non-stoichiometric defects /Intrinsic/Thermodynamic defects Arises when foreign atoms are Disturb stoichiometry of solid. Do not disturb stoichiometry present at lattice site. of solid. E.g., SrCl2, CdCl2, AgCl, etc. Vacancy defects Interstitial defects Frenkel defects Schottky defects Arises when lattice Arises when constituent Smaller ion (usually Arises when equal sites are vacant. particles occupy cation) dislocates number of cations & Decreases density, interstitial site. from its normal site to anions are missing. shown by non-ionic Increases an interstitial site. Decreases density, solids. The density, shown by Des not change shown by ionic concentration of non-ionic solids. density, shown by solids. vacancies increases Vacancy & interstitials ionic solids. E.g., NaCl, KCl, with increasing are inverse E.g., ZnS, AgCl, CsCl, AgBr, etc. temperature. phenomenon. AgBr, AgI, etc. Metal excess defects Metal deficiency defects Arises when metal shows variable valency. E.g., Fe2+, Fe3+. Due to ionic vacancies Due to the presence of A negative ion is missing from its extra cations lattice site, leaving a ‘hole’ which Usually arises when metal oxides are is occupied by an electron to heated, e.g., ZnO. maintain electrical neutrality. E.g., LiCl, NaCl, etc.  Solid State Properties Of Solids Some remarkable properties of solids are described below Electrical Properties of Solids Depending upon conduction power, solids are divided into the following types: 1. Conductors: The solids whose conductivity is in the order of 107 (Ω m) −1 are termed as conductors. Conductivity is due to the presence of free electrons. E.g., metals. 2. Semiconductors: The solids whose conductivity range from 10−6 to 104 (Ω m) −1 are called semiconductors. E.g., Semi-metals. (a) n-type semiconductors: Group 14 elements when doped with group 15 elements from n-type semiconductors. (b) p-type semiconductors: Group 14 elements when doped with group 13 elements form p-type semiconductors. 3. Insulators: The solids having conductivity range 10−20 to 10−10 (Ω m) −1 are known as insulators. E.g., Non-metals. 4. Super conductors: The conductors which allow electricity to pass through them without any resistant. The temperature at which a substance starts behaving as superconductor is called transition temperature. E.g., Ba2Cu3O7 at 90K. Dielectric Properties of Solids These are seen in insulators. Insulators show generation of dipoles in them, when they are placed in an electric field. There are the following terms related to dielectric properties of solids. 1. Piezoelectricity: It is produced on applying mechanical stress on polar crystals. E.g., Rochelle’s salt. 2. Pyroelectricity: It is produced either by heating or cooling some polar crystals. E.g., Cane sugar, Lithium sulphate. 3. Ferroelectricity: Piezoelectric crystals with permanent dipoles are said to have ferroelectricity. E.g., BaTiO3 , KH2PO4 , etc. 4. Anti-ferroelectricity: In anti-ferromegnetic crystals, the dipoles in alternate polyhedral points up & down in such a way that the crystal does not possess any net dipole moment. E.g., PbZrO3. Used in transistor, telephones & computers to prepare small size capacitor of high capacitance. Dielectric Properties of Solids Properties Description Alignment of Examples Application magnetic dipole Diamagnetic These are feebly repelled by the All electrons are TiO2 , V2O5 NaCl, Insulator magnetic fields due to the paired. C6H6 existence of paired electron. ↑↓↑↓↑↓ Paramagnetic Attracted by magnetic field due to At least one O2, Cu2+, Fe3+, TiO, Electronic the presence of unpaired electrons, unpaired electron. Ti2O3, VO, VO2, applications acts as a tiny bar magnet. ↑↓↑↑ CuO Ferromagnetic Acts as a permanent magnet even All unpaired Fe, Ni, Co, CrO2 CrO2 is used in after removing magnetic field. electrons are in audio & vedio Above curie temperature, no same direction. tapes. ferromagnetism occurs. ↑↑↑↑↑ Anti-ferromagnetic Net dipole moment becomes zero ↑↓↑↓↑↓ CoO, Co3O4, MnO, due to equal & opposite alignment. MnO2, M2O3, Ferromagnetic This arises when there is net dipole ↑↓↓↓↑↓ Fe3O4, Ferrites moment.  Solid State Allotropy  Allotropy is a phenomena, in which two or more different forms of an element exist in the same physical state.  It is a common phenomena shown by metals, metalloids & non-metals.  In the gaseous or liquid state, allotropes most often differ from one another in molecular formula while solid allotropes may differ from one another in their bonding pattern.  Diamond, graphite & Buckminster fullerene are allotropes of solids carbon.  White tin, the stable form of solid tin at 250C, shows metallic bonding & has properties we associate with metals, when it is kept at low temperatures for long periods of time, it crumbles to powdery grey tin, which has the covalent network structure of diamond.  Phosphorous forms a variety of allotropic forms in solid state. E.g., red phosphorous (PX, X = very large number), white phosphorous (P4).  Sulphur has two allotropes named as rhombic & monoclinic, both allotropes have same molecular formula (S8), but differ in crystal structure. Bragg’s Law According to this law, when a beam of monochromatic X-ray strikes two planes of atoms in a crystal at a certain angle ′𝜃′ & reflected, the intensity of the reflected beam is maximum when, 𝑛𝜆 = 2𝑑 𝑠𝑖𝑛𝜃 (for constructive interference) 𝑛𝜆 = 4𝑑 𝑠𝑖𝑛𝜃 (for destructive interference) When, n = order of different (𝑛𝜀𝐼 + ) When, n = 1 ⇒ first order diffraction n=2 ⇒ second order diffraction 𝑑 = distance between two planes 𝜆 = wavelength of X-rays 𝜃 = angle between X-rays & plane of crystal (angle of diffraction) Determination of structures of solids by X-ray crystallography  X-ray crystallography is a powerful method for determining the structure of a crystalline solid.  This technique relies on the fact that the spacing between atoms is similar in magnitude to the wavelength of X-rays.  The electron clouds of the atoms in the crystal scatter the X-rays, with the degree of scattering proportional to the atomic number Z.  In a crystal, the individual atoms, ions or molecules all lie in identical positions, so they scatter the X-rays in an identical manner.  In a modern instrument, the scattering leads to a diffraction pattern, i.e., observed as a series of spots on an image plate.  From this diffraction pattern, the electron density in the crystal can be reconstructed using a computer. This allows the bond length & bond angle data to be determined. Crystal Image plate NOTE ON GRAPHENE & ITS BIOLOGICAL APPLICATIONS  Solid State Graphene is an allotrope of carbon in the form of a two dimensional, atomic scale, hexagonal lattice in which one atom form each vertex. It is about 207 times stronger than steel by weight, it conducts heat & electricity efficiently & is nearly transparent.  Solid State 1. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB....... Any packing of spheres leaves out voids in the lattice. The percentage by volume of empty space of this is (a) 26% (b) 21% (c) 18% (d) 16 % Solution :(a) The hexagonal base consists of six equilateral triangles, each with side 2r and altitude 2r sin 60°. 1  Hence, area of base = 6  (2r) (2r sin 60 o )  6 3.r 2 2  The height of the hexagonal is twice the distance between closest packed layers. The latter can be determined to a face centred cubic lattice with unit cell length a. In such a 3a lattice, the distance between closest packed layers is one third of the body diagonal, i.e. , 3  3 a  2a Hence, Height (h)  2    3  3 Now, in the face centred lattice, atoms touch one another along the face diagonal, Thus, 4r  2. a 2  4r   2 With this, the height of hexagonal becomes : Height (h)     4 .r 3  2   3  4 2  Volume of hexagonal unit is, V  (base area)  (height)  (6 3 r 2 ) . r   24 2. r 3  3  In one hexagonal unit cell, there are 6 atoms as described below :  3 atoms in the central layer which exclusively belong to the unit cell.  1 atom from the centre of the base. There are two atoms of this type and each is shared between two hexagonal unit cells.  2 atoms from the corners. There are 12 such atoms and each is shared amongst six hexagonal unit cells. 4  Now, the volume occupied by atoms = 6  r 3  3  Volume occupied by atoms Fraction of volume occupied by atoms  Volume of hexagonal unit cell 4  6  r 3  3   3   / 3 2  0. 74. 24 2. r Fraction of empty space = (1.00  0.74 )  0.26 Percentage of empty space = 26% 2. Lithium borohydride (LiBH 4 ), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are : a = 6.81Å, b= 4.43Å, c=717Å. If the molar mass of LiBH 4 is 21.76 g mol 1. The density of the crystal is – (a) 0.668 g cm 3 (b) 0.585 g cm 2 (c) 1.23 g cm 3 (d) None 1 ZM 4  (21. 76 gmol ) Solution : (a)We know that,     0.668 g cm 3 N 0 V (6. 023  10 mol 1 )(6. 81  4. 43  7. 17  10  24 cm 3 ) 23 3. Atomic radius of silver is 144.5 pm. The unit cell of silver is a face centred cube. Calculate the density of silver. (a) 10.50 g/cm3 (b) 16.50 g/cm3 (c) 12.30 g/cm3 (d) 15.50 g/cm3  Solid State Solution : (a) For (fcc) unit cell, atoms touch each other along the face diagonal. a 2 Hence, Atomic radius (R)  4 4R 4 a   144.5 pm  408.70 pm  408.70  10 10 cm 2 2 ZM Density (D)  , V  a3 VN 0 ZM D= 3 ; where Z for (fcc) unit cell = 4 , Avagadro’s number (N 0 )  6.023  10 23 , Volume a N0 of cube ( V )  (408.70  10 10 ) 3 cm 3 and M (Mol. wt.) of silver = 108, 4  108 D   10.50 g / cm 3 (408.70  10 10 )3  6.023  10 23 4. A metallic elements exists as a cubic lattice. Each edge of the unit cell is 2.88Å. The density of the metal is 7.20 g cm 3. How many unit cells will be present in 100 gm of the metal. (a) 5.82  10 23 (b) 6.33  10 23 (c) 7.49  10 24 (d) 6.9  10 24 Solution : (a)The volume of unit cell (V) = a3 = (2.88Å)3 = 23.9  10 24 cm 3 Mass 100 Volume of 100 g of the metal =   13. 9 cm 2 Density 7. 20 13.9 cm 3 Number of unit cells in this volume =  5.82  10 23 23.9  10  24 cm 3 5. The diffraction of barium with X-radiation of wavelength 2.29Å gives a first – order reflection at 30°. What is the distance between the diffracted planes. (a) 3.29 Å (b) 4.39 Å (c) 2.29 Å (d) 6.29 Å Solution :(c) Using Bragg's equation 2d sin   n  n d , where d is the distance between two diffracted planes,  the angle to have 2 sin  maximum intensity of diffracted X-ray beam, n the order of reflection and  is the wavelength 1  2.29 Å  1  d  2.29 Å  sin 30   o 2  sin 30 o  2 6. When an electron in an excited Mo atom falls from L to the K shell, an X-ray is emitted. These X-rays are diffracted at angle of 7.75 o by planes with a separation of 2.64Å. What is the difference in energy between K-shell and L-shell in Mo assuming a first-order diffraction. (sin 7.75 o  0.1349 ) (a) 36.88 10 15 J (b) 27.88  10 16 J (c) 63.88  10 17 J (d) 64.88  10 16 J Solution : (b) 2d sin   n    2d sin  2  2.64  10 10  sin 7.75 o  0.7123  10 10 m hc 6.62  10 34  3  10 8 E   27.88  10 16 J  0.7123  10 10 7. CsBr has a (bcc) arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl. (a) 346.4 pm (b) 643 pm (c) 66.31 pm (d) 431.5 pm Solution:(a) The (bcc) structure of CsBr is given in figure The body diagonal AD  a 3 , where a is the length of edge of unit cell C On the basis of figure, AD  2(rCs   rCl  ) A B a 3 3 O a 3  2(rCs   rCl  ) or (rCs   rCl  )   400  2 2 D  200  1.732  346.4 pm  Solid State 8. Fraction of total volume occupied by atoms in a simple cube is  3 2  (a) (b) (c) (d) 2 8 6 6 Solution:(d) In a simple cubic system, number of atoms a = 2r 4 3 4 3 r r Volume occupied by one atom 3 3   Packing fraction     Volume of unit cell a3 (2r)3 6 9. A solid AB has the NaCl structure. If radius of cation A  is 120 pm, calculate the maximum possible value of the radius of the anion B  (a) 240 pm (b) 280 pm (c) 270 pm (d) 290 pm Solution:(d) We know that for the NaCl structure rA  radius of cation/radius of anion = 0.414;  0. 414 ; rB  rA  120 rB     290 pm 0.414 0.414 10. Silver crystallizes in a face centred cubic system, 0.408 nm along each edge. The density of silver is 10.6 g / cm 3 and the atomic mass is 107.9 g / mol. Calculate Avogadro's number. (a) 6.00  10 23 atom/mol (b) 9.31  10 23 atom/mol (c) 6.23  10 23 atom/mol (d) 9.61  10 23 atom/mol Solution:(a) The unit cell has a volume of (0.408  10 9 m )3  6.79  10 29 m 3 per unit cell and contains four atoms. The volume of 1 mole of silver is,  (1  10 2 m )3  5   1.02  10 m / mol ; 3 107.9 g / mol   10. 6 g  where 107.9 g/mol is the molecular mass of the silver The number of unit cells per mol. is,  1unit cell  1.02  10 5 m 3 / mol   29   1.50  10 unit cells per mol. 23  6.79  10 m  3  4 atoms   1.50  10 unit cell  23 and the number of atoms per mol. is,     6.00  10 atom/mol. 23  unit cell   mol  11. Silver metal crystallises in a cubic closest – packed arrangement with the edge of the unit cell having a length a  407 pm.. What is the radius of silver atom. (a) 143.9 pm (b) 15.6 pm (c) 11.59 pm (d) 13.61 pm Solution :(a) AC 2  AB 2  BC 2 here AC  AB  a, BC  4 r Cr a 2  a 2  (4 r) 2 2a 2  16 r 2 2r 407 pm a2  r  2 8 r A B a 407 407 pm  r   143.9 pm. 2 2 2 2  Solid State Multiple Choice Questions: 1. How many kinds of space lattices are possible in a crystal? (a) 23 (b) 7 (c) 230 (d) 14 2. If ‘a’ is the length of unit cell, then which one is correct relationship? 𝑎 (a) For simple cubic lattice, Radius of metal atom = 2 √3𝑎 (b) For bcc lattice, Radius of metal atom = 4 𝑎 (c) For fcc lattice, Radius of metal atom = 2√2 (d) All of these 3. Close packing is maximum in the crystal lattice is: (a) simple-cubic (b) face-centred (c) body-centred (d) none of these 4. CaF2 possesses: (a) face centred cubic (b) body centred cubic (c) simple cubic (d) hexagonal close packing 5. The yellow colour of ZnO & conducting nature produced in heating is due to: (a) metal excess defects due to interstitial cation (b) extra positive ions present in an interstitial site (c) trapped electrons (d) all of these 6. Which substance shows anti-ferromagnetism? (a) ZrO2 (b) CdO (c) CrO2 (d) Mn2O3 7. The melting point of RbBr is 6820C, while that of NaF is 9880C. The principle reason that melting point of NaF is much higher than that of RbBr, is that: (a) the two crystals are not isomorphous (b) the molar mass of NaF is smaller than that of RbBr (c) the inter-nuclear distance 𝑟𝑐 + 𝑟𝑎 is greater for RbBr than for NaF (d) the bond in RbBr has more covalent character than the bond in NaF 8. The contribution of particle at the edge of centre of a particular unit cell is: (a) 1/2 (b) 1/4 (c) 1 (d) 1/8  Solid State 9. If the unit cell of a mineral has cubic closed packed (ccp) array of oxygen atoms with ‘m’ fraction of octahedral holes occupied by aluminium ions &’n’ fraction of tetrahedral holes occupied by magnesium ions, ‘m’ & ‘n’, respectively 1 1 (a) , 2 8 1 (b) 1, 4 1 1 (c) 2 , 2 1 1 (d) 4 , 8 10. The composition of sample of wustite is Fe0.93 O1.00. The percentage of iron in the form of Fe(III) is: (a) 15% (b) 25% (c) 50% (d) none of these 11. Which of the following properties of a crystalline solid changes with change of direction? (a) refractive index (b) electrical conduction (c) electrical resistance (d) all of these 12. Iodine molecules are held in the crystal lattice by: (a) London force (b) dipole-dipole interactions (c) covalent bonds (d) coulombic forces 13. The solid NaCl is a bad conductor of electricity since,: (a) in solid NaCl, there are no ions (b) solid NaCl is covalent (c) in solid NaCl, there is no movement of ions (d) in solid NaCl, there are no electrons. 14. In which of the following substances, the carbon atom is arranged in a regular tetrahedral structure? (a) Diamond (b) Graphite (c) Fullerene (d) All of these 15. The geometry of the lattice can be obtained by joining the ________ with _________ lines. (a) lattice points, straight (b) lattice point, curved (c) unit cells, straight (d) unit cells, curved  Solid State 16. Calculate the distance ‘d’ between two successive planes when a first order diffraction of X-rays of wavelength 1.54 Å is reflected at an angle of 11.290. (a) 3.93 Å (b) 3.13 Å (c) 2.57 Å (d) 2.07 Å 17. The unit cell with crystallographic dimensions, 𝑎 ≠ 𝑏 ≠ 𝑐, 𝛼 = 𝛽 = 900 & 𝛾 = 900 is: (a) triclinic (b) monoclinic (c) orthorhombic (d) tetragonal 18. The crystal system of a compound with unit cell dimensions 𝑎 = 0.387, 𝑏 = 0.387 & 𝑐 = 0.504𝑛𝑚 & 𝛼 = 𝛽 = 900 & 𝛾 = 1200 is: (a) cubic (b) hexagonal (c) orthorhombic (d) rhombohedral 19. An element occurring in the bcc structure has 12.08 × 1023 unit cells. The total number of atoms of the element in these cells will be: (a) 24.16 × 1023 (b) 36.18 × 1023 (c) 6.04 × 1023 (d) 12.08 × 1023 20. A compound is formed by elements A & B. This crystallizes in the cubic structure when atoms A are at the corners of the cube & atoms B are at the centre of the body. The simplest formula of the compounds is (a) AB (b) AB2 (c) A2B (d) AB4 21. The packing efficiency in simple cubic unit cell is: (a) 52.4% (b) 68% (c) 74% (d) 80% 22. The edge lengths of the unit cells in terms of the radius of spheres constituting fcc, bcc & simple cubic unit cell are respectively: 4𝑟 (a) 2√2𝑟 , , 2𝑟 √3 4𝑟 (b) , 2√2𝑟 , 2𝑟 √3 4𝑟 (c) 2𝑟, 2√2𝑟 , √3 4𝑟 (d) 2𝑟 , ,2√2𝑟 √3  Solid State 23. TlCl crystallises in simple cubic lattice with density 7.00 𝑔 𝑐𝑚−3. The molecular weight of TlCl is 239.87 𝑔 𝑚𝑜𝑙 −1. The edge length of the unit cell is ______ × 10−8 𝑐𝑚. (a) 2.5 (b) 3.0 (c) 3.84 (d) 5.0 24. An element (atomic mass 100 g/mol) having bcc structure has unit cell edge 400 pm. Then, the density of the element is: (a) 10.376 g/cm3 (b) 5.189 g/cm3 (c) 7.289 g/cm3 (d) 2.144 g/cm3 25. Which of the following is true about point defects? (a) Due to valency defect, the density of substance decreases & due to interstitial defect, the density of the substance increases. (b) Due to valency defect, the density of substance increases & due to interstitial defect, the density of the substance decreases. (c) The density of the substance increases in both vacancy defect & interstitial defect. (d) The density of the substance decreases in both vacancy defect & interstitial defect. 26. In the solid lattice, the cation has left a lattice site & is located at an interstitial position. The lattice defect is: (a) interstitial defect (b) vacancy defect (c) Frankel defect (d) Schottky defect 27. Assertion: Insulators are poor conductor of electricity. Reason: According to band theory, electrons cannot move from valence band to conduction band. (a) Assertion & reason are true. Reason is correct explanation of assertion. (b) Assertion & reason are true. Reason is not the correct explanation of assertion. (c) Assertion is true, reason is false. (d) Assertion is false, reason is true. 28. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because: (a) all the domains get oriented in the direction of magnetic field. (b) all the domains get oriented in the direction opposite to the direction of magnetic field. (c) domains get oriented randomly. (d) domains are not affected by magnetic field. 29. How many chloride ions are there around sodium chloride crystal? (a) 3 (b) 8 (c) 4 (d) 6  Solid State 30. Match the items given in the column I with the item given in column II. Column I Column II (i) Aluminium solid state (A) Electrolytic conductor (ii) Germanium doped with Indium (B) n-type semiconductor (iii) Silicon doped with Bismuth (C) p-type semiconductor (iv) Al(NO3)3 in molten state (D) Metallic conductor (a) (i) – (C), (ii) – (A), (iii) – (B), (iv) – (D) (b) (i) – (D), (ii) – (C), (iii) – (B), (iv) – (A) (c) (i) – (A), (ii) – (C), (iii) – (B), (iv) – (D) (d) (i) – (B), (ii) – (D), (iii) – (A), (iv) – (C) 31. A hard, crystalline solid with a high melting point does not conduct electricity in any phase. The solid is most likely: (a) an ionic solid (b) a metallic solid (c) a molecular solid (d) a network covalent solid 32. The crystal system of a compound with unit cell dimensions 𝑎 = 0.387 𝑛𝑚, 𝑏 = 0.387 𝑛𝑚 & 𝑐 = 0.504 𝑛𝑚 & 𝛼 = 𝛽 = 900 & 𝛾 = 1200 is: (a) cubic (b) hexagonal (c) orthorhombic (d) rhombohedral 33. The density of KCl is 1.9893 𝑔𝑐𝑚−3 & the edge length of unit cell (𝑎) = 6.29082Å as determined by X-rays diffraction. The value of Avogadro’s number calculated from these data is (a) 6.021 × 1023 (b) 6.023 × 1023 (c) 6.03 × 1023 (d) 6.017 × 1019 34. An oxide of rhenium crystallises with eight rhenium atoms at the corners of the unit cell & 12 oxygen atoms on the edges between them. What is the formula of this oxide? (a) ReO (b) Re2O3 (c) ReO2 (d) ReO3 35. Platinum crystallises in a fcc (face centred cubic crystal) with a unit cell length of 3.9231Å. The density & atomic radius of platinum are respectively (a) 45.25 𝑔 𝑐𝑚−3 , 2.516Å (b) 21.45 𝑔 𝑐𝑚−3 , 1.387Å (c) 29.46 𝑔 𝑐𝑚−1 , 1.48Å (d) None of the above  Solid State 36. Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are (a) 6.022 × 1023 (b) 3.011 × 1023 (c) 9.034 × 1023 (d) 4.516 × 1023 37. Sodium metal crystallizes in a body centred cubic lattice with the cell length (𝑎) = 4.29Å. The radius of sodium atom (in Å) is (a) 1.8576 (b) 2.8576 (c) 3.8576 (d) None of these 38. Conduction in a p-type semiconductor is increased by (a) increasing the band gap (b) decreasing the temperature (c) adding appropriate electron deficient impurities (d) adding appropriate electron rich impurities 39. A non-stoichiometric compound Cu1.8S is formed due to incorporation of Cu2+ ions in the lattice of cuprous sulphide. What percentage of Cu2+ ion in the total copper content is present in the compound? (a) 88.88 (b) 11.11 (c) 99.8 (d) 89.8 40. If sphere of radius ‘r’ are arranged in a ccp fashion (ABCABC...), then vertical distance between any two consecutive ‘A’ layer is 2 (a) 4𝑟√3 3 (b) 4𝑟√2 (c) 6r (d) 𝑟√6 41. Which of the following statements is/are correct? (I) NaCl & KCl show metal excess defect due to anionic vacancies. (II) F-centres are the anionic sites occupied by unpaired electrons. (III) F-centres impart yellow colour to the NaCl crystals. (a) II & III (b) Only III (c) I & II (d) All of these  Solid State 42. A metal crystallizes into two cubic phases, face-centred cubic (fcc) & body-centred cubic (bcc) whose unit cell length are 3.5Å & 3.0Å, respectively. Calculate the ratio of densities of fcc to bcc. (a) 1.259 (b) 1.871 (c) 2.112 (d) 0.115 43. In which of the following crystal alternate tetrahedral voids are occupied? (a) NaCl (b) ZnS (c) CaF2 (d) Na2O 44. A compound MpXq has ccp arrangement of X. Its unit cell structure shown below. The empirical formula of the compound is M= X= (a) MX (b) MX2 (c) M2X (d) M5X14 45. For N-atoms in a crystal with Ni-interstitial position in its structure is ‘n’. If there are ‘n’ Frankel defects in the crystal. The value of ‘n’ should be 𝑁 1/2 (a) 𝑛 = ( 𝑁𝑖 ) 𝑒 𝐸𝑖 /2𝐾𝑇 𝑁 1/2 (b) 𝑛 = (𝑁 ) 𝑒 (𝐸𝑖 /2𝐾𝑇) 𝑖 𝑁 1/2 (c) 𝑛 = (𝑁 ) 𝑒 (𝐸𝑖 /2𝐾𝑇) 𝑖 𝑁 1/2 (d) 𝑛 = ( 𝑁𝑖 ) 𝑒 (𝐸𝑖 /2𝐾𝑇) 46. Which of the following defects shows decrease in the mass of the crystal lattice? (a) Frankel defect (b) Schottky defect (c) Edge dislocation (d) Constitution of F-centres.  Solid State 47. Among the following, which pair is considered to be ferromagnetic? (a) MgFe2O4 , ZnFe2O4 (b) H2O , NaCl (c) Fe, Co (d) Ni, MnO 48. The ionic radius of 𝐶𝑙 − ion is 1.81Å. The inter-ionic distances of NaCl & NaF are 2.79Å & 2.31Å respectively. The ionic radius of 𝐹 − ions will be (a) 0.78Å (b) 1.33Å (c) 0.23Å (d) 2.34Å 49. In fcc lattice of NaCl structure, if the diameter of Na+ is ‘x’ & the radius of 𝐶𝑙 − is ‘y’, then the edge length of NaCl in the crystal is (a) 3𝑥 + 2𝑦 𝑥 𝑦 (b) + 2 3 (c) 𝑥 + 2𝑦 1 (d) 2 (𝑥 + 𝑦) 50. Which of the following statements is/are correct about metals? (I) Valence band overlaps with conduction band. (II) The gap between valence band & conduction band is negligible. (III) The gap between these bands cannot be determined. (IV) Valence band may remain partially filled. (a) I, II & III (b) I, II & IV (c) II & III (d) Al of these  Solid State Answer set of MCQ Types Questions 1. (d) 2. (d) 3. (b) 4. (a) CaF2 has fcc structure with 8:4 co-ordination & has 4 units of CaF2 per unit cell. 5. (d) 6. (d) 7. (c) 8. (b) Each particle at edge centre is surrounded by 4 unit cells, thus its contribution to each unit cell is 1/4. 9. (a) For ccp lattice: Number of O-atoms = Z = 4 Number of octahedral voids = 4 Number of tetrahedral voids = 8 Number of Al3+ ions = 4 × 𝑚 Number of Mg2+ ions = 8 × 𝑚 ∴ The formula of mineral is Al4mMg8nO4. 4𝑚(+3) + 8𝑛(+2) + 4(−2) = 0 12𝑚 + 16𝑛 − 8 = 0 4(3𝑚 + 4𝑛 − 2) = 0 1 1 𝑚 =2 &𝑛 = 8 10. (a) Oxidation number of Fe in wustite = 200/93. (∴ charge of 100 oxide is balanced by 93 Fe2+ & Fe3+ ions) It is an intermediate value between two oxidation states of Fe as, Fe (II) & Fe (III). Let a% Fe ions are present as Fe(III). 200 Then 2 × (100 − 𝑎) + 3 × 𝑎 = 93 × 100  𝑎 = 15.05% 11. (d) 12. (a) Solid I2 is a non-polar molecular solid in which constituent particles are non-polar molecules held by weak dispersion forces or London forces. 13. (c) 14. (a) 15. (a) 16. (a) 𝑛 = 1, 𝜆 = 1.54Å, 𝜃 = 11.290 According to the Bragg’s equation, 𝑛𝜆 = 2𝑑 sin 𝜃 𝑛𝜆 1×1.54 1.54 𝑑 = 2 sin 𝜃 = 2×sin 11.290 = 2×0.1957 = 3.93 Å 17. (b) 18. (b) 19. (a) There are two atoms in a bcc unit cell. So, number of atoms in 12.08 × 1023 unit cells = 2 × 12.08 × 1023 = 24.16 × 1023 atoms.  Solid State 20. (a) ‘A’ atoms are at eight corners of the cube. 8 Therefore, the number of ‘A’ atoms in the unit cell = 1, 8 ‘B’ atoms are at the centre of the cube. Therefore, ‘B’ atoms per unit cell = 1. Hence, the formula is AB. 21. (a) 22. (a) 𝑍 ×𝑀 23. (c) Density = 𝑎3 × 𝑁 𝑔 𝑐𝑚−3 0 1 ×239.87 𝑔 𝑚𝑜𝑙−1 7.00 𝑔 𝑐𝑚−3 = 𝑎3 ×(6.022 × 1023 𝑚𝑜𝑙−1 ) 𝑎3 = 5.690 × 10−23 𝑎 = 3.84 × 10−8 𝑐𝑚 𝑍 ×𝑀 24. (b) Density = 𝑎3 × 𝑁 −30 𝑔 𝑐𝑚−3 0 × 10 2 ×100 (When ‘a’ is in pm) = (400)3 ×(6.022×1023)×10−30 = 5.189 𝑔/𝑐𝑚3 25. (a) 26. (c) Frankel defect is observed in ionic solids, when ions are displaced from their normal crystal sites & occupy interstitial sites. 27. (a) 28. (a) 29. (d) In the NaCl crystal, 𝐶𝑙 − ions from ccp arrangement & 𝑁𝑎 + ions occupy octahedral holes. Thus each 𝑁𝑎 + ion is surrounded by 6𝐶𝑙 − ions. 30. (b) 31. (d) a network covalent solid 32. (b) hexagonal 33. Density of KCl is 1.9893 𝑔𝑐𝑚−3 ; 𝑎 = 6.29082Å = 6.29082 × 10−8 𝑐𝑚 Molecular weight of KCl = 74.55 ; NA = ? Since, KCl crystallises in the NaCl type structure, it has fcc lattice. Hence, there are four atoms per unit cell. 𝑍 ×𝑀 𝑑 = 𝑎3 × 𝑁 𝐴 4 × 74.55 ⇒ 1.9893 = (6.29082)3 × 10−24 × 𝑁𝐴 23 NA = 6.021 × 10 34. (d) ReO3 𝑍 ×𝑀 4 × 195 35. 𝑑 = 𝑎3 × 𝑁 ⇒ 1.9893 = (3.9231 × 10−8 )3 × 6.023 × 10−24 = 21.45𝑔 𝑐𝑚−3 𝐴 𝑎 3.9231 𝑟 = 2√2 = = 1.387Å. 2√2  Solid State 36. Number of closed packed particles = 0.5 × 6.023 × 1023 ∴ Number of octahedral voids = 3.0115 × 1023 Number of tetrahedral voids = 2 × 3.0115 × 1023 = 6.023 × 1023 Therefore, total no. voids = 6.023 × 1023 + 3.0115 × 1023 = 9.0345 × 1023. 37. Radius of Na (for bcc lattice) √3 √3 ×4.29 = 𝑎 = = 1.8576Å 4 4 38. Adding electron deficient impurities creates an abundance of holes. These holes are majority carriers in p-type semiconductors & are responsible for conduction. 39. Let x Cu2+ & (1.8 – x)Cu+ ions are present in the compound Cu1.8S. Compound is electrically neutral. ∴ 2𝑥 + (1.8 − 𝑥) − 2 = 0 ⇒ 𝑥 = 0.2 0.2 %𝐶𝑢2+ = × 100 = 11.11 1.8 3 40. Because 𝑎√2 = 4𝑟 ⇒ 𝑎√3 = 4𝑟√. 2 41. (d) All of these. 42. fcc unit cell length = 3.5Å bcc unit cell length = 3.0Å 𝑍 × 𝑎𝑡.𝑤𝑡. 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑉 × 𝑁𝐴 𝑑𝑓𝑐𝑐 𝑍1 𝑉2 𝑑𝑏𝑐𝑐 =𝑍 ×𝑉 2 1 Now, Z1 for bcc = 4; Also, V1 = a3 = (3.5 × 10−8 )3 Z2 for bcc = 2; Also, V2 = a3 = (3.0 × 10−8 )3 3 𝑑𝑓𝑐𝑐 4 × (3.5×10−8 ) =2 × (3.0×10−8 )3 = 1.259 𝑑𝑏𝑐𝑐 43. (b) ZnS 44. (b) MX2 𝑁 1/2 45. (c) 𝑛 = (𝑁 ) 𝑒 (𝐸𝑖 /2𝐾𝑇) 𝑖 46. (b) Schottky defect. 47. (c) Fe, Co is a pair which is considered to be ferromagnetic. 48. (b) 1.33Å 49. (c) x + 2y 50. (b) I, II & IV.  Chapter – 2  SOLUTION & COLL