Chemistry Class 12 Syllabus PDF

 Syllabus CENTRAL BOARD OF SECONDARY EDUCATION, NEW DELHI CHEMISTRY (Theory) CLASS–XII Total Per...

 Syllabus CENTRAL BOARD OF SECONDARY EDUCATION, NEW DELHI CHEMISTRY (Theory) CLASS–XII Total Periods (Theory 160+ Practical 60) One paper Time: 3 hours 70 marks Unit Title No. of Periods Marks I Solid State 10 II Solutions 10 III Electrochemistry 12 23 IV Chemical Kinetics 10 V Surface Chemistry 08 VI General Principles and Processes of Isolation of Elements 08 VII p-Block Elements 12 19 VIII d- and f-Block Elements 12 IX Coordination Compounds 12 X Haloalkanes and Haloarenes 10 XI Alcohols, Phenols and Ethers 10 XII Aldehydes, Ketones and Carboxylic Acids 10 XIII Amines 10 28 XIV Biomolecules 12 XV Polymers 08 XVI Chemistry in Everyday Life 06 Total 160 70 Unit I: Solid State (10 Periods) Classification of solids based on different binding forces: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea). Unit cell in two dimensional and three dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects, electrical and magnetic properties. Band theory of metals, conductors, semiconductors and insulators and n and p type semiconductors. Unit II: Solutions (10 Periods) Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, Raoult’s law, colligative properties—relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, Van’t Hoff factor. Unit III: Electrochemistry (12 Periods) Redox reactions, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and law of electrolysis (elementary idea), dry cell-electrolytic cells and Galvanic cells, lead accumulator, fuel cells, corrosion. Unit IV: Chemical Kinetics (10 Periods) Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, no mathematical treatment), activation energy, Arrhenius equation. Unit V: Surface Chemistry (8 Periods) Adsorption–physisorption and chemisorption, factors affecting adsorption of gases on solids, catalysis: homogenous and heterogenous, activity and selectivity of solid catalysts; enzyme catalysis, colloidal state: distinction between true solutions, colloids and suspension; lyophilic, lyophobic, multi-molecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation, emulsion—types of emulsions. Unit VI: General Principles and Processes of Isolation of Elements (8 Periods) Principles and methods of extraction-concentration, oxidation, reduction—electrolytic method and refining; occurrence and principles of extraction of aluminium, copper, zinc and iron. Unit VII: p-Block Elements (12 Periods) Group 15 Elements: General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; Nitrogen preparation properties and uses; compounds of Nitrogen: preparation and properties of Ammonia and Nitric Acid, Oxides of Nitrogen (Structure only); Phosphorus- allotropic forms, compounds of Phosphorus: Preparation and properties of Phosphine, Halides and Oxoacids (elementary idea only). Group 16 Elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties, dioxygen: preparation, properties and uses, classification of Oxides, Ozone, Sulphur-allotropic forms; compounds of Sulphur: preparation properties and uses of Sulphur-dioxide, Sulphuric Acid: industrial process of manufacture, properties and uses; Oxoacids of Sulphur (Structures only). Group 17 Elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens, Preparation, properties and uses of Chlorine and Hydrochloric acid, interhalogen compounds, Oxoacids of halogens (structures only). Group 18 Elements: General introduction, electronic configuration, occurrence, trends in physical and chemical properties, uses. Unit VIII: d- and f-Block Elements (12 Periods) General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4. Lanthanoids—Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences. Actinoids—Electronic configuration, oxidation states and comparison with lanthanoids. Unit IX: Coordination Compounds (12 Periods) Coordination compounds—Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner’s theory, VBT, and CFT; structure and stereoisomerism, importance of coordination compounds (in qualitative analysis, extraction of metals and biological system). Unit X: Haloalkanes and Haloarenes (10 Periods) Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties, mechanism of substitution reactions, optical rotation. Haloarenes: Nature of C-X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only). Uses and environmental effects of—dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT. Unit XI: Alcohols, Phenols and Ethers (10 Periods) Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special reference to methanol and ethanol. Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols. Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses. Unit XII: Aldehydes, Ketones and Carboxylic Acids (10 Periods) Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses. Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties, uses. Unit XIII: Amines (10 Periods) Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines. Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry. Unit XIV: Biomolecules (12 Periods) Carbohydrates: Classification (aldoses and ketoses), monosaccahrides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates. Proteins: Elementary idea of—amino acids, peptide bond, polypeptides, proteins, structure of proteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones—Elementary idea excluding structure. Vitamins: Classification and functions. Nucleic Acids: DNA and RNA. Unit XV: Polymers (8 Periods) Classification—natural and synthetic, methods of polymerization (addition and condensation), copolymerization, some important polymers: natural and synthetic like polythene, nylon polyesters, bakelite, rubber. Biodegradable and non-biodegradable polymers. Unit XVI: Chemistry in Everyday Life (6 Periods) Chemicals in medicines: analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines. Chemicals in food: preservatives, artificial sweetening agents, elementary idea of antioxidants. Cleansing agents: soaps and detergents, cleansing action.  Design of Question Paper CHEMISTRY (Theory) CLASS–XII (2020-21) S. No. Domains Total Marks % 1. Remembering and Understanding: Exhibit memory of previously learned material by recalling facts, terms, basic concepts and answers. 28 40 Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions and stating main ideas. 2. Applying: Solve problems to new situations by applying 21 30 acquired knowledge, facts, techniques and rules in a different way. 3. Analysing, Evaluating and Creating: Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations. Present and defend opinions by making 21 30 judgments about information, validity of ideas or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions. Note: 1. No chapter wise weightage. Care to be taken to cover all the chapters. 2. Suitable internal variations may be made for generating various templates. Choice(s): OO There will be no overall choice in the question paper. OO However, 33% internal choices will be given in all the sections. The changes for classes XI-XII (2021-22) internal year-end/Board Examination are as under: Year-end (2020-21) (2021-22) Examination/Board Existing Modified Examination (Theory) OO  bjective type Questions including O OO Competency Based Questions will be 20% Multiple Choice Question-20% OO  hese can be in the form of Multiple- T OO Case-based/Source-based Choice Questions, Case-based Questions, Integrated Questions-10% Source Based Integrated Questions or Composition OO  hort Answer/Long Answer S any other types Questions- Remaining 70% OO Objective Questions will be 20 % OO emaining 60% Short Answer/Long R Answer Questions—(as per existing pattern) Part-A NCERT Textbook Questions Multiple Choice Questions Assertion-Reason Questions Passage-based/Case-based Questions Very Short Answer Questions Short Answer Questions–I Short Answer Questions–II Long Answer Questions Self-Assessment Test  Chapter–1 The Solid State The particles in solid are closely packed and held together by strong intermolecular forces. The building constituents have fixed positions and can only oscillate about their mean positions. They have definite shape and definite volume. The density of solids is high and they have low compressibility. 1. Classes of Solids: Two types of solids are known: B D (i) Amorphous solids, (ii) Crystalline solids. (i) Amorphous solids: In amorphous solids, the arrangement of building constituents is not regular but haphazard. They may have a short range order. Their melting points are not sharp. They are isotropic in nature, i.e., their properties such as mechanical strength, electrical conductivity, etc. are same in all directions. Examples: rubber, quartz glass, etc. (ii) Crystalline solids: In crystalline solids, the arrangement of C building constituents is regular throughout the entire three- A dimensional network. A crystalline solid has sharp melting point Fig. 1.1: Anisotropic behaviour and is anisotropic in nature i.e., some of their physical properties of crystal such as electrical resistance or refractive index show different values when measured along different directions in the same crystal. It has a definite geometrical shape with flat faces and sharp edges. Examples: sodium chloride, quartz, etc. Table 1.1: Distinction between Crystalline and Amorphous Solids Property Crystalline Solids Amorphous Solids Shape Definite characteristics and geometrical Irregular shape. shape. Melting point Melt at a sharp and characteristic Gradually soften over a range of temperature. temperature. Cleavage property When cut with a sharp edged tool, they When cut with a sharp edged tool, split into two pieces and the newly they cut into two pieces with irregular generated surfaces are plain and smooth. surfaces. Heat of fusion They have a definite and characteristic They do not have a definite heat of heat of fusion. fusion. Isotropy Anisotropic in nature. Isotropic in nature. Nature True solids. Pseudo solids or super cooled liquids. Order in arrangement of Long range order. Only short range order. constituent particles The Solid State 7  Table 1.2: Different Types of Solids Type of Solid Constituent Bonding/ Examples Physical Electrical Melting Particles Attractive Nature Conductivity Point Forces 1. Molecular solids (i) Non-polar Molecules Dispersion Ar, CCl4, H2, I2, Soft Insulator Very low or London CO2 forces (ii) Polar Dipole- HCl, SO2 Soft Insulator Low dipole interactions (iii) Hydrogen Hydrogen H2O (ice) Hard Insulator Low bonded bonding 2. Ionic solids Ions Coulombic NaCl, MgO, ZnS, Hard but Insulators in High or CaF2 brittle solid state but electrostatic conductors in molten state and in aqueous solutions 3. Metallic solids Positive Metallic Fe, Cu, Ag, Mg Hard but Conductors Fairly high ions in bonding malleable in solid state a sea of and ductile as well as in delocalised molten state electrons 4. Covalent or Atoms Covalent SiO2 (quartz), SiC, Hard Insulators Very high network solids bonding C (diamond), AlN, C (graphite) Soft Conductor (exception) 2. Space Lattice and Unit Cell Space Lattice: It is the three- dimensional arrangement of identical points in the space which represent how the constituent particles (atoms, c ions, molecules) are arranged in a β α b crystal. Each particle is depicted as a a γ point. Fig. 1.3: Illustration of Unit Cell: A unit cell is the smallest parameters of portion of a space lattice which, a unit cell Fig. 1.2: A portion of a three dimensional when repeated in different directions, cubic lattice and its unit cell generates the entire lattice. A unit cell is characterised by six parameters, i.e., axial angles a, b and g and axial lengths a, b and c. Thus, unit cell of a crystal possesses all the structural properties of a given crystal. 3. Crystal Systems: On the basis of the axial distances and the axial angles between the edges, the various crystals can be divided into seven systems. These are listed in Table 1.3. Table 1.3: Seven Primitive Unit Cells and their Possible Variations as Centred Unit Cells Crystal Possible variations Axial distances or Axial angles Examples system edge lengths Cubic Primitive, Body-centred, a=b=c a = b = g = 90° NaCl, Zinc blende, Cu, Face-centred KCl, Diamond Tetragonal Primitive, Body-centred a=b≠c a = b = g = 90° White tin, SnO2, TiO2, CaSO4 8 Xam idea Chemistry–XII  Orthorhombic Primitive, Body-centred, a≠b≠c a = b = g = 90° Rhombic sulphur, KNO3, or Rhombic Face-centred, End- BaSO4 centred Hexagonal Primitive a=b≠c a = b = 90°, Graphite, ZnO, CdS, Mg, g = 120° Zn Rhombohedral Primitive a=b=c a = b = g ≠ 90° Calcite (CaCO3) HgS or Trigonal (cinnabar), ICl, As, Sb, Bi Monoclinic Primitive, End-centred a≠b≠c a = g = 90°, Monoclinic sulphur, b ≠ 90° PbCrO4, Na2SO4.10H2O Triclinic Primitive a≠b≠c a ≠ b ≠ g ≠ 90° K2Cr2O7, CuSO4.5H2O, H3BO3 There can be 14 different ways in which similar points can be arranged in a three-dimensional space. These are called Bravais lattices. a a a Primitive Body-centred Face-centred (or simple) The three cubic lattices: all sides of same length, angles between faces all 90° Primitive Body-centred The two tetragonal: one side different in length from the other, two angles between faces all 90° Primitive End-centred Body-centred Face-centred The four orthorhombic lattices: unequal sides, angles between faces all 90° More than 90° Less than 90° Primitive End-centred The two monoclinic lattices: unequal sides, two faces have angles other than 90° The Solid State 9  60° a less than 90° a a a Hexagonal lattice: one side different in Rhombohedral lattice: length to the other two, all sides of equal the marked angles on length, angles on two two faces are 60° and faces are less than 90° the remaining angle is 120°. a b A Triclinic lattice: B C unequal sides a, b, c, A, B, C are unequal c angles with none equal to 90° Fig. 1.4: Unit Cells of 14 Types of Bravais Lattices 4. Number of Atoms in a Unit Cell (i) An atom lying at the corner of a unit cell is shared equally by eight unit cells and therefore, only one- eighth (1/8) of an atom belongs to the given unit cell. (ii) An atom present on an edge is distributed among the four unit cells, therefore only one-fourth (1/4) of an atom belongs to the given unit cell. (iii) A face-centred atom is shared between two adjacent unit cells. Therefore, one-half (1/2) of an atom lies in each unit cell. (iv) A body-centred atom belongs entirely to one unit cell since it is not shared by any other unit cell. Therefore, its contribution to the unit cell is one. Applying above stated points, let us calculate the number of atoms in the different cubic unit cells. 1 Simple cubic: 8 (corner atoms) × atom per unit cell = 1 atom 8 1 Body-centred cubic: 8 (corner atoms) × atom per unit cell + 1 (body centre atom) × 1 atom per unit cell 8   = 1 + 1 = 2 1 Face-centred cubic: 8 (corner atoms) × atom per unit cell 8 1 + 6 (face atoms) × atom per unit cell = 1 + 3 = 4 2 1/8 Atom 1/8 Atom 1/8 Atom 1 Atom 1/2 Atom Simple cubic Body-centred cubic Face-centred cubic Fig. 1.5: Space-filling structures and actual portions of atoms belonging to one unit cell 10 Xam idea Chemistry–XII  Table 1.4: Number of Atoms per Unit Cell Type of cell Number of atoms Number of atoms Number of atoms Total at corners in faces in the body of cube Simple or primitive cubic 1 0 0 1 8× =1 8 Body-centred cubic (bcc) 1 0 1×1=1 2 8× =1 8 Face-centred cubic (fcc) 1 1 0 4 8× =1 6× =3 8 2 5. Density of Unit Cell: Suppose edge of a unit cell of a cubic crystal is a, d is the density of the substance and M is the molar mass, then in case of cubic crystal, Mass of unit cell = Number of atoms in unit cell × Mass of each atom = z × m Molar mass Mass of each atom (m) = Avogadro number M m = NA z#M Mass of unit cell = NA Volume of a unit cell = a3 Therefore, density of the unit cell, Mass of unit cell d= Volume of unit cell z×M d= , where d is in g/cm3 and a is in cm. a3 × N A 6. Other Parameters of a Cubic System (a) Atomic radius: It is defined as half of the distance between nearest neighbouring atom in a crystal. It is expressed in terms of length of the edge (a) of unit cell of the crystal. a (i) Simple cubic structure (sc): Radius of atom ‘r’ = , as atoms touch each other along the edges. 2 3a (ii) Body-centred cubic structure (bcc): Radius of atom ‘r’ = , as the atoms touch each other 4 along the cross diagonal of the cube. a (iii) Face-centred cubic structure (fcc): Radius of atom ‘r’ = , as the atoms touch each other 2 2 along the face diagonal of the cube. (b) Coordination number: It is defined as the number of nearest neighbours that a particle has in a unit cell. It depends upon the structure of unit cell of the crystal. (i) Simple cubic structure (sc): Coordination number (C.N.) = 6 (ii) Body-centred cubic structure (bcc): C.N. = 8 (iii) Face-centred cubic structure (fcc): C.N. = 12 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: Let ‘a’ be the cube edge and ‘r’ the atomic radius. The Solid State 11  As the particles touch each other along the edge, therefore a = 2r Volume of the unit cell = a3 Since one atom is present in a unit cell, its volume G B rr = r b l = 4 3 4 a 3ra 3 H A v = 3 3 2 6 v ra 3 6 F \ Packing efficiency = # 100 = # 100 C V a3 E D = r # 100 = 3.14 # 100 6 6 Fig. 1.6: Simple cubic unit cell. The spheres = 52.36% = 52.4% are in contact with each other along Therefore, 52.4% of unit cell is occupied by atoms and the the edge of the cube. rest 47.6% is empty space. (b) Packing efficiency in ccp and hcp structures: The efficiencies of both types of packing, ccp and 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 the unit cell length be ‘a’ and face diagonal be ‘b’ (represented as AC in Fig. 1.7). In this figure other sides are not shown for the sake of clarity. In triangle ABC, ∠ABC is 90°, therefore, G B AC2 = b2 = BC2 + AB2 A H = a2 + a2 = 2a2 \ b = 2a b If r is the radius of the sphere, we find b = 4r = 2 a F C 4r a or a= = 2 2r or, r= 2 2 2 E D As ccp structure has 4 atoms per unit cell, therefore the total volume Fig. 1.7: Cubic close packing 4 of 4 spheres (v) is = 4 × rr3 3 Total volume of the unit cell (V) = a3 = (2 2 r) 3 v Packing efficiency = × 100 V 4 # (4/3) # rr3 = # 100 (2 2 r ) 3 (16/3) # rr3 r = # 100 = # 100 = 74% 16 # 2 r 3 3 2 Therefore, 74% of unit cell is occupied by atoms and the rest 26% is empty space. (c) Efficiency of packing in bcc structures: In this case the atom at the centre is in touch with other two atoms which are diagonally arranged (see Fig. 1.8). The spheres along the body diagonal are shown with solid boundaries. In DEFD, b2 = a2 + a2 = 2a2 \ b = 2a In DAFD, c2 = a2 + b2 = a2 + 2a2 = 3a2 \ c= 3a 12 Xam idea Chemistry–XII  The length of the body diagonal c is equal to 4r, r being the radius of G B the sphere (atom). As all the three spheres along the diagonal touch each other, A H c = 4r Therefore, c = 4r = 3a c a 4r 3 a= or r = a F C 3 4 b a As already calculated, the total number of atoms associated with a E a D bcc unit cell is 2, the volume (v) is, therefore, 4 8 Fig. 1.8: Body-centred cubic unit cell          2 × rr3 = rr3 3 3 4r 3 64r3 Volume of the unit cell (V) = a3 = e o = 3 3 3 v (8/3) rr3 3 Packing efficiency = # 100 = # 100 = r # 100 = 68% V (64/3 3 ) # r 3 8 Therefore, 68% of unit cell is occupied by atoms and the rest 32% is empty space. 8. Close Packing of Constituents (a) Close packing in one dimension There is only one way of arranging spheres in a one-dimensional close packed structure, that is to arrange them in a row and touching each other. In one-dimensional close packed arrangement, the coordination number is 2. Fig. 1.9: Close packing of spheres in one dimension (b) Close packing in two dimensions Two-dimensional close packed structure can be generated by stacking (placing) the rows of close packed spheres. This can be done in two different ways as shown in Figs. 1.10(a) and (b). (i) Square close packing [Fig. 1.10(a)] (ii) Hexagonal close packing [Fig. 1.10(b)] A B A A A B A A Fig. 1.10(a): AAA..... type arrangement, C.N. = 4, Fig. 1.10(b): ABAB.... type arrangement, C.N. = 6, Square close packing in 2-D Hexagonal close packing in 2-D (c) Close packing in three dimensions (i) Hexagonal close packing (hcp): The first layer is formed utilizing maximum space, thus wasting minimum space. In every second row the particles occupy the depressions (also called voids) between the particles of the first row (Fig. 1.11). In the third row, the particles are vertically aligned with those in the first row giving AB AB AB... arrangement. This structure has hexagonal symmetry and is known as hexagonal close packing (hcp). This packing is more efficient and leaves small space which is unoccupied by spheres. In hcp arrangement, the coordination number is 12 and only 26% space is free. A single unit cell has 4 atoms. The Solid State 13  2-fold axis A B A B A     (a) (b) (c) Fig. 1.11: Hexagonal close packing (hcp) of spheres in 3-D (ii) Cubic close packing (ccp): Again, if we start with hexagonal layer of spheres and second layer of spheres is arranged by placing the spheres over the voids of the first layer, half of these holes can be filled by these spheres. Presume that spheres in the third layer are arranged to cover octahedral holes. This arrangement leaves third layer not resembling with either first or second layer, but fourth layer is similar to first, fifth layer to second, sixth to third and so on giving pattern ABCABCABC.... This arrangement has cubic symmetry and is known as cubic closed packed (ccp) arrangement. This is also called face-centred cubic (fcc) arrangement [Fig. 1.12(a) and (b)]. 3-fold axis A C B A (a) (b ) (c) Fig. 1.12: Cubic close packing (ccp) of spheres in 3-D The free space available in this packing is 26% and coordination number is 12. 9. Voids or holes: The empty spaces left between closed packed spheres are called voids or holes. Fig. 1.13: Tetrahedral (T) and octahedral (O) voids Voids are of three types: (a) Octahedral voids: This void is surrounded by six spheres and formed by a combination of two triangular voids of the first and second layer. There is one octahedral void per atom in a crystal. The rvoid radius ratio e o is 0.414. rsphere 14 Xam idea Chemistry–XII  Tetrahedral hole Tetrahedral hole Tetrahedron Octahedral Octahedral hole hole Octahedron (a) (b) (c ) Fig. 1.14: Tetrahedral and octahedral voids (a) top view, (b) exploded side view, and (c) geometrical shape of the void (b) Tetrahedral voids: These voids are surrounded by four spheres which lie at the vertices of a regular tetrahedron. There are 2 tetrahedral voids per atom in a crystal and the radius ratio is 0.225. (c) Trigonal voids: The void, enclosed by three spheres in contact is called a trigonal void. There are 8 trigonal voids per atom in crystal and the radius ratio is 0.155. Fig. 1.15: Trigonal void 10. Locating Tetrahedral and Octahedral Voids: All closed packed structures have both octahedral and tetrahedral voids. In a ccp pattern, there is one octahedral void at the centre of body and 12 octahedral voids on each of the 12 edges of the cube. Each void on the edge is shared by four other unit cells. Octahedral void at centre of cube = 1 1 Effective number of voids at edges = 12 × =3 4 Total number of octahedral voids = 1 + 3 = 4 In ccp structure, there are 8 tetrahedral voids. These are located at the body diagonals, two on each body diagonal at one-fourth of the distance from each end. 11. Radius Ratio: For ionic solids, the ratio of the radius of cation to that of anion is called radius ratio. Radius of the cation r + Radius ratio = = – Radius of the anion r 12. Crystal Defects: The defects are basically irregularities in the arrangement of constituent particles. Broadly, crystal defects are of two types, namely, point defects and line defects. Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance, whereas the line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points. These irregularities are called crystal defects. 13. Point Defects Interstitials: Atoms or ions which normally occupy 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, (B) Impurity defects, and (C) Non-stoichiometric defects. (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 and interstitial defects. The Solid State 15  (a) Vacancy defect: When some of the lattice sites are vacant, the crystal is said to have vacancy defect. It results in decrease in density of the substance. This defect can arise when a substance is heated. (b) 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 and interstitial defects are generally shown by non-ionic solids because ionic solids must always maintain electrical neutrality. Ionic solids show these defects as Schottky and Frenkel defects as explained below: (i) Schottky defect: This defect arises when equal number of cations and anions are missing from the lattice. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect, density of crystal decreases and it begins to conduct electricity to a smaller extent [Fig. 1.16(a)]. (ii) Frenkel 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 and there is no change in overall chemical composition of the crystal [Fig. 1.16(b)]. A+ B– A+ A+ A+ B– A+ B– A+ B– A+ B– A+ B– B– B– A+ B– A+ B– B– A+ A+ B– A+ B– A+ A+ – – – – – B A+ B A+ B B A + B A+ B–    Fig. 1.16(a): Schottky defect Fig. 1.16(b): Frenkel defect (B) Impurity Defects: These defects arise when foreign + Cl– Na+ Cl– atoms or ions are present in the lattice site (substitutional Na solid solutions) or in the interstitial sites (interstitial solid solutions). For example, when molten NaCl containing Cl– Sr2+ Cl– Na+ 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 and the other site Na+ Cl– Cl– remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions. (C) Non-stoichiometric Defects: These defects arise when Cl– Na+ Cl– Na+ stoichiometry of a substance is disturbed. These are of two types. Fig. 1.17: Introduction of cation vacancy in (a) Metal excess defect: This may occur in either of the NaCl by substitution of Na+ by Sr2+. following two ways: (i) 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 which is occupied by the electron originally associated with the anion. In this way crystal remains neutral. Alkali halides like NaCl and KCl show this type of defect. °° 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 and 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. 16 Xam idea Chemistry–XII  (ii) Metal excess defect due to interstitial cation: In this defect an extra positive ion occupies interstitial position in the lattice and 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 and turns yellow. Heating 1 ZnO Zn2+ + O + 2e– 2 2 The excess of Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites. (b) 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 and 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– B– B– A2+ A+ B– A+ B– B– A+ B– A+ Fig. 1.18: Metal deficiency defect due to missing cations 14. (a) 13-15 Compounds: When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc. (b) 12-16 Compounds: Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character. 15. Magnetic Moments: The magnetic properties Magnetic Magnetic of substances result from their magnetic moment moment moments associated with individual electrons. Electron Each electron has a magnetic moment, origin of which lies in two sources. It is a known fact + Electron that an electron shows two types of motions, i.e., it rotates (spins) around its own axis and Atomic nucleus Direction of spin simultaneously revolves around the nucleus (a ) (b) (orbital motion). An electron in motion is just Fig. 1.19: Magnetic moment associated with (a) orbital like a small current loop. Two types of motions motion and (b) spin motion of an electron give rise to two types of magnetic moments— spin and orbital magnetic moments. Spin moment is directed along the spin axis and is shown in up or down direction [Fig. 1.19(b)]. Orbital motion also generates a magnetic field and thus gives rise to orbital moment along its axis of rotation [Fig. 1.19(a)]. In this way each electron of the atom behaves like a small bar magnet having permanent orbital and spin magnetic moments. Magnetic moments are measured in Bohr magneton (µB) unit (B.M.). eh 1 B.M. = = 9.27 × 10–24 Am2 or 9.27 × 10–21 erg/gauss 4r mc where e is charge on electron; h is Planck’s constant; m is the mass of electron and c is the velocity of light. Depending upon two spin motions (clockwise and anticlockwise), spin magnetic moment may acquire two values ! MB. Contribution of the orbital magnetic moment is equal to ML. MB, where ML is magnetic quantum number of electron. The Solid State 17  16. Magnetic Properties of Solids: On the basis of their magnetic properties, substances can be classified into five categories. (a) Diamagnetic: Diamagnetic substances are weakly repelled by the external magnetic field. The atoms of these substances have all paired up electrons. As pairing of electrons cancel their magnetic moments, they lose their magnetic character. NaCl, H2O, TiO2 and C6H6 are some examples of diamagnetic substances. (b) Paramagnetic: Paramagnetic substances are weakly attracted by the external magnetic field. The atoms of these substances have one or more unpaired electrons. Paramagnetism is temporary and is present as long as external magnetic field is present. O2, Fe3+, Cr3+, TiO, VO2, Cu2+ are some examples of paramagnetic substances. (c) Ferromagnetic: Ferromagnetic substances are strongly attracted by the external magnetic field. In solid state, the metal ions of these substances are grouped together into small regions called domains. Each domain acts as a tiny magnet. When such a substance is placed in a magnetic field all the domains get oriented in the direction of magnetic field and a strong magnetic effect is produced. This ordering of domains persists even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet. Thus, besides strong attractions, these substances can be permanently magnetised. (d) Antiferromagnetic: These substances have domain structure similar to that of ferromagnetic substances but their domains are oppositely oriented and cancel out each other’s magnetic moment. MnO is an antiferromagnetic substance. (e) Ferrimagnetic: In ferrimagnetic substances due to unequal number of magnetic moment in parallel and antiparallel directions, the net magnetic moment is small. These substances lose ferrimagnetism on heating and become paramagnetic. Fe3O4 and ferrites like MgFe2O4 and ZnFe2O4 are examples of such substances. (a) Ferromagnetic (b) Antiferromagnetic (c) Ferrimagnetic Fig. 1.20: Schematic alignment of magnetic moments °° Curie Temperature: The temperature at which a ferromagnetic substance loses its ferromagnetism and attains paramagnetism only is called curie temperature. For iron, the curie temperature is 1033 K, for Ni it is 629 K and for Fe3O4 it is 850 K. Below this temperature paramagnetic substances behave as ferromagnetic substances. 17. Electrical Properties: Solids are classified into three groups on the basis of their electrical conductivities: (a) Conductors: These generally include metals. Their conductivity is of the order of 104–107 ohm–1 m–1. (b) Semiconductors: Those solids which have intermediate conductivities ranging from 10–6 to 104 ohm–1 m –1 are classified as semiconductors. As the temperature rises there is a rise in conductivity because electrons from the valence band jump to conduction band. (c) Insulators: These are solids which have very low conductivity values ranging from 10–20 to 10–10 ohm–1 m–1. °° Causes of conductance in solids: In most of the solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic compounds, it is the ions that are responsible for the conducting behaviour due to their movement. In metals, conductivity strongly depends upon the number of valence electrons available in an atom. A band is formed due to closeness of molecular orbitals which are formed from atomic orbitals. If this band is partially filled or it overlaps the higher energy unoccupied conduction band, the electrons can flow easily under applied electric field and the solid behaves as conductor [Fig. 1.21(a)]. If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator. [Fig. 1.21(b)] 18 Xam idea Chemistry–XII  Conduction band Empty band Empty band Forbidden zone Small energy gap (Large energy gap) Energy Filled band Partially Overlapping filled bands band (a) Metal (b) Insulator (c) Semiconductor Fig. 1.21: Distinction among metals, insulators and semiconductors If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor [Fig. 1.21(c)]. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump from valence to conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. (d) Doping: It is a process by which impurity is introduced in semiconductors to enhance their conductivity. °° n-type semiconductor: When silicon or germanium crystal is doped with a Group 15 element like P or As, the dopant atom forms four covalent bonds like a Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and contributes its share towards electrical conduction. Thus, silicon or germanium doped with P or As is called n-type semiconductor, n indicates negative charge of electron since it is the electron that conducts electricity [Fig. 1.22(b)]. °° p-type semiconductor: When silicon or germanium is doped with a group 13 element like B or Al, the dopant atom forms three covalent bonds, but at the place of fourth electron a hole is created. This hole moves through the crystal like a positive charge giving rise to electrical conductivity. Thus, Si or Ge doped with B or Al is called p-type semiconductor ( p stands for positive hole), since it is the positive hole that is responsible for conduction [Fig. 1.22(c)]. Silicon atom Mobile electron Positive hole (no electron) As B (a) Perfect crystal (b) n-type (c) p-type Fig. 1.22: Creation of n-type and p-type semiconductors °° Diode: Diodes are made by the combination of n-type and p-type semiconductors. They are used as rectifiers. °° Transistors: These are used to detect or amplify radio or audio signals. They consist of pnp or npn sandwich semiconductors. °° Photodiode: These are diodes which are capable of converting light energy into electrical energy and are used in solar cells. The Solid State 19  18. Structure of Some Ionic Solids (a) Ionic solids of the type AB (i) NaCl (fcc): Cl– in ccp, Na+ occupy all the octahedral voids. Coordination number is 6 : 6 and r+ = 0.52 r– Number of formula units per unit cell = 4. (ii) CsCl (bcc): Cl– ions at the corners of a cube, Cs+ ion at the body centre and vice-versa. Coordination number is 8 : 8 and r+ = 0.93 r– Number of formula units per unit cell = 1. (iii) ZnS (Zinc blende): S2– ions form ccp structure and Zn2+ ions occupy alternate tetrahedral voids, coordination number is 4 : 4 and r+ = 0.4 r– Number of formula units per unit cell = 4. (iv) ZnS (Wurtzite): S2– ions form hcp structure and Zn2+ ions occupy alternate tetrahedral voids. Coordination number is 4 : 4 and r+ = 0.4 r– Number of formula units per unit cell = 4. (b) Ionic solid of the type AB2 CaF2 (Fluorite): Ca2+ ions form ccp structure and F – ions occupy all tetrahedral voids. Coordination number is 8 : 4 and r+ = 0.73 r– Number of formula units per unit cell = 4. (c) Ionic solid of the type A2B Na2O (Antifluorite structure): O2– ions form ccp structure and Na+ ions occupy all tetrahedral voids. Coordination number is 4 : 8. Important Formulae Mass of unit cell z#M 1. Density of unit cell (d) = = Volume of unit cell a3 # N A 2. Table 1.5: Different Parameters of Cubic System Unit cell No. of atoms per Distance between C.N. Radius unit cell nearest neighbour (r) (d) a Simple cubic 1 a 6 2 a a Face-centred cubic 4 12 2 2 2 Body-centred cubic 2 3 8 3 a a 2 4 20 Xam idea Chemistry–XII  Volume occupied by atoms in unit cell (v) 3. Packing efficiency = # 100 Total volume of the unit cell (V) Table 1.6: Packing efficiency of different crystals S.No. Crystal system Packing efficiency (i) Simple cubic 52.4% (ii) Body-centred cubic 68% (iii) Face-centred cubic 74% (iv) Hexagonal close-packed 74% Radius of the cation r + 4. Radius ratio = = – Radius of the anion r Table 1.7: Structural arrangement of different radius ratios of ionic solids Radius ratio Possible coordination Structural Examples (r+/r–) number arrangement    0.155 – 0.225 3   Trigonal planar B2O3    0.225 – 0.414 4   Tetrahedral ZnS, SiO4– 4    0.414 – 0.732 6   Octahedral NaCl    0.732 – 1.0 8   Body-centred cubic CsCl 5. If R is the radius of the spheres in the close packed arrangement, then (i) Radius of octahedral void, r = 0.414 R (ii) Radius of tetrahedral void, r = 0.225 R 6. In a close packed arrangement: (i) Number of octahedral voids = Number of atoms present in the close packed arrangement. (ii) Number of tetrahedral voids = 2 × Number of atoms present in the close packed arrangement. NCERT Textbook Questions NCERT Intext Questions Q. 1. Why are solids rigid? Ans. In solids, the constituent particles (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. This imparts rigidity. Q. 2. Why do solids have a definite volume? Ans. The constituent particles in solids are bound to their mean positions by strong forces of attraction. The interparticle distances remain unchanged at a given temperature and thus solids have a definite volume. Q. 3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper. Ans. Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride and fibre glass. Crystalline solids: Naphthalene, benzoic acid, potassium nitrate and copper. Q. 4. Why is glass considered a super cooled liquid? Ans. Glass is an amorphous solid. Like liquids, it has a tendency to flow, though very slowly. This is evident from the fact that the glass panes in the windows of old buildings are invariably found to be slightly thicker at the bottom than at the top. This is because the glass flows down very slowly and makes the bottom portion slightly thicker. The Solid State 21  Q. 5. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property? Ans. 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 cleavage when cut with a knife. Instead, it would break into pieces with irregular surfaces. Q. 6. Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide. Ans. Ionic solids: Potassium sulphate, zinc sulphide. Covalent solids: Graphite, silicon carbide. Molecular solids: Benzene, urea, ammonia, water, argon. Metallic solids: Rubidium, tin. Q. 7. Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it? Ans. Covalent. Q. 8. Ionic solids conduct electricity in molten state but not in solid state. Explain. Ans. In the molten state, ionic solids ionise to give free ions and hence can conduct electricity. However, in the solid state, since the ions are not free to move about but remain held together by strong electrostatic forces of attraction, they behave as insulators. Q. 9. What type of solids are electrical conductors, malleable and ductile? Ans. Metallic solids. Q. 10. Give the significance of a ‘lattice point’. Ans. Each lattice point represents one constituent particle of the solid. This constituent particle may be an atom, an ion, or a molecule. Q. 11. Name the parameters that characterise a unit cell. Ans. A unit cell is characterised by (i) its dimensions along the three edges, a, b and c. (ii) angles between the edges, which are a (between b and c), b (between a and c) and g (between a and b). Thus, a unit cell is characterised by six parameters, a, b, c, a, b and g. Q. 12. Distinguish between (i) Hexagonal and monoclinic unit cells (ii) Face-centred and end-centred unit cells Ans. (i) Hexagonal and monoclinic unit cells System Hexagonal Monoclinic γ = 120° β ≠ 90° b c α = 90° α = 90° β = 90° γ = 90° c b a a No. of space lattices 1 2 Possible variations Primitive Primitive, End-centred Axial distances a=b≠c a≠b≠c Axial angles a = b = 90°, g = 120° a = g = 90°, b ≠ 90° Examples Graphite, ZnO, CdS Na2SO4.10H2O, monoclinic sulphur 22 Xam idea Chemistry–XII  (ii) Face-centred and end-centred unit cells Unit cell Face-centred End-centred Position of lattice points At the corners and at the centre of At the corners and at the centres of each face two end faces 1 1 1 1 No. of atoms per unit cell 8× +6× =4 8× +2× =2 8 2 8 2 Q. 13. Explain how much portion of an atom is located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell. 1 Ans. (i) A point lying at the corner of a unit cell is shared equally by eight unit cells and therefore, only of 8 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. 14. What is the two dimensional coordination number of a molecule in square close packed layer? [CBSE (F) 2013] Ans. 4. Q. 15. 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 tetrahedral voids? Ans. No. of atoms in the 0.5 mol close-packed structure = 0.5 × 6.022 × 1023 = 3.011 × 1023 No. of octahedral voids = 1 × No. of atoms in the close-packed structure = 3.011 × 1023 No. of tetrahedral voids = 2 × No. of atoms in the close-packed structure = 2 × 3.011 × 1023        = 6.022 × 1023 23 23 23 Total number of voids = 3.011 × 10 + 6.022 × 10 = 9.033 × 10 Q. 16. A compound is formed by two elements, M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound? Ans. Suppose atoms of element N present in ccp = x \ Number of tetrahedral voids = 2x 1 Since rd of the tetrahedral voids are occupied by atoms of element M, 3 1 2x ` Mumber of atoms of element, M = × 2x = 3 3 2x       Ratio of M : N = : x = 2:3 3 Hence, the formula of the compound = M2N3. Q. 17. 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. 18. An element with molar mass 2.7 × 10–2 kg mol–1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m–3, what is the nature of the cubic unit cell? z#M d # a3 # N A Ans. Density, d = 3 or    z =...(i) a # NA M Here, M = 2.7 × 10–2 kg mol–1 a = 405 pm = 405 × 10–12 m = 4.05 × 10–10 m d = 2.7 × 103 kg m–3 NA = 6.022 × 1023 mol–1 The Solid State 23  Substituting these values in expression (i), we get (2.7 # 103 kg m –3) (4.05 # 10 –10 m) 3 (6.022 # 10 23 mol –1) z= 2.7 # 10 –2 kg mol –1 =4 As there are 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face-centred. Q. 19. What type of defect can arise when a solid is 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. 20. What type of stoichiometric defect is shown by: (i) ZnS (ii) AgBr? Ans. (i) ZnS shows Frenkel defect because its ions have a large difference in size. (ii) AgBr shows both Frenkel and Schottky defects. Q. 21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it. Ans. When a cation of higher valence is added as an impurity in an ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.  Number of cations of higher valency     Cationic vacancies produced =  Difference in valencies of the original cation     and cation of higher valency  Q. 22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example. Ans. In ionic solids with anionic vacancies due to metal excess defect, when the metal atoms deposit on the surface, they diffuse into the crystal and after ionisation, the metal ion occupies cationic vacancy while electron occupies anionic vacancy. Such anionic sites occupied by an electron are known as F-centres. These electrons get excited to higher energy levels by adsorption of suitable wavelengths from the visible white light and therefore appear coloured. When Na vapours are passed over NaCl crystals such defect is created and the crystals become yellow due to excess Na+ and presence of F-centres. Q. 23. A group 14 element 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 element into n-type semiconductor, it should be doped with group 15 element e.g., As. Q. 24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic? Justify your answer. Ans. Ferromagnetic substances make better permanent magnets. This is because the metal ions of a ferromagnetic substance are grouped into small regions called ‘domains’. Each domain acts as a tiny magnet. These domains are randomly oriented. When a ferromagnetic substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic field is produced. Such order of domains persists even when the external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet. On the other hand, the net or resultant magnetic moment of ferrimagnetic substances is small. 24 Xam idea Chemistry–XII NCERT Textbook Exercises Q. 1. Define the term ‘amorphous’. Give a few examples of amorphous solids. Ans. An amorphous solid consists of particles of irregular shape. The arrangement of constituent particles in such a solid has only short range order. In such an arrangement, a regular and periodically repeating pattern is observed over short distances only. Examples: Glass, rubber and plastics. Q. 2. 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 (SiO4 tetrahedra) have only a short range order and there is no long range order. In quartz, the constituent particles (SiO4 tetrahedra) have both short range as well as long range orders. On melting quartz and then cooling it rapidly, it is converted into glass. Q. 3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous. (i) Tetra phosphorus decoxide (P4O10) (ii) Ammonium phosphate (NH4)3PO4 (iii) SiC (iv) I2 (v) P4 (vi) Plastic (vii) Graphite (viii) Brass (ix) Rb (x) LiBr (xi) Si Ans. Ionic: (NH4)3PO4 and LiBr; Metallic: Brass, Rb; Molecular: P4O10, I2, P4; Network (covalent): Graphite, SiC, Si; Amorphous: Plastic. Q. 4. (i) What is meant by the term ‘coordination number’? (ii) What is the coordination number of atoms: (a) in a cubic close packed structure? (b) in a body-centred cubic structure? Ans. (i) Coordination number is defined as the number of nearest neighbours in a close-packed structure. In ionic crystals, coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion. (ii) (a) 12 (b) 8. Q. 5. How can you determine the atomic mass of an unknown metal if know its density and the dimension of its unit cell? Explain your answer. d × a3 × N A Ans. Refer to Basic Concepts Point 5. e Atomic mass, M = z o Q. 6. ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules? Ans. Higher the melting point, stronger are the forces holding the constituent particles together and hence greater is the stability. In other words, stronger the lattice structure, higher will be lattice energy and more will be the stability of crystal, hence higher the melting point. 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 and the intermolecular forces present in it is dipole–dipole attraction. Whereas methane is a non- polar molecule and the only forces present in it is the weak van der Waals’ forces. Q. 7. How will you distinguish between the following pairs of terms: [CBSE (AI) 2014] (i) Hexagonal close packing and cubic close packing? (ii) Crystal lattice and unit cell? (iii) Tetrahedral void and octahedral void? Ans. (i) Refer to Basic Concepts Points 8(c) (i) and (ii). (ii) The regular three dimensional arrangement of identical points in the space which represent how the constituent particles (atoms, ions, molecules) are arranged in a crystal is called a crystal lattice. A unit cell is the smallest portion of a crystal lattice, which when repeated over and again in different directions produces the complete crystal lattice. The Solid State 25  (iii) A void surrounded by four spheres occupying the corners of tetrahedron is called a tetrahedral void. It is much smaller than the size of spheres in the close packing. A void surrounded by six spheres along the corners of an octahedral is called octahedral void. The size of the octahedral void is smaller than that of the spheres in the close packing but larger than the tetrahedral void. Q. 8. 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. Ans. (a) 14 (b) 14 (c) 9. Q. 9. Explain: (a) The basis of similarities and differences between metallic and ionic crystals. (b) Ionic solids are hard and brittle. Ans. (a) Similarities: (i) Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions. In metals, these are among the valence electrons and the kernels. (ii) In both cases, the bond is non-directional. Differences: (i) In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the solid state. They can do so only in the molten state or in aqueous solution. In metals, the valence electrons are not bound but are free to move. Hence, they can conduct electricity in the solid state. (ii) Ionic bond is strong due to electrostatic forces of attraction. Metallic bond may be weak or strong depending upon the number of valence electrons and the size of the kernels. (b) Ionic solids are hard because there are strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because ionic bond is non-directional. Q. 10. Calculate the efficiency of packing in case of a metal crystal for (a) simple cubic (b) body-centred cubic (c) face-centred cubic (with the assumption that atoms are touching each other). Ans. Refer to Basic Concepts Point 7. Q. 11. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 10.5 g cm–3, calculate the atomic mass of silver. [CBSE (AI) 2010; (F) 2010] d # a # N A 10.5 g cm # (4.07 # 10 cm) # (6.02 # 10 mol ) 3 –3 –8 3 23 –1 Ans. M = z = 4 10.5 g cm –3 # 67.419 # 10 –24 cm3 # 6.02 # 10 23 mol –1 = = 106.54 g mol–1 4 Q. 12. A cubic solid is made up of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body centre. What is the formula of the compound? What are the coordination numbers of P and Q? Ans. Number of P atoms per unit cell = 1 (at the body centre) × 1 = 1 1 Number of Q atoms per unit cell = 8 (at the corners) × = 1 8 Hence the formula is PQ. Coordination number of each of P and Q = 8. Q. 13. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93u. [CBSE (AI) 2008] 3 Ans. d = 8.55 g/cm , M = 93 g/mol For bcc, z = 2, a = ? NA = 6.02 × 1023 mol–1 z×M d= a3 × N A 26 Xam idea Chemistry–XII  Substituting the values, 2 × 93   8.55 = a3 × 6.02 × 10 23 2 × 93 a3 =    8.55 × 6.02 × 10 23 1/3 a =d n 10 –8    930 8.55 × 3.01 1/3 x=d n 930 Let    8.55 × 3.01 1/3 log x = log d n 930 ` 8.55 × 3.01 1 = (log 930 – log 8.55 – log 3.01) 3 1 = (2.9685 – 0.9320 – 0.4786) 3 1 log x = (1.5579) = 0.5193 3 x = Antilog (0.5193) x = 3.306 ` a = 3.306 × 10 –8 cm 3 1.732 × 3.306 × 10 –8 Now r= × 3.306 × 10 –8 = 4 4 –8 r = 1.4315 × 10 cm = 143.15 pm Q. 14. If the radius of the octahedral void is r and the radius of the atoms in close packing is R, derive relationship between r and R. Ans. A sphere fitting into the octahedral void is shown by shaded circle. The spheres present above and below the void are not shown in the figure. As ABC is a right-angled triangle, Pythagoras theorem is applied. AC2 = AB2 + BC2 (2R)2 = (R + r)2 + (R + r)2 = 2(R + r)2 4R2 = 2(R + r)2 2R2 = (R + r)2 ( 2 R)2 = (R + r)2 B r r 2R =R+r R R r = 2R–R A R R C r = ( 2 – 1) R = (1.414 – 1)R r = 0.414 R Q. 15. Copper crystallises into a fcc lattice with edge length 3.61 × 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3. z#M Ans. d = 3 a # NA For fcc lattice of copper, z = 4 Atomic mass of copper, M = 63.5 g mol–1 4 × 63.5 g mol –1 \ d = = 8.97 g cm–3 (3.61 # 10 –8 cm) 3 # (6.022 # 10 23 mol –1) which is in agreement with the measured value. The Solid State 27 Q. 16. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions? Ans. 98 Ni atoms, are associated with 100 O atoms. Out of 98 Ni atoms, suppose Ni present as Ni2+ = x. Then Ni present as Ni3+ = 98 – x. Total charge on x Ni 2+ and (98 – x) Ni3+ should be equal to charge on 100 O2– ions. Therefore, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94 94 \ Fraction of Ni present as Ni2+ = × 100 = 96% 98 4 Fraction of Ni present as Ni3+ = × 100 = 4% 98 Q. 17. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanisms. Ans. Refer to Basic Concepts Point 17. Q. 18. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. Can you account for the fact that this substance is a p-type semiconductor? Ans. The ratio less than 2 : 1 in Cu2O shows that some cuprous (Cu+) ions have been replaced by cupric (Cu2+) ions. For maintaining electrical neutrality, every two Cu+ ions will be replaced by one Cu2+ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor. Q. 19. Ferric oxide crystallises in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide. Ans. Let the number of oxide ions (O2–) in the close packing be x. \ Number of octahedral voids = x As 2/3rd of the octahedral voids are occupied by ferric ions, number of ferric ions present = 2 # x = 2x 3 3 2x \ Ratio of Fe3+: O2– = :x=2:3 3 Hence, the formula of ferric oxide is Fe2O3. Q. 20. Classify each of the following as being either a p-type or a n-type semiconductor: (i) Ge doped with In (ii) Si doped with B. Ans. (i) Ge is Group 14 element and In is Group 13 element therefore, an electron defic