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The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4, Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on, 30.01.2020 and it has been decided to implement it from academic year 2020-21., , CHEMISTRY, STANDARD TWELVE, , Download DIKSHA App on your, smartphone. If you scan the Q.R.Code on, this page of your textbook, you will be able, to access full text and the audio-visual, study material relevant to each lesson, provided as teaching and learning aids., , 2020, , Maharashtra State Bureau of Textbook Production and, Curriculum Research, Pune.
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The Constitution of India, , Preamble, WE, THE PEOPLE OF INDIA, having, solemnly resolved to constitute India into a, SOVEREIGN, SOCIALIST, SECULAR, DEMOCRATIC REPUBLIC and to secure to, all its citizens:, JUSTICE, social, economic and political;, LIBERTY of thought, expression, belief, faith, and worship;, EQUALITY of status and of opportunity;, and to promote among them all, FRATERNITY assuring the dignity of, the individual and the unity and integrity of the, Nation;, IN OUR CONSTITUENT ASSEMBLY this, twenty-sixth day of November, 1949, do HEREBY, ADOPT, ENACT AND GIVE TO OURSELVES, THIS CONSTITUTION.
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NATIONAL ANTHEM
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Preface, Dear Students,, We welcome you all to std. XII. For the first time, you were introduced to the subject, of chemistry discipline in std XI., Chemistry is very vast subject that covers many aspects of our everyday experience., This textbook aims to create awareness and to understand certain essential aspects by, the state curriculum framework (NCF) which was formulated in 2005, followed by, the national curriculum framework (SCF) in 2010. Based on these two frameworks,, reconstruction of the curriculum and prepartion of a revised syllabus has been done and, designed now., Three major branches of chemistry, namely organic chemistry, inorganic chemistry, and physical chemistry are presented in separate units in this book. Care is taken to, have an integrated approach in deliberation of their contents. Chemistry is highly, applied subject. In the unit "applied chemistry" along with the application of chemistry, in life science and materials in everyday life, the upcoming applied branches, namely, nanochemistry and green chemistry are also introduced. You can learn basic principles,, understand facts and put them into practice by learning in the classroom and laboratory., The textbook is presented in a simple language with relevant diagrams, graphs, tables,, photographs. This will help you to understand various terminology, concepts with more, clarity. All the illustrations are in colour form. The new syllabus focuses on the basic, principles, concepts, laws based on precise observations, their applications in everyday, life and ability to solve different types of problems. The general teaching - learing, objectives of the revised syllabus are further determined on the basis of the ‘Principle of, constructivism’ i.e. self learning., The curriculum and syllabus is designed to make the students to think independently., The students are encouraged to read, study more through the additional information given, in the colored boxes. Activities have been introduced in each chapter. These activities, will help to understand the content knowledge on your own efforts. QR code has been, introduced for gaining the additional information, abstracts of chapters and practice, questions/ activities., The efforts taken to prepare the text book will help the students think about more, than just the content of the chemical concepts. Teachers, parents as well as the aspiring, condidates preparing for the competitive examinations will be benefited., We look forward to a positive response from the teachers and students., Our best wishes to all !, , Pune, Date : 21 February 2020, Bharatiya Saur : 2 Phalguna 1941, , (Vivek Gosavi), Director, Maharashtra State Bureau of Textbook, Production and Curriculum Research, Pune 4
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- For Teachers -, , Dear Teachers,, We are happy to introduce the revised textbook, of chemistry for std. XII. This book is a sincere, attempt in continuation with the standard XI book, to follow the maxims of teaching as well as develop, a ‘constructivist’ approach to enhance the quality, of learning. The demand for more activity based,, experiential and innovative learning opportunities, is the need of the time. The present curriculum has, been restructured so as to bridge the credibility gap, that exists in the experience in the outside world., Guidelines provided below will help to enrich the, teaching - learning process and achieve the desired, learning outcomes., • To begin with, get familiar with the textbook, yourself., • The present book has been prepared for, constructivism and activity based learning., • Teachers must skilfully plan and organize the, activities provided in each chapter to develop, interest as well as to stimulate the thought process, among the students., • Always teach with proper planning., • Use teaching aids as required for the proper, understanding of the subject., • Do not finish the chapter in short., • Follow the order of the chapters strictly as listed, in the contents because the units are introduced in, a graded manner to facilitate knowledge building., • Each unit is structured in a definite manner. It, starts from the basic concepts of general chemistry, , Can you recall ?, , Observe and discuss..., , Internet my friend, , •, , •, •, •, , •, •, , •, , required for each branch of chemistry. Application, of this knowledge will help students to understand, further chapters in each unit., Each chapter provides solved problems on each, and every concept and various laws. The solved, problems are put into boxes. Teachers should, explain each step of the problems to the student, and give them practice., Invite students' participation by making use of the, boxes like ‘Can You Recall’, ‘Do you know?’, Encourage the students to collect related, information by providing them the websites., Teaching- learning interactions, processes and, participation of all students are necessary and so is, your active guidance., Do not use the content of the boxes titled ‘Do you, know’? for evaluation., Exercises include parameters such as co-relation,, critical thinking, analytical reasoning etc., Evaluation pattern should be based on the given, parameters. Equal weightage should be assigned, to all the topics. Use different combinations of, questions., Illustrative figures are included to help assimilation, of new concepts. Teachers should note that students, are not expected to draw all the figures included in, this book. As a part of evaluation students can be, asked to draw schematic diagrams/structures and, interpret or label other complicated figures., , Remember..., , Do you know ?, , Use your brain power, , Try this..., , Activity :, , Can you tell ?, , Can you think ?, , Cover page :, • (+) - Limonene and (-) Limonene : major contributors to the characteristic flavours of peelings of oranges and lemons respectively., • Image of gold surface : Au (111), obtained by Scanning Tunneling Microscope (STM)., • Structures of coordination complexes : haemoglobin, chlorophyll and metal -EDTA complex., , DISCLAIMER Note : All attempts have been made to contact copy right/s (©) but we have not heard from them., We will be pleased to acknowledge the copy right holder (s) in our next edition if we learn from them.
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Competency Statements - Standard XII, Area/ Unit/, Lesson, , After studying the contents in Textbook students....., •, •, •, •, •, •, •, •, •, •, •, •, •, •, , Physical, Chemistry •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, , Inorganic •, chemistry •, •, •, •, •, •, •, •, •, , Distinguish crystal structures illustrating unit cell and packing efficiency in cubic systems., Gain information on point defects and band theory in relation to electric and magnetic behavior., Define solubility and rationalise its dependence on various factors., Explain Henry and Raoult's laws., Derive expressions for colligative properties., Learn van’t Hoff factor and its correlation with dissociation constant., Catagorize strong and weak acid bases., Learn Ostwald’s dilution law., Derive Henderson Balch Hassel equation., Explain the role of buffer solutions in controlling of pH., Understand spontaneity of reactions., Know reversible/irreversible processes and PV work., Understand first and second laws of thermodynamics., Work out change in enthalpy, entropy and Gibbs' functions in physical and chemical, transformations., Apply Hess’s Law in thermochemical equations., State what are strong and weak electrolytes., Define Kohlrausch law and state its importance., Understand functioning of electrolytic and galvanic cells. Write half cell reactions there in., Describe type of electrodes., Derive Nernst equation and understand its importance., Know what are dry cell, lead strong batteries and fuel cells., Describe the electrochemical series and its implications., Define average and instantaneous rate, order and molecularity in kinetics, Formulate differential and integral rate laws for zero and first order reactions., Understand basis of collision theory of reaction rates, Sketch qualitatively potential energy curve. Understand acceleration of reactions in the presence, of catalyst., Solve relevant numerical problems., Write electronic configuration of groups 16, 17, 18 and those of d and f blocks., Correlate atomic properties of elements with electron configuration., Explain the anomalous behaviour of ‘O’ and ‘F’., Understand allotropy in ‘O’ and ‘S’., Draw structures of oxyacids of ‘sulfur’ and ‘halogens’., Write reactions for preparation, chemical properties of O2, O3, SO2, H2SO4, Cl2, HCl, KMnO4,, K2Cr2O7., Draw structures of interhalogen and xenon compounds and illustrate their properties. State the, methods of preparation with reaction., Know chemistry of the elements belonging to groups 16, 17, 18., Understand the principles of metallurgy in extraction of iron., Compare lanthanoides and actinoides., Enlist properties of the manmade post actinoide elements., Understand Werner theory of coordination compounds., Understand and apply EAN rule for stability of coordination compounds., Understand diverse isomerism in coordination compounds., Use the concept of hybridization for predicting structures and magnetic behaviour of complexes, based on the V.B.T., Understand C.F.T. Sketch qualitatively d-orbital splitting diagrams in octahedral and tetrahedral, ligand field environments., Distinguish between high spin and low spin complexes., Predict structure, colour and magnetic properties of the complexes based on the C.F.T.
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•, •, , Organic •, Chemistry •, •, •, •, •, •, •, •, •, •, •, , Applied •, Chemistry •, •, •, •, •, •, •, , State common and IUPAC names of compounds and methods of preparation of halogen, derivatives, alcohols, phenols, ethers, aldehydes, ketones, carboxylic acid and amines., Understand structure, chemical properties, laboratory tests and reactions of the above functional, groups., Explain acid or base strength of alcohols, phenols, carboxylic acids and amines., Explain trends in boiling point and solubility of compounds of above functional groups in terms, of intermolecular forces., Understand optical activity, recognize chiral molecules and represent with Fischer projection, and wedge formulae., Understand mechanism of nucleophilic substitution reactions and influencing factors., Classify carbohydrates, amino acids, nucleic acids., Represent monosaccharides using the Fischer projection formula., Represent monosaccharides, disaccharides and polysaccharides using the Haworth formula., Correlate properties of carbohydrates to the presence or absence of potential aldehyde group., Learn four level structure of proteins and primary structures nucleic acid., Represent primary structure of dipeptide and tripeptide from data on the terminals., Understand enzyme catalysis and double strand DNA structure., Understand classification of polymers on the basis of source, structure, intermolecular forces,, polymerization, number of monomers and biodegradability., Understand addition and condensation polymerization., Know properties, structure and preparation of natural rubber, vulcanized rubber, Buna-S,, viscose, LDP, HDP, teflon, polyacrylo nitrile, polyamide, polyesters, phenol-formaldehyde, resin and PHBV., Understand scope of green chemistry with reference to sustainable development., Recognize twelve principles of green chemistry and their implementation., Correlate the Chemistry knowledge gained so far as pro or counter to the principles of green, chemistry., Understand scope and applications of nanochemistry., Gain knowledge of a synthetic method and properties of nanoparticles., Know instrumental techniques for characterization of nanomaterials., , CONTENTS, Sr. No., 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, , Title, Solid State, Solutions, Ionic Equilibria, Chemical Thermodynamics, Electrochemistry, Chemical Kinetics, Elements of Groups 16, 17 and 18, Transition and Inner transition Elements, Coordination Compounds, Halogen Derivatives, Alcohols, Phenols and Ethers, Aldehydes, Ketones and Carboxylic acids, Amines, Biomolecules, Introduction to Polymer Chemistry, Green Chemistry and Nanochemistry, , Page No., 01-27, 28-46, 47-62, 63-89, 90-119, 120-137, 138-164, 165-191, 192-209, 210-233, 234-253, 254-281, 282-297, 298-321, 322-339, 340-352
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1. SOLID STATE, Can you recall ?, • What are the three most common, states of matter?, , Try this..., , • How does solid state differ from the, other two states ? (Answer with reference, to volume, shape, effect of temperature, and pressure on these and the motion, of constituent particles and interparticle, forces.), 1.1 Introduction : As studied earlier, the solid, state of matter is characterised by strong, interparticle forces of attraction. As a result, most solids have definite shape and volume,, which change only slightly with change, in temperature and pressure. The smallest, constituent particles of various solids are, atoms, ions or molecules. All such smallest, constituent particles of solids will be referred, to as 'particles' in this chapter., , Observe the above figure carefully., The two types of circles in this figure, represent two types of constituent particles, of a solid., • Will you call the arrangement of particles, in this solid regular or irregular ?, • Is the arrangement of constituent, particles same or different in directions, → →, →, AB, CD, and EF ?, , 1.2 Types of solids : There are two types, of solids, namely crystalline solids and, amorphous solids., , 1.2.1 Crystalline solids : Study of many, crystalline solids indicates that they possess, the following characteristic properties., , Observe and discuss..., • Collect the following solids :, grannular sugar, common salt,, blue vitriol., • Observe a few grannules of these solids, under a magnifying lens or microscope., • Discuss your observations with reference, to the following points : (i) Shape of the, grannules, (ii) Smoothness of faces of, the grannules and (iii) Angles between, various edges of the grannules., • All the above solids are crystalline solids., Name the properties of crystals that you, observed in this activity., , i., , There is a regularity and periodicity in, the arrangement of constituent particles in, crystalline solids. The ordered arrangement, of particles extends over a long range., , ii. Crystalline solids have sharp melting, points, that is, they melt at a definite, temperature., iii. All crystalline substances except those, having cubic structure are anisotropic. In, other words their properties like refractive, index, thermal and electrical conductivity,, etc, are different in different directions., Ice, salts such as NaCl, metals such, as sodium, gold, copper and materials such, as diamond, graphite, ceramics are examples, of crystalline solids., , 1
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Glass, plastic, rubber, tar, and metallic glass, (metal-metalloid alloy) are a few examples of, amorphous solids., , Do you know ?, • A single crystal has ordered, (regular, and, periodic), arrangement of constituent particles, throughout its bulk., • Majority of crystalline solids, including, metals, are polycrystalline in nature., Single grannule of a polycrystalline, solid is made of many single crystals or, crystallites packed together with different, orientations., • Single crystals are difficult to obtain., Diamond is an example of naturally, formed single crystal., , Use your brain power, Identify the arrangements A and B, as crystalline or amorphous., , 1.2.3 Isomorphism and polymorphism, Similarity or dissimilarity in crystal, structure of different solids is described as, isomorphism and polymorphism., i. Isomorphism : Two or more substances, having the same crystal structure are said to be, isomorphous. In these substances the chemical, composition has the same atomic ratio. For, example (i) NaF and MgO (ii) NaNO3 and, CaCO3 are isomorphous pairs, and have the, same atomic ratios, 1:1 and 1:1:3, respectively,, of the constituent atoms., , 1.2.2 Amorphous solids : The particles of, a liquid are in constant motion. The stop, action photograph of a liquid describes the, amorphous state. In fact, they are supercooled, liquids. Amorphous solids have the following, characteristics., i. The constituent particles in amorphous, solids are randomly arranged. The, particles do not have long range ordered, structure, but they do have a short range, order., , ii. Polymorphism : A single substance that, exists in two or more forms or crystalline, structures is said to be polymorphous., Polymorphs of a substance are formed under, different conditions. For example : Calcite and, aragonite are two forms of calcium carbonate;, α-quartz, b-quartz and cristobalite are three, of the several forms of silica. Polymorphism, occuring in elements is called allotropy. For, example: three polymorphic (allotropic) forms, of carbon are diamond, graphite and fullerene., , ii. Amorphous solids do not have sharp, melting points. They melt gradually, over a temperature interval. On, heating, amorphous solids gradually and, continuously soften and start to flow., iii. These solids are isotropic. In other, words, their properties such as refractive, index, conductivity are all independent of, direction of measurement. They exhibit the, same magnitude for any property in every, direction., , 2
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1.3.2 Covalent network crystals, , Do you know ?, Many crystalline forms of, silica (SiO2) are found in nature., Three of them are α-quartz, b-quartz and, cristobalite, , Characteristics of covalent network crystals, are as follows :, i. The constituent particles in covalent, network solids are atoms., ii. The atoms in these crystals are linked by a, continous system of covalent bonds. The, result is a rigid three dimensional network, that forms a giant molecule. The entire, crystal is a single molecule., , cristobalite, , iii. As a result of rigid and strongly bonded, structure, covalent network crystals are, very hard. In fact they are the hardest and, most incompressible of all the materials., These crystals have high melting and, boiling points., , α-quartz, , iv. The electrons are localised in covalent, bonds and hence are not mobile. As a, result, covalent solids are poor conductors, of heat and electricity., , b-quartz, , 1.3 Classification of crystalline solids :, Crystalline solids are further classified into, four categories : ionic solids, covalent network, solids, molecular solids and metallic solids., , For example : diamond, quartz (SiO2), boron, nitride, carborandum are covalent network, solids., , 1.3.1 Ionic crystals : Ionic crystals have the, following characteristics :, , Do you know ?, Diamond is the hardest known, material., , i. The constituent particles of ionic crystals, are charged ions. The cations and anions, may differ in size., , Try this..., , ii. Each ion of a given sign of charge is, bonded to ions of opposite charge around, it by coulomb force. In other words, the, particles of ionic crystals are held by, electrostatic force of attraction between, oppositely charged ions., , Graphite is a covalent solid, yet soft and good conductor of, electricity. Explain., Can you recall ?, , iii. Ionic crystals are hard and brittle. They, have high melting points., , •, , What are structures, diamond and graphite ?, , iv. These are nonconductors of electricity, in solid state. However, they are good, conductors when melted or dissolved in, water., , •, , What are the types of covalent bonds, those link carbon atoms in diamond, and graphite ?, , •, , Are all the valence electrons of, carbon atoms in graphite localized to, specific covalent bonds ?, , For example : NaCl, K2SO4, CaF2, KCl are, ionic crystals., , 3, , of
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c. Intermolecular hydrogen bonds in, solids such as H2O (ice), NH3, HF and, so forth., , Remember..., Both ionic and covalent, crystals are hard and have high, melting and boiling points. We can, use electrical properties to distinguish, between them. Both are insulators at low, temperature., , iii. Because of weak intermolecular attractive, forces, molecular solids are usually soft, substances with low melting points., iv. These solids are poor electrical conductors, and are good insulators., , Ionic solids become good conductors, only at high temperature, above their, melting points., , 1.3.4 Metallic crystals : These are crystalline, solids formed by atoms of the same metallic, element, held together by a metallic bond., , The conductivity of covalent solids, is in general low and increases with, temperature. However, there is no abrupt, rise in conductivity when substance is, melted., , Metallic bond : In a solid metal, the valence, electrons are delocalised over the entire crystal, leaving behind positively charged metal ions., Therefore, metallic crystals are often described, as an array of positive ions immersed in a sea, of mobile electrons. The attractive interactions, between cations and mobile electrons constitute, the metallic bonds. (For more details refer to, section 1.9.2), , 1.3.3 Molecular crystals : Substances such, as Cl2, CH4, H2, CO2, O2 on solidfication, give molecular crystals. Crystalline organic, compounds are also molecular solids., i. The constituent particles of molecular, solids are molecules (or unbonded single, atoms) of the same substance., , Metallic crystals have the following properties:, , Can you recall ?, , ii. Metals are ductile, that is, they can be, drawn into wires., , i. Metals are malleable, that is, they can be, hammered into thin sheets., , What is a hydrogen bond ?, , iii. Metals have good electrical and thermal, conductivity., , ii. The bonds within the molecules are, covalent. The molecules are held together, by various intermolecular forces of, attraction. (Refer to XI Std. Chemistry, Textbook, Chapter 10). For example :, , Examples : metals such as Na, K, Ca, Li, Fe,, Au, Ag, Co, etc., The properties of different types of crystalline, solids are summarized in Table 1.1., , a. Weak dipole-dipole interactions in, polar molecules such as solid HCl,, H2O, SO2, which possess permanent, dipole moment., , 1.4 Crystal structure : The ordered three, dimensional arrangement of particles in a, crystal is described using two terms, namely,, lattice and basis., , b. Very weak dispersion or London, forces in nonpolar molecules such as, solid CH4, H2. These forces are also, involved in monoatomic solids like, argon, neon. (These substances are, usually gases at room temperature.), , 1.4.1 Crystal, lattice and basis : Lattice is a, geometrical arrangement of points in a three, dimensional periodic array. A crystal structure, is obtained by attaching a constituent particle, to each of the lattice points. Such constituent, particles that are attached to the lattice points, form the basis of the crystal lattice. Crystal, , 4
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Table 1.1 : Properties of four types of crystalline solids, Type Ionic solids, , Covalent network Molecular, solids, solids, , Property, 1. Particles of unit Cations, cell, anions, , and Covalently bonded, atoms, , Monoatomic, or, polyatomic, molecules, London, dipoledipole forces and/, or hydrogen bonds, , 2. Interparticle, forces, , Electrostatic, , Covalent bonds, , 3. Hardness, , Hard and brittle, , Very hard, , 4. Melting points, , High, 6000C to 30000C, Poor electrical, conductors in, solid state. Good, conductors, when melted, or dissolved in, water, NaCl, CaF2, , High, 12000C to 40000C, Poor conductors, Exceptions :, i. Graphite : good, conductor, of, electricity., ii. Diamond : good, conductor of heat, diamond, silica, ice, benzoic acid, , 5. Thermal and, electrical, conductivity, , 6. Examples, , ., ., ., ., , Lattice, , ., ., ., ., , +, , Metallic ions in a, sea of electrons, , Metallic, bonds, (attraction between, cations and mobile, valence electrons), Soft, Variable from soft, to very hard, Low, Wide range, (-2720C to 4000C) (-390C to 34000C), poor conductor of good conductor of, heat and electricity heat and electricity, , Na, Mg, Cu, Au, , The dimensions of unit cell along the three, axes are denoted by the symbols a, b and c., The angles between these axes are represented, by the symbols ∝, β and γ as shown in Fig 1.1., , lattice is also called space lattice of crystal., Thus, crystal is the structure that results by, attaching a basis to each of the lattice points., It is represented by the following equation., , ., ., ., ., , Metallic solids, , =, , + Basis =, , b, , Crystal, , 1.4.2 Unit Cell : The space lattice of a crystal, is built up of a three dimensional basic, pattern. This basic pattern is repeated in three, dimensions to generate the entire crystal., The smallest repeating structural unit of a, crystalline solid is called unit cell., , a, g, , Fig. 1.1 : Unit cell parameters, , 1.4.3 Types of unit cell : There are four types, of unit cells., , When the unit cells are stacked together, to generate the crystal, each unit cell shares, its faces, edges and corners with neighbouring, unit cells. It is important to understand that the, geometric shape of a unit cell is same as that, of the macroscopic crystal. For example, if the, crystal has cubic shape the unit cell will also, have its constituent particles arranged to form, a tiny cube., , i. Primitive or simple unit cell : In primitive, unit cell, the constituent particles are present at, its corners only., ii. Body-centred unit cell : In this type of, unit cell, one constituent particle is present at, the centre of its body in addition to the corner, particles., , 5
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iii. Face-centred unit cell : This unit cell, consists of particles at the centre of each of the, faces in addition to the corner particles., iv. Base-centred unit cell : This unit cell, consists of particles at the centre of any two, of its opposite faces in addition to the corner, particles., 1.4.4 Crystal systems : By mathematical, analysis, it has been proved that only fourteen, different kinds of space lattices are possible., In other words, there are only 14 ways in, which similar points can be arranged in a, three dimensional order. These 14 lattices,, which describe the crystal structure, are called, Bravais lattices., Fourteen Bravais lattices are divided, into seven crystal systems. The possible, combinations of lattice point spacings (a, b, and c) along three axes and the angles (∝, β, and γ) between these axes give rise to seven, crystal systems. In other words, seven crystal, systems are associated with 14 Bravais lattices, also called 14 unit cells., , sc, , bcc, , fcc, , sc, , bcc, , fcc, , sc, , bcc, , fcc, , Fig. 1.2 : Cubic unit cell, , 1.5.1 Number of particles in cubic unit cells, i. Primitive or simple cubic unit cell (sc) :, A simple cubic cell has particles at its eight, corners. When these unit cells are stacked, together, particle at each corner of a given unit, cell is shared with seven other neighbouring, cubes that come together at that corner. As a, result the corner particle contributes its 1/8th, part to the given unit cell. Thus, a simple cubic, cell has 1/8 × 8 = 1 particle per unit cell., ii. Body-centred cubic unit cell (bcc) : A, bcc unit cell has eight corner particles and an, additional particle at the centre of the cube., One eighth of each particle from eight corners, belongs to the given unit cell as mentioned in, simple cubic unit cell., The particle at the centre of a given, cube is not shared by any other cube. Hence, it, belongs entirely to the given unit cell. Thus bcc, unit cell has one particle from eight corners, plus one particle in the centre of the cube,, making total of 2 particles per bcc unit cell., iii. Face-centred cubic unit cell (fcc) : A, fcc unit cell has particles at the eight corners, plus particles at the centre of its six faces. As, described in simple cubic unit cell, one particle, , The seven crystal systems are, named as cubic, tetragonal, orthorhombic,, rhombohedral, monoclinic, triclinic and, hexagonal system. Cubic system will be, discussed in the following section., 1.5 Cubic system : There are three kinds of, unit cells in cubic system : primitive or simple, cubic (sc), body-centred cubic (bcc) and facecentred cubic (fcc) (Fig. 1.2)., i. Simple cubic unit cell (sc) has a particle at, each of the eight corners of a cube., ii. Body-centred cubic unit cell (bcc) has, particles at its eight corners and an additional, particle in the center of the cube., iii. Face-centred cubic unit cell (fcc) has, particle at the centre of each of six faces in, addition to the particles at eight corners of the, cube., , 6
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Do you know ?, The names of fourteen Bravais lattices (unit cells) for each of the seven crystal system, are shown below., Crystal system, 1. Cubic, , Bravais lattices, Name, i. simple or primitive, ii. body-centred, iii. face-centred, , unit cell structure, , 2. Orthorhombic, , i. simple or primitive, ii. body-centred, iii. face-centred, iv. base-centred, i., , 3. Tetragonal, , ii., , iii., , iv., , i. simple or primitive, ii. body-centred, , i., , 4. Monoclinic, , iii., , ii., , i., , ii., , i. simple or primitive, ii. base centred, , i., , 5. Rhombohedral i. simple or primitive, , 6. Triclinic, , i. simple or primitive, , 7. Hexagonal, , i. simple or primitive, , 7, , ii.
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from eight corners belongs to the given unit, cell., , Problem 1.1 : When gold crystallizes, it, forms face-centred cubic cells. The unit cell, edge length is 408 pm. Calculate the density, of gold. Molar mass of gold is 197 g/mol., , Each particle at the centre of the six, faces is shared with one neighbouring cube., Thus,1/2 of each face particle belongs to the, given unit cell. From six faces, 1/2 × 6 = 3, particles belong to the given unit cell., , Solution :, , Mn, a3 NA, M = 197 g mol-1, n = 4 atoms for fcc,, NA = 6.022 ×1023 atoms mol-1,, a = 408 pm = 408 ×10-12m = 4.08×10-8 cm, ρ=, , Therefore, fcc unit cell has one corner, particle plus 3 face particles, total 4 particles, per unit cell., , Substitution of these quantities in the, equation gives, , Remember..., Each corner particle of a cube, is shared by 8 cubes, each face, particle is shared by 2 cubes and each edge, particle is shared by 4 cubes., , ρ=, 197 g mol-1× 4 atom, (4.08×10-8)3 cm3× 6.022×1023 atom mol-1, , 1.5.2 Relationship between molar mass,, density of the substance and unit cell edge, length, is deduced in the following steps, i. If edge length of cubic unit cell is a, the, volume of unit cell is a3., , = 19.27 g/cm3, 19.27 × 103 kg/m3., 1.6 Packing of particles in crystal lattice, Constituent particles of a crystalline, solid are close packed. While describing, the packing of particles in a crystal, the, individual particles are treated as hard, spheres. The closeness of particles maximize, the interparticle attractions., , ii. Suppose that mass of one particle is m and, that there are n particles per unit cell., Mass of unit cell = m × n, , (1.1), , iii. The density of unit cell (ρ), which is same, as density of the sub-substance is given by, m×n, mass of unit cell, ρ = volume of unit cell = a3 = density of, substance, (1.2), , The number of neighbouring spheres, that touch any given sphere is its coordination, number. Magnitude of the coordination, number is a measure of compactness of, spheres in close-packed structures. The larger, the coordination number, the closer are the, spheres to each other., , iv. Molar mass (M) of the substance is given, by, , 1.6.1 Close packed structures : The three, dimensional close packed structure can be, understood conveniently by looking at the, close packing in one and two dimensions., , M = mass of one particle × number of particles, per mole, = m×NA, , (NA is Avogadro number), , Therefore, m= M/NA , , (1.3), , a. Close packing in one dimension : A close, packed one dimensional structure results by, arranging the spheres to touch each other in a, row (Fig. 1.3 (a))., , v. Combining Eq. (1.1) and (1.3), gives, nM, , (1.4), a3 NA, By knowing any four parameters of Eq. (1.4),, the fifth can be calculated, ρ=, , b. Close packing in two dimensions : A close, packed two dimensional (planar) structure, results by stacking the rows together such, , 8
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the free space in this arrangement is less than, in square packing, making it more effecient, packing than square packing. From Fig.1.3(c), it is evident that the free spaces (voids) are, triangular in shape. These triangular voids are, of two types. Apex of the triangular voids in, alternate rows points upwards and downwards., , that they are in contact with each other. There, are two ways to obtain close packing in two, dimensions., i. Square close packing : One dimensional, rows of close packed spheres are stacked over, each other such that the spheres align vertically, and horizontally (Fig. 1.3 (b)). If the first row, is labelled as 'A' type, being exactly same as, the first row, the second row is also labelled as, 'A' type. Hence this arrangement is called A,, A, A, A..... type two dimensional arrangement., In this arrangement, every sphere touches four, neighbouring spheres. Hence, two dimensional, coordination number, here, is 4. A square is, obtained by joining the centres of these four, closest neighbours (Fig. 1.3(b)). Therefore,, this two dimensional close packing is called, square close packing in two dimension., , (a), , (b), , (c), , ii. Hexagonal close packing : Close packed, one dimensional row (Fig. 1.3 (a)) shows that, there are depressions between the neighbouring, spheres. If the second row is arranged in such, a way that its spheres fit in the depressions of, the first row, a staggered arrangement results., If the first row is called 'A' type row, the, second row, being different, is called 'B' type., Placing the third row in staggered manner in, contact with the second row gives rise to an, arrangement in which the spheres in the third, row are aligned with the spheres in the first, row. Hence the third row is 'A' type. Similarly, spheres in the fourth row will be alligned with, the spheres in the second row and hence the, fourth row would be 'B' type. The resulting two, dimensional arrangement is 'ABAB...' type, (Fig. 1.3 (c)). In this arrangement each sphere, touches six closest neighbours. Thus, the two, dimensional coordination number in this, packing is 6. A regular hexagon is obtained, by joining the centres of these six closest, spheres (Fig. 1.3 (c)). Hence, this type of two, dimensional close packing is called hexagonal, close packing in two dimensions. Compared to, the square close packing in two dimensions,, the coordination number in hexagonal close, packing in two dimensions is higher. Moreover, , Fig. 1.3 : (a) Close packing in one dimension, (b) square close packing, (c) Hexagonal close packing in two dimension, , c. Close packing in three dimensions :, Stacking of two dimensional layers gives rise, to three dimensional crystal structures. Two, dimensional square close packed layers are, found to stack only in one way to give simple, cubic lattice. Two dimensional hexagonal, close packed layers are found to stack in, two distinct ways. Accordingly two crystal, structures, namely, hexagonal close packed, (hcp) structure and face centred cubic (fcc), structure are formed., i. Stacking of square close packed layers:, Stacking of square close packed layers, generates a three dimensional simple cubic, structure. Here, the second layer is placed, over the first layer so as to have its spheres, exactly above those of the first layer (Fig. 1.4)., Subsequent square close packed layers are, placed one above the other in the same manner., In this arrangement, spheres of all the layers, are perfectly aligned horizontally as well as, vertically. Hence, all the layers are alike, and, are labelled as 'A' layers. This arrangement, of layers is described as 'AAAA... ' type. The, , 9
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structure that results on stacking square close, packed layers is simple cubic. Its unit cell is, the primitive cubic unit cell (Fig. 1.4)., , Fig. 1.5 : Two layers of closed packed spheres, , Fig. 1.4 : Stacking of square closed packed, layers, , It can be seen that in the simple cubic, structure, each sphere touches six neighbouring, spheres, four in its own layer, one in the layer, above and one in the layer below. Hence,, coordination number of each sphere is 6., Polonium is the only metal that crystallizes, in simple cubic closed packed structure., , Fig. 1.6 : Tetrahedral void, , ii. Stacking of two hexagonal close packed, layers : To generate a close packed three, dimensional structure, hexagonal close, packed layers are arranged in a particular, manner. In doing so, spheres of the second, layer are placed in the depression of the first, layer (Fig. 1.5). If the first layer is labelled, as 'A' layer, the second layer is labelled as, 'B' layer because the two layers are aligned, differently. It is evident from the Fig. 1.5, that all triangular voids of the first layers, are not covered by the spheres of the second, layer. The triangular voids that are covered, by spheres of the second layer generate, tetrahedral void(Fig. 1.6). A tetrahedral void, is surrounded by four spheres. On joining the, centres of these four spheres a tetrahedron is, formed which encloses the tetrahedral voids, (Fig. 1.6). The remaining triangular voids of, the first layer have above them the triangular, voids of the second layer. The overlapping, triangular voids from the two layers together, form an octahedral void which is surrounded, by six spheres (Fig. 1.7)., , Fig. 1.7 : Octahedral void, , Remember..., It is important to note that the, triangular shapes of depressions in, A and B layer do not overlap. The apices of, two triangular depressions in A and B layer, point in opposite directions., The depressions in which spheres, of second layer rest are tetrahedral voids, while the depressions in which no sphere, rests are octahedral voids., , 10
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iii. Placing third hexagonal close packed, layer : There are two ways of placing the, third hexagonal close packed layer on the, second., , 1 above and 1 below. Hence coordination, number of any sphere in sc is 6., b. In both hcp and ccp/fcc structures that result, from stacking of hexagonal close packed, layers in two different ways, each sphere is, surrounded by 12 neighbouring spheres, 6 in, its own layer, 3 above and 3 below. Hence, the, coordination number of any sphere in hcp or, ccp/fcc structure is 12., , One way of doing this is to align the, spheres of the third layer with the spheres, of the first layer. The resulting pattern of the, layers will be 'ABAB....'. This arrangement, results in hexagonal close packed (hcp), structure (Fig. 1.8(a)). Metals such as Mg,, Zn, have hcp crystal structure., , 1.6.3 Number of voids per atom in hcp and, ccp : The tetrahedral and octahedral voids, occur in hcp and ccp/fcc structures. There are, two tetrahedral voids associated with each, atom. The number of octahedral voids is half, that of tetrahedral voids. Thus, there is one, octahedral void per atom., , The second way of placing the third, hexagonal close packed layer on the second, is to cover the octahedral voids by spheres of, the third layer. In such placing, the spheres of, the third layer do not align with the spheres, of the second or the spheres of the first layer., The third layer is, therefore, called 'C' layer., The spheres of the fourth layer get aligned, with the spheres of the first layer. Hence, the, fourth layer is called 'A' layer. This pattern, of stacking hexagonal close packed layers, is called 'ABCABC....'. This arrangement, results in cubic close packed (ccp) structure, (Fig. 1.8(b)). This is same as fcc structure., Metals such as copper and Ag have ccp (or, fcc) crystal structure., , Remember..., If N denotes number of particles,, then number of tetrahedral voids is, 2N and that of octahedral voids is N., 1.7 Packing efficiency : Like coordination, number, the magnitude of packing efficiency, gives a measure of how tightly particles are, packed together., Packing efficiency is the fraction or a, percentage of the total space occupied by the, spheres (particles)., Packing efficiency =, , (a), , (b), , Expanded, view (a), , volume occupied by particles in unit cell, ×100, total volume of unit cell, , (1.5), 1.7.1 Packing efficiency of metal crystal, in simple cubic lattice is obtained by the, following steps., , Expanded, view (b), , Step 1 : Radius of sphere : In simple cubic, unit cell, particles (spheres) are at the corners, and touch each other along the edge. A face of, simple cubic unit cell is shown in Fig. 1.9. It is, evident that, a = 2r, or, r = a/2, (1.6), , Fig. 1.8 : Formation of hexagonal closed packed, structures, , 1.6.2 Coordination number in close packed, structure, a. In the simple cubic (sc) crystal structure,, that results from stacking of square close, packed layers, each sphere is surrounded by, 6 neighbouring spheres, 4 in its own layer,, , where r is the radius of atom and ‘a’ is the, length of unit cell edge., , 11
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a, r, , r, , Fig. 1.9 : Face of simple cubic unit cell, , Step 2 : Volume of sphere : Volume of a, sphere = (4/3π)(r3). Substitution for r from, Eq. (1.6) gives, , Fig. 1.10 : bcc unit cell, , •, , ∴ FD2 = FE2 + ED2 = a2 +a2 = 2a2 (because, FE = ED = a) , (1.8), , Volume of one particle, , πa3, (1.7), 6, Step 3 : Total volume of particles : Because, simple cubic unit cell contains only one, particle,, πa3, volume occupied by particle in unit cell =, 6, Step 4 : Packing efficiency, = (4/3π) (a/2)3 =, , •, , For triangle AFD, ∠ ADF = 900, ∴ AF2 = AD2 + FD2, , (1.9), , Substitution of Eq. (1.8) into Eq. (1.9) yields, AF2 = a2 + 2a2 = 3a2 (because AD = a), or AF = 3 a, , (1.10), , The Fig. 1.10 shows that AF = 4r., , Packing efficiency, =, , For triangle FED, ∠ FED = 900., , Substitution for AF from equation (1.10) gives, 3, 3a = 4r and hence, r =, a, (1.11), 4, Step 2 : Volume of sphere : Volume of sphere, particle = 4/3 π r3. Substitution for r from, Eq. (1.11), gives, 4, volume of one particle = 3 π ( 3/4a)3, , volume occupied by particle in unit cell, × 100, total volume of unit cell, , 100π 100 × 3.142, πa3/6, = 52.36%, = a3 × 100 = 6 =, 6, Thus, in simple cubic lattice, 52.36 % of, total space is occupied by particles and 47.64%, is empty space, that is, void volume., , , , 1.7.2 Packing efficiency of metal crystal in, body-centred cubic lattice, , 4, ( 3)3 3, = 3 π×, a, 64, , 3 π a3, , = 16, Step 3 : Total volume of particles : Unit, cell bcc contains 2 particles. Hence, volume, occupied by particles in bcc unit cell, 2 3 πa3, , =2×, 16, , 3 πa3, =, (1.12), 8, , Step 1 : Radius of sphere (particle) :, In bcc unit cell, particles occupy the corners, and in addition one particle is at the centre of, the cube. Figure 1.10 shows that the particle, at the centre of the cube touches two corner, particles along the diagonal of the cube., To obtain radius of the particle (sphere), Pythagorus theorem is applied., , 12
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Step 4 : Packing efficiency, , =, , Packing efficiency, , 4, 1 3, 3, 3 π a × ( 2 2), , π a3, 12 2, Step 3 : Total volume of particles : The unit, cell of fcc lattice contains 4 particles. Hence,, volume occupied by particles in fcc unit cell, 3, π a3, =4× πa =, 32, 12 2, , volume occupied by particles in unit cell, =, ×100, total volume of unit cell, , , , πa3, = 3 3 × 100 = 68 %, 8a, Thus, 68% of the total volume in bcc unit, lattice is occupied by atoms and 32 % is empty, space or void volume., , =, , Step 4 : Packing, efficiency, , 1.7.3 Packing efficiency of metal crystal, in face-centred cubic lattice (or ccp or hcp, lattice), , =, , efficiency : Packing, , volume occupied by particles in unit cell, total volume of unit cell, , × 100, , πa3, × 100 = π × 100 = 74%, 3, 3 2a, 32, Thus in fcc/ccp/hcp crystal lattice, 74% of the, total volume is occupied by particles and 26%, is void volume or empty space., =, , Step 1 : Radius of particle/sphere : The corner, particles are assumed to touch the particle at, the centre of face ABCD as shown in Fig. 1.11., The triangle ABC is right angled, with ∠ABC = 900. According to Pythagorus, theorem,, AC2 = AB2 + BC2 = a2 +a2 = 2a2, , Table 1.3 shows the expressions for, various parameters of particles in terms of unit, cell dimension for cubic systems., Use your brain power, Which of the three lattices, sc,, bcc and fcc has the most efficient, packing of particles? Which one has the, least efficient packing?, Table 1.3 : Edge length and particle parameters, in cubic system, , Fig. 1.11 : fcc unit cell, , Unit, cell, , (because AB = BC = a), Hence, AC = 2 a , , (1.13), , Figure 1.11 shows that AC = 4 r. Substitution, for AC from Eq. (1.13) gives, a, 2a = 4r or r = 2 a =, (1.14), 2 2, 4, Step 2 : Volume of sphere : Volume of one, 4, particle = 3 πr3. Substitution for r from, Eq. (1.14) gives, 4, a 3, ), Volume of one particle = 3 π (, 2 2, , 1. sc, 2. bcc, 3. fcc/, ccp, , 13, , Relation, between a, and r, , Volume, of one, particle, , Total, volume, occupied, by, particles, in unit, cell, 3, r = a/2 =, πa /6 =, πa3/6 =, 0.5000a, 0.5237 a3 0.5237 a3, 3, 3, r = 3a/4 = 3 πa /16 3 πa /8, 3, 0.4330a, = 0.34a3 = 0.68a, 3, 3, r = 2a/4 = πa /12 2 πa /3 2, 0.3535a, = 0.185a3 = 0.74a3
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Table 1.4 shows the summary of coordination, number of particles and packing efficiency in, various cubic systems., , substitution of M gives, , xNA, Number of particles in 'x' g = ρa3N /n, A, xn, =, ρa3, • Number of unit cells in 'x' g metal :, , Table 1.4 : Coordination number and packing, efficiency in systems, Lattice, , Coordination, number of atoms, , Packing, efficiency, , ∴, , 1. sc, , 6 : four in the, 52.4 %, same layer, one, directly above, and one directly, below, 2. bcc, 8 : four in the, 68 %, layer below and, one in the layer, above, 3. fcc/ccp/ 12 : six in its, 74 %, hcp, own layer, three, above and three, below, , ∴, , 'n' particles correspond to 1 unit cell, xn, xn, particles correspond to, ×, 3, ρa, ρa3, , 1, n, , unit cells., x, ρa3, Number of unit cells in volume 'V' of, V, metal = 3, a, , ∴Number of unit cells in 'x' g metal =, •, , Problem 1.2 A compound made of elements, C and D crystallizes in fcc structure. Atoms, of C are present at the corners of the cube., Atoms of D are at the centres of faces of the, cube. What is the formula of the compound?, , 1.7.4 Number of particles and unit cells in, x g of metallic crystal :, The number of particles and the number of, unit cells in given mass of a metal can be, calculated from the known parameters of unit, cell, namely, number of particles 'n' per unit, cell and volume 'a3' of unit cell. Density (ρ), and molar mass (M) of a metal are related, to each other through unit cell parameters as, shown below :, mass, ρ=, volume, number of particles in unit cell, M, =, ×, volume of unit cell, NA, n, M, ∴ρ = 3 ×, a, NA, 3, a NA, ∴M = ρ, n, where 'n' is the number of particles in unit cell, and 'a3' is the volume of unit cell., , Solution:, i. C atoms are present at the 8 corners. The, contribution of each corner atom to the unit, cell is 1/8 atom. Hence, the number of C, atom that belongs to the unit cell = 8×(1/8), =1, ii. D atoms are present at the centres of, six faces of unit cell. Each face-centre, atom is shared between two cubes. Hence,, centribution of each face centre atom to the, unit cell is 1/2 atom., The number of D atoms that belong, to unit cell = 1/2×6 =3, There are one C atom and three D atoms in, the unit cell., ∴ Formula of compound = CD3, , • Number of particles in 'x' g metal :, ∴, Molar mass, M, contains NA particles, xNA, ∴ x g of metal contains, particles., M, , 14
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Solution : The atoms of element B form, ccp structure. The number of tetrahedral, voids generated is twice the number of B, atoms., , Problem 1.3 : The unit cell of metallic, silver is fcc. If radius of Ag atom is 144.4, pm, calculate (a) edge length of unit cell(b), volume of Ag atom, (c) the percent of the, volume of a unit cell, that is occupied by Ag, atoms, (d) the percent of empty space., , Thus, number of tetrahedral voids = 2B, The atoms A occupy (1/3) of these, tetrahedral voids., , Solution:, (a) For fcc unit cell, r = 0.3535 a, , Hence, number of A atoms = 2B×1/3, , r = 144.4 pm = 144.4 × 10 m, , Ratio of A and B atoms = 2/3 B: 1B, , = 144.4 × 10 cm, r, 144.4 × 10-10cm, a = 0.3535 =, 0.3535, , , , -12, , -10, , = 4.085 × 10-8 cm, (b) Volume of Ag atom =, =, , 4, 3, , = 2/3:1 = 2:3, , Formula of compound = A2B3, , Problem 1.5 : Niobium forms bcc structure., The density of niobium is 8.55 g/cm3 and, length of unit cell edge is 330.6 pm. How, many atoms and unit cells are present in, 0.5 g of niobium?, , π r3, , 4, × 3.142 × (144.4 × 10-10 cm)3, 3, , = 1.261 × 10-23 cm3, (c) In fcc unit cell, there are 4 Ag atoms, , Solution:, , Volume occupied by 4 Ag atoms, , i. Number of atoms in x g niobium =, , = 4×1.26×10-23 cm3, , x = 0.5 g, n = 2 (for bcc structure),, , = 5.044 × 10-23 cm3, , ρ = 8.55 g/cm3,, , Total volume of unit cell = a3, , a = 330.6pm = 3.306×10-8cm., , = (4.085×10-8 cm)3, , , = 6.817×10, , -23, , Number of atoms in 0.5 g of niobium, 0.5 g × 2, = 8.55 g cm-3 × (3.306×10-8 cm)3, , cm, , 3, , Percent of volume occupied by Ag atoms, =, =, , volume occupied by atoms in unit cell, total volume of unit cell, , 5.044×10 cm, 6.817×10-23cm3, -23, , 3, , xn, ρa3, , × 100, , = 3.25×1021, , x, ii. Number of unit cells in x g = ρa3, , = 74%, , Number of unit cells in 0.5 g of niobium, , (d) Percent empty space = 100 - 74 = 26%, , =, , 0.5 g × 2, 8.55 g cm × (3.306×10-8cm)3, -3, , Problem 1.4 : A compound is formed, by two elements A and B. The atoms of, element B forms ccp structure. The atoms, of A occupy 1/3rd of tetrahedral voids., What is the formula of the compound ?, , = 1.62 ×1021, , 15
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There are three types of defects: point, defects, line defects and plain defects. Only, point defects will be discussed in this chapter., , Problem 1.6 : A compound forms hcp, structure. What is the number of (a), octahedral voids (b) tetrahedral voids (c), total voids formed in 0.4 mol of it., , 1.8.1 Point defects : These defects are, irregularities produced in the arrangement of, basis at lattice points in crystalline solids., , Solution :, Number of atoms in 0.4 mol = 0.4 × NA, , There are three major classes of point, defects: stoichiometric point defects, impurity, defects and nonstoichiometric point defects., , = 0.4 × 6.022 × 1023 = 2.4098 × 1023, (a) Number of octahedral voids = number, of atoms = 2.4098 × 1023, , a. Stoichiometric point defects : Chemical, formula of a compound shows fixed ratio of, number of atoms or number of cations and, anions. This fixed ratio is the stoichiometry of, the compound., , (b) Number of tetrahedral voids, = 2×number of atoms, = 2×2.4098×1023, = 4.818×1023, , In stoichiometric defect, the stoichiometry, remains unchanged. In other words, the ratio, of number of atoms or number of cations and, anions of compound remains the same as, represented by its chemical formula., , (c) Total number of voids, = 2.409×1023+ 4.818×1023, = 7.227 × 1023, , There are four types of stoichiometric, point defects: vacancy defect, self interstitial, defect, Schottky defect and Frenkel defect., , 1.8 Crystal defects or imperfections : The, real, naturally occurring crystalline substances, do not have perfect crystal structures. They, have some disorders or irregularities in the, stacking of atoms. Such irregularities in the, arrangement of constituent particles of a solid, crystal are called defects or imperfections., , i. Vacancy defect : During crystallization of, a solid, a particle is missing from its regular, site in the crystal lattice. The missing particle, creates a vacancy in the lattice structure. Thus,, some of the lattice sites are vacant because of, missing particles as shown in Fig. 1.12. The, crystal is, then, said to have a vacancy defect., The vacancy defect can also be developed, when the substance is heated., , Defects are created during the process, of crystallization. The imperfections are more, if the crystallization occurs at a faster rate. It, means that the defects can be minimized by, carrying out crystallization at a slower rate., In fact ideal crystals with no, imperfections are possible only at the absolute, zero of temperature. Above this temperature, no crystalline materials are 100 % pure. They, contain defects., Whatever be the nature of a crystal defect,, electrical neutrality of the solid is maintained., It is important to note that sometimes, defects are to be intentionally created for, manipulating the desired properties in, crystalline solids., , Fig. 1.12 : Vacancy defect, , 16
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Due to the absence of particles, the, mass of the substance decreases. However,, the volume remains unchanged. As a result the, density of the substance decreases., , It is interesting to note that in this second, case, because of the displacement of a particle a, vacancy defect is created at its original regular, lattice site. At the same time interstitial defect, results at its new position. We can, therefore,, say that in this defect there is a combination of, vacancy defect and self interstitial defect., , ii. Self interstitial defect in elemental solid, Interstitial sites in a crystal are the, spaces or voids in between the particles at, lattice points. When some particles of a, crystalline elemental solid occupy interstitial, sites in the crystal structure, it is called self, interstitial defect., , This defect preserves the density of the, substance because there is neither loss nor, gain in mass of a substance., , Firstly, an extra particle occupies an empty, interstitial space in the crystal structure as, shown in Fig. 1.13. This extra particle is same, as those already present at the lattice points., , iii. Schottky defect : In an ionic solid, equal, number of cations and anions are missing, from their regular positions in the crystal, lattice creating vacancies as shown in Fig., 1.15. It means that a vacancy created by a loss, of cation is always accompanied by a vacancy, formed by a loss of anion., , Fig. 1.13 : Self interstitial defect, , Fig. 1.15 : Schottky defect, , The extra particles increase the total, mass of substance without increasing volume., Hence its density increases., , Thus, there exist two holes per ion pair, lost, one created by missing cation and the, other by a missing anion. Such a paired cationanion vacancy defect is a Schottky defect., , This defect occurs in the following two ways :, , Secondly, in an elemental solid a particle, gets shifted from its original lattice point and, occupies an interstitial space in the crystal as, shown in the Fig. 1.14., , Conditions for the formation of Schottky, defect, i. Schottky defect is found in ionic compounds, with the following characteristics :, , Fig. 1.14 : Self interstitial defect, , 17, , •, , High degree of ionic character., , •, , High coordination number of anion, , •, , Small difference between size of cation, and anion. The ratio rcation/ranion is not far, below unity.
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Consequences of Schottky defect, • As the number of ions decreases, mass, decreases. However, volume remains, unchanged. Hence, the density of a, substance decreases., , Conditions for the formation of Frenkel, defect, •, , Frenkel defect occurs in ionic compounds, with large difference between sizes of, cation and anion., , •, , •, , The ions of ionic compounds must be, having low coordination number., , The number of missing cations and anions, is equal, the electrical neutrality of the, compound is preserved., This defect is found in ionic crystals such as, NaCl, AgBr and KCl., , Consequences of Frenkel defect, , iv. Frenkel defect : Frenkel defect arises when, an ion of an ionic compound is missing from, its regular lattice site and occupies interstitial, position between lattice points as shown in, Fig. 1.16., , •, , As no ions are missing from the crystal, lattice as a whole, the density of solid and, its chemical properties remain unchanged., , •, , The crystal as a whole remains electrically, neutral because the equal numbers of, cations and anions are present., , This defect is found in ionic crystals like ZnS,, AgCl, AgBr, AgI, CaF2., , The cations are usually smaller than, anions. It is, therfore, more common to find, the cations occupying interstitial sites. It is, easier for the smaller cations to accomodate, the interstitial spaces., , b. Impurity defect : Impurity defect arises, when foreign atoms, that is, atoms different, from the host atoms, are present in the crystal, lattice. There are two kinds of impurity defects :, Substitutional and interstitial impurity defects., i. Substitutional impurity defect : In this, defect, the foreign atoms are found at the, lattice sites in place of host atoms. The regular, atoms are displaced from their lattice sites by, impurity atoms., For example :, •, , Fig. 1.16 : Frenkel defect, , Do you know ?, Frenkel defect is not found in, pure alkali metal halides because, cations of allkali metals due to large size, cannot occupy interstitial space., It is important to note that the smaller, cation is displaced from its normal site to, an interstitial space. It, therefore, creates a, vacancy defect at its original position and, interstitial defect at its new location in the, same crystal. Frenkel defect can be regarded, as the combination of vacancy defect and, interstitial defect., , Solid solutions of metals (alloys) : Brass, is an alloy of Cu and Zn. In brass, host, Cu atoms are replaced by impurity of Zn, atoms. The Zn atoms occupy regular sites, of Cu atoms as shown in Fig. 1.17., , Fig. 1.17 : Brass, , 18
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•, , Vacancy through aliovalent impurity :, , c. Nonstoichiometric defects :, , Vacancies are created by the addition, of impurities of aliovalent ions (that is, ions, with oxidation state (o.s.) different from that, of host ions) to an ionic solid., , Nonstoichiometric defect arises when, the ratio of number of atoms of one kind to, that of other kind or the ratio of number of, cations to anions becomes different from that, indicated by its chemical formula. In short,, stoichiometry of the compound is changed., It is important to note that the change, in stoichiometry does not cause any change in, the crystal structure., There are two types of nonstoichiometric, defects, i. Metal deficiency defect : This defect is, possible only in compounds of metals that, show variable oxidation states., , Fig. 1.18 : Vacancy through aliovalent ion, , Suppose that a small amount of, SrCl2 impurity is added to NaCl during its, crystallization. The added Sr2⊕ ions (O.S. + 2), occupy some of the regular sites of Na⊕ host, ions (O.S.+1)., , In some crystals, positive metal ions, are missing from their original lattice sites., The extra negative charge is balanced by the, presence of cation of the same metal with, higher oxidation state than that of missing, cation., , In order to maintain electrical, neutrality, every Sr2⊕ ion removes two Na⊕, ions. One of the vacant lattice sites created by, removal of two Na⊕ ions is occupied by one, Sr2⊕ ion. The other site of Na⊕ ion remains, vacant as shown in Fig. 1.18., , For example, in the compound NiO, one Ni ion is missing creating a vacnacy at, its lattice site. The deficiency of two positive, charges is made up by the presence of two Ni3⊕, ions at the other lattice sites of Ni2⊕ ions as, shown in Fig. 1.20. The composition of NiO, then becomes Ni0.97O1.0, 2⊕, , ii. Interstitial impurity defect : In this defect,, the impurity atoms occupy interstitial spaces, of lattice structure. For example in steel, Fe, atoms occupy normal lattice sites. The carbon, atoms are present at interstitial spaces, as, shown in Fig. 1.19., , Fig. 1.20 : Nonstoichiometric Ni0.97O1.0, , ii. Metal excess defect : There are two types of, metal excess defects., •, Fig. 1.19 : Stainless steel, , 19, , A neutral atom or an extra positive, ion occupies interstitial position : ZnO, presents two ways of metal excess defect., In the first case in ZnO lattice one neutral
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Zn atom is present in the interstitial space, as shown in Fig. 1.21(a), , Cl ions diffuse to the crystal surface, creating vacancies at their regular sites. These, Cl ions combine with Na atoms on the surface, to form NaCl, by releasing electron from, sodium atom., Na + Cl, NaCl +e, The electrons released diffuse into the, crystal and occupy vacant sites of anions as, shown in Fig. 1.22. The anion vacant sites, occupied by electrons are F-centres or colourcentres., , Fig. 1.21 (a) : Neutral Zn atom at interstitial, site, , Fig. 1.21 (b) : Zn2+ ions and electrons at, interestitial sites, , Fig. 1.22 : An F-centre in a crystal, , In the second case, when ZnO is heated it, decomposes as :, ZnO, , NaCl shows yellow colour due to the, formation of F-centre. The crystal of NaCl, has excess Na. The nonstoichiometric formula, of NaCl is the Na1+xCl1.0, , Zn2⊕ + 1/2 O2 +2e, , The excess Zn2⊕ ions are trapped in, interstitial site in the lattice. The electrons also, diffuse in the crystal to occupy interstitial sites, as shown in Fig. 1.21(b)., , 1.9 Electrical properites of solids, Can you recall ?, , In both the cases, nonstoichiometric, formula of ZnO is Zn1+xO1.0, , • What is electrical conductivity ?, • What is meant by the terms electrical, 'insulator' and 'semiconductor' ?, , Can you think ?, , When ZnO is heated it turns, yellow and returns back to original, white colour on cooling. What, could be the reason ?, , Solids show very wide range of electrical, conductivity. Accordingly solids are classified, into the following three categories : conductors,, insulators and semiconductors., , •, , i. Conductors : Solids having electrical, conductivities in the range 104 to 107 Ohm-1m-1, are called conductors. Metals and electrolytes, (ionic solids) are examples of electrical, conductors. Metals conduct electricity by, movement of electrons while electrolytes, conduct electricity by movement of ions. In, , By anion vacancies (Colour or F-centres), , This type of defect imparts colour, to the colourless crystal. For example, when, NaCl crystal is heated in the atmosphere of, sodium vapour, sodium atoms are deposited, on the crystal surface., , 20
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this section we will study some aspects of, electronic conduction of electricity., , They conduct electricity when electrical, potential is applied., , ii. Insulators : Solids having low electrical, conductivities in the range 10-20 to, 10-10 Ohm-1m-1 are called insulators. Most, nonmetals and molecular solids belong to this, category., , ii. Valence band : The band having lower, energy than conduction band is the valence, band., The electrons in valence band are not, free to move because they are tightly bound to, the respective nuclei., , iii. Semiconductors : Solids having electrical, conductivities in the range 10-6 to 104 Ohm-1 m-1, are semiconductors. This range is intermediate, between conductors and insulators. Metalloids, like silicon, germanium belong to this category., , iii. Band gap : The energy difference between, valence band and conduction band is called, band gap. Size of the band gap decides whether, electrons from valence band can be promoted, to vacant conduction band or not when band, gap is too large to promote electrons from, valence band to vacant conduction band by, thermal energy, it is called forbidden zone., When band gap is small, electrons from higher, energy levels in valence band can be promoted, to conduction band by absorption of energy, (such as thermal, electromagnetic)., , 1.9.1 Band theory : Electrical conductivities, of solid metals, nonmetals and metalloids are, explained in terms of band theory. A band, is made of closely spaced electronic energy, levels. Band formation can be correlated to, formation of molecular orbitals (MOs) by, interaction of atomic orbitals. (Refer to Std., XI Chemistry Textbook, chapter 5)., Can you recall ?, • How many molecular orbitals, are formed by interaction of two, atomic orbitals ?, • What is metallic bond ?, , Do you know ?, The band gap energy values of, a few solids are as shown here., Solid, , According to MO theory interaction of, atomic orbitals of combining atoms results, in formation of equal number of MOs which, spread over the entire molecule. Similar to, this, interaction of energy levels of electrons in, the closely spaced constituent atoms in solids, result in formation of bands. Band theory, considers formation of two types of bands,, namely, conduction band and valence band., Another important concept of band theory is, the band gap., , Egap eV, , Diamond, , 5.47, , Sodium, , 0, , Silicon, , 1.12, , Germanium, , 0.67, , The electrical properties of metallic, conductors, insulators and semiconductors are, explained in terms of band theory as follows :, 1.9.2 Metals : Metals are good conductors, of electricity. The outermost electrons of, all the atoms in the metallic crystal occupy, conduction band. The number of electrons, in conduction band of metals is large. Hence, metals are good conductors of electricity. The, conduction bands in metals can be further, labelled as 's' band (Fig. 1.23 (a)), overlapping, s and p bands (Fig. 1.23(b)) and so on. This, , i. Conduction band : The highest energy band, containing electrons is the conduction band., It is formed by interaction of the outermost, energy levels of closely spaced atoms in solids., Conduction band may be partially occupied, or vacant. Electrons in conduction band are, mobile and delocalized over the entire solid., , 21
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depends on the atomic orbitals involved in, band formation. Band formation in metallic, conductors, thus, results in delocalization of, the outermost electrons of all the metal atoms, leaving behind metal ions. This is described as, 'cations of metal are immersed in the sea of, electrons'., , between these two halves. The 3s band, in sodium is the conduction band which, contains same number of electrons as the, sodium atoms. This is responsible for the, high conductivity of metallic sodium., • Metallic magnesium is an example, of conductor with overlapping bands., Electronic configuration of Mg is, [Ar]3s23p0. Interaction of completely, filled 3s AOs of all the Mg atoms gives, rise to the same number of MOs all of, which are filled. These together form the, 3s band which is a completely filled band., Interaction of vacant 3p AOs of all the Mg, atoms gives rise to the same number of, vacant MOs together called 3p band. This, is the vacant band. The filled 3s band and, vacant 3p band overlap each other. As a, result, higher energy electrons move from, 3s band to 3p band., , p-band, , s-band, s-band, (a), , overlapping bands, (b), , Fig. 1.23: Metalic conductor, , 1.9.3 Insulators : In insulators the valence, band is completely filled with electrons and, the conducation band is empty., , The cations of metal atoms occupying, lattice sites vibrate about their mean positions., At higher temperatures, metal cations undergo, increased vibrational motion about their lattice, sites. The flow of electrons is interrupted by, increased vibrational motion. As a result, conductivity of metals decreases with increase, in temperature. (Refer to Std. XI Physics, chapter 11)., Do you know ?, • Metallic sodium is an example of, conductor where the conduction, band is partially filled and there is, no band gap. Electronic configuration of, Na is [Ar]3s1. Interaction of the partially, filled 3s AOs of all the Na atoms gives, rise to same number of MOs. All these, closely spaced MOs together form a, continuous band of energies which is, called 3s band. Lower half of 3s band, corresponds of BMOs and is filled while, the upper half of 3s band corresponds, to AMOs and is empty. There is no gap, , Fig. 1.24: Insulators, , The valence band and conduction, band in insulators are separated by a large, energy gap called forbidden zone as shown in, Fig. 1.24. Here, thermal energy is insufficient, to promote electrons from valence band to, conduction band., As a result the conduction band, remains vacant. The material is, therefore, an, insulator., , 22
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1.9.4 Semiconductors : Electrical conductivity, of a semiconductor material is intermediate, between that of metals and insulators. The, metalloids Si and Ge are semiconductors., , the valence band than at lower temperature., In fact semiconductors are insulators at, low temperatures and conductors at high, temperatures., , Like insulators, the valence band in, semiconductor is completely filled with, electrons and conduction band is empty., However, the energy gap between the two, bands is smaller than that in an insulator. (Fig., 1.25), , Remember..., Electrical conductivity of, metals decreases and that of, semiconductor increases with increasing, temperature., 1.9.5 Extrinsic semiconductors and doping :, The conductivity of a semiconductor, can be increased by doping. The process of, addition of minute quantity of impurities to a, semiconductor to increase its conductivity is, called doping. The added impurity is called, dopant., A doped semiconductor, having higher, conductivity than pure intrinsic semiconductor,, is an extrinsic semiconductor., There are two types of extrinsic, semiconductors, namely, n-type and p-type, semiconductors., , Fig. 1.25 : Semiconductors, , At a temperature above absolute zero a, few electrons in the valence band have enough, thermal energy to jump through the small band, gap and occupy higher energy conduction, band. The conduction band, thus, becomes, partially filled and the valence band becomes, partially empty., , i., n-type, semiconductor, :, n-type, semiconductor contains increased number of, electrons in the conduction bond., An n-type semiconductor is obtained, by adding group 15 element to intrinsic, semiconductor which belongs to group 14., , The electrons in conduction band are free, to move. When electric potential is applied to, a semiconductor, it conducts a small amount, of electicity., , Can you tell ?, Let a small quantity of, phosphorus be doped into pure, silicon., , Such a pure semiconductor material which, has a very low but finite electrical conductivity, is called intrinsic semiconductor., , • Will the resulting material contain, the same number of total number of, electrons as the original pure silicon ?, , The electrical conductivity of a, semiconductor increases with increasing, temperature. This is because, the number, of electrons with sufficient energy so as, to get promoted to the conduction band, increases as temperature rises. Thus, at higher, temperatures, there are more mobile electrons, in the conduction band and more vacancies in, , • Will the material be electrically, neutral or charged ?, Consider, for example, doping of Si, with phosphorus. Si has a crystal structure in, which each Si atom is linked tetrahedrally to, four other Si atoms. When small quantity of, , 23
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contain less number of valence electrons than, that of the pure semiconductor., , phosphorous is added to pure Si, the P atoms, occupy some vacant sites in the lattice in place, of Si atoms, as shown in Fig. 1.26. The overall, crystal structure of Si remains unchanged., , Consider, for example, pure Si doped, with boron. The B atoms occupy normal, positions of some of the Si atoms in the lattice, as shown in Fig. 1.28. Boron atom has only, three valence electrons. It does not have, enough electrons to form bonds with its four Si, neighbours., , Fig. 1.26 : P atom occupying regular site of Si, atom, , Four of the five valence electrons of P are, utilized in bonding the closest to four Si atoms., Thus, P has one extra electron than needed for, bonding. Therefore, Si doped with P has more, number of electrons in the conduction band, than those in the conduction band in pure Si, as shown in Fig. 1.27. It is thus transperent, that the conductivity of Si doped with P is, higher than that of pure Si. The electrons in, conduction band move under the influence of, an applied potential and conduct electricity., , Fig. 1.28 : B atom occupying regular site of Si, atom, , B atom forms bonds with three Si atoms, only. The missing fourth electron creates an, electron vacancy. It is called a hole., Fig. 1.28 shows the holes in the valence, band of p-type semiconductor., A hole has a tendency to accept electron, from its close vicinity. Thus, a hole behaves, as if it has a positive charge. The electrons in, partially filled valence band move under the, influence of an applied potential. The holes, move in the opposite direction., Remember..., •, , Whether intrinsic or extrinsic, semiconductor, the material is, electrically neutral., , •, , An n-type semiconductor such as Si, doped with P has more electrons than, those needed for bonding and thus has, electrons in the partially filled conduction band., A p-type semiconductor such as Si, doped with B has the less electrons, than needed for bonding and thus has, vacancies (holes) in the valence band., , Fig. 1.27 : n-type and p-type semiconductor, , Because the charge carriers are the, increased number of electrons, Si or Ge doped, with group 15 elements such as P, As, Sb or Bi, is an n-type semiconductor., ii. p-type semiconductor : A p-type, semiconductor is produced by doping a pure, semiconductor material (Si or Ge) with an, impurity of group 13 elements. These elements, , •, , 24
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Because the charge carriers are holes, which behave like positive charge, the Si or, Ge doped with group 13 elements like B, Ga, or In, is a p-type semiconductor., , ii. Paramagnetic solids : The substances with, unpaired electrons are weakly attracted by, magnetic field. These substances are called, paramagnetic substances., , 1.10 Magnetic properties of solids : Magnetic, properties of solids can be understood easily, in terms of classical picture of electron. The, electrons spin about their own axis. The, spinning electrons act like tiny magnets, because their spinning action generates, induced magnetic field., , The spinning of unpaired electron gives, rise to a magnetic moment. The substance, is attracted by magnetic field because of, magnetic moment. It is important to understand, that these substances exhibit magnetism in, presence of external magnetic field only. They, lose magnetism when the external magnetic, field is removed., , If an orbital contains one electron,, the unbalanced spin exhibits magnetism., However, when electrons are paired their, spin is balanced and no magnetic property is, observed. On the basis of magnetic properties, solids are classified into three major, classes : diamagnetic, paramagnetic and, ferromagnetic., , Oxygen, Cu2⊕, Fe3⊕, Cr3⊕ are some, examples of paramagnetic substances., iii. Ferromagnetism : The substances, containing large number of unpaired electrons, are attracted strongly by magnetic field. These, substances are said to be ferromagnetic., These substances can be permanently, magnetised. They retain magnetism even after, the removal of external magnetic field., , i. Diamagnetic solids : The substances with, all electrons paired, are weakly repelled by, magnetic fields. These substances are said to, be diamagnetic., , Some, example, of, ferromagnetic, substances are Fe, Co, Ni, Gd, CrO2., , Pairing of electrons balances the spins, and hence, cancels their magnetic moments., N2, F2, NaCl, H2O and benzene are some, examples of diamagnetic substances., , Exercises, 1. Choose the most correct answer., i., , iii., , Molecular solids are, a. crystalline solids, b. amorphous solids, c. ionic solids, d. metallic solids, ii., Which of the follwong is n-type, semiconductor?, a. Pure Si, b. Si doped with As, c. Si doped with Ga, d. Ge doped with In, , iv., , 25, , In Frenkel defect, a. electrical neutrality of the, substance is changed., b. density of the substance is, changed., c. both cation and anion are missing, d. overall electical neutrality is, preserved., In crystal lattice formed by bcc unit, cell the void volume is, a. 68 %, b. 74 %, c.32 %, d. 26 %
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v., , vi., , The coordination number of atoms in, bcc crystal lattice is, a. 2 , , b. 4, , c. 6 , , d. 8, , having different electrical properties., Classify, them, as, insulator,, semiconductor or a metal., , Which of the following is not, correct?, a. Four spheres are involved in the, formation of tetrahedral void., b. The centres of spheres in octahedral, voids are at the apices of a regular, tetrahedron., , iv., , What is the unit cell?, , v., , How does electrical conductivity, of a semiconductor change with, temperature? Why?, , vi., , The picture represents bands of, MOs for Si. Label valence band,, conduction band and band gap., , vii., , A solid is hard, brittle and electrically, nonconductor. Its melt conducts, electricity. What type of solid is it?, , c. If the number of atoms is N the, number of octahedral voids is 2N., d. If the number of atoms is N/2, the, number of tetrahedral voids is N., vii., , A compound forms hcp structure., Number of octahedral and tetrhedral, voids in 0.5 mole of substance is, respectively, a. 3.011×1023, 6.022×1023, b. 6.022×1023, 3.011×1023, c. 4.011×1023, 2.011×1023, d. 6.011×1023, 12.022×1023, , vii., , Pb has fcc structure with edge length, of unit cell 495 pm. Radius of Pb, atom is, a. 205 pm , , b. 185 pm, , c. 260 pm , , d. 175 pm, , viii. Mention two properties that are, common to both hcp and ccp lattices., , 2. Answer the following in one or two, sentences, i., , ii., , iii., , ix., , Sketch a tetrahedral void., , x., , What are ferromagnetic substances?, , 3. Answer the following in brief., , What are the types of particles in, each of the four main classes of, crystalline solids ?, Which of the three types of packing, used by metals makes the most, efficient use of space and which, makes the least efficient use?, The following pictures show, population of bands for materials, , 26, , i., , What are valence, conduction band?, , ii., , Distinguish between ionic solids and, molecular solids., , iii., , Calculate the number of atoms in fcc, unit cell., , iv., , How are the spheres arranged in first, layer of simple cubic close-packed, structures? How are the successive, , band, , and
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layers of spheres placed above this, layer?, v., , Calculate the packing efficiency of, metal crystal that has simple cubic, structure., , vi., , What are paramagnetic substances?, Give examples., , vii., , What are the consequences of, Schottky defect?, , g/cm3. What is the nature of cubic unit, cell? (fcc or ccp), 10. An element has a bcc structure with, unit cell edge length of 288 pm. How, many unit cells and number of atoms, are present in 200 g of the element?, (1.16×1024, 2.32×1024), 11. Distinguish with the help of diagrams, metal conductors, insulators and, semiconductors from each other., 12. What are n-type semiconductors? Why, is the conductivity of doped n-type, semiconductor higher than that of pure, semiconductor? Explain with diagram., 13. Explain, with, diagram,, Frenkel, defect. What are the conditions for its, formation? What is its effect on density, and electrical neutrality of the crystal?, 14. What is an impurity defect? What are, its types? Explain the formation of, vacancies through aliovalent impurity, with example., , viii. Cesium chloride crystallizes in cubic, unit cell with Cl ions at the corners, and a Cs⊕ ion in the centre of the, cube. How many CsCl molecules are, there in the unit cell?, ix., , Cu crystallizes in fcc unit cell with, edge length of 495 pm. What is the, radius of Cu atom?, , x., , Obtain the relationship between, density of a substance and the edge, length of unit cell., , 4. The density of iridium is 22.4 g/cm3. The, unit cell of iridium is fcc. Calculate the, radius of iridium atom. Molar mass of, iridium is 192.2 g/mol. (136 pm), 5. Aluminium crystallizes in cubic close, packed structure with unit cell edge, length of 353.6 pm. What is the radius of, Al atom? How many unit cells are there, in 1.00 cm3 of Al? (125 pm, 2.26×1022), 6. In an ionic crystalline solid atoms of, element Y form hcp lattice. The atoms, of element X occupy one third of, tetrahedral voids. What is the formula, of the compound ? (X2Y3), 7. How are tetrahedral and octahedral, voids formed?, 8. Third layer of spheres is added to second, layer so as to form hcp or ccp structure., What is the difference between the, addition of third layer to form these, hexagonal close-packed structures?, 9. An element with molar mass 27 g/mol, forms cubic unit cell with edge length of, 405 pm. If density of the element is 2.7, , Activity :, •, , With the help of plastic balls,, prepare models of, 1. tetrahedral, voids., , and, , octahedral, , 2. simple cubic, bcc and fcc unit, cells., 3. ccp and hcp lattices., •, , Draw structures of network of, carbon atoms in diamond and, graphite. Discuss with reference to, the following points :, 1. Are the outermost electrons of, carbons in diamond localized or, delocalized ?, 2. Is the energy gap between BMOs, and AMOs in diamond expected, to be large or small ?, , 27
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2. SOLUTIONS, 2.1 Introduction, , combination of these three states of their, components. This gives rise to nine types of, solutions depending on the states of solute, and solvent. These are summarised in table, 2.1., , You are familiar with mixtures. The, mixture is a combination of two or more, substances. Air is a mixture of gases, rock, is a mixture of two or more minerals and so, forth., , Table 2.1 : Types of solutions, , Recall that the mixtures are (a), homogeneous in which mixing of components, is uniform or (b) heterogeneous which have, nonuniform mixing of components., , No. State of State of Some examples, solute solvent, 1, Solid, Liquid Sea water, benzoic, acid in benzene,, sugar in water, 2, Solid, Solid, Metal alloys such, as brass, bronze., 3, Solid, Gas, Iodine in air, 4, Liquid Liquid Gasoline, ethanol, in water., 5, Liquid Solid, Amalgams of, mercury with, metals as mercury, in silver, Chloroform in, 6, Liquid Gas, nitrogen, 7, Gas, Liquid Carbonated water, (CO2 in water),, oxygen in water., 8, Gas, Solid, H2 in palladium, 9, Gas, Gas, Air (O2, N2, Ar and, other gases), , We studied in standard XI, that, homogeneous mixtures are classified, according to the size of their constituent, particles as colloids or as true solutions. In, this chapter we deal with the properties of, homogeneous mixtures especially of true, solutions., Can you recall ?, The size of particles of colloids, and those of true solutions., The solutions are commonly found, in all life processes. The body fluids are, solutions. The solutions are also important for, industrial processes, and many other areas., Can you recall ?, The terms solute and solvent., , Our focus, in this chapter, will be, on the solution of solid in liquid with some, attention to a solution of gas in liquid. The, solvent in most of the cases will be water., , The solution is a homogeneous, mixture of two or more pure substances. A, true solution consists of a solvent and one, or more solutes. We explore the properties, of binary true solutions containing only one, solute., , Can you recall ?, The different units used to express, the concentrations of solutions., , 2.2 Types of solutions, We generally think that a solution is a, either solid dissolved in liquid or a mixture of, two liquids. There are other types of solutions, as well. The solute and the solvent may be, in any of three states namely, solid, liquid, or gas. The solutions thus may involve any, , 2.3 Capacity of solution to dissolve solute, Chemists need to know the capacity of, solutions to dissolve a solute. Suppose that a, solute is added to a solvent. It dissolves readily, at first. The dissolution then slows down as, more solute is added. If we continue the, , 28
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and ion-ion attractions between Na⊕ and Cl, ions are comparable., , addition of solute, the dissolution stops. The, solution at this point is said to be saturated., A dynamic equilibrium can be reached where, the number of solute molecules leaving the, crystal to pass into solution is equal to the, number returning from the solution to the, crystal. Thus,, Solute + solvent, , dissolution, crystallization, , Can you recall ?, The types of force, molecules., , between, , Nonpolar organic compounds like, cholesterol dissolves in nonpolar solvent such, as benzene., , Solution, , A, saturated solution contains, maximum amount of solute dissolved, in a given amount of solvent at a given, temperature. A solution containing greater, than the equilibrium amount of solute is said, to be supersaturated solution. Such solutions, are unstable. The precipitation occurs by the, addition of a tiny crystal of solute and the, supersaturated solution changes to saturated, solution., , Can you tell ?, Why naphthalene dissolves in, benzene but not in water?, Sugar dissolves in water! The, dissolution of sugar in water is due to, intermolecular hydrogen bonding between, sugar and water., ii. Effect of temperatute on solubility :, How solubility of substance changes with, temperature depends on enthalpy of solution., Many solids, for example KCl, dissolve in, water by endothermic process, that is, with, the absorption of heat. When temperature is, increased by adding heat to the system, the, solubility of substance increases according to, the Le-Chatelier principle. Addition of heat, causes a stress on the saturated solution. This, stress favours endothermic process., , 2.4 Solubility : The solubility of a solute is its, amount per unit volume of saturated solution, at a specific temperature. The solubility of, a solute is its maximum concentration and, expressed in the concentration units mol L-1., 2.4.1 Factors affecting solubility : The extent, to which a substance dissolves in a solvent, varies greatly with different substances. It, depends on the nature of solute and solvent,, temperature and pressure., , Can you recall ?, , i. Nature of solute and solvent : Generally the, compounds with similar chemical character, are more readily soluble in each other, than those with entirely different chemical, characters. The saying that ‘like dissolves, like’ guides to predict the solubility of a solute, in a given solvent., , Le-Chatelier principle, exothermic, and endothermic processes., On the other hand, when the substance, dissolves in water by an exothermic process, its solubility decreases with an increase of, temperature. The substances such as CaCl2, and Li2SO4.H2O dissolve in water releasing, heat., , Thus, substances having similar, intermolecular forces are likely to be soluble, in each other. Generally polar solutes, dissolve in polar solvents. This is because, in these, solute-solute, solute-solvent and, solvent-solvent interactions are all of similar, magnitude. For example, NaCl dissolves, in water. The strong ion-dipole interactions, of Na⊕ and Cl ions with water molecules,, hydrogen bonding between water molecules, , Can you tell ?, Anhydrous sodium sulphate, dissolves in water with the, evolution of heat. What is the effect, of temperature on its solubility ?, , 29
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It is important to understand that there, is no direct correlation between solubility and, exothermicity or endothermicity. For example,, dissolution of CaCl2 in water is exothermic and, that of NH4NO3 is endothermic. The solubility, of these increases with the temperature., , iii. Effect of pressure on solubility, Pressure has no effect on the, solubilities of solids and liquids as they are, incompressible. However pressure greatly, affects solubility of gases in liquids. The, solubility of gases increases with increasing, pressure. The quantitative relationship between, gas solubility in a liquid and pressure is given, by Henry’s law., , Figure 2.1 shows the result of, experimental determination of solubilities, of some ionic solids in water at various, temperatures., Following, are, some, experimental observations from Fig 2.1, , Henry’s law: It states that the solubility of a, gas in a liquid is directly proportional to the, pressure of the gas over the solution. Thus,, S ∝ P or S = KHP, , (2.1), , where, S is the solubility of the gas in, mol L-1, P is the pressure of the gas in bar, over the solution. KH, the proportionality, constant is called Henry’s law constant., S, mol L-1, = bar, Units of KH : KH =, P, = mol L-1 bar-1, When P = 1 bar, KH = S. Thus, KH is, the solubility of the gas in a liquid when its, pressure over the solution is 1 bar., , Fig. 2.1 : Variation of solubilities of some ionic, solids with temperature, , i. Solubilities of NaBr, NaCl and KCl change, slightly with temperature., ii. Solubilities of KNO3, NaNO3 and, KBr increase appreciably with increasing, temperature., , Demonstration of Henry’s law., Before sealing the bottle of soft, drink, it is pressurised with a mixture of air,, CO2 saturated with water vapour. Because, of high partial pressure of CO2, its amount, dissolved in soft drink is higher than the, solubility of CO2 under normal conditions., When the bottle of soft drink is, opened, excess dissolved CO2 comes out, with effervescence., , iii. Solubility of Na2SO4 decreases with, increase of temperature., The solubility of gases in water usually, decreases with increase of temperature., When gases are dissolved in water, the gas, molecules in liquid phase are condensed. The, condensation is an exothermic process. The, solubility of gases in water must decrease as, temperature is raised., , Exceptions to Henry’s law, Gases like NH3 and CO2 do not obey, Henry’s law. The reason is that these gases, react with water., , In united states about 1000000, billion galons of water from rivers and, lakes are used for industrial cooling., The cooling process heats water. The hot, water then returns to rivers and lakes. The, solubility of oxygen decreases in hot water, thereby affecting the life of cold blooded, animals like fish., , NH3+ H2O, CO2 + H2O, , NH4⊕ + OH, H2CO3, , Because of these reactions, NH3 and, CO2 gases have higher solubilities than, expected by Henry’s law., , 30
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2.5 Vapour pressure of solutions of liquids, in liquids : Consider a binary solution of two, volatile liquids A1 and A2. When the solution, is placed in a closed container, both the, liquids vaporize. Eventually an equilibrium, is established between vapor and liquid, phases. Both the components are present in, the vapour phase. The partial pressure of the, components are related to their mole fractions, in the solution. This realationship is given by, Raoult’s law., , Do you know ?, O2 gas has very low solubility in, water. However, its solubility in, blood is exceedingly high. This is because, of binding of O2 molecule to haemoglobin, present in blood., Hb + 4O2, , Hb(O2)4, , Problem 2.1 : The solubility of N2 gas in, water at 25 0C and 1 bar is 6.85 × 10-4 mol, L-1. Calculate (a) Henry’s law constant, (b) molarity of N2 gas dissolved in water, under atmospheric conditions when partial, pressure of N2 in atmosphere is 0.75 bar., , 2.5.1 Raoult’s law : It states that the partial, vapour pressure of any volatile component of, a solution is equal to the vapour pressure of, the pure component multiplied by its mole, fraction in the solution., , Solution :, , Suppose that for a binary solution, of two volatile liquids A1 and A2, P1 and P2, are their partial vapour pressures and x1 and, x2 are their mole fractions in solution. Then, according to Raoult’s law,, , S 6.85 × 10-4 mol dm-3, a. KH = =, 1 bar, P, = 6.85 × 10-4 mol L-1 bar-1, b. S = KHP = 6.85 × 10-4 mol L-1 bar-1, , we write P1 = x1P10 and P2 = x2P20, , × 0.75 bar, , where P10 and P20 are vapour pressures of, pure liquids A1 and A2, respectively., , = 5.138 × 10 mol L-4, , (2.2), , 1, , According to Dalton’s law of partial, pressures, the total pressure P above the, solution is,, , Problem 2.2 : The Henry’s law constant, of methyl bromide (CH3Br), is 0.159 mol, L-1 bar-1 at 250C. What is the solubility of, methyl bromide in water at 250C and at, pressure of 130 mmHg?, , P = P1 + P2, = P10x1 + P20x2, , (2.3), , Solution :, , Since x1 = 1- x2, the Eq. (2.2) can also be, written as, , According to Henry’s law, , P = P10 (1-x2) + P20x2, , S = K HP, , = P10 - P10x2 + P20x2, = (P20 - P10) x2 + P10, , 1. KH = 0.159 mol L-1 bar-1, 1, 760 mm Hg/atm, = 0.171 atm × 1.013 bar/atm, , (2.4), , Because P10 and P20 are constants, a plot, of P versus x2 is a straight line as shown in, the Fig 2.2. The figure also shows the plots, of P1 versus x1 and P2 versus x2 according, to the equations 2.2. These are straight lines, passing through origin., , 2. P = 130 mm Hg ×, , = 0.173 bar, Hence, S = 0.159 mol L-1 bar-1 × 0.173 bar, = 0.271 M, , 31
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Ideal solution, , P2, , Vapour Pressure, , PTotal = P1 + P2, , P10, , 600 - 450 = 150 = 250x2 or x2 =, , 0, , x1 = 1-x2 = 1-0.6 = 0.4, , ii. Compositions of A and B in vapour are y1, and y2 respectively., , P2, P1, Mole fraction, x2, , x1 = 1, x1 = 0, , 150, = 0.6, 250, , P1 = y1P and P2 = y2P, P1 = P10x1 and, P2 = P20x2, P P 0x, 450 mm Hg × 0.4, y1 = 1 = 1 1 =, P, P, 600 mmHg, = 0.3, , x2 = 0, x2 = 1, , Fig. 2.2 : Variation of vapour pressure with, mole fraction of solute, , y2 = 1 - y1 = 1 - 0.3 = 0.7, , i. For P10 versus x2 straight line,, P = P10 at x2 = 0 and P = P20 at x2 = 1, , 2.5.2 Ideal and nonideal solutions, , ii. For P1 versus x1 straight line,, , 1. Ideal solutions, , P1 = 0 at x1 = 0 and P1 = P10, at x1 = 1, , i. Ideal solutions obey Raoult’s law over, entire range of concentrations., , iii. For P2 versus x2 straight line,, , ii. No heat is evolved or absorbed when two, components forming an ideal solution are, mixed. Thus, the enthalpy of mixing is, zero. ∆mixH = 0, , P2 = 0 at x2 = 0 and P2 = P20 at x2 = 1, Composition of vapour phase :, The composition of vapour in equilibrium, with the solution can be determined by, Dalton’s law of partial pressures., , iii. There is no volume change when two, components forming an ideal solution are, mixed. Thus volume of an ideal solution, is equal to the sum of volumes of two, components taken for mixing., , If we take y1 and y2 as the mole, fractions of two components in the vapour,, then P1 = y1P and P2 = y2P, , ∆mixV = 0, , where P1 and P2 are the partial pressures of, two components in the vapour and P is the, total vapour pressure., , iv. In an ideal solution solvent-solute, solutesolute and solvent-solvent molecular, interactions are comparable., , Problem 2.3 : The vapour pressures of, pure liquids A and B are 450 mm Hg and, 700 mm Hg, respectively at 350 K. Find the, composition of liquid and vapour if total, vapour pressure is 600 mm., , v. The vapour pressure of ideal solution, always lies between vapour pressures of, pure components, as shown in Fig. 2.2., It is important to understand that, perfectly ideal solutions are uncommon and, solutions such as benzene + toluene behave, nearly ideal., , Solution : i. Compositions of A and B in the, solution are x1 and x2, P = (P20 - P10) x2 + P10, , 2. Nonideal solutions, , P10 = 450 mmHg, P20 = 700 mmHg,, , i., , P = 600 mmHg, Hence, 600 mm Hg = (700 mm Hg -, , ii. The vapour pressures of these solutions, can be higher or lower than those of pure, components., , 450 mm Hg)x2 + 450 mm Hg, , , These solutions do not obey Raoult’s law, over the entire range of concentrations., , = 250x2 + 450, , 32
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iii. Deviation from the Raoult's law : These, solutions show two types of deviation, from the Raoult's law., , vapour pressures of such solutions are lower, than those of pure components as shown in, Fig. 2.4. The solutions of phenol and aniline,, chloroform and acetone exhibit negative, deviations from the Raoult’s law., , A. Positive deviations from Raoult’s law -, , Vapour Pressure, , Vapour Pressure, of solution, , P1, , 2.6 Colligative properties of nonelectrolyte, solutions : The physical properties of solutions, that depend on the number of solute particles, in solutions and not on their nature are called, colligative properties. These are, , P2, , 1. vapour pressure lowering, 2. boiling point elevation, , Mole fraction, x1, x2, , x1 = 0, x2 = 1, , 3. freezing point depression, , x1 = 1, x2 = 0, , 4. osmotic pressure, , Fig. 2.3 : Positive derivations from, Raoult's law, , While dealing with colligative properties of, nonelectrolyte solutions, the relatively dilute, solutions with concentrations 0.2 M or less, are considered., , The solutions in which solute-solvent, intermolecular attractions are weaker than, those between solute-solute molecules and, solvent-solvent molecules, exhibit positive, deviations. The vapour pressures of such, solutions are higher than those of pure, components as shown in Fig. 2.3. The solutions, of ethanol and acetone, carbon disulphide and, acetone show positive deviations from the, Raoult’s law., , 2.7 Vapour pressure lowering : When a, liquid in a closed container is in equilibrium, with its vapours, the pressure exerted by the, vapour on the liquid is its vapour pressure., Can you recall ?, Vapour pressure of a liquid, i., , B. Negative deviations from Raoult’s law, , Vapour Pressure, , Vapour Pressure, of solution, , x1 = 0, x2 = 1, , P2, , P1, , Mole fraction, x1, x2, , x1 = 1, x2 = 0, , Experiments have shown that when, a nonvolatile, nonionizable solid is, dissolved in a liquid solvent, the vapour, pressure of the solution is lower than, that of pure solvent. In other words the, vapour pressure of a solvent is lowered, by dissolving a nonvoltile solute into it., When the solute is nonvolatile it does, not contribute to the vapour pressure, above the solution. Therefore, the vapour, pressure of solution is equal to the vapour, pressure of solvent above the solution., , ii. If P10 is the vapour pressure of pure, solvent and P1 is the vapour pressure of, solvent above the solution, then, , Fig. 2.4 : Negative derivations from, Raoult's law, , The solutions in which the interactions, between solvents and solute molecules are, stronger than solute-solute or solvent-solvent, interactions, exhibit negative deviations.The, , P1 < P10, The vapour pressure lowering is, 0, , ∆P = P1 - P1 , , 33, , (2.5)
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iii. Why the vapour pressure of solution, containing nonvolatile solute is lower, than that of pure solvent? Vapour, pressure of a liquid depends on the ease, with which the molecules escape from, the surface of liquid. When nonvolatile, solute is dissolved in a solvent, some, of the surface molecules of solvent are, replaced by nonvolatile solute molecules., These solute molecules do not contribute, to vapour above the solution. Thus, the, number of solvent molecules available, for vaporization per unit surface area, of solution is less than the number at, the surface of pure solvent. As a result, the solvent molecules at the surface of, solution vaporize at a slower rate than, pure solvent. This results in lowering of, vapour pressure., , For a binary solution containing one solute,, , 2.7.1 Raoult’s law for solutions of nonvolatile, solutes : We saw in section 2.5.1 that Raoult’s, law expresses the quantitative relationship, between vapour pressure of solution and, vapour pressure of solvent., , The ratio of vapour pressure lowering, of solvent divided by the vapour pressure of, pure solvent is called relative lowering of, vapour pressure. Thus, Relative vapour pressure lowering, , In solutions of nonvolatile solutes, the, law is applicable only to the volatile solvent., The law states that the vapour pressure of, solvent over the solution is equal to the vapour, pressure of pure solvent multiplted by its mole, fraction in the solution. Thus, , P 1 - P1, ∆P, =, =, 0, 0, P1, P1, The Eq. (2.6) shows that, , x1 = 1 - x2, It therefore, follows that, 0, , P1 = P1 x1, 0, , = P1 (1 - x2), 0, , or, , 0, , 0, , P1 - P1 = P1 x2, 0, , The Eq. (2.5) defines P1 - P1 as ∆P,, the lowering of vapour pressure. Hence, 0, , ∆P = P1 x2 , , (2.6), , The Eq. (2.6) shows that ∆P depends, on x2 that is on number of solute particles., Thus, ∆P, the lowering of vapour pressure is, a colligative property., 2.7.2 Relative lowering of vapour pressure, , 0, , (2.7), , 0, , P 1 - P1, ∆P, = x2 =, (2.8), 0, 0, P1, P1, Thus, relative lowering of vapour, pressure is equal to the mole fraction of solute, in the solution. Therefore, relative vapour, pressure lowering is a colligative property., , 0, , P 1 = P 1 x1, A plot of P1 versus x1 is a straight line, as shown in Fig. 2.5, , 2.7.3 Molar mass of solute from vapour, pressure lowering : We studied that the, relative lowering of vapour pressure is equal, to the mole fraction x2 of solute in the solution., From Eq. 2.6, it follows that :, ∆P = x, (2.8), 0, 2, P1, Recall (Chapter 10, sec 10.5.6) that, the mole fraction of a component of solution, is equal to its moles divided by the total, moles in the solution. Thus,, n2, x2 = n +n, 1, 2, , Vapour Pressure of solvent, x = 0 Mole fraction of solvent, , 0, , = P 1 - P 1 x2, , x=1, , Fig. 2.5 : Variation of vapour pressure of, solution with mole fraction of solvent, , 34
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where n1 and n2 are the moles of solvent and, solute respectively, in the solution., , Problem 2.5 : The vapour pressure of, pure benzene (molar mass 78 g/mol) at, a certain temperature is 640 mm Hg. A, nonvolatile solute of mass 2.315 g is added, to 49 g of benzene. The vapour pressure, of the solution is 600 mm Hg. What is the, molar mass of solute?, 0, W2 × M1, P 1 - P1, =, 0, M2 × W1, P1, , We are concerned only with dilute, solutions hence n1 >> n2 and n1+n2 ≈ n1. The, mole fraction x2 is then given by, n2, x2 = n, and, 1, ∆P = n2, (2.9), 0, n1, P, 1, , Suppose that we prepare a solution, by dissolving W2 g of a solute in W1 g of, solvent. The moles of solute and solvent in, the solution are then,, W2, W1, n2 = M and n1 = M, (2.10), 2, 1, where M1 and M2 are molar masses of, solvent and solute, respectively. Substitution, of Eq. (2.10) into Eq. (2.9) yields, W2/M2, ∆P, x2 =, 0 =, W1/M1, P1, 0, W2M1, P 1 - P1, ∆P, = 0 = MW, (2.11), 0, P1, P, 2, 1, , 0, , P1 = 640 mm Hg, P1 = 600 mm Hg,, W1 = 49 g, W2 = 2.315 g, 640 mm Hg- 600 mm Hg, 640 mm Hg, 2.315 g - 78 g/mol, =, 40 g × M2, 40 mm Hg, 2.315 g - 78 g/mol, 640 mm Hg =, 40 g × M, , Hence,, , 2, , 2.315 g - 78 g/mol × 640 mm Hg, M2=, 40 mm Hg × 40 g, , = 72.23 g mol, 2.8 Boiling point elevation : Recall that the, boiling point of liquid is the temperature at, which its vapour pressure equals the applied, pressure. For liquids in open containers the, applied pressure is atmospheric pressure., , 1, , Knowing all other quantities, molar, mass of solute M2 can be calculated., Problem 2.4 : A solution is prepared by, dissolving 394 g of a nonvolatile solute, in 622 g of water. The vapour pressure of, solution is found to be 30.74 mm Hg at, 30 0C. If vapour pressure of water at 30 0C, is 31.8 mm Hg, what is the molar mass of, solute?, Solution :, 0, , P 1 - P1, ∆P, = 0, 0, P1, P1, , =, , It has been found that the boiling, point of a solvent is elevated by dissolving, a nonvolatile solute into it. Thus, the, solutions containing nonvolatile solutes boil, at temperatures higher than the boiling point, of a pure solvent., , W2M1, M2W1, , 0, , If T b is the boiling point of a pure, 0, solvent and Tb that of a solution, Tb > T b, . The difference between the boiling point of, solution and that of pure solvent at any given, constant pressure is called the boiling point, elevation., , W2 = 394 g, W1 = 622 g, M1 = 18 g mol-1,, 0, , P1 = 30.74 mm Hg, P1 = 31.8 mm Hg, Substitution of these quantities into the, equation gives, 31.8 mm Hg - 30.74 mm Hg, 31.8 mm Hg, , 0.0333 =, M2 =, , =, , ∆Tb = Tb - T0b, , 394 g ×18 gmol, M2 × 622 g, , -1, , 11.4 g mol-1, M2, , 11.4 g mol-1, 0.0333, , = 342 g mol-1, , 35, , (2.12)
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2.8.1 Boiling point elevation as a consequence, of vapour pressure lowering : To understand, the elevation of boiling point, let us compare, the vapour pressures of solution and those of, pure solvent. The vapour pressures of solution, and of pure solvent are plotted as a function, of temperature as shown in Fig. 2.6., , A, , Vapour Pressure, , 760 mm, , higher temperature to cause it to boil than, the pure solvent., 2.8.2 Boiling point elevation and, concentration of solute : The boiling point, elevation is directly proportional to the, molality of the solution. Thus,, ∆Tb ∝ m, , C, , If m = 1,, , (2.13), , ∆Tb = Kb, , Thus, ebullioscopic constant is the boiling, point elevation produced by 1 molal solution., , Solution, 0, , Temperature, , ∆Tb = Kbm, , where m is the molality of solution. The, proportionality constant Kb is called boiling, point elevation constant or molal elevation, constant or ebullioscopic constant., , Solvent, B, D, , or, , Tb, , ∆T, , Tb, , K, , Units of Kb : Kb = m b = mol kg-1 =K kg mol-1, , Fig. 2.6 : Vapour pressure-temperature of pure, solvent and solution, , Remember..., Kb and ∆Tb are the differences, between two temperatures. Hence,, their values will be the same in K or in 0C., , As stated in section 2.7 that at any, temperature the vapour pressure of solution, is lower than that of pure solvent. Hence, the, vapour pressure-temperature curve of solution, (CD) lies below that of solvent (AB). The, difference between the two vapour pressures, increases as temperature and vapour pressure, increase as predicted by the equation, , We are dealing with the systems whose, temperature is not constant. Therefore, we, cannot express the concentration of solution, in molarity which changes with temperature, whereas molality is temperature independent., , ∆P = x2 P10, , Therefore the concentration of solute is, expressed in mol/kg (molality) rather than, mol/L (molarity)., , The intersection of the curve CD with, the line corresponding to 760 mm is the boiling, point of solution. The similar intersection of, the curve AB is the boiling point of pure, solvent. It is clear from the figure that the, boiling point (Tb) of the solution is higher, than that of pure solvent (Tb0), , 2.8.3 Molar mass of solute from boiling, point elevation, The Eq. (2.13) is ∆Tb = Kbm, Suppose we prepare a solution by, dissolving W2 g of solute in W1 g of solvent., , At the boiling point of a liquid, its, vapour pressure is equal to 1 atm. In order to, reach boiling point, the solution and solvent, must be heated to a temperature at which, their respective vapour pressures attain 1 atm., At any given temperature the vapour pressure, of solution is lower than that of pure solvent., Hence, the vapour pressure of solution needs, higher temperature to reach 1 atm than that, needed for vapour pressure of solvent. In, other words the solution must be heated to, , W, , 2, Moles of solute in W1 g of solvent =, M2, where M2 is the molar mass of solute., , W1g, , W1, , Mass of solvent = W1 g = 1000 g/kg =1000 kg, Recall the expression of molality, m., m=, , moles of solute, mass of solvent in kg, W2/M2 mol, , = W /1000 kg =, 1, , 36, , 1000 W2, mol kg-1, M2W1, , (2.14)
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Substitution of this value of m in Eq. (2.13), gives, 1000 W2, ∆Tb = Kb M W, 2, 1, 1000 KbW2, , ∆TbW1, , 2.9 Depression in freezing point, , (2.15), , Recall that freezing point of a liquid, is the temperature at which liquid and solid, are in equilibrium and the two phases have, the same vapour pressure., The general experimental observation, is that the freezing point of a solvent is, lowered by dissolving a nonvolatile solute into, it. Thus, freezing point of solution containing, a nonvolatile solute is lower than that of pure, solvent., , Solution :, 1000 KbW2, , ∆TbW1, , If Tf 0 is the freezing point of pure, solvent and Tf that of a solution in which, nonovolatile solute is dissolved, Tf 0 > Tf. The, difference between the freezing point of pure, solvent and that of the solution is depression, in freezing point., , W2 = 50 g, W1 = 150 g,, ∆Tb = Tb-T0b = 84.27 0C - 77.06 0C, = 7.210C = 7.21 K, Kb = 2.77 0C kg mol-1 = 2.77 K kg mol-1, Substitution of these in above equation, , Thus, , 1000 g Kg-1 × 2.77 K kg mol-1 × 50 g, M2 =, 7.21 K × 150 g, , = 128 g mol, , The effect of dissolution of a, nonvolatile solute into a solvent on freezing, point of solvent can be understood in terms, of the vapour pressure lowering., , Problem 2.7 : 3.795 g of sulphur is dissolved, in 100 g of carbon disulfide. This solution, boils at 319.81 K. What is the molecular, formula of sulphur in solution? The boiling, point of the solvent is 319.45 K., (Given that Kb for CS2 = 2.42 K kg mol-1 and, atomic mass of S = 32 u, , id, , 1000 g kg-1 × 2.42 K kg mol-1 × 3.795 g, 0.36 K × 100 g, , = 255.10 g mol-1, , t, en, so, lv, , B, , qu, Li, , E, , t, en D, , lv, so, , F, , n, , tio, , lu, So, , lid, , ∆Tb = (319.81 - 319.45)K = 0.36 K, , C, , So, , Vapour Pressure, , 1000 KbW2, , ∆TbW1, W1 = 100 g, W2 = 3.795 g, M2=, , (2.16), , ∆Tf = T f0 - Tf, , 2.9.1 Freezing point depression as a, consequence of vapour pressure lowering, , -1, , M2 =, , 32, , The molecular formula would be S8 in CS2, , Problem 2.6 : The normal boiling point of, ethyl acetate is 77.06 0C. A solution of 50, g of a nonvolatile solute in 150 g of ethyl, acetate boils at 84.27 0C. Evaluate the, molar mass of solute if Kb for ethyl acetate, is 2.77 0C kg mol-1., M2 =, , = 255.1, , = 7.92 ≈ 8, , Hence,, , M2 =, , molar mass of S, atomic mass of S, , =, , A, , T, T0, Temperature, , Fig. 2.7 : Variation of vapour pressure with, temperature of pure solvent, solid solvent and, solution, , Atomic mass of S = 32 u, Therefore number of atoms in a molecule of, sulphur, , 37
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∆Tf ∝ m, , Consider the vapoure pressuretemperature diagram as shown in Fig. 2.7., The diagram consists of three curves. AB is, the vapour pressure curve of solid solvent, while CD is the vapour pressure curve of, pure liquid solvent. EF is the vapour pressure, curve of solution that always lies below the, pure solvent., , or, , ∆Tf = Kfm, , (2.17), , The proportionality constant Kf is, called freezing point depression constant or, cryoscopic constant., If m = 1, ∆Tf = Kf . The cryoscopic, constant thus is the depression in freezing, point produced by 1 molal solution of a, nonvolatile solute., , It is important to note that solute does, not dissolve in solid solvent., , Unit of Kf :, , ∆Tf, , K or 0C, , m = mol kg-1, = K kg mol-1 or 0C kg mol-1, , The curves AB and CD intersect at, point B where solid and liquid phases of pure, solvent are in equilibrium. The two phases, have the same vapour pressure at B. The, temperature corresponding to B is the freezing, point of solvent ( T f0)., , 2.9.3 Molar mass of solute from freezing, point depression, Refer to Eq. (2.17), ∆Tf = Kfm, The molality m of the solution is given by, Eq. (2.14) as, 1000W2, m=, M2W1, Substitution of Eq. (2.14) into Eq. (2.17) gives, 1000W, ∆Tf = Kf M W 2, 2, 1, , Similarly at E, the point of intersection, of EF and AB, the solid solvent and solution, are in equilibrium. They have the same vapour, pressure at E. The temperature corresponding, to E is the freezing point of solution, Tf ., It is clear from the figure that freezing, point of solution Tf is lower than that of pure, solvent Tf0. It is obvious because the vapour, pressure curve of solution lies below that of, solvent., , Hence, M2 =, , 1000KfW2, ∆Tf W1, , (2.18), , Problem 2.8 : 1.02 g of urea when, dissolved in 98.5 g of certain solvent, decreases its freezing point by 0.211K., 1.609 g of unknown compound when, dissolved in 86 g of the same solvent, depresses the freezing point by 0.34 K., Calculate the molar mass of the unknown, compound., , Why freezing point of solvent is lowered, by dissolving a nonvolatile solute into it?, At the freezing point of a pure liquid the, attractive forces among molecules are large, enough to cause the change of phase from, liquid to solid., In a solution, the solvent molecules are, separated from each other because of solute, molecules. Thus, the separation of solvent, molecules in solution is more than that in, pure solvent. This results in decreasing the, attractive forces between solvent molecules., Consequently, the temperature of the solution, is lowered below the freezing point of solvent, to cause the phase change., 2.9.2 Freezing point depression and, concentration of solute : As varified, experimentally for a dilute solution the, freezing point depression (∆Tf) is directly, proportional to the molality of solution. Thus,, , (Molar mass of urea = 60 g mol-1), Solution :, Urea , , Unknown compound, , W2 = 1.02 g , , W2' = 1.609 g, , W1 = 98.5 g , , W1' = 86 g, , ∆Tf = 0.211 K , , ∆Tf ' = 0.34 K, , M2 = 60 gmol-1, M2' =?, M × ∆T × W1, 1000Kf= 2 W f, 2, , 38
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the vapour pressure of solvent is greater than, that of solution. The net spontaneous flow of, solvent molecules into the solution or from, more dilute solution to more concentrated, solution through a semipermeable membrane, is called osmosis. See Fig. 2.8., , M2' × ∆Tf' × W1', 1000Kf =, W2', , -1, = 60 g mol × 0.211 K × 98.5 g, , 1.02 g, , ', = M2 × 0.34K × 86 g, 1.609 g, , Semipermeable, membrane, , Semipermeable, membrane, , 1247.01 g K mol-1, M2' × 29.24 K, =, 1.02, 1.609, Net flow of solvent, , 1222.55 g K mol-1 = M2' × 18.173, M2' =, , 1222.55 g K mol-1, = 67.3 g mol-1, 18.175 K, , Solvent, , 2.10 Osmotic pressure : Besides the boiling, point elevation and freezing point depression,, the osmotic pressure is associated with, vapour pressure lowering and can be used to, determine molar masses of dissolved solutes., , Solution, , Net flow of solvent, , Solution of Solution of, lower, higher, concentration concentration, , Fig. 2.8 : Osmosis, , As a result of osmosis, the amount of, liquid on the pure solvent side or more dilute, solution side decreases. Consequently, the, amount of liquid on the other side increases., This results in decrease of the concentration, of solution., , Semipermeable membrane : The osmotic, pressure phenomenon involves the use of, semipermeable membrane., It is a film such as cellophane which, has pores large enough to allow the solvent, molecules to pass through them. These pores, are small enough not to allow the passage, of large solute molecules or ions of high, molecular mass through them. The semipermeable membrane selectively allows, passage of solvent molecules., , 2.10.2 Osmotic pressure : Osmosis can be, demonstrated with experimental set up shown, in Fig. 2.9, Semipermeable membrane is firmly, fastened across the mouth of thistle tube., The solution of interest is placed inside an, inverted thistle tube. This part of the tube, and the membrane are then immersed in a, container of pure water., , 2.10.1 Osmosis : When a solution and, pure solvent or two solutions of different, concentrations are separated by a semipermeable membrane, the solvent molecules, pass through the membrane., , Hydrostatic, pressure (hg), , What is the direction of flow of solvent, molecules? It is important to understand that, the passage of solvent molecules through the, semipermeable membrane takes place in both, directions, since solvent is on both sides of, the membrane. However, the rate of passage, of solvent molecules into the solution or from, more dilute solution to more concentrated, solution is found to be greater than the rate in, the reverse direction. This is favourable since, , Sugar solution, Semipermeable, membrane, Water, , Fig. 2.9 : Osmosis and osmotic pressure, , 39
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As a result, some of the solvent passes, through the membrane into the solution. It, causes the liquid level in the tube to rise. The, liquid column in the tube creates hydrostatic, pressure that pushes the solvent back through, the membrane into the container. The column, of liquid in the tube continues to rise and, eventually stops rising. At this stage, hydrostatic pressure developed is sufficient, to force solvent molecules back through the, membrane into the container at the same rate, they enter the solution., , ii. Hypertonic and hypotonic solutions :, If two solutions have unequal osmotic, pressures, the more concentrated solution, with higher osmotic pressure is said to be, hypertonic solution., The more dilute solution exhibiting, lower osmotic pressure is said to be hypotonic, solution., For example, if osmotic pressure of, sucrose solution is higher than that of urea, solution, the sucrose solution is hypertonic, to urea solution, and the urea solution is, hypotonic to sucrose solution., , An equilibrium is thus established, where rates of forward and reverse passages, are equal. The height of liquid column in the, tube remains constant. This implies that the, hydrostatic pressure has stopped osmosis., , 2.10.4 Osmotic pressure and concentration, of solution, For very dilute solutions, the osmotic, pressure follows the equation,, n RT, π= 2, , (2.19), V, where V is the volume of a solution in dm3, containing n2 moles of nonvolatile solute. R, is the gas constant equal to 0.08206 dm3 atm, K-1mol-1 and π is osmotic pressure in atm., , Remember..., It is important to note that, osmotic pressure is not the pressure, produced by a solution. It exists only when, the solution is separated from the solvent by, a suitable kind of semipermeable membrane., , The term n2/V is concentration in, molarity (M). Eq. (2.19) thus can be written, as, , The hydrostatic pressure that stops, osmosis is an osmotic pressure (π) of the, solution. The hydrostatic pressure is equal to, hρg where h is the height of liquid column, in the tube, ρ is density of solution and g is, acceleration due to gravity., , π = MRT , , (2.20), , Note that the solute concentration, is expressed in molarity while calculating, osmotic pressure rather than molality. The, reason is that osmotic pressure measurements, are made at a specific constant temperature., It is not necessary to express concentration in, a temperature independent unit like molality., , 2.10.3 Isotonic, hypertonic and hypotonic, solutions, i. Isotonic solutions : Two or more solutions, having the same osmotic pressure are said to, be isotonic solutions., , 2.10.5 Molar mass of solute from osmotic, pressure, n RT, Consider Eq. (2.19), π= 2, V, If the mass of solute in V litres of, solution is W2 and its molar mass is M2 then, n2 = W2/M2. With this value of n2, Eq. (2.19), becomes, W2RT, W2RT, π=, or M2 =, (2.21), πV, M2V, , For example, 0.1 M urea solution and, 0.1 M sucrose solution are isotonic because, their osmotic pressures are equal. Such, solutions have the same molar concentrations, but different concentrations in g/L. If these, solutions are separated by a semipermeable, membrane, there is no flow of solvent in, either direction., , 40
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Remember..., , W2 = 0.822 g, , Osmotic, pressure, is, much larger and therefore more, precisely measurable property than other, colligative properties. It is therefore,, useful to determine molar masses of, very expensive substances and of the, substances that can be prepared in small, quantities., , R = 0.08205 dm3 atm K-1 mol-1, T = 298K, 149 (mmHg), π = 149 (mmHg) = 760 (mmHg/atm), = 0.196 atm, V = 300 mL = 0.3dm3, M2 =, , 2.10.6 Reverse osmosis : As mentioned, earlier osmosis is a flow of solvent through, a semipermeable membrane into the solution., , = 342 g mol-1, 2.11 Collgative properties of electrolytes, , The direction of osmosis can be, reversed by applying a pressure larger than, the osmotic pressure., , Pressure > π, Fresh water, Water, outlet, , Can you recall ?, Electrolytes and nonelectrolytes., Solutions of nonelectrolytes in water, exhibit colligative properties as described, in the preceeding sections. These solutions, also give, for the dissolved substances, the, molar masses expected from their chemical, formulae., , Piston, , Salt water, , The study of colligative properties, of electrolytes, however, require a different, approach than used for colligative properties, of nonelectrolytes. Following are the, experimental observations for the colligative, behavior of electrolytes., , Semipermeable membrane, , Fig. 2.10 : Reverse osmosis, , The pure solvent then flows from, solution into pure solvent through semipermeable membrane. This phenomenon is, called reverse osmosis. Fig. 2.10 shows the, schematic set up for reverse osmosis. Fresh, water and salty water are separated by a, semipermeable membrane. When the pressure, larger than the osmotic pressure of solution, is applied to solution, pure water from salty, water passes into fresh pure water through, the membrane., , i. The solutions of elctrolytes also exhibit, colligative properties which do not obey the, relations of nonelectrolytes., ii. The colligative properties of the solutions, of electrolytes are greater than those to be, expected for solutions of nonelectrolytes of, the same concentration., iii. The molar masses of electrolytes in, aqueous solutions determined by colligative, properties are found to be considerably lower, than the formula masses., , Problem 2.9 : What is the molar mass of a, solute if a solution prepared by dissolving, 0.822 g of it in 300 mdm3 of water has an, osmotic pressure of 149 mm Hg at 298 K?, , Why the colligative properties of, electrolyte solutions are greater than those, for nonelectrolyte solutions of the same, concentration? Recall that electrolytes, dissociate into two or more ions when, , Solution :, M2 =, , 0.822 g × 0.08205 L atm K-1 mol-1 × 298K, 0.196 atm × 0.3dm3, , W2RT, πV, , 41
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Mtheoretical, , (2.24), Mobserved, Thus, i is equal to 1 for nonelctrolyte,, 2 for KNO3 and NaCl, 3 for Na2SO4 and, CaCl2 and so forth. The colligative properties, of these electrolytes are, therefore, twice and, thrice respectively as those of noneletrolytes, of the same concentration., , dissolved in water whereas nonelctrolytes do, not. Stated differently, one formula unit of, electrolyte dissolved in water produces two or, more ions. Consequently number of particles, in solution increases., , =, , The, colligative, properties, of, electrolyte solutions are thus higher than the, nonelectrolyte solutions., , The foregoing arguments are valid, only for infinitely dilute solutions where the, dissociation of electrolytes is complete., , If 1.25 m sucrose solution has ∆Tf of 2.320C,, what will be the expected value of ∆Tf for, 1.25m CaCl2 solution?, , In reality especially at high, concentrations, the colligative properties of, strong electrolytes and their i values are, usually smaller than expected. The reason is, that the electrostatic forces between oppositely, charged ions bring about the formation of ion, pairs. Each ion pair consists of one or more, cations and one or more anions held together, by electrostatic attractive forces. This results, in decrease in the number of particles in, solution causing reduction in the expected i, value and colligative properties., , For example, when NaCl is dissolved, in water, it produces two ions, Na⊕ and, Cl , whereas sucrose does not dissociate. It, is expected then that the colligative property, of 0.1 m NaCl is twice that of 0.1 m sucrose, solution., 2.11.1 van’t Hoff factor(i), To obtain the colligative properties, of electrolyte solutions by using relations, for nonelectrolytes, van’t Hoff suggested a, factor i. It is defined as the ratio of colligative, property of a solution of electrolyte divided, by the collogative property of nonelectrlyte, solution of the same concentration. Thus, i=, , 2.11.2 Modification of expressions of, colligative properties : The expressions of, colligative properties mentioned earlier for, nonelectrolytes are to be modified so as to, make them applicable for electrolyte solutions., The modified equations are, WM, i. ∆P = i P10x2 = i 2 1, M 2 W1, , colligative property of elctrolyte solution, colligative property of nonelectrolyte solution, of the same concentration, , (π), (∆T ), (∆T ), (∆P), = (∆T f = (∆T b) =, = (π), (∆P)0, ), 0, b 0, f 0, , (2.22), ii. ∆Tb = iKbm = i, , where quantities without subscript, refer to electrolytes and those with subscript, to nonelectrolytes., , iii. ∆Tf = i Kf m = i, , The van’t Hoff factor i is also defined, in an alternative but exactly equivalent, manner as, i=, , =, , 1000 KfW2, M2W1, , W2RT, M2V, 2.11.3 van’t Hoff factor and degree of, dissociation : A discussion of colligative, properties of electrolytes in the preceeding, sections is based on the fact that the, elctrolytes are completely dissociated in their, aqueous solutions. This is approximately true, iv. π = i MRT = i, , actual moles of particles in solution after, dissociation, moles of formula units dissolved in solution, , , , 1000 KbW2, M2W1, , (2.23), , formula mass of substance, observed molar mass of substance, , 42
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for solutions of strong electrolytes and not for, weak electrolytes that dissociate to a small, extent. The weak electrolytes involve the, concept of degree of dissociation (∝), that, changes the van’t Hoff factor., , (∆Tf)0= 1.86 K kg mol-1 × 0.2 mol kg-1, = 0.372 K, (∆Tf), 0.680 K, = 0.372 K = 1.83, i=, (∆Tf)0, , Relation between van’t Hoff factor and, degree of dissociation, , (π)0 = MRT, n, = V2 RT, , Consider an elctrolyte AxBy that, dissociates in aqueous solution as, AxBy, Initially : 1 mol, , x Ay⊕ + y Bz, 0, , =, 0.2 mol × 0.08205 L atm.mol-1K-1 × 273K, 1L, = 4.48 atm, π, i = 1.83 = π, 0, π = 1.83 × 4.48 atm, , (2.25), , 0, , At equilibrium :, (1-∝) mol, , (x ∝ mol) (y ∝) mol, , If ∝ is the degree of dissociation of, elctrolyte, then the moles of cations are ∝x, and those of anions are ∝y at equilibrium., We have dissolved just 1 mol of electrolyte, initially. ∝ mol of eletrolyte dissociates and (1∝) mol remains undissociated at equilibrium., , π = 8.2 atm, Problem 2.11 : 0.01m aqueous formic acid, solution freezes at -0.021 0C. Calculate its, degree of dissociation. Kf = 1.86 K kg mol-1, ∆Tf = i Kf m, , Total moles after dissociation, = (1- ∝) + (x∝) + (y∝), , ∆Tf = 0 0C - (-0.021 0C) = 0.021 0C, m = 0.01 mol kg-1, 0.021 = i × 1.86 K kg mol-1 ×0.01 mol kg-1, 0.021, i = 1.86 × 0.01 = 1.13, i-1, ∝ = n - 1 = i -1 because n = 2, Hence, ∝ = 1.13 - 1 = 0.13 = 13%, , = 1+∝(x+y-1), = 1+∝(n-1), , (2.26), , where, n = x+y = moles of ions obtained from, dissociation of 1 mole of electrolyte., The van’t Hoff factor given by Eq. (2.23) is, i=, actual moles of particles in solution after dissociation, moles of formula units dissolved in solution, , 1 + ∝(n-1), 1, i-1, Hence i = 1 + ∝(n-1) or ∝ = n -1, , Problem 2.12 : 3.4 g of CaCl2 is dissolved, in 2.5 L of water at 300 K. What is the, osmotic pressure of the solution? van’t, Hoff factor for CaCl2 is 2.47., Solution :, W RT, π = iMRT = i 2, M2V, , =, , (2.27), , Problem 2.10 : 0.2 m aqueous solution, of KCl freezes at -0.680 0C. Calculate, van’t Hoff factor and osmotic pressure of, solution at 0 0C.(Kf = 1.86 K kg mol-1), Solution :, , i = 2.47, W2 = 3.4 g, R = 0.08206 dm3atm, K-1 mol-1, T = 300 K, M2 = 40+71 = 111 g, mol-1, V = 2.5 dm3, π = 2.47 ×, 3.4 g × 0.08206 dm3atm K-1mol-1 × 300 K, 111 g mol-1 × 2.5 dm3, , ∆Tf = Kf.m, ∆Tf = 0.680 K, m = 0.2 mol kg-1, , = 0.745 atm, , 43
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Arrange the following solutions in order, of increasing osmotic pressure. Assume, complete ionization. a) 0.5m Li2SO4 b), 0.5m KCl c) 0.5m Al2(SO4)3 d) 0.1m BaCl2., , Problem 2.13 : Which of following, solutions will have maximum boiling point, elevation and which have minimum freezing, point depression assuming the complete, dissociation? (a) 0.1 m KCl (b) 0.05 m NaCl, (c) 1m AlPO4 (d) 0.1 m MgSO4, , Problem 2.14 : Assuming complete, dissociation, calculate the molality of an, aqueous solution of KBr whose freezing, point is -2.950C. Kf for water is 1.86 K kg, mol-1, Solution :, KBr = K⊕ + Br ,, , Solution : Boiling point elevation and, freezing point depression are colligative, properties that depend on number of particles, in solution. The solution having more, number of particles will have large boiling, point elevation and that having less number, of particles would show minimum freezing, point depression., (a) KCl, 0.1m, , moles of particles after dissociation, moles of particles dissolved, 2, = =2, 1, , Total particles in, 0.1m 0.1m solution = 0.2 mol, , (b) NaCl, 0.05m, , (c) AlPO4, 1m, , (d) MgSO4, 0.1m, , i=, , K⊕ + Cl, , Na⊕ + Cl, , Total particles in, 0.05m 0.05m solution = 0.1 mol, , ∆Tf = iKfm, ∆Tf = 0 0C- (-2.95 0C) = 2.95 0C, , Al3⊕ + PO43 Total particles in, 1m, 1m solution = 2.0mol, , ∆Tf, 2.95 K, m = iK =, 2 × 1.86 K kg mol-1, f, , Mg2⊕ + SO42 Total particles in, 0.1m, 0.1m solution = 0.2mole, , = 0.793 mol kg-1, , AlPO4 solution contains highest, moles and hence highest number particles, and in turn, the maximum ∆Tb. NaCl solution, has minimum moles and particles. It has, minimum ∆Tf., , Exercises, 1. Choose the most correct option., i. The vapour pressure of a solution, containing 2 moles of a solute in 2, moles of water (vapour pressure of, pure water = 24 mm Hg) is, a. 24 mm Hg , b. 32 mm Hg, c. 48 mm Hg , d. 12 mm Hg, ii. The colligative property of a solution is, a. vapour pressure, b. boiling point, c. osmotic pressure d. freezing point, iii. In calculating osmotic pressure the, concentration of solute is expressed in, , a. molarity , b. molality, c. mole fraction, d. mass percent, iv. Ebullioscopic constant is the boiling, point elevation when the concentration, of solution is, a. 1m , b. 1M, c. 1 mass%, d. 1 mole fraction of solute., v. Cryoscopic constant depends on, a. nature of solvent, b. nature of solute, c. nature of solution, d. number of solvent molecules, , 44
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xii. Which of the following statement is, NOT correct for 0.1 M urea solution, and 0.05 M sucrose solution?, a. osmotic pressure exhibited by urea, solution is higher than that exhibited, by sucrose solution, b. urea solution is hypertonic to sucrose, solution, c. they are isotonic solutions, d. sucrose solution is hypotonic to urea, solution, 2. Answer the following in one or two, sentences, i., What is osmotic pressure?, ii. A solution concentration is expressed, in molarity and not in molality while, considering osmotic pressure. Why?, iii. Write the equation relating boiling, point elevation to the concentration, of solution., iv. A 0.1 m solution of K2SO4 in water, has freezing point of -4.3 0C. What is, the value of van’t Hoff factor if Kf for, water is 1.86 K kg mol-1?, v., What is van’t Hoff factor?, vi. How is van’t Hoff factor related to, degree of ionization?, vii. Which of the following solutions will, have higher freezing point depression, and why ?, a. 0.1 m NaCl b. 0.05 m Al2(SO4)3, viii. State Raoult’s law for a solution, containing a nonvolatile solute, ix. What is the effect on the boiling point, of water if 1 mole of methyl alcohol, is added to 1 dm3 of water? Why?, x., Which of the four colligative, properties is most often used for, molecular mass determination? Why?, 3. Answer the following., i. How vapour pressure lowering is, related to a rise in boiling point of, solution?, ii. What are isotonic and hypertonic, solutions?, , vi. Identify the correct statement, a. vapour pressure of solution is higher, than that of pure solvent., b. boiling point of solvent is lower than, that of solution, c. osmotic pressure of solution is lower, than that of solvent, d. osmosis is a colligative property., vii. A living cell contains a solution which, is isotonic with 0.3 M sugar solution., What osmotic pressure develops when, the cell is placed in 0.1 M KCl solution, at body temperature?, a. 5.08 atm , b. 2.54 atm, c. 4.92 atm , d. 2.46 atm, viii. The osmotic pressure of blood is 7.65, atm at 310 K. An aqueous solution of, glucose isotonic with blood has the, percentage (by volume), a. 5.41 % , b. 3.54 %, c. 4.53 % , d. 53.4 %, ix. Vapour pressure of a solution is, a. directly proportional to the mole, fraction of the solute, b. inversely proportional to the mole, fraction of the solute, c. inversely proportional to the mole, fraction of the solvent, d. directly proportional to the mole, fraction of the solvent, x. Pressure cooker reduces cooking time, for food because, a. boiling point of water involved in, cooking is increased, b. heat is more evenly distributed in the, cooking space, c. the higher pressure inside the cooker, crushes the food material, d. cooking involves chemical changes, helped by a rise in temperature., xi. Henry’s law constant for a gas CH3Br, is 0.159 moldm-3 atm at 250 0C. What, is the solubility of CH3Br in water at 25, 0, C and a partial pressure of 0.164 atm?, a. 0.0159 mol L-1, b. 0.164 mol L-1, c. 0.026 M , d. 0.042 M, , 45
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iii. A solvent and its solution containing, a nonvolatile solute are separated by, a semipermable membrane. Does the, flow of solvent occur in both directions?, Comment giving reason., iv., , The osmotic pressure of CaCl2 and urea, solutions of the same concentration at, the same temperature are respectively, 0.605 atm and 0.245 atm. Calculate, van’t Hoff factor for CaCl2, , v., , Explain reverse osmosis., , 9. The vapour pressure of water at 20 0C is, 17 mm Hg. What is the vapour pressure, of solution containing 2.8 g urea in 50 g of, water? (16.17 mm Hg), 10. A 5% aqueous solution (by mass) of cane, sugar (molar mass 342 g/mol) has freezing, point of 271K. Calculate the freezing point, of 5% aqueous glucose solution. (269.06, K), 11. A solution of citric acid C6H8O7 in 50 g, of acetic acid has a boiling point elevation, of 1.76 K. If Kb for acetic acid is 3.07 K, kg mol-1, what is the molality of solution?, (0.573 m), 12. An aqueous solution of a certain organic, compound has a density of 1.063 gmL-1,, an osmotic pressure of 12.16 atm at 250C, and a freezing point of -1.030C. What is, the molar mass of the compound? (334 g/, mol), 13. A mixture of benzene and toluene contains, 30% by mass of toluene. At 300C, vapour, pressure of pure toluene is 36.7 mm Hg, and that of pure benzene is 118.2 mm Hg., Assuming that the two liquids form ideal, solutions, calculate the total pressure and, partial pressure of each constituent above, the solution at 300C. (86.7 mm, P = 96.5, mm), 14. At 25 0C a 0.1 molal solution of CH3COOH, is 1.35 % dissociated in an aqueous, solution. Calculate freezing point and, osmotic pressure of the solution assuming, molality and molarity to be identical., (-0.189 0C, 2.48 atm), 15. A 0.15 m aqueous solution of KCl freezes, at -0.510 0C. Calculate i and osmotic, pressure at 0 0C. Assume volume of, solution equal to that of water (1.83, 6.15, atm), , vi. How molar mass of a solute is, determined by osmotic pressure, measurement?, vii. Why vapour pressure of a solvent is, lowered by dissolving a nonvolatile, solute into it?, viii. Using Raoult’s law, how will you show, 0, that ∆P = P1 x2 ? Where x2 is the mole, 0, fraction of solute in the solution and P1, vapour pressure of pure solvent., ix. While considering boiling point, elevation and freezing point depression, a solution concentration is expressed in, molality and not in molarity. Why?, 4. Derive the relationship between degree, of dissociation of an electrolyte and van’t, Hoff factor., 5. What is effect of temperature on solubility, of solids in water? Give examples., 6. Obtain the relationship between freezing, point depression of a solution containing, nonvolatile nonelctrolyte and its molar, mass., 7. Explain with diagram the boiling point, elevation in terms of vapour pressure, lowering., 8. Fish generally needs O2 concentration in, water at least 3.8 mg/L for survival. What, partial pressure of O2 above the water is, needed for the survival of fish? Given the, solubility of O2 in water at 00C and 1 atm, partial pressure is 2.2 × 10-3 mol/L (0.054, atm), , Activity :, Boil about 100 mL of water, in a beaker. Add about 10 to 15 g of, salt (NaCl) to the boiling water. Write, your observations and conclusions., , 46
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3. IONIC EQUILIBRIA, Can you recall ?, • What is chemical equilibrium ?, •, , The weak electrolytes dissociate only, partially in dilute aqueous solutions. An, equilibrium thus can be established between, the ions and nonionized molecules. The, ionization reaction therein is represented, as double arrow( ) between the ions and, nonionized molecule., , What are electrolytes ?, , 3.1 Introduction : The equilibrium between, ions and unionized molecules in solution is, called ionic equilibrium. The principles of, chemical equilibrium we studied in standard, XI will be applied to ionic equilibria. In this, chapter with the help of these principles, we determine equilibrium constants and, concentrations of ions and unionized species., In particular examine the following ionic, equilibria :, • H⊕ and OH, molecules., , Use your brain power, Which of the following is a, strong electrolyte ?, HF,, H3PO4., , AgCl,, , CuSO4,, , CH3COONH4,, , 3.2.3 Degree of dissociation (∝) : The degree, of dissociation of an electrolyte is defined, as a fraction of total number of moles of the, electrolyte that dissociates into its ions when, the equilibrium is attained. It is denoted by, symbol ∝ and given by, , ions and unionized water, , • Ionization of weak acids and weak bases., • Reactions between ions of salt and ions of, water., • Solid salt and its ions in water., , number of moles dissociated, total number of moles, Percent dissociation = ∝ × 100, ∝=, , 3.2 Types of electrolyte : The substances, which give rise to ions when dissolved in water, are electrolytes. The non electrolytes are those, which do not ionize and exist as molecules in, aqueous solutions., , (3.1), (3.2), , If 'c' is the molar concentration of an, electrolyte the equilibrium concentration of, cation or anion is (∝ × c) mol dm-3., , The electrolytes are classified into strong, and weak electrolytes. This classification is, based on their extent of ionisation in dilute, aqueous solutions., , 3.3 Acids and Bases : Acids and bases are, familiar chemical compounds. Acetic acid, is found in vinegar, citric acid in lemons,, magnesium hydroxide in antacids, ammonia, in household cleaning products. The tartaric, acid is present in tamarind paste. These are, some acids and bases we come across in, everyday life., , 3.2.1 Strong electrolyte : The electrolytes, ionizing completely or almost completely are, strong electrolytes. For example : strong acids,, strong bases and salts., 3.2.2 Weak electrolyte : The electrolytes, which dissociate to a smaller extent in aqueous, solution are weak electrolytes. Weak acids, and weak bases belong to this class., , 47
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3.3.1 Arrhenius theory of acids and bases, , Base : Base is a substance that accepts a, proton (H⊕) from another substance., , According to this theory acids and bases are, defined as follows :, , For example :, , Acid : Acid is a substance which contains, hydrogen and gives rise to H⊕ ions in aqueous, solution. For example :, HCl (aq), , water, , CH3COOH(aq), , water, , H (aq) + Cl (aq), CH3COO (aq)+ H⊕(aq), , Acid1, , Acid2, , Base2, , Base1, , The base produced by accepting the, proton from an acid is the conjugate base of, that acid. Likewise the acid produced when a, base accepts a proton is called the conjugate, acid of that base. A pair of an acid and a, base differing by a proton is said to be a, conjugate acid-base pair., , Do you know ?, Hydrochloric acid, HCl, present in the gastric juice is, secreted by our stomach and is essential, for digestion of food., , HCl(aq) + H2O(l), Acid1, , Base2, , H3O⊕ (aq) + Cl (aq), , Acid2, , Base1, , conjugate acid-base pair2, , Base : Base is a substance that contains OH, group and produces hydroxide ions (OH ), ions in aqueous solution. For example,, NH4OH(aq), , NH4⊕ + Cl, , In the above reaction HCl and NH4⊕ are, proton donors and act as acids. The NH3 and, Cl are proton acceptors and act as bases., Further it follows that the products of the, Bronsted-Lowry acid-base reactions are acids, bases., , ⊕, , Arrhenius described H⊕ ions in water as, bare ions; they hydrate in aqueous solutions, and thus represented as hydronium ions H3O⊕., We herewith conveniently represent them as, H⊕., , NaOH (aq), , HCl + NH3, , conjugate acid-base pair1, , 3.3.3 Lewis theory : A more generalized acidbase concept was put forward by G.N. Lewis, in 1923. According to this theory acids and, bases are defined as follows., , Na⊕(aq) + OH (aq), NH4⊕(aq) + OH (aq), , Arrhenius theory accounts for properties, of different acids and bases and is applicable, only to aqueous solutions. It does not account, for the basicity of NH3 and Na2CO3 which do, not have OH group., , Acid : Any species that accepts a share in an, electron pair is called Lewis acid., Base : Any species that donates a share in an, electron pair is called Lewis base., , 3.3.2 Bronsted - Lowry theory : J. N., Bronsted and T. M. Lowry (1923) proposed, a more general theory known as the BronstedLowry proton transfer theory. According to, this theory acids and bases are defined as, follows., , For example :, H, ⊕, H + N-H, H, , Acid : Acid is a substance that donates a, proton (H⊕) to another substance., , F H, F-B+ N-H, F H, , (acid) (base), , (acid) (base), , 48, , H, H←N-H, H, , ⊕, , H, F, F-B←N-H, H, F
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3.4.1 Dissociation constant of weak acids, and weak bases : The dissociation of a weak, acid HA in water is expressed as, , Use your brain power, • All Bronsted bases are also, Lewis bases, but all Bronsted, acids are not Lewis acids. Explain., , HA(aq), , The equilibrium constant called aciddissociation constant for this equilibrium is :, [H⊕][A ], Ka =, (3.3), [HA], , Amphoteric nature of water : Water has, the ability to act as an acid as well as a base., Such behaviour is known as amphoteric nature, of water. For example :, H2O(l) + NH3(aq), , Similarly the dissociation of weak base, BOH in water is represented as :, , OH (aq) + NH4⊕(aq), , BOH(aq), , Acid, , H2O(l) + HCl(aq), , [B⊕][OH ], (3.4), [BOH], Thus, the dissociation constant of a, weak acid or a weak base is defined as, the equilibrium constant for dissociation, equilibrium of weak acid or weak base,, respectively., Kb =, , H2O acts as an acid towards NH3 and as a base, towards HCl. Therefore H2O is amphoteric., 3.4 Ionisation of acids and bases, Acids and bases are classified as strong acids, and strong bases, weak acids and weak bases, on the basis of their extent of dissociation., Strong acids and bases are almost completely, dissociated in water. For example :, NaOH (aq), , 3.4.2 Ostwald's dilution law : Arrhenius, concept of acids and bases was expressed, quantitatively by F. W. Ostwald in the form of, the dilution law in 1888., , H⊕(aq) + Cl (aq), Na⊕(aq) + OH (aq), , a. Weak acids : Consider an equilibrium of, weak acid HA that exists in solution partly as, the undissociated species HA and partly H⊕, and A ions. Then, , Typical strong acids are HCl, HNO3,, H2SO4, HBr and HI while typical strong bases, may include NaOH and KOH., Weak acids and weak bases are partially, dissociated in water. The solution of a weak, acid or a weak base contains undissociated, molecules along with a small number of ions, at equilibrium. For example :, CH3COOH(aq), NH4 OH(aq), , B⊕(aq) + OH (aq), , The equilibrium constant called basedissociation constant for this equilibrium is,, , H3O⊕(aq) + Cl (aq), , Base, , HCl (aq), , H⊕(aq) + A (aq), , HA(aq), , H⊕(aq) + A (aq), , The acid dissociation constant is given by, Eq. (3.3), [H⊕][A ], Ka =, [HA], , CH3COO (aq) + H⊕(aq), , Suppose 1 mol of acid HA is initially, present in volume V dm3 of the solution. At, equilibrium the fraction dissociated would, be ∝, where ∝ is degree of dissociation of, the acid. The fraction of an acid that remains, undissociated would be (1 - ∝)., , NH4⊕(aq) + OH (aq), , Note that HCOOH, HF, H2S are examples, of weak acids while Fe(OH)3, Cu(OH)2 are, examples of weak bases., , 49
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Amount, present at, equilibrium/, mol, concentration, at, equilibrium/, mol dm-3, , HA(aq), (1-∝), , 1- ∝, V, , H⊕(aq) + A (aq), ∝, ∝, , ∝, V, , BOH(aq), (1-∝), , Amount, present at, equilibrium, concentration, at equilibrium, , ∝, V, , (3.5), , Kb =, , For the weak acid HA, ∝ is very small,, or (1 - ∝) ≅ 1. With this Eq. (3.5) and (3.6), reduce., , ∝=, , Ka, or ∝ = Ka.V, c, , (∝/V) (∝/V), ∝2, =, (1- ∝)/V, (1- ∝)V, , (3.9), , The degree of dissociation of a weak, base is inversely proportional to square root of, its concentration and is directly proportional, to square root of volume of the solution, containing 1 mol of weak base., , (3.7), (3.8), , Problem 3.1 : A weak monobasic acid, is 0.05% dissociated in 0.02 M solution., Calculate dissociation constant of the acid., Solution : The dissociation constant of acid, is given by Ka = ∝2 c. Here,, percent dissociation, ∝=, 100, 0.05, = 5 × 10-4, =, 100, , The Eq. (3.8) implies that the degree, of dissociation of a weak acid is inversely, proportional to the square root of its, concentration or directly proportional to, the square root of volume of the solution, containing 1 mol of the weak acid., b. Weak base : Consider 1 mol of weak base, BOH dissolved in V dm3 of solution. The base, dissociates partially as, BOH (aq), , ∝, V, , Similar arguments in the case of weak, acid, led to, ∝2 c, Kb =, (3.10), (1- ∝), Kb, ∝ = Kb.V , ∝ =, (3.11), c, , (3.6), , Ka = ∝2/V and Ka = ∝2c, , ∝, V, , Substitution of these concentrations in Eq., (3.4), gives, , If c is the initial concentration of an acid, in mol dm-3 and V is the volume in dm3 mol-1, then c = 1/V. Replacing 1/V in Eq. (3.5) by c, we get, 2, Ka = ∝ c, 1- ∝, , 1- ∝, V, , A equilibrium,, [BOH] = 1- ∝ mol dm-3,, V, [B⊕] = [OH ] = ∝ mol dm-3., V, , Thus, at equilibrium [HA] = 1- ∝ , mol dm-3,, V, [H⊕] = [A ] = ∝ mol dm-3., V, Substituting these in Eq. (3.3), ∝2, (∝/V) (∝/V), = (1- ∝)V, Ka =, (1- ∝)/V, , B⊕(aq) +OH (aq), ∝, ∝, , c = 0.02 M = 2 × 10-2 M, , B⊕(aq) + OH (aq), , Hence Ka = (5 × 10-4)2 × 2 × 10-2, , The base dissociation constant is, [B⊕][OH ], Kb =, [BOH], , = 25 × 10-8 × 2 × 10-2, = 50 × 10-10 = 5 × 10-9, , Let the fraction dissociated at equilibrium, is ∝ and that remains undissociated is (1 - ∝)., , 50
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3.5 Autoionization of water : Pure water, ionizes to a very small extent. The ionization, equilibrium of water is represented as,, , Problem 3.2 : The dissociation constant of, NH4OH is 1.8 × 10-5. Calculate its degree, of dissociation in 0.01 M solution., Solution : The degree of dissociation is, given by ∝ = Kb/c . Here,, , H2O (l) + H2O (l), , The equilibrium constant (K) for the, ionization of water is given by, [H O⊕][OH ], K= 3, (3.12), [H2O(l)]2, , Kb = 1.8 × 10-5; c = 0.01 = 1 × 10-2 M, Hence, ∝ =, , 1.8 × 10-5 = 1.8 × 10-3, 1 × 10-2, , or K[H2O]2 = [H3O⊕][OH ], , = 18 × 10-4 = 4.242 × 10-2 = 0.04242, , (3.13), , A majority of H2O molecules are, undissociated, consequently concentration of, water [H2O] can be treated as constant. Then, , Problem 3.3 : A weak monobasic acid is, 12% dissociated in 0.05 M solution. What, is percent dissociation in 0.15 M solution., Solution : If ∝1 and ∝2 are the values of, degree of dissociation at two different, concentrations c1 and c2 respectively, then, , [H2O]2 = K'. Substituting this in Eq. (3.13), we get,, K×K' = [H3O⊕][OH ], , Ka = ∝12c1 = ∝22c2 Therefore ∝12c1 = ∝22c2, 12, ∝1 =, c = 0.05 M, c2 = 0.15 M,, 100 1, ∝2 = ?, , (3.14), , Kw = [H3O⊕][OH ], where Kw = KK' is called ionic product of, water. The product of molar concentrations of, hydronium (or hydrogen) ions and hydroxyl, ions at equilibrium in pure water at the given, temperature is called ionic product of water., , Substituting of these values in the equation, gives, (0.12)2 × 0.05 = ∝22 × 0.15, (12)2 × 0.05, = 0.0048, ∝22 =, 0.15, Hence ∝2 = 0.0693 %, ∴ percent dissociation = 6.93 %, , In pure water H3O⊕ ion concentration, always equals the concentration of OH ion., Thus at 298 K this concentration is found to be, 1.0 × 10-7 mol/L., Kw = (1.0 × 10-7) (1.0 × 10-7), , Problem 3.4 : Calculate [H3O⊕] in, 0.1 mol dm3 solution of acetic acid., , Kw = 1.0 × 10-14, , (3.15), , Internet my friend, Find out the values of ionic, product Kw of water at various, temperatures., , Given : Ka [CH3COOH] = 1.8 × 10-5, Solution : Let ∝1 be the degree of, dissociation. Concentrations of various, species involved at equilibrium are as, follows., , 273 K, 283K, 293K, 303K, 313K, 323 K, 3.6 pH Scale : Instead of writing concentration, of H3O⊕ ions in mol dm-3, sometimes it is, convenient to express it on the logarithmic, scale. This is known as pH scale., , CH3COOH + H2O, CH3COO + H3O⊕, (1- ∝)c, ∝c, ∝c, 1.8 × 10-5, Ka, =, ∝=, 0.1, c, = 1.34 × 10-2, , Sorensen in 1909 defined the pH of a, solution as the negative logarithm to the base, 10, of the concentration of H⊕ ions in solution, in mol dm-3. Expressed mathematically as, , [H3O ] = ∝ × c = 1.34 × 10 × 0.1, ⊕, , H3O⊕(aq) + OH (aq), , -2, , = 1.34 × 10-3 mol/L, , 51
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pH = - log10[H⊕], , Problem 3.5 : Calculate pH and pOH of, 0.01 M HCl solution., , Similarly pOH of a solution can be, defined as the negative logarithm to the base, 10, of the molar concentration of OH ions in, solution., Thus, pOH = -log10[OH ], , Solution : HCl is a strong acid. It dissociates, almost completely in water as, , 3.6.1 Relationship between pH and pOH, , Hence, [H3O⊕] = c = 0.01M = 1 × 10-2 M, , The ionic product of water is, , pH = -log10[H3O+] = -log10[1 × 10-2] = 2, , ⊕, , Kw = [H3O ][OH ], , We know that pH + pOH = 14, , Now, Kw = 1 × 10-14 at 298 K and thus, , ∴ pOH = 14 - pH = 14 - 2 = 12, , [H3O⊕][OH ] = 1.0 × 10-14, , Problem 3.6 : pH of a solution is 3.12., Calculate the concentration of H3O⊕ ion., , Taking logarithm of both the sides, we write, log10[H3O⊕] + log10[OH ] = -14, , Solution : pH is given by, , -log10[H3O⊕] + {- log10[OH ]} = 14, , pH = -log10[H3O⊕], , From Eq. (3.16) and (3.17), pH + pOH = 14, , H3O⊕(aq) + Cl, , HCl (aq) + H2O(l), (aq), , (3.16), , log10[H3O⊕] = -pH, , (3.18), , = - 3.12, , 3.6.2 Acidity, basicity and neutrality of, aqueous solutions, , = - 3 - 0.12 + 1 - 1, = (- 3 - 1) + 1 - 0.12, , 1. Neutral solution : For pure water or any, aqueous neutral solution at 298 K, , = - 4 + 0.88 = 4.88, , [H3O⊕] = [OH ] = 1.0 × 10-7 M, , Thus [H3O⊕] = antilog [4.88], , Hence,pH = -log10[H⊕] = -log10[1 × 10-7] = 7, , = 7.586 × 10-4 M, , 2. Acidic solution : In acidic solution, there is, excess of H3O⊕ ions, or [H3O⊕] > [OH ] Hence,, [H3O⊕] > 1 × 10-7 and pH < 7, , Problem 3.7 : A weak monobasic acid is, 0.04 % dissociated in 0.025M solution., What is pH of the solution ?, , 3. Basic solution : In basic solution, the excess, of OH ions are present that is [H3O⊕] < [OH ], or [H3O⊕] < 1.0 × 10-7 with pH > 7., , Solution : A weak monobasic acid HA, dissociates as :, HA + H2O(l), , H3O⊕(aq) + A (aq), , Percent dissociation = ∝ × 100, percent dissociation, or ∝ =, 100, = 0.04 = 4 × 10-4, 100, , Fig. 3.1 : pH scale, , Now [H3O⊕] = ∝ × c, = 4 × 10-4 × 0.025 M = 10-5 M, ∴ pH = -log10[H3O⊕] = -log10[10-5] = 5, , 52
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3.7 Hydrolysis of salts, 3.7.1 Types of salts, These are of four types, , Use your brain power, • Suppose that pH of monobasic, and dibasic acid is the same. Does, this mean that the molar concentrations, of both acids are identical ?, , I., , • How pH of pure water vary with, temperature ? Explain., , Salts derived from strong acid and, strong base. For example : NaCl,, Na2SO4, NaNO3, KCl, KNO3., , II. Salts derived from strong acids and, weak bases. For example : NH4Cl,, CuSO4, NH4NO3, CuCl2., , Problem 3.8 : The pH of monoacidic, weak base is 11.2. Calculate its percent, dissociation in 0.02 M solution., , III. Salts derived from weak acids and, strong bases. For example : CH3COONa,, KCN, Na2CO3., , Solution : pOH of the solution is given as :, , IV. Salts derived from weak acids and weak, bases. For example : CH3COONH4,, NH4CN., , pOH = 14 - pH = 14 - 11.2 = 2.8, pOH = -log10[OH ], , 3.7.2 Concept of hydrolysis : When a salt is, dissociated in water, it dissociates completely, into its constituent ions. The solvent water, dissociates slightly as,, , log10[OH ] = - pOH, = - 2.8 = - 2 - 0.8 - 1 + 1, = - 3 + 0.2 = 3.2, [OH ] = antilog 3.2 = 1.585 × 10-3 mol/dm3, , H2O (l) + H2O (l), , For monoacidic base,, , Pure water is neutral and [H3O⊕] = [OH ]., If the ions of the salt do not interact with water,, the hydronium and hydroxyl ion concentrations, remain equal and the solution is neutral. When, one or more of the salt ions react with water, the, equality of concentrations of H3O⊕ and OH, ions is disturbed. The solution, does not remain, neutral and becomes acidic or basic depending, on the type of the salt. Such a reaction between, the ions of salt and the ions of water is called, hydrolysis of salt. Hydrolysis of salt is defined, as the reaction in which cations or anions or, both ions of a salt react with ions of water to, produce acidity or alkalinity (or sometimes, even neutrality)., , BOH (aq), , B⊕(aq) + OH (aq), , [OH ] = ∝c, [OH ], 1.585 × 10-3, =, = 0.07925, ∝=, 0.02, c, Percent dissociation = ∝ × 100, = 0.07925 × 100, = 7.925 %, Do you know ?, • pH is crucial for digestion of, food and other biochemical, reactions in our body., , H3O⊕(aq) + OH (aq), , 3.7.3 Salts of strong acids and strong bases, NaCl is a salt of strong acid HCl and a, strong base NaOH. When it is dissolved in, water, it dissociates completely into its ions., , • pH of gastric juice is about 2., • pH of blood is maintained within range, 7.36 to 7.42., • Enzymes function effectively only at, a certain pH. For example trypsin acts, best for alkaline pH., , NaCl (aq), , Na⊕(aq) + Cl (aq), , The ions Na⊕ and Cl have no tendency to, react with water. This is because the possible, products, NaOH and HCl of such reactions are, , 53
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CH3COONa(aq), Na⊕(aq), , strong electrolytes and dissociate completely, in aqueous solutions., In other words,, Na⊕(aq) + Cl (aq) + H2O, , H2O (l) + H2O (l), , H3O⊕(aq) + OH (aq), , Solution of CH3COONa contains Na⊕,, H3O , CH3COO , OH . The Na⊕ ions of salt, have no tendency to react with OH ions of, water since the possible product of the reaction, is NaOH, a strong electrolyte., , ⊕, , HCl(aq) + NaOH(aq) + H2O, H3O (aq) +, Cl (aq)+Na⊕(aq)+OH (aq), , ⊕, , Thus the reactants and the products are the, same. This implies that neither the cation nor, anion of the salt reacts with water or there is, no hydrolysis. Equality H3O⊕ = OH produced, by ionization of water is not disturbed and, solution is neutral. It may be concluded that, salt of strong acid and strong base does not, undergo hydrolysis., , On the other hand the reaction of, CH3COO ions of salt with the H3O⊕ ions from, water produces unionized CH3COOH., CH3COO (aq) + H2O (l), Thus, the hydrolytic, CH3COONa is,, , 3.7.4 Salts of strong acids and weak bases :, CuSO4 is salt of strong acid H2SO4 and, weak base Cu(OH)2. When CuSO4 is dissolved, in water, it dissociates completely as,, , CH3COOH(aq), + OH (aq), equilibrium, , for, , CH3COONa(aq) + H2O (l), CH3COOH(aq), ⊕, + Na (aq) + OH (aq), As a result of excess OH ions produced, the solution becomes basic. The solution of, CH3COONa is therefore basic., , Cu2⊕(aq) + SO42 (aq), , SO42 ions of salt have no tendency to, react with water because the possible product, H2SO4 is strong electrolyte. The reaction of, Cu2⊕ ions with OH ions form unionized, Cu(OH)2. The hydrolytic equilibrium for, CuSO4 is then written as,, Cu2⊕(aq)+4H2O(l), , +, , Water dissociates slightly as,, , HCl(aq) +NaOH (aq), , (strong acid) (strong base), [Possible products], , CuSO4 (aq), , CH3COO (aq), , Can you tell ?, Why an aqueous solution, of NH4Cl is acidic while that of, HCOOK basic ?, , Cu(OH)2(aq)+2H3O⊕(aq), , Remember..., , Due to the presence of excess of H3O⊕, ions, the resulting solution of CuSO4 becomes, acidic and turns blue litmus red., Formation of sparingly soluble Cu(OH)2, by hydrolysis makes the aqueous solution of, CuSO4 turbid. If H2SO4, that is H3O⊕ ions, are added, the hydrolytic equilibrium shifts, to the left. A turbidity of Cu(OH)2 dissolves, to give a clear solution. To get clear solution, of CuSO4, the addition of H2SO4 would be, required., , As a general rule the solutions, of salts of strong acids and strong, bases are neutral, the solutions of salts of, strong acids and weak bases are acidic, and the solutions of salts of strong bases, and weak acids are basic., 3.7.6 Salts of weak acids and weak bases:, When salt BA of weak acid HA and weak, base BOH is dissolved in water, it dissociates, completely as, , 3.7.5 Salts of weak acids and strong bases, , BA(aq), B⊕(aq) + A (aq), The hydrolysis reaction involves the, interaction of both the ions of the salt with, water,, , CH3COONa is a salt of weak acid, CH3COOH and strong base NaOH, when, dissolved in water, it dissociates completely., , 54
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CH3COONa(aq), , Can you think ?, Home made jams and gellies, without any added chemical, preservative additives spoil in a few, days whereas commercial jams and jellies, have a long shelf life. Explain. What role, does added sodium benzoate play?, , On the other hand since the acetic acid is a, weak acid, the concentration of undissociated, CH3COOH molecules is usually high. If a, strong acid is added to this solution the added, H⊕ ions will be consumed by the conjugate, base CH3COO present in large concentration., Similarly, if small amount of base is added,, the added OH ions will be neutralized by the, large concentration of acetic acid as shown in, the following reactions :, CH3COO (aq) + H⊕(aq), CH3COOH(aq), , 3.8.1 Types of buffer solutions, There are two types of buffer solutions., Acidic buffer used to maintain an acidic pH,, while basic buffer maintains alkaline pH., , (large concentraion) (added acid), , a. Acidic buffer solution : A solution, containing a weak acid and its salts with, strong base is called an acidic buffer solution., , CH3COOH(aq) +OH (aq), (large concentraion) (added acid), , pH of acidic buffer is given by the equation, [salt], [acid], , where pKa = - log10Ka, , CH3COO (aq), + H2O(l), , The acid or base added thus can not change, the [H⊕] or [OH ] concentrations and, pH of, the buffer remains unchanged. Dilution does, not have any effect on pH of buffer. This is, because the concentration ratio term in Eq., (3.23) and Eq. (3.25 ) remains the same. The, dilution does not change this ratio., 3.8.3 Properties of buffer solution, , For example : A solution containing weak, acid such as CH3COOH and its salt such as, CH3COONa is an acidic buffer solution., , pH = pKa + log10, , CH3COO (aq) + Na⊕(aq), , (3.23), (3.24), , The pH of a buffer solution does not, change appereciably, , and Ka is the dissociation constant of the acid., b. Basic buffer solution : A solution, containing a weak base and its salt with, strong acid is the basic buffer solution., , i. by addition of small amount of either, strong acid or strong base, ii. on dilution or iii., when it is kept for long time., , For example : A solution containing a, weak base such as NH4OH and its salt such as, NH4Cl is a basic buffer solution., , Can you tell ?, It is enough to add a few mL, of a buffer solution to maintain its, pH. Which property of buffer is used here?, , The pOH of basic buffer is given by,, [salt], pOH = pKb + log10, (3.25), [base], (3.26)and, where pKb = - log10Kb, Kb is the dissociation constant for the base., Equations (3.26) and (3.25) are known to, Henderson Hasselbalch equation., , 3.8.4 Applications of buffer solution, Buffer, solution, finds, extensive, applications in a variety of fields. Some of its, applications are given., , 3.8.2 Buffer action : Let us consider, sodium acetate - acetic acid buffer. Here, sodium acetate is a strong electrolyte which, dissociates completely in water producing, large concentration of CH3COO as follows :, , i. In biochemical system : pH of blood in, our body is maintained at 7.36 - 7.42 due to, (HCO3 + H2CO3) buffer. A mere change of 0.2, pH units can cause death. The saline solution, used for intravenous injection must contain, , 56
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buffer system to maintain the proper pH of the, blood., , Problem 3.10 : Calculate the pH of buffer, solution composed of 0.1 M weak base, BOH and 0.2 M of its salt BA. [Kb = 1.8×, 10-5 for the weak base], Solution : pOH of basic buffer is given by, Henderson-Hasselbalch equation, pOH = pKb + log10 [salt], [base], , ii. Agriculture : The soils get buffered, due to presence of salts such as carbonate,, bicarbonate, phosphates and organic acids., The choice of fertilizers depends upon pH of, soil., iii. Industry : Buffers play an important role, in paper, dye, ink, paint and drug industries., , ∴ pKb = - log10Kb, = - log10 (1.8 × 10-5) = 5 - log10 1.8, = 5 - 0.2553 = 4.7447, [salt] = 0.02 M,, [acid] = 0.1M, Substitution of these in the above equation, gives, pOH = 4.7447 + log 0.02 = 4.7447 + log 2, 0.1, = 4.7447 + 0.3010 = 5.0457, pH = 14 - pOH, = 14 - 5.0457, = 13.9543 ≈ 13.95, , iv. Medicine : Penicillin preparations are, stabilized by addition of sodium citrate as, buffer. When citric acid is added to milk of, magnesia (Mg(OH)2), magnesium citrate is, formed, which is a buffer., v. Analytical chemistry : In qualitative, analysis, a pH of 8 to 10 is required for, precipitation of cations IIIA group. It is, maintained with the use of (NH4OH + NH4Cl), buffer., , 3.9 Solubility product, , Problem 3.9 : Calculate the pH of buffer, solution containing 0.05 mol NaF per litre, and 0.015 mol HF per litre. [Ka = 7.2 × 10-4, for HF], , Can you recall ?, • What is solubility, compound ?, , Solution : The pH of acidic buffer is given, by Henderson-Hasselbalch equation, pH = pKa + log10 [salt], [acid], ∴ pKa = - log10Ka = - log10 7.2 × 10-4, , • What is meant by the sparingly soluble, salt ?, , a, , • What is saturated solution ?, , Do you know ?, The process of dissolution, and precipitation of sparingly, soluble ionic compounds are of important, in our everyday life, industry and medicine., Kidney stone is developed due to the, precipitation of insoluble calcium oxalate,, CaC2O4. The process of tooth decay occurs, due to dissolution of enamel composed of, hydroxyapatite, Ca5(PO4)3OH in acidic, medium., , = 4 - log10 7.2 = 4 - 0.8573 = 3.1427, [salt] = 0.05 M,, , of, , [acid] = 0.015M, , Substitution in the above equation gives, pH = 3.1427 + log10 0.05, 0.015, = 3.1427 + log 3.33, = 3.1427 + 0.5224 = 3.6651 ≈ 3.67, , 3.9.1 Solubility equilibria : Hereafter we, confine our attention to sparingly soluble, compounds that is, compounds those dissolve, only slightly in water., , 57
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Consider following examples., , Suppose some powdered sparingly, soluble salt such as AgCl is put into water and, stirred vigorously. A very small amount of, AgCl dissolves in water to form its saturated, solution. Most of the salt remains undissolved., Thus, solid AgCl is in contact with its saturated, solution. AgCl is a strong electrolyte. Hence, the quantity of AgCl that dissolves in water, dissociates completely into its constituent, ions, Ag⊕ and Cl . A dynamic equilibrium, exists between undissolved solid AgCl and the, dissolved ions, Ag⊕ and Cl , in the saturated, solution. This equilibrium, called solubility, equilibrium, is represented as :, AgCl(s), , i. BaSO4(s), , Ksp = [Ba2⊕][SO42 ], ii. CaF2 (s), iii. Bi2S3 (s), , 2Bi3⊕ (aq) + 3S2 (aq), , Ksp = [Bi3⊕]2[S2 ]3, iv. Ca3(PO4)2 (s), , 3Ca2⊕ (aq) + 2PO43 (aq), , Ksp = [Ca2⊕]3[PO43 ]2, 3.9.2 Relationship between solubility, and solubility product : The solubility of, a compound is the amount in grams that, dissolves per unit volume (which may be 100, mL or 1L of its saturated solution)., , Ag⊕(aq) + Cl (aq), , Molar solubility : The number of moles, of a compound that dissolve to give one, litre of saturated solution is called its molar, solubility., , The concentration of undissolved solid, AgCl is constant we may write, [AgCl] = constant = K', , solubility in g/L, , molar solubility (mol/L) = molar mass in g/mol, , Substituting in Eq. (3.27) we write, [Ag⊕][Cl ], K=, K', , Consider once again the solubility, equilibrium for BxAy,, , K × K' = [Ag⊕][Cl ], , BxAy(s), , The product of K × K' is another constant, and is called solubility product, that is the, product of concentrations of ions in a saturated, solution. It is denoted by Ksp., , xBy⊕ (aq) + yAx (aq), , The solubility product is given by Eq., (3.28) :, Ksp = [By⊕]x[Ax ]y, , Ksp = [Ag⊕][Cl ], , If S is the molar solubility of the, compound, the equilibrium concentrations of, the ions in the saturated solution will be, , For the general salt solubility equilibrium, xBy⊕ (aq) + yAx (aq), , [By⊕] = xS mol/L, , The solubility product is, Ksp = [By⊕]x[Ax ]y, , Ca2⊕ (aq) + 2F (aq), , Ksp = [Ca2⊕][F ]2, , The expression for its equilibrium constant is :, [Ag⊕][Cl ], K= [AgCl], (3.27), , BxAy(s), , Ba2⊕ (aq) + SO42 (aq), , [Ax ] = xS mol/L, , (3.28), From Eq. (3.28), , Thus, in the saturated solution, of sparingly soluble salt the product, of equilibrium concentrations of the, constituent ions raised to the power equal, to their respective coefficients in the, balanced equilibrium expression at a given, temperature is called solubility product., , Ksp = [xS]x[yS]y = xxyySx+y, For example :, i. For AgBr,, AgBr(s), Ag⊕ (aq) + Br (aq), Here, x = 1, y = 1, , 58, , (3.29)
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∴ Ksp = S × S = S2, , 0.3, [C2O42-] = 2 M = 0.15 mol/L, , ii. For PbI2,, PbI2(s), , Pb2⊕ (aq) + 2I (aq), x = 1, y = 2, , These ions would react to form sparingly, soluble salt MgC2O4 in accordance with, reaction, , Therefore, Ksp = (1)1(2)2S1+2 = 4S3, iii. Al(OH)3,, , Mg2⊕ (aq) + C2O42 (aq), , Al(OH)3(s), , Al3⊕ (aq) + 3OH (aq), x = 1, y = 3, , Ionic product in the solution is given by, [Mg2+][C2O42-(aq)] = 0.05 × 0.15, , Ksp = (1)1(3)3S1+3 = 27S4, , = 0.0075 = 7.5 × 10-3, the Ksp value for MgC2O4 at 293 K is, 8.56 × 10-5. As ionic product is greater than, Ksp precipitation will take place., , Use your brain power, What is the relationship, between molar solubility and, solubility product for salts given below, i. Ag2CrO4, , ii. Ca3(PO4)2, , MgC2O4 (s), , Problem 3.12 : The solubility product of, AgBr is 5.2 × 10-13. Calculate its solubility, in mol dm-3 and g dm-3(Molar mass of AgBr, = 187.8 g mol-1), , iii. Cr(OH)3., , 3.9.3 Condition of precipitation : Ionic, product (IP) of an electrolyte is defined in the, same way as solubility product (Ksp). The only, difference is that the ionic product expression, contains concentration of ions under any, condition whereas expression of Ksp contains, only equilibrium concentrations. If,, , Solution : The solubility equilibrium of, AgBr is :, AgBr(s), , Ag⊕(aq) + Br (aq), , x = 1, y = 1, Ksp = [Ag⊕][Br ] = S2, , a. IP = Ksp ; the solution is saturated and, solubility equilibrium exists., , S = Ksp = 5.2 × 10-13, = 7.2 × 10-7 mol dm-3, , b. IP > Ksp ; the solution is supersaturated, and hence precipitation of the compound will, occur., , The solubility in g dm-3 = molar solubility, in mol dm-3 × molar mass g mol-1, , c. If IP < Ksp, the solution is unsaturated and, precipitation will not occur., , S = 7.2 × 10-7 mol dm-3 × 187.8 g mol-1, = 1.35 × 10-4 g dm-3, , Problem 3.11 : A solution is prepared by, mixing equal volumes of 0.1M MgCl2 and, 0.3M Na2C2O4 at 293 K. Would MgC2O4, precipitate out ? Ksp of MgC2O4 at 293 K is, 8.56 × 10-5., , Problem 3.13 : If 20.0 cm3 of 0.050 M, Ba(NO3)2 are mixed with 20.0 cm3 of 0.020, M NaF, will BaF2 precipitate ? Ksp of BaF2, is 1.7 × 10-6 at 298 K., Solution : Final volume of solution is, , Solution : When solution is prepared by, mixing equal volumes, volume gets doubled, and hence effective concentration of ions, would be half of initial concentration,, , 20 + 20 = 40 cm3,, 0.050 × 20, = 0.025 M, [Ba(NO3)2] =, 40, 0.020 × 20, [NaF] =, = 0.010M, 40, , 0.1, [Mg2+] = 2 = 0.05 mol/L, , 59
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Hence ionic product of BaF2 is, , The common ion effect states that the, ionisation of a weak electrolyte is supressed, in presence of a strong electrolyte containing, an ion common to the weak electrolyte., , IP = [Ba2⊕][F ]2, , Remember..., , Therefore [Ba2⊕] = 0.025 M and, [F ] = 0.010M, , = 0.025 × (0.01)2, , Common ion effect is a special, case of Le-Chatelier's principle in, which the stress applied to an equilibrium, system is an increase in the concentration, of one of the product (ions). The effect, of this stress is reduced by shifting the, equilibrium to the reactant side., , = 2.5 × 10-6, Ksp (BaF2) = 1.7 × 10-6 Thus, Ksp < IP, Ionic product in the solution is greater than, Ksp. Hence BaF2 will precipitate from the, solution., 3.10 Common ion effect :, , Can you tell ?, How does the ionization of, NH4OH suppressed by addition of, NH4Cl to the solution of NH4OH ?, , Can you recall ?, Which reagents are used to, precipitate (i) group II, (ii) group III, B, (iii) group III A of basic radicals/, cations ?, , 3.10.1 Common ion effect and solubility, Do you know ?, , Consider a solution of weak acid, CH3COOH and its soluble ionic salt, CH3COONa., , The hardness of water is due to, presence of Ca2⊕ ions. It is surprising, to know that Ca2⊕ ions can be removed by, adding more Ca2⊕ ions in the form of lime, Ca(OH)2, to the hard water. The OH ions, of lime react with HCO3 ions present in the, hard water to form CO32 ions., OH (aq) + HCO3 (aq), CO32 (aq) + H2O(l), Solubility product of CaCO3 is very low, (Ksp = 4.5 × 10-9). Addition of lime makes, IP >>Ksp which results in the precipitation, of CaCO3 and thereby removal of hardness., , CH3COOH is weak acid, dissociates only, slightly in solution, CH3COOH, , CH3COO (aq) + H⊕ (aq), , CH3COONa being a strong electrolyte, dissociates almost completely in solution., CH3COONa(aq), CH3COO + Na⊕, Both the acid and the salt produce, CH3COO ions in solution. CH3COONa, dissociates completely. Therefore it provides, high concentration of CH3COO ions., According to Le-Chatelier principle, the, addition of CH3COO from CH3COONa to, the solution of CH3COOH, shifts equilibrium, of dissociation of CH3COOH to left. Thus, reverse reaction is favoured in which, CH3COO combines with H⊕ to form unionised, CH3COOH. Hence dissociation of CH3COOH, is supressed due to presence of CH3COONa, containing a common CH3COO ion., , The presence of a common ion also, affects the solubility of a sparingly soluble, salt. Consider, the solubility equilibrium of, AgCl,, AgCl(s), , Ag⊕(aq) + Cl (aq), , The solubility product of AgCl is, Ksp = [Ag⊕][Cl ], , 60
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to Le-chatelier's principle the addition of Ag⊕, ions from AgNO3 to the solution of AgCl shifts, the solubility equilibrium of AgCl from right, to left. The reverse reaction in which AgCl, precipitates, is favoured until the solubility, equilibrium is re-established. The value of, Ksp however, remains the same since it is, an equilibrium constant. The solubility of a, sparingly soluble compound, thus decreases, with the presence of a common ion in solution., , Suppose AgNO3 is added to the saturated, solution of AgCl. The salt AgNO3 being a, strong electrolyte dissociates completely in the, solution., AgNO3(aq), , Ag⊕ (aq) + NO3 (aq), , The dissociation of AgCl and AgNO3, produce a common Ag⊕ ion. The concentration, of Ag⊕ ion in the solution increases owing to, complete dissociation of AgNO3. According, , Exercises, 1. Choose the most correct answer :, i. The pH of 10-8 M of HCl is, a. 8, b. 7, c. less than 7, d. greater than 7, ii. Which of the following solution will, have pH value equal to 1.0 ?, a. 50 mL of 0.1M HCl + 50mL of 0.1M, NaOH, b. 60 mL of 0.1M HCl + 40mL of, 0.1M NaOH, c. 20 mL of 0.1M HCl + 80mL of, 0.1M NaOH, d. 75 mL of 0.2M HCl + 25mLof 0.2M, NaOH, iii. Which of the following is a buffer, solution ?, a. CH3COONa + NaCl in water, b. CH3COOH + HCl in water, c. CH3COOH+CH3COONa in water, d. HCl + NH4Cl in water, iv. The solubility product of a sparingly, soluble salt AX is 5.2×10-13. Its, solubility in mol dm-3 is, a. 7.2 × 10-7 , b. 1.35 × 10-4, c. 7.2 × 10-8 , d. 13.5 × 10-8, v. Blood in human body is highly buffered, at pH of, a. 7.4 , b. 7.0, c. 6.9 , d. 8.1, , vi. The conjugate base of [Zn(H2O)4]2⊕ is, a. [Zn(H2O)4]2 NH3, b. [Zn(H2O)3]2, c. [Zn(H2O)3OH]⊕, d. [Zn(H2O)H]3⊕, vii. For pH > 7 the hydronium ion, concentration would be, a. 10-7M, b. < 10-7M, c. > 10-7M d. ≥ 10-7M, 2. Answer the following in one sentence :, i., Why cations are Lewis acids ?, ii., Why is KCl solution neutral to, litmus?, iii. How are basic buffer solutions, prepared?, iv. Dissociation constant of acetic, acid is 1.8 × 10-5. Calculate percent, dissociation of acetic acid in 0.01 M, solution., v., Write one property of a buffer, solution., vi. The pH of a solution is 6.06. Calculate, its H⊕ ion concentration., vii. Calculate the pH of 0.01 M sulphuric, acid., viii. The dissociation of H2S is suppressed, in the presence of HCl. Name the, phenomenon., ix. Why is it necessary to add H2SO4, while preparing the solution of, CuSO4?, , 61
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x., , Classify the following buffers into, different types :, a. CH3COOH + CH3COONa, b. NH4OH + NH4Cl, c. Sodium benzoate + benzoic acid, d. Cu(OH)2 + CuCl2, 3. Answer the following in brief :, i., What are acids and bases according, to Arrhenius theory ?, ii. What is meant by conjugate acidbase pair?, iii. Label the conjugate acid-base pair in, the following reactions, a. HCl + H2O, H3O⊕ + Cl, b. CO32 + H2O, OH + HCO3, iv. Write a reaction in which water acts, as a base., v., Ammonia serves as a Lewis base, whereas AlCl3 is Lewis acid. Explain., vi. Acetic acid is 5% ionised in its, decimolar solution. Calculate the, dissociation constant of acid, (Ans : 2.63 × 10-4), vii. Derive the relation pH + pOH = 14., viii. Aqueous solution of sodium, carbonate is alkaline whereas, aqueous solution of ammonium, chloride is acidic. Explain., ix. pH of a weak monobasic acid is 3.2, in its 0.02 M solution. Calculate its, dissociation constant., x. In NaOH solution [OH ] is 2.87 ×, 10-4. Calculate the pH of solution., 4. Answer the following :, i., Define degree of dissociation., Derive Ostwald's dilution law for the, CH3COOH., ii. Define pH and pOH. Derive, relationship between pH and pOH., iii. What is meant by hydrolysis ? A, solution of CH3COONH4 is neutral., why ?, , iv., , Dissociation of HCN is suppressed, by the addition of HCl. Explain., vi. Derive the relationship between, degree, of, dissociation, and, dissociation constant in weak, electrolytes., vii. Sulfides of cation of group II are, precipitated in acidic solution (H2S, + HCl) whereas sulfides of cations, of group IIIB are precipitated, in ammoniacal solution of H2S., Comment on the relative values of, solubility product of sulfides of these., viii. Solubility of a sparingly soluble salt, get affected in presence of a soluble, salt having one common ion. Explain., ix. The pH of rain water collected in a, certain region of Maharashtra on, particular day was 5.1. Calculate, the H⊕ ion concentration of the rain, water and its percent dissociation., x., Explain the relation between ionic, product and solubility product to, predict whether a precipitate will, form when two solutions are mixed?, , Activity :, Take two test tubes and label them, as A and B. Add Zinc filings in both, the test tubes. In the test tube labelled A, add 5 mL of 1M HCl and in test B 5 mL, of acetic acid. Keep the test tubes on the, stand. Note down your observations., a. Do you see any effervescence coming, from the two test tubes ?, b. Which gas is evolved ?, c. How do you identify the gas ?, d. What is the relative rate at which the, gas is evolved in the two test tubes, e. Based on your observations comment, on the strength of acids used., , 62
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4. CHEMICAL THERMODYNAMICS, Can you recall ?, 1. How do you define energy?, 2. What are the different forms of, energy?, , putting certain mass on the piston. In this case,, a gas under study is called the system., , Surrounding, , 4.1 Introduction : You know transformation, of liquid water into vapour, solid ice into liquid, water or burning of carbon forming carbon, dioxide, CO2, are accompanied by a change, in energy. In dry cell, the chemical energy is, converted into electrical energy. On the other, hand, in electroplating of metals electrical, energy is converted into chemical energy., Thus it may be realized that the energy can be, transformed from one form into another., , Surrounding, , Surrounding, , Gas, (system), , Fig. 4.1 : System and surroundings, , A part of the universe under, thermodynamic investigation is called, the system.All other parts of the universe, outside the system such as cylinder, room, and others, are surroundings. The universe, is made of system plus surroundings., , Do you know ?, At the top of dam, water is, stored in a reservoir. It has certain, potential energy due to its height from ground, level and its kinetic energy is negligible as it, is not in motion. As the water starts to fall, down through an outlet its potential energy, decreases and kinetic energy increases due, to the downward velocity. It means that, potential energy of falling water is converted, into kinetic energy., , 4.2.2 Types of system :, Observe and discuss..., Observe Fig. 4.2 and discuss, with reference to transaction of, energy and matter., , Matter, , Thermodynamics is concerned with the, energy changes in physical and chemical, transformations. Thermodynamics, however, gives no information on the rates of physical or, chemical processes or underlying mechanisms, involved in these., , Energy, , Energy, , Insulator, Matter, , Energy, Open system, , (a), , 4.2 Terms used in thermodynamics, , Matter, , Closed system, , Isolated system, , (b), , (c), , Fig. 4.2 : Kinds of systems, , Three types of systems are shown in Fig. 4.2., , 4.2.1 System and surrounding : Consider, a gas enclosed in a cylinder equipped with a, movable piston as shown in Fig. 4.1. Suppose, we undertake study of change in volume of, a gas and the amount of energy released or, gained by a gas when the pressure is varied by, , i. Open system : Fig. 4.2(a) shows an open, cup containing hot coffee placed in a room., You observe coffee cools down releasing, heat to the surroundings. The water vapour, from coffee simultaneously passes into, , 63
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surroundings. Such a system (coffee) which, exchanges both energy and matter with the, surroundings is called an open system., ii. Closed system : In Fig. 4.2(b), a cup, containing hot coffee is covered with a saucer., Coffee cools down by giving away heat to the, surroundings. The water vapour from coffee, now does not pass into surroundings. Such, a system that exchanges energy and not the, matter with the surroundings is called a closed, system., , Fig. 4.3 : Change of state, , Suppose the pressure of the system is, increased to 2 bar, (P2 ) volume changes to 0.5, dm3 (V2 ) and the temperature is maintained at, 300 K (T1 ). This is the final state of the system, which is different from the initial state. A, change in state functions of the system brings, forth a change of its state. This is shown in Fig., 4.3., , iii. Isolated system : As you see in Fig. 4.2(c),, a cup containing hot coffee covered with a, saucer is insulated from the surroundings., Coffee does not cool down. Moreover,, there is no escape of water vapour into the, surroundings. Such a system that does not, allow exchange of either energy or matter with, the surroundings is an isolated system., , The final state of the system in Fig. 4.3. is, described by pressure 2 bar (P2 ), volume 0.5, dm3(V2 ) and temperature 300 K(T1 ). A system, continues to be in such state as long as the state, functions are unchanged. How the pressure 2, bar is attained whether by increasing from 1, bar to 2 bar or decreasing from 5 bar to 2 bar,, would not matter., , 4.2.3 Properties of system, i. Extensive property :, A property which depends on the amount, of matter present in a system is called an, extensive property., , The property which depends on the state, of a system and independent of a path followed, to attain it, is called the state function., , Examples : Mass, volume, internal energy,, heat capacity, number of moles., ii. Intensive property :, , The term process means a physical or, chemical change in a system on going from, one state to another. This can be achieved by a, number of paths by some operation. A path here, refers to a sequence of situations the system, undergoes during the accomplishment of the, change. In other words the process in general, may not necessarily determine the change in, unique way. Only isothermal and adiabatic, reversible processes follow the unique path to, bring about the change of state of the system., , A property which is independent of the amount, of matter in a system is called intensive, property., Examples : Pressure, temperature, surface, tension, viscosity, melting point, boiling point,, specific heat., 4.2.4 State functions : As shown in Fig., 4.1, certain amount of a gas is enclosed in a, cylinder fitted with a movable piston. Suppose, the pressure of the gas is 1 bar (P1 ), volume is, 1 dm3 (V1 ) and temperature is 300 K (T1 ) in, the beginning. This initial state of the system, is fully defined by specifying the values of, these properties. Such properties defining the, state of a system, are state functions., , 4.2.5 Path Functions : The properties which, depend on the path are called path functions., For example, work (W) and heat (Q)., , 64
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4.2.6 Thermodynamic equilibrium :, Consider a gas enclosed in a cylinder fitted, with a movable piston shown in Fig. 4.1., The gas has temperature T1, pressure P1 and, volume V1. These state functions continue to, be constant as long as piston is motionless,, and no heat exchange takes place. This is an, equilibrium state., , carried out in a closed container is isochoric., For isochoric process ∆V = 0., iv. Adiabatic process : A process in which, there is no exchange of heat between system, and surroundings is an adiabatic process., (Q = 0). In adiabatic process the system is, completely insulated from the surroundings., For an exothermic process the heat is released, which rises temperature of the system. If the, process is endothermic the temperature falls., This results in either increase or decrease of, internal energy., , Now move the piston in upward direction, so that the gas expands. It passes through states, for which pressure, volume and temperature, are not specified and vary continuously during, the movement of the piston. The gas would, then be in nonequilibrium state., , v. Reversible process : Consider a gas enclosed, in a cylinder fitted with a movable piston., Let the external pressure be Pext on the outer, surface of the piston be set equal to pressure P, of the gas. Neither expansion nor compression, of the gas occurs. A system is then said to be, in mechanical equilibrium with surroundings., , Stop the movement of the piston. Suppose, at this stage the pressure and volume of the gas, are respectively P2 and V2 and the temperature, is constant at T1. The state functions are, constant since the piston is motionless. The, gas is then in another equilibrium state., , Consider Pext is reduced by an infinitesimal, amount. Now it the Pext is infinitesimally, smaller than P the piston moves out slowly, allowing gas to expand., , A system is said to be in thermodynamic, equilibrium when its state functions do not, vary with time. Thermodynamics considered, here is limited to equilibrium states., , If Pext is slightly increased so that it, becomes infinitesimally greater than P, the, piston moves inward with a compression of, the gas., , 4.2.7 Process and its types : A, transition, from one equilibrium state to another is called, a process. They are of different types., i. Isothermal process : It is the process in, which temperature of the system remains, constant throughout the transformation., , For the system in mechanical equilibrium, with its surroundings, infinitesimal change, may cause the process to occur in the reverse, direction. The process is then said to be, thermodynamically reversible. A process, conducted in such a way so that at every, stage the driving force due to pressure (P) is, infinitesimally greater than the opposing force, due to external pressure (Pext) and which can, be reversed by a slight change of the opposing, force is reversible process., , In such process heat flows from the system, to surroundings and vice versa so as to keep the, temperature constant. For a given temperature, the internal energy (U) of the system remains, constant. Thus, ∆T = 0 and ∆U = 0., ii. Isobaric process : In isobaric process, the pressure remains constant during the, transformation. In the laboratory chemical, reactions are carried out in open containers at, constant atmospheric pressure or ∆P = 0, , Features of reversible process, i. The driving and opposing forces differ by an, infinitesimal amount., , iii. Isochoric process : It is a process during, which volume of the system remains constant, during the transformation. A chemical reaction, , ii. The process can be reversed by an, infinitesimal change in conditions., , 65
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surroundings. With no heat being transferred, a loss of energy by the system is equal to work, done by the system on the surroundings. This, is PV expansion., , iii. A reversible process proceeds infinitely, slowly and takes place in infinite number of, steps., iv. At the end of every step of the process, the, system attains mechanical equilibrium with, the surroundings., , ii. Reaction between NH3 gas and HCl gas, Now, consider, NH3(g) + HCl(g), NH4Cl(s), As the reaction progresses the gases are, consumed resulting in a decrease of volume., The piston moves down. A decrease in the, height of the mass is shown in Fig. 4.5., , 4.3 Nature of heat and work, 4.3.1 Nature of work (W) : In mechanics the, work is defined as the energy by which body, is displaced through a distance d with an, application of force. Thus,, W=f×d, In thermodynamics the type of work, involved is pressure-volume or PV work, that, is, work is done when the system (gas) expands, or contracts against the external opposing, force., It may be realized that the product of, pressure and volume is equal to work. Pressure, is defined as force per unit area. If d is the, distance, area A = d2 and volume V = d3. Then, f, f, PV =, × V = 2 × d3 = f d = W, A, d, Now let us explore the PV work with, two chemical reactions in a cylinder equipped, with frictionless movable piston attached with, a certain mass on its outer surface., i. Decomposition of H2O2, Consider 2 H2O2(l), 2 H2O(l) + O2(g), , Fig. 4.5 : Reaction between NH3(g) and HCl(g), , In the process the surroundings lose, energy to the system and perform work on the, system. If no heat transfer occurs work done, by the surroundings is equal to gain in energy, by the system. This is PV work., Thus the work refers to a way by which, a system exchanges energy with surroundings., 4.3.2 Nature of heat (Q) : Like heat is a form of, energy by which the system exchanges energy, with its surroundings. When the system and its, surroundings are at different temperatures heat, either flows in or let out of the system., 4.3.3 Sign conventions of W and Q : The, energy changes for the system are considered, hereafter., , O2(g), H2O2, , The energy entering the system from the, surroundings has positive value. While the, energy leaving the system and flowing into the, surroundings is negative. This is shown in Fig., 4.6., , Fig. 4.4 : Decomposition of H2O2, , The gas produced in above reaction, pushes the piston upwards so that the mass in, the surroundings is raised as shown in Fig. 4.4., In lifting the mass the system loses energy to, the surroundings or it performs work on the, , +Q : Heat is absorbed by the system from the, surroundings., , 66
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W = f × d , , (4.2), , Substitution from Eq. (4.1) gives, W = - Pext × A × d, , (4.3), , The product of area of the piston and, distance it moves is the volume change (∆V), in the system., , Fig. 4.6 : Sign conventions, , -Q : Heat is released by the system to the, surroundings., , ∆V = A × d, , (4.4), , Combining equations (4.3) and (4.4) we write, , +W : Work is done on the system by the, surroundings., , W = - Pex ∆V , , -W : Work is done by the system on the, surroundings., , (4.5), , W = - Pex (V2 - V1), , Note W and Q are path functions., , where V2 is final volume of the gas., , 4.4 Expression for pressure-volume (PV), work : Consider a certain amount of gas at, constant pressure P is enclosed in a cylinder, fitted with frictionless, rigid movable piston of, area A. This is shown in Fig. 4.7., , When the gas expands, work is done by the, system on the surroundings. Since V2 > V1, W, is negative. When the gas is compressed, work, is done on the system by surroundings. In this, case V1< V2, and -Pext ∆V or W is positive., Eq. (4.5) shows the external pressure, determines the work during expansion (or, compression) of the gas. A volume change, does no work unless the system is linked to the, surroundings by external pressure., Remember..., Remember during expansion, of a gas, work is done by the, system on the surroundings and during, compression work is done on the system by, the surroundings., , Fig. 4.7 : Pressure-volume work, , Let volume of the gas be V1 at temperature T., On expansion the force exerted by a, gas is equal to area of the piston multiplied, by pressure with which the gas pushes against, piston. This pressure is equal in magnitude and, opposite in sign to the external atmospheric, pressure that opposes the movement and has, its value -Pext. Thus,, f = -Pext × A, , 4.4.1 Free expansion : A free expansion, means expansion against zero opposing force., Such expansion occurs in vacuum. The work, done by a system during such expansion is, given by Eq. (4.5), W = - Pext ∆V. When the, gas expands in vacuum, there is no opposing, force that is Pext and hence, W = 0. In other, words no work is done when the gas expands, freely in vacuum., , (4.1), , where Pext is the external atmospheric pressure., If the piston moves out a distance d,, then the amount of work done is equal to the, force multiplied by distance., , 67
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4.4.2 Units of energy and work, , Problem 4.3 : 200 mL ethylene gas and 150, mL of HCl gas were allowed to react at 1, bar pressure according to the reaction, , 1 J = 1 kg m s = 1 Pa m, 2, , -2, , 3, , 1 Pa = 1 kg m-1 s-2, , C2H4(g) + HCl(g), , From to Eq. (4.5), W = - Pext ∆V, if pressure is, expressed in bar and ∆V in dm3, then the work, has the units of bar dm3., , Calculate the PV work in joules., Solution :, , 1 bar = 105 Pa = 105 kg m-1 s-2, , W = - Pext ∆V = - Pext (V2 - V1), , 1 dm3 bar = dm3 × 105 kg m-1 s-2, , According to the equation of reaction 1, mole of C2H4 reacts with 1 mole of HCl to, produce 1 mole of C2H5Cl. Hence, 150 mL, of HCl would react with only 150 mL of, C2H4 to produce 150 mL of C2H5Cl., , = m3 × 10-3 × 105 kg m-1 s-2, , , = 100 kg m2 s-2 = 100 J, , Problem 4.1 : Three moles of an ideal gas, are expanded isothermally from 15 dm3 to, 20 dm3 at constant external pressure of 1.2, bar. Estimate the amount of work in dm3, bar and J., Solution :, W = - Pext ∆V = - Pext (V2 - V1), Pext = 1.2 bar, V1 = 15 dm3, V2 = 20 dm3, Substitution of these quantities into the, equation gives, W = -1.2 bar (20 dm3 - 15 dm3), = -1.2 bar × 5dm3 = -6 dm3 bar, 1 dm3 bar = 100 J, Hence, W = -6 dm3 bar × 100 J/dm3, bar = -600 J, , V1 = 150 mL + 150 mL = 300 mL = 0.3 dm3, V2 = 150 mL = 0.15 L, Pext = 1 bar, Substitution of these quantities in above, W = -1 bar (0.15 dm3 - 0.3 dm3), = 0.15 dm3 bar, = 0.15 dm3 bar × 100, , J, dm bar, 3, , = 15.0 J, 4.5 Concept of maximum work : Eq. (4.5), shows the amount of work performed by a, system is governed by the opposing force (Pext)., Larger the opposing force more work is done, by the system to overcome it., If the opposing force is zero no work is, involved. With an increase of the opposing, force from zero, more work will be needed by, the system. When the opposing force reaches, its maximum the system performs maximum, work. With an opposing force being greatest, more effort would be needed to overcome it., , Problem 4.2 : Calculate the constant, external pressure required to compress 2, moles of an ideal gas from volume of 25, dm3 to 13 dm3 when the work obtained is, 4862.4 J., Solution :, W = - Pext ∆V = - Pext (V2 - V1), V1 = 25 dm3, V2 = 13 dm3, W = 4862.4 J, dm3 bar, W = 4862.4 J × 100 J = 48.62 dm3 bar, Substitution of these into the equation gives, 48.62 dm3 bar = - Pext (13 dm3-25 dm3), = - Pext × 12 dm3, Hence, Pext =, , C2H5Cl(g), , Thus when the opposing force (Pext), becomes greater than the driving force (P), the process gets reversed. Since the opposing, force cannot be greater than the driving force, it should be the maximum., If the pressure P of the gas differs from, Pext by a quantity ∆P then P - Pext = ∆P and, Pext = P - ∆P. The eq. (4.5) then becomes, , 48.62 dm3 bar, = 4.052 bar, 12 dm3, , W = -(P - ∆P) ∆V, , 68
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When the volume of a gas increases by, an infinitesimal amount dV in a single step, the, small quantity of work done, , The work (W) would be maximum, when ∆P is smallest. This means the opposing, force (Pex) must be infinitesimally smaller, than the driving force (P) for the work to be, maximum. This is required for the process to, be reversible. The maximum work is obtained, from the change which is thermodynamically, reversible., , dW = -Pext dV , , As the expansion is reversible, P is greater by, a very small quantity dp than pex. Thus,, P -Pext = dP or Pext = P - dP, , 4.5.1 Expression for the maximum work :, , Step 2, , dW = - (P - dP)dV = - PdV + dP dV, Neglecting the product dpdV which is very, small, we get, dW = - PdV , , dv, dv, , dv, V1, , Gas, , Gas, , Gas, , (4.8), , The total amount of work done during, entire expansion from volume V1 to V2 would, be the sum of infinitesimal contributions of, all the steps. The total work is obtained by, integration of Eq. (4.8) between the limits of, initial and final states. This is the maximum, work, the expansion being reversible. Thus,, v2, final, , Continued, , Step 3, , (4.7), , Combining equations (4.6) and (4.7),, , Consider n moles of an ideal gas enclosed, in a cylinder fitted with frictionless movable, rigid piston. It expands isothermally and, reversibly from the initial volume V1 to final, volume V2 at temperature T. The expansion, takes place in a number of steps illustrated in, Fig. 4.8. , , Step 1, , (4.6), , ∫ dW, , Gas, , =-, , initial, , v2, , ∫ PdV, , v1, , ∫ PdV, v1, Using the ideal gas law, Hence Wmax = -, , Fig. 4.8 : Reversible expansion, , During each step the external pressure, Pext is made infinitesimally smaller than the, pressure P of the gas, with a gradual removal, of masses from the piston. The gas expand, slowly and its pressure P would decrease., The expansion continues until the pressure, of the gas falls to Pext. Beyond this no further, expansion occurs and the system attains, mechanical equilibrium with its surroundings., The volume of a gas is increased by an, infinitesimal quantity dv in each single step., , (4.9), , PV = nRT, v2, Wmax = -∫ nRT dV, V, v1, v2, = -nRT ∫ dV because T is constant., v1 V, v, = - nRT ln(V) 2, v1, = - nRT (ln V2 - ln V1), V, = - nRT ln 2, V1, V2, = -2.303 nRT log10, V1, , The process is repeated in such a way, that every time Pext is lowered infinitesimally, the gas undergoes a series of infinitesimal, increments in volume until the volume V2 is, attained., , (4.10), , At tconstant temperature, P1V1 = P2V2 or, P, V2, = 1, P2, V1, , 69
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Replacing V2/V1 in Eq. (4.10) by P1/P2, We, have, P, Wmax = -2.303 nRT log 1, (4.11), P2, , Problem 4.6 : 300 mmol of an ideal gas, occupies 13.7 dm3 at 300 K. Calculate the, work done when the gas is expanded until, its volume has increased by 2.3 dm3 (a), isothermally against a constant external, pressure of 0.3 bar (b) isothermally and, reversibly (c) into vacuum., , Problem 4.4 : 2 moles of an ideal gas are, expanded isothermally and reversibly from, 20 L to 30 L at 300 K. Calculate the work, done (R= 8.314 J K-1 mol-1), V, Solution : Wmax = -2.303 nRT log10 2, V1, n = 2 mol, T = 300 K, V1 = 20 L, V2 = 30 L,, R = 8.314 J/K mol, , Solution :, a. W = -Pex ∆V, Pext = 0.3 bar, ∆V = 2.3 dm3, W = -0.3 bar × 2.3 dm3, = -0.69 dm3 bar, 100 J, = -0.69 dm3 bar ×, dm3 bar, = - 69 J, V, b. Wmax = - 2.303 nRT log10 2, V1, -3, n = 300 mmol = 300 × 10 mol = 0.3 mol,, T = 300 K, , Substitution of these quantities into the, equation gives, Wmax = -2.303 × 2 mol × 8.314 J/K mol ×, , 300K × log10 30 L, 20 L, = -2.303 × 2 × 8.314 J×300 × log10 1.5, = -2.303 × 2 × 8.314 J×300 × 0.1761, = -2023 J = -2.023 kJ, , Wmax = - 2.303 × 0.3 mol × 8.314 J K-1mol-1, , 16, × 300K × log10, 13.7, = -2.303 ×0.3 ×8.314 J ×300 ×0.0674, = - 116.1 J, c. W = - Pex ∆V, When gas is expanded to vaccum, Pext = 0, and W = 0, , Problem 4.5 : 22 g of CO2 are compressed, isothermally and reversibly at 298 K from, initial pressure of 100 kPa when the work, obtained is 1.2 kJ. Find the final pressure., Solution :, , P1, P2, =0.5 mol, T = 298 K,, , W = -2.303 nRT log10, n=, , 22 g, 44 g mol-1, , 4.6 Internal energy (U) : Every substance is, associated with a definite amount of energy., This energy stored in a substance is internal, energy denoted by U., , P1= 100 kPa, W = 1.2 kJ = 1200 J, Hence, 1200 J = -2.303 × 0.5 mol ×8.314 J, 100 kPa, K-1 mol-1 × 298K × log10, P2, or log10 100 kPa =, P2, -1200, 2.303 ×0.5 ×8.314 ×298, , The internal energy of a system is made up, of kinetic and potential energies of individual, particles of the system., ∆U = U2 - U1, where U1 and U2 are internal energies of initial, and final states, respectively. U is a state, function and extensive property., , , = -0.4206, 100 kPa, = antilog (-0.4206) = 0.3797, P2, 100 kPa, Therefore, P2 = 0.3797 = 263.4 kPa, , 70
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4.7.1 Formulation of first law of, thermodynamics : A system exchange energy, with its surroundings either by transfer of, heat or by doing work. An energy supplied, to the system increases its internal energy., On the other hand, removal of heat or work, from the system decreases its internal energy., , Try this..., 25 kJ of work is done on the, system and it releases 10 kJ of, heat. What is ∆U?, A transfer of energy (as heat or work), from the system would change its internal, energy. To know ∆U the energy supplied to or, removed from the system need to be monitored., , Suppose (Q) is heat supplied to the, system and W work done on the system by, the surroundings. The internal energy of the, system would increase., , i. The energy transferred to the system by, heating it or performing work on it is added, to the system., , Increase in internal energy of the system, is equal to sum of the quantity of heat supplied, to the system and amount of work done on the, system or, , ii. The energy transferred from the system, by cooling or by performing work on the, surroundings is removed from the system., , ∆U = Q + W , , The following examples illustrate how to, determine ∆U., , (4.12), , where ∆U is an increase in internal energy, of the system. Eq. (4.12) is the first law of, thermodynamics. For infinitesimal changes., dU = dQ +dW, (4.13), , i. 30 kJ of heat supplied to the system. It, would be added to internal energy of the, system and ∆U = +30 kJ., , 4.7.2 First law of thermodynamics for, various processes, , ii. If 20 kJ of work is done on the system, it, is added to internal energy of the system., Consequently, ∆U = + 20 kJ., , i. Isothermal process : Temperature is constant, in such process, internal energy is constant., Hence, ∆U = 0, For isothermal process, , iii. Suppose a system releases 10 kJ of heat and, performs 15 kJ of work on the surroundings., These quantities are removed from internal, energy of the system and ∆U = - 25 kJ, , 0 = Q +W or W = -Q, , (4.14), , The above equation implies that heat, absorbed by the system is entirely used for, doing work on the surroundings. When work, is done on the system by the surroundings it, results in release of heat., , 4.7 First law of thermodynamics : First law, of thermodynamics is simply the conservation, of energy. According to this law the total, energy of a system and surroundings remains, constant when the system changes from an, initial state to final state. The law is stated in, different ways as follows., , ii. Adiabatic process : In adiabatic process,, there is no exchange of heat between system, and its surroundings that is, Q = 0. then, (4.15), -∆U = -W, Thus an increase in internal energy of, the system is the work done on it. If the work, is done by the system on the surroundings at, the expense of its internal energy, the internal, energy accompanying the adiabatic process, would decrease., , i. Energy of the universe remains constant, ii. The total internal energy of an isolated, system is constant, iii. Energy is neither created nor destroyed, and can only be converted from one form to, another., All above statements are equivalent., , 71
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From Eq. (4.19), we write, , iii. Isochoric process : Substitution of, W = -Pext ∆V into Eq. (4.12), ∆U = Q - Pext ∆V, , H1 = U1 + P1V1 and H2 = U2 + P2V2, With these, , (4.16), , ∆H = U2 + P2V2 - U1 + P1V1, , As the reaction is carried out in a closed, container, volume of the system is constant or, ∆V = 0 and, ∆U = Qv, , = (U2 - U1) + (P2V2 - P1V1), = ∆U + ∆(PV), , (4.17), , For constant pressure, P1 = P2 = P and, , Equation (4.17) shows a change in, internal energy of the system is due to heat, transfer at constant volume. The subscript ‘V’, indicates that heat is transferred at the constant, volume. Further U being a state function, Qv is, also a state function., , ∆H = ∆U + P∆V, , Qp = ∆U + P ∆V, ∆H = Qp , , (4.24), , Thus change in enthalpy of a system is, equal to heat transferred from it at the constant, pressure. H and Qp are state functions., 4.8.1 Relationship between ∆H and ∆U for, chemical reactions : At constant pressure, ∆H, and ∆U are related as, , (4.18), , The reactions carried out in open, containers under constant atmospheric, pressure are common in chemistry, a special, symbol ∆H, the enthalpy change, is given to, indicate heat changes occurring at constant, pressure., , ∆H = ∆U + P∆V, i. For reactions involving solids and liquids, ∆V, usually is very small (solids or liquids do not, show volume change with change of pressure), and ∆H = ∆U, ii. For reactions involving gases, ∆V cannot be, neglected and, , Remember..., q is not a state function., Whereas Qv and Qp are state, functions., , ∆H = ∆U + P∆V, = ∆H + P(V2 - V1), , 4.8 Enthalpy (H) : Enthalpy of a system is, sum of internal energy of a system and the, energy equivalent to PV work., , ∆H = ∆U + PV2 - PV1, , (4.25), , where V1 is the volume of gas phase reactants, and V2 that of the gaseous products., , (4.19), , We assume reactant and product behave, ideally. Applying ideal gas equation PV = nRT., When n1 moles of gaseous reactants produce, n2 moles of gaseous products. The ideal gas, equation give,, , Change in enthalpy, ∆H, is also state, function given by, ∆H = H2 - H1 , , (4.23), , From equations (4.22) and (4.23), , Replacing Q by Qp and ∆U by Qp - Pext ∆V in, equation (4.16) gives, , H = U + PV , , (4.22), , If the pressure inside and outside is the same or, Pext = P, Eq. (4.18) gives, , iv. Isobaric process : Usually, chemical, reactions are carried out in the open containers, under constant atmospheric pressure. In such, reactions, ∆V ≠ 0, , Qp = ∆U + Pext ∆V, , (4.21), , (4.20), , where H1 and H2 are the enthalpies of initial, and final states, respectively., , PV1 = n1RT and PV2 = n2RT, , 72, , (4.26)
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Substitution of Eq. (4.26) into Eq. (4.25) yields, , Problem 4.9 : Calculate the work done in, oxidation of 4 moles of SO2 at 250C if, , ∆H = ∆U + n2RT - n1RT, , 2 SO2(g) + O2(g), , = ∆U + (n2- n1) RT, = ∆U + ∆ng RT, , 2 SO3(g), , R = 8.314 J K-1mol-1, , (4.27), , State whether work is done on the system or, by the system., , where ∆ng is difference between the number of, moles of products and those of reactants., , Solution :, , Problem 4.7 : ∆H for the reaction,, , For oxidation of 4 moles of SO2, the reaction, is, , 2C(s) + 3H2(g), C2H6(g) is -84.4, 0, kJ at 25 C. Calculate ∆U for the reaction, at 25 0C. (R = 8.314 J K-1 mol-1), , 4 SO2(g) + 2 O2(g), , 4 SO3(g), , W = -∆ng RT, , Solution :, , ∆ng = 4 - 6 = - 2 mol, T = 298 K, , ∆H = ∆U + ∆ng RT, , Hence,, , ∆ng = (moles of product gases) - (moles of, reactant gases), , W = -2 mol × -8.314 J K-1 mol-1 × 298 K, = 4955 J = 4.955 J, , ∆ng = 1 - 3 = -2 mol, , Work is done on the system (since W > 0)., , ∆H = -84.4 kJ, R = 8.314 J K-1 mol-1, , 4.8.2 Work done in chemical reaction :, , = 8.314 × 10-3 kJ K-1 mol-1, , The work done by a system at constant, temperature and pressure is given by, W = Pext ∆V. Assuming Pext = P,, , Substitution of these in above, -84.4 kJ = ∆U + 8.314 × 10-3 kJ K-1 mol-1 ×, 298 K × (-2 mol), , W = - P∆V, , = ∆U - 4.96 kJ, , = - P (V2 - V1), , Hence, ∆U = -84.4 kJ + 4.96 kJ = - 79.44 kJ, , = - PV2 + PV1, II the gases were ideal, using Eq. (4.26), , Under what conditions ∆H = ∆U?, , PV1 = n1RT and PV2 = n2RT, At constant temperature and pressure., , Problem 4.8 : In a particular reaction 2 kJ, of heat is released by the system and 6 kJ, of work is done on the system. Determine, of ∆H and ∆U?, , W = - n2RT + n1RT, = - (n2 - n1) RT, = - ∆ng RT , , Solution : According to the first law of, thermodynamics, , The above equation gives the work done by, the system in chemical reactions. The sign of, W depends on ∆V. We consider the following, cases:, , ∆U = Q + W, Q = -2 kJ,, , (4.28), , W = +6 kJ, , i. If n2> n1, ∆ng is positive and W < 0, or work is done by the system., , ∆U = -2 kJ + 6 kJ = + 4 kJ, Qp = ∆H = - 2kJ, , ii. If n1> n2, ∆ng is negative and W > 0, or work is done on the system., , 73
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l (liquid), g (gas) and aq (aqueous). ∆rH0, value refers to physical states of substances, those appear in the equation., iv. The given value of ∆rH0 assumes that the, reaction occurs in a given direction. ∆rH0 for, the reverse reaction equals in magnitude and, opposite in the sign to that of the forward, reaction. A exothermic reaction on reversal, becomes endothermic and vice versa., v. When the coefficients indicating the, number of moles of all substances in, thermochemical equation are multiplied or, divided by a certain numerical factor, the, corresponding ∆rH0 need to be multiplied or, divided by the same., Example of thermochemical equation, , On the other hand, if, ∑ Hproducts < ∑ Hreactants, ∆rH is negative, which means that heat is released and the, reaction is exothermic., For example,, N2(g) + O2(g), , 2 NO2(g),, , ∆rH = 66.4 kJ (endothermic), 2 KClO3(s), , 2 KCl(s) + 3O2(g),, , ∆rH = -78 kJ (exothermic), 4.10.3 Standard enthalpy of reaction(∆rH0), To compare enthalpy changes of different, reactions they have to be reported under similar, set of conditions., Thermodynamic standard state : The, standard state of a substance is the form, in which the substance is most stable at a, pressure of 1 bar and at temperature 298 K., If the reaction involves species in solution its, standard state refers to 1 M concentration., , CH4(g) + 2 O2(g), CO2(g) + 2H2O(l), , ∆rH0 = -890 kJ, The equation signifies that when 1 mole, of gaseous CH4 and 2 moles of O2 in their, standard states produce 1 mole of CO2 gas and, 2 moles of liquid water also in their standard, states the enthalpy change would be -890 kJ., , Standard states of certain elements and, compounds are H2(g), Hg(g), Na(s) or, C(graphite), C2H5OH(l), CaCO3(s), CO2(g), C2H5OH(l), H2O(l), CaCO3(s), CO2(g) refer to, 1 bar and 25 0C., , Try this..., Given the thermochemical equation,, C2H2(g)+ 5/2 O2(g), , 2CO2(g)+ H2O(l),, ∆rH0 = -1300 kJ, Write thermochemical equations when, i. Coefficients of substances are multiplied, by 2., ii. equation is reversed., , The standard enthalpy (∆rH ) of, reaction is the enthalpy change accompanying, the reaction when the reactants and products, involved are in their standard states., 0, , 4.10.4 Thermochemical equation : It is, the balanced chemical equation in which, the enthalpy change, physical states and the, number of moles of reactants and products,, have been specified. Here follow the guidelines, for writing thermochemical equations :, , 4.10.5 Standard enthalpy of formation, (∆fH0), Consider, 1, , H2(g) + 2 O2(g), , i. Consider the balanced equation for, reactants and products., ii. The value and appropriate sign of enthalpy, change is given on the right hand side. This, value is ∆rH0., iii. The physical states of reactants and, products are specified by letter, s (solid),, , H2O(l), ∆rH0 = -286 kJ, , For the reaction where one mole of, liquid water in standard state is formed from, H2 and O2 gases in their standard states, the, enthalpy changes for the reaction would be the, standard enthalpy of formation of water. ∆fH, of water is -286 kJmol-1., , 76
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4.11 Spontaneous (irreversible) process :, , processes tend to occur in a direction, that leads to equilibrium., , Spontaneous processes have a natural, tendency to occur and do not require any, external influence for their occurrence., , 4.11.1 Energy and spontaneity :, The spontaneous reaction takes place, in a direction in which energy of the system, is lowered. It is accompanied by release of, energy. The reaction between NaOH and, HCl is exothermic (∆rH° = -57 kJ) and is, spontaneous., , Do you know ?, i. The aqueous NaOH and, HCl solutions mixed together., NaOH immediately combines with, HCl to form NaCl and water., NaOH(aq)+HCl(aq), , , , On the other hand :, , NaCl(aq)+ H2O(l),, ∆rH0 = -57 kJ, , i. Ice melts spontanoeusly above 0 °C by, absorbing heat from the surroundings. It, is endothermic., , No external force or energy is required, for the reaction to occur. This is, spontaneous. The process stops when, HCl or NaOH is consumed., , ii. Likewise, NaCl dissolves spontaneously in, water with the absorbtion of heat from the, surroundings., , NaCl is dissolved in water, it does not, react with water to produce NaOH, and HCl., , NaCl(s) + aq, , , Na⊕(aq) + Cl (aq), ∆H0 = +3.9 kJ mol-1, , These are endothermic and spontaneous. It, is therefore, clear that the exothermicity is, not the sufficient criterion for deciding of, spontaneity. There needs to be an another, factor to describe spontaneity., , ii. Water flows from higher level to lower, level. It is not necessary to apply, external force. It is a spontaneous, process. The flow ceases when two, levels become equal or when the, equilibrium is reached., , 4.11.2 Entropy :, To know what entropy consider the following, processes:, , iii. Ice melts spontaneously above 0 °C., iv. Hot coffee in a cup placed in a room, cools down releasing heat to the, surroundings. This is spontaneous., , i., , In solid state water molecules in ice are, arranged in a definite order., , ii. When ice melts, this highly crystalline, arrangements of water molecules collapse., The molecules become free in liquid state., An ordered state thus tends to become, more disordered., , Key points of spontaneous process, i. It occurs of its own and does not require, any external agency to occur., ii. It proceeds in one direction and cannot, take place in the opposite direction unless, the external stimulant is present., iii. The spontaneous processes can be rapid or, slow or spontaneity is not concerned with, the rate of the reaction., , Ice highly, ordered, state, , iv. The processes continues till equilibrium, is reached. The spontaneous (natural), , H2O(l), disordered, state, , H2O (g), highly, disordered, state, , Fig. 4.10 Increasing disorder, , 81
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Quantitative definition of entropy, , iii. When liquid water vaporises, gaseous water, molecules move freely and randomly in, the available space. A less disordered, state becomes highly disordered as shown, in Fig. 4.10., , Entropy is a measure of molecular, disorder or randomness. An entropy change, of a system is equal to the amount of heat, transferred (Qrev) to it in a reversible manner, divided by the temperature in kelvin T at, which the transfer takes place. Thus, Q, ∆S = rev , (4.32), T, the ∆S is thus expressed in J K-1., , During melting of ice or the vaporisation, of liquid water the disorder or randomness, increases. The disorder or randomness is, measured by entropy, denoted by S. Greater, the disorder of a system larger is its entropy., The melting of ice and vaporisation of liquid, water show that disorder and hence, entropy, of substance increases as it passes from solid, to liquid to gas., , Entropy or its change ∆S is a state, function and depends on the initial and final, states of the system and not on the path, connecting two states., , In both processes entropy change, ∆S > 0. Look at the following processes :, , i. When heat is added to a system the molecular, motions increase owing to increase of their, kinetic energies. This results in increased, molecular disorder and thus entropy of the, system. ∆S is proportional to Qrev., , i. Dissolution of solid I2 in water :, I2(s) + aq., , I2 (aq), , ordered state, , disordered state, , (∆S is positive), , ii. The effectiveness of the addition of, heat to increase randomness depends on, temperature., , When solid iodine dissolves in water I2, molecules move randomly. Thus disorder and, hence, entropy of the system increases or ∆S, is positive for the dissolution process., , If a certain amount of heat is added to, system at the higher temperature then the, disorder caused is lesser than that caused, by adding the same amount of heat is added, to system at the lower temperature Thus,, ∆S relates reciprocally to temperature at, which the of heat is added., , ii. Dissociation of H2 molecule into atoms, H2(g), , 2H(g) , , (∆S is positive), , One mole of H2 gas is converted into two, H atoms. Larger disorder is associated with, separated H atoms than with H2 molecule., Thus, disorder and hence entropy increases, or ∆S is positive., , 4.11.3 Entropy and spontaneity, (Second law of Thermodynamics), Look at the following examples :, , Try this..., , i. The entropy increases when ice melts, above 0 °C and water vaporizes at 100, °C. Both are spontaneous., , State whether ∆S is positive,, negative or zero for the following, reactions., i. 2H2(g) + O2(g), ii. CaCO3(s), , ∆, , ii. Consider the spontaneous reaction at room, temperature, , 2H2O(l), , 2H2O2 (l), , CaO(s) + CO2(g), , 2H2O (l) + O2 (g),, , ∆S = +126 J K-1, Entropy increases due to the formation of, O2 gas., , 82
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From above,, , From above examples, it is clear that, the entropy of the system increases in the, spontaneous processes. Consider the reaction., 2H2(g) + O2(g), , i. ∆Stotal > 0,, , ii. ∆Stotal < 0, the process is nonspontaneous, , 2H2O(l),, , iii. ∆Stotal = 0, the process is at equilibrium, , ∆S = -327 J K-1., , 4.11.5 Gibbs energy, , The entropy of the system decreases., Note the reaction is spontaneous., , As pointed out in the preceding section, it, is necessary to determine, ∆Ssys and ∆Ssurr, for, predicting the spontaneity of a reaction. We, are more interested in the system (reaction, mixture) . It. is, therefore convenient to, consider the criterion of spontaneity in terms, of the thermodynamic properties of a system., This problem was solved by American, theoretician J. W. Gibbs. He introduced a new, thermodynamic property called Gibbs energy, usually denoted by G., , 4.11.4 Second law of thermodynamics :, The second law of thermodynamics, states that total entropy of a system and, its surroundings increases in a spontaneous, process. For the process to be spontaneous, ∆Stotal = ∆Ssys + ∆Ssurr > 0, , (4.33), , Consider, 2H2(g) + O2(g), , the process is spontaneous, , 2H2O(l), , ∆S = -327 J K-1, and ∆H = -572 kJ (both at, 298 K), , The Gibbs energy is defined as, , To find ∆Stotal, we need to know ∆Ssurr., ∆H for the reaction is -572 kJ. When 2 moles, of H2 and 1 mole of O2 gas combine to form, 2 moles of liquid water, 527 kJ of heat is, released which is received by surroundings at, constant pressure (and 298 K). The entropy, change of surroundings is, 572 × 103 J, Q, = 1919 J K-1, ∆Ssurr = rev =, 298 K, T, ∆Stotal = ∆Ssys + ∆Ssurr, , where H is enthalpy and S entropy of the, system. Since H, T and S are state functions,, G is state function. A change in Gibbs energy, depends on initial and final states of the, system and not on a path connecting the two, states., , G = H - TS , , (4.34), , The change in Gibbs energy at constant, temperature and constant pressure is given by, ∆G = ∆H - T ∆S, , = -327 J K-1 + 1919 J K-1, , 4.11.6 Gibbs energy and spontaneity, , = + 1592 J K-1, , The total entropy change, accompanies a process is given by, , ∆Stotal > 0., , (4.35), that, , ∆Stotal = ∆Ssys + ∆Ssurr, , The reaction is thus spontaneous. It, follows that to decide spontaneity of reactions,, we need to consider the entropy of system and, its surroundings., , = ∆S + ∆Ssurr , , (4.36), , The subscript sys that refers to the system, is dropped hereafter., , The total entropy increases during a, spontaneous process that finally reaches, equilibrium. The equilibrium corresponds to, maximum total entropy. The total entropy, change, ∆Stotal must be zero for a process at, equilibrium., , Relation between ∆G and ∆Stotal, According, to, second, law, thermodynamics for a process to, spontaneous, ∆ Stotal > 0, , 83, , of, be
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3. For ∆H negative and ∆S is positive it, follows that ∆G is negative regardless of, temperature., , If ∆H is the enthalpy change, accompanying a reaction (system) the, enthalpy change of the surroundings is -∆H., With, ∆Ssurr = - ∆H, (4.37), T, Substituting above into Eq. (4.36),, ∆Stotal = ∆S - ∆H, T, , 4. For ∆H positive and ∆S is negative then, ∆G is positive regardless of temperature., Such reactions are nonspontaneous at all, temperatures., 4.11.8 Temperature of equilibrium, , (4.38), , For equilibrium, ∆G = ∆H - T∆S gives, , Thus ∆Stotal is expressed in terms of the, properties of the system only. Rearranging, T ∆Stotal = ∆H - T ∆S, , ∆H, (4.41), ∆S, T is the temperature at which the change, over from spontaneous to nonspontaneous, behavior occurs. ∆H and ∆S are assumed to, be independent of temperature in Eq. (4.41)., Introducing of temperature dependence of ∆H, or ∆S would not cause significant error for, the moderate temperature range., ∴T=, , (4.39), , Substituting in Eq. (4.35), ∆G = - T ∆Stotal, , (4.40), , For a spontaneous reaction Stotal > 0 and, hence, ∆G < 0. At constant temperature and, pressure Gibbs energy of the system decreases, in a spontaneous process., , 4.11.9 Gibbs function and equilibrium, constant : Gibbs energy change for a chemical, reaction is given by, , The second law leads to the conditions, of spontaneity which are summarised here., i. ∆Stotal > 0 and ∆G < 0, the process is, spontaneous., , ∆G = ∆G0 + RT ln Q, , (4.42), , where ∆G0 is standard Gibbs energy change, that is, the Gibbs energy change when the, reactants and products in a reaction are in their, standard states. Q is called reaction quotient Q, is analogus to that of the equilibrium constant., and involves nonequilibrium concentrations or, partial pressures in case of gaseous reaction., , ii. ∆Stotal < 0 and ∆G > 0, the process is, nonspontaneous., iii. ∆Stotal = 0 and ∆G = 0, the process is at, equilibrium., 4.11.7 Sponaneity and ∆H or ∆S, From ∆G = ∆H - T ∆S (at constant T, and P)., , Consider, aA + bB, , The temperature term determines relative, contributions of ∆H and ∆S to ∆G., , cC + dD, , ∆G = ∆G0 + RT ln Qc, , 1. ∆H and ∆S are both negative then ∆G, will be negative only when ∆H is more, negative than T∆S. This is possible at low, temperatures only., , [C]c[D]d, = ∆G0 + RT ln [A]a [B]b, or, , 2. ∆H amd ∆S both positive ∆G will be, negative only if T∆S > ∆H. This is possible, only at high temperatures., , ∆G = ∆G0 + RT ln Qp, PCc×PDd, = ∆G + RT ln P a× P b, A, B, 0, , 84, , (4.43), , (4.44)
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When the reaction reaches equilibrium, ∆G0 = 0, and Qc and Qp become Kc and Kp, respectively., Thus,, , Problem 4.17, For a certain reaction ∆H0 is -224 kJ, and ∆S is -153 J K-1. At what temperature, the change over from spontaneous to, nonspontaneous will occur?, 0, , 0 = ∆G0 + RT ln Kc and 0 = ∆G0 + RT ln Kp, or, , Solution -, , ∆G0 = -RT ln Kc and ∆G0 = -RT ln Kp, , ∆H0, T = ∆S0, ∆H0 = -224 kJ, ∆S0 = -153 JK-1 = -0.153, , kJ K-1, -224 kJ, Therefore, T = -0.153 J K-1 = +1464 K, , (4.45), or, , ∆G0 = -2.303 RT log10Kc, , and, ∆G0 = -2.303 RT log10Kp, , (4.46), , Problem 4.16 : State whether following, reactions are spontaneous or not. Further, state whether they are exothermic or, endothermic., a. ∆H = -110 kJ and ∆S = +40 JK-1 at 400 K, b. ∆H = +50 kJ and ∆S = -130 JK-1 at 250 K, Solution :, a. ∆G = ∆H - T∆S, ∆H = -110 kJ, ∆S = +40 J K-1, = +40 × 10-3 kJ K-1, T = 400 K, Therefore, ∆G = -110 kJ -400 K × 40, ×10-3 kJ K-1, , = -110 kJ - 16 kJ = -126 kJ, Since ∆G is negative, the reaction is, spontaneous. It is exothermic since ∆H is, negative, b. ∆H = +50 kJ, ∆S = -130 ×J K-1, = -130 ×10-3 kJ K-1 T = 250 K, ∆G = +50 kJ - 250 K ×(-130 ×10-3 kJ K-1), = 50 kJ + 32.5 kJ = +82.5 kJ, As ∆G is positive, the reaction is, nonspontaneous. It is endothermic since, ∆H is positive., , Since ∆H0 and ∆S0 are both, negative, the reaction is spontaneous at low, temperatures. A change over will occur at, 1464 K. The reaction is spontaneous below, 1464 K ., Problem 4.18, For the reaction,, CH4(g) + H2(g), , C2H6(g),, , Kp = 3.356 × 1017, Calculate ∆G0 for the reaction at 25 0C., Solution :, ∆G0 = -2.303 RT log10 Kp, R = 8.314 J K-1mol-1, T = 298 K,, Kp = 3.356 × 1017, ∆G0 = -2.303×8.314 × 298 ×, log10(3.356 × 1017), = -2.303 × 8.314 J mol-1 × 298 × 17.526, = -100,000 J mol-1, = -100 kJ mol-1, , 85
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Problem 4.20 : Calculate ∆G for the, reaction at 25 0C, , Problem 4.19 : Calculate ∆Stotal and state, whether the reaction is spontaneous or, nonspontaneous at 25 0C., HgS(s) + O2(g), , CO(g) + 2 H2(g), CH3OH(g), ∆G0, -1, = -24.8 kJ mol The partial pressures of, gases are PCO = 4 bar, PH2 = 2 bar and, PCH3OH = 2 bar, , Hg(l) + SO2(g),, , ∆H0 = -238.6 kJ, ∆S0 = +36.7 J K-1, Solution :, ∆Ssurr = -, , =, , Solution : ∆G = ∆G0 + RT ln Qp, , ∆H, T, , 0, , = ∆G0 + 2.303 RT log10, , (-238.6 kJ), 298 K, , ∆G0 = -24.8 kJ mol-1, R =8.314 ×10-3 kJ, K-1 mol-1, T = 298 K, , = +0.8007 kJ K-1 = +800.7 J K-1, , Calculate Qp,, PCH3OH, 2, Qp =, =, =, PCo×P2H2, 4×4, , ∆Stotal = ∆Ssys + ∆Ssurr, = +36.7 JK-1 + 800.7 JK-1, , , PCH3OH, PCO×P2H2, , = +837.4 J K-1, , 1, 8, , = 0.125, , ∆G = -24.8 kJ mol-1 +2.303 ×8.314× 10-3, kJ K-1 mol-1 × 298 K × log100.125, = -24.8 kJ mol-1 + 5.706 × (-0.903) kJ, mol-1, = -24.8 kJ mol-1 - 5.153 kJ mol-1, = -29.953 kJ mol-1, , ∆Stotal > 0, the reaction is spontaneous at, 25 0C., , Exercises, 1. Select the most apropriate option., i., , ii., , The, correct, thermodynamic, conditions for the spontaneous, reaction at all temperatures are, a., , ∆H < 0 and ∆S > 0, , b., , ∆H > 0 and ∆S < 0, , c., , ∆H < 0 and ∆S < 0, , d., , ∆H < 0 and ∆S = 0, , a. -500 J, , b. + 500 J, , c. -1013 J d. + 1013 J, iii., , In which of the following, entropy of, the system decreases?, a. Crystallization of liquid into, solid, b. Temperature of crystalline solid is, increased from 0 K to 115 K, , A gas is allowed to expand in a, well insulated container against a, constant external pressure of 2.5 bar, from an initial volume of 2.5 L to a, final volume of 4.5 L. The change in, internal energy, ∆U of the gas will, be, , c. H2(g), , 2H(g), , d. 2 NaHCO3(s), Na2CO3(s) + CO2(g) + H2O(g), , 86
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iv., , x., , The enthalpy of formation for all, elements in their standard states is, a. unity, b. zero, c. less than zero, d. different elements, , v., , vi., , ii., , State the first law of thermodynamics., , H2O(l), , iii., , What is enthalpy of fusion?, , iv., , What is standard state of a substance?, , v., , State whether ∆S is positive, negative, or zero for the reaction 2H(g), H2(g). Explain., , vi., , State second law of thermodynamics, in terms of entropy., , vii., , If the enthalpy change of a reaction, is ∆H how will you calculate entropy, of surroundings?, , C(g), , c. 2 Cl(g), d. H2O(s), , 6.24 g of ethanol are vaporized by, supplying 5.89 kJ of heat. Enthalpy, of vaporization of ethanol will be, , c. 38.9 kJ mol-1, d. 20.4 kJ mol-1, If the standard enthalpy of formation, of methanol is -238.9 kJ mol-1 then, entropy change of the surroundings, will be, a. -801.7 J K-1, , b. 801.7 J K-1, , c. 0.8017 J K, , d. -0.8017 J K, , ix., , viii. Comment on spontaneity of reactions, for which ∆H is positive and ∆S is, negative., 3. Answer in brief., , -1, , Which of the following are not state, functions?, 2. Q, , d. 93 kJ mol-1, , Cl2(g), , b. C(s), , 1. Q + W, , c. -245 kJ mol-1, , Comment on the statement: no work, is involved in an expansion of gas in, vacuum., , 2H(g), , b. 60.2 kJ mol-1, , viii., , b. -93 kJmol-1, , i., , a. H2(g), , -1, , a. 245 kJ mol-1, , 2. Answer the following in one or two, sentences., , Which of the following reactions is, exothermic?, , a. 43.4 kJ mol-1, , vii., , Bond enthalpies of H-H, Cl-Cl and, H-Cl bonds are 434 kJ mol-1, 242 kJ, mol-1 and 431 kJ mol-1, respectively., Enthalpy of formation of HCl is, , 3. W, , 4. H-TS, , a. 1,2 and 3, , b. 2 and 3, , c. 1 and 4 , , d. 2,3 and 4, , For vaporization of water at 1 bar,, ∆H = 40.63 kJ mol-1 and ∆S = 108.8, J K-1 mol-1. At what temperature,, ∆G = 0 ?, a. 273.4 K , , b. 393.4 K, , c. 373.4 K , , d. 293.4 K, , 87, , i., , Obtain the relationship between ∆G0, of a reaction and the equilibrium, constant., , ii., , What is entropy? Give its units., , iii., , How will you calculate reaction, enthalpy from data on bond, enthalpies?, , iv., , What is the standard enthalpy of, combustion ? Give an example., , v., , What is the enthalpy of atomization?, Give an example., , vi., , Obtain the expression for work done, in chemical reaction.
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vii., , ix. Calculate standard enthalpy of reaction,, , Derive the expression for PV work, , viii. What are intensive properties?, Explain why density is intensive, property., , Fe2O3(s) + 3CO(g), 2 Fe(s) + 3CO2(g),, from the following data., , ix., , ∆fH0(CO) = -110 kJ/mol,, , ∆fH0(Fe2O3) = -824 kJ/mol,, , How much heat is evolved when 12 g, of CO reacts with NO2 ? The reaction, is :, , ∆fH0(CO2) = -393 kJ/mol , , Ans. : (-25 kJ), , 4 CO(g) + 2 NO2(g), , x. For a certain reaction ∆H0 =219 kJ and, ∆S0 = -21 J/K. Determine whether the, reaction is spontaneous or nonspontaneous., , 4 CO2(g) + N2(g), ∆rH0 = -1200 kJ, 4. Answer the following questions., , xi. Determine whether the following reaction, is spontaneous under standard state, conditions., , i. Derive the expression for the maximum, work., ii. Obtain the relatioship between ∆H and ∆U, for gas phase reactions., , 2 H2O(l) + O2(g), , 2H2O2(l), , if ∆H0 = 196 kJ, ∆S0 = -126 J/K, , iii. State Hess’s law of constant heat, summation. Illustrate with an example., State its applications., , Does it have a cross-over temperature? , , (Nonspontaneous, No), , iv. Although ∆S for the formation of two, moles of water from H2 and O2 is, -327JK-1, it is spontaneous. Explain., (Given ∆H for the reaction is -572 kJ)., , xii. Calculate ∆U at 298 K for the reaction,, C2H4(g) + HCl(g), C2H5Cl(g), , ∆H = -72.3 kJ, , v. Obtain the relation between ∆G and ∆Stotal, Comment on spontaneity of the reaction., , How much PV work is done? , Ans. : (-69.8 kJ, 2.48 kJ), , vi. One mole of an ideal gas is compressed, from 500 cm3 against a constant external, pressure of 1.2 × 105 Pa. The work involved, in the process is 36.0 J. Calculate the final, volume. (200 cm3), , xiii. Calculate the work done during synthesis, of NH3 in which volume changes from, 8.0 dm3 to 4.0 dm3 at a constant external, pressure of 43 bar. In what direction the, work energy flows?, , vii. Calculate the maximum work when 24, g of O2 are expanded isothermally and, reversibly from the pressure of 1.6 bar to, 1 bar at 298 K. , , Ans. : (17.2 kJ, work energy flows into, system), , , , xiv. Calculate the amount of work done in the, (a) oxidation of 1 mole HCl(g) at 200 0C, according to reaction., , Ans. : (-873.4 J), , 4HCl(g) + O2(g), , viii. Calculate the work done in the, decomposition of 132 g of NH4NO3 at, 100 0C., NH4NO3(s), , (b) decomposition of one mole of NO at, 300 0C for the reaction, , N2O(g) + 2 H2O(g), , 2 NO(g), , State whether work is done on the system, or by the system., , , 2 Cl2(g) + 2 H2O(g), , , , Ans. : (-18.6 kJ), , 88, , N2(g) + O2, Ans. : (a = + 983 kJ ; b = 0 kJ)
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5. ELECTROCHEMISTRY, makes possible the manufacture of essential, chemicals. You have learnt preparation of, NaOH, widely used in the manufacture of, soaps, detergents and paper, by electrolysis of, NaCl. Electrolysis is possibly the only means, to produce fluorine. The processes such as, electro-refining (for purification of metals),, electroplating (for coating one metal is on the, surface of another) are also electrochemical, processes., , Can you recall ?, • What is a redox reaction ?, • Which form of energy is, converted into electrical energy in dry, cells ?, , In standard XI, you learnt redox reactions., Redox reaction forms the basis for the, generation of electricity by chemical reactions, and also for chemical reactions brought out, by means of electricity. These processes, are carried out in an electrochemical cell., Electrochemistry deals with the design and, operation of such cells., , • How is NaOH manufactured from, NaCl ?, 5.1 Introduction : Dry cell is used to power, our electrical and electronic equipments, because it generates electricity. Do you, know how does a dry cell generate electricity, ? A chemical reaction occurs in it which, generates electricity. Thus in a dry cell, chemical energy is converted into electrical, energy., , The current research in electrochemistry, is focused on the design of fuel cells. The, fuel cells are being explored as convenient, and compact source of electricity., 5.2 Electric conduction : We know that the, electric current represents a charge transfer., A charge transfer or flow of electricity occurs, through substances called conductors. There, are two types of conductors which give rise, to two types of conduction of electricity., , You are familiar with the electrolysis, of solutions of ions. Electrolysis is breaking, down of an ionic compound by the passage, of eletricity. Breaking down of an electrolyte, during electrolysis is a chemical reaction, that takes place by the passage of electricity., Electrical energy is, thus, converted into, chemical energy., , 5.2.1 Metallic conduction :, Can you recall ?, , Electrochemistry is the area of chemistry, which is concerned with interconversion of, chemical and electrical energy., , • What is the origin of electrical, conductivity of metals ?, Electrical conduction through metals, involves a direct flow of electrons from one, point to the other. The outermost electrons of, metals form conduction bond. The electrons in, conduction band are free to move and hence, flow under the influence of applied electrical, potential (Chapter 1). Metallic conductors, are, thus, electronic conductors., , It also deals with the resistance and, conductance of aqueous electrolytic solutions., The determination of conductivities of aqueous, electrolytic solutions provide an information, on the extent of ionization of electrolytes in, water. (Refer to Chapter 3)., The study of electrochemical cells is, important in science and technology. It, , 90
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5.2.2 Electrolytic or ionic conduction :, Electrolytic conduction involves conduction, of electric current by the movement of ions, of the electrolytes. In this type of conduction, the charge transfer occurs in the form of, movement of ions through molten electrolytes, or the aqueous solutions of electrolytes., Substances such as ionic salts, strong or weak, acids and bases are the electrolytes. These, dissociate into ions when dissolved in polar, solvents such as water. Ionic solids dissociate, into ions in molten state as well., , i., , On the other hand, substances like, potassium chloride, acetic acid, sodium, hydroxide, HCl dissociate in their aqueous, solutions. The conductivities of their, aqueous solutions are higher than that, of water. These are called electrolytes., Electrolytes conduct electricity in molten, state or when dissolved in water., , Conduction, through, electrolytic, conductors involves transfer of matter from, one part of the conductor to the other. It, means that the current flowing through an, electrolytic conductor is accompanied by a, chemical change., , ii. On the basis of high or low electrical, conductivity electrolytes are classified, into strong and weak electrolytes. The, substances such as ionic salts, strong, acids or bases are almost completely, dissociated in aqueous solutions. These, are strong electrolytes. The solutions, of strong electrolytes exhibit high, conductivities., , 5.2.3 Information provided by measurement, of conductivities of solutions :, Try this..., , Switch, , Non conductive, frame, , Battery, , The weak acids and weak bases are, weak electrolytes. They dissociate to a, very small extent in aqueous solutions, and show lower conductivities than those, of strong electrolytes., , Lamp, , Electrodes, Solution, , •, , Arrange a simple set up as shown in, the diagram above., , •, , The lamp will glow when circuit is, complete., , •, , Prepare 5 % (mass/volume) solutions, of cane sugar, acetic acid, sodium, chloride and urea in distilled water., , •, , Check the electrical conductivity, of these solutions using the above, assembly. Compare these with that, observed with distilled water., , The, conducting and nonconducting, nature of solutions can be identified, by measurement of their conductivity., Sucrose and urea do not dissociate in their, aqueous solutions. The conductivities of, these solutions are nearly the same as, that of water. These substances are called, nonelectrolytes., , Remember..., Electrolyte is a compound that, conducts electricity when molten or, in aqueous solution and breaks down into, ions during electrolysis., 5.3 Electrical conductance of solution :, According to Ohm's law, the electrical, resistance R of a conductor is equal to the, electric potential difference V divided by the, electric current, I :, V, R=, , (5.1), I, The SI unit of potential is volt (V) and, that of current is ampere (A). The unit of, , 91
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From Eq. (5.2) and Eq. (5.4), we write, 1 l, l, k=G, =, (5.5), R a, a, Combination of Eq. (5.3) and Eq. (5.5), shows that k = 1/ρ., , electrical resistance is ohm denoted by the, symbol Ω (omega). Thus, Ω = VA-1., The electrical conductance, G, of a, solution is reciprocal of resistance., 1, G=, (5.2), R, The SI unit of G is siemens, denoted by, S, which is equal to Ω-1. Therefore, we write, S = Ω-1 = AV-1 = CV-1s-1 where C represents, coulomb, the unit of electricity related to, current strength in ampere and time in, seconds as C = A s., , Units of electrolytic conductivity, , The electrical resistance of a conductor, is proportional to length l and inversely, proportional to cross sectional area a. Thus,, l, l, R∝, or R = ρ, (5.3), a, a, where ρ, the proportionality constant is, called resistivity of the conductor. It is the, resistance of conductor of unit length and unit, cross sectional area., , Quantity, , SI unit, , Length, Area, Resistance, Conductivity, , m, m2, Ω, -1, Ω m-1 or, S m-1, , Common unit, , cm, cm2, Ω, Ω-1 cm-1, , 5.3.2 Molar conductivity (∧) : The electrolytic, conductivity is not suitable for comparing, conductivities of different solutions. The, conductivity of a solution depends on number, of ions present in unit volume of solution., The solution of higher concentration contains, more ions and exhibits higher conductivity, than the solution of lower concentration. To, compare conductivities of different solutions,, they must have the same concentration., In 1880, the German physicist F.W.G., Kohlrausch introduced the term molar, conductivity denoted by ∧ (lambda)., The molar conductivity of an electrolytic, solution is the electrolytic conductivity, k,, divided by its molar concentration c., k, ∧=, (5.6), c, SI units of k are S m-1 and that of, c are mol m-3. Hence SI units of ∧ are, S m2 mol-1. Common units employed for, molar conductivity are Ω-1 cm2 mol-1., Significance of molar conductivity : To, understand the significance of ∧, consider, volume of a solution containing 1 mole of, dissolved electrolyte. Suppose the solution is, placed between two parallel electrodes 1 cm, apart and large enough to accommodate it., The electrical conductance exhibited by this, solution is the molar conductivity. The molar, conductivity is the electrical conductance, generated by all the ions in 1 mole of the, electrolyte., , Can you recall ?, What is the SI unit of, resistivity ?, 5.3.1 Conductivity (k) : We have seen that, G = 1/R and R is directly proportional to, length and inversely proportional to its cross, sectional area. It, therefore, follows that G, is directly proportional to a and inversely, proportional to the length l. Thus, a, a, G∝, or G = k, (5.4), l, l, The proportionality constant k is called, conductivity. G = k if length and cross, sectional area of conductor are unity., Thus, conductivity is the electrical, conductance of a conductor of unit length, and unit area of cross section. In other words,, the conductivity is the electrical conductance, of unit cube of material. Conductivity of, solution of an electrolyte is called electrolytic, conductivity which refers to the electrical, conductance of unit volume (1 m3 or 1 cm3), of solution., , 92
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∧ = 223 Ω-1 cm2 mol-1, c = 0.05 mol L-1, , Remember..., Conductivity is electrical, conductance due to all the ions in, 1 cm3 of given solution. Molar conductivity, is the electrical conductance due to the ions, obtained from 1 mole of an electrolyte in, a given volume of solution., , Hence, 223 Ω-1 cm2 mol-1 × 0.05 mol L-1, k=, 1000 cm3L-1, -1, = 0.01115 Ω cm-1, 5.3.4 Variation, concentration, , 5.3.3 Relation between k and ∧ : Conductivity, k is the electrical conductance of 1 cm3 of, solution. If V is volume of solution in cm3, containing 1 mole of dissolved electrolyte,, its electrical conductance is ∧. Each 1 cm3, portion in the volume V has conductance, k. Hence, total conductance of V cm3 is kV, which is molar conductivity., Thus, we have ∧ = k V, , i., , of, , conductivity, , with, , The electrolytic conductivity is electrical, conductance of unit volume (1 cm3) of, solution. It depends on the number of, current carrying ions present in unit, volume of solution., , ii. On dilution total number of ions increase, as a result of increased degree of, dissociation., , (5.7), , iii. An increase in total number of ions is, not in proportion of dilution. Therefore,, the number of ions per unit volume, of solution decreases. This results in, decrease of conductivity with decrease, in concentration of solution., , Concentration of solution, = c mol L-1, c mol L-1, c, mol cm-3, 3 -1 =, 1000 cm L, 1000, Volume, V of solution in cm3 containing, 1 mole of an electrolyte is reciprocal of, concentration. Therefore,, 1, 1000, =, cm3 mol-1, V=, concentraion, c, (5.8), =, , Suppose 100 cm3 of solution of an, electrolyte contains 8 × 1020 ions. The, number of ions per cm3 is 8 × 1018., If the solution is diluted to 1000 cm3, the total number of ions will increase but, not by a factor of 10. Assume that the, number of ions increases from 8 × 1020, to 64 × 1020 on dilution. After dilution the, number of ions per cm3 is 6.8 × 1018., It is evident that the number of ions, per cm3 decreases from 8 × 1018 to 6.8 ×, 1018 on dilution from 100 cm3 to 1000 cm3, and in turn, the conductivity decreases., , Substitution for V in Eq. (5.7) yields, 1000k, ∧=, (5.9), c, Try this..., What, must, be, the, concentration of a solution of silver, nitrate to have the molar conductivity of, 121.4 Ω-1 cm2 mol-1 and the conductivity of, 2.428 × 10-3 Ω-1 cm-1 at 25 0C ?, , 5.3.5 Variation of molar conductivity with, concentration, , Problem 5.1 : The molar conductivity of, 0.05 M BaCl2 solution at 250C is 223 Ω-1, cm2 mol-1. What is its conductivity ?, , i., , Solution :, 1000k, ∧c, ∧=, or k =, c, 1000, , ii. The increasing number of ions produced, in solution by 1 mole of the electrolyte, lead to increased molar conductivity., , 93, , The molar conductivity is the electrical, conductance of 1 mole of an electrolyte, in a given volume of solution.
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5.3.6 Variation of molar conductivity with, concentration : The variation of molar, conductivity with concentration in case of, strong and weak electrolytes is qualitatively, different., , Molar conductivity of strong electrolytes, at zero concentration can be determined by, extrapolation of linear part of ∧ versus c, curve as shown in Fig. 5.1. This method, cannot be used for weak electrolytes since ∧, versus c curve does not approach linearity., Kohlrausch law is useful for calculating ∧0, of weak electrolytes., , i. Strong electrolytes : The molar conductivity, of solution of strong electrolyte increases, rapidly with dilution. It approaches the limiting, value for 0.001 M or 0.0001 M solution. The, dilution has no effect on molar conductivity, thereafter. The maximum limiting value of, molar conductivity is the molar conductivity, at zero concentration or at infinite dilution., It is denoted by ∧0. The zero concentration, or infinite dilution means the solution is so, dilute that further dilution does not increase, the molar conductivity., , 5.3.7 Kohlrausch law of independent, migration of ions : The law states that at, infinite dilution each ion migrates independent, of co-ion and contributes to total molar, conductivity of an elctrolyte irrespective of the, nature of other ion to which it is associated., Both cation and anion contribute to, molar conductivity of the electrolyte at zero, concentration and thus ∧0 is sum of molar, conductivity of cation and that of the anion, at zero concentration. Thus,, , During nineteenth century F. Kohlrausch, with repeated experiments showed that the, molar conductivity of strong electrolytes varies, linearly with square root of concentration as :, ∧ = ∧0 - a c, , ∧0 = n⊕ λ0⊕ + n λ0, , (5.11), , where λ⊕ and λ are molar conductivities of, cation and anion, respectively, and n⊕ and, n are the number of moles of cation and, anion, specified in the chemical formula of, the electrolyte., , (5.10), , where a, is constant. For strong, electrolytes a plot of ∧ versus c is linear as, shown in Fig. 5.1., , Applications of Kohlrausch theory, 1., Strong, electrolyte, , ∧0 (KCl) = λ0K⊕+ λ0Cl, ∧0 [Ba(OH)2] = λ0Ba2⊕+ 2 λ0OH, , Weak, electrolyte, , Knowing the molar conductivites of ions, at infinite dilution, ∧0 values of electrolyte, can be obtained., , Fig. 5.1 : Variation of ∧ with c, , ii. Weak electrolytes : The molar conductivity, of weak electrolytes increases rapidly on, dilution. For concentrations of 0.001M or, 0.0001 M, the ∧ value is lower than ∧0 the, molar conductivity at zero concentration., with, , The theory can be used to calculate the, molar conductivity of an electrolyte at, the zero concentration. For example,, , 2., , For weak electrolytes the variation of ∧, c shown in Fig. 5.1 is not linear., , The theory is particularly useful in, calculating ∧0 values of weak electrolytes, from those of strong electrolytes. For, example, ∧0 of acetic acid can be, calculated by knowing those of HCl, NaCl, and CH3COONa as described below :, , ∧0 (HCl) + ∧0 (CH3COONa) - ∧0 (NaCl), = λ0H⊕ + λ0Cl + λ0CH3COO + λ0Na⊕ - λ0Na⊕ - λ0Cl, , 94
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1. Determination of cell constant : The cell, constant is determined using the 1 M, 0.1 M, or 0.01 M KCl solutions. The conductivity, of KCl solution is well tabulated at various, temperatures. The resistance of KCl solution, is measured by Wheatstone bridge. (Refer to, standard XII Physics Textbook Chapter 9), , Try this..., Obtain the expression for, dissociation constant in terms of ∧c, and ∧0 using Ostwald's dilution law., 5.3.9 Measurement of conductivity : The, conductivity of a solution can be determined, from the resistance measurements by, Wheatstone bridge., , In Fig. 5.3 AB is the uniform wire. Rx is, the variable known resistance placed in one, arm of Wheatstone bridge., , Conductivity Cell : The conductivity cell, consists of a glass tube with two platinum, plates coated with a thin layer of finely, divided platinium black. This is achieved by, the electrolysis of solution of chloroplatinic, acid. The cell is dipped in a solution whose, resistance is to be measured as shown in Fig., 5.2., , conductivity cell, , solution of unknown, resistance, , A.C., , Fig. 5.3 : Measurement of resistance, , The conductivity cell containing KCl, solution of unknown resistance is placed in, the other arm of Wheatstone bridge. D is a, current detector. F is the sliding contact that, moves along AB. A.C. represents the source, of alternating current., The sliding contact is moved along AB, until no current flows. The detector D shows, no deflection. The null point is, thus, obtained, at C., , Fig. 5.2 : Conductivity cell, , Cell constant : The conductivity of an, electrolytic solution is given by Eq. (5.5),, 1 l, k=, R a, For a given cell, the ratio of separation, (l) between the two electrodes divided by the, area of cross section (a) of the electrode is, called the cell constant. Thus,, l, Cell constant =, (5.13), a, SI unit of cell constant is m-1 which is, conveniently expressed in cm-1. The Eq. (5.5), then becomes, cell constant, k=, (5.14), R, The determination of conductivity consists, of three steps :, , According to Wheatstone bridge principle,, Rx, Rsolution, =, l (BC), l (AC), Hence, Rsolution =, , l(AC), × Rx, l(BC), , (5.15), , By measuring lengths AC and BC and, knowing Rx, resistance of KCl solution can, be calculated. The cell constant is given by, Eq. (5.13)., Cell constant = kKCl × Rsolution, The conductivity of KCl solution is, known. The cell constant, thus, can be, calculated., , 96
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2. Determination of conductivity of given, solution : KCl solution in the conductivity, cell in step (1) is replaced by the given, solution whose conductivity is to be measured., Its resistance is measured by the process, described in step (1). The conductivity of, given solution is then calculated as :, Cell constant, k=, Rsolution, 3. Calculation of molar conductivity : The, molar conductivity of the given solution is, then calculated using Eq. (5.9)., 1000 k, ∧= c, , 5.4.1 Electrochemical reactions :, Can you recall ?, What is the reaction involving, transfer of electrons from one, chemical species to another called?, The chemical reaction occuring in, electrochemical cells involves transfer, of electrons from one species to the, other. It is a redox reaction, we learnt in, (Std. XI, Chapter 6)., Electrochemical reactions, are made of, oxidation and reduction half reactions. The, oxidation half reaction occurs at one electrode, and the reduction half reaction occurs at the, other electrode. The net cell reaction is the, sum of these half reactions., 5.4.2 Electrodes : Electrodes are the surfaces, on which oxidation and reduction half, reactions take place. Electrodes may or may, not participate in the reactions. The electrodes, which do not take part in reactions are inert, electrodes., Cathode : It is an electrode at which the, reduction takes place. At this electrode the, species undergoing reduction gains electrons., Anode : It is an electrode at which oxidation, takes place. At this electrode, the species, undergoing oxidation loses electrons., 5.4.3 Types of electrochemical cells : There, are two types of electrochemical cells., 1. Electrolytic cell : In this type of cell,, a nonspontaneous reaction, known as, electrolysis, is forced to occur by passing a, direct current from an external source into, the solution. In such cells electrical energy, is converted into chemcial energy. The anode, of electrolytic cell is positive and cathode is, negative., 2. Galvanic or voltaic cell : In galvanic, (voltaic) cell a spontaneous chemical reaction, produces electricity. In these cells chemical, energy is converted into electrical energy., The anode of galvanic cell is negative and, cathode is positive., , Problem 5.5 : A conductivity cell containing, 0.01M KCl gives at 250C the resistance, of 604 ohms. The same cell containing, 0.001M AgNO3 gives resistance of 6530, ohms. Calculate the molar conductivity of, 0.001M AgNO3. [Conductivity of 0.01M, KCl at 25 0C is 0.00141 Ω-1 cm-1], Solution :, i. Calculation of cell constant, Cell constant = kKCl × RKCl, = 0.00141 Ω- cm-× 604 Ω, = 0.852 cm-1, ii. Calculation of conductiviy of AgNO3', Cell constant, k=, where R = 6530 Ω, R, 0.852 cm-1, 6530 Ω, = 1.3 × 10-4 Ω-1 cm-1, iii. Calculation of molar conductivity of, AgNO3, 1000k, ∧=, where c = 0.001 M, c, 1000 cm3 L-1 × 1.3 × 10-4 Ω-1 cm-1, =, 0.001 mol L-1, =, , = 130 Ω-1 cm2 mol-1, 5.4 Electrochemical cells : An electrochemical, cell consists of two metal plates or carbon, (graphite) rods. These electronic conductors, are dipped into an electrolytic or ionic, conductor., , 97
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The carbon electrode connected to, terminal electrode of the battery is anode, and that connected to negative terminal of, the battery is cathode., , Use your brain power, Distinguish, between, electrolytic and galvanic cells., 5.5 Electrolytic cell, , Remember..., In electrolysis the electrodes, are usually inert, Pt or graphite., , Do you know ?, Michael Faraday was the, first person to explain electrolysis, nearly 200 years ago., , Reactions occuring in the cell : Fused NaCl, contains Na⊕ and Cl ions which are freely, mobile. When potential is applied, cathode, attracts Na⊕ ions and anode attracts Cl ions., As these are charged particles, their migration, results in an electric current. When these, ions reach the respective electrodes, they, are discharged according to the following, reactions., , Electrolytic cell consists of a container, in which electrolyte is placed. Two electrodes, are immersed in the electrolyte and connected, to a source of direct current., At anode (+) a species oxidises with, the removal of electrons. These electrons are, pulled from anode and pushed to cathode, through an external circuit. The electrons, are supplied to species at cathode which are, reduced., , Oxidation half reaction at anode :, Cl ions migrate to anode. Each Cl, ion, that reaches anode, gives one electron, to anode. It oxidises to neutral Cl atom in the, primary process. Two Cl atoms then combine, to form chlorine gas in the secondary process., , Remember..., Electrolysis is the process, of breaking down of an ionic, compound in molten state or in aqueous, solution by the passage of electricity., 5.5.1 Electrolysis of molten NaCl, Construction of cell : The electrolytic cell, consists of a container in which fused NaCl is, placed. Two graphite electrodes are immersed, in it. They are connected by metallic wires to, a source of direct current that is battery. This, is shown in Fig. 5.4., , Cl2 gas, , Cl (g) + Cl (g) + 2e, (primary process), , Cl (g) + Cl (g), , Cl2 (g), (secondary process), , 2Cl (l), , Cl2 (g) + 2e, (overall oxidation), , The battery sucks electrons so produced, at the anode and pushes them to cathode, through a wire in an external circuit. The, battery thus serves as an electron pump. The, electrons from the battery enter into solution, through cathode and leave the solution, through anode., , Battery, (D.C. source), Carbon anode, , 2 Cl (l), , Reduction half reaction at cathode : The, electrons supplied by the battery are used, in cathodic reduction. Each Na⊕ ion, that, reaches cathode accepts an electron from the, cathode and reduces to metallic sodium., , Carbon cathode, , Fused Na, Fused NaCl, , Na⊕ (l) +e, , Fig. 5.4 : Electrolysis of fused NaCl, , 98, , Na (l)
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Net cell reaction, , The other is the reduction of water to, hydrogen gas., , The net cell reaction is the sum of two, electrode reactions., 2 Cl (l), 2 Na⊕ (l) + 2e, 2 Na⊕ (l) + 2 Cl (l), , ii. 2 H2O (l) + 2e, , Cl2 (g) + 2e, (oxidation half reaction), , The standard potential (section 5.7.1) for, the reduction of water is higher than that for, reduction of Na⊕. This implies that water has, much greater tendency to get reduced than the, Na⊕ ion. Hence reaction (ii), that is, reduction, of water is the cathode reaction when the, aqueous NaCl is electrolysed., , 2 Na (l), (reduction half reaction), 2 Na (l) + Cl2(g), (overall cell reaction), , Results of electrolysis of molten NaCl, i., , Oxidation half reaction at anode : At anode, there will be competition between oxidation, of Cl ion to Cl2 gas as in case of molten, NaCl and the oxidation of water to O2 gas., , A pale green Cl2 gas is released at anode., , ii. A molten silvery-white sodium is formed, at cathode., Decomposition of NaCl into metallic, sodium and Cl2(g) is nonspontaneous. The, electrical energy supplied by the battery, forces the reaction to occur., , i. 2 Cl (aq), , Cl2 (g) +2e ,E0oxi = - 1.36 V, , ii. 2H2O (l), , O2 (g) + 4H⊕ (aq) + 2e, E0oxi = - 0.4 V, , Standard electrode potential for the, oxidation of water is greater than that of Cl, ion or water has greater tendency to undergo, oxidation. It is, therefore, expected that anode, half reaction would be oxidation of water., The experiments have shown, however, that, the gas produced at the anode is Cl2 and, not O2. This suggests that anode reaction is, oxidation of Cl to Cl2 gas. This is because, of the overvoltage, discussion of which is, beyond the scope of the present book., , Remember..., When molten ionic compound, is electrolysed, a metal is formed, at the negative electrode and a nonmetal, at the positive electrode., 5.5.2 Electrolysis of aqueous NaCl :, Electrolysis of an aqueous NaCl can be, carried out in the cell used for the electrolysis, of molten NaCl using inert electrodes shown, in Fig. 5.4. The fused NaCl is replaced by, moderately concentrated aqueous solution of, NaCl. The water involved in electrolysis of, aqueous NaCl, leads to electrode reactions, that differ from electrolysis of molten NaCl., , It has been found experimentally, that the actual voltage required for, electrolysis is greater than that calculated, using standard potentials. This additional, voltage required is the overpotential., , Reduction half reaction at cathode : At, cathode, two reduction reactions compete., One is the reduction of Na⊕ ions as in case, of molten NaCl., i. Na⊕ (aq) + e, , H2 (g) + 2 OH (aq),, E0 = - 0.83 V, , Do you know ?, Refining of metal and, electroplating are achieved by, electrolysis., , Na (s), E0 = -2.71 V, , 99
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Overall cell reaction, , Q (C) = I (A) × t (s), , It is the sum of electrode reactions., 2 Cl (aq), 2 H2O (l) + 2e, , Total charge passed is Q(C). The charge, of one mole electrons is 96500 coulombs (C)., It is referred to as one faraday (IF). Hence,, , H2 (g) + 2 OH (aq), (reduction at cathode), Cl2 (g) + H2(g), + 2 OH (aq), (overall cell reaction), , Results of electrolysis of aqueous NaCl, i., , ii. Calculation of moles of electrons passed, , Cl2 (g) + 2e, (oxidation at anode), , 2 Cl (aq) + 2 H2O (l), , H2 gas is liberated at cathode., , ii. Cl2 gas is released at anode., iii. Because Na⊕ ions remain unreacted and, OH ions are formed at cathode, NaCl, solution is converted to NaOH solution., Do you know ?, Sea water is the source of, 300000 tones of Mg produced, every year by electrolysis., Electrochemical art : Al, Cr and Sn, can be coloured by an electrochemical, process called anodizing. In this process, metal anode oxidizes to give metal oxide, coat. When an organic dye is added to the, electrolyte, dye molecules soak forming, spongy surface of coating and become, trapped with the hardening of the metal, oxide surface., 5.5.3 Quantitative aspects of electrolysis :, a. The mass of reactant consumed or, the mass of product formed at an electrode, during electrolysis can be calculated by, knowing stoichiometry of the half reaction, at the electrode., i. Calculation of quantity of electricity, passed : To calculate the quantity of, eletricity (Q) passed during electrolysis, the, amount of current, I, passed through the cell, is measured. The time for which the current, is passed is noted., , (5.16), , Moles of electrons actually passed, Q(C), =, (5.17), 96500 (C/mol e ), iii. Calculation of moles of product formed, The balanced equation for the half, reaction occuring at the electrode is devised., The stoichiometry of half reaction indicates, the moles of electrons passed and moles of, the product formed. For the reaction,, Cu2⊕ (aq) + 2e, Cu (s), two moles of, electrons are required for the production of, one mole of Cu. So we can calculate the moles, of product formed. The moles of electrons, actually passed are given by Eq. (5.16)., To simplify further we introduce the, entity mole ratio given by, Mole ratio =, moles of product formed in the half reaction, moles of electrons required in the half reaction, 1, For the reaction of Cu, mole ratio =, 2, Therefore,, Moles of product formed, = moles of electrons actually passed × mole, ratio, Q(C), =, × mole ratio, (5.18), 96500 (C/mol e ), =, , I (A) × t (s), × mole ratio, 96500 (C/mol e ), , (5.19), , iv. Calculation of mass of product :, Mass of product, W = moles of product × molar mass of product, I (A) × t (s), =, × mole ratio × molar mass, 96500 (C/mol e ), of product, (5.20), , 100
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b. Suppose two cells containing different, electrolytes are connected in series. The same, quantity of electricity is passed through them., The masses of the substances liberated at the, electrodes of the two cells are related as given, below :, The mass of the substance produced at, the electrode of first cell is given by, Q(C), W1 =, × (mole ratio)1 × M1, 96500 (C/mol e ), Hence,, =, , Q(C), 96500 (C/mol e ), W1, (mole ratio)1 × M1, , Similarly mass of substance liberated, at the electrode of second cell is W2 in the, equation,, W2, Q(C), = (mole ratio), × M2, 96500 (C/mol e ), 2, M1 and M2 are the molar masses of, substances produced at the electrodes of cells, 1 and 2., Q(C), Because, is the same, 96500 (C/mol e ), for both,, We have, , ii. Mass of Cu formed,, W=, I (A) × t (s), × mole ratio × molar mass, 96500 (C/mol e ), of Cu, 1 mol, 5 A × 100 × 60 s, ×, 2, mol e96500 (C/mol e ), , × 63.5 g mol-1, = 9.87 g, , =, , Problem 5.7 : How long will it take to, produce 2.415 g of Ag metal from its salt, solution by passing a current of 3 ampere ?, Molar mass of Ag is 107.9 g mol-1., Solution :, i. Stoichiometry :, Ag⊕ (aq) + e, mole ratio =, , Ag (s), 1 mol, 1 mol e, , ii. W =, I (A) × t (s), × mole ratio × molar mass, 96500 (C/mol e ), of Ag, 1 mol, 3A×t, 2.415 g =, ×, 96500 (C/mol e ) 1 mol e, , W1, W2, =, (mole ratio)1 × M1, (mole ratio)2 × M2, , × 107.9 g mol-1, , (5.21), , Do you know ?, Names, galvanic or voltaic, are given in honour of Italian, scientists L. Galvani and A. Volta for their, work in electrochemistry., , Solution :, i. Stoichiometry for the formation of Cu is, Cu2⊕ (aq) + 2 e = Cu (s), mole ratio =, , 2.415 × 96500 (C = As), 3 A × 107.9, , = 720 s = 12 min., , Problem 5.6 : What is the mass of Cu, metal produced at the cathode during, the passage of 5 ampere current through, CuSO4 solution for 100 minutes. Molar, mass of Cu is 63.5 g mol-1., , Hence,, , t=, , 1 mol, 2 mol e, , 101
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Substitution of the quantities gives, , Problem 5.8 : How many moles of, electrons are required for reduction of 3, moles of Zn2⊕ to Zn ? How many Faradays, of electricity will be required ?, , 4.36g, =, 1 mol/2mol e × 65.4 g mol-1, W2, 1 mol/3mol e- × 27 g mol-1, , Solution :, i. The balanced equation for the reduction, of Zn2⊕ to Zn is, Zn2⊕ (aq) + 2e, Zn (s), The equation shows that 1 mole of Zn2⊕, is reduced to Zn by 2 moles of electrons., For reduction of 3 moles of Zn2⊕ 6 moles, of electrons will be required., Faraday (96500 Coulombs) is the amount, of charge on one mole of electrons., Therefore, for 6 moles of electrons, 6 F, electricity will be required., Problem 5.9 : In a certain electrolysis, experiment 4.36 g of Zn are deposited, in one cell containing ZnSO4 solution., Calculate the mass of Al deposited in, another cell containing AlCl3 solution, connected in series with ZnSO4 cell. Molar, masses of Zn and Al are 65.4 g mol-1 and, 27 g mol-1, respectively., Solution :, Cell 1 :, Zn2⊕ (aq) + 2e, (mole ratio)1 =, , Zn (s), 1 mol, 2 mol e, , Cell 2 :, Al3⊕ (aq) + 3e, (mole ratio)2 =, , Al (s), 1 mol, 3 mol e, , W1, W2, =, (mole ratio)1 × M1, (mole ratio)2 × M2, W1 = 4.36 g, M1 = 65.4 g mol-1,, M2 = 27 gmol-1, , or, , 4.36 g × 2 W2 × 3, =, 65.4, 27, , Hence, W2 =, , 4.36 g × 2 × 27, = 1.2 g, 65.4 × 3, , 5.6 Galvanic or voltaic cell : In galvanic or, voltaic cells, electricity is generated through, the use of spontaneous chemical reactions., A galvanic (or voltaic) cell is made of, two half cells. Each half cell consists of a, metal strip immersed in the solution of its own, ions of known concentration. For example, a, strip of zinc metal immersed in 1 M aqueous, solution of zinc ions forms an half cell., It follows that two metal plates and, the solutions of their ions with known, concentrations are required for the construction, of a galvanic (voltaic) cell. Two half cells, are constructed by immersing the two metal, plates in solutions of their respective ions, placed in separate containers. The two half, cells so constructed are combined together, to form the galvanic cell. The metal plates, called electrodes are connected through, voltmeter by a conducting wire for transfer, of electrons between them. To complete the, circuit the two solutions are connected by, conducting medium through which cations, and anions move from one compartment to, the other. This requirement is fulfilled by a, salt bridge., 5.6.1 Salt bridge : In a galvanic cell, the two, solutions are connected by a salt bridge. It is, an U tube containing a saturated solution of, an inert electrolyte such as KCl or NH4NO3, and 5 % agar solution. The ions of electrolyte, do not react with the ions of electrode, solutions or the electrodes., , 102
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Salt bridge is prepared by filling a U, tube with hot saturated solution of the salt, and agar agar solution allowing it to cool., The cooled solution sets into a gel which does, not come out on inverting the tube. The salt, bridge is kept dipped in distilled water when, not in use as shown in Fig. 5.5., Saturated KCl, solution + agar gel, , i., , ii. The insoluble species if any or gases are, placed in the interior position adjacent to, the metal electrodes., iii. The aqueous solutions of ions are placed, at the middle of the cell formula., , U tube, , iv. A single vertical line between two phases, indicates the phase boundary. It indicates, the direct contact between them., , Beaker, Distilled water, , v., , Glass, wool plugs, , Fig. 5.5 : Salt bridge, , A double vertical line between two, solutions indicates that they are connected, by salt bridge., , vi. The additional information such as, concentration of solutions and gas, pressures is also given., , Try this..., Salt bridge can be prepared, with a laminated long strip of good, quality filter paper. Cut the two ends of, a laminated strip. Dip the two ends in a, saturated solution of KCl for 24 hours., This strip can be used as salt bridge by, dipping the two ends in two solutions., Functions of salt bridge, The salt bridge serves the following, functions :, i., , The metal electrodes or the inert electrodes, are denoted by vertical lines placed at the, ends of the formula or the short notation., The anode (-) is written at the extreme, left and cathode (+) at extreme right., , vii. A single half cell is written in the order:, aqueous solution of ions first and then, the solid electrode., For example Zn2⊕(1M) Zn (s). This, order is reversed when the electrode, acts as anode in the cell. The following, example illustrates these conventions., The cell composed of Mg (anode) and, Ag (cathode) consists of two half cells,, Mg2⊕ (1M) Mg (s) and Ag⊕ (1M)Ag(s)., The cell is represented as :, , It provides an electrical contact between, two solutions and thereby completes the, electrical circuit., , Mg (s) Mg2⊕ (1M) Ag⊕(1M) Ag(s)., Can you tell ?, You have learnt Daniel cell, in XI th standard. Write notations, for anode and cathode. Write the cell, formula., , ii. It prevents mixing of two solutions., iii. It maintains electrical neutrality in both, the solutions by transfer of ions., 5.6.2 Formulation or short notation of, galvanic cells : A galvanic cell is represented, by a formula or short notation that includes, electrodes, aqueous solutions of ions and other, species which may or may not be involved in, the cell reaction. The following conventions, are used to write the cell notation., , 5.6.3 Writing of cell reaction : The cell, reaction corresponding to the cell notation is, written on the assumption that the right hand, side electrode is cathode (+) and left hand, side electrode is anode (-)., , 103
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As mentioned in section 5.4.2, oxidation, half reaction occurs at anode and reduction, half reaction at cathode. It, therefore, follows, that in galvanic cell oxidation half reaction, takes place on the left hand side electrode, and reduction half reaction on the right hand, side electrode., The following steps are followed to write, the cell reaction., i. Write oxidation half reaction for the left, hand side electrode and reduction half, reaction for the right hand side electrode., ii. Add two electrode half reactions to get, the overall cell reaction. While adding, the electrons must be cancelled. For this, purpose, it may be necessary to multiply, one or both the half reactions by a suitable, numerical factor (s). No electrons should, appear in the overall reaction., iii. It is important to note that the individual, half reactions may be written with one, or more electrons. For example, half, reactions for H2 gas, whether written as, 2H⊕ (aq) + 2e, H2 (g) or H⊕(aq) +, e, 1/2 H2 (g) makes no difference., In writing the overall cell reaction, the, electrons must be balanced., Consider the cell,, e, 2⊕, Ni (s)Ni (1M)Al3⊕ (1M)Al (s), The oxidation at anode is, Ni (s), Ni2⊕ (1M) + 2e, The reduction half reaction at cathode is, Al3⊕ (1M) + 3 e, Al (s)., To balance the electrons, oxidation, reaction is multiplied by 3 and reduction, reaction by 2. The two half reactions so, obtained when added give the overall cell, reaction. Thus,, 3 Ni (s), 3 Ni2⊕ (1M) + 6 e, (oxidation half reaction), 3⊕, 2 Al (1M) + 6 e, 2 Al (s), (reduction half reaction), 3 Ni (s) + 2Al3⊕ (1M), , 3Ni2⊕ (1M) + 2 Al (s), (overall cell reaction), , Try this..., Write electrode reactions and, overall cell reaction for Daniel cell, you learnt in standard XI., 5.7 Electrode potential and cell potential, : A galvanic cell is composed of two half, cells, each consisting of electronic (metal, plates) and electrolytic (solution of ions), conductors in contact. At the surface of, separation of solid metal and the solution,, there exists difference of electrical potential., This potential difference established due, to electrode half reaction occurring at the, electrode surface, is the electrode potential., The potential is associated with each of, the half reaction, be it oxidation or reduction., The potential associated with oxidation, reaction is oxidation potential while that, associated with reduction gives the reduction, potential. The overall cell potential, also, called electromotive force (emf), is made of, the contributions from each of the electrodes., In other words, the cell potential is algebraic, sum of the electrode potentials,, Ecell = Eoxi (anode) + Ered (cathode), (5.22), where Eoxi is the oxidation potential of, anode (-) and Ered is the reduction potential, of cathode (+)., When galvanic cell operates, electrons, are generated at the anode. These electrons, move through external circuit to the cathode., The cell potential is the force that pushes, electrons away from anode (-) and pulls them, toward cathode where they are consumed., 5.7.1 Standard potentials : The electrode, potential and the cell potential depend on, concentrations of solutions, pressures of gases, and the temperature. To facitilate comparison, of different galvanic cells, it is necessary to, measure the cell voltage under given set of, standard conditions of concentration and, temperature., , 104
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The standard conditions chosen are 1, M concentration of solution, 1 atm pressure, for gases, solids and liquids in pure form, and 250C. The voltage measured under, these conditions is called standard potential, designated as E0., The standard cell potential is the, algebraic sum of the standard electrode, potentials similar to Eq. (5.22)., E0cell = E0oxi (anode) + E0red (cathode), (5.23), Here E0oxi is standard oxidation potential, and E0red is the standard reduction potential., According to IUPAC convention,, standard potential of an electrode is taken, as the standard reduction potential., It must be realised that standard oxidation, potential of any electrode is numerically, equal to its standard reduction potential with, the reversal of sign. For example standard, oxidation potential of Zn2⊕ (1M) Zn electrode, is 0.76V. Its standard reduction potential will, be -0.76 V. Hereafter the standard reduction, potential will be called standard potential, the, voltage associated with a reduction reaction., It follows that the standard cell potential, (emf) is written in terms of the standard, potentials of the electrodes. In Eq. (5.23),, E0oxi(anode) is replaced by - E0red (anode)., We then write,, E0cell = - E0red (anode) + E0red (cathode), Omitting the subscript red, we have, E0cell = E0 (cathode, +) - E0 (anode, -), , 5.7.2 Dependence of cell potential on, concentration (Nernst equation) : The, standard cell potential tells us whether or, not the reactants in their standard states, form the products in their standard states, spontaneously. To predict the spontaneity of, reactions for anything other than standard, concentration conditions we need to know, how voltage of galvanic cell varies with, concentration., Dependence of cell voltage on, concentrations is given by Nernst equation., For any general reaction,, aA + bB, , The cell voltage is given by, Ecell = E0cell = E0cell -, , • The difference in electrical potential, between anode and cathode is cell, voltage., , [C]c [D]d, RT, 1n, [A]a [B]b, nF, [C]c [D]d, 2.303RT, log10, [A]a [B]b, nF, (5.25), , where n = moles of electrons used in, the reaction, F = Faraday = 96500 C, T =, temperature in kelvin, R = gas constant =, 8.314 J K-1mol-1, 2.303RT, At 25 0C,, = 0.0592 V, F, , Therefore at 25 0C, eq. (5.24) becomes, Ecell = E0cell -, , (5.24), , Remember..., • While constructing a galvanic, cell from two electrodes, the, electrode with higher standard, potential is cathode (+) and that with, lower standard potential is anode (-)., , cC + dD, , [C]c [D]d, 0.0592V, log10, [A]a [B]b, n, (5.26), , The Eq. (5.25) or Eq. (5.26) is the, Nernst equation. The first term on the right, hand of Nernst equation represents standard, state electrochemical conditions. The second, term is the correction for non standard state, conditions. The cell potential equals standard, potential if the concentrations of reactants, and products are 1 M each. Thus,, , 105
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if [A] = [B] = [C] = [D] = 1M,, Ecell = E cell, 0, , If a gaseous substance is present in the, cell reaction its concentration term is replaced, by the partial pressure of the gas., The Nernst equation can be used to, calculate cell potential and electrode potential., i. Calculation of cell potential : Consider the, cell, Cd(s)Cd2⊕(aq) Cu2⊕ (aq) Cu., Let us first write the cell reaction, Cd2⊕ (aq) + 2 e, , Cd (s), , (oxidation at anode), Cu2⊕ (aq) + 2 e, , Cu (s), (reduction at cathode), , Cd (s) + Cu2⊕ (aq), , Cd2⊕ (aq) + Cu (s), (overall cell reaction), , Here n = 2, The potential of cell is given by Nernst, equation,, [Cd2⊕], 0.0592, log10, Ecell = E0cell [Cu2⊕], 2, at 25 0C., , (Concentration of solids and pure liquids, are taken to be unity.), ii. Calculation of electrode potential, Consider Zn2⊕(aq)Zn(s), , Sn (s) + 2 Ag⊕ (0.01M), Sn2⊕ (0.02 M) + 2 Ag (s), (overall cell reaction), The cell potential is given by, [Sn2⊕], 0.0592V, 0, Ecell = E cell log10, [Ag⊕]2, 2, E0cell = E0Ag - E0Sn = 0.8 V + 0.136 V, = 0.936 V, Hence,, 0.02, 0.0592V, Ecell = 0.936V log10, (0.01)2, 2, 0.0592V, = 0.936V log10 200, 2, 0.0592V, = 0.936V × 2.301, 2, = 0.936 V - 0.0681V = 0.8679 V, Problem 5.11 : The standard potential, of the electrode, Zn2⊕ (0.02 M) Zn (s), is - 0.76 V. Calculate its potential., Solution :, , The reduction reaction for the electrode is, Zn2⊕ (aq) + 2 e, , Problem 5.10 : Calculate the voltage, of the cell, Sn (s)Sn2⊕ (0.02M)Ag⊕, (0.01M)Ag (s) at 25 0C., E0Sn = - 0.136 V,, E0Ag = 0.800 V., Solution :, First we write the cell reaction., Sn (s), Sn2⊕ (0.02M) + 2 e, (oxidation at anode), [Ag⊕ (0.01M) + e, Ag (s)] × 2, (reduction at cathode), , Electrode reaction :, , Zn (s), , Applying Nernst equation, electrode potential, is given by, 1, 0.0592, log10, EZn = E0Zn 2⊕, [Zn, ], 2, , Zn2⊕ (0.02M) + 2 e, , 0.0592, log10 [Zn2⊕] at 25 0C, 2, , = - 0.76 V +, , = E0Zn +, , EZn = E0Zn -, , Zn (s), , 1, 0.0592V, log10, 2⊕, [Zn, ], n, 0.0592V, log10 (0.02), 2, , 0.0592V, × (-1.6990), 2, = - 0.76 V - 0.0503V = - 0.81 V, , = - 0.76 V +, , 106
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5.8 Thermodynamics of galvanic cells, 5.8.1 Gibbs energy of cell reactions and, cell potential : The electrical work done in a, galvanic cell is the electricity (charge) passed, multiplied by the cell potential., Electrical work, = amount of charge passed × cell potential., Charge of one mole electrons is F, coulombs. For the cell reaction involving n, moles of electrons., , Remember..., For chemical reaction to be, spontaneous, ∆G must be negative., Because ∆G = - nFEcell, Ecell must be positive, for a cell reaction if it is spontaneous., 5.8.2 Standard cell potential and equilibrium, constant : The relation between standard, Gibbs energy change of cell reaction and, standard cell potential is given by Eq. (5.27)., , charge passed = nF coulombs, , - ∆G0 = nFE0cell, , Hence, electrical work = nFEcell, W. Gibbs in 1878 concluded that, electrical work done in galvanic cell is equal, to the decrease in Gibbs energy, - ∆G, of cell, reaction. It then follows that, Electrical work = - ∆G, , ∆G0 = - RT ln K, , ∆G = -nFEcell, , -nFE0cell = - RT ln K, , (5.27), , Under standard state conditions, we write, ∆G0 = -nFE0cell, , (5.28), , or, , E0cell =, , The Eq. (5.28) explains why E0cell is an, intensive property., We know that ∆G0 is an extensive, property since its value depends on the, amount of substance. If the stoichiometric, equation of redox reaction is multiplied by, 2 that is the amounts of substances oxidised, and reduced are doubled, ∆G0 doubles. The, moles of electrons transferred also doubles., The ratio,, ∆G0, E0cell = then becomes, nF, E0cell = -, , 2∆G, ∆G, =2 nF, nF, 0, , (5.29), , Combining Eq. (5.28) and Eq. (5.29), we, have, , and thus - ∆G = nFEcell, or, , The relation between standard Gibbs, energy change of a chemical reaction, and its equilibrium constant as given in, thermodynamics is :, , 0, , RT, ln K, nF, , =, , 2.303 RT, log10K, nF, , =, , 0.0592, log10K, n, , Try this..., Write expressions to calculate, equilibrium constant from, i. Concentration data, ii. Thermochemical data, iii. Electrochemical data, , Thus, E0cell remains the same by, multiplying the redox reaction by 2. It, means E0cell is independent of the amount of, substance and the intensive property., , 107, , at 25 0C, (5.29)
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What would happen if potential of one of, the electrodes in a galvanic cell is zero ? Can, we measure the potential of such a galvanic, cell ? There are two electrodes combined, together and a redox reaction results. The, measured cell potential is algebraic sum of two, electrode potentials. One electrode potential is, zero. Therefore, the measured cell potential is, equal to the potential of other electrode., , Problem 5.12 : Calculate standard Gibbs, energy change and equilibrium constant at, 250C for the cell reaction,, Cd (s) + Sn2⊕ (aq), , Cd2⊕ (aq) + Sn (s), , Given :, E0Cd = -0.403V and, E0Sn = - 0.136 V. Write formula of the cell., Solution :, i. The cell is made of two electrodes,, Cd2⊕ (aq) Cd (s) and Sn2⊕ (aq) Sn (s)., E0 value for Sn2⊕ (aq) Sn (s) electrode, is higher than that of Cd2⊕(aq) Cd, (s) electrode. Hence, Sn2⊕ (aq) Sn (s), electrode is cathode and Cd2⊕ (aq) Cd (s), anode. Cell formula is Cd(s) Cd2⊕ (aq), Sn2⊕ (aq) Sn (s), , From foregoing arguments, it follows that, it is necessary to choose an arbitrary standard, electrode as a reference point. The chemists, have chosen hydrogen gas electrode consisting, of H2 gas at 1 atm pressure in contact with, 1 M H⊕ ion solution as a primary reference, electrode. The potential of this electrode has, arbitrarily been taken as zero. The electrode, is called standard hydrogen electrode (SHE)., , ii. ∆G0 = - nF E0cell, E0cell = E0Sn - E0Cd = - 0.136 V - (-0.403V), , = - 51531 V C = - 51531 J = - 51.53 kJ, 0.0592 V, iii. E0cell =, log10K, 2, , We will see later that SHE is not the most, convenient electrode. Several other electrodes, namely calomel, silver-silver chloride and, glass electrodes with known potentials are, used as secondary reference electrodes. The, potentials of these electrodes are determined, using SHE., , 0.0592 V, 0.267 V =, log10K, 2, , A reference electrode is then defined as, an electrode whose potential is arbitrarily, taken as zero or is exactly known., , = 0.267 V., n = 2 mol e, ∆G0 = -2 mol e 96500 C/mol e × 0.267 V, , 0.267 × 2, log10K =, = 9.0203, 0.0592, K = antilog 9.0203 = 1.05 × 109, 5.9 Reference electrodes : Every oxidation, needs to be accompanied by reduction. The, occurrence of only oxidation or only reduction, is not possible. (Refer to Std. XI Chemistry, Textbook Chapter 6), In a galvanic cell oxidation and reduction, occur simultaneously. The potential associated, with the redox reaction can be experimentally, measured. For the measurement of potential, two electrodes need to be combined together, where the redox reaction occurs., , 5.9.1 Standard hydrogen electrode (SHE) :, Construction : SHE consists of a platinum, plate, coated with platinum black used as, electrodes. This plate is connected to the, external circuit through sealed narrow glass, tube containing mercury. It is surrounded by, an outer jacket., The platinum electrode is immersed in, 1 M H⊕ ion solution. The solution is kept, saturated with dissolved H2 by bubbling, hydrogen gas under 1 atm pressure through, the side tube of the jacket as shown in Fig.5.6., Platinum does not take part in the electrode, reaction. It is inert electrode and serves as the, site for electron transfer., , 108
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Cu wire, , e, , Pure and, dry H2 gas, at 1 atm, Glass jacket, , Salt bridge, Zn, anode, , Vessel, Pt wire, , H2, (g, 1atm), SHE, , Mercury, 1M, ZnSO4, soution, , Platinised, platinum, plate, , Solution, H⊕ ions, (1M), , Fig. 5.6 : Standard hydrogen electrode, , Formulation : Standard hydrogen electrode, is represented as, H⊕ (1M) H2 (g, 1atm)Pt, Electrode reaction : The platinum black, capable of adsorbing large quantities of H2, gas, allows the change from gaseous to ionic, form and the reverse process to occur., The reduction half reaction at the, electrode is, 2H⊕ (1M) + 2e, , e, , H2 (g, 1atm), E0H2 = 0.000 V, , Application of SHE, SHE is used as a primary reference, electrode to determine the standard potentials, of other electrodes., To determine the standard potential of, 2⊕, Zn (1M)Zn (s), it is combined with SHE, to form the cell,, ZnZn2⊕(1M)H⊕ (1M)H2 (g,, This is shown in Fig. 5.7., , 1atm)Pt, , The standard cell potential, E0cell, is measured., E0cell = E0H2 - E0Zn = - E0Zn , because E0H2 is zero., Thus, the measured emf of the cell is equal, to standard potential of Zn2⊕(1M)Zn (s), electrode., Difficulties in setting SHE, i. It is difficult to obtain pure and dry, hydrogen gas., ii. The pressure of hydrogen gas cannot be, maintained exactly at 1 atm throughout, the measurement., , 1M, H⊕ ion, solution, , Fig. 5.7 : Determination of standard potential, using SHE, , iii. The concentration of H⊕ ion solution, cannot be exactly maintained at 1 M., Due to bubbling of gas into the solution,, evaporation of water may take place., This results in changing the concentration, of solution., Hydrogen gas electrode, For, hydrogen, gas, electrode,, ⊕, H (aq)H2(g,PH ) Pt, [H ] and pressure of, 2, hydrogen gas differ from unity., ⊕, , Electrode reaction :, 2H⊕ (aq) + 2e, , H2 (g, PH2), , From the Nernst equation, 0.0592, P, log10 H⊕2 2, 2, [H ], 0.0592, P, log10 H⊕2 2, = 2, [H ], , EH = E0H22, , Because E0H2 = 0, 5.10 Galvanic cells useful in day-to-day life, Voltaic (or galvanic) cells in common use can, be classified as primary and secondary cells., i. Primary voltaic cells : When a galvanic, cell discharges during current generation,, the chemicals are consumed. In primary, voltaic cell, once the chemicals are, completely consumed, cell reaction stops., The cell reaction cannot be reversed even, after reversing the direction of current, flow or these cells cannot be recharged., The most familiar example is dry cell., , 109
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ii. Secondary voltaic cells : In secondary, voltaic cell, the chemicals consumed, during current generation can be, regenerated. For this purpose an external, potential slightly greater than the cell, potential is applied across the electrodes., This results in reversal of the direction of, current flow causing the reversal of cell, reaction This is recharging of cell. The, voltaic cells which can be recharged are, called secondary voltaic cells., It is amazing to see that secondary, cells are galvanic cells during discharge and, electrolytic cells during recharging. Examples, of secondary cells are lead storage battery,, mercury cell and nickel-cadmium cell., 5.10.1 Dry cell (Leclanche' cell) : It is a cell, without liquid component, but the electrolyte, is not completely dry. It is a viscous aqueous, paste., Construction : The container of the cell is, made of zinc which serves as anode (-). It, is lined from inside with a porous paper to, separate it from the other material of the cell., An inert graphite rod in the centre of the, cell immersed in the electrolyte paste serves, as cathode. It is surrounded by a paste of, manganese dioxide (MnO2) and carbon black., , Brass cap, Paper spacer, Zn container, Paste of MnO2 +, carbon, Graphite rod, Paste of NH4Cl, + ZnCl2, Bottom of Zn container, , Fig. 5.8 : Dry cell (Leclanche' cell), , Cell reactions:, i., , Oxidation at anode : When the cell, operates the current is drawn from the, cell and metallic zinc is oxidised to zinc, ions., Zn (s), , Zn2⊕ (aq) + 2e, , ii. Reduction at cathode : The electrons, liberated in oxidation at anode flow along, the container and migrate to cathode. At, cathode NH4⊕ ions are reduced., 2NH4⊕ (aq) + 2e, , 2NH3 (aq) + H2 (g), , Hydrogen gas produced in reduction, reaction is oxidised by MnO2 and, prevents its collection on cathode., H2(g)+2MnO2(s), , Mn2O3(s)+H2O(l), , The net reduction reaction at cathode is, combination of these two reactions., , The rest of the cell is filled with an, electrolyte. It is a moist paste of ammonium, chloride (NH4Cl) and zinc chloride (ZnCl2)., Some starch is added to the paste to make it, thick so that it cannot be leaked out., , 2NH4⊕(aq) + 2 MnO2(s) + 2e, Mn2O3 (s) + 2 NH3 (aq) + H2O (l), , The cell is sealed at the top to prevent, drying of the paste by evaporation of moisture., See Fig. 5.8., , Zn (s) + 2 NH4⊕ (aq) + 2 MnO2(s), , iii. Net cell reaction : The net cell reaction is, sum of oxidation at anode and reduction, at cathode., Zn2⊕ (aq) + Mn2O3 (s) + 2 NH3 (aq) + H2O(l), The ammonia produced combines with, Zn2⊕ to form soluble compound containing, complex ion., Zn2⊕ (aq) + 4 NH3 (aq), [Zn (NH3)4]2⊕(aq), , 110
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The electrodes are immersed in, an electrolytic aqueous solution of, 38 % (by mass) of sulphuric acid of density, 1.2 g/mL., , Do you know ?, Alkaline dry cell : The Leclanche', dry cell works under acidic, conditions due to the presence of NH4Cl., The difficulty with this dry cell is that Zn, anode corrodes due to its actions with H⊕, ions from NH4⊕ ions., This results in shortening the life of, dry cell. To avoid this a modified or of, the dry cell called alkaline dry cell has, been proposed. In alkaline dry cell NaOH, or KOH is used as electrolyte in place of, NH4Cl., The alkaline dry cell has longer life, than acidic dry cell since the Zn corrodes, more slowly., , Notation of the cell : The cell is formulated, as Pb(s)PbSO4(s)38%H2SO4(aq)PbSO4(s), PbO2(s) Pb(s), a. Cell reactions during discharge, i., , Uses of dry cell : Dry cell is used as a source, of power in flashlights, portable radios, tape, recorders, clocks and so forth., 5.10.2 Lead storage battery (Lead, accumulator) : Lead accumulator stores, electrical energy due to regeneration of original, reactants during recharging. It functions as, galvanic cell and as electrolytic cell, as well., Construction : A group of lead plates packed, with spongy lead serves as anode (-). Another, group of lead plates bearing lead dioxide, (PbO2) serves as cathode (+)., Anode (-), , Cathode (+), , Pb (s), , Pb2⊕ (aq) + 2 e, (oxidation), , Pb2⊕ (aq) + SO42 (aq), Pb (s)+SO42 (aq), , PbSO4 (s), (precipitation), PbSO4 (s) + 2 e ...(i), (overall oxidation), , ii. Reduction at cathode (+) : The electrons, produced at anode travel through external, circuit and re-enter the cell at cathode. At, cathode PbO2 is reduced to Pb2⊕ ions in, presence of H⊕ ions. Subsequently Pb2⊕, ions so formed combine with SO42 ions, from H2SO4 to form insoluble PbSO4 that, gets coated on the electrode., PbO2 (s) + 4H⊕ (aq) + 2 e, , 38 %, H2SO4, Pb, plates, , Oxidation at anode (-) : When the, cell provides current, spongy lead is, oxidised to Pb2⊕ ions and negative charge, accumulates on lead plates. The Pb2⊕, ions so formed combine with SO42 ions, from H2SO4 to form insoluble PbSO4. The, net oxidation is the sum of these two, processes., , Pb (s) + SO42 (aq), , Pb2⊕ (aq), +2H2O(l), (reduction), , PbSO4 (s), (precipitation), , PbO2 (s) + 4H⊕ (aq) + SO42 (aq)+ 2 e, PbSO4 (s) + 2H2O (l) ...(ii), , Pb plates with, PbO2, , Fig. 5.9 : Lead storage cell, , (overall reduction), , To provide large reacting surface, the, cell contains several plates of each type. The, two types of plates are alternately arranged, as shown in Fig. 5.9., , 111, , iii. Net cell reaction during discharge: The, net cell reaction is the sum of overall, oxidation at anode and overall reduction, at cathode.
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Pb (s) + PbO2 (s) + 4H⊕ (aq) + 2SO42 (aq), 2PbSO4 (s) + 2H2O (l), or, Pb (s) + PbO2 (s) + 2H2SO4 (aq), 2PbSO4(s) + 2H2O (l) ...(iii), As the cell operates to generate current,, H2SO4 is consumed. Its concentration (density), decreases and the cell potential is decreased., The cell potential thus depends on sulphuric, acid concentration (density)., b. Cell reactions during recharging :, The potential of lead accumulator is 2V. It, must be recharged with the falling of the, cell potential to 1.8 V. To recharge the cell, external potential slightly greater than 2 V, needs to be applied across the electrodes., During recharging the cell functions as, electrolytic cell. The anode and cathode are, interchanged with PbO2 electrode being anode, (+) and lead electrode cathode (-)., iv. Oxidation at anode (+) : It is reverse, of reduction reaction (ii) at cathode that, occurs during discharge., PbSO4 (s) + 2H2O (l), PbO2 (s) + 4H⊕(aq) + SO42 (aq) + 2 e, , ...(iv), , v. Reduction at cathode (-) : It is reverse, of oxidation reaction (i) at anode that occurs, during discharge., PbSO4 (s)+2 e, , Pb (s)+SO42 (aq) ...(v), , vi. Net cell reaction : It is the sum of, reaction (iv) and (v) or the reverse of, net cell reaction (iii) that occurs during, discharge, 2PbSO4 (s) + 2H2O (l), Pb (s) +, PbO2 (s) + 2 H2SO4 (aq), The above reaction shows that H2SO4 is, regenerated. Its concentration (density), and in turn, the cell potential increases., Applications of lead accumulator, i., , ii. A 12 V lead storage battery constructed, by connecting six 2 V cells in series is, used in automobiles and inverters., 5.10.3 Nickel-Cadmium or NICAD storage, cell : Nickel-cadmium cell is a secondary dry, cell. In other words it is a dry cell that can, be recharged., Anode of the NICAD storage cell is, cadmium metal. The cathode is nickel (IV), oxide, NiO2 supported on Ni. The electrolyte, solution is basic., The electrode reactions and overall cell, reaction are as follows :, Cd (s) + 2OH (aq), , Cd(OH)2 (s) + 2 e, (anodic oxidation), NiO2 (s) + 2 H2O (l) + 2 e, Ni(OH)2 (s) + 2OH (aq), (cathodic reduction), Cd (s) + NiO2 (s) + 2 H2O (l), Cd(OH)2 (s) + Ni(OH)2 (s), (overall cell reaction), The reaction product at each electrode, is solid that adheres to electrode surface., Therefore the cell can be recharged. The, potential of the cell is about 1.4 V. The cell, has longer life than other dry cells. It can, be used in electronic watches, calculators,, photographic equipments, etc., 5.10.4 Mercury battery : Mercury battery is, a secondary dry cell and can be recharged., The mercury battery consists of zinc anode,, amalgamated with mercury. The cathode is, a paste of Hg and carbon. The electrolyte, is strongly alkaline and made of a paste of, KOH and ZnO. The electrode ractions and, net cell reaction are :, Zn(Hg)+2OH (aq), HgO(s)+ H2O(l)+2e, , It is used as a source of direct current in, the laboratory., , 112, , Zn (Hg) + HgO(s), , ZnO(s) +H2O(l) + 2 e, (anode oxidation), Hg(l) + 2 OH (aq), (cathode reduction), ZnO(s) + Hg(l), (overall cell reaction)
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The overall reaction involves only solid, substances. There is no change in electrolyte, composition during operation. The mercury, battery, therefore, provides more constant, voltage (1.35V) than the Leclanche' dry cell., It also has considerably higher capacity and, longer life than dry cell., , Hydrogen gas is continuously bubbled,, through anode and oxygen gas through, cathode into the electrolyte., , The mercury dry cell finds use in hearing, aids, electric watches, pacemakers, etc., , 2H2 (g) + 4OH (aq), , 5.11 Fuel cells : The functioning of fuel cells, is based on the fact that combustion reactions, are of redox type and can be used to generate, electricity., The fuel cells differ from ordinary, galvanic cells in that the reactants are not, placed within the cell. They are continuously, supplied to electrodes from a reservoir., In these cells one of the reactants is a, fuel such as hydrogen gas or methanol. The, other reactant such as oxygen, is oxidant., The simplest fuel cell is hydrogen-oxygen, fuel cell., 5.11.1 Hydrogen-oxygen fuel cell : In H2 O2 fuel cell, the fuel is hydrogen gas. Oxygen, gas is an oxidising agent. The energy of the, combustion of hydrogen is converted into, electrical energy., H2O, (anode), , (cathode), , H2 (g), , O2 (g), Aqueous, KOH, , Fig. 5.10 : H2 - O2 fuel cell, , Construction : The anode and cathode are, porous carbon rods containing small amount, of finely divided platinum metal that acts, as a catalyst. The electrolyte is hot aqueous, solution of KOH. The carbon rods immersed, into electrolyte are shown in Fig. 5.10., , Cell reactions, i., , Oxidation at anode (-) : At anode, hydrogen gas is oxidised to H2O., 4H2O (l) + 4 e, , ii. Reduction at cathode (+) : The electrons, released at anode travel, through external, circuit to cathode. Here O2 is reduced to, OH-., O2 (g) + 2H2O (aq)+ 4 e, , 4OH (aq), , iii. Net cell reaction : The overall cell, reaction is the sum of electrode reactions, (i) and (ii)., 2H2 (g) + O2 (g), , 2H2O (l), , The overall cell reaction is combustion, of H2 to form liquid water. Interestingly, the, fuel H2 gas and the oxidant O2 do not react, directly., The chemical energy released during the, formation of O-H bond is directly converted, into electrical energy accompanying in above, combustion reaction. The cell continues, to operate as long as H2 and O2 gases are, supplied to electrodes., The cell potential is given by, E0cell = E0cathode - E0anode = 0.4V - (-0.83V), = 1.23 V., Advantages of fuel cells, i. The reacting substances are continuously, supplied to the electrodes. Unlike, conventional galvanic cells, fuel cells do, not have to be discarded on consuming, of chemicals., ii. They are nonpolluting as the only reaction, product is water., iii. Fuel cells provide electricity with an, efficiency of about 70 % which is twice, as large when compared with efficiency, of thermal plants (only 40 %)., , 113
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Drawbacks of fuel cell, H2 gas is hazardous to handle and the, cost of preparing H2 is high., Internet my friend, Fuel cells are also used in cell, phones and laptop computers. The, cell proposed for use in these products, is direct methanol fuel cell (DMFC)., Collect information of this cell., Applications of fuel cells, i., , The fuel cells are used on experimental, basis in automobiles., , ii. The fuel cell are used for electrical power, in the space programme., , ii. Below hydrogen electrode the negative, standard potential increases and above, hydrogen electrode the positive standard, potential increases., iii. E0 values apply to the reduction half, reactions that occur in the forward, direction as written., iv. Higher (more positive) E0 value for a half, reaction indicates its greater tendency to, occur in the forward direction and in, turn greater tendency for the substance, to reduce. Conversely, the low (more, negative) E0 value of a half reaction, corresponds to its greater tendency to, occur in the reverse direction or for the, substance to oxidise., , iii. In space crafts the fuel cell is operated, at such a high temperature that the, water evaporates at the same rate as it, is formed. The vapour is condensed and, pure water formed is used for drinking, by astronauts., , The half reactions are listed in order of, their decreasing tendency in the forward, direction., Remember..., The left side of half reaction, has cations of metals or nonmetallic molecules (oxidants). There are, free metals or anions of non metals on the, right side (reductants)., , iv. In future, fuel cells can possibly be, explored as power generators in hospitals,, hotels and homes., Can you tell ?, In what ways are fuel cells, and galvanic cells similar and in, what ways are they different ?, 5.12 Electrochemical series (Electromotive, series) : The standard potentials of a number, of electrodes have been determined using, standard hydrogen electrode. These electrodes, with their half reactions are arranged according, to their decreasing standard potentials as, shown in Table 5.1. This arrangement is, called electrochemical series., Key points of electrochemical series, i., , The half reactions are written as, reductions. The oxidizing agents and, electrons appear on the left side of half, reactions while the reducing agents, are shown on the right side in the half, reaction., , Applications of electrochemical series, i. Relative strength of oxidising agents:, The species on the left side of half, reactions are oxidizing agents. E0 value, is a measure of the tendency of the species, to accept electrons and get reduced., In other words, E0 value measures the, strength of the substances as oxidising, agents. Larger the E0 value greater is, the oxidising strength. The species in the, top left side of half reactions are strong, oxidising agents. As we move down the, table, E0 value and strength of oxidising, agents decreases from top to bottom., Use your brain power, Indentify the strongest and the, weakest oxidizing agents from the, electrochemical series., , 114
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ii. Relative strength of reducing agents:, The species on the right side of half, reactions are reducing agents., , From Table 5.1 of electrochemical series we, have, E0Mg = -2.37 V and E0Ag = 0.8 V. For the, cell having Mg as anode and Ag cathode., , The half reactions at the bottom of the, table with large negative E0 values have, a little or no tendency to occur in the, forward direction as written. They tend, to favour the reverse direction. It follows,, that the species appearing at the bottom, right side of half reactions associated, with large negative E0 values are the, effective electron donors. They serve as, strong reducing agents. The strength of, reducing agents increases from top to, bottom as E0 values decrease., , E0Cell = E0Ag - E0Mg = 0.8V - (-2.37V), = 3.17 V., EMF being positive the cell reaction, is spontaneous. Thus Ag⊕ ions oxidise to, metallic Mg., General rules, i., , Use your brain power, Identify the strongest and the, weakest reducing agents from the, electrochemical series., , ii. An reducing agent can reduce the, oxidising agent located above it in the, electrochemical series., , iii. Spontaneity of redox reactions : A redox, reaction in galvanic cell is spontaneous, only if the species with higher E0 value, is reduced (accepts electrons) and that, with lower E0 value is oxidised (donates, electrons)., , Use your brain power, From E0 values given in, Table 5.1, predict whether Sn can, reduce I2 or Ni2⊕., , The standard cell potential must be positive, for a cell reaction to be spontaneous under, the standard conditions. Noteworthy, application of electromotive series is, predicting spontaneity of redox reactions, from the knowledge of standard potentials., , Do you know ?, The fuel cells for power, electric vehicles incorporate the, proton conducting plastic membrane., These are proton exchange membranes, (PEM) fuel cells., , Suppose, we ask a question : At standard, conditions would Ag⊕ ions oxidise, metallic magnesium ? To answer this, question, first we write oxidation of Mg, by Ag⊕., Mg (s), , An oxidizing agent can oxidize any, reducing agent that appears below it,, and cannot oxidize the reducing agent, appearing above it in the electrochemical, series., , Mg2⊕ (aq) + 2 e (oxidation), , 2Ag2⊕ (aq) + 2 e, , 2Ag (s)(reduction), , Mg (s) +2Ag2⊕ (aq), , Mg2⊕ (aq) + 2Ag (s), (overall reaction), , 115
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x., , Which of the following expressions, represent molar conductivity of, Al2(SO4)3?, , ii., , What is a salt bridge ?, , iii., , a. 3 λ0Al3⊕ + 2 λ0SO42, , Write electrode reactions for the, electrolysis of aqueous NaCl., , iv., , How many moles of electrons are, passed when 0.8 ampere current is, passed for 1 hour through molten, CaCl2 ?, , v., , Construct a galvanic cell from, the electrodes Co3⊕Co and, Mn2⊕ Mn. E0Co = 1.82 V,, E0Mn = - 1.18V. Calculate E0cell., , vi., , Using the relationsip between ∆G0, of cell reaction and the standard, potential associated with it, how will, you show that the electrical potential, is an intensive property ?, , viii., , Derive the relationship between, standard, cell, potential, and, equilibrium constant of cell reaction., , ix., , It is impossible to measure the, potential of a single electrode., Comment., , b. 2 λ Al3⊕ + 3 λ SO42, 0, , 0, , c. 1/3 λ0Al3⊕ + 1/2 λ0SO42, d. λ0Al3⊕ + λ0SO42, , 2. Answer the following in one or two, sentences., i., , What is a cell constant ?, , ii., , Write the relationship between, conductivity and molar conductivity, and hence unit of molar conductivity., , iii., , Write the electrode reactions during, electrolysis of molten KCl., , iv., , Write any two functions of salt, bridge., , v., , What is standard cell potential for, the reaction, 3Ni (s) +, 3Ni2⊕(1M), , 2Al3⊕(1M), , + 2Al(s) if E0Ni = - 0.25 V and E0Al =, -1.66V?, vi., , Write Nerst equation. What part of, it represents the correction factor for, nonstandard state conditions ?, , vii., , Under what conditions the cell, potential is called standard cell, potential ?, , viii. Formulate a cell from the following, electrode reactions :, Au3⊕(aq) + 3e, Mg(s), , Au(s), Mg2⊕(aq) + 2e, , ix., , How many electrons would have a, total charge of 1 coulomb ?, , x., , What is the significance of the single, vertical line and double vertical line, in the formulation galvanic cell., , 3. Answer the following in brief, i., , Explain the effect of dilution of, solution on conductivity ?, , x., , Why do the cell potential of lead, accumulators decrease when it, generates electricity ? How the cell, potential can be increased ?, xi., Write the electrode reactions and net, cell reaction in NICAD battery., 4. Answer the following :, i., What is Kohrausch law of, independent migration of ions?, How is it useful in obtaining molar, conductivity at zero concentration of, a weak electrolyte ? Explain with an, example., ii., Explain electrolysis of molten NaCl., iii. What current strength in amperes, will be required to produce 2.4, g of Cu from CuSO4 solution in, 1 hour ? Molar mass of Cu = 63.5 g, mol-1. (2.03 A), iv. Equilibrium constant of the reaction,, 2Cu⊕(aq), Cu2⊕(aq) + Cu(s), 6, is 1.2 × 10 . What is the standard, , 118
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potential of the cell in which the, reaction takes place ?, (0.36 V), v., , Calculate emf of the cell, Zn(s)Zn 2⊕ (0.2M)H ⊕ (1.6M), H2(g, 1.8 atm) Pt at 250C., (0.785V), , vi., , Calculate emf of the following cell at, 250C., Zn (s) Zn2⊕(0.08M)Cr3⊕(0.1M)Cr, , E0Zn = - 0.76 V, E0Cr = - 0.74 V, , vii., , (0.0327 V), , What is a cell constant ? What are, its units? How is it determined, experimentally?, , viii. How will you calculate the moles, of electrons passed and mass of, the substance produced during, electrolysis of a salt solution using, reaction stoichiometry., ix., , x., , Write the electrode reactions when, lead storage cell generates electricity., What are the anode and cathode and, the electrode reactions during its, recharging?, What are anode and cathode of H2O2 fuel cell ? Name the electrolyte, used in it. Write electrode reactions, and net cell reaction taking place in, the fuel cell., , xi., , What are anode and cathode for, Leclanche' dry cell ? Write electrode, reactions and overall cell reaction, when it generates electricity., , xii., , Identify oxidising agents and arrange, them in order of increasing strength, under standard state conditions., The standard potentials are given in, parenthesis., Al, (-1.66V),, Al3⊕(-1.66V),Cl2, (1.36V), Cd2⊕(-0.4V), Fe(-0.44V),, I2(0.54V), Br (1.09V)., , xiii. Which of the following species are, reducing agents? Arrange them in, , 119, , order of increasing strength under, standard state conditions. The, standard potentials are given in, parenthesis., K (-2.93V), Br2(1.09V), Mg(-2.36V),, Ce3⊕(1.61V), Ti2⊕(-0.37V), Ag⊕(0.8, V), Ni (-0.23V)., xiv. Predict whether the following, reactions would occur spontaneously, under standard state conditions., a. Ca (s) + Cd2⊕ (aq), Ca2⊕(aq) + Cd(s), b. 2 Br (s) + Sn2⊕ (aq), Br2(l) + Sn(s), c. 2Ag(s) + Ni2⊕ (aq), 2 Ag⊕ (aq) + Ni (s), (use information of Table 5.1), , Activity :, 1. Write electrode reactions net, cell reaction in the electrolysis, of molten barium chloride., 2. Prepare the salt bridge and set up, the Daniel cell in your laboratory., Measure its emf using voltmeter and, compare it with the value calculated, from the information in Table 5.1, 3. k1 and k2 are conductivities of two, solutions and c1 and c2 are their, concentrations., Establish, the, relationship between k1, k2, c1, c2 and, molar conductivities ∧1 and ∧2of the, two solutions., 4. Find and search working of power, inverters in day-to-day life., 5. Collect information of pollution free, battery.
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6. CHEMICAL KINETICS, Can you recall ?, , Average rate =, , • What is the influence of particle, size of reacting solid on rate of a, chemical reaction ?, , change in concentration of a species, change in time, , ∆c, = ∆t, Consider the reaction A, B in which A is, consumed and B is produced., average rate of consumption of A = - ∆[A], ∆t, ∆[B], Average rate of formation of B = +, ∆t, , 6.1 Introduction : Three important, characteristics of chemical reactions include, : extent of reaction, feasibility and its rate., In standard XI, we learnt how equibrium, constants predict the extent of reaction. In unit, 3 of this text, we learnt how thermodynamic, properties such as change in entropy or, enthalpy tell us whether under the given set, of conditions chemical reaction represented, by chemical equation occurs or not. Chemical, kinetics is a branch of chemistry which deals, with the rate of chemical reactions and the, factors those affect them., A chemist wants to know the rates of, reactions for different reasons. One the study, of reaction rates help us to predict how rapidly, the reaction approaches equilibrium. Secondly, it gives information on the mechanism of, chemical reactions., , Therefore, average rate of reaction = - ∆[A], , ∆t, , ∆[B], ∆t, The rate of reaction represents a, decrease in concentration of the reactant, per unit time or increase in concentration of, product per unit time. The dimensions of rate, are concentration divided by time, that is,, mol dm-3 sec-1., , =+, , 6.2.2 Instantaneous rate of : To determine the, instantaneous rate of a reaction the progress, of a reaction is followed by measuring, the concentrations of reactant or product, for different time intervals. The changes, in concentration are relatively fast in the, , A number of reactions occur as a sequence, of elementary steps constituting the mechanism, of reaction., 6.2 Rate of reactions : The rate of reaction, describes how rapidly the reactants are, consumed or the products are formed., 6.2.1 Average rate of chemical reaction : The, average rate of a reaction can be described by, knowing change in concentration of reactant, or product divided by time interval over which, the change occurs. Thus,, , reactant concerntration, , • What is effect of change of temperature, on the rate of a chemical reaction ?, , product concerntration, , • Why is finely divided nickel used in, hydrogenation of oil ?, , Fig. 6.1 : Determination of instantaneous rate, , 120
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beginning which later become slow. The, concentration of a reactant or a product plotted, against time are shown in Fig. 6.1 (a) and, 6.1 (b). A tangent drawn to the curve at time, t1 gives the rate of the reaction. The slope thus, obtained gives the instantaneous rate of the, reaction at time t1. The instantaneous rate dc/, dt, is represented by replacing ∆ by derivative, dc/dt in the expression of average rate. In, chemical kinetics we are concerned with, instantaneous rates., For the reaction, A, , In general, For, aA + bB, rate = -, , B,, , Rate of consumption of A at any time t = -, , d[A], dt, , Rate of formation of B at any time t = d[B], , ∆t, d, [B], d, [A], Rate of reaction at time t = =, dt, dt, For the reaction involving one mole, of A and B each, the rate of consumption of, A equals the rate of formation of B. This is, not true for the reactions involving different, stoichiometries. Consider, for example, a, reaction :, A + 3B, , 2C, , When one mole of A is consumed three, moles of B are consumed and two moles of, C are formed. The stoichiometric coefficients, of the three species are different. Thus the, rate of consumption of B is three times the, rate of consumption of A. Likewise the rate of, formation of C is twice the rate of consumption, of A. We write,, , d[C], d[A], d[A], = -3, and, = -2, dt, dt, dt, dt, With this, , -, , d[B], , d[A], , 1 d[B], b dt, 1 d[D], d dt, , Problem 6.1 : For the reaction, 2 N2O5(g), 4 NO2(g) + O2(g) in liquid, bromine, N2O5 disappears at a rate of 0.02, moles dm-3 sec-1. At what rate NO2 and O2, are formed? What would be the rate of, reaction?, Solution :, d[N2O5], Given :, = 0.02, dt, d[O ], = dt 2, , 1 d[NO2], 4, dt, , 1 d[N2O5], dt, 2, 1 d[N2O5], = dt, 2, d[O ], Rate of formation of O2 = dt 2, 1, = × 0.02, 2, d[N, O, ], 1, 1, 2 5, = = × 0.02 moles dm-3 sec -1, dt, 2, 2, , Rate of reaction = -, , = 0.01 moles dm-3 sec -1, , d[NO5], dt, d[N2O5], = 4 dt, 2, , Rate of formation of NO2 =, , = 2 × moles dm3 sec -1, = 0.04 moles dm3 sec -1, , =-, , Try this..., For the reaction,, 3I (aq)+S2O82 (aq), , I3 (aq) + 2 SO42 (aq), , Calculate the rate of formation of, I3 , the rates of consumption of I- and, S2O82 and the overall rate of reaction, if the rate of formation of SO42 is, 0.022 moles dm-3 sec -1, , Write the expression for:, 2 N2O(g), , 1 d[A], =a dt, 1 d[C], =, =, c dt, , , , 1 d[C], 1 d[B], =, dt, 2 dt, 3 dt, d[A], d[B], or rate of reaction = =- 1, dt, 3 dt, 1 d[C], =, 2 dt, -, , cC + dD,, , 4 NO2(g) + O2(g) ?, , 121
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6.3 Rate of reaction and reactant, concentration : The rate of a reaction at a, given temperature for a given time instant, depends on the concentration of reactant. Such, rate-concentration relation is the rate law., 6.3.1 Rate law : Consider the general reaction,, aA + bB, , cC + dD, , (6.1), , The rate of reaction at a given time is, proportional to its molar concentration at that, time raised to simple powers or, , (iii) If x = 0, the rate is independent of, concentration of A., (iv) If x < 0 the rate decreases as [A] increases., 6.3.2 Writing the rate law, Consider the reaction,, 2H2O2(g), , If the rate of the reaction is proportional, to concentration of H2O2. The rate law is given, by, rate = k[H2O2], , Rate of reaction ∝ [A]x [B]y or, rate , , = k [A]x[B]y, , 2 H2O(l) + O2(g)., , (6.2), , Try this..., For the reaction,, , where k the proportionality constant is called, the rate constant, which is independent of, concentration and varies with temperature., For unit concentrations of A and B, k is equal, to the rate of reaction. Equation (6.2) is called, differential rate law., , NO2(g) + CO(g), NO(g) + CO2(g),, the rate of reaction is experimetally found, to be proportional to the squre of the, concentration of NO2 and independent, that of CO. Write the rate law., , The powers x and y of the concentration, terms A and B in the rate law not necessarily, equal to stoichiometric coefficients (a and, b) appearing in Eq. (6.1). Thus x and y may, be simple whole numbers, zero or fraction., Realize that x and y are experimentally, determined. The rate law in Eq. (6.2) is, determined experimentally and expresses, the rate of a chemical reaction in terms of, molar concentrations of the reactants and, not predicted from the stoichiometries of the, reactants., , Problem 6.2 : Write the rate law for the, reaction, A + B, P from the following, data :, , The exponents x and y appearing in the rate, law tell us how the concentration change, affects the rate of the reaction., (i) For x = y = 1, Eq. (6.2) gives, rate = k[A][B], The equation implies that the rate of a, reaction depends linearly on concentrations of, A and B. If either of concentration of A or B is, doubled, the rate would be doubled., (ii) For x = 2 and y = 1. The Eq. (6.2) then, leads to rate = k[A]2[B]. If concentration of A, is doubled keeping that of B constant, the rate, of reaction will increase by a factor of 4., , 122, , [A] moles [B] moles, dm-3 sec -1 dm-3 sec -1, (Initial), (Initial), (i) 0.4, 0.2, (ii) 0.6, 0.2, (iii) 0.8, 0.4, , Initial rate/, moles dm-3 sc -1, 4.0 × 10-5, 6.0 × 10-5, 3.2 × 10-4, , Solution :, a. From above data (i) and (ii), when [A], increases by a factor 1.5 keeping [B] as, constant, the rate increases by a factor, 1.5. It means rate ∝ [A] and x = 1, b. From observations (i) and (iii), it can be, seen that when concentrations of A and, B are doubled, the rate increases by a, factor 8. Due to doubling of [A] the rate, is doubled (because x = 1) that is rate, increases by a factor 2., This implies that doubling [B], the, rate increases by a factor 4. or rate ∝ [B]2, and y = 2. Therefore, rate = k[A] [B]2, contd....
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Problem 6.2 contd...., Alternatively, , Try this..., •, , The rate law gives rate = k [A] [B] ., x, , y, , 2C +D, if, 2A + 2B, concentration of A is doubled at constant, [B] the rate increases by a factor of 4. If, the concentration of B is doubled with [A], being constant the rate is doubled. Write, the rate law of the reaction., , a. From above observations (i) and (ii), (i) 4 × 10-5 = (0.4)x(0.2)y, (ii) 6 × 10-5 = (0.6)x(0.2)y, , (, , (0.6)x(0.2)y, 0.6, 6 × 10-5, =, 1.5, =, x, y =, -5, (0.4), (0.2), 0.4, 4 × 10, = (1.5)x, , (, , Dividing (ii) by (i), we have, , •, , x, , C is found to be, , rate = k[A]2[B]., The rate constant of the reaction at 25 0C, is 6.25 M-2s-1. What is the rate of reaction, when [A] = 1.0 moles dm-3 sec -1 and, [B] = 0.2 moles dm-3 sec -1?, , b. From observations (i) and (iii) separately, in the rate law gives, iii) 4 × 10-5 = (0.4) × (0.2)y, , since x = 1, , iv) 3.2 × 10-4 = 0.8 × (0.4)y, Dividing (iv) by (iii) we write, 0.8 (0.4) y, 3.2 × 10-4, =, 0.4 (0.2)y, 4 × 10-5, y, 0.6, or 8 = 2 ×, 0.2, , (, , or 4 = 22 = 2y, , The rate law for the reaction, , A+B, , Hence x = 1, , (, , For the reaction, , Therefore y = 2., The rate law is then rate = k[A][B]2., , 6.3.3 Order of the reaction : For the reaction,, aA + bB, , cC + dD is, , If the rate of the reaction is given as, , , rate = k[A]x[B]y., , Then the sum x + y gives overall, order of the reaction. Thus overall order of, the chemical reaction is given as the sum of, powers of the concentration terms in the rate, law expression. For example :, i. For the reaction,, , Problem 6.3 : For the reaction,, 2 NOBr(g), , 2 NO2(g) + Br2(g),, , the rate law is rate = k[NOBr]2. If the rate, of the reaction is 6.5 × 10-6 mol L-1 s-1 when, the concentration of NOBr is 2 × 10-3 mol, L-1. What would be the rate constant for the, reaction?, Solution :, rate, rate = k[NOBr]2 or k = [NOBr]2, =, , , 6.5 × 10-6 mol L-1s-1, (2 × 10-3mol L-1)2, , = 1.625 mol-1 L1 s-1, , 2H2O(l), +, 2H2O2(g), experimentally determined rate law is, , O2(g), , rate = k[H2O2]., The reaction is of first order., ii. If the experimentally determined rate law, for the reaction, H2(g) + I2(g), , 2 HI(g) is, , rate = k[H2][I2]., The reaction is of first order in H2 and I2 each, and hence overall of second order., , 123
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Key points about the order of reaction, a. The order of chemical reaction is, experimentally determined., , Solution : The reaction is first order in A, and second order in B. Hence, the rate law, gives, rate = k[A][B]2, rate, or k =, [A][B]2, , b. The order can be integer or fractional., Look at the reaction,, CH3CHO(g), CH4(g) + CO(g)., The rate law for the reaction was found to be, rate = k[CH3CHO]3/2., Here the order of the reaction is 3/2., , rate = 3.6 × 10-2 mol/s,, [A] = 0.2 mol dm-3 and [B] = 0.1 mol dm-3, Substitution gives, 3.6 × 10-2 mol dm-3s-1, k=, 0.2 mol dm-3 × (0.1 mol dm-3)2, , c. The order of the reaction, can be zero for :, NO2(g) + CO(g), NO(g) + CO2(g), The rate expression for this is : rate = k[NO2]2., This shows that order of reaction with respect, to NO2 is 2 and with CO is zero or the rate, is independent of concentration of CO. The, overall order of reaction is 2., d. Only a few reactions of third order are, known. Reactions with the orders higher than, three are scanty., Problem 6.4 : For the reaction, 2NO(g) + 2H2(g), N2(g) + 2 H2O(g),, the rate law is rate = k[NO]2 [H2]. What is, the order with respect to NO and H2 ? What, is the overall order of the reaction ?, , 3.6 × 10-2 s-1, = 0.2 × 0.01 mol2 dm-6 sec-1, = 18 mol-2 dm-6 sec-1, Use your brain power, The rate of the reaction, 2A + B, 2C + D is 6 × 10-4 mol, -3 -1, dm s when [A] = [B] = 0.3 mol dm-3. If, the reaction is of first order in A and zeroth, order in B, what is the rate constant?, Problem 6.6 : Consider,, A+ B, P. If the concentration of A is, doubled with [B] being constant, the rate of, the reaction doubles. If the concentration, of A is tripled and that of B is doubled,, the rate increases by a factor 6. What is, order of the reaction with respect to each, reactant ? Determine the overall order of, the reaction., Solution :, The rate law of the reaction :, rate = k[A]x[B]y , (i), If [A] is doubled, the rate doubles., Hence 2 × rate = k [2A]x[B]y, = k 2x [A]x[B]y, (ii), x, y, 6 × rate = k [3A] [2B], (iii), y, y, (iii), 6 × rate 3 k[A]2 [B], (i) gives rate = k [A][B]y, or 6 = 3 × 2y or 2 = 2y and y =1, The reaction is first order in A and first, order in B. The overall reaction is of the, second order., , Solution : In the rate law expression, the, exponent of [NO] is 2 and that of [H2] is 1., Hence, reaction is second order in NO, first, order in H2 and the reaction is third order., Try this..., The reaction, CHCl3(g) + Cl2(g), CCl4(g) + HCl(g) is first, order in CHCl3 and 1/2 order in Cl2., Write the rate law and overall order of, reaction., Problem 6.5 : The rate of the reaction,, A+B, P is 3.6 × 10-2 mol dm-3 s-1, when [A] = 0.2 moles dm-3 and [B] = 0.1, moles dm-3. Calculate the rate constant if, the reaction is first order in A and second, order in B., , 124
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6.4 Molecularity of elementary reactions :, Complex reactions are those which, constitute a series of elementary reactions., , these steps is slower than others. The slowest, step is the rate determining step., , 6.4.1 Elementary reaction, , The slowest step determines the rate of overall, reaction., , Consider,, , Consider, 2NO2Cl(g), , O3(g), , O2(g) + O(g), , C2H5I(g), , The reaction takes place in two steps:, , C2H4(g) + HI(g), , i. NO2Cl(g), , k1, , NO2(g) + Cl(g) (slow), , These reactions occur in a single step and, cannot be broken down further into simpler, reactions. These are elementary reactions., , ii. NO2Cl(g) + Cl (g), , 6.4.2 Molecularity of reaction : The, molecularity refers to how many reactant, molecules are involved in reactions. In the, above reactions there is only one reactant, molecule. These are unimolecular reactions or, their molecularity is one., , Overall 2NO2Cl(g), , O3(g) + O(g), 2 NO2(g), , 2NO2(g) + Cl2(g)., , 2 O2(g), , k2, , NO2(g) + Cl2(g)(fast), 2NO2(g) + Cl2(g), , The first step being slower than the second it is, the rate determining step., The rate law is, rate = k[NO2Cl], This also represents the rate law of the overall, reaction. The reaction thus is of the first order., , 2 NO(g) + O2(g), , The elementary reactions involving two, reactant molecules are bimolecular reactions, or they have molecularity as two., The molecularity of an elementary, reaction is the number of reactant molecules, taking part in it., 6.4.3 Order and molecularity of elementary, reactions:, The rate law for the elementary reaction, 2NO2(g), 2NO2(g) + O2(g) is found to, 2, be rate = k[NO2] . The reaction is second order, and bimolecular. The order of reaction is 2 and, its molecularity is also 2., , Reaction intermediate:, In the above reaction Cl is formed in the, first step and consumed in the second. Such, species represents the reaction intermediate., The concentration of reaction intermediate, does not appear in the rate law., Distinction between order and molecularity, of a reaction :, , For the elementary reaction,, C2H5 I(g), C2H4(g) + HI(g), rate = k[C2H5I], It is unimolecular and first order. However the, order and molecularity of the reaction may or, may not be the same., 6.4.4 Rate determining step : A number of, chemical reactions are complex. They take, place as a series of elementary steps. One of, , 125, , Order, Molecularity, 1. It is experimentally i. It is theoretical, determined property. entity., 2. It is the sum, of powers of the, concerntration terms, of reactants those, appear in the rate, equation., , ii. It is the number of, reactant molecules, taking part in an, elementary reaction., , 3. It may be an iii. It is integer., integer, fraction or, zero.
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The differential rate law is given by, d[A], rate = = k [A], (6.4), dt, where [A] is the concentration of reactant at, time t., Rearranging Eq. (6.4),, , Problem 6.7 : A reaction occurs in the, following steps, i. NO2(g) + F2(g), , NO2F(g) + F(g) (slow), , ii. F(g) + NO2(g), , NO2F(g) (fast), , a. Write the equation of overall reaction., , d[A], = -k dt , (6.5), [A], Let [A]0 be the initial concentration of the, reactant A at time t = 0. Suppose [A]t is the, concentration of A at time = t., , b. Write down rate law., c. Identify the reaction intermediate., Solution :, a. The addition of two steps gives the overall, reaction as, 2NO2(g) + F2(g), , The equation (6.5) is integrated between limits, [A] = [A]0 at t = 0 and [A] = [A]t at t = t, , 2 NO2F(g), , [A]t, , b. Step (i) is slow. The rate law of the reaction, is predicted from its stoichiometry. Thus,, , ∫, [A], , rate = k[NO2] [F2], , On integration,, , c. F is produced in step (i) and consumed in, step (ii) or F is the reaction intermediate., , [A]t, , [ln[A]] [A], , 0, , [A]t, = -kt , [A]0, [A]0, 1, or k = ln, t, [A]t, or, , i) NO(g) + O3(g), , NO3(g) + O(g), , ii) NO3(g) + O(g), , NO2(g) + O2(g), , The predicted rate law is rate = k[NO][O3]., , ln, , Converting ln to log10, we write, [A]0, 2.303, k = t log10 [A], t, , Identify the rate determining step. Write, the overall reaction. Which is the reaction, intermediate? Why?, 6.5 Integrated rate law : We introduced the, differential rate law earlier. It describes how, rate of a reaction depends on the concentration, of reactants in terms of derivatives., The differential rate laws are converted, into integrated rate laws. These tell us the, concentrations of reactants for different times., 6.5.1 Integrated rate law for the first order, reactions in solution : Consider first order, reaction,, product, , t, 0, , = -k (t), , Substitution of limits gives, ln [A]t - ln [A]0 = -k t, , Try this..., A complex reaction takes place in, two steps:, , A, , 0, , t, d[A], = -k ∫ dt, [A], 0, , (6.3), , (6.6), , (6.7), , Eq. (6.7) gives the integrated rate law for the, first order reactions., The rate law can be written in the following, forms, [A]t, i. Eq. (6.6) is ln, = -kt, [A]0, By taking antilog of both sides we get, [A]t, = e-kt or [A]t = [A]0e-kt, [A]0, , (6.8), , ii. Let ‘a’ mol dm-3 be the initial concentration, of A at t = 0, Let x mol dm-3 be the concentration of, A that decreases (reacts) during time t. The, , 126
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concentration of A that remains unreacted at, time t would be (a - x) mol/dm3, Substitution of [A]0 and [A]t = (a - x), k=, , a, 2.303, log10 (a-x), t, , (6.9), , Equations (6.7), (6.8) and (6.9) represent the, integrated rate law of first order reactions., , 0.693, t1/2, 0.693, t1/2 =, k1/2, , k=, , (6.10), , Eq. (6.10) shows that half life of the, first order reaction is independent of initial, reactant concentration. This is shown in Fig, (6.2) as a plot of [A]t versus t., , 6.5.2 Units of rate constant for the first order, reaction:, , [A], , The integrated rate law is, [A]0, 2.303, k = t log10 [A], t, [A]0, , Because log10 [A] is unitless quantity, the, t, dimensions of k will be (time)-1. The units of k, will be s-1, min-1or (hour)-1, , time, , 6.5.3 Half life of the first order reactions (t1/2), Radioactive processes follow the first, order kinetics. The half life of reaction is time, required for the reactant concentration to fall, to one half of its initial value., 6.5.4 Half life and rate constant of the first, order reaction :, The integrated rate law for the first order, reaction is, [A]0, 2.303, k=, log10, [A]t, t, where [A]0 is the initial concentration of, reactant at t = 0. It falls to [A]t at time t after, the start of the raction. The time required for, [A]0 to become [A]0/2 is denoted as t1/2, , Fig. 6.2 : Half life period of first order reaction, , 6.5.5 Graphical representation of the first, order reactions, i. The differential rate law for the first order, reaction A, P is, d[A], rate = - dt = k [A]t + 0, , m x c, y, The equation is of the form y = mx + c. A plot, of rate versus [A]t is a straight line passing, through origin. This is shown in Fig. 6.3. The, slope of straight line = k., , or, [A]t = [A]0/2, , at t = t1/2, , =, , rate, , Putting this condition in the integrated rate, law we write, 2.303, [A]t, k= t, log10, [A]0/2, 1/2, , intitial Concerntration, , 2.303, log10 2, t1/2, , Fig. 6.3 : Variation of rate with [A], , 2.303, = t, × 0.3010, 1/2, , 127
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ii. From Eq. (6.7) the integrated rate law is, [A]0, 2.303, k=, log10, [A]t, t, On rearrangement, the equation becomes, , 6.5.6 Examples of first order reactions, Some examples of reactions of first order are :, i. 2 H2O2(l), rate = k[H2O2], , kt, = log10 [A]0 - log10 [A]t, 2.303, k, Hence, log10 [A]t = t + log10 [A]0, 2.303, m, , y, , x, , ii. 2 N2O5(g), , [A]0, [A]t, , 4 NO2(g) + O2(g),, , rate = k [N2O5], 6.5.7 Integrated rate law for gas phase f, reactions, , c, , The equation is of the straight line. A graph of, log10, , 2 H2O(l) + O2(g),, , For the gas phase reaction,, A(g), , versus t yields a stright line with, , slope -k/2.303 and y-axis intercept as log10[A]0, This is shown in Fig. 6.4, , B(g) + C(g), , Let initial pressure of A be Pi that decreases by, x within time t., Pressure of reactant A at time t, PA = Pi - x, (6.11), , [A]0, log, [A]t, , The pressures of the products B and C at time, t are, PB = PC = x, , time, , The total pressure at time t is then, , Fig. 6.4 : A plot showing log [At]t/[A]0 vs time, , P = Pi - x + x + x = Pi + x, , iii. Eq. (6.7) gives, k, [A], log10 0 =, t, 2.303, [A]t, v, , m, , Hence, x = P - Pi , , Pressure of A at time t is obtained by substitution, of Eq. (6.12) into Eq. (6.11). Thus, , x, , PA = Pi - (P - Pi) = Pi - P + Pi = 2Pi - P, , The equation has a straight line form y = mx ., Hence, the graph of, , (6.12), , [A], log10 0, [A]t, , The integrated rate law turns out to be, , versus t is, , straight line passing through origin as shown, in Fig. 6.5., , [A]0, 2.303, log10, [A]t, t, The concentration now expressed in terms of, pressures., k=, , Thus, [A]0 = Pi and [A]t = PA = 2 Pi - P, , log, , Substitution gives in above, Pi, 2.303, k=, log10, t, 2 Pi - P, , [A]0, [A]t, , (6.13), , P is the total pressure of the reaction mixture, at time t., conc, Fig. 6.5 :, , 128
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Problem 6.8 : The half life of first order, reaction is 990 s. If the initial concentration, of the reactant is 0.08 mol dm-3, what, concentration would remain after 35, minutes?, Solution :, 0.693 0.693, k= t, = 990 s = 7 × 10-4 s-1, 1/2, [A]0, 2.303, k=, log10, [A]t, t, , Problem 6.10 : Following data were, obtained during the first order decomposition, of SO2Cl2 at the constant volume., , [A]0 = 0.08 mol dm-3, t = 35 min or 2100 s,, [A]t = ?, [A], kt, 7 × 10-4 s-1 × 2100 s, log10 0 = 2.303 =, 2.303, [A]t, , = 0.6383, [A]0, = antilog 0.6383 = 4.35, [A]t, [A]0, 0.08, Hence, [A]t =, =, 4.35 4.35, = 0.0184 mol dm-3, , Solution :, , SO2Cl2(g), Times/s, , k=, , 100 0.6, Calculate the rate constant of the reaction., k=, , (, 2.303, =, log (, 100, 10, , ) = 2.23 × 10, , -3, , s-1, , Integrated rate law for zero order reactions, : For zero order reaction,, A, , P, , the differntial rate law is given by, d[A], rate = = k [A]0 = k, [A], Rearrangement of Eq. (6.14) gives, , Substitution of these in above, 2.303, 100, log10, t, 40, 2.303, = 45 log10 2.5, 2.303, = 45 × 0.3979 = 0.0204 min-1, , k=, , 0.693, 0.0204 min-1, , 0.5, 0.4, , ), , 6.5.8 Zero order reactions: The rate of zero, order reaction is independent of the reactant, concentration., , [A]0, 2.303, log10, [A]t, t, , 0.693, k =, , Pi, 2.303, log10, t, 2 Pi - P, , Pi = 0.5 bar, P = 0.6 bar, t = 100 s, 2.303, 0.5 bar, k = 100 log10 2 × 0.5 bar - 0.6 bar, , [A]0 = 100, [A]t = 100 - 60 = 40, t = 45 min, , t1/2 =, , Toatl pressure/bar, , 0 0.5, , Problem 6.9 : In a first order reaction 60%, of the reactant decomposes in 45 minutes., Calculate the half life for the reaction, Solution :, , SO2(g) + Cl2(g), , .....(6.14), , d[A] = -k dt, Integration between the limits, [A] = [A]0 at t = 0 and [A] = [A]t at t = t gives, [A]t, , ∫, [A], , = 34 min, , 0, , t, , d[A] = -k ∫ dt, 0, , or [A]t - [A]0 = - kt, Hence, k t = [A]0 - [A]t, , Try this..., The half life of a first order, reaction is 0.5 min. Calculate time, needed for the reactant to reduce to 20% and, the amount decomposed in 55 s., , (6.15), , Units of rate constant of zero order reactions, [A]0 - [A]t, mol L-1, k=, =, = mol dm-3 t-1, t, t, The units of rate constant of zero order reaction, are the same as the rate., , 129
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Half life of zero order reactions : The rate, constant of zero order reaction is given by eqn, (6.15), [A]0 - [A]t, k=, t, Using the conditions t = t1/2, [A]t = [A]1/2,, Eq. (6.15) becomes, [A]0 - [A]0/2, [A], k=, = 2t 0, t1/2, 1/2, [A], , Hence, t1/2 = 2 k 0 , , 6.16, , The half life of zero order reactions is, proportional to the initial concentration of, reactant., Graphical representation of zero oder, reactions : The rate law in Eq. (6.15) gives, [A]t = -k t + [A]0, , 6.17, , mx c, y, Which is straight line given by, y = mx + c., , The metals surface gets completely, covered by a layer of NH3 molecules. A number, of NH3 molecules attached on platinum, surface is small compared to ammonia. A large, number of the NH3 molecules tend to remain as, gas which do not react. The molecules present, on the metal surface only react. The rate of, a reaction was thus independent of the total, concentration of NH3 and remains constant., ii. Decomposition of nitrous oxide in the, presence of Pt catalyst., 2 N2O(g), , Pt, , 2 N2(g) + O2(g), , iii. The catalytic decomposition of PH3 on hot, tungsten at high pressure., 6.5.9 Pseudo-first order reactions : Certain, reactions which are expected to be of higher, order follow the first order kinetics. Consider, hydrolysis of methyl acetate., CH3COOCH3(aq) + H2O(l), , CH3COOH(aq) + CH3OH(aq), The rate law is, , A plot of [A]t versus t is a straight line as, shown in Fig 6.6., , rate = k' [CH3COOCH3] [H2O], The reaction was expected to follow, the second order kinetics, however, obeys the, first order., The reason is that solvent water is, present in such large excess that the change, in its concentration is negligible compared, to initial one or its concentration remains, constant., , [A]t, , time, , Thus [H2O] = constant = k''. The rate law, becomes, , Fig. 6.6 : [A]t vs t for zero order reaction, , The slope of straight line is -k and its intercept, on y-axis is [A]0., , rate = k' [CH3COOCH3] k'', , The t1/2 of zero order reaction is, directly proportional to the initial concentration., , where k = k'k'', , Examples of zero order reactions :, , = k [CH3COO CH3], The reaction is thus of first order., The reaction C12H22O11(aq) + H2O (l) (excess), , Here follow some examples, , C6H12O6(aq) + C6H12O6(aq), , Decomposition of NH3 on platinum metal, 2 NH3(g), , glucose, , fructose, , Can it be of pseudo-first order type ?, , N2(g) + 3 H2(g), , 130
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6.6 Collision theory of bimolecular reactions, , 6.6.4 Potential energy barrier, , 6.6.1 Collision between reactant molecules, , Consider again the reaction, , Chemical reactions occur as a result, of collisions between the reactant species. It, may be expected that the rate of the reaction is, equal to the rate of collision. For the gas-phase, reactions the number of collisions is far more, and typically many powers of tens compared, to the observed rate., 6.6.2 Activation : For the reaction to occur, the colliding reactant molecules must possess, the minimum kinetic energy. This minimum, kinetic energy is the activation energy., The reaction would occur only if colliding, molecules possess kinetic energies equal to or, greater than the activation energy., 6.6.3 Orientation of reactant molecules, The requirement for successful collision, described above is sufficient for reactions, involving simple molecules (or ions) however, not for those involving complex molecules., Besides the above considerations, the colliding molecules must have proper, orientation. The molecules need to be so, oriented relative to each other that the reacting, groups approach closely., Consider, A + C - B, , A+C-B, , During a course of collision, new bond, A - B developes. At the same time bond C - B, breaks. A configuration in which all the three, atoms are weakly connected together is called, activated complex., A+B-C, , B, , A, , C, , +, A, , B, , A-B+C, , C, , Energy, , ∆H, , B, , No reaction will takes place. The reactant, molecules would collide and separate owing, to the improper orientation of C - B., +, ii., A, B C, , , C, , Ea, , +, C, , B, , The energy barrier between reactants, and products is shown in Fig. 6.7. The reactant, molecules need to climb up and overcome this, before they get converted to products. The, height of the barrier is called as activation, energy (Ea). Thus the reactant molecules, transform to products only if they possess, energy equal to or greater than such activation, energy. A fraction of molecules those possess, energy greater than Ea is given by f = e-Ea/RT., , i. The collision of A with C approaching toward, A would not lead to reaction., A, , A, , To attain the configuration A, B, C, atoms need to gain energy, which comes from, the kinetic energy of colliding molecules., , A-B+C, , +, , A-B+C, , , , The reaction is successful as a result of, proper orientation of C - B. A fraction of such, collisions bring forth conversion of reactants, to products., , Fig. 6.7 : Potential energy barrier, , As a result only a few collisions lead to, products. The number of successful collisions, are further reduced by the orientation, requirement already discussed., , 131
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6.7.2 Graphical determination of activation, energy : Taking logarithm of both sides of eqn, (6.18) we obtain, E, ln k = - a + ln A, (6.19), RT, Converting natural base to base 10 we write, Ea, 1, log10k = + log10 A, (6.20), 2.303 R T, , Do you know ?, For a gaseous reaction at, 298 K, Ea = 75 kJ/mol. The fraction, of successful collisions is given by f = e-Ea/RT, = e-75000/8.314 × 298 = 7 × 10-14 or only 7 collisions, in 1014 collisions are sufficiently energetic, to lead to the reaction., , y, , m, , x, , c, , This equation is of the form of straight, line y = mx + c., , Remember..., All collisions of reactant, molecules do not lead to a chemical, reaction. The colliding molecules need to, possess certain energy which is greater, than the activation energy Ea and proper, orientation., , The Arrhenius plot of log10 k versus, 1/T giving a straight line is shown in fig., (6.8). A slope of the line is -Ea /2.303R with its, intercept being log10 A., , 6.7 Temperature dependence of reaction, rates, Do you know ?, It has been observed that, the rates of most of the chemical, reactions usually increase with temperature., In everyday life we see that the fuels such as, oil, coal are inert at room temperature but, burn rapidly at higher temperatures. Many, foods spoil rapidly at room temperature and, lasts longer in freezer., , Fig. 6.8 : Variation of log10k with 1/T, , The concentrations change only a little, with temperature. The rate constant shows a, strong dependence on the temperature., 6.7.1 Arrhenius equation: Arrhenius, suggested that the rate of reactions varies with, temperature as, k = A e-Ea /RT, , 6.18, , where k is the rate constant, Ea is the activation, energy, R molar gas constant, T temperature in, kelvin, and A is the pre-exponential factor. Eq., (6.18) is called as the Arrhenius equation., The pre exponential factor A and the, rate constant have same unit in case of the, first order reactions. Besides A is found to be, related to frequency of collisions., , From a slope of the line the activation, energy can be determined. Eq. (6.18) shows, that with an increase of temperature, -Ea/RT, and in turn, the rate of reaction would increase., 6.7.3 Determination of activation energy :, For two different temperatures T1 amd T2, Ea, log10 k1 = log10A (6.21), 2.303 RT1, Ea, log10k2 = log10A (6.22), 2.303 RT2, where k1 and k2 are the rate constants, at temperatures T1 and T2 respectively., Subtracting Eq. (6.21) from Eq. (6.22),, Ea, Ea, 1, 1, +, log10k2 - log10 k1 = 2.303 R T2, 2.303 R T1, , 132
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T2 - T1, T1T2, , ..........(6.23), , 6.7.4 Graphical description of effect of, temperature : It has been realized that average, kinetic energy of molecules is proportional to, temperature. The collision theory suggested a, bimolecular reaction occurs only if the reacting, molecules have sufficient kinetic energies (at, least Ea) and proper orientation when they, collide., At a given temperature, the fraction, of molecules with their kinetic energy equal, to or greater than Ea may lead to the product., With an increase of temperature the fraction, of molecules having their energies (Ea) would, increases. The rate of the reaction thus would, increase. This is depicted by plotting a fraction, of molecules with given kinetic energy versus, kinetic energy for two different temperatures, T1 and T2 (T2 being > T1) in Fig. 6.9., , Problem 6.12 : The rate constants for a first, order reaction are 0.6 s-1 at 313 K and 0.045, s-1 at 293 K. What is the activation energy?, Solution Ea, k, log10 2 =, k1 2.303 R, , (, , T2 - T1, T1T2, , k1 = 0.045 s-1, k2 = 0.6 s-1, T1 = 293 K,, T2 = 313 K, R = 8.314 J K-1mol-1, Substituting, Ea, 2.303 × 8.314 ×, 313 - 293, 293 × 313, , Ea, log10 13.33 =, ×, 2.303 × 8.314, 20, 293 × 313, Ea, 1.1248 =, × 2.18 × 10-4, 19.15, log10, , 0.6, =, 0.045, , [, , [, , (, , (, , Ea, 2.303 R, , (, , =, , increase with temperature. The rate of reaction, increases accordingly., , (, , (, , Ea, k, 1, 1, Hence, log10 k2 =, T2, 2.303 R T1, 1, , Ea = 1.1248 × 19.15 J mol-1/ 2.18 × 10-4, = 98810 J/mol-1 = 98.8 kJ/mol-1, Problem 6.13 : A first order gas phase, reaction has activation energy of 240 kJ, mol-1. If the pre-exponential factor is 1.6, × 1013 s-1. what is the rate constant of the, reaction at 600 K?, Solution : Arrhenius equation, k = A e-Ea/RT is written as, A, Ea, log10 k =, 2.303 RT, , Fig. 6.9 : Comparison of fraction of molecules, activated at T1 and T2, , Ea = 240 kJ mol-1 = 240 × 103 J mol-1,, , The shaded areas is proportional to, total number of molecules. The total area, is the same at T1 and T2. The area (a + b), represents a fraction of molecules with kinetic, energy exceeding Ea is larger at T2 than at T1, (since T2 > T1). This indicates that a fraction of, molecules possessing energies larger than Ea, , 133, , T = 600 K, A = 1.6 × 1013 s-1, A, =, k, 240 × 103 J mol-1, , Hence log10, , 2.303 × 8.314 J mol-1 K-1× 600 K, , = 20.89, contd....
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A, k, , 6.8 Effect of a catalyst on the rate of reaction, , = antilog 20.89, , A catalyst is a substance added to the, reactants that increases the rate of the reaction, without itself being consumed in the reaction., , = 7.78 × 1020, A, and k = 7.78 × 1020, , Consider, , 1.6 × 10 s, 7.78 × 1020, 13, , =, , -1, , 2 KClO3(s)+ 3O2(g), , = 2.055 × 10-8 s-1, , Problem 6.14 : The half life of a first order, reaction is 900 min at 820 K. Estimate its, half life at 720 K if the activation energy is, 250 kJ mol-1., Solution :, 0.693, t1/2 = k, Rate constants at two different, temperatures, T1 and T2 are k1 and k2, respectively, and the corresponding half, lives (t1/2)1 and (t1/2)2., 0.693, 0.693, (t1/2)1 =, and (t1/2)2 =, k, k, k, (t ), Hence, 1/2 1 = k2, (t1/2)2, 1, Ea, k, The equation, log10 2 = 2.303 RT ×, k1, T2 - T1, , T1T2, , [ TT -TT, 2, , 1 2, , Ea = 250 kJ mol , T1 = 720 K,, -1, , 1, , (t1/2)1, =, (t1/2)2, , 2.303 × 8.314 J K mol, -1, , -1, , 820 K - 720 K, [820, K × 720 K, , [, , 250 × 103 J mol-1, , Fig. 6.10 : Potential energy barriers for, catalyzed and uncatalyzed reactions, , 2H2O2(l), , T2 = 820 K, (t1/2)2 = 900 min, Thus, log10, , Fig. 6.10 compares the potential energy, barriers for the catalysed and uncatalysed, reactions. The barrier for uncatalysed reaction, (Ea) is larger than that for the same reaction in, the presence of a catalyst Ea., , Consider the decomposition of H2O2 in, aqueous solution catalysed by I- ions., , [, , (t ), Ea, log10 (t1/2)1 =, 2.303 RT, 1/2 2, , 2 KCl(s), , Here MnO2 is the catalyst. It has been observed, that the decomposition rate increases with, the addition of catalyst. A catalyst provides, alternative pathway associated with lower, activation energy., , [, , [, , MnO2, , I-, , 2 H2O(l) + O2(g), , At room temperature the rate of reaction is, slower in the absence of catalyst with its, activation energy being 76 kJ mol-1. In the, presence of iodide ion/ the catalyst I- the, reaction is faster since the activation energy, decreases to 57 kJ mol-1., , , = 2.212, (t1/2)1, = antilog 2.212 = 162.7, (t1/2)2, (t1/2)1 = (t1/2)2 × 162.7 = 900 × 162.7 , , = 1.464 × 105 min, , 134
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Fig 6.11 shows a plot of fraction of molecules, as a function of energy. A catalyst lowers, the the threshold energy. Consequently more, molecules acquire the minimum amount of, energy and tend to cross the energy barrier. A, fraction of activated molecules is greater for, the catalyzed reaction. The rate of catalyzed, reaction thus is larger than the reaction with, no catalyst., Fig. 6.11 : Comparison of fraction of molecules, for catalyzed and uncatalyzed reactions, , Exercises, 1. Choose the most correct option., i., , The rate law for the reaction, aA + bB, P is rate = k[A] [B]., The rate of reaction doubles if, a. concentrations of A and B are both, doubled., b. [A] is doubled and [B] is kept, constant, c. [B] is doubled and [A] is halved, d. [A] is kept constant and [B] is, halved., , ii. The order of the reaction for which the, units of rate constant are mol dm-3 s-1, is, a. 1 , , b. 3, , c. 0 , , d. 2, , iii. The rate constant for the reaction, 2 N2O5(g), 2 N2O4(g) + O2(g) is, -4 -1, 4.98 × 10 s . The order of reaction is, a. 2 , , b. 1, , c. 0 , , d. 3, , b. 2t, , c. t/2 , , d. 3t, , Slope of the graph ln[A]t versus t for, first order reaction is , a. -k , b. k, c. k/2.303, , d. -k/2.303, , vi. What is the half life of a first order, reaction if time required to decrease, concentration of reactant from 0.8 M to, 0.2 M is 12 h? , a. 12 h , , b. 3 h, , c. 1.5 h, , d. 6 h, , vii. The reaction, 3 ClO, ClO3 +2 Cl, occurs in two steps,, (i) 2 ClOClO2, (ii) ClO2 + ClO, ClO3 + Cl, The reaction intermediate is , a. Cl , , b. ClO2, , c. ClO3, , d. ClO, , viii. The elementary reaction, O3(g) + O(g), 2 O2(g) is , a. unimolecular and second order, , iv. Time required for 90 % completion of, a certain first order reaction is t. The, time required for 99.9 % completion, will be , a. t , , v., , b. bimolecular and first order , c. bimolecular and second order , d. unimolecular and first order, ix. Rate law for the reaction,, 2 NO + Cl2, 2 NOCl is , 2, rate = k[NO2] [Cl2]. Thus of k would, increase with, , 135
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a. increase of temperature , b. increase of concentration of NO , c. increase of concentration of Cl2 , d. increase of concentrations of , both Cl2 and NO, x. For an endothermic reaction, X, Y., If Ef is activation energy of the forward, reaction and Er that for reverse reaction,, which of the following is correct?, a. Ef = Er b. Ef < Er, , if time is expressed in seconds and, concentration of reactants in mol/L?, vii. Write Arrhenius equation and explain, the terms involved in it., viii. What is the rate determining step?, ix. Write the relationships between rate, constant and half life of first order and, zeroth order reactions., x., , c. Ef > Er, d. ∆H = Ef - Er is negative, , 3. Answer the following in brief., , 2. Answer the following in one or two, sentences., i., , For the reaction,, N2(g) + 3 H2(g), , i., , How instantaneous rate of reaction is, determined?, , ii. Distinguish between order, molecularity of a reaction., , 2 NH3(g),, , what is the relationship among d[N2] ,, dt, d[NH3], d[H2], and dt ?, dt, ii. For the reaction,, CH3OH (aq) +Br (aq), rate law is, rate = k[CH3Br][OH ], a. How does reaction rate changes if, [OH ] is decreased by a factor of 5 ?, b. What is change in rate if, concentrations of both reactants are, doubled?, iii. What is the relationship between, coefficients of reactants in a balanced, equation for an overall reaction and, exponents in rate law. In what case the, coefficients are the exponents?, iv. Why all collisions between reactant, molecules do not lead to a chemical, reaction?, What is the activation energy of a, reaction?, , vi. What are the units for rate constants for, zero order and second order reactions, , 136, , and, , iii. A reaction takes place in two steps,, 1. NO(g) + Cl2(g), , NOCl2(g), , 2. NOCl2(g) + NO(g), NOCl(g), , 2, , a. Write the overall reaction. b. Identify, reaction intermediate. c. What is the, molecularity of each step?, , CH3Br(aq) + OH-(aq), , v., , How do half lives of the first order, and zero order reactions change with, initial concentration of reactants?, , iv. Obtain the relationship between the, rate constant and half life of a first, order reaction., v., , How will you represent zeroth order, reaction graphically?, , vi. What are pseudo-first order reactions?, Give one example and explain why it, is pseudo-first order., vii. What are requirements for the colliding, reactant molecules to lead to products?, viii. How catalyst increases the rate of, reaction? Explain with the help of, potential energy diagram for catalyzed, and uncatalyzed reactions., ix. Explain with the help of Arrhenius, equation, how does the rate of reaction, changes with (a) temperature and (b), activation energy.
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x., , Derive the integrated rate law for first, order reaction., , xi., , How will you represent first order, reactions graphically., , xii. Derive the integrated rate law for the first, order reaction, A(g), B(g) + C(g), in terms of pressure., xiii. What is zeroth order reaction? Derive its, integrated rate law. What are the units of, rate constant?, xiv. How will you determine activation, energy: (a) graphically using Arrhenius, equation (b) from rate constants at two, different temperatures?, xv. Explain graphically the effect, temperature on the rate of reaction., , of, , v., , The rate constant of a reaction at 5000C, is 1.6 × 103 M-1s-1. What is the frequency, factor of the reaction if its activation, energy is 56 kJ/mol. (9.72 × 106 M-1s-1), , vi., , Show that time required for 99.9%, completion of a first order reaction is, three times the time required for 90%, completion., , vii. A first order reaction takes 40 minutes, for 30% decomposition. Calculate its half, life. (77.66 min), viii. The rate constant for the first order reaction, is given by log10 k = 14.34 - 1.25 × 104, T. Calculate activation energy of the, reaction. (239.3 kJ/mol), ix., , xvi. Explain graphically the effect of catalyst, on the rate of reaction., xvii. For the reaction 2A + B, products,, find the rate law from the following data., [A]/M, 0.3, 0.6, 0.6, , [B]/M, 0.05, 0.05, 0.2, , rate/M s-1, 0.15, 0.30, 1.20, , 4. Solve, i., , In a first order reaction, the concentration, of reactant decreases from 20 mmol dm-3, to 8 mmol dm-3 in 38 minutes. What is, the half life of reaction? (28.7 min), , ii., , The half life of a first order reaction is 1.7, hours. How long will it take for 20% of, the reactant to react? (32.9 min), , iii. The energy of activation for a first order, reaction is 104 kJ/mol. The rate constant, at 25 0C is 3.7 × 10-5 s-1. What is the rate, constant at 300C? (R = 8.314 J/K mol), (7.4 × 10-5), iv., , What is the energy of activation of a, reaction whose rate constant doubles, when the temperature changes from 303, K to 313 K? (54.66 kJ/mol), , 137, , What fraction of molecules in a gas at, 300 K collide with an energy equal to, activation energy of 50 kJ/mol ? (2 × 10-9), , Activity :, 1. You wish to determine, the reaction order and rate, constant for the reaction, 2AB2, A2 + 2B2. a) What data would, you collect? b) How would you use, these data to determine whether the, reaction is zeroth or first order?, 2. The activation energy for two, reactions are Ea and E'a with Ea > E'a., If the temperature of reacting system, increase from T1 to T2, predict which, of the following is correct?, a., , c., , k', k'1, = 2, k2, k1, , b., , k'1, k'2, >, k1, k2, , k'1, k'2, k'1, <, d., <2, k1, k2, k1, ks are rate constants at, temperature and k's at, temperature., , k'2, k2, lower, higher
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7. ELEMENTS OF GROUPS 16, 17 AND 18, Fluorine (9F), chlorine (17Cl), bromine, (35Br), iodine (53I) and astatine (85At) constitute, Group 17. These are collectively known as, halogens (Greek halo means salt, gene means, born), that is, salt producing element., , Can you recall ?, • How does the valence shell, electronic configuration of the, elements vary in the p-block of periodic, table ?, • Name the first element of groups 16,, 17 and 18., 7.1 Introduction : You have learnt in, Std. XI that in the p-block elements the, differentiating electron (the last filling, electron) enters the p-orbital of the outermost, shell. Since maximum of six electrons can be, accommodated in a p-subshell it gives rise to, groups 13 to 18, in the p-block. In this chapter, we shall study the properties of elements of, groups 16, 17 and 18., 7.2 Occurence : The elements oxygen (8O),, sulfur (16S), selenium (34Se), tellurium (52Te), and polonium (84Po) constitute Group 16,, called the oxygen family. Large number of, metal ores are oxides or sulfides. Group 16, elements are also called chalcogens or ore, forming elements., Oxygen is the most abundant of all the, elements on earth. Oxygen forms 20.95 %, by volume of air and about 46.6 % by mass, of earth's crust. Sulfur forms 0.034% by, mass of the earths crust. It occurs mainly in, combined forms as sulfates such as gypsum, (CaSO4.2H2O), epsom salt (MgSO4.7H2O),, baryte (BaSO4) and sulfides such as galena, (PbS), zinc blende (ZnS), copper pyrites, (CuFeS2)., Selenium and tellurium are also found as, metal selenides and tellurides in sulfide ores., Polonium which is radioactive is a decay, product of thorium and uranium., , Halogens are very reactive due to high, electronegativities and hence they are not, found in free sate. They occur in the form, of compounds., Fluorine occurs mainly as insoluble, fluorides (fluorspar CaF2, cryolite Na3AlF6,, fluorapatite 3Ca3(PO4)2.CaF2) and small, quantities are present in soil, fresh water, plants, and bones and teeth of animals. Sea, water contains chlorides, bromides and iodides, of Na, K, Mg and Ca. However it mainly, contains NaCl (2.5 % by mass). The deposits, of dried up sea beds contain sodium chloride, and carnallite, KCl.MgCl2.6H2O. Marine life, also contains iodine in their systems. For, example, sea weed contains upto 0.5 % iodine, and chile saltpetre contains upto 0.2 % of, sodium iodate. Astatine, the last member of, halogen family is radioactive and has a half, life of 8.1 hours., The elements helium (2He), neon (10Ne),, argon (18Ar), krypton (36Kr), xenon (54Xe) and, radon (86Rn) constitute the Group 18., All the noble gases except radon occur, in the atmosphere. Their abundance in dry air, is ∼ 1% (by volume) with argon as the major, constituent. The main commercial source of, helium is natural gas. Helium and neon are, found in minerals of radioactive origin e.g., pitchblende, monazite, cleveite. Xenon and, radon are the rarest elements of the group., Radon is a decay product of 226Ra., , 138
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Table 7.1 : Condensed electronic configuration of elements of group 16, 17 and 18, Group 16 (Oxygen family), Element, , Condensed, Electronic, Configuration, , Group 17 (Halogen family), Element, , Condensed, Electronic, Configuration, , Group 18 (Noble gases), , 2, , O, , [He]2s22p4, , 9, , 16, , S, , [Ne]3s23p4, , 17, , 34, , Se, , [Ar]3d104s24p4, , 35, , Te, , [Kr]4d105s25p4, , 53, , Po, , [Xe]4f145d106s26p4, , 85, , 8, , 52, 84, , Condensed, Electronic, Configuration, , Element, , He, , 1s2, , F, , [He]2s22p5, , 10, , Ne, , [He]2s22p6, , Cl, , [Ne]3s23p5, , 18, , Ar, , [Ne]3s23p6, , Br, , [Ar]3d104s24p5, , 36, , Kr, , [Ar]3d104s24p6, , I, , [Kr]4d105s25p5, , 54, , Xe, , [Kr]4d105s25p6, , At, , [Xe]4f145d106s26p5, , 86, , Rn, , [Xe]4f145d106s26p6, , 7.3 Electronic configuration of elements of, group 16, 17 and 18: : The general electronic, configuration of the group 16 elements is, ns2np4 while that of group 17 elements is, ns2np5. The group 18 elements are shown by, ns2np6 configuration., The elements of groups 16 and 17, repectively have two and one electrons less, than the stable electronic configuration of the, nearest noble gas., Table 7.1 shows the condensed electronic, configuration of the elements of group 16, 17, and 18., 7.4 Atomic and physical properties of, elements of group 16, 17 and 18., 7.4.1 Atomic properties of Group 16, 17, and 18 elements : These properties are given, in Tables 7.2, 7.3 and 7.4., i. Atomic and Ionic radii : In group 16,, 17 and 18 atomic and ionic radii increase, , down the group, as a result of increase in the, number of quantum shells., Across a period atomic or ionic radii decrease, with increasing atomic number, consequent, to increase in (Zeff) effective nuclear charge., Group 17 elements (Halogens) have the, smallest atomic radii in their respective, periods., ii. Ionisation enthalpy : The group 16, 17 and, 18 elements have high ionisation enthalpy., The ionisation enthalpy decreases down the, group due to increase in the atomic size., Across a period ionisation enthalpy increases, with increase of atomic number. This is due, to addition of electrons in the same shell., However the elements of group 16 have lower, ionisation enthalpy values compared to those, of group 15 in the corresponding periods,, owing to extra stable half filled electronic, configuration of p-orbitals in elements of, group 15., , Table 7.2 : Atomic and physical properties of group 16 elements., Element, , Atomic, number, , Atomic, mass, g/mol, , Atomic, radius, (pm), , Ionic, radius, E2, (pm), , Ionization, enthalpy, (∆iH1), kJ/mol, , Electronegativity, , Electron, gain, enthalpy, kJ/mol, , Density, g /cm3, , M.P., (K), , B.P., (K), , O, S, Se, Te, Po, , 8, 16, 34, 52, 84, , 16.00, 32.06, 78.96, 127.60, 210.00, , 66, 104, 117, 137, 146, , 140, 184, 198, 221, 230, , 1314, 1000, 941, 869, 813, , 3.50, 2.44, 2.48, 2.01, 1.76, , -141, -200, -195, -190, -174, , 1.32, 2.06, 4.19, 6.25, -, , 55, 393, 490, 725, 520, , 90, 718, 958, 1260, 1235, , 139
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Table 7.3 : Atomic and physical properties of group 17 elements, Element, , Atomic, number, , Atomic, mass, g/mol, , Atomic, radius, (pm), , Ionic, radius, E (pm), , Ionization, enthalpy, (∆iH1), kJ/mol, , Electro, negativity, , Electron, gain, enthalpy, kJ/mol, , Density, g/cm3, , M.P., (K), , B.P., (K), , F, Cl, Br, I, At, , 9, 17, 35, 53, 85, , 19.00, 35.45, 79.90, 126.90, 210, , 64, 99, 114, 133, -, , 133, 184, 196, 220, -, , 1680, 1256, 1142, 1008, -, , 4.0, 3.2, 3.0, 2.7, 2.2, , -333, -349, -325, -296, -, , 1.5, 1.66, 3.19, 4.94, -, , 54.4, 172.0, 265.8, 386.6, -, , 84.9, 239.0, 332.5, 458.2, -, , Table 7.4 : Atomic and physical properties of group 18 elements., Element, , Atomic, number, , Atomic, mass, g/mol, , Atomic, radius, (pm), , Ionization, enthalpy, (∆iH1), kJ/mol, , Electron, gain, enthalpy, kJ/mol, , Density, g/cm3, , M.P., (K), , B.P., (K), , Atmospheric, content, (% by volume), , He, , 2, , 4.00, , 120, , 2372, , 48, , 1.8 × 10-4, , -, , 4.2, , 5.24 × 10-4, , Ne, , 10, , 20.18, , 160, , 2080, , 116, , 9.0 × 10-4, , 24.6, , 27.1, , -, , Ar, , 18, , 39.95, , 190, , 1520, , 96, , 1.8 × 10-3, , 83.8, , 87.2, , 1.82 × 10-3, , Kr, , 36, , 83.80, , 200, , 1351, , 96, , 3.7 × 10-3, , 115.8, , 119.7, , 0.934, , Xe, , 54, , 131.30, , 220, , 1170, , 77, , 5.9 × 10, , 161.3, , 165.0, , 1.14 × 10-4, , Rn, , 86, , 222.00, , -, , 1037, , 68, , 9.7 × 10-3, , 202, , 211, , 8.7 × 10-6, , iii. Electronegativity : In a group (16, 17 and, 18) the electronegativity decreases down the, group., * Oxygen has the highest electronegativity, next to fluorine amongst all the elements., * Halogens have very high electronegativity., Fluorine is the most electronegative element, in the periodic table., iv. Electron gain enthalpy : In the groups 16, and 17 electron gain enthalpy becomes less, negative down the group., However in group 16, oxygen has less, negative electron gain enthalpy than sulfur, due to its small atomic size., * In group 17, fluorine has less negative, electron gain enthalpy than that of chlorine., This is due to small size of fluorine atom., * Group 18 elements (noble gases) have, no tendency to accept electrons because of, their stable electronic configuration (ns2np6), and thus have large positive electron gain, enthalpy., , 140, , -3, , Try this..., •, , Observe Table no 7.3 and, explain the trend in following, atomic properties of group 17 elements., i. Atomic size, ii. Ionisation enthalpy,, iii. electronegativity, iv. electron gain, enthalpy, •, , Oxygen has less negative electron gain, enthalpy than sulfur. Why ?, , Problem 7.1 : Elements of group 16, generally show lower values of first, ionisation enthalpy compared to the, elements of corresponding period of group, 15. Why ?, Solution : Group 15 elements have extra, stable, half filled p-orbitals with electronic, configuration (ns2np3). Therefore more, amount of energy is required to remove an, electron compared to that of the partially, filled orbitals (ns2np4) of group 16 elements, of the corresponding period.
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b. Group 17 elements (Halogen family) :, Problem 7.2 : The values of first ionisation, enthalpy of S and Cl are 1000 and 1256 kJ, mol-1, respectively. Explain the observed, trend., , Fluorine, chlorine are gases, bromine is a liquid, and iodine is a solid at room temperature. F2, is yellow, Cl2 greenish yellow, Br2 red and I2, is violet, in colour., , Solution : The elements S and Cl belong, to second period of the periodic table., , Fluorine and chlorine react with water., Bromine and iodine are only sparingly soluble, in water and are soluble in various organic, solvents such as chloroform, carbon disulfide,, carbon tetrachloride, hydrocarbons which, give coloured solutions. Bond dissociation, enthalpies of halogen molecules follow the, order : Cl - Cl > Br - Br > F - F > I - I., , Across a period effective nuclear charge, increases and atomic size decreases with, increase in atomic number., Therefore the energy required for the, removal of electron from the valence, shell (I.E.) increases in the order, S < Cl., , Problem 7.4 : Fluorine has less negative, electron gain affinity than chlorine. Why ?, Solution : The size of fluorine atom is, smaller than chlorine atom. As a result, there, are strong inter electronic repulsions in the, relatively small 2p orbitals of fluorine and, therefore, the incoming electron does not, experience much attraction. Thus fluorine, has less negative electron gain affinity than, chlorine., , 7.4.2 Physical properties of group 16, 17, and 18 elements :, a. Group 16 elements (Oxygen family or, chalcogens) : Oxygen is a gas while other, elements are solids at room temperature., Oxygen and sulfur are nonmetals, selenium, and tellurium are metalloids, while polonium, is a metal. Polonium is radioactive with its, half life of 13.8 days., , Problem 7.5 : Bond dissociation enthalpy, of F2 (158.8 KJ mol-1) is lower than that, of Cl2 (242.6 KJ mol-1) Why ?, Solution : Fluorine has small atomic size, than chlorine. The lone pairs on each F, atoms in F2 molecule are so close together, that they strongly repel each other, and, make the F - F bond weak (fig. 7.1) Thus, it requires less amount of energy to break, the F - F bond. In Cl2 molecule the lone, pairs on each Cl atom are at a larger, distance and the repulsion is negligible., Thus Cl - Cl bond is comparitively stronger., Therefore bond dissociation enthalpy of F2, is lower than that of Cl2., , Melting and boiling points increase with, increasing atomic number., All the elements of group 16 exhibit allotropy., Problem 7.3 : Why is there a large, difference between the melting and boiling, points of oxygen and sulfur ?, Solution : Oxygen exists as diatomic, molecule (O2) where as sulfur exists as, polyatomic molecule (S8)., The van der Waals forces of attraction, between O2 molecules are relatively weak, owing to its much smaller size., The large van der Waals attractive forces in, the S8 molecules can be noticed because of, large molecular size. Therefore oxygen has, low m.p. and b.p. as compared to sulfur., , lone pair - lone pair, repulsion, F-F, , Fig. 7.1, , 141
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c. Group 18 elements (Noble gases) :, , iv. Nature of hydrides : Hydride of oxygen, (H2O) is liquid at room temperature while, hydrides of other members of the group are, gases., , Noble gases are monoatomic., They are sparingly soluble in water., Noble gases have very low melting and, boiling points. Helium has the lowest boiling, point (4.2 K) of any known substances., , Use your brain power, , Problem 7.6 : Noble gases have very low, melting and boiling points. Why ?, Solution : Noble gases are monoatomic,, the only type of inter atomic interactions, which exist between them are van der, Waals forces. Therefore, they can be, liquified at very low temperatures and, have very low melting or boiling points., , •, , Which of the following possess, hydrogen bonding ? H2S, H2O,, H2Se, H2Te, , •, , Show hydrogen bonding in the above, molecule with the help of a diagram., , v. Common covalency of oxygen is 2. In rare, cases it is four. But for the other members of, the group 16 the covalency can exceed four., The anomalous behaviour of oxygen is due, to the following reasons., , Can you tell ?, The first member of a group, usually differs in properties from, the rest of the members of the group., Why ?, , i. small atomic size, ii. high electronegativity., iii. absence of inner d-orbitals., , 7.5 Anamalous Behaviour, 7.5.1 Anomalous behaviour of oxygen :, Oxygen shows the following anomalous, properties compared to other members of, group 16 :, i. Atomicity : Oxygen is a diatomic molecule, (O2) while others are polyatomic molecules., For example P4, S8., , 7.5.2 Anomalous behaviour of fluorine :, Fluorine, the first member of group 17, differs, in properties from the other members of the, group. The anomalous behaviour of fluorine, is due to the following reasons., i. small atomic size, ii. high electronegativity, iii. absence of d-orbitals in valence shell, , ii. Magnetic property : Oxygen is, paramagnetic while others are diamagnetic., , iv. low F-F bond dissociation enthalpy, , iii. Oxidation state : Oxygen shows -2, -1,, and +2 oxidation states while other elements, show, -2, +2, +4, +6 oxidation states. Oxygen, can not exhibit higher oxidation state due to, absence of vacant d orbitals., , i. Ionisation enthalpy, electronegativity,, electrode potential are all higher for fluorine, than expected trends shown by other halogens., , Use your brain power, •, , Oxygen forms only OF2 with, fluorine while sulfur forms SF6., Explain. Why ?, , Some anomalous properties of fluorine :, , ii. Ionic and covalent radii, m.p., b.p. and, electron gain enthalpy are quite lower than, expected., iii. Most of the reactions of fluorine are, exothermic (due to the short and strong bond, formed by it with other elements)., , 142
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iv. It forms only one oxoacid (HOF) while, other halogens form a number of oxoacids., v. Hydrogen fluoride is a liquid (b.p. 293K), due to strong hydrogen bonding while other, hydrogen halides are gases., 7.6 Chemical Properties of elements of, groups 16, 17 and 18, 7.6.1 Oxidation state : i. The group 16, elements have the valence shell electronic, configuration ns2np4. They attain a noble gas, configuration either by gaining two electrons,, forming E2 ions or by sharing two electrons,, forming two covalent bonds. These elements,, thus, show -2 and +2 oxidation states in their, compounds., Oxygen being highly electronegative,, shows common oxidation state of -2 except, two cases. In the case of OF2, its oxidation, state is +2 and in peroxides, it shows oxidation, state -1 (H2O2, Na2O2). Other elements of the, group exhibit +2, +4, +6 oxidation states with, +4 and +6 being more common. The stability, of higher (+6) oxidation state decreases down, the group while the stability of the lower, oxidation state (+4) increases down the group, due to inert pair effect. Bonding in +4 and +6, oxidation states are primarily covalent., Try this..., Complete the following tables, Element, , O, , O, , S, , F, , compound, , H 2O, , OF2, , H2S, , HF, , Oxidation, state, Element, compound, , -2, Se, SeO3, , Te, TeF6, , Cl, HOCl, , Oxidation, state, , Se, SeO2, , All halogens exhibit -1 oxidation state., However Cl, Br and I exhibit +1, +3, +5, and +7 oxidation states as well. This is, because they are less electronegative than F, and possess empty d-orbitals in the valence, shell and therefore, can expand the octet. The, oxidation states +4 and +6 occur in the oxides, and oxoacids of Cl and Br., The fluorine atom has no d - orbitals, in its valence shell and therefore cannot, expand its octet. Thus fluorine being most, electronegative exhibits mostly -1 oxidation, state., iii. Group 18 elements (noble gases), have stable valence shell electronic, configuration ns2np6 with completely filled, orbitals. Thus they have no tendency to, gain or lose electrons, that is, they are zero, valent and mostly exist as monoatomic gases., However, xenon has large atomic size and, lower ionisation enthalpy compared to He,, Ne, Ar and Kr. Hence xenon exhibits higher, oxidation states. Its outermost shell has, d-orbitals. The paired electrons of the valence, shell can be unpaired and promoted to empty, d-orbitals. The unpaired electrons are shared, with fluorine or oxygen atoms and covalent, compounds showing higher oxidation state, such as XeF2(+2), XeF4 (+4), XeF6 (+6), XeO3, (+6) and XeOF4 (+6) are formed., 7.6.2, Chemical, hydrogen:, , Reactivity, , towards, , i. Group 16 elements : The elements of group, 16 react with hydrogen to form hydrides of, the type H2E. (Where E = O, S, Se, Te, Po)., For example, H2O, H2S, H2Se, H2Te and H2Po., Some properties of hydrides of group 16 are, given in Table 7.5., , +6, , ii. The group 17 elements are represented, by their valence shell electronic configuration, as ns2np5. They attain noble gas configuration, either by gaining one electron forming E, ions or by sharing one electron forming one, covalent bond., , H2O is a colourless, odourless liquid, while, H2S, H2Se, H2Te and H2Po are colourless, bad smelling, poisonous gases at ambient, conditions., All hydrides have angular structures which, involve sp3 hybridisation of central atom (E)., , 143
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The hydrides of group 16 elements are weakly, acidic. The acidic character of the hydrides, increases, while thermal stability decreases, from H2O to H2Te. This is due to decrease, in the bond dissociation enthalpy of the H-E, bond down the group (Table 7.5)., , Table 7.6 Properties of hydrides of group, 17 elements., , All hydrides except H2O possess reducing, property which increases in the order H2S <, H2Se < H2Te., ii. Group 17 elements : The elements of, group 17 react with hydrogen to give hydrogen, halides., H2 + X2, (Where X = F, Cl, Br, I), , 2HX, , Some of the properties of hydrogen halides, are given in Table 7.6., Acidic strength of halogen acids increases in, the order :, HF < HCl < HBr < HI, It is due to decreasing bond dissociation, enthalpy of H-X bond in the order, HF > HCl > HBr > HI., Thermal stability of hydrogen halides, decreases in the order HF>HCl>HBr>HI. It is, due to decrease in bond dissociation enthalpy, of H-X bond down the group., Table 7.5 Properties of hydrides of group, 16 elements, , Property, , HF, , HCl, , HBr, , HI, , m.p (K), , 190, , 159, , 185, , 222, , b.p (K), , 293, , 189, , 206, , 238, , Bond length, (H-X) pm, , 91.7, , 127.4, , 141.4, , 160.9, , ∆dissH0, kJ/mol, , 574, , 432, , 363, , 295, , pKa, , 3.2, , -7.0, , -9.5, , -10.0, , iii. Group 18 elements (Noble gases) :, Noble gases are chemically inert towards, hydrogen due to their stable electronic, configuration., 7.6.3 Reactivity towards oxygen :, i. Group 16 elements : All the elements of, group 16 form oxides of the type EO2 and, EO3 where E = S, Se, Te, Po., EO2 type oxides, Ozone (O3) and sulfur, dioxide (SO2) are gases, while selenium, dioxide (SeO2) is solid. They are acidic in, nature and react with water to form acids., SO2 + H2O, SeO2 + H2O, , H2SO3 (Sulfurous acid), H2SeO3 (Selenious acid), , Reducing property of dioxides decreases from, SO2 to TeO2. SO2 is reducing while TeO2, serves as an oxidising agent., , Property, , H2O, , H2S, , H2Se, , H2Te, , m.p (K), , 273, , 188, , 208, , 222, , EO3 type oxides, SO3, SeO3, TeO3 are also, acidic in nature. They dissolve in water to, form acids., , b.p (K), , 373, , 213, , 232, , 269, , SeO3 + H2O, TeO3 + 3H2O, , H-E, Bond length, (pm), , 96, , 134, , 146, , 169, , ∆dissH (H-E), kJ/mol, , 463, , 347, , 276, , 238, , ∆fH kJ/mol, , -286, , -20, , 73, , 100, , HEH angle (°), , 104, , 92, , 91, , 90, , pKa, , 14.0, , 7.0, , 3.8, , 2.6, , H2SeO4 (Selenic acid), H6TeO6 (Telluric acid), , ii. Group 17 elements : Elements of group 17, (Halogens) form many oxides with oxygen,, but most of them are unstable., Fluorine forms two oxides OF2 and O2F2., However, only the OF2 is thermally stable at, 298 K. Both are strong fluorinating agents., O2F2 oxidises plutonium to PuF6 and the, reaction is used in removing plutonium as, PuF6 from spent nuclear fuel., , 144
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Chlorine oxides, Cl2O, ClO2, Cl2O6 and Cl2O7, are highly reactive oxidising agents and tend, to explode., ClO2 is used as bleaching agent for paper, pulp and textiles and in water treatment., Bromine oxides, Br2O, BrO2, BrO3 are the, least stable halogen oxides (middle row, anomally). They are very powerful oxidising, agents., Iodine oxides, I2O4, I2O5 and I2O7 are insoluble, solids and decompose on heating. I2O5 is a, very good oxidising agent and used for the, estimation of carbon monoxide., The higher oxides of halogens are more stable, than the lower ones., iii. Group 18 elements : Noble gas elements, are chemically inert and do not directly react, with oxygen., , Internet my friend, Find and draw the structures of, SeF4 and SCl2., ii. Group 17 elements :, Halogens (Group 17 elements) combine, amongst themselves to form a number of, compounds known as interhalogen compounds., These are of following types : XX', XX'3,, XX'5, XX'7,, Where X is the halogen atom with larger size, and X', is the halogen atom with smaller size., More details of interhalogen compounds are, included in section 7.12., iii. Group 18 elements : Group 18 elements, (Noble gases) are chemically inert. Krypton, and xenon, however react directly with, fluorine to give their fluorides. For example,, , 7.6.4 Reactivity towards halogens :, , Xe(g) + F2(g), , 673K, 1 atm, , XeF2(s), , i. Group 16 elements : Elements of group 16, react with halogens to give a large number, of halides of the types EX6 , EX4 and EX2., (Where E = S, Se, Te), , Xenon fluorides XeF2, XeF4 and XeF6 are, crystalline and colourless which sublime, readily at 298 K. They are powerful, fluorinating agents., , Hexahalides, SF6, SeF6 and TeF6 are formed by, direct combination. They are colourless gases., They have sp3d2 hybridisation and possess, octahedral structure. SF6 is exceptionally, stable halide for steric reasons., , 7.6.5 Reactivity towards metals :, , Stability of halides decreases in the order, fluorides > chlorides > bromides > iodides, , e.g., , Tetrahalides, SF4, SeF4, TeF4, TeCl4 have sp3, hybridisation and thus trigonal bipyramidal, geometry with one equatorial position, occupied by a lone pair., , i. Group 16 elements : Elements of group, 16 react with metals to form corresponding, compounds., 4Al + 3O2, Cu + S, Mg + Se, , 2Al2O3, CuS, MgSe, , magnesium selenide, , Dihalides, SCl2, SeCl2, TeCl2 have sp3, hybridisation and thus possess tetrahedral, structure with two equatorial positions, occupied by lone pairs., Monohalides are dimeric in nature. For, example, S2F2, S2Cl2, Se2Cl2 and SeBr2. These, dimeric halides undergo disproportionation., 2 Se2Cl2, SeCl4 + 3Se, , 145, , Do you know ?, Tellurium has the unusal property, of combining with gold metal to, form telluride., 2Au + 3Te, , Au2Te3, (gold telluride)
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Internet my friend, , Do you know ?, The photocopying process., , www.chemistry.explained.com, , A selenium-coated rotating, drum is given a uniform positive charge, (step 1) and is then exposed to an image, (step 2). Negatively charged toner particles, are attracted to the charged area of the, drum (step 3) and the image is transferred, from the drum to a sheet of paper (step, 4). Heating then fixes the image and the, drum is flooded with light and cleaned to, ready the machine for another cycle (step, 5). Figure of photocopying process using, Se is as shown below :, , ii. Group 17 elements :, Elements of group 17 (Halogens) react with, metals instantly to give metal halides., 2Na (s) + Cl2 (l), , 2NaCl (s), , Mg (s) + Br2 (l), , MgBr2 (s), magnesium bromide, , Ionic character of halides decreases in the, order MF > MCl > MBr > MI, where M is a, monovalent metal., The metal halides having metals in their higher, oxidation states are more covalent than the, ones having metals in lower oxidation state., For example, SnCl4, PbCl4, SbCl5 and UF6 are, more covalent than SnCl2, PbCl2, SbCl3 and, UF4 respectively., iii. Group 18 elements : Noble gases do not, directly react with metals., 7.7 Allotropy :, , • Tellurium exists in two allotropic forms (i), crystalline and (ii) amorphous., , Can you tell ?, •, , What is allotropy ?, , •, , What is the difference between, allotropy and polymorphism ?, , • Polonium reveals two allotropic forms α, and β (both metallic)., 7.7.1 Allotropes of sulfur :, , Elements of the group 16 exhibit, allotropy. Oxygen has two allotropes O2 and, O3 (ozone). Sulfur exists in a number of, allotropic forms. Rhombic and monoclinic, sulfur are the important allotropes of sulfur., Both are non metallic., , Sulfur exhibits numerous allotropic, forms. However rhombic sulfur (α- sulfur), and monoclinic sulfur (β - sulfur) are the most, important allotropes of sulfur (Table 7.7)., , • Selenium exists in two allotropic forms red, (non metallic) and grey (metallic)., Do you know ?, Grey selenium allotrope of is a, photoconductor used in photocells., , 146, , Problem 7.7 : Which form of sulfur shows, paramagnetic behaviour ?, Solution : In vapour state, sulfur partly, exists as S2 molecule, which has two, unpaired electrons in the antibonding, π* orbitals like O2. Hence it exhibits, paramagnetism.
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Table 7.7 Allotropes of sulfur, , Rhombic Sulfur (α - Sulfur), Pale Yellow, Orthorhombic crystals, 385.8 K, 2.069/ cm3, Insoluble in water and soluble in, CS2, Stable below 369 K and transforms, to β - Sulphur above this, temperature., S8 molecules having puckered ring, structure, , Color, Shape, M. P., Density, Solubility, Stability, Structure, , Method of, preparation, , It is prepared by evaporation of, roll sulphur in CS2., , Soluble in CS2, Stable above 369 K and transforms, into α - Sulphur below this, temperature., S8 molecules with puckered ring, structure, Rhomic sulphur melted in a dish, and cooled till crust is formed. Two, holes are made in the crust and, remaining liquid is poured out to, give needle shaped crystals of, β - Sulphur, , 7.8 Oxoacids, , Remember..., , Several modifications of sulfur, containing 6-20 sulfur atoms per ring, have, been synsthesised. In the S8 molecule the, ring is puckered and has a crown shape., In cyclo - S6, the ring adopts the chair, form. At elevated temperature (~ 1000 K),, S2 is the dominant species which like O2, is paramagnetic., S, , Monoclinic Sulfur (β - Sulfur), Bright yellow solid, Needle shaped monoclinic crystals, 393 K, 1.989 / cm3, , S, , 2.04, pm S, , S, , S, , Some important oxoacids of sulfur and, their structures are given below., i. Sulfurous acid, H2SO3, , HO, HO, , S, , S, , 107°, , 7.8.1 Oxoacids of sulfur : Sulfur forms, a number of oxoacids. Some of them are, unstable and cannot be isolated. They are, known to exist in aqueous solutions or in the, form of their salts., , S, , ii. Sulfuric Acid, H2SO4, , S, , O, , Structure of S8 ring in rhombic sulfur, HO, , S, S, S, , S, , O, HO, iii. Di or pyrosulfuric acid, H2S2O7, , 205.7°c S, , 02.2°, , O, , S, S, , O, , Chair form of S6, , O, , S, OH, , 147, , O, O O, , S, HO, , O
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7.9.2 Simple Oxides : A binary compound, of oxygen with another element is called an, oxide., , The layer of ozone protects the earth’s surface, from harmful ultraviolet (U.V) radiations., Hence, it is called as 'ozone umbrella'., , Oxides can be classified into, , a. Preparation of Ozone : Ozone is prepared, in the laboratory by passing silent electric, discharge through pure and dry oxygen in an, apparatus called ozoniser. As the conversion, of oxygen to ozone is only 10%, the product, is known as ozonised oxygen. It is an, endothermic process., , a. Acidic oxides, , b. Basic oxides, , c. Amphoteric oxides d. Neutral oxides, a. Acidic oxides : An oxide which dissolves, in water to give an acid or reacts with a, base to give a salt is called acidic oxide. For, example, SO2, SO3, CO2, N2O5, Cl2O7 etc., SO2 + H2O, , H2SO3, , SO3 + 2NaOH, , 3O2(g), , Generally, oxides of nonmetals are acidic, oxides., b. Basic oxides : An oxide which dissolves in, water to give a base or reacts with an acid to, give salt is called basic oxide. For example,, , i. Pure Ozone is a pale - blue gas, dark blue, liquid and violet - black solid., ii. Ozone has a characteristic smell. When, inhaled in concerntration above 100 ppm, it, causes nausea and headache., iii. It is diamagnetic in nature., , Na2O, CaO, BaO etc., , Problem 7.9 : High concerntration of ozone, can be dangerously explosive. Explain., , Ca(OH)2, , BaO + 2HCl, , BaCl2 + H2O, , Solution : i. Thermal stability : Ozone is, thermodynamically unstable than oxygen, and decomposes into O2. The decomposition, is exothermic and results in the liberation, of heat (∆H is -ve) and an increase in, entropy (∆S is positive). This results in, large negative Gibbs energy change (∆G)., Therefore high concerntration of ozone can, be dangerously explosive., , c. Amphoteric oxides : The oxide which, reacts with a base as well as with an acid to, give salt is called an amphoteric oxide. For, example, Al2O3, Al2O3 + 6NaOH(aq) + 3H2O(l), , 2Na3[Al(OH)6], , (Acidic), , Al2O3(s)+ 6HCl(aq)+ 9H2O(l), , 2[Al(H2O)6]3⊕, , Eq. O3, , (Basic), , + 6Cl, , (aq), , d. Neutral oxides : The oxides which are, neither acidic nor basic, are called as neutral, oxides. For example, CO, NO, N2O etc., 7.9.3 Ozone : Ozone (O3) is an allotrope, of oxygen. Oxygen in the upper atmosphere, absorbs energy in the form of ultra-violet, light and changes to atomic oxygen, which, combines with molecular oxygen to form O3., O2 U.V. light, , O+O, , O2 + O, , O3, , ∆H = +142 kJ/mol, , b. Physical properties of ozone :, , Na2SO4 + H2O, , CaO + H2O, , 2O3, , O2 + O, , c. Chemical Properties :, i. Oxidising property :, Ozone is a powerful oxidising agents as it, easily decomposes to liberate nascent oxygen., (O3, O2 + O)., Ozone oxidises lead sulfide to lead sulfate, and iodide ions to iodine., PbS(s) + 4O3(g), , PbSO4(s) + 4O2(g), , 2KI(aq) + H2O(l) + O3(g), , 150, , 2KOH(aq) +, I2(g) + O2(g)
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Ozone oxidises nitrogen oxide and gives, nitrogen dioxide., NO2(g) + O2(g), , Hence the nitrogen oxide emmited from the, exhaust systems of supersonic jet aeroplanes, can bringforth depletion of ozone layer in the, upper atmosphere., , The depletion of ozone layer has been most, pronounced in polar regions, especially, over Antarctica., , •, , Ozone depletion is a major environmental, problem because it increases the amount, of ultraviolet (UV) radiation that reaches, earth’s surface, thus causing an increase, in rate of skin cancer, eye cataracts and, genetic as well as immune system damage, among people., , ii. Bleaching property : Ozone acts as a good, bleaching agent due to its oxidising nature., O3, O + O2, Coloured matter + O, colourless matter, Ozone bleaches in absence of moisture so it, is also known as dry bleach., , Do you know ?, Ozone reacts with unsaturated, compounds containing double, bonds to form addition products called, ozonides. Ozonides are decomposed by, water or dilute acids to give aldehydes, or ketones. This reaction is termed as, ozonolysis., , iii. Reducing property : Ozone reduces, peroxides to oxides., H2O + 2O2, , BaO2 + O3, , BaO + 2O2, , Try this..., , Internet my friend, , a. Ozone is used as bleaching, agent. Explain., , www.britannica.com, ozone depletion., , b. Why does ozone act as a powerful, oxidising agent ?, iv. Ozone depletion : Thinning of ozone, layer in upper atmosphere is called ozone, depletion., •, , •, , The ozone (O3) layer in the upper, atmosphere, absorbs harmful UV, radiations from the sun, thus protecting, people on the earth., Depletion of ozone layer in the upper, atmosphere is caused by nitrogen oxide, released from exhausts system of car or, supersonic jet aeroplanes., NO (g) + O3 (g), , •, , NO2 (g) + O2 (g), , Depletion (thining) of ozone layer can, also be caused by chlorofluoro carbons, (freons) used in aerosol and refrigerators, and their subsequent escape into the, atmosphere., , d. Structure of Ozone : Ozone (O3) is an, angular molecule. The two O, O bond, lengths in the ozone molecule are identical,, 128 pm and the O, O, O bond angle of, about 117°. It is a resonance hybrid of two, canonical forms., O, , O, O, , O, , O, , pm, , H2O2 + O3, , O, , 12, , 8, , e.g., , O, , 12, , NO(g) + O3(g), , •, , O, , 117°, , 8p, , m, , O, , e. Uses of ozone :, • Ozone is used for air purification at, crowded places like cinema halls, tunnels,, railways, etc., • In sterilizing drinking water by oxidising, all germs and bacteria., • For bleaching ivory, oils, starch, wax and, delicate fabrics such as silk., • In the manufacture of synthetic camphor,, potassium permanganate, etc., , 151
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7.10 Compounds of sulfur :, , iv. Reaction with Na2SO3 : Sulfur dioxide, reacts with sodium sulfite solution to form, sodium hydrogen sulfite., , 7.10.1 Sulfur dioxide, a. Preparation :, i. From sulfur : Sulfur dioxide gas can be, prepared by burning of sulfur in air., S(s) + O2 (g), , SO2 (g), , ii. From sulfite : In the laboratory sulfur, dioxide is prepared by treating sodium sulfite, with dilute sulfuric acid., Na2SO3 + H2SO4(aq), , Na2SO4+H2O(l)+ SO2(g), , iii. From sulfides : (Industrial method), Sulfur dioxide can be prepared by roasting, zinc sulfide and iron pyrites., 2ZnS(s) + 3O2(g), 4FeS2(s) + 11O2 (g), , ∆, , 2ZnO(s) + 2SO2(s), ∆, , 2Fe2O3(s) + 8SO2(g), , b. Physical properties of SO2, i. Sulfur dioxide is a colourless gas with a, pungent smell., , Na2SO3 + H2O(l) + SO2, , v. Reducing property : Sulfur dioxide acts as, a reducing agent in the presence of moisture., •, , Moist sulfur dioxide reduces ferric salts, into ferrous salts., , 2Fe3⊕ + SO2 + 2H2O, •, , 2KMnO4 + 5SO2 + 2H2O, K2SO4 +, 2MnSO4 + 2H2SO4, •, , Moist sulfur dioxide reduces halogens to, halogen acids., , I2 + SO2 + 2H2O, , S, , iv. It liquifies at room temperature under a, pressure of 2 atm and boils at 263 K., c. Chemical Properties :, i. Reaction with Cl2 : Sulfur dioxide reacts, with chlorine in the presence of charcoal, (catalyst) to form sulfuryl chloride., charcoal, , SO2Cl2(l), , ii. Reaction with O2 : Sulfur dioxide is, oxidised by dioxygen in presence of vanadium, (V) oxide to sulfur trioxide., V 2O 5, , 2SO3(g), , iii. Reaction with NaOH : Sulfur dioxide, readily reacts with sodium hydroxide solution, to form sodium sulfite., 2NaOH + SO2, , H2SO4 + 2HI, , d. Structure of SO2 : Sulfur dioxide is, angular with O, S, O bond angle of, 119.5°., , iii. SO2 is highly soluble in water and its, solution in water is called sulfurous acid., , 2SO2(g) + O2(g), , 2Fe2⊕ + SO42 + 4H⊕, , Moist sulfur dioxide decolourises acidified, potassium permangnate (VII) solution., , ii. It is poisonous in nature., , SO2(g) + Cl2(g), , 2NaHSO3, , Na2SO3 + H2O, , O, , S, O, , O, , O, , O double bond arises from dπ The S, pπ bonding. It is a resonance hybrid of two, canonical forms., e. Uses : Sulfur dioxide is used, •, •, •, •, •, •, , In refining of petroleum and sugar., In bleaching wool and silk., As an anti-chlor, disinfectant., As a preservative., In the manufacture of H2SO4, NaHSO3., Liquid SO2 is used as a solvent to dissolve, a number of organic and inorganic, chemicals., 7.10.2 Sulfuric acid, H2SO4, a. Preparation : Sulfuric acid is, manaufactured by Contact process, which, involves the following three steps., , 152
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Fig. 7.1 : Flow diagram for manufacture of Sulfuric acid, , i. Sulfur or sulfide ore (iron pyrites) on, burning or roasting in air produces sulfur, dioxide., ∆, , S(s) + O2(g), 4FeS2(s) + 11O2(g), , ∆, , iii. It freezes at 283 K and boils at 611 K., iv. It is highly corrosive and produces severe, burns on the skin., , SO2 (g), , Do you know ?, , 2Fe2O3(s) + 8SO2(g), , Sulfuric acid dissolves in water, with the evolution of a large quantity, of heat. Hence care must be taken while, preparing solution of sulfuric acid from, concentrated sulfuric acid. Concentrated, H2SO4 must be added slowly to water with, constant stirring by keeping the beaker in, water bath., , ii. Sulfur dioxide is oxidised catalytically, with oxygen to sulfur trioxide, in the presence, of V2O5 catalyst., 2SO2(g) + O2, , V2O5, , 2SO3(g), , The reaction is exothermic and reversible, and the forward reaction leads to decrease, in volume. Therefore low temperature (720K), and high pressure (2 bar) are favourable, conditions for maximum yield of SO3., iii. Sulfur trioxide gas (from the catalytic, converter) is absorbed in concentrated, H2SO4 to produce oleum., Dilution of oleum with water gives sulfuric, acid of desired concentration., SO3(g) + H2SO4, , i. Acidic Property : Sulfuric acid ionises in, aqueous solution in two steps., H2SO4(aq)+H2O(l), , , H3O⊕(aq)+HSO4 (aq), Ka > 10, , HSO 4(aq)+ H2O(l), , H3O⊕(aq) + SO42 (aq), , Ka = 1.2 × 10-2, , H 2S 2 O 7, , oleum, H2S2O7 + H2O, , c. Chemcial Properties :, , 2H2SO4, , The sulfuric acid obtained by contact process, is 96 - 98 % pure., b. Physical properties of H2SO4 :, i. Sulfuric acid is a colourless, dense, oily, liquid., , The greater value of Ka (Ka>10) means that, H2SO4 is largely dissociated into H⊕ and, HSO4 ions. Thus H2SO4 is a strong acid., ii. Reaction with metals and nonmetals, (oxidising property) : Metals and nonmetals, both are oxidised by hot, concentrated sulfuric, acid which itself gets reduced to SO2., Cu + 2H2SO4, , ii. It has a density (specific gravity) of 1.84, g/cm3 at 298 K., , 153, , (Conc.), , CuSO4 + SO2 + 2H2O
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viii. Bleaching Property : Chlorine requires, the presence of moisture (water) for bleaching., It liberates nascent oxygen from water which, is responsible for its oxidising and bleaching, property, , 7.11.2 Hydrogen Chloride : Hydrogen, chloride was prepared by Glauber in 1648, by heating common salt with concentrated, sulfuric acid. Davy in 1810 showed that it is, a compound of hydrogen and chlorine., , Cl2 + H2O, , a. Preparation : In the laboratory, hydrogen, chloride is prepared by heating sodium, chloride (common salt) with concentrated, sulfuric acid., , HOCl, , HCl + HOCl, HCl + [O], , Chlorine bleaches vegetable matter or coloured, organic matter in the presence of moisture to, colourless matter., Coloured organic matter + [O], organic matter, , NaCl + H2SO4, , Colourless, , NaHSO4 + NaCl, , 420 K, 420 K, , NaHSO4 + HCl, Na2SO4 + HCl, , e. Uses :, , HCl gas can be dried by passing it, through concentrated sulfuric acid., , Chlorine is used, , a. Physical properties of HCl, , •, , For purification (sterilizing) of drinking, water., , i. Hydrogen chloride is a colourless and, pungent smelling gas., , •, , For bleaching wood pulp required for, manufacture of paper and rayon, bleaching, cotton and textiles., , ii. It can be easily liquified to a colourless, liquid (b.p. 189 K) which freezes to a white, crystalline solid (m.p. 159 K), , •, , For extraction of metals like gold and, platinum., , iii. It is highly soluble in water., , •, , In the manufacture of dyes, drugs and, organic compounds such as CCl4, CHCl3,, DDT, refrigerants, etc., , •, , In the preparation of poisonous gases such, as phosgene (COCl2), tear gas (CCl3NO2),, mustard gas (ClCH2CH2SCH2CH2Cl)., Try this..., i. Give the reasons for bleaching, action of chlorine., ii. Name the two gases used in war., Do you know ?, Bleaching by chlorine is permanent., It bleaches cotton fabrics, wood, pulp, litmus, etc. However chlorine is not, used to bleach delicate materials such as, silk, wool etc. as it is a strong bleaching, and oxidising agent. This dual action will, damage the base material., , Chemical properties :, i. Acidic property : Hydrogen chloride is, highly soluble in water and ionises as follows, :, HCl (g) + H2O (l), , H3O⊕ (aq) + Cl (aq), Ka = 107, , The aqueous solution of HCl gas is called, hydrochloric acid. High value of dissociation, constant (Ka) indicates that it is a strong acid, in water., ii. Reaction with NH3 : Hydrochloric acid, reacts with ammonia and gives white fumes, of ammonium chloride., NH3 + HCl, NH4Cl, iii. Reaction with noble metals : When three, parts of concentrated HCl and one part of, concentrated HNO3 are mixed, aqua regia is, formed., Noble metals like gold, platinum get dissolved, in aqua regia., , 156
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In the general formula XX'n. X is, the halogen having larger size and is more, electropositive. X' is the halogen having, smaller size and more electronegativity., 7.12.1, General, characteristics, interhalogen compounds, , Use your brain power, •, , of, , 1. The compound is considered as the, halide of X. For example, ClF. Here the, halogen having larger size is chlorine, it is, more electropositive than F and hence the, interhalogen compound is named as chlorine, monofluoride. (n) is the number of atoms of, X' attached to X., As the ratio [radius of X : radius of X'], increases the value of n also increases., 2. Interhalogen compounds have even number, of atoms 2, 4, 6, 8. For example, ClF3 has 4, atoms., 3. The properties of interhalogen compounds, are generally intermediate between those of, the halogens from which they are made., 4. The central halogen exhibits different, oxidation states in different interhalogen, compounds., 5. Number of X' atoms in the compounds is, always odd., 6. They are all diamagnetic ., , Which halogen has tendency, to form more interhalogen, compounds?, , Use your brain power, • Which will be more reactive ?, a. ClF3 or ClF,, •, , Complete the table, , Formula, ClF, ClF3, , Do you know ?, XX' compounds are more reactive, than X2 or X'2. In the X-X' bond X', is more electronegative than X, while, in X2 and X'2 both atoms have same, electronegativity, hence the X-X' bond, energy is less than the X-X or X'-X' bond, energy., , Name, Chlorine monofluoride, Chlorine penta fluoride, , BrF, Bromine penta fluoride, ICl, ICl3, Table 7.10 : States of Interhalogen compounds, at 25°C, , XX', ClF, BrF, BrCl, ICl, IBr, , Use your brain power, • What will be the names of, the following compounds :, ICl, BrF., • Which halogen (X) will have maximum, number of other halogen (X' ) attached?, , b. BrF5 or BrF, , Colorless gas, Pale brown gas, Gas, Ruby red solid (α- form), Brown red soid (β - form), Black solid, XX3', , ClF3, BrF3, , Colorless gas, Yellow green liquid, , IF3, , Yellow powder, , ICl3, , Orange solid dimerises, to form (I2Cl6 having Cl bridges), XX5', , IF5, BrF5, , Colorless gas at R. T. but, solid below 77 K, Colorless liquid, , ClF5, , Colorless liquid, XX7', , IF7, , 158, , Colorless gas
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Table 7.11 Structures of some Interhalogen Compounds, Name, Structure, , Formula, ICl, , Iodine monochloride, , ClF3, , Chloride trifluoride, , BrF3, , Bromine trifluoride, , BrF5, , Bromine pentafluoride, , Cl I Cl, F, , 89°, , Cl, , F, F, , Br, F, , ClF5, , F, , Chlorine pentafluoride, , F, , Cl, F, , Table 7.12 Oxidation states of central halogen, atom in interhalogen compounds., , O.S., central, Halogen, , No. of lone, pairs of, electrons, , +7, , 0, , IF7, , +5, , 1, , ClF5, BrF5,, IF5, , +3, , 2, , ClF3, BrF3,, IF3, I2Cl6, , 3, , ClF, BrF, IF,, BrCl, ICl,, IBr, , +1, , Examples, , Linear, F, , F, , Br, , Shape, , F, F, F, , 86°, , F, F, F, F, , Bent T- shaped, Bent T- shaped, , Square Pyramidal, , Square pyramidal, , We have studied earlier in this chapter, that group 18 elements have very high, ionisation energies and due to this property, they are unreactive. Each noble gas atom, has a completely filled valence electron shell, which makes it inert., The first ionization potential decreases, down the group, hence heavier noble gases, Kr, Xe and Rn can form compounds due to, low ionization energy., Do you know ?, First true compound of noble, gas was made in 1962 by Neil, Bartelt and Lohman., , 7.13 Compounds of Xenon, , Xe + 2[PtF6], , Can you recall ?, , [XeF]⊕ [Pt2F11], , Only Xenon reacts directly with, fluorine to form Xenon fluorides., , • What is the correlation, between ionization energies, and reactivity of elements?, , Remember..., XeF2, XeF4, XeF6 are stable, fluoride of xenon., , • Trends in ionization energy down a, group., , Xenon also forms compounds with, oxygen, such as XeO3, XeOF2, XeOF4,, XeO2F2., , 160
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a. Preparation, i., , Partial hydrolysis of Xenon fluorides, yields different oxyfluorides., XeF4 + H2O, , 80 C, 0, , XeF6 + H2O, , XeOF2 + 2HF, XeOF4 + 2HF, , ii. Reaction of Xenon oxyfluoride with SiO2, or hydrolysis yields xenon dioxydifluoride., 2XeOF4 + SiO2, XeOF4 + H2O, , 2XeO2F2 + SiF4, , Use your brain power, What are the missing entries ?, Formula, Name, XeOF2, Xenon mono, oxyfluoride, Xenon dioxydifluoride, XeO3F2, XeO2F4, , XeO2F2 + 2HF, , Table 7.14 : Uses of helium, neon and argon, Element, Helium, , i., , Uses, Mixture of He and O2 is used for artificial breathing of asthma patients., , ii. Mixture of He and O2 is used for respiration by sea divers., iii. For filling balloons, a mixture of helium (85%) and hydrogen (15 %) is, used., iv. Helium is used for producing inert atmosphere required for welding, purpose and metallury of some metals., v., , Liquid helium is used for producing low temperature required for research., , vi. In low temperature gas thermometry, for production of lasers., vii. Used to pressurise fuel tanks of liquid fueled rockets., viii. Used as shielding gas for arc welding., ix. In supersonic wind tunnels., x., , Helium nucleus is used as a bombarding particle for disintegration of, atoms., xi. Used for magnetic resonance imaging., Neon, , i., , In Neon discharge lamps and signs. These signs are visible from the long, distances and also in mist or fog., , ii. Mixture of Ne and He is used in certain protective electrical devices such, as voltage stabilizers and current rectifiers., iii. For production of lasers., Argon, , iv. In fluorescent tubes., i. For producing inert atmosphere in welding and steel production., ii. Mixture of 85 % Ar and 15 % N2 is filled in electric bulb to increase life, of filament., iii. In filling fluorescent tubes and radio valves., iv. It is mixed with neon to get lights of various colors., , 162
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x., , Write the reaction of conc. H2SO4 with, sugar., , xi. Give two uses of chlorine., , ix. How is hydrogen chloride prepared from, sodium chloride ?, , xii. Complete the following., 1. ICl3 + H2O, ........ + ...... + ICl, 2. I2 + KClO3, ....... + KIO2, 3. BrCl + H2O, ....... + HCl, 4. Cl2 + ClF3, ........, 5. H2C = CH2 + ICl, ......., 6. XeF4 + SiO2, ....... + SiF4, 7. XeF6 + 6H2O, ........ + HF, 8. XeOF4 + H2O, ....... + HF, , x., , xiv. What is the oxidation state of xenon in the, following compounds., XeOF4, XeO3, XeF6, XeF4, XeF2., 3., , Answer the following., , i., , The first ionisation enthalpies of S, Cl, and Ar are 1000, 1256 and 1520 kJ/mol-1,, respectively. Explain the observed trend., “Acidic character of hydrides of group, 16 elements increases from H2O to H2Te”, Explain., , xii. What is the action of hydrochloric acid on, the following ?, a. NH3 , b. Na2CO3, xiii. Give two uses of HCl., xiv. Write the names and structural formulae, of oxoacids of chlorine., xv. What happens when, a. Cl2 reacts with F2 in equal volume at, 437 K., b. Br2 reacts with excess of F2., xvi. How are xenon fluorides XeF2, XeF4 and, XeF6 obtained ? Give suitable reactions., xvii. How are XeO3 and XeOF4 prepared ?, xviii. Give two uses of neon and argon., xix. Describe the structure of Ozone. Give two, uses of ozone., xx. Explain the trend in following atomic, properties of group 16 elements., , iii. How is dioxygen prepared in laboratory, from KClO3 ?, iv. What happens when, a. Lead sulfide reacts with ozone (O3)., b. Nitric oxide reacts with ozone., v., , Draw structures of XeF6, XeO3, XeOF4,, XeF2., , xi. What are inter-halogen compounds ? Give, two examples., , xiii. Match the following, A B, XeOF2, Xenon trioxydifluoride, XeO2F2, Xenon monooxydifluoride, XeO3F2, Xenon dioxytetrafluoride, XeO2F4, Xenon dioxydifluoride, , ii., , viii. What is the action of chlorine on the, following, a. Fe, b. Excess of NH3, , Give two chemical reactions to explain, oxidizing property of concentrated H2SO4., , i. Atomic radii ii. Ionisation enthalpy, iii. Electronegativity., 4., , Answer the following., , i., , Distinguish between rhombic sulfur and, monoclinic sulfur., , ii., , Give two reactions showing oxidising, property of concentrated H2SO4., , vi. Discuss the structure of sulfure dioxide., vii. Fluorine shows only -1 oxidation state, while other halogens show -1, +1, +3, +5, and +7 oxidation states. Explain., , iii. How is SO2 prepared in laboratory, from sodium sulfite? Give two physical, properties of SO2., iv. Describe the manufacturing of H2SO4 by, contact process., , 164
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8. TRANSITION AND INNER TRANSITION ELEMENTS, 8.2 Position in the periodic table, , Do you know ?, In which block of the modern, perodic table are the transition and, inner transition elements placed ?, , The transition elements are placed in, the periods 4 to 7 and groups 3 to 12 those, constitute 3d, 4d, 5d and 6d series (Fig.8.1)., , 8.1 Introduction, , They are placed at the centre with, s block on one side and p on the other. The, electropositivity, reactivity and other properties, show a gradual change from s block to p block, through those of the d block elements., , The transition elements belong to, d block of the periodic table. As per IUPAC, convention the transition metal atom has an, incomplete d-subshell or it give cations with, incomplete d subshell. They exhibit properties, 8.3 Electronic configuration, between those of s and p block elements. The, The 3d series begins with Sc (Z = 21) and, transition elements of the modern periodic, ends with Zn (Z=30). Argon, Ar is the noble, table appear as groups 3 to 12 or as four long, gas preceding to 3d series and its electronic, periods. The (n-1) d-orbital is comprised of, configuration is 1s2 2s2 2p6 3s2 3p6. Calcium, successively filled in each element, where ‘n’, (Z = 20) belonging to ‘s’ block of 4th period, is the ultimate or valence shell. The 3d series, has electronic configuration 1s2 2s2 2p6 3s2 3p6, is comprised of elements from scandium, 4s2. Hence 21st electron in scandium (Z = 21), (Z=21) to zinc (Z=30), 4d series has elements, enters in the available 3d orbital. Electronic, from yttrium (Z=39) to cadmium (Z=48), 5d, configuration of Sc is written as 1s2 2s2 2p6 3s2, series from lanthanum (Z=57) to mercury, 3p6 3d1 4s2 or also can be represented as [Ar], (Z=80) without those from cerium to lutecium,, 3d1 4s2., and 6d series has actinium to curium without, those fromthorium to lawrentium. The general, electronic configuration of transition elements, is (n-1)d1-10 ns1-2., Table 8.1 Shows the four transition series elements, Group, d series, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 3d, , Sc(21), , Ti(22), , V(23), , Cr(24), , Mn(25), , Fe(26), , Co(27), , Ni(28), , Cu(29), , Zn(30), , 4d, , Y(39), , Zr(40), , Nb(41), , Mo(42), , Tc(43), , Ru(44), , Rh(45), , Pd(46), , Ag(47), , Cd(48), , 5d, , La(57), , Hf(72), , Ta(73), , W(74), , Re(75), , Os(76), , Ir(77), , Pt(78), , Au(79), , Hg(80), , 6d, , Ac(89), , Rf(104) Db(105) Sg(106) Bh(107) Hs(108) Mt(109) Ds(110) Rg(111), , Cn(112), , 165
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s - block, , p - block, , 1, 1, , 18, 13, , 2, , 2, 3, , 14, , 15, , 16, , 17, , d - block, 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 4, 5, 6, 7, Series, and, f - block, , Fig. 8.1 Position of d-block elements in the modern periodic table, , The electronic configuration of the, elements of 3d series is given in Table 8.2., , 8.3.1 Electronic configuration of chromium, and copper, , Since zinc has completely filled, (n - 1)d orbital in the ground state (3d10,4s2), and (3d10) in its common oxidation state +2,, it is not regarded as transition element. On the, same ground, cadmium and mercury from 4d, and 5d series are not considered as transition, elements. Copper in the elementary state (3d10, 4s1) contains filled 3d orbitals but in the +2, oxidation state it has partly filled 3d orbital, (3d9), hence copper is a transition element., , Table 8.2 indicates that the expected, electronic configuration of chromium (Z = 24), differs from the observed configuration., , General electronic configuration of, four series of d-block elements of periodic, table can be represented as given below:, i. 3d series : [Ar] 3d1-10 4s2, ii. 4d series : [Kr] 4d1-10 5s0-2, iii. 5d series : [Xe] 5d1-10 6s2, iv. 6d series : [Rn] 6d1-10 7s2, , This can be explained on the basis of, the concept of additional stability associated, with the completely filled and half filled, subshells., Remember..., Any subshell having a half filled, or completely filled electronic, configuration has extra stability., The general electronic configuration, of the elements of the 3d series is 3d1-10 4s2, with the exceptions of Cr and Cu. The 3d and, 4s orbitals are close in energy and in order to, gain extra stability the last electron instead of, occupying 4s orbital occupies the 3d orbital, that assigns Cr the 3d5, 4s1 and Cu 3d10, 4s1, configuration., , 166
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Remember..., Electronic configuration of Cr, is [Ar] 3d5 4s1 and, Cu is : [Ar]3d10, 4s1., 8.4 Oxidation states of first transition, series, One of the notable features of transition, elements is the great variety of oxidation states, they show in their compounds. Table 8.3 lists, the common oxidation states of the first row, transition elements., Can you tell ?, Which of the first transition series, elements shows the maximum, number of oxidation states and why?, Which elements in the 4d and 5d series will, show maximum number of oxidation states?, Loss of 4s and 3d electrons progressively, leads to formation of ions. The transition, elements display a variety of oxidation states, in their compounds. Loss of one 4s electron, leads to the formation of M⊕ ion, loss of two, 4s electrons gives a M2⊕ ion while loss of, unpaired 3d and 4s electrons gives M3⊕, M4⊕, ions and so on. Some examples are as shown, in Table 8.4, , From Table 8.3 it is clear that as the number, of unpaired electrons in 3d orbitals increases,, the number of oxidation states shown by the, element also increases. Scandium has only one, unpaird electron. It shows two oxidation states, while manganese with 5 unpaired d electrons, shows six different oxidation states., The elements which give the greatest, number of oxidation states occur in or near the, middle of the series. Manganese, for example,, shows oxidation states from +2 to +7., 8.5 Physical properties of first transition, series : All transition elements are metals, and show properties that are characteristic of, metals. They are hard, lustrous, malleable,, ductile and form alloys with other metals. They, are good conductors of heat and electricity., Except Zn, Cd, Hg and Mn, all the other, transition elements have one or more typical, metallic structures at ambient temperature., These transition metals (with the exception, of Zn, Cd and Hg) are very hard and have, low volatility. They possess high melting and, boiling points., , Try this..., Write the electronic configuration, of Mn6⊕, Mn4⊕, Fe4⊕, Co5⊕, Ni2⊕, , Fig. 8.2 : Trends in melting points of transition, elements, , Table 8.4: Electronic configuration of various ions of 3d elements, Elements, Atomic no :, Species, M, , Sc, 21, , Ti, 22, , V, 23, , Cr, 24, , Mn, 25, , Fe, 26, , Co, 27, , Ni, 28, , Cu, 29, , Zn, 30, , Valence shell Electronic Configuration, , M⊕, , 3d1 4s2, 3d1 4s1, , 3d2 4s2, 3d2 4s1, , 3d3 4s2, 3d3 4s1, , 3d5 4s2, 3d5, , 3d5 4s2, 3d5 4s1, , 3d6 4s2, 3d6 4s1, , 3d1 4s2, 3d7 4s1, , 3d8 4s2, 3d8 4s1, , 3d10 4s1, 3d10 4s0, , 3d10 4s2, 3d10 4s1, , M2⊕, , 3d2, , 3d2, , 3d3, , 3d4, , 3d5, , 3d6, , 3d7, , 3d8, , 3d9, , 3d10, , M3⊕, , [Ar], , 3d1, , 3d2, , 3d3, , 3d4, , 3d5, , 3d6, , 3d7, , -, , -, , 168
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atomic radii in nm, , 8.6 Trends in atomic properties of the first, transition series, 8.6.1 Atomic and ionic radii, Atomic radii of the elements of the, transition series decrease gradually from left, to right (Fig. 8.3 and Table 8.5). As we move, across a transition series from left to right the, nuclear charge increases by one unit at a time., The last filled electron enters a penultimate, (n-1)d subshell. However, d orbitals in an, atom are less penetrating or more diffused and,, therefore d electrons offer smaller screening, effect. The result is that effective nuclear, charge also increases as the atomi number, increases along a transition series. Hence, the atomic radii decrease gradually across a, transition series from left to right., The explanation for the minor variation, in atomic radii within a particular transition, series is out of the scope of this textbook., , Fig. 8.3 : Trends in atomic radii of d block, elements, , Ionic radii of transition elements show, the same trend as of the atomic radii (Table, 8.5), The elements of first transition series, show variable oxidation states. The trends, in ionic radii, thus, can be studied with (i), elements having same oxidation state or (ii), considering various oxidation states of the, same element., (i) For the same oxidation state, with an, increase of nuclear charge a gradual decrease, in ionic radii was observed. The trend is, pronounced for the divalent ions of the first, transition series (Cr2⊕ - 82 pm, Cu2⊕ - 72 pm)., , Table 8.5 Atomic properties of first transition series elements, , Element Atomic, (M), number, (Z), Sc, 21, Ti, 22, V, 23, Cr, 24, Mn, 25, Fe, 26, Co, 27, Ni, 28, Cu, 29, Zn, 30, , Density, (g/cm3), 3.43, 4.1, 6.07, 7.19, 7.21, 7.8, 8.7, 8.9, 8.9, 7.1, , Atomic/ionic radius (pm), M, M2⊕, M3⊕, 164, 147, 135, 129, 127, 126, 125, 125, 128, 137, , 169, , 79, 82, 82, 77, 74, 70, 73, 75, , 73, 67, 64, 62, 65, 65, 61, 60, -, , Ionisation, enthalpy (kJ/mol), 631, 656, 650, 653, 717, 762, 758, 736, 745, 906
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(ii) The oxidation states of the same element, shows difference of one unit such as M⊕, M2⊕,, M3⊕, M4⊕ and so on. With higher oxidation state, the effective nuclear charge also increases and, hence, decrease of ionic radii can be observed, from M2⊕ to M3⊕ (Table 8.5). Ionic radii of, transition elements are smaller than ionic radii, of representative elements of the same period., 8.6.2 Ionisation Enthalpy : The ionisation, enthalpies of transition elements are, intermediate between those of s-block or, p-block elements. This suggests that transition, elements are less electropositive than elements, of group 1 and 2. Depending on the conditions,, they form ionic or covalent bonds. Generally in, the lower oxidation states these elements form, ionic compounds while in the higher oxidation, states they form covalent compounds., Ionisation enthalpies shown in, Table 8.6 reveal that for a given element there, is substantial increase from the first ionisation, enthalpy IE1 to the third ionisation enthalpy, IE3., As we move across the transition series,, slight variation is observed in the successive, enthalpies IE1, IE2, IE3 of these elements (Table, 8.6)., , The atoms of elements of third, transition series possess filled 4f- orbitals. 4f, orbitals show poor shielding effect on account, of their peculiar diffused shape. As a result,, the valence electrons experience greater, nuclear attraction. A greater amount of energy, is required to ionize elements of the third, transition series. The ionisation enthalpies of, the elements of the third transition series are,, therefore much higher than the first and second, series (Fig.8.4)., 8.6.3 Metallic character : Low ionization, enthalpies and vacant d orbitals in the, outermost shell are responsible for the metallic, character of the transition elements. These, favour the formation of metallic bonds and, thus these elements show typical metallic, properties. The hard nature of these elements, suggests the formation of covalent bonds in, them. This is possible due to the presence of, unpaired (n-1)d electrons in these elements., Nearly all transition metals have simple, hexagonal closed packed (hcp), cubic closed, packed (ccp) or body centered cubic (bcc), lattices which are characteristic of true, metals (You have learnt more about this in, Chapter 1)., Remember..., Hardness, high melting points, and metallic properties of the, transition elements indicate that the, metal atoms are held strongly by metallic, bonds with covalent character., , Fig. 8. 4: Trends in first ionisation enthalpies of, d block elements, Table 8.6 Ionisation enthalpies of first transition series elements, , Element Sc, IE, IE1, 632, , Ti, , V, , Cr, , Mn, , Fe, , Co, , Ni, , Cu, , Zn, , 659, , 650, , 652, , 717, , 762, , 756, , 736, , 744, , 906, , IE2, , 1245 1320 1376, , 1635, , 1513, , 1563 1647, , 1756, , 1961 1736, , IE3, , 2450 2721 2873, , 2994, , 3258, , 2963 3237, , 3400, , 3560 3838, , (IE = Ionisation Enthalpy in kJ/mol), , 170
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In all the transition series the melting, points steadily increase upto d5 configuration., Cr, Mo and W show highest melting points, in their respective series. Mn and Tc display, anomalous values of melting points. After this, with increasing atomic number the melting, point decreases regularly., 8.6.4 Magnetic Properties :, , Each unpaired electron gives rise to a, small magnetic field (magnetic moment) due, to its spin angular momentum and orbital, angular momentum. In case of the first row, transition elements, the contribution from, the orbital angular momentum is quenched, and hence, can be neglected. The spin-only, formula for magnetic moment is :, m = n(n + 2) BM, , Can you recall ?, 1. What happens when magnetic, field is applied to substances ?, 2. What is meant by the terms, paramagnetism and diamagnetism ?, The compounds of transition metals, exhibit magnetic properties due to the unpaired, electrons present in their atoms or ions. When, a magnetic field is applied, substances which, are attracted towards the applied magnetic, field are called paramagnetic, while the ones, which are repelled are called diamagnetic., Some substances are attracted very, strongly and these are called ferromagnetic, substances., , where n is the number of unpaired, electrons and m is the magnetic moment, expressed in Bohr Magneton (BM). A single, unpaired electron has magnetic moment, m = 1.73 BM., From the magnetic moment (m), measurements of the metal complexes of the, first row transition elements, the number of, unpaired electrons can be calculated, with the, use of spin-only formula. As magnetic moment, is directly related to number of unpaired, electrons, value of m will vary directly with, the number of unpaired electrons., Try this..., What will be the magnetic, moment of transition metal having, 3 unpaired electrons ?, a. equal to 1.73 BM,, b. less than 1.73 BM or, c. more than 1.73 BM ?, , Remember..., Paramagnetism, and, ferromagnetism arises due to, presence of unpaired electrons in a species., When all electron spins are paired, the, compound becomes diamagnetic., Among transition metals Fe, Co, Ni, are ferromagnetic. When magnetic field is, applied externally all the unpaired electrons in, these metals and their compounds align in the, direction of the applied magnetic field. Due to, this the magnetic susceptibility is enhanced., These metals can be magnetized, that is, they, aquire permanent magnetic moment., , Magnetic moments are determined, experimentally in solution or in solid state, where the central metal is hydrated or bound, to ligands. A slight difference in the calculated, and observed values of magetic moments thus, can be noticed., Use your brain power, A metal ion from the first transition, series has two unpaired electrons., Calculate the magnetic moment., , Try this..., Pick up the paramagnetic species, Cu1⊕, Fe3⊕, Ni2⊕, Zn2⊕, Cd2⊕, Pd2⊕., , Table 8.7 gives the calculated and, observed magnetic moments of cations of 3d, series., , 171
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Table 8.7 Magnetic moments of ions of first transition series elements (values in BM), , Ion, , Outer electronic, configuration, , Sc3⊕, Ti3⊕, V3⊕, Cr3⊕, Cr2⊕, Mn2⊕, Fe2⊕, Co2⊕, Ni2⊕, Cu2⊕, Zn2⊕, , 3d0, 3d1, 3d2, 3d3, 3d4, 3d5, 3d6, 3d7, 3d8, 3d9, 3d10, , Number of, unpaired, electrons, 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, , Problem : Calculate the spin only magnetic, moment of divalent cation of a transition, metal with atomic number 25., , Experimental, value, 0, 1.75, 2.76, 3.86, 4.80, 5.96, 5.3-5.5, 4.4-5.2, 2.9-3.0, 4.0, 1.8-2.2, 0, , Can you tell ?, Compounds of s and p block, elements are almost white. What, could be the absorbed radiation : uv or, visible ?, , Solution : For element with atomic number, 25, electronic configuration for its divalent, cation will be, 3d, , Calculated value, of magnetic, moment, 0, 1.73, 2.84, 3.87, 4.90, 5.92, 4.90, 3.87, 2.84, 1.73, 0, , 4s, , [Ar], There are 5 unpaired electrons, so n = 5., ∴ m = 5(5 + 2) = 5.92 BM, Try this..., Calculate the spin only, magnetic moment of divalent cation, of element having atomic number 27., In second and third transition series,, orbital angular moment is significant., Therefore, the simple spin only formula is not, useful and more complicated equations have to, be employed to determine magnetic moments., The magnetic moments further are found to be, temperature dependent., 8.6.5 Colour : A substance appears coloured if, it absorbs a portion of visible light. The colour, depends upon the wavelength of absorption in, the visible region of electromagnetic radiation., , The ionic and covalent compounds, formed by the transition elements are coloured., Transition elements contain unpaired electrons, in their d orbitals. When the atoms are free or, isolated, the five d orbitals are degenerate;, or have the same energy. In complexes, the, metal ion is surrounded by solvent molecules, or ligands. The surrounding molecules affect, the energy of d orbitals and their energies are, no longer the same [You will learn more about, this in Chapter 9]. As the principal quantum, number of ‘d’ orbitals is the same, the amount, of energy required for transition of electron, from one d orbital to another is quite small., The small energy required for this transition, is available by absorption of radiation, having certain wavelength from the visible, region. Remaining light is transmitted and the, observed colour of the compound corresponds, to the complimentary colour of light absorbed., That means, if red light is absorbed then the, transmitted light contains excess of other, colours in the spectrum, in particular blue, so, , 172
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the compound appears blue. The ions having, no unpaired electrons are colorless for example, Cu⊕(3d10); Ti4⊕(3d0). Table 8.8 enlists colours, of 3d transition metal ions., 620 nm, 800 nm, 400 nm, , red, , violet, , blue, , orange, , 580 nm, , Thus the colour of a transition metal ion, relates to, 1. presence of unpaired d electrons, 2. d - d transitions, 3. nature of ligands attached to the metal ion, 4. geometry of the complex formed by the, metal ion, Do you know ?, , yellow, , green, , 560 nm, , 430 nm, 490 nm, , Let us see how colour of the transition, metal ion depends upon ligand and geometry, of the complex formed by metal ion., When cobalt chloride (Co2⊕) is dissolved, in water, it forms a pink solution of the, complex [Co(H2O)6]2⊕ which has octahedral, geometry. But when this solution is treated, with concentrated hydrochloric acid, it turns, deep blue. This change is due to the formation, of another complex [CoCl4]2 which has a, tetrahedral structure., , Colour of transition metal ions, may arise due to a charge transfer., For example, MnO4 ion has an intense purple, colour in solution. In MnO4 , an electron is, momentarily transferred from oxygen (O) to, metal, thus momentarily changing O2 to O, and reducing the oxidation state of manganese, from +7 to +6. For charge transfer transition, to take place, the energy levels of the two, different atoms involved should be fairly close., Colours of Cr2O72 , CrO4 , Cu2O and Ni-DMG, (where DMG = dimethyl glyoxime) complex, thus can be explained through charge transfer, transitions., , 8.6.6 Catalytic Properties : Transition metals, and their compounds exhibit good catalytic, properties. They have proven to be good, homogeneous and heterogeneous catalysts., Partly because of their ability to participate in, [Co(H2O)6]2⊕ + 4Cl, [CoCl4]2 + 6H2O, different oxidation-reduction steps of catalytic, reactions., Table 8.8 Colour of 3d transition metal ions, Ion, Outer electronic configuration Number of unpairedelectrons, Colour, 3⊕, 0, Sc, 3d, 0, Colourless, 3⊕, 1, Ti, 3d, 1, Purple, 4⊕, 0, Ti, 3d, 0, Colourless, 3⊕, 2, V, 3d, 2, Green, 3⊕, 3, Cr, 3d, 3, violet, Mn2⊕, 3d5, 5, Light pink, 3⊕, 4, Mn, 3d, 4, Violet, 2⊕, 6, Fe, 3d, 4, Pale green, 3⊕, 5, Fe, 3d, 5, Yellow, 2⊕, 7, Co, 3d, 3, Pink, 2⊕, 8, Ni, 3d, 2, Green, 2⊕, 9, Cu, 3d, 1, Blue, ⊕, 10, Cu, 3d, 0, Colourless, 2⊕, 10, Zn, 3d, 0, Colourless, , 173
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These steps involve changes in the, oxidation states of these metal ions., Compounds of Fe, Co, Ni, Pd, Pt, Cr, etc. are, used as catalysts in a number of reactions., Their compounds enhance the rate of the, chemical reactions., In homogeneous catalysis reactions, the, metal ions participate by forming unstable, intermediates. In heterogeneous catalysis, reactions on the other hand, the metal provides, a surface for the reactants to react., Examples :, 1. MnO2 acts as a catalyst for decomposition, of KClO3., 2. In manufacture of ammonia by Haber’s, process Mo/Fe is used as a catalyst., 3. Co-Th alloy is used in Fischer Tropsch, process in the synthesis of gasoline., 4. Finely divided Ni, formed by reduction of, the heated oxide in hydrogen is an extremely, efficient catalyst in hydrogenation of ethene, to ethane at 140 0C., H2C = CH2 + H - H, , ∆, , Ni, 140 0C, , formed are called interstitial compounds., Sometimes sulphides and oxides are also, trapped in the crystal lattice of transition, elements. Steel and cast iron are examples of, interstitial compounds of carbon and iron. Due, to presence of carbon, the malleability and, ductility of iron is reduced while its tenacity, increases., Some properties of interstitial compounds, i. They are hard and good conductors of heat, and electricity., ii. Their chemical properties are similar to the, parent metal., iii. Their melting points are higher than the, pure metals., iv. Their densities are less than the parent, metal., v. The metallic carbides are chemically inert, and extremely hard as diamond., vi. Hydrides of transition metals are used as, powerful reducing agents., Remember..., • Tungsten carbide is used for, cutting tools., , H3C - CH3, , Commercially, hydrogenation with nickel as, catalyst is used to convert inedible oils into, solid fat for the production of margarine., 5. In the contact process of industrial, production of sulfuric acid; sulphur dioxide, and oxygen from the air react reversibly, over a solid catalyst of platinised asbestos., 2SO2 + O2, , platinised, asbestos, , (steam), , Fe-Cr, catalyst, , CO2 + H2, , 8.6.7 Formation of interstitial compounds, When small atoms like hydrogen, carbon, or nitrogen are trapped in the interstitial spaces, within the crystal lattice, the compounds, , Iron carbide is used in manufacture, of steel., , 8.6.8 Formation of Alloys, Can you recall ?, • What is an alloy ?, , 2SO3, , 6. Carbon dioxide and, hydrogen are, formed by reaction of the carbon monoxide, and steam at about 500 0C with an Fe-Cr, catalyst., CO + H2O, , •, , •, , Do atomic radii of 3d transition, elements differ largely ?, , Transition metals form alloys where, atoms of one metal are distributed randomly, in the lattice of another metal. The metals with, similar radii and similar properties readily, form alloys., Alloys are classified into ferrous and nonferrous., , 174
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Ferrous alloys have atoms of other elements, distributed randomly in atoms of iron in the, mixture. As percentage of iron is more, they, are termed ferrous alloys eg. nickel steel,, chromium steel, stainless steel etc. All steels, have 2% carbon., Non-ferrous alloys are formed by mixing, atoms of transition metal other than iron with, a non transition element. eg. brass, which is, an alloy of copper and zinc. Ferrous and nonferrous alloys are of industrial importance., Uses of alloys, •, , Bronze, an alloy of copper and tin is tough,, strong and corrosion resistant. It is used for, making statues, medals and trophies., , •, , Cupra-nickel, an alloy of copper and nickel, is used for making machinary parts of, marine ships, boats. For example, marine, condenser tubes., , •, , Stainless steels are used in the construction, of the outer fuselage of ultra-high speed air, craft., , •, , •, , Nichrome an alloy of nickel and chromium, in the ratio 80 : 20 has been developed, specifically for gas turbine engines., Titanium alloys withstand stress up to, high temperatures and are used for ultrahigh speed flight, fire proof bulkheads and, exhaust shrouds., , 8.7 Compounds of Mn and Cr (KMnO4 and, K2Cr2O7), Remember..., Both KMnO4 and K2Cr2O7 are, strong oxidising agents., , 3K2MnO4 + 4CO2+ 2H2O, 3MnO42, , +, , 2KMnO4 +, MnO2 + 4KHCO3, , 4H⊕, , 2MnO4, +, MnO2 + 2H2O, , The liquid is filtered through glass, wool or sintered glass and evaporated, until crystallisation occurs. Potassium, permanganate forms small crystals which are, almost black in appearance., ii. Electrolytic oxidation, In electrolytic oxidation, alkaline solution, of manganate ion is electrolysed between iron, electrodes separated by a diaphragm. Overall, reaction is as follows :, 2K2MnO4 + H2O + [O], , 2KMnO4 + 2KOH, , The oxygen evolved at the anode converts, manganate to permanganate., The solution is filtered and evaporated to, get deep purple black coloured crystals of, KMnO4., 8.7.2 Chemical properties of KMnO4 :, a. In acidic medium :, The oxidizing reactions of KMnO4 in acidic, medium, i. Oxidation of iodide to iodine :, + 10 I, , + 16H⊕, , 2Mn2⊕ +, 8H2O + I2, , ii. Oxidation of Fe2⊕ to Fe3⊕, MnO4 + 5Fe2⊕ + 8H⊕, , i. Chemical oxidation, When a finely divided manganese dioxide, (MnO2) is heated strongly with fused mass, of caustic potash (KOH) and an oxidising, agent, potassium chlorate (KClO3), dark green, potassium manganate, K2MnO4 is formed., , 3K2MnO4 +, KCl + 3H2O, , In neutral or acidic medium the green potassium, manganate disproportionates to KMnO4 and, MnO2., , 2MnO4, , 8.7.1 Preparation of potassium permaganate, , ∆, , 3MnO2 + 6KOH + KClO3, , 5Fe3⊕ + Mn2⊕, +4H2O, , iii. Oxidation of H2S, H2S, , 175, , 2H⊕ + S2, , 5S2 + 2MnO4 + 16H⊕, , 2Mn2⊕ + 5S + 8H2O
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Table 8.9 : List of minerals and ores of some, transition metals., , Differences : Although most properties, exhibited by d block elements are similar, the, elements of first row differ from second and, third rows in stabilization of higher oxidation, states in their compounds., , Metals, Mineral, Ore, Haematite, Iron, Haematite Fe2O3, Magnetite Fe3O4, Limonite 2Fe2O3, 3H2O, Iron pyrites FeS2, Siderite FeCO3, Chalcopyrite, Copper Chalcopyrite CuFeS2, Chalcocite, Chalcocite, Cuprite Cu2O, , For example, Mo(V) and W (VI) compounds, are more stable than Cr(VI) and Mn (VIII)., Highest oxidation state for elements of first row, is +7, and in the case of 3rd row +8 oxidation, state as in (RuO4) and (OsO4)., , Zinc, , Can you recall ?, •, , How are metals found in nature?, , •, , Name two the salts of metals that, are found in nature., , Zinc blende ZnS, Zinccite ZnO, Calamine ZnCO3, , Zinc blende, , Metallurgy, Pyrometallurgy, , Internet my friend, 1. Collect the information on, different steps involved in the, extraction of metals from their ores., 2. Collect information about place where, deposits of iron ores are found., 8.9 Extraction of metals, Most metals are found in the earth’s, crust in the form of their salts, such as, carbonates, sulphates, sulphides and oxides., A few metals are nonreactive and occur in, the free state in the earth’s crust, for example,, silver, gold, and platinum., Mineral : A naturally occuring substance, found in the earth’s crust containing inorganic, salts, solids, siliceous matter etc, is called a, mineral., , Hydrometallurgy, , Electrometallurgy, , 8.9.1 Metallurgy : Commercial extraction of, metals from their ores is called metallurgy., Different methods are used for their extraction, depending on the nature of a metal and its ore., a. Pyrometallurgy: A process in which the, ore is reduced to metal at high temperature, using reducing agents like carbon, hydrogen,, aluminium, etc. is called pyrometallurgy., b. Hydrometallurgy : The process of, extracting metals from the aqueous solution, of their salts using suitable reducing agent is, called hydrometallurgy., c. Electrometallurgy : A process in which, metal is extracted by electrolytic reduction of, molten (fused) metallic compound is called, electrometallurgy., Steps Involved in Process of Extraction, , The mineral which contains high, percentage of the metal and from which the, metal can be extracted economically is called, an ore., , 177, , Do you know ?, 1. Extraction of iron has been, known to Indians since 700 BC., Indian blacksmiths also knew the, thermo-mechnical process for forging., During Archaeological studies in Harappa,, Madhya Pradesh different iron objects, belonging to middle iron age were found., 2. Famous iron pillar in Delhi is 1300 years, old and is free of rust till to date.
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Concentration : After mining the ore from the, earth’s crust it is subjected to concentration., In this step, impurities termed as gangue, are removed from the ore and the ore gets, concentrated., , Iron is extracted from haematite by its, reduction using coke and limestone. Carbon in, the limestone is reduced to carbon monoxide., Carbon and carbon monoxide together reduce, Fe2O3 to metallic iron., , The sand, mud and other unwanted, impurities which remain mixed with the ore, deposit are called gangue., , Limestone acts as flux, it combines with, the gangue material to form molten slag., Extraction of iron from haematite ore involves, the following steps., , During the process of concentration,, the ore is separated from the gangue material, using different methods such as washing,, hydraulic classification, magnetic separation,, froth floatation, etc., The method chosen for concentration, depends upon the nature of the ore., 8.9.2 Extraction of Iron from Haematite ore, using Blast furnace, Do you know ?, Iron is the fourth most, abundant element in the earth’s, crust, Composition of Haematite ore :, , {, , Fe2O3 + SiO2 + Al2O3 + phosphates, , Gangue, , Internet my friend, Find percentage of oxygen, silicon,, aluminium and iron in earth’s crust., Iron ore, , i. Concentration : The powdered ore is washed, in a powerful current of water introduced into, the hydraulic classifier., The lighter gangue particles are, separated and the concentrated ore is collected, at the bottom., ii. Roasting : The concentrated ore is heated, in a current of air. The sulfur and arsenic, impurties present in the ore get converted into, their oxides and escape as vapour. Ferrous, oxide in the ore is converted to Fe2O3., 4FeO + O2, , 2Fe2O3, , The roasted ore is converted into lumps, by sintering., iii. Reduction (Smelting) : This step is carried, out in a blast furnace. Blast furnace is a tall, cylindrical steel tower which is lined with, refractory bricks., The height of a typical blast furnace is, 25 m and its diameter varies between 5 and, 10 m. The furnace works on counter current, principle where the charge comes down and, Reduction by coke, , Crushing and grinding, , Reduction by CO, Reduction by heat, , Concentration, , Reduction, , Reduction by Al, Reduction by electrolysis, , Leaching Froth Flotation, Magnetic Gravity, separation, separation, , Refining, , Pure iron, , Liquification Distillation Oxidation Electro-refining, , 178
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Burden charging, 3Fe2O3 + CO, Fe3O4 + CO, , C + O2, , 2Fe3O4 + CO2, 3FeO + CO2, , FEO + C, , Fe + CO, , MnO + C, , Mn + CO, , P2O5 + 5C, , P2 + 5CO, , SiO2 + 2C, , Si + 2CO, , Hot blast air, CO2, CO2 + C, , 2CO, , Slag tap hole, 1/2 SiO2, 1/3CaO, 1/3Al2O3, MgO, MnO, , 2500C, , Charge distribution, Low temperature, distribution, , CO + N2, Stack, , Indirect reduction, , 12000C, Bosh, , Coke solution, direct reduction, primary slag, formation, , 17000C, Tuyeres, 15000C, , Food phenomena, , Swelling and, softening, , Effect of raceway, conditions, Pig iron tap hole Fe, C, Si, S, P,, Mn - 93% Iron, , 13500C, Hearth, , Fig. 8.5 : Blast furnace, , hot gases move up the tower. The furnace is, comprised of 3 parts - 1. Hearth, 2. Bosh and, 3. Stack, The charge contaning ore and lime, stone is introduced into the furnace through a, cup and cone arrangement. In this arrangement, the cone enables uniform distribution of charge, and the cup prevents the loss of gases. A blast, of preheated air is introduced into the furnance, below the bosh. The charge and hot air come in, contact with each other and various reactions, take place., , Chemical reactions taking place in different, zones of the blast furnace, 1. Zone of combustion : This is 5 - 10 m from, the bottom. The hot air blown through the, tuyers reacts with coke from the charge to form, CO., 1, C+, O2, CO, ∆H = -220 kJ, 2, The reaction is highly exothermic; thus, the temperature of this zone is around 2000 K., Some of the CO formed dissociates to form, finely divided carbon., , Reactions in the blast furnace : There are, different temperature zones in the blast furnace., The temperature goes on increasing from, top to bottom in the furnace. At the top, the, temperature is 500 K. Maximum temperature, of the furnace is 2000 K above the tuyers., There are 3 temperature zones in the furnace., , The hot gas rich in CO rises upwards, in the blast furnace. The charge coming down, gets heated and reacts with CO. Thus CO acts, as a fuel and also a reducing agent., , 1. Zone of combustion - Combustion of coke, with O2 in the air., , Here, the temperature is around 900 K., Fe2O3 is reduced to spongy iron by CO, , 2. Zone of reduction - Reduction of Fe2O3 to, metallic iron, 3. Zone of slag formation - Formation of slag, by reaction of gangue with limestone, , 2 CO, , 2 C + O2, , 2. Zone of Reduction (22-25 m near the top), , Fe2O3+ 3 CO, , 2Fe + 3 CO2, , some amount of Fe2O3 is reduced to iron by, carbon, Fe2O3 + 3C, , 179, , 2 Fe + 3 CO
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Table 8.9 : Summary of reactions taking place in blast furnace at different temperature zones, , Temp K, Changes taking place, 500, loss of moisture from ore, 900, Reduction of ore by CO, , Reaction, Fe2O3 + 3CO, ∆, , 2 Fe + 3CO2, , 1200, , Decomposition of lime, , CaCO3, , 1500, , Reduction of ore by C, , Fe2O3 + 3C, , 2Fe + 3CO, , Fusion of iron, , CaO + SiO2, , CaSiO3, , Slag formation, , 12CaO + 2Al2O3, , 2000, , 3. Zone of slag formation (20 m unit) :, The gangue present in the ore is converted, to slag. This slag can be used for making, road foundation. Temperature of this zone is, 1200 K. The gangue contains silica, alumina, and phosphates. A removal of this gangue is, effected by adding lime-stone in the charge,, which acts as flux. Limestone decomposes to, give CaO (quick lime), CaCO ∆ CaO + CO, 3, , 2, , CaO combines with gangue to form, molten slag of calcium silicate and calcium, aluminate., CaO + SiO2, 12 CaO + 2Al2O3, , CaSiO3, , CaO + CO2, , 4Ca3AlO3 + 3 O2, , 5. Refining : Pure iron can be obtained, by electrolytic refining of impure iron or, other methods given in the flow chart. The, choice of extraction technique is governed, by the following factors. 1. Nature of ore, 2. Availability and cost of reducing agent,, generally cheap coke is used. 3. Availability of, hydraulic power. 4. Purity of product (metal), required. 5. Value of byproducts for example,, SO2 obtained during roasting of sulphide ores, is vital for manufacture of H2SO4. Knowledge, of electrochemical series provides solutions to, many problems., Commercial forms of Iron, Iron, , 4Ca3 AlO3 + 3 O2, , 4. Zone of fusion (15 m ht) : MnO2 and, Ca3(PO4)2 present in the iron ore are reduced, to Mn and P. Some of the silica is also reduced, to Si., The spongy iron coming down in the, furnace melt absorbs impurities like C, Si, Mn,, P and S. This molten iron collects at the bottom, in the furnace. The slag which is lighter floats, on the surface of molten iron. Molten slag and, iron are collected through separate outlets., Molten iron is poured into moulds., These solid blocks are called pigs. This iron, contains about 4% of carbon. When pig iron, is remelted, run into moulds and cooled, it, becomes cast iron. The waste gases containing, N2, CO and CO2 escape through the outlet at, the top. These hot gases are used for preheating, the blast of air., , 180, , Cast, , Wrought, , Steel, , Remember..., 1., , Iron melts at a very high, temperature (1800 K). On, addition of carbon its melting, point decreases depending upon, percentage of carbon., , 2., , Mechanical properties of steel can, be modified by addition of small, amounts of suitable elements such, as manganese, chromium, sulfur,, nickel etc. These elements are, called alloying elements and steels, are alloy steels.
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Differences between cast iron, wrought iron, and steel, Cast iron, 1. Hard and, brittle, 2. Contains, 4% carbon., , Wrought, iron, 1. Very soft, , 2.Contains, less than, 0.2%, carbon, 3. Used for, 3. Used for, making pipes, making, manufacturing pipes,, automotive, bars for, stay bolts,, parts, pots,, pans, utensils engine bolts, and rivetts., , and (n-1)f orbitals are very similar and are, sensitive to electronic configurations., , Steel, , Do you know ?, , 1. Neither too, hard nor too, soft., 2. Contains, 0.2 to 2%, carbon, , Glenn Seaborg first proposed a, revised design of periodic table with, a whole new series of elements. When, he showed his design to two prominent, inorganic chemists of the time, they warned, him against publishing it. They told him, that tampering with the established periodic, table will affect his career. Seaborg went, ahead and published it. He later remarked, “I did not have any scientific reputation, so, I published it anyway”. Now we see that, elements 89 -102 (filling of 5f orbitals) fit, in Seaborg’s proposed order., , 3. Used in, buildings, infrastructure,, tools, ships,, automobiles,, weapons etc., , Do you know ?, Iron possesses a high degree, of magnetism below 1042 K. This is, known as ferromagnetism., , 8.11 Properties of f-block elements, , 8.10 Inner Transition (f-block) Elements:, Lanthanoids and Actinoids : Elements whose, f orbitals get filled up by electrons, are called, f block elements. These elements are placed, separately at the bottom of the periodic table., They are a subset of 6th and 7th periods., , i. Properties are similar to d block elements, ii. Electrons are added to f subshells of (n-2), level, iii. Placed between (n-1)d and ns block, elements, , Can you tell ?, •, •, , Why f-block elements are, called inner transition metals?, Are there any similarities between, transition and inner transition metals?, , Since f orbital lies much inside the d orbital,, in relation to the transition metals the f block, elements are called inner transition elements., These elements have 1 to 14 electrons in their, f orbital, 0 or 1 in the penultimate energy level, and 2 electrons in the outermost orbital. The, lanthanoids are characterized by gradual, filling up of 4f and actinides by the 5 f orbitals., There are 14 elements filling the f orbital in, each series. The relative energies of the nd, , Lanthanoids begin with atomic number 57 and, end at 71. Although, historically, lanthanoids, are termed as rare earth elements, they are, fairly abundant in earth’s crust. For example,, thulium is found more in abundance than silver, (4.5 x 10-5 vs 0.79 x 10-5 percent by mass). The, name rare earth elements was coined because, of difficulty in extracting them economically, in pure form from other lanthanoids having, similar chemical properties. Now, due to newer, separation methods like ion exchange resins,, the separation of these elements has become, easier and more economical., , 181
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These metals are soft with moderate, densities of about 7 g cm-3. They have high, melting (~1000 0C) and boiling points (~3000, 0, C). Similar to groups 1 and 2, lanthanoids, in the metallic state are very reactive and, resemble alkali and alkaline earth metals in, their reactivities than transition metals. For, example, they react with water to give the, metal hydroxide and hydrogen gas, 2M(s) + 6 H2O (l), , 2M(OH)3(s) + 3H2(g), , Although, the common oxidation state, for lanthanoids is +3, the +2 oxidation state, is also important. They all form stable oxides, of the type M2O3 where M is metal ion. Eu2+, and Yb2+ are the most stable dipositive metal, ions. Higher oxidation states are unusual for, lanthanoids with the only exception of cerium, which forms a stable +4 species. The energy, required to break up the metal lattice is heat, of atomization. Lanthanoids have lower heat, of atomization than transition metals. This, is because with d electrons, transition metals, are much harder and require high heat of, atomization. Europium and ytterbium have the, lowest enthalpies of vaporization and largest, atomic radii of lanthanoids, resemble barium., These two elements resemble alkaline earth, elements; they dissolve in liquid ammonia to, give blue conducting solutions., Their ionic radii decrease from 117 pm of, La to 100 pm for Lu. This is because 5f orbitals, do not shield the outer 5s and 5p electrons, effectively, leading to increase in effective, nuclear charge and decrease in the ionic size., Such large ions have higher coordination, number that varies from 6 (most common) to, 9, 10 and upto12 in some cases. For example,, hydrated lanthanum ion is a nonahydrate,, [La(H2O)9]3+., All the lanthanoids form hydroxides of, the general formula Ln(OH)3(Ln represents, any elements of lanthanoid series). These are, ionic and basic. Since the ionic size decreases, from La3+ to Lu3+, the basicity of hydroxides, , decreases. La(OH)3 is the strongest base, while Lu(OH)3 is the weakest base., Lanthanoids react with nitrogen and, halogens to give nitrides and halides of the, formulae LnN and LnX3 repectively. While, doing so, lanthanoids lose their outermost, 3 electrons to form stable compound in +3, oxidation state. When lanthanoids are heated, at elevated temperatures (~ 2800 K) with, carbon, the carbides with general formula, LnC2 are obtained., In +3 oxidation state, many of the, lanthanoids are coloured, mostly green, pink, and yellow. This is attributed to the electronic, transitions among the f orbitals. Like transition, metals the electronic spectra of lanthanoids, however, do not get affected with different, ligands., 8.12 Properties of Lanthanoids, i. Soft metals with silvery white colour and, moderate densities of ~ 7 g cm-3. Colour and, brightness reduces on exposure to air, ii. Good conductors of heat and electricity., iii. Except promethium (Pm), all are nonradioactive in nature., iv. The atomic and ionic radii decrease from, lanthanum (La) to lutetium (Lu). This is known, as lanthanoid contraction., v. Binding to water is common (i.e.) such that, H2O is often found in products when isolated, from aqueous solutions., vi. Coordination numbers usually are greater, than 6, typically 8, 9,... (up to 12 found)., vii. The, lanthanoides are strongly, paramagnetic., Gadolinium, becomes, 0, ferromagnetic below 16 C (Curie point). The, other heavier lathanoids terbium, dysprosium,, holmium, erbium, thulium, and ytterbium, – become ferromagnetic at much lower, temperatures., viii. Magnetic and optical properties are, largely independent of environment (similar, spectra in gas/solution/solid)., , 182
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Table 8.12: First, second, third and fourth, ionization enthalpies of lanthanoids in kJ/mol, , 8.12.1 Electronic configuration : The, electronic configuration of lanthanoids, is [Xe] 4f0-14 5d 0-2 6s2. This is because, 1s22s22p63s23p64s23d104p65s24d105p6 is the, electronic configuration of xenon and we, can simplify the electronic configuration, of lanthanoids by [Xe] 4f0-145d0-26s2. The, lanthanoids involve gradual filling of, f-orbitals. The energies of 5d and 4f orbitals, are very close. For lanthanum 4f is slightly, higher in energy than 5d. The lanthanum has, electronic configuration [Xe]6s25d1 and not, [Xe]6s24f1. Gadolinium (Gd) and lutetium, (Lu) have 5d1 electron to make f-orbital, half-filled and full-filled which render them, extra stability. The electronic configuration, of lanthanoids have variable occupancy in 4f, (0 to 14) orbitals. This can be noticed from, Table 8.10. Number of electrons in 6s orbitals, remains constant in the ground state. The, valence shell electronic configuration of these, elements, thus can be represented as: (n-2)f 0,214, (n-1)d0,1,2 ns2, , Lanthanoid, , The electronic distribution in different, orbitals of elements in their ground and excited, states are shown in Table 8.11., , IE1, , IE2, , IE3, , La, , 538.1 1067 1850.3, , Ce, , 528.0 1047, , 1949, , Pr, , 523.0 1018, , 2086, , Nd, , 530.0 1034, , 2130, , Pm, , 536.0 1052, , 2150, , Sm, , 543.0 1068, , 2260, , Eu, , 547.0 1085, , 2400, , Gd, , 592.0 1170, , 1990, , Tb, , 564.0 1112, , 2110, , Dy, , 572.0 1126, , 2200, , Ho, , 581.0 1139, , 2200, , Er, , 589.0 1151, , 2190, , Tm, , 596.7 1163, , 2284, , Yb, , 603.4 1175, , 2415, , Lu, , 523.5 1340, , 2022, , Problem : Which of the following will, have highest forth ionization enthalpy IE4?, La4⊕, Gd4⊕, Lu4⊕., , Try this..., , Solution : First write electronic, configuration of that element/ion. Check, for any unpaired electrons present. The, energy required for removal of that electron, will be less as compared to the energy, required to remove an electron from an, electron pair.Also compare the energies of, the orbitals occupying those electrons. It, will be easier to remove an electron from, an orbital that is lower in energy than the, one with higher in energy. First ionization, enthalpy generally decrease across the, period., , Fill in the blanks in Table 8.11., Ionization Enthalpies, Can you recall ?, , • What is ionization enthalpy?, • Some elements have variable, , oxidation states and some have only, two. Can this be justified based on, their ionization enthalpies?, , The ionization enthalpies of lanthanoids are, given in Table 8.12, , 184
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8.12.2 Oxidation state : +3 oxidation state is, common to all elements in which 2 electrons, of s- subshell and one from d or f- subshell, are removed. The 4f electrons are strongly, screened by inner electrons of 5s and 5p, electrons. Thus, 4f electrons are not involved in, the bonding. Besides these, some lanthanoids, show oxidation states +2 and +4. They are, formed in case of f0, f7 , f14 configurations or, resulting ions., , 4f subshell. Thus the 4f electrons shield each, other from the nuclear charge poorly owing to, their diffused nature. With increasing atomic, number and nuclear charge, the effective, nuclear charge experienced by each 4f, electrons increases. As a result, the whole of, 4f electron shell contracts at each successive, element., , For example : Ce4⊕ (f 0) ; Eu2⊕ and Tb4⊕ (f 7) ;, Yb2⊕ (f 14) Refer Table 8.11., 8.12.3 Colour and Spectra :, Some trivalent ions (M3⊕) are coloured in, solid state as well as in solution. The colour of, lanthanoid ion is due to f-f transitions which, correspond to energy in the visible region of, the electromagnetic spectrum. The colour of, ions having nf electrons is about the same as, those having (14 – n)f electrons. (where n is an, integer 1-13)., Ln, ion, , Pr, , 3⊕, , No. of, f-electrons, , Colour, , 4f, , green, , 2, , Fig. 8.6 : Ionic radii of lanthanoids in +3, oxidation state, , In section 8.3.4 we have learnt about, the magnetic behaviour of transition metal, complexes., Use your brain power, , (14 -n), f-electrons, =14-2 =12, , • Do you think that lanthanoid, complexes, magnetism?, , would, , show, , Tm3⊕, , 4f 12, , green, , n f-electrons, =12, , Nd3⊕, , 4f 3, , pink, , (14 -n), f-electrons, =14-3 =11, , magnetic moment of lanthanoid, complexes using the same formula, that you used for transition metal, complexes?, , Er3⊕, , 4f 11, , pink, , nf-electrons, =11, , • Calculate the spin only magnetic, , • Can you calculate the spin only, , 8.12.4 Atomic and ionic radii (Lanthanoid, Contraction) : As we move along the, lanthanoid series, there is a decrease in, atomic and ionic radii (Fig.8.6). This steady, decrease in the atomic and ionic radii is called, Lanthanoide contraction. As we move from, one element to another the nuclear charge, increases by one unit and one electron is added., The new electrons are added to the same inner, , 185, , moment of La3+. Compare the value, with that given in Table 8.13 Is it same, or different?
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For example, mixed oxide of europium and, yttrium (Eu,Y)2O3 releases an intense red, colour when bombarded with the high energy, electrons.The optoelectronics applications use, lanthanoid ions as active ions in luminescent, materials. The most notable application is the, Nd: YAG laser (Nd: YAG = neodymium doped, yttrium aluminium garnet). Erbium-doped fibre, amplifiers are significant devices in the opticalfibre communication systems. Lanthanoids, are used in hybrid cars, superconductors and, permanent magnets., , Table 8.13: Effective magnetic moments of, lanthanoids in +3 oxidation state, Ln, , Ln3+, oxidation, state, , No. of, unpaired, electrons, , Observed, magnetic, moment,, μeff B.M, , La, , 4f0, , 0, , 0, , Ce, , 4f1, , 1, , 2.3-2.5, , Pr, , 4f2, , 2, , 3.4-3.6, , Nd, , 4f, , 3, , 3, , 3.5-3.6, , Pm, , 4f4, , 4, , --, , Sm, , 5, , 4f, , 5, , 1.4-1.7, , Eu, , 4f6, , 6, , 3.3-3.5, , Gd, , 4f7, , 7, , 7.9-8.0, , Tb, , 4f, , 8, , 6, , 9.5-9.8, , Dy, , 4f9, , 5, , 10.4-10.6, , Ho, , 4f10, , 4, , 10.4-10.7, , Er, , 4f11, , 3, , 9.4-9.6, , Tm, , 4f, , 12, , 2, , 7.1-7.6, , Yb, , 4f13, , 1, , 4.3-4.9, , Lu, , 4f14, , 0, , 0, , 8.13 Applications, The lanthanoid compounds are present, in every household. It is inside the colour, television tubes. When electrons are bombarded, on certain mixed lanthanoid compounds, they, emit visible light over a small wavelength, range. Therefore, the inside surface of a, television tube or computer monitor is coated, with tiny patches of three different lanthanoid, compositions to give three colours that make, the colour image., , Promethium, , Europium, , Nobelium, , Actinium, , Mendelenium, , Brekelium, , Sebargium, , 8.14 Actinoids : The last row of elements in the, periodic table is the actinoid series. It begins, at thorium (Z =72) and ends at lawrencium, (Z=103). Most of these elements are not, found in nature. They are all radioactive and, man-made. The half-lives of the isotopes, of thorium (Th-232=1.4 x 1010 years) and, uranium (U-238=4.5 x 109 years) are so long, that these elements exist in rocks on earth., The long lived isotopes such as thorium,, protactinium, uranium, neptunium, plutonium, and americium, are studied in more details., These elements have high densities (~ 15-20, g cm-3), high melting points (~1000 oC) and, high boiling points (~3000 oC). Actinoids are, less reactive than lanthanoids. For example,, they react with hot, but not cold water to, give the hydroxide and hydrogen gas. Unlike, lanthanoids, they exhibit a range of oxidation, numbers in their compounds which varies, from +2 to +8. The most common oxidation, numbers of the actinoids are shown in Fig. 8.7., As can be seen from Fig.8.7, the most, common oxidation state of early actinoids, reflects the loss of all outer electrons which, is similar to transition metals than the, lanthanoids. A ready loss of 5f electrons by, early actinoids indicates that these electrons, are much closer in energy to 7s and 6d electrons, than the 4f electrons to 6s and 5d electrons as, in lanthanoids. All three sets of orbitals that, is 6d, 5f and 7s have similar energies. For Th,, Pa and Np difference in energy levels is small, , 186
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Internet my friend, With the help of internet find out the applications of the elements listed in, table below. Share this information with your friends, Element, Applications, Lanthanum, Ytterbium, Erbium, Praseodymium, Samarium, Promethium, so electrons occupy either 6d or 5f oribitals. In, actinoids series the 5f are orbitals appreciablly, lower in energy, thus from Pu onwards 5f shell, gets filled in a regular way., , Oxidation numbers, , The electronic configuration of actinoids is, [Rn] 5f0-14 6d0-2 7s2, where Rn is the electronic, configuration of radon. As seen from Fig. 8.7,, the most stable oxidation state in actinoids, is +3. The highest common oxidation states, of early actinoids reflect the loss of all outer, electrons which is similar to transition metals, than lanthanoids. For example, uranium has, electronic configuration of [Rn]7s25f 36d1. The, formation of +6 oxidation state corresponds, to an electronic configuration of [Rn]. Similar, to lanthanoids, loss of s and d electrons occur, before f electrons, in formation of 3+ ions. A, ready loss of 5f electrons by early actinoids, indicates that these electrons are much closer, in energy to 7s and 6d electrons than the 4f, electrons are to 6s and 5d electrons in the, lanthanoids. In turn, 5f and 6s orbitals expand, as they are partially shielded from the nuclear, , attraction by 7s electrons. As a result all three, sets of orbitals i.e. 6d, 5f and 7s have very, similar energies. The ionic radius decreases, as we move across the series which is known, as ‘Actinoid contraction’. This is attributed to, poor shielding offered by f electrons., , Fig. 8. 8 Figure depicting contraction of ionic, radii of lanthanoides and actinoids, , 8.15 Properties of Actinoids, i. Similar to lanthanoids, they appear silvery, white in colour., ii. These are highly reactive radioactive, elements, , 6, 5, 4, 3, 2, 1, Ac Th, , Pa U, , Np Pu Am Cm Bk Cf, , Es Fm Md No Lr, , Fig. 8. 7 The most common oxidation numbers of actinoids, , 187
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iii. Except promethium (Pm), all are nonradioactive in nature, , Do you know ?, Uranium is another actinoid which, is in great demand as it is used in, the nuclear reactors. One of the extraction, methods for uranium has a very interesting, chemistry. The ore containing U(IV) oxide,, UO2, is first treated with Fe(III) ion to give, U(VI) oxide, UO3, UO2 (s) + H2O(l) UO3 (s) + 2H⊕ (aq), + 2e, 3⊕, 2⊕, Fe (aq) + e Fe (aq), Addition of H2SO4 to this solution produces, uranyl sulphate containing UO22⊕ cation:, UO3 (s) + H2SO4 (aq) UO2SO4 (aq) +, H2O (l), After purification, ammonia is added to the, solution giving bright yellow precipitate of, ammonium diuranate, (NH4)2U7O7:, 2 UO2SO4 (aq) + 6 NH3 (aq) + 3 H2O (l), (NH4)2U7O7 (s) + 2 (NH4)2SO4 (aq), This yellow cake is the marketable form of, uranium!, , iv. They experience decrease in the atomic and, ionic radii from actinium (Ac) to lawrencium, (Lw), known as actinoid contraction, v. They usually exhibit +3 oxidation state., Elements of first half of the series usually, exhibit higher oxidation states., 8.16 Applications of actinoids : We have, seen that the half-lives of natural thorium, and uranium isotopes are so long that we get, very negligible radiation from these elements., We find them in everyday use. For example,, Th(IV) oxide, ThO2 with 1% CeO2 was used, as a major source of indoor lighting before, incandescent lamps came into existence only, because these oxides convert heat energy from, burning natural gas to an intense light. Even, today, there is a great demand for these lights, for outdoor camping., , Similarities and differences between lanthanides and actinoids, Similarities, Both the series show a +3 oxidation state, , Differences, Lanthanoids show a maximum oxidation state of +4, while actinoids show oxidation states of +3, +4, +5,, +6 and +7, , In both the series, the f-orbitals are filled Lanthanoids do not form complexes easily. Actinoids, gradually, have a greater tendency to form complexes with, ligands such as thio-ethers, Ionic radius of the elements in both series All lanthanoids are non-radioactive except, decreases with an increase in atomic number, promethium but actinides are radioactive in nature, The electronegativity of all the elements in Lanthanoids do not form oxocations, but actinides, both the series is low and are said to be highly form oxocations such as UO+, PuO+, NpO2+, reactive, The nitrates, perchlorates and sulphates of all Most of the lanthanoids are colourless in nature, the elements are soluble while the hydroxides, whereas the actinoids are coloured ions, fluorides and carbonates are insoluble, , 188
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Table 8.14 Electronic configuration of actinoids and their ionic radii in +3 oxidation state, Element, , Symbol, , Atomic, number, , Electronic configuration, ground state, , +3, oxidation, state, , *Atomic, radii, pm, , *Ionic radii, (Ac3⊕), pm, , Actinium, , Ac, , 89, , [Rn]5f 06d17s2, , 5f 0, , 203, , 126, , Thorium, , Th, , 90, , [Rn]5f 06d27s2, , 5f 1, , 180, , -, , Protactinium, , Pa, , 91, , [Rn]5f 6d 7s, , 5f, , 2, , 162, , 118, , Uranium, , U, , 92, , [Rn]5f 36d17s2, , 5f 3, , 153, , 118, , Neptunium, , Np, , 93, , [Rn]5f 46d17s2, , 5f 4, , 150, , 116, , Plutonium, , Pu, , 94, , [Rn]5f 6d 7s, , 5f, , 5, , 162, , 115, , Americium, , Am, , 95, , [Rn]5f 76d07s2, , 5f 6, , 173, , 114, , Curium, , Cm, , 96, , [Rn]5f 6d 7s, , 5f, , 7, , 174, , 112, , Berkelium, , Bk, , 97, , [Rn]5f 96d07s2, , 5f 8, , 170, , 110, , Californium, , Cf, , 98, , [Rn]5f 6d 7s, , 5f, , 9, , 186, , 109, , Einsteinium, , Es, , 99, , [Rn]5f 116d07s2, , 5f 10, , 186, , 98, , Fermium, , Fm, , 100, , [Rn]5f 126d07s2, , 5f 11, , 198, , 91, , Mendelevium, , Md, , 101, , [Rn]5f 6d 7s, , 5f, , 194, , 90, , Nobelium, , No, , 102, , [Rn]5f146d07s2, , 5f 13, , 197, , 95, , Lawrencium, , Lr, , 103, , [Rn]5f 6d 7s, , 5f, , 171, , 88, , 2, , 6, , 7, , 1, , 0, , 1, , 10, , 13, , 14, , 2, , 2, , 2, , 0, , 0, , 1, , 2, , 2, , 2, , 12, , 14, , Table 8.15 : Some comparison between Pre-Transition, Lanthanoids and Transition Metals, , Pre-Transition Metals, Essentially monovalent show group (n+) oxidation, state, , Lanthanoids, Essentially in (+3), oxidation state, (+2/+4 for certain, configurations), , Transition Metals, Show variable oxidation, states, , Periodic trends dominated, by effective nuclear charge, at noble gas configuration, , Lanthanoid contraction of, Ln3⊕, , Similar properties for a, given group, , Similar properties, , Substantial changes in, properties, , Always 'hard' (O, X,, N donors, preferably, negatively charged), , Always 'hard' (O, X,, N donors, preferably, negatively charged), , heavier metals, (increasingly from Fe-Cu), may show a 'soft' character, , No ligand field effects, , Insignificant ligand field, effects, , Substantial ligand field, effects, , Poor coordination, properties(C.N. determined, by size), , High coordination, numbers (C.N. determined, by size), , Coordination number 6, is typical maximum (many, exceptions), , Flexibility in geometry, , Flexibility in geometry, , Fixed geometries (ligand, field effects), , No magnetism, , Show magnetism, , Show magnetism, , 189, , Size changes of Mn⊕, less, marked
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8.17 Postactinoid elements : You have seen, that elements with atomic number greater than, 92 are called ‘Transuranium’. Elements from, atomic number 93 to 103 now are included, in actinoid series and those from 104 to 118, are called as postactinoid elements. The, postactinoid elements that are known so far, are transition metals. They are included as, postactinoids because similar to actinoid, elements, they can be synthesized in the, nuclear reactions. So far, nine postactinoid, elements have been synthesized. It is difficult, to study their chemistry owing their short halflives. For example, element 112 has a half-life, of only 2.8 x 10-4 seconds., , With half-lives of miliseconds only a little is, known about the chemistry of these elements., Rutherfordium forms a chloride, RfCl4, similar, to zirconium and hafnium in the +4 oxidation, state. Dubnium resembles to both, group 5, transition metal, niobium(V) and actinoid,, protactinium(V)., Do you know ?, Traditionally, no element was, named after a still-living scientist., This principle was put to an end with, naming the element 106 as ‘Seaborgium’., , Exercises, 1. Choose the most correct option., i., , ii., , iii., , iv., , v., , vi., , Which one of the following is, dimagnetic, a. Cr2⊕, b. Fe3⊕, c. Cu2⊕, d. Sc3⊕, Most stable oxidation state of, Titanium is, a. +2 , b. +3, , c. +4, d. +5, Components of Nichrome alloy are, are, a. Ni, Cr, Fe b. Ni, Cr, Fe, C, c. Ni, Cr, d. Cu, Fe, Most stable oxidation state of, Ruthenium is, a. +2 , b. +4, c. +8 , d. +6, Stable oxidation states for chromiom, are, a. +2, +3, b. +3, +4, c. +4, +5, d. +3, +6, Electronic configuration of Cu and, Cu+1, a. 3d10, 4s0; 3d9, 4s0, b. 3d9, 4s1; 3d94s0, , 190, , c. 3d10, 4s1; 3d10, 4s0, d. 3d8, 4s1; 3d10, 4s0, vii. Which of the following have d0s0, configuration, a. Sc3⊕, b. Ti4⊕, c. V5⊕ , d. all of the above, viii. Magnetic moment of a metal, complex is 5.9 B.M. Number of, unpaired electrons in the complex is, a. 2 , b. 3, c. 4 , d. 5, ix. In which of the following series, all the elements are radioactive in, nature, a. Lanthanides , b. Actinides, c. d-block elements, d. s-block elements, x., Which of the following sets of ions, contain only paramagnetic ions, a. Sm3⊕, Ho3⊕, Lu3⊕ , b. La3⊕, Ce3⊕, Sm3⊕, c. La3⊕, Eu3⊕, Gd3⊕ , d. Ce3⊕, Eu3⊕, Yb3⊕
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xi., , Which actinoid, other than uranium,, occur in significant amount, naturally?, a. Thorium , b. Actinium, c. Protactinium, d. Plutonium, xii. The flux added during extraction of, Iron from teamatite are its?, a. Silica , b. Calcium carbonate, c. Sodium carbonate, d. Alumina, 2. Answer the following, i What is the oxidation state of, Manganese in (i) MnO42- (ii) MnO4- ?, ii. Give uses of KMnO4, iii. Why salts of Sc3⊕, Ti4⊕, V5⊕ are, colourless ?, iv. Which steps are involved in, manufacture of potassium dichromate, from chromite ore ?, v. Balance the following equation, (i) KMnO4 + H2C2O4, MnSO4 +, K2SO4 + H2O + O2, (ii) K2Cr2O7 + KI + H2SO4, K2SO4 +, Cr2(SO4)3 + 7H2O + 3I2, vi. What are the stable oxidation states, of plutonium, cerium, manganese,, Europium ?, vii. Write probable electronic configuration, of chromium and copper., viii. Why nobelium is the only actinoid, with +2 oxidation state?, ix. Explain with the help of balanced, chemical equation, why the solution of, Ce(IV) is acidic., x. What is meant by ‘shielding of, electrons’ in an atom?, xi. The atomic number of an element is, 90. Is this element diamagnetic or, paramagnetic?, 3. Answer the following, i. Explain the trends in atomic radii of d, block elements, , 191, , ii. Name different zones in the Blast, furnace. Write the reactions taking, place in them., iii. What are the differences between cast, iron, wrought iron and steel., iv. Iron exhibits +2 and +3 oxidation states., Write their electronic configuration., Which will be more stable ? Why ?, v. Give the similarites and differences in, elements of 3d, 4d and 5d series., vi. Explain trends in ionisation enthalpies, of d block elements., vii. What is meant by diamagnetic and, paramagnetic? Give one example, of diamagnetic and paramagnetic, transition metal and lanthanoid metal., viii.Why the ground-state electronic, configurations of gadolinium and, lawrentium are different than expected?, ix. Write steps involved in metallugical, process, x. Cerium and Terbium behaves as good, oxidising agents in +4 oxidation state., Explain., xi. Europium and xtterbium behave as, good reducing agents in +2 oxidation, state explain., , Activity :, Make groups and each group, prepares a powerpoint presentation, on properties and applications of one, element. You can use your imagination, to create some innovative ways of, presenting data., You can use pictures, images, flow, charts, etc. to make the presentation, easier to understand. Don’t forget to, cite the reference(s) from where data, for presentation is collected (including, figures and charts). Have fun!
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9. COORDINATION COMPOUNDS, 9.1 Introduction : A coordination compound, consists of central metal atom or ion surrounded, by atoms or molecules. For example, a, chemotherapy drug, cisplatin, Pt(NH3)2Cl2, is, a coordination compound in which the central, platinum metal ion is surrounded by two, ammonia molecules and two chloride ions. The, species surrounding the central metal atom or, ion are called ligands. The ligands are linked, directly to central metal ion through coordinate, bonds. A formation of coordinate bond occurs, when the shared electron pair is contributed, by ligands. A coordinate bond is conveniently, represented by an arrow →, where the arrow, head points to electron acceptor. The central, metal atom or ion usually an electron deficient, species, accepts an electron pair while the, ligands serve as electron donors. Coordination, compounds having a metal ion in the centre are, common. In cisplatin two ammonia molecules, and two chloride ligands utilize their lone pairs, of electrons to form bonds with the Pt(II)., NH3, Cl Pt NH3, Cl, , ligands depending on the number of electron, donor atoms they have., 9.2.1 Monodentate ligands : A monodentate, ligand is the one where a single donor atom, shares an electron pair to form a coordinate, bond with the central metal ion. For example:, the ligands Cl , OH or CN attached to metal, have electron pair on Cl, O and N, respectively, which are donor atoms :, O-H C≡N, , Cl, , Use your brain power, Draw Lewis structure of the, following ligands and identify the, donor atom in them : NH3, H2O, 9.2.2 Polydentate ligands : A polydentate, ligand has two or more donor atoms linked, to the central metal ion. For example,, ethylenediamine and oxalate ion. Each of these, ligands possesses two donor atoms. These are, bidentate ligands., i. Ethylenediamine binds to metal using, electron pair on each of its two nitrogens., , The donor nitrogen and chlorine atoms of, the ligands are directly attached to and form, bonds with platinum., Can you recall ?, , H 2N, , CH2 - CH2, , Similarly oxalate ion (C2O4)2 utilizes, electron pair on each of its negatively charged, oxygen atoms upon linking with the metal., , What are Lewis acids and bases ?, , O, , Formation of a coordination compound, can be looked upon as the Lewis acid-base, interaction. The ligands being electron pair, donors are Lewis bases. The central metal ion, being electron pair acceptor serves as Lewis, acid., 9.2 Types of ligands : The ligands can, be classified as monodentate and polydentate, , NH2, , O, C-C, , O, O, ii., Ethylenediaminetetraacetate, ion, (EDTA)4 binds to metal ion by electron pairs, of four oxygen and two nitrogen atoms. It is a, hexadentate ligand., O, O, O - C - H2C, CH2 - C - O, N - CH2 - CH2 - N, CH2 - C - O, O - C - H2C, O, O, , 192
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9.2.3 Ambidentate ligand : The ligands which, have two donor atoms and use the electron, pair of either donor atoms to form a coordinate, bond with the metal ion, are ambidentate, ligands. For example, the ligand NO2 links to, metal ion through nitrogen or oxygen., , [Fe(CN)6]4 has charge number of -4. It can be, utilised to calculate O.S. of Fe. Thus,, charge number of complex = -4, = (O.S. of Fe + charge of ligands), = (O.S. of Fe + 6 × charge of CN ion), = (O.S. of Fe + 6 × (-1)), , O, M, , N, , or, , M, , Therefore, O.S. of Fe = -4 + 6 = +2., , O-N=O, , O, , Can you tell ?, , Similarly, SCN has two donor atoms, nitrogen and sulfur either of which links to, metal depicted as M ← SCN or M← NCS., 9.3 Terms used in coordination chemistry:, The following terms are used for describing, coordination compounds., 9.3.1 Coordination sphere : The central metal, ion and ligands linked to it are enclosed in a, square bracket. This is called a coordination, sphere, which is a discrete structural unit. When, the coordination sphere comprising central, metal ion and the surrounding ligands together, carry a net charge, it is called the complex, ion. The ionisable groups shown outside the, bracket are the counter ions. For example,, the compound K4[Fe(CN)6] has [Fe(CN)6]4coordination sphere with the ionisable K⊕, ions representing counter ions. The compound, ionizes as :, K4[Fe(CN)6], , 4K⊕ + [Fe(CN)6]4, , Try this..., , A complex is made of Co(III), and consists of four NH3 molecules, and two Cl ions as ligands. What is the, charge number and formula of complex ion ?, 9.3.3 Coordination number (C.N.) of central, metal ion : Look at the complex [Co(NH3)4Cl2]⊕., Here four ammonia molecules and two chloride, ions, that is, total six ligands are attached to, the cobalt ion. All these are monodentate since, each has only one donor atom. There are, six donor atoms in the complex. Therefore,, the coordination number of Co3⊕ ion in the, complex is six. Thus, the coordination number, of metal ion attached to monodentate ligands, is equal to number of ligands bound to it., Consider the bidentate ligand C2O42, or ethylenediamine (en). The complexes,, [Fe(C2O4)3]3 and [Co(en)3]3⊕, have three, bidentate ligands each. The total donor atoms, in three of ligands is six and the C.N. of Fe3⊕, and Co3⊕ in these complexes is six., C.N. of metal ion in a complex is the, number of ligand donor atoms directly, attached to it or the number of electron, pairs involved in the coordinate bond., , Can you write ionisation of, [Ni(NH3)6]Cl2? Identify coordination, sphere and counter ions., 9.3.2 Charge number of complex ion and, oxidation state of metal ion :, The net charge residing on the complex, ion is its charge number. It is algebraic sum, of the charges carried by the metal ion and, the ligands. The charge carried by the metal, ion is its oxidation state (O.S.). The complex, , 193, , Use your brain power, Coordination number used, in coordination of compounds in, somewhat different than that used in, solid state. Explain
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The secondary valencies for a metal ion are, fixed and satisfied by either anions or neutral, ligands. Number of secondary valencies is, equal to the coordination number., , Can you tell ?, What is the coordination, number of Co in [CoCl2(en)2]⊕,, of Ir in [Ir(C2O4)2Cl2]3⊕ and of Pt in, [Pt(NO2)2(NH3)2]?, 9.3.4 Double salt and coordination complex, Combination of two or more stable, compounds in stochiometric ratio can give two, types of substances, namely, double salt and, coordination complexes., , Postulate (iv) The secondary valencies have, a fixed spatial arrangement around the metal, ion. Two spheres of attraction in the complex, [Co(NH3)6]Cl3 are shown., , Double salt : A double salt dissociates in water, completely into simple ions. For example (i), Mohr's salt, FeSO4(NH4)2SO4.6H2O dissociates, as :, water, FeSO4(NH4)2SO4.6H2O, Fe2⊕(aq), + 2NH4⊕(aq) +2SO42 (aq), ii. Carnalite KCl.MgCl2.6H2O dissociates as, KCl.MgCl2.6H2O, , water, , K⊕(aq), +, 2⊕, Mg (aq) + 3Cl (aq), , Coordination complex : A coordination, complex dissociates in water with at least, one complex ion. For example, K4[Fe(CN)6], dissociates as the complex ion and counter ion., K4[Fe(CN)6], , 4K⊕(aq) + [Fe(CN)6]4, (counter ion) (complex ion), , 9.3.5 Werner theory of coordination, complexes : The first attempt to explain nature, of bonding in coordination compounds was, put forth by Werner. The postulates of Werner, theory are as follows., Postulate (i) Unlike metal salts, the metal in, a complex possesses two types of valencies, : primary (ionizable) valency and secondary, (nonionizable) valency., Postulate (ii) The ionizable sphere consists, of entities which satisfy the primary valency, of the metal. Primary valencies are generally, satisfied by anions., Postulate (iii) The secondary coordination, sphere consists of entities which satisfy the, secondary valencies and are non ionizable., , 194, , H 3N, , NH3, , NH3, Cl3, , Co, H 3N, , NH3, , NH3, , coordination, (secondary) sphere, , ionization, (primary) sphere, , Remember..., When a complex is brought, into solution it does not dissociate, into simple metal ions. When [Co(NH3)6], Cl3 is dissolved in water it does not give the, test for Co3⊕ or NH3. However, on reacting, with AgNO3 a curdy white precipitate of, AgCl corresponding to 3 moles is observed., Problem 9.1 : One mole of a purple coloured, complex CoCl3 and NH3 on treatment with, excess AgNO3 produces two moles AgCl., Write the formula of the complex if the, coordination number of Co is 6., Solution : One mole of the complex gives, 2 moles of AgCl. It indicates that two Cl, ions react with Ag⊕ ions. The complex has, two ionisable Cl ions. The formula of the, complex is then [Co(NH3)5Cl]Cl2., Can you tell ?, One mole of a green coloured, complex of CoCl3 and NH3 on, treatment with excess of AgNO3, produces 1 mole of AgCl. What is the, formula of the complex ? (Given : C.N. of, Co is 6)
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A complex with coordination number six has, octahedral structure. When four coordinating, groups are attached to the metal ion the, complex would either be with square planar, or tetrahedral structure., 9.4 Classification of complexes: The, coordination complexes are classified, according to types of ligands and sign of, charge on the complex ion., 9.4.1 Classification on the basis of types of, ligands, , anionic sphere complex., For example,, 2, [Ni(CN)4] and K3[Fe(CN)6] have anionic, coordination sphere;, [Fe(CN)6]3, and, ⊕, three K ions make the latter electrically, neutral., iii. Neutral sphere complexes : A neutral, coordination complex does not possess cationic, or anionic sphere. [Pt(NH3)2Cl2] or [Ni(CO)4], have neither cation nor anion but are neutral, sphere complexes., , Use your brain power, Classify the complexes, as cationic, anionic or neutral., Na4[Fe(CN)6],, Co(NH3)6Cl2,, Cr(H2O)2(C2O4)23 , PtCl2(en)2 and, Cr(CO)6., , i. Homoleptic complexes : Consider, [Co(NH3)6]3⊕. Here only one type of ligands, surrounds the Co3⊕ ion. The complexes in, which metal ion is bound to only one type of, ligands are homoleptic., ii. Heteroleptic complexes : Look at the, complex [Co(NH3)4Cl2]⊕. There are two types, of ligands, NH3 and Cl attached to Co3⊕, ion. Such complexes in which metal ion is, surrounded by more than one type of ligands, are heteroleptic., Use your brain power, Classify, the, complexes, as homoleptic and heteroleptic, [Co(NH3)5Cl]SO4, [Co(ONO)(NH3)5]Cl2, [CoCl(NH3)(en)2]2⊕ and [Cu(C2O4)3]3 ., 9.4.2 Classification on the basis of charge on, the complex, i. Cationic complexes : A positively charged, coordination sphere or a coordination, compound having a positively charged, coordination sphere is called cationic sphere, complex., For example: the cation [Zn(NH3)4]2⊕ and, [Co(NH3)5Cl]SO4 are cationic complexes. The, latter has coordination sphere [Co(NH3)5Cl]2⊕;, the anion SO42 makes it electrically neutral., ii. Anionic sphere complexes : A, negatively charged coordination sphere or a, coordination compound having negatively, charged coordination sphere is called, , 9.5 IUPAC nomenclature of coordination, compounds : Tables 9.1, 9.2 and 9.3 summarize, the IUPAC nomenclature of coordination, compounds., Rules for naming coordination compounds, recommended by IUPAC are as follows:, 1. In naming the complex ion or neutral, molecule, name the ligand first and then the, metal., 2. The names of anionic ligands are obtained, by changing the ending -ide to -o and -ate to, -ato., 3. The name of a complex is one single word., There must not be any space between, different ligand names as well as between, ligand name and the name of the metal., 4. After the name of the metal, write its, oxidation state in Roman number which, appears in parentheses without any space, between metal name and parentheses., 5. If complex has more than one ligand of the, same type, the number is indicated with, prefixes, di-, tri-, tetra-, penta-, hexa- and, so on., 6. For the complex having more than one type, of ligands, they are written in an alphabetical, order. Suppose two ligands with prefixes, , 195
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are tetraaqua and dichloro. While naming, in alphabetical order, tetraaqua is first and, then dichloro., 7. If the name of ligand itself contains, numerical prefix then display number, by prefixes with bis for 2, tris for 3,, tetrakis for 4 and so forth. Put the ligand, name in parentheses. For example,, (ethylenediamine)3 or (en3) would appear, as tris(ethylenediamine) or tris(ethane-1,, 2-diamine)., 8. The metal in cationic or neutral complex, is specified by its usual name while in the, anionic complex the name of metal ends, with 'ate'., Try this..., Write the representation of, • Tricarbonatocobaltate(III) ion., , rule states that a metal ion continues to accept, electrons pairs till it attains the electronic, configuration of the next noble gas. Thus if the, EAN is equal to 18 (Ar), 36 (Kr), 54 (Xe), or, 86 (Rn) then the EAN rule is obeyed., EAN can be calculated with the following., Formula, EAN = number of electrons of metal ion + total, number of electrons donated by ligands, = atomic number of metal (Z) - number, of electrons lost by metal to form the, ion (X) + number of electrons donated, by ligands (Y)., =Z-X+Y, Cosider Co[NH3]63⊕, Oxidation state of Cobalt is +3, six ligands, donate 12 electrons., Z = 27; X = 3; Y = 12, EAN of Co3⊕ = 27 - 3 + 12 = 36., Try this..., , • Sodium hexacyanoferrate(III)., , Find, out, the, EAN, 4, 2⊕, Zn(NH3)4 ,[Fe(CN)6], , • Potassium hexa cyanoferrate (II), • Aquachlorobis(ethylenediamine), cobalt(III)., • Tetraaquadichlorochromium(III), chloride., • Diamminedichloroplatinum(II)., 9.6 Effective Atomic Number (EAN) Rule :, An early attempt to explain the stability, of coordination compounds was made by, Sidgwick who proposed an empirical rule, known as effective atomic number (EAN) rule., EAN equals total number of electrons around, the central metal ion in the complex. EAN, , of, , Cr(CO)6 and [Fe(CN)6]4 are some examples, of coordination compounds which, obey, the EAN rule. Certain other coordination, compounds however do not obey the EAN, rule. For example, [Fe(CN)6]3 and Cu[NH3]42⊕, have EAN 35., Use your brain power, Do the following complexes follow, the EAN rule ? Cr(CO)4, Ni(CO)4,, Mn(CO)5, Fe(CO)5., Isomers, Constitutional or structural, isomers, , Stereo isomers, , Geometric or, Distereoisomers, , Optical isomers or, enantiomers, Linkage isomers, , Coordination isomers Ionization isomers, , Fig. 9.1 : Classification of isomers in coordination compounds, , 197, , Solvate isomers
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9.7 Isomerism in coordination compounds :, One of the interesting aspects of, coordination chemistry is existence of isomers., Isomers are different compounds that have, the same molecular formula. Their chemical, reactivities and physical properties such, as colour, solubility and melting point are, different., Broadly speaking, isomers are classified, into two types namely stereoisomers and, constitutional (or structural) isomers as, displayed in Fig. 9.1., 9.7.1 Stereoisomers : Stereoisomers have the, same links among constituent atoms however, the arrangements of atoms in space are, different., There are two kinds of stereoisomers in, coordination compounds: (a) geometric, isomers or distereoisomers and (b) enantiomers, or optical isomers., a. Geometric isomers or distereoisomers :, These are nonsuperimposable mirror image, isomers. These are possible in heteroleptic, complexes. In these isomers, there are cis and, trans types of arrangements of ligands., Cis-isomers : Identical ligands, adjacent positions., , Here the cis isomer is more soluble in, water than the trans isomer. The cis isomer, named cisplatin is an anticancer drug while, the trans isomer is physiologically inactive., The cis isomer is polar with non-zero dipole, moment. The trans isomer has zero dipole, moment as a result of the two opposite Pt - Cl, and two Pt-NH3 bond moments, which cancel, each other., [Pt(NH3)(H2O)Cl2] (MA2BC type), H2O, Cl, Cl, H2O, Pt, Pt, Cl, H3N, Cl, NH3, cis isomer, trans isomer, Four coordinate tetrahedral complexes do, not show cis and trans isomers., iii. Cis and trans isomers in octahedral, complexes : The octahedral complexes of, the type MA4B2, MA4BC and M(AA)2B2 exist, as cis and trans isomers. (AA) is a bidentate, ligand., [Co(NH3)4Cl2]⊕, (MA4B2 type), , occupy, , Trans-isomer : Identical ligands occupy the, opposite positions., Cis and trans isomers have different, properties. Cis trans isomerism is observed in, square planar and octahedral complexes., i. Cis and trans isomers in square planar, complexes : The square planar complexes of, MA2B2 and MA2BC type exist as cis and trans, isomers, where A, B and C are monodentate, ligands, M is metal. For example : Pt(NH3)2Cl2,, (MA2B2 type), Cl, , NH3, , Cl, , Pt, Cl, NH3, cis isomer, , Cl, , Co, , 198, , NH3, , Co, , H3N, , [Pt(NH3)4ClBr]2⊕, (MA4BC type), , Cl, , 2⊕, , NH3, , NH3, , H3N, , NH3, , H 3N, , Pt, Br, , NH3, , cis isomer, , 2⊕, , Cl, Pt, Br, , NH3, NH3, , trans isomer, , [Co(en)2Cl2]⊕, (M(AA)2B2 type), Cl, , Cl, , ⊕, , ⊕, , Cl, en, , cis isomer, , trans isomer, , ⊕, , Cl, , NH3, Cl, trans isomer, , NH3, NH3, cis isomer, , en, , Cl, , H3N, , H3N, , en Co, , NH3, Pt, , H 3N, , H3N, , ⊕, , Cl, , Co, , en, , Cl, trans isomer
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en, , Draw structures of the cis and, trans isomers of [Fe(NH3)2(CN)4], , Enantiomers have identical properties, however differ in their response to the planepolarized light. The enantiomer that rotates, the plane of plane-polarized light to right, (clockwise) is called the dextro (d) isomer,, while the other that rotates the plane to left, (anticlockwise) is called laevo (l) isomer., , Co, , Co, , en, d, , 3⊕, , en, en, , en, mirror, , l, , Remember..., Our, hands, are, non, superimposable mirror images., When you hold your left hand upto a, mirror the image looks like right hand., , en, , en, l, , cis isomer, , 2⊕, , Cl, en, , Pt, , en, , Cl, trans isomer, Square planar complexes do not show, enantiomers since they have mirror plane and, axis of symmetry., Try this..., 1. Draw enantiomers of, [Cr(ox)3]3, 2. Draw enantiomers and cis and trans, isomers, of, [Cr(H2O)2(ox)2] (where, 2, ox = C2O4 ), , [Co(en)3]3⊕, 3⊕, , 2⊕, , Cl, Pt, , Pt, d, , i. Optical iomers in octahedral complexes, , en, , Cl, , Cl, en, , b. Optical isomers (Enantiomers) :, The complex molecules or ions that are, nonsuperimposable mirror images of each, other are enantiomers. The nonsuperimposable, mirror images are chiral. (A more elaborate, discussion on chirality and optical isomerism, is included in Chapter 10.), , en, , 2⊕, , Cl, , Try this..., , 9.7.2 Structural isomers (Constitutional, isomers) : Structural isomers possess different, linkages among their constituent atoms and, have, their chemical formulae to be the same., They can be classified as linkage isomers,, ionization isomers, coordination isomers and, solvate isomers., a. Linkage isomers : These isomers are formed, when the ligand has two different donor atoms., It coordinates to the metal via different donor, atoms. Thus the nitrite ion NO2 having two, donor atoms show isomers as :, [Co(NH3)5(NO2)]2⊕ and [Co(NH3)5(ONO)]2⊕, , ii. Octahedral complexes existing as both, geometric and optical isomers, , The nitro complex has Co-N bond and the, nitrito complex is linked through Co-O bond., These are linkage isomers., , [PtCl2(en)2]2⊕, , Can you tell ?, , In this type of complex, only the cis isomer, exists as pair of enantiomers, , Write linkage isomers of, [Fe(H2O)5SCN]⊕. Write their IUPAC, names., , 199
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b. Ionization isomers : Ionization isomers, involve exchange of ligands between, coordination and ionization spheres. For, example:, [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4, (I), , (II), , In compound I, anion SO42 , bonded to, Co is in the coordination sphere while Br is, in the ionization sphere. In compound II, anion, Br is in the coordination sphere linked to Co, while SO42 is in the ionisation sphere. These, complexes in solution ionize to give different, ions., [Co(NH3)5SO4]Br, , [Co(NH3)5SO4]⊕ + Br, , [Co(NH3)5Br]SO4, , [Co(NH3)5Br]2⊕+ SO42, , the free solvent molecule. I and II represent, solvate (hydrate) isomers., 9.8 Stability of the coordination compounds:, The stability of coordination compounds, can be explained by knowing their stability, constants. The stability is governed by metalligand interactions. In this the metal serves, as electron-pair acceptor while the ligand as, Lewis base (since it is electron donor). The, metal-ligand interaction can be realized as the, Lewis acid-Lewis base interaction. Stronger, the interaction greater is stability of the, complex., Consider the equilibrium for the metal, ligand interaction :, Ma⊕ + nLx, , I and II are examples of ionization isomers., , [MLn]a⊕ + nx, , where a, x, [a⊕ + nx ] denote the charge on the, metal, ligand and the complex, respectively., Now, the equilibrium constant K is given by, [MLn]a⊕ + nx, K=, [Ma⊕][Lx ]n, , Can you tell ?, Can you write IUPAC names of, isomers I and II?, c. Coordination isomers : Coordination, isomers show interchange of ligands between, cationic and anionic spheres of different metal, ions. For example :, [Co(NH3)6] [Cr(CN)6] [Cr(NH3)6] [Co(CN)6], (cationic) (anionic) (cationic) (anionic), (I) , (II), In isomer I, cobalt is linked to ammine, ligand and chromium to cyanide ligand. In, isomer II the ligands coordinating to metals, are interchanged. Cobalt coordinates with, cyanide ligand and chromium to NH3 ligand., I and II are examples of coordination isomers., d. Solvate isomers (Hydrate isomers when, water is solvent) : These are similar to, ionization isomers. Look at the complexes., [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 .H2O, (I), (II), In compound I the solvent water is directly, bonded to Cr. In compound II, H2O appears as, , Stability of the complex can be explained in, terms of K. Higher the value of K larger is the, thermodynamic stability of the complex., The equilibria for the complex formation, with the corresponding K values are given, below., Ag⊕ + 2CN, , [Ag(CN)2], , K = 5.5 ×1018, , Cu2⊕ + 4CN, , [Cu(CN)4]2, , K = 2.0 ×1027, , Co3⊕ + 6NH3, , [Co(NH3)6]3⊕ K = 5.0 ×1033, , From the above data, [Co(NH3)6]3⊕ is more, stable than [Ag(CN)2] and [Cu(CN)4]2 ., 9.8.1 Factors which govern stability of the, complex : Stability of a complex is governed, by (a) charge to size ratio of the metal ion and, (b) nature of the ligand., a. charge to size ratio of the metal ion, Higher the ratio greater is the stability., For the divalent metal ion complexes their, stability shows the trend : Cu2⊕ > Ni2⊕ > Co2⊕ >, Fe2⊕ > Mn2⊕ > Cd2⊕. The above stability order, , 200
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is called Irving-William order. In the above list, both Cu and Cd have the charge +2, however,, the ionic radius of Cu2⊕ is 69 pm and that of, Cd2⊕ is 97 pm. The charge to size ratio of Cu2⊕, is greater than that of Cd2⊕ . Therefore the Cu2⊕, forms stable complexes than Cd2⊕., b. Nature of the ligand., A second factor that governs stability, of the complexes is related to how easily the, ligand can donate its lone pair of electrons to, the central metal ion that is, the basicity of the, ligand. The ligands those are stronger bases, tend to form more stable complexes., Use your brain power, The stability constant K of the, [Ag(CN)2] is 5.5 × 1018 while that, for the corresponding [Ag(NH3)2]⊕ is 1.6 ×, 107. Explain why [Ag(CN)2]2 is more stable., 9.9 Theories of bonding in complexes :, The metal-ligand bonding in coordination, compounds has been described by Valence, bond theory (VBT) and Crystal field theory, (CFT)., 9.9.1 Valence bond theory (VBT), Can you recall ?, What is valence bond theory and, the concept of Hybridisation?, The hybridized state is a theoretical step that, describes how complexes are formed. VBT, is based on the concept of hybridization. The, hybrid orbitals neither exist nor can be detected, spectroscopically. These orbitals, however,, help us to describe structure of coordination, compounds. The steps involved in describing, bonding in coordination compounds using the, VBT are given below., i. Metal ion provides vacant d orbitals, for formation of coordinate bonds with, ligands., ii. The vacant d orbitals along with s and, p orbitals of the metal ion take part in, hybridisation., , iii. The number of vacant hybrid orbitals, formed is equal to the number of ligand, donor atoms surrounding the metal ion, which equals the coordination number of, metal., iv. Overlap between the vacant hybrid, orbitals of the metal and filled orbitals of, the ligand leads to formation of the metalligand coordinate bonds., v. The hybrid orbitals used by the metal ion, point in the direction of the ligand., vi. The (n-1)d, or nd orbitals used in, hybridisation allow the complexes to be, classified as (a) outer orbital and (b) inner, orbital complexes., vii. For hybridisation in the outer orbital, complex nd orbitals are used, whereas, in the inner orbital complexes (n-1)d, orbitals are used., Type of hybridisation decides the structure, of the complex. For example when the, hybridisation is d2sp3 the structure is octahedral., Steps to understand the metal-ligand bonding, include :, i. Find oxidation state of central metal ion, ii. Write valence shell electronic configuration, of metal ion., iii. See whether the complex is low spin or, high spin. (applicable only for octahedral, complexes with d4 to d8 electronic, configurations)., iv. From the number of ligands find the, number of metal ion orbitals required for, bonding., v. Identify the orbitals of metal ion, available for hybridisation and the type of, hybridisation involved., vi. Write the electronic configuration after, hybridisation., vii Show filling of orbitals after complex, formation., viii.Determine the number of unpaired, electrons and predict magnetic behaviour, of the complex., , 201
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vi. Six orbitals available for hybridisation, are two 3d, one 4s, three 4p orbitals, , Remember..., Complete the missing entries., Coordination Geometry Hybridization, number, of complex, 2, sp, Tetrahedral, 4, 4, , 3d, , 4p, , d2sp3, , The orbitals for hybridization are decided, from the number of ammine ligands which, is six. Here (n-1)d orbitals participate in, hybridization since it is the low spin complex., vii. Electronic configuration after complex, formation., , Square, planar, , 6, , 4s, , d2sp3/ sp3d2, , Try this..., , 3d, , Give VBT description of, bonding in each of following, complexes. Predict their magnetic, behavior., , 4s, , 4p, , d2sp3, , a. [ZnCl4]2, , 3⊕, , b. [Co(H2O)6]2⊕ (high spin), c. [Pt(CN)4]2 (square planar), , H3N, , d. [CoCl4]2 (tetrahedral), , H3N, , e. [Cr(NH3)6]3⊕, , NH3, Co, NH3, , NH3, NH3, , 9.9.2 Octahedral, complexes, a. [Co(NH3)6]3 ⊕ low spin, i. Oxidation state of Cobalt:3⊕, ii. Valence shell electronic configuration, of Co3⊕ is represented in box diagram as, shown below :, 3d, 4s, 4p, , viii. As all electrons are paired the complex is, diamagnetic., b. [CoF6]3 high spin, i.Oxidation state of central metal Co is 3+, ii.Valence shell electronic configuration of, Co3⊕ is, 3d, 4s, 4p, , iii. Number of ammine ligands is 6, number, of vacant metal ion orbitals required for, bonding with ligands must be six., iv. Complex is low spin, so pairing of electrons, will take place prior to hybridisation., v. Electronic configuration after pairing would, be, 3d, 4s, 4p, , iii. Six fluoride F ligands, thus the number of, vacant metal ion orbitals required for bonding, with ligands would be six., iv. Complex is high spin, that means pairing, of electrons will not take place prior to, hybridisation. Electronic configuration would, remain the same as in the free state shown, above., , 202
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v. Six orbitals available for the hybridisation., Those are one 4s, three 4p, two of 4d orbitals, 3d, 4s, 4p, 4d, , one 4s, three 4p. The complex is tetrahedral., 3d, 4s, 4p, , sp3, sp3d2, , Six metal orbitals after bonding with six, F ligands led to the sp3d2 hybridization. The d, orbitals participating in hybridisation for this, complex are nd., vi. Six vacant sp3d2 hybrid orbital of Co3+, overlap with six orbitals of fluoride forming, Co - F coordinate bonds., vii. Configuration after complex formation., 3d, , 4s, , 4p, , The four metal ion orbitals for bonding with, Cl ligands, are derived from the sp3, hybridization., vi. Four vacant sp3 hybrid orbital of Ni2⊕, overlap with four orbitals of Cl ions., vii. Configuration after complex formation, would be., 3d, 4s, 4p, , 4d, sp3, , viii.The complex has four unpaired electrons, and hence, paramagnetic., , sp3d2, , viii. The complex is octahedral and has four, unpaired electrons and hence, is paramagnetic., , F, F, , Cl, , 3, , F, , 2, , Cl, Ni, , Cl, , Cl, , F, Co, , 9.9.4 Square planar complex, [Ni(CN)4]2, i. Oxidation state of nickel is +2, ii. Valence shell electronic configuration of, Ni2⊕, 3d, 4s, 4p, , F, F, , 9.9.3 Tetrahedral complex, [Ni(Cl)4]2, i. Oxidation state of nickel is +2, ii. Valence shell electronic configuration of, Ni2+, 3d, 4s, 4p, , iii. number of Cl ligands is 4. Therefore, number of vacant metal ion orbitals, required for bonding with ligands must be, four., iv. Four orbitals on metal available for, hybridisation are, , iii. Number of CN ligands is 4, so number, of vacant metal ion orbitals required for, bonding with ligands would be four., iv. Complex is square planar so Ni2⊕ ion uses, dsp2 hybrid orbitals., v. 3d electrons are paired prior to the, hybridisation and electronic configuration, of Ni2⊕ becomes :, , 203, , 3d, , 4s, , 4p
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vi. Orbitals available for hybridisation are, one 3d, one 4s and two 4p which give dsp2, hybridization., vii. Four vacant dsp2 hybrid orbitals of Ni2⊕, overlap with four orbitals of CN ions to form, Ni - CN coordinate bonds., vii. Configuration after the complex formation, becomes., 4p, 3d, 4s, , dsp2, , viii.The complex has no unpaired electrons, and hence, dimagnetic., 2, , NC, , CN, Ni, , NC, , CN, , Try this..., Based on the VBT predict, structure and magnetic behavior of, the [Ni(NH3)6]3⊕ complex., 9.9.5 Limitations of VBT, i. It does not explain the high spin or low spin, nature of the complexes. In other words, strong, and weak field nature of ligands can not be, distinguished., ii. It does not provide any explanation for the, colour of coordination compounds., iii. The structure of the complexes predicted, from the VBT would not always match, necessarily with those determined from the, experiments., To overcome these difficulties in VBT, the, Crystal field theory has been proposed which, has widely been accepted., 9.9.6 Crystal Field theory (CFT), C.F.T. is based on following assumptions, i. The ligands are treated as point charges., The interaction between metal ion and ligand, is purely electrostatic or there are no orbital, interactions between metal and ligand., , In an isolated gaseous metal ion the five d, orbitals, d x2-y2 ,dz2, dxy,dyzd,zx have the same, energy i.e. they are degenerate., ii. When ligands approach the metal ion they, create crystal-field around the metal ion. If, it were symmetrical the degeneracy of the d, orbitals remains intact., Usually the field created is not symmetrical, and the degeneracy is destroyed. The d orbitals, thus split into two sets namely, (dxy, dyz, dxz), usually refered by t2g and ( d x2-y2 ,dz2) called, as eg. These two sets of orbitals now have, different energies. A separation of energies of, these two sets of d orbitals is the crystal field, splitting parameter. This is denoted by Δo (O, for octahedral)., iii. The Δo depends on strength of the ligands., The ligands are then classified as (a) strong, field and (b) weak field ligands. Strong field, ligands are those in which donor atoms are, C,N or P. Thus CN , NC , CO, NH3, EDTA,, en (ethylenediammine) are considered to be, strong ligands. They cause larger splitting of, d orbitals and pairing of electrons is favoured., These ligands tend to form low spin complexes., Weak field ligands are those in which donor, atoms are halogens, oxygen or sulphur. For, example, F , Cl , Br , I , SCN , C2O42 . In, case of these ligands the Δo parameter is, smaller compared to the energy required for, the pairing of electrons, which is called as, electron pairing energy. The ligands then can, be arranged in order of their increasing field, strength as, I < Br < Cl < S2 < F < OH < C2O42, <H2O<NCS <EDTA< NH3,< en< CN < CO., Let us understand splitting of d orbitals and, formation of octahedral complexes, In octahedral environment the central metal, ion is surrounded by six ligands., Ligands approach the metal ion along the x, y,, z axes. As the ligands approach the metal ion, the degeneracy of d orbitals is resolved., , 204
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Energy, , eg, eg - The higher energy set of orbitals (dz2 and dx2 - y2 ), t2g - The lower energy set of orbitals (dxy,dyz and dzx), Δ0 or 10 Dq - The energy separation between the two, levels, The eg orbitals are repelled by an amount of 0.6 Δ0., The t2g orbitals to be stabilized to the extend of 0.4 Δ0., , 3/5 Δ0, Δ0, 2/5 Δ0, , t2g, , Fig. 9.2 : Crystal field Splitting in an octahedral complex, , With closer approach of ligands along the, axes, the doubly degenerate dx2-y2 and dz2 (eg), orbitals experience larger repulsion than the, triply degenerate, t2g orbitals. As shown in Fig., 9.2, the eg set has higher energy than the t2g set, by the amount Δo. The Δo parameter is equal, to 10 Dq units of splitting of t2g and eg levels., For electronic configurations d1, d2, d3 the, electrons occupy t2g orbitals and obey the, Hund’s rules. For electronic configurations d1,, d2, d3 and d8, d9, d10 the high spin and low spin, configurations cannot be distinguished. Only, the electronic configurations d4 to d7 render, the high and low spin complexes.These are, depiciated in Fig 9.5., Table 9.5 : d orbital diagrams for high spin, and low spin complexes., d orbital, electronic, High spin, Low spin, configuration, , d4, , eg, t2g, , d5, , eg, t2g, , d6, , eg, t2g, , d7, , eg, , 9.9.7 Factors affecting Crystal Field Splitting, parameter (∆0), a. The magnitude of crystal field splitting, depends on strength of the ligands., The strong ligands those appear in, spectrochemical series approach closer to, the central metal which results in a large, crystal field splitting., b. Oxidation state of the metal : A metal with, the higher positive charge is able to draw, ligands closer to it than that with the lower, one. Thus the metal in higher oxidation, state results in larger separation of t2g and, eg set of orbitals. The trivalent metal ions, cause larger crystal field splitting than, corresponidng divalent ones., 9.9.8 Colour of the octahedral complexes, As discussed above, the formation of, an octahedral complex is accompanied by, splitting of d orbitals into t2g and eg sets. A, separation of these two sets of orbitals is Δo,, which can be measured from experiments., The Δo corresponds to a certain frequency, of electromagnetic radiation usually in the, visible region. A colour complementary to the, absorbed frequency is thus observed. Consider, the [Ti(H2O)6]3⊕ complex. The central metal, ion titanium has electronic configuration 3d, and the electron occupies one of the t2g orbitals, (Figure 9.3)., , t2g, , 205
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eg, , Energy, , t2g, , Try this..., , eg, , Sketch qualitatively crystal, field d orbital energy level diagrams, for each of the following complexes:, , t2g, Ground state, , Excited state, , Fig. 9.3 : d - d transition in d' system, , a. [Ni(en)3]2⊕ b. [Mn(CN)6]3, , The absorption of the wavelength of light, corresponding to Δo parameter promotes an, electron from the t2g level. Such energy gap in, case of the [Ti(H2O)6]3⊕ complex is 20,300 cm-1, (520 nm, 243 kJ/mol) and a complimentary, colour to this is imparted to the complex. A, violet color of the [Ti(H2O)6]3⊕ complex arises, from such d-d transition., 9.9.9 Tetrahedral complexes, A pattern of splitting of d orbitals,, which is a key in the crystal field theory, is, dependent on the ligand field environment., This is illustrated for the tetrahedral ligand, field environment., z, , y, , Fig. 9.4 : Tetrahedral structure, , The tetrahedral structure having the, metal atom M at the centre and four ligands, occupying the corners is displayed along with, in Fig. 9.4., Tetrahedral, t2g, eg, , Octahedral, eg, Δo, , Δtet, Free-ion, , Predict whether each of the complexes is, diamagnetic or paramagnetic., The dxy, dyz, dzx orbitals with their lobes, lying in between the axes point toward the, ligands. On the other hand, dx2-y2 and dz2, orbitals lie in between metal-ligand bond, axes. The dxy, dyz and dzx orbitals experience, more repulsion from the ligands compared to, that by dx2-y2 and dz2 orbitals., Due to larger such repulsions the dxy, dyz, and dzx orbitals are of higher energy while the, dx2-y2 and dz2 orbitals are of relatively lower, energy., Each electron entering in one of the, dxy,dyz and dzx orbitals raises the energy by, 4 Dq whereas that accupying d x2-y2 and dz2, orbitals lowers it by 6 Dq compared to the, energy of hypothetical degenerate d orbitals, in the ligand field., , x, , M, , c. [Fe(H2O)6]2⊕, , t2g, , A splitting of d orbitals in tetrahedral, crystal fields (assumed to be 10 Dq) thus is, much less (typically 4/9) compared to that, for the octahedral environment. The crystal, field splitting of d orbitals in a tetrahedral, ligand field is compared with the octahedral, one in Fig. 9.5. Thus the pairing of electrons, is not favoured in tetrahedral structure. For, example, in d4 configuration an electron, would occupy one of the t2g orbitals. The low, spin tetrahedral complexes thus are not found., Typically metal complexes possessing, the cetral metal ion with d8 electronic, configuration, for example, Ni(CO)4, favours, the tetrahedral structure., , Fig. 9.5 : Splitting of d orbitals in tetrahedral, and octahedral complexes, , 206
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9.10, Applications, compounds, , of, , coordination, , a. In biology : Several biologically important, natural compounds are metal complexes. They, play important role in a number of processes, occuring in plants and animals., For example, chlorophyll present in, plants is a complex of Mg. Haemoglobin, present in blood is a complex of iron., b. In medicines, i. Pt complex cisplatin is used in the treatment, of cancer., ii. EDTA is used for treatment of lead, poisoning., , c. To estimate hardness of water : Hardness, of water is due to the presence of Ca2+ and, Mg2⊕ ions. The ligand EDTA forms stable, complexes with Ca2⊕ and Mg2⊕. It can,, therefore, be used to estimate hardness., d. In electroplating :, Usually stable, coordination complexes on dissolution, dissociate to small extent and furnish a, controlled supply of metal ions. The metal, ions when reduced clump together to form, the clusters or nanoparticles. When the, coordination complexes are used the ligands, in the complex keep the metal atoms well, separated from each other. These metal atoms, tend to form a protective layer on the surface., Certain cyanide complexes K[Ag(CN)2], and K[Au(CN)2] find applications in the, electroplating of these noble metals., , Exercises, 1., , Choose the most correct option., , i., , The oxidation state of cobalt ion in the, complex [Co(NH3)5Br]SO4 is, a. +2 , b. +3, c. +1 , d. +4, , 3. [PtCl2Br2]2 (square planar), a. 1 and 3 , , b. 2 and 3, , IUPAC, name, of, the, complex, 2+, [Pt(en)2(SCN)2] is, a. bis (ethylenediamine, dithiocyanatoplatinum (IV) ion, b. bis (ethylenediamine), dithiocyantoplatinate (IV) ion, c. dicyanatobis (ethylenediamine), platinate IV ion, d. bis (ethylenediammine)dithiocynato, platinate (IV) ion, , c. 1 and 3 , , d. 4 only, , ii., , iii. Formula for the compound sodium, hexacynoferrate (III) is, a. [NaFe(CN)6], , b. Na2[Fe(CN)6], , c. Na[Fe(CN)6], , d. Na3[Fe(CN)6], , iv. Which of the following complexes exist, as cis and trans isomers ?, 1. [Cr(NH3)2Cl4], , 4. [FeCl2(NCS)2]2 (tetrahedral), , v., , Which of the following complexes are, chiral ?, 1. [Co(en)2Cl2]⊕ 2. [Pt(en)Cl2], 3. [Cr(C2O4)3]3, , 4. [Co(NH3)4Cl2]⊕, , a. 1 and 3 , , b. 2 and 3, , c. 1 and 4 , , d. 2 and 4, , vi. On the basis of CFT predict the number of, unpaired electrons in [CrF6]3 ., a. 1, , b. 2, , c. 3, , d. 4, , vii. When an excess of AgNO3 is added to the, complex one mole of AgCl is precipitated., The formula of the complex is, a. [CoCl2(NH3)4]Cl, b. [CoCl(NH3)4] Cl2, , 2. [Co(NH3)5Br]2⊕, , 207
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c. [CoCl3(NH3)3], , v., , d. [Co(NH3)4]Cl3, viii. The sum of coordination number and, oxidation number of M in [M(en)2C2O4]Cl, is, a. 6, , b. 7, , c. 9, , d. 8, , Write formulae of the following complexes, a. Potassium amminetrichloroplatinate, (II), b. Dicyanoaurate (I) ion, , vi. What are ionization isomers ? Give an, example., , 2., , Answer the following in one or two, sentences., , vii. What are the high-spin and low-spin, complexes ?, , i., , Write, the, formula, tetraammineplatinum (II) chloride., , ii., , Predict whether the [Cr(en)2(H2O)2]3+, complex is chiral. Write structure of its, enantiomer., , viii. [CoCl4]2 is tetrahedral complex. Draw its, box orbital diagram. State which orbitals, participate in hybridization., , for, , iv. Name the Lewis acids and bases in the, complex [PtCl2(NH3)2]., v., , What is the shape of a complex in which, the coordination number of central metal, ion is 4 ?, , ix. What are strong field and weak field, ligands ? Give one example of each., x., , With the help of crystal field energylevel diagram explain why the complex, [Cr(en)3]3⊕ is coloured ?, , 4., , Answer the following questions., , i., , Give valence bond description for the, bonding in the complex [VCl4] . Draw, box diagrams for free metal ion. Which, hybrid orbitals are used by the metal ?, State the number of unpaired electrons., , ii., , Draw, qualitatively, energy-level, diagram showing d-orbital splitting in, the octahedral environment. Predict, the number of unpaired electrons in the, complex [Fe(CN)6]4 . Is the complex, diamagnetic or paramagnetic? Is it, coloured? Explain., , vi. Is the complex [CoF6] cationic or anionic, if the oxidation state of cobalt ion is +3 ?, vii. Consider the complexes [Cu(NH3)4][PtCl4], and [Pt(NH3)4] [CuCl4]. What type of, isomerism these two complexes exhibit?, viii. Mention two applications of coordination, compounds., 3., , Answer in brief., , i., , What are bidentate ligands ? Give one, example., , ii., , What is the coordination number and, oxidation state of metal ion in the complex, [Pt(NH3)Cl5] ?, , iii. What is difference between a double salt, and a complex ? Give an example., iv. Classify, following, complexes, homoleptic and heteroleptic, , as, , [Cu(NH3)4]SO4,, [Cu(en)2(H2O)Cl]2⊕,, [Fe(H2O)5(NCS)]2⊕, tetraammine zinc (II), nitrate., , 208, , iii. Draw isomers in each of the following, a. Pt(NH3)2ClNO2, b. Ru(NH3)4Cl2, c. [Cr(en2)Br2]⊕, iv. Draw geometric isomers and enantiomers, of the following complexes., a. [Pt(en)3]4⊕, , b. [Pt(en2)ClBr]2⊕
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v., , What are ligands ? What are their types ?, Give one example of each type., , vi. What are cationic, anionic and neutral, complexes? Give one example of each., , vii. How stability of the coordination, compounds can be explained in terms of, equilibrium constants ?, viii. Name the factors governing the, equilibrium constants of the coordination, compounds., , Activity :, 1. The reaction of chromium metal with H2SO4 in the absence of air gives blue, solution of chromium ion., Cr(s) + 2H⊕(aq), , Cr2⊕(aq) + H2(s), , Cr2⊕ forms octahedral complex with H2O ligands., a. Write formula of the complex, b. Describe bonding in the complex using CFT and VBT., Draw crystal field splitting and valence bond orbital diagrams., 2. Reaction of complex [Co(NH3)3(NO2)3 with HCl gives a complex, [Co(NH3)3H2OCl2]⊕ in which two chloride ligands are trans to one another., a. Draw possible stereoisomers of starting material, b. Assuming that NH3 groups remain in place, which of two starting isomers would give, the observed product ?, , 209
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10. HALOGEN DERIVATIVES, Can you recall ?, , CH3 - CH2 - X, , Identify the functional group, in the following compounds., , (Haloalkane), , i., , (Benzyl bromide), , iii., , Cl, , Cl, , Cl, , (Haloarene), b. On the basis of number of halogen atoms,, halogen derivatives are classified as mono,, di, tri or poly halogen compounds., , iv. Cl - CH = CCl2, (Westrosol), , X, , CH3 - CH2 - X, , Cl, , Monohalogen compounds, , (Hexachlorobenzene), , v., , X, , (Haloalkyne), , (Freon - 12), , Cl, Cl, , (Haloalkene), , HC ≡ C - X, , ii. CCl2F2, , Br, , CH2 = CH - X, , Cl, Cl, , Cl, , Cl, , Cl, , CH3 - CH - X, X, , CH2 - CH2, , X, , X, , X, , X, , Dihalogen compounds, , Cl, , X, , (Benzene hexachloride), (Hexachlorocyclohexane), , The parent family of organic compounds, is hydrocarbon. Replacement of hydrogen, atom/s in aliphatic or aromatic hydrocarbons, by halogen atom/s results in the formation of, halogen derivatives of hydrocarbons., In this chapter we will study halogen, derivatives in a systematic way., Internet my friend, Find out the structures, of two thyroid hormones T3, (triiodothyronine) and T4 (thyroxine)., How do these help our body ?, 10.1 Classification of halogen derivatives :, Halogen derivatives of hydrocarbons are, classified mainly in two ways., a. On the basis of hydrocarbon skeleton to, which halogen atom is bonded, the halogen, derivatives are classified as haloalkanes,, haloalkenes, haloalkynes and haloarenes., , CH2 - CH - CH2, X X, X, , X, , X, Trihalogen compounds, We will consider classification of mono, halogen derivatives in more detail., 10.1.1 Classification of monohalogen, compounds : Monohalogen compounds are, further classified on the basis of position of, halogen atom and the type of hybridization of, carbon to which halogen is attached., a. Alkyl halides or haloalkanes : In alkyl, halides or haloalkanes the halogen atom is, bonded to sp3 hybridized carbon which is, a part of saturated carbon skeleton. Alkyl, halides may be primary, secondary or tertiary, depending on the substitution state of the, carbon to which halogen is attached : (Refer, to Std. XI Chemistry Textbook, section 14.3)., , 210
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10.2 Nomenclature of halogen derivatives, , a wide variety. The hydroxyl group may be, replaced by halogen atom using (a) halogen, acid, (b) phosphorous halide or (c) thionyl, chloride., , Can you recall ?, •, , •, , In, IUPAC, system, of, nomenclature, does, the, functional group 'halogen' appear, as a suffix or prefix ?, What are the trivial names of laboratory, solvents CHCl3 and CCl4 ?, , The common names of alkyl halides are, derived by naming the alkyl group followed, by the name of halogen as halide. For, example, methyl iodide, tert-butyl chloride., According to IUPAC system of nomenclature, (Std. XI Chemistry Textbook Chapter 14,, section 14.4.7) alkyl halides are named as, haloalkanes. Aryl halides are named as, haloarenes in common as well as IUPAC, system. For dihalogen derivative of an arene,, prefix o-, m-, p- are used in common name, system but in IUPAC system the numerals, 1,2 ; 1,3 and 1,4 respectively are used., Common and IUPAC names of some halogen, derivatives are given in Table 10.1., Use your brain power, Write IUPAC names, following., , of, , a. By using halogen acid or hydrogen, halide (HX) : The conditions for reaction, of alcohol with halogen acid (HX) depend, on the structure of the alcohol and particular, halogen acid used. The order of reactivity of, alcohols with a given haloacid is 30>20>10., (Refer to section 11.2.1 a), suitable, condition, , R - OH + HX, , R - X + H2O, , (Alcohol) (Alkyl halide), , Hydrogen chloride is used with zinc, chloride (Grooves' process) for primary and, secondary alcohols, but tertiary alcohols, readily react with concentrated hydrochloric, acid in absence of zinc chloride., R - OH + HCl, , anhydrous, ZnCl2, , R - Cl + H2O, , Do you know ?, Zinc chloride is a Lewis acid, and consequently can coordinate, with the alcohol, weakening R - O bond., Mixture of concentrated HCl and anhydrous, ZnCl2 is called Lucas reagent., , the, , iii. CH3 - CH = CH - CH2Cl, , Constant boiling hydrobromic acid, (48%) is used for preparing alkyl bromides., Primary alkyl bromides can also be prepared, by reaction with NaBr and H2SO4. Here HBr, is generated in situ., , iv. CH3 - C ≡ C - CH2 - Br, , R - OH + HBr, , i., , CH3 - CH - CH3, Br, , ii. CH3 - CH - CH2I, CH3, , Br, , R - Br + H2O, , Good yield of alkyl iodides may be, obtained by heating alcohols with sodium or, potassium iodide in 95 % phosphoric acid., Here HI is generated in situ., , Br, v., , NaBr, H2SO4, heat, , vi., Br, , 10.3 Methods of preparation of alkyl halides, 10.3.1 From alcohol : The most widely used, method of preparation of alkyl halide is, replacement of hydroxyl group of an alcohol, by halogen atom. Alcohols are available in, , 212, , R - OH + HI, , NaI/H3PO4, , R - I + H2O, , Can you tell ?, Why phosphoric acid is preferred to, H2SO4 to prepare HI in situ ?
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When toluene is brominated in presence, of iron, a mixture of ortho and para bromo, toluene is obtained., , Do you know ?, When alkenes are heated, with Br2 or Cl2 at high temperature,, hydrogen atom of allylic carbon is, substittued with halogen atom giving, allyl halide., , CH3, + Br2, , R - I + NaCl ↓, , 10.3.5 Sandmeyer's reaction : Aryl halides, are most commonly prepared by replacement, of nitrogen of diazonium salt. (For details, refer to Chapter 13 section 13.6)., , Alkyl fluorides are prepared by heating, alkyl chlorides or bromides with metal, fluorides such as AgF, Hg2F2, AsF3, SbF3 etc., , 10.4 Physical properties : Physical properties, of alkyl halides are considerably different, from those of corresponding alkanes. The, boiling point of alkyl halides is determined, by polarity of the C-X bond as well as the, size of halogen atoms., , R - F + AgCl ↓, , The reaction is known as Swartz reaction., 10.3.4 Electrophilic substitution :, Can you recall ?, , 10.4.1 Nature of intermolecular forces:, Halogens (X = F, Cl, Br and I) are more, electronegative than carbon., , • Identify the product of the, following reaction., + Cl2, , +, , + HBr, Aromatic, electrophilic, substitution, with iodine is reversible. In this case use, of HNO3/HIO4 removes HI by oxidation to, I2, equilibrium is shifted to right and iodo, product is formed. F2 being highly reactive,, fluoro compounds are not prepared by this, method., , The reaction is known as Finkelstein reaction., , R - Cl + AgF, , Br, , (o - Bromotoluene) (p-Bromotoluene), , 10.3.3 Halogen exchange : Alkyl iodides, are prepared conveniently by treating alkyl, chlorides or bromides with sodium iodide in, methanol or acetone solution. The sodium, bromide or sodium chloride precipitates from, the solution and can be separated by filtration., acetone, , Fe, dark, , CH3, , Br, , CH2 = CH - CH3 + Cl2, CH2 = CH - CH2Cl + HCl, , R - Cl + NaI, , CH3, , Carbon atom that carries halogen, develops a partial positive charge while the, halogen carries a partial negative charge., Thus carbon-halogen bond in alkyl halide is a, polar covalent bond. Therefore alkyl halides, are moderately polar compounds., , anhyd., AlCl3, , • Name the type of halide produced in, the above reaction., • What type of reactions are shown by, benzene ?, Aryl chlorides and bromides can be, prepared by direct halogenation of benzene, and its derivatives through electrophilic, substitution. It may be conveniently carried, out in dark at ordinary temperature in presence, of suitable Lewis acid catalyst like Fe, FeCl3, or anhydrous AlCl3., , δ⊕ δ, C, X, Size of the halogen atom increases from, fluorine to iodine. Hence the C-X bond length, increases. The C-X bond strength decreases, with an increase in size of halogen. This is, because as the size of p-orbital of halogen, increases the p-orbital becomes more diffused, , 214
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and the extent of overlap with orbital of, carbon decreases. Some typical bond lengths,, bond enthalpies and dipole moments of C-X, bond are given in Table 10.2., , CH3 - F, , Bond, length/, (pm), 139, , Bond enthalpy/, (kJ mol-1), 452, , Dipole, moment/, debye, 1.847, , CH3 - Cl, , 178, , 351, , 1.860, , CH3 - Br, , 193, , 293, , 1.830, , CH3 - I, , 214, , 234, , 1.636, , 10.4.2 Boiling point : Boiling points of alkyl, halides are considerably higher than those of, corresponding alkanes due to higher polarity, and higher molecular mass. Within alkyl, halides, for a given alkyl group, the boiling, point increases with increasing atomic mass, of halogen, because magnitude of van der, Waals force increases with increase in size, and mass of halogen., Thus boiling point of alkyl halide, decreases in the order RI > RBr > RCl > RF, For example, :, Haloalkane, Boiling, point (K), , CH3F CH3Cl, 194.6, , 248.8, , CH3Br, , CH3I, , 276.6, , 315.4, , For the given halogen, boiling point rises, with increasing carbon number., , Boiling point (K), , CH3CH2CH2CH2Br, , 375, , CH3 - CH - CH2 - CH3, Br, CH3, CH3− C − CH3, Br, , Table 10.2 : Bond parameters of C-X bond, Bond, , Haloalkane, , 364, 346, , 10.4.3 Solubility : Though alkyl halides, are moderately polar, they are insoluble in, water. It is due to inability of alkyl halides to, form hydrogen bonds with water. Attraction, between alkyl halide molecules is stronger, than attraction between alkyl halide and, water. Alkyl halides are soluble in non-polar, organic solvents., Aryl halides are also insoluble in water, but soluble in organic solvents. If aryl halides, are not modified by presence of any other, functional group, they show properties similar, to corresponding alkyl halides. The isomeric, dihalobenzenes have nearly the same boiling, points, but melting points of these isomers, show variation. Melting point of para isomer, is quite high compared to that of ortho or meta, isomer. This is because of its symmetrical, structure which can easily pack closely in, the crystal lattice. As a result intermolecular, forces of attraction are stronger and therefore, greater energy is required to overcome its, lattice energy., Cl, , For example,, , Cl, , Cl, , Cl, , Haloalkane, , Boiling point (K), , CH3Cl, , 248.8, , CH3CH2Cl, , 285.5, , b.p./K, , 453, , 446, , 448, , CH3CH2CH2Cl, , 320.0, , m.p./K, , 256, , 249, , 323, , CH3CH2CH2CH2Cl, , 351.5, , Cl, , For isomeric alkyl halides, boiling point, decrease with increased branching as surface, area decreases on branching and van der, Waals forces decrease. For example :, , 215, , Cl
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Let us, now, jot down the atoms/groups, attached to each carbon in 2 - chlorobutane., , Problem 10.2 Arrange the following, compounds in order of increasing boiling, points : bromoform, chloromethane,, dibromomethane, bromomethane., , 1, , Thus increasing order of boiling point is,, CH3Cl < CH3Br < CH2Br2 < CHBr3, isomerism, , in, , halogen, , Can you recall ?, • What is the relationship between, two compounds having the same, molecular formula?, • What is meant by stereoisomerism ?, Isomers having the same bond connectivities,, that is, structural formula are called, stereoisomers., Knowledge, of, optical, isomerism, which is a kind of stereoisomerism, will be useful to understand nucleophilic, substitution reactions of alkyl halides (see, 10.6.3)., , 3, , C-1 : - H, -H,, , Solution : The comparative boiling points, of halogen derivatives are mainly related, with van der Waals forces of attraction, which depend upon the molecular size. In, the present case all the compounds contain, only one carbon. Thus the molecular size, depends upon the size of halogen and, number of halogen atoms present., , 10.5 Optical, derivatives :, , 2, , -H,, 1, , 3, , 2, , 4, , -CHCl-CH2-CH3, 4, , 3, , C-2 : -H,, , -Cl, -CH3,, , C-3 : -H,, , -H, -CH3,, , -CHCl-CH3, , C-4 : -H,, , -H,, , -CH2-CHCl-CH3, , 4, , -H,, , -CH2-CH3, 2, , 1, , 2, , 3, , 1, , It can be seen that the four groups bonded, to C-2 are all different from each other., Carbon atom in a molecule which carries, four different groups/atoms is called chiral, carbon atom. Thus, the C-2 in 2-chlorobutane, is a chiral carbon. Chiral atom in a molecule, is marked with asterisk (*). For example,, CH3-*CHCl-CH2-CH3., When a molecule contains one chiral, atom, it acquires a unique property. Such a, molecule can not superimpose perfectly on its, mirror image. It is called chiral molecule. A, chiral molecule and its mirror image are not, identical (see Fig. 10.1)., CH3, , CH3, C*, H, , *C, Cl, , C2H5, , Cl, mirror plane, , H, C2H5, , CH3 CH3, , 10.5.1 Chiral atom and molecular chirality, Try this..., • Make a three - dimensional, model of 2 - chlorobutane., • Make another model which is a mirror, image of the first model., • Try to superimpose the two models on, each other., • Do they superimpose on each other, exactly ?, • Comment on whether the two models, are identical or not., , 4, , CH3 - CHCl - CH2 - CH3, , H, , C * *C, Cl Cl H, C2H5 C2H5, , Fig. 10.1 : Nonsuperimposable mirror images, , A chiral molecule and its mirror image, both have the same structural formula and,, of course, the same molecular formula. The, spatial arrangement of the four different, groups around the chiral atom, however, is, different. In other words, a chiral molecule, and its mirror image are stereoisomers of each, , 216
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other. (Refer to Std. XI Chemistry Textbook,, Chapter 14)., , Do you know ?, Nicol prism is a special type of, prism made from pieces of calcite,, a crystalline form of CaCO3, arranged, in a specific manner. Nicol prism is also, called polarizer., , The relationship between a chiral, molecule and its mirror image is similar to, the relationship between left and right hands., Therefore it is called handedness or chirality., (Origin : Greek word : Cheir means hand), The stereoisomerism in which the isomers, have different spatial arrangements of groups/, atoms around a chiral atom is called optical, isomerism. The optical isomers differ from, each other in terms of a measurable property, called optical activity., To understand optical activity, we must, know what is plane polarized light., Remember..., The phenomenon of optical, isomerism in organic compounds, was observed first and its origin in, molecular chirality was recognized later., 10.5.2 Plane polarized light : An ordinary, light consists of electromagnetic waves having, oscillations of electric and magnetic field in, all possible planes perpendicular to direction, of propagation of light., When ordinary light is passed through, Nicol's prism, oscillations only in one plane, emerge out. Such a light having oscillations, only in one plane perpendicular to direction, of propagation of light is known as plane, polarized light., , 10.5.3 Optical activity : When an aqueous, solution of certain organic compounds like, sugar, lactic acid is placed in the path of, plane polarized light, the transmitted light, has oscillations in a different plane than, the original. In other words, the incident, light undergoes rotation of its plane of, polarization. The plane of polarization rotates, either to the right (clockwise) or to the left, (anticlockwise). This property of a substance, by which it rotates plane of polarization, of incident plane polarized light is known, as optical activity. The compounds which, rotate the plane of plane polarized light are, called optically active compounds and those, which do not rotate it are optically inactive, compounds. Optical activity of a substance, is expressed numerically in terms of optical, rotation. The angle through which a substance, rotates the plane of plane polarized light on, passing through it is called optical rotation., In accordance with the direction of optical, rotation an optically active substance is either, dextrorotatory or laevorotatory. A compound, which rotates the plane of plane polarized, light towards right is called dextrorotatory, and designated by symbol d- or by (+), sign. A compound which rotates plane of, plane polarized light towards left is called, laevorotatory and designated by symbol lor by (-) sign., Isomerism in which isomeric compounds, have different optical activity is known as, optical isomerism. French scientist Louis, Pasteur first recognized that optical activity is, associated with certain type of 3-dimensional, structure of molecules. Pasteur introduced, the term enantiomers for the optical isomers, having equal and opposite optical rotation., , 217
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Figure 10.2 indicates a few objects in our, day to day life which exhibit superimposable, and non-superimposable mirror image, relationship., , Non superimposable, , Enantiomers have identical physical, properties (Such as melting point, boiling, points, densities, refractive index) except the, sign of optical rotation. The magnitude of, their optical rotation is equal but the sign, of optical rotation is opposite. They have, identical chemical properties except towards, optically active reagent., An equimolar mixture of enantiomers, (dextrorotatory and laevorotatory) is called, racemic modification or racemic mixture., A racemic modification is optically inactive, because optical rotation due to molecules, of one enatiomer is cancelled by equal and, opposite optical rotation due to molecules of, the other enantiomer. A racemic modification, is designated as (dl) or by (±) sign., , Superimposable, , Fig. 10.2 : Superimposable and, nonsuperimposable mirror image, , 10.5.5 Representation of configuration of, molecules :, Can you recall ?, , Remember..., • Optical, activity, is, an, experimentally, observable, property of compounds. Chirality is, a description of molecular structure., Optical activity is the consequence of, chirality., •, , •, , The two non-superimposable mirror, image structures are called pair of, enantiomers., , •, , Enantiomers have equal and opposite, optical rotation. Thus, enantiomers are, a kind of optical isomers., , W, , W, X, , Molecules which contain one chiral, atom are chiral, that is, they are, nonsuperimposable on their mirror, image., , •, , Identify the type of following, 3-D representation (I) and, (II) of a molecule and state, significance of the lines drawn., , Y, , X, , C, Z, , Y, , Z, , (I), , (II), , a. Fischer projection formula (cross, formula) : Two representations are used to, represent configuration of chiral carbon and, the 3-dimensional structure of optical isomers, on plane paper. These are (a) wedge formula, and (b) Fischer projection formula (also called, cross formula) (Std. XI Chemistry Textbook, Chapter 14 section 14.2.3)., , 10.5.4 Enantiomers : The optical isomers, which are non-superimposable mirror image, of each other are called enantiomers or, enantiomorphs or optical antipodes. For, example, 2 - chlorobutane exists as a pair of, enantiomers (Fig. 10.1)., , 218, , Cl Bonds below, , Cl Chiral carbon, Br, , I, , the plane, , I, , C, , Br, , Bonds above, H, the plane, Fischer projection, Convention of vertical, and horizontal lines, , H, , Fig. 10.3 Fischer projection formula
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a. Wedge formula : When a tetrahedral, carbon is imagined to be present in the plane, of paper all the four bonds at this carbon, cannot lie in the same plane. The bonds in, the plane of paper are represented by normal, lines, the bonds projecting above the plane, of paper are represented by solid wedges, (or simply by bold lines) while bonds going, below the plane of paper are represented by, broken wedges (or simply by broken lines)., , Can you recall ?, • What is meant by substitution, reaction ?, • Can you identify substitution reaction, from the following ?, ZnCl2, , (i) CH3 - CH2 - OH + HCl, CH3 - CH2 - Cl + H2O, (ii) CH2 = CH2 + HI, , Br, below the plane, , I, , C, , • Is the carbon carrying halogen in alkyl, halide, an electrophilic or a nucleophilic, centre ?, , In the plane, , H, , Cl, Above the plane, , Try this..., 1. Draw structures of enantiomers, of lactic acid (CH3-CH-COOH), OH, using Fischer projection formulae., 2. Draw structures of enantiomers of, 2-bromobutane using wedge formula., 10.6 Chemical properties :, 10.6.1 Laboratory test of haloalkanes :, Haloalkanes are of neutral type in aqueous, medium. On warming with aqueous sodium, or potassium hydroxide the covalently bonded, halogen in haloalkane is converted to halide, ion., R - X + OH, , ∆, , R - OH + X, , When this reaction mixture is acidified, by adding dilute nitric acid and silver nitrate, solution is added a precipitate of silver halide, is formed which confirms presence of halogen, in the original organic compound., Ag⊕ (aq) + X (aq), , CH3 - CH2 - I, , AgX↓ (s), , 10.6.2 Nucleophilic substitution reactions of, haloalkanes :, , hybridization of that carbon the reaction is, called substitution reaction. The C-X bond, in alkyl halides is a polar covalent bond, and the carbon in C-X bond is positively, polarized. In other words, the C-X carbon is, an electrophilic centre. It has, therefore, a, tendency to react with a nucleophile. (Refer to, Std. XI Chemistry Textbook Chapter 14.) Alkyl, halides react with a variety of nucleophiles, to give nucleophilic substitution reactions, (SN). The reaction is represented in general, form as shown below., δ⊕ δ, Nu + − C − X, , − C − Nu + X, , When a substrate reacts fast it is said to, be reactive. The reactivity of alkyl halides, in SN reaction depends upon two factors,, namely, the substitution state (10, 20 or 30), of the carbon and the nature of the halogen., The order of reactivity influenced by these, two factors is as shown below., tertiary alkyl halide (30) > secondary alkyl, halide (20) >primary alkyl halide (10) and, R - I > R - Br > R - Cl, Examples of some important nucleophilic, substitution reactions of alkyl halides are, shown in Table 10.3., , When a group bonded to a carbon in, a substrate is replaced by another group to, get a product with no change in state of, , 219
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10.3 Nucleophilic substitution reactions of alkyl halides, , Sr. No., 1., , Alkyl halide, R-X, , +, , Reagent, NaOH(aq) ∆, , R-X, , +, , 3., , R-X, , +, , (alcohol), , (or KOH), ⊕, , 2., , Substitution product, R - OH + NaX, , NaOR', , R - O - R' + NaX, , ∆, , (ether), , (sodium alkoxide), , O, ⊕, R' - C - OAg, , O, R' - C - OR + AgX ↓, , ∆, , (silver carboxylate), , 4., , R-X, , +, , NH3(alc.), , (ester), , ∆, pressure, , R - NH2, , (excess), , 5., , R-X, , +, , KCN (alc.), , 6., , R-X, , +, , AgCN (alc.), , (or KX), , +, , HX, , (primary amine), , R - CN, , ∆, , R-N C, , ∆, , 7., , R-X, , +, , 8., , R-X, , +, , Ag - O - N = O, , + AgX ↓, , (isocyanide), , ⊕, , KO - N = O, , + RX, , (nitrile)(alkyl cyanide), , R-O-N=O, , + KX, , (alkyl nitrite), , (potassium nitrite), , ⊕, , R N, , (silver nitrite), , O, O, , + AgX ↓, , (nitroalkane), , Can you tell ?, , Do you know ?, , Alkyl halides when treated with, alcoholic solution of silver nitrite give, nitroalkanes whereas with sodium nitrite, they give alkyl nitrites Explain., , Cyanide ion is capable of, attacking through more than one, site (atom)., , C≡N, , C=N, , Such nucleophiles are called ambident, nucleophiles. KCN is predominantly ionic, (K⊕C ≡ N) and provides cyanide ions. Both, carbon and nitrogen are capable of donating, electron pair. C-C Bond being stronger than, C-N bond, attack occurs through carbon atom, of cyanide group forming alkyl cyanides as, major product. However AgCN (Ag-C ≡ N), is mainly covalent compound and nitrogen, is free to donate pair of electron. Hence, attack occurs through nitrogen resulting in, formation of isocyanide., Another ambident nucleophile is nitrite, ion, which can attack through ‘O’ or ‘N’., O-N=O, , 10.6.3 Mechanism of SN reaction :, Can you recall ?, • What is meant by order and, molecularity of a reaction ?, • What is meant by mechanism of, chemical reaction ?, It can be seen from the Table 10.3 that in, a nucleophilic substitution reactions of alkyl, halides the halogen atom gets detached from, the carbon and a new bond is formed between, that electrophilic carbon and nucleophile., The covalently bonded halogen is converted, into halide ion (X ). It means that the two, electrons constituting the original covalent, bond are carried away by the halogen along, with it. The halogen atom of alkyl halide, is, therefore, called ‘leaving group’ in the, , 220
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context of this reaction. Leaving group is, the group which leaves the carbon by taking, away the bond pair of electrons. The substrate, undergoes two changes during a SN reaction., The original C-X bond undergoes heterolysis, and a new bond is formed between the carbon, and the nucleophile using two electrons of the, nucleophile. These changes may occur in one, or more steps. The description regarding the, sequence and the way in which these two, changes take place in SN reaction is called, mechanism of SN reaction. The mechanism is, deduced from the results of study of kinetics, of SN reactions. Two mechanisms are observed, in various SN reactions. These are denoted as, SN1 and SN2 mechanisms., a. SN2 Mechanism : The reaction between, methyl bromide and hydroxide ion to give, methanol follows a second order kinetics,, that is, the rate of this reaction depends on, concentration of two reacting species, namely,, methyl bromide and hydroxide. Hence it is, called subtitution nucleophilic bimolecular,, SN2., CH3Br + OH, , CH3OH + Br, , rate = k [CH3Br] [OH ], Rate of a chemical reaction is influenced by, the chemical species taking part in the slowest, step of its mechanism. In the above reaction, only two reactants are present and both are, found to influence the rate of the reaction., This means that the reaction is a single step, reaction which can also be called the slow, step. This further implies that the two changes,, namely, bond breaking and bond forming at, the carbon take place simultaneously. This, SN2 mechanism is represented as shown in, Fig. 10.4., , H, HO, , C, H, , H, , Br, , [ [, H, , 1, 2, , 1, 2, , HO C Br, H, , H, , Transition state, (T.S.), , H, HO, , C, H, , + Br, H, , Fig. 10.4 : SN2 mechanism, , i., , Salient features of SN2 mechanism :, Single step mechanism with simultaneous, bond breaking and bond forming., , ii. Backside attack of nucleophile : The, nucleophile attacks the carbon undergoing, substitution from the side opposite to that, of the leaving group. This is to avoid steric, repulsion (repulsion due to bulkyness of, the groups) and electrostatic repulsion, between the incoming nucleophile and, the leaving group., iii. In the transition state (T.S.) the nucleophile, and leaving groups are bonded to the, carbon with partial bonds and carry, partial negative charge. (Thus, the total, negative charge is diffused.), iv. The T.S. contains pentacoordinate, carbon having three σ (sigma) bonds in, one plane making bond angles of 1200, with each other and two partial covalent, bonds along a line perpendicular to this, plane., v., , 221, , When SN2 reaction is brought about at, chiral carbon (in an optically active, substrate), the product is found to have, opposite configuration compared to, that of the substrate. In other words,, SN2 reaction is found to proceed with, inversion of configuration. This is like, flipping of an umbrella (See Fig. 10.4)., It is known as Walden inversion. The, inversion in configuration is the result of, backside attack of the nucleophile.
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b. SN1 Mechanism : The reaction between, tert-butyl bromide and hydroxide ion to give, tert-butyl alcohol follows a first-order kinetics,, that is the rate of this reaction depends on, concentration of only one species, which is, the substrate molecule, tert-butyl bromide., Hence it is called substitution nucelophilic, unimolecular, SN1., CH3, CH3, CH3− C − Br + OH, CH3− C − OH + Br, CH3, CH3, rate = k [(CH3)3CBr], , iv. When SN1 reaction is carried out at chiral, carbon in an optically active substrate,, the product formed is nearly racemic., This indicates that SN1 reaction proceeds, mainly with racemization. This means, both the enantiomers of product are formed, in almost equal amount. Racemization in, SN1 reaction is the result of formation, of planar carbocation intermediate (Fig., 10.5). Nucleophile can attack planar, carbocation from either side which results, in formation of both the enantiomers of, the product., , It can be seen in this reaction that, concentration of only substrate appears, in the rate equation; concentration of the, nucleophile does not influence the reaction, rate. In other words, tert-butyl bromide reacts, with hydroxide by a two step mechanism., In the slow step C-X bond in the substrate, undergoes heterolysis and in the subsequent, fast step the nucleophile uses its electron, pair to form a new bond with the carbon, undergoing change. This SN1 mechanism is, represented as shown in Fig. 10.5., Step I, CH3, (CH3)3C - Br, , slow, , C, , ⊕, , H 3C, , Use your brain power, • Draw the Fischer projection, formulae of two products, obtained when compound (A) reacts, with OH by SN1 mechanism., C2H5, H3C− C − Br (A), n-C3H7, • Draw the Fischer projection formula of, the product formed when compound (B), reacts with OH by SN2 mechanism., CH3, H− C − Cl, C2H5, , + Br, CH3, , (carbocation intermediate), , 10.6.4 Factors influencing SN1 and SN2, mechanism :, , Step II, CH3, C⊕, H3C, , + OH, CH3, , a. Nature of substrate : SN2 : The T.S. of, SN2 mechanism is pentacoordinate and thus, crowded (See Fig. 10.4). As a result SN2, mechanism is favoured in primary halides, and least favoured in tertiary halides., , (CH3)3C - OH, , Fig. 10.5 : SN1 mechanism, , Salient features of SN1 mechanism :, i., , (B), , Two step mechanism., , ii. Heterolyis of C-X bond in the slow, and reversible first step to form planar, carbocation intermediate., iii. Attack of the nucleophile on the, carbocation intermediate in the fast, second step to form the product., , SN1 : A planar carbocation intermediate, is formed in SN1 reaction. It has no steric, crowding. Bulky alkyl groups can be easily, accommodated in planar carbocation See, (Fig. 10.5). As a result SN1 mechanism is, most favoured in tertiary halides and least, favoured in primary halides. (Formation of, planar carbocation intermediate results in a, , 222
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carbocation is relatively poor and solvation, of anion is particularly important. Anions are, solvated by hydrogen bonding solvents, that, is, protic solvents. Thus SN1 reaction proceeds, more rapidly in polar protic solvents than in, aprotic solvents., , b. Nucleophilicity of the reagent :, Can you recall ?, •, , Give some examples of, nucleophiles that are electrically, neutral., , •, , Give some examples, nucleophiles., , •, , What is the difference between a base, and a nucleophile ?, , of, , anionic, , A nucleophile is a species that uses its, electron pair to form a bond with carbon., Nucleophilic character of any species is, expressed in its electron releasing tendency,, which can be corelated to its strength as, Lewis base., , Polar protic solvents usually decrease the, rate of SN2 reaction. In the rate determining, step of SN2 mechanism substrate as well, as nucleophile is involved. A polar solvent, stabilizes nucleophile (one of the reactant), by solvation. Thus solvent deactivates the, nucleophile by stabilizing it. Hence aprotic, solvents or solvents of low polarity will, favour SN2 mechanism., , A more powerful nucleophile attacks the, substrate faster and favours SN2 mechanism., The rate of SN1 mechanism is independent of, the nature of nucleophile. Nucleophile does, not react in slow step of SN1. It waits till, the carbocation intermediate is formed, and, reacts fast with it., Do you know ?, 1. A, negatively, charged, nucleophile is more powerful, than its conjugate acid. For example, R-O is better nucleophile than R-OH., 2. When donor atoms are from same, period of periodic table, nucleophilicity, decreases from left to right in a period., For example H2O is less powerful, nucleophile than NH3., , Problem 10.5 : Which of the following, two compounds would react faster by SN2, mechanism and Why ?, CH3-CH2-CH2-CH2Cl, 1-Chlorobutane, , CH3-CH-CH2-CH3, Cl, , 2-Chlorobutane, , Solution : In SN2 mechanism, a, pentacoordinate T.S. is involved. The order, of reactivity of alkyl halides towards SN2, mechanism is,, Primary > Secondary > Tertiary, (due to, increasing crowding in T.S. from primary, to tertiary halides. 1-Chlorobutane being, primary halide will react faster by SN2, mechanism, than the secondary halide, 2-chlorobutane., Can you recall ?, , 3. When donor atoms are from same group, of the periodic table, nucleophilicity, increases down the group. For example,, I is better nucleophile than Cl ., , •, , How are alkenes prepared, from alkyl halides ?, , •, , Which is stronger base from the, following ?, i. aq. KOH, , c. Solvent polarity : SN1 mechanism proceeds, via formation of carbocation intermediate., A good ionizing solvent, polar solvent,, stabilizes the ions by solvation. Solvation of, , 224, , ii. alc. KOH
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10.6.5, Elimination, Dehydrohalogenation, , reaction, , :, , When alkyl halide having at least one, β-hydrogen is boiled with alcoholic solution of, potassium hydroxide, it undergoes elimination, of hydrogen atom from β-carbon and halogen, atom from α - carbon resulting in the formation, of an alkene., Remember..., The carbon bearing halogen is, commonly called α-carbon (alpha, carbon) and any carbon attached to α-carbon, is β-carbon (beta carbon). Hydrogens, attached to β-carbon are β-hydrogens., This reaction is called β-elimination, (or 1,2 - elimination) reaction as it involves, elimination of halogen and a β - hydrogen, atom., H, B+, , β, , C, , α, , alc. KOH, , C, , The different products of elimination do, not form in equal proportion. After studying, a number of elimination reactions, Russian, chemist Saytzeff formulated an empirical rule, given below., In dehydrohalogenation reaction, the, preferred product is that alkene which has, greater number of alkyl groups attached to, doubly bonded carbon atoms., Therefore, in the above reaction but-2-ene, is the preferred product, and is formed as the, major product. It turned out that more highly, substituted alkenes are also more stable alkenes., Hence Saytzeff elimination is preferred, formation of more highly stabilized alkene, during an elimination reaction. The stability, order of alkyl substituted alkenes is :, , C = C + B⊕H +X, , ∆, , X, (base) (alkyl halide), , (alkene), , As hydrogen and halogen is removed, in this reaction it is also known as, dehydrohalogenation reaction., If there are two or more non-equivalent, β-hydrogen atoms in a halide, then this, reaction gives a mixture of products. Thus,, 2-bromobutane on heating with alcoholic, KOH gives mixture of but-1-ene and but-2ene., β1, , β2, , α, , CH3 − CH2 − CH − CH3, Br, (2-bromobutane), alc. KOH, 2, loss of β - hydrogen, , ∆, , loss of β1 -hydrogen, , HC3 − CH2 − CH = CH2, (But-1-ene), , CH3− CH = CH - CH3, (But-2-ene), , 225, , R2C = CR2 > R2C = CHR > R2C = CH2,, RCH = CHR > RCH = CH2, Do you know ?, Elimination versus substitution:, Alkyl halides undergo sunstitution as, well as elimination reaction. Both reactions, are brought about by basic reagent, hence, there is always a competition between these, two reactions. The reaction which actually, predominates depends upon following, factors., a. Nature of alkyl halides : Tertiary alkyl, halides prefer to undergo elimination, reaction where as primary alkyl halides, prefer to undergo substitution reaction., b. Strength and size of nucleophile :, Bulkier electron rich species prefers to act, as base by abstracting proton, thus favours, elimination. Substitution is favoured in the, case of comparatively weaker bases, which, prefer to act as nucleophile, c. Reaction conditions : Less polar solvent,, high temperature fovours elimination where, as low tempertaure, polar solvent favours, substitution reaction.
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b. Nucleophilic substitution SN of haloarenes:, Can you recall ?, • What is resonance?, •, , Draw resonance structures of, bromobenzene., , •, , Identify the type of hybridization of, carbon to which halogen is attached in, haloarene., , Aryl halides show low reactivity, towards nucleophilic substitution reactions., The low reactivity of aryl halides is due to :, i. Resonance effect and, ii. sp2 hybrid state of C ., , Thus nuclophilic substitution reaction, involving cleavage of C-X bond in haloarene, proceeds with difficulty. However, the presence, of certain groups at certain positions of the, ring, markedly activate the halogen of aryl, halides towards substitutuion. For example,, presence of electron withdrawing group at, ortho and/or para postion greatly increases the, reactivity of haloarenes towards subsitution of, halogen atom. Greater the number of electron, withdrawing groups at o/p position, greater, is the reactivity. Electron withdrawing group, at meta position has practically no effect on, reactivity., i., , NO2, , NO2, (p-nitrophenol), , (p-nitrochlorobenzene), , ii., , :, , :, , OH, (i)NaOH 433 K, (ii)H3O⊕, , :, , :, , i. One of the lone pairs of electrons, on halogen atom is in conjugation with, π -electrons of the ring. For example the, following different resonance structures can, be written for chlorobenzene., +Cl :, + Cl :, : Cl, , Cl, , Cl, , OH, , NO2, , NO2, , (i)aq.Na2CO3 403 K, (ii)H3O⊕, , III, , II, :, , + Cl :, , :, , I, , : Cl :, , NO2, , NO2, , (2,4-dinitrochlorobenzene), , (2,4 - dinitrophenol), , iii., NO2, , Cl, , NO2, , NO2, , OH, , NO2, , warm, , IV, , V, , H2O, , NO2, , Resonance structures II, III and IV show, double bond character to carbon-chlorine bond., Thus carbon-chlorine bond in chlorobenzene, is stronger and shorter than chloroalkane, molecule, C-Cl bond length in chlorobenzene, is 169 pm as compared to C-Cl bond length, in alkyl chloride 178 pm. Hence it is difficult, to break. Phenyl cation produced due to selfionization of haloarene will not be stabilised, by resonance, which rules out possibility of SN1, mechanism. Back side attack of nucleophile is, blocked by the aromatic ring, which rules out, SN2 mechanism., , (2,4,6-trinitrochlorobenzene), , 227, , NO2, (2,4,6 - trinitrophenol), , Can you tell ?, Conversion of chlorobenzene, to phenol by aqueous sodium, hydroxide requires high temperture of, about 623K and high pressure. Explain., OH, , Cl, (i) 623K, OH 300 atm, (ii) H3O⊕, , Chlorobenzene, , Phenol
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Do you know ?, , •, , Occurrence of nucleophilic, substitution in p-nitrochlorobenzene, can be explained on the basis of resonance, stabilization of the intermediate., , + OH, , O, , N, O ⊕ O, (II), , N, ⊕ O, , N, O ⊕ O, (III), , Cl OH, , OH, fast, step, , N, O ⊕ O, (IV), , In resonance structures of chlorobenzene, (see section 10.6.5) elelctron density is, relatively more at ortho and para position., Therefore incoming electrophilic group is, more likely to attack at these positions. But, due to steric hinderance at ortho position, para, product usually predominates. In haloarenes,, halogen atom has strong electron withdrawing, inductive effect (-I). This deactivates the ring, and electrophilic substitution reaction occurs, slowly., , (I), Cl OH, , Cl OH, , Remember..., The -I effect of Cl is more, powerful than its +R effect., Therefore Cl is o-/p- directing but ring, deactivating group., , + Cl, N, O ⊕ O, , i. Halogenation : It is carried out by reacting, haloarene with halogen in presence of ferric, salt as Lewis acid catalyst., , The resonance structure (III) shows, that the electron withdrawing nitro group, (-NO2) in the p-position extends the, conjugation. As a result, the intermediate, carbanion is better stabilized which favours, nucleophilic substitution reaction., c. Electrophilic, arylhalides, , substitution, , (SE), , Cl, + Cl2, , in, , (Chlorobenzene), anhydrous Cl, 2, FeCl3, , Can you recall ?, •, , What is an electrophile?, , •, , Give some, electrophiles, , •, , A, , Aryl halides undergo electrophilic, substitution reaction slowly as compared to, benzene., , slow, step, , N, O ⊕ O, , conc. H2SO4, , + HNO3, , Cl OH, , Cl, , Identify the product A of following, reaction., , examples, , Cl, , Cl, , of, , +, , What type of reactions are observed in, benzene?, , Cl, + HCl, , Cl, (1,4 - Dichlorobenzene), (major), , 228, , (1,2 - Dichlorobenzene), (minor)
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ii. Nitration : It is carried out by heating, haloarene with conc. HNO3 in presence of, conc. H2SO4., , Cl, , +, , Cl, , (Chlorobenzene), anhyd. AlCl3, , + HNO3, conc., , Cl, , (Chlorobenzene), , ∆, , (1-Chloroacetophenone), , Cl, , Cl, +, , (1-Chloro-2-nitrobenzene), , iii. Sulfonation : It is carried out by heating, haloarene with fuming H2SO4., Cl, +, , H2SO4, (fuming), , (chlorobenzene), ∆, , Cl, , SO3H, +, , SO3H, (1 - Chlorobenzene, sulfonic acid), , (4 - Chlorobenzene sulfonic, acid), , iv. Friedel Craft’s reaction : It is carried out, by treating haloarene with alkyl chloride or, acyl chloride in presence of anhydrous AlCl3, as a catalyst., Cl, , Cl, , + CH3-Cl, (Chlorobenzene), , anhyd., AlCl3, , O C CH3, , (4 -Chloroacetophenone), , 10.7.1 Dichloromethane/ methylene chloride, (CH2Cl2) : It is a colourless volatile liquid, with moderately sweet aroma. It is used as a, solvent, and used as a propellant in aerosols., Over exposure to dichloromethane causes, dizziness, fatigue, nausea, headaches,, numbness, weakness. It is highly dangerous, if it comes in direct contact with eyes as it, damages cornea., 10.7.2 Chloroform / trichloromethane, (CHCl3) : It is a colourless liquid with, peculiar sweet smell. It is used to prepare, chlorofluromethane, a refrigerant R-22. It, is used as a solvent for extraction of natural, products like gums, fats, resins. It is used as a, source of dichlorocarbene. Chloroform causes, depression of central nervous system. Inhaling, chloroform for a short time causes fatigue,, dizziness and headache. Long exposure to, chloroform may affect liver. Chloroform when, exposed to air and light forms a poisonous, compound phosgene so it is stored in dark, coloured air tight bottles., 10.7.3 Carbon tetrachloride /, tetrachloromethane (CCl4) :, , Cl, CH3, , (1-Chlorotoluene), , +, , 10.7 Uses and Environmental effects of some, polyhalogen compounds, , NO2, , NO2, (1-Chloro-4-nitrobenzene), , Cl, , O, C CH, 3, , conc. H2SO4, , Cl, , O, CH3-C -Cl, , +, , CH3, (4-Chlorotoluene), , It is a colourless liquid with sweet smell. It is, very useful solvent for oils, fats, resins. It serves, as a source of chlorine. It is used as a cleaning, agent. It is highly toxic to liver. Exposure to, high concentration of CCl4 can affect central, , 229
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nervous system and it is suspected to be, carcinogenic. Prolonged exposure may cause, death. It is a green house gas., 10.7.4 Idoform or triiodomethane (CHI3):, It is a yellow crystalline substance with, disagreeable smell. It is used in medicine as, a healing agent and antiseptic in dressing of, wounds, however its use is limited., It causes irritation to skin and eyes. It may, cause respiratory irritation or breathing, difficulty, dizziness, nausea, depression of, central nervous system, visual disturbance., 10.7.5 Freons : These are organic compounds, of chlorine and fluorine, chlorofluorocarbons,, CFC's are commonly used as refrigerants., The most common representative is, dichlorodifluromethane (Freon-12) others, include, chlorodifluromethane, (R-22),, trichlorofluromethane (R-11) and so on., , 10.7.6, Dichlorodiphenyltrichloroethane, (DDT) : It is colourless, tasteless and odorless, crystalline compound having insecticidal, property., It kills insects such as houseflies,, mosquitoes and body lice. It was used for, controlling maleria and typhus., Exposure to high doses of DDT may cause, vomiting, tremors or shakiness. Laboratory, animal studies showed adverse effect of DDT, on liver and reproduction. DDT is a pressistent, organic pollutant, readily absorbed in soils, and tends to accumulate in the ecosystem., When dissolved in oil or other lipid, it is, readily absorbed by the skin. It is resistant to, metabolism. It accumulates in fatty tissues., There is a ban on use of DDT due to all these, adverse effects ., Cl, , They are used as refrigerants in fridge and, airconditioning, propellants in aerosol and, solvents. They are used as blowing agents in, making foams and packing materials., , Cl, , Cl, , Cl, , Cl, , Do you know ?, , Chloroflurocarbons are responsible for ozone, depletion of ozone in stratosphere. Regular, large inhalation of freons results in breathing, problems, organ damage, loss of consciousness., , DDT, the first chlorinated, organic insecticides, was originally, prepared in 1873, but it was not until 1939, that Paul Muller of Geigy Pharmaceuticals, in Switzerland discovered the effectiveness, of DDT as an insecticide. Paul Muller, was awarded the Nobel Prize in Medicine, and Physiology in 1948 for this discovery., The use of DDT increased enormously, on a worldwide basis after World War, II, primarily because of its effectiveness, against the mosquito that spreads malaria, and lice that carry typhus. Many species, of insects developed resistance to DDT,, and it was also discovered to have a, high toxicity towards fish. DDT is not, metabolised very rapidly by animals;, instead, it is deposited and stored in the, fatty tissues. The use of DDT was banned, in the United States in 1973, although it is, still in use in some other parts of the world., , Do you know ?, How do CFC distroy the ozone, layer in the atmosphere ?, When ultraviolet radiation (UV), strikes CFC (CFCl3) molecules in the, upper atmosphere, the carbon-chlorine, bond breaks and produces highly reactive, chlorine atom (Cl)., CFCl3, CFCl2 + Cl, This reactive, chlorine atom, decomposes ozone (O3) molecule into, oxygen molecule (O2)., O3 + Cl, O2 + ClO, O2 + Cl, ClO + O, One atom of chlorine can destroy upto, 100,000 ozone molecules., , 230
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vi., , Name the reagent used to bring about, the following conversions., a. Bromoethane to ethoxyethane, b.1-Chloropropane to 1 nitropropane, c. Ethyl, bromide, isocyanide, , to, , ethyl, , i., , HCl is added to a hydrocarbon 'A', (C4H8) to give a compound 'B' which, on hydrolysis with aqueous alkali, forms tertiary alcohol 'C' (C4H10O)., Identify 'A' , 'B' and 'C'., , ii., , Complete the following reaction, sequences by writing the structural, formulae of the organic compounds, 'A', 'B' and 'C' ., , Arrange the following in the increase, order of boiling points, a. 1-Bromopropane, b. 2- Bromopropane, c. 1- Bromobutane, d. 1-Bromo-2-methylpropane, , viii. Match the pairs., , a. 2-Bromobutane, , Column I , , Column II, , a. CH3CH-CH3, , i. vinyl halide, , alc.KOH, , b. Isopropyl alcohol, iii., , X, , b. CH2=CH-CH2X ii. alkyl halide, c. CH2=CH-X, , But-1-ene to n-butyl iodide, 2-Chloropropane to propan-1-ol, tert-Butyl bromide to isobutyl bromide, Aniline to chlorobenzene, Propene to 1-nitropropane, , 7. Answer the following, , d. Chlorobenzene to biphenyl, vii., , iv., v., vi., vii., viii., , iii. allyl halide, , , , iv. benzyl halide, , , , v. aryl halide, , 3. Give reasons, i., , Haloarenes are less reactive than halo, alkanes., ii., Alkyl halides though polar are, immiscible with water., iii. Reactions involving Grignard reagent, must be carried out under anhydrous, condition., iv. Alkyl halides are generally not, prepared by free radical halogenation, of alkanes., 4. Distinguish between - SN1 and SN2, mechanism of substitution reaction ?, 5. Explain, Optical, isomerism, in, 2-chlorobutane., 6. Convert the following., i., Propene to propan-1-ol, ii., Benzyl alcohol to benzyl cyanide, iii. Ethanol to propane nitrile, , 233, , A, , ∆, PBr3, , Br2, , B NaNH C, 2, , A NH excess B, 3, , Observe the following and answer the, questions given below., ⊕, , CH2=CH-X, , CH2-CH=X, , a. Name the, derivative, , type, , of, , halogen, , b. Comment on the bond length of, C-X bond in it, c. Can react by SN1 mechanism?, Justify your answer., , Activity :, 1. Collect detailed information, about Freons and their uses., 2. Collect information about DDT as a, persistent pesticide., Reference books, i. Organic chemistry by Morrison, Boyd,, Bhattacharjee, 7th edition, Pearson, ii. Organic chemistry by Finar, Vol 1, 6th, edition, Pearson
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11. ALCOHOLS, PHENOLS AND ETHERS, Can you recall ?, , Do you know ?, , 1. What is the name and formula, of 2nd member of homologous, series of alcohols?, , Epoxide are cyclic ethers in which, the ethereal oxygen is a part of a, three membered ring., , 2. What is the structural formula of, functional group of ether?, , C, O, , 3. What is the name of the compound, having -OH group bonded to benzene, ring?, , H 2C, , O, , CH2, , (Ethylene oxide) (1,2 - Epoxyethane), , 11.1 Introduction : Alcohols are organic, compounds whose molecules contain hydroxyl, group, (-OH) attached to a saturated carbon, atom., C, , C, , OH, , Hydroxyl group can also be present in aromatic, compounds. There are two types of aromatic, hydroxy compounds: phenols and aromatic, alcohols. Phenols contain a hydroxyl group, directly attached to the carbon atom of benzene, ring. When the hydroxyl group is present in the, side chain of aromatic ring, the compound is, termed as aromatic alcohol., , 11.2 Classification : Let us first consider, classification of alcohols, phenols and then, ethers., 11.2.1 Mono, di, tri and polyhydric, compounds : Alcohols and phenols are, classified as mono, di-, tri, or polyhydric, compounds on the basis of one, two, three, or more hydroxyl groups present in their, molecules as :, CH3 - CH2 - OH, , H2C - OH, H2C - OH, , H2C - OH, HC - OH, H2C - OH, , OH, , OH, OH, , OH, , OH, , OH, , OH, , OH, Monohydric Dihydric alcohols/ Trihydric, alcohols/phenols, alcohols/phenols, phenols, , Phenols, , CH2-OH, Aromatic alcohol, , Ethers are compounds which contain, an oxygen atom bonded to two alkyl groups, or two aryl groups or one alkyl and one aryl, group. Ethers are organic oxides. Ethers are, considered as anhydrides of alcohols., R-O-R', R-O-Ar, Ar-O-Ar'., , Monohydric alcohols are further classified on, the basis of hybridisation state of the carbon, atom to which hydroxyl group is attached., a. Alcohols containing sp3C - OH bond : In, these alcohols -OH group is attached to a sp3, hybridised carbon atom of alkyl group. These, alcohols are further classified as primary, (10), secondary (20) and tertiary (30) alcohols, in which -OH group is attached to primary,, secondary and teriary carbon atom respectively., (see Fig. 11.1), also refer to Std. XI Chemistry, Textbook Chapter 14, sec. 14.3.2), , 234
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H, R, , C, , Primary, carbon, , R, , OH, , R, , H, 10 alcohol, , C, , b. Alcohols contianing sp2C -OH bonds :, , Secondary, carbon, , In these alcohols -OH group is attached, to a sp2 hybridised carbon atom which is part of, a carbon-carbon double bond. These alcohols, are known as vinylic alcohols. For example, , OH, , H, 20 alcohol, , CH2 = CH - OH (Vinyl alcohol), , tertiary, , R, , R carbon, C OH, , 11.2.2 Classification of Ethers : Ethers are, classified as symmetrical ethers (simple, ethers) or unsymmetrical ethers (mixed, ethers) depending on whether the two alkyl/, aryl groups bonded to oxygen atom are same, or different respectively. For example :, , R, 30 alcohol, Fig. 11.1 : Primary, secondary and tertiary, alcohols, , Each of these three types of alcohols can also, be either allylic or benzylic if the sp3 carbon, carrying -OH is further bonded to sp2 carbon., •, , •, , R - O - R/Ar - O - Ar, CH3 - O - CH3, symmetrical ethers (simple ethers), , Allylic alcohols : In this type of alcohols, -OH group is attached to sp3 hybridised, carbon atom which is further bonded to a, carbon-carbon double bond. Allylic alcohol, may be primary, secondary or tertiary., Benzylic alcohols : In this type of alcohols, -OH group is attached to sp3 hybridised, carbon atom which is further bonded to an, aromatic ring. Benzylic alcohol may be, primary, secondary or tertiary., , Use your brain power, Classifiy the following, alcohols as 10 / 20 / 30 and allylic/, benzylic, CH3, H2C = CH - CH2 - OH,, H2C = CH - CH - OH,, CH3, , R - O - R'/Ar - O - Ar', CH3 - O - C2H5, C6H5 - O - CH3, unsymmetrical ethers (mixed ethers), , 11.3 Nomenclature :, 11.3.1 Alcohols : There are three systems of, nonmenclature of monohydric alcohols., a. Common/trivial names : The common, or trivial names of alcohols are obtained by, adding word alcohol after the name of alkyl, group bonded to -OH. Names of higher alkyl, groups also include prefixes like normal, iso,, secondary, tertiary (see. Table 11.1)., b. Carbinol system : In this system alcohols, are considered as derivatives of methyl, alcohol which is called carbinol. The alkyl, group attached to the carbon carrying -OH, group are named in alphabetical order. Then, the suffix carbinol is added. For example :, , CH3, , H Carbinol carbon, , H2C = CH - C - OH,, CH3, , H 3C, , OH, , Methyl carbinol, , CH3, CH2 OH, , C, H, , CH OH, , 235
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alcohol. (Refer to Std. XI Chemistry Textbook, section 15.2.4). This is an antimarkownikoff, hydration of alkene., , Problem 11.1 : Draw structures of, following compounds., i. 2,5-Diethylphenol ii. Prop-2-en-1-ol, , Do you know ?, The mechanism of acid catalyzed, hydration of alkene involves the, following three steps:, , iii. 2-methoxypropane iv. Phenylmethanol, Solutuion :, i. OH, 6, 5, , H5C2, , 1, , 2, 3, , C2H5, , ii. , 3, 2, 1, H2C = CH - CH2 - OH, , Step 1: Formation, intermediate., H, H-O-H+ C=C, ⊕, , 4, , iii. CH3-CH-CH3, O-CH3, , iv., , CH2 - OH, , of, , carbocation, , H 2O + C - C, ⊕, H, , Step 2: Nucleophilic attack of H2O on C⊕, Try this..., Write IUPAC names, following compounds., OH OH, i. , , the, , C-C +H-O, ⊕, H, , ii. H3C-CH-CH2- CH3, OCH2- CH3, , Step 3: Deprotonation, , of, , C-C, , iii., , CH2 - CH2 - OH, , H, , iv. H C, 3, , H, , OH, , C-C, H, , H2O, , O⊕, H H, , + H3O⊕, , C-C, , O⊕, H H, , H, , OH, , Use your brain power, , 11.4 Alcohols and Phenols :, , Predict the major product of the, following reactions :, , 11.4.1 Prepartion of alcohols :, a. From alkyl halide by hydrolysis with, aqueous alkali or moist silver oxide (refer to, section 10.6.2), , •, •, , b. By acid catalyzed hydration of alkenes :, Alkene reacts with sulfuric acid to produce, alkyl hydrogen sulfate, which on hydrolysis, gives alcohol (Refer to Std XI Chemistry, Textbook, section 15.2.4). This reaction, follows Markownikoff’s rule., c. Hydroboration - Oxidation of alkenes :, With diborane (B2H6) alkene undergoes, addition reaction (Hydroboration) to give to, trialkylborane (R3B), which on oxidation with, hydrogen peroxide in alkaline medium gives, , CH3 - CH = CH2, CH3, , (i)B2H6 - THF, , ?, , (ii)H2O2, OH, , (i) con. H2SO4, , ?, , (ii) H2O, , d. By reduction of carbonyl compounds :, i. By reduction of aldehydes and ketones :, Aldehydes on reduction by H2/Ni or, LiAlH4 give primary alcohols (10). Similarly, ketones on reduction with H2/Ni or LiAlH4, give secondary alcohols (20)., , 238, , R - CHO, , H2/Ni or Pd, ∆, (i) LiAlH4, (ii) H3O⊕, , R- CH2 - OH, 10 alcohol
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O, C, , R, , H2/Ni or Pd, ∆, (i)LiAlH4, , R, , (ii) H3O⊕, , R, , CH, , Problem 11.2 : Predict the products for the, following reaction., , OH, , R, 20 alcohol, , CH3 - CH = CH - CH2 - CHO, , ii. By reduction of carboxylic acids :, Caboxylic acids require strong reducing agent, LiAlH4 to form primary alcohols., R, , O, C, , (i) LiAlH4, , OH, , (ii) H3O⊕, , (A), , H⊕, , RCOOR' + 2H2, , Ni/Pd, ∆, , R - CH2 - OH, , R - COOR' + H2O, , R - CH2OH + R'OH, , ?, , (i) LiAlH4, (ii) H3O⊕, , Solution : The substrate (A) contains an, isolated C = C and an aldehyde group., H2/Ni can reduce both these functional, groups while LiAlH4 can reduce only -CHO, of the two, Hence, , However LiAlH4 is an expensive reagent., Therefore, commercially acids are first, transformed into esters which on catalytic, hydrogenation give primary alcohols., R - COOH + R'OH, , ?, , H2/Ni, , Ni, , (A) (i, , H 2/, , CH3-CH2-CH2-CH2-CH2-OH, , )L, (ii) iAlH, H O⊕ 4, , CH3-CH=CH-CH2-CH2-OH, , 3, , This reaction is useful in synthesis of a variety, of alcohols (see Table 11.4)., Table 11.4 Preparation of alcohols by Grignard, reagent, , Remember..., The advantage of LiAlH4 over, H2/Ni is that it does not reduce the, isolated olefinic bond and hence it can, reduce unsaturated aldehyde and ketones to, unsaturated alcohols., , Aldehyde/, ketone, , H - CHO, R - Mg Br, (formaldehyde), , e.By addition of Grignard reagent to, aldeheydes and ketones : Grignard reagent, reacts with aldehyde or ketone to form an, adduct which on hydrolysis with dilute acid, gives the corresponding alcohols., δ, , O, δ δ⊕, δ⊕C + R - Mg - X, , dry, ether, , R' - CHO, (aldehyde), , R - Mg Br, , R' - CO - R'', (ketone), , R - Mg Br, , Final product, , Type of, alcohol, , R - CH2OH, , 10, , R - CH - OH, R', R'', R - C - OH, R', , 20, , 30, , Do you know ?, , OMgX, -CR, , Epoxide reacts with Girgnard, reagent followed by acidic hydolysis, to give primary alcohols, , (adduct), H3O⊕, , Grignard, reagent, , H2C - CH2 + R Mg X, , OH, X, - C - + Mg, OH, R, , dry, ether, , O, [R - CH2- CH2 - OMgX], , H3O⊕, , R-CH2CH2-OH + Mg, , 239, , X, OH
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Problem 11.4 : The solubility of o-nitrophenol, and p-nitrophenol is 0.2 g and 1.7 g/100 g of, H2O respectively Explain the difference., , Ar - OH + NaOH(aq), , Ar-O Na⊕(aq), + H2O(l), , Ar - O Na⊕(aq) + HCl(aq), , Ar - OH↓, + NaCl(aq), , Solution :, δO δ⊕, H (intramolecular hydrogen, ⊕ O - bonding in o-nitrophenol), N, Oδ δ⊕, , H, , δ-, , O, δ⊕, , O-H, , O, , ⊕, , N, -, , δ⊕, , O-H, , δ-, , O, , Do you know ?, Sodium bicarbonate, sodium, hydroxide, sodium metal are, increasingly strong bases. Weak and strong, acids can be distinguished from each other, qulitatively by testing their reactivity, towards bases of different strengths. A, weak acid does not react with a weak, base, it requires a stronger base instead., Hence phenols react with NaOH but not, with NaHCO3. A strong acid shows high, reactivity towards weak as well as strong, base. For example : HCl is a strong acid., Its reacts with both NaHCO3 and NaOH as, shown below:, , δ⊕, , H, , H, , δ⊕, , (intermolecular hydrogen bonding in p-nitrophenol, and water), , p-Nitrophenol has strong intermolecular, hydrogen bonding with solvent water. On, the other hand, o-nitrophenol has strong, intramolecular hydrogen bonding and, therefore the intermolecular attraction, towards solvent water is weak. The stronger, the intermolecular attraction between solute, and solvent higher is the solubility. Hence, p-nitrophenol has higher solubility in water, than that of o-nitrophenol., , HCl(aq)+NaHCO3(aq), H2O(l)+ NaCl(aq)+ CO2↑, HCl(aq)+NaOH(aq), , 11.4.4 Chemical properties of Alcohols and, Phenols, , •, , a. Laboratory tests of alcohols and phenols :, i. Litmus test : Water soluble alcohols and, phenols can be tested with litmus paper., Aqueous solution of alcohols is neutral to, litmus (neither blue nor red litmus change, colour). Aqueous solutions of phenols turn, blue litmus red. Thus, phenols have acidic, character., ii. Reaction with bases :, •, , Acid strength of phenols being very low,, phenols cannot react with NaHCO3 but, react with NaOH., , Ar - OH + NaHCO3(aq), , No reaction, , Phenols dissolve in aqueous NaOH by forming, water soluble sodium phenoxide and are, reprecipitated/regenerated on acidification, with HCl., , NaCl(aq)+H2O(l), , Alcohols show no acidic character in, aqueous solution, thus, alcohols do, not react with either aqueous NaHCO3, or aqueous NaOH. Very weak acidic, character of alcohol is revealed in the, reaction with active metal. When alcohols, are treated with very strong base like alkali, metal Na or K they react to give sodium, or potassium alkoxide with liberation of, hydrogen gas., , 2R - OH + 2Na, , 2R-O Na⊕ + H2↑, , Liberation of H2 gas is used to detect the, presence of alcoholic -OH group in a molecule., iii. Characteristic test for phenols : Phenols, reacts with neutral ferric chloride solution to, give deep (purple/violet/green) colouration of, ferric phenoxide., , 3Ar - OH + FeCl3, , 242, , (neutral), , (Ar -O)3 Fe + 3HCl, (deep colour)
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iv. Distinguishing test for alcohols (Lucas, test) : Primary, secondary and tertiary, alcohols can be distinguished from each other, in the laboratory using Lucas reagent (conc., HCl and ZnCl2). The reaction involved is :, R - OH, , HCl, ZnCl2, , R - Cl, , Alcohols are soluble in Lucas reagent but, the product alkyl chloride is not. Hence, the, clear solution becomes turbid when product, starts forming. Tertiary alcohols reacts fast, and the reagent turns turbid instantaneously., Secondary alcohols turn the reagent turbid, slowly. Primary alcohols turn the reagent, turbid only on heating., , Electron donating inductive effect (+I effect) of, alkyl group destabilizes the alkoxide ion (the, conjugate base of alcohol). As a result alcohol, does not ionize much in water, and behaves, like neutral compound in aqueous medium., • Ionization of phenol is represented by the, equilibrium shown in Fig. 11.1., O, , O-H, + H2O, , + H3O⊕, (phenoxide), , Phenol, O, , O, , O, , I, , II, , III, , b. Reactions due to breaking of O -H bond., i. Acidic character of alcohols and phenols :, From the laboratory tests it is, understood that in aqueous medium phenols, show weak acidic character while alcohols are, neutral. It is clear, therefore, that the reactivity, of alcohols and phenols towards ionization, of O-H bond in them is different. The reason, behind this difference lies in the extent of, stabilization of their respective conjugate, bases by electronic effects as shown below., •, , Ionization of alcohols is represented by the, following equilibrium, R - OH + H2O, (alcohol), , R O + H3O⊕, (alkoxide), , O, , O, , IV, , V, , Fig. 11.1 Ionization of phenol and resonance, stabilization phenoxide ion, , Phenoxide ion, the conjugate base of, phenol, is resonace stabilized by delocalization, of the negative charge.Therefore phenol ionizes, in aqueous medium to a moderate extent, and, thereby shows a weak acidic character., , Problem 11.5 : Arrange the following compounds in decreasing order of acid strength and, justify., i. CH3-CH2-OH, , ii. (CH3)3C-OH, , iii. C6H5-OH , , iv. p-NO2-C6H4-OH, , Solution : Compounds (iii) and (iv) are phenols and therefore are more acidic than the alcohols, (i) and (ii). The acidic strenghts of compounds depend upon stabilization of the corresponding, conjugate bases. Hence let us compare electronic effects in the conjugate bases of these, compounds :, Alcohols :, CH3, , CH2, , H3C, O (Conjugate base of (i)) and H3C, H3C, , 243, , C, , O, , (Conjugate base of (ii))
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The conjugate base of the alcohol (i) is destabilized by +I effect of one alkyl group,, where as conjugate base of the alcohol (ii) is destabilized by +I effect of three alkyl groups., Hence (ii) is weaker acid than (i), O, , O, , O, , O, , O, , O, , N, O ⊕ O, I, , N, O ⊕ O, II, , N, O ⊕ O, III, , N, O ⊕ O, IV, , N, O ⊕ O, V, , N, O ⊕ O, VI, , Phenols : The conjugate base of p-nitrophenol (iv) is better resonance stabilized due to six, resonance structures compared to the five resonance structures of conjugate base of phenol (iii), (see Fig. 11.1). The resonance structure VI has -ve charge on only electronegative oxygens., Hence the phenol (iv) is stronger acid than (iii). Thus the decreasing order of acid strength is, (iv) > (iii) > (i) > (ii), O, H⊕, R-OH + HO-C -R', , Use your brain power, What are the electronic effects, exerted by -OCH3 and -Cl ? predict, the acid strength of H3C-O-OH and, Cl -OH relative to parent phenol, -OH ., ii. Esterification : Alcohols and phenols form, esters by reaction with carboxylic acid, acid, halides and acid anhydrides. The reaction, between alcohol or phenol with a carboxylic, acid to form an ester is called esterification., , (alcohol), , (acid), , O, R-O-C -R' + H2O, (ester), , O, O, H⊕, Ar-OH + HO-C -R', Ar-O-C -R' + H2O, (phenol), , (acid), , (ester), , Alcohols and phenols react with acid, anhydrides in presence of acid catalyst to form, ester., O O, O, H⊕, R-OH+R'-C -O-C -R', R-C -OR'+R'-COOH, (alcohol) (anhydride), , Esterification of alcohol or phenol is carried, out in the presence of concentrated sulphuric, acid. The reaction is reversible and can be, shifted in the forward direction by removing, water as soon as it is formed., , (ester), , (acid), , O O, O, H⊕, Ar-OH+R'-C -O-C -R', Ar-O-C -R'+R'-COOH, (phenol), , 244, , (anhydride), , (ester), , (acid)
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The reaction of alcohol and phenols with, acid chloride is carried out in the presence of, pyridine (base), which neutralizes HCl., O, O, pyridine, R'-C -OR + HCl, R-OH + Cl-C -R', , ii. Reaction with phosphorous halide :, Alcohols react with phosphorous pentahalide, (PX5) and phosphorous trihalide (PX3) to form, alkyl halides. (refer to Chapter 10 section, 10.3.1)., , (alcohol) (acid chloride), , iii. Dehydration of alcohols to alkenes :, Alcohol when treated with concerntrated, sulphuric acid or phosphoric acid or alumina, undergoes dehydration to form alkene and, water. (refer to Std. XI Chemistry Textbook, section 15.2) The reaction gives more, substituted alkene as the major product, in, accordance with Saytzeff rule., , O, Ar-OH+ Cl-C -R', , pyridine, , (ester), , O, R'-C -O-Ar + HCl, , (phenol) (acid chloride), , (ester), , Acetyl derivatives : The CH3-CO- group, is called acetyl group. The acetate esters of, alcohols or phenols are also called ‘acetyl, derivatives’ of alcohols or phenols repectively., The number of alcoholic or phenolic -OH, groups in the given compound can be deduced, from the number of acetyl groups introduced, in it as a result of acetylation. Aspirin, a, well known generic medicine, is an acetyl, derivative of salicylic acid formed by its, acetylation using acetic anhydride., COOH, OH, , Problem 11.6 : Write the reaction showing, major and minor products formed on heating, butan-2-ol with concentrated sulfuric acid., Solution : In the reaction described, butan-2-ol undergoes dehydration to give, but-2-ene (major) and but-1-ene (minor) in, accordance with Saytzeff rule., , O O, H⊕, + CH3-C -O-C -CH3, , OH, CH3-CH -CH2-CH3Conc H2SO4, , (Salicylic acid) (Acetic anhydride), , CH3-CH=CH-CH3, But-2-ene (major), , ∆, , (Butan-2-ol), , CH2=CH-CH2-CH3, , COOH, , But-1-ene (minor), , O, -O-C -CH3 + CH3COOH, (Acetic acid), , Do you know ?, , (Aspirin/, Acetyl salicyclic acid), , According to the common accepted, mechanism dehydration involves, following three steps., , c. Reaction due to breaking of C-O bond in, alcohols :, i. Reaction with hydrogen halides :, Alcohols reacts with hydrogen halides to form, alkylhalides (refer to Chapter 10 section 10.3.1), In general, tertiary alcohols react rapidly with, hydrogen halides; secondary alcohols react, somewhat slower; and primary alcohols,, even more slowly. The order of reactivity of, hydrogen halides is, HI > HBr > HCl, , ⊕, , 1. Formation of protonated alcohols R-OH2, 2. Its slow dissociation into carbocation, 3. Fast removal of hydrogen ion to form, alkene., ⊕, -C -C- H, , H OH, , fast, , (Alcohol), , HCl reacts only in the presence of anhydrous, ZnCl2. No catalyst is required in the case of, HBr and HI., , 245, , -C - C-, , -H2O, , H ⊕OH2, , (Protonated, alcohol), , slow, , -C -C -, , H, , ⊕, , -H⊕, -C = Cfast, , (Carbocation) (Alkene)
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The intermolecular hydrogen bonding that, holds alcohol molecules together strongly, is, not present in ethers and alkanes. However,, solubility/miscibility of ethers in water is, similar to that of alcohols of comparable, molecular mass. This is because ethers can, form hydrogen bonds with water through the, ethereal oxygen., δ, , R-O-R + H-O-H, , H 3O ⊕, ∆, ⊕, , 2R-OH, , R-O-R' + H-O-H, , H, ∆, , R-OH + R'-OH, , Ar-O-R + H-O-H, , H⊕, ∆, , Ar-OH + R-OH, , ii. Reaction with PCl5 : Ethers react with PCl5, to give alkyl chlorides, R-O-R' + PCl D R-Cl + R'-Cl + POCl, 5, , δ⊕, , R- O H -O, H, R, , 3, , For example diethyl ether and n-butyl alcohol, have respective miscibilities of 7.5 and 9g per, 100 g of water., , iii. Reaction with hot concentrated acid :, Alkyl ethers react with hot and concentrated, HI and HBr to give an alcohol and an alkyl, halide., R-O-R + HX, R-X + R-OH HX R-X, , 11.5.3 Chemical properties of ethers :, , R-OH HX, , a. Laboratory test for ethers : Ethers are, neutral compounds in aqueous medium., Ethers do not react with bases, cold dilute, acids, reducing agents, oxidizing agents and, active metals. However, ethers dissolve in, cold concerntrated H2SO4 due to formation of, oxonium salts., H, R-O-R' + H2SO4, R-O-R' HSO4, , The order of reactivity of HX is HI>HBr>HCl, , ⊕, , This property distinguishes ethers from, hydrocarbons., b. Reaction involving alkyl group of ether :, i. Formation of peroxide : Ethers combine, with atmospheric oxygen to form peroxide., O-OH, CH3-CH2-O-C2H5 + O2 Long CH3-CH-O-C2H5, (diethyl ether), , contact, with air, , (oxygen), , (peroxide of, diethyl ether), , R-X + H2O, , Do you know ?, Mechanism of first stage : Reaction, of ether with hot concentrated, HI involves formation of oxonium ion by, protonation in the first step and subsequent, nucleophilic substitution reaction brought, about by the powerful nucleophile I . The, least substituted carbon in oxoinium ion is, attacked by I following SN2 mechanism., ⊕, ∆, CH3-O-CH2-CH3, CH3-O-CH2-CH3 + H-I, H, + I, I, , ⊕, CH3-O-CH2-CH3, H, , 1, 2, , 1, ⊕, 2, , I CH3 O-CH2-CH3, H, CH3- I + CH3-CH2-OH, , For example :, • Use of excess HI converts the alcohol into, alkyl iodide., • In case of ether having one tertiary alkyl, group the reaction with hot HI follows SN1, mechanism, and tertiary iodide is formed, rather than tertiary alcohol., ⊕, slow, Step 1 : (CH3)3C-O-CH3, (CH3)3C⊕ +, H, , CH3OH, , All ethers which have been exposed to, the atmosphere contain peroxide. This is very, undesirable reaction. Peroxides are hazardous, because they decompose violently at high, temperature., c. Reaction involving C-O bond, i. Reaction with hot dilute sulphuric acid, (Hydrolysis) : Ethers when heated with, dilute sulfuric acid undergo hydrolysis to give, alcohols/phenols., , Step 2 : (CH3)3C⊕ + I, , 250, , fast, , (CH3)3C-I
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ix. Ethers are kept in air tight brown bottles, because, , 4., , An ether (A), C5H12O, when heated, with excess of hot HI produce two, alkyl halides which on hydrolysis form, compound (B)and (C), oxidation of (B), gave and acid (D), whereas oxidation, of (C) gave a ketone (E). Deduce the, structural formula of (A), (B), (C), (D), and (E)., , 5., , Write structural formulae for, , A. Ethers absorb moisture, B. Ethers evaporate readily, C. Ethers oxidise to explosive peroxide, D. Ethers are inert, x., , Ethers reacts with cold and concentrated, H2SO4 to form, A. oxonium salt , , B. alkene, , C. alkoxides , , D. alcohols, , b. Methyl vinyl ether, c. 1-Ethylcyclohexanol, , 2., , Answer in one sentence/ word., , i., , Hydroboration-oxidation, gives....., , ii., , Write the IUPAC name of alcohol having, molecular formula C4H10O which is, resistant towards oxidation., , of, , a. 3-Methoxyhexane, , d. Pentane-1,4-diol, , propene, , e. Cyclohex-2-en-1-ol, 6., , HO, , i., , iii. Write structure of optically active alcohol, having molecular formula C4H10O, , Answer in brief., , i., , Explain why phenol is more acidic than, ethyl alcohol., , ii., , Explain why p-nitrophenol is a stronger, acid than phenol., , CH3, , ii. CH3-CH -CH -CH2-OH, OH CH3, , iv. Write name of the electrophile used in, Kolbe’s Reaction., 3., , Write IUPAC names of the following, , iii., , OH, , iv., , O-CH3, , NO2, , iii. Write two points of difference between, properties of phenol and ethyl alcohol., , Activity :, , iv. Give the reagents and conditions necessary, to prepare phenol from, , •, , Collect information about, production of ethanol as byproduct, in sugar industry and its, importance in fuel economy., , •, , Collect information about phenols, used as antiseptics and polyphenols, having antioxidant activity., , a. Chlorobenzene, b. Benzene sulfonic acid., v., , Give the equations of the reactions for, the preparation of phenol from isopropyl, benezene., , vi. Give a simple chemcial test to distinguish, between ethanol and ethyl bromide., , 253
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12. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS, Can you recall ?, •, , •, , 12.2.1, Classification of aldehydes :, (Aldehydes are classified as aliphatic and, aromatic aldehydes), , Draw the structures of the, following compounds and, classify them on the basis of, C-O single bond and C = O double, bond present in them, Ethyl alcohol,, acetaldehyde, o-nitrophenol, Diethyl, ether, isopropyl alcohol, acetone., What are carbonyl compounds?, , 12.1 Introduction : In the previous chapter,, you learnt about the organic compounds, which contain carbon –oxygen single bond. In, this chapter, we are going to study the organic, compounds containing carbon - oxygen double, bond (>C=O ) called carbonyl group, which, is one of the most important functional group, in organic chemistry., O, , a. Aliphatic aldehydes : The compounds, in which the –CHO group (formyl group) is, attached directly to sp3 hybridized carbon, atom that is saturated carbon atom are, called aliphatic aldehydes. (Exception :, Formaldehyde, H-CHO is also classified as, aliphatic aldehyde though –CHO group is not, attached to any carbon ). For example :, CH3 - CHO, , CH3 - CH2 - CHO, , (Acetaldehyde), , ( Propionaldehyde), , O, R, , C, , H, , General formula, (R= H or alkyl group), , carbonyl oxygen, , carbonyl carbon, C, ', R, R, Both aldehydes and ketones contain, a carbon –oxygen double bond (-C-) as their, O, functional group. Therefore they are called, carbonyl compounds. In aldehydes , carbonyl, carbon is bonded to at least one hydrogen apart, from an alkyl or aryl group. The functional, group of aldehydes, therefore, is - CHO which, is called formyl group or aldehydic carbonyl, group. On the other hand in ketones, carbonyl, carbon is bonded to two alkyl or aryl groups, either identical (R = R) or different (R ≠ R')., It is called ketonic carbonyl group. The, functional group of carboxylic acids is -COOH, called carboxyl group. Due to the -OH group, bonded to (>C=O ) group, carboxylic acids are, distinct from aldehydes and ketones., , b. Aromatic aldehydes : The compounds in, which –CHO group is attached directly to an, aromatic ring are called aromatic aldehydes., For example :, H O, H O, H O, OH, NO2, (Benzaldehyde) (Salicylaldehyde)(p-Nitrobenzaldehyde), , 12.2 Classification of aldehydes, ketones, and carboxylic acids : Aldehydes, ketones, and carboxylic acids are classified as per the, nature of carbon skeleton bonded to (>C=O )., , 254, , Use your brain power, Classify the following as aliphatic, and aromatic aldehydes., O, , O, H, , H, CHO, , CHO, CH3
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12.2.2 Classification of ketones : Ketones are, classified as aliphatic and aromatic ketones:, , Use your brain power, •, , a. Aliphatic ketones : The compounds in, which >C=O group is attached to two alkyl, groups are called aliphatic ketones., O, R, , C, , •, , Oxo or ketonic carbonyl group, , Do you know ?, • Aldehydes and ketones are, responsible for many flavours and, odours that you will readily recognize :, CHO, CHO, , R', , General formula, (Where R , R’ = alkyl group, identical or different), , On the basis of types of alkyl groups bonded to, carbonyl carbon, aliphatic ketones are further, classified as simple and mixed ketones., , OH, , i. Simple or symmetrical ketones : The, ketones in which both the alkyl groups bonded, to carbonyl carbon are identical, are called, simple ketones or symmetrical ketones. For, example :, O, O, H5C2 - C - C2H5, H3C - C - CH3, (Dimethyl ketone) ( Acetone), , Benzaldehyde, (Bitter almond flavour), , (Diethyl ketone), , Benzophenone, (Diphenyl ketone), , Do you know ?, , CH3, , Acetophenone, (Methyl phenyl ketone), , Camphor, (Camphor fragance), , • Structures of many important biological, compounds contain carbonyl moiety. For, example progesterone and testosterone,, the female and male sex hormones, respectively., , (Ethyl n-propyl ketone), , O, C, , Vanillin, (Vanilla flavour), , O, Cinnamaldehyde, (Cinnamon flavour), , b. Aromatic ketones : The compounds in, which a >C=O group is attached to either two, aryl groups or one aryl and one alkyl group are, called aromatic ketones. For example :, O, C, , OCH3, , CH = CH - CHO, , ii. Mixed or unsymmetrical ketones : The, ketones in which two alkyl groups bonded, to carbonyl carbon are different, are called, mixed ketones or unsymmetrical ketones. For, example :, O, O, H5C2 - C - CH2 - CH2 - CH3, H5C2 - C - CH3, (Ethyl methyl ketone), , Classify the followings as, simple and mixed ketones., Benzophenone, acetone, butanone,, acetophenone., , •, , Butyraldehyde is used in, margarine and food for its, buttery odour., , •, , Acetophenone has smell of pistachio, and is used in ice-cream. Muscone has, musky aroma and is used in perfumes., Popcorn has butter flavour which, contains butane-2,3-dione., , 12.2.3 Classification of carboxylic acids :, Carboxylic acids are classified as aliphatic, and aromatic carboxylic acids :, , 255
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carbonyl groups. In case of polyfunctional, ketones, higher priority group is given lower, number. When ketonic carbonyl is a lower, priority group it is named as 'oxo', preceded, by the locant. In alicyclic ketones, carbonyl, carbon is numbered as 1. (Refer Table 12.2)., , ii. By dehydrogenation of alcohols : This, method has industrial application. Aldehydes, and ketones are prepared by passing the, vapours of primary and secondary alcohols, respectively over hot copper powder. (See, Chapter 11), b. From hydrocarbons :, , Use your brain power, Write IUPAC names for the, following compounds., i., , H, Br, , ii., , Br, , O, , H, , iii., O, CHO, , •, , What is ozonolysis ?, , •, , What is the role of zinc dust in, ozonolysis process?, , i. By ozonolysis : Alkene reacts with ozone to, give ozonide which on decomposition with zinc, dust and water gives aldehyde and/or ketones., (See Std. XI Chemistry Textbook, Chapter 15), , O, O, , iv., , Can you recall ?, , v., , O, , Try this..., Draw structures for the following, a. 2-Methylpentanal, b. Hexan-2-one, , ii. By hydration of alkynes : Alkynes react, with water in presence of 40% sulfuric acid, and 1% mercuric sulfate to give aldehydes, or ketones. (See Std. XI Chemistry Textbook,, Chapter 15), 12.4.2 Other methods of preparation of, aldehydes and ketones : Some methods of, preparation of aldehydes and ketones involve, common starting functional groups but, different types., a. From acyl chlorides (Acid chlorides) :, Aldehydes and ketones both can be obtained, from acyl chloride, but the reactions involved, are different., , 12.4 Preparation of aldehydes and ketones :, 12.4.1 General methods of preparation of, aldehydes and ketones :, , •, , a. By oxidation of alcohols : i. Aldehydes, and ketones are prepared by the oxidation of, primary and secondary alcohols respectively., (See Chapter 11 ), , Acyl chloride is reduced to corresponding, aldehyde by hydrogen using a palladium, catalyst poisoned with barium sulfate. This, reaction is known as Rosenmund reduction., , Can you tell ?, What is the reagent which, oxidizes primary alcohols to, only aldehydes and does not oxidize, aldehydes further into carboxylic acid ?, , Preparation of aldehyde from acyl, chloride, , O, R - C - Cl, , H2, Pd-BaSO4, , (Acyl chloride), , 260, , O, R - C - H + HCl, (Aldehyde)
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b. Anhydrides on hydrolysis with water give, carboxylic acids., , Try this..., Draw the structure of the product, formed by the combination of, carbon monoxide and HCl., , O, O, R - C - O - C - R + H2O, , O, 2R - C - O - H, , (Anhydride), , Use your brain power, , (Carboxylic acid), , 12.5.3 From esters : Carboxylic acids can be, obtained from esters either by acid hydrolysis, or alkaline hydrolysis., , Identify the reagents necessary, to achieve each of the following, transformation, O, O, Cl, , a. Acid hydrolysis of ester : Esters on, hydrolysis with dilute mineral acid like dilute, HCl or dilute H2SO4 give the corresponding, carboxylic acid ., O, O, SO, R - C - O - R' + H2O dil.H, R, C, -O-H, ∆, , H, O, , 2, , 4, , (Ester), , + R' - OH, , O, , O, , (Alcohol), , H, , OCH3, O2N, , O2N, , 12.5 Preparation of carboxylic acids :, 12.5.1 From nitriles and amides : Alkyl, nitriles or aryl nitriles on acid hydrolysis give, amides . Amides on further acid hydrolysis give, corresponding carboxylic acids. Hydrolysis is, carried out by using dilute mineral acids like, dilute sulfuric acid or dilute hydrochloric acid., R - C ≡ N + H2O, , OH, [R - C = NH], , O, R - C - NH2, , H 2O, dil.HCl, , O, H5C2 - C - O - CH3 + dil.NaOH, , R - C ≡ N + 2H2O + dil.HCl, , O, ⊕, H5C2 - C - O - Na + CH3 - OH, O, ⊕, H5C2 - C - O - Na + H2O, , R - COOH + NH3, ∆, , R - COOH, + NH4Cl, , 12.5.2 From acyl chloride and anhydrides :, a. Acyl chlorides on hydrolysis with water give, carboxylic acids. This method is useful for, preparation of aliphatic as well as aromatic, acid., R - COOH + H - Cl, (Carboxylic acid), , ∆, , (Methyl propanoate), , (Sodium propanoate), , (carboxylic acid), , (Amide), , (Acyl chloride), , b. Alkaline hydrolysis of ester using dilute, alkali like dilute NaOH or dilute KOH form, solution of water soluble sodium or potassium, salt of the acid (carboxylate). On acidification, with concentrated HCl, free acid is formed., , (Sodium propanoate), , (Nitrile), , R - COCl + H2O, , (Carboxylic acid), , H⊕, Conc. HCl, , O, H5C2 - C - O - H + NaOH, (Propanoic acid), , The sodium or potassium salts of higher fatty, acids are known as soaps. Hence alkaline, hydrolysis of esters is called saponification, (Std XI Chemistry Textbook, Chapter 16)., 12.5.4 From alkyl benzene : Aromatic, carboxylic acids can be prepared by oxidation, of alkyl benzene with dilute HNO3 or alkaline, /acidic KMnO4 or chromic acid. The entire, , 263
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Table 12.4, , Number of carbon atoms, 1, 2, 3, 4, 5, 6, , Boiling points of aldehydes and ketones, , Aldehyde, Methanal, Ethanal, Propanal, Butanal, Pentanal, Hexanal, , Boiling point, 252 K, 294 K, 319 K, 348 K, 376 K, 392 K, , 12.6.2 Physical state and boiling points of, aldehydes and ketones :, Formaldehyde is a gas at room temperature, and has irritating odour. Acetaldehdye is, extremely volatile, colourless liquid. Higher, aldehydes have pleasant odour. Acetone is a, liquid at room temperature and has pleasant, odour but most of the higher ketones have, bland odours., Increasing boiling points in the homologous, series of aldehydes and ketones are listed in, Table 12.4., 12.6.3 Solubility of aldehydes and ketones :, The oxygen atom of ( C=O) can involve, in hydrogen bonding with water molecule, (Fig 12.2). As a result of this, the lower, aldehydes and ketones are water soluble (For, example : acetaldehyde, acetone). As the, molecular mass increases, the proportion of, hydrocarbon part of the molecule increases, which cannot form hydrogen bond; and the, water solubility decreases., δ, δ⊕ C = O, , δ⊕ δ, H -O, , Ketone, Boiling point, -------------------------Propanone, 329 K, Butan -2-one, 353 K, Pentan-2-one, 375 K, Hexan-2-one, 400 K, , 12.6.4 Physical state, boiling points and, solubilities of carboxylic acids : Lower, aliphatic carboxylic acids upto nine carbon, atoms are colourless liquids with irritating, odours. The higher homologues are colourless,, odourless wax like solids, have low volatility., Boiling points of lower carboxylic acids are, listed in Table 12.5., Carboxylic acids have higher boiling, points than those of alkanes, ethers, alcohols, aldehydes and ketones of comparable mass, (Table 12.6). The reason is that , in liquid, phase, carboxylic acids form dimer in which, two molecules are held by two hydrogen, bonds. Acidic hydrogen of one molecule form, hydrogen bond with carbonyl oxygen of the, other molecule (Fig.12.3). This doubles the, size of the molecule resulting in increase in, intermolecular van der Waals forces, which in, turn results in high boiling point. In the case of, acetic acid dimers exist even in the gas phase, (Fig.12.3)., δ⊕, δ, O, H O, H3 C - C, , δ⊕, H, , C - CH3, O, , Fig. 12.2 : Hydrogen bonding in carbonyl, compound and water, , H O, δ⊕ δ, , Fig. 12.3 : Dimer of acetic acid (Two molecules, held by two hydrogen bonds), Table 12.5 Increasing boiling points of carboxylic acids, , Name, Formic acid, Acetic acid, , Formula, HCOOH, CH3COOH, , Boiling point in K, 373 K, 391 K, , Propionic acid, , CH3CH2COOH, , 414 K, , Butyric acid, , CH3CH2CH2COOH, , 437 K, , Valeric acid, , CH3CH2CH2CH2COOH, , 460 K, , 265
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Table 12.6 : Variation of boiling point with functional group, Family, , Molecular mass, , Boiling point, , CH3-CH2-CH2-CH3, , Alkane, , 58, , 272 K, , CH3-O-CH2-CH3, , Ether, , 60, , 281 K, , CH3-CH2-CHO, , Aldehyde, , 58, , 322 K, , CH3-CO-CH3, , Ketone, , 58, , 329 K, , CH3-CH2-CH2-OH, , Alcohol, , 60, , 370 K, , CH3-COOH, , Carboxylic acid, , 60, , 391 K, , Remember..., Relative, strength, of, intermolecular force : H-Bond, > dipole-dipole attraction > van der, Waals force. Hence, Boiling points :, Carboxylic acids > Alcohols > Ketones >, Aldehydes > ether > Alkanes, , Strength of intermolecular, forces, increases, , Compound, , solubility of carboxylic acids in water decreases, with increase in molecular mass. Higher, carboxylic acids are practically insoluble, in water due to the increased hydrophobic, (water hating) interaction of hydrocarbon part, with water. Aromatic acids like benzoic acid, are also practically insoluble in water at room, temperature. Water insoluble carboxylic acids, are soluble in less polar organic solvents like, ether, alcohol, benzene, and so on., , Do you know ?, Commercially available forms of, formaldehyde and acetaldehyde:, , 12.7 Polarity of carbonyl group : The, polarity of a carbonyl group originates from, higher electronegativity of oxygen relative to, carbon as well as resonance effects as shown, in Fig. 12.4., , i. Formaldehyde is available commercially, as solid polymer called paraformaldehyde, HO [CH2 - O]n H and trioxane (CH2O)3, (Trioxane has cyclic structure). These are, convenient for use in chemical reactions as, source of formaldehyde., ii. Aqueous solution of formaldehyde, gas is called formalin, which is used for, preservation of biological and anatomical, specimens., , O, , Oδ, , C, , C δ⊕, (A), , iii. When dry formaldehyde is required, it, is obtained by heating paraformaldehyde, or trioxane., , O, , O, , C, , C, ⊕, , major, , iv. Acetaldehyde is also conveniently used, as solid trimer (paraldehyde) and tetramer, (metaldehyde)., , minor, , (B), , (C), , Oδ, , Lower aliphatic carboxylic acids, containing upto four carbons are miscible, with water due to formation of intermolecular, hydrogen bonds between carboxylic acid, molecules and solvent water molecules. The, , 266, , C δ⊕, resonance, hybrid, (D), , Fig. 12.4 : Polarity of carbonyl group
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The carbonyl carbon has positive polarity, (see structures (A) and (D)). Therefore, it is, electron deficient. As a result, this carbon, atom is electrophilic (electron loving) and is, susceptible to attack by a nucleophile (Nu: )., 12.7.1 Reactivity of aldehydes and ketones, : Reactivity of aldehydes and ketones is, due to the polarity of carbonyl group which, results in electrophilicity of carbon. In general,, aldehydes are more reactive than ketones, toward nucleophilic attack. This can be well, explained in terms of both the electronic effects, and steric effect., 1. Influence of electronic effects : Alkyl, groups have electron donating inductive effect, (+I). A ketone has two electron donating alkyl, groups bonded to carbonyl carbon which are, responsible for decreasing its positive polarity, and electrophilicity. In contrast, aldehydes, have only one electron donating group bonded, to carbonyl carbon. This makes aldehydes, more electrophilic than ketones., 2. Steric effects : Two bulky alkyl groups, in ketone come in the way of incoming, nucleophile. This is called steric hindrance to, nucleophilic attack., On the other hand, nucleophile can easily, attack the carbonyl carbon in aldehyde because, it has one alkyl group and is less crowded or, sterically less hindered . Hence aldehyde are, more easily attacked by nucleophiles., Oδ, C, δ⊕, R, R, (Ketone), , Oδ, , Try this..., Draw structure of propanone and, indicate its polarity., 12.8 Chemical properties of aldehydes and, ketones :, 12.8.1 Laboratory tests for aldehydes and, ketones : Aldehydes are easily oxidized to, carboxylic acids and therefore, act as reducing, agents toward mild oxidizing agents. Ketones,, do not have hydrogen atom directly attached to, carbonyl carbon. Hence, they are not oxidized, by mild oxidizing agents. On the basis of this, difference in the reactivity, aldehydes and, ketones are distinguished by the following, tests:, a. Tests given by only aldehydes :, 1. Schiff test : When alcoholic solution of, aldehyde is treated with few drops of Schiff 's, reagent, pink or red or magenta colour appears., This confirms the presence of aldehydic, (-CHO) group., 2. Tollens' test or silver mirror test : When, an aldehyde is boiled with Tollens' reagent, (ammonical silver nitrate), silver mirror, is formed., The aldehyde is oxidized to, carboxylate ion by Tollens' reagent and Ag⊕, ion is reduced to Ag., R - CHO + 2 Ag (NH3)2⊕ + 3OH, (aldehyde), , Remember..., Aromatic aldehydes are less, reactive than aliphatic aldehydes, in nucleophilic addition reactions. This is, due to electron-donating resonance effect, of aromatic ring which makes carbonyl, carbon less electrophilic., , Tollens reagent, , R - COO + 2 Ag↓ + 4NH3↑ + 2H2O, , (carboxylate), , C, δ⊕, R, H, (Aldehyde), , ∆, , (Silver mirror), , 3. Fehling test : When a mixture of an aldehyde, and Fehling solution is boiled in hot water, a, red precipitate of cuprous oxide is formed., An aldehyde is oxidized to carboxylate ion, by Fehling solution and Cu2⊕ ion is reduced, to Cu⊕ ion. It may be noted that α-hydroxy, ketone also gives this test positive., , 267
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Table 12.8 : pKa values of haloacetic acids, , Acid, , Can you recall ?, What is the numerical parameter, to express acid strength?, , pKa, , F-CH2-COOH, Cl-CH2-COOH, , 2.56, 2.86, , Br-CH2-COOH, , 2.90, , I-CH2-COOH, , 3.18, , CH3-COOH, , 4.76, , Acid, strength, decreases, , Carboxylate ion has two resonance structures, (i) and (ii) and both of them are equivalent to, each other (Refer to Std. XI Chemistry Textbook, Chapter 14). This gives good resonance, stabilization to carboxylate ion, which in turn, gives acidic character to carboxylic acids., , Halogens are electronegative atoms and exert, electron withdrawing inductive effect (-I, effect). The negatively charged carboxylate ion, in the conjugate base of haloacetic acid gets, stabilized by the -I effect of halogen. Which is, responsible to diffuse the native charge., , Remember..., Lower Ka value, higher pKa:, Weaker acid., Higher Ka value, lower pKa : stronger acid., Influence of electronic effects on acidity, of carboxylic acids : All the carboxylic, acids do not have the same pKa value. The, structure of 'R' in R-COOH has influence on, the acid strength of carboxylic acids. Various, haloacetic acids illustrate this point very well, (Tables 12.8 and 12.9)., , O, X, , CH2, , C, , O, Higher the electronegativity of halogen greater, is the stabilization of the conjugate base,, stronger is the acid and smaller is the pKa, value., , Problem 12.1, Alcohols (R-OH), phenols (Ar-OH) and carboxylic acids (R-COOH) can undergo ionization, of O-H bond to give away proton H⊕; yet they have different pKa values, which are 16, 10 and, 4.5 respectively. Explain, Solution : pKa value is indicative of acid strength. Lower the pKa value stronger the acid., Alcohols, phenols and carboxylic acids, all involve ionization of an O-H bond. But their, different pKa values indicate that their acid strength are different. This is because the resulting, conjugate bases are stabilized to different extent., Acid(HA), R-O-H, , Conjugate base(A ) Electronic effect, Stabilization/destabilization, R O, +I effect of R destabilization of conjugate base, group, Ar-O-H, Ar-O, -R effect or Ar stabilization of conjugate base is, group, moderate because all the resonance, structures are not equivalent to each, other, O, O, -R effect of C = O stabilization is good because all the, R - C - O-H, R-C-O, group, resonance structures are equivalent, to each other, As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids, are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized,, alcohols are weakest acids and have highest pKa value. As conjugate base of phenols is, moderately stabilized phenols are moderately acidic and have intermediate pKa value., , 276
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ring exerts electron withdrawing inductive, effect (-I effect) which stabilizes the conjugate, base and increases the acid strength of, aromatic acids., , Try this..., Compare the following, conjugate bases and answer., , two, , O, CH3 C, , O, , Table 12.9 illustrates that more the number of, electron withdrawing substituents higher is the, acid strength., , O, Cl - CH2 C, , (a), , O, , (b), , Try this..., , • Indicate the inuctive effects of CH3 -, , Arrange the following acids in, order of their decreasing acidity., , group in (a) and Cl - group in (b) by, putting arrowheads in the middle of, appropriate covalent bonds., , CH3-CH-CH2-COOH, CCl3-CH2-COOH,, Cl, CH3COOH, , • Which species is stabilized by, inductive effect, (a) or (b) ?, , • Which species is destabilized by, , Electron–withdrawing groups like -Cl, -CN,, and -NO2 increase the acidity of substituted, benzoic acids while electron –donating group, like –CH3, - OH , - OCH3 and -NH2 decrease, the acidity of substituted benzoic acids ., , inductive effect, (a) or (b) ?, Use your brain power, , • Compare the pKa values and, , COOH, , COOH, , arrange the following in an, increasing order of acid strength., , Cl3CCOOH, ClCH2COOH, CH3COOH,, Cl2CHCOOH, , COOH, , NO2, , • Draw structures of conjugate bases, , (4-Nitrobenzoic, acid), (pKa = 3.41), , of, monochloroacetic, acid, and, dichloroacetic acid. Which one is more, stabilized by -I effect ?, Acidity of aromatic carboxylic acids :, Benzoic acid is the simplest aromatic acid., From the pKa value of benzoic acid (4.2) we, understand that it is stronger than acetic acid, (pKa 4.76). The sp2 hybrid carbon of aromatic, , CH3, (Benzoic acid), (pKa = 4.2), , (4-Methylbenzoic, acid), (pKa = 4.4), , Try this..., Arrange the following carboxylic, acids in order of increasing acidity., m-Nitrobenzoic acid, Trichloroacetic acid,, benzoic acid, α-Chlorobutyric acid., , Table 12.9 pKa values of chloroacetic acids, , Structure, , pKa, , Monochloroacetic acid, , Cl - CH2 - COOH, , 2.86, , Dichloroacetic acid, , Cl - CH - COOH, Cl, , 1.26, , Trichloroacetic acid, , Cl, Cl C - COOH, Cl, , 0.6, , 277, , Acid strength, , increases, , Name
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Exercises, 1. Choose the most correct option., , d., , i. In the following resonating structures, A and B, the number of unshared, electrons in valence shell present on, oxygen respectively are, O, , O, , C, , C⊕, , (A), a. 2, 4 , , (B), b. 2, 6, , c. 4, 6 , , d. 6, 4, , OCH3, v. Diborane reduces, , b. alcohols, , c. hydrazones , , d. alkenes, , a. ester group , , b. nitro group, , c. halo group, , d. acid group, , vi. Benzaldehyde does NOT show positive, test with, a. Schiff reagent, , ii. In the Wolf -Kishner reduction, alkyl, aryl ketones are reduced to alkyl, benzenes. During this change, ketones, are first converted into, a. acids , , COOH, , b. Tollens' ragent, c. Sodium bisulphite solution, d. Fehling solution, 2. Answer the following in one sentence, i. What are aromatic ketones?, , iii. Aldol condensation is, , ii. Is phenyl acetic acid an aromatic, carboxylic acid ?, , a. electrophilic substitution reaction, b. nucleophilic substitution reaction, , iii. Write reaction showing conversion of, ethanenitrile into ethanol., , c. elimination reaction, d. addition - elimination reaction, iv. Which one of the following has lowest, acidity ?, , iv. Predict the product of the following, reaction:, CH3 CH2 COOCH3, , a. , COOH, , i. AlH (i-Bu)2, ii. H3O⊕, , ?, , v. Name the product obtained by reacting, toluene with carbon monoxide and, hydrogen chloride in presence of, anhydrous aluminium chloride., , NO2, b., COOH, , vi. Write reaction showing conversion of, Benzonitrile into benzoic acid., , Cl, , vii.Name, the, product, obtained, by, the, oxidation, of, 1,2,3,4-tetrahydronaphthalene, with, acidified potassium permanganate ., , c. , COOH, , viii.What is formalin ?, , 280
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ix. Arrange the following compounds in the, increasing order of their boiling points :, , 4. Answer the following, , Formaldehyde, ethane, methyl alcohol., , i. Write a note on –, a. Cannizaro reaction, , x. Acetic acid is prepared from methyl, magnesium bromide and dry ice, in presence of dry ether. Name the, compound which serves not only reagent, but also as cooling agent in the reaction., 3. Answer in brief., , b. Stephen reaction., ii. What is the action of the following, reagents on toluene ?, a. Alkaline KMnO4 , dil. HCl and heat, b. CrO2Cl2 in CS2, , i. Observe the following equation of, reaction of Tollens', reagent with, aldehyde. How do we know that a redox, reaction has taken place. Explain., R CHO + 2 Ag(NH3)2+, R COO- + 2 Ag, , +, , OH-, , ∆, , c. Acetyl chloride, anhydrous AlCl3., , presence, , of, , iii. Write the IUPAC names of the following, structures :, O, a., b. COOH, , + 4 NH3 + 2 H2O, , COOH, , ii. Formic acid is stronger than acetic acid., Explain., iii. What is the action of hydrazine on, cyclopentanone in presence of ---., KOH in ethylene glycol ?, iv. Write reaction showing conversion, of Acetaldehyde into acetaldehyde, dimethyl acetal., v. Aldehydes are more reactive toward, nucleophilic addition reactions than, ketones. Explain., vi. Write reaction showing the action of the, following reagent on propanenitrile –, a. Dilute NaOH, , iv. Write reaction showing conversion, of p- bromoisopropyl benzene into, p-Isopropyl benzoic acid ( 3 steps)., v. Write, reaction, showing, aldol, condensation of cyclohexanone., , Activity :, Draw and complete the, following reaction scheme which, starts with acetaldehyde. In each empty, box, write the structural formula of, the organic compound that would be, formed., HCN, , b. Dilute HCl ?, vi. Arrange the following carboxylic acids, with increasing order of their acidic, strength and justify your answer., COOH, , COOH, , in, , CH3CHO, , dilute H2SO4, heat, , reduction, , Tollens' reagents, , Conc. H2SO4 heat, , O, , CH2 = CHCO2H, , COOH, O, , 281, , Cold dilute KMnO4/H+
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13 AMINES, Can you recall ?, , Use your brain power, , •, , Write some examples of nitrogen, containing organic compounds., , •, , What are the types of amines?, , Classify the following amines, as simple/mixed; 1°, 2°, 3° and, aliphatic or aromatic., (C2H5)2NH, (CH3)3N,, , C2H5-NH-CH3,, NH-C6H5, , CH3, C6H5-NH2, CH3-CH- NH2,, , Amines are nitrogen containing organic, compounds having basic character. Amines, are present in structure of many natural, compounds like proteins, vitamins, hormones, and many plant products like nicotine., , CH3, CH3-C-NH2,, CH3, , 13.1 Classification of Amines : Amines are, classified as primary (1°), secondary (2°), and tertiary (3°) amines. Their structures are, obtained in simple way by replacing one, two, or three hydrogen atoms of NH3 molecule by, alkyl/aryl groups (see Table 13.1)., , N(CH3)2, , ,, , ,, , N, , Remember..., Other organic compopunds, like alkyl halides or alcohols are, classified as 1°, 2°, 3° depending upon, the nature of the carbon atom to which, functional group is attached where as, amines are classified depending upon the, number of alkyl or aryl groups directly, attached to the nitrogen atom. Thus,, isopropyl amine is 10 amine, but isopropyl, alcohol is 20 alcohol., , Secondary and tertiary amines are further, classified as simple / symmetrical amines and, mixed / unsymmetrical amines. When all the, alkyl or aryl groups on nitrogen are same, it is, a simple amine. If these groups are different,, then the amine is a mixed amine., Amines are also divided into two major, classes, namely, aliphatic and aromatic amines, on the basis of nature of the groups attached to, the nitrogen atom., , Table 13.1 Types of amines, , Functional group, , Type, , Examples, , Name, , Formula, , Formula, , Common Name, , Primary amine, (10), , Amino, , -NH2, , C2H5-NH2, , Ethylamine, , Secondary amine, (20), , Imino, , NH, , Tertiary amine, (3 ), , Tertiary, nitrogen, , 0, , N, , 282, , CH3, CH3, , NH, , CH3, CH3 N, CH3, , Dimethylamine, , Trimethylamine
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13.2 Nomenclature of Amines :, , 13.3 Preparation of Amines :, , 13.2.1 Common names : Common names, of aliphatic amines are given by writing the, name of alkyl group followed by suffix-amine,, that is, ‘alkyl amine’. In the case of mixed, amines, the names of alkyl groups are written, in alphabetical order. If two or three identical, alkyl groups are attached to nitrogen atom, the, prefix ‘di-’ or ‘tri-’ is added before the name of, alkyl group. The parent arylamine, C6H5-NH2,, is named as aniline. Other aromatic amines, are named as derivatives of aniline (see Table, 13.2)., , 13.3.1 : By ammonolysis of alkyl halides :, , 13.2.2 IUPAC names : In IUPAC system,, primary amines are named by replacing, the ending ‘e’ of the parent alkane by suffix, -amine (alkanamine). A locant indicating the, position of amino group is added before the, suffix amine. When two or more amino groups, are present, the prefix ‘di-’, ‘tri-’ etc. are used, with proper locant. In this case the ending ‘e’, of parent alkane is retained., , When alkyl halide is heated with alcoholic, solution of excess ammonia it undergoes, nucleophilic substitution reaction in which the, halogen atom is replaced by an amino (-NH2), group to form primary amine. This process of, breaking of C-X bond by ammonia is known, as ammonolysis. The reaction is also known, as alkylation of ammonia. The reaction is, carried out in a sealed tube at 373 K. It may be, noted that the primary amine obtained in the, 1st step is stronger nucleophile than ammonia., Hence, it further reacts with alkyl halide to, form secondary and tertiary amines and finally, quaternary ammonium salt if NH3 is not used, in large excess., R-X +NH3(alc). ∆, (excess), , R-NH2, 10 amine, , The order of reactivity of alkyl halides, with ammonia is R-I > R-Br > R-Cl., , Secondary or tertiary amines are named, as N-substituted derivatives of primary, amines. The largest alkyl group attached to, nitrogen is taken as the parent alkane and other, alkyl groups as N-substituents. While naming, arylamines ending ‘e’ of arene is replaced by, ‘amine’. The common name of aniline is also, accepted by IUPAC (see Table 13.2)., Remember..., The, name, of, amine, (common or IUPAC) is always, written as one word. For example :, C2H5-NH2 Ethylamine (Ethanamine), , Use your brain power, • Write chemical equations for, 1. reaction of alc. NH3 with C2H5I., 2. Ammonolysis of benzyl chloride, followed by the reaction with two, moles of CH3-I., • Why is ammonolysis of alkyl halide not, a suitable method for the preparation of, primary amine ?, , Try this..., Draw possible structures of all, the isomers of C4H11N. Write their, common as well as IUPAC names., , 283
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Solution : Methyl bromide can be converted, into ethyl amine in two stage reaction, sequence as shown below., , Do you know ?, When tert-butyl bromide, is treated with alcoholic NH3,, isobutylene is formed. This is the result, of elimination reaction preferred over, nucleophilic substitution through the stable, tertiary butyl carbocation intermediate., CH3, -Br, H3C-C-CH3 NH (alc), 3, Br, , CH3-Br + KCN, CH3-CN, , Na/C2H5OH, (reduction), , CH3-CH2-NH2, , The starting compound methyl bromide, contains one carbon atom while the product, ethylamine contains two carbon atoms., A reaction in which number of carbons, increases involves a step up reaction. The, overall conversion of methyl bromide into, ethyl amine is a step up conversion., , CH3, H3C-C-CH3, ⊕, , (tert-Butyl bromide), , CH3-CN + KBr, , -H⊕, , CH2, H3C-C-CH3, (isobutylene), , Use your brain power, 13.3.2 Reduction of nitrocompounds :, , Use your brain power:, , Aliphatic and aromatic nitrocompounds, can be reduced to primary amines by using, metal-acid mixture (Sn/HCl or Fe/HCl or Zn/, HCl) or catalytic hydrogenation (H2/Ni or Pt, or Pd) or LiAlH4 in ether., , Identify ‘A’ and ‘B’ in the following, conversions., Na/C2H5OH, (i) CH3-I KCN A, , (ii) CH3-Br, , of, , alkyl, , cyanide, , • How is alkyl halide converted, into alkyl cyanide ?, , 5, , 4, , Problem 13.1 : Write reaction to convert, methyl bromide into ethyl amine ? Also,, comment on the number of carbon atoms, in the starting compound and the product., , B, , 13.3.4 By reduction of amides :, , O, , Primary amines can be obtained by the, reduction of alkyl cyanide with sodium and, ethanol. This is known as Mendius reduction., The reaction can also be brought about by, lithium aluminium hydride., 2, , A Sn/HCl B, , Primary amines having same number of, carbon atoms can be obtained by the reduction, of amides by LiAlH4 in ether or by Na/C2H5OH., , Can you recall ?, , H OH, R-C N + 4[H] Na/C, R-CH2-NH2, or LiAlH, 1° amine, , B, , Na/C2H5OH, (iii) C2H5-I AgCN A, , R-NO2 + 6[H] Sn/HCl R-NH2 + 2H2O, 13.3.3 Reduction, (alkanenitriles) :, , AgNO2, , CH3-C-NH2 + 4[H], (Acetamide), , LiAlH4 / ether, or Na/C2H5OH, , CH3-CH2-NH2, (Ethylamine), , 13.3.5 Gabriel phthalimide synthesis : This, method is used for the synthesis of primary, amine. It involves the following three stages., i. Formation of potassium salt of phthalimide, from phthalimide on reaction with alcoholic, potassium hydroxide., ii. Formation of N-alkyl phthalimide from, the potassium salt by reaction with alkyl, halide., iii. Alkaline hydrolysis of N-alkyl phthalimide, to form the corresponding primary amine., , 285
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O, C, , i., , C, O, , N-H alc.KOH, -H O, 2, , (Potassium, phthalimide), , (Phthalimide), , O, C, , ii., , C, O, iii., , O, C, , O, C, , ⊕, , NK, , R-X, -KX, , C, O, , O, , ⊕, , NK, , salt, , O, C, , of, , Use your brain power, Write the chemical equations for, the following conversions :, , N-R, , C, O, , i. Methyl chloride to ethylamine., , (N-Alkylphthalimide), , N-R, , NaOH (aq), , C, O, , ii. Benzamide to aniline., , O, C-ONa, ⊕, C-ONa, O, , iii. 1, 4 - Dichlorobutane to hexane - 1, 6 diamine., iv. Benzamide to benzylamine., , (sodium phthalate), , + R-NH2, , (1° amine), , Aromatic amines cannot be prepared, by this method because aryl halides do not, undergo nucleophilic substitution with the, anion formed by phthalimide., 13.3.6 By Hofmann degradation (Hofmann, rearrangement / Hofmann bromamide, degradation / Hofmann hypobromite, degradation ) :, This is a good laboratory method for the, conversion of an amide into primary amine, containing one carbon less. The reaction is, brought about by warming the amide with, bromine and concentrated aqueous KOH, solution., O, R-C-NH2 + Br2 + 4KOH(aq), , (Amide), , The overall result is removal of the -Cgroup from the amide. As the product contains, one carbon atom less than the original amide., It is a step down reaction., , 13.4 Physical properties of Amines :, 13.4.1 Intermolecular forces, boiling points, and solubility : The N-H bond in amines, is polar because the electronegativities of, Nitrogen (3.0) and Hydrogen (2.1) are different., Due to the polar nature of N-H bond primary, and secondary amines have intermolecular, hydrogen bonding. The intermolecular, hydrogen bonding is to greater extent in, primary amine than in secondary amines,, because primary amines have two hydrogen, atoms bonded to nitrogen for hydrogen bond, formation (see Fig 13.1)., , ∆, , δ⊕, δ⊕, H δ⊕ R δ⊕ H, R-N-H δ N-H δ N-R, Hδ⊕, H, δ, Hydrogen bond, H-N-R, H, Fig. 13.1 : Intermolecular hydrogen bonding in, primary amines, , R-NH2 + 2KBr+ K2CO3 + 2H2O, , (1° amine), For example :, , O, ∆, CH3-NH2, CH3-C-NH2 + Br2 + 4KOH (aq), (methylamine), (Acetamide), , +2KBr + K2CO3 + 2H2O, , 286, , Tertiary amines do not have intermolecular, hydrogen bonding as there is no hydrogen, atom on nitrogen of tertiary amine. But due, to polar N-C bonds, tertiary amines are polar, molecules, and have intermolecular dipoledipole attractive forces. Thus intermolecular, forces of attraction are strongest in primary, amines and weakest in tertiary amines.
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Table 13.3 Boiling points of alkane, alcohol, and amines of similar molar masses, , The observed order of boiling points of, isomeric amines is : primary amine > secondary, amine > tertiary amine (see Table 13.3 serial, numbers 1, 2, 3). It can be explained on the, basis of the intermolecular forces in them., , Sr., No, , The lower aliphatic amines are gases, with fishy odour, middle members are liquids, and higher members are solids under ordinary, temperature and pressure., Aniline and other arylamines are usually, colourless liquids but get coloured as they are, easily oxidised by air., Due to their ability to form hydrogen, bond with water molecule, lower aliphatic, amines are soluble in water (see Fig. 13.2)., Solubility of amines decreases with increase, in molar mass of amines due to increase in size, of hydrophobic alkyl group. Aromatic amines, and higher aliphatic amines are insoluble in, water., Hydrogen bond, , δ, H-O, H, , δ⊕ R δ, H-N, H, , δ⊕ δ, H-O, H, , Fig. 13.2 : Hydrogen bonding between, amine and water molecule, , Since N-H bonds in amines are less polar, than O-H bond in alcohol, water solubilities of, alcohols, amines and alkanes of comparable, molar mass in water are in the decreasing, order: alcohols > amines > alkanes., The order of boiling points of alkanes,, amines, alcohols and carboxylic acid of, comparable molar mass is as follows :, Alkanes < Amines < Alcohols < Carboxylic, acid. (Table 13.3, serial number 4, 5, 6, 7), , Compound, , Molar, mass, , B.P. (K), , 1, , n-C4H9NH2, , 73, , 350.8, , 2, , (C2H5)2NH, , 73, , 329.3, , 3, , C2H5N(CH3)2, , 73, , 310.5, , 4, , C2H5COOH, , 74, , 414.4, , 5, , n-C4H9OH, , 74, , 390.3, , 6, , (CH3)3C-NH2, , 73, , 318.15, , 7, , C2H5CH(CH3)2, , 72, , 300.8, , 13.5 Basicity of Amines, The basic nature of amines is due to, presence of a lone pair of electrons on the, nitrogen atom. In terms of Lewis theory, amines, are bases because they can share a lone pair of, electrons on ‘N’ atom with an electron deficient, species. For example : Trimethylamine shares, its lone pair of electrons with the electron, deficient boron trifluoride., ⊕, , Me3N + BF3, , Me3N-BF3, , Basic strength of amines is expressed, quantitatively as Kb or pKb value. In terms of, Lowry-Bronsted theory, the basic nature of, amines is explained by writing the following, equilibrium., N + H2O, , (amine), , ⊕, , N-H + OH ........... (13.1), (conjugate acid), , In this equilibrium amine accepts H⊕,, hence an amine is a Lowry-Bronsted base., For stronger base, this equilibrium shifts, towards right, thereby the Kb value is larger, and pKb value is smaller and vice versa (refer, to Chapter 3). Table 13.4 gives pKb values of, some amines., , Use your brain power, Arrange the following :, a. In decreasing order of the b.p., C2H5-OH, C2H5-NH2, (CH3)2NH, b. In increasing order of solubility in water:, C2H5-NH2, C3H7-NH2, C6H5-NH2, , 287
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H, , H, O, H⊕, R-N-R, H, O, H H, , a. Influence of +I effect on stabilization of, conjugate acids of aliphatic amines and NH3, can be represented as shown below :, H, H, H, H, ⊕, ⊕, ⊕, ⊕, HNH, R NH, R N R, R N R, H, H, H, R, An alkyl group exerts electron releasing, inductive effect (+I) which stabilizes positive, charge on atom bonded to it. As we move, from conjugate acid of ammonia (NH4⊕) to, that of tertiary amine (R3NH⊕), the number of, alkyl groups (R) bonded to Nitrogen goes on, increasing steadily. This results in increasing, stabilization of the conjugate acids and, thereby an increasing order of basic strength, is expected., Order of stabilization :, ⊕, , ⊕, , ⊕, , ⊕, , NH4 < R-NH3 < R2NH2 < R3N-H, , ,, , The solvent water stabilizes the conjugate, acid by hydrogen bonding through the ‘H’, bonded to the ‘N⊕’. The number of ‘H’ atoms, bonded to the ‘N⊕’ decreaes from 4 in NH4⊕, to 1 in R3NH⊕. As a result NH4⊕ is best, stabilized by solvation while the stabilization, by solvation is very poor in R3NH⊕., c. Combined influence of +I effect and, solvation on stabilization if conjugate acids, of aliphatic amines decides the observed basic, strength and pKb value. These two influencing, factors operate in opposite directions., , Expected order of basic strength :, NH3 < R-NH2 < R2NH < R3N, The expected order of basic strength on the, basis of +I effect differs from the observed order, (Eq.13.2). It is seen that the observed increasing, basic strength from ammonia to amine and, from 1° amine to 2° amine is explained on the, basis of increased stabilization of conjugate, acids by +I effect of increased number of, alkyl (R) groups. However, decreased basic, strength of 3° amine implies that the conjugate, acid of 3° amine is less stabilized even, though, the +I effect of three alkyl groups in, ⊕, R3NH is expected to be large. This is suggestive, of existance of another influencing factor in, stabilization of conjugate acids of amines., b. Influence of solvation by water on, stabilization of conjugate acids of aliphatic, amines and ammonia can be represented as, shown below :, H, H, H, H, O, O, H, H⊕, H, H, H, O H-N-H, O, R-N-H, O, ⊕, H, H ,, H,, H, H, O, O, H H, H H, , R⊕, R-N-R, H, O, H H, , increases, , Solvation, ⊕, , NH4, +I effect, , ⊕, , ⊕, , R-NH3 R2NH, , ⊕, , R3NH, , increases, , The net results is that as we move from, NH3 to RNH2 to R2NH, the basic strength, increases due to better stabilization of the, corresponding conjugate acids. But 3° amine, is weaker base than 2° amine because the, stabilization of conjugate acid of 3° amine by, solvation is very poor., 13.5.2 Basicity of arylamines :, , 289, , Can you recall ?, Refer to Table 13.4 and answer :, Are the pKb values of aniline,, N-methylaniline and N, N-dimethylaniline, larger or smaller than those of NH3 and, CH3NH2 ?, Which one of the two, aniline or CH3NH2, is, stronger base ?
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From the pKb values we understand that, arylamines in general are weaker bases than, ammonia and aliphatic amines. Strength, of arylamines is explained in accordance, with Lowery Bronsted theory by writing the, following equilibrium (Eq. 13.3) For aniline, (similar to eq. 13.1)., ⊕, , NH2, , As a result the equilibrium (13.3) is, shifted towards left side. This makes aniline, (and also other arylamines) weaker bases than, aliphatic amines and ammonia., Use your brain power, Arrange the following amines, in decreasing order of their basic, strength -, , NH3, , + H 2O, , + OH ....... (13.3), , (Base), , NH3, CH3-NH2, (CH3)2NH, C6H5NH2., , (Conjugate acid), , 13.6 Chemical properties of amines, , Here, both the species base and conjugate, acid, are resonance stabilized but to different, extent., In arylamines, the -NH2 group is attached, directly to an aromatic ring. The lone pair, of electrons on nitrogen is conjugated to, the aromatic ring and is less available for, protonation. Aniline is resonance stabilized by, the following five resonance structures., ⊕, NH2, , NH2, , 13.6.1 Laboratory test for amines :, a. Test for amines as the ‘base’ : All amines, 1°, 2° and 3° are basic compounds. Aqueous, solution of water soluble amines turns red, litmus blue., The ‘basic’ nature of amines is detected in, laboratory by reaction with aqueous solution, of strong mineral acid HCl., N + HCl(aq), (amine), , ⊕, NH2, , ⊕, , N-H(aq) + Cl (aq), (a substituted, ammonium chloride), , ⊕, , N-H(aq)+ Cl (aq) + NaOH(aq), (I), , (II), ⊕, NH2, , (IV), , , , (III), , NH3, , (V), , (I), , ⊕, , NH3, , (II), , , , N +, , NaCl(aq) + H2O, , Water insoluble amine dissolves in, aqueous HCl due to formation of water, soluble substituted ammonium chloride,, which on reaction with excess aqueous NaOH, regenerates the original insoluble amine., , NH2, , On the other hand anilinium ion obtained, by accepting a proton does not have lone, pair of electrons on nitrogen. Hence it can be, stabilized by only two resonance structures, and therefore less stabilized than aniline., ⊕, , (excess), , b. Diazotization reaction/ Orange dye test:, In a sample of aromatic primary amine,, 1-2 mL of conc. HCl is added. The aqueous, solution of NaNO2 is added with cooling. This, solution is transfered to a test tube containing, solution of b naphthol in NaOH. Formation, of orange dye indicates presence of aromatic, primary amino group. (It may be noted that, temperature of all the solutions and reaction, mixtures is maintined near 0°C throughout the, reaction)., , 290
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13.6.4 Acylation of amines :, , primary amines. Secondary and tertiary, amines do not give this test., , Can you recall ?, , R-NH2 + CHCl3 + 3KOH, (1° amine), ∆, R-NC + 3KCl + 3H2O, , • What is an acyl group ?, • How are alcohols acylated ?, , (Alkyl isocyanide), , Aliphatic and aromatic primary and, secondary amines undergo acylation reaction., These amines contain replaceable hydrogen, atoms (positively polarised H) on the nitrogen, atom. These hydrogen atoms are replaced, by acyl groups such as acetyl group. On, reaction of amines with acetyl chloride or, acetic anhydride, acetyl derivative of amine, is obtained. It is also called amide. Amide, is less basic than the amine. Acylation is a, nucleophilic substitution reaction. The reaction, is carried out in presence of strong base like, pyridine, which neutralizes the acid produced, during the reaction. For example :, (i), , O, C2H5-N-H + C-CH3, H, Cl, , Pyridine, , (Ethanamine) (Ethanoyl, chloride), , Use your brain power, Write the carbylamine reaction by, using aniline as starting material., 13.6.6 Reaction with nitrous acid : Primary,, secondary and tertiary amines react differently, with nitrous acid. Reactions of only primary, amines will be considered here., Can you tell ?, • What is the formula of nitrous, acid ?, • Can nitrous acid be stored in bottle ?, , H O, C2H5-N⊕-C-CH3, H Cl, , Nitrous acid is an unstable compound., Hence it is prepared in situ by adding aqueous, sodium nitrite to hydrochloric acid already, mix with the substrate, that is amine., , O, C2H5-N-C-CH3 + HCl, H, (N-Ethylethanamide), , (ii), , H, O, O, Pyridine, H3C-N-C-CH3 + HCl, H3C-N + C-CH3, C6H5, C6H5 Cl, (N-Methylaniline) (N-methyl-N-phenylethanamide), , Benzoyl chloride also gives similar reaction, with amines., , a. Aliphatic primary amines on reaction with, nitrous acid form aliphatic diazonium salts as, very unstable intermidiates which decompose, immediately by reaction with solvent water., Corresponding alcohol is formed as the product, of the reaction and nitrogen gas is liberated ., R-NH2 + HNO2, , Use your brain power, •, , CH3-NH2 + Ph-CO-Cl, , •, , (CH3)3N + Ph-CO-Cl, , 273-278 K, (NaNO2 + HCl), , H 2O, , ?, , ⊕, , R-N2 Cl, , (alkyl diazonium, chloride), , R-OH + N2 + HCl, , b. Aromatic primary amines react with nitrous, acid to form diazonium salts which have, reasonable stability at 273 K., , ?, , 13.6.5 Carbylamine reaction :, Aliphatic or aromatic primary amines on, heating with chloroform give foul (offensive), smelling products called alkyl/aryl isocyanides, or carbylamines. This reaction is a test for, , 292, , NH2 + HNO2, , 273-278 K, (NaNO2 + HCl), , ⊕, , N NCl, , (benzene diazonium, chloride)
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Azo coupling with b-naphthol in NaOH, is used as a confirmatory test for primary, aromatic amines. Benzenediazonium chloride, reacts with aniline in mild alkaline medium to, give p-aminoazo-benzene (yellow dye.), ⊕, , N NCl +, (Benzenediazonium, chloride), , N=N, , NH2, , O, S-Cl + H-N-C2H5, O, C2H5, (2° amine), , O, S-N-C2H5 + HCl, O C2H5, , OH, , (N, N-diethylbenzene sulfonamide), , (Aniline), , N,N-diethylbenzenesulfonamide does not, contain any H-atom attached to nitrogen atom., Hence it is not acidic and does not dissolve in, alkali., , NH2 + HCl, , (p-Aminoazobenzene), , Do you know ?, , Can you tell ?, , The acid-base indicator methyl, orange is an azo dye., (CH3)N, , • Do tertiary amines have ‘H’, bonded to ‘N’ ?, , ⊕, , SO3Na, , N=N, , • Why do tertiary amines not react with, benzene sulfonyl chloride ?, , 13.8 Reaction with arenesulfonyl chloride :, (Hinsberg’s test) : Benzenesulfonyl, chloride (C6H5SO2Cl) is known as Hinsberg’s, reagent., , Use your brain power, How will you distinguish between, methylamine, dimethylamine and, trimethylamine by Hinsberg’s test ?, , a. Ethyl amine (primary amine) reacts with, benzenesulfonyl chloride to form N-ethyl, benzenesulfonyl amide., O, S-Cl + H-N-C2H5, O, H, (Benzenesulfonyl, chloride), , (1° amine), , O, S-N-C2H5 + HCl, OH, (N-Ethylbenzene sulfonamide), (Soluble in alkali), , 13.9 Electrophilic aromatic substitution in, aromatic amines : Amino group is ortho and, para directing and powerful ring activating, group. As a result aromtic amines readily, undergo electrophilic substitution reactions., a. Bromination : Aniline reacts with bromine, water at room temperature to give a white, precipitate of 2,4,6- tribromoaniline., NH2, , The hydrogen attached to nitrogen in, sulfonamide ethanamine (a primary amine) is, strongly acidic. Hence it is soluble in alkali., , Br, + 3Br2, , (Aniline), , b. Diethyl amine reacts with benzene-sulfonyl, chloride to give N, N- diethyl benzene, sulfonamide., , NH2, , Br2/H2O, , Br, + 3HBr, , Br, (2,4,6-tribromoaniline), , 294
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14. BIOMOLECULES, Can you recall ?, , Try this..., , •, , What are the constituents of, balanced diet ?, , •, , What are the products of digestion of, carbohydrates?, , •, , Which constituent of diet is useful for, building muscles?, , Observe, the, following, structural formulae carefully and, answer the questions., CH2OH, CHO, CHO, CO, (CHOH)3, (CHOH)4, CH2OH, (CHOH)3, CH OH, , •, , Which constituent of diet is a source, of high energy?, , •, , What is the genetic material of, organisms?, , 14.1 Introduction : Principal molecules of, the living world : Bodies of living organisms, contain large number of different molecules, which constitute their structure. They are, also part of various physiological processes, taking place in them. Primary structural, materials of organisms are proteins and, cellulose. By means of the unique process of, photosynthesis plants produce carbohydrates., Plants utilize the minerals absorbed by their, roots to produce proteins. Lipids are the main, ingredient of vegetable oils and milk fats., Nucleic acids constitute the genetic material, of organisms., In this chapter we are going to study, some aspects of three principal biomolecules,, namely, carbohydrates, proteins and nucleic, acids., 14.2 Carbohydrates : From the simple, chemical reactions of many carbohydrates it is, understood that carbohydrates are polyhydroxy, aldehydes or ketones or compounds which, give rise to such units on hydrolysis. Some, carbohydrates like glucose, fructose are, sweet in taste, and are called sugars. The, most commonly used sugar is sucrose which, is obtained from sugarcane or sugar beet., The sugar present in milk is called lactose., , 2, , (glucose), , CH2OH, , (ribose), , (fructose), , 1. How many OH groups are present in, glucose, fructose and ribose respectively?, 2. Which other functional groups are, present in these three compounds?, Greek word for sugar is sakkharon. Hence, carbohydrates are also called saccharides., Origin of the term carbohydrate lies in the, finding that molecular formulae of many of, them can be expressed as Cx(H2O)y(hydrates, of carbon). For example: glucose (C6H12O6 Or, C6(H2O)6, sucrose (C12H22O11 or C12(H2O)11),, starch [(C6H10O5)n or [C6(H2O)5]n]., 14.2.1 Classification of carbohydrates :, Carbohydrates, are, clssified, into, three broad groups in accordance with, their behaviour on hydrolysis. These are, monosaccharides,, oligosaccharides, and, polysaccharides (Fig. 14.1)., Monosaccharides do not hydrolyse, further into smaller units of polyhydroxy, aldehydes or ketones. Oligosaccharides, on hydrolysis yield two to ten units of, monosaccharides and accordingly they, are further classified as disaccharides,, trisaccharides and so on. Polysaccharides, give very large number of monosaccharide, units on complete hydrolysis., , 298
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Carbohydrates (Saccharides), Monosaccharides, (Do not hydrolyse into smaller units, Examples: glucose, fructose, ribose, , Oligosaccharides, Polysaccharides, (Yield two to ten monosaccharide, (Yield large number of, units on hydrolysis), monosaccharide units on hydrolysis), Example : starch, glycogen, cellulose, , Disaccharides, (Yield two monosaccharide units on, hydrolysis), Examples:, Sucrose : (One glucose unit + one fructose, unit), Maltose : (two glucose units), Lactose : (one glucose unit + one galactose unit), , Trisaccharides, (Yield three, monosaccharide units on, hydrolysis), Example : Raffinose :, (one unit each of glucose,, fructose and galactose), , Tetrasaccharides, (Yield four monosaccharide, units on hydrolysis), Examples : Stachyose : (one, glucose unit + one fructose, unit + two galactose units), , Fig. 14.1 : Classification of carbohydrates, , Remember..., • About, twenty, different, monosaccharides are found in, carbohydrates., •, , •, , Use your brain power, Give IUPAC names to the following monosaccharides., 1. CHO, 2. CHO 3. CH2OH, , Disaccharides, are, the, most, common oilgosacchrides. The two, monosaccharide units in disaccharides, may be same or different., , CHOH, , (CHOH)3, , CO, , CH2OH, , CH2OH, , (CHOH)4, CH2OH, , Polysaccharides : Starch is common, ingredient of food grains. Cellulose is, constituent of cell wall of plant cells., Animals store in their body in the form, of glycogen., , 14.2.3 Glucose : Glucose occurs in nature in, free as well as in combined state. Glucose, can be obtained from sucrose or starch by, acid catalysed hydrolysis as shown below., , 14.2.2 Nomenclature of monosaccharides :, , a. Prepartion of glucose from sucrose :, , According, to, IUPAC, system, of nomenclature, general name for, monosaccharide is glycose. Monosaccharide, with one aldehydic carbonyl group is called, aldose while that with one ketonic carbonyl, group is called ketose. These names are further, modified in accordance with the total number, of carbon atoms in the monosaccharide. For, example, glucose (C6H12O6) is an aldose with, six carbons, and is thereby, an aldohexose., Fructose (C6H12O6) is a ketose with six, carbons, and is, thereby, a ketohexose., , Sucrose is hydrolysed by warming, with dilute hydrochloric acid or sulfuric acid, for about two hours. This hydrolysis converts, sucrose into mixture of glucose and fructose., Glucose is separated from fructose by adding, ethanol during cooling. Glucose being almost, insoluble in alcohol crystallizes out first., The solution is filltered to obtain crystals of, glucose., C12H22O11 + H2O, (Sucrose), , 299, , H⊕, ∆, , C6H12O6 + C6H12O6, (Glucose), , (Fructose)
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b. Prepartion of glucose from starch :, Commercially glucose is obtained by, hydrolysis of starch by boiling it with dilute, sulfuric acid at 393K under 2 to 3 atm, pressure., (C6H10O5)n + n H2O, , H⊕, 393K, 2-3 atm, , 4. The carbonyl group in glucose is in the, form of aldehyde. This was inferred from the, observation that glucose gets oxidised to a six, carbon monocarboxylic acid called gluconic, acid on reaction with bromine water which is, a mild oxidizing agent., , n C6H12O6, , (Starch), , CHO, (CHOH)4, CH2OH, , (Glucose), , 14.2.4 Structure and properties of glucose, , (Glucose), , Glucose has an aldohexose structure., In other words, glucose molecule contains, one aldehydic, that is, formyl group and the, remaining five carbons carry one hydroxyl, group (-OH) each. The six carbons in glucose, form one straight chain. This aldohexose, structure of glucose was established on the, basis of the following chemical properties., , 2. The six carbons in glucose molecule form, a straight chain. This was inferred from, the following observation : Glucose gives, n-hexane on prolonged heating with HI., CHO - (CHOH)4 - CH2OH, (Glucose), CH3 - (CH2)4 - CH3, (n-Hexane), , 3. Glucose molecule contains one carbonyl, group. This was inferred from the observation, that glucose forms oxime by reaction with, hydroxylamine and gives cyanohydrin on, reaction with hydrogen cyanide., , NH, , CH=N-OH, (CHOH)4, CH2OH, (Oxime), , OH, , 2, , CHO, (CHOH)4, CH2OH, (Glucose), , Br2 water, , COOH, (CHOH)4, CH2OH, , (Gluconic acid), , Problem 14.1 :, An alcoholic compound was found to have, molcular mass of 90 u. It was acetylated., Molecular mass of the acetyl derivative, was found to be 174 u. How many alcoholic, (-OH) groups must be present in the original, compound?, Solution : In acetylation reaction H atom, of an (-OH) group is replaced by an acetyl, group (-COCH3). This results in an increase, in molcular mass by [(12+16+12+3×1)-1],, that as, 42 u., In the given alcohol,, increase in molecular mass = 174 u - 90 u, = 84 u, 84 u, =2, ∴ Number of -OH groups =, 42 u, , 1. Molecular formula of glucose was found, to be C6H12O6, on the basis of its elemental, compostion and colligative properties., , HI, ∆, , (O), , Can you recall ?, What are the products of reaction, of, i. CH3 - CO - CH3 with NH2 - OH ?, ii. CH3 - CHO with HCN?, iii. CH3 - OH with CH3 - CO - O - CO- CH3?, 5. Glucose contains five hydoxyl groups :, , HC, , N, , OH, CH CN, (CHOH)4, CH2OH, , (Cyanohydrin), , This was inferred from the observation that, Glucose reacts with acetic anhydride to form, glucose pentaacetate. As glucose is a stable, compound, it was further inferred that the five, hydroxyl groups are bonded to five different, carbon atoms in glucose molecule., , 300
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CHO, (CHOH)4, CH2OH, (Glucose), , CHO O, (CH-O-C-CH3)4, Acetic, anhydride, CH -O-C-CH +CH3COOH, 2, , 14.2.5 Optical isomerism in glucose :, Structural formula of glucose shows that it, contains four chiral carbon atoms. You have, learnt that every chiral carbon can have two, distinct spatial arrangements of groups around, it (section 10.5.1). In other words, two distinct, configurations are possible for each of the four, chiral carbons of glucose. Stereostructure of, glucose is therefore one out of several possible, stereostructures of an aldohexose., , 3, , O, , (glucose pentacetate), , (acetic acid), , 6. Glucose contains one primary alcoholic, (- CH2OH) group : This was inferred from, the following observation : Glucose and, gluconic acid both on oxidation with dilute, nitric acid give the same dicarboxylic acid, called saccharic acid., CHO, COOH, (CHOH)4, (CHOH)4, CH2OH, CH2OH, (Glucose), , (O, , ), , HN, , (Gluconic acid), , ), , (O, , O, , Do you know ?, A structural formula containing, ‘n’ number of chiral carbon can, have maximum ‘2n’ numbers of, stereostructures or optical isomers. An, aldohexose therefore, can exist as sixteen, (24 = 16) optical isomers, and glucose is, one of them., , O3, , HN, , 3, , COOH, (CHOH)4, COOH, , Can you recall ?, , (Saccharic acid), , Use your brain power, •, , •, •, •, , Write structural formula of, glucose showing all the bonds, in the molecule., Number all the carbons in the molecules, giving number 1 to the (-CHO) carbon., Mark the chiral carbons in the molecule, with asterisk (*)., How many chiral carbons are present, in glucose?, 1, , H, , *2, , HO, , *3, , H, , * 4, , H, , * 5, 6, , •, , How is a Fischer projection, formula drawn?, , 1, , COOH, , H, , 2, , OH, , H, , HO, , 3, , H, , OH, , H, , 4, , OH, , H, , 5, , OH, , OH, , OH, , CH2OH, , (Glucose), I, , What are the ways to, represent three dimensional, structure of an organic molecule?, , On the basis of very elaborate chemical, evidence and measurement of optical activity, of various chemicals involved, Emil Fischer,, a German Nobel laureate (1902), determined, the configuration of the four chiral carbons, (C-2, C-3, C-4, C-5) in glucose., 1, , CHO, , •, , 6, , CH2OH, (Gluconic acid), II, , COOH, , H, , 2, , OH, , HO, , 3, , H, , H, , 4, , OH, , H, , 5, , OH, , 6, , COOH, , (Saccharic acid), III, , Fig 14.2 : Fischer projection formulae of glucose, gluconic acid and saccharic acid, , 301
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Figure 14.2 shows the Fisher projection, formulae of glucose (I), gluconic acid (II) and, saccharic acid (III)., , Do you know ?, Optical, rotation, is, an, experimentally measurable property, of a compound. Configuration of chiral, carbon, on the other hand, is difficult to, observe by simple experiment. In 1951, X-ray crystallographic studies of (+) sodium rubidium tartarate established its, configuration as :, COO, , Glucose is an optically active compound, and has its specific rotation, [∝]20, , equal to, D, 0, +52.7 . Due to its dextrorotation glucose is, also called dextrose. The designations (+)glucose or d-glucose imply the dextrorotatory, nature of glucose. D-glucose is another, designation of glucose, which is more common., This designation indicates the configuration, of glucose rather than the sign of its optical, rotation., D/L configuration system : The prefix D- or, L- in the name of a compound indicates relative, configuration of a stereoisomer. It refers to, a particular enantiomer of glyceraldehyde., Glyceraldehyde has one chiral carbon(C-2), and exists as two enantiomers. These are, represented by two Fischer projection formulae, (see Fig. 14.3)., CHO, H, , *, , CHO, OH, , CH2OH, D-(+)- Glyceraldehyde, IV, , HO, , *, , H, , OH, , HO, , H, , COO, This was the first instance of determining, absolute configuration., A monosaccharide is assigned D/L, configuration on the basis of the configuration, of the lowest chiral carbon in its Fischer, projection formula. Figure 14.4 illustrates the, D-configuration of (+) - glucose., , H, , CHO, , CH2OH, , H, , L -(-)- Glyceraldehyde, V, , 1, , OH, CH2OH, , Fig. 14.3 : Enantiomers of glyceraldehyde, , Conventionally (+)-glyceraldehyde is, represented by the Fischer projection formula, having OH group attached to C-2 on right, side (IV) and this configuration is denoted, by symbol ‘D’. Similarly, configuration of, (-) glyceraldehyde (V) is denoted by symbol, ‘L’. All the compounds which can be, correlated by a series of chemical reactions, to (+) - glyceraldehyde are said to have, D-configuration. The compounds which are, chemically correlated to (-) - glyceraldehyde, are said to have L- configuration. This is the, system of relative configuration of chiral, compounds., , IV, D-(+)-glyceraldehyde, , CHO, , H, , 2, , OH, , HO, , 3, , H, , H, , 4, , OH, , H, , 5, , OH, , 6, , CH2OH, , I, D-(+)-glucose, , Fig. 14.4 : Relative configuration of, (+) - glucose, , 14.2.6 Ring structure of glucose : On the, basis of chemical evidence stereostructure, of D-glucose was represented by the Fischer, projection formula I (Fig. 14.2 and Fig. 14.4)., Glucose, however, was found to exhibit some, more chemical properties which could not be, explained on the basis of the structure I. It was, necessary to write another structure for glucose, which will explain all the properties. Ring, structure of glucose fulfils this requirement., , 302
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Glucose is found to have two cyclic structures, (VI and VII) which are in equilibrium with, each other through the open chain structure, (I) in aqueous solution (Fig.14.5)., , Problem 14.2 : Assign D/L configuration, to the following monosaccharides., i., , ii., , CHO, H, , HO, H, , CHO, , The ring structure of glucose is formed, by reaction between the formyl (-CHO) group, and the alcoholic (-OH) group at C-5. Thus,, the ring structure is a hemiacetal structure, (section 12.8.2 c). The two hemiacetal, structures (VI and VII) differ only in the, configuration of C-1 (Fig. 14.5), the additional, chiral centre resulting from ring closure., The two ring structures are called ∝- and, β- anomers of glucose and C-1 is called, the anomeric carbon. The ring of the cyclic, structure of glucose contains five carbons and, one oxygen. Thus, it is a six membered ring., It is called pyranose structure, in analogy, with the six membered heterocyclic compound, pyran (Fig. 14.6). Hence glucose is also, called glucopyranose. Haworth formula is, a better way than Fischer projection formula, to represent structure of glucopyranose (Fig., 14.6). In the Haworth formula the pyranose, ring is considered to be in a perpendicular, plane with respect to the plane of paper., The carbons and oxygen in the ring are in, the places as they appear in Fig. 14.6. The, lower side of the ring is called ∝-side and, the upper side is the β-side. The ∝-anomer, has its anomeric hydroxyl (-OH) group (at, C-1) on the ∝-side, whereas the β-anomer has, , HO, , H, H, HO, H, HO, CH2OH, , OH, CH2OH, , (Ribose), , (Threose), , Solution :, D/L configuration is assigned to Fischer, projection formula of monosaccharide on, the basis of the lowest chiral carbon., 1 CHO, Threose has two chiral, i., 2, carbons C-2 and C-3., H, HO, The given Fischer, 3, H, OH projection formula of, 4, CH2OH threose has -OH groups, at the lowest C-3 chiral, (Threose), carbon on right side., ∴ It is D-threose., 1, , ii., HO, , Ribose has three chiral, carbons C-2, C-3 and, C-4. The given Fischer, projection formula of, ribose has -OH group, at the lowest C-4 chiral, carbon on left side., , CHO, 2, , H, H, HO, 4, H, HO, 5 CH OH, 2, 3, , (Ribose), , ∴ It is L-ribose, 1, 1, , H, , 2, , H, , 3, , HO, H, , 4, , H, , 5, , OH, OH, H, OH, , 2, , H, O, , (∝-D-(+) Glucose), VI, , OH, , H, , 4, , OH, , H, , H, , 5, , OH, , H, , H, , CH2OH, I, , 2, , H, , HO, , CH2OH, , 1, , HO, , 3, , 6, 6, , CHO, , 3, , HO, , 4, , H, OH, H, OH, , 5, 6, , CH2OH, , β-D-(+) glucose, VII, , Fig. 14.5 Ring structures of glucose: Fischer projection formulae, , 303, , O
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14.2.8 Representation of Fructose structure, Fructose (C6H12O6) is a laevorotatory, ketohexose. Fructose is also called laevulose, 20, due to its laevorotation [∝]D = -92.40. Being, an ∝-hydroxy keto compound fructose is, a reducing sugar. In free state it exists, as mixture of fructopyranose (major) and, fructofuranose. In combined state fructose, is found in the form of fructofuranose ring, structre (as in sucrose, see section 14.2.9). The, name furanose is given by analogy with furan,, a five membered heterocyclic compound., Figure 14.7 shows representations of open, chain structure of fructose and ring structures, of ∝- and β- anomers of fructofuranose. Ring, structure of fructose is a hemiketal (section, 12.8.2 c)., 14.2.9 Disaccharides : Disaccharides, give rise to two units of same or different, monosaccharides, on, hydrolysis, with, dilute acids or specific enzymes. The two, monosaccharide units are linked together by an, ether oxide linkage (-O-), which is termed as, glycosidic linkage in carbohydrate chemistry., Glycosidic linkage is formed by removal of a, water molecule by reaction of two hydroxyl, , O, , (Pyran), 6, , H, 4, , HO, , 5, , CH2OH, , H, OH, 3, , O, H, , 2, , H, , OH, , 6, , b - side, H, H, 1, , 5, , CH2OH, H, OH, , 4, , OH HO, , 1, , H, , 2, , 3, , (α- D - (+) - Glucopyranose), , OH, , H, , H, , α- side, , O, , OH, , (b - D - (+) Glucopyranose), , Fig. 14.6 : Haworth formula of anomers of, glucopyranose, , its anomeric hydroxyl (-OH) group (at C-1), on the β-side. The groups which appear on, right side in the Fischer projection formula, appear on a-side in the Haworth formula, and, viceversa., 14.2.7 Reducing nature of glucose :, Hemiacetal group of glucopyranose structure, is a potential aldehyde group. It imparts, reducing properties to glucose. Thus, glucose, gives positive Tollens test and positive, Fehling test (Section 12.8.1 a)., 1, 2, , CH2OH, C=O, , 3, , HO, H, H, , 4, , 1, , HOH2C, , 2, , HO, , 3, , H, OH, , H, , OH, 6, CH2OH, , H, , 5, , 2, , H, , HO, , 3, , H, , 4, , H, , 5, , O, , OH, , 5, , 6, , CH2OH, , 1, , HO- H2C, , O, , b- side, , H, , H, 4, , HO, , HO, 3, , H, , OH, , O, , OH, , CH2OH, , 1, , O, , b- side, OH, 2, , 5, , H, , H, 4, , α- side, , H, , 6, , CH2 - OH HO- H C, 2, 2, , 5, , 1, , CH2OH, , b -D-(-) - Fructofuranose, , ∝-D-(-) - Fructofuranose, , 6, , Furan, , HO, , 6, , Open chain structure of fructose, , O, , 4, , OH, , HO, 3, , OH, , H, , C H2 - OH, a- side, , α- D - (-) - Fructofuranose, b - D - (-) - Fructofuranose, Fig. 14.7 : Representations of fructose structure, , 304
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6, , H, 4, , HO, , 5, , CH2-OH, O, H, OH H, 3, , H, , 2, , H, OH, α- D - glucose unit, 6, , H, 4, , O, , 2, 1, , 5, , H, , α, or, , 1, , CH2-OH, 5, O, H, H, 4, OH H, HO, 3, 2, , HO-H2C, H, , H, , O, , 5, 6, , 2, , 1, , α, , OH, H, α- D - glucose unit, , b, , O, , HO-H2C, , 6, , 1, , O, , H, , b, , CH2 - OH, , HO, 4, , 3, , HO, H, b- D - fructose unit, , CH2 - OH, , HO, 3, , OH, H, b- D - fructose unit, , α, b - 1, 2- glycosidic linkage, Fig. 14.8 : Haworth formula of sucrose, , b. Maltose : Maltose (C12H22O11) is a, disaccharide made of two units of D-glucose., The glycosidic bond in maltose is formed, between C-1 of one glucose ring and C-4, of the other. The glucose ring which uses its, hydroxyl group at C-1 is ∝-glucopyranose., Hence the linkage is called ∝-1,4-glycosidic, , (-OH) groups from two monosaccharide units., At least one of the two monosaccharide, units must use its anomeric hydroxyl group, in formation of the glycosidic linkage. Three, most common disaccharides are sucrose,, maltose and lactose., a. Sucrose : Sucrose (C12H22O11) is, dextrorotatory (+66.50). On hydrolysis with, dilute acid or an enzyme called invertase, sucrose gives equimolar mixture of D-(+), glucose and D-(-) fructose., H, C12H22O11 + H2O or, C6H12O6 + C6H12O6, , Do you know ?, Invert sugar is commerically, available as invert syrup. It is, used as sweetene in bakery and, confectionary products and also in fruit, preserves and beverages. It is sweeter, than sucrose and glucose. It is resistant to, crystallization and promotes retention of, moisture, enhances flavour and texture and, also prolongs shelf life., , ⊕, , invertase, , (Sucrose), , (D-(+) glucose) (D-(-)fructose), , Since the laevoratotion of fructose (-92.40), is larger than the dextrorotation of glucose, (+52.70), the hydrolysis product has net, laevorotation. Hence hydrolysis of sucrose, is also called inversion of sucrose, and the, product is called invert sugar. Structure of, sucrose contains glycosidic linkage between, C-1 of ∝-glucose and C-2 of β-fructose (Fig., 14.8)., Try this..., , linkage. The hemiacetal group at C-1 of the, second ring is not involved in glycosidic, linkage. Hence maltose is a reducing sugar., Maltose gives glucose on hydrolysis with, dilute acids or the enzyme maltase. Figure, 14.9 shows Haworth formula of maltose., 6, , 6, , Make models corresponding to the, two Haworth formulae of sucrose, in Fig. 14.8. Check that both are identical., , H, 4, , HO, , Since the potential aldehyde and ketone, groups of both the monosaccharide units are, involved in formation of the glycosidic bond,, sucrose is a non reducing sugar., , 305, , 5, , CH2OH, H, OH, 3, , H, , O, H, , 2, , OH, , H, , H, , 4, , 1, , α, , O, , 5, , CH2OH, H, OH, 3, , H, , α- D-glucose, , O, H, , 2, , H, 1, , OH, , OH, glucose, , α- 1, 4- glycosidic bond, Fig. 14.9 : Haworth formula of maltose
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6, , 6, , HOCH2, 5, , HO, , O, , H, OH, , 4, , H, , HOCH2, , 3, , O, , 1, , H, , 5, , H, , H, OH, , 4, , H, , 2, , 3, , b- D - galactose, , OH, , •, , 1, , H, , •, , OH, , b-D-glucose, , •, , b- 1, 4- glycosidic linkage, Fig. 14.10 : Haworth formula of lactose, , •, , c. Lactose : Lactose (C12H22O11) is a, disaccharide present in milk. It is formed, from two monosaccharide units, namely, D-galactose and D-glucose. The glycosidic, linkage is formed between C-1 of β-Dgalactose and C-4 of glucose. Therefore the, linkage in lactose is called β-1,4-glycosidic, linkage. The hemiacetal group at C-1 of the, glucose unit is not involved in glycosidic, linkage but is free. Hence lactose is a, reducing sugar. Figure 14.10 shows Haworth, formula of lactose., , •, , Starch, cellulose and glycogen are the most, common natural polysaccharides. Starch is, storage carbohydrate of plants and important, nutrient for humans and other animals., Cellulose is the main constituent of cell wall, of plant and bacterial cells. It is also main, constituent of wood and cotton. Glycogen, constitutes storage carbohydrate of animals, and is present in liver, muscles and brain. It, is also found in yeast and fungi., , 14.2.10 Polysaccharides : Polysaccharides, are formed by linking large number of, monosaccharide units by glycosidic linkages., 6, , 6, , H, 4, , 5, , CH2OH, , H, OH, , O, , CH2OH, H, OH, , 4, , 1, , 2, , 6, , O, , H, , 5, , CH2OH, H, OH, , 4, , 2, , 3, , O, , OH, , H, , H, , 1, , H, , 3, , O, , OH, , H, , 5, , H, , H, , H, , 3, , -O, , Is galactose an aldohexose or, a ketohexose?, Which carbon in galactose has different, configuration compared to glucose?, Draw, Haworth, formulae, of, ∝-D-galactose and β-D-galactose., Which disaccharides among sucrose,, maltose and lactose is/are expected to, give positive Fehling test?, What are the expected products of, hydrolysis of lactose?, , H, , 2, , H, , OH, , H, , Use your brain power, , O, , 6, , O, , H, 1, , H, , 3, , O, , OH, , CH2OH, , H, OH, , 4, , 2, , H, , H, , 5, , O, , H, 1, , H, 2, , OH, , H, , α- 1, 4 - glycosidic linkages, Fig. 14.11 : Amylose, 6, , 6, , H, 4, , -O, , 5, , CH2OH, H, OH, 3, , O, , H, , H, , 2, , CH2OH, , H, OH, , O, , OH, , 6, , H, 4, , -O, , 5, , H, OH, 3, , H, , 1, , 2, , H, , CH2OH, , H, , H, , 3, , O, , OH, , H, , 4, , 1, , H, , 5, , 6, , O, H, , 2, , OH, , H, , H, , 4, , 1, , O, , α- 1, 6 - glycosidic linkage (branch), , O, , 5, , 6, , CH2, , O, , H, OH, 3, , H, , H, , H, , 4, , H, 2, , OH, , O, , α- 1, 4 - glycosidic linkage, Fig. 14.12 : Amylopectin, , 306, , 5, , CH2OH, H, OH, 3, , H, , O, H, , 2, , OH, , H, 1, , O-, , O-
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6, , H, O, , 4, , 5, , 3, , HOCH2, , O, , H, OH, H, , 6, , 6, , HOCH2, , H, 2, , 1, , H, , O, , H, 4, , 5, , O, , H, OH, 3, , 1, , H, , O, , H, 4, , H, , 2, , OH, , H, , OH, , HOCH2, 5, , O, , H, OH, 3, , H, , H, 2, , OH, , 1, , O, , H, , b- 1, 4 - glycosidic link, Fig. 14.13 : Cellulose, , c. Glycogen : Glycogen has its structure, similar to that of amylopectin, but it is more, highly branched., , Can you think ?, When you chew plain bread,, chapati or bhaakari for long time, it, tastes sweet. What could be the resason ?, a. Starch : Starch is a polymer of, ∝-D-glucose.Starch has two components,, namely, amylose (15-20%) and amylopectin, (80-85%). Amylose is soluble in water and, forms blue coloured complex with iodine., It contains 200-1000 ∝-glucose units linked, by ∝-1,4- glycosidic linkages giving rise to, unbranched chain of variable length (Fig., 14.11). Amylopectin is water insoluble, component of starch which forms blue-violet, coloured complex with iodine. It is a branched, chain polysaccharide. In amylopectin, chains, are formed by ∝-1,4- glycosidic linkages, between ∝-glucose units, where as branches, are formed by ∝-1,6- glycosidic linkages, (Fig. 14.12)., b. Cellulose : Cellulose is a straight chain, polysaccharide of β-glucose units linked by, β-1,4- glycosidic bonds. Chemical hydrolysis, of cellulose requires use of concerntrated strong, acids at high temperature and pressure. This, implies that the β-1,4- glycosidic bond is very, strong and difficult to hydrolyse. Humans do, not have enzymes which can hydrolyse this, linkage. Hence cellulose cannot be digested, by human beings; it serves as the fibrous, content of food useful for bowel movement., Figure 14.13 shows the Haworth formula of, cellulose., , Do you know ?, The symbiotic bacteria in guts, of insects called termites have, enzymes that can hydrolyse, β-1,4- glycosidic linkage in cellulose., 14.3 Proteins, Can you recall ?, • What is the product of reaction, of acetic acid with ammonia ?, •, , Write the structural formula of N-methyl, acetamide. What is the name of the, functional group in this compound?, , •, , What are the nitrogenous nutrients in, human diet?, , Proteins are the fundamental structural, materials of animal bodies. Proteins in the, form of enzymes play prime role in all the, physiological reactions. The name protein is, derived form the Greek word, ‘proteios’ which, means ‘primary’ or ‘of prime importance’., Nutritional sources of proteins are milk,, pulses, nuts, fish, meat, etc. Chemically, proteins are polyamides which are high, molecular weight polymers of the monomer, units called ∝-amino acids., 14.3.1 ∝-Amino acids : Proteins on complete, hydrolysis give rise to a mixture of ∝-amino, acids. ∝-Amino acids are carboxylic acids, having an amino (-NH2) group bonded to, , 307
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for example, zwitter ion and the other forms, of alanine (Fig. 14.16)., , Lone pair can, bond to a proton, , H, :, , types as neutral, acidic or basic. Ten ∝-amino, acids from this list cannot be synthesised in, human body and have to be obtained through, diet. These are called essential amino acids, and are marked with asterisk (*) in Table, 14.1., Use your brain power, Tryptophan and histidine, have the structures (I) and (II), respectively. Classify them into neutral/, acidic/basic ∝-amino acids and justify, your answer.(Hint : Consider involvement, of lone pair in resonance)., , ⊕, , H3N, , CH2 - CH - COOH, , ⊕, , H 3N, , COOH, , C5H11 - NH2, , 87, , -550C, , C3H7 - COOH, , 88, , -7.90C, , (A), Zwitter ion of alanine, (No net charge), pH ~, ~6, , COO, , CH3, (C), Overall -1 charge, pH < 10, , Fig. 14.16 : Three forms of alanine, , Molecular mass Melting point, 293.50C, , COO, , H, H 2N, , 89, , H, CH3, , NH2, , Compare the molecular masses of, the following compounds and explain, the observed melting points., , NH2, , Zwitter ion, , H, , (B), overall +1 charge, pH < 2, , Can you think ?, , CH3 - CH - COOH, , O, , R, , carboxyl group, can donate proton, , CH3, , (II), , Formula, , C, , Fig. 14.15 : Zwitter ion, , N, N, H, , H O, , ⊕, H3N, , proton trasfer, , O H, , R, , NH2, , (I), , C, , H 2N, , CH2 - CH - COOH, N, H, , O, , Do you know ?, At the physiological pH of 7.4,, neutral ∝-amino acids are primarily, in their zwitterionic forms. On the other, hand, at this pH acidic ∝-amino acids, exist as anion (due to deprotonation of, the carboxyl group), while basic ∝-amino, acids exist as cation (due to protonation, of the amino groups). Ionic structures of, constituent ∝-amino acids result in ionic, nature of proteins., , ∝-Amino acids are high melting, water, soluble crystalline solids, unlike simple, amines or carboxylic acids. These properties, are due to a peculiar structure called zwitter, ion structure of ∝-amino acids. An ∝-amino, acid molecule contains both acidic carboxyl, (-COOH) group as well as basic amino (-NH2), group. Proton transfer from acidic group to, basic group of amino acid forms a salt, which, is a dipolar ion called zwitter ion (Fig. 14.15)., , Can you recall ?, •, •, , Amino acid can exist in different forms, depending upon the pH of the aqueous, solution in which it is dissolved. Consider,, , 309, , What does the enzyme pepsin, do?, What are the initial and final products, of digestion of proteins?
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14.3.2 Peptide bond and protein :, Proteins are known to break down into, peptides in stomach and duodenum under, the influence of enzymes, pepsin being, one of them which is secreted by stomach., Polypeptides are further broken down to, ∝-amino acids. This implies that proteins, are formed by connecting ∝-amino acids to, each other. The bond that connects ∝-amino, acids to each other is called peptide bond., Consider, for example, linking of a molecule, of glycine with that of alanine. One way of, doing this is to combine carboxyl group of, glycine with ∝-amino group of alanine. This, results in elimination of a water molecule and, formation of a dipeptide called glycylalanine, in which the two amino acid units are linked, H2N - CH2 - COOH, (glycine), , + H2N - CH - COOH, CH3, , 14.3.3 Types of proteins : Depending upon, the molecular shape proteins are classified, into two types., a. Globular proteins : Molecules of globular, proteins have spherical shape. This shape, results from coiling around of the polypeptide, chain of protein. Globular proteins are usually, soluble in water. For example : insulin, egg, albumin, serum albumin, legumelin (protein, in pulses), , (alanine), , -H2O, , H2N - CH2 - CO - NH - CH - COOH, CH3, (peptide bond), (glycylalanine), , Fig. 14.17 : Peptide bond, , by a peptide bond (Fig. 14.17). It can be, seen that a peptide bond or peptide linkage, is same as what is described as secondary, amide in organic chemistry. Combination of, a third molecule of an ∝-amino acid with, a dipeptide would result in formation of a, tripeptide. Similarly linking of four, five, or six ∝-amino acids results in formation, of tetrapeptide, pentapeptide or hexapeptide, Use your brain power, • Write the structural formula of, dipeptide formed by combination of carboxyl group of alanine and, amino group of glycine., • Name the resulting dipeptide., • Is this dipeptide same as glycyalanine, or its structural isomer?, , respectively. When the number of ∝-amino, acids linked by peptide bonds is more than, ten, the products are called polypeptides., The -CHR- units linked by peptide bonds are, referred to as ‘amino acid residues’. Proteins, are polypeptides having more than hundred, amino acid residues linked by peptide bonds., It may be, however, noted that distinction, between proteins and polypeptides is not, sharp. The two ends of a polypeptide chain of, protein are not identical. The end having free, carboxyl group is called C-terminal while the, other end having free amino group is called, N-terminal. In the dipeptide glycylalanine, glycine residue is N-terminal and alanine, residue is C-terminal., , b. Fibrous proteins : Molecules of fibrous, proteins have elongated, rod like shape. This, shape is the result of holding the polypeptide, chains of protein parallel to each other., Hydrogen bonds and disulfide bonds are, responsible for this shape. Fibrous proteins, are insoluble in water. For example : keratin, (present in hair, nail, wool), myosin (protein, of muscles)., The shapes of protein molecules are the, result of four level structure of proteins., 14.3.4 Structure of proteins : Proteins are, responsible for a variety of functions in, organisms. Proteins of hair, muscles, skin, give shape to the structure, while enzymes, are proteins which catalyze physiological, reactions. These diverse functions of proteins, can be understood by studying the four, , 310
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O, O, O, O, - NH - CH - C - NH - CH - C - NH - CH - C - NH - CH - C N - terminal R', , R'', , R''', , R' C - terminal, , a. Representation by structural formula, Ala - Gly - Ser - Tyr - Gly - Gly - Lys, N - terminal, C - terminal, , b. Representation with amino acid symbols, Fig. 14.18 : Representation of primary structure of protein, , level structure of proteins, namely primary,, secondary, tertiary and quaternary structure, of proteins., a. Primary structure of proteins : Primary, structure of proteins is the sequence of, constituent ∝-amino acid residues linked, by peptide bonds. Any change in the, sequence of amino acid residuce results, in a different protein. Primary structure of, proteins is represented by writing the three, letter symbols of amino acid residuces as per, their sequence in the concerned protein. The, , symbols are separated by dashes. According, to the convention, the N-terminal amino acid, residue as written at the left end and the, C-terminal amino acid residue at the right, end (Fig. 14.18)., Problem 14.4 : Write down the structures, of amino acids constituting the following, peptide., CH3-CH-CO-NH-CH-CO-NH-CH-COOH, NH2, CH2OH, CH2SH, Solution : The given peptide has two, amide bonds linking three amino acids., The structures of these amino acids are, obtained by adding one H2O molecule, across the amide bond as follows :, , Problem 14.3, Chymotrypsin is a digestive enzyme, that hydrolyzes those amide bonds for, which the carbonyl group comes from, phenylalanine, tyrosine or tryptophan., Write the symbols of the amino acids and, peptides smaller than pentapeptide formed, by hydrolysis of the following hexapeptide, with chymotrypsin., , HO H, HO H, CH3-CH-CO-NH-CH-CO-NH-CH-COOH, NH2, CH2SH, CH2OH, → CH3-CH-COOH + H2N-CH-COOH +, NH2, , Gly-Tyr-Gly-Ala-Phe-Val, Solution : In the given hexapeptaide, hydroylsis by chymotripsin can take, place at two points, namely, Phe and Tyr., The carbonyl group of these residuces, is towards right side, that is, toward, the C-terminal. Therefore the hydrolysis, products in required range will be :, Gly-Tyr, Gly-Ala-Phe and Val, (a dipeptide) (a tripeptide) (∝-amino acid), , CH2SH, H2N-CH-COOH, CH2OH, , b. Secondary structure of proteins :, The three-dimensional arrangement of, localized regions of a protein chain is called, the secondary structure of protein. Hydrogen, bonding between N-H proton of one amide, linkage and C=O oxygen of another gives, rise to the secondary structure. Two types, of secondary structures commonly found in, proteins are ∝-helix and β-pleated sheet., , 311
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β-Pleated sheet : The secondary structure, is called β-pleated sheet when two or more, polypeptide chains, called strands, line, up side-by-side (Fig. 14.20). The β-pleated, sheet structure of protein consists of extended, strands of polypeptide chains held together, by hydrogen bonding. The characteristics of, β-pleated sheet structure are :, , R, H, H, , R, , H, , R, , H, , R, , R, , Hydrogen, bond, , H, H, , R, , R, , H, , 3.6 residues, H, , R, , R, , H, , R, , N, H, , Fig. 14.19 : Backbone of ∝ - Helix, , O, , ∝-Helix : The ∝-helix forms when a, polypeptide chain twists into a right handed, or clockwise spiral (Fig. 14.19). Some, characteristic features of ∝-helical structure, of protein are:, •, •, , Each turn of the helix has 3.6 amino acids., A C=O group of one amino acid is, hydrogen bonded to N-H group of the, fourth amino acid along the chain., • Hydrogen bonds are parallel to the axis, of helix while R groups extend outward, from the helix core., Myosin in muscle and ∝-keratin in hair, are proteins with almost entire ∝-helical, secondary structure., Do you know ?, In collagen, the protein of, connective tissue, the polypeptide, chains have unusual left-handed helix, structure. Three strands of these chains, wind around each other in a right-handed, triple helix., , R, , O, , H, N, , R, , O, , R, , N, H, , H, N, , R, , O, , O, , N, H, , R, , H, N, O, , R, , H, N, , R, , O, , Fig. 14.20 : β - pleated sheet, , •, , The C=O and N-H bonds lie in the planes, of the sheet., , •, , Hydrogen bonding occurs between the, N-H and C=O groups of nearby amino, acid residues in the neighbouring chains., , •, , The R groups are oriented above and, below the plane of the sheet., , The β-pleated sheet arrangement is favoured, by amino acids with small R groups., Most proteins have regions of ∝-helix, and β-pleated sheet, in addition to other, random regions that cannot be characterised, by either of these secondary structures. For, example: Spider dragline silk protein is strong, due to β-pleated sheet region, yet elastic due, to ∝-helical regions in it., , 312
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c. Tertiary structure of proteins : The, three-dimensional shape adopted by the, entire polypeptide chain of a protein is, called its tertiary structure. It is the result of, folding of the chain in a particular manner, that the structure is itself stabilized and also, has attractive interaction with the aqueous, environment of the cell. The globular and, fibrous proteins represent two major molecular, shapes resulting from the tertiary structure., The forces that stabilize a particular tertiary, structure include hydrogen bonding, dipoledipole attraction (due to polar bonds in the, side chains), electrostatic attraction (due to, the ionic groups like -COO , NH3⊕ in the side, chain) and also London dispersion forces., Finally, disulfide bonds formed by oxidation, of nearby -SH groups (in cysteine residues), are the covalent bonds which stabilize the, tertiary structure (Fig. 14.21)., , Haemoglobin can do its function of oxygen, transport only when all the four subunits are, together. Figure 14.22 summerizes the four, levels of protein structure., a - amino, acids, , b - pleated, sheet, , Primary structure, (sequence of, a - amino acids), , a - helix, , Secondary structure, , Tertiary structure, , b - pleated, sheet, a - helix, , Quaternary structure, (multiple units of, tertiary structure), , hydrogen, bond, , Fig. 14.22 : Four levels of protein structure, , Use your brain power, A protein chain has the following, amino acids residues. Show and, label the interactions that can be present, in various pairs from these giving rise to, tertiary level structure of protein., , a-Helical structure, , b-Pleated, structure, , disulfide, bond, , electrostatic, attraction, , London, dispersion, forces, , - HN - CH - CO CH2OH, , Fig. 14.21 : Tertiary structure of protein, , d. Quaternary structure of proteins : When, two or more polypeptide chains with folded, tertiary structures come together into one, protein complex, the resulting shape is, called quaternary structure of the protein., Each individual polypeptide chain is called, a subunit of the overall protein. For example:, Haemoglobin consists of four subunits, called haeme held together by intermolecular, forces in a compact three dimensional shape., , 313, , - HN - CH - CO - ,, CH2 - Ph, ,, , - HN - CH - CO - HN - CH - CO - ,, CH2-CO-NH2 ,, CH2-CH-Me2, , Can you tell ?, What is the physical change, observed when (a) egg is boiled, (b), milk gets curdled on adding lemon juice ?
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14.3.5 Denaturation of proteins, High temperature, acid, base and, even agitation can disrupt the noncovalent, interactions responsible for a specific shape, of protein. This is denaturation of protein., Denaturation is the process by which the, molecular shape of protein changes without, breaking the amide/peptide bonds that, form the primary structre., , Chemically enzymes are proteins. Every living, cell contains at least 1000 different enzymes., Most enzymes catalyse only one reaction or, one group of similar reactions. Thus, enzyme, catalysis is highly specific. You have learnt, that a mineral acid can catalyse hydrolysis, of many types of compounds such as esters,, acetals and amides. In contrast, an enzyme, that catalyses hydrolysis of amide will not, work on ester or acetal., , Do you know ?, Globular proteins are typically, folded with hydrophobic side chains, in the interior and polar residues on the, outside, and thereby are water soluble., Denaturation exposes the hydrophobic, region of globular proteins and makes, them water insoluble., , substrate, , active site, enzyme, , enzyme substrate, complex, , denaturation, , coiled globular protein, , loose coils and loops, , Denaturation results in disturbing the, secondary, tertiary or quaternary structure of, protein. This causes change in properties of, protein and the biological activity is often, lost., , product, , enzyme, , 14.3.6 Enzymes :, Fig. 14.23 : Enzyme catalysis, , Can you recall ?, • Which parameter, equilibrium, constant or activation energy,, decides the rate of a chemical, reaction?, • What is the influence of a catalyst on, activation energy?, , Mechanism of enzyme catalysis, , A very large number of chemical, reactions take place in our bodies. These are, brought about at the physiological pH of 7.4, and the body temperature of 370 C with the, help of biological catalysts called enzymes., For example : insulin, an enzyme secreted by, pancreas, controls blood sugar level; amylase,, an enzyme present in saliva, hydrolyzes starch., , Action of an enzyme on a substrate, is described as lock-and-key mechanism, (Fig.14.23). Accordingly, the enzyme has, active site on its surface. A substrate molecule, can attach to this active site only if it has the, right size and shape. Once in the active site,, the substrate is held in the correct orientation, to react and forms the products of reaction., , 314
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The products leave the active site and the, enzyme is then ready to act as catalyst again., Formation of enzyme-substrate complex has, very low activation energy. That is how, the rate of the reaction is very high. Some, enzymes are so efficient that one enzyme, molecule can catalyse the reaction of 10000, substrate molecules in one second., , Do you know ?, Synthesis of protein, the, fundamental structural material, of body, is the process in which genetic, information is transferred. DNA governs this, process. DNA is present in the chromosomes, of the cell nucleus. Each chromosome has, a different type of DNA. An individual, chromosome is composed of many genes., Gene is a portion of DNA molecule, responsible for synthesis of a single protein., DNA stores the genetic information,, while RNA translates this into synthesis of, proteins needed by cells for proper function, and development., , Several enzymes have been isolated, from organisms (such as bacteria), purified, and crystallised,and amino acid sequences of, many of them have been determined. In many, industrial processes specific reactions are, carried out by use of enzymes extracted from, organisms, and also by use of new enzymes, made using genetic engineering., Some examples of industrial application of, enzyme catalysis are :, •, , Conversion of glucose to sweet-tasting, fructose, using glucose isomerase., , •, , Manufacture of new antibiotics, using, pencillin G acylase., , •, , Manufacture of laundry detergents, using, proteases., , •, , Manufacture of esters used in cosmetics,, using genetically engineered enzyme., , Can you tell ?, , •, •, •, , Cell, , Chromosome, , DNA, , of information is passed unchanged from one, generation to the next. Such information is, called genetic information and its transfer to, new cells is accomplished by nucleic acids., There are two types of nucleic, acids : ribonucleic acids (RNA) and, deoxyribonucleic acids (DNA). RNA are, found mainly in the fluid of living cells, (cytoplasm) while DNA are found primarily, in the nuclei of living cells., , 14.4 Nucleic acids :, •, , Gene, , What is the single term that, answers all the following, questions ?, What decides whether you are blue eyed, or brown eyed ?, Why does wheat grain germinate to, produce wheat plant and not rice plant ?, Which acid molecules are present in, nuclei of living cells ?, , Knowledge of structure of nucleic acids, is essential to understand their biological, functions. In this chapter we are going to look, at the structural aspects of nucleic acids., , One of the most remarkable properties, of living cells is their ability to produce their, replicas through thousands of generations., This becomes possible because certain type, , 14.4.1 Nucleotides : Nucleic acids are, unbranched polymers of repeating monomers, called nucleotides. In other words, nucleic, acids have a polynucleotide structure. DNA, molecules contain several million nucleotides, while RNA molecules contain a few thousand, , 315
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HO-CH2, , HO-CH2, OH, O, O OH, H H, H H, H, H, H, H, HO H, OH OH, , or, , Remember..., RNA contains : D-ribose, A, G, C, U, DNA contains : D-2-deoxyribose, A, G, C, T, , or, 5, , 5, , HO-CH2, , OH, , O, , HO-CH2, , 1, , 4, 3, , 1, , 4, 3, , 2, , 2, , OH, , OH OH, , bases with two rings (adenine and guanine), are derived from the parent compound, purine. Each base in designated by a oneletter symbol (Fig. 14.25). Uracil (U) occurs, only in RNA while thymine (T) occurs only, in DNA., , OH, , O, , No OH, at C-2, , 2 - Deoxy-D-ribose, (present in DNA), , D - Ribose, (present in RNA), , A nucleoside is formed by joining, the anomeric carbon of the furanose with, nitrogen of a base. While numbering the, atoms in a nucleoside, primes (') are used, for furanose numbering to distinguish them, from the atoms of the base (Fig. 14.26). With, pyrimidine bases, the nitrogen atom at the 1, position bonds with the 1' carbon of the sugar., With purine bases, the nitrogen atom at the, 9 position bonds with 1' carbon of the sugar., , Fig 14.24 : Sugar Components of nucleic acids, , nucleotides., The nucleotide monomers consist of three, components : a monosaccharide, a nitrogencontaining base and a phosphate group., Nucleotides of both RNA and DNA, contain five-membered ring monosaccharide, (furanose), often called simply sugar, component. In RNA, the sugar component of, nucleotide unit is D-ribose, while in DNA, it, is 2-deoxy-D-ribose (Fig. 14.24)., , Nucleotides, are formed by adding, a phosphate group to the 5'-OH of a, nucleoside (Fig. 14.27). Thus, nucleotides are, monophosphates of nucleosides. Abridged, names of some nucleotides are AMP, dAMP,, UMP, dTMP and so on. Here, the first capital, letter is derived from the corresponding base., , Total five nitrogen - containing bases are, present in nucleic acids. Three bases with one, ring (cytosine, uracil and thymine) are derived, from the parent compound pyrimidine. Two, , 5, , N3, , 6, , 2, , N, , N, H, , 7, , 8, , 5, , 9, , N, H, , 4, , 6, , N1, N, , N, H, , O, , Cytosine, C, , Pyrimidine, (Parent compound), , 2, , 3, , Purine, (Parent compound), , O, H 3C, , NH, , N, , 1, , N, , O, , NH2, , 4, , N, N, H, , NH, , O, , N, H, , Uracil, U, , NH2, N, N, , Thymine, T, , O, N, N, H, , Adenine, A, , NH, N, Guanine, G, , Fig 14.25 : Bases in nucleic acids, , 316, , O, , NH2
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HO-CH2, , O, , NH2, , NH2, , OH, , N, , +, N, H, , OH OH, , N, , 5', , HO-CH2, , O, , N, , O, , O, , 1', , 4', 2', , 3', , OH OH, (D - ribose), , HO-CH2, , O, , (a ribonucleoside), , (Cytosine), , OH, , NH2, , N, , +, OH, , N, , N, , 5', , N, , N, H, , NH2, , HO-CH2, , N, , O, , 9, , N, , N, , 1', , 4', 2', , 3', , OH, , (Adenine), , (D - 2 - deoxyribose), , (a deoxyribonnucleoside), , Fig 14.26 : Formation of nucleoside, , O, O-P-O-CH2, O, , N, , O, , N, , NH2, , NH2, , N, , O, O-P-O-CH2, O, O, , N, , OH OH, , N, N, , O, , OH, , (AMP), , (dCMP), , Fig 14.27 : Structures of nucleotides, , other. One end having free phosphate group, of 5' position is called 5' end. The other end is, 3' end and has free OH- group at 3' position., , Use your brain power, Draw structural formulae, of nucleosides formed from the, following sugars and bases., i. D-ribose and guanine, ii. D-2-deoxyribose and thymine, MP stands for monophosphate. Small letter, 'd' in the beginning indicates deoxyribose in, the nucleotide., 14.4.2 Structure of nucleic acids : Nucleic, acids, both DNA and RNA, are polymers of, nucleotides, formed by joining the 3' - OH, group of one nucleotide with 5'-phosphate of, another nucleotide (Fig. 14.28). Two ends of, polynucleotide chain are distinct from each, , The polynucleotide structure of nucleic, acids can be represented schematically as in, Fig. 14.29 (a and b)., The primary structure of nucleic acids, is the sequence of the nucleotides in it. This,, in turn, is determined by the identity of the, bases in the nucleotides. Different nucleic, acids have distinct primary structure. It is, the sequence of bases in DNA which carries, the genetic information of the organism., The polynucleotide chains of nucleic acids, are named by the sequence of the bases,, beginning at the 5' end and using the one, letter symbols of the bases. For example the, , 317
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NH2, , O, 5', O-P-O-CH2, O, , N, N, , O, , +, , O, , 3`, , O 5', O-P-O-CH2, O, , OH, , NH2, , N, , N, , N, , O, , N, , 3', , OH, (dAMP), , (dCMP), , NH2, , O, 5', O-P-O-CH2, O, , 5' end, , Phosphodiester, linkage, , N, , O, , N, , O, , 3', , NH2, , N, , {, , O, 5', O P-O-CH2, O, , N, , N, , O, , N, , 3', , 3' end, , OH, Fig 14.28 : Formation of a dinucleotide, , }, , Sugar-phosphate, backbone, (a), , (b), , 5', , Sugar, , Phosphate, , Sugar, , Phosphate, , Sugar, , Phosphate, , 3', , base, , base, , P, , 5', , S, B, , P, , S, , P, , S, , B, , B, , P, , base, , S 3', B, , Fig 14.29 : Polynucleotide structure of nucleic acids :, Schematic representations (a) and (b), , Problem 14.5 : Draw a schematic representation of trinucleotide segment 'ACT' of a DNA, molecule., Solution : In DNA molecule sugar is deoxyribose. The base 'A' in the given segment is at 5', end while the base 'T' at the 3' end. Hence the schematic representation of the given segment of, DNA is, Phosphate, , deoxyribose, , Phosphate, , deoxyribose, , Phosphate, , deoxyribose, , 5', , 3', , A, , C, , 318, , T
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base pairs and the two strands of the, double helix are complementary to each, other., , name CATG means there are four nucleotides, in the segment containing the bases cytosine,, adenine, thymine and guanine, in the indicated, order from the 5' end., , It may be noted that RNA exists as single, stranded structure., , Remember..., • A nucleic acid contains, a backbone consisting of, alternating sugar and phosphate groups., • Backbone of all types of DNA contains, the sugar 2-deoxy-D-ribose while that, of RNA contains the sugar D-ribose., • The identity and sequence of bases, distinguish one polynucleotide from the, other., • A polynucleotide has one free phosphate, group at the 5' end., • A polynucleotide has a free OH group, at the 3' end., , 50, , 30, G, , C, T, C, , A, G, , T, , A, , Hydrogen bond, G, , C, A, T, , A, , G, , C, C, , C, , 14.4.3 DNA double helix : James Watson, and Francis Crick put forth in 1953 a double, helix model for DNA structure, which was, later verified by electron microscopy. Salient, features of the Watson and Crick model of, DNA are :, , •, , •, , •, , •, , DNA consists of two polynucleotide, strands that wind into a right-handed, double helix., The two strands run in opposite directions;, one from the 5' end to the 3' end, while, the other from the 3' end to the 5' end., The sugar- phosphate backbone lies on, the outside of the helix and the bases lie, on the inside, perpendicular to the axis, of the helix., The double helix is stabilized by hydrogen, bonding between the bases of the two, DNA strands. This gives rise to a ladderlike structure of DNA double helix., Adenine always forms two hydrogen, bonds with thymine, and guanine forms, three hydrogen bonds with cytosine. Thus, A - T and C - G are complementary, , 319, , Sugar phosphate, backbone, , T, , G, , •, , Axis of helix, , T, , A, , G, , Base, T, , A, , T, , G, , C, A, , A, G, C, T, , 50, , 30, , Fig. 14.30 : DNA double helix, , Do you know ?, Hydrogen bonding between, complementary base pairs., N, G, , O, , N, , N H, , Sugar, , N, Sugar, , C, N, , H, , N, , H, , N, , N, H, , A, , H N, , N H, , O, , N H, , O, , N, , Sugar, , CH3, , H N, , T, N, , N, O, , Sugar
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Problem 14.6 : Write the sequence of the complementary strand of the following portion of a, DNA molecule : 5' - ACGTAC-3', Solution : The complementary strand runs in opposite direction from the 3' end to the 5' end. It, has the base sequence decided by complementary base pairs A - T and C - G., Original strand, , 5' - A C G T A C - 3', , Complementary strand, , 3' - T G C A T G - 5', , Exercises, 1. Select the most correct choice., i., , CH2OH-CO-(CHOH)4-CH2OH is an, example of, a. Aldohexose , c. Ketotetrose , , v., , b. Aldoheptose, d. Ketoheptose, , a. Primary, b. Secondary, , ii. Open chain formula of glucose does, not contain, , c. Tertiary, d. Quaternary, , a. Formyl group, , vi. RNA has, , b. Anomeric hydroxyl group, , a. A - U base pairing, , c. Primary hydroxyl group, , b. P-S-P-S backbone, , d. Secondary hydroxyl group, , c. double helix, , iii. Which of the following does not apply, to CH2NH2 - COOH, , d. G - C base pairing, 2. Give scientific reasons :, , a. Neutral amino acid, , i., , b. L - amino acid, c. Exists as zwitter ion, d. Natural amino acid, iv. Tryptophan is called essential amino, acid because, a. It contains aromatic nucleus., b. It is present in all the human, proteins., c. It cannot be synthesised by human, body., d., , It is essential, enzymes., , A disulfide link gives rise to the, following structure of protein., , constituent, , of, , ii. On complete hydrolysis DNA gives, equimolar quantities of adenine and, thymine., iii. α - Amino acids have high melting, points compared to the corresponding, amines or carboxylic acids of, comparable molecular mass., iv. Hydrolysis, inversion., v., , 320, , The disaccharide sucrose gives, negative Tollens test while the, disaccharide maltose gives positive, Tollens test., , of, , sucrose, , is, , called, , On boiling egg albumin becomes, opaque white.
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3. Answer the following, i., , 4. Draw a neat diagram for the following:, , Some of the following statements, apply to DNA only, some to RNA, only and some to both. Lable them, accordingly., , i., , a. The polynucleotide, stranded. (, ), , iv. Secondary structure of protein, , is, , Haworth formula of glucopyranose, , ii. Zwitter ion, iii. Haworth formula of maltose, , double, , v., , AMP, , b. The polynucleotide contains uracil., (, ), , vi. dAMP, , c. The, polynucleotide, D-ribose (, )., , contains, , viii. Enzyme catalysis, , d. The, polynucleotide, Guanine (, )., , contains, , ii. Write, the, sequence, complementary, strand, following segments of, molecule., , of, the, for, the, a DNA, , a. 5' - CGTTTAAG - 3', b. 5' - CCGGTTAATACGGC - 3', iii. Write the names and schematic, representations of all the possible, dipeptides formed from alanine,, glycine and tyrosine., iv. Give two evidences for presence of, formyl group in glucose., , 321, , vii. One purine base from nucleic acid, , Activity :, •, , Draw structure of a segment, of DNA comprising at least ten, nucleotides on a big chart paper., , •, , Make a model of DNA double, stranded structure as group, activity.
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15. INTRODUCTION TO POLYMER CHEMISTRY, Can you recall ?, i. Classify, the, following, materials as bio-degradable, and, non-bio-degradable, :, Thermocol, glass, wood, cotton, clothes, paper bags, polythene bags,, nylon ropes, fruit peels., , aspects of organic polymers. You have learnt, in Chapter 14 about carbohydrates, proteins, and nucleic acids which are important organic, biopolymers playing crucial role in living, world., In this chapter we will consider mainly, man made organic polymers with reference, to aspects such as types, preparation and, applications., , ii. Give examples of man made materials, we use in our daily life., iii. Which material is used in manufacture, of toys, combs ?, , Do you know ?, Nobel prizes for pioneering work, in 'Polymers' :, • The polymeric substances, that we, know today as macromolecules,, were considered hundred years ago, as associated molecules. Staudinger, received Nobel prize (1953) for his work, which established macromolecular, nature of polymers., • In 1963 Natta received Nobel prize for, recognizing stereospecific regularity in, vinyl polymers., • In 1974 Flory received Nobel prize for, elucidating the three step mechanism, of chain-reaction in polymerization, involving initiation, propagation and, termination., , iv. Write examples of thermosetting, plastic articles., v., , List various properties of plastic., , 15.1 Introduction : Today the overall, development in polymer science and, technology has enriched human life. The world, would be at totally different place without, polymers such as artificial fibres, plastics, and elastomers. From the throwaway candy, wrapper to the artificial heart, polymers touch, our lives as does no other class of material., In short we are living in the world of, polymers. Polymer chemistry emerged as, a separate branch of chemistry during the, last several decades due to the voluminous, knowledge built up in this field and the ever, increasing applications in everyday life., Chemically polymers are complex, giant, macromolecules made from the repeating units, which are derived from small molecules called, 'monomers'. The term 'polymer' originates, from Greek word 'poly' meaning many and, 'mer' meaning part or unit. Interlinking of, many units constitutes polymers., Polymers are high molecular mass, macromolecules (103 - 107 u)., Both inorganic as well as organic polymers, are known. In this chapter we will study some, , 15.2 Classification of polymers : Polymers, are classified in number of ways on the basis, of their source, chemical structures, mode of, polymerization, molecular forces, type of, monomers and biodegradability., 15.2.1 Classification of polymers on the basis, of source or origin : Poylmers are divided, into three categories : a. Natural b. Synthetic, c. Semisynthetic, a. Natural polymers : The polymers obtained, from natural source are said to be natural, polymers. They are further subdivided into two, types., , 322
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bifunctional monomers or alkenes. (Fig., 15.1(a)). For example : PVC, high density, polythene., , i. Plant polymers : These are obtained from, plants. For example, cotton and linen are, obtained from cotton plant and flax plant, respectively. Natural rubber is another example, of natural polymer which is manufactured, from the latex obtained from bark of rubber, trees., , b. Branched chain polymers : The second, most common arrangement is that of branched, chain. Monomer having 3 functional groups, or already having side chains give rise to, branched chain polymers. (Fig. 15.1 (b)). For, example : low density polythene., , ii. Animal polymers : These are derived from, animal sources. For example, wool is obtained, from hair of sheep. Silk is obtained from, silkworm., , c. Cross-linked polymers : Third type of, arrangement is said to be cross linked or, network polymers where cross links are, produced between linear chains as shown, in Fig. 15.1 (c). Cross linking results from, polyfunctional monomers. For example,, bakelite, melamine., , b. Synthetic Polymers : These are man- made, polymers. These polymers are artificially, prepared by polymerization of one monomer, or copolymerization of two or more monomers., Nylon, terylene, neoprene are synthetic, polymers. These are further divided into three, subtypes, namely, fibres, synthetic rubbers and, plastics., c. Semisynthetic polymers : These are derived, from natural polymers. These are also called, regenerated fibres.Cellulose derivatives such, as cellulose acetate rayon, cellulose nitrate,, viscose rayon, cuprammonium rayon are a, few examples of this category., Semisynthetic polymers are used in, preparation of non-inflammable photographic, films, cinema films, varnishes, etc., , Fig. 15.1 : Different chain configurations of, polymers, , Use your brain power, • Differentiate between natural, and synthetic polymers., 15.2.2 Classification of polymers on the, basis of structure : Depending upon how, the monomers are linked together, that is, the, chain configuration, polymers are classified, in three general types : a. linear b. branched, and c. three dimensional cross - linked, polymers (Fig. 15.1). The nature of linking, the monomers depends upon the nature and, number of functional groups in them., , 15.2.3 Classification of polymers on the basis, of mode of polymerization : Polymerization, is the fundamental process by which low, molecular mass compounds are converted, into high molecular weight compounds by, linking together of repeating structural units, with covalent bonds. This process is illustrated, below., Low molecular, mass material, (Possessing, reactive groups), , a. Linear or straight chain polymers : When, the monomer molecules are joined together, in a linear arrangement the resulting polymer, is straight chain polymer. It is obtained from, , 323, , High temprature, and/or pressure, and/or catalyst, , High molecular, mass material, , There are three modes of polymerization, according to the types of reactions taking place, between the monomers.
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a. Addition polymerization (or chain growth, polymerization), b. Condensation polymerization (or step, growth polymerization), c. Ring opening polymerization, a. Addition polymerization : Addition, polymerization is a process of formation of, polymers by addition of monomers without, loss of any small molecules. The repeating unit, of an addition polymer has the same elemental, composition as that of original monomer., Compounds containing double bond, undergo addition polymerization. It is also, referred as vinyl polymerization, since, majority of monomers are from vinyl category., For example : vinyl chloride (CH2=CHCl),, acrylonitrile (CH2=CHCN). Formation of, polyethylene from ethylene is well known, example of addition polymerization. Addition, polymerization produces high molecular mass, polymeric materials without formation of any, intermediate low molecular mass polymeric, materials., Free radical mechanism is most common, in addition polymerisation. It is also called, chain reaction which involves three distinct, steps chain initiation, chain propagation and, chain termination., Step 1 : Chain initiation : The chain reaction, is initiated by a free radical. An initiator, (catalyst) such as benzoyl peroxide, acetyl, peroxide, tert-butyl peroxide, etc. can be used, to produce free radical. For example acetyl, peroxide generates methyl radical as shown, below :, O, O, CH3 - C - O - O - C - CH3, (acetyl peroxide), , O, 2CH3 - C - O, - CO2, , R + CH2 = CHY, (free, radical), , R - CH2 - CHY, , (vinyl, monomer), , (new radical), , Step 2 : Chain propagation : The new radical, formed in the initiation step reacts with, another molecule of vinyl monomer, forming, another still bigger sized radical, which in turn, reacts with another monomer molecule. The, repetition of this sequence takes place very, rapidly. It is called chain propagation., R - CH2 - CHY + nCH2 = CHY, R (CH2 - CHY)n CH2 - CHY, This step is very rapid and leads to high, molecular mass radical., Step 3 : Chain termination : Ultimately,, at some stage,termination of the growing chain, takes place. It may occur by several processes., One mode of termination is by combination of, two growing chain radicals., 2 R (CH2 - CHY)n CH2 - CHY, R (CH2CHY)n+ 1 (CHYCH2 )n+ 1 R, , , (polymer), , Internet my friend, Study, audiovisual, free, radical mechanism of addition, polymerization. (Refer/search for free, radical polymerization.Animation (IQOGCSIC) on youtube channel), b. Condensation polymerization :, Consider the formation of terylene, a, poly ester polymer, from ethylene glycol and, terephthalic acid., n HO - CH2 - CH2 - OH + n HOOC, , CH3, , (methyl radical), , The free radical (say R) so formed, attaches itself to the olefin (vinyl monomer), and produces a new radical, made up of two, parts, namely, the attached radical and the, monomer unit., , 324, , (ethylene glycol), - nH2O, , HO, , COOH, , (terephthalic acid), , O, , O, , [CH - CH - O - C, , ]C, , 2, , 2, , n, , OH, , ester link, (terylene or dacron)
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In this reaction an alcoholic OH group, in ethylene glycol condenses with a carboxyl, group in terephthalic acid by eliminating a, water molecule to form an ester linkage., , Use your brain power, What is the type of, polymerization in the following, examples ?, (i), CH, 2n CH2 = CH - CH3, , The process of formation of polymers from, polyfunctional monomers with the elimination, of some small molecules such as water,, hydrochloric acid, methanol, ammonia is, called condensation polymerization., , 3, , H3C (CH2- CH)n (CH - CH2 )n CH3, CH3 CH3, (ii), nHO-(CH2)x-OH + nHOOC - (CH2)y - COOH, , In this type of polymerization the chain, growth occurs by a series of condensation, steps. Therefore it is also referred to as step, growth polymerization. This process is, continued until a high molecular mass polymer, is obtained., , HO [ (CH2)x-O-CO - (CH2)y - CO ]n OH, 15.2.4 Classification of polymers on the, basis of intermolecular forces : Mechanical, properties of polymers such as tensile strength,, toughness, elasticity differ widely depending, upon the intermolecular forces. Polymers are, classified into various categories on the basis, of intermolecular forces as follows., , Remember..., Repeating, units, of, condensation polymer do not have, the same elemental composition as, that of monomer., c. Ring opening polymerization : The, third type of polymerization is ring opening, polymerization. Cyclic compounds like, lactams, cyclic ethers, lactones, etc. polymerize, by ring opening polymerization. Strong acid, or base catalyze this reaction. For example :, polymerization of e-caprolactam. (For more, details see section 15.3.5 (b)., O, NH, e, NH - (CH2)5 - C, CO, n, b a, 2, , [, , ], , Elemental composition of the repeating, unit in the polymer resulting from ring opening, polymerization is same as that of the monomers,, as in the case of addition polymerization., Addition polymerizations are often very rapid., But ring opening polymerization proceeds by, addition of a single monomer unit (but never of, larger units) to the growing chain molecules., In this sense, ring opening polymerization, is a step growth polymerization similar to, condensation polymerization., , a. Elastomers : Elasticity is a property by, which a substance gets stretched by external, force and restores its original shape on, release of that force. Elastomers, the elastic, polymers, have weak van der Waals type of, intermolecular forces which permit the polymer, to be stretched. A few crosslinks between the, chains help the stretched polymer to retract, to its original position on removal of applied, force. For example : vulcunized rubber, bunaS, buna-N, neoprene, etc., , b. Fibres : Polymeric solids which form threads, are called fibres. The fibres possess high tensile, strength which is a property to have resistance, to breaking under tension. This characteristic, is due to the strong intermolecular forces like, hydrogen bonding and strong dipole-dipole, forces. Due to these strong intermolecular, , 325
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forces the fibres are crystalline in nature. For, example : polyamides (nylon 6, 6), polyesters, (terylene), etc., , c. Thermoplastic polymers : Plasticity is a, property of being easily shaped or moulded., Thermoplastic polymers are capable of, repeated softening on heating and hardening, on cooling. These polymers possess, moderately strong intermolecular forces that, are intermediate between elastomers and, fibres. For example : polythene, polystyrene,, polyvinyls, etc. :, , 15.2.5 Classification of polymers on the basis, of type of different monomers : Polymers are, divided into two classes :, a. Homopolymers : The polymers which, have only one type of repeating unit are called, homopolymers. Usually they are formed from, a single monomer. In some cases the repeating, unit is formed by condensation of two, distinct monomers. For example : polythene,, polypropene, Nylon 6, polyacrylonitrile,, Nylon 6, 6., b. Copolymers : The polymers which have, two or more types of repeating units are called, copolymer. They are formed by polymerization, of two or more different types of monomers in, presence of each other. The different monomer, units are randomly sequenced in the copolymer., For example : Buna-S, Buna-N., , d. Thermosetting polymers : Themosetting, polymers are rigid polymers. During their, formation they have property of being shaped, on heating; but they get hardened while hot., Once hardened these become infusible; cannot, be softened by heating and therefore cannot, be remoulded. This characteristic is the result, of extensive cross linking by covalent bonds, formed in the moulds during hardening/setting, process while hot. For example : bakelite, urea, formaldehyde resin, etc., , 326, , Problem 15.1 : Refer to the following table, listing for different polymers formed from, respective monomer. Identify from the list, whether it is copolymer or homopolymer., Sr., Monomer, Polymers, No., 1., Ethylene, Polyethylene, 2., Vinyl chloride, Polyvinyl chloride, 3., Isobutylene, Polyisobutylene, 4., Acrylonitrile, Polyacrylonitrile, 5., Caprolactum, Nylon 6, 6., Hexamethylene Nylon 6, 6, diammonium, adipate, 7., Butdiene +, Buna-S, styrene, Solution : In each of first five cases,, there is only one monomer which gives, corresponding homopolymer. In the sixth, case hexamethylene diamine reacts with, adipic acid to form the salt hexamethylene, diammonium adipate which undergoes, condensation to form Nylon 6, 6. Hence, nylon 6, 6 is homopolymer. The polymer, Buna-S is formed by polymerization of, the monomers butadiene and styrene in, presence of each other. The repeating units, corresponding to the monomers butadiene, and styrene are randomly arranged in the, polymer. Hence Buna-S in copolymer.
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Fig. 15.2 Shows all the classes of polymers in the form of a tree diagram., Polymers, Based on source, or origin, Natural, , Based on mode of, polymerization, , Based on inter, molecular forces, , Based on, structure, Linear, , No. of, monomers, Biodegradability, , Elastomers, , Addition, , Homopolymers, Copolymers, , Synthetic, , Branched chain, , Fibres, , Condensation, , Semisynthetic, , Cross - linked, , Thermoplastic, , Ring opening, , Biodegradable, Non-biodegradable, , Thermosetting, , Fig. 15.2 : Classification of polymers, , 15.2.6 Classification of polymers on the basis, of biodegradability : Most of the synthetic, polymers are not affected by microbes. These, are called non-biodegradable polymers. These,, in the form of waste material which stays, in the environment for very long time and, pose pollution hazards. Most natural fibres, in contrast are biodegradable. In attempt to, tackle the environmental problem, scientists, have developed bio-degradable synthetic, polymers. More details will be described in, section 15.5., 15.3 Some important polymers :, , H3C, , H, C=C, CH2, H2C, CH2, , Natural rubber : Monomer of natural, rubber is isoprene (2-methyl - 1, 3-butadiene)., , Natural rubber is a high molecular mass, linear polymer of isoprene. Its molecular mass, varies from 130, 000u to 340, 000u (that is, number of monomer units varies from 2000, to 5000)., , CH2, , H, C=C, , CH2, , CH2, , H, , Natural rubber, , Reaction involved in formation of, natural rubber by the process of addition, polymerization is as follows., nH2C = C - C = CH2, CH3 H, , Polymerization, , H, H, C-C=C-C, H, , CH3 H, , H n, , (polyisoprene/rubber), , Properties of Natural rubber :, 1., , Polyisoprene, molecule, has, cis, configuration of the C = C double bond., It consists of various chains held together, by weak van der Waals forces and has, coiled structure., , 2., , It can be stretched like a spring and, exhibits elastic property., , CH3, CH2 = C-CH = CH2, (isoprene), , C=C, , H3C, , (isoprene), , 15.3.1 Rubber : Elastomers are popularly, known as rubbers. For example, balloons,, shoesoles, tyres, surgeon's gloves, garden, hose, etc. are made from elastomeric material, or rubber., , H3C, , Vulcanization of rubber : To improve, the physical properties of natural rubber,, a process of vulcanization is carried out., In 1839 Charles Goodyear, an American, inventor invented the process of vulcanization., , 327
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The process by which a network of cross, links is introduced into an elastomers is, called vulcanization. The profound effect, of vulcanization enhances the properties like, tensile strength, stiffness, elasticity, toughness;, etc. of natural rubber. Sulfur vulcanization, is the most frequently used process. Sulfur, forms crosslinks between polyisoprene chains, which results in improved properties of, natural rubber., , Use your brain power, • From, the, cis-polyisoprene, structure of natural rubber, explain, the, low, strength of, van der Waals forces in it., • Explain how vulcanization of natural, rubber improves its elasticity ? (Hint :, consider the intermolecular links.), Do you know ?, Natural Rubber first came into, the market in early 19 th century. It, was entirely recovered from wild Hevea, brasiliensis trees which usually grew on the, banks of Amazon river and its tributaries in, South America. The amount of hydrocarbon, present in Hevea Tree is very high (35%)., As per the demand the production of natural, rubber increased by leaps and bounds and, at present 1.5 million tons of natural rubber, is sent to the market., The latex is collected from a mature, Hevea tree by making deep cuts on the, bark and by allowing the latex stream in a, pot attached below the cut. The latex is an, emulsion like milk., When a coagulant like acetic acid is, added to the latex the rubber hydrocarbon, gets coagulated in the amorphous solid, form., , Do you know ?, Vulcanizing is carried out, by heating raw rubber with sulfur, powder in presence of some organic, compounds called accelerators at about, 150 0C. (The most common accelerator, is ZBX or zincbutyl xanthate). By, increasing amount of sulfur the rubber, can be hardened. For example when the, amount of sulfur is raised to 40-45 %, a non-elastic hard material known as, ebonite is obtained., One or more sulfur atoms cross-link, two polyisoprene chains. Cross-linking, takes place by opening of a double, bond and produces three dimensional, vulcanised rubber., Probable 3-D structure of vulcanized, rubber is, CH3 S, CH - C - CH - CH2, , 15.3.2 Polythene :, Can you recall ?, , S, , How is ethylene prepared ?, , S, S, H2C - HC - C - CH2- CH2, S CH3, S, S, , CH3, H2C - CH - C - CH2- CH2, S, , Polythene is the simplest and most, commonly used hydrocarbon thermoplastic, and has following structure., [CH2 − CH2]n, The IUPAC name of polyethylene is, polythene. Polythene is of two kinds, namely, low density polythene (LDP) and high density, polythene (HDP) ., , 328
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a. Low density polyethylene (LDP) :, LDP is obtained by polymerization of ethylene, under high pressure (1000 - 2000 atm) and, temperature (350 - 570 K) in presence of, traces of O2 or peroxide as initiator., CH2 = CH2, , Traces of O2, , peroxide at 350 - 370K,, 1000 - 2000 atm, , LDP, , The mechanism of this reaction, involves free radical addition and H-atom, abstraction. The latter results in branching., As a result the chains are loosely held and, the polymer has low density., Do you know ?, In the H-atom abstraction,, process involved in formation, of LDP; the terminal carbon radical, abstracts H-atom from an internal carbon, atom in the form of an internal carbon, radical. Termination step in addition, polymerization gives rise to branching, of these internal carbons., , b. High density polyethylene (HDP) : It is, essentially a linear polymer with high density, due to close packing., CH2 = CH2, , 333 K - 343 K, 6 - 7 atm, catalyst, , HDP, , HDP is obtained by polymerization of, ethene in presence of Zieglar-Natta catalyst, which is a combination of triethyl aluminium, with titanium tetrachloride at a temperature, of 333K to 343K and a pressure of 6-7 atm., Properties of HDP : HDP is crystalline,, melting point in the range of 144 - 150 0C. It, is much stiffer than LDP and has high tensile, strength and hardness. It is more resistant to, chemicals than LDP., Uses of HDP : HDP is used in manufacture, of toys and other household articles like, buckets, dustbins, bottles, pipes etc. It is used, to prepare laboratory wares and other objects, where high tensile strength and stiffness is, required., , H, H, , intramolecular, H-atom abstraction, , R, Termination, , R, , Properties of LDP : LDP films are extremely, flexible, but tough, chemically inert and, moisture resistant. It is poor conductor of, electricity with melting point 110 0C., Uses of LDP : LDP is mainly used in, preparation of pipes for agriculture, irrigation,, domestic water line connections as well as, insulation to electric cables. It is also used, in submarine cable insulation. It is used in, producing extruded films, sheets, mainly, for packaging and household uses like in, preparation of squeeze bottles, attractive, containers etc., , Internet my friend, Where, is, polythene, manufactured in India ?, 15.3.3 Teflon : Chemically teflon is, polytetrafluoroethylene. The monomer used, in preparation of teflon is tetrafluoroethylene,, (CF2 = CF2) which is a gas at room temperature., Tetrafluoroethylene is polymerized by, using free radical initiators such as hydrogen, peroxide or ammonium persulphate at high, pressure., nCF2 = CF2, , Polymerization, Peroxide, , (Tetrafluoroethene), , [CF2 − CF2]n, (Teflon), , Properties : Telflon is tough, chemically inert, and resistant to heat and attack by corrosive, reagents., , 329
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C - F bond is very difficult to break, and remains unaffected by corrosive alkali,, organic solvents., Uses : Telflon is used in making non-stick, cookware, oil seals, gaskets, etc., , Polyamides contain - CO - NH - groups, as the inter unit linkages. Two important, polyamide polymers are nylon 6, 6 and nylon, 6., a. Nylon 6,6 : The monomers adipic acid and, hexamethylendiamine on mixing forms nylon, salt, which upon condensation polymerization, under conditions of high temperature and, pressure give the polyamide fibre nylon 6,6., n HOOC-(CH2)4-COOH + n H2N-(CH2)6-NH2, (adipic acid) , (hexamethylene, diamine), , Internet my friend, Collect the information of, Teflon coated products used in, daily life and in industries., , O, O, ⊕, ⊕, n O - C ( CH2 )4 C - OH3N ( CH2 )6 NH3, (nylon salt), , 553K high, -nH2O pressure, , 15.3.4 Polyacrylonitrile : Polyacrylonitrile, is prepared by addition polymerization of, acrylonitrile by using peroxide initiator., nCH2 = CHCN, , Polymerization, Peroxide, , (Acrylonitrile), , [CH2 − CH]n, CN, , (Polyacrylonitrile), , Polyacrylonitrile resembles wool and is, used as wool substitute and for making orlon, or acrilan., , H, O, O, [ C-(CH2)4-C-NH-(CH2)6-N]n, (nylon 6,6), The numerals 6,6 in the name of this, polymer stand for the number of carbon atoms, in the two bifunctional monomers, namely,, adipic acid and hexamethylenediamine., Nylon 6,6 is high molecular mass (1200050000 u) linear condensation polymer. It, possesses high tensile strength. It does not, soak in water. It is used for making sheets,, bristles for brushes, surgical sutures, textile, fabrics, etc., , Do you know ?, , Do you know ?, , Orlon is used to make blankets,, shawls, sweat shirts, sweaters., 15.3.5 Polyamide polymers : Polyamide, polymers are generally known as nylons. Nylon, is the generic name of the synthetic linear, polyamides obtained by the condensation, polymerization between dicarboxylic acids and, diamines, the self condensation of an amino, acid or by the ring opening polymerization, of lactams., , 330, , •, , When an amino group and, carboxyl group present in the, same molecule react intramolecularly, the resulting amide is cyclic and is, called lactam., , •, , A cyclic ester formed by intramolecular, reaction of hydroxyl and carboxyl, groups is called lactone.
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b. Nylon 6 : When epsilon (ε)-caprolactam, is heated with water at high temperature it, undergoes ring opening polymerization to give, the polyamide polymer called nylon 6., e, , n, , H2C, , δ, , NH, , C, , O, , α, , CH2, , H 2O, 533 - 543 K, , H2C, CH2 CH2, β, γ, (e - caprolactam), , Polycarbonates are also a kind of, polyester polymers. These are high melting, thermosetting resins., 15.3.7 Phenol - formaldehyde and related, polymers :, , [NH ( CH2 )5 CO ]n, , a. Bakelite : Bakelite, the thermosetting, polymer obtained from reaction of phenol and, formaldehyde is the oldest synthetic polymer., Phenol and formaldehyde react in presence of, acid or base catalyst to form thermosetting/, moulding powder (novolac) in two stages. In, the third stage, various articles are shaped from, novolac by putting it in appropriate moulds and, heating at high temperature (1380C to 1760C), and at high pressure. The reactions involved, are represented in the Fig. 15.3., , (Nylon 6), , The name nylon 6 is given on the basis of six, carbon atoms present in the monomer unit., Due to its high tensile strength and luster nylon, 6 fibres are used for manufacture of tyre cords,, fabrics and ropes., 15.3.6 Polyesters : The polyester polymers, have ester linkage joining the repeating units., Commercially the most important polyester, fibre is 'terylene' (also called dacron). It is, obtained by condensation polymerization, of ethylene glycol and terephthalic acid in, presence of catalyst at high temperature., n HO -CH2 - CH2 - OH + n H - O - C, O, (ethyleneglycol), , During the third stage of thermosetting, in the moulds, many crosslinks are formed, which results in formation of rigid polymeric, material, called bakelite which is insoluble, and infusible and has high tensile strength. It, can also serve as substitute for glass. Bakelite, is used for making articles like telephone, instrument, kitchenware, electric insulators., , C-O-H, O, , (terephthalic acid), 420 - 460 K, -nH2O, , zinc acetate-antimony, trioxide catalyst, , [ O - CH2 - CH2 - O - C, , O, , CO, , ]n, , (terylene or dacron), , Terylene has relatively high melting, point (2650C) and is resistant to chemicals, and water. It is used for making wrinkle, free fabrics by blending with cotton (terycot), and wool (terywool), and also as glass, reinforcing materials in safety helmets. PET, is the most common thermoplastic which, is another trade name of the polyester, polyethyleneterephthalate. It is used for, making many articles like bottles, jams,, packaging containers., , b. Melamine-formaldehyde polymer :, Decorative table tops like formica and plastic, dinner-ware are made from heat and moisture, resistant themosetting plastic called melamine, - formaldehyde resin. The reactions are shown, in Fig. 15.4. Melamine and formaldehyde, undergo condensation polymerisation to give, cross linked melamine formaldehyde., , 331, , Internet my friend, Find applications of bakelite, in day to day life.
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Cl, nCH2 = C - CH = CH2, , Cl, [CH2 - C = CH - CH2 ]n, , Polymerization, , (Neoprene), , (chloroprene), (a) : Polymerization, , - CH2 - C = CH - CH2 - CH2 - C = CH - CH2 Cl, Cl, + MgO, , + MgO, , - CH2 - C = CH - CH2 - CH2 - C = CH - CH2 Cl, Cl, vulcanization, , - CH2 - C = CH - CH2 - CH2 - C = CH - CH2 O, O, - CH2 - C = CH - CH2 - CH2 - C = CH - CH2 (b) : Vulcanization of neoprene, , Fig. 15.6 Neoprene and vulcanization, , 15.3.8 Buna-S rubber : Buna-S is an, elastomer which is a copolymer of styrene with, butadiene (Fig.15.5). Its trade name is SBR, (styrene-butadiene rubber). The copolymer is, usually obtained from 75 parts of butadiene, and 25 parts of styrene subjected to addition, polymerization by the action of sodium. It is, vulcanized with sulfur., , Neoprene is particularly resistant to, petroleum, vegetable oils, light as well as, heat. Neoprene is used in making hose pipes, for transport of gasoline and making gaskets., It is used for manufacturing insulator cable,, jackets, belts for power transmission and, conveying., , Buna-S is superior to natural rubber with, regard to mechanical strength and has abrasion, resistance. Hence it is used in tyre industry., , 15.3.9 Neoprene : Neoprene, a synthetic, rubber,, is, a, condensation, polymer, of chloroprene (2-chloro-1,3-butadiene)., Chloroprene polymerizes rapidly in presence, of oxygen. Vulcanization of neoprene takes, place in presence of magnesium oxide. The, reactions involved can be represented in Fig., 15.6., , 15.3.10 Viscose rayon : Viscose rayon is, a semisynthetic fibre which is regenerated, cellulose. Cellulose in the form of wood pulp, is transformed into viscose rayon. Cellulose, is a linear polymer of glucose units and has, molecular formula (C6H10O5)n. A modified, representation of the molecular formula of, cellulose Cell-OH, is used in the reactions, involved in viscose formation, as shown in, Fig. 15.7. Cellulose in the form of wood pulp, is treated with concentrated NaOH solution to, get fluffy alkali cellulose. It is then converted, to xanthate by treating with carbon disulphide., On mixing with dilute NaOH it gives viscose, solution which is extruded through spinnerates, , 333
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⊕, , Cell - ONa + H2O, , Cell - OH + NaOH, (cellulose pulp wood), , (alkali cellulose), , ⊕, , ⊕, , Cell - ONa + CS2, , Cell - O - C - SNa, S, , (cellulose xanthate), , ⊕, , Cell - O - C - SNa + H2O, S, , Cell - O - C - SH + NaOH, S, , Cell - O - C - SH + H2O, S, , Cell - OH + CS2, , (viscose rayon), (Regenerated cellulose filaments), , Fig. 15.7 : Formation of viscose rayon, , of spining machine into acid bath when, regenerated cellulose fibres precipitate., Use your brain power, •, , Write structural formulae of, styrene and polybutadiene., , 15.4 Molecular mass and degree of, polymerization of polymers : A polymer is, usually a complex mixture of molecules of, different molecular masses. Hence, molecular, mass of a polymer is an average of the, molecular masses of constituent molecules., Molecular mass of polymer depends upon, the degree of polymerization (DP). DP is, the number of monomer units in a polymer, molecule., , Most of the mechanical properties, of polymers depend upon their molecular, mass. Low molecular mass polymers are, liable to be brittle and have low mechanical, strength. If a polymer is allowed to attain, very high molecular mass it becomes tough, and unmanageable. Both these ends are, undesirable. A polymer must possess a, molecular mass more than certain minimum, value in order to exhibit the properties, needed for a particular application. This, minimum molecular mass corressponds to, the critical degree of polymerization. But the, polymerization process has to be controlled, after certain stage. For polymers containing, hydrogen bonding the critical degree of, polymerization is lower than those containing, weak intermolecular forces., , Can you tell ?, 1., , Classify the following polymers as addition or condensation., i. PVC ii. Polyamides iii. Polystyrene iv. Polycarbonates v. Novolac, , 2., , Complete the following table :, Condensation, polymers, , Repeating unit, , Name of, monomer, , 1. Nylon 6, 2. Nylon 6, 6, 3. Terylene, 4. Melamine, , 334, , Formula of, monomer, , Uses
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Problem 15.2 : The critical degree of, polymerization is low for nylon 6 while, high for polythene. Explain., , Can you recall ?, •, , Solution : Nylon 6 is a polyamide, polymer, and has strong intermolecular, hydrogen bonding as inter molecular, forces. On the other hand polythene chains, have only weak van der Waals forces as, intermolecular interaction. Because of the, stronger intermolecular forces the critical, DP is lower for nylon 6 than polythene., , this problem biodegradable polymers are being, developed. These polymers contain functional, groups similar to those in biopolymers such, as proteins. Aliphatic polyesters are also an, important class of biodegradable polymers., Use your brain power, • Represent the copolymerization, reaction between glycine and, e - amino caproic acid to form the, copolymer nylon 2- nylon 6., , 15.5 Biodegradable polymers :, Can you recall ?, •, , Name some materials which, undergo degradation after use., , •, , List the materials which do not decay, even after a long time., , •, , How is the environment affected by non, decaying substances ?, , •, , Which bonds are broken during digestion, of proteins and carbohydrates ?, , •, , What happens to disposed natural wastes, such as stale food, fruit peels, torn cotton, cloth ?, , What are the structural formulae, of glycine and e - amino caproic, acid ?, , •, , Inspite of large number of useful, applications, polymers are blamed for creating, environmental pollution. To strike the golden, mean, certain new biodegradable synthetic, polymers have been developed., , 15.5.1 PHBV : PHBV is a copolymer of, two bifunctional b- hydroxy carboxylic, acids, namely, b- hydroxybutyric acid (3 hydroxybutanoic acid) and b- hydroxyvaleric, acid (3 - hydroxypentanoic acid). Hydroxyl, group of one monomer forms ester link by, reacting with carboxyl group of the other. Thus, PHBV is an aliphatic polyester with name poly, b- hydroxy butyrate - co - b- hydroxy valerate, (PHBV). PHBV is degraded by microbes in the, environment., b, , a, , n(HO - CH - CH2 - COOH) + n(HO - CH - CH2 - COOH), , CH2 CH3, CH3, (b - hydroxy butyric acid) (b - hydroxy valeric acid), , Aliphatic polyesters and polyamides, with large proportion of polar linkages are, one of the important classes of biodegradable, polymers. Some important examples are, discussed below., Disposed natural wastes are usually, attacked by soil microbes and get degraded, to humus. But most synthetic polymers and, plastics cannot be degraded by microbes and, stay in the environment for very long period of, time posing pollution problems. To overcome, , What is the origin of the numbers 2 and, 6 in the name of this polymer ?, , - nH2O, , O, [ O - CH - CH2 - C - O - CH - CH2- C ]n, CH3, O, CH2 - CH3, (PHBV), , 15.5.2 Nylon 2 - nylon 6 : Nylon 2 - nylon 6, is a polyamide copolymer of two amino acids,, namely, glycine and e - amino caproic acid. It, is a biodegradable polymer., , 335
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15.6 Commercially important polymers : Apart from the polymers already discussed in this, chapter, many more polymers are used extensively. Structures and applications of some of them, are given in the Table 15.1., Table 15.1 : Commercially important polymers, Trade name, Perspex/acrylic, glass, , Buna N, , Monomer, methyl, methacrylate, , Polymer structure, CH3, , lenses, paint, security, barrier, LCD screen,, shatter resistant glass, , [ CH2 - C ]n, COOMe, , COOMe, , Butadiene and, acrylonitrile, , Applications, , [H2C-CH=CH-CH2-CH2-CH ]n, CN, , CN, , PVC (polyvinyl, chloride), , Polyacrylamide, , vinyl chloride, Cl, , acrylamide, , Ureaformaldehyde, resin, , a. urea, b. formaldehyde, , Glyptal, , a. ethyleneglycol, b. phthalic acid, , Polycarbonate, , a. bisphenol, b. phosgene, , Thermocol (made Styrene, from airfilled thin, walled beads of, polystyrene, , water pipes, rain coats,, flooring, , Cl, [ CH2 - CH ]n, , [ CH2 - CH ]n, , Polyacrylamide gel, used in electrophoresis, , CONH2, , CONH2, , [ NH - CO - NH - CH2 ]n, , [ O - CH2 - CH2 - OOC, O, C-O, , [ CH2 - CH ]n, , 336, , CO ]n, , CH3, C, CH3, , adhesives, rubber belts,, shoe soles, O-rings,, gaskets, , O, n, , unbreakable dinner, ware, decorative, laminates, , paints and lacquers, , electrical and, telecommunication, hardware, food grade, plastic containers, non-biodegradable, styrene can leach when, heated. Therefore it is, banned.
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Exercises, 1. Choose the correct option from the given, alternatives., i., , viii. PET is formed by ----, , Nylon fibres are ---A. Semisynthetic fibres, , ix., , B. Polyamide fibres, , iii., , C. PVC, , D. Protein, , x., , i., , C. Animal, D. Vegetable, , vii., , A. C-H bond , , B. C-F bond, , C. H- bond , , D. C=C bond, , Identify 'A' and 'B' in the following , reaction ----, , C. Novolac D. Terylene, , a. HO-CH2-CH2-OH, 533K, +, ∆, O, O, H-O-C, C-O-H, , Which of the following is made up of, polyamides ?, , b. H2N-(CH2)6-NH2+HOOC-(CH2)4COOH, , Dacron is another name of ---A. Nylon 6 B. Orlon, , vi., , Teflon is chemically inert, due to, presence of ..........., , 2. Answer the following in one sentence, each., , B. Synthetic, , v., , Chemically pure cotton is ----, , D. Cellulose, , Silk is a kind of ---- fibre, A. Semisynthetic , , iv., , D. Hydration, , C. Cellulose nitrate, , Which of the following is naturally, occurring polymer ?, B. Polyethylene, , C. Alkylation, , B. Viscose rayon, , D. Cellulose fibres, , A. Telfon, , B. Condensation, , A. Acetate rayon, , C. Polyester fibres, ii., , A. Addition , , A. Dacron, , B. Rayon, , C. Nylon, , D. Jute, , N2, 533K, , The number of carbon atoms present, in the ring of e - caprolactam is, A. Five, , B. Two, , C. Seven, , D. Six, , 'A', , ii., , Complete the following statements, a., b., c., , Terylene is ---A. Polyamide fibre, B. Polyester fibre, , d., , C. Vegetable fibre, D. Protein fibre, , e., , 337, , ‘B’, , Caprolactam is used to, prepare-------Novolak is a copolymer of ------- and --------Terylene is ----------polymer of, terephthalic acid and ethylene, glycol., Benzoyl peroxide used in, addtion polymerisation acts as, ---------Polyethene, consists, of, polymerised ----------
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iii., , vi., , Draw the flow chart diagram to show, classification of polymers based on, type of polymerisation., , Match the following pairs :, , Name of polymer , , Monomer, , 1. Teflon , , a. CH2=CH2, , 2. PVC , , b. CF2=CF2, , v. Name some chain growth polymers., , 3. Polyester, , c) CH2=CHCl, , vi. Define the terms :, , 4. Polythene, , d) C6H5OH and HCHO, , 5. Bakelite , , , e) Dicarboxylic acid , and polyhydoxyglycol, , iv., , Write examples of Addition polymers, and condensation polymers., , 1) Monomer, 2) Vulcanisation, , vii., , 3) Synthetic fibres, vii., , What type of intermolecular force, leads to high density polymer ?, , 1. nHOOC-R-COOH + nHO-R'-OH, 2. H2N-(CH2)5-COOH, , viii. Give one example each of copolymer, and homopolymer., ix., , viii. Name and draw structure of the, repeating unit in natural rubber., , Identify, Thermoplastic, and, Thermosetting Plastics from the, following -----, , ix., , a. Cellulose b. Polystyrene, , 2. Urea formaldehyde resin, , c. Terylene d. Starch, , 3. Polythene, , e. Protein, , 4. Phenol formaldehyde, , g. Orlon (Polyacrylonitrle), , iii., , iv., , v., , h. Phenol-formedehyde resins, , Write the names of classes of polymers, formed according to intermolecular, forces and describe briefly their, structural characteristics., , x., , What are synthetic resins? Name, some natural and synthetic resins., , xi., , Distinguish between thermosetting, and thermoplastic resins. Write, example of both the classes., , xii., , Write name and formula of raw, material from which bakelite is made., , Write reactions of formation of :, a. Nylon 6, , f. Silicones, , b. Terylene, , Write structure of natural rubber and, neoprene rubber along with the name, and structure of thier monomers., , 4. Attempt the following :, , Name the polymer type in which, following linkage is present., -C-OO, Write structural formula of the, following synthetic rubbers :, a. SBR rubber, , i. Identify condensation polymers and, addition polymers from the following., a. -(CH2-CH-)n, −, , ii., , Classify the following polymers as, natural and synthetic polymers, , 1. PET, , 3. Answer the following., i., , Draw the structures of polymers, formed, from, the, following, monomers, , C6H5, b. -(CH2-CH=CH-CH2)n, c. -(CO(CH2)4-CONH(CH2)6NH-)n, d. -(OCH2-CH2-O-CO, , b. Buna-N rubber, c. Neoprene rubber, , 338, , CO-)n
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ii., , Write the chemical reactions involved, in manufacture of Nylon 6,6, , iii., , 5. Answer the following., , Explain vulcanisation of rubber., Which vulcanizing agents are used, for the following synthetic rubber., a. Neoprene b. Buna-N, , iv., , Write reactions involved in the, formation of --1) Teflon, , 2) Bakelite, v., , What is meant by LDP and HDP?, Mention the basic difference between, the same with suitable examples., , vi., , Write preparation, properties and uses, of Teflon., , vii., , Classify the following polymers as, straight chain, branched chain and, cross linked polymers., , How is polythene manufactured ?, Give their properties and uses., , ii., , Is synthetic rubber better than natural, rubber ? If so, in what respect?, , iii., , Write main specialities of Buna-S,, Neoprene rubber?, , iv., , Write the structure of isoprene and, the polymer obtained from it., , v., , Explain in detail free radical, mechanism, involved, during, preparation of addition polymer., , Activity :, i., , Collect the information of, the process like extrusion and, moulding in Textile Industries., , ii., , Make a list of polymers used to, make the following articles, , − −, , −, , a. -(CH2-CH-)n, CN, b. -(CH2-CH2-CH-CH2-CH2-)n, CH2, CH2, , i., , c., , a. Photographic film, b. Frames of spectacles, , OH, , OH, , CH2, , CH2, , H2C, , OH, , c. Fountain pens, , CH2, , d. Moulded plastic chains, e. Terywool or Terycot fabric, , CH2, CH2, OH, CH2, , H2C, CH2, , OH, , CH2, , CH2, , OH, , CH2, , (Bakelite), , 339, , iii., , Prepare a report on factors, responsible for degradation of, polymers giving suitable example., , iv., , Search and make a chart/note, on silicones with reference to, monomers, structure, properties, and uses., , v., , Collect the information and data, about Rubber industry, plastic, industry and synthetic fibre (rayon), industries running in India.
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16. GREEN CHEMISTRY AND NANOCHEMISTRY, need to implement 12 principles of green, chemistry enunciated by Paul Anastas, whereever possible., , Can you recall ?, 1. What do you mean by, environment?, , 16.2 Sustainable development : Green, chemistry plays an important role in sustainable, development. We can achieve sustainable, development by adapting the twelve principles, of green chemistry. Sustainable development, is development that meets the needs of the, present, without compromising the ability of, future generations to meet their own need., Sustainable development has been continued to, evolve as that protecting the world’s resources., , 2. Which are the factors affecting the, environment?, 3. What is pollution? Which are the types, of pollution?, 4. Why it occurs?, 16.1 Introduction, Chemistry plays an important role to, improve the quality of our life. Unfortunately,, due to this achievement our health and, global environment are under threat. Also,, due to increase in human population and, the industrial revolution, energy crisis and, environmental pollution are highlighted, major global problems in the 21st century. To, minimize the problems of energy crisis and, pollution, we have to adapt green chemistry., , 16.3 Principles of green chemistry :, 1. Prevention of waste or by products :, To give priority for the prevention of, waste rather than cleaning up and treating, waste after it has been created., , Do you know ?, Paul T. Anastas (Born on May 16,, 1962) is the director of Yale university’s Center for green chemistry and green, engineering. He is known as father of green, chemistry., Green Chemistry is an approach to, chemistry that aims to maximize efficiency, and minimize hazardous effects on human, health and environment. The concept of green, chemistry was coined by Paul T. Anastas., Definition : Green Chemistry is the, use of chemistry for pollution prevention by, environmentally conscious design of chemical, products and processes that reduce or eliminate, the use or generation of hazardous substances., , Illustration : To develop the zero waste, technology (ZWT). In terms of ZWT, in a, chemical synthesis, waste product should be, zero or minimum. It also aims to use the waste, product of one system as the raw material, for other system. For example : 1. bottom, ash of thermal power station can be used as, a raw material for cement and brick industry., 2. Effluent coming out from cleansing of, machinery parts may be used as coolant water, in thermal power station., 2. Atom economy : Atom economy is a, measure of the amount of atoms from the, starting materials that are present in the useful, products at the end of chemical process. Good, atom economy means most of the atoms of, the reactants are incorporated in the desired, products and only small amounts of unwanted, byproducts are formed and hence lesser, problems of waste disposal., , To reduce the impact of energy crisis,, pollution and to save natural resources, we, , 340
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Illustration : The concept of atom economy, gives the measure of the unwanted product, produced in a particular reaction., % atom economy =, Formula weight of the desired product, sum of formula weight of all the reactants used in, the reaction, , × 100, , 5. Use Safer solvent and auxilaries :, , For example : conversion of Butan-1-ol to 1 bromobutane, CH3CH2CH2CH2OH + NaBr + H2SO4, CH3CH2CH2CH2-Br + NaHSO4 + H2O, % atom economy =, mass of (4C + 9H + 1Br) atoms, mass of (4C + 12H + 5O + 1Br + 1Na + 1S)atoms, × 100, , =, , For example : Adipic acid is widely used in, polymer industry. Benzene is the starting, material for the synthesis of adipic acid but, benzene is carcinogenic and benzene being, volatile organic compound (VOC) pollutes air., In green technology developed by Drath and, Frost, adipic acid is enzymatically synthesised, from glucose., , 137 u, × 100, 275 u, , = 49.81 %, 3. Less hazardous chemical synthesis :, Designed chemical reactions and, synthesis routes should be as safe as possible., So that we can avoid formation of hazardous, waste from chemical processes., Illustration : Earlier DDT (Dichlorodiphenyl, trichloroethane) was used as insecticide and, which was effective in controlling diseases like, typhoid and malaria carrying mosquitos. It was, realized that DDT is harmful to living things., Nowadays benzene hexachloride (BHC), is used as insecticide. One of the ϒ-isomer, (gamma) of BHC is called gammexane or, lindane., 4. Desigining Safer Chemicals : This principle, is quite similar to the previous one. To develop, products that are less toxic or which require, less toxic raw materials., Illustration : In Chemical industries workers, are exposed to toxic environment. In order to, prevent the workers from exposure to toxicity,, we should think of designing safer chemicals., , Choose the safer solvent available for, any given step of reaction. Minimize the total, amount of solvents and auxilary substances, used, as these make up a large percentage of, the total waste created., Illustration : The main aim behind this, principle is to use green solvents. For example,, water, supercritical CO2 in place of volatile, halogenated organic solvents, for example,, CH2Cl2, CHCl3, CCl4 for chemical synthesis, and other purposes. Solvents as chemicals that, dissolve solutes and form solutions, facilitate, many reactions. Water is a safe benign solvent, while dichloromethane is hazardous. Use of, toxic solvent affects millions of workers every, year and has implications for consumers and, the environment as well. Many solvents are, used in high volumes and many are volatile, organic compounds. Their use creates large, amounts of waste, air pollution and other, health impacts. Finding safer, more efficient, alternatives or removing solvents altogether is, one of the best ways to improve a process or, product., 6. Design for energy efficiency : Chemical, synthesis should be designed to minimize the, use of energy. It is better to minimize the energy, by carrying out reactions at room temperature, and pressure., This can be achieved by use of proper, catalyst, use of micro organisms for organic, synthesis, use of renewable materials, ... ,etc., Illustration : The biocatalyst can work at, the ambient condition. Similarly, in chemical, synthesis, refluxing conditions require less, , 341
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energy, improving the technology of heating, system, use microwave heating, .... , etc., , 8. Reduce derivatives : [Minimization of, steps], , 7. Use of renewable feedstocks : The, perspective of this principle is largely toward, petrochemicals. Use chemicals which are, made from renewable (plant based) sources, rather than other (for example : Crude Oil)., , A commonly used technique in, organic synthesis is the use of protecting or, blocking group. Unnecessary derivatization, (for example installation / removal of use of, protecting groups) should be minimized or, avoided if possible, because such steps require, additional reagents and can generate waste., , Illustration, :, Overexploitation, of, nonrenewable feed stocks will deplete the, resources and future generation will be, deprived. Moreover, use of these nonrenewable, resources puts burden on the environment., , Illustration : In organic synthesis, we need, very often protection of some functional, groups. Finally, we again need their, deprotection. It is explained in the following, example of synthesis of m-hydroxybenzoic, acid from m-hydroxy benzaldehyde., , On the other hand, use of renewable resources, for example agricultural or biological product, ensures the sharing of resources by future, generation. This practice generally does not, put much burden on environment. The products, and waste are generally biodegradable., , CHO, , Obviously, in such cases, atom economy, is also less. The green chemistry principle, aims to develop the methodology where, unnecessary steps should be avoided, if, practicable biocatalytic reactions very often, need no protection of selective group., , CHO, OH, , C6H5CH2Cl, protection of, -OH group, , COOH, [O], , OCH2C6H5, , (m-hydroxybenzaldehyde), , OCH2C6H5, Deprotection of, OH group, , COOH, OH, (m-hydroxybenzoic acid), , 9. Use of catalysis : Use of catalyst in the chemical reaction speeds up its rate. Catalyst helps, to incease selectivity, minimize waste and reduce reaction times and energy demands., Complete the chart, Reaction, , Name of Catalyst used, , 1. Hydrogenation of oil (Hardening) , 2. Haber’s process of manufature of ammonia, 3. Manufacture of HDPE polymer, 4. Manufacture of H2SO4 by contact process , 5. Fischer-Tropsch process (synthesis of gasoline), , 342
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10. Design for degradation : Design chemicals, that degrade and can be discarded easily. Ensure, that both chemicals and their degradation, products are not toxic, bioaccumulative or, environmentally persistent., , Do you know ?, Does plastic packaging impact the, food they wrap ?, Phthalates leach into food through, packaging so you should avoid, microwaving food or drinks in plastic and, not use plastic cling wrap and store your, food in glass container whenever possible., Try to avoid prepackaging, processed food, so that you will reduce exposure to harmful, effect of plastic., , Illustration : The aim behind this principle, is that the waste product should degrade, automatically to clean the environment. Thus,, the biodegradable polymers and pesticides are, always preferred. To make separation easier, for the consumer an international plastic, recycle mark is printed on larger items., , 1, , 2, , PETE, , HDPE, , Number 1 : PETE or PET, (Polyethylene terephthalate), Used In : microwavable food trays;, salad dressing, soft drink, water and, beer bottles, Status : hard to clean; absorbs bacteria, and flavors; avoid reusing, Is Recycled To Make : carpet,, furniture, new containers, polar, fleece, , Number 2 : HDPE (high, density polyethylene), Used In : household cleaner and, shampoo bottles, Status : transmits no known chemicals, into food, Is Recycled To Make : detergent, bottles, fencing, floor tiles, pens, , 7, , Other, , Number 7 : Other, (Miscellaneous), Used In : 3- and 5- gallon waterjugs,, nylon, some food containers, Status : Contains bisphenol A, which, has been linked to heart disease and, obesity; avoid, Is Recycled To Make : custommade products, , 3, , V, , HOUSEHOLD, PLASTICS, , Number 3 : V or PVC, (vinyl), Used In : cooking oil bottles, clear, food packaging, mouthwash bottles, Status : is belived to contain, phalates that interfere with hormonal, development; avoid, Is Recycled To Make : cables., mudflaps, paneling, roadway, gutters, , 6, PS, , Number 6 : PS, (Polystyrene), Used In : disposable cups and plates,, egg cartons, take-out containers, Status : is belived to leach styrene,, a possible human carcinogen into, food; avoid, Is Recycled To Make : foam, packaging, insulation, light, switchplates, rulers, , 4, , 5, , PP, Number 5 : PP, (Polypropylene), Used In : ketchup bottles, medicine,, and syrup bottle, drinking straws, Status : transmits no known chemicals, into food, Is Recycled To Make : battery, cables, brooms, ice scrapers, rakes, , Use this chart to sort plastic materials in daily life, , 343, , LDPE, , Number 4 : LDPE (Low, density polyethylene), Used In : bread and shopping bags,, carpet, clothing, furniture, Status : Transmits no known, chemicals into food, Is Recycled To Make : envelopes,, floor tiles, lumber, trash-can liners
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11. Real-time Analysis Pollution Prevention:, Analytical methods need to be further, developed to allow for real-time, in process, monitoring and control prior to the formation, of hazardous substances., , negative environmental effects. Right now, the green chemistry revolution is beginning, and it is an exciting time with new challenges, for chemists involved with the discovery,, manufacture and use of chemicals., , Illustration : Analytical methodologies, should be developed or modified, so that, continuous monitoring of the manufacturing, and processing units is possible. This is very, much important for the chemical industries, and nuclear reactors., , Green chemistry helps to protect the, presence of ozone in the stratosphere essential, for the survival of life on the earth. Green, chemistry is useful to control green house, effect (Global warming). So we should think, about save environment and save earth., , 12. Safer chemistry for Accident prevention:, , Can you recall ?, , We need to develop chemical processes, that are safer and minimize the risk of, accidents., Illustration : The substances to be used in a, chemical reaction should be selected in such, a way that they can minimize the occurrence, of chemical accidents, explosions, fire and, emission. For example, if the chemical process, works with the gaseous substances, then the, possibility of accidents including explosion, is relatively higher compared to the system, working with non volatile liquid and solid, substances., 16.4 The role of Green chemistry : The green, chemistry approach recognizes that the Earth, does have a natural capacity for dealing with, much of the waste and pollution that society, generates, it is only when that capacity is, exceeded that we become unsustainable., To promote innovative chemical, technologies that reduce or eliminate the use, or generation of hazardous substances in the, design, manufacture and use of chemical, products., The green chemistry helps to reduce, capital expenditure, to prevent pollution., Green chemistry incorporates pollution, prevention practices in the manufacture of, chemicals and promotes pollution prevention, and industrial ecology. Green chemistry is a, new way of looking at chemicals and their, manufacturing process to minimize any, , 1. What are the shapes of a, bacillus and coccus? (Refer to, chapter from Biology, Std. XI), 2. Which instrument is used to observe the, cells ? (Refer to chapter 5 from Biology,, Std. XI), 3. What is the size range of molecules of, lipids and proteins ?, 16.5 Introduction to nano chemistry : From, clothes, sunglasses you wear to computer, hard drives and even cleaning products,, nanotechnology plays a big part in the, manufacture of many materials. We have been, using Lasers in DVD, CD players for a long, time which contain nanosize components., Look at Fig. 16.1 which shows comparative, scales from macro-materials to atoms., Also observe another Fig. 16.2 which depicts, the materials in nature, as well as devices that, are man made. In both figures some objects, like tennis ball (Fig. 16.1), ant, human hair, (Fig. 16.2) we can see with our own eyes, whereas bacteria, virus, red blood cell, we can, not observe with naked eye. These are known, as nanomaterials., a. What is nanoscience ?, Nanoscience is the study of phenomena and, manipulation of materials at atomic, molecular, and macromolecular scales where properties, differ significantly from those at a larger scale., , 344
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Fig. 16.1 Macro-materials to atoms, ant, 5 mm, , microelectromechanical, 10-2 metre =, devices 10-100µm, 1 cm =, 10 mm, 10-3 metre =, 1 mm =, 1.000 microns (µm), 10-4 metre =, 0.1 mm =, 100 µm, , dust mite, 200 µm, , 10-5 metre =, 0.01 mm =, 10 µm, 10-8 metre =, 1 µm =, 1,000 (nm), , human hair, 10-50 µm wide, , red blood cells, 2-5 µm wide, , 10-7 metre =, 0.1 µm =, 1,000 nm =, 1,000 angstroms, (A), , DNA 2-12 nm diameter, , red blood, cells, , quantum corral of 48 iron, atoms on copper surface, positioned one at a time, with a scanning tunneling, microscope tip 14nm corral, diameter, , carbon nanotube 2nm, diameter, , Fig. 16.2 Scale of nanomaterials, , b. How do we define nanotechnology ?, Nanotechnology is the design, characterization,, production and application of structures,, device and system by controlling shape and, size at nanometer scale., c. Why Nano ?, The nanometer scale : ‘Nano’ in Greek means, dwarf but in actual case ‘nano’ is even smaller, than dwarf. Conventionally, the nanometer, scale is defined as 1-100 nm. One nanometer is, one billionth of a meter. (that is 1nm = 10-9m)., The materials we see around us are bulk, materials that possess macroscopic physical, properties. Grain of sand that is micron-sized, , material also possesses same bulk properties., But material synthesized at nanoscale (1nm 100nm) possesses unique optical, structural,, thermal, catalytic, magnetic and electrical, properties. These properties change as a, function of size and are very different from, their bulk materials., d. What is a nanomaterial ?, The nanomaterial is a material having, structural components with atleast one, dimension in the nanometer scale that is 1-100, nm. Nanomaterials are larger than single atoms, but smaller than bacteria and cells. These may, , 345
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be nanoparticles, nanowires and nanotubes, according to dimensions. Nanostructured, materials may be large organic molecules,, inorganic cluster compounds and metallic or, semiconductor particles., , materials science as well as engineering,, biological and medical applications., Do you know ?, A very highly useful application, of nanochemistry is ‘medicine’. A, simple skin care product of nanochemistry, is, sunscreen., Sunscreen, contains, nanoparticles of Zinc oxide, (ZnO) and, Titanium dioxide, (TiO2). These chemicals, protect the skin against harmful UV, (ultraviolet) rays by absorbing or reflecting, the light and prevent the skin from damage., , What are zero, one and two dimensional, nanoscale system ?, i. Zero-Dimensional Nanostructures : For, example, Nanoparticles., A zero dimensional structure is one in which, all three dimensions are in the nanoscale., ii. One-Dimensional Nanostructures : For, example, Nanowires and Nano rods., , Internet my friend, Find out similar applications in, medicine related to wounds, healing, process. Also find out applications of TiO2, and ZnO in other areas., , A one dimensional nanostructure is one in, which two dimensions are in the nanoscale., iii. Two-Dimensional Nanostructures : For, example, Thin films., A two-dimensional nanostructure is one in, which one dimension is in the nanoscale., Nanomaterial Nanomaterial, Dimension, Type, , Example, , Nanoparticles,, All three, Quantum dots,, dimensions < nanoshells,, 100 nm, nanorings,, microcapsules, Two dimensions < 100, nm, , Nanotubes, fibres,, nanowires, , One dimension < 100, nm, , Thin films, layers, and coatings, , 16.6 Characteristic features of Nanoparticles, : What makes the science at nanoscale special, is that at such a small scale, different laws, dominate over those that we experience in our, everyday life., 16.6.1 Colour : It is an optical property that, is different at nanoscale. Elemental gold as, we know, has nice shining yellow colour., However, if you had only 100 gold atoms, arranged in cube, its colour would be much, more red., , Fig. 16.3 Illustration of zero, one, two, dimensions, , e. Definition of Nanochemistry : It is the, combination of chemistry and nanoscience., It deals with designing and synthesis of, materials of nanoscale with different size and, shape, structure and composition and their, organization into functional architectures., Nanochemistry is used in chemical, physical,, , Fig. 16.4 Formation of gold nanoparticles, solution, , 16.6.2 Surface area : High surface-to-volume, ratio is a very important characteristic of, nanoparticles. If a bulk material is sub divided, into a group of individual nanoparticles,, the total volume remains the same, but the, , 346
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collective surface area is largely increased., With large surface area for the same volume,, these small particles react much faster because, more surface area provides more number, of reaction sites, leading to more chemical, reactivity. Explanation of increase in surface, area with decrease in particle size., , sodium clusters (Nan) of 1000 atoms appeared, to melt at 288 K while cluster of 10,000 atoms, melted at 303 K and bulk sodium melts at, 371K., 16.6.5 Mechanical properties, Mechanical strength : Nanosized copper and, palladium clusters with diameter in the size, range of 5-7 nm can have hardness upto 500%, greater than bulk metal., 16.6.6 Electrical conductivity : Electrical, conductivity is observed to change at nanoscale., For example, carbon nanotube can act as a, conductor or semiconductor in behaviour., , Area = 6 × 1m2 = 6 m2 Area = 6×(1/2m)2 ×8 = 12 m2, , 16.7 Synthesis of nanomaterials, 16.7.1 : There are two approaches to the, synthesis of nanomaterials. Bottom up and Top, down. Fig. 16.6 shows schematic illustration, of the preparation methods of nanoparticles., , Area = 6 × (1/3m)2 × 27 = 18 m2, , Fig. 16.5 : Surface area of nanoparticles, , Fig. 16.5 shows the surface areas when a cube, of 1m3 were progressively cut into smaller, cube until cube of 1nm3 formed., 16.6.3 Catalytic activity : Due to increase, in surface area with decrease in particle, size, nanomaterial-based catalysts show, increased catalytic acitivity. Usually they are, heterogeneous catalysts that means catalysts, are in solid form and the reactions occur on the, surface of the catalyst. Nanoparticle catalysts, can be easily separated and can be recycled., Example, Pd, Pt metal nanoparticles used in, hydrogenation reactions., TiO2, ZnO are used in photocatalysis. Gold in, bulk form is unreactive, but gold nanoparticles, are found to be very good catalyst for various, organic reactions., Internet my friend, Find out various applications or, use of gold nanoparticles., 16.6.4 Thermal properties : melting point, The melting point of nanomaterial changes, drastically and depends on size. For example,, , 347, , Top-Down, , Bottom-up, , Fig. 16.6 : Schematic illustration of the, preparation of nanoparticles, , In the bottom approach, molecular, components arrange themselves into more, complex assemblies atom by atom, molecule, by molecule and cluster by cluster from the, bottom. Example : synthesis of nanoparticles, by colloidal dispersion., In the top-down approach, nanomaterials, are synthesized from bulk material by breaking, the material. The bulk solids are dis-assembled, into finer pieces until they are constituted of, only few atoms., 16.7.2 Wet chemical synthesis of, Nanomaterials : Sol-gel process : Sols are, dispersions of colloidal particles in a liquid., Colloids are solid particles with diameters of, 1-100nm. A gel is interconnected rigid network
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with pores of submicrometer dimensions and, polymeric chains whose average length is, greater than a micrometer., , 2. Gelation resulting from the formation of an, oxide or alcohol-bridged network. (gel) by a, polycondensation reaction., , A sol-gel process is based on inorganic, polymerization reactions. It is generally carried, out at room temperature and includes four, steps : hydrolysis, polycondensation, drying, and thermal decomposition. This method is, widely employed to prepare oxide materials., , 3. Aging of the gel means during that period, gel transforms into a solid mass., 4. Drying of the gel : In this step, water and, other volatile liquids are removed from the, gel network., 5. Dehydration : The material is heated at, temperatures upto 800 0C., 16.7.3 Analysis or characterization of, nanomaterials :, The synthesized material is analyzed by, various analytical tools or techniques. The, name of the technique and its use is described, in the following Table 16.1., 16.7.4 Photographs of instruments, , Fig. 16.7 : Schematic representation of sol-gel, process of synthesis of nanoparticles, , The rections involved in the sol-gel process, can be described as follows :, MOR + H2O, , MOH + ROH (hydrolysis), , metal alkoxide, MOH + ROM, , M-O-M + ROH, (condensation), , Fig. 16.8 Photograph of X-ray diffractometer, , 1. Formation of different stable solution of the, alkoxide or solvated metal precursor., , Table 16.1, , Name of Technique, 1. UV-visible spectroscopy, , Instrument used, Information, UV-visible spectrophotometer Preliminary confirmation of, formation of nanoparticles, , 2. Xray Diffraction (XRD), , Xray diffractometer, , 3. Scanning electron, microscopy, , Scanning electron microscope Structure of surface of, (SEM), material that is morphology, , 4. Transmission electron, microscopy, , Transmission electron, microscope (TEM), , particle size, , 5. FTIR Fourier transform, infrared spectroscopy, , Fourier transform infrared, spectrophotometer, , Absorption of functional, groups, Binding nature., , 348, , particle size, crystal structure,, geometry
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16.8 History of nanotechnology :, Electron Gun, Condenser Lens, , Vaccum Chamber, Anode, , Condenser Lens, , Electron Beam, , Sample, chamber, X-ray detector, Sample, , Objective Lens, Backscatter, Detector, Secondary, Detector, , Watson, and Crick, 1953, , Richard, Feynman, 1959, , Von, Neumann, Machines, 1966, , Center for, responsible, Nanotechnology, , Fig. 16.9 Schematic diagram of scanning, electron microscope, , National, Nanotechnology, Initiative, (INR) 1999, , 1989, Foresight, Institute 89, , Eric, Drexler, 1986, , Fig. 16.13 : Scientists contributed to, nanotechnology, , Nanomaterials have been produced and used, by humans for hundreds of years. However,, understanding of certain materials as, nanostructured materials is relatively recent., Due to the development of advanced tools, that is sophisticated instruments, it has been, possible to reveal the information at nanoscale., , Fig. 16.10 Scanning electron microscope, , Filament, Wehnelt, Anode, Fixed Aperture, Condenser stigmator, Condenser Aperture, Objective stigmator, Objective Aperture, Selective Area Aperture, or intermediate Aperture, Different stigmator, , Gun, Gun Deflectors, Condenser lens 1, Condenser lens 2, Beam Deflectors, Objective lens upper, Specimen holder, Objective lens lower, Image Deflectors, Intermediate Lens, Projector Lens 1, Projector Lens 2, , CCD Camera, Viewing screen, , Beam axis, , Fig. 16.11Transmission electron microscope (TEM), , Fig. 16.12 FTIR specroscope, Fig. 16.14 Ruby red colour, , 349
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a. Beautiful ruby red colour of some ancient, glass paintings is due to gold and silver, nanoparticles trapped in the glass matrix., , Can you think ?, Visualize the size effect : Size, difference between the earth and, an apple is equal to the size difference, between atoms (30 nm) and an apple., , b. The decorative glaze or metallic film known, as lustre found on some medieval pottery is, due to certain spherical metallic nanoparticles., c. Carbon black is a nanostructured material, that is used in tyres of car to increase the life of, tyre. (Discovery in 1900). Carbon nanotubes, are made up of graphite sheets with nanosized, diameter. They have highest strength., d. Fumed silica, a component of silicon rubber,, coatings, sealants and adhesives is also a, nanostructured material., Internet my friend, Find out more number of, nanostructured materials in day to, day used products., Do you know ?, 1. The term ‘nanotechnology’, was defined by Tokyo science, University Professor, Nario Taniguchi in, 1974., 2. Invention of Scanning Tunneling, Microscope (STM) in 1980, led to the, discovery of fullerenes in 1986 and, carbon nanotubes a few years later., Internet my friend, Collect the information about the, scientists who discovered SEM,, STM, TEM instruments., , (a) 0D spheres and clusters, (b) 1D nanofibers,, wires and rods (c) 2D films, plates and networks,, (d) 3D nanomaterials, , 16.9 Applications of nanomaterials :, Nanochemistry has already contributed to, number of innovative products in various, disciplines because of their unique physical,, chemical, optical, structural, catalytic, properties and so on. Few applications are, given below :, a. Nanoparticles can contribute to stronger,, lighter, cleaner and smarter surfaces and, systems. They are used in the manufacturing of, scratchproof eyeglasses, transport, sunscreen,, crack resistant paints and so on., b. Used in electronic devices. For example,, Magnetoresistive Random Acess memory, (MRAM), c. Nanotechnology plays an important role in, water purification techniques., Water contains waterborne pathogens like, viruses, bacteria. 1.1 billion people are without, access to an improved water supply. The, provision of safe drinking water is currently, high priority. Recently, cost effective filter, materials coated with silver nanoparticles, (AgNps) is an alternative technology. (For, example : water purifier) Silver nanoparticles, act as highly effective bacterial disinfectant,, remove E.Coli from water., d. Self cleaning materials : Lotus is an example, of self cleaning. The lotus plant (Nelumbo, nucifera) although grows in muddy water, its, leaves always appear clean. The plants’ leaves, are superhydrophobic. Nanostructures on lotus, leaves repel water which carries dirt as it rolls, off. Lotus effect is the basis of self cleaning, windows., , Fig. 16.15 Classification of nanomaterials, , 350
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Disadvantages : Despite the possibilities and, the advancements that the nanotechnology, offers to the world, there also exist certain, potential risks involved with the disadvantages, of it., , Do you know ?, Sol-gel processes are used in the, motor vechicle industry to produce, water repellent coatings for wind screens, or exterior mirrors., , Nanotechnology has raised the standard, of living but at the same time, it has increased, the pollution which includes air pollution. The, pollution caused by nanotechnology is known, as nano pollution. This kind of pollution is, very dangerous for living organisms., , 16.10 Nanoparticles and Nanotechnology :, Advantages :, 1. Revolution in electronics and computing., 2. Energy sector - nanotechnology will make, solar power more economical. Energy storage, devices will become more efficient., , Nanoparticles can cause lung damage., Inhaled particulated matter can be deposited, throughout the human respiratory tract and, then deposit in lungs., , 3. Medical field :, Manufacturing of smart drugs, helps cure, faster and without side effects. Curing of life, threatening diseases like cancer and diabetes., , The characteristics of nanoparticles, that are relevant for health effects are size,, chemical composition and shape., , Exercises, 1. Choose the most correct option., i., , The development that meets the needs, of present without compromising the, ability of future generations to meet, their own need is known as, , iv., , a. Absorption of functional groups, b. Particle size, , a. Continuous development, , c. Confirmation of formation of, nanoparticles, , b. Sustainable development, c. True development, , d. Crystal structure, , d. Irrational development, ii., , Which of the following is ϒ-isomer, of BHC?, a. DDT , , b. lindane, , v., , The concept of green chemistry was, coined by, a. Born Haber, b. Nario Taniguchi, , c. Chloroform , , c. Richard Feynman, , d. Chlorobenzene, iii., , Which of the following information, is given by FTIR technique ?, , d. Paul T. Anastas, , The prefix 'nano' comes from, , 2. Answer the following, , a. French word meaning billion, b. Greek word meaning dwarf, , i. Write the formula to calculate % atom, economy., , c. Spanish word meaning particle, , ii. Name the ϒ-isomer of BHC., , d. Latin word meaning invisible, , 351
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iii. Ridhima wants to detect structure, of surface of materials. Name the, technique she has to use., , vi., , vii. How will you illustrate the use of, safer solvent and auxiliaries ?, , iv. Which nanomaterial is used for tyres, of car to increase the life of tyres ?, v. Name the scientist who discovered, scanning tunneling microscope (STM), in 1980., , Which nanomaterial is used in, sunscreen lotion ? Write its use., , viii. Define catalyst. Give two examples., 4. Answer the following, i. Explain any three principles of green, chemistry., , vi. 1 nm = .....m ?, , ii. Explain atom economy with suitable, example., , 3. Answer the following, i., , Define (i) Green chemistry (ii), sustainable development., , iii. How will you illustrate the principle,, minimization of steps ?, , ii., , Explain the role of green chemistry., , iii., , Give the full form (long form) of the, names for following instruments., , iv. What do you mean by sol and gel?, Describe the sol-gel method of, preparation for nanoparticles., , iv., , a. XRD, , b. TEM., , d. FTIR, , e. SEM, , c. STM, , v. Which flower is an example of self, cleaning ?, , Define the following terms :, , Activity :, , a. Nanoscience, b. Nanotechnology, , •, , c. Nanomaterial, d. Nanochemistry, v., , How nanotechnology plays an, important role in water purification, techniques?, , 352, , Collect, information, about, application of nanochemistry in, cosmetics and pharmaceuticals
