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www.tntextbooks.in, , GOVERNMENT OF TAMIL NADU, , HIGHER SECONDARY SECOND YEAR, , CHEMISTRY, VOLUME - I, , A publication under Free Textbook Programme of Government of Tamil Nadu, , Department of School Education, Untouchability is Inhuman and a Crime, , Introduction Pages.indd 1, , 2/19/2020 4:36:24 PM

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www.tntextbooks.in, , Government of Tamil Nadu, First Edition, , -, , 2019, , Revised Edition, , -, , 2020, , (Published under new syllabus), , NOT FOR SALE, , Content Creation, , The wise, possess all, , State Council of Educational, Research and Training, © SCERT 2019, , Printing & Publishing, , Tamil NaduTextbook and Educational, Services Corporation, www.textbooksonline.tn.nic.in, , II, , Introduction Pages.indd 2, , 2/19/2020 4:36:24 PM

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www.tntextbooks.in, , Key features …, , Scope of Chemistry, Learning objectives, Do you know, Example Problems, , Awareness about higher education avenues in, the field of Chemistry, Describe the specific competency /, performance capability acquired by the learner, Additional information provided to relate the, content to day-to-day life / development in the, field, Model problems worked out for clear-cut, comprehension by the learners, , Evaluate yourself, , To help the students to assess their own, conceptual understanding, , Q.R code, , Quick access to concepts, videos, animations, and tutorials, , ICT, , opens up resources for learning; enables the, learners to access, extend transform ideas /, informations, , Summary, , A glance on the substance of the unit, , Concept map, Evaluation, , Inter relating the concepts for enabling learners, to visualize the essence of the unit, To assess the level of understanding through, multiple choice question,numerical problems, etc…, , Books for Reference List of relevant books for further reading, Key answers, Glossary, , To help the learners confirm the accuracy of, the answers arrived and remedy the gaps in, learning, Important terms are enlisted with equivalent, Tamil words, , III, , Introduction Pages.indd 3, , 2/19/2020 4:36:25 PM

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www.tntextbooks.in, , INSTITUTE, Central Drug research, institute (CDRI), , RESEARCH AREAS, , WEBSITE, , Drug design, Pharmaceuticals and Biomedical Research, , www.cdri.res.in, , Salt and Marine Chemicals, Inorganic Materials and Catalysis, Electro Membrane Processes, Reverse Osmosis, , www.csmcri.org, , Lucknow,Uttar Pradesh., , Central Salt & Marine, Chemicals Research, Institute, Bhavnagar, Gujarat, , National Institute, of Pharmaceutical, Education and Research, (NIPER), Mohali, Punjab., , Institute of Nano, Science and, Technology(INST), Mohali, Punjab, , Laboratory of Advanced, Research in Polymeric, Materials (LARPM), Bhubaneswar, disha., , Tata institute, of fundamental, research (TIFR), Mumbai, Maharashtra, , Development of oral anti diabetic drugs, Nano crystalline solid, Xanthine oxidase inhibitors, Microbial and marine origin for, therapeutic purposes, Standardization and quality control of, herbal drugs and products, Development of chemical process, technologies for important natural, products involving isolation, , Bio-inspired soft nanostructures, Bio-sensors and online diagnostics, Bio-targeting and therapeutics, Microfluidics based devices, Materials and devices for energy storage, and harvesting, Nanotechnology in Agriculture and Rural, Development, Nano toxicology, Biopolymer, Fuel Cells, Polymer Nano composite, Carbon Nanotubes, Polymer Blends & Alloys, E Waste Recycling, , Molecular biophysics and imaging, Chemical biology and synthetic, chemistry, Bioinorganic and biomimetic chemistry, Nano science and catalysis, Chemical physics and dynamics, , www.niper.gov.in, , www.inst.ac.in, , www.larpm.gov.in, , www.tifr.res.in, , V, , Introduction Pages.indd 5, , 2/19/2020 4:36:25 PM

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www.tntextbooks.in, , CONTENTS, CHEMISTRY, S.No., , Topic, , Page.No., , Month, , I, , Metallurgy, , 01, , June, , 2, , p-Block Elements-I, , 26, , June, , 3, , p-Block Elements - II, , 56, , July, , 4, , Transition and Inner Transition Elements, , 100, , August, , 5, , Coordination Chemistry, , 130, , October, , 6, , Solid State, , 176, , June, , 7, , Chemical Kinetics, , 204, , July, , Answers, , 234, , Practicals, , 256, , Glossary, , 280, , E-book, , Assessment, , DIGI links, , Let’s use the QR code in the text books!, •, , Download DIKSHA app from the Google Play Store., , •, , Tap the QR code icon to scan QR codes in the textbook., , •, , Point the device and focus on the QR code., , •, , On successful scan, content linked to the QR code gets listed., , Note: For ICT corner, Digi Links QR codes use any other QR scanner., , IX, , Introduction Pages.indd 9, , 2/19/2020 4:36:26 PM

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www.tntextbooks.in, , UNIT, , 1, , METALLURGY, , Learning Objectives, After studying this unit, students will be, able to, , Harold Johann Thomas Ellingham, (1897–1975), , Ellingham, was, a, British, physical chemist, best known, for his Ellingham diagrams., Ellingham diagram summarizes a, large amount of information about, extractive metallurgy, and are, useful in predicting the favourable, thermodynamic conditions under, which an ore will be reduced to, its metal. Ellingham was able to, compare the temperature stability, of many different oxides. The, phenomenon of reduction of, metal oxides into free metal by, carbon or carbon monoxide was, known before Ellingham's time,, but Ellingham demonstrated it in, a scientific manner., , , , , , , , , , , , , , , , , , describe, various, concentrating ores, , methods, , of, , explain various methods of extraction, of crude metals, apply thermodynamic principles to, metallurgical processes, predict the favourable conditions for, the reduction process using Ellingham, diagram, describe the electrochemical principles, of metallurgy, apply the electrochemical principles in, the extraction of metals, explain the electrode reactions in, electrolytic refining., list the uses of Al, Zn, Fe, Cu and Au, , 1, , XII U1 Metallurgy - Jerald.indd 1, , 2/19/2020 4:37:56 PM

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www.tntextbooks.in, , INTRODUCTION, Metallurgy relate to the science and technology of metals. In nature, only a few metals, occur in their native state, all other metals occur in a combined state as their oxides,, sulphides, silicates etc... The extraction of pure metals from their natural sources, is linked, to the history of human civilisation. Ancient people used the available materials in their, environment which includes fire and metals, and they were limited to the metals available, on the earth's surface. In the modern world, we use a wide range of metals in our daily life,, which is the result of the development of metallurgical knowledge over thousands of years., Our need for the materials with specific properties have led to production of many metal, alloys. It is essential to design an eco-friendly metallurgical process that would minimize, waste, maximize energy efficiency. Such advances in metallurgy is vital for the economic, and technical progress in the current era. In this unit we will study the various steps, involved in the extraction of metals and the chemical principles behind these processes., , 1.1 Occurrence of metals, In general, pure metals are shiny and malleable, however, most of them are found in, nature as compounds with different properties. Metals having least chemical reactivity, such as copper, silver, gold and platinum occur in significant amounts as native elements., Reactive metals such as alkali metals usually occurs in their combined state and are, extracted using suitable metallurgical process., 1.1.1 Mineral and ore, A naturally occurring substance obtained by mining which contains the metal in free, state or in the form of compounds like oxides, sulphides etc... is called a mineral. In, most of the minerals, the metal of interest is present only in small amounts and some of, them contains a reasonable percentage of metal. For example iron is present in around, 800 minerals. However, some of them such as hematite magnetite etc., containing high, percentage of iron are commonly used for the extraction of iron. Such minerals that, contains a high percentage of metal, from which it can be extracted conveniently and, economically are called ores. Hence all ores are minerals but all minerals are not ores., Let us consider another example, bauxite and china clay (Al2O3.2SiO2.2H2O). Both are, minerals of aluminium. However, aluminium can be commercially extracted from bauxite, while extraction from china clay is not a profitable one. Hence the mineral, bauxite is an, ore of aluminium while china clay is not., The extraction of a metal of interest from, its ore consists of the following metallurgical, processes., (i) concentration of the ore, (ii) extraction of crude metal, (iii) refining of crude metal, , Malachite – copper mineral, 2, , XII U1 Metallurgy - Jerald.indd 2, , 2/19/2020 4:37:56 PM

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www.tntextbooks.in, , Table 1.1 List of some metals and their common ores with their chemical formula, Metal, , Aluminum, , Iron, , Copper, , Ore, , Composition, , Bauxite, , Al2O3.nH2O, , Diaspore, , Al2O3.H2O, , Kaolinite, , Metal, , Ore, , Composition, , Zinc blende or, Sphalerite, , ZnS, , Calamine, , ZnCO3, , Al2Si2O5(OH)4, , Zincite, , ZnO, , Haematite, , Fe2O3, , Galena, , PbS, , Magnetite, , Fe3O4, , Anglesite, , PbSO4, , Siderite, , FeCO3, , Cerrusite, , PbCO3, , Iron pyrite, , FeS2, , Limonite, , Fe2O3.3H2O, , Copper pyrite, , CuFeS2, , Copper glance, , Cu2S, , Cuprite, , Cu2O, , Stefinite, , Ag5SbS4, , Malachite, , CuCO3.Cu(OH)2, , Proustite, , Ag3AsS3, , Azurite, , 2CuCO3.Cu(OH)2, , Zinc, , Lead, , Tin, , Silver, , Cassiterite, (Tin stone), Silver glance, (Argentite), Pyrargyrite, (Ruby silver), Chlorargyrite, (Horn Silver), , SnO2, Ag2S, Ag3SbS3, AgCl, , 1.2 Concentration of ores, Generally, the ores are associated with nonmetallic impurities, rocky materials and siliceous, matter which are collectively known as gangue. The preliminary step in metallurgical process is, removal of these impurities. This removal process is known as concentration of ore. It increases the, concentration of the metal of interest or its compound in the ore. Several methods are available for, this process and the choice of method will depend on the nature of the ore, type of impurity and, environmental factors. Some of the common methods of ore concentration are discussed below., 1.2.1 Gravity separation or Hydraulic wash, In this method, the ore having high specific gravity is separated from the gangue that has, low specific gravity by simply washing with running water. Ore is crushed to a finely powdered, form and treated with rapidly flowing current of water. During this process the lighter gangue, particles are washed away by the running water. This method is generally applied to concentrate, the native ore such as gold and oxide ores such as haematite (Fe2O3), tin stone (SnO2) etc., 3, , XII U1 Metallurgy - Jerald.indd 3, , 2/19/2020 4:37:56 PM

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www.tntextbooks.in, , 1.2.2 Froth flotation, This method is commonly used to concentrate sulphide ores such as galena (PbS), zinc, blende (ZnS) etc... In this method, the metallic ore particles which are preferentially wetted by, oil can be separated from gangue., In this method, the crushed, ore is suspended in water and, mixed with frothing agent such, motor, as pine oil, eucalyptus oil etc., controller, A small quantity of sodium, ethyl xanthate which acts as, air supply, froth concena collector is also added. A, layer, trate, froth is generated by blowing, air through this mixture. The, feed, collector molecules attach to, valve, the ore particle and make them, sensor, water repellent. As a result, ore, particles, wetted by the oil, rise, stirrer, tailings, to the surface along with the, froth. The froth is skimmed, off and dried to recover the, Figure 1.1 Froth Flotation, concentrated ore. The gangue, particles that are preferentially, wetted by water settle at the bottom., When a sulphide ore of a metal of interest contains other metal sulphides as impurities,, depressing agents such as sodium cyanide, sodium carbonate etc are used to selectively, prevent other metal sulphides from coming to the froth. For example, when impurities, such as ZnS is present in galena (PbS), sodium cyanide (NaCN) is added to depresses the, flotation property of ZnS by forming a layer of zinc complex Na2[Zn(CN)4] on the surface, of zinc sulphide., 1.2.3 Leaching, This method is based on the solubility of the ore in a suitable solvent and the reactions, in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable, solvent, the metal present in the ore is converted to its soluble salt or complex while the, gangue remains insoluble. The following examples illustrate the leaching processes., Cyanide leaching, Let us consider the concentration of gold ore as an example. The crushed ore of, gold is leached with aerated dilute solution of sodium cyanide. Gold is converted into a, soluble cyanide complex. The gangue, aluminosilicate remains insoluble., 4Au (s) + 8CN (aq) + O2 (g) + 2H2O (l), , 4[Au(CN)2] (aq) + 4OH (aq), , 4, , XII U1 Metallurgy - Jerald.indd 4, , 2/19/2020 4:37:58 PM

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www.tntextbooks.in, , Recovery of metal of interest from the complex by reduction:, Gold can be recovered by reacting the deoxygenated leached solution with zinc. In this, process the gold is reduced to its elemental state (zero oxidation sate) and the process is called, cementation., [Zn(CN) ] 2(aq) + 2Au (s), Zn (s) + 2[Au(CN) ] (aq), 4, , 2, , Ammonia leaching, When a crushed ore containing nickel, copper and cobalt is treated with, aqueous ammonia under suitable pressure, ammonia selectively leaches these metals by, forming their soluble complexes viz. [Ni(NH3)6]2+, [Cu(NH3)4]2+, and [Co(NH 3)5H2O]3+, respectively from the ore leaving behind the gangue, iron(III) oxides/hydroxides and, aluminosilicate., Alkali leaching, In this method, the ore is treated with aqueous alkali to form a soluble complex., For example, bauxite, an important ore of aluminum is heated with a solution of sodium, hydroxde or sodium carbonate in the temperature range 470 - 520 K at 35 atm to form, soluble sodium meta-aluminate leaving behind the impurities, iron oxide and titanium, oxide., Al2O3 (s) + 2NaOH (aq) + 3H2O (l), , 2Na[Al(OH)4] (aq), , The hot solution is decanted, cooled, and diluted. This solution is neutralised by passing, CO2 gas, to the form hydrated Al2O3 precipitate., 2Na[Al(OH)4] (aq) + 2CO2 (g), , Al2O3.3H2O (s) + 2NaHCO3 (aq), , The precipitate is filtered off and heated around 1670 K to get pure alumina Al2O3., Acid leaching, Leaching of sulphide ores such as ZnS, PbS etc., can be done by treating them with hot, aqueous sulphuric acid., 2ZnS (s) + 2H2SO4 (aq) + O2(g), , 2ZnSO4 (aq) + 2S (s) + 2 H2O, , In this process the insoluble sulphide is converted into soluble sulphate and elemental sulphur., Evaluate yourself 1, 1. Write the equation for the extraction of silver by leaching with sodium cyanide and, show that the leaching process is a redox reaction., , 5, , XII U1 Metallurgy - Jerald.indd 5, , 2/19/2020 4:37:58 PM

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www.tntextbooks.in, , 1.2.4 Magnetic separation, This method, is, applicable, to, ferromagnetic, ores, and it is based on, Powdered, wdered, ore, the difference in the, magnetic properties, of the ore and the, Magnetic, impurities., For, Wheel, example tin stone can, be separated from the, Moving belt, wolframite impurities, Magnetic, which is magnetic., ore, Non-Magnetic, Similarly, ores such as, ore, chromite, pyrolusite, Figure 1.2 Magnetic separation, having, magnetic, property, can, be, removed from the, non magnetic siliceous impurities. The crushed ore is poured on to an electromagnetic, separator consisting of a belt moving over two rollers of which one is magnetic. The, magnetic part of the ore is attracted towards the magnet and falls as a heap close to the, magnetic region while the nonmagnetic part falls away from it as shown in the figure 1.2., , 1.3 Extraction of crude metal, The extraction of crude metals from the concentrated ores is carried out in two steps, namely, (i) conversion of the ore into oxides of the metal of interest and (ii) reduction of the, metal oxides to elemental metals. In the concentrated ore, the metal exists in positive oxidation, state and hence it is to be reduced to its elemental state. We can infer from the principles of, thermodynamics, that the reduction of oxide is easier when compared to reduction of other, compounds of metal and hence, before reduction, the ore is first converted into the oxide of, metal of interest., Let us discuss some of the common methods used to convert the concentrated ore into the, oxides of the metal of interest., 1.3.1 Conversion of ores into oxides, Roasting, Roasting is the method, usually applied for the conversion of sulphide ores into their, oxides. In this method, the concentrated ore is oxidised by heating it with excess of oxygen in, a suitable furnace below the melting point of the metal., 6, , XII U1 Metallurgy - Jerald.indd 6, , 2/19/2020 4:37:59 PM

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www.tntextbooks.in, , Fe2O3.3H2O, , ∆, , Fe2O3 (s) + 3H2O (g), , Al2O3.2H2O, , ∆, , Al2O3 (s)+ 2H2O (g), , Evaluate yourself 2, 2. Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to, make refractory bricks. Write the decomposition reaction., 1.3.2 Reduction of metal oxides, Metal oxide can be reduced to crude metal by using a suitable reducing agent like carbon,, carbon monoxide, hydrogen, aluminium and other reactive metals such as sodium etc...The, choice of reducing agent depends on the nature of the metal. For example, carbon cannot, be used as a reducing agent for the reactive metals such as sodium, potassium, aluminium, etc...Similarly CO cannot be used to reduce oxides such as ZnO, Al2O3. Later in this,we study, selection of suitable reducing agents by applying Ellingham diagram., Smelting, In this method, a flux (a chemical substance that forms an easily fusible slag with gangue), and a reducing agent such as carbon, carbon monoxide (or) aluminium is added to the, concentrated ore and the mixture is melted by heating at an elevated temperature (above the, melting point of the metal) in a smelting furnace. For example the oxide of iron can be reduced, by carbon monoxide as follows., Fe2O3 (s) + 3CO (g), , 2Fe (s) + 3CO2 (g), , In this extraction, a basic flux, limestone (CaO) is used. Since the silica gangue present in, the ore is acidic in nature, the limestone combines with it to form calcium silicate (slag)., CaO (s) + SiO2 (s), Flux, , CaSiO3 (s), , Gangue, , Slag, , In the extraction of copper from copper pyrites, the concentrated ore is heated in a, reverberatory furnace after mixing with silica, an acidic flux. The ferrous oxide formed due, to melting is basic in nature and it combines with silica to form ferrous silicate (slag). The, remaining metal sulphides Cu2S and FeS are mutually soluble and form a copper matte., 2CuFeS2 (s)+ O2 (g), , 2FeS (l)+ Cu2S (l)+ SO2 (g), , 2FeS (l) + 3O2 (g), , 2FeO (l) + 2SO2 (g), , FeO (s) + SiO2 (s), Gangue, , FeSiO3 (s), , Flux, , Slag, , 8, , XII U1 Metallurgy - Jerald.indd 8, , 2/19/2020 4:38:00 PM

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www.tntextbooks.in, , The matte is separated from the slag and fed to the converting furnace. During conversion,, the FeS present in the matte is first oxidised to FeO. This is removed by slag formation with, silica. The remaining copper sulphide is further oxidised to its oxide which is subsequently, converted to metallic copper as shown below., 2Cu2S (l,s) + 3O2 (g), , 2Cu2O (l,s) + 2SO2 (g), , 2Cu2O (l) + Cu2S (l), , 6Cu (l) + SO2 (g), , The metallic copper is solidified and it has blistered appearance due to evolution of SO2, gas formed in this process. This copper is called blistered copper., Reduction by carbon:, In this method the oxide ore of the metal is mixed with coal (coke) and heated strongly in, a furnace (usually in a blast furnace). This process can be applied to the metals which do not, form carbides with carbon at the reduction temperature., Examples:, ZnO (s)+ C (s), , Zn (s) + CO (g), , Mn3O4 (s) + 4C (s), , 3Mn (s) + 4CO (g), , Cr2O3 (s) + 3C (s), , 2Cr (s) + 3CO (g), , Reduction by hydrogen:, This method can be applied to the oxides of the metals (Fe, Pb, Cu) having less electropositive character than hydrogen., Ag2O (s)+ H2 (g), , 2Ag (s) + H2O (l), , Fe3O4 (s) + 4H2 (g), , 3Fe (s) + 4H2O (l), , Nickel oxide can be reduced to nickel by using a mixture of hydrogen and carbon monoxide, (water gas), 2NiO (s) + CO (g) + H2 (g), , 2Ni (s) + CO2 (g) + H2O (l), , Reduction by metal:, Metallic oxides such as Cr2O3 can be reduced by an aluminothermic process. In this, process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible. To, initiate the reduction process, an ignition mixture (usually magnesium and barium peroxide), is used., 9, , XII U1 Metallurgy - Jerald.indd 9, , 2/19/2020 4:38:00 PM

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www.tntextbooks.in, , BaO2 + Mg, , BaO + MgO, , During the above reaction a large amount of heat is evolved (temperature up to 2400°C, is, generated and the reaction enthalpy is : 852 kJ mol-1) which facilitates the reduction of Cr2O3, by aluminium power., Cr2O3 + 2Al, , Δ, , 2Cr + Al2O3, , Active metals such as sodium, potassium and calcium can also be used to reduce the, metal oxide, B2O3 + 6Na, , 2B + 3Na2O, , Rb2O3 + 3Mg, , 2Rb + 3MgO, , TiO2 + 2Mg, , Ti + 2MgO, , ThO2 + 2Ca, , 1250 K, , Th + 2CaO, , Auto-reduction:, Simple roasting of some of the ores give the crude metal. In such cases, the use of, reducing agents is not necessary. For example, mercury is obtained by roasting of its ore, cinnabar (HgS), HgS (s) + O2 (g), , 1.4, , Hg (l) + SO2 , , Thermodynamic principle of metallurgy, , As we discussed, the extraction of metals from their oxides can be carried out by using, different reducing agents. For example, consider the reduction of a metal oxide MxO y., 2, y, , MxOy (s), , 2x, y, , M (s) + O2 (g) ------ (1), , The above reduction may be carried out with carbon. In this case, the reducing agent, carbon may be oxidised to either CO or CO2., C + O2, , CO2 (g) ------ (2), , 2C + O2, , 2CO (g) ------ (3), , If carbon monoxide is used as a reducing agent, it is oxidised to CO2 as follows,, 2CO + O2, , 2CO2 (g) ------ (4), , A suitable reducing agent is selected based on the thermodynamic considerations. We, know that for a spontaneous reaction, the change in free energy (ΔG) should be negative., Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given, 10, , XII U1 Metallurgy - Jerald.indd 10, , 2/19/2020 4:38:01 PM

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www.tntextbooks.in, , The change in Gibbs free energy (ΔG) for a reaction is given by the expression., ΔG = ΔH - TΔS ------ (1), where, ΔH is the enthalpy change , T the temperature in kelvin and ΔS the entropy, change. For an equilibrium process, ΔG⁰ can be calculated using the equilibrium constant by, the following expression, ΔG⁰ =-RT lnKp, Harold Ellingham used the above relationship to calculate the ΔG⁰ values at various, temperatures for the reduction of metal oxides by treating the reduction as an equilibrium, process., He has drawn a plot by considering the temperature in the x-axis and the standard free, energy change for the formation of metal oxide in y-axis. The resultant plot is a straight line with, ΔS as slope and ΔH as y-intercept. The graphical representation of variation of the standard, Gibbs free energy of reaction for the formation of various metal oxides with temperature is, called Ellingham diagram, Observations from the Ellingham diagram., 1. For most of the metal oxide formation, the slope is positive. It can be explained as follows., Oxygen gas is consumed during the formation of metal oxides which results in the decrease, in randomness. Hence, ΔS becomes negative and it makes the term, TΔS positive in the, straight line equation., 2. The graph for the formation of carbon monoxide is a straight line with negative slope. In, this case ΔS is positive as 2 moles of CO gas is formed by the consumption of one mole of, oxygen gas. It indicates that CO is more stable at higher temperature., 3. As the temperature increases, generally ΔG value for the formation of the metal oxide, become less negative and becomes zero at a particular temperature. Below this temperature,, ΔG is negative and the oxide is stable and above this temperature ΔG is positive. This, general trend suggests that metal oxides become less stable at higher temperature and their, decomposition becomes easier., 4. There is a sudden change in the slope at a particular temperature for some metal oxides like, MgO, HgO. This is due to the phase transition (melting or evaporation)., 1.4.2 Applications of the Ellingham diagram:, Ellingham diagram helps us to select a suitable reducing agent and appropriate temperature, range for reduction. The reduction of a metal oxide to its metal can be considered as a, competition between the element used for reduction and the metal to combine with oxygen., If the metal oxide is more stable, then oxygen remains with the metal and if the oxide of, element used for reduction is more stable, then the oxygen from the metal oxide combines, with elements used for the reduction. From the Ellingham diagram, we can infer the relative, stability of different metal oxides at a given temperature., 12, , XII U1 Metallurgy - Jerald.indd 12, , 2/19/2020 4:38:02 PM

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www.tntextbooks.in, , 1. Ellingham diagram for the formation of Ag2O and HgO is at upper part of the diagram and, their decomposition temperatures are 600 and 700 K respectively. It indicates that these, oxides are unstable at moderate temperatures and will decompose on heating even in the, absence of a reducing agent., 2. Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides, of one metal by another metal. Any metal can reduce the oxides of other metals that are, located above it in the diagram. For example, in the Ellignham diagram, for the formation, of chromium oxide lies above that of the aluminium, meaning that Al2O3 is more stable, than Cr2O3. Hence aluminium can be used as a reducing agent for the reduction of chromic, oxide. However, it cannot be used to reduce the oxides of magnesium and calcium which, occupy lower position than aluminium oxide., 3. The carbon line cuts across the lines of many metal oxides and hence it can reduce all, those metal oxides at sufficiently high temperature. Let us analyse the thermodynamically, favourable conditions for the reduction of iron oxide by carbon. Ellingham diagram for the, formation of FeO and CO intersects around 1000 K. Below this temperature the carbon, line lies above the iron line which indicates that FeO is more stable than CO and hence, at this temperature range, the reduction is not thermodynamically feasible. However,, above 1000 K carbon line lies below the iron line and hence, we can use coke as reducing, agent above this temperature. The following free energy calculation also confirm that the, reduction is thermodynamically favoured., From the Ellingham Diagram at 1500 K,, 2Fe (s) + O2 (g), , 2FeO (g), , ΔG1 = -350 kJ mol–1 ------ (1), , 2C (s) + O2 (g), , 2CO (g), , ΔG2 = -480 kJ mol–1------ (2), , Reverse the reaction (1), 2FeO (s), , 2Fe (s)+ O2 (g) – ΔG1 = +350 kJ mol–1 ------ (3), , Now couple the reactions (2) and (3), 2FeO (s) + 2C, , 2Fe (l,s)+ 2CO (g) ΔG3 = -130 kJ mol–1 ------ (4), , The standard free energy change for the reduction of one mole of FeO is, ΔG3/2 = -65 kJ mol-1, Limitations of Ellingham diagram, 1. Ellingham diagram is constructed based only on thermodynamic considerations. It gives, information about the thermodynamic feasibility of a reaction. It does not tell anything, about the rate of the reaction. More over, it does not give any idea about the possibility of, other reactions that might be taking place., 2. The interpretation of ΔG is based on the assumption that the reactants are in equilibrium, with the products which is not always true., , 13, , XII U1 Metallurgy - Jerald.indd 13, , 2/19/2020 4:38:02 PM

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www.tntextbooks.in, , Evaluate yourself 3, 3. Using Ellingham diagram (fig 1.4) indicate the lowest temperature at which ZnO, can be reduced to Zinc metal by carbon. Write the overall reduction reaction at this, temperature., , 1.5 Electrochemical principle of metallurgy, Similar to thermodynamic principles, electrochemical principles also find applications, in metallurgical process. The reduction of oxides of active metals such as sodium, potassium, etc., by carbon is thermodynamically not feasible. Such metals are extracted from their ores, by using electrochemical methods. In this technique, the metal salts are taken in a fused form, or in solution form. The metal ion present can be reduced by treating it with some suitable, reducing agent or by electrolysis., Gibbs free energy change for the electrolysis process is given by the following expression, ΔG° = -nFE°, Where n is number of electrons involved in the reduction process, F is the Faraday and E0, is the electrode potential of the redox couple., If E0 is positive then the ΔG is negative and the reduction is spontaneous and hence a, redox reaction is planned in such a way that the e.m.f of the net redox reaction is positive., When a more reactive metal is added to the solution containing the relatively less reactive, metal ions, the more reactive metal will go into the solution. For example,, Cu (s) + 2Ag+ (aq), , Cu 2+ (aq) + 2Ag (s), , Cu2+ (aq) + Zn (s), , Cu (s) + Zn 2+ (aq), , 1.5.1 Electrochemial extraction of aluminium - Hall-Heroult process:, In this method, electrolysis is carried out in an iron tank lined with carbon which acts, as a cathode. The carbon blocks immersed in the electrolyte act as a anode. A 20% solution, of alumina, obtained from the bauxite ore is mixed with molten cryolite and is taken in the, electrolysis chamber. About 10% calcium chloride is also added to the solution. Here calcium, chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a, temperature of above 1270 K. The chemical reactions involved in this process are as follows., Ionisaiton of alumina, , Al2O3, , Reaction at cathode, , 2Al3+ (melt) + 6e, , Reaction at anode, , 6O2- (melt), , 2Al3+ + 3O22Al (l), 3O2 + 12e, , Since carbon acts as anode the following reaction also takes place on it., 14, , XII U1 Metallurgy - Jerald.indd 14, , 2/19/2020 4:38:02 PM

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www.tntextbooks.in, , -, , C (s) + O2- (melt), , CO + 2e, , C (s) + 2O2- (melt), , CO2 + 4e, , -, , Due to the above two reactions, anodes are slowly consumed during the electrolysis. The, pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction, can be written as follows., 4Al3+ (melt) + 6O2- (melt) + 3C (s), , 4Al (l) + 3CO2 (g), , Evaluate yourself 4, 4. Metallic sodium is extracted by the electrolysis of brine (aq. NaCl). After electrolysis, the electrolytic solution becomes basic in nature. Write the possible electrode, reactions., , 1.6 Refining process, Generally the metal extracted from its ore contains some impurities such as unreacted, oxide ore, other metals, nonmetals etc...Removal of such impurities associated with the isolated, crude metal is called refining process. In this section, let us discuss some of the common, refining methods., 1.6.1 Distillation, This method is employed for low boiling volatile metals like zinc (boiling point 1180 K), and mercury (630 K). In this method, the impure metal is heated to evaporate and the vapours, are condensed to get pure metal., 1.6.2 Liquation, This method, is employed to remove the impurities with high melting points from metals, having relatively low melting points such as tin (Sn; mp= 904 K), lead (Pb; mp=600 K), mercury, (Hg; mp=234 K), and bismuth (Bi; mp=545 K). In this process, the crude metal is heated to, form fusible liquid and allowed to flow on a sloping surface. The impure metal is placed on, sloping hearth of a reverberatory furnace and it is heated just above the melting point of the, metal in the absence of air, the molten pure metal flows down and the impurities are left, behind. The molten metal is collected and solidified., 1.6.3 Electrolytic refining:, The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing, aqueous solution of the salts of the metal of interest. The rods of impure metal are used as, anode and thin strips of pure metal are used as cathode. The metal of interest dissolves from, the anode, pass into the solution while the same amount of metal ions from the solution will be, deposited at the cathode. During electrolysis, the less electropositive impurities in the anode,, settle down at the bottom and are removed as anode mud., 15, , XII U1 Metallurgy - Jerald.indd 15, , 2/19/2020 4:38:03 PM

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www.tntextbooks.in, , Let us understand this process by considering electrolytic refining of silver as an example., Cathode, , : Pure silver, , Anode, , : Impure silver rods, , Electrolyte, , : Acidified aqueous solution of silver nitrate., , When a current is passed through the electrodes the following reactions will take place, Reaction at anode, Ag (s), Ag+ (aq) + 1e, Reaction at cathode, , -, , Ag+ (aq) + 1e, , Ag (s), , During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The, positively charged silver cations migrate towards the cathode and get discharged by gaining, electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be, refined by this process in a similar manner., 1.6.4 Zone Refining, This method is based on the principles of fractional crystallisation. When an impure, metal is melted and allowed to solidify, the impurities will prefer to be in the molten region. i.e., impurities are more soluble in the melt than in the solid state metal. In this process the impure, metal is taken in the form of a rod. One end of the rod is heated using a mobile induction, heater which results in melting of the metal on that portion of the rod. When the heater is, slowly moved to the other end the pure metal crystallises while the impurities will move on, to the adjacent molten zone formed due to the movement of the heater. As the heater moves, further away, the molten zone containing impurities also moves along with it. The process is, repeated several times by moving the heater in the same direction again and again to achieve, the desired purity level. This process is carried out in an inert gas atmosphere to prevent the, oxidation of metals . Elements such as germanium (Ge), silicon (Si) and galium (Ga) that are, used as semiconductor are refined using this process., 1.6.5 Vapour phase method, In this method, the metal is treated with a suitable reagent which can form a volatile, compound with the metal. Then the volatile compound is decomposed to give the pure metal., We can understand this method by considering the following process., Mond process for refining nickel:, The impure nickel is heated in a stream of carbon monoxide at around 350 K. The nickel, reacts with the CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left, behind., [Ni(CO)4] (g), Ni (s) + 4 CO (g), On heating the nickel tetracarbonyl around 460 K, the complex decomposes to give pure, metal., Ni (s) + 4 CO (g), [Ni(CO)4] (g), 16, , XII U1 Metallurgy - Jerald.indd 16, , 2/19/2020 4:38:03 PM

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www.tntextbooks.in, , Van-Arkel method for refining zirconium/titanium:, This method is based on the thermal decomposition of metal compounds which lead to, the formation of pure metals. Titanium and zirconium can be purified using this method., For example, the impure titanium metal is heated in an evacuated vessel with iodine at a, temperature of 550 K to form the volatile titanium tetra-iodide.(TiI4). The impurities are left, behind, as they do not react with iodine., 550K, Ti (s) + 2I2 (s), TiI4 (vapour), The volatile titanium tetraiodide vapour is passed over a tungsten filament at a temperature, aroud 1800 K. The titanium tetraiodide is decomposed and pure titanium is deposited on the, filament. The iodine is reused., TiI4 (vapour), , 1800 K, , Ti (s) + 2I2 (s), , 1.7 Applications of metals, 1.7.1 Applications of Al, Aluminium is the most abundant metal and is a good conductor of electricity and heat. It, also resists corrosion. The following are some of its applications., ÂÂ Many heat exchangers/sinks and our day to day cooking vessels are made of aluminium., ÂÂ It is used as wraps (aluminium foils) and is used in packing materials for food items,, ÂÂ Aluminium is not very strong, However , its alloys with copper, manganese, magnesium, and silicon are light weight and strong and they are used in design of aeroplanes and other, forms of transport., ÂÂ As Aluminium shows high resistance to corrosion, it is used in the design of chemical, reactors, medical equipments,refrigeration units and gas pipelines., ÂÂ Aluminium is a good electrical conductor and cheap, hence used in electrical overhead, electric cables with steel core for strength., 1.7.2 Applications of Zn, ÂÂ Metallic zinc is used in galvanising metals such as iron and steel structures to protect them, from rusting and corrosion., ÂÂ Zinc is also used to produce die-castings in the automobile, electrical and hardware, industries, ÂÂ Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics,, 17, , XII U1 Metallurgy - Jerald.indd 17, , 2/19/2020 4:38:03 PM

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www.tntextbooks.in, , pharmaceuticals, plastics, inks, batteries, textiles and electrical equipment. Zinc sulphide, is used in making luminous paints, fluorescent lights and x-ray screens., ÂÂ Brass an alloy of zinc is used in water valves and communication equipment as it is highly, resistant to corrosion., 1.7.3 Applications of Fe, ÂÂ Iron is one of the most useful metals and its alloys are used everywhere including bridges,, electricity pylons, bicycle chains, cutting tools and rifle barrels., ÂÂ Cast iron is used to make pipes, valves and pumps stoves etc..., ÂÂ Magnets can be made from iron and its alloys and compounds., ÂÂ An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in, architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for, making cables, automobiles and aeroplane parts. Chrome steels are used for maufacturing, cutting tools and crushing machines, 1.7.4 Applications of Cu, Copper is the first metal used by the human and extended use of its alloy bronze resulted in a, new era,'Bronze age', Copper is used for making coins and ornaments along with gold and other metals., Copper and its alloys are used for making wires, water pipes and other electrical parts, 1.7.5 Applications of Au, ÂÂ Gold, one of the expensive and precious metals. It is used for coinage, and has been used as, standard for monetary systems in some countries., ÂÂ It is used extensively in jewellery in its alloy form with copper. It is also used in electroplating, to cover other metals with a thin layer of gold which are used in watches, artificial limb, joints, cheap jewellery, dental fillings and electrical connectors., ÂÂ Gold nanoparticles are also used for increasing the efficiency of solar cells and also used, an catalysts., , 18, , XII U1 Metallurgy - Jerald.indd 18, , 2/19/2020 4:38:03 PM

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www.tntextbooks.in, , The Iron Pillar – Delhi:, The Iron pillar, also known as Ashoka Pillar,, is 23 feet 8 inches high, 16 inches wide and, weighs over 6000 kg., The surprise comes in knowing its age, some 1600 years, old,an iron column should have turned into a pile of dust long ago., Despite that, it has avoided corrosion for over the last 1600 years, and stands as an evidence of the exquisite skills and knowledge of, ancient Indians., A protective film was created through a complicated, combination of the presence of raw and unreduced iron in the pillar and cycles of the weather,, which helped to create a thin, uniform layer of misawite on the pillar. Misawite is a compound of, iron, oxygen and hydrogen which does not rust and gives corrosion resistance., , Summary, „„ Metallurgy relates to the science and technology of metals., „„ A naturally occurring substance obtained by mining which contains the metal, , in free state or in the form of compounds like oxides, sulphides etc... is called a, mineral., , „„ minerals that contains a high percentage of metal, from which it can be extracted, , conveniently and economically are called ores., , „„ The extraction of, , a metal of interest from its ore consists of the following, metallurgical processes., (i) concentration of the ore, (ii) extraction of crude metal, (iii) refining of crude metal, , „„ The extraction of crude metals from the concentrated ores is carried out in two steps, , namely, (i) conversion of the ore into oxides of the metal of interest and (ii) reduction, of the metal oxides to elemental metals., , „„ The graphical representation of variation of the standard Gibbs free energy of reaction, , for the formation of various metal oxides with temperature is called Ellingham diagram, , „„ Ellingham diagram helps us to select a suitable reducing agent and appropriate, , temperature range for reduction., , 19, , XII U1 Metallurgy - Jerald.indd 19, , 2/19/2020 4:38:04 PM

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www.tntextbooks.in, , „„ Similar to thermodynamic principles, electrochemical principles also find applications, , in metallurgical process., , „„ If E0 is positive then the ΔG is negative and the reduction is spontaneous and hence, , a redox reaction is planned in such a way that the e.m.f of the net redox reaction is, positive. When a more reactive metal is added to the solution containing the relatively, less reactive metal ions, the more reactive metal will go into the solution., , „„ Generally the metal extracted from its ore contains some impurities such as unreacted, , oxide ore, other metals, nonmetals etc...Removal of such impurities associated with the, isolated crude metal is called refining process., , EVALUATION, Choose the correct answer:, 1. Bauxite has the composition, a) Al 2O3, , b) Al 2O3 .nH2O c) Fe2O3 .2H2O, , d)None of these, , 2. Roasting of sulphide ore gives the gas (A).(A) is a colourless gas. Aqueous solution of (A), is acidic. The gas (A) is, a) CO2, , b) SO3, , c) SO2, , d) H2S, , 3. Which one of the following reaction represents calcinations?,  2ZnO, a) 2Zn + O2 →, ,  2ZnO + 2SO2, b) 2ZnS + 3O2 →, ,  MgO + CO2, c) MgCO3 →, , d)Both (a) and (c), , 4. The metal oxide which cannot be reduced to metal by carbon is, a) PbO, , b) Al 2O3, , c) ZnO, , d) FeO, , 5. Which of the metal is extracted by Hall-Heroult process?, a) Al, , b) Ni, , c) Cu, , d) Zn, , 6. Which of the following statements, about the advantage of roasting of sulphide ore before, reduction is not true?, a) Δ G f 0 of sulphide is greater than those for CS2 and H2S ., b) Δ Gr 0 is negative for roasting of sulphide ore to oxide, c) Roasting of the sulphide to its oxide is thermodynamically feasible., d) Carbon and hydrogen are suitable reducing agents for metal sulphides., 20, , XII U1 Metallurgy - Jerald.indd 20, , 2/19/2020 4:38:07 PM

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www.tntextbooks.in, , 7. Match items in column - I with the items of column – II and assign the correct code., Column-I, , Column-II, , A, , Cyanide process, , (i), , Ultrapure Ge, , B, , Froth floatation process, , (ii), , Dressing of ZnS, , C, , Electrolytic reduction, , (iii), , Extraction of Al, , D, , Zone refining, , (iv), , Extraction of Au, , (v), , Purification of Ni, , A, , B, , C, , D, , (a), , (i), , (ii), , (iii), , (iv), , (b), , (iii), , (iv), , (v), , (i), , (c), , (iv), , (ii), , (iii), , (i), , (d), , (ii), , (iii), , (i), , (v), , 8. Wolframite ore is separated from tinstone by the process of, a) Smelting, , b) Calcination, , c) Roasting, , d) Electromagnetic separation, , 9. Which one of the following is not feasible,  Cu(s) + Zn 2+ (aq), a) Zn(s) + Cu 2+ (aq) →, b) Cu(s) + Zn 2+ (aq) →,  Zn(s) + Cu 2+ (aq),  2Ag(s) + Cu 2+ (aq), c) Cu(s) + 2Ag + (aq) →, , d) Fe(s) + Cu 2+ (aq) →,  Cu(s) + Fe 2+ (aq), 10. Electrochemical process is used to extract, a) Iron, , b) Lead, , c) Sodium, , d) silver, , 11. Flux is a substance which is used to convert, a) Mineral into silicate , , b) Infusible impurities to soluble impurities, , c) Soluble impurities to infusible impurities d) All of these, 12. Which one of the following ores is best concentrated by froth – floatation method?, a) Magnetite, , b) Haematite, , c) Galena, , d) Cassiterite, , 21, , XII U1 Metallurgy - Jerald.indd 21, , 2/19/2020 4:38:07 PM

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www.tntextbooks.in, , 13. In the extraction of aluminium from alumina by electrolysis, cryolite is added to, a) Lower the melting point of alumina b) Remove impurities from alumina, c) Decrease the electrical conductivity d) Increase the rate of reduction, 14. Zinc is obtained from ZnO by, a) Carbon reduction, , b) Reduction using silver, , c) Electrochemical process, , d) Acid leaching, , 15. Extraction of gold and silver involves leaching with cyanide ion. silver is later recovered, by (NEET-2017), a) Distillation, , b) Zone refining, , c) Displacement with zinc, , d) liquation, , 16. Considering Ellingham diagram, which of the following metals can be used to reduce, alumina? (NEET-2018), a) Fe, , b) Cu, , c) Mg, , d) Zn, , 17. The following set of reactions are used in refining Zirconium, K, Zr (impure) + 2I2 523, , → ZrI4, , ZrI4 1800K, , → Zr (pure) + 2I2, a) Liquation, , This method is known as, , c) Zone refining, , b) van Arkel process, d) Mond’s process, , 18. Which of the following is used for concentrating ore in metallurgy?, a) Leaching, , b) Roasting, , c) Froth floatation, , d) Both (a) and (c), , 19. The incorrect statement among the following is, a) Nickel is refined by Mond’s process, b) Titanium is refined by Van Arkel’s process, c) Zinc blende is concentrated by froth floatation, d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution, 20. In the electrolytic refining of copper, which one of the following is used as anode?, a) Pure copper, , b) Impure copper, , c) Carbon rod, , d) Platinum electrode, 22, , XII U1 Metallurgy - Jerald.indd 22, , 2/19/2020 4:38:07 PM

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www.tntextbooks.in, , 21. Which of the following plot gives Ellingham diagram, a) Δ S Vs T, c) Δ G 0 Vs, , 1, T, , b) Δ G0 Vs T, 0, 2, d) Δ G Vs T, , 22. In the Ellingham diagram, for the formation of carbon monoxide,  ∆ S0 , is negative, a) ,  ∆ T , ,  ∆ G0 , b) , is positive,  ∆ T , ,  ∆ G0 , is negative, c) ,  ∆ T , ,  ∆T , d) initially , is positive, after 7000C ,,  ∆ G0 ,  ∆ G0 ,  ∆ T  is negative, , 23. Which of the following reduction is not thermodynamically feasible?,  Al 2O3 + 2Cr, a) Cr2O3 + 2Al →, ,  Cr2O3 + 2Al, b) Al 2O3 + 2Cr →, ,  2 Al 2O3 + 3Ti, c) 3TiO2 + 4Al →, , d) none of these, , 24. Which of the following is not true with respect to Ellingham diagram?, a) Free energy changes follow a straight line. Deviation occurs when there is a phase, change., b) The graph for the formation of CO2 is a straight line almost parallel to free energy axis., c) Negative slope of CO shows that it becomes more stable with increase in temperature., d) Positive slope of metal oxides shows that their stabilities decrease with increase in, temperature., Answer the following questions:, 1. What are the differences between minerals and ores?, 2. What are the various steps involved in extraction of pure metals from their ores?, 3. What is the role of Limestone in the extraction of Iron from its oxide Fe2O3 ?, 4. Which type of ores can be concentrated by froth floatation method? Give two examples for, such ores., 5. Describe a method for refining nickel., 6. Explain zone refining process with an example., , 23, , XII U1 Metallurgy - Jerald.indd 23, , 2/19/2020 4:38:10 PM

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www.tntextbooks.in, , 7. Using the Ellingham diagram,, (A) Predict the conditions under which, (i) Aluminium might be expected to reduce magnesia., (ii) Magnesium could reduce alumina., (B) it is possible to reduce Fe2O3 by coke at a temperature around 1200K, 8. Give the uses of zinc., 9. Explain the electrometallurgy of aluminium., 10. Explain the following terms with suitable examples., (i) Gangue, , (ii) slag, , 11. Give the basic requirement for vapour phase refining., 12. Describe the role of the following in the process mentioned., (i) Silica in the extraction of copper., (ii) Cryolite in the extraction of aluminium., (iii) Iodine in the refining of Zirconium., (iv) Sodium cyanide in froth floatation., 13. Explain the principle of electrolytic refining with an example., 14. The selection of reducing agent depends on the thermodynamic factor: Explain with an, example., 15. Give the limitations of Ellingham diagram., 16. Write a short note on electrochemical principles of metallurgy., , 24, , XII U1 Metallurgy - Jerald.indd 24, , 2/19/2020 4:38:10 PM

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www.tntextbooks.in, , Metallurgy, minerals, gravity separation, , ores, , froth floatation, , concentration of ores, , magnetic separation, cyanide, leaching, , acid, alkali, , extraction of crude metal, Roasting, conversion of ore into oxides, smelting, reduction by C, or H or metal, , calcination, reduction of metal oxides, , Auto reduction, Distillation, Liquation, , Refining process, , principles of metalurgy, , zone refining, thermodynamic, principles, DG = DH–TDS, , vapour phase, method, , Electrochemical, principles, DG = –nFE0, , Ellingham diagram, Pure metal, , Applications of Al ,Cu ,Zn , Fe and Au, , 25, , XII U1 Metallurgy - Jerald.indd 25, , 2/19/2020 4:38:10 PM

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www.tntextbooks.in, , UNIT, , 2, , p-BLOCK, ELEMENTS-I, , Kenneth wade, (1932–2014), , Learning Objectives, After studying this unit, the students will, be able to, , Kenneth, Wade,, was, a, British chemist, and professor, emeritus at Durham University., He developed a method for the, prediction of shapes of borane, clusters. Wade’s rules are used to, rationalize the shape of borane, clusters by calculating the total, number of skeletal electron pairs, (SEP) available for cluster bonding., For his substantial contribution,, Kenneth Wade was granted FRS, award from royal society, London, In 1989.He received the Tilden, prize award in 1999 from the Royal, Society of Chemistry for advances, in chemistry., , , , , , , , , , , , describe the general trends in the, properties of p-block elements, explain the anomalous properties of the, first element of p-block groups, discuss the preparation, properties and, uses of boron, discuss the preparation of important,, compounds of boron and aluminium, discuss the preparation and properties, of important compounds of carbon and, silicon, , 26, , XII_U2-P-Block.indd 26, , 2/19/2020 4:38:39 PM

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www.tntextbooks.in, , INTRODUCTION, We have already learnt the classification of elements into four blocks namely s, p, d and, f. We have also learnt the properties of s-block elements and their important compounds in, the XI standard. This year we learn the elements of remaining blocks, starting with p-block, elements., The elements in which their last electron enters the 'p' orbital, constitute the p-block, elements. They are placed in 13th to 18th groups of the modern periodic table and the first, member of the groups are B, C, N, O, F and He respectively. These elements have quite, varied properties and this block contains nonmetals, metals and metalloids. Nonmetallic, elements of this group have more varied properties than metals. The elements of this block, and their compounds play an important role in our day to day life, for example, without, molecular oxygen we cannot imagine the survival of living system. The most abundant, metal aluminium and its alloys have plenty of applications ranging from household, utensils to parts of aircraft. The semi conducting nature of elements such as silicon and, germanium made a revolutionary change in the field of modern electronics. In this unit, we discuss the properties of first three groups (Group 13 - 15) of p-block elements namely, boron, carbon and nitrogen family elements and their important compounds., , 2.1 General trends in properties of p-block elements:, We already learnt that the properties of elements largely depends on their electronic, configuration, size, ionisation enthalpy, electronegativity etc... Let us discuss the general, trend in such properties of various p-block elements., 2.1.1 Electronic configuration and oxidation state:, The p-block elements have a general electronic configuration of ns2, np1-6. The, elements of each group have similar outer shell electronic configuration and differ only, in the value of n (principal quantum number). The elements of group 18 (inert gases), have completely filled p orbitals, hence they are more stable and have least reactivity., The elements of this block show variable oxidation state and their highest oxidation state, (group oxidation state) is equal to the total number of valance electrons present in them., Unlike s-block elements which show only positive oxidation state, some of the p-block, elements show negative oxidation states also. The halogens have a strong tendency to, gain an electron to give a stable halide ion with completely filled electronic configuration, and hence -1 oxidation state is more common in halogens. Similarly, the other elements, belonging to pnictogen and chalcogen groups also show negative oxidation states., Evaluate yourself :, Why group 18 elements are called inert gases? Write the general electronic configuraton, of group 18 elements, 27, , XII_U2-P-Block.indd 27, , 2/19/2020 4:38:41 PM

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www.tntextbooks.in, , Table 2.1 General electronic configurations and oxidation states of p-block elements, Group No., , 13, , Group Name, , 14, , 15, , 16, , 17, , 18, , Icosagens Tetragens Pnictogens Chalcogens Halogens, , General outer, electronic, configuration, , Inert, gases, , ns2 np1, , ns2 np2, , ns2 np3, , ns2 np4, , ns2 np5, , ns2 np6, , Highest oxidation, state (Group, oxidation state), , +3, , +4, , +5, , +6, , +7, , +8, , Other observed, oxidation states, , +1, , +2, -4, , +3, -3, , +4, +2, -2, , +5, +3,, +1, -1, , +6. +4,, +2, , 2.1.2 Metallic nature:, The tendency of an element to form a cation by loosing electrons is known as electropositive, or metallic character. This character depends on the ionisation energy. Generally on descending, a group the ionisation energy decreases and hence the metallic character increases., 18, , Figure 2.1 p-block elements with their ionisation, enthalpies, electronegativity and metallic nature., Group No, , 13, , 14, , 15, , 16, , IE1-800.63, , IE1-800.63, , IE1-1402.33, , IE1-1313.94, , EN-2.04, , EN-2.55, , EN-3.04, , EN-3.44, , IE1-577.54, , IE1-786.52, , IE1-1011.81, , IE1-999.59, , IE1-2372.32, , 17, IE1-1681.04, , EN-3.98, IE1-1251.19, , EN-1.61, , EN-1.90, , EN-2.19, , EN-2.58, , IE1-578.84, , IE1-762.18, , IE1-944.47, , IE1-940.96, , IE1-1139.86, , EN-2.55, , EN-2.96, , EN-1.81, , EN-2.01, , EN-2.18, , IE1-558.3, , IE1-708.58, , IE1-830.58, , EN-1.78, , EN-1.96, , IE1-589.35, , IE1-715.57, , EN-1.8, , EN-1.8, IE1-, , EN-, , IE1-, , EN-, , EN-3.16, , ENIE1-2080.67, , ENIE1-1520.57, , EN-, , EN-, , IE1-1008.39, , IE1-1170.35, , EN-2.1, , EN-2.66, , EN-2.60, , IE1-702.94, , IE1-811.82, , IE1-, , IE1-1037.07, , EN-1.9, , EN-2.0, , EN-2.2, , EN-, , IE1-, , IE1-, , IE1-, , EN-, , EN-, , IE1-, , EN-, , EN-, , Metalloids, , IE1-1350.76, , IE1-869.29, , EN-2.1, , Metals, , Non Metal, , Radio, active, IE1- First ionisattion, energy, EN- Electro negativity, , 28, , XII_U2-P-Block.indd 28, , 2/19/2020 4:38:41 PM

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www.tntextbooks.in, , In p-block, the elements present in lower left part are metals while the elements in the, upper right part are non metals. Elements of group 13 have metallic character except the first, element boron which is a metalloid, having properties intermediate between the metal and, nonmetals. The atomic radius of boron is very small and it has relatively high nuclear charge, and these properties are responsible for its nonmetallic character. In the subsequent groups, the non-metallic character increases. In group 14 elements, carbon is a nonmetal while silicon, and germanium are metalloids. In group 15, nitrogen and phosphorus are non metals and, arsenic & antimony are metalloids. In group 16, oxygen, sulphur and selenium are non metals, and tellurium is a metalloid. All the elements of group 17 and 18 are non metals., 2.1.3 Ionisation Enthalpy:, We have already learnt that as we move down a group, generally there is a steady, decrease in ionisation enthalpy of elements due to increase in their atomic radius. In, p-block elements, there are some minor deviations to this general trend. In group 13, from, boron to aluminium the ionisation enthalpy decreases as expected. But from aluminium, to thallium there is only a marginal difference. This is due to the presence of inner d and, f-electrons which has poor shielding effect compared to s and p-electrons. As a result, the, effective nuclear charge on the valance electrons increases. A similar trend is also observed, in group 14. The remaining groups (15 to 18) follow the general trend. In these groups,, the ionisation enthalpy decreases, as we move down the group. Here, poor shielding effect, of d- and f-electrons are overcome by the increased shielding effect of the additional, p-electrons. The ionisation enthalpy of elements in successive groups is higher than the, corresponding elements of the previous group as expected., 2.1.4 Electronegativity, As we move down the 13th group, the electronegativity first decreases from boron to, aluminium and then marginally increases for Gallium, thereafter there is no appreciable, change. Similar trend is also observed in 14 th group as well. In other groups, as we move down, the group, the electro negativity decreases. This observed trend can be correlated with their, atomic radius., 2.1.5 Anomalous properties of the first elements:, In p-block elements, the first member of each group differs from the other elements of, the corresponding group. The following factors are responsible for this anomalous behaviour., 1. Small size of the first member, 2. High ionisation enthalpy and high electronegativity, 3. Absence of d orbitals in their valance shell, The first member of the group 13, boron is a metalloid while others are reactive metals., Moreover, boron shows diagonal relationship with silicon of group 14. The oxides of boron, and silicon are similar in their acidic nature. Both boron and silicon form covalent hydrides, that can be easily hydrolysed. Similarly, except boron trifluoride, halides of both elements are, readily hydrolysed., 29, , XII_U2-P-Block.indd 29, , 2/19/2020 4:38:41 PM

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www.tntextbooks.in, , In group 14, the first element carbon is strictly a nonmetal while other elements are, metalloids (silicon & germanium) or metals (tin & lead). Unlike other elements of the group, carbon can form multiple bonds such as C=C, C=O etc... Carbon has a greater tendency to, form a chain of bonds with itself or with other atoms which is known as catenation. There, is considerable decrease in catenation property down the group (C>>Si>Ge≈Sn>Pb)., In group 15 also the first element nitrogen differs from the rest of the elements of the, group. Like carbon, the nitrogen can from multiple bonds (N=N, C=N, N=O etc...). Nitrogen is, a diatomic gas unlike the other members of the group. Similarly in group 16, the first element,, oxygen also exists as a diatomic gas in that group. Due to its high electronegativity it forms, hydrogen bonds., The first element of group 17, fluorine the most electronegative element, also behaves, quiet differently compared to the rest of the members of group. Like oxygen it also forms, hydrogen bonds. It shows only -1 oxidation state while the other halogens have +1, +3, +5 and, +7 oxidation states in addition to -1 state. The fluorine is the strongest oxidising agent and the, most reactive element among the halogens., 2.1.6 Inert pair effect:, We have already learnt that the alkali and alkaline earth metals have an oxidation, state of +1 and +2 respectively, corresponding to the total number of electrons present in, them. Similarly, the elements of p-block also show the oxidation states corresponding to, the maximum number of valence electrons (group oxidation state). In addition they also, show variable oxidation state. In case of the heavier post-transition elements belonging to, the groups (13 to 16), the most stable oxidation state is two less than the group oxidation, state and there is a reluctance to exhibit the group oxidation state. Let us consider group, 13 elements. As we move from boron to heavier elements, there is an increasing tendency, +3, to have +1 oxidation state, rather than the group oxidation state, +3. For example Al is, +1, +1, +3, more stable than Al while Tl is more stable than Tl . Aluminium(III)chloride is stable, whereas thallium(III)chloride is highly unstable and disproportionates to thallium(I), chloride and chlorine gas. This shows that in thallium the stable lower oxidation state, corresponds to the loss of np electrons only and not ns electrons. Thus in heavier posttransition metals, the outer s electrons (ns) have a tendency to remain inert and show, reluctance to take part in the bonding, which is known as inert pair effect. This effect is, also observed in groups 14, 15 and 16., 2.1.7 Allotropism in p-block elements:, Some elements exist in more than one crystalline or molecular forms in the same physical, state. For example, carbon exists as diamond and graphite. This phenomenon is called, allotropism (in greek 'allos' means another and 'trope' means change) and the different forms, of an element are called allotropes. Many p-block elements show allotropism and some of the, common allotropes are listed in the table., 30, , XII_U2-P-Block.indd 30, , 2/19/2020 4:38:41 PM

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www.tntextbooks.in, , Table 2.2 : Some of common allotropes of p-block elements, Element, , Most common allotropes, , Boron, , Amorphous boron, α-rhombohedral boron, β-rhombohedral boron,, γ-orthorhombic boron, α-tetragonal boron, β-tetragonal boron, , Carbon, , Diamond, Graphite, Graphene, Fullerenes, Carbon nanotubes, , Silicon, , Amorphous silicon, crystalline silicon, , Germanium, , α-germanium, β-germanium, , Tin, , Grey tin, white tin, rhombic tin, sigma tin, , Phosphorus, , White phosphorus, Red phosphorus, Scarlet, phosphorus, Black phosphorus., , Arsenic, , Yellow arsenic, gray arsenic & Black arsenic, , Anitimony, , Blue-white antimony, Yellow, Black, , Oxygen, , Dioxygen, ozone, , Sulphur, , Rhombus sulphur, monoclinic sulphur, , Selenium, , Red selenium, Gray selenium, Black selenium, Monoclinic selenium,, , Tellurium, , Amorphous & Crystalline, , phosphorus, Violet, , 2.2 Group 13 (Boron group) elements:, 2.2.1 Occurrence:, The boron occurs mostly as borates and its important ores are borax - Na2[B4O5(OH)4].8H2O, and kernite - Na2[B4O5(OH)4].2H2O.. Aluminium is the most abundant metal and occurs as, oxides and also found in aluminosilicate rocks. Commercially it is extracted from its chief, ore, bauxite (Al2O3.2H2O). The other elements of this group occur only in trace amounts. The, other elements Ga, In and Tl occur as their sulphides., 2.2.2 Physical properties:, Some of the physical properties of the group 13 elements are listed below, Table 2.3 Physical properties of group 13 elements, Property, , Boron, , Aluminum, , Gallium, , Indium, , Thallium, , Physical state at, 293 K, , Solid, , Solid, , Solid, , Solid, , Solid, , Atomic Number, , 5, , 13, , 31, , 49, , 81, , Isotopes, , B, , 11, , Al, , Ga, , 27, , 69, , In, , 115, , Tl, , 205, , 31, , XII_U2-P-Block.indd 31, , 2/19/2020 4:38:41 PM

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www.tntextbooks.in, , Property, , Boron, , Aluminum, , Gallium, , Indium, , Thallium, , Atomic Mass, (g.mol-1 at 293 K), , 10.81, , 26.98, , 69.72, , 114.81, , 204.38, , Electronic, configuration, , [He]2s 2p, , [Ne]3s 3p, , [Ar]3d10 4s2, 4p1, , [Kr]4d10 5s2, 5p1, , [Xe] 4f14, 5d10 6s2 6p1, , Atomic radius (Å), , 1.92, , 1.84, , 1.87, , 1.93, , 1.96, , Density (g.cm-3 at, 293 K), , 2.34, , 2.70, , 5.91, , 7.31, , 11.80, , Melting point (K), , 2350, , 933, , 302.76, , 429, , 577, , Boiling point (K), , 4273, , 2792, , 2502, , 2300, , 1746, , 2, , 1, , 2, , 1, , 2.2.3 Chemical properties of boron:, Boron is the only nonmetal in this group and is less reactive. However, it shows reactivity, at higher temperatures. Many of its compounds are electron deficient and has unusual type, of covalent bonding which is due to its small size, high ionisation energy and similarity in, electronegativity with carbon and hydrogen., Formation of metal borides:, Many metals except alkali metals form borides with a general formula MxBy (x ranging, upto 11 and y ranging upto 66 or higher), Direct combination of metals with boron:, 1500 K, , Cr + nB, , CrBn, , Reduction of borontrihalides:, Reduction of borontrichloride with a metal assisted by dihydrogen gives metal borides., 2BCl 3 + 2W 1500 K, , 2WB + 2Cl 2 +2HCl, , H2, , Formation of hydrides:, Boron does not react directly with hydrogen. However, it forms a variety of hydrides, called boranes. The simplest borane is diborane - B2H6. Other larger boranes can be prepared, from diborane. Treatment of gaseous boron trifluoride with sodium hydride around 450 K, gives diborane. To prevent subsequent pyrolysis, the product diborane is trapped immediately., 2BF3 + 6NaH, , 450 K, , B2H6 + 6NaF, , Formation of boron trihalides:, Boron combines with halogen to form boron trihalides at high temperatures., 32, , XII_U2-P-Block.indd 32, , 2/19/2020 4:38:42 PM

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www.tntextbooks.in, , 2.3.2 Physical properties:, Some of the physical properties of the group 14 elements are listed below, Table 2.4 Physical properties of group 14 elements, Property, , Carbon, , Silicon, , Germanium, , Tin, , Lead, , Physical state at, 293 K, , Solid, , Solid, , Solid, , Solid, , Solid, , Atomic Number, , 6, , 14, , 32, , 50, , 82, , Isotopes, , 12, , 28, , 73, , 120, , 208, , Atomic Mass, (g.mol-1 at 293 K), , 12.01, , 28.09, , 72.63, , 118.71, , 207.2, , Electronic, configuration, , [He]2s2 2p2 [Ne]3s2 3p2 [Ar]3d10 4s2 4p2, , [Kr]4d10 5s2, 5p2, , [Xe] 4f14, 5d10 6s2 6p2, , C, 13C, 14C, , Si, 30Si, , Ge, 74Ge, , Sn, , Pb, , Atomic radius (Å) 1.70, , 2.10, , 2.11, , 2.17, , 2.02, , Density (g.cm-3 at, 293 K), , 2.33, , 5.32, , 7.29, , 11.30, , Sublimes at 1687, 4098, 3538, , 1211, , 505, , 601, , 3106, , 2859, , 2022, , Melting point (K), Boiling point (K), , 3.51, , 2.3.3 Tendency for catenation, Catenation is an ability of an element to form chain of atoms. The following conditions are, necessary for catenation. (i) the valency of element is greater than or equal to two, (ii) element, should have an ability to bond with itself (iii) the self bond must be as strong as its bond, with other elements (iv) kinetic inertness of catenated compound towards other molecules., Carbon possesses all the above properties and forms a wide range of compounds with itself, and with other elements such as H, O, N, S, and halogens., 2.3.4 Allotropes of carbon, Carbon exists in many allotropic forms., Graphite and diamond are the most common, allotropes. Other important allotropes are, graphene, fullerenes and carbon nanotubes., Graphite is the most stable allotropic, form of carbon at normal temperature and, pressure. It is soft and conducts electricity. It, is composed of flat two dimensional sheets of, carbon atoms. Each sheet is a hexagonal net, of sp2 hybridised carbon atoms with a C-C, bond length of 1.41 Å which is close to the, , Figure 2.4 Structure of graphite, , 41, , XII_U2-P-Block.indd 41, , 2/19/2020 4:38:52 PM

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www.tntextbooks.in, , C-C bond distance in benzene (1.40 Å). Each carbon atom forms three σ bonds with three, neighbouring carbon atoms using three of its valence electrons and the fourth electron present, in the unhybridised p orbital forms a π-bond. These π electrons are delocalised over the entire, sheet which is responsible for its electrical conductivity. The successive carbon sheets are held, together by weak vander Waals forces. The distance between successive sheet is 3.40 Å. It is, used as a lubricant either on its own or as a graphited oil., Unlike graphite the other allotrope, diamond is very hard. The carbon atoms in, diamond are sp3 hybridised and bonded to four, neighbouring carbon atoms by σ bonds with a, C-C bond length of 1.54 Å. This results in a, tetrahedral arrangement around each carbon, atom that extends to the entire lattice as shown, in figure 2.5. Since all four valance electrons, of carbon are involved in bonding there is, no free electrons for conductivity. Being the, hardest element, it used for sharpening hard, tools, cutting glasses, making bores and rock, drilling., Fullerenes are newly synthesised allotropes, of carbon. Unlike graphite and diamond, these, allotropes are discrete molecules such as C32,, C50, C60, C70, C76 etc.. These molecules have cage, like structures as shown in the figure. The C60, molecules have a soccer ball like structure and, is called buckminster fullerene or buckyballs., It has a fused ring structure consists of 20 six, membered rings and 12 five membered rings., Each carbon atom is sp2 hybridised and forms, three σ bonds & a delocalised π bond giving, aromatic character to these molecules. The, C-C bond distance is 1.44 Å and C=C distance, 1.38 Å., Carbon nanotubes, another recently, discovered allotropes, have graphite like tubes, with fullerene ends. Along the axis, these, nanotubes are stronger than steel and conduct, electricity. These have many applications in, nanoscale electronics, catalysis, polymers and, medicine., , Figure 2.5 Structure of diamond, , Figure 2.6 Structure of Fullerenes, , Figure 2.7 Structure of carbon nanotubes, , 42, , XII_U2-P-Block.indd 42, , 2/19/2020 4:38:55 PM

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www.tntextbooks.in, , Another allotrophic form of carbon is, graphene. It has a single planar sheet of sp2, hybridised carbon atoms that are densely, packed in a honeycomb crystal lattice., 2.3.5 Carbon monoxide [CO]:, Preparation:, Carbon monoxide can be prepared by, the reaction of carbon with limited amount of, oxygen., 2C + O2, , Figure 2.8 Structure of graphene, 2CO, , On industrial scale carbon monoxide is produced by the reaction of carbon with air., The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and, carbon monoxide is called producer gas., 2C + O2/N2 (air), , 2CO + N2, , Producers Gas, , The producer gas is then passed through a solution of copper(I)chloride under, pressure which results in the formation of CuCl(CO).2H2O. At reduced pressures this, solution releases the pure carbon monoxide., Pure carbon monoxide is prepared by warming methanoic acid with concentrated, sulphuric acid which acts as a dehydrating agent., HCOOH + H SO, CO + H SO . H O, 2, , 4, , 2, , 4, , 2, , Properties, It is a colourless, odourless, and poisonous gas. It is slightly soluble in water., It burns in air with a blue flame forming carbon dioxide., 2CO2, , 2CO + O2, , When carbon monoxide is treated with chlorine in presence of light or charcoal, it forms, a poisonous gas carbonyl chloride, which is also known as phosgene. It is used in the synthesis, of isocyanates., COCl2, CO + Cl2, Carbon monoxide acts as a strong reducing agent., 2Fe + 3CO2, , 3CO + Fe2O3, , Under high temperature and pressure a mixture of carbon monoxide and hydrogen, (synthetic gas or syn gas) gives methanol., CO + 2H2, CH3OH, In oxo process, ethene is mixed with carbon monoxide and hydrogen gas to produce, propanal., 43, , XII_U2-P-Block.indd 43, , 2/19/2020 4:38:56 PM

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www.tntextbooks.in, , CO + C2H4 + H2, , CH3CH2CHO, , Fischer Tropsch synthesis:, The reaction of carbon monoxide with hydrogen at a pressure of less than 50 atm using, metal catalysts at 500 - 700 K yields saturated and unsaturated hydrocarbons., nCO + (2n+1)H2, CnH(2n+2) + nH2O, nCO + 2nH2, , CnH2n + nH2O, , Carbon monoxide forms numerous complex compounds with transition metals in which, the transition meal is in zero oxidation state. These compounds are obtained by heating the, metal with carbon monoxide., Eg. Nickel tetracarbonyl [Ni(CO)4], Iron pentacarbonyl [Fe(CO)5], Chromium, hexacarbonyl [Cr(CO)6]., Structure:, It has a linear structure. In carbon monoxide, three electron pairs are shared between, carbon and oxygen. The bonding can be explained using molecular orbital theory as discussed, in XI standard. The C-O bond distance is 1.128Å. The structure can be considered as the, resonance hybrid of the following two canonical forms., , C, , C, , O, , O, , C, , O, , Figure 2.9 Structure of carbon monoxide, Uses of carbon monoxide:, 1. Equimolar mixture of hydrogen and carbon monoxide - water gas and the mixture of, carbon monoxide and nitrogen - producer gas are important industrial fuels, 2. Carbon monoxide is a good reducing agent and can reduce many metal oxides to metals., 3. Carbon monoixde is an important ligand and forms carbonyl compound with transition, metals, 2.3.6 Carbon dioxide:, Carbon dioxide occurs in nature in free state as well as in the combined state. It is a, constituent of air (0.03%). It occurs in rock as calcium carbonate and magnesium carbonate., Production, On industrial scale it is produced by burning coke in excess of air., C + O2, , CO2, , -1, ∆H = -394 kJ mol, , Calcination of lime produces carbon dioxide as by product., CaO + CO2, , CaCO3, 44, , XII_U2-P-Block.indd 44, , 2/19/2020 4:38:57 PM

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www.tntextbooks.in, , Carbon dioxide is prepared in laboratory by the action of dilute hydrochloric acid on, metal carbonates., CaCO3 + 2HCl, , CaCl2 + H2O + CO2, , Properties, It is a colourless, nonflammable gas and is heavier than air. Its critical temperature is 31⁰, C and can be readily liquefied., Carbon dioxide is a very stable compound. Even at 3100 K only 76 % decomposes to, form carbon monoxide and oxygen. At still higher temperature it decomposes into carbon, and oxygen., 3100 K, , CO2, CO2, , high temperature, , CO + ½O2, C + O2, , Oxidising behaviour:, At elevated temperatures, it acts as a strong oxidising agent. For example,, CO2 + 2Mg, 2MgO + C, Water gas equilibrium:, The equilibrium involved in the reaction between carbon dioxide and hydrogen, has many, industrial applications and is called water gas equilibrium., CO2 + H2, , CO + H2O, Water gas, , Acidic behaviour:, The aqueous solution of carbon dioxide is slightly acidic as it forms carbonic acid., +, CO2 + H2O, H2CO3, H + HCO3, Structure of carbon dioxide, Carbon dioxide has a linear structure with equal bond distance for the both C-O bonds., In this molecule there is two C-O sigma bond. In addition there is 3c-4e bond covering all the, three atoms., , O, , C, , O, , O, , C, , O, , O, , C, , O, , Figure 2.10 Structure of carbon dixide, Uses of carbon dioxide, 1. Carbon dioxide is used to produce an inert atomosphere for chemical processing., 2. Biologically, it is important for photosynthesis., 45, , XII_U2-P-Block.indd 45, , 2/19/2020 4:38:58 PM

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www.tntextbooks.in, , Preparation:, Generally silicones are prepared by the hydrolysis of dialkyldichlorosilanes (R2SiCl2) or, diaryldichlorosilanes Ar2SiCl2, which are prepared by passing vapours of RCl or ArCl over, silicon at 570 K with copper as a catalyst., Cu / 570 K, , 2RCl+ Si, , R2SiCl2, , The hydrolysis of dialkylchloro silanes R2SiCl2 yields to a straight chain polymer which, grown from both the sides, R, R, Cl, , +2H2O, , Si Cl, , HO Si OH, , -2HCl, , R, , R, R, , R, , R, , HO Si OH + HO Si OH, R, , R, , R, , R, , HO Si O Si OH + HO Si OH, , -H2O, , R, , R, , R, -H2O, , R, Etc, , R, , R, , HO Si O Si O, , -H2O, , R, , Si OH, R, , R, , The hydrolysis of monoalkylchloro silanes RSiCl3 yields to a very complex cross linked, polymer.. Linear silicones can be converted into cyclic or ring silicones when water molecules, is removed from the terminal –OH groups., , Me, Me, , Si, , Me, O, , O, Me, , Si, , Si Me, O, , O, , Si Me, , Me, , Me, , Me, Me, , Si, , O, , O, Me, , Si, O, , Si, , Me, , Me, Me, , O, R Si, , R, O, , O, O, , Si, R, , Si O, R, , O, O, , Si, R, , O, , Si O, O, , Types of silicones:, (i) Liner silicones:, They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl, silicon chlorides., 47, , XII_U2-P-Block.indd 47, , 2/19/2020 4:38:59 PM

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www.tntextbooks.in, , a) Silicone rubbers: These silicones are bridged together by methylene or similar groups, b) Silicone resins: They are obtained by blending silicones with organic resins such as acrylic, esters., (ii) Cyclic silicones, These are obtained by the hydrolysis of R2SiCl2., (iii) Cross linked silicones, They are obtained by hydrolysis of RSiCl3, Properties, The extent of cross linking and nature of alkyl group determine the nature of polymer. They, range from oily liquids to rubber like solids. All silicones are water repellent. This property, arises due to the presence of organic side groups that surrounds the silicon which makes the, molecule looks like an alkane. They are also thermal and electrical insulators. Chemically they, are inert. Lower silicones are oily liquids whereas higher silicones with long chain structure are, waxy solids. The viscosity of silicon oil remains constant and doesn’t change with temperature, and they don't thicken during winter, Uses:, 1. Silicones are used for low temperature lubrication and in vacuum pumps, high temperature, oil baths etc..., 2. They are used for making water proofing clothes, 3. They are used as insulting material in electrical motor and other appliances, 4. They are mixed with paints and enamels to make them resistant towards high temperature,, sunlight, dampness and chemicals., 2.3.9 Silicates, 4-, , The mineral which contains silicon and oxygen in tetrahedral [SiO ] units linked, 4, together in different patterns are called silicates. Nearly 95 % of the earth crust is composed, of silicate minerals and silica. The glass and, ceramic industries are based on the chemistry, silicates., , , , Types of Silicates:, Silicates are classified into various types, based on the way in which the tetrahedral, 4units, [SiO ] are linked together., 4, , Ortho silicates (Neso silicates): The simplest, 4silicates which contain discrete [SiO ], 4, tetrahedral units are called ortho silicates or, neso silicates., , , , , , , , , , Figure 2.11, Structure of Ortho silicates, , 48, , XII_U2-P-Block.indd 48, , 2/19/2020 4:38:59 PM

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www.tntextbooks.in, , Examples : Phenacite - Be2SiO4 (Be2+ ions are tetrahedrally surrounded by O2- ions), Olivine (Fe/Mg)2SiO4 ( Fe2+ and Mg2+ cations are octahedrally surrounded by O2- ions),, Pyro silicate (or) Soro silicates):, Silicates which contain [Si2O7]6ions are called pyro silicates (or), Soro silicates. They are formed by, 4joining two [SiO ] tetrahedral, 4, units by sharing one oxygen atom, at one corner.(one oxygen is, removed while joining). Example :, Thortveitite - Sc2Si2O7, , , , , , , , , , , , , , , , , , , , , , , , , , Figure 2.12 Structure of Pyro silicate, , Cyclic silicates (or Ring silicates), Silicates which contain (SiO3)n2n- ions, which are formed by linking three or more, tetrahedral SiO44- units cyclically are called, cyclic silicates. Each silicate unit shares two, of its oxygen atoms with other units., , , , , , , Example: Beryl [Be3Al2 (SiO3)6] (an, aluminosilicate with each aluminium, is surrounded by 6 oxygen atoms, octahedrally), , , , , , , , , , , , , , , , , , , , , , , , , , , Figure 2.13 Structure of Cyclic silicates, , Inosilicates :, Silicates which contain 'n' number of silicate units linked by sharing two or more oxygen, atoms are called inosilicates. They are further classified as chain silicates and double chain, silicates., Chain, silicates, (or, pyroxenes):, These, silicates contain [(SiO3), 2n], ions formed by, n, linking ‘n’ number of, 4tetrahedral [SiO ] units, 4, linearly. Each silicate unit, shares two of its oxygen, atoms with other units., , , , , , , , , , , , , , , , , , , , , , Figure 2.14 Structure of Chain silicates, , Example: Spodumene - LiAl(SiO3)2., Double chain silicates (or amphiboles): These silicates contains [Si4O11]n6n- ions. In these, silicates there are two different types of tetrahedra : (i) Those sharing 3 vertices (ii) those, sharing only 2 vertices., 49, , XII_U2-P-Block.indd 49, , 2/19/2020 4:39:00 PM

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www.tntextbooks.in, , Examples:, 1) Asbestos: These are, fibrous and noncombustible silicates., Therefore they are used, for thermal insulation, material,, brake, linings, construction, material and filters., Asbestos, being, carcinogenic silicates,, their applications are, restricted., Sheet or phyllo silicates, , Figure 2.15 Structure of Double chain silicates, , , , , , , , , , , , Silicates, which, 2nare, contain (Si2O5)n, called sheet or phyllo, , , , , silicates. In these, Each, 4[SiO ] tetrahedron, 4, , , , unit shares three oxygen, atoms with others and, Figure 2.16 Structure of Sheet or phyllo silicates, thus by forming twodimensional sheets. These, sheets silicates form layered structures in which silicate sheets are stacked over each other. The, attractive forces between these layers are very weak, hence they can be cleaved easily just like, graphite., Example: Talc, Mica etc..., Three dimensional silicates (or tecto silicates):, 4-, , Silicates in which all the oxygen atoms of [SiO ] tetrahedra are shared with other, 4, tetrahedra to form three-dimensional network are called three dimensional or tecto silicates., They have general formula (SiO2)n ., Examples: Quartz, These tecto silicates can be converted into Three dimensional aluminosilicates by replacing, [SiO4]4- units by [AlO4]5- units. E.g. Feldspar, Zeolites etc.,, 2.3.10 Zeolites:, Zeolites are three-dimensional crystalline solids containing aluminium, silicon, and, oxygen in their regular three dimensional framework. They are hydrated sodium alumino, silicates with general formula Na2O.(Al2O3).x(SiO2).yH2O (x=2 to 10; y=2 to 6)., 50, , XII_U2-P-Block.indd 50, , 2/19/2020 4:39:00 PM

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www.tntextbooks.in, , Zeolites have porous structure in which the monovalent sodium ions and water molecules, are loosely held. The Si and Al atoms are tetrahedrally coordinated with each other through, shared oxygen atoms. Zeolites are similar to clay minerals but they differ in their crystalline, structure., Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting, of a network of interconnected tunnels and cages. Water molecules moves freely in and out of, these pores but the zeolite framework remains rigid. Another special aspect of this structure is, that the pore/channel sizes are nearly uniform, allowing the crystal to act as a molecular sieve., We have already discussed in XI standard, the removal of permanent hardness of water using, zeolites., Boron Neutron Capture Therapy:, The affinity of Boron-10 for neutrons is the basis of a technique known as, boron neutron capture therapy (BNCT) for treating patients suffering from brain, tumours., It is based on the nuclear reaction that occurs when boron-10 is irradiated with low-energy, thermal neutrons to give high linear energy α particles and a Li particle., Boron compounds are injected into a patient with a brain tumour and the compounds collect, preferentially in the tumour. The tumour area is then irradiated with thermal neutrons and results, in the release of an alpha particle that damages the tissue in the tumour each time a boron-10, nucleus captures a neutron. In this way damage can be limited preferentially to the tumour, leaving, the normal brain tissue less affected. BNCT has also been studied as a treatment for several other, tumours of the head and neck, the breast, the prostate, the bladder, andthe liver., Summary, „„ The elements in which their last electron enters the 'p' orbital, constitute the p-block, , elements., , „„ The p-block elements have a general electronic configuration of ns2, np1-6. The, , elements of each group have similar outer shell electronic configuration and differ, only in the value of n (principal quantum number)., , „„ Generally on descending a group the ionisation energy decreases and hence the, , metallic character increases., , „„ The ionisation enthalpy of elements in successive groups is higher than the, , corresponding elements of the previous group as expected., , „„ As we move down the 13th group, the electronegativity first decreases from boron, , to aluminium and then marginally increases., , 51, , XII_U2-P-Block.indd 51, , 2/19/2020 4:39:00 PM

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www.tntextbooks.in, , „„ In p-block elements, the first member of each group differs from the other elements, , of the corresponding group., , „„ In heavier post-transition metals, the outer s electrons (ns) have a tendency to, , remain inert and show reluctance to take part in the bonding, which is known as, inert pair effect., , „„ Some elements exist in more than one crystalline or molecular forms in the, , same physical state. For example, carbon exists as diamond and graphite. This, phenomenon is called allothropism, , „„ Borax is a sodium salt of tetraboric acid. It is obtained from colemanite ore by, , boiling its solution with sodium carbonate., , „„ Boric acid can be extracted from borax and colemanite., „„ Boric acid has a two dimensional layered structure., „„ The name alum is given to the double salt of potassium aluminium sulphate [K2SO4., , Al2(SO4)3.24.H2O]., , „„ Carbon is found in the native form as graphite., „„ Silicon occurs as silica (sand and quartz crystal). Silicate minerals and clay are, , other important sources for silicon., , „„ Catenation is an ability of an element to form chain of atoms, „„ Carbon nanotubes, another recently discovered allotropes, have graphite like tubes, , with fullerene ends., , „„ Silicones or poly siloxanes are organo silicon polymers with general empirical, , formula (R2SiO). Because of their very high thermal stability they are called high, –temperature polymers., , „„ The mineral which contains silicon and oxygen in tetrahedral [SiO ], , together in different patterns are called silicates., , 4, , 4-, , units linked, , „„ Types of Silicates:, ▶▶ Ortho silicates (Neso silicates), Pyro silicate (or) Soro silicates), Cyclic, , silicates (or Ring silicates), , ▶▶ Inosilicates : Chain silicates (or pyroxenes), Double chain silicates, , (or amphiboles):, , ▶▶ Sheet or phyllo silicates, ▶▶ Three dimensional silicates (or tecto silicates), „„ Zeolites are three-dimensional crystalline solids containing aluminium, silicon,, , and oxygen in their regular three dimensional framework., , „„ zeolites act as a molecular sieve for the removal of permanent hardness of water, , 52, , XII_U2-P-Block.indd 52, , 2/19/2020 4:39:01 PM

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www.tntextbooks.in, , EVALUATION, Choose the correct answer:, 1. An aqueous solution of borax is, a) neutral, , b) acidic, , c) basic, , d) amphoteric, , 2. Boric acid is an acid because its molecule (NEET), a) contains replaceable H+ ion, b) gives up a proton, c) combines with proton to form water molecule, d) accepts OH- from water ,releasing proton., 3. Which among the following is not a borane?, a) B2 H6, , b) B3H6, , c) B4 H10, , d) none of these, , 4. Which of the following metals has the largest abundance in the earth’s crust?, a) Aluminium, , b) calcium, , c) Magnesium, , d) sodium, , 5. In diborane, the number of electrons that accounts for banana bonds is, a) six, , b) two, , c) four, , d) three, , 6. The element that does not show catenation among the following p-block elements is, a) Carbon, , b) silicon, , c) Lead, , d) germanium, , 7. Carbon atoms in fullerene with formula C60 have, a) sp3 hybridised, , b) sp hybridised, , c) sp2 hybridised, , d) partially sp2 and partially sp3 hybridised, , 8. Oxidation state of carbon in its hydrides, a) +4, , b) -4, , c) +3, , d) +2, , 9. The basic structural unit of silicates is (NEET), b) ( SiO4 ), , a) ( SiO3 )2−, , 2−, , c) ( SiO ), , 10. The repeating unit in silicone is, , 4−, , R, , b), , a) SiO2, , d) ( SiO4 ), , −, , Si, , O, , R, , c), , R, , O, , Si, , O, , d), , Si, , O, , O, , R, , R, , R, , 53, , XII_U2-P-Block.indd 53, , 2/19/2020 4:39:02 PM

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www.tntextbooks.in, , 11. Which of these is not a monomer for a high molecular mass silicone polymer?, a) Me3SiCl, , b) PhSiCl3, , c) MeSiCl3, , d) Me2SiCl2, , 12. Which of the following is not sp2 hybridised?, a) Graphite, , b) graphene, , c) Fullerene, , d) dry ice, , 13. The geometry at which carbon atom in diamond are bonded to each other is, a) Tetrahedral, , b) hexagonal, , c) Octahedral, , d) none of these, , 14. Which of the following statements is not correct?, a) Beryl is a cyclic silicate, b) Mg2SiO4 is an orthosilicate, c) SiO4 4− is the basic structural unit of silicates, d) Feldspar is not aluminosilicate, 15. Match items in column - I with the items of column – II and assign the correct code., Column-I, , A, , B, , C, , D, , (a), , 2, , 1, , 4, , 3, , (b), , 1, , 2, , 4, , 3, , (c), , 1, , 2, , 4, , 3, , Column-II, , A, , Borazole, , 1, , B(OH)3, , B, , Boric acid, , 2, , B3N3H6, , C, , Quartz, , 3, , Na2[B4O5(OH)4] 8H2O, , D, , Borax, , 4, , SiO2, , (d), , None of these, , 16. Duralumin is an alloy of, a) Cu,Mn, , b) Cu,Al,Mg, , c) Al,Mn, , d) Al,Cu,Mn,Mg, , 17. The compound that is used in nuclear reactors as protective shields and control rods is, a) Metal borides, , b) metal oxides c) Metal carbonates d) metal carbide, , 18. The stability of +1 oxidation state increases in the sequence, a) Al < Ga < In < Tl, , b) Tl < In < Ga < Al, , c) In < Tl < Ga < Al, , d) Ga< In < Al < Tl, , Answer the following questions:, 1., 2., 3., 4., 5., , Write a short note on anamolous properties of the first element of p-block., Describe briefly allotropism in p- block elements with specific reference to carbon., Give the uses of Borax., What is catenation ? describe briefly the catenation property of carbon., Write a note on Fisher tropsch synthesis., 54, , XII_U2-P-Block.indd 54, , 2/19/2020 4:39:03 PM

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www.tntextbooks.in, , 6. Give the structure of CO and CO2., 7. Give the uses of silicones., 8. Describe the structure of diborane., 9. Write a short note on hydroboration., 10. Give one example for each of the following, (i) icosogens, , (ii) tetragen, , (iii) prictogen, , (iv) chalcogen, , 11. Write a note on metallic nature of p-block elements., 12. Complete the following reactions, a. B ( OH )3 +NH3 →, , , , b. Na 2 B4O7 + H2SO4 + H2O →, c. B2 H6 + 2NaOH + 2H, H2O →, , d. B2 H6 + CH3OH →, , e. BF3 + 9 H2O →, , f . HCOOH + H2SO4 →, , g . SiCl 4 + NH3 →, , h. SiCl 4 + C 2 H5OH →, , i. B + NaOH →, , hot, j. H2 B 4O7 Red, , →, , 13. How will you identify borate radical?, 14. Write a note on zeolites., 15. How will you convert boric acid to boron nitride?, 16. A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a, reducing agent (C). identify A, B and C., 17. A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B)., aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound, with alizarin. Identify A and B., 18. CO is a reducing agent. justify with an example., , 55, , XII_U2-P-Block.indd 55, , 2/19/2020 4:39:03 PM

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www.tntextbooks.in, , UNIT, , 3, , p-BLOCK, ELEMENTS - II, , Learning Objectives, , Sir William Ramsay,, (1852 – 1916), , After studying this unit, the students will, be able to, , Sir William Ramsay was a Scottish, chemist who discovered the noble, gases. During the years 1885–1890, he published several important, papers on the oxides of nitrogen.In, August 1894, Ramsay had isolated, a new heavy element of air, and he, named it "argon", (the Greek word, meaning "lazy").In the following, years, he worked with Morris Travers, and discovered neon, krypton,, and xenon. In 1910 he isolated and, characterized radon. In recognition, of his services in the discovery of the, inert gases, he was awarded a noble, prize in chemistry in 1904. His work, in isolating noble gases led to the, development of a new section of the, periodic table., , , , , , , , , , , , , , discuss the preparation and properties, of important compounds of nitrogen, and phosphorus, describe the preparation and properties, of important compounds of oxygen and, sulphur, describe the preparation, properties of, halogens and hydrogen halides, explain the chemistry of inter-halogen, compounds, describe the occurrence, properties and, uses of noble gases, appreciate the importance of p-block, elements and their compounds in day, today life., , 56, , XII U3-P-block.indd 56, , 2/19/2020 4:39:45 PM

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www.tntextbooks.in, , INTRODUCTION, We have already learnt the general characteristics of p-block elements and the first two, group namely icosagens (boron group) and tetragens (carbon group) in the previous unit. In, this unit we learn the remaining p-block groups, pnictogens, chalcogens, halogens and inert, gases., , 3.1 Group 15 (Nitrogen group) elements:, 3.1.1 Occurrence:, About 78 % of earth atmosphere contains dinitorgen (N2) gas. It is also present in earth, crust as sodium nitrate (Chile saltpetre) and potassium nitrate (Indian saltpetre). The 11th, most abundant element phosphorus, exists as phosphate (fluroapatite, chloroapatite and, hydroxyapatite). The other elements arsenic, antimony and bismuth are present as sulphides, and are not very abundant., 3.1.2 Physical properties:, Some of the physical properties of the group 15 elements are listed below, Table 3.1 Physical properties of group 15 elements, Property, Physical state at, 293 K, Atomic Number, Isotopes, Atomic Mass, (g.mol-1 at 293 K), Electronic, configuration, Atomic radius (Å), Density (g.cm-3 at, 293 K), Melting point (K), Boiling point (K), , Nitrogen, , Phosphorus Arsenic, , Antimony, , Bismuth, , Gas, , Solid, , Solid, , Solid, , 7, , 15, P, , N, N, , 14, , Solid, , 15, , 14, , 33, As, , 1.14 x 10-3, 63, 77, , 83, Bi, , 75, , 121, , 209, , 30.97, , 74.92, , 121.76, , 209.98, , [Ar]3d10 4s2, 4p3, 1.85, , [Kr]4d10 5s2, 5p3, 2.06, , [Xe] 4f14, 5d10 6s2 6p3, 2.07, , 6.68, , 9.79, , 904, 1860, , 544, 1837, , [He]2s2 2p3 [Ne]3s2 3p3, 1.55, , 51, Sb, , 31, , 1.80, 1.82 (white, 5.75, phosphorus), 317, Sublimes at, 889, 554, , 3.1.3 Nitrogen:, Preparation:, Nitrogen, the principal gas of atmosphere (78 % by volume) is separated industrially from, liquid air by fractional distillation, Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about, 575 K, 2NaN3, , 2 Na + 3N2, 57, , XII U3-P-block.indd 57, , 2/19/2020 4:39:46 PM

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www.tntextbooks.in, , It can also be obtained by oxidising ammonia using bromine water, 8NH3 + 3Br2, , 6NH4Br + N2, , Properties, Nitrogen gas is rather inert. Terrestrial nitrogen contains 14.5% and 0.4% of nitrogen-14, and nitrogen-15 respectively. The later is used for isotopic labelling. The chemically inert, character of nitrogen is largely due to high bonding energy of the molecules 225 cal mol-1 (946, kJ mol-1). Interestingly the triply bonded species is notable for its less reactivity in comparison, with other iso-electronic triply bonded systems such as -C≡C-, C≡O, X-C≡N, X-N≡C, -C≡C-,, and -C≡N. These groups can act as donor where as dinitrogen cannot. However, it can form, complexes with metal (M← N≡N) like CO to a less extent, The only reaction of nitrogen at room temperature is with lithium forming Li3N. With, other elements, nitrogen combines only at elevated temperatures. Group 2 metals and Th, forms ionic nitrides., 6Li + N 2 →,  2Li 3 N, hot, 3Ca + N 2 red, , → Ca 3 N 2, red hot, 2 B + N 2 bright, , → 2BN, , Direct reaction with hydrogen gives ammonia. This reaction is favoured by high pressures, and at optimum temperature in presence of iron catalyst. This reaction is the basis of Haber’s, process for the synthesis of ammonia., 1, 3, −1, , N2 + H2 , ,  NH 3 ∆ H f = −46.2 kJ mol, 2, 2, With oxygen, nitrogen produces nitrous oxide at high temperatures. Even at 3473 K, nitrous oxide yield is only 4.4%., 2N 2 + O2 , → 2N 2O, Uses of nitrogen, 1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc., 2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in, biological preservation ., 3.1.4 Ammonia (NH3), Preparation:, Ammonia is formed by the hydrolysis of urea., NH2CONH2 +H2O, , 2NH3 +CO2, , Ammonia is prepared in the laboratory by heating an ammonium salt with a base., 2 NH 4 + + OH − , → 2 NH 3 + H 2 O, 2NH 4 Cl + CaO , → CaCl 2 + 2NH 3 + H 2 O, , 58, , XII U3-P-block.indd 58, , 2/19/2020 4:39:49 PM

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www.tntextbooks.in, , It can also be prepared by heating a metal nitrides such as magnesium nitride with water., Mg 3 N 2 + 6H 2 O , → 3Mg(OH) 2 + 2NH 3, , It is industrially manufactured by passing nitrogen and hydrogen over iron catalyst (a, small amount of K2O and Al2O3 is also used to increase the rate of attainment of equilibrium), at 750 K at 200 atm pressure. In the actual process the hydrogen required is obtained from, water gas and nitrogen from fractional distillation of liquid air., Properties, Ammonia is a pungent smelling gas and is lighter than air. It can readily liquefied by, at about 9 atmospheric pressure. The liquid boils at -38.4°C and freezes at -77° C. Liquid, ammonia resembles water in its physical properties. i.e. it is highly associated through strong, hydrogen bonding. Ammonia is extremely soluble in water (702 Volume in 1 Volume of water), at 20°C and 760mm pressure., At low temperatures two soluble hydrate NH3.H2O and 2NH3.H2O are isolated. In these, molecules ammonia and water are linked by hydrogen bonds. In aqueous solutions also, ammonia may be hydrated in a similar manner and we call the same as (NH3.H2O), −, +, , NH 3 + H 2 O , ,  NH 4 + OH, , The dielectric constant of ammonia is considerably high to make it a fairly good ionising, solvent like water., −, +, , 2NH 3 , ,  NH 4 + NH 2, , K −500C = [NH 4 + ][NH 2 − ] = 10−30, +, −, , 2H 2O , ,  H 3O + OH, , K 250 C = [H 3O + ][OH − ] = 10−14, , Chemical Properties, Action of heat: Above 500°C ammonia decomposes into its elements. The decomposition may, be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on, continuous sparking., 500 C, 2NH 3 >, → N 2 + 3H 2, 0, , Reaction with air/oxygen: Ammonia does not burn in air but burns freely in free oxygen with, a yellowish flame to give nitrogen and steam., , , 4NH 3 + 3O 2 , ,  N 2 + 6H 2 O, , In presence of catalyst like platinum, it burns to produce nitric oxide. This process is used for, the manufacture of nitric acid and is known as ostwalds process., , , 4NH 3 + 5O 2 , ,  4NO + 6H 2 O, , Reducing property: Ammonia acts as a reducing agent. It reduces the metal oxides to metal, when passed over heated metallic oxide., 59, , XII U3-P-block.indd 59, , 2/19/2020 4:39:53 PM

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www.tntextbooks.in, , 3PbO + 2NH 3 , → 3Pb + N 2 + 3H 2 O, , Reaction with acids: When treated with acids it forms ammonium salts. This reaction shows, that the affinity of ammonia for proton is greater than that of water., Reaction with chlorine and chlorides: Ammonia reacts with chlorine and chlorides to give, ammonium chloride as a final product. The reactions are different under different conditions, as given below., With excess ammonia, 2 NH 3 + 3 Cl2 , → N 2 + 6 HCl, 6 HCl + 6 NH3 , → 6 NH 4 Cl, , With excess of chlorine ammonia reacts to give nitrogen, trichloride, an explosive substance., 2NH 3 + 6Cl 2 , → 2NCl3 + 6 HCl, , 2NH 3 ( g ) + HCl(g) , → NH 4Cl (s), , Formation of amides and nitrides: With strong electro, positive metals such as sodium, ammonia forms amides, while it forms nitrides with metals like magnesium., 2Na + 2NH 3 , → 2NaNH 2 + H 2, , Reaction of ammonia with HCL, , 3Mg + 2NH 3 , → Mg 3 N 2 + 3H 2, , With metallic salts: Ammonia reacts with metallic salts to give metal hydroxides, (in case of Fe) or forming complexes (in case Cu), −, , OH, Fe3+ + 3NH 4 + 3, → Fe(OH)3 + 3NH 4 +, , Cu 2+ + 4NH 3 , → [Cu(NH 3 ) 4 ]2+, Tetraamminecopper(II)ion, (a coordinattion complex), , Formation of amines: Ammonia forms ammonated compounds by ion dipole attraction., Eg. [CaCl2.8NH3]. In this, the negative ends of ammonia dipole is attracted to Ca2+ ion., It can also act as a ligand and form coordination compounds such as [Co(NH3)6]3+, [Ag(NH3)2]+., For example when excess ammonia is added to, aqueous solution copper sulphate a deep blue colour, compound [Cu(NH3)4]2+ is formed., sp 3, , 6Å, , 1.01, , Structure of ammonia, Ammonia molecule is pyramidal in shape N-H, bond distance is 1.016 Å and H-H bond distance, is 1.645 Å with a bond angle 107°. The structure, of ammonia may be regarded as a tetrahedral with, , 107, , 0, H, N, , Figure 3.1 Structure of ammonia, , 60, , XII U3-P-block.indd 60, , 2/19/2020 4:39:58 PM

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www.tntextbooks.in, , one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as, shown in the figure., 3.1.5 Nitric acid, Preparation, Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with, concentrated sulphuric acid., KNO3 + H 2SO 4 , → KHSO 4 + HNO3, , The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid, condenses to a fuming liquid which is coloured brown by the presence of a little nitrogen, dioxide which is formed due to the decomposition of nitric acid., 4HNO3 , → 4NO 2 + 2H 2 O + O 2, , Commercial method of preparation, Nitric acid prepared in large scales using Ostwald's process. In this method ammonia, from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into, the catalyst chamber where they come in contact with platinum gauze. The temperature rises, to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia, resulting in the formation of NO, which then oxidised to nitrogen dioxide., 4NH3 + 5O 2 , → 4NO + 6H 2 O + 120 kJ, 2NO + O 2 , → 2NO 2, , The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts, with water to give nitric acid. Nitric oxide formed is bleached by blowing air., 3NO 2 + H 2 O , → 2HNO3 + NO, , Properties, Pure nitric acid is colourless. It boils at 86 °C. The acid is completely miscible with water, forming a constant boiling mixture (98% HNO3, Boiling point 120.5 °C). Fuming nitric acid, contains oxides of nitrogen. It decomposes on exposure to sunlight or on being heated, into, nitrogen dioxide, water and oxygen., 4HNO3 , → 4NO 2 + 2H 2 O + O 2, , Due to this reaction pure acid or its concentrated solution becomes yellow on standing., In most of the reactions, nitric acid acts as an oxidising agent. Hence the oxidation state, changes from +5 to a lower one. It doesn’t yield hydrogen in its reaction with metals. Nitric, acid can act as an acid, an oxidizing agent and an nitrating agent., As an acid: Like other acids it reacts with bases and basic oxides to form salts and water, ZnO + 2HNO3 , → Zn(NO3 ) 2 + H 2 O, 3FeO + 10HNO3 , → 3Fe(NO3 )3 + NO + 5 H 2 O, , 61, , XII U3-P-block.indd 61, , 2/19/2020 4:40:03 PM

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www.tntextbooks.in, , As an oxidising agent: The nonmetals like carbon, sulphur, phosphorus and iodine are, oxidised by nitric acid., C + 4HNO3 , → 2H 2 O + 4NO 2 + CO 2, S + 2HNO3 , → H 2SO 4 + 2NO, P4 + 20HNO3 , → 4H 3 PO 4 + 4H 2 O + 20NO 2, 3I 2 + 10HNO3 , → 6HIO3 + 10NO + 2H 2 O, HNO3 + F2 , → HF + NO3 F, 3H 2S + 2HNO3 , → 3S + 2NO + 4H 2 O, , As an nitrating agent: In organic compounds replacement of a –H atom with –NO2 is often, referred as nitration. For example., 2SO 4, C6 H 6 + HNO3 H, → C6 H 5 NO 2 + H 2 O, , Nitration takes place due to the formation of nitronium ion, HNO3 + H 2SO 4 , → NO 2 + + H 2O + HSO4 −, , Action of nitric acid on metals, All metals with the exception of gold, platinum, rhodium, iridium and tantalum reacts, with nitric acid. Nitric acid oxidises the metals. Some metals such as aluminium, iron, cobalt,, nickel and chromium are rendered passive in concentrated acid due to the formation of a layer, of their oxides on the metal surface, which prevents the nitric acid from reacting with pure, metal., With weak electropositive metals like tin, arsenic, antimony, tungsten and molybdenum,, nitric acid gives metal oxides in which the metal is in the higher oxidation state and the acid, is reduced to a lower oxidation state. The most common products evolved when nitric acid, reacts with a metal are gases NO2, NO and H2O. Occasionally N2, NH2OH and NH3 are also, formed., +5, , +4, , HNO3, , NO 2, , +3, , HNO 2, , +2, , +1, , NO N 2 O, , 0, , N2, , −3, , NH 3, , The reactions of metals with nitric acid are explained in 3 steps as follows:, Primary reaction: Metal nitrate is formed with the release of nascent hydrogen, M + HNO3 , → MNO3 + (H), , Secondary reaction: Nascent hydrogen produces the reduction products of nitric acid., HNO3 + 2H , → HNO 2 + H 2 O, Nitrous acid, , HNO3 + 6H , → NH 2 OH + 2H 2 O, Hydroxylamine, , → NH 3 + 3H 2 O, HNO3 + 8H , Ammonia, , 2HNO3 + 8H , → H 2 N 2 O 2 + 4H 2 O, Hypo nitrous acid, , 62, , XII U3-P-block.indd 62, , 2/19/2020 4:40:10 PM

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www.tntextbooks.in, , Structures of oxides of nitrogen:, , Name, , Formula, , Nitrous oxide, , N2O, , Nitric oxide, , NO, , Dinitrogen, trioxide (or), Nitrogen, sesquoxide, Nitrogen, dioxide, , N2O3, , Structure, , N, , , , , , , , , , N, , O, , N, , N, , N, , O, , 115 pm, , O, , O−, , O, , O, , N, , N+, , N, , N+, O−, , O, , , NO2, , O, , N, –, , O, Nitrogen, tetraoxide, , N2O4, , N, , O, O, , +, , N, , +, –, , O, , O, , O, Nitrogen, pentoxide, , O, , O, , N+, , N2O5, , N+, , O, , −, , O−, , O, , Structures of oxoacids of nitrogen:, , Name, , Hyponitrous, acid, , Formula, , Structure, , H2N2O2, , HO, , N, , N, , OH, , HO, Hydronitrous, acid, , N, , H4N2O4, , HO, , OH, , N, OH, , 65, , XII U3-P-block.indd 65, , 2/19/2020 4:40:15 PM

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www.tntextbooks.in, , 3.1.7 Allotropic forms of phosphorus:, Phosphorus has several allotropic modification, of which the three forms namely white, red and black, phosphorus are most common., The freshly prepared white phosphorus is, colourless but becomes pale yellow due to formation, of a layer of red phosphorus upon standing. Hence it, is also known as yellow phosphorus. It is poisonous, in nature and has a characteristic garlic smell. It, glows in the dark due to oxidation which is called, phosphorescence. Its ignition temperature is very low, and hence it undergoes spontaneous combustion in air, at room temperature to give P2O5., , P, , P, , P, P, , Figure 3.2 Structure of white, phosphorus, , The white phosphorus can be changed into red phosphorus by heating it to 420 ⁰C in, the absence of air and light. Unlike white phosphorus it is not poisonous and does not show, Phosphorescence. It also does not ignite at low temperatures. The red phosphorus can be, converted back into white phosphorus by boiling it in an inert atmosphere and condensing, the vapour under water., The phosphorus has a layer structure and also acts as a semiconductor. The four atoms in, phosphorus have polymeric structure with chains of P4 linked tetrahedrally. Unlike nitrogen, P≡P is less stable than P-P single bonds. Hence, phosphorus atoms are linked through single, bonds rather than triple bonds. In addition to the above two more allotropes namely scarlet, and violet phosphorus are also known for phosphorus., , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, , P, P, , Figure 3.3 Structure of red phosphorus, 3.1.8 Properties of phosphorus, Phosphorus is highly reactive and has the following important chemical properties, Reaction with oxygen: Yellow phosphorus readily catches fire in air giving dense white, fumes of phosphorus pentoxide. Red phosphorus also reacts with oxygen on heating to give, phosphorus trioxide or phosphorus pentoxide., 67, , XII U3-P-block.indd 67, , 2/19/2020 4:40:19 PM

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www.tntextbooks.in, , pm which is shorter than the single bond distance of, P-O (184 pm) due to pπ-dπ bonding and results in, considerable double bond character., , O, P1020, , O, , In P4O10 each P atoms form three bonds to oxygen, atom and also an additional coordinate bond with an, oxygen atom., , m, 143 p, , O, , Terminal P-O bond length is 143 pm, which is less, than the expected single bond distance. This may be, due to lateral overlap of filled p orbitals of an oxygen, atom with empty d orbital on phosphorous., , 160 pm, , O, , P, , O 1230, P, O, O, O, , O, P, , O, , Figure 3.7 Structure of P4O10, , Oxoacids of Phosphorous-Structure:, , Name, , Structure, , Formula, , H, Hypophosphorous, acid, , H3PO2, , P, , H, , OH, , O, O, Orthophosphrous, acid, , H3PO3, , P, , HO, , OH, , H, Hypophosphoric, acid, , H4P2O6, , HO, , O, , O, , P, , P, , HO, , OH, , OH, , O, Orthophosphoric, acid, , H3PO4, , HO, , P, , OH, , OH, O, Pyrophosphoric, acid, , H4P2O7, , HO, , P, HO, , O, O, , P, , OH, , OH, , 72, , XII U3-P-block.indd 72, , 2/19/2020 4:40:44 PM

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www.tntextbooks.in, , Atomic radius (Å) 1.52, , 1.80, , 1.90, , 2.06, , 1.97, , Density (g.cm-3 at, 1.3 x 10-3, 293 K), , 2.07, , 4.81, , 6.23, , 9.20, , Melting point (K), , 54, , 388, , 494, , 723, , 527, , Boiling point (K), , 90, , 718, , 958, , 1261, , 1235, , 3.2 Oxygen:, Preparation: The atmosphere and water contain 23% and 83% by mass of oxygen respectively., Most of the world’s rock contain combined oxygen. Industrially oxygen is obtained by, fractional distillation of liquefied air. In the laboratory, oxygen is prepared by one of the, following methods., The decomposition of hydrogen peroxide in the presence of catalyst (MnO2) or by, oxidation with potassium permanganate., , , 2H 2 O 2 , ,  2H 2 O + O 2, , 5H 2 O 2 + 2MnO 4 − + 6H + , → 5O 2 + 8H 2 O + 2Mn 2+, , The thermal decomposition of certain metallic oxides or oxoanions gives oxygen., ∆, 2HgO , → 2Hg + O 2, ∆, 2BaO 2 , → 2BaO + O 2, ∆, 2KClO3 MnO, , → 2KCl + 3O 2, 2, , 2KNO3 ∆, → 2KNO 2 + O 2, , Properties, Under ordinary condition oxygen exists as a diatomic gas. Oxygen is paramagnetic. Like, nitrogen and fluorine, oxygen form strong hydrogen bonds. Oxygen exists in two allotropic, forms namely dioxygen (O2) and ozone or trioxygen (O3). Although negligible amounts of, ozone occurs at sea level it is formed in the upper atmosphere by the action of ultraviolet, light. In the laboratory ozone is prepared by passing electrical discharge through oxygen. At, a potential of 20,000 V about 10% of oxygen is converted into ozone it gives a mixture known, as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of, liquefied ozonised oxygen., , , O , , , , 2, Oxygen, , 2(O), atomic oxygen, , , , O 2 + (O) , ,  O3, , Ozone, , The ozone molecule has, a bent shape and symmetrical, with delocalised bonding, between the oxygen atoms., , Figure 3.8, , Structure of ozone, , 74, , XII U3-P-block.indd 74, , 2/19/2020 4:40:51 PM

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www.tntextbooks.in, , Chemical properties:, The chemical properties of oxygen and ozone differ vastly. Oxygen combines with, many metals and non-metals to form oxides. With some elements such as s-block elements, combination of oxygen occurs at room temperature. Some of less reactive metals react when, powdered finely and made to react exothermically with oxygen at room temperature but a, lump of metal is unaffected under same condition. These finely divided metals are known as, pyrophoric and when set the powder on fire, heat is liberated during a reaction., On the other hand ozone is a powerful oxidising agent and it reacts with many substances, under conditions where oxygen will not react. For example, it oxidises potassium iodide to, iodine. This reaction is quantitative and can be used for estimation of ozone., O3 + 2KI + H 2 O , → 2KOH + O 2 + I 2, , Ozone is commonly used for oxidation of organic compounds. In acidic solution ozone, exceeds the oxidising power of fluorine and atomic oxygen. The rate of decomposition of, ozone drops sharply in alkaline solution., Uses:, 1. Oxygen is one of the essential component for the survival of living organisms., 2. It is used in welding (oxyacetylene welding), 3. Liquid oxygen is used as fuel in rockets etc..., 3.2.1 Allotrophic forms of sulphur, Sulphur exists in crystalline as well as amorphous allotrophic forms. The crystalline, form includes rhombic sulphur (α sulphur) and monoclinic sulphur (β sulphur). Amorphous, allotropic form includes plastic sulphur (γ sulphur), milk of sulphur and colloidal sulphur., Rhombic sulphur also known as α sulphur, is the only thermodynamically stable allotropic, form at ordinary temperature and pressure. The crystals have a characteristic yellow colour, and composed of S8 molecules. When heated slowly above 96 ⁰C, it converts into monoclinic, sulphur. Upon cooling below 96 ⁰C the β form converts back to α form. Monoclinic sulphur, also contains S8 molecules in addition to small amount of S6 molecules. It exists as a long, needle like prism and is also called as prismatic sulphur. It is stable between 96 ⁰C - 119 ⁰C, and slowly changes into rhombic sulphur., When molten sulphur is poured into cold water a yellow rubbery ribbon of plastic sulphur, is produced. They are very soft and can be stretched easily. On standing (cooling slowly) it, slowly becomes hard and changes to stable rhombic sulphur., Sulphur also exists in liquid and gaseous states. At around 140 ⁰C the monoclinic sulphur, melts to form mobile pale yellow liquid called λ sulphur. The vapour over the liquid sulphur, consists of 90 % of S8, S7 & S6 and small amount of mixture of S2, S3, S4, S5 molecules., 75, , XII U3-P-block.indd 75, , 2/19/2020 4:40:52 PM

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www.tntextbooks.in, , Name, , Molecular Formula, , Structure, , O, Disulphuric acid or, pyrosulphuric acid, , O, , HO S O S OH, , H2S2O7, , O, , O, , O, Peroxymono sulphuric acid, (Caro's acid), , HO S O OH, , H2SO5, , O, O, Peroxodisulphuric acid., Marshall’s acid, , O, , HO S O O, , H2S2O8, , O, , S OH, O, , O O, Dithionic acid, , HO S S OH, , H2S2O6, , O O, O, Polythionic acid, , O, , HO S (S)n S OH, O, O, , H2Sn+2O6, , 3.3 Group 17 (Halogen group) elements:, 3.3.1 Chlorine, Occurrence:, The halogens are present in combined form as they are highly reactive. The main source, of fluorine is fluorspar or fluorite. The other ores of fluorine are cryolite, fluroapatite. The main, source of chlorine is sodium chloride from sea water. Bromides and iodides also occur in sea, water., Physical properties:, The common physical properties of the group 17 elements are listed in the table., , 81, , XII U3-P-block.indd 81, , 2/19/2020 4:41:21 PM

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www.tntextbooks.in, , 3.3.1 Manufacture of chlorine:, Chlorine is manufactured by the electrolysis of brine in electrolytic process or by oxidation, of HCl by air in Deacon’s process., Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na+ and Cl ions are, formed. Na+ ion reacts with OH ions of water and forms sodium hydroxide. Hydrogen and, chlorine are liberated as gases., NaCl , → Na + + Cl−, H 2O , → H + OH, +, , −, , −, , Na + OH , → NaOH, +, , At the cathode,, , At the anode,, , H + + e − , →H, , Cl− , → Cl + e −, , H + H , → H2, , Cl + Cl , → Cl2, , Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up, a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are, placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber., 0, , 400 C, 4HCl + O2 Cu, , → 2H 2O + 2Cl2 ↑, Cl, 2, , 2, , The chlorine obtained by this method is dilute and is employed for the manufacture of, bleaching powder. The catalysed reaction is given below,, 2Cu 2 Cl2 + O 2 , → 2Cu 2 OCl2, Cuprous oxy chloride, , → 2CuCl2 + H 2 O, Cu 2 OCl2 + 2HCl , Cupric chloride, , 2CuCl2 , → Cu 2 Cl2 + Cl2, Cuprous chloride, , Physical properties:, Chlorine is a greenish yellow gas with a pungent irritating odour. It produces headache, when inhaled even in small quantities whereas inhalation of large quantities could be fatal. It, is 2.5 times heavier than air., Chlorine is soluble in water and its solution is referred as chlorine water. It deposits, greenish yellow crystals of chlorine hydrate (Cl2.8H2O). It can be converted into liquid (Boiling, point – 34.6° C) and yellow crystalline solid (Melting point -102° C), Chemical properties:, Action with metals and non-metals: It reacts with metals and non metals to give the, corresponding chlorides., 2Na + Cl2 , → 2NaCl, 2Fe + 3Cl2 , → 2FeCl3, 2Al + 3Cl2 , → 2AlCl3, Cu + Cl2 , → CuCl2, H 2 + Cl2 , → 2HCl ; ∆H = − 44kCal, , 83, , XII U3-P-block.indd 83, , 2/19/2020 4:41:35 PM

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www.tntextbooks.in, , AuCl4 + NO + 2H2O, , Au + 4H+ + NO3 + 4 Cl, 3Pt + 16H+ + 4NO3 + 18Cl, , 3[PtCl6]2- + 4NO + 8H2O, , Uses of hydrochloric acid:, 1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose, from corn starch etc.,, 2. It is used in the extraction of glue from bone and also for purification of bone black, 3.3.3 Trends in physical and chemical properties of hydrogen halides:, Preparation:, Direct combination is a useful means of preparing hydrogen chloride. The reaction, between hydrogen and fluorine is violent while the reaction between hydrogen and bromine, or hydrogen and iodine are reversible and don’t produce pure forms., Displacement reactions:, Concentrated sulphuric acid displaces hydrogen chloride from ionic chlorides. At higher, temperatures the hydrogen sulphate formed react with further ionic chloride. Displacement, can be used for the preparation of hydrogen fluorides from ionic fluorides. Hydrogen bromide, and hydrogen iodide are oxidised by concentrated sulphuric acid and can’t be prepared in this, method., Hydrolysis of phosphorus trihalides:, Gaseous hydrogen halides are produced when water is added in drops to phosphorus tri, halides except phosphorus trifluoride., PX 3 + 3H 2 O , → H 3 PO3 + 3HX, , Hydrogen bromide may be obtained by adding bromine dropwise to a paste of red, phosphorous and water while hydrogen iodide is conveniently produced by adding water, dropwise to a mixture of red phosphorous and iodine., 2P + 3X 2 , → 2PX 3, 2PX 3 + 3H 2 O , → H 3PO3 + 3HX, (where X=Br or I), , Any halogen vapours which escapes with the hydrogen halide is removed by passing the, gases through a column of moist red phosphorous., From covalent hydrides:, Halogens are reduced to hydrogen halides by hydrogen sulphide., H 2S + X 2 , → 2HX + S, , Hydrogen chloride is obtained as a by-product of the reactions between hydrocarbon of, halogens., 87, , XII U3-P-block.indd 87, , 2/19/2020 4:41:57 PM

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www.tntextbooks.in, , Table 3.4: General Properties:, HF, , HCl, , HBr, , HI, , Bond dissociation enthalphy(KJmol-1) +562, , +431, , +366, , +299, , % of ionic character, , 17, , 13, , 7, , 43, , In line with the decreasing bond dissociation enthalpy, the thermal stability of hydrogen, halides decreases from fluoride to iodide., For example, Hydrogen iodide decomposes at 400° C while hydrogen fluoride and, hydrogen chloride are stable at this temperature., At room temperature, hydrogen halides are gases but hydrogen fluoride can be readily, liquefied. The gases are colourless but, with moist air gives white fumes due to the production, of droplets of hydrohalic acid. In HF, due to the presence of strong hydrogen bond it has high, melting and boiling points. This effect is absent in other hydrogen halides., Acidic properties:, The hydrogen halides are extremely soluble in water due to the ionisation., HX + H 2 O , → H 3O + + X −, , (X – F, Cl, Br, or I), Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids., Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are, therefore strong acids but HF is a weak acid i.e. 0.1mM solution is only 10% ionised, but in, 5M and 15M solution HF is stronger acid due to the equilibrium., +, −, , HF + H 2 O , ,  H 3O + F, −, , HF + F− , ,  HF2, , At high concentration, the equilibrium involves the removal of fluoride ions is important., Since it affects the dissociation of hydrogen fluoride and increases and hydrogen ion, concentration Several stable salts NaHF2, KHF2 and NH4HF2 are known. The other hydrogen, halides do not form hydrogen dihalides., Hydrohalic acid shows typical acidic properties. They form salts with acids, bases and, reacts with metals to give hydrogen. Moist hydrofluoric acid (not dry) rapidly react with silica, and glass., SiO 2 + 4HF , → SiF4 + 2H 2 O, , Na 2SiO3 + 6HF , → Na 2SiF6 + 3H 2 O, , Oxidation: Hydrogen iodide is readily oxidised to iodine hence it is a reducing agent., +, −, , 2HI , ,  2H + I 2 + 2e, , Acidic solution of iodides is readily oxidised. A positive result is shown by liberation of, iodine which gives a blue-black colouration with starch., 88, , XII U3-P-block.indd 88, , 2/19/2020 4:42:01 PM

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www.tntextbooks.in, , Hydrogen bromide is more difficult to oxidise than HI. HBr reduces slowly H2SO4 into SO2, 2HBr + H 2SO 4 , → 2H 2 O + Br2 + SO 2, , But hydrogen iodide and ionic iodides are rapidly reduced by H2SO4 into H2S and not into SO2., 8HI + H 2SO 4 , → 4H 2 O + 4I 2 + H 2S, , Reducing property of hydrogen iodide can be also explained by using its reaction with, alcohols into ethane. It converts nitric acid into nitrous acid and dinitrogen dioxide into, ammonium., Hydrogen chloride is unaffected by concentrated sulphuric acid but affected by only strong, oxidising agents like MnO2, potassium permanganate or potassium chloride., To summarize the trend,, , Table 3.5, , Property, , Order, , Reactivity of hydrogen, , Decreases from fluorine to iodine, , Stability, , Decreases from HF to HI, , Volatility of the hydrides, , HF < HI < HBr < HCl, , Thermal stability, , HF > HI > HBr > HCl, , Boiling point, , HCl < HBr < HI, , Acid strength, , Increases from HF to HI, , 3.3.4 Inter halogen compounds:, Each halogen combines with other halogens to form a series of compounds called inter, halogen compounds. In the given table of inter halogen compounds a given compound A is, less electronegative than B., Table 3.6, , AB, , AB3, , AB5, , AB7, , ClF, , ClF3, , IF5, , IF7, , BrF, , BrF3, , BrF5, , IF, , IF3, , BrCl, , ICl3, , ICl, IBr, Properties of inter halogen compounds:, i. The central atom will be the larger one, ii. It can be formed only between two halogen and not more than two halogens., 89, , XII U3-P-block.indd 89, , 2/19/2020 4:42:02 PM

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www.tntextbooks.in, , Br, , Br2O, , BrO2, , -, , -, , -, , I, , -, , -, , I2O5, , -, , -, , I4O9, I2O4 (+4), , 3.3.6 Oxoacids of halogens:, Chlorine forms four types of oxoacids namely hypochlorus acid, chlorous acid, chloric, acid and perchloric acid. Bromine and iodine forms the similar acids except halous acid., However, flurine only forms hypofulric acid. The oxidizing power oxo acids follows the order:, HOX > HXO2 > HXO3 > HXO4, Table 3.9, Type, , HOX, , HXO2, , HXO3, , HXO4, , Halous acid, , Halic acid, , Perhalic acid, , X = Cl, Br and I, Common Name Hypohalous acid, Oxidation state, , +1, , +3, , +5, , +7, , F, , HOF, , -, , -, , -, , Cl, , HOCl, , HClO2, , HClO3, , HClO4, , Br, , HOBr, , HBrO3, , HBrO4, , I, , HOI, , HIO3, , HIO4, , 3.4 Group 18 (Inert gases) elements:, 3.4.1 Occurrence:, , All the noble gases occur in the atmosphere., Physical properties:, As we move along the noble gas elements, their atomic radius and boiling point increases, from helium to radon. The first ionization energy decreases from helium to radon. Noble gases, have the largest ionisation energy compared to any other elements in a given row as they have, completely filled orbital in their outer most shell. They are extremely stable and have a small, tendency to gain or lose electrons. The common physical properties of the group 18 elements, are listed in the Table., Property, , Neon, , Argon, , Krypton, , Xenon, , Radon, , Physical state at, 293 K, , Gas, , Gas, , Gas, , Gas, , Gas, , Atomic Number, , 10, , 18, , 36, , 54, , 86, , 91, , XII U3-P-block.indd 91, , 2/19/2020 4:42:05 PM

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www.tntextbooks.in, , Xenon reacts with PtF6 and gave an orange yellow solid [XePtF6] and this is insoluble in, CCl4., Xenon difluoride forms addition compounds XeF2.2SbF5 and XeF2.2TaF5. Xenon hexa fluorides, forms compound with boron and alkali metals. Eg : XeF6.BF3, XeF6MF, M-alkali metals., There is some evidence for existence of xenon dichloride XeCl2., Krypton form krypton difluoride when an electric discharge is passed through Kr and fluorine, at 183° C or when gases are irradiated with SbF5 it forms KrF2.2SbF3., Table 3.11 Structures of compounds of Xenon:, Compound, , Hybridisation Shape / Structure, , XeF, , sp3d, , Linear, , XeF4, , sp3d2, , Square planar, , XeF6, , sp3d3, , Distorted octahedron, , XeOF2, , sp3d, , T Shaped, , XeOF4, , sp3d2, , Square pyramidal, , XeO3, , sp3, , Pyramidal, , Uses of noble gases:, The inertness of noble gases is an important feature of their practical uses., Helium:, 1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents, the painful dangerous condition called bends., 2. Helium is used to provide inert atmosphere in electric arc welding of metals, 3. Helium has lowest boiling point hence used in cryogenics (low temperature science)., 4. It is much less denser than air and hence used for filling air balloons, Neon:, Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing, electric current through neon gas under low pressure., Argon:, Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs, Krypton:, Krypton is used in fluorescent bulbs, flash bulbs etc..., Lamps filed with krypton are used in airports as approaching lights as they can penetrate, through dense fog., 93, , XII U3-P-block.indd 93, , 2/19/2020 4:42:11 PM

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www.tntextbooks.in, , Xenon:, Xenon is used in fluorescent bulbs, flash bulbs and lasers., Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed, electronic flash bulbs used by photographers, Radon:, Radon is radioactive and used as a source of gamma rays, Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e., cancer growth, Summary, „„ Occurrence: About 78 % of earth atmosphere contains dinitorgen (N2) gas. It is also, , present in earth crust as sodium nitrate (Chile saltpetre) and potassium nitrates, (Indian saltpetre)., , „„ Nitrogen, the principle gas of atmosphere (78 % by volume) is separated industrially, , from liquid air by fractional distillation, , „„ Ammonia is formed by the hydrolysis of urea., „„ Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate, , with concentrated sulphuric acid., , „„ In most of the reactions, nitric acid acts as an oxidising agent. Hence the oxidation, , state changes from +5 to a lower one. It doesn’t yield hydrogen in its reaction with, metals., , „„ The reactions of metals with nitric acid are explained in 3 steps as follows:, ▶▶ Primary reaction: Metal nitrate is formed with the release of nascent hydrogen, ▶▶ Secondary reaction: Nascent hydrogen produces the reduction products of, , nitric acid., , ▶▶ Tertiary reaction: The secondary products either decompose or react to give, , final products, , „„ Phosphorus has several allotropic modification of which the three forms namely, , white, red and black phosphorus are most common., , „„ yellow phosphorus is poisonous in nature and has a characteristic garlic smell. It, , glows in the dark due to oxidation which is called phosphorescence., , „„ Yellow phosphorus readily catches fire in air giving dense white fumes of phosphorus, , pentoxide., , „„ Phosphine is prepared by action of sodium hydroxide with white phosphorous in an, , inert atmosphere of carbon dioxide or hydrogen., 94, , XII U3-P-block.indd 94, , 2/19/2020 4:42:11 PM

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www.tntextbooks.in, , „„ Phosphine is used for producing smoke screen as it gives large smoke., „„ When a slow stream of chlorine is passed over white phosphorous, phosphorous, , trichloride is formed., , „„ phosphorus trichloride: and Phosphorous pentachloride are used as a chlorinating, , agent, , „„ Oxygen is paramagnetic. It exists in two allotropic forms namely dioxygen (O2), , and ozone or trioxygen (O3)., , „„ Ozone is commonly used for oxidation of organic compounds., „„ Sulphur exists in crystalline as well as amorphous allotrophic forms. The crystalline, , form includes rhombic sulphur (α sulphur) and monoclinic sulphur (β sulphur)., Amorphous allotropic form includes plastic sulphur (γ sulphur), milk of sulphur, and colloidal sulphur., , „„ Sulphuric acid can be manufactured by lead chamber process, cascade process or, , contact process., , „„ When dissolved in water, it forms mono (H2SO4.H2O) and dihydrates (H2SO4.2H2O), , and the reaction is exothermic., , „„ Halogens are present in combined form as they are highly reactive., „„ Chlorine is manufactured by the electrolysis of brine in electrolytic process or by, , oxidation of HCl by air in Deacon’s process., , „„ Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen., „„ When three parts of concentrated hydrochloric acid and one part of concentrated, , nitric acid are mixed, Aquaregia (Royal water) is obtained. This is used for, dissolving gold, platinum etc..., , „„ Hydrogen halides are extremely soluble in water due to the ionisation., „„ Each halogen combines with other halogens to form a series of compounds called, , inter halogen compounds., , „„ Fluorine reacts readily with oxygen and forms difluorine oxide (F2O) and difluorine, , dioxide (F2O2) where it has a -1 oxidation state., , „„ All the noble gases occur in the atmosphere., „„ They are extremely stable and have a small tendency to gain or lose electrons., „„ Sodium per xenate is very much known for its strong oxidizing property., „„ The inertness of noble gases is an important feature of their practical uses., , 95, , XII U3-P-block.indd 95, , 2/19/2020 4:42:11 PM

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www.tntextbooks.in, , 8. The molarity of given orthophosphoric acid solution is 2M. its normality is, a) 6N, , b) 4N, , c) 2N, , d) none of these, , 9. Assertion : bond dissociation energy of fluorine is greater than chlorine gas, Reason: chlorine has more electronic repulsion than fluorine, a) Both assertion and reason are true and reason is the correct explanation of assertion., b) Both assertion and reason are true but reason is not the correct explanation of, assertion., c) Assertion is true but reason is false., d) Both assertion and reason are false., 10. Among the following, which is the strongest oxidizing agent?, a) Cl2, , b) F2, , c) Br2, , d) l2, , 11. The correct order of the thermal stability of hydrogen halide is, a) HI > HBr > HCl > HF, , b) HF > HCl > HBr > HI, , c) HCl > HF > HBr > HI, , d) HI > HCl > HF > HBr, , 12. Which one of the following compounds is not formed?, a) XeOF4, , b) XeO3, , c) XeF2, , d) NeF2, , 13. Most easily liquefiable gas is, a) Ar, , b) Ne, , c) He, , d) Kr, , 14. XeF6 on complete hydrolysis produces, a) XeOF4, , b) XeO2F2, , c) XeO3, , d) XeO2, , 15. Which of the following is strongest acid among all?, a) HI, , b) HF, , c) HBr, , d) HCl, , 97, , XII U3-P-block.indd 97, , 2/19/2020 4:42:12 PM

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www.tntextbooks.in, , UNIT, , 4, , TRANSITION AND, INNER TRANSITION, ELEMENTS, , Martin Heinrich Klaproth,, (1743— 1817), , Learning Objectives, After studying this unit, the students will, be able to, , Martin Heinrich Klaproth,, German, chemist, who, discovered, uranium,, zirconium and cerium . He, described them as distinct, elements, though he did, not obtain them in the pure, metallic state. He verified, the discoveries of titanium,, tellurium, and strontium, His, role is the most significant, in systematizing analytical, chemistry and mineralogy., , , , , , , , , , , , , , , , , , recognise the position of d and f block, elements in the periodic table, describe the general trend in properties, of elements of 3d series, discuss the trends in Mn+/M standard, electrode potential, predict the oxidising and reducing, property based in Eo values, explain the tendencies of d-block, elements towards the formation of alloy,, complex and interstitial compounds, describe the preparation and properties, of potassium permanganate and, potassium dichromate, describe the properties of f-block, elements, compare the properties of lanthanoides, and actinides, , 100, , XII U4-D-Block-Jerald Folder.indd 100, , 2/19/2020 4:40:15 PM

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www.tntextbooks.in, , INTRODUCTION, Generally the metallic elements that have incompletely filled d or f sub shell in the, neutral or cationic state are called transition metals. This definition includes lanthanoides, and actinides. However, IUPAC defines transition metal as an element whose atom has an, incomplete d sub shell or which can give rise to cations with an incomplete d sub shell. They, occupy the central position of the periodic table, between s and p block elements, and their, properties are transitional between highly reactive metals of s block and elements of p block, which are mostly non metals. Except group- 11 elements all transition metals are hard and, have very high melting point., Transition metals, iron and copper play an important role in the development of human, civilization. Many other transition elements also have important applications such as tungsten, in light bulb filaments, titanium in manufacturing artificial joints, molybdenum in boiler, plants, platinum in catalysis etc. They also play vital role in living system, for example iron in, hemoglobin, cobalt in vitamin B12 etc.,, , In this unit we study the general trend in properties of d block elements with specific reference, to 3d series, their characteristics, chemical reactivity, some important compounds KMnO4 and, K2Cr2O7, we also discuss the f-block elements later in this unit., , 4.1 Position of d- block elements in the periodic table:, We have already learnt the periodic classification of elements in XI std. the transition, metals occupy from group –3 to group-12 of the modern periodic table., p-Block, , s-Block, , hydrogen, , helium, , H, , He, , 1, , 2, , 1.0079, , 4.0026, , lithium, , beryllium, , boron, , carbon, , nitrogen, , oxygen, , fluorine, , neon, , Li, , Be, 9.0122, , B, , 10.811, , C, , 12.011, , N, , 14.007, , O, , 15.999, , F, , 18.998, , Ne, , sodium, , magnesium, , aluminium, , silicon, , phosphorus, , sulfur, , chlorine, , argon, , Al, , Si, 28.086, , P, , 30.974, , S, , 32.065, , gallium, , germanium, , arsenic, , selenium, , 3, , 6.941, , 11, , 4, , 5, , 12, , Na Mg, 22.990, , 24.305, , potassium, , calcium, , 19, , K, , 39.098, rubidium, , 37, , d-Block, scandium, , 20, , 21, , Ca Sc, 40.078, , 44.956, , strontium, , yttrium, , 38, , 39, , Rb Sr, , Y, , 85.468, , 87.62, , 88.906, , caesium, , barium, , lanthanum, , 55, , 56, , titanium, , vanadium, , Ti, , V, , 22, , 23, , 47.867, , 50.942, , zirconium, , niobium, , 40, , 41, , 132.91, , 137.33, , radium, , 87, , 88, , 24, , manganese, , 25, , iron, , 26, , 26.982, , cobalt, , nickel, , 27, , 28, , copper, , 29, , zinc, , 30, , 31, , [226], , 15, , 16, , 33, , 34, , 51.996, , 54.938, , 55.845, , 58.933, , 58.693, , 63.546, , 65.38, , 69.723, , molybdenum, , technetium, , ruthenium, , rhodium, , palladium, , silver, , cadmium, , indium, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 72.64, , 74.922, , 78.96, , tin, , antimony, , tellurium, , 50, , 51, , 52, , Sn Sb Te, , 9, , 17, , 10, , 20.180, , 18, , Cl Ar, 35.453, , 39.948, , bromine, , krypton, , 35, , 36, , Br Kr, 79.904, , 83.798, , iodine, , xenon, , I, , Xe, radon, , 53, , 92.906, , 95.96, , [98], , 101.07, , 102.91, , 106.42, , 107.87, , 112.41, , 114.82, , tantalum, , tungsten, , rhenium, , osmium, , iridium, , platinum, , gold, , mercury, , thallium, , W Re Os, , Ir, , Pt Au Hg Tl Pb Bi Po At Rn, , 73, , 138.91, actinium, , 178.49, , 180.95, , rutherfordium, , dubnium, , 104, , 105, , 74, , 75, , 76, , 77, , 183.84, , 186.21, , 190.23, , 192.22, , seaborgium, , bohrium, , hassium, , meitnerium, , 106, , 107, , 108, , 109, , 78, , 79, , 80, , 81, , 200.59, , 204.38, , 207.2, , 208.98, , [209], , Copernicium, , Nahonium, , Flerovium, , Mascovium, , Livermorium, , 111, , 112, , 113, , [277], , [268], , [271], , [272], , [285], , [286], , cerium, , praseodymium, , neodymium, , promethium, , samarium, , europium, , gadolinium, , terbium, , dysprosium, , holmium, , 62, , 63, , astatine, , 84, , 196.97, , [264], , 61, , 126.90, , polonium, , 83, , roentgenium, , 110, , [266], , 60, , 127.60, , bismuth, , 82, , 195.08, , [262], , 59, , 121.76, , lead, , darmstadtium, , [261], , 58, , 118.71, , 54, , 91.224, , 72, , [227], , 32, , 8, , Cr Mn Fe Co Ni Cu Zn Ga Ge As Se, , Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh, [223], , 14, , 7, , hafnium, , 57, , 89, , chromium, , Zr Nb Mo Tc Ru Rh Pd Ag Cd In, , Cs Ba La Hf Ta, francium, , 13, , 6, , 64, , 65, , 66, , 67, , 114, , 115, , 116, , Fl Mc Lv, [289], , [289], , [293], , erbium, , thulium, , ytterbium, , 68, , 69, , 70, , 85, , [210], , Tennessine, , 117, , 131.29, , 86, , [222], , Oganessom, , 118, , Ts Og, [294], , [294], , lutetium, , 71, , Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu, 140.12, , 140.91, , 144.24, , [145], , 150.36, , 151.96, , 157.25, , 158.93, , 162.50, , 164.93, , 167.26, , 168.93, , 173.05, , 174.97, , thorium, , protactinium, , uranium, , neptunium, , plutonium, , americium, , curium, , berkelium, , californium, , einsteinium, , fermium, , mendelevium, , nobelium, , lawrencium, , 90, , 91, , Th Pa, 232.04, , 231.04, , 92, , U, , 238.03, , 93, , 94, , 95, , 96, , 97, , 98, , Np Pu Am Cm Bk Cf, [237], , [244], , [243], , [247], , [247], , [251], , 99, , 100, , 101, , 102, , 103, , Es Fm Md No Lr, [252], , [257], , [258], , [259], , [262], , f-Block, , Figure 4.1-Position of d- block elements in the periodic table, 101, , XII U4-D-Block-Jerald Folder.indd 101, , 2/19/2020 4:40:17 PM

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www.tntextbooks.in, , d- Block elements composed of 3d series (4th period) Scandium to Zinc ( 10 elements), 4d, series ( 5th period) Yttrium to Cadmium ( 10 elements) and 5d series ( 6th period) Lanthanum,, Haffinium to mercury. As we know that the group-12 elements Zinc, Cadmium and Mercury, do not have partially filled d-orbital either in their elemental state or in their normal oxidation, states. However they are treated as transition elements, because their properties are an, extension of the properties of the respective transition elements. As per the IUPAC definition,, the seventh period elements, starting from Ac, Rf to Cn also belong to transition metals. All of, them are radioactive. Except Actinium; all the remaining elements are synthetically prepared, and have very low half life periods., , 4.2 Electronic configuration:, We have already learnt in XI STD to write the electronic configuration of the elements, using Aufbau principle, Hund’s rule etc. According to Aufbau principle, the electron first fills, the 4s orbital before 3d orbital. Therefore filling of 3d orbital starts from Sc, its electronic, configuration is [Ar]3d1 4s2 and the electrons of successive elements are progressively filled, in 3d orbital and the filling of 3d orbital is complete in Zinc, whose electronic configuration, is [Ar] 3d10 4s2. However, there are two exceptions in the above mentioned progressive filling, of 3d orbitals; if there is a chance of acquiring half filled or fully filled 3d sub shell, it is given, priority as they are the stable configuration, for example Cr and Cu., The electronic configurations of Cr and Cu are [Ar] 3d5 4s1 and [Ar] 3d10 4s1 respectively., The extra stability of half filled and fully filled d orbitals, as already explained in XI STD, is due, to symmetrical distribution of electrons and exchange energy., Note: The extra stability due to symmetrical distribution can also be visualized as follows., When the d orbitals are considered together, they will constitute a sphere. So the half filled and, fully filled configuration leads to complete symmetrical distribution of electron density. On, the other hand, an unsymmetrical distribution of electron density as in the case of partially, filled configuration will result in building up of a potential difference. To decrease this and to, achieve a tension free state with lower energy, a symmetrical distribution is preferred., With these two exceptions and minor variation in certain individual cases, the general, , electronic configuration of d- block elements can be written as [Noble gas] ( n −1) d1−10 ns1−2,, Here, n = 4 to 7 . In periods 6 and 7, (except La and Ac) the configuration includes, ((n −2 ) f orbital ; [Noble gas] ( n −2 ) f 14 ( n −1) d1−10ns1−2 ., , 4.3 General trend in properties:, 4.3.1 Metallic behavior:, All the transition elements are metals. Similar to all metals the transition metals are good, conductors of heat and electricity. Unlike the metals of Group-1 and group-2, all the transition, metals except group 11 elements are hard. Of all the known elements, silver has the highest, electrical conductivity at room temperature., 102, , XII U4-D-Block-Jerald Folder.indd 102, , 2/19/2020 4:40:19 PM

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www.tntextbooks.in, , Most of the transition elements are hexagonal close packed, cubic close packed or body, centrered cubic which are the characteristics of true metals., 21, Sc, HCP, , 22, Ti, HCP, , 23, V, BCC, , 24, Cr, BCC, , 25, Mn, BCC, , 26, Fe, BCC, , 27, Co, HCP, , 28, Ni, FCC, , 29, Cu, FCC, , 30, Zn, HCP, , 39, Y, HCP, , 40, Zr, HCP, , 41, Nb, BCC, , 42, Mo, BCC, , 43, Tc, HCP, , 44, Ru, HCP, , 45, Rh, FCC, , 46, Pd, FCC, , 47, Ag, FCC, , 48, Cd, HCP, , 57*, La, DHCP, , 72, Hf, HCP, , 73, Ta, BCC/, TETR, , 74, W, BCC, , 75, Re, HCP, , 76, Os, HCP, , 77, Ir, FCC, , 78, Pt, FCC, , 79, Au, FCC, , 80, Hg, RHO, , 89**, Ac, FCC, , 104, Rf, [HCP], , 105, Db, [BCC], , 106, Sg, [BCC], , 107, Bh, [HCP], , 108, Hs, [HCP], , 109, Mt, [FCC], , 110, Ds, [BCC], , 111, Rg, [BCC], , 112, Cn, [BCC], , Figure 4.2 lattice structures of 3d, 4d and 5d transition metals, As we move from left to right along the transition metal series, melting point first, increases as the number of unpaired d electrons available for metallic bonding increases, reach, a maximum value and then decreases, as the d electrons pair up and become less available for, bonding., For example, in the first series the melting point increases from Scandium (m.pt 1814K), to a maximum of 2183 K for vanadium, which is close to 2180K for chromium. However,, manganese in 3d series and Tc in 4d series have low melting point. The maximum melting, point at about the middle of transition metal series indicates that d5 configuration is favorable, for strong interatomic attraction. The following figure shows the trends in melting points of, transition elements., , V, Sc, , Cr, Fe, , Ti, , Co, Mn, , Ni, , Cu, , (K), , Zn, , Figure 4.3-Variation in melting point of 3d series elements, 103, , XII U4-D-Block-Jerald Folder.indd 103, , 2/19/2020 4:40:21 PM

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www.tntextbooks.in, , 4.3.2 Variation of atomic and ionic size:, , Atomic radus (in Å), , It is generally expected a steady decrease in atomic radius along a period as the nuclear, charge increases and the extra electrons are added to the same sub shell. But for the 3d transition, elements, the expected decrease in atomic radius is observed from Sc to V , thereafter up to, Cu the atomic radius nearly remains the same. As we move from Sc toZn in 3d series the extra, electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased, nuclear charge and hence the effective nuclear charge increases slightly. However, the extra, electrons added to the 3d, 2.5, sub shell strongly repel, 2.3, the 4s electrons and these, Sc, Ti, two forces are operated, V, Cr, Mn, Fe, 2.1, Zn, Co, Ni, in opposite direction and, Cu, as they tend to balance, 1.9, each other, it leads to, 1.7, constancy in atomic radii., , Generally as we move, down a group atomic, radius increases, the, same trend is expected, in d block elements also., As the electrons are, added to the 4d sub shell,, the atomic radii of the, 4d elements are higher, than the corresponding, elements of the 3d series., However there is an, unexpected observation, in the atomic radius of 5d, , 1.5, , Atmoic Number, , Figure 4.4 (a) Atomic radius of 3d Elements, , 2.5, , Atomic radus (in Å), , 2.3, , Y, Zr, , Nb, , Mo, , 2.1, , Tc, , Ru, , Cd, , Ag, , Pd, , Rh, , 1.9, 1.7, 1.5, , Atmoic Number, , Figure 4.4 (b) Atomic radius of 4d Elements, 2.5, 2.3, Atomic radus (in Å), , At the end of the, series, d – orbitals of, Zinc contain 10 electrons, in which the repulsive, interaction between the, electrons is more than, the effective nuclear, charge and hence, the, orbitals slightly expand, and atomic radius slightly, increases., , Lu, , Hf, , Ta, , 2.1, , W, , Re, , Os, , Hg, Ir, , Pt, , Au, , 1.9, 1.7, 1.5, , Atmoic Number, , Figure 4.4 (c) Atomic radius of 5d Elements, , 104, , XII U4-D-Block-Jerald Folder.indd 104, , 2/19/2020 4:40:23 PM

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www.tntextbooks.in, , elements which have nearly same atomic radius as that of corresponding 4d elements. This is, due to lanthanoide contraction which is to be discussed later in this unit under inner transition, elements., 4.3.3 Ionization enthalpy:, Ionization energy of transition element is intermediate between those of s and p, block elements. As we move from left to right in a transition metal series, the ionization, enthalpy increases as expected. This is due to increase in nuclear charge corresponding to, the filling of d electrons. The following figure show the trends in ionisation enthalpy of, transition elements., , The increase in first ionisation enthalpy with increase in atomic number along a particular, series is not regular. The added electron enters (n-1)d orbital and the inner electrons act as a, shield and decrease the effect of nuclear charge on valence ns electrons. Therefore, it leads to, variation in the ionization energy values., The ionisation enthalpy values can be used to predict the thermodynamic stability of their, compounds. Let us compare the ionisation energy required to form Ni2+ and Pt2+ ions., For Nickel,, , IE1 + IE2 = ( 737 + 1753 ), = 2490 kJmol, , −1, , For Platinum, IE1 + IE2 = ( 864 + 1791), = 2655 kJmol −1, Since, the energy required to form Ni2+ is less than that of Pt2+, Ni(II) compounds are, thermodynamically more stable than Pt(II) compounds., 105, , XII U4-D-Block-Jerald Folder.indd 105, , 2/19/2020 4:40:25 PM

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www.tntextbooks.in, , Evaluate yourself:, Compare the stability of Ni4+ and Pt4+ from their ionisation enthalpy values., IE, I, , Ni, , Pt, , 737, 1753, 3395, 5297, , II, III, IV, , 864, 1791, 2800, 4150, , 4.3.4 Oxidation state:, The first transition metal Scandium exhibits only +3 oxidation state, but all other transition, elements exhibit variable oxidation states by loosing electrons from (n-1)d orbital and ns, orbital as the energy difference between them is very small. Let us consider the 3d series; the, following table summarizes the oxidation states of the 3d series elements., Sc, , +3, , Ti, , +4, +3, +2, , V, , Cr, , +5, +4, +3, +2, , +6, +5, +4, +3, +2, , Mn, +7, +6, +5, +4, +3, +2, , Fe, , Co, , Ni, , +4, +3, +2, , +4, +3, +2, , Cu, , +6, , +4, +3, +2, , +2, +1, , At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation, state becomes stable., The number of oxidation states increases with the number of electrons available, and it, decreases as the number of paired electrons increases. Hence, the first and last elements show, less number of oxidation states and the middle elements with more number of oxidation states., For example, the first element Sc has only one oxidation state +3; the middle element Mn has, six different oxidation states from +2 to +7. The last element Cu shows +1 and +2 oxidation, states only., The relative stability of different oxidation states of 3d metals is correlated with the, extra stability of half filled and fully filled electronic configurations. Example: Mn2+ ( 3d 5 ) is, more stable than Mn 4+ ( 3d 3 ), The oxidation states of 4d and 5d metals vary from +3 for Y and La to +8 for Ru and Os., The highest oxidation state of 4d and 5d elements are found in their compounds with the, higher electronegative elements like O, F and Cl. for example: RuO4, OsO4 and WCl6. Generally, in going down a group, a stability of the higher oxidation state increases while that of lower, oxidation state decreases.It is evident from the Frost diagram (ΔG0 vs oxidation number) as, shown below,For titanium,vanadium and chromium, the most thermodynamically stable, oxidation state is +3. For iron, the stabilities of +3 and +2 oxidation states are similar.Copper, 106, , XII U4-D-Block-Jerald Folder.indd 106, , 2/19/2020 4:40:27 PM

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www.tntextbooks.in, , is unique in 3d series having a stable +1 oxidation state. It is prone to disproportionate to the, +2 and 0 oxidation states., Evaluate yourself:, , Ti, , +7, , 4.3.5 Standard electrode potentials of, transition metals, , Cr, , Mn, , +6, , Fe, , Co, , Ni, , Cu, , +6, , +7, , +5, , +6, , +4, ΔGo/F or -nEo (V . mol e-, , Why iron is more stable in +3, oxidation state than in +2 and the reverse, is true for Manganese?, , V, , +3, +6, , +2, , +3, , +2, , +1, 0, , 0, , 0, , 0, +5, , -1, , +4, +3, , 0, , 0, +3, +2, , +1, , 0, , 0, , 0, , +2, , +2, , +2, Redox reactions involve transfer of, -2, +4, +2, +2, +3, +3, -3 +4, electrons from one reactant to another. Such, +2, +3, -4, reactions are always coupled, which means, 0, 0, 0, 0, 0, 0, 0, 0, that when one substance is oxidised, another, Oxidation number, must be reduced. The substance which is, Figure 4.6 Frost diagram, oxidised is a reducing agent and the one, which is reduced is an oxidizing agent. The oxidizing and reducing power of an element is, measured in terms of the standard electrode potentials., , Standard electrode potential is the value of the standard emf of a cell in which molecular, hydrogen under standard pressure ( 1atm) and temperature (273K) is oxidised to solvated, protons at the electrode., If the standard electrode potential (E0), of a metal is large and negative, the metal is a, powerful reducing agent, because it loses electrons easily. Standard electrode potentials (reduction, potential) of few first transition metals are given in the following table., Reaction, Ti 2+ + 2e − →,  Ti, , Standard reduction, potential ( V ), −1.63, , −, , V + 2e →, V, , –1.19, , Cr 2+ + 2e − →,  Cr, , –0.91, , Mn2+ + 2e − →,  Mn, , –1.18, , 2+, , −, , Fe + 2e →,  Fe, , –0.44, , Co2+ + 2e − →,  Co, , –0.28, , Ni + 2e →,  Ni, , –0.23, , Cu + 2e →,  Cu, , +0.34, , 2+, , 2+, , −, , 2+, , −, , 2+, , −, , 0.5, , Cu, , 0, , Co, Fe, , 0.5, Cr, , Zn, , 1, V, , Ni, , Mn, , 1.5, Ti, , 2, Sc, , 2.5, , 3d -Series, ,  0, , Figure 4.7 (a)  E M 2+  -3d series, Zn + 2e →,  Zn, , –0.76, M, , , In 3d series as we move from Ti to Zn, the standard reduction potential  E 0 2+  value,  M M, is approaching towards less negative value and copper has a positive reduction potential. i.e.,, 107, , XII U4-D-Block-Jerald Folder.indd 107, , 2/19/2020 4:40:37 PM

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www.tntextbooks.in, , elemental copper is more stable than Cu2+. There are two deviations., In the general trend,, , , Fig shows that  E 0 2+  value for manganese and zinc are more negative than the regular,  M M, trend. It is due to extra stability which arises due to the half filled d5 configuration in Mn2+ and, completely filled d10 configuration in Zn2+., Transition metals in their high oxidation states tend to be oxidizing . For example,, Fe3+ is moderately a strong oxidant, and it oxidises copper to Cu2+ ions. The feasibility of the, reaction is predicted from the following standard electrode potential values., Fe 3+ (aq) + e −  Fe2+ E0 = 0.77V, Cu2+ (aq) + 2e −  Cu(s) E0 = +0.34 V, 3+, The standard electrode potential for the M, , half-cell gives the relative stability, M 2+, between M3+ and M2+. The reduction potential values are tabulated as below., 2.5, , Standard reduction, potential ( V ), , Reaction, , 2, , Ti 3+ + e − →,  Ti 2+, , –0.37, , , , V 3+ + e − →,  V 2+, , –0.26, , , , –0.41, , , , Cr + e →,  Cr, 3+, , −, , 2+, , Mn + e →,  Mn, 3+, , −, , Fe + e →,  Fe, 3+, , −, , 2+, , 2+, , Co3+ + e − →,  Co2+, , +1.51, , , , +0.77, , 0.5, , +1.81, , Co, Mn, , Fe, , Ti, , V, , Cr, , Figure 4.7 (b), , 3d-Series, , M3+, , M 2+, , -3d series, , The negative values for titanium, vanadium and chromium indicate that the higher, oxidation state is preferred. If we want to reduce such a stable Cr3+ ion, strong reducing agent, which has high negative value for reduction potential like metallic zinc ( E0 = − 0.76 V ) is, required., 3+, The high reduction potential of Mn Mn2+ indicates Mn2+ is more stable than Mn3+. For, Fe3+ 2+ the reduction potential is 0.77V, and this low value indicates that both Fe3+ and Fe2+ can, Fe, , exist under normal conditions. The drop from Mn to Fe is due to the electronic structure of the ions, concerned.Mn3+ has a 3d4 configuration while that of Mn2+ is 3d5. The extra stability associated with a, half filled d sub shell makes the reduction of Mn3+ very feasible (E0 = +1.51V)., , 4.3.6 Magnetic properties, Most of the compounds of transition elements are paramagnetic. Magnetic properties, are related to the electronic configuration of atoms. We have already learnt in XI STD that, 108, , XII U4-D-Block-Jerald Folder.indd 108, , 2/19/2020 4:40:47 PM

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www.tntextbooks.in, , the electron is spinning around its own axis, in addition to its orbital motion around the, nucleus. Due to these motions, a tiny magnetic field is generated and it is measured in terms, of magnetic moment. On the basis of magnetic properties, materials can be broadly classified, as (i) paramagnetic materials (ii) diamagnetic materials, besides these there are ferromagnetic, and antiferromagnetic materials., Materials with no elementary magnetic dipoles are diamagnetic, in other words a, species with all paired electrons exhibits diamagnetism. This kind of materials are repelled, by the magnetic field because the presence of external magnetic field, a magnetic induction is, introduced to the material which generates weak magnetic field that oppose the applied field., Paramagnetic solids having unpaired electrons possess magnetic dipoles which are, isolated from one another. In the absence of external magnetic field, the dipoles are arranged, at random and hence the solid shows no net magnetism. But in the presence of magnetic field,, the dipoles are aligned parallel to the direction of the applied field and therefore, they are, attracted by an external magnetic field., Ferromagnetic materials have domain structure and in each domain the magnetic dipoles, are arranged. But the spin dipoles of the adjacent domains are randomly oriented. Some, transition elements or ions with unpaired d electrons show ferromagnetism., 3d transition metal ions in paramagnetic solids often have a magnetic dipole moments, corresponding to the electron spin contribution only. The orbital moment L is said to be, quenched. So the magnetic moment of the ion is given by, µ= g, , S (S + 1) µB, , = g S (and, S + 1is, ) µB Bohr, Where S is the total spin quantum number of the unpaired µelectrons, Magneton., n, For an ion with n unpaired electrons S = and for an electron g=2, 2, Therefore the spin only magnetic moment is given by, , µ= 2, ,  n  n ,  2   2 + 1 µB, , µ= 2, ,  n ( n + 2) , ,  µB, 4, , µ=, , n ( n + 2 ) µB, , The magnetic moment calculated using the above equation is compared with the, experimental values in the following table. In most of the cases, the agreement is good., , 109, , XII U4-D-Block-Jerald Folder.indd 109, , 2/19/2020 4:40:49 PM

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www.tntextbooks.in, , Ion, , Configuration, , n, , µ=, , n ( n + 2 ) µB, , μ (observed), , Sc3+ ,Ti4+ ,V5+, , d0, , 0, , µ=, , 0 ( 0 + 2 ) = 0 µB, , diamagnetic, , Ti3+, V4+, , d1, , 1, , µ=, , 1 (1 + 2 ) = 3 = 1.73 µB, , 1.75, , Ti2+, V3+, , d2, , 2, , µ=, , 2 ( 2 + 2 ) = 8 = 2.83 µB, , 2.76, , Cr3+, Mn4+, V2+ d3, , 3, , µ=, , 3 ( 3 + 2 ) = 15 = 3.87 µB, , 3.86, , Cr2+, Mn3+, , d4, , 4, , µ=, , 4 ( 4 + 2 ) = 24 = 4.89 µB, , 4.80, , Mn2+, Fe3+, , d5, , 5, , µ=, , 5 ( 5 + 2 ) = 35 = 5.91 µB, , 5.96, , Co3+, Fe2+, , d6, , 4, , µ=, , 4 ( 4 + 2 ) = 24 = 4.89 µB, , 5.3-5.5, , Co2+, , d7, , 3, , µ=, , 3 ( 3 + 2 ) = 15 = 3.87 µB, , 4.4-5.2, , Ni2+, , d8, , 2, , µ=, , 2 ( 2 + 2 ) = 8 = 2.83 µB, , 2.9-3.4, , Cu2+, , d9, , 1, , µ=, , 1(1 + 2 ) = 3 = 1.732 µB, , 1.8-2.2, , Cu+ , Zn2+, , d10, , 0, , µ=, , 0 ( 0 + 2 ) = 0 µB, , diamagnetic, , 4.3.7 Catalytic properties, The chemical industries manufacture a number of products such as polymers, flavours,, drugs etc., Most of the manufacturing processes have adverse effect on the environment so, there is an interest for eco friendly alternatives. In this context, catalyst based manufacturing, processes are advantageous, as they require low energy, minimize waste production and, enhance the conversion of reactants to products., Many industrial processes use transition metals or their compounds as catalysts. Transition, metal has energetically available d orbitals that can accept electrons from reactant molecule or, metal can form bond with reactant molecule using its d electrons. For example, in the catalytic, hydrogenation of an alkene, the alkene bonds to an active site by using its π electrons with an, empty d orbital of the catalyst., The σ bond in the hydrogen molecule breaks, and each hydrogen atom forms a bond with, a d electron on an atom in the catalyst. The two hydrogen atoms then bond with the partially, broken π -bond in the alkene to form an alkane., H, , H, , H, , C, , Ni /H2, , C, , H, , H, , C, , H, , C, , H, , H, , H, H, , Alkane, , Alkene, , In certain catalytic processes the variable oxidation states of transition metals find, applications. For example, in the manufacture of sulphuric acid from SO3, vanadium pentoxide, 110, , XII U4-D-Block-Jerald Folder.indd 110, , 2/19/2020 4:40:58 PM

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www.tntextbooks.in, , (V2O5) is used as a catalyst to oxidise SO2. In this reaction V2O5 is reduced to vanadium (IV), Oxide (VO2)., Some more examples are discussed below,, (i) Hydroformylation of olefins, CHO, , +, , CO + H2, , Co2(CO)8, , CHO, , +, , Butan-1-al, , Propene, , 2-methylpropan-1-al, , (ii) Preparation acetic acid from acetaldehyde., CH3- CHO + CO, , Rh / Ir complex, , CH3- COOH, Acetic acid, , Acetaldehyde, , (iii) Zeigler – Natta catalyst, A mixture of TiCl4 and trialkyl aluminium is used for polymerization., H3C, , n, , CH3, , TiCl4 + Al(C2H5)3, CH, , CH2, , *, , Propylene, , CH, , CH2, , poly propylene, , *, , n, , 4.3.8 Alloy formation, An alloy is formed by blending a metal with one or more other elements. The elements, may be metals or non-metals or both. The bulk metal is named as solvent, and the other, elements in smaller portions are called solute. According to Hume-Rothery rule to form a, substitute alloy the difference between the atomic radii of solvent and solute is less than 15%., Both the solvent and solute must have the same crystal structure and valence and their electro, negativity difference must be close to zero. Transition metals satisfying these mentioned, conditions form a number of alloys among themselves, since their atomic sizes are similar and, one metal atom can be easily replaced by another metal atom from its crystal lattice to form an, alloy. The alloys so formed are hard and often have high melting points. Examples: Ferrous, alloys, gold – copper alloy, chrome alloys etc.,, 4.3.9 Formation of interstitial compounds, An interstitial compound or alloy is a compound that is formed when small atoms like, hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They, are usually non-stoichiometric compounds.Transition metals form a number of interstitial, compounds such as TiC, ZrH1.92 , Mn 4 N etc . The elements that occupy the metal lattice, provide them new properties., (i), , They are hard and show electrical and thermal conductivity, , (ii) They have high melting points higher than those of pure metals, 111, , XII U4-D-Block-Jerald Folder.indd 111, , 2/19/2020 4:40:59 PM

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www.tntextbooks.in, , (iii) Transition metal hydrides are used as powerful reducing agents, (iv) Metallic carbides are chemically inert., 4.3.10 Formation of complexes, Transition elements have a tendency to form coordination compounds with a species, that has an ability to donate an electron pair to form a coordinate covalent bond. Transition, metal ions are small and highly charged and they have vacant low energy orbitals to accept an, electron pair donated by other groups. Due to these properties, transition metals form large, number of complexes. Examples: [Fe(CN)6]4- , [Co(NH3)6]3+ , etc.., The chemistry of coordination compound is discussed in unit 5., , 4.4 important compound of Transition elements, Oxides and Oxoanions of Metals, Generally, transition metal oxides are formed by the reaction of transition metals with, molecular oxygen at high temperatures. Except the first member of 3d series, Scandium, all, other transition elements form ionic metal oxides. The oxidation number of metal in metal, oxides ranges from +2 to +7. As the oxidation number of a metal increases, ionic character, decreases, for example, Mn2O7 is covalent. Mostly higher oxides are acidic in nature, Mn2O7, dissolves in water to give permanganic acid (HMnO4 ) , similarly CrO3 gives chromic acid, (H2CrO4) and dichromic acid (H2Cr2O7). Generally lower oxides may be amphoteric or basic,, for example, Chromium (III) oxide - Cr2O3, is amphoteric and Chromium(II) oxide, CrO, is, basic in nature., Potassium dichromate K2Cr2O7, Preparation:, Potassium dichromate is prepared from chromate ore. The ore is concentrated by gravity, separation. It is then mixed with excess sodium carbonate and lime and roasted in a reverbratory, furnace., 0, , - 1000 C, 4 FeCr2O4 + 8 Na 2CO3 + 7 O2 900, , → 8 Na 2CrO4 + 2 Fe2O3 + 8 CO2 ↑, , The roasted mass is treated with water to separate soluble sodium chromate from, insoluble iron oxide. The yellow solution of sodium chromate is treated with concentrated, sulphuric acid which converts sodium chromate into sodium dichromate., 2 Na 2 CrO4 + H2 SO4 →,  Na 2 Cr2 O7 + Na 2 SO4 + H2 O, sodium chromate, (yellow), , sodium dichromate, (orange red), , The above solution is concentrated to remove less soluble sodium sulphate. The resulting, solution is filtered and further concentrated. It is cooled to get the crystals of Na2SO4.2H2O., The saturated solution of sodium dichromate in water is mixed with KCl and then, concentrated to get crystals of NaCl. It is filtered while hot and the filtrate is cooled to obtain, K2Cr2O7 crystals., 112, , XII U4-D-Block-Jerald Folder.indd 112, , 2/19/2020 4:41:01 PM

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www.tntextbooks.in, , Na Cr O + 2KCl →,  K 2Cr2O7, , 2, 2 7, sodium dichromate, (orange red), , potassium dichromate, (orange red), , + 2NaCl, , Physical properties:, Potassium dichromate is an orange red crystalline solid which melts at 671K and it is, moderately soluble in cold water, but very much soluble in hot water. On heating it decomposes, and forms Cr2O3 and molecular oxygen. As it emits toxic chromium fumes upon heating, it is, mainly replaced by sodium dichromate., → 4 K 2CrO4 + 2 Cr2O3 + 3O2 ↑, 4 K 2Cr2O7 ∆, potassium, dichromate, , potassium, chromate, , chromium(III), oxide, , Structure of dichromate ion:, , Figure 4.8 (b) Structure of, dichromate ion, , Figure 4.8 (a) Structure of, chromate ion, , Both chromate and dichromate ion are oxo anions of chromium and they are moderately, strong oxidizing agents. In these ions chromium is in +6 oxidation state. In an aqueous solution,, chromate and dichromate ions can be interconvertible, and in an alkaline solution chromate ion, is predominant, whereas dichromate ion becomes predominant in acidic solutions. Structures, of these ions are shown in the figure., Chemical properties:, 1. Oxidation, Potassium dichromate is a powerful oxidising agent in acidic medium. Its oxidising action, +, in the presence of H ions is shown below. You can note that the change in the oxidation, state of chromium from Cr6+ to Cr3+.Its oxidising action is shown below., Cr2O7 2− + 14H+ + 6e− →,  2C r 3+ + 7 H 2O, The oxidising nature of potassium dichromate (dichromate ion) is illustrated in the, following examples., 113, , XII U4-D-Block-Jerald Folder.indd 113, , 2/19/2020 4:41:02 PM

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www.tntextbooks.in, , Uses of potassium permanganate:, Some important uses of potassium permanganate are listed below., 1. It is used as a strong oxidizing agent., 2. It is used for the treatment of various skin infections and fungal infections of the foot., 3. It used in water treatment industries to remove iron and hydrogen sulphide from well, water., 4. It is used as Bayer’s reagent for detecting unsaturation in an organic compound., 5. It is used in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen, peroxide and iodides., Note HCl cannot be used for making the medium acidic since it reacts with KMnO4 as follows., 2MnO4 − + 10 Cl − + 16H+ →,  2Mn2+ + 5Cl 2 + 8H2O, HNO3 also cannot be used since it is good oxidising agent and reacts with reducing agents, in the reaction., However,H2SO4 is found to be most suitable since it does not react with potassium, permanganate., Note, Molecular weight of KMnO4, 158, Equivalent weight of KMnO4 in, =, =, = 31.6, acid medium no of mols of electrons transferred, 5, Molecular weight of KMnO4, 158, Equivalent weight of KMnO4 in, =, =, = 158, basic medium no of mols of electrons transferred, 1, Molecular weight of KMnO4, 158, Equivalent weight of KMnO4 in, =, =, = 52.67, neutral medium no of mols of electrons transferred, 3, f-block elements – Inner transition elements, In the inner transition elements there are two series of elements., 1) Lanthanoids ( previously called lanthanides), 2) Actinoids ( previously called actinides), Lanthanoid series consists of fourteen elements from Cerium (58Ce) to Lutetium (71Lu), following Lanthanum (57La).These elements are characterised by the preferential filling of, 4f orbitals, Similarly actinoids consists of 14 elements from Thorium (90Th) to Lawrencium, (103Lr) following Actinium (89Ac).These elements are characterised by the preferential filling, of 5f orbital., The position of Lanthanoids in the periodic table, The actual position of Lanthanoids in the periodic table is at group number 3 and period, 118, , XII U4-D-Block-Jerald Folder.indd 118, , 2/19/2020 4:41:33 PM

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www.tntextbooks.in, , number 6.However, in the sixth period after lanthanum, the electrons are preferentially filled, in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical, properties. Therefore these elements are grouped together and placed at the bottom of the, periodic table. This position can be justified as follows., 1. Lanthanoids have general electronic configuration [Xe] 4 f 1−14 5d 0−1 6s 2, 2. The common oxidation state of lanthanoides is +3, 3. All these elements have similar physical and chemical properties., Similarly the fourteen elements following actinium resemble in their physical and, chemical properties. If we place these elements after Lanthanum in the periodic table below, 4d series, the properties of the elements belongs to a group would be different and it would, affect the proper structure of the periodic table. Hence a separate position is provided to the, inner transition elements as shown in the figure., p-Block, , s-Block, hydrogen, , helium, , H, , He, , 1, , 2, , 1.0079, , 4.0026, , lithium, , beryllium, , boron, , carbon, , nitrogen, , oxygen, , fluorine, , neon, , Li, , Be, 9.0122, , B, , 10.811, , C, , 12.011, , N, , 14.007, , O, , 15.999, , F, , 18.998, , Ne, , sodium, , magnesium, , aluminium, , silicon, , phosphorus, , sulfur, , chlorine, , argon, , Al, , Si, 28.086, , P, , 30.974, , S, , 32.065, , gallium, , germanium, , arsenic, , selenium, , 3, , 6.941, , 11, , 4, , 5, , 12, , Na Mg, 22.990, , 24.305, , potassium, , calcium, , 19, , K, , 39.098, rubidium, , 37, , 20, , d-Block, scandium, , 21, , Ca Sc, 40.078, , 44.956, , strontium, , yttrium, , 38, , Rb Sr, , 39, , Y, , 85.468, , 87.62, , 88.906, , caesium, , barium, , lanthanum, , 55, , 56, , titanium, , vanadium, , Ti, , V, , 22, , 23, , 47.867, , 50.942, , zirconium, , niobium, , 40, , 41, , 132.91, , 137.33, , radium, , 87, , 88, , 24, , manganese, , 25, , 26.982, , iron, , cobalt, , 26, , nickel, , 27, , 28, , copper, , 29, , zinc, , 30, , 31, , [226], , 33, , 16, , 34, , 51.996, , 54.938, , 55.845, , 58.933, , 58.693, , 63.546, , 65.38, , 69.723, , molybdenum, , technetium, , ruthenium, , rhodium, , palladium, , silver, , cadmium, , indium, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 72.64, , 74.922, , 78.96, , tin, , antimony, , tellurium, , 50, , 51, , 52, , Sn Sb Te, , 9, , 17, , 10, , 20.180, , 18, , Cl Ar, 35.453, , 39.948, , bromine, , krypton, , 35, , 36, , Br Kr, 79.904, , 83.798, , iodine, , xenon, , I, , Xe, radon, , 53, , 92.906, , 95.96, , [98], , 101.07, , 102.91, , 106.42, , 107.87, , 112.41, , 114.82, , tantalum, , tungsten, , rhenium, , osmium, , iridium, , platinum, , gold, , mercury, , thallium, , W Re Os, , Ir, , Pt Au Hg Tl Pb Bi Po At Rn, , 73, , 138.91, actinium, , 178.49, , 180.95, , rutherfordium, , dubnium, , 104, , 105, , 74, , 75, , 76, , 77, , 183.84, , 186.21, , 190.23, , 192.22, , seaborgium, , bohrium, , hassium, , meitnerium, , 106, , 107, , 108, , 109, , 78, , 79, , 80, , 81, , 200.59, , 204.38, , 207.2, , 208.98, , [209], , Copernicium, , Nahonium, , Flerovium, , Mascovium, , Livermorium, , 111, , 112, , 113, , [277], , [268], , [271], , [272], , [285], , [286], , cerium, , praseodymium, , neodymium, , promethium, , samarium, , europium, , gadolinium, , terbium, , dysprosium, , holmium, , 62, , 63, , astatine, , 84, , 196.97, , [264], , 61, , 126.90, , polonium, , 83, , roentgenium, , 110, , [266], , 60, , 127.60, , bismuth, , 82, , 195.08, , [262], , 59, , 121.76, , lead, , darmstadtium, , [261], , 58, , 118.71, , 54, , 91.224, , 57, , [227], , 32, , 15, , 8, , Cr Mn Fe Co Ni Cu Zn Ga Ge As Se, , Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh, [223], , 14, , 7, , hafnium, , 72, , 89, , chromium, , Zr Nb Mo Tc Ru Rh Pd Ag Cd In, , Cs Ba La Hf Ta, francium, , 13, , 6, , 64, , 65, , 66, , 67, , 114, , 115, , 116, , Fl Mc Lv, [289], , [289], , [293], , erbium, , thulium, , ytterbium, , 68, , 69, , 70, , 85, , [210], , Tennessine, , 117, , 131.29, , 86, , [222], , Oganessom, , 118, , Ts Og, [294], , [294], , lutetium, , 71, , Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu, 140.12, , 140.91, , 144.24, , [145], , 150.36, , 151.96, , 157.25, , 158.93, , 162.50, , 164.93, , 167.26, , 168.93, , 173.05, , 174.97, , thorium, , protactinium, , uranium, , neptunium, , plutonium, , americium, , curium, , berkelium, , californium, , einsteinium, , fermium, , mendelevium, , nobelium, , lawrencium, , 90, , 91, , Th Pa, 232.04, , 231.04, , 92, , U, , 238.03, , 93, , 94, , 95, , 96, , 97, , 98, , Np Pu Am Cm Bk Cf, [237], , [244], , [243], , [247], , [247], , [251], , 99, , 100, , 101, , 102, , 103, , Es Fm Md No Lr, [252], , [257], , [258], , [259], , [262], , f-Block, , Figure 4.10 position of inner transition elements, Electronic configuration of Lanthanoids:, We know that the electrons are filled in different orbitals in the order of their increasing, energy in accordance with Aufbau principle. As per this rule after filling 5s,5p and 6s and, 4f level begin to fill from lanthanum, and hence the expected electronic configuration of, Lanthanum(La) is [Xe] 4 f 1 5d 0 6s 2 but the actual electronic configuration of Lanthanum is, 119, , XII U4-D-Block-Jerald Folder.indd 119, , 2/19/2020 4:41:37 PM

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www.tntextbooks.in, , [Xe] 4 f 0 5d1 6s 2 and it belongs to d block. Filling of 4f orbital starts from Cerium (Ce) and its, electronic configuration is [Xe] 4 f 1 5d1 6s 2 . As we move from Cerium to other elements the, additional electrons are progressively filled in 4f orbitals as shown in the table., Table : electronic configuration of Lanthanum and Lanthanoids, Atomic, number, , Symbol, , Lanthanum, , 57, , La, , [Xe] 4 f 0 5d1 6s 2, , Cerium, , 58, , Ce, , [Xe] 4 f 1 5d1 6s 2, , Praseodymium, , 59, , Pr, , [Xe] 4 f 3 5d 0 6s 2, , Neodymium, , 60, , Nd, , [Xe] 4 f 4 5d 0 6s 2, , Promethium, , 61, , Pm, , [Xe] 4 f 5 5d 0 6s 2, , Samarium, , 62, , Sm, , [Xe] 4 f 6 5d 0 6s 2, , Europium, , 63, , Eu, , [Xe] 4 f 7 5d 0 6s 2, , Gadolinium, , 64, , Gd, , [Xe] 4 f 7 5d1 6s 2, , Terbium, , 65, , Tb, , [Xe] 4 f 9 5d 0 6s 2, , Dysprosium, , 66, , Dy, , [Xe] 4 f 10 5d 0 6s 2, , Holmium, , 67, , Ho, , [Xe] 4 f 11 5d 0 6s 2, , Erbium, , 68, , Er, , [Xe] 4 f 12 5d 0 6s 2, , Thulium, , 69, , Tm, , [Xe] 4 f 13 5d 0 6s 2, , Ytterbium, , 70, , Yb, , [Xe] 4 f 14 5d 0 6s 2, , Lutetium, , 71, , Lu, , [Xe] 4 f 14 5d1 6s 2, , Name of the element, , Electronic configuration, , In Gadolinium (Gd) and Lutetium (Lu) the 4f orbitals, are half-filled and completely filled,, and one electron enters 5d orbitals. Hence the general electronic configuration of 4f series of, elements can be written as [Xe] 4 f 1−14 5d 0−1 6s 2, Oxidation state of lanthanoids:, The common oxidation state of lanthanoids is +3. In addition to that some of the, lanthanoids also show either +2 or +4 oxidation states., Gd3+ and Lu3+ ions have extra stability, it is due to the fact that they have exactly half filled, and completely filled f-orbitals respectively.their electronic c onfigurations are, Gd 3+ : [ Xe ]4 f 7, , Lu 3+ : [ Xe ]4 f 14, 120, , XII U4-D-Block-Jerald Folder.indd 120, , 2/19/2020 4:41:53 PM

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www.tntextbooks.in, , Similarly Cerium and terbium attain 4f 0 and 4f 7 configurations respectively in the +4, oxidation states. Eu2+ and Yb2+ ions have exactly half filled and completely filled f orbitals, respectively., The stability of different oxidation states has an impact on the properties of these elements., the following table shows the different oxidation states of lanthanoids., Ce, , Pr, , Nd, , Pm, , +2, +3, , +3, , +3, , +4, , +4, , +4, , +3, , Atomic and ionic radii:, As we move across, 4f series, the atomic and, ionic radii of lanthanoids, show gradual decrease, with increse in atomic, number. This decrese, in ionic size is called, lanthanoid contraction., , Sm, , Eu, , +2, , +2, , +3, , +3, , 1.04, , Gd, , Tb, , +3, , La, , 1.02, , Dy, , +3, , +3, , +4, , +4, , Ho, , Er, , +3, , +3, , Tm, , Yb, , +2, , +2, , +3, , +3, , Pr, , 0.98, , Nd, , Pm, , 0.96, , Sm, Eu, , 0.94, , Gd, Tb, , 0.92, , Dy, Ho, , 0.9, , Er, , 0.88, , Tm, Yb, , 0.86, , 55, , 56, , 57, , +3, , Ce, , 1, , 0.84, , Lu, , 58, , 59, , 60, , 61, , 62, , 63, , 64, , 65, , 66, , 67, , 68, , 69, , 70, , Lu, 71, , 72, , 73, , 74, , 75, , Figure 4.11 Variation of atomic radii of lanthanoids, Cause of lanthanoid contraction:, As we move from one element to another in 4f series ( Ce to Lu) the nuclear charge, increases by one unit and an additional electron is added into the same inner 4f sub shell. We, know that 4f sub shell have a diffused shapes and therefore the shielding effect of 4f elelctrons, relatively poor.hence, with increase of nuclear charge, the valence shell is pulled slightly towards, nucleus. As a result, the effetive nuclear charge experienced by the 4f elelctorns increases and, the size of Ln3+ ions decreases. Lanthanoid contraction of various lanthanoids is shown in the, graph, Consequences of lanthanoid contraction:, 1. Basicity differences, As we from Ce3+ to Lu3+ , the basic character of Ln3+ ions decrease. Due to the decrease, in the size of Ln3+ ions, the ionic character of Ln −OH bond decreases (covalent character, increases) which results in the decrease in the basicity., 2. Similarities among lanthanoids:, In the complete f - series only 10 pm decrease in atomic radii and 20 pm decrease in ionic, radii is observed. because of this very small change in radii of lanthanoids, their chemical, properties are quite similar., 121, , XII U4-D-Block-Jerald Folder.indd 121, , 2/19/2020 4:41:53 PM

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www.tntextbooks.in, , The elements of the second and third transition series resemble each other more closely, than the elements of the first and second transition series. For example, Series, , Element, , Atomic radius, , 3d Series, , Ti, , 132 pm, , 4d Series, , Zr, , 145 pm, , 5d Series, , Hf, , 144 pm, , Actinoids:, The fourteen elements following actinium ,i.e., from thorium (Th) to lawrentium (Lr) are, called actinoids. Unlike the lanthanoids, all the actinoids are radioactive and most of them, have short half lives. Only thorium and uranium(U) occur in significant amount in nature, and a trace amounts of Plutonium(Pu) is also found in Uranium ores.Neptunium(Np) and, successive heavier elements are produced synthetically by the artificial transformation of, naturally occuring elements by nuclear reactions., Similar to lanthanoids, they are placed at the bottom of the periodic table., Electronic configuration:, The electronic configuration of actinoids is not definite. The general valence shell electronic, configuration of 5f elements is represented as [Rn]5f 0-146d0-27s2. The following table show the, electronic configuration of actinoids., Table : electronic configuration of actinoids, Name of the element, , Atomic, number, , Symbol, , Electronic configuration, , Actinium, , 89, , Ac, , [Rn] 5f 0 6d1 7s 2, , Thorium, , 90, , Th, , [Rn] 5f 0 6d 2 7s 2, , Protactinium, , 91, , Pa, , [Rn] 5f 2 6d1 7s 2, , Uranium, , 92, , U, , [Rn] 5f 3 6d1 7s 2, , Neptunium, , 93, , Np, , [Rn] 5f 4 6d1 7s 2, , Plutonium, , 94, , Pu, , [Rn] 5f 6 6d 0 7s 2, , Americium, , 95, , Am, , [Rn] 5f 7 6d 0 7s 2, , Curium, , 96, , Cm, , [Rn] 5f 7 6d1 7s 2, , Berkelium, , 97, , Bk, , [Rn] 5f 9 6d 0 7s 2, , Californium, , 98, , Cf, , [Rn] 5f 10 6d 0 7s 2, , 122, , XII U4-D-Block-Jerald Folder.indd 122, , 2/19/2020 4:42:01 PM

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www.tntextbooks.in, , Name of the element, , Atomic, number, , Symbol, , Electronic configuration, , Einstenium, , 99, , Es, , [Rn] 5f 11 6d 0 7s 2, , Fermium, , 100, , Fm, , [Rn] 5f 12 6d 0 7s 2, , Mendelevium, , 101, , Md, , [Rn] 5f 13 6d 0 7s 2, , Nobelium, , 102, , No, , [Rn] 5f 14 6d 0 7s 2, , Lawrentium, , 103, , Lr, , [Rn] 5f 14 6d1 7s 2, , Oxidation state of actinoids:, Like lanthanoids, the most common state of actinoids is +3. In addition to that actinoids, show variable oxidation states such as +2 , +3 , +4 ,+5,+6 and +7., The elements Americium(Am) and Thorium (Th) show +2 oxidation state in some, compounds , for example thorium iodide (ThI2). The elements Th , Pa, U ,Np , Pu and Am, show +5 oxidation states. Np and Pu exhibit +7 oxidation state., , Th, , Pa, , U, , Np, , Pu, , +2, , Am Cm, , Bk, , Cf, , Es, , Fm, , Md, , No, , Lr, , +3, , +3, , +3, , +3, , +3, , +3, , +2, , +3, , +3, , +3, , +3, , +3, , +3, , +3, , +3, , +4, , +4, , +4, , +4, , +4, , +4, , +4, , +4, , +5, , +5, , +5, , +5, , +5, , +5, , +6, , +6, , +6, , +6, , +7, , +7, , +7, , Differences between lanthanoids and actinoids:, s.no, , Lanthanoids, , Actinoids, , 1, , Differentiating electron enters in 4f Differentiating electron eneters in 5f, orbital, orbital, , 2, , Binding energy of 4f orbitals are higher, , 3, , They show less tendency to form They show greater tendency to form, complexes, complexes, , 4, , Most of the actinoids are coloured. For, example., U3+ (red), U4+ (green) , UO2 2+ (yellow), , Most of the lanthanoids are colourless, , Binding energy of 5f orbitals are lower, , 123, , XII U4-D-Block-Jerald Folder.indd 123, , 2/19/2020 4:42:06 PM

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www.tntextbooks.in, , s.no, , Lanthanoids, , Actinoids, They do form oxo cations such as, UO2 2+ , NpO2 2+ etc, , 5, , They do not form oxo cations, , 6, , Besides +3 oxidation states lanthanoids Besides +3 oxidation states actinoids, show +2 and +4 oxidation states in few show higher oxidation states such as +4,, cases., +5, +6 and +7., Summary, „„ IUPAC defines transition metal as an element whose atom has an incomplete d, , sub shell or which can give rise to cations with an incomplete d sub shell. They, occupy the central position of the periodic table, between s and p block elements,, , „„ d- Block elements composed of 3d series (4th period) Scandium to Zinc ( 10, , elements), 4d series ( 5th period) Yttrium to Cadmium ( 10 elements) and 5d, series ( 6th period) Lanthanum, Haffinium to mercury., , „„ the general electronic configuration of d- block elements can be written as, , [Noble gas] ( n −1) d, , 1−10, , 1−2, ns ,, , Here, n = 4 to 7 . In periods 6 and 7, the configuration includes, ((n −2 ) f orbital ; [Noble gas] ( n −2 ) f 14 ( n −1) d1−10ns1−2 ., „„ All the transition elements are metals. Similar to all metals the transition metals, , are good conductors of heat and electricity. Unlike the metals of Group-1 and, group-2, all the transition metals except group 11 elements are hard., , „„ As we move from left to right along the transition metal series, melting point first, , increases as the number of unpaired d electrons available for metallic bonding, increases, reach a maximum value and then decreases, as the d electrons pair up, and become less available for bonding., , „„ Ionization energy of transition element is intermediate between those of s and, , p block elements. As we move from left to right in a transition metal series, the, ionization enthalpy increases as expected., , „„ The first transition metal Scandium exhibits only +3 oxidation state, but all other, , transition elements exhibit variable oxidation states by loosing electrons from, (n-1)d orbital and ns orbital as the energy difference between them is very small., , , , 2+,  M M , value is approaching towards less negative value and copper has a positive, , „„ In 3d series as we move from Ti to Zn, the standard reduction potential  E 0, , reduction potential. i.e., elemental copper is more stable than Cu2+., 124, , XII U4-D-Block-Jerald Folder.indd 124, , 2/19/2020 4:42:07 PM

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www.tntextbooks.in, , „„ Most of the compounds of transition elements are paramagnetic. Magnetic, , properties are related to the electronic configuration of atoms., , „„ Many industrial processes use transition metals or their compounds as catalysts., , Transition metal has energetically available d orbitals that can accept electrons, from reactant molecule or metal can form bond with reactant molecule using its, d electrons., , „„ Transition metals form a number of interstitial compounds such as, , TiC, ZrH1.92 , Mn 4 N etc ., , „„ Transition elements have a tendency to form coordination compounds with a species, , that has an ability to donate an electron pair to form a coordinate covalent bond., , „„ In the inner transition elements there are two series of elements. 1) Lanthanoids, , ( previously called lanthanides) 2) Actinoids ( previously called actinides), , „„ Lanthanoids have general electronic configuration [Xe] 4 f 1−14 5d 0−1 6s 2, „„ The common oxidation state of lanthanoides is +3, „„ As we move across 4f series, the atomic and ionic radii of lanthanoids show, , gradual decrease with increse in atomic number. This decrese in ionic size is, called lanthanoid contraction., , „„ The electronic configuration of actinoids is not definite. The general valence, , shell electronic configuration of 5f elements is represented as [Rn]5f 0-146d0-27s2., , „„ Like lanthanoids, the most common state of actinoids is +3. In addition to that, , actinoids show variable oxidation states such as +2 , +3 , +4 ,+5,+6 and +7., , EVALUATION, Choose the best answer:, 1. Sc( Z=21) is a transition element but Zinc (z=30) is not because, , a) both Sc3+ and Zn2+ ions are colourless and form white compounds. , b) in case of Sc, 3d orbital are partially filled but in Zn these are completely filled, c) last electron as assumed to be added to 4s level in case of zinc, d) both Sc and Zn do not exhibit variable oxidation states, 2. Which of the following d block element has half filled penultimate d sub shell as well as half, filled valence sub shell?, a) Cr, , b) Pd, , c) Pt, , d) none of these, 125, , XII U4-D-Block-Jerald Folder.indd 125, , 2/19/2020 4:42:08 PM

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www.tntextbooks.in, , (, , 2+, 3. Among the transition metals of 3d series, the one that has highest negative M, M, standard electrode potential is, , ), , a) Ti    b) Cu     c) Mn    d) Zn, 4. Which one of the following ions has the same number of unpaired electrons as present in, V3+?, a) Ti3+, , b) Fe3+, , c) Ni2+, , d) Cr3+, , 5. The magnetic moment of Mn2+ ion is, a) 5.92BM, , b) 2.80BM, , c) 8.95BM, , d) 3.90BM, , 6. the catalytic behaviour of transition metals and their compounds is ascribed, mainly due to, a) their magnetic behaviour , b) their unfilled d orbitals, c) their ability to adopt variable oxidation states , d) their chemical reactivity, 7. The correct order of increasing oxidizing power in the series, +, 2−, −, a) VO2 < Cr2O7 < MnO4, , −, 2−, +, b) Cr2O7 < VO2 < MnO4, , −, 2−, +, c) Cr2O7 < MnO4 < VO2, , −, 2−, +, d) MnO4 < Cr2O7 < VO2, , 8. In acid medium, potassium permanganate oxidizes oxalic acid to, a) oxalate, , b) Carbon dioxide, , c) acetate, , d) acetic acid, , 9. Which of the following statements is not true?, a) on passing H2S, through acidified K2Cr2O7 solution, a milky colour is observed., b) Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis, c) K2Cr2O7 solution in acidic medium is orange in colour, d) K2Cr2O7 solution becomes yellow on increasing the PH beyond 7, 10. Permanganate ion changes to ________ in acidic medium, 2−, a) MnO4, , b) Mn2+, , c) Mn3+, , d) MnO2, 126, , XII U4-D-Block-Jerald Folder.indd 126, , 2/19/2020 4:42:15 PM

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www.tntextbooks.in, , 11. How many moles of I2 are liberated when 1 mole of potassium dichromate react with, potassium iodide?, , a) 1, , b) 2, , c) 3, , d) 4, , 12. The number of moles of acidified KMnO4 required to oxidize 1 mole of ferrous, oxalate(FeC2O4) is, a) 5   b) 3   c) 0.6   d) 1.5, 13. Which one of the following statements related to lanthanons is incorrect?, a) Europium shows +2 oxidation state., b) The basicity decreases as the ionic radius decreases from Pr to Lu., c) All the lanthanons are much more reactive than aluminium., d) Ce4+ solutions are widely used as oxidising agents in volumetric analysis., 14. Which of the following lanthanoid ions is diamagnetic?, a) Eu2+, , b) Yb2+, , c) Ce2+, , d) Sm2+, , 15. Which of the following oxidation states is most common among the lanthanoids?, a) 4, , b) 2, , c) 5, , d) 3, , 16. Assertion : Ce4+ is used as an oxidizing agent in volumetric analysis., Reason: Ce4+ has the tendency of attaining +3 oxidation state., a) Both assertion and reason are true and reason is the correct explanation of assertion., b) Both assertion and reason are true but reason is not the correct explanation of, assertion., c) Assertion is true but reason is false., d) Both assertion and reason are false., 17. The most common oxidation state of actinoids is, a) +2, , b) +3, , c) +4, , d) +6, , 18. The actinoid elements which show the highest oxidation state of +7 are, a) Np, Pu ,Am, , b) U, Fm, Th, , c) U, Th, Md, , d) Es, No, Lr, 127, , XII U4-D-Block-Jerald Folder.indd 127, , 2/19/2020 4:42:15 PM

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www.tntextbooks.in, , 19. Which one of the following is not correct?, a) La(OH)3 is less basic than Lu(OH)3 , b) In lanthanoid series ionic radius of Ln3+ ions decreases, c) La is actually an element of transition metal series rather than lanthanide series, d) Atomic radii of Zr and Hf are same because of lanthanide contraction, Answer the following questions:, 1. What are transition metals? Give four examples., 2. Explain the oxidation states of 4d series elements., 3. What are inner transition elements?, 4. Justify the position of lanthanides and actinides in the periodic table., 5. What are actinides? Give three examples., 6. Describe the preparation of potassium dichromate., 7. What is lanthanide contraction and what are the effects of lanthanide contraction?, 8. complete the following, a. MnO4 2− + H+ →, ?, → ?, b. C 6 H5CH3 acidified, KMnO, 4, , c. MnO4 + Fe →, ?, −, , 2+, , ∆, d. KMnO4 Red, , →?, hot, , e. Cr2O7 2− + I− + H+ →, ?, f. Na 2Cr2O7 + KCl →, ?, 9. What are interstitial compounds?, 10. Calculate the number of unpaired electrons in Ti3+ , Mn2+ and calculate the spin only, magnetic moment., 11. Write the electronic configuration of Ce4+ and Co2+., 12. Explain briefly how +2 states becomes more and more stable in the first half of the first row, transition elements with increasing atomic number., 13. Which is more stable? Fe3+ or Fe2+ - explain., 14. Explain the variation in E0, , M, , 3+, , /M, , 2+, , 3d series., 128, , XII U4-D-Block-Jerald Folder.indd 128, , 2/19/2020 4:42:16 PM

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www.tntextbooks.in, , 15. Compare lanthanides and actinides., 16. Explain why Cr2+ is strongly reducing while Mn3+ is strongly oxidizing., 17. Compare the ionization enthalpies of first series of the transition elements., 18. Actinoid contraction is greater from element to element than the lanthanoid contraction,, why?, 19. Out of Lu(OH)3 and La(OH)3 which is more basic and why?, 20. Why europium (II) is more stable than Cerium (II)?, 21. Why do zirconium and Hafnium exhibit similar properties?, 22. Which is stronger reducing agent Cr2+ or Fe2+?, 23. The E0, , M, , 2+, , /M, , value for copper is positive. Suggest a possible reason for this., , 24. Describe the variable oxidation state of 3d series elements., 25. Which metal in the 3d series exhibits +1 oxidation state most frequently and why?, 26. Why first ionization enthalpy of chromium is lower than that of zinc?, 27. Transition metals show high melting points why?, , 129, , XII U4-D-Block-Jerald Folder.indd 129, , 2/19/2020 4:42:17 PM

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www.tntextbooks.in, , UNIT, , 5, , COORDINATION, CHEMISTRY, , Alfred Werner, (1866 –1919), , Learning Objectives, After studying this unit, students will be, able to, , Alfred Werner was a Swiss chemist, who explainedthe bonding in, coordination complexes. Werner, proposed his coordination theory, in 1893. It must be remembered, that this imaginative theory was, proposed before the electron had, been discovered by J.J. Thompson, in 1896. Werner did not have any, modern instrumental techniques, at his time and all his studies, were made using simple reaction, chemistry. Complexes must have, been a complete mystery without, any knowledge of bonding or, structure. This theory and his, painstaking work over the next 20, years won Alfred Werner the Nobel, Prize for Chemistry in 1913.He was, the first inorganic chemist to win the, Nobel Prize., , , , , , , , , , , , , , , , , , , , define important terms in coordination, chemistry, nomenclate, the, coordination, compounds in accordance with the, guidelines of IUPAC, describe different types of isomerism in, coordination compounds, discuss the postulates of Werner's, theory of coordination compounds, predict the geometry of coordination, compounds using valence bond theory, apply crystal field theory to explain, the colour and magnetic properties of, coordination compounds, differentiate high spin and low spin, coordination compounds, explain the stability of coordination, compounds interms of stability, constants., explain the applications of coordination, compounds in day to day life, , 130, , XII U5 Coordination jagan.indd 130, , 2/19/2020 4:40:41 PM

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www.tntextbooks.in, , INTRODUCTION, We have already learnt in the previous unit that the transition metals have a tendency, to form complexes (coordination compounds). The name is derived from the Latin, words 'complexus' and 'coordinate' which mean 'hold' and 'to arrange' respectively. The, complexes of transition metals have interesting properties and differ from simple ionic, and covalent compounds. For example, chromium(III)chloride hexahydrate, CrCl3.6H2O,, exists as purple, pale green or dark green compound. In addition to metals, certain non, metals also form coordination compounds but have less tendency than d block elements., Coordination compounds play a vital role in the biological functions, and have wide range, of catalytic applications in chemical industries. For example, haemoglobin, the oxygen, transporter of human is a coordination compound of iron, and cobalamine, an essential, vitamin is a coordination compound of cobalt. Chlorophyll, a pigment present in plants, acting as a photo sensitiser in the photosynthesis is also a coordination compound. Various, coordination compounds such as Wilkinson's compound, Ziegler Natta compound are, used as catalysts in industrial processes. Hence, it is important to understand the chemistry, of coordination compounds. In this unit we study the nature, bonding, nomenclature,, isomerism and applications of the coordination compounds., , 5.1 Coordination compounds and double salts:, When two or more stable compounds in solution are mixed together and allowed, to evaporate, in certain cases there is a possibility for the formation of double salts or, coordination compounds. For example when an equimolar solution of ferrous sulphate, and ammonium sulphate are mixed and allowed to crystallise, a double salt namely, Mohr's salt (Ferrous ammonium sulphate, FeSO4.(NH4)2SO4.6H2O) is formed. Let, us recall the blood red colour formation in the inorganic qualitative analysis of ferric, ion, the reaction between ferric chloride and potassium thiocyanate solution gives a, blood red coloured coordination compound, potassium ferrithiocyanate K3[Fe(SCN)6]., If we perform a qualitative analysis to identify the constituent ions present in both the, compounds, Mohr's salt answers the presence of Fe2+ ,NH4+ and SO42-ions, whereas the, potassium ferrithiocyanate will not answer Fe3+ and SCN ions. From this we can infer that, the double salts lose their identity and dissociates into their constituent simple ions in, solutions , whereas the complex ion in coordination compound, does not loose its identity, and never dissociate to give simple ions., , 5.2 Werner's theory of coordination compounds:, Swiss chemist Alfred Werner was the first one to propose a theory of coordination, compounds to explain the observed behaviour of them., Let us consider the different coloured complexes of cobalt(III) chloride with ammonia, which exhibit different properties as shown below., 131, , XII U5 Coordination jagan.indd 131, , 2/19/2020 4:40:42 PM

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www.tntextbooks.in, , Complex, , Colour, , CoCl3.6NH3, , Yellow, , No. of moles of AgCl precipitated on reaction of, one mole of complex with excess Ag+, 3, , CoCl3.5NH3, , Purple, , 2, , trans - CoCl3.4NH3, , Green, , 1, , cis - CoCl3.4NH3, , Violet, , 1, , In this case, the valences of the elements present in both the reacting molecules, cobalt(III), chloride and ammonia are completely satisfied. Yet these substances react to form the above, mentioned complexes., To explain this behaviour Werner postulated his theory as follows, 1. Most of the elements exhibit, two types of valence namely primary valence and secondary, valence and each element tend to satisfy both the valences.In modern terminology, the, primary valence is referred as the oxidation state of the metal atom and the secondary, valence as the coordination number. For example, according to Werner, the primary and, secondary valences of cobalt are 3 and 6 respectively., 2. The primary valence of a metal ion is positive in most of the cases and zero in certain, cases. They are always satisfied by negative ions. For example in the complex CoCl3.6NH3,, The primary valence of Co is +3 and is satisfied by 3Cl ions., 3. The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the, combination of these. For example, in CoCl3.6NH3 the secondary valence of cobalt is 6, and is satisfied by six neutral ammonia molecules, whereas in CoCl3.5NH3 the secondary, valence of cobalt is satisfied by five neutral ammonia molecules and a Cl ion., 4. According to Werner, there, are two spheres of attraction, around a metal atom/ion in a, complex. The inner sphere is, known as coordination sphere, and the groups present in this, sphere are firmly attached to, the metal. The outer sphere is, called ionisation sphere. The, groups present in this sphere, are loosely bound to the, central metal ion and hence, can be separated into ions, upon dissolving the complex, in a suitable solvent., , Inner sphere or coordination sphere, NH3, , M, , Cl, , NH3, , NH3, , Cl, , Co, , NH3, , NH3, , NH3, Cl, , outer sphere or ionization sphere, , Figure 5.1 inner and outer spheres of attraction in, coordination compounds, 132, , XII U5 Coordination jagan.indd 132, , 2/19/2020 4:40:43 PM

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www.tntextbooks.in, , 1. The primary valences are non-directional while the secondary valences are directional. The, geometry of the complex is determined by the spacial arrangement of the groups which, satisfy the secondary valence. For example, if a metal ion has a secondary valence of six, it, has an octahedral geometry. If the secondary valence is 4, it has either tetrahedral or square, planar geometry., , The following table illustrates the Werner's postulates., Groups satisfy the, secondary valence, (non-ionaisable, inner, coordination sphere), , No. of ionisable Cl, ions in the complex, (outer coordination, sphere), , CoCl3.6NH3, , 6 NH3, , 3 Cl-, , 3 AgCl, , CoCl3.5NH3, , 5 NH3 & 1 Cl-, , 2 Cl-, , 2 AgCl, , CoCl3.4NH3, , 4 NH3 & 2 Cl-, , 1 Cl-, , 1 AgCl, , CoCl3.4NH3, , 4 NH3 & 2 Cl-, , 1 Cl-, , 1 AgCl, , Complex, , No. of moles of, AgCl formed =, no. of moles of, ionisable Cl, , 5.2.1 Limitations of Werner’s theory:, Even though, Werner’s theory was able to explain a number of properties of coordination, compounds, it does not explain their colour and the magnetic properties., Evaluate yourself 1:, When a coordination compound CrCl3.4H2O is mixed with silver nitrate solution, one, mole of silver chloride is precipitated per mole of the compound. There are no free solvent, molecules in that compound. Assign the secondary valence to the metal and write the, structural formula of the compound., , 5.3 Definition of important terms pertaining to co-ordination compounds, 5.3.1 Coordination entity:, Coordination entity is an ion or a neutral molecule, composed of a central atom, usually, a metal and the array of other atoms or groups of atoms (ligands) that are attached to it. In, the formula, the coordination entity is enclosed in square brackets. For example, in potassium, 4ferrocyanide, K4[Fe(CN)6], the coordination entity is [Fe(CN)6] . In nickel tetracarbonyl, the, coordination entity is [Ni(CO)4]., 5.3.2 Central atom/ion:, The central atom/ion is the one that occupies the central position in a coordination, entity and binds other atoms or groups of atoms (ligands) to itself, through a coordinate, 133, , XII U5 Coordination jagan.indd 133, , 2/19/2020 4:40:43 PM

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www.tntextbooks.in, , covalent bond. For example, in K4[Fe(CN)6], the central metal ion is Fe2+. In the, 4coordination entity [Fe(CN)6] , the Fe2+ accepts an electron pair from each ligand, CN, and thereby forming six coordinate covalent bonds with them. since, the central metal ion, has an ability to accept electron pairs, it is referred to as a Lewis acid., 5.3.3 Ligands:, The ligands are the atoms or groups of atoms bound to the central atom/ion. The atom in a, ligand that is bound directly to the central metal atom is known as a donor atom. For example,, in K4[Fe(CN)6], the ligand is CN ion, but the donor atom is carbon and in [Co(NH3)6]Cl3 the, ligand is NH3 molecule and the donor atom is nitrogen., Coordination sphere:, The complex ion of the coordination compound containing the central metal atom/ion, and the ligands attached to it, is collectively called coordination sphere and are usually enclosed, in square brackets with the net charge. The other ionisable ions, are written outside the bracket, are called counter ions. For example, the coordination compound K4[Fe(CN)6] contains the, 4complex ion [Fe(CN)6] and is referred as the coordination sphere. The other associated ion, K+ is called the counter ion., Coordination polyhedron:, The three dimensional spacial arrangement of ligand atoms/ions that are directly attached, to the central atom is known as the coordination polyhedron (or polygon). For example,, in K4[Fe(CN)6], the coordination polyhedra is octrahedral. The coordination polyhedra of, [Ni(CO)4] is tetrahedral., Coordination number:, The number of ligand donor atoms bonded to a central metal ion in a complex is called, the coordination number of the metal. In other words, the coordination number is equal to the, number of σ-bonds between ligands and the central atom., For example,, i. In K4[Fe(CN)6], the coordination number of Fe2+ is 6., ii. In [Ni(en)3]Cl2, the coordination number of Ni2+ is also 6. Here the ligand 'en' represents, ethane-1,2-diamine (NH2-CH2-CH2-NH2) and it contains two donor atoms (Nitrogen), Each ligand forms two coordination bonds with nickel. So,totally there are six coordination, bonds between them., Oxidation state (number):, The oxidation state of a central atom in a coordination entity is defined as the, charge it would bear if all the ligands were removed along with the electron pairs that, were shared with the central atom. In naming a complex, it is represented by a Roman, 4numeral. For example, in the coordination entity [Fe(CN) 6] , the oxidation state of, iron is represented as (II). The net charge on the complex ion is equal to the sum of, , 134, , XII U5 Coordination jagan.indd 134, , 2/19/2020 4:40:43 PM

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www.tntextbooks.in, , the oxidation state of the central metal and the charge the on the ligands attached to, it. Using this relation the oxidation number can be calculated as follows, Net charge = (oxidation state of the central metal) + [(No. of ligands) X (charge on the ligand)], , Example 1:, , 4-, , In [Fe(CN)6], , ,, , let the oxidation number of iron is x :, , The net charge: -4 = x + 6 (-1) => x = +2, Example 2:, , 2+, , In [Co(NH3)5Cl], , ,, , let the oxidation number of cobalt is x :, , The net charge: +2 = x + 5 (0) + 1 (-1) => x = +3, Evaluate yourself 2:, 2. In the complex, [Pt(NO2)(H2O)(NH3)2]Br , identify the following, i. Central metal atom/ion, ii. Ligand(s) and their types, iii. Coordination entity, iv. Oxidation number of the central metal ion, v. Coordination number, Types of complexes:, The coordination compounds can be classified into the following types based on (i) the, net charge of the complex ion, (ii) kinds of ligands present in the coordination entity., Classification based on the net charge on the complex:, A coordination compound in which the complex ion, +, , i. carries a net positive charge is called a cationic complex. Examples: [Ag(NH3)2] ,, 3+, 2+, [Co(NH3)6] , [Fe(H2O)6] , etc, ii. carries a net negative charge is called an anionic complex. Examples: [Ag(CN)2] ,, 34[Co(CN)6] , [Fe(CN)6] , etc, iii. bears no net charge, is called a neutral complex. Examples: [Ni(CO)4], [Fe(CO)5] ,, [Co(NH3)3(Cl)3],, Classification based on kind of ligands:, A coordination compound in which, i. the central metal ion/atom is coordinated to only one kind of ligands is called a homoleptic, complex. Examples: [Co(NH3)6]3+ , [Fe(H2O)6]2+,, 135, , XII U5 Coordination jagan.indd 135, , 2/19/2020 4:40:43 PM

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www.tntextbooks.in, , ii. the central metal ion/atom is coordinated to more than one kind of ligands is called a, heteroleptic complex. Example, [Co(NH3)5Cl]2+, [Pt(NH3)2Cl2)], , 5.4 Nomenclature of coordination compounds, In the earlier days, the compounds were named after their discoverers. For example,, K[PtCl3(C2H4)] was called Zeise’s salt and [Pt(NH3)4][PtCl4] is called Magnus’s green salt etc..., There are numerous coordination compounds that have been synthesised and characterised., The International Union of Pure and Applied Chemistry (IUPAC) has developed an elaborate, system of nomenclature to name them systematically. The guidelines for naming coordination, compounds based on IUPAC recommendations (2005) are as follows:, 1. The cation is named first, followed by the anion regardless of whether the ion is simple or, complex. For example, +, , 4-, , • In K4[Fe(CN)6], the cation K is named first followed by[Fe(CN)6] ., • In [Co(NH3)6]Cl3, the complex cation [Co(NH3)6]3+ is named first followed by the, anion Cl, 2+, , • In [Pt(NH3)4][PtCl4], the complex cation [Pt(NH3)4] is named first followed by the, 2complex anion [PtCl4], 2. The simple ions are named as in other ionic compounds. For example,, Simple cation, , Symbol, , Simple anion, , Symbol, , Sodium, , Na+, , Chloride, , Cl-, , Potassium, , K+, , Nitrate, , NO3-, , Copper, , Cu2+, , Sulphate, , SO42-, , 3. To name a complex ion, the ligands are named first followed by the central metal atom/ion., When a complex ion contains more than one kind of ligands they are named in alphabetical, order., a. Naming the ligands:, i. The name of anionic ligands ends with the letter 'o' and the cationic ligand ends with, 'ium'. The neutral ligands are usually called with their molecular names with fewer, exceptions namely, H2O (aqua), CO (carbonyl), NH3 (ammine) and NO (nitrosyl)., ii. A κ-term is used to denote an ambidendate ligand in which more than one coordination, mode is possible. For example, the ligand thiocyanate can bind to the central atom/, ion, through either the sulfur or the nitrogen atom. In this ligand, if sulphur forms a, coordination bond with metal then the ligand is named thiocyanato-κS and if nitrogen, is involved, then it is named thiocyanato-κN., 136, , XII U5 Coordination jagan.indd 136, , 2/19/2020 4:40:43 PM

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www.tntextbooks.in, , Evaluate yourself 3:, 14. Write the IUPAC name for the following compounds., (i), , K 2 Fe ( CN )3 ( Cl )2 ( NH3 ), , (ii) Cr ( CN )2 ( H2O ) 4  Co ( ox )2 ( en ) , (iii) Cu ( NH3 ) Cl 2 , , , 2, (iv) Cr ( NH3 )3 ( NC )2 ( H2O ), 4−, (v) Fe ( CN )6 , , +, , 15. Give the structure for the following compounds., (i), , diamminesilver(I) dicyanidoargentate(I), , (ii) Pentaammine nitrito-κNcobalt (III) ion, (iii) hexafluorido cobaltate (III) ion, (iv) dichloridobis(ethylenediamine) Cobalt (IV) sulphate, (v) Tetracarbonylnickel (0), , 5.5 Isomerism in coordination compounds, We have already learnt the concept of isomerism in the context of organic compounds,, in the previous year chemistry classes. Similarly, coordination compounds also exhibit, isomerism. Isomerism is the phenomenon in which more than one coordination compounds, having the same molecular formula have different physical and chemical properties due to, different arrangement of ligands around the central metal atom. The following flow chart gives, an overview of the common types of isomerism observed in coordination compounds,, arises due to the, difference in the, structures of, coordination compounds, , Isomerism, , Structural, Isomerism, , Ionisation, isomerism, , Hydration, isomerism, , Linkage, isomerism, , arises due to different, spatial orientation of, ligands around the, metal ion, , Stereo, Isomerism, , Coordination, isomerism, , Geometrical, isomerism, , Optical, isomerism, , Figure 5.2 Isomerism in coordination compounds, 5.5.1 Structural isomers, The coordination compounds with same formula, but have different connections among, their constituent atoms are called structural isomers or constitutional isomers. Four common, types of structural isomers are discussed below., 141, , XII U5 Coordination jagan.indd 141, , 2/19/2020 4:40:51 PM

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www.tntextbooks.in, , Linkage isomers:, This type of isomers arises when an ambidentate ligand is bonded to the central metal, atom/ion through either of its two different donor atoms. In the below mentioned examples, the, nitrite ion is bound to the central metal ion Co3+ through a nitrogen atom in one complex,and, through oxygen atom in other complex., [Co(NH3)5(NO2)]2+, O, 2+, O, , 2+, , O, , N, , N, H3N, , O, , Co, , H 3N, , NH3, , H3N, , NH3, , H 3N, , NH3, , Co, , NH3, NH3, , NH3, , Figure 5.3 Linkage isomers, Coordination isomers:, This type of isomers arises in the coordination compounds having both the cation and, anion as complex ions. The interchange of one or more ligands between the cationic and the, anionic coordination entities result in different isomers., For example, in the coordination compound, [Co(NH3)6][Cr(CN)6] the ligands ammonia, and cyanide were bound respectively to cobalt and chromium while in its coordination isomer, [Cr(NH3)6][Co(CN)6] they are reversed., Some more examples for coordination isomers, 1. [Cr(NH3)5CN][Co(NH3)(CN)5] and [Co(NH3)5CN][Cr(NH3)(CN)5], 2. [Pt(NH3)4][Pd(Cl)4] and [Pd(NH3)4][Pt(Cl)4], Ionisation isomers:, This type of isomers arises when an ionisable counter ion (simple ion) itself can act as, a ligand. The exchange of such counter ions with one or more ligands in the coordination, entity will result in ionisation isomers. These isomers will give different ions in solution. For, example, consider the coordination compound [Pt(en)2Cl2]Br2. In this compound, both Br, and Cl have the ability to act as a ligand and the exchange of these two ions result in a different, isomer [Pt(en)2Br2]Cl2. In solution the first compound gives Br ions while the later gives Cl, ions and hence these compounds are called ionisation isomers., Some more example for the isomers,, 1. [Cr(NH3)4ClBr]NO2 and [Cr(NH3)4Cl NO2]Br, 2. [Co(NH3)4Br2]Cl and [Co(NH3)4Cl Br]Br, 142, , XII U5 Coordination jagan.indd 142, , 2/19/2020 4:40:51 PM

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www.tntextbooks.in, , Evaluate yourself 4:, 3. A solution of [Co(NH3)4l2]Cl when treated with AgNO3 gives a white precipitate. What, should be the formula of isomer of the dissolved complex that gives yellow precipitate, with AgNO3. What are the above isomers called?, Solvate isomers., The exchange of free solvent molecules such as water , ammonia, alcohol etc.. in the, crystal lattice with a ligand in the coordination entity will give different isomers. These type, of isomers are called solvate isomers. If the solvent molecule is water, then these isomers are, called hydrate isomers. For example, the complex with chemical formula CrCl3.6H2O has three, hydrate isomers as shown below., [Cr(H2O)6]Cl3, , a violet colour compound and gives three chloride, ions in solution,, , [Cr(H2O)5Cl]Cl2.H2O, , a pale green colour compound and gives two chloride, ions in solution and,, , [Cr(H2O)4Cl2]Cl.2H2O, , dark green colour compound and gives one chloride, ion in solution, , 5.5.2 Stereoisomers:, Similar to organic compounds, coordination compounds also exhibit stereoisomerism., The stereoisomers of a coordination compound have the same chemical formula and, connectivity between the central metal atom and the ligands. But they differ in the spatial, arrangement of ligands in three dimensional space. They can be further classified as geometrical, isomers and optical isomers., Geometrical isomers:, Geometrical isomerism exists in heteroleptic complexes due to different possible three, dimensional spatial arrangements of the ligands around the central metal atom. This type of, isomerism exists in square planer and octahedral complexes., In square planar complexes of the form [MA2B2]n± and [MA2BC]n± (where A, B and C are, mono dentate ligands and M is the central metal ion/atom), Similar groups (A or B) present, either on same side or on the opposite side of the central metal atom (M) give rise to two, different geometrical isomers, and they are called, cis and trans isomers respectively., The square planar complex of the type [M(xy)2]n± where xy is a bidentate ligand with two, different coordinating atoms also shows cis-trans isomerism. Square planar complex of the, form [MABCD]n± also shows geometrical isomerism. In this case, by considering any one, of the ligands (A, B, C or D) as a reference, the rest of the ligands can be arranged in three, different ways leading to three geometrical isomers., 143, , XII U5 Coordination jagan.indd 143, , 2/19/2020 4:40:51 PM

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www.tntextbooks.in, , Octahedral complexes:, Octahedral complexes of the type [MA2B4]n±, [M(xx)2B2]n± shows cis-trans isomerism., Here A and B are monodentate ligands and xx is bidentate ligand with two same kind of, donor atoms. In the octahedral complex, the position of ligands is indicated by the following, numbering scheme., Y, , 1, L, , Z′, , 5, 4, , L, , X′, , 2, , M, , L, , L, , Mn+, , X, , L, , 3, Z, , Figure 5.5 Position of ligands in, octahedral complex, , 6, , L, , Y′, , In the above scheme, the positions (1,2), (1,3), (1,4), (1,5), (2,3), (2,5), (2,6), (3,4), (3,6),, (4,5), (4,6), and (5,6) are identical and if two similar groups are present in any one of these, positions, the isomer is referred as a cis isomer. Similarly, positions (1,6), (2,4), and (3,5) are, identical and if similar ligands are present in these positions it is referred as a trans-isomer., Octahedral complex of the type [MA3B3]n± also shows geometrical isomerism. If the three, similar ligands (A) are present in the corners of one triangular face of the octahedron and the, other three ligands (B) are present in the opposing triangular face, then the isomer is referred, as a facial isomer (fac isomer)- Figure 5.6 (a)., If the three similar ligands are present around the meridian which is an imaginary, semicircle from one apex of the octahedral to the opposite apex as shown in the figure 5.6(b),, the isomer is called as a meridional isomer (mer isomer). This is called meridional because, each set of ligands can be regarded as lying on a meridian of an octahedron., Cl, , Cl, , CN, Cl, , Co3+, , CN, CN, , CN, , Cl, , Co3+, , CN, , Cl, , CN, , Figure 5.6 (a) Facial isomer, , Cl, , Figure 5.6 (b) Meridional isomer, 145, , XII U5 Coordination jagan.indd 145, , 2/19/2020 4:40:56 PM

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www.tntextbooks.in, , As the number of different ligands increases, the number of possible isomers also, increases. For the octahedral complex of the type [MABCDEF]n±, where A, B, C, D, E and, F are monodentate ligands, fifteen different orientation are possible corresponding to 15, geometrical isomers. It is difficult to generate all the possible isomers., Evaluate yourself 5:, 5. Three compounds A ,B and C have empirical formula CrCl3.6H2O. they are kept in a, container with a dehydrating agent and they lost water and attaining constant weight, as shown below., Compound, , Initial weight, compound(in g), , of, , the Constant weight, dehydration (in g), , A, , 4, , 3.46, , B, , 0.5, , 0.466, , C, , 3, , 3, , after, , 6. Indicate the possible type of isomerism for the following complexes and draw their, isomers, (i) [Co(en)3][Cr(CN)6], , (ii) [Co(NH3)5(NO2)]2+ (iii) [Pt(NH3)3(NO2)]Cl, , 5.5.3 Optical Isomerism, Coordination, compounds, which possess chairality exhibit, optical isomerism similar to organic, compounds. The pair of two optically, active isomers which are mirror, images of each other are called, enantiomers. Their solutions rotate, the plane of the plane polarised light, either clockwise or anticlockwise and, the corresponding isomers are called, 'd' (dextro rotatory) and 'l' (levo, rotatory) forms respectively. The, octahedral complexes of type [M(xx)3], n±, , [M(xx)2AB]n± and [M(xx)2B2]n±, exhibit optical isomerism., , , , , , , , Co3+, , , , , , , , , , , , , , , Co3+, , , , , , , , , , [()]+, , , , H2N-CH2-CH2-NH2, (en), , Figure 5.7 - Optical isomer, , Examples:, The optical isomers of [Co(en)3]3+ are shown in figure 5.7., , 146, , XII U5 Coordination jagan.indd 146, , 2/19/2020 4:40:57 PM

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www.tntextbooks.in, , The coordination complex [CoCl2(en)2]+ has three isomers, two optically active cis forms, and one optically inactive trans form. These structures are shown below., , , , , , , , Co3+, , , , , , , , , , , , , , , Co3+, , , , , , , , , , [()]+, , , , H2N-CH2-CH2-NH2, (en), , Figure 5.8 - Optical isomers, Evaluate yourself 6:, 10. Draw all possible stereo isomers of a complex Ca[Co(NH3)Cl(Ox)2], , 5.6 Theories of coordination compound, Alfred Werner considered the bonding in coordination compounds as the bonding, between a lewis acid and a lewis base. His approach is useful in explaining some of the observed, properties of coordination compounds. However, properties such as colour, magnetic property, etc.. of complexes could not be explained on the basis of his approach. Following werner theory,, Linus pauling proposed the Valence Bond Theory (VBT) which assumes that the bond formed, between the central metal atom and the ligand is purely covalent. Bethe and Van vleck treated, the interaction between the metal ion and the ligands as electrostatic and extended the Crystal, Field Theory (CFT) to explain the properties of coordination compounds. Further, Ligand, field theory and Molecular orbital have been developed to explain the nature of bonding in the, coordination compounds. In this porton we learn the elementry treatment of VBT and CFT to, simple coordination compounds., 5.6.1 Valence Bond Theory, According to this theory, the bond formed between the central metal atom and the ligand, is due to the overlap of filled ligand orbitals containing a lone pair of electron with the vacant, hybrid orbitals of the central metal atom., Main assumptions of VBT:, 1. The ligand → metal bond in a coordination complex is covalent in nature. It is formed by sharing, of electrons (provided by the ligands) between the central metal atom and the ligand., 147, , XII U5 Coordination jagan.indd 147, , 2/19/2020 4:40:58 PM

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www.tntextbooks.in, , 2. Each ligand should have at least one filled orbital containing a lone pair of electrons., 3. In order to accommodate the electron pairs donated by the ligands, the central metal ion, present in a complex provides required number (coordination number) of vacant orbitals., 4. These vacant orbitals of central metal atom undergo hybridisation, the process of mixing, of atomic orbitals of comparable energy to form equal number of new orbitals called, hybridised orbitals with same energy., 5. The vacant hybridised orbitals of the central metal ion, linearly overlap with filled orbitals, of the ligands to form coordinate covalent sigma bonds between the metal and the ligand., 6. The hybridised orbitals are directional and their orientation in space gives a definite, geometry to the complex ion., Coordination, number, , Hybridisation, , 2, , sp, , Linear, , [CuCl2] , [Ag(CN)2], , 3, , sp2, , Trigonal, planar, , [HgI3], , 4, , sp3, , Tetrahedral, , [Ni(CO)4], [NiCl4]2-, , 4, , dsp2, , Square, planar, , [Ni(CN)4]2-, [Pt(NH3)4]2+, , Trigonal, bipyramidal, , Fe(CO)5, , Octahedral, , [Ti(H2O)6]3+, [Fe(CN)6]2-, [Fe(CN)6]3-,, [Co(NH3)6]3+, , dsp3, 5, , (dx2-y2 orbital is, involved), , Geometry, , d2sp3, 6, , (dz2 and dx2-y2, orbitals of inner shell, , sp3d2, 6, , -, , (Inner orbital complexes), , are involved), , (dz2 and dx2-y2, orbitals of the outer, , Examples, , Octahedral, , [FeF6]4-,[CoF6]4-, [Fe(H2O)6]2+, (Outer orbital complexes), , shell are involved), , 7. In the octahedral complexes, if the (n-1) d orbitals are involved in hybridisation, then they, are called inner orbital complexes or low spin complexes or spin paired complexes. If the, nd orbitals are involved in hybridisation, then such complexes are called outer orbital or, high spin or spin free complexes. Here n represents the principal quantum number of the, outermost shell., 8. The complexes containing a central metal atom with unpaired electron(s) are paramagnetic., If all the electrons are paired, then the complexes will be diamagnetic., 148, , XII U5 Coordination jagan.indd 148, , 2/19/2020 4:40:58 PM

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www.tntextbooks.in, , 3-, , Complex, Hybridised orbitals of, the metal atom in the, complex, , [CoF6], ,     , 3d6, ,      , sp3d2 Hybridised orbitals, , 4d0, , Octahedral, Geometry, , Magnetic property, , In this complex outer d orbitals are involved in the hybridisaion, and hence the complex is called outer orbital complex, No. of unparied electrons = 4;, Hence paramagnetic, , Magnetic moment, (Using spin only, formula), , µs = n(n+ 2) = 4(4 + 2) = 4.899 BM, , Limitations of VBT, Eventhough VBT explains many of the observed properties of complexes, it still has following, limitations, 1. It does not explain the colour of the complex, 2. It considers only the spin only magnetic moments and does not consider the other, components of magnetic moments., 3. It does not provide a quantitative explanation as to why certain complexes are inner orbital, complexes and the others are outer orbital complexes for the same metal. For example,, 44[Fe(CN)6] is diamagnetic (low spin) whereas [FeF6] is paramagnetic (high spin)., Evaluate yourself 7:, 7. The spin only magnetic moment of Tetrachloridomanganate(II)ion is 5.9 BM. On the, basis of VBT, predict the type of hybridisation and geometry of the compound., 8. Predict the number of unpaired electrons in [CoCl4]2- ion on the basis of VBT., 9. A metal complex having composition Co(en)2Cl2Br has been isolated in two forms, A and B. (B) reacted with silver nitrate to give a white precipitate readily soluble in, ammonium hydroxide. Whereas A gives a pale yellow precipitate. Write the formula of, A and B. state the hybridization of Co in each and calculate their spin only magnetic, moment., 152, , XII U5 Coordination jagan.indd 152, , 2/19/2020 4:40:59 PM

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www.tntextbooks.in, , 5.6.2 Crystal Field Theory, Valence bond theory helps us to visualise the bonding in complexes. However, it has, limitations as mentioned above.Hence Crystal Field Theory to expalin some of the properties, like colour, magnetic behavior etc.,This theory was originally used to explain the nature of, bonding in ionic crystals. Later on, it is used to explain the properties of transition metals and, their complexes. The salient features of this theory are as follows., 1. Crystal Field Theory (CFT) assumes that the bond between the ligand and the central metal, atom is purely ionic. i.e. the bond is formed due to the electrostatic attraction between the, electron rich ligand and the electron deficient metal., 2. In the coordination compounds, the central metal atom/ion and the ligands are considered, as point charges (in case of charged metal ions or ligands) or electric dipoles (in case of, metal atoms or neutral ligands)., 3. According to crystal field theory, the complex formation is considered as the following, series of hypothetical steps., Step 1: In an isolated gaseous state, all the five d orbitals of the central metal ion are degenerate., Initially, the ligands form a spherical field of negative charge around the metal. In this field, the, energies of all the five d orbitals will increase due to the repulsion between the electrons of the, metal and the ligand., L, Step 2: The ligands, are approaching the, metal atom in actual, bond, directions., L5, To illustrate this, let us consider an, octahedral field, in, which the central, metal ion is located at, the origin and the six, L6, ligands are coming, from the +x, -x, +y, -y,, +z and -z directions, as shown below., , 4, , Z, dz2, , -X, , dyz, dxz, , Y, , dxy, , L3, , dx2 - y2, , -Y, X, L2, , -Z, L1, Figure 5.9 octahedral ligand field, 153, , XII U5 Coordination jagan.indd 153, , 2/19/2020 4:41:01 PM

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www.tntextbooks.in, , As shown in the figure, the orbitals lying along the axes dx2-y2 and dz2 orbitals will, experience strong repulsion and raise in energy to a greater extent than the orbitals with lobes, directed between the axes (dxy, dyz and dzx). Thus the degenerate d orbitals now split into two, sets and the process is called crystal field splitting., Step 3: Up to this point the complex formation would not be favoured. However, when the, ligands approach further, there will be an attraction between the negatively charged electron, and the positively charged metal ion, that results in a net decrease in energy. This decrease in, energy is the driving force for the complex formation., Crystal field splitting in octahedral complexes:, During crystal field splitting in octahedral field, in order to maintain the average energy of the, orbitals (barycentre) constant, the energy of the orbitals dx2-y2 and dz2 (represented as eg orbitals), will increase by 3/5Δo while that of the other three orbitals dxy, dyz and dzx (represented as t2g orbitals), decrease by 2/5Δo. Here, Δo represents the crystal field splitting energy in the octahedral field., dx2 y2, dz2, -, , eg, , Energy, , + 3 ∆o, 5, , Average energy of, the d orbitals in a, sphercial crystal, field, , d orbitals, in free ion, (dxy , dyz , dxz, dx2-y2 and dz2), , ∆o, , - 2 ∆o, 5, , dxy , dyz , dxz, , t2g, , Splitting of d-orbitals, in an octahedral, crystal field, , Figure: 5.10 - Crystal field splitting in octahedral field, Crystal field splitting in tetrahedral, complexes:, The approach of ligands in, tetrahedral field can be visualised as, follows. Consider a cube in which, the central metal atom is placed at its, centre (i.e. origin of the coordinate, axis as shown in the figure). The, four ligands approach the central, metal atom along the direction of, the leading diagonals drawn from, alternate corners of the cube., , L3, , L4, Z, , X', Y, , Y', , X, , L1, , L2, Z', , Figure 5.11 tetrahedral ligand field, 154, , XII U5 Coordination jagan.indd 154, , 2/19/2020 4:41:01 PM

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www.tntextbooks.in, , In this field, none of the d, orbitals point dirctly towards the, ligands,however the t2 orbitals (dxy,, dyz and dzx) are pointing close to, the direction in which ligands are, approaching than the e orbitals, (dx2-y2 and dz2)., , L3, , L4, Z, , dz2, dyz, dxz, , X', , As a result, the energy of t2, Y, d, orbitals increases by 2/5Δt and that, d, of e orbitals decreases by 3/5Δt as, Y', X, shown below. when compared to, the octahedral field, this splitting is, inverted and the spliting energy is, less. The relation between the crystal, L2, field splitting energy in octahedral L1, and tetrahedral ligand field is given, Z', by the expression; ∆ t = 4 ∆ 0, 9, Figure 5.12 d-orbitals in tetrahedral ligand field, xy, , x2 - y2, , dxy , dyz , dxz, Energy, , 2 ∆, 5 t, - 3 ∆t, 5, , t2, ∆t = 94 ∆o, , Average energy of, e, the d orbitals in a, dx2 y2, dz2, sphercial crystal, orbitals, d, Splitting of d-orbitals, field, in free ion, in an tetrahedral, (dxy , dyz , dxz, dx2-y2 and dz2), crystal field, , Figure: 5.13 - Crystal field splitting in tetrahedral field, Crystal filed splitting Energy and nature of ligands:, The magnitude of crystal field splitting energy not only depends on the ligand field, as discussed above but also depends on the nature of the ligand, the nature of the central, metal atom/ion and the charge on it. Let us understand the effect of the nature of ligand, on crystal field splitting by calculating the crystal field splitting energy of the octahedral, complexes of titanium(III) with different ligands such as fluoride, bromide and water, 155, , XII U5 Coordination jagan.indd 155, , 2/19/2020 4:41:04 PM

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www.tntextbooks.in, , using their absorption spectral data. The absorption wave numbers of complexes [TiBr6]3-,, [TiF6]3- and [Ti(H2O)6]3+ are 12500, 19000 and 20000 cm-1 respectively. The energy, associated with the absorbed wave numbers of the light, corresponds to the crystal field, splitting energy (Δ) and is given by the following expression,, Δ = hν =, , hc, = hcν, λ, , where h is the Plank' s constant; c is velocity of light, υ is the wave number of absorption, maximum which is equal to 1/λ, [TiBr6]3-, , [TiF6]3Δ = hcν, , [Ti(H2O)6]3+, Δ = hcν, , Δ = hcν, , = (6.626 × 10–34 Js ), × (3 × 108 ms–1 ), × (12500 × 102 m–1), , = (6.626 × 10–34 Js ), × (3 × 108 ms–1 ), × (19000 × 102 m–1), , = (6.626 × 10–34 Js ), × (3 × 108 ms–1 ), × (20000 × 102 m-1), , = 248475 × 10–24 J, , = 377682 × 10–24 J, , = 397560 × 10–24 J, , = 2.48 × 10–22 kJ, , = 3.78 × 10–22 kJ, , = 3.98 × 10–22 kJ, , To express Δ on a per, mole basis, multiply it by, Avogadro number, , To express Δ on a per, mole basis, multiply it by, Avogadro number, , To express Δ on a per, mole basis, multiply it by, Avogadro number, , = (2.48 × 10–22 kJ), × (6.023 × 1023mol–1), , = (3.78 × 10–22 kJ), × (6.023 × 1023mol-1), , = (3.98 × 10–22 kJ), × (6.023 × 1023mol-1), , = 149.4 kJ mol–1, , = 227.7 kJ mol–1, , = 239.7 kJ mol–1, , From the above calculations, it is clear that the crystal filed splitting energy of the Ti3+ in, complexes,the three ligands is in the order; Br < F < H2O. Similarly, it has been found form, the spectral data that the crystal field splitting power of various ligands for a given metal ion,, are in the following order, 24- - - 2I <Br <SCN <Cl <S <F-<OH ≈urea< ox < H2O< NCS <EDTA <NH3<en<NO2 <CN < CO, The above series is known as spectrochemical series. The ligands present on the right, side of the series such as carbonyl causes relatively larger crystal field splitting and are, called strong ligands or strong field ligands, while the ligands on the left side are called, weak field ligands and causes relatively smaller crystal field splitting., 156, , XII U5 Coordination jagan.indd 156, , 2/19/2020 4:41:04 PM

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www.tntextbooks.in, , Distribution of d electrons in octahedral complexes:, The filling of electrons in the d orbitals in the presence of ligand field also follows Hund's, rule. In the octahedral complexes with d2 and d3 configurations, the electrons occupy different, degenerate t2g orbitals and remains unpaired. In case of d4 configuration, there are two, possibilities. The fourth electron may either go to the higher energy eg orbitals or it may pair, with one of the t2g electrons. In this scenario, the preferred configuration will be the one with, lowest energy., If the octahedral crystal field splitting energy (Δo) is greater than the pairing energy, (P), it is necessary to cause pairing of electrons in an orbital, then the fourth electron will pair, up with an the electron in the t2g orbital. Conversely, if the Δo is lesser than P, then the fourth, electron will occupy one of the degenerate higher energy eg orbitals., For example, let us consider two different iron(III) complexes [Fe(H2O)6]3+ (weak field, complex; wave number corresponds to Δo is 14000 cm-1) and [Fe(CN)6]3- (Strong field complex;, wave number corresponds to Δo is 35000 cm-1). The wave number corresponds to the pairing, energy of Fe3+ is 30000 cm-1. In both these complexes the Fe3+ has d5 configuration. In aqua, complex, the Δo < P hence, the fourth & fifth electrons enter eg orbitals and the configuration is, t2g3, eg2. In the cyanido complex Δo > P and hence the fourth & fifth electrons pair up with the, electrons in the t2g orbitals and the electronic configuration is t2g5, eg0., The actual distribution of electrons can be ascertained by calculating the crystal field, stabilisation energy (CFSE). The crystal field stabilisation energy is defined as the energy, difference of electronic configurations in the ligand field (ELF) and the isotropic field/barycentre, (Eiso)., CFSE (ΔEo) = {ELF } - {Eiso }, = {[nt2g(-0.4)+neg(0.6)] Δo + npP} - {n'p P}, Here, nt2g is the number of electrons in t2g orbitals; neg is number of electrons in eg orbitals;, np is number of electron pairs in the ligand field; & n'p is the number of electron pairs in the, isotropic field (barycentre)., Calculating the CFSE for the Iron complexes, Complex: [Fe(H2O)6]3+, High Spin Complex, , Low spin complex, , Electronic configuration in isotropic field : d5, , , , , , , , , , , , No. of paired electrons (n'p)= 0 ; Therefore, Eiso = 0, 157, , XII U5 Coordination jagan.indd 157, , 2/19/2020 4:41:04 PM

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www.tntextbooks.in, , High Spin Complex, , Low spin complex, , Electronic configuration( for high spin com- Electronic configuration ( for low spin complex) : t 5 2g e0g, plex) : t 32g e2g, CFSE = {[3(-0.4)+2 (0.6)] Δo + 0 × P} - {0}, , CFSE = {[5(-0.4)+0 (0.6)] Δo + 2 × P} - {0}, , =0, , = -2 Δo + 2P, = (-2 ×14000) + (2× 30000), = 32000 cm-1, High positive CFSE value indicates that low, spin complex is not a favourable one., , Actual nature of [Fe(H2O)6]3+, , High spin (Spin free), , Electronic configuration of central metal ion, , t 32g e2g, No. of unparied electrons = 5;, , Magnetic property, , Hence paramagnetic, , Magnetic moment, , µs = n(n+ 2) = 5(5 + 2) = 5.916 BM, , (Using spin only formula), Complex: [Fe(CN)6]3High Spin Complex, , Low spin complex, , Electronic configuration in isotropic field : d5 , , , , , , , , , , No. of paired electrons (n'p)= 0 ; Therefore, Eiso = 0, Ligand field:, , Ligand field Electronic configuration : t5 2g e0g, , Electronic configuration : t 32g e2g, , CFSE = {[5(-0.4)+0 (0.6)] Δo + 2 × P} - {0}, = -2 Δo + 2P, , CFSE = {[3(-0.4)+2 (0.6)] Δo + 0 × P} - {0}, =0, , = (-2 ×35000) + (2× 30000), = -10000 cm-1, Negative CFSE value indicates that low spin, complex is favoured, , 158, , XII U5 Coordination jagan.indd 158, , 2/19/2020 4:41:05 PM

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www.tntextbooks.in, , Nature of the complex, , Low spin (Spin paired), , Electronic configuration of central metal ion, , t 5 2g e0g, No. of unparied electrons = 1;, Hence paramagnetic, , Magnetic property, Magnetic moment, (Using spin only formula), , µs = n(n+ 2) = 1(1+ 2) = 1.732 BM, , Colour of the complex and crystal field splitting, energy:, , Blue, 450-480 nm, , Most of the transition metal complexes are, Indigo, 400-450 nm, Green, coloured. A substance exhibits colour when it absorbs, 480-560 nm, the light of a particular wavelength in the visible region, Violet, and transmit the rest of the visible light. When this, Yellow, 400-450 nm, 560-600 nm, transmitted light enters our eye, our brain recognises, orange, 600-640 nm, its colour. The colour of the transmitted light is given, Red, 640-700 nm, by the complementary colour of the absorbed light., For example, the hydrated copper(II) ion is blue, Figure 5.15 Colour Wheel in colour as it absorbs orange light, and transmit, Complementary colours are shown, its complementary colour, blue. A list of absorbed, on opposite sides., wavelength and their complementary colour is given, in the following table., Wave length(λ) of, absorbed light (Å), , Wave number(ν) of the Colour of, absorbed light (cm-1) absorbed light, , Observed, Colour, , 4000, , 25000, , Violet, , Yellow, , 4750, , 21053, , Blue, , Orange, , 5100, , 19608, , Green, , Red, , 5700, , 17544, , Yellow, , Violet, , 5900, , 16949, , Orange, , Blue, , 6500, , 15385, , Red, , Green, , The observed colour of a coordination compound can be explained using crystal field, theory. We learnt that the ligand field causes the splitting of d orbitals of the central metal atom, into two sets (t2g and eg). When the white light falls on the complex ion, the central metal ion, absorbs visible light corresponding to the crystal filed splitting energy and transmits rest of the, light which is responsible for the colour of the complex., 159, , XII U5 Coordination jagan.indd 159, , 2/19/2020 4:41:06 PM

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www.tntextbooks.in, , This absorption causes excitation of d-electrons of central metal ion from the lower energy, t2g level to the higher energy eg level which is known as d-d transition., Let us understand the d-d transitions, by considering [Ti(H2O)6]3+ as an example., In this complex the central metal ion is, Ti3+, which has d1 configuration. This, single electron occupies one of the t2g, orbitals in the octahedral aqua ligand field., When white light falls on this complex, the d electron absorbs light and promotes, itself to eg level. The spectral data show, the absorption maximum is at 20000 cm-1, corresponding to the crystal field splitting, energy (Δo) 239.7 kJ mol-1. The transmitted, colour associated with this absorption is, purple and hence ,the complex appears, purple in colour., , dx2 y2, dz2, -, , eg, , ∆o=239.7 kJ mol-1, , dxy , dyz , dxz, , t2g, , Figure 5.16 d-d Transition, , The octahedral titanium(III) complexes with other ligands such as bromide and fluoride, have different colours. This is due to the difference in the magnitude of crystal field splitting, by these ligands (Refer page 156). However, the complexes of central metal atom such as of, Sc3+, Ti4+, Cu+, Zn2+, etc... are colourless. This is because the d-d transition is not possible in, complexes with central metal having d0 or d10 configuration., Evaluate yourself 8:, 11. The mean pairing energy and octahedral field splitting energy of [Mn(CN)6]3- are, 28,800 cm-1 and 38500 cm-1 respectively. Whether this complex is stable in low spin or, high spin?, 12. Draw energy level diagram and indicate the number of electrons in each level for the, complex [Cu(H2O)6]2+. Whether the complex is paramagnetic or diamagnetic?, 13. For the [CoF6]3- ion the mean pairing energy is found to be 21000 cm-1 . The magnitude, of Δ0 is 13000cm-1. Calculate the crystal field stabilization energy for this complex ion, corresponding to low spin and high spin states., Metallic carbonyls, Metal carbonyls are the transition metal complexes of carbon monoxide, containing MetalCarbon bond. In these complexes CO molecule acts as a neutral ligand. The first homoleptic, carbonyl Ni ( CO ) 4  nickel tetra carbonyl was reported by Mond in 1890.These metallic, carbonyls are widely studied because of their industrial importance, catalytic properties and, their ability to release carbon monoxide., 160, , XII U5 Coordination jagan.indd 160, , 2/19/2020 4:41:07 PM

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www.tntextbooks.in, , Classification:, Generally metal carbonyls are classified in two different ways as described below., (i) Classification based on the number of metal atoms present., Depending upon the number of metal atoms present in a given metallic carbonyl, they are, classified as follows., a. Mononuclear carbonyls, These compounds contain only one metal atom, and have comparatively simple structures., For example, Ni ( CO ) 4  - nickel tetracarbonyl is tetrahedral, Fe ( CO )5  - Iron pentacarbonyl, , is trigonalbipyramidal, and Cr ( CO )6  - Chromium hexacarbonyl is octahedral., b. Poly nuclear carbonyls, , Metallic carbonyls containing two or more metal atoms are called poly nuclear carbonyls. Poly, , (, , nuclear metal carbonyls may be Homonuclear Co2 ( CO )8  , Mn2 ( CO )10  , Fe3 ( CO )12 , or hetero nuclear MnCo ( CO )9  , MnRe ( CO )10  etc., , (, , ), , ), , (ii) Classification based on the structure:, The structures of the binuclear metal carbonyls involve either metal–metal bonds or, bridging CO groups, or both. The carbonyl ligands that are attached to only one metal atom, are referred to as terminal carbonyl groups, whereas those attached to two metal atoms, simultaneously are called bridging carbonyls. Depending upon the structures, metal carbonyls, are classified as follows., a. Non-bridged metal carbonyls:, These metal carbonyls do not contain any bridging carbonyl ligands. They may be of two, types., (i) Non- bridged metal carbonyls which contain only terminal carbonyls. Examples:, Ni ( CO ) 4  , Fe ( CO )5  and Cr ( CO )6 , , CO, , O, , CO, , O, Ni, , C, C, , CO, OC, , O, , CO, ,   , , C, , OC, , C, , Fe, , CO, Cr, , O, OC, , C, ,   , , O, , CO, , CO, , (ii) Non- bridged metal carbonyls which contain terminal carbonyls as well as Metal-Metal, bonds. For examples,The structure of Mn2(CO)10actually involve only a metal–metal, bond, so the formula is more correctly represented as (CO)5Mn−Mn(CO)5., 161, , XII U5 Coordination jagan.indd 161, , 2/19/2020 4:41:18 PM

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www.tntextbooks.in, , CO, , CO, CO, , OC, Mn, , OC, , CO, , Mn, , 2.79A0, , OC, , CO, CO, , CO, , Other examples of this type are,Tc2(CO)10, and Re2(CO)10., b. Bridged carbonyls:, These metal carbonyls contain one or more bridging carbonyl ligands along with terminal, carbonyl ligands and one or more Metal-Metal bonds. For example,, , CO, , CO, OC, , CO, , (i) The structure of Fe2(CO)9, di-iron nona carbonyl molecule consists of three bridging CO ligands,, six terminal CO groups, , CO, Fe, , CO, , Fe, CO, , CO, , CO, , CO, , CO, , CO, , (ii) For dicobaltoctacarbonylCo2(CO)8two isomers are possible. The one has a metal–metal, bond between the cobalt atoms, and the other has two bridging CO ligands., CO, , CO, , CO, , OC, Co, , 0, , 2.52 A, , OC, , Co, , OC, , CO, , CO, , CO, , CO, , Co, , CO, , CO, , Co, , CO, , Bonding in metal carbonyls, In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of, two components. The first component is an electron pair donation from the carbon atom, of carbonyl ligand into a vacant d-orbital of central metal atom. This electron pair donation, σ bond, forms M ←,  CO sigma bond. This sigma bond formation increases the electron density, in metal d orbitals and makes the metal electron rich. In order to compensate for this increased, electron density, a filled metal d-orbital interacts with the empty π* orbital on the carbonyl, ligand and transfers the added electron density back to the ligand. This second component, 162, , XII U5 Coordination jagan.indd 162, , 2/19/2020 4:41:19 PM

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www.tntextbooks.in, , π-back, is called π-back bonding . Thus in metal carbonyls, electron, density moves from ligand to, M, C, O, bond, metal through sigma bonding and from metal to ligand through pi bonding, this, synergic, π-back, M, C, O, is shown, effect accounts for strong M ← CO bond in metal carbonyls. This phenomenonbond, π, ,π, diagrammatically as follows., x y, , πx,πy, , M, , C, , π-back, bond, , OM, , O, , πx,πy, M, , Mπ-forwardC, , O, , bond, , πx,πy, , 5.7 Stability of metal complexes:, , M, πx,πy, O, M, σs, , π-forward, bond, , π-forward, bond, , O, , πx,πy, σ-forward, bond, , C, , σs, , O, , σ-forward, bond, , The stability of coordination complexes can be interpreted in two different ways. The first, σ-forward, M, C stability, O and second, one is thermodynamic, one is kinetic stability. Thermodynamic stability, bond, of a coordination complex, σsrefers to the free energy change (∆G) of a complex formation, reaction. Kinetic stability of a coordination complex refers to the ligand substitution. In some, cases, complexes can undergo rapid ligand substitution; such complexes are called labile, complexes. However, some complexes undergo ligand substitution very slowly (or sometimes, no substitution), such complexes are called inert complexes., Stability constant:(β), The stability of a coordination complex is a measure of its resistance to the replacement of, one ligand by another. The stability of a complex refers to the degree of association between two, species involved in an equilibrium. Let us consider the following complex formation reaction, , , Cu 2+ + 4 NH3 , ,  Cu ( NH3 ) 4 , Cu ( NH3 ) 4 , β=, 4, Cu 2+  [NH3 ], , 2+, , 2+, ,    ---------( 1 ), , So, as the concentration of Cu ( NH3 ) 4  increases the value of stability complexes also, increases. Therefore the greater the value of stability constant greater is the stability of the, 2+, , complex., Generally coordination complexes are stable in their solutions; however, the complex ion, can undergo dissociation to a small extent. Extent of dissociation depends on the strength of, the metal ligand bond, thus Stronger the M ← L , lesser is the dissociation., 163, , XII U5 Coordination jagan.indd 163, , 2/19/2020 4:41:25 PM

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www.tntextbooks.in, , In aqueous solutions, when complex ion dissociates, there will be equilibrium between, undissociated complex ion and dissociated ions. Hence the stability of the metal complex can, be expressed in terms of dissociation equilibrium constant or instability constant ( α) . For, 2+, example let us consider the dissociation of Cu ( NH3 ) 4  in aqueous solution., 2+, , , Cu ( NH3 ) 4  , ,  Cu + 4 NH3, 2+, , The dissociation equilibrium constant or instability constant is represented as follows,, Cu 2+  [NH3 ], α =, 2+, Cu ( NH3 ) 4     ---------( 2 ), 4, , From (1) and (2) we can say that, the reciprocal of dissociation equilibrium constant ( α), is called as formation equilibrium constant or stability constant ( β) .,  1, β=  ,  α, Significance of stability constants, , The stability of coordination complex is measured in terms of its stability constant ( β) ., Higher the value of stability constant for a complex ion, greater is the stability of the complex, ion. Stability constant values of some important complexes are listed in table, Complex ion, , Fe ( SCN ), , 2+, , Cu ( NH3 ) 4 , Ag ( CN )2 , , −, , Co ( NH3 )6 , Hg ( CN ) 4 , , 2+, , 3+, , 2−, , Instability constant value ( α), , stability constant value ( β), , −3, , 1.0 × 103, , 1.0 × 10, , −12, , 1.0 × 1012, , 1.8 × 10, , −19, , 5.4 × 1018, , 6.2 × 10, , −36, , 1.6 × 1035, , 4.0 × 10, , − 42, , 2.5 × 10 41, , 1.0 × 10, , By comparing stability constant values in the above table, we can say that among the five, 2−, complexes listed, Hg ( CN ) 4  is most stable complex ion and Fe ( SCN ), 5.7.1. Stepwise formation constants and overall formation constants, , 2+, , is least stable., , When a free metal ion is in aqueous medium, it is surrounded by (coordinated with), water molecules. It is represented as [MS6]. If ligands which are stronger than water are added, to this metal salt solution, coordinated water molecules are replaced by strong ligands., 164, , XII U5 Coordination jagan.indd 164, , 2/19/2020 4:41:50 PM

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www.tntextbooks.in, , Let us consider the formation of a metal complex ML6 in aqueous medium.(Charge on the, metal ion is ignored) complex formation may occur in single step or step by step., If ligands added to the metal ion in single step, then, , [MS ], , , , + 6L , ,  [ML 6 ] + 6 S, , 6, , [ML ] [S ], [MS ] [L ], , 6, , βoverall =, , 6, , 6, , 6, , βoverall is called as overall stability constant. As solvent is present in large excess, its, concentration in the above equation can be ignored., ∴ βoverall =, , [ML ], [MS ] [L ], 6, , 6, , 6, , If these six ligands are added to the metal ion one by one, then the formation of complex, [ML6] can be supposed to take place through six different steps as shown below. Generally step, wise stability constants are represented by the symbol k., , [MS L ], [MS ][L ], [MS L ], =, [MS L ][L ], [MS L ], =, [MS L ][L ], [MS L ], =, [MS L ][L ], [MSL ], =, [MS L ][L ], [ML ], =, [MSL ][L ], , [MS ], , , , + L, ,  [MS5L ] + S, , k1 =, , [MS L ], , , , +L , ,  [MS 4 L 2 ] + S, , k2, , [MS L ], , , , + L, ,  [MS3L 3 ] + S, , k3, , [MS L ], , , , + L, ,  [MS2 L 4 ] + S, , k4, , , , L, ,  [MSL5 ] + S, , k5, , 6, , 5, , 4, , 2, , 3, , 3, , [MS L ] +, 2, , 4, , [MSL ], 5, , , , + L, ,  [ML 6 ], , 5, , 6, , 4, , 2, , 3, , 3, , 5, , 4, , 2, , 2, , 3, , k6, , 3, , 5, , 2, , + S, , 4, , 4, , 6, , 5, , In the above equilibrium, the values k1 , k 2 , k 3 , k 4 , k 5 and k 6 are called step wise stability, constants. By carrying out small a mathematical manipulation, we can show that overall, stability constant β is the product of all step wise stability constants k1 , k 2 , k 3 , k 4 , k 5 and k 6 ., β = k1 × k 2 × k 3 × k 4 × k 5 × k 6, , On taking logarithm both sides, log ( β) = log ( k1 ) + log ( k 2 ) + log ( k 3 ) + log ( k 4 ) + log ( k 5 ) + log ( k 6 ), 165, , XII U5 Coordination jagan.indd 165, , 2/19/2020 4:41:57 PM

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www.tntextbooks.in, , 5.8 Importance and applications of coordination complexes:, The coordination complexes are of great importance. These compounds are present in, many plants, animals and in minerals. Some Important applications of coordination complexes, are described below., 1. Phthalo blue – a bright blue pigment is a complex of Copper (II) ion and it is used in, printing ink and in the packaging industry., 2. Purification of Nickel by Mond’s process involves formation [Ni(CO)4], which Yields 99.5%, pure Nickel on decomposition., 3. EDTA is used as a chelating ligand for the separation of lanthanides,in softening of hard, water and also in removing lead poisoning., 4. Coordination complexes are used in the extraction of silver and gold from their ores by, forming soluble cyano complex. These cyano complexes are reduced by zinc to yield metals., This process is called as Mac-Arthur –Forrest cyanide process., 5. Some metal ions are estimated more accurately by complex formation. For example, Ni2+, ions present in Nickel chloride solution is estimated accurately for forming an insoluble, complex called [Ni(DMG)2]., 6. Many of the complexes are used as catalysts in organic and inorganic reactions. For example,, (i), , Wilkinson’s catalyst - ( PPh 3 )3 RhCl  is used for hydrogenation of alkenes., , (ii) Ziegler-Natta catalyst - [TiCl 4 ]+ Al ( C 2 H5 )3 is used in the polymerization of ethene., 7. In order to get a fine and uniform deposit of superior metals (Ag, Au, Pt etc.,) over, −, , −, , base metals, Coordination complexes Ag ( CN )  and Au ( CN )  etc., are used in, 2, 2, , , electrolytic bath., 8. Many complexes are used as medicines for the treatment of various diseases. For example,, (1) Ca-EDTA chelate, is used in the treatment of lead and radioactive poisoning. That is for, removing lead and radioactive metal ions from the body., (2) Cis-platin is used as an antitumor drug in cancer treatment., 9. In photography, when the developed film is washed with sodium thio sulphatesolution, (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called, sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with, water., AgBr + 2 Na 2S2O3 →,  Na 3 Ag ( S2O3 )2  + 2 NaBr, , 166, , XII U5 Coordination jagan.indd 166, , 2/19/2020 4:42:00 PM

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www.tntextbooks.in, , 10. Many biological systems contain metal complexes. For example,, (i), , A red blood corpuscles (RBC) is composed of heme group, which is Fe2+- Porphyrin, complex.it plays an important role in carrying oxygen from lungs to tissues and, carbon dioxide from tissues to lungs., , (ii), , Chlorophyll, a green pigment present in green plants and algae, is a coordination complex, containing Mg2+ as central metal ion surrounded by a modified Porphyrin ligand called, corrin ring. It plays an important role in photosynthesis, by which plants converts CO2 and, water into carbohydrates and oxygen., , (iii), , Vitamin B12(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination, complex in which the central metal ion is Co+ surrounded by Porphyrin like ligand., , (iv), , Many enzymes are known to be metal complexes, they regulate biological processes. For, example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in, digestion, contains a zinc ion coordinated to the protein., , Cisplatin:, Cisplatin is a square planar coordination, complex (cis- [Pt (NH3)2Cl2]), in which two, similar ligands are in adjacent positions., It is a Platinum-based anticancer drugThis drug undergoes, hydrolysis and reacts with DNA to produce various, crosslinks. These crosslinks hinder the DNA replication, and transcription, which results in cell growth inhibition and ultimately cell death., It also crosslinks with cellular proteins and inhibits mitosis., Summary, , „ When two or more stable compounds in solution are mixed together and allowed, to evaporate, in certain cases there is a possibility for the formation of double, salts or coordination compounds. The double salts loose their identity and, dissociates into their constituent simple ions in solutions , whereas the complex, ion in coordination compound, does not loose its identity and never dissociate to, give simple ions., „ According to werner, most of the elements exhibit, two types of valence namely, primary valence and secondary valence and each element tend to satisfy both the, valences.In modern terminology, the primary valence is referred as the oxidation, state of the metal atom and the secondary valence as the coordination number., „ Coordination entity is an ion or a neutral molecule, composed of a central atom,, usually a metal and the array of other atoms or groups of atoms (ligands) that, are attached to it., 167, , XII U5 Coordination jagan.indd 167, , 2/19/2020 4:42:02 PM

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www.tntextbooks.in, , „„ The central atom/ion is the one that occupies the central position in a coordination, entity and binds other atoms or groups of atoms (ligands) to itself, through a, coordinate covalent bond., „„ The ligands are the atoms or groups of atoms bound to the central atom/ion. The, atom in a ligand that is bound directly to the central metal atom is known as a, donor atom., „„ The complex ion of the coordination compound containing the central metal, atom/ion and the ligands attached to it, is collectively called coordination sphere, and are usually enclosed in square brackets with the net charge., „„ The three dimensional spacial arrangement of ligand atoms/ions that are, directly attached to the central atom is known as the coordination polyhedron, (or polygon)., „„ The number of ligand donor atoms bonded to a central metal ion in a complex, is called the coordination number of the metal., „„ The oxidation state of a central atom in a coordination entity is defined as the, charge it would bear if all the ligands were removed along with the electron pairs, that were shared with the central atom., „„ This type of isomers arises when an ambidentate ligand is bonded to the central, metal atom/ion through either of its two different donor atoms., „„ This type of isomers arises in the coordination compounds having both the cation, and anion as complex ions. The interchange of one or more ligands between the, cationic and the anionic coordination entities result in different isomers., „„ Ionisation isomers arises when an ionisable counter ion (simple ion) itself can, act as a ligand. The exchange of such counter ions with one or more ligands in, the coordination entity will result in ionisation isomers., „„ Geometrical isomerism exists in heteroleptic complexes due to different possible, three dimensional spatial arrangements of the ligands around the central metal, atom. This type of isomerism exists in square planer and octahedral complexes., „„ Coordination compounds which possess chairality exhibit optical isomerism, similar to organic compounds. The pair of two optically active isomers which, are mirror images of each other are called enantiomers., „„ Linus pauling proposed the Valence Bond Theory (VBT) which assumes that the, bond formed between the central metal atom and the ligand is purely covalent., Bethe and Van vleck treated the interaction between the metal ion and the, ligands as electrostatic and extended the Crystal Field Theory (CFT) to explain, the properties of coordination compounds., , 168, , XII U5 Coordination jagan.indd 168, , 2/19/2020 4:42:02 PM

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www.tntextbooks.in, , EVALUATION, Choose the correct answer:, 1. The sum of primary valence and secondary valence of, the metal M in the complex M ( en )2 ( Ox )  Cl is L, a) 3 b) 6 , , c) ­-3 , , d) 9, , 2. An excess of silver nitrate is added to 100ml of a 0.01M solution of, pentaaquachloridochromium(III)chloride. The number of moles of AgCl precipitated, would be, a)0.02 , , b) 0.002, , c) 0.01 , , d) 0.2, , 3. A complex has a molecular formula MSO4Cl. 6H2O .The aqueous solution of it gives white, precipitate with Barium chloride solution and no precipitate is obtained when it is treated, with silver nitrate solution. If the secondary valence of the metal is six, which one of the, following correctly represents the complex?, a) M ( H2O ) 4 Cl  SO4 .2H2O , , b) M ( H2O )6  SO4, , c) M ( H2O )5 Cl  SO4 .H2O d) M ( H2O )3 Cl  SO4 .3H2O, , 4. Oxidation state of Iron and the charge on the ligand NO in Fe ( H2O )5 NO SO4 are, a) +2 and 0 respectively, , b) +3 and 0 respectively, , c) +3 and -1 respectively, , d) +1 and +1 respectively, , 5. As per IUPAC guidelines, the name of the complex Co ( en )2 ( ONO ) Cl  Cl is, a) chlorobisethylenediaminenitritocobalt(III) chloride, b) chloridobis(ethane-1,2-diamine)nitro, , -Ocobaltate(III) chloride, , de, c) chloridobis(ethane-1,2-diammine)nitrito -Ocobalt(II) chlorid, , d) chloridobis(ethane-1,2-diammine)nitrito κ -Ocobalt(III)chloride, 6. IUPAC name of the complex K 3 Al ( C 2O4 )3  is, a) potassiumtrioxalatoaluminium(III), b) potassiumtrioxalatoaluminate(II), c) potassiumtrisoxalatoaluminate(III), d) potassiumtrioxalatoaluminate(III), 169, , XII U5 Coordination jagan.indd 169, , 2/19/2020 4:42:16 PM

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www.tntextbooks.in, , 7. A magnetic moment of 1.73BM will be shown by one among the following (NEET), a) TiCl 4, , b) [CoCl 6 ], , 4−, 2−, , c) Cu ( NH3 ) 4 , d) Ni ( CN ) 4 , 8. Crystal field stabilization energy for high spin d5 octahedral complex is, 2+, , a) −0.6∆ 0, , b) 0, , c) 2(P −∆ 0 ), , d) 2(P + ∆ 0 ), , 9. In which of the following coordination entities the magnitude of Δ0 will be maximum?, a) Co ( CN )6 , , b) Co ( C 2O4 )3 , , 3−, , 3−, , c) Co ( H2O )6 , d) Co ( NH3 )6 , 10. Which one of the following will give a pair of enantiomorphs?, 3+, , 3+, , a) Cr ( NH3 )6  Co ( CN )6 , c) Pt ( NH3 ) 4  [PtCl 4 ], , b) Co ( en )2 Cl2  Cl, , a) Coordination isomerism, , b) Linkage isomerism, , c) Optical isomerism, , d) Geometrical isomerism, , d) Co ( NH3 ) 4 Cl2  NO2, 11. Which type of isomerism is exhibited by Pt ( NH3 )2 Cl2  ?, , 12. How many geometrical isomers are possible for Pt ( Py ) ( NH3 ) ( Br )( Cl )  ?, a) 3, , b) 4, , c) 0, , d) 15, , 13. Which one of the following pairs represents linkage isomers?, a) Cu ( NH3 ) 4  [PtCl 4 ] and Pt ( NH3 ) 4  [CuCl 4 ], b) Co ( NH3 )5 ( NO3 ) SO4 and Co ( NH3 )5 ( ONO ) , c) Co ( NH3 ) 4 ( NCS )2  Cl and Co ( NH3 ) 4 ( SCN )2  Cl, d) both (b) and (c), 14. Which kind of isomerism is possible for a complex Co ( NH3 ) 4 Br2  Cl ?, a) geometrical and ionization, , b) geometrical and optical, , c) optical and ionization, , d) geometrical only, , 15. Which one of the following complexes is not expected to exhibit isomerism?, a) Ni ( NH3 ) 4 ( H2O )2 , c) Co ( NH3 )5 SO4  Cl, , 2+, , b) Pt ( NH3 )2 Cl2 , d) FeCl6, , 3, , 170, , XII U5 Coordination jagan.indd 170, , 2/19/2020 4:42:33 PM

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www.tntextbooks.in, , 16. A complex in which the oxidation number of the metal is zero is, a) K 4 Fe ( CN )6 , c) Fe ( CO ) , 5, , , b) Fe ( CN )3 ( NH3 )3 , d) both (b) and (c), , a) Fe ( CH3 -CH(NH2 )2 )3  ( PO4 )3, , b) Fe ( H2 N-CH2 -CH2 -NH2 )3  ( PO4 ), , 17. Formula of tris(ethane-1,2-diamine)iron(II)phosphate, , c) Fe ( H2 N-CH2 -CH2 -NH2 )3  ( PO4 )2 d) Fe ( H2 N-CH2 -CH2 -NH2 )3 3 ( PO4 )2, 18. Which of the following is paramagnetic in nature?, b) Co ( NH3 )6 , 2−, d) Ni ( CN ) 4 , , 3+, , a) Zn ( NH3 ) 4 , 2+, c) Ni ( H2O )6 , , 2+, , 19. Fac-mer isomerism is shown by, 3+, , a) Co ( en )3  b), c) Co ( NH3 )3 (Cl )3  , d), , Co ( NH3 ) 4 (Cl )2 , , +, , Co ( NH3 )5 Cl  SO4, , 20. Choose the correct statement., a) Square planar complexes are more stable than octahedral complexes, 2−, , b) The spin only magnetic moment of Cu ( Cl ) 4  is 1.732 BM and it has square planar, structure., c) Crystal field splitting energy ( ∆ 0 ) of [FeF6 ] is higher than the ( ∆ 0 ) of Fe ( CN )6 , 4−, , d) crystal field stabilization energy of V ( H2O )6 , 2+, stabilization of Ti ( H2O ) , 6, , , 2+, , 4−, , is higher than the crystal field, , Answer the following questions:, 1. Write the IUPAC names for the following complexes., i) Na 2 Ni ( EDTA ), ii) Ag ( CN )2 , , −, , iii) Co ( en )  ( SO4 ), 3 2, , 3, iv) Co ( ONO ) ( NH3 ) , 5, , , 2+, , v) Pt ( NH3 )2 Cl(NO2 ), , 171, , XII U5 Coordination jagan.indd 171, , 2/19/2020 4:42:42 PM

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www.tntextbooks.in, , 2. Write the formula for the following coordination compounds., a) potassiumhexacyanidoferrate(II), b) pentacarbonyliron(0), c) pentaamminenitrito −κ −N -cobalt(III)ion, d) hexaamminecobalt(III)sulphate, e) sodiumtetrafluoridodihydroxidochromate(III), 3. Arrange the following in order of increasing molar conductivity, , i) Mg Cr ( NH3 ) ( Cl )5      ii) Cr ( NH3 )5 Cl  [CoF6 ]2, 3, , iii) Cr ( NH3 )3 Cl 3 , 4. Give an example of coordination compound used in medicine and two examples of, biologically important coordination compounds., 5. Based on VB theory explain why Cr ( NH3 )6 , diamagnetic., , 3+, , 2−, , is paramagnetic, while Ni ( CN ) 4  is, , 6. Draw all possible geometrical isomers of the complex Co ( en )2 Cl 2  and identify the, optically active isomer., +, , 7. Ti ( H2O )6  is coloured, while Sc ( H2O )6  is colourless- explain., 8. Give an example for complex of the type [Ma 2 b2 c2 ] where a, b, c are monodentate ligands, and give the possible isomers., 3+, , 3+, , 9. Give one test to differentiate Co ( NH3 )5 Cl  SO4 and Co ( NH3 )5 SO4  Cl ., 10. In an octahedral crystal field, draw the figure to show splitting of d orbitals., 11. What is linkage isomerism? Explain with an example., 12. Classify the following ligand based on the number of donor atoms., a) NH3  b) en   c) ox2-   d) pyridine, 13. Give the difference between double salts and coordination compounds., 14. Write the postulates of Werner’s theory., 15. Why tetrahedral complexes do not exhibit geometrical isomerism., 16. Explain optical isomerism in coordination compounds with an example., 17. What are hydrate isomers? Explain with an example., 18. What is crystal field splitting energy?, 19. What is crystal field stabilization energy (CFSE) ?, 20. A solution of Ni ( H2O )6 , Explain, , 2+, , is green, whereas a solution of Ni ( CN ) 4 , , 2−, , is colorless -, , 172, , XII U5 Coordination jagan.indd 172, , 2/19/2020 4:42:47 PM

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www.tntextbooks.in, , 21. Discuss briefly the nature of bonding in metal carbonyls., 22. What is the coordination entity formed when excess of liquid ammonia is added to an, aqueous solution of copper sulphate?, 3−, , 23. On the basis of VB theory explain the nature of bonding in Co ( C 2O4 )3  ., 24. What are the limitations of VB theory?, 25. Write the oxidation state, coordination number , nature of ligand, magnetic property and, electronic configuration in octahedral crystal field for the complex K 4 Mn ( CN )6  ., , 173, , XII U5 Coordination jagan.indd 173, , 2/19/2020 4:42:48 PM

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www.tntextbooks.in, , ICT Corner, CRYSTAL FIELD THEORY, , By using this tool you can understand, the crystal field splitting of different, metal ions in octahedral and tetrahedral, ligand field and also calculate the Crystal, Field Stabilisation Energy (CFSE) of a, complex using crystal field theory., , Please go to the URL, http://vlab.amrita.edu/index., php?sub=2&brch=193&sim=, 610&cnt=4, (or) Scan the QR code on the, right side, , Steps, •, , Open the browser and type the URL given (or) Scan the QR Code. You will see the webpage as, shown in the figure., Note: One time sign up is needed to access this webpage. Login using your username and password., Once logged in click the simulation tab., , •, , You can select a suitable ligand field splitting using the drop down menu (box 1). Select a metal of, interest and a ligand using the drop down menu (box 2). Now crystal field splitting for the selected, complex appears on the screen., , •, , Apply crystal field theory to the selected complex and fill the d-electrons in the t2g and eg orbitals by, clicking each orbital. Click on the orbitals thrice to remove electrons. After completion, click submit, button (box 4). Now you can check the correctness of the electron distribution. If wrong try again., , •, , Enter the number of electrons in the t2g & eg orbitlals in the work sheet at the bottom of the page, (box 6), The calculated Crystal Field Stabilisation Energy (CFSE) will be displayed., , 1, 2, , 3, 4, 5, , 6, , 175, , XII U5 Coordination jagan.indd 175, , 2/19/2020 4:42:49 PM

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www.tntextbooks.in, , UNIT, , 6, , Sir William Henry Bragg, (1862 –1942), , SOLID STATE, , Sir Lawrence Bragg, (1890 –1971), , Learning Objectives, , Sir William Henry Bragg was a, British physicist, chemist, and a, mathematician. Sir William Henry, Bragg and his son Lawrence Bragg, worked on X-rays with much, success. They invented the X-ray, spectrometer and founded the new, science of X-ray crystallography,, the analysis of crystal structure, using X-ray diffraction. Bragg, was joint winner (with his son,, Lawrence Bragg) of the Nobel, Prize in Physics in 1915, for their, services in the “analysis of crystal, structure by means of ray”. The, mineral Braggite (a sulphide ore of, platinum, palladium and Nickel) is, named after him and his son., , After studying this unit, the students will, be able to, , , , , , , , , , , , , , , describe general characteristics of solids, distinguish amorphous and crystalline, solids, define unit cell, describe different types of voids and, close packed structures, calculate the packing efficiency of, different types of cubic unit cell, solve numerical problems involving, unit cell dimensions, explain point defects in solids, , 176, , XII U6 Solid State - Jerald.indd 176, , 2/19/2020 4:41:27 PM

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www.tntextbooks.in, , INTRODUCTION, , molecules) have fixed positions and, can only oscillate about their mean, positions, , Matter may exist in three different, physical states namely solid, liquid and gas., If you look around, you may find mostly, solids rather than liquids and gases. Solids, differ from liquids and gases by possessing, definite volume and definite shape. In the, solids the atoms or molecules or ion are, tightly held in an ordered arrangement, and there are many types of solids such, as diamond, metals, plastics etc., and, most of the substances that we use in, our daily life are in the solid state. We, require solids with different properties for, various applications. Understanding the, relation between the structure of solids, and their properties is very much useful, in synthesizing new solid materials with, different properties., , 6.2 Classification of solids:, Classification of solids, , Crystalline solids, , Ionic crystals Ex: NaCl ,KCl, Covalent crystals Ex: Diamond, SiO2, Molecular crystals Ex: naphthalene, anthracene, glucose, Metallic crystals Ex: All metallic elements, (Na, Mg,Cu,Au,Ag etc..), Atomic solids - ex: frozen elements of Group 18, , We can classify solids into the, following two major types based on the, arrangement of their constituents., (i) Crystalline solids, (ii) Amorphous solids., The term crystal comes from the, Greek word “krystallos” which means, clear ice. This term was first applied to the, transparent quartz stones, and then the, name is used for solids bounded by many, flat, symmetrically arranged faces., , In this chapter, we study the, characteristics of solids, classification,, structure and their properties; we also, discuss the crystal defects and their, significance., , 6.1 General characteristics of solids, We have already learnt in XI STD that, gas molecules move randomly without, exerting reasonable forces on one another., Unlike gases, in solids the atoms , ions or, molecules are held together by strong force, of attraction. The general characteristics, of solids are as follows,, (i), , Amorphous solids, Ex: Glass, rubber etc, , Solids have definite volume and, shape., , (ii) Solids are rigid and incompressible, (iii) Solids have strong cohesive forces., (iv) Solids have short inter atomic, ionic, or molecular distances., (v) Their constituents ( atoms , ions or, 177, , XII U6 Solid State - Jerald.indd 177, , 2/19/2020 4:41:28 PM

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www.tntextbooks.in, , A crystalline solid is one in which its constituents (atoms, ions or molecules), have, an orderly arrangement extending over a long range. The arrangement of such constituents, in a crystalline solid is such that the potential energy of the system is at minimum. In, contrast, in amorphous solids (In Greek, amorphous means no form) the constituents are, randomly arranged., The following table shows the differences between crystalline and amorphous solids., S.no, , Crystalline solids, Long range orderly arrangement of, constituents., , Amorphous solids, Short range, random arrangement of, constituents., , 2, , Definite shape, , Irregular shape, , 3, , Generally crystalline solids are, anisotropic in nature, , They are isotropic* like liquids, , 4, , They are true solids, , They are considered as pseudo solids (or), super cooled liquids, , 5, , Definite Heat of fusion, , Heat of fusion is not definite, , 6, , They have sharp melting points., , Gradually soften over a range of, temperature and so can be moulded., , 7, , Examples: NaCl , diamond etc.,, , Examples: Rubber , plastics, glass etc, , 1, , Table 6.1 differences between crystalline and amorphous solids, *Isotropy, Isotropy means uniformity in all directions. In solid state isotropy means having identical, values of physical properties such as refractive index, electrical conductance etc., in all, directions, whereas anisotropy is the property which depends on the direction of measurement., Crystalline solids are anisotropic and they show different values of physical properties when, measured along different directions. The following figure illustrates the anisotropy in crystals, due to different arrangement of their constituents along different directions., A, , A, A, , B, , Anisotropy in Crystals, , Isotropy in Amorphous solids, , 178, , XII U6 Solid State - Jerald.indd 178, , 2/19/2020 4:41:30 PM

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www.tntextbooks.in, , 6.3 Classification of crystalline, solids:, , 6.3.2 Covalent solids:, In covalent solids, the constituents, (atoms) are bound together in a three, dimensional network entirely by covalent, bonds. Examples: Diamond, silicon, carbide etc. Such covalent network crystals, are very hard, and have high melting, point. They are usually poor thermal and, electrical conductors., , 6.3.1 Ionic solids:, The structural units of an ionic crystal, are cations and anions. They are bound, together by strong electrostatic attractive, forces. To maximize the attractive force,, cations are surrounded by as many anions, as possible and vice versa. Ionic crystals, possess definite crystal structure; many, solids are cubic close packed. Example:, The arrangement of Na+ and Cl- ions in, NaCl crystal., , 6.3.3 Molecular solids:, In molecular solids, the constituents, are neutral molecules. They are held, together by weak van der Waals forces., Generally molecular solids are soft and, they do not conduct electricity. These, molecular solids are further classified into, three types., Graphite is used, inside pencils. It slips, easily off the pencil, onto the paper and, leaves a blackmark. Graphite is also, a component of many lubricants , for, example bicycle chain oil , because it, is slippery, , Characteristics:, 1. Ionic solids have high melting points., 2. These solids do not conduct electricity,, because the ions are fixed in their, lattice positions., 3. They do conduct electricity in molten, state (or) when dissolved in water, because, the ions are free to move in, the molten state or solution., , (i) Non-polar molecular solids:, , 4. They are hard as only strong external, force can change the relative positions, of ions., , In non polar molecular solids, constituent molecules are held together by, weak dispersion forces or London forces., 179, , XII U6 Solid State - Jerald.indd 179, , 2/19/2020 4:41:31 PM

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www.tntextbooks.in, , They have low melting points and are, usually in liquids or gaseous state at room, temperature. Examples: naphthalene,, anthracene etc.,, , molecules, relative to one another in a, three dimensional pattern. The regular, arrangement of these species throughout, the crystal is called a crystal lattice. A basic, repeating structural unit of a crystalline, solid is called a unit cell. The following, figure illustrates the lattice point and the, unit cell., , (ii) Polar molecular solids, T h e, c ons t itu e nt s, are molecules, formed, by, polar covalent, bonds. They, are, held, together, by, relatively, strong dipole-dipole interactions. They, have higher melting points than the nonpolar molecular solids. Examples are solid, CO2 , solid NH3 etc., (iii), , – Unit cell, , – Lattice points, , A crystal may be considered to, consist of large number of unit cells,, each one in direct contact with its nearer, neighbour and all similarly oriented in, space. The number of nearest neighbours, that surrounding a particle in a crystal is, called the coordination number of that, particle., , Hydrogen bonded molecular solids, , The constituents are held together by, hydrogen bonds. They are generally soft, solids under room temperature. Examples:, solid ice (H2O), glucose, urea etc.,, 6.3.4 Metallic solids:, You have already studied in XI STD, about the nature of metallic bonding., In metallic solids, the lattice points are, occupied by positive metal ions and a, cloud of electrons pervades the space., They are hard, and have high melting, point. Metallic solids possess excellent, electrical and thermal conductivity. They, possess bright lustre. Examples: Metals, and metal alloys belong to this type of, solids, for example Cu,Fe,Zn, Ag ,Au, CuZn etc., , A unit cell is characterised by the, three edge lengths or lattice constants a ,b, and c and the angle between the edges α,, β and γ, , c, , a, , 6.4 Crystal lattice and unit cell:, Crystalline solid is characterised by, a definite orientation of atoms, ions or, , , , b, , , Unit cell, , 180, , XII U6 Solid State - Jerald.indd 180, , 2/19/2020 4:41:33 PM

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www.tntextbooks.in, , 6.5 , Primitive and non-primitive, unit cell, There are two types of unit cells:, primitive and non-primitive. A unit cell, that contains only one type of lattice point, is called a primitive unit cell, which is, made up from the lattice points at each of, the corners., , Primitive, , In case of non-primitive unit cells,, there are additional lattice points, either, on a face of the unit cell or with in the unit, cell., , Non-Primitive, , There are seven primitive crystal systems; cubic, tetragonal, orthorhombic, hexagonal,, monoclinic, triclinic and rhombohedral. They differ in the arrangement of their crystallographic, axes and angles. Corresponding to the above seven, Bravis defined 14 possible crystal systems, as shown in the figure., , Cubic, abc, 90, , Rhombohedral, abc, 90, , Tetragonal, Hexagonal, abc, abc, 90120 90, , c, , , , a, , , , b, , Orthorhombic, abc, 90, , Monoclinic, abc, 9090, , Triclinic, abc, , , 181, , XII U6 Solid State - Jerald.indd 181, , 2/19/2020 4:41:36 PM

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www.tntextbooks.in, , Number of atoms in a cubic unit cell:, , 6.5.2 Body centered cubic unit cell., (BCC), , 6.5.1 Primitive (or) simple cubic unit, cell.(SC), , H, E, , G, , r, , F, , a, , 2rr, D, , A, , r, , 2a, , a, , C, , a, , B, , r, a, , r, , In a body centered cubic unit cell,, each corner is occupied by an identical, particle and in addition to that one atom, occupies the body centre. Those atoms, which occupy the corners do not touch, each other, however they all touch the, one that occupies the body centre. Hence,, each atom is surrounded by eight nearest, neighbours and coordination number, is 8. An atom present at the body centre, belongs to only to a particular unit cell i.e, unshared by other unit cell., , In the simple cubic unit cell, each, corner is occupied by an identical atoms, or ions or molecules. And they touch, along the edges of the cube, do not touch, diagonally. The coordination number of, each atom is 6., Each atom in the corner of the cubic, unit cell is shared by 8 neighboring unit, cells and therefore atoms `per unit cell is, N, equal to c , where N c is the number of, 8, atoms at the corners., , ∴ Number of atoms,  Nc   N b , in a bcc unit cell =  8  +  1 ,  8 1, = + ,  8 1, , ∴ Number of atoms,  Nc , ∴ no of atoms in, inaaSC, SCunit, unitcell, cell =  ,  8 ,  8, = =1,  8, , = (1 + 1), =2, , 183, , XII U6 Solid State - Jerald.indd 183, , 2/19/2020 4:41:44 PM

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www.tntextbooks.in, , 6.5.3 Face centered cubic unit cell.(FCC), , is not an easy task. The constituents in, a unit cell touch each other and form a, three dimensional network. This can be, simplified by drawing crystal structure, with the help of small circles (spheres), corresponding constituent particles and, connecting neighbouring particles using a, straight line as shown in the figure., 6.5.4 Calculations involving unit cell, dimensions:, X-Ray diffraction analysis is the most, powerful tool for the determination of, crystal structure. The inter planar distance, (d) between two successive planes of atoms, can be calculated using the following, equation form the X-Ray diffraction data, 2dsinθ = nλ, , C, , r, a, 2r, a, , r, A, , The above equation is known as, Bragg’s equation., , B, , a, , Where, In a face centered cubic unit cell,, , λ is the wavelength of X-ray used for, diffraction., , identical atoms lie at each corner as well, as in the centre of each face. Those atoms, , θ is the angle of diffraction, , in the corners touch those in the faces, , n is the order of diffraction, , By knowing the values of θ,λ and n, we can calculate the value of d., , but not each other. The atoms in the face, centre is being shared by two unit cells,,  1, each atom in the face centers makes  ,  2, contribution to the unit cell., , d=, , Using these values the edge length of, the unit cell can be calculated., , ∴ Number of atoms, N  N , in a fcc unitcell =  c  +  f ,    , , 8,  8 6, = + ,  8 2, = (1 + 3), , nλ, 2sinθ, , 6.5.5 Calculation of density:, , 2, , Using the edge length of a unit cell, we, can calculate the density ( ρ) of the crystal, by considering a cubic unit cell as follows., Density of, mass of the unit cell, the unit cell ρ = volume of the unit cell, , =4, Drawing the crystal lattice on paper, , ...(1), 184, , XII U6 Solid State - Jerald.indd 184, , 2/19/2020 4:41:53 PM

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www.tntextbooks.in, , total number of , ,  mass of , mass of the unit cell= atoms belongs to  × , , one, atom, , , that unit cell, , , , mass of one atom =, m=, , ...(2), , molar mass (gmol −1 ), Avagadro number (mol −1 ), , M, NA, , ...(3), , Substitute (3) in (2), mass of the unit cell= n ×, , M, NA, , ...(4), , For a cubic unit cell, all the edge lengths are equal i.e , a=b=c, volume of the unit cell = a × a × a = a 3, nM, ∴ Density of the unit cell ρ = a 3 N, A, , ...(5), ...(6), , Equation (6) contains four variables namely ρ , n , M and a . If any three variables are, known, the fourth one can be calculated., Example 2, Barium has a body centered cubic unit cell with a length of 508pm along an edge., What is the density of barium in g cm-3?, Solution:, ρ =, , nM, a 3NA, , In this case,, n=2 ; M=137.3 gmol-1 ; a = 508pm= 5.08X10-8cm, ρ =, ρ =, , 2 atoms × 137.3 g mol −1, , (5.08 × 10, (5.08 ), , −8, , cm ) ( 6.023 × 1023 atoms mol −1 ), 3, , 2 × 137.3, , 3, , × 10, , −24, , × 6.023 × 10, , 23, , g cm −3, , ρ = 3.5 g cm −3, , 185, , XII U6 Solid State - Jerald.indd 185, , 2/19/2020 4:41:59 PM

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www.tntextbooks.in, , Evaluate yourself -1, 1. An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The, density of the element is 8.9 gcm-3. How many atoms are present in 100 g of an element?, 2. Determine the density of CsCl which crystallizes in a bcc type structure with an edge, length 412.1 pm., 0, , 3. A face centered cubic solid of an element (atomic mass 60) has a cube edge of 4 A ., Calculate its density., , 6.6 Packing in crystals:, Let us consider the packing of fruits for, display in fruit stalls. They are in a closest packed, arrangement as shown in the following fig. we can, extend this analogy to visualize the packing of, constituents (atoms / ions / molecules) in crystals,, by treating them as hard spheres. To maximize, the attractive forces between the constituents, they generally tend to pack together as close as, possible to each other. In this portion we discuss how to pack identical spheres to create cubic, and hexagonal unit cell. Before moving on to these three dimensional arrangements, let us first, consider the two dimensional arrangement of spheres for better understanding., 6.6.1 Linear arrangement of spheres in one direction:, In a specific direction, there is only, one possibility to arrange the spheres in, one direction as shown in the fig. in this, arrangement each sphere is in contact with two, neighbouring spheres on either side., 6.6.2 Two dimensional close packing:, Two dimensional planar packing can be, done in the following two different ways., (i) AAA… type:, Linear arrangement of spheres in one, direction is repeated in two dimension i.e., more, number of rows can be generated identical to, the one dimensional arrangement such that all, spheres of different rows align vertically as well, as horizontally as shown in the fig. If we denote, the first row as A type arrangement, then the, above mentioned packing is called AAA type, because all rows are identical as the first, one. In this arrangement each sphere is in contact with four of its neighbours., 186, , XII U6 Solid State - Jerald.indd 186, , 2/19/2020 4:42:01 PM

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www.tntextbooks.in, , (i) ABAB.. Type:, In this type, the second row spheres are, arranged in such a way that they fit in the, depression of the first row as shown in the, figure. The second row is denoted as B type., The third row is arranged similar to the first, row A, and the fourth one is arranged similar, to second one. i.e., the pattern is repeated as, ABAB….In this arrangement each sphere is in, contact with 6 of its neighbouring spheres., On comparing these two arrangements, (AAAA...type and ABAB….type) we found, that the closest arrangement is ABAB…type., 6.6.3 Simple cubic arrangement:, This type of three dimensional packing, arrangements can be obtained by repeating the, AAAA type two dimensional arrangements, in three dimensions. i.e., spheres in one layer, sitting directly on the top of those in the, Simple Cubic (SC), previous layer so that all layers are identical. All spheres of different layers of crystal are, perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement, as simple cubic structure as shown in fig., In simple cubic packing, each sphere is in contact with 6 neighbouring spheres Four in its own layer, one above and one below and hence the coordination number of the, sphere in simple cubic arrangement is 6., Packing efficiency:, There is some free space between the spheres of a single layer and the spheres of, successive layers. The percentage of total volume occupied by these constituent spheres, gives the packing efficiency of an arrangement. Let us calculate the packing efficiency in, simple cubic arrangement,, Packing fraction , , =, (or), efficiency, , , , Total volume occupied by , , , , spheres in a unit cell, Volume of the unit cell, , × 100, , Let us consider a cube with an edge length ‘a’ as shown in fig., , r, , Volume of the cube with edge length a is = a × a × a = a 3, , Let ‘r’ is the radius of the sphere. From the figure, a=2r ⇒ r =, , a, , r, , a, 2, , 187, , XII U6 Solid State - Jerald.indd 187, , 2/19/2020 4:42:05 PM

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www.tntextbooks.in, , ∴ Volume of the sphere with radius ‘r’, =, , coordination number of 8, four neighbors, in the layer above and four in the layer, below., , 4 3, πr, 3, , 4  a, = π  , 3  2, , 3, , Layer a, , 4  a3 , = π  , 3  8, =, , πa 3, 6, , Layer b, Layer a, ,   , , ... (1), , Body Certered Cubic (BCC), , In a simple cubic arrangement, number of, spheres belongs to a unit cell is equal to one, ∴ Total volume, 3, occupied by the, spheres in sc unit cell, ,  πa , = 1× ,  6 , , Packing efficiency:, Here, the spheres are touching along the, leading diagonal of the cube as shown in, the fig., , ... (2), , Dividing (2) by (3), , H, ,  πa 3 ,  6 , 100 π, Packing fraction =, × 100 =, 3, 6, (a ), , E, , G, , r, , F, , =52.38%, , a, , 2rr, , i.e., only 52.38% of the available, volume is occupied by the spheres in simple, cubic packing, making inefficient use of, available space and hence minimizing the, attractive forces., , D, , A, , r, , 2a, , a, , C, , a, , B, , More to know Of all the metals, , In ∆ ABC, , in the periodic table, only polonium, crystallizes in simple cubic pattern., , AC 2 = AB2 + BC 2, AC = AB2 + BC 2, , 6.6.4 Body centered cubic arrangement, , AC = a 2 + a 2 = 2a 2 = 2 a, , In this arrangement, the spheres, in the first layer ( A type ) are slightly, separated and the second layer is, formed by arranging the spheres in the, depressions between the spheres in layer, A as shown in figure. The third layer is a, repeat of the first. This pattern ABABAB, is repeated throughout the crystal. In, this arrangement, each sphere has a, , In ∆ ACG, AG2 = AC 2 + CG 2, AG = AC 2 + CG 2, AG =, , ( 2a ), , 2, , + a2, , AG = 2a 2 + a 2 = 3a 2, AG = 3 a, 188, , XII U6 Solid State - Jerald.indd 188, , 2/19/2020 4:42:13 PM

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www.tntextbooks.in, , i.e.,, ,  3 π a3 , =2 × ,  =,  16 , , 3a = 4r, r=, , 3, a, 4, , 3 π a3, 8, , Dividing (2) by (3), , ∴ Volume of the sphere with radius ‘r’, ,  3 π a3 , , 8 , , Packing fraction =, × 100, (a 3 ), , 4 3, πr, 3, 3, 4  3 , a, = π, 3  4 , =, , 3π, × 100, 8, = 3 π × 12.5, = 1.732 × 3.14 × 12.5, = 68 %, =, , 3 3, πa, ...(1), 16, Number of spheres belong to a unit, cell in bcc arrangement is equal to two, and hence the total volume of all spheres, =, , i.e., 68 % of the available volume is, occupied. The available space is used more, efficiently than in simple cubic packing., , 6.6.5 The hexagonal and face centered, cubic arrangement:, , , , , Formation of first layer:, , , , , In this arrangement, the first layer, is formed by arranging the spheres as, in the case of two dimensional ABAB, arrangements i.e. the spheres of second, row fit into the depression of first row., Now designate this first layer as ‘a’. The, next layer is formed by placing the spheres, in the depressions of the first layer. Let, the second layer be ‘b’., , , , , , , , , , , , , , , , , , , , , , , , , Formation of second layer:, , , In the first layer (a) there are two, types of voids (or holes) and they are, designated as x and y. The second layer, (b) can be formed by placing the spheres, either on the depression (voids/holes) x, (or) on y. let us consider the formation of, second layer by placing the spheres on the, depression (x)., , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,  , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 189, , XII U6 Solid State - Jerald.indd 189, , , , 2/19/2020 4:42:18 PM

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www.tntextbooks.in, , Wherever a sphere of second layer, (b) is above the void (x) of the first layer, (a), a tetrahedral void is formed. This, constitutes four spheres – three in the, lower (a) and one in the upper layer (b)., When the centers of these four spheres, are joined, a tetrahedron is formed., , , , , , , , , , , , At the same time, the voids (y) in the, first layer (a) are partially covered by the, spheres of layer (b), now such a void in (a), is called a octahedral void. This constitutes, six spheres – three in the lower layer (a), and three in the upper layer (b). When, the centers of these six spheres are joined,, an octahedron is formed. Simultaneously, new tetrahedral voids (or holes) are also, created by three spheres in second layer, (b) and one sphere of first layer (a), , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The number of voids depends on, the number of close packed spheres., If the number of close packed spheres, be ‘n’ then, the number of octahedral, voids generated is equal to n and the, number of tetrahedral voids generated, , , , , is equal to, 2n. , , , , , , , Formation of third layer:, , , , , , , , , , , , , , , , , ,  closest,  packing, , The third layer of spheres can be formed in two ways to achieve, , (i) aba arrangement - hcp structure, , , , , , , , , , , , , , , , , , , , (ii) abc arrangement – ccp structure, The spheres can be arranged so as to fit into the depression in such a way that the third, layer is directly over a first layer as shown in the figure. This “aba’’ arrangement is known, as the hexagonal close packed (hcp) arrangement. In this arrangement, the tetrahedral, voids of the second layer are covered by the spheres of the third layer., Alternatively, the third layer may be placed over the second layer in such a way that all, the spheres of the third layer fit in octahedral voids. This arrangement of the third layer is, different from other two layers (a) and (b), and hence, the third layer is designated (c). If, the stacking of layers is continued in abcabcabc… pattern, then the arrangement is called, cubic close packed (ccp) structure., In both hcp and ccp arrangements, the coordination number of each sphere is 12 – six, neighbouring spheres in its own layer, three spheres in the layer above and three sphere in, the layer below. This is the most efficient packing., 190, , XII U6 Solid State - Jerald.indd 190, , 2/19/2020 4:42:19 PM

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www.tntextbooks.in, , Radius ratio:, , The cubic close packing is based on, the face centered cubic unit cell. Let us, calculate the packing efficiency in fcc unit, cell., , a, , crystals the bigger anions are present in, , 4r = a 2, , 2r, , the close packed arrangements and the, , a 2, r=, 4, , B, , a, , cations occupy the voids. The ratio of,  rC+ ,  plays,  rA− , , radius of cation and anion , , , In ∆ ABC, AC 2 = AB2 + BC 2, , an important role in determining the, , 2, , structure. The following table shows the, , AC = a + a = 2a = 2 a, 2, , 2, , Volume of the sphere, with radius r is, , rleation between the radius ratio and the, , 4  2a , = π, 3  4 , , =, =, , 3, , structural arrangement in ionic solids., , 2 πa 3, 24, , ∴ the volume, ,  2 πa 3 , = 4×, , of all spheres in a fcc,  24 , unit cell, , 2r, a, , r, , a, , =, , B2O3, , 0.225 –, 0.414, , 4, , Tetrahedral, , ZnS, , 0.414 –, 0.732, , 6, , Octahedral, , NaCl, , 0.732 – 1.0, , 8, , Cubic, , CsCl, , According to the law of nature, nothing is perfect, and so crystals need, not be perfect. They always found to have, some defects in the arrangement of their, constituent particles. These defects affect, the physical and chemical properties, of the solid and also play an important, role in various processes. For example, a, process called doping leads to a crystal, , 2π, × 100, 6, 1.414 × 3.14 × 100, , = 74%, , Trigonal, planar, , 6.7 Imperfection in solids:, ,  2 π a3 , , 6 , , packing efficiency =, × 100, (a 3 ), , =, , 3, , , , , , Table 6.3 Radius ratio, ,  2 πa 3 , =, ,  6 , , a, , 0.155 –, 0.225, ,  rC+, ,  rA−, , 4  2 2a 3 , π, 3  64 , , Total number of spheres belongs to a, single fcc unit cell is 4, , r, , Example, , 2, , Structure, , AC = AB + BC, 2, , Coordination, number, , r, , the size of the ions.generally in ionic, , AC = 4r, , r, , A, , depends upon the stoichiometry and, , From the figure, , C, , a, , The structure of an ionic compound, , 6, 192, , XII U6 Solid State - Jerald.indd 192, , 2/19/2020 4:42:28 PM

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www.tntextbooks.in, , imperfection and it increases the electrical conductivity of a semiconductor material such, as silicon. The ability of ferromagnetic material such as iron, nickel etc., to be magnetized, and demagnetized depends on the presence of imperfections. Crystal defects are classified, as follows, 1) Point defects, 2) Line defects, 3) Interstitial defects, 4) Volume defects, In this portion, we concentrate on point defects, more specifically in ionic solids., Point defects are further classified as follows, Point defects, , non- stiochiometric, defects, , stiochiometric, defects, , Schottky defect, , Frenkel defect, , metal excess, defect, , impurity defects, , metal deficiency, defec, , Stoichiometric defects in ionic solid:, This defect is also called intrinsic (or) thermodynamic defect. In stoichiometric ionic, crystals, a vacancy of one ion must always be associated with either by the absence of, another oppositely charged ion (or) the presence of same charged ion in the interstitial, position so as to maintain the electrical neutrality., 6.7.1 Schottky defect:, Schottky defect arises due to, the missing of equal number of, cations and anions from the crystal, lattice. This effect does not change, the stoichiometry of the crystal., Ionic solids in which the cation, and anion are of almost of similar, size show schottky defect. Example:, NaCl., Presence of large number of, schottky defects in a crystal, lowers, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Na+, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Cl, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Schottky Defect, 193, , XII U6 Solid State - Jerald.indd 193, , 2/19/2020 4:42:29 PM

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www.tntextbooks.in, , its density. For example, the theoretical density of vanadium monoxide (VO) calculated, using the edge length of the unit cell is 6.5 g cm-3, but the actual experimental density, is 5.6 g cm-3. It indicates that there is approximately 14% Schottky defect in VO crystal., Presence of Schottky defect in the crystal provides a simple way by which atoms or ions, can move within the crystal lattice., 6.7.2 Frenkel defect:, Frenkel defect arises, due to the dislocation of ions, from its crystal lattice. The, ion which is missing from, the lattice point occupies, an interstitial position. This, defect is shown by ionic solids, in which cation and anion, differ in size. Unlike Schottky, defect, this defect does not, affect the density of the crystal., For example AgBr, in this case,, small Ag+ ion leaves its normal, site and occupies an interstitial, position as shown in the figure., , Ag+ Missing, , , , , , , , , , , , , , , , , , , , +,  Ag in interstitial, , , , , , , , , , , , , , , , , , , , Ag+, , , , , , , , , , Br-, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , position, , Ag+ Missing, , Ag+ in interstitial position, Frenkel Defect, , 6.7.3 Metal excess defect:, Metal excess defect, arises due to the presence, of more number of metal, Na+, ions as compared to anions., e, Cl, Alkali metal halides NaCl,, KCl show this type of defect., F center, The electrical neutrality of, Metal Excess Defect, the crystal can be maintained, by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of, extra cation and electron present in interstitial position., For example, when NaCl crystals are heated in the presence of sodium vapour, Na+, ions are formed and are deposited on the surface of the crystal. Chloride ions (Cl-) diffuse, to the surface from the lattice point and combines with Na+ ion. The electron lost by the, sodium vapour diffuse into the crystal lattice and occupies the vacancy created by the, Cl- ions. Such anionic vacancies which are occupied by unpaired electrons are called F, centers. Hence, the formula of NaCl which contains excess Na+ ions can be written as, Na1+ x Cl ., 194, , XII U6 Solid State - Jerald.indd 194, , 2/19/2020 4:42:32 PM

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www.tntextbooks.in, , ZnO is colourless at room temperature. When it is heated, it becomes yellow in colour., On heating, it loses oxygen and thereby forming free Zn2+ ions. The excess Zn2+ ions move, to interstitial sites and the electrons also occupy the interstitial positions., 6.7.4 Metal deficiency defect:, Metal deficiency defect, arises due to the presence of, less number of cations than the, anions. This defect is observed, in a crystal in which, the cations, have variable oxidation states., , _, , , , +, , For example, in FeO, +, crystal, some of the Fe2+ ions are, missing from the crystal lattice., To maintain the electrical, neutrality, twice the number of, Metal Deficiency Defect, other Fe2+ ions in the crystal is, oxidized to Fe3+ ions. In such cases, overall number of Fe2+ and Fe3+ ions is less than the, O2- ions. It was experimentally found that the general formula of ferrous oxide is FexO,, where x ranges from 0.93 to 0.98., 6.7.5 Impurity defect:, A general method of introducing defects in ionic solids is by adding impurity ions. If, the impurity ions are in different valance state from that of host, vacancies are created in, the crystal lattice of the host. For example, addition of CdCl2 to AgCl yields solid solutions, where the divalent cation Cd2+ occupies the position of Ag+. This will disturb the electrical, neutrality of the crystal. In order to maintain the same, proportional number of Ag+ ions, leaves the lattice. This produces a cation vacancy in the lattice, such kind of crystal defects, are called impurity defects., Energy harvesting by piezoelectric crystals:, Piezoelectricity (also called the piezoelectric effect) is the appearance of, an electrical potential across the sides of a crystal when you subject it to, mechanical stress. The word piezoelectricity means electricity resulting, from pressure and latent heat. Even the inverse is possible which is known, as inverse piezoelectric effect., If you can make a little amount of electricity by pressing one piezoelectric crystal, once, could youmake a significant amount by pressing many crystals over and over, again? What happens if we bury piezoelectric crystals under streets to capture energy, as vehicles pass by?, 195, , XII U6 Solid State - Jerald.indd 195, , 2/19/2020 4:42:33 PM

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www.tntextbooks.in, , This idea, known as energy harvesting, has caught many people's interest., Even though there are limitations for the large-scale applications, you can produce, electricity that is enough to charge your mobile phones by just walking. There are, power generating footwears that has a slip-on insole with piezoelectric crystals that, can produce enough electricity to charge batteries/ USB devices.`, , Summary, „„ Solids have definite volume and shape., „„ solids can be classified into the following two major types based on the arrangement, , of their constituents. (i) Crystalline solids (ii)Amorphous solids., , „„ A crystalline solid is one in which its constituents (atoms, ions or molecules),, , have an orderly arrangement extending over a long range., , „„ In contrast, in amorphous solids (In Greek, amorphous means no form) the, , constituents are randomly arranged., , „„ Crystalline solid is characterised by a definite orientation of atoms, ions or, , molecules, relative to one another in a three dimensional pattern. The regular, arrangement of these species throughout the crystal is called a crystal lattice., , „„ A crystal may be considered to consist of large number of unit cells, each one in, , direct contact with its nearer neighbour and all similarly oriented in space., , „„ A unit cell is characterised by the three edge lengths or lattice constants a ,b and, , c and the angle between the edges α, β and γ, , „„ There are seven primitive crystal systems; cubic, tetragonal, orthorhombic,, , hexagonal, monoclinic, triclinic and rhombohedral. They differ in the, arrangement of their crystallographic axes and angles. Corresponding to the, above seven, Bravis defined 14 possible crystal systems, , „„ In the simple cubic unit cell, each corner is occupied by an identical atoms or, , ions or molecules. And they touch along the edges of the cube, do not touch, diagonally. The coordination number of each atom is 6., , „„ In a body centered cubic unit cell, each corner is occupied by an identical particle, , and in addition to that one atom occupies the body centre. Those atoms which, occupy the corners do not touch each other, however they all touch the one, that occupies the body centre. Hence, each atom is surrounded by eight nearest, neighbours and coordination number is 8., , „„ In a face centered cubic unit cell, identical atoms lie at each corner as well as in, , the centre of each face. Those atoms in the corners touch those in the faces but, not each other.The coordination number is 12., 196, , XII U6 Solid State - Jerald.indd 196, , 2/19/2020 4:42:33 PM

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www.tntextbooks.in, , „„ X-Ray diffraction analysis is the most powerful tool for the determination of crystal, , structure. The inter planar distance (d) between two successive planes of atoms, can be calculated using the following equation form the X-Ray diffraction data, 2dsinθ = nλ, , „„ The structure of an ionic compound depends upon the stoichiometry and the, , size of the ions.generally in ionic crystals the bigger anions are present in the, close packed arrangements and the cations occupy the voids. The ratio of radius,  rC+ ,  plays an important role in determining the structure, r, −,  A , , of cation and anion , , „„ Crystals always found to have some defects in the arrangement of their constituent, , particles., , „„ Schottky defect arises due to the missing of equal number of cations and anions, , from the crystal lattice., , „„ Frenkel defect arises due to the dislocation of ions from its crystal lattice. The ion, , which is missing from the lattice point occupies an interstitial position., , „„ Metal excess defect arises due to the presence of more number of metal ions as, , compared to anions., , „„ Metal deficiency defect arises due to the presence of less number of cations than, , the anions., , EVALUATION, Choose the best answer:, 1. Graphite and diamond are, a) Covalent and molecular crystals, , b) ionic and covalent crystals, , c) both covalent crystals, , d) both molecular crystals, , 2. An ionic compound AxBy crystallizes in fcc type crystal structure with B ions at the, centre of each face and A ion occupying corners of the cube. the correct formula of, AxBy is, a) AB, , b) AB3, , c) A3B, , d) A8B6, , 3. The ratio of close packed atoms to tetrahedral hole in cubic packing is, a) 1:1, , b) 1:2, , c) 2:1, , d) 1:4, 197, , XII U6 Solid State - Jerald.indd 197, , 2/19/2020 4:42:33 PM

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www.tntextbooks.in, , 4. Solid CO2 is an example of, a) Covalent solid, , b) metallic solid, , c) molecular solid, , d) ionic solid, , 5. Assertion : monoclinic sulphur is an example of monoclinic crystal system, Reason: for a monoclinic system, a≠b≠c and α = γ = 900 , β ≠ 900, a) Both assertion and reason are true and reason is the correct explanation of assertion., b) Both assertion and reason are true but reason is not the correct explanation of assertion., c) Assertion is true but reason is false., d) Both assertion and reason are false., 6. In calcium fluoride, having the flurite structure the coordination number of Ca2+ ion, and F- Ion are (NEET), a) 4 and 2, , b) 6 and 6, , c) 8 and 4, , d) 4 and 8, , 7. The number of unit cells in 8 gm of an element X ( atomic mass 40) which crystallizes, in bcc pattern is (NA is the Avogadro number), a) 6.023 X 1023 , , b) 6.023 X 1022, ,  6.023 × 1023 , c) 60.23 X 10 d) ,  8 × 40 ,  1, 8. In a solid atom M occupies ccp lattice and   of tetrahedral voids are occupied by,  3, 23, , atom N. find the formula of solid formed by M and N., a) MN, , b) M3N, , c) MN­3, , d) M3N2, , 9. The ionic radii of A+ and B− are 0.98 × 10−10 m and 1.81 × 10−10 m . the coordination, number of each ion in AB is, a) 8, , b) 2, , c) 6, , d) 4, , 10. CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance, is,  3, a) 400pm, b) 800pm, c) 3 × 100pm, d)   × 400pm,  2 , 11. A solid compound XY has NaCl structure. if the radius of the cation is 100pm , the, radius of the anion will be,  0.414 ,  0.732 , a)  100 , b) , c) 100 × 0.414, d) , ,  100 ,  100 ,  0.414 , 198, , XII U6 Solid State - Jerald.indd 198, , 2/19/2020 4:42:38 PM

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www.tntextbooks.in, , 12. The vacant space in bcc lattice unit cell is, a) 48%, , b) 23%, , c) 32%, , d) 26%, , 13. The radius of an atom is 300pm, if it crystallizes in a face centered cubic lattice, the, length of the edge of the unit cell is, a) 488.5pm, , b) 848.5pm, , c) 884.5pm, , d) 484.5pm, , 14. The fraction of total volume occupied by the atoms in a simple cubic is,  π ,  π ,  π,  π, b), c), d), a) , , ,  4 ,  6 ,  4 2 , 3 2, , 15. The yellow colour in NaCl crystal is due to, a) excitation of electrons in F centers, , b) reflection of light from Cl- ion on the surface, c) refraction of light from Na+ ion, d) all of the above, 16. if ‘a’ stands for the edge length of the cubic system; sc , bcc, and fcc. Then the ratio of, radii of spheres in these systems will be respectively., 1, 3, 2 , a:, a, a)  a :, 2, 2 , 2, , b), , 1, 3, 1 , a:, a, c)  a :, 4, 2 2 , 2, , 1 , 1, a, d)  a : 3a :, 2, 2 , , (, , 1a : 3a : 2a, , ), , 17. If ‘a’ is the length of the side of the cube, the distance between the body centered atom, and one corner atom in the cube will be,  4 ,  2 , b)   a, a)   a,  3,  3,  3, d)   a,  2 , ,  3, c)   a,  4 , , 18. Potassium has a bcc structure with nearest neighbor distance 4.52 A0 . its atomic weight, is 39. its density will be, a) 915 kg m-3, , b) 2142 kg m-3, , c) 452 kg m-3, , d) 390 kg m-3, , 19. Schottky defect in a crystal is observed when, a) unequal number of anions and cations are missing from the lattice, b) equal number of cations and anions are missing from the lattice, c) an ion leaves its normal site and occupies an interstitial site, d) no ion is missing from its lattice., 199, , XII U6 Solid State - Jerald.indd 199, , 2/19/2020 4:42:42 PM

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www.tntextbooks.in, , 20. The cation leaves its normal position in the crystal and moves to some interstitial, position, the defect in the crystal is known as, a) Schottky defect, , b) F center, , c) Frenkel defect, , d) non-stoichiometric defect, , 21. Assertion: due to Frenkel defect, density of the crystalline solid decreases., Reason: in Frenkel defect cation and anion leaves the crystal., a) Both assertion and reason are true and reason is the correct explanation of assertion., b) Both assertion and reason are true but reason is not the correct explanation of assertion., c) Assertion is true but reason is false., d) Both assertion and reason are false, 22. The crystal with a metal deficiency defect is, a) NaCl, , b) FeO, , c) ZnO, , d) KCl, , 23. A two dimensional solid pattern formed by two different atoms X and Y is shown, below. The black and white squares represent atoms X and Y respectively. the simplest, formula for the compound based on the unit cell from the pattern is, , a) XY8, , b) X4Y9, , c) XY2, , d) XY4, , Answer the following questions:, 1. Define unit cell., 2. Give any three characteristics of ionic crystals., 3. Differentiate crystalline solids and amorphous solids., 4. Classify the following solids, b. Brass, a. P4, , c. diamond, , d. NaCl, , e. Iodine, , 5. Explain briefly seven types of unit cell., 6. Distinguish between hexagonal close packing and cubic close packing., 7. Distinguish tetrahedral and octahedral voids., 200, , XII U6 Solid State - Jerald.indd 200, , 2/19/2020 4:42:42 PM

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www.tntextbooks.in, , 8. What are point defects?, 9. Explain Schottky defect., 10. Write short note on metal excess and metal deficiency defect with an example., 11. Calculate the number of atoms in a fcc unit cell., 12. Explain AAAA and ABABA and ABCABC type of three dimensional packing with the, help of neat diagram., 13. Why ionic crystals are hard and brittle?, 14. Calculate the percentage efficiency of packing in case of body centered cubic crystal., 15. What is the two dimensional coordination number of a molecule in square close packed, layer?, 16. What is meant by the term “coordination number”? What is the coordination number, of atoms in a bcc structure?, 17. An element has bcc structure with a cell edge of 288 pm. the density of the element is, 7.2 gcm-3. how many atoms are present in 208g of the element., 18. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm., calculate the edge length of unit cell., 19. if NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration, of cation vacancy?, 20. KF crystallizes in fcc structure like sodium chloride. calculate the distance between K +, −3, and F − in KF.( given : density of KF is 2.48 g cm ), −3, 21. An atom crystallizes in fcc crystal lattice and has a density of 10 gcm with unit cell, edge length of 100pm. calculate the number of atoms present in 1 g of crystal., , 22. Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the, cube and Y is at the centre of the cube. What is the formula of the compound?, 23. Sodium metal crystallizes in bcc structure with the edge length of the unit cell, 4.3 × 10−8 cm . calculate the radius of sodium atom., 24. Write a note on Frenkel defect., , 201, , XII U6 Solid State - Jerald.indd 201, , 2/19/2020 4:42:44 PM

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www.tntextbooks.in, , Solids, , Crystalline solid, , Amorphous solid, , Ionic crystals, Covalent crystals, Crystalline structure, Molecular crystals, , Crystal lattice, , Close packing, Sc, , Schottky defect, , Bcc, , Frenkel defect, , Hcp, , Metal excess defect, , Unit cell, , Simple Cubic, Body Centered Cubic, Face Centered Cubic, , Imperfection in crystals, , Ccp, , Metal deficiency defect, , Packing fraction, , 202, , XII U6 Solid State - Jerald.indd 202, , 2/19/2020 4:42:45 PM

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www.tntextbooks.in, , ICT Corner, CRYSTAL SYSTEMS, By using this tool, you will be, able to visualize different crystal, systems and know their unit cell, parameters., , Please go to the URL, http://vlab.amrita.edu, (or) Scan the QR code on the, right side, , Steps, •, , Open the Browser and type the URL given (or) Scan the QR Code. In the webpage click physical, science tab and then click solid state virtual lab. Then go to crystal structure and then click, simulator., Note: One time sign up is needed to access this webpage. Login using your username and password., Once logged in click the simulator tab., , •, , Now the using the menu (box 1) select any one of the seven crystal systems and the lattice type., Now the unit cell of the selected crystal system will appear on screen (box 2) and the unit cell, parameters will also be displayed in the measurement tab (box 3), , 1, , 2, , 3, , 203, , XII U6 Solid State - Jerald.indd 203, , 2/19/2020 4:42:46 PM

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www.tntextbooks.in, , UNIT, , 7, , CHEMICAL, KINETICS, , Svante August Arrhenius, (1859 –1927), , Learning Objectives, After studying this unit, the students will, be able to, , Svante August Arrhenius was, a Swedish scientist. Arrhenius, was one of the founders of the, science of physical chemistry., He focused his attention on the, conductivities of electrolytes., He proposed that crystalline, salts dissociate into paired, charged ions when dissolved, in water, for which he received, the Nobel Prize for Chemistry, in 1903. He also proposed, definitions for acids and bases., He formulated the concept of, activation energy., , , , , , define the rate and order of a reaction,, derive the integrated rate equations for, zero and first order reactions,, , , , describe the half life period,, , , , describe the collision theory,, , , , , , discuss the temperature dependence of, the rate of a reaction, and, explain various factors which affect the, rate of a reaction., , 204, , XII U7 kinetics - Jerald Folder.indd 204, , 2/19/2020 4:43:29 PM

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www.tntextbooks.in, , 7.1 Rate of a chemical reaction:, , INTRODUCTION, , A rate is a change in a particular, variable per unit time. You have already, learnt in physics that change in the, displacement of a particle per unit time, gives its velocity. Similarly in a chemical, reaction, the change in the concentration, of the species involved in a chemical, reaction per unit time gives the rate of a, reaction., , We have already learnt in XI standard that, the feasibility of a chemical reaction under, a given set of conditions can be predicted,, using the principles of thermodynamics., However, thermodynamics does not, provide an answer to a very important, question of how fast a chemical reaction, takes place. We know from our practical, experience that all chemical reactions take, some time for completion. Reaction speeds, ranging from extremely fast (in femto, seconds) to extremely slow (in years). For, example, when the reactants BaCl2 solution, and dilute H2SO4 are just mixed, a white, precipitate of BaSO4 is immediately formed;, on the other hand reactions such as rusting, of Iron take many years to complete. The, answers to the questions such as (i) how fast, a chemical change can occur and (ii) What, happens in a chemical reaction during the, period between the initial stage and final, stage are provided by the chemical kinetics., The word kinetics is derived from the Greek, word “kinesis” meaning movement., , Let us consider a simple general reaction, A , →B, , The concentration of the reactant, ([A]) can be measured at different time, intervals. Let the concentration of A at two, different times t 2 and t 2 , (t2>t1) be [A1] and, [A2] respectively. The rate of the reaction, can be expressed as, , Rate=, , – [Change in the concentration, of the reactants], , i.e., Rate =, , Chemical kinetics is the study of the rate, and the mechanism of chemical reactions,, proceeding under given conditions of, temperature, pressure, concentration etc., , (Change in time), - ([A 2 ] - [A1 ]), (t 2 -t1 ), ,  ∆[A], = -, ...(7.1),  ∆ t , , During the reaction, the concentration, of the reactant decreases i.e. [A 2 ] < [A1 ], and hence the change in concentration, [A 2 ] - [A1 ] gives a negative value. By, convention the reaction rate is a positive, one and hence a negative sign is introduced, in the rate expression (equation 7.1), , The study of chemical kinetics not only, help us to determine the rate of a chemical, reaction, but also useful in optimizing, the process conditions of industrial, manufacturing processes, organic and, inorganic synthesis etc., , If the reaction is followed by measuring, the product concentration, the rate is given, , In this unit, we discuss the rate of a, chemical reaction and the factors affecting, it. We also discuss the theories of the, reaction rate and temperature dependence, of a chemical reaction., ,  ∆[B], since [B2 ]>[B1 ] , no minus sign, by ,  ∆ t , is required here., 205, , XII U7 kinetics - Jerald Folder.indd 205, , 2/19/2020 4:43:32 PM

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www.tntextbooks.in, , *, , 1.0, , Concentraon(M), , 0.8, 0.6, 0.4, 0.2, 0, , 0, , 20, , 40, , 60, Time (mins), * Schematic representation- Not to scale, , 80, , 100, -A, , 120, -B, , Fig 7.1 change in concentration of A and B for the reaction A → B, Unit of rate of a reaction:, , Now, let us consider a different reaction, , unit of concentration, unit of time, Usually, concentration is expressed, in number of moles per litre and time is, expressed in seconds and therefore the, unit of the rate of a reaction is mol L-1s -1 ., Depending upon the nature of the reaction,, minute, hour, year etc can also be used., , A , → 2B, , unit of rate =, , In this case, for every mole of A, that, disappears two moles of B appear, i.e., the, rate of formation of B is twice as fast as the, rate of disappearance of A. therefore, the rate, of the reaction can be expressed as below, + d[B],  -d[A], = 2,  dt , dt, In other words,, , Rate =, , For a gas phase reaction, the, concentration of the gaseous species is, usually expressed in terms of their partial, pressures and in such cases the unit of, reaction rate is atm s-1 ., , Rate =, , For a general reaction, the rate of the, reaction is equal to the rate of consumption, of a reactant (or formation of a product), divided by its coefficient in the balanced, equation, xA + yB , → lC + mD, , 7.1.1 Stoichiometry and rate of a reaction:, In a reaction A , → B , the, stoichiometry of both reactant and product, are same, and hence the rate of disappearance, of reactant (A) and the rate of appearance of, product (B) are same., , -d[A] 1 d[B], =, dt, 2 dt, , =, Rate, , -1 d[A] -1 d[B] 1 d[C] 1 d[D], = = =, x dt, y dt, l dt, m dt, , 206, , XII U7 kinetics - Jerald Folder.indd 206, , 2/19/2020 4:43:35 PM

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www.tntextbooks.in, , 7.1.2 Average and instantaneous rate:, , Let us calculate the average rate for an, initial and later stage over a short period., , Let us understand the average rate, and instantaneous rate by considering the, isomerisation of cyclopropane., 780 K, cyclopropane, , (Rate)later, , propene, , stage, , The kinetics of the above reaction is, followed by measuring the concentration, of cyclopropane at regular intervals and the, observations are shown below. (Table 7.1), , From the above calculations, we come, to know that the rate decreases with time, as the reaction proceeds and the average, rate cannot be used to predict the rate of, the reaction at any instant. The rate of the, reaction, at a particular instant during the, reaction is called the instantaneous rate., The shorter the time period, we choose, the, closer we approach to the instantaneous, rate,, , Table 7.1 Concentration of cyclopropane, at various times during its isomerisation at, 780K, Time ( min), , [cyclopropane], ( mol L-1 ), , 0, , 2.00, , 5, , 1.67, , 10, , 1.40, , 15, , 1.17, , 20, , 0.98, , 25, , 0.82, , 30, , 0.69, , Rate of the reaction=, , As ∆ t → 0;, -∆[cyclopropane] -d[cyclopropane], =, ∆t, dt, A plot of [cyclopropane] Vs (time), gives a curve as shown in the figure 7.2., Instantaneous rate at a particular instant ‘t ’, , – ∆ [cyclopropane], , -d [cyclopropane], is obtained by calculating, dt, , ∆t, , the slope of a tangent drawn to the curve at, , The rate over the -(0.69-2) molL-1, =, entire 30 min, (30-0) min, =, , − (1.4 −2 ), (10 −0), 0.6, =, = 6 x 10 -2 mol L-1 min-1, 10, -( 0.69 - 0.98), =, (30-20), 0.29, =, = 2.9 x 10-2 mol L-1min-1, 10, , =, ( Rate)initial, stage, , that instant., , 1.31, −2, -1, -1, = 4.36 × 10 molL min, 30, , In general, the instantaneous, reaction rate at a moment of mixing the, reactants (t = 0) is calculated from the slope, of the tangent drawn to the curve. The rate, calculated by this method is called initial, rate of a reaction., , It means that during the first 30 minutes, of the reaction, the concentration of the, reactant ( cyclo propane) decreases as an, average of 4.36 × 10-2 mol L-1 each minute., , 207, , XII U7 kinetics - Jerald Folder.indd 207, , 2/19/2020 4:43:37 PM

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www.tntextbooks.in, , 2.00, , [Cyclopropane] (M), , 1.50, , 1.00, , 0.50, , 0.00, 0, , 10, , 20, , 30, , 40, , 50, , 60, , 70, , 80, , 90, , 100, , Time (mins), , Fig 7.2 Concentration of cyclopropane vs time - graph, Let us calculate the instantaneous, rate of isomerisation cyclopropane at, different concentrations: 2 M, 1M and 0.5 M, from the graph shown in fig 7.2, the results, obtained are tabulated below., [cyclopropane], mol L-1, , Rate mol L-1min-1, , 2, , 6.92 × 10 –2, , 1, , 3.46 × 10 –2, , 0.5, , 1.73 × 10 –2, , The rate law for the above reaction is, generally expressed as, Rate = k [A]m[B]n, Where k is proportionality constant, called the rate constant. The values of m and, n represent the reaction order with respect, to A and B respectively. The overall order of, the reaction is given by (m+n). The values, of the exponents (m and n) in the rate law, must be determined by experiment. They, cannot be deduced from the Stoichiometry, of the reaction. For example, consider the, isomerisation of cyclopropane, that we, discussed earlier., , Table 7.2 Rate of isomerisation, , The results shown in table 7.2 indicate, that if the concentration of cyclopropane, is reduced to half, the rate also reduced to, half. It means that the rate depends upon, [cyclopropane] raised to the first power, , 7.3 Rate law and rate constant:, We have just learnt that, the rate of the, reaction depends upon the concentration of, the reactant. Now let us understand how the, reaction rate is related to concentration by, considering the following general reaction., , i.e.,, ⇒, , xA + yB , → products, , Rate = k[cyclopropane]1, Rate, =k, [cyclopropane], , 208, , XII U7 kinetics - Jerald Folder.indd 208, , 2/19/2020 4:43:38 PM

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www.tntextbooks.in, , mol L-1min-1, , mol L-1, , Rate, , [cyclopropane], , Table 7.3 Rate constant for isomerisation, , k=, , 38.40 X10-2 = k [1.3]m[2.2]n, For experiment 3, , Rate, [cyclopropane], , ...(2), , Rate3 = k [NO]m [O 2 ]n , , 76.8 × 10-2 = k [2.6]m[1.1]n, , ...(3), , 6.92 × 10 –2, , 2, , 3.46 × 10 –2, , 38.40 x 10-2, k [1.3]m[2.2]n, (2), ⇒, =, (1), 19.26 x 10-2, k [1.3]m[1.1]n, , 3.46 × 10 –2, , 1, , 3.46 × 10 –2, , 1.73 × 10 –2, , 0.5, , 3.46 × 10 –2, ,  2.2 , 2= ,  1.1 , , 2=2n i.e., n=1, , Let us consider an another example, the, oxidation of nitric oxide (NO), , Therefore the reaction is first order with, respect to O 2, , 2NO(g) + O2 ( g ) , → 2NO2 (g), , k [2.6]m [1.1]n, 76.8 x 10-2, (3), =, ⇒, k [1.3]m [1.1]n, 19.26 x 10-2, (1), , Series of experiments are conducted, by keeping the concentration of one of the, reactants constant and the changing the, concentration of the others., Experiment, , n, ,  2. 6 , 4=  ,  1. 3 , , m, , 4=2m i.e., m=2, Therefore the reaction is second order, with respect to NO, , [NO] X 10-2 [O 2 ] X 10-2 Initial rate x 10-2, (mol L-1 ), , (mol L-1 ), , (mol L-1min -1 ), , 1.3, , 1.1, , 19.26, , 2, , 1.3, , 2.2, , 38.40, , Differences between rate and rate constant, of a reaction:, , 3, , 2.6, , 1.1, , 76.80, , s.no, , 1, , The rate law is Rate1 = k [NO]2 [O 2 ]1, The overall order of the reaction = (2 + 1) = 3, , Rate = k [NO]m[O2 ]n, For experiment 1, the rate law is, Rate1 = k [NO]m[O2 ]n, , 19.26 X10-2 = k [1.3]m[1.1]n, , 1, , ...(1), , Similarly for experiment 2, Rate 2 = k [NO]m [O 2 ]n, , Rate of a, reaction, , Rate constant of a, reaction, , It represents the, speed at which, the reactants are, converted into, products at any, instant., , It is a, proportionality, constant, , 209, , XII U7 kinetics - Jerald Folder.indd 209, , 2/19/2020 4:43:44 PM

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s.no, , www.tntextbooks.in, , 2, , 3, , H 2 O 2 and I− ,which indicates that I− is also, , Rate of a, reaction, , Rate constant of a, reaction, , involved in the reaction. The mechanism, involves the following steps., , It is measured as, decrease in the, concentration of, the reactants or, increase in the, concentration of, products., , It is equal to the, rate of reaction,, when the, concentration, of each of the, reactants is unity, , Step:1, , It depends, on the initial, concentration of, reactants., , It does not, depend on, the initial, concentration of, reactants., , H 2 O 2 (aq) + I−(aq) , → H 2 O(l ) + OI−(aq), , Step : 2, , H 2 O 2 (aq) + OI−(aq) → H 2 O(l ) + I−(aq) + O2 (g), , Overall reaction is, 2H2O2 (aq) , → 2H2O(l ) + O2 (g), These two reactions are elementary, reactions. Adding equ (1) and (2) gives the, overall reaction. Step 1 is the rate determining, step, since it involves both H 2 O 2 and I−, the, overall reaction is bimolecular., , 7.4 Molecularity:, , Differences between order and, molecularity:, Order of a, reaction, , Molecularity of a, reaction, , 1, , It is the sum of, the powers of, concentration, terms involved in, the experimentally, determined rate, law., , It is the total, number of, reactant species, that are involved, in an elementary, step., , 2, , It can be zero (or) It is always a, fractional (or), whole number,, integer, cannot be zero, or a fractional, number., , 3, , It is assigned for a It is assigned, overall reaction., for each, elementary step, of mechanism., , s.no, , Kinetic studies involve not only, measurement of a rate of reaction but, also proposal of a reasonable reaction, mechanism. Each and every single step in a, reaction mechanism is called an elementary, reaction., An elementary step is characterized, by its molecularity. The total number of, reactant species that are involved in an, elementary step is called molecularity of that, particular step. Let us recall the hydrolysis, of t-butyl bromide studied in XI standard., Since the rate determining elementary step, involves only t-butyl bromide, the reaction, is called a Unimolecular Nucleophilic, 1, substitution (SN ) reaction., Let us understand the elementary, reactions by considering another reaction,, the decomposition of hydrogen peroxide, catalysed by I− ., 2H2O2 (aq) , → 2H2O(l ) + O2 (g), It is experimentally found that the, reaction is first order with respect to both, 210, , XII U7 kinetics - Jerald Folder.indd 210, , 2/19/2020 4:43:48 PM

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www.tntextbooks.in, , 5Br −(aq ) + BrO3−(aq ) + 6H+ (aq ) , → 3Br2 (l ) + 3H 2 O(l ), , Example 1, , The experimental rate law is, , Consider the oxidation of nitric, oxide to form NO 2, , Rate = k [Br −][BrO3−][H+ ]2, ∆, → CH 4 ( g ) + CO( g ), (b). CH 3CHO( g ) , the experimental rate law is, , 2NO(g) + O2 (g) , → 2NO 2 (g), , (a). Express the rate of the reaction in, terms of changes in the concentration of, NO,O 2 and NO 2 ., , 3, , Rate = k [CH 3CHO] 2, , Solution:, , (b). At a particular instant, when [O 2 ] is, decreasing at 0.2 mol L−1s −1 at what rate is, [NO 2 ] increasing at that instant?, , a) First order with respect to Br − , first, order with respect to BrO3− and second, order with respect to H+ . Hence the, overall order of the reaction is equal to, , Solution:, =, a) Rate, , -1 d[NO] - d[O 2 ] 1 d[NO 2 ], =, =, 2 dt, dt, 2 dt, , 1+1+2=4, , b) Order of the reaction with respect to, , -d[O2 ] 1 d[NO2 ], b), =, dt, 2 dt, , acetaldehyde is, , d[NO2 ],  -d[O2 ], = 2x , = 2 x 0.2 mol L−1s −1, ,  dt , dt, , also, , 3, 2, , 3, and overall order is, 2, , Example 3, , −1 −1, , = 0.4 mol L s, , 2. The, , Evaluate yourself 1, , rate, , of, , the, , reaction, , x + 2y → product is 4 x 10 mol L−1s −1, , if [x]=[y]=0.2 M and rate constant, at 400K is 2 x 10-2 s -1 , What is the, −3, , 1). Write the rate expression for the, following reactions, assuming them as, elementary reactions., , overall order of the reaction., , → 4CD, i) 3A + 5B2 , , Solution :, , → 2XY, ii) X 2 + Y2 , , Rate = k [x]n [y]m, , 2). Consider the decomposition of N 2 O5 (g), to form NO 2 (g) and O 2 (g) . At a, particular instant N 2 O5 disappears at a, rate of 2.5x10-2 mol dm -3s-1 . At what rates, are NO 2 and O 2 formed? What is the, rate of the reaction?, , 4 x 10-3 mol L-1s -1 = 2 x 10-2 s -1 (0.2mol L-1 ) n (0.2mol L-1 ) m, n+m, 4 x 10−3 mol L−1s −1, = (0.2) n + m ( mol L−1 ), −2 −1, 2 x 10 s, , 0.2 ( mol L-1 ) = (0.2) n + m ( mol L-1 ), , n+m, , Comparing the powers on both sides, The overall order of the reaction n + m = 1, , Example 2, 1. What is the order with respect to each, of the reactant and overall order of the, following reactions?, , Evaluate yourself 2, , 1). For a reaction, X + Y , → product ;, quadrupling [x] , increases the rate by a, −, −, +, → 3Br2 (l ) + 3H, (a). 5Br (aq ) + BrO3 (aq ) + 6H (aq) , factor, 2 O(l )of 8. Quadrupling both [x] and [y],, 211, , XII U7 kinetics - Jerald Folder.indd 211, , 2/19/2020 4:44:10 PM

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www.tntextbooks.in, , reactant after a time ‘ t ’?. To answer such, , increases the rate by a factor of 16. Find the, order of the reaction with respect to x and, y. what is the overall order of the reaction?, , questions, we need the integrated form of, the above rate law which contains time as a, , 2). Find the individual and overall order of, the following reaction using the given, data., , variable., 7.5.1 Integrated rate law for a first order, reaction, , 2NO(g) + Cl2 (g) , → 2NOCl(g), , Initial rate, , NO, , Cl2, , NOCl mol L-1s -1, , A , → product, , 1, , 0.1, , 0.1, , 7.8 x10, , −5, , 2, , 0.2, , 0.1, , 3.12 x10−4, , 3, , 0.2, , 0.3, , 9.36 x10−4, , Experiment, number, , Initial, concentration, , A reaction whose rate depends on, the reactant concentration raised to the first, power is called a first order reaction. Let us, consider the following first order reaction,, Rate law can be expressed as, Rate = k [A]1, , Where, k is the first order rate constant., -d[A], = k [A]1, dt, -d[A], ⇒, = k dt, [A], , ...(1), Integrate the above equation between, the limits of time t = 0 and time equal to t,, while the concentration varies from the initial, concentration [A 0 ] to [A] at the later time., , 7.5 The integrated rate equation:, We have just learnt that the rate of change, of concentration of the reactant is directly, proportional to that of concentration of the, reactant. For a general reaction,, , t, -d[A], k, =, ∫[A0 ] [A], ∫0 dt, [A], , ( −ln [A ])[ ], , A , → product, , = k ( t )0, A , A, , The rate law is, − d[A], Rate =, = k [A]x, dt, , , , t, , 0, , - ln [A]- ( - ln [A0 ]) = k (t-0), , - ln [A] + ln [A 0 ]= kt, , Where k is the rate constant, and x is the, ,  [A ] , ln  0  = kt,  [A] , , order of the reaction. The above equation is, -d[A], a differential equation,, , so it gives the, dt, rate at any instant. However, using the above, , ...(2), , This equation is in natural logarithm., To convert it into usual logarithm with base, 10, we have to multiply the term by 2.303., , expression, we cannot answer questions, such as how long will it take for a specific, ,  [A ] , 2.303 log  0  = kt,  [A] , , concentration of A to be used up in the, reaction? What will be the concentration of, 212, , XII U7 kinetics - Jerald Folder.indd 212, , 2/19/2020 4:44:20 PM

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www.tntextbooks.in, , 0, , -0.5, , ln [A], , -1, , -1.5, , -2, , -2.5, , 0, , 10, , 20, , 30, Time (in mins), , 40, , 50, , 60, , Fig: 7.3 A plot of ln[A] Vs t for a first order reaction, A , → product with initial, concentration of [A] = 1.00 M and k = 2.5 x10−2 min -1 ., k=, ,  [A ] , 2.303, log  0  ----- (3),  [A] , t, , Equation (2) can be written in the form y = mx + c as below, ln [A 0 ] −ln [A] = kt, ln [A] = ln [A 0 ] −kt, ⇒ y = c + mx, , If we follow the reaction by measuring the concentration of the reactants at regular time, interval‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope.From this, the, rate constant is calculated., Examples for the first order reaction, (i) Decomposition of dinitrogen pentoxide, 1, N 2 O5 (g) , → 2NO 2 (g) + O 2 (g), 2, → SO 2 (g) + Cl 2 (g), (ii) Decomposition of sulphurylchloride; SO 2 Cl2 (l) , , (iii) Decomposition of the H2O2 in aqueous solution; H 2 O 2 (aq) , → H 2 O(l) +, , 1, O 2 (g), 2, , (iv) Isomerisation of cyclopropane to propene., , 213, , XII U7 kinetics - Jerald Folder.indd 213, , 2/19/2020 4:44:26 PM

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www.tntextbooks.in, , Pseudo first order reaction:, Kinetic study of a higher order reaction is difficult to follow, for example, in a study of a second, order reaction involving two different reactants; the simultaneous measurement of change in the, concentration of both the reactants is very difficult. To overcome such difficulties, A second order, reaction can be altered to a first order reaction by taking one of the reactant in large excess, such, reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,, +, , CH 3COOCH 3 (aq) + H 2 O (l ) H→ CH 3COOH (aq) + CH 3OH (aq), Rate = k [CH3COOCH 3 ] [H 2 O], , If the reaction is carried out with the large excess of water, there is no significant change, in the concentration of water during hydrolysis. i.e.,concentration of water remains almost a, constant., Now, we can define k [H 2 O] = k' ; Therefore the above rate equation becomes, Rate = k' [CH3COOCH 3 ], , Thus it follows first order kinetics., 7.5.2 Integrated rate law for a zero order reaction:, A reaction in which the rate is independent of the concentration of the reactant over a, wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us, consider the following hypothetical zero order reaction., A , → product, , The rate law can be written as,, Rate = k [A]0 , −d[A], =k (1), dt, , , (∴[A], , 0, , = 1), , ⇒ −d[A] = k dt, , Integrate the above equation between the limits of [A 0 ] at zero time and [A] at some later, time 't',, −∫, , [ A], , [ A0 ], , t, , d[A] = k ∫ dt, 0, , −([ A])[ A ] = k ( t )0, [ A], , t, , 0, , [A 0 ] −[A] = kt, k=, , [A 0 ] −[A], t, , Equation (2) is in the form of a straight line y = mx + c, Ie., [A] = −kt + [A 0 ], ⇒ y = c + mx, , A plot of [A] Vs time gives a straight line with a slope of −k and y - intercept of [A 0 ] ., 214, , XII U7 kinetics - Jerald Folder.indd 214, , 2/19/2020 4:44:38 PM

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www.tntextbooks.in, , [A] in M, , 0.75, , 0.5, , 0.25, , 0, 0, , 5, , 10, , 15, Time (in mins), , 20, , 25, , 30, , → product with, Fig 7.4 : A plot of [A] Vs time for a zero order reaction A , -2, initial concentration of [A] = 0.5M and k = 1.5x10 mol−1L−1min −1, , Examples for a zero order reaction:, , 7.6 Half life period of a reaction:, , 1. Photochemical reaction between H2 and I2, H2 (g)+Cl 2 (g) hv, → 2HCl(g), 2. Decomposition of N2O on hot platinum, surface, , The half life of a reaction is defined, as the time required for the reactant, concentration to reach one half its initial, value. For a first order reaction, the half life, is a constant i.e., it does not depend on the, initial concentration., , 1, , N 2 O(g) , ,  N 2 (g) + O 2 (g), 2, , 3. Iodination of acetone in acid medium is, zero order with respect to iodine., , The rate constant for a first order, reaction is given by, , +, , CH 3COCH3 + I 2 H→ ICH 2 COCH 3 + HI, , k=, , +, , Rate = k [CH3COCH3 ] [H ], , [A ], 2.303, log 0, [A], t, , =, at t t=, ; [A], 1, 2, , General rate equation for a nth order, reaction involving one reactant [A]., , k=, , A , → product, , Rate law, , k=, , Consider the case in which n≠1,, [A 0 ] and [A] at time t = 0 and t = t, 1, 1, = (n-1)kt, respectively gives, n -1, [A] [A 0 ]n -1, , [A 0 ], 2.303, log, [, A0 ], t1, 2, , −d[A], = k[A]n, dt, , integration of above equation between, , [A 0 ], 2, , 2, , 2.303, log 2, t1, 2, , =, k, , 2.303x0.3010 0.6932, =, t1, t1, 2, , t1 =, 2, , 2, , 0.6932, k, , 215, , XII U7 kinetics - Jerald Folder.indd 215, , 2/19/2020 4:44:50 PM

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www.tntextbooks.in, , Let us calculate the half life period for a, zero order reaction., , k=, , [A 0 ] −[A], t, [A 0 ], =, at t t=, ; [A], 1, 2, 2, , Rate constant, k =, , k=, , t 80% =, , 2.303, log ( 5) −−−−−(2), k, , Find the value of k using the given data, , [A 0 ], , [A 0 ] −, , 2.303,  100 , log ,  20 , t 80%, , 2, , k=, , 2.303,  100 , log ,  10 , t 90%, , k=, , 2.303, log 10, 8 hours, , k=, , 2.303, (1), 8 hours, , t1, 2, , [A 0 ], k=, 2t 1, 2, , t1 =, 2, , [A 0 ], 2k, , Substitute the value of k in equation (2), , Hence, in contrast to the half life of a, first order reaction, the half life of a zero, order reaction is directly proportional to the, initial concentration of the reactant., , 2.303, log ( 5), 2.303, 8 hours, t 80% = 8hours x 0.6989, t 80% =, , t 80% = 5.59hours, , More to know, , Example 5, , Half life for an nth order reaction involving, reactant A and n ≠ 1, t1 =, 2, , (ii) , The half life of a first order, reaction, x , → products, is, 6.932 x 104s at 500K . W h a t, percentage of x would be, decomposed on heating at 500K, for 100 min. (e0.06 = 1.06), , n −1, , 2 −1, (n −1)k [ A0 ]n−1, , Example 4, (i) A first order reaction takes 8 hours, for 90% completion. Calculate the, time required for 80% completion., , Solution:, Given t 1 = 0.6932 x 104 s, 2, To solve : when t=100 min,, , (log 5 = 0.6989 ; log10 = 1), , Solution:, , [A 0 ] −[A], x 100 = ?, [A 0 ], We know that, , For a first order reaction,,  [A ] , 2.303, log  0 ,  [A]  , t, Let [A 0 ] = 100M, k=, , ....(1), , For a first order reaction, t 1 =, 2, , When, t = t 90% ; [A]=10M (given that t90 % =8hours), , 0.6932, k, , 0.6932, k=, 6.932 x 104, k = 10−5 s −1, , t = t 80% ; [A]=20M, 216, , XII U7 kinetics - Jerald Folder.indd 216, , 2/19/2020 4:45:06 PM

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www.tntextbooks.in, ,  [A ] ,  1, k =   ln  0 ,  t,  [A] , , t 99.9% =, ,  [A ] , 10 s x 100 x 60 s = ln  0 ,  [A] , , t 99.9%  10 x, , −5 −1, , t 99.9%  10 t 1, ,  [A ] , 0.06= ln  0 ,  [A] , , [A 0 ] - [A], × 100 %, [A 0 ], , (2) The rate constant for a first order, reaction is 2.3 X 10−4 s −1 If the initial, concentration of the reactant is 0.01M ., What concentration will remain after 1, hour?, ,  [A] , × 100 %, =  1 [A 0 ] , , 1 , , × 100 %, = 1 1.06 , , (3) Hydrolysis of an ester in an aqueous, solution was studied by titrating the, liberated carboxylic acid against sodium, hydroxide solution. The concentrations, of the ester at different time intervals are, given below., , = 5.6 %, , Example 6, Show that in case of first order, reaction, the time required for 99.9%, completion is nearly ten times the time, required for half completion of the, reaction., Let, , 2.303, log 1000, k, , t 99.9% =, , 2.303, (3), k, , 0, , 30, , 60, , 90, , Ester, concentration, mol L−1, , 0.85, , 0.80, , 0.754 0.71, , 7.7 Collision theory :, ,  [A ] , 2.303, k=, log  0 ,  [A] , t, , t 99.9% =, , Time (min), , Show that, the reaction follows first, order kinetics., , [A 0 ] = 100;, when t = =, t 99.9% ; [A] (100, =, -99.9) 0.1, , 2.303,  100 , log ,  0.1 , k, , 2, , (1) In a first order reaction A , → products, 60% of the given sample of A decomposes, in 40 min. what is the half life of the, reaction?, , [A 0 ], = 1.06, [A], , t 99.9% =, , 0.69, k, , Evaluate yourself:, , [A 0 ], = e0.06, [A], , ∴, , 6.909, k, , Collision Theory was proposed, independently by Max Trautz in 1916 and, William Lewis in 1918. This theory is based, on the kinetic theory of gases. According, to this theory, chemical reactions occur as, a result of collisions between the reacting, molecules. Let us understand this theory by, considering the following reaction., A 2 (g) + B2 (g) , → 2AB(g), 217, , XII U7 kinetics - Jerald Folder.indd 217, , 2/19/2020 4:45:19 PM

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Potential energy Æ, , www.tntextbooks.in, , Ea, , Reactants, Products, Reaction progress Æ, , Fig 7.5 progress of the reaction, If we consider that, the reaction, between A 2 and B2 molecules proceeds, through collisions between them, then the, rate would be proportional to the number of, collisions per second., , are not effective to lead to the reaction. In order, to react, the colliding molecules must possess, a minimum energy called activation energy., The molecules that collide with less energy, than activation energy will remain intact and, no reaction occurs., , Rate ∝ number of molecules colliding, per litre per second (collision rate), , Fraction of effective collisions (f) is, given by the following expression, , The number of collisions is directly, proportional to the concentration of both, A 2 and B2 ., , f =, , −Ea, e RT, , To understand the magnitude of, collision factor (f), Let us calculate the, collision factor (f) for a reaction having, −1, activation energy of 100 kJ mol at 300K., , Collision rate ∝ [A 2 ][B2 ], Collision rate = Z [A 2 ][B2 ], , Where, Z is a constant., The collision rate in gases can be, calculated from kinetic theory of gases., For a gas at room temperature (298K) and, 1 atm pressure, each molecule undergoes, approximately 109 collisions per second, i.e., 1, collision in 10-9 second. Thus, if every collision, resulted in reaction, the reaction would be, complete in 10-9 second. In actual practice this, does not happen. It implies that all collisions, , f =e, , , , 100 × 103 J mol-1, -, ,  8.314J K -1 mol-1 × 300 K , , f = e-40 ≈ 4 x 10-18, , Thus, out of 1018 collisions only four, collisions are sufficiently energetic to, convert reactants to products. This fraction, of collisions is further reduced due to, 218, , XII U7 kinetics - Jerald Folder.indd 218, , 2/19/2020 4:45:27 PM

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www.tntextbooks.in, , Proper allignment, Effective collission, , A, A, , B, , +, , B, , A, , B, , A, , B, , B, , A, , Products, , A, , Reactants, , B, , improper allignment, , A, , A + B, , B, , Reactants, , A, , A, , B, , B, , ineffective collision, , A, , A + B, , B, , Reactants, , Fig 7.6 - Orientation of reactants - schematic representation, , 7.8 Arrhenius equation – The effect, of temperature on reaction rate, , orientation factor i.e., even if the reactant, collide with sufficient energy, they will not, react unless the orientation of the reactant, molecules is suitable for the formation of, the transition state., , Generally, the rate of a reaction, increase with increasing temperature., However, there are very few exceptions. The, magnitude of this increase in rate is different, for different reactions. As a rough rule, for, many reactions near room temperature,, reaction rate tends to double when the, temperature is increased by 100 C ., , The figure 7. 6 illustrates the importance, of proper alignment of molecules which, leads to reaction., The fraction of effective collisions (f), having proper orientation is given by the, steric factor p., , Activity, , ⇒ Rate = p x f x collision rate, , i.e., Rate = p x, , −Ea, e RT x, , Let us understand the effect of, temperature on reaction rate by doing, this activity., , Z [A 2 ][B2 ] ...(1), , As per the rate law,, , i. Take two test tubes, label them as A, and B, , Rate = k [A2] [B2] ...(2), , ii. Take 5 ml of cold water in A, add a, drop of phenolphthalein indicator and, then add Magnesium granules., , Where k is the rate constant, On comparing equation (1) and (2), the, rate constant k is ‘, k=pZe, , iii. Repeat the above with 5 ml of hot, water in test tube B., , −Ea, RT, , 219, , XII U7 kinetics - Jerald Folder.indd 219, , 2/19/2020 4:45:30 PM

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www.tntextbooks.in, , Where A the frequency factor,, , iv. Observe the two test tubes., , R the gas constant,, , v. The observation shows that the, solution in test tube B changes to pink, colour and there is no such colour, change in test tube A. That is, hot water, reacts with magnesium according to, the following reaction and there is no, such reaction in cold water., , E a the activation energy of the reaction, , and,, T the absolute temperature (in K), , The frequency factor (A) is related, to the frequency of collisions (number, of collisions per second) between the, reactant molecules. The factor A does not, vary significantly with temperature and, hence it may be taken as a constant., , Mg + 2H 2 O , → Mg 2+ + 2OH − + H 2 ↑, , E a is the activation energy of the, , reaction, which Arrhenius considered as the, minimum energy that a molecule must have, to posses to react., Taking logarithm on both side of the, equation (1), ln k = ln A + ln e, , Ea , -  RT, , , E , ln k = ln A − a ,  RT , , (∴ ln e =1), ,  E   1, ln k = ln A − a   ,  R   T, , vi. The resultant solution is basic and it is, indicated by phenolphthalein., , ....(2), , y=c+mx, , A large number of reactions are, known which do not take place at room, temperature but occur readily at higher, temperatures. Example: Reaction between, H 2 and O 2 to form H 2 O takes place only, when an electric spark is passed., , The above equation is of the form of a, straight line y = mx + c., A plot of ln k Vs 1 T gives a straight, line with a negative slope − E a . If the rate, R, , constant for a reaction at two different, , Arrhenius suggested that the rates of, , temperatures is known, we can calculate the, , most reactions vary with temperature in, , activation energy as follows., , such a way that the rate constant is directly,  Ea , , At temperature T = T1 ; the rate constant, , proportional to e−  RT  and he proposed, a relation between the rate constant and, , k = k1, ,  E , ln k1 = ln A-  a ,  RT1 , , temperature., k=Ae, , E , − a ,  RT , , ....(1), , ....(3), , 220, , XII U7 kinetics - Jerald Folder.indd 220, , 2/19/2020 4:45:41 PM

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www.tntextbooks.in, , At temperature T = T2 ; the rate constant, , log ( 2 ) =, , k = k2, ,  E , ln k 2 = ln A-  a ,  RT2 , , Ea,  1 , , −1, −1 , 2.303 x 8.314 JK mol  400K , , E a = log ( 2 ) x 2.303 x 8.314 JK −1mol−1 x 400K, E a = 2305 J mol−1, , ....(4), , (4) –(3), , Example 8, ,  E   E , ln k 2 - ln k1 = -  a  +  a ,  RT2   RT1 , , Rate constant k of a reaction varies, with temperature T according to the, following Arrhenius equation, , k  E  1 1 , ln  2  = a  - ,  k1  R  T1 T2 , , log k = log A −, ,  k  E  T −T , 2.303 log  2  = a  2 1 ,  k1  R  T1T2 , , Ea  1 ,  , 2.303R  T , , Where E a is the activation energy., , 1, T, , k , E a  T2 −T1 , log  2  =,  k1  2.303R  T1T2 , , When a graph is plotted for log k Vs, ,  E   E , ln k 2 −ln k1 = − a  +  a ,  RT2   RT1 , , is obtained. Calculate the activation, , a straight line with a slope of -4000K, , energy, , This equation can be used to calculate, E a from rate constants k1 and k 2 at, temperatures T1 and T2 ., , Solution, , Example 7, , log k = log A-, , The rate constant of a reaction, at 400 and 200K are 0.04 and 0.02 s-1, respectively. Calculate the value of, activation energy., , Ea  1 ,  , 2.303R  T , , y = c + mx, m= -, , Ea, 2.303R, , E a = − 2.303 R m, E a = − 2.303 x 8.314 J K −1 mol−1 x (− 4000K ), , Solution, According to Arrhenius equation, , Ea = 76, 589J mol −1, , k , E a  T2 −T1 , log  2  =,  k1  2.303R  T1T2 , , E a = 76.589 kJ mol−1, , T2 = 400K ;, , k 2 = 0.04 s −1, , Evaluate yourself, , T1 = 200K ;, , k1 = 0.02 s −1, , For a first order reaction the rate, constant at 500K is 8 X 10−4 s −1 . Calculate the, frequency factor, if the energy of activation, for the reaction is 190 kJ mol-1 ., ,  0.04 s −1 , Ea,  400K −200K , =, log , , ,  0.02 s −1  2.303 x 8.314 JK −1mol−1  200K x 400K , , 221, , XII U7 kinetics - Jerald Folder.indd 221, , 2/19/2020 4:45:56 PM

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www.tntextbooks.in, , 7.9 Factors affecting the, reaction rate:, , than the reaction between solid sodium and, solid iodine., , The rate of a reaction is affected by the, following factors., , Let us consider another example that, you carried out in inorganic qualitative, analysis of lead salts. If you mix the, aqueous solution of colorless potassium, iodide with the colorless solution of lead, nitrate, precipitation of yellow lead iodide, take place instantaneously, whereas if, you mix the solid lead nitrate with solid, potassium iodide, yellow coloration will, appear slowly., , 1. Nature and state of the reactant, 2. Concentration of the reactant, 3. Surface area of the reactant, 4. Temperature of the reaction, 5. Presence of a catalyst, 7.9.1 Nature and state of the reactant:, We know that a chemical reaction, involves breaking of certain existing bonds, of the reactant and forming new bonds, which lead to the product. The net energy, involved in this process is dependent on the, nature of the reactant and hence the rates, are different for different reactants., Let us compare the following two, reactions that you carried out in volumetric, analysis., 1). Redox reaction between ferrous, ammonium sulphate (FAS) and KMnO 4, 2). Redox reaction between oxalic acid and, KMnO 4, , The oxidation of oxalate ion by, KMnO 4 is relatively slow compared to the, reaction between KMnO 4 and Fe2+ . In fact, heating is required for the reaction between, KMnO 4 and Oxalate ion and is carried out, at around 600 C ., , 7.9.2 Concentration of the reactants:, The rate of a reaction increases with, the increase in the concentration of the, reactants. The effect of concentration is, explained on the basis of collision theory, of reaction rates. According to this theory,, the rate of a reaction depends upon the, number of collisions between the reacting, molecules. Higher the concentration,, greater is the possibility for collision and, hence the rate., , The physical state of the reactant also, plays an important role to influence the rate, of reactions. Gas phase reactions are faster, as compared to the reactions involving solid, or liquid reactants. For example, reaction of, sodium metal with iodine vapours is faster, 222, , XII U7 kinetics - Jerald Folder.indd 222, , 2/19/2020 4:45:59 PM

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www.tntextbooks.in, , Activity, , 1. Take three conical flasks and label them as A, B, and C., 2. using a burette, add 10, 20 and 40 ml of 0.1M sodium thiosulphate solution to the flask, A, B and C respectively. And then add 40, 30 and 10 ml of distilled water to the respective, flasks so that the volume of solution in each flask is 50ml., 3. Add 10 ml of 1M HCl to the conical flask A. Start the stop watch when half of the HCl, has been added. Shake the contents carefully and place it on the tile with a cross mark as, shown in the figure. Observe the conical flask from top and stop the stops watch when, the cross mark just becomes invisible. Note the time., 4. Repeat the experiment with the contents on B and C. Record the observation., Flask, , Volume of Na 2S2 O3, , Volume of water Strength of Na 2S2 O3, , A, , 10, , 40, , 0.02, , B, , 20, , 30, , 0.04, , C, , 40, , 10, , 0.08, , Time taken ( t), , Vs, , 1, is a direct measure of rate of, t, , concentration of sodium thiosulphate. A, , reaction and therefore, the increase in the, , graph like the following one is obtained., , concentration of reactants i.e Na 2S2 O3 ,, , Draw, , a, , graph, , between, , 1, t, , 1 1, (s ), t, , increases the rate., , 0, , 0.02, , 0.04, 0.08, Concentraon(M), , 7.9.3 Effect of surface area of the reactant:, In heterogeneous reactions, the surface areas of the solid reactants play an important role, in deciding the rate. For a given mass of a reactant, when the particle size decreases surface, area increases. Increase in surface area of reactant leads to more collisions per litre per second,, and hence the rate of reaction is increased.For example, powdered calcium carbonate reacts, much faster with dilute HCl than with the same mass of CaCO3 as marble, 223, , XII U7 kinetics - Jerald Folder.indd 223, , 2/19/2020 4:46:03 PM

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www.tntextbooks.in, , Activity, , A Known mass of marble chips are taken in a flask and a known volume of dilute HCl is, added to the content, a stop clock is started when half the volume of HCl is added. The mass, is noted at regular intervals until the reaction is complete. Same experiment is repeated with, the same mass of powdered marble chips and the observations are recorded., Reaction, , CaCO3 (s) + 2HCl (aq) , → CaCl2 (aq) + H 2 O (l ) + CO 2 ( g ), , Since, carbon dioxide escapes during, reaction, the mass of the flask gets lighter as, the reaction proceeds. So, by measuring the, flask, we can follow the rate of the reaction., A plot of loss in mass Vs time is drawn and, it looks like the one as shown below., , Mass, , Pow, der, ed, , ma, r, , s, hip, ec, bl, , For the powdered marble chips, the, reaction is completed in less time. i.e., rate, of a reaction increases when the surface, area of a solid reactant is increased., , 0, , 10, , ble, ar, m, , 20, , ps, chi, , 30, , 40, , 50, , 60, , Time, , 7.9.4 Effect of presence of catalyst:, , Potential energy Æ, , So far we have learnt, that rate of reaction can be increased to certain extent by increasing, the concentration, temperature and surface area of the reactant. However significant changes, in the reaction can be brought out by the addition of a substance called catalyst. A catalyst, is substance which alters the rate of a reaction without itself undergoing any permanent, chemical change. They, may participate in the, reaction, but again, regenerated at the end, of the reaction. In the, presence of a catalyst,, the energy of activation, Ea Without, catalyst, is lowered and hence,, Ea With, greater, number, of, catalyst, molecules can cross, Reactants, the energy barrier and, Products, change over to products,, Reaction progress Æ, thereby increasing the, rate of the reaction., 224, , XII U7 kinetics - Jerald Folder.indd 224, , 2/19/2020 4:46:05 PM

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www.tntextbooks.in, , Activity, , Take two test tubes and label them as A and B. Add 7 ml of 0.1N oxalic acid solution, 5, ml of 0.1N KMnO 4 solution and 5 ml of 2N dilute H 2SO 4 in both the test tubes. The colour, of the solution is pink in both the test tubes., Now add few crystals of manganese sulphate to the content in test tube A. the pink, colour fades up and disappears. In this case, MnSO4 acts as a catalyst and increases the rate, 2−, of oxidation of C2O 4 by MnO 4, Chemical kinetics in pharmaceuticals, Chemical kinetics has many applications in the field of pharmaceuticals. It is used, to study the lifetimes and bioavailability of drugs within the body and this branch, of study is called pharmacokinetics Doctors usually prescribe drugs to be taken at, different times of the day. i.e.some drugs are to be taken twice a day, while others, are taken three times a day, or just once a day. Pharmacokinetic studies is used to determine the, prescription (dosage and frequency). For example, Paracetamol is a well known anti-pyretic and, analgesic that is prescribed in cases of fever and body pain. Pharmacokinetic studies showed that, Paracetamol has a half-life of 2.5 hours within the body i.e.the plasma concentration of a drug, is halved after 2.5 hrs. After 10 hours (4 half-lives)only 6.25 % of drug remains. Based on such, studies the dosage and frequency will be decided. In case of paracetamol, it is usually prescribed, to take once in 6 hours depending upon the conditions., Summary, „„ Chemical kinetics is the study of the rate and the mechanism of chemical reactions,, , proceeding under given conditions of temperature, pressure, concentration etc., , „„ The change in the concentration of the species involved in a chemical reaction per, , unit time gives the rate of a reaction., , „„ The rate of the reaction, at a particular instant during the reaction is called the, , instantaneous rate. The shorter the time period, we choose, the closer we approach, to the instantaneous rate,, , „„ The rate represents the speed at which the reactants are converted into products at, , any instant., , „„ The rate constant is a proportionality constant and It is equal to the rate of reaction,, , when the concentration of each of the reactants is unity., , „„ Molecularity of a reaction is the total number of reactant species that are involved, , in an elementary step., , „„ The half life of a reaction is defined as the time required for the reactant concentration, , to reach one half its initial value. For a first order reaction, the half life is a constant, 225, , XII U7 kinetics - Jerald Folder.indd 225, , 2/19/2020 4:46:07 PM

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www.tntextbooks.in, , i.e., it does not depend on the initial concentration., „„ According, , to collision theory, chemical reactions occur as a result of collisions, between the reacting molecules., , „„ Generally, the rate of a reaction increase with increasing temperature. However,, , there are very few exceptions. The magnitude of this increase in rate is different, for different reactions. As a rough rule, for many reactions near room temperature,, reaction rate tends to double when the temperature is increased by 100 C ., , „„ According to Arrhenius, activation energy of the reaction is the minimum energy, , that a molecule must have to posses to react., , „„ The rate of a reaction is affected by the following factors., , 1. Nature and state of the reactant, , 4. Temperature of the reaction, , 2. Concentration of the reactant, , 5. Presence of a catalyst, , 3. Surface area of the reactant, a), , EVALUATION, , ln k, , B, 1. For a first order reaction A →, −1, the rate constant is x min . If the, initial concentration of A is 0.01M , the, concentration of A after one hour is, given by the expression., a) 0.01 e−x, , c) ­(1 × 10, , −2, , )e, , 1/T, , b), log k, , b) 1 × 10−2 (1 −e−60 x ), , −60 x, , 1/T, , d) none of these, , c), , 2. A zero order reaction X →,  Product ,, with an initial concentration 0.02M has, a half life of 10 min. if one starts with, concentration 0.04M, then the half life, is, , ln k, , 1/T, , d) both (b) and (c), , a) 10 s    b) 5 min    c) 20 min, ,  product, 4. For a first order reaction A →, , d) cannot be predicted using the given, information, , with initial concentration x mol L−1 , has, , 3. Among the following graphs showing, variation of rate constant with, temperature (T) for a reaction, the one, that exhibits Arrhenius behavior over, the entire temperature range is, , a half life period of 2.5 hours . For the, same reaction with initial concentration,  x, −1,  2  mol L the half life is, 226, , XII U7 kinetics - Jerald Folder.indd 226, , 2/19/2020 4:46:09 PM

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www.tntextbooks.in, , a) ( 2.5 × 2 ) hours, , (, , c) mol, ,  2. 5 , b)   hours,  2 , , 1, , 2, , ), , L 2 s −1 , ( mol L−1s −1 ), 1, , (, , d) ( mol L s −1 ) , mol, , c) 2.5 hours, d) Without knowing the rate constant,, t1/2 cannot be determined from the, given data, , −d [NH3 ], = k1 [NH3 ] ,, dt, , b) k1 = 3 k 2 = 2 k 3, , b)Activation, energy, , c) Entropy, , d) Internal energy, , d) 2k1 = k 2 = 3 k 3, , (iv) a plot of ln ( k ) vs T is a straight line., , 6. The decomposition of phosphine (PH3), on tungsten at low pressure is a first, order reaction. It is because the (NEET), ,  1, (v) a plot of ln ( k ) vs   is a straight,  T, line with a positive slope., , a) , rate is proportional to the surface, coverage, , Correct statements are, , b) rate is inversely proportional to the, surface coverage, , d) rate of decomposition is slow, 3, , 7. For a reaction Rate = k [acetone ] 2 then, unit of rate constant and rate of reaction, respectively is, , (, , 2, , ), , −1, , 2, , 1, , L 2 s −1, , a) (ii) only, , b) (ii) and (iv), , c) (ii) and (v), , d) (i), (ii) and (v), , 10. In a reversible reaction, the enthalpy, change and the activation energy in, the forward direction are respectively, −x kJ mol −1 and y kJ mol −1 . Therefore ,, the energy of activation in the backward, direction is, , c) rate is independent of the surface, coverage, , b) mol, , increase in concentration of the, reactant increases the rate of a zero, order reaction., , (iii) rate constant k is equal to collision, frequency A if Ea=∞, , c) 1.5 k1 = 3 k 2 = k 3, , −1, a) ( y −x ) kJ mol, , ), , −1, b) ( x + y ) J mol, , c) ( x − y ) kJ mol −1, , L 2 s −1 , ( mol L−1s −1 ), 1, , ), , (ii) rate constant k is equal to collision, frequency A if Ea = 0, , a) k1 = k 2 = k 3, , −1, , 2, , a) Enthalpy, , (i), , then the relation between k1 ,k 2 and k 3 is, , (, , 1, , L s, , 9. Consider the following statements :, , d [N 2 ], d H, = k 2 [NH3 ], [ 2 ] = k 3 [NH3 ], dt, dt, , a) ( mol L−1s −1 ) , mol, , 2, , 8. The addition of a catalyst during a, chemical reaction alters which of the, following quantities? (NEET), ,  N 2 + 3H2, 5. For the reaction, 2NH3 →, , if, , 1, , d) ( x + y ) × 103 J mol −1, 227, , XII U7 kinetics - Jerald Folder.indd 227, , 2/19/2020 4:46:14 PM

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www.tntextbooks.in, , 11. What is the activation energy for a, reaction if its rate doubles when the, temperature is raised from 200K to, 400K? (R = 8.314 JK-1mol-1), , 2A + B , → C + 3D, Reaction, number, , a) 234.65 kJ mol −1K −1, −1, , b) 434.65 kJ mol K, , −1, , c) 2.305 kJ mol, , -1, −1, , d) 334.65 J mol K, 12., , −1, , ;, , This reaction, , 2, , 0.2, , 0.1, , 2x, , 3, , 0.1, , 0.2, , 4x, , 4, , 0.2, , 0.2, , 8x, 2, , 3, , Reason: rate constant also doubles, , b) 0.215M, , c) 0.25 × 2.303M d) 0.05M, 13. For a first order reaction, the rate, constant is 6.909 min-1.the time taken, for 75% conversion in minutes is, , a), , Both assertion and reason are true, and reason is the correct explanation, of assertion., , b), , Both assertion and reason are, true but reason is not the correct, explanation of assertion., , c), , Assertion is true but reason is false., , d), , Both assertion and reason are false., , 17. The rate constant of a reaction is, −2 −1, 5.8 × 10 s . The order of the reaction is, ,  y ; if, 14. In a first order reaction x →, k is the rate constant and the initial, concentration of the reactant x is 0.1M,, then, the half life is,  0.693 ,  log 2 , b) , a) , , ,  k ,  ( 0.1) k ,  ln2 , c) ,  k , , 0.1, , 16. Assertion: rate of reaction doubles when, the concentration of the reactant is, doubles if it is a first order reaction., , ( log 2 = 0.3010 ), ,  2,  4, d)   log  ,  3,  3, , 1, , 1, , of cyclopropane after 1806 minutes?, ,  3,  3, c)   log  ,  2,  4, , (min) (M s-1), x, 0.1, , d) rate = k [A ] 2 [B ] 2, , What will be the concentration, ,  2, b)   log 2,  3, , (min), , c) rate = k [A ][B ], , concentration of cyclopropane is 0.25, ,  3, a)   log 2,  2, , Initial rate, , 2, , constant at particular temperature, −2, −1, is 2.303 × 10 hour . The initial, , a) 0.125M, , [B], , a) rate = k [A ] [B ] b) rate = k [A ][B ], , follows first order kinetics. The rate, , M., , [A], , 18., , a) First order, , b) zero order, , c) Second order, , d) Third order, , For, , the, , reaction, , 1, O (g) , the, 2 2, value of rate of disappearance of N2O5, N 2O5 (g) →,  2NO2 (g) +, , d) none of these, , is given as 6.5 × 10−2 mol L−1s −1 . The, , 15. Predict the rate law of the following, reaction based on the data given below, , rate of formation of NO2 and O2 is given, 228, , XII U7 kinetics - Jerald Folder.indd 228, , 2/19/2020 4:46:18 PM

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www.tntextbooks.in, ,  2P0 ,  2.303 , log , d) k = , ,  t ,  3P0 −2P , −2, −1 −1, −2, −1 −1, a) ( 3.25 × 10 mol L s ) and (1.3 × 10 mol L s ), 22. If 75% of a first order reaction was, −2, −1 −1, × 10 mol L s ) and (1.3 × 10−2 mol L−1s −1 ), completed in 60 minutes , 50% of the, same reaction under the same conditions, b) (1.3 × 10−2 mol L−1s −1 ) and ( 3.25 × 10−2 mol L−1s −1 ), would be completed in, 3 × 10−2 mol L−1s −1 ) and ( 3.25 × 10−2 mol L−1s −1 ), a) 20 minutes, b) 30 minutes, −1, −1 −1, −2, −1 −1, c) (1.3 × 10 mol L s ) and ( 3.25 × 10 mol Lc)s 35, ) minutes, d) 75 minutes, respectively as, , × 10−1 mol L−1s −1 ) and ( 3.25 × 10−2 mol L−1s −1 ), , 23. The half life period of a radioactive, element is 140 days. After 560 days , 1 g, of element will be reduced to, , d) None of these, 19. During the decomposition of H2O2 to, give dioxygen, 48 g O2 is formed per, minute at certain point of time. The rate, of formation of water at this point is, ,  1, a)   g,  2, c)  1  g,  8, , a) 0.75 mol min−1, b) 1.5 mol min−1, ,  1, b)   g,  4,  1, d)   g,  16 , , 24. The correct difference between first and, second order reactions is that (NEET), , c) 2.25 mol min−1, d) 3.0 mol min−1, 20. If the initial concentration of the reactant, is doubled, the time for half reaction, is also doubled. Then the order of the, reaction is, a) Zero, , b) one, , c) Fraction, , d) none, , 21. In, a, homogeneous, reaction, A →,  B + C + D , the initial pressure, was P0 and after time t it was P. expression, for rate constant in terms of P0 , P and t, will be, , a), , A first order reaction can be, catalysed; a second order reaction, cannot be catalysed., , b), , The half life of a first order reaction, does not depend on [A0]; the half, life of a second order reaction does, depend on [A0]., , c), , The rate of a first order reaction, does not depend on reactant, concentrations; the rate of a second, order reaction does depend on, reactant concentrations., , d) The rate of a first order reaction does, depend on reactant concentrations;, the rate of a second order reaction, does not depend on reactant, concentrations., ,  2P0 ,  2.303 , a) k = , log,  3P −P ,  t , 0,  2P0 ,  2.303 , log , b) k = , ,  t ,  P0 −P , , 25. After 2 hours, a radioactive substance, th,  1, becomes   of original amount.,  16 , ,  3P0 −P ,  2.303 , c) k = , log,  2P ,  t , 0, 229, , XII U7 kinetics - Jerald Folder.indd 229, , 2/19/2020 4:46:22 PM

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www.tntextbooks.in, , Then the half life ( in min) is, a) 60 minutes, , b) 120 minutes, , c) 30 minutes, , d) 15 minutes, , (iv) Concentration of [A] is reduced to, 1, 3 and concentration of [L] is, , ( ), , quadrupled., , Answer the following questions:, , 11. The rate of formation of a dimer in a second, −3, −1 −1, order reaction is 7.5 × 10 mol L s at, 0.05 mol L−1 monomer concentration., Calculate the rate constant., , 1. Define average rate and instantaneous, rate., 2. Define rate law and rate constant., 3. Derive integrated rate law for a zero,  product ., order reaction A →, ,  products, 12. For a reaction x + y + z →, 3, 1, the rate law is given by rate = k [x ] 2 [ y ] 2, , 4. Define half life of a reaction. Show, that for a first order reaction half life is, independent of initial concentration., , what is the overall order of the reaction, , 5. What is an elementary reaction? Give, the differences between order and, molecularity of a reaction., , 13. Explain briefly the collision theory of, bimolecular reactions., , and what is the order of the reaction with, respect to z., , 14. Write Arrhenius equation and explains, the terms involved., , 6. Explain the rate determining step with, an example., , 8. Write the rate law for the following, reactions., , 15. The decomposition of Cl2O7 at 500K in, the gas phase to Cl2 and O2 is a first order, reaction. After 1 minute at 500K, the, pressure of Cl2O7 falls from 0.08 to 0.04, atm. Calculate the rate constant in s-1., , (a) A reaction that is 3/2 order in x and, zero order in y., , 16. Give two exapmles for zero order, reaction, , (b) A reaction that is second order in, NO and first order in Br2., , 17. Explain pseudo first order reaction with, an example., , 7. Describe the graphical representation of, first order reaction., , 9. Explain the effect of catalyst on reaction, rate with an example., , 18. Identify the order for the following, reactions, , 10. The rate law for a reaction of A, B and 3C, 2, has been found to be rate = k [A ] [B ][L ] 2, . How would the rate of reaction change, when, , (i) Rusting of Iron, 238, (ii) Radioactive disintegration of 92 U, ,  products ; rate = k [A ] 2 [B ], (iii) 2A + 3B →, 1, , 2, , 19. A gas phase reaction has energy of, activation 200 kJ mol-1. If the frequency, factor of the reaction is 1.6 × 1013 s −1, . Calculate the rate constant at 600 K., (e −40.09 = 3.8 × 10−18 ), , (i) Concentration of [L] is quadrupled, (ii) Concentration of both [A] and [B], are doubled, (iii) Concentration of [A] is halved, 230, , XII U7 kinetics - Jerald Folder.indd 230, , 2/19/2020 4:46:24 PM

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www.tntextbooks.in, , 28. Benzene diazonium chloride in aqueous, solution decomposes according to the,  C 6 H5Cl + N 2 ., equation C 6 H5 N 2Cl →, Starting with an initial concentration of, 10 g L−1 , the volume of N2 gas obtained, at 50 °C at different intervals of time was, found to be as under:, ,  L find the, 20. For the reaction 2x + y →, rate law from the following data., [x], (M), , [y], (M), , rate, (M s-1), , 0.2, , 0.02, , 0.15, , 0.4, , 0.02, , 0.4, , 0.08, , 0.30, 1.20, , t (min):, , 6, , 12, , 18, , 24, , ∞, , 30, , Vol. of N2 19.3 32.6 41.3 46.5 50.4 58.3, (ml):, , 21. How do concentrations of the reactant, influence the rate of reaction?, 22. How do nature of the reactant influence, rate of reaction., 23. The rate constant for a first order reaction, is 1.54 × 10-3 s-1. Calculate its half life, time., 24. The half life of the homogeneous gaseous, reaction SO2Cl2 → SO2 + Cl2 which obeys, first order kinetics is 8.0 minutes. How, long will it take for the concentration of, SO2Cl2 to be reduced to 1% of the initial, value?, 25. The time for half change in a first order, decomposition of a substance A is 60, seconds. Calculate the rate constant., How much of A will be left after 180, seconds?, 26. A zero order reaction is 20% complete, in 20 minutes. Calculate the value of, the rate constant. In what time will the, reaction be 80% complete?, , Show that the above reaction follows the, first order kinetics. What is the value of, the rate constant?, 29. From the following data, show that the, decomposition of hydrogen peroxide is, a reaction of the first order:, t (min), , 0, , 10, , 20, , V (ml), , 46.1, , 29.8, , 19.3, , Where t is the time in minutes and V is, the volume of standard KMnO4 solution, required for titrating the same volume of, the reaction mixture., 30. A first order reaction is 40% complete, in 50 minutes. Calculate the value of, the rate constant. In what time will the, reaction be 80% complete?, , 27. The activation energy of a reaction, is 22.5 k Cal mol-1 and the value of, rate constant at 40°C is 1.8 × 10−5 s −1 ., Calculate the frequency factor, A., , 231, , XII U7 kinetics - Jerald Folder.indd 231, , 2/19/2020 4:46:25 PM

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XII U7 kinetics - Jerald Folder.indd 232, , Order, , Instantaneous rate, , Average rate, , Initial rate, , Collision theory, , Molecularity, , Rate law, , Reaction rate, , Chemical kinetics, , Temperature, dependence of, rate ( Arrhenius, theory), , Rate constant, , factors affecting, reaction rate, , k=, , k=, , [A 0 ] −[A], t, , [A ], 2.303, log 0, t, [A], , First order, reaction, , Zero order, reaction, , Surface area, , Catalyst, , Temperature, , Concentration, , www.tntextbooks.in, , 232, , 2/19/2020 4:46:25 PM

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www.tntextbooks.in, , ICT Corner, CHEMICAL KINETICS, By using this tool you can, understand collision and the, activation energy. You can also, perform virtual kinetic experiment, to understand the effect of, temperature on reaction rate., , Please go to the URL, https://phet.colorado.edu/en/, simulation/legacy/reactionsand-rates, (or) Scan the QR code on the, right side, , Steps, •, , Open the Browser and type the URL given (or) Scan the QR Code. You can see a webpage, which displays the java applet called reactions and rates. You can click it and you will get a, window as shown in the figure. This applet contains three modules which can be selected by, clicking the appropriate tab (box1)., , •, , Select single collision tab (tab 1) to visualise collision between two molecules. You can visualise, the progress of the reaction (box 5) by varying the temperature using the slider (box 2). You, can visualise that the raise in temperature, will raise the energy of the system and allows the, reactants to cross the energy barrier to form products. You can repeat this simulation with, more molecules in the many collisions tab (box 1)., , •, , You can also perform virtual kinetic experiment, using rate experiments mode. Choose the types, reacting molecules and their stoichiometry using the options provided (box 2). The number of, reactant and product molecules at a given time will be displayed in panel (box-3). You can see, the effect of temperature on reaction rate by varying the temperature (box 4)., 1, 3, 2, , 5, 4, , 233, , XII U7 kinetics - Jerald Folder.indd 233, , 2/19/2020 4:46:26 PM

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www.tntextbooks.in, , ANSWERS, , VOLUME, , I, , UNIT-I, Choose the correct answer, 1. b) Al 2O3 .nH2O 2. c) SO2   3. c) MgCO3 →,  MgO + CO2, 4. b) Al 2O3 5. a) Al, 6. d) Carbon and hydrogen are suitable reducing agents for metal sulphides., 7. c) A-iv , B-ii , C-iii , D-i, 8. d)Electromagnetic separation, 9. b) Cu(s) + Zn 2+ (aq) →,  Zn(s) + Cu 2+ (aq), 10. c) sodium, , 11. b) Infusible impurities to soluble impurities, , 12. c) Galena, , 13. a) Lower the melting point of alumina, , 14. a) Carbon reduction, , 15. c) Displacement with zinc, , 16. c) Mg, , 17. b) van Arkel process, , 18. d) Both (a) and (c), 19. d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution, 20. b) Impure copper, , 21. b) Δ G0 Vs T, ,  ∆ G0 ,  Cr2O3 + 2Al, 22. c) , is negative, 23. b) Al 2O3 + 2Cr →,  ∆ T , 24. b) The graph for the formation of CO2 is a straight line almost parallel to free energy axis., , UNIT-2, Choose the correct answer:, 1. c) basic, , , Na 2 B4O7 + 7H2O , + 4H3BO3 , ,  2NaOH, Strong base, Weak acid, , 234, , Answers.indd 234, , 2/19/2020 4:34:17 PM

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www.tntextbooks.in, , 2., , 18. a) Al < Ga < In < Tl, , d) accepts OH- from water ,releasing, proton., −, +, , , B ( OH )3 + H2O , ,  B ( OH ) 4  + H, 3. b) B3H6, nido borane, , stability of +1 oxidation state increases, down the group due to inert pair effect, , : BnH4+n, , UNIT-3, , aracno borane : BnH6+n, B3H6 is not a borane, , Choose the correct answer:, , 4. a) Aluminium, , 1. a) Nessler’s reagent , , 5. c) four, , 2. d) , ability to form pπ − pπ bonds with, itself, , There are two 3c – 2e- bonds i.e., the, bonding in the bridges account for 4, electrons., , 3. d) 1s2 2s2 2p6 3s2 3p3, 4. b) P4(white) and PH3, , 6. c) Lead, , 5. a) H3PO3, , 7. c) sp2 hybridised, , 6. a) H3PO3, , 8. a) +4 , , 7. b) 2, 8. a) 6N, , Example : CH4+ in which the oxidation, state of carbon is +4, 9. d) ( SiO4 ), , 9. d) Both assertion and reason are wrong., The converse is true., , 4−, , 10. b) F2, , R, , 10. b), , Si, , 11. b) HF > HCl > HBr > HI, , O, , 12. d) NeF2, , R, , 11. a) Me3SiCl, , 13. c) He, , 12. d) dry ice, , 14. c) XeO3, , dry ice – solid CO2 in which carbon is, in sp hybridized state, , 15. a) HI, 16. d) Cl2 > Br2 > F2 > I2, , 13. a) Tetrahedral, , 17. d) HClO < HClO2 < HClO3 < HClO4, , 14. d) Feldspar is a three dimensional silicate, , 18. c) Cu(NO3)2 and NO2, , 15. a) A-b , B-1 , C-4 , D-3, 16. d) Al,Cu,Mn,Mg, Al-95% , Cu-4% , Mn-0.5% , Mg-0.5%, 17. a) Metal borides, 235, , Answers.indd 235, , 2/19/2020 4:34:19 PM

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www.tntextbooks.in, , 15. d) 3, 16. a) Both assertion and reason are true and reason is the correct explanation of assertion., 17. b) +3, 18. a) Np, Pu ,Am, 19. a) La(OH)2 is less basic than Lu(OH)3, , UNIT-5, Choose the correct answer:, 1. In the complex M ( en )2 ( Ox )  Cl For the central metal ion M3+, The primary valance is = +3, The secondary valance = 6, sum of primary valance and secondary valance = 3+6 = 9, Answer : option (d), 2. The complex is M ( H2O )5 Cl  Cl 2, 1000 ml of 1M solution of the complex gives 2 moles of Cl − ions, 1000 ml of 0.01M solution of the complex will give, 100 ml × 0.01M × 2Cl −, = 0.002 moles of Cl −ions, 1000 ml × 1M, , Answer : option (b), 3. Molecular formula: MSO4Cl. 6H2O ., Formation of white precipitate with Barium chloride indicates that SO4 2− ions are outside, the coordination sphere, and no precipitate with AgNO3 solution indicates that the Cl −, −, ions are inside the coordination sphere.Since the coordination number of M is 6, Cl and, 2−, 5 H2O are ligands, remaining 1 H2O molecular and SO4 are in the outer coordination, sphere., Answer : option (c), 2+, , +, +, , 2−, 4. Fe ( H2O )5 NO SO4, , , +1 and +1 respectively, , Answer : option(d), 5. Answer : option(d), 6. Answer : option(d), , 237, , Answers.indd 237, , 2/19/2020 4:34:23 PM

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www.tntextbooks.in, , 7. Answer : option(c), , Ti 4+ ( d 0 ⇒ 0BM ), , (, (d Low spin ⇒ t, (d Low spin ⇒ t, , Co2+ d 7spin free ⇒ t 2 g 5 , e g 2 ; n = 3 ; µ = 3.9 BM, Cu 2+, Ni 2+, , 9, , 8, , 2g, , 2g, , 6, , 6, , ), , , e g 3 ; n = 1 ; µ = 1.732 BM, , , e g 2 ; n = 2 ; µ = 2.44 BM, , ), , ), , 8. Answer : option(b), The electronic configuration t 2 g 3 , e g 2, 3 × ( −0.4 ) + 2 ( 0.6 )  ∆ 0, , [−1.2 + 1.2 ]∆, , =0, 9. Answer : option(a), 0, , In all the complexes, the central metal ion is Co3+, among the given ligands CN − is the, strongest ligand, which causes large crystal field splitting i.e maximum ∅0, 10. Answer : option(b), en, , en, N, , M, , en, N, , N, , Cl, , N, , Cl, , Co, , N, , N, , Cl, Co, , Co, Cl, , N, , N, , N, , Cl, , N, , Cl, N, , N, , en, , en, , en, inactive, , Active forms ( enantiomorphs), , Complexes given in other options (a), (c) and (d) have symmetry elements and hence they, are optically inactive., 11. Answer : option(d), Cl, , NH3, , H3N, Pt, , Pt, Cl, , Cl, , NH3, , Cl, , NH3, , 12. Three isomers. If we consider any one of the ligands as reference ( say Py), the arrangement, of other three ligands ( NH3, Br- and Cl-) with respect to (Py) gives three geometrical, isomers., 13. Answer : option(c), (a)coordination isomers, (b) no isomerism ( different molecular formula), (c) ← NCS , ← SCN coordinating atom differs : linkage isomers, , 238, , Answers.indd 238, , 2/19/2020 4:34:25 PM

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www.tntextbooks.in, , 14. Answer : option(a), , For [MA 4 B2 ] complexes-geometrical isomerism is possible, n+, ,  Co ( NH3 ) Br2  Cl, , , 4, ,  Co ( NH3 ) 4 Br Cl  Br, , ,  → ionisation isomers, , , 15. Answer : option(d), Option (a) and (b) – geometrical isomerism is possible, Option (c) – ionization isomerism is possible, Option (d) – no possibility to show either constitutional isomerism or stereo isomerism, 16. Answer : option(c), (b) Fe3+, , (a) Fe2+, , (c) Fe0, , 17. Answer : option(d), , Fe ( en )3  ( PO4 3− ), 18. Answer : option(c), 2+, , (a) Zn2+ ( d10 ⇒ diamagnetic ), (b) Co3+ d 6 Low spin ⇒ t 2 g 6 , e g 0 ; diamagnetic, (c) Ni, , 2+, , (, (d Low spin ⇒ t, 8, , 2g, , 6, , ), , , e g ; paramagnetic, 2, , ), , (d) [Ni(CN)4 ] ( dsp ; square planar, diamagnetic ), 2−, , 2, , 19. Answer : option(c), , Co ( NH3 )3 (Cl )3 , 20. Answer : option(d), V 2+ t 2 g 3 , e g 0 ; CFSE = 3 × (-0.4)∆ 0 = −1.2∆ 0, Ti, , 2+, , (, (t, , 2g, , 2, , , e g ; CFSE = 2 × (-0.4)∆ 0, 0, , ), = −0.8∆ ), 0, , Statements given in option (a) ,(b), and (c) are wrong., The current statements are, (a) since, the crystal field stabilization is more in octahedral field , octahedral complexes, are more stable than square planar complexes., (b), , [FeF ], , 4−, , 6, ,  Fe2+ −d 6 High Spin −t 2 g 4 , e g 2 ), , −, ,  CFSE = 4 × (-0.4) + ( 0.6 ) × 2 + P, ,  Fe2+ −d 6 Low Spin −t 2 g 6 , e g 0 ), Fe ( CN )6  −, ,  CFSE = 6 × (-0.4) + 3P, , 4−, , 239, , Answers.indd 239, , 2/19/2020 4:34:27 PM

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www.tntextbooks.in, , UNIT-6, Choose the correct answer:, 1. c) both covalent crystals , 2. b) AB3, ,  N   8, number of A ions =  c  =   = 1,  8   8,  Nf   6, number of B ions =   =   = 3,  2   2, simplest formula AB3, , 3. b) 1:2, if number of close packed atoms =N; then,, The number of Tetrahedral holes formed = 2N, number of Octahedral holes formed = N, therefore N:2N = 1:2, 4. c) molecular solid, lattice points are occupied by CO2 molecules, , 5. a) Both assertion and reason are true and reason is the correct explanation of assertion., 6. c) 8 and 4, CaF2 has cubical close packed arrangement, , Ca2+ ions are in face centered cubic arrangement, each Ca2+ ions is surrounded by 8 F − ions, and each F − ion is surrounded by 4 Ca2+ ions., Therefore coordination number of Ca2+ is 8 and of F − is 4, 7. b) 6.023 × 1022, in bcc unit cell,, 2 atoms ≡ 1 unit cell, Number of atoms in 8g of element is ,, 8g, = 0.2 mol, Number of moles =, 40 g mol −1, 1 mole contains 6.023 × 1023 atoms, 0.2 mole contains 0.2 × 6.023 × 1023 atoms,  1unit cell , 23,  2 atoms  × 0.2 × 6.023 × 10, 6.023 × 1022 unit cells, , 240, , Answers.indd 240, , 2/19/2020 4:34:29 PM

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www.tntextbooks.in, , 8. d) M3N2, , (100-68) = 32%, 13. b) 848.5pm, , if the total number of M atoms is n,, then the number of tetrahedral voids, =2n,  1, given that   rd of tetrahedral voids,  3, 1, are occupied i.e.,   × 2n are occupied,  3, by N atoms, , let edge length =a, 2a = 4r, 4 × 300, a=, 2, a = 600 × 1.414, a = 848.4 pm,  π, 14. b)  ,  6, ,  2, ∴ M:N ⇒ n :   n,  3,  2, 1:  ,  3, 3 : 2 ⇒ M3 N 2, , 3,  , 4,  43 πr 3   3 π  a2    π ,  a 3  =  a 3  =  6 , , , , 9. c) 6, rC, , +, , rA, , −, , 15. a) excitation of electrons in F centers, =, , 0.98 × 10−10, = 0.54, 1.81 × 10−10, , 1, 3, 1 , a:, a, 16. c)  a :, 4, 2 2 , 2, , it is in the range of 0.414 - 0.732 ,, hence the coordination number of each, ion is 6, , sc ⇒ 2r = a ⇒ r =, , 3a, 4, 2a, a, =, fcc ⇒ 4r = 2a ⇒ r =, 4, 2 2,  a   3a   a ,  2  :  4  : , , ,  2 2, , bcc ⇒ 4r = 3a ⇒ r =, ,  3, 10. d)  2  × 400pm,  , 3 a = rCs + 2rCl + rCs, +, , −, ,  3,  2  a = rCs + rCl,  , , (, , +, , −, , a, 2, , +, , ), ,  3, 17. d)   a,  2 , if a is the length of the side, then the, length of the leading diagonal passing, through the body centered atom is 3a,  3, Required distance =   a,  2 , ,  3,  2  400 = inter ionic distance,  ,  100 , 11. a)  0.414 , rX, = 0.414, for an fcc structure, ry, given that rX = 100 pm, +, , −, , +, , 18. a) 915 kg m-3, , 100 pm, ry =, 0.414, 12. c) 32%, , ρ=, , −, , n × M, a 3NA, , for bcc, , packing efficiency = 68%, therefore empty space percentage =, 241, , Answers.indd 241, , 2/19/2020 4:34:33 PM

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www.tntextbooks.in, , n, n=, = 22, M=39, M=39, nearest, nearest distance, distance 2r, 2r =, = 4.52, 4.52, , −10, −10, 44rr 22 ×, 44..52, ×, 10, ×, ×, 52, 10, 10, 100−−10, =, 55..21, ×, aa =, = 3=, =, =, ×, 1, 21, 3, 3, 3, 22 ×, 39, × 39, ρ, =, ρ=, −10 33, 23 ), 21 ×, 10−10 ) ×, 6..023, 023 ×, 1023, ((55..21, (, × 10, 6, 10, ×, ×, ), (, ), −3, −3, ρ, =, Kg, m, 915, ρ = 915 Kg m, , NA , , 19. b) equal number of cations and anions are missing from the lattice, 20. c) Frenkel defect , 21. d) Both assertion and reason are false, 22. b) FeO, 23. a) XY8, , UNIT-7, Choose the correct answer:, 1. option (c),  [A 0 ],  2.303 , log, k= ,  A ,  t ,  [ ],  1  [A ], k =   ln  0 ,  t   [A ] , ,  [A ], e kt =  0 ,  [A ] , , [A ]= [A ]e, 0, , −kt, , In this case, k = x min−1 and [A 0 ] = 0.01M = 1 × 10−2 M, t = 1 hour = 60 min, , [A ] = 1, , × 10−2 ( e−60 x ), , 2. option (c), , 242, , Answers.indd 242, , 2/19/2020 4:34:34 PM

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www.tntextbooks.in, , for, for n, n≠, ≠ 11, for, = 00, for n, n=, , [[AA0 ]], =, = 0, , tt 1, 12, , tt 1 =, =, 12, 2, , tt 1, 12, 2, , 22nn−−11 −, −11, n−1, n−1, (( n-1, n-1)) kk [[A, A 0 ]], , 11, =, −1, =, −1, 22 kk [[A, A 0 ]], , 0, , 0, , 22 kk, , 2, , tt 1 α, A 00 ]] −−−−−, −−−−−((11)), α [[A, 12, 2, , given, given, [[AA00 ]] == 00..02, 02M, M ;; tt 112 =, = 10, 10 min, min, , [[AA ]] == 00..04, 04M, M ;; tt, 0, 0, , 2, , 1, 12, 2, , =, = ??, , substitute, substitute in, in (1), (1), 10 min, min α, 10, α 0.02M, 0.02M −−−−−, −−−−−((22)), tt 1 α, α 0.04M, 0.04M −−−−−, −−−−−((33)), 12, 2, , ((33)), , ((22 )), , tt 1, 12, , 00..04, M, 04 M, 2, ⇒, =, ⇒ 10 min, = 0.02 M, 10 min 0.02 M, , tt 1 =, = 22 ×, × 10, 10 min, min =, = 20, 20 min, min, 12, 2, , 3. option (b), k= Ae, , E , − a ,  RT , ,  E  1, ln k = ln A − a   ,  R  T, this equation is in the form of a straight line equatoion, y=c+mx,  1, a plot of lnk vs   is a straight line with negative slope,  T, 4. option(c), For a first order reaction, 0.693, t1 =, 2, k, t1, 2 does not depend on the initial concentration and it remains constant (whatever may, be the initial concentration), t 1 = 2.5 hrs, 2, , 243, , Answers.indd 243, , 2/19/2020 4:34:35 PM

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www.tntextbooks.in, , 5. option(c), ,  −1 d [NH3 ] d [N 2 ]  1  d [H2 ], Rate =  , =, = ,  2  dt,  3  dt, dt, ,  1,  1,  2  k1 [NH3 ] = k 2 [NH3 ] =  3  k 3 [NH3 ],  3,  2  k1 = 3k 2 = k 3, 1.5 k1 = 3k 2 = k 3, 6. option(a), Given :, At low pressure the reaction follows first order, therefore, Rate α[reactant ], , 1, , Rate α ( surface area ), At high pressure due to the complete coverage of surface area, the reaction follows zero, order., , Rate α[reactant ], , 0, , Therefore the rate is independent of surface area., 7. option(b), rate = k [A ], , n, , rate =, , −d [A ], dt, , mol L−1, = mol L−1s −1, s, mol L−1s −1 ), (, unit of rate constant =, = mol1−n Ln−1s −1, −1 n, ( mol L ), , unit of rate =, , in this case, 3, , rate = k [Acetone ] 2, n= 3, 2, , ( 2 ) ( 3 2 )−1, , 1− 3, , L s −1, −( 1 ), (1 ), mol 2 L 2 s −1, , mol, , 8. option(b), A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers, the activation energy., 244, , Answers.indd 244, , 2/19/2020 4:34:37 PM

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www.tntextbooks.in, , 9. option(a), In zero order reactions, increase in the concentration of reactant does not alter the rate., So statement (i) is wrong., E , − a ,  RT , , k= Ae, if Ea = 0, k= Ae, k= A, , so, statement (ii) is correct, and statement (iii) is wrong, , 0, ,  E  1, ln k = ln A − a   ,  R  T, this equation is in the form of a straight line equatoion, y=c+mx,  1, a plot of lnk vs   is a straight line with negative slope,  T, so statements (iv) and (v) are wrong., , (Ea)f = y kJ mol-1, , 10. option(d), , (Ea)b = (x + y) kJ mol-1, , Reactant, x kJ mol-1, , ( x + y ) kJmol −1, ( x + y ) × 103 Jmol −1, 11. option(c), T1 = 200K ; k = k1, T2 = 400K ; k = k 2 = 2k1, k , Ea  T2 −T1 , log  2  =,  k1  2.303R  T1 T2 , 2k , Ea,  400 K −200K , log  1  =, , −1 , −1,  k1  2.303 × 8.314 J K mol  200K × 400K , 0.3010 × 2.303 × 8.314 J K-1 mol −1 × 200 K × 400 K, Ea =, 200 K, –1, Ea = 2305 J mol, Ea = 2.305 kJ mol-1, , 245, , Answers.indd 245, , 2/19/2020 4:34:39 PM

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www.tntextbooks.in, , 12. option(a),  [[A,  2.303, A 0 ]], 2.303 , kk =, =  t  log, log  A0 ,  [[A ]] ,  t , ,  2.303, -2, −1, 2.303  log  00..25, 25 , -2 hour −1 = , ×, 2.303, 10, 2.303 × 10 hour =  1806 min  log  A ,  [[A ]] ,  1806 min , -2, −1,  2.303, 25 , × 10, × 1806, 2.303 ×, 10-2 hour, hour −1 ×, 1806 min, min  = log  00..25, =, log, , , , ,  [A, , 2., 03, 2.3303, ,  [A ]] , , -2,  1806, 1806 ×, × 10, 10-2  = log  00..25, 25 , ,  = log  A , 60,  [[A ]] , 60, , ,  00..25, 25 , = log, 0.301, log  A , 0.301 =,  [[A ]] ,  00..25, 25 , log, =, log, 2, log 2 = log  A ,  [[A ]] ,  00..25, 25 , 22 =, =  A ,  [[A ]] ,  0.25 , [[AA ]] ==  0.225  == 00..125, 125M, M,  2 , , 13. option(b),  [A 0 ],  2.303 , k= , log,  A ,  t ,  [ ], , [A ]= 100 ; [A ]=25, 0, ,  2.303 ,  100 , 6.909 = , log , ,  t ,  25 ,  2.303 , log ( 4 ), t= ,  6.909 ,  1, t =   log 22,  3,  2, t =   log 2,  3, , 246, , Answers.indd 246, , 2/19/2020 4:34:39 PM

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www.tntextbooks.in, , 14. option(c),  1  [A ], k =   ln  0 ,  t   [A ] , , [A ]=0.1 ; [A ]=0.05, 0, ,  , 1,  0. 1 , k =   ln , ,  t 1   0.05 , 2,  , 1, k =   ln ( 2 ),  t 1 , 2, t1 =, 2, , ln ( 2 ), k, , 15. option(b), rate1 = k [0.1] [0.1] −−−−−(1), n, , m, , rate2 = k [0.2 ] [0.1] −−−−−(2), n, , (2), , m, , (1), , 2 x k [0.2 ] [0.1], =, x k [0.1]n [0.1]m, 2x, = 2n ∴ n =1, x, n, , m, , rate3 = k [0.1] [0.2 ] −−−−−(3), n, , m, , rate4 = k [0.2 ] [0.2 ] −−−−−(4), n, , (4), , m, , (2), , 8 x k [0.2 ] [0.2 ], =, 2 x k [0.2 ]n [0.1]m, 8, = 2m ∴ m =2, 2, 1, 2, ∴ rate = k [A ] [B ], n, , m, , 16. option(c), For a first reaction, If the concentration of reactant is doubled, then the rate of reaction, also doubled., Rate constant is independent of concentration and is a constant at a constant temperature,, 17. option(a), The unit of rate constant is s-1 and it indicates that the reaction is first order., 247, , Answers.indd 247, , 2/19/2020 4:34:41 PM

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www.tntextbooks.in, ,  1, 2=  ,  2, , n−1, , 2 = ( 2−1 ), , n−1, , 21 = ( 2−n+1 ), , n= 0, , 21. Answer : option(a), A, , →, , , B, , C, , D, , Initial, , a, , 0, , 0, , 0, , Reacted at time t, , x, , –, , –, , –, , (a −x ), , x, , x, , x, , After time t, Total number of moles, , = (a + 2x ), , a α P0, , (a + 2x ) α P, P, a, = 0, (a + 2x ) P, x=, , ( P −P ) a, 0, , 2P0,  ( P −P0 ) a ,  2P , , (a −x ) = a −, , 0, ,  3P0 −P , ,  2P0 , , (a −x ) = a , , [A ],  2.303 , k= , log 0, ,  t , [A ],  2.303 ,  a , k= , log , ,  t ,  a −x , , , , , a,  2.303 , , , log, k= ,  t ,   3P0 −P  , ,  a,   2P0  ,  2P0 ,  2.303 , log, k= ,  3P −P ,  t , 0, 249, , Answers.indd 249, , 2/19/2020 4:34:44 PM

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www.tntextbooks.in, , 22. Answer : option(b)30min, t 75% = 2t 50%,  t   60 , t 50% =  75%  =   = 30 min,  2   2, 23. Answer : option(d), in140, in140 days, days ⇒, ⇒ initial, initial concentration, concentration reduced, reduced to, to, ,  11 ,  2  gg,  2, ,  11 , in 2280, in, 80 days, days ⇒, ⇒ initial, initial concentration, concentration reduced, reduced to, to  4  gg,  4,  11 , in, in 442200 days, days ⇒, ⇒ initial, initial concentration, concentration reduced, reduced to, to  8  gg,  8,  1, in, days ⇒, ⇒ initial, initial concentration, concentration reduced, reduced to, to  1  gg, in 56, 5600 days,  16, 16 , 24. Answer : option(b), For a first order reaction, t1 =, 2, , 0.6932, k, , For a second order reaction, t1 =, 2, , 2n−1 −1, , ( n-1) k [A0 ], , n−1, , n= 2, t1 =, 2, , t1 =, 2, , 22−1 −1, , (2-1) k [A0 ], , 2 −1, , 1, k [A 0 ], , 25. Answer : option(c), t1,  1 t 1  1  t 1  1 t 1  1 , 1  2 →    2 →    2 →    2 →  ,  2,  4,  8,  16 , ∴ 4 t 1 = 2 hours, 2, , t 1 = 30 min, 2, , 250, , Answers.indd 250, , 2/19/2020 4:34:45 PM

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www.tntextbooks.in, , Answer the following:, 10 solution, 3, , Rate = k [A ] [B ][L ] 2 -----(1), 2, , (i) when [L ] = [4L ], , 3, , Rate = k [A ] [B ][4L ] 2, 2, , (, , 3, , ), , Rate = 8 k [A ] [B ][L ] 2 -----(2), 2, , Comparing (1) and (2) ; rate is increased by 8 times., (ii) when [A ] = [2A ]and [B ] = [2B ], 3, , Rate = k [2A ] [2B ][L ] 2, 2, , (, , 3, , ), , Rate = 8 k [A ] [B ][L ] 2 -----(3), 2, , Comparing (1) and (3) ; rate is increased by 8 times., A, (iii) when [A ] =  , 2, 2, , 3, A, Rate = k   [B ][L ] 2, 2, 3, 2,  1, Rate =   k [A ] [B ][L ] 2 -----(4),  4, , (, , ), , Comparing (1) and (4) ; rate is reduced to ¼ times., A, (iv ) when [A ] =   and [L ] = [4L ], 3, 2, , 3, A, Rate = k   [B ][4L ] 2, 3, 3, 2,  8, Rate =   k [A ] [B ][L ] 2 -----(5),  9, , (, , ), , Comparing (1) and (5) ; rate is reduced to 8/9 times., 11. solution, Let us consider the dimerisation of a monomer M, 2M →,  ( M )2, Rate= k [M ], , n, , Given that n=2 and [M] = 0.05 mol L-1, Rate = 7.5 X 10-3 mol L-1s-1, , 251, , Answers.indd 251, , 2/19/2020 4:34:47 PM

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www.tntextbooks.in, , k=, k=, , Rate, , [M ], , n, , 7.5 × 10−3, , (0.05), , 2, , = 3 mol −1Ls −1, , 12. Solution,  3,  2 , , rate = k [x ], ,  1,  2 , , [y], ,  3 1, overall order =  +  = 2,  2 2, i.e., second order reaction., Since the rate expression does not contain the concentration of z , the reaction is zero order, with respect to z., 15. Solution:, 2.303, [A ], k=, log 0, t, [A ], , 2.303, [0.08 ], log, 1 min, [0.04 ], k = 2.303 log 2, k = 2.303 × 0.3010, k=, , k = 0.6932 min−1,  0.6932  −1, s, k= ,  60 , k = 1.153 × 10−2 s −1, 19. Solution, k= Ae, , E , − a ,  RT ,  200 ×103 J mol −1 , −, , −1, −1, 13 −1  8.314 JK mol × 600 K , , k = 1.6 × 10 s e, , k = 1.6 × 1013s −1 e −( 40.1), k = 1.6 × 1013s −1 × 3.8 × 10−18, k = 6.21 × 10−5s −1, , 252, , Answers.indd 252, , 2/19/2020 4:34:49 PM

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www.tntextbooks.in, , 20. Solution, , For a first order reaction,, , rate = k [x ] [ y ], n, , m, , t=, , 0.15 = k [0.2 ] [0.02 ] −−−−−(1), n, , m, , 0.30 = k [0.4 ] [0.02 ] −−−−−(2), n, , m, , 2.303,  100 , log,  1 , 0.087 min −1, t = 52.93 min, t=, , 1.20 = k [0.4 ] [0.08 ] −−−−−(3), n, , (3), , m, , (2 ), , 1.2 k [0.4 ] [0.08 ], =, 0.3 k [0.4 ]n [0.02 ]m, , m, , n, ,  [0.08 ], 4= , ,  [0.02 ], 4 = (4), , 2.303, [A ], log 0, k, [A ], , 25 Solution:, i) Order of a reaction = 1; t1/2 = 60 ;, seconds, k = ?, , m, , 0.6932, We know that, k =, t1/2, , m, , 2.303, = 0.01155 s −1, 60, , ∴ m =1, , k=, , (2), , ii) [A 0 ] = 100% t = 180 s , k = 0.01155, seconds-1, [A ] = ?, , (1), , 0.30 k [0.4 ] [0.02 ], =, 0.15 k [0.2 ]n [0.02 ]m, n, ,  [0.4 ], 2= , ,  [0.2 ], , m, , For, , n, , 1, , 0.15 = k [0.2 ] [0.02 ], , [0.2 ] [0.02], 1, , 1, , 1, , 0.9207 = log100 −log [A ], log [A ] = log100 −0.9207, , =k, , k = 37.5 mol L s, −1, , log [A ] = 2 −0.9207, , −1, , log [A ] = 1.0973, , [A ] = antilog of (1.0973), [A ] = 12.5%, , 23Solution:, We know that, t1/2 = 0.693/ k, , 26 Solution:, , t1/2 = 0.693/1.54x 10-3 s-1 = 450 s, , i) Let A = 100M, [A0]–[A] = 20M,, , For the zero order reaction, , 24.Solution:, We know that,, , reaction, ,  100 , 0.01155 × 180, = log , , 2.303,  [A ] , , Rate = k [x ] [ y ], 0.15, , order, , 2.303, [A ], log 0, t, [A ],  100 , 2.303, 0.01155 =, log , , 180,  [A ] , , 2 = (2), ∴ n =1, , 1, , first, , k=, , n, , 1, , the, ,  [A ] −[A ], k=  0, , t, , , k = 0.693/ t1/2, , k = 0.693/ 8.0 minutes = 0.087 minutes-1, 253, , Answers.indd 253, , 2/19/2020 4:34:55 PM

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www.tntextbooks.in, ,  20M , k= , = 1 Mmin−1, ,  20min , Rate constant for a reaction = 1Mmin-1, ii) To calculate the time for 80% of, completion, k = 1Mmin-1, [A0] = 100M, [A0]-[A] =, 80M, t = ?, ,  E, , a, log A = logk + , , 2.303RT, , , 22500, , , log A = log (1.8 × 10−5 ) + ,  2.303 × 1.987 × 313 , , log A = log (1.8 ) −5 + (15.7089 ), log A = 10.9642, , Therefore,, , A = antilog (10.9642 ), ,  [A ] −[A ]  80M , t=  0,  =  1Mmin-1  = 80min, k, , , 28.Solution:, , 27 Solution:, , A = 9.208 × 1010 collisions s −1, For a first order reaction, , Here, we are given that, , k=, , 2.303, [A ], log 0, t, [A ], , k=, , V∞, 2.303, log, t, V∞ - Vt, , Ea = 22.5 kcal mol-1 = 22500 cal mol-1, , T = 40°C = 40 + 273 = 313 K, k = 1.8 × 10-5 sec-1, Substituting, equation, , the, , values, , in, , the, , In the present case, V∞ = 58.3 ml., , The value of k at different time can be, calculated as follows:, , t (min), , Vt, , V∞ - Vt, , V°, 2.303, log, t, V° - Vt, , 6, , 19.3, , 58.3-19.3 = 39.0, , k=, , 2.303,  58.3 , log , = 0.0670 min-1, , , , 6, 39, , 12, , 32.6, , 58.3-32.6 = 25.7, , k=, , 2.303,  58.3 , log , = 0.0683 min-1,  25.7 , 12, , 18, , 41.3, , 58.3-41.3 = 17.0, , k=, , 2.303,  58.3 , log , = 0.0685 min-1,  17 , 18, , 24, , 46.5, , 58.3-46.5 = 11.8, , k=, , 2.303,  58.3 , log , = 0.0666 min-1,  11.8 , 24, , 254, , Answers.indd 254, , 2/19/2020 4:34:59 PM

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www.tntextbooks.in, , Since the value of k comes out to be nearly constant, the given reaction is of the first, order. The mean value of k = 0.0676 min-1, 29.Solution:, k=, , 2.303, [A ], log 0, t, [A ], , V ,  2.303 , k= , log  0 , ,  t ,  Vt , In the present case, Vo = 46.1 ml., , The value of k at each instant can be calculated as follows:, , t (min), , Vt, ,  V0 ,  2.303 , k= , log,  V ,  t , t, , 10, , 29.8, , k=, , 2.303,  46.1, log , = 0.0436 min-1, , , , 10, 29.8, , 20, , 19.3, , k=, , 2.303,  46.1, log , = 0.0435 min-1,  19.3 , 20, , Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first, order., 30. Solution:, i) For the first order reaction k =, , 2.303, [A ], log 0, t, [A ], , Assume, [A0] = 100 %, t = 50 minutes, Therefore, [A] = 100 – 40 = 60, k = (2.303 / 50) log (100 / 60), k = 0.010216 min-1, Hence the value of the rate constant is 0.010216 min-1, ii) t = ?, when the reaction is 80% completed,, [A] = 100 – 80 = 20 %, From above, k = 0.010216 min-1, t = (2.303 / 0.010216) log (100 / 20), t = 157.58 min, The time at which the reaction will be 80% complete is 157.58 min., 255, , Answers.indd 255, , 2/19/2020 4:35:02 PM

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www.tntextbooks.in, , +2, , PRACTICALS, , I-ORGANIC QUALITATIVE ANALYSIS, S.no, , Experiment, , Observation, , Inference, , Preliminary tests, 1, , Odour:, , (i) Fish odour, , (i) May be an amine, , Note the Odour of the organic (ii) Bitter almond odour, compound., (iii) Phenolic odour, (iv) Pleasant, odour, 2, , Test with litmus paper:, , 3, , Action, with, bicarbonate:, , (iii) May be phenol, , fruity (iv) May be an ester, , (i) Blue litmus turns red (i) M,  ay be a carboxylic acid, or phenol, Touch the Moist litmus paper, with an organic compound., (ii) , Red litmus turns (ii) May be an amine, blue, (iii) , Absence of carboxylic, (iii) No colour change is, acid , phenol and amine, noted,  risk effervescence, sodium (i) B, , Take 2 ml of saturated sodium, bi carbonate solution in a test, tube. Add 2 or 3 drops (or a, pinch of solid) of an organic, compound to it., 4, , (ii) May be benzaldehyde, , (ii) N,  o brisk, effervescence, , (i) P,  resence of a carboxylic, acid., (ii) A,  bsence of a carboxylic, acid., , Action with Borsche’s reagent: yellow or orange or red Presence of an aldehyde or, ketone, Take a small amount of an precipitate, organic compound in a test, tube. Add 3 ml of Borsche’s, reagent, 1 ml of Conc HCl to it,, then warm the mixture gently, and cool it., , 256, , organic analysis.indd 256, , 2/19/2020 4:36:46 PM

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www.tntextbooks.in, , 5, , Charring test:, , Charring takes place with Presence of carbohydrate, Take a small amount of an smell of burnt sugar, organic compound in a dry test, tube. Add 2 ml of conc H2SO4, to it, and heat the mixture., Tests for Aliphatic or Aromatic nature:, , 6, , Ignition test:, , (i) B,  urn with sooty, flame, Take small amount of the, organic compound in a Nickel, spatula and burn it in Bunsen (ii) Burns with non, flame., sooty flame, , (i) P,  resence of an aromatic, compound, (ii) P,  resence of an aliphatic, compound, , Tests for an unsaturation:, 7, , 8, , Test with bromine water:, , (i) O,  range - yellow, colour of, Take small amount of the, bromine water is, organic compound in a test, decolourised, tube add 2 ml of distilled water, to dissolve it. To this solution (ii) N,  o Decolourisation, add few drops of bromine water, takes place, and shake it well., (iii) D,  ecolourisation, with formation of, white precipitate., , (i) S ubstance is, unsaturated., , Test with KMnO4 solution:, , (i) S ubstance is, unsaturated., , (i) P,  ink colour of, KmnO4 solution is, Take small amount of the, decolourised, organic compound in a test, tube add 2 ml of distilled water (ii) N,  o Decolourisation, to dissolve it. To this solution, takes place, add few drops of very dilute, alkaline KmnO4 solution and, shake it well., , (ii) Substance is saturated., (iii) P,  resence of an aromatic, amine or phenol., , (ii) Substance is saturated., , TEST FOR SELECTED ORGANIC FUNCTIONAL GROUPS, Test For Phenol, 9, , Neutral FeCl3 test:, , (i) V,  iolet colouration is (i) Presence of phenol., seen, Take 1 ml of neutral ferric, chloride solution is taken in a (ii) ,  resence of α–naphthol, violet, –, blue (ii) P, dry clean test tube. Add 2 or, colouration is seen, (iii) P,  resence of β– naphthol, 3 drops (or a pinch of solid), oforganic compound to it. If no (iii) green colouration is, seen, colouration occurs add 3 or 4, drops of alcohol., 257, , organic analysis.indd 257, , 2/19/2020 4:36:47 PM

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www.tntextbooks.in, , TEST FOR CARBOXYLIC ACIDS, 10, , Esterification reaction:, , A Pleasant fruity odour is Presence of carboxylic group., Take 1 ml (or a pinch of solid) noted., of an organic compoundin, a clean test tube. Add 1 ml of, ethyl alcohol and 4 to 5 drops, of conc. sulphuric acid to it., Heat the reaction mixture, strongly for about 5 minutes., Then pour the mixture into a, beaker containing dil. Sodium, carbonate solution and note the, smell., Test for aldehydes., , 11, , Tollen’s reagent test:, , 12, , Fehling’s test:, , Shining silver mirror is Presence of an aldehyde, Take 2 ml of Tollen’s reagent in formed., a clean dry test tube. Add 3-4, drops of an organic compound, (or 0.2 g of solid) to it, and, warm the mixture on a water, bath for about 5 minutes., Red precipitate is formed. Presence of an aldehyde, , Take 1 ml each of Fehling’s, solution A and B are taken in, a test tube. Add 4-5 drops of, an organic compound (or 0.2g, of solid) to it, and warm the, mixture on a water bath for, about 5 minutes., Test for ketones, 13, , Legal’s test:, , Red colouration., , Presence of a ketone., , A small amount of the, substance is taken in a test tube., 1 ml sodium nitro prusside, solution is added. Then sodium, hydroxide solution is added, dropwise., Test for an amine., , 258, , organic analysis.indd 258, , 2/19/2020 4:36:47 PM

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www.tntextbooks.in, , 14, , Dye test:, , Scarlet red, Take A small amount of an obtained., organic substance in a clean, test tube, add 2 ml of HCl to, dissolve it. Add few crystals of, NaNO2, and cool the mixture, in ice bath. Then add 2 ml of ice, cold solution of β-naphtholin, NaOH., , dye, , is Presence of an, primary amine, , aromatic, , Test for diamide, 15, , Biuret test:, , Violet colour is appeared. presence of a diamide, , Take A small amount of an, organic compound in a test, tube. Heat strongly and then, allow to cool. Dissolve the, residue with 2 ml of water., To this solution Add 1 ml of, dilute copper sulphate solution, and few drops of 10% NaOH, solution drop by drop., Test for carbohydrates, 16, , Molisch’s test:, , 17, , Osazone test:, , Violet or purple ring is Presence of carbohydrate, Take A small amount of an formed at the junction of, organic compound in a test the two liquids., tube. It is dissolved in 2 ml of, water. Add 3-4 drops of alpha, naphthol to it.Then add conc, H2SO4 through the sides of test, tube carefully., Yellow, crystals, Take A small amount of an obtained, organic compound in a test, tube. Add 1 ml of phenyl, hydrazine solution and heat the, mixture for about 5 minutes on, a boiling water bath., , are Presence of carbohydrate, , Report:, The given organic compound contains /is, (i), , Aromatic / aliphatic, , (ii), , Saturated / unsaturated, , (iii), , __________ functional group, 259, , organic analysis.indd 259, , 2/19/2020 4:36:47 PM

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www.tntextbooks.in, , 5.Charring test:, When carbohydrates are treated with concentrated sulphuric acid, dehydration of, carbohydrates results in charring., H SO, C X ( H2O ) y , → x C + yH2O, ∆, 2, , 4, , 6. Ignition test, Aromatic compounds burn with a strong sooty yellow flame because of the high, carbon–hydrogen ratio. Aliphatic compounds burn with non-sooty flame., 7.Test with bromine water:, In this test, the orange-red colour of bromine solution disappears when it is added, to an unsaturated organic compound., , C, , Br, , +, , C, , unsaturated, compound, , Br, , Orange-yellow, , Br, , Br, , C, , C, , Colour less, , 8. Test with KMnO4 (Baeyer’s Test ), In this test, pink colour of KMnO4 disappears, when alkaline KMnO4 is added to, an unsaturated hydrocarbon. The disappearance of pink colour may take place with or, without the formation of brown precipitate of MnO2., 2KMnO4 + H2O, , C, , C, , 2KOH + 2MnO2 + 3(O), , + H2O + (O), , unsaturated, compound, , OH, , OH, , C, , C, , Colour less, , 9. Neutral FeCl3 test:, Phenol reacts with ferric ions to form violet coloured complex., Aqueous solution Naphthols do not give any characteristic colour with neutral, ferric chloride. But alcoholic solution of α and β naphtholsgiveblue-violet and green, colouration respectively due to the formation of binaphthols., , 261, , organic analysis.indd 261, , 2/19/2020 4:36:48 PM

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www.tntextbooks.in, , 10. Esterification test:, Alcohols react with carboxylic acids to form fruity smelling compounds called, esters. This esterification is catalysed by an acid such as concentrated sulphuric acid., [Fe(OC6H5)3]3+ + 3HCl, , FeCl3 +O 3C6H5OH, Ferric, R, C, chloride, , Phenol, OH, , +, , HO, , Carboxylic acid, , violet complex, H2SO4, R', , ∆, , R, , O, C, , O, , R', , +, , H 2O, , Ester, (pleasent, OH fruity odour), , OH, Alcohol, , 2FeCl3 +, , 11. Tollen’s reagent test:, , OH, , + 2FeCl2 + 2HCl, , Aldehydes react, with Tollen’s reagent to form elemental silver, accumulated onto the, β-naphthol, inner surface of the test tube. Thus silver mirror is produced on the inner walls of the, Green complex, test tube., , R-CHO  2  Ag  NH 3 2  OH ,  2Ag   R  COONH 4 + H 2 O + 3NH 3, Aldehyde, Metallic silver, Tollen'sreagent, , (Silver mirror), , Tollen’s reagent preparation:, Tollen’s reagent is ammoniacal silver nitrate. It is prepared as follows. About 1 g, of silver nitrate crystals are dissolved in distilled water in a clean dry test tube. To this, aqueous solution of silver nitrate, add 2 ml of dilute NaOH solution to it. A brown, precipitate of silver oxide is formed. This precipitate is dissolved by adding dilute, ammonia solution drop wise., 12. Fehling’s Test, Fehling’s solution A is an aqueous solution of copper sulphate., Fehling’s solution B is a clear solution of sodium potassium tartrate (Rochelle salt), and strong alkali (NaOH)., The Fehling’s solution is obtained by mixing equal volumes of both Fehling’s solution, A and Fehling’s solution B that has a deep blue colour. In Fehling’s solution, copper (II), ions form a complex with tartrate ions in alkali. Aldehydes reduces the Cu(II) ions in the, Fehling’s solution to red precipitate of cuprous oxide(copper (I) oxide)., RCHO + 2Cu 2  5OH  ,  Cu 2 O   RCOO   3H 2 O, Aldehyde, , , , Fehling's solution, , (Cuprous oxide), (Red colour), , Note: Benzaldehyde may not give this test as the reaction is very slow., , 262, , organic analysis.indd 262, , 2/19/2020 4:36:49 PM

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www.tntextbooks.in, , 16. Molisch’s test:, Disaccharides, and polysaccharidesare hydrolysed to Monosaccharides by strong, mineral acids. Pentoses are then dehydrated to furfural, while hexoses are dehydrated to, 5-hydroxymethylfurfural. These aldehydes formed will condense with two molecules of, α-Naphthol to form a purple-coloured product, as shown below., H OH, H, , O, , H3O+, , HO, HO, , H, , H, , O, , -3H2O, , 5-(hydroxymethyl)furfural, , OH, , H, , O, , HO, , OH, , OH, HO, , O, , H3O+, , +2, , O, , HO, , [O], , HO, , -H2O, , 5-(hydroxymethyl), furfural, , O, , OH, O, , H3O+, , O, , α-naphthol, OH, , OH, , A purple dye, , 17.Osazone test:, Phenyl hydrazine in acetic acid, when boiled with reducing sugars forms Osazone., The first two carbon atoms are involved in this reaction. The sugars that differ in their, configuration on these carbon atoms give the same type of Osazone. Thus glucose,, fructose and mannose give the same needle type yellow crystals., H, , O, , H, , N, , NH, , C, , C, H, , C, , OH, , HO, , C, , H, , H, , C, , OH, , H, , C, , OH, , H2N, , +, , HN, , phenyl hydrazine, , C, , N, , HO, , C, , H, , H, , C, , OH, , H, , C, , OH, , NH, , CH2OH, , CH2OH, , Glucosazone, (Yellow crystals), , Glucose, , 264, , organic analysis.indd 264, , 2/19/2020 4:36:51 PM

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www.tntextbooks.in, , Titration –II, (Link KMnO 4 )Vs (Unknown oxalic acid), s.no, , Volume of, Unknown oxalic, acid (ml), , 1, , 20, , 2, , 20, , 3, , 20, , Burette readings, , Concordant value, , Initial, , Final, , (Volume of KMnO 4 ), , (ml), , (ml), , (ml), , Calculation :, Volume of Unknown oxalic acid solution, , V1, , =, , 20 ml, , Normality of Unknown oxalic acid solution, , N1, , =, , ?N, , Volume of KMnO4 (link) solution, , V2, , =, , ml, , Normality KMnO4 (link) solution, , N2 =, , N, , According to normality equation:, V1 x N1 = V2 x N 2, , N1 =, , V2 x N 2, V1, , Y, Normality of Unknown oxalic acid solution N1 = _______________, N, Weight calculation:, The amount of oxalic acid dissolved in 1 lit, =(Normality) x (equivalent weight), of the solution, The amount of oxalic acid dissolved in 500 Y × 63 × 500, ml of the solution =, 1000, =, , x 63 x 500, 1000, , = g, Report :, , The amount of oxalic acid dissolved in 500 ml of given the solution = g, , 273, , organic analysis.indd 273, , 2/19/2020 4:36:54 PM

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www.tntextbooks.in, , 4. Estimation of sodium hydroxide, Aim :, To estimate the amount of sodium hydroxide dissolved in 250 ml of the given unknown, solution volumetrically. For this you are given with a standard solution of sodium carbonate, solution of normality 0.0948 N and hydrochloric acid solution as link solution., Principle:, Neutralization of Sodium carbonate by HCl is given below. To indicate the end point, methyl, orange is used as an indicator., Na 2CO3 + 2HCl , → 2NaCl + CO2 + H2O, Neutralization of Sodium hydroxide by HCl is given below. To indicate the end point,, phenolphthalein is used as an indicator., NaOH + HCl , → NaCl + H 2 O, , Short procedure:, s.no, , Content, , Titration-I, , Titration-II, , 1, , Burette solution, , HCl ( link solution), , HCl ( link solution), , 2, , Pipette solution, , 20 ml of standard Na2CO3 20 ml of unknown NaOH, solution, solution, , 4, , Temperature, , Lab temperature, , Lab temperature, , 5, , Indicator, , Methyl orange, , Phenolphthalein, , 6, , End point, , Colour change from straw, Disappearance of pink colour, yellow to pale pink, , 7, , Equivalent weight of NaOH = 40, , Procedure :, Titration–I, (Link HCl )Vs (standard Na2CO3), Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up to, the zero mark. Exactly 20 ml of standard Na2CO3solution is pipetted out into the clean, washed, conical flask. To This solution 2 to 3 drops of methyl orange indicator is added and titrated, against HCl link solution from the burette. HCl is added drop wise till the colour change from, straw yellow to pale pink. Burette reading is noted and the same procedure is repeated to get, concordant values., , 274, , organic analysis.indd 274, , 2/19/2020 4:36:54 PM

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www.tntextbooks.in, , Titration –II, (Link HCl )Vs (Unknown NaOH solution), Volume of, Unknown NaOH, (ml), , s.n, 1, , 20, , 2, , 20, , 3, , 20, , Burette readings, , Concordant value, , Initial, , Final, , (Volume of HCl), , (ml), , (ml), , (ml), , Calculation :, Volume of Unknown NaOH solution, , V1, , = 20 ml, , Normality of Unknown NaOH solution, , N1, , = ?N, , Volume of HCl (link) solution, , V2, , = ml, , Normality HCl (link) solution, , N2, , = N, , According to normality equation:, V1 x N1 = V2 x N 2, , N1 =, , V2 x N 2, V1, , Y, Normality of Unknown HCl solution N1 = _______________, N, Weight calculation:, The amount of NaOH dissolved in 1 lit of the, = (Normality) x (equivalent weight), solution, The amount of NaOH dissolved in 250 ml of the, Normality x equivalentweight x 250, =, solution, 1000, = Y × 40 × 250, 1000, x 40 x 250,   g, 1000, , =   , Report :, The amount of NaOH dissolved in 750 ml of the solution, , =, , g, , 276, , organic analysis.indd 276, , 2/19/2020 4:36:54 PM

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www.tntextbooks.in, , 5. Estimation of oxalic acid, Aim :, To estimate the amount of oxalic acid dissolved in 1250 ml of the given unknown solution, volumetrically. For this you are given with a standard solution of HCl solution of normality, 0.1010 N and sodium hydroxide solution as link solution., Principle:, Neutralization of Sodium hydroxide by HCl is given below. To indicate the end point,, phenolphthalein is used as an indicator., NaOH + HCl , → NaCl + H 2 O, , Neutralization of Sodium hydroxide by oxalic acid is given below. To indicate the end point,, phenolphthalein is used as an indicator., 2NaOH +, , → ( COONa )2, ( COOH )2 , Oxalic acid, , + 2H 2 O, , Sodium oxalate, , Short procedure:, s.no, , Content, , Titration-I, , Titration-II, , 1, , Burette solution, , HCl (standard solution), , Oxalic acid ( unknown, solution), , 2, , Pipette solution, , 20 ml of NaOH link, solution, , 20 ml of NaOH link solution, , 4, , Temperature, , Lab temperature, , Lab temperature, , 5, , Indicator, , Phenolphthalein, , Phenolphthalein, , 6, , End point, , Disappearance of pink, colour, , Disappearance of pink colour, , 7, , Equivalent weight of oxalic acid = 63, , Procedure :, Titration–I, (standard HCl )Vs (link NaOH), Burette is washed with water, rinsed with HCl solution and filled with same HCl solution up, to the zero mark. Exactly 20 ml of NaOH is pipetted out into the clean, washed conical flask., To This solution 2 to 3 drops of phenolphthalein indicator is added and titrated against HCl, solution from the burette. HCl is added drop wise till the pink colour disappears completely., Burette reading is noted and the same procedure is repeated to get concordant values., , 277, , organic analysis.indd 277, , 2/19/2020 4:36:55 PM

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www.tntextbooks.in, , Titration –I, (standard HCl )Vs (link NaOH), Burette readings, , Volume of, NaOH(ml), , s.no, 1, , 20, , 2, , 20, , 3, , 20, , Concordant value, , Initial, , Final, , (Volume of std HCl), , (ml), , (ml), , (ml), , Calculation :, Volume of NaOH(link) solution, , V1 = 20 ml, , Normality NaOH(link) solution, , N1 = ? N, , Volume of standard HCl solution, , V2 = ml, , Normality of standard HCl solution, , N 2 = 0.1010 N, , According to normality equation:, V1 x N1 = V2 x N 2, , N1 =, , × 0.1010, =, 20, , X, Normality NaOH (link) solution N1 =_______________, N, Titration–II, (Unknown oxalic acid ) Vs (Link NaOH), Burette is washed with water, rinsed with oxalic acid solution and filled with same, oxalic acid solution up to the zero mark. Exactly 20 ml of NaOH solution is pipetted out into, the clean, washed conical flask. To This solution 2 to 3 drops of phenolphthalein indicator is, added and titrated against oxalic acid solution from the burette. oxalic acid is added drop wise, till the pink colour disappears completely. Burette reading is noted and the same procedure is, repeated to get concordant values., , 278, , organic analysis.indd 278, , 2/19/2020 4:36:55 PM

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www.tntextbooks.in, , Titration –II, (Link NaOH )Vs (Unknown oxalic acid solution), Volume of NaOH, link (ml), , s.no, 1, , 20, , 2, , 20, , 3, , 20, , Burette readings, , Concordant value, , Initial, , Final, , (Volume of oxalic acid), , (ml), , (ml), , (ml), , Calculation :, Volume of Unknown oxalic acid solution , , V1, , =, , ml, , Normality of Unknown oxalic acid solution, , N1, , =, , ?N, , Volume of NaOH solution , , V2, , =, , 20 ml, , Normality NaOH solution , , N2, , =, , N, , N1, , Y, = _______________N, , According to normality equation:, V1 x N1 = V2 x N 2, , N1 =, , V2 x N 2, V1, , Normality of Unknown oxalic acid solution, Weight calculation:, , The amount of oxalic acid dissolved in 1 lit of, = (Normality) x (equivalent weight), the solution, The amount of oxalic acid dissolved in 1250, = Normality x equivalentweight x 1250, ml of the solution, 1000, =, , Y × 63 × 1250, 1000, , =, , x 63 x 1250, 1000, , =      g, Report :, The amount of oxalic acid dissolved in 1250 ml of the solution, , = g, , 279, , organic analysis.indd 279, , 2/19/2020 4:36:55 PM