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1, , STANDAR XII, , CHEMISTRYQUESTION BANK, 2019-20, (BASED ON NEW SYLLABUS), UNIT NO, UNIT 1:, , TOPICS, METALLURGY, , UNIT 2:, , p-BLOCK ELEMENTS - I, , PAGE NO, 3, , 14, UNIT 3:, , p-BLOCK ELEMENTS-II, 22, , UNIT 4:, , TRANSITION & INNER TRANSITION ELEMENTS, 40, , UNIT 5:, , COORDINATION CHEMISTRY, , UNIT 6:, , SOLID STATE, , 55, 67, , UNIT 7:, , CHEMICAL KINETICS, , 81

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3, , UNIT – 1 METALLURGY, , I. Text Book Questions and Answers, 1., , What is the difference between minerals and ores?, , SNo., , Minerals, , 1, , Naturally occurring substances obtained, by mining which contain the metals in, free state or in the form of compounds, like oxides, sulphides, etc. are called, minerals., All the minerals are not ores, , 2, 3, 2., , 3., , 4., , 5., , Ores, Minerals that contain high, percentage of metal from which it, can be extracted conveniently and, economically are called ores., All the ores are minerals, , Mineral of Al is Bauxite (Al2O3 nH2O ) Ore of Al is Bauxite (Al2O3 nH2O ), and China clay (Al2O3 SiO2 .2H2O ), What are the various steps involved in extraction of pure metals from their ores?, Steps involved in extraction of pure metals from their ores are, i) Concentration of the ore, ii) Extraction of the crude metal., iii) Refining of the crude metal., What is the role of Limestone in the extraction of Iron from its oxide Fe2O3?, Lime stone (CaO3) is used as a basic flux in the extraction of iron from its oxide Fe2O3., Limestone decomposes to form CaO, , CaCO3 ⎯⎯, → CaO + CO2, Impurity silica (SiO2)react with CaOform fusible slagcalcium silicate., CaO(s) + SiO2(s)→CaSiO3(s), Flux, Gaugue Slag, Which type of ores can be concentrated by froth flotation method? Give two, examples for such ores., Sulphide ores can be concentrated by froth flotation method., (eg) Galena (PbS), Zinc blende (ZnS), Out of coke and CO, which is better reducing agent for the reduction of ZnO? Why?, , Out of coke and CO, coke is better reducing agent than CO for the reduction of ZnO., ZnO(s)+C →Zn(s)+ CO(g), •, In Ellingham diagram formation ZnO line lies above the formation C→CO at low, temperature (T1), ZnO line also lies above the CO → CO2 but at high temperature.

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4, , •, , Hence carbon can be used as a better reducing agent than CO for the reduction of, ZnO. Below the temperature T1 both Coke and COcannot reduce ZnO, 6. Describe a method for refining nickel., Impure nickel is heated in a stream of carbon monoxide at around 350K. Nickel reacts, with CO to form a highly volatile nickel tetracarbonyl. The solid impurities are left, behind., 350 K, → Ni[CO]4(g), Ni(s) +4CO(g) ⎯⎯⎯, On heating nickel tetra carbonyl around 460K, decomposes to give pure nickel., 460 K, → Ni(s)+ 4CO(g), Ni[CO]4(g) ⎯⎯⎯, 7. Explain zone refining process with an example, The principle is fractional crystallisation., • When an impure metal is melted and allowed to solidify, the impurities will prefer to, remain in the molten region. 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, melting the metal on that portion of the rod., • When the heater is slowly moved to the other end pure metal crystallises while, 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., • This 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., • Germanium, Silicon and Gallium which are used as semiconductor are refined by this, process., 8. Using the Ellingham diagram given below., (A) Predict the conditions under which, i) Aluminium might be expected to reduce magnesia., ii) Magnesium could alumina., B) Carbon monoxide is more effective reducing agent than carbon below, 983K but, above this temperature, the reverse is true - Explain., C) It is possible to reduce Fe2O3 by coke at a temperature around 1200K, , A) i) Ellingham diagram for the formation of Al2O3 and MgO intersects around 1600K., Above this temperature aluminium line lies below the magnesium line. Hence we can use, aluminium to reduce magnesia above 1600K.

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5, , ii) In Ellingham diagram below 1600K magnesium line lies below aluminium line. Hence, below 1600K magnesium can reduce alumina., , B)The two lines for CO→CO2 and C →CO cross at about 983K. Below this temperature, the reaction to form CO2 is energetically more favourable hence CO is more effective, reducing agent than carbon. But above 983K the formation of CO is preferred, hence, carbon is more effective reducing agent than CO above this temperature., C)In Ellingham diagram above 1000K carbon line lies below the iron line. Hence it is, possible to reduce Fe2O3 by coke at a temperature around 1200K., 9. Give uses of zinc., 1. Metallic zinc is used in galvanisation to protect iron and steel structures from rusting, and corrosion., 2. Zinc is used to produce die - castings in the automobile, electrical and hardware industries., 3. Zinc oxide is used in the manufacture of paints, rubber, cosmetics, pharmaceuticals,, plastics, inks, batteries, textiles and electrical equipment., 4. Zinc sulphide is used in making luminous paints, fluorescent lights and x - ray screens., 5. Brass an alloy of zinc is used in water valves and communication equipment as it is, highly resistant to corrosion, 10. Explain the electrometallurgy of aluminium., Hall – Herold Process, Cathode: Iron tanked lined with carbon, Anode: Carbon blocks, Electrolyte: 20% solution of alumina obtained from bauxite + Molten Cryolite +, 10% calcium chloride (lowers the melting point of the mixture), Temperature: Above 1270K, Ionisation of Alumina Al2O3→ 2Al3+ + 3O2Reaction at cathode: 2Al3+(melt) + 6e−→2Al(l), Reaction at anode:, 6O2- (melt) →3O2 + 12e−, Since carbon acts as anode the following reaction also takes place on it., C(s) + O2− (melt) → CO + 2e−, C(s) + 2O2− (melt) → CO2 + 4e−, During electrolysis anodes are slowly consumed due to the above two reactions. Pure, aluminium is formed at the cathode and settles at the bottom., Net electrolysis reaction is, 4Al3+ (melt) + 6O2- (melt) + 3C(s)→4Al(l) + 3CO2(g), 11. Explain the following terms with suitable examples. i) Gangue ii) Slag, i) Gangue:, The non metallic impurities, rocky materials and siliceous matter present in the ores are, called gangue. (eg): SiO2 is the gangue present in the iron ore Fe2O3 ., ii) Slag: Slag is a fusible chemical substance formed by the reaction of gangue with a flux., CaO(s)+ SiO2(s)→CaSiO3(s), Flux, gangue slag, 12. Give the basic requirement for vapour phase refining., The metal is treated with a suitable reagent to form a volatile compound.

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6, , Then the volatile compound is decomposed to give the pure metal at high temperature., 13. 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., i) In the extraction of copper, silica acts as an acidic flux to remove FeO as slag FeSiO3., FeO(s) + SiO2(s)→ FeSiO3(s), Flux, Slag, ii) As Al2O3 is a poor conductor, cryolite improves the electrical conductivity., In addition, cryolite serves as an added impurity and lowers the melting point of the, electrolyte., iii) First Iodine forms a Volatile tetraiodide with impure metal, which decomposes to give, pure metal. Impure zirconium metal is heated in an evacuated vessel with iodine to, form the volatile zirconium tetraiodide (ZrI4). The impurities are left behind, as they, do not react with iodine., Zr(s) + 2I2(s)→ZrI4(Vapour), On passing volatile zirconium tetraiodide vapour over a tungsten filament, it is, decomposed to give pure zirconium., ZrI4(Vapour)→ Zr(s) + 2I2(s), iv) Sodium cyanide acts as a depressing agent in froth floatation process. When a, sulphide ore of a metal contains other metal sulphides, the depressing agent sodium, cyanide selectively prevent other metal sulphides coming to the froth. eg: NaCN, depresses the floatation property ZnS present in Galena (PbS) by forming a layer of, Zinc complex Na2 [Zn(CN)4]on the surface of Zinc sulphide., 14. Explain the principle of electrolytic refining with an example., Crude metal is refined by electrolysis carried out in an electrolytic cell., Cathode: Thin strips of pure metal., Anode: Impure metal to be refined., Electrolyte: Aqueous solution of the salt of the metal with dilute acid., As current is passed, the metal of interest dissolves from the anode and pass into the, electrolytic solution., At the same time same amount of metal ions from the electrolytic solution will be, deposited at the cathode., Less electro positive impurities in the anode settle down as anode mud., eg : Electro refining of silver:, Cathode: Pure silver, Anode: Impure silver rods., Electrolyte: Acidified aqueous solution of silver nitrate., When current passed, the following reactions will take place., Reaction at anode: Ag(s)→ Ag+(aq) + e−, Reaction at cathode: Ag+(aq) + e−→ Ag(s), At anode silver atoms lose electrons and enter the solution. From the solution, silver ions, (Ag+)migrate towards the cathode. At cathode silver ions get discharged by gaining, electrons and deposited on the cathode., 15. The selection of reducing agent depends on the thermodynamic factor: Explain with, an example., • A suitable reducing agent is selected based on the thermodynamic considerations., • For a spontaneous reaction G should be negative.

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7, , •, , Thermodynamically, the reduction of metal oxide with a given reducing agent can, occur if G for the coupled reaction is negative., • Hence the reducing agent is selected in such a way that it provides a large negative G, value for the coupled reaction., • Ellingham diagram is used to predict thremodynamic 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., • Ellingham diagram for the formation of FeO and CO intersects around 1000K. Below, this temperature the carbon line lies above the iron line., • Hence FeO is more stable than CO and the reduction is not thermodynamically, feasible., • However above 1000K carbon line lies below the iron line. Hence at this condition FeO, is less stable than CO and the reduction is thermodynamically feasible. So coke can be, used as a reducing agent above this temperature., • Following free energy calculation also confirm that the reduction is thermodynamically, favoured., From the Ellingham diagram at 1500K, 2Fe(s)+O2(g)→ 2FeO(g) G1 = − 350 kJmol-1 ............1, 2C(s) + O2(g)→ 2CO(g) G2 = − 480 kJmol-1 ............2, Reverse the reaction 1, 2FeO(s)→ 2Fe(s) + O2(g)G1 = 350 kJmol-1 ............3, Couple the reactions 2 and 3, 2FeO(s)+2C(s)→2Fe(s) + 2CO(g)G3 = − 130 kJmol-1 ............4, • The standard free energy change for the reduction of one mole of FeO is, G3, = - 65 Jmol-1, 2, 16. Give the limitations of Ellingham diagram., Ellingham diagram is constructed based only on thermodynamic considerations., 1. It gives information about the thermodynamic feasibility of a reaction., 2. It does not tell anything about the rate of the reaction., 3. More over it does not give any idea about the possibility of other reactions that might be, taking place., 4. The interpretation of G is based on the assumption that the reactants are in, equilibrium with the product which is not always true., 17. Write a short note on electrochemical principles of metallurgy., • 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 method the metal salts are taken in fused form or in solution form., • The metal ion present can be reduced by treating the solution with suitable reducing, agent or by electrolysis., • Gibbs free energy change for the electrolysis is, Go = - nFEo, n = number of electrons involved in the reduction, F = Faraday = 96500 coulombs, Eo = electrode potential of the redox couple., • If Eo is positive, Go is negative and the reduction is spontaneous., • Hence a redox reaction is planned in such a way that the e.m.f of the net redox reaction, is positive.

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8, , •, , II., 1., , 2., , 3, , A more reactive metal displaces a less reactive metal from its salt solution., eg: Cu2+(aq) + Zn(s)→ Cu(s) + Zn2+(aq), • Zinc is more reactive than copper and displaces copper from its salt solution., Evaluate yourself, Write the equation for the extraction of silver by leaching with sodium cyanide and, show that the leaching process is a redox reaction., In the metallurgy of silver metal is leached with a dilute solution of NaCN in the presence, of air (O2), 4Ag + 8CN- +2H2O +O2→ 4[Ag(CN)2] + 4OH In this reaction, Ag →Ag+ oxidation number of Ag increases from 0 to +1, hence oxidation, O2→OH- (oxidation number of oxygen decreases from 0 to -2, hence reduction), Hence Leaching of silver is a redox reaction., Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to, make refractory bricks. Write the decomposition reaction, Magnesite (Magnesium carbonate) is heated in the absence of oxygen decomposes to form, Magnesium oxide ( Magnesia), MgCO3→MgO+ CO2, Using Ellingham diagram indicate the lowest temperature at which ZnO can, be reduced to Zinc metal by carbon. Write the overall reduction reaction at, this temperature, , Ellingham diagram for the formation of ZnO and CO intersects around 1233K Below this, temperature, Carbon line lies above Zinc line. Hence ZnO is more stable than CO so the, reduction is thermodynamically not feasible at this temperature range., However above 1233K carbon line lies below the zinc line, hence carbon can be used as a, reducing agent above 1233K., 2Zn +O2→2ZnO ...........1, 2C + O2→2CO ..........2, Reversing 1 and adding with equation 2, 2ZnO →2Zn+O2, 2C + O2→2CO, 2ZnO +2C →2Zn + 2CO, 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., Sodium metal is prepared by Down’s process. This involves the electrolysis of fused, NaCland CaCl2 at 873K During electrolysis sodium is discharged at the cathode and Cl2 is, obtained at the anode., NaCl(l)→Na+(melt ) + Cl−(melt), Cathode: Na+(melt + e−→ Na (s), Anode: 2Cl−(aq)→ Cl2(g) + 2e−, If an aqueous solution of NaCl is electrolysed, H2 is evolved at cathode and Cl2 is evolved, at anode. NaOH is obtained in the solution.., Electrolysis, NaCl(aq) ⎯⎯⎯⎯→ Na+(aq) + Cl−(aq), Cathode: 2H2O(l) + 2e−→ H2(g) + 2OH−(aq)

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9, , 1, Anode: Cl−(aq)→ Cl2(g) + 2e−, 2, +, −, Na and OH ions to form NaOH ., Hence solution is basic in nature., , IV. ADDITIONAL QUESTIONS AND ANSWERS, 1. What is concentration of ores?, The removal of non-metallic impurities, rocky materials and siliceous matter (called as, gangue) from the ores is known as concentration of ores., 2. What is leaching?, In this method crushed ore is allowed to dissolve in a suitable solvent to form a soluble, metal salt or complex leaving the gangue undissolved is called leaching., 3. What is ammonia leaching?, 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 namely, [Ni(NH3)6]2+, [Cu(NH3)4]2+ and [Co(NH3)5H2O]3+ from the ore., The gangue left behind are iron (III) oxides / hydroxides and alumino silicate., 4. What is acid leaching?, Sulphide ores ZnS, PbS can be leached with hot aqueous sulphuric acid., In this process the insoluble sulphide is converted into soluble sulphate and elemental, sulphur., 2ZnS(s) +2H2SO4(aq) + O2(g)→2ZnSO4(aq) + 2S(s) + 2H2O, 5. What are the steps involved in the extraction of crude metal?, 1.Conversion of the ore into metal oxide either by roasting or calcination., 2.Reduction of the metal oxide into metal., 6. In the extraction of metal, ore is first converted into metal oxide before reduction into, metal. Why?, • In the concentrated ore the metal exists in positive oxidation state and hence it is to be, reduced to elemental state., • From the principles of thermodynamics, the reduction of oxide is easier compared to, the reduction of other compounds of metal., • Hence before reduction the ore is first converted into metal oxide., 7. Write about the extraction of metal by the process of reduction by hydrogen., This method can be applied to the oxides of the metals (Fe, Pb, Cu) which are less, electropositive than hydrogen., Ag2O(s) + H2(g)→ Ag(s) + H2O(l)↑, Nickel oxide is reduced to nickel by a mixture of hydrogen and carbon monoxide(water gas), 2NiO(s) + CO(g) + H2(g) → 2Ni(s) + CO2(g) + H2O(l)↑, 8. What is auto reduction of metallic ores?, Simple roasting of some of the metallic ores give the crude metal., Use of reducing agent is not necessary because of low thermal stability, (eg) Cinnabar is roasted to give mercury., HgS(s) + O2(g) → Hg(l) + SO2(g)↑, 9. Write the applications of copper., 1. Copper is the first metal used by humans and extended use of its alloy bronze resulted, in a new era, ‘Bronze age’., 2.Used for making coins and ornaments along with gold and other metals., 3.Copper and its alloys are used for making wires, water pipes and other electrical parts., 10. Write the applications of gold., 1.Gold is one of the expensive and precious metals.

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10, , 2.Used for coinage and has been used as standard for monetary systems in some countries., 3.Extensively used in jewellery in its alloy form with copper., 4. 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., 5.Gold nanoparticles are used for increasing the efficiency of solar cells., 6.Used as a catalyst., 11. Write about alumino thermite process., • In this method a metal oxide such as Cr2O3 is reduced to metal by aluminium., • Metal oxide (Cr2O3) is mixed with aluminium powder in a fire clay crucible.The, reduction process is initiated by ignition mixture of Magnesium power and barium, peroxide., BaO2+ Mg→BaO + MgO, • It is an exothermic process where heat is liberated., • Temperature = 2400oC Heat liberated = 852kJmol-1.This heat initiate the reduction of, Cr2O3 by Al., , → 2Cr + Al2O3, Cr2O3 + 2Al ⎯⎯, 12. What is refining process of a metal?, Metals extracted from its ore contains impurities such as unreacted oxide ore, other, metals, non metals etc,. Removal of such impurities from crude metal is known as refining, process of a metal., 13. Write about distillation process of refining a metal?, This method is used for low boiling volatile metals like zinc and mercury. In this method, impure metal is heated to evaporate and the vapours are condensed to get pure metal., 14. Write about liquation process of refining a metal?, • This method is used to remove the impurities with high melting points from metals, having relatively low melting points.(eg) Tin, lead, mercury, bismuth., • 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 metal flows, down and impurities are left behind .The molten metal is collected and solidified., 15. Give example for the following, 1. Frothing agent, 2. Collector 3. Depressing agent, Frothing agent, :Pine oil , eucalyptus oil, Collector, :Sodium ethyl xanthate, Depressing agent, :Sodium cyanide , sodium carbonate, 16. What is cementation ?, Gold can be recovered by reacting the deoxygenated leached solution with Zinc. In this, process Gold is reduced to its elemental state ( zero oxidation state) and the process is, called cementation, Zn (S)+ 2[ Au (CN)2] –(aq) → [Zn(CN)4] 2-(aq) + 2Au (S), 17. Why Fe, Pb , Cu are reduced by hydrogen ?, The oxides of metal Fe ,Pb , Cu having less electropositive character than hydrogen , these, metal oxide can be reduced by hydrogen., Ag2O (S) + H2 (g) → Ag (S) + H2O (l) , Fe2O3 (S) + 4H2(g) → 4Fe (S) + 4H2O (l) , 18. Write about gravity separation or hydraulic wash?, • Ore with high specific gravity is separated from gaugue with low specific gravity by, simply washing with running water., • Finely powdered ore is treated with rapidly flowing current of water., • Lighter gaugue particles are washed away by the running water.

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11, , •, , This method is used for concentrating native ore such as gold and oxide ores such as, haematite(Fe2O3), tin stone(SnO2)., 19. Write about magnetic separation., • This method is applicable to ferromagnetic ores., • It is based on the difference in the magnetic properties of the ore and the impurities., • Non-magnetic tin stone can be separated from the magnetic impurities wolframite., • Similarly magnetic ores chromite, pyrolusite can be removed from non magnetic, siliceous impurities., • The crushed ore is poured to an electromagnetic separator with a belt moving over two, rollers of which one is magnetic., • Magnetic part of the ore is attached towards the magnet and falls as a heap close to the, magnetic region., • Non- magnetic part falls away from it., , 20. Write about calcination., • Calcination is the process in which the concentrated ore is strongly heated in the, absence of air., • During this process water of crystallisation present in the hydrated oxide escapes as, moisture., • Any organic matter present also get expelled leaving the ore porous., • This method can also be carried out with a limited supply of air., • During calcination of carbonate ore is decomposed to metal oxide and carbon dioxide, is liberated., , , → PbO + CO2↑, → CaO + CO2↑, PbCO3 ⎯⎯, CaCO3 ⎯⎯, , → Al2O3(s) + 2H2O(g)↑, Al2O3.2H2O ⎯⎯, 21. Write about Van – Arkel method for refining zirconium/titanium?, • This method is based on the thermal decomposition of gaseous metal compounds to, metals.(eg) Titanium and Zirconium., • Impure titanium is heated in an evacuated vessel with iodine at 550K to form, volatiletitanium tetra iodide., • The impurities do not react with iodine., 550 K, → TiI4(vapour), Ti(s) + 2I2(s) ⎯⎯⎯, • Volatile titanium tetraiodide is passed over a tungsten filament at 1800K., • Titanium tetraiodide is decomposed to pure titanium which is deposited over the, filament.Iodine is reused., 1800 K, TiI4(vapour) ⎯⎯⎯→ Ti(s) + 2I2(s), 22. Write the applications of aluminium., 1. Used for making heat exchangers/sinks.

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12, , 2. Used for making our day to day cooking vessels., 3. Used for making aluminium foils for packing, food items., 4. Alloys of aluminium with copper, manganese, magnesium, silicon are light weight and, strong hence used in design of aeroplanes and other forms of transport., 5. Due to its high resistance to corrosion, it is used in the design of chemical reactors,, medical equipment’s, refrigeration units and gas pipelines., 5. It is a good electrical conductor and cheap, hence used in electrical over head cables, with steel core for strength., 23. Write the applications of iron., 1.Iron is one the most useful metals and its alloys are used everywhere including, bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels., 2.Cast iron is used to make pipes, valves and pump stoves etc., 3.Magnets can be made from iron , its alloys and compounds., 4.Important alloy of iron is stainless steel which is very resistant to corrosion., 5.It is used in architecture, bearings, cutlery, surgical instruments and jewellery., 6.Nickel steel is used for making cables, automobiles, and aeroplane parts., 7.Chrome steels are used for manufacturing cutting tools and crushing machines., 24. Explain froth floatation method., • This is used to concentrate sulphide ores such as galena (PbS) Zinc blende (ZnS) etc., • Metallic ore particles preferentially wetted by oil can be separated from gangue., • Crushed ore is mixed with water and a frothing agent like pine oil or eucalyptus oil., • A small amount of sodium ethyl xanthate is added as a collector., • A froth is formed by blowing air through the mixture., • The collector molecules attach to the ore particles and make them water repellent., • As a result ore particles wetted by the oil rise to the surface along with the froth., • The froth is skimmed off and dried to recover the concentrated ore., • Gangue particles preferentially wetted by water settle at the bottom., • When sulphide ore contains other metal sulphides as impurities, depressing agents, such as sodium cyanide, sodium carbonate etc. are used to selectively prevent other, 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 ZnS., , ., , UNIT-2

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13, , UNIT 2, p – Block element I, Text Book Exercise Questions and Answers:, I Answer the following question:, Write a short note on anamolous properties of the first element of p-block., (i) Small size of the first member, (ii) High ionization enthalpy and high electronegativity, (iii) Absences of d-orbital in their valence shell, , 1., , 2., , Describe briefly allotropism in p-block elements with specific reference to carbon., Some elements exist in more than one crystalline or molecular forms in the same, physical state. This phenomenon is called allotropism and the different forms of an, element are called allotropes., Eg: carbon exist as diamond, graphite, graphene, fullerenes and carbon nanotubes, , ., 3., , Boron does not react directly with hydrogen. Suggest one method to prepare, diboranefrom BF3., 2BF3 + 6NaH 450K, B2H6 + 6NaF, Boron trifluoride, Diborane, , 4., , Give the uses of borax., (i) Borax is used for the identification of coloured metal ions, (ii) It is used as a flux in metallurgy., (iii) It act as a preservative., , 5., , What is catenation? Describe briefly the catenation property of carbon., 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, 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., , 6., , Write a note on Fisher tropsch synthesis., The reaction of carbon monoxide with hydrogen at pressure of less than 50 atm using, metal catalysts at 500 – 700K yields saturated and unsaturated hydrocarbons, nCO + (2n+1)H2 → CnH(2n+2) + nH2O, nCO + 2nH2, CnH2n, + nH2O, →, Carbon monoxide forms complex compounds with transition metals. Eg : Nickel, tetracarbonyl., , 7., , Give the structure of CO and CO2., Structure of CO, •, , It has a linear structure., , Structure of CO2, •, , It has a linear structure.

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14, , •, , 8., , The C—O bond distance is, 1.128Ao., , •, , It has equal bond distance for both C—O, bonds., , Give the uses of silicones.(i) Silicones are used for low temperature lubrication., (ii) Silicones are used for making water proofing clothes., (iii)Silicones are used as insulating material in electric motor., AlCl3 behaves like a lewis acid. Substantiate this statement., AlCl3 is electron deficient. Al forms three bonds with chloride and hence outer shell has, six electron. It needs two more electrons to complete its octet. So readily accept a pair of, electrons. Hence AlCl3 is a Lewis acid and forms addition compounds with ammonia and, phosphate., , 9., , 10., , Describe the structure of diborane., , •, , In diborane twoBH2 units are linked by two bridged hydrogens. Therefore, it has, eight B-H bonds. However, diborane has only 12 valence electrons and are not, sufficient to form normal covalent bonds., • The four terminal B-H bonds (two centre – two electron bond or 2c-2e bond) are, normal covalent bonds., • The remaining four electrons have to be used for the bridged bonds. i.e. two three, centred B-H-B bonds utilize two electrons each. Hence, these bonds are three, centre –two electron bonds (3c-2e), • The bridging hydrogen atoms are in a plane. In diborane, the boron is sp3, hybridised., , 11., , Write a short note on hydroboration., Diborane adds on alkenes and alkynes in ether solvent at room temperature. This, reaction is called hydroboration., B2H6 + 6RCH = CHR → 2B(RCH – CH2R)3, Diborane, , 12., , trialkyl borane, , Give one example for each of the following:, (i) Icosagens --- Boron family (group 13), --- Eg : aluminium, (ii) Tetragen, --- Carbon family (group 14) --- Eg : Silicon, (iii)Pnictogen ---Nitrogen family(group 15) --- Eg : Phosphorous, (iv)Chalcogen --- Oxygen family (group 16) --- Eg : Sulphur

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16, , 19., , 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., K2SO4.Al2(SO4)3.24H2O, Potash alum (A), , 20., , 500K, , K2SO4.Al2(SO4)3 + 24H2O, (B) Burnt alum, , COis a reducing agent.Justify with an example., Carbon monoxide is a strong reducing agent. It reduce the oxides of the less active, metals to theirrespective metals and itself gets oxidised to carbon dioxide in the, process., Eg: 3CO + Fe2O3, , Heat, , 2Fe + 3CO2, , II.Additional Question with Answers:, 1., , Why boron does not form B3+ ions?, Because of its small size and sum of first three ionization enthalpies is high, boron does, not lose all its valence electrons to form B3+ ions., , 2., , Why the ionization enthalpy from aluminium to thallium 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., , 3., , What isInert pair effect?, Inert pair effect: In heavier post transition metals, the outer s-electrons (ns) have a, tendency to remain inert andshow reluctance to take part in the bonding is known as, inert pair effect, , 4., , What happen when boron burns with nitrogen and oxygen (or) air?, (i) 2B + N2, 2BN, 900K, (ii)4B + 3O2, 2B2O3, , 5., , Write any two methods for the preparation of metal borides, 1500K, (i) Cr + nB, CrBn, 1500K, (ii) BCl3 + W, WB + Cl2 + HCl, , 6., , Write any three uses of boron., a. Amorphous boron is used as a rocket fuel igniter., b. Boron is essential for the cell walls of plants., c. Isotope of boron10B5 is used as a moderator in nuclear reactors., 7.How borax is prepared from colemanite ore?, 2Ca2B6O11 + 2Na2CO3 + H2Oheat3Na2B4O7 + 3CaCO3 + Ca(OH)2, Colemanite, borax, 8. Why Borax solution is basic in nature?, Borax solution in hot-water is alkaline as it dissociates into boric acid and sodium, hydroxide, Na2B4O7 + 7H2O→ 4H3BO3 + 2NaOH

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19, , (ii)Pyro silicate (or) Soro silicates:, •, Silicates which contain [Si2O7]6- ions,, •, They are formed by joining two [SiO4] 4- tetrahedral units by sharing one, oxygen atom, at one corner., •, Eg : Thortveitite – Sc2Si2O7., (iii)Cyclic silicate (or) Ring silicates:, • Silicates which contain (SiO3 )n2n- ions., • Which are formed by linking three or more tetrahedral SiO44-units., • Each silicate unit shares two of its oxygen atoms with other units, • Eg : Beryl [Be3Al2 (SiO3)6], (iv)Inosilicates:, • Silicates which contain ‘n’ number of silicate units liked by sharing two or, more oxygen atoms are called inosilicates., • They are further classified as chain silicates and double chain silicates., (v)Chain silicates (or) Pyroxenes:, • These silicates contain [SiO3)]2n- ions formed by linking ‘n’ number of, tetrahedral [SiO4]4- units linearly., • Each silicate unit shares two of its oxygen atoms with other units., • Eg : Spodumene – LiAl(SiO3)2., (vi) Double chain silicates(or) amphiboles:, • These silicates contain [Si4O11]6n- ions., • In these silicates there are two different types of tetrahedral :, (i) Those sharing 3 vertices, (ii) those sharing only 2 vertices., • Eg : Asbestos., (vii) Sheet (or) phyllo silicates:, • Silicates which contain (Si2O5)n2n- ions., • Each [SiO4]4- tetrahedron unit share three oxygen atoms with others and thus by, forming, two dimensional sheets., • Eg : Talc, Mica etc..., (viii)Three dimensional silicates (or) tecto silicates:, • Silicates in which all the oxygen atoms of [SiO4]4- tetrahedral are shared with, other tetrahedralto form three-dimensional network are called three dimensional, silicate., • They have general formula (SiO2) n., • Eg : Quartz.

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20, , UNIT - 3., p - Block Elements II, I Text Book Question & Answer, 1., , What is inert pair effect?, In heavier post transition metals, the outermost electrons (ns) have a, tendency to remain inert and show reductance to take part in the bonding, which is known as inert pair effect., , 2., , Chalcogens belongs to p - block give reason?, 1 Their outer electronic configuration isns2np4., 2 In these elements, the last electron enter np orbital., 3 Hence they belong to p-block elements., 4 Since the outer most electron of Chalcogens enter into ‘p’ orbital it belongs, to ‘p’ block. Its group number is16., , 3., , Explain why fluorine always exhibit an oxidation state of -1?, 1. Since fluorine is the most electronegative element, it exhibits only a, negative oxidation state of -1., 2. Due to the absence of d-orbital, fluorine does not show positive oxidation state., , 4., , Give the oxidation state of halogen in the following, a) OF2, b) O2F2, c) Cl2O3, d) I2O4, Fluorine shows only -1 oxidation state. Hence, a) Oxidation state of‘F’ inOF2 is-1, b) Oxidation state of ‘F’ in O2F2 is -1, , c) Cl2O3, , 2x + 3(-2) = 0, 2x - 6 = 0, 2x = 6, X=3, Oxidation state of Cl is +3, d) I2O4, 2x + 4(-2) = 0, 2x - 8 = 0, 2x = 8, X=4, Oxidation state of Iodine is +4

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21, , 5., , What are interhalogen compounds? Give example?, Each halogen combines with other halogen to form a series of compounds, called interhalogen compounds, Eg : ClF, BrCl, IF7, , 6., , Why fluorine is more reactive than other halogens?, Fluorine is the most reactive element among halogen. This is due to the low, value of F-F bond dissociation energy., , 7., , Give the uses of helium., 1. It is much less denser than air and hence used for filling air balloons., 2.Helium has lowest boiling paint and hence used in cryogenics., 3.Helium is used to provide inert atmosphere in electric arc welding metals., 4.Helium and oxygen mixture is used by the divers in place of air oxygen, mixture. This prevents the painful dangerous condition called bends., What is the hybridisation of iodine in IF7? Give its structure?, , 8., , Hybridisation of IF7 is sp3d3, , Structure of IF7, , Structure - Pentagonal Bipyramidal, , 9., , 10., , 11., , Give the balanced equation for the reaction between chlorine with cold, NaOH and hot NaOH., 1. Chlorine reacts with cold NaOH to give sodium chloride and sodium, hypochlorite., Cl2 + 2 NaOH → NaOCl + NaCl + H2O, Sodium hypochlorite, 2. Chlorine reacts with hot NaOH to give sodium chloride and sodium, chlorate., 3Cl2 + 6NaOH → NaClO3 + 5 NaCl +3H2O, Sodium chlorate, How will you prepare chlorine in the laboratory?, In the laboratory, chlorine is prepared by the oxidation of hydrochloric acid, by KMnO4., 2KMnO4 + 16HCl→ 2 KCl + 2 MnCl2 +8H2O + 5Cl2↑, Give the uses of sulphuric acid ?, i. Sulphuric acid is used in the manufacture of fertilizers, ammonium, sulphate and super phosphates and other chemicals such as hydrochloric, acid, nitric acid etc., ii. It is used as a drying agent and also used in the preparation of pigments,, explosives etc.

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22, , 12., , 13., , 14., , 15., , 16., , Give a reason to support that sulphuric acid is a dehydrating agent?, It is highly soluble in water and has strong affinity towards water and hence it, can be used as a dehydrating agent., The dehydrating property can also be illustrated by its reaction with organic, compounds such as sugar, oxalic acid and formic acid., C 12H 22 O 11+H2SO4→ 12C + H2SO4. 11H2O, (COOH)2+ H2SO4 → CO + CO2 + H2SO4. H2O, HCOOH + H2SO4→ CO + H2SO4. H2O, Write the reason for the anamolous behaviour of nitrogen?, 1. Small size, 2. High ionisation enthalpy and high electronegativity., 3. Absence of d-orbitals in their valence shell., 4. Nitrogen is a diatomic gas unlike the other members of the group., Write the molecular and structural formula for the following molecules?, a) Nitric Acid b) Dinitrogen Pentoxide c) Phosphoric acid d) Phosphine, , Give the uses of Argon?, Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs, Write the valence shell electronic configuration of group 15elements?, Valence shell electronic configuration of group 15 elements ns2np3

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23, , Nitrogen, , 17., , [He]2s2p3, , Phosphorous, , [Ne] 3s23p3, , Arsenic, , [Ar]3d104s24p3, , Antimony, , [Kr] 4d105s25p3, , Bismuth, , [Xe] 5d106s26p3, , Give two equations to illustrate the chemical behaviour of phosphine?, (i), Basic nature : Phosphine is weakly basic and forms phosphonium, salts with halogen acids., , PH3 + HI ⎯⎯, → PH4I., ii) Reducing property : Phosphine precipitates some metal from their salt, solutions, 3AgNO3 + PH3→Ag3P + 3HNO3, , 18., , Give a reaction between nitric acid and a basic oxide?, Nitric acid reacts with basic oxides to form salt and water., ZnO + 2HNO3 → Zn (NO3)2 + H2O, 3FeO + 10 HNO3 →3Fe (NO3)3 + NO + 5 H2O, , 19., , What happens when PCl5 is heated?, On heating phosphorous penta chloride, it decomposes into phosphorous, trichloride and chlorine., PCl5(g) → PCl3(g) + Cl2(g), , 20., , Suggest a reason why HF is a weak acid, whereas binary acids of the all, other halogens are strong acids?, Hydrofluoric acid is a weak acid due to the presence of intermolecular, hydrogen bonding in it and it cannot be completely ionized and hydrogen ion, concentration will not be increased. But other hydrohalic acids are, completely ionized and therefore they are strong acids., , 21., , Deduce the Oxidation number of Oxygen in Hypofluorous acid -HOF?, HOF, +1 + x - 1= 0, x =0, Hence oxidation number of oxygen in HOF is zero., , 22., , What type of hybridisation occur in a) BrF5 b)BrF3, a) BrF5 -sp3d2, b) BrF3 -sp3d

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26, , 6), , 7), , Write the reaction of hydrolysis of urea., Hydrolysis of Urea gives ammonia., NH2CONH2+H2O→2NH3+CO2, How is ammonia prepared in the laboratory?, , Ammonia is prepared in the laboratory by heating ammonium salts with a, base., 2NH4+ + OH--→ 2NH3 + H2O, 2NH4Cl + CaO → CaCl2 + 2NH3 + H2O, 8), , How is ammonia prepared from magnesium nitride?, , Mg3N2 + 6 H2O →3Mg(OH)2+2NH3, By heating magnesium nitride with water ammonia is formed., 9), , What happens when ammonia is heated above500°C?, , When ammonia is heated above 500°C, it decomposes into Nitrogen and hydrogen., >500oC, , 2NH3, 10), , →, , N2+3H2, , Illustrate the reducing property of ammonia with an example, When Ammonia is passed over heated lead oxide, it is reduced into lead, 3PbO + 2NH3 → 3Pb +N2+3H2O, , 11), , 12), , The Affinity of ammonia for proton is greater than that of water .Justify it., When treated with acids ammonia forms ammonium salts. This reaction, shows that the affinity of ammonia for proton is greater than that of water., What happens when nitric acid is exposed to sunlight (or) heating? (or), , Colourless pure concentrated nitric acid turns yellow on standing. why?, Nitric acid decomposes on exposure to sunlight (or) on being heated., 4HNO3→4NO2 + 2H2O + O2`, 13), , Mention the uses of nitricacid, , i) Nitric acid is used as a oxidizing agent in the preparation of aquaregia, ii) Salts of nitric acid are used in photography (AgNO3) and gun powder for, firearms(NaNO3), 14), , Mention the three common allotropic forms of phosphorous., , The three common allotropic forms of phosphorous are white, red and black, phosphorous., 15), , How do you convert, , i) White phosphorous → Red Phosphorous, ii) Red phosphorous → White phosphorous, i) The White phosphorous can be changed into red phosphorous by heating it to, 420°C in the absence of air and light, ii) The red phosphorous can be converted back into white phosphorous by boiling it, in an inert atmosphere and condensing the vapour underwater.

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29, 35), , How does ammonia react with metallic salts?, , i) Fe3+ ii)Cu2+, Ammonia reacts with metallic salts to give metal hydroxides (in case of Fe) (or), forming complexes (in case of Cu)., Fe 3+ + 3NH4+ → Fe(OH)3 + 3NH4+, , Cu2+ + 4NH3→ [(Cu(NH3)4]+2, Tetraamminecopper(II)ion, 36), , Prove that nitric acid is an oxidizing agent., , C+4HNO3 → 2H2O + 4 NO2 + CO2, S + 2HNO3 → H2SO4 + 2 NO, 37), , Write a note on nitration of benzene., , In organic compounds replacement of a - H atom with NO2 is referred to as nitration., H2SO4, , C6H6 + HNO3 → C6H5NO2 + H2O, Nitration takes place due to the formation of nitronium ion., +, +, HNO3 + H2SO4→ NO, 3 2 + H3O + HSO4, 38), , Write the reaction of conc.HNO3 with copper., , Cu + 4HNO3→ Cu (NO3)2+ 2 NO2+2H2O, 39), , What is called phosphorescence?, , The freshly prepared white phosphorous is colourless but becomes pale yellow due, to the formation of a layer of red phosphorous on standing. It is poisonous and has a, characteristic garlic smell. It glows in the dark due to oxidation which is called, phosphorescence., 40), , Give an account on structure of phosphorous., , Phosphorous has a layer structure. The four atoms in phosphorous have, polymeric structure with chains of P4 linked tetrahedrally. P≡P is less stable than P-P, single bonds. Hence phosphorous atoms are linked through single bonds rather than, triple bonds., 41), , How is phosphine obtained from yellow phosphorous?, , Yellow phosphorous reacts with alkali on boiling in an inert atmosphere and it, liberates phosphine., P4 + 3NaOH + 3H2O → 3NaH2PO2 + PH3↑, 42), , How do you prepare orthophosphoric acid from phosphorous?, , When phosphorous is treated with conc.nitric acid in presence of iodine catalyst, ortho phosphoric acid is formed., P4 + 20HNO3→ 4H3PO4 + 20NO2 + 4H2O, 43), , Mention the hydride of phosphorous and give its hybridization and structure. The, hydride of phosphorous is phosphine (PH3).The hybridization of phosphorous in, phosphine is sp3.Three orbitals are occupied by bond pair with fourth corner, occupied by lone pair of electrons. Hence bond angle is reduced to 94°. It has, pyramidal shape., , 44), , How do you prepare phosphine from metallic phosphides?, , Metallic phosphides on hydrolysis with water (or) dilute mineral acids give, phosphine., Ca3P2 + 6H2O →2PH3↑+3Ca(OH)2, AlP + 3HCl → PH3↑+ AlCl3

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33, , 71), , 72), , Show that chlorine is a strong oxidizing agent, Chlorine is a strong oxidizing agent because of the nascent oxygen. It oxidizes ferrous, sulphate to ferric sulphate, 2FeSO4 + H2SO4 + Cl2→ Fe2 (SO4)3 +2HCl, How is bleaching powder prepared?, Bleaching Powder is prepared by passing chlorine gas through dry slaked time, (Calcium hydroxide), Ca(OH)2 + Cl2→ CaOCl2 + H2O, , 73), , Write two displacement redox reactions of chlorine, Chlorine displaces bromine from bromides and iodine from iodide salts, Cl2 + 2KBr → 2KCl + Br2, Cl2 + 2KI → 2KCl + I2, , 74), , Give the uses of chlorine, Chlorine is used in, i) Purification of drinking water, ii) Bleaching of cotton textiles paper and rayon, iii) Extraction of gold and platinum, , 75), , What is aqua regia (Royal water) mention it’s use, When three parts of concentrated hydrochloric acid and one part of concentrated, Nitric acid are mixed aquar egia is obtained. It is used for gold, platinum etc.,, What is the action of gold on aquaregia, , 76), , Au + 4H+ + NO3- + 4Cl-→ AuCl4- + NO + 2H2O, 77), , Give the reaction of platinum on aquaregia, Pt + 8H+ + 4NO3- + 6Cl-→ [Pt Cl6]2- + 4 NO2 + 4 H2O, , 78), , Give the uses of hydrochloric acid, i) Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride,, glucose from corn starch etc.,, ii) It is used in the extraction of glue from bone and also for purification of bone, black, , 79), , Thermal stability of hydrogen halide decreases from hydrogen fluoride to iodide, Give reason, Bond dissociation enthalpy decreases from hydrogen fluoride to hydrogen iodide and, hence thermal stability decreases. For eg, hydrogen iodide decomposes at 400°C, while hydrogen fluoride and hydrogen chloride are stable at this temperature., , 80), , Hydrofluoric acid is a weak acid whereas other hydrohalic acids are strong acids, Give reason, Hydrofluoric acid is a weak acid due to the presence of intermolecular hydrogen, bonding in it and it cannot be completely ionized and hydrogen ion concentration, will not be increased. But other hydrohalic acids are completely ionized and, hydrogen therefore they are strong acids.

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34, , 81), , Hydrofluoric acid is a weak acid at low concentration, but becomes stronger as, the concentration increases why?, 0.1M HF is only 10% ionized and hence it is a weak acid but 5M and 15M solution, of HF is stronger acid due to the equilibrium, , HF+H2O, , H3O++F-, , HF+F-, , HF2-, , At high concentration, the equilibrium involves the removal of fluoride ions. It affects, the dissociation of hydrogen fluoride and increases the hydrogen ion concentration, and hence HF becomes stronger acid., 82), , Hydrofluoric acid cannot be stored in silica and glass bottles why?, , 83), , Moist hydrofluoric acid rapidly reacts with silica and glass and thus it cannot be, stored init, SiO2 + 4HF → SiF4 + 2H2O, Na2SiO3 + 6HF → Na2 SiF6 + 3H2O, Show that hydrogen iodide is a good reducing agent, and how is it tested?, Hydrogen iodide is readily oxidized to iodine hence it is a good reducing, agent, 2HI → 2H+ + I2 + 2CLiberated iodine gives blue-black colouration with starch (Test for Iodine)., , 84), , Give the conditions for formation of interhalogen compounds, i) The central atom must be less electronegative and larger insize, ii) It can be formed only between two halogens and not with more than two, halogens., , 85), , Fluorine cannot act as central atom in inter halogen compoundsWhy?, Fluorine cannot act as central atom in inter halogen compounds because it is highly, electronegative and smallest among halogens., , Group 18 (Inert Gases) Elements, 86), , How are xenon fluorides prepared?, Xenon fluorides are prepared by direct reaction of xenon and fluorine under, different conditions., Ni, , Xe + F2 → XeF2, 4000 C, Ni/ acetone, , Xe + F2 ____ →, , XeF4, , 4000 C, Ni/ 200atm, , Xe +3F2 →, 4000 C, , XeF6

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35, , 87), , What happens when XeF6 is hydrolysed?, On hydrolysis of X6F6 with water vapour it gives, XeO3 XeF6 + 3H2O → XeO3 + 6HF, , 88), , How is sodium per xenate obtained fromXeF6?, When XeF6 reacts with 2.5m NaOH, sodium perxenate is obtained, 2XeF6 + 16 NaOH → Na4XeO6 + Xe+O2 + 12 NaF + 8H2O, (Sodium per xenate), , 89), , Show that sodium per xenate is has strong oxidizing property?, Sodium per xenate oxidises manganese(II) ion into permanganate ion even in, the absence of catalyst, 5XeO6 4- + 2Mn2+ 14H+ → 2MnO4-+ 5XeO3 + 7 H2O, , 90), , Xenon is used in high speed electronic flash bulbs used by photographers Why?, Xenon emits an intense light in discharge tubes instantly. Due to this it is used in, high speed electronic flash bulbs used by photographers., , ADDITIONAL 5 MARK QUESTIONS, 1), , Explain commercial method of preparation of nitric acid by ostwald’s process., , 2), , Write the primary, secondary and tertiary reactions of metals with nitric acid., , 3), , Give the various steps involved in the reaction of dilute nitric acid with, i. copper ii. magnesium, , 4), , Mention the oxides of nitrogen and give their preparation., , 5), , Give the structures of oxides of nitrogen., , 6), , Give the structures of oxoacids of nitrogen., , 7), , Mention the oxoacids of nitrogen and give their method of preparation., , 8), , Give one method of preparation for each oxyacids of phosphorous., , 9), , Write briefly on allotropic form of Sulphur., , 10), , Explain any three methods of preparation of sulphur dioxide with equations., , 11), , Phosphorous compound ‘A’ which is poisonous and has the smell of rotten fish, reacts with chlorine and gives ‘B’ ‘B’ reacts with water to give an oxyacid of, phosphorous ‘C’ which is tribasic in nature Identify the compounds A, B and C and, explain the reactions.

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36, , 12), , How is chlorine manufactured by the electrolysis of brine solution?, , 13), , Explain deacon’s process of manufacture of Chlorine, , 14), , Mention the hybridization, geometry and number of bond pair and lone pairs of, electrons present in different types of inter halogen compounds.

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37, , UNIT – 4, TRANSITION AND INNER TRANSITION ELEMENTS, I., 1., , Text book question and Answer:, What are transition metals? Give four examples, 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. Example : Iron, Cobalt,, Nickel, Copper., , 2., , Explain the oxidation states of 4d series elements., •, , At the beginning of the series, +3 oxidation state is stable but towards the, end +2 oxidation state become stable., , •, , Hence the first and last elements show less number of oxidation states, and the middle elements with more number of oxidation states., 4d series elements, , 3., , Oxidation states, , Y, , +3, , Ru, , From +2 to +8, , Cd, , +2, , What are inner transition elements?, •, , The elements in which the extra electron enters (n-2)f orbitals are called fblock elements., , •, , These elements are also called as inner transition elements because they, form a transition series within the transition elements., , •, , In the inner transition elements there are two series of elements., , 1) Lanthanoids (Previously called lanthanides), 2) Actinoids (Previously called actinides), 4., , Justify the position of lanthanides and actinides in the periodic table., •, , The actual position of Lanthanoids in the periodic table is at group number 3, and period number6., , •, , In sixth period after lanthanum, the electrons are preferentially filled in inner, 4f sub shell and these fourteen elements following lanthanum show similar, chemical properties. Similarly, the fourteen elements following actinium, resemble in their physical and chemical properties.

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40, , 11., , What are 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. Example : TiC, ZrH1.92,Mn4N, , 12., , Calculate the numbers of unpaired electrons in Ti 3+, Mn2+ and calculate the, spin only magnetic moment., Ti3+, , 13., , 1) Electronic Configuaration of Ti3+ is, , Mn2+, 1) Electronic Configuration of Mn2+ is, , [Ar] 3d1 4s0, , [Ar] 3d54s0, , 2) No. of unpaired electron = 1, , 2) No. of unpaired electron = 5, , 3) Spin only magnetic moment, , 3) Spin only magnetic moment, s = √𝑛(𝑛 + 2), , s = √𝑛(𝑛 + 2), = √1(1 + 2) = √3, = 1.732 B, , 13, , = √5(5 + 2) = √35, , = 5.92B, , Write the electronic configuration of Ce4+ and Co2+, Electronic Configuration of Ce4+ = [Xe] 4f 05d06s0, Electronic configuration of Co2+ = [Ar]3d74s0, , 14., , Explain briefly how +2 states become more and more stable in the first half, of the first row transition elements with increasing atomic number., E0 M2+/M for 3d series upto Mn is highly negative. Therefore +2 states become more stable in, the first half of the first row transition elements., , 15., 1)lect, , Which is more stable? Fe3+ or Fe2+ -Explain., Fe3+, , Fe2+, , 54s0, Electro, Electronic configuration =[Ar] 3dElectronic, configuration =[Ar] 3d64s0, 1, , It, 2)consiIt consists of 5 unpaired electrons, Half filled d sub shell and more stable., , Hence Fe3+ is more stable than Fe2+., , It consists of 4 unpaired electrons., Partially filled d sub shell and less stable

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41, , Explain the variation EOM3+/M2+in 3dseries, , 16., , ▪, , The standard electrode potential for the M3+/M2+ half cell gives the relative, stability between M3+ and M2+., , ▪, , 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 (E=0.76V) is required., , o The high reduction potential of Mn3+ / Mn2+ indicates Mn2+ is more stable than Mn3+., For Fe3+ / Fe2+ the reduction potential is 0.77V, and this low value indicates that both, Fe3+ and Fe2+ can exist under normal conditions., 17., S.No., , Compare lanthanides and actinides., , Lanthanoids, Actinoids, orbital, 1, Differentiating electrons enters in 4forbital Differentiating electrons enters in 5f orbital., orbital., higher, 2, Binding energy of 4f orbitals are higher., lower Binding energy of 5f orbitals are lower., 3, 4, , They show less tendency to form, complexes, Most of the lanthanoids are colourless., , They show greater tendency to form complexes., Most of the actinoids are coloured. For Eg. U3+, (Red), U4+(Green),UO22+ (Yellow), 2, , 5, They do not form oxo cations, Beside, Besides +3 oxidation states lanthanoids, 6, show +2 and +4oxidationstates in few, , They do not form oxo cations such U022+,NpO22+, Besides +3 oxidation states actinoids show, higher oxidationstates such as +4, +5, +6 and +7., , cases., , 18., , Explain why Cr2+ is strongly reducing while Mn3+ is strongly oxidizing., , Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a, half filled t2g level. On the other hand, the change from Mn2+ to Mn3+ results in, the half – filled (d5) configuration which has extra stability., 19., , Compare the ionization enthalpies of first series of the transition elements., Ionization energy of transition elements 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 change, corresponding to the filling of d electrons.

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42, , IONIZATION ENTHALPIES OF FIRST SERIES, , Actinoid contraction is greater from the elements to element than the lanthanoid, , 20., , contraction, why?, This is due to poor shielding effect by 5f electrons in actinoids as compared to, that by 4f electrons in lanthanoids., Out of Lu(OH)3and La(OH)3which is more basic and why?, , 21., , La (OH)3, 1)1)Size of La3+ is large, 2), , 2)Ionic characters of La-OH bond is, high, , 3), , is low., , 22., , 1) Size of Lu3+ is small, 2) Ionic characters of Lu-OH bond is, low, , 3)Covalent character of La-OH bond, , 4), , Lu(OH)3, , 4)La(OH)3 is more basic., , 3) Covalent character of Lu-OH bond is, high., 4) Lu(OH)3 is less basic., , Why Europium (II) is more stable than Cerium(II)?, Europium (II), 1), , Electronic configuration of Europium, 1), Electronic configuration of Cerium, , (II) is [Xe]4f 75d06s0, 2), , Cerium (II), , In Eu2+, 4f sub shell is half filled, , (II) is [Xe]4f 15d16s0, In Ce2+, 4f and 5d sub shells are, partially filled., , Hence, 3), , Eu2+ is, , more stable, , Hence Ce2+ is less stable

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44

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45, , Describe the variable oxidation state of 3d series elements., , 27., , The first transition metal scandium exhibits only +3 oxidation state, but all other, transition elements exhibit variable oxidation states by losing electrons from (n-1)d, orbital and ns orbitals as the energy difference between them is very small., The first and last elements show less number of oxidation states and the middle, elements with more number of oxidation states., 3d Series Elements, Sc, Mn, Cu, 28., , Oxidation States, +3, From +2 to +7, +1 and +2, , Which metal in the 3d series exhibits +1 oxidation state most frequently and why?, Copper exhibits +1 oxidation state in the 3d series., Reason : Electronic configuration of Cu = [Ar] 3d104s1, It can easily lose 4s1 electron to give stable 3d10 configuration. Hence, it exhibits +1, oxidation state., , 29., , Why first ionization enthalpy of chromium is lower than that of zinc?, Chromium, , Zinc, , 1) Chromium (Z=24), 2)Electronic, , Configuration, , 1) Zinc (Z=30), of 2) Electronic configuration of Zinc is, , Chromium is [Ar] 3d5 4s1, , [Ar] 3d104s2, , 3) It can easily lose 4s1 electron to give, , 3) It is difficult to remove one electron, , stable half filled (3d5) configuration., , from 4s2 (completely filled). Hence, , Hence first ionization enthalpy of, , first ionization enthalpy of zinc is, , chromium is less, , comparatively more., , 30., , Transition metals show high melting points.Why?, , The high melting points of transition metals are attributed to the involvement of, greater number of electrons in the inter atomic metallic bonding from (n-1)d, electrons in addition to ns electrons.

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46, , 1., , II Evaluate yourself, Why iron is more stable in +3 oxidation state than in +2 and the reverse is true, for manganese?, Electronic Configuration., Fe3+ : [Ar] 3d5, Fe2+ : [Ar] 3d6, 2+, 5, Mn :[Ar] 3d, Mn3+ : [Ar] 3d4, , Fe3+ ion has half filled d orbital which is more sable than partially filled d orbital of Fe2+, Mn2+ ion has half filled d orbital which is more stable than partially filled d orbital of Mn3+, 2., Compare the stability of Ni4+ and Pt4+ from their ionization enthalpy values, IE, Ni, Pt, I, 737, 864, II, 1753, 1791, III, 3395, 2800, IV, 5297, 4150, For Ni, IE1+IE2+IE3+IE4=11,182 kJ, For Pt, IE1+IE2+IE3+IE4= 9605 kJ, Pt4+ is thermodynamically more stable than Ni4+ .Smaller the Ionisation enthalpy, greater will be the thermodynamic stability of its compounds, , i., ii., iii., iv., , III.ADDITIONAL QUESTIONS, 1.How d block elements are classified?, 3d series (4th period), Scandium to Zinc, 4d series (5th period), Yttrium to Cadmium, th, 5d serried (6 period), Lanthanum,Haffinium to Mercury, 6d series (7th period), Actinium, Rutherfordium to Copernicium., 2.Which transition series contains radioactive elements?, 6dseries(7th period), Ac,Rf to Cn., In this period all the elements are radioactive and have very low half –life period., 3.Write the electronic configuration of Cr and Cu, Cr :, Z=24, Cr : [Ar] 3d54s1, , Cu :, Z=29, Cu :[Ar] 3d104s1, 4.a In transition metals, which group elements are not hard?, b. Which metal has highest electrical conductivity at room temperature., a. Group 11 elements are not hard, b. Silver, 5 Give reason for the slight increase in atomic radius of Zn, The d orbitals of Zn 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., 6. Ionisation enthalpy to form Ni2+ =2490kJ, Ionization enthalpy to form Pt2+ =2655kJ, Which is thermodynamically stable Ni2+ (or) Pt2+? Why?, The energy required to form Ni2+ is less than that of Pt2+ so Ni2+compounds are, thermodynamically more stable than Pt2+compounds., 7. Calculate the number of unpaired electrons for the following ions?, a., Cu+, b. Ni2+, a), For, ↑↓, ↑↓, ↑↓, ↑↓, ↑↓, Cu+ : Ar] 3d10

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47, , Number of unpaired electrons are zero (n=0), ↑↓, ↑↓, ↑↓, ↑, For, Number of unpaired electrons are two (n = 2), Name the catalyst used for the following reaction?, , b), 8., , ↑, , Ni2+ : [Ar] 3d8, , CHO, a., Propene, , + CO + H2, , ?, , CHO, , Butan-1- al, , +, 2-methylpropan-1-al, , b., CH3CHO + CO, ?, CH3-COOH, Acetaldehyde, Acetic acid, c., In the manufacture of Sulphuric acid from SO3, a.Co2(CO)8, b.Rh / Ir Complex, c.Vanadium pentoxide (V2O5), 9. Why transition metals form number of alloys ?, Atomic sizes of transition metals are similar and one metal atom can be easily replaced by, another metal atom from its crystal lattice to form an alloy., 10. Why transition elements form complexes?, i. Transition metal ions are small and highly charged, ii. They have vacant low energy orbitals to accept an electron pair donated by other group, 11. Under what oxidation state do transition metals form ionic oxide and covalent oxides?, Metals in lower oxidation state form ionic oxides., Metals in higher oxidation state form covalent oxides, Example Mn2O7 is covalent. Oxidation state is +7., 12. Zn, Cd and Hg do not have partially filled d-orbitals why they are treated as transition, elements?, They are treated as transition elements because their properties are an extension of the, properties of the respective transition elements., 13. When potassium dichromate acts as oxidizing agent in the presence of H+ ion, what is the, change in oxidation state of Chromium? Give equation ?, Cr2O72- + 14H+ + 6eCr3+ + 7 H2O, Oxidation state of chromium changes from +6 to +3, 14.What is the action of heat on K 2Cr2 O7?, On heating it decomposes and forms Cr2O3 and molecular oxygen, 4K2Cr2O7, 4K2CrO4 + 2 Cr2O3+3O2, Red orange, yellow, 15.How neutral KMnO4 oxidizes thio sulphate ion? Give equation?, It oxidizes thio sulphate into sulphate, 8MnO4- + 3S2O32- + H2O, 6SO42- + 8MnO2 + 2 OH16.a. Which element in 3d series does not form ionic metal oxide ?, b. What is the acid formed when Mn2O7 dissolved in water?, a. Scandium, b .Mostly higher oxide are acidic. Mn2O7 dissolved in water to give permanganic acid, (HMnO4), , 17. Why d orbitals containing symmetrical distribution of electron is more stable?, 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. An unsymmetrical distribution of electron density 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.

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48, , 18. In first transition series from Sc to V atomic radius decreases, thereafter up to Cu atomic, radius nearly the same. Why ?, The added 3d electrons only partially shield the increased nuclear charge and hence the effective, nuclear change increases slightly. At the same time the extra electrons added to the 3d sub shell, strongly repel the 4s electron, These two forces are operated in opposite direction, As they tend to balance each other it leads to constancy in atomic radius, 19.Which is more acidic in nature MnO(Mn2+) or Mn2O7 (Mn7+) why?, Mn2O7 is more acidic in nature. Acidic strength increases with increase in oxidation state of the, element. In higher oxidation state there is no scope for further loss of electron, rather it can accept, electrons. So it is more acidic in nature., 20.Classify the following oxides as acidic, basic and amphoteric oxides?, i.CrO ii. Cr2O3 iii.CrO3, i.CrO, –Basic oxide, ii.Cr2O3, – Amphoteric oxide, iii. CrO3 – Acidic oxide, 21. Calculate spin magnetic moment of Co2+?, Electronic configuration of Co2+ [Ar] 3d7, Number of unpaired electrons n=3, Magnetic momentµ =√𝑛(𝑛 + 2), =√3(3 + 2), = √15= 3.87 µB, , 22.Why transition metal and its compound act as good catalyst?, Transition metal has energetically available d orbitals., d-orbitals can accept electrons from reactant molecule and form bond with reactant molecule using, its d electron., 23. Draw the structure of chromate ion and dichromate ion ?, , 24.What is zeigler – Natta Catalyst?Give a reaction in which it catalysis?, zeigler – Natta Catalyst ( TiCl4 + Al (C2H5)3, Mixture of TiCl4 and trialkyl aluminium., CH3, TiCl4+Al(C2H5) 3, n, CH = CH2, , CH3, CH-CH2, n, , propylene, , polypropylene

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50, , v., 31., , It is used in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen, peroxide and iodides., Complete the following reaction and give the balanced equation?, i MnO4- +, , COO-, , + H+, , ?, , COO-, , ii MnO4- + H2S, i. 2MnO-4 + 5 COO- + 16H+, , ?, 2Mn2+ + 10CO2+ 8H2O, , COO, , ii.2MnO4-+ 3H2S, 32., , 33., , 2MnO2 + 3S + 2 OH- + 2H2O, , What is the hybridization in Mn7+ ion of KMnO4 and what is the structure of permanganate, ion?, Permanganate ion has tetrahedral geometry in which the central Mn7+ is sp3 hybridised., Compound A is an orange red crystalline solid which on heating with NaCl and conc H2SO4, evolves red orange vapours B. On passing the vapors of B into a solution of NaOH and then, adding the solution of acetic acid and lead acetate, yellow precipitate C is obtained. Identify, A,B, andC. Give chemical equations for these reactions, What is the name of this test?, , Chromylchloride, , This is chromyl chloride Test., 34.What are the properties of Interstitial compound?, They are hard and show electrical and thermal conductivity, i., They have high melting points higher than those of pure metals, ii., Transition metal hydrides are used as powerful reducing agents, iii. Metallic carbides are chemically inert., 35. Complete the following reactions and give the balanced equations?, Cr2O72- + S2-+ H+, ?, +, MnO4 + I + H, ?, Cr2O72- + 3S2- + 14H+, 2Cr3+ + 3S + 7H2O, 2MnO4 + 10 I+ 16H+, 2Mn2+ + 5 I2 + 8H2O, 36. What is Hume- Rothery rule to form a substitute alloy?, i., The difference between the atomic radius of solvent and solute is less than 15%, ii., Both the solvent and solute must have the same crystal structure and valence., iii., The eletronegativity difference between solvent and solute must be close to zero., 37.What are the uses of potassium dichromate?, i. It is used as a strong oxidizing agent., ii. It is used in dyeing and printing., iii.It used in leather tanneries for chrome tanning., iv.It is used in quantitative analysis for the estimation of iron compounds and iodides.

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53, , 9. Draw all possible geometrical isomers of the complex [Co(en)2Cl2]+and identify the optically active, isomer., , Trans isomer [Co(en)2Cl2]+ does not show optical isomerism, because of plane of symmetry, , Cis isomer shows optical isomerism because it lacks symmetry, , 10. [Ti(H2O)6]3+ is coloured, while [Sc(H2O)6]3+ is colourless- explain., CONTENT, Central metal ion, , [Ti(H2O)6]3+, Ti3+, , Electronic Configuration, , 3d1, , Number of unpaired electron, , 1, , [Sc(H2O)6]3+, Sc3+, 3d0, , 0, , Ti3+ has one unpaired electron No unpaired electron, so, for d-d transition, hence it is, d-d transition is not possible, coloured, hence it is not coloured, , 11. Give an example for complex of the type. [Ma2b2c2] where a, b, c are monodentate ligands and, give the possible isomers., [PdI2(ONO)2(H2O)2] Diaquadiiododinitrito-K -O Palladium (IV), [PdI2(NO2)(H2O)2] Diaquadiiododinitrito-K -N palladium (IV), They will exhibit linkage isomerism, , 12. Give one test to differentiate [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl.

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55, , 17. Give the difference between double salts and coordination compounds., S.NO, Doublesalt, Coordinationcompound, 1., Double salts lose their identity in aqueous They don’t lose their identity in aqueous, solution by completely dissociating in to, solution as they do not ionize completely, ions in the solvent, (the complex ion further doesnot get, ionized), 2., They give test for all the constituent ions They do not show test for all their, constituent ions for example in, K4[Fe(CN)6], it does not show the test for, Fe2+ and CNExample : K2SO4.Al2(SO4)3.24H2O, Example : K4[Fe(CN)6], 18. Write the postulates of Werner’s theory., 1. Most of the elements exhibit, two types of valence., •, , Primary valence, , •, , Secondary valence, , 2. Primary valence is referred as the oxidation state of the metal atom and the Secondary, , valence as the coordination number., 3. 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., 4. The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the, , combination of these., 5. 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., , •, , The outer sphere is called ionisation sphere., , 6. The primary valences - non-directional, the secondary valences - directional., , 7. The geometry of the complex is determined by the spacial arrangement of the groups which, satisfy the secondary valence., Six - octahedral geometry., Four -either tetrahedral or square planar geometry., , 19. [Ni(CN)4]2- is diamagnetic, while [Ni(Cl)4]2- is paramagnetic using crystal field theory?, Square planar geometry of [Ni(CN)4]2-experience undergo Jahn Teller Distortion

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56, [Ni(CN)4]2−:, , /, Jahn Teller distrotion, The d orbitals fill with 8 electrons, then, with a low spin configuration. The even number of d orbitals will get filled, (dyz,dxz,dz2,dxy) with an even number of 3d electrons., This gives rise to a diamagnetic configuration, as expected., [NiCl4]2−:, Splitting in the tetrahedral field, , The d orbitals here fill with 8 electrons, but instead, high spin. So, the dz2 and dx2−y2fill with one electron each, then, the dxy, dxz, and dyz with one electron each, and then pairing occurs only after that, filling the dz2, dx2−y2,, and dxy completely. This leaves two unpaired electrons in the t2 orbitals, and thus this complex is paramagnetic with, two unpaired electrons, as expected., , 20. Why tetrahedral complexes do not exhibit geometrical isomerism., • All the four ligands are adjacent or equidistant to one another in tetrahedral complex., • As the relative positions of donor atoms of ligands attached to the central atom are same with respect, to each other., 21. Explain optical isomerism in coordination compounds with an example., Optical Isomerism, • Coordination compounds which possess chirality exhibit optical isomerism., • 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., , Example:

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57, , 22. What are hydrate isomers? Explain with an example., The exchange water molecules in the crystal lattice with a ligand in the coordination entity will give different isomers., These type of isomers are called hydrate isomers., , Example :, CrCl3.6H2O has three hydrate isomers as shown below., [Cr(H2O)6]Cl3, violet colour, [Cr(H2O)5Cl]Cl2.2H2O, pale green colour, [Cr(H2O)5Cl2]Cl.H2O, dark green colour, , gives three chloride ions in solution, two chloride ions in solution, one chloride ion in solution., , 23. What is crystal field splitting energy?, The splitting of five d-orbitals of the metal ion in the presence of ligand field into two sets having different, energies is called crystal field splitting or energy level splitting. The difference in the energy of the two sets, is called crystal field splitting energy., 24. What is crystal field stabilization energy (CFSE) ?, The crystal field stabilization energy is defined as the energy difference of electronic configurations in the, ligand filed (ELF) and the isotropic field/barycentre (Eiso)., CFSE (ΔEo) = {ELF } - {Eiso }, = {[nt2g(-0.4)+neg(0.6)] Δo + npP} - {n'p P}, nt2g is the number of electrons in t2g orbitals, negis number of electrons in eg orbitals, npis number of electron pairs in the ligand field, n'pis the number of electron pairs in the isotropic field (barycentre)., , 25. A solution of [Ni(H2O)6]2+ is green, whereas a solution of [Ni(CN)4]2- is colorless – Explain., , 26. Discuss briefly the nature of bonding in metal carbonyls., In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components., • An electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of central, metal atom., σ, , • This electron pair donation forms M←CO sigma bond., This sigma bond formation increases the electron density in metal d orbitals and makes the, metal electron rich.

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58, , 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 is called π-back bonding ., Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from, metal to ligand through pi bonding, this synergic effect accounts for strong, M ←CObond in metal carbonyls., 27. What is the coordination entity formed when excess of liquid ammonia is added to an aqueous, solution copper sulphate?, When excess of liquid ammonia is added to an aqueous solution of copper sulphate it gives tetraammine, copper (II) sulphate., CuSO4 + 4NH3→ [Cu(NH3)4]SO4, So, the coordination entity is : [Cu(NH3)4]2+, 28. On the basis of VB theory explain the nature of bonding in [Co(C2O4)3]3[Co(C2O4)3]3Central metal /atom, , Co(III), , Electronic configuration, , Co:, , 3d7, , 4s2, , 4p, , C2O42- is a strong field ligand., 3-, , [Co(C2O4)3], , Geometry, Hybridization, Number of unpaired, electron, , 3d6, x, x, , 4s0, x, x, , x, x, , 4p, x, x, , x, x, , x, x, , Inner orbital (low spin) complex, Octahedral, d2sp3, , n = 0 ; DIAMAGNETIC, 29. What are the limitations of VB theory?, 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, [Fe(CN)6]4- is diamagnetic (low, spin) whereas [FeF6]4- is paramagnetic (high spin)., 30. Write the oxidation state, coordination number , nature of ligand, magnetic property and, electronic configuration in octahedral crystal field for the complex K4[Mn(CN)6].

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60, (iv) dichloridobis(ethylenediamine)Cobalt(III) sulphate - [Co(en)2Cl2]2SO4, (v) Tetracarbonylnickle(0), , - [Ni(CO)4], , 5. A solution of [Co(NH3)4I2]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?, Silver nitrate reacts with chloride ions to give silver chloride white precipitate, therefore in one isomer the Cl- is the, counter ion. Whereas silver nitrate reacts with Iodide ion to form yellow precipitate therefore the counter ion is I-., Therefore the isomers are called ionization isomer, ISOMERS, [Co(NH3)4Cl I]I - Yellow precipitate, , [Co(NH3)4I2] Cl – white precipitate, , 6. Three compounds A,B and C have the molecular formulaCrCl3.6H2o they are kept in a container with a, dehydrating agent and they lost water and attaining constant weight as shown below., Compound, Initial weight of compound (in g), A, 4, B, 0.5, C, 3, Compound A: [Cr(H2O)4 Cl2]Cl. 2H2O, , Constant weight after dehydration (in g), 3.46, 0.466, 3, , Compound B: [Cr(H2O)5 Cl]Cl2. H2O, Compound C: [Cr(H2O)3Cl3]. 3H2O, 7. Indicate the possible type of isomerism for the following complexes and draw their isomers, Given Compound, [Co(en)3][Cr(CN)6], , Isomer, [Cr(en)3][Co(CN)6], , [Co(NH3)5(NO2)]2+– N attached, [Co(NH3)5(ONO)]2+ - O attached, [Pt(NH3)3(NO2)]Cl, [Pt(NH3)3Cl]NO2, 8. Draw all possible isomers of a complex Ca[Co(NH3)Cl(ox)2], , Type, Coordiantion Isomerism, Linkage Isomerims, Ionization Isomerism

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61, 9. The spin only magnetic moment of tetrachloridomanganate(II) ion is 5.9 BM. On the basis of VBT predict, the type of hybridization and geometry of the compound, Given complex [MnCl4]2When magnetic moment is 5.9BM it means it has 5 unpaired electron i.e., n=5, 3d5, ↑ ↑ ↑ ↑ ↑, Chlorido is a weak ligand therefore no pairing takes place ., [MnCl4]24s2, , 4p6, , X, , Clx, , X, , X, , X, , x, , x, , x, , Cl- Cl-Cl-, , x, x, , Hybridization sp3, Shape : Tetrahedral, 10. Predict the number of unpaired electrons in [CoCl4]2- ion on the basis of VBT., Co : 3d7, , 4s2, , 4p, , 10. Predict the number of unpaired electrons in [CoCl4]2- ion on the basis of VBT., Co : 3d7, , 4s2, , 4p, , Co2+:3d7, , 4s0, , 4p, , Cl- is a weak ligand therefore pairing does not takeplace, [CoCl4]2X, , X, , X, , x, , x, , x, , x, , Cl-, , Cl- Cl-Cl-, , X, , Number of unpaired electron, n=3, 11. 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. Where as 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., Compound A pale yellow precipitate counter ion Br [Co(en)2Cl2]Br, , Compound B – white precipitate – counter ion Cl[Co(en)2ClBr]Cl, , 12. The mean pairing energy and octahedral field splitting energy [Mn(CN)6]4- are 28,800 cm-1 and 38500 cm-1, respectively. Whether this complex is stable in low spin or high spin?

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62, , Given mean pairing energy 28000 cm-1 and Octahedral splitting energy is 38500 cm-1, For [Mn(CN)4]4- the electronic configuration Mn3+ is 3d4, ↑ ↑ ↑ ↑, , Case I: High Spin complex t2g3 eg1, CFSE: {[nt2g(-0.4) + neg (0.6)] ∆ₒ + np P} – { np’P}, ={3(-0.4) + 1(0.6)] ∆ₒ+ 0xP} – 0, = {-1.2 +0.6}∆ₒ, = -0.6 ∆ₒ, = -0.6 x 38500, = -23100 cm-1, , Case II: low Spin complex t2g4 eg0, CFSE: {[nt2g(-0.4) + neg (0.6)] ∆ₒ + np P} – { np’P}, ={4(-0.4) + 0(0.6)] ∆ₒ+ 1xP} – 0, = {-1.6 }∆ₒ + P, = -1.6(38500) + 28800, = - 61600 + 28800, = - 32800 cm-1, , Large negative CFSE indicate in case II that the complex is stable in low spin, , 13. Draw energy level diagram and indicate the number of electron in each level for the complex [Cu(H 2O)6]2+ ., Whether the complex is paramagnetic or diamagnetic?, [Cu(H2O)6]2+ ion is a stretched octahedron. The electronic configuration of copper(II) ion is [Ar} 3d9. When the, complex ion formed by Cu2+ has an octahedral structure, the d orbitals will split in to two different orbitals t2g and eg., The electronic configuration is (t2g)6 (eg)3 , the 3 electrons in eglevel can be arranged in two different ways t2g6 eg3 (, dz2)2 (dx2-y2)1 or t2g6 eg3 ( dz2)1 (dx2-y2)2 . To break the degeneracy there is a distortion of the octahedral splitting, The presence of one unpaired electron in the eg level makes [Cu(H2O)6]2+ as paramagnetic, 14. For the [CoF6]3- ion in the mean pairing is found to be 21000 cm-1. The magnitude of ∆ₒ is 13000 cm-1., Calculate the crystal field stabilization energy for this complex ion corresponding to low spin and high spin, states., Given mean pairing energy 21000 cm-1 and Octahedral splitting energy is 13000 cm-1, Forthe [CoF6]3 electronic configuration Co3+ is 3d6, ↑↑ ↑ ↑ ↑ ↑, Case I: High Spin complex t2g4 eg2, CFSE: {[nt2g(-0.4) + neg (0.6)] ∆ₒ + np P} – { np’P}, ={4(-0.4) + 2(0.6)] ∆ₒ+ 1xP} – 1P, = {-1.6 +1.2}∆ₒ, = -0.4 ∆ₒ, = -0.4 x13000, = -5200 cm-1, , Case II: low Spin complex t2g6 eg0, CFSE: {[nt2g(-0.4) + neg (0.6)] ∆ₒ + np P} – { np’P}, ={6(-0.4) + 0(0.6)] ∆ₒ+ 3xP} – 1P, = {-2.4}∆ₒ +2 P, = -2.4(13000) + 2x21000, = 10,800 cm-1, , High positive CFSE indicate in case II that low spin complex is not favorable one, , EXTRA QUESTIONS, 1. Write the coordination number and oxidation state of platinum in the complex [Pt(en)2Cl2], Coordination number and oxidation state of Pt in the complex [Pt(en)2Cl2] are 6 and +2 because en is a bidentate and, neutral ligand, 2. What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4 and [Co(NH3)6][Cr(CN)6] ?, Ionization isomerism and coordination isomerism respectively., 3.Write the postulates of Valence bond theory., 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.

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63, 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, , Geometry, , Geometry, , 2, , Sp, , Linear, , [CuCl2]-, [Ag(CN)2], , UNIT 6

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64, , SOLID STATE, I, , TEXT BOOK QUESTION AND ANSWERS, 1. Define unit cell., A basic repeating structural unit of a crystalline solid is called a unit cell., 2. Give any three characteristics of ionic crystals., • Ionic solids have high melting points., • They do not conduct electricity in solid state., • They conduct electricity in molten state (or) when dissolved in water., • They are hard and brittle., 3. Differentiate crystalline solids and amorphous solids., S.NO, , CRYSTALLINE SOLIDS, , AMORPHOUS SOLIDS, , 1, , Long range orderly arrangement of, constituents, , Short range random arrangement of, constituents, , 2, , Definite shape, , Irregular shape, , 3, , Anisotropic in nature, , Isotropic in nature, , 4, , They are true solids, , 5, , Definite Heat of fusion, , They are pseudo solids (or) super, cooled liquids, Heat of fusion is not definite, , 6, , They have sharp melting points, , They do not have sharp melting points, , 4 Classify the following solids, a. P4, b. Brass, c. diamond, , d. NaCl, a, b, c, d, e, , e. Iodine, P4, Brass, diamond, NaCl, Iodine, , Covalent solid, Metallic solid, Covalent solid, Ionic solid, Molecular solid, , 5. Explain briefly seven types of unit cell., There are seven primitive crystal systems. They differ in the arrangement of their crystallographic, axes and angles., S.NO, NAME OF THE UNIT CELL EDGE LENGTH, ANGLES, 1, Cubic, a=b=c, α=β=γ=90o, 2, Rhombohedral, a=b=c, α=β=γ≠90o, 3, Hexagonal, a=b≠c, α=β= 90o, γ=120o, 4, Tetragonal, a=b≠c, α=β=γ=90o, 5, Orthorhombic, a≠b≠c, α=β=γ=90o, 6, Monoclinic, a≠b≠c, α= γ=90o, β≠90o, 7, Triclinic, a≠b≠c, α≠β≠γ≠90o, 6. Distinguish between hexagonal close packing and cubic close packing.

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65, , S.NO, 1, 2, , 3, , 4, , HEXAGONAL CLOSE PACKING, , CUBIC CLOSE PACKING, , “ABA’’ arrangement is known asthe, “ABC’’ arrangement is known as the, hexagonal close packed (hcp)arrangement cubic close packed (ccp) arrangement., The arrangement of spheres in the third, The arrangement of spheres inthe third, layer exactly resembles the first layer., layer dose not resembles with those of, either the first or second layer., The hexagonal close packing is based, The cubic close packing is based on the, hexagonal unit cells with sides of, face centered cubic unit cell., equal length., Tetrahedral voids of the second, Octahedral voids of the second layer are, layer are exactly covered by the sphere of partially covered by the sphere of the third, the third layer., layer., , 7. Distinguish tetrahedral and octahedral voids., S.NO, 1, , TETRAHEDRAL VOIDS, When a sphere of second layer (b) is, above the void in the first layer (a),, tetrahedral void is formed, , OCTAHEDRAL VOIDS, When a sphere of second layer (b), partially covers the void in the first layer, (a), octahedral void is formed, , 2, , If the number of close packed spheres, be ‘n’ then, the number of tetrahedral, voids generated is equal to 2n., , If the number of close packed spheres, be ‘n’ then, the number of octahedral, voids generated is equal to n, , 3, , This constitutes four spheres, three in, the lower layer (a) and one in the upper, layer (b)., When the centers of these four spheres, are joined, a tetrahedron is formed., , 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., , 4, , 8. What are point defects?, Point defects are the deviations from ideal arrangement that occurs at some points or atoms in a, crystalline substance., 9. Explain Schottky defect.

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66, , •, , Schottky defect arises due to the missing of equal number of cations and anions, from the crystal lattice .Hence stoichiometry of the crystal is not changed., 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 its density., , •, •, , 10. Write short note on metal excess and metal deficiency defect with an example.(or) Write a, short note on non-stoichiometric defects., Metal excess defect:, , •, •, , Metal excess defect arises due to the presence of more number of metal ions than the, anions. Ex: NaCl, KCl, The electrical neutrality of the crystal is 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 ., , The anionic vacancies, which are occupied by unpaired electrons are called F centers., , •, , 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 state. 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 other Fe2+oxidized to Fe3+ ions., In such cases, overall number of Fe2+ and Fe3+ ions is less than the O2- ions., , 11. Calculate the number of atoms in an fcc unit cell., , Number of atoms in fcc unit cell =, , Nc, 8, , +, , Nf, 2, , =, , 8, 8, , 6, , + =1+3=4, 2

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67, , 12. Explain AAAA and ABABA and ABCABC type of three dimensional packing with the, help of neat diagram., i., AAAA three dimensional packing:, , •, •, •, •, •, , ii, , •, •, •, •, •, , iii, , It occurs in simple cubic arrangement., This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two, dimensional arrangements in three dimensions., Spheres in one layer sitting directly on the top of those in the previous layer so that all layers are, identical., All spheres of different layers of crystal are perfectly aligned horizontally and also vertically., 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 is 6., ABABA three dimensional packing:, , It occurs in body centeredcubic arrangement., In this arrangement, the spheres in the first layer(a) are slightly separated and the second layer (b) is, formed by arranging the spheres in the depressions between the spheres in first layer (a) ., The third layer is a repeat of the first., This pattern ABABAB is repeated throughout the crystal., In this arrangement, each sphere is in contact with 8 neighbouring spheres- Four in the above layer and, four in the below layer and hence the coordination number is 8., ABCABC three dimensional packing:

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68, , •, •, •, •, •, •, •, , The first layer (a) is formed by arranging the spheres as in the case of two dimensional ABAB, arrangements., The second layer (b)is formed by placing the spheres in the depressions of the first layer., A tetrahedral void(x) and octahedral voids(y) are formed in the first layer., The third layer is 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) . Hence, the third, layer is designated (c)., If the stacking of layers is continued in abcabc...pattern, then the arrangement is called cubic close, packed (ccp) structure., In 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., , 13. Why ionic crystals are hard and brittle?, • Ionic crystal are hard due to strong electrostatic force of attraction between cations and anions., • They are brittle because ionic bonds are non directional., 14. Calculate the percentage efficiency of packing in case of body centered cubic crystal., In bcc, the spheres are touching along the leading diagonal of the cube as shown in the fig., , 𝐼𝑛 ∆𝐴𝐵𝐶, 𝐴𝐶2 = AB2 + BC2, 𝐴𝐶 = √𝐴𝐵 2 + 𝐵𝐶 2, 𝐴𝐶 = √𝑎2 + 𝑎2 = √2𝑎2 = √2 𝑎, In ∆ACG, AG2 = AC2 + CG2, AG = √AC2 + CG 2, 2, , AG = √(√2a) + a2 = √2a2 + a2 = √3a2 = √3 a, But √3 a = 4 r :, , r =, , √3, a, 4, 4, , Volume of the sphere with radius ‘r’ = 3 𝜋𝑟, Number of spheres in bcc unit cell = 2, , 3, , =, , 4, , √3, 𝜋[4, 3, √3, , 3, , a] =, , Total volume of all spheres in bcc unit cell= 2 x 16 𝜋 a3 =, Volume of the cube with edge length a, = a x a x a = a3, Packing fraction =, , √3, 𝜋, 8, , Total volume occupied by all spheres in a unit cell, Volume of unit cell, , √3, 𝜋a3, 16, , a3, , x 100

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69, √3 3, 𝜋a, =8 3 x, a, , 100=, , √3𝜋, 8, , x 100 = √3𝜋 x 12.5= 1.732 x 3.14 x 12.5 = 68%, , 15. What is the two dimensional coordination number of a molecule in square close packed layer?, The two dimensional coordination number of a molecule in square close packed layer is 4. In this, arrangement each sphere is in contact with four of its neighbours., 16. Experiment shows that Nickel oxide has the formula Ni0.96 O1.00 . What fractionof Nickel exists as, of Ni2+ and Ni3+ ions?, Number of Ni2+ ion = x, Number ofNi3+ ion = y, x + y = 0.96, x = 0.96 –y, The charge on Ni2+ ion and charge on Ni3+ ion are balanced by the charge on O2- ion., Therefore,, 2x + 3y = 2, 2(0.96 – y) + 3y = 2, 1.92 – 2y + 3y = 2, 1.92+ y = 2, y = 2 – 1.92 = 0.08, x = 0.96– 0.08= 0.88, Percentage of Ni2+ inNi0.96 O1.00 = 0.88 x 100 / 0.96 =91.67 %, Percentage of Ni3+ inNi0.96 O1.00 = 0.08 x 100 / 0.96 =8.33 %, 17. What is meant by the term “coordination number”? What is the coordination number of atoms in, a bcc structure?, The neighbouring atoms surrounded by each atom is called coordination number. In the, bodycentre, each atom is surrounded by eight nearest neighbours and coordination number is 8., 18. 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., Given:, Edge of bcc (a)= 288pm = 2,88 x 10-8 cm (since density is given as gcm-3, the edge length should, be converted to cm). For bcc, Z = 2, Density = 7.2 gcm-3, Mass of element = 208g, a3 𝜌 x NA, (2.88 x 10−8 )3 x 7.2 x 6.023 x 1023, M =, =, n, 2, 1035.9 x 10−24 x 1023, , =, , 2, mass, , Number of atoms of an element =, , =51.795gmol-1, , x 6.023 x 1023, , atomic mass, , Number of atoms present in 208g =, , 208, 51.795, , x 6.023 x 1023 = 2.418 x 1024 atoms, , 19. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm.Calculate, the edge length of unit cell., Given: r = 125pm., For ccp, , r=, , a√ 2, 4, , =, , a √2, 2x√2x√2, , a = 2√2 r, , =, , 𝑎, 2√2

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70, , = 2 x 1.414 x 125 pm = 353.5 pm, 20. If NaCl is doped with 10-2 mol percentage of strontium chloride, what is the, concentration of cation vacancy?, 100 moles of NaCl is dopped with 10-2 moles of SrCl2, Therefore 1 mole of NaCl is dopped with 10-2 / 100 = 10-4 moles of SrCl2, Each Sr2+ ion will create 1 cation vacancy in NaCl., Numberof cationic vacancy produced by 10-4 mol Sr2+ ion = 6.023 x 1023 x 10-4, = 6.023 x 1019, Number of cation vacancies produced by SrCl2 = 6.023 × 1019 per mol, 21. KF crystallizes in fcc structure like sodium chloride. calculate the distance between K+ and F− in, KF.( given : density of KF is 2.48 g cm−3), Molar mass of KF = 39.1 + 19 = 58.1g/mol, a3=, , nxM, , NA x 𝜌, , =, , 4 x 58.1, , = 15.56 x 10-23 = 1.56 x 10-24, , 6.023 x 1023 x 2.48, , 3, , a = √1.56x 10-8 = 1.1597 x 10-8 cm, Inter ionic distance (d)=, , a, √2, , =, , 1.1597 x10− 8, 1.414, , = 0.8202 x 10-8 = 8.202 x 10-9cm, , 22. An atom crystallizes in fcc crystal lattice and has a density of 10 gcm-3, with unit cell edge length of 100pm. Calculate the number of atoms present in 1 g of crystal., Given:, Density =10 gcm-3: a = 100 pm = 100 x 10 -10cm : Mass = 1g, Number of atoms in fcc unit cell = 4, a3 𝜌 x NA, (100 x 10−10 )3 x 10 x 6.023 x 1023, M =, =, n, 4, −8, (1 x 10 )3 x 10 x 6.023x 1023 6.023, =, =, = 1.505 gmol−1, 4, 4, Number of atoms of an element=, , mass, , atomic mass, , Number of atoms in 1 g of crystal ==, , 1, 1,505, , x 6.023 x 1023, , x 6.023 x1023= 4 x 10 23atoms, , 23. 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?, Number of corner atoms (X) =, , N𝐶, 8, , 8, , = =1, , Number of body centre atoms(Y) =, , 8, N𝑏, 1, , 1, , = =1, 1, , Formula of the compound = XY, 24. Sodium metal crystallizes in bcc structure with the edge length of the unit cell, 4.3 ×10−8cm. Calculate the radius of sodium atom., Given: a = 4.3 ×10−8cm., For bcc, r=, , √3, 4, , 𝑎=, , 1.732 x 4.3 ×10−8, 4, , 25. Explain Frenkel defect, , =1.86 ×10−8 cm

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71, , •, •, •, •, , 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 occurs when cation and anion differ in size., Unlike Schottky defect, this defect does not affect the density of the crystal.Ex: AgBr, , II EVALUATE YOURSELF:, 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?, Given: Density =8.9 gcm-3: a = 352.4 pm = 3.524 x 10 -8cm , Mass = 100g , Z = 4, a3 𝜌 x NA, M =, n, , (3.524x 10−8 )3 x 8.9 x 6.023 x 1023, =, 4, 2345.9x 10−1, =, = 586.5x10−1 = 58.65 gmol−1, 4, , Number of atoms of an element=, , mass, , atomic mass, , Number of atoms in 100g of an element, , x 6.023 x 1023, , =, , 100, , x 6.023 x1023= 10.27x 10 23atoms, , 58.65, , 2. Determine the density of CsCl which crystallizes in a bcc type structure with an edge length 412.1, pm., Given: M=133 + 35.5 =168.5 gmol-1, a= 412.1 pm= 4.121 x 10-8 cm , Z=1, , 𝜌=, , Mxn, a3 x N A, , =, , 168.5 x 1, (4.121x 10−8 )3 x 6.023 x 1023, , Density of CsCl crystal, , =, , 168.5, 42.15, , =3.99, , = 4gcm-3, , 3. A face centered cubic solid of an element (atomic mass 60) has a cube edge of 4A0. Calculate its, density., Given: M=60 gmol-1, a = 4 A0= 4 x 10-8 cm , Z=4, , 𝜌=, , Mxn, a3, , x NA, , =, , 60 x 4, (4 x 10−8 )3 x 6.023 x 1023, , =, , 240, 38.54, , = 6.227gcm-3, , Density of an element = 6.227gcm-3, , III TEXTBOOK EXAMPLES:, 1. 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?, Given: M=137.3 gmol-1, a= 508pm = 5.08x 10-8cm , Z=2, , 𝜌=, , Mxn, a3 x N A, , =, , 137.3x 2, (5.08 x 10−8 )3 x 6.023 x 1023, , =, , 274.6, 78.96, , IV ADDITIONAL QUESTIONS:, 1. How solids are classified?, Solids are classified into two types, (i) Crystalline solids (ii) Amorphous solids..Ex: glass, rubber, , = 3.478gcm-3 = 3.5gcm-3

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72, , Crystalline solids are further classified depending upon nature of their constitutents., (a) Ioniccrystals. Ex: NaCl, KCl, (b) Covalent crystals Ex: Diamond, SiO2, (c) Molecular crystals. Ex: Naphthalene, Anthracene, Glucose, (d) Metallic crystals. Ex: Na,Mg,Au,Ag, (e) Atomic solids. Ex: Frozen elements of group 18, 2. Explain isotropy and anisotropy ?, • Isotropy means having identical values of physical properties (refractive index, electrical, conductance) in all directions. Ex-amorphous solids., • Anisotropy means having different values of physical properties when measured along different, directions. Ex- crystalline solids, 3. Give a note on 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, , .4. What are molecular crystals? How are they classified?, •, •, •, •, , In molecular solids, the constituents are neutral molecules., They are held together by weak Vanderwaals forces., Generally, molecular solids are soft and they do not conduct electricity, Molecular solids are further classified into three types., , 5. Explain metallic solids., • 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,, Cu- Zn etc., , 6. Name the parameters that characterize a unit cell, 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.

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73, , 7. What are primitive and non-primitive unit cell?, A unit cell that contains only one lattice point is called a primitive unit cell,, A unit cell that contains additional lattice points, either on a face or within the unit cellis called anonprimitive unit cells., 8. Sketch i.sc ii.bcc iii.fcc&calculate its numberof atoms per unit cell, i .sc, , Number of atoms in scunit cell =, , ii., , Nc, 8, , =, , 8, 8, , =1, , bcc, , Number of atoms in bcc unit cell =, , Nc, 8, , +, , N𝑏, 1, , 8, , 1, , = +, 8, 1, =1+1=2, , iii.fcc, , Number of atoms in fcc unit cell =, , Nc, 8, , +, , Nf, 2, , 8, , 6, , = +, 8, 2, =1+3=4, , 9. How inter planar distance are calculated in unit cell?, The inter planar distance (d) between two successive planes of atoms can be calculated using the, following equation form the X-Ray diffraction data, 2dsin θ= n λ, The above equation is known as Bragg’s equation, where, λ-wavelength of X-ray used for diffraction. θ-angle of diffraction n – order of refraction, By knowing the values of λ, θ and n we can calculate the value of d., d = nλ/2dsin θ, 10 .Explain the relationship between atomic radius and edge length of fcc unit cell and calculate its, packing efficiency

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74, , In ∆ABC, AC2 = AB2 + BC2, AC = √AB2 + BC2, AC = √𝑎2 + 𝑎2 = √2𝑎2 = √2 a, From the figure, AC = 4r, 4r = a√2, , r=, , a√2, 4, , Volume of the sphere with radius ‘r’ =, 4, , 2√2, , 4, , 𝜋𝑟 3 =, 3, , 4, , =4, , Total volume of all spheres in fcc unit cell = 4 x, Volume of the cube with edge length a = axax a = a3, Packing fraction =, , =, , =, , √2𝜋, 6, , 3, , √2, , =3 𝜋 64 𝑎3 = 24 𝜋 a3, Total number of spheres in fcc unit cell, , √2, 𝜋a3, 6, x, a3, , √2, , 𝜋[4 a], 3, , √2, 𝜋, 24, , a3, , Total volume occupied by all spheres in a unit cell, Volume of unit cell, , =, , √2, 𝜋a3, 6, , x 100, , 100, , x 100 =, , 1.414 x 3.14 x 100, 6, , = 74 %, , 11. Calculate the packing efficiency of polonium., Polonium crystallizes in simple cubic pattern. So let us consider simple cubic system., Let us consider a cube with an edge length ‘a’ as shown in fig.

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75, , Volume of the cube with edge length a is = a x a x a = a3, Let ‘r’ is the radius of the sphere., From the figure, a=2r, r=a/2, 4, , Volume of the sphere with radius ‘r’ =, , 3, , Total number of spheres in sc unit cell, , πr 3 =, , 3, , a 3 4, , a3, , 2, , 8, , 𝜋[ ] = 𝜋, 3, , πa3, 6, , x 1=, , =, , 100π, 6, , 6, , 6, , Volume of unit cell, , 100=, , πa3, , πa3, , Total volume occupied by all spheres in a unit cell, , πa3, 6, x, a3, , =, , =1, , Total volume of all spheres insc unit cell =, Packing fraction =, , 4, , =, , 100 x 3.14, 6, , x 100, , = 52.33%, , 12. Radius ratio of an ionic solid is found to be 0.415. Where the cations are occupied?, Cations occupies the octahedral voids., {Hint: radius ratio <0.4 – tetrahedral voids, radius ratio >0.4 – octahedral voids}, 13. Write the possible octahedral voids and tetrahedral voids exist per atom in a crystal., Number of tetrahedral voids = 2, Number of octahedral voids = 1, 14. How electrical neutrality is maintained in stoichiometric ionic crystals?, 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., 15. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as impurity, in it. (Impurity defect), , •, , Due to the presence of impurity, ions in ionic solids causes vacancies in the crystal, lattice of the host., • For example, addition of CdCl2 to silver chloride yields solid solutions where the, divalent cation Cd 2+ occupies the position of Ag+., • This disturbs 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., 16. What happens when ZnO is heated?, • 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., 17. Why Frenkel defect is not found in alkali halides?, Larger size of alkali metal ions dose not allow them to fit in interstitial sites., 18. Schottky defect lowers the density of the ionic solid. Why?, The total number of ions in a crystal with this defect is less than the theoretical value of ions. Thus the, density of the solid crystal is less than expected., 19. How F centers are formed when NaCl crystals are heated in the presence of sodium vapour, excess, of Na+ ions are observed. Justify your answer., • When NaCl crystals are heated in the presence of sodium vapour, Na+ ions are, formed and are deposited on the surface of the crystal.

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76, , •, •, , Chloride ions (Cl-) diffuse to the surface from the lattice point and combines with Na+, 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 asNa1+ x Cl, 20. Give the relationship between atomic radius (r), inter atomic distance (d), and packing efficiency of the unit cell of a cubic crystal., UNIT, CELL, SIMPLE, CUBIC, BODY, CENTERED, CUBIC, FACE, CENTERED, CUBIC, , NO; OF, ATOMS, , COORDINATION, NUMBER, , 1, , 6, , 2, , 8, , INTER ATOMIC, DISTANCE (d), d=a, d =, , √3, 𝑎, 2, , = 0.866 a, , 4, , 12, , d=, = 0.7076 a, , 𝑎, √2, , ATOMIC, RADIUS (r), r=, r=, , 𝑎, , PACKING, EFFICIENCY, 52.33%, , 2, , √3, 𝑎, 4, , 68%, , = 0.433 a, r=, , 𝑎, 2√2, , = 0.3536 a, , 74%

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77, UNIT-7, CHEMICAL KINETICS, , I., , ANSWER THE FOLLOWING QUESTIONS, , 1., , Define average rate and instantaneous rate., Average rate, , Instantaneous rate, , It is the rate between the intervals of time, , It is the rate at a particular time., Average rate becomes instantaneous rate when t→ 0, , 2., , Define rate law and rate constant., Rate Law, , → products, aA + bB ⎯⎯, rate [A]x [B]y, rate = k [A]x [B]y, Rate is given in terms of molar concentration of reactants raised to the power which may or, may not equal to stoichiometric coefficient., Rate constant, Rate constant is same as rate of reaction when concentration of all the reactants is unity., 3., , → product., Derive integrated rate law for a zero order reaction A ⎯⎯, Rate is independent of the concentration of reactant is called zero order reaction., Consider a general zero order reaction.A, , ⎯⎯, → Product, , At t = 0 concentration of reactant, , =, , At time t, concentration of reactant left =, , [A0], [A], , rate = k[A]0, , −, , d[A], = k, dt, , – d[A] = k dt, Integrate between limits of [A0] at t = 0 and [A] at time 't'., [A], , −, , , , [A0 ], , t, , d[A] = k  dt, 0

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78, , − ([A])[A, , [A], , = k(t)0t, , 0], , [A0] – [A] = k, , k=, 4., , [A]0 − [A], t, , Define half life of a reaction. Show that for a first order reaction half life is independent, of initial concentration., Time required for the reactant concentration to reach one half its of initial value is called, half life of a reaction., First order reaction half-life, , k=, , [A ], 2.303, log 0, t, [A], , when t = t1/2 then [A] =, , k=, , [A 0 ], 2, , [A ], 2.303, log 0, [A0 ], t1/2, 2, , t1/2 =, , 2.303, log 2, k, , t1/2 =, , 2.303,  0.3010, k, , t1/2 =, , 0.693, k, , No concentration terms involved, so half life is independent on initial concentration of reactant., 5., , What is an elementary reaction? Give the differences between order and molecularity, of a reaction., Each and every single step in a reaction mechanism is called an elementary reaction.

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79, No, 1., , 2., , Order of a reaction, , Molecularity of a reaction, , 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, , It can be zero (or) fraction (or) Integer, , It is always whole number, cannot be, , that are involved in an elementary step., , zero or fraction., 3., , It is assigned for a overall reaction, , It is assigned for each elementary step, of mechanism., , 6., , Explain the rate determining step with an example., I−, , Consider a reaction 2H 2O2 ⎯⎯→ 2H 2O + O2, This reaction takes place in two steps, Step – 1, , → H2O(l) + OI –(aq), H2O2(aq)+ I –(aq) ⎯⎯⎯, , Step – 2, , → H2O(l) + I –(aq) + O2(g), H2O2(aq)+ OI –(aq) ⎯⎯, , slow, RDS, , → 2H2O(l) + O2(g), Overall reaction 2H2O2(aq) ⎯⎯, , Step – 1 is a slow step and rate determining step. In this step H2O2 and I– are involved,, hence it is bimolecular reaction., 7., , Describe the graphical representation of first order reaction., , 1 [A ], k = ln 0, t [A], kt = ln [A0] – ln[A], , Intercept =ln [A0], ln [A], , slope = – k, , ln [A] = – kt + ln [A0], y = – mx + C, Plot of ln [A] Vs t gives straight line with negative slope., Slope equal to –k and intercept equals to ln[A]., 8., , Write the rate law for the following reactions., (a) A reaction that is 3/2 order in x and zero order in y., , t

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80, (b) A reaction that is second order in NO and first order in Br2., (a) rate = k [X]3/2 [Y]0, rate = k [X]3/2, (b) rate = k [NO]2 [Br2]1, 9., , Explain the effect of catalyst on reaction rate with an example., 1. A catalyst is a substance which alters the rate of a reaction without itself undergoing, any permanent chemical change., 2. It may participate in a reaction but regenerated at the end of reaction., 3. A catalyst increases rate of reaction by lowering the activation energy., MnO2, ⎯⎯⎯→, , Example : 2KClO3, , 2KCl +3O2, , Potential, , MnO2 acts as catalyst., , Ea, Without, , Energy, , catalyst, , Reactant, s, , 10., , Ea, With, catalyst, , Product, , Reaction progress, →, , The rate law for a reaction of A, B and L has been found to be rate = k [A]2 [B] [L]3/2., How would the rate of reaction change when, (i) Concentration of [L] is quadrupled, (ii) Concentration of both [A] and [B] are doubled, (iii) Concentration of [A] is halved, (iv) Concentration of [A] is reduced to (1/3) and concentration of [L] is quadrupled., , (i), , → (1), rate = k [A]2 [B] [L]3/2 ⎯⎯, , (ii), , → (3), (ii) x rate = k [2A]2 [2B] [ L]3/2 ⎯⎯, , → (2), x rate = k [A]2 [B] [4L]3/2 ⎯⎯, 2, , k [A]2 [B][4 L ]3/2, (2), x rate, , =, 3/2, (1), rate, k [A]2 [B] [L], , k [2 A] [2 B] [L]3/2, (3), x rate, , =, (1), rate, k [A]2 [B] [L]3/2, x = (2)2 (2), x=8, , x=4, , 3/2, , 3, , x = (4), , 1, 2, , Rate increase by 8 times

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81, , x=8x, , 1, = (64) 2, , Rate increase by 8 times, (iii), , 2, , A, 3/ 2, ⎯⎯, →(4), x rate = k   [B][L], 2, 2, , A, k   [ B ][ L]3/2, 2, (4), x rate, , =  , (1), rate, k [ A ]2[ B ][ L]3/2, 1, x = , 2, , Rate decrease by, , 11., , 2, , A, 3/ 2, ⎯⎯, →(5), x rate = k   [B][4L], 3, 2, , A, k   [ B ][ 4L]3/2, 3, (5), x rate, , =  , (1), rate, k [ A ]2[ B ][ L]3/2, 2, , 1, x =   (4)3/2, 3, , 2, , x=, , (iv), , 1,  (8), 9, , 1, 4, , 1, times, 4, , x =, , Rate decrease by, , 8, times, 9, , The rate of formation of a dimer in a second order reaction is 7.5 × 10 – 3 mol L – 1 s – 1, at 0.05 mol L – 1 monomer concentration. Calculate the rate constant., rate = k [monomer]2, , 7.5 10−3 = k  (0.05)2, 1, 75, 7.5  10−3, 7.5  10−3, 7.5  10 −3 7.5  10, =, = 3 molL−1s −1, =, k=, =, =, 2, −2 2, −4, 25, 25, 25  10, (0.05), (5  10 ), , 12., , → products the rate law is given by rate = k [X3/2 [Y]1/2 ., For a reaction X + Y + Z ⎯⎯, What is the overall order of the reaction and what is the order of the reactionwith respect to z., , → product, X + Y + Z ⎯⎯, rate = k [X]3/2 [Y]1/2, order w.r.to Z = 0, overall order =, , 3 1, + =2, 2 2

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82, 13.Explain briefly the collision theory of bimolecular reactions. (or) Derive k = pz, , − Ea, e RT, , Consider a bimolecular reaction between A2 and B2 proceeds through collision betweenthem, which is proportional to number of collision per second, Rate Collision rate, Collision rate [A2] [B2], Collision rate =Z [A2] [B2], , Z is the collision Frequency, , Collision rate in gas phase reaction is calculated from kinetic theory of gas and its value is, 109collision per seconds at 298K and 1 atm pressure., All these collisions are not effective and in order for reaction to takes place, the colliding, molecules must possess activation energy., , Fraction of effective collision (f) is given by, , f =e, , −, , Ea, RT, , In addition to activation energy, orientation factor(p) also necessary for reaction to take place., rate = p  f  collision rate, rate = p  e, , −, , Ea, RT  Z [A2] [B2], , ⎯⎯, → (1), , → (2), As per rate lawrate = k [A2] [B2] ⎯⎯, comparing (1) and (2), , k = pze, 14., , −, , Ea, RT, , Write Arrhenius equation and explains the terms involved., , Arrhenius equation, , k = Ae, , −, , Ea, RT, , k is rate constant, A is Frequency factor, Ea is Activation energy( Jmol-1), T is Temperature in Kelvin, R is gas constant (8.314 JK–1 mol–1)

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83, 15., , The decomposition of Cl2O7 at 500 K 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., , k=, , [A ], 2.303, log 0, t, [A], , k=, , 2.303, 2.303, 0.08 2.303, 0.6932,  0.3010 =, log, =, log 2 =, 60, 60, 0.04, 60, 60, , = 0.011553 s – 1, , k = 1.153 10 – 2 s –1, 17., , Explain pseudo first order reaction with an example., A second order reaction can be altered to a first order reaction by taking one of the reactants in, , large excess, such reaction is called pseudo first order reaction., Example :, , Acid hydrolysis of ester, +, , CH3COOCH3 + H2O, 18., , H, CH3COOH + CH3OH, ⎯⎯→, , Identify the order for the following reactions, (i) Rusting of Iron, (ii) Radioactive disintegration of 92U238, , → products ; rate = [A]1/2 [B]2, (iii) 2A + 3B ⎯⎯, (i) First order, (ii) First order, (iii), 19., , 1, 1+ 4 5, +2 =, =, 2, 2, 2, , 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, Ea = 200 kJ mol – 1, A = 1.6  1013 s – 1, T = 600 K andR = 8.314 J K –1 mol–1, k=?, , −40.09, , = 3.8 × 10 – 18)

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84, , log k = log A −, , Ea, 2.303RT, , = log1.6 1013 −, , 2000 0 0, 2.303  8.314  6 0 0, , = log1.6  1013 −, , 2000, 2.303  8.314  6, , = log1.6  1013 −, , 2000, 114.88, , = log1.6 1013 − 17.409, = log 1.6 + 13 – 17.409, = 0.2041 + 13 – 17.409, = 13.2041 – 17.409, log k = – 4.2049, = – 4 – 0.2049 + 1 – 1, = – 5 + 0.7951, = = 5.7951, , k = Antilog 5.7951, k = 6.238 10−5 s−1, 20., , → L. Find the rate law from the following data., For the reaction 2x + y ⎯⎯, [x], , [y], , rate, , (min), , (min), , (M s – 1), , 0.2, , 0.02, , 0.15, , 0.4, , 0.02, , 0.30, , 04, , 0.08, , 1.20, , rate = k [X]a [Y]b, , → (1), 0.15 = k (0.2)a (0.02)b ⎯⎯, → (2), 0.30 = k (0.4)a (0.02)b ⎯⎯

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85, , → (3), 1.20 = k (0.4)a (0.08)b ⎯⎯, , (2), 0.30 k(0.4)a (0.02) b, , =, (1), 0.15 k(0.2)a (0.02) b, , 2 = 2a, order w.r.to X = 1, , (3), 1.20 k(0.4)a (0.08) b, , =, (2), 0.30 k(0.4)a (0.02) b, 4 = 4b, b=1, order w.r.to y = 1, rate law is rate = k [X]1 [Y]1, 21., , How do concentrations of the reactant influence the rate of reaction?, Rate of reaction increases with increases of concentration of reactant., rate[Reactant], As reactant concentration is more, which leads to more collision of reactant molecules which, increases the rate of reaction., , 22., , How do nature of the reactant influence rate of reaction. (or), , Titration between potassium per mangate and oxalic acid is carried out at 60C where as, titration between potassium per manganate and ferrous ammonium sulphate at room, temperature. Give reason., Chemical reaction involves bond breaking and bond formation. The net energy involved in, this process depends on nature of reactants and hence rate differs for different reactants., For example, titration between KMnO4 vs FAS takes place at room temperature whereas, titration between KMnO4 vs Oxalic acid is heated to 60C. This is because oxidation of oxalate ion, by KMnO4 is slow compared to reaction between KMnO4 and Fe2+., 23., , The rate constant for a first order reaction is 1.54 × 10 – 3 s – 1. Calculate its half life time., k = 1.54  10–3 s–1, t1/2 = ?

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86, , 0.693, 0.693  103 693, 0.693, =, =, =, =, 1.54, k, 1.54, 1.54 10−3, , t1/2, , t1/2 = 450 sec, 24., , The half life of the homogeneous gaseous reaction SO2Cl2 ⎯⎯, → SO2 + Cl2 which obeys first, orderkinetics is 8.0 minutes. How long will it take for the concentration of SO2Cl2 to be, reduced to 1% of the initial value?, , t1/2 = 8 min, [A0] = 100, [A] = 1% of initial value, [A] =, , 1,  100 = 1, 100, , k=, , 0.693, t1/2, , k=, , 0.693, min −1, 8, , k=, , [A ], 2.303, log 0, t, [A], , 0.693 2.303, 100, =, log, 8, t, 1, , 0.693 2.303, =, 2, 8, t, 0.693 4.606, =, 8, t, t=, , 4.606  8 36.848, =, 0.693, 0.693, t = 53.17 min, , t=?

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87, 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., (i) t1/2 = 60sec, (ii)t = 180 sec, , k=?, , [A] = ?, , [A0] = 100, (i) t1/2 =, , k=, , 0.693, k, , 0.693 0.693, =, = 0.01155 sec−1, t t1/2, 60, , (ii) k =, , [A ], 0.693, log 0, t t1/2, [A], , 0.01155 =, , [A ], 2.303, log 0, 180, [A], , log, , 100 0.01155 180, =, [A], 2.303, , log, , 100 2.079, =, [A] 2.303, , log, , 100, = 0.9027, [A], , 100, = Anti log 0.9027, [A], , 100, = 7.993, [A], [A] =, , 100, 7.993, [A] = 12.5%, , 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?, (i) [A0] = 100, t = 20 min, , [A] = 80, k=?

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88, (ii) t = ?, , (i) k =, , [A] = 20, , [A 0 ] − [A] 100 − 80 20, =, =, 20, 20, t, , k = 1 mol L– 1min – 1, , (ii) k =, , 1=, , [A 0 ] − [A], t, , 100 − 20, t, , t = 80 min, 27., , The activation energy of a reaction is 225 k cal mol – 1 and the value of rate constant at 40C, is1.8 × 10 – 5 s – 1. Calculate the frequency factor A., , E a = 225 k cal.mol−1 = 225000 cal .mol–1, k = 1.8  10–5 s–1, R = 1.987 cal K –1 mol –1, T = 40C = 40 + 273 = 313 K, A=?, log k = log A –, , Ea, 2.303RT, , log 1.8  10 – 5 = log A –, , 22500, 2.303  1.987  313, , log 1.8  10 – 5 = log A –, , 22500, 1432, , log 1.8  10 – 5= log A – 15.7089, log 1.8  10 – 5= log A – 15.7089, log 1.8 – 5 = log A – 15.7089, 0.2553 – 5 = log A – 15.7089

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89, log A = 15.7089 + 0.2553 – 5, log A = 10.9642, A = Antilog 10.9642, , A = 9.208 1010 collisons s –1, 28., , Benzene diazonium chloride in aqueous solution decomposes according to the equation, , → C6H5Cl + N2. Starting with an initial concentration of 10 g L – 1,, C6H5N2Cl ⎯⎯, the volume of N2 gas obtained at 50C at different intervals of time was found to be as under:, t (min) :, Vol. of N2 (ml ):, , 6, , 12, , 18, , 24, , 30, , , , 19.3 32.6 41.3 46.5 50.4 58.3, , Show that the above reaction follows the first order kinetics. What is the value of the rate constant?, , k=, , 2.303, V, log, t, V − Vt, , (i) k =, , 2.303, 58.3 2.303, 2.303, 2.303, 58.3, =, log, =, log1.495 =,  0.1746 = 0.067 min −1, log, 6, 39, 6, 6, 6, 58.3 − 19.3, , (ii) k =, , 2.303, 2.303, 58.3 2.303, 2.303, 58.3, =, log, =, log 2.268 =, log,  0.3556 = 0.0682 min −1, 12, 25.7, 12, 12, 58.3 − 32.6, 12, , (iii) k =, , 2.303, 2.303, 58.3 2.303, 2.303, 58.3, =, log, =, log 3.429 =, log,  0.5351 = 0.0685min −1, 18, 17, 18, 18, 58.3 − 41.3, 18, , (iv) k =, , 2.303, 58.3 2.303, 2.303, 2.303, 58.3, =, log, =, log 4.9407 =,  0.6938 = 0.666 min −1, log, 24, 11.8, 24, 24, 24, 58.3 − 46.5, , Average k = 0.0676 min –1 and as follows first order kinetics., 29., , From the following data, show that the decomposition of hydrogen peroxide is a reaction of the, first order :, t (min), V (ml ), , 0, , 10, , 20, , 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.

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90, , k=, , V, 2.303, log 0, t, Vt, , (i) k =, , 2.303, 2.303, 46.1 2.303, =, log1.5469 =,  0.1895 = 0.0436 min −1, log, 10, 10, 10, 29.8, , (ii) k =, , 2.303, 46.1 2.303, 2.303, log, =, log 2.3886 =,  0.3781 = 0.0435min −1, 20, 19.3, 20, 20, , Average k = 0.04355min –1 and as follows first order kinetics., 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?, (i) t = 50 min, , [A0] = 100, , (ii) [A] = 20, , t=?, , (i) k =, , [A] = 60, , k=?, , [A ] 2.303, 2.303, 2.303, 100 2.303,  0.2219 =, log, log 0 =, =, log1.667 =, 50, 50, 60, 50, t, [A], , 0.01022 min – 1, (ii) k =, , 0.01022 =, , 2.303, 2.303, 100, 0.01022 =, log, log 5, t, t, 20, , 0.01022 =, , 2.303,  0.6990, t, , t=, , II., , [A ], 2.303, log 0, t, [A], , 2.303, 1.61,  0.6990 =, t = 157.51 min, 0.01022, k, , EXAMPLES, Example-1 (Page 211), Consider the oxidation of nitric oxide to form NO2, , → 2NO2 (g), 2NO(g) + O2(g) ⎯⎯, (a) Express the rate of the reaction in terms of changes in the concentration of NO, O2 and NO2., (b) At a particular instant, when [O2] is decreasing at 0.2 mol L–1 s–1 at what rate is [NO2], increasing at that instant?

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91, (a), , −, , 1 d[NO], d[O 2 ] 1 d[NO 2 ], =−, =, 2 dt, dt, 2 dt, , (b), , −, , d[O 2 ] 1 d[NO 2 ], =, dt, 2 dt, , 0.2 =, , 1 d[NO 2 ], 2 dt, , d[NO 2 ], = 0.2  2 = 0.4 mol  L−1s −1, dt, Example-2 (Page 211), What is the order with respect to each of the reactant and overall order of the following reaction?, −, −, +, → 3Br2(l) + 3H2O(l), (a) 5Br (aq) + BrO3 (aq) + 6H (aq) ⎯⎯, , The experimental rate law is Rate = k [Br–] [BrO3] [H+]2, , , → CH 4 (g) + CO(g) the experimental rate law is Rate = k[CH3CHO]3/2, (b) CH 3CHO(g) ⎯⎯, (a)Order w.t.to Br – is 1, −, , Order w.t.to BrO 3 is 1, Order w.t.to H+ is 2, Overall order is 1 + 1 + 2 = 4, (b)order = 3/2, , Example-3 (Page 211), The rate of the reaction x + 2y → product is 4  10 –3 mol L –1 s –1, if [x] = [y] = 0.2 M and, rate constant at 400 K is 2  10 –2 s –1, What is the overall order of the reaction., rate = k [X]9 [Y]6, 4  10 –3 = 2  10 –2 (0.2)a (0.2)b, , 4 10−3, 2 10, , −2, , = (0.2)a + b, , 2 10−1 = (0.2)a + b, 0.22 = (0.2)a + b

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92, , Comparing the powers a + b = 1, Overall order = a + b = 1, Example-4 (Page 216), A first order reaction takes 8 hours for 90% completion. Calculate the time required for, 80% completion. (log 5 = 0.6989 ; log 10 = 1), t = 8 hr, [A0] = 100, [A] = 10, t = ?for [A] = 20 (80% completion), k=, , [A ] 2.303, 2.303, 100 2.303, log 0 =, =, log, log10, 8, 8, 10, t, [A], , k=, , 2.303 −1, hr, 8, , k=, , [A ], 2.303, log 0, t, [A], , 2.303 2.303, 100, =, log, 8, t, 20, 2.303 2.303, =, log 5, 8, t, , 2.303 2.303, =,  0.6989, 8, t, 2.303 1.6096, =, 8, t, , t=, , 1.6096 12.876, =, 2.303, 2.303, , t = 5.59 hr, , Example-5 (Page 216), The half life of a first order reaction x → products is 6.932  104 s at 500 K. What, percentage, of x would be decomposed on heating at 500 K for 100 min. (e0.06 = 1.06)

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93, , t1/2 = 6.932  104 s, t = 100 min = 100  60 = 6000 s, t1/ 2 =, , 0.693, k, , k=, , 0.693 10−1 0.693 10−1, 0.693, 0.693, = 10−5 s −1, =, =, =, 4, 4, 4, t1/ 2, 6.932  10, 6.932 10, 6.932 10, , k=, , [A ], 2.303, log 0, t, [A], , 10−5 =, , [A ], 2.303, log 0, 6000, [A], , log, , [A 0 ] 6000 10−5 6  103  10 −5 6 10−2, =, =, =, = 2.605  10 –2, 2.303, 2.303, [A], 2.303, , log, , [A 0 ], = 0.02605, [A], , [A0 ], = Anti log(0.02605), [A], [A 0 ], = 1.0618, [A], 100, = 1.0618, [A], [A] =, , 100, 1.0618, , [A] = 94179, Left out concentration of reactant x = 94.179 mol .L –1. Reacted amount = 5.82%, Example-6 (Page 217), 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., t99.9%, k=, , [A ], 2.303, log 0, t, [A]

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94, t99.9% =, , 100, 2.303, log, 0.1, k, , t99.9% =, , 2.303, log 1000, k, , t99.9% =, , 2.303, ×3, k, , t 99.9% =, , t 50% =, , From, , 6.909, ⎯⎯, → (1), k, , 0.693, k, , ⎯⎯, → (2), , (1) t 99.9% 6.909, k, ,, =, ×, 0.693, k, t 50%, (2), , t 99.9%, = 10, t 50%, , t 99.9% = 10 t 50%, Hence proved., , Example-7 (Page 221), The rate constant of a reaction of a reaction at 400 and 200 K are 0.04 and 0.02 s –1, respectively. Calculate the value of activation energy., T1 = 200 K, , k1 = 0.02 s –1, , T2 = 400 K, , k1 = 0.04 s –1, , R = 8.314 JK –1mol –1, E a= ?, , log, , Ea  T2 − T1 , k2, =, , , k1 2.303 R  T1T2 , , log, , Ea, 0.04,  400 − 200 , =, , , 0.02 2.303  8.314  400  200 , , log 2 =, , Ea, 20 0, , 2.303  8.314 800 0 0

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95, , 0.3010 =, , Ea, 2, , 2.303  8.314 800, , 0.3010 =, , Ea, 1, , 2.303  8.314 400, , Ea = 2.303  8.314  400  0.3010, E a = 2.305 J mol−1, Example-8 (Page 221), Rate constant k of a reaction varies with temperature T according to the following, Arrhenius equation, log k = log A, , Ea  1 ,  , 2.303R  T , , Where Ea is the activation energy. When a graph is plotted for log k Vs, , 1, astraight line, T, , with a slope of –4000 K is obtained. Calculate the activation energy, Slope = –4000 K, E a= ?, slope = −, , Ea, 2.303R, , − 4000K = −, , Ea, , 2.303  8.314 JK −1mol−1, , Ea = 2.303  8.314 J K −1 mol−1  4000 K, E a = 76589 J mol−1, Solved Problem in page 209, For the reaction 2NO(g) + O 2(g) ⎯⎯, → 2NO(g) the following data were obtained., Expt., , [NO]  10 –2 (mol, L –1), , [O2]  10 –2 (mol, L –1), , Initial rate  10 –2 (mol L –, 1 –1, s ), , 1., , 1.3, , 1.1, , 19.26

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96, , 2., , 1.3, , 2.2, , 38.40, , 3., , 2.6, , 1.1, , 76.80, , Determine order with respect to NO, O2 and overall order., rate = k[NO]x [O 2 ]y, , 19.26  10−2 = k(1.3  10−2 ) x (1.1 10−2 ) y ⎯⎯, → (1), 38.40  10−2 = k(1.3  10−2 ) x (2.2  10−2 ) y ⎯⎯, → (2), 76.80  10−2 = k(2.6  10−2 ) x (1.1 10−2 ) y ⎯⎯, → (3), −2, , (2), 38.40  10, , =, (1), 19.26  10−2, , k (1.3  10−2 ) x (2.2  10−2 ), , y, , k (1.3  10−2 ) x (1.1 10−2 ), , y, , 2 = 2x, x=1, order w.r.to, , NO = 1, y, , k (2.6  10−2 ) x (1.1 10−2 ), (3), 76.80  10−2, , =, y, −2, (1), 19.26  10, k (1.3  10−2 ) x (2.1 10−2 ), 4 = 2x, x=2, order w.r.to, , O2 = 2, , rate = k [NO]1 [O2]2, Overall order = 1 + 2 = 3, , III, , EVALUATE YOURSELF, , Evaluate Yourself-1 (Page 211), (1) Write the rate expression for the following reactions, assuming them as elementary, reactions., (i) 3A + 5B2 ⎯⎯, → 4CD, , (ii) X2 + Y2 ⎯⎯, → 2XY

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97, , (2) Consider the decomposition of N2O5(g) to form NO2(g) and O2(g). At a particular, instant N2O5 disappears at a rate of 2.5  10 –2 mol dm –3 s –1. At what rate are NO2, and O2 formed? What is the rate of the reaction?, (1) (i) rate = k [A]3 [B2]5, (ii)rate = k [X2]1 [Y2]1, (2) 2N2O5 ⎯⎯, → 4NO2 + O2, −, , 1 d[N 2O5 ] 1 d[NO 2 ], =, 2, dt, 4 dt, , 1, 1 d[NO 2 ],  2.5  10−2 =, 2, 4 dt, , d[NO 2 ] 4, =  2.5  10−2 = 2  2.5  10 –2 = 5  10 –2 mol L –1 s –1, dt, 2, −, , 1 d[N 2O5 ] d[O 2 ], =, 2, dt, dt, , 1, d[O 2 ],  2.5  10−2 =, 2, dt, d[O2 ], = 1.25  10−2 mol L−1s −1, dt, , Evaluate Yourself-2 (Page 212), → product ; quadrupling [x], increases the rate by a factor of 8., (1) For a reaction, X + Y ⎯⎯, , Quadrupling both [x] and [y], increases the rate by a factor 16. Find the order of the, reaction with respect to x and y. What is the overall order of the reaction?, (2) Find the individual and overall order of the following reaction using the given date, 2NO(g) + Cl2(g) ⎯⎯, → 2NOCl(g) ., , Expt., , [NO], , [Cl2], , Initial, rate, , 1., , 0.1, , 0.1, , 7.8  10 –, 5, , 2., , 0.2, , 0.1

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98, , 3., , 0.2, , 0.3, , 3.12  10, –4, , 9.36  10, –4, , → (1), (1)rate = k [X]a [Y]b ⎯⎯, , → (2), 8 rate = k [4X]a [Y]b ⎯⎯, → (3), 16 rate = k [4X]a [Y]b ⎯⎯, a, b, (2) 8 rate k [4 X ] [ Y], , =, (1), rate, k [ X ]a [ Y]b, , 8 = 4a, 23 = 22a, 3 = 2a, a = 3/2, order w.r.to X = 3/2, a, b, (3) 16 rate k [ 4X ] [ 4Y], , =, (2), 8rate, k [ 4X ]a [ Y]b, , 2 = 4b, b = 1/2, order w.r.to Y = 1/2, (2) rate = k[NO] [Cl2 ], x, , y, , 7.8  10−5 = k(0.1) x (0.1) y ⎯⎯, → (1), , 3.12  10−4 = k(0.2) x (0.1) y ⎯⎯, → (2), 9.36  10−4 = k(0.2) x (0.3) y ⎯⎯, → (3), y, , x, (2), 3.12 10−4 k (0.2) ( 0.1), , =, y, (1), 7.8 10−5, k (0.1) x ( 0.1), , 4 = 2x

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99, , x=2, order w.r.to, , NO = 2, , x, y, (3), 9.36  10−4 k (0.2) (0.1), , =, (2), 3.12  10−4 k (0.2) x (0.3) y, , 1, 3= , 3, , y, , 3 = (3−1 ) y, , 3=3, Equating the powers, 1 = –y, y = –1, order w.r.to O2 = 1, rate = k [NO]2 [O2]–1, overall order = 2 – 1 = 1, Evaluate Yourself-3 (Page 217), (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?, (2) The rate constant for a first order reaction is 2.3  10 –4 s –1. If the initial concentration, of the reactant is 0.01 M. What concentration will remain after 1 hour?, (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., Time (min), , 0, , 30, , 60, , 90, , Ester concentration mol L –1, , 0.85, , 0.80, , 0.754, , 0.71, , Show that, the reaction follows first order kinetics.

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100, , (1), , t = 40 min, [A0] = 100, [A] = 40, , t1/2 = ?, k=, , [A ] 2.303, 2.303, 2.303, 0.9163, 100 2.303, log 0 =, =,  0.3979 =, log, log 2.5 =, 40, 40, 40, 40, 40, t, [A], , k = 0.0229 min −1, , t1/2 =, , 0.693 0.693, =, 0.0229, k, , t1/2 = 3.026 min, , t1/2 = 3min, (2) k = 2.3  10 –4 s –1, [A0] = 0.01 M, t = 1 hour = 3600 s, [A] = ?, k=, , [A ], 2.303, log 0, t, [A], , 2.3 10−4 =, , 2.303, 0.01, log, 3600, [A], , log, , 0.01 2.3 10−4  3600, =, = 3600  10 –4 = 36  10210 –4 = 36  10 –2, [A], 2.303, , log, , 0.01, = 0.36, [A], , 0.01, = Anti log 0.36, [A], 0.01, = 2.29, [A], [A] =, , 0.01, = 0.0044 M = 4.4  10 –3 M, 2.29

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101, , (3) k =, , V, 2.303, 2.303, 2.303, 0.0606, 0.85 2.303, log 0 =, =, log1.0625 =,  0.0263 =, log, = 0.002019 min –, 30, 30, 30, 30, 0.80, t, Vt, , 1, , k = 2.019  10−3 min −1, , k=, , 2.303, 2.303, 2.303, 0.05918, 0.85, =, log1.061 =,  0.0257 =, log, 60, 60, 60, 30, 0.754, , = 0.001972, k = 1.972 10−3 min −1, k=, , 2.303, 0.18009, 2.303, 0.85 2.303, =, log1.1972 =,  0.782 =, log, = 0.002001min −1, 90, 90, 90, 90, 0.71, , k = 2.00110−3 min −1, , As all k values are nearly same, ester hydrolyzing follows first order kinetics., Evaluate Yourself-4(Page 221), For a first order reaction the rate constant at 500 K is 8  10 –4 s –1. Calculate the frequency, factor, if the energy of activation for the reaction is 190 kJ mol –1., T = 500 K, k = 8  10 –4 s –1, A=?, Ea = 190 kJ mol –1 = 190000 J mol –1, R = 8.314 JK–1 mol –1, log k = log A −, , Ea, 2.303 RT, , log8 10−4 = log A −, , 1900 0 0, 2.303  8.314  5 0 0, , log 8 10−4 = log A −, , 1900, 95.735, , log 8  10 –4 = log A – 19.846, log 8 + log 10 –4 = log A – 19.846, log 8 – 4 = log A – 19.846

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102, , 0.9031 – 4 = log A – 19.846, log A = 0.9031 – 4 + 19.846, = 20.7491 – 4, log A = 16.7491, A = Antilog 16.7491, , A = 5.6117 1016 collision s −1, , IV, , ADDITIONAL QUESTIONS WITH ANSWERS, , 1., , Define rate of a reaction., Rate is defined as change in concentration of reactant or product with respect to time., , 2., , Give the unit of rate of reaction for (i) aqueous reaction, (i) mol L–1 s–1, , 3., , (ii) Gaseous reaction, , (ii) atm s–1, , How to determine rate of a reaction., Draw a plot of concentration vs time graph. The slope of the curve between the interval of time, gives the average rate., Instantaneous rate is the rate at a particular time which is calculated by drawing a tangent at that, point on the concentration vs time graph. The slope of the tangent gives the instantaneous rate., 0.5, , 0.5, , 0.4, , Slope, , Conc 0.3, 0.2, , Conc 0.3, 0.2, , Average Rate, , 0.1, , 4., , Instantaneous Rate, , 0.1, 0, , S, , Slope, , 0.4, , 5, , 10 15 20, Time, 0, , 25, , Differentiate rate and rate constant., , 0, , 5, , 10 15 20, Time, 0, , No, , Rate of a reaction, , 1., , It represents the speed at which the, reactants are converted intoproducts at any, instant., , It is a proportionality, constant, , 2., , It is measured as decrease in the, concentration of the reactants or increase in, the concentration of, , It is equal to the rate of, reaction,, , 25, , Rate constant of a reaction, , when the concentration of

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103, products., , each, of the reactants in unity., , 3., , It depends on the initial on the, concentrations of reactants., , It does not depend on the, initial, concentration of reactants., , 5., , Define molecularity., It is the total number of reactant speciesthat are involved in an elementary step., , 6., , Define order of a reaction., , → products, aA + bB ⎯⎯, rate = k [A]x [B]y, overall order = x + y, It is the sum of the powers of concentration terms involved in the experimentally determined rate law., 7., , Derive an expression of half life of a zero order reaction., , k=, , [A 0 ] − [A], t, , t = t1/2 then [A] =, , k=, , [A 0 ] −, , [A 0 ], 2, , t1/2, , t1/2 =, 8., , [A 0 ], 2, , 2[A 0 ] − [A 0 ], 2k, , t1/2 =, , Give general expression for half life of nth order reaction., Half life of nth order reaction is t1/2 =, , 9., , [A0 ], 2k, , 2n −1 − 1, (n − 1)k[A 0 ]n −1, , Draw a plot of concentration vs time for zero order reaction., , k=, , [A 0 ] − [A], t, , [A], , Intercept = [A0], slope = – k, , [A] = −kt + [A0 ], t

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104, y = – mx + C, Plot of [A] Vs t gives straight line with negative slope., Slope equal to –k and intercept equals to [A]., 10., , Give examples of first order reaction., , → 2NO 2 (g) +, (i) N 2O5 (g) ⎯⎯, (ii), , 1, O 2 (g), 2, , SO2Cl2 (l ) ⎯⎯, →SO2 (g) + Cl2 (g), , → H 2O(l ) +, (iii) H 2O 2 (aq) ⎯⎯, , 1, O 2 (g), 2, , (iv) All radioactive decay, (v) Isomerisation of cyclopropane to propene, 11., , Give two examples of zero order reaction., (i), , h, , H2 (g) + Cl2 (g) ⎯⎯→ 2HCl(g), , ⎯⎯, → N 2 (g) +, (ii) N 2O(g) ⎯, ⎯, 12., , 1, O 2 (g), 2, , Give units of rate constant of, (a) first order reaction (b) zero order reaction, (a) sec–1, , 13., , (b) mol L–1s–1, , What is collision frequency?, Number of collisions per second per unit volume is called collision frequency., , 14., , Why molecularity can never be more than 3?, Probability of simultaneous collision of more than three reactants is rare., So molecularity can never be more than three., , 15., , Define activation energy?, In order to react, the colliding molecules must possess a minimum energy called activation energy., , 16., , What is the usefulness of Arrhenius equation., With the help of Arrhenius equation, we can calculate activation energy of the reaction provided, rate constants at two different temperatures are given., , 17., , Mention the factors affecting rate of reaction.

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105, 1. Nature and state of reactant, , 2. Concentration of reactant., , 3. Surface area of reactant., , 4. Temperature of reaction., , 5. Presence of catalyst., 18., , What does the slope represent in the following graphs., , [R], , log[R], , t, Slope = -k, 19., , t, Slope = -k/2.303, , Which of the following reaction is fast. Give reason., , → 2NaI(s), (i) 2Na (s) + I2(s) ⎯⎯, → 2NaI(s), (ii) 2Na (s) + I2(g) ⎯⎯, Second reaction is fast because the state of reactant is gas. Gaseous reaction is faster than, solid state reactants., 20., , Which of the two reacts faster? Why?, (i) Powdered CaCO3 with dil. HCl, (ii) Lump of CaCO3 as marble with dil. HCl, First reaction is faster because powdered form of reactant has more surface area., , 21., , → product., Derive integrated rate law for a first order reaction A ⎯⎯, Rate is directly proportional to the concentration of one reactant is called first order reaction., Consider a general first order reactionA, , ⎯⎯, → Product, , At t = 0 concentration of reactant, At time t, concentration of reactant left =, , rate = k[A]1, , −, , d[A], = k[A], dt, , −, , d[A], = k dt, [A], , =, [A], , [A0]

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106, Integrate between limits of [A0] at t = 0 and [A] at time 't'., [A], , t, , d[A], − , = k  dt, [A], [A ], 0, 0, , − ( l n [A])[A, , [A], 0], , = k(t) 0t, , –ln [A] – (–ln [A0] ) = k (t – 0), –ln [A] + ln [A0] = kt, ,  [A ] , ln  0  = kt,  [A] , , 1 [A ], k = ln 0, t, [A], k=, , [A ], 2.303, log 0, t, [A], ***, , Page 219 Computing e–40 = 4  10–18, f = e −40, 1, = e 40, , f, , Let x = e40, Take log(ln) on both sides, ln x = ln e40, ( ( l n e = 1), , ln x = 40 ln e, ln x = 40, 2.303 log x = 40, log x =, , 40, 2.303, , log x = 17.3686, x = Antilog 17.3686, x = 2.34  1017, , f=, , 1, 2.34 10, , 17, , = 0.4  10 –17 = 4  10−18

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www.kalviexpress.in, , www.kalviexpress.in, , 7. A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25 C0 to get a, solution with [H3O+ ] = 4×10-5 M. Is the solution neutral (or) acidic (or) basic., GIVEN: [H3 O+ ] = 4 x 10−5 M, pH = −log [H3 O+ ], = − log [ 4 x 10−5 ], = 5 – log 4 = 5 – 0.6021 = 4.3979, pH = 4.3979. Since the pH is lesser than 7, the solution is acidic, 8. Calculate the pH of 0.04 M HNO3, GIVEN:, , [H + ] = 0.04M, pH = − log[H + ] = − log 0.04, = − log [ 4 x 10−2 ], = 2 – log 4 = 2 – 0.6021 = 1.3979, pH = 1.3979, , 9. Define Solubility Product., The solubility product of a compound is defined as the product of the molar concentration, of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced, equilibrium equation., , www.kalviexpress.in, XmYn ⇌ mX n+, , +, , nY m−, , Ksp = [X n+]m× [Y m−]n, , 10. Define Ionic product of water. Give its value at room temperature., • The product of concentration of hydrogen ion and hydroxyl ion of pure water is known as ionic, product of water (Kw )., Kw = [H3O++][OH¯], Kw = [H3O ][OH¯], Kw = 1 ×10-7× 1 × 10-7, Kw = 1 × 10-14mol2dm-6, Kw = Ionic product of water, •, , At room temperature, the value of Kw is 1 × 10-14mol2dm-6, , ., 11. Explain Common Ion effect with an example., • When the salt of the weak acid is added to the acid, the dissociation of the weak acid decreases., This is known as common ion effect., • Ex. When sodium acetate is added to acetic acid, the dissociation of acetic acid decreases. Here, CH3COO¯ is the common ion present., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 3

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www.kalviexpress.in, 3., , Identify the Lewis acid and Lewis base in the following reactions., i) CaO + CO2 → CaCO3, ii) CH3 –O--CH3 + AlCl3 → (CH3)2O→AlCl3, , 1, 2., 4., , www.kalviexpress.in, , Compound, CaCO3, (CH3)2O→AlCl3, , Lewis Acid, CO2, AlCl3, , Lewis Base, CaO, (CH3)2O, , H3BO3 accepts hydroxide ion from water as shown below, H3BO3(aq) + H2O(l) → [B(OH)4 ]- + H+. Predict the nature of H3BO3 using Lewis concept., H3BO3 is a Lewis acid. It accepts an electron pair from H2O as follows., , HO, HO, , B, , OH, , HO, , 5., , At a particular temperature, the Kw of a neutral solution was equal to 4 x 10-14. Calculate the, concentration of [H3O+] and [OH¯], Since the solution is neutral, [H3O+] = [OH¯]. Let their concentration be x, Kw = [H3O+] [OH¯], 4 x 10-14 = (x). (x), x 2 = 4 x 10−14, x = √4 x 10−14 = 2 x10−7, [H3O+] = [OH¯] = 𝟐 𝐱𝟏𝟎−𝟕, , www.kalviexpress.in, 6. a) Calculate the pH of 10 -8 M H2SO4, Given: [H + ]= 2 x Concentration of H2SO4 = 2 x 10-8M, [H3O+] = 10-7 (from water) + 2 x 10-8 (from H2SO4), = 10 x 10-8 + 2 x 10-8 = (10 + 2 ) x 2 x 10-8, = 12 x 10-8 M, pH = -log (12 x 10-8) = 8 – log 12 = 8 – 1.0792 = 6.9208, pH= 6.9208, b) Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4., Given: pH = 5.4., [H3O+] = antilog (-pH), = antilog (-5.4), = antilog (-5.4 +6 -6), = antilog (-6 +0.6), = 3.981 x 10-6, [H3O+] = 3.981 x 10-6M, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 11

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www.kalviexpress.in, , www.kalviexpress.in, , [H + ] =, , (1.8 x 10−5 )(0.42), (0.38), , = 1.99 x10−5, , pH = − log[1.99 x10−5 ], = 5 – log 1.99 = 5 – 0.2989= 4.7011, 𝐩𝐇 = 𝟒. 𝟕𝟎𝟏𝟏, The addition of strong acid 0.01 mol HCl increased the pH only slightly ie., from 4.7447 to, 4.7011. So the buffer action is verified., 9. a) How can you prepare a buffer solution of pH 9. You are provided with 0.1M NH4OH solution and, ammonium chloride crystals. (𝐩𝐊 𝐛 = 𝟒. 𝟕), [salt], pOH = pK b + log, [base], pH + pOH = 14, 9 + pOH = 14, pOH = 14 − 9 = 5, [NH Cl], , 5 = 4.7 + log [NH 4OH], 4, , [NH Cl], , 0.3 = log 0.14, [NH4 Cl], = antilog of (0.3), 0.1, , www.kalviexpress.in, [NH4 Cl] = 0.1M x 1.995 = 0.2M, , Amount of NH4Cl required to prepare 1 litre of 0.2M solution, = Strength x molar mass of NH4Cl = 0.2 x 53.5 = 10.70g, Amount of NH4Cl required to prepare 1 litre of 0.2M solution= 10.70g, , b) What volume of 0.6M sodium formate solution is required to prepare a buffer solution of pH 4.0 by, mixing it with 100ml of 0.8M formic acid (pKa = 3.75), [sodium formate], [formic acid], [sodium formate] = 0.6 x V, [formic acid] = 0.8 x 100 = 80, [0.6 V], 4 = 3.75 + log, [80], [0.6 V], 4 − 3.75 = log, [80], [0.6 V], antilog of 0.25 =, [80], pH = pK a + log, , 0.6V = 1.778 x 80 = 142.24, V=, , 142.24, = 237.08, 0.6, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 14

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www.kalviexpress.in, , www.kalviexpress.in, , 𝐕𝐨𝐥𝐮𝐦𝐞 𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝 = 𝟐𝟑𝟕. 𝟎𝟖 𝐦𝐋, 10. Calculate the i) hydrolysis constant ii) degree of hydrolysis and iii) pH of 0.05M sodium carbonate, (pK a= 10.26), pK a= 10.26, pK a= -log Ka, Ka= antilog (-pK a) = antilog (-10.26), = antilog (-10.26 +11 -11), = antilog (-11 + 0.74) = 5.495 x 10-11, i) hydrolysis constant (Kh), Kh =, , K𝑤, K𝑎, , =, , 1 x10−14, 5.5 x10−11, , = 1.8 x 10−4, , hydrolysis constant (Kh) = 𝟏. 𝟖 𝐱 𝟏𝟎−𝟒, ii) degree of hydrolysis (h), 1 x 10−14, x 0.05, , K, , h = √ K wC = √5.5 x 10−11, 𝑎, , 1 x 10−3, 0.275, , = √, , = √3.63 x 10−3 = √36.3 x 10−4 = 6.025 x 10−2, , degree of hydrolysis (h) = 𝟔. 𝟎𝟐𝟓 𝐱 𝟏𝟎−𝟐, iii) pH of the solution:, , www.kalviexpress.in, 1, , 1, , pH = 7 + 2 p K a + 2 log C = 7 +, pH= 𝟏𝟏. 𝟒𝟕𝟗𝟓, , 10.26, 2, , +, , log 0.05, 2, , = 7 + 5.13 − 0.6505 = 11.4795, , III TEXT BOOK EXAMPLES:, Example: 1 (page no 8), Calculate the concentration of OH- in a fruit juice which contains 2 x 10-3M, H3O+ ion. Identify, the nature of solution., Given: [H3O+] = 2 x 10-3M, Kw = [H3O+][OH¯], K𝑤, 1 x 10−14, [OH − ] =, =, = 5 x 10−12 M, [H3 O+ ], 2 x 10−3, 2 x 10-3 >> 5 x 10-12 i.e. [H3O+] >>[OH¯], hence the juice is acidic in nature., Example: 2 (page no 11), Calculate the pH of 0.001M HCl solution., pH = −log [H3 O+ ] = −log(0.001), = −log 10−3 =3, 𝐩𝐇 = 𝟑, Example:3 (page no 11), Calculate the pH of 10 -7M HCl solution, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 15

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www.kalviexpress.in, , www.kalviexpress.in, , Given: Concentration of HCl = 10-7, [H3O+] = 10-7 (from water) + 10-7 (from HCl), = 10-7 + 10-7 = (1 + 1) x 10-7 = 2 x 10-7, pH = −log [H3 O+ ] = − log (2 x 10−7 ) = 7 − log 2 = 7 − 0.3010 = 6.699 = 6.70, pH= 6.70, Example: 4 (page no 14), A solution of 0.10M of a weak electrolyte is found to be dissociated to the extent of 1.20% at 25oC., Find the dissociation constant of the acid, Given: α = 1.20% =, , 1.20, 100, , = 1.2 x 10−2, , K 𝑎 = α2 C = (1.2 x 10−2 )2 (0.1) = 1.44 x 10−4 x 10−1 = 1.44x 10−5, , Example:5 (page no 15), Calculate the pH of 0.1M CH3COOH solution. Dissociation constant of acetic acid is 1.8 x 10-5, Given: Ka = 1.8 x 10-5 , C = 0.1M, For weak acids, [H+] = √ K a x C, [H+] = √1.8 x 10−5 x 0.1 = √1.8 x 10−6, = 1.34 x 10−3 M, pH = -log (1.34 x 10-3), = 3 – log 1.34, = 3- 0.1271 = 2.8729, pH= 2.8729, , www.kalviexpress.in, Example:6 (page no 19), Find the pH of the buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per, litre acetic acid. Ka = 1.8 x 10-5, Given: Ka = 1.8 x 10-5, [acid] = 0.18 mole per litre, [salt] = 0.20 mole per litre, [salt], [acid], 0.20, = 4.7447 + log, 0.18, 10, = 4.7447 + log, 9, = 4.7447 + log10 − log 9 = 4.7447 + 1 − 0.9542, = 4.7905, , pH = pK a + log, , pH= 4.7905, Example:7 (page no 20), What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and 8.2 gram of, sodium acetate and making the volume equal to 500ml. (Ka = 1.8 x 10-5), pH = pK a + log, , [sodium acetate], [acetic acid], , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 16

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www.kalviexpress.in, , www.kalviexpress.in, , pK a = 4.7447, Number of moles of sodium acetate =, [sodium acetate] =, , number of moles, volume of soluition, , Number of moles of acetic acid =, [acetic acid] =, , number of moles, volume of soluition, , mass, 8.2, =, = 0.1mole, molar mass 82, , =, , 0.1, 0.5, , = 0.2M, , mass, 6, =, = 0.1mole, molar mass 60, , =, , 0.1, 0.5, , = 0.2M, , 0.2, , pH = 4.7447 + log 0.2, = 4.7447, pH= 4.7447, , Example: 8 (page no 24), Calculate the i) hydrolysis constant ii) degree of hydrolysis and iii) pH of 0.1M sodium acetate, (pK a of acetic acid= 4.74), pK a= 4.74, , www.kalviexpress.in, pK a= -log Ka, , Ka= antilog (-pK a) = antilog (-4.74), = antilog (-4.74 +5 -5), = antilog (-5 + 0.26) = 1.8 x 10-5, , i) hydrolysis constant (Kh), Kh =, , K𝑤, K𝑎, , =, , 1 x10−14, 1.8 x x10−5, , = 5.6x 10−10, , hydrolysis constant (Kh) = 𝟓. 𝟔𝐱 𝟏𝟎−𝟏𝟎, ii) degree of hydrolysis (h), 1 x 10−14, , K, , h = √ K wC = √1.8 x 10−5 x 0.1 = √, 𝑎, , 1 x 10−8, 1.8, , = √0.5555x 10−8 = 0.75 x 10−4, , degree of hydrolysis (h) = 𝟕. 𝟓 𝐱 𝟏𝟎−𝟓, iii) pH of the solution:, 1, , 1, , pH = 7 + 2 p K a + 2 log C = 7 +, , 4.74, 2, , +, , log 0.1, 2, , = 7 + 2.37 − 0.5 = 8.87, , pH= 𝟖. 𝟖𝟕, Example:9 (page no 25), Find out whether lead chloride gets precipitated or not when 1mL of 0.1M lead nitrate and, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 17

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www.kalviexpress.in, , www.kalviexpress.in, , IV ADDITIONAL QUESTIONS AND ANSWERS:, 1. Discuss Arrhenius concept of acids and bases with suitable example. Give its limitations., • An acid is a substance that dissociates to give hydrogen ion in water. Ex: HCl, H2SO4, • A base is a substance that dissociates to give hydroxyl ion in water. Ex: NaOH, KOH, Limitations:, • It does not explain the behavior of acids and bases in non aqueous solvents like acetone,, tetrahydrofuran etc.., • It does not account the basicity of the substance like ammonia which do not posses hydroxyl group., 2. Distinguish Lewis acids and Lewis bases., Lewis acids, Electron deficient molecules ., Ex: BF3, AlCl3, All metal ions or atoms. Ex: Fe2+, Fe3+, Cr3+, Molecules with polar double bond. Ex: SO2, CO2, Molecules in which the central atom can expand its, octet due to the availability of empty d-orbitals. Ex:, SiF4, SF4, Carbonium ion. Ex: (CH3)3C+, , Lewis bases, Molecules with one or more lone pairs of electrons., Ex: NH3, H2O, All anions. Ex: F-,Cl-, CNMolecules with carbon – carbon double bond Ex:, CH2=CH2, CH ≡ CH, All metal oxides. Ex: CaO, MgO, Carbanion. Ex: CH3-, , www.kalviexpress.in, , 3.What are strong acids and bases. Give examples., S.No Strong Acids, , Strong Bases, , 1., , A strong base is one that is almost completely, dissociated in water, Ex: NaOH, KOH, , 2., , A strong acid is one that is almost completely, dissociated in water, Ex: HCl,HNO3, , 4.What are weak acids and bases. Give examples, S.No Weak Acids, , Weak Bases, , 1., , A weak base is one which are partially dissociated in, water., Ex: NH4OH, , 2., , A weak acid is one which are partially, dissociated in water., Ex: CH3COOH, , 5. The K a value of Acid (A) is 2 x 106 and Acid (B) is 1.8 x 10-5 at 250C. Identify the strong acid., Ka is called the ionization constant or dissociation constant of the acid. It measures the strength of the acid., Higher the Ka value, stronger the acid is. So Acid (A) is stronger than Acid (B)., 6. Conjugate base of a strong acid is a weak base. Justify your answer, Consider the dissociation of HCl in aqueous solution., HCl + H2O, Acid1, , Base2, , ⇌ H3O + + ClAcid2, , Base1, , Due to complete dissociation, the equilibrium lies almost to the right i.e., the Cl- ion has only a negligible, tendency to accept proton from H3O + . It means that conjugate base of a strong acid is a weak base., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 19

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www.kalviexpress.in, , www.kalviexpress.in, , 7. At 250C, the value of Kw is 1.00 x 10-14 and at 400C, it is 2.71 x 10-14. Why such variation in Kw value, is observed?, Kw is a constant at a particular temperature. The dissociation of water is an endothermic reaction. With, increase in temperature, the concentration of H3O + and OH- also increases, and hence ionic product of, water also increases., 8. Aqueous solution of NaCl is neutral. Comment on it., In aqueous solution of NaCl, both Na+ and Cl- do not undergo hydrolysis. So the concentration, of both H3O + and OH- are equal. Hence the solution is neutral., 9. Aqueous solution of HCl is acidic and solution of NaOH is basic. Why?, The following equilibrium exist in aqueous solution of HCl., HCl + H2O ⇌ H3O+ + ClHCl molecules also produces H3O+ by donating proton to water and so [H3O+] > [OH-]., Hence, aqueous solution of HCl is acidic. Similarly, in basic solutions, [OH-] > [H3O+]., Hence, aqueous solution of NaOH is basic., 10. Express the concentration of H3O+ and OH- present in a solution, when concentration of, acid and base is less than 10-6., If the concentration of acid and base is less than 10 -6, then the concentration of H3O+ produced due, to the auto ionisation of water cannot be neglected., [H3O+] = 10-7 (from water) + [H3O+] (from the acid), [OH-] = 10-7 (from water) + [OH-] (from the base), , www.kalviexpress.in, , 11. From Ostwald dilution law, how the concentration of H+( H3O+) can be calculated from Ka, value., [H + ] = α C, K𝑎, [H + ] = √, x C = √K 𝑎 C, C, [𝐇 + ] = √𝐊 𝒂 𝐂, 12 What are buffer solution? What are its types.?, •, •, •, , Buffer solution is a mixture of a weak acid and its conjugate base (or) a weak base and its, conjugate acid., It resists drastic changes in its pH upon addition of a small amount of acids or bases., There are two types of buffer solutions., 1. Acidic buffer solution: a solution containing weak acid and its salt. Ex: solution containing, acetic acid and sodium acetate., 2. Basic buffer solution : a solution containing weak base and its salt. Ex: solution containing, ammonium hydroxide and ammonium chloride., , 13. Explain the buffer action in a acidic buffer containing equimolar acetic acid and sodium acetate., Consider the buffer action in a solution containing CH3COOH and CH3COONa. The dissociation, of buffer components occur below., CH3COOH (aq), , ⇌ CH3COO-(aq)+ H+(aq), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 20

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www.kalviexpress.in, , www.kalviexpress.in, , CH3COONa(s), , H2 O(l), , →, , Na+ (aq) + CH3COO-(aq), , If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO- to form, the undissociated weak acid i.e. the increases in the concentration of H+ does not reduce the pH, significantly., H+(aq) + CH3COO-, , (aq), , → CH3COOH (aq), , If a base is added, it will be neutralised by H+, and the acetic acid is dissociated to maintain the, equilibrium. Hence the pH is not significantly altered., H+ (aq) + OH-(aq) → H2O(l), H2 O(l), , CH3COOH (aq) ⇔, H+(aq) + CH3COO____________________________________, OH-(aq) + CH3COOH (aq) →, , (aq), , CH3COO- (aq) + H2O(l), , 14. What is buffer capacity?, It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer solution to, change its pH by unity., , www.kalviexpress.in, 𝛃=, , 𝐝𝐁, 𝐝(𝐩𝐇), , 15. Derive Henderson – Hasselbalch equation to calculate pH of acidic buffer., The concentration of hydronium ion in an acidic buffer solution depends on the ratio of the, concentration of the weak acid to the concentration of its conjugate base present in the solution., [H3 O+ ] = K 𝒂, , [acid], [salt], , Due to common ion effect and very less extent of dissociation, the concentration of weak acid is, nearly equal to the initial concentration of unionized acid. Similarly, the concentration of the, conjugate base is nearly equal to the initial concentration of added salt., [H3 O+ ] = K 𝒂, , [acid], [salt], , Taking logarithm on both sides of the equation, [acid], , log [H3 O+ ] = logK 𝒂 + log [salt], Reverse the sign on both sides,, − log [H3 O+ ] = −logK 𝒂 − log, pH = pK a − log, , [acid], [salt], , [acid], [salt], , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 21

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www.kalviexpress.in, , www.kalviexpress.in, , 19. How is solubility product is used to decide the precipitation of ions?, Ionic product > Ksp, , Precipitation occurs., , Solution is super saturated, , Ionic product < Ksp, , No precipitation, , Solution is unsaturated, , Ionic product = Ksp, , Equilibrium exist, , Solution is saturated, , 20. What is molar solubility?, The maximum number of moles of solute that can be dissolved in one litre of the solution., , www.kalviexpress.in, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 25

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www.kalviexpress.in, , www.kalviexpress.in, UNIT – 9 ELECTRO CHEMISTRY, , I. Text book question and answer., 1. Define anode and cathode., Anode: The electrode at which the oxidation occurs is called the anode., Cathode: The electrode at which the reduction occurs is called the cathode., 2. Why does conductivity of a solution decrease on dilution of the solution., Conductivity of a solution decreases on dilution because the number of ions per unit volume that, carry the current in a solution decreases on dilution., 3. State Kohlrausch Law. How is it useful to determine the molar conductivity of weak, electrolyte at infinite dilution., Kohlrausch Law:, At infinite dilution, the limiting molar conductivity of an electrolyte is equal to the sum of the, limiting molar conductivities of its constituent ions., The molar conductance of CH3 COOH, can be calculated using the experimentally determined, molar conductivities of strong electrolytes HCl, NaCl and CH3 COONa., , www.kalviexpress.in, 4. Describe the electrolysis of molten NaCl using inert electrodes.,  The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they, are connected to an external DC power supply via a key.,  The electrode which is attached to the negative end of the power supply is called the cathode,, and the one which attached to the positive end is called the anode.,  Once the key is closed, the external DC power supply drives the electrons to the cathode and at, the same time pull the electrons from the anode., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 26

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www.kalviexpress.in, , www.kalviexpress.in, , Cell reactions, Na+ ions are attracted towards cathode, where they combine with the electrons and reduced to liquid, sodium., , The negative E° value shows that the above reaction is a non spontaneous one. Hence, we have to, supply a voltage greater than 4.07V to cause the electrolysis of molten NaCl., 5. State Faraday’s Laws of electrolysis., Faraday’s Laws of electrolysis, First Law, The mass of the substance (m) liberated at an electrode during electrolysis is directly, proportional to the quantity of charge (Q) passed through the cell., m𝛼Q, m = Z It, , www.kalviexpress.in, Faraday’s Second Law, When the same quantity of charge is passed through the solutions of different electrolytes, the, amount of substances liberated at the respective electrodes are directly proportional to their, electrochemical equivalents., , 6. Describe the construction of Daniel cell. Write the cell reaction., The separation of half reaction is the basis for the construction of Daniel cell.It consists of two, half cells., Oxidation half cell, A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker., Reduction half cell, A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker., Joining the half cells, The zinc and copper strips are externally connected using a wire through a switch (k) and a load, (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are, connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolytes such as, KCl, Na2SO4 etc.,, When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This, is due to the following redox reactions which are taking place at the respective electrodes., Anodic oxidation, The electrode at which the oxidation occurs is called the anode. In Daniel cell, the oxidation, takes place at zinc electrode, i.e., zinc is oxidised to Zn2+ ions by losing its electrons., Zn(S)  Zn2+ (aq) + 2e(loss of electron-oxidation), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 27

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www.kalviexpress.in, , www.kalviexpress.in, , Cathodic reduction, The electrons flow through the circuit from zinc to copper, where the Cu2+ ions in the solution, accept the electrons, get reduced to copper., Cu2+ (aq) + 2e-  Cu(S), (gain of electron - reduction), Salt bridge,  The electrolytes present in two half cells are connected using a salt bridge.,  To maintain the electrical neutrality in both the compartments, the non reactive anions Cl (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic, compartment, at the same time some of the K+ ions move from the salt bridge into the cathodic, compartment., , www.kalviexpress.in, Completion of circuit, Electrons flow from the negatively charged zinc anode into the positively charged copper, cathode through the external wire, at the same time, anions move towards anode and cations move, towards the cathode compartment. This completes the circuit., Consumption of Electrodes, As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of, the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is, converted in to Zn2+or the entire Cu2+ ions are converted in to metallic copper., Daniel cell is represented as, , 7. Why is anode in galvanic cell considered to be negative and cathode positive electrode?, At anode, oxidation occurs and electrons are liberated. Hence it is negative (-ve). At cathode,, reduction occurs and electrons are consumed. Hence it is positive (+ve)., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 28

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www.kalviexpress.in, , www.kalviexpress.in, , 11. Why is AC current used instead of DC in measuring the electrolytic conductance?,  If we apply DC current through the conductivity cell, it will lead to the electrolysis of the, solution taken in the cell.,  So, AC current is used for this measurement to prevent electrolysis., 12. 0.1M NaCl solution is placed in two different cells having cell constant 0.5 and 0.25cm -1, respectively. Which of the two will have greater value of specific conductance., 1 𝑙, ,  Specific conductance k=𝑅 𝐴, ஃ specific conductance ∝ cell constant,  When cell constant value increases then specific conductance also increases.,  ஃ The cell with cell constant value 0.5cm-1 will have greater value of specific conductance., 13. A current of 1.608A is passed through 250 mL of 0.5M solution of copper sulphate for 50, minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and, the current efficiency is 100%., Answer:, , www.kalviexpress.in, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 30

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www.kalviexpress.in, , www.kalviexpress.in, , 19. In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidized at the, anode and 2 O at cathode. If 44.8 litre of H2 at 25°C and 1atm pressure reacts in 10 minutes,, what is average current produced? If the entire current is used for electro deposition of Cu, from Cu2+ , how many grams of deposited?, Answer:, , www.kalviexpress.in, , 20. The same amount of electricity was passed through two separate electrolytic cells containing, solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in, the first cell. The amount of Cr deposited in the another cell? Give : molar mass of Nickel and, chromium are 58.74 and 52gm-1 respectively., Answer:, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 32

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www.kalviexpress.in, , www.kalviexpress.in, , www.kalviexpress.in, 23. 8.2 x 1012litres of water is available in a lake. A power reactor using the electrolysis of water in, the lake produces electricity at the rate of 2 x 106 Cs-1 at an appropriate voltage. How many, years would it like to completely electrolyse the water in the lake. Assume that there is no loss, of water except due to electrolysis., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 34

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www.kalviexpress.in, , www.kalviexpress.in, , 25. Write a note on sacrificial protection., In this technique, unlike galvanising the entire surface of the metal to be protected need not be, covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than, iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but, Mg or Zn is corroded. Hence, this process is otherwise known as sacrificial protection., 26. Explain the function of H2 –O2 fuel cell., The galvanic cell in which the energy of combustion of fuels is directly converted into electrical, energy is called the fuel cell. It requires a continuous supply of reactant to keep functioning. The, general representation of a fuel cell is follows, Fuel | Electrode |, Electrolyte | Electrode |, Oxidant, Let us understand the function of fuel cell by considering, hydrogen – oxygen, fuel cell. In this case, hydrogen act as a fuel and oxygen as an, oxidant and the electrolyte is aqueous KOH maintained at, 200oCand 20 – 40 atm. Porous, graphite electrode containing Ni and NiO serves as the inert, electrodes., Hydrogen and oxygen gases are bubbled through the anode and, cathode, respectively., , www.kalviexpress.in, The above reaction is the same as the hydrogen combustion reaction, however, they do not react, directly ie., the oxidation and reduction reactions take place separately at the anode and cathode, respectively like H - O 2 2 fuel cell. Other fuel cells like propane –O2 and methane O2 have also been, developed., 27. Ionic conductance at infinite dilution of Al3+ and SO4 2- are 189 and 160 mho cm2 equiv-1., Calculate the equivalent and molar conductance of the electrolyte Al2(SO4 )3 at infinite, dilution., Equivalent conductance, (Λ ∝ ) Al2(SO4 )3, , 1, 3, 189, 160, + 2, 3, , 1, 2, , = (λ∞ ) Al3+ + (λ∞ ) SO42=, = 63+80, = 143 mho cm2 (g equiv)-1, , Molar conductance, (Λ°M ) Al2(SO4 )3 = 2((λ°m )Al3++3 (λ°m ) SO42= (2 x 189) + (3 x 160), = 378 + 480, = 858 mho cm2 mole-1, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 36

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www.kalviexpress.in, , www.kalviexpress.in, , °, 𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙, −, , °, = 𝐸𝑐𝑒𝑙𝑙, −, , 0.0591, 1, 𝑙𝑜𝑔, 2, 1, , 0.0591, 𝑋0, 2, , 𝐸𝑐𝑒𝑙𝑙 = 𝐸 °, , When anode compartment increased by a factor of 10., 0.0591, 10, 𝐸𝑐𝑒𝑙𝑙 = 𝐸 ° −, 𝑙𝑜𝑔, 2, 1, 𝐸𝑐𝑒𝑙𝑙 = 𝐸 ° − 0.02955 𝑋 1, ஃ Cell voltage decrease by 0.02955V., 7. A solution of salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes., The mass of metal deposited at the cathode is 0.783g. calculate the equivalent mass of the, metal., , t = 150 min = 150 X 60 sec, I=0.15 amperes, m=0.783g, Faraday’s First law,, m=Zit, 𝑒𝑞.𝑚𝑎𝑠𝑠, m= 𝐹, 𝑋𝐼𝑋𝑡, , www.kalviexpress.in, equivalent mass =, , 𝑚𝑋𝐹, 𝐼𝑋𝐹, , 0.783 𝑋 96500, =, 0.15 𝑋 150 𝑋 60, 75559.5, 1350, =55.97 g.equv-1, =, , Additional Questions:, 1. Check the feasibility of the following redox reaction with the help of electrochemical series., +, 𝑁𝑖(𝑠) + 2𝐴𝑔(𝑎𝑞),  𝑁𝑖 2+ + 2𝐴𝑔(𝑠), The E° value of 𝑁𝑖 2+ /𝑁𝑖 is -0.25V while that of Ag+ / Ag is +0.80V. This means that nickel is, placed below silver in the series and can easily reduce Ag+ ions to silver by releasing electrons. The, redox reaction is therefore, feasible., 2. Rusting of iron becomes quicker in saline medium. Explain., Saline medium has extra salts such as sodium chloride dissolved in water. This means that it, has a greater concentration of electrolyte than ordinary medium. The ions present will favour the, formation of more electrochemical cells and will thus promote rusting or corrosion., 3. Explain the factors affecting electrolytic conductance., If the interionic attraction between the oppositely charged ions of solutes increases, the, conductance will decrease., • Solvent of higher dielectric constant show high conductance in solution., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 39

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www.kalviexpress.in, •, , www.kalviexpress.in, , Conductance is inversely proportional to the Viscosity of the medium. i.e., conductivity increases, with the decrease in viscosity., If the temperature of the electrolytic solution increases, conductance also increases. Increase in, temperature increases the kinetic energy of the ions and decreases the attractive force between the, oppositely charged ions and hence conductivity increases., Molar conductance of a solution increases with increase in dilution. This is because, for a strong, electrolyte, interionic forces of attraction decrease with dilution. For a weak electrolyte, degree of, dissociation increases with dilution., , •, •, , 4. Explain the variation of molar conductivity with concentration., •, •, , •, •, •, •, , The molar reconductance of an electrolytic solution increases, with dilution., The relationship between the molar conductance and the, concentration is given by, , ∧m =∧° m − k√C, The plot of ∧m Vs C gives a straight line with a negative slope., The slope value is –k and the intercept is ∧° m, For strong electrolytes it gives a straight line., For weak electrolytes it gives a non linear curve., , Variation of molar conductance of strong electrolytes with, concentration, • For a strong electrolyte, at high concentration, the number of, ions at a given volume is very high., • The force of attraction between the ions is also very high., • Viscous drag is very high., • Hence molar conductance decrease at high concentration., • But on dilution the molar conductance increase., • On dilution the ions are far apart and the force of attraction between the ions decrease., Various of molar conductance of weak electrolytes with concentration., • For the weak electrolytes when the concentrations approaches zero, there is a sudden increase, in the molar conductance and curve becomes parallel to the axis., • On dilution the dissociation of the weak electrolyte increases., , www.kalviexpress.in, 5. Explain the construction of Leclanche cell and Mercury button cell., Leclanche Cell, Mercury button Cell, Anode, , Zinc container, , Zinc amalgamated with mercury., , Cathode, , Graphite rod in contact with MnO2, , HgO mixed with graphite., , Electrolyte, Emf of the, cell, Cell, reaction, oxidation, at anode., Reduction, at cathode, , Ammonium chloride and zinc chloride in water, , Paste of KOH and ZnO, , 2𝑁𝐻4+ (𝑎𝑞) + 2𝑒 −  2𝑁𝐻3 (𝑎𝑞) + 𝐻2(𝑔), H2 gas is oxidized to H2O by MnO2., 𝐻2 (𝑔) + 2𝑀𝑛𝑂2 (𝑠) 𝑀𝑛2 𝑂3 (𝑠) + 𝐻2 𝑂(𝑙), , 𝐻𝑔𝑂(𝑠) + 𝐻2 𝑂(𝑙) + 2𝑒− 𝐻𝑔(𝑙), + 2𝑂𝐻 − (𝑎𝑞), , Overall, redox, reaction, , 𝑍𝑛(𝑠) + 2𝑁𝐻4+ (𝑎𝑞) + 2𝑀𝑛𝑂2 (𝑆) 𝑍𝑛2+ (𝑎𝑞), + 𝑀𝑛2 𝑂3 (𝑠) + 𝐻2 𝑂(𝑙) + 2𝑁𝐻3, , 𝑍𝑛(𝑠) + 𝐻𝑔𝑂(𝑠)𝑍𝑛𝑂(𝑠) + 𝐻𝑔(𝑙), , 1.5 V, , 1.35 V, , 𝑍𝑛(𝑠) 𝑍𝑛2+ (𝑎𝑞) + 2𝑒 −, , 𝑍𝑛° (𝑠) + 2𝑂𝐻 − (𝑎𝑞)𝑍𝑛𝑂 + 𝐻2 𝑂(𝑙) + 2𝑒−, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 40

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www.kalviexpress.in, , www.kalviexpress.in, , 4. How will you measure the conductivity of ionic solution by using wheat stone bridge circuit?, 5. Write Debye – Huckel and Onsagar equation., 6. How will you measure single electrode potential?, 7. How will you relate emf with Gibb’s free energy?, 8. Give IUPAC definition for the following, (i), , Electrode potential (ii), , Standard electrode potential, , 9. Explain the electrochemical mechanism of corrosion., 10. Name some process to protect metals from corrosion., 11. Define equivalent conductance., , www.kalviexpress.in, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 42

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www.kalviexpress.in, , UNIT 10, , www.kalviexpress.in, , SURFACE CHEMISTRY, , I. Text Book Questions:, 1. Give two important characteristics of physisorption.,  Physisorption decreases with increase in temperature.,  In physisorption multilayer of the adsorbate is formed on the adsorbent., 2. Differentiate physisorption and chemisorption., S.No, 1, 2, 3, , Chemical adsorption (Activated, adsorption), It is very slow., It is very specific depends on nature of, adsorbent and adsorbate., Chemical adsorption is fast with, increase pressure, it can not alter the, amount., When temperature is raised, chemisorption first increases and then, decreases., Chemisorption involves transfer of, electrons between the adsorbent and, adsorbate., Heat of adsorption is high i.e., from, 40-400KJ/mole., Monolayer of the adsorbate is formed., , Physical adsorption (van der waals, adsorption), It occurs immediately., It is non-specific, In Physisorption, when pressure, increases the extent of adsorption, increases., Physisorption decreases with increase in, temperature., , www.kalviexpress.in, 4, , 5, , 6, 7, 8, , 9, , No transfer of electrons, , Heat of adsorption is low in the order of, 40kJ/mole., Multilayer of the adsorbate is formed on, the adsorbent., It occurs on all sides., , Adsorption occurs at fixed sites called, active centres. It depends on surface, area, Chemisorption involves the formation Activation energy is insignificant., of activated complex with appreciable, activation energy., , 3. In case of chemisorption, why adsorption first increases and then decreases with, temperature?, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 43

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www.kalviexpress.in, , www.kalviexpress.in, , In chemisorption, adsorption first increases with rise in temperature due to the fact, that formation of activated complex requires certain energy., The decrease at high temperature is due to desorption, as the kinetic energy of the, adsorbate increases., 4. Which will be adsorbed more readily on the surface of charcoal and why? NH 3, or CO2?,  Gases with high critical temperature are adsorbed readily.,  Critical temperature of ammonia (406K) is higher than the critical temperature of, carbondioxide (304K). Hence ammonia (NH3) will be adsorbed more readily on, the surface of charcoal., 5. Heat of adsorption is greater for chemisorptions than physisorption. Why?, In chemical adsorption, gas molecules are held to the surface by formation of, strong chemical bonds and hence heat of adsorption is high., In physical adsorption, weak physical forces such as Vander Waals force of, attraction, dipole-dipole interaction etc exist between adsorbent and adsorbate and hence, heat of adsorption is low., , www.kalviexpress.in, 6. In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water, and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found, that all solutions containing more than 6.6 mL of AB coagulate within 5, minutes. What is the flocculation values of AB for sol (X)?, Flocculation value is the minimum concentration of an electrolyte, millimoles/lit required to cause precipitation of a sol in 2 hrs., 20 ml of sol (x) contains, 0.1 mole of AB (electrolyte), 6.6ml of sol (x) contains, 0.1 x 6.6 = 0.33 moles of AB, 20, , 33 millimoles of AB(electrolyte)is required for coagulating 1litre of electrolyte sol(x), Floculation value of AB for sol(x) = 33, [I mole = 1000 millimoles], 7. Peptising agent is added to convert precipitate into colloidal solution. Explain, with anexample., Due to common ion effect, when peptising or dispersing agent like HCl is added to, AgCl precipitate, Cl- ion is adsorbed on the precipitate and it is converted into, colloidal solution., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 44

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www.kalviexpress.in, , www.kalviexpress.in, , AgCl, (Precipitate), , HCl, , AgCl, (Colloid), , 8. What happens when a colloidal sol of Fe(OH)3 and As2O3 are mixed?, When colloidal sols with opposite charges are mixed mutual coagulation takes place., It is due to migration of ions form the surface of the particles., When positively charged Fe(OH)3 colloid is mixed with negativity charged As2O3, colloid mutual coagulation takes place., 9. What is the difference between a sol and a gel?, Sol is a colloidal solution of solid in liquid. Eg: Ink, Paint., Gel is a colloidal solution of liquid in solid. Eg: Butter, Cheese., 10. Why are lyophillic colloidal sols are more stable than lyophobic colloidal sols?,  In lyophillic colloidal sols , definite attractive force exists between dispersion, medium and dispersed phase. E.g: sols of starch.,  In lyophobic colloidal sols, no attractive force exists between the dispersed phase, and dispersion medium. E.g: sols of gold , silver.,  Hence, lyophillic colloidal sols are more stable than lyophobic colloidal sol., , www.kalviexpress.in, 11. Addition of Alum purifies water. Why?, Water containing suspended impurities are negatively charged. The Al3+ was present, in alum coagulates the suspended impurities in water purifies water., 12. What are the factors which influence the adsorption of a gas on a solid?, The factors which influence the adsorption of a gas on a solid are, i) Nature of adsorbent, ii) Nature of adsorbate, iii), Pressure iv) concentration at a given temperature. v) Temperature., 13. What are enzymes? Write a brief note on the mechanism of enzyme catalysis., Enzymes are complex protein molecules with three dimensional structures. They, catalyse the chemical reaction in living organisms. They are often present in colloidal, state and extremely specific in catalytic action., Mechanism of enzyme catalysed reaction, The following mechanism is proposed for the enzyme catalysis, E + S ⇌ ES → P + E, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 45

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www.kalviexpress.in, , www.kalviexpress.in, , Where E is the enzyme, S the substrate (reactant), ES represents activated complex and P, the products., 14. What do you mean by activity and selectivity of catalyst?, Activity :,  Ability of a catalyst to alter the rate of a reaction is called activity of catalyst,  Ability of a catalyst depends on chemisorption., Selectivity:,  Ability of a catalyst to catalyse a specific reaction to form particular products is, selectivity of a catalyst., 𝑁𝑖, , e.g: 𝐶𝑂(𝑔) + 3𝐻2 (𝑔) → 𝐶𝐻4 + 𝐻2 𝑂, 𝐶𝑢, , 𝐶𝑂(𝑔) + 𝐻2 (𝑔) → 𝐻𝐶𝐻𝑂, 15. Describe some feature of catalysis by Zeolites., i) Zeolites are microporous, crystalline, hydrated alumino silicates., ii) They are made of silicon and aluminium tetrahedron., iii) There are about 50 natural zeolites and 150 synthetic zeolites., iv) As silicon is tetravalent and aluminum is trivalent, the zeolite matrix carries extra, negative charge., v) To balance the negative charge, there are extra framework cations for example, H +, or Na+ ions., vi) Zeolites carrying protons are used as solid acids and in catalysis., vii) They are extensively used in the petrochemical industry for cracking heavy, hydrocarbon fractions into gasoline, diesel, etc.,, viii) Zeolites carrying Na+ ions are used as basic catalysis., ix) One of the most important applications of zeolite is their shape selectivity., x) In zeolites, the active sites namely protons are lying inside their pores. So,, reactions occur only inside the pores of zeolites., Reactant selectivity:, When bulkier molecules in a reactant mixture are prevented from reaching the active sites, with in the zeolite crystal, this selectivity is called reactant shape selectivity., Transition state selectivity:, If the transition state of a reaction is large compared to the pore size of the zeolite, then, no product will be formed., Product selectivity:, It is encountered when certain product molecules are too big to diffuse out of the zeolite, pores., , www.kalviexpress.in, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 46

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www.kalviexpress.in, , www.kalviexpress.in, , 16. Give three uses of emulsions., Food:, Food stuffs like milk, cream, butter, etc are present in colloidal form., In washing:, The cleansing action of soap is due to the formation of emulsion of soap molecules with, dirt and grease., Rubber industry:, Latex is the emulsion of natural rubber with negative particles. By heating rubber with, sulphur, vulcanized rubbers are produced for tyres, tubes, etc., 17. Why does bleeding stop by rubbing moist alum?, Ions present in moist alum (peptizing agent) neutralizes the colloidal protein, present in blood and coagulate it. Due to coagulation of blood, bleeding stops by rubbing, with moist alum., , www.kalviexpress.in, 18. Why is desorption important for a substance to act as good catalyst?, To create free surface on the catalyst for more reactants to adsorb and react, the, products already formed on the surface of the catalyst should be desorbed., Hence desorption is important for a substance to act as a good catalyst., 19. Comment on the statement: Colloid is not a substance but it is a state of, substance.,  Any substance can be converted into a colloid by reducing its particle size, between 1nm and 200 nm.,  Hence we can say that colloid is not a substance but it is a state of the substance,, which is dependent on the size of the particle.,  A colloidal state is intermediate between a true solution and suspension., 20. Explain any one method for coagulation, The flocculation and settling down of the sol particles is called coagulation., Electrophoresis is a method used for coagulation., Electrophoresis:, In the electrophoresis, charged particles migrate to the electrode of opposite sign., It is due to neutralization of the charge of the colloids. The particles are discharged and, so they get precipitated., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 47

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www.kalviexpress.in, , www.kalviexpress.in, , 21. Write a note on electro osmosis., A sol is electrically neutral. Hence the medium carries an equal but opposite, charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field, the medium moves in a direction opposite to that of, the sol particles. This movement of dispersion medium under the influence of electric, potential is called electro osmosis., , 22. Write a note on catalytic poison., Certain substances when added to a catalysed reaction either decreases or, completely destroys the activity of a catalyst and they are often known as catalytic, poisons., , www.kalviexpress.in, 2𝑆𝑂2(𝑔) + 𝑂2, , Pt, , → 2SO3(g), , As2 O3, , In this reaction,, Catalyst - Pt, Catalytic poison – As2O3., 23. Explain intermediate compound formation theory of catalysis with an example.,  A catalyst acts by providing a new path with low energy of activation.,  In homogeneous catalysed reactions a catalyst may combine with one or more, reactant to form an intermediate which reacts with other reactant or decompose to, give products and the catalyst is regenerated., Consider the reactions:, A+B → AB, ….. (1), A+C → AC (intermediate), ……(2), (where C is the catalyst), AC+B → AB+ C (catalyst), …….(3), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 48

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www.kalviexpress.in, , www.kalviexpress.in, , Activation energies for the reactions (2) and (3) are lowered compared to that of, (1). Hence the formation and decomposition of the intermediate accelerate the rate of the, reaction., Example:, 𝐶𝑢, , 1, , 𝐻2 + 𝑂2 → 𝐻2 𝑂, , …..(1), , 1, , ……(2), , 2, , 2𝐶𝑢 +, , 2, , 𝑂2 𝐶𝑢2 𝑂 (Intermediate), , 𝐶𝑢2 𝑂 + 𝐻2 𝐻2 𝑂 + 2𝐶𝑢 (Catalyst), ……(3), This theory describes, (i) the specificity of a catalyst and (ii) the increase in the, rate of the reaction with increase in concentration of a catalyst., 24. What is the difference between homogenous and heterogenous catalysis?, s.no, Homogenous catalysis, Heterogeneous catalysis, 1, In a homogenous catalysed reaction, In a heterogeneous catalysis the catalyst, the reactants, products and catalyst is present in a different phase ie, it is not, are present in the same phase., present in the same phase as that of, reactants or products. It is also referred as, contact catalysis., 2, 𝑃𝑡(𝑠), 𝑁𝑂(𝑔), 2𝑆𝑂2(𝑔) + 𝑂2(𝑔) → 2𝑆𝑂3(𝑔), 2𝑆𝑂2(𝑔) + 𝑂2(𝑔) →, 2𝑆𝑂3(𝑔), In this reaction the catalyst (Pt) is in solid, In this reaction, all the reactants,, phase, where as the reactants and the, catalyst and the products are in, gaseous phase., product are in gaseous phase., , www.kalviexpress.in, 25. Describe adsorption theory of catalysis., Adsorption theory explains the action of catalyst in heterogeneous catalysis. It can, also be called as contact catalysis., According to this theory, the reactants are adsorbed on the catalyst surface to form, an activated complex, which subsequently decomposes and gives the product., The various steps involved in a heterogeneous catalysed reaction are given as follows:, 1. Reactant molecules diffuse from the bulk to the catalyst surface., 2. The reactant molecules are adsorbed on the surface of the catalyst., 3. The adsorbed reactant molecules are activated to form activated complex, which is, decomposed to form the products., 4. The product molecules are desorbed., 5. The products diffuse away from the surface of the catalyst., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 49

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www.kalviexpress.in, , www.kalviexpress.in, , Additional questions:, 1. Distinguish between adsorption and absorption., S.no, Adsorption, Absorption, 1, Adsorption is a surface, Absorption is a bulk phenomenon., phenomenon., 2, Adsorbate molecules are adsorbed Adsorbate molecules are distributed, on the surface of the adsorbent., throughout the adsorbent., 3, Adsorption is a quick process., Absorption is a slow process., 2. What is called positive adsorption and negative adsorption.,  In adsorption, if the concentration of a substance in the interface is high, then it is, called positive adsorption.,  If it is less then it is called negative adsorption., 3. Give the characteristics of adsorption., i) Adsorption can occur in all interfacial surfaces i.e. the adsorption can occur in, between gas-solid, liquid-solid, liquid-liquid, solid- solid and gas-liquid., ii) Adsorption is a spontaneous process and it is always accompanied by decrease in, free energy. When Δ G reaches zero, the equilibrium is attained., iii) When molecules are adsorbed, there is always a decrease in randomness of the, molecules., , www.kalviexpress.in, 4. What is called Adsorption isobar? Give an account on adsorption isobars of, physisorption and chemisorption., A plot of “amount of adsorption” versus temperature at constant pressure is called, adsorption isobar., Physical Adsorption, , Chemical Adsorption, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 50

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www.kalviexpress.in, , www.kalviexpress.in, ,  In physical adsorption,, , 𝑥, 𝑚, 𝑥, ,  In chemical adsorption,,  The increase of, , 𝑥, 𝑚, , decreases with increase in Temperature., , 𝑚, , increases with rise in temperature and then decreases., , with rise in temperature is due to the fact that formation of, , activated complex requires certain energy.,  The decrease at high temperature is due to desorption, as the kinetic energy of the, adsorbate increases., 5. What are called adsorption isotherms?, A plot between the amount of adsorbate adsorbed and pressure (or concentration, of adsorbate) at constant temperature is called adsorption isotherms., 6. Write note on Freundlich adorption isotherm., According to Freundlich,, 1, 𝑥, = 𝑘𝑝𝑛, 𝑚, For adsorption of gases in solutions, with ‘C’ as concentration., 1, 𝑥, = 𝐾𝐶 𝑛, 𝑚, , www.kalviexpress.in, where ‘x’ is the amount of adsorbate adsorbed on ‘m’ gm of adsorbent at a pressure of, ‘p’. K and n are constants., This equation quantitively predicts the effect of pressure(or concentration) on the, adsorption of gases(or adsorbates) at constant temperature., Taking log on both sides of equation,, , 𝑥, 𝑚, , 1, , = 𝑘𝑝𝑛, , 𝑥, 1, log = log 𝐾 + log 𝑝, 𝑚, 𝑛, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 51

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www.kalviexpress.in, , www.kalviexpress.in, , without itself undergoing any chemical change., 12. What are positive and negative catalysis?, Positive catalysis is a process in which, the rate of a reaction is increased by the, presence of catalyst., 𝑃𝑡, 1, 𝐻2 𝑂2 → 𝐻2 𝑂 + 𝑂2, 2, Negative catalysis is a process in which, the rate of reaction is decreased by the, presence of a catalyst., 𝑔𝑙𝑦𝑐𝑒𝑟𝑖𝑛𝑒, 1, 𝐻2 𝑂2 →, 𝐻2 𝑂 + 𝑂2, 2, 13. Give the characteristics of catalysts., i) For a chemical reaction, catalyst is needed in very small quantity. Generally, a, pinch of catalyst is enough for a reaction in bulk., ii) There may be some physical changes, but the catalyst remains unchanged in mass, and chemical composition in a chemical reaction., iii) A catalyst itself cannot initiate a reaction. It means it cannot start a reaction which, is not taking place. But, if the reaction is taking place in a slow rate it can increase, its rate., iv) A solid catalyst will be more effective if it is taken in a finely divided form., v) A catalyst can catalyse a particular type of reaction, hence they are said to be, specific in nature., vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of, Equilibrium and hence it does not affect the position of equilibrium and the value, of equilibrium constant., vii) A catalyst is highly effective at a particular temperature called as optimum, temperature., viii) Presence of a catalyst generally does not change the nature of products, , www.kalviexpress.in, , 𝑃𝑡, , For example. 2SO2 + O2 → 2SO3, This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst., 14. What are promoters?, In a catalysed reaction the presence of a certain substance increases the activity of, a catalyst. Such a substance is called a promoter., Fe, , 𝑁2 + 3𝐻2 → 2𝑁𝐻3, Mo, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 53

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www.kalviexpress.in, , www.kalviexpress.in, , In Haber’s process, the activity of the iron catalyst is increased by the presence of, molybdenum. Hence molybdenum is called a promoter., 15. Illustrate auto catalysis(or) Ester hydrolysis reaction is slow in the beginning, and becomes faster after sometime. Give reason., In certain reactions one of the products formed acts as a catalyst to the reaction., Initially the rate of the reaction will be very slow but with the increase in time the rate of, the reaction increases. Such reactions are called auto catalysis., e.g: 𝐶𝐻3 𝐶𝑂𝑂𝐶2 𝐻5 + 𝐻2 𝑂 𝐶𝐻3 𝐶𝑂𝑂𝐻 + 𝐶2 𝐻5 𝑂𝐻, In this reaction acetic acid acts as auto catalyst., 16. Give the limitations of intermediate compound formation theory., i) The intermediate compound theory fails to explain the action of catalytic poison, and promoters., ii) This theory is unable to explain the mechanism of heterogeneous catalysed, reactions., , www.kalviexpress.in, 17. Write notes on active centers.,  The surface of a catalyst is not smooth.,  It bears steps, cracks and corners.,  Atoms on such locations of the surface are co-ordinatively unsaturated.,  So, they have much residual force of attraction. Such sites are called active centres.,  The presence of such active centres increases the rate of reaction by adsorbing and, activating the reactants., , 18. Explain active centers on the basis of adsorption theory., The adsorption theory explains the following, i) Increase in the surface area of metals and metal oxides by reducing the particle size, increases acting of the catalyst and hence the rate of the reaction., ii) The action of catalytic poison occurs when the poison blocks the active centres of, the catalyst., iii) A promoter or activator increases the number of active centres on the surfaces., 19. Give three examples for Enzyme catalysis., i) The enzyme diastase, hydrolyses starch into maltose, 𝑑𝑖𝑎𝑠𝑡𝑎𝑠𝑒, , 2(C6H10O5)n+nH2O→, nC12H22O11, ii) The enzyme micoderma aceti oxidises alcohol into acetic acid., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 54

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www.kalviexpress.in, , www.kalviexpress.in, , 𝑚𝑖𝑐𝑜𝑑𝑒𝑟𝑚𝑎 𝑎𝑐𝑒𝑡𝑖, , C2H5OH+O2 →, CH3COOH+H2O, iii) The enzyme urease, present in soya beans hydrolyses the urea., 𝑢𝑟𝑒𝑎𝑠𝑒, , NH2 -CO-NH2+H2O→, , 2NH3+CO2, , 20. Explain the general characteristics of enzyme catalysed reactions., i) Effective and efficient conversion is the special characteristic of enzyme catalysed, reactions., An enzyme may transform a million molecules of reactant in a minute., ii) Enzyme catalysis is highly specific in nature., iii) Enzyme catalysed reaction has maximum rate at optimum temperature., iv) The rate of enzyme catalysed reactions varies with the pH of the system. The rate is, maximum at a pH called optimum pH., v) Enzymes can be inhibited. Activity of an enzyme is decreased and destroyed by a, poison., Example: Penicillin inhibits the action of bacteria and used for curing diseases like, pneumonia and other infectious diseases., vi) Catalytic activity of enzymes is increased by coenzymes or activators., A small non protein (vitamin) called a coenzyme promotes the catalytic activity of, enzyme., , www.kalviexpress.in, 21. Describe phase transfer catalysis.,  If the reactants of a reaction are present in two different solvents which are, immiscible, then the reaction between them is very slow.,  As the solvents form separate phases, the reactants have to migrate across the, boundary to react. But it is not easy.,  For such situations a third solvent is added which is miscible with both and hence, phase boundary is eliminated, reactants freely mix and react fast.,  For large scale production of any product, use of a third solvent is not convenient, as it may be expensive.,  Phase transfer catalysis provides a solution for it.,  Phase transfer catalysis facilitate transport of a reactant in one solvent to the other, solvent where the second reactant is present.,  As the reactants are now brought together, they rapidly react and form the product., R – Cl + NaCN  R – CN + NaCl, Organic, phase, , Aquous, phase, , Organic, phase, , Aquous, phase, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 55

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www.kalviexpress.in, , www.kalviexpress.in, , R – Cl is 1 – Chlorooctane, R – CN is 1- Cyanooctane,  By direct heating of organic 1-chlorooctane with aqueous sodium cyanide for, several days, 1-cyanooctane is not obtained.,  If a small amount of quaternary ammonium salt like tetra alkyl ammonium chloride, is added, a rapid transition, of 1-cyanooctane occurs in about 100% yield after 1 or 2 hours.,  In this reaction, the tetra alkyl ammonium cation, which has hydrophobic and, hydrophilic ends, transports CN- from the aqueous phase to the organic phase using, its hydrophilic end facilitates the reaction with 1-chloro octane, Where,, , NaCN, , +, , R4N+Cl-, , R4N+ CN-, , + Cl-, , -, , -, , CN moved to organic phase, , CN in aquous phase, , R4N+ CN-, , , , +, , R - Cl, , Both in organic phase, , , , R – CN, , + R4N+Cl-, , Organic phase, , www.kalviexpress.in, , So phase transfer catalyst, speeds up the reaction by transporting one reactant from, one phase to another., , 22. Give an account on Nano catalysis.,  Nano materials such a metallic nano particles, metal oxides, are used as catalyst in, many chemical transformations.,  Nano catalysts carry the advantages of both homogeneous and heterogeneous, catalyses.,  Like homogeneous catalysts, the nano catalysts give 100% selective transformations, and excellent yield and show extremely high activity.,  Like the heterogenous catalysts nano catalysts can be recovered and recycled,  Nano, catalysis, are, soluble, heterogenous, catalysts, ., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 56

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www.kalviexpress.in, , www.kalviexpress.in, , 23. Define colloids., Colloid is a homogeneous mixture of two substances in which one substance is, dispersed in another substance., Size of the colloidal particle is 1-200 nm., 24. Explain dispersion methods of preparation of colloids in detail., There are 4 methods of dispersion they are,, i) Mechanical dispersion, ii) Electro dispersion, iii) Ultrasonic dispersion, iv) Peptisation, Mechanical dispersion:, Using a colloid mill, the solid is ground to colloidal dimension. The colloid mill, consists of two metal plates rotating in opposite direction at very high speed of nearly, 7000 revolution / minute., The colloidal particles of required colloidal size is obtained by adjusting the, distance between two plates., , www.kalviexpress.in, Colloidal Mill, By this method, colloidal solutions of ink and graphite are prepared., Electro Dispersion:, An electrical arc is struck between electrodes dispersed in water surrounded by, ice. When a current of 1 amp /100 V is passed, an arc produced forms vapours of metal, which immediately condense to form colloidal solution. By this method colloidal, solution of many metals like copper, silver, gold, platinum, etc. can be prepared., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 57

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www.kalviexpress.in, , www.kalviexpress.in, , Bredig’s arc method, Ultrasonic dispersion:, Sound waves of frequency more than 20kHz (audible limit) could cause, transformation of coarse suspension to colloidal dimensions., , www.kalviexpress.in, Mercury sol can be obtained by subjecting mercury to sufficiently high frequency, ultrasonic vibrations., The Ultrasonic vibrations produces by generator spread the oil and transfer the vibration, to the vessel with mercury in water, Peptisation:, By addition of suitable electrolytes, precipitated particles can be brought into, colloidal state. The process is termed as peptisation and the electrolyte added is called, peptising or dispersing agent., 𝐴𝑔𝐶𝑙, (precipitate), , 𝐻𝐶𝑙, , →, , 𝐴𝑔𝐶𝑙, (colloid), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 58

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www.kalviexpress.in, , www.kalviexpress.in, , 25. Explain chemical methods of the preparation of colloids., Oxidation:, Sols of some non metals are prepared by this method., When hydroiodic acid is treated with iodic acid, I2 sol is obtained., 𝐻𝐼𝑂3 + 5𝐻𝐼, , , , 3𝐻2 𝑂 + 𝐼2 (𝑠𝑜𝑙), , Reduction:, Many organic reagents like phenyl hydrazine, formaldehyde, etc are used for the, formation of sols. For example: Gold sol is prepared by reduction of auric chloride using, formaldehyde., 2AuCl3+3HCHO+3H2O  2Au(sol)+6HCl+3HCOOH, Hydrolysis:, Sols of hydroxides of metals like chromium and aluminium can be produced by, this method., FeCl3+3H2O  Fe(OH)3+3HCl, , www.kalviexpress.in, Double decomposition, For the preparation of water insoluble sols this method can be used. When, hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow coloured, arsenic sulphide is obtained as a colloidal solution., As2O3+3H2S  As2S3(sol)+3H2O, , Decomposition:, When few drops of an acid is added to a dilute solution of sodium thiosulphate, the, Colloidal sulphur is prepared, S2O3 2- + 2H+  S+H2O+SO2, 26. What is tyndall effect., When light passes through colloidal solution, it is scattered in all directions. This, effect is called as Tyndall effect., 27. What is Brownian movement?, When a colloidal solution is viewed through ultra microscope, continuous, bombardment of colloidal sol particles with the molecules of the dispersion medium, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 59

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www.kalviexpress.in, , www.kalviexpress.in, , can be seen and they show a random, zigzag, ceaseless motion. This is called, Brownian movement., , 28. Mention and brief the property of colloid, which explains its stability (or), What is Helmholtz double layer?, The surface of colloidal particle adsorbs one type of ion due to preferential, adsorption. This layer attracts the oppositely charged ions in the medium and hence at the, boundary separating the two electrical double layers are setup. This is called as, Helmholtz electrical double layer., As the particles nearby are having similar charges they cannot come close and, condense. Hence this helps to explain the stability of colloid, 29. What is coagulation? Mention the various methods of coagulation., The flocculation and settling down of the sol particles is called coagulation., Various methods of coagulation are:, (i) Addition of electrolytes, (ii) Electrophoresis, (iii) Mixing oppositively charged sols., (iv) Boiling, , www.kalviexpress.in, 30. What is flocculation value?, The precipitation power of electrolyte is determined by finding the minimum, concentration (millimoles/lit) required to cause precipitation of a sol in 2 hours. This, value is called flocculation value. The smaller the flocculation value greater will be, precipitation., 31. Define Gold number., Gold number is defined as the number of milligrams of hydrophilic colloid that, will just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl, solution., Smaller the gold number greater the protective power., 32. What are Emulsions? What are the two types of Emulsions?, Emulsions are colloidal solution in which a liquid is dispersed in another liquid., Generally there are two types of emulsions., (i) Oil in water (O/W), (ii) Water in oil (W/O), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 60

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www.kalviexpress.in, , www.kalviexpress.in, , 33. What is inversion of phases? Give example., The change of W/O emulsion into O/W emulsion is called inversion of phases., For example:, An oil in water emulsion containing potassium soap as emulsifying agent can be, converted into water in oil emulsion by adding CaCl2 or AlCl3., , 34. Give an account on dispersion medium and dispersed phase., Colloid is a homogeneous mixture of two substances in which one substance, (smaller proportion) is dispersed in another substance ( large proportion)., In a colloid, the substance present in larger amount is called dispersing medium, and the substance present in less amount is called dispersed phase., 35. Mention the dispersion medium of the colloids., i), , Hydrosols or aquasols, , ii) alcosols iii) benzosol, , www.kalviexpress.in, S.no, , 1, 2, 3, , Colloid, Hydrosols (aquasols), Alcosols, Benzosol, , Dispersion medium, Water, Alcohol, Benzene, , 36. Why gas in gas colloid is not following formed?,  Gases are completely miscible with each other and form a homogeneous mixture.,  A gas in gas is a true solution and not a colloidal solution., 37. Give the principle involved in the dispersion and condensation methods of, preparation of colloids., Dispersion methods: In this larger particles are broken into colloidal dimensions., Condensation methods: In this small sized particles or molecules or ions are converted, into larger colloidal sized particles., 38. How is colloid of phosphorous or sulphur prepared?, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 61

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www.kalviexpress.in, , www.kalviexpress.in, ,  Colloid of phosphorous or sulphur is prepared by the method of exchange of, solvent.,  Solution of phosphorus or sulphur is prepared in alcohol and then it is poured into, water. As they are insoluble in water, they form colloidal solution., Other important questions:, 1. Explain the methods of purification of colloids by, i), , Dialysis ii) Electrodialysis iii) ultrafiltration, , 2. Give the advantages of Brownian movement., 3. Illustrate the method of detection of charge on colloidal particles., Or, Write a note on Electrophoresis., 4. Give an account on protective action of gold sol., , www.kalviexpress.in, 5. What is called emulsification? Mention three types of emulsifications., , 6. Discuss the different tests to indentify two types of emulsions, oil in water (O/W), and water in oil (W/O)., 7. What is de-emulsification? Give various de-emulsification techniques., 8. Give the application of cortrell’s precipitator?, 9. Write short note on Delta?, 10. Distinguish between lyophilic and lyophobic colloids., 11. Give the uses of colloids in medicine., 12. What is tanning of leather?, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 62

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www.kalviexpress.in, , www.kalviexpress.in, UNIT - 11, , 63, , HYDROXY DERIVATIVES, , I, , TEXT BOOK QUESTIONS AND ANSWERS, , 1., , Identify the product (s) is / or formed when 1 – methoxy propane is heated with excess of HI . Name, the mechanism involved in the reaction., , CH3CH2CH2OH + HI CH, , 3CH2CH2 I + H2, The final products are methyl iodide ( CH3I ) and n- propyl iodidie (CH3CH2CH2 I), Mechanism:, Nucleophilic substitution (SN2 mechanism), , www.kalviexpress.in, Halide ion attack the alkyl group contain less number of carbon atom (Because it is less crowded) by SN2, mechanism., 2., , Draw the major product formed when 1 – ethoxy prop 1 ene is heated with one equivalent of HI, , 3., , Suggest a suitable reagent to prepare sec – alcohol with identical group using Grignard reagent., When Grignard reagent reacts with aldehydes gives secondary alcohol. If aldehyde contains same alkyl or, aryl group as Grignard reagent, it gives secondary alcohol with identical groups., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com

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www.kalviexpress.in, 4., , www.kalviexpress.in, , What is the major product obtained when two moles of ethyl magnesium bromide is treated with, methyl benzoate followed by acid hydrolysis., , 5., , Predict the major product, when 2-methyl but -2-ene is converted into an alcohol in each of the, following methods., (i.) Acid catalysed hydration, , www.kalviexpress.in, (Addition takes place according to Markownikoff’s rule), , (ii) Hydroboration, , (Addition takes place according to Anti Markownikoff’s rule), (No carbocation intermediate is formed), (iii)Hydroxylation using Baeyer’s reagent, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 64

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www.kalviexpress.in, , www.kalviexpress.in, , 72, , www.kalviexpress.in, , NO2 group will enter the ring that is attached to O – atom as, , It will not enter the ring attached to the carbonyl group as, , is activating group., , is deactivating group., , is O, P directing group. Due to steric hinderence, P isomer is only formed., , ii), , Conc.H SO, , 2 4, , C6H5 – CH2 – CH(OH) CH (CH3)    , , Product is based on Saytzeff’s rule during intra molecular dehydration ,if there is a possibility to form a, carbon – carbon double bond at different locations, the preferred location is the one that gives the more, highly substituted alkene ie stable alkene ., Conc.H SO, , 2 4, , C6H5 – CH2 – CH(OH) CH – CH3    , , CH3, , C6H5 – CH2 – CH  C – CH3, , CH3, , 20. Phenol is distilled with Zn dust gives A followed by friedel – crafts alkylation with propyl chloride to, give a compound B, B on oxidation gives (c) Identify A, B and C., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com

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www.kalviexpress.in, , www.kalviexpress.in, , 84, , The reaction involves protonation of oxygen which is followed by SN1 mechanism, , III., 1., , ADDITONAL QUESTIONS AND ANSWERS, Write the uses of ethylene glycol., , www.kalviexpress.in, , i. It is used as an antifreeze in automobile radiator., , ii. Its dinitrate is used as an explosive in the name of DNG., 2., , How will you prepare nitroglycerine (TNG)?, Glycerol is treated with mixture of con.HNO3 and con. H2SO4 nitroglycerine is formed.Use: an explosive., , 3., , Give the uses of glycerol., i. It is used as a sweetening agent in confectionary and beverages., ii. It is used in the manufacture of cosmetics and transparent soaps., iii. It is used in making printing inks and stamp pad ink and lubricant for watches and clocks., iv. It is used in the manufacture of explosive like dynamite and cordite by mixing it with china clay., , 4., , Write the uses of methanol., i. It is used as a solvent for paints, varnishes shellac, gums and cement., ii. In the manufacture of dyes, drugs, perfumes and formaldehyde., , 5., , Write the uses of ethanol., i. It is used as an important beverage., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com

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www.kalviexpress.in, , www.kalviexpress.in, , 85, , ii. It is also used in the preparation of, a. Paints and varnishes., b. Organic compounds like ether, CHCl3, CHI3., c. Dyes, transparent soaps., iii. As a substitute for petrol under the name power alcohol used as fuel for aeroplane, iv. It is used as a preservative for biological specimens., 6., , What happens when phenol is heated with Zinc dust?, Phenol is converted to benzene on heating with zinc dust., Zn, Δ, C6H5OH  , C6H6 + ZnO, Benzene, , 7., , How to prepare the following from phenol?, a. 2, 4, 6 – tri bromo phenol: When phenol is treated with bromine water, it gives white precipitate which, is 2, 4, 6 – tri bromophenol., , www.kalviexpress.in, , b. Picric acid: When phenol reacts with nitrating mixture con. H2SO4 and con. HNO3 the product formed is, picric acid., , 8., , Write note on Riemer – Tiemann reaction., When phenol is heated with CHCl3/NaOH the product formed is salicylaldehyde., , 9., , Write note on coupling reaction., P – hydroxy azobenzene: When phenol, benzene diazonium chloride and NaOH solution are mixed, coupling reaction take place to give p - hydroxy azobenzene., , This is called dye test., 10. Explain Phthalein reaction., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com

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www.kalviexpress.in, , www.kalviexpress.in, , Phthalein fusion reaction: Phenols are heated with phthalic anhydride and con. H2SO4 to give, phenolphthalein., , 11. Write any three tests to differentiate alcohols and phenols., a. Phenols + neutral FeCl3 → Purple colour, b. Phenol + C6H5N2+Cl- → Red orange colour., c. Phenol + NaOH → Sodium phenoxide, Alcohols do not answer all the above tests., 12. Write the uses of phenol., i. It is used for making phenol formaldehyde resin (Bakelite)., ii. Starting material for the preparation of, a. Drugs such as phenacetin, Salol, aspirin., b. Phenolphthalein indicator., c. explosive like picric acid., iii. It is used as an antiseptic – carbolic lotion and carbolic soaps., , www.kalviexpress.in, , 13. Give the uses of diethyl ether and anisole., a. Diethylether:, , i. It is used as a surgical anaesthetic agent., ii. Good solvent for organic reactions and extraction., iii. A volatile starting fluid for diesel and gasoline engine., iv. As a refrigerant., b. Anisole:, i. It is a precursor to the synthesis of perfumes and insecticide pheromones., ii. As a pharmaceutical agent., 14. What happens when Anisole reacts with HI?, Anisole reacts with HI to give phenol and methyl iodide., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 86

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www.kalviexpress.in, , www.kalviexpress.in, , (ii) Fehlings solution test: When aldehyde is warmed with Fehlings solution deep blue, colour solution is changed to red precipitate, (iii) Benedict’s solution test: Benedicts solution is reduced by aldehyde to give red, precipitate, (iv) Schiff’s reagent test: Dilute solution of aldehydes when added to Schiff’s reagent yields, its red colour, , 16. Mention the Tests for Carboxylic acid:, , (i) Turns blue litmus red, (ii) Gives brisk effervescence with sodium bicarbonate, (iii) Warmed with alcohol and conc H2SO4 it forms an ester, which is detected by its fruity, odour, , 17. Why acetic acid is less acidic than Formic acid?, , In acetic acid, the electron releasing group (+I group) increases the negative charge on the, carboxylate ion and destabilize it and hence the loss of proton becomes difficult., , 18. Why Carboxylic acids have higher boiling point than aldehyeds, ketones and alcohols of, comparable molecular masses?, , www.kalviexpress.in, This is due to more association of carboxylic acid molecules through intermolecular, hydrogen bonding. They exist as dimer in its vapour state., , O, , R, , H O, , C, , C, O, , H, , R, , O, , 19. What is glacial acetic acid? How is it obtained?, , Pure acetic acid is called glacial acetic acid. Because it forms ice like crystal when cooled., When aqueous acetic acid is cooled at 289.5K, acetic acid solidifies and forms ice like crystals,, where as water remains in liquid state and removed by filtration., , 20. Why formic acid reduces Tollens reagent and Fehlings solution? (or) Account for reducing, property of formic acid?, , Formic acid contains both an aldehyde as well as an acid group., Like other aldehydes, formic acid can easily be oxidised and thereore acts as a strong, reducing agent., , O, , H, , C, , O, , OH, , Aldehyde group, , H, , C, , OH, , Carboxylic acid group, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 103

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www.kalviexpress.in, , www.kalviexpress.in, , c) Uses of Aetone, (i), (ii), (iii), (iv), , Used as solvent in the manufacture of smokeless powder(cordite), Used as a nail polish remover, Used in the preparation of sulphonal, a hypnotic, Used in the manufacture of thermosoftening plastic Perspex, , d) Uses of Benzaldehyde, (i), (ii), (iii), (iv), , As a flavoring agent, Used in perfumes, Used in dye intermediates, As starting materials for the synthesis of cinnamaldehyde, Cinnamic acid, benzoyl, chloride etc.,, , e) Uses of Acetophenone, , (i) Used in perfumary, (ii) As a hypnotic under the name hypnone, , f) Uses of Benzophenone, , www.kalviexpress.in, (i) Used in perfumary, (ii) In the preparation of benzhydrol drop, , g) Uses of Formic acid, , It is used, (i) For the dehydration of hides, (ii) As a coagulating agent for rubber latex, (iii) In medicine for treatment of gout, (iv) As an anticeptic in the preservation of fruit juice, , h) Uses of Acetic acid, , (i) As table vinegar, (ii) For coagulating rubber latex, (iii) For manufacture of cellulose acetate and poly vinyl acetate, , i) Uses of benzoic acid:, , (i) As food preservative either in pure form or in the form of Sodium benzoate, (ii) In medicine as an urinary antiseptic, (iii) Manaufacture of dyes, , j) Uses of acetyl chloride:, , (i) Acylating agent in organic analysis, (ii) In detection and estimation of –OH, -NH2 groups in organic compounds, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 105

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www.kalviexpress.in, , www.kalviexpress.in, , (iii) Acetaldehyde and Acetone, , Acetaldehyde, 1. on oxidation gives acetic acid, 2. Reduces Tollen’s reagent, 3. Reduces Fehling’s solution, 4. on reduction with NaBH4 gives ethanol, (primary Alcohol), , Acetone, on oxidation gives acetic acid with loss of one, carbon atom, Does not reduce Tollen’s reagent, Does not reduce Fehling’s solution, on reduction with NaBH4 gives, Iso propyl Alcohol (secondry Alcohol), , (iv) Formic acid and Acetic acid, , Formic acid, 1. Reduces Tollen’s reagent, 2. Calcium salt of formic acid on dry, distillation gives formaldehyde, 3. It contains both aldehyde and carboxylic, acid group, , Acetic acid, Does not reduce Tollen’s reagent, Calcium salt of acetic acid on dry distillation gives, acetone, It contains only carboxylic acid group, , (v) Acetophenone and benzophenone, , www.kalviexpress.in, Acetophenone gives idoform test but benzophenone does not gives idoform test, , (vi) Phenol and Benzoic acid, Phenol, 1. Phenol does not react with NaHCO3, 2. Phenol gives violet colour with neutral FeCl3, , Benzoic acid, Benzoic acid reacts with NaHCO3 gives CO2, effervescence, Benzoic acid does not give violet colour with, neutral FeCl3, , (vii) Pent - 2 - one and Pent - 3 - one, Pent - 2 - one undergoes idoform reaction, but Pent - 3 -one does not undergo idoform reaction., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 109

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www.kalviexpress.in, , www.kalviexpress.in, , 119, , ii. The stability of arene diazonium salt is due to the dispersal of the positive charge over the, benzene ring., , N:, , N, , .., , .., N, , N:, , N, , .., , N:, , N:, , N, , .., N, , N:, , iii. 1) In aniline the lone pair of electron on the N-atom is delocalized over the benzene, ring due do resonance. 2) In CH3 NH2, +I effect of CH3 increases the electron density on, the N - atom. Therefore aniline is a weaker base than methylamine and hence its pkb value, is more than that of methylamine, iv. Gabriel phthalimide synthesis is preferred only for 10 amines., O, , O, , alcoholic, C, KOH, NH, –H2O, C, , O, , O, aqueous, , C, , R-X, , C, , KOH, , C, , OK, , www.kalviexpress.in, O, , Phthalimide, , C, , NK, , (SN2), , O, Potassium phthalimide, , +R, , N R, , C, , O, N - alkyl, phthalimide, , C, , OK, , NH2, , (10 amine), , O, Potassium pthalati, , The 10 amine thus formed does not undergo further reaction to form 20 & 30 amines. Thus this, method is used for preparation of 10 amines, , V. When Ethylanmine is added to water forms intermolecular H - bonds with water. Aniline, does not form H - bond due to the presence of a large hydrophobic - C6H5 group, , Vi. Amines are more basic than amides because in amines the lone pair of electrons is on, nitrogen are available for protonation whereas in amides the electron pair on nitrogen is, involved in resonance with C = O Group., O, , .., , O, +, , R - C - NH2, , R - C = NH2, , (Amide resonance Structure), +, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com

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www.kalviexpress.in, , www.kalviexpress.in, , BIOMOLECULES, I. Text Book Questions:, 1. What type of linkages hold together monomers of DNA?, ❖ Monomers of DNA are linked together by phospho diester bond, between 5’OH group of one nucleotide and 3’OH group on another, nucleotide., 2. Give the difference between primary and secondary structure of, proteins., S.no, Primary, Secondary, 1 It is the relative arrangement of The amino acids in the polypeptide, amino acids in the polypeptide chain forms highly regular shapes, chain, through the hydrogen bond between, carbonyl, oxygen and amine, hydrogen., 2 It is essential as even small ∝ - helix and 𝛽 – strands or sheets, changes can alter the overall are two most common sub –, structure and function of a structures formed by proteins., protein., , www.kalviexpress.in, 3. Name the Vitamins whose deficiency cause i) rickets ii) scurvy, i), Rickets, Vitamin D, ii), Scurvy, Vitamin C, 4. Write the Zwitter ion structure of alanine., H3+N – CH – COO–, CH3, , 5. Give any three difference between DNA and RNA., S.no, DNA, 1, It is mainly present in nucleus,, mitochondria and chloroplast, 2, It contains deoxyribose sugar, 3, Base pair A=T G ≡ C, 4, Double stranded molecules, 5, It’s life time is high, 6, It is stable and not hydrolysed easily, by alkalies., 7, It can replicate itself, , RNA, It is mainly present in cytoplasm,, nucleolus and ribosomes., It contains ribose sugar, Base pair A=U C ≡ G, Single stranded molecules, It is short lived., It is unstable and hydrolysed easily, by alkalies., It cannot replicate itself.It is formed, from DNA, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 139

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www.kalviexpress.in, , www.kalviexpress.in, , 6. Write a short note on peptide bond., The carboxyl group of the first amino acid react with the amino group of, the second amino acid to give an amide linkage between these amino, acids. This amide linkage is called peptide bond, , 7. Give two difference between Hormones and Vitamins., S.no, Hormones, 1 Hormone is an organic substance, that is secreted by one tissue. It, limits the blood stream and induces a, biological response in other tissues., 2 Endocrine glands, which are special, groups of cells, make hormones., , Vitamins, Vitamins, are, organic, compounds that cannot be, synthesized by our body and, must be obtained through diet., They are essential for the, normal growth and maintenance, of our health, Eg. Vitamin A,B,C,D,E and K., , 3 Eg. Insulin, 8. Write a note on denaturation of Proteins., ❖ Each protein has a unique three dimensional structure formed by, interactions., ❖ These interactions can be disturbed when the protein is exposed to, a higher temperature., ❖ The process of losing its higher order structure without losing the, primary structures is called denaturation., ❖ When a protein denatures, its biological function is lost., Eg. Coagulation of egg white by action of heat., , www.kalviexpress.in, , 9. What are reducing and non- reducing sugars., Reducing sugars:, ❖ These are carbohydrates which contain free aldehyde or ketonic, group., ❖ Reduces Fehling’s solution and Tollen’s reagent. Eg. Glucose., Non - reducing sugars:, ❖ They do not have free aldehyde group., ❖ They do not reduce Tollen’s reagent and Fehling’s solution., Eg. Sucrose, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 140

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www.kalviexpress.in, , www.kalviexpress.in, , 10. Why carbohydrates are generally optically active., ❖ Carbohydrates are optically active as they have one or more chiral, carbons., ., 11. Classify the following into monosaccharides, oligosaccharides and, polysaccharides., i), Starch, - polysaccharides, ii), Fructose, - monosaccharides, iii) Sucrose, - oligosaccharides (disaccharides), iv) Lactose, - oligosaccharides (disaccharides), v), Maltose, - oligosaccharides (disaccharides), 12. How are vitamins classified., Vitamins are classified into two groups based on their solubility., i), Fat soluble Vitamins – They do not dissolve in water., ❖ Vitamin A, D, E and K., ii), Water soluble Vitamins - They are readily soluble in water, ❖ Vitamins B(B1, B2, B3, B5, B6, B7, B9 and B12) and, Vitamin C, 13. What are hormones? Give examples., ❖ Hormone is an organic substance that is secreted by one tissue., ❖ It limits the blood stream and induces a physiological response in, other tissues., ❖ Endocrine glands, which are special groups of cells make, hormones, ❖ It is an intercellular signaling molecule., ❖ Eg. Insulin, estrogen., , www.kalviexpress.in, , 14. Write the structure of all possible dipeptides which can be obtained, from glycine and alanine., They form two dipeptides namely glycylalanine and alanylglycine., i), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 141

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www.kalviexpress.in, , www.kalviexpress.in, , ii), , 15. Define enzymes., ❖ All biochemical reactions occur in the living systems are, catalysed by the catalytic proteins called Enzymes., ❖ Enzymes are biocatalysts that accelerate the reaction rate in the orders, of 105 and also make them highly specific. Eg. Sucrase, 16.Write the structure of 𝜶 – D(+) glucophyranose., , www.kalviexpress.in, 17. What are the different types of RNA which are found in cell?, RNA molecules are classified into three major types., 1. Ribosomal RNA ( rRNA), 2. Messenger RNA ( mRNA), 3. Transfer RNA (tRNA), , 18. Write a note on formation of 𝜶-helix., ❖ In the 𝛼-helix sub-structure, the amino acids are arranged in a right, handed helical structure., ❖ They are stabilised by the hydrogen bond between the carbonyl, oxygen of one amino acid with amino hydrogen of the fifth residue., ❖ The side chains of the residues protrude outside of the helix., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 142

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www.kalviexpress.in, , www.kalviexpress.in, , ❖ Each turn contains 3.6 residues and is about 5.4 A° long., ❖ The amino acid proline produces a kink in the helical structure and, often called as helix breaker due to its rigid cyclic structure., 19. What are the functions of lipids in living organisms., ❖ Lipids are the integral component of cell membrane., ❖ The main function of triglycerides(lipids) in animals is as an, energy reserve., ❖ They act as protective coating in aquatic organisms., ❖ Lipids of connective tissues give protection to internal organs., ❖ Lipids help in the absorption and transport of fat soluble vitamins., ❖ Essential for activation of enzymes such as lipases., ❖ Act as emulsifier in fat metabolism., 20., , Is the following sugar, D- sugar or L – sugar?, CHO, OH, , H, , OH, , H, , OH, , H, , www.kalviexpress.in, CH2OH, , L - Sugar, , Additional questions and Answers., 1. What are monosaccharides? Give example., Monosaccharides are carbohydrates that cannot be hydrolysed further, and are also called simple sugars., General formula Cn(H2O)n, Eg: glucose, fructose, 2. What are disaccharides? Give example., Disaccharides are sugars that yield two molecules of monosaccharides, on hydrolysis catalysed by dilute acid or enzyme., General formula Cn(H2O)n-1., Eg: Sucrose, Lactose, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 143

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www.kalviexpress.in, , www.kalviexpress.in, , 3. What are polysaccharide? Give example., Polysaccharide consists of large number of monosaccharide units, bonded together by glycosidic bonds.Since, they do not have sweet taste, polysaccharides are called as non-sugars., Eg: starch, cellulose, 4. What is mutarotation, ❖ The specific rotation of pure α- and β-(D) glucose are 112° & 18.7°, ❖ When a pure form of any one of these sugars is dissolved in water, slow, interconversion of α -D glucose and β -D glucose via open chain form, occurs until equilibrium is established giving a constant specific, rotation + 53°, ❖ This phenomenon is called mutarotation., 5. What is epimerization., ❖ Sugar differing in configuration at an asymmetric centre is known as epimers., ❖ The process by which one epimer is converted into other is called, epimerisation and it requires the enzymes epimerase., ❖ Galactose is converted to glucose by this manner in our body., , 6. Sucrose is called as invert sugar? Why?, , www.kalviexpress.in, ❖ Sucrose (+66.6°) and glucose (+52.5°) are dextrorotatory compounds while, fructose is levo rotatory (-92.4°)., ❖ During hydrolysis of sucrose the optical rotation of the reaction mixture, changes from dextro to levo., ❖ Hence, sucrose is also called as invert sugar., , 7. Write a short note on the structure of sucrose (or) sucrose is a nonreducing sugar. Justify., ❖ In sucrose, C1 of α-D-glucose is joined to C2 of β-D-fructose., ❖ The glycosidic bond thus formed is called α-1,2 glycosidic bond., ❖ Since, both the carbonyl carbons (reducing groups) are involved in the, glycosidic bonding, sucrose is a non-reducing sugar., , 8. What is glycosidic linkage?, ❖ In disaccharides two monosaccharide’s are linked by oxide linkage called, ‘glycosidic linkage’., ❖ It is formed by the reaction of the anomeric carbon of one monosaccharide, with a hydroxyl group of another monosaccharide., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 144

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www.kalviexpress.in, , www.kalviexpress.in, , 9. Lactose is a reducing sugar? Justify., ❖ In lactose the β-D–galactose and β-D–glucose are linked by β-1,4 glycosidic, bond., ❖ The aldehyde carbon is not involved in the glycosidic bond, ❖ It retains its reducing property and is called a reducing sugar., 10. Maltose acts as a reducing sugar justify., ❖ Maltose consists two molecules of α -D-glucose units linked by an α-1,4, glycosidic bond between anomeric carbon of one unit and C-4 of the other, unit., ❖ Since one of the glucose has the carbonyl group intact, it also acts as a, reducing sugar., 11. Write a note on a starch., ❖ Starch is used for energy storage in plants., ❖ It is a polymer of glucose in which glucose molecules are linked by α (1,4), glycosidic bonds, ❖ They are separated into two fractions,, 1. water soluble amylose - 20 %, 2. water insoluble amylopectin - 80%, , www.kalviexpress.in, 12. Write two difference between Amylose and Amylopectin, S.No Amylose, Amylopectin, 1, , Amylose is composed of, unbranched chains upto 4000 α -Dglucose molecules joined by α, (1,4) glycosidic bonds., , 2, , Gives blue colour with iodine, solution., Starch contains 20% amylose, which is water soluble, , 3, , Amylopetin contains chains upto, 10000 α -D-glucose molecules linked, by, α (1,4)glycosidic bonds. At branch, points, new chains of 24 to 30 glucose, molecules are linked by, α (1,6)glycosidic bonds., Gives purple colour with iodine, solution., Starch contains 80% amylopectin, which is water insoluble, , 13.Write the importance of carbohydrates?, ❖ Carbohydrates, widely distributed in plants and animals, act mainly as energy, sources and structural polymers, ❖ Carbohydrate is stored in the body as glycogen and in plant as starch., ❖ Carbohydrates such as cellulose which is the primary components of plant cell, wall, is, ❖ used to make paper, furniture and cloths., ❖ Simple sugar glucose serves as an instant source of energy., ❖ Ribose sugars are one of the components of nucleic acids., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 145

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www.kalviexpress.in, , www.kalviexpress.in, , ❖ Modified carbohydrates such as hyaluronate (glycosaminoglycans) act as, shock absorber and lubricant., ., 14. What is isoelectric point., ❖ At a specific pH the net charge of an amino acid is neutral and this pH is, called isoelectric point., ❖ At a pH above the isoelectric point the amino acid will be negatively charged, and positively charged at pH values below the isoelectric point., 15.What are Zwitter ions?, ❖ In aqueous solution the proton from carboxyl group can be transferred to the, amino group of an amino acid leaving these groups with opposite charges., ❖ Despite having both positive and negative charges this molecule is neutral and, has amphoteric behaviour., ❖ These ions are called zwitter ions., , www.kalviexpress.in, CH3, , 16.How are proteins classified? Explain., Proteins are classified into two major types., 1. Fibrous proteins, 2. Globular proteins, 1.Fibrous proteins, , ❖ Fibrous proteins are linear molecules similar to fibres., ❖ Generally insoluble in water and are held together by disulphide bridges and weak, intermolecular hydrogen bonds., ❖ The proteins are often used as structural proteins. Example: Keratin, Collagen, 2.Globular proteins, ❖ They have an overall spherical shape., ❖ The polypeptide chain is folded into a spherical shape., ❖ These proteins are usually soluble in water and have many functions including, catalysis Example: myoglobin, 17. Give the importance of proteins., ❖ All biochemical reactions occur in the living systems are catalysed by the catalytic, proteins called enzymes., ❖ Proteins such as keratin, collagen act as structural back bones., ❖ Antibodies help the body to fight various diseases., ❖ They are used as messengers to coordinate many functions. Insulin and glucagon, control the glucose level in the blood., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 146

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www.kalviexpress.in, , www.kalviexpress.in, , ❖ They act as receptors that detect presence of certain signal molecules and activate, the proper response., ❖ They are also used to store metals such as iron (Ferritin)., 18. Give the catalytic activity of the following enzymes i) Carbonic anhydrase ii) Sucrase ii), Lactase., (i) Carbonic anhydrase - Catalyses the interconversion of carbonic acid to water and carbon, dioxide., (ii) Sucrase - Catalyses the hydrolysis of sucrose to fructose and glucose., (iii) Lactase enzyme - Hydrolyses the lactose into its constituent monosaccharides, glucose and, galactose., , 19. What are the components of nucleic acids?, The three components of nucleic acids, (i) Nitrogenous base, (ii) Pentose sugar, (iii) Phosphate group, , 20. Human cannot use cellulose as food? Why?, Human cannot use cellulose as food because our digestive system do not contain the, necessary enzymes (glycosidases or cellulases) that can hydrolyse the cellulose., , www.kalviexpress.in, 21. What are nucleoside and nucleotide., Sugar + Base, Nucleoside, Nucleoside + Phosphate, Nucleotide, 22. Give the Biological functions of nucleic acids., (i) Energy carriers (ATP), (ii) Components of enzyme cofactors (Eg. FAD), (iii) Chemical messengers. (Eg. Cyclic AMP), , 23. What are the types of RNA? Write its functions. Explain., Types of RNA, i. Ribosomal RNA (rRNA), ❖ rRNA is mainly found in cytoplasm and in ribosomes., ❖ It contains 60% RNA and 40% protein., ❖ Protein synthesis takes place at this site., ii. Messenger RNA (mRNA), ❖ It is present in small quantity and very short lived., ❖ The synthesis of mRNA from DNA strand is called transcription., ❖ It carries genetic information from DNA to the ribosomes for protein synthesis., iii. Transfer RNA (tRNA), ❖ Molecules have lowest molecular weight of all nucleic acids., ❖ They carry amino acids to the sites of protein synthesis on ribosomes., 24.Elucidate the structure of glucose., (i) Elemental analysis and molecular weight determination show that the molecular, formula of glucose is C6H12O6, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 147

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www.kalviexpress.in, , www.kalviexpress.in, , (ii) On reduction with concentrated HI and red phosphorus at 373K, glucose gives a mixture of, n hexane and 2–iodohexane indicating that the six carbon atoms are bonded linearly., (iii) Glucose reacts with hydroxylamine to form oxime and with HCN to form cyanohydrins., The above reactions indicate the presence of carbonyl group in glucose., (iv)Glucose gets oxidized to gluconic acid with mild oxidizing agents like bromine water, It shows that the carbonyl group is an aldehyde group and it occupies one end of the carbon, chain., (v)When oxidised using strong oxidising agent such as conc. nitric acid gives glucaric acid, (saccharic acid).It shows that the other end is occupied by a primary alcohol group., (vi)Glucose is oxidised to gluconic acid with ammonical silver nitrate (Tollen’s reagent) and, alkaline copper sulphate (Fehling’s solution).Tollen’s reagent is reduced to metallic silver, and Fehling’s solution to cuprous oxide which appears as red precipitate., These reactions further confirm the presence of an aldehyde group., (vii)Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol, groups., (viii), Glucose is a stable compound and does not undergo dehydration easily., ❖ It indicates that not more than one hydroxyl group is bonded to a single carbon atom., ❖ Thus the five hydroxyl groups are attached to five different carbon atoms., ❖ Sixth carbon is an aldehyde group., (ix) The glucose is referred to as D(+) glucose as it has D configuration and is dextrorotatory., Structure of D(+) glucose, , www.kalviexpress.in, 25. What are anomers,  In the formation of cyclic structure of glucose, the achiral aldehyde carbon in it is converted, to a chiral one leading to the possibility of two isomers.,  These two isomers differ only in the configuration of C1 carbon.,  These isomers are called anomers.,  The two anomeric forms of glucose are called 𝛼 and β-forms., 26. Elucidate the structure of fructose., 1) Elemental analysis and molecular weight determination of fructose show that it has the, molecular formula C6H12O6., 2) On reduction with concentrated HI and red phosphorus, fructose gives a mixture of, n hexane and 2–iodohexane indicating that the six carbon atoms are bonded linearly, 3) Fructose reacts with NH2OH and HCN. It shows the presence of a carbonyl groups in the, fructose., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 148

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www.kalviexpress.in, , www.kalviexpress.in, , 4) Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This, reaction indicates the presence of five hydroxyl groups in a fructose molecule., 5) Fructose is not oxidized by bromine water. This rules out the possibility of presence of an, aldehyde (-CHO) group., 6) Partial reduction of fructose with sodium amalgam and water produces mixtures of Sorbitol and, Mannitol which are epimers at the second carbon.In the above reaction new asymmetric carbon is, formed at C-2. This confirms the presence of a keto group., 7) On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller, number of carbon atoms than in fructose.This shows that a keto group is present in C-2. It, also shows that 1° alcoholic groups arepresent at C- 1 and C- 6., , The structure of fructose is, , www.kalviexpress.in, 27. Write a note on DNA finger printing., ❖ DNA fingerprinting is also called DNA typing or DNA profiling., ❖ The DNA finger print is unique for every person., ❖ It can be extracted from traces of samples from blood, saliva, hair etc…, ❖ By using this method we can detect the individual specific variation in human DNA., , 28 Explain the method of DNA finger printing., ❖ The extracted DNA is cut at specific points along the strand with restriction of enzymes., ❖ It resulting in the formation of DNA fragments of varying lengths which were analysed by, technique called gel electrophoresis., ❖ This method separates the fragments based on their size., ❖ The gel containing the DNA fragments is then transferred to a nylon sheet using a technique, called blotting., ❖ Then, the fragments will undergo autoradiography in which they were exposed to DNA, probes., ❖ A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at, any point where a radioactive probe had become attached., ❖ The resultant pattern of marks could then be compared with other samples., ❖ DNA fingerprinting is based on slight sequence differences between individuals, , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 149

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www.kalviexpress.in, , www.kalviexpress.in, UNIT 15, , I., , CHEMISTRY IN EVERDAY LIFE, , EVALUATION – SHORT ANSWER QUESTIONS, 1. Which chemical is responsible for the antiseptic properties of dettol?, (i) Chloroxylenol,, (ii) terpineol, 2. What are antibiotics?, The medicines that have the ability to kill the pathogenic bacteria are grouped as antibiotics., Example: (i) Penicillins (ii) amoxicillin (iii) cefpodoxime, 3. Name one substance which can act as both analgesic and antipyretic., (i) Aspirin (ii) paracetamol, 4. Write a note on synthetic detergents., Synthetic detergents are formulated products containing either sodium salts of alkyl hydrogen, sulphates or sodium salts of long chain alkyl benzene sulphonic acids. There are three types of, detergents., Detergent type, Example, (i), Anionic detergent, Sodium Lauryl sulphate ( SDS ), (ii), Cationic detergent, n- hexaadecyltrimethyl ammonium chloride, (iii), Non – ionic detergent, pentaerythrityl stearate, 5. How do antiseptics differ from disinfectants?, Antiseptics,  Stop or slow down the growth of micro, organisms.,  Applied to living tissue, Example, (i), Hydrogen peroxide, (ii), Povidine – Iodine, (iii), Benzalkonium chloride, , Disinfectants,  Stop or slow down the growth of micro, organisms.,  Generally used on inanimated objects., Example, (i), Chlorine compounds, (ii), Alcohol, (iii), Hydrogen peroxide, , www.kalviexpress.in, 6. What are food preservatives?, Preservatives are capable of inhibiting retarding or arresting the process of fermentation, acidification or other decomposition of food by growth of microorganisms., Examples:, ACETIC ACID, i., Acetic acid is used as a preservative for the preparation of pickles., ii., Sodium metasulphite is used as a preservative for fresh vegetables and fruits., iii., Benzoic acid, sorbic acid and their salts are potent inhibitors of a number of fungi, yeast, and bacteria., 7. Why do soaps not work in hard water?, Ca2+ and Mg2+ ions present in hard water reacts with soaps to produce insoluble calcium or, magnesium salts of fatty acids., These insoluble salts separate as scum, and get on the fabrics of the clothes., 8. What are drugs? How are they classified?, A drug is substance that is used to modify or explore physiological systems or pathological, states for benefit of the recipient. It is used for the purpose of diagnosis, prevention cure or relief of a, disease., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 150

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www.kalviexpress.in, , www.kalviexpress.in, , Classification of drugs:, Classification based on, i., ii., iii., iv., , The chemical structure, Pharmacological effect, Target system (drug action), Site of action (molecular target), , 9. How the tranquilizers work in body?,  They are neurologically active drugs.,  They acts on the central nervous system by blocking the neuro transmitter dopamine in the, brain.,  They are used in the treatment of stress, anxiety, depression, sleep disorders and severe mental, diseases like schizophrenia., 10. Write the structural formula of aspirin., , www.kalviexpress.in, 11. Explain the mechanism of cleansing action of soaps of detergents,  The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate, ion) present in soap. The structure of palmitate exhibit dual polarity. The hydrocarbon portion is, non polar and the carboxyl portion is polar.,  The non polar portion is hydrophobic while the polar end is hydrophilic. The hydrophobic, hydrocarbon portion is soluble in oils and greases, but not in water.,  The hydrophilic carboxylate group is soluble in water.,  When the soap is added to an oily or greasy part of the cloth. The hydrocarbon part of the soap, dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease, surface.,  At the same time the negatively charged carboxylate groups are strongly attracted by water,, thus leading to the formation of small droplets called micelles and grease is floated away from, the solid object,  When the water is rinsed away the grease goes with it. As a result, the cloth gets free from dirt, and the droplets are washed away with water., Eg. Sodium Palmitate, Detergents:,  Detergents are superior to soaps as they can be used even in hard water and in acidic conditions., The cleansing action of detergents are similar to the cleansing action of soaps., 12. Which sweetening agents are used to prepare sweets for a diabetic patient?, i) Saccharin ii) Aspartarme iii) Sucralose iv) Alitame are artificial sweeteners., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 151

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www.kalviexpress.in, , www.kalviexpress.in, , 13. What are narcotic and non-narcotic drugs. Give examples, Narcotic drugs relieve pain and produce sleep. These drugs are addictive. In poisonous dose,, these produces coma and ultimately death., i), Narcotic drugs Uses:,  Used for either short – term or long term relief of severe pain.,  Mainly used for post operative pain, pain of terminal cancer.,  Example: Morphine, Codeine.,  Non narcotic drugs are Analgesics reduce the pain without causing impairment of, consciousness.,  They alleviate pain by reducing local inflammatory responses., ii), i., ii., , Non-narcotic drugs Uses:, Used for short – term pain, Relief and for modest pain like headache, muscle strain, bruising (arthritis), Example: paracetamol, Asprin, , 14. What are antifertility drugs? Give examples, Antifertility drugs are synthetic hormones that suppresses ovulation (or) fertilisation., Uses: used in birth control pills., Example:, ❖ Synthetic oestrogen, i., Ethynylestradiol, ii., Menstranol, , www.kalviexpress.in, ❖ Synthetic progesterone, i., Norethindrone, ii., Norethynodrel, , 15. Write a note on co- polymer, A polymer containing two or more different kinds of monomer units is called a co-polymer., Example:, i., Buna –S(SBR rubber) contains styrene and butadiene monomer units., ii., Buna –N, Nylon – 6, 6, 16. What are bio degradable polymers? Give examples.,  The materials that are readily decomposed by microorganisms in the environment are called, biodegradable., Examples:, ❖ Poly hydroxy butyrate (PHB), ❖ Polyglycolic acid ( PGA), ❖ Polylactic acid ( PLA), Uses:, In medical field such as, ❖ Surgical sutures, ❖ Plasma substitute, 17. How is terylene prepared?, The monomers are ethylene glycol and terephthalic acid these monomers are mixed and heated, at 500K in the presence of zinc acetate and antimony trioxide catalyst, terylene is formed., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 152

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www.kalviexpress.in, , www.kalviexpress.in, , Uses:,  blending with cotton or wool fibres.,  glass reinforcing materials in safety helmets., 18. Write a note on vulcanization of rubber?,  Natural rubber is not so strong or elastic, the properties of natural rubber can be modified by the, process called vulcanization., , www.kalviexpress.in,  Natural rubber is mixed with 3-5% sulphur and heated at 100-150˚C causes cross linking of the, cis-1,4-polyisoprene chains through disulphide (-S-S-) bonds. The physical properties of rubber, can be altered by controlling the amount of sulphur that is used for vulcanization.,  Ratio of sulphur controls the properties of rubber., Sulphur Ratio, Properties of rubber, 1-3%, soft and stretchy, 3 -10%, harder but flexible, 19. Classify the following as linear, branched or cross linked polymers., a) Bakelite b) Nylon c) polythene, a) Bakelite, - cross linked polymers, b) Nylon, - linear polymers, c) Polyethene, - linear polymers., 20. Differentiate thermoplastic and thermosetting., S.No, 1, , Thermoplastic, Linear polymers, , 2, , They become soft on heating and hard on cooling., , 3, , They can be remoulded, , 4, , Example: Polyethene, PVC, Polystrene, , Thermosetting, Cross linked polymers, Don’t become soft on heating but, set to an infusible mass upon, heating., They cannot be remoulded, Example: Bakelite, Melamine,, formaldehyde., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 153

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www.kalviexpress.in, , www.kalviexpress.in, , Additional questions:, 1. Explain the terms (i) drug (ii) medicine (iii) chemotherapy, (i), Drug:, A drug is a substance that is used to modify or explore physiological systems or pathological states, for the benefit of the recipient. It is used for the purpose of diagnosis, prevention, cure/relief of a, disease., (ii), Medicine:, The drug which interacts with macromolecular targets such as proteins to produce a therapeutic and, useful biological response is called medicine., (iii), Chemotherapy:, The specific treatment of a disease using medicine is known as chemotherapy., 2. Define the term therapeutic index?, Therapeutic index is defined as the ratio between the maximum tolerated dose of a drug (above, which it becomes toxic) and the minimum curative dose (below which the drug is ineffective). Higher, the value of therapeutic index, safer is the drug., 3. Write short notes on (i) antagonists (ii) agonists,  Antagonist is drug which block the message by binding to the receptor side and inhibits to, natural function.,  Agonists are drugs which mimic the natural messenger by switching on the receptor., , www.kalviexpress.in, 4. Explain the action of antagonist and agonist with examples., When adenosine binds to the adenosine receptors, it induces sleepiness. On, the other hand, the antagonist drug caffeine binds to the adenosine receptor and makes it inactive. This, results in the reduced sleepiness (wakefulness)., Antagonist is a drug, morphine, which is used as a pain killer, binds to the opioid receptors and, activates them. This suppress the neuro transmitters that causes pain., 5. Explain Anaesthetics with example., Anaesthetics are two types. They are (i) Local Anaesthetics (ii) General Anaesthetics., , Types, , Local, anaesthetics, , General, Anaesthetics, , Mode of action, It, causes, loss, of, sensation, in the area in, which it is applied, without, losing, consciousness., They, block pain perception that, is transmitted via, peripheral nerve fibres to, the brain., Cause a controlled, and reversible loss, of consciousness by, affecting central nervous, system, , Uses, They are often used, during minor surgical, procedures., , Example, (i) Procaine, (Ester-linked local), (ii)Lidocaine, (Amide-linked), , They are often used, for major surgical, procedures., , (i) Propofol, (Intravenous ), (ii) Isoflurane, (Inhalational ), , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 154

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www.kalviexpress.in, , www.kalviexpress.in, , 6. What are antacids? Give examples?, Antacids neutralize the acid in the stomach that causes acidity., Uses:, To relieve symptoms such as burning sensation in the chest/ throat area (heart burns) caused by, acid reflux., Examples:, (i) Milk of Magnesia,, (ii) calcium bicarbonate,, (iii) Aluminium hydroxide, 7. What are Antihistamines? Give examples., Antihistamines block histamine release from histamine-1 receptors., Uses:, To provide relief from the allergic effects., Examples:, (i)Cetirizine, (ii) levocetirizine, , 8. Write short notes on Antioxidant with example.,  Antioxidants are substances which retard the oxidative deteriorations of food. Food containing, fats and oils are easily oxidised and turn rancid.,  To prevent the oxidation of the fats and oils, chemical BHT(butylhydroxy toluene),, BHA(Butylated hydroxy anisole) are added as food additives., Example:, Sulphur dioxide and sulphites are used as as antioxidants and enzyme inhibitors., , www.kalviexpress.in, 9. Define saponificaiton., Soaps are made from animal fats or vegetable oils. They contain glyceryl esters of long chain, fatty acids. When the glycerides are heated with a solution of sodium hydroxide they become soap and, glycerol.., 10. Define TFM value., The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the, total amount of fatty matter that can be separated from a sample after splitting with mineral acids.,, Higher the TFM quantity in the soap better is its quality.,  As per BIS standards, Grade-1 soaps 76% TFM,, Grade-2 soaps 70% TFM, Grade-3 soaps 60% TFM, 11. Explain the types of polyethene (LDPE, HDPE), LDPE:, It is formed by heating ethene at 200° to 300°C under oxygen as a catalyst. The reaction follows, free radical mechanism. The peroxides formed from oxygen acts as a free radical initiator., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 155

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www.kalviexpress.in, , www.kalviexpress.in, , Uses:, (i) insulators for cables, making toys etc…, HDPE:, The polymerization of ethylene is carried out at 373K and 6 to 7 atm pressure using Zeiglar –, Natta catalyst [TiCl4 +(C2H5 )3 Al] HDPE has high density and melting point and it is used to make, bottles, pipes etc..,, 12. How Teflon is prepared?, The monomer is tetrafluroethylene when heated with oxygen (or) ammonium persulphate under, high pressure, Teflon is obtained., , Uses:, coating articles and preparing non – stick utensils., 13. How orlon (PAN) is prepared?, It is prepared by the addition polymerisation of vinylcyanide (acrylonitrile) using a peroxide, initiator., , www.kalviexpress.in, Uses:, It is used as a substitute of wool for making blankets, sweaters etc.,, 14. How Nylon 6,6 is prepared? Give its use., Nylon – 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene – diamine in, equimolar proportion to form a nylon salt which on heating eliminate a water molecule to form amide, bonds., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 156

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www.kalviexpress.in, , www.kalviexpress.in, , Uses:, It is used in textiles, manufacture of cards etc…, , www.kalviexpress.in, 15. How Nylon-6 is prepared? Give its uses., Capro lactum (monomer) on heating at 533K in an inert atmosphere with traces of water, gives ∈ amino caproic acid which polymerises to give nylon – 6, , Uses:, It is used in the manufacture of tyrecords fabrics etc…., 16. How is melamine prepared? Give its uses, The monomers are melamine and formaldehyde. These monomers undergo, condensation polymerisation to form melamine formaldehyde resin., , Usage:, It is used for making unbreakable crockery., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 157

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www.kalviexpress.in, , www.kalviexpress.in, , 17. How Buna-N rubber prepared? Mention its uses., It is a co-polymer of acrylonitrile and buta-1,3-diene., , Uses:, It is used in the manufacture of hoses and tanklinings., 18. How PHBV polymer is prepared? Mention its uses., It is the co – polymer of the monomers 3 – hydroxybutanoic acid and 3-hydroxypentanoic acid., In PHBV, the monomer units are joined by ester linkages., , www.kalviexpress.in, Uses:, It is used in orthopaedic devices, and in controlled release of drugs., , 19. How Nylon -2 - Nylon-6 is prepared?,  It is a co – polymer which contains polyamide linkages.,  It is obtained by the condensation polymersiation of the monomers, glycine and ∈ - amino, caproic acid., , 20. How Buna-S rubber is prepared?, It is a co-polymer. It is obtained by the polymerisation of buta-1,3-diene and styrene in the ratio, 3:1 in the presence of sodium., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 158

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www.kalviexpress.in, , www.kalviexpress.in, , 21. Explain the mechanism followed in free radical polymerization., When alkenes are heated with free radical initiator such as benzyl peroxide, they undergo, polymerisation reaction., For example preparation of polystyrene from styrene., , www.kalviexpress.in, Chain growth will continue with the successive addition of several thousands of monomer units., Termination:, In termination step coupling of two monomers, to form polystyrene polymer., , 22., , Explain the preparation of Bakelite.,  The monomers are phenol and formaldehyde., , , The polymer is obtained by the condensation polymerization of these monomers in presence of, either an acid or a base catalyst., , , , In this preparation phenol reacts with methanal to form ortho or para hydroxyl methylphenols, which on further reaction with phenol gives linear polymer called novolac. Novalac on further, heating with formaldehyde undergo cross linkages to form bakelite., , Send Your Study Material & model question - our email ID - kalviexpress@gmail.com, , 159