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CHEMISTRY, EVALUATION PLAN, 1. (a) Theory / Written examination ( 3 hours )  :   70 marks, (b) Practical examination, ( 3 hours )  :   30 marks,               , Total, , : 100 marks, , 2. Question paper pattern for the theory / written examination :, Section, , Question Type, , Question No., , Internal, Choice, , Total, Marks, , Marks, with, Option, , Multiple Choice Questions ( MCQ ), , Q. 1. [(i) to (x)], , –, , 10, , 10, , Very Short Answer Questions (VSA), , Q. 2. [(i) to (viii)], , –, , 8, , 8, , B, , Short Answer Questions (SA) – I, , Q. 3. to Q. 14., , 8 out of, 12 Qs., , 16, , 24, , C, , Short Answer Questions (SA) – II, , Q. 15. to Q. 26., , 8 out of, 12 Qs., , 24, , 36, , D, , Long Answer Questions (LA), , Q. 27. to Q. 31., , 3 out of, 5 Qs., , 12, , 20, , 70, , 98, , A, , 3. Chapterwise distribution of marks in the question paper :, Chapter, No., , Name of the Chapter, , Marks, , Marks with, Option, , 1, , Solid State, , 3, , 5, , 2, , Solutions, , 4, , 6, , 3, , Ionic Equilibria, , 4, , 6, , 4, , Chemical Thermodynamics, , 6, , 8, , 5, , Electrochemistry, , 5, , 7, , 6, , Chemical Kinetics, , 4, , 6, , 7, , Elements of Groups 16, 17 and 18, , 6, , 8, , 8, , Transition and Inner Transition Elements, , 6, , 8, , 9, , Coordination Compounds, , 5, , 7, , 10, , Halogen Derivatives, , 5, , 7, , 11, , Alcohols, Phenols and Ethers, , 4, , 6, , 12, , Aldehydes, Ketones and Carboxylic Acids, , 6, , 8, , 13, , Amines, , 3, , 4, , 14, , Biomolecules, , 3, , 4, , 15, , Introduction to Polymer Chemistry, , 3, , 4, , 16, , Green Chemistry and Nanochemistry, , 3, , 4, , Total, , 70, , 98, , , , 1, 166, , Youtube.com/smfreedigestchannel

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NON-EVALUATIVE PORTION FOR THE ACADEMIC YEAR 2020-21, AS DECLARED ON 22-07-2020, Chapter No. & Name, 1. Solid State, , Non-evaluative portion, 1.2.1 : Crystalline solids, 1.2.2 : Amorphous solids, 1.7.3 : Packing efficiency of metal crystal in fcc lattice, Table 1.3 : Edge length and particle parameters in cubic system, Table 1.4 : Point no. 3 fcc/hcp only, 1.9 : Electrical properties of solids, 1.10 : Magnetic properties of solids, , 2. Solutions, , Demonstration and Exceptions to Henry’s Law, 2.11.1 : Van’t Hoff factor(i), 2.11.2 : Modification of expressions of colligative property, 2.11.3 : Van’t Hoff factor and degree of dissociation, Problems : 2.10 to 2.12 and 2.14, , 3. Ionic Equilibria, , 3.1 : Introduction, 3.6.2 : Acidity, basicity and neutrality of aqueous solutions, 3.8.3 : Properties of buffer solution, 3.10.1 : Common ion effect and solubility, , 4. Chemical Thermodynamics, , 4.1 : Introduction, 4.2.6 : Thermodynamic Equilibrium, Key points of spontaneous process, 4.11.3 : Entropy and spontaneity, 4.11.4 : Second law of thermodynamics, 4.11.5 : Gibbs energy, 4.11.6 : Gibbs energy and spontaneity, 4.11.7 : Spontaneity and  H or  S, 4.11.8 : Temperature of equilibrium, 4.11.9 : Gibbs function and equilibrium constant, Problems : 4.16 to 4.20, , 5. Electrochemistry, , 5.1 : Introduction, 5.2.2 : Ionic conduction, 5.2.3 : Measurement of conductivity of solution, Significance of molar conductivity, 5.4.1 : Electrochemical reactions, 5.4.2 : Electrodes, 5.10.1 : Dry cell, 5.10.2 : Lead accumulator, 5.11 : Fuel cells, , 2, 167, , Youtube.com/smfreedigestchannel

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6. Chemical Kinetics, , 6.1 : Introduction, 6.6 : Collision theory of bimolecular reactions, 6.7 : Temperature dependence of reaction rate, Problems : 6.12 to 6.14, , 7. Elements of Groups 16, 17 and, 18, , Introduction, Table 7.2 : Atomic and physical properties of group 16 elements, Table 7.3 and 7.4 : Atomic and physical properties of groups 17 and 18, elements, Table 7.5 and 7.6 : Properties of hydrides of groups 16 and 17 elements, 7.9 : Oxygen and compounds of oxygen, Fig. 7.1 : Flow diagram for manufacture of sulphuric acid, 7.11.2 : Hydrogen chloride, 7.13 : Compounds of Xenon (Excluding Table no. 7.14), , 8. Transition and Inner Transition 8.1.1 : General Introduction, Elements, Table 8.5 : Atomic properties of first transition series, Table 8.6 : Ionisation enthalpies of first transition series elements, 8.3 : Compounds of Mn and Cr, 8.6.1 : Metallurgy, 8.6.2 : Extraction of iron from haematite ore using blast furnace, Table 8.12 : I.E. of Lanthanoids, Problem on textbook page no. 184, Table 8.13 : Effective magnetic moments of Lanthanoids, 9. Coordination Compounds, , 9.9.6 : CFT, 9.9.7 : Factors affecting Crystal Field splitting parameters, 9.9.8 : Colour of the octahedral complexes, 9.9.9 : Tetrahedral complexes, , 10. Halogen Derivatives, , 10.3.5 : Sandmeyer’s reaction, 10.5.5 : Representation of configuration of molecule, 10.6.1 : Laboratory test of haloalkane, 10.7 : Uses and environmental effect of some polyhalogen compound, , 11. Alcohols, Phenols and Ethers, , Preparation of alcohols (a) From alkyl halide (b) By acid catalyzed, hydration of alkenes, (a) Laboratory test of alcohols and phenol (i) Litmus test, (ii) Reaction, with bases, (ii) Reaction with Phosphorus Halide, (iii) Dehydration of alcohols to, alkenes, (a) Laboratory test of ether, , 12. Aldehydes, Ketones and, Carboxylic Acids, , 12.1 : Introduction, (a) By oxidation of alcohol, (b) from hydrocarbons, (ii) Preparation of aromatic ketones from acyl chloride, (b) Laboratory test for ketonic group, 12.9.2 : Laboratory tests for carboxyl group, , 3, 168, , Youtube.com/smfreedigestchannel

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13. Amines, , 13.6.1 : Laboratory test for amines, Reaction with fluoroboric acid, (b) Reactions involving retention of diazonium group, , 14. Biomolecules, , 14.1 : Introduction, 14.2.10 : Polysaccharides (Starch, cellulose and glycogen), (b) Secondary structure of protein, (c) Tertiary structure of protein,, (d) Quaternary structure of protein, Fig. 14.26 : Formation of nucleoside, Fig. 14.27 : Structure of nucleotide,, Fig. 14.28 : Formation of dinucleotide, 14.4.3 : DNA double helix, , 15. Introduction to Polymer, Chemistry, , Fig. 15.2 : Classification of polymers, 15.3.6 : Phenol-formaldehyde and related polymers, Fig. 15.3 : Preparation of Bakelite, Fig. 15.4 : Formation of crosslinked malemine formaldehyde resin, 15.3.9 : Viscose rayon, Fig. 15.7 : Formation of viscose rayon, 15.4 : Molecular mass and degree of polymerization of polymers, , 16. Green Chemistry and, Nanochemistry, , Fig. 16.1 : Macro-materials to atoms, Fig. 16.2 : Scale of nanomaterials, 16.6.4 : Thermal properties, 16.6.5 : Mechanical properties, 16.6.6 : Electrical conductivity, 16.7 : Synthesis of nanomaterial, 16.7.4 : Photographs of instruments, , , 4, 169, , Youtube.com/smfreedigestchannel

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NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), CHEMISTRY, , CHEMISTRY, , PART, , MODEL QUESTION PAPER, , 1, , (WITH SOLUTION AND MARKING SCHEME), , CHEMISTRY, Time : 3 Hours], , [Max. Marks : 70, , General Instructions :, 1. Question paper consists of 31 questions divided into FOUR sections, namely A, B, C and D., (1) Section – A : Q. No. 1 contains 10 multiple choice type questions carrying one mark each., Q. No. 2 contains 8 very short answer type questions carrying one mark each., (2) Section – B : Q. No. 3 to Q. No. 14 are 12 short answer – I type questions carrying two marks, each. Attempt any eight questions., (3) Section – C : Q. No. 15 to Q. No. 26 are 12 short answer–II type questions carrying three marks, each. Attempt any eight questions., (4) Section – D : Q. No. 27 to Q. No. 31 are 5 long answer type questions carrying four marks each., Attempt any three questions., 2. Figures to the right indicate full marks., 3. Start each section on a new page., 4. For each MCQ, the correct answer must be written along with its alphabet :, e.g. (a) …… / (b) …… / (c) …… / (d) …… , etc., 5. Evaluation of each MCQ would be done for the first attempt only., 6. Draw neat, labelled diagrams and write balanced chemical equations wherever necessary., 7. Use log table if necessary. Use of calculator is not allowed., , SECTION – A, Q. 1. Select and write the correct answer :, (i), , The reaction, 3ClO\ ; ClO\ ;2Cl\ occurs in two steps, , \, (1) 2ClO\ ; ClO ;Cl\, , (2) ClO\ ;ClO\ ; ClO\ ;Cl\, , , The reaction intermediate is, (a) Cl\, , 170, , [10], , (b) ClO\, , , (c) ClO\, , , (d) ClO\, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , (1)

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(ii) Which one of the following is dimagnetic ?, (a) Cr ;, (iii), , (b) Fe ;, , (c) Cu ;, , (d) Sc ;, , The oxidation state of cobalt ion in the complex [Co(NH ) Br]SO is, , , (a) ;2, (b) ;3, (c) ;1, (d) ;4, , (1), , (1), , (iv) A gas performs the work of expansion equal to 1000 J by absorbing the heat, 1600 J. The internal energy change for the process is, (b) 9600 J, , (a) 2600 J, , (c) 600 J, , (d) 1300 J, , (1), , (v) Resorcinol on distillation with Zn dust gives, (a) cyclohexane, , (b) benzene, , (c) toluene, , (d) benzene–1, 3-diol, , (1), , (vi) The colligative property of a solution is, (a) vapour pressure, , (b) boiling point, , (c) osmotic pressure, , (d) freezing point, , (1), , (vii) pH of 0.01 M HCl solution is, (a) 2, (viii), , (b) 3, , (c) 1.8, , (d) 2.1, , (1), , In the Wolf-Kishner reduction, alkyl aryl ketones are reduced to alkyl, benzenes. During this change, ketones are first converted into, (a) acids, , (b) alcohols, , (c) hydrazones, , (d) alkenes, , (1), , (ix) Which of the following is not correct ?, (a) Four spheres are involved in the formation of tetrahedral void., (b) The centres of spheres in octahedral voids are at the apices of a regular, tetrahedron., (c) If the number of atoms is N, the number of octahedral voids is 2N., (d) If the number of atoms is N/2, the number of tetrahedral voids is N., , (1), , (x) Which one of the following compounds does not react with acetyl chloride ?, (a) CH –CH –NH, (b) (CH –CH ) NH, , , , , , (c) (CH –CH ) N, (d) C H –NH, , ,  , , Q. 2. Answer the following questions :, , (1), , (i) Write any two functions of a salt bridge., , (1), , [8], , (ii) What is the oxidation state of S in H S O ?, (1),   , (iii) How many moles of acetic anhydride will be required to form glucose, pentaacetate from 1 mole of glucose ?, (iv) What is an isolated system ?, , (1), (1), , CHEMISTRY, , Youtube.com/smfreedigestchannel, , 171

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(v) What type of intermolecular force leads to high density polymer ?, , (1), , (vi) Write probable electronic configuration of chromium., , (1), , (vii) The vapour pressure of pure water at 25 °C is 18 mm Hg. When a, nonelectrolyte is dissolved, its vapour pressure becomes 17.32 mm Hg., What is the relative lowering of vapour pressure of the solution ?, (viii) Name the -isomer of BHC., , (1), (1), , SECTION – B, Attempt any eight of the following questions :, , [16], , Q. 3. What happens when :, (a) Lead sulphide reacts with ozone, (b) Nitric oxide reacts with ozone ?, , (2), , Q. 4. Write the sequence of the complementary strand for the following segments of a, DNA molecule., 59CGTTTAAG93, , (2), , Q. 5. Using Raoult’s law, show that P:x P where P is vapour pressure of a solvent, , and x is the mole fraction of a solute., (2), , Q. 6. Write electronic configuration of iron with oxidation states ;2 and ;3. Which, is more stable and why ?, , (2), , Q. 7. Phenol is more acidic than ethyl alcohol. Explain., , (2), , Q. 8. Derive an expression for maximum work., , (2), , Q. 9. Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation, constant of acid., , (2), , Q. 10. Write a note on Hofmann bromamide degradation., , (2), , Q. 11. Explain atom economy with suitable example., , (2), , Q. 12. What is standard cell potential for the reaction,, 3Ni(s) ;2Al>(1M) IIIIIJ 3NI>(1 M) ;2Al(s), if ENi :90.25 V and EAl :91.66 V ?, Q. 13. Distinguish between order and molecularity of the reaction., , (2), (2), , Q. 14. What is the difference between a double salt and a coordination compound, (complex) ?, , 172, , (2), , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel

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SECTION – C, Attempt any eight of the following questions :, , [24], , Q. 15. Explain the variation of molar conductivity with concentration for strong and, weak electrolytes., , (3), , Q. 16. Explain the trend in the following atomic properties of group 16 elements :, (1) Atomic radii, (2) Ionisation enthalpy, (3) Electronegativity., , (3), , Q. 17. What are the salient features of valence bond theory (VBT) ?, , (3), , Q. 18. An element with molar mass 27 g/mol forms cubic unit cell with edge length of, 405 pm. If density of the element is 2.7 g/cm, what is the nature of cubic unit, cell ? (fcc or ccp), , (3), , Q. 19. Explain the following :, (a) Why are cations Lewis acids ?, (b) Why is ammonia a Lewis base ?, (c) Why is AlCl a Lewis acid ?, , Q. 20. Give reasons : Haloarenes are less reactive than haloalkanes., , (3), (3), , Q. 21. Answer the following :, (a) What are the types particles in the unit cells of crystalline solids ?, (b) What are intensive properties ?, (c) What is enthalpy of fusion ?, , (3), , Q. 22. Describe the action of PCl on (1) ethanol (2) Propan-1-ol (3) Propan-2-ol., , Q. 23. Explain Stephen reaction with suitable examples., , (3), (3), , Q. 24. What are meant by diamagnetic and paramagnetic properties ? Give one example, of diamagnetic and paramagnetic transition metal and lanthanoid metal., , (3), , Q. 25. In a first order reaction, the concentration of reactant decreases from 20 mmol dm\, to 8 mmol dm\ in 38 minutes. What is the half-life of reaction ?, Q. 26., , (3), , (i) What is the shape of a complex in which the coordination number of central, metal ion is 4., (ii) Write the reaction to prepare ethanamine from acetonitrile., (iii) What is a peptide bond (peptide linkage) ?, , CHEMISTRY, , Youtube.com/smfreedigestchannel, , (3), , 173

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SECTION – D, Attempt any three of the following questions :, , [12], , Q. 27. (a) Obtain the relationship between freezing point depression of a solution, containing nonvolatile nonelectrolyte and its molar mass., (b) Calculate the work done during synthesis of NH in which volume changes, , from 8.0 dm to 4.0 dm at a constant external pressure of 43 bar. In what, direction the work energy flows ?, Q. 28. Discuss SN mechanism of methyl bromide using aqueous KOH., , (4), (4), , Q. 29. (a) Explain the oxidation states of halogens., (b) (i) Formulate a cell from the following electrode reactions :, > ; 3e\, Au(aq), Mg(s), , IIIIJ, , IIIIJ, , Au(s), , Mg>, (aq) ;2e\, , (ii) Explain the variation in atomic or ionic radii of 3d-series elements., Q. 30. Write a note on aldol condensation., , (4), (4), , Q. 31. (1) What is HDP ? How is it prepared ? Give its properties and uses., (2) Name the scientist who discovered scanning tunneling microscope (STM) in, 1980., , 174, , (4), , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel

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SOLUTION : MODEL QUESTION PAPER – CHEMISTRY, SECTION – A, Note : There are two questions in Section-A. Q. 1 consists of ten Multiple Choice, Questions (MCQs) carrying one mark each including two numerical problems. Q. 2, consists of eight Very Short Answer type questions carrying one mark each including one, numerical problem., , Q. 1., , (i) (b) ClO \, , , (1 mark), , (ii) (d) Sc>, , (1 mark), , (iii) (b), , ;3, , (1 mark), , (iv) (c) 600 J, , (1 mark), , (v) (b) Benzene, , (1 mark), , (vi) (c) osmotic pressure, , (1 mark), , (vii) (a) 2, , (1 mark), , (viii) (c) hydrazones, , (1 mark), , (ix) (c) If the number of atoms is N, the number of octahedral voids is, 2 N., , (1 mark), , (x) (c) (CH 9CH ) N, , , , (1 mark), , CHEMISTRY, , Youtube.com/smfreedigestchannel, , 175

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Q. 2., , (i) The functions of a salt bridge are :, (1) It maintains the electrical contact between the two electrode, solutions of the half cells., (2) It prevents the mixing of electrode solutions., , 1, , ( 2 mark), 1, , ( 2 mark), , (ii) > > \, H S O,  , ,  The oxidation state of S atom: ;5., , (1 mark), , (iii) 10 moles of acetic anhydride., , (1 mark), , (iv) Isolated system : It is defined as a system which can neither, exchange energy nor matter with its surroundings, e.g. hot water filled, in a thermally insulated closed vessel like thermos flask., , (1 mark), , (v) High density polymers have low degree of branching along the hydrocarbon chain. The molecules are closely packed together during, crystallization. This closer packing means that the van der Waals, attraction between the chains are greater and so the plastic (high, density polymer) is stronger and has a melting point., , (vi) Chromium ( Cr) has electronic configuration,, , Cr (Expected) : 1s 2s 2p 3s 3p 3d 4s, , (Observed) : 1s 2s 2p 3s 3p 3d 4s, , (vii) Given : P :18 mm Hg; P:17.32 mm Hg, , P 9P, , :?, P, , P 9P 18917.32, , :, P, 18, , :0.03788, , (1 mark), , (1 mark), , 1, , ( 2 mark), 1, , ( 2 mark), ,  Relative lowering of vapour pressure:0.03788 mm Hg, , (viii) Lindane, 176, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , (1 mark)

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SECTION – B, Note : There are 12 Short Answer Type Questions from Q. 3 to Q. 14 carrying two marks, each. You have to attempt any eight questions. This section includes ten theory questions, and two numerical problems. Write appropriate units for each calculated value., , Q. 3. (a), , It oxidises lead sulphide (PbS) to lead sulphate (PbSO ) changing, , oxidation state of S from 92 to ;6., PbS(s) ;4O, , (b), , IIIIIJ, , PbSO(s) ;4O (g), , , (1 mark), , Ozone oxidises nitrogen oxide (nitric oxide) to nitrogen dioxide., NO(g) ;O, , Q. 4., , (g), , (g), , IIIIJ, , NO, , (g), , ;O (g), , , (1 mark), , DNA molecule : 59CGTTTAAG93, The complementary strand runs in opposite direction from the 3 end, to the 5 end. It has the base sequence decided by complementary, base pairs A9T and C9G., Original strand :, , 59C G T T T A A G93,        , , Complementary strand : 39G C A A A T T C95, , Q. 5., , (1 mark), , (1 mark), , If ~ and ~ are the mole fractions of solvent and solute respectively,, , , then, ~ ;~, , , By Raoult's law,, 1, , P:~ ;P, ( 2 mark), , , where P is the vapour pressure of a pure solvent and P is the vapour, pressure of the solution at given temperature., P, :~, , , P, , P, 19 :19~, , P, , P 9P, , :~, , P, , P, If P 9P:P, then, :~ ., , , P, , , , , 7/Navneet Practice Papers : Std. XII (Science) D0685 (2021), , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark), (1 mark), 177

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Q. 6., , The electronic configuration of Fe2; and Fe3; :, 1, , Fe2; : 1s 2s 2p 3s 3p 3d, , ( 2 mark), , Fe3; : 1s 2s 2p 3s 3p 3d, , ( 2 mark), , 1, , Due to loss of two electrons from the 4s-orbital and one electron from, 3d-orbital, iron attains 3 ; oxidation state. Since in Fe3; , the, 3d-orbital is half filled, it gets extra stability, hence Fe3; is more, stable than Fe, , Q. 7. (1), , 2;, , ., , (1 mark), , In ethyl alcohol, the – OH group is attached to sp-hybridised carbon, while in phenols, it is attached to sp-hybridised carbon., , (2), , Due to higher electronegativity of sp-hybridised carbon, electron, density on oxygen decreases. This increases the polarity of O-H bond, and results in more ionization of phenol than that of alcohols., , 1, , ( 2 mark), (3), , Electron donating inductive effect (;I effect) of the alkyl group, destabilizes alkoxide ion. As a result alcohol does not ionize much in, water, therefore alcohol is neutral compound in aqueous medium., , (4), , In alkoxide ion, the negative charge is localized on oxygen, while in, phenoxide ion the negative charge is delocalized. The delocalization, of the negative charge (structure I to V) makes phenoxide ion more, stable than that of phenol., , 178, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel

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(1 mark), , The delocalization of charge in phenol (structures VI to X), the, resonating structures have charge separation (where oxygen atom of, OH group to be positive and delocalization of negative charge over the, ortho and para positions of aromatic ring) due to which phenol molecule, is less stable than phenoxide ion. This favours ionization of phenol., Thus phenols are more acidic than ethyl alcohol., , Q. 8., , 1, , ( 2 mark), , Consider 'n' moles of an ideal gas enclosed in an ideal cylinder., Let V be the volume of the gas at a pressure P and a temperature T., When pressure changes from P to P9dP and volume increases from, V to V;dV, then the work obtained is, dW:9(P9dP) dV, 1, , :9PdV;dPdV, , ( 2 mark), , Since dP.dV is negligibly small relative to PdV, dW:9PdV, Let the state of the system change from A(P , V ) to B (P , V ),  ,  , isothermally and reversibly, at temperature T involving number of, infinitesimal steps., Then the total work or maximum work in the process is obtained by, integrating above equation., B, , Wmax :  dW, A, B, , :  9PdV, A, 1, , ( 2 mark), , B PV:nRT,  P:, , nRT, V, CHEMISTRY, , Youtube.com/smfreedigestchannel, , 179

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V, , , Wmax :  9nRT, V, , V, , , : 9nRT , , V, , , dV, V, , 1, , ( 2 mark), , dV, V, , : 9nRT (lnV 9lnV ), , , V, , V, , V, ,  Wmax : 9 2.303 nRT log,  V, , At constant temperature,, P, V, B P ;V :P ;V,  : , , , , , P, V, , , : 9nRT log e, , P, ,  Wmax : 9 2.303 nRT log,  P, , Q. 9., , 1, , ( 2 mark), , Given : C:0.1 M, Dissociation:5%, Ka :?, :, , :, , Percent dissociation, 100, , 1, , ( 2 mark), , 5, :0.05, 100, , Ka :, :, , C 0.1;(0.05), :, 19, 190.05, , (1 mark), , 0.1;(0.05), 0.95, , :2.63;10\, , 1, , ( 2 mark), , Ans. Dissociation constant of acid:Ka :2.63;10\, , Q. 10., , The conversion of amides into amines in the presence of bromine and, alkali is known as Hofmann degradation of amides. An important, characteristic of this reaction is that an amine with one carbon less, than those in the amide is formed. Thus, decreasing the length of, carbon chain. This reaction is an example of molecular rearrangement, and involves the migration of an alkyl or aryl group from the carbonyl, carbon to the adjacent nitrogen atom., , 180, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , (1 mark)

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For example,, (1) When propanamide is treated with bromine and aqueous or, alcoholic sodium hydroxide, ethanamine is obtained which has one, carbon atom less., O, CH – CH – C– NH ;Br ;4NaOH IIIIIJ, , , , , Propanamide, CH – CH – NH ;Na CO ;2NaBr;2H O, , , ,  , , 1, Ethanamine, ( 2 mark), (2) When benzamide is treated with bromine and aqueous or alcoholic, sodium hydroxide, aniline is obtained., , 1, , ( 2 mark), , (1), , Atom economy is a measure of the amount of atoms from the starting, material that are present in the final product at the end of a chemical, process. Good atom economy means most of the atoms of the, reactants are incorporated in the desired products. Only small amount, , Q. 11., , of waste is produced, hence lesser is problem of waste disposal. (1 mark), (2), , The atom economy of a process can be calculated using the following, formula., % atom economy:, Formula weight of the desired product, ;100, Sum of formula weight of all the reactants used in the reaction, Consider the conversion of Butan-1-ol to 1-bromobutane, CH –CH –CH –CH OH;NaBr;H SO IIIIIIIJ, , , , ,  , CH –CH –CH –CH –Br ;NaHSO ;H O, , , , , , , % atom economy:, :, , mass of 1-Bromobutane, ;100, sum of mass of 1-Butanol;sodium bromide, , mass of (4C;9H;1Br) atoms, ;100, mass of (4C;12H;5O;1Br;1Na;1S) atoms, CHEMISTRY, , Youtube.com/smfreedigestchannel, , 181

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137 u, ;100:49.81%, 275 u, The atom economy of the above reaction is less than 50% and hence, :, , waste produced is higher., , Q. 12., , (1 mark), ,  :?, Given : E0Ni2>/Ni :90.25 V; E0Al3>/Al :91.66 V; Ecell, Since Ni is oxidised and Al3> is reduced,, 0, , 1, , Ecell :E0Al3>/Al 9E0Ni2>/Ni, , ( 2 mark), 1, , :91.669(90.25), , ( 2 mark), , :91.41 V, , (1 mark), , 0, , Ans. Ecell :91.41 V, Order, , Q. 13., , Molecularity, , 1. It is the sum of the exponents 1. It is the number of molecules, to, , concentration, , (or atoms or ions) of the, , terms in rate law expression, , reactants taking part in the, , are raised., , elementary reaction., , 2. It, , which, , is, , the, , experimentally 2. It is theoretical property and, , determined and indicates the, , indicates, , the, , number, , of, , dependence of the reaction, , molecules of reactant in each, , rate on the concentration of, , step of the reaction., , particular reactants., 3. It may have values which are 3. It is always an integer., integer, fractional or zero., 4. Its, , value, , depends, , upon 4. Its value does not depend upon, , experimental conditions., , experimental conditions., , 5. It is the property of elemen- 5. It is the property of elementary and complex reactions., , tary reactions only., , 6. Rate law expression describes 6. Rate law does not describe, the order of the reaction., , 182, , molecularity., , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel

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Q. 14., , Double salt, , Coordination compound (complex), , 1. Double salts exist only in the 1. Coordination compounds exist, solid state and dissociate into, , in the solid state as well as in, , their constituent ions in the, , the aqueous or non-aqueous, , aqueous solutions., , solutions., , 2. Double, , salts, , lose, , their 2. They, , identity in the solution., , do, , not, , lose, , their, , identity completely., , 3. The properties of double salts 3. The, , properties, , of, , are same as those of their, , coordination compounds are, , constituents., , different, , from, , their, , constituents., 4. Metal ions in the double salts 4. Metal ions in the coordishow their normal valence., , nation compounds show two, valences, , namely, , primary, , valence and secondary valence, satisfied by anions or neutral, molecules called ligands., 5. In K SO . Al (SO ) . 24H O. 5. In K [ Fe(CN) ], the ions K>,  , , , , , , the ions K>, Al> and SO\, and [ Fe(CN) ]\ show their, , , show their properties., properties., (Any four points :, , CHEMISTRY, , Youtube.com/smfreedigestchannel, , 1, 2, , mark each), , 183

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SECTION– C, Note : There are 12 Short Answer Type questions from Q. 15 to Q. 26 carrying three, marks each. You have to attempt any eight questions. This section includes ten theory, questions and two numerical problems. Write appropriate units for each calculated value., , Q. 15. (i), , As the dilution of an electrolytic solution increases, the dissociation of, the electrolyte increases, hence the total number of ions increases,, 1, , (2 mark), , therefore, the molar conductivity increases., , Fig. : Variation of molar conductivity with (c, , (1 mark for figure), , (ii) The increase in molar conductivity with increase in dilution or decrease, in concentration is different for strong and weak electrolytes., 1, , ( 2 mark), (iii), , On dilution, the molar conductivity of strong electrolytes increases, rapidly and approaches to a maximum limiting value at infinite dilution, or zero concentration and represented as u- or u or um . In case of, , weak electrolytes which dissociate less as compared to strong, electrolytes, the molar conductivity is low and increases slowly in high, concentration region, but increases rapidly at low concentration or, high dilution. This is because the extent of dissociation increases with, dilution rapidly., , (iv), , 1, , ( 2 mark), , u values for strong electrolytes can be obtained by extrapolating the, , linear graph to zero concentration (or infinite dilution). However u for, , the weak electrolytes cannot be obtained by this method, since the, graph increases exponentially at very high dilution and does not, intersect um axis at zero concentration., , 184, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark)

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Q. 16. (1), , Atomic and ionic radii :, (i) As compared to group 15 elements, the atomic and ionic radii of, group 16 elements are smaller due to higher nuclear charge., 1, , ( 2 mark), (ii) The atomic and ionic radii increase down the group from oxygen to, polonium. This is due to the addition of a new shell at each, successive elements on moving down the group., The atomic radii increases in the order O, , S, , Se, , Te, , Po, 1, , ( 2 mark), (2), , Ionisation enthalpy :, (i) The ionisation enthalpy of group 16 elements has quite high values., (ii) Ionisation enthalpy decreases down the group from oxygen to, polonium. This is due to the increase in atomic volume down the, 1, , ( 2 mark), , group., , (iii) The first ionisation enthalpy of the lighter elements of group 16, (O, S, Se) have lower values than those of group 15 elements in, the corresponding periods. This is due to difference in their, electronic configurations., Group 15 : (valence shell) ns npx npy npz, Group 16 : (valence shell) ns npx npy npz, Group 15 elements have extra stability of half filled and more, symmetrical orbitals, while group 16 elements acquire extra, stability by losing one of paired electrons from npx-orbital, forming half filled p-orbitals., Hence group 16 elements have lower first ionisation enthalpy than, group 15 elements., , (3), , 1, , ( 2 mark), , Electronegativity :, (i) The electronegativity values of group 16 elements have higher, values than corresponding group 15 elements in the same periods., (ii) Oxygen is the second most electronegative elements after, fluorine. (O:3.5, F:4), CHEMISTRY, , Youtube.com/smfreedigestchannel, , 185

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(iii) On moving down the group electronegativity decreases from, oxygen to polonium., (iv) On moving down the group atomic size increases, hence nuclear, attraction decreases, therefore electronegativity decreases., (1 mark), , Q. 17., , The salient features of valence bond theory (VBT) are as follows :, (1) According to this theory, a central metal atom or ion present in a, complex provides a definite number of vacant orbitals (s, p, d and f), to accommodate the electrons from the ligands for the formation, coordinate bonds with the metal ion / atom., (2) The number of vacant orbitals provided by the central metal atom, or ion is the same as the coordination number of the metal., For example : Cu> provides 4 vacant orbitals in the complex., [Cu(NH ) ]>., , (3) The vacant orbitals of metal atom or ion undergo hybridisation, forming the same number of hybridised orbitals, since the bonding, with the hybrid orbitals is stronger., (4) Each ligand has one or more orbitals containing one or more lone, pairs of the electrons., (5) The shape or geometry of the complex depends upon the type of, hybridisation of the metal ion / atom., (6) When inner orbitals namely (n91) d orbitals in transition metal, atom or ion hybridise, the complex is called inner complex and, when outer orbitals i.e., nd orbitals hybridise then the complex is, called outer complex., (7) When the central metal atom or ion in the complex contains one or, more unpaired electrons the complex is paramagnetic while if all, the electrons are paired, the complex is diamagnetic., (Any six points :, , 186, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , 1, 2, , mark each)

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Given : Molar mass:M:27 g mol 91, , Q. 18., , Nature of crystal:cubic unit cell, Edge length:a:405 pm:4.05;10 98 cm, Density: :2.7 g cm 93, Nature of unit cell:?, :, , n;M, a;NA, ,  n:, , 1, , ( 2 mark), , ;a;NA, M, , 2.7;(4.05;10 98);6.022;1023, 27, :3.997, , 1, , :, , ( 2 mark), , X4, , (1 mark), , Hence the nature of unit cell:face-centred cubic unit cell, , (1 mark), , Ans. The nature of cubic unit cell is fcc., , Q. 19. (a), , Since cations are deficient of electrons they accept a pair of, electrons, hence they are Lewis acids., , (1 mark), , (b), , Since ammonia molecule, NH has a lone pair of electrons to donate it, , acts as a Lewis base., (1 mark), , (c), , AlCl is a molecule with incomplete octet hence it is electron deficient, , and acts as a Lewis acid., (1 mark), , CHEMISTRY, , Youtube.com/smfreedigestchannel, , 187

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Q. 20., , Haloarenes (Aryl halides) are less reactive than (alkyl halides), haloalkanes due to the following reasons :, (1) Resonance effect : In haloarenes, the electron pairs on halogen, atom are in conjugation with -electrons of the benzene ring. The, delocalization of these electrons C-Cl bond acquires partial double, bond character., , Due to partial double bond character of C-Cl bond in aryl halides,, the bond cleavage in haloarene is difficult and are less reactive., On the other hand, in alkyl halides, carbon is attached to chlorine, by a single bond and it can be easily broken., , (1 mark), , (2) Aryl halides are stabilized by resonance but alkyl halides are not., Hence, the energy of activation for the displacement of halogen, from aryl halides is much greater than that of alkyl halides., (1 mark), (3) Different hybridization state of carbon atom in C-X bond :, (i) In alkyl halides, the carbon of C-X bond is sp-hybridized with, less s-character and greater bond length of 178 pm, which, requires less energy to break the C-X bond., (ii) In aryl halides, the carbon of C-X bond is sp-hybridized with, more s-character and shorter bond length which requires, more energy to break C-X bond. Therefore, aryl halides are, less reactive than alkyl halides., (iii) Polarity of the C-X bond : In aryl halide C-X bond is less polar, than in alkyl halides. Because sp-hybrid carbon of C-X bond, has less tendency to release electrons to the halogen than a, sp-hybrid carbon in alkyl halides. Thus halogen atom in aryl, halides cannot be easily displaced by nucleophile., 188, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , (1 mark)

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Given : [ A ] :20 mmol dm\; [ A ]t :8 mmol dm\; t:38 min; t / :?, , , , Q. 25., , k:, , 2.303, [A], , log,  [A ]t, t, , :, , 2.303, 20, log, , 38, 8, , :, , 2.303, ;0.3979, 38, , 1, , ( 2 mark), , :0.02411 min\, , (1 mark), , 0.693, k, , ( 2 mark), , t / :, , :, , 1, , 0.693, :28.74 min, 0.02411, , Ans. Half-life period:28.74 min, Q. 26. (i), , (1 mark), , A complex with the coordination number of central metal ion equal to, 4 may be tetrahedral or square planar., , (ii), , CH – C Y N;4 [ \H ], , , Na/C H OH, ,  , IIIIIIIIIIIIIIIIIIIJ, , Acetonitrile, , (iii), , CH – CH – NH, , , , , (1 mark), , (1 mark), , Ethanamine, , Proteins are the polymers of -amino acids and they are connected to, each other. The bond that connects -amino acids to each other is, called peptide bond (peptide linkage, 9CONH9)., , CHEMISTRY, , Youtube.com/smfreedigestchannel, , (1 mark), , 191

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SECTION – D, Note : This section has five long Answer Type Questions from Q. 27 to Q. 31, carrying four marks each. This section includes theory questions and one numerical, problem. You have to attempt any three questions., , Q. 27. (a), , The freezing point depression, Tf of a solution is directly, proportional to molality (m) of the solution.,  Tf . m,  Tf :Kf m, where Kf is a molal depression constant., , 1, , ( 2 mark), , The molality of a solution is given by,, m:, , Number of moles of the solute, Weight of the solvent in kg, , If W grams of a solvent contain W grams of a solute of the molar, , , mass M , then the molality m of the solution is given by,, , m:, , W ;1000, W, , , :, mol kg\, W ;M, W M,  , , , ,  Tf :Kf ;, , W ;1000, , W M,  , , 1, , ( 2 mark), 1, , ( 2 mark), , If the weights are expressed in kg then,, W, , W M,  , The unit of Kf is K kg mol 91., Tf :Kf ;, , (b), , 1, , ( 2 mark), , Given : V :8.0 dm; V :4.0 dm; Pex :43 bar, , , W:? What direction work energy flows ?, 1, , W:9Pex (V 9V ), , , :943 (498), , ( 2 mark), 1, , ( 2 mark), , :172 dm bar, :172;100 J, :17200 J, , :17.2 kJ, , (1 mark), , In this compression process, the work is done on the system and work, energy flows into the system., 192, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel

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Q. 28. (1), , Consider alkaline hydrolysis of methyl bromide (Bromomethane),, CH Br with aqueous NaOH or KOH., , ;, CH 9OH, CH 9Br, ;, OH 9, , , Bromomethane, , nucleophile, , ;, , Br 9, , Methanol, , substrate, , (2), , Stereochemistry and Kinetics of the reaction (R.D.S.) : This hydrolysis, reaction takes place only in one step which is a rate determining step, i.e. R.D.S. The rate of hydrolysis reaction depends on the concentration, of CH Br and OH 9 which are present in the R.D.S. of the reaction., , 1, Rate:R:k [ CH Br ] [ OH 9 ], ( 2 mark), , where k is rate constant of the reaction., SN reaction : The reaction between methyl bromide and hydroxide, ion to form methanol follows a second order kinetics, since the rate of, the reaction depends on the concentrations of two reacting species,, namely methyl bromide and hydroxide ion it is bimolecular second, order (2nd) Nucleophilic Substitution reaction denoted by SN., 1, , ( 2 mark), (3), , Mechanism of the reaction :, (i) It is a single step mechanism. The reaction takes place in the, following steps :, , Fig. : Backside attack of nucleophile in SN mechanism, , (1 mark), , (ii) Backside attack of the nucleophile : Nucleophile, OH 9 attacks, carbon atom of CH Br from back side i.e. from opposite side to, , that of the leaving group i.e. Br\ to experience minimum steric, repulsion and electrostatic repulsion between the incoming, nucleophile (OH\) and leaving Br\., CHEMISTRY, , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark), 193

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(iii) Transition state : When a nucleophile, OH 9 approaches carbon, atom of CH Br, the potential energy of the system increases until, , a transition state (T.S.) of maximum potential energy is formed in, which C9Br bond is partially broken and C9OH bond is partially, formed. The negative charge is equally shared by both incoming, nucleophile – OH\ and outgoing, leaving group – Br\. (Thus, the, total negative charge is diffused.), , 1, , ( 2 mark), , (iv) In CH Br, carbon atom is sp-hybridized and CH Br molecule is, , , tetrahedral. The hybridisation of carbon atom changes to, sp hybridization. The transition state contains pentacoordinate, carbon having three, , (sigma) bonds in one plane making bond, , angles of 120° with each other i.e., H , H and H atoms lie in one,  , , plane while two partial covalent bonds containing Br and OH lie, 1, , collinear and on opposite sides perpendicular to the plane. ( 2 mark), , (v) Inversion of configuration : The transition state decomposes fast, by the complete breaking of the C– Br bond and the new C– OH, bond is formed on the other side. The breaking of C– Br bond and, the formation of C– OH bond take place simultaneously. The, energy required to break the C– Br bond is partly obtained from, the energy released when C– OH bond is formed. The formation of, product CH OH is accompanied by complete or 100% inversion of, , configuration forming again sp-hybridized carbon atom giving, tetrahedral CH OH molecule. But in this structure the positions, , of H and H atoms in the reactant (CH Br) and in product are on, , , , the opposite side. This inversion of configuration is called Walden, inversion., , 194, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark)

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Q. 29. (a), , (i) Halogens have outer electronic configuration ns np., (ii) Halogens have tendency to gain or share one electron to attain the, stable configuration of nearest inert element with configuration, 1, , nsnp., , ( 2 mark), , (iii) Hence they are monovalent and show oxidation state 91., (iv) Since fluorine does not have vacant d-orbital, it shows only one, oxidation state of 91 while all other halogens show variable, oxidation states from 91 to ;7. These oxidation states are,, 91, ;1, ;3, ;5 and ;7. Cl and Br also show oxidation states, ;4 and ;6 in their oxides and oxyacids., 1, , ( 2 mark for each point), , (b), , (i) The electrochemical cell from above reactions is, > ; 3e\, Au(aq), Mg(s), , IIIIJ, , IIIIJ, , Au(s), , (Cathode reduction reaction), , > ; 2e\, Mg(aq), , (Anode oxidation reaction), , > 1 M Au(aq), Mg(s)  Mg(aq),  > 1 M  Au, (ii) (1) The atomic or ionic radii of 3-d series transition elements are, smaller than those of representative elements, with the same, 1, , ( 2 mark), , oxidation states., , (2) For the same oxidation state, there is an increase in nuclear, charge and a gradual decrease in ionic radii of 3d-series, elements is observed., Thus ionic radii of ions with oxidation state ;2 decreases, 1, , ( 2 mark), , with increase in atomic number., , (3) There is slight increase observed in Zn2; ions. With the higher, oxidation, , states,, , effective, , nuclear, , charge, , increases., , Therefore ionic radii decrease with increase in oxidation state, of the same element. For example, Fe2; ion has ionic radius, 77 pm whereas Fe3; has 65 pm., , CHEMISTRY, , Youtube.com/smfreedigestchannel, , (1 mark), , 195

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Q. 30. (1) The carbon atom adjacent to carbonyl carbon atom is called -carbon, atom (9C) and the hydrogen atom attached to -carbon atom is called, -hydrogen atom (9H)., O, , , , , , , 9 C9 C9 C9C9, , H, , 1, , ( 2 mark), , (2) The -hydrogen of aldehydes and ketones is acidic in nature due to, (i) the strong– I effect of carbonyl group, (ii) resonance stabilization of the carbanion., , 1, , ( 2 mark), , (3) Aldol condensation reaction is characteristic reaction of aldehydes and, ketones containing active -hydrogen atom., , 1, , ( 2 mark), , (4) When aldehydes or ketones containing 9H atoms are warmed with a, dilute base or dilute acid, two molecules of them undergo self, condensation to give -hydroxy aldehyde (aldol) or -hydroxy ketone, (ketol) respectively. The reaction is known as Aldol addition Reaction., 1, , ( 2 mark), (5) In aldol condensation, the product is formed by the nucleophilic, addition of -carbon atom of a second molecule which gets attached to, carbonyl carbon atom of the first molecule and -hydrogen atom of the, second molecule gets attached to carbonyl oxygen atom of the first, molecule forming (9OH) group to give -hydroxy aldehyde or ketone., 1, , ( 2 mark), (6) This is a reversible reaction, establishing an equilibrium favouring aldol, formation to a greater extent than ketol formation., , 196, , NAVNEET PRACTICE PAPERS : STANDARD XII (SCIENCE), , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark)

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(7) For aldehyde :, , 1, , ( 2 mark), Acetaldol on heating undergoes subsequent elimination of water giving, rise to ,  unsaturated aldehyde., , OH,  , CH 9 C9 CH 9C, ,  , H, , Q. 31., , O, , , , H, , , J, , , , CH 9 C: C9 C,  , , , O, , ; H O, , H, , 1, , ( 2 mark), , H H, , Crotonaldehyde (But-2-enal), , (1), , HDP means high density polyethylene. It is a linear polymer with high, density due to close packing., , (1 mark), , 333 K9343 K, , nCH :CH IIIIIIIIIIIIIIIIIIIIIJ –( CH 9CH )– n, ,  6-7 atm catalyst, , , , (1 mark), , HDP, , HDP is obtained by polymerization of ethene in presence of ZieglarNatta catalyst which is a combination of triethyl aluminium with, titanium tetrachloride at a temperature of 333K to 343K and a, pressure of 6-7 atm., Properties of HDP :, (i) HDP is crystalline, melting point in the range of 1449150 °C., (ii) It is much stiffer than LDP and has high tensile strength and, hardness., (iii) It is more resistant to chemicals than LDP., , 1, , ( 2 mark), , Uses of HDP :, (i) HDP is used in manufacture of toys and other household articles, like buckets, dustbins, bottles, pipes, etc., (ii) It is used to prepare laboratory wares and other objects where, high tensile strength and stiffness is required., (2) Gerd Bining and Heinrich Rohrer. (Nobel Prize 1986), CHEMISTRY, , Youtube.com/smfreedigestchannel, , 1, , ( 2 mark), (1 mark), 197