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FIRST PRE BOARD EXAM CLASS XII BLUE PRINT SL NO 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 UNITS VSA 1 SA II 3M M( NO. Q) -3(1) 3(1) -3(1) 3(1) LA 5M M( NO. Q) ---5(1) --- TOTAL ----1(1) -- SA 1 2M M( NO. Q) 4(2) 2(1) 2(1) ---- SOLID STATE SOLUTIONS ELECTROCHEM CHEM KINETICS SURFACE CHEM ISOLATION OF ELEMENTS P BLOCK D AND F BLOCK CO-ORDINATION COMPOUNDS HALO ALKANES AND HALO ARENES ALCOHOLS,PHENOLS AND ETHERS ALDEHEYDES,KETONES AND CARB ACIDS ORGANIC COMPS CONT NITROGEN BIOMOLECULES POLYMERS CHEMISTRY IN EVERYDAY LIFE TOTAL --1(1) -2(1) 2(1) 3(1) 3(1) -- 5(1) --- 8(2) 5(2) 3(2) -- 4(2) -- -- 4(2) 1(1) -- 3(1) -- 4(2) 1(1) -- -- 5(1) 6(2) 1(1) -- 3(1) -- 4(2) 1(1) 1(1) 1(1) -2(1) 2(1) 3(1) --- ---- 4(2) 3(2) 3(2) 8(8) 20(10) 27(9) 15(3) 70(30) 4(2) 5(2) 5(2) 5(1) 4(2) 3(1) KENDRIYA VIDYALAYA GANESHKHIND PUNE Time: Three Hours CLASS – XII CHEMISTRY I PRE BOARD EXAM Max. Marks: 70 General Instructions 1. All questions are compulsory. 2. Question nos. 1 to 8 are very short answer questions and carry 1 mark each. 3. Question nos. 9 to 18 are short answer questions and carry 2 marks each. 4. Question nos. 19 to 27 are also short answer questions and carry 3 marks each 5. Question nos. 28 to 30 are long answer questions and carry 5 marks each 6. Use log tables if necessary, use of calculators is not allowed. Q1.Define peptisation.What is colloidon? The process of converting the precipitate in to colloidal particles on adding a suitable electrolyte. 1/2M Colloidon is a sol of cellulose nitrate in ethyl alcohol 1/2M Q2. How do the co-ordination compounds show colour? Co-ordination compounds impart colour due to the presence of unpaired electrons in the central metal atom or ion. 1/2M The unpaired electrons show d-d transition by jumping between t2g and eg orbitals. 1/2M Q3. Write the structures of the each of the following alcohols. 1)3,3-dimethyl cyclopentanol 2)3-chloro-2-methyl-1-butanol 1) 1/2M 2) CH3-CH(Cl)-CH(CH3)-CH2OH 1/2M Q4.What does PCC stand for?Give its one use. PCC stands for pyridunuum chlorochromate. 1/2M It is used to oxidise 10 alcohol to aldehydes and 20 alcohols to ketones. 1/2M Q5. Why are alkyl cyanides more volatile than the corresponding acids? Alkyl cyanides are more volatile because they can not associate via intermolecular hydrogen bonding. 1M Q6. What do you understand by reducing sugars?Give one example. Carbohydrates which contain free aldehyde group and reduce Fehling solution or Tollens reagent are called reducing sugars. 1/2M Eg Glucose 1/2M Q7. What are co-polymers? Give one example. Co-Polymers are obtained by the polymerisation of two or more than two different monomers. 1/2M Eg Nylon 66. 1/2M Q8. Why aspartame has a limited use? Aspartame has a limited use because it is unstable at cooking temperture. 1M Q9. Aluminium crystallises in a cubic close packed structure.Its metallic radius is 125pm.What is the length of the side of the unit cell? CCP means fcc. Z= 4 r=125pm 1/2M For FCC 4r = 2 a 1M a= 4x125/2 a= 353.6pm 1/2M Q10.Explain the following terms a)Schottky defect b)Frenkel defect Schottky defect:- It is a stoichiometric defect and is formed when one cation and one anion is missing from the lattice.The lattice remains electrically neutral. 1M Frenkel defect:- This also is a stoichiometric defect in which one ion ( preferably cation) occupies an interstitial site instead of its correct crystal site.Electrical neutrality and stoichiometry is maintained. 1M Q11. When two solutions X and Y are mixed the solution becomes hot.When Y and Z are mixed the solution becomes cold.Which of these solutions will exhibit negative deviation from ideal behaviour? Solution of X and Y exhibit negative deviation.Inthis X—Y interaction is much stronger than the X—X and Y—Y interaction hence heat is evolved. 2M Q12.What happens to conductivity of solution with dilution?Why? The conductivity of solution decreases on dilution 1M because the number of ions per unit volume that carry the current in a solution decreases. 1M Q13. Give oxidising property of KmnO4 in neutral medium with example. KmnO4 is a moderate oxidising agent in neutral medium because it gives three oxygen atom in water. 1M 2KmnO4 + H2O 2KOH + 2MnO2 + 3 [O] In water KmnO4 oxidises manganese sulphate to manganese dioxide. 3MnSO4 3H2O + 3O 3MnO2 + 3H2SO4 1M Q14.Write IUPAC names of the following co-ordination compounds. a) [PtCl(NH2CH3)(NH3)2]Cl b) [Co(en)3]3+ a)Diamminechloro (methylamine) platinum (II) chloride 1M b)Tris (1,2-ethanediamine) cobalt (III) ion 1M Q15. How would you convert 2-methyl propene to a)Tertiary butyl bromide b)iso-butyl bromide a) CH3—C (CH3) = CH2 + HBr CH3 – C (CH3) (Br) – CH3 1M b) CH3 – C (CH3) = CH2 + HBr (Benzoyl peroxide) CH3—CH ( CH3) – CH2Br 1M Q16.Convert a)Chlorobenzene to bromobenzene b)Chlorobenzene to toluene C6H5—Cl + NH3 C6H5—NH2 NaNO2/HCl C6H5N2Cl CuBr/HBr C6H5—Br. 1M C6H5—Cl NaOH/High Temp/High Press C6H5—OH Zn C6H6 C6H6 + CH3 Cl/FeCl3 C6H5—CH3 1M Q17.Explain the structure of natural rubber.Name its monomeric unit. The natural rubber is a linear 1,4-polymer of isopropene. 1M Monomer is isoprene, 1M Q18. What is an antibiotic? Name the first antibiotic discovered.Is this antibiotic narrow spectrum or broad spectrum antibiotic? Antibiotics are chemical substances which are produced by microorganisms inorder to inhibit the growth or even to destroy other microorganisms. 1M The first antibiotic discovered was penicillin. 1/2M It is a narrow spectrum antibiotic. 1/2M Q19.Calculate the mass of ascorbic acid (Vitamine C = C6H8O6) to be dissolved in 75 g acetic acid to lower its melting point by1.50C .Kf=3.9K Kg mol—1. Tf=1.5K W1 = 75g Molar mass of solute Mb = 72+8+96 =176 g mol—1 Kf = 3.9 K kg mol—1 Tf = kf x Wb x 1000/Wa x Mb 1+1/2M Wb = Mb x Wa x Tf/1000 x kf 176 x 75 x 1.5/1000 x 3.9 = 5.077 1+ 1/2 Q20.How much electricity in coulombs is required for the oxidation of a) 1 mol of H2O to O2 b) 1 mol of FeO to Fe2O3 H2O 2H+ + ½ O2 + 2e— 1M Quantity of electricity required is 2F = 2 x 96500 C 1/2M Fe 2+ Fe3+ e—1 ( FeO ½ Fe2O3) Quantity of electrons required is 1F = 96500 C 1M 1/2M Q 21. Why a) True solutions do not show Tyndall effect? b) Medicines are more effective in colloidal state. c) Alum is added to purify muddy water a) Because the particle size is so small that no scattering of light is possible. 1M b) A colloidal state has a larger surface area. Thus medicines in colloidal state are effectively adsorbed and assimilated and give better results. 1M c) Since the clay particles present are negatively charged alum provides Al3+ ions which neutralise the negatively charge of the particles which then coagulate and get settled at the bottom. 1M Q22.The value of fG0 for the formation of Cr2O3 is –540kj/mol and that of Al2O3 is –827 kj/mol.Is the reduction of Cr2O3 possible with Al? 2Cr + 3/2 O2 Cr2O3 fG0 = -540kj/mol Cr2O3 2Cr + 3/2 O2 fG0 = +540kj/mol………1 2Al + 3/2O2 Al2O3 fG0 = -827kj/mol……………..2 1+ 2 represents reduction of Cr2O3 with Al. 1M 1M Since fG0 of this reaction is negative the reduction is possible. 1M Q23.Starting from elemental sulphur how would you prepare the following a)H2SO4 b)SCl2 c)SF6 a) 1/8 S8 + O2 SO2 2SO2 + O2 V2O5 2SO3 SO3 + H2SO4 H2S2O7H2O H2SO4 1M b)1/8S8 + Cl2 SCl2 1M c)1/8S8 + 3F2 SF6 1M Q24.Describe the preparation of potassium dichromate from ferrochrome. I Step Fusion of chomite ore with sodium carbonate and O2 4FeCr2O4 + 8Na2CO3 + 7O2 8Na2CrO4 + 2Fe2O3 + 8CO2 II Step Acidification of chromate ore 2Na2CrO4 + 8H2SO4 Na2Cr2O7 + Na2SO4 + H2O III Step Conversion of sodium dichromate to potassium dichromate Na2Cr2O7 + 2KCl K2Cr2O7 + 2NaCl A Q25.How are the following conversions carried out? 1M 1M 1M a)Toluene to benzyl alcohol b)Ethyl magnesium chloride to propan-1-ol c)Methyl magnesium bromide to 2-methyl propane-2-ol a)C6H5—CH3 Cl2/UV LIGHT C6H5—CH2Cl Aq KOH C6H5—CH2 OH 1M b)CH3—CH2 MgCl + H—CHO H—CH(CH2—CH3)—OMgCl H2O CH3—CH2-CH2—OH 1M c)CH3 Mg Br + CH3—CO—CH3 H2O CH3—CH(CH3)(OH) – CH3 1M Q26.Explain why does nitrobenzene on nitration with nitric acid and sulphuric acid forms only meta dinitro benzene? NO2 group is electron withdrawing .The electron displacement is away from the benzene.As a result O and P carbon acquires positive charge.So meta position is more electron rich and is favourable for an electrophile (NO2+). 1M Show resonance hybrids. 2M Q27.Give any three points of differences between DNA and RNA. 3M DNA RNA 1.It is mainly found in chromosomes It is mainly found in cytoplasm but also found in chromosomes 2.Has de oxy ribose sugar 2.Has ribose sugar 3.Four nitrogenous bases in DNA are 3.The four nitrogenous bases are A,G,U and A,G,T,C C. 4.It has double helix structure. 4.It has single helix structure. Q28.From the following data for the reaction between A + B A+BM+N Expt no [A] mol/l [B] mol/l Initial Rate mol/l/sec At 300K 1 2.5 x 10--4 3.0 x 10--5 5.0 x 10—4 2 5.0 x 10--4 6.0 x 10--5 4.0 x 10—3 3 1.0 x 10--3 6.0 x 10--5 1.6 x 10—2 1.Calculate order of reaction with respect to A and B 2.Rate constant at 300K 3.Energy of activation if the rate constant at 320 K is 1.07 x 109 Assume the rate law to be r= k[A]x [B]y……………………1 5.4 x 10—4 = )x (3.0 x 10—5 )y…………………2 4.0 x 10 –3 = (5.0 x 10—4)x (6.0x 10—5)y……………………3 1.6 x 10—2 = (1.0 x 10—3)x (6.0 x 10—5)y……………………4 4/3 …………………1.6x 10—2/4.0 x 10—3 = (1.0 x 10—3)x/(5.0 x 10—4)y 2x = 4 x= 2 1M 3/2 , 4.0 x 10 –3/5.4 x 10—4 = 2x 2y 4 x 2y =8 y=1 1M order with respect to A = 2 and with respect to B = 1. Rate law is r = k [A]2 [B] K300 = r/[A]2[B] K300 = 5.4 x 10—4/5.4 x 10—4 x3.0 x 10—5 = 2.7 x 108 1M 1M k320 = 1.07 x 109 Ea = 2.303 x 8.314 x 300x 320 /20 log (10.7/2.7) Ea = 54962j/mol Ea = 54.962 1M OR The activation energy for the reaction 2HI (g) H2(g) + I2 (g) is 209.5 kj/mol. At 581 K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy? Y = e –Ea/RT 1M Logy = --Ea / 2.303RT --209500/2.303x8.314 x581 = --18.8323 2M y = antilog(--18.8323) y = 1.47 x 10—19 2M Q29.Discuss oxides of nitrogen.Give structure and calculate the oxidation number of nitrogen in each of them.Arrange the oxides in order of increasing acid strength. Nitrogen forms a wide range of oxided .The oxidation number of nitrogen in these oxides varies from +1 to +5.The oxides are 1)N2O dinitrogen oxide It is formed by gently heating ammonium nitrate. NH4NO3 N2O + 2H2O N2O is the oside of nitrogen with the lower oxidation number +1. The structure of N2 O is 1M 2)Nitrogen oxide NO. NO is also known as nitric oxide.The oxidation number of nitrogen is +2.NO can be prepared by in the laboratory by reducing anitrite with mild reducing agent such as I—1 2HNO2 + 2HI 2NO2 +I2 + H2O The structure of NO is 1M 3)Dinitrogen trioxide N2O3. This can be obtained by condensing NO and NO2 together Its structure is 1M 4)Nitrogen dioxide NO2.It is obtained by thermal decomposition of lead nitrate. 2Pb(NO3)2 4000C 4NO2 + PbO + O2 The oxidation number of nitrogen in NO2 is +4. Its structure is 5)Dinitrogen pentoxide N2O5 Oxidation number of nitrogen in +5. It is obtained by the reaction of P2O5 on HNO3. 2HNO3 + P2O5 N2O5 + 2HPO3 Its structure is 1M The order of increasing acidic strength is N2O NO,N2O3,NO2,N2O5 1M OR Illustrate how nitrogen compounds provide good examples of multiple bonding and resonance. Show the resonance hybrids 5M Q30.a)Explain the mechanism of a nucleophilic aattack on carbonyl group of an aldehyde or a ketone. b)An organic compound A molecular formula C8H16O2 was hydrolysed with dilute sulphuric acid to give a carboxylic acid B and an alcohol C.Oxidation of C with chromic acid also produced B.On dehydration C gives but-1-ene.Write equations of reactions involved. a) =C=O + Nu— =C(Nu) – O -- E+ Nu—C—OE 2M b)There are two possible structures of A CH3-CH2-CH2-CO-O-CH2-CH2-CH2-CH3 CH3-CH(CH3)-CO-O-CH2-CH(CH3)-CH3 CH3-CH2-CH2-CO-O-CH2-CH2-CH2-CH3 + H2O/H+ CH3-CH2-CH2-COOH + (B) CH3-CH2-CH2-CH2-OH ( (C) CH3-CH2-CH2-CH2-OH Chromic acid CH3-CH2-CH2-COOH (C) CH3CH(CH3)COOCH2CH(CH3)CH3 H2O/H+ CH3CH(CH3)CH2OH + (C ) CH3CH(CH3)COOH (B) CH3—CH(CH3) CH2OH Chromic acid CH3 CH(CH3) COOH OR a)Give chemical test to distinguish between the following pairs of compounds. 1) ethanal and propanal 2) Phenol and benzoic acid. b)How will you bring about the following conversions 1.Benzoic acid to benzaldehyde 2.Ethanal to but-2-ene 3.Propanone to propene. Give complete reaction in each case. 3M a) Ethanal answers iodoform test where as propanal does not give iodoform test. 1M CH3CH2OH I2/NaOH CHI3 + Na2CO3 b)C6H5-OH + NaHCO3 No effervescence 1M C6H5-COOH + NaHCO3 Brisk effervescence due to liberation of CO2. 1. Benzoic acid to benzaldehyde C6H5-COOH + SOCl2 C6H5-COCl H2/Pd/BaSO4 C6H5—CHO 1M 2.CH3 CHO + CH3—CH2MgBr/H2O CH3—CH(OH)—CH2—CH3 Alc KOH CH3—CH=CH—CH3 1M 3. CH3—CO—CH3 [H] CH3—CH(OH)—CH3Alc KOH CH3-CH=CH2 1M …………………………………………………………………………………………………….