
	Trending ▼ ICSE CBSE 10th ISC CBSE 12th CTET GATE UGC NET Vestibulares ResFinder

CBSE 12th Board Class 12 2017 : Chemistry

18 pages, 18 questions, 0 questions with responses, 0 total responses,

Shailesh Shahabadkar
D. A. V. Public School
XII SCIENCE STREAM

+Fave Message

Profile Timeline Uploads

Home > shaileshs >

Formatting page ...

Physics class XII C.B.S.E Exam Paper 2017 Solved Page 1 of 18 Chemistry Class XII C.B.S.E Exam Paper 2017 Solved Physics class XII C.B.S.E Exam Paper 2017 Solved Page 2 of 18 General Instructions : (i) All questions are compulsory. (ii) Questions number 1 to 5 are very short answer questions and carry 1 mark each. (iii) Questions 6 to 10 are short answer questions and carry 2 marks each. (iv) Question number 11 to 22 are also short-answer questions and carry 3 marks each. (v) Question number 23 is a value based questions and carry 4 marks. (vi) Question number 24 to 26 are long-answer questions and carry 5 marks each. (vii) Use Log Tables, if necessary. Use of calculators is not allowed Chapter:Haloalkanes and Haloarenes Topic: Allylic halides Question 1: Out of the and which is an example of allylic halide? Solution: allylic halides Chapter: Chemical Kinetics Topic: Activation Energy Question 2: What is the effect of adding a catalyst on: (a) Activation energy (Ea), and (b) Gibbs energy ( G) of a reaction? Solution: (a) Catalyst provide a new reaction pathway in which a lower activation is offered. Hence, catalyst increases the rate of reaction by lowering the activation energy. (b) Gibbs free energy will remain same as for catalyzed & un catalyzed reaction, the equilibrium constant is not affected which is a function of Gibbs free energy. Chapter: P-Block elements Topic:Compund formed by halogens Question 3: Write the formula of the compound of Iodine which is obtained when conc. HNO3 oxidises I2. Solution: HIO3 (Iodic Acid) is obtained when conc. HNO3 oxidises I2. Chapter: Surface chemistry Topic: Aerosol Question 4: What type of colloid is formed when a gas is dispersed in a liquid? Give an example. Solution: Aerosol eg. Fog, mist. Physics class XII C.B.S.E Exam Paper 2017 Solved Page 3 of 18 Chapter: Alcohols, phenols and ethers Topic: IUPAC nomenclature Question 5: Write the IUPAC name of the following compound : Solution: 2-methoxy-2-methyl Propane. Chapter: p-block elements Question 6: Draw the structure of the following: (a) XeF4 (b) BrF5 Topic: Fluorides Solution: Chapter: Electrochemistry Topic: Dry cell Question 7: Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and cathode of this cell. Solution: Zinc carbon batteries were the first commercial dry batteries, developed from the technology of the wet Leclanche cell. They made flash lights and other portable devices possible, because the battery can function in any orientation. They are still useful in low drain Physics class XII C.B.S.E Exam Paper 2017 Solved Page 4 of 18 or intermittent use devices such as remote controls, flashlights, clocks or transistor radios. Reaction at ANODE: Zn(s) Zn2+(aq) + 2 e Reaction at Cathode: 2 MnO2(s) + 2 e + 2 NH4Cl(aq) Mn2O3(s) + 2 NH3(aq) + H2O(l) + 2 Cl Chapter: Alcohols phenols and ethers Topic: Stability and Mechanism Question 8: (a) Arrange the follwowing compounds in the increasing order of their strength: p-cresol, p-nitrophenol, phenol (b) Write the mechanism (using curved arrow notation) of the following reaction: Solution: (a)Below is the increasing order of the compounds according to their strength: o-Cresol < Phenol < o-Nitrophenol. (b) Or Write the structures of the products when Butan-2-ol reacts with the following: (a) CrO3 (b) SOCl2 (Alcohols phenols and ethers) Solution: (a) (b) Physics class XII C.B.S.E Exam Paper 2017 Solved Page 5 of 18 Chapter: Coordination compounds Topic: IUPAC nomenclature Question 9: Using IUPAC norms write the formulae for the following: (a) Potassium trioxalatoaluminate(III) (b) Dichlorido bis (ethane- 1,2 -diamine) cobalt(II) Solution: (a)K3[Al(C2O4)] (b)[COCl2(en)2]+ Chapter:Solid state Topic: Number of unit cells Question 10: Calculate the number of unit cells in 8.1 g of aluminum if it crystallizes in a facecentered cubic (f.c.c.) structure. (Atomic mass of Al = 27g mo1-1) Solution: = nAl = 8.1 2.7 = 0.3 We know that one unit of f.c.c., No. of atoms = 4 4 - atoms are there in unit cell = 1 1 - atoms are there in unit cell = (1 mole) NA atoms are found in unit cell = NA/4 0.3 moles atoms are found in unit cell = NA/4 0.3 .075 NA Chapter: Coordination compounds Topic: Isomerism and Hybridization Question 11: (a) What type of isomerism shown by the complex[Co(NH3)5 (SCN)]2+ ? (b) Why is [NiCl4]2- paramegnatic while [Ni(CN)4]2- is diamagnetic? (Atomic number of Ni = 28) (c) Why are low spin tetrahedral complexes rarely observed? Solution: (a) This complex is Linkage isomerism (b) Because Ni[Cl4]2- has 2 unpaired electrons and is Paramagnetic. While Ni[CN4]2- has paired electrons and is Diamagnetic. (c) There are only 4 ligands in a tetrahedral complex. So,the ligand field is roughly 2/3 of the octahedral field. The direction of DC ligand approach in tetrahedral complex does not coincide with the d-orbitals. This reduces the field by a factor of 2/3. Hence, t is roughly 2/3 x 2/3 = 4/9 of o. As a result, all tetrahedral complexes are high-spin since the CFSE is normally smaller than the paring energy. Physics class XII C.B.S.E Exam Paper 2017 Solved Page 6 of 18 Chapter: Surface chemistry Topic: Coagulation and peptization Question 12: Write one difference in each of the following: (a) Multi-molecular colloid and associated colloid (b) Coagulation and Peptization (c) Homogeneous catalysis and Heterogeneous catalysis Or (a) Write the dispersed phase and dispersion medium of milk. (b) Write one similarity between Physisorption and chemisorption. (c) Wrtie the chemical method by which Fe(OH)3 sol is prepared from FeCl3. Solution: (a) In Multimolecular colloid the atoms or molecules are held by weak vander walls forces and In associated colloid. They behave as normal electrolytes at low concentrations and colloidal only at high concentrations. (b) The process of Changing the colloid particle in a solution into insoluble precipitate by addition of suitable electrolyte is known as cogulation and conversion of freshly precipitated substance in to colloidal solution by shaking with the suitable electrolyte is called peptization. (c) In Homogenous catalysis the catalyst and reactant are in the same phase and In Hetrogeneous catalysis the catalyst is in different phase than are the reactants. Or (a)Milk is an emulsion in which both the dispersed phase and the dispersion medium are in liquid state. In milk, liquid fat is dispersed in water. So fat is the dispersed phase and water is the dispersion medium. (b)Both are surface phenomenon. (c)A colloidal solution of Fe[OH]3 is prepared when concentrated solution of ferric chloride is added drop wise to hot water. FeCl3 + 3H2O Fe[OH]3 + 3HCl Chapter : Electrochemistry energy Question 13: (a) The cell in which the following reactions occurs: Topic: Gibbs free Eocell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F = 96,500 C mol-1) (b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F = 96,500 C mol-1) Physics class XII C.B.S.E Exam Paper 2017 Solved Page 7 of 18 Solution: (a) E0 cell = 0.236 V G = -NFE n=2, F=96500C G0 = 2 (96500 C) (0.236 V) = - 45548 J Or = - 45.55 KJ G = -2303 RT log KC logKc = G0 / 2.303 RT = - 45.55 / (2.303 8.314 10-3 298) = 7.983 Kc = antilog (7.983) = 9.62 107 (b) Current, I = 0.5 A Time, t = 2 hours In sec we get Time, t = 2 60 60 s = 7200 s Charge: Q = I t = 0.5 A 7200 s = 3600 Coulombs Number of electrons = total charge / charge on 1 electrons = 3600 / (1.6 1019) = 2.25 1022 electrons Chapter: d & f block elements Topic: Structure and chemical properties Question 14: (a) Based on the nature of intermolecular forces, classify the following solids : Silicon carbide, Argon (b) ZnO turns yellow on heating. Why ? (c) What is meant by groups 12-16 compounds ? Give an example. Solution: (a) On the basis of intermolecular forces : (i) Silicon carbide : - Covalent or network solid (Covalent Bonding) (ii) Argon : - Non-polar molecular solid which posses dispersion or london forces. (b) zinc oxide is white in colour at room temperature. On heating it loses oxygen & turns Physics class XII C.B.S.E Exam Paper 2017 Solved Page 8 of 18 yellow 1 ZnO Zn+2 + 2 + 2e 2 the excess zn+2 ions move to interstitial sites and the electron to neighbouring interstitial sites. (c) Some of the compound like Zns, CdSe and HgTe are example of group 12 16 compound. In these compound bonds are having same ionic character along with covalent. Chapter : General principles and processes of isolation of elements Topic: Electrolytic refining and froth floatation process Question 15: (a) Write the principle of electrolytic refining? (b) Why does copper obtained in the extraction from copper pyrites have a blistered appearance? (c) What is the role of depressants in the froth floation process? Solution: (a) Electrolytic refining: It is the process of refining impure metals by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. (b) Copper is extracted from its principal ore copper pyrites (CuFeS2). The ore is concentrated by froth flotation process. The molten copper is poured out and allowed to cool. The sulphur dioxide escaping from the melt gets trapped in the cooler parts of the surface giving a blistery appearance for copper and hence it is called blister copper. (c) In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. Chapter : Compounds containing nitrogen Topic: Conversion Question 16: Write the structures of compounds A, B and C in the following reactions : NH3 / (a) CH3 COOH Br2 /KOH(aq) CHCl3 +alc.KOH A NaNO2 / Cu / B C Fe/HCl CH3 COCl/pyridine (b) C6 H5 N2+ BF4 A Solution: (a) A: CH3CONH2 (Ethanamide) B: CH3NH2 (Methenamine) C: CH3NC (Methyl isocyanide) (b) A: C6H5NO2 (Nitrobenzene) B C Physics class XII C.B.S.E Exam Paper 2017 Solved Page 9 of 18 B: C6H5NH2 (Aniline) C: Chapter : Compounds containing nitrogen Topic: Chemical properties Question 17: Give reasons for the following: (a) Acetylation of aniline reduces its activation effect. (b) CH3NH2 is more basic than C6H5NH2. (c) Although -NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. Solution: (a) Because with acetylation of aniline is result in decrease of electron density on nitrogen so activation effect reduces. (b) CH3NH2 is more basic than C6H5NH2 because of positive I effect. (c) Although amino group is o, p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline: Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is metadirecting). For this reason, aniline on nitration gives a substantial amount of m-nitroaniline. Chapter: P-block elements Topic: Chemical properties Question 18. Give reasons for the following : (a) Red phosphorus is less reactive than white phosphorus. (b) Electron gain enthalpies of halogens are largely negative. (c) N2O5 is more acidic than N2O3. Solution: (a) Red phosphorus are less reactive than white phosphorus as the white phosphorous posses angle strain in the P4 molecule where the angle are only 60 & also they have low M.P. (b) Electron gain enthalpy of halogen are largely negativity it is due to the fact that they have high effective nuclear charge & smallest size among period. Although they contain 7e in Physics class XII C.B.S.E Exam Paper 2017 Solved Page 10 of 18 valence shell & required one electron to attain their nearest noble gas configuration. (c) N2O5 is more acidic then N2O3 as in N2O5 the N is in +5 O.S. while in N2O3 it is in +3 O.S. So higher the oxidation state of central atom in a given oxide, higher will be acidic character" Chapter : Chemistry in everyday life Topic: Detergents & Medicines Question 19: Define the following: (a) Cationic detergent (b) Broad spectrum antibiotics (c) Tranquillizer Solution: (a) Cationic detergent: a type of detergent in which the active part of the molecule is a positive ion (cation). Cationic detergents are usually quaternary ammonium salts and often also have bactericidal properties. (b) Broad spectrum antibiotics: The term broad-spectrum antibiotic refers to an antibiotic that acts against a wide range of disease-causing bacteria. A broad-spectrum antibiotic acts against both Gram-positive and Gram-negative bacteria eg. Amoxicillin, Streptomycin (c) Tranquillizers: a drug used to reduce stress or tension without reducing mental clarity. Chapter: Polymers Topic: Teflon, neoprene Question 20: Write the structures of the monomers used for getting the following polymers: (a) Teflon (b) Melamine-formaldehyde polymer (c) Neoprene Solution: (a) Teflon (PTFE) - The monomeric unit present is tertrafluoroethene (n F2C = CF2). Teflon is used as a material resistant to heat and chemical attack. It is also used for making gaskets, pumps packing, valves, oil seals, non lubricated bearings. (b) Melamine-formaldehyde polymer Physics class XII C.B.S.E Exam Paper 2017 Solved Page 11 of 18 (c) Neoprene - It is prepared by the free radical polymerization of chloroprene. It is used as insulator for making belts , gaskets , hoses etc Chapter: Haloalkanes and Haloarenes Topic: SN1 & SN2 reaction Question 21: The following compounds are given to you: 2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane (a) Write the compound which is most reactive towards SN2 reaction. (b) Write the compound which is optically active. (c) Write the compound which is most reactive towards -elimination reaction. Solution: (a) 1- Bromopentane is most reactive towards SN2 reaction. (b) 2-Bromo-2-methylbutane (c) 2- Bromopentane is most reactive towards -elimation reaction with product of pent-1-ene pent-2-ene. Chapter: Chemical kinetics Topic: First order reaction Question 22: A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed? (Given: Log 2 = 0.3010, Log 3 = 0.4771, Log 4 = 0.6021) Solution: For the first order reaction = (2.303 / ) ( / ) When x = (25 / 100) a = 0.25 a t = 20 minutes (given) Therefore, k = (2.303 / 20) log (a / a 0.25 a) k = (2.303 / 20) log (1 / 0.75) = 0.013120 min-1 Hence the value of the rate constant is 0.013120 min-1 Now we need to find Time for rest 75% reaction t= ?, when x = 0.75 a From above, k = 0.013120 min-1 Therefore, t = (2.303 / 0.013120) log (a / a 0.75 a) = (2.303 / 0.013120 ) log (1 / 0.25 ) = 105.6 min. Physics class XII C.B.S.E Exam Paper 2017 Solved Page 12 of 18 The time at which the reaction will be 75% complete is 105.6 min. Chapter: Biomolecules Topic: Polysaccharide Question 23: After watching a programme on TV about the presence of carcinogens (cancer causing agents) Potassium bromate and Potassium iodate in bread and other bakery products, Rupali a class XII student decided to make others aware about adverse affect of these carcinogens in foods. She consulted the school principal and requested him to instruct canteen contactor to stop selling sandwiches, pizzas, burgers and other bakery products to the students. The Principal took an immediate action and instructed to canteen contractor to replace the bakery products with some protein and vitamin rich food like fruit, salads , sprouts etc .The decision was welcomed by the parents and the students. After reading the above passage, answer the following questions: (a) What are the values (at least two) displayed by the Rupali? (b) Which Polysaccharide component of carbohydrates is commonly present in bread? (c) Write the two types of secondary structures of proteins. (d) Give two examples of water solube vitamins. Solution: (a) 1.To aware about adverse effect of carcinogen present in bread. 2.To Replace the bakery product with some protein and vitamin rich food like fruits, salads etc. (b) Starch, Glycogen (c) Alpha heix, Beta pleated sheet (d) Riboflavin, Thiamine Chapter:Aldehyde, ketones and Carboxyclic acid Question 24: (a) Write the product(s) in the following reactions : Topic: Conversion Physics class XII C.B.S.E Exam Paper 2017 Solved Page 13 of 18 (b) Give simple chemical tests to distinguish between the following pairs of compounds : (i) Butanal and Butan-2-one (ii) Benzoic acid and Phenol Or (a) Write the reactions involved in the following : (i) Etard reaction (ii) Stephen reduction (b) How will you convert the following in not more than two steps : (i) Benzoic acid to Benzaldehyde (ii) Acetophenone to Benzoic acid (iii) Ethanoic acid to 2-Hydroxyethanoic acid Solution: (a) Physics class XII C.B.S.E Exam Paper 2017 Solved Or (a).(i) (ii) (b) (i) Page 14 of 18 Physics class XII C.B.S.E Exam Paper 2017 Solved Page 15 of 18 (ii) (iii) Chapter: D & f block elements Topic: Chemical properties Question 25: (a) Account for the following: (i) Transition metals show variable oxidation states. (ii) Zn, Cd and Hg are soft metals. (iii) E value for the Mn3+/Mn2+ couple is highly positive (+ 1.57 V) as compared to Cr3+/Cr2+. (b) Write one similarity and one difference between the chemistry of lanthanoid and actionoid elements. Or (a) Following are the transition metal ions of 3d series: Ti4+. V2+, Mn3+, Cr+3 (Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24) Answer the following: (i) Which ion is most stable in an aqueous solution and why? (ii) Which ion is a strong oxidising agent and why? (iii) Which ion is colourless and why? (b) Complete the following equations: + 2 (i) 2MnO 4 + 16H + 5S heat (ii) KMnO4 Solution: (a) Physics class XII C.B.S.E Exam Paper 2017 Solved Page 16 of 18 (i) The variable oxidation states of transition elements are due to the participation of ns and (n1)d-electrons in bonding. Lower oxidation state is exhibited when ns-electrons take part in bonding. Higher oxidation states are exhibited when (n - 1)d-electrons take part in bonding. In each group, the highest oxidation state increases with the increase in atomic number; reaches the maximum in the middle, and then starts decreasing. (ii) Zn, Cd and Hg are soft metals due to absence of any unpaired electrons resulting weak metal-metal bonding. (iii) Mn2+ exists in half-filled d5 state which is highly stable while Mn3+ is d4 which is not very stable. Conversion from d4 to d5 will be quick and have negative AGvalue. Hence, because of the stability factor the E value is high for this process. While, Cr3+ is d3 is half filled (t293) is stable in nature and Cr2+ is d4 has one extra electron which it would like to donate to attain the stable half filled (t293 ) configuration. Hence for the process Cr3+ to Cr2+ the value of E is less. (b) Similarity: They both show contraction of radii, the progressive decrease in the radii of atoms of the lanthanoid and actionoid elements as the atomic number increases. Difference: Actinoids display a large number of oxidation states while lanthanoids primarily show only three oxidation states. Or (ii) An examination of the E values for the redox couple M3+/M2+ (from electrode potential table) shows that Mn3+ ion are the strongest oxidising agents in aqueous solutions. (iii) Only the ions that have electrons in d-orbital and in which d-d transition is possible will be coloured. The ions in which d-orbitals are empty or completely filled will be colourless as no dd transition is possible in those configurations. From the given transition metal ions, it can be easily observed that only Ti4+ has an empty d-orbital, so, it is colourless ion. Physics class XII C.B.S.E Exam Paper 2017 Solved Page 17 of 18 (b) Chapter: Solutions Topic: Coagulative properties. Question 26: (a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K. Given : (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) (b) Define the following terms: (i) Molality (m) (ii) Abnormal molar mass Or -1 (a) 30g of Urea (M = 60 g mol ) is dissolved in 846 g of water. calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. (b) Write two difference between ideal solution and non-ideal solutions. Solution: (a) For Sucrose: n = 10 / 342 n = 0.292 mol Now, m = 0.0292*1000/90 = 0.3244 We know that, Kf = Tf / M = 4 / 0.3244 = 12.33 Kg / Mol Now for Glucose, n = 10 / 180 = 0.055 mol Now, m = 0.055 * 1000 /90 = 0.6160 mol / kg Again we know that, Tf = Kf * m = 12.33 * 0.616 Physics class XII C.B.S.E Exam Paper 2017 Solved Page 18 of 18 = 7.60 Tf = 273.15 -7.60 = 265.55 = 0.055 mol So, freezing point of 10% glucose in water is 0.055 mol. (b) (i) Molality is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol/kg. (ii) Abnormal molar masses and colligative properties. For the solute which undergo association or dissociation, observed value of colligative property is different from the calculated value of colligative property. Or (a) Vapour pressure of water, p1 = 23.8 mm of Hg Weight of water = 846 g Weight of urea = 30 g Molecular weight of water (H2O) = 1 2 + 16 = 18 g mol 1 Molecular weight of urea (NH2CONH2) = 2N + 4H + C + O = 2 14 + 4 1 + 12 + 16 = 60 g mol 1 Number of moles of water n1 = 846 / 18 = 47 Number of mole of urea n2 = 30 / 60 = 0.5 Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1. Use the formula of Raoult s law (P10 - P1) / P10 = n2 / (n1-n2) Plug the values we get (23.8 - p1) / 23.8 = 0.5 / (47+0.5) (23.8 - p1) / 23.8 = 0.5106 after cross multiply 23.8 p1 = 23.8 0.5106 Solve it we get p1 = 11.6 mm Hg So, Vapour pressure of water in the given solution = 11.6 mm of Hg (b) Ideal Solution Non-ideal Solution Obey Raoult's law at every range of concentration. Do not obey Raoult's law. Hmin = 0 Neither heat is evolved nor absorbed during dissolution. Hmin > 0 Endothermic dissolution; heat is absorbed.

Related ResPapers
CBSE 12th Board Class 12 2020 : Chemistry
by abhishekkmr0025

CBSE 12th Board Class 12 2017 : Chemistry
by shaileshs

CBSE 12th Board Class 12 2020 : Chemistry
by sifan

CBSE 12th Board Class 12 2020 : Chemistry
by 9369655917r

Formatting page ...

shaileshs chat

Horizontal lines at:
Guest Horizontal lines at:
AutoRM Data:
Box geometries: