CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2014

Time allowed: 3 hours                                                                                      Maximum Marks: 70

General Instructions:

  1. All questions are compulsory.
  2. Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
  3. Questions number 6 to 10 are short-answer questions and carry 2 marks each.
  4. Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
  5. Questions number 23 is a value based question and carry 4 marks.
  6. Questions number 24 to 26 are long-answer questions and carry 5 marks each.
  7. Use log tables, if necessary. Use of calculators is not allowed.

SET I

Note : Except for the following questions, all the remaining question have been asked in previous sets.

Question.1. Give one example each of ‘oil in water’ and Water in oil’ emulsion.
Answer : Oil in water — Milk
Water in oil – Butter

Question.2. Which reducing agent is employed to get copper from the leached low grade copper ore ?
Answer : Scrap iron is used as reducing agent to obtain copper metal from the solution containing copper. Cu2+(aq) + Fe(s) —-> Cu(s) + Fe2+(aq)

Question.3. Which of the following is more stable complex and why ?
[Co(NH3)6]3+and [Co(en)3]3+
Answer: NH3 is a unidentate ligand and H2NCH2CH2NH?(en) is a bidentate ligand. Chelating ligands form more stable complexes compared to non-chelating ligands. Thus, [Co(en)3]‘ is more stable.

Question.4. Write the IUPAC name of the compound.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-1

Question.5. Which of the following isomers is more volatile: o-nitrophenol or p-nitrophenol ?
Answer: o-nitropenol is more volatile because of the presence of intra-molecular hydrogen bonding.

Question.6. What are isotonic solutions ?
Answer: Two or more solutions having same osmotic pressure are called isotonic solutions, e.g. 0.5 M NaCl, 0.5 M KCl and 1M glucose are isotonic.

Question.7. Arrange the following compounds in increasing order of solubility in water :
C6H5NH2, (C2H5)2NH, C2H5NH2
Answer : C6H5NH2 < (C2H5)2NH < C2H5NH2

Question.8. Which of the two components of starch is water soluble ?
Answer : Amylose is the water soluble content of starch between Amylose and Amylopectin.

Question.9. An element with density 11.2 gm cm-3 forms f.c.c. lattice with edge length 4 x 10-8cm. Calculate the atomic mass of the element. (Given NA = 6.022 x 1023 mol-1)
Answer: Given,
d= 11.2 g cm-1 NA = 6.022 x 1023 mol-1
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Question.10. Examine the given defective crystal
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Answer the following questions :
(i) What type of stoichiometric defect is shown by crystal ?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect ?
Answer : (i) Schottky defect, as equal number of cations and anions are missing.
(ii) The density of the crystal decreases as ions are missing from crystal lattice.
(iii) Highly ionic substances with almost comparable size of cation and anion show this defect, e.g. KCl.

Question.11. Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf=5.12 K kg mol-1).
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Question.12. Define an ideal solution and write, one of its characte-ristics.
Answer : A solution which obeys Raoult’s law at all temperatures and concentrations is called an ideal solution. For an ideal solution,
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Question.13. Write two differences between ‘order of reaction’ and ‘molecularity of reaction’.
Answer:
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Question.14. Outline the principles behind the refining of metals by the following methods:
(i) Zone refining method
(ii) Chromatographic method .
Answer : (i) Zone refining Method : This method is employed when impurities are more soluble in the melt than in solid form of the metal. It is used to obtain metals such as Germanium (Ge), Silicon (Si), Gallium (Ga), etc in their purest form.
(ii) Chromatographic Method : This method is used when the impurities are not very different in chemical properties from the element to be purified. The components of the mixture show different mobility on the stationary phase, i.e., the components are adsorbed differently on the adsorbent. Gas chromatography, liquid chromatography and paper chromatography are various chromatographic methods based on ” different mobile and stationary phase.

Question.15. Complete the following chemical equations:
(i) Ca3P2 + H20 —>
(ii) Cu + H2S04 (cone.) —>
OR
Arrange the following in order of property indicated against each set:
(i) HF, HCl, HBr, HI : increasing bond dissociation enthalpy
(ii) H20, H2S, H2Se, H2Te,: increasing acidic character.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-7

Question.16. Write the IUPAC name of the complex (Cr (NH3)4Cl2]+. What type of isomerism does it exhibit ?
Answer : Tetraammine dichloridochromium(III) ion. The complex exhibits geometric (cis-trans) isomerism.
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Question.17. (i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction ?
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(ii) Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration ^
(b) Racemisation
Answer: (i) Compound (b), i.e., CH3CHBrCH2CH3 possesses chiral centre and undergoes faster SN2 reaction.
(ii) (a) Inversion of configuration results in SN2 reaction as there is formation of intermediate transition state in which there is simultaneous attack and migration of leaving group.
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(b) Racemisation in SN1 reaction is due to attack of nucleophile on both sides of the planar carbocation.
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Question.18. Draw the structure of major monohalo product in each of the following reactions :
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Question.19. (a) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(b) Write an important characteristic  of lyophilic sols.
(c) Based on type of particles of dispersed phase, give one example each of associated colloid and multi molecular colloid.
Answer : (a) Expression for Freundlich adsorption isotherm- x/m =kp 1/n (Where n>1)
Where, x = mass of the gas adsorbed (adsorbate)
m = mass of the adsorbent (solid)
p = pressure of the gas
n and k are constants, which depend on the nature of the adsorbate and adsorbent.
(b) Lyophilic sols are stable and reversible in nature
(c) Associated colloid: Soap solution/detergent solution Multimolecular colloid: Sulphur sol/Gold sol.

Question.20. (a) Draw the structures of the following molecules :
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus
Answer: (a)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-13
(b) Structural Difference : White Phosphorus : In white phosphorus, P4 molecules are held by weak Vander Waal’s forces and exist as a discrete tetrahedral.
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Red Phosphorus — In red phosphorus, P4 molecules are held by covalent bonds in polymeric structure.
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Question.21. Account for the following:
(i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two 0-0 bond lengths in the ozone molecule are equal.
Answer : (i) According to Fajan’s rule higher the oxidation state of the central atom, more is its polarising power, thus more is the covalent character of the bond formed. In the PCl5 molecule the oxidation state of P is +5 while in PCl3 it is +3. Thus, PCl5 is more covalent in nature than PCl3.
(ii) The hydrogen gas produced prevents further oxidation of Fe2+ to Fe3+.
Fe + 2HCl —-> FeCl2 + H2.
(iii) An ozone molecule a is resonance hybrid of the molecule in which there is single bond with one terminal oxygen and double bond with other terminal oxygen of the central oxygen atom.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-16
So, neither single or double bond is pure. Thus both 0 — 0 bond length are equal.

Question.22. The following data were obtained during the first order thermal decomposition of S02Cl2 at a constant volume :
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Calculate the rate constant. (Given : log 4 = 0.6021, log 2 = 0.3010)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-18

Question.23. (i) Give two examples of macromolecules that are chosen as drug targets.
(ii) What are antiseptics? Give an example.
(iii) Why is use of aspartame limited to cold foods and soft drinks ?
Answer : (i) Proteins, enzymes, nucleic acids and lipids are called drug targets as drug interacts with these macro-molecules.
(ii) Antiseptics are the chemical substances which prevent the growth of micro-organism and are capable of killing them without harming the human tissues. These are applied on wounds, ulcer, cuts and diseased skin surfaces, e.g. savlon, 0.2% solution of phenol, dettol, iodine tincture etc.
(iii) Aspartaime decomposes on heating hence” it is used as an artificial sweetner for foods and soft drinks at low temperatures.

Question.24. (i) Deficiency of which vitamin causes night-blindness ?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What
does it suggest about the structure of glucose?
Answer : (i) Vitamin A.
(ii) Uracil
(iii) Glucose exists in acyclic straight six membered carbon chain => open structure of Glucose.
CHO – (CHOH)4 – CH2OH

Question.25. After the ban on plastic bags, students of a school decided to make the people aware of the harmful effects of plastic bags on the environment and Yamuna River. To make the awareness more impact full, they organized rally by joining hands with other schools and distributed paper bags to Vegetable vendors, shopkeepers and departmental stores. All the students pledged not to use polythene bags in the future to save the Yamuna River.
After reading the above passage, answer the following questions:
(i) What values are shown by the students?
(ii) What are bio-degradable polymers? Give one example.
(iii) Is polythene a condensation or the addition polymer?
Answer : (i) The students show awareness about the environment and its protection. There contribution towards the cleaner pollution free environment. They also understand the value of team work as rally organization with other school students imbibe the quality of cooperation.
(ii) Polymers That are decomposed over a period of time either by itself or by the action of micro-organisms are called biodegradable polymers. PHBV (Poly β-hydroxy butrate Co-β-hydroxy valerate)
(iii) Polythene is an addition polymer.

Question.26. (a) Write the mechanism of the following reaction :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-19
(b) Write the equation involved in Reimer-Tiemann reaction.
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Question.27. Give the structures of A, B and C in the following reactions:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-22
OR
How will you convert the following?
(i) Nitrobezene into aniline.
(ii) Ethanoic acid into methanamine.
(iii) Aniline into N-phenylethanamide.
(Write the chemical equations involved.)
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-23
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-24

Question.28. (a) Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
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OR
(a) The mass of substance deposited or liberated at any eletrode is proportional to the quantity of electric charge passed through an electrolysis cell.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-27

Question.29.
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(iv) Name a member of the Lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation
MnO4 + 8H+ + 5e ——->
Answer : (a) (i) Pyrolusite is fused with KOH in presence of atmospheric oxygen or an oxidizing agent (KNO3 or KClO3) to give potassium permanganate.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-29
(ii) Unpaired electrons account for stronger metallic bond. Zinc lacks unpaired electrons as it has electronic configuration of [Ar] 3d10 4s2, thus metallic bonding is the weakest. So, Zn has the lowerst enthalpy of atomization.
(iii) Due to comparable energies of 5 f 6d and 7s orbitals, actinoids show wide range of oxidation states.
OR
(i) Manganese shows maximum number of oxidation states in 3d transition series. This is because all the five d-orbital electrons are unpaired i.e [Ar] 3d54s2. Thus, Mn shows oxidation states from +2 to +7.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-30

Question.30. (a) Write the products of the fallowing reactions:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-31
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic add
(ii) Propanal and Propanone 
OR
(a) Account for the following:
(i) CH3CHO is more reactive than CH3COCH3 towards reaction with HCN.
(ii) Carboxylic acid is a stronger acid than phenol.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Wolff-Kishner reduction
(ii) Aldol condesnation
(iii) Cannizzaro reaction 
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-32
(a) (i) CH3CHO is more reactive than CH3COCH3 because the reactivity of compound depends on the steric hindrance due to the groups present around the carbonyl group. More the steric hindrance, less will be the reactivity of the compound. Therefore, due to the presence of more electrophilic carbonyl carbon in CH3CHO, it is more reactive than CH3COCH3.
(ii) Carboxylic acid is stronger than phenol because of resonance, stabilisation of more electronegative oxygen atom in carboxylate ion than carbon atom in phenoxide ion.
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The, reason for greater stability of carboxylate ion is that it involves equivalent resonance in which negative charge is present on oxygen atom all the time whereas, in phenoxide ion, out of total four resonating structures, in three structures, negative charge is present on carbon atom which makes it less stable.
(b) (i) Reduction of aldehyde or ketone to respective, hydrocarbon.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-35

SET II

Note : Except for the following questions, all the remaining question have been asked in previous sets.

Question.1. Give one example each of sol and gel.
Answer: Sol-paint, gel-butter

Question.3. Write the IUPAC name of the compound
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-36

Question.5. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?
Answer : Azeotropes are mixtures with fixed concentrations of components such that they boil at constant temperatures. e.g. 95% ethanol and 5% water by mass mixture.

Question.7. Which component of starch is a branched polymer of α-glucose and insoluble in water?
Answer : Amylopectin is a branched polymer of α-glucose and insoluble in water.

Question.9. State Henry’s law. What is the effect of temperature on the solublity of a gas in a liquid.
Answer: Henry’s law states that “the partial pressure of the gas in vapour phase (p) is directly proportional to the mole fraction of gas (x) in the solution.”
p = KHx   (KH—Henry S law constant)
The solubility of a gas in liquid decreases with rise in temperature as dissolution of a gas in a liquid is an exothermic process.

Question.10. Define the following terms :
(i) Pseudo first-order reaction
(ii) Half-life period of reaction (t1/2).
Answer : (i) Reactions which are actually first order but behave as first order under certain conditions like excess of one of the reactants, is a psuedo first order reaction, e.g. Acid hydrolysis of ethyl acetate.
(ii) The time taken for the concentration of reactants to be reduced to half of its initial value is called the half life of a reaction.

Question.11. Write the principle behind the following methods of refining:
(i) Hydraulic washing (ii) Vapour-phase refining
Answer : (i) Hydraulic washing based on the differences in densities or gravities of the are and the gangue particles. The lighter gangue particles are washed away and the heavier ores are left behind.
(ii) Vapour phase refining method the metal is converted into its volatile compound and convected elsewhere. It is then thermally decomposed to get the pure metal. E.g. Mond’s process.

Question.22. (a) Draw the structures of the following:
(i) XeF2 (ii) BrF3
(b) Write the structural difference between white phosphorus and red phosphorus.
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(b) White phosphorus consist of discrete tetrahedral P4 molecule with six P-P covalent bonds.
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Red Phosphorus has polymeric structure in which P4 tetrahedral are linked together through P-P covalent bond to form chain.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-39

Question.23. Account for the following:
(i) Bi(V) is a stronger oxidizing agent than Sb(V).
(ii) N – N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points.
Answer : (i) Due to inert pair effect +3 oxidation state of Bi is more stable than its +5 oxidation state while oxidation state of Sb is more stable than its +3 oxidation state. Therefore, Bi (V) can accept a pair of electrons to form more stable Bi (III) more easily than Sb (V).
(ii) N has small size so lone pair is more concentrated over N hence repulsion takes place and bond becomes weak but P has large size therefore, no repulsion takes place and Hence, bond becomes strong. ‘
(iii) Noble gas are monoatomic with weak Vander Waals forces of attraction, Hence, noble has has very low boiling point.

Question.24. (i) Name the sweetening agents used in the preparation of sweets for a diabetic patient.
(ii) What are antibioties? Give an example.
(iii) Give two example of macro molecules that are chosen as drug targets. 
Answer : (i) Sucrolose.
(ii) Antibiotics are the chemicals substances produced wholly or partly by chemical synthesis which in low concentrations inhibit the growth or destroy micro¬organisms by intervening in their metabolic processes, e.g. Penicillin, Tetracyline.
(iii) Carbohydrate, lipid, proteins, enzymes, nucleic acid.

Question.27. Deficiency of which vitamin causes rickets?
(ii) Give an example for each of fibrous protein and globular protein.
(iii) Write the product formed on reaction of D-Glucose With Br2 water.
Answer: (i) Rickets.
(ii) Fibrous protein : Keratin or myosin Globular protein : Insulin or albumin.
(iii) Gluconic acid i.e. HOOC (CH2OH)4 CH2OH.

SET III

Note : Except for the following questions, all the remaining question have been asked in previous sets.

Question.1. Give one example each of lyophobic sol and lyophillic sol.
Answer : Lyophobic sol: Metal sol or metal sulphide. Lyophilic sol: Gum, Starch, gelatin.

Question.2. Write the IUPAC name of the compound.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-40

Question.3. What type of inter molecular attractive interaction exists in the pair of methanol and acetone?
Answer : Hydrogen bonding (intermolecular)

Question.6. Name the products of hydrolysis of sucrose.
Answer : Glucose and fructose.

Question.9. State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law?
Answer : Raoult’s law states that the partial pressure of the vapour of a volatile component in a solution is directly proportional to its mole fraction in the solution. Raoult’s law is a special Case of Henry’s law.

Question.10. Explain the following terms :
(i) Rate constant (k)
(ii) Half life period of reaction (t1/2).
Answer: (i) Rate constant (k) is the rate of reaction when the concentration of reactants is unity.
(ii) Half life period (t1/2) of reaction is the time in which the concentration of reactant is reduced to half of its initial concentration.

Question.11. Write the principles of the following methods:
(i) Froth floatation method (ii) Electrolytic refining
Answer : (i) It is based on the difference in wetting qualities of gangue and the sulphide are particles with water and oil. Whereas the ore particles are wetted by oil, the gangue or the earthy particles are wetted by water.
(ii) In this process a slab of impure copper is used as anode and a thin sheet of pure copper as cathode. The copper sulphate is used as electrolytic solution. By passing electricity through the cell copper is dissolved from the anode and deposited on cathode. The impurities either remain in solution or collect as an insoluble gangue.

Question.20. (a) Draw structure of the following compounds :
(i) XeF4 (ii) N2O5
(b) Write the structural difference between white phos-phorus and red phosphorus.
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(b) White phosphorus consist of discrete tetrahedral P4 molecule with six P-P covalent bonds.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-42
Red Phosphorus has polymeric structure in which P4 tetrahedral are linked together through P-P covalent bond to ‘ form chain.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2014-43

Question.22. Account for the following :
(i) Sulphur in vapour form exhibit paramagnetic behaviour. ,
(ii) SnCl4 is more covalent than SnCl2.
(iii) H3P02 is stronger reducing agent than H3PO3.
Answer: (i) In vapour form sulphur partly exits as S2 molecule which have two unpaired electrons in the antibonding p molecular orbitals like O2 molecule and hence exhibits paramagnetism.
(ii) It is due to higher oxidation state (+4) of Sn is SnCl4, or because of it high polarising, power which, As turn increases the covalent character of bond formation between the central atom and the atoms around it.
(iii) It is because of two P-H bonds in H3PO2 whereas there is only on P-H bond in H3PO3.

Question.23. (i) What are disinfectants? Give an example.
(ii) Give two examples of macro molecules that are chosen as drug targets.
(iii) What are anionic detergents? Give an example.
Answer: (i) Chemicals which are used to kill micro-organisms and applied on non-living objects like floors and drains are called disinfectants e.g. 1% phenol solution.
(ii) Proteins, amino acids and enzymes.
(iii) Detergents in which the anionic part of the molecule is responsible for cleansing action are called anionic detergents eg. Sodium laurylsulphate.

Question.24. (i) Deficiency of which vitamin causes scurvy?
(ii) What type of linkage is responsible for the formation of proteins?
(iii) Write the product formed when glucose is treated with HI.
Answer: (i) Vitamin-C
(ii) Peptide linkage
(ii) n-hexane : CH3CH2CH2CH2CH2CH3