CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2010

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 questions have been asked in previous sets.

Question.1. Write a feature which will distinguish a metallic solid from
an ionic solid.
Answer : A metallic solid does not ionize in water.

Question.2. Define ‘order of a reaction’.
Answer : The sum of power of the concentration terms on which rate of reaction depend^ according to the rate law is called order of the reaction.

Question.3. What is an emulsion?
Answer : A colloid in which both dispersed phase and dispersing medium are liquid is called an emulsion, e. g. fat or oil in milk.

Question.4. Why does N02 dimerise?
Answer : Due to the presence of unpaired electrons in N02 it dimerises to form  N2O4.

Question.5. Give an example of linkage isomerism.
Answer : The example of linkage isomerism are : [Cr(SCN) (H2O)5]3+ and [Cr(NCS)(H2O)5]3+

Question.6. A solution KOH hydrolyses CH3CHClCH2CH3and CH3CH2CH2CH2Cl which one of these in more easily hydrolysed?
Answer : CH3CHCICH2CH3 is easily hydrolysed due to more stability of secondary carbocation.

Question.7. Draw the structural formula of 1-phenyl Propan-l-one molecule.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-1

Question.8.Give the IUPAC name of
H2N – CH2 – CH2 – CH = CH2.
Answer: But-3-ene-l-amine

Question.9. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused ? Explain with one example for each type.
Answer : Negative deviation from Raoult’s law : For any composition of the non-ideal solution the partial vapour pressure of each component and total vapour pressure of the solution is less than expected from Raoult’s law. Such solutions show negative deviation.
Example : Mixture of CHCl3 and acetone, water and nitric acid etc.
Non-ideal solutions show positive deviations from Raoult’s law on mixing of two volatile components of the solution the partial pressure of each component and total vapour pressure of the solution is more than expected from Raoult’s Law, such solution show positive deviation.
Example : Mixture of acetone and benzene solutions show., positive deviation, water and ethanol, etc.

Question.10. A reaction is of first order in reactant A and of second order in reactants B. How is the rate of this reaction affected when
(i) the concentration of B alone is increased to three times
(ii)the concentrations of A as well as B are doubed?
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Question.11. The rate constant for a reaction of zero order in A is 0.0030 mol L-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-4

Question.12. Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-4

Question.13. Explain the following observations:
(i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(ii)Transition elements and their compounds are generally found to be good catalysts in chemical reactions.
Answer :
(i) Due to decrease in atomic radii from titanium to copper, the corresponding density increases.
(ii) The catalytic activity of transition elements is attributed to the following reasons.

(1) Because of their variable valencies transition metal
sometimes form unstable intermediate compound and provide a new path with lower activation energy for the reaction.
(2) In some cases, transition metals provide a suitable surface ‘ for the reaction to take place. The reactants are adsorbed on
the surface of the catalysts where the reaction occurs.

Question.14. Name the following coordination compounds according to IUPAC system of nomenclature:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-6
Answer: (i) Tetraamineaquachloridocobalt (III) chloride (ii) Dichlorido-bi$(ethane-l, 2-diamine)-Chromium(III) chloride

Question.15. Illustrate the following reactions giving a chemical equation for each:
(i) Kolbe’s reaction,
(ii)Williamson synthesis..
Answer: (i) Kolbe’s reaction :
br-lazy"

Question.16. How are the following conversions carried out?
(i) Benzyl chloride to benzyl alcohol.
(ii)Methyl magnesium bromide to 2-Methylpropan-2-ol.
Answer:(i)
br-lazy"

Question.17. Explain the following terms :
(i) Invert sugar (ii) Polypeptides .
OR
Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar ?
Answer : (i) Sucrose is dextrorotatory but after hydrolysis it gives an equimolar mixture of Q (+) -glucose and q- (—)—fructose which is laevorotatory. This change in specific rotation from dextrorotation to laevorotation is called inversion of sugar and the mixture obtained is called invert sugar.
(ii)Polypeptide is a polyamide formed by large number of ‘a-amino acids joined together by peptide bonds.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-9
The hydrolysis product of sucrose are D-(+)-glucose D-(-)-fructose. It is not considered as a reducing sugar , because the anomeric carbon of both units are involved in glycoside formation. Also it is unable to reduce Tollen’s reagent or Fehling solution.

Question.18. What are essential and non-essential Amino acids in human food? Give one example of each.
Answer : Amino acids which the body cannot synthesise like phenylalanine, valine are essential amino acids and those which can be synthesised by the body are non-essential like glucose and serine.

Question.19. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F ions and that Ca2+ ions are arranged in a fee lattice. The Fions fill all the tetrahedral holes in the face centred cubic lattice ofCa2+ jions. The edge of the unit cell is 5.46 x 10-8 cm in length. The density of the solid is 3.18 g cm 3. Use this information to calculate Avogadro’s number (Molar mass of CaF2 = 78.08 g mol-1)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-10

Question.20.A solution prepared by dissolving 1.25g of oil of winter green (methyle salicylate) in 99.0g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and Kb for benzene = 2.53°C kg mol-1)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-11

Question.21. What is the difference between multimolecular and macromolecular colloids? Give one example of each type. How are associated colloids different from these two types of colloids?
Answer : Multimolecular colloids : In this type of colloids, colloidal particles are aggregates of atoms or molecules each having size less than 1 am. e. g. Sulphur sol, gold sol. Macromolecular colloids : In this type of colloids, colloidal particles are themselves large molecules of colloidal dimensions, e. g. Starch, proteins, polythene etc.
Associated colloids : There are certain substances which at low concentration behave as normal electrolyte, but at higher concentration exhibit colloidal behaviour due to the formation of aggregates. Such colloids are known as associated colloids, e. g. Soaps and detergents.

Question.22. Describe how the following changes are brought about:

  1. Pig iron into steel.
  2. Zinc oxide into metallic zinc.
  3. Impure titanium into pure titanium.

OR
Describe the role of

  1.  NaCN in the extraction of gold from gold ore.
  2. SiO2 in the extraction of copper from copper matter.
  3. Iodine in the refining of zirconium.

Answer :

  1. For the conversion of pig iron into steel, basic oxygen process (BOP) is used. In this process, the furnance is charged with molten pig iron and lime. The pure O2 is blown over the surface of the metal at a great speed through water-cooled retractable lances. The O2 penetrates through the metal and oxidizes the impurities rapidly. When all the impurities are removed, this required elements (Cr, Ni etc.) are added to produce steel of desirable properties.
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-12
  2. Calcination of zinc oxide reduces it to metallic zinc using coke this can be explained as follows – above 1270 k Δf  G° for ZnO is higher than that of CO2 and CO from carbon therefore, above 1270 K Δ2G° ΔlG° for reduction of ZnO by carbon is negative and hence ZnO is easily reduced by coke. The ZnO is made into brickettes with coke and clay ‘ and heated above 1270 K so that the reduction process goes to complete
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-13
  3. The impure titanium is converted to pure titanium by , Van Arkel method. In this method, the impure titanium
    is heated in an evacuated vessel with iodine at 870 K. The volatile titanium tetraiodide thus formed is separated which is then decomposed by heating over a tungsten filament at 2075 * K to give pure titanium.
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-14

(i) NaCN is used to leach the metal present in the ore, when an ore of gold is leached with NaCN in a current of air, gold particle pass into solution as Au+, which combines with CN’ ions, to form soluble complex of aurocyanides. The gold metal is then recovered from the complex by precipitating it with electropositive reducing agent i.e., zinc.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-15
To remove the basic impurities i. e. FeO, an acidic flux silica is added during smelting to remove it as iron silicate slag (FeSiO3) which floats over molten matte and hence can be easily removed.
(iii) Iodine is used as a reducing agent in refining of zirconium to form a volatile iodide which on decomposition at high temperature gives pure metal.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-16

Question.23. How would you account for the following?

  1. The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
  2. The E° value for the Mn3+/Mn2+ couple is much more positive than that of Cr3+/Cr2+ couple or Fe3+/Fe2+ couple.
  3.  The highest oxidation state of a metal is exhibited in its oxide of fluoride.

Answer :

  1.  Due to lanthanide contraction in second series i. e. Ad after lanthanum, and poor shielding of electrons, the atomic radii of elements of second (Ad) and third (5d) transition series become almost same.
  2.  This is mainly due to much greater third ionization energy of Mn when d5 configuration changes to d4. 
  3. Due to small size, low ionization energy and higher electronegativity the highest oxidation state of a metal is exhibited in its oxide or fluoride.

Question.24. (i) State one use of DDT and iodoform.
(ii) Which compound in the following couples will react fester in SN2 displacement and why?
(a) 1-Bromopentane or 2-bromopentane
(b) l-Bromo-2 methylbutane or 2-Bromo-2 methyl- butance.
Answer : (i) DDT is used as an insecticide to prevent sugarcane and fodder crops by killing lice and mosquitoes which carry pathogens.
Iodoform dissolved in alcohol is used as an antiseptic for dressing wounds.
(ii) (a) 1-bromopentane being primary alkyl halide have less steric hindrance than 2-bromopentane, therefore, it will be more reactive and hence undergoes SN2 reaction faster.
(b) l-bromo-2-methylbutane, experience less steric hindrance as it has -Br at primary carbon as compared to 2-bromo-2- methylbutane which has -Br at secondary carbon. Therefore, l-bromo-2methyl butane will be more reactive and hence undergoes SN2 reaction faster. –

Question.25. Arrange the following in order of property indicated for each set.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-17

Question.26. Give one example each of
(i) Addition polymers
(ii)Condensation polymers
(iii)Copolymers.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-18

Question.27. What are analgesic medicines? How are they classified and when are they commonly recommended for use?
Answer: Analgesics are drugs, which-relieve or decrease pain without causing unconsciousness, paralysis or coordination and mental confusion. They are classified into the following two categories:
(i) Non-Narcotic (non-addictive) drugs, e. g. aspirin, ibuprofen etc. These drug are used to relief skeletal pain.
(ii)Narcotic (addictive) drugs, e. g. Heroin, morphine etc. These are used to relieve postoperative pain, cardiac pain.

Question.28. (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic add at infinite dilution according to Kohlrauch law.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-19
(a) Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.
(b) A copper silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-20
Answer : (a) Kohlrausch law : It states that “at infinite dilution, the molar conductance of an electrolyte is equal to the sum of the molar conductances of the two ions i. e. cation and anions.
Mathematically, A°m= A°+ + A°_.
Expression for the molar conductivity of acetic acid.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-21
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-22

Question.29. (a) Complete the following chemical equations :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-23
(b) How would you account for the following?
(i) The value of electron gain enthalpy with negative . sign for sulphur is higher than that for oxygen.
(ii)NF3 is an exothermic compound but NCl3   is endothermic compound.
(iii)ClF3   molecule has a T-shaped structure and not a trigonal planar one.
OR
(a) Complete the following chemical reaction equations:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-24
(b) Explain the following observations giving appropriate reasons:
(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.
(ii)Solid phosphorus pentachloride behaves as an ionic compound.
(iii)Halogens are strong oxidizing agents.
Answer: (a) (i)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-201025
(b) (i) Because enthalpy of S-S bond is higher then 0-0 bond and the hydration energy of S2- is less than that of O2- ion.
(ii) N F3   is an exothermic compound whereas NCl3   is an endothermic compound because in case of NF3 , N-F bond strength is greater than the F-F bond strength while in case of NCl3  , N-Cl bond strength is lower than the Cl-Cl bond strength. Thus, the formation NF3 is spontaneous while energy has to be supplied during the formation of NCl3. In other words, the formation of NF3 is an exothermic reaction while formation of NCl3 is an endothermic reaction.
(iii) ClF3 molecule has a T-shaped structure. This is due to the presence of two lone pairs in the outer shell of chlorine in ClF3 molecule which repel the bond pairs.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-26
state becomes more and more common-on moving down the group from N to Bi. This is because of inert pair effect.
(ii) Solid PCl behaves as ionic compound because it is a salt containing the tetrahedral cation [PCl4 ]+ and octahedral anion [PCl]
(iii) Halogens are strong oxidizing agents because they have high electron affinities, so they easily accept electrons from other substances and change into negative ions i.e. they undergo reduction. Therefore, they are strong oxidizing agents.

Question.30. (a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone.
(b) An organic compound (A) (molecular formulaC8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (G). Oxidation of (C) with chromic add also produced (B). Oh dehydration (C) gives but-l-ene. Write the equations for the reactions involved.
OR
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal
(ii)Phenol and Benzoic add
(b) How will you bring about the following conversions?
(i) Benzoic add to benzaldehyde 
(ii)Ethanal to but-2-enal
(iii)Propanone to propene
Give complete reaction in each case.
Answer: (a)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-201027
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-28
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-29
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(b) The hydrolysis of given compound (A) with molecular formula C8H16O2 upon hydrolysis with dil H2SO4 gives an alcohol (C) and carboxylic acid (B) it suggests that the compound is an ester further since oxidation of (C) with chronic acid produces the acid (B), therefore, both the carboxylic acid (B) and the alcohol must contain the same number of carbon atom.

SET II

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

Question.1.Which point defect in the crystals of a solid does not change the density of the solid?
Answer: Frenkel defect.

Question.4. What is the oxidation number of phosphorus in H3POmolecule?
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-48

Question.5. Give an example of coordination isomerism.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-49

Question.12. Draw the structural formulae of molecules of following compounds:
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-31

Question.14.Describe the shapes and magnetic behaviour of following complexes:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-32

Question.15. Explain the following reactions with an example for each:

  1.  Reimer-Tiemann reaction
  2.  Friedel-Crafts reaction

Answer:

  1.  Reimer-Tiemann reaction: Phenol on treatment with chloroform in aqueous sodium hydroxide at 340 K give salicyldehyde.
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-33
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-34
    Friedel-crafts reaction : This reaction is used for introducing an alkyl or an acyl group into an aromatic compound in presence of Lewis acid catalyst (AlCl3).
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-35

Question.16. How are the following conversions carried out?

  1. Propane to propan-2-ol
  2. Etylmagnesium chloride to propan- l-ol

Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-36

Question.22. A solution of glycerol (C3H8O3; molar mass = 92 g mol-1) in water was prepared by dissolving some glycerol 500 g of water. This solution has a boiling point of 100.42 C. , What mass of glycerol was dissolved to make this solution?
Kb for water = 0.512K kg mol-1.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-37

Question.24. Complete the following chemical equations
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-38
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-39

Question.25. Write the name and structure of the monomer of each of the following polymers :
(i) Neoprene
(ii)Buna-S
(iii)Teflon 3.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-40

SET III

Note: Except for the following questions, all the remaining question have been asked in previous sets.
Question.1. Which point defect in the crystals of a solid decreases the density of the solid.
Answer: Schottky defect

Question.2. Define‘rate of reaction’.
Answer : The rate of decrease of reactant concentration or increase in product concentration per unit time is called rate of reation.

Question.3. Give an example of‘shape-selective catalyst’.
Answer : Zeolites with its honey-comb structure is an example of shape selective catalysts.

Question.4. Draw the structure of 03 molecule.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-41

Question.5. Give an example of ionization isomerism.
Answer :[CO(NH3)5Br]SO4 and [CO(NH3)5SO4]Br

Question.13. Explain the following observations :

  1. Transition elements generally form coloured compounds.
  2. Zinc is not regarded as a transition element.

Answer:

  1. Because of the presence of incomplete d-orbitals, transition elements, undergo d-d transition by four coloured absorption of energy from visible region and then they emit the light of complementary colour compounds.
  2.  Zinc has completely filled d-subsheil in its common oxidation state. Hence, it is considered as a non-transition element.

Question.18. State clearly what are known as nucleotides and nucleosides?
Answer : Nucleoside : A nucleoside contains only two ’ basic components of nucleic acid i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 1-position of the purine moiety is linked to C1 of the sugar (ribose or deoxyribose) by a β-linkage
br-lazy"
Nucleotides : A nucleotide contrain all the three basic components of nucleic acid a phosphoric acid group, at 5′-position of pentose sugar and a nitrogenous base
br-lazy"

Question.19. The density of copper metal is 8.95 g cm-3. If the radius of copper is 127.8 pm, is the copper unit cell a simple cubic, a body centred cubic or a face centred cubic structure. (Givens At. Mass of Cu = 63.54 g mol-1and NA = 6.02 x 1023 mol-1)
Answer : If copper atom were simple cubic:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-44
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-45

Question.20. How are the following colloids different from each other in respect of their dispersion medium and dispersed phase? Give one example of each.

  1.  Aerosol
  2. Emulsion
    (iii)Hydrosol
    Answer:
    cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-46

Question.26. Differentiate between thermoplastic and thermosetting polymers. Give one example of each.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2010-47

Question.27. Explain the following terms with one suitable example in each case.

  1. Cationic detergents
  2. Enzymes
  3. Antifertility drugs

Answer :

  1.  Cationic detergents are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Therefore, they are called as cationic detergents. e.g. Cetyltrimethyl ammonium bromide.
  2. Enzymes are biological catalysts which are chemically made up of globular proteins. They have high molecular mass and are highly specific in their actions due to presence of active sites of definite shape and size on their surfaces so that only specific substrate can fit in them. e.g. Pepsin, Trypsin etc.
  3. Antifertility drugs are oral contraceptives which are used to check pregnancy in women. It control the female menstrual cycle and ovulation, e.g. N.ovestrol, Norethindrone etc.