CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2013

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. How many atoms constitute one unit cell of a face-centered cubic crystal ?
Answer : 4 atoms constitute one unit cell of a face centered cubic crystal. It can be determined from the number of atoms contributed from the faces and the corners of the unit cell as:
(8 corners x 1/8 atom per corner = 1 atom) + (6 faces x 1/2 atom per unit face = 3 atoms)
Total number of atoms per unit cell =1+3 = 4 atoms.

Question.2. Name the method used for refining of Nickel metal.
Answer: Method used for refining of nickel metal is Mond’s process.

Question.3. What is the covalency of nitrogen in N205?
Answer: In N205, the covalency of N is restricted to ‘4’ due to sp2 hybridisation of nitrogen atom involving one 2s and three 2p orbitals.

Question.4.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-1

Question.5. What happens when CH3-Br is treated with KCN?
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-2
It is a nucleophilic substitution reaction.

Question.6. Write the structure of 3-methyl butanal.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-3

Question.7. Arrange the following in increasing order of their basic strength in aqueous solution: CH3NH2, (CH3)3N, (CH3)2 NH.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-4

Question.8. What are the types of RNA molecule which perform different functions?
Answer: Three types of RNA molecules which perform different functions are :
(i) Messenger RNA (m-RNA)
(ii) Transfer RNA (t-RNA)
(iii) Ribosomal RNA (r-RNA)

Question.9. 18 g of glucose, C6H12O6 (Molar Mass = 180g/mol) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?
(Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-5

Question.10. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-6

Question.11. Write the dispersed! phase and dispersion medium of the following colloidal system ?
(i) Smoke (ii) Milk.
OR
What are lyophilic and lyophobic colloids? Which of these , sols can be easily coagulated on addition of small amounts of electrolytes?
Answer: (i) Dispersed phase in smoke : Solid, dispersion medium in smoke : Gas
(ii) Dispersed phase in milk : Fats (liquid), dispersion medium in milk: Water (liquid)
OR
Answer : Lyophilic colloids (Liquid Loving) : These are the colloidal solutions in which dispersed phase has great affinity for dispersion medium. Such solutions are quite stable and are reversible in nature. E.g. starch, proteins, etc.
Lyophobic Colloids (Liquid Hating): These are the colloidal solutions in which dispersed phase has very little affinity for the dispersion medium. Such solutions are unstable and are irreversible in nature. Eg. (As2S3 Sol).
Lyophobic colloids can be easily coagulated because on addition of small amount of electrolyte, the charge on colloidal particles is removed, as a result the particles will come closer to each other and then aggregate to form a cluster which settle down under the force of gravity.

Question.12. Write the differences between physisorption and chemisorption with respect to the following:
(i) Specificity (ii) Temperature dependence
(iii) Reversibility and (iv) Enthalpy change
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-7

Question.13. (a) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver?
(b) Out of ‘C’ and ‘CO’, which is a better reducing agent . at the lower temperature range in the blast furnace to extract iron from the oxide ore?
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-8

Question.14. What happens when
(i) PCl5 is heated? (ii) H3P03 is heated?
Write the reaction involved.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-9

Question.15. (a) Which metal in the first transition series (3d series) exhibits + 1 oxidation state most frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+ (Atomic Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer: (a) Cu is the only metal in the first transition series (3d series) which exhibits +1 oxidation state more frequently. This is because the electronic configuration of Cu is 3d10 4s1 and after losing one s electron it acquires a stable 3d10 configuration.
(b) The colour of cations depend upon the number of unpaired electrons present in d-orbital. The electronic configuration of the following cations is as follows :
Sc (Atomic number 21) = Sd1s2 and Sc3+ = 3d°4s°. As d-orbital is empty, it is colourless.
V (atomic number 23) = 3d34s2 and V3+ = 3d24s°. As d-orbital is having 2 unpaired electrons, it undergoes d-d transition and depicts green colour.
Ti (Atomic number 22) = 3d24s2 and Ti+4 = 3d°4s°. As ‘d’ orbital is empty, it is colourless.
Mn(Atomic number 25) = 3d54s2 and Mn+2=3d54s°. As ‘d’ orbital has 5 unpaired electrons, it depicts pink colour.
Thus, V3+ and Mn2+ ions are coloured in their aqueous solution due to presence of unpaired electron.

Question.16. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same.
Answer: Chlorobenzene is extremely less reactive towards a’ nucleophilic substitution reaction because of the following reasons:
1. Resonance effect: The electron pair on chlorine atom is in conjugation with the benzene-electrons of the’ benzene ring which results in the following resonating structures :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-10
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-11
This’results in delocalization of electrons of C-Cl bond and a partial double bond character develops in the bond, which makes it difficult for the nucleophile to cleave the C-Cl bond.
2. The nucleophile suffers repulsion from the increased electron density on the benzene ring as a result the nucleophile is unable to make a close approach for the attack on the molecule.

Question.17. Explain the mechanism of the following reaction:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-12
Answer: The mechanism of the reaction is given below:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-13

Question.18. How will you convert:
(i) Propene to Propan-2-ol ?
(ii) Phenol to 2, 4, 6 – trinitrophenol?
Answer: When H2S04 is added to propene, propan-2-ol is formed. The addition of H2S0takes place in accordance with Markovnikov’s rule.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-14
(ii) When concentrated nitric acid is added to phenol in the presence of sulphuric acid it gives 2, 4,6 – trinitrophenol.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-15

Question.19. (a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments?
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-16
(c) What type of point defect is produced when AgCl is doped with CdCl2?
Answer: (a) When silicon is doped with boron, p-type semiconductor is obtained.
(b) Ferromagnetism is shown in this alignment of magnetic moments.
(c) Impurity defect of ionic solids is produced when AgCl is doped with CdCl2.

Question.20. Determine the osmotic pressure of a solution prepared by dissolving 2.5 x 10-2g of K2S04 in 2L of water at 25°C, assuming that it is completely dissociated.
(R = 0.0821 L atm K-1 mol-1, Molar mass of K2S04 = 174 g mol-1)
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-17

Question.21. Calculate the emf of the following cell at 298 K:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-18
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-19

Question.22. How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst.
OR
Complete the following chemical equations:
(i) Cr2O72- + 6Fe2+ + 14H+ —>
(ii) 2CrO42- + 2H+ —>
(iii) 2MnO4  + 5C2O42-  + 16H+ —>
Answer: (i) The variable oxidation states of transition elements is due to the participation of ns and (n-1)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.
(ii) This is because atomic radii of 4d and 5d transition elements are nearly same. This similarity in size is a consequence of lanthanide contraction which is due to weak shielding of d-electrons. As a result, the radii of Hf becomes nearly equal to that of Zr.
(iii) The catalytic activity of transition elements and their compound is due to the following reasons:
(i) Due to their tendency to shown variable oxidation state transition metal form instable intermediate compounds and provideds a new path for the reaction with lower activation energy.
(ii) In some cases, the transition metals provide a suitable large surface area with free valencies ion which reactant are adsorbed.
OR
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-20

Question.23. Write the IUPAC names of the following coordination compounds:
(i) [Cr(NH3)3Cl3]
(ii) K3[Fe(CN)6]
(iii) [CoBr2(en)2]+, (en = ehylenediamine)
Answer: (i) Triammine trichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromidobis-(ethylene-diammine) cobalt (III) ion .

Question.24. Give the structures of A, B and C in the following reactions:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-21

Question.25. Write the names and structures of the monomers of the following polymers:
(i) Buna – S (ii) Neoprene (iii) Nylon-6, 6
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-22

Question.26. After watching a programme on TV about the adverse effects of junk food and soft drinks on the health of school children, Sonali, a student of Class XII, discussed the issue with the principal. Principal immediately instructed the canteen contractor to replace the fast food with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision was welcomed by the parents and the students.
After reading the above passage, answer the following questions:
(a) What values are expressed by Sonali and the Principal of the school?
(b) Give two examples of water-soluble vitamins.
Answer : (a) The values showed by Sonali are awareness regarding adverse effect of junk food and concern for the health of her school mates. The value showed by the Principal is responsible behaviour in listening to Sonali’s views and taking prompt action in replacing junk food with healthy food.
(b) The two water soluble vitamins are vitamin B2 (Riboflavin) and Vitamin C (Ascorbic acid).

Question.27. (a) Which one of the following is a food preservative? Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
Answer : (a) Sodium benzoate is used as a food preservative whereas equanil is a tranquilizer and morphine is a narcotic analgesic.
(b) Bithional is an antiseptic so it is added to soaps to reduce
the odours produced by bacterial decomposition of organic matter on the skin.
(c) Tranquilizers relieve stress, fatigue by inducing sense of well being, so they are used in sleeping pills.

Question.28. (a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given: log 1.428 = 0.1548)
OR
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
(b) Rate constant ‘If of a reaction varies with temperature ‘T’ according to the equation:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-23
(ii) On increasing the concentration of‘A’ three times i.e. 3A; the rate of reaction becomes 9 times of the initial rate.
Rate = K[3A]2 [B]=9K[A]2 [B] = 9 times rate.
(iii) On increasing the concentration of A and B as 2A and 2B. The rate of reaction becomes 8 times of the initial rate. Rate = K[2A]2[2B] = 8K[A]2[B] = 8 times rate
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-24
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-25
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-26

Question.29. (a) Give reasons for the following:
(i) Bond enthalpy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH3.
(b) Draw the structures of the following molecules:
(i) BrF3 (ii) (HP03)3 (iii)XeF4 OR
(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Flourine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(b) Draw the structures of the following molecules.
(i) XeF2 (ii) H2S2OS
Answer: (a) (i) Bond enthalpy of F2 is lower than that of Cl2 because ‘F’ atom is small in size and due to this the electron- electron repulsions between the lone pairs of F-F electrons are very large. Thus, the bond dissociation energy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH3 because NH3 molecule possess intermolecular hydrogen bondings which binds them strongly, whereas, PH3 has weaker van der Waals forces. Thus PH, has a lower boiling point than NH3.
(b) (i) BrF3, Bent T-shape
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-27
OR
(a) (i) Helium mixed with oxygen under pressure is given to sea- divers for artificial respiration because of its very low solubility in blood. Air is not given because nitrogen present in air being soluble in blood will give a painful sensation called bends by bubbling out blood on moving from high pressure to the atmospheric pressure. Thus, oxygen-helium mixture is used.
(ii) Fluorine being the most electronegative atom does not exhibit positive oxidation state because, it does not have d-orbitals for octet expansion and therefore, it shows only a negative oxidation state of -1.
(iii) Oxygen shows catenation behaviour less than sulphur because the oxygen atom is smaller in size as compared to -> sulphur due to this the lone pair of electrons in 0-0 bonds in oxygen experiences more repulsions as compared to the S-S bonds and thus, S-S forms strong bond.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-28
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-29

Question.30. (a) Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring about the following conversions?
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal.
OR
(a) Complete the following reactions :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-30
(b) Give simple chemical tests to distinguish between the . following pairs of compounds :
(i) Ethanal and Propanal
(ii) Benzoic acid and Phenol.
Answer : (a) On losing a proton, carboxylic acids forms carboxylate ion and phenol forms phenoxide ion as follows :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-31
The conjugate base of carboxylic acid has two resonance structures in which negative charge is declocalized over two oxygen atoms which stabilizes the carboxylate ion. On the other hand, in phenoxide ion the charge is delocalized over entire molecule on the less electronegative atom, thus resonance of phenoxides is not important in comparison to resonance in carboxylate ion.
Further, in carboxylate ion the negative charge is effectively delocalized over two oxygen atoms whereas it is less effectively delocalized over one oxygen atom and less electronegative carbon atom.
Thus, phenol is less acidic than carboxylic acids. In other words, carboxylic acids are stronger acids than phenol.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-32
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-33
(b) (i) Ethanol and Propanal:
Iodoform Test : When ethanol is treated with sodium liydroxide and ammonia, iodoform is obtained.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-34
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-35

SET II

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

Question.1. What type of stoichiometric defect is shown by AgCl?
Answer: Frenkel defect is shown by AgCl.

Question.2.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-36

Question.4. What type of bonding helps in stabilizing the a-helix structure of proteins?
Answer: Hydrogen bonding between the -NH group of each amino acid residue and the > C = O of an adjacent turn of the helix helps in stabilizing the <-helix- structure of proteins.

Question.6. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
Answer: N. Bartlett observed that the first ionization enthalpy of molecular oxygen is almost identical with that of xenon. So after preparing red coloured compound 02+PtF6- he made efforts to prepare Xe+PtF6– by mixing PtF6 and Xe.

Question.7. What happens when ethyl chloride is treated with aqueous KOH?
Answer: When C2H5Cl reacts with aq. KOH, substitution Nucleophilic bimolecular (SN2) reaction takes place and Ethanol is formed.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-37

Question.8. Write the structure of 4-chloropentan-2-one.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-38

Question.9. How will you convert the following?
(i) Propan-2-ol to propanone.
(ii) Phenol to 2,4,6 – tribromophenol?
Answer: (i) Propan-2-ol to propanone :
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-39

Question.11. What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions? Give an example of each type.
Answer: Emulsion of oil-in-water has oil has dispersed phase and water as dispersion medium. For example, Milk etc. Emulsion of water-in-oil has water as dispersed phase and oil as dispersion medium. For example, Cod liver oil etc.

Question.17. (a) Which of the following ores can be concentrated by froth floatation method and why?
Fe203, ZnS, Al203
(b) What is the role of silica in the metallurgy of Copper?
Answer: (a) Only sulphide ores are concentrated by this process because pine oil selectively wets the sulphide ore and hence bring it to the froth.
(b) Silica is added in the reverberatory furnance during the extraction of copper to remove iron oxide present in the ore. * Iron oxide reacts with silica and is removed as slag of iron silicate.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-40

Question.18. (a)Why does p-dichlorobenzene have a higher m p than its o- and m- isomers? (b) Why is (—) — Butan-2-ol is optically inactive?
Answer: (a) p-dichlorobenzene have higher- melting point than ortho and meta isomer. This is because the para isomer is having a symmetrical structure and therefore, its packing is more efficient as compared to the ortho and meta isomer, therefore, it shows higher melting point.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-41
(b) The (+) – Butan-2-ol is optically inactive because it is racemic mixture and exists in two enantiomeric forms which are non-superimposable mirror images of each other. Both the isomers are present in equal amounts therefore, it does not rotate the plane of polarized light and is optically inactive.
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-42

Question.23. Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
(ii) Dacron
(iii) Teflon
Answer: Polymers with their monomers and their structures:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-43

Question.27. Write the types of isomerism exhibited by the following complexes:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-44

SET III

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

Question.1. What type of substances would make better Permanent Magnets, Ferromagnetic or Ferrimagnetic?
Answer: Ferromagnetic substance would make better permanent magnets because when the ferromagnetic substance is placed in a magnetic field, all domains get oriented in the direction of magnetic field and strong a magnetic effect is produced.

Question.3. What is the composition of ‘Copper matte’?
Answer: Composition of‘Copper matte’ is Cu2S and FeS.

Question.5. What is a glycosidic linkage?
Answer: The linkage between the two monosaccharide units through oxygen atom accompanied by the loss of a water molecule is called glycosidic linkage.

Question.6. Write the IUPAC name of (CH3)2CH.CH(Cl)CH3
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-45

Question.7. Which compound in the following pair undergoes faster SN1 reaction?
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-46

Question.8. Write the structure of p-Methylbenzaldehyde molecule.
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-47

Question.9. What is the difference between multi-molecular and macromolecular colloids? Give one example”of each.
Answer: Difference between Multi-molecular and Macromolecular colloids:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-48

Question.14. (a) Give an example of zone refining of metals.
(b) What is the role of cryolite in the metallurgy of aluminum?
Answer: (a) Zone Refining of metals : This method is used for production of semiconductor and other metals of very high purity like germanium, silicon, boron, gallium and indium.
(b) Role of cryolite in metallurgy of Aluminium : Cryolite is added to.lower the melting point of mixture and to increase the conductivity of electrolyte.

Question.17. Account for the following:
(i) The C-Cl bond length in chlorobenzene is shorter than that in CH3-Cl.
(ii) Chloroform is stored in closed dark brown bottles.
Answer: (i) This is due to partial double bond character to C-Cl bond (due to resonance in C6H5Cl).
(ii) Chloroform in the presence of air gets oxidized to phosgene. Phosgene is carbonyl chloride and is represented as COCl2. To prevent the formation of phosgene, chloroform is stored in dark coloured bottles. The reaction represented as CHCl3 + 1/2 O2 –> COCl2 + HCl

Question.18. How will you convert:
(i) Propene to Propan- 1-ol? (ii) Ethanal to Propan-2-ol?
Answer:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-49

Question.23. Give the structures of products A, B and C in the following reactions:
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-50
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-51

Question.27. Write the names and structures of the monomers of the following polymers:
(i) Bakelite (ii) Nylon-6 (iii) Polythene
cbse-previous-year-solved-papers-class-12-chemistry-delhi-2013-52

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