CBSE Previous Year Solved Papers Class 12 Chemistry Outside Delhi 2013

CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside 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

Question.1. Of physisorption or chemisorption, which has a higher enthalpy of adsorption ?
Answer: Chemisorption.

Question.2. Name the method used for refining of copper metal.
Answer: Electrolytic fefining of copper metal.

Question.3. Name two poisonous gases which can be prepared from chlorine gas.
Answer: Chloropicrin or tear gas (CCl₃N0₂) and phosgene gas (COCl₂).

Question.4. Write the IUPAC name of the following compound :

Question.5. Rearrange the following compounds in the increasing order of their boiling points: CH₃ – CHO, CH₃ – CH₂ – OH₂ -CH₃ – CH₂ – CHl₃
Answer: CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question.6. Write the structure of n-methylethanmine.

Question.7. What are the products of hydrolysis of sugar?
Answer: Glucose and fructose.

Question.8.

Question.9. Account for the following:
(i) Schottky defects lower the density of related solids.
(ii) Conductivity of silicon increases on doping it with phosphorus.
Answer: (i) Due to the missing of equal number of cations and anions from lattice, the density of lattice solid decreases in Schottky defect.
(ii) Presence of one unshared electron in phosphorous increases its conductivity and make silicon an n-type semiconductor.

Question.10. Aluminium crystallizes in an FCC structure. Atomic radius of the metal is 125 pm, what is the length of the side of the unit cell of the metal ?

Question.11.

Question.12. (a) For a reaction A + B —> P , the rate law is given by [2] r = k [A]^1/2 [B]²
(a) What is the order of this reaction?
(b) A first order reaction is found to have a rate constant k = 5.5 x 10-14 s-1. Find the half life of the reaction.

Question.13. (a) Name the method used for removing gangue from sulphide ores.
(b) How is wrought iron different from steel?
Answer: (a) Froth floatation method.
(b) Wrought iron is the purest form of iron with 0.2-0.5% carbon and steel is an alloy of iron. Wrought iron is produced from cast iron.
Steel is an alloy of iron and other elements. It has carbon content of 0.1-1.5%.
They also have different properties, industrial and decorative applications.

Question.14. Draw the structures of the following molecules:
(i) XeOF₄ (ii)H₃PO₃

Question.15. How are interhalogen compounds formed? What general compositions can be assigned to them?
Answer: Halogens react with each other to form interhalogen compounds.
Their general composition is XY_n where X is the less electronegative halogen of the two. They can be expressed by the formulas AB₃ to AB₇.

Question.16. Explain the mechanism of the following reaction :

Question.17. Write the equations involved in the following reactions :
(i) Reimer-Tiemann reaction
(ii) Williamson ether synthesis

Question.18. Define thermoplastic and thermosetting polymers. Give one example of each.
OR
What is a biodegradable polymer? Give an example of biodegradable aliphatic polyester.
Answer : Thermoplastic polymers : Intermolecular forces of attraction are intermediate between elastomers and fibres. Eg. Polythene.
Thermosetting polymers : These polymers are cross linked or heavily branchted forming 3-D chain, which can’t be moulded on heating. Eg. Bakelite.
OR
Natural polymers such as starch and cellululose that naturally disintegrate themselves, over a period of time are called biodegradable polymers. Example of biodegradable aliphatic polyester is PHBV (Poly-β-hydroxybutyrate-co-β-hydroxy Vaterate).

Question.19. The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature. [R = 8.314 J/K mol^-1, log 4 = 0.6021]

Question.20. What are the characteristics of the following colloids? Give one example of each.
(i) Multimolecular colloids (ii) Lyophobic sols (iii) Emulsions.
Answer: (i) On dissolution, a large number of atoms or molecules of a substance aggregate to form colloidal particles. This colloid is called a multimolecular colloid. E.g. Sulphur sol.
(ii) The colloids in which there is no affinity between the particles of dispersed phase and dispersion medium are called lyophobic colloids. They are not stable, that is easily coagulated and irreversible. They are produced only by special methods. Eg. AS₂S₃S0l, Fe(OH)₃sol.
(iii) Emulsions are colloids in which both dispersed phase and dispersing medium are liquid and immiscible with each other Eg. Milk, cod liver oil, etc.

Question.21. Give reasons for the following:
(i) When R is an alkyl group R₃ P = 0 exist but R; N = 0 doesn’t.
(ii) PbCl₄ is more covalent than PbCl₂.
(iii) N₂ is much less reactive at room temperature.
Answer: (i) Due to absence of d-orbitals, N cannot form pπ -dπ multiple bonds. Thus, it cannot expand its covalency beyond 4. In R₃ N = 0, N has covalency 5 so it does not exist. On the other hand, due to presence of d-orbitals, P forms pπ —dπ multiple bonds and expand its covalency beyond 4. In R₃ P=0 covalency of P is 5 hence it exists.
(ii) Because ‘Pb’ is in +4 oxidation state in PbCl₄ and has high charge/size ratio than Pb²⁺. According to Fazan’s rule, a higher charge on cation or anion makes compound more covalent, +4 state is more stable than +2 state. Hence PbCl₄ is more covalent than PbCl₂
(iii) N₂ has. strong pπ -dπ overlap resulting to triple bond (N = N). Due to short bond length of (N = N) triple bond, N₂ has high bond energy and hence it is less reactive at room temperature.

Question.22. For the complex [NiCl₄]^2-, write
(i) the IUPAC name (ii) the hybridization type
(iii) the shape of the complex
(Atomic no. of Ni = 28)
OR
‘What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration or d⁴ in terms of t_2g and e_g in an octahedral field when
(i) ∆₀ > P
(ii) ∆₀ < P

Question.23. Give reasons for the following;
(i) Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C — X bond length in halobenzene is smaller than C — X bond length in CH₃ — X.
Answer: (i) I^– ion is better leaving group than Br~ion, therefore, ethyl iodide reacts faster than ethyl bromide in S_N2 reaction.
(ii) (±) 2-butanol is optically inactive because it is racemic mixture i.e. it has enantiomers in equal amount and hence they cancel each other’s rotation of plane polarised light.
(iii) In halobenzene halogen atom is attached to sp³ hybridized carbon while in CH₃-X sp² hybridized carbon which is smaller in size than sp³ hybridized carbon attached to the halogen. Therefore C-X bond in halobenzene is shorter than in CH₃-X.

Question.24. Complete the following reactions :

Question.25. (i) What class of drug is Ranitidine?
(ii) If water contains dissolved Ca²⁺ ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
(iii) Which of the following is an antiseptic? 0.2% phenol, 1% phenol
Answer: (i) Antacid.
(ii) Soap gets precipitated in hard water hence it can’t be used to wash clothes. On the other hand, synthetic detergents do not precipitate in hard water because its calcium salt is soluble in water. Therefore, it can be used to wash clothes in hard water.
(iii) 0.2% phenol.

Question.26. Calculate the emf of the following cell at 25°C :

Question.27. Shanti, a domestic helper of Mrs. Anuradha, fainted while mopping the floor. Mrs. Anuradha immediately took her to* the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron rich diet and multivitamins supplement to her. Mrs. Anuradha supported her financially to get the medicines. After a month, Shanti was diagnosed to be normal.
(i) What values are displayed by Mrs. Anuradha?
(ii) Name the vitamin whose deficiency causes ‘pernicious anaemia’.
(iii) Give an example of a water soluble vitamin.
Answer: (i) Mrs. Anuradha was caring, helpful and compassionate for her domestic help.
(ii) Pernicious anaemia is caused due to deficiency of Vitamin B₁₂.
(iii) Vitamin C.

Question.28. (a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K_f for benzene = 5.12 K kg/ mol)
OR
(a) Define the following terms :
(i) Ideal solution (ii) Azeotrope (iii) Osmotic pressure.
(b) A solution of glucose (C₆H₁₂0₆) in water is labelled , as 10% by weight. What would be the molality of the solution? (Molar mass of glucose = 180 g/mol)
Answer : (a) Raoult’s law states that the partial pressure of a vapour of a component in the solution in directly proportional to its mole fraction in the solution.
Wauuiis’jaw’necomes a special case of Yienry s’law as it states that the partial pressure of gas in vapour phase is directly proportional to the mole fraction of the gas in the solution.

(a) (i) A solution that obeys Raoult’s law. over all ranges of temperature and concentration and shows no attractive forces between components, is called as ideal solution.
(ii) A liquid mixture which distill at constant temperature without undergoing any change in composition is called Azeotropes.
(iii) The minimum external pressure required to prevent osmosis is known as osmotic pressure.

Question.29. (a) Give reasons for the following:
(i) Mn³⁺ is a good oxidising agent.
(ii) E°_M2/M values are not regular for fist row transition metals (3d series).
(iii) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF₄, whereas the highest oxide is Mn₂ G ₇
(b) Complete the following equations :

(a) Why do transition elements show variable oxidation states?
(i) Name the element -showing maximum number of
oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
(ii) Name the element which shows only +3 oxidation state.
(b) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals.
Answer : (i) Mn⁺³ has electronic configuration 3d⁴ 40. It gains one electron on reduction and become 3d³40 which is half filled stable configuration. Hence it is a good oxidizing agent.
(ii) Due to extra stability of half and fully filled d-orbitals and variations in ionization energies E°_M2/M values are not regular.
(iii) Due to the ability of oxygen to form multi bonds with, metals and because oxygen stabilizes the highest oxidation state even more than fluorine.

OR
(a) Due to the presence of incomplete d-orbitals.
(i) Manganese shows oxidation states from +2 to +7.
(ii) Lanthanum.
(b) Lanthonoid contraction refers to the steady and regular decrease in atomic size along the period from La³⁺ to Lu³⁺ eg. Misch metal alloy which contains 95% lanthanoids and 5% iron.

Question.30. (a) How will you convert the following:
(i) Propanone to Propan-2-ol
(ii) Ethanal to 2-hydroxy propanoic acid
(iii) Toluene to benzoic acid
(b) Give simple chemical test to distinguish between:
(i) Pentan-2 -one and Pentan-3 -one
(ii) Ethanal and Propanal
OR
(a) Write the products of the following reactions:

(b) Which acid of each pair shown here would you expect to be stronger?


(b) (i) F-CH₂-COOH is a stronger acid than Cl-CH₂– COOH, because F is more electronegative than Cl, so it will favour release of H⁺ ion faster by dragging electron density towards itself more as compared to Cl.
(ii) Acetic acid is stronger acid than phenol. Acetic acid forms carboxylate ion and phenol forms phenoxide ion. Carboxylate ion is more stable than phenoxide ion due to resonance

SET II

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

Question.1. Write the structure of 2-aminotoluene.
Answer:

Question.2. Which aerosol depletes ozone layer?
Answer: Chloroflurocarbon (CFCs) present in aerosols depletes ozone layer.

Question.4. Ethanal is soluble in water . why?
Answer: Ethanal is soluble in water due to hydrogen bonding.

Question.5. Write the IUPAC name of the following compound:

Question.8. Give one example of a condensation polymer.
Answer: Dacron or Nylon-6, 6.

Question.9. (a) Why does presence of excess of lithium makes LiCl crystals pink?
(b) A solid with cubic crystal is made of two elements P and Q Atoms of Q are at the comers of the cube and P at the body-centre. What is the formula of the compound?
Answer: (a) Excess of Li in LiCl causes metal excess defect due to anionic vacancy in the crystal. Pink colour results due to excitation of electrons on absorbing visible light.
(b) Atoms of Qare present at the corners of the cube. Therefore, member of atoms of Q in one unit cell =8 x 1/8= 1. Atoms of P are present at the body centre. Therefore, number of atoms of P in one unit cell = 1. Hence, the formula of compound is PQ.

Question.14. Draw the structures of the following molecules :
(i)XeF₆ (ii)H₂S₂0₇

Question.18. Outline the principles of refining of metals by the following methods:
(i) Zone refining (ii) Vapour phase refining
Answer: (i) This method is based on the principle that the impurities are more soluble in the molten state than in thesolid state of metal.
(ii) In this process the metal is converted to its volatile compound and collected elsewhere and then decomposed to give pure metal.

Question.19. Define the following terms giving an example of each:
(i) Associated colloids (ii) Lyophilic sol
(iii) Adsorption.
Answer: (i) Associated colloids are colloidal substances which at low concentrations behave as normal electrolytes but at higher concentration they aggregate to form colloids. Eg. Soaps and detergents.
(ii) Lyophilic sols are the colloids in which the particles of dispersed phase have a strong affinity for the dispersion medium. They are reversible in nature because on precipitation they can be easily converted back to colloidal form by adding dispersion medium, eg. Starch sol
(iii) The aggregation of a substance on the surface of liquid or solid is known as adsorption eg. Platinum adsorbs H₂ gas.

Question.22. Write the main products of the following reactions:

Question.27. Give reasons for the following:
(i) Oxygen is a gas but sulphur is a solid.
(ii) O₃ acts as a powerful oxidising agent.
(iii) BiH₃ is the strongest reducing agent amongst all the hydrides of Group 15 elements.
Answer: (i) Oxygen is smaller in size than sulphur. Due to small size, it can effectively form pπ — pπ bonds, and forms diatomic 0₂ molecule. The intermolecular forces in oxygen are weak van der Waals forces, which causes it to exit as gas. On the other hand, sulphur does not form strong S=S double bonds and exists as puckered structure held together by covalent bonds and exists as polyatomic molecule. So, it exists as solid.
(ii) O₃ act as powerful oxidising agent because it decompose to give nascent oxygen.
(iii) On moving down the group, the bond length, increases and its strength decreases. Bi-H bond is the weakest. It can break easily and evolves H₂ gas which is a strong reducing agent. Hence, BiH₃ is the strongest reducing agent among group 15 elements.

SET III

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

Question.1. What is especially observed when a beam of light is passed through a colloidal solution ?
Answer: Tyndall effect is observed due to scattering of light.

Question.2. What is the basicity of H₃PO₃ and why?

Third hydrogen is directly attached to phosphorous is not acidic.

Question.3. Write the IUPAC name of the following compound:

Question.8. Write the structure of prop-2-en-1-amine.
Answer: H₂C=CH-CH₂-NH₂

Question.12. Draw the structures of the following molecules :
(i) N₂O₅ (ii) XeF₂

Question.13. (a) What change occurs when AgCl is doped with CdCl₂? (b) What type of semiconductor is produced when silicon is doped with boron ?
Answer: (a) When AgCl is doped with CdCl₂ impurity defect is produced one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical conductivity. One position is occupied by Cd²⁺ ion and other will be left as a hole similar to schottky defect.
(b) n-type semiconductor.

Question.18. Name the principal ore of aluminium. Explain the significance of leaching in extraction as aluminium.
Answer: Bauxite (Al₂0₃.2H₂0) is the principal ore of aluminium. The significance of leaching in extraction of aluminium is to prepare pure alumina from the bauxite ore in the following steps :

Question.19, Define the following terms with an example in each case:
(i) Macromolecular Sol (ii) Peptization
(ii) Emulsion.
Answer: (i) Macromolecular sol: They are molecules of large size having high molecular masses. Due to long chain, the van der Waals forces are stronger. Eg. rubber, nylon, etc.
(ii) Peptization : It is the process of converting a precipitate into colloidal solution by shaking it with dispersion medium in the presence of small amount of electrolyte. E.g. A preciptate of Agl can be peptized by shaking with a dilute solution of silver nitrate.
(iii) Emulsion : A type of colloidal solution in which both the dispersed phase and dispersion medium are liquid and are immiscible with each other is called emulsion. Ex. milk.

Question.21. Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does – not form pentahalide
(ii) Electron gain enthalpy with negative sign of flourine is less than that of chlorine.
(iii) The two oxygen-oxygen bond lengths in ozone molecules are identical.
Answer: (i) Due to lack of empty d-orbitals in nitrogen, it does not form pentahalide.
(ii) Due to small size, fluorine atom has high electro negativity and strong electron – electron repulsions in its compact 2p orbitals, its electron gain enthalpy is less than that of chlorine.
(iii) The two oxygen bond lengths in ozone are identical due to resonance.