CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2011

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.


Question.1.Define ‘order of a reaction’.
Answer : The sum of powers of the concentrations of the reactants in the rate law expression is called the order of a reaction.

Question.2.What is meant by ‘shape selective catalysis’?
Answer : Catalysis using selective absorbents like zeolites as catalyst is called shape selective catalysis. In this catalysis small sized molecules are absorbed in the pores and cavities of zeolites.

Question.3.Differentiate between a mineral and an ore.

Question.4. What is meant by ‘lanthanoid contraction’?
Answer : The lanthanoid contraction refers to the steady and regular decrease in atomic radii along the period from La+3 to Lu+3.

Question.5. Write the IUPAC name of the following compound:
CH2 = CHCH2 Br
Answer: 3-Bromo-l-propene

Question.6. Draw the structure of 4-chloropentan-2-one.

Question.7. How would you convert ethanol to ethane?

Question.8. Rearrange the following in an increasing order of their basic strengths:

Question.9. Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal.

Question.10.Calculate the packing efficiency of ametal crystal for a simple cubic lattice.
Question.11. State the following:

  1.  Raoult’s law in its general form in reference to solutions.
  2. Henry’s law about partial pressure of a gas in a mixture.

Answer :

  1.  Raoult’s law : The partial vapour pressure of each component of a solution is equal to the vapour pressure of pure component multiplied by its mole fraction in the solution
  2.  Henry’s law : It states that the partial pressure of gas in vapour phase (P) is direedy proportional to mole fraction of gas (x) in the solution and is expressed as

Question.12. What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction if the units of its rate constant are:

  1. L-1mol s-1
  2. L mol s-1

Answer : Rate law of a chemical reaction is the expression relating the rate of reaction to the concentrations or pressures of various reactants taking part in the reaction.
The rate of reaction at unit concentration of all reactants is known as the rate constant (K)

  1.  Zero order
  2. Second order

Question.13. The thermal decomposition of HCO2 H is a first order reaction with a rate constant of 2.4 x 10-3 s-1 at a certain temperature. Calculate how long will it take for three- fourths of initial quantity of HCO2H to decompose, (log 0.25 = -0.6021)

Question.14. Describe the principle controlling each of the following

  1.  Vapour phase refining of titanium metal
  2. Troth floatation method of concentration of a sulphide ore

Answer :

  1. Titanum is converted to its volatile form which is evaporated and then decomposed to give pure titanium.
  2.  The ore particles get adsorbed on oil droplets and come to the surface where they can be collected as froth gangue is wetted by water and gets settledown. ,

Question.15. How would you account for the following:

  1.  Cr2+ is reducing in nature while with the same d-orbital configuration (d4) Mn3+ is an oxidizing agent.
  2. In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.

Answer :

  1.  Cr2+has configuration d4 which easily changes to d3 due to stable half filled ftg orbitals. Therefore, Cr2+ is reducing agent and Mn3+ easily changes to Mn2+ and acts as an oxidizing agent.
  2. Due to presence of more unpaired electrons and more number of partially filled orbitals in the middle of a transition series,these metals exhibit the greatest number of oxidation states.

Question.16. Complete the following chemical equations :
State reasons for the following :
(i) Cu (I) ion is not stable in an aqueous solution.
(ii)Unlike Cr3+, Mn2+, Fe3+ and the subsequent other M2+ ions of the 3d series of elements, the 4d and the 5d series metals generally do not form stable cationic species.
(ii) Because of Lanthanoid contraction, expected increase in size does not occur. That is why they do not form stable cations.

Question.17. Explain what is meant by the following:

  1.  Peptide linkage
  2. Pyranose structure of glucose

Answer :

  1. Peptide linkage is the peptide bond formed between amino acids. It is a covalent bond formed between amino group of one molecule and carboxylic acid group of another molecule, causing the release of one molecule of water.
  2.  Cyclic structure of glucose is called pyranose because it resembles pyran ring. Its structure includes a six membered ring with 5-carbon atoms and one oxygen atom having no double bonds.

Question.18. Write the main structural difference between DNA and RNA. Of the four bases, name those which are common to both DNA and RNA.

Question.19. A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

Question.20. Classify colloids where the dispersion medium is water. State their characteristics and write an example of each of these classes.
Explain what is observed when

  1. an electric current is passed through a soln.
  2. a beam of light is passed through a soln.
  3. an electrolyte (say NaCl) is added to ferric hydroxide sol

Answer : Colloids can be classified into two types where the dispersion medium is water. They are as follows :

  1.  Hydrophillic colloids or Lyophillic: The substances when mixed with dispersion medium form colloidal solution direcdy are called hydrophilic colloids. They are quite stable, reversible
    solutions and can’t get precipated easily, e.g., Gum starch, etc.
  2. Hydrophobic colloids or Lyophobic : The substances which do not form colloidal solution with dispersion medium are called hydrophobic colloids. They are unstable, irreversible solutions and can be easily precipitated, e.g., Metals and their sulphides.


  1. When electric current is passed through a solution, then positively charged ions move towards cathode and negatively charged ions move towards anpde. Then they get coagulate. This is known as Electrophoresis.
  2. When a beam of strong light is passed through solution, light gets scattered by the colloidal particles and path of light becomes visible. This is known as Tyndall effect.
  3. When NaCl is added to ferric hydroxide sol then a negatively charged solution is obtained with absorption of OH ion.

Question.21. How would you account for the following:

  1.  H2S is more acidic than H2O.
  2. The N-O bond in NO2 is shorter than the N-O bond in NO3 
  3. Both O2 and F2 stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine.


  1. Size of sulphur is large than oxygen due to which S-H bond length increase and hence the bond dissociation energy of S-H is less than O-H. Therefore S-H easily loses H+ and is more acidic than H2O.
  2. Due to tendency of nitrogen to form pπ-pπ  multiple
    bonds, there is a difference in N-O bond lengths of NO2 and NO2 .
  3. Due to the property of oxygen to form double bonds with the metal atoms, oxygen stabilizes the higher oxidation state even more than fluorine.

Question.22. Explain the following terms giving a suitable example in each case:

  1.  Ambident ligand
  2. Denticity of a ligand
  3. Crystal field splitting in an octahedral field

Answer :

  1. Ligands which can ligate to the central atom in two places ligands such as SCN which can. attach at either S atom or N atom are called ambidentate ligand.
  2. The number of donor atoms of a ligand when bound with central atom in a coordination complex is called its denticity.
  3. The splitting of five degenerate d-orbitals in two sets, one with three orbitals and another with two orbitals is known as crystal field splitting.

Question.23. Rearrange the compounds of each of the following sets in order of reactivity towards SN2 displacement:

  1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bro- mopentane
  2. 1 -Bromo-3-methylbutane, 2-Bromo-2-methyl-butane, 2-Bromo-3-methylbutane
  3. 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, 1- Bromo-2-methylbutane, l-Bromo-2, 2-dimethylpropane.


  1. 1-bromopentane > 2-bromopentane > 2-bromo- 2-methylbutane.
  2. 1-bromo-3 methylbutane > 2-bromo-3-methylbutane > 2-bromo-2-methyl butane.
  3. 1-bromobutane>1 -bromo-3-methylbutane> 1-bromo-2-, 2-dimethylpropane >1-bromo-2, 2-dimethyl propane.

Question.24. How would you obtain the following:
(i) Benzoquione from phenol
(ii)2-Methylpropan-2-ol from methylmagnesium bromide
(iii)Propan-2-ol from propene

Question.25. State reasons for the following:

  1. pKbvalue for aniline is more than that for methylamine.
  2. Ethyiamine is soluble in water whereas aniline is not soluble in water.
  3.  Primary amines have higher boiling points than tertiary amines.

Answer :

  1. Higher value of  pKb means lower basicity, therefore, aniline is less basic than methylamine because in aniline the lone pair of electrons on N atom gets delocalized over the benzene ring and remain unavailable for protonation due to resonance but this is absent in methylamine.
  2. Ethyiamine forms H-bonds with water therefore, it is soluble in water but aniline does not form H-bonds with water due to its larger hydrocarbon part and is insoluble in water.
  3. In primary amines, two H-atoms are attached to N-atom and they undergo intermolecular H-bonding, but tertiary amines due to the absence of H-atom on N-atom do not undergo H-bonding . Therefore, primary amines have higher boiling point than tertiary amines.

Question.26. Draw the structures of the monomers of the following polymers:

  1. Polythene
  2. PVC
  3. Teflon

Answer :

  1. ethene CH2 = CH2.
  2. Vinyl chloride CH2 = CHCl
  3. Tetraflouroethylene CF2 = CF2

Question.27. What are the following substances? Give one example of each.

  1.  Food preservatives
  2. Synthetic detergents
  3. Antacids

Answer :

  1.  Chemicals added to food to prevent its spoilage by killing or preventing the growth of microorganisms like bacteria, yeasts and moulds. e.g. sodium benzoate.
  2. Synthetic detergents are sodium or potassium salts of long chain sulphonic acids. They don’t precipitate in hard water. e.g. Sodium laurylsulphate.
  3. Antacids are chemicals consumed to get relief from acidity in the stomach by neutralizing excess acid, e.g, milk of magnesia.

Question.28. (a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
(b) Calculate the potential for half-cell containing 0.10 M K2Cr2O7 (aq), 0.20 M Cr3+ (aq) and 1.0 x 10-4 M H+ (aq) The half-cell reaction is and the standard electrode potential is given as E° = 1.33 V
(a) How many moles of mercury will be produced by electrolyzing 1.0 M Hg(N03)2 solution with a current of 2.00 A for 3 hours? [Hg(NO3)2 = 200.6 g mol1]
(b) A voltaic cell is set up at 25°C with the following half-cells Al3+ (0.001 M) and Ni2+ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
Answer :
(a) Lead storage battery is a secondary cell (rechargable). The electrode reaction is as follows :

Question.29. (a) Draw the structures of the following molecules:
(a) What happens when
(i) Chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) Sulphur dioxide gas is passed through an aqueous solution of a Fe(III) salt?
(b) Answer the following :
(i) What is the basicity of H3PO3and why?
(ii)Why does fluorine not play and role of a central atom in interhalogen compounds?
(iii)Why do noble gases have very low boiling points?
(b)(i) Two, because H3PO3 has two replaceable H+ ions as one hydrogen is attached directly to phosphorus and can’t be released as a proton.
(ii)Since fluorine is the most electro-negative halogen, it does not acts as a central atom in interhalogen compounds.
(iii)Noble gases are monoatomic with weak Vander Waals forces of attraction and hence have low boiling points.

Question.30. (a) Illustrate the following name reactions :
(i) Cannizzaro’s reaction
(ii)Clemmensen reduction
(b) How would you obtain the following :
(i) But-2-enal from ethanal
(ii)Butanoic acid from butanol
(iii)Benzoic acid from ethylbenzene
(a) Give chemical tests to distinguish between the following:
(i) Benzoic acid and ethyl benzoate
(ii)Benzaldehyde and acetophenone
(b) Complete each synthesis by giving missing reagents or products in the following :
Answer : (a)
(i) Aldehydes having no a-hydrogen atom undergoes self oxidation and reduction on treatment with concentrated alkali and produce alocohol and carboxylic acid salt.


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

Question.2.What are lyophobic colloids? Given one example for them.
Answer : Colloids in which the colloidal particles have no affinity for the dispersion medium and they do not form colloidal solution are lyophobic colloids, like, Al(OH)3 and  AS2S3 sols.

Question.3. Why is it that only sulphide ores are concentrated by ‘froth floatation process’?
Answer : Due to the affinity of heavy oil droplets to adsorb sulphide particles. The ore particles come on the surface as froth, gangue particles are wetted by water and get settle down.

Question.5. Write the IUPAC name of the following compound:
Answer: 3-Bromo-2-methyl prop-l-ene

Question.6. Draw the structure of 2, 6-Dimethylphenol.

Question.9. Define the following terms in relation to crystalline solids:

  1.  Unit Cell
  2. Coordination number

Answer :

  1.  Unit Cell: The smallest three dimensional unit of a lattice which when repeated in three dimensions generates the entire lattice is called the unit cell, e.g., Cubic unit cell.
  2.  Co-ordination number : Coordination number of an atom in a crystal is the number of immediate neighbours of the atom in that crystal, e.g., Coordination number of hexagonal (hep) structure isT2.

Question.12. A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ? What is the unit of rate constant for such a reaction ?

Question.14. Describe the principle controlling each of the following processes:

  1.  Zone refining of metals
  2. Electrolytic refining of metals

Answer :

  1.  Zone refining of metals : It is based on the principle that impurities are more soluble in molten state of metal than in the solid state,
  2.  Electrolytic refining : Impure metal is made of anode, thin sheet of pure metal is made of cathode and a salt of the metal is used as an electrolyte. On passing current, metal from anode goes into the solution and ions in the solution reduce on cathode leading to deposition of pure metal.

Question.15. Explain giving a suitable reason for each of the following :

  1.  Transition metals and their compounds are generally found to be good catalysts.
  2. Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.

Answer :

  1.  Due to presence of vaccant orbitals and tendency to form large number of oxidation states, transition metals have a high tendency to form complexes and hence acts as a catalysts.
  2.  In transition metals of 4d and 5series the 4d and 5d electrons are at greater distance from the nucleus therefore they are less tighdy held to . the atom by the nucleus and hence contribute more to metallic bonding , as compared to transition metals of 3d series.

Question.19. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50 °C? The freezing point depression constant (Kf) for water is 1.86°C/m. Assume Van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g)

Question.22. Write the structures and names of all the stereoisomers of the following compounds :

Question.27. (a) Differentiate between a disinfectant and an antiseptic. Give one example of each.
(b) What is tincture of iodine and what is it used fori
Answer: (a)
Answer : (b) 2-3% solution of iodine in alcohol and water is called tincture iodine and are widely is used as an antiseptic


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

Question.1. Define ‘activation energy’ of a reaction
Answer: Activation energy is the minimum energy that need to be provided to the reactants for the reaction to take place.

Question.2. What is meant by ‘reverse osmosis’?
Answer : When a pressure larger than osmotic pressure is applied the solvent will flow from the solution into the pure solvent through the semipermeable membrane, called reverse osmosis.

Question.3. What type of ores can be concentrated by magnetic separation method?
Answer : Ores with different magnetic properties and impurities magnetic in nature can be concentrated by magnetic separation method.
e.g. Chromite (Fe O. Cr2O3), Magnetite (Fe3O4)

Question.14. Describe the principle controlling each of the following processes:

  1. Preparation of cast iron from pig iron.
  2. Preparation of pure alumina (Al2O3) from bauxite ore.

Answer :

  1.  Preparation of cast iron from pig iron : The iron obtained from blast furnace is called pig iron. Cast iron is prepared by meltiilg pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%). It is extremely hard and brittle.
  2.  Preparation of pure alumina from bauxite ore : The principal ore of aluminium is bauxite Al2O3.xH20. Bauxite is concentrated by digesting the powdered ore with concentrated solution of NaOH at 473-523K. The Al2O3 is leached out as sodium aluminate. The sodium aluminate is neutralized by passing CO2gas and hydrated Al2O3is precipitated which is filtered, dried and heated to give pure Al2O3.

Question.15. Explain giving reasons :

  1.  Transition metals and their compounds generally  exhibit a paramagnetic behaviour.
  2. The chemistry of actinoids is not so smooth as that of lanthanoids.

Answer :

  1. Transition metals and their compounds exhibit a paramagnetic behaviour due to presence of unpaired electrons in the Penultimate shell of d-orbital.
  2.  Lanthanoids show limited number of oxidation states, +2, +3, +4 because of large energy gap between 4f and 5d subshells. Actinoids show a number of oxidation states +4,+5, +6, +7 due to small energy difference between 5f, 6d and 7s subshells.

Question.18. Write such reactions and facts about glucose which cannot be explained by its open chain structure.
Answer : Limitations of open chain structure of glucose :

  1.  Glucose does not form NaHS03 as addition product.
  2. Glucose penta-acetate does not react with NH2OH due to absence of aldehyde group.

Question.21. How would you account for the following:

  1.  NF3 is an exothermic compound but NCl3is not.
  2. The acidic strength of compounds increase in the order:
  3.  SF6 is kinetically inert.

Answer :

  1. In NF3, fluorine molecule, F-F bond enthalpy is very high, so large amount of energy is released when NF3 is formed but this is not in the case of NCl3.
  2. As electronegativity increases in the same period from left to right so their electronegativity is in the increasing order P < S < Cl. Therefore, acidic strength increases in the order,
    PH3 < H2S < HCl
  3. SF6 is protected by six F atoms and hence does not allow to attack on sulphur atom.

Question.22 Write the state of hybridization, the shape and the magnetic behaviour of the following complex entitles :
Question.26. Write the names and structures of the monomers of the following polymers:

One thought on “CBSE Previous Year Solved Papers Class 12 Chemistry Outside Delhi 2011

Comments are closed.