CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2016

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. What type of magnetism is shown by a substance if moments of domains are arranged same direction ?
Answer : If magnetic moments of domains are arranged in same directions then the ferromagnetic type of magnetism is shown by the substance.


Question.3. On adding NaOH to ammonium sulphate a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu2+ ion. Identify the gas.

Question.4. Write the main reason for the stability of colloidal solutions.
Answer: All the colloidal particles in a given solution carry the same charge and the dispersion medium has an opposite and equal charge; the system as a whole being electrically neutral. This is the main reason for the stability of the colloidal solution.

Question.5. Write the IUPAC name of the given compound.

Question.6. When a coordination compound CrCl3.6H2O is mixed with AgNO3, 2 mole of AgCl are precipitated per mole of the compound. Write
(i) Structural formula of the complex.
(ii) IUPAC name of the complex
Answer: (i) Structural formula of the complex is [Cr (H20)5 Cl] Cl2. H20, because two mole of chlorine are outside coordination entity to form two mole of AgCl from per mole compound.
(ii) Pentaaquachloride chromium(III) chloride monohydrate.

Question.7. From the given cells :
Lead storage cell, Mercury cell, Fuel cell and Dry cell Answer the following;
(i) Which cell is used in hearing aids ?
(ii) Which cell was used in Apollo Space Programme ?
(iii) Which cell is used in auto mobiles and inverters ?
(iv) Which cell does not have long life ?
Answer : (i) Hearing aid—Mercury cell.
(ii) Apollo Space Programme—Fuel cell.
(iii) Automobile and inverters—Lead storage cell.
(iv) Cell does not have long life—Dry cell.

Question.8. When chromite ore FeCr204 is (used with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B), compound (B) on reaction with KCl forms a orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C).
Complete the following chemical equations :

Question.9. Write the mechanism of the following reaction :
Answer : Formation of ether from alcohol is a nucleophilic bimolecular reaction (SN2). A protonated alcohol is attacked by another alcohol molecule.

(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer: (i) For any reaction; Rate = K [A] order
[A] = concentration of reactant.
Hence its a zero order reactions and its molecularity is two.
(ii) Unit of K for a zero order reaction is mol L-1 sec-1.

Question.11. The rate constant for the first order decomposition of H2O2 is given by the following equation :
Calculate Ea for this reaction and rate constant K if its half life period be 200 minute. (Given : R = 8.314 J K-1 mol-1)

Question.12. (i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl, which one is more effective in causing coagulation of negatively charged sol and why ?
(iii) Out of sulphur sol and proteins, which one forms multi molecular colloids ?
(ii) According to Hardy-Schulze law the ions carrying opposite ‘ charge to that on sol are responsible for coagulation of the sol. These are called active ions. Hence as the sol is negative, Mg2+ and Al5+ ions will cause coagulation. As coagulation power of electrolyte is proportional to the valency of oppositely charged ion, so AlClwill be more effective than MgCl2 .
(iii) Out of sulphur sol & protein, sulphur sol will form multi molecular colloids. As the particles are similar than 103 pm. So they form aggregate to form sol. sulphur sol consists of S8 molecules.

Question.13. Give reasons:
(i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH3—Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) SN1 reactions are accompanied by racemization in optically active alkyl halides
Answer: (i) C — Cl bond length in chlorobenzene is shorter than CH3 – Cl, C – Cl bond as in chlorobenzene due to resonance C-Cl bond has partial double bond character which reduces the bond length.
(ii)In cyclohexyl chloride, carbon in C – Cl bond is sp3 hybridised whereas in chlorobenzene C – Cl bond carbon is sp2 hybridised sp2 is more electronegative than sp3 carbon. Hence C — Cl bond of chlorobenzene less polar.
(iii) In SN1 reaction a carbocation intermediate is formed. In case of optically active albyl halide the attack of nucleophile in the next step to carbocation can occur from both the faces of the trigonal planar species in equal probability. Thus 50 : 50 racemic mixture is obtained.

Question.14. An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contain 2 x 1024 atoms.

Question.15. Give reasons :
(i) Mn shows the highest oxidation state of + 7 with oxygen but with fluorine it shows the highest oxidation state of + 4
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.
Answer : (i) oxygen has a greater capacity to stabilised the highest oxidation state of a particular metal than fluorine because it can form multiple bonds in oxide (double bond) which is not possible in case of fluoride. ‘
(ii) Transition metals show variable oxidation state because of use of ns and (n-1) d shell electrons while bonding as the shells have similar energy.
(iii) Actinoids show irregularities in their electronic configurations because 6d, Is and 5f electrons/shells have less energy difference and electrons can be accommodated in any of them.

Question.16. Write the main product(s) in each of the following reactions :

Question.17. (i) Name the method of refining of metals such as Germanium.
(ii) In the extraction of Al, impure Al2O3 is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of his process ?
(iii) What is the role of coke in the extraction of iron from its oxides ?
Answer : (i) Zone refining method used for refining of metals such as germanium which is based on the principle that the impurities are more soluble in the meat than in the solid state of the metal.
(ii) Leaching: This method consists of treating the powdered ore with a suitable reagent which can selectively dissolve the ore but not the impurities.
(iii) Coke act as a reducing agent and it reduces the iron ore hematite.

Question.18. Calculate e.m.f. of the following cell at 298 K.

Question.19. (i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar ?
(ii) Why Vitamin C cannot be stored in our body ?
(iii) What is the difference between a nucleoside and nucleotide ?
Answer: (i) The two monosaccharides are β-D-galactose and β-D- glucose.
(ii) Vitamin C is, an water soluble,, vitamin and hence get excreted by urine. So it cannot be stored in body and needs to be supplemented regularly.
(iii) When a base (purine or pyriamidine) get attached to T position of a pentose sugar a nucleoside is formed. When a nucleoside is further linked to phosphoric acid at 5′ position of sugar moeity, we get a nucleotide.

Question.20, (a) For the complex [Fe(CN)6]3- write the hybridization type, magnetic character and spin nature of the complex, (pt number Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt (en)2 Cl2]2+, which is optically active ?

Question.21. Write the structure of A, B and C in the following :

Question.22. (i) What is the role of t-butyl peroxide in the polymerization of ethane ?
(ii) Identify the monomers in the following polymer :
(iii) Arrange the following polymers in the increasing order of their inter molecular forces. Polystyrene, Terylene, Buna-S.
Write the mechanism of free radical polymerisation of ethene. 
Answer : (i) Polymerisation of ethene to low density polyethene (L.D.P.) needs presence of a free radical generating initiator (catalyst), t-butyl peroxide helps in starting the chain of radical formations.

Question.23. Due to hectic and busy schedule Mr. Angad made his life full of tensions and anxiety. He started taking sleeping pills to overcome the depression without consulting the doctor. Mr. Deepak a close friend of Mr. Angad advised him to stop taking sleeping pills and suggested to change his lifestyle by doing Yoga, meditation and some physical exercise. Mr. Angad followed his friend s advice and after few days he started feeling better. After reading the above passage answer the following:
(i) What are the values (at least two) displayed by Mr. Deepak.
(ii) Why is it not advisable to the sleeping pills without consulting doctor.
(iii) What are tranquilizers ? Give two examples.
Answer : (i) Values displayed by Mr. Deepak ;
(a) Knowledge of side effects of pills without prescription.
(b) Friendship
(ii) Sleeping pills are tranquilizers and may cause harmful side effects and act as poison. Hence a doctor must be consulted to regularise the doses of such drugs.
(iii) Tranquilizers are a class of druge or chemicals which are used to treat stress and mental disease. Example : proniazid and Equanil.

Question.24. (a) Write the structures of A, B, C, D and E in the following reactions:
(a) Write the chemical equation for the reaction involved in Cannizzaro reaction.
(b) Draw the structure of the semicarbazpne of ethanal.
(c) Why pKa of F-CH2-COOH is lower than that of CI-CH2-COOH ?
(d) Write the product in the following reaction ,
(e) How can you distinguish between propanal and proanone ?
(a) For aldehydes which do not have a-hydrogen atom self oxidation and reduction takes place in presence of concentrated alkali. This produces one mole of alcohol and one. mole of salt of carboxylic acid. This is called Cannizzaros reaction.
(e) Tollens reagent will give a positive test of silver mirror formation with propanal.
Aldehyde can oxidise Tollens’ reagent to metallic silver but ketones cannot.

Question.25. (a) Calculate the freezing point of solution when 1.9 g (of MgCl2 (M = 95 g/mol) was dissolved in 50 g of water, assuming MgCl2. Undergoes complete ionization. (Kf for water = 1.86 K kg mol-1)
(i) Out of 1 M glucose and 2 M glucose, which one has a higher boilling point and why ?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of so lution ?
When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mole-1, Atomic mass of sulphur = 32 g/mol-1).
Blood cells are isotonic with 0.9% sodium chloride solution what happens if we place blood cells in a solution containing.
(i) 1.2% sodium chloride solution ?
(ii) 0.4% sodium chloride solution ?
(b) (i) 2M glucose will have higher boiling point because boiling point of a solution of a non-volatile liquid increases with increase in concentration
(ii)When external pressure applied is more than the osmotic pressure then reverse osmosis takes place. The solvent will then flow from the solution to the five solvent by semipermeable membrane.
(b) (i) 1.2 % Sodium chloride is hypertonic than blood cells, hence cells will shrink. Plasmolyis will take place.
(ii) 0.4% Sodium chloride solution is hypotonic than blood cell, so cells will swell. Endo osmosis will take place.

Question.26. (a) Account for the following:
(i) Ozone is Thermodynamically unstable.
(ii) Solid PCl5 is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(b) Draw the structure of: (i) Br5 (ii) XeF4
(i) Compare the oxidising action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write the conditions to maximize the yield of H2SO4 by contact person.
(iii) Arrange the following in the increasing order of property mentioned.
(a) H3PO3, H3PO4, H3PO2 (Reducing Characters)
(b) NH3, PH3, ASH3, SbH3, BiH3 (Base strength)
Answer : (a) (i) Ozone easily decompose to give nascent oxygen:
O2 —> O2 + [O], because to reaction is exothermic (ΔH —> -ve) and results in increase in entropy (Δs time). Over all Gibbs energy change in quite high and negative.
(ii) In solid state PCI5 exist as [PCl4]+ [PCl6]-. The cation is tetrahedral and in formed by breaking of weaker P—Cl axial bonds of trigonal bipyramidal structure to form [PCl4]+ and it Bond ionically with octahedral [PCl6] .
(iii) Because of high electro negativity and small size fluorine forms only one oxO acid, HOF.
(i) (a) F2 is a stronger oxidising agent than chlorine parameters;
(b) Bond dissociation enthalpy of F2 is less than chlorine.
(c) Electron gain enthalpy of Cl2 is more negative than F2, because of bigger size and less repulsion. Also because of small size the hydration enthalpy of F2 is quite high than Cl2.
(d) Altogether bond dissociation enthalpy and high hydration enthalpy compensates the smaller electron stronger oxidising agent than Cl2.
(ii) (a) A moderately low temperature of about 720 K and high
pressure of about 2 bar yields maximum H2SO4 acid.
(b) Its an exothermic reaction and forward reaction causes decrease in pressure.
(iii) (a) H3PO4 < H3PO3 < H3PO2.
(b) BiH3 < SbH3 < ASH3 < PH3 < NH3.

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