CBSE Previous Year Solved Papers Class 12 Chemistry Delhi 2009
Time allowed: 3 hours Maximum Marks: 70
- All questions are compulsory.
- Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
- Questions number 6 to 10 are short-answer questions and carry 2 marks each.
- Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
- Questions number 23 is a value based question and carry 4 marks.
- Questions number 24 to 26 are long-answer questions and carry 5 marks each.
- Use log tables, if necessary. Use of calculators is not allowed.
Note : Except for the following questions, all the remaining questions have been asked in previous sets.
Question.1. Which point defect in crystals does not affect the density of the relevant solid?
Answer : Frenkel defect does not alter the density of the relevant solids ?
Question.2. Define the term ‘Tyndall effect’.
Answer : Tyndall effect is the phenomenon due to which the yaath perpendicular to the path of the incident light becomes visible, when a beam pf light , is passed through a colloidal solution.
Question.3. Why is the froth flotation method selected for the concentration of sulphide ores?
Answer : The surface of sulphide ores is preferentially wetted by heavy oils like turpentine or pine oil while the gangue is wetted by water hence, froth floatation method is selected for concentration of sulphide ores.
Question.4. Why is Br(V) a stronger oxidant than Sb (V)?
Answer : Br (V) has a strong metallic character compared to Sb(V) which is a metalloid. The inert pair effect also contributes to Br (V) being a stronger oxidizing agent compound to Sb(V).
Question.5. Give the IUPAC name of the following compound:
Question.6. Write t! s e structure of 3-oxopentanal.
Question.7. Why is an alkylamine more basic than ammonia?
Answer : Since alkyl groups are electron donating groups, the electron density on nitrogen»,atom in case of alkylamines increases making it to donate the lone pair more easily than ammonia. Hence, alkylamines are more basic than ammonia.
Question.8.Given an example of elastomers.
Answer : The example of elastomers are Buna-S and Buna-N.
Question.9. A reacti on is of second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is
(ii) Reduced to half ?
Answer : For second order reaction,
So rate becomes 4 times.
If [A] is reduced to half i.e, a/2
Question.10. Explain the role of
- Carbon electrolytic reduction of alumina.
- Carbon monoxide in the purification of nickel.
- In the electrolytic reduction of alumina, cryolite is used to make it a good conductor of electricity and to lower the melting point of the mixture.
- Carbon monoxide acts as a reducing agent in the process of purification of nickel. When nickel is heated in a stream of CO gas, it forms a volatile complex, nickel tetracarbonyl, which when subjected to higher temperature decomposes to give pure nickel.
Question.11. Draw the structures of the following molecules :
Question.12. Complete the following chemical reaction equations :
Question.13. Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity?
Answer : Molality is the number of moles of the solute dissolve per kg of the solvent.
Molarity is the number of moles of the solute dissolved per litre of the solution.
Molality does not change with temperature. On the other hand, molarity increases with rise in temperature but only to a certain limit.
Question.14. Which ones in the following pairs of substances undergoes Sn2 substitution reaction faster and why?
Question.15. Complete the following reaction equations :
Question.16. Explain what is meant by
- A peptide linkage
- A glycosidic linkage
- Peptide linkage is the amide linkage formed by the condensation of an amino group of an a-amino acid with the carboxyl group of another molecule of the same or a different a-amino acid with dehydration.
- The linkage between two monosaccharides through oxygen atom in an oligosaccharide or a polysaccharide is known as glycosidic linkage.
Question.17. Name two water soluble vitamins, their sources and the disease caused due to their deficiency in diet.
Question.18. Draw the structures of the monomers of the following polymers:
Question.19.Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm.The density of iron is 7.87 g cm”3. Use this information to calculate Avogadro’s number.(At. Mass of Fe = 56 g mol-1)
Question.20. 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?
Question.21. A first-order reaction has a rate constant of0.0051 min-1. If we begin with 0.10 M concentration of the reactant, what concentration of reactant will remain in the solution after 3 hours?
Question.22. How are the following colloids differ from each other with respect to dispersion medium and dispersed phase ? Give on example of each type. (i) An aerosol (ii) A hydrosol (iii) An emulsion
- An aerosol is a colloidal solution in which dispersion medium is gas whereas dispersed phase can be liquid or solid eg. Perfume, smoke, dust particles in air.
- A hydrosol is a colloidal solution in which the dispersion medium is water and dispersed phase is mostly solid, e.g. Carbonated drinks, starch sol.
- An emulsion is a colloidal solution in which both the dispersed phase and dispersing medium are in liquid state. e.g. Cold cream, milk.
Question.23. Account for the following
- NH3 is a stronger base than PH3.
- Sulphur has a greater tendency for catenation than
- Bond dissociation energy of F2 is less than that of Cl2.
Explain the following situations:
- In the structure of HNO3molecule, the N-O bond (121 pm) is shorter than the N-OH bond (140 pm).
- SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
- XeF2 has a straight linear structure and not a bent angular structure.
- Due to the small size and greater electronegativity of nitrogen compared to phosphorus, it has greater electron density on its surface and hence NH3 is a stronger base than PH3.
- Due to presence of d-orbitals in sulphur and lower electronegativity of sulphur compared to oxygen, it has greater tendency for catenation as compared to oxygen.
- Due to small size and high electronegativity of F compared to Cl, the inter-electronic repulsion between lone pair of electrons is very large, hence bond dissociation energy of F2 is less than Cl2.
- In gaseous state the HNO3 molecule has a planar structure which is a resonance hybrid in which nitrogen atom is sp2 hybridized in nitrate ions. Hence, N-O bond is shorter than N-OH bond in HNO3.
- SF4 has sp3d hybridization with a lone pair of electrons in equatorial position which makes it unstable hence, it is easily hydrolyzed.
SF6 does not have a lone pair and is therefore, exceptionally stable.
- XeF2 has three lone pairs of electrons and two bond pairs and thus show sp3d hybridization. But according to VSEPR theory due to presence of l.p-l.p and l.p-b p. repulsion its geometry will in order to minimize repulsion and to have maximum stability.
Question.24. For the complex [Fe(en)2Cl2] Cl, (en = enthyiene diamine) identify
(i) The oxidation number of iron,
(ii)The hybrid orbitals and the shape of the complex,
(iii)The magnetic behaviour of the complex,
(iv)The number of geometrical isomers,
(v) Whether there is an optical isomer also, and
(vi)Name of the complex. (At. No. of Fe = 26)
Question.25. Explain the mechanism of the following reactions :
(i) Addition of Grignard’s reagent to the carboxyl group of a compound forming an adduct followed by hydrolysis.
(ii)Acid catalysed dehydration of an alcohol forming an alkene.
(iii)Acid catalysed hydration of an alkene forming an alcohol.
(i) Carboynl group under goes nucleophilic addi¬tion reaction with Grignard reagent to form an adduct which undergoes hydrolysis to give alcohol in the following manner
Question.26. Giving an example for each, describe the following reactions
- Hofmann’s bromamide reaction
- Gatterman reaction
- A coupling reaction
- Hofmann’s bromamide reaction : This reaction involves the conversion of a primary amide to a primary amine on heating with a mixture of bromine in presence of
- Gattermann Reaction : When Benzene diazonium chloride is treated with halo-acid in presence of Cu powder, it form arylhalide.
- Coupling reaction : when benzene diazonium chloride reacts with phenol at 0.5°C, it form an azo compound i. e.,p -hydroxyazobenzene (orange dye).
Question.27. Explain the following types of substances with one suitable example, for each case s
- Cationic detergents
- Food preservatives
- Cationic detergents are chloride, bromides or acetates of quarternary ammonium salts where the cationic part possesses a long hydrocarbon chain e.g. (Cetyltrimethyl ammonium bromide.)
- Food preservatives are substances used to preveht food spoilage due to microbial growth and help extend the life span of an edible substance without any decline in its nutritional value, e.g. Sugar, salt, sodium benzoate.
- The drugs that are used to rexluce or abolish pain without causing impairment of the consciousness mental confusion in co ordination or paralysis or some, other disturbance of nervous system, e.g. Novalgin, Butazolidine etc.
Question.28. (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
(b) A voltaic cell is set up at 25°C with the following half cells:
(a) State the relationship amongst the cell constant of a cell, the resistance of the solution in the cell and the conductivity of the solution. How is molar conductivity of a solute related to the conductivity of its solution?
(b) A voltaic cell is set up at 25°C with the following half-cells :
Answer : (a) Molar conductivity of a solution at a given concentration is the conductance of a volume ‘V’ of the solution containing a mole of electrolyte placed in between two electrodes with area of cross-section ‘A’ at distance of unit length Y, that is
Effect of change of concentrations on molar conductivity : Molar conductivity increases with decrease in concentration. This is because both number of ions as well as mobility of ions increases with dilution.
In case of strong electrolyte number of ions do not increase appreciably only mobility of ions increases therefore, Km increases as shown in graph as straight line.
Ir» case of weak electrolytes both no. of ions as well as mobility of ions increases therefore, Km increases sharply as shown by curve in the figure.
Question.29. (a) Complete the following equations :
(b) Explain the fallowing observations about the transition/ inner transition elements:
(i) There is in general an increase in density of element from titanium (Z = 22) to copper (Z = 29)
(ii)There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3rd series).
(iii)The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
(a) Complete the following equations :
(b) Give an explanation for each of the following observations:
(i) The gradual decrease in sire (actinoid contraction) from element to element is greater among the actinoids than among the lanthanoids (lanthanoid contraction).
(ii)The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
(iii)With the same d-orbital configuration (d4) Cr2+ ion is a reducing agent but Mn3+ ion is an oxidizing agent.
Answer: (a) (i)
(b) (i) The density increases from titanium to copper due to decrease in atomic radii with increase in nuclear charge Hence the atomic volume decreases. At the same time, atomic mass increases, therefore, density increases.
(ii) It is due to more number of unpaired electrons and smaller size due to poor shielding effect of/orbitals and high enthalpy of atomization of heavy transition elements.
(iii)Actinoids are larger in size and have lower ionization energy, therefore it can show greater range of oxidation states. Secondary, 5/ 6dand7s orbitals have comparable energies in actinoid series.
(i) This is due to poor shielding of 4f and 5f electrons in actions from element to element whereas in lanthanoids, there is poor shielding effect of 4f electrons only, that is why nuclear charge increases from element to element in actinoids than lanthanoids.
(ii)The numbers in the middle of transition series elements posses one unpaired electron per d-orbital and hence exhibit greater number of oxidation states.
(iii) Mn3+ is strongly oxidizing as it gets reduced to much more stable Mn2+ ion. This involves change from 3d4to 3d5 configurations which is very stable being half filled.
Cr2+ is a reducing agent as it gets oxidized to more stable Cr3+ ion with half-filled 3d3 or t2g orbital.
Question.30. (a) Illustrate the following name reactions by giving example:
(i) Cannizzaros reaction
(b) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce ToUen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous i oxidation it gives ethanoic and propanoic acids. Derive the possible structure of compound A.
(a) How are the following obtained?
(i) Benzoic acid from ethyl benzene
(ii)Benzaldehyde from toluene
(b)Complete each synthesis by giving the missing material reagent or products:
Answer: (a) (i) Cannizzaro reaction: When aldehydes which do not have a- hydeogen atom undergo disproportionation reaction, potassium salt of acid and alcohol are formed.
Since hydrogen atoms are double than carbon atoms, therefore, it is likely to be aldehyde or ketone. It does not reduce Tollen’s reagent therefore, it is not an aldehyde. It is a ketone. It reacts with NaHSO3 and gives iodoform test therefore, it is a methyl ketone. On vigorous oxidation it gives ethanoic acid and propanoic acid.
Therefore, it must be pentan-2-one, CH3COCH2CH2CH3
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question.21. What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer : Multimolecular colloids : They are aggregates of atoms or molecules each having size less than lnm. e.g. Sulphur sol. Their molecular masses are not very high and they are held together by weak Vander Waals forces.
Micromolecular colloids : They themselves are large mole-cules of colloidal dimensions e.g. Starch sol. They have high molecular masses and Vander Waals forces holding the molecules are comparatively stronger.
Associated colloids: They are colloids which behaves as normal electrolytes at low concentration and as colloid at higher concentration, e.g. soaps, detergents etc. Their molecular masses are generally high and higher in the concentration, greater are the Vander Waals forces concentration, they exhibit colloidal properties.
Question.24. Explain the following observations :
- Fluorine does not exhibit any positive oxidation state..
- The majority of known noble gas compounds are those of xenon.
- Phosphorus is much more reactive than nitrogen.
- Fluorine is the most electronegative element, has the maximum reduction potential and has no d-orbitals for octet expansion, therefore it shows only a negative oxidation state of-1.
- Due to low ionization energy and large atomic radius, Xenon readily forms compounds, particularly with fluorine and oxygen.
- Phosphorus is more reactive than nitrogen because bond dissociation energy to break N = N triple bond is very large as compared to the energy required to break P-P single bond, which makes nitrogen inert and unteactive.
Question.27. How do antiseptics differ from disinfectants? Give one example of each type.
Answer : Antiseptics are the chemical substance that are applied to living tissues to kill or prevent the growth of micro-organisms. They are safe to be applied to living tissues and are generally applied on wounds, ulcers and diseased skin surfaces e.g. Dettol, furacin, soframycin etc.
Disinfectants, on the other hand, are chemical substances which are applied to non-living objects to kill micro-organisms. They are not safe to be applied to the living tissues and are generally used to kill micro-organisms present in the drains, toilets, floors etc. eg., 1% phenol solution or SO2 in low concentration.
Question.28. (a) Complete the following equations :
(b) Explain the following observations :
(i) Transition elements are known to form many interstitial compounds.
(ii)With the same d4 d-orbital configuration Cr2+ ion is reducing while Mn3+ion is oxidizing.
(iii)The enthalpies of atomization of transition’ elements are quite high.
(b)(i) Because of their ability to accommodate small non- metallic atoms owing to spaces or voids present between atoms due to their variable valency transition elements form many interstitial compounds.
(ii)Cr2+ has the configuration which easily changes to d^ due to stable half-filled t2g orbitals. Therefore Cr2+ is reducing agent. While Mn2+ has stable half-filled d^ configuration. Hence, Mn3+easily changes to Mn2+ and acts as an oxidizing agent.
(iii)As transition metals certain a large number of unpaired electrons, they have strong interatomic attractions (metallic) bonds. Hence, they have high enthalpies of atomization.
(b)(i) Transition elements generally have unpaired electrons in their d-orbitals and hence due to d → d transition of electrons, their compounds are coloured.
(ii)Due to incompletely filled d-orbitals, the oxidation states of transition metals vary.
(iii)Due to small energy gaps between, 5f,6dand 7s subshells, actinoids exhibit a greater range of oxidation states.
Question.29.(a) What type of a cell is lead storage battery? Write the anode and cathode reactions and the overall reaction occurring in the lead storage battery while operating.
(b) A voltaic cell is set up at 25°C with the half cells, Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M). Write the equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(a) Express the relation amongst the cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to the conductivity of the solution ?
(b) Calculate the equilibrium constant for the reaction
Answer : (a) A lead storage battery is a secondary cell and it can be recharged. During discharging, the cell reactions at anode and cathode are as follows :
(a) The relationship between cell constant of a cell (G*) resistance of the solution in the cell (R) and conductivity (K) is given by –
Note : Except for the following questions, all the remaining question have been asked in previous sets.
Question.8. What does the part ‘6,6’ mean lh the name nylon-6,6?
Answer : ‘6,6’ refers to the 6 carbon atoms in monomers molecules of nylon-6,6, i.e. adipic acid and hexamethylene diamine.
Question.19. Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2S04. If this solution actually freezes at -0.320°C, what would be the value of Van’t Hoff factor?
Question.24. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved :
Question.27. What are the following substances?.Give an example of each
- Non-ionic detergent
- Antacids are drugs that help to overcome the problem of hyperacidity by neutralizing the excess acid and raising the pH to. an appropriate level in stomach eg. Rantidine, sodium bicarbonate etc.
- Non-ionic detergents do not contain any ion in their constitution as they are esters by high molecular mass alcohol which are obtained by reaction between polyethylene glycol and steric acid. e.g. Polyethylene glycol sterate, lawyrl alcohol ethoxylate.
- Antiseptics are the drugs applied to living tissues in order to prevent microbial growth, e.g, iodoform for wounds, boric acid for eyes.
Question.28. (b) (i) In general the atomic radii of transition elements decreases with atomic number in a given series.
(ii)The E°Mn2+ |M for copper is positive (+0.24 V). It is the only metal in the first transition series showing this type of behavior.
(a) What is meant by lanthanoid contraction? What is it due to and what consequences does it have on the chemistry of elements following lanthanoids in the periodic table?
(b) Explain the following observations:
(i)Cu2+ is unstable in aqueous solution
(ii) Although CO2+ appears to be stable, it is easily oxidized to CO3+ ion in the presence of a strong ligand.
(iii) The E°Mn2+|Mn value for manganese is much more than expected from the trend of the other elements in the series.
(iii) The E° value for Mn3+|Mn2+ is much more positive than for Fe3+|Fe2+ or Cr3+|Cr2+.
Answer : (b)
(i) When moving from left to right across a period, because of increase in effective nuclear charge and weak shielding effect of d-electrons the atomic radii decreases.
(ii)Because of its high enthalpy of atomization and low hydration enthalpy. Hence E°Mn2+|M for copper is positive.
(iii)This is because of the much larger third ionization
enthalpy of Mn (where the required charge is d5to d4 )
(a) Lanthanoid contraction is the steady decrease in the
atomic radii of lanthanoids with increasing atomic number due to mutual imperfect or poor shielding of the electrons in the Af orbital – As we move along the lanthanoid series, the effective nuclear charge increases on addition of electrons and the electrons thus added in f-subshell causes imperfect shielding which is unable to counter balance the effect of the increased nuclear charge. As a result, the contraction in size occurs.
(i) Due to lanthanoid contraction, zirconium (Zr) and Hafnium (Hf) have a comparable size. They are also known as chemical twins due to their similar radii.
(ii) Due to lanthanoid contraction, the chemical properties of lanthanoids are very similar due to which their separation becomes very difficult.
(b) (i) Since Cu+ in aqueous solution undergoes dispropertio- nate i.e.
2Cu+ (aq)–> 2 Cu2+(aq)+Cu (s)
The E° value for this is favourable.
(ii)CO(III) is stablised because of higher crystal field splitting energy [or strong ligand causes pairing of electron to give more stable CO(III) ion]
(iii)The comparatively high value for Mn shows that Mn2+( d5) is particularly stable.
Question.29. (a) Corrosion is essentially an electrochemical phenomenon. Explain the reaction occurring during the corrosion of iron kept in an open atmosphere.
(b) One half-cell in a voltalic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3 )2. A voltage of 1.48 V is measured for this cell.
Use this information to calculate the concentration of silver nitrate solution used.
(a) Corrosion is the wearing away of a metal due to gases and water vapour present in the atmosphere. It results generally in the formation of oxides, sulphides or carbonates.
In the atmosphere, pure iron surface behaves as a small electro chemical cell in the presence of Oxygen and water vapour.