Chemistry cbse solution_2011-12


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Chemistry cbse solution_2011-12

  1. 1. Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 56/1/1 Candidates must write the Code onRoll No. the title page of the answer-book.  Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate.  Please check that this question paper contains 30 questions.  Please write down the Serial Number of the questions before attempting it.  15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not write any answer on the answer script during this period. CHEMISTRY[Time allowed : 3 hours] [Maximum marks : 70]General Instructuions:(i) All questions are compulsory.(ii) Marks for each question are indicated against it.(iii) Questions numbered 1 to 8 are very short-answer questions and carry 1 mark each.(iv) Questions numbered 9 to 18 are short-answer questions and carry 2 marks each.(v) Questions numbered 19 to 27 are also short-answer questions and carry 3 marks each.(vi) Questions numbered 28 to 30 are long-answer questions and carry 5 marks each.(vii) Use Log Tables, if necessary. Use of calculators is not allowed. -(1)-
  2. 2. STUDYmate1. What is meant by ‘doping’ in a semiconductor?Ans. Introducing impurities into elements of Group 13 {B} or Group 15 {P} into the elements of Group 14 {Si or Ge}. This improves conductivity of Group 14 elements.2. What is the role of graphite in the electrometallurgy of aluminium?Ans. Graphite is used as anode and cathode during electrolysis of fused alumina in molten cryolite.3. Which one of PCl +4 and PCl –4 is not likely to exist and why?Ans. PCl 4 can not exist, because ‘P’ can not withdraw e– from more electronegative Cl. –4. Give the IUPAC name of the following compound. CH2 = C – CH2Br CH3Ans. 3-Bromo-2-methyl propene.5. Draw the structural formula of 2-methylpropanl-2-ol molecule. CH3Ans. CH3 – C – CH3 OH6. Arrange the following compounds in an increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone.Ans. butanone < propanone < propanal < ethanal.7. Arrange the folllwing in the decreasing order of their basic strength in aqueous solution: CH3NH2, (CH3)2NH, (CH3)3N and NH3Ans. (CH3)2NH > CH3NH2 > (CH3)3N > NH3.8. Define the term, ‘homopolymerisation’ giving an example.Ans. Homopolymerization is defined as a polymerization reaction which involves polymerization of single monomeric species. Example: polythene from ethene.9. A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18 °C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol–1) OR Define the following terms: (i) Mole fraction (ii) Isotonic solutions (iii) Van’t Hoff factor (iv) Ideal solutionAns. m = 1 Tb = 100.18 °C Tob = 100 °C Tb = 0.18 °C i=? Kb = 0.512 K Kg/mol -(2)-
  3. 3. STUDYmate Tb(theo) = Kb × m Tb(theo) = 0.512 × 1 = 0.512 Tbobs i = T btheo 0.18 i= = 0.35156 0.512 ORAns. (i) Mole fraction: It is the ratio of number of moles of one component to the total number of moles present in solution/mixture. (ii) Isotonic solutions: Two solutions having same osmotic pressure at same temperature are isotonic solutions. (iii) Van’t Hoff factor: It is a ratio of observed colligative property and calculated colligative property. So, Van’t Hoff factor ‘i’ is, observed colligative property i= calculated colligative property Total number of moles of particle after association /dissociation i= Number of moles of particles before association /dissociation normal molar mass i= abnormal molar mass (iv) Ideal solution: The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.10. What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant: (i) L–1 mol s–1 (ii) L mol–1 s–1Ans. It is the sum of powers of concentrations of the reactants in the rate law expression. Order of a reaction can be 0, 1, 2, 3 or a fraction. (i) Zero (ii) Second11. Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved.Ans. The two groups in which catalysis can be divided are (i)  NO(g )  Homogeneous catalysis: Eg: SO2 (g) + O2 (g)  SO3 (g) or other examples. Ni/Pt (ii) Heterogeneous catalysis: Eg: C2H4 (g)  H2 (g)  C2H6 (g) or other  Ethene Ethane examples.12. What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.Ans. The process of settling of colloidal particles is called coagulation or precipitation of the sol. -(3)-
  4. 4. STUDYmate Methods by which coagulation of lyophobic sols can be carried out. (i) By electrophoresis: The colloidal particles move towards oppositely charged electrodes, get discharged and precipitated under an applied electric field. (ii) By mixing two oppositely charged sols: Oppositely charged sols when mixed in almost equal proportions, neutralise their charges and get partially or completely precipitated. Mixing of hydrated ferric oxide (+ve sol) and arsenious sulphide (–ve sol) bring them in the precipitated forms. This type of coagulation is called mutual coagulation. (iii) By boiling: When a sol is boiled, the adsorbed layer is disturbed due to increased collisions with the molecules of dispersion medium. This reduces the charge on the particles and ultimately lead to settling down in the form of precipitate.13. Describe the principle involved in each of the following processes. (i) Mond process for refining of Nickel. (ii) Column chromatography for purification of rare elements.Ans. (i) Mond process for refining of Nickel: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel carbonyl which decomposes to form pure nickel. 330–350 K 450–470 K Ni + 4CO Ni(CO)4 Ni + 4CO Impure Tetracarboxyl Pure nickel (ii) Column chromatography: Different components of mixture are adsorbed to different extents, depending upon their polarity.14. Explain the following giving an appropriate reason in each case. (i) O2 and F2 both stabilize higher oxidation state of metals but O2 exceeds F2 in doing so. (ii) Structures of Xenon fluorides cannot be explained by Valance Bond approach.Ans. (i) The ability of O2 to stabilise higher oxidation states exceeds that of fluorine because oxygen can form multiple bonds with metals. (ii) According to VBT, if electrons in an orbital are paired (especially of noble gases) then they do not participate in bond formations (or undergo hybridisation).15. Complete the following chemical equations: (i) Cr2O2  H  I –  7 (ii) MnO4  NO2  H   Ans. (i) Cr2O2  14H  6I–  2Cr 3   3I2  7H2O 7 (ii) 2MnO4  5NO2  6H  2Mn2  5NO3  3H2O   –16. What is meant by (i) peptide linkage (ii) biocatalysts ?Ans. (i) Peptide linkage: Peptide is an amide formed between –COOH group and –NH2 group of -amino acids in proteins. (ii) Biocatalysts: Biocatalysts are enzymes, which are complex nitrogenous compounds produced by living species and catalyse certain biological reactions. -(4)-
  5. 5. STUDYmate17. Write any two reactions of glucose which cannot be explained by the open chain structure of glucose molecule.Ans. Any two (i) Despite having the aldehyde group, glucose does not give 2,4–DNP test, Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3. (ii) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free –CHO group. (iii) Glucose is found to exist in two different crystalline forms which are named as  and . The -form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the -form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.18. Draw the structure of the monomer for each of the following polymers: (i) Nylon 6 (ii) Polypropene H N H 2C C=OAns. (i) -caprolactam H2C CH2 H 2C CH2 (ii) H3C – CH = CH2 Propene19. Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom? OR Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm–3. Use this information to calculate Avogadro’s number. (At. mass of Fe = 55.845u)Ans. For a body centred cubic, unit cell arrangement, ( 3) a  4rW ( 3)  316.5  rW 4 rW = 137.04 pm OR ZMAns. d = NA  a 3 ZM NA = d  a3 2  55.845 NA = 7.874  (286.65  10 10 cm)3 NA = 5.931 × 1023 atoms per mol20. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2K. (Kf for water = 1.86 K kg mol–1)Ans. For KCl, presuming complete dissociation, (i = 2) Tf = i × kf × m -(5)-
  6. 6. STUDYmate mass of KCl = 2 × 1.86 × 1 74.5 2  1.86  mass of KCl  2 = 74.5 74.5 Mass of KCl = = 40.05 gm 1.8621. For the Reaction 2NO(g) + Cl2(g)  2 NOCl(g) the following data were collected. All the measurements were taken at 263 K : Experiment Initial [NO] Initial [Cl2] (M) Initial rate of disaappearance No of Cl2 (M/min) 1 0.15 0.15 0.60 2 0.15 0.30 1.20 3 0.30 0.15 2.40 4 0.25 0.25 ? (a) Write the expression for rate law. (b) Calculate the value of rate constant and specify its units. (c) What is the initial rate of disappearance of Cl2 in exp. 4 ? d[R]Ans. (a) R  k[NO]2 [Cl2 ]1 dt (b) R = k [NO]2 [Cl2]1 R 0.6 0.6 k = [NO]2 [Cl ]1  [0.15]2 [0.15]1  (0.15)3 = 177.75 [Using experiment (1)] 2 (c) R = k [NO]2 [Cl2] R = 177.78 × (0.25)2 × (0.25) R = 2.78 M / min.22. How would you account for the following? (i) Many of the transition elements are known to form interstitial compounds. (ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series. (iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even + 6 being typical.Ans. (i) They have interstitial spaces on their surface, in which small atoms such as H, C or N can be trapped For e.g. steel. (ii) This is due to lanthanoid contraction. (iii) This is attributed to the fact that 5f, 6d and 7s levels are of comparable energies in case of actinoids. -(6)-
  7. 7. STUDYmate23. Give the formula of each of the following coordination entities. (i) Co3+ ion is bound to one Cl–, one NH3 molecule and two bidentate ethylene diamine (en) molecules. (ii) Ni2+ ion is bound to two water molecules and two oxalate ions. Write the name and magnetic behaviour of each of the above coordination entities. (At Nos.Co = 27, Ni = 28)Ans. (i) [CO (NH 3 ) (en) 2 Cl] +2 – IUPAC name Amminechloridobis(ethane-1,2- diamine)cobalt (III) ion (ii) [Ni (H2O) (C2O4)2]2–  Diaquadioxalatonickelate (II) ion Magnetic behaviour (i) Diamagnetic (d2sp3) hybridisation (ii) Paramagnetic (sp3d2) hybridisation24. Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why it is so?Ans. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution. Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho– and para– positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonanace effect tends to oppose the inductive effect for the attack at ortho– and para– positions and hence makes the deactivation less for ortho– and para– attack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.25. Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used. (i) CH3CH2CH2CH2OH (ii) 2-Butenol (iii) 2-Methyl-l-propanol -(7)-
  8. 8. STUDYmateAns. (i) K 2Cr2O7 /H CH3CH2CH2CH2OH  CH3CH2CH2COOH  Butanoic acid OH PCC (ii) CH3 – CH = CH – CH2 CH3 – CH = CH – CHO But-2-en-1-al + K2 Cr2O7/H (iii) CH3 – CH – CH2OH CH3 – CH – COOH CH3 CH3 2-Methylpropanoic acid26. Write chemical equations for the following conversions: (i) Nitrobenzene to benzoic acid. (ii) Benzyl chloride to 2-phenylethanamine. (iii) Aniline to benzyl alcohol. NO2 NH2 N2Cl CN COOHAns. (i) + Sn/HCl NaNO 2 + HCl CuCN H 3O O°C CH2Cl CH2CN CH2CH2NH2 (ii) KCN (aq) LiAlH 4 NH2 N2Cl CN CH2NH2 CH2OH (iii) NaNO 2, HCl CuCN LiAlH4 HNO2 0°C27. What are the following substances? Give one example of each one of them. (i) Tranquilizers (ii) Food preservatives (iii) Synthetic detergentsAns. (i) Tranquilizers are a class of compounds used for the treatment of stress, mild or even severe mental diseases. (e.g.), veronal / luminal / seconal. (ii) Food preservatives prevent spoilage of food due to microbial growth (e.g.) sodium benzoate. (iii) Synthetic detergents are cleansing agents which have all the properties of soaps but which actually do not contain any soap. (e.g.)Sodium laurylsulphate.28. (a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it. (b) In the button cell, widely used in watches, the following reaction takes place 2  Zn(s) + Ag2O(s) + H2O(l)  Zn(aq) + 2 Ag(s) + 2OH(aq ) Determine E° and G° for the reaction. (given: Eo   0.80V, Eo 2 /Zn  0.76V ) Ag /Ag Zn -(8)-
  9. 9. STUDYmate OR (a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 . What is the cell constant if the conductivity of 0.001 M KC1 solution at 298 K is 0.146 × 10–3S cm–1?Ans. (a) It is a secondary cell. Anode: Pb + SO2–  PbSO4 + 2e– 4 Cathode: PbO2 + 4H+ + SO2– + 2e–  PbSO4 + 2H2O 4 Overall reaction: Pb + PbO2 + 4H+ + 2SO2–  2PbSO4 + 2H2O 4 (b) Zn is oxidized and Ag2O is reduced. E = E   E cell Ag 2 O, Ag (r eduction) Zn / Zn 2 (oxidation) = 0.8 – (–0.76) = 1.56 V G° = – nFE = – 2×96500 × 1.56 = 301080 J. cell ORAns. (a) Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross-section A and distance of unit length. K  1000 m = S cm2 mol1 C Molar conductivity increases marginally with decrease in concentration/increase in dilution for a strong electrolyte. m KCl CH3COOH (C)½ For a weak electrolyte, m increases steeply with decrease in concentration/ increase in dilution. This is in accordance with Le chatelier principle. 1  (b) K=  R A 1 or K= G* R  G* = Cell constant A G* = R × K = 1500 ohms × 0.146 × 10–3 Ohm–1 cm–1 = 1500 × 0.146 × 10–3 cm–1 = 219 × 10–3 cm–1. -(9)-
  10. 10. STUDYmate29. (a) Complete the following chemical equations: (i) P4 + SO2Cl2  (ii) XeF6 + H2O  (b) Predict the shape and the asked angle (90° or more or less) in each of the following cases (i) SO2 and the angle O – S – O 3 (ii) ClF3 and the angle F – Cl – F (iii) XeF2 and the angle F – Xe – F OR (a) (i) NaOH  Cl 2  (hot and conc.) (ii) XeF4 + O2F2  (b) Draw the structures of the following molecules: (i) H3PO2 (ii) H2S2O7 (iii) XeOF4Ans. (a) (i) P4 + 8SO2Cl2  4PCl3 + 4SO2 + 2S2Cl2 (ii) XeF6 + H2O  2HF + XeOF4 (b) (i) Shape of SO2– – pyramidal 3 Bond angle – More than 90°. (ii) ClF3  Bent T-shape Bond angle – Less than 90° (iii) XeF2  Linear shape Bond angle – More than 90° ORAns. (a) (i) 6NaOH + 3Cl2  5NaCl + NaClO3 + 3H2O (hot and conc.) (ii) XeF4 + O2F2  XeF6 + O2 O (b) (i) H3PO2 P H H OH O O S S (ii) H2S2O7 O O O OH OH O F F Xe (ii) XeOF4 F F30. (a) Illustrate the following name reactions giving suitable example in each case: (i) Clemmensen reduction (ii) Hell-Volhard-Zelinsky reaction -(10)-
  11. 11. STUDYmate (b) How are the following conversions carried out? (i) Ethylcyanide to ethanoic acid (ii) Butan-1-ol to butanoic acid (iii) Benzoic acid to m-bromobenzoic acid OR (a) Illustrate the following reactions giving a suitable example for each. (i) Cross aldol condensation (ii) Decarboxylation (b) Give simple tests to distinguish between the following pairs of compounds (i) Pentan-2-one and Pentan-3-one (ii) Benzaldehyde and Acetophenone (iii) Pehnol and Benzoic acidAns. (a) (i) Clemmensen Reduction O Zn(Hg) in CH3 – C – H H3C – CH3 conc. HCl Ethanal Ethane O Zn(Hg) in H3C – C – CH3 H3C – CH2 – CH3 conc. HCl Propanone Propane (ii) Hell-Volhard-Zelinsky reaction (i) Cl 2/Red P. H3C – CH2 – COOH H3C – CH – COOH (ii) H2O Cl Propanoic acid -Chloropropanoic acid NH3,  + H3O (b) (i) CH3CH2CN CH3CH2COOH CH3CH2CONH2 NaOH, Br2 HONO CH3CH2OH CH3CH2NH2 [O] CH3COOH O K2Cr2O7/dil. H2SO4 (ii) CH3 – CH2 – CH2 – CH2 – OH H3C – CH2 – CH2 – C – OH oxidation Butan-1-ol Butanoic acid O O C – OH C – OH AlBr3 or FeBr3 (iii) + Br2  Br Benzoic acid m-Bromobenzoic acid ORAns. (a) (i) Cross Aldol condensation: When aldol condensation is carried out between two different aldehydes and / or ketones, it is called cross aldol condensation. If both of them contain -hydrogen atoms, it gives a mixture of four products. This is illustrated below by aldol reaction of a mixture of ethanal and propanal. -(11)-
  12. 12. STUDYmate CH3CHO (i) NaOH + CH3 – CH = CH – CHO + CH3CH2 – CH = C – CHO (ii)  CH3CH2CHO But-2-enal CH3 from two molecules of ethanal 2-Methylpent-2-enal from two molecules of propanal simple or self aldol products + CH3 – CH = C – CHO + CH3CH2 – CH = CHCHO CH3 2-Methylbut-2-enal Pent-2-enal from one molecule of ethanal and one molecule of propanal cross aldol products (ii) Decarboxylation: Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO in the ratio of 3 : 1). The reaction is known as decarboxylation. NaOH & CaO R–COONa  R–H + Na2CO3 Heat  Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. The reaction is known as Kolbe electrolysis. electrolysis 2CH3COONa (aq) + 2H2O  CH3CH3 + 2CO2 + 2NaOH + H2  O (b) (i) CH3 – C – CH2CH2 – CH3 gives yellow precipitate of Iodoform with NaOH Pentan-2-one and iodine. O H3C – CH2 – C – CH2 – CH3 does not give yellow precipitate. Pentan-3-one O O C – OH C – CH3 (ii) Acetophenone Benzaldehyde No reaction Gives yellow precipitate of Iodoform with NaOH and iodine. OH COOH (iii) (I) Gives violet colouration No reaction with neutral FeCl3 solution. (II) No reaction Gives brisk effervescence with NaHCO3. ×·×·×·×·× -(12)-
  13. 13. STUDYmate Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 56/1/2 UNCOMMON QUESTIONS ONLY1. Write a point of distinction between a metallic solid and an ionic other than metallic lustre.Ans. Metallic solids are good conductors but an ionic solid is a a bad conductor of electricity.11. Describe a conspicuous change observed when (i) a solution of NaCl is added to a sol of hydrated ferric oxide. (ii) a beam of light is passed through a solution of NaCl and then through a sol.Ans. (a) Addition of NaCl to a sol of hydrated Fe2O3 causes coagulation. (b) When a beam of light is passed through NaCl solution, no tyndall effect is produced. But when it is passed through a sol tyndal effect is produced.13. Describe the following : (i) The role of cryolite in electro metallurgy of aluminium. (ii) The role of carbon monoxide in the refining of crude nickel.Ans. (i) In the metallurgy of Aluminium, purified Al2O3 is mixed with cryolite. Cryolite lowers melting point of the mix and brings conductivity. (ii) CO forms a volatile complex. (Nickle tetralarbonyl) 300  350 K Ni  4CO  Ni [CO]4  The carbonyl is subjected to higher temperture so that it is decomposed giving a pure metal. 450  470 K Ni [ CO]4  Ni  4CO 18. Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in DNA ?Ans. The main structural difference between DNA and RNA is (i) DNA has double helical structural RNA has single strand structure (ii) DNA contains cytosine and thymine as pyrimidine bases RNA contains cytosine and uracil as pyrimidine bases.20. A solution of glycerol (C3H8O3) in water prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C while pure water boils at 100 °C. What mass of glycerol was dissolved to make the solution ? (Kb for water = 0.512 K kg mol–1). -(13)-
  14. 14. STUDYmateAns. WH O = 500 g = 0.5 kg 2 Tb = 100.42°C Tb = 100°C 0 Tb 0.42 ºC Wglycerol = ? Tb = Kb . m Tb = Kb. m Wglycerol Tb = W Mglycerol H2 O Wglycerol 0.42 = 0.512 × 0.5  62 0.42  0.5  62 Wglycerol = 0.512 Wglycerol = 25.429 gm22. State a reaon for each of the following situations : (i) Co2+ is easily oxidized to Co3+ in presence of a strong ligand. (ii) CO is a stronger complexing reagent than NH3. (iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2–Ans. (i) Co (III) has greater tendency to from coordination complexes than CO (II). Hence, in the presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidized. (ii) As co creates a synergic effect which strengthens the bond between Co and the metal. NH3 doesn’t exhibit any synergic effect (iii) In Ni(CO)4, Ni si in zero oxidation state NiG.S. (28) 4s 4p 3 sp hydbridisation Ni0 (ES ) ×× ×× ×× ×× tetrahedral shape In (Ni(CN)4 ]2– Ni(28) 3d 4s 4p 2 Ni (G.S.) Ni 2 ES dsp2 (Square planar shape) Thus the shapes are different.23. How would you account for the following ? (i) With the same d-orbital configuration (d4) Cr2+ is a reducing agent while Mn3+ is an oxidizing agent. (ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series. (iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions. -(14)-
  15. 15. STUDYmateAns. (i) Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn2+ to Mn3+ results in the half-filled (d5) configuration which has extra stability. (ii) There is a greater range if oxidation states, which is attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. (iii) Due to presence of unpaired electrons in d-orbitals.30. (a) Give a plausible explanation for each one of the following : (i) There are two – NH2 groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones. (ii) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6–trimethylcyclohexanone does not. (b) An organic compound with molecular formula C9H10O forms 2, 4, – DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1, 2-benzene-di-carboxylic acid. Identify the compound.Ans. (a) (i) In semicarbazides, out of two – NH2 groups, one of the –NH2 group bonded O to is involved in conjugation through its lone pair. –C– O : : H 2N – C – NH – NH 2 :O: :  : : H2N = C – NH – NH2 O H3C CH3 (ii) highly sterically hindered ketone. CH3 2, 4, 6 – Trimethylcyclohexanone Due to steric hindrance posed by three methyl groups around carbonyl group in 2, 4, 6-Trimethyl cyclohexanone, it will not give good yields of cyanohydrin. (b) (i) As C9H10O forms 2, 4-DNP recation it is an aldehyde or ketone. (ii) As C9H10O reduces Tollen’s reagent, it is an aldehyde (iii) As C9H10O undergoes cannizaro’s reaction it is substitued benzaldehyde. (iv) As C9H10O gives 1, 2-Benzene-di-carboxylic acid, the compound is O O C–H C – OH CH2 – CH3 C – OH O 2-Ethylbenzaldehyde 1, 2-Benzene di-carboxylicalid OR -(15)-
  16. 16. STUDYmate (a) Give chemical tests to distinguish between (i) Phenol and Benzoic acid (ii) Benzophenone and Acetophenone (b) Write the structures of the main products of following reactions : (i)  C6H5 COCl  Anhydrous AlCl3 CS2  2 (ii) H3C – C  C – H  Hg ,H2SO4 CH3 (iii)  1. CrO2Cl2 2. H O  3 NO2Ans. (a) (i) Same as Set 1 Q 30 (b) (iii) (ii) Acetophenone gives yellow crystals of iodoform on reaction with I2 and NaOH Benzophenone doesn’t give yellow crystals of iodoform on reaction with I2 and NaOH O C (b) (i) O (ii) H 3C – C – CH 3 CHO (iii) NO2 ×·×·×·×·× -(16)-
  17. 17. STUDYmate Studymate Solutions to CBSE Board Examination 2011-2012 Series : SMA/1 Code No. 56/1/3 UNCOMMON QUESTIONS ONLY12. Explain the following terms giving one example for each: (i) Micelles (ii) AerosolAns. (i) Micelles : These are associated colloids. For e.g. soap. cleansing action of soap is explained through formation of micelles around the oil droplet on a fabric, in such a way that hydrophobic part of the soap molecule is attached to oil droplet and hydrophilic part projects out of the oil droplet. O C + ONa hydrophobic hydrophillic Since the polar groups can interact with water, the oil droplet is pulled in water and removed from the dirty surface. Thus soap helps in emulsification and wasing away of oils from the fabric. (ii) Aerosol : It is a colloidal state. It may be of two types : Aerosol of solid (e.g., smoke) and Aerosol of liquid (e.g. fog).20. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at – 0.34 °C. what is the molar mass of this material (Kf for water = 1.86 K kg mol–1).Ans. Tf= Kf × molality 15g  1000 0 – (– 0.34) = 1.86 K kg mol–1 ×  M  450 1.86 K kg mol 1  15g  1000 M= 0.34 K  450 kg = 182.35 gm/mol22. Explain the following observations giving an appropriate reason for each. (i) The enthalpies of atomization of transition elements are quite high. (ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e., 3rd series). (iii) Mn2+ is much more resistant than Fe2+ towards oxidation.Ans. (i) This is due to strong metallic bonding. (ii) Greater the number of unpaired electrons, stronger and more frequent is the resultant bonding. -(17)-
  18. 18. STUDYmate (iii) Mn2+ has stable half filled (3d5) configuration. Hence does not get oxidised. But Fe2+ has 3d6 configuration hence it can lose 1 electron to become the more stable Fe3+ of 3d5 configuration (or) Fe2+ easily gets oxidised.23. Write the name, the structure and the magnetic behaviour of each one of the following complexes: (i) [Pt(NH3)Cl(NO2)] (ii) [Co(NH3)4Cl2]Cl (iii) Ni(CO)4 (Atomic numbers : Co = 27, Ni = 28, Pt = 78)Ans. (i) [Pt(NH3)Cl(NO2)] IUPAC : Amminechloridonitrito-N-platinum (II) Structure : Trigonal planar Magnetic behaviour : diamagnetic (ii) [Co(NH3)4Cl2]Cl IUPAC : Tetraamminedichloridocobalt (III) chloride Structure : Octahedral Magnetic behaviour : diamagnetic (iii) Ni(CO)4 IUPAC : Tetracarbonylnickel (0) Structure : Tetrahedral Magnetic behaviour : diamagnetic27. Explain the following terms giving one example of each type: (i) Antacids (ii) Disinfectants (iii) EnzymesAns. (i) Antacids : Antacids are mild bases which neutralise the excess production of acid in the stomach. E.g., NaHCO3 (ii) Disinfectants : Disinfectants are chemical substances which either kill or present the growth of microorganisms. These are applied to inanimate objects such as floor, drainage system. E.g., chlorine (iii) Enzymes : Enzymes are proteins/macromolecules of biological origin which perform various functions in the metabolic reactions. They are biological catalysts. E.g., maltase, amylase30. (a) Draw the molecular structures of following compounds: (i) XeF6 (ii) H2S2O8 (b) Explain the following observations: (i) The molecules NH3 and NF3 have dipole moments which are of opposite direction. (ii) All the bonds in PCl5 molecule are not equivalent. (iii) Sulphur in vapour state exhibits paramagnetism. -(18)-
  19. 19. STUDYmate OR (a) Complete the following chemical equations: (i) XeF4 + SbF5  (ii) Cl2 + F2 (excess)  (b) Explain each of the following: (i) Nitrogen is much less reactive than phosphorus. (ii) The stability of + 5 oxidation state decreases down group 15. (iii) The bond angles (O – N – O) are not of the same value in NO2– and NO2+. FAns. (a) (i) XeF6 : F F Xe F F F O O (ii) H2S2O8 : S S O O O O OH HO Peroxodisulphuric acid (H2S2O8) (b) (i) N N H H F F H F Since ‘N’ is more electronegative than H hence the dipole is pointing towards ‘N’ in NH3. In NF3, F is more electronegative than ‘N’ hence the dipole is pointing towards ‘F’. (ii) The three equatorial P–Cl bonds are equivalent, while the other two axial bonds are longer than equatorial bonds. This is due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. Cl Cl 240 pm P Cl Cl Cl -(19)-
  20. 20. STUDYmate (iii) In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding * orbitals like O 2 and, hence, exhibits paramagnetism. OR (a) (i) XeF4 + SbF5 [XeF3]+ [SbF6]– (ii) Cl2 + 3F2 (excess) 2ClF3 (b) (i) Nitrogen is associated with high bond dissociation energy and hence unreactive. (ii) This is due to inert pair effect. (iii) + – In NO2 , due to linear geometry the bond angle is 180°, but in NO2, due to bent shape, it is greater than 90° and thus different. ×·×·×·×·× -(20)-