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ISC Class XII Board Exam 2019 : Chemistry

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ISC SOLVED PAPER - 2019 Class XII CHEMISTRY (THEORY) (Maximum Marks : 70) (Time allowed : Three hours) (Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.) All question are compulsory Question 1 is of 20 marks having four sub parts, all of which are compulsory. Question number 2 to 8 carry 2 marks each, with any two questions having internal choice. Question numbers 9 to 15 carry 3 marks each, with any two questions having an internal choice Question numbers 16 to 18 carry 5 marks each, with an internal choice. All working, including rough work, should be done on the same sheet as, and adjacent to the rest of the answer. The intended marks for questions or parts of questions are given in brackets []. Balanced equations must be given wherever possible and diagrams where they are helpful. When solving numerical problems, all essential working must be shown. In working out problems, use the following data : Gas constant R = 1.987 cal deg 1 mol 1 = 8.314 JK 1 mol 1 = 0.0821 dm3 atm K 1 mol 1 1 l atm = 1 dm3 atm = 101.3 J. 1 Faraday = 96500 coulombs. Avogadro's number = 6.023 1023. Question 1 (a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets : [4 1] (more than, primary, cathode, Lucas reagent, two, four, less than, Grignard's reagent, tertiary, anode, zero, equal to, three) (i) The elevation of boiling point of 0.5 M K2SO4 solution is __________ that of 0.5 M urea solution. The elevation of boiling point of 0.5 M KCl solution is __________ that of 0.5 M K2SO4 solution. (ii) A mixture of conc. HCl and anhydrous ZnCl2 is called __________ which shows maximum reactivity with ________ alcohol. (iii) In electrolytic refining the impure metal is made __________ while a thin sheet of pure metal is used as __________. (iv) When the concentration of a reactant of first order reaction is doubled, the rate of reaction becomes __________ times, but for a __________ order reaction, the rate of reaction remains the same. (b) Complete the following statements by selecting the correct alternative from the choices given : [4 1] (i) The cell reaction is spontaneous or feasible when emf of the cell is : (1) negative (2) positive (3) zero (4) either positive or negative (ii) Which, among the following polymers, is a polyester : (1) melamine (2) bakelite (3) terylene (4) polythene (iii) The correct order of increasing acidic strength of the oxoacids of chlorine is : (1) HClO3 < HClO4 < HClO2 < HClO (2) HClO < HClO2 < HClO3 < HClO4 (3) HClO2 < HClO < HClO4 < HClO3 (4) HClO3 < HClO4 < HClO < HClO2 (iv) A catalyst is a substance which : (1) changes the equilibrium constant of the reaction. (2) increases the equilibrium constant of the reaction. To know about more useful books for class-12 click here 2 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII (3) supplies energy to the reaction. (4) shortens the time to reach equilibrium. (c) Match the following : [4 1] (i) Diazotization (1) Anisotropic (ii) Crystalline solid (2) Reimer-Tiemann reaction (iii) Phenol (3) Diphenyl (iv) Fittig reaction (4) Aniline (d) Answer the following question : [4 2] (i) (1) Which trivalent ion has maximum size in the Lanthanoid series i.e., Lanthanum ion (La3+) to Lutetium ion (Lu3+) ? (at. no. of Lanthanum = 57 and Lutetium = 71) (2) Explain why Cu2+ is paramagnetic but Cu+ is diamagnetic. (at. no. of Cu = 29) (ii) When a coordination compound CoCl3.6NH3 is mixed with AgNO3, three moles of AgCl are precipitated per mole of the compound. Write the structural formula and IUPAC name of the coordination compound. (iii) Calculate the boiling point of urea solution when 6 g of urea is dissolved in 200 g of water. (Kb for water = 0.52 K kg mol 1, boiling point of pure water = 373 K, mol. wt. of urea = 60) (iv) Identify the compounds A, B, C and D in the given reaction : H O [O ] heat 2 ( OH )2 HC CH A B Ca D C dry K Cr O + H SO distillation Hg 2 + / H SO 2 2 4 2 7 2 4 Question 2 (a) For the reaction A + B C + D, the initial rate for different reactions and initial concentration of reactants are given below : [2] S.No. Initial Conc. [A] mole L 1 Initial rate [B] mole L 1 (mole L 1 sec 1) 1 1.0 1.0 2 10 3 2 2.0 1.0 4 10 3 3 4.0 1.0 8 10 3 4 1.0 2.0 2 10 3 5 1.0 4.0 2 10 3 (i) What is the overall order of reaction ? (ii) Write the rate law equation. OR (b) 25% of a first order reaction is completed in 30 minutes. Calculate the time taken in minutes for the reaction to go to 90% completion. Question 3 (a) Name the type of drug which lowers the body temperature in high fever condition. [2] (b) What are tranquilizers ? Given one example of a tranquilizer. Question 4 Write the balanced chemical equation for each of the following : [2] (a) Chlorobenzene treated with ammonia in the presence of Cu2O at 475 K and 60 atm. (b) Ethyl chloride treated with alcoholic potassium hydroxide. Question 5 (a) Name the monomer and the type of polymerisation that takes place when PTFE is formed. [2] (b) Name the monomers of nylon 6, 6. Question 6 Name two water soluble vitamins and the diseases caused by their deficiency in the diet of an individual. [2] Question 7 (a) How will you obtain the following (give balanced chemical equations) : [2] (i) Benzene from phenol. (ii) Iodoform from ethanol. OR To know about more useful books for class-12 click here Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII 3 (b) How will you obtain the following (give balanced chemical equations) : (i) Salicylaldehyde from phenol. (ii) Propan 2 ol from Grignard's reagent. Question 8 [2] Show that for a first order reaction the time required to complete 75% of reaction is about 2 times more than that required to complete 50% of the reaction. Question 9 [3] (a) When 0.4 g of acetic acid is dissolved in 40 g of benzene, the freezing point of the solution is lowered by 0.45 K. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene. (Kf for benzene = 5.12 K kg mol 1, at. wt. C = 12, H = 1, 0 = 16) OR (b) A solution is prepared by dissolving 9.25 g of non-volatile solute in 450 ml of water. It has an osmotic pressure of 350 mm of Hg at 27 C. Assuming the solute is non-electrolyte, determine its molecular mass. (R = 0.0821 lit atm K 1 mol 1) Question 10 [3] An element occurs in body centered cubic structure. Its density is 8.0 g/cm3. If the cell edge is 250 pm, calculate the atomic mass of an atom of this element. (NA = 6.023 1023) Question 11 [3] Describe the role of the following : (a) Cryolite in the extraction of aluminium from pure alumina. (b) NaCN in the extraction of silver from a silver ore. (c) Coke in the extraction of iron from its oxides. Question 12 [3] (a) Write the IUPAC names of the following : (i) K3[Fe(C2O4)3] (ii) [Co(NH3)5Cl]SO4 (b) [Fe(CN)6]4 is a coordination complex ion. (i) Calculate the oxidation number of iron in the complex. (ii) Is the complex ion diamagnetic or paramagnetic ? (iii) What is the hybridisation state of the central metal atom ? (iv) Write the IUPAC name of the complex ion. Question 13 (a) Explain why : (i) Transition elements form alloys. (ii) Zn2+ salts are white whereas Cu2+ salts are coloured. (iii) Transition metals and their compounds act as catalyst. OR (b) Complete and balance the following chemical equations. (i) KMnO4 + H2SO4 + H2C2O4 _________ + _________ + _________ + _________ (ii) K2Cr2O7 + H2SO4 + KI _________ + _________ + _________ + _________ (iii) K2Cr2O7 + H2SO4 + FeSO4 _________ + _________ + _________ + _________ [3] Question 14 [3] Give balanced equations for the following : (i) Aniline is treated with bromine water. (ii) Ethylamine is heated with chloroform and alcoholic solution of potassium hydroxide. (iii) Benzene diazonium chloride is treated with ice cold solution of aniline in acidic medium. Question 15 [3] Define the following terms with suitable examples : (i) Peptization (ii) Electrophoresis. (iii) Dialysis To know about more useful books for class-12 click here 4 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII Question 16 [5] (a) (i) Calculate the mass of silver deposited at cathode when a current of 2 amperes is passed through a solution of AgNO3 for 15 minutes. (at. wt. of Ag = 108, 1 F = 96,500 C) (ii) Calculate the emf and DG for the cell reaction at 298 K Mg(s) | Mg2+(0.1M) || Cu2+(0.01M)|Cu(s) Given, E cell = 2.71V 1F = 96,500 C OR (b) (i) Define the following terms : (1) Specific conductance (2) Kohlrausch's Law (ii) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 ohm. What is the cell constant and molar conductivity of 0.001 M KCl solution, if the conductivity of this solution is 0.146 10 3 S cm 1 at 298 K ? Question 17 [5] (a) (i) Explain why : (1) Fluorine has lower electron affinity than chlorine. (2) Red phosphorus is less reactive than white phosphorous. (3) Ozone acts as a powerful oxidising agent. (ii) Draw the structures of the following : (1) XeF6 (2) IF7 OR (b) (i) Explain why : (1) Interhalogen compounds are more reactive than the related elemental halogens. (2) Sulphur exhibits tendency for catenation but oxygen does not. (3) On being slowly passed through water, PH3 forms bubbles but NH3 dissolves. (ii) Complete and balance the following reactions : (1) P4 + H2SO4 _________ + _________ + _________ (2) Ag + HNO3 _________ + _________ + _________ (dilute) Question 18 (a) (i) Give balanced chemical equations for the following reactions : (1) Acetaldehyde reacts with hydrogen cyanide. (2) Acetone reacts with phenyl hydrazine. (3) Acetic acid is treated with ethanol and a drop of conc. H2SO4. (ii) Give one chemical test each to distinguish between the following pairs of compounds : (1) Acetone and benzaldehyde. (2) Phenol and benzoic acid. OR (b) (i) Write chemical equations to illustrate the following name reactions : (1) Aldol condensation. (2) Cannizzaro's reaction. (3) Benzoin condensation. (ii) Identify the compounds A and B in the given reactions : CH Cl [O ] 3 A A K B lCl ( anhy ) Cr O + H SO (1) [5] 3 2 2 7 2 4 Benzene [O] 5 A PCl (2) CH3COCH3 HNO B 3 ( conc.) qqq To know about more useful books for class-12 click here 5 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII ANSWERS Answer 1. (a) (i) more than, less than (ii) Lucas reagent, tertiary (iii) anode, cathode (iv) two, zero (b) (i) (2) positive (ii) (3) terylene (iii) (2) HClO < HClO2 < HClO3 < HClO4 (iv) (4) shortens the time to reach equilibrium. (c) (i) Diazotization (4) Aniline (ii) Crystalline solid (1) Anisotropic (iii) Phenol (2) Reimer-Tiemann reaction (iv) Fitting reaction (3) Diphenyl (d) (i) (1) La3+, as size decreases with increase in atomic number (lanthanoid contraction). (2) Electronic configuration of Cu+: 3d10 3 2 10 Electronic configuration of Cu2+: 3d9 As, Cu+ contains no unpaired electron therefore, it is diamagnetic whereas Cu2+ contains one unpaired electron therefore, it is paramagnetic. (ii) Structural formula : [Co(NH3)6] Cl3 IUPAC name : Hexa amine cobalt (III) chloride (iii) Given : Kb = 0.52 K kg mol 1, W2=6g, W1=200g, Tb = 373 K M2 = 60 Tb = Kb W2 1000 M 2 W1 Tb Tb = Kb W2 1000 M 2 W1 B Answer 2. (a) Rate = k [A]x [B]y 2 10-3 = k (1)x (1)y 4 10-3 = k (2)x (1)y ........................................ (5) From equation (1) and (2) 2 10 3 3 4 10 \ k (1)x (1)y = y x k ( 2 ) (1) 1 1 = 2 2 x x = 1 From equation (1) and (4) 2 10 3 k (1)x (1)y = k (1)x ( 2 ) y 1 1 = 1 2 \ y = 0 y (i) Order of reaction = x + y = 1 + 0 = 1 (ii) Rate law equation = k [A]1 [B]0 OR (b) For the first order reaction : For 25% completion of reaction. 2.303 a log t a x 2.303 = 0.1248 30 A CH3COCH3 heat dry distillation Acetone 2 10-3 = k (1)x (4)y 2.303 k= log 1.333 30 CH3CHO [O ] Acetaldehyde K 2 Cr2 O7 + H 2 SO4 ........................................ (4) 2.303 100 k= log 30 75 CH3COOH (CH3COO )2 Ca Ca( OH )2 Acetic acid Calcium acetate 2 10-3 = k (1)x (2)y 2.303 100 k= log 30 100 25 (iv) H O ........................................ (3) k= 0.52 6 1000 Tb 373 = 60 200 Tb 373 = 0.26 Tb = 0.26 + 373 = 373.26 K 2 HC CH Hg 2 + / H 2 SO4 8 10-3 = k (4)x (1)y C D ........................................ (1) ........................................ (2) = 0.00958 = 9.58 10-3 minutes For 90% completion of reaction k= 2.303 a log t a x 9.58 10-3 = 2.303 100 log t 100 90 9.58 10-3 = 2.303 100 log t 10 9.58 10-3 = 2.303 log 10 t To know about more useful books for class-12 click here 6 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII t = 2.303 9.58 10 3 1 NaOH 0.240396 103 Answer 3. (a) Antipyretic (b) Tranquilizers are the chemical substances used in the treatment of stress and severe mental stress. Example : Equanil, Valium or Luminal (a) NH2 Cl 475 K 60 atm +2NH3+Cu2O +Cu2Cl2+H2O 2 Aniline Chlorobenzene (b) C2H5Cl+KOH(alc) CH2=CH2+KCl+H2O Ethyl chloride Ethene Answer 5. (a) Monomer : CF2 = CF2 Tetrafluoroethene Type of polymerisation : Addition (b) Monomer : NH2 (CH2)6 NH2; Hexamethylenediamine and COOH (CH2)4 COOH; Adipic acid Answer 6. Vitamin B and Vitamin C are two water soluble vitamins. Deficiency of Vitamin B cause Beriberi, Paralysis, Convulsions and deficiency of Vitamin C causes Scurvy. Answer 7. (a) (i) Benzene from phenol : OH +ZnO (ii) Iodoform from ethanol : D CHI3 + CH3CH2OH + 4I2 + 6NaOH Ethanol Iodoform HCOONa + 5NaI + 5H2O OR (b) (i) Salicylaldehyde from phenol; (Reimer-Tiemann reaction) OH O Na+ CHCl2 CHCl +NaOH(aq) 3 Phenol OH Answer 8. For a first order reaction, 340K OH Propan 2 ol 2.303 [ R]o t= log k [ R]t For 75% completion of reaction, 2.303 100 t75%= log k 25 t75%= 2.303 log 4 k t75%= 2.303 log (2)2 k 2.303 2.303 t75%= 2 log 2 = 2 log 2 k k For 50% completion of reaction, 2.303 100 t50%= log k 50 ...(i) 2.303 t50%= log 2 ...(ii) k From equation (i) and (ii), we get t75%= 2 t50% Answer 9. (a) w=0.4g, W=40g, T=0.45k, kg=5.12 K kg mol-1 Benzene Phenol X CH3 CH CH3+Mg Given, T = Zn Distillation CHO (ii) Propan-2-ol from Grignard's reagent: Dry CH3 CHO + CH3MgX ether Acetaldehyde (Any one) Answer 4. OH Salicylaldehyde =240.396 minutes 2 ONa CHO H O+ 3 0.45 = 1000 k g m W 1000 5.12 0.4 m 40 1000 5.12 0.4 m = 40 0.45 mobserved = 113.77 Molecular weight of acetic acid = 60 2CH3COOH (CH3COOH)2 Initial. 1 0 After association 1 /2 Degree of association, i =1 + 2 Normal molecular mass 60 i = Observed molecular mass = 113 .77 To know about more useful books for class-12 click here 7 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII 60 = 1 + 113.77 2 2 2 + 0.527 = 2 1.054 = 2 = 2 1. 054 = 0.946. OR 350 (b) W = 9.25g, p = 350 mm Hg = =0.46 atm, 760 R = 0.0821 m atm k-1 mol-1 , V = 450 ml p = crt n p = RT V 0.46 = n = n 0.0821 300 1000 450 0.46 450 0.0821 300 1000 = 0.0084 W 9.25 Molar mass = = = 1101g n 0.0084 Answer 10. For BCC z = 2, d = 8.0 g/cm3, a = 250 pm = 250 10 10 cm NA = 6.023 1023 V = a3 = (250 10 10)3 cm3 M Density, d = V M = d V = 8.0 (250 10 10)3 = 8.0 (250)3 10 30 = 125 10 24g Therefore, 6.023 1023 atoms will have mass : 6.023 10 23 125 10 24 = 2 = 376 10 1 = 37.6g. Answer 11. (a) Cryolite lowers the melting point of alumina and makes it a good conductor of electricity. (b) Dilute NaCN acts as a leaching agent forming cyano compound complex, Na[Ag(CN)2]. It is further treated with strong metal Zn which replaces silver. (c) Coke acts as a reducing agent, which combines with silica impurities to remove them as slag. Answer 12. (a) (i) Potassium trioxalatoferrate (III) (ii) Pentaaminechlorocobalt (III) sulphate (b) (i) Oxidation number : + 2 (ii) Diamagnetic (iii) Fe2+ 3d 4s 4p 3d 4s 4p 4d CN [Fe(CN)6]4 Hybridisation d2sp3 (iv) Hexacyanoferrate (III) Answer 13. (a) (i) Transition elements forms alloy due to almost similar size of the metal ions, their high ionic size and availability of d-orbitals. Therefore, they can mutually substitute their position in crystal lattice to form alloys. (ii) Electronic Configuration : Zn2+ : 3d10 Cu2+ : 3d9 2+ 10 Zn has completely filled d-orbitals (3d ) whereas Cu2+ has incompletely filled d-orbital (3d9). Therefore, in Zn2+ d-d transition does not take place imparting white colour whereas in Cu2+ d-d transition takes place causing electrons to emit light in visible range which imparts blue colour. (iii) Because they have variable valencies and show multiple oxidation state leading to formation of unstable intermediate compounds. This provides a new path with lower activation energy for the reaction. Transition metals and their compounds also provides suitable surface for a reaction to take place. OR (b) (i) 2KMnO4 + 3H2SO4 + 5H2C2O4 2MnSO4 + 10CO2 + K2SO4 + 8H2O (ii) K2Cr2O7 + 7H2SO4 + 6KI 4K2SO4 + Cr2(SO4)3 + 3I2 + 7H2O (iii) K2Cr2O7 + 7H2SO4 + 6FeSO4 Cr2(SO4)3 Answer 14. (i) + 3Fe2 (SO4)3 + K2SO4 + 7H2O NH2 NH2 Br Br + 3HBr Hydrogen bromide +3 Br2 Br 2,4,6 Tribromoaniline Aniline (ii) C2H5NH2 + Ethylamine + 3KOH(alc.) CHCl3 Chloroform C2H5NC + 3KCl + 3H2O Ethyl iso-cyanide (iii) N =NCl + Benzene diazonium chloride To know about more useful books for class-12 click here NH2 Aniline OH 8 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII NH2+Cl +H2O N=N p Aminazobenzene (yellow dye) Answer 15. (i) Peptization can be defined as the process of converting a precipitate into colloidal form by shaking it with dispersion medium in the presence of small amount of electrolyte. It is used to convert a freshly prepared precipitate into a colloidal sol. (ii) Electrophoresis can be defined as the movement of colloidal particles towards a particular electrode under the influence of an electric field. Positively charged particles move towards cathode whereas negatively charged particles move towards anode. The gel electrophoresis is used to separate the macromolecules such as the DNA, RNA and proteins. (iii) Dialysis can be defined as the process of removing a dissolved substance from a colloidal solution by means of diffusion through which fresh water is continuously flowing. The molecules and ions diffuse into the water and pure colloidal solution is left behind. Dialysis is used to remove the waste from the blood in human body when the kidney of human gets failed. Answer 16. (a) (i) Molar mass of silver = 108 g mol 1, t = 15 minutes = 15 60 seconds, 1F = 96500 C, i = 2 Amperes Q = it = 2 15 60 = 1800 C w = ZQ M Z = nF Here, n = 1 So, Z = 108 1 96500 w = 108 1800 1 96500 = 2.014g Mass of silver deposited at cathode is 2.014g (ii) Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s) 0.0591 [ Mg 2 + ] Ecell = E cell log n [Cu 2 + ] = 2.71 0.0591 = 2.71 log 10 2 = 2.71 0.0591 [0.1] log 2 [0.01] 0.0591 1 2 = 2.68 V DG = nFEcell = 2 96500 2.68 = 517240 J = 517.240 kJ OR (b) (i) (1) Specific conductance can be defined as the reciprocal of specific resistance or conductance of a solution of definite dilution enclosed in a cell having two electrodes of unit area separated by 1 cm. (2) Kohlrausch's law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anions and cations of the electrolyte. L = v+l+ + v l (ii) k = 0.146 10 3 Scm 1, R = 1500 ohm, C = 0.001 M Cell constant = k R = 0.146 10 3 1500 = 0.219 cm 1 k 1000 Molar conductivity Lm = C = 0.146 10 3 1000 0.001 = 146 S cm2 mol 1. Answer 17. (a) (i) (1) Because the atomic size of fluorine is smaller than chlorine due to which there is larger electronic repulsion between the electrons of fluorine. Therefore, in fluorine atom in the incoming electron there is lesser attraction towards the nucleus than chlorine. (2) Red phosphorus is less reactive than white phosphorus because white contains tetrahedral units which are under strain. While red exists in polymeric chain form bonded through single covalent bond and are more stable and thus less reactive. (3) Under normal conditions, ozone is not very stable and decomposes readily on heating into a molecule of oxygen. As nascent oxygen is a free radical, it becomes very reactive and acts as a powerful oxidising agent. O3 Ozone (ii) (1) XeF6 O2 + [O] Oxygen Nascent oxygen F F F Xe F = 2.71 0.02955 To know about more useful books for class-12 click here F F 9 Oswaal ISC Solved Paper - 2019, CHEMISTRY, Class-XII (2) IF7 F C6H5CHO + 2 [Ag (NH3)2]+ + 2OH Benzaldehyde C6H5COOH + 2 Ag + 4NH3 + H2O (2) Benzoic acid reacts with NaHCO3 to give CO2 gas with effervescence whereas phenol does not. F F F I COOH F + NaHCO3 F NH=NH2 (2) CH2 C = O + Acetone Phenylhydrazine H2O CH3 C CH3 C=N NH OR (b) (i) (1) Aldol condensation : OH CHO | | dil.NaOH 2CH3 CHO CH3 CH CH2 Ethanal 3-Hydroxybutanal H2O CH3 CH = CH CHO But 2 enal (Aldol condensation product) (2) Cannizzaro reaction : H H C=O+ C = O + conc. KOH H H (ii) (1) Benzaldehyde being aldehyde reduces Tollen's reagent to form silver mirror but acetone being ketone does not. O H C OH+H C OK H Methanol Potassium formate (3) Benzoin condensation : O C H 2 Benzaldehyde O C CN H2O/C2H5OH OH (ii) (1) Benzoin COOH CH3 AlCl3(anhy) Ethanol O || CH3 C O CH2CH3 + H2O Ethyl acetate Foomaldehyde H conc.H 2 SO4 Acetic acid (Aldol) (3) CH3 COOH + CH3CH2OH Sodium Benzoate [O] CH3Cl Acetone phenyl hydrazone + H2O+CO2 Benzoic acid = OR (b) (i) (1) Because X X' bond in inter halogens is weaker than X X bond in halogens except F-F bond. (2) Oxygen is small in size and the lone pair on oxygen repel the bond pairs of O O bond to larger extent than the lone pairs on sulphur in S S bond. The S S bond energy i.e., bond strength is more compared to O O bond energy. Thus, sulphur exhibits tendency for catenation but oxygen does not. (3) When passed slowly in water NH3 forms hydrogen bonds with water and hence becomes soluble in water. Whereas, PH3 is not able to form hydrogen bonds with water due to its larger size in comparison to nitrogen and hence is insoluble in water and forms bubbles. (ii) (1) P4 + 10H2SO4 4H3PO4 + 10SO2 + 4H2O (2) Ag + 2HNO3 AgNO3 + NO2 + H2O (dilute) Answer 18. (a) (i) (1) H3C C H + HCN CH3 CH CN || | O OH Acetaldehyde Acetaldehyde cyanohydrin CH3 COONa = F K2Cr2O7+H2SO4 (A) Toluene (B) Benzoic acid [ O] (2) CH3COCH3 conc.HNO CH3COOH 3 Acetic acid (A) PCl 5 CH3COCl Acetyl chloride (B) qqq To know about more useful books for class-12 click here

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