AIEEE- 08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 1 KELWIN AIEEE CLASSES CHEMISTRY KELWIN CLASSES SOLUTIONS TO AIEEE-2008 1. The organic chloro compound which shows complete stereochemical inversion during a SN2 reaction is (1) (CH3)2CHCl (2) CH3Cl (3) (C2H5)2CHCl (4) (CH3)3CCl Sol. (2) For SN2 reaction, the C atom is least hindered towards the attack of nucleophile in the case of (CH3Cl). 2. Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazzotised and then heated with cuprous bromide. The reaction mixture so formed contains (1) mixture of o-and p-bromoanlines (2) mixture of o-and m-bromotoluenes (3) mixture of o-and p-bromotoluenes (4) mixture of o-and p-dibromobenzenes Sol. (3) 3. The coordination number and the oxidation state of the element ‘E’ in the complex [E (en)2 (C2O4)] NO2 (where (en) is ethylene diamine) are, respectively (1) 4 and 3 (2) 6 and 3 (3) 6 and 2 (4) 4 and 2 Sol. (2) Coordination no. = 6 and Oxidation no. = 3.AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 2 KELWIN AIEEE CLASSES 4. Identify the wrong statements in the following (1) Ozone layer does not permit infrared radiation from the sun to reach the earth (2) Acid rain is mostly because of oxides of nitrogen and sulphur (3) Chlorofluorocarbons are responsible for ozone layer depletion (4) Greenhouse effect is responsible for globol warming Sol. (1) Ozone layer does not allow ultraviolest radiation from sun to reach earth. 5. Phenol when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives (1) p-nitrophenol (2) nitrobenzene (3) 2, 4, 6-trinitrobenzene (4) o-nitrophenol Sol. (4) 6. In the following sequence of reactions, the alkene affords the compound ‘B’ CH3CH = CHCH3 3 O A ZnO H2 B The compound B is (1) CH3CH2COCH3 (2) CH3CHO (3) CH3CH2CHO (4) CH3COCH3 Ans. (2) 7. Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (1) more energy difference between 5f and 6d than between 4f and 5d orbitals (2) more reactive nature of the actinoids than the lanthanoids (3) 4f orbitals more diffused than the 5f orbitals (4) lesser energy difference between 5f and 6d than between 4f and 5d orbitals Sol. (4) Being lesser energy difference between 5f and 6d than 4f and 5d orbitals.AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 3 KELWIN AIEEE CLASSES 8. In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of 0 be the highest ? (1) [Co(H2O)6]3+ (2) [Co(NH3)6]3+ (3) [Co(CN)6]3– (4) [Co(C2O4)3]3– Sol. (3) CN is stronger ligand hence 0 is highest. 9. At 80º C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80º C and 1 atm pressure the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) (1) 48 mol present (2) 50 mol present (3) 52 mol present (4) 34 mol present Sol. (2) PT = B o B A oA X P X P 760 = 520 XA + o B P (1–XA) XA = 0.5 Thus, mole% of A = 50% 10. For a reaction 21 A 2B, rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression (1) dt ] B [ d dt ] A [ d (2) dt ] B [ d 4 dt ] A [ d (3) dt ] B [ d 21 dt ] A [ d (4) dt ] B [ d 41 dt ] A [ d Sol. (4) B 2 A 21 dt 2 ] B [ d dt ] A [ d 2 dt ] B [ d 41 dt ] A [ d 11. The equilibrium constants 1 P K and 2 P K for the reactions X 2Y and Z P + Q, respectively are in the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at these equilibrium is (1) 1 : 3 (2) 1 : 9 (3) 1 : 36 (4) 1 : 1 Sol. (3) X 2Y 1 0 (1-x) 2x kp1 = 1 1 2 x 1P ) x 1 ( ) x 2 ( Z P +Q 1 0 0 (1-x) x x kp2 = 1 2 2 x 1P ) x 1 ( x 36 1 PP 91 PP 4 21 2 1 AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 4 KELWIN AIEEE CLASSES 12. Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below 21 Cl2(g) H 21 dise Cl (g) H eg Cl– (g) H hyd Cl– (aq) The energy involved in the conversion of 21 Cl2 (g) to Cl– (aq) (using the data, 2 Cl dissH = 240 kJ mol–1 Cl egH = – 349 kJ mol–1 Cl hydH = – 381 kJ mol–1) will be (1) – 850 kJ mol–1 (2) + 120 kJ mol–1 (3) + 152 kJ mol–1 (4) – 610 kJ mol–1 Sol. (4) For the process 21 Cl2 (g) aq Cl H = 21 Hdiss of Cl2 + Cl Cl hyd eg = + 381 349 2 240 = – 610 kJ mol–1 13. Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly ? (1) Metal sulphides are less stable than the corresponding oxides (2) CO2 is more volatile than CS2 (3) Metal sulphides are thermodynamically more stable than CS2 (4) CO2 is thermodynamical more stable than CS2 Sol. (2) CO2 is more volatile than CS2 14. Bakelite is obtained from phenol by reacting with (1) CH3COCH3 (2) HCHO (3) (CH2OH)2 (4) CH3CHO Sol. (2) 15. For the following three reactions a, b and c equilibrium constants are given a. CO (g) + H2O (g) CO2 (g) + H2 (g); K1 b. CH4 (g) + H2O (g) CO (g) + H2 (g); K2 c. CH4 (g) +2 H2O (g) CO2 (g) + 4 H2 (g); K3 Which of the following relations is correct (1) K3 = K1 K2 (2) K3. 32 K = 21 K (3) K1 2 K = K3 (4) K2 K3 = K1 Sol. (1) Equation (c) = equation (a) + equation (b) Thus K3 = K1.K2AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 5 KELWIN AIEEE CLASSES 16. The absolute configuration of is (1) R, S (2) S, R (3) S, S (4) R, R Sol. (4) Both C1 and C2 have R – configuration 17. The electrophile, E attacks the benzene ring to generate the intermidiate -complex. Of the following which -complex is of lowest energy ? (1) (2) (3) (4) Sol. (4) – NO2 is electron withdrawing which will destabilize – complex 18. -D-(+) glucose and -D-(+) glucose are (1) anomers (2) enantiomers (3) conformers (4) epimers Sol. (1) -D (+) -glucose and -D (+) glucose are anomers. 19. Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 J K–1 mol–1 , respectively. For the reaction, 21 X2 + 23 Y2 XY3, H = – 30 kJ, to be at equilibrium, the temperature will be (1) 750 K (2) 1000 K (3) 1250 K (4) 500 K Sol. (1) 21 X2 + 23 Y2 XY3 Sreaction = 50 – 60 21 40 23 = – 40 J mol–1 G = H – TS at equilibrium G = 0 H = TS 30 × 103 = T × 40 T = 750 KAIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 6 KELWIN AIEEE CLASSES 20. Four species are listed below (i) HCO3 (ii) H3O+ (iii) 4 HSO (iv) HSO3F Which one of the following is the correct sequence of their acid strength ? (1) i < iii < ii < iv (2) iii < i < iv < ii (3) iv < ii < iii < i (4) ii < iii < i < iv Sol. (1) i < iii < ii < iv 21. Which one of the following constitutes a group of the isoelectronic species ? (1) CN– , N2, 22 O , 22 C (2) N2, 3 O , NO+, CO (3) 22 C , 2 O , CO, NO (4) NO+, 22 C , CN–, N2 Sol. (4) NO+, 22 C , CN– and N2 all have fourteen electrons. 22. Which one of the following pairs of species have the same bond order ? (1) 2 O and CN– (2) NO+ and CN+ (3) CN– and NO+ (4) CN– and CN+ Sol. (3) Both are isoelectronic and have same bond order. 23. The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excited the electron in the atom from n = 1 to n = 2 is (1) 7.56 × 105 J mol–1 (2) 9.84 × 105 J mol–1 (3) 8.51 × 105 J mol–1 (4) 6.56 × 105 J mol–1 Sol. (2) E = E2 – E1 = 1 10 312 . 1 2 10 312 . 1 6 2 6 = 9.84 × 105 J mol–1 24. Which one of the following is the correct statement ? (1) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase (2) B2H6. 2NH3 is known as inorganic benzene (3) Boric acid is a protonic acid (4) Beryllium exhibits coordination number of six Sol. (1) AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 7 KELWIN AIEEE CLASSES 25. Given 0 Cr /Cr3 E = – 0.72 V, 0 Fe /Fe2 E = – 0.42 V. The potential for the cell Cr | Cr3+ (0.1 M) | | Fe2+ (0.01 M) | Fe is (1) – 0.339 V (2) – 0.26 V (3) 0.26 V (4) 0.339 V Sol. (3) As V 42 . 0 E and V 72 . 0 E 0 Fe /Fe 0 Cr /Cr 2 2 2Cr + 3Fe2+ 3Fe + 2Cr3+ Ecell = 3 2 2 3 0cell ) Fe ( ) Cr ( log 6 0591 . 0 E = (–0.42 + 0.72) 3 2 3 2 ) 01 . 0 ( ) 1 . 0 ( log 6 0591 . 0 30 . 0 ) 01 . 0 ( ) 1 . 0 ( log 6 0591 . 0 = 0.30 6 0591 . 0 30 . 0 10 10 log 6 0591 . 0 62 log 104 Ecell = 0.2606 V 26. Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl, because HCl (1) reduces permanganate to Mn2+ (2) oxidises oxalic acid to carbon dioxide and water (3) gets oxidised by oxalic acid to chlorine (4) furnishes H+ ions in addition to those from oxalic acid Sol. (1) HCl being stronger reducing agent reduces 4 MnO to Mn2+ and result of the titration becomes unsatisfactory. 27. The vapour pressure of water at 20º C is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20ºC, the vapour pressure of the resulting solution will be (1) 16.500 mm Hg (2) 17.325 mm Hg (3) 17.675 mm Hg (4) 15.750 mm Hg Sol. (2) solute s s 0 X P P P 101 . 0 P P 5 . 17 s s 01 . 0 P P 5 . 17 s s Ps = 17.325 mm HgAIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 8 KELWIN AIEEE CLASSES 28. Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is (1) R2SiCl2 (2) R3SiCl (3) R4Si (4) RSiCl3 Sol. (4) 29. In context with the industrial preparation of hydrogen form water gas (CO + H2) which of the following is the correct statement ? (1) H2 is removed through occlusion with Pd (2) CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 is alkali (3) CO and H2 are fractionally separated using difference in their densities (4) CO is removed by absorption is aqueous Cu2Cl2 solution Sol. (2) CO + H2 O H2 CO2 + 2H2 30. In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be (1) X2Y (2) X3Y4 (3) X4Y3 (4) X2Y3 Sol. (3) No. of atoms of Y = 4 No. of atoms of X = 8 32 Formula of compound will be X4Y3 31. Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is (1) A < C < B < D (2) B < D < A < C (3) D < A < C < B (4) C < B < D < A Sol. (1) Highter the gold number lesser will be the protective power of colloid.AIEEE-08 PAPER WITH SOLUTION CODE -C DATE -27-04-08 Head Office : A/2 Jawahar Nagar-Kota. Ph : 0744-3291239, Fax No. : 0744-2426495 9 KELWIN AIEEE CLASSES 32. The hydrocarbon which can react with sodium in liquid ammonia is (1) CH3CH = CHCH3 (2) CH3CH2C CCH2CH3 (3) CH3CH2CH2C CCH2CH2CH3 (4) CH3CH2C CH Sol. (4) CH3CH2– C CH 3 NH . Liq /Na CH3CH2C Na C It is a terminal alkyne, having acidic hydrogen. Note : Solve it as a case of terminal alkynes, otherwise all alkynes react with Na in liq. NH3. 33. The treatment of CH3MgX with CH3C C – H produces (1) 3 3 CH C C CH | | H H (2) CH4 (3) CH3 – CH = CH2 (4) CH3 C C – CH3 Sol. (2) CH3 – MgX + CH3 – C C – H CH4 34. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (1) – CHO, – COOH, – SO3H, – CONH2 (2) – CONH2, – CHO, – SO3H, – COOH (3) – COOH, – SO3H, – CONH2, – CHO (4) – SO3H, – COOH, – CONH2, – CHO Sol. (3) – COOH, – SO3H, – CONH2, – CHO 35. The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be (1) 7.01 (2) 9.22 (3) 9.58 (4) 4.79 Sol. (1) It is a salt of weak acid and weak base b a WK K K H pH = 7.01