PAPER-1 Select the correct alternative. (Only one is correct) [27 × 3 = 81] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.25 If decomposition reaction X (g) Y (g) + Z (g) follows first order kinetics then the graph of rate of formation (R) of Y against time t will be (A) (B) (C*) (D) Q.26 Which one of following represents different molecules? (A) and (B*) and (C) and (D) and [Sol. (B) is non planar so double bonds are localised ] Q.27 Which of the following is an oxidizing agent? (A*) Mn(CO)5 (B) Fe(CO)5 (C) Mn2(CO)10 (D) Fe2(CO)9 Q.28 The equilibrium constant for the ionization of R–NH2(g) in water as R–NH2(g) + H2O(l) l R–NH3+(aq) + OH–(aq) is 8 × 10–6 at 25°C. Find the pH of a solution at equilibrium with pressure of R–NH2(g) is 0.5 bar. [Take log 2.82 = 0.45] (A) 12.3 (B*) 11.3 (C) 11.45 (D) None x 2 [Sol. 8 × 10–6 = 0.5 4 × 10–6 = x2 x = 2 × 10–3 pOH = 2.7 So, pH = 11.3 ] Q.29 Which statement(s) is/are true about the relation between the following compounds? (a) (b) (c) (I) a and b are tautomers (II) b and c are resonating structures (III) a and c are resonating structures (IV) a and c are tautomers (A*) I & III (B) II & IV (C) I, II & III (D) I & IV Q.30 Select correct statement about given square planar complex. (A) It has no geometrical isomer. (B*) It is optically active because it does not have plane of symmetry. (C) It is optically inactive because square planar complexes have plane of symmetry. (D) It is optically active because it has symmetric carbon. Q.31 XeF2 (g) + H2(g) 2HF(g) + Xe(g) H° = – 430 kJ using the following bond energies : H–H = 435 kJ/mol H–F = 565 kJ/mol Calculate the average bond energy of Xe–F in XeF2. (A) 267 kJ/mol (B) 562.5 kJ/mol (C*) 132.5 kJ/mol (D) None [Sol. rH° = [ 2 × x + 435] – 2 × 565 –430 = (2x + 435) – 2 × 565 x = 132.5 kJ/mol ] Q.32 Arrange the following carbocations in the increasing order of their stability. (I) (II) (III) (A*) I > II > III (B) I > II = III (C) I > III > II (D) III > I > II [Sol. (I) Two benzenoid RS & Two H (II) Two benzenoid RS & but zero H (III) Only one benzenoid RS zero H ] Q.33 The total possible co-ordination isomers for the following compounds respectively are [Co(en)3] [Cr(C2O4)3] [Cu(NH3)4] [CuCl4] [Ni(en)3] [Co(NO2)6] (A) 4, 4, 4 (B) 2, 2, 2 (C) 2, 2, 4 (D*) 4, 2, 4 Q.34 Calculate f H° (in kJ/mol) for Cr2O3 from the rG° and the S° values provided at 27° 4Cr(s) + 3O2(g) 2Cr2O3(s) rG° = – 2093.4 kJ/mol S°(J/K mole) : S°(Cr, s) = 24 ; S°(O2, g) = 205 ; S° (Cr2O3, s) = 81 (A) –2258.1 (B) –946.35 (C*) –1129.05 (D) None [Sol. rG° = rH° – T·rS° –2093.4 = rH° – 300 × [ 2 × 81 – 4 × 24 – 3 × 205] rH° = –2093.4 – 164.7 = –2258.1 rH° = 2 × f H°(Cr2O3, s) H° of Cr O 2258.1 = – = –1129.05 kJ/mol ] f 2 3 2 Q.35 Select correct code about complex [Cr(NO2)(NH3)5][ZnCl4] (I) IUPAC name of the compound is Pentaammine nitrito-N chromium (I