NJC 2018 SH1 H1 Promo Paper 1 Answer

1 2018 SH1 H1 Promotional Exam Answer Paper 1 1 B 6 D 11 C 16 D 2 A 7 A 12 A 17 B 3 B 8 A 13 C 18 D 4 C 9 B 14 C 19 C 5 A 10 D 15 B 20 C 1 Ans: B Relative atomic mass of Zn = 100 210 × 64 + 80 210 × 65 + 20 210 × 67 + 10 210 × 68 = 64.86 ≈ 64.9 (1dp) 2 Ans: A Isotope No. of protons No. of neutrons 10B 5 5 32S 16 16 32P 31 17 40K 19 21 40Ar 18 22 3 Ans: B Let the molecular formula of the hydrocarbon be CxHy. CxHy + (x + 𝑦 4 ) O2 → x CO2 + 𝑦 2 H2O Initial 15 75 - - Change –15 –45 +30 End 0 30 30 Amt reacted 1 3 2 x = 2 and x + 𝑦 4 = 3  y = 4 molecular formula = C2H4 4 Ans: C MnO4– + 5e + 8H+ → Mn2+ + 4H2O Amt of electrons released during [O] = Amt of electrons gained during [R] = (0.006) × 5 = 0.030 mol Amount of Z2Ox = 0.005 mol Amt of electrons lost per mole of Z2Ox = 0.030/0.005 = 6 mol Amt of electrons lost per mole of Z = 3 mol During oxidation, oxidation state of Z increases by 3 units from +3 (in Z2Ox) to +6 (in product) Hence x = 3 5 Ans: A Angle of deflection ∝ 𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 Protons would be attracted to negatively charged plate while electron would be attracted to positively charged plate.

2 NJC SH1 Promotional Examination 8873/01/18 Since charge of protons and electrons are the same but mass of electrons is much smaller, electron would be deflected more. 6 Ans: D Second ionisation is the removal of an electron from 1 mole gaseous singly charged cation to form 1 mole of gaseous doubly charged cation. 7 Ans: A A dative bond is formed when N atom in NH 2CN donates the lone pair of electrons to the empty 2p orbital of B atom in BF3. Since there are four bond pairs, the geometry about B and N central atoms is tetrahedral. The C atom in NH2CN does not have any lone pair of electrons. The two covalent molecules do not form ions when reacted. 8 Ans: A Bond angle Electrons regions Shape Bond angle x 3b.p. 0 l.p. Trigonal planar 120° y 3b.p. 1 l.p. Trigonal pyramidal 107° z 4b.p. 0 l.p. Tetrahedral 109.5° 9 Ans: A After bonding, there are 6 electrons around Al (3b.p. 0l.p.). The shape is trigonal planar and it is a non -polar molecule with intermolecular forces of temporary dipole – induced dipole interaction. 10 Ans: D Both Si and O are non -metal that form a giant covalent lattice for SiO 2 which has structure similar to that of diamond. 11 Ans: C Enthalpy change of formation is the formation of 1 mol of product from its constituent elements in their standard state (C(s) and O2(g)). Enthalpy change of combustion is the complete combustion of 1 mol of reactant to give oxidized product (e.g. CO2(g)) 12 Ans: A By Hess’s Law, H1 = 601+ 58 + ½ (−1700) = −191 kJ mol−1 13 Ans: C Formation of ionic bonds from gaseous Al3+ and O2− is an exothermic process. Breaking of Br−Br covalent bond is an endothermic process. First I.E. of Na is an endothermic process. 14 Ans: C From the graph, at low [HCl], as [HCl] increases, rate increases proportionally. Hence it is first order w.r.t. HCl at low [HCl]. At high [HC l], as [HC l] increases, rate remains constant. Hence it is zero order w.r.t. HC l at high [HCl]. [sodium thiosulfate] is kept constant throughout all reactions, hence we are unable to determine the effect of [sodium thiosulfate] on rate of reaction.

3 15 Ans: B Since we are monitoring time taken for a fixed amount of I 2 to be formed, rate ∝ 1 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Comparing expt 1 & 2, [KI] remain constant and [(NH 4)2S2O8] × 2, rate also × 2. Hence it is first order w.r.t (NH4)2S2O8. Comparing expt 1 & 3, [(NH 4)2S2O8] remain constant and [KI] × 3, rate also × 3. Hence it is first order w.r.t KI. Rate = k[(NH4)2S2O8][KI] Using expt 1 data, 1/35 = k (0.1)(0.2), k = 1.429 Using the value of k for expt 4 (since same temperature condition), 1/(time taken) = 1.429(0.02)(0.750), Time taken = 46.7s 16 Ans: D Catalyst increases the rate of reaction by lowering the activation energy. Rate constant increases. H is not affected by the presence of a catalyst. 17 Ans: B Reactants: 100% → 50% → 25% → 12.5% To achieve 20% of the initial amount, time taken is between 2-3 half-lifes. ( 1 2 )n = final concentration initial concentration , where n is the number of half-life ( 1 2 )n= 20 100 n = 2.32 The time required = 2.32 × 5500 = 12760 years 18 Ans: D The compounds are chain-branch isomers. Straight chain isomer has greater surface area for intermolecular interactions of temporary dipole-induced dipole. Hence higher boiling point as compared to the branched isomers. 19 Ans: C C C Br H H Br C C Br H Br H C C H H Br Br 20 Ans: C For C=C, there are 1σ+ 1π bond For C≡N, there are 1σ+ 2π bonds Total 8σ + 6π bonds There are 2 lone pairs electrons (1 for each N atom) The shapes around the C atoms are linear and trigonal planar. Thus the molecule is planar.