NJC 2016 A Level Chem Solution

H2 Chemistry TYS 2016 suggested answers Questions that are no longer in syllabus Paper 1 Paper 2 5(b)(ii) Paper 3 Paper 1 Answer Key 1 B 11 C 21 B 31 B 2 D 12 A 22 B 32 A 3 C 13 C 23 B 33 C 4 D 14 B 24 A 34 D 5 A 15 D 25 D 35 C 6 C 16 A 26 C 36 B 7 C 17 A 27 A 37 C 8 D 18 A 28 B 38 B 9 B 19 D 29 B 39 A 10 D 20 D 30 B 40 C 2016 A level Paper 1 suggested answers 1) Amount of molecule = 1.0 𝑥 mol Amount of atom = (1.0 𝑥 × 2)mol No of atoms = 2.0 𝑥 × L Ans: B 2) 2H2S + CS2 + 6O2  CO2 + 4SO2 + 2H2O Ans: D 3) From the data, the isotopes of copper have mass number of 63 and 65. Note: mass number 197 is Au Ar of Cu = 63×65+65×29 94 = 63.6 Ans: C 4) No. of protons No. of neutrons No. of electrons 36S2 16 20 18 37Cl 17 20 18 Nucleons = protons + neutrons (different nucleon no. for the 2 ions) The ions have the same outer electronic configuration of 3s23p6 Ans: D

5) Ans: A 6) Possible structures of C2N Ans: C 7) pV = nRT = 𝑚 𝑀𝑟 RT (Note: Vol of gas collected = 782 = 76 cm3) Mr = 𝑚𝑅𝑇 𝑝𝑉 = 0.293×8.31×(97+273) (101000)×(76×10−6) = 117 Ans: C 8) Lattice Energy is the amount of energy released when 1 mole of ionic solid is formed from its gaseous ions. Ans: D 9) By Hess' Law, Hf KCl(s) = Hat K(s) + Hat Cl2(g) + 1st I.E.of K(g) + 1st E.A. of Cl(g) + L.E. of KCl(s) Hf KCl(s) = +90 + ½ ×(+242) + (418) + (355) + (710) = 436 kJmol1 Ans: B Energy / kJ mol1 0 K (s) + ½ Cl2 (g) KCl (s) (s)KCl ΔH θ f K (g) + ½ Cl2 (g) (s)K ΔHθ at (g)K of I.E. 1st K (g) + Cl (g) K+ (g) + e + Cl (g) K+ (g) + Cl − (g) (g) Cl E.A. 1st (s)KC of L.E. l (g) Cl H 2 θ at

10) S > 0 as there is an increase in number of gaseous molecules after the reaction G = H  TS For G to be > 0 at 378 K, H must be a positive value. Ans: D 11) Eocell = Eo (Ag+/Ag)  Eo (Fe3+/ Fe2+) = + 0.03V. When [Fe3+] increases, position of eqm Fe3+ + e Fe2+ shifts to the right, E(Fe3+/ Fe2+) increases and thus Eocell could become 0.00V. Ans: C (Note: Surface area of electrode would not affect Eo values) 12) [O] 2O2  O2 + 4e Q = It = neF ne = 8×60×100 96500 = 0.4974 mol nO2 = 0.4974 4 = 0.12435 mol VO2 = 0.12435 × 22.7 = 2.82 dm3 Ans: A 13) Let solubility of ZnF2(s) be s mol dm3 ZnF2(s) + aq Zn2+(aq) + 2 F (aq) Initial / mol dm3  0 0 Change / mol dm3 s +s +2s Eqm / mol dm3  s 2s Ksp = [Zn2+][F]2 = 4s3 s = 0.2 mol dm3 [F] = 2s = 0.4 mol dm3 For precipitation of BaF2, [Ba2+][F]2 = Ksp and [F] = 0.4 mol dm3 [Ba2+] = 1.6×10−7 (0.4)2 = 1×106 mol dm3 Ans: C 14) For weak acid, [H+] = √𝐾𝑎×𝑐 = 1×103 mol dm3 pH = lg [H+] = 3 Ans: B 15) Rate = k[IO3][I]2[H+]2 Expt 1: y = k (a)(a)2(a)2  eqn (1) Expt 2: rate 2 = k (1 2 a)(2a)2(3a)2  eqn (2) 𝑒𝑞𝑛 2 𝑒𝑞𝑛 1 : 𝑟𝑎𝑡𝑒 2 𝑦 = 𝑘×18𝑎5 𝑘×𝑎5 Rate 2 = 18y Ans: D

16) Both Al2O3 and MgO exist as giant ionic lattice. Since Al3+ has a higher charge density than Mg2+, Al2O3 has the most covalent character. Ans: A 17) Mg(NO3)2(s) MgO(s) + 2NO2(g) + ½ O2(g) Amount of Mg(NO3)2 = 10.4 24.3+2×14+6×16 = 0.07013 mol Amount of O2 = ½ × 0.07013 = 0.03506 mol Mass of O2 = 0.03506 × 32.0 = 1.12 g Ans: A 18) MgCl2 is soluble in water to give a solution of pH = 6.5 Mg has a metallic lattice structure with strong metallic bond between Mg2+ and sea of delocalised electrons. Sulfur exists as S8 with simple molecular structure with weak intermolecular forces of attraction between S8 molecules. Hence melting point of Mg is higher than sulfur. Mg(OH)2 is a white ppt that is only sparingly soluble in water. Option D is not in syllabus. Mg solid reacts slowly with cold water. Ans: A 19) AgCl (white ppt), AgBr (cream ppt) and AgI (yellow ppt) are formed. Both AgCl and AgBr are soluble in excess concentrated NH3, only AgI remains insoluble due to the extremely small Ksp value. A small [Ag+] will easily lead to precipitation of AgI. Note: can refer to Data Booklet for the reactions of AgX with NH3. Ans: D 20) As the element contains a partially filled 3d subshells, it is a transition element. Ans: D 21) Redox reaction between SO2 and VO2+, Eocell = 1.00  0.17 = +0.83 V > 0, rxn is feasible Redox reaction between SO2 and VO2+, Eocell = 0.34  0.17 = +0.17 V > 0, rxn is feasible Redox reaction between SO2 and V3+, Eocell = 0.26  0.17 = 0.43V < 0, rxn is not feasible VO2+ can be reduced to V3+ in the presence of excess SO2. Ans: B 22) As Pt has a oxidation number of +4, there are total of 4 Cl− in the compound. Complex ion has a +2 charge and hence there are two free Cl− anion. The other 2 Cl− are acting as ligands.The coordination number is 6, there are 4 other NH3 ligands. The compound is [Pt(NH3)4Cl2]2+ + 2 Cl− Ans: B

23) CC bond consists of 1  + 2  bonds. C−C and C−H bonds consist of 1  bond. Ans: B 24) Termination step for free radical substitution mechanism involves the reaction of two radicals. H is not produced during the free radical substitution mechanism. Ans: A 25) The slow step of SN1 mechanism involves the breaking of C−X bond to give carbocation intermediate. Since C−Cl bond (BE 340 kJ mol−1) is stronger than C−Br bond (BE 280 kJ mol−1), the activation energy for reaction 1 is lower than reaction 2. Reaction 1 is faster than reaction 2 under identical conditions. Ans: D 26) Na(s) undergoes redox reaction with −OH groups in alcohol, phenol and carboxylic acids to give H2(g) and −O−Na+. Ans: C 27) Compound Z has empirical formula of CH2O and - contains the following structure [gives yellow ppt CHI3 with alkaline I2(aq)] - contains a −OH group from alcohol or carboxylic acid [gives white fume HCl with PCl5] Ans: A 28) H atoms in the phenol −OH and carboxylic acid −COOH can easily be replaced by deuterium, D, due to the weak acid dissociation equilibrium of H+ from phenol and carboxylic acid group in aqueous medium. RCOOH RCOO− + H+ . RCOO− + D2O RCOOD + OD− Ans: B 29) Compound X - does not contain the following structure [no reaction with alkaline I2(aq)] - does not contain aldehyde functional group [no reaction with Tollens’ reagent] - contain a ketone functional group [reduced by NaBH4] Compound X, CH3CH2COCH2CH3, reacts with NaBH4 to give CH3CH2CH(OH)CH2CH3 Ans: B

30) Reaction A and D is the hydrolysis of ester, which requires heating and acid/alkaline catalyst. Reaction B is the hydrolysis of acyl chloride, which occurs readily at room temperature. Reaction C is the hydrolysis of amide, which requires prolonged heating and acid/alkaline catalyst. Ans: B 31) 4He 12C 24Mg 14N 20Ne 30P 28Si 34S 40Ca Protons 2 6 12 7 10 15 14 16 20 Neutrons 2 6 12 7 10 15 14 18 20 Ans: B 32) 1 SO2 Bent (polar) CO2 Linear (non-polar) 2 PF3 Trigonal pyramidal (polar) BF3 Trigonal planar (non-polar) 3 BrF5 Square pyramidal (polar) SiF4 Tetrahedral (non-polar) Ans: A 33) A Bronsted-Lowry base is a proton (H+) acceptor. In reaction 1, NH3 acts as a nucleophile (electron pair donor/Lewis base) Ans: C 34) Comparing experiments 1 and 3, when [C2H5ONa] doubles while keeping [CH3I] constant, rate of reaction doubles. Hence it is first order with respect to C2H5ONa. Comparing experiments 2 and 3, when [CH3I] doubles while keeping [C2H5ONa] constant, rate of reaction doubles. Hence it is first order with respect to CH3I. Overall order of reaction is 2. Ans: D 35) A Bronsted-Lowry acid is a proton (H+) donor. In reaction 1, Be2+ acts as a Lewis acid (electron pair acceptor) In reaction 2, [Be(H2O)4]2+ donates H+ to give Be(H2O)2(OH)2 In reaction 3, Be(H2O)2(OH)2 donates H+ to give [Be(OH)4]2− Ans: C 36) Oxidation number of Cl changes from 0 in Cl2 to −1 in NaCl and +5 in NaClO3 Ans: B 37) The structure