NJC 2019 A Level Chem Solution

H2 Chemistry TYS 2019 suggested answers Paper 1 Answer Key 1 B 11 A 21 D 2 C 12 B 22 C 3 B 13 D 23 A 4 C 14 D 24 C 5 C 15 D 25 B 6 A 16 A 26 B 7 D 17 B 27 C 8 B 18 A 28 C 9 D 19 A 29 B 10 D 20 D 30 B/C 2019 A level Paper 1 suggested answers 1) Sr2+ W X─ Y2─ No of electrons *36 *36 *36 *36 No of protons 38 36 35 34 No of neutrons *46 *46 *46 *46 Nucleon Number (no of protons + neutrons) *84 82 81 80 * told in the question Ans: B 2) Angle of deflection  𝐶ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 Mass of electron is approximately 1 2000 the mass of a proton, hence electron has a larger angle of deflection. The charged particles are attracted to the oppositely charged plate. Ans: C 3) For Group 15 and Group 16 element Electronic configuration where electrons are lost are as such: Group 15 element Group 16 element 1st IE ns2 np3 ns2 np4 2nd IE ns2 np2 ns2 np3 3rd IE ns2 np1 ns2 np2 4th IE ns2 ns2 np1 4th IE in group 15 element will experience a slightly large spike as electrons are removed from an inner subshell. Comparing the values given in the table, 4th IE of option A and C experience a slightly larger spike than that of B and D => A and C belong to Group 15, B and D belong to Group 16. Between Se and Te, Te is in a period further down the periodic table. This means that it has lower IE (since increase in no. of electron shell and shielding effect outweighs increase in nuclear charge) => Te must be B Ans: B

4) BaO2: ionic compound : Ba2+ O22─ (peroxide ion) For O22─ dot and cross: Each O─ is bonded to another O─ by a single covalent bond (told in question). Ans: C 5) Molecule No of bonds 1 Cl Al ClCl Cl Al Cl Cl 6 Sigma bonds 2 COO 2 Sigma bonds 2 Pi bonds 3 C H H H C H O 6 Sigma bonds 1 Pi bond 4 CC H H H CH HH 8 Sigma bonds 1 Pi bond Ans: C 6) Assuming the compounds behave as ideal gases, apply pV = nRT, rearranging, p = 𝑛𝑅𝑇 𝑉 = 𝑚𝑅𝑇 𝑀𝑟𝑉 Since T, V and m is constant, the smaller the Mr, the higher the pressure. Molecule CH4 HCHO CH3Cl HCO2H Mr 16 30 50.5 46 Ans: A 7) Follow the question closely for the reaction that happens: NaN3 (s) ⎯→ Na (s) + 3 2 N2 (g) Amount of NaN3 = 5.00 65.0 = 0.07692 mol Amount of N2 = 1.5 × 0.07692 = 0.1154 mol pV = nRT V = 𝑛𝑅𝑇 𝑝 = 0.1154 × 8.31 × (30+273) 9.85 × 104 = 0.00295 m3 = 2.95 dm3 Ans: D 8) Context of question: Period 3 and Period 4 elements behave similarly. As such, Ge behaves similarly to Silicon, which has high melting point due to its giant covalent lattice structure with strong covalent bonds and is also a semiconductor. Ans: B

9) Clue 1: X dissolves in HCl to give Solution Y, which gives a white ppt insoluble in excess NaOH (aq) Possible Metal Solution Y it forms White ppt it forms Al Al3+ Al(OH)3 White ppt dissolves in excess NaOH to give [Al(OH)4]− Mg Mg2+ Mg(OH)2 White ppt insoluble in excess NaOH Clue 2: Give off gas Z, which reacts with O2(g) to form H2O and a white solid oxide that is insoluble in dilute acid or alkali. Possible Non-Metal Oxides it forms Reaction with acid Reactions with alkali Si SiO2 No rxn Insoluble in dilute alkali due to strong covalent bonds P P4O10 No rxn React with bases to give a soluble salt. Ans: D 10) A Not relevant, electron affinity is for non-metal forming anion. B False, electronegativity decreases down the group C True statement, but does not explain why chemical reactivity increases down the group D True, easier to lose the valence electrons as a result of increase in electronic shell, M is more easily oxidised to give M2+ . Ans: D 11) Assume 100g of solder glass, of which 16g is B2O3 and 84g is PbO. Amount of B2O3 = 16 10.8 × 2 + 16.0 × 3 = 0.2298 mol Amount of B = 0.2298 × 2 = 0.4596 mol Amount of PbO = 84 207.2 + 16.0 = 0.3763 mol Amount of Pb = 0.3763 mol 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑃𝑏 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐵 = 0.3763 0.4596 = 0.82 Ans: A 12) A False, for ∆Hꝋcombustion, H2O should be in liquid state and not gaseous B True, ∆Hꝋformation involves element in the natural state. C False, ∆Hꝋatomisation is for per mol of atom formed, and hence, it should be 12 × ∆Hꝋatomisation D False, ∆Hꝋformation involves element in the natural state, hydrogen exist as H2(g), not H(g) atom. Ans: B 13) 1 False, ∆H = −ve, ∆S = +ve (increase in no. of mol of gas) => ∆G < 0 at all temperature. 2 True, since H2O2 can be stored for several week with little decomposition at rtp, suggesting rate is very slow, and hence high Ea. 3 False, ∆S = +ve (increase in no. of mol of gas) Ans: D

14) 1 False. When [substrate]= x, initial rate of reaction remains constant despite any further increase in concentration of substrate. 2 False, since the graph shows that initial rate of reaction remains constant despite any further increase in concentration of substrate. order of reaction = 0 wrt to substrate. 3 True, as enzyme active sites are saturated, increasing substrate conc has no effect on the rate of reaction (as oppose to before [substrate]= x, where increasing conc of substrate increases rate of reaction by allowing more substrate to bind to the active sites of the enzymes). Ans: D 15) 1 False. Increasing concentration may increase rate (if order of reaction wrt reactant is not zero), but only temperature and catalyst increases rate constant. Note : rate = k[reactant]x 2 False, increasing Temperature, not conc, increases the number of particles with energy greater than or equal to activation energy. 3 True. Refer explanation for statement 1. 4 True. Refer explanation for statement 2. Ans: D 16) Recall hybridisation: Atom in a molecule undergoes hybridization depending on no. of electron regions after bonding. Type of hybridization: Type of hybridisation No of eln region around atom (incl. bond pairs and lone pairs) sp3 4 sp2 3 sp 2 Note : H atom does not undergo hybridisation as it has only 1 electron pair region after bonding. Ans: A 17) Chiral C => carbon with 4 different groups attached. OCH3CH CH3 C2H5 Ans: B

18) Compound 2 is a disubstituted product. Compounds 1, 2 and 3 can be formed in termination steps of FRS. Compounds 1 and 2 can also be formed in the propagation steps. Ans: A 19) Reaction is electrophilic substitution. E+ must be electron deficient and is generated using a Lewis acid catalyst, AlCl3. CH3CO H COCH3 Note : The benzene C that is bonded to 4 groups is sp3 hybridised and hence it does not take part in the delocalization of pi electron, resulting in the “opening” in the benzene ring. Ans: A 20) In order to for a racemic mixture from nucleophilic substitution/hydrolysis with aq NaOH, heat 1st criteria: must undergo SN1 mechanism => must be tertiary RX. 2nd criteria: the product must have a chiral C (C atom bonded to 4 different groups). 1st criteria 2nd criteria A CH3(CH2)6Cl ⎯→ CH3(CH2)6OH   B (CH3CH2)3Cl ⎯→ (CH3CH2)3OH   C (CH3)2CHCH2C(CH3)2Cl ⎯→(CH3)2CHCH2C(CH3)2OH   D (CH3)2CHC(CH3)(CH2CH3)Cl ⎯→(CH3)2CHC(CH3)(CH2CH3)OH   Ans: D

21) Recap reaction with Na, NaOH, Na2CO3: Na NaOH Na2CO3 or NaHCO3 Alcohol    Phenol    Carboxylic acid    • All -OH group react with Na(s). • Acids (phenol and carboxylic acid) react with strong base like NaOH. • Phenol (a very weak acid) cannot react with Na2CO3 (a very weak base) Ans: D 22) Type of alcohol Oxidation product when heated under reflux with excess heat acidified potassium dichromate 1 Carboxylic acid (heating with reflux ensures any aldehyde produced is further oxidized to carboxylic acid) 2 Ketone 3 Does not oxidise Ans: C 23) Based on the given options, Y is an amide, formed from the reaction between acid chloride (X) and the amine given in the question. C and D cannot be the answer as the N is not part of the ring, and hence contradict the amine given. Molecules of W are c