NJC 2021 A Level Chem Solution

H2 Chemistry TYS 2021 suggested answers Paper 1 Answer Key 1 B 11 C 21 A 2 D 12 A 22 A 3 C 13 B 23 A 4 B 14 B 24 B 5 C 15 A 25 B 6 D 16 A 26 D 7 B 17 C 27 B 8 D 18 B 28 C 9 D 19 D 29 C 10 A 20 C 30 D 2021 A level Paper 1 suggested worked solutions 1 Ans: B Statement 1: Correct, as electrons are lighter than protons and angle of deflection ∝charge mass Statement 2: Correct, as electron and proton are attracted towards oppositely charged plate due to electrostatic forces of attraction. Statement 3: Incorrect, as the beam do not travel in a straight line but curved path. 2 Ans: D Since the formula of compound is QNO3, cation is Q+. Hence since there is 80 electrons, there should be 81 protons in element Q.  Element is Tl in group 13. As such, nucleon number = 81 + 122= 203. 3 Ans: C Statement A: Incorrect F+: 1s2 2s2 2p4 Al2+: 1s2 2s2 2p6 3s1 Electron is removed from second electronic shell of F which is much closer to the nucleus (vs from 3rd electronic shell for Al) hence more energy is required to overcome the stronger attraction between the nucleus and valence electron. Statement B: Incorrect, 3rd I.E. is removing electron from gaseous X2+ to form X3+ F2+: 1s2 2s2 2p3 Ne2+: 1s2 2s2 2p4 Na2+: 1s2 2s2 2p5 Mg2+: 1s2 2s2 2p6 Al2+: 1s2 2s2 2p6 3s1 Al 3rd IE involve electron removed from electronic shell 3 while the rest are removed from electronic shell 2. Statement C: Correct, Na3+ and Ne2+ are isoelectronic (1s2 2s2 2p4 ) with same shielding effect. Na3+ has a greater proton number, hence a greater nuclear charge and stronger nuclear attraction for the most loosely held electron, resulting in a greater ionisation energy value. Statement D: Incorrect, successive I.E. involves removal of electrons from the outer filled electronic shell towards the inner filled electronic shell, and hence are not all taken from the same shell.

2021 A level Paper 1 suggested worked solutions 4 Ans: B For element W, there is a big jump in IE between 7th and 8th IE, implying 8th electron is in inner electronic shell and hence 7 valence electrons, thus W is a group 17 element. Since the elements are consecutive, it means X is in group 18 of the same period, n. Y and Z are in group 1 and group 2 of the next period (n+1). As 1st IE generally increases across a period and decreases down the group, X has the highest 1st I.E. E.g. W is F, X is Ne, Y is Na, Z is Mg, Ne has the highest first I.E. 5 Ans: C From C2H6 to C2H4 to C2H2, Bond order of the C-C changes from 1 (C−C) to 2 (C=C) to 3 (C≡C).  Bond energy increases  Bond length decreases. 6 Ans: D molecule Molecular shape Polarity A BCl3 Trigonal planar Non-polar B NCl3 Trigonal pyramidal Polar C SO2 Bent polar D CHCl3 tetrahedral polar 7 Ans: B Predominant IMF Average number of H bond per molecule M H-bond 2 N Pd-pd NA P H-bond 1 Q H-bond 1 Between P and Q, Q has the higher boiling point as carboxylic acid has an additional electronegative C=O which increases the polarity of the O−H bond and strengthen the hydrogen bond, compared to that of P. Furthermore, Q also has stronger id-id due to larger electron cloud. Hence, order of boiling point, N < P < Q < M 8 Ans: D Dalton’s Law of partial pressure: PT = PA + PB + PC +…. Where PA, PB, PC is the partial pressure of gas A, gas B, gas C etc…

2021 A level Paper 1 suggested worked solutions 9 Ans: D SiO2 is insoluble in water as it has a giant covalent lattice and has strong covalent bond, hence it remains as solid when shaken with water. SiO2 is an acidic oxide, hence it does not react with acidic, and remains as solid when shaken with HCl (aq). SiO2 reacts only with hot concentrated NaOH and thus remains as solid when shaken with NaOH (aq). 10 Ans: A No. of mol No of molecules 2.00 g of HCOOCH2CH3 2 74= 0.027 0.027 × 6.02 × 1023 = 1.62 × 1022 4.00g of Br2(l) 4 158= 0.0252 0.025 × 6.02 × 1023 = 1.51 × 1022 550cm3 of H2 550 24000= 0.0229 0.0229 × 6.02 × 1023 = 1.38 × 1022 1.55 × 1022 molecules of water 1.55 × 1022 11 Ans: C H2SO4 + 2 NaOH ⎯→ Na2SO4 + 2H2O Amount of H2SO4 = 20 1000  5 = 0.1 Amount of NaOH = 20 1000  5 = 0.1 NaOH is limiting reagent. Amount of NaOH = Amount of water formed ∆Hn = − q amount of water formed= −(40)(4.18)(50 − 25) 0.1 = −41800 J mol−1 = −41.8 kJ mol-1 12 Ans: A The rate is constant at the start as an equilibrium has been established. At time t, as the pressure is lowered, there is a reduced frequency of effective collision, hence the rate of forward reaction decreased. As the pressure slowly returns to atmospheric pressure, the rate of reaction is restored to the original rate of reaction.

2021 A level Paper 1 suggested worked solutions 13 Ans: B Statement 1: Correct, looking at the rate equation for reaction 1, as [HBr] is in the denominator of the rate equation, formation of HBr slows down the rate of reaction. Statement 2: Correct, as the same number of moles of reactants are observed in the rate equation as well as in the overall stoichiometric equation, this implies a single step reaction. Statement 3: Incorrect, since the order of reaction is NOT 1 wrt Br2 in reaction 1, doubling its concentration will not double the rate of reaction. 14 Ans: B • * Chiral centre • Circled in blue: double bond that gives cis trans Note: double bond in ring does NOT give cis trans isomers. Total number of stereoisomer = 2(no. of chiral C + no. of cis/trans C=C) Z: no of stereoisomer = 2(1+1) = 4 Q: no of stereoisomer = 2(3+0) = 8 15 Ans: A A: Correct, propagation step produces (CH3)3C• that reacts with X2 to give (CH3)3CX. B: Incorrect, in termination step, two radicals react to form molecule. C: Incorrect, enthalpy change of reaction also affected by energy released when forming of C−X and H−X bonds. For reaction with Br2, ∆H = (410 + 193) – (280 + 366) = −43 kJ mol−1 For reaction with Cl2, ∆H = (410 + 244) – (340 + 431) = −117 kJ mol−1 D: Incorrect, Initiation produces halogen radical, not halide ion. 16 Ans: A For electrophilic addition of alkene, pi electrons in the C=C bond of electron rich alkene are donated to the electron-deficient species (electrophile). Statement D is incorrect,  bond is strong while  bond is weaker.

2021 A level Paper 1 suggested worked solutions 17 Ans: C In absence of sunlight, FRS do not occur, In presence of iron-containing catalyst (a halogen carrier), an electrophile is generated. Since Cl is more electronegative than Br in Cl−Br, Br+ electrophile is produced, and it react with benzene via electrophilic substitution. 18 Ans: B Statement 1: incorrect, ionisation energy has to do with gaseous species, so it does not explain the difference in rate of reaction. Statement 2: correct, the bond pair electron in C−Br bond experiences weaker attraction from the nucleus, leading to smaller BE of C−Br, and rate of reaction is faster. Statement 3: correct, as the radius of bromine is bigger, the orbital size gets larger and the degree of orbital overlap gets less effective, resulting in a longer and weaker C−Br bond and thus a faster rate of reaction. Statement 4: incorrect, proton number does not directly affect the strength of C−X bond. C−Br bond (280 kJ mol−1) is weaker than C−Cl bond (340 kJ mol−1). 19 Ans: D Student P: incorrect, the negative test only indicates the absence of chloroalkane, bromoalkene and iodoalkane, the compound may not be a halogenoarene. Other organic functional group also has no reaction with warm silver nitrate. Student Q: incorrect, it may be a fluoroalkane with strong C−F bond, and hence inert towards nucleophilic substitution. 20 Ans: C Ester is present in DOTP, which can be formed from the reaction between an a