RI 2023 Y6 H2 Chemistry Common Test Suggested Solutions with Examiner Comments

1 2023 Year 6 H2 Chemistry Common Test Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A C D A D B B A A D C B B D B Question 1 (A) Angle of deflection ⍺ charge mass Since the electron has the smallest mass ( 1 1840 that of proton), it has the largest angle of deflection. Question 2 (C) There is a large increase from the 7th to 8th ionisation energy. Significantly more energy is required to remove the 8th electron as it is located in an inner shell. Hence, the element has 7 valence electrons i.e. chlorine. Question 3 (D) Since the dipole moment of the highly polar C=O is not completely cancelled out in X and Y, X and Y have greater polarity than W and Z, which have no highly polar bonds i.e. polarity of W, Z < X, Y. Since the dipole moment of the C=O in Y is partially cancelled out by the dipole moments of the 2 C–Cl bonds, the overall dipole moment in Y is lower than in X i.e. polarity of Y < X. The dipole moments in Z are exactly cancelled out, while the dipole moments in Y are not. Hence, Z is less polar than W i.e. polarity of Z < W. Question 4 (A) Comparing expts 2 and 3, when [H+] x 2, rate x 4 i.e. order of reaction wrt H+ = 2. Comparing expts 3 and 4, when [BrO3–] x 2, rate x 2 i.e. order of reaction wrt BrO3– = 1. Comparing expts 1 and 3, when [Br–] x 2 , rate x 2 i.e. order of reaction wrt Br– = 1.

2 Question 5 (D) 1 Correct. As HBr is formed, [HBr] in the denominator of the rate equation increases, causing the rate to decrease i.e. rate slows down. 2. Incorrect. In a propagation step, a radical is consumed as a reactant and another radical is produced as a product. Hence, step 2 is also a propagation step, in addition to steps 3 and 4. 3. Incorrect. [Br2] appears in both the numerator and denominator . Do ubling its concentration does not double the rate due to the powers which each [Br2] is raised to. Question 6 (B) From the initial buffer provided, 4.46 = pKa + lg [ethanoate] [ethanoic acid] 4.46 = pKa + lg( 1 2) --- (1) Let x be the no. of mols of NaOH added to the buffer to achieve pH 4.60. Amt / mol CH3COOH + NaOH CH3COO– Na+ + H2O Initial 2(2) x 2(1) -- Change –x –x +x -- final 4–x 0 2+x After adding x mol of NaOH, 4.60 = pKa + lg( 2 + x 4 - x ) --- (2) Equation (2) – (1) gives: 4.60 – 4.46 = lg( 2 + x 4 - x ) – lg( 1 2) 0.14 = lg( 2 + 𝑥𝑥 4 − 𝑥𝑥 ÷ 1 2 ) 100.14 = 2(2 + x) 4 - x x = 0.4503 mol mass of NaOH = 0.4503 x 40.0 = 18.0 g. Question 7 (B) A Incorrect. If HA is a weak acid, then Ka(HA) x Kb(conjugate base of HA) = Kw. This is because Ka = [H+][A−] [HA] and Kb(conjugate base of HA) = Kb(A–) = [HA][OH−] [A−] . Ka(HA) x Kb(conjugate base of HA) = [H+][A−] [HA] x [HA][OH−] [A−] = [H+][OH–] = Kw. For this statement involving water to be correct, it should read either • Ka(H2O) x Kb(OH–) = Kw, or • Ka(H3O+) x Kb(H2O) = Kw. B Correct. When an excess of a weak base reacts with a strong acid, the resultant solution contains a mixture of unreacted weak base and the conjugate acid of the weak base, resulting in a buffer solution. C Incorrect. When large volumes of water are added to an acidic buffer, the pH gradually approaches the pH of water. D Incorrect. Sulfuric acid is considered a strong acid. When titrated against NaOH(aq), only one region of rapid pH change (i.e. one equivalence point) will be observed.

3 Question 8 (A) A Correct. While the addition of concentrated HC l increases [Cl–] and should cause more AgCl to precipitate, the AgCl ppt dissolves instead. Students need to recognise that this increase in solubility of the ppt must be due to the formation of a soluble complex, even if they do not know the formula of the complex. In this case, the soluble complex formed is [AgCl2]–. B Incorrect. Cations, such as Ag+, form complexes with neutral species (e.g. NH 3) or anionic species (e.g. Cl–), not cations. C Incorrect. The common ion effect, exerted by the presence of the C l– common ion, would have caused more ppt to form. D Incorrect. AgCl does not react with H+. Question 9 (A) A With the lowest 1st IE, this suggests that element R is Na. B Since the chloride of R reacts with water to give an acidic solution, element R could be Mg, Al, Si or P. C Since the oxide of R is a solid at room temperature, element R could be Na, Mg, A l, Si or P. D The chlorides of Na, Mg and Al dissolve in water to form the corresponding cations and chloride ions. For SiCl4, the following reaction occurs to form its corresponding oxide: SiCl4 + 2H2O → SiO2 + 4HCl. PCl 5 reacts with water to form H3PO4 and HCl. This option implies that element R is Si. Hence, option A does not fit with the other three. Question 10 (D) To obtain the structure of the octapeptide, align the overlapping sections of the dipeptide and tripeptide fragments. Gly-Asn-Tyr Tyr-Asn-Tyr Tyr-Leu-Tyr Tyr-Arg Gly-Asn-Tyr-Asn-Tyr-Leu-Tyr-Arg Question 11 (C) Reaction with excess concentrated H2SO4 causes some alcohols to undergo elimination. The carbon next to the carbon bearing the -OH group needs to contain a hydrogen for elimination to take place.

4 Question 12 (B) The products of reaction with the stated reagents and conditions are shown. 1 Correct. The aldehyde and primary alcohol are oxidised by hot acidified K2Cr2O7 to form –COOH. 2. Correct. The aldehyde and carboxylic acid are reduced by LiAlH4 to form the corresponding primary alcohols. 3. Incorrect. NaBH4 reduces carbonyl compounds only and hence the aldehyde is reduced to the corresponding primary alcohol. However, NaBH4 is unable to reduce the carboxylic acid. Question 13 (B) The energy profile diagram has two peaks, indicating that the reaction proceeds via a two-step mechanism. 1 Correct. (CH3)3CCl is a tertiary chloroalkane. Hence, its reaction with KOH proceeds via the two-step SN1 mechanism. 2. Incorrect. CH3CH2Cl is a primary chloroalkane. Hence, its reaction with KOH proceeds via the SN2 reaction which is a one-step mechanism. 3. Correct. The reaction between CH 3CHO and HCN is a nucleophilic addition reaction which proceeds via a two-step mechanism. Question 14 (D) 1 Incorrect The 2° alcohol and alkene undergoes oxidation, decolourising the purple KMnO4. The ethyl side chain undergoes oxidation, decolouring the purple KMnO4. 2 Correct The 2° alcohol undergoes oxidation, causing the orange K 2Cr2O7 to turn green. The ethyl side chain is not oxidised by K2Cr2O7. Hence, there is no colour change. 3 Correct There is no phenol group present. Hence, there is no colour change. The neutral FeC l3 reacts with the phenol group, producing a violet colouration. 4 Incorrect The alkene undergoes electrophilic addition with Br 2(aq), decolourising the orange Br2(aq). The phenol group undergoes electrophilic substitution with Br2(aq), decolourising the orange Br2(aq). Question 15 (B) 1 Incorrect. To form an ester with phenol, an acyl chloride needs to be used. Phenol does not react directly with carboxylic acids. 2. Correct. Amides are formed from the reaction between an acyl chloride and a primary or secondary amine. 3. Incorrect. Acyl chlorides do not react with tertiary amines.

5 Section B 1 (a) (i) The half -life (t 1/2) of a reaction is the time taken for the concentration of a reactant to decrease to half its initial value / to be halved. Examiners’ Comments • This was generally well done. • For definitions, candidates should follow what is given in notes, as more often than not, when candidates rephrase, a different meaning to what is being described surfaces. (ii) time taken for [X] to decrease from 0.010 mol dm–3 to 0.005 mol dm–3 = 182.