2023 SOTA Yr 6 Chem HL Prelim Exam MS1 2

Mark scheme 2023 HL Prelim P1 P2 Question 1 2 3 4 5 6 7 8 9 10 Answer D C D A B D B B B A Question 11 12 13 14 15 16 17 18 19 20 Answer A A D B A D D B C D Question 21 22 23 24 25 26 27 28 29 30 Answer B C D C C D D C B D Question 31 32 33 34 35 36 37 38 39 40 Answer A C C A C B B A C C Qn Mark scheme Remark 1a chromium: n(Cr) = !".$%&'.(( % *+,!" = 1.32 mol OR oxygen: n(O) = -..!.!.(( = 1.98 mol ✔ ratio n(Cr) : n(O) = 1.32 : 1.98 = 1 : 1.5 = 2 : 3 empirical formula: Cr2O3 ✔

Let the molecular formula be Cr2xO3x. 156 = 2(52.00)x + 3(16.00)x x = 1 molecular formula: Cr2O3 ✔ Allow ecf for empirical and molecular formula. Award [3] for correct final answer. 1b i «promoted» electrons fall back to lower energy level ✔ Energy difference between levels is different ✔ Accept “Cu, Ca and Cr have different nuclear charge” for M2 1bii energy levels are closer together at high energy / high frequency / short wavelength ✔ 1biIi IE = ΔE = hν = 6.63 × 10–34 J s × 1.64 × 1015 s–1 = 1.09 × 10–18 J ✔ 1ci 1s2 2s2 2p6 3s2 3p6 3d5 4s1 OR 1s2 2s2 2p6 3s2 3p6 4s1 3d5 ✔ Do not accept condensed electron configuration like [Ar] 3d5 4s1 1cii Both are paramagnetic. ✔ «both» contain unpaired electrons OR each chromium atom has 6 unpaired electrons and each chromium(III) ion has 3 unpaired electrons. ✔ Accept orbital diagrams showing unpaired electrons.

1ciii H2O is LB as it donates electron pair OR Cr3+ is LA as it accepts electron pair iv Any THREE from: partially filled d-orbitals ligands/water cause d-orbitals to split red light is absorbed as electrons move to a higher energy orbital «in d–d transitions» OR light is absorbed as electrons are promoted «green» colour observed is the complementary colour «of red» 2ai. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ✔ aii. [(-393.50 + (2 x – 285.8 ) – ( -74.0)] ; ✔ -891.1 <<kJmol-1>> ✔ Award 2m for correct final answer; Award 1 m for <<+>> 891.1 kJmol-1 aiii. BE assume all bonds are in gaseous state / H2O formed is in liquid form for literature value/ OWTTE ✔ bi. Methane AND tetrachloromethane is non-polar AND Dichloromethane is polar; ✔ C-Cl bond is polar <<C-H bond is non-polar>>; ✔ In dichloromethane, dipole moment/bond polarities does not cancel out OR In tetrachloromethane, dipole moment/bond polarities cancels out; ✔ bii. London dispersion forces is stronger in tetrachloromethane due to greater number of electrons <<larger molecular mass>> ✔ biii. Methane cannot form hydrogen bond/favourable intermolecular forces of attraction/interaction with water. ✔ ci. (3 x 498) – (4 x O-O in ozone) = <<+>> 285.4; ✔ (O-O in ozone) = << [(3 x 498) - 285.4] / 4 >> = <<+>> 302 <<kJ>> ✔ Award 2m for correct final answer

cii. ; ✔ FC ( -1 +1 0) ✔ ciii. <<Bond energy is greater than O-O single but weaker than O=O double bond>> Both bonds are identical in bond length << due to resonance>>, longer than O=O single bond AND shorter than O-O single bond; ✔ Bond order is 1.5/ between 1 and 2 ✔ di. ΔH1¡ : enthalpy change of formation <<of MgO>> ✔ ΔH2¡: lattice energy << of MgO>> ✔ dii. 3791 = 602 + 249 + 148 + 738 + 2nd IE + 612; ✔ 2nd IE = <<+>> 1442 <<kJ mol-1>> ✔ Award 2m for correct final answer Award 1m for -1442 diii. 2nd IE more endothermic than 1st IE as it is more difficult/ energy needed to remove an electron from a cation/ OWTTE ✔ div. Down the group, lattice enthalpy decreases/less endothermic; ✔ Radius of M2+ increases; ✔ O-O=O

3a (i) (ii) b Pressure/Concentration of NO will double ; ✔ rate increases by 4 times; ✔ c Rate = k [NO]2 mol dm-3 s-1 = k (mol dm-3)2 units of k = mol-1dm3s-1 ; ✔ d Increases in <<average>> kinetic energy of the particles; ✔ More particles have energy > Ea ; ✔ Increase in frequency of effective collisions ; ✔ 4ai C5H5N + HCl à C5H5NHCl/ C5H5NH+Cl- ✔ aii pyridinium ion is a Bronsted-Lowry acid as it the conjugate acid of the weak Bronsted base, pyridine. / pyridinium ion is a Bronsted-Lowry acid as it can act as a proton donor. ✔ [NO] [H2]

bi C5H5NH+Cl- + NaOH à C5H5N + NaCl + H2O Amount of C5H5N = 25.00/1000 x 0.100 = 2.50 x 10-3 <<mol>> ✔ Concentration of C5H5N at the equivalence point = 2.50 x 10-3/ (50.00/1000) = 0.0500 <<mol dm-3>> ✔ C5H5N + H2O ⇌ C5H5NH+ + OH- Kb = [C5H5NH+ ][OH-] [C5H5N] 1.4 x 10-9 = [OH-]2 0.0500 [OH-] = 8.367 x 10-6<<mol dm-3>> ✔ pOH =5.08 pH = 14 -pOH= 8. 92 ✔ Award [4] for correct final answer.

bii non-symmetrical sigmoidal curve, starting pH 2–7 AND terminating pH>12 ✔ equivalence point pH approximately 8.92 AND at a volume 25.00 cm3 ✔ ECF (acid or base) from a(ii) c (I) The position of the equilibrium will shift to the right hand side with the smaller number of moles of gaseous particles to decrease the total pressure ✔ (II) The position of the equilibrium remains unchanged as the rates of the forward and backward reactions are increased by the same extent. ✔ di Yield is very small as POE lies to the left/equilibrium constant is <<1/very small number. ✔ dii DG = -RT ln K = - 8.31 x 500 x ln (3.80 x 10-16 ) ✔ = <<+>>148 000 J mol-1 = <<+>>148 kJ mol-1 ✔ [1] for correct working [1] for conversion to kJ 25.00 Volume of NaOH added/ cm3

5a. Left electrode is anode AND right electrode is cathode ✔ b. Electrons through the wires/ cables; ✔ Ions towards the electrodes in the electrolyte/cation towards cathode and anion towards anode ✔ ci. Cathode: Pb2+ + 2e- → Pb ✔ Anode: H2O → ½ O2 + 2H+ + 2e- ✔ [1] for reversed equations at electrodes Accept if student shows reduction of OH- instead of H2O cii. Pb2+ + H2O → ½ O2 + 2H+ + Pb OR 2Pb2+ + 2H2O → O2 + 4H+ + 2Pb ✔ d. Hydrogen gas would be formed at cathode/ water will be reduced at cathode ; ✔ Reduction potential of water is less negative (-0.83V) than manganese (-1.18V) ✔

6a Signals : 2 ✔ Ratio: 9:3 or 3:9 or 3: 1 or 1:3 ✔ b Any one of the following: ✔ ci HNO3 + 2H2SO4 ⇌ H3O+ + NO2+ + 2HSO4– ✔ Accept: HNO3 + H2SO4 ⇌ NO2+ + HSO4– + H2O. Accept: HNO3 + H2SO4 ⇌ H2NO3+ + HSO4–. Accept single arrow instead of equilibrium sign. Accept equivalent two step reactions in which sulfuric acid first behaves as strong acid and protonates nitric acid, before behaving as dehydrating agent removing water from it.

cii curly arrow going from benzene ring to N «of +NO2/NO2 + » ✔ carbocation with correct formula and positive charge on ring ✔ curly arrow going from C–H bond to benzene ring of cation ✔ formation of organic product nitromesitylene AND H+ ✔ Accept mechanism with corresponding Kekulé structures. Do not accept a circle in M2 or M3. Accept first arrow starting either inside the circle or on the circle. If Kekulé structure used, first arrow must start on the double bond. M2 may be awarded from correct diagram for M3. M4: Accept “C6H5(CH3)3NO2 +H2SO4” if HSO4− used in M3. 7ai Cinnamaldehyde AND Absence of << strong, very broad band >>peak at 2500–3000 «cm–1»/peak due to O–H/hydroxyl in carboxylic acids✔ CH3 CH3 H3C CH3 CH3 CH3 CH3 CH3 CH3 H3C H3C H3C