2025 JC2 Prelims H2 Chem Paper 1 Solution TJC

H2 P1 REVIEW 1 2 3 4 5 6 7 8 9 10 A C B B C B D B B C 11 12 13 14 15 16 17 18 19 20 D A D A B B C C D C 21 22 23 24 25 26 27 28 29 30 A B B C D C D A A B 1 Answer: A Angle of deflection  charge/mass Option A : 2/4 = 0.5 Option B : 1/19 = 0.05 Option C : 1/28 = 0.036 Option D : 2/32 = 0.06 2 Answer: C Cu: 1s2 2s2 2p6 3s2 3p6 3d104s1 Mn: 1s2 2s2 2p6 3s2 3p6 3d5 4s2 Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 Fe2+: 1s2 2s2 2p6 3s2 3p6 3d6 3 Answer: B Option A : [F --- H – F ]–. There is hydrogen bond formed between H in HF and F– as indicated by ---. Option B : Both compounds have hydrogen bonding between molecules. However, volatility depends on the electron cloud size (inferred from Mr) which affects the strength of id-id. Option C : 2-nitrophenol can form intramolecular hydrogen bonding giving rise to less extensive H bonding between molecules and lower melting point. Option D : Ethanoic acid can dimerise in benzene by forming two H bonds between two acid molecules.

4 Answer: B A simple molecular solid should have a low er melting point (due to weak intermolecular forces) and does not conduct electricity (due to lack of mobile charge carriers – ions or free electrons). It may be soluble or insoluble in water depending on the type of intermolecular forces that it can form with water. Since the unknown compound is a solid at room temperature, the answer cannot be A due to the melting point. 5 Answer: C Option 1 has a dative bond from O to C while option 3 has a dative bond from N to O. Option 2 has no dative bond present. 6 Answer: B Vol of original gas mixture = 10 + 50 cm3 Vol of residual gas = ¼ x 60 = 15 cm3 Vol of unreacted O2 = 15 cm3 Vol of reacted O2 = 50 – 15 = 35 cm3 CxHy + (x + y/4)O2 xCO2 + y/2 H2O 10 35 x + y/4 = 35/10 4x + y = 14 x = 2, y = 6 7 Answer: D A: 17 + 4(8) + 1 = 50 e B: 2 + 16 + 4(8) = 50 e C: 16 + 4(8) + 2 = 50 e D: 50 – 2 = 48 e 8 Answer: B During thermal decomposition, heat is absorbed to the reaction, so the reaction is endothermic (option A or B). The activation energy is twice that of the enthalpy change, so option A is incorrect.

9 Answer: B Na(g) → Na+ (g) + e → Na2+(g) + 2e 10 Answer: C Rate = k[L][N] = k[L]2[M] 1: incorrect as Units for k = mol dm–3 s–1 / (mol dm–3)3 = mol–2 dm6 s–1. 2: correct as adding up the 3 elementary steps gives 2L + N → 2P 3: correct. Catalyst is consumed first then regenerated at the end of the reaction while an intermediate is produced and consumed at the end of the reaction. 11 Answer: D Rate = k[W][X]2 Using run 1 data, 0.0064 = k(0.02)(0.015)2, k = 1420 12 Answer: A Using pV = nRT 101000(400x10-6) = n(8.31)(300) n = 0.0162 mol Ar of gas = 40.1 (Ar) 13 Answer: D Option D: By Le Chatelier’s principle, when the experiment is conducted at a lower temperature, the position of equilibrium will shift to the right to release heat as forward reaction is exothermic (mole fraction of I2(g) should be lower). Option A: Incorrect as catalyst only speeds up the reaction (equilibrium reached in shorter time), there should not be changes in composition (mole fraction of I2(g) should remain the same). Option B: Incorrect as the mole fraction of I 2(g) at time = 0 is the same (0.5) as the first experiment. Option C: Incorrect as change in pressure does not affect the equilibrium position. w x – w x

14 Answer: A Options C and D: Incorrect as first IE of X is greater than that of Y, X should be above Y in the Periodic Table. Option B: Incorrect. Since phosphorus has simple molecular structure, the melting point of P is lesser than aluminium which has giant metallic structure. 15 Answer: B Eo(Cl2/Cl–) = + 1.36 V Eo(Br2/Br–) = + 1.07 V Eo(I2/I–) = + 0.54 V Option A: Chloride ion is less readily oxidised and thus chloride ions are weaker reducing agent. Option B: Ecello = Eredo – Eoxo = +1.07 – 1.36 < 0  reaction is not feasible Option C: A yellow precipitate that is insoluble (not partially soluble) in aqueous ammonia is formed when iodide (not iodine) is added to silver nitrate solution. Option D: Incorrect. When one drop of manganate(VII) is added to bromide, manganate(VII) decolourises and orange aqueous bromine is formed. 16 Answer: B Ionic product of ZnS = Ionic product of CuS = [M2+][S2−] = (0.1)(10−35) = 10−36 mol2 dm−6 For ZnS, ionic product < Ksp  No ppt observed For CuS, ionic product > Ksp  Ppt observed (Option B is correct) H2S ⇌ 2H+ + S2− ⎯ (1) M2+(aq) + S2−(aq) ⇌ MS(s) ⎯ (2) When pH is lowered, [H +] is increased, by LCP, position of equilibrium (1) shifts to the left. [S2−] decreases. Thus position of equilibrium (2) shifts to the left and MS(s) dissolves (less ppt formed). 17 Answer: C As the equivalence pH is less than 7, this is a reaction between strong acid (HCl) and weak base (NH3). [NH3] = 20×0.1 10 = 0.2 mol dm−3

18 Answer: C At 25 C, [H+] = [OH−] = 10−7 mol dm−3 pH = −log [H+] = 7 Kw = [H+][OH−] = 10−14 mol2 dm−3 pKw = -lg Kw = 14 Ka of H2O = [H+][OH−] [H2O] = 10−14 55.6 = 1.80 x 10−16 mol dm−3 pKa = −lg Ka = 15.7 (Statement 1 is incorrect) pH < pKw < pKa (Statement 2 is correct) When temperature increases, by Le Chatelier’s Principle, position of equilibrium shifts right to absorb heat since forward reaction is endothermic. [H+] and [OH −] will increase and Kw will increase and p Kw will decrease. (Statement 3 is correct) 19 Answer: D A CH2=C=CHCH3 contains one sp hybridised and one sp 3 hybridised C atoms. B CH2=CHCN contains one sp hybridised C atom but no sp 3 hybridised C atoms. C HOCH2C≡CH contains two sp hybridised and one sp3 hybridised C atoms. D CH3CH2CH=CHCN contains one sp hybridised and two sp 3 hybridised C atoms. 20 Answer: C The reaction involves breaking of C–X bond to release free X– ions which can then combine with Ag+ ions to form a precipitate of AgX. The rate of reaction depends on the strength of C–X bond. The time taken to observe precipitate for compound 1, 2 and 3 depends on strength of C–Cl, C–Br and C–I respectively and independent of C–F bond. Since C–I bond is the weakest of the three bonds, it is the easiest to break and will take the shortest time for the formation of precipitate. Since C– Cl bond is the strongest of the three bonds, it is the hardest to break and will take the longest time for the formation of precipitate.

21 Answer: A 22 Answer: B Step 1 is electrophilic substitution (Friedel Crafts alkylation) using halogenoalkane, (CH3)3CCl. Step 2 involves converting benzene to cycloalkane hence this is reduction (inferred from gain of H). Step 3 involves formation of ester from either the use of CH3COOH, warm with conc. H2SO4 or CH3COCl. 23 Answer: B 1. The ring contains more than 8 carbon atoms hence the C=C bond in the ring can exhibit cis-trans isomerism. Statement 1 is correct.2. The alkene functional group can react with Br 2 via electrophilic addition reaction. However, the molecular formula of the product should be C17H30OBr2 not C17H32OBr2 as only Br atoms are added across the C=C bond and there should be no change in number of H atoms. Statement 2 is incorrect. 3. Civetone does not exhibit intermolecular hydrogen bonding since there are no H atoms directly bonded to F, O or N atoms in the molecule. The dominant type of intermolecular forces should be permanent dipole - permanent dipole and instantaneous dipole-induced dipole attractions. Note: the presence of O atom in civetone can still allow it to form hydrogen bonding with other types of molecules. Statement 3 is correct.

24 Answer: C The three compounds undergo elimination to form alkenes. Only one of the compounds can form two final products with molecular formula C7H10. 25 Answer: D A: Incorrect. –NH2 group is 2 - and/or 4 - directing and