2025 JC2 Prelims H1 Chem Paper 1 Solutions TJC

2025 TJC H1 Chemistry Prelim Paper 1 Worked Solutions 1 2 3 4 5 6 7 8 9 10 B A C B A B B C A D 11 12 13 14 15 16 17 18 19 20 B B C D D D D A A B 21 22 23 24 25 26 27 28 29 30 C A D C B A D C D C 1 Answer: B Al : 1s22s22p63s23p1 Al3+ : 1s22s22p6 Ca : 1s22s22p63s23p64s2 Ca2+ : 1s22s22p63s23p6 Cu : 1s22s22p63s23p63d104s1 Cu2+ : 1s22s22p63s23p63d9 S : 1s22s22p63s23p4 S2– : 1s22s22p63s23p6 Only options A, B and D have noble gas configuration. Only options A and B have more protons than electrons Option A involves removing electron from 3p and 3s orbitals 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 hydrogen bonds between two acid molecules. The apparent M r (120.0) is double than of its actual Mr (60.0). 5 Answer: A A Sr: giant metallic structure Al2O3: giant ionic lattice structure I2: simple molecular structure B Be: giant metallic structure Al2Cl6: simple molecular structure SO3: simple molecular structure C Al: giant metallic structure C: giant molecular structure SiO2: giant molecular structure D Ni : giant metallic structure PCl3: simple molecular structure SiCl4: simple molecular structure 2 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 3 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 4 Answer: B Option A : Water molecules are attracted to other water molecules via hydrogen bonding resulting in a regular three dimensional tetrahedral structure. The arrangement of water molecules in ice create an ‘open cage’ structure resulting in ice to be less dense than water. Option B : Both compounds have hydrogen bonding between molecules. However, volatility depends on the electron cloud size (inferred from M r) which affects the strength of id-id. 6 Answer: B A simple molecular solid should have a lower 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.

7 Answer: B Option 1 has a dative bond from O to C while option 3 has a dative bond from Cl to Al. Option 2 has no dative bond present. 12 8 Answer: C At standard temperature and pressure, graphite is thermodynamically more stable than diamond. However, the conversion from diamond to graphite, has a very high activation energy and therefore, takes place at an extremely slow rate. 13 Answer: C Y: standard enthalpy change of combustion is always negative Z: standard enthalpy change of combustion of C and H2 are negative X: by Hess’ Law, X = –890 – (–964) = +74 9 Answer: A CxHy + (x+y/4) O2 → x CO2 + (y/2) H2O 10 excess 0 0 0 40 H2O is a liquid at room temperature, hence does not contribute to gaseous volume. CO2 in the residual gas is acidic and is absorbed by KOH. x = 40/10  x = 4 Volume of unreacted O2 = 80 – 40 cm3 = 40 cm3 Volume of reacted O2 = 100 – 40 cm3 = 60 cm3 x + y/4 = 60/10 4 + y/4 = 6 y = 8 14 Answer: D 1: correct as there will be more molecules with energy greater than or equals to Ea. 2: incorrect as Ea is only affected by presence of catalyst. 3: incorrect as total number of molecules remain the same when temperature changes. 15 Answer: D Rate = k[W][X]2 Using run 1 data, 0.0064 = k(0.02)(0.015)2, k = 1420 16 16 Answer: D A is incorrect Rate = k[Br2][HCOOH] k = Rate / [Br2][HCOOH] For units: mol dm-3 s-1/(mol dm-3)2 = mol–1 dm3 s–1 B is incorrect. For overall second order reaction, half-life is not constant. C is incorrect as the mole ratio of 1Br 2≡2H+, so rate of formation of H + should be twice that of disappearance of Br2. D is correct. Rate = k[Br2][HCCOH] The rate doubles when concentration of methanoic acid doubles so the rate of product formed will also be doubled. 10 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 11 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.

17 Answer: D Options 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. 21 Answer: C No of moles of excess NaOH = 0.0040 – 0.0025 = 0.0015 mol [NaOH] = 0.0015/2 = 7.5 x 10−4 mol dm−3 pOH = −lg(7.5 x 10−4) = 3.13 pH = 14 – 3.13 = 10.88 (10.9) 22 Answer: A H2O ⇌ H+ + OH− When temperature increases, by Le Chatelier’s Principle, position of equilibrium shifts towards endothermic reaction to absorb heat. Since Kw increases with temperature, [H+] and [OH−] increase implying that forward reaction is favoured. Therefore, forward reaction (dissociation of water) is endothermic. (Statement 1 is correct) At 62 C, Kw = [H+][OH−] = 10−13 mol2 dm−6 [H+] = [OH −] = 3.16 x 10 −7 mol dm−3 (Statement 3 is incorrect, water remains neutral) pH = −log [H+] = 6.5 (Statement 2 is correct) 18 Answer: A Down the group, atomic radius increases and I.E. decreases  attraction of the nucleus for the valence electrons decreases  electrons can be removed more easily through ionisation (i.e. more readily oxidised)  reducing power increases 19 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. 23 Answer: D All three compounds are non -polar in nature, and they are constitutional isomers with the same Mr. The difference in their boiling points is not dependent on the size of electron cloud which is based on Mr. From the data, increased branching reduces the boiling point. The polarity of the alkane does not change due to branching. In fact, increased branching reduces the surface area of contact hence reducing the strength of intermolecular id-id attractions and thereby reducing the boiling point. 20 Answer: B Down the group, oxidising power of the halogens decreases  halogens are less easily reduced  halides are more easily oxidised Option A: Incorrect. Chloride ion is less readily oxidised and thus chloride ions are weaker reducing agent. Option B: Correct. Bromine is less reactive than chlorine and less easily reduced. It is unable to displace chloride. Only the more reactive halogen can displace the less reactive halogen from its aqueous halide ions. Option C: Incorrect. Down the group, size of electron cloud for the halogens increases, more energy required to overcome the stronger id -id attractions and boiling point increases. Thus volatility of halogens decreases down the group. Option D: Incorrect. Down the group, the H -X bon