2025 RI H1Chem Prelims P1 P2 Answers

1 Q1 no. of protons no. of neutrons no. of electrons 32S2− 16 16 18 34S 16 18 16 Ans: A Q2 Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶  1 The third shell has no unpaired electrons. 2 The valence electronic configuration of an isolated atom is 3s2 3p6. 3 The 2p orbitals have the same energy (degenerate) and same shape (dumbbell). Ans: B (1 only) Q3 electronic configuration A Ti2+ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d²  B Zn2+ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ C Cu2+ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ D Fe2+ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ Ans: B Q4 Ans: D Q5 The reaction involves the cleavage of the C−X bond. The weaker the C−X bond, the easier it can be broken and the faster is the rate of reaction. 1 While permanent dipole -permanent dipole forces between chloroethane molecules are stronger than those between bromoethane molecules, it does not contribute to the ease of breaking the C−X bond. species bonding A Al2O3 ionic B SO3 covalent only C NH4Cl ionic, covalent and dative  D H3O+ covalent and dative Suggested Answers for 2025 Y6 H1 Chemistry Preliminary Examination Paper 1: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A B B D D D C C C C D C C B C B C A A C 21 22 23 24 25 26 27 28 29 30 B B C B C D A B C C Worked solutions for Paper 1 Al 3+ 2− O 2 3 S O O O N H H H H + Cl − O H H H +

2 2 Though C−Cl bond is more polar than C −Br bond, it does not increase the reactivity as the rate of reaction is observed to be more rapid with bromoethane than chloroethane.  3 Since C−Cl bond is stronger than C−Br bond, more energy is required to break the C −Cl bond, this leads to a slower rate of reaction in chloroethane. Ans: D (3 only) Q6 Liquefaction of gases can be achieved by: • increasing pressure on a gas to bring the gas molecules closer together so that the strength of the intermolecular forces becomes more significant. • decreasing temperature so that sufficient energy is removed for the gas molecules to move at a slower speed and to make the gas change from the gaseous to the liquid state. The ease of liquefaction of gases generally depends on the strength of intermolecular forces . The stronger the intermolecular forces between the gas molecules, the easier it will be to liquefy a gas. Ans: D Q7 physical state room temperature conductivity solubility in water ionic solid non- conductors in solids, conductors in molten / aq usually soluble in polar solvents, insoluble in non– polar solvents metallic solid, except for Hg conductors in solids / liquids insoluble in polar and non– polar solvents giant molecular solid non- conductors, except graphite insoluble in all solvents. simple molecular mostly gases/ volatile liquids non- conductors when solid, liquid and usually in aq solutions. A few (e.g.HCl) ionise in water so that they conduct electricity in aq solutions. many are insoluble in polar solvents, soluble in non–polar solvents Ans: C Q8 Steps to take: 1. convert mg to g (divide mg by 1000) 2. divide by Mr (divide by 71) to get amount (no. of moles) of Cl2 molecules 3. multiply the amount by Avogadro’s constant to obtain number of Cl2 molecules ( 6  1023) Since density is 0.005 mg dm-3 In 1 dm3, the mass of Cl2 in air = 0.005g Hence: 230.005 1 × 6 × 101000 71  Ans: C Q9 CxHy(g) + (x + y/4)O2(g) ⎯→ xCO2(g) + (y/2)H2O(l) Amt of O2 used = 0.720/24 = 0.03 mol Amt of H2O produced = 0.36/18.0 = 0.02 mol Taking mole ratio: Q : water = 1 : y/2 = 0.0050 : 0.02  y = 8 O2 : Q = x + 8/4 : 1 = 0.03 : 0.0050  x = 4  Q is C4H8. Ans: C Q10 Heat released = (100)(4.18)(15.0) = 6270 J 1.0Amt of Mg = 24.3 = 0.0411 mol Amt of HCl = 100/1000 x 1 = 0.1 Since 1 mole of Mg reacts with 2 moles of HCl, Mg is limiting. No. of moles of H2 formed = no. of moles of Mg reacted = 0.0411 Heat released = − 6270 0.041152 = − 152361 J mol−1 = − 152 kJ mol−1 Ans: C

3 Q11 A The activation energy for the forward reaction of step 1 is lower than that of step 2. B Step 2 is less exothermic than step 1. C The activation energy for the backward reaction of step 1 is +121 kJ mol–1.  D The enthalpy change of reaction for the conversion of W to Y is 55-168 = −113 kJ mol−1. Ans: D Q12 By Hess’ Law, 4BE(S-F) + 158 = -434 + 6BE(S-F) BE (S-F) = 296 kJ mol−1 Ans:C Q13 Hr = nHf (products) − mHf (reactants) ∆Hr = 6(−394) + 6(−286) – (−1273) = −2807 kJ mol−1 Ans: C Q14 100 µg cm-3 ⎯⎯→ 50 µg cm-3 ⎯⎯→ 25 µg cm-3 (after 4 hours) 12 t = 2 hours 200 µg cm-3 → 100 µg cm-3 → 50 µg cm-3 → 25 µg cm-3 3 x 12 t = 6 hours Ans: B Q15 A Rate constant is increased by • increase in temperature • lowering of activation energy B A catalyst is a substance which increases the rate of a reaction by providing an alternative pathway with lower activation energy without itself undergoing any permanent chemical change.  C The kinetic energy of the molecules is affected by the temperature, not catalyst. Hence the introduction of catalyst does not make the particles move more quickly. D Since more particles have E greater or equal to Ea, the frequency of successful collision increases Ans: C Q16 N2(g) + 3H2(g) ⇌ 2NH3(g) H = –92 kJ mol–1 Statement 1 incorrect: a catalyst increases rate of reaction but does not affect the yield Statement 2 incorrect: a higher pressure favours the side with fewer (not greater) number of molecules Statement 3 correct: an increase in temperature always leads to increased rate Statement 4 incorrect: Haber process is exothermic. Applying Le Chatelier’s principle, a lower temperature (not higher) will give greater yield of ammonia. Ans: B (3 only) Q17 Given that the volume of the flask is 1 dm3, N2O4(g) ⇌ 2NO2(g) Initial conc/ mol dm-3 0 4/1 = 4 Change in conc. / mol dm-3 +x 4 – 2x Eqm conc. / mol dm-3 1.0 4 – 2(1.0) = 2.0 (Since x = 1.0) Kc = [NO2]2 / [N2O4] = 2.02 / 1.0 = 4.0 Incorrect answers: • Reverse K expression, resulting in incorrect K = 1.0/2.02 = 0.25 12 t 12 t 12 t 12 t 12 t

4 • Failed to square the numerator, resulting in incorrect K = 2.0/1.0 = 2.00 • Failed to −2x for NO2, resulting in eqm amount of NO2 = 3.0, incorrect K = 3.02/1.0 = 9.00 Ans: C Q18 Dynamic equilibrium refers to a state in a reversible system in which the forward and backward reactions are continuing at the same rate, resulting in no net change in the macroscopic properties of the reactants and products. Ans: A Q19  A + -pH -4.0 -4 -3 [H ] = 10 = 10 = 1 x 10 mol dm Since [HX] > [H+], HX undergoes partial dissociation. Hence it is a weak acid. B 4.0 2 -7 [H ] A ][ [HA] (10 ) 0.100 =1 x 10 aK +− − = = C 4.0 -4 -3 [H ] 10 10 =1 x 10 mol dm pH+− − = = D When 25.0 cm3 of 0.100 mol dm−3 NaOH(aq) is added to the solution, HX is completely neutralised to give salt and water. Ans: A Q20 Recall the following definitions Arrhenius acid hydrogen-containing substance that releases hydrogen ions Arrhenius base hydroxide-containing substance that releases hydroxide ions Brønsted- Lowry Acid proton donor Brønsted- Lowry Base proton acceptor Ans: C Q21 Statement 1 incorrect: electronegative of X decreases down the group, so the bonding pair of electrons gets further away (not closer) from the