2025 DHS H2 Chem Prelim Paper 1 Worked Solutions

2025 DHS Preliminary Examination H2 Chemistry 9729 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 6 Answer Key 1 2 3 4 5 6 7 8 9 10 D B A C D A B D A C 11 12 13 14 15 16 17 18 19 20 D C B A C B B D D A 21 22 23 24 25 26 27 28 29 30 D C D C A C D B A C 1 D  A Since both particles have a positive angle of deflection, they are deflected in the same direction towards the same plate. Similar to 4He2+, X must be positively charged.  B The beam of positively charged particles X travels in a curved path towards the negatively charged plate.  C Charge to mass ratio of 12C+ = + 1 12 Charge to mass ratio of 4He2+ = + 2 4 = + 1 2 Since angle of deflection is proportional to charge to mass ratio, the angle of deflection of 12C+ is expected to be 1 6(1) = +0.17  D None of the above options are correct. 2 B 1. Determine the number of neutrons in Sr (find proton number of Sr using Data Booklet.) ⇒ 84 – 38 = 46 2. Find number of electrons present in Sr 2+ (using proton number of Sr given in Data Booklet.) ⇒ 38 – 2 = 36 ⇒ W has 36 electrons 3. Find number of protons present in W W: 36 protons 4. Calculate the nucleon number of W W: 36 + 46 = 82 The outer electronic configuration of W is 4s24p6 and hence a total of four orbitals (one 4s and three 4p) are occupied in the valence shell. 3 A (1, 2 and 3 only) The molecules have the following linear structures: O=C=O, S=C=S and O=C=S  1 C=O and C=S are polar covalent bonds due to the difference in electronegativity between the atoms.  2 Each C=O and C=S bond contains one sigma and one pi bond so each molecule contains two sigma and two pi bonds.  3 The dipole moments of the polar bonds are equal and opposite in O=C=O and S=C=S molecules so they cancel out and the molecules are non -polar. There exists only instantaneous dipole –induced dipole ( id–id) interactions between these non –polar molecules. As O is more electronegative than S, C=O bond is more polar than C=S bond and there is a net dipole moment in O=C=S. Hence, COS molecule is polar and it has both id –id and permanent dipole –permanent dipole interactions between molecules.  4 O=C=S has a smaller, less polarisable electron cloud than S=C=S. Hence, O=C=S has weaker id-id interactions between molecules than S=C=S. 5 D 2H2S + 3O2 → 2SO2 + 2H2O CS2 + 3O2 → CO2 + 2SO2 Hence, SO2 : CO2 will be 4 : 1.  Options A & C are incorrect. In the 60 cm 3 mixture, there is 40 cm 3 of H 2S and 20 cm3 of CS2. 40 cm3 of H2S will form 40 cm3 of SO2. 20 cm3 of CS2 will form 20 cm3 of CO2 and 40 cm3 of SO2. Hence, total volume of acidic gases, CO2 and SO2, = 40 + 20 + 40 = 100 cm3 These acidic gases will react with NaOH(aq) and cause a reduction in gas volume. 6 A  A pV = nRT = mRT/M  density = m/V = pM/RT Since density = pM/RT and M, T and R are constants, density is directly proportional to p  graph of density against p is a n upward sloping straight line starting from the origin This graph does not represent the behaviour of a fixed mass of an ideal gas. 4 C  A Ice has a simple molecular structure with hydrogen bonds between water molecules. There are covalent bonds between atoms in the water molecules.  B Iodine, I2, has a simple molecular structure and is a non -polar molecule. There exist s instantaneous dipole –induced dipole (id –id) interactions between molecules and covalent bond between iodine atoms in a molecule.  C Sodium nitrate, NaNO 3, has a giant ionic structure with ionic bonds between oppositely charged Na + and NO 3− ions. Within the NO 3− ion, there are covalent bonds between N and O atoms.  D Manganese has a giant metallic structure with metallic bonds between the positively charged metal cations and sea of delocalised electrons.

Dunman High School 2025 DHS Preliminary Examination – H2 Chemistry 9729/01 Solutions © DHS Chemistry Unit Page 2 of 6  B Since pV = nRT and n, p and R are constants, V is directly proportional to T (in K)  graph of V against T is an upward sloping straight line starting from the origin  C Since pV = nRT and n and R are constants, pV/T = constant  graph of pV/T against p is a horizontal straight line.  D Since pV = nRT and n, V and R are constants, p/T = constant  graph of p against p/T is a vertical straight line. 7 B Oxidation half-equation: Fe2+ → Fe3+ + e– Fe2+ ≡ e– amount of electrons transferred = 30.0 1000  0.200 = 0.00600 mol amount of electrons amount of nitrate ions = 0.00600 0.00200 = 3 1 The oxidation state of nitrogen decreases by 3 from +5 in NO3– to +2.  A The oxidation state of N in N2 is 0.  B The oxidation state of N in NO is +2.  C The oxidation state of N in NO2– is +3.  D The oxidation state of N in NO2 is +4. 8 D K is phosphorus as the P3– has the largest ionic radii amongst the Period 3 ions. The cations are smaller has they have one less shell of electrons compared to the anions. Amongst the isoelectronic anions, P 3– has the lowest nuclear charge hence the weakest electrostatic forces of attraction for its valence electrons. L is silicon. Large amount of energy is required to break the strong and extensive covalent bonds between Si atoms in the 3 -dimensional network structure. M is magnesium. It has the second highest number of delocalised electrons in its solid lattice, hence the second highest electrical conductivity behind aluminium. In order of increasing atomic number: M (magnesium) < L (silicon) < K (phosphorus) 9 A (1 and 2 only)  1 Electronegativity decreases down every group as the number of electron shells increases. This leads to an increase in screening effect. Despite the increase in nuclear charge, the ability of the atom to attract bonding electrons decreases.  2 Reducing power of Group 2 elements increases as E(M2+/M) gets increasingly negative.  3 As the size of electron cloud increases down the group, polarisability increases. More energy is required to overcome the stronger instantaneous dipole –induced dipole interactions between X 2 molecules. Boiling point increases and volatility (ease of vaporisation) decreases. 10 C  A 2Na(s) + ½O2(g) → Na2O(s) The number of moles of gases decreases from ½ to 0 per mole of Na 2O(s) formed from its constituent elements at standard conditions.  B Mg(s) + ½O2(g) → MgO(s) The number of moles of gases decreases from ½ to 0 per mole of MgO(s) formed from its constituent elements at standard conditions. The Sf of MgO(s) is expected to be similar to that of Na 2O(s) since the change in number of gaseous particles is the same for both reactions.  C Si(s) + O2(g) → SiO2(s) The number of moles of gases decreases from 1 to 0 per mole of SiO 2(s) formed from its constituent elements at standard conditions. The Sf of SiO2(s) is the most negative as its formation results in the largest decrease in number of gaseous particles.  D S(s) + O2(g) → SO2(g) The number of moles of gases does not change as one mole of SO2(g) is formed from its constituent elements at standard conditions. 11 D The magnitude of lattice energy of an ionic compound is dependent on the product of ionic charge and the sum of ionic radii: |L.E.| | q+q– r++ r– |  A For TiF3, |L.E.| | (+3)(–1) 0.067 + 0.136| = 14.8  B For FeF3, |L.E.| | (+3)(–1) 0.055 + 0.136| = 15.7  C For TiO, |L.E.| | (+2)(–2) 0.086 + 0.140| = 17.7  D For FeO, |L.E.| | (+2)(–2) 0.061 + 0.140| = 19.9 12 C heat transferred to water = mc|T| = 500  4.18  7.5 = 15675 J heat released by combustion = 100 70  15675 = 22392 J = 22.392 kJ energy released per gram of fuel burnt = 22.392  0.7 = 32.0 kJ g–1

Dunman High School 2025 DHS Preliminary Examination – H2 Chemistry 9729/01 Solut