2024 Prelims RVHS H2 Chem P1 (Ans)

2024 JC 2 H2 Chemistry Prelim Exam Paper 1 Worked Solutions 1 D The decomposition involves: n → p+ + e− where the newly -formed proton stays in the nucleus. As the proton number of the atom increases by 1 following the decomposition, the element will no longer be the same. Options A and C are eliminated. Option D involves an increase in proton number by 1 going from K to Ca, hence it is the correct answer. 2 C Option A: By definition, one mole of a substance contains exactly 6.02  1023 (or Avogadro number) elementary entities. Option A is incorrect because of the phrase “same number of atoms as there are in 12.000 g of carbon-12”. Option B: By definition, relative isotopic mass (Ar) is the mass of one mole of atoms of an isotope (of a certain element) relative to 1 12 the mass of one mole of 12C atoms. Option B is incorrect because in the formula given, the numerator used is “average mass of all isotopes of lithium”. Option C: By definition, relative atomic mass ( Ar) is the average mass of one mole of atoms of an element relative to 1 12 the mass of one mole of 12C atoms. Note that some textbooks define relative atomic mass as “the average mass of one atom of an element relative to 1 12 the mass of one 12C atom”. Hence, option C is correct. Option D: By definition, relative molecular ( Mr) mass is the average mass of one mole of molecules relative to 1 12 the mass of one mole of 12C atoms. Option D is incorrect because in the formula given, the numerator used is “average mass of one atom of E”.

3 C Amount of CO2 = 48.0 24 000 = 0.00200 mol Amount of sodium percarbonate = 10.00.100 1000 = 0.00100 mol (Na2CO3)xy(H2O2) : CO2 x : 1 0.00200 : 0.00100 2 : 1 Therefore, x = 2 Amount of H2O2 = 0.00100y mol Amount of KMnO4 = 24.00.0500 1000 = 0.00120 mol Given the ratio is 2 : 5 0.00120 2 0.00100y 5= y = 3 Therefore, y3 x2= 4 D A: Due to the overlap of unhybridised 2p orbitals, mobile delocalised electrons are found in the lattice structure. B: Each carbon atom forms 3 sigma bonds with 3 other carbon atoms. C: Instantaneous dipole -induced dipole interactions exist between each graphite plane. D: Conduction of electricity occurs due to the overlap of unhybridised 2p orbitals. The overlapping occurs perpendicular to the axis of the unhybridised 2p orbitals.

5 A A: There is an unpaired electron on N atom in NO2 molecule. B: The Cu+ ion has an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 4s0. C: The Li+ ion has an electronic configuration of 1s2 2s0. D: After heterolytic fission, Cl+ and Cl ions are formed, which do not have a single unpaired electron. 6 D Using Boyle’s Law, p1V1 = p2V2. Upon connecting, the total volume of the container becomes 3 dm3. Helium: Neon: (2)(1) = p2(3) (1)(2) = p2(3) p2 = 2 3 kPa p2 = 2 3 kPa Final pressure = 2 3 + 2 3 = 4 3 kPa 7 B Option B is correct: Using Hess’ Law, −1561 = − (−85) + 2(−394) + 3(−243) − 3Hvap (H2O) Hvap (H2O) = +43 kJ mol−1

8 B Option A is incorrect: Hsoln does not accurately predict the solubility of a substance. Generally, the more negative Hsoln is, the more soluble the substance; the more positive Hsoln is, the less soluble the substance. Option B is correct: Hsoln = − LE + 2Hhyd (Hg+) + Hhyd (SO42−) = −(−2127) + 2(−625) + (−1160) = −238 kJ mol−1 Option C is incorrect: Since ionic radius of Hg + > Cd+, the lattice energy of Hg 2SO4 should be less exothermic than that of Cd2SO4. Option D is incorrect: The magnitude of Hhyd is dependent on the charge density of the ion. Since ionic radius of Hg+ > Cd+, Hg+ has a lower charge density and less exothermic Hhyd than Cd+. 9 A Option 1 is correct: S > 0 due to mixing of particles. When ionic solids dissolve in water , entropy increases because the ordered ionic lattice is broken up and the ions are then free to move in solution. Option 2 is correct: S > 0 due to change in phase. When the solvent evaporates (a liquid becomes a gas), particles move randomly in all directions, and there are more ways to arrange the particles and distribute the energy. Furthermore, the large increase in volume going from a liquid to gas also leads to an increase in entropy. Option 3 is correct: S > 0 due to mixing of particles. When pure solvent particles pass through the membrane into the solution, the particles mix together towards increased disorder. Chemical Bonding Teacher’s copy River Valley High School 2015 Year 5 H2 Chemistry 3.3 Dot-and-cross diagrams (Lewis diagrams) for ionic bonding ‘Dot-and-cross’ diagrams involve using dots ‘•’ or crosses ‘ ’ to represent electrons. Only the valence electrons of the species are shown. NaCl: MgCl2: Checkpoint 3 Draw a ‘dot-and-cross’ diagram for Na2O. *Note: there is no need to draw a legend for dot-and-cross diagram at A-Levels as long as it is clear which electrons belong to which elements. 3.4 Factors affecting strength of ionic bonds Ionic bond strength can be deduced from the lattice energy of an ionic compound. Lattice energy is the energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions (refer to topic of Chemical Energetics). The magnitude of lattice energy is dependent on the charges of the cations and anions, as well as the ionic radii of the cations and anions. It can be described with the relationship below: Magnitude of lattice energy (LE) + + ZZ rr where, Z+ charge of the cation Z charge of the anion r + radius of the cation r radius of the anion In other words, ionic bonds formed between ions with high charge and small size (i.e high charge density) are the strongest. Correspondingly, the lattice energy of such ionic compounds will be the most exothermic. Lattice energy is exothermic because gaseous ions need to lose energy before being brought together into an ordered arrangement in the giant ionic lattice structure. Mg Cl 2+ 2 Na Cl + Inter-ionic distance, r + + r Fig. 3.3 Illustrating inter-ionic distance r+ r

10 A Let the initial [A] and [B] be x: Rate = k[A][B]2 = k(x3) = y After compression to 0.25V: Initial [A] = [B] = 4x Rate = k(4x)(4x)2 = k(64x3) = 64y 11 C A – (Theoretically possible, but not the best answer) This could theoretically be used to determine change in [H+] over time, and when [H+] value is plotted against time, this would allow us to detect if it’s 0 th order (straight line graph) or 1st order (curve with constant half-life) but will not be able to determine 2nd or higher orders of reaction. B – (Incorrect) This can let us determine the order of reaction with respect to ethanamide, and not H+ since [H+] is the same for all experiments C – (Best answer) This would allow us to determine initial rate of reaction at different [H+] and this data will allow us to determine order of reaction with respect to H+ D – (Theoretically possible, but not the best answer) Similar to option A, this can be used to determine change in [H+] over time. 12 B Option A is correct: Anionic size decreases in the order of P3− > S2− > Cl−. Option B is incorrect: The element with the highest melting point is silicon, which has giant covalent structure. Option C is correct: Aluminium has giant metallic lattice structure. Due to the highest number of delocalised valence electrons (per atom), aluminium has the highest electrical conductivity. Option D is correct: Sulfur exists as S8 molecules. 13 A Statement 2 is correct and Statement 3 is incorrect. Since Cl− and Br − are spectator ions, the ionic equation is the same for both, Ag+(aq) + 2NH3(aq) = [Ag(NH3)2]+(aq). Therefore, G2 = G4 Statement 4 is correct. Since AgC l is soluble in NH 3(aq) while AgBr is only soluble in concentrated