2022 TMJC H2 Chem Prelim P1 (Ans)

2022 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry 1 In which species are the numbers of protons, neutrons and electrons all different? A F− 9 19 B Na+ 11 23 C P15 31 D S2− 16 32 Answer: B F− 9 19 9 protons, 10 neutrons, 10 electrons Na+ 11 23 11 protons, 12 neutrons, 10 electrons (all different) P15 31 15 protons, 16 neutrons, 15 electrons S2− 16 32 16 protons, 16 neutrons, 18 electrons 2 Beams of charged particles are deflected by an electric field. When a beam of protons passes through an electric field of constant strength, the angle of deflection is +12 . In another experiment under identical conditions, particle Y is deflected by an angle of – 4 . What could be the composition of particle Y? protons neutrons electrons 1 1 2 2 2 3 3 5 3 4 5 1 A 1, 2 and 3 B 1 and 2 C 1 only D 3 only Answer: B Magnitude of angle deflected  charge/mass charge/mass for 1H+ = 1  deflected by 12 o for 1 unit of charge/mass To be deflected through angle of – 4 o (i.e. opposite side of protons), particle Y should be negatively-charged, with charge/mass = – 1/3 Option 1: charge/mass = – 1/3 ✓ Option 2: charge/mass = – 2/6 = – 1/3 ✓ Option 3: charge/mass = + 3/9 = + 1/3 

2 2022 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry 3 Use of the Data Booklet is relevant to this question. In which pair of compounds does the first molecule have a smaller bond angle than the second molecule? A BF3, NH3 B H2O, H2S C BeCl2, SCl2 D XeF4, SiCl4 Answer: D A: BF3 (trigonal planar; bond angle 120o) > NH3 (trigonal pyramidal; bond angle 107o) B: Both H2O and H2S have bent shape. As O is more electronegative than S, bond pairs of electrons are nearer to the central O atom. There is greater repulsion between bond pairs in H2O and thus bond angle of H2O > bond angle of H2S. C: BeCl2: 2 bond pairs and 0 lone pairs around Be atom  180o SnCl2: 2 bond pairs and 1 lone pairs round Sn atom  118o D: XeF4 (square planar; bond angle ~90 o) < SiCl4 (tetrahedral; bond angle 109.5 o) 4 To produce decaffeinated coffee, pure liquid CO2 is sometimes used to extract caffeine from coffee beans. It was discovered that the solubility of caffeine greatly increased when a mixture of ethanol and liquid CO2 was used. Which interaction best explains why caffeine is more soluble in the ethanol -CO2 mixture as compared to liquid CO2? A instantaneous dipole - induced dipole interactions B permanent dipole - permanent dipole interactions C hydrogen bonding D dative covalent bond

3 Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry [Turn over Answer: C Ethanol can form hydrogen bond to both caffeine and CO2, allowing greater solubility. 5 Which graph does not share the same general shape as the other three graphs according to the ideal gas law for a fixed mass of gas with pressure p, volume V and temperature T in Kelvin? A p against 1 V (at constant T) B pV against p (at constant T) C pV against V (at constant T) D V T against T (at constant p) Answer: A From pV = nRT, A: p = nRT ( 1 V)  straight line through origin B & C: pV = nRT = constant at constant T  horizontal straight line D: V T = nR/p = constant at constant p  horizontal straight line 6 Which statements about Group 2 elements are correct? 1 The charge density of cations increases down the Group. 2 The reducing strength of the elements increases down the Group. 3 The minimum temperature needed for the thermal decomposition of Group 2 carbonates increases down the Group. 4 The melting point of MgO is higher than CaO due to the higher polarising power of Mg2+. A 1 and 3 B 1 and 4 C 2 and 3 D 2 and 4 Answer: C 1 is incorrect: Down the group, ionic charge remains constant while the cationic radius increases. Hence, charge density of Group 2 cations decreases down the group.

4 2022 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry 2 is correct: Down the group, Eo value becomes more negative.  Group 2 element becomes more easily oxidised (i.e. loses valence electrons more easily)  Reducing strength increases down the group. 3 is correct: Down the group, charge density and polarising power of the cation decreases. Hence, the ability of the cation to distort the electron cloud and break the C−O bond in CO32− decreases. Therefore, Group 2 carbonates become thermally more stable down the group i.e. more energy (higher temperature) needed for thermal decomposition to occur. 4 is incorrect: lattice energy - - qq rr + +  + . Mg2+ has a higher ionic charge and smaller ionic radius than Ca2+ which results in a greater magnitude of lattice energy. Ionic bonds in MgO are stronger than that in CaO and require more energy to break. Concept of charge density and polarising power is not applied here. Hence, only statements 2 and 3 are correct. 7 Due to its radioactive nature, the properties of astatine, At, have to be estimated based on its position in the Periodic Table. Which prediction concerning At or its compounds is correct? A Astatine is a weaker oxidising agent than iodine. B Astatine is a liquid at room temperature. C Astatine forms diatomic molecules which dissociate into atoms less readily than iodine molecules. D Hydrogen astatide has a higher decomposition temperature than hydrogen iodide. Answer: A Option A is correct: Oxidising power decreases down Group 17, so astatine should be a weaker oxidising agent than iodine. Option B is incorrect: Boiling and melting point increases down the group. Iodine is already a solid at room temperature, hence astatine, with a higher melting point, would also exist as a solid at room temperature. Option C is incorrect: Due to astatine’s larger atomic radius, the extent of orbital overlap between two At atoms would be smaller as compared to between two I atoms. Hence, the At–At bond would be weaker and hence At2 would dissociate more readily than I2. Option D is incorrect: Down the Group, ease of thermal decomposition of the Group 17 hydrides increases.  thermal stability: HF > HCl > HBr > HI

5 Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry [Turn over 8 Sodium thiosulfate (Na 2S2O3) is used in the textile industry to remove an y excess chlorine from bleaching processes by reducing it to chloride ions. 10 cm 3 of 0.20 mol dm −3 of sodium thiosulfate requires 192 cm 3 of chlorine gas for complete reaction at room temperature and pressure. Which of the following is a possible formula of the sulfur-containing product? A H2S B S C SO2 D HSO4– Answer: D Amount of S2O32– = 0.20 × 10 1000 = 0.00200 mol Amount of Cl2 = 192 24000 = 0.00800 mol Cl2 + 2e– ⎯⎯ → 2Cl– Amount of e– gained by Cl2 = 2(0.00800) = 0.0160 mol = Amount of electrons lost by S2O32– 2 23 0.0160 80.00200 e SO n n − − == ⇒ S2O32– ≡ 8e– ≡ 2S Since there are 2 S atoms per S2O32–, each of the S atom would lose 4 electrons. Initial oxidation state of S in S2O32– = +2 Final oxidation state of S in product = +2 + 4 = +6 Only HSO4– has sulfur with the oxidation state of +6

6 2022 H2 Chem Prelim P1 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry 9 Aqueous solutions of P, Q and R react according to the following equation: P + 3Q + 2R ⎯⎯ → T + U The kinetics of the above reaction was studied and the experimental results obtained are shown in the table below. experiment volume of P / cm3 volume of Q / cm3 volume of R / cm3 volume of water / cm3 relative initial rate 1 20 20 20 20 16 2 2