2023 VJC H2 Chem Prelim P1 (Ans)

1 VICTORIA JUNIOR COLLEGE 2023 JC2 PRELIMINARY EXAMINATION H2 CHEMISTRY PAPER 1 ANSWERS 1 D 6 D 11 C 16 B 21 D 26 B 2 C 7 A 12 D 17 C 22 A 27 B 3 B 8 D 13 B 18 D 23 C 28 C 4 C 9 D 14 B 19 D 24 A 29 A 5 A 10 B 15 C 20 C 25 B 30 A 1 D No of protons No of e– No of neutrons A: Wrong For Al13 27 13 13 27 – 13 = 14 B: Wrong For Cl– 17 35 17 17 + 1 = 18 35 – 17 = 18 C: Wrong For S2–16 32 16 16 + 2 = 18 32 – 16 = 16 D: Correct For K+19 39 19 19 – 1 = 18 39 – 19 = 20 2 C Since central C atom is sp hybridised and there are no unpaired electrons, C2O molecule has the following structure: Hence, C2O has 3 lone pairs of electrons. 3 B (1 and 2 only) Option 1: Correct It shows the structure of diamond which has a giant molecular structure. Option 2: Correct It shows the structure of graphite which has a layered giant molecular structure. Option 3: Wrong Sodium chloride has a giant ionic structure. Option 4: Wrong Iodine has a simple molecule structure. The molecules are held by weak intermolecular forces of attraction. It has a solid lattice structure. 4 C PV = nRT PV = (m/Mr)RT Mr = mRT/(PV) Given the same mass of gases in the identical flasks of same volume, hence, m, V and R are constant, Mr ∝ T/P For gas E For gas F For gas G Mr ∝ t/p Mr ∝ t/2p Mr ∝ 2t/p Hence, Mr(G) > Mr(E) > Mr(F) 5 A Option A: Correct Real gases behave less ideally at high pressure. At high pressure, real gases behave less ideally as the gas particles are close to each other and size of the particle is not negligible. Option B: Wrong Real gases behave more ideally at high temperature. Intermolecular forces of attraction is negligible at high temperature since gas particles have sufficient kinetic energy to overcome it. Option C: Correct The presence of a catalyst increases the rates of both the forward and backward reactions. It does not have any effect on the ideal behaviour of gas. Option D: Wrong Strong covalent bonds between the nitrogen atoms does not explain if the gas is behaving ideally. We should consider the intermolecular attractions between the molecules instead. 6 D For element X: Chloride and oxide dissolve to give strong acidic solution  P or S For element Y: Low conductivity at room temperature  Si Only 1 chloride with formula SiCl4  Si Chloride dissolves giving acid solution  Si For element Z: Chloride and oxide have high m.p.  Na or Mg Oxide reacts readily in water  Na Chloride dissolves giving neutral solution  Na 7 A Down Group 2, charge density of cation ( ∝ q+/r+) decreases, causing the polarising power of cation to decrease. As a result, there is less distortion (or polarisation) of the anionic charge cloud and so the iodate becomes more stable to heat down the group. This means that iodate is decomposed at a higher temperature. With the mass of Q( IO3)2 being unchanged with time upon heating, that means Q( IO3)2 is the most stable to heat. While P( IO3)2 is decomposed at a faster rate than R(IO3)2, meaning that P(IO3)2 is the least stable to heat. Hence, order of the Group 2 elements: P, R, Q Option A: Correct Down the group, nuclear charge increases. However, the distance between the nucleus and the valence electrons increases due to an increase in number of electron shells. This results in weaker attraction between the nucleus and the valence electrons, le ss energy is required to remove the

2 valence electron and first ionisation energy (IE) decreases down the group. Hence, order of first IE of Group 2 elements: P, R, Q, with P having the most endothermic first IE. Option B: Wrong Down the group, with the weaker attraction between the nucleus and the valence electrons, valence electrons are more readily lost and hence reducing strength increases down the group. Hence, order of reducing power (reactivity) of Group 2 elements: P, R, Q. Q being the most reactive among all, reacts the most readily with water to give hydrogen gas. Option C: Wrong Melting point of an ionic solid depends on the strength of ionic bond, which is measured by lattice energy as follows: |Lattice Energy|  | 𝑞+×𝑞− 𝑟++𝑟− | Since size of Group 2 cations increases down the group, with P 2+ having the smallest cationic size, P(IO3)2 has the most endothermic lattice energy and highest melting point. Option D: Wrong All Group 2 oxides are basic, hence they will react with acids to form salts. 8 D Option A: Wrong Astatine has more electron shells than bromine. The orbital overlap between hydrogen and astatine atoms will be less effective than the orbital overlap between hydrogen and bromine atoms. Therefore, the bond strength of H–At will be weaker than H–Br implying a lower decomposition temperature. Option B: Wrong Going down Group 17, the oxidising power of halogen decreases. As such, At2 will not be able to oxidise I– to I2, which is a brown solution. Option C: Wrong Going down Group 17, the solubility product of silver halides decreases. Since the position of astatine is below iodine in the periodic table, when NH3(aq) is added, the ionic product of silver astatide will still exceed its solubility product, making silver astatide insoluble in NH3(aq). Option D: Correct Going down Group 17, the melting point increases due to the stronger instantaneous dipole -induced dipole interactions between the non -polar halogen molecules. Since At 2 has a higher melting point than I2 which exists as solid at room temperature, At2 should exist as solid as well. 9 D Oxidation half equation: Tl+ → Tl3+ + 2e ---(1) Reduction half equation: 6H+ + VO3– + 3e → V2+ + 3H2O ---(2) Hence, overall redox equation: (1) x 3 + (2) x 2 gives 3Tl+ + 12H+ + 2VO3– → 3Tl3+ + 2V2+ + 6H2O Since only VO 3– is present at the end of the reaction, Tl+ is the limiting reagent. At the end of the reaction, let the a mount of VO3– (aq) = amount of V2+(aq) = a 3Tl+ + 2VO3– + … → 3Tl3+ + 2V2+ + … I / mol 3a/2 2a 0 C / mol –3a/2 –a +a E / mol 0 a a Hence, at the start of the reaction, amount of Tl+ : amount of VO3– = 3a/2 : 2a = 3 : 4 3 mol of Tl+ and 4 mol of VO3– are required at the start. 10 B Na(g) → Na+(g) + e– ∆H = w = 1st IE Na(g) → Na2+(g) + 2e– ∆H = x ∆H = x = 1st IE + 2nd IE = w + 2nd IE Hence, 2nd IE = x – w (Option B) OR Na(s) → Na(g) ∆H = y = enthalpy change of atomisation Na(s) → Na2+(g) + 2e– ∆H = z = enthalpy change of atomisation + 1st IE + 2nd IE 2nd IE = z – enthalpy change of atomisation – 1st IE = z – y – w (This option is not given) 11 C Option A: Wrong Six water ligands have been displaced by one edta ligand. As the number of free ligands increases, entropy also increases. Option B: Wrong When gases are mixed at constant pressure, the volume of the mixture increases, leading to more ways to arrange particles . Hence , entropy increases. Option C: Correct A de crease in temperature of liquid ethanol decreases entropy. Option D: Wrong A change in state from solid to gas (sublimation) should result in an increase in entropy.

3 12 D Option A: Wrong Since the gradient of tangent at t = 0 min (initial rate) increases by two times when the initial [ Y] doubles, the reaction is first order wrt Y. Since the t1/2 of X is constant at 20 min, the reaction is first order wrt X. Hence, rate = k[ X][Y] and the unit of k is mol–1 dm3 min–1. Option B: Wrong Based on the rate equation, 1 molecule of X and 1 molecule of Y are involved in the slow step of the mechanism. The reaction will mostly likely proceed in more than one step as follows: X + Y → D