VJC 2022 JC1 Promo H2 Chem P1 MCQ Final Solutions

1 VICTORIA JUNIOR COLLEGE 2022 JC1 PROMOTIONAL EXAM H2 CHEMISTRY PAPER 1 ANSWERS 1 C 6 A 11 B 16 B 21 C 26 C 2 B 7 B 12 B 17 A 22 C 27 D 3 A 8 B 13 A 18 D 23 B 28 D 4 A 9 C 14 C 19 A 24 D 29 A 5 A 10 A 15 D 20 B 25 C 30 D 1 C (1 and 3 only) Option 1: Correct Lv+ has 116 – 1=115 electrons while Fl– has 114 + 1 = 115 electrons Option 2: Wrong Angle of deflection α |charge / mass| Angle of deflection of Lv3+ α |+3/292| = 0.0103 Angle of deflection of Fl2– α |–2/289| = 0.00692 Option 3: Correct From the Data Booklet, Fl having 114 proton no. is in Group 14, Lv having 116 proton no. is in Group 16. Hence, outer electronic configuration of Fl: ns2np2 ; Lv: ns2np4 ; Fl2–: ns2np4 In addition, being in the same period, they have the same number of core electrons , similar to that of the noble gas at the end of the previous period. Hence, one Lv atom has the same electronic configuration as one Fl 2– ion. 2 B Large increase between the 2nd and 3rd IE ⇒ element X is in Group 2. Hence, X forms the cation X2+. To maintain overall charge neutrality, the formula of compound Y should thus be XO2. 3 A The N atom in –CN group of NH 2CN has a lone pair of electrons for donation, while the B atom in BF 3 has a vacant orbital to accept the lone pair of electrons . This will allow the formation of dative bond when the shared pair of electrons is provided by only N. 4 A Option A: Correct POCl3 (4bp) Tetrahedral 109.5o CCl4 (4bp) Tetrahedral 109.5o Option B: Wrong AlCl3 (3bp) Trigonal planar 120o NCl3 (3bp + 1lp) Trigonal pyramidal 107o Option C: Wrong SO2 (2bp + 1lp) Bent shape 118o CO2 (2bp) Linear 180o Option D: Wrong ClF3 (3bp + 2lp) T-shaped 90o SF4 (4bp + 1lp) See-saw 120o & 90o 5 A Option A: Correct Both substances have intermolecular hydrogen bonding but the second substance has larger electron cloud size which is more easily polarised. Hence, there are stronger instantaneous dipole - induced dipole (id-id) interactions between CH3CH2CH2CH2OH molecules. Hence, the first substance has a lower bp. Option B: Wrong I2 is a solid whereas HC l is a gas. This indicates that I2 has a higher bp. This is because I2 has a much larger electron cloud size which is more easily polarised. Hence, larger amount of energy is required to overcome stronger id-id interactions between I2 molecules than weaker permanent dipole-permanent dipole interactions between HCl molecules. Hence, the first substance has a higher bp. Option C: Wrong Larger amount of energy is required to overcome the stronger hydrogen bonding between CH3CH2NHCH3 molecules than the weaker permanent dipole -permanent dipole interactions between CH 3CH2OCH3 molecules. Hence, first substance has a higher boiling point. Option D: Wrong CH3COOH exists as dimers, whereby two CH3COOH molecules bond to each other via intermolecular hydrogen bonds. As such, higher extent of hydrogen bonds is formed between CH3COOH molecules than that between CH3CH2OH molecules. Hence, the first substance has a higher boiling point. 6 A Since silicon carbide has a high m p, it should have a giant molecular structure ⇒ options A or C Since it is also hard, its structure is expected to represent the tetrahedral structure of diamond rather than the layer structure of graphite ⇒ option A 7 B Let the percentage yield for each step be y%. For first step: Theoretical nNa2CO3 produced from Na2S = nNa2S reacted = (50 x 1000) / 78.1 = 640 mol Actual nNa2CO3 produced = y% x 640 = (y / 100) x 640 = 6.40y mol For second step: Theoretical nNaHCO3 produced from Na2CO3 = 2 x nNa2CO3 reacted = 2 x 6.40y = 12.8y mol Actual nNaHCO3 produced = y% x 12.8y = (y / 100) x 12.8y = 0.128y2 mol Hence, mass of NaHCO3 = 20.5 x 1000 = 0.128y2 x 84.0 y = 43.7%

2 8 B C4H10 + 13/2O2 → 4CO2 + 5H2O 3/4w (13/2)(3/4)w C4H10 + 9/2O2 → 4CO + 5H2O 1/4w (9/2)(1/4)w Vol. of O2 = (13/2)(3/4)w + (9/2)(1/4)w = (48/8)w = 6w dm3 9 C Replacing all the alcohol groups bonded directly to the ring carbon atoms with chlorine gives the structure Molecular formula of the product = C6H8Cl4O2. Dividing by 2 gives the empirical formula C3H4Cl2O. 10 A Zn → Zn2+ + 2e- Amt of Zn = 0.5/65.4 = 7.65 x 10–3 mol Amt of e – lost = 2 x 7.65 x 10–3 = 1.53 x 10–2 mol = Amt of e- gained Amt of VO2+ = 10.2/1000 x 0.500 = 5.10 x 10-3 mol  1 mol VO2+ will gain (1.53 x 10–2)/(5.10 x 10-3) = 3 mol e- for reduction  O.N. of V will decrease by 3 units from +5 in VO2+ to +2 in the reduced product 11 B By Hess’ law: Hr = −283 + 2(−286) − (−715) = −140 kJ mol−1 12 B Hsol = Hhyd(Mg2+) + 2Hhyd(Cl–) –LE(MgCl2) = (–1890) + 2(–384) – (−2526) = – 132 kJ mol–1 Amt of MgCl2 = 2.00/95.3 = 0.0210 mol Assuming 100% heat transfer, heat gained by water = heat evolved in expt = 132  0.0210 = 2.77 kJ = 2770 J q = mcT T= q mc = (2770) (50  4.18) = +13.3 °C (exothermic reaction, ∆T is positive) 13 A (1, 2 and 3) Option 1: Correct ∆Hrxn = ∑∆Hf(products) − ∑∆Hf(reactants) = −1273 – [6(−394) + 6(−286)] = +2807 kJ mol−1 Option 2: Correct The process has no change in number of particles of gases . However, comparing 6 mol of H 2O(l) reactant with 1 mol of C 6H12O6(s) product, the number of particles decreases and a liquid reactant is changed to a solid product with a more ordered structure. Hence, there is a decrease in overall entropy and ∆S has a negative sign ( i.e. less disordered). Option 3: Correct Since ∆G = ∆H (+ve) – T∆S (–ve), the ∆G for the reaction will always be positive at all temperatures as ∆H and – T∆S is always positive. 14 C Vol. of KMnO4 required  [H2O2] remaining Graph shows constant half-life of 14 min  Order of reaction wrt H2O2 is 1. [H2O2] in 10cm3 sample at 0 min = ( 5 2 x 30 1000 x 0.05 ) / (10 x 10-3) = 0.375 mol dm-3 3.00 → 1.50 → 0.75 → 0.375 Contamination period = 3t1/2 = 3 x 14 = 42 min 15 D (3 and 4 only) From the Arrhenius equation, k = A 𝑒−𝐸𝑎 𝑅𝑇, rate constant is affected only by temperature or catalyst. Hence, changing the concentration does not affect the rate constant. ⇒ Option 1 wrong, Option 3 correct The amount of kinetic energy possessed by the reactant particles is affected by temperature only. Hence, increasing the temperature increases the proportion of particles having energy greater than the activation energy but increasing the concentration has no effect. ⇒ Option 2 wrong, Option 4 correct 16 B Based on Step II, the slow step, rate equation is rate = k1 [N2O2][H2] However, N2O2 is an intermediate, not a reactant. Hence it will have to be re -expressed in terms of its reactants based on Step I: [N2O2] ∝ [NO]2 ⇒ [N2O2] = k2 [NO]2 Substituting, Rate = k1 k2 [NO]2 [H2] = k [NO]2 [H2] (k = k1ꞏk2) 17 A Moles before opening = moles after opening (1 x 105) x (V) R x (273+25) = (p) x (4V) R x (273+100) p = 3.13 x 104 Pa Not replaced since –OH not bonded to a ring carbon Hrxn t1/2 t1/2 t1/2

3 18 D When pressure increases, position of equilibrium shifts right to favour the side with less number of moles of gas to reduce the pressure. Hence, more NH3 is formed and yield increases. 19 A N2O4(g) ⇌ 2NO2(g) Initial moles 1 0.2 Change in mole –0.24 +0.48 Eqm moles 0.76 0.68 Kc = [NO2]2 [N2O4] = (0.68 / 4)2 (0.76 / 4) = 0.15 mol dm–3 20 B (1 and 2 only) A Bronsted Lowry acid donates a proton to form its conjugate base, e.g. HA → H+ + A– acid conjugate base Option 1: Correct H2O → H+ + OH– a