RI 2021 Prelim P1 Answers

1 2021 H2 Chemistry Y6 Prelim Paper 1 Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B B D C A A C D C D B A C B D Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C C A D A C C D A B A B D A B Question 1 (B) Electronic configuration of Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (not 1s2 2s2 2p6 3s2 3p6 3d9 4s2 – see Atomic Structure notes if you are unsure why) Number of electrons in d orbitals with 4 lobes = 8 (dz2 does not have 4 lobes) Question 2 (B) In Period 2, there are two irregularities in the first ionisation energies of the elements. One irregularity occurs at Group 13 (i.e. B) and the other occurs at Group 16 (i.e. O). Two irregularities are also observed in the second ionisation energies at G roup 14 (removing e – from C+) and at Group 17 (removing e– from F+). A quick way to realise is to check the Data Booklet. Therefore, K, the element with the lower 2nd IE can be either from either Group 14 or Group 17, corresponding to carbon and fluorine respectively. Question 3 (D) Q is non -volatile (does not vapourise easily ) eliminates nitrogen dioxide which has a simple covalent structure with weak instantaneous dipole- induced dipole interactions and is a gas at rtp. Q does not conduct electricity in its standard state eliminates sodium as metals can conduct electricity. Q dissolves in water eliminates silicon dioxide is insoluble in water due to its giant covalent structure. Sodium oxide is the only option that • is non -volatile (due to strong ionic bonds holding the giant ionic lattice), • does not conduct electricity in its standard state (no mobile ions as charge carriers in solid state), and • dissolves in water (by reacting with water to form NaOH(aq)). Question 4 (C) BeF2 is the simplest compound of beryllium and fluorine. F–Be–F In BeF 2, Be is sp hybridised and contain two unhybridised p-orbitals which, in this molecule, are empty. Hence Be in BeF 2 can accept 2 pairs of electrons into its two unhybridised p-orbitals. 1 Incorrect. • F donates a lone pair to Be which accepts the pair of electrons from F, thus the arrow representing the dative bond should point from F to Be. 2 Correct. • With 2 empty unhybridised p -orbitals, Be can accept 2 pairs of electrons from 2 F i.e. Be in BeF 2 forms 2 dative bonds with 2 fluorines from other BeF 2, resulting in the polymeric structure. 3 Correct. • Similarly, BeF 2 can form 2 dative bonds with two F– to give BeF42–. Question 5 (A) X, Y, and Z are Period 3 elements. Oxide of X is amphoteric ⇒ X is Al Oxide of Y is basic ⇒ Y is Na or Mg Oxide of Z is acidic ⇒ Z is non-metal (P, S, Cl)

2 Since Z is a non-metal, it forms anions. Since X and Y are metals, they form cations. Since ionic radii of cations are smaller than that of anions (formed from non -metals) and ionic radii decreases from Na+ to Al3+, ionic radii increases in this order X < Y < Z. Question 6 (A) Highest Ka implies strongest weak acid. A cation with higher charge density has stronger polarising power and distort the electron cloud of water molecules to a greater extent and weaken O–H bonds to a larger extent. element ionic radii / nm A Co3+ 0.055 B Mg+ 0.065 C Mn2+ 0.083 D V3+ 0.064 Since Co3+ has the highest char ge and smallest ionic radius, [Co(H2O)6]3+ is the most acidic and has the highest Ka. Question 7 (C) At constant T and constant number of moles of gas, pV = nRT = constant. CH3OH and SiH 4 have a molar mass of 32 g/mol which is lower than that of 81 g/mol for both HBr and H 2Se. When equal masses of each gas is used, there are more number of moles of CH3OH and SiH4 than HBr and H2Se. Thus, CH3OH and SiH4 will have a higher pV value (Options C or D). CH3OH can form strong hydrogen bonds and hence deviates more from ideal gas behaviour than SiH 4 which can only form weak instantaneous -dipole induced-dipole attractions. Hence Option C is correct. Question 8 (D) You need to check whether the following reaction is spontaneous by checking whether it has a positive Ecell. 2Fe2+ + X2 → 2Fe3+ + 2X– From Data Booklet, F2 + 2e− ⇌ 2F− E = +2.87 V Cl2 + 2e− ⇌ 2Cl− E = +1.36 V Br2 + 2e− ⇌ 2Br − E = +1.07 V I2 + 2e− ⇌ 2I− E = +0.54 V Fe3+ + e− ⇌ Fe2+ E = +0.77 V For the reaction between Fe2+ and I2, Ecell = Ecathode − Eanode = 0.54 – (0.77) = –0.23 V Since Ecell for reaction between I2 is less than zero, the reaction is not spontaneous and I2 cannot oxidise Fe2+ to Fe3+. Question 9 (C) Using algebraic method: (1) N2(g) + 5/2 O2(g) → N2O5(g) Hf = ??? kJ mol−1 (2) 2NO(g) + O2(g) → 2NO2(g) H = −114.1 kJ mol−1 (3) 4NO2(g) + O2(g) → 2N2O5(g) H = −110.2 kJ mol−1 1/2 × (3) to get (4) (4) 2NO2(g) + 1/2 O2(g) → N2O5(g) H = −55.1 kJ mol−1 (5) N2(g) + O2(g) → 2NO(g) H = +180.5 kJ mol−1 Add (2), (4) and (5) to get (1) (1) N2(g) + 5/2 O2(g) → N2O5(g) Hf = −114.1 + (−55.1) + 180.5 = +11.3 kJ mol−1 Question 10 (D) This question tests students on heterogeneous catalysis (by iron, in the Haber Process), which involves the understanding that there are active sites on the catalyst surface for reactants to adsorb onto for reaction. At low pressures of N 2, the active sit es on the catalyst surface are not saturated with N 2, hence initial rate depends on the partial pressure of N2. At moderate or high pressures of N2, most, if not all, of the active sites present on the catalyst surface are taken up by N 2. Consequently, any increase in the partial pressure of N2 has no effect on the initial rate of reaction.

3 Question 11 (B) Since concentration of compound P decreases to 25% (i.e. ¼)of its initial concentration in 1 hour, C0 ½C0 ¼C0 1 h 2 half-lives ⇒ 1 h 1 half-life ⇒ 30 minutes Question 12 (A) By considering the slow step, the rate equation for step 2 is rate = 𝑘2[O •][O3] but O• is an intermediate, thus [O•] cannot be in the overall rate equation. Using the equilibrium constant of step 1, 𝐾1 = [O •][O2] O3 [O •] = 𝐾1[O3] [O2] Substituting this into the rate equation from step 2, rate = 𝑘2 𝐾1[O3] [O2] [O3] rate = 𝑘2𝐾1 [O3]2 [O2] Thus, statements 1 and 2 are correct. Statement 3 is correct because [O 2] is the denominator – when [O 2] increases, rate decreases. Question 13 (C) Rate is a measure of how the fast the reaction proceeds. Yield is about how much product is formed and is related to position of equilibrium. 1 Correct. • At high temperature, more reactant particles move faster and have energy higher than activation energy, resulting in a faster reaction. 2 Incorrect. • At high temperature, position of equilibrium of the reaction shifts to the left to favour the endothermic backward reaction to remove excess heat, resulting in a lower yield. 3 Incorrect. • Pressure does not affect rate constant as rate constant is only affected by activation energy and temperature. 4 Correct. • Presence of a catalyst lowers the activation energy, thus rate constant increases, resulting in a faster reaction. Question 14 (B) G = H − TS G = (−S)T + H y = (m)x + c The correct graph should have a positive y - intercept since the reaction is endothermic (H > 0). Options B and C are possible answers. At high T, the ratio of [products]/[reactants] at equilibrium is lower than 1 means that there are more reactants than products. Therefore, the position of equilibrium lies to the left, thus G must be greater than 0 at high T. Question 15 (D) H2O(l) ⇌ H+(aq) + OH−(aq) equilibrium 1 Option A is incorrect as water is neutral at all temperature since [ H+] is still equals to [ OH−] regardless of temperature. Increasing temperature increases K w i.e. the position of equilib