MI 2021 PU1 H2 EOY P1 answers

2021 End-of-Year Examination Pre-University 1 H2 CHEMISTRY 9729/01 Paper 1 Multiple Choice & Structured Questions 15 Oct 2021 2 hours Section A – Multiple Choice 1 2 3 4 5 B B C D D 6 7 8 9 10 A C B D B 11 12 13 14 15 A C A C C 1 1 Correct. Empirical formula is the lowest whole number ratio of the number of atoms of each element. 2 Correct. Molecular formula is the number of atoms of each element present in one molecule. 3 Not correct. The relative molecular mass of C6H12 is 72.0 + 12.0 = 84.0, not 48.0. Relative molecular masses are calculated by summing the relative atomic masses of each element found in the Periodic Table, multiplied by the number of atoms.

2 2 CxHy + (𝑥 + 𝑦 4) O2 ⟶ 𝑥 CO2 + 𝑦 2 H2O initial volume / cm3 10 90 0 change in volume / cm3 -10 -50 +30 final volume / cm3 0 40 30 volume of oxygen gas unreacted = 40 cm3 volume of carbon dioxide gas formed = 70 – 40 = 30 cm3 volume of oxygen reacted = 90 – 40 = 50 cm3 𝜂𝐶𝑥𝐻𝑦 : 𝜂𝑂2 : 𝜂𝐶𝑂2 = 10 : 50 : 30 = 1 : 5 : 3 = 1 : 𝑥 + 𝑦 4 : 𝑥 hence 𝑥 = 3 𝑥 + 𝑦 4 = 5 𝑦 = 8 Molecular formula of hydrocarbon is C3H8. 3 A Not correct. BeCl2 is a linear molecule. Be only has 2 valence electrons, and forms two bond pairs with Cl. Be has no lone pairs of electrons. Hence the two electron domains around Be form a 180° from each other. B Not correct. BeCl2 is a molecule. Be 2+ ion has a high charge density, strongly polarising Cl- ion electron clouds towards itself, forming Be–Cl covalent bonds. C Correct. Be has an electronic configuration of 1s 2 2s2. To form 2 bonds, its valence orbitals (2s and 2p) undergo sp hybridisation, forming 2 sp orbitals which form covalent bonds with C l atoms. Be hence has 2 energetically -accessible vacant unhybridised 2p orbitals which can accept lone pairs of electrons from chlorine atoms, forming a polymer. D Not correct. The 1s orbital of Be is fully-filled and cannot accept additional electrons.

3 [Turn over 4 2Zn(NO3)2(s) ⟶ 2ZnO(s) + 4NO2(g) + O2(g) +5 -2 +4 0 Zn(NO3)2 was oxidised to form O2. The oxidation state of O increased from -2 (in Zn(NO3)2) to 0 (in O2). Zn(NO3)2 was also reduced to form NO2. The oxidation state of N decreased from +5 (in Zn(NO3)2) to +4 (in NO2). 5 Ground-state electronic configuration of Cl: 1s2 2s2 2p6 3s2 3p5 Excited-state electronic configuration of Cl: 1s2 2s2 2p6 3s1 3p6 One electron must have transferred from 3s (lower energy) to 3p (higher energy). 6 element electronic configuration number of half- filled orbitals A C 2 B F 1 C Mg 0 D Al 1 7 The first bond between two atoms is a sigma (σ) bond. Any bond formed in excess of the first one is a pi (π) bond. C C C N H H H 8 melting point of ionic compound ∝ LE ∝ | 𝑞+×𝑞− 𝑟++𝑟− | Since O2– and F– ions have similar ionic radii (F – ion 0.136 nm is much larger than half the radius of O 2– ion 0.140 nm), only the magnitudes of 𝑞+ and 𝑞− will result in significant differences in lattice energies. 1s 2s 2p 1s 2s 2p 1s 2s 2p 3s 1s 2s 2p 3s 3p 1 σ, 1 π 1 σ, 2 π

4 compound 𝑞+ × 𝑞− melting point A CaF2 +2 × –1 = –2 B CaO +2 × –2 = –4 highest C K2O +1 × –2 = –2 D KF +1 × –1 = –1 9 A Not correct. All boron and nitrogen atoms in the network use up all of their available valence electrons to form covalent bonds. Hence there are no delocalised electrons to behave as mobile charge carriers. B Not correct. Only metals are ductile and malleable. Boron nitride resembles graphite in structure, which has a giant molecular structure. C Not correct. Boron nitride is a soft material. Between planes of boron and nitrogen atoms, there are weak instantaneous dipole-induced dipole forces of attraction, which require little energy to overcome and allows the planes to slide over one another easily. D Correct. The atoms of boron and nitrogen are held together in a network by strong covalent bonds. These bonds require much energy to break before the substance can melt. 10 Metals are ductile and malleable. Only metals are able to conduct electricity in solid state. This is achieved by the ‘sea’ of delocalised electrons surrounding the metal ion lattice, acting as mobile charge carriers. melting point / °C electrical conductivity in different states identity of substance (s) (l) (aq) A 122    C6H5CO2H B 181   insoluble Li C 373   insoluble PbBr2 D 802    NaCl

5 [Turn over 11 A Correct. The volume of the gas measured is assumed to only consist of the space containing the gas, and does not include the volume that the gas molecules themselves occupy. B Not correct. A gas is most ideal under low pressure and high temperature , so that molecules are sufficiently far apart and can overcome the forces of attraction between molecules. C Not correct. Ideal gas molecules collide perfectly elastically – kinetic energy is conserved in all the molecules. D Not correct. The correct ideal gas assumption is ‘forces of attraction between molecules are insignificant’, and makes no mention about the type of force. 12 Standard temperature and pressure refers to 273 K (0 °C) and 100000 Pa (1 bar). V1 = 3.00 dm3 P1 = 300 kPa P2 = 100 kPa Since only p and V varied while n and T remained constant, the relationship 𝑝1𝑉1 = 𝑝2𝑉2 should be used. (300 𝑘𝑃𝑎)(3.00 𝑑𝑚3) = (100 𝑘𝑃𝑎)(𝑉2) Hence 𝑉2 = 9.00 dm3 13 A Since 𝑝𝑉 = 𝑛𝑅𝑇 and 𝑝 ∝ 𝑇, the graph showing a direct proportional relationship is correct. B Since 𝑝𝑉 = 𝑛𝑅𝑇 and 𝑝𝑉 ∝ 𝑇, 𝑝𝑉 is therefore inversely proportional to 1 𝑇 and the graph of a direct proportional relationship is not correct. The graph should be an inverse relationship. C Since 𝑝𝑉 = 𝑛𝑅𝑇 and 𝑉 = 𝑛𝑅𝑇( 1 𝑝), 1 𝑝 = 1 𝑛𝑅𝑇 𝑉. Therefore 1 𝑝 ∝ 𝑉 and the graph of an inverse relationship is not correct. The graph should be a direct proportional relationship. D Since 𝑝𝑉 = 𝑛𝑅𝑇 and 𝑉 ∝ 𝑇, the graph of an inverse relationship is not correct. The graph should be a direct proportional relationship.

6 14 Since 𝑝𝑉 = 𝑛𝑅𝑇, 𝑛𝑅 is the gradient of the graphs. gradient = 𝑛𝑅 = 𝑚 𝑀𝑟 𝑅 gas 𝑀𝑟 rank in 𝑀𝑟 rank in 𝑚 𝑀𝑟 𝑅 N2 28.0 3rd 2nd O2 32.0 2nd 3rd Ar 39.9 1st 4th CH4 16.0 4th 1st 15 H O OH OH OH OH OH There are 4 chiral carbon atoms in the D-glucose molecule. Therefore, number of enantiomers in total = 24 = 16. * * * *

7 [Turn over Section B – Structured Questions 1 (a) MnO4– + 8 H+ + 5 e– ⟶ Mn2+ + 4 H2O 2 I– ⟶ I2 + 2 e– (b) I2 + 2 S2O32– ⟶ 2 I– + S4O62– (c) 𝜂𝑆2𝑂32− = 24.90 1000 dm3 × 1.50 mol/dm3 = 0.03735 mol since 𝜂𝐼2 𝜂𝑆2𝑂32− = 1 2 ecf (b) coefficients 𝜂𝐼2 = 1 2 × 0.03735 mol = 0.018675 mol = 0.0187 mol (d) since 𝜂𝑒− 𝜂𝐼2 = 2 1 ecf (a) coefficients 𝜂𝑒− = 2 × 0.018675 mol = 0.03735 mol ecf (c) since 𝜂𝑀𝑛𝑂4− 𝜂𝑒− = 1 5 ecf (a) coefficients 𝜂𝑀𝑛𝑂4− = 1 5 × 0.03735 mol = 0.00747 mol accept 0.00748 if 0.0187 used from (c) (e) [MnO4–] = 0.00747 𝑚𝑜𝑙 10.0 1000 𝑑𝑚3 = 0.747 mol dm–3 ecf (d) accept 0.748 if 0.0187 used from (c)

8 2 (a) endothermic energy / cosmic ray is absorbed / required to break the covalent bond (b) (i) (ii) 3 sets of axes: 1 point each 4 shapes + label: 1 point each 7 / 7 points: [2] ecf (i) 3s, 3p 4 / 7 points: [1] (c) (i) 𝐶6 14 accept X (ii) no they have different number of protons / are of different elements ecf from (c)(i): yes, they have the same number of p