RI 2021 Y5 Timed Practice Solutions

© Raffles Institution 2021 9729/J/21 2021 Y5 H2 Chemistry July Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D B A C A D D B C D B A A B C MCQ worked solutions 1 Amount of CO2 formed = 39.6 / 44 = 0.9 mol Amount of H2O formed = 16.2 / 18 = 0.9 mol Since X and Y have the same empirical formula, the mole ratio of C : H in the products is the same as the mole ratio of C : H in X and Y.  mole ratio of C : H = 0.9 : 2(0.9) = 1 : 2  empirical formula of X and Y is CH2. Let the molecular formula of X be CnH2n. Then the molecular formulae of Y will be C2nH4n. By considering the total amount of CO2 formed from the combustion of 0.1 mol of CnH2n and 0.1 mol of C2nH4n, 0.1 × (n + 2n) = 0.9 n = 3  Y has molecular formula C6H12. 2 Let the other species be X. Oxidation number of Mn in MnO42− = +6 Oxidation number of Mn in MnO4− = +7 oxidation: MnO42− ⎯→ MnO4− + e− 1 mol of MnO42− loses 1 mol of electron to form 1 mol of MnO4−. Since this is a disproportionation reaction, MnO42− undergoes oxidation to form MnO4− and reduction to form X.  Oxidation number of Mn in X should be lower than +6.  Option D is incorrect as oxidation number of Mn in Mn2O7 = +7 X Oxidation number of Mn No. of electrons gained by 1 mol of MnO42− to form 1 mol of X Since number of electrons gained = number of electrons lost, ratio of X : MnO4− A Mn(OH)2 +2 4 mol 1 : 4 B MnO2 +4 2 mol 1 : 2 C Mn2O3 +3 3 mol 1 : 3 Only option B gives the correct ratio between the two manganese-containing products.

© Raffles Institution 2021 9729/J/21 3 HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O(l) Let the volume of each solution used be V dm3. Amount of HCl = V mol Amount of NaOH = 1.5V mol Since the mole ratio of HCl : NaOH = 1 : 1, HCl is limiting. In the resulting solution: Amount of NaOH unreacted = 0.5V mol Amount of NaCl formed = V mol Final total volume = 2V dm3 [NaOH] = 0.5V / 2V = 0.25 mol dm−3 [NaCl] = V / 2V = 0.50 mol dm−3 4 Species Electronic Configuration Electrons-in-boxes diagram (for highest energy subshell) Number of unpaired electrons N− 1s22s22p4 2 F2+ 1s22s22p3 3 Cr2+ 1s22s22p63s23p63d4 4 Co2+ 1s22s22p63s23p63d7 3 5 Option A is correct (as the statement is incorrect). A neutron has no charge while an electron is negatively charged. Hence, they are not oppositely charged. Option B incorrect (as the statement is correct). A neutron is found in the nucleus at the centre of the atom while an electron is found around the nucleus. Hence, they are found in different parts of the atom. Option C is incorrect (as the statement is correct). A neutron has a relative mass of 1 while an electron has a relative mass of approximately 1/1840. Hence, the neutron has a greater mass than the electron. Option D is incorrect (as the statement is correct). The number of electrons in an atom must equal to the number of protons in an atom (for charge neutrality). The number of neutrons need not be the equal to the number of electrons in an atom. For example, H has 1 electron and 1 proton, but 0 neutron.

© Raffles Institution 2021 9729/J/21 6 B: 1s2 2s2 2p1 The valence shell of B refers to the electronic shell containing the 2s and 2p subshell. Option A is incorrect. B only contains one half-filled orbital in the 2p subshell. Option B is incorrect. The 2s orbital is spherical in shape while the 2p orbitals are dumbbell in shape. Option C is incorrect. The 2p electron is at a higher energy level than the 2s electrons. Option D is correct. There is one orbital in the 2s subshell and three orbitals in the 2p subshell. Since each of the four orbitals can hold a maximum of 2 electrons, the valence shell of B can accommodate a maximum of eight electrons. 7 In the molecule, there are • 15 single bonds  15 σ bonds • 2 double bonds  2 σ bonds and 2 π bonds • 1 triple bond  1 σ bond and 2 π bonds Hence, there are a total of 18 σ bonds and 4 π bonds. 8 Option A is incorrect. SiO2 has a giant molecular structure with strong covalent bonds. Boiling SiO2 requires overcoming these strong covalent bonds. Option B is correct. Greater number of valence electrons increases the strength of metallic bonds. Option C is incorrect. AlCl3 has a simple molecular structure with id-id interactions.

© Raffles Institution 2021 9729/J/21 Option D is incorrect. In the first compound, the close proximity of the two -OH groups allows for i ntramolecular hydrogen bonding, which results in less extensive intermolecular hydrogen bonding . Since boiling involves overcoming the intermolecular hydrogen bonding, the first compound has a lower boiling point. 9 Given that the electronegativity values of C and H are similar, the C –H bond can be considered as non-polar. Hence, A is non-polar. In B, the C–Cl bond dipoles cancel out each other. Hence, B is also non-polar. Both C and D are polar molecules. In D, the dipole moment of C=O is partially offset by the dipole moments of the C–Cl bonds, resulting in a smaller overall dipole. C D 10 PCl5 SF6 XeF4 trigonal bipyramidal (90° between axial and equatorial bonds) octahedral (all bond angles are 90°) square planar (all bond angles are 90°) 11 pV = nRT = mRT M M = mRT pV = m V RT p = density × RT p Average Mr = density × RT p = (5.00 × 103) × 8.31 × 298 100 × 103 = 123.82 Let the mole fraction of He and Xe be 𝑥 and (1 – 𝑥) respectively. 123.82 = 4𝑥 + 131.3(1 – 𝑥) 𝑥 = 0.059 Note the conversion of units: • density to g m−3 • temperature to K • pressures to Pa

© Raffles Institution 2021 9729/J/21 12 Using pV = nRT and making the y-axis the subject: graph 1 (correct) pV = nRT = constant at a particular T (since n, R and T are constants)  graph is similar to y = k (horizontal straight line)  pV is larger at higher T graph 2 (correct) V = ( nR p )(T) = (k)(T) at a particular p (since n, R and p are constants)  graph is similar to y = kx (straight line that passes through origin)  gradient (= nR p ) is smaller at higher pressure graph 3 (incorrect) p = (nRT)( 1 V) = (k)( 1 V) at a particular T (since n, R and T are constants)  graph is similar to y = kx (straight line that passes through origin)  gradient (= nRT) is larger at higher T 13 Hhyd is directly proportional to the charge density (or charge ÷ radius) of the ion. The higher the charge density of the ion, the greater the magnitude of its hydration energy. |𝐻hyd o [M+(g)] |  | q+ r+ | ion q+ r+ |q+ / r+| Cr2+ 2+ 0.073 2 ÷ 0.073 = 27.4 Fe2+ 2+ 0.061 2 ÷ 0.061 = 32.8 Ti3+ 3+ 0.067 3 ÷ 0.067 = 44.8 14 Option A is correct. Hr = BE of bonds in reactants − BE of bonds in products Since Hr < 0, energy released during bond forming is more than energy absorbed during bond breaking. Option B is incorrect. Hr = nHc (reactants) − mHc (products) Hence, Hc (H2(g)) is also required. Note: If formation enthalpies are given, Hr = nHf (products) − mHf (reactants), only Hf (C2H4(g)) and Hf (C2H6(g)) are required as Hf (H2(g)) = 0. Option C is correct. As 2 mol of gaseous reactants react to form only 1 mol of gaseous product, the disorder of the system decreases and hence S < 0. Option D is correct. G = H – TS Since H < 0 and S < 0, G < 0 only at low T.

© Raffles Institution 2021 9729/J/21 15 equation 1 is incorrect. P = Hatomisation(C(s)) + 2Hatomisation(O2(g)) equation 2 is correct. By Hess’ law, S = – P + Q + R Since Q + R = Hformation(CO2(g)) Hence, S = – P + Hformation(CO2(g)) equation 3 is correct. Q + R = Hcombustion(C(s)) Hence, Q = Hcombustion(C(s)) – R Note: The equation for Hcombustion(C(s)) is