RVHS H1 Chemistry P2 Soln 2024

River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination [Turn over River Valley High School 2024 JC 2 H1 Chemistry 8873 Prelim Exam Paper 2 Suggested Solution 1 (a) (b) Shape: Trigonal pyramidal Bond angle: 107° Explanation: There are 3 bond pairs and one lone pair of electrons around the central S atom. The electron pairs are arranged as far apart as possible to minimise repulsion. (c) Dative bond is formed when the N atom donates a lone pair of electrons into an empty 1s orbital of an electron deficient H atom for sharing. As a result, the electron deficient H atom achieves a stable duplet configuration. (d) Ammonium sulfite has a higher boiling point than ammonia. Ammonium sulfite has a giant ionic lattice structure while ammonia has a simple covalent structure. More energy is needed to overcome the stronger electrostatic forces of attraction between NH 4+ and SO32− ions than the weaker hydrogen bonding between NH3 molecules.

2 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination 2 (a) Ca2+(g) + 2Cl–(g) → CaCl2(s) H = lattice energy (b) Ionic radius of Ca2+ = 0.099 nm Ionic radius of Mg2+ = 0.065 nm |𝐿𝐸| | 𝑞+𝑞− 𝑟+ + 𝑟−| Magnitude of lattice energy of MgC l2 is higher as the ionic radius of Mg 2+ is smaller than that of Ca2+ (and both ions have the same charge), resulting in a (stronger electrostatic force of attraction between Mg2+ and Cl– ions in MgCl2). (c) [Ca(H2O)6]2+ + H2O = [Ca(H2O)5(OH)]+ + H3O+ Ca2+ ions will polarise H2O molecule, causing partial hydrolysis to occur. This will cause a small amount of H+ to form, resulting in a solution of pH 6.5.

3 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination [Turn over (d) Amount of CaCO3 = 11.375 100.1 = 0.1136 mol Amount of HCl = 250 1000 × 2.00 = 0.5000 mol Hence, CaCO3 is limiting. Energy released = mcΔT = (250)(4.2)(1.58) = 1659 J = 1.659 kJ x = − 1.659 0.1136 = −14.6 kJ mol−1 (e) (i) To prevent the solution from spraying out of the conical flask due to bubbling. (ii) CaCO3  CO2 Mass loss = mass of CO2 produced = 0.1136 × 44.0 = 4.998 g = 5.00 g (iii) Rate = k[HCl] (iv)

4 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination (v) An increase in temperature will lead to a n increase in the average kinetic energy of reactant molecules. More particles will have energy greater than or equal to Ea. The frequency of effective collisions increases. Hence, the rate of reaction increases.

5 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination [Turn over 3 (a) Nanomaterial is a material with at least 1 dimension on nanoscale from 1 nm to 100 nm. (b) High tensile strength. It has strong covalent bonds between atoms in a giant network of carbon atoms. (c) It is due to the (black) colour of CNT material which does not allow the driver to see through the window. Or CNT material is opaque. (d) It has delocalised valence electrons which act as mobile charge carriers and will conduct static charges/conduct the accumulated charges away. (e) structures names Armchair Zigzag 4 (a) Reaction 1: LiAlH4 in dry ether Reject heat for LiAlH4 or H2, Ni, heat or NaBH4 Reaction 2: Excess concentrated H2SO4, heat Rejected Al2O3 and heat because alcohol given is not a vapour (b) (c) CH3CH2CH2COOH (d)

6 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination 5 (a) Polymers are macromolecules with at least 100 repeat units of monomers and average molar mass of at least 1000 g mol–1. (b) (i) Condensation polymerisation (ii) Thermosetting polymer There are strong covalent bonds as cross-linkages between the chains. (iii) Bakelite cannot be recycled. A large amount of energy is needed to overcome the strong covalent bond cross linkages which will result in the thermal decomposition of Bakelite.

7 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination [Turn over (c) PVA dissolves in water by forming hydrogen bonds between its protonic hydrogen and the lone pair of electrons on the oxygen atom of water. 6 (a) Amount of S2O32– reacted = 0.00150 × 25.60 1000 = 3.840 × 10−5 mol 4S2O32–  2I2  4Mn(OH)3  O2 Amount of oxygen = 1 4 × 3.840 × 10−5 = 9.60 × 10−6 mol (b) (i) Half-equation: NO2– + H2O + e– → NO + 2OH– Overall: 2I– + 2NO2– + 2H2O → I2 + 2NO + 4OH–

8 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination (ii) Calculated DOC value is larger than the expected value.[✓] Presence of NO2– will cause more I2 to form[✓], which in turn causes a larger titre value of Na2S2O3(aq) [✓], resulting in a larger calculated value of dissolved oxygen[✓]. (iii) 2NO2–  I2  2S2O32– Amount of S2O32– reacted with I2 formed from dissolved oxygen = 3.840 × 10–5 – 5.70 × 10–6 = 3.270 × 10–5 mol 4S2O32–  O2 Amount of oxygen = 1 4 × 3.270 × 10−5 = 8.18 × 10–6 mol (iv) DOC = 8.18 × 10−6 × 1000 30 × 32.0 × 1000 = 8.72 mg dm–3 (v) There is enough oxygen to support small aquatic animals but not the larger fishes.

9 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination [Turn over (c) (i) Mn O Mass % 72.0 100 – 72.0 = 28.0 Mole ratio 72.0 54.9 = 1.311 28.0 16.0 = 1.75 Adjusted Mole ratio 1 1.335 Simplest ratio 3 4 Empirical formula: Mn3O4 (ii) Average oxidation state = +(8/3) = +2.667 +3 oxidation state: Mn2O3 The other oxide can only have oxidation state of either +2 or +1 Assuming +2 oxidation state: MnO and letting the proportion of Mn2O3 be x: (x)(3) + (1−x)(2) = 2.667 x = 0.667 Mn2O3 : MnO ratio of 2:1 OR Assuming +1 oxidation state: Mn 2O and letting the proportion of Mn2O3 be x: (x)(3) + (1−x)(1) = 2.667 x = 0.833 Mn2O3 : Mn2O ratio of 5:1

10 River Valley High School 8873/02/PRELIMS/24 2024 Preliminary Examination 7 (a) (i) 1s2 2s2 2p6 3s2 3p6 3d8 4s2 (ii) For all atoms, there is an increase in successive ionisation energies as the electron is removed from an ion of increasing positive charge. The significant jump from the 2nd to the 3rd ionisation energy for element A and B is due the removal of the 3 rd electron from an inner principal quantum shell which is more strongly attracted to the positive nucleus. (iii) Element B is Mg. (Both element A and B are from Group 2. Ionisation energies decrease down the group.) [1] (b) Ar of nickel = ( .. .. 18 157 98 18 1 44 9 + ) + ( .. .. 44 959 02 18 1 44 9 + ) [1] = 58.72 [1] [2] (c) (i) From fluorine to iodine, the number of electrons increases, hence more energy is required to overcome the stronger instantaneous dipole - induced dipole interactions between the halogen molecules . This results in an increase in boiling point and a decrease in volatility. [2] (ii) The thermal stability of hydrogen halides decreases down the group due to decreasing HX bond energy. As the size of halogen atoms increases, their valence orbitals become more diffuse. This results in less effective overlap of orbitals between the H atom and the halogen atom . Thus, less energy is required to break the weaker HX bond. [2] (d) (i) Table 7.2 X Y Z nucleon number 121 86 35 charge +1 +1 0 [1]