2022 HCI H2 Chemistry Promo Mark Scheme

2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme & Teaching Solutions HWA CHONG INSTITUTION 2022 C1 H2 CHEMISTRY PROMOTIONAL EXAM MARK SCHEME & TEACHING SOLUTIONS Paper 1 1 2 3 4 5 6 7 8 9 10 C A B A D A D B D D 11 12 13 14 15 16 17 18 19 20 C D C B C A C A B C Comments 1 C Obtain the proton number based on the identity of the element in the Data Booklet. Calculate the no. of neutrons by taking mass number (indicated as superscript on the element) minus proton number: A No. of protons = 2; no. of neutrons = 2 – 2 = 0 B No. of protons = 4; no. of neutrons = 8 – 4 = 4 C No. of protons = 20; no. of neutrons = 40 – 20 = 20 D No. of protons = 82; no. of neutrons = 210 – 82 = 128 Only 40Ca has both numbers matching the given nuclear magic numbers. 2 A 1 Each nitrogen atom in cyanogen has one lone pair of electrons, hence statement is correct. 2 Both carbon atoms in cyanogen are sp hybridized – there are two electron groups around each carbon atom and the shape around each carbon is linear, hence statement is correct. 3 There should be three σ bonds: C–N, C–N and C–C, hence statement is incorrect. 3 B A The zig-zag chains are flat and unlikely to form regular 3D spaces in the lattice structure like the tetrahedral 3D arrangement for ice, hence statement is incorrect. B Each hydrogen bond in HF is stronger than that in H 2O since F is more electronegative than O, and hence draw away more electron density from H atom, resulting in a greater partial positive charge on H so attraction between the H and the lone pair of another HF molecule is stronger. Statement is correct.

2 2022 HCI C1 H2 Chemistry Promotional Exam / Mark Scheme & Teaching Solutions C The question states that the bond angle around F atom is similar to that around O in ice, hence bond angle around each F atom should be 104.5 o. Statement is incorrect. D Solid HF exists as discrete HF molecules held together by intermolecular hydrogen bonding. It is still a simple molecular structure, not giant covalent. Statement is incorrect. 4 A 1 More gas added into a flask of fixed volume, hence total pressure in the flask will increase. Or consider PV = nRT, where n increases and V is constant, so P will increase. Statement is correct. 2 Mole fraction of a gas in a mixture is “no. of moles of that gas / total no. of moles of gas”, since total moles of gas has increased, the mole fraction of each gas will decrease. Statement is correct. 3 Partial pressure of each gas is defined as the pressure that each gas exerts on the flask as though it is alone in the flask. Since the no. of moles of N2 and O2 did not change and the volume of the flask did not change, their partial pressures will be unchanged. Or consider partial pressure = mole fraction × total pressure = (ngas / ntotal) × Ptotal. Ptotal/ntotal is a constant (use PV = nRT), so if ngas remain the same, partial pressure of that gas remain