NJC 2025 H2 Chemistry Prelim P1 Ans

NJC/H2 Chem Prelim/01/2025 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Optical Answer Sheet Data Booklet 9729/01 23 September 2025 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, subject class and registration number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. Instructions on how to fill in the Optical Mark Sheet Shade the index number in a 5 digit format on the optical mark sheet: 2nd digit and the last 4 digits of the Registration Number. Example: Student Examples of Registration No. Shade: 2405648 45648 This document consists of 16 printed pages.

2 NJC/H2 Chem Prelim/01/2025

3 NJC/H2 Chem Prelim/01/2025 [Turn over Answer Key for P1 (MCQ) 1 A 6 C 11 C 16 C 21 C 26 D 2 D 7 C 12 C 17 B 22 A 27 C 3 A 8 C 13 B 18 D 23 D 28 C 4 B 9 B 14 B 19 C 24 C 29 C 5 B 10 D 15 A 20 C 25 C 30 A 1 Sodium chromate(VI), Na2CrO4, is manufactured by heating chromite, FeCr2O4, with sodium carbonate in an oxidising atmosphere. Chromite contains Cr2O42– ions. 4FeCr2O4 + 8Na2CO3 + 7O2 ⎯→ 8Na2CrO4 + 2Fe2O3 + 8CO2 What happens in this reaction? A Chromium and iron are the only elements oxidised. B Chromium, iron and carbon are oxidised. C Only chromium is oxidised. D Only iron is oxidised. Ans: A Reactants: FeCr₂O4 (Fe = +2 and Cr = +3), Na2CO3 (C = +4), O2 (O = 0). Products: Fe₂O₃ (Fe = +3), Na₂CrO₄ (Cr = +6), CO₂ (C = +4). Changes: Fe: +2 → +3 (oxidised) Cr: +3 → +6 (oxidised) C: +4 → +4 (no change) O: 0 → –2 (reduced) So, only Fe and Cr are oxidised. 2 Which species has two unpaired electrons? A B+ B Cu+ C Mg D S Ans: D B+: 1s2 2s2 No unpaired electrons Cu+: 1s2 2s2 2p6 3s2 3p6 3d10 No unpaired electrons Mg: 1s2 2s2 2p6 3s2 No unpaired electrons S: 1s2 2s2 2p6 3s2 3p4 2 unpaired 3p electrons

4 NJC/H2 Chem Prelim/01/2025 3 The variation in the second ionisation energy of eight consecutive elements in the Periodic Table with atomic numbers ≤ 20 is shown in the graph. Which element is a Group 13 element? Ans: A The large drop in 2 nd I.E. shows that D+ ion has the ns 1 electronic configuration. Hence, the electronic configuration of element D is ns2 and belongs to Group 2. Since the graph is showing consecutive elements in the Periodic Table, the configuration of the other elements can be derived. A − Group 13 B – Group 15 C – Group 17 D – Group 2

5 NJC/H2 Chem Prelim/01/2025 [Turn over 4 What do the ions 15N3− and 14C4− have in common? A They have 10 neutrons in their nuclei. B They have more electrons than neutrons. C They have a valence electronic configuration of 3s2 3p6. D They contain the same number of nucleons in their nuclei. Ans: B species 15N3− 14C4− number of protons 7 6 number of neutrons 8 8 number of electrons 10 10 number nucleons 15 14 valence electron configuration 2s2 2p6 2s2 2p6 5 What is the order of decreasing enthalpy change for the three reactions shown? Rb+(g) ⎯→ Rb2+(g) + e− ∆H1 Br−(g) ⎯→ Br(g) + e− ∆H2 Kr(g) ⎯→ Kr+(g) + e− ∆H3 A ∆H1 > ∆H2 > ∆H3 B ∆H1 > ∆H3 > ∆H2 C ∆H2 > ∆H1 > ∆H3 D ∆H2 > ∆H3 > ∆H1 Ans: B All 3 equations are for ionisation energy. Rb+, Br− and Kr are isoelectronic with 36 electrons. Nuclear charge increases from Br− < Kr < Rb+ while shielding effect remains constant. Nuclear attraction for the most loosely held electron increases from Br− < Kr < Rb+. Largest amount of energy is required to remove the most loosely held electron from Rb+.

6 NJC/H2 Chem Prelim/01/2025 6 Barium dithionate, BaS2O6.2H2O, is soluble in water. S2O62– ions slowly decompose in acidic solution. S2O62–(aq) ⎯→ SO2(g) + SO42–(aq) 3.513 g of BaS2O6•2H2O is dissolved in some water and the solution made up to the mark with HCl(aq) in a 100 cm3 volumetric flask. At time x min, a white precipitate of mass 0.661 g is present in the flask. What is the concentration of BaS2O6 in the volumetric flask at time x min? [Ar: Ba, 137.3; S, 32.1; O, 16.0; H, 1.0] A 0.0077 mol dm–3 B 0.0090 mol dm–3 C 0.077 mol dm–3 D 0.090 mol dm–3 Ans: C Amount of BaS2O6.2H2O = 3.513𝑔 (137.3+32.1×2+16.0×6+18.0×2) = 0.01053 𝑚𝑜𝑙 The white ppt is BaSO4. Amount of BaSO4 = 0.661𝑔 (137.3+32.1+16.0×4) = 0.002831 𝑚𝑜𝑙 Amount of BaS2O6 at x min = 0.01053 – 0.002831 = 0.007698 mol [BaS2O6] at x min = 0.007698 100 1000 = 0.077 mol dm–3

7 NJC/H2 Chem Prelim/01/2025 [Turn over 7 NH4Fe(SO4)2•12H2O is a hydrated ‘double salt’. A student analyses this double salt using the following chemical tests. Which row gives the correct result for the stated test? Test Results 1 Reaction with cold NaOH(aq) Green ppt 2 Reaction with Ba(NO3)2(aq) White ppt 3 Reaction with warm NaOH(aq) Red-brown ppt and an alkaline gas A 1, 2 and 3 B Only 1 and 2 C Only 2 and 3 D Only 1 Ans: C In the double salt, oxidation state of Fe is +3. 1 is a wrong result. With cold NaOH(aq), Fe(OH)3 red-brown ppt is formed. 2 is a correct result. BaSO4 white ppt is formed. 3 is a correct result. With warm NaOH(aq), as NH4+ is present, NH3 gas will be given off. At the same time, Fe(OH)3 red-brown ppt is formed. 8 Which statements about BF3 and NF3 are correct? 1 The shape of BF3 is trigonal planar while that of NF3 is trigonal pyramidal. 2 Both BF3 and NF3 are polar molecules. 3 BF3 can act as a Lewis acid because the boron atom has empty low -lying orbitals. A 1 and 2 B 2 and 3 C 1 and 3 D 1, 2 and 3 Ans: C Statement 1: Shapes ● BF₃: central B, 3 bonding pairs, 0 lone pairs → trigonal planar. ● NF₃: central N, 3 bonding pairs, 1 lone pair → trigonal pyramidal. So statement 1 is correct. ● Statement 2: Polarity ● BF₃: although each B–F bond is polar, the symmetry (trigonal planar) cancels dipoles → non-polar molecule. ● NF₃: asymmetric, lone pair on N, net dipole → polar molecule. So statement 2 is incorrect (only NF₃ is polar). Statement 3: Lewis acidity BF₃: central B has an empty 2p orbital (low-lying, energetically accessible) → can accept lone pair electron from a donor. So statement 3 is correct.

8 NJC/H2 Chem Prelim/01/2025 9 Silicon carbide has a similar structure to diamond. Silicon carbide can be used as A a lubricant. B a tip for cutting tools. C a substitute for pencil ‘lead’. D an electrical conductor. Ans: B Diamond: each sp3 hybridised C atom forms strong covalent bond with 4 other C atoms in a tetrahedral arrangement in the giant covalent lattice structure. Properties of diamond include: very high melting point, non-electrical conductor, hard, insoluble in all solvents. 10 Which statement is not a basic assumption of the kinetic theory of gases? A The atoms or molecules have negligible size in comparison with the space they occupy. B There are negligible intermolecular forces between the gas particles. C Collisions between the individual particles and the vessel are perfectly elastic.