EJC 2026 Prelim P1 Ans

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2026 General Certificate of Education Advanced Level Higher 2 CHEMISTRY Paper 1 Multiple Choice 9476/01 18 September 2026 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST 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 multiple choice answer sheet. Write in soft pencil. Write your name, civics group and registration number on the Answer Sheet in the spaces provided unless this has been done for you. Do not use correction fluid or tape. The use of an approved scientific calculator is expected, where appropriate. The total mark for this paper is 30. Each correct answer will score one mark. Any rough working should be done in this question paper. This document consists of 12 printed pages.

2 © EJC 9476/01/J2PRELIM/26 1 The shape of two atomic orbitals is shown. s p Which statement correctly describes these orbitals? A All s orbitals are circular in shape. B There are five occupied p orbitals in an atom with atomic number 15. C A  bond can be formed by the sideway overlap of 2px orbital on one atom and a 2py orbital on an adjacent atom. D A  bond can be formed by the head -on overlap of two p orbitals and has electron density along the internuclear axis. Answer: D  A: S orbitals are spherical in shape and 3-dimensional (a region of space where there is a high probability of finding an electron).  B: Electron configuration of an atom with atomic number 15: 1s 22s22p63s23p3, so 6 p orbitals are occupied.  C: 2px and 2py orbitals are perpendicular to each other. A  bond requires two parallel orbitals of the same orientation (2px with 2px and 2py with 2py).  D: A  bond is defined as a bond formed by the head -on (axial) overlap of atomic orbitals (such as s-s, s-p, or p-p). Its electron density is concentrated directly along the internuclear axis connecting the two nuclei. 2 Use of the Data Booklet is relevant to this question. Which compound consists of a cation and an anion that does not contain the same number of electrons as each other? A LiH B NaOH C NH4F D ZnCO3 Answer: D  A: Number of electrons in Li+: 3 – 1 = 2; Number of electrons in H−: 1 + 1 = 2  B: Number of electrons in Na+: 11 – 1 = 10; Number of electrons in OH−: 8+ 1 + 1 = 10  C: Number of electrons in NH4+: 7 + 4 – 1 = 10; Number of electrons in F−: 9 + 1 = 10  D: Number of electrons in Zn2+: 30 – 2 = 28; Number of electrons in CO32−: 6 + 3 × 8 + 2 = 32

3 © EJC 9476/01/J2PRELIM/26 [Turn over 3 For the following chlorine-containing species, ClO3−, ICl2−, PCl4+ Which statements are correct? 1 All atoms achieve stable octet configuration. 2 ClO3− is trigonal pyramidal, ICl2− is linear, while PCl4+ is tetrahedral. 3 The bond angles increase in the order: ClO – 3 < PCl4+ < ICl2−. A 1, 2 and 3 B 2 and 3 only C 1 only D 2 only Answer: B  1: In all three species, the central atom has an expanded octet (10 electrons). ICl2⁻ has 5 paired-electron domains (AX2E3), PCl4+ has 4 paired-electron domains (AX4), and ClO3⁻ has 4 paired-electron domains (AX3E). Chlorine and iodine as the central atom exceed 8 electrons in the stable resonance structures in ClO3− and ICl2⁻. Only the atoms (Cl, F, O), as terminal atoms, satisfy the octet rule.  2: Using VSEPR theory: • ClO3⁻ has 3 bonding pairs + 1 lone pair (4 electron domains). The parent geometry is tetrahedral, making the molecular shape trigonal pyramidal. → Trigonal pyramidal. • PCl4+ has 4 bonding pairs + 0 lone pairs (4 electron domains). The 4 electron pairs occupy tetrahedral positions, resulting in a tetrahedral molecule. → Tetrahedral. • ICl2⁻ has 2 bonding pairs + 3 lone pairs (5 electron domains total). The geometry is trigonal bipyramidal, with lone pairs occupying the equatorial positions to minimize repulsion, resulting in a linear molecular shape. → Linear.  3: The bond angles match the molecular shapes perfectly: • ClO3⁻ is Trigonal pyramidal - bond angles of approximately 107° (~106°). • PCl4+ is Tetrahedral - bond angles are 109.5°. • ICl2⁻ is Linear - bond angle of exactly 180°. • The lone pair compresses bond angles slightly in ClO3⁻ (one lone pair) compared to PCl4+ which is tetrahedral, and not at all in the linear ICl2⁻. • Bond angles increase in the order: ClO3⁻ < PCl4+ < ICl2⁻

4 © EJC 9476/01/J2PRELIM/26 4 Which of the following substances is least soluble in dilute NaOH(aq)? A SiO2 B P4O10 C Na2O D Al2O3 Answer: A Option Substance Behavior with NaOH Dissolves? A SiO2 SiO2 is acidic by nature. No reaction with dilute NaOH(aq). Reacts with hot conc. NaOH to form Na2SiO3(aq). SiO2(s) + 2NaOH(conc) → Na2SiO3(aq) + H2O(l) ❌ No B P4O10 Acidic oxide. Reacts with NaOH to form salt and water. P4O10(s) + 12NaOH(aq) → 4Na3PO4(aq) + 6H2O(l) ✅ Yes C Na2O Basic oxide. Reacts vigorously with water (present in the aqueous NaOH) to form NaOH: Na2O + H₂O → 2NaOH ✅ Yes D Al2O3 Amphoteric oxide. Reacts with NaOH to form soluble sodium aluminate: Al2O3 + 2NaOH + 3H2O → 2Na[Al(OH)4] ✅ Yes 5 20.0 cm 3 of 0.02 mol dm –3 aqueous sodium bromate( V), NaBrO 3, was found to react completely with 80.0 cm3 of 0.01 mol dm–3 hydroxylamine, NH2OH. The half equation for the reduction of bromate(V) ion is shown. BrO – 3(aq) + 6H+(aq) + 6e− → Br−(aq) + 3H2O(l) What could be the nitrogen-containing product of this reaction? A NO B NO2 C N2O D NO – 3 Answer: A n(BrO3−) = (0.02)(0.02) = 0.0004 mol n(NH2OH) = (0.08)(0.01) = 0.0008 mol BrO3− ≡ 2NH2OH ≡ 6e−  Every 1 mol of NH2OH lose 3 mol of e– ( )2O.N. of N in NH OH=3 -1 +2=-1 O.N. of N in product 1 3 2=− + =+

5 © EJC 9476/01/J2PRELIM/26 [Turn over 6 Which factor is the most significant in explaining the non -ideal behaviour of the gases present in the reaction chamber in the Haber process? A the presence of a catalyst B the high pressure of 250 atm C the high temperature of 450 oC D strong bonds between the atoms in nitrogen molecules Answer: B Real gases deviate most from ideal gas behaviour at high pressure and low temperature. At high pressure, the volume of the gas molecules compared to the volume occupied by the gas becomes more significant as the volume occupied by the gas becomes smaller. 7 Which equation shows the first reactant acting as a Bronsted–Lowry acid? A O2– + H2O → 2OH– B H2S + SO2 → 2H2O + 3S C C6H5OH + OH– → C6H5O− + H2O D HNO3 + H2SO4 → H2NO3+ + HSO4– Answer: C Bronsted–Lowry acid is a proton donor.  A: Oxide anion is a proton acceptor, hence a Bronsted–Lowry base.  B: This is not a Bronsted–Lowry acid–base reaction as H2S did not lose a proton (H+) to form HS−.  C: C6H5OH loses a proton to form C6H5O−.  D: HNO3 is a proton acceptor, hence a Bronsted–Lowry base.

6 © EJC 9476/01/J2PRELIM/26 8 Use of the Data Booklet is relevant to this question. Which property decreases from sodium to aluminium and also from aluminium to phosphorus? A melting point of the oxide B melting point of the element C electrical conductivity of the element D pH of the resulting solution when the chloride is added to water Answer: D  A: Melting point of oxide increases from Na to Al, then falls sharply. (The high charge density of Al3+ gives very strong ionic bonding, while P4O10 is simple molecular)  B: Melting point of the element increases from Na to Al, then falls sharply. (The metallic bonding strengthens from Na to Al due to more delocalised electrons per metal ion, while P4 is simple molecule with id-id between molecules only)  C: Electrical condu