ASRJC 2026 Prelim P3 Ans

ASRJC JC2 PRELIM 2026 9476/03/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2026 JC 2 PRELIMINARY EXAMINATION NAME:______________________________ ( ) CLASS: 26 / _____ CHEMISTRY 9476/03 Paper 3 Structured Questions SUGGESTED SOLUTIONS 1 (a) Nickel is found in the d-block of the Periodic Table. Explain why nickel is described as a transition element. [1] Ni: [Ar] 3d8 4s2 Ni2+: [Ar] 3d8 • Ni atom has an incomplete/partially filled d subshell and it forms ions with an incomplete/partially filled d subshell. [1] (b) Transition elements form stable complex ions in aqueous solution. When water ligands in a hydrated metal ion are substituted by other ligands, the equilibrium constant for the reaction is referred to as stability constant, Kstab, of the new complex. For example: [Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l) Kc = Kstab = 2+ 36 2+ 6 2 6 3 [Ni(NH ) ] [Ni(H O) ][NH ] Table 1.1 lists the Kstab of three nickel(II) complexes at 25 °C. Table 1.1 complex ligand Kstab [Ni(NH3)6]2+ NH3 5.0 × 107 [Ni(EDTA)]2− EDTA4− 4.0 × 1018 [Ni(en)3]2+ en 2.0 × 1018 “en” is the abbreviation for ethylenediammine, H2NCH2CH2NH2.

2 ASRJC JC2 PRELIM 2026 9476/03/H2 (i) Predict, with the aid of an equation, how entropy will change when [Ni(EDTA)]2− is formed from [Ni(H2O)6]2+. Explain your answer. [2] [Ni(H2O)6]2+(aq) + EDTA4−(aq) [Ni(EDTA)]2-(aq) + 6H2O(l) [1] For the formation of [Ni(EDTA)]2-, there is an increase in the number of particles in the solution. This increases the number of ways to distribute particles and their energies . Hence, there is greater disorder, and entropy will increase. [1] (ii) Write an expression for the Kstab of the [Ni(EDTA)]2- complex ion. [1] [Ni(H2O)6]2+(aq) + EDTA4−(aq) [Ni(EDTA)]2-(aq) + 6H2O(l) Kstab = [ Ni(EDTA)2− ] [ Ni(H2O)62+ ] [EDTA4−] [1] correct expression (iii) Deduce an expression for the Kc for the following equilibrium in terms of Kstab of the complexes involved. Hence, calculate the value of Kc and include the units. [Ni(NH3)6]2+(aq) + EDTA4−(aq) [Ni(EDTA)]2−(aq) + 6NH3(aq) [2] [Ni(NH3)6]2+(aq) + EDTA4−(aq) [Ni(EDTA)]2−(aq) + 6NH3(aq) Kc = [ Ni(EDTA)2− ] [ NH3 ]6 [ Ni(NH3)62+ ] [EDTA4−] = [ Ni(EDTA)2− ] [ Ni(H2O)62+ ] [EDTA4−] × [ Ni(H2O)62+ ] [ NH3 ]6 [ Ni(NH3)62+ ] = Kstab(EDTA complex) × 1 Kstab(NH3 complex) [1] = 4.0 × 1018 5.0 × 107 = 8.0 × 1010 mol5 dm−15 [1] correct answer and units

3 ASRJC JC2 PRELIM 2026 9476/03/H2 [Turn over (iv) Name the type of isomerism displayed by [Ni(en)3]2+. Draw relevant 3 -dimensional structures to explain your answer, showing clearly the atoms involved in bonding. You can use N N to represent the ligand, en. [2] Enantiomerism Ni N N N N. II N N 2+ Ni N N N N. II N N 2+ Or N CH2 CH2 H2 NH2 H2N CH2 CH2 NH2 H2N CH2 H2C 2+ Ni NH2 II NCH2 CH2 H2 NH2 NH2H2C CH2 NH2 NH2 CH2 CH2 H2N 2+ Ni II [1] Name [1] Diagrams

4 ASRJC JC2 PRELIM 2026 9476/03/H2 (c) Copper forms numerous compounds such as copper(II) sulfate and copper oxides which are widely used in agriculture as fungicides and soil additives. When copper(II) sulfate, CuSO4, dissolves in water, a blue solution A is formed. Fig. 1.1 shows some reactions of solution A. Fig. 1.1 (i) Identify the species present in A, B and C. [2] species A [Cu(H2O)6]2+ B Cu(OH)2(H2O)4 C [Cu(NH3)4(H2O)2]2+ [2] (ii) Explain how B and C are formed. For each reaction that occurs, identify the type of reaction and write relevant chemical equation(s). [3] When dilute NH3 is added gradually, acid-base reaction occurs to form B. NH3 (aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) [Cu(H2O)6]2+ (aq) + 2OH– (aq) ⇌ Cu(H2O)4(OH)2 (s) + 2H2O(l) [1] In excess NH 3, NH 3 ligands replace H 2O ligands in a ligand exchange reaction to form C. [Cu(H2O)6]2+ (aq) + 4NH3 (aq) ⇌ [Cu(NH3)4(H2O)2]2+ (aq) + 4H2O(l) [1] [1] both types of reaction (iii) Suggest why there are different numbers of ligands surrounding the Cu2+ ion in A and [CuCl4]2−. [1] Cl− ions are bigger than H2O and hence less Cl− ligands can surround Cu2+. [1] blue solution A blue solid B dark blue solution C yellow solution [CuCl4]2− NH3(aq) excess NH3(aq) concentrated HCl

5 ASRJC JC2 PRELIM 2026 9476/03/H2 [Turn over (d) Fig. 1.2 shows the energy levels of the five 3d orbitals in Cu2+ as an isolated gaseous species and Cu2+ in the tetrahedral complex ion CuCl42−. Fig. 1.2 (i) Explain the relative energy levels of the 3d orbitals of Cu2+ in [CuCl4]2−. [2] In a tetrahedral complex, the four C l− ligands approach the central metal cation in between the x, y and z axes. [1] 3dxy, 3dxz and 3dyz orbitals are at higher energy level as there is greater electrostatic repulsion occurring between the lone pairs of the ligands and electr ons in these orbitals. 3dz2 and 3d x2–y2 orbitals are at lower energy level as there is smaller electrostatic repulsion occurring between the lone pairs of the ligands and electrons in these orbitals. [1] (ii) Explain the origin of the yellow colour of [CuCl4]2−. [3] Cu2+: [Ar] 3d9 • In the presence of Cl− ligands, the partially filled 3d orbitals of Cu2+ are split into two levels with a small energy gap, E. • When a 3d electron absorbs energy from the visible light region corresponding to E, • this electron is promoted from the d orbital of a lower energy level to a d orbital of a higher energy level. • The yellow colour of CuCl42− ion observed is the complementary of the purple (violet) colour absorbed. [3]

6 ASRJC JC2 PRELIM 2026 9476/03/H2 (e) Iron(II) hydroxide, Fe(OH)2, is sparingly soluble in water and has a Ksp value of 7.9 × 10−15 at 298 K. (i) Write the expression for Ksp Fe(OH)2. [1] Fe(OH)2(s) Fe2+(aq) + 2OH–(aq) Ksp Fe(OH)2 = [Fe2+][OH–]2 [1] (ii) Calculate the pH of a saturated solution of Fe(OH)2. [2] Let solubility of Fe(OH)2 be x mol dm−3 Ksp Fe(OH)2 = [Fe2+][OH–]2 7.9 × 10−15 = ( x )( 2x )2 = 4x3 x = √7.9 x 10-15 43 = 1.2546 × 10−5 [1] [OH–] = 2 × 1.2546 × 10−5 = 2.5092 × 10−5 mol dm-3 pH = 14 – pOH = 14 – (– lg (2.5092 × 10−5)) = 9.40 [1] [Total: 22]

7 ASRJC JC2 PRELIM 2026 9476/03/H2 [Turn over 2 (a) Compound N can be synthesised from bromobenzene using the route shown in Fig. 2.1. Br step 1 Br step 2 Br O [O] step 3 J K L step 4 M step 5 Br OH OO N CH3CH2OH, conc. H2SO4, heat Fig. 2.1 (i) Suggest the type of reaction and the reagents and conditions for step 1. [2] Electrophilic substitution [1] CH3Cl, anhydrous AlCl3 [1] (ii) Step 2 is an oxidation reaction. Write an equation for the reaction in step 2. Use [O] to represent oxygen from the oxidising agent. [1] Br Br O 2[O]+ + H2O [1] (iii) Suggest structures for the organic compounds L and M. [2] Br CN OH [1] Br COOH OH [1] L M (iv) Suggest reagents and conditions for step 3. [1] HCN with NaCN catalyst [1]

8 ASRJC JC2 PRELIM 2026 9476/03/H2 (b) Separate samples of three organobromine compounds are added to warm AgNO3(aq). Br Br Br O P Q R State the expected observations, if any. Explain your answer. [4] P: no precipitate Q: slower appearance of cream precipitate R: cream precipitate appears immediately [1] for observations Ppt is formed faster for R because the carbon atom of the acyl group is bonded to two highly electronegative atoms (O and Br ) while the carbon atom in Q is bonded to only one e