2025 ASRJC Prelim H2Chem P3 MS

1 9729/03/H2 ANDERSON SERANGOON JUNIOR COLLEGE 2025 JC 2 PRELIMINARY EXAMINATION PAPER 3 SOLUTIONS Section A 1 (a) Transition elements have characteristic physical and chemical properties. (i) Explain what is meant by the term transition element. [1] A transition element is a d block element that can form one or more stable ions with partially filled d subshells. [1] (ii) Explain why the melting point and density of nickel is higher than that of calcium. [3] • [A] In Ni, both the 3d and 4s electrons are available for metallic bonds since the energy level difference between the 3d and 4s orbitals is small. In Ca, only two 3s electrons are available for metallic bonding. • [B] The greater number of 3d and 4s delocalised electrons available for metallic bonding in Ni results [A or B - 1m] • in stronger electrostatic attraction between the cations and the ‘sea’ of delocalised electrons. More energy is required to overcome the stronger metallic bonds in transition elements as compared to Ca. [1m] • Ni has smaller atomic radii and higher atomic mass compared to Ca. • Also, Ni has more closely packed [1m] structure due to their stronger metallic bonding. • Thus, transition elements are denser than Ca. (b) (i) A solution containing the [Ni(H2O)6]2+ complex ion is green. When 1,2 -diaminoethane, en, H 2NCH2CH2NH2, is added, the colour of the solution changes to purple. This is due to the formation of the [Ni(en)2(H2O)2]2+ complex ion. Explain why the two solutions are coloured, and why the colours are different. [3] In the presence of the ligands, the partially filled 3d orbitals of Ni2+ are split into two levels with a small energy gap, E (d orbital splitting). 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 (d–d transition). The colour observed is the complementary of the colour absorbed. The colours are different as different ligands split the partially filled 3d orbitals to different extents with different E. Red is absorbed for [Ni(H2O)6]2+ complex and yellow is absorbed for [Ni(en)2(H2O)2]2+ complex. [1]: presence of ligand; partially filled 3d–orbital for Fe2+(aq); small energy gap [1]: d–d transition explanation; colour is the complement of the colour absorbed [1]: different ligands, different E

2 9729/03/H2 (ii) [Ni(en)2(H2O)2]2+ complex ion can exist in three different forms where the ions differ in the spatial arrangement of the ligands around the central metal ion. Fig. 1.1 shows one of the isomers. Ni N N OH2 N H2O N 2+ H2 H2 H2 H2 Fig. 1.1 Draw another isomer and state the type of isomerism. [2] 2+ Ni NH2O N N N H2O H2 H2 H2 H2 [1] cis-trans isomerism [1]

3 9729/03/H2 (c) When a sample of aqueous copper(II) sulfate was added to a small amount of aqueous ammonia, a blue precipitate A was formed. Upon adding excess aqueous ammonia, A dissolved, and a deep blue solution containing a complex ion, B, was formed. When another sample of aqueous copper( II) sulfate was added to concentrated hydrochloric acid, a yellow-green solution was formed. (i) Identify A and B and explain their formation. Include relevant equations. [4] [1] correct ID of A and B A is Cu(OH)2(H2O)4 [accept Cu(OH)2] and B is [Cu(NH3)4(H2O)2]2+. To explain formation of A [1] for mention of acid-base reaction with correct relevant equation Either In aqueous solution, the weak base NH3 undergoes partial ionisation, producing OH– ions. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) When a small amount of NH3(aq) is added, the OH–(aq) undergoes acid-base reaction with [Cu(H2O)6]2+(aq) to produce a blue precipitate of Cu(OH)2. [Cu(H2O)6]2+(aq) + 2OH–(aq) [Cu(OH)2(H2O)4](s) + 2H2O(l) ......(1) blue solution blue ppt OR Acid-base reaction occurs [Cu(H2O)6]2+SO42– (aq) + 2NH3(aq) [Cu(OH)2(H2O)4](s) + (NH4)2SO4 (l) .......(1) To explain formation of B [1] for mention of ligand exchange reaction with correct relevant equation [1] explanation for ppt dissolving The increasing addition of NH3(aq) shifts the position of equilibrium (2) to the right, forming a deep blue solution of [Cu(NH 3)4(H2O)2]2+(aq). As the concentration of [Cu(H2O)6]2+(aq) decreases, the position of equilibrium (1) shifts left, causing the blue precipitate of Cu(OH)2 to dissolve. When excess NH 3(aq) is added, ligand exchange reaction occurs and the [Cu(NH3)4(H2O)2]2+ ion is formed. [Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) ......(2) blue solution deep blue solution (ii) Write an equation to explain the observation of the yellow-green solution. [1] [Cu(H2O)6]2+ + 4Cl– [CuCl4]2– + 6H2O [1]

4 9729/03/H2 (d) Ruthenium can form complexes with ligands like H2NCH2CH2NH2 (represented as en) and NH3. Three half equations involving rubidium complex ions are shown in Table 1.1. Table 1.1 half-cell half equation A [Ru(H2O)6]3+ + e- ⇌ [Ru(H2O)6]2+ B [Ru(NH3)6]3+ + e- ⇌ [Ru(NH3)6]2+ C [Ru(en)3]3+ + e- ⇌ [Ru(en)3]2+ (i) Two electrochemical cells are set up to compare the standard electrode potential, Eo, of the three half-cells. Fig 1.2 shows the relative potential of each electrode. half-cell C half-cell B half-cell A half-cell C Fig 1.2 Using this information, state and explain the order of standard reduction potential, Eθ, for the three half-cells from the least negative to the most negative. [2] In the first electrochemical cell, half-cell C has a positive electrode (cathode) and half-cell B has a negative electrode (anode). Hence, reduction take place in half -cell C and oxidation take place in half-cell B. Therefore half-cell C will have a less negative Eθ then half-cell B. In the second electrochemical cell, half-cell A has a positive electrode (cathode) and half- cell C has a negative electrode (anode). Hence , reduction take place in half -cell A and oxidation take place in half-cell C. Therefore half-cell A will have a less negative Eθ then half-cell C. [1] Hence, from the order of Eθ from the least negative to the most negative is A, C, B. [1]

5 9729/03/H2 (ii) The standard electrode potential of the half-cell A is +0.25 V. An electrochemical cell was set up using half-cell A and a [Cu(NH3)4]2+/Cu half cell. Use data from the Data Booklet to calculate the Eo cell for this cell [1] [Ru(H2O)6]3+ + e ⇌ [Ru(H2O)6]2+ +0.25V (Ered) [Cu(NH3)4]2++ 2e ⇌ Co + 4NH3 – 0.05 V (Eoxd) Eocell = 0.25 – (– 0.05) = +0.30 V [1] (iii) Write the overall equation for the reaction that occurs in the cell in (d)(ii). Using the Eocell you have calculated in (d)(ii), calculate a value of ∆Go for the cell reaction represented by your overall equation. [2] [R]: [Ru(H2O)6]3+ + e → [Ru(H2O)6]2+ [O]: Cu + 4NH3 → [Cu(NH3)4]2+ + 2e Cu + 4NH3 + 2[Ru(H2O)6]3+ → [Cu(NH3)4]2+ + 2[Ru(H2O)6]2+ [1] ∆Go = −nFEocell The no. of electrons exchanged in the overall balanced equation is 2. Hence, n = 2. ∆Go = −2 x 96500 x (+0.300) = –57900 J mol–1 or –57.9 kJ mol–1 [1] [Total: 19]

6 9729/03/H2 2 (a) The reactivity of three nitrogen -containing compounds with aqueous bromine is shown in the Table 2.1. Table 2.1 NO2 nitrobenzene NHCOCH3