RI 2026 Transition Metals Tut Ans

-1- Suggested Answers to Transition Elements Practice Questions Question 1 N08/2/4 (a) Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1 ; Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 (b) (i) [Ag(NH3)2]Cl Note: The question requires the candidate to give the formula of compound B. Hence, [Ag(NH 3)2]+ is not acceptable. Do check if the question asks for the formula of compound or complex ion and give the answer accordingly. (ii) Cation in CuCl2(aq): [Cu(H2O)6]2+ Cation in C: [Cu(NH3)4(H2O)2]2+ or [Cu(NH3)4]2+ (c) (i) CuCl2 + 2HCl → H2[CuCl4]; D CuCl2 + Cu + 2HCl → 2H[CuCl2] E (c) (ii) Complex ion in D: [CuCl4]2− Complex ion in E: [CuCl2]− (charged, thus soluble in water to give colourless solution) (c) (iii) ligand exchange (c) (iv) D is tetrahedral in shape. Also accepted: square planar (c) (v) +1 (d) (i) Reduction Reduction: [CuCl4]2− + e−  CuCl + 3 Cl− Oxidation: SO2 + 2 H2O  SO42− + 4 H+ + 2e− Overall: 2[CuCl4]2− + SO2 + 2 H2O  2 CuCl + 6 Cl− + SO42− + 4H+ (d) (ii) CuCl (Equation for conversion of E to F: H[CuCl2](aq) + H2O (l) → CuCl(s) + H3O+(aq) + Cl−(aq) The expected answer for identity of F is CuCl, since it is stated in the question that F contains only Cu and Cl. Extra information: However the complex form, F, could be [CuCl(H 2O)], a 2-coordinate neutral complex, thus insoluble in water. The addition of H2O to [CuCl2]– causes a shift in the position of equilibrium resulting in Cl− being replaced by H2O: [CuCl2]– + H2O ⇌ [CuCl(H2O)] + Cl−) (e) (i) Molar ratio of Cu to F to K = : : = 0.338 : 2.036 : 1.018 = 1 : 6 : 3 Empirical formula of G: K3CuF6 (e) (ii) +3 (f) (i) Both E and F contain Cu(I) which has electronic configuration of [Ar]3d 10. Cu(I) has a fully filled 3d subshell, i.e. absence of partially filled d subshell. Consequently d-d transitions are not possible and hence both E and F are colourless. Note: A colourless compound transmits all wavelengths of visible light while a white compound reflects all wavelengths of visible light. 5.63 5.21 0.19 7.38 1.39 8.39

-2- (f) (ii) G contains CuF63− with the central Cu(III) which has electronic configuration [Ar] 3d8. • presence of F− ligands causes the • splitting of the five 3d orbitals in Cu3+ ion into two sets of slightly different energy levels. • Since the 3d subshell is partially filled, • electrons from the lower -energy d orbitals can absorb energy corresponding to certain wavelengths from the visible spectrum and get promoted to the higher -energy d orbitals. (d-d transitions) • The colour observed is the complement of the colour absorbed. • This is illustrated in the diagrams below: Question 2 (i) dxz (dxy, dyz are also acceptable) 𝐝𝐝𝐳𝐳𝟐𝟐 𝐝𝐝𝐱𝐱𝟐𝟐−𝐲𝐲𝟐𝟐 (ii) 3dx2-y2 and 3dz2 : These orbitals have their greatest electron density along the co-ordinate axes on which the ligands are situated. Hence electrons in these orbitals are pointing towards the lone pairs of ligands, and will be repelled by them. L S P E C lower d-orbitals have the same energy in a n isolated Cu3+ ion In the presence of ligands, d -orbitals of Cu3+ are split into 2 sets of different energy, with energy gap corresponding to energies in the visible light spectrum. An e– in the lower energy level absorbs visible light energy corresponding to ∆ E and is promoted to the higher energy level. Energy gap (∆E)

-3- 3dxy, 3d yz, 3d xz : These orbitals have their greatest electron density in between the co - ordinate axes. Hence the repulsion between electrons in these orbitals and those of the approaching ligands will be less compared to electrons in 3dx2-y2 or 3dz2 orbitals. Hence the d-orbitals are split into two different energy levels. (iii) • Electronic configuration of Fe2+ : [Ar]3d6 • As shown in Fig. 1, in haemoglobin, the presence of ligands causes the splitting of the five 3d orbitals in Fe2+ into two sets of slightly different energy levels. • Since the 3d subshell in Fe2+ is partially filled, • electrons from the lower-energy d orbitals can absorb energy corresponding to certain wavelengths from the visible spectrum and get promoted to the higher-energy d orbitals. (d-d transitions) • The colour observed (red) is the complement of the colour absorbed (green). (iv) High Spin State Low Spin State (v) Electrons are negatively charged and experience electronic repulsion when paired together in the same orbital. Occupying an orbital singly minimises inter-electron repulsion. (vi) Oxyhaemoglobin has a larger energy gap. Due to the larger energy gap in oxyhaemoglobin, it is energetically more favourable for electrons to pair up in d-orbitals in the lower energy level (despite inter-electronic repulsion) than to occupy a d-orbital in the higher energy level singly. Hence electronic configuration of Fe2+ in oxyhaemoglobin is the low spin state. (Conversely, the energy gap between the 2 different levels in deoxyhaemoglobin is small so that it is energetically more favourable for electrons to occupy d -orbitals in higher energy level singly before pairing up). d orbitals of an isolated Fe2+ ion d orbitals of Fe2+ in haemoglobin energy gap E d orbitals of an isolated Fe2+ ion d orbitals of Fe2+ in haemoglobin energy gap E

-4- Question 3 (RI Prelim 2017/P3/Q1(c) and (d) (i) A ligand exchange reaction has taken place. [Fe(H2O)6]2+ + 6 CN− ⇌ [Fe(CN)6]4− + 6 H2O Examiner’s Comments • Ligand exchange/ligand displacement were acceptable as possible answers. • A handful of students who wrote the charge on the complex incorrectly were not given credit. • Note that in the question, it was already stated that the complex was [Fe(CN) 6]4−, thus ALL 6 H 2O ligands had to be exchanged for CN− ligands and not only one of them. (ii) The complexes contain different ligands, H2O and CN−, and these ligands split the d- orbitals of Fe2+ to a different extent. Thus, different wavelengths of light are absorbed to promote the electrons from the lower energy d-orbitals to the higher energy d-orbitals in the two complexes, leading to different complementary colours observed. Examiner’s Comments • Students who only mentioned that a different wavelength of light was absorbed without explaining why were not awarded any credit for their answers. (iii) Ecell = Ecathode − Eanode = +0.54 – (+0.36) = +0.18 V > 0. Oxidation of yellow [Fe(CN)6]4− to red [Fe(CN)6]3− by I2 is spontaneous. Thus, I2 can be used. Examiner’s Comments • It was important to calculate the Ecell for the reaction AND conclude that it was spontaneous before concluding that I2 can be used. A positive Ecell calculation is not sufficient to substantiate the possibility for the use of I2 if there is no link provided that the reaction is spontaneous/feasible. • A significant number of students used +0.77 instead of +0.36 and came to the incorrect conclusion that I 2 cannot be used. Note that the complex involved is [Fe(CN) 6]4- and not [Fe(H2O)6]2+. The value of +0.77 which is for Fe3+/Fe2+ is for the aqua complex. Question 4 (a) Thioglycolic acid acts as a reducing agent as it reduces Fe 3+ to Fe 2+ as shown in equation 5.1. (b) , tetrahedral or square planar (c) Ammonia was added to react with thioglycolic acid to form the corresponding anion, which can then complex with Fe2+ to form the pink-coloured complex, M. OR Ammonia was added to provide the basic conditions necessary for thioglycolic acid to complex with Fe2+ to form the pink-coloured complex, M. (d) Fe(SCH2COOH)2 + 2NH3 → [Fe(SCH2COO)2]2– + 2NH4+ (e)(i) [Fe(H2O)6]2+(aq) + 2NH3(aq) → [Fe(OH)2(H2O)4](