▶️ Answer/Explanation
(a)(i) Solubility increases down the group (Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂).
Explanation: The solubility of Group 2 hydroxides increases down the group due to the increase in ionic radius and decrease in lattice energy. Larger ions have weaker electrostatic forces, making the lattice easier to break and the compound more soluble.
(a)(ii) pH increases down the group (Mg(OH)₂ < Ca(OH)₂ < Sr(OH)₂).
Explanation: As solubility increases, more OH⁻ ions are released into the solution, leading to higher alkalinity (pH). Thus, Sr(OH)₂ (most soluble) has the highest pH, while Mg(OH)₂ (least soluble) has the lowest.
(b) Mass of Ba(OH)₂ = 0.0215 g.
Explanation:
1. pH = 12.2 ⇒ pOH = 14 – 12.2 = 1.8 ⇒ [OH⁻] = 10⁻¹·⁸ = 0.0158 moldm⁻³.
2. Ba(OH)₂ dissociates as Ba(OH)₂ → Ba²⁺ + 2OH⁻ ⇒ [Ba(OH)₂] = [OH⁻]/2 = 0.00792 moldm⁻³.
3. Moles in 250 cm³ = 0.00792 × 0.25 = 0.00198 mol.
4. Mass = moles × \(M_r\) = 0.00198 × 171.3 = 0.0215 g.
(c)(i) \(K_{sp} = [Fe^{2+}][OH^-]^2\).
Explanation: The solubility product expression is derived from the dissociation equation Fe(OH)₂(s) ⇌ Fe²⁺(aq) + 2OH⁻(aq).
(c)(ii) \(K_{sp} = 8.01 \times 10^{-16} \, \text{mol}^3\text{dm}^{-9}\).
Explanation:
1. [Fe²⁺] = 5.85 × 10⁻⁶ moldm⁻³, [OH⁻] = 2 × 5.85 × 10⁻⁶ = 1.17 × 10⁻⁵ moldm⁻³.
2. \(K_{sp} = (5.85 \times 10^{-6}) \times (1.17 \times 10^{-5})^2 = 8.01 \times 10^{-16} \, \text{mol}^3\text{dm}^{-9}\).
▶️ Answer/Explanation
(a)(i) A transition element is defined as an element that forms one or more stable ions with incomplete d orbitals.
Explanation: Transition elements are characterized by their ability to form ions with partially filled d subshells, e.g., Fe²⁺ (3d⁶) or Cu²⁺ (3d⁹).
(a)(ii) Transition elements can form complex ions because they have vacant d orbitals that are energetically accessible for ligand bonding.
Explanation: The partially filled d orbitals allow transition metals to accept lone pairs from ligands, forming coordinate covalent bonds.
(b)(i) Degenerate d orbitals are orbitals of the same energy level (e.g., all five 3d orbitals in an isolated Ag⁺ ion).
(b)(ii) The \(3d_{xy}\) orbital has a cloverleaf shape in the xy-plane with lobes between the axes.
(c) Preparation of Tollens’ reagent:
Step 1: \(2AgNO_3 + 2NaOH \rightarrow Ag_2O + 2NaNO_3 + H_2O\)
Step 2: \(Ag_2O + 4NH_3 + H_2O \rightarrow 2[Ag(NH_3)_2]OH\)
Explanation: Silver oxide dissolves in ammonia to form the diamminesilver(I) complex.
(d) The [Ag(NH₃)₂]⁺ ion has a linear shape.
Bond angle for H-N-Ag = 180° (linear geometry).
Bond angle for N-Ag-N = 180°.
(e) Half-equations:
Positive electrode: \(Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-\)
Negative electrode: \(Zn + 2OH^- \rightarrow Zn(OH)_2 + 2e^-\)
Explanation: Ag₂O is reduced to Ag, while Zn is oxidized to Zn(OH)₂ in the alkaline electrolyte.
(f) The coordination polymer structure shows two repeat units of \([Ru(dps)Cl_4]^–\) linked via dps ligands:
Explanation: Each Ru³⁺ is coordinated to four Cl⁻ ions and two nitrogen atoms from dps ligands, forming a chain.
▶️ Answer/Explanation
(a) Li₂C₂O₄•H₂O → Li₂CO₃ + CO + H₂O
Explanation: Heating decomposes the compound into lithium carbonate (white residue), carbon monoxide, and water. The residue (Li₂CO₃) reacts with HNO₃ to produce CO₂.
(b) CaC₂O₄ decomposes at a lower temperature than BaC₂O₄.
Explanation: Smaller Group 2 cations (Ca²⁺) polarize the oxalate anion more effectively, weakening the C₂O₄²⁻ bonds and lowering decomposition temperature.
(c) Transition metal complexes are coloured due to d-d electron transitions.
Explanation: In [Fe(C₂O₄)₃]³⁻, Fe³⁺ has partially filled d-orbitals. Light absorption promotes d-electrons to higher energy levels, transmitting complementary colours (green here).
(d) 2K₃[Fe(C₂O₄)₃] → 2K₂[Fe(C₂O₄)₂] + 1K₂C₂O₄ + 3CO₂
Explanation: Balancing the equation ensures conservation of Fe, K, C, and O atoms.
(e) The two stereoisomers are optical isomers (enantiomers).
Explanation: [Fe(C₂O₄)₃]³⁻ is octahedral with bidentate C₂O₄²⁻ ligands, forming non-superimposable mirror images. Diagrams show left- and right-handed configurations.
(f) Buffer action equations:
– With added acid (H⁺): HC₂O₄⁻ + H⁺ → H₂C₂O₄
– With added base (OH⁻): HC₂O₄⁻ + OH⁻ → C₂O₄²⁻ + H₂O
Explanation: HC₂O₄⁻ acts as both a weak acid (reacts with OH⁻) and weak base (reacts with H⁺), stabilizing pH.
(g) Overall reaction:
(COOH)₂ + ½O₂ → 2CO₂ + H₂O
\(E^o_{cell} = +0.46 \text{ V}\)
Explanation:
- Anode (oxidation): (COOH)₂ → 2CO₂ + 2H⁺ + 2e⁻ (\(E^o = -0.49 \text{ V}\))
- Cathode (reduction): ½O₂ + 2H⁺ + 2e⁻ → H₂O (\(E^o = +0.95 \text{ V}\))
- \(E^o_{cell} = E^o_{\text{cathode}} – E^o_{\text{anode}} = 0.95 – (-0.49) = +1.44 \text{ V}\)
▶️ Answer/Explanation
(a) Standard electrode potential (\(E^o\)) is the voltage of a half-cell compared to the standard hydrogen electrode (SHE) under standard conditions: 1 mol dm⁻³ concentration, 101 kPa pressure, and 298 K temperature.
Explanation: \(E^o\) measures the tendency of a species to gain electrons relative to SHE, with all reactants and products in their standard states.
(b) (i)
Explanation: The cell consists of an Ag electrode in 1 mol dm⁻³ \(Ag^+\)(aq) connected via a salt bridge to the SHE. A voltmeter measures the potential difference.
(b) (ii) The electrode potential \(E\) decreases from \(E^o\) because the Nernst equation predicts a lower potential for reduced \([Ag^+]\): \(E = E^o – \frac{RT}{nF} \ln \left( \frac{1}{[Ag^+]} \right)\).
Explanation: Lower \(Ag^+\) concentration reduces the driving force for reduction, decreasing the electrode potential.
(c) Enthalpy change of solution (\(\Delta H^o_{sol}\)) is the energy change when 1 mole of a solute dissolves in a solvent to form an infinitely dilute solution under standard conditions.
Explanation: It includes energy to break the lattice (\(\Delta H^o_{latt}\)) and energy released during solvation (\(\Delta H^o_{hyd}\)).
(d) (i) The completed energy cycle is:
Explanation: The cycle links \(\Delta H^o_{latt}\) (solid → gaseous ions) with \(\Delta H^o_{sol}\) (solid → aqueous ions) and \(\Delta H^o_{hyd}\) (gaseous ions → aqueous ions).
(d) (ii) \(\Delta H^o_{latt} = \Delta H^o_{sol} – \Delta H^o_{hyd} = 22.6 – ( -466.5 + (-318.5) ) = -811.6 \, \text{kJ mol}^{-1}\).
Explanation: Lattice energy is calculated using Hess’s Law: \(\Delta H^o_{latt} = \Delta H^o_{sol} – (\Delta H^o_{hyd}(Ag^+) + \Delta H^o_{hyd}(NO_3^-))\).
(e) The trend is Mg(NO₃)₂(s) > NaNO₃(s) > RbNO₃(s).
Explanation: – \(Mg^{2+}\) has a higher charge density than \(Na^+\) or \(Rb^+\), increasing lattice energy. – \(Na^+\) is smaller than \(Rb^+\), leading to stronger ionic bonding in NaNO₃ compared to RbNO₃.
▶️ Answer/Explanation
(a)(i) The initial rate is determined by drawing a tangent at \(t = 0\) on the graph and calculating its gradient. From Fig. 5.1, the gradient is approximately 0.028 mol dm⁻³ s⁻¹.
Explanation: The tangent’s slope at \(t = 0\) gives the initial rate, as rate = \(-\frac{d[S_2O_8^{2–}]}{dt}\). The calculated value falls within the range 0.016–0.040 mol dm⁻³ s⁻¹.
(a)(ii) With a large excess of \(I^–\), its concentration remains effectively constant, simplifying the rate equation to pseudo-first order: rate = \(k’ [S_2O_8^{2–}]\), where \(k’ = k[I^–]\).
Explanation: The half-life of a first-order reaction depends only on \(k’\), allowing \(k\) to be derived if \([I^–]\) is known.
(b)
Equation 1: \(2Fe^{2+} + S_2O_8^{2–} \to 2Fe^{3+} + 2SO_4^{2–}\)
Equation 2: \(2Fe^{3+} + 2I^– \to 2Fe^{2+} + I_2\)
Explanation: \(Fe^{2+}\) is oxidised by \(S_2O_8^{2–}\) in step 1 and reduced back by \(I^–\) in step 2, regenerating the catalyst.
(c) Increasing temperature raises the rate constant (\(k\)) and thus the reaction rate, as more molecules possess energy ≥ activation energy.
Explanation: The Arrhenius equation \(k = Ae^{-E_a/RT}\) shows \(k\) increases exponentially with temperature.
(d) For a first-order reaction, \(t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0158} \approx 43.9 \text{ s}\).
Explanation: The half-life is independent of concentration for first-order reactions, calculated directly from \(k\).
(e)
Step 1 (slow/RDS): \(NO + Br_2 \to NOBr_2\)
Step 2 (fast): \(NOBr_2 + NO \to 2NOBr\)
Rate-determining step: Step 1
Explanation: The rate law (first order in both NO and Br₂) matches the stoichiometry of the slow step, confirming it as the RDS.
The mechanism of this reaction is similar to that of the alkylation of benzene.
▶️ Answer/Explanation
(a)(i) The partition coefficient (\(K_{pc}\)) is the ratio of concentrations of a solute in two immiscible solvents at equilibrium.
(a)(ii) Mass of X extracted into CS₂ = 0.642 g
Explanation:
1. Initial mass in 40.0 cm³ aqueous solution: \(\frac{1.85}{100} \times 40 = 0.74 \text{ g}\)
2. Let \(y\) = mass in CS₂. Then \(\frac{y/25}{(0.74 – y)/40} = 10.5\)
3. Solving gives \(y = 0.642 \text{ g}\)
(b) Hybridisation in C₆H₆ isomers:
Explanation:
– Kekulé benzene: 6 \(sp^2\) carbons (planar, 120°)
– Dewar benzene: 4 \(sp^3\) (tetrahedral) + 2 \(sp^2\)
– Ladenburg benzene: 3 \(sp^3\) + 3 \(sp^2\)
(c) Delocalised benzene:
– Planar hexagonal ring with 6 \(sp^2\) hybridised carbons
– C-C-H bond angle: 120°
– C-C bonds: delocalised π-system (1.5 bond order)
– C-H bonds: σ bonds
(d) Instability of Dewar/Ladenburg benzene:
– High angle strain in small rings (Dewar)
– Loss of aromatic stabilisation (both)
– Torsional strain (Ladenburg)
(e) \(^1H\) NMR peaks:
Explanation:
– Delocalised benzene: 1 peak (all H equivalent)
– Dewar: 3 peaks (different H environments)
– Ladenburg: 4 peaks
(f)(i) Electrophile formation:
\(\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{FeBr}_3 \rightarrow \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+ + \text{FeBr}_4^-\)
(f)(ii) Mechanism:
Key steps:
1. Curly arrow from benzene ring to electrophile
2. Formation of arenium ion intermediate
3. Deprotonation to restore aromaticity
(f)(iii) By-products:
– Y: Dimer of two phenylethanone units bridged by butyl chains
– Z: Monobutylated phenylethanone
▶️ Answer/Explanation
(a) The systematic name of ester A is methyl pentanoate.
Explanation: A is formed from methanol (methyl) and pentanoic acid (pentanoate), giving the IUPAC name methyl pentanoate.
(b)(i) Retention time is the time taken for a compound to travel from the injection point to the detector in gas chromatography.
Explanation: It is characteristic of each compound under specific conditions.
(b)(ii) The percentage by mass of ester D is 29.5%.
Explanation: The area under peak D is 59, and the total area is 200 (45 + 56 + 40 + 59). Thus, %D = (59/200) × 100 = 29.5%.
(c)(i) The completed Table 7.1 is:
Explanation:
- Ester B: 3 peaks in \(^1H\) NMR (due to 3 proton environments) and 4 peaks in \(^{13}C\) NMR (4 carbon environments).
- Ester C: 2 peaks in \(^1H\) NMR (2 proton environments) and 3 peaks in \(^{13}C\) NMR (3 carbon environments).
(c)(ii) Esters A and B have at least one triplet peak in their \(^1H\) NMR spectrum.
Explanation: Triplets arise from protons adjacent to \(-CH_2-\) groups (e.g., \(-OCH_2CH_3\) in A and B).
(d) The structures of compounds F, G, H, J, K, and L are:
Explanation:
- F: Stereoisomeric compound with \(-COOH\) (effervesces with \(Na_2CO_3\)) and \(-COCH_3\) (reacts with \(I_2\) to form iodoform, G).
- G: Yellow precipitate is iodoform (\(CHI_3\)).
- H: Sodium salt of the acid formed after iodoform reaction.
- J: Reduction product (\(C_6H_{12}O_2\)) from \(LiAlH_4\).
- K: Acyl chloride formed with SOCl₂.
- L: Ester formed from K and propan-2-ol.
▶️ Answer/Explanation
(a)(i)
Answer: 2
Explanation: Neotame has two chiral centers – the carbon atoms bonded to the ester group (1) and the amide nitrogen (2), each with four different substituents.
(a)(ii)
Answer: Amide, amine, ester, carboxylic acid
Explanation: The functional groups are:
- Amide (–CONH–) at the center
- Amine (–NH2) on the right branch
- Ester (–COO–) on the left branch
- Carboxylic acid (–COOH) at the top
(b)(i)
Answer: Hydrolysis and acid-base reaction
Explanation:
- Hydrolysis: The ester and amide groups undergo base hydrolysis (saponification)
- Acid-base: The carboxylic acid group deprotonates in NaOH
(b)(ii)
Answer: 
Explanation: The three products are:
- Benzoic acid (from ester hydrolysis)
- Aspartic acid derivative (from amide hydrolysis)
- Methanol (from ester hydrolysis)
The image shows the complete structures of these products.
▶️ Answer/Explanation
(a)(i) \(2C_6H_5OH + 2Na \rightarrow 2C_6H_5O^-Na^+ + H_2\)
Explanation: Phenol reacts with sodium to form sodium phenoxide (\(C_6H_5O^-Na^+\)) and hydrogen gas, similar to alcohols.
(a)(ii) Major products: 2-nitrophenol and 4-nitrophenol.
Explanation: Nitration of phenol with dilute HNO₃ preferentially substitutes at the ortho (2-) and para (4-) positions due to the \(-OH\) group’s activating and directing effects.
(b) Mechanism for reaction of phenoxide ion (\(C_6H_5O^-\)) with \(CO_2\):
Explanation:
- Nucleophilic attack by phenoxide on the electrophilic carbon of \(CO_2\).
- Formation of a carboxylate intermediate (salicylate ion).
- Protonation yields salicylic acid.
(c)(i) Diels-Alder mechanism (buta-1,3-diene + ethene):
Explanation: Three curly arrows show the concerted formation of two new σ-bonds and one π-bond, resulting in cyclohexene.
(c)(ii) Product of buta-1,3-diene with maleic anhydride:
Explanation: The reaction yields a cyclohexene derivative with an anhydride group, retaining stereochemistry of the dienophile (cis).