▶️ Answer/Explanation
(a)(i) Strontium hydroxide > Calcium hydroxide > Magnesium hydroxide (most soluble to least soluble).
Explanation: Solubility of Group 2 hydroxides increases down the group due to decreasing lattice energy and increasing ionic size, making hydration more favorable.
(a)(ii) pH increases from magnesium hydroxide to strontium hydroxide.
Explanation: Higher solubility means more OH⁻ ions in solution, resulting in a higher pH for strontium hydroxide compared to magnesium hydroxide.
(b) Mass of Ba(OH)₂ = 0.0219 g.
Explanation:
- pH = 12.2 ⇒ pOH = 1.8 ⇒ [OH⁻] = 10⁻¹·⁸ = 0.0158 mol/dm³.
- Since Ba(OH)₂ dissociates as Ba(OH)₂ → Ba²⁺ + 2OH⁻, [Ba(OH)₂] = 0.0158/2 = 0.0079 mol/dm³.
- Moles in 250 cm³ = 0.0079 × 0.25 = 0.001975 mol.
- Mass = 0.001975 × 171.3 = 0.0219 g.
(c)(i) Ksp = [Fe²⁺][OH⁻]².
Explanation: The solubility product expression is derived from the dissociation equation Fe(OH)₂ ⇌ Fe²⁺ + 2OH⁻.
(c)(ii) Ksp = 8.00 × 10⁻¹⁶ mol³dm⁻⁹.
Explanation:
- [Fe²⁺] = 5.85 × 10⁻⁶ mol/dm³, [OH⁻] = 2 × 5.85 × 10⁻⁶ = 1.17 × 10⁻⁵ mol/dm³.
- Ksp = (5.85 × 10⁻⁶)(1.17 × 10⁻⁵)² = 8.00 × 10⁻¹⁶ mol³dm⁻⁹.
▶️ Answer/Explanation
(a)(i) A transition element is 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, such as Fe²⁺ (3d⁶) or Cu²⁺ (3d⁹).
(a)(ii) Transition elements can form complex ions because they have vacant d orbitals that are energetically accessible for bonding with lone pairs from ligands.
Explanation: The presence of empty d orbitals allows transition metals to accept electron pairs from ligands, forming coordinate bonds in complex ions like \([Fe(CN)_6]^{4-}\).
(b)(i) Degenerate d orbitals are orbitals of the same energy level.
Explanation: In an isolated Ag⁺ ion, all five 3d orbitals (dxy, dxz, dyz, dx²-y², dz²) have identical energy.
(b)(ii)
Explanation: The 3dxy orbital has four lobes lying in the xy plane between the axes, with a node at the nucleus.
(c)
Step 1: \(2AgNO_3 + 2NaOH → Ag_2O + 2NaNO_3 + H_2O\)
Step 2: \(Ag_2O + 4NH_3 + H_2O → 2[Ag(NH_3)_2]OH\)
Explanation: In step 1, silver oxide precipitates, while in step 2, it dissolves in ammonia to form the diamminesilver(I) complex.
(d) Linear; H-N-Ag angle: 180°; N-Ag-N angle: 180°.
Explanation: The \([Ag(NH_3)_2]^+\) ion has two ammonia ligands arranged linearly around the central Ag⁺ ion, giving bond angles of 180°.
(e)
Positive electrode: \(Ag_2O + H_2O + 2e^- → 2Ag + 2OH^-\)
Negative electrode: \(Zn + 2OH^- → Zn(OH)_2 + 2e^-\)
Explanation: At the positive electrode, Ag₂O is reduced to Ag, while at the negative electrode, Zn is oxidized to Zn(OH)₂.
(f)
Explanation: The dps ligand bridges between Ru³⁺ ions, with each nitrogen coordinating to a different metal center, forming a polymeric chain with \([Ru(dps)Cl_4]^-\) repeating units.
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▶️ Answer/Explanation
(a) \(\text{Li}_2\text{C}_2\text{O}_4 \cdot \text{H}_2\text{O} \rightarrow \text{Li}_2\text{CO}_3 + \text{CO} + \text{H}_2\text{O}\)
Explanation: Heating lithium ethanedioate releases CO and H₂O gases, leaving Li₂CO₃ as residue. When HNO₃ is added, CO₂ is produced (turns limewater milky).
(b) BaC₂O₄ decomposes at a lower temperature than CaC₂O₄.
Explanation: Larger cations (Ba²⁺) stabilize the carbonate ion less effectively than smaller cations (Ca²⁺), so BaC₂O₄ decomposes more easily.
(c) Transition elements form coloured complexes due to d-d electron transitions.
Explanation: The partially filled d-orbitals in transition metals split in the ligand field, absorbing specific wavelengths of light and transmitting the complementary colour (green in this case).
(d) \(\text{2K}_3[\text{Fe}(\text{C}_2\text{O}_4)_3] \rightarrow \text{2K}_2[\text{Fe}(\text{C}_2\text{O}_4)_2] + \text{K}_2\text{C}_2\text{O}_4 + \text{2CO}_2\)
Explanation: The decomposition breaks down the complex into a simpler iron(II) complex, potassium oxalate, and CO₂ gas.
(e) 
Explanation: The two stereoisomers are the Λ (lambda) and Δ (delta) forms, which are non-superimposable mirror images (enantiomers).
(f)
For acid: \(\text{HC}_2\text{O}_4^- + \text{OH}^- \rightarrow \text{C}_2\text{O}_4^{2-} + \text{H}_2\text{O}\)
For base: \(\text{HC}_2\text{O}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{C}_2\text{O}_4\)
Explanation: HC₂O₄⁻ acts as a buffer by neutralizing added OH⁻ or H⁺, maintaining pH stability.
(g) Overall reaction: \((\text{COOH})_2 + \text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}\)
\(E^{\Theta}_{cell} = 1.23 – (-0.49) = 1.72 \text{ V}\)
Explanation: The cell potential is calculated by subtracting the anode potential from the cathode potential (oxygen reduction).
▶️ Answer/Explanation
(a) The standard electrode potential (\(E^{\Theta}\)) 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).
(b)(i)
Explanation: The diagram shows the \(Ag^+(aq)/Ag(s)\) half-cell connected to the SHE. The cell includes a salt bridge, voltmeter, and electrodes (Ag and Pt).
(b)(ii) The electrode potential (\(E\)) would become more negative. Explanation: Lower \(Ag^+\) concentration shifts the equilibrium \(Ag^+ + e^- \leftrightarrow Ag\) to the left (Le Chatelier’s Principle), reducing the tendency for reduction.
(c) The enthalpy change of solution (\(\Delta H^{\Theta}_{sol}\)) is the energy change when one mole of a solute dissolves in water to form an infinitely dilute solution.
(d)(i)
Explanation: The cycle links lattice energy (\(\Delta H^{\Theta}_{latt}\)), enthalpy of hydration (\(\Delta H^{\Theta}_{hyd}\)), and enthalpy of solution (\(\Delta H^{\Theta}_{sol}\)).
(d)(ii) \(\Delta H^{\Theta}_{latt} = -811.6 \ \text{kJ mol}^{-1}\). Explanation: Using Hess’s Law: \(\Delta H^{\Theta}_{latt} = \Delta H^{\Theta}_{sol} – \Delta H^{\Theta}_{hyd} = +22.6 – (+834.2) = -811.6 \ \text{kJ mol}^{-1}\).
(e) The trend is Mg(NO₃)₂(s) > NaNO₃(s) > RbNO₃(s). Explanation: Mg²⁺ has a higher charge density than Na⁺ or Rb⁺, leading to stronger ionic attraction. Na⁺ is smaller than Rb⁺, so NaNO₃ has a higher lattice energy than RbNO₃.
▶️ Answer/Explanation
(a)(i)
Answer: The initial rate is determined from the tangent at \( t = 0 \). From the graph, the slope is approximately \(-0.0125 \, \text{mol dm}^{-3} \, \text{s}^{-1}\).
Explanation: The initial rate is the change in concentration of \( S_2O_8^{2-} \) per unit time at \( t = 0 \). The tangent at \( t = 0 \) gives a slope of \(-0.0125\), so the rate is \( 0.0125 \, \text{mol dm}^{-3} \, \text{s}^{-1} \).
(a)(ii)
Answer: A large excess of \( I^- \) ensures its concentration remains approximately constant, making the reaction pseudo-first order with respect to \( S_2O_8^{2-} \). The half-life then depends only on \( k \).
Explanation: Since \( [I^-] \) is constant, the rate equation simplifies to \( \text{rate} = k’ [S_2O_8^{2-}] \), where \( k’ = k[I^-] \). The half-life of a first-order reaction is \( t_{1/2} = \frac{\ln 2}{k’} \), allowing \( k \) to be determined.
(b)
Answer: \[ Fe^{2+} + S_2O_8^{2-} \rightarrow Fe^{3+} + 2SO_4^{2-} \] \[ Fe^{3+} + I^- \rightarrow Fe^{2+} + \frac{1}{2} I_2 \]
Explanation: \( Fe^{2+} \) is oxidised to \( Fe^{3+} \) by \( S_2O_8^{2-} \), and \( Fe^{3+} \) is reduced back to \( Fe^{2+} \) by \( I^- \), regenerating the catalyst.
(c)
Answer: An increase in temperature increases the rate constant \( k \) (Arrhenius equation) and thus the rate of reaction.
Explanation: The Arrhenius equation \( k = Ae^{-E_a/RT} \) shows that \( k \) increases with temperature. Since rate \( = k[S_2O_8^{2-}][I^-] \), the rate also increases.
(d)
Answer: The half-life \( t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{1.58 \times 10^{-2}} \approx 43.9 \, \text{s} \).
Explanation: For a first-order reaction, \( t_{1/2} = \frac{\ln 2}{k} \). Substituting \( k = 1.58 \times 10^{-2} \, \text{s}^{-1} \) gives \( t_{1/2} \approx 43.9 \, \text{s} \).
(e)
Answer: Step 1 (slow, rate-determining): \( NO + Br_2 \rightarrow NOBr_2 \)
Step 2 (fast): \( NOBr_2 + NO \rightarrow 2NOBr \)
Explanation: The rate-determining step involves one NO and one \( Br_2 \) molecule, matching the observed rate law. The second step is fast and does not affect the rate.
▶️ Answer/Explanation
(a)(i) The partition coefficient, \(K_{pc}\), is the ratio of the concentrations of a solute in two immiscible solvents at equilibrium.
Explanation: It describes how a solute distributes itself between two phases, such as organic and aqueous layers.
(a)(ii) 0.642 g
Explanation: Using the partition coefficient formula \(K_{pc} = \frac{[X]_{CS_2}}{[X]_{H_2O}}\), we set up the equation \(10.5 = \frac{(y/25)}{(0.74 – y)/40}\), where \(y\) is the mass extracted into CS₂. Solving gives \(y = 0.642 \, \text{g}\).
(b)
Explanation: Kekulé benzene has 6 \(sp^2\) carbons. Dewar benzene has 4 \(sp^2\) and 2 \(sp^3\) carbons. Ladenburg benzene has 2 \(sp\), 2 \(sp^2\), and 2 \(sp^3\) carbons.
(c) Delocalised benzene is planar with each carbon having trigonal planar geometry. The C-C-H bond angle is 120°. The C-C bonds are delocalised \(\pi\)-bonds, and C-H bonds are sigma (\(\sigma\)) bonds.
Explanation: The ring is flat due to \(sp^2\) hybridisation, with equal bond lengths from electron delocalisation.
(d) Dewar benzene and Ladenburg benzene are unstable due to ring strain and lack of aromaticity (delocalisation).
Explanation: Their structures introduce angle strain and prevent the stability conferred by aromatic \(\pi\)-electron delocalisation in benzene.
(e)
Explanation: Dewar benzene has 2 peaks (2 types of H), Ladenburg benzene has 3 peaks (3 types of H), and delocalised benzene has 1 peak (all H equivalent).
(f)(i) \(\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{FeBr}_3 \rightarrow \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+ + \text{FeBr}_4^-\)
Explanation: FeBr₃ acts as a Lewis acid, polarising the C-Br bond to generate the electrophile.
(f)(ii)
Explanation: The mechanism involves electrophilic attack on the benzene ring, forming a carbocation intermediate, followed by deprotonation to restore aromaticity.
(f)(iii) Y is a dimer (two phenylethanone units bridged by the dibromobutane chain). Z is a monosubstituted product (one phenylethanone with a butyl group).
Explanation: Y forms from two electrophilic attacks, while Z results from a single substitution.
▶️ Answer/Explanation
(a) methyl pentanoate
Explanation: Ester A has a 5-carbon chain (pentanoate) and a methyl group (\(-OCH_3\)) attached to the carbonyl carbon, hence the name methyl pentanoate.
(b)(i) Time between injection and detection.
Explanation: Retention time is the time taken for a compound to travel through the chromatography column from injection to detection.
(b)(ii) 29.5%
Explanation: The area under peak D is 59, and the total area is 200. Thus, the percentage of D is \(\frac{59}{200} \times 100 = 29.5\%\).
(c)(i)
Explanation: Ester B has 4 proton environments (peaks in \(^1H\) NMR) and 5 carbon environments (peaks in \(^{13}C\) NMR). Ester C has 3 proton environments and 4 carbon environments.
(c)(ii) A and B
Explanation: Triplet peaks arise from protons adjacent to \(-CH_2\) groups. Esters A and B have such environments, while C and D do not.
(d)
Explanation: Compound F is a \(\beta\)-ketoacid (\(CH_3COCH_2COOH\)) due to its reactions. G is iodoform (\(CHI_3\)), H is sodium salt (\(CH_3COONa\)), J is reduced to a diol (\(HOCH_2CH(OH)CH_3\)), K is the acyl chloride (\(CH_3COCH_2COCl\)), and L is the ester (\(CH_3COCH_2COOCH(CH_3)_2\)).
▶️ Answer/Explanation
(a)(i) Two.
Explanation: Chiral carbons are those attached to four different groups. In neotame, the two carbons marked with asterisks (*) in the structure are chiral because they each have four distinct substituents.
(a)(ii) Amide, amine, ester, carboxylic acid.
Explanation: The functional groups in neotame are identified as follows: the amide (–CONH–), amine (–NH2), ester (–COO–), and carboxylic acid (–COOH) groups are all present in addition to the arene (benzene ring).
(b)(i) Hydrolysis and acid-base reaction.
Explanation: The ester and amide groups undergo hydrolysis in hot NaOH, breaking into smaller molecules. The carboxylic acid group participates in an acid-base reaction with NaOH, forming a carboxylate ion.
(b)(ii)
Explanation: The three organic products are: (1) sodium carboxylate (from the carboxylic acid group), (2) an amine (from the amide hydrolysis), and (3) an alcohol (from the ester hydrolysis). The structures are correctly drawn in the image above.
▶️ Answer/Explanation
(a)(i) The reaction of phenol with sodium produces sodium phenoxide and hydrogen gas:
\[ 2C_6H_5OH + 2Na \rightarrow 2C_6H_5O^-Na^+ + H_2 \]
Explanation: Phenol reacts with sodium similarly to alcohols, where the acidic hydrogen is replaced by sodium, forming a phenoxide ion and releasing \( H_2 \).
text
(a)(ii) The two major isomeric products from nitration of phenol are 2-nitrophenol and 4-nitrophenol.
Explanation: Dilute \( HNO_3 \) nitrates phenol at the ortho (2-) and para (4-) positions due to the electron-donating effect of the hydroxyl group.
(b) The mechanism involves electrophilic attack of \( CO_2 \) on the phenoxide ion, forming a carboxylate intermediate.
Explanation: \( CO_2 \) acts as an electrophile, attacking the electron-rich ortho position of the phenoxide ion. The intermediate is stabilised by resonance, leading to salicylic acid after protonation.
(c)(i) The Diels-Alder mechanism involves three curly arrows showing the movement of π-electrons to form new σ-bonds.
(c)(ii) The product is a cyclohexene derivative formed by [4+2] cycloaddition.
Explanation: Buta-1,3-diene reacts with ethene to form cyclohexene. The reaction proceeds via a concerted mechanism, where π-electrons rearrange to form a six-membered ring.