Flashcards · YCH11 · A Level Topic 5

Unit 5 – Transition Metals and Organic Nitrogen Chemistry (A Level) flashcards

38 cards on A Level Topic 5, subtopics 16 to 20. Pick a subtopic or study them all.

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16 Redox Equilibria

Standard electrode potential, E⦵: The voltage of a half-cell measured against a standard hydrogen electrode, both under standard conditions: 298 K, 100 kPa for any gas, 1.00 mol dm⁻³ for each ion.
Standard hydrogen electrode: Platinum coated with platinum black, H₂ at 100 kPa, H⁺(aq) at 1.00 mol dm⁻³, 298 K. E⦵ = 0.00 V by definition.
Al³⁺/Al −1.66 V and Cu²⁺/Cu +0.34 V. Cell diagram and E⦵cell: Al(s) | Al³⁺(aq) || Cu²⁺(aq) | Cu(s). E⦵cell = E⦵(right) − E⦵(left) = +0.34 − (−1.66) = +2.00 V. Overall 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.
Why a salt bridge of potassium nitrate, and a high-resistance voltmeter?: The salt bridge completes the circuit by ion movement without mixing the solutions; nitrate forms no precipitates. The high-resistance voltmeter draws almost no current, so it reads the true emf.
Limitations of using E⦵ to predict whether a reaction happens: E⦵ predicts thermodynamics, not rate: a feasible reaction can be too slow if its activation energy is large. Non-standard conditions change E and can reverse a prediction when E⦵cell is close to zero.
Iron(II) with manganate(VII): ionic equation, ratio and end point: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Fe²⁺ : MnO₄⁻ = 5 : 1. Self-indicating: first permanent pale pink. Use dilute sulfuric acid (HCl would be oxidised to chlorine).
Iodine–thiosulfate titration: equation, ratio and indicator: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻. S₂O₃²⁻ : I₂ = 2 : 1. Starch added near the end: blue-black to colourless.
Hydrogen fuel cell electrode equations with an alkaline electrolyte: Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻. Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall 2H₂ + O₂ → 2H₂O.

17 Transition Metals and their Chemistry

Transition metal, and why scandium and zinc are not: A d-block element with at least one stable ion with incompletely filled d-orbitals. Sc forms only Sc³⁺ (d⁰) and Zn only Zn²⁺ (d¹⁰).
Electron configurations of Cr, Cu and Fe³⁺: Cr: [Ar] 3d⁵ 4s¹. Cu: [Ar] 3d¹⁰ 4s¹. Fe³⁺: [Ar] 3d⁵ (4s electrons are removed first).
Ligand, and coordination number: Ligand: a molecule or ion that bonds to a metal ion by donating a lone pair (a dative bond). Coordination number: the total number of dative bonds to the metal ion.
Shapes of [Fe(H₂O)₆]³⁺, [CoCl₄]²⁻ and cis-platin: Octahedral (90°), coordination number 6. Tetrahedral (109.5°), 4: large Cl⁻ ligands. Square planar (90°), 4: [Pt(NH₃)₂Cl₂] with the two Cl on the same side.
Why are transition metal complexes coloured?: Ligands split the 3d orbitals into two levels. A d electron absorbs visible light of frequency f where ΔE = hf. The complementary colour is transmitted and seen. d⁰ and d¹⁰ ions are colourless.
Why is carbon monoxide toxic (haemoglobin)?: CO replaces O₂ on the Fe²⁺ by ligand exchange and binds far more strongly, so the change is not easily reversed and less oxygen is carried.
Colours of vanadium in oxidation states +5, +4, +3, +2: VO₂⁺ (+5) yellow; VO²⁺ (+4) blue; V³⁺ (+3) green; V²⁺ (+2) violet.
Cr³⁺(aq) with NaOH(aq) and with NH₃(aq), then excess: Grey-green precipitate with both. Excess NaOH: dark green solution (amphoteric). Excess NH₃: purple solution (ligand exchange).
Why is an EDTA⁴⁻ complex more stable than the aqua complex it replaces?: [M(H₂O)₆]ⁿ⁺ + EDTA⁴⁻ → 7 particles from 2, so ΔSsystem is large and positive. ΔH ≈ 0, so ΔStotal is positive and K is large (the chelate effect).
Heterogeneous vs homogeneous catalysts: Heterogeneous: different phase; reaction at the surface (V₂O₅ in the Contact process). Homogeneous: same phase; goes via an intermediate species (Fe²⁺ with S₂O₈²⁻ and I⁻).
Catalyst vs autocatalyst: An autocatalyst forms during the reaction and then catalyses it (Mn²⁺ in the manganate(VII)–ethanedioate reaction).

18 Organic Chemistry – Arenes

Three lines of evidence against the Kekulé structure of benzene: Thermochemical: benzene is more stable than predicted for three C=C. X-ray: all six C–C bonds equal, about 0.140 nm. IR: no alkene C=C band near 1650 cm⁻¹.
The delocalised model of benzene: Each carbon forms three σ bonds at 120° and has one electron in a p-orbital perpendicular to the ring. The p-orbitals overlap sideways all round, so six π-electrons are delocalised above and below the ring.
Why does benzene need a halogen carrier to react with bromine, unlike cyclohexene?: Its π-electrons are delocalised, so electron density between any two carbons is too low to polarise Br₂. FeBr₃ makes Br⁺. It substitutes rather than adds, keeping the stable delocalised ring.
Nitration of benzene: reagents, conditions and electrophile: Concentrated HNO₃ + concentrated H₂SO₄, at or below about 55 °C. Electrophile NO₂⁺: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Product nitrobenzene.
Friedel-Crafts acylation: reagents and product: An acyl chloride with anhydrous AlCl₃, warm under reflux. RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻; product an aromatic ketone C₆H₅COR + HCl.
Phenol with bromine water: observation, product and reason: Orange decolourised and a white precipitate of 2,4,6-tribromophenol (C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr). An O lone pair is partly delocalised into the ring, raising its electron density.

19 Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins

Order of base strength: ammonia, butylamine, phenylamine: Butylamine > ammonia > phenylamine. The alkyl group releases electron density onto N; in phenylamine the N lone pair is delocalised into the ring and is less available.
Making phenylamine from nitrobenzene: Heat under reflux with tin and concentrated HCl, then add NaOH(aq) until alkaline to free the amine from C₆H₅NH₃⁺. ArNO₂ + 6[H] → ArNH₂ + 2H₂O.
Making an azo dye from phenylamine: Diazotisation: NaNO₂ + HCl below 10 °C → benzenediazonium chloride. Coupling: add to phenol in NaOH(aq) → a yellow-orange azo dye. Too warm and the diazonium ion decomposes to N₂ and phenol.
Amine + ethanoyl chloride: CH₃COCl + 2RNH₂ → CH₃CONHR + RNH₃⁺Cl⁻. An N-substituted amide forms fast at room temperature; two moles of amine are needed.
Nylon-6,6: monomers and small molecule lost: H₂N(CH₂)₆NH₂ + HOOC(CH₂)₄COOH (or the diacyl dichloride). Repeat unit –[NH(CH₂)₆NHCO(CH₂)₄CO]–; water (or HCl) is lost at each link.
Forms of 2-aminopropanoic acid in the solid, in acid and in alkali: Solid / neutral: zwitterion H₃N⁺CH(CH₃)COO⁻. Acid: H₃N⁺CH(CH₃)COOH. Alkali: H₂NCH(CH₃)COO⁻.
Why does poly(ethenol) dissolve in water, and one use?: Its many –OH groups hydrogen bond with water. Used for soluble hospital laundry bags and liquid-detergent capsules.

20 Organic Synthesis

Grignard reagent: how it is made, and why everything must be dry: RX + Mg → RMgX in dry ether. Water reacts with RMgX to give the alkane RH, destroying the reagent.
Grignard reagent with CO₂, methanal, another aldehyde and a ketone: Then hydrolyse with dilute acid. CO₂ → carboxylic acid. Methanal → primary alcohol. Other aldehyde → secondary alcohol. Ketone → tertiary alcohol.
Recrystallisation: Dissolve the solid in the minimum volume of hot solvent, filter hot, cool to crystallise, filter, wash with a little cold solvent and dry.
Steam distillation is used for…: A volatile, water-immiscible compound that would decompose at its boiling point: it distils with steam below 100 °C.
Making aspirin (Core Practical 16): how is purity checked?: Measure the melting temperature: pure aspirin melts sharply at about 135–136 °C. A lower, broad range shows impurity (often unreacted 2-hydroxybenzoic acid) or water.
Reagents and conditions: halogenoalkane → amine, and nitrile → amine: Halogenoalkane → amine: excess ammonia in ethanol, heat in a sealed tube. Nitrile → amine: LiAlH₄ in dry ether.