Sketch and explain the Maxwell–Boltzmann distribution of molecular energies at two different temperatures (T2 > T1) and with a catalyst.

The curve starts at the origin (0,0), peaks at the most probable energy (Emp), has mean energy slightly to the right of Emp, and approaches the energy axis asymptotically; the total area under the curve equals the total number of molecules. At higher temperature T2, the peak shifts to the right and lowers, exponentially increasing the area past Ea. A catalyst shifts the Ea line to the left without changing the shape of the curve.

Explain why transition metal complex ions are coloured, and state the equation ΔE = hν = hc/λ.

When ligands coordinate to a central transition metal ion, the degenerate 3d orbitals split into two different energy levels (crystal field splitting). Visible light hits the complex and electrons absorb a specific frequency of light (ΔE = hν = hc/λ) to become excited from the lower 3d orbitals to the higher unfilled 3d orbitals. The complementary wavelengths not absorbed are transmitted/reflected, giving the observed colour.

How do you select the right acid–base indicator (Methyl Orange vs Phenolphthalein) from pH titration curves?

The indicator's pH colour-change range must lie entirely within the vertical equivalence section of the titration curve: 1) Strong acid + Strong base (vertical pH 3–11): either indicator works; 2) Strong acid + Weak base (vertical pH 3–7): use Methyl Orange (range 3.1–4.4); 3) Weak acid + Strong base (vertical pH 7–11): use Phenolphthalein (range 8.3–10.0); 4) Weak acid + Weak base: no sharp vertical section, use a pH meter.

Compare the reactions of aqueous [Fe(H2O)6]²⁺, [Cu(H2O)6]²⁺, [Fe(H2O)6]³⁺, and [Al(H2O)6]³⁺ with aqueous NaOH, NH3, and Na2CO3 (AQA).

2+ ions ([Fe(H2O)6]²⁺ green, [Cu(H2O)6]²⁺ blue) are less acidic than 3+ ions ([Fe(H2O)6]³⁺ purple/yellow-brown, [Al(H2O)6]³⁺ colourless, which have high charge density that polarises O–H bonds in H2O ligands). With Na2CO3: 2+ ions form simple insoluble carbonate precipitates (FeCO3 green, CuCO3 blue-green) with NO gas; 3+ ions are strongly acidic enough to protonate CO3²⁻, forming M(H2O)3(OH)3 precipitates AND effervescence of CO2 gas. Excess NH3 dissolves Cu(H2O)4(OH)2 to form deep blue [Cu(NH3)4(H2O)2]²⁺; excess NaOH dissolves Al(H2O)3(OH)3 to form colourless [Al(OH)4]⁻.

How do you determine the order of reaction (0, 1, or 2) with respect to a reactant from concentration–time and rate–concentration graphs, and state the Arrhenius equation?

Zero order: [A] vs t is a straight downward line (gradient = −k); Rate vs [A] is horizontal. First order: [A] vs t is exponential decay with constant half-life (t₁/₂ = ln 2 / k); Rate vs [A] is a straight line through origin. Second order: Rate vs [A] is an upward parabola (Rate vs [A]² is linear). Arrhenius equation: k = A e^(−Ea/RT) or ln k = −(Ea/R)(1/T) + ln A (gradient of ln k vs 1/T is −Ea/R).

State the Ideal Gas Equation with its strict SI units, and show how to calculate the Mr of a volatile liquid.

pV = nRT, where p = pressure in Pascals (Pa; 1 kPa = 1,000 Pa), V = volume in cubic metres (m³; 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³), n = moles (mol), R = 8.31 J K⁻¹ mol⁻¹, and T = temperature in Kelvin (K = °C + 273). Since n = m / Mr, substituting gives Mr = mRT / (pV).

describe the four stages of Time-of-Flight (ToF) Mass Spectrometry: Ionisation, Acceleration, Ion Drift, and Detection.

1) Ionisation: Electrospray ionisation (sample dissolved in volatile solvent, pushed through high-voltage needle to gain H⁺ → MH⁺(g), no fragmentation) OR Electron impact (vaporised sample bombarded with high-energy electrons: X(g) + e⁻ → X⁺(g) + 2e⁻). 2) Acceleration: positive ions accelerated by an electric field so all ions have the same kinetic energy (Ek = ½mv²). 3) Ion Drift: lighter ions (lower m/z) travel faster through the flight tube (t = d√(m/2Ek)). 4) Detection: ions hit negative detector plate, gaining electrons and generating a current proportional to abundance.

How do you construct a Born–Haber Cycle for an ionic solid (e.g., NaCl), and why do experimental and theoretical lattice enthalpies sometimes differ?

ΔfH°(NaCl) = ΔatH°(Na) + IE1(Na) + ΔatH°(Cl) + EA1(Cl) + ΔLEH°(formation of NaCl). Theoretical lattice enthalpy assumes a purely 100% ionic model of spherical ions. If experimental lattice enthalpy is more exothermic (larger magnitude) than the theoretical value (e.g., in AgCl or MgI2), the compound has significant covalent character caused by a small, highly charged cation polarising a large anion.

How does the rate equation relate to the Rate-Determining Step (RDS) of a multi-step reaction mechanism?

Only the species involved in the slowest step (the Rate-Determining Step, RDS) or in fast equilibrium steps prior to the RDS appear in the rate equation, and the order with respect to each reactant equals the stoichiometric number of moles of that species involved up to and including the RDS.

Compare the cells and electrode reactions in non-rechargeable primary cells, rechargeable Lithium-ion cells, and alkaline Hydrogen Fuel Cells.

Lithium-ion cell: Negative electrode: Li → Li⁺ + e⁻; Positive electrode: Li⁺ + CoO2 + e⁻ → Li⁺[CoO2]⁻ (during charging, an external current reverses both reactions). Alkaline Hydrogen Fuel Cell: Negative anode: H2 + 2OH⁻ → 2H2O + 2e⁻; Positive cathode: O2 + 2H2O + 4e⁻ → 4OH⁻; Overall: 2H2 + O2 → 2H2O (E°cell = +1.23 V; operates continuously as long as H2 and O2 fuel are supplied).

Define Standard Enthalpy of Formation (ΔfH°) and Standard Enthalpy of Combustion (ΔcH°), and state their Hess's Law cycle equations.

ΔfH°: enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, 298 K). ΔcH°: enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions. Using Formation data: ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants). Using Combustion data: ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products).

How do you calculate the pH of a weak acid (HA) given its acid dissociation constant Ka and concentration [HA]?

For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻] / [HA]. Making the two standard approximations that [H⁺] ≈ [A⁻] (negligible water dissociation) and [HA]_eqm ≈ [HA]_initial (minimal dissociation), we get Ka ≈ [H⁺]² / [HA], so [H⁺] = √(Ka × [HA]), and pH = −log₁₀[H⁺]. Also pKa = −log₁₀(Ka).

Compare the three types of intermolecular forces (London dispersion forces, permanent dipole–dipole, and hydrogen bonding) and explain why ice is less dense than liquid water.

1) Induced dipole–dipole (London dispersion/van der Waals): present in all molecules due to temporary fluctuating electron clouds; increases with number of electrons/surface area. 2) Permanent dipole–dipole: between polar molecules with unsymmetrical electronegativity. 3) Hydrogen bonding: strongest IMF, occurs between a δ+ H bonded to N, O, or F and a lone pair on N, O, or F of another molecule. In ice, each H2O forms 4 tetrahedral H-bonds, holding molecules in an open lattice with large spaces.

Write the equations for the disproportionation of Chlorine (Cl2) with cold water (in water treatment) and with cold dilute NaOH (making bleach).

Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. 1) With water: Cl2(g) + H2O(l) ⇌ HCl(aq) [Cl = -1] + HClO(aq) [chloric(I) acid, Cl = +1, kills bacteria]. In bright sunlight: 2Cl2 + 2H2O → 4HCl + O2. 2) With cold dilute NaOH: Cl2(aq) + 2NaOH(aq) → NaCl(aq) [Cl = -1] + NaClO(aq) [sodium chlorate(I) bleach, Cl = +1] + H2O(l).

Compare the reducing power of halide ions (Cl⁻, Br⁻, I⁻) when solid sodium halides react with concentrated sulfuric acid (H2SO4).

Reducing power increases down Group 7 (I⁻ > Br⁻ > Cl⁻) as ionic radius and shielding increase. 1) NaCl + H2SO4 → NaHSO4 + HCl (steamy white fumes; acid-base only, Cl⁻ cannot reduce S(+6)). 2) NaBr + H2SO4 → HBr (steamy fumes), then Br⁻ reduces S(+6) to S(+4) in SO2 (choking gas) + orange Br2 vapour. 3) NaI + H2SO4 → HI, then I⁻ reduces S(+6) all the way to S(+4) (SO2), S(0) (yellow sulfur solid), and S(-2) in H2S (rotten egg smell) + purple I2 vapour.

What is the Chelate Effect in ligand substitution reactions (e.g., replacing 6 H2O ligands with 3 bidentate en or 1 hexadentate EDTA⁴⁻ ligand)?

In [Cu(H2O)6]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H2O, 2 reactant particles form 7 product particles in solution (and 6 Cu–O bonds are broken while 6 new coordinate bonds are formed, so ΔH ≈ 0). This large increase in the number of free molecules in solution creates a large positive entropy change (+ΔS), making ΔG = ΔH − TΔS strongly negative and the multidentate complex thermodynamically very stable.

Explain why enthalpies of hydration (ΔhydH°) of ions become more exothermic as ionic radius decreases and ionic charge increases.

Enthalpy of hydration is the enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions at infinite dilution. Smaller ions with higher charges (higher charge density, e.g., Mg²⁺ vs Na⁺ or Al³⁺ vs Mg²⁺) exert a much stronger electrostatic attraction on the δ− oxygen or δ+ hydrogen atoms of polar water molecules, releasing more energy when ion–dipole bonds form. Also: ΔsolH° = ΔLEH°(dissociation) + ΣΔhydH°.

Describe the acid–base characters and reactions with water of the Period 3 oxides (Na2O, MgO, Al2O3, SiO2, P4O10, SO2, SO3).

Ionic basic oxides: Na2O + H2O → 2NaOH (pH 13–14); MgO + H2O → Mg(OH)2 (sparingly soluble, pH 9–10). Amphoteric oxide (ionic with covalent character): Al2O3 is insoluble in water, but reacts with BOTH acids (forming Al³⁺ salts) and alkalis (Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4). Giant covalent acidic oxide: SiO2 is insoluble in water but reacts with hot concentrated NaOH. Simple molecular acidic oxides: P4O10 + 6H2O → 4H3PO4 (pH 1–2); SO2 + H2O → H2SO3 (pH 2–3); SO3 + H2O → H2SO4 (pH 0–1).

How does an iodine–sodium thiosulfate (I2 / S2O3²⁻) redox titration work to determine the concentration of an oxidising agent (e.g., Cu²⁺, ClO⁻, or IO3⁻)?

1) Add excess potassium iodide (KI) to a measured volume of the oxidising agent so it oxidises I⁻ into brown aqueous I2 in a known molar ratio; 2) Titrate the liberated I2 with standard sodium thiosulfate (Na2S2O3): 2S2O3²⁻ + I2 → S4O6²⁻ + 2I⁻ (ratio 2 mol S2O3²⁻ : 1 mol I2); 3) When the solution fades to pale straw-yellow, add starch indicator (turns deep blue-black) and continue adding S2O3²⁻ dropwise until the blue-black colour disappears to colourless.

What is a buffer solution, and how does an acidic buffer (e.g., ethanoic acid CH3COOH + sodium ethanoate CH3COONa) resist pH changes when small amounts of H⁺ or OH⁻ are added?

An acidic buffer contains a large reservoir of weak acid (CH3COOH) and its conjugate base salt (CH3COO⁻). ForCH3COOH(aq) ⇌ H⁺(aq) + CH3COO⁻(aq): 1) On adding acid (H⁺), H⁺ reacts with the large reservoir of CH3COO⁻ ions to form undissociated CH3COOH (equilibrium shifts left); 2) On adding alkali (OH⁻), OH⁻ reacts with H⁺ to form H2O, and more CH3COOH dissociates (shifts right) to restore [H⁺]. Buffer pH is calculated using [H⁺] = Ka × [HA] / [A⁻].

Explain how heterogeneous catalysts (e.g., V2O5 in the Contact Process) and autocatalysis (Mn²⁺ in MnO4⁻ + C2O4²⁻) work.

In the Contact Process (SO2 to SO3), V2O5 changes oxidation state from +5 to +4 and back: SO2 + V2O5 → SO3 + V2O4, then V2O4 + ½O2 → V2O5. In the reaction between negative MnO4⁻ and C2O4²⁻ ions, initial rate is slow due to electrostatic repulsion between two anions; once a trace of Mn²⁺ product forms, it acts as an autocatalyst (reacting with MnO4⁻ to form Mn³⁺, which then oxidises C2O4²⁻), rapidly accelerating the reaction.

Using VSEPR theory, state the shapes and bond angles of molecules/ions with 2, 3, 4, 5, and 6 bonding pairs and 0 lone pairs.

2 bonding pairs (e.g., BeCl2, CO2): Linear, 180°. 3 bp (e.g., BF3): Trigonal planar, 120°. 4 bp (e.g., CH4, NH4⁺): Tetrahedral, 109.5°. 5 bp (e.g., PCl5): Trigonal bipyramidal, 90° and 120°. 6 bp (e.g., SF6): Octahedral, 90°.

What are the standard conditions for a Standard Hydrogen Electrode (SHE), and how do you calculate E°cell and predict redox feasibility?

SHE standard conditions: H2(g) at 100 kPa, 1.00 mol dm⁻³ H⁺(aq) solution, 298 K, inert porous Platinum electrode (defined as E° = 0.00 V). E°cell = E°(most positive / reduced right-hand electrode) − E°(most negative / oxidised left-hand electrode). A redox reaction is thermodynamically feasible if E°cell > 0 V.

How do lone pairs of electrons affect bond angles in VSEPR theory? State the shapes and bond angles of NH3, H2O, SF4, and XeF4.

Lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion, reducing bond angles by ~2.5° per lone pair. NH3 (3 bp, 1 lp): Trigonal pyramidal, 107°. H2O (2 bp, 2 lp): Bent / non-linear, 104.5°. SF4 (4 bp, 1 lp): Seesaw, 87° and 102°. XeF4 (4 bp, 2 lp): Square planar, 90°.

Define Brønsted–Lowry acids and bases, write the expression for the ionic product of water (Kw), and explain why pure water at 50°C has pH < 7 but is still neutral.

Brønsted–Lowry acid = proton (H⁺) donor; base = proton (H⁺) acceptor. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. The dissociation of water (H2O ⇌ H⁺ + OH⁻) is endothermic; heating water shifts equilibrium right, increasing Kw and [H⁺] so pH = −log₁₀[H⁺] drops below 7, but it remains completely neutral because [H⁺] still equals [OH⁻].

State the Gibbs Free Energy equation (ΔG = ΔH − TΔS) and explain the four combinations of ΔH and ΔS that determine reaction feasibility.

A reaction is thermodynamically feasible (spontaneous) when ΔG ≤ 0 (where ΔS must be divided by 1,000 to convert J K⁻¹ mol⁻¹ into kJ K⁻¹ mol⁻¹). 1) ΔH 0: feasible at ALL temperatures; 2) ΔH > 0, ΔS 0, ΔS > 0: feasible only at HIGH temperatures (T > ΔH/ΔS).

Explain why there is a slight drop in First Ionisation Energy between Magnesium and Aluminium, and between Phosphorus and Sulfur, across Period 3.

Mg (3s²) to Al (3s² 3p¹): Al's outer electron is in a higher-energy 3p subshell, which is slightly further from the nucleus and shielded by the 3s² electrons, so it requires less energy to remove. P (3p³) to S (3p⁴): in P, each of the three 3p orbitals is singly occupied, whereas in S, one 3p orbital contains a spin-paired pair of electrons; mutual repulsion between the paired electrons makes one easier to remove.

Define a transition metal in exact A-Level terms, and explain why Scandium (Sc) and Zinc (Zn) are d-block elements but NOT transition metals.

A transition metal is a d-block element that forms at least one stable ion with a partially filled (incomplete) d-subshell. Scandium only forms Sc³⁺ ([Ar] 3d⁰, empty d-subshell) and Zinc only forms Zn²⁺ ([Ar] 3d¹⁰, full d-subshell); therefore neither is a transition metal and their compounds are white/colourless.

How do you perform a redox titration using acidified potassium manganate(VII) (KMnO4) to determine Fe²⁺ content in iron tablets?

Dissolve crushed iron tablets in dilute H2SO4 (provides H⁺ ions without being oxidised by MnO4⁻, unlike HCl) and make up to 250 cm³ in a volumetric flask. Titrate 25.0 cm³ aliquots against standard KMnO4 from the burette; KMnO4 is self-indicating, turning from colourless to the first permanent pale pink colour at the endpoint. Half-equations give stoichiometric ratio 1 MnO4⁻ : 5 Fe²⁺ (MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O).

Write the expressions for Kc and Kp for the general gas reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), and state what factor changes Kc or Kp.

Kc = ([C]^c [D]^d) / ([A]^a [B]^b). Kp = (p(C)^c × p(D)^d) / (p(A)^a × p(B)^b), where partial pressure p(X) = mole fraction χ(X) × total pressure P_total. ONLY temperature changes the numerical value of Kc or Kp (concentration, pressure, and catalysts have zero effect on Kc or Kp).

State and explain the trends down Group 2 (Alkaline Earth Metals: Mg to Ba) for reactivity, solubility of hydroxides, and solubility of sulfates.

Reactivity with water/oxygen increases down Group 2 because atomic radius and shielding increase, lowering 1st and 2nd ionisation energies. Solubility of Group 2 Hydroxides M(OH)2 INCREASES down the group (Mg(OH)2 is sparingly soluble, used as Milk of Magnesia; Ca(OH)2 is used in agriculture; Ba(OH)2 is strongly alkaline). Solubility of Group 2 Sulfates MSO4 DECREASES down the group (BaSO4 is completely insoluble, used as a 'barium meal' in X-rays and in the sulfate ion test).

Write the full electron configurations of Chromium (Z=24) and Copper (Z=29) and explain why they are exceptions to the Aufbau rule.

Chromium (Z=24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (NOT 3d⁴ 4s²). Copper (Z=29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (NOT 3d⁹ 4s²). Half-filled (3d⁵) and completely filled (3d¹⁰) d-subshells provide extra thermodynamic stability due to symmetrical electron distribution and exchange energy. When transition metals form positive ions, 4s electrons are ALWAYS lost before 3d electrons.

A-Level Chemistry I (Physical & Inorganic): Amount of Substance, Bonding, Energetics, Kinetics, Equilibria & Periodicity

Core AS & A-Level Physical and Inorganic Chemistry across AQA, OCR, and Edexcel: s/p/d electron configurations, ionisation energies, VSEPR shapes, intermolecular forces, Hess's Law, Born-Haber cycles, Kp/Kc, pH/Ka/buffers, electrode potentials, and Group 2/7 trends.

Level: basic
Languages: en → en
Cards: 32
Markdown: A-Level Chemistry I (Physical & Inorganic): Amount of Substance, Bonding, Energetics, Kinetics, Equilibria & Periodicity (.md)

Flashcards (32)

#	Front (Term / Question)	Back (Definition / Answer)
1	Sketch and explain the Maxwell–Boltzmann distribution of molecular energies at two different temperatures (T2 > T1) and with a catalyst.	The curve starts at the origin (0,0), peaks at the most probable energy (Emp), has mean energy slightly to the right of Emp, and approaches the energy axis asymptotically; the total area under the curve equals the total number of molecules. At higher temperature T2, the peak shifts to the right and lowers, exponentially increasing the area past Ea. A catalyst shifts the Ea line to the left without changing the shape of the curve.
2	Explain why transition metal complex ions are coloured, and state the equation ΔE = hν = hc/λ.	When ligands coordinate to a central transition metal ion, the degenerate 3d orbitals split into two different energy levels (crystal field splitting). Visible light hits the complex and electrons absorb a specific frequency of light (ΔE = hν = hc/λ) to become excited from the lower 3d orbitals to the higher unfilled 3d orbitals. The complementary wavelengths not absorbed are transmitted/reflected, giving the observed colour.
3	How do you select the right acid–base indicator (Methyl Orange vs Phenolphthalein) from pH titration curves?	The indicator's pH colour-change range must lie entirely within the vertical equivalence section of the titration curve: 1) Strong acid + Strong base (vertical pH 3–11): either indicator works; 2) Strong acid + Weak base (vertical pH 3–7): use Methyl Orange (range 3.1–4.4); 3) Weak acid + Strong base (vertical pH 7–11): use Phenolphthalein (range 8.3–10.0); 4) Weak acid + Weak base: no sharp vertical section, use a pH meter.
4	Compare the reactions of aqueous [Fe(H2O)6]²⁺, [Cu(H2O)6]²⁺, [Fe(H2O)6]³⁺, and [Al(H2O)6]³⁺ with aqueous NaOH, NH3, and Na2CO3 (AQA).	2+ ions ([Fe(H2O)6]²⁺ green, [Cu(H2O)6]²⁺ blue) are less acidic than 3+ ions ([Fe(H2O)6]³⁺ purple/yellow-brown, [Al(H2O)6]³⁺ colourless, which have high charge density that polarises O–H bonds in H2O ligands). With Na2CO3: 2+ ions form simple insoluble carbonate precipitates (FeCO3 green, CuCO3 blue-green) with NO gas; 3+ ions are strongly acidic enough to protonate CO3²⁻, forming M(H2O)3(OH)3 precipitates AND effervescence of CO2 gas. Excess NH3 dissolves Cu(H2O)4(OH)2 to form deep blue [Cu(NH3)4(H2O)2]²⁺; excess NaOH dissolves Al(H2O)3(OH)3 to form colourless [Al(OH)4]⁻.
5	How do you determine the order of reaction (0, 1, or 2) with respect to a reactant from concentration–time and rate–concentration graphs, and state the Arrhenius equation?	Zero order: [A] vs t is a straight downward line (gradient = −k); Rate vs [A] is horizontal. First order: [A] vs t is exponential decay with constant half-life (t₁/₂ = ln 2 / k); Rate vs [A] is a straight line through origin. Second order: Rate vs [A] is an upward parabola (Rate vs [A]² is linear). Arrhenius equation: k = A e^(−Ea/RT) or ln k = −(Ea/R)(1/T) + ln A (gradient of ln k vs 1/T is −Ea/R).
6	State the Ideal Gas Equation with its strict SI units, and show how to calculate the Mr of a volatile liquid.	pV = nRT, where p = pressure in Pascals (Pa; 1 kPa = 1,000 Pa), V = volume in cubic metres (m³; 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³), n = moles (mol), R = 8.31 J K⁻¹ mol⁻¹, and T = temperature in Kelvin (K = °C + 273). Since n = m / Mr, substituting gives Mr = mRT / (pV).
7	describe the four stages of Time-of-Flight (ToF) Mass Spectrometry: Ionisation, Acceleration, Ion Drift, and Detection.	1) Ionisation: Electrospray ionisation (sample dissolved in volatile solvent, pushed through high-voltage needle to gain H⁺ → MH⁺(g), no fragmentation) OR Electron impact (vaporised sample bombarded with high-energy electrons: X(g) + e⁻ → X⁺(g) + 2e⁻). 2) Acceleration: positive ions accelerated by an electric field so all ions have the same kinetic energy (Ek = ½mv²). 3) Ion Drift: lighter ions (lower m/z) travel faster through the flight tube (t = d√(m/2Ek)). 4) Detection: ions hit negative detector plate, gaining electrons and generating a current proportional to abundance.
8	How do you construct a Born–Haber Cycle for an ionic solid (e.g., NaCl), and why do experimental and theoretical lattice enthalpies sometimes differ?	ΔfH°(NaCl) = ΔatH°(Na) + IE1(Na) + ΔatH°(Cl) + EA1(Cl) + ΔLEH°(formation of NaCl). Theoretical lattice enthalpy assumes a purely 100% ionic model of spherical ions. If experimental lattice enthalpy is more exothermic (larger magnitude) than the theoretical value (e.g., in AgCl or MgI2), the compound has significant covalent character caused by a small, highly charged cation polarising a large anion.
9	How does the rate equation relate to the Rate-Determining Step (RDS) of a multi-step reaction mechanism?	Only the species involved in the slowest step (the Rate-Determining Step, RDS) or in fast equilibrium steps prior to the RDS appear in the rate equation, and the order with respect to each reactant equals the stoichiometric number of moles of that species involved up to and including the RDS.
10	Compare the cells and electrode reactions in non-rechargeable primary cells, rechargeable Lithium-ion cells, and alkaline Hydrogen Fuel Cells.	Lithium-ion cell: Negative electrode: Li → Li⁺ + e⁻; Positive electrode: Li⁺ + CoO2 + e⁻ → Li⁺[CoO2]⁻ (during charging, an external current reverses both reactions). Alkaline Hydrogen Fuel Cell: Negative anode: H2 + 2OH⁻ → 2H2O + 2e⁻; Positive cathode: O2 + 2H2O + 4e⁻ → 4OH⁻; Overall: 2H2 + O2 → 2H2O (E°cell = +1.23 V; operates continuously as long as H2 and O2 fuel are supplied).
11	Define Standard Enthalpy of Formation (ΔfH°) and Standard Enthalpy of Combustion (ΔcH°), and state their Hess's Law cycle equations.	ΔfH°: enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, 298 K). ΔcH°: enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions. Using Formation data: ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants). Using Combustion data: ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products).
12	How do you calculate the pH of a weak acid (HA) given its acid dissociation constant Ka and concentration [HA]?	For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻] / [HA]. Making the two standard approximations that [H⁺] ≈ [A⁻] (negligible water dissociation) and [HA]_eqm ≈ [HA]_initial (minimal dissociation), we get Ka ≈ [H⁺]² / [HA], so [H⁺] = √(Ka × [HA]), and pH = −log₁₀[H⁺]. Also pKa = −log₁₀(Ka).
13	Compare the three types of intermolecular forces (London dispersion forces, permanent dipole–dipole, and hydrogen bonding) and explain why ice is less dense than liquid water.	1) Induced dipole–dipole (London dispersion/van der Waals): present in all molecules due to temporary fluctuating electron clouds; increases with number of electrons/surface area. 2) Permanent dipole–dipole: between polar molecules with unsymmetrical electronegativity. 3) Hydrogen bonding: strongest IMF, occurs between a δ+ H bonded to N, O, or F and a lone pair on N, O, or F of another molecule. In ice, each H2O forms 4 tetrahedral H-bonds, holding molecules in an open lattice with large spaces.
14	Write the equations for the disproportionation of Chlorine (Cl2) with cold water (in water treatment) and with cold dilute NaOH (making bleach).	Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced. 1) With water: Cl2(g) + H2O(l) ⇌ HCl(aq) [Cl = -1] + HClO(aq) [chloric(I) acid, Cl = +1, kills bacteria]. In bright sunlight: 2Cl2 + 2H2O → 4HCl + O2. 2) With cold dilute NaOH: Cl2(aq) + 2NaOH(aq) → NaCl(aq) [Cl = -1] + NaClO(aq) [sodium chlorate(I) bleach, Cl = +1] + H2O(l).
15	Compare the reducing power of halide ions (Cl⁻, Br⁻, I⁻) when solid sodium halides react with concentrated sulfuric acid (H2SO4).	Reducing power increases down Group 7 (I⁻ > Br⁻ > Cl⁻) as ionic radius and shielding increase. 1) NaCl + H2SO4 → NaHSO4 + HCl (steamy white fumes; acid-base only, Cl⁻ cannot reduce S(+6)). 2) NaBr + H2SO4 → HBr (steamy fumes), then Br⁻ reduces S(+6) to S(+4) in SO2 (choking gas) + orange Br2 vapour. 3) NaI + H2SO4 → HI, then I⁻ reduces S(+6) all the way to S(+4) (SO2), S(0) (yellow sulfur solid), and S(-2) in H2S (rotten egg smell) + purple I2 vapour.
16	What is the Chelate Effect in ligand substitution reactions (e.g., replacing 6 H2O ligands with 3 bidentate en or 1 hexadentate EDTA⁴⁻ ligand)?	In [Cu(H2O)6]²⁺ + EDTA⁴⁻ → [Cu(EDTA)]²⁻ + 6H2O, 2 reactant particles form 7 product particles in solution (and 6 Cu–O bonds are broken while 6 new coordinate bonds are formed, so ΔH ≈ 0). This large increase in the number of free molecules in solution creates a large positive entropy change (+ΔS), making ΔG = ΔH − TΔS strongly negative and the multidentate complex thermodynamically very stable.
17	Explain why enthalpies of hydration (ΔhydH°) of ions become more exothermic as ionic radius decreases and ionic charge increases.	Enthalpy of hydration is the enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions at infinite dilution. Smaller ions with higher charges (higher charge density, e.g., Mg²⁺ vs Na⁺ or Al³⁺ vs Mg²⁺) exert a much stronger electrostatic attraction on the δ− oxygen or δ+ hydrogen atoms of polar water molecules, releasing more energy when ion–dipole bonds form. Also: ΔsolH° = ΔLEH°(dissociation) + ΣΔhydH°.
18	Describe the acid–base characters and reactions with water of the Period 3 oxides (Na2O, MgO, Al2O3, SiO2, P4O10, SO2, SO3).	Ionic basic oxides: Na2O + H2O → 2NaOH (pH 13–14); MgO + H2O → Mg(OH)2 (sparingly soluble, pH 9–10). Amphoteric oxide (ionic with covalent character): Al2O3 is insoluble in water, but reacts with BOTH acids (forming Al³⁺ salts) and alkalis (Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4). Giant covalent acidic oxide: SiO2 is insoluble in water but reacts with hot concentrated NaOH. Simple molecular acidic oxides: P4O10 + 6H2O → 4H3PO4 (pH 1–2); SO2 + H2O → H2SO3 (pH 2–3); SO3 + H2O → H2SO4 (pH 0–1).
19	How does an iodine–sodium thiosulfate (I2 / S2O3²⁻) redox titration work to determine the concentration of an oxidising agent (e.g., Cu²⁺, ClO⁻, or IO3⁻)?	1) Add excess potassium iodide (KI) to a measured volume of the oxidising agent so it oxidises I⁻ into brown aqueous I2 in a known molar ratio; 2) Titrate the liberated I2 with standard sodium thiosulfate (Na2S2O3): 2S2O3²⁻ + I2 → S4O6²⁻ + 2I⁻ (ratio 2 mol S2O3²⁻ : 1 mol I2); 3) When the solution fades to pale straw-yellow, add starch indicator (turns deep blue-black) and continue adding S2O3²⁻ dropwise until the blue-black colour disappears to colourless.
20	What is a buffer solution, and how does an acidic buffer (e.g., ethanoic acid CH3COOH + sodium ethanoate CH3COONa) resist pH changes when small amounts of H⁺ or OH⁻ are added?	An acidic buffer contains a large reservoir of weak acid (CH3COOH) and its conjugate base salt (CH3COO⁻). ForCH3COOH(aq) ⇌ H⁺(aq) + CH3COO⁻(aq): 1) On adding acid (H⁺), H⁺ reacts with the large reservoir of CH3COO⁻ ions to form undissociated CH3COOH (equilibrium shifts left); 2) On adding alkali (OH⁻), OH⁻ reacts with H⁺ to form H2O, and more CH3COOH dissociates (shifts right) to restore [H⁺]. Buffer pH is calculated using [H⁺] = Ka × [HA] / [A⁻].
21	Explain how heterogeneous catalysts (e.g., V2O5 in the Contact Process) and autocatalysis (Mn²⁺ in MnO4⁻ + C2O4²⁻) work.	In the Contact Process (SO2 to SO3), V2O5 changes oxidation state from +5 to +4 and back: SO2 + V2O5 → SO3 + V2O4, then V2O4 + ½O2 → V2O5. In the reaction between negative MnO4⁻ and C2O4²⁻ ions, initial rate is slow due to electrostatic repulsion between two anions; once a trace of Mn²⁺ product forms, it acts as an autocatalyst (reacting with MnO4⁻ to form Mn³⁺, which then oxidises C2O4²⁻), rapidly accelerating the reaction.
22	Using VSEPR theory, state the shapes and bond angles of molecules/ions with 2, 3, 4, 5, and 6 bonding pairs and 0 lone pairs.	2 bonding pairs (e.g., BeCl2, CO2): Linear, 180°. 3 bp (e.g., BF3): Trigonal planar, 120°. 4 bp (e.g., CH4, NH4⁺): Tetrahedral, 109.5°. 5 bp (e.g., PCl5): Trigonal bipyramidal, 90° and 120°. 6 bp (e.g., SF6): Octahedral, 90°.
23	What are the standard conditions for a Standard Hydrogen Electrode (SHE), and how do you calculate E°cell and predict redox feasibility?	SHE standard conditions: H2(g) at 100 kPa, 1.00 mol dm⁻³ H⁺(aq) solution, 298 K, inert porous Platinum electrode (defined as E° = 0.00 V). E°cell = E°(most positive / reduced right-hand electrode) − E°(most negative / oxidised left-hand electrode). A redox reaction is thermodynamically feasible if E°cell > 0 V.
24	How do lone pairs of electrons affect bond angles in VSEPR theory? State the shapes and bond angles of NH3, H2O, SF4, and XeF4.	Lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion, reducing bond angles by ~2.5° per lone pair. NH3 (3 bp, 1 lp): Trigonal pyramidal, 107°. H2O (2 bp, 2 lp): Bent / non-linear, 104.5°. SF4 (4 bp, 1 lp): Seesaw, 87° and 102°. XeF4 (4 bp, 2 lp): Square planar, 90°.
25	Define Brønsted–Lowry acids and bases, write the expression for the ionic product of water (Kw), and explain why pure water at 50°C has pH < 7 but is still neutral.	Brønsted–Lowry acid = proton (H⁺) donor; base = proton (H⁺) acceptor. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. The dissociation of water (H2O ⇌ H⁺ + OH⁻) is endothermic; heating water shifts equilibrium right, increasing Kw and [H⁺] so pH = −log₁₀[H⁺] drops below 7, but it remains completely neutral because [H⁺] still equals [OH⁻].
26	State the Gibbs Free Energy equation (ΔG = ΔH − TΔS) and explain the four combinations of ΔH and ΔS that determine reaction feasibility.	A reaction is thermodynamically feasible (spontaneous) when ΔG ≤ 0 (where ΔS must be divided by 1,000 to convert J K⁻¹ mol⁻¹ into kJ K⁻¹ mol⁻¹). 1) ΔH < 0, ΔS > 0: feasible at ALL temperatures; 2) ΔH > 0, ΔS < 0: NEVER feasible at any temperature; 3) ΔH < 0, ΔS < 0: feasible only at LOW temperatures (T < ΔH/ΔS); 4) ΔH > 0, ΔS > 0: feasible only at HIGH temperatures (T > ΔH/ΔS).
27	Explain why there is a slight drop in First Ionisation Energy between Magnesium and Aluminium, and between Phosphorus and Sulfur, across Period 3.	Mg (3s²) to Al (3s² 3p¹): Al's outer electron is in a higher-energy 3p subshell, which is slightly further from the nucleus and shielded by the 3s² electrons, so it requires less energy to remove. P (3p³) to S (3p⁴): in P, each of the three 3p orbitals is singly occupied, whereas in S, one 3p orbital contains a spin-paired pair of electrons; mutual repulsion between the paired electrons makes one easier to remove.
28	Define a transition metal in exact A-Level terms, and explain why Scandium (Sc) and Zinc (Zn) are d-block elements but NOT transition metals.	A transition metal is a d-block element that forms at least one stable ion with a partially filled (incomplete) d-subshell. Scandium only forms Sc³⁺ ([Ar] 3d⁰, empty d-subshell) and Zinc only forms Zn²⁺ ([Ar] 3d¹⁰, full d-subshell); therefore neither is a transition metal and their compounds are white/colourless.
29	How do you perform a redox titration using acidified potassium manganate(VII) (KMnO4) to determine Fe²⁺ content in iron tablets?	Dissolve crushed iron tablets in dilute H2SO4 (provides H⁺ ions without being oxidised by MnO4⁻, unlike HCl) and make up to 250 cm³ in a volumetric flask. Titrate 25.0 cm³ aliquots against standard KMnO4 from the burette; KMnO4 is self-indicating, turning from colourless to the first permanent pale pink colour at the endpoint. Half-equations give stoichiometric ratio 1 MnO4⁻ : 5 Fe²⁺ (MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O).
30	Write the expressions for Kc and Kp for the general gas reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), and state what factor changes Kc or Kp.	Kc = ([C]^c [D]^d) / ([A]^a [B]^b). Kp = (p(C)^c × p(D)^d) / (p(A)^a × p(B)^b), where partial pressure p(X) = mole fraction χ(X) × total pressure P_total. ONLY temperature changes the numerical value of Kc or Kp (concentration, pressure, and catalysts have zero effect on Kc or Kp).
31	State and explain the trends down Group 2 (Alkaline Earth Metals: Mg to Ba) for reactivity, solubility of hydroxides, and solubility of sulfates.	Reactivity with water/oxygen increases down Group 2 because atomic radius and shielding increase, lowering 1st and 2nd ionisation energies. Solubility of Group 2 Hydroxides M(OH)2 INCREASES down the group (Mg(OH)2 is sparingly soluble, used as Milk of Magnesia; Ca(OH)2 is used in agriculture; Ba(OH)2 is strongly alkaline). Solubility of Group 2 Sulfates MSO4 DECREASES down the group (BaSO4 is completely insoluble, used as a 'barium meal' in X-rays and in the sulfate ion test).
32	Write the full electron configurations of Chromium (Z=24) and Copper (Z=29) and explain why they are exceptions to the Aufbau rule.	Chromium (Z=24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (NOT 3d⁴ 4s²). Copper (Z=29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (NOT 3d⁹ 4s²). Half-filled (3d⁵) and completely filled (3d¹⁰) d-subshells provide extra thermodynamic stability due to symmetrical electron distribution and exchange energy. When transition metals form positive ions, 4s electrons are ALWAYS lost before 3d electrons.