In IB Chemistry Mass Spectrometry, what common mass fragments correspond to losses of m/z = 15, 17, 18, 28, 29, 31, and 45?

m/z = 15: •CH3 (methyl radical); m/z = 17: •OH (hydroxyl); m/z = 18: H2O (loss of water from alcohol); m/z = 28: CO or C2H4; m/z = 29: •CH2CH3 (ethyl) or •CHO (aldehyde acylium); m/z = 31: •OCH3 or •CH2OH; m/z = 43: •C3H7 or CH3CO⁺; m/z = 45: •COOH (carboxylic acid) or •OCH2CH3; m/z = 77: C6H5⁺ (phenyl ring).

Summarize the Characteristic Properties of First-Row Transition Elements (Sc to Cu, excluding Zn) and explain paramagnetism vs diamagnetism (HL Structure 3.2).

Transition elements have an incomplete d-subshell in at least one stable oxidation state (so Zn is not a transition metal). Five characteristic properties: 1) Variable oxidation states (because 4s and 3d energy levels are very close in energy; all show +2 state); 2) Formation of complex ions with monodentate/polydentate ligands via coordinate bonds; 3) Coloured compounds (d–d orbital splitting); 4) Catalytic activity (heterogeneous: Fe in Haber, V2O5 in Contact, Ni in hydrogenation; homogeneous: Fe²⁺ in heme); 5) Magnetic properties: Paramagnetic = contains unpaired d-electrons (attracted into magnetic field); Diamagnetic = all electrons spin-paired (weakly repelled).

Why does First Ionisation Energy show large jumps when plotting successive log₁₀(IE) values for a single element (e.g., Sodium or Aluminum)?

Successive ionisation energies always increase because an electron is being removed from an increasingly positive ion (greater electrostatic attraction). A large jump in log₁₀(IE) occurs when an electron is removed from a new inner principal energy level (shell) that is much closer to the nucleus and experiences far less electron shielding. Counting the number of electrons removed before the first big jump reveals the Group number (number of valence electrons).

Explain the Bonding Triangle (van Arkel–Ketelaar diagram) in IB Chemistry 2023+ syllabus (Structure 2.4).

The Bonding Triangle plots Average Electronegativity (∑χ / 2) on the horizontal x-axis against Electronegativity Difference (Δχ) on the vertical y-axis to show that ionic, covalent, and metallic bonding form a continuous spectrum rather than rigid categories: Bottom-left (low ∑χ, low Δχ) = Metallic; Bottom-right (high ∑χ, low Δχ) = Covalent (non-polar → polar as Δχ rises); Top apex (large Δχ > 1.8) = Ionic. Allows prediction of intermediate properties (e.g., AlCl3, SiO2, metalloids, alloys).

How does Reaction Quotient (Q) compared to Equilibrium Constant (Kc) predict the direction of shift to reach equilibrium?

The Reaction Quotient Q has the exact same algebraic expression as Kc, but is calculated using non-equilibrium concentrations at any given instant: 1) If Q Kc: ratio of products to reactants is too large, so the net reverse reaction proceeds (shifts left) until Q = Kc.

For HL Chemistry, state the VSEPR Electron Domain and Molecular Geometries for 5 and 6 electron domains.

5 Electron Domains (Trigonal Bipyramidal domain geometry, 90° axial & 120° equatorial; lone pairs ALWAYS occupy equatorial positions to minimise 90° repulsions): 5 bp + 0 lp = Trigonal bipyramidal (PCl5); 4 bp + 1 lp = Seesaw (SF4); 3 bp + 2 lp = T-shaped (ClF3); 2 bp + 3 lp = Linear (180°, XeF2, I3⁻). 6 Electron Domains (Octahedral 90° domain geometry): 6 bp + 0 lp = Octahedral (SF6); 5 bp + 1 lp = Square pyramidal (BrF5); 4 bp + 2 lp = Square planar (90°, XeF4).

Explain the reactions of Period 3 Oxides (Na2O, MgO, Al2O3, SiO2, P4O10, SO2, SO3, Cl2O7) with water and the chemical cause of Acid Deposition vs Ocean Acidification.

Across Period 3, oxides change from Giant Ionic Basic (Na2O, MgO) → Giant Ionic Amphoteric (Al2O3) → Giant Covalent Weakly Acidic (SiO2) → Simple Molecular Acidic (P4O10, SO2, SO3). Normal rainwater is naturally slightly acidic (pH ~5.6) due to dissolved CO2 forming weak carbonic acid (H2CO3). 'Acid rain' (pH < 5.6) is caused by anthropogenic sulfur oxides (SO2 + H2O → H2SO3; SO3 + H2O → H2SO4) and nitrogen oxides (2NO2 + H2O → HNO2 + HNO3).

How do you distinguish between Empirical Formula, Molecular Formula, and Structural Formula, and determine the formula of a hydrated salt (e.g., CuSO4·xH2O)?

Empirical formula: simplest whole-number ratio of atoms of each element in a compound. Molecular formula: actual number of atoms of each element in one molecule (a whole-number multiple of the empirical formula). For hydrated CuSO4·xH2O: weigh a crucible, add blue hydrated crystals and weigh, heat strongly until white anhydrous CuSO4 reaches constant mass, and calculate the mole ratio x = n(H2O evaporated) ÷ n(anhydrous CuSO4 remaining).

Using VSEPR Theory, state the Electron Domain Geometry, Molecular Geometry, and Bond Angles for 2, 3, and 4 electron domains (SL & HL Structure 2.2).

Count total Electron Domains (single, double, or triple bond = 1 domain; each lone pair = 1 domain): 1) 2 domains: Linear (180°, e.g., CO2, HCN, BeCl2). 2) 3 domains (Trigonal Planar 120° domain geometry): 3 bp + 0 lp = Trigonal planar (120°, BF3, CO3²⁻); 2 bp + 1 lp = Bent / V-shaped (~117°, SO2, O3, NO2⁻). 3) 4 domains (Tetrahedral 109.5° domain geometry): 4 bp + 0 lp = Tetrahedral (109.5°, CH4, NH4⁺); 3 bp + 1 lp = Trigonal pyramidal (107°, NH3, H3O⁺); 2 bp + 2 lp = Bent / V-shaped (104.5°, H2O).

In IB Chemistry Structure 2, why is Silicon Dioxide (SiO2, quartz) a high-melting giant covalent solid whereas Carbon Dioxide (CO2) is a discrete molecular gas?

Carbon has a small atomic radius, allowing effective sideways overlap of 2p orbitals with oxygen 2p orbitals to form strong C=O double bonds (O=C=O discrete linear molecules with weak London dispersion forces between them). Silicon has a larger atomic radius (3p orbitals), making π-overlap with oxygen ineffective; instead, each Si atom forms 4 strong single covalent σ-bonds to 4 oxygen atoms in a giant 3D tetrahedral network lattice (empirical formula SiO2).

In HL Born–Haber Cycles, compare the Lattice Enthalpies (ΔH°_lat) of NaCl, MgCl2, and MgO, and define Electron Affinity vs Electronegativity.

IB defines Lattice Enthalpy endothermically: MX(s) → M⁺(g) + X⁻(g) (ΔH°_lat > 0). Magnitude of lattice enthalpy increases with HIGHER ionic charge and SMALLER ionic radius: ΔH°_lat(MgO: Mg²⁺/O²⁻) > ΔH°_lat(MgCl2: Mg²⁺/Cl⁻) > ΔH°_lat(NaCl: Na⁺/Cl⁻). First Electron Affinity (1st EA, exothermic) is the enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons to form 1 mole of gaseous 1− ions (X(g) + e⁻ → X⁻(g)), whereas 2nd EA (e.g., O⁻(g) + e⁻ → O²⁻(g)) is always endothermic due to repulsion between two negative species.

How do you calculate Percentage Uncertainty and propagate uncertainties in addition/subtraction vs multiplication/division in IB Chemistry IA and Paper 1B?

1) When ADDING or SUBTRACTING measurements (e.g., ΔT = T_final − T_initial or burette titre = V_final − V_initial): ADD the ABSOLUTE uncertainties (Δy = Δa + Δb). 2) When MULTIPLYING or DIVIDING measurements (e.g., q = mcΔT or ΔH = −q/n): ADD the PERCENTAGE uncertainties (%Δy = %Δa + %Δb + %Δc), where % uncertainty = (Absolute uncertainty ÷ Measured value) × 100%, then convert back to absolute uncertainty at the end rounded to 1 significant figure.

How do you apply Beer–Lambert's Law (A = εcl) in Colorimetry / UV-Vis Spectrophotometry to find the concentration of a coloured transition metal solution?

1) Select the wavelength (λ_max) of maximum absorbance (using a complementary-colour filter to the solution's observed colour, e.g., orange/red ~650 nm filter for blue Cu²⁺(aq)); 2) Zero the colorimeter with a cuvette of distilled water (blank); 3) Measure the absorbance (A = log₁₀(I₀/I)) of a series of standard solutions of known concentration to plot a linear calibration curve (A = εcl); 4) Measure the absorbance of the unknown solution and interpolate its concentration from the linear graph.

State the Degree of Unsaturation / Index of Hydrogen Deficiency (IHD) formula and calculate the IHD of Aspirin (C9H8O4) and Benzene (C6H6).

IHD measures how many molecules of H2 would be needed to convert the molecule into a saturated, acyclic alkane: IHD = ½(2C + 2 + N − H − X), where X = halogens and O/S are ignored (each double bond = 1 IHD, each ring = 1 IHD, each triple bond = 2 IHD). For Benzene (C6H6): IHD = ½(12 + 2 − 6) = 4 (1 ring + 3 π-bonds). For Aspirin (C9H8O4): IHD = ½(18 + 2 − 8) = 6 (1 benzene ring [=4] + 2 C=O carbonyl double bonds [=2]).

Distinguish between Sigma (σ) and Pi (π) covalent bonds, and determine the Hybridisation (sp³, sp², sp) of carbon/nitrogen/oxygen atoms (HL Structure 2.2).

Sigma (σ) bond: formed by head-on axial overlap of atomic/hybrid orbitals, concentrating electron density along the internuclear axis (all single bonds are σ bonds). Pi (π) bond: formed by sideways parallel overlap of unhybridised p-orbitals above and below the internuclear axis (a double bond = 1σ + 1π; a triple bond = 1σ + 2π). Hybridisation matches the number of electron domains around the atom: 4 domains = sp³ (109.5°); 3 domains = sp² (120°); 2 domains = sp (180°).

How do intermolecular forces explain the boiling point trends of Group 16 hydrides (H2O, H2S, H2Se, H2Te) and organic homologous series?

In Group 16 hydrides, boiling points increase from H2S → H2Se → H2Te because increasing molar mass/electrons increases London dispersion forces, but H2O has an anomalously high boiling point (100°C) due to strong intermolecular hydrogen bonding. For organic compounds of similar molar mass, boiling point follows IMF strength: Carboxylic acids (dimer H-bonds) > Alcohols (H-bonds) > Aldehydes/Ketones/Esters (permanent dipole–dipole) > Alkanes/Alkenes (London dispersion forces only; branched isomers have smaller surface area and lower b.p. than straight-chain isomers).

Contrast using Average Bond Enthalpies vs Standard Enthalpies of Formation/Combustion in Hess's Law calculations.

1) Bond Enthalpies (DEFINED IN THE GAS PHASE ONLY!): ΔH° = Σ(BE of bonds broken in reactants) − Σ(BE of bonds formed in products). Less accurate than formation data because average bond enthalpies are mean values across many different compounds and neglect intermolecular forces in liquids/solids. 2) Formation Data: ΔH° = ΣΔfH°(products) − ΣΔfH°(reactants). 3) Combustion Data: ΔH° = ΣΔcH°(reactants) − ΣΔcH°(products).

In HL Energetics (Reactivity 1.2 & 1.3), how do you predict the sign of Entropy Change (ΔS°) from a chemical equation and connect ΔG°, ΔH°, ΔS°, and the Equilibrium Constant K?

Entropy (S°, in J K⁻¹ mol⁻¹) measures the dispersal of energy and matter (number of microstates: S(gas) ≫ S(liquid) > S(solid)). ΔS° is positive when the number of moles of GAS increases or a solid dissolves. Standard entropy change: ΔS° = ΣS°(products) − ΣS°(reactants). Gibbs Free Energy: ΔG° = ΔH° − TΔS° (spontaneous when ΔG° 0 and K > 1, so products predominate at equilibrium).

Compare Systematic Errors (Accuracy) and Random Uncertainties (Precision) in IB Chemistry experiments.

Random uncertainties arise from the finite precision of measuring apparatus (e.g., ±0.05 cm³ on a burette) and fluctuate equally in both directions; they reduce precision (spread of replicates) and can be reduced by repeating trials and taking a mean. Systematic errors arise from flaws in experimental design or miscalibrated equipment (e.g., heat loss to surroundings in calorimetry), shifting all measurements in the same direction and reducing accuracy (% error = |Experimental − Literature| ÷ Literature × 100%).

Explain Formal Charge (FC), Resonance Structures, Delocalisation, and Bond Order in O3, CO3²⁻, NO3⁻, and Benzene (HL Structure 2.2).

Formal Charge: FC = (Valence electrons in free atom) − ½(Bonding electrons) − (Non-bonding lone pair electrons). The most stable Lewis structure has FC values closest to zero and any negative FC on the most electronegative atom. Resonance hybrids occur when π-electrons delocalise over >2 atoms: Ozone (O3) has 2 resonance structures (bond order = 1.5, intermediate length/strength); Carbonate (CO3²⁻) and Nitrate (NO3⁻) have 3 resonance structures (bond order = 1.33); Benzene has bond order = 1.5.

In HL Transition Metal Chemistry, list the Spectrochemical Series of ligands and state the 3 factors that determine the magnitude of d-orbital splitting (ΔE) and colour.

When ligands approach a central metal ion, the 5 degenerate d-orbitals split into lower and higher energy sets; absorbed light promotes an electron across the gap ΔE = hf = hc/λ, and the complementary colour (from the IB Data Booklet colour wheel) is observed. Three factors affect ΔE: 1) Identity and oxidation state of the central metal ion (higher charge = larger ΔE); 2) Geometry of the complex (octahedral vs tetrahedral); 3) Position of the ligand in the Spectrochemical Series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H2O < NH3 < CO ≈ CN⁻ (strong-field ligands cause larger ΔE → absorb shorter λ / higher frequency light).

How do metallic properties (electrical conductivity, thermal conductivity, malleability, and melting point) vary across Period 3 (Na, Mg, Al)?

Moving from Na → Mg → Al: 1) Ionic charge increases (+1 → +2 → +3); 2) Number of delocalised valence electrons donated per atom increases (1 → 2 → 3); 3) Cationic radius decreases. Therefore, the electrostatic attraction between the lattice of cations and the sea of delocalised electrons becomes progressively stronger, so melting point and electrical conductivity increase from Na to Mg to Al.

Explain the principle of Green Chemistry Atom Economy and how E-factor or carbon efficiency is evaluated in IB Chemistry.

Atom Economy = (Molar mass of desired product ÷ Total molar mass of all reactants in balanced stoichiometric equation) × 100%. Addition reactions and rearrangements have 100% atom economy (zero by-products), whereas substitution and elimination reactions have < 100% atom economy. High atom economy minimises chemical waste and improves sustainability independently of experimental percentage yield.

Compare the structures, bonding, and physical properties of the four covalent allotropes of Carbon: Diamond, Graphite, Graphene, and Buckminsterfullerene (C60).

1) Diamond: giant covalent network, sp³ hybridised C (109.5°), no delocalised electrons (insulator), hardest natural substance, high thermal conductivity. 2) Graphite: giant covalent layered structure, sp² hybridised C (120° hexagonal rings) with London forces between layers (soft lubricant) and 1 delocalised p-electron per C (electrical conductor along layers). 3) Graphene: single 2D atomic layer of sp² hexagons (flexible, transparent, highest electrical/thermal conductivity). 4) C60 Fullerene: simple molecular substance (truncated icosahedron cage of 20 hexagons + 12 pentagons, sp² C, dissolves in non-polar solvents, semiconductor).

Explain how the Hydrogen Atomic Emission Spectrum provides evidence for quantized Bohr electron energy levels and how to find First Ionisation Energy from convergence (Structure 1.3).

When excited electrons fall from higher energy levels (n) to lower energy levels, they emit photons of discrete frequencies (ΔE = hf = hc/λ), producing a line emission spectrum rather than a continuous spectrum (transitions to n=1 give the UV Lyman series; to n=2 give the visible Balmer series; to n=3 give the IR Paschen series). The spectral lines converge at high frequency because energy levels get closer together at higher n; the convergence limit (n = ∞, where the electron leaves the atom) gives the First Ionisation Energy: IE₁ = L × h × f_limit.

What is the Winkler Method for determining Biochemical Oxygen Demand (BOD) in water samples via redox titration?

BOD is the amount of dissolved oxygen (in mg dm⁻³ or ppm) required by aerobic microorganisms to decompose organic matter in a water sample over 5 days in the dark at 20°C. In the Winkler titration, dissolved O2 oxidises Mn²⁺ to Mn(IV) (MnO2), which in acid oxidises I⁻ to I2, and the liberated I2 is titrated with standard sodium thiosulfate (S2O3²⁻). Stoichiometric ratio across the 3 equations: 1 mol O2 : 2 mol MnO2 : 2 mol I2 : 4 mol S2O3²⁻.

Explain how Ozone (O3) is continuously formed and depleted in the stratosphere (Chapman Cycle), why O3 absorbs longer-wavelength UV than O2, and how CFCs/NOx catalyse ozone depletion.

O2 has a double bond (bond order 2, BE = 498 kJ mol⁻¹), requiring high-energy short-wavelength UV-C (< 242 nm) to dissociate, whereas O3 has weaker delocalised bonds (bond order 1.5, BE = 364 kJ mol⁻¹), so it is dissociated by longer-wavelength, lower-energy UV-B (< 330 nm). CFCs (e.g., CF2Cl2) undergo C–Cl homolytic fission in UV light to release chlorine free radicals (Cl•): Cl• + O3 → ClO• + O2, then ClO• + O• → Cl• + O2 (net: O3 + O• → 2O2, regenerating Cl• to destroy thousands of O3 molecules).

How do you calculate the Equilibrium Constant Kc using an ICE (Initial, Change, Equilibrium) table in IB Chemistry?

1) Write the balanced reversible equation; 2) Convert all given quantities into moles (or concentrations in mol dm⁻³ by dividing by container volume V); 3) Fill in the Initial row, use the stoichiometric coefficients (±x) for the Change row, and sum to get the Equilibrium row; 4) Divide equilibrium moles by V (if exponents don't cancel) and substitute into Kc = [Products]^coeff / [Reactants]^coeff. If Kc ≫ 10³, reaction goes almost to completion; if Kc ≪ 10⁻³, reaction barely proceeds.

State the rules for drawing orbital box diagrams (Aufbau Principle, Hund's Rule, Pauli Exclusion Principle) and define the maximum electron capacity of main energy level n.

1) Aufbau Principle: electrons fill lowest-energy subshells first (1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p; note Cr = [Ar] 3d⁵ 4s¹ and Cu = [Ar] 3d¹⁰ 4s¹). 2) Hund's Rule: degenerate orbitals of equal energy are occupied singly with parallel spins before pairing up. 3) Pauli Exclusion Principle: an orbital can hold at most 2 electrons, and they must have opposite spins (+½ and −½). Main shell n holds at most 2n² electrons and n² orbitals.

State the Ideal Gas Equation in IB Chemistry Data Booklet units and specify the two conditions under which real gases deviate most from ideal behaviour (Structure 1.5).

PV = nRT, where P is in kPa (or Pa), V is in dm³ (when P is in kPa) or m³ (when P is in Pa), R = 8.31 J K⁻¹ mol⁻¹, T is in K, and Molar Volume of an ideal gas at STP (273.15 K, 100 kPa) = 22.7 dm³ mol⁻¹. Real gases deviate most from ideal gas assumptions at LOW temperature (intermolecular forces become significant as kinetic energy drops) and HIGH pressure (the finite volume of the gas particles themselves is no longer negligible compared to container volume).

In IB Energetics (Reactivity 1.1), how do you calculate ΔH from calorimetry (q = mcΔT), why are experimental combustion values less exothermic than Data Booklet values, and how do you extrapolate cooling curves?

Heat energy transferred to/from water: q = mcΔT (m = mass of water/solution in g, c = 4.18 J g⁻¹ K⁻¹), and ΔH = −q / (1000 × n_limiting) in kJ mol⁻¹. Experimental ΔcH° values using a spirit burner and copper calorimeter are much less exothermic than Data Booklet values due to: 1) Heat loss to surroundings and apparatus; 2) Incomplete combustion (forming soot C and CO); 3) Non-standard conditions and evaporation of volatile fuel. In solution calorimetry, extrapolate the linear cooling curve back to the time of mixing to correct for heat loss.

Explain the Periodic Trends across Period 3 and down a Group for Atomic Radius, Ionic Radius, First Ionisation Energy, Electron Affinity, and Electronegativity (Structure 3.1).

Across a Period (left → right): nuclear charge (protons) increases while electrons are added to the same outer shell (similar shielding), increasing effective nuclear charge (Z_eff) → Atomic radius DECREASES, IE₁ INCREASES (with dips at Group 13 p-subshell and Group 16 paired p-orbital), and Electronegativity INCREASES. Down a Group: number of occupied electron shells and shielding increase → Atomic radius INCREASES, IE₁ DECREASES, and Electronegativity DECREASES. Cations are smaller than parent atoms; isoelectronic anions (N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺) shrink as proton number rises.

IB Chemistry HL/SL: Stoichiometry, Atomic Structure, Periodicity, Chemical Bonding (VSEPR) & Energetics

IBDP Chemistry (Structure 1–3 & Reactivity 1) core and HL revision: mole concept & gas laws, emission spectra & ionisation energies, VSEPR electron domain geometries, hybridisation (sp/sp2/sp3), Period 3 oxides, transition metal complexes, Born-Haber cycles, and entropy/Gibbs free energy.

Level: basic
Languages: en → en
Cards: 32
Markdown: IB Chemistry HL/SL: Stoichiometry, Atomic Structure, Periodicity, Chemical Bonding (VSEPR) & Energetics (.md)

Flashcards (32)

#	Front (Term / Question)	Back (Definition / Answer)
1	In IB Chemistry Mass Spectrometry, what common mass fragments correspond to losses of m/z = 15, 17, 18, 28, 29, 31, and 45?	m/z = 15: •CH3 (methyl radical); m/z = 17: •OH (hydroxyl); m/z = 18: H2O (loss of water from alcohol); m/z = 28: CO or C2H4; m/z = 29: •CH2CH3 (ethyl) or •CHO (aldehyde acylium); m/z = 31: •OCH3 or •CH2OH; m/z = 43: •C3H7 or CH3CO⁺; m/z = 45: •COOH (carboxylic acid) or •OCH2CH3; m/z = 77: C6H5⁺ (phenyl ring).
2	Summarize the Characteristic Properties of First-Row Transition Elements (Sc to Cu, excluding Zn) and explain paramagnetism vs diamagnetism (HL Structure 3.2).	Transition elements have an incomplete d-subshell in at least one stable oxidation state (so Zn is not a transition metal). Five characteristic properties: 1) Variable oxidation states (because 4s and 3d energy levels are very close in energy; all show +2 state); 2) Formation of complex ions with monodentate/polydentate ligands via coordinate bonds; 3) Coloured compounds (d–d orbital splitting); 4) Catalytic activity (heterogeneous: Fe in Haber, V2O5 in Contact, Ni in hydrogenation; homogeneous: Fe²⁺ in heme); 5) Magnetic properties: Paramagnetic = contains unpaired d-electrons (attracted into magnetic field); Diamagnetic = all electrons spin-paired (weakly repelled).
3	Why does First Ionisation Energy show large jumps when plotting successive log₁₀(IE) values for a single element (e.g., Sodium or Aluminum)?	Successive ionisation energies always increase because an electron is being removed from an increasingly positive ion (greater electrostatic attraction). A large jump in log₁₀(IE) occurs when an electron is removed from a new inner principal energy level (shell) that is much closer to the nucleus and experiences far less electron shielding. Counting the number of electrons removed before the first big jump reveals the Group number (number of valence electrons).
4	Explain the Bonding Triangle (van Arkel–Ketelaar diagram) in IB Chemistry 2023+ syllabus (Structure 2.4).	The Bonding Triangle plots Average Electronegativity (∑χ / 2) on the horizontal x-axis against Electronegativity Difference (Δχ) on the vertical y-axis to show that ionic, covalent, and metallic bonding form a continuous spectrum rather than rigid categories: Bottom-left (low ∑χ, low Δχ) = Metallic; Bottom-right (high ∑χ, low Δχ) = Covalent (non-polar → polar as Δχ rises); Top apex (large Δχ > 1.8) = Ionic. Allows prediction of intermediate properties (e.g., AlCl3, SiO2, metalloids, alloys).
5	How does Reaction Quotient (Q) compared to Equilibrium Constant (Kc) predict the direction of shift to reach equilibrium?	The Reaction Quotient Q has the exact same algebraic expression as Kc, but is calculated using non-equilibrium concentrations at any given instant: 1) If Q < Kc: ratio of products to reactants is too small, so the net forward reaction proceeds (shifts right) until Q = Kc; 2) If Q = Kc: system is at dynamic equilibrium; 3) If Q > Kc: ratio of products to reactants is too large, so the net reverse reaction proceeds (shifts left) until Q = Kc.
6	For HL Chemistry, state the VSEPR Electron Domain and Molecular Geometries for 5 and 6 electron domains.	5 Electron Domains (Trigonal Bipyramidal domain geometry, 90° axial & 120° equatorial; lone pairs ALWAYS occupy equatorial positions to minimise 90° repulsions): 5 bp + 0 lp = Trigonal bipyramidal (PCl5); 4 bp + 1 lp = Seesaw (SF4); 3 bp + 2 lp = T-shaped (ClF3); 2 bp + 3 lp = Linear (180°, XeF2, I3⁻). 6 Electron Domains (Octahedral 90° domain geometry): 6 bp + 0 lp = Octahedral (SF6); 5 bp + 1 lp = Square pyramidal (BrF5); 4 bp + 2 lp = Square planar (90°, XeF4).
7	Explain the reactions of Period 3 Oxides (Na2O, MgO, Al2O3, SiO2, P4O10, SO2, SO3, Cl2O7) with water and the chemical cause of Acid Deposition vs Ocean Acidification.	Across Period 3, oxides change from Giant Ionic Basic (Na2O, MgO) → Giant Ionic Amphoteric (Al2O3) → Giant Covalent Weakly Acidic (SiO2) → Simple Molecular Acidic (P4O10, SO2, SO3). Normal rainwater is naturally slightly acidic (pH ~5.6) due to dissolved CO2 forming weak carbonic acid (H2CO3). 'Acid rain' (pH < 5.6) is caused by anthropogenic sulfur oxides (SO2 + H2O → H2SO3; SO3 + H2O → H2SO4) and nitrogen oxides (2NO2 + H2O → HNO2 + HNO3).
8	How do you distinguish between Empirical Formula, Molecular Formula, and Structural Formula, and determine the formula of a hydrated salt (e.g., CuSO4·xH2O)?	Empirical formula: simplest whole-number ratio of atoms of each element in a compound. Molecular formula: actual number of atoms of each element in one molecule (a whole-number multiple of the empirical formula). For hydrated CuSO4·xH2O: weigh a crucible, add blue hydrated crystals and weigh, heat strongly until white anhydrous CuSO4 reaches constant mass, and calculate the mole ratio x = n(H2O evaporated) ÷ n(anhydrous CuSO4 remaining).
9	Using VSEPR Theory, state the Electron Domain Geometry, Molecular Geometry, and Bond Angles for 2, 3, and 4 electron domains (SL & HL Structure 2.2).	Count total Electron Domains (single, double, or triple bond = 1 domain; each lone pair = 1 domain): 1) 2 domains: Linear (180°, e.g., CO2, HCN, BeCl2). 2) 3 domains (Trigonal Planar 120° domain geometry): 3 bp + 0 lp = Trigonal planar (120°, BF3, CO3²⁻); 2 bp + 1 lp = Bent / V-shaped (~117°, SO2, O3, NO2⁻). 3) 4 domains (Tetrahedral 109.5° domain geometry): 4 bp + 0 lp = Tetrahedral (109.5°, CH4, NH4⁺); 3 bp + 1 lp = Trigonal pyramidal (107°, NH3, H3O⁺); 2 bp + 2 lp = Bent / V-shaped (104.5°, H2O).
10	In IB Chemistry Structure 2, why is Silicon Dioxide (SiO2, quartz) a high-melting giant covalent solid whereas Carbon Dioxide (CO2) is a discrete molecular gas?	Carbon has a small atomic radius, allowing effective sideways overlap of 2p orbitals with oxygen 2p orbitals to form strong C=O double bonds (O=C=O discrete linear molecules with weak London dispersion forces between them). Silicon has a larger atomic radius (3p orbitals), making π-overlap with oxygen ineffective; instead, each Si atom forms 4 strong single covalent σ-bonds to 4 oxygen atoms in a giant 3D tetrahedral network lattice (empirical formula SiO2).
11	In HL Born–Haber Cycles, compare the Lattice Enthalpies (ΔH°_lat) of NaCl, MgCl2, and MgO, and define Electron Affinity vs Electronegativity.	IB defines Lattice Enthalpy endothermically: MX(s) → M⁺(g) + X⁻(g) (ΔH°_lat > 0). Magnitude of lattice enthalpy increases with HIGHER ionic charge and SMALLER ionic radius: ΔH°_lat(MgO: Mg²⁺/O²⁻) > ΔH°_lat(MgCl2: Mg²⁺/Cl⁻) > ΔH°_lat(NaCl: Na⁺/Cl⁻). First Electron Affinity (1st EA, exothermic) is the enthalpy change when 1 mole of gaseous atoms gains 1 mole of electrons to form 1 mole of gaseous 1− ions (X(g) + e⁻ → X⁻(g)), whereas 2nd EA (e.g., O⁻(g) + e⁻ → O²⁻(g)) is always endothermic due to repulsion between two negative species.
12	How do you calculate Percentage Uncertainty and propagate uncertainties in addition/subtraction vs multiplication/division in IB Chemistry IA and Paper 1B?	1) When ADDING or SUBTRACTING measurements (e.g., ΔT = T_final − T_initial or burette titre = V_final − V_initial): ADD the ABSOLUTE uncertainties (Δy = Δa + Δb). 2) When MULTIPLYING or DIVIDING measurements (e.g., q = mcΔT or ΔH = −q/n): ADD the PERCENTAGE uncertainties (%Δy = %Δa + %Δb + %Δc), where % uncertainty = (Absolute uncertainty ÷ Measured value) × 100%, then convert back to absolute uncertainty at the end rounded to 1 significant figure.
13	How do you apply Beer–Lambert's Law (A = εcl) in Colorimetry / UV-Vis Spectrophotometry to find the concentration of a coloured transition metal solution?	1) Select the wavelength (λ_max) of maximum absorbance (using a complementary-colour filter to the solution's observed colour, e.g., orange/red ~650 nm filter for blue Cu²⁺(aq)); 2) Zero the colorimeter with a cuvette of distilled water (blank); 3) Measure the absorbance (A = log₁₀(I₀/I)) of a series of standard solutions of known concentration to plot a linear calibration curve (A = εcl); 4) Measure the absorbance of the unknown solution and interpolate its concentration from the linear graph.
14	State the Degree of Unsaturation / Index of Hydrogen Deficiency (IHD) formula and calculate the IHD of Aspirin (C9H8O4) and Benzene (C6H6).	IHD measures how many molecules of H2 would be needed to convert the molecule into a saturated, acyclic alkane: IHD = ½(2C + 2 + N − H − X), where X = halogens and O/S are ignored (each double bond = 1 IHD, each ring = 1 IHD, each triple bond = 2 IHD). For Benzene (C6H6): IHD = ½(12 + 2 − 6) = 4 (1 ring + 3 π-bonds). For Aspirin (C9H8O4): IHD = ½(18 + 2 − 8) = 6 (1 benzene ring [=4] + 2 C=O carbonyl double bonds [=2]).
15	Distinguish between Sigma (σ) and Pi (π) covalent bonds, and determine the Hybridisation (sp³, sp², sp) of carbon/nitrogen/oxygen atoms (HL Structure 2.2).	Sigma (σ) bond: formed by head-on axial overlap of atomic/hybrid orbitals, concentrating electron density along the internuclear axis (all single bonds are σ bonds). Pi (π) bond: formed by sideways parallel overlap of unhybridised p-orbitals above and below the internuclear axis (a double bond = 1σ + 1π; a triple bond = 1σ + 2π). Hybridisation matches the number of electron domains around the atom: 4 domains = sp³ (109.5°); 3 domains = sp² (120°); 2 domains = sp (180°).
16	How do intermolecular forces explain the boiling point trends of Group 16 hydrides (H2O, H2S, H2Se, H2Te) and organic homologous series?	In Group 16 hydrides, boiling points increase from H2S → H2Se → H2Te because increasing molar mass/electrons increases London dispersion forces, but H2O has an anomalously high boiling point (100°C) due to strong intermolecular hydrogen bonding. For organic compounds of similar molar mass, boiling point follows IMF strength: Carboxylic acids (dimer H-bonds) > Alcohols (H-bonds) > Aldehydes/Ketones/Esters (permanent dipole–dipole) > Alkanes/Alkenes (London dispersion forces only; branched isomers have smaller surface area and lower b.p. than straight-chain isomers).
17	Contrast using Average Bond Enthalpies vs Standard Enthalpies of Formation/Combustion in Hess's Law calculations.	1) Bond Enthalpies (DEFINED IN THE GAS PHASE ONLY!): ΔH° = Σ(BE of bonds broken in reactants) − Σ(BE of bonds formed in products). Less accurate than formation data because average bond enthalpies are mean values across many different compounds and neglect intermolecular forces in liquids/solids. 2) Formation Data: ΔH° = ΣΔfH°(products) − ΣΔfH°(reactants). 3) Combustion Data: ΔH° = ΣΔcH°(reactants) − ΣΔcH°(products).
18	In HL Energetics (Reactivity 1.2 & 1.3), how do you predict the sign of Entropy Change (ΔS°) from a chemical equation and connect ΔG°, ΔH°, ΔS°, and the Equilibrium Constant K?	Entropy (S°, in J K⁻¹ mol⁻¹) measures the dispersal of energy and matter (number of microstates: S(gas) ≫ S(liquid) > S(solid)). ΔS° is positive when the number of moles of GAS increases or a solid dissolves. Standard entropy change: ΔS° = ΣS°(products) − ΣS°(reactants). Gibbs Free Energy: ΔG° = ΔH° − TΔS° (spontaneous when ΔG° < 0; at equilibrium ΔG = 0, so T_eqm = ΔH° / ΔS°). Relating ΔG° to the equilibrium constant K: ΔG° = −RT ln K (if ΔG° < 0, then ln K > 0 and K > 1, so products predominate at equilibrium).
19	Compare Systematic Errors (Accuracy) and Random Uncertainties (Precision) in IB Chemistry experiments.	Random uncertainties arise from the finite precision of measuring apparatus (e.g., ±0.05 cm³ on a burette) and fluctuate equally in both directions; they reduce precision (spread of replicates) and can be reduced by repeating trials and taking a mean. Systematic errors arise from flaws in experimental design or miscalibrated equipment (e.g., heat loss to surroundings in calorimetry), shifting all measurements in the same direction and reducing accuracy (% error = \|Experimental − Literature\| ÷ Literature × 100%).
20	Explain Formal Charge (FC), Resonance Structures, Delocalisation, and Bond Order in O3, CO3²⁻, NO3⁻, and Benzene (HL Structure 2.2).	Formal Charge: FC = (Valence electrons in free atom) − ½(Bonding electrons) − (Non-bonding lone pair electrons). The most stable Lewis structure has FC values closest to zero and any negative FC on the most electronegative atom. Resonance hybrids occur when π-electrons delocalise over >2 atoms: Ozone (O3) has 2 resonance structures (bond order = 1.5, intermediate length/strength); Carbonate (CO3²⁻) and Nitrate (NO3⁻) have 3 resonance structures (bond order = 1.33); Benzene has bond order = 1.5.
21	In HL Transition Metal Chemistry, list the Spectrochemical Series of ligands and state the 3 factors that determine the magnitude of d-orbital splitting (ΔE) and colour.	When ligands approach a central metal ion, the 5 degenerate d-orbitals split into lower and higher energy sets; absorbed light promotes an electron across the gap ΔE = hf = hc/λ, and the complementary colour (from the IB Data Booklet colour wheel) is observed. Three factors affect ΔE: 1) Identity and oxidation state of the central metal ion (higher charge = larger ΔE); 2) Geometry of the complex (octahedral vs tetrahedral); 3) Position of the ligand in the Spectrochemical Series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H2O < NH3 < CO ≈ CN⁻ (strong-field ligands cause larger ΔE → absorb shorter λ / higher frequency light).
22	How do metallic properties (electrical conductivity, thermal conductivity, malleability, and melting point) vary across Period 3 (Na, Mg, Al)?	Moving from Na → Mg → Al: 1) Ionic charge increases (+1 → +2 → +3); 2) Number of delocalised valence electrons donated per atom increases (1 → 2 → 3); 3) Cationic radius decreases. Therefore, the electrostatic attraction between the lattice of cations and the sea of delocalised electrons becomes progressively stronger, so melting point and electrical conductivity increase from Na to Mg to Al.
23	Explain the principle of Green Chemistry Atom Economy and how E-factor or carbon efficiency is evaluated in IB Chemistry.	Atom Economy = (Molar mass of desired product ÷ Total molar mass of all reactants in balanced stoichiometric equation) × 100%. Addition reactions and rearrangements have 100% atom economy (zero by-products), whereas substitution and elimination reactions have < 100% atom economy. High atom economy minimises chemical waste and improves sustainability independently of experimental percentage yield.
24	Compare the structures, bonding, and physical properties of the four covalent allotropes of Carbon: Diamond, Graphite, Graphene, and Buckminsterfullerene (C60).	1) Diamond: giant covalent network, sp³ hybridised C (109.5°), no delocalised electrons (insulator), hardest natural substance, high thermal conductivity. 2) Graphite: giant covalent layered structure, sp² hybridised C (120° hexagonal rings) with London forces between layers (soft lubricant) and 1 delocalised p-electron per C (electrical conductor along layers). 3) Graphene: single 2D atomic layer of sp² hexagons (flexible, transparent, highest electrical/thermal conductivity). 4) C60 Fullerene: simple molecular substance (truncated icosahedron cage of 20 hexagons + 12 pentagons, sp² C, dissolves in non-polar solvents, semiconductor).
25	Explain how the Hydrogen Atomic Emission Spectrum provides evidence for quantized Bohr electron energy levels and how to find First Ionisation Energy from convergence (Structure 1.3).	When excited electrons fall from higher energy levels (n) to lower energy levels, they emit photons of discrete frequencies (ΔE = hf = hc/λ), producing a line emission spectrum rather than a continuous spectrum (transitions to n=1 give the UV Lyman series; to n=2 give the visible Balmer series; to n=3 give the IR Paschen series). The spectral lines converge at high frequency because energy levels get closer together at higher n; the convergence limit (n = ∞, where the electron leaves the atom) gives the First Ionisation Energy: IE₁ = L × h × f_limit.
26	What is the Winkler Method for determining Biochemical Oxygen Demand (BOD) in water samples via redox titration?	BOD is the amount of dissolved oxygen (in mg dm⁻³ or ppm) required by aerobic microorganisms to decompose organic matter in a water sample over 5 days in the dark at 20°C. In the Winkler titration, dissolved O2 oxidises Mn²⁺ to Mn(IV) (MnO2), which in acid oxidises I⁻ to I2, and the liberated I2 is titrated with standard sodium thiosulfate (S2O3²⁻). Stoichiometric ratio across the 3 equations: 1 mol O2 : 2 mol MnO2 : 2 mol I2 : 4 mol S2O3²⁻.
27	Explain how Ozone (O3) is continuously formed and depleted in the stratosphere (Chapman Cycle), why O3 absorbs longer-wavelength UV than O2, and how CFCs/NOx catalyse ozone depletion.	O2 has a double bond (bond order 2, BE = 498 kJ mol⁻¹), requiring high-energy short-wavelength UV-C (< 242 nm) to dissociate, whereas O3 has weaker delocalised bonds (bond order 1.5, BE = 364 kJ mol⁻¹), so it is dissociated by longer-wavelength, lower-energy UV-B (< 330 nm). CFCs (e.g., CF2Cl2) undergo C–Cl homolytic fission in UV light to release chlorine free radicals (Cl•): Cl• + O3 → ClO• + O2, then ClO• + O• → Cl• + O2 (net: O3 + O• → 2O2, regenerating Cl• to destroy thousands of O3 molecules).
28	How do you calculate the Equilibrium Constant Kc using an ICE (Initial, Change, Equilibrium) table in IB Chemistry?	1) Write the balanced reversible equation; 2) Convert all given quantities into moles (or concentrations in mol dm⁻³ by dividing by container volume V); 3) Fill in the Initial row, use the stoichiometric coefficients (±x) for the Change row, and sum to get the Equilibrium row; 4) Divide equilibrium moles by V (if exponents don't cancel) and substitute into Kc = [Products]^coeff / [Reactants]^coeff. If Kc ≫ 10³, reaction goes almost to completion; if Kc ≪ 10⁻³, reaction barely proceeds.
29	State the rules for drawing orbital box diagrams (Aufbau Principle, Hund's Rule, Pauli Exclusion Principle) and define the maximum electron capacity of main energy level n.	1) Aufbau Principle: electrons fill lowest-energy subshells first (1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p; note Cr = [Ar] 3d⁵ 4s¹ and Cu = [Ar] 3d¹⁰ 4s¹). 2) Hund's Rule: degenerate orbitals of equal energy are occupied singly with parallel spins before pairing up. 3) Pauli Exclusion Principle: an orbital can hold at most 2 electrons, and they must have opposite spins (+½ and −½). Main shell n holds at most 2n² electrons and n² orbitals.
30	State the Ideal Gas Equation in IB Chemistry Data Booklet units and specify the two conditions under which real gases deviate most from ideal behaviour (Structure 1.5).	PV = nRT, where P is in kPa (or Pa), V is in dm³ (when P is in kPa) or m³ (when P is in Pa), R = 8.31 J K⁻¹ mol⁻¹, T is in K, and Molar Volume of an ideal gas at STP (273.15 K, 100 kPa) = 22.7 dm³ mol⁻¹. Real gases deviate most from ideal gas assumptions at LOW temperature (intermolecular forces become significant as kinetic energy drops) and HIGH pressure (the finite volume of the gas particles themselves is no longer negligible compared to container volume).
31	In IB Energetics (Reactivity 1.1), how do you calculate ΔH from calorimetry (q = mcΔT), why are experimental combustion values less exothermic than Data Booklet values, and how do you extrapolate cooling curves?	Heat energy transferred to/from water: q = mcΔT (m = mass of water/solution in g, c = 4.18 J g⁻¹ K⁻¹), and ΔH = −q / (1000 × n_limiting) in kJ mol⁻¹. Experimental ΔcH° values using a spirit burner and copper calorimeter are much less exothermic than Data Booklet values due to: 1) Heat loss to surroundings and apparatus; 2) Incomplete combustion (forming soot C and CO); 3) Non-standard conditions and evaporation of volatile fuel. In solution calorimetry, extrapolate the linear cooling curve back to the time of mixing to correct for heat loss.
32	Explain the Periodic Trends across Period 3 and down a Group for Atomic Radius, Ionic Radius, First Ionisation Energy, Electron Affinity, and Electronegativity (Structure 3.1).	Across a Period (left → right): nuclear charge (protons) increases while electrons are added to the same outer shell (similar shielding), increasing effective nuclear charge (Z_eff) → Atomic radius DECREASES, IE₁ INCREASES (with dips at Group 13 p-subshell and Group 16 paired p-orbital), and Electronegativity INCREASES. Down a Group: number of occupied electron shells and shielding increase → Atomic radius INCREASES, IE₁ DECREASES, and Electronegativity DECREASES. Cations are smaller than parent atoms; isoelectronic anions (N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺) shrink as proton number rises.