IB Chemistry · Unit 1 · Foundations

The smallest
thing
that matters.

Where everything starts. Electrons in orbitals, protons in nuclei, and a periodic table that explains chemistry from a few simple rules.

10Lessons
3HL extensions
42Key terms
SL+HLLevel
Unit 1 · Standard Level

7 lessons to work through.

The required syllabus content for Unit 1, in order. Each card is one lesson-sized checkpoint.

Lesson 0

Introduction to DP Chemistry

Lesson 0 of Unit 1.

Lesson 1

Separating techniques

Lesson 1 of Unit 1.

Lesson 2

Introduction to the particulate nature of matter

Distinguish between the properties of elements, compounds and mixtures.

Lesson 3

The Nuclear Atom

Conceptual understanding: The mass of an atom is concentrated in its minute, positively charged nucleus.

Lesson 4

Isotopes and RAM

Lesson 4 of Unit 1.

Lesson 6

The emission spectrum

1) Emission and absorption spectra provide both evidence for: A. the existence of neutrons B. the existence of isotopes C. the existence of atomic energy levels D. the nuclear model of an atom

Lesson 7

Electron Configurations

Conceptual understanding: The electron configuration of an atom can be deduced from its atomic number.

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

Separating techniques

Structure 1.1.1Structure 1.1.2

Lesson outcomes

Chemistry begins with classifying matter. The opening lesson establishes the three-way taxonomy of elements, compounds and mixtures — and the laboratory techniques used to separate the last category.

Six separation techniques (Tool 1)

Worked example

The separation challenge — Cu, S and NH₄Cl

Problem. You have a mixture of copper filings (Cu), sulfur (S) and ammonium chloride (NH₄Cl). Design a method to separate and collect each component in pure form. Cu is metallic and non-magnetic; S is non-metal and insoluble in water; NH₄Cl sublimates on heating.
Solution. Step 1 — gentle heat in a closed vessel. NH₄Cl sublimes and re-deposits on a cool surface above; collect it. Cu and S remain in the original vessel.
Step 2 — add water and stir. Neither dissolves (S is insoluble; Cu is unreactive), but the densities differ. Decant or use density.
Step 3 — alternative for the Cu/S step: heat in carbon disulfide (CS₂) which dissolves S but not Cu. Filter to retain Cu on the paper; evaporate the filtrate to recover S.
Each technique here exploits a physical property (sublimation T, solubility) — never a chemical reaction. That's the hallmark of a separable mixture.

Try these

  1. Suggest suitable methods for separating each mixture: (a) salt and pepper, (b) several water-soluble dyes, (c) sugar and water, (d) iron and copper filings.
    Show answer
    (a) Dissolve in water → filter (pepper insoluble, salt in solution); evaporate filtrate to recover salt. (b) Paper chromatography. (c) Evaporation/crystallisation. (d) Magnetism — iron is magnetic, copper is not.
  2. Summarise each separation technique: state how it works, what is removed, and what is left behind. (a) Solvation, (b) Filtration, (c) Recrystallisation, (d) Evaporation, (e) Distillation, (f) Paper chromatography.
    Show answer
    (a) Solvation: dissolve soluble component in chosen solvent → soluble part removed as solute; insoluble part remains. (b) Filtration: pour mixture through filter paper → liquid (filtrate) passes through; insoluble solid (residue) retained on paper. (c) Recrystallisation: dissolve in hot solvent then cool slowly → pure crystals form (the solid); soluble impurities remain in cold liquid. (d) Evaporation: gently heat solution → solvent boils away; dissolved solute left as residue. (e) Distillation: heat mixture, condense vapour, collect → component with lower b.p. comes off first; higher-b.p. component remains. (f) Paper chromatography: place spot on paper, lower into solvent → soluble components travel up paper at different rates based on affinity; separates components by their Rf values.
  3. Data-based: in a chromatography experiment, which colour dot had the strongest affinity for both solvent 1 and solvent 2? Which dots had a stronger affinity for solvent 1 than for solvent 2? Which had a stronger affinity for solvent 2 than solvent 1?
    Show answer
    Affinity for a solvent is inferred from how far up the paper the dot moves in that solvent. The dot that moved the highest in both solvents has strongest affinity for both. Dots that moved high in solvent 1 but stayed low in solvent 2 have stronger affinity for solvent 1. Vice versa for solvent 2. (Specifics depend on the chromatography figure in the deck.)
  4. Why does an impurity in a recovered solid lower and broaden its melting point?
    Show answer
    Impurities disrupt the regular crystal lattice, so lattice forces can be overcome at lower temperature. A variety of impurity sites gives a range of melting temperatures — so the m.p. is both lower and less sharp than the pure literature value. A sharp m.p. at the literature value is the chemist's confirmation of high purity.
  5. Why is the recrystallisation step considered most successful when the substance is very soluble in hot solvent but barely soluble in cold solvent?
    Show answer
    All the impure substance dissolves on heating (so impurities go into solution too). On cooling, the substance crystallises out cleanly because the cold solvent can hold very little — the impurities, present in small quantities, remain dissolved in the small amount they can occupy. Filtering off the crystals gives pure product.
Lesson 2

Introduction to the particulate nature of matter

Structure 1.1.1Structure 1.1.2

Lesson outcomes

Everything is made of matter — but not all matter is the same. Elements contain only one type of atom; compounds contain two or more elements chemically joined; mixtures are elements/compounds together in the same vessel without chemical bonding.

Pure substances (elements and compounds) have a uniform chemical composition. Mixtures can be homogeneous (same state, uniform composition — e.g. dissolved salt in water) or heterogeneous (different states/non-uniform — e.g. sand in water, oil and water).

Kinetic molecular theory

Particles in a solid vibrate about fixed positions. Heating increases their kinetic energy — temperature is a measure of average kinetic energy. Adding enough energy lets particles overcome the forces holding them in place: solid → liquid → gas. The reverse: gas → liquid → solid. Direct s → g is sublimation; g → s is deposition.

Colligative property

Adding an impurity (any soluble solute) depresses the melting point and elevates the boiling point of a solvent. The magnitude depends on the number of solute particles, not their identity — that's what "colligative" means.

State symbols

Every species in a chemical equation should carry a state symbol: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

Worked example

Solvent vs solute

Problem. When 5 g of NaCl is dissolved in 100 g of water, identify the solute, the solvent, and the type of mixture formed.
Solution. Solute = NaCl (the substance dissolved, smaller amount).
Solvent = water (the substance doing the dissolving, larger amount).
Type of mixture = solution — a homogeneous mixture: uniform composition, single visible phase.

Try these

  1. List 5 elements, 5 compounds, and 5 mixtures.
    Show answer
    Elements: Na, O₂, Au, He, C. Compounds: H₂O, NaCl, CO₂, C₆H₁₂O₆, NH₃. Mixtures: air, sea water, salad dressing, bronze, granite.
  2. Of elements, compounds and mixtures, which are pure? Why?
    Show answer
    Elements and compounds — they have uniform, fixed chemical composition. Mixtures vary.
  3. What types of mixtures can we have?
    Show answer
    Homogeneous (one visible phase, uniform — e.g. salt water) and heterogeneous (multiple visible phases, non-uniform — e.g. sand in water).
  4. Find out the meaning of solute, solvent and solution.
    Show answer
    Solute: substance that dissolves in a solvent (often present in smaller amount). Solvent: substance doing the dissolving (often present in larger amount; usually water in chemistry). Solution: homogeneous mixture of solute and solvent.
  5. Explain what happens to particles as you go from solid → liquid → gas, in terms of internal energy, kinetic energy and temperature.
    Show answer
    Internal energy increases. Particles gain kinetic energy (vibrate faster in solids; move past one another in liquids; fly freely in gases). Temperature is a measure of average kinetic energy — heating a solid increases the vibration speed of its particles.
  6. Why does adding a soluble solute (e.g. salt) to water depress the melting point and elevate the boiling point? What is the property called?
    Show answer
    Solute particles disrupt the regular structure of the solvent's solid lattice (depressing m.p.) and increase the energy needed for solvent molecules to escape from the liquid (elevating b.p.). The effect depends on the number of solute particles, not their identity — this is a colligative property. (Pure water boils at exactly 100 °C; salt water above 100 °C.)
  7. Name the changes of state, including sublimation and deposition. Which involve energy absorbed and which energy released?
    Show answer
    Melting (s→l, absorbed), freezing (l→s, released), evaporation/boiling (l→g, absorbed), condensation (g→l, released), sublimation (s→g, absorbed — e.g. CO₂, I₂, NH₄Cl), deposition (g→s, released).
  8. Write state symbols for all species in the equation: hydrochloric acid + calcium carbonate → calcium chloride + water + carbon dioxide.
    Show answer
    2 HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g).
Lesson 3

The nuclear atom

Structure 1.2.1

Lesson outcomes

Through the 19th and 20th centuries, models of the atom became progressively more refined: Dalton's indivisible spheres → Thomson's "plum pudding" → Rutherford's nuclear atom → Bohr's quantised shells → the modern quantum-mechanical model.

Rutherford's gold-foil experiment (1909)

α-particles fired at a thin gold foil. Three observations and what each tells us:

Subatomic particles

ParticleLocationRelative massRelative charge
ProtonNucleus1+1
NeutronNucleus10
ElectronSurrounds nucleus1/1836−1

Atomic number, mass number, nuclear symbol

Atomic number Z = number of protons. Defines the element. Mass number A = protons + neutrons. Written in nuclear notation as AZX. For neutral atoms, # electrons = # protons. For an ion of charge q, # electrons = Z − q.

Worked example

Counting subatomic particles

Problem. How many protons, neutrons and electrons are in 5626Fe³⁺?
Solution. Protons = Z = 26. Neutrons = A − Z = 56 − 26 = 30. Electrons = Z − charge = 26 − 3 = 23.

Try these

  1. Starter: briefly outline the models of the atom in chronological order. Which model do we use now? Is it 'accurate'?
    Show answer
    Dalton (~1808): atoms are indivisible spheres. Thomson (1897): 'plum pudding' — electrons embedded in positive matter. Rutherford (1909): nuclear atom — dense positive nucleus, electrons in space around. Bohr (1913): electrons in fixed circular shells with quantised energies. Modern quantum model (1920s+): orbitals as probability distributions. We use the quantum-mechanical model. It is the best predictive model we have — but no model is the 'truth'; each is a useful approximation in its regime.
  2. What are the three conclusions from Rutherford's gold-foil experiment?
    Show answer
    (1) Most α-particles passed straight through → atoms are mostly empty space. (2) A small fraction were deflected through small angles → there is a positive region within the atom. (3) A very few were deflected by more than 90° (or bounced straight back) → the positive region is small, dense and highly concentrated (the nucleus).
  3. Match each gold-foil observation to the property of the nucleus it reveals: (i) most α pass through; (ii) some are deflected; (iii) very few bounce back.
    Show answer
    (i) → atom is mostly empty space. (ii) → there is a positive charge in the atom (repels positive α). (iii) → the positive region is small, dense and concentrated (nucleus has enough mass to stop a moving α).
  4. State the relative mass and relative charge of a proton, neutron and electron. Where is each particle in the atom?
    Show answer
    Proton: mass ≈ 1 u, charge +1, in the nucleus. Neutron: mass ≈ 1 u, charge 0, in the nucleus. Electron: mass ≈ 1/1836 u (negligible), charge −1, outside the nucleus.
  5. What does the atomic number Z tell you? What does the mass number A tell you? How is the nuclear-symbol notation written?
    Show answer
    Z = number of protons; defines the element. A = total nucleons = protons + neutrons; identifies the isotope. Nuclear symbol: AZX (e.g. 2311Na).
  6. State the number of protons, neutrons and electrons in: (a) 23Na, (b) 35Cl⁻, (c) 40Ca²⁺.
    Show answer
    (a) 11 p, 12 n, 11 e. (b) 17 p, 18 n, 18 e (gained 1 e to become −1). (c) 20 p, 20 n, 18 e (lost 2 e to become +2).
Lesson 4

Isotopes and relative atomic mass

Structure 1.2.2

Lesson outcomes

The periodic table lists chlorine with Ar = 35.45 — but you can't have half a proton. The value is a weighted average across all the naturally occurring isotopes of chlorine: 35Cl (≈ 75%) and 37Cl (≈ 25%).

Isotopes

Isotopes = atoms of the same element (same Z) with different numbers of neutrons (different A). They have identical chemical properties (same electrons, same bonding) but slightly different physical properties — mass, density, rates of effusion, and any property depending on inertia. This is why isotopes can only be separated by mass-based techniques (mass spec, centrifugation), never by chemistry.

Calculating Ar

The formula:

Ar = Σ (isotopic mass × fractional abundance)

where each fractional abundance is the % divided by 100. Use percentages straight from a mass spectrum if you prefer.

Worked example

A<sub>r</sub> of chlorine

Problem. Chlorine has two natural isotopes: 35Cl (75.00%) and 37Cl (25.00%). Calculate the relative atomic mass.
Solution. Ar(Cl) = (35 × 0.7500) + (37 × 0.2500) = 26.25 + 9.25 = 35.50 (≈ 35.45 in the data booklet).
Worked example

A<sub>r</sub> of iron from a 4-isotope table

Problem. Iron exists as 54Fe (5.85%), 56Fe (91.75%), 57Fe (2.12%), 58Fe (0.28%). Calculate Ar(Fe).
Solution. Ar(Fe) = (54 × 0.0585) + (56 × 0.9175) + (57 × 0.0212) + (58 × 0.0028)
= 3.159 + 51.380 + 1.208 + 0.162 = 55.91 (matches data booklet 55.85).
Worked example

Abundance from A<sub>r</sub>

Problem. Indium has two stable isotopes 113In and 115In. From the periodic table Ar(In) = 114.82. Calculate the percentage abundance of each.
Solution. Let x = fraction of 113In, so (1−x) is the fraction of 115In.
113x + 115(1−x) = 114.82
113x + 115 − 115x = 114.82
−2x = −0.18 → x = 0.090
113In = 9.0%, 115In = 91.0%.

Try these

  1. How do the chemical and physical properties of isotopes differ? Give an example.
    Show answer
    Chemical properties are identical (same electrons → same bonding). Physical properties differ — e.g. D₂O (heavy water) boils at 101.4 °C vs 100 °C for H₂O; density is about 11% higher. Slight differences in m.p., b.p. and reaction rates (kinetic isotope effect).
  2. In nature, silicon exists as 28Si, 29Si and 30Si. The lightest has an abundance of 92.23%. Using Ar(Si) = 28.09, calculate the abundances of the other two.
    Show answer
    Let y = % of 29Si, z = % of 30Si. Then y + z = 7.77, and (28×0.9223) + (29×y/100) + (30×z/100) = 28.09. Solving: y ≈ 4.68%, z ≈ 3.09%.
  3. Are there any unstable isotopes? Name one and its use.
Lesson 6

The emission spectrum

Structure 1.3.1Structure 1.3.2

Lesson outcomes

White light contains all visible wavelengths → a continuous rainbow. Hot atomic hydrogen emits only certain wavelengths → discrete bright lines on a black background. This line emission spectrum is the direct experimental evidence that electrons in atoms have only certain allowed energies — energy is quantised.

The mechanism

Heat or electrical excitation promotes electrons to higher energy levels. As they fall back down, photons are emitted. Photon energy = (Ehigh − Elow) = h·ν. Each gap is fixed, so each line is fixed.

Three named series for hydrogen

What you'd see in the lab

656 nm n=3→2 486 nm n=4→2 434 nm n=5→2 410 nm n=6→2 convergence → 400 700 nm Hydrogen Balmer series · visible emission

Why those lines · the energy-level picture

ENERGY → n = 1 (ground state) n = 2 n = 3 n = 4 n = 5 n = ∞ · ionised 656 nm 486 nm 434 nm BALMER (visible) → n = 2 LYMAN (UV) → n = 1 Energy levels of the hydrogen atom

Useful equations

c = ν λ · speed of light = frequency × wavelength. (c = 3.00 × 10⁸ m s⁻¹.)
E = h ν · photon energy = Planck constant × frequency. (h = 6.626 × 10⁻³⁴ J s.)
Combined: E = hc / λ.

Convergence → ionisation

Energy levels get closer as n increases. The spectral lines converge to a limit; beyond that limit, the electron is no longer bound. The energy at this convergence limit is the ionisation energy for that series.

Worked example

Photon energy from wavelength

Problem. Calculate the energy (J) of a photon with wavelength 486 nm — the blue-green Balmer line.
Solution. E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / (486 × 10⁻⁹)
= 1.988 × 10⁻²⁵ / 4.86 × 10⁻⁷ = 4.09 × 10⁻¹⁹ J.
(Per mole: × NA = 246 kJ mol⁻¹.)
Worked example

Photon energy from frequency

Problem. Calculate the energy of a photon of frequency 3.00 × 10¹⁵ Hz in (a) J and (b) kJ mol⁻¹. Which region of the electromagnetic spectrum does it fall in?
Solution. (a) E = hν = 6.626 × 10⁻³⁴ × 3.00 × 10¹⁵ = 1.99 × 10⁻¹⁸ J per photon.
(b) Per mole: 1.99 × 10⁻¹⁸ × 6.022 × 10²³ = 1.20 × 10⁶ J mol⁻¹ = 1.20 × 10³ kJ mol⁻¹.
Region: λ = c/ν = 3.00 × 10⁸ / 3.00 × 10¹⁵ = 1.00 × 10⁻⁷ m = 100 nm → ultraviolet (UV).
Worked example

Wavelength from photon energy

Problem. A photon has energy 120.4 kJ mol⁻¹. Calculate its wavelength.
Solution. Per photon: E = 120 400 / 6.022 × 10²³ = 2.00 × 10⁻¹⁹ J.
λ = hc/E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / 2.00 × 10⁻¹⁹ = 1.988 × 10⁻²⁵ / 2.00 × 10⁻¹⁹ = 9.94 × 10⁻⁷ m ≈ 994 nm (near-infrared).

Try these

  1. Components of the electromagnetic spectrum, lowest to highest energy?
    Show answer
    Radio < Microwave < Infrared < Visible < Ultraviolet < X-ray < γ-ray.
  2. What is the difference between absorption and emission of light?
    Show answer
    Emission: an electron drops from a higher to a lower energy level, releasing a photon of energy equal to the gap. Absorption: an electron absorbs a photon of the right energy and jumps to a higher level.
  3. Explain the difference between a continuous spectrum and a discrete (line) spectrum, with an example of each.
    Show answer
    Continuous: all wavelengths present, blending together (e.g. the rainbow from a hot tungsten filament, or sunlight). Discrete (line): only certain specific wavelengths present, showing as bright lines (or dark, in absorption) on a black/continuous background (e.g. hydrogen emission).
  4. Write the equation linking c, ν and λ.
    Show answer
    c = νλ.
  5. Write the equation linking the energy of a photon to its frequency. Name the constant.
    Show answer
    E = hν. h is the Planck constant (6.626 × 10⁻³⁴ J s).
  6. Write the equation linking the energy of a photon to its wavelength.
    Show answer
    E = hc/λ.
  7. What is the difference between an emission spectrum and an absorption spectrum?
    Show answer
    Emission: bright lines on a black background — atoms emit photons as electrons fall to lower levels. Absorption: dark lines on a continuous (rainbow) background — atoms absorb photons of specific energies as electrons jump up. Same line positions for the same element.
  8. Sodium flames are orange; lithium flames are red. Outline the source of the colours and why they differ.
    Show answer
    When metal compounds are heated in a flame, the metal's electrons are excited to higher energy levels. As they fall back, they emit photons at characteristic frequencies. The energy gaps in Na and Li are different (different elements, different nuclear charges, different orbitals), so they emit at different wavelengths → different colours.
  9. Multiple choice: Emission and absorption spectra provide evidence for: (A) the existence of neutrons, (B) the existence of isotopes, (C) the existence of atomic energy levels, (D) the nuclear model of the atom.
    Show answer
    (C). Discrete spectral lines can only occur if electrons have specific, fixed energies — i.e. quantised energy levels.
  10. How would an emission line from an n=3 → n=2 transition differ from an n=3 → n=1 transition?
    Show answer
    n=3 → n=1 has a larger energy gap → higher-frequency / shorter-wavelength photon. It's in the UV (Lyman series). n=3 → n=2 is in the visible (Balmer).
  11. How does the hydrogen emission spectrum help us understand other one-electron systems like He⁺? How would the energy levels differ between He⁺ and H?
    Show answer
    He⁺ has the same single-electron architecture as H, so it shows a similar spectrum mathematically. But He⁺ has Z = 2 (twice the nuclear charge), so each energy level is 4× more negative (En ∝ −Z²/n²). He⁺ lines are at much higher frequencies than the corresponding H lines.
Lesson 7

Electron configurations

Structure 1.3.3Structure 1.3.4Structure 1.3.5

Lesson outcomes

The Bohr model only worked for one-electron atoms. The modern model treats electrons probabilistically — an atomic orbital is a region of space where an electron is likely (~90% probability) to be found. Shells subdivide into sublevels (s, p, d, f), each containing a fixed number of orbitals.

Orbital geometry

The three p-orbitals are degenerate (equal in energy). Similarly the five d's, the seven f's. Maximum electrons per shell = 2n² (so n=1→2, n=2→8, n=3→18, n=4→32).

Three rules for filling

  1. Aufbau principle: build from lowest energy upward. Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s …
  2. Hund's rule: in a set of degenerate orbitals, place one electron per orbital with parallel spin before pairing.
  3. Pauli exclusion: an orbital holds at most 2 electrons, with opposite spin.

Periodic-table blocks reveal the valence sublevel

Groups 1–2 fill s; groups 13–18 fill p; the d-block (groups 3–12, periods 4+) fills d; lanthanides/actinides fill f.

Condensed notation

Replace inner-core electrons with the previous noble gas symbol in brackets. Na = [Ne] 3s¹. Fe = [Ar] 4s² 3d⁶.

Two anomalies: Cr and Cu

Half-full and fully-full d-subshells are unusually stable, so the configurations are not what aufbau predicts:

Configurations of ions

To form a cation, remove electrons from the highest occupied energy level first. For transition metals this means remove 4s before 3d — counter-intuitive but a recurring exam catch.

Electrons in boxes

Each box is one orbital; each arrow is one electron. ↑ for spin-up, ↓ for spin-down. Pauli: max 2 per box, opposite spin. Hund: in a degenerate sublevel, fill each box singly first.

Nitrogen · Z=7 · 1s² 2s² 2p³ Hund: three 2p electrons unpaired with parallel spin 1s 2s 2p unpaired ×3
Oxygen · Z=8 · 1s² 2s² 2p⁴ One 2p orbital pairs; two remain unpaired 1s 2s 2p unpaired ×2

The aufbau filling order

Follow the arrows to get the filling order 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f Reading order: 1s · 2s · 2p · 3s · 3p · 4s · 3d · 4p · 5s · 4d · 5p · 6s · 4f · 5d · 6p · 7s · 5f · 6d 4s fills before 3d (slightly lower energy)

First ionisation energy across a period · evidence for sub-shells

If atoms had only shells (no sub-shells), first IE would rise smoothly across a period. It doesn't — there are two systematic dips in every period. Each dip exposes a sub-shell boundary.

First ionisation energy · Li (Z=3) → Na (Z=11) ATOMIC NUMBER (Z) 1st IE / kJ mol⁻¹ 0 500 1000 1500 2000 s-block (2s) p-block (2p) Li 520 Be 900 B 800 C 1086 N 1402 O 1314 F 1681 Ne 2081 Na 496 DIP 1 · s → p B's 2p¹ electron is at higher energy than Be's 2s² DIP 2 · half-fill O's 4th 2p electron must pair — repulsion makes it easier to lose NEW SHELL · n=3 Na's 3s¹ is much further out & shielded
Worked example

Electron configuration of oxygen

Problem. Write the full electron configuration for oxygen (Z = 8).
Solution. Oxygen is in period 2, so n = 1 and n = 2 are involved. First shell: 1s². Second shell: 2s holds 2 e⁻; 2p (since O is in the p-block, 2 positions from B) holds 4 e⁻. Configuration = 1s² 2s² 2p⁴.
Worked example

Electron configuration of Fe³⁺

Problem. Write the electron configuration of (a) Fe and (b) Fe³⁺.
Solution. (a) Fe (Z = 26) = [Ar] 4s² 3d⁶.
(b) For Fe³⁺ we remove 3 e⁻. Take them from the highest energy occupied level: first the 4s electrons (2 of them), then one 3d. Fe³⁺ = [Ar] 3d⁵ — half-filled d, which contributes to the stability of Fe³⁺.

Try these

  1. How many electrons can the first four shells each hold?
    Show answer
    Maximum electrons per shell = 2n². n=1 → 2; n=2 → 8; n=3 → 18; n=4 → 32.
  2. Name the four types of orbital and rank them from lowest to highest energy.
    Show answer
    s < p < d < f (within the same principal energy level). s is spherical; p is dumbbell-shaped (3 orientations: px, py, pz); d has 5 different shapes; f has 7.
  3. How many electrons can a single orbital hold? How many orbitals are in each sub-shell? How many electrons per sub-shell?
    Show answer
    Each orbital holds at most 2 electrons (Pauli). s sub-shell: 1 orbital, 2 e⁻. p: 3 orbitals, 6 e⁻. d: 5 orbitals, 10 e⁻. f: 7 orbitals, 14 e⁻.
  4. How many sub-shells can you have in a given shell?
    Show answer
    Equal to n. Shell n=1 has only 1s. Shell n=2 has 2s and 2p. Shell n=3 has 3s, 3p, 3d. Shell n=4 has 4s, 4p, 4d, 4f. (You cannot have more sub-shells than n.)
  5. Which orbitals are present in each electron shell, and comment on their relative energies.
    Show answer
    Shell 1: 1s only. Shell 2: 2s < 2p. Shell 3: 3s < 3p < 3d. Shell 4: 4s < 4p < 4d < 4f. Within a shell, s is lowest energy then p < d < f. The three p-orbitals are degenerate (same energy). 4s is slightly lower than 3d (note: 4s fills first).
  6. For the energy level n = 4, state: the sub-level types, the number of atomic orbitals in each, the total number of orbitals, and the maximum number of electrons.
    Show answer
    Sub-levels: 4s, 4p, 4d, 4f. Orbitals: 1 + 3 + 5 + 7 = 16 orbitals. Maximum electrons = 2 × 16 = 32 (= 2n²). ✓
  7. Where are the four blocks (s, p, d, f) of the periodic table, and what does the block tell us about the valence subshell?
    Show answer
    s-block: groups 1–2 on the left (valence electrons in ns). p-block: groups 13–18 on the right (valence in np). d-block: transition metals in the middle (valence in (n−1)d). f-block: lanthanides and actinides at the bottom (valence in (n−2)f). The block tells you which subshell the highest-energy valence electron occupies.
  8. State the three rules for filling orbitals.
    Show answer
    Aufbau: build from lowest energy upward. Hund's rule: within a degenerate sub-shell, fill each orbital singly before pairing (minimises repulsion). Pauli exclusion: within an orbital, two electrons must have opposite spin.
  9. Write the full electron configurations for sodium, nitrogen and argon.
    Show answer
    Na (Z=11): 1s² 2s² 2p⁶ 3s¹. N (Z=7): 1s² 2s² 2p³. Ar (Z=18): 1s² 2s² 2p⁶ 3s² 3p⁶.
  10. Draw orbital (electrons-in-boxes) diagrams for aluminium and chlorine.
    Show answer
    Aluminium (Z=13, 1s² 2s² 2p⁶ 3s² 3p¹): 1s ↑↓   2s ↑↓   2p ↑↓ ↑↓ ↑↓   3s ↑↓   3p ↑ _ _.
    Chlorine (Z=17, 1s² 2s² 2p⁶ 3s² 3p⁵): 1s ↑↓   2s ↑↓   2p ↑↓ ↑↓ ↑↓   3s ↑↓   3p ↑↓ ↑↓ ↑. (Pauli: paired electrons opposite spin. Hund: 3p fills singly then pairs.)
  11. Looking at the aufbau diagonal, what is the order of filling in the region containing 4s, 3d, 4p? Which fills first?
    Show answer
    4s before 3d before 4p. The 4s sub-shell is slightly lower in energy than 3d (despite higher n), so it fills first. Then 3d, then 4p. This is why period 4 contains the first row of transition metals (3d filling).
  12. The sodium configuration is 1s² 2s² 2p⁶ 3s¹. Write its condensed (noble-gas-core) configuration. Then give the condensed configuration for oxygen.
    Show answer
    Na = [Ne] 3s¹. Oxygen (Z=8, 1s² 2s² 2p⁴) = [He] 2s² 2p⁴.
  13. Why does chromium have configuration [Ar] 4s¹ 3d⁵ rather than [Ar] 4s² 3d⁴? Apply the same logic to copper.
    Show answer
    A half-filled d-subshell (d⁵, all 5 spins parallel by Hund) is unusually stable — promoting one 4s electron to 3d gives a more stable arrangement. Copper similarly: Cu = [Ar] 4s¹ 3d¹⁰ rather than 4s² 3d⁹ — a fully-filled d is also extra-stable.
  14. Write the full electron configurations for Na⁺, O²⁻, Cu⁺ and Ti³⁺. What does the d-electron behaviour tell you?
    Show answer
    Na⁺ = 1s² 2s² 2p⁶ ([Ne]) — lost one 3s. O²⁻ = 1s² 2s² 2p⁶ ([Ne]) — gained two 2p. Cu⁺ = [Ar] 3d¹⁰ — removed only 4s (full d retained). Ti³⁺ = [Ar] 3d¹ — 4s² and one 3d removed. Pattern: for cations, remove 4s electrons before 3d. The 4s is filled last but emptied first.
HL extension

Inside the
ionisation
energies.

HL pushes deeper: mass spectrometry data, successive ionisation energies, and the limit of convergence in emission spectra.

Lesson 5 HL only

Using mass spectra to determine A_r

Structure 1.2.3 (HL)

Lesson outcomes

  • Interpret a mass spectrum: identify m/z and relative intensity axes.
  • Calculate Ar from peak heights and m/z values on a mass spectrum.
  • Explain why diatomic molecules (e.g. Cl₂) give multiple molecular-ion peaks.

A mass spectrometer ionises a sample, accelerates the ions through a magnetic field, and separates them by mass-to-charge ratio (m/z). The output is a "stick" plot with one line per ion species — height proportional to relative abundance.

Reading a single-element spectrum

For an atomic sample (e.g. Cl atoms), each m/z value corresponds to one isotope. Peak heights give the abundances directly. Ar is then the weighted mean.

Why Cl2 shows three peaks

If chlorine is fed in as Cl₂, the molecular ions are 35Cl−35Cl (mass 70), 35Cl−37Cl (mass 72) and 37Cl−37Cl (mass 74). With ~3:1 ratio of 35Cl:37Cl, the probabilities give relative peak heights of 9 : 6 : 1.

Worked example

Identifying an unknown element

Problem. A mass spectrum shows m/z peaks at 204 (2%), 206 (24%), 207 (22%) and 208 (52%). Calculate Ar and identify the element.
Solution. Ar = (204×0.02) + (206×0.24) + (207×0.22) + (208×0.52)
= 4.08 + 49.44 + 45.54 + 108.16 = 207.22.
This matches lead (Pb), Ar = 207.2.

Try these

  1. Why are there three molecular-ion peaks in the mass spectrum of Cl₂, and what is the expected height ratio?
    Show answer
    Combinations 35Cl-35Cl (m/z 70), 35Cl-37Cl (m/z 72), 37Cl-37Cl (m/z 74). With abundances 3:1, the probabilities (3×3):(2×3×1):(1×1) = 9 : 6 : 1.
  2. Use a database to find the mass spectra of three different elements. From each, calculate Ar and compare to the data booklet value.
Lesson 8 HL only

The limit of convergence

Structure 1.3.6 (HL)

Lesson outcomes

  • Define ionisation energy and write a general ionisation equation.
  • Identify the convergence limit on an emission spectrum and link it to ionisation energy.
  • Calculate ionisation energy from a convergence frequency (E = hν, per mol via NA).
  • Calculate the wavelength of the convergence limit from an ionisation energy.
  • Identify and explain four factors affecting ionisation energy.

The first ionisation energy (IE₁) is the energy required to remove the most loosely held electron from one mole of gaseous atoms:

X(g) → X⁺(g) + e⁻    ΔH = IE₁

On a hydrogen emission spectrum (Lyman series), the spectral lines converge at high frequency. The line at the convergence limit corresponds to a transition from n = ∞ down to n = 1 — meaning the electron has come back from the just-ionised state. Therefore the energy of a photon at the convergence frequency, multiplied by Avogadro's number, equals the molar ionisation energy.

IE (J mol⁻¹) = h · νlimit · NA

Four factors affecting ionisation energy

  • Nuclear charge — more protons pull electrons in more tightly → higher IE.
  • Distance from nucleus — outer-shell electrons are further away → lower IE.
  • Shielding — inner electrons screen outer electrons from full nuclear charge → lower effective pull on outer e⁻ → lower IE.
  • Electron pairing — paired electrons in the same orbital repel each other slightly → easier to remove → lower IE than expected (the N→O dip).
Worked example

Wavelength of convergence for sodium

Problem. The first ionisation energy of Na is 496 kJ mol⁻¹. Calculate the wavelength of the convergence limit for sodium in Å (1 Å = 10⁻¹⁰ m).
Solution. Convert IE to J per atom: 496 000 / NA = 496 000 / 6.022 × 10²³ = 8.236 × 10⁻¹⁹ J.
E = hc / λ, so λ = hc / E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) / 8.236 × 10⁻¹⁹
= 1.988 × 10⁻²⁵ / 8.236 × 10⁻¹⁹ = 2.413 × 10⁻⁷ m
= 2413 × 10⁻¹⁰ m = ≈ 2413 Å (UV).
Worked example

Helium ionisation energy

Problem. Use the data booklet (IE₁(He) = 2372 kJ mol⁻¹). Calculate the frequency and wavelength of a photon at the convergence limit of helium.
Solution. Per atom: E = 2 372 000 / NA = 3.939 × 10⁻¹⁸ J.
ν = E / h = 3.939 × 10⁻¹⁸ / 6.626 × 10⁻³⁴ = 5.945 × 10¹⁵ Hz.
λ = c / ν = 3.00 × 10⁸ / 5.945 × 10¹⁵ = 5.05 × 10⁻⁸ m (≈ 50.5 nm, far UV).
Compare to hydrogen (IE₁ = 1312 kJ mol⁻¹): He's IE is much higher because the +2 nuclear charge attracts each electron much more strongly, while the second electron provides only weak shielding.

Try these

  1. Define ionisation energy and write a generic chemical equation for any element X.
    Show answer
    Energy required to remove one mole of electrons (the highest in energy) from one mole of gaseous atoms or ions, forming gaseous positive ions. X(g) → X⁺(g) + e⁻. Always endothermic.
  2. How is the ionisation energy of hydrogen represented on its emission spectrum?
    Show answer
    At the high-frequency end of the Lyman series, the spectral lines converge. The frequency at this convergence limit corresponds to the photon emitted when an electron falls from n = ∞ to n = 1 — equivalent in energy to fully removing the electron from the ground state. So the convergence frequency × h × NA = IE₁.
  3. How can you use data from an emission spectrum to determine the ionisation energy?
    Show answer
    Plot frequencies of lines in the Lyman series and extrapolate to find the convergence frequency (where line spacing approaches zero). Multiply by h to get energy per atom, then by NA for energy per mole.
  4. List four factors that affect the ionisation energy of an element.
    Show answer
    1. Nuclear charge — more protons pull electrons in more tightly → higher IE. 2. Distance from nucleus — outer-shell electrons are further away → lower IE (atoms get bigger down a group). 3. Shielding — inner electrons screen the outer electrons from the full nuclear charge → lower effective pull → lower IE. 4. Electron pairing — paired electrons in the same orbital repel each other → easier to remove (gives the N → O dip).
  5. What is the general trend in first IE across a period and down a group? Justify.
    Show answer
    Across a period: IE generally increases — nuclear charge grows, shielding stays similar, so outer electrons are held more tightly. Down a group: IE decreases — outer electrons are in higher shells, further from the nucleus and more shielded.
  6. What accounts for the two systematic dips in a first-IE plot across a period (e.g. across period 2)?
    Show answer
    Dip 1 (Be → B): removing a 2p¹ electron is easier than removing one of Be's 2s² electrons because 2p is higher in energy. Evidence for s/p sub-shell split. Dip 2 (N → O): nitrogen's 2p³ has three unpaired electrons (Hund-stable). Oxygen's 2p⁴ must have one pair, and that paired electron experiences e-e repulsion, making it easier to remove. Evidence for the pair-vs-unpair distinction in p orbitals.
  7. Use the data booklet to find IE₁(He) = 2372 kJ mol⁻¹. Compare it to IE₁(H) = 1312 kJ mol⁻¹. Why is He's so much larger?
    Show answer
    He has nuclear charge +2 (vs +1 for H), pulling each of its two 1s electrons twice as strongly. The second electron in He provides only weak shielding (it's in the same shell, not an inner shell). The net effective nuclear charge on each He electron is much greater than 1+ → significantly higher IE.
Lesson 9 HL only

Successive ionisation energies

Structure 1.3.7 (HL)Practical 1.2

Lesson outcomes

  • Define successive ionisation energy.
  • Explain why IE₂ > IE₁ > … always.
  • Use a log(IE) vs ionisation number graph to identify jumps and deduce a group/configuration.
  • Complete Practical 1.2: plot successive IE data and identify the element.

Successive ionisation energies are the energies needed to remove the 1st, 2nd, 3rd … electrons in turn:

X⁺(g) → X²⁺(g) + e⁻   (IE₂)   ;   X²⁺(g) → X³⁺(g) + e⁻   (IE₃)   …

Each successive removal is from an increasingly positive ion, with fewer electrons sharing the same nuclear pull. So IE₂ > IE₁ always — and the increases get steeper.

Why big jumps reveal shells

Removing the last electron of a given shell is much easier than removing the first electron of the next-inner shell, which is closer to the nucleus and less shielded. So a jump in successive IE plotted on a log scale signals a shell boundary. Counting electrons up to each jump reveals the electron configuration directly.

Sodium · Z=11 · successive ionisation energies (log scale) IONISATION NUMBER (electrons removed) log(IE / kJ mol⁻¹) 1 2 3 4 5 6 7 8 9 10 11 2.5 3.5 4.5 5.5 large jump (n=3 → n=2) huge jump (n=2 → n=1) eight 2s/2p electrons two 1s 3s¹
Worked example

Identifying an element from successive IEs

Problem. The first five successive ionisation energies of an unknown element are 801, 2427, 3660, 25 026 and 32 827 kJ mol⁻¹. Deduce the group of the periodic table to which the element belongs.
Solution. Successive: 801 → 2427 → 3660 → 25 026 → 32 827. The big jump is from IE₃ to IE₄ (3660 → 25 026, a factor of ~7).
This means the first 3 electrons come out from one shell (relatively easy), and the 4th must come from an inner shell (much harder).
Three electrons in the outer shell → Group 13 (e.g. boron). For boron, IE₁ ≈ 801 kJ mol⁻¹ ✓ — the unknown is most likely boron.

Try these

  1. Why does IE₂ > IE₁ always?
    Show answer
    Removing the 2nd electron means pulling from a positive ion (X⁺) with the same nuclear charge but one fewer electron — less e-e repulsion, stronger effective nuclear pull on each remaining electron.
  2. Sodium has IE values 496, 4562, 6912, 9544, 13 353, 16 613, 20 117, 25 496, 28 932, 141 362, 159 076 kJ mol⁻¹. Where are the jumps and what configuration does this confirm?
    Show answer
    Big jump from IE₁ to IE₂ (1 → 8 electrons removed from the same shell would be tiny; here 496 → 4562 is a 9× jump because we now cross from n=3 to n=2). Another huge jump from IE₉ to IE₁₀. Pattern 1, 8, 2 confirms Na: 1s² 2s² 2p⁶ 3s¹.
  3. Practical 1.2 — use a database to find successive IE data for an unknown element of your choice. Plot log(IE) vs ionisation number, identify the jumps, and predict the element's group.
Vocabulary

42 terms to own.

If you can't define one of these in a sentence, that's where to revise next. Click any term for its definition.

elementcompoundmixturehomogeneousheterogeneousfiltrationdistillationchromatographyrecrystallisationkinetic molecular theorysublimationdepositionatomprotonneutronelectronatomic numbermass numberisotoperelative atomic massemission spectrumexcited stateenergy levelconvergenceorbitalsublevelelectron configurationaufbau principlepauli exclusionionisation energysuccessive ionisation energieslimit of convergencemolesolutesolventsolutiongraphmass spectrometryinternal assessmenttemperaturesurface areaendothermic
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