IB Chemistry · Unit 7 · Structure

The bonds
that build
matter.

What holds atoms together is what determines what they do. Ionic, covalent, metallic — three answers to the same problem.

Lessons Key terms
10Lessons
0HL extensions
42Key terms
SL+HLLevel
A B SHARED ELECTRONS
Unit 7 · Standard & Higher Level

10 lessons to work through.

The required syllabus content for Unit 7, in order. Each card is one lesson-sized checkpoint.

Lesson 1

Ionic Bonding

The electrostatic attraction between two oppositely charged ions

Lesson 2

Covalent Bonding

Compare and contrast ionic, metallic, and covalent bonding.

Lesson 3

Metallic bonding

Metallic bonding can be defined as electrostatic attractions between a bed of metal cations and delocalized electrons.

Lesson 4+5

From models to materials

What role do bonding and structure have in the design of materials?

Lesson 6

Shapes of molecules (VSEPR Theory)

Lesson 6 of Unit 7.

Lesson 7

Molecular Shapes Simulation

Lesson 7 of Unit 7.

Lesson 8+9

Bond Polarity and Intermolecular forces

Lesson 8+9 of Unit 7.

Lesson 10

Practical 7.1 Isolation of caffeine from coffee

Lesson 10 of Unit 7.

Lesson 11

Thin Layer Chromatography

Chromatography is a technique to separate and identify components in a mixture

Lesson 12

Giant Covalent Structures

Lesson 12 of Unit 7.

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

Ionic bonding

Structure 2.1.1Structure 2.1.2

Lesson outcomes

An ionic bond forms when electrons are transferred from a metal to a non-metal, creating a cation and an anion held together by electrostatic attraction.

Predicting ionic formulas

Charges must balance: Na⁺ + Cl⁻ → NaCl. Mg²⁺ + 2 Cl⁻ → MgCl₂. Al³⁺ + 3 O²⁻... cross-multiply: Al₂O₃.

Ionic lattice and properties

Worked example

Predicting an ionic formula

Problem. Predict the formula of the ionic compound formed between magnesium and nitrogen.
Solution. Mg → Mg²⁺ (group 2). N → N³⁻ (group 15). Charge balance: 3 Mg²⁺ + 2 N³⁻ → Mg₃N₂.

Try these

  1. Define an ionic bond.
    Show answer
    Electrostatic attraction between oppositely-charged ions, formed when one or more electrons are transferred from a metal atom to a non-metal atom.
  2. Predict the formula of the ionic compound formed between magnesium and chlorine. Show the electron transfer.
    Show answer
    Mg → Mg²⁺ + 2 e⁻. Each Cl + e⁻ → Cl⁻. Need 2 Cl per Mg. Formula: MgCl₂.
  3. Explain why ionic compounds are brittle.
    Show answer
    Striking the crystal displaces a layer of ions, bringing like charges into contact (cation next to cation). Strong repulsion → the lattice cleaves cleanly along the displaced plane.
  4. Why do ionic compounds conduct electricity when molten or aqueous but not when solid?
    Show answer
    Conduction requires charge carriers that can move. In the solid lattice, ions are locked in place — no mobility. Melting (or dissolving) frees the ions to move, enabling current flow.
  5. Why does MgO have a much higher melting point (~2850 °C) than NaCl (~801 °C)?
    Show answer
    Coulomb's law: F ∝ (q₁q₂)/r². MgO has higher charges (2+ and 2−) than NaCl and smaller ionic radii. Both factors give MgO a much larger lattice enthalpy.
  6. Why are ionic compounds usually soluble in water but not in non-polar solvents like hexane?
    Show answer
    Water is polar — its dipoles can solvate (surround) ions, providing energy comparable to lattice enthalpy. Hexane is non-polar; weak London dispersion forces with ions cannot break the lattice. So ionic compounds dissolve readily in water but not in hexane.
Lesson 2

Covalent bonding

Structure 2.2.1Structure 2.2.2

Lesson outcomes

A covalent bond forms when two non-metal atoms share electrons to gain stable noble-gas-like configurations.

Single, double, triple bonds

Single bond: 1 shared pair (e.g. H–H). Double: 2 (O=O). Triple: 3 (N≡N). Higher bond order → shorter, stronger bonds. C–C 154 pm; C=C 134 pm; C≡C 120 pm.

Coordinate (dative) bonds

Both shared electrons come from the same atom. The donor has a lone pair; the acceptor has an empty orbital. Examples: NH₄⁺, H₃O⁺, CO.

Worked example

Lewis structure of CO₂

Problem. Draw the Lewis structure of CO₂.
Solution. Total valence e⁻: C (4) + 2 × O (6) = 16. C in centre, two O's outside. C–O–O? No, both O attached to C. Distribute 12 e⁻ as lone pairs on outer O's. C has only 4 e⁻ — need double bonds. Move one lone pair from each O into a π bond: O=C=O. Each atom has 8 e⁻. ✓

Try these

  1. Define a covalent bond. What is the octet rule?
    Show answer
    A pair of electrons shared between two atoms (usually non-metals). Octet rule: atoms tend to share/transfer electrons to achieve a full outer shell of 8 (or 2 for H/He) — a stable noble-gas-like arrangement.
  2. Compare the bond lengths and strengths of C–C, C=C, and C≡C.
    Show answer
    Higher bond order → shorter, stronger bond. C–C: 154 pm, ~346 kJ mol⁻¹. C=C: 134 pm, ~614 kJ mol⁻¹. C≡C: 120 pm, ~839 kJ mol⁻¹.
  3. Draw the Lewis structure of H₂O. How many lone pairs are on O?
    Show answer
    H–O–H with two lone pairs on O. Each H makes one bond (its 1s electron paired with one of O's 2p electrons). O has 4 + 2 + 2 = 8 e⁻ total: 4 in bonds, 4 in 2 lone pairs.
  4. Draw the Lewis structure of N₂. State the number and type of bonds.
    Show answer
    N≡N with one lone pair on each N. Triple bond: 1 σ + 2 π. Each N has 5 valence e⁻: 3 in the triple bond + 2 in the lone pair = 5 ✓.
  5. Draw the Lewis structure of NH₄⁺. Identify the coordinate bond.
    Show answer
    N central, 4 N–H bonds, no lone pairs on N. One of the four N–H bonds is dative — both electrons came from N (which originally had a lone pair, donated to H⁺).
  6. Why does carbon usually form 4 bonds?
    Show answer
    C has 4 valence electrons (2s² 2p²). Forming 4 covalent bonds gives it a share of 8 electrons (full octet) → most stable configuration. Hence C makes 4 σ bonds in alkanes, 3 σ + 1 π in alkenes, 2 σ + 2 π in alkynes.
Lesson 3

Metallic bonding

Structure 2.3.1

Lesson outcomes

Metallic bonding: a regular array of metal cations sits in a "sea" of delocalised electrons. Electrons came from the outer shells; now belong to the whole lattice.

Properties explained

Bond strength scales with charge: Na (+1) is soft and low-melting; Fe (+2/+3) is hard; W (+6) has m.p. 3422 °C.

Try these

  1. Define metallic bonding.
    Show answer
    Electrostatic attraction between a regular lattice of metal cations and a 'sea' of delocalised electrons that came from the outer shells of those atoms.
  2. Explain why metals are good thermal conductors.
    Show answer
    The delocalised 'sea' of electrons carries kinetic energy quickly through the lattice. A small region heated up transfers thermal energy via electron motion much faster than through phonon (lattice vibration) transport alone.
  3. Why are metals malleable and ductile (whereas ionic crystals are brittle)?
    Show answer
    When you strike a metal, layers of cations can slip past one another without breaking — the delocalised electron sea adjusts and continues to hold them. In an ionic crystal, the same displacement brings like charges into contact → repulsion → cleavage.
  4. Why are metals lustrous (shiny)?
    Show answer
    Delocalised electrons can absorb and immediately re-emit visible photons across a broad range. This gives high reflectivity at the metal surface — the characteristic 'metallic' shine.
  5. Predict the relative metallic-bonding strength of Na, Mg, Al.
    Show answer
    Bond strength: Al > Mg > Na. Each Al contributes 3 valence electrons (higher charge density); Mg 2; Na 1. M.p.: Na 98 °C; Mg 650 °C; Al 660 °C. (Al's atomic radius is also smaller.)
  6. Why is mercury liquid at room temperature, while most other metals are solid?
    Show answer
    Relativistic effects on the 6s shell contract it close to the nucleus, weakening Hg's metallic bonding. Weak bonds → can be overcome by thermal energy at room T (m.p. −39 °C).
Lesson 4-5

From models to materials · the bonding continuum

Structure 2.4.1

Lesson outcomes

Real bonds aren't always purely one type. The van Arkel-Ketelaar triangle (see diagram below in section 7.2) plots compounds by average electronegativity vs ΔEN, mapping the three bonding extremes (ionic, covalent, metallic) as vertices. Most real compounds lie inside.

ΔEN thresholds

Polar bonds vs polar molecules

A molecule can have polar bonds but be non-polar overall — if the dipoles cancel by symmetry. CO₂ (linear, two C=O dipoles cancel) vs H₂O (bent, dipoles add).

Worked example

Classify bonds by ΔEN

Problem. Classify each bond: (a) C–H (ΔEN 0.4), (b) H–Cl (ΔEN 0.9), (c) Na–Cl (ΔEN 2.1), (d) Cs–F (ΔEN 3.3).
Solution. (a) Non-polar covalent. (b) Polar covalent. (c) Predominantly ionic. (d) Very strongly ionic (top of van Arkel triangle).

Try these

  1. Why is the van Arkel-Ketelaar triangle useful?
    Show answer
    It maps real compounds on a continuum between three idealised bonding extremes (ionic, covalent, metallic). Most bonds aren't 'pure' — the triangle quantifies the blend using χ̄ (average electronegativity) and Δχ (difference).
  2. Place each on the triangle: Cs, F₂, CsF, HCl, H₂O.
    Show answer
    Cs: low χ̄, Δχ = 0 → metallic vertex. F₂: high χ̄, Δχ = 0 → covalent vertex. CsF: middle χ̄, very high Δχ → ionic vertex. HCl: high χ̄, Δχ ≈ 0.9 → polar covalent (right side, near the bottom). H₂O: high χ̄, Δχ ≈ 1.24 → polar covalent (right side, slightly above HCl).
  3. What is the difference between a polar bond and a polar molecule?
    Show answer
    Polar bond: a bond between atoms with ΔEN > 0.4, creating partial charges. Polar molecule: a molecule with a net dipole moment overall. A molecule can have polar bonds but be non-polar if symmetry causes the dipoles to cancel (e.g. CO₂).
  4. Predict polarity: (a) CO₂, (b) NH₃, (c) CCl₄, (d) CH₃Cl.
    Show answer
    (a) Non-polar (linear, dipoles cancel). (b) Polar (trigonal pyramidal — lone pair makes it asymmetric). (c) Non-polar (tetrahedral symmetric). (d) Polar (asymmetric tetrahedral).
  5. Predict whether SF₄ is polar (assuming see-saw shape).
    Show answer
    Polar. SF₄ has 5 electron domains (4 bonding + 1 lone pair on S), giving a see-saw shape. The lone pair makes the molecule asymmetric, and the S–F dipoles don't cancel.
Lesson 6

VSEPR · shapes of molecules

Structure 2.2.3

Lesson outcomes

VSEPR (Valence Shell Electron Pair Repulsion) theory: electron pairs around a central atom arrange themselves to minimise repulsion.

Lone pairs repel more strongly than bonding pairs (their cloud is held by one nucleus rather than two), squeezing bonding angles smaller than the idealised geometry would predict — that's why H₂O's 104.5° is less than 109.5°.

See the interactive 3D molecule viewers below — drag and zoom each shape.

Worked example

Predict the shape of NH₃

Problem. Predict the shape and bond angle of NH₃ using VSEPR.
Solution. N has 5 valence e⁻. Three in N–H bonds (3 bonding pairs); one is a lone pair. Total = 4 domains. Tetrahedral electron geometry; trigonal pyramidal molecular shape. Lone pair squeezes H–N–H from 109.5° to ~107°.

Try these

  1. Predict the shape, bond angle and polarity of: (a) BF₃, (b) NH₃, (c) H₂O, (d) CH₄, (e) PCl₅, (f) SF₆.
    Show answer
    (a) Trigonal planar, 120°, non-polar. (b) Trigonal pyramidal, ~107°, polar. (c) Bent, 104.5°, polar. (d) Tetrahedral, 109.5°, non-polar. (e) Trigonal bipyramidal, 90°/120°, non-polar (Cl symmetric around P). (f) Octahedral, 90°, non-polar.
  2. Why is the bond angle in H₂O (104.5°) smaller than in NH₃ (~107°)?
    Show answer
    Lone pair – lone pair repulsion is the strongest, followed by lone pair – bond pair, then bond pair – bond pair. H₂O has 2 lone pairs squeezing the bond angle. NH₃ has only 1 lone pair, so less squeezing. (Idealised tetrahedral is 109.5°.)
  3. Predict the shape of SF₆ and the F–S–F bond angle.
    Show answer
    6 electron domains, no lone pairs. Octahedral, 90°.
  4. Why is H₂O bent but BeCl₂ linear?
    Show answer
    H₂O has 4 domains (2 bonds + 2 lone pairs) → tetrahedral electron geometry → bent shape. BeCl₂ has only 2 domains and no lone pairs → linear.
  5. Why is the central atom in BF₃ deviating from the octet rule? Is this a problem?
    Show answer
    B has only 6 electrons in BF₃ (3 bonds × 2 e⁻). It's an 'electron-deficient' species and a Lewis acid (can accept a lone pair). Not a 'problem' — boron is small and stable with fewer than 8 electrons; it just makes BF₃ very reactive.
Lesson 7

Molecular shapes simulation · 3D practice

Structure 2.2.3

Lesson outcomes

Use the interactive 3D models in the diagrams section below to verify VSEPR predictions. Drag to rotate; zoom in and out to confirm bond angles.

Tasks: identify the shape, count lone pairs, predict polarity, and confirm against the viewer.

Try these

  1. Open the SO₂ 3D viewer below. Identify its shape, the position of the lone pair, and predict whether SO₂ is polar.
    Show answer
    Bent shape (3 domains, 1 lone pair). Lone pair on top of central S. Polar — the bent shape doesn't allow the two S=O dipoles to cancel.
  2. Compare CO₂ and SO₂ in shape and polarity. Why are they different?
    Show answer
    CO₂: linear (2 domains, no lone pair). Polar bonds cancel by symmetry → non-polar molecule. SO₂: bent (3 domains, 1 lone pair). Bent geometry can't cancel the dipoles → polar molecule. Same general formula AB₂, different geometry, different polarity.
  3. Why is CH₃Cl polar but CCl₄ is non-polar?
    Show answer
    CH₃Cl: tetrahedral but asymmetric (3 C–H and 1 C–Cl). The C–Cl dipole isn't cancelled by the three C–H dipoles → net dipole moment, polar. CCl₄: tetrahedral and symmetric (4 identical C–Cl bonds). All four C–Cl dipoles cancel by symmetry → non-polar.
Lesson 8-9

Bond polarity and intermolecular forces

Structure 2.2.4Structure 2.2.5

Lesson outcomes

The three intermolecular forces (weakest → strongest)

See the diagrams below for visual representations of each force, including the O–H...O linear geometry of the hydrogen bond.

Worked example

Predicting b.p. order from IMFs

Problem. Predict the order of boiling points for CH₄, NH₃, H₂O, HF.
Solution. CH₄: only London — lowest. NH₃: H-bonding. HF: H-bonding (stronger per bond but only 1 H to donate). H₂O: H-bonding (multiple H-bonds per molecule). Experimental: CH₄ (−162 °C) < NH₃ (−33) < HF (+20) < H₂O (+100).

Try these

  1. Why do all molecules experience London dispersion forces?
    Show answer
    Electrons in a molecule are constantly moving, creating instantaneous, temporary dipoles at any moment. These induce dipoles in neighbouring molecules. Even noble gases (single atoms) show this — it's a universal phenomenon.
  2. Why does I₂ (s) have higher b.p. than F₂ (g)?
    Show answer
    Both are non-polar diatomics with only London forces between molecules. I₂ has many more electrons (and a much larger, more polarisable electron cloud) than F₂ → stronger London forces → higher b.p. (I₂ solid at room T; F₂ gas).
  3. What conditions are needed for hydrogen bonding to occur?
    Show answer
    (1) Hydrogen must be bonded directly to N, O or F (highly electronegative, small atoms). (2) The H carries δ⁺. (3) That H interacts with a lone pair on another N, O or F atom on a neighbouring molecule.
  4. Why is water much higher boiling than methane, despite both being small molecules?
    Show answer
    Water has hydrogen bonding (strong dipole-dipole between O–H and lone pairs on O). Methane has only weak London dispersion.
  5. Compare boiling points of ethanol (C₂H₆O) and dimethyl ether (also C₂H₆O). Explain.
    Show answer
    Ethanol has –OH → can hydrogen-bond. Dimethyl ether has no O–H → only dipole-dipole. Ethanol b.p. 78 °C; DME b.p. −24 °C. Ethanol is far higher.
  6. Why is H₂O denser as a liquid than as a solid (ice floats)?
    Show answer
    Hydrogen bonding in ice forces molecules into an open hexagonal lattice with lots of empty space — lower density than liquid water (where H-bonding is more dynamic and packs more closely). Most other liquids contract on freezing; water expands.
Lesson 10

Practical 7.1 · isolation of caffeine from coffee

Structure 2.2Tool 1

Lesson outcomes

Caffeine is a non-polar molecule, weakly soluble in water but readily soluble in non-polar organic solvents like dichloromethane (DCM).

Procedure (overview)

  1. Brew strong coffee.
  2. Add Na₂CO₃ to deprotonate any acidic compounds (keeping them in the aqueous layer).
  3. Add DCM. Caffeine partitions into DCM.
  4. Drain the lower (DCM) layer; dry over anhydrous Na₂SO₄.
  5. Evaporate the DCM → solid caffeine.

Confirm purity by m.p. — pure caffeine: 235–237 °C.

Try these

  1. Why is DCM chosen rather than ethanol or hexane?
    Show answer
    DCM is immiscible with water (so a separating funnel works), has low b.p. (40 °C — easy to evaporate later), and dissolves caffeine well. Ethanol is miscible with water (no separation). Hexane is too non-polar.
Lesson 11

Thin-layer chromatography (TLC)

Tool 1

Lesson outcomes

TLC uses a thin layer of silica gel or alumina bonded to a glass/plastic plate (stationary phase). Drops of sample are spotted near the bottom; the plate is placed in a covered tank with mobile-phase solvent.

Capillary action carries the solvent up. Components migrate at different rates based on their affinity for the two phases.

Rf value

Rf = distance moved by spot / distance moved by solvent front

Rf is between 0 and 1. Characteristic for a given compound in a given solvent system.

Worked example

Calculate R_f

Problem. A spot moves 4.2 cm on a TLC plate while the solvent front moves 6.0 cm. Calculate Rf.
Solution. Rf = 4.2 / 6.0 = 0.70.

Try these

  1. Why must the solvent level in the tank be below the spotted line?
    Show answer
    Otherwise the solvent dissolves the spots directly into the reservoir, rather than letting them migrate up the plate.
Lesson 12

Giant covalent structures

Structure 2.2.6

Lesson outcomes

A giant covalent structure (network solid) has covalent bonds extending throughout the whole crystal — billions of atoms, one giant molecule.

Typical properties

Four IB examples

Interactive 3D models of diamond and graphite are in the diagrams section below.

Try these

  1. Define a giant covalent (network) solid. Give two examples.
    Show answer
    A solid in which covalent bonds extend throughout the whole crystal — no discrete molecules, just one continuous network of atoms held together by strong bonds. Examples: diamond, graphite, graphene, silicon, SiO₂ (quartz).
  2. Compare the m.p. and hardness of diamond vs graphite.
    Show answer
    Diamond: very high m.p. (~3550 °C), extremely hard (one of the hardest known materials). Graphite: very high m.p. (~3650 °C), but soft and slippery in the direction perpendicular to its layers (lubricant). The hardness difference comes from the structure, not the bond strength.
  3. Why is diamond an excellent insulator while graphite conducts electricity?
    Show answer
    Diamond: every C uses all 4 valence electrons in σ bonds. No free electrons. Graphite: each C uses 3 in σ bonds; the 4th sits in a delocalised π orbital across the layer — these electrons can move along the plane.
  4. Why is graphite useful as a lubricant but diamond is not?
    Show answer
    Graphite has weak London forces between its layers — layers slip easily. Diamond is one rigid 3D network; no weak directions.
  5. What makes graphene so industrially interesting?
    Show answer
    A single layer of graphite. Strongest known material per weight; excellent electrical and thermal conductor; nearly transparent; only one atom thick. Potential applications: flexible electronics, ultra-strong composites, batteries, sensors.
  6. Why is SiO₂ a giant covalent network rather than a small molecule like CO₂?
    Show answer
    Si has larger atomic radius than C and forms weaker π bonds. So instead of Si=O double bonds (which would give a hypothetical molecule O=Si=O analogous to CO₂), each Si bonds to 4 O atoms by single bonds (sp³), and each O bridges 2 Si atoms — a 3D network. The smaller, π-bond-strong C can form O=C=O cleanly.
7.1 · Three types of bond

How atoms stay together.

Atoms bond to reach a more stable arrangement (typically a full outer shell — the octet rule). Three strategies: transfer, share, or pool electrons.

Ionic

Electrons transferred

Metal donates 1+ electrons to a non-metal. Resulting cations and anions are held in a regular lattice by electrostatic attraction. Strong but brittle; high melting points; conduct only when molten or aqueous.

Covalent

Electrons shared

Two non-metals share one or more pairs of electrons. Discrete molecules with strong intramolecular bonds but variable intermolecular forces.

Metallic

Electrons pooled

Metal atoms release their outer electrons into a delocalised sea. The lattice of cations sits in this sea. Conductive, malleable, ductile, lustrous.

7.2 · Where the bond type comes from

Electronegativity difference.

There's no sharp line — bond type is a continuum. The electronegativity difference (ΔEN) between two atoms tells you where on it the bond lies.

⚖️

ΔEN < 0.4

Essentially non-polar covalent. C–C, C–H. Electrons shared roughly equally.

📊

0.4 < ΔEN < 1.7

Polar covalent. The more electronegative atom gets δ⁻. C–O, O–H, H–Cl.

⚡

ΔEN > 1.7

Predominantly ionic. NaCl (ΔEN ≈ 2.1). Electrons effectively transferred.

🌫️

Watch the symmetry

A molecule can contain polar bonds but be non-polar overall — if the dipoles cancel. CO₂ (linear, polar bonds cancel) vs H₂O (bent, dipoles add).

The van Arkel-Ketelaar triangle · placing a bond on the map

There are three pure bond types — ionic, covalent, metallic — but most real bonds lie somewhere between. Plot a compound by the average of the two atoms' electronegativities (x-axis) and the difference between them (y-axis), and the position reveals the bond character.

ΔEN ≈ 1.7 AVERAGE ELECTRONEGATIVITY (χ̄) DIFFERENCE (Δχ) 0.8 1.6 2.4 3.2 4.0 0 1.0 2.0 3.0 METALLIC COVALENT IONIC Cs Na Mg Fe Cu F₂ O₂ Cl₂ N₂ CH₄ HCl H₂O NH₃ CO₂ SiO₂ AlCl₃ NaCl KCl MgO CsF KF LiCl KEY metals · delocalized e⁻ covalent · shared e⁻ ionic · transferred e⁻ non-polar diatomic
Read it 1

x-axis · χ̄

Average electronegativity. Low (left): metals, electrons easily lost / pooled. High (right): non-metals, electrons held tightly / shared.

Read it 2

y-axis · Δχ

Difference between the two atoms. Small Δχ = symmetric sharing (or pooling). Large Δχ = electrons effectively transferred from low-χ to high-χ → ionic.

Read it 3

The zones blur

The triangle has no sharp lines — there is no "ionic / not ionic" cutoff in nature. NaCl is mostly ionic but a bit covalent; HCl is mostly covalent but quite polar. Position on the triangle quantifies the blend.

7.3 · Drawing Lewis structures

Dot-and-cross, step by step.

A Lewis structure shows every valence electron. Bonded pairs are drawn as lines (or pairs of dots between atoms); lone pairs are drawn on individual atoms.

📝

The procedure

1) Count total valence electrons (add/subtract for charge).
2) Place the least electronegative atom in the centre (usually) — H is never central.
3) Draw single bonds between central and outer atoms.
4) Distribute the remaining electrons as lone pairs around outer atoms first (to give octets), then to the central atom.
5) If the central atom doesn't have an octet, form double or triple bonds by moving lone pairs onto bonds.

7.4 · VSEPR and molecular shape

Electron pairs repel.

Electron domains (bonding pairs, lone pairs, and multiple bonds counted as one domain each) arrange themselves around a central atom to maximise the angle between them.

Domains Lone pairs Shape Angle Example
20Linear180°CO₂, BeCl₂
30Trigonal planar120°BF₃, CH₂O
31Bent~118°SO₂, O₃
40Tetrahedral109.5°CH₄, NH₄⁺
41Trigonal pyramidal~107°NH₃, H₃O⁺
42Bent~104.5°H₂O
50Trigonal bipyramidal90° / 120°PCl₅, PF₅
60Octahedral90°SF₆, [Fe(H₂O)₆]³⁺

Lone pairs repel slightly more strongly than bonding pairs (their electron cloud is held by only one nucleus, not two). This squeezes bonding-pair angles smaller than the "ideal" geometry would predict — that's why H₂O's 104.5° is less than the tetrahedral 109.5°.

The shapes · interactive 3D models

Drag to rotate · scroll to zoom · the geometry is the chemistry

Linear · CO₂

180° · 2 domains · 0 lone pairs

Trigonal planar · BF₃

120° · 3 domains · 0 lone pairs

Tetrahedral · CH₄

109.5° · 4 domains · 0 lone pairs

Trigonal pyramidal · NH₃

~107° · 4 domains · 1 lone pair (pink)

Bent · H₂O

104.5° · 4 domains · 2 lone pairs (pink)

Bent · SO₂

~118° · 3 domains · 1 lone pair (pink)

Trigonal bipyramidal · PCl₅

90° / 120° · 5 domains · 0 lone pairs

Octahedral · SF₆

90° · 6 domains · 0 lone pairs

📐

Reading the perspective

Solid wedge (filled triangle) = bond pointing out of the page toward you. Dashed line = bond pointing into the page away from you. Solid line = bond lying in the plane of the page. This convention lets a 2-D drawing show real 3-D geometry. In trigonal bipyramidal PCl₅, the two axial atoms (top, bottom) are at 90° to the three equatorial atoms (which are at 120° to each other). In octahedral SF₆, every adjacent bond is at 90°.

7.5 · Intermolecular forces

The forces between molecules.

Three flavours, strongest to weakest. The dominant one determines boiling point, solubility, and most physical properties.

Type 1

London dispersion

Universal — happens between any two molecules. Strength rises with molecular size and surface area. Why I₂ is solid at room T but F₂ is gas.

Type 2

Dipole-dipole

Between polar molecules. δ⁺ end of one attracts δ⁻ end of next. Stronger than London for molecules of similar size.

Type 3

Hydrogen bond

A strong dipole-dipole specifically between H bonded to F, O or N and a lone pair on F, O or N. Up to ~10% the strength of a covalent bond. Responsible for water's anomalously high boiling point.

💧

Why water is weird

Compare boiling points of group 16 hydrides: H₂S (−60 °C), H₂Se (−42 °C), H₂Te (−2 °C). Extrapolating, H₂O should boil around −90 °C. It boils at +100 °C. That extra 190 K is hydrogen bonding.

The three intermolecular forces · drawn

London (dispersion) forces · instantaneous induced dipole t = 0 · symmetric Ar Ar no net charge · no attraction t = δt · instant dipole forms Ar δ⁻ δ⁺ random e⁻ fluctuation Ar δ⁻ δ⁺ dipole is induced attract Random electron fluctuation in one atom creates a temporary dipole, which induces an opposite dipole in its neighbour. Weak, but universal.
Dipole-dipole forces · permanent partial charges align H Cl δ⁺ δ⁻ HCl molecule 1 dipole-dipole attraction H Cl δ⁺ δ⁻ HCl molecule 2 δ⁻ end of one attracts δ⁺ end of next
Hydrogen bonding · the strongest IMF (H bonded to N, O, or F) : : H H O δ⁻ δ⁺ δ⁺ hydrogen bond O—H ⋯ O collinear (≈180°) : : H H O δ⁻ δ⁺ δ⁺ Three rules H bonded to N, O, or F + that H carries δ⁺ + that H interacts with a lone pair on another N, O, or F atom. Geometry O—H ⋯ O is ideally linear. Strength drops fast as the angle bends away from 180°. ~10% of a covalent bond's strength. : = lone pair
Weakest

London dispersion

~1–10 kJ mol⁻¹. Universal — between any molecules. Strength rises with size and surface area. Why I₂ is a solid at room T but F₂ is a gas.

Stronger

Dipole-dipole

~5–25 kJ mol⁻¹. Between polar molecules. The δ⁺ of one molecule attracts the δ⁻ of another. Adds to the London dispersion that's always there too.

Strongest IMF

Hydrogen bonding

~10–40 kJ mol⁻¹. A special, very strong dipole-dipole. Requires H on N/O/F (highly polar bond) interacting with a lone pair on N/O/F. Responsible for water's high b.p., DNA pairing, protein folding.

7.6 · Giant covalent structures

One huge molecule.

Some non-metals form covalent bonds that extend throughout the whole crystal — billions of atoms, one network.

Diamond

3-D tetrahedral network

Each carbon bonded covalently to four neighbours in a tetrahedral arrangement. The whole crystal is one giant molecule. Extremely hard (each bond must break to deform). Doesn't conduct (all electrons localised).

Graphite

2-D layers

Each carbon bonded to three neighbours in flat hexagonal sheets. The fourth electron is delocalised between layers. Layers slip past each other (lubricant; pencil "lead") and conduct along the plane.

Silicon dioxide

Tetrahedral network

Each Si bonded to 4 O; each O bridges 2 Si. Quartz, sand, glass. Hard, high-melting (~1700 °C), insulating.

Graphene

One layer of graphite

A single hexagonal sheet of carbon. Strongest known material per weight. Electrically conductive. Industrially important.

Diamond vs graphite · interactive 3D

Drag to rotate · scroll to zoom · the structure is the property

Diamond

sp³ · tetrahedral · 109.5° · 1.54 Å C–C · 3-D network

Graphite

sp² · trigonal planar · 120° · 1.42 Å in-plane · 3.35 Å between layers

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Why the same element gives such different properties

Diamond — every C uses all 4 valence electrons in σ bonds to 4 neighbours. No free electrons. Every direction in the crystal is locked by strong σ bonds → hard, insulating, very high melting (~3550 °C).
Graphite — every C uses 3 valence electrons in σ bonds to 3 in-plane neighbours; the 4th sits in a π orbital delocalised across the layer. Strong bonds within a layer (hard to break) but only weak London forces between layers (easy to slip → lubricant; layers flake off onto paper → pencil "lead"). Delocalised π electrons conduct along the plane.

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.

elementcompoundmixturechromatographyatomprotonelectronenergy levelorbitalelectron configurationmolemolar massempirical formulasolutesolventsolutiongraphtemperatureendothermicenthalpy changebond enthalpylattice enthalpyperiodic tableperiodgrouptransition metalsmetallic characterelectronegativityoxidecomplex ionligandionic bondcovalent bondsingle bonddouble bondtriple bondcoordinate bondmetallic bonddelocalised electronselectronegativity differencepolar moleculedipole