IB DP Chemistry Structure 2.4: From models to materials

This study guide teaches Structure 2.4, From models to materials, following the IB Chemistry guide (Diploma Programme; published February 2023; first assessment 2025). It works through understandings Structure 2.4.1 to 2.4.6. The first five are for SL and HL; Structure 2.4.6 is additional higher level, so it is HL only.

The unit asks how bonding and structure decide what a material is good for. When you finish, check your recall with the matching revision notes, then attempt the unit’s practice questions.

What this unit covers

Understanding	What you must be able to do	Level
Structure 2.4.1	Treat bonding as a continuum between ionic, covalent and metallic; use bonding models to explain a material’s properties	SL and HL
Structure 2.4.2	Use electronegativity values to locate a binary compound on the triangle; predict its properties from where it sits	SL and HL
Structure 2.4.3	Explain the properties of alloys using non-directional bonding	SL and HL
Structure 2.4.4	Link typical plastic properties to chain structure; discuss natural and synthetic polymers	SL and HL
Structure 2.4.5	Draw an addition polymer’s repeating unit from a supplied monomer, and deduce a monomer from a polymer	SL and HL
Structure 2.4.6	Draw polyamide and polyester repeating units from supplied monomers; link biological macromolecules to condensation and hydrolysis	HL only

Bonding as a continuum (Structure 2.4.1)

The three bonding models are idealised end points that few real substances match perfectly.

Model	Particles	What holds them	Typical properties
Ionic	Cations and anions in a lattice	Electrostatic attraction between oppositely charged ions	High melting point, brittle, conducts only when molten or in solution
Covalent	Atoms sharing electron pairs	Attraction of two nuclei to a shared pair	Molecular: low melting point, non-conducting. Network: very high melting point, hard
Metallic	Cations in a sea of delocalized electrons	Attraction between cations and delocalized electrons	Conducts as a solid, malleable, ductile, lustrous

A bond between atoms of slightly different electronegativity is polar covalent: covalent with some ionic character. A lattice where a small, highly charged cation distorts the anion’s electron cloud is ionic with some covalent character. So ask “how far towards each end?” rather than “ionic or covalent?”. This is a limitation of neat categories.

Worked example 1: explaining a material’s use

A furnace uses magnesium oxide bricks, copper cable and silicon carbide cutting discs. Explain each choice.

1. Magnesium oxide is ionic. Small, doubly charged Mg²⁺ and O²⁻ ions attract very strongly, so much energy is needed to separate them and the bricks stay solid.
2. Copper is metallic. Its delocalized electrons move under a potential difference, so it conducts; non-directional bonding lets layers slide, so it can be drawn into wire.
3. Silicon carbide is a covalent network. Each atom has four strong covalent bonds in three dimensions, and many must break to scratch it, so it is extremely hard.

The bonding triangle (Structure 2.4.2)

A bonding triangle (also called a van Arkel–Ketelaar triangle) turns the continuum into a graph. For a binary compound of elements A and B:

• the horizontal axis is the average electronegativity, (χA + χB) / 2
• the vertical axis is the electronegativity difference, Δχ = |χA − χB|

According to the guide, the data booklet contains both a bonding triangle diagram and electronegativity values.

The corners are the pure models. With Pauling values, caesium (0.79, Δχ = 0) is the metallic corner, fluorine (3.98, Δχ = 0) the covalent corner, and caesium fluoride (Δχ = 3.19, average 2.385) the ionic corner.

• Higher up (large Δχ): more ionic character.
• Low down and to the left (small Δχ, low average): metallic.
• Low down and to the right (small Δχ, high average): covalent; both elements hold electrons tightly.
• Between regions: intermediate character, such as polar covalent.

The boundaries are gradual, not sharp lines. You will not have to calculate a percentage ionic character.

Worked example 2: placing compounds

Use these Pauling electronegativity values: K 0.82, Cl 3.16, Al 1.61, Si 1.90, C 2.55, Ni 1.91. Place KCl, AlCl₃, SiC and NiAl in the triangle.

Compound	Δχ	Average χ	Position	Bonding
KCl	3.16 − 0.82 = 2.34	(0.82 + 3.16) / 2 = 1.99	High up, near the middle	Mainly ionic
AlCl₃	3.16 − 1.61 = 1.55	(1.61 + 3.16) / 2 = 2.385	Middle height, right of centre	Polar covalent
SiC	2.55 − 1.90 = 0.65	(1.90 + 2.55) / 2 = 2.225	Low, right of centre	Covalent
NiAl	1.91 − 1.61 = 0.30	(1.61 + 1.91) / 2 = 1.76	Very low, left of centre	Metallic

Only the two elements matter: AlCl₃ has three chlorines, but you use one value for each element.

Worked example 3: predicting properties

Predict and compare the properties of KCl and AlCl₃ from their positions.

• KCl lies well into the ionic region: predict a high melting point, a brittle solid, and conduction only when molten or dissolved.
• AlCl₃ lies lower, between ionic and covalent: predict a much lower melting point and little conduction. Aluminium chloride in fact sublimes at about 450 K, and its vapour is made of Al₂Cl₆ molecules.
• NiAl should conduct as a solid; SiC should be a hard, high-melting non-conductor.

Structure 3.1 uses the same reasoning for the bonding trend in the period 3 oxides.

Alloys (Structure 2.4.3)

An alloy is a metal mixed with one or more further elements, metal or non-metal. It is a mixture, not a compound, because its composition can vary. Alloys usually have enhanced properties.

A metal can take in foreign atoms because metallic bonding is non-directional: the delocalized electrons attract cations equally in all directions, so a different-sized atom can sit in the lattice without breaking a particular bond.

• Harder and stronger. Different-sized atoms distort the regular layers of cations, so the layers cannot slide as easily.
• Still conducts. The delocalized electrons remain mobile.
• Other changes. Some alloys resist corrosion better; some melt lower.

Alloy	Main components	Improved property
Bronze	Copper with tin	Harder than copper; resists corrosion
Brass	Copper with zinc	Harder than copper; easy to machine; attractive colour
Stainless steel	Iron with chromium (often nickel), a little carbon	Resists rusting; strong

You do not need to learn specific alloys, only the explanation.

Worked example 4

Explain why brass is harder than copper, yet both conduct. Copper’s identical cations form regular layers that slide easily. In brass, zinc atoms of a different size fit into the lattice, which non-directional bonding allows. They disrupt the layers, so more force is needed to make them slide. Both keep mobile delocalized electrons, so both conduct.

Polymers and plastics (Structure 2.4.4)

A polymer is a macromolecule made from many small repeating subunits called monomers. The section of chain that repeats is the repeating unit.

Natural polymers	Synthetic polymers
Proteins (from 2-amino acids)	Poly(ethene)
Starch and cellulose (from glucose)	Poly(chloroethene), PVC
DNA (from nucleotides)	Polyamides such as nylons
Natural rubber	Polyesters

Plastics are synthetic polymers that can be moulded. Their common properties follow from their structure:

Property	Structural reason
Low density	Light atoms (mostly C and H), loosely packed chains
Electrical insulators	Electrons held in localized bonds; no mobile charges
Can be softened and moulded	Warming overcomes the intermolecular forces between chains; chain bonds stay intact
Strong and flexible	Long chains tangle, with many intermolecular forces along each one
Unreactive, slow to biodegrade	Strong, non-polar C–C and C–H bonds

Longer, less branched chains pack closer, so the plastic is stronger and denser. Covalent cross-links between chains stop a plastic softening on heating. Low reactivity also makes plastic waste persist; Structure 3.2 asks what makes some plastics biodegradable.

Addition polymers (Structure 2.4.5)

In addition polymerization, the π bond of each monomer’s C=C breaks and each carbon forms a new C–C bond to a neighbouring monomer. Nothing else forms, so the atom economy is 100%.

To draw the repeating unit:

1. Draw the C=C carbons joined by a single bond, keeping all their groups.
2. Add a continuation bond on each side through square brackets, and write n after.

Worked example 5: from monomer to repeating unit

Propenoic acid, CH₂=CHCOOH, polymerizes to make water-absorbing polymers. Draw its repeating unit and work out the molar mass of a chain of 2000 units.

Repeating unit: –[–CH₂–CH(COOH)–]–ₙ, a two-carbon backbone with continuation bonds through the brackets.

COOH is a side group, not part of the backbone. Each repeating unit has the same atoms as one monomer, C₃H₄O₂. Taking C = 12.01, H = 1.01 and O = 16.00: M = 3(12.01) + 4(1.01) + 2(16.00) = 72.07 g mol⁻¹. A chain of 2000 units has M ≈ 2000 × 72.07 = 1.44 × 10⁵ g mol⁻¹ (ignoring the end groups).

Worked example 6: from polymer to monomer

A polymer has the repeating unit –CH₂–C(CH₃)Cl–. Deduce the monomer.

Put a double bond between the two backbone carbons and remove the continuation bonds: CH₂=C(CH₃)Cl, 2-chloropropene.

Condensation polymers (Structure 2.4.6, HL only)

In condensation polymerization, each monomer has two functional groups. When two groups react, they form a link and release a small molecule, usually water. Because a by-product forms, the atom economy is below 100%.

• Polyesters: a carboxyl group (–COOH) reacts with a hydroxyl group (–OH) to give an ester link, –COO–.
• Polyamides: a carboxyl group reacts with an amino group (–NH₂) to give an amide link, –CONH–.

The groups can be on two different monomers or on the two ends of one monomer.

Worked example 7: a polyester

Butane-1,4-diol, HO(CH₂)₄OH, reacts with butanedioic acid, HOOC(CH₂)₂COOH. Draw the repeating unit and find the atom economy.

1. H from the diol’s –OH and OH from the acid’s –COOH leave as H₂O; the diol’s O joins the acid’s C=O, at both ends.
2. Repeating unit: –[–O(CH₂)₄O–CO(CH₂)₂CO–]–ₙ, with two ester links and two water molecules released per unit.
3. Atom economy = 172.20 / (90.14 + 118.10) × 100 = 82.7%.

Worked example 8: polyamides

(a) 4-Aminobutanoic acid, H₂N(CH₂)₃COOH, has both groups on one molecule. Each monomer links head-to-tail: –[–NH(CH₂)₃CO–]–ₙ, releasing one H₂O per unit.

(b) Pentane-1,5-diamine and hexanedioic acid give –[–NH(CH₂)₅NH–CO(CH₂)₄CO–]–ₙ, releasing two H₂O per unit.

Biological macromolecules

All biological macromolecules form by condensation and break down by hydrolysis, the reverse reaction, in which water splits each link.

• Proteins: 2-amino acids join through amide (peptide) links. A chain of 120 amino acids releases 119 water molecules.
• Polysaccharides such as starch and cellulose: glucose units join, releasing water at each link.
• Nucleic acids (DNA, RNA): nucleotides join by condensation.

Worked example 9

Starch is fully hydrolysed to glucose. Find the mass of glucose from 8.10 g of starch.

Each chain unit is C₆H₁₀O₅ (162.16 g mol⁻¹); adding water gives glucose, C₆H₁₂O₆ (180.18 g mol⁻¹). Moles of units = 8.10 / 162.16 = 0.04995 mol, so mass of glucose = 0.04995 × 180.18 = 9.00 g. The mass rises because water is added.

Common errors

• Multiplying by a subscript: Δχ for AlCl₃ uses one Cl value.
• Calling an alloy a compound.
• Saying alloys are harder “because the bonds are stronger” instead of “layers cannot slide”.
• Putting a side group into the backbone of an addition polymer, or leaving out continuation bonds and n.
• (HL) Forgetting to remove H₂O at each link, or confusing the amide link –CONH– with an amine.

Next steps

• Recap with the condensed notes for this unit
• Test yourself on the worked practice set
• Other units: course page; progress: syllabus checklist
• Spot weak areas quickly with a free diagnostic

Official syllabus

Document followed: Chemistry guide, Diploma Programme, International Baccalaureate Organization (February 2023; first examined 2025). Scope here: Structure 2.4, From models to materials; 2.4.1 to 2.4.5 for SL and HL, 2.4.6 additional higher level.