Educerie
Level

1 higher-level section hidden.

Educerie · IB Diploma · Chemistry

Structure 2 Models of bonding and structure · S2.4 From models to materials

Level
SL and HL. Section 8 is HL only. If you are SL, skip it; nothing in your papers tests it.
Themes (key concepts)
structure, and models. The three bonding models of S2.1 to S2.3 are idealized ends of one scale; this subtopic puts them on a single diagram and uses them to explain the structure, and so the properties, of real materials: alloys, plastics and composites.
The question this unit answers
what role do bonding and structure have in the design of materials?
Where it is examined
Paper 1A multiple choice (1 mark: a position on the bonding triangle, a repeating unit, a monomer); Paper 1B data on the properties of materials; Paper 2 parts of 2 to 4 marks where you place a compound on the triangle and predict its properties, explain an alloy's hardness, or draw a repeating unit. HL adds condensation polymers, usually as a 2- or 3-mark drawing.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
Describe bonding as a continuum between ionic, covalent and metallic, shown by a bonding triangleSL, HL"Outline why bonding is described as a continuum" (2 marks)
Use bonding models to explain the properties of a materialSL, HL"Explain why reinforced concrete is used in…" (2–3 marks)
Determine the position of a binary compound on the bonding triangle from electronegativity dataSL, HL"Use the bonding triangle in the data booklet to determine the bonding in MgCl₂" (2 marks)
Predict properties from a compound's position on the triangleSL, HL"Predict two physical properties of…" (2 marks)
Explain the properties of alloys in terms of non-directional bondingSL, HL"Explain why brass is harder than copper" (2–3 marks)
Describe the common properties of plastics in terms of their structure; give natural and synthetic polymersSL, HL"Describe two properties of plastics and explain one" (3 marks)
Represent the repeating unit of an addition polymer from its monomer, and deduce the monomer from the polymerSL, HL"Draw the repeating unit of…" (1–2 marks)
Represent the repeating unit of polyamides and polyesters from their monomersHL only"Draw the repeating unit of the polymer formed from…" (2 marks)

Before you start

You need all three models: the ionic lattice (S2.1), molecules, networks and intermolecular forces (S2.2), and the metallic bond (S2.3), because this subtopic is about how they blend. You need electronegativity values from the data booklet, used in S2.2 for bond polarity, and the C=C double bond from S2.2.


1The idea in one paragraph

Real bonds are rarely purely ionic, purely covalent or purely metallic. A bond between two atoms sits somewhere on a continuum, and two numbers tell you where: how different the two atoms' electronegativities are, and how high they are on average. Plot those two numbers and you land on a bonding triangle with ionic, covalent and metallic at its corners; your position predicts how the substance behaves. Materials scientists use the same thinking in reverse. They mix metals to make alloys harder than either metal alone, link small molecules into long chains to make polymers with the properties they want, and combine different materials into composites that do what no single one can.

2Bonding is a continuum

The three models are useful because they are simple, and simple models are ideals. Consider the bond in hydrogen chloride. The pair is shared, so it is covalent; but chlorine pulls it strongly, so the bond has a large δ+ and δ−, some of the character of an ionic bond. Or consider aluminium chloride: from a metal and a non-metal, so "ionic" by the S2.1 rule of thumb, yet it turns to vapour at about 180 °C, which no ionic lattice does. Most real bonds are a blend.

Bonding is a continuum: most bonds have partly ionic, partly covalent and partly metallic character.

The bonding triangle (a version is in the data booklet) turns that blend into a position. The bottom axis is the average electronegativity of the two elements; the vertical axis is the difference in electronegativity, Δχ. Figure 1 draws it with examples.

Figure 1 · The bonding triangle: where a binary substance sits Figure 1 · The bonding triangle: where a binary substance sits 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Average electronegativity, (χ₁ + χ₂) ÷ 2 0 1 2 3 Electronegativity difference, Δχ IONIC METALLIC COVALENT polar covalent Δχ ≈ 1.8 CsF NaCl MgO Al₂O₃ AlCl₃ SiO₂ H₂O CCl₄ Na Mg Al Si Cl₂ Electronegativities from the data booklet. The zone boundaries are gradual, not sharp: bonding is a continuum.
Figure 1 · The bonding triangle: where a binary substance sits

The three corners are the three pure models, and the reasoning behind each is short.

  • Top: ionic. A large Δχ means one atom takes the electrons almost completely. Caesium fluoride, from the least and the most electronegative elements, is at the apex.
  • Bottom left: metallic. Both atoms have low electronegativity, so neither holds its electrons tightly and the electrons are delocalized. Pure metals sit on the base, where Δχ = 0.
  • Bottom right: covalent. Both atoms have high electronegativity, so both hold on to electrons, and they share them. Cl₂ and F₂ sit on the base at the right.

In between, the character changes gradually. Moving up from the bottom-right corner, bonds become more polar, then polar covalent, then predominantly ionic. The dashed lines in Figure 1 mark rough boundaries only; above about Δχ = 1.8 bonding is mostly ionic, but nothing special happens at that value.

3Placing a compound on the triangle

The guide asks you to determine the position of a binary compound (two elements) from electronegativity data. It is two sums. Take magnesium oxide, with χ(Mg) = 1.3 and χ(O) = 3.4.

average χ = (1.3 + 3.4) ÷ 2 = 2.35
Δχ = 3.4 − 1.3 = 2.1
position (2.35, 2.1): well into the ionic region

Do the same for aluminium chloride, χ(Al) = 1.6 and χ(Cl) = 3.2: average 2.4, Δχ 1.6. It sits below the ionic zone, in the polar covalent region, which is exactly why it behaves as a covalent substance. For a pure element, Δχ is zero and the average is just its own electronegativity: silicon sits on the base at 1.9, between the metallic and covalent corners, as a metalloid should.

You do not calculate a percentage of ionic character; the guide says so. You read the region off the triangle and say what kind of bonding dominates.

4From position to properties

Once you know which corner a substance is nearest, the properties follow from the model for that corner.

RegionTypical propertiesExample
Ionichigh melting point; conducts when molten or dissolved, not as a solid; brittle; often soluble in waterNaCl, MgO
Polar covalent, molecularlow melting and boiling points; does not conduct; solubility depends on polarityHCl, H₂O, AlCl₃
Covalent, non-polarlow melting and boiling points (unless a network); does not conduct; insoluble in waterCl₂, CCl₄
Metallicconducts as a solid; malleable; melting point variesNa, Mg, Al

The chlorides of period 3 show the continuum at work. Following them from sodium to phosphorus, the average electronegativity rises and Δχ falls, so the bonding moves steadily from ionic towards covalent, and the melting points collapse.

ChlorideAverage χΔχMelting point / °CBonding and structure
NaCl2.052.3801ionic lattice
MgCl₂2.251.9714ionic lattice
AlCl₃2.41.6sublimes at about 180covalent character, molecules in the vapour
SiCl₄2.551.3−69covalent molecules
PCl₃2.71.0−94covalent molecules

The triangle has limits, and the guide wants you to know them. It uses only two numbers per compound, so it cannot tell a covalent network from covalent molecules: SiO₂ (average 2.65, Δχ 1.5) sits near AlCl₃ and H₂O, yet melts above 1600 °C because it is a network. It says nothing about structure, only about the bond. Discrete labels such as "ionic" and "covalent" are useful precisely because they are simplifications, and the continuum is the more honest description.

Composites show the same design thinking at a larger scale. A composite is made of two or more materials with different properties, combined so that each makes up for the other's weakness. Concrete is mostly ionic and covalent-network solids: very strong when squeezed, but it cracks when pulled or bent. Steel is metallic: strong when pulled. Set steel bars inside concrete and you have reinforced concrete, which resists squeezing and bending and holds up bridges and buildings that neither material could hold alone.

5Alloys

An alloy is a mixture of a metal with other metals or with non-metals. Some common examples:

AlloyMain componentsWhy it is used
Bronzecopper and tinharder than copper; resists corrosion
Brasscopper and zincharder than copper; easily machined; looks like gold
Steeliron and carbonmuch harder and stronger than iron
Stainless steeliron, chromium (and often nickel)strong and resists rusting

You do not have to learn specific alloys; you do have to explain their properties.

Why alloys form at all. Metallic bonding is non-directional: the cations are held by the electron sea around them, not by bonds to particular neighbours. So a cation of a different element, or a small non-metal atom, can take a place in the lattice or sit in the gaps between cations without breaking the bonding. The alloy still has delocalized electrons, so it still conducts and is still metallic.

Why alloys are mixtures, not compounds. The composition can vary continuously: brass can be 60% copper or 70% copper. There is no fixed ratio and no new chemical formula, so an alloy is a mixture, even though it is held together by metallic bonding.

Why alloys are harder and stronger. In a pure metal all the cations are the same size, so the layers are regular and slide over each other easily, which is what makes a pure metal soft and malleable. Atoms of a different size distort the layers, as Figure 2 shows. The distorted layers can no longer slide past each other easily, so a larger force is needed to change the metal's shape: the alloy is harder and stronger, and less malleable.

Figure 2 · Why an alloy is harder than the pure metal Figure 2 · Why an alloy is harder than the pure metal (a) Pure metal layers slide easily (b) Alloy atoms of a different size distort the layers, so they no longer slide past each other easily The bonding is still metallic and non-directional, so the foreign atoms fit in. They just get in the way of slipping.
Figure 2 · Why an alloy is harder than the pure metal

Other properties can change too. Chromium in stainless steel forms a thin oxide layer on the surface that stops the iron beneath from rusting. The bonding is the same idea; the design is choosing what to add.

6Polymers and plastics

A polymer is a very large molecule, a macromolecule, made of many small repeating subunits called monomers, joined by covalent bonds. A chain may contain thousands of monomer units.

Polymers are everywhere in living things. Natural polymers include proteins (from amino acids), starch and cellulose (from glucose), DNA (from nucleotides) and natural rubber. Synthetic polymers include poly(ethene), PVC, PTFE, polystyrene, nylon and PET. Plastics are synthetic polymers that can be moulded into shape.

Figure 3 shows the structure behind the properties of plastics: long chains, with strong covalent bonds along each one and weaker intermolecular forces between chains.

Figure 3 · Plastics are tangles of long chains Figure 3 · Plastics are tangles of long chains (a) Chains, held by intermolecular forces strong covalent bonds along each chain; weaker forces between chains (b) Warm it: chains slide apart a thermoplastic softens and can be reshaped; the chains are not broken Longer chains and closer packing mean stronger forces between chains: a stiffer, higher-melting plastic.
Figure 3 · Plastics are tangles of long chains
Property of most plasticsWhy, from the structure
Low densitymade of light atoms (C, H) in chains that do not pack tightly
Electrical insulatorevery electron is held in a localized covalent bond; there are no ions and no delocalized electrons
Poor conductor of heatno mobile electrons to carry energy
Unreactive and durablestrong C–C and C–H bonds, which are non-polar and not easily attacked
Not biodegradable, so persists in the environmentthe same strong, non-polar chains that microorganisms cannot readily break
Softens on heating and can be moulded (thermoplastics)the chains are held by intermolecular forces, which are overcome on warming so the chains slide
Flexible, but strongchains can move and bend, and long tangled chains resist being pulled apart

Properties can be tuned by changing the structure. Longer chains mean more intermolecular forces per chain, so a higher melting point and a stronger plastic. Chains with few branches pack closely, so their intermolecular forces are stronger and the plastic is denser and more rigid; branched chains cannot pack closely, giving a softer, lower-density plastic. Polar side groups add dipole–dipole forces: the C–Cl bonds in PVC make it stiffer than poly(ethene). Plastics whose chains contain ester or amide links can be broken down by hydrolysis, which is one route to biodegradable plastics (Structure 3.2).

7Addition polymers

An addition polymer forms when monomers containing a C=C double bond join together. In each monomer the double bond breaks (strictly, its π bond does), and the two carbon atoms each use the freed electron to bond to a neighbouring monomer. Nothing else is lost, so the polymer is the only product and the atom economy is 100%. Figure 4 shows ethene becoming poly(ethene).

Figure 4 · Addition polymerization: the double bond opens Figure 4 · Addition polymerization: the double bond opens (a) Three ethene molecules join; each C=C becomes C–C C C H H H H C C H H H H C C H H H H C C C C C C H H H H H H H H H H H H … … (b) The equation, with the repeating unit in brackets n C C H H H H C C H H H H [ ] n ethene → poly(ethene) the bonds that stick out through the brackets join to the next unit nothing is lost: atom economy 100% Two carbon atoms from the C=C, and everything attached to them, make one repeating unit.
Figure 4 · Addition polymerization: the double bond opens

The polymer is too long to draw, so you draw its repeating unit: the smallest section that repeats along the chain, in square brackets with a subscript n and a bond sticking out through each bracket. For an addition polymer made from one monomer, the repeating unit is always the two carbons of the old C=C with everything attached to them, now joined by a single bond.

The same rule works for any alkene. Figure 5 shows three more.

Figure 5 · Monomers and their repeating units Figure 5 · Monomers and their repeating units C C H H H CH₃ propene C C H H H CH₃ [ ] n poly(propene) C C H H H Cl chloroethene C C H H H Cl [ ] n poly(chloroethene), PVC C C F F F F tetrafluoroethene C C F F F F [ ] n poly(tetrafluoroethene), PTFE To go backwards from a polymer: 1. find two carbons in the main chain that repeat, with their side groups 2. draw them once, in brackets 3. put the C=C back between them: that is the monomer The main chain is always the carbons of the old C=C. Side groups such as CH₃ or Cl hang off it, never in it. The side groups keep their places: a CH₃ on the monomer's second carbon is a CH₃ on every second chain carbon.
Figure 5 · Monomers and their repeating units

The monomer structures are given in the exam, or you deduce them, so the skill is the drawing. Three checks catch almost every mistake.

  1. The main chain contains only the two carbon atoms of the C=C. Side groups such as CH₃ or Cl hang off the chain; they are never in it.
  2. There is no double bond in the repeating unit.
  3. There are bonds through both brackets, showing the chain continues.

To go backwards, from a section of polymer to its monomer, find the two-carbon unit that repeats, draw it once, and put the C=C back. A chain reading –CH₂–CH(CH₃)–CH₂–CH(CH₃)– has the repeating unit –CH₂–CH(CH₃)–, so the monomer is CH₂=CH–CH₃, propene.

8HLCondensation polymers

SL students can skip to section 9.

A condensation polymer forms when monomers with two reactive functional groups each join, and every link made releases a small molecule, usually water. Each monomer needs a functional group at each end, so the chain can keep growing in both directions.

Polyamides. A monomer with two amine groups, a diamine, reacts with a monomer with two carboxylic acid groups, a dicarboxylic acid. An H from –NH₂ and the OH from –COOH leave as water, and the nitrogen bonds to the carbon: an amide link, –CONH–. Figure 6 shows the result. Nylon-6,6 is made this way from hexane-1,6-diamine, H₂N(CH₂)₆NH₂, and hexanedioic acid, HOOC(CH₂)₄COOH.

Figure 6 · A polyamide: diamine + dicarboxylic acid (HL) Figure 6 · A polyamide: diamine + dicarboxylic acid (HL) H₂N chain NH₂ + HOOC chain COOH a diamine a dicarboxylic acid an H from –NH₂ and the OH from –COOH leave together as H₂O The polymer: the link repeats, and so do both blocks [ NH chain NH C O chain C O ] n example: nylon-6,6, from hexane-1,6-diamine and hexanedioic acid Each amide link, –CONH–, forms as an –NH₂ and a –COOH join and lose one water molecule. The repeating unit contains one of each monomer, so n units release 2n − 1 water molecules.
Figure 6 · A polyamide: diamine + dicarboxylic acid (HL)

Polyesters. A diol, with two –OH groups, reacts with a dicarboxylic acid. The H of –OH and the OH of –COOH leave as water, and an ester link, –COO–, forms. PET, used for drinks bottles and polyester fibres, is made from ethane-1,2-diol and benzene-1,4-dicarboxylic acid.

Figure 7 · A polyester: diol + dicarboxylic acid (HL) Figure 7 · A polyester: diol + dicarboxylic acid (HL) HO chain OH + HOOC chain COOH a diol a dicarboxylic acid the H from –OH and the OH from –COOH leave together as H₂O The polymer: the link repeats, and so do both blocks [ O chain O C O chain C O ] n example: PET, from ethane-1,2-diol and benzene-1,4-dicarboxylic acid Each ester link, –COO–, forms as an –OH and a –COOH join and lose one water molecule. Biological macromolecules are made the same way, and hydrolysis reverses it.
Figure 7 · A polyester: diol + dicarboxylic acid (HL)

Drawing a repeating unit follows one method. Write each monomer with its reactive groups at the ends. Remove H from one end and OH from the other. Join them with the link (–CONH– or –COO–). The repeating unit contains one of each monomer, with the link in the middle and a bond through each bracket. For nylon-6,6:

n H2N(CH2)6NH2 + n HOOC(CH2)4COOH → –[NH(CH2)6NHCO(CH2)4CO]n– + (2n − 1) H2O

If the acid is replaced by a diacyl chloride (–COCl at each end), the small molecule released is HCl instead of water; the polymer is the same.

Biological macromolecules are all condensation polymers. Proteins form from amino acids joined by amide (peptide) links; starch and cellulose from glucose; nucleic acids from nucleotides. Every link releases a water molecule, and the reverse reaction, hydrolysis, uses water to break the links again: that is what digestion does.

9Where marks are lost

Placing a compound using Δχ alone. The triangle needs two coordinates. Δχ tells you how ionic; the average tells you whether low-Δχ bonding is metallic or covalent.

Saying a metal–non-metal compound must be ionic. AlCl₃ is metal plus non-metal and has strong covalent character. Use the data, not the rule of thumb.

Calling an alloy a compound. Alloys have variable composition and no formula: they are mixtures, held together by metallic bonding.

Explaining alloy hardness by "stronger bonds". The point is that different-sized atoms disrupt the regular layers so they cannot slide easily.

Putting a side group in the main chain. In poly(propene) the chain is –CH₂–CH– and the CH₃ hangs off it. A three-carbon chain unit is wrong.

Leaving the double bond in the repeating unit, or leaving off the continuation bonds and n. The C=C is gone in the polymer, and the brackets need bonds through them.

Saying plastics are unreactive "because they are covalent". Name the reason: strong, non-polar C–C and C–H bonds.

HL: including both link-forming groups in the repeating unit, or drawing two of one monomer. One diamine unit, one diacid unit, one pair of amide links shared across the brackets.

10Draw it right

  1. Bonding triangle: give both coordinates with working, (χ₁ + χ₂) ÷ 2 and |χ₁ − χ₂|, then name the region.
  2. Alloy diagrams: regular layers of one size of atom for the pure metal; atoms of a different size breaking up the layers for the alloy; label "layers cannot slide easily".
  3. Addition repeating unit: two carbons in the chain, single bond between them, all four substituents drawn, square brackets, a bond through each bracket, and n.
  4. Equation: n in front of the monomer, n after the bracket, and no other product.
  5. HL condensation repeating unit: one of each monomer, the amide –CONH– or ester –COO– link drawn out with its C=O, bonds through both brackets, n; and the small molecule named in the equation.

11Try it

Marks in brackets. Answers and marker's notes are at the end.

Q1. Which is the repeating unit of poly(propene)? 1 mark

A. –CH₂–CH₂–CH₂– · B. –CH₂–CH(CH₃)– · C. –CH=CH(CH₃)– · D. –CH₂–CH₂–

Q2. Magnesium chloride is a binary compound. χ(Mg) = 1.3, χ(Cl) = 3.2. 4 marks

(a) Determine the position of MgCl₂ on the bonding triangle and state the type of bonding that dominates. 2 marks

(b) Predict two physical properties of magnesium chloride. 2 marks

Q3. Brass is an alloy of copper and zinc. Explain why brass is harder than pure copper, and why it is described as a mixture rather than a compound. 3 marks

Q4. Two forms of poly(ethene) are compared. 3 marks

FormChainsDensity / g cm⁻³Softens at about / °C
low-density poly(ethene), LDPEmany side branches0.92110
high-density poly(ethene), HDPEvery few branches0.95130

Explain the differences in density and softening temperature. 3 marks

Q5. A section of a polymer chain is –CH₂–CH(CN)–CH₂–CH(CN)–CH₂–CH(CN)–. 2 marks

(a) Draw the repeating unit. 1 mark

(b) Deduce the structure of the monomer. 1 mark

Q6 (HL). Ethane-1,2-diol, HOCH₂CH₂OH, reacts with butanedioic acid, HOOCCH₂CH₂COOH, to form a condensation polymer. 3 marks

(a) State the type of polymer formed and the small molecule released. 1 mark

(b) Draw the repeating unit of the polymer. 2 marks

12In one breath

Bonding is a continuum, not three boxes: most bonds are partly ionic, partly covalent, partly metallic. The bonding triangle places a binary compound by two numbers, the average electronegativity along the bottom and the electronegativity difference up the side: large Δχ is ionic at the top, low Δχ with low average is metallic at the bottom left, low Δχ with high average is covalent at the bottom right, and position predicts properties. The period 3 chlorides slide from ionic NaCl to covalent PCl₃, and their melting points collapse with it. The triangle cannot tell networks from molecules, which is one limit of discrete categories. Composites such as reinforced concrete combine materials so that each covers the other's weakness. Alloys are mixtures of a metal with other elements; non-directional metallic bonding lets foreign atoms fit, and their different size disrupts the layers so they cannot slide, making alloys harder and stronger. Polymers are macromolecules of repeating monomers, natural (proteins, starch, cellulose, DNA) or synthetic; plastics are light, insulating, unreactive and persistent because of strong non-polar covalent chains held together by weaker intermolecular forces. In addition polymerization each C=C opens and the repeating unit is the two old C=C carbons with their side groups, in brackets, with 100% atom economy. HL: condensation polymers form from monomers with two functional groups, releasing water at each link: diamine plus diacid gives a polyamide, diol plus diacid a polyester, and biological macromolecules form the same way and are broken by hydrolysis.


Answers

Q1. B. The chain contains the two carbons of propene's C=C, with the CH₃ as a side group. A puts the CH₃ carbon into the chain, C keeps a double bond, and D is poly(ethene). B only.

Q2. (a) Average χ = (1.3 + 3.2) ÷ 2 = 2.25; Δχ = 3.2 − 1.3 = 1.9. The point lies in the ionic region of the triangle, so the bonding is predominantly ionic. (b) Any two, each linked to ionic bonding: a high melting point (it is a lattice held by strong electrostatic attractions); does not conduct as a solid but conducts when molten or in aqueous solution; soluble in water; brittle. 1 for both coordinates with working, 1 for predominantly ionic, 1 for each correct property up to 2. "Ionic, because Mg is a metal" with no data scores 0 for (a).

Q3. Brass contains zinc atoms of a different size from copper atoms. These distort the regular layers of the lattice, so the layers can no longer slide over each other easily, and a larger force is needed to change the shape: brass is harder. It is a mixture because its composition can vary, with no fixed ratio of copper to zinc and no chemical formula. 1 for different-sized atoms disrupting the layers, 1 for layers not sliding easily, 1 for variable composition. "Zinc makes the bonds stronger" scores 0 for the hardness marks.

Q4. HDPE has very few branches, so its chains can pack closely together, giving more mass in a given volume, a higher density. Closely packed chains have stronger intermolecular (London) forces between them, so more energy is needed to overcome them and let the chains move: HDPE softens at a higher temperature. LDPE's branches keep its chains further apart, so it is less dense and its intermolecular forces are weaker. 1 for branching preventing close packing, 1 for close packing giving a higher density, 1 for stronger intermolecular forces giving a higher softening temperature. Any mention of covalent bonds breaking on softening scores 0 for the last mark.

Q5. (a) –CH₂–CH(CN)–, drawn in square brackets with a bond through each bracket and n. (b) CH₂=CH–CN (propenenitrile). 1 for each. A repeating unit with the CN in the main chain, or a monomer without the C=C, scores 0.

Q6 (HL). (a) A polyester; the small molecule released is water. (b) –O–CH₂–CH₂–O–CO–CH₂–CH₂–CO–, drawn with each C=O shown, in square brackets with a bond through each bracket and n. The unit contains one diol and one diacid, joined by an ester link, with the second ester link shared across the brackets. 1 for (a); in (b) 1 for the ester link correctly drawn, 1 for one of each monomer with continuation bonds and n. Leaving OH or H on the ends of the unit loses the last mark.


Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section S2.4 From models to materials. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!