SL & HL Structure · Models of bonding and structure BETA S2.4

Bonding and materials

IB syllabus topic Structure 2.4 — From models to materials

The bonding triangle (van Arkel-Ketelaar triangle) treats ionic, covalent and metallic bonding as a continuum rather than three separate boxes, and from a compound's position you can predict its structure and properties. This revision note covers the bonding continuum and triangle, predicting properties, alloys, polymers and plastics, addition polymers, and (HL) condensation polymers.

Part of IB Chemistry (2025-2030 syllabus) — Standard and Higher Level.

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Key points

  • Bonding is a continuum, not three separate categories: most compounds have character somewhere between purely ionic, purely covalent and purely metallic.
  • The van Arkel-Ketelaar (bonding) triangle plots the electronegativity difference Δχ\Delta\chi (y-axis) against the average electronegativity χˉ\bar\chi (x-axis); the triangle itself is given in the data booklet (section 17), and electronegativity values are in section 9.
  • Position on the triangle predicts bonding type and therefore properties: large Δχ\Delta\chi = ionic (top), small Δχ\Delta\chi with high χˉ\bar\chi = covalent (bottom right), small Δχ\Delta\chi with low χˉ\bar\chi = metallic (bottom left).
  • An alloy is a mixture of a metal with one or more other elements (usually metals); the different-sized atoms disrupt the regular lattice, making layers harder to slip, so alloys are generally harder and stronger than the pure metal.
  • Addition polymers form from monomers containing a C=C double bond; the bond opens and the units join with no atoms lost, so the repeat unit contains exactly the same atoms (the same molecular formula) as the monomer.
  • Plastics are synthetic polymers; their bulk properties (rigidity, flexibility, melting behaviour) depend on chain length, branching and intermolecular forces between chains.
  • HL(HL) Condensation polymers form when monomers each carrying two functional groups join with the loss of a small molecule (usually water or HCl) at each link, producing polyesters (ester links) and polyamides (amide links).

Subtopic by subtopic

The bonding continuum and bonding triangle

Earlier topics treated bonding as three neat boxes: ionic, covalent and metallic. In reality bonding is a continuum - most substances sit somewhere between the extremes.

The van Arkel-Ketelaar triangle (given in data booklet section 17) maps this continuum. Each substance is plotted using two numbers worked out from electronegativity values (data booklet section 9):

  • x-axis = the average electronegativity χˉ=χA+χB2\bar\chi = \dfrac{\chi_A + \chi_B}{2}
  • y-axis = the electronegativity difference Δχ=χA−χB\Delta\chi = \chi_A - \chi_B (always take the larger minus the smaller, so Δχ≥0\Delta\chi \ge 0)

The three corners are:

  • Metallic - bottom left (low χˉ\bar\chi, small Δχ\Delta\chi): e.g. Na, Cu.
  • Covalent - bottom right (high χˉ\bar\chi, small Δχ\Delta\chi): e.g. Cl2\text{Cl}_2, SiO2\text{SiO}_2.
  • Ionic - top (large Δχ\Delta\chi): e.g. NaCl, MgO.

A large Δχ\Delta\chi pushes a substance up towards ionic; a small Δχ\Delta\chi keeps it along the bottom edge, with χˉ\bar\chi deciding metallic vs covalent. Substances near the edges and middle (e.g. some semiconductors and polar covalent compounds) have intermediate character.

Predicting properties from the bonding triangle

Once you have plotted a substance, its region lets you predict structure and physical properties.

  • Ionic region (top): giant ionic lattice. High melting point, brittle, conducts only when molten or aqueous (ions free to move), often soluble in water.
  • Covalent region (bottom right): either small molecules (low melting point, non-conducting, e.g. CO2\text{CO}_2) or covalent networks (very high melting point, hard, e.g. diamond, SiO2\text{SiO}_2). The triangle alone does not tell you which - you also need the formula and structure.
  • Metallic region (bottom left): giant metallic lattice. Conducts as solid and liquid, malleable and ductile, lustrous.

Working method:

1. Look up each element's χ\chi (section 9).

2. Calculate χˉ\bar\chi and Δχ\Delta\chi.

3. Plot on the triangle and read off the region.

4. State the structure, then the expected properties.

Watch out: the triangle gives the dominant bonding type, not a yes/no answer. A compound such as AlCl3\text{AlCl}_3 sits in an intermediate zone (significant covalent character despite a metal + non-metal), which is why it has a low sublimation point - exactly the kind of nuance the triangle is designed to reveal.

Alloys

An alloy is a mixture of a metal with one or more other elements (usually other metals). It is not a compound - there is no fixed ratio and no new chemical bond type.

In a pure metal, identical positive ions sit in regular layers within a sea of delocalised electrons. Applying a force lets layers slip over each other, so pure metals are relatively soft.

In an alloy, atoms of different sizes are mixed in. These disrupt the regular layers, so the layers can no longer slide easily. The metallic bonding (delocalised electrons) is retained, but the alloy is:

  • harder and stronger than the pure metal
  • often more resistant to corrosion

Examples:

  • Bronze = copper + tin (harder than copper)
  • Brass = copper + zinc
  • Steel = iron + carbon (small non-metal atoms wedge into the lattice)
  • Solder = tin + lead (low melting point)

Because the electron sea is unchanged, alloys still conduct electricity and heat and remain malleable, just less easily deformed than the pure metal. A simple particle diagram shows two sizes of cation in a single delocalised electron sea.

Polymers and plastics

A polymer is a very large molecule (a macromolecule) made by linking many small monomers into long chains. Plastics are synthetic polymers.

The repeating unit is the part that recurs along the chain; we write polymers with this unit in brackets and a subscript nn to show it repeats many times.

Properties depend on:

  • Chain length - longer chains tangle more, giving stronger, higher-melting plastics.
  • Branching - straight (linear) chains pack closely, so they are denser and stronger (e.g. HDPE); branched chains pack loosely, giving softer, lower-density plastics (e.g. LDPE).
  • Intermolecular forces between chains - stronger forces (e.g. hydrogen bonding in nylon) raise the melting point and strength.

Many plastics are thermoplastics: the chains are held together only by intermolecular forces, so they soften on heating and can be remoulded and recycled.

Polymers can be natural (proteins, cellulose, DNA) or synthetic (poly(ethene), PVC, nylon). The big environmental issue is that many synthetic plastics are non-biodegradable because their strong C-C backbones resist breakdown.

Addition polymers

Addition polymers are made from unsaturated monomers - molecules containing a C=C\text{C=C} double bond. The double bond opens up and the monomers join end to end with no atoms lost.

General pattern for an ethene-type monomer:

n CH2=CH2⟶−(CH2−CH2)−nn\,\text{CH}_2\text{=CH}_2 \longrightarrow {-}(\text{CH}_2{-}\text{CH}_2){-}_n

Key features:

  • The repeat unit contains exactly the same atoms as the monomer (the same molecular formula) - nothing is eliminated.
  • The product backbone is a saturated C-C chain.

How to draw the repeat unit:

1. Draw the monomer with the C=C\text{C=C} horizontal.

2. Change the double bond to a single bond.

3. Extend a bond from each carbon out through the brackets.

4. Add the subscript nn.

Common examples (monomer -> polymer):

  • ethene -> poly(ethene)
  • chloroethene -> PVC (poly(chloroethene))
  • tetrafluoroethene -> PTFE (Teflon)
  • propene -> poly(propene)

Given a polymer you can also work backwards to the monomer: find the repeat unit, then put a C=C\text{C=C} back between the two backbone carbons.

Condensation polymersHL

Condensation polymers form when each monomer carries two reactive functional groups, so chains can grow from both ends, and a small molecule is eliminated at every new link (usually H2O\text{H}_2\text{O}).

Polyesters form from a diol + a dicarboxylic acid (or a single monomer with both an -OH and a -COOH). The link is an ester group:

-COOH+HO-⟶-COO-+H2O\text{-COOH} + \text{HO-} \longrightarrow \text{-COO-} + \text{H}_2\text{O}

Example: ethane-1,2-diol + benzene-1,4-dicarboxylic acid -> PET (Terylene).

Polyamides form from a diamine + a dicarboxylic acid. The link is an amide (peptide) group, and water is again released:

-COOH+H2N-⟶-CONH-+H2O\text{-COOH} + \text{H}_2\text{N-} \longrightarrow \text{-CONH-} + \text{H}_2\text{O}

Example: nylon-6,6 from a 6-carbon diamine + a 6-carbon diacid.

Contrast with addition polymers:

  • Addition: monomer has C=C; no small molecule lost; repeat unit = monomer atoms.
  • Condensation: monomers have two functional groups; a small molecule is lost each link; repeat unit is lighter than the combined monomers.

Proteins (amino acids joined by amide/peptide bonds) and polysaccharides are natural condensation polymers.

Formulae

Δχ=χA−χB\Delta\chi = \chi_A - \chi_B

Electronegativity difference between two bonded elements; the yy-coordinate when plotting on the bonding triangle. You CONSTRUCT this from electronegativity values (data booklet section 9); the value Δχ\Delta\chi itself is not listed.

χˉ=χA+χB2\bar\chi = \dfrac{\chi_A + \chi_B}{2}

Average (mean) electronegativity of the two bonded elements; the xx-coordinate on the bonding triangle. Constructed by the student from data booklet section 9 electronegativity values.

Definitions

Bonding continuum
The idea that ionic, covalent and metallic bonding are extremes of a smooth range, and that real substances usually have a blend of bonding character rather than belonging to one pure type.
van Arkel-Ketelaar (bonding) triangle
A triangular diagram (in data booklet section 17) that places a compound by its average electronegativity and its electronegativity difference, with ionic, covalent and metallic bonding at the three corners.
Electronegativity difference, Δχ\Delta\chi
The difference in electronegativity between the two bonded elements; a large value indicates more ionic character (the y-axis of the bonding triangle).
Alloy
A mixture (not a compound) of a metal with one or more other elements, in which the host metal's delocalised sea of electrons is retained but the lattice is disrupted by atoms of different size.
Addition polymer
A long-chain molecule built by repeated joining of unsaturated monomers (containing C=C) with no loss of atoms, so the repeat unit contains exactly the same atoms (the same molecular formula) as the monomer.
Condensation polymer (HL)HL
A polymer formed from monomers with two reactive functional groups that link together with the elimination of a small molecule (such as water) at each new bond.
Monomer
A small molecule, many of which react together (each contributing once) to build the repeating units of a polymer chain.

Worked examples

1

(SL) Using electronegativity values χ(Mg)=1.3\chi(\text{Mg}) = 1.3 and χ(O)=3.4\chi(\text{O}) = 3.4, calculate the average electronegativity and the electronegativity difference for MgO, and use the bonding triangle to predict its bonding type and one physical property.

  1. 1
    Average electronegativity: χˉ=(1.3+3.4)/2=4.7/2=2.35\bar\chi = (1.3 + 3.4)/2 = 4.7/2 = 2.35.
  2. 2
    Electronegativity difference: Δχ=3.4−1.3=2.1\Delta\chi = 3.4 - 1.3 = 2.1.
  3. 3
    Plot the point: moderate χˉ\bar\chi (about 2.35) and a large Δχ\Delta\chi (about 2.1) places it high up on the triangle.
  4. 4
    A large Δχ\Delta\chi is the ionic corner (top) of the van Arkel-Ketelaar triangle.
  5. 5
    Ionic substances form giant lattices, so MgO has a very high melting point and conducts only when molten or in solution.

Answer: χˉ=2.35\bar\chi = 2.35, Δχ=2.1\Delta\chi = 2.1. The large Δχ\Delta\chi places MgO in the ionic region, so it is predominantly ionic: a giant ionic lattice with a very high melting point that conducts electricity only when molten or aqueous.

2

(SL) The monomer chloroethene is CH2=CHCl\text{CH}_2\text{=CHCl}. Draw/describe the repeat unit of the addition polymer it forms (PVC) and explain why no small molecule is released.

  1. 1
    Identify the C=C double bond - this is the reactive site for addition polymerisation.
  2. 2
    Open the double bond to a single C-C bond.
  3. 3
    Extend a bond from each carbon outward (these continue the chain on both sides).
  4. 4
    Keep all the original atoms: one carbon carries two H, the other carries one H and one Cl.
  5. 5
    Because the C=C simply opens and the monomers add on, every atom of the monomer ends up in the chain - nothing is eliminated.

Answer: Repeat unit: −(CH2−CHCl)−n-(\text{CH}_2-\text{CHCl})-_n. No small molecule is lost because addition polymerisation works by opening the C=C double bond, so the repeat unit contains exactly the same atoms (the same molecular formula) as the monomer.

3

(HL) A polyester is made from ethane-1,2-diol, HO-CH2-CH2-OH\text{HO-CH}_2\text{-CH}_2\text{-OH}, and ethanedioic acid, HOOC-COOH\text{HOOC-COOH}. Identify the link formed, the small molecule released, and explain why this is condensation rather than addition polymerisation.HL

  1. 1
    The diol provides two -OH groups; the diacid provides two -COOH groups.
  2. 2
    An -OH reacts with a -COOH to form an ester link: -COO-\text{-COO-}.
  3. 3
    Forming each ester link eliminates one molecule of water (-COOH+HO-→-COO-+H2O\text{-COOH} + \text{HO-} \rightarrow \text{-COO-} + \text{H}_2\text{O}).
  4. 4
    Each monomer has two functional groups, so chains grow from both ends.
  5. 5
    Because a small molecule (water) is lost at each new bond, this is condensation, not addition (addition needs a C=C and loses nothing).

Answer: An ester link (-COO-\text{-COO-}) forms and water (H2O\text{H}_2\text{O}) is eliminated at each link. It is condensation polymerisation because the monomers each have two functional groups that join with loss of a small molecule, unlike addition polymerisation which opens a C=C with no atoms lost.

Common mistakes

  • ×Confusing the two axes of the bonding triangle: Δχ\Delta\chi (difference) is the y-axis and decides ionic character, while χˉ\bar\chi (average) is the x-axis and decides metallic vs covalent.
  • ×Treating the triangle as giving a pure category - many substances are intermediate, and the triangle shows the dominant character, not an absolute label.
  • ×Calling an alloy a compound. An alloy is a mixture with no fixed ratio; the metallic bonding (electron sea) is retained, which is why alloys still conduct.
  • ×(HL) Saying water is lost in addition polymerisation. Addition (C=C opening) loses nothing; only condensation polymerisation eliminates a small molecule such as H2O\text{H}_2\text{O}.

Exam tips

  • ✓When asked to place a substance on the bonding triangle, always show your working: state each χ\chi from data booklet section 9, then calculate χˉ\bar\chi and Δχ\Delta\chi before reading off the region.
  • ✓To explain why an alloy is harder than the pure metal, mention that different-sized atoms disrupt the regular layers so they cannot slip easily - do not just say 'it is stronger'.
  • ✓For addition polymers, draw the repeat unit with the double bond changed to a single bond AND a bond extending out through each bracket; forgetting the extending bonds is a common mark loss.
  • ✓(HL) For condensation polymers, name BOTH the link (ester or amide) and the small molecule eliminated (usually water), and identify the two functional groups on the monomers.

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