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Structure 2.4 · Bonding as a continuum

1Bonding as a continuum 2.4.1 SL + HL

The ionic, covalent and metallic models are idealizations. Real substances rarely fit one model perfectly: the bond in HCl is covalent but polar; aluminium chloride has properties between those of an ionic salt and a molecular compound; brass is a metal made of two different elements. Materials science uses the models together to explain, and then to design, the properties of real materials.

The syllabus statement: Bonding is best described as a continuum between the ionic, covalent and metallic models, and can be represented by a bonding triangle. The skill assessed is Use bonding models to explain the properties of a material — so the triangle is a means, and the properties are the end.

Electronegativity controls the type of bonding between two elements:

• a large difference in electronegativity (Δχ) → electrons transferred → ionic bonding (a metal with a reactive non-metal);
• small Δχ, high average electronegativity → electrons shared and held between the atoms → covalent bonding (non-metals);
• small Δχ, low average electronegativity → electrons loosely held and delocalized → metallic bonding (metals and alloys).

Between these extremes lie compounds with mixed character, for example polar covalent bonds with partial ionic character. A bonding triangle (van Arkel–Ketelaar triangle) displays this continuum: each binary substance is plotted using the average electronegativity of its two elements (horizontal axis) and their electronegativity difference (vertical axis). The three corners represent the most metallic (caesium), the most covalent (fluorine) and the most ionic (caesium fluoride) substances.

Exam focus · what the published papers show

2Locating a compound in the bonding triangle and predicting its properties 2.4.2 SL + HL

The syllabus statement: The position of a compound in the bonding triangle is determined by the relative contributions of the three bonding types to the overall bond. You are asked to determine the position of a compound in the bonding triangle from electronegativity data and to predict the properties of a compound based on its position.

Exam focus · what the published papers show

Structure 2.4 · Materials

3Alloys 2.4.3 SL + HL

The syllabus statement: Alloys are mixtures of a metal and other metals or non-metals. They have enhanced properties. You are asked to explain the properties of alloys in terms of non-directional bonding. Two words in that statement carry the whole outcome: mixtures, and non-directional.

Because metallic bonding is non-directional, atoms of another element can be introduced into a metal lattice without breaking a particular set of bonds: the delocalized electrons simply bind the new atoms into the lattice as well. Alloys are therefore mixtures of variable composition, not compounds with a fixed formula. There are two ways the added atoms fit in (Figure 2.24):

• Substitutional alloys: atoms of similar size replace some host atoms. In brass about one-third of the copper atoms are replaced by zinc; bronze is copper with tin.
• Interstitial alloys: much smaller atoms occupy the gaps between host atoms. Steel is iron containing a small percentage of carbon; stainless steel also contains chromium and nickel, which improve corrosion resistance.

Explaining the enhanced properties. In a pure metal the layers of identical cations are regular and slide over one another easily, so pure metals are often soft. Atoms of a different size distort the regular arrangement of the lattice and act as obstacles that prevent the layers sliding. The alloy is therefore harder and stronger (higher tensile strength) but less malleable and ductile than the pure metal. Alloying can also change other physical properties such as density, melting point and electrical conductivity; the irregular lattice scatters moving electrons, so alloys usually conduct electricity less well than the pure host metal. Specific alloys need not be memorized; the explanation must be.

Exam focus · what the published papers show

4Polymers and the properties of plastics 2.4.4 SL + HL

The syllabus statement: Polymers are large molecules, or macromolecules, made from repeating subunits called monomers. The skill is to describe the common properties of plastics in terms of their structure — so again, a property with a structural reason attached.

Polymers occur naturally and are made synthetically:

Properties of plastics and their structural explanation

“Plastics” are synthetic polymers that can be moulded. Their common properties follow from their structure: very long molecules held together by covalent bonds along the chain, with intermolecular forces between the chains.

• Low density and light weight: made of light atoms (C, H, O, Cl) and the chains do not pack as closely as metal ions.
• Electrical insulators: no mobile ions or delocalized electrons.
• Durable and chemically unreactive: strong C–C and C–H covalent bonds; this is also why many plastics are not biodegradable and persist in the environment.
• Mouldable when heated (thermoplastics): heating gives the chains enough energy to overcome the intermolecular forces between chains so that they can slide past each other; the covalent bonds in the chains are not broken.
• Solids despite being molecular: each chain has a very large number of electrons, so the total London forces between long chains are large.

Properties can be tuned by changing the structure. Longer chains give stronger intermolecular forces and tougher materials (ultra-high molecular weight polyethene is used in body armour). Less branching lets chains pack closely into crystalline regions: high-density polyethene (HDPE) is harder and stronger than branched low-density polyethene (LDPE). Polar side groups add dipole–dipole forces: the C–Cl dipoles in PVC make it more rigid than polyethene. Cross-links (covalent bonds between chains) produce thermosetting polymers that do not soften on heating but decompose.

Exam focus · what the published papers show

5Addition polymers 2.4.5 SL + HL

The syllabus statement: Addition polymers form by the breaking of a double bond in each monomer. The skill is to represent the repeating unit of an addition polymer from given monomer structures. You are told that Structures of monomers do not have to be learned but will be provided or will need to be deduced from the polymer — so this outcome is entirely about drawing, in either direction.

Alkenes (and other monomers with a C=C double bond) undergo addition polymerization: the π bond of each monomer breaks and the carbon atoms form new single σ bonds to neighbouring monomers, building a long saturated chain. No other product forms, so the atom economy is 100%.

Exam focus · what the published papers show

Past-paper practice · Practice set M · Polymers and addition polymerization

No supplied IB paper asks directly about the general properties of plastics (2.4.4); the items below assess addition polymers (2.4.5).

M1IB · May 2022 · SL Paper 1 · TZ2 · Q25 · [1]

Which structure represents a repeating unit of a polymer formed from propene?

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

M2IB · November 2019 · SL Paper 1 · Q28 · [1]

M3IB · May 2017 · SL Paper 2 · TZ1 · Q5(d) · [1]

Chloroethene, C₂H₃Cl, can undergo polymerization. Draw a section of the polymer with three repeating units. [1]

M4IB · May 2018 · SL Paper 2 · TZ1 · Q3(a)(ii) · [1]

Outline the formation of polyethene from ethene by drawing three repeating units of the polymer. [1]

Solutions and mark-scheme guidance · Set M

M1 A

Propene, CH₂=CH–CH₃: the C=C opens and the CH₃ group becomes a side branch, so the repeating unit is –CH₂–CH(CH₃)–. The main chain has two carbons per unit, not three.

M2 C

Addition polymerization requires a C=C double bond outside the benzene ring. Ethylbenzene has only a saturated ethyl side chain; the ring does not undergo addition.

M3 [1]

–CH₂–CHCl–CH₂–CHCl–CH₂–CHCl– with continuation bonds at both ends ✓. All C–C bonds in the chain must be single; brackets and n are ignored; a head-to-head arrangement is also accepted.

M4 [1]

Six carbon atoms in a chain, each carrying two H atoms, with continuation lines at both ends: –CH₂–CH₂–CH₂–CH₂–CH₂–CH₂– ✓. The continuation lines must be shown.

6Condensation polymers 2.4.6 HL only

The syllabus statement: Condensation polymers form by the reaction between functional groups in each monomer with the release of a small molecule. The skill is to represent the repeating unit of polyamides and polyesters from given monomer structures, and the syllabus adds the idea that carries far beyond this topic: All biological macromolecules form by condensation reactions and break down by hydrolysis.

In condensation polymerization, each monomer has two reactive functional groups. A group on one monomer reacts with a group on the next, forming a new covalent linkage and releasing a small molecule, usually water. Because each monomer reacts at both ends, a long chain forms.

• Polyamides: a dicarboxylic acid (–COOH at both ends) and a diamine (–NH₂ at both ends) form amide links, –CO–NH–. Nylon-6,6 is made from hexanedioic acid and hexane-1,6-diamine. Kevlar is an aromatic polyamide.
• Polyesters: a dicarboxylic acid and a diol (–OH at both ends) form ester links, –CO–O–. PET is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Polar amide and ester links give condensation polymers strong intermolecular forces between chains: hydrogen bonds between the N–H of one chain and the C=O oxygen of a neighbouring chain in polyamides, and dipole–dipole forces in polyesters. This explains the high tensile strength of nylon and PET fibres. Unlike most addition polymers, the chains contain links that can be hydrolysed (split by reaction with water, catalysed by acid, base or enzymes), which is the basis of some biodegradable polymers.

Exam focus · what the published papers show

Past-paper practice · Practice set N · Condensation polymers HL only

Condensation polymers appear only in papers set under earlier guides; these items remain aligned with 2.4.6.

N1IB · May 2014 · HL Paper 1 · TZ1 · Q38 · [1]

N2IB · November 2014 · HL Paper 1 · Q38 · [1]

Which combination of monomers produces a condensation polymer with the repeating unit –[OC–C₆H₄–COOCH₂CH₂O]_n–?

A. C₆H₅COOH and HOCH₂CH₂OHB. C₆H₅COOH and CH₃CH₂OHC. C₆H₄(COOH)₂ and CH₃CH₂OHD. C₆H₄(COOH)₂ and HOCH₂CH₂OH

N3IB · May 2015 · HL Paper 1 · TZ2 · Q38 · [1]

Which pairs of compounds can react together to undergo condensation polymerization reactions?

I. HOOC–C₆H₄–COOH and C₂H₅OH
II. H₂N–(CH₂)₆–NH₂ and HOOC–(CH₂)₄–COOH
III. H₂N–CH₂–COOH and H₂N–CH(CH₃)–COOH

A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III

N4IB · May 2015 · HL Paper 2 · TZ1 · Q5(e)(i) · [2]

A polyester can be formed when ethane-1,2-diol reacts with benzene-1,4-dicarboxylic acid. Deduce the structure of the repeating unit and state the other product formed. [2]

N5IB · May 2009 · HL Paper 2 · TZ1 · Q9(c) · [5]

One form of nylon has the repeating unit:

(i) Identify the circled functional group. [1]
(ii) Deduce the structures of the two monomers used to make this form of nylon. [2]
(iii) Nylon is a condensation polymer. Another condensation polymer can be formed by reacting ethane-1,2-diol with benzene-1,4-dicarboxylic acid. Deduce the equation for this reaction using n moles of each reactant. [2]

Solutions and mark-scheme guidance · Set N

N1 B

A polyester needs two monomers with two reactive groups each: a dicarboxylic acid (benzene-1,4-dicarboxylic acid) and a diol (ethane-1,2-diol). A monofunctional alcohol (A, C) would end the chain; D has an aldehyde group rather than a second –COOH.

N2 D

The repeat unit contains –OC–C₆H₄–CO– (from benzene-1,4-dicarboxylic acid, C₆H₄(COOH)₂) and –OCH₂CH₂O– (from ethane-1,2-diol, HOCH₂CH₂OH).

N3 C

Pair I contains ethanol, which has only one –OH group, so the chain cannot grow. Pair II (diamine + dicarboxylic acid) forms a polyamide; pair III (two amino acids, each with –NH₂ and –COOH) forms a polypeptide.

N4 [2]

Repeating unit –O–CH₂–CH₂–O–CO–C₆H₄–CO– with continuation lines (condensed formulae acceptable) ✓; other product: water ✓.

N5 [5]

(i) Amide (peptide) ✓.
(ii) H₂N(CH₂)₆NH₂ ✓ and HOOC(CH₂)₈COOH (or the acyl chloride ClOC(CH₂)₈COCl) ✓.
(iii) n HOOC–C₆H₄–COOH + n HOCH₂CH₂OH → HO–[OC–C₆H₄–COO–CH₂CH₂O]_n–H + (2n − 1) H₂O. One mark for the polymer ✓ and one for (2n − 1) H₂O ✓. The water count is 2n − 1 because the two chain ends keep an H and an OH; –[OC–C₆H₄–COOCH₂CH₂O]_n– is also accepted for the polymer.

Review · Structure 2.4

7The examiner’s view, and two habits

The two halves of this topic are answered by two different habits.

8Quick check

9Summary and knowledge organiser

Essential knowledge

• Real compounds lie on a continuum: the bonding triangle uses average electronegativity and electronegativity difference to locate a substance between ionic, covalent and metallic.
• Alloys are harder and stronger than the pure metal because atoms of a different size disrupt the regular lattice and stop layers sliding.
• Addition polymers form from alkenes by opening the C=C bond, with no other product.
• HL Condensation polymers (polyesters, polyamides) form from monomers with two functional groups each, releasing a small molecule such as water.

Essential definitions and relationships

Examination checklist

• For alloys: different-sized atoms disrupt the regular arrangement and prevent layers sliding.
• Show continuation bonds on polymer sections; keep side groups off the main chain.

Knowledge organiser · from models to materials