S2.4 From models to materials | IB DP Chemistry

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🎯What you need to be able to do

Use bonding models to explain the properties of a material, and describe bonding as a continuum between the ionic, covalent and metallic models.
Determine the position of a binary compound in the bonding triangle from electronegativity data, and predict its properties from that position.
Explain the properties of alloys in terms of non-directional bonding, and explain why alloys are mixtures rather than compounds.
Describe polymers as macromolecules made from repeating monomers, and describe the common properties of plastics in terms of their structure.
Represent the repeating unit of an addition polymer from a given monomer, and deduce the monomer from a polymer.
AHL Represent the repeating unit of polyamides and polyesters formed by condensation polymerization.

📚The chemistry

Bonding is a continuum

Structure 2 has presented three separate models — ionic, covalent, metallic. The honest position, and the one the syllabus takes, is that bonding is best described as a continuum between them, and the three “types” are the idealised corners of a space in which real substances lie somewhere in between.

Evidence for this is everywhere once you look. Aluminium chloride has an ionic formula but sublimes at 180 °C and dissolves in organic solvents — it has substantial covalent character, because the small, highly charged \( \mathrm{Al^{3+}} \) ion distorts the electron cloud of the chloride ion. Silicon is classified as a metalloid because it is genuinely intermediate between a covalent network and a metal. Describing a substance as “100% ionic” is almost never true.

The bonding triangle

The triangular bonding diagram (in the data booklet) places a binary compound using two quantities calculated from electronegativity values:

\( \Delta\chi \) — the difference in electronegativity between the two elements. Plotted on the y-axis. It measures how ionic the bonding is.

\( \chi_{\text{av}} \) — the average electronegativity of the two elements. Plotted on the x-axis. Low average means metallic, high average means covalent.

The three corners are then: top (large difference) ionic; bottom right (small difference, high average) covalent; bottom left (small difference, low average) metallic. Everything else sits somewhere between, and the regions shade into one another rather than having hard borders.

Two things the guide states explicitly: only binary compounds need be considered, and calculations of percentage ionic character are not required. You read a position off the diagram; you do not compute a percentage.

Having placed a compound, predict its properties from the region it falls in — high melting point and conduction only when molten or aqueous near the ionic corner; low melting point and no conduction in the molecular covalent region; conduction as a solid, malleability and lustre in the metallic region; very high melting point and no conduction for a covalent network.

Alloys

An alloy is a mixture of a metal with other metals or non-metals. It is a mixture and not a compound because the components are present in no fixed ratio, are not chemically bonded to one another in a fixed way, and the composition can be varied continuously to tune the properties.

Alloys form readily because metallic bonding is non-directional: the electron sea does not care very much which cations it is surrounding, so an atom of a different size or charge can simply take a place in the lattice.

Alloys usually have enhanced properties compared with the pure metal, and the standard explanation is worth learning: the added atoms are of a different size, so they disrupt the regular layers of the lattice and make it harder for layers to slide over one another. The alloy is therefore harder and stronger, and less malleable, than the pure metal. Bronze (copper and tin), brass (copper and zinc) and stainless steel (iron, chromium and nickel) are the standard examples; the syllabus says specific examples do not have to be learned.

Polymers

A polymer is a very large molecule, a macromolecule, built from many repeating sub-units called monomers. Natural polymers include starch, cellulose, proteins and DNA; synthetic ones include poly(ethene), PVC, nylon and PET.

The common properties of plastics follow from that structure. The chains are long, so London (dispersion) forces between them are cumulatively substantial even though each individual interaction is weak — which is why a material made only of hydrocarbon chains can be a solid at all. Because the chains can slide and uncoil, plastics are flexible and can be moulded; because there are no ions and no delocalized electrons, they are electrical insulators; because the chains are non-polar, most are unreactive and not biodegradable, which is precisely the environmental problem. Cross-linking between chains gives a rigid thermoset that cannot be remelted.

Addition polymers

An addition polymer forms when the double bond in each monomer breaks and the monomers join end to end. No other product is formed, so the atom economy is 100% — a point Reactivity 2.1 returns to.

To draw the repeating unit from a monomer: open the double bond, draw the resulting two-carbon backbone with all its substituents, put a bond extending from each end through the brackets, and write \( n \) outside:

\[ n\ \mathrm{CH_2{=}CHCl} \;\longrightarrow\; \mathrm{-\!\!\left[CH_2{-}CHCl\right]\!\!-}_n \]

To go the other way — deduce the monomer from a polymer — find the repeating unit, take two backbone carbons, remove the bonds through the brackets and put the double bond back. The syllabus says monomer structures do not have to be learned but must be deducible, so practise the reverse direction, which is the one that appears in exams.

Leaving the double bond in the repeating unit. The whole point of addition polymerization is that the \( \pi \) bond is used up joining the monomers together. A repeating unit drawn with C=C inside it is wrong and scores nothing. The second half of the same mistake is forgetting the bonds extending through the brackets, which are what show the unit continues in both directions.

AHL Condensation polymers

A condensation polymer forms when functional groups on the monomers react, joining them and releasing a small molecule — usually water, sometimes HCl. Unlike addition polymerization, the atom economy is therefore below 100%.

Each monomer must carry two reactive functional groups, one at each end, or the chain cannot grow.

Polyester
diol + dicarboxylic acid
links by an ester group, –COO–
releases \( \mathrm{H_2O} \) · e.g. PET

Polyamide
diamine + dicarboxylic acid
links by an amide group, –CONH–
releases \( \mathrm{H_2O} \) · e.g. nylon

The syllabus makes the biological connection explicitly: all biological macromolecules form by condensation reactions and break down by hydrolysis. Proteins are polyamides of amino acids, joined by peptide (amide) links; polysaccharides are condensation polymers of sugars. Reversing the reaction with water — hydrolysis — is what digestion does, and it is also why polyesters and polyamides are more readily broken down than poly(ethene): the ester and amide links are polar and attackable, while a saturated carbon chain is not.

✏️Worked example

Electronegativity values: Na 0.9, Mg 1.3, Al 1.6, Si 1.9, Cl 3.2, O 3.4.
(a) Calculate \( \Delta\chi \) and \( \chi_{\text{av}} \) for NaCl, \( \mathrm{SiO_2} \) and an Mg–Al alloy, and state which region of the bonding triangle each falls in.
(b) Predict, for each, whether it conducts electricity as a solid and whether it has a high or low melting point.
(c) \( \mathrm{AlCl_3} \) has \( \Delta\chi = 1.6 \) and sublimes at 180 °C. Comment on what this suggests about its bonding.
(d) Deduce the monomer of the polymer whose repeating unit is –[CH₂–CH(CH₃)]–_n, and state the atom economy of its formation.

(a) Take the difference and the mean of the two electronegativities in each case.

NaCl: \( \Delta\chi = 3.2 - 0.9 = 2.3 \); \( \chi_{\text{av}} = \dfrac{3.2 + 0.9}{2} = 2.05 \). A large difference puts it at the top of the triangle: the ionic region.
SiO₂: \( \Delta\chi = 3.4 - 1.9 = 1.5 \); \( \chi_{\text{av}} = \dfrac{3.4 + 1.9}{2} = 2.65 \). A moderate difference with a high average puts it on the right, in the covalent region — but well up towards the middle, consistent with a polar covalent network rather than a purely covalent molecule.
Mg–Al: \( \Delta\chi = 1.6 - 1.3 = 0.3 \); \( \chi_{\text{av}} = \dfrac{1.6 + 1.3}{2} = 1.45 \). A very small difference with a low average puts it at the bottom left: the metallic region.

(b)

NaCl — does not conduct as a solid (ions fixed in the lattice), conducts when molten or aqueous; high melting point, because a giant ionic lattice must be broken down.
SiO₂ — does not conduct in any state (no ions, no delocalized electrons); very high melting point, because it is a giant covalent network in which strong covalent bonds must be broken.
Mg–Al alloy — conducts as a solid, because of delocalized electrons; high melting point, and it will be harder and less malleable than either pure metal, because the differing atomic sizes disrupt the regular layers and prevent them sliding.

(c) An electronegativity difference of 1.6 would ordinarily place \( \mathrm{AlCl_3} \) near the boundary of the ionic region, and a truly ionic chloride would have a high melting point and conduct when molten. Instead it sublimes at only 180 °C, which is behaviour characteristic of a simple molecular covalent substance held by weak intermolecular forces. This tells us the bonding has substantial covalent character: the \( \mathrm{Al^{3+}} \) ion is small and highly charged, so it polarises the large chloride ion, drawing electron density back between the nuclei. It is a direct illustration of the continuum — the compound sits between the ionic and covalent corners rather than at either.

(d) The repeating unit has a two-carbon backbone, \( \mathrm{-CH_2-CH(CH_3)-} \). Remove the bonds passing through the brackets and restore the double bond between those two carbons: the monomer is \( \mathrm{CH_2{=}CH(CH_3)} \), propene, and the polymer is poly(propene). Because this is an addition polymerization, the monomer is the only reactant and the polymer is the only product, so every atom of the reactant appears in the desired product and the atom economy is 100%.

Check it. For the triangle, remember that \( \Delta\chi \) can never exceed the range of the electronegativity scale and \( \chi_{\text{av}} \) must lie between the two values you averaged — if either fails, you have subtracted where you should have added. As a coarse guide, \( \Delta\chi \) above about 1.8 is usually ionic and below about 0.4 with a low average is metallic, but treat those as regions and not thresholds, which is the whole point of (c). For the polymer, check the monomer by counting: propene is \( \mathrm{C_3H_6} \) and the repeating unit is also \( \mathrm{C_3H_6} \) — in an addition polymer the two formulas must match exactly.

Treating the bonding triangle as three boxes with hard edges. Students calculate \( \Delta\chi = 1.6 \) for \( \mathrm{AlCl_3} \), announce “ionic”, and then cannot explain why it sublimes below 200 °C. The triangle exists to show that the categories shade into one another; a compound near a boundary should be described as having “significant covalent character” or “polar covalent”, and the physical evidence — melting point, conductivity, solubility — outranks the calculation whenever the two disagree.

📝Practise

Work through these on paper, then reveal the answer.

1. Explain why alloys are correctly described as mixtures rather than as compounds.

A compound contains its elements chemically bonded in a fixed ratio, has a fixed composition by mass, and can only be separated chemically. An alloy satisfies none of these. Its components are present in no fixed ratio — brass can be made with anything from about 5% to 45% zinc, and the properties vary continuously as the proportion changes. The added atoms simply occupy positions in the metallic lattice; there is no new chemical bond of fixed stoichiometry between copper and zinc, only the same non-directional metallic bonding between all the cations and the shared electron sea. And the properties of an alloy are intermediate and predictable from those of its components, as one expects for a mixture, rather than being unrelated to them as a compound’s are. The reason mixing works so readily is precisely that metallic bonding is non-directional, so it does not require a particular partner or geometry.

2. Using electronegativity values (Li 1.0, F 4.0, C 2.6, H 2.2, Cu 1.9, Zn 1.7), calculate \( \Delta\chi \) and \( \chi_{\text{av}} \) for LiF, CH₄ and a Cu–Zn alloy, and assign each to a region of the bonding triangle.

LiF: \( \Delta\chi = 4.0 - 1.0 = 3.0 \), \( \chi_{\text{av}} = 2.5 \). The largest possible difference on this data set, so it sits at the top of the triangle: strongly ionic. Predicted properties: very high melting point (it is 845 °C), non-conducting as a solid but conducting when molten, soluble in polar solvents. CH₄: \( \Delta\chi = 2.6 - 2.2 = 0.4 \), \( \chi_{\text{av}} = 2.4 \). Small difference, high average → bottom right, the covalent region; and since it is a small discrete molecule rather than a network, it is a gas with a very low boiling point. Cu–Zn: \( \Delta\chi = 1.9 - 1.7 = 0.2 \), \( \chi_{\text{av}} = 1.8 \). Small difference, low average → bottom left, the metallic region; conducts as a solid, malleable, and harder than either pure metal.

3. Draw the repeating unit of the addition polymer formed from (a) ethene, (b) chloroethene, (c) tetrafluoroethene, and name each polymer.

In every case the C=C opens and the two carbons become the backbone of the repeating unit, with bonds extending through the brackets at each end. (a) From \( \mathrm{CH_2{=}CH_2} \): –[CH₂–CH₂]–_n, poly(ethene). (b) From \( \mathrm{CH_2{=}CHCl} \): –[CH₂–CHCl]–_n, poly(chloroethene), better known as PVC. (c) From \( \mathrm{CF_2{=}CF_2} \): –[CF₂–CF₂]–_n, poly(tetrafluoroethene) or PTFE. In each the empirical formula of the repeating unit is identical to that of the monomer, confirming 100% atom economy. PTFE is a useful case to think about: the very strong C–F bonds and the non-polar surface make it extremely unreactive and give it the low friction it is used for.

4. Explain, in terms of structure and bonding, why poly(ethene) is flexible and does not conduct electricity, and why it is not biodegradable.

Flexible: poly(ethene) consists of long non-polar hydrocarbon chains held to one another only by London (dispersion) forces. These are weak individually, though substantial in total because the chains are long, so the chains can slide and uncoil past one another when a force is applied, and the material deforms and can be moulded rather than shattering. Non-conducting: conduction requires charged particles that are free to move. Poly(ethene) contains no ions and no delocalized electrons — every valence electron is localised in a C–C or C–H \( \sigma \) bond — so there are no mobile charge carriers and it is an insulator. Not biodegradable: the chain is a saturated hydrocarbon with strong, essentially non-polar C–C and C–H bonds. There is no \( \delta+ \) site for a nucleophile or an enzyme to attack, and no functional group that can be hydrolysed, so no organism has enzymes that recognise it. Contrast a polyester or polyamide, whose polar ester and amide links can be hydrolysed.

5. AHL Ethane-1,2-diol reacts with benzene-1,4-dicarboxylic acid to form PET. Identify the type of polymerization, the link formed, the small molecule released, and explain why each monomer must have two functional groups.

This is condensation polymerization. The –OH group of the diol reacts with the –COOH group of the dicarboxylic acid to form an ester link, –COO–, releasing a molecule of water. The product is therefore a polyester. Each monomer must carry two functional groups — two –OH on the diol, two –COOH on the diacid — because a molecule with only one reactive group would form a single ester link and then have no site left to react further, terminating the chain. With two groups apiece, every molecule added to the chain still presents a reactive end, so the polymer can grow indefinitely in both directions. This is also why the atom economy is below 100%: a molecule of water is lost at every link, so some of the reactant mass ends up in an unwanted product.

6. Compare addition and condensation polymerization under four headings: the monomer required, the bond broken or formed, the by-product, and the atom economy. Give one example of each.

Monomer: addition requires a monomer containing a C=C double bond (an alkene or a substituted alkene); condensation requires monomers each carrying two functional groups such as –OH, –COOH or –NH₂. Bond change: in addition, the \( \pi \) bond of the C=C breaks and a new C–C \( \sigma \) bond forms between monomers; in condensation, a new ester or amide link is formed between the functional groups. By-product: addition produces none — the polymer is the only product; condensation releases a small molecule at every link, usually water. Atom economy: addition is 100%; condensation is less than 100%, since the mass of the released water is not in the desired product. Examples: ethene → poly(ethene) for addition; a diamine plus a dicarboxylic acid → nylon, or a diol plus a dicarboxylic acid → a polyester, for condensation. A useful consequence: condensation polymers can be hydrolysed back to their monomers by reversing the reaction with water, which addition polymers cannot.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Your data booklet — the triangular bonding diagram and the electronegativity table are on facing pages for a reason. Plot half a dozen familiar compounds on it now, before you need to do it under time pressure.
RSC Learn Chemistry — classroom preparations of nylon (the “nylon rope trick”) and of a simple polyester, which make the condensation mechanism physical rather than theoretical.
The RSC and IUPAC material on plastics and sustainability, useful background for a collaborative sciences project or an Extended Essay on polymer degradation.