DP Chemistry HL S2.4 From Models to Materials

DP Chemistry HL S2.4 From Models to Materials - Full notes

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DP Chemistry HL | S2.4 From Models to Materials

DP CHEMISTRY · STRUCTURE

S2.4

From Models to Materials

Full notes for HL · IB-style

Contents

1	Bonding as a continuum
2	The bonding triangle
3	Alloys
4	Polymers
5	Addition polymers
6	Plastics and the environment
7	Condensation polymers (HL only)
8	Using the bonding triangle
9	Predicting properties from bonding type
10	Composite materials
11	Monomers and addition polymers
12	Properties of addition polymers
13	Plasticizers and polymer modification
14	Condensation polymers
15	Plastics and the environment
16	Recycling and resin identification codes
17	Biodegradable and bio-based polymers
18	Nanomaterials
19	Choosing materials for a purpose
20	Mixed review: worked problems
21	Additional exam-style practice
22	Polymer calculations
23	Materials and sustainability goals
24	Common misconceptions and links
25	Exam-style question with markscheme
26	Check your understanding
27	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
explain that bonding is a continuum between ionic, covalent and metallic
use electronegativity values and the bonding triangle to deduce bonding type
explain the properties of alloys compared with pure metals
describe the structure of polymers and relate properties to structure
draw repeating units of addition polymers from monomers
evaluate the environmental impact of plastics
explain condensation polymerization to form polyesters and polyamides
deduce repeating units and monomers of condensation polymers

Sections marked HL only (Section 7) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How can bonding models be used to explain and design materials? What are the properties of alloys and polymers?

How this topic is examined

S2.4 is examined through the bonding triangle, explanations of alloy properties, and drawing polymer repeating units. HL questions include condensation polymers such as nylon and PET.

When drawing repeating units, include continuation bonds at both ends and show all substituent groups.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Bonding as a continuum

Few compounds are purely ionic or purely covalent.

Think about it

Aluminium chloride is a compound of a metal and a non-metal, yet it sublimes at 180 °C, far lower than typical ionic compounds. What could explain this?

The ionic, covalent and metallic models are useful but simplified. Real bonding often lies between these models: it is a continuum. The type of bonding depends on the electronegativity of the elements involved.

Bonding between elements with a large difference in electronegativity is mainly ionic. Bonding between elements with high, similar electronegativities is covalent. Bonding between elements with low, similar electronegativities is metallic. Many substances have intermediate character: polar covalent bonds have partial ionic character; ionic compounds with small, highly charged cations (such as AlCl3) have significant covalent character.

2 The bonding triangle

Two electronegativity values place a substance on the triangle.

The van Arkel-Ketelaar bonding triangle plots the difference in electronegativity (Δχ, vertical axis) against the average electronegativity (Σχ/2, horizontal axis) of the two elements. The corners represent ionic (top), metallic (bottom left) and covalent (bottom right) bonding.

Ionic: high Δχ (above about 1.8), for example NaCl (Δχ = 2.3, average 2.1) and CsF. Covalent: low Δχ and high average, for example Cl2, CH4. Metallic: low Δχ and low average, for example Na, Cu. Substances near boundaries have intermediate bonding: AlCl3 (Δχ = 1.6, average 2.4) is polar covalent.

Figure 1. A simplified bonding triangle.

WORKED EXAMPLE 1 Using the bonding triangle

Electronegativities: Mg 1.3, O 3.4, Si 1.9, Cl 3.2. Deduce the bonding in MgO and SiCl4.

Solution

MgO: Δχ = 2.1, average 2.35: ionicA1

SiCl4: Δχ = 1.3, average 2.55: polar covalentA1

Quick check

1. What is plotted on the axes of the bonding triangle?

2. Where would a metallic substance appear?

Answers: 1. Δχ vertically; average χ horizontally 2. bottom left: low difference and low average

3 Alloys

Mixtures that are stronger than pure metals.

An alloy is a mixture of a metal with one or more other elements (usually metals, sometimes non-metals such as carbon). Alloys have metallic bonding.

In a pure metal, all the cations are the same size and arranged in regular layers that slide easily over each other, so pure metals are relatively soft. In an alloy, atoms of different sizes disrupt the regular arrangement of layers, making it harder for layers to slide. So alloys are usually harder and stronger than the pure metal. Alloys may also be more resistant to corrosion and have lower melting points.

Alloy	Composition	Property and use
Steel	iron + carbon (0.2 to 1.5%)	strong: buildings, cars
Stainless steel	iron + chromium + nickel	corrosion resistant: cutlery, surgical tools
Brass	copper + zinc	hard, golden, corrosion resistant: instruments, fittings
Bronze	copper + tin	hard: statues, bells
Solder	tin + other metals	low melting point: joining electrical components

Quick check

1. Define alloy.

2. Why are alloys harder than pure metals?

Answers: 1. mixture of a metal with other elements 2. different-sized atoms disrupt layers so they cannot slide easily

4 Polymers

Giant molecules made of repeating units.

Polymers are very large molecules (macromolecules) made of many repeating units joined by covalent bonds, formed from small molecules called monomers. They include natural polymers (proteins, cellulose, DNA, rubber) and synthetic polymers (plastics such as poly(ethene), PVC, nylon).

Properties depend on structure. Long chains held together by intermolecular forces; longer chains and stronger forces give higher melting points and strength. Branching reduces how closely chains pack, lowering density and strength: low-density poly(ethene) (LDPE) is branched and flexible; high-density poly(ethene) (HDPE) has few branches and is stronger. Cross-links (covalent bonds between chains) make polymers rigid and heat-resistant (thermosets, which do not melt on heating); polymers without cross-links soften on heating (thermoplastics).

5 Addition polymers

Monomers with C=C double bonds join together.

Addition polymerization occurs when monomers containing a C=C double bond join together. The π bond breaks and the monomers link to form a long saturated chain. No other product is formed.

The repeating unit is drawn by: taking the two carbons of the C=C; changing the double bond to a single bond; keeping the substituent groups; and adding continuation bonds at each end, usually within square brackets with n as a subscript.

Monomer	Polymer	Use
ethene, CH2=CH2	poly(ethene)	bags, bottles
propene, CH2=CHCH3	poly(propene)	ropes, crates
chloroethene, CH2=CHCl	poly(chloroethene), PVC	pipes, window frames
tetrafluoroethene, CF2=CF2	PTFE	non-stick coatings
phenylethene, CH2=CHC6H5	poly(phenylethene), polystyrene	packaging, insulation

Figure 2. Monomers for addition polymers.

WORKED EXAMPLE 2 Drawing a repeating unit

Describe the repeating unit of poly(propene).

Solution

Two carbons joined by a single bondM1

One carbon has two H atoms; the other has one H and one CH3 groupA1

Continuation bonds at both ends: –[CH2–CH(CH3)]n–A1

Common mistake

Leaving the double bond in the repeating unit or drawing the CH3 group as part of the main chain. The chain only contains the two carbons from the C=C.

Quick check

1. What feature must a monomer have for addition polymerization?

2. Write the repeating unit of PVC.

Answers: 1. a C=C double bond 2. –[CH2–CHCl]n–

6 Plastics and the environment

Durable materials, persistent problems.

Most synthetic polymers are made from crude oil, a finite resource. Many addition polymers are non-biodegradable because their strong C–C and C–H bonds and non-polar chains are not broken down by microorganisms. They persist in landfill and the environment for hundreds of years, and break into microplastics that enter food chains (D4.2 Biology).

Solutions: reducing use; reusing; recycling (mechanical recycling requires sorting by resin identification code; chemical recycling breaks polymers into monomers); incineration with energy recovery (but releases CO2 and possibly toxic gases such as HCl from PVC); and developing biodegradable and bio-based polymers.

Quick check

1. Why are most addition polymers non-biodegradable?

2. Give one disadvantage of incinerating PVC.

Answers: 1. strong non-polar C–C and C–H bonds not broken by microorganisms 2. releases HCl and CO2

7 Condensation polymers (HL only)

Monomers with two functional groups join, releasing a small molecule.

HL only

This section is Additional Higher Level content. SL students do not need it.

Condensation polymerization occurs between monomers that each have two functional groups. Each time two monomers join, a small molecule (usually water or HCl) is eliminated.

Polyesters form from a dicarboxylic acid and a diol, joined by ester links (–COO–). PET (polyethylene terephthalate), used in drinks bottles and polyester fibres, forms from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Polyamides form from a dicarboxylic acid (or diacyl chloride) and a diamine, joined by amide links (–CONH–). Nylon-6,6 forms from hexanedioic acid and 1,6-diaminohexane. Kevlar is a polyamide with aromatic rings; hydrogen bonds between chains make it extremely strong. Proteins are natural polyamides of amino acids.

Condensation polymers can be hydrolysed back into monomers, because the ester and amide links can be broken by water (with acid or alkali), so many are more biodegradable than addition polymers.

Figure 3. Monomers for condensation polymers.

	Addition	Condensation
Monomer feature	C=C double bond	two functional groups
By-product	none	small molecule (H2O or HCl)
Linkage	C–C	ester or amide
Examples	poly(ethene), PVC	PET, nylon-6,6, Kevlar

Quick check

1. What is eliminated in condensation polymerization?

2. Which two functional groups form a polyamide?

Answers: 1. a small molecule, usually water 2. carboxylic acid (or acyl chloride) and amine

8 Using the bonding triangle

Locating a substance.

The van Arkel–Ketelaar bonding triangle plots the average electronegativity (x-axis) against the electronegativity difference (y-axis). Ionic compounds lie near the top (large difference); covalent substances at the bottom right (high average, small difference); metals at the bottom left (low average, small difference).

WORKED EXAMPLE 3 Locating (standard)

Locate MgO (Mg 1.3, O 3.4) on the bonding triangle.

Solution

Average 2.35; difference 2.1M1

Upper region: predominantly ionicA1

WORKED EXAMPLE 4 Covalent (standard)

Locate Cl2.

Solution

Average 3.2; difference 0M1

Bottom right: covalentA1

Quick check

1. Locate SiO2 (Si 1.9, O 3.4).

2. Locate an Al–Cu alloy (Al 1.6, Cu 1.9).

Answers: 1. average 2.65, difference 1.5: polar covalent 2. average 1.75, difference 0.3: metallic

9 Predicting properties from bonding type

Structure determines behaviour.

Bonding/structure	Melting point	Conductivity	Solubility in water
ionic lattice	high	molten or aqueous only	often soluble
simple molecular	low	none	polar ones soluble
giant covalent	very high	none (except graphite)	insoluble
metallic	variable, often high	solid and liquid	insoluble (some react)

WORKED EXAMPLE 5 Identifying a structure (exam level)

Substance W melts at 1610 °C, does not conduct as a solid or liquid, and is insoluble in water. Suggest its structure.

Solution

Very high melting point without conduction when moltenR1

Giant covalentA1

Quick check

1. Substance V melts at 80 °C and does not conduct. Suggest its structure.

2. Substance U conducts when solid and melts at 1085 °C. Suggest its structure.

Answers: 1. simple molecular 2. metallic

10 Composite materials

Combining materials.

A composite combines two or more materials with different properties, typically a reinforcing phase (fibres) embedded in a matrix. Examples: reinforced concrete (steel in concrete), fibreglass (glass fibres in polymer), carbon-fibre composites, and bone (collagen and calcium phosphate).

WORKED EXAMPLE 6 Reinforced concrete (standard)

Explain why steel rods are placed in concrete.

Solution

Concrete is strong in compression but weak in tensionR1

Steel is strong in tensionR1

In the real world

Carbon-fibre reinforced polymers are used in aircraft such as the Boeing 787, where they make up about half the structure by weight, reducing fuel use.

Quick check

1. Identify the matrix and reinforcement in fibreglass.

Answers: 1. matrix: polymer resin; reinforcement: glass fibres

11 Monomers and addition polymers

Opening the double bond.

Addition polymers form from alkene monomers: the C=C double bond opens and monomers join into long chains, with no other product. The repeating unit is drawn with the former double-bond carbons in brackets with extension bonds and the subscript n.

Monomer	Polymer	Use
ethene	poly(ethene)	bags, bottles
propene	poly(propene)	ropes, containers
chloroethene	poly(chloroethene), PVC	pipes, window frames
tetrafluoroethene	PTFE	non-stick coatings
phenylethene	poly(phenylethene)	packaging (polystyrene)

WORKED EXAMPLE 7 Repeating unit (standard)

Draw the repeating unit of poly(propene).

Solution

–[CH2–CH(CH3)]–nA2

Quick check

1. Identify the monomer of –[CH2–CHCl]–n.

2. What is the molar mass of a poly(ethene) chain with n = 5000?

Answers: 1. chloroethene 2. about 140 000 g mol−1

12 Properties of addition polymers

Chains and forces.

Polymer properties depend on chain length, branching and intermolecular forces. Long, unbranched chains (HDPE) pack closely, giving higher density and strength; branched chains (LDPE) pack loosely, giving flexibility. Polar side groups (Cl in PVC) increase intermolecular forces.

WORKED EXAMPLE 8 HDPE vs LDPE (exam level)

Explain why HDPE is stronger and has a higher melting point than LDPE.

Solution

HDPE chains are unbranched and pack closelyR1

Stronger London forces between chainsR1

Quick check

1. Why is PVC more rigid than poly(ethene)?

Answers: 1. polar C–Cl bonds give dipole–dipole forces between chains

13 Plasticizers and polymer modification

Tuning properties.

Adding plasticizers (small molecules) between polymer chains increases spacing, weakening intermolecular forces and making the polymer flexible (for example, PVC in cables). Cross-linking chains with covalent bonds makes polymers harder and heat-resistant (thermosets).

WORKED EXAMPLE 9 Plasticizer (standard)

Explain how a plasticizer makes PVC flexible.

Solution

Molecules fit between polymer chains, pushing them apartR1

Weaker forces allow chains to slideR1

Quick check

1. Distinguish thermoplastics and thermosets.

Answers: 1. thermoplastics soften on heating (no cross-links); thermosets do not (cross-linked)

14 Condensation polymers

A small molecule is lost.

Condensation polymers form when monomers with two functional groups join, releasing a small molecule such as water. Polyesters (diol + dicarboxylic acid, ester links) and polyamides (diamine + dicarboxylic acid, amide links) are examples. Proteins are natural polyamides.

This content is examined in detail at HL (S2.4 AHL); SL students should recognize the idea.

WORKED EXAMPLE 10 Polyamide (standard)

Identify the link and small molecule formed when hexane-1,6-diamine reacts with hexanedioic acid.

Solution

Amide link –CONH–A1

Water is releasedA1

Quick check

1. Name the type of polymer formed from ethane-1,2-diol and benzene-1,4-dicarboxylic acid.

Answers: 1. a polyester (PET)

15 Plastics and the environment

Persistence and pollution.

Most synthetic polymers are not biodegradable because microorganisms cannot break their strong C–C bonds. Disposal options: landfill (persists, uses land), incineration (energy recovery but CO2 and toxic gases such as HCl from PVC), and recycling (requires sorting by resin code).

WORKED EXAMPLE 11 Evaluating disposal (exam level)

Discuss incineration of plastic waste.

Solution

Advantage: energy recovery; reduces landfill volumeR1

Disadvantage: CO2 emissions; toxic gases need removalR1

In the real world

It is estimated that over 8 million tonnes of plastic enter the oceans each year; microplastics have been found in deep-sea sediments and in human blood.

Quick check

1. Why is PVC difficult to dispose of by incineration?

Answers: 1. forms HCl and possibly dioxins

16 Recycling and resin identification codes

Sorting plastics.

Recycling plastics requires sorting by type, because different polymers have different melting points and properties. Resin identification codes (1 PET, 2 HDPE, 3 PVC, 4 LDPE, 5 PP, 6 PS, 7 other) help sorting. Infrared spectroscopy is used in automated sorting.

WORKED EXAMPLE 12 Recycling challenges (standard)

Give two reasons why plastic recycling rates are low.

Solution

Different polymers must be separated; mixing lowers qualityR1

Contamination and low economic value of recycled plasticR1

17 Biodegradable and bio-based polymers

Greener alternatives.

Biodegradable polymers, such as polylactic acid (PLA, made from corn starch), contain bonds (esters) that microorganisms can hydrolyse. Bio-based polymers come from renewable resources but are not always biodegradable.

WORKED EXAMPLE 13 Biodegradability (standard)

Explain why polyesters can be biodegradable but poly(ethene) is not.

Solution

Ester links can be hydrolysed by enzymesR1

Poly(ethene) has only non-polar C–C and C–H bondsR1

18 Nanomaterials

Properties at the nanoscale.

Nanomaterials have structures between 1 and 100 nm. Their large surface-area-to-volume ratio gives new properties: nanoparticles of gold appear red, titanium dioxide nanoparticles are transparent yet block UV in sunscreens, and silver nanoparticles are antibacterial.

WORKED EXAMPLE 14 Surface area (standard)

Explain why nanoparticles are effective catalysts.

Solution

Very large surface area per unit massR1

More active sites availableR1

Quick check

1. Suggest one risk of nanoparticles.

Answers: 1. small enough to enter cells; unknown long-term effects

19 Choosing materials for a purpose

Properties to applications.

Materials scientists choose materials by matching properties (strength, density, conductivity, corrosion resistance, cost, recyclability) to requirements. A life-cycle assessment considers environmental impact from extraction to disposal.

WORKED EXAMPLE 15 Choosing (exam level)

Suggest a material for a kitchen saucepan base and justify.

Solution

Copper or aluminium (often stainless steel with copper base)A1

High thermal conductivity; stainless steel resists corrosionR1

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem (exam level)

Compare poly(ethene) and silicon dioxide in terms of structure, bonding and melting point.

Solution

Poly(ethene): long covalent chains, London forces between chains: low melting (130 °C)R1 R1

SiO2: giant covalent network: very high melting pointR1

Quick check

1. Use the bonding triangle to explain why AlCl3 has a low melting point for a metal chloride.

Answers: 1. intermediate difference (1.6): significant covalent character; molecular (Al2Cl6)

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 17 Paper 1 question (standard)

Which can undergo addition polymerization? A. CH4 B. C2H6 C. C3H6 D. CH3OH

Solution

C (propene has C=C)A1

WORKED EXAMPLE 18 Paper 2 question (exam level)

(a) Draw two repeating units of PTFE. (b) Explain why PTFE is used for non-stick pans. (c) State one environmental problem associated with PTFE.

Solution

(a) –CF2–CF2–CF2–CF2–A1

(b) very unreactive; low friction; heat resistantR1

22 Polymer calculations

Masses and chain lengths.

Average chain length n = M(polymer)/M(repeating unit). Percentage atom economy for addition polymerization is 100%; for condensation, it is less because a small molecule is lost.

WORKED EXAMPLE 19 Chain length (standard)

A sample of PVC has M = 62 500 g mol−1. Find n.

Solution

M(C2H3Cl) = 62.50A1

n = 1000A1

Quick check

1. Find the atom economy of making a polyester when water (M = 18) is lost for each repeating unit of M = 192 formed from monomers of total M = 210.

Answers: 1. 192/210 = 91.4%

23 Materials and sustainability goals

Green chemistry.

Green chemistry principles applied to materials include using renewable feedstocks, designing for degradation, reducing waste, and improving energy efficiency. These link materials chemistry with the UN Sustainable Development Goals.

Quick check

1. Give one green chemistry principle relevant to plastics.

Answers: 1. design for degradation (or use renewable feedstocks)

24 Common misconceptions and links

Students often think…	Actually…
bonding is either ionic or covalent	it is a continuum
alloys are compounds	alloys are mixtures
the repeating unit keeps the double bond	it becomes a single bond
condensation polymers form without by-products	a small molecule is eliminated

How S2.4 connects to the rest of the course

Topic	Connection
S2.1, S2.2, S2.3	the three bonding models
S3.2 Functional groups	alkenes, esters, amides
R3.4 Electron-pair sharing	addition reactions
S3.1 Periodic table	electronegativity trends

25 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. Brass is an alloy of copper and zinc. PVC is made by addition polymerization of chloroethene.

(a) Deduce the bonding type in KBr (K 0.8, Br 3.0) using the bonding triangle. [2]

(b) Explain why brass is harder than copper. [2]

(d) Discuss two methods of dealing with plastic waste. [4]

(e) State the by-product and linkage formed when nylon-6,6 is made. [2]

Markscheme

Part	Answer	Marks
(a)	Δχ = 2.2, average 1.9	M1
	ionic	A1
(b)	zinc atoms different size	R1
	disrupt layers so they cannot slide	R1
(c)	two-carbon chain with one Cl	A1
	continuation bonds, brackets and n	A1
(d)	recycling saves resources but requires sorting	R1 R1
	incineration recovers energy but releases CO2 / HCl	R1 R1
(e)	water	A1
	amide link (–CONH–)	A1

Examiner's comments. In (c), the most common error was leaving a double bond in the repeating unit.

26 Check your understanding

Section A: multiple choice

1. An alloy is

A. a mixture of a metal with other elements B. a compound C. a pure metal D. a polymer

2. In the bonding triangle, the vertical axis is

A. average electronegativity B. atomic number C. electronegativity difference D. melting point

3. Addition polymers form from monomers with

A. two OH groups B. C=C bonds C. amine groups D. rings only

4. Branching in poly(ethene)

A. increases density B. adds cross-links C. makes it a thermoset D. lowers density

5. The monomer of PTFE is

A. CH2=CHCl B. CF2=CF2 C. CH2=CH2 D. CH2=CHCH3

6. Steel is iron alloyed with

A. carbon B. copper C. tin D. zinc

7. PET is a

A. polyamide B. addition polymer C. protein D. polyester

Section B: structured

8. Explain why bonding is described as a continuum. [3]

9. Compare the properties of LDPE and HDPE. [3]

10. Evaluate the use of plastics. [4]

11. Compare addition and condensation polymerization. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D

8. Electronegativity differences vary continuously; partial ionic character; triangle positions.

9. LDPE branched, packs loosely, flexible, low density; HDPE unbranched, packs closely, stronger, higher density.

10. Cheap, durable, light, versatile; but finite resources, non-biodegradable, microplastics, disposal problems.

11. Addition: C=C monomers, no by-product; condensation: two functional groups, water eliminated, ester/amide links.

27 Summary and topic map

Idea	Remember
Continuum	ionic, covalent, metallic as extremes
Bonding triangle	Δχ vs average χ
Alloys	different-sized atoms stop layers sliding
Polymers	chains; branching; cross-links
Addition polymers	C=C monomers; repeating units
Plastics	non-biodegradable; recycling, incineration
HL	condensation polymers: polyesters, polyamides

Before the exam, make sure you…

•use both axes of the bonding triangle
•explain alloy hardness with layers
•draw repeating units with continuation bonds
•evaluate plastics both ways

From Models to Materials on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for From Models to Materials.

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