# Bonding continuum and materials (polymers)

> IB Chemistry Higher Level · IB Chem HL 2023+
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u2-bonding-continuum-and-materials/

This module maps the full bonding continuum across ionic, covalent, and metallic regimes, then connects intermediate bonding behaviour to the structure, properties, and real-world uses of synthetic polymers.

**Prerequisites:** [Understanding of ionic, covalent and metallic primary bonding types](https://www.owlsprep.com/study/ib-chemistry-hl-u2-primary-bonding-types/); [Knowledge of intermolecular forces (van der Waals, hydrogen bonding)](https://www.owlsprep.com/study/ib-chemistry-hl-u2-intermolecular-forces/)

## Learning objectives

- Explain the bonding continuum between ionic, covalent, and metallic bonding with no sharp dividing lines
- Relate addition and condensation polymer structure to physical properties like tensile strength and melting point
- Predict the effect of plasticizer and cross-linking on polymer mechanical characteristics
- Evaluate real-world polymer applications based on bonding and structural data

## The Bonding Continuum Model

The introductory model of discrete, mutually exclusive bonding types is a simplification. In reality, all bonding exists on a continuous spectrum, with most real compounds exhibiting mixed character that does not fit neatly into a single category. Fajans' rules quantify this by describing how high charge density cations can polarize anions to introduce partial covalent character to otherwise ionic compounds.

**Bonding Continuum** — A qualitative and quantitative scale that describes bonding as a continuous spectrum from 100% ionic, through partially polar covalent, to 100% non-polar covalent and metallic, with no sharp dividing lines between categories.

**Worked example:** Classify the bonding character of tin(IV) chloride, SnCl₄, given its electronegativity difference ΔEN = 1.3, and melting point of -33°C

1. First recall that the old discrete rule classifies ΔEN > 1.8 as ionic, 0.5-1.8 as polar covalent, but the continuum accounts for additional factors like cation polarizing power.
2. Sn⁴⁺ has very high charge density, so it strongly polarizes Cl⁻ anions, introducing significant covalent character that overrides the ionic classification.
3. The low melting point confirms no extended ionic lattice exists, so SnCl₄ sits far on the covalent side of the continuum, despite its relatively high electronegativity difference.
4. Final classification: predominantly covalent, with negligible ionic behaviour.

> **info**
>
> Virtually no real-world compound sits at the extreme 100% ionic or 100% covalent ends of the continuum.

> **Exam tip:** Always reference Fajans' rules when justifying a compound's position on the bonding continuum for 2+ mark questions to get full marks.

## Polymer Chain Structure and Primary Bonding

Addition and condensation polymers are built from covalently bonded monomer units, forming long linear chains with a strong covalent backbone. The length of these chains is defined by the degree of polymerization, which directly impacts bulk material properties.

**Degree of Polymerization** — The total number of repeating monomer units covalently linked in a single polymer chain.

*Notation:* $n_{DP}$

$$M_{\text{polymer}} = n_{DP} \times M_{\text{monomer}}$$

**Worked example:** Calculate the approximate molar mass of a polyethylene chain with a degree of polymerization of 2500, given ethene monomer molar mass is 28.0 g mol⁻¹.

1. Identify the given values: $n_{DP} = 2500$, $M_{\text{monomer}} = 28.0$ g mol⁻¹.
2. Substitute into the polymer molar mass formula: $M_{\text{polymer}} = 2500 \times 28.0$ g mol⁻¹.
3. Compute the final result: 70 000 g mol⁻¹, or 70 kg mol⁻¹.

**Check your understanding**

Test your understanding of basic polymer structure

1. Which of the following is an addition polymer?

   - Nylon 6,6
   - Polyethylene terephthalate (PET)
   - Polytetrafluoroethene (PTFE)
   - Urea-formaldehyde resin

   *Why:* PTFE forms from addition polymerization of tetrafluoroethene monomers, with no small elimination product during synthesis.

## Intermolecular Forces and Polymer Bulk Properties

The physical properties of non-crosslinked thermoplastics are almost entirely determined by the strength of intermolecular forces between adjacent polymer chains, not the strong covalent bonds in the polymer backbone. Polar side groups on the chain can drastically increase intermolecular attraction, raising melting point and tensile strength.

> **Polymer Melting Point Mnemonic**
>
> SSS: Side group polarity, chain Stiffness, chain Size (degree of polymerization) are the three factors that increase polymer melting point.

**Worked example:** Explain why polyvinyl chloride (PVC, -CH₂-CHCl- repeating unit) has a much higher melting point than polyethylene (PE, -CH₂-CH₂- repeating unit) of the same degree of polymerization.

1. Compare the side groups on each polymer backbone: PE has only non-polar H atoms, PVC has electronegative Cl atoms on every second carbon.
2. The C-Cl bond is polar, so permanent dipole-dipole interactions form between adjacent PVC chains, while PE only has weak London dispersion forces.
3. More energy is required to overcome the stronger intermolecular forces in PVC, leading to a significantly higher melting temperature.

**Exam command terms**

IB exam questions on polymer properties often use these command terms with specific expectations:

- **Deduce** — You must use given structural information to predict a property, a memorized list alone will not earn full marks

- **Evaluate** — You must explicitly link bonding characteristics directly to a real-world application of the polymer to score maximum points

## Modifying Polymer Properties: Cross-Linking and Plasticizers

Polymer properties can be tuned for specific industrial uses via two common modification strategies: cross-linking and plasticizer addition. These two interventions have opposite effects on chain mobility and bulk material behaviour.

**Worked example:** Predict and explain how adding 10% by mass of a plasticizer to rigid PVC (used for water pipes) changes its flexibility and melting point.

1. Recall the function of a plasticizer: small, non-volatile molecules insert between adjacent PVC polymer chains.
2. The plasticizer molecules separate the chains, reducing the strength of dipole-dipole interactions between them.
3. Less force is required to make chains slide past each other, so flexibility increases, and the melting point / glass transition temperature decreases. This is how flexible PVC products like shower curtains are manufactured.

## Common pitfalls

- **Wrong:** Stating that a compound is '100% ionic' or '100% covalent'
  - Why it fails: Virtually all real bonds have some mixed character, no substance sits exactly at the extreme end of the bonding continuum.
  - Correct: Describe the dominant bonding character, and note any partial covalent/ionic contributions from Fajans' rules.
- **Wrong:** Claiming polymer melting points depend on breaking strong covalent backbone bonds
  - Why it fails: Thermoplastics melt when intermolecular forces between chains are overcome, the covalent backbone bonds remain fully intact.
  - Correct: Explicitly state that only intermolecular forces are disrupted during melting of non-crosslinked thermoplastics.
- **Wrong:** Confusing cross-linking effect with plasticizer effect
  - Why it fails: Cross-linking increases rigidity and melting point, while plasticizers decrease both, they have exactly opposite effects.
  - Correct: Memorize that cross-links are covalent connections between chains, plasticizers are spacers that reduce chain interaction.
- **Wrong:** Classifying all polymers as covalent network solids
  - Why it fails: Most thermoplastics are made of separate long polymer chains held by intermolecular forces, not a single continuous covalent network.
  - Correct: Reserve 'covalent network' classification only for fully crosslinked thermoset polymers or giant covalent substances like diamond.
- **Wrong:** Forgetting to reference electronegativity difference when justifying position on the bonding continuum
  - Why it fails: IB exam markers require explicit use of electronegativity or Fajans' rules to support continuum claims, not arbitrary classification.
  - Correct: Always cite ΔEN or cation/anion charge density to justify your placement of a compound on the bonding spectrum.

## Cheatsheet

| Factor | Effect on polymer melting point | Effect on tensile strength |
| --- | --- | --- |
| Increasing degree of polymerization | Increase | Increase |
| Polar side groups on chain | Significant increase | Significant increase |
| Cross-linking between chains | Large increase (no melting for high cross-link) | Large increase |
| Added plasticizer | Decrease | Decrease |

## What's next

You now have a full framework to map bonding character across the entire spectrum, and connect polymer structure to real-world material properties. This knowledge will be critical for upcoming topics on nanotechnology, environmental impact of polymer waste, and organic reaction pathways for monomer synthesis. You will also use the bonding continuum model to explain unexpected properties of ionic compounds like aluminium oxide in the acids and bases unit. Mastering these links will help you score maximum marks on the Paper 2 long answer questions that combine bonding, structure and material applications.

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