MYP integrated science

C4.1 - Carbon’s Bonding Rules and Molecular Geometry

In the previous unit, you learned about atomic structure and how electrons are arranged in shells and subshells. In this lesson, we focus on carbon — an element with unique bonding properties that allow it to form an enormous range of compounds. Carbon is the foundation of organic chemistry because it is tetravalent: it can form four covalent bonds.

Why is carbon tetravalent?

Carbon has four valence electrons in its outer shell. To achieve stability, it shares these electrons with others, forming four covalent bonds. These bonds arrange themselves to minimise repulsion, which gives rise to predictable molecular shapes.

Activity 1: Map Pin Molecular Geometry

Map Pin Challenge

In groups, use large polystyrene spheres and map pins:

Place 2, 3, 4 and 5 pins as far apart as possible on the surface.
Identify the shapes adopted (linear, trigonal planar, tetrahedral, trigonal bipyramidal).
Estimate bond angles using geometry and trigonometry.

Extension: use coloured pins to represent non-bonding electron pairs and compare electron-domain geometry with molecular shape.

Activity 2: Introducing Molymods

Molymod Models

Explore molecular geometry further by building methane, ammonia, and water with molymods.

Compare your ball-and-stick models with space-filling representations in MolView.
Discuss: are molymods a good representation of molecules? What do they show well, and what do they miss?

Drawing Molecules

We also need a way to represent molecules on paper. Chemists use Lewis structures (dot-cross diagrams) to show bonding pairs and lone pairs of electrons. These diagrams connect the visual models with the abstract electron count.

Examples

CH₄ (Methane)

four single bonds, tetrahedral shape.

NH₃ (Ammonia)

three single bonds and one lone pair, trigonal pyramidal.

H₂O (Water)

two single bonds and two lone pairs, bent shape.

Summary

Carbon forms four covalent bonds; electron pairs arrange to minimise repulsion.
Electron-domain geometry (pairs) is not always the same as the molecular shape (atoms).
Models (pins, molymods, MolView) and Lewis structures connect counting electrons to 3D shape.

Check your understanding

Explain why carbon forms four covalent bonds (tetravalency) in terms of valence electrons.
What is the difference between electron-domain geometry and molecular shape? Give an example.
State the typical bond angle for a tetrahedral arrangement and outline (in words) why it arises.

C4.2 - Allotropes Across the Elements (with a focus on Carbon)

Many elements exist in more than one structural form called allotropes. Changes in bonding, dimensionality (0D/1D/2D/3D), and packing explain dramatic differences in properties such as hardness, conductivity, and stability.

Carbon allotropes

Model & Compare

Diamond – 3D network (each C is tetrahedral, sp³); very hard, electrical insulator.
Graphite – 2D layers (sp² trigonal planar) with delocalised electrons; soft, conducts along layers.
Graphene – single graphite layer; very strong, excellent conductor.
Fullerenes / nanotubes – curved 0D/1D cages/tubes; unusual mechanical/electronic properties.

Build small lattice fragments with molymods (diamond tetrahedra; graphite hexagonal sheet). Sketch graphene, fullerene (C₆₀), and a nanotube.

Allotropy across other elements

Compare & Explain

Oxygen: O₂ (stable) vs O₃ ozone (metastable; decomposes more readily).
Sulfur: S₈ crown rings (stable at room temperature) vs polymeric chains at high temperature.
Phosphorus: white P (P₄ tetrahedra; strained, reactive) → red P (polymeric, more stable) → black P (layered, most stable).
Tin: β-Sn (white, metallic, stable above ~13 °C) ⇄ α-Sn (grey, brittle, “tin pest”, stable below ~13 °C).

For each element, note bonding type (molecular, network, metallic), dimensionality, and a key property or use.

Case Study: Napoleon’s Buttons (Myth-busting)

The story claims tin buttons on uniforms crumbled in the 1812 Russian winter due to tin pest (β-Sn → α-Sn). Evaluate plausibility: did buttons actually contain tin? Were temperatures low enough? What alternative explanations exist?

Why are some allotropes more stable?

Stability depends on bond strength, electron delocalisation, and strain:

Stronger, more directional covalent networks (diamond) are often very stable.
Delocalised π systems (graphite/graphene) stabilise 2D layers but are weakly held between layers.
Small-ring strain (P₄) raises energy → less stable, more reactive.
Temperature/pressure can shift which allotrope is favoured (tin’s α/β transition near 13 °C).

Summary

Allotropy shows how the same element can display radically different behaviours when its atoms connect in different ways. Understanding structure (bonding and dimensionality) explains properties and stability trends across elements.

Check your understanding

Why does graphite conduct electricity while diamond does not?
What structural feature makes white phosphorus less stable than red/black phosphorus?
Why does tin undergo the α ⇄ β transition around 13 °C, and what consequence does this have in cold climates?

C4.3 Organic Nomenclature Basics (Part 1)

In this lesson you will learn the core language of organic chemistry: how to name simple molecules using roots and functional group suffixes. We focus on straight-chain compounds (C1–C6) before moving to branching and locants.

Roots and homologous series

Build the backbone

Using molymods, build methane → hexane (C1–C6). Say the names out loud as you build.
Record the general formula for the alkane homologous series.
Discuss: how does adding one CH₂ unit change properties (mass, b.p.)?

Functional group suffixes

Spot the group

Add a double bond to make an alkene (–ene). Add a triple bond for an alkyne (–yne).
Swap an H for –OH to make an alcohol (–ol). Swap an H for –Cl / –Br to make haloalkanes (chloro–, bromo–).
Extend the chain and add –COOH to make a carboxylic acid (–oic acid).

For each build, write the name and draw a condensed structural formula.

Examples

Ethane → Ethene → Ethyne; Propane → 1-propanol; Butane → Chlorobutane; Ethanoic acid.

Summary

You can now name straight-chain C1–C6 molecules and recognise common functional groups. This vocabulary prepares you for branching, locants, and isomerism next lesson.

Check your understanding

Name these: CH₃–CH₂–CH₃; CH₂=CH–CH₃; CH≡C–CH₃.
Which suffix would you use for an alcohol? For a carboxylic acid?
Give the general formula for alkanes and explain what “homologous series” means.

C4.4 Organic Nomenclature Basics (Part 2): Substitution, Locants & Branching

This lesson adds precision to your naming: numbering the chain (locants), mono/di-substitution, and simple branching. You’ll practise rapidly with a Molymod Challenge.

Locants and substitution

Number it right

Choose the longest continuous carbon chain as the parent (e.g. butane, pentane).
Number the chain to give the lowest set of locants to double/triple bonds and substituents.
Apply mono-/di- prefixes (e.g. 1-chloropropane vs 1,2-dichloroethane).

Write names for teacher-provided structures, then swap: build each other’s named molecules to check accuracy.

Branching (alkyl substituents)

Branching is when the carbon chain forks into two different chains. A branch may have as few as one carbon atom.

The longest unbroken chain provides the root of the name of the compound.

Add side chains

Introduce methyl– and ethyl– substituents. Example: 2-methylpropane, 3-methylpentane.
Alphabetise different substituents; use di-/tri- for repeats (e.g. 2,3-dimethylbutane).
Redraw neatly as condensed structural formulae after checking with molymods.

Examples

1-chloropropane; 2-chloropropane; 1,2-dichloroethane; 2-methylpropane; 2,3-dimethylbutane.

Molymod Challenge

Name ↔ Build Race

Round A (Name → Build): Teacher calls “2-methylbutane”, “1-bromopropane”, “2-butanol”. Teams build fast, then hold up.
Round B (Build → Name): Teacher shows a model; teams write the IUPAC name with correct locants.
Round C (Isomer Sprint): “Make a different isomer of C₅H₁₂.” Bonus for neat, unambiguous condensed formulas.

Summary

You can now number chains, position functional groups, and name simple branched molecules accurately. Next, you’ll measure boiling points and connect structure to intermolecular forces.

Check your understanding

Give correct names (with locants): CH₃–CH(Cl)–CH₃ and CH₃–CH₂–CH₂Cl.
Name a different isomer of C₅H₁₂ than pentane, and draw its condensed formula.
Explain what “lowest set of locants” means and apply it to 2,3-dimethylbutane.

C4.5 Experimental Determination of Boiling Points

This lesson introduces experimental techniques for measuring boiling points of small organic molecules. Students use the Siwolobov micro-method to measure methanol and ethanol. The data will later be extended to a homologous series of alkanes.

The Siwolobov micro-method

Practical setup

A small capillary tube, sealed at one end, is inverted into the liquid sample inside a thin-walled melting point tube.
The tube is heated gently in a water bath until a continuous stream of bubbles emerges.
The temperature is recorded when the last bubble escapes and the liquid begins to be drawn back.

Work in pairs, and record careful temperature readings for methanol and ethanol.

Accuracy and safety

Ensure steady heating — avoid superheating or rapid boiling.
Both methanol and ethanol are flammable; keep open flames away, use a hot water bath.
Compare results with literature values (methanol 65 °C, ethanol 78 °C).

Worked example

If a student records methanol b.p. as 63 °C, the % error compared to literature is:
(|63–65| ÷ 65) × 100 = 3.1 %.

Summary

You have learned how to use the Siwolobov micro-method to measure boiling points of small organic molecules. This technique provides accurate results with only a small volume of liquid and is safer than bulk boiling experiments.

Check your understanding

Explain why the capillary tube is sealed at one end in the Siwolobov method.
What is the literature boiling point of ethanol, and how does it compare to your measured value?
Suggest two possible sources of error when using this method to determine boiling points.

C4.6 Boiling Points of Alkanes and Intermolecular Forces

This lesson extends the boiling point investigation from PS7.5 to a homologous series of alkanes (C1–C8). Students collect literature values, plot a graph in Excel, and use the trend to introduce London dispersion forces (LDF).

Data collection

Student activity

Work in groups to collect boiling point data for alkanes from methane (C1) to octane (C8).
Enter values into Excel as a table: “Relative mass” vs. “Boiling point (°C)”.
Create a scatter plot with a best-fit line.
Ensure the graph has correctly labelled axes and a descriptive title.

Observing the trend

Boiling point increases with chain length.
This is due to stronger London dispersion forces as the number of electrons increases.
Branching reduces surface contact and lowers boiling point.

Worked example

Compare butane (C₄H₁₀, b.p. –0.5 °C) with octane (C₈H₁₈, b.p. 126 °C). Octane has twice as many carbons, more electrons, and stronger dispersion forces → higher boiling point.

Summary

You have observed how boiling points of alkanes increase with chain length and decrease with branching. This reflects the role of intermolecular London dispersion forces, which are stronger in larger, more extended molecules.

Check your understanding

Why do boiling points increase as the alkane chain length increases?
How does branching affect the boiling point of alkanes, and why?
Explain why London dispersion forces are stronger in octane than in methane.

C4.7 Isomerism and Molecular Design

In this lesson you will explore how molecules with the same molecular formula can have different structures (isomers). You will use molymods to build and compare isomers, and discover how small changes in structure affect physical properties.

Structural isomers

Isomers are compounds with the same molecular formula but different structural arrangements.
Example: Butane (C₄H₁₀) exists as n-butane (straight chain) and isobutane (branched chain).
Even though the formula is the same, their boiling points differ due to branching.

Molymod challenge

In pairs, race to build all possible isomers of C4H10 (butane) and C5H12 (pentane).
Check your models against classmates — are there duplicates or missed structures?
Discuss how branching changes surface area and affects boiling points.

Designing molecules

Chemists design molecules by altering chain length, branching, and functional groups.
This structural control allows tuning of properties like volatility, solubility, and reactivity.
Isomerism is a foundation for drug design, fuels, and materials science.

Worked example

Pentane (C₅H₁₂) has three isomers: pentane, methylbutane, and dimethylpropane. Their boiling points are 36 °C, 28 °C, and 9 °C respectively. Increased branching → lower boiling point.

Summary

You have learned that isomers share the same molecular formula but differ in structure. Branching changes physical properties such as boiling point, and molecular design is a key tool in applied chemistry.

Check your understanding

What is the difference between butane and methylpropane?
Why does dimethylpropane have a much lower boiling point than pentane?
How does isomerism provide opportunities for molecular design in industry?

C4.8 Combustion Analysis and Empirical Formula Determination

In this lesson you will perform a classic combustion analysis to determine the empirical formula of a hydrocarbon (candle wax). By measuring mass loss and the volume of carbon dioxide produced, you will calculate the simplest whole-number ratio of carbon to hydrogen in the fuel.

Practical procedure

Student experiment

Weigh a candle before burning.
Light the candle and allow it to burn under a gas jar until extinguished.
Extract 80 cm³ of the gas using a gas syringe.
Add 20 cm³ of concentrated NaOH solution into the syringe, stopper, and shake gently to absorb CO₂.
Record the new gas volume in the syringe to determine CO₂ produced.
Measure the total volume of the gas jar by filling it with water.
Reweigh the candle to calculate mass lost.

Calculations

Use the decrease in gas syringe volume to calculate moles of CO₂ in the 80 cm³ sample.
Scale up to the full gas jar volume to determine total moles of CO₂.
From moles of CO₂, deduce moles of carbon.
From mass lost by the candle, deduce moles of hydrogen (by subtraction).
Calculate the empirical formula of candle wax.

Worked example

Gas jar volume = 2.00 L (2000 cm³). After shaking the 80 cm³ sample with NaOH, the volume drops by 36 cm³ → the CO₂ fraction is 36/80 = 0.45.

Total CO₂ volume in jar = 0.45 × 2000 cm³ = 900 cm³.
Assuming molar volume ≈ 24 000 cm³ mol⁻¹ (room temperature): n(CO₂) = 900 / 24 000 = 0.0375 mol.
n(C) = n(CO₂) = 0.0375 mol → m(C) = 0.0375 × 12.0 = 0.450 g.

Candle mass loss measured = 0.520 g, which is the mass of fuel consumed = m(C) + m(H). So m(H) = 0.520 − 0.450 = 0.070 g → n(H) = 0.070 / 1.008 ≈ 0.0694 mol.

Mole ratio C:H = 0.0375 : 0.0694 → divide by 0.0375 → 1 : 1.85 ≈ 1 : 2 → empirical formula ≈ CH₂.

Summary

You have learned how to determine the empirical formula of a hydrocarbon using combustion analysis. Measuring CO₂ produced and mass lost allows you to calculate mole ratios of C and H in the fuel.

Check your understanding

Why is it necessary to absorb CO₂ in the syringe with NaOH?
How can the moles of carbon be determined from the amount of CO₂ collected?
What is the difference between an empirical formula and a molecular formula?

Now test yourself

Click on the button below to access the self-tests for MYP9 and MYP10.

MYP Self-test

Life Science Enhancement - Chemistry 4 - Carbon the element of life

Content

Scheme of work