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Giant Covalent Structures (Diamond, Graphite, Silicon Dioxide)

Giant covalent structures: link structure → property → use for diamond, graphite, and silicon dioxide (conductivity, hardness, melting point).

  • SEC G3 Pure Chemistry 2027
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Learning objectives

  • compare the structures of the following substances in order to deduce their properties: — simple molecular substances, e.g. methane, iodine
  • compare the structures of the following substances in order to deduce their properties: — macromolecules, e.g. poly(ethene)
  • compare the structures of the following substances in order to deduce their properties: — giant covalent substances, e.g. sand (silicon dioxide), diamond, graphite (see also 3.4(g))
  • compare the bonding and structures of diamond and graphite in order to deduce their properties such as electrical conductivity, lubricating or cutting action (candidates will not be required to draw the structures)
  • deduce the physical and chemical properties of substances from their structures and bonding and vice versa (see also 3.1(d), 3.2(d), 3.3(b) and 3.4(e)).

Giant covalent structure questions are comparison questions. If you cannot link structure → property → use, you will not score.

1. Definition

Giant covalent structures are structures with many atoms joined by strong covalent bonds in a giant network.

Common O-Level examples: diamond (C), graphite (C), and silicon(IV) oxide (SiO₂).

Syllabus wording: “macromolecules”

In the syllabus, macromolecules often refers to polymers like poly(ethene), not diamond/graphite. If you need polymers, see Polymers. For side-by-side bonding comparisons, revise Covalent Bonds before you attempt structure-property questions.

2. Key Ideas

  • Giant covalent structures have very high melting points because many strong covalent bonds must be broken.
  • Diamond: each carbon bonds to 4 (tetrahedral) → rigid 3D network → very hard; does not conduct.
  • Graphite: each carbon bonds to 3 (layers) + 1 delocalised electron → conducts; layers slide → soft.
  • Silicon(IV) oxide (SiO₂): giant covalent network → hard and high melting point; does not conduct.

3. Detailed Explanations

Quick Recall (structure → property)
  • Giant covalent = many strong covalent bonds in a giant network → very high melting point.
  • Diamond: 4 bonds per carbon → rigid 3D network → hard, does not conduct.
  • Graphite: 3 bonds per carbon + delocalised electrons → conducts; weak forces between layers → soft.

A. Diamond

In diamond, each carbon atom is covalently bonded to four other carbon atoms in a rigid tetrahedral arrangement.

Diamond giant covalent structureDiamond: every carbon atom forms four strong covalent bonds in a rigid three-dimensional giant network. It has no mobile charged particles.Diamond: four bonds from each carbonrigid 3D networkstrong covalent bonds4 bonds per C atomno mobile electronsdoes not conduct electricity
Diamond: every carbon atom forms four strong covalent bonds in a rigid three-dimensional giant network. It has no mobile charged particles.
  • Hardness: rigid 3D lattice; atoms cannot slide because bonds hold the network together.
  • Melting point: very high because many strong covalent bonds must be broken.
  • Electrical conductivity: does not conduct because there are no delocalised electrons (all 4 valence electrons are used in bonds).

B. Graphite

In graphite, each carbon atom is covalently bonded to three other carbon atoms, forming hexagonal layers.

Graphite layered giant covalent structureGraphite: every carbon atom bonds to three others in hexagonal layers. Weak forces act between layers, while one delocalised electron per carbon can move within the layers.Graphite: hexagonal carbon layerse⁻e⁻e⁻e⁻each C bonds to 3 othersdelocalised electronsmove within each layerweak forces between layerslayers slide → soft and slipperymobile electrons → conducts
Graphite: every carbon atom bonds to three others in hexagonal layers. Weak forces act between layers, while one delocalised electron per carbon can move within the layers.
  • Softness: layers are held together by weak forces between layers, so layers can slide over each other.
  • Electrical conductivity: the 4th electron is delocalised and can move along the layers, carrying charge.
  • Melting point: very high because covalent bonds within layers are strong and numerous.
Key term: weak forces between layers

Graphite is not made of separate molecules. The weak forces are between layers of the giant structure (often called Van der Waals forces).

C. Silicon(IV) Oxide (SiO₂)

Silica is the main component of sand. Each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. This results in a giant tetrahedral structure similar to diamond.

Silicon dioxide network diagramA two-dimensional connectivity map of the three-dimensional silicon dioxide network. Each silicon has four oxygen neighbours, each interior oxygen bridges two silicons, and dashed ends continue beyond the fragment. Bond angles are not represented.Silicon(IV) oxide giant covalent networkSiSiSiSiSiSiBlue circles: O. Dashed bonds continue beyond this fragment.Connectivity only; tetrahedral bond angles are not shown.How to write the explanationGiant covalent network- Each Si bonds to 4 O atoms.- Each O bridges 2 Si atoms.Strong Si–O bonds throughoutTherefore:- very high melting point- hard structure- does not conduct electricityNo mobile charged particles
Silicon dioxide giant covalent network: each silicon bonds to four oxygens and each oxygen bridges two silicon atoms, giving a rigid high-melting non-conducting structure.

D. Comparison Table (Diamond vs Graphite vs Silicon(IV) oxide)

FeatureDiamond (C)Graphite (C)Silicon(IV) oxide (SiO₂)
StructureGiant 3D tetrahedral latticeGiant layers of hexagonsGiant network (each Si to 4 O; each O to 2 Si)
HardnessVery hardSoft/slipperyHard
Electrical conductivityDoes not conductConductsDoes not conduct
Typical usesCutting tools, jewelleryLubricant, pencil leads, electrodesSand; glass/ceramics (as silica)

E. Syllabus Comparison: Simple Molecular vs Polymers vs Giant Covalent

In structure questions, the syllabus often wants you to compare three structure types, not just “simple vs giant”.

Structure typeExampleWhat holds it together?Melting/boiling pointElectrical conductivity
Simple molecularmethane (CH₄), iodine (I₂)strong covalent bonds within molecules; weak intermolecular forces between moleculesusually lowdoes not conduct
Polymer (macromolecule)poly(ethene)strong covalent bonds along the chain; weak forces between chainsoften high (softens over a range)does not conduct
Giant covalentdiamond, graphite, SiO₂many strong covalent bonds in a giant networkvery highgraphite conducts; most others do not

4. Common Mistakes

  • Calling these “large molecules” and then using intermolecular-force explanations for melting points. These are giant covalent networks.
  • Saying graphite is soft because “covalent bonds are weak”. Wrong: covalent bonds are strong; weak forces between layers allow sliding.
  • Saying diamond conducts because it is carbon (wrong). Diamond has no delocalised electrons.
  • Forgetting to use the keyword delocalised electrons for graphite conductivity.
  • Mixing up “silicon” and “silicon(IV) oxide” (different substances with different formulas).

5. Exam Tips

One sentence that scores for melting point

“Very high melting point because many strong covalent bonds must be broken.”

Graphite conductivity line

“One electron per carbon is delocalised and can move along the layers to carry charge.”

6. Worked Examples

Modelled example 1

Lubrication (Structure → Property → Use)

Core

Problem

Explain why graphite is a suitable lubricant for high-temperature machinery, whereas a simple molecular oil may not be.
Study the worked solution

  1. Explain graphite's slipperiness

    Method

    Identify its layered giant structure.

    Reason

    Weak forces between layers allow the layers to slide over one another.

    Working

    Layered structure → sliding layers → lubricating action.

  2. Explain high-temperature stability

    Method

    Identify many strong covalent bonds within the layers.

    Reason

    Much energy is required to break enough of these bonds for graphite to melt.

    Working

    Giant covalent bonding → very high melting point.

  3. Compare with oil

    Method

    Contrast the giant structure with simple molecules.

    Reason

    A simple molecular oil is more likely to evaporate or decompose at high operating temperatures.

    Working

    Graphite retains its useful solid, sliding-layer structure under conditions where the oil may not.

Guided practice 2

Melting Point (Network Explanation)

About 5 min

Problem

Both diamond and silicon(IV) oxide have very high melting points. Identify the shared structural feature and explain the energy requirement.

Link structure, bond and energy

Shared structure
What melting must overcome

Hints

Hint 1: decide whether molecules exist
Neither diamond nor silicon(IV) oxide consists of separate small molecules.
Hint 2: identify the repeated connection
Strong covalent bonds extend throughout each giant network.
View solution step by step

  1. State the shared structure

    Method

    Identify both as giant covalent structures.

    Reason

    Atoms are joined by covalent bonds through a continuous network.

    Working

    Diamond and SiO₂: giant covalent networks.

  2. Link bonding to melting point

    Method

    Break many strong covalent bonds.

    Reason

    A large energy input is required to disrupt enough of the network for melting.

    Working

    Many strong bonds → much energy → very high melting point.

Common misconception 3

Error Analysis (Fix the Claim)

Find and correct the mistake

Learner response

A student writes: “Graphite has a low melting point because its layers are held together by weak forces.” Locate the property error and separate the roles of the weak and strong forces.

Match each force to the correct property

Graphite melting point
Weak forces explain

View solution step by step

  1. Locate the property error

    Method

    Correct “low melting point” to “very high melting point.”

    Reason

    Melting the giant structure requires many strong covalent bonds within layers to be broken.

    Working

    Many strong covalent bonds → very high melting point.

  2. Assign the weak forces correctly

    Method

    Use weak forces between layers to explain sliding.

    Reason

    The layers can move relative to one another without breaking the strong bonds within a layer.

    Working

    Weak interlayer forces → softness and slipperiness.

  3. Write the corrected distinction

    Working

    Strong intralayer covalent bonds explain graphite’s high melting point; weak interlayer forces explain its softness.

Examiner practice 4

Conductivity (Why Graphite Conducts but Diamond Doesn’t)

4 marks

Examination question

Explain why graphite conducts electricity but diamond does not. [4 marks]

Compare bonding and charge carriers

View solution step by step

  1. Explain graphite bonding

    1 mark

    Method

    State that each carbon bonds to three other carbon atoms.

    Reason

    Only three of carbon’s four valence electrons are used in these bonds.

    Working

    One electron per carbon remains delocalised.

  2. Explain graphite conduction

    1 mark

    Method

    Make the delocalised electrons mobile along the layers.

    Reason

    Moving electrons carry electrical charge.

    Working

    Graphite conducts.

  3. Explain diamond bonding

    1 mark

    Method

    State that each carbon bonds to four other carbon atoms.

    Reason

    All four valence electrons are used in covalent bonds.

    Working

    Diamond has no delocalised electrons.

  4. Explain diamond's result

    1 mark

    Method

    Identify the absence of mobile charged particles.

    Reason

    No electrons are available to move through the structure and carry charge.

    Working

    Diamond does not conduct electricity.

Challenge 5

Data Interpretation (Identify the Substance)

Minimal support

Property-evidence transfer

Substance X is a solid that conducts electricity, feels slippery and has a very high melting point. Decide whether X is most likely diamond, graphite or silicon(IV) oxide, and account for all three observations.

Use every observation

Most likely substance
Conduction evidence

Hints

Hint 1: use the conducting exception
Of the three listed giant covalent substances, only one has delocalised electrons.
Hint 2: account for softness separately
Slipperiness requires layers that can slide; high melting point requires many strong covalent bonds.
View solution step by step

  1. Use conductivity

    Method

    Identify graphite from mobile delocalised electrons.

    Reason

    Diamond and silicon(IV) oxide lack mobile charge carriers.

    Working

    Conducts → graphite.

  2. Use slipperiness

    Method

    Identify graphite’s layered structure.

    Reason

    Weak forces between layers allow them to slide.

    Working

    Sliding layers → slippery feel.

  3. Use melting point

    Method

    Identify strong covalent bonds within the giant layers.

    Reason

    Many such bonds require much energy to break.

    Working

    Very high melting point confirms a giant covalent structure; X is graphite.

7. Mind Stretchers

Mind stretcher 1: Choosing an Electrode MaterialExtension

Question: An electrode is needed for electrolysis at high temperature. Choose diamond or graphite and explain.

Show Answer

Choose graphite:

  • conducts electricity due to delocalised electrons
  • high melting point due to strong covalent bonds

Diamond does not conduct electricity.

Mind stretcher 2: “Both are carbon” TrapExtension

Question: Diamond and graphite are both C. Why do they have different properties?

Show Answer

They have different structures (different bonding/arrangement of atoms):

  • diamond: 3D tetrahedral network, no delocalised electrons
  • graphite: layered structure with delocalised electrons

Different structure → different properties.

8. Quiz

Quiz time

Ready to check your understanding? Try the practice links, then review anything you missed.

Practice next

Lesson quizCheck this lesson quickly with targeted MCQs.Structured practiceSwitch to exam-style written responses after the quiz.