Structure and Bonding - Giant Covalent Structures

1 What is a Giant Covalent Structure?

A giant covalent structure (sometimes called a macromolecular crystal) is like a massive 3D network where every atom is joined to others by strong covalent bonds. Think of it as one enormous structure rather than lots of separate molecules stuck together.

• Every covalent bond has the same strength and points in the same direction

• There are no weak forces between molecules to break - the only way to melt or boil the substance is to break many strong covalent bonds

• The most common examples you'll meet are diamond, graphite and silicon dioxide (silica)

1.1 Recognising Giant Structures in Diagrams

When you see a bonding diagram in an exam, look for these clues:

• Each atom is bonded to three or four neighbouring atoms

• The same pattern repeats in all directions

• There are no small, separate groups of atoms

• The structure appears to continue beyond the edges of the diagram

A network that looks like it goes on forever almost certainly represents a giant covalent structure.

2 Physical Properties and Their Explanations

1. Very high melting and boiling points

• Many strong covalent bonds must be broken, which needs huge amounts of energy

2. Hardness varies depending on structure

• Diamond is the hardest natural substance because every carbon atom forms four strong bonds in a rigid 3D network

• Graphite is soft and slippery because it has layers that can slide over each other, even though each layer is very strong

3. Electrical conductivity depends on available electrons

• Diamond and silica don't conduct electricity because all electrons are used up in bonding

• Graphite conducts electricity because each carbon atom has one spare electron that can move freely through the layers

4. Insoluble in water and other solvents

• The covalent bonds are too strong for water molecules to break apart the giant structure

2.1 Energy Considerations

Breaking lots of covalent bonds needs enormous amounts of energy. We can estimate the energy needed using:

Q=mcΔTQ=mcΔT

where QQ is the energy supplied, mm is the mass, cc is the specific heat capacity, and ΔTΔT is the temperature change.

3 Case Studies of Common Giant Covalent Substances

3.1 Diamond

• Each carbon atom bonds to four others at angles of 109.5°109.5° (tetrahedral shape)

• Creates a rigid 3D network that makes diamond incredibly strong

• Density of about 3.5 g cm−33.5 g cm−3

• No free electrons, so it's an electrical insulator

3.2 Graphite

• Carbon atoms form flat layers of hexagons (like chicken wire)

• Each carbon atom bonds to only three others

• The fourth electron from each carbon is free to move, creating mobile electrons

• Weak forces between layers allow them to slide - this is why graphite works as pencil "lead" and as a lubricant

3.3 Silicon Dioxide (SiO₂)

• Each silicon atom bonds to four oxygen atoms

• Each oxygen atom connects two silicon atoms together

• Found naturally as quartz and sand

• Very high melting point (about 1700°C1700°C) makes it useful for glassware and furnace linings

3.4 Recognising Differences in Diagrams

Look at how many bonds each atom makes:

• Diamond and SiO₂: four bonds around the central atoms

• Graphite: three bonds and you can see the layered structure

Comparison table

4 Interpreting Exam Graphs and Diagrams

You'll often see:

• Space-filling models: Look for patterns that repeat endlessly

• Line diagrams: Covalent bonds extending beyond the edge show the lattice continues

• Melting point data: Values above 1000°C suggest giant covalent or ionic structures - check electrical conductivity to tell them apart

Worked example

Question: A sample of diamond needs 2.8×105 J2.8×105 J to heat it from 25°C25°C to its melting point at 3550°C3550°C and then melt it completely. If the mass is 15 g15 g and the specific heat capacity is 0.5 J g−1 °C−10.5 J g−1 °C−1, how much energy was used just to break covalent bonds?

Solution:

1. Calculate energy needed to heat the diamond: Q1=mcΔT=15×0.5×(3550−25)=15×0.5×3525=26,438 JQ1​=mcΔT=15×0.5×(3550−25)=15×0.5×3525=26,438 J

2. Calculate energy used to break bonds (melting): Q2=Qtotal−Q1=2.8×105−2.64×104=2.54×105 JQ2​=Qtotal​−Q1​=2.8×105−2.64×104=2.54×105 J

Therefore, about 2.5×105 J2.5×105 J were needed to break covalent bonds in 15 g15 g of diamond.

Testing electrical conductivity: When graphite is connected to a simple circuit with a battery and bulb, the bulb lights up, showing that graphite conducts electricity. When the same test is done with diamond, the bulb doesn't light because diamond has no free electrons to carry current.

This simple test helps distinguish between different types of giant covalent structures.

Investigating properties of carbon allotropes:

Aim: To compare the hardness and electrical conductivity of different carbon structures

Apparatus:

• Graphite sample (pencil lead works well)

• Diamond-tipped tool or cubic zirconia

• Glass slide

• 6V battery, bulb, connecting wires with crocodile clips

• Safety spectacles

Method:

1. Hardness test: Try to scratch a glass slide with graphite, then with the diamond tool

2. Conductivity test: Connect the battery and bulb in a circuit with crocodile clips

3. Place graphite between the clips and observe the bulb

4. Replace with diamond simulant and observe again

Safety: Wear safety spectacles and handle sharp materials carefully

Expected results:

• Diamond scratches glass easily; graphite doesn't

• Graphite conducts electricity (bulb lights); diamond doesn't (bulb stays off)

Conclusion: This shows diamond is harder but graphite conducts electricity due to its different structure

Key terms

Giant covalent structure - A continuous 3D network where all atoms are held together by covalent bonds, with no separate molecules

Allotrope - Different structural forms of the same element in the same physical state (e.g. diamond and graphite are both carbon)

Delocalised electron - An electron that isn't tied to one particular atom or bond and can move freely through a structure to conduct electricity

Specific heat capacity - The energy needed to raise the temperature of 1 kg of a substance by 1°C