How to Distinguish Molecular Solids from Covalent Networks Using Data

A white crystalline solid melts at 355 °C. Another white crystalline solid does not melt until temperatures above 1,500 °C. Both contain only carbon and oxygen. Which one is more likely to be a molecular solid, and which is more likely to be a covalent network?

The melting points give you a strong clue, but they don’t tell the whole story. To distinguish a molecular solid from a covalent network solid, you need to connect three things: the way atoms are bonded, the structure represented by the chemical model, and the physical properties measured in experiments.

The key distinction is this: a molecular solid contains separate molecules held together by intermolecular forces, whereas a covalent network solid contains an extended structure in which atoms are joined by covalent bonds throughout the solid.

The challenge in an HSC Chemistry question is recognising that distinction from data, rather than simply memorising examples.

01Start with the structure, not the melting point

Imagine comparing two buildings.

One is made from individual rooms that are connected to one another using relatively weak clips. The other is made from a continuous framework in which every section is joined by strong structural connections.

Pulling the first building apart means overcoming the connections between its rooms. Pulling the second apart means breaking the framework itself.

This is a useful starting model for molecular and covalent network solids.

Molecular solids: separate molecules in a crystal

A molecular solid consists of individual molecules arranged in an organised crystal structure.

Within each molecule, atoms are held together by covalent bonds. Between neighbouring molecules, intermolecular forces hold the crystal together.

These intermolecular forces include:

• London dispersion forces, which occur between all molecules.
• Dipole-dipole forces, which occur between permanent dipoles.
• Hydrogen bonding, which occurs when hydrogen is bonded to nitrogen, oxygen, or fluorine and interacts with an appropriate lone pair.

When a molecular solid melts, its molecules generally remain intact. The intermolecular attractions are overcome sufficiently for the molecules to move past one another.

For example, solid iodine, \(\mathrm{I_2(s)}\), consists of separate \(\mathrm{I_2}\) molecules. Each molecule contains a covalent I-I bond, while London dispersion forces attract neighbouring molecules.

Covalent network solids: one extended structure

A covalent network solid does not consist of separate molecules.

Instead, atoms are connected by covalent bonds in an extended arrangement. The bonding continues through the solid, rather than stopping at the boundary of an individual molecule.

Diamond is a familiar example. Each carbon atom forms four covalent bonds to four neighbouring carbon atoms in a three-dimensional network.

To melt diamond, you cannot simply separate intact molecules. There are no individual carbon molecules to separate. The extended covalent structure must be disrupted.

Graphite is another covalent network solid, although its structure is different. Each carbon atom forms three covalent bonds to other carbon atoms in a flat hexagonal layer. Strong covalent bonds hold each layer together, while weaker attractions act between layers.

This distinction matters because graphite can be relatively soft and can conduct electricity, even though it is a covalent network solid.

The building analogy has a limit: real crystal structures are three-dimensional arrangements of atoms and electrons, not miniature buildings with literal connections. Its purpose is to help you distinguish separate molecular units from continuous covalent bonding.

The opening numbers are a hypothetical data prompt, not reference values for named compounds. In an actual question, elemental analysis tells you which atoms are present, but it does not tell you how those atoms are connected.

02Read the bonds, not just the chemical formula

A common mistake is to look at a chemical formula and assume it tells you how many atoms are joined together in one molecule.

It doesn’t always.

You need to distinguish between a molecular formula and an empirical formula.

• A molecular formula tells you the actual number of each type of atom in one molecule.
• An empirical formula gives the simplest whole-number ratio of atoms in a substance.

For a molecular solid, the formula usually describes the individual molecules.

For a covalent network solid, the formula describes the repeating ratio of atoms throughout the extended structure.

Comparing CO₂ and SiO₂

Consider carbon dioxide and silicon dioxide.

Both have a 1:2 ratio between their central atom and oxygen. Both are neutral substances. However, their structures are completely different.

Carbon dioxide, CO₂, is molecular.

Each carbon atom forms two double covalent bonds with oxygen atoms.

The resulting structure is a discrete molecule:

\[
\mathrm{O=C=O}
\]

Carbon has four covalent bonds in total, while each oxygen has two. These are consistent with their usual valencies.

Each molecule has a definite boundary. In solid carbon dioxide, separate CO₂ molecules are held together by intermolecular forces.

Silicon dioxide, SiO₂, is a covalent network solid.

Each silicon atom forms four covalent bonds to four oxygen atoms. Each oxygen atom forms two covalent bonds, connecting two neighbouring silicon atoms.

This means that the bonding continues throughout the solid.

There are no separate SiO₂ molecules in the ordinary silica network structure.

Why does the network have the formula SiO₂?

This is where valency becomes useful.

Imagine starting with one silicon atom. It has four oxygen neighbours, giving four Si-O connections.

However, each oxygen is shared between two silicon atoms.

Therefore, each silicon effectively accounts for:

\[
4\times\frac{1}{2}=2
\]

oxygen atoms.

The resulting ratio is:

\[
\mathrm{Si:O}=1:2
\]

So the formula is SiO₂.

Notice what happened: the formula describes the ratio of atoms in the network. It does not mean that the structure consists of individual SiO₂ molecules.

This gives you a useful rule:

If the covalent bonds continue from atom to atom throughout the crystal, you have a covalent network. If the covalent bonds stop at the boundaries of individual molecules, you have a molecular solid.

Formal charge and oxidation state are not the same thing

There is another trap in questions involving silicon dioxide.

You might calculate the oxidation states of silicon and oxygen and obtain:

• Silicon: \(+4\)
• Oxygen: \(-2\)

A student might then conclude that SiO₂ must be an ionic solid containing \(\mathrm{Si}^{4+}\) and \(\mathrm{O}^{2-}\) ions.

That conclusion is incorrect.

An oxidation state is an electron-bookkeeping assignment. It assumes that bonding electrons belong to the more electronegative atom when assigning oxidation states.

It does not automatically tell you that free ions exist in the solid.

Formal charge is a different calculation. It considers how electrons are assigned in a Lewis structure, with bonding electrons shared equally.

For a neutral CO₂ molecule, the Lewis structure \(\mathrm{O=C=O}\) gives formal charges of zero on all three atoms.

A simple covalent representation of the local Si-O bonding in silica also gives zero formal charges on the atoms. Its oxidation states are nevertheless \(+4\) for silicon and \(-2\) for oxygen.

The important distinction is:

None of these should be used in isolation.

A correct HSC response needs to connect the electron arrangement to the actual bonding pattern.

03Use physical data as supporting evidence

What if a question doesn’t give you a complete structural representation?

You can use physical properties to help identify the likely structure.

The most useful starting point is usually melting behaviour.

Melting point: what actually needs to be overcome?

When a molecular solid melts, the molecules gain enough energy to move past one another.

The intermolecular forces holding the crystal together are overcome sufficiently for melting to occur.

The covalent bonds within the individual molecules generally remain intact.

For a covalent network solid, the situation is different.

Covalent bonds extend throughout the structure. Melting requires substantial disruption and rearrangement of this extended covalent bonding.

This generally requires much more energy.

This explains the general trend:

• Molecular solids usually have relatively low to moderate melting points.
• Covalent network solids usually have very high melting points.

However, don’t turn this into an absolute rule.

Molecular solids with extensive hydrogen bonding or large molecules can have comparatively high melting or decomposition temperatures. Some covalent network materials, such as graphite, also have unusual physical properties.

A comparison using real substances

The following values are approximate reference values.

The melting-point difference supports the structural classifications.

Iodine consists of individual I₂ molecules. Its relatively low melting point reflects the comparatively weak intermolecular attractions between those molecules.

Silica has an extended covalent structure. Much more energy is needed to disrupt that structure.

But notice something important: both substances are poor electrical conductors.

This means that electrical conductivity alone cannot distinguish a molecular solid from a covalent network solid.

Which physical properties are actually useful?

Let’s look at the two most common misconceptions.

Misconception 1: Every covalent network solid is hard.

Diamond is extremely hard because its carbon atoms form a rigid three-dimensional network.

Graphite is also a covalent network solid, but it is relatively soft. Its carbon atoms form strong covalent bonds within flat layers, while weaker attractions act between the layers.

The layers can slide over one another.

Misconception 2: Every solid that does not conduct electricity is molecular.

This is also incorrect.

Most molecular solids are poor conductors because they do not contain freely moving charged particles.

However, many covalent network solids are also poor conductors. Diamond and silica are examples.

Graphite is an important exception because it contains delocalised electrons that can move along its layers.

Therefore, poor electrical conductivity is consistent with both classifications. It is rarely the deciding piece of evidence.

What about solubility?

Solubility can provide another clue, but you need to interpret it carefully.

Molecular substances may dissolve in solvents with suitable intermolecular interactions.

For example, non-polar molecular substances often dissolve more readily in non-polar solvents than in water.

Covalent network solids are generally insoluble in common solvents because dissolving them would require disrupting an extended covalent structure.

However, insolubility does not prove that a solid is a covalent network. Some molecular substances are also poorly soluble.

Likewise, if a molecular substance dissolves in water and its solution conducts electricity, that does not automatically make the original solid ionic. The substance may form ions after dissolving.

Always distinguish between the properties of the solid and the properties of a solution made from it.

04What do the experiments actually measure?

Suppose you are given an unknown solid and asked to identify its structure.

You cannot directly look at the atoms and see whether the substance is molecular or network covalent. Instead, you collect evidence from experiments and structural models.

Two useful measurements are melting behaviour and electrical conductivity.

Measuring melting behaviour

A typical melting-point measurement uses a small amount of powdered solid inside a capillary tube.

The tube is placed in a controlled heating apparatus, close to a temperature sensor.

The apparatus heats the sample gradually. The observer records the temperature at which melting first appears and the temperature at which the sample has completely melted.

The difference between these readings gives the melting range.

For a molecular solid, you would generally expect a relatively low melting temperature.

For a covalent network solid, you would generally expect a much higher temperature or decomposition before a conventional melting point can be measured.

There is an important experimental limitation.

A standard school melting-point apparatus cannot measure every high-temperature solid. If a sample remains solid up to the maximum temperature of the equipment, you cannot claim to have measured its melting point.

You can only report that no melting was observed within the tested range.

Similarly, if the substance decomposes before melting, record the decomposition rather than incorrectly reporting a melting point.

Measuring electrical conductivity

A simple conductivity experiment uses two electrodes connected to a low-voltage power supply.

The solid sample is placed between the electrodes, ideally as a compact pellet.

An ammeter is connected in series with the sample to measure current. A voltmeter is connected in parallel across the sample to measure the potential difference.

The current passes from the power supply, through the ammeter and first electrode, through the sample, and back through the second electrode.

If the solid contains mobile charge carriers, current can flow.

The original prediction might be that a molecular solid and a covalent network solid behave differently.

However, the experiment often produces a less helpful result: both samples show very little current.

This is not a failed experiment. It tells you that neither sample has significant electrical conductivity under the test conditions. It simply doesn’t distinguish the two structures.

There is another possible problem. If you place loose powder between electrodes, poor electrical contact can produce an artificially high resistance.

A better design uses compacted pellets with consistent dimensions and electrode pressure. Repeat the measurement and compare samples under the same conditions.

Worked measurement: calculate resistance

Suppose a compacted solid sample is connected to a power supply.

The voltmeter reads 1.50 V, and the ammeter reads 0.30 mA.

Calculate the electrical resistance of the sample.

Use Ohm’s law:

\[
R=\frac{V}{I}
\]

Where:

• \(R\) is resistance in ohms (\(\Omega\)).
• \(V\) is potential difference in volts (V).
• \(I\) is current in amperes (A).

Convert the current into amperes:

\[
0.30\text{ mA}=0.00030\text{ A}
\]

Substitute:

\[
\begin{aligned}
R&=\frac{1.50}{0.00030}\\
&=5000\ \Omega\\
&=5.0\text{ k}\Omega
\end{aligned}
\]

The sample has a resistance of 5.0 kΩ under these test conditions.

Suppose another sample has a resistance of 30 kΩ using the same apparatus and sample geometry.

The second sample has a higher resistance, but that still does not establish that it is a covalent network solid.

Both molecular and covalent network solids can have high resistance.

The calculation tells you about electrical behaviour, not the connectivity of the covalent bonds.

05A reliable decision process for HSC questions

When a question asks you to distinguish a molecular solid from a covalent network using data, work through the following sequence.

Step 1: Identify the structural units.

Does the representation show separate molecules, or does it show covalent bonds continuing between neighbouring atoms throughout the solid?

This is your most important question.

Step 2: Check the valencies.

Check whether the number of covalent bonds around each atom is consistent with its usual valency.

For example:

• Carbon commonly forms four covalent bonds.
• Oxygen commonly forms two covalent bonds.
• Silicon commonly forms four covalent bonds.

Do not assume that satisfying valency means a molecule must exist. The same valencies can produce an extended network.

Step 3: Check the charges.

Distinguish formal charges from oxidation states.

A positive oxidation state does not automatically mean that a free positive ion exists in the structure.

Check whether the representation is consistent with the proposed neutral molecule or extended covalent structure.

Step 4: Compare the physical properties.

Use melting point, volatility, hardness, solubility, and conductivity as supporting evidence.

Give more weight to the overall pattern than to any single property.

Step 5: Write a conclusion that links structure to properties.

Avoid writing something vague such as:

“The solid is covalent because it has a high melting point.”

A stronger answer is:

“The solid is a covalent network because its structural representation shows silicon atoms connected to oxygen atoms through continuous covalent bonding. Each silicon forms four covalent bonds, and each oxygen bridges two silicon atoms. The high melting point is consistent with the need to disrupt the extended covalent structure. Its poor electrical conductivity is also consistent with the absence of mobile charge carriers, although this property alone cannot distinguish it from a molecular solid.”

That answer connects the representation, valency, and experimental evidence.

06Worked examples

07Questions and solutions

Question 1

An unknown crystalline solid contains separate molecules. Each molecule contains two atoms joined by a covalent bond. The molecules are held together by London dispersion forces.

The solid has a melting point of 90 °C and does not conduct electricity.

Classify the solid and explain which interactions are overcome when it melts.

Solution 1

The substance is a molecular solid.

Each molecule is a separate structural unit, and the covalent bonds do not extend between neighbouring molecules.

The molecules are held together by London dispersion forces. When the solid melts, these intermolecular attractions are overcome sufficiently for the molecules to move past one another.

The covalent bonds within each molecule remain intact.

The relatively low melting point supports the molecular classification. Its poor electrical conductivity is also consistent with the absence of mobile charge carriers.

Question 2

A solid contains two elements, A and B.

Experimental and structural data indicate that:

• Each A atom forms four covalent bonds to four B atoms.
• Each B atom forms two covalent bonds to two A atoms.
• The covalent bonding continues throughout the solid.
• The melting point is approximately 1,600 °C.
• The solid is a poor electrical conductor.

(a) Determine the simplest formula of the solid.

(b) Classify the structure.

(c) Explain why the formula does not indicate that the solid contains individual AB₂ molecules.

Solution 2

(a) The simplest formula is AB₂.

Each A atom forms four covalent connections to B atoms.

However, each B atom is shared between two A atoms.

Therefore:

\[
\begin{aligned}
\text{B per A}&=4\times\frac{1}{2}\\
&=2
\end{aligned}
\]

The simplest ratio is:

\[
\mathrm{A:B}=1:2
\]

So the formula is AB₂.

(b) The substance is a covalent network solid.

The covalent bonds extend continuously throughout the structure.

There are no separate molecular units.

The high melting point is consistent with the need to disrupt an extended covalent structure.

(c) AB₂ describes the atom ratio, not the number of atoms in a molecule.

A molecular formula would describe the number of atoms in one discrete molecule.

Here, A and B form an extended structure. The formula AB₂ simply represents the simplest repeating ratio throughout that structure.

This is the key distinction between an empirical formula and a molecular formula.

Question 3

Consider two neutral substances, solid X and solid Y.

(a) Calculate the oxidation state of carbon in CO₂ and silicon in SiO₂.

(b) Classify both solids.

(c) Explain why their identical oxidation-state values and similar electrical conductivity do not mean they have the same structure.

Solution 3

(a) Carbon and silicon both have an oxidation state of +4.

Oxygen has an oxidation state of −2 in both substances.

For CO₂, let the oxidation state of carbon be \(x\).

The molecule is neutral, so:

\[
\begin{aligned}
x+2(-2)&=0\\
x-4&=0\\
x&=+4
\end{aligned}
\]

For SiO₂, let the oxidation state of silicon be \(y\).

\[
\begin{aligned}
y+2(-2)&=0\\
y-4&=0\\
y&=+4
\end{aligned}
\]

Therefore, both central atoms have an oxidation state of +4.

(b) Solid X is molecular, while solid Y is a covalent network.

Solid X contains separate CO₂ molecules. The covalent bonds stop within each O=C=O unit, and intermolecular attractions hold the solid together.

Solid Y has continuous Si-O covalent bonding. Each silicon connects to four oxygen atoms, and each oxygen connects two silicon atoms.

(c) Oxidation states and electrical conductivity do not determine structural connectivity.

Oxidation states are assigned through electron bookkeeping. They do not establish whether a substance contains separate molecules or an extended network.

Similarly, both molecular CO₂ and network SiO₂ are poor electrical conductors because neither contains significant numbers of mobile charge carriers in the solid state.

The distinguishing evidence is the structural representation.

The thermal behaviour provides further support: solid CO₂ sublimes readily because its molecules are held together by relatively weak intermolecular forces, whereas silica has a highly extended covalent structure.

Question 4

Graphite has the following properties:

• It is relatively soft.
• It conducts electricity.
• Each carbon atom forms three covalent bonds to neighbouring carbon atoms.
• The carbon atoms are arranged in hexagonal layers.
• Weaker attractions act between adjacent layers.

A student concludes that graphite cannot be a covalent network solid because it is soft and conducts electricity.

Evaluate this conclusion. Explain how graphite’s bonding and structure account for its properties.

Solution 4

Graphite is a covalent network solid, despite being relatively soft and electrically conductive.

The student’s conclusion is incorrect because hardness and conductivity are not universal defining properties of covalent network solids.

Each carbon atom forms three covalent bonds with neighbouring carbon atoms within a hexagonal layer.

These bonds create an extended covalent network within each layer. There are no separate carbon molecules.

Each carbon also contributes an electron to a delocalised electron system. These electrons can move along the layers, allowing graphite to conduct electricity.

The layers are held together by weaker attractions. They can slide over one another relatively easily, which explains why graphite is soft.

Therefore, graphite is a covalent network solid with unusual electrical and mechanical properties.

This example demonstrates why you should identify the bonding structure before using physical properties to support your classification.

Question 5

Two crystalline solids, P and Q, both have structures that appear to form continuous three-dimensional arrangements.

Both are poor electrical conductors.

The following information is available.

Solid P:

• Each oxygen atom forms two covalent bonds with hydrogen atoms.
• Each water molecule is attracted to neighbouring water molecules through hydrogen bonding.
• These hydrogen bonds create an extended arrangement throughout the crystal.
• The solid melts at 0 °C under atmospheric pressure.

Solid Q:

• Each silicon atom forms four covalent bonds with oxygen atoms.
• Each oxygen atom connects two silicon atoms.
• The covalent bonding continues throughout the crystal.
• The solid has a melting point of approximately 1,710 °C.

A student argues:

“Both P and Q must be covalent network solids because both structures form continuous three-dimensional networks. They also both have poor electrical conductivity.”

Evaluate this argument.

In your answer:

(a) Classify both solids.

(b) Explain why a continuous arrangement of attractions does not necessarily mean a covalent network.

(c) Describe which interactions are disrupted when each solid melts.

(d) Explain why electrical conductivity cannot distinguish the two structures.

Solution 5

Solid P is a molecular solid, while solid Q is a covalent network solid. The student’s argument is incorrect because an extended arrangement of intermolecular attractions is not the same as an extended covalent network.

(a) Classify both solids.

Solid P is ordinary ice, which consists of discrete H₂O molecules.

Solid Q is silica, which consists of an extended covalent network.

(b) Explain the difference between their connectivity.

In ice, each oxygen atom forms two covalent bonds to hydrogen atoms.

These bonds create an individual H₂O molecule.

Hydrogen bonds then attract neighbouring water molecules. In ordinary ice, these intermolecular attractions form an extended three-dimensional arrangement.

However, the hydrogen bonds do not turn the separate water molecules into one continuous covalent structure.

In silica, the situation is different.

Each silicon atom forms four covalent bonds to oxygen atoms. Each oxygen atom connects two silicon atoms.

The covalent bonds themselves extend throughout the solid.

The important distinction is therefore the type of bonding that creates the extended arrangement, not simply whether the diagram looks connected.

(c) Explain what happens during melting.

When ice melts, sufficient hydrogen bonds are disrupted to allow the water molecules to move past one another.

The covalent O-H bonds within individual water molecules generally remain intact.

The substance remains composed of H₂O molecules.

When silica melts, the extended Si-O covalent framework must be disrupted and rearranged.

There are no pre-existing SiO₂ molecules that simply separate from one another.

This difference in the structural units explains why their melting behaviours are so different.

(d) Explain why conductivity does not distinguish them.

Both substances are poor electrical conductors in the solid state.

Ice does not contain significant concentrations of mobile charge carriers under ordinary conditions. Silica also lacks freely moving electrons or ions that would produce substantial electrical conductivity.

Therefore, poor conductivity is consistent with both structures.

The decisive evidence is the bonding connectivity:

• Ice contains discrete covalently bonded H₂O molecules connected by intermolecular hydrogen bonds.
• Silica contains an extended network of covalent Si-O bonds.

This is the subtle point: a solid can have an extended network of intermolecular forces and still be a molecular solid.

08Where to go next

Once you can distinguish discrete molecules from extended covalent structures, the next step is understanding how the bonding within and between molecules determines physical properties.

For a closer look at molecular geometry, covalent bonding, and intermolecular forces, read Covalent Molecular Structures and Intermolecular Forces.

If you want to practise identifying unknown substances using a broader range of experimental evidence, continue with How to Identify an Unknown Solid from Chemistry Data.

The skill to carry into your next data question is simple: identify what the covalent bonds actually connect before deciding what the physical properties mean.