Period 3 Trends: Melting Points and Bonding Revision Notes | A Level AQA Chemistry

1. Why Period 3 Melting Points Tell a Bigger Story

Melting point rises from Na to Al because metallic bonding becomes stronger, then rises dramatically at Si because its giant covalent structure requires many strong covalent bonds to be broken. After the large drop at P, there is a slight rise at S before values fall again at Cl and Ar.

This lesson connects three trends — atomic radius, first ionisation energy, and melting point — into one coherent picture. Once you see how nuclear charge drives all three, the trends stop being a list to memorise and become something you can reason through in an exam.

2. Atomic Radius Across Period 3

Moving from Na to Cl, atomic radius decreases steadily. At sodium the outer electron sits in the third shell, shielded from the nucleus by the two inner shells. As you move right, each successive element adds one more proton to the nucleus and one more electron to the same third shell.

Electrons in the same shell shield each other very poorly. The nuclear charge therefore increases without a proportional increase in shielding, so the effective nuclear charge — the net pull felt by the outer electrons — rises across the period. A stronger pull contracts the electron cloud, and the atomic radius shrinks.

Argon is included for completeness, but its radius is defined differently (van der Waals radius) because it forms no bonds. For exam purposes, the key trend to state is: atomic radius decreases across Period 3 from Na to Cl due to increasing effective nuclear charge with constant shielding.

3. First Ionisation Energy Across Period 3

First ionisation energy (IE₁) is the energy needed to remove one mole of electrons from one mole of gaseous atoms: $\text{X}(g) \rightarrow \text{X}^+(g) + e^-(g)$. The general trend across Period 3 is an increase from Na to Ar, but two anomalies interrupt that rise and they are exactly the kind of detail AQA examiners test.

The general increase makes sense from what we already know about atomic radius. A smaller atom with higher effective nuclear charge holds its outer electrons more tightly, so more energy is needed to remove one. Sodium, with its lone outer electron in a large third shell, has the lowest IE₁ in the period.

Two dips break the upward trend. Aluminium has a lower IE₁ than magnesium, and sulphur has a lower IE₁ than phosphorus. These are not errors — they reveal something important about subshell energy levels and electron repulsion.

3a. The Anomaly at Aluminium

Magnesium’s outer electron is in a 3s subshell (configuration: [Ne] 3s²). Aluminium’s outer electron is in a 3p subshell (configuration: [Ne] 3s² 3p¹). The 3p subshell is at a slightly higher energy than 3s and is partially shielded by the 3s electrons below it. This makes the 3p electron easier to remove than either of magnesium’s 3s electrons, so IE₁ drops from Mg to Al.

3b. The Anomaly at Sulphur

Phosphorus has the configuration [Ne] 3s² 3p³. Each 3p orbital contains exactly one electron — a half-filled subshell with no electron pairing. Sulphur adds a fourth 3p electron, which must pair up in an already-occupied orbital. The repulsion between the two electrons sharing that orbital makes it easier to remove one of them, so IE₁ at sulphur is lower than at phosphorus.

These two anomalies — Al lower than Mg, and S lower than P — are a favourite AQA target. Knowing the subshell explanation is what separates a grade 7 answer from a grade 9 one.

4. Structure and Bonding Across Period 3

Now for the melting point trend, and this is where the chemistry becomes genuinely interesting. The elements of Period 3 do not all bond in the same way. Three distinct structure types appear across the period, and each produces a characteristic range of melting points.

4a. The Metals: Na, Mg, and Al

Sodium, magnesium, and aluminium all adopt giant metallic structures. Picture a regular lattice of positive metal ions immersed in a delocalised ‘sea’ of electrons. The melting point depends on the strength of the electrostatic attraction between the positive ions and the electron sea.

Three factors strengthen that attraction as you move from Na to Al: the ionic charge increases (Na⁺, Mg²⁺, Al³⁺), the number of delocalised electrons per atom increases (1, 2, 3), and the ionic radius decreases. All three factors act together, which is why melting point rises sharply from Na (98°C) to Mg (650°C) and then only slightly further to Al (660°C).

4b. Silicon: The Giant Covalent Peak

Silicon’s melting point of 1414°C dwarfs everything else in the period. The reason is its giant covalent structure: every silicon atom forms four strong covalent bonds to four neighbouring silicon atoms in a three-dimensional network. To melt silicon, you must break a large number of these strong covalent bonds, which requires an enormous amount of energy.

This is the single most important structural contrast in Period 3. The jump from Al (metallic, 660°C) to Si (giant covalent, 1414°C) reflects a complete change in bonding type, not a continuation of the metallic trend.

4c. The Sharp Drop: P, S, Cl, and Ar

Phosphorus, sulphur, chlorine, and argon all have low melting points because they exist as simple molecules (P₄, S₈, Cl₂) or as individual atoms (Ar). The covalent bonds within each molecule are strong, but the forces between molecules are only weak London dispersion forces (also called van der Waals forces or instantaneous dipole–induced dipole forces).

Melting these substances only requires overcoming the weak intermolecular forces, not breaking covalent bonds. That is why the melting points are so low. The variation within this group — S₈ (113°C) higher than P₄ (44°C), which is higher than Cl₂ (−101°C) — reflects the relative sizes of the molecules. Larger molecules have more electrons, stronger London dispersion forces, and therefore higher melting points.

5. The Full Melting Point Trend: Putting It Together

Earlier we established that atomic radius decreases and effective nuclear charge increases across Period 3. Now we can see how those atomic-level changes feed into the macroscopic melting point trend. The metals get stronger lattices, silicon builds an unbreakable covalent network, and then the non-metals revert to weak intermolecular forces.

The chart makes the structural story visible. The rise from Na to Si reflects increasingly strong bonding (metallic then covalent). The cliff-edge drop from Si to P marks the transition from giant covalent to simple molecular. The slight rise from P to S reflects the larger S₈ molecule having stronger London dispersion forces than P₄, before the trend falls again at Cl₂ and Ar.

Worked example: Explaining the Melting Point Trend in an Exam Answer

Explain why silicon has a much higher melting point than phosphorus. [4 marks]

1. Silicon has a giant covalent structure. — State the structure type first — this is the foundation of the explanation and earns a mark.
2. Each silicon atom is covalently bonded to four other silicon atoms in a three-dimensional lattice. — Specify the number of bonds and the 3D nature of the network. Vague answers ('lots of bonds') do not score.
3. Phosphorus exists as simple molecular P₄ molecules. — Name the molecular formula P₄ for precision. AQA mark schemes often require this.
4. The forces between P₄ molecules are weak London dispersion forces (van der Waals forces). — Name the force type explicitly. 'Weak forces' alone is insufficient at A Level.
5. Melting silicon requires breaking many strong covalent bonds, requiring a large amount of energy. Melting phosphorus only requires overcoming weak intermolecular forces, requiring much less energy. — The comparison must be explicit. Contrast 'breaking covalent bonds' with 'overcoming intermolecular forces' in the same sentence or adjacent sentences.

Answer: Silicon has a giant covalent structure in which each Si atom forms four strong covalent bonds to neighbouring Si atoms throughout a 3D lattice. A large amount of energy is needed to break these bonds on melting. Phosphorus exists as simple molecular P₄ molecules held together by weak London dispersion forces. Much less energy is needed to overcome these forces, so phosphorus has a far lower melting point.

6. Connecting All Three Trends

Atomic radius, first ionisation energy, and melting point are not three separate topics — they are three windows onto the same underlying change: increasing effective nuclear charge across Period 3. A smaller atom with a higher effective nuclear charge is harder to ionise (higher IE₁) and, in the metals, forms a stronger metallic lattice (higher melting point).

The melting point trend diverges from the IE₁ trend at silicon because melting point depends on the type of structure, not just on atomic properties. Silicon’s giant covalent structure gives it a melting point far above what its position in the period would suggest if it were metallic. This is why you must always identify the structure type before explaining the melting point.

7. Exam Technique and Common Mistakes

AQA questions on Period 3 trends tend to use the command words ‘explain’ and ‘compare’. An ‘explain’ answer must always give a cause-and-effect chain, not a list of facts. A ‘compare’ answer must address both similarities and differences explicitly — writing about only one element scores at most half the available marks.

For melting point questions, the single most common error is describing the forces within a molecule rather than the forces between molecules. The covalent bonds inside Cl₂ are strong, but they are irrelevant to melting point. What matters is the London dispersion forces between Cl₂ molecules. Always ask yourself: what must be broken or overcome when this substance melts?

The three trends covered in this lesson — atomic radius decreasing, first ionisation energy generally increasing (with anomalies at Al and S), and melting point following a pattern dictated by structure type — all trace back to the same root cause: increasing effective nuclear charge across Period 3 with constant inner-shell shielding. Keeping that single idea at the centre of your reasoning will allow you to reconstruct any part of this topic under exam conditions.