Periodicity and Trends

This lesson examines the key periodic trends across Period 3 of the Periodic Table: atomic radius, first ionisation energy, electronegativity, and melting point. Understanding these trends — and the reasons behind them — is fundamental to inorganic chemistry at A-Level. The AQA specification (3.2.1) requires you to explain these patterns in terms of structure, bonding, nuclear charge, and shielding.

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Atomic Radius across Period 3

Atomic radius decreases across Period 3 from sodium to chlorine.

Element	Na	Mg	Al	Si	P	S	Cl	Ar
Atomic radius / pm	186	160	143	117	110	104	99	71
Protons	11	12	13	14	15	16	17	18
Electron config	[Ne]3s¹	[Ne]3s²	[Ne]3s²3p¹	[Ne]3s²3p²	[Ne]3s²3p³	[Ne]3s²3p⁴	[Ne]3s²3p⁵	[Ne]3s²3p⁶

Explanation

• Across the period, each successive element has one more proton in the nucleus and one more electron in the same principal energy level (n = 3).
• The additional electron enters the same shell, so there is no significant increase in shielding (inner electrons remain the same: 1s²2s²2p⁶).
• The increased nuclear charge pulls the outer electrons closer to the nucleus, reducing the atomic radius.
• Argon has the smallest atomic radius, but as a noble gas with a full outer shell, it is not usually compared directly in bonding contexts.

Exam Tip: When asked to explain the trend in atomic radius, always refer to (1) increasing nuclear charge, (2) same number of inner shielding electrons, and (3) greater attraction for outer electrons. Never say "more protons attract more electrons" without specifying that shielding remains roughly constant.

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First Ionisation Energy across Period 3

The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions:

X(g) → X⁺(g) + e⁻

Element	Na	Mg	Al	Si	P	S	Cl	Ar
1st IE / kJ mol⁻¹	496	738	578	786	1012	1000	1251	1521

General Trend

First ionisation energy generally increases across Period 3. The reasons are the same as for atomic radius:

• Increasing nuclear charge across the period.
• Negligible change in shielding (electrons added to the same shell).
• Outer electrons are held more tightly, so more energy is required to remove them.

The Two Dips

There are two important departures from the general upward trend:

Dip 1: Mg to Al

• Magnesium: outer electron is in the 3s sub-shell (3s²).
• Aluminium: outer electron is in the 3p sub-shell (3s²3p¹).
• The 3p electron is at a slightly higher energy than the 3s, and the 3s² electrons provide some additional shielding for the 3p electron.
• Therefore, less energy is required to remove the 3p electron from aluminium than the 3s electron from magnesium.

Dip 2: P to S

• Phosphorus: 3p³ — each p orbital contains one electron (no pairing).
• Sulfur: 3p⁴ — one p orbital now contains a pair of electrons.
• The paired electrons in sulfur experience spin-pair repulsion, which raises their energy slightly and makes one easier to remove.
• Therefore, sulfur has a slightly lower first ionisation energy than phosphorus.

Exam Tip: These two dips are examined frequently. For the Mg → Al dip, emphasise sub-shell energy levels. For the P → S dip, emphasise spin-pair repulsion in the doubly occupied 3p orbital.

Worked Example

Q: Explain why the first ionisation energy of sodium (496 kJ mol⁻¹) is much lower than that of neon (2081 kJ mol⁻¹).

A: Sodium's outer electron is in the 3s sub-shell, which is much further from the nucleus than neon's outer 2p electrons. Sodium has an extra shell of inner electrons providing substantial shielding. Although sodium has one more proton, the combined effect of greater distance and increased shielding far outweighs the extra nuclear charge, so sodium's first ionisation energy is much lower.

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Electronegativity across Period 3

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself (Pauling scale).

Element	Na	Mg	Al	Si	P	S	Cl
Electronegativity	0.93	1.31	1.61	1.90	2.19	2.58	3.16

Trend

Electronegativity increases across Period 3 from sodium to chlorine.

• Increasing nuclear charge with similar shielding means the nucleus attracts bonding electrons more strongly.
• Decreasing atomic radius means bonding electrons are closer to the nucleus.
• Argon is excluded because it does not normally form bonds.

Common Misconception: Students sometimes confuse electronegativity with electron affinity. Electronegativity is a relative scale for atoms in covalent bonds; electron affinity is the enthalpy change when a gaseous atom gains an electron.

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Melting Points across Period 3

Melting points across Period 3 show a distinctive pattern that reflects changes in structure and bonding.

Element	Na	Mg	Al	Si	P₄	S₈	Cl₂	Ar
Structure	Giant metallic	Giant metallic	Giant metallic	Giant covalent	Simple molecular	Simple molecular	Simple molecular	Monatomic
Melting point / °C	98	649	660	1414	44	115	−101	−189

Metals: Na, Mg, Al

Melting point increases Na → Mg → Al because:

1. Number of delocalised electrons increases: Na donates 1, Mg donates 2, Al donates 3.
2. Cation charge increases: Na⁺, Mg²⁺, Al³⁺.
3. Ionic radius of cation decreases (higher charge/radius ratio).
4. All three factors strengthen the metallic bonding (electrostatic attraction between cations and the sea of delocalised electrons).

Giant Covalent: Si

Silicon has the highest melting point in Period 3 (1414 °C). It has a diamond-like structure with each Si atom covalently bonded to four others in a tetrahedral arrangement. Many strong covalent bonds must be broken to melt silicon.

Simple Molecular: P₄, S₈, Cl₂

These exist as discrete molecules with weak London (dispersion) forces between them. Melting point depends on the number of electrons per molecule (which determines the strength of London forces):

• S₈ has 128 electrons per molecule → strongest London forces → highest melting point of the three.
• P₄ has 60 electrons → intermediate.
• Cl₂ has 34 electrons → weakest London forces → very low melting point.

Monatomic: Ar

Argon has the lowest melting point. Individual atoms with only 18 electrons have very weak London forces.

Exam Tip: The key to explaining melting points is identifying the correct structure type first. A common mistake is trying to explain the trend using only one factor (e.g. nuclear charge) when the real explanation requires discussing the type of bonding.

Worked Example

Q: Explain why silicon has a much higher melting point than phosphorus.

A: Silicon has a giant covalent structure in which each atom is bonded to four others by strong covalent bonds in a tetrahedral lattice. A very large amount of energy is needed to break these bonds. Phosphorus exists as P₄ molecules held together by weak London forces. Only these weak intermolecular forces need to be overcome to melt phosphorus, so its melting point is much lower.

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Summary

Property	Trend across Period 3	Key explanation
Atomic radius	Decreases	Increasing nuclear charge, same shielding
1st ionisation energy	Generally increases (dips at Al and S)	Increasing nuclear charge; sub-shell and pairing effects
Electronegativity	Increases	Increasing nuclear charge, smaller radius
Melting point	Rises (metals), peaks at Si, drops sharply (non-metals)	Change from metallic to giant covalent to simple molecular structures