Classification and Trends in Atomic Properties — AQA A-level Chemistry

SPEC 3.2.1.1–3.2.1.2 • PERIODICITY

Classification & Physical Properties of Period 3

s, p, d, f block classification and trends in atomic radius, first ionisation energy, and melting point across Na to Ar

Why Study Periodicity?

The Periodic Table is one of the most powerful tools in chemistry. It organises elements not randomly, but according to their electronic structure, which determines their chemical and physical properties. Periodicity refers to the repeating patterns (trends) we observe as we move across periods or down groups.

Understanding periodicity allows chemists to predict how elements will behave without having to test every single one. For example, if you know how sodium reacts, you can predict how potassium and lithium will react too. This predictive power is what makes the Periodic Table so valuable.

Learning Objectives

By the end of this lesson, you will be able to:

Classify elements as s, p, d or f block based on their electron configuration

Explain trends in atomic radius, first ionisation energy, and melting point across Period 3

Link these trends to electronic structure, nuclear charge, and bonding

Exam Focus

Periodicity questions typically ask you to explain trends, not just describe them. Always link your answer to electron configuration, nuclear charge, shielding, or bonding.

Block Classification

Definition

An element is classified as s, p, d or f block according to the sub-shell that contains its highest energy (outermost) electron.

The Periodic Table can be divided into four distinct regions called blocks. The block an element belongs to tells you which type of orbital its outermost electron occupies. This classification is directly linked to electron configuration:

BlockHighest Energy ElectronGroupsExamples

s-blocks sub-shellGroups 1 and 2Na (1s² 2s² 2p⁶ 3s¹), Mg, Ca

p-blockp sub-shellGroups 13–18Al, Si, P, S, Cl, Ar

d-blockd sub-shellGroups 3–12Fe, Cu, Zn (transition metals)

f-blockf sub-shellLanthanides & ActinidesCe, U (rare earths)

Common Mistake

Helium (He) is in Group 18 but is an s-block element because its electron configuration is 1s². Its position in Group 18 reflects its chemical inertness, not its electron configuration.

Periodic Table Block Diagram

The Periodic Table is divided into four blocks based on which sub-shell the outermost electron occupies:

Period 3 Classification

Na, Mg are s-block (3s sub-shell). Al, Si, P, S, Cl, Ar are p-block (3p sub-shell). Period 3 has no d-block elements because the 3d sub-shell fills in Period 4. Note: He is s-block (1s²) despite its position in Group 18.

Worked Example01

Identify which block each of the following elements belongs to, and explain your reasoning:

(a) Sulfur (S, Z = 16)

(b) Calcium (Ca, Z = 20)

(c) Iron (Fe, Z = 26)

a)

Sulfur (Z = 16): Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁴

The highest energy electron is in the 3p sub-shell.

Therefore, sulfur is a p-block element.

b)

Calcium (Z = 20): Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

The highest energy electron is in the 4s sub-shell.

Therefore, calcium is an s-block element.

c)

Iron (Z = 26): Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

The highest energy electron is in the 3d sub-shell (3d fills after 4s but 3d is higher in energy for transition metals).

Therefore, iron is a d-block element.

Atomic Radius

Definition

Atomic radius is defined as half the distance between the nuclei of two bonded atoms of the same element.

We cannot measure the "size" of an isolated atom directly because the electron cloud has no sharp boundary. Instead, we measure the distance between nuclei in a diatomic molecule (like Cl₂) and halve it. This gives us a consistent way to compare atomic sizes.

Terminology

In metals, this measurement is called the metallic radius. In non-metals, it is called the covalent radius. Both represent half the internuclear distance between bonded atoms.

The atomic radius depends on two competing factors:

FactorEffect on Atomic RadiusWhy?

Nuclear chargeMore protons → smaller radiusGreater attraction pulls electrons closer to nucleus

Electron shieldingMore inner shells → larger radiusInner electrons repel outer electrons, reducing nuclear attraction

Key Insight

Across Period 3, the number of electron shells stays the same (all have 3 shells), but the nuclear charge increases. This means shielding is roughly constant while nuclear charge increases, so atoms get progressively smaller.

Atomic Radius Across Period 3

Trend: Atomic radius decreases across Period 3 from sodium to chlorine (argon is not typically included as it doesn't form covalent bonds for measurement).

Explanation

As we move Na → Cl: the number of protons increases (11 → 17), so nuclear charge increases. The number of electron shells remains constant (all have 3 shells), so shielding is roughly constant. Therefore, the increased nuclear charge pulls the outer electrons closer, reducing atomic radius.

Worked Example02

Explain why the atomic radius of aluminium (143 pm) is smaller than that of sodium (186 pm).

1

Compare electron configurations:

Na: 1s² 2s² 2p⁶ 3s¹ (11 protons, 11 electrons)

Al: 1s² 2s² 2p⁶ 3s² 3p¹ (13 protons, 13 electrons)

2

Identify shielding:

Both elements have the same number of inner shells (1s², 2s², 2p⁶ = 10 inner electrons). Therefore, shielding is approximately the same.

3

Compare nuclear charge:

Al has 13 protons compared to Na's 11 protons. This greater nuclear charge creates a stronger attraction on the outer electrons.

4

Conclusion:

The increased nuclear charge in Al (without additional shielding) pulls the outer electrons closer to the nucleus, resulting in a smaller atomic radius.

First Ionisation Energy

Definition

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 1+ ions.

X (g) \to X^{+} (g) + e^{-}

First ionisation energy measures how tightly an atom holds onto its outermost electron. A high ionisation energy means the electron is difficult to remove (strongly attracted to the nucleus), while a low ionisation energy means the electron is easily removed.

FactorEffect on IEReason

Increased nuclear chargeIncreases IEMore protons attract electrons more strongly

Increased atomic radiusDecreases IEOuter electrons are further from the nucleus

Increased shieldingDecreases IEInner electrons reduce nuclear attraction

Important Conditions

The definition specifies gaseous atoms. This is because we want to measure the energy to remove an electron without any interference from intermolecular forces that exist in solids or liquids.

First IE Trend Across Period 3

General Trend: First ionisation energy increases across Period 3 from sodium to argon. However, there are two important anomalies (dips) at aluminium and sulfur.

Why the General Increase?

As nuclear charge increases across the period (with constant shielding), the outer electrons are held more tightly, requiring more energy to remove them. The decreasing atomic radius also means electrons are closer to the nucleus.

Anomaly 1: Mg → Al Drop

Despite aluminium having a higher nuclear charge than magnesium, its first ionisation energy is lower (578 kJ mol⁻¹ vs 738 kJ mol⁻¹). This appears to contradict the general trend, but it can be explained by examining the electron configurations:

Electron Configurations

Mg: 1s² 2s² 2p⁶ 3s²

Al: 1s² 2s² 2p⁶ 3s² 3p¹

The key difference is which sub-shell the outermost electron occupies:

ElementOuter ElectronSub-shellReason for Difference

Mg3s²s sub-shells electrons penetrate closer to nucleus, strongly attracted

Al3p¹p sub-shellp electrons are further from nucleus on average, less attracted

Explanation Summary

The 3p electron in Al is higher in energy and further from the nucleus on average than the 3s electron in Mg. Additionally, the 3p electron in Al is shielded by the 3s² electrons, making it easier to remove despite Al having more protons.

Anomaly 2: P → S Drop

Sulfur has a slightly lower first ionisation energy than phosphorus (1000 vs 1012 kJ mol⁻¹), even though sulfur has a higher nuclear charge. This anomaly is explained by examining the arrangement of electrons in the 3p sub-shell:

3p Orbital Diagrams

P (3p³):↑↑↑S (3p⁴):↑↓↑↑

In phosphorus, each of the three 3p orbitals contains one unpaired electron. In sulfur, one orbital contains a pair of electrons. When two electrons occupy the same orbital, they experience spin-pair repulsion (mutual repulsion between the two negatively charged electrons).

Explanation Summary

The paired electrons in one of sulfur's 3p orbitals repel each other. This repulsion makes it easier to remove one of the paired electrons compared to removing an unpaired electron from phosphorus. The energy saved by eliminating this repulsion partially offsets the increased nuclear charge.

Successive Ionisation Energies

Electrons can be removed one at a time from an atom. The energy required to remove each successive electron is called the successive ionisation energies. These provide evidence for electron shell structure.

Definitions

Second ionisation energy: The energy required to remove one mole of electrons from one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions.

Mg^{+} (g) \to Mg^{2 +} (g) + e^{-} Δ H_{I E 2} = + 1451 kJ mol^{- 1}

What the Pattern Shows

Within a shell: IE increases gradually as remaining electrons experience greater attraction to the increasingly positive ion.

Between shells: Large jumps occur when moving to an inner shell (electrons are closer to the nucleus and experience less shielding). For Mg, the big jump between IE₂ and IE₃ shows that the 3rd electron removed comes from a different (inner) shell.

Worked Example03

Explain why the first ionisation energy of sulfur (1000 kJ mol⁻¹) is lower than that of phosphorus (1012 kJ mol⁻¹), despite sulfur having a higher nuclear charge.

1

Write electron configurations:

P: 1s² 2s² 2p⁶ 3s² 3p³ (15 protons)

S: 1s² 2s² 2p⁶ 3s² 3p⁴ (16 protons)

2

Compare 3p sub-shell occupancy:

P has three unpaired electrons in 3p (one in each orbital).

S has four electrons in 3p, meaning one orbital contains a pair of electrons.

3

Explain the effect of pairing:

The two electrons in the same orbital experience spin-pair repulsion (electron-electron repulsion). This repulsion destabilises the electron and makes it easier to remove.

4

Conclusion:

Although sulfur has a higher nuclear charge, the spin-pair repulsion in the 3p⁴ configuration outweighs this effect, making it easier (less energy required) to remove an electron from sulfur than from phosphorus.

Melting Points Across Period 3

Unlike atomic radius and ionisation energy, the melting point trend across Period 3 is not a simple increase or decrease. Instead, it reflects the changing structure and bonding of the elements as we cross the period.

The melting point depends on the energy required to break the forces holding particles together. These forces vary dramatically across Period 3:

Structure Types in Period 3

Element(s)StructureBonding/ForcesNa, Mg, AlGiant metallicMetallic bonding (electrostatic attraction between cations and delocalised electrons)SiGiant covalentStrong covalent bonds throughout structure (tetrahedral, like diamond)P₄, S₈, Cl₂Simple molecularWeak van der Waals (London dispersion) forces between moleculesArMonatomicIndividual atoms with very weak van der Waals forces

Comparing P, S, Cl, and Ar

S₈ > P₄ > Cl₂ > Ar in terms of melting point. Larger molecules have more electrons, creating stronger London dispersion forces. Chlorine (Cl₂) has fewer electrons than S₈, and Argon exists as individual atoms (monatomic) with a full outer shell, giving it the weakest intermolecular forces and lowest melting point.

Why S Has Higher Melting Point Than P

Sulfur exists as S₈ molecules while phosphorus exists as P₄ molecules. The larger S₈ molecule has more electrons, creating stronger London dispersion forces between molecules. This is why sulfur (392 K) has a higher melting point than phosphorus (317 K).

Why Metallic Bonding Strengthens Na → Al

In metallic bonding, metal cations form a giant lattice held together by a "sea" of delocalised electrons. The electrons in this sea come from the valence shell of each atom:

ElementElectrons DonatedCation FormedIonic RadiusRelative Bonding Strength

Na1 electronNa⁺102 pm (largest)Weakest

Mg2 electronsMg²⁺72 pmStronger

Al3 electronsAl³⁺53 pm (smallest)Strongest

Why Melting Points Increase Na → Al

Three factors strengthen metallic bonding from Na to Al:

1. Higher cation charge: Al³⁺ has greater electrostatic attraction than Na⁺

2. More delocalised electrons: Al contributes 3 electrons per atom vs 1 for Na

3. Smaller ionic radius: Al³⁺ (53 pm) is much smaller than Na⁺ (102 pm), so delocalised electrons are closer to the positive nuclei, strengthening the attraction

Melting Point Trend

Pattern Summary

Na → Al: Melting point increases (stronger metallic bonding: more delocalised electrons, higher charge on cation).

Si: Highest melting point (giant covalent structure with strong Si-Si bonds).

P → Ar: Low melting points (simple molecules/atoms held by weak van der Waals forces).

Why Silicon Has the Highest Melting Point

Silicon has a giant covalent structure (like diamond). Each Si atom is covalently bonded to four other Si atoms in a tetrahedral arrangement. To melt silicon, you must break strong covalent bonds throughout the structure, requiring very high temperatures (1687 K). This is fundamentally different from metals where melting only requires overcoming electrostatic attractions.

Worked Example04

Explain why silicon has a much higher melting point (1687 K) than phosphorus (317 K).

1

Identify the structure of silicon:

Silicon has a giant covalent structure (like diamond). Each silicon atom is bonded to four other silicon atoms by strong covalent bonds in a tetrahedral arrangement.

2

Identify the structure of phosphorus:

Phosphorus exists as P₄ molecules (simple molecular structure). Each molecule contains four phosphorus atoms bonded together, but the molecules are held to each other only by weak van der Waals forces.

3

Compare the forces that must be overcome:

To melt silicon, many strong covalent bonds must be broken.

To melt phosphorus, only weak intermolecular forces between P₄ molecules need to be overcome.

4

Conclusion:

Much more energy is required to break the strong covalent bonds in silicon than to overcome the weak van der Waals forces between P₄ molecules, so silicon has a much higher melting point.

Practice

Question 1: Chlorine is in the p-block of the Periodic Table. State the electron configuration of chlorine and explain why it is classified as a p-block element.

Electron configuration of Cl: 1s² 2s² 2p⁶ 3s² 3p⁵

Explanation: Chlorine is a p-block element because its highest energy (outermost) electron is in a p sub-shell (specifically the 3p sub-shell). The block classification is determined by which sub-shell receives the final electron when building up the electron configuration.

Practice

Question 2: Explain why the first ionisation energy of magnesium (738 kJ mol⁻¹) is higher than that of aluminium (578 kJ mol⁻¹).

Electron configurations:

Mg: 1s² 2s² 2p⁶ 3s²

Al: 1s² 2s² 2p⁶ 3s² 3p¹

Explanation: In magnesium, the electron being removed is from the 3s sub-shell. In aluminium, the electron being removed is from the 3p sub-shell. The 3p electron is higher in energy and on average further from the nucleus than the 3s electron. Additionally, the 3p electron in Al experiences some shielding from the 3s² electrons. Despite Al having a higher nuclear charge, these factors make it easier to remove an electron from Al than from Mg, so Mg has the higher first ionisation energy.

Practice

Question 3: Explain why the melting point of aluminium (933 K) is higher than that of sodium (371 K).

Both elements have giant metallic structures, but the strength of metallic bonding differs:

Sodium: Na⁺ ions with 1 delocalised electron per atom. Relatively weak metallic bonding.

Aluminium: Al³⁺ ions with 3 delocalised electrons per atom. The higher charge on the cation and the greater number of delocalised electrons create much stronger electrostatic attraction between the cations and the "sea" of electrons.

Additionally: Al³⁺ ions are smaller than Na⁺ ions, so the delocalised electrons are closer to the nuclei, strengthening the metallic bonding further.

Conclusion: More energy is required to overcome the stronger metallic bonding in aluminium, so it has a higher melting point.

Summary

Key Trends Across Period 3 (Na → Ar)

PropertyTrendExplanation

Atomic RadiusDecreasesIncreasing nuclear charge with constant shielding pulls electrons closer

First Ionisation EnergyGenerally increases (with dips at Al and S)Al: 3p electron easier to remove than 3s. S: Spin-pair repulsion in 3p⁴

Melting PointRises Na→Al, peaks at Si, drops sharply for P→ArReflects structure change: metallic → giant covalent → simple molecular

Exam Tips

Always link trends to nuclear charge, shielding, and atomic radius

For ionisation energy anomalies, mention sub-shell differences (Mg→Al) or spin-pair repulsion (P→S)

For melting points, identify the structure type first, then explain the bonding/forces

Remember: block classification is based on the highest energy electron's sub-shell