The Periodic Table

The Periodic Table becomes much more useful once we stop treating its trends as disconnected facts. Atomic radius, ionisation energy, melting point, oxide acidity and oxidising power may appear to be separate topics, but each can be traced back to a small number of underlying ideas.

Across a period, the number of protons increases while electrons are added to the same principal quantum shell. Down a group, additional occupied electron shells are introduced, so the outer electrons become farther from the nucleus and more strongly shielded.

In this chapter, we will use these ideas to understand Period 3, Group 2 and Group 17. The aim is not simply to memorise whether a property increases or decreases. It is to explain why the trend occurs and to connect microscopic changes in particles with the properties that we observe.

The Periodic Table as a map of electronic structure

Elements are arranged in order of increasing proton number. Their chemical properties repeat periodically because similar outer-shell electronic structures recur at regular intervals.

Most periodic trends can therefore be understood by asking how strongly the nucleus attracts the electrons involved.

Nuclear charge, shielding and effective nuclear attraction

The nuclear charge is determined by the number of protons in the nucleus. As proton number increases, the nucleus becomes more positively charged.

However, an outer electron does not experience the full attraction of the nucleus. Electrons between it and the nucleus repel it and reduce the net attraction that it experiences. This is called shielding.

The resulting attraction is often discussed in terms of the effective nuclear charge experienced by the electron.

At this level, the important skill is normally to compare these factors rather than calculate an exact effective nuclear charge.

Atomic and ionic radii across Period 3

Period 3 provides a clear example of how increasing nuclear charge changes atomic size. From sodium towards chlorine, the atomic radius generally decreases.

Why atomic radius decreases across Period 3

Three changes must be considered.

1. The proton number increases. Each successive element has an additional proton, so the nuclear charge becomes more positive.
2. Electrons are added to the same principal quantum shell. From Na to Cl, the additional electrons enter the third principal shell.
3. Inner-shell shielding changes very little. The number of electrons in the first two principal shells remains unchanged.

The increase in nuclear charge therefore outweighs the relatively small change in shielding. The outer electrons experience a stronger effective nuclear attraction and are pulled closer to the nucleus.

Why cations are smaller than their parent atoms

Sodium, magnesium and aluminium form Na⁺, Mg²⁺ and Al³⁺ respectively. These ions are substantially smaller than their neutral atoms.

• Na has [Ne]3s¹, while Na⁺ has [Ne].
• Mg has [Ne]3s², while Mg²⁺ has [Ne].
• Al has [Ne]3s²3p¹, while Al³⁺ has [Ne].

Formation of each ion removes all the electrons from the third principal shell. The ion therefore has one fewer occupied shell than the parent atom.

Comparing Na⁺, Mg²⁺ and Al³⁺

Na⁺, Mg²⁺ and Al³⁺ are isoelectronic: each contains 10 electrons.

Since the electron arrangements are the same, the important difference is nuclear charge.

• Na⁺ has 11 protons attracting 10 electrons.
• Mg²⁺ has 12 protons attracting 10 electrons.
• Al³⁺ has 13 protons attracting 10 electrons.

Al³⁺ therefore attracts the same electron arrangement most strongly and has the smallest radius.

Why anions are larger than their parent atoms

When an atom gains electrons to form an anion, the number of protons remains unchanged but there are more electrons in the electron cloud.

Electron–electron repulsion increases and the same nuclear charge is now attracting more electrons. The electron cloud expands, so the anion is larger than its parent atom.

P³⁻, S²⁻ and Cl⁻ are another isoelectronic series. Each contains 18 electrons.

Cl⁻ has the greatest nuclear charge and therefore attracts the same 18-electron arrangement most strongly.

First ionisation energy across Period 3

The first ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions.

Across Period 3, the first ionisation energy generally increases.

The outer electron is removed from the same principal quantum shell, while nuclear charge increases and atomic radius generally decreases. Inner-shell shielding changes very little.

The outer electron is therefore held increasingly strongly and more energy is required to remove it.

Why the first ionisation energy decreases from Mg to Al

Magnesium and aluminium have the outer electronic configurations:

Mg: 3s²
Al: 3s²3p¹

The electron removed from magnesium is a 3s electron, whereas the electron removed from aluminium is a 3p electron.

A 3p electron is higher in energy and experiences slightly greater shielding than a 3s electron. It is therefore easier to remove despite aluminium having the greater nuclear charge.

Why the first ionisation energy decreases from P to S

Phosphorus and sulfur have the outer electronic configurations:

P: 3s²3p³
S: 3s²3p⁴

In phosphorus, the three 3p electrons occupy separate 3p orbitals.

In sulfur, one of the 3p orbitals contains a pair of electrons. Repulsion between these paired electrons makes one of them easier to remove.

Electronegativity across Period 3

Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself.

Electronegativity generally increases from sodium towards chlorine.

Across the period, nuclear charge increases and atomic radius decreases while inner-shell shielding changes very little.

The nucleus therefore attracts a bonding pair of electrons increasingly strongly.

Melting points and electrical conductivity across Period 3

The melting points of Period 3 elements cannot be explained simply by saying that their atoms become smaller. Melting requires the attractions holding the solid structure together to be overcome.

We therefore need to identify the structure and bonding of each element first.

Why melting points generally increase from Na to Al

Sodium, magnesium and aluminium contain positive metal ions surrounded by delocalised electrons.

From Na to Mg to Al:

• the charge on the metal ion increases;
• the metal ion becomes smaller;
• the charge density of the metal ion increases; and
• more delocalised electrons are contributed per atom.

The electrostatic attraction between the positive metal ions and the sea of delocalised electrons therefore becomes stronger overall. More energy is required to disrupt the metallic lattice.

Why silicon has the highest melting point

Silicon forms a giant covalent structure. Each silicon atom is joined to other silicon atoms by strong covalent bonds throughout a three-dimensional network.

Melting therefore requires many strong Si–Si covalent bonds to be broken. This requires a large amount of energy.

Why the melting point falls sharply from Si to P₄

Phosphorus, sulfur and chlorine consist of discrete molecules rather than giant structures.

During melting, the covalent bonds within P₄, S₈ and Cl₂ molecules remain intact. It is the much weaker intermolecular attractions between molecules that are overcome.

Why S₈ has a higher melting point than P₄

An S₈ molecule contains more electrons and has a larger, more polarisable electron cloud than a P₄ molecule.

Stronger temporary dipoles can therefore form. The London dispersion forces between S₈ molecules are stronger than those between P₄ molecules.

London dispersion forces are the instantaneous dipole–induced dipole component of van der Waals forces.

Electrical conductivity across Period 3

Sodium, magnesium and aluminium conduct electricity because their metallic structures contain mobile delocalised electrons.

Silicon is a semiconductor. Its conductivity is much lower than that of a metal, but it can conduct when sufficient charge carriers are available.

Phosphorus, sulfur and chlorine do not conduct electricity because they contain no mobile ions or delocalised electrons.

Period 3 oxides: bonding and acid–base behaviour

The oxides of Period 3 show another important transition. Moving from left to right, the bonding becomes increasingly covalent and the oxides change from basic through amphoteric to acidic.

Basic oxides

Sodium oxide contains O²⁻ ions. The oxide ion is strongly basic and reacts readily with water.

Magnesium oxide is also basic, although it reacts much less readily with water. Its basic character is clear from its reaction with acids.

An oxide does not need to be highly soluble in water to be basic. Basicity is demonstrated by its reaction with an acid.

Why Al₂O₃ is amphoteric

An amphoteric oxide reacts with both acids and bases.

Aluminium oxide reacts with acids:

It also reacts with strong aqueous alkali:

Aluminium lies close to the transition between strongly metallic and non-metallic behaviour across Period 3, and its oxide shows both basic and acidic behaviour.

Acidic oxides

Silicon dioxide is an acidic oxide. Its giant covalent structure means that it does not simply dissolve or react with water, but it reacts with hot concentrated bases.

Phosphorus(V) oxide and sulfur trioxide react with water to form acids.

Period 3 chlorides and their reactions with water

The chlorides also show a progression from predominantly ionic to increasingly covalent behaviour. Their reactions with water reveal an important distinction between dissolving and hydrolysis.

Sodium chloride simply dissolves

Sodium chloride separates into hydrated Na⁺ and Cl⁻ ions.

Neither ion hydrolyses water significantly, so the resulting solution is approximately neutral.

Why MgCl₂ solution is slightly acidic

A Mg²⁺ ion becomes surrounded by water molecules to form a hydrated ion such as [Mg(H₂O)₆]²⁺.

The positively charged Mg²⁺ centre attracts electron density from the O–H bonds in the coordinated water molecules. This weakens an O–H bond sufficiently for slight hydrolysis to occur.

Only a relatively small concentration of H₃O⁺ is produced, so the solution is only slightly acidic.

Why AlCl₃ solution is more acidic

Al³⁺ is smaller and more highly charged than Mg²⁺. It therefore has a much greater charge density.

It polarises the O–H bonds of coordinated water molecules much more strongly, encouraging greater hydrolysis.

A greater concentration of H₃O⁺ is therefore produced than in magnesium chloride solution.

Hydrolysis of SiCl₄

Silicon tetrachloride reacts vigorously with water. The Si–Cl bonds are hydrolysed and hydrogen chloride is produced.

In moist air, this reaction can produce white fumes because of the tiny droplets and particles formed.

Hydrolysis of PCl₅

Phosphorus(V) chloride is hydrolysed in stages. With excess water, the overall reaction can be represented as:

Hydrogen chloride and phosphoric acid are formed, so the resulting mixture is acidic.

Group 2: reactivity and reducing power

Group 2 metals form M²⁺ ions by losing two outer-shell electrons.

Down the group, each element has one additional occupied principal quantum shell. The outer electrons therefore become farther from the nucleus and experience greater shielding.

Although nuclear charge also increases, the combined effects of greater distance and greater shielding outweigh the increase in nuclear charge.

The attraction between the nucleus and the outer electrons becomes weaker. The first and second ionisation energies therefore decrease overall, so the two outer electrons are lost more readily.

A stronger reducing agent loses electrons more readily. The increase in Group 2 reducing power therefore follows directly from the increasing ease with which the metals are oxidised to M²⁺.

Thermal stability of Group 2 carbonates

Group 2 carbonates decompose on heating to form the corresponding metal oxide and carbon dioxide.

The carbonates become more thermally stable down Group 2. This means that progressively stronger heating is required to decompose them.

How cation charge density explains the trend

1. Every Group 2 cation has a charge of +2.
2. The ionic radius of M²⁺ increases down the group.
3. The charge density and polarising power of M²⁺ therefore decrease.
4. A small cation such as Mg²⁺ strongly distorts the electron cloud of CO₃²⁻.
5. This polarisation weakens bonding within the carbonate ion.
6. A larger cation such as Ba²⁺ polarises CO₃²⁻ less strongly.
7. The carbonate ion is therefore more resistant to decomposition.

Worked comparison: MgCO₃ and BaCO₃

Mg²⁺ and Ba²⁺ have the same ionic charge, but Mg²⁺ has the smaller radius.

Mg²⁺ therefore has the greater charge density and polarising power. It distorts the carbonate ion more strongly, making MgCO₃ easier to decompose.

MgCO₃ is less thermally stable than BaCO₃ because Mg²⁺ has the greater charge density and polarises CO₃²⁻ more strongly.

Group 17: volatility and boiling points

The halogens exist as non-polar diatomic molecules, X₂. Between their molecules, the important attractions are London dispersion forces.

These are the instantaneous dipole–induced dipole component of van der Waals forces.

Down the group, the molecules contain more electrons and their electron clouds become larger and more polarisable.

Temporary dipoles can therefore be produced more easily, giving stronger London dispersion forces between neighbouring molecules.

More energy is required to separate the molecules, so boiling point increases and volatility decreases down the group.

Oxidising power of the halogens

A halogen acts as an oxidising agent by accepting electrons and being reduced to halide ions.

The overall oxidising power of the halogens decreases down Group 17.

This means that chlorine gains electrons more readily than bromine under comparable aqueous conditions, while bromine gains electrons more readily than iodine.

It is safest to treat this as an overall redox tendency. The trend reflects several energetic factors rather than just one simple change in atomic radius.

Halogen displacement reactions

The trend can be demonstrated using displacement reactions. A stronger halogen oxidising agent oxidises the halide ion of a weaker halogen.

I₂ cannot oxidise Br⁻ or Cl⁻ under these conditions because iodine is the weaker oxidising agent.

Thermal stability of the hydrogen halides

The thermal stability of the hydrogen halides decreases down Group 17.

Down the group, the halogen atom becomes larger. The H–X bond therefore becomes longer.

Orbital overlap becomes less effective, so the H–X bond becomes weaker and its bond enthalpy decreases.

Less energy is therefore required to break the H–X bond, so the hydrogen halide decomposes more readily on heating.

A reliable method for explaining periodic trends

Periodic-trend explanations become much easier when you build a logical chain rather than jumping directly from position in the Periodic Table to the final property.

1. Identify the particles involved. Are you comparing atoms, ions, molecules or lattices?
2. State what changes. Consider nuclear charge, shielding, radius, charge density, electron-cloud size or bond length.
3. State how the relevant attraction changes. Is an electron, molecule or ion held more or less strongly?
4. Connect that attraction to the observed property. Explain why more or less energy is required, or why electron transfer becomes easier or harder.

Worked example: comparing Mg²⁺ and Al³⁺

Both ions contain 10 electrons, but Al³⁺ has 13 protons whereas Mg²⁺ has 12.

The greater nuclear charge in Al³⁺ attracts the same electron arrangement more strongly.

Same number of electrons + greater nuclear charge → stronger attraction → smaller ion

Worked example: comparing Br₂ and I₂ boiling points

I₂ contains more electrons and has a larger, more polarisable electron cloud than Br₂.

Stronger instantaneous dipoles can therefore form, producing stronger London dispersion forces. More energy is required to separate I₂ molecules, so I₂ has the higher boiling point.

Worked example: comparing MgCO₃ and CaCO₃

Mg²⁺ is smaller than Ca²⁺ and therefore has greater charge density and polarising power.

It distorts CO₃²⁻ more strongly, making MgCO₃ easier to decompose. MgCO₃ is therefore less thermally stable than CaCO₃.

Common mistakes when explaining periodic trends

Periodic Table checklist