Full chapter
The Periodic Table
Use electron arrangements and structure to explain trends, reactions and unknown elements.
A-Level 8873, revised syllabus (2026-2027)
Read the electron pattern across and down
A period adds electrons to one main shell; a group repeats an outer pattern.
| Element | Outer configuration | Group |
|---|---|---|
| Na | 3s1 | 1 |
| Mg | 3s2 | 2 |
| Al | 3s2 3p1 | 13 |
| Si | 3s2 3p2 | 14 |
| P | 3s2 3p3 | 15 |
| S | 3s2 3p4 | 16 |
| Cl | 3s2 3p5 | 17 |
From Na to Cl the nuclear charge increases. Added electrons enter the third shell, so the shielding of these outer electrons changes relatively little. Stronger effective attraction generally draws the outer shell closer: atomic radius decreases, while first ionisation energy and electronegativity generally increase.
Explain the two first-ionisation-energy exceptions using orbitals. Al loses a higher-energy 3p electron rather than Mg's 3s electron. S loses a paired 3p electron, with extra electron-electron repulsion, whereas P has three singly occupied 3p orbitals. The resulting dips do not mean the general nuclear-charge trend is wrong.
| Element | Outer configuration | Relative pattern down the group |
|---|---|---|
| Cl | 3s2 3p5 | Smaller atom; stronger attraction to outer electrons |
| Br | 4s2 4p5 | An additional principal shell |
| I | 5s2 5p5 | Larger atom; more shielding |
Down Cl to I, the outer shell is farther from the nucleus and is more shielded. These effects outweigh the increased nuclear charge: atomic radius increases, first ionisation energy decreases and electronegativity decreases. The number of outer electrons stays seven.
For ionic radii, compare species with similar electron arrangements. Na+, Mg2+ and Al3+ each have ten electrons: increasing nuclear charge contracts this isoelectronic series. P3-, S2- and Cl- each have eighteen; their radii also decrease in that order. Crossing from a cation series to an anion series introduces a large jump, so ionic radius does not form one smooth decreasing line across Period 3. Down Cl-, Br-, I-, more occupied shells give larger ions.
Check your understandingPut Na+, Mg2+ and Al3+ in decreasing radius and explain.Think it through, then reveal the answer
A melting-point trend can change its cause
First classify the structure; then compare the attractions.
| Elements | Structure | Explanation |
|---|---|---|
| Na, Mg, Al | Metallic | Melting requires overcoming metallic attraction. More delocalised electrons and greater positive-ion charge help explain the increase from Na to Mg/Al. |
| Si | Giant covalent | Many strong covalent bonds extend through the solid, so its melting point is high. |
| P, S, Cl | Simple molecular: P4, S8, Cl2 | Intermolecular attractions, not the bonds inside the molecules, are overcome. These melt far below silicon. |
The molecular comparison uses white phosphorus, P4. S8 has a larger, more easily polarised electron cloud than P4, so sulfur has stronger intermolecular attractions and a higher melting point. Cl2 is much smaller and melts much lower. Other allotropes can have different structures; an element name alone does not specify an allotrope.
Na, Mg and Al conduct electricity through mobile delocalised electrons. Conductivity generally increases from Na to Al: each atom supplies one, two and three valence electrons respectively, giving a greater density of mobile charge carriers. The ions remain in the solid lattice. Silicon's covalent network has far fewer mobile carriers at room temperature, so it is a semiconductor with much lower conductivity than the metals. P4, S8 and Cl2 lack mobile charged particles and are electrical insulators in their pure forms.
| Halogen | Appearance near room conditions | Volatility |
|---|---|---|
| Chlorine | Greenish-yellow gas | Highest of these three |
| Bromine | Red-brown liquid | Intermediate |
| Iodine | Dark solid; purple vapour on heating | Lowest of these three |
Down the group, the molecules have more electrons and more polarisable electron clouds. Instantaneous dipole-induced dipole attractions become stronger, so more energy is needed to separate molecules and volatility decreases. This is an intermolecular explanation; do not use the X-X covalent bond strength to explain boiling.
Check your understandingSulfur atoms have more protons than silicon atoms. Why does sulfur still melt much lower?Think it through, then reveal the answer
Move from basic oxides to acidic oxides
Track oxidation number, bonding and the reactions with water, acid and alkali.
In these highest oxides, oxygen has oxidation number -2. The element's highest oxidation number rises across the period: Na +1, Mg +2, Al +3, Si +4, P +5 and S +6. More outer electrons can participate in bonding across this sequence. With oxygen fixed, the electronegativity difference generally decreases, and bonding changes from predominantly ionic to covalent.
| Oxide; element oxidation number | Bonding/structure | With water |
|---|---|---|
| Na2O; +1 | Giant ionic | Na2O(s) + H2O(l) → 2NaOH(aq) |
| MgO; +2 | Giant ionic | MgO(s) + H2O(l) → Mg(OH)2(s); slow reaction, sparingly soluble hydroxide makes the water weakly alkaline |
| Al2O3; +3 | Predominantly ionic giant structure | No appreciable reaction with water |
| SiO2; +4 | Giant covalent | No reaction with water |
| P4O10; +5 | Covalent, molecular | P4O10(s) + 6H2O(l) → 4H3PO4(aq) |
| SO3; +6 | Covalent | SO3(g) + H2O(l) → H2SO4(aq); rapid and strongly exothermic when gaseous SO3 contacts water |
Na2O and MgO are basic: their oxide ions accept protons. For example, MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l). Their corresponding hydroxides neutralise acids: NaOH(aq) + H+(aq) → Na+(aq) + H2O(l), and Mg(OH)2(s) + 2H+(aq) → Mg2+(aq) + 2H2O(l). Magnesium hydroxide is only sparingly soluble, so an excess solid does not make an arbitrarily concentrated solution.
Al2O3 and Al(OH)3 are amphoteric: each reacts with acid and with strong alkali. The specified alkali is sodium hydroxide; heat the oxide with concentrated aqueous NaOH, whereas freshly precipitated Al(OH)3 dissolves in excess aqueous NaOH. With acid, Al2O3 + 6H+ → 2Al3+ + 3H2O; Al(OH)3 + 3H+ → Al3+ + 3H2O. In aqueous hydroxide, soluble aluminate species form.
| Starting solid | Balanced ionic equation |
|---|---|
| Al(OH)3 | Al(OH)3(s) + OH-(aq) → [Al(OH)4]-(aq) |
| Al2O3 | Al2O3(s) + 2OH-(aq) + 3H2O(l) → 2[Al(OH)4]-(aq) |
SiO2, P4O10 and SO3 are acidic oxides and react with alkali. Insoluble SiO2 does not need to react with water to be acidic: SiO2 + 2NaOH → Na2SiO3 + H2O with hot concentrated NaOH (or fused NaOH). With sufficient NaOH, P4O10 + 12NaOH → 4Na3PO4 + 6H2O; SO3 + 2NaOH → Na2SO4 + H2O.
Worked example
Predict from a pair of observations
A white oxide does not react with water. It dissolves in acid and also in aqueous NaOH. Which specified Period 3 oxide fits?
- No reaction with water alone leaves more than one candidate.
- Reaction with both acid and alkali establishes amphoteric behaviour.
- Of the named Period 3 oxides, Al2O3 fits.
Al2O3. The two chemical tests are more diagnostic than colour or insolubility alone.
Dissolving and hydrolysing are different processes
The pH after adding water reveals more than whether the solid disappears.
| Chloride; highest oxidation number | Bonding | Behaviour with water |
|---|---|---|
| NaCl; Na +1 | Ionic | Dissolves to Na+ and Cl-; approximately neutral solution. |
| MgCl2; Mg +2 | Ionic | Dissolves; slight hydrolysis of hydrated Mg2+ can make the solution mildly acidic. |
| AlCl3; Al +3 | Covalent when anhydrous | Dissolves and forms hydrated Al3+; marked hydrolysis produces an acidic solution. |
| SiCl4; Si +4 | Covalent molecular | Rapid hydrolysis: SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq). |
| PCl5; P +5 | Covalent in the molecular representation used for reactions | With excess water: PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq). |
Across the series the highest element oxidation number rises with the available outer electrons. Increasing electronegativity of the Period 3 element generally reduces the difference from chlorine and favours covalent bonding. AlCl3 is the stated exception to the simple electronegativity classification; the small, highly charged aluminium centre strongly polarises chloride electron density. Its molecular dimer Al2Cl6 is discussed in Bonding.
For NaCl, separating ions into water is principally dissolution. A hydrated Mg2+ or Al3+ ion attracts electron density from its water ligands, weakening O-H bonds and allowing proton release. The effect is stronger for Al3+, with greater charge density. It is the hydrated cation, not Cl- simply splitting into HCl, that explains this acidity.
For example, [Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5(OH)]2+ + H3O+. The equation makes proton release and charge conservation explicit. By contrast, SiCl4 and PCl5 undergo extensive hydrolysis of their covalent chlorides.
Check your understandingA Period 3 chloride reacts with water, forming an acidic solution and a white insoluble oxide. Which chloride is suggested?Think it through, then reveal the answer
Separate losing electrons from gaining them
Group 1 metals reduce other species; halogens oxidise them.
A reducing agent supplies electrons and is itself oxidised. Group 1 atoms have one outer electron: M → M+ + e-. From lithium through sodium, potassium, rubidium and caesium, the outer electron is farther from the nucleus and more shielded. It is generally lost more easily, so the familiar chemical reactivity increases down the group.
| Element | Outer configuration | Shared behaviour |
|---|---|---|
| Li | 2s1 | Forms a 1+ ion by losing the outer electron |
| Na | 3s1 | Same outer pattern |
| K | 4s1 | Same outer pattern |
| Rb | 5s1 | Same outer pattern |
| Cs | 6s1 | Same outer pattern |
A common reaction is 2M(s) + 2H2O(l) → 2MOH(aq) + H2(g). The metal is oxidised and water is reduced. For this H1 comparison, explain the trend through ease of electron loss; do not import an H2 electrode-potential ranking, which includes additional energetic factors.
An oxidising agent accepts electrons and is itself reduced. X2 + 2e- → 2X-. From chlorine to bromine to iodine, larger size and greater shielding weaken attraction for an incoming electron. Oxidising ability decreases in the order Cl2 > Br2 > I2.
| Mixture | Prediction | Reason |
|---|---|---|
| Cl2 + Br- | Cl2(aq) + 2Br-(aq) → 2Cl-(aq) + Br2(aq) | Chlorine is the stronger oxidant. |
| Br2 + I- | Br2(aq) + 2I-(aq) → 2Br-(aq) + I2(aq) | Bromine is stronger than iodine. |
| I2 + Cl- | No displacement reaction | Iodine is too weak an oxidant to oxidise chloride under these conditions. |
Thermal stability of the hydrogen halides is a different comparison. HCl is more thermally stable than HBr, which is more stable than HI. Down the group, the halogen atom becomes larger, the H-X bond longer and overlap poorer, so bond energy decreases. Less energy is needed to decompose HI: 2HI(g) ⇌ H2(g) + I2(g). This concerns covalent H-X bond breaking, not the intermolecular forces used to explain halogen volatility.
Check your understandingWhy does iodine have a higher boiling point but HI lower thermal stability than the chlorine comparison?Think it through, then reveal the answer
Use several clues to identify an unknown
Predict with a periodic pattern, then check the observations.
Worked example
Identify a Period 3 element
Element X conducts electricity as a solid. Its highest oxide contains X in oxidation state +3 and reacts with both acid and NaOH. Identify X.
- Conductivity suggests a metal among Na, Mg and Al.
- The +3 highest oxide selects aluminium from those candidates.
- The oxide amphoterism agrees with Al2O3, checking the first inference.
X is aluminium. State the independent clues rather than guessing from just one property.
Worked example
Predict an unfamiliar group member
An unfamiliar element has outer configuration ns2 np5 and lies below chlorine. Predict two properties relative to chlorine.
- Seven outer electrons place it in Group 17.
- A higher occupied shell gives greater size and shielding, so it attracts an incoming electron less strongly.
- A larger molecular electron cloud gives stronger instantaneous dipole-induced dipole attractions.
It is a weaker oxidising agent and its diatomic molecules are less volatile than chlorine. These are separate electronic and intermolecular explanations.
When inferring bonding from an oxide or chloride, combine melting behaviour, conductivity in different states and chemical reactions with water. High melting point does not uniquely establish ionic bonding: SiO2 is a covalent network. An acidic chloride solution also does not uniquely establish molecular hydrolysis: hydrated metal ions can release protons.
Worked example
Use an oxide and a chloride to test the same structural model
A Period 3 element forms an oxide that has a very high melting point, does not conduct when molten and does not react with water. Its chloride is a volatile liquid that reacts with water to produce an acidic solution and a white insoluble oxide. Suggest the element and both structures.
- The oxide has strong bonding throughout its structure, because melting needs much energy. Its non-conducting melt argues against an ordinary ionic lattice: mobile ions would carry charge.
- A giant covalent oxide fits these observations. SiO2 is the specified Period 3 example; its extended Si-O network also explains its failure to dissolve in water.
- The chloride is volatile, so the attractions separating its particles are relatively weak. A simple molecular chloride fits; giant ionic chlorides such as NaCl have high melting points.
- SiCl4 hydrolysis gives insoluble SiO2 and HCl, agreeing with both observed products. AlCl3 can also give acidic water, but that single observation cannot explain the entire set.
The evidence supports silicon: giant covalent SiO2 and simple molecular SiCl4. Bonding is inferred from several independent observations, then checked against the reaction products.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Property | Main cause |
|---|---|
| Atomic radius / first IE | Nuclear charge, shielding and occupied shell; pairing/subshell exceptions. |
| Halogen boiling/volatility | Intermolecular attractions increase with polarisability. |
| Hydrogen-halide thermal stability | H-X covalent bond energy falls down the group. |
| Group 1 reducing behaviour | Ease of losing the outer electron increases down Li-Cs. |
| Group 17 oxidising behaviour | Ease of accepting electrons decreases down Cl-I. |
| Period 3 oxide character | Basic → amphoteric → acidic; insolubility is not neutrality. |
| Chloride solution pH | Distinguish simple dissolution, hydrated-cation hydrolysis and molecular hydrolysis. |
Scope and references
Learning outcomes and sources
4. The Periodic Table (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
4(a) Recognise electronic patterns.
- Period 3 Na-Cl and Group 17 Cl-I.
4(b) Explain radius, first-ionisation and electronegativity trends.
- (i) Across a period: nuclear charge/shielding.
- (ii) Down a group: shells/shielding/nuclear charge; atomic and ionic radii.
4(c) Explain melting and conductivity across Period 3.
- Metallic, giant molecular and simple molecular structures.
4(d) Explain halogen volatility.
- Cl2, Br2, I2; instantaneous dipole-induced dipole attraction.
4(e) Connect Period 3 compounds to their behaviour.
- (i) Highest oxidation numbers in Na2O, MgO, Al2O3, SiO2, P4O10, SO3 and NaCl, MgCl2, AlCl3, SiCl4, PCl5.
- (ii) Bonding versus electronegativity; AlCl3 exception.
- (iii) All six oxides with water.
- (iv) Acid/base behaviour of six oxides and NaOH, Mg(OH)2, Al(OH)3; amphoterism with NaOH and acids.
- (v) All five chlorides with water.
- (vi) Infer oxide/chloride structure from chemical and physical evidence.
Move from basic oxides to acidic oxidesDissolving and hydrolysing are different processesUse several clues to identify an unknown
4(f) Compare reducing and oxidising behaviour.
- (i) Group 1 Li-Cs: ease of electron loss.
- (ii) Group 17 Cl-I: ease of electron gain.
4(g) Explain hydrogen-halide thermal stability.
- HCl, HBr, HI bond energies.
4(h) Predict properties from group membership.
- Use periodic patterns with explanations.
4(i) Deduce unknown element identity/position.
- Combine physical and chemical observations.
- SEAB H1 Chemistry 8873, 2026 revision
Official scope: section 4, printed pages 14-15. All lettered outcomes and nested requirements checked.
- SEAB H1 Chemistry 8873, 2027
Section 4: same substantive outcomes as the revised 2026 course.
- RI 2022 Periodic Table I (Grail)
Consulted printed pp. 5, 8-11 and chloride comparison questions for conductivity and oxide/chloride properties. Authored an original combined-evidence deduction; did not adopt the source claim that SiCl4 hydrolysis requires low-lying 3d orbitals.