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The Periodic Table

Full chapter

The Periodic Table

Use electron arrangements and structure to explain trends, reactions and unknown elements.

A-Level 8873, revised syllabus (2026-2027)

01

Read the electron pattern across and down

A period adds electrons to one main shell; a group repeats an outer pattern.

Period 3: the outer shell fills
ElementOuter configurationGroup
Na3s11
Mg3s22
Al3s2 3p113
Si3s2 3p214
P3s2 3p315
S3s2 3p416
Cl3s2 3p517

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.

Group 17: same outer pattern, a different shell
ElementOuter configurationRelative pattern down the group
Cl3s2 3p5Smaller atom; stronger attraction to outer electrons
Br4s2 4p5An additional principal shell
I5s2 5p5Larger 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
Na+ > Mg2+ > Al3+. They all have ten electrons, but the proton numbers are 11, 12 and 13. The larger nuclear charge attracts the same electron population more strongly.
02

A melting-point trend can change its cause

First classify the structure; then compare the attractions.

Why Period 3 melting points do not form a smooth trend
ElementsStructureExplanation
Na, Mg, AlMetallicMelting requires overcoming metallic attraction. More delocalised electrons and greater positive-ion charge help explain the increase from Na to Mg/Al.
SiGiant covalentMany strong covalent bonds extend through the solid, so its melting point is high.
P, S, ClSimple molecular: P4, S8, Cl2Intermolecular 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 volatility: Cl2 to I2
HalogenAppearance near room conditionsVolatility
ChlorineGreenish-yellow gasHighest of these three
BromineRed-brown liquidIntermediate
IodineDark solid; purple vapour on heatingLowest 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
Silicon is a continuous covalent network, whereas ordinary sulfur consists of S8 molecules. Sulfur melting overcomes weaker intermolecular forces; comparing proton counts misses the change in structure.
03

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.

Every specified oxide
Oxide; element oxidation numberBonding/structureWith water
Na2O; +1Giant ionicNa2O(s) + H2O(l) → 2NaOH(aq)
MgO; +2Giant ionicMgO(s) + H2O(l) → Mg(OH)2(s); slow reaction, sparingly soluble hydroxide makes the water weakly alkaline
Al2O3; +3Predominantly ionic giant structureNo appreciable reaction with water
SiO2; +4Giant covalentNo reaction with water
P4O10; +5Covalent, molecularP4O10(s) + 6H2O(l) → 4H3PO4(aq)
SO3; +6CovalentSO3(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.

Amphoteric reactions in aqueous NaOH
Starting solidBalanced ionic equation
Al(OH)3Al(OH)3(s) + OH-(aq) → [Al(OH)4]-(aq)
Al2O3Al2O3(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?

  1. No reaction with water alone leaves more than one candidate.
  2. Reaction with both acid and alkali establishes amphoteric behaviour.
  3. Of the named Period 3 oxides, Al2O3 fits.
Answer

Al2O3. The two chemical tests are more diagnostic than colour or insolubility alone.

04

Dissolving and hydrolysing are different processes

The pH after adding water reveals more than whether the solid disappears.

Every specified chloride
Chloride; highest oxidation numberBondingBehaviour with water
NaCl; Na +1IonicDissolves to Na+ and Cl-; approximately neutral solution.
MgCl2; Mg +2IonicDissolves; slight hydrolysis of hydrated Mg2+ can make the solution mildly acidic.
AlCl3; Al +3Covalent when anhydrousDissolves and forms hydrated Al3+; marked hydrolysis produces an acidic solution.
SiCl4; Si +4Covalent molecularRapid hydrolysis: SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq).
PCl5; P +5Covalent in the molecular representation used for reactionsWith 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
SiCl4 is suggested: hydrolysis gives SiO2 and HCl. Support the inference with the products, rather than saying only that the original chloride is covalent.
05

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.

The required Group 1 range
ElementOuter configurationShared behaviour
Li2s1Forms a 1+ ion by losing the outer electron
Na3s1Same outer pattern
K4s1Same outer pattern
Rb5s1Same outer pattern
Cs6s1Same 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.

Displacement is a redox test
MixturePredictionReason
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 reactionIodine 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.

06

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.

  1. Conductivity suggests a metal among Na, Mg and Al.
  2. The +3 highest oxide selects aluminium from those candidates.
  3. The oxide amphoterism agrees with Al2O3, checking the first inference.
Answer

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.

  1. Seven outer electrons place it in Group 17.
  2. A higher occupied shell gives greater size and shielding, so it attracts an incoming electron less strongly.
  3. A larger molecular electron cloud gives stronger instantaneous dipole-induced dipole attractions.
Answer

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.

  1. 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.
  2. 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.
  3. 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.
  4. 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.
Answer

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.

Choose the right explanation
PropertyMain cause
Atomic radius / first IENuclear charge, shielding and occupied shell; pairing/subshell exceptions.
Halogen boiling/volatilityIntermolecular attractions increase with polarisability.
Hydrogen-halide thermal stabilityH-X covalent bond energy falls down the group.
Group 1 reducing behaviourEase of losing the outer electron increases down Li-Cs.
Group 17 oxidising behaviourEase of accepting electrons decreases down Cl-I.
Period 3 oxide characterBasic → amphoteric → acidic; insolubility is not neutrality.
Chloride solution pHDistinguish 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
  1. 4(a) Recognise electronic patterns.

    • Period 3 Na-Cl and Group 17 Cl-I.

    Read the electron pattern across and down

  2. 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.

    Read the electron pattern across and down

  3. 4(c) Explain melting and conductivity across Period 3.

    • Metallic, giant molecular and simple molecular structures.

    A melting-point trend can change its cause

  4. 4(d) Explain halogen volatility.

    • Cl2, Br2, I2; instantaneous dipole-induced dipole attraction.

    A melting-point trend can change its cause

  5. 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

  6. 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.

    Separate losing electrons from gaining them

  7. 4(g) Explain hydrogen-halide thermal stability.

    • HCl, HBr, HI bond energies.

    Separate losing electrons from gaining them

  8. 4(h) Predict properties from group membership.

    • Use periodic patterns with explanations.

    Use several clues to identify an unknown

  9. 4(i) Deduce unknown element identity/position.

    • Combine physical and chemical observations.

    Use several clues to identify an unknown

  • 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.