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8873 / 2027
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 understanding

Put Na+, Mg2+ and Al3+ in order of decreasing radius.

  1. Al3+ > Mg2+ > Na+, because more protons means a bigger nucleus and ion
  2. Na+ > Mg2+ > Al3+, because nuclear charge rises with the same 10 electrons
  3. The three are the same size, because they have the same 10 electrons
Show the answer

Na+ > Mg2+ > Al3+, because nuclear charge rises with the same 10 electrons

All three ions have 10 electrons, but 11, 12 and 13 protons. A larger nuclear charge pulls the same electrons in 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 understanding

Sulfur atoms have more protons than silicon atoms. Why does sulfur melt at a much lower temperature?

  1. Silicon is a giant covalent network; sulfur is made of S8 molecules held by weak forces.
  2. Sulfur is a metal with weak metallic bonds, while silicon is a non-metal with ionic bonds.
  3. Sulfur's extra protons pull its electrons in, so its bonds become weaker and easier to break.
Show the answer

Silicon is a giant covalent network; sulfur is made of S8 molecules held by weak forces.

Melting silicon breaks strong covalent bonds throughout the network. Melting sulfur overcomes only weak forces between S8 molecules. The structure matters, not the proton count.

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 examplePredict 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?

Try it first. Then show the worked stepsHide the worked steps
  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.

Check your understanding

Which Period 3 oxide reacts with both dilute acid and aqueous sodium hydroxide?

  1. Aluminium oxide, Al2O3
  2. Sulfur trioxide, SO3
  3. Magnesium oxide, MgO
Show the answer

Aluminium oxide, Al2O3

Al2O3 is amphoteric: it reacts with acids to give Al3+ salts and with NaOH to give aluminate ions. Na2O and MgO are basic; SiO2, P4O10 and SO3 are acidic.

04

Dissolving and hydrolysing are different processes

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

Make a guess. You are not marked.

Sodium chloride and phosphorus(V) chloride are each added to water. How do the pH values compare?

  1. NaCl gives about pH 7; PCl5 gives about pH 2
  2. Both give about pH 7, because both are chlorides
  3. Both give about pH 2, because both contain chlorine
Show the answer

NaCl gives about pH 7; PCl5 gives about pH 2

NaCl is ionic and simply dissolves. PCl5 is covalent and is hydrolysed, producing HCl and H3PO4, which make the solution strongly acidic.

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 understanding

A Period 3 chloride reacts with water to give an acidic solution and a white insoluble oxide. Which chloride is it?

  1. PCl5, which hydrolyses to give P4O10 as a solid
  2. NaCl, which dissolves to give a neutral solution
  3. SiCl4, which hydrolyses to SiO2 and HCl
Show the answer

SiCl4, which hydrolyses to SiO2 and HCl

SiCl4 + 2H2O -> SiO2 + 4HCl. The white solid is SiO2 and the HCl makes the solution strongly acidic.

05

Separate losing electrons from gaining them

Group 1 metals reduce other species; halogens oxidise them.

Make a guess. You are not marked.

Chlorine water is added to colourless aqueous potassium iodide. What do you see?

  1. A white precipitate of KCl forms
  2. No change, because both are halogens
  3. The solution turns brown as iodine forms
Show the answer

The solution turns brown as iodine forms

Chlorine is a stronger oxidising agent than iodine, so it oxidises iodide to iodine. Iodine in water is brown.

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.

Check your understanding

Which mixture shows no displacement reaction?

  1. Chlorine water added to aqueous bromide ions
  2. Iodine solution added to aqueous chloride ions
  3. Bromine water added to aqueous iodide ions
Show the answer

Iodine solution added to aqueous chloride ions

Iodine is too weak an oxidising agent to oxidise Cl- to Cl2, so nothing happens. Oxidising power falls down Group 17: Cl2 > Br2 > I2.

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 exampleIdentify 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.

Try it first. Then show the worked stepsHide the worked steps
  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 examplePredict an unfamiliar group member

An unfamiliar element has outer configuration ns2 np5 and lies below chlorine. Predict two properties relative to chlorine.

Try it first. Then show the worked stepsHide the worked steps
  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 exampleUse 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.

Try it first. Then show the worked stepsHide the worked steps
  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.

Exam-style question

Paper 2 Section B style

This question draws on Atomic Structure and the Periodic Table.

Q is an element in Period 3. Its first four ionisation energies are shown below.

Table 1. Successive ionisation energies of Q
ionisation1st2nd3rd4th
energy / kJ mol-15771820274011600

The first ionisation energy of magnesium is 736 kJ mol-1.

  1. (a) Define first ionisation energy. [1]
  2. (b) Explain why each successive ionisation energy of Q is larger than the one before. [2]
  3. (c) Use Table 1 to deduce the group of Q. Explain your answer. [2]
  4. (d) Write the full electronic configurations of a Q atom and of the Q3+ ion. [2]
  5. (e) Q has a larger nuclear charge than magnesium, yet its first ionisation energy is lower. Explain why. [2]
  6. (f) Explain why atomic radius decreases from sodium to chlorine. [2]
  7. (g) Describe how sodium oxide and phosphorus(V) oxide react with water. Give an equation for each and the approximate pH of the solution formed. [4]
  8. (h) The oxide of Q is amphoteric. Write ionic equations for its reactions with aqueous acid and with aqueous sodium hydroxide. [2]
  9. (i) Element R is also in Period 3. Its chloride is a volatile liquid that reacts with water to give a white solid and an acidic solution. Its oxide has a very high melting point. Identify R, write the equation for the reaction of its chloride with water, and explain the high melting point of its oxide. [3]
[20]

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Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Atomic radiusHalf the distance between the nuclei of two neighbouring atoms of the same element.
ShieldingThe repulsion of the outer electrons by inner-shell electrons, which reduces the attraction the outer electrons feel from the nucleus.
First ionisation energyThe energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous singly charged positive ions: M(g) → M+(g) + e-.
ElectronegativityThe ability of an atom in a molecule to attract the shared pair of electrons in a covalent bond towards itself.
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

Learning outcomes

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