The Periodic Table

Learn and apply The Periodic Table in the published Chemistry course sequence.

GCE A-Level H2 Chemistry 9476-2027

Learning goals

Periodic Trends (Radius, IE, EN)
Period 3 Elements: Melting Point and Conductivity
Period 3 Oxides and Chlorides
Group 2 Chemistry Trends
Group 17 Chemistry Trends
Deducing Unknown Elements from Data

H2 Periodic Table learning outcomes

This hub covers every Topic 5 outcome: physical periodicity, Period 3 chemistry, Group 2 and Group 17 trends, and evidence-based deductions.

The 9476 periodicity focus: sodium to chlorine across Period 3, magnesium to barium down Group 2, and chlorine to iodine down Group 17.

Before you start

Beginner path: follow the Lessons (Recommended Order) from top to bottom.

Revision: Jump to: Quick Reference · What You Must Memorise · Common Exam Traps · Hub Quiz.

Prerequisites: skim the Prerequisites list first if you’re rusty.

Useful links: A Level portal · Exam Skills

What You’ll Learn

Explain trends with the correct drivers (nuclear charge, shielding, distance, subshell/pairing where needed).
Link Period 3 structure and bonding to melting point, conductivity, oxidation number, oxide/hydroxide behaviour, and chloride hydrolysis.
Use supplied E⦵ values to compare Group 2 reducing strength and Group 17 oxidising strength; explain carbonate and hydrogen-halide thermal stability.
Deduce identity/position from unfamiliar data in a structured chain of reasoning.

Lessons (Recommended Order)

Periodic Trends (Radius, IE, EN)
Trend drivers plus ionic radius comparisons.
Open lesson
Period 3 Elements: Melting Point and Conductivity
Explain physical-property patterns using structure and bonding.
Open lesson
Period 3 Oxides and Chlorides
Oxide acidity, amphoterism, and chloride hydrolysis.
Open lesson
Group 2 Chemistry Trends
Reducing strength from electrode potentials and carbonate thermal stability.
Open lesson
Group 17 Chemistry Trends
Volatility, oxidising strength from electrode potentials, and hydrogen-halide stability.
Open lesson
Deducing Unknown Elements from Data
“Unknown element” workflows (data → inference → conclusion).
Open lesson

Quick Reference

Explain a trend: state the change in Z, shielding and distance; link the resulting attraction to the property.
Compare ionic radii: check electron count first; within an isoelectronic series, higher Z means smaller radius.
Classify an oxide or hydroxide: state basic, amphoteric or acidic, then prove it with the requested reaction.
Use E⦵ data: remember that tables show reductions. More positive means the left-hand oxidant is reduced more readily; more negative E⦵(M²⁺/M) means the metal is oxidised more readily.
Deduce an element: quote the clue, make one inference, cross-check it, then state the conclusion.

What You Must Memorise

Across a period: Z increases; electrons added to same shell; shielding changes little → attraction increases.
Down a group: more shells → distance and shielding increase → attraction decreases.
Isoelectronic ions: same electrons; higher Z → smaller radius.
Oxide acidity across Period 3: basic → amphoteric → acidic (support with equations).
Highest oxidation number: rises from +1 in Na₂O/NaCl to +6 in SO₃; the specified chlorides stop at +5 in PCl₅.
Thermal stability: Group 2 carbonates increase down the group; hydrogen halides decrease from HCl to HI.

Common Exam Traps

Writing “it increases” with no driver (nuclear charge / shielding / distance) and no attraction link.
Treating shielding as the main across-period driver (it changes little across a period).
Mixing atomic radius vs ionic radius rules (cation smaller; anion larger; isoelectronic → higher Z is smaller).
Period 3 oxides: claiming acidic/basic/amphoteric without a supporting reaction.
Group 2: treating a more negative E⦵(M²⁺/M) as a weaker reducing metal, or omitting charge density and polarisation for carbonates.
Group 17: treating a less positive E⦵(X₂/X⁻) as a stronger oxidising halogen, or confusing H–X bond strength with intermolecular forces.
Halogens are diatomic: write Cl₂, Br₂, I₂ (not Cl).
Displacement ionic equations must conserve atoms and charge (and show the correct electron transfer logic).

Hub Quiz and Check Your Understanding

Use the quiz without notes first. If a trend explanation fails, identify whether the missing link is electronic structure, attraction, bonding, redox language or evidence selection before revisiting the lesson.

Periodic Table Hub Quiz
Check Period 3, Group 2, Group 17 and unfamiliar-data deductions across the 9476 scope.
Start hub quiz
H2 Periodic Table Knowledge Check
Find the idea that needs more work, use the feedback, then try a fresh question independently.
Practice onlyCheck understanding

Practise

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Electronegativity increases from sodium (0.9) to chlorine (3.0). Which explanation is correct?

Nuclear charge increases while the outer electrons stay in the third shell with similar shielding, so a bonding pair is attracted more strongly.
Chlorine has more electrons, so it attracts other electrons more strongly.
The atomic radius increases across the period, so bonding electrons are held more strongly.
Chlorine has more electron shells than sodium, so it attracts bonding electrons more.

Electronegativity decreases down Group 16 from oxygen to tellurium. Which explanation is correct?

The bonding pair is in a shell further from the nucleus and is more shielded by inner shells, so it is attracted less strongly despite the higher nuclear charge.
Tellurium has fewer protons than oxygen.
Tellurium has fewer outer electrons than oxygen.
Larger atoms have stronger covalent bonds, so they attract bonding electrons less.

The melting points increase from sodium (98 °C) to magnesium (650 °C) to aluminium (660 °C). Which explanation is correct?

Each atom contributes more delocalised electrons (1, 2, 3) and the cations are smaller with higher charge, so metallic bonding becomes stronger.
Instantaneous dipole–induced dipole attractions increase as the atoms get more electrons.
More covalent bonds form between the atoms from sodium to aluminium.
The atoms become larger, so the metallic bonding becomes stronger.

Which statement about NaCl and PCl₅ is correct? (Electronegativities: Na 0.9, P 2.1, Cl 3.0.)

P is +5 in PCl₅, using all five outer electrons; NaCl is ionic because of the large electronegativity difference, while PCl₅ is covalent because the difference is small.
P is +3 in PCl₅, because phosphorus has three unpaired electrons, and PCl₅ is ionic.
P is −5 in PCl₅, because phosphorus gains five electrons.
P is +5 in PCl₅, and PCl₅ is ionic because chlorine is very electronegative.

Phosphorus(V) oxide, P₄O₁₀, is added to excess water. Which equation and pH are correct?

P₄O₁₀ + 6H₂O → 4H₃PO₄; pH about 2
P₄O₁₀ + 6H₂O → 4H₃PO₄; pH about 13
P₄O₁₀ does not react with water; pH 7
P₄O₁₀ + 6H₂O → 4PH₃ + 8O₂; pH 7

Aluminium chloride dissolves in water to give a solution of pH about 3. Which explanation is correct?

The small, highly charged Al³⁺ ion in [Al(H₂O)₆]³⁺ polarises its water ligands, so a proton is released: [Al(H₂O)₆]³⁺ ⇌ [Al(H₂O)₅(OH)]²⁺ + H⁺.
The chloride ions react with water to form HCl.
AlCl₃ dissolves without reaction, and the solution is acidic because aluminium is a metal.
The large Al³⁺ ion has a low charge density, so it releases H⁺ from water.

Bromine is a liquid and iodine a solid at room temperature, and hydrogen iodide decomposes more readily on heating than hydrogen bromide. Which pair of explanations is correct?

I₂ has more electrons, so its instantaneous dipole–induced dipole attractions are stronger; the H–I bond is weaker than the H–Br bond.
The I–I bond is stronger than the Br–Br bond; the H–I bond is weaker than the H–Br bond.
I₂ has stronger instantaneous dipole–induced dipole attractions; HI has weaker attractions between its molecules, so it decomposes.
I₂ has permanent dipoles; the H–I bond is stronger than the H–Br bond but HI is more polar.

Which equation represents a reaction that takes place in aqueous solution?

Br₂ + 2I⁻ → 2Br⁻ + I₂
I₂ + 2Br⁻ → 2I⁻ + Br₂
Br₂ + 2Cl⁻ → 2Br⁻ + Cl₂
I₂ + 2Cl⁻ → 2I⁻ + Cl₂

Element Z is in Period 3. It is a yellow solid that melts at about 115 °C and does not conduct electricity. It burns in oxygen to form ZO₂, a gas at room temperature. What is Z, and what structures do Z and ZO₂ have?

Sulfur; both the element (S₈) and its oxide SO₂ are simple molecular.
Silicon; the element and its oxide are giant covalent.
Phosphorus; the element is simple molecular and ZO₂ is giant covalent.
Sulfur; the element is simple molecular but SO₂ has a giant ionic lattice.

E⊖(Ca²⁺/Ca) = −2.87 V and E⊖(Sr²⁺/Sr) = −2.89 V. Which is the stronger reducing agent, and which carbonate decomposes at the lower temperature?

Strontium is the stronger reducing agent; CaCO₃ decomposes at the lower temperature.
Calcium is the stronger reducing agent; CaCO₃ decomposes at the lower temperature.
Strontium is the stronger reducing agent; SrCO₃ decomposes at the lower temperature.
Calcium is the stronger reducing agent; SrCO₃ decomposes at the lower temperature.

Practise after feedback

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About 10 minutes

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Electronegativity increases from sodium (0.9) to chlorine (3.0). Which explanation is correct?

Nuclear charge increases while the outer electrons stay in the third shell with similar shielding, so a bonding pair is attracted more strongly.
Chlorine has more electrons, so it attracts other electrons more strongly.
The atomic radius increases across the period, so bonding electrons are held more strongly.
Chlorine has more electron shells than sodium, so it attracts bonding electrons more.

Electronegativity decreases down Group 16 from oxygen to tellurium. Which explanation is correct?

The bonding pair is in a shell further from the nucleus and is more shielded by inner shells, so it is attracted less strongly despite the higher nuclear charge.
Tellurium has fewer protons than oxygen.
Tellurium has fewer outer electrons than oxygen.
Larger atoms have stronger covalent bonds, so they attract bonding electrons less.

The melting points increase from sodium (98 °C) to magnesium (650 °C) to aluminium (660 °C). Which explanation is correct?

Each atom contributes more delocalised electrons (1, 2, 3) and the cations are smaller with higher charge, so metallic bonding becomes stronger.
Instantaneous dipole–induced dipole attractions increase as the atoms get more electrons.
More covalent bonds form between the atoms from sodium to aluminium.
The atoms become larger, so the metallic bonding becomes stronger.

Which statement about NaCl and PCl₅ is correct? (Electronegativities: Na 0.9, P 2.1, Cl 3.0.)

P is +5 in PCl₅, using all five outer electrons; NaCl is ionic because of the large electronegativity difference, while PCl₅ is covalent because the difference is small.
P is +3 in PCl₅, because phosphorus has three unpaired electrons, and PCl₅ is ionic.
P is −5 in PCl₅, because phosphorus gains five electrons.
P is +5 in PCl₅, and PCl₅ is ionic because chlorine is very electronegative.

Phosphorus(V) oxide, P₄O₁₀, is added to excess water. Which equation and pH are correct?

P₄O₁₀ + 6H₂O → 4H₃PO₄; pH about 2
P₄O₁₀ + 6H₂O → 4H₃PO₄; pH about 13
P₄O₁₀ does not react with water; pH 7
P₄O₁₀ + 6H₂O → 4PH₃ + 8O₂; pH 7

Aluminium chloride dissolves in water to give a solution of pH about 3. Which explanation is correct?

The small, highly charged Al³⁺ ion in [Al(H₂O)₆]³⁺ polarises its water ligands, so a proton is released: [Al(H₂O)₆]³⁺ ⇌ [Al(H₂O)₅(OH)]²⁺ + H⁺.
The chloride ions react with water to form HCl.
AlCl₃ dissolves without reaction, and the solution is acidic because aluminium is a metal.
The large Al³⁺ ion has a low charge density, so it releases H⁺ from water.

Bromine is a liquid and iodine a solid at room temperature, and hydrogen iodide decomposes more readily on heating than hydrogen bromide. Which pair of explanations is correct?

I₂ has more electrons, so its instantaneous dipole–induced dipole attractions are stronger; the H–I bond is weaker than the H–Br bond.
The I–I bond is stronger than the Br–Br bond; the H–I bond is weaker than the H–Br bond.
I₂ has stronger instantaneous dipole–induced dipole attractions; HI has weaker attractions between its molecules, so it decomposes.
I₂ has permanent dipoles; the H–I bond is stronger than the H–Br bond but HI is more polar.

Which equation represents a reaction that takes place in aqueous solution?

Br₂ + 2I⁻ → 2Br⁻ + I₂
I₂ + 2Br⁻ → 2I⁻ + Br₂
Br₂ + 2Cl⁻ → 2Br⁻ + Cl₂
I₂ + 2Cl⁻ → 2I⁻ + Cl₂

Sulfur; both the element (S₈) and its oxide SO₂ are simple molecular.
Silicon; the element and its oxide are giant covalent.
Phosphorus; the element is simple molecular and ZO₂ is giant covalent.
Sulfur; the element is simple molecular but SO₂ has a giant ionic lattice.

E⊖(Ca²⁺/Ca) = −2.87 V and E⊖(Sr²⁺/Sr) = −2.89 V. Which is the stronger reducing agent, and which carbonate decomposes at the lower temperature?

Strontium is the stronger reducing agent; CaCO₃ decomposes at the lower temperature.
Calcium is the stronger reducing agent; CaCO₃ decomposes at the lower temperature.
Strontium is the stronger reducing agent; SrCO₃ decomposes at the lower temperature.
Calcium is the stronger reducing agent; SrCO₃ decomposes at the lower temperature.

Check what I know

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Atomic radius decreases from sodium to chlorine across Period 3. Which explanation is correct?

Electrons are added to the same third shell (3s, then 3p) while nuclear charge increases and shielding by the inner electrons stays about the same, so the outer electrons are pulled closer.
A new electron shell is added for each element, and the shells are packed more tightly.
Shielding increases greatly across the period, so the outer electrons are pulled closer.
The extra electrons repel each other and compress the outer shell.

Atomic radius increases down Group 15 from nitrogen to bismuth, even though nuclear charge increases. Which explanation is correct?

Each element has an extra occupied shell, so its outer electrons are further from the nucleus and more shielded; this outweighs the increase in nuclear charge.
Nuclear charge decreases down the group, so the electron cloud spreads out.
Shielding decreases down the group, so the outer electrons move further out.
The number of outer electrons increases down the group, so the atoms become larger.

Why does silicon have the highest melting point of the Period 3 elements?

Silicon has a giant covalent lattice, so many strong covalent bonds must be broken to melt it.
Silicon has the strongest metallic bonding in Period 3.
Silicon molecules have the strongest instantaneous dipole–induced dipole attractions.
Silicon has ionic bonding between Si⁴⁺ and Si⁴⁻ ions.

Consider the oxides Na₂O, MgO, Al₂O₃, SiO₂, P₄O₁₀ and SO₃. Which statement is correct?

The oxidation number of the Period 3 element rises from +1 to +6, equal to its number of outer electrons; bonding changes from ionic to covalent as the electronegativity difference with oxygen decreases.
The oxidation number rises from +1 to +6, but bonding changes from covalent to ionic as the element’s electronegativity increases.
The oxidation number is +2 in every oxide, because oxygen is always −2.
The oxidation number of phosphorus is +10 in P₄O₁₀ because it has ten oxygen atoms, and all the oxides are ionic.

Which statement about aluminium oxide, Al₂O₃, is correct?

It does not react with water, but it reacts with both hydrochloric acid and sodium hydroxide, so it is amphoteric: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄.
It reacts with water to give a solution of pH 7, because it is amphoteric.
It reacts only with acids, like magnesium oxide.
It reacts only with sodium hydroxide, like silicon dioxide.

What happens when silicon tetrachloride, SiCl₄, is added to water?

It reacts vigorously, releasing fumes of HCl: SiCl₄ + 2H₂O → SiO₂ + 4HCl, giving a solution of pH about 1–2.
It dissolves without reaction to give a neutral solution, like sodium chloride.
It does not react, like tetrachloromethane, because both are Group 14 chlorides.
It forms a solution of pH about 13 containing silicate ions.

From chlorine to iodine, volatility of the halogens decreases, and the thermal stability of the hydrogen halides decreases from HCl to HI. Which pair of explanations is correct?

Volatility falls because larger electron clouds give stronger instantaneous dipole–induced dipole attractions; stability falls because the H–X bond energy decreases.
Volatility falls because the X–X covalent bond gets stronger; stability falls because the H–X bond energy decreases.
Volatility falls because of stronger instantaneous dipole–induced dipole attractions; stability falls because the attractions between HX molecules decrease.
Volatility falls because the molecules become more polar; stability falls because HI has the strongest attractions between molecules.

Chlorine water is added to KBr(aq) and to KI(aq), and bromine water is added to KCl(aq). Which observations are correct?

Chlorine turns KBr(aq) orange-yellow and KI(aq) brown; bromine water does not react with KCl(aq).
Chlorine reacts with both, and bromine turns KCl(aq) pale green as chlorine forms.
Chlorine reacts with KBr(aq) only; there is no reaction with KI(aq).
None of the mixtures react, because all halogens have similar reactivity.

A Period 3 element X forms a white oxide that melts at 2852 °C and reacts slightly with water to give a solution of pH about 9. Its chloride melts at 714 °C and dissolves to give a solution of pH about 6.5. What is X, and what structures do its oxide and chloride have?

Magnesium; both the oxide and the chloride have giant ionic lattices.
Aluminium; the oxide is ionic and the chloride is simple molecular.
Sodium; both have giant ionic lattices.
Silicon; the oxide is giant covalent and the chloride simple molecular.

E⊖(Mg²⁺/Mg) = −2.38 V and E⊖(Ba²⁺/Ba) = −2.90 V. Which statement about magnesium and barium is correct?

Barium is the stronger reducing agent, and BaCO₃ is more thermally stable than MgCO₃ because the larger Ba²⁺ ion polarises the carbonate ion less.
Magnesium is the stronger reducing agent, and BaCO₃ is more thermally stable.
Barium is the stronger reducing agent, and MgCO₃ is more thermally stable because the small Mg²⁺ holds the carbonate tightly.
Magnesium is the stronger reducing agent, and MgCO₃ is more thermally stable.

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The Periodic Table

H2 Periodic Table learning outcomes

What You’ll Learn

Lessons (Recommended Order)

Periodic Trends (Radius, IE, EN)

Period 3 Elements: Melting Point and Conductivity

Period 3 Oxides and Chlorides

Group 2 Chemistry Trends

Group 17 Chemistry Trends

Deducing Unknown Elements from Data

Quick Reference

What You Must Memorise

Hub Quiz and Check Your Understanding

Periodic Table Hub Quiz

H2 Periodic Table Knowledge Check

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

Your answers

Check what I know

Starting check complete

Your answers

Check my progress

Progress check complete

Your answers

Check again

Progress check complete

Your answers

Review

Review complete

Your answers