The Periodic Table

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

GCE A-Level H1 Chemistry 8873-2027

Learning goals

recognise variation in the electronic configurations across a Period and down a Group
describe and explain qualitatively the general trends and variations in atomic radius, ionic radius, first ionisation energy and electronegativity: — across a Period in terms of shielding and nuclear charge
describe and explain qualitatively the general trends and variations in atomic radius, ionic radius, first ionisation energy and electronegativity: — down a Group in terms of increasing number of electron shells, shielding and nuclear charge
interpret the variation in melting point and in electrical conductivity across a Period in terms of structure and bonding in the elements (metallic, giant molecular, or simple molecular)
describe and explain the trend in volatility of the Group 17 elements in terms of instantaneous dipole- induced dipole attraction Trends and variations in chemical properties For elements in the third period (sodium to chlorine), candidates should be able to:
— state and explain the variation in the highest oxidation number of the elements in oxides (for Na2O; MgO; Al 2O3; SiO2; P4O10; SO3) and chlorides (for NaCl; MgCl 2; AlCl 3; SiCl 4; PCl 5)
— state and explain the variation in bonding in oxides and chlorides in terms of electronegativity (with the exception of AlCl 3)
— describe the reactions of the oxides with water (for Na2O; MgO; Al 2O3; SiO2; P4O10; SO3)
— describe and explain the acid/base behaviour of oxides (for Na2O; MgO; Al 2O3; SiO2; P4O10; SO3) and hydroxides (for NaOH; Mg(OH)2; Al (OH)3), including, where relevant, amphoteric behaviour in reaction with sodium hydroxide (only) and acids
— describe and explain the reactions of the chlorides with water (for NaCl; MgCl 2; AlCl 3; SiCl 4; PCl 5)
— suggest the types of structure and bonding present in the oxides and chlorides from observations of their chemical and physical properties For elements in Group 1 (lithium to caesium) and Group 17 (chlorine to iodine), candidates should be able to:
describe and explain the relative reactivity of elements of: — Group 1 as reducing agents in terms of ease of loss of electrons
describe and explain the relative reactivity of elements of: — Group 17 as oxidising agents in terms of ease of gain of electrons
describe and explain the trend in thermal stability of Group 17 hydrides in terms of bond energies In addition, candidates should be able to:
predict the characteristic properties of an element in a given Group by using knowledge of chemical periodicity
deduce the nature, possible position in the Periodic Table, and identity of unknown elements from given information of physical and chemical properties

H1 Periodic Table learning outcomes

The Periodic Table turns electronic structure into predictions about size, bonding, physical properties, acid–base behaviour and redox strength. Learn the causal models separately, then combine them when unfamiliar evidence asks you to identify an element.

Follow the six lessons in order for first learning. During revision, classify an error as an electron-attraction explanation, a structure–property link, a Period 3 equation, a group redox trend or an evidence-chain failure.

What You’ll Learn

Explain atomic radius, ionic radius, first ionisation energy and electronegativity using nuclear charge, shielding and electron-shell evidence.
Interpret Period 3 melting-point and electrical-conductivity patterns using metallic, giant molecular and simple molecular structures.
Explain the bonding, water reactions and acid–base behaviour of the specified Period 3 oxides, hydroxides and chlorides.
Explain why Group 1 reducing strength increases down the group through increasingly easy outer-electron loss.
Explain Group 17 volatility, oxidising strength and hydrogen-halide thermal stability using the correct distinct causal model.
Deduce an unknown element or compound from unfamiliar electronic, physical, acid–base and redox evidence.

Lessons (Recommended Order)

The sequence moves from general periodic drivers to Period 3 physical and chemical patterns, then Group 1, Group 17 and unfamiliar-data deduction.

Periodic Trends: Radius, Ionisation Energy and Electronegativity
Explain radius, ionisation energy, electronegativity and ionic-radius patterns through attraction and electron structure.
Open lesson
Period 3 Elements: Melting Point and Conductivity
Relate the Period 3 melting-point and conductivity pattern to metallic, giant and molecular structures.
Open lesson
Period 3 Oxides and Chlorides
Explain the bonding, water reactions and acid–base behaviour of the specified oxides, hydroxides and chlorides.
Open lesson
Group 1 as Reducing Agents
Explain why Group 1 reducing strength increases down the group through easier outer-electron loss.
Open lesson
Group 17 Chemistry Trends
Keep volatility, oxidising strength and hydrogen-halide stability tied to their three distinct causal models.
Open lesson
Deducing Unknown Elements from Data
Build evidence chains from unfamiliar electron, physical, acid–base and redox information.
Open lesson

Quick Reference

Idea	Exam-ready reminder
Across Period 3	effective nuclear attraction generally increases while shielding changes little
Isoelectronic ions	the species with more protons has the smaller radius
Period 3 structure	metallic → giant molecular → simple molecular
Oxides	basic → amphoteric → acidic across Period 3
Chlorides with water	distinguish no reaction, aquation and hydrolysis from the specified evidence
Group 1	reducing strength increases down the group as electron loss becomes easier
Group 17 volatility	larger, more polarisable molecules have stronger instantaneous dipole–induced dipole attractions
Group 17	oxidising strength decreases down the group as electron gain becomes harder
Hydrogen halides	thermal stability decreases as H–X bond energy decreases
Deduction	quote clue → state inference → cross-check → identify

Identify the changing structural driver, state how attraction or bonding changes, then connect that cause to the measured property or reaction. Quote supplied evidence when the command word is “deduce”.

Common Exam Traps

Describing a trend without linking nuclear charge, distance and shielding to attraction.
Explaining a Period 3 melting-point change without first naming the structure and particles.
Forcing aluminium chloride into a simplistic ionic-to-covalent electronegativity trend when the syllabus states it as the exception.
Confusing an oxide’s reaction with water with its separate acid–base reactions.
Calling a Group 1 metal reduced when it acts as a reducing agent; it donates electrons and is oxidised.
Using H2 Group 2 carbonate or standard-electrode-potential treatment in an H1 response.
Using intermolecular attraction to explain hydrogen-halide decomposition, which requires breaking an H–X covalent bond.
Naming an unknown element from one clue without testing the conclusion against the remaining evidence.

Practice and Check Your Understanding

H1 Periodic Table Topic Practice
Course-specific 8873 questions across Period 3, Group 1, Group 17 and unfamiliar evidence.
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H1 Periodic Table: Check Your Understanding
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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.

First ionisation energies: Na 496, K 419, Cs 376 kJ mol⁻¹. Which is the strongest reducing agent, and why?

Caesium, because its outer 6s electron is furthest from the nucleus and most shielded, so it is lost most easily.
Sodium, because its small atoms hold their electrons tightly.
They are equally strong, because all lose one electron to form M⁺.
Caesium, because it has the greatest nuclear charge, which pushes out the electron.

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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

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.

First ionisation energies: Na 496, K 419, Cs 376 kJ mol⁻¹. Which is the strongest reducing agent, and why?

Caesium, because its outer 6s electron is furthest from the nucleus and most shielded, so it is lost most easily.
Sodium, because its small atoms hold their electrons tightly.
They are equally strong, because all lose one electron to form M⁺.
Caesium, because it has the greatest nuclear charge, which pushes out the electron.

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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.

Predict how rubidium reacts with cold water compared with potassium, and explain the difference.

More vigorously, forming RbOH(aq) and H₂, because rubidium’s outer 5s electron is further from the nucleus and more shielded, so it is lost more easily and Rb is the stronger reducing agent.
Less vigorously, because rubidium has a greater nuclear charge that holds its outer electron more tightly.
At the same rate, because both have one outer electron.
More vigorously, forming Rb₂O₂ and H₂O, because rubidium gains electrons more easily.

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The Periodic Table

H1 Periodic Table learning outcomes

What You’ll Learn

Lessons (Recommended Order)

Periodic Trends: Radius, Ionisation Energy and Electronegativity

Period 3 Elements: Melting Point and Conductivity

Period 3 Oxides and Chlorides

Group 1 as Reducing Agents

Group 17 Chemistry Trends

Deducing Unknown Elements from Data

Quick Reference

Practice and Check Your Understanding

H1 Periodic Table Topic Practice

H1 Periodic Table: Check Your Understanding

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