Launching your learning experience…
Detailed notes on The Periodic Table for Cambridge IGCSE Coordinated Science, covering key concepts, explanations, examples, and exam-focused revision points.
Properties change in regular patterns across periods and down groups. Cambridge tests trends in reactivity, atomic radius, ionisation energy, melting point, and metallic/non-metallic character.
Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).
Atomic size, ionisation energy, and reactivity all follow predictable patterns explained by electron shell structure and nuclear charge.
Atomic radius:
Ionisation energy (energy to remove 1 outer electron from gaseous atom):
Reactivity of metals (Group I):
Reactivity of non-metals (Group VII):
Metallic to non-metallic character across Period 3: Na (metal) → Mg (metal) → Al (metal) → Si (metalloid) → P (non-metal) → S (non-metal) → Cl (non-metal) → Ar (noble gas)
Verbatim phrases and definitions Cambridge mark schemes credit.
Paper 4: 'Explain why potassium is more reactive than sodium' (3 marks — outer electron in shell further from nucleus, more shielding, less energy needed to remove it). 'State the trend in ionisation energy across Period 3' (1 mark — increases from left to right). MCQ tests which element is more reactive in displacement reactions.
Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (C9); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.
Step-by-step solutions to past-paper-style questions on periodic trends, written exactly the way a tutor would explain them at the board.
Question
Explain why atomic radius decreases from sodium to argon across Period 3.
Step-by-step solution
Step 1
Going from Na (Z=11) to Ar (Z=18), the atomic number (and therefore number of protons) increases from 11 to 18.
Step 2
All elements in Period 3 have the same number of electron shells (3 shells); the outer electrons are all in the 3rd shell.
Step 3
As nuclear charge increases (more protons), the nucleus attracts the outer electrons more strongly; the outer shell is pulled closer to the nucleus.
Step 4
Radius decreases from Na to Ar across Period 3.
Answer
Number of protons increases (11 → 18); same number of shells (3); stronger nuclear attraction pulls outer electrons closer → atomic radius decreases across Period 3.
Examiner tip
Key phrases: 'increasing nuclear charge', 'same number of shells (shielding constant)', 'stronger attraction on outer electrons'.
Question
Explain why ionisation energy decreases down Group I (from Li to Cs).
Step-by-step solution
Step 1
Going down Group I, each element has one more electron shell than the element above.
Step 2
The outer (valence) electron is in a shell further from the nucleus and is more shielded by inner shells from the nuclear charge.
Step 3
The effective nuclear attraction on the outer electron decreases; it requires less energy to remove the outer electron.
Step 4
Therefore, ionisation energy decreases down Group I.
Answer
More shells → outer electron further from nucleus → greater shielding → weaker nuclear attraction → less energy to remove outer electron → lower ionisation energy.
Question
State the trend in reactivity going down Group VII and explain the trend in terms of atomic structure.
Step-by-step solution
Step 1
Reactivity DECREASES going down Group VII (F > Cl > Br > I).
Step 2
Halogens react by gaining one electron to complete their outer shell.
Step 3
Going down the group, the outer shell is further from the nucleus and more shielded by inner electrons.
Step 4
It becomes harder for the nucleus to attract an incoming electron to the outer shell → decreasing reactivity.
Answer
Reactivity decreases down Group VII; the outer shell is further from the nucleus and more shielded → weaker attraction for an extra electron → less reactive.
Examiner tip
Contrast with Group I (reactivity increases down): metals react by losing electrons; easier to lose further from nucleus. Halogens react by gaining electrons; harder to attract further from nucleus.
Definitions to memorise and the exact keywords mark schemes credit for periodic trends answers — sharpened from recent examiner reports for the 2026 0654 sitting.
The energy required to remove one mole of electrons from one mole of atoms in the gaseous state to form one mole of positive ions: X(g)→X+(g)+e−. Increases across a period; decreases down a group.
A measure of the size of an atom; decreases across a period (increasing nuclear charge) and increases down a group (additional electron shells).
The reduction in the effective nuclear attraction on outer electrons caused by inner shell electrons. Greater shielding → weaker effective nuclear charge on outer electrons.
The positive charge of the nucleus, equal to the number of protons (atomic number). Increasing nuclear charge across a period increases attraction on outer electrons.
The tendency of an element to lose electrons and form positive ions; decreases across a period (left to right) and increases down a group.
The traps other students keep falling into on periodic trends questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.
0654 Examiner Report 2023
Why it happens
Students apply the same reactivity trend to both groups without considering whether the reaction involves gaining or losing electrons.
How to avoid it
Group I: reactivity INCREASES down (easier to LOSE outer electron). Group VII: reactivity DECREASES down (harder to GAIN electron). Always state whether electrons are lost or gained.
Why it happens
Students state the trend without linking it to the mechanism.
How to avoid it
Explain: inner electrons repel the nuclear attraction on outer electrons → reduced effective nuclear charge on outer electrons → weaker attraction.
Why it happens
Students see more electrons and assume a larger atom, ignoring the stronger nuclear charge.
How to avoid it
More electrons across a period are all in the SAME shell; the increasing nuclear charge pulls them all closer → SMALLER radius across a period.
The things students keep getting wrong in this sub-topic, answered.