Cambridge IGCSE Chemistry 0620: Elements, compounds and mixtures; atomic structure and the Periodic Table; isotopes – Study Guide

This study guide teaches sub-topics 2.1 (Elements, compounds and mixtures), 2.2 (Atomic structure and the Periodic Table) and 2.3 (Isotopes) of topic 2 in the Cambridge IGCSE Chemistry 0620 syllabus for examination in 2026, 2027 and 2028 (version 2). Everything in 2.1 and 2.2 is Core, so every candidate needs it. In 2.3, outcomes 1 and 2 are Core, and outcomes 3 and 4 are Supplement content, labelled Extended only below. Core content is examined on Paper 1 (multiple choice) and Paper 3 (theory); Core and Supplement content is examined on Paper 2 (multiple choice) and Paper 4 (theory).

Use it alongside the Cambridge IGCSE Chemistry hub and the printable 0620 checklist. For a short recap of the same content, read the atomic structure revision notes; to test yourself, use the atomic structure practice questions.

What these sub-topics cover

0620 outcome	What you must be able to do	Tier
2.1.1	Describe the differences between elements, compounds and mixtures	Core
2.2.1	Describe the atom as a central nucleus of protons and neutrons, surrounded by electrons in shells	Core
2.2.2	State the relative charges and relative masses of a proton, a neutron and an electron	Core
2.2.3	Define proton number / atomic number	Core
2.2.4	Define mass number / nucleon number	Core
2.2.5	Determine the electronic configuration of elements and their ions, proton number 1 to 20	Core
2.2.6	State the links between electronic configuration, group and period	Core
2.3.1	Define isotopes	Core
2.3.2	Interpret and use symbols for atoms and ions	Core
2.3.3	State why isotopes of an element have the same chemical properties	Extended only
2.3.4	Calculate relative atomic mass from isotope masses and abundances	Extended only

2.1 Elements, compounds and mixtures

• An element is a substance made of only one type of atom. It cannot be broken down into anything simpler by a chemical reaction.
• A compound is a substance made of two or more elements chemically combined in a fixed ratio.
• A mixture is two or more substances that are not chemically combined. The proportions can vary.

Feature	Element	Compound	Mixture
Types of atom	one	two or more, bonded together	can be any; substances not bonded to each other
Composition	fixed	fixed ratio (one formula)	variable
Properties	its own	different from the elements it is made from	each substance keeps its own properties
How to separate	cannot be split	only by a chemical reaction	by physical methods, e.g. filtration, distillation, magnet

Two points cause confusion.

• Molecules can be elements. Oxygen gas is made of O₂ molecules. Each molecule has two atoms, but both are oxygen atoms, so oxygen is an element, not a compound.
• “Pure” does not mean “element”. Water is a pure substance, but it is a compound, because each molecule contains hydrogen and oxygen atoms bonded together.

Worked example 1: iron, sulfur and iron(II) sulfide

A student mixes iron filings with powdered sulfur, then heats some of the mixture strongly. A black solid, iron(II) sulfide (FeS), forms. Describe how the unheated mixture differs from the product.

Mixture: not bonded; any proportions; a magnet pulls the iron out (physical)
FeS:     Fe and S chemically bonded; fixed ratio 1 : 1; new properties;
         iron recovered only by a chemical reaction

Worked example 2: classifying substances

Classify each substance as an element, a compound or a mixture: argon, carbon dioxide, air, sea water, nitrogen gas, sodium chloride.

elements:  argon; nitrogen gas (N2 molecules, but only N atoms)
compounds: carbon dioxide (CO2); sodium chloride (NaCl) - fixed ratios
mixtures:  air; sea water - substances not bonded, composition varies

In a particle diagram, look at each separate particle. If every particle is identical and contains one kind of atom, it is an element. If every particle is identical but contains more than one kind of atom, it is a compound. If there is more than one kind of particle, it is a mixture.

2.2 Atomic structure and the Periodic Table

The structure of the atom

An atom has a tiny central nucleus containing protons and neutrons. Electrons are arranged in shells around the nucleus. Almost all the mass of the atom is in the nucleus.

Particle	Where it is	Relative charge	Relative mass
proton	nucleus	+1	1
neutron	nucleus	0	1
electron	shells	−1	1/1840 (negligible)

The electron’s mass is roughly 1/1840 of a proton’s, which is why it is usually called negligible.

Proton number and mass number

• Proton number (atomic number) = the number of protons in the nucleus of an atom.
• Mass number (nucleon number) = the total number of protons and neutrons in the nucleus of an atom.

Use the definitions word for word in an exam. “Number of protons” alone is not enough for mass number, and “number of particles” is too vague.

From these two numbers you can work out everything else:

protons   = proton number
neutrons  = mass number − proton number
electrons = proton number   (for a neutral atom)
electrons = proton number − charge   (for an ion)

Worked example 3: counting particles in atoms and ions

Complete the table for an aluminium atom (proton number 13, mass number 27), an aluminium ion Al³⁺, a sulfide ion S²⁻ (proton number 16, mass number 32) and a calcium ion Ca²⁺ (proton number 20, mass number 40).

particle   protons   neutrons        electrons
Al         13        27 − 13 = 14    13
Al3+       13        14              13 − 3 = 10
S2−        16        32 − 16 = 16    16 + 2 = 18
Ca2+       20        40 − 20 = 20    20 − 2 = 18

Forming an ion changes only the number of electrons. The protons and neutrons stay the same, so Al³⁺ is still aluminium.

Electronic configuration (proton number 1 to 20)

Electrons fill shells from the nucleus outwards. For proton numbers 1 to 20:

first shell    up to 2 electrons
second shell   up to 8 electrons
third shell    up to 8 electrons (for this syllabus range)
fourth shell   the remaining electrons (only K and Ca in this range)

Write the configuration as numbers separated by commas, one per shell, as the syllabus does: 2,8,3 for aluminium.

Worked example 4: atoms

Write the electronic configurations of nitrogen (7), silicon (14), argon (18) and calcium (20).

N   7 electrons:  2 then 5         -> 2,5
Si  14 electrons: 2, 8, then 4     -> 2,8,4
Ar  18 electrons: 2, 8, then 8     -> 2,8,8
Ca  20 electrons: 2, 8, 8, then 2  -> 2,8,8,2

Worked example 5: ions

Write the electronic configurations of N³⁻, F⁻ and S²⁻.

Step 1: find the number of electrons in the ion
  N3−   7 + 3 = 10
  F−    9 + 1 = 10
  S2−   16 + 2 = 18
Step 2: fill the shells with that number
  N3−   2,8
  F−    2,8
  S2−   2,8,8

Notice that N³⁻, F⁻ and Al³⁺ all end up as 2,8, the same as neon. S²⁻ and Ca²⁺ are 2,8,8, the same as argon. Why ions form is sub-topic 2.4.

Group, period and the noble gases

The syllabus asks you to state three facts:

1. Group VIII noble gases have a full outer electron shell (He 2; Ne 2,8; Ar 2,8,8).
2. The number of outer shell electrons equals the group number in Groups I to VII.
3. The number of occupied electron shells equals the period number.

Fact 2 does not apply to Group VIII. Helium has only 2 outer electrons, yet it is in Group VIII because its only shell is full.

Worked example 6: from configuration to position, and back

(a) An element has the configuration 2,8,5. State its group, its period and its proton number.

outer shell has 5 electrons   -> Group V
3 shells occupied             -> Period 3
total electrons 2 + 8 + 5 = 15 -> proton number 15 (phosphorus)

(b) An element is in Period 2 and Group VI. Write its configuration and identify it.

Period 2   -> 2 shells
Group VI   -> 6 outer electrons
configuration 2,6 ; total 8 electrons -> proton number 8 (oxygen)

Use the configuration of the atom for this, not of an ion. The configuration of O²⁻ (2,8) would wrongly suggest Group VIII.

2.3 Isotopes

Definition

Isotopes are different atoms of the same element that have the same number of protons but different numbers of neutrons. Because the neutron number differs, the mass number differs too.

Interpreting symbols for atoms and ions

In a symbol such as ¹⁸₈O, the top number is the mass number and the bottom number is the proton number. For an ion, the charge goes at the top right, as in ³⁵₁₇Cl⁻.

Worked example 7: oxygen isotopes

Oxygen has three naturally occurring isotopes: ¹⁶₈O, ¹⁷₈O and ¹⁸₈O. Give the numbers of protons, neutrons and electrons in each atom.

isotope   protons   neutrons       electrons
16-O      8         16 − 8 = 8     8
17-O      8         17 − 8 = 9     8
18-O      8         18 − 8 = 10    8

Worked example 8: are these isotopes?

Which of these pairs are isotopes of each other? (i) ⁴⁰₁₈Ar and ⁴⁰₂₀Ca (ii) ¹⁹₉F and ¹⁹₉F⁻ (iii) ¹⁶₈O and ¹⁸₈O.

(i)   proton numbers 18 and 20 -> different elements -> NOT isotopes
      (same mass number does not matter; neutrons are 22 and 20)
(ii)  same protons (9) and same neutrons (10); only electrons differ
      -> an atom and its ion, NOT isotopes
(iii) same protons (8), neutrons 8 and 10 -> isotopes

Why isotopes have the same chemical properties (Extended only)

Isotopes of the same element have the same number of electrons and therefore the same electronic configuration. Chemical reactions depend on electrons, so isotopes react in the same way. A full answer needs both links: same number of electrons, so same configuration, so same chemical properties. Saying only “same number of protons” does not explain it.

Calculating relative atomic mass (Extended only)

Relative atomic mass, Aᵣ, is an average of the masses of an element’s isotopes, weighted by how common each one is. (Sub-topic 3.2 describes Aᵣ as the average mass of the isotopes compared with 1/12 of the mass of an atom of carbon-12; see formulae, equations and the mole.)

Ar = sum of (relative isotopic mass × abundance) ÷ total abundance

When abundances are percentages, the total is 100.

Worked example 9: boron (Extended only)

Boron is about 20% boron-10 and 80% boron-11. Calculate its relative atomic mass.

Ar = (10 × 20 + 11 × 80) ÷ 100
   = (200 + 880) ÷ 100
   = 1080 ÷ 100
   = 10.8

The answer lies between 10 and 11 and closer to 11, because boron-11 is more common. That is a quick sense check. A plain average, (10 + 11) ÷ 2 = 10.5, ignores abundance and scores no credit.

Worked example 10: three isotopes (Extended only)

Magnesium is about 79% magnesium-24, 10% magnesium-25 and 11% magnesium-26. Calculate Aᵣ to one decimal place.

check abundances: 79 + 10 + 11 = 100
Ar = (24 × 79 + 25 × 10 + 26 × 11) ÷ 100
   = (1896 + 250 + 286) ÷ 100
   = 2432 ÷ 100
   = 24.32
   = 24.3 (1 d.p.)

Common errors

• Calling a substance made of molecules a compound automatically. O₂, N₂ and Cl₂ are elements.
• Giving the electron a relative mass of exactly 0. Its mass is very small, not zero: write 1/1840 or “negligible”.
• Defining mass number as “the mass of the atom” or “the number of protons”. It is the total number of protons and neutrons.
• Changing the proton number when writing an ion. Only the electrons change.
• Reading the group from an ion’s configuration instead of the atom’s.
• Treating ⁴⁰Ar and ⁴⁰Ca as isotopes because the mass numbers match.
• (Extended) Taking a simple average of isotope masses, or forgetting to divide by the total abundance.

Where to go next

Recap with the atomic structure revision notes and test yourself with the atomic structure practice questions. Sub-topic 2.4 onwards is in the bonding study guide, bonding revision notes and bonding practice questions. Check where you stand with the free 10-minute 0620 diagnostics: Core or Extended.

Official syllabus

Cambridge IGCSE Chemistry 0620 syllabus for examination in 2026, 2027 and 2028, version 2 (August 2026), Cambridge University Press & Assessment. Sub-topics 2.1, 2.2 and 2.3.