4.1 Atomic structure and the periodic table — revision question pack

15 specification points · notes, questions, answers and worked methods

Checked against AQA 8462 section 4.1. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

Loading your tier…

Answer all questions in the spaces provided.

4.1.1.1 · Atoms, elements and compounds

Explanation

All substances are made of atoms, and an atom is the smallest part of an element that can exist. Required chemical symbols include the first 20 elements, Groups 1 and 7, and other elements named in the specification.
Compounds form when elements react and become chemically combined in fixed proportions, so a formula records the elements and their ratio.
A compound has different properties from its constituent elements and can be separated into elements only by chemical reactions.
Exam questions require names, symbols and formulae to be distinguished, then specified reactions to be represented using word equations or balanced symbol equations.
Higher tier: write balanced half equations and ionic equations where appropriate.

Worked example

Classify oxygen, sodium chloride and air as an element, compound or mixture, giving one reason for each.

1.Oxygen contains only oxygen atoms, so it is an element.
2.Sodium chloride contains sodium and chlorine chemically combined in a fixed ratio, so it is a compound.
3.Air contains several gases not chemically combined, so it is a mixture.

Answer: Oxygen is an element, sodium chloride is a compound and air is a mixture.

Common mistakes

Don't call a compound a mixture because it contains more than one element, ignoring that its elements are chemically combined in fixed proportions.
Don't change a formula when balancing an equation, which changes the substance instead of changing the number of particles.

Exam tip

For a ‘state the difference’ question, contrast both chemical combination and the method of separation.

Tier 1 · Easy

A sealed jar contains only argon atoms. State whether its contents are an element, a compound or a mixture, and give one reason.
[2 marks]
Total for this question: 2
Water is a compound of hydrogen and oxygen. State whether filtration can separate water into these elements and give one reason.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Magnesium burns in oxygen to form magnesium oxide. Write the word equation, then complete and balance the symbol equation Mg + O₂ → MgO.
[3 marks]
Total for this question: 3
A container holds nitrogen molecules, N₂, mixed with ammonia molecules, NH₃. Classify the whole sample and explain why ammonia is a compound rather than an element.
[3 marks]
Total for this question: 3
One formula unit of aluminium sulfate contains two aluminium atoms and three sulfate groups. Each sulfate group contains one sulfur atom and four oxygen atoms. Write the formula using brackets, state the number of different elements present, and explain why it is a compound.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Molecule E has formula $\mathrm{C}_2\mathrm{H}_6\mathrm{O}$ . It reacts fully with oxygen to produce carbon dioxide and water. Balance $\mathrm{C}_2\mathrm{H}_6\mathrm{O}+\mathrm{O}_2\rightarrow\mathrm{CO}_2+\mathrm{H}_2\mathrm{O}$ , then explain why chemical products are not merely the starting materials mixed together.
[5 marks]
Total for this question: 5
Aluminium reacts with bromine to form aluminium bromide. Complete and balance $\mathrm{Al}+\mathrm{Br}_2\rightarrow\mathrm{AlBr}_3$ , then explain why changing a subscript is not a valid way to balance the equation.
[5 marks]
Total for this question: 5
Sample P contains iron powder and sulfur powder. A magnet removes the iron from P. Sample Q forms when iron and sulfur react, and a magnet does not separate it. Compare P and Q and explain the evidence.
[5 marks]
Total for this question: 5
Two clear drinks contain only sugar and water. Drink A contains twice as much sugar per 100 cm³ as drink B. Distillation separates water from each drink and leaves sugar behind. A student claims that each drink is a compound because it looks uniform. Evaluate the claim.
[5 marks]
Total for this question: 5
A student says the labels Co, CO and 2CO all represent the same substance. Evaluate the statement, explain the meaning of each label, and state how a sample containing both Co and CO should be described.
[6 marks]
Total for this question: 6

4.1.1.2 · Mixtures

Explanation

A mixture contains two or more elements or compounds that are not chemically combined, so each substance keeps its chemical properties and no new substance is formed.
Separation therefore uses physical properties.
Filter an insoluble solid from a liquid; crystallise a dissolved solid from solution; use simple distillation to recover a solvent; use fractional distillation for miscible liquids with different boiling points; and use chromatography when substances move differently because of their solubilities and attractions.
When asked to suggest a method, identify the relevant property and describe what is collected.
Examiners reward a linked explanation, not just the name of a technique.

Worked example

A mixture contains sand, salt and water. Describe how to obtain dry salt crystals.

1.Filter the mixture: insoluble sand remains as the residue and salt solution passes through.
2.Heat the filtrate to evaporate some water until the solution is concentrated.
3.Cool the solution so crystals form, then filter and dry the crystals.

Answer: Filter off the sand, crystallise the salt from the filtrate, then filter and dry the salt crystals.

Common mistakes

Don't use filtration to separate a dissolved solid, although dissolved particles pass through the filter paper.
Don't say fractional distillation separates liquids by density instead of by their different boiling points.
Don't evaporate a solution to complete dryness when the question asks for crystals.

Exam tip

In a method question, name the physical property that makes the chosen separation work.

Tier 1 · Easy

A beaker contains chalk powder suspended in water. Name the physical process that collects the chalk and identify what passes through the paper.
[2 marks]
Total for this question: 2
A student needs to collect pure water from sodium chloride solution. Name the separation method and state the change of state that forms liquid water in the receiver.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student needs dry copper sulfate crystals from a copper sulfate solution. Describe a suitable sequence after the solution has been placed in an evaporating basin.
[4 marks]
Total for this question: 4
A student pours copper sulfate solution through filter paper and predicts that copper sulfate crystals will remain as the residue. Evaluate the prediction and name the process needed to obtain crystals.
[3 marks]
Total for this question: 3
Describe how to use paper chromatography to determine whether a water-soluble black ink contains more than one dye.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A liquid mixture contains propanone, which boils at 56 °C, water, which boils at 100 °C, and a dissolved non-volatile blue solid. Design a separation that obtains all three components.
[6 marks]
Total for this question: 6
Potassium nitrate decomposes if heated strongly. Method A boils its solution to dryness. Method B evaporates some water gently, cools the concentrated solution, then filters and dries the solid. Evaluate both methods for obtaining dry potassium nitrate crystals.
[5 marks]
Total for this question: 5
A paper chromatogram has reference spots R at 2.0 cm, S at 4.5 cm and T at 6.0 cm above the baseline. Sample X gives spots at 2.0 cm and 6.0 cm. Sample Y gives one spot at 4.5 cm and another at 5.2 cm. Determine which reference substances are present in X and Y, then evaluate the claim that Y is pure.
[5 marks]
Total for this question: 5
A student tests a green food colouring by paper chromatography. The baseline is drawn in green ink, the solvent starts above the sample spot, and the student concludes that the colouring is pure because one green mark is visible after the run. Evaluate the method and conclusion, then describe the key corrections.
[6 marks]
Total for this question: 6
Sample Z gives one spot when water is used as the chromatography solvent but two separated spots when ethanol is used. Evaluate the claim that the first chromatogram proves Z is pure, and explain why the two results can differ.
[4 marks]
Total for this question: 4

4.1.1.3 · The development of the model of the atom (common content with physics)

Explanation

Atomic models changed when new experimental evidence could not be explained by the existing model. Atoms were first pictured as indivisible spheres.
Discovering the electron produced the plum pudding model: negative electrons embedded in a ball of positive charge. In alpha-particle scattering, most particles passed straight through, a few were deflected and very few rebounded.
This showed that the atom is mostly empty space and that nearly all its mass and positive charge are concentrated in a tiny nucleus, so the nuclear model replaced plum pudding.
Bohr then proposed electrons at specific distances; later work identified protons and Chadwick supplied evidence for neutrons.
Examiners expect evidence linked explicitly to each conclusion.

The sequence of atomic models from the solid sphere to electrons in shells around a nucleus.

Worked example

Most alpha particles passed through gold foil, but a very small number rebounded. Explain what each observation showed.

1.Most particles passing through showed that most of an atom is empty space.
2.Rare large deflections showed that positive charge is concentrated in a very small region.
3.Particles rebounding showed that this small nucleus is dense and contains most of the atom’s mass.

Answer: The observations support a mostly empty atom with a tiny, dense, positively charged nucleus.

Common mistakes

Don't claim that every alpha particle was deflected, instead of recognising that most passed through the mostly empty atom.
Don't say Rutherford’s scattering experiment discovered electrons, although it provided evidence for a small, charged nucleus.

Exam tip

For ‘explain why the model changed’, pair each observation with the conclusion it supports.

Tier 1 · Easy

Place these developments in chronological order: Chadwick's neutron evidence, the plum pudding model, Bohr's shells, the nuclear model.
[2 marks]
Total for this question: 2
Identify the atomic model that described negative electrons embedded in a sphere of positive charge.
[1 mark]
Total for this question: 1

Tier 2 · Standard

In an alpha-scattering trial, nearly every alpha particle crossed a thin metal sheet without changing direction, but a tiny proportion returned towards the source. Explain the conclusions about atomic structure.
[4 marks]
Total for this question: 4
Experiments showed that atoms could emit negatively charged electrons. Explain why this evidence caused the solid-sphere model to be replaced and describe the model proposed next.
[3 marks]
Total for this question: 3
Bohr proposed that electrons orbit the nucleus at specific distances. Explain why agreement between his theoretical calculations and experimental observations supported this change to the nuclear model.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Compare the plum pudding and nuclear models, then explain how scattering evidence and later discoveries produced the modern GCSE model of the atom.
[6 marks]
Total for this question: 6
Measurements showed that the known protons in a nucleus could not account for all its mass. Chadwick later obtained evidence for particles with about the same mass as a proton but no electrical charge. Explain how this evidence changed the atomic model.
[5 marks]
Total for this question: 5
Experiments found that the positive charge of a nucleus could be divided into a whole number of identical units. Each unit had the same charge as the nucleus of a hydrogen atom. A nitrogen nucleus had seven of these units. Explain how this evidence refined the nuclear model.
[6 marks]
Total for this question: 6
Compare the plum pudding model with the nuclear model. State one feature of the atom that the nuclear model kept, one feature that it rejected, and explain why the superseded plum pudding model was still scientifically useful.
[5 marks]
Total for this question: 5
Before a scattering experiment, two atomic models both explain why atoms can release electrons. One places the remaining positive charge throughout the atom; the other places it in a small centre. Explain why the electron evidence alone cannot choose between the models, then state a prediction and result that would distinguish them.
[5 marks]
Total for this question: 5

4.1.1.4 · Relative electrical charges of subatomic particles

Explanation

Protons have relative charge $+1$ , neutrons have charge $0$ and electrons have charge $-1$ .
Protons and neutrons are in the nucleus; electrons occupy energy levels around it.
An atom has no overall electrical charge because it has equal numbers of protons and electrons, so the positive and negative charges cancel.
The atomic number is the number of protons, and it identifies the element: every atom of one element has the same proton number, while atoms of different elements have different proton numbers.
In questions, use the atomic number to find protons and, for a neutral atom, electrons; do not infer neutron number unless a mass number is also supplied.

Worked example

A neutral atom has atomic number $13$ . State its numbers of protons and electrons and explain its overall charge.

1.Atomic number $=13$ , so the nucleus contains $13$ protons.
2.A neutral atom has equal numbers of protons and electrons, so it has $13$ electrons.
3.The charges sum to $13(+1)+13(-1)=0$ .

Answer: It has $13$ protons and $13$ electrons, giving no overall charge.

Common mistakes

Don't use the atomic number as the number of neutrons rather than the number of protons.
Don't give a neutral atom one more electron than proton, which would describe a negative ion.

Exam tip

When the particle is an atom, explicitly use ‘protons = electrons’ before concluding that its charge is zero.

Tier 1 · Easy

State the relative charge of a proton, a neutron and an electron.
[3 marks]
Total for this question: 3
Identify the subatomic particle with no electrical charge and state where it is found in an atom.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A particle contains 13 protons, 14 neutrons and 10 electrons. Determine its overall charge and explain whether it is an atom or an ion.
[3 marks]
Total for this question: 3
A neutral sulfur atom contains 16 protons. State its electron number, then determine the electron number and charge after it gains two electrons.
[3 marks]
Total for this question: 3
A collection contains four X²⁺ ions. Determine the number of Y⁻ ions needed to make the collection electrically neutral and explain the charge balance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Species X has 26 protons, 30 neutrons and 23 electrons. Give its atomic number, mass number and ionic charge, then identify X using a periodic table.
[5 marks]
Total for this question: 5
Ion Q has mass number 31, charge 3+ and 10 electrons. Determine its numbers of protons and neutrons, identify the element, and write the ion symbol.
[4 marks]
Total for this question: 4
A student states that a neutral oxygen atom becomes O²⁻ by gaining two neutrons. Evaluate the statement and describe the particle change that forms the ion.
[4 marks]
Total for this question: 4
Ion T has proton number 35 and an overall charge of 1−. Determine its number of electrons. A student says that T must contain 44 neutrons. Explain why the neutron number cannot be determined from the information given.
[4 marks]
Total for this question: 4
Four neutral atoms each contain 13 protons. Each atom loses three electrons. Determine the charge on each resulting ion, the total charge of the four ions, the total number of electrons transferred, and explain why the proton number is unchanged.
[4 marks]
Total for this question: 4

4.1.1.5 · Size and mass of atoms

Explanation

An atom has a radius of about $0.1\ \mathrm{nm}$ , or $1\times10^{-10}\ \mathrm{m}$ . Its nucleus has a radius below one ten-thousandth of the atom’s radius, about $1\times10^{-14}\ \mathrm{m}$ , yet contains almost all the mass.
Protons and neutrons each have relative mass $1$ ; an electron’s relative mass is very small.
The mass number is the total number of protons and neutrons, so neutron number equals mass number minus atomic number.
Isotopes are atoms of the same element with the same proton number but different neutron numbers.
Exam questions may require scale comparisons, particle-count calculations or an explanation of why isotopes remain the same element.

Worked example

An isotope is written as $^{37}_{17}\mathrm{Cl}$ . Determine its numbers of protons, neutrons and electrons.

1.The lower number is the atomic number, so there are $17$ protons.
2.Neutrons $=37-17=20$ .
3.The symbol represents a neutral atom, so there are $17$ electrons.

Answer: $17$ protons, $20$ neutrons and $17$ electrons.

Common mistakes

Don't subtract the mass number from the atomic number and obtain a negative neutron count.
Don't define isotopes as having different proton numbers, which would make them different elements.

Exam tip

Write ‘neutrons = mass number − atomic number’ before substituting values.

Tier 1 · Easy

Convert an atomic radius of $0.12\,\text{nm}$ into metres and write the result in standard form.
[2 marks]
Total for this question: 2
State the relative mass of a neutron and describe the relative mass of an electron.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An oxygen-18 atom has atomic number 8. Calculate its numbers of protons, neutrons and electrons, and state where almost all its mass is located.
[4 marks]
Total for this question: 4
A row is 1.0 mm long. Treat each atom in the row as having diameter 0.20 nm. Calculate the number of atoms that fit side by side along the row.
[3 marks]
Total for this question: 3
The ion $^{31}_{15}\mathrm{P}^{3-}$ is shown in isotope notation. Determine its numbers of protons, neutrons and electrons.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An atom has radius $9.0\times10^{-11}\,\text{m}$ and its nucleus has radius $7.5\times10^{-15}\,\text{m}$ . Calculate how many times larger the atomic radius is. A related ion has 26 protons, 30 neutrons and 24 electrons; write its nuclide symbol and charge.
[6 marks]
Total for this question: 6
Atoms A and B each have 12 protons. A has 12 neutrons and B has 14 neutrons. Compare the atoms, explain why they are the same element but have different masses, and give both mass numbers.
[5 marks]
Total for this question: 5
A scale model gives an atom a radius of 50 cm. The nucleus has a radius less than 1/10 000 of the atomic radius. Calculate the upper limit for the nuclear radius in millimetres, then explain how the model represents the distribution of mass in an atom.
[5 marks]
Total for this question: 5
Treat an atom and its nucleus as spheres. Take the nuclear radius as 1/10 000 of the atomic radius for this model, and take the volume of a sphere as proportional to radius cubed. A student claims that the nucleus occupies 1/10 000 of the atom's volume. Evaluate the claim and relate the result to the distribution of mass.
[5 marks]
Total for this question: 5
A neutral particle contains 12 protons, 12 neutrons and 12 electrons. Take the relative mass of each proton and neutron as 1 and of each electron as $1/1840$ . Calculate the total relative mass contributed by the nucleus and by the electrons, then evaluate the statement that electrons make no contribution to atomic mass.
[5 marks]
Total for this question: 5

4.1.1.6 · Relative atomic mass

Explanation

Relative atomic mass, $A_r$ , is the weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.
It accounts for both the masses of the isotopes and their relative abundances, so the periodic-table value is often not a whole number.
Multiply each isotope mass by its abundance, add the products, then divide by the total abundance: $A_r=\dfrac{\sum(\text{isotope mass}\times\text{abundance})}{\sum\text{abundances}}$ .
Percentages total $100$ , but ratios may have another total.
In an exam, show the complete weighted calculation because an unsupported rounded answer can lose method marks.

Worked example

A sample contains $72\%$ of isotope $^{63}\mathrm{Cu}$ and $28\%$ of isotope $^{65}\mathrm{Cu}$ . Calculate $A_r$ .

1.Multiply each isotope mass by its percentage abundance: $63\times72$ and $65\times28$ .
2.Add the weighted masses: $4536+1820=6356$ .
3.Divide by the total percentage: $A_r=6356\div100=63.56$ .

Answer: $A_r=63.56$ .

Common mistakes

Don't take the simple mean of the isotope masses even though their abundances are unequal.
Don't divide by the number of isotopes instead of the total abundance.
Don't round the relative atomic mass to a whole number without being asked.

Exam tip

Keep the abundance products visible so the examiner can award the weighted-mean method mark.

Tier 1 · Easy

A sample of copper contains 69% copper-63 and 31% copper-65. Calculate its relative atomic mass.
[2 marks]
Total for this question: 2
The relative atomic mass shown for chlorine is 35.5. Explain why this does not mean that one chlorine atom has a mass number of 35.5.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Element Z has isotopes Z-24, Z-25 and Z-26. Their abundances are 79%, 10% and 11% respectively. Determine the relative atomic mass of Z.
[3 marks]
Total for this question: 3
An element contains 20% of an isotope with mass 10 and 80% of an isotope with mass 11. A student reports $(10+11)\div2=10.5$ . Evaluate the method and calculate the relative atomic mass.
[3 marks]
Total for this question: 3
Element Q has isotopes Q-50, Q-53 and Q-57 with relative abundances in the ratio 18 : 4 : 3. Calculate the relative atomic mass of Q to 1 decimal place.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An element has only isotopes of mass 79 and 81. Its relative atomic mass is 79.90. Calculate the percentage abundance of the mass-79 isotope.
[4 marks]
Total for this question: 4
Element J contains 25% J-62 and 75% of one other isotope. Its relative atomic mass is 63.50. Calculate the mass number of the other isotope and explain why the result is consistent with its abundance.
[4 marks]
Total for this question: 4
A sample contains 240 atoms of element G: 180 are G-24, 24 are G-25 and the rest are G-26. Calculate the relative atomic mass to 1 decimal place. Determine the new relative atomic mass if 12 of the G-24 atoms are replaced by 12 G-26 atoms, and explain the change.
[5 marks]
Total for this question: 5
A hypothetical new survey reports Mg-30 as a fourth naturally occurring magnesium isotope, at 0.1% abundance. The revised abundances are Mg-24: 78.9%, Mg-25: 10.0%, Mg-26: 11.0% and Mg-30: 0.1%. Calculate the relative atomic mass using all four isotopes. State whether the periodic-table value 24.3 changes when printed to 1 decimal place.
[3 marks]
Total for this question: 3
Element R has only two isotopes, R-41 and R-44. Three laboratories report relative atomic masses of 40.8, 42.5 and 44.1 for pure samples containing only these isotopes. Determine which report could be valid and explain why the other two are impossible without calculating any isotope abundances.
[5 marks]
Total for this question: 5

4.1.1.7 · Electronic structure

Explanation

Electrons occupy the lowest available energy levels, also called shells. For the first 20 elements, use the simple filling pattern $2,8,8,2$ : fill an inner shell before starting the next.
Electronic structure may be written as numbers separated by commas or drawn as electrons on concentric shells.
For example, sodium has atomic number $11$ , so a neutral atom has $11$ electrons and structure $2,8,1$ .
The number of occupied shells gives the period, while the outer-shell electron count explains the group and similar chemical properties of main-group elements.
Examiners expect the correct total and distribution, not merely the right number of shells.

A sodium atom with two electrons in the first shell, eight in the second and one in the third.

Worked example

Give the electronic structure and period of a neutral calcium atom, atomic number $20$ .

1.A neutral calcium atom has $20$ electrons.
2.Fill the shells from the inside: $2,8,8,2$ .
3.Four shells are occupied, so calcium is in period $4$ .

Answer: Electronic structure $2,8,8,2$ ; period $4$ .

Common mistakes

Don't place more than two electrons in the first shell.
Don't use the mass number rather than the atomic number to choose the total electrons in a neutral atom.

Exam tip

After drawing shells, add the electrons and check that the total equals the atomic number.

Tier 1 · Easy

Give the electronic structure of an aluminium atom, which has atomic number 13.
[2 marks]
Total for this question: 2
An atom has electronic structure 2,8,5. Determine its total number of electrons.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Calcium has atomic number 20. State the electronic structure of a calcium atom and of a Ca²⁺ ion.
[3 marks]
Total for this question: 3
Two students suggest electronic structures 2,7,8 and 2,8,7 for a neutral chlorine atom. Choose the correct structure and explain the choice.
[3 marks]
Total for this question: 3
A shell diagram shows 2 electrons in the first shell, 8 in the second shell and 6 in the third shell. Write the electronic structure, determine the atomic number of the neutral atom, and state how many electrons would fill its outer shell.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Neutral atom X is one of the first 20 elements. It has three occupied shells and seven electrons in its outer shell. Identify X, give its atomic number and electronic structure, and state the structure after it gains one electron.
[5 marks]
Total for this question: 5
Ion R has 18 electrons and charge 2−. Determine the proton number of R, identify the element, and give the electronic structures of its neutral atom and the ion.
[4 marks]
Total for this question: 4
Species A has 19 protons and electronic structure 2,8,8. Species B has 17 protons and the same electronic structure. Identify both elements, determine the charge on each species, give both neutral-atom electronic structures, and explain why A and B have different charges despite having the same number of electrons.
[6 marks]
Total for this question: 6
Four neutral atoms from the first 20 elements are assigned the electronic structures A 2,8,9,1; B 2,5,7; C 2,8,6; and D 2,9,2. Identify every impossible structure, state the shell-capacity and inside-out filling rules, and correct each impossible structure without changing its total number of electrons.
[6 marks]
Total for this question: 6
A ground-state diagram for a neutral atom places its 15 electrons in the structure 2,7,6. State the error, give the correct ground-state electronic structure, and identify the element.
[4 marks]
Total for this question: 4

4.1.2.1 · The periodic table

Explanation

The modern periodic table arranges elements in increasing atomic, or proton, number. Similar properties recur at regular intervals, so elements with related chemical behaviour are placed in vertical columns called groups.
Main-group elements in the same group have the same number of outer-shell electrons, which explains their similar reactions.
The period indicates the number of occupied electron shells.
Use an electronic structure to locate an element, then use its position to predict likely reactions and reactivity.
The modern ordering is not by relative atomic mass: atomic number gives each element one unambiguous position and explains apparent mass-order anomalies caused by isotopes.

Worked example

An element has electronic structure $2,8,7$ . State its period and group and predict whether it has properties similar to fluorine.

1.Three occupied shells place the element in period $3$ .
2.Seven outer-shell electrons place it in Group $7$ .
3.Fluorine is also in Group $7$ , so the elements have similar chemical properties.

Answer: Period $3$ , Group $7$ ; yes, it should have properties similar to fluorine.

Common mistakes

Don't state that the modern table is ordered by relative atomic mass rather than atomic number.
Don't use the total number of electrons as the group number instead of the outer-shell count.

Exam tip

For a prediction, cite the shared group or outer-electron arrangement before stating the similar property.

Tier 1 · Easy

An atom has electronic structure 2,8,2. Give its atomic number, period and group.
[3 marks]
Total for this question: 3
State the quantity used to arrange elements in the modern periodic table and state what elements in the same group share.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Elements P and Q have electronic structures 2,1 and 2,8,1. Explain why they have similar chemical properties and identify which one has the larger atomic number.
[4 marks]
Total for this question: 4
Element Y is directly below sulfur in the periodic table. Predict the number of electrons in Y's outer shell and explain why Y should react in ways similar to sulfur.
[3 marks]
Total for this question: 3
Elements A, B and C have atomic numbers 9, 10 and 11 respectively. Write their electronic structures and explain why C starts a new period after B.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Element R has atomic number 16 and element S has atomic number 19. Write both electronic structures, locate each by period and group, and predict which is more likely to form a positive ion.
[6 marks]
Total for this question: 6
Elements T and U are both in period 3. T is in Group 7 and U is in Group 0. Give both electronic structures, predict which element is more reactive, and explain the prediction.
[5 marks]
Total for this question: 5
Evaluate these three statements: V has atomic number 12 and is in period 2, Group 2; W has electronic structure 2,8,6 and is in period 3, Group 6; X has electronic structure 2,8,8 and is a reactive Group 7 element. Write each corrected placement or property.
[6 marks]
Total for this question: 6
Element M is one of the first 20 elements. Its atoms have three occupied electron shells, and each atom forms M²⁺ with electronic structure 2,8. Identify M, give its neutral electronic structure, period and group, and name one element expected to have similar chemical properties.
[5 marks]
Total for this question: 5
A student claims that sodium, electronic structure 2,8,1, and chlorine, electronic structure 2,8,7, should have similar chemical properties because they are in the same period. Evaluate the claim and explain the different information given by a period and a group.
[5 marks]
Total for this question: 5

4.1.2.2 · Development of the periodic table

Explanation

Before subatomic particles were known, scientists tried to classify elements mainly in order of atomic weight. Early tables were incomplete, and keeping a strict weight order sometimes placed elements with different properties together.
Mendeleev prioritised repeating chemical properties: he left gaps for elements not yet discovered and changed the order in a few places.
Newly discovered elements filled those gaps and had properties close to his predictions, providing strong evidence for his arrangement.
Later knowledge of isotopes explained why atomic-weight order can appear inconsistent, while atomic number gives the correct modern sequence.
Examiners want a chronological account showing how predictions were tested by later evidence.

Worked example

Explain why the later discovery of an element with Mendeleev’s predicted properties supported his periodic table.

1.Mendeleev had left a gap because the repeating pattern suggested an undiscovered element.
2.He predicted properties for the element from neighbouring elements.
3.The discovered element fitted the gap and matched those predictions, so the table made a successful testable prediction.

Answer: The close match between prediction and later evidence supported Mendeleev’s arrangement.

Common mistakes

Don't say Mendeleev simply arranged every element in strict atomic-weight order, ignoring his deliberate gaps and reversals.
Don't claim Mendeleev used proton number, although protons had not yet been discovered.
Don't state that gaps weakened the table without explaining that accurate predictions later supported it.

Exam tip

In a history question, use the chain ‘prediction → discovery → matching properties → support’.

Tier 1 · Easy

Give two decisions Mendeleev made that improved the arrangement of the elements known in his time.
[2 marks]
Total for this question: 2
Give one reason why the earliest periodic tables were incomplete.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Mendeleev predicted that an empty position would be filled by an element forming an oxide X₂O₃. Years later, a new element was found and its oxide had that formula. Explain why this strengthened his periodic table.
[3 marks]
Total for this question: 3
A strict atomic-weight table places element A among elements with which it does not react similarly. Moving A one position puts it with elements that form similar compounds. Explain the decision Mendeleev would make and why.
[4 marks]
Total for this question: 4
Chlorine-35 and chlorine-37 have the same atomic number but different mass numbers. Explain why knowledge of isotopes supports arranging the modern periodic table by atomic number rather than by atomic weight.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two elements have relative atomic masses 39.1 and 40.0. Their chemical properties place the 40.0 element before the 39.1 element in the modern table. Explain why this ordering troubled early tables, how Mendeleev could respond, and how later atomic theory resolved the issue.
[5 marks]
Total for this question: 5
Mendeleev predicted that a missing element would be a grey solid with density close to 5.5 g/cm³. The element later discovered was grey with density 5.3 g/cm³. Evaluate the claim that the match was only luck, and explain how the modern table improved the ordering.
[5 marks]
Total for this question: 5
A critic states that Mendeleev's successful predictions gave no support to his periodic table because scientists could have fitted the facts after each element was discovered. Evaluate the statement.
[5 marks]
Total for this question: 5
A gap in an early periodic table lies between related elements whose element densities are 3.8 g/cm³ and 5.2 g/cm³. The missing element is predicted to have an element density from 4.1 to 4.9 g/cm³ and to form an oxide with formula J₂O₃. Candidate element Z forms Z₂O₃, but the measured element density of Z is 5.8 g/cm³. Evaluate the candidate for this gap and state what should be concluded about the periodic table.
[5 marks]
Total for this question: 5
Table P lists every known element strictly in increasing atomic weight and leaves no gaps, but several columns contain elements with very different reactions. Table Q changes the order of a few elements and leaves gaps so that each column contains elements with similar reactions. Compare the scientific strengths and risks of the two tables, and explain what later evidence could decide between them.
[6 marks]
Total for this question: 6

4.1.2.3 · Metals and non-metals

Explanation

Metals are elements that react to form positive ions; non-metals do not form positive ions. Most elements are metals, found mainly on the left and towards the bottom of the periodic table, while non-metals lie towards the right and top.
Typical metals conduct heat and electricity, are strong and malleable, and often have high melting points.
Solid non-metals are generally brittle and poor conductors, although the specification includes important structure-based exceptions such as graphite.
Link chemical behaviour to electron arrangement: metal atoms tend to lose outer electrons, while non-metal atoms tend to gain or share electrons.
In comparisons, state a property and then explain it using particles or electrons where requested.

Worked example

An element conducts electricity, is malleable and forms a $2+$ ion. Decide whether it is a metal or non-metal.

1.Electrical conduction and malleability are characteristic metallic properties.
2.Forming a positive ion means the atom loses electrons, which is characteristic of a metal.
3.Both the physical and chemical evidence therefore support the same classification.

Answer: The element is a metal.

Common mistakes

Don't define a metal only as shiny, instead of using its formation of positive ions as the chemical distinction.
Don't claim every non-metal is a gas or every metal is magnetic.
Don't say metals gain electrons when explaining the formation of positive ions.

Exam tip

For ‘explain the difference’, pair a characteristic property with the relevant electron behaviour.

Tier 1 · Easy

Element T reacts by losing two electrons from each atom. Classify T as a metal or non-metal and state the charge on the ion formed.
[2 marks]
Total for this question: 2
Element E gains one electron when it reacts. Classify E as a metal or non-metal and give one reason.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Magnesium has electronic structure 2,8,2, while sulfur has 2,8,6. Use these structures and periodic-table positions to explain why magnesium is a metal but sulfur is a non-metal.
[4 marks]
Total for this question: 4
Graphite conducts electricity even though carbon is a non-metal. Explain why conductivity alone is not enough to classify an element and state the chemical test used in the specification.
[3 marks]
Total for this question: 3
Compare a typical metal with a typical non-metal in terms of electrical conduction, response to hammering, the ions formed in reactions and position in the periodic table.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Unknown A is shiny, bends without snapping, conducts electricity and forms A³⁺. Unknown B is dull, breaks when hammered and does not conduct as a solid. Compare the evidence and predict where each lies in the periodic table.
[6 marks]
Total for this question: 6
Solid C conducts electricity but forms no positive ions in its reactions. Solid D is a poor conductor in one form but its atoms form D²⁺ ions. Evaluate the classifications using both physical and chemical evidence.
[5 marks]
Total for this question: 5
Element J is near the lower left of the periodic table and has one outer-shell electron. Element K is near the upper right and has six outer-shell electrons. Predict the type of ion J forms, state whether K forms positive ions, and compare their likely physical properties.
[5 marks]
Total for this question: 5
The oxide ion has charge 2−. An unknown element E forms an ionic oxide with formula E₂O₃. Determine the charge on each E ion, state the electron change that forms it, state whether E is a metal or non-metal, and predict its broad position in the periodic table.
[5 marks]
Total for this question: 5
A student proposes three rules: every electrical conductor is a metal; every metal is hard with a very high melting point; and every non-metal gains electrons to form a negative ion. Evaluate all three rules using graphite, Group 1 metals and Group 0 elements, then state how elements should be classified.
[6 marks]
Total for this question: 6

4.1.2.4 · Group 0

Explanation

Group 0 elements are the noble gases. Their atoms have stable outer-electron arrangements: eight outer electrons, except helium, whose only shell is complete with two.
This stability makes them very unreactive, so they do not easily form molecules and exist as single atoms.
Down the group, relative atomic mass increases and boiling point increases.
A data trend can be used to predict the boiling point or physical state of another noble gas, while recognising that a prediction is an estimate.
For explanation marks, low reactivity must be connected directly to the complete outer shell rather than merely describing noble gases as ‘stable’.

Worked example

Neon boils at $-246\,^{\circ}\mathrm{C}$ and argon at $-186\,^{\circ}\mathrm{C}$ . Predict whether krypton’s boiling point is above or below $-186\,^{\circ}\mathrm{C}$ .

1.Krypton is below argon in Group $0$ .
2.Boiling point increases down Group $0$ as relative atomic mass increases.
3.Krypton should therefore boil at a temperature above $-186\,^{\circ}\mathrm{C}$ .

Answer: Krypton’s boiling point should be above $-186\,^{\circ}\mathrm{C}$ .

Common mistakes

Don't say helium needs eight outer electrons, although its first shell is complete with two.
Don't describe noble gases as diatomic molecules rather than single atoms.
Don't predict that boiling point decreases down Group 0, reversing the specified trend.

Exam tip

When explaining unreactivity, write ‘complete outer shell’ before stating that atoms do not readily react.

Tier 1 · Easy

Explain why neon is unreactive using its electronic structure 2,8.
[2 marks]
Total for this question: 2
State how Group 0 particles exist and give the name for this particle arrangement.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The boiling points of neon, argon and krypton are −246 °C, −186 °C and −153 °C. Choose the most plausible boiling point for xenon from −260 °C, −170 °C and −108 °C, and justify your choice.
[3 marks]
Total for this question: 3
Arrange neon, argon and xenon in order of increasing boiling point and explain the trend.
[3 marks]
Total for this question: 3
Helium has two electrons, both in its first shell. Explain why helium is unreactive even though the other Group 0 elements have eight outer-shell electrons.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An unknown gas is monatomic, has electronic structure 2,8,8 and boils at a higher temperature than neon. Identify the gas and explain all three observations.
[5 marks]
Total for this question: 5
Neon melts at −249 °C and boils at −246 °C; argon melts at −189 °C and boils at −186 °C; krypton melts at −157 °C and boils at −153 °C. Determine the state of each element at −155 °C and explain how the data support a Group 0 trend.
[5 marks]
Total for this question: 5
Argon is placed around a hot metal filament inside a lamp. Its electronic structure is 2,8,8. Explain why argon is suitable and why it exists as single atoms rather than molecules under these conditions.
[4 marks]
Total for this question: 4
A liquefied mixture contains helium (boiling point −269 °C), argon (−186 °C) and xenon (−108 °C). Describe the order in which the gases are collected as the mixture is warmed in a fractional-distillation apparatus, and explain why this is a physical separation rather than a chemical reaction.
[5 marks]
Total for this question: 5
Helium, neon, argon, krypton, xenon and radon have boiling points of −269 °C, −246 °C, −186 °C, −153 °C, −108 °C and −62 °C respectively. A student claims, ‘Radon must be reactive because it is the heaviest and densest noble gas.’ Use the boiling-point trend and electronic structure to evaluate the claim.
[5 marks]
Total for this question: 5

4.1.2.5 · Group 1

Explanation

Group 1 elements are alkali metals with one electron in their outer shell. They react by losing this electron to form $+1$ ions, so their reactions are similar.
Lithium, sodium and potassium react with oxygen to form oxides, with chlorine to form chlorides, and with water to form a metal hydroxide plus hydrogen.
The water reactions become more vigorous down the group.
Reactivity increases because the outer electron is farther from the nucleus and experiences more shielding, so the attraction to the nucleus is weaker and the electron is lost more easily.
Examiners may ask for observations, products, balanced equations or a prediction for a lower element.

Worked example

Predict the products when potassium reacts with water and explain why potassium reacts more vigorously than lithium.

1.A Group 1 metal reacting with water forms the metal hydroxide and hydrogen.
2.The products are potassium hydroxide and hydrogen.
3.Potassium’s outer electron is farther from the nucleus and more shielded, so it is less strongly attracted and lost more easily.

Answer: Potassium hydroxide and hydrogen form; potassium is more reactive because its outer electron is lost more easily.

Common mistakes

Don't write metal oxide as the product of a Group 1 metal reacting with water.
Don't reverse the trend and say lithium is the most reactive of the first three alkali metals.
Don't explain the trend only by saying atoms are larger, without linking distance and shielding to weaker attraction.

Exam tip

A reactivity-trend explanation needs distance, shielding, nuclear attraction and ease of electron loss.

Tier 1 · Easy

Name the two products when sodium reacts with water, and state one visible observation.
[3 marks]
Total for this question: 3
Complete the word equation for lithium burning in oxygen: lithium + oxygen → _____.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Complete and balance the equation Na + Cl₂ → NaCl. Explain why sodium and potassium both form compounds with one metal atom for each chlorine atom.
[4 marks]
Total for this question: 4
Complete and balance Li + H₂O → LiOH + H₂, then state one observation when lithium reacts with water.
[4 marks]
Total for this question: 4
Equal-sized pieces of three Group 1 metals react separately with water. A fizzes steadily and moves slowly; B melts into a ball and moves rapidly; C ignites with a lilac flame. Identify A, B and C, then state the reactivity trend shown.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A teacher compares lithium, sodium and potassium in water using equal-sized pieces. Predict the order from least to most vigorous, explain the trend using atomic structure, and predict how rubidium would behave.
[6 marks]
Total for this question: 6
Equal numbers of lithium atoms and caesium atoms react separately with excess water. Each reaction follows 2M + 2H₂O → 2MOH + H₂. Compare the reaction times and final amounts of hydrogen, then explain both predictions.
[5 marks]
Total for this question: 5
A teacher compares lithium, sodium and potassium in water. The pieces have different exposed surface areas, the water temperatures are 18 °C, 25 °C and 31 °C, and different volumes of water are used. Evaluate the method, describe improvements, and predict the order of reaction speed after improvement.
[6 marks]
Total for this question: 6
A Group 1 metal M forms M⁺ ions. Oxide ions are O²⁻, chloride ions are Cl⁻ and hydroxide ions are OH⁻. Determine the formulae of the metal oxide, chloride and hydroxide, and explain how the electron change in M produces the common pattern.
[5 marks]
Total for this question: 5
Metal A has electronic structure 2,8,1 and metal B has electronic structure 2,8,8,1. A student predicts that A reacts faster with water because its outer electron is in a shell closer to the nucleus. Identify A and B, evaluate the prediction, and compare the products of their reactions with water.
[5 marks]
Total for this question: 5

4.1.2.6 · Group 7

Explanation

Group 7 elements are halogens: non-metals whose atoms have seven outer-shell electrons. They exist as diatomic molecules such as $\mathrm{Cl}_2$ , and form ionic halides with metals or covalent compounds with non-metals.
Down the group, relative molecular mass, melting point and boiling point increase, but reactivity decreases.
A halogen reacts by gaining one electron; farther down the group, the outer shell is farther from the nucleus and more shielded, so attracting an incoming electron is harder.
A more reactive halogen displaces a less reactive halogen from an aqueous halide solution.
Examiners expect correct trends, observations or equations and an electron-based explanation.

Worked example

Chlorine water is added to potassium bromide solution. Predict the products and explain whether a reaction occurs.

1.Chlorine is above bromine in Group $7$ , so chlorine is more reactive.
2.The more reactive chlorine displaces bromine from bromide ions.
3.The equation is $\mathrm{Cl}_2+2\mathrm{KBr}\rightarrow2\mathrm{KCl}+\mathrm{Br}_2$ .

Answer: Potassium chloride and bromine form because chlorine is more reactive than bromine.

Common mistakes

Don't say Group 7 reactivity increases down the group, confusing the trend with Group 1.
Don't write a halogen atom as a $-1$ ion before it has gained an electron.
Don't predict that iodine displaces chlorine from chloride solution.

Exam tip

For displacement, compare the two halogens’ positions before writing the products.

Tier 1 · Easy

State two features shared by chlorine, bromine and iodine atoms or molecules that explain their placement in Group 7.
[2 marks]
Total for this question: 2
State the physical states of bromine and iodine at room temperature.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Chlorine water is added to aqueous potassium bromide. Predict the products, write a balanced equation and explain why the reaction occurs.
[5 marks]
Total for this question: 5
Astatine lies below iodine in Group 7. Predict how its boiling point and reactivity compare with iodine, and explain the reactivity prediction.
[4 marks]
Total for this question: 4
A chlorine atom has electronic structure 2,8,7. State the formula of a chlorine molecule, determine the electron change and ion charge when chlorine forms chloride ions, and explain why bromine reacts similarly.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Bromine water is tested separately with sodium chloride and sodium iodide solutions. Predict each result, write any equation that occurs, and explain the different outcomes using the Group 7 reactivity trend.
[6 marks]
Total for this question: 6
Halogens P, Q and R are chlorine, bromine and iodine in an unknown order. P displaces both Q and R from their halide solutions; Q displaces R but not P. Identify P, Q and R, then predict their order of increasing boiling point and explain the reactivity evidence.
[6 marks]
Total for this question: 6
Bromine reacts separately with magnesium and hydrogen. Write the formula of magnesium bromide and hydrogen bromide, state the type of bonding in each compound, and explain the difference.
[5 marks]
Total for this question: 5
The melting and boiling points are: chlorine, −101 °C and −34 °C; bromine, −7 °C and 59 °C; iodine, 114 °C and 184 °C. State the physical state of each halogen at 50 °C. Choose the most reactive halogen that is liquid at this temperature, then predict whether it reacts with potassium chloride solution and potassium iodide solution.
[6 marks]
Total for this question: 6
Solution U contains potassium chloride, potassium bromide or potassium iodide. Bromine water causes no reaction with U, but chlorine water causes a displacement reaction. Identify the halide in U, explain how both observations determine it, and write the balanced equation for the chlorine reaction.
[5 marks]
Total for this question: 5

4.1.3.1 · Comparison with Group 1 elements (chemistry only)

Explanation

Transition elements are metals, including chromium, manganese, iron, cobalt, nickel and copper. Compared with Group 1 metals, they generally have higher melting points and densities and are stronger and harder.
They are also much less reactive with water, oxygen and halogens.
These differences make transition metals suitable where a material must keep its shape, withstand force or avoid rapid reaction, whereas Group 1 metals are soft and highly reactive.
Questions often provide unfamiliar data and require an element to be classified or compared.
Conclusions must use the data and stated general patterns, with a named transition element supporting broad statements when requested.

Worked example

Compare iron with sodium using density, hardness and reaction with water.

1.Iron, a transition metal, is generally denser than sodium, a Group $1$ metal.
2.Iron is harder and stronger, while sodium is soft.
3.Iron is much less reactive with water than sodium.

Answer: Iron is denser, harder and much less reactive with water than sodium.

Common mistakes

Don't claim transition metals are more reactive than Group 1 metals because they are stronger.
Don't use sodium or potassium as an example of a transition element.
Don't treat every transition metal as having exactly the same melting point, density and reactivity.

Exam tip

In a comparison, use paired language such as ‘higher density but lower reactivity than Group 1’.

Tier 1 · Easy

Give three ways in which iron typically differs from sodium in its physical or chemical properties.
[3 marks]
Total for this question: 3
Choose the two transition elements from chromium, potassium, calcium and copper.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A manufacturer needs a metal component that remains solid above 900 °C, resists deformation and does not react rapidly with water. Explain why nickel is a better choice than potassium.
[4 marks]
Total for this question: 4
Sodium is stored under oil, whereas an iron nail can be handled in air. Explain the different storage using their reactions with water and oxygen.
[4 marks]
Total for this question: 4
A chromium sample has mass 35.7 g and volume 5.0 cm³. Calculate its density using density = mass ÷ volume, then compare it with sodium, whose density is 0.97 g/cm³.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Metal U melts at 98 °C, has low density, cuts easily and reacts violently with water. Metal V melts at 1495 °C, is dense and hard, and reacts slowly with oxygen when heated. Classify U and V, justify each classification, and name one specified transition element in V's class.
[6 marks]
Total for this question: 6
Metal X is dense, melts at 1270 °C and does not react with cold water. A student concludes from density alone that X is a transition metal. Evaluate the conclusion and identify further evidence that would strengthen it.
[5 marks]
Total for this question: 5
Fresh potassium reacts immediately with chlorine, whereas copper reacts with chlorine only when heated. Compare the metals using this evidence, then predict how their reactions with water and their strength and hardness differ.
[5 marks]
Total for this question: 5
Metal A has mass 24.0 g, volume 30.0 cm³, melts at 63 °C and reacts rapidly with cold water. Metal B has mass 21.0 g, volume 2.50 cm³, melts at 1085 °C and shows no reaction when placed in cold water. Calculate both densities, state whether each is a Group 1 metal or transition element, and evaluate the claim that A is denser because its sample has greater mass.
[6 marks]
Total for this question: 6
Plan comparisons that would distinguish an unknown Group 1 metal from an unknown transition metal using reaction with water and hardness. State the controls, measurements and expected pattern needed for a valid conclusion.
[6 marks]
Total for this question: 6

4.1.3.2 · Typical properties (chemistry only)

Explanation

Transition elements show characteristic chemical properties as well as typical metallic ones. Many form ions with different charges, form coloured compounds and are useful catalysts.
For example, iron can form $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ ions, and iron is the catalyst in the Haber process.
Compounds of chromium, manganese, iron, cobalt, nickel and copper provide examples of coloured transition-metal compounds.
A catalyst increases reaction rate without being used up overall; being coloured does not itself make a compound a catalyst.
In exams, identify the property shown by information in the question and give a named example rather than assuming every transition element shows every property in the same way.

Worked example

Iron forms $\mathrm{FeCl}_2$ and $\mathrm{FeCl}_3$ . State the transition-metal property shown and give one other typical property.

1.Chloride ions have charge $-1$ , so the formulae contain iron ions with different charges.
2.The compounds therefore show that iron forms ions with different charges.
3.Another typical property is forming coloured compounds or acting as a catalyst.

Answer: Iron forms ions with different charges; transition elements also form coloured compounds or act as catalysts.

Common mistakes

Don't say transition elements form only one ion charge, overlooking variable-charge ions such as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ .
Don't define a catalyst as a substance that is used up to provide energy.
Don't assume any coloured compound must be acting as a catalyst.

Exam tip

When asked for a typical property, state it precisely and attach it to a named transition element or compound.

Tier 1 · Easy

State three typical chemical properties of transition elements or their compounds.
[3 marks]
Total for this question: 3
Copper sulfate solution is blue. State the typical transition-element property demonstrated by this observation.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Iron forms pale-green FeCl₂ and yellow-brown FeCl₃. Chloride ions have charge 1−. Determine the charge on iron in each compound and explain what the colours demonstrate.
[4 marks]
Total for this question: 4
Iron forms two oxides. One has formula FeO; the other contains two iron atoms for every three oxygen atoms. Name both oxides and state the typical transition-element property shown. Oxide ions have charge 2−.
[4 marks]
Total for this question: 4
State the typical property of transition elements shown by each observation: a cobalt compound is pink; iron forms Fe²⁺ and Fe³⁺ ions; a nickel compound speeds up a reaction.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Equal samples react under identical conditions. Without a catalyst the reaction takes 240 s; with an iron compound it takes 80 s; with a copper compound it takes 60 s. Choose the more effective catalyst, calculate how many times faster its trial is than the uncatalysed trial using reciprocal time as rate, and explain two catalyst features.
[6 marks]
Total for this question: 6
Solid A is orange and shortens a reaction from 150 s to 90 s, but its mass falls from 2.0 g to 1.2 g. Solid B is black, shortens the reaction to 60 s and is recovered with a mass of 2.0 g. Evaluate the evidence for A and B being catalysts.
[6 marks]
Total for this question: 6
A reaction produces 18 g of product after 40 s without a catalyst and 31 g after 40 s with a copper compound. Both trials produce 36 g when allowed to finish, and the copper compound is recovered unchanged. Evaluate the claim that the copper compound increases the final amount of product.
[5 marks]
Total for this question: 5
Element P forms only colourless P⁺ compounds in the data supplied. Element Q forms blue QCl₂ and green QCl₃, and a Q oxide speeds up a reaction while being recovered unchanged. Evaluate which element has the stronger evidence for being a transition element and link each observation to a typical property.
[6 marks]
Total for this question: 6
A coloured transition-metal compound makes a reaction finish sooner. Design further checks needed to support the claim that the compound is a catalyst rather than a reactant or merely a coloured additive.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.1.1.1 · Atoms, elements and compounds

Tier 1 · Easy

Mark scheme for 4.1.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Element. It contains only one type of atom.	Identify the number of atom types present. Every particle is an argon atom, so only one type of atom is present; the substance is therefore an element.	2
Total Question 1			2
02.1	Filtration cannot separate water into hydrogen and oxygen. The elements are chemically combined, so separating them requires a chemical reaction.	Filtration is a physical separation and does not break chemical bonds. The elements in a compound can be recovered only through chemical reactions.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	magnesium + oxygen → magnesium oxide 2Mg + O₂ → 2MgO	Write the named reactant and product first. In the symbol equation oxygen enters as O₂, so place 2 before MgO to give two oxygen atoms on the right, then place 2 before Mg to balance magnesium.	3
Total Question 1			3
02.1	The whole sample is a mixture. Nitrogen and ammonia are present together but are not chemically combined with each other. Ammonia is a compound because each molecule contains nitrogen and hydrogen atoms chemically bonded in a fixed ratio.	Classify the container from all the particles present: it contains two substances, so it is a mixture. Then classify NH₃ itself from the different elements bonded within each molecule.	3
Total Question 2			3
03.1	The formula is Al₂(SO₄)₃. Three different elements are present. Aluminium, sulfur and oxygen are chemically combined. The elements are combined in fixed proportions.	Write Al₂ for the two aluminium atoms and put the sulfate group in brackets with an outside subscript 3. A compound contains different elements chemically combined in fixed proportions.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	C₂H₆O + 3O₂ → 2CO₂ + 3H₂O Atoms have been rearranged and new chemical bonds have formed. The products have different properties from the reactants.	Balance carbon first by placing 2 before CO₂, then hydrogen by placing 3 before H₂O. The products now contain seven oxygen atoms in total; one comes from ethanol, so six must come from 3O₂. A chemical reaction rearranges atoms into substances with new bonding and properties, unlike physical mixing.	5
Total Question 1			5
02.1	2Al + 3Br₂ → 2AlBr₃ The equation has two aluminium atoms and six bromine atoms on each side. Changing a subscript would change the formula and therefore the identity of a substance. Balancing must change only the coefficients in front of formulae.	The least common multiple of 2 bromine atoms in Br₂ and 3 in AlBr₃ is 6. Use 3Br₂ and 2AlBr₃, then balance aluminium with 2Al. Formula subscripts describe fixed particle composition, so only coefficients may be adjusted.	5
Total Question 2			5
03.1	P is a mixture. Its substances are not chemically combined, so the iron keeps its magnetic property. P can be separated by a physical process. Q is a compound because a new substance has formed from chemically combined elements. Separating Q into its elements would require a chemical reaction.	Use the separation evidence to distinguish the samples. Retaining magnetism and physical separation identify a mixture; loss of the constituents' separate properties after reaction identifies a compound.	5
Total Question 3			5
04.1	The claim is incorrect: each drink is a mixture. Sugar and water are not chemically combined. The different sugar-to-water proportions show that the composition is not fixed. Distillation is a physical process, so its success supports the mixture classification. Looking uniform does not prove that a substance is a compound.	Test the claim against the defining features rather than appearance. Compounds have chemically combined elements in fixed proportions and require chemical reactions for separation. The variable composition and successful physical separation both show that the drinks are mixtures.	5
Total Question 4			5
05.1	The statement is incorrect because capital letters and coefficients have different meanings. Co is the chemical symbol for the element cobalt. CO is the formula of carbon monoxide, a compound containing carbon and oxygen. The capital O in CO is the symbol for oxygen, whereas the lower-case o in Co is part of cobalt's symbol. 2CO means two carbon monoxide particles; the coefficient does not change the compound's formula. A sample containing both cobalt and carbon monoxide is a mixture because two substances are present without being chemically combined with each other.	Read chemical notation exactly. Letter case distinguishes the element symbol Co from the two element symbols C and O in CO, while a coefficient counts unchanged particles. Finally describe the whole sample from the two different substances present together.	6
Total Question 5			6

4.1.1.2 · Mixtures

Tier 1 · Easy

Mark scheme for 4.1.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Filtration. Water passes through as the filtrate.	Chalk is an insoluble solid with particles too large to pass through filter paper, so filtration traps it as the residue while the liquid water becomes the filtrate.	2
Total Question 1			2
02.1	Simple distillation. Condensation.	Heat the solution so water vaporises while the dissolved salt remains. Cooling the vapour condenses it to liquid water in the receiver.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Heat gently to evaporate some water and make a concentrated solution. Leave the solution to cool so crystals form. Filter the crystals from the remaining solution. Dry the crystals, for example between sheets of filter paper.	Concentrate rather than boil the solution dry. Cooling reduces the amount of solute that remains dissolved, so crystals form. Separate them by filtration and remove surface solution by drying.	4
Total Question 1			4
02.1	The prediction is incorrect because dissolved copper sulfate particles pass through the filter paper. The filtrate remains copper sulfate solution rather than leaving copper sulfate as the residue. Use crystallisation: concentrate the solution and then cool it so crystals form.	Filtration retains an insoluble solid, not a dissolved solute. Copper sulfate must first be brought out of solution by evaporating some water and cooling; the resulting crystals can then be filtered.	3
Total Question 2			3
03.1	Draw a pencil baseline on the chromatography paper and place a small spot of the ink on it. Stand the paper in water with the solvent level below the ink spot. Allow the solvent to move up the paper, then remove the paper before the solvent reaches the top. More than one separated spot shows that the ink contains more than one dye.	Keep the sample above the solvent so it travels with the solvent rather than dissolving into the beaker. Different soluble dyes can move different distances and appear as separate spots.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Use fractional distillation and collect propanone near 56 °C. The fractionating column allows repeated condensation and evaporation, improving separation of the liquids. Then use simple distillation to collect most of the water near 100 °C, stopping before the flask is dry. The non-volatile solid remains in the concentrated solution in the flask. Cool the concentrated solution, then filter and dry the solid crystals.	Exploit the two different boiling points first. Fractional distillation separates the more volatile propanone from water; subsequent simple distillation collects most of the water. Because the blue solid does not vaporise, stop before dryness and cool the concentrated residue so it crystallises.	6
Total Question 1			6
02.1	Method A is unsuitable because strong heating may decompose the potassium nitrate, so the product would not all be the required substance. Boiling to dryness can also cause hot solution or solid to spit. Method B is suitable because gentle evaporation concentrates the solution without taking it to dryness. Cooling allows potassium nitrate crystals to form. Filtering separates the crystals from the remaining solution and drying removes solution from their surfaces.	Use the stated thermal decomposition risk to reject heating to dryness. Controlled concentration followed by cooling produces crystals; filtration and drying then isolate the solid safely.	5
Total Question 2			5
03.1	X contains R because both give a spot at 2.0 cm. X contains T because both give a spot at 6.0 cm. Y contains S because both give a spot at 4.5 cm. Y is not pure because it produces two spots. The 5.2 cm spot does not match R, S or T, so Y also contains an unidentified substance.	Compare spots produced under the same conditions. Matching heights support a match to a reference, while each additional separated spot is evidence for another component.	5
Total Question 3			5
04.1	The ink baseline may dissolve and produce extra marks, so it should be drawn in pencil. Starting with the spot below the solvent can wash the sample into the solvent instead of carrying it up the paper. The solvent level should begin below the sample spot. A suitable solvent in which the colouring dissolves should be used. The solvent should be allowed to rise up the paper before the chromatogram is removed and dried. One visible mark from this flawed run is not reliable evidence that the colouring is pure.	Check each part of the setup against what must move and what must stay fixed. Pencil does not dissolve to contaminate the chromatogram, while the sample must start above the solvent so it travels with the rising solvent. Because both controls were wrong, the observed mark cannot support the purity conclusion.	6
Total Question 4			6
05.1	The claim is incorrect because the ethanol result shows that Z contains at least two substances. The substances may have travelled the same distance in water, so their spots overlapped. Alternatively, one substance may not have dissolved or moved in water. Using a different suitable solvent can separate substances because their solubilities and attractions to the paper differ.	A single spot is evidence of purity only if the method can separate the possible components. The second solvent reveals two components, so the first result must have hidden them through overlap or failure of one component to move.	4
Total Question 5			4

4.1.1.3 · The development of the model of the atom (common content with physics)

Tier 1 · Easy

Mark scheme for 4.1.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	plum pudding model → nuclear model → Bohr's shells → Chadwick's neutron evidence	The electron prompted the plum pudding model. Scattering evidence then produced the nuclear model, Bohr refined electron positions, and the neutron was identified last.	2
Total Question 1			2
02.1	The plum pudding model.	The model with electrons spread through a ball of positive charge is the plum pudding model.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.1.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Most of an atom is empty space because nearly all particles passed through. Positive charge is concentrated in a very small nucleus because some positive alpha particles were strongly repelled. Most of the atom's mass is concentrated in the nucleus.	Match each observation to a structural inference. Easy passage requires mostly empty space. Rare, very large deflections require a small concentration of charge and mass capable of exerting a strong repulsive force.	4
Total Question 1			4
02.1	The atom could not be an indivisible solid sphere because smaller particles came from it. The new model included negative electrons. The electrons were embedded in a sphere of positive charge, giving the plum pudding model.	Use the evidence to challenge the old assumption of indivisibility. The replacement model had to contain electrons while retaining an overall neutral atom, so it also included spread-out positive charge.	3
Total Question 2			3
03.1	The proposal made predictions that could be compared with observations. The observations agreed with the theoretical calculations. This evidence supported electrons occupying specific distances rather than being placed arbitrarily around the nucleus.	A model gains support when predictions derived from it agree with experimental observations. Bohr's calculated model passed that comparison, so scientists had evidence for the adaptation.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The plum pudding model had diffuse positive charge with embedded electrons and no nucleus. The nuclear model concentrated positive charge and most mass in a tiny central nucleus, with electrons outside it. Large alpha deflections contradicted diffuse positive charge and supported a concentrated nucleus. Bohr proposed electrons at specific distances or energy levels. Positive nuclear charge was divided into protons. Chadwick supplied evidence for neutrons in the nucleus.	Begin with one direct contrast between the models. Link the rare large deflections to the replacement of spread-out positive charge by a nucleus. Then add the later refinements in sequence: fixed electron levels, protons and finally neutrons.	6
Total Question 1			6
02.1	The measured nuclear mass was greater than the mass explained by its protons. The extra nuclear particles could add mass without adding positive charge. Chadwick provided evidence for neutrons. Neutrons were added to the model as neutral particles in the nucleus. This accounted for nuclear mass without increasing the positive charge.	Combine the mass evidence with the absence of charge. A neutral particle with about the same relative mass as a proton explains additional nuclear mass while leaving the nucleus's charge unchanged; this particle is the neutron.	5
Total Question 2			5
03.1	The nucleus was no longer treated as one undivided region of positive charge. Its positive charge could be subdivided into identical smaller particles. Each particle had the same positive charge as a hydrogen nucleus. These positively charged particles were called protons. Seven charge units meant that the nitrogen nucleus contained seven protons. This added the proton to the nuclear model without moving the electrons from outside the nucleus.	Use the whole-number charge evidence to identify identical positive particles within the nucleus. One hydrogen-nucleus charge corresponds to one proton, so seven units correspond to seven protons; this refines the contents of the existing nucleus.	6
Total Question 3			6
04.1	Both models include negatively charged electrons. Both models account for an atom being neutral by including positive charge as well as electrons. The nuclear model rejected the idea that positive charge is spread diffusely throughout the atom. It placed the positive charge in a small central nucleus instead. The plum pudding model was useful because it incorporated the evidence for electrons into a testable model that could be compared with later evidence and improved.	Separate continuity from change. The nuclear model retained electrons and overall neutrality, but replaced diffuse positive charge with a concentrated nucleus. A model can remain scientifically useful after being superseded if it organises the evidence available at the time and provides ideas that later evidence can test and refine.	5
Total Question 4			5
05.1	Both models contain electrons, so electron emission is consistent with either model. Evidence that fits two models cannot by itself decide where the positive charge is located. The spread-out model predicts no rare, very large deflections of positive alpha particles. The central-charge model predicts that most alpha particles pass through but a very small number are strongly deflected or rebound. Observing those rare large deflections would support the small central positive region and reject the spread-out model.	First identify the feature shared by the models: both already contain electrons. A useful test must instead target their different distributions of positive charge. Comparing their contrasting scattering predictions with the observed result provides evidence that can discriminate between them.	5
Total Question 5			5

4.1.1.4 · Relative electrical charges of subatomic particles

Tier 1 · Easy

Mark scheme for 4.1.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Proton: +1. Neutron: 0. Electron: −1.	Recall the relative charge table: the proton is positive, the neutron is neutral and the electron has an equal-magnitude negative charge.	3
Total Question 1			3
02.1	Neutron. It is found in the nucleus.	A neutron has relative charge 0 and is located with protons in the nucleus.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Overall charge: +3. It is a positive ion.	Neutrons contribute no charge. Add the charges from protons and electrons: $13(+1)+10(-1)=+3$ . Because proton and electron numbers are unequal, the particle is an ion rather than a neutral atom.	3
Total Question 1			3
02.1	The neutral atom has 16 electrons. After gaining two electrons it has 18 electrons. Its charge is 2−.	A neutral atom has equal proton and electron numbers. Adding two negative electrons without changing the 16 protons gives two more negative charges than positive charges, so the ion is 2−.	3
Total Question 2			3
03.1	The four X²⁺ ions have a total charge of 8+. Eight Y⁻ ions are needed. Their total charge is 8−, so the positive and negative charges cancel.	Multiply the charge on each X ion by four, then supply the same magnitude of negative charge. Eight ions of charge 1− balance a total charge of 8+.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Atomic number: 26. Mass number: 56. Charge: 3+. X is Fe³⁺, an iron ion.	Atomic number equals protons, so it is 26. Mass number is protons plus neutrons: $26+30=56$ . There are three fewer electrons than protons, so the charge is 3+. Element 26 in the periodic table is iron, Fe.	5
Total Question 1			5
02.1	13 protons. 18 neutrons. Q represents aluminium, Al. The ion is Al³⁺.	A 3+ charge means there are three fewer electrons than protons, so the proton number is $10+3=13$ . Neutrons equal mass number minus proton number: $31-13=18$ . Atomic number 13 identifies aluminium, giving Al³⁺.	4
Total Question 2			4
03.1	The statement is incorrect. Neutrons have no electrical charge, so gaining neutrons would not produce a 2− charge. The oxygen atom becomes O²⁻ by gaining two electrons. It then has two more electrons than protons, giving an overall charge of 2−.	Only a change in charged particles can change overall charge. The nucleus remains unchanged when an atom forms this ion; adding two negative electrons produces the 2− charge.	4
Total Question 3			4
04.1	T has 36 electrons. Its 35 protons contribute 35+ and its 36 electrons contribute 36−, giving 1− overall. Neutron number equals mass number minus proton number. No mass number is given, and natural isotopes with proton number 35 have different neutron numbers, so the student's value cannot be determined.	A particle with charge 1− has an electron excess of one, so the electron number is $35+1=36$ . Charge data do not give neutron number because neutrons are uncharged. A mass number is needed to use neutron number $=$ mass number $-35$ ; atoms with 35 protons occur naturally with more than one mass number, so 44 neutrons is only one possibility.	4
Total Question 4			4
05.1	Each ion has charge 3+. The four ions have a total charge of 12+. Twelve electrons are transferred in total. Ion formation changes electrons outside the nucleus; it does not remove protons from the nucleus.	Losing three negatively charged electrons leaves each atom with three more protons than electrons, so each ion is 3+. Multiply both the charge and the electrons lost by four. The nucleus is not changed during ordinary ion formation, so the proton number remains 13.	4
Total Question 5			4

4.1.1.5 · Size and mass of atoms

Tier 1 · Easy

Mark scheme for 4.1.1.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$1.2\times10^{-10}\,\text{m}$	The prefix nano means $10^{-9}$ , so $0.12\,\text{nm}=0.12\times10^{-9}\,\text{m}=1.2\times10^{-10}\,\text{m}$ .	2
Total Question 1			2
02.1	Neutron: 1. Electron: very small compared with 1.	A neutron has relative mass 1. An electron's relative mass is negligible at this level.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	8 protons. 10 neutrons. 8 electrons. Almost all the mass is in the nucleus.	Atomic number gives 8 protons. Neutrons equal $18-8=10$ . A neutral atom has the same number of electrons as protons, so it has 8 electrons. Protons and neutrons carry almost all the mass and both are in the nucleus.	4
Total Question 1			4
02.1	$5.0\times10^6$ atoms (or 5 000 000 atoms).	Convert both lengths to metres: $1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}$ and $0.20\,\text{nm}=2.0\times10^{-10}\,\text{m}$ . Divide the row length by the atom diameter: $(1.0\times10^{-3})/(2.0\times10^{-10})=5.0\times10^6$ .	3
Total Question 2			3
03.1	15 protons. 31 − 15 = 16 neutrons. 18 electrons.	The lower number gives 15 protons and the upper number gives 31 total protons and neutrons. Subtract to obtain 16 neutrons. A 3− ion has three more electrons than protons, so it has 18 electrons.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The atomic radius is $1.2\times10^4$ times the nuclear radius. The ion has mass number 56 and charge 2+. $\,^{56}_{26}\mathrm{Fe}^{2+}$	Divide the radii: $(9.0\times10^{-11})/(7.5\times10^{-15})=1.2\times10^4$ . The mass number is $26+30=56$ . Two electrons are missing relative to the proton count, so the ion is 2+. Atomic number 26 identifies Fe, giving $\,^{56}_{26}\mathrm{Fe}^{2+}$ .	6
Total Question 1			6
02.1	A and B are isotopes of the same element. They are the same element because both have 12 protons. They have different masses because they contain different numbers of neutrons. A has mass number 24 and B has mass number 26. Almost all atomic mass comes from protons and neutrons in the nucleus.	Element identity is fixed by proton number, so both atoms are magnesium isotopes. Add protons and neutrons for each mass number: $12+12=24$ and $12+14=26$ . The extra neutrons increase the nuclear mass.	5
Total Question 2			5
03.1	50 cm = 500 mm. 500 ÷ 10 000 = 0.05 mm. The nuclear radius is less than 0.05 mm in this model. The tiny nucleus contains the protons and neutrons. Because electrons have very little relative mass, almost all the atomic mass is concentrated in that tiny nucleus.	Convert the model radius to millimetres before dividing by 10 000. Treat 0.05 mm as an upper limit because the specification states that the nuclear radius is less than this fraction of the atomic radius.	5
Total Question 3			5
04.1	The nuclear-to-atomic radius ratio is $1/10\,000=10^{-4}$ . The volume ratio is $(10^{-4})^3=10^{-12}$ . The nucleus therefore occupies about one trillionth of the atom's volume in this model. The student's claim is incorrect because it treats a radius ratio as a volume ratio. Despite this tiny volume, the nucleus contains the protons and neutrons and therefore almost all the atomic mass.	Cube the radius scale factor because the objects are treated as similar spheres. Cubing $10^{-4}$ gives $10^{-12}$ , far smaller than $10^{-4}$ . Then separate the amount of space occupied from the mass distribution: the tiny nucleus still contains the massive subatomic particles.	5
Total Question 4			5
05.1	The nucleus contributes a relative mass of 12 + 12 = 24. The electrons contribute $12/1840=0.00652$ to 3 significant figures. The total relative mass from these values is about 24.0065. The electrons contribute about 0.0272% of this total. The statement is not exactly correct: electron mass is not zero, but its contribution is negligible compared with the nucleus.	Add the proton and neutron contributions because both particles are in the nucleus. Multiply the mass of one electron by 12, then compare this with the total: $0.00652\div24.00652\times100\approx0.0272\%$ . This quantifies why almost all atomic mass is concentrated in the nucleus without treating electron mass as exactly zero.	5
Total Question 5			5

4.1.1.6 · Relative atomic mass

Tier 1 · Easy

Mark scheme for 4.1.1.6 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A_r = 63.62	Weight each isotope by its percentage abundance: $A_r=(63\times69+65\times31)/100=(4347+2015)/100=63.62$ .	2
Total Question 1			2
02.1	Mass number is a whole number because it counts protons and neutrons. The relative atomic mass is a weighted mean for a mixture of chlorine isotopes.	Distinguish one atom from a natural sample. Each isotope has a whole-number mass number, but averaging their masses according to abundance can give a non-integer relative atomic mass.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.1.6 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	A_r = 24.32	Use all three weighted contributions: $A_r=(24\times79+25\times10+26\times11)/100=(1896+250+286)/100=24.32$ .	3
Total Question 1			3
02.1	The method is incorrect because it ignores the unequal percentage abundances. A_r = 10.8	Use a weighted mean rather than a simple mean: $A_r=(10\times20+11\times80)/100=(200+880)/100=10.8$ .	3
Total Question 2			3
03.1	Total relative abundance = 18 + 4 + 3 = 25. Weighted total = (50 × 18) + (53 × 4) + (57 × 3) = 1283. A_r = 1283 ÷ 25 = 51.32, which is 51.3 to 1 decimal place.	Multiply each isotope mass by its ratio part, add the three products, then divide by the total number of ratio parts: $A_r=[(50\times18)+(53\times4)+(57\times3)]/25=51.32$ , so $A_r=51.3$ to 1 decimal place.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.6 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Mass-79 abundance: 55%	Let the mass-79 abundance be x%, so mass-81 has abundance (100 − x)%. Then $[79x+81(100-x)]/100=79.90$ . Expanding gives $79x+8100-81x=7990$ , so $-2x=-110$ and $x=55$ . Therefore the mass-79 isotope is 55% abundant.	4
Total Question 1			4
02.1	The other isotope has mass number 64. The relative atomic mass lies between 62 and 64. It is closer to 64 because J-64 is the more abundant isotope.	Let the unknown mass number be $x$ . Then $[62(25)+75x]/100=63.50$ . This gives $1550+75x=6350$ , so $75x=4800$ and $x=64$ . Its 75% abundance pulls the weighted mean nearer 64 than 62.	4
Total Question 2			4
03.1	There are 240 − 180 − 24 = 36 G-26 atoms initially. Initial total isotope mass = (180 × 24) + (24 × 25) + (36 × 26) = 5856. Initial relative atomic mass = 5856 ÷ 240 = 24.4. Replacing 12 G-24 atoms by 12 G-26 atoms adds 24 to the total, so the new relative atomic mass is 5880 ÷ 240 = 24.5. The mean increases because some lighter isotopes have been replaced by heavier isotopes while the total number of atoms stays the same.	Find the missing isotope count, calculate the weighted total and divide by 240. Each replacement increases the isotope-mass total by 2, so 12 replacements increase it by 24 without changing the denominator.	5
Total Question 3			5
04.1	Weighted total = $(24\times78.9)+(25\times10.0)+(26\times11.0)+(30\times0.1)=1893.6+250+286+3=2432.6$ . $A_r=2432.6\div100=24.326$ , which rounds to 24.3 to 1 decimal place. The periodic-table value therefore does not change at the printed precision.	Check the abundances sum to $100\%$ , then use every abundance in the weighted mean: $A_r=[(24\times78.9)+(25\times10.0)+(26\times11.0)+(30\times0.1)]\div100=24.326$ . Compare values only after completing the calculation: both the recalculated value and the stated periodic-table value print as 24.3 to 1 decimal place.	3
Total Question 4			3
05.1	A weighted mean of masses 41 and 44 must lie from 41 to 44 inclusive. The report of 42.5 could be valid because it lies between the two isotope masses. The report of 40.8 is impossible because it is below both isotope masses. The report of 44.1 is impossible because it is above both isotope masses. Changing isotope abundances can move the mean within the interval but cannot move it outside the lightest and heaviest isotope masses.	Use a consistency bound rather than a weighted-mean calculation. An average made only from 41 and 44 cannot be smaller than 41 or larger than 44, whatever non-negative abundances are used. Therefore only 42.5 is possible.	5
Total Question 5			5

4.1.1.7 · Electronic structure

Tier 1 · Easy

Mark scheme for 4.1.1.7 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	2,8,3	A neutral aluminium atom has 13 electrons. Fill 2 into the first shell and 8 into the second, leaving 3 for the third shell: 2,8,3.	2
Total Question 1			2
02.1	15 electrons.	Add the electrons in the occupied shells: $2+8+5=15$ .	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.1.1.7 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Calcium atom: 2,8,8,2. Ca²⁺ ion: 2,8,8.	Distribute the atom's 20 electrons as 2,8,8,2. A 2+ ion has lost two electrons from its outer shell, leaving 18 electrons arranged 2,8,8.	3
Total Question 1			3
02.1	2,8,7 is correct. A chlorine atom has 17 electrons. The lower energy levels fill first, so the second shell fills with eight electrons before the third shell contains seven.	Both suggestions total 17, so the total alone does not decide. Apply inside-out filling: 2 fill the first shell and 8 fill the second, leaving 7 for the third.	3
Total Question 2			3
03.1	The electronic structure is 2,8,6. The atomic number is 2 + 8 + 6 = 16. Two more electrons would fill the outer shell to 8.	Translate the shell counts directly into comma-separated notation. A neutral atom has equal numbers of electrons and protons, and atomic number is the proton number.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.1.7 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	X is chlorine, Cl. Atomic number: 17. Atom: 2,8,7. After gaining one electron: 2,8,8.	Three occupied shells place X in period 3, and seven outer electrons place it in Group 7. The period-3 Group-7 element is chlorine. Counting $2+8+7$ gives atomic number 17. Adding one electron completes the outer shell, producing 2,8,8.	5
Total Question 1			5
02.1	Proton number: 16. R is sulfur, S. Neutral atom: 2,8,6. Ion: 2,8,8.	A 2− ion has two more electrons than protons, so $18-2=16$ protons. Atomic number 16 is sulfur. The atom has 16 electrons arranged 2,8,6; gaining two electrons produces the 2,8,8 ion.	4
Total Question 2			4
03.1	A is potassium and B is chlorine. A is K⁺ because it has 19 protons and 18 electrons. B is Cl⁻ because it has 17 protons and 18 electrons. A neutral potassium atom has electronic structure 2,8,8,1. A neutral chlorine atom has electronic structure 2,8,7. The ions have equal electron numbers but different proton numbers, so their overall charges differ.	Use proton number to identify each element. Compare each proton count with the 18 electrons to obtain the charge, then restore electron number to equal proton number for each neutral atom.	6
Total Question 3			6
04.1	A is impossible: for the first 20 elements the third shell fills to 8 before the fourth shell; its 20 electrons should be 2,8,8,2. B is impossible: electrons must fill the second shell to 8 before the third; its 14 electrons should be 2,8,4. C is valid and represents 16 electrons arranged as 2,8,6. D is impossible because the second shell cannot hold 9 electrons; its 13 electrons should be 2,8,3. The first shell holds up to 2 electrons and the second holds up to 8. Electrons occupy the lowest available inner shells first; for the first 20 elements, the third shell holds up to 8 before the fourth shell is occupied.	Keep each proposed total fixed, then rebuild the structure from the inside out. Fill shell 1 to 2 and shell 2 to 8 before using shell 3; in the first 20 elements, fill shell 3 to 8 before placing electrons in shell 4. This gives A 2,8,8,2; B 2,8,4; C unchanged; and D 2,8,3.	6
Total Question 4			6
05.1	The diagram places electrons in shell 3 before shell 2 is full. The eighth electron in shell 2 must be placed there before the remaining electrons occupy shell 3. The correct ground-state structure is 2,8,5. A neutral atom with 15 electrons has proton number 15, so the element is phosphorus.	Count 15 electrons in the proposed diagram, but apply the inside-out filling rule before identifying the atom. Moving one electron from shell 3 into the vacancy in shell 2 gives 2,8,5. A neutral atom then has 15 protons, which identifies phosphorus.	4
Total Question 5			4

4.1.2.1 · The periodic table

Tier 1 · Easy

Mark scheme for 4.1.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Atomic number: 12. Period 3. Group 2.	The electron total is $2+8+2=12$ , so a neutral atom has atomic number 12. Three occupied shells mean period 3, and two outer electrons mean Group 2.	3
Total Question 1			3
02.1	Elements are arranged in order of increasing atomic number or proton number. Elements in the same group have the same number of outer-shell electrons and similar chemical properties.	The modern ordering variable is atomic number. A vertical group collects elements with matching outer-electron patterns, which produces similar reactions.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Both have one electron in the outer shell. Both are in Group 1 and tend to react in similar ways by losing that electron. Q has the larger atomic number: 11 compared with 3 for P.	Compare the last number in each structure: both atoms have one outer electron, so they occupy the same group and react similarly. Add all electrons to obtain atomic numbers: P has $2+1=3$ and Q has $2+8+1=11$ .	4
Total Question 1			4
02.1	Y has six electrons in its outer shell. Y and sulfur are in the same group. Their matching outer-shell electron numbers give them similar chemical properties.	Moving vertically within a group changes the number of occupied shells but preserves the outer-electron count. Sulfur is in Group 6, so Y also has six outer electrons and similar chemistry.	3
Total Question 2			3
03.1	A has electronic structure 2,7. B has electronic structure 2,8. C has electronic structure 2,8,1. B's second shell is full, so the next electron in C occupies a new third shell and C begins a new period.	Fill the lowest available shells in order. Atomic number 10 completes the second shell; electron 11 must occupy the third shell, creating a new row of the periodic table.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	R: 2,8,6; period 3, Group 6. S: 2,8,8,1; period 4, Group 1. S is more likely to form a positive ion because it can lose one outer electron to obtain a stable arrangement.	Distribute 16 electrons as 2,8,6 and 19 electrons as 2,8,8,1. The number of occupied shells gives periods 3 and 4; the outer electron counts give Groups 6 and 1. Losing S's single outer electron is the simpler route to a stable shell, so S is the likely positive-ion former.	6
Total Question 1			6
02.1	T has electronic structure 2,8,7. U has electronic structure 2,8,8. T is more reactive than U. T can gain one electron to complete its outer shell. U already has a stable full outer shell, so it is unreactive.	Period 3 gives three occupied shells. The group supplies the outer-electron number: seven for T and a full shell of eight for U. Compare their tendency to change: T can gain one electron, while U already has a stable arrangement.	5
Total Question 2			5
03.1	V has electronic structure 2,8,2. V is in period 3, not period 2, because it has three occupied shells. V is correctly placed in Group 2 because it has two outer-shell electrons. W is correctly placed in period 3 and Group 6 because it has three occupied shells and six outer-shell electrons. X is in Group 0, not Group 7, because its outer shell is full. X is unreactive rather than reactive because it has a stable outer-shell arrangement.	For each species, use occupied shells for the period and outer-shell electrons for the group. A full outer shell identifies Group 0 and explains low reactivity.	6
Total Question 3			6
04.1	The ion has 10 electrons, so the neutral atom has 12 electrons because it lost two. Atomic number 12 identifies M as magnesium. The neutral electronic structure is 2,8,2. M is in period 3 and Group 2. Calcium has similar chemical properties because it is also in Group 2.	Reverse the ion formation first: a 2+ ion with 10 electrons came from a neutral atom with 12 electrons. Use the resulting structure to identify the element and locate it: occupied shells give the period, while outer-shell electrons give the group. A same-group element provides the similarity prediction.	5
Total Question 4			5
05.1	Sodium and chlorine are both in period 3 because each has three occupied shells. Being in the same period does not mean that their chemical properties are similar. Sodium has one outer-shell electron and is in Group 1. Chlorine has seven outer-shell electrons and is in Group 7. Elements in the same group have the same number of outer-shell electrons and therefore similar chemical properties.	Read the structures in two different ways. The number of occupied shells places both elements in period 3, but their outer-shell electron numbers place them in different groups. Chemical similarity follows the group pattern, so the student's conclusion does not follow from sharing a period.	5
Total Question 5			5

4.1.2.2 · Development of the periodic table

Tier 1 · Easy

Mark scheme for 4.1.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	He left gaps for elements that had not yet been discovered. He changed the strict atomic-weight order where properties showed an element fitted a different group.	Recall the two deliberate departures from simply listing known elements by weight: preserve a gap when evidence suggested a missing element, and prioritise matching properties over strict weight order.	2
Total Question 1			2
02.1	Some elements had not yet been discovered.	Early classifiers could arrange only the elements known at the time, so undiscovered elements were absent.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.1.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The prediction was made before the element was discovered. The observed compound matched the predicted property. This agreement supplied evidence that the gap and group placement were valid.	Treat the prediction as a test of the model. An independently discovered element displaying the forecast property is unlikely to be explained by chance alone and supports both the missing-element prediction and its position.	3
Total Question 1			3
02.1	Mendeleev would move A away from the strict atomic-weight position. He prioritised similar chemical properties when choosing groups. The new position places A with elements that form similar compounds. He accepted occasional reversals of weight order to preserve the repeating property pattern.	Compare the two kinds of evidence. Mendeleev used atomic weight as a guide but treated chemical similarity as stronger evidence for group placement.	4
Total Question 2			4
03.1	Both isotopes have the same proton number, so both are atoms of chlorine. Their different neutron numbers give them different mass numbers. Atomic number keeps all isotopes of one element in the same position, whereas atomic weight is affected by the masses and abundances of its isotopes.	Separate element identity from isotope mass. Proton number fixes the element and its periodic-table position; neutron number can vary without producing a different element.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Strict atomic-weight order would place 39.1 before 40.0. That order could put elements into groups with the wrong chemical properties. Mendeleev could reverse the order to preserve property patterns. The modern table uses atomic number, not relative atomic mass. Isotopes explain why relative atomic masses do not always rise in the same order as atomic numbers.	State the conflict between numerical weight order and repeating properties. Mendeleev prioritised the property pattern even without knowing why. Proton discovery supplied atomic number as the correct ordering variable, while isotope mixtures explained the apparently reversed average masses.	5
Total Question 1			5
02.1	The colour and density both closely matched predictions made before the discovery. Agreement of more than one predicted property provides evidence for the gap and its position. A close match could occur by chance, but matching independent properties makes luck less convincing. The modern table orders elements by atomic number rather than relative atomic mass. Atomic number gives each element a position consistent with its proton number and recurring properties.	Evaluate rather than accept or reject the claim absolutely: the evidence strongly supports the prediction because two properties match, although evidence never proves chance impossible. Finish by contrasting Mendeleev's mass-based evidence with modern proton-number order.	5
Total Question 2			5
03.1	Mendeleev left gaps before the missing elements were discovered. He predicted properties for the elements expected to fill those gaps. Later discoveries could therefore be compared with predictions recorded in advance, rather than fitted after the event. Agreement between predicted and observed properties supported the arrangement of the table. A serious disagreement would instead have counted against the proposed arrangement, so the predictions provided a genuine test.	Distinguish a prediction made before new evidence from an explanation invented afterwards. A prior prediction creates a test that can either support or refute the scientific idea.	5
Total Question 3			5
04.1	The oxide formula Z₂O₃ matches the predicted oxide pattern. The element density of Z, 5.8 g/cm³, is outside the predicted range 4.1 to 4.9 g/cm³. The candidate therefore fails one of the advance predictions for the gap. Z should be rejected as the element for this particular gap unless further evidence changes the data or prediction. This result does not by itself reject the periodic table: Z may belong elsewhere, and the gap can remain for a different element that fits the predictions.	Test the candidate against every stated prediction. Its oxide supplies one match, but its explicitly measured element density lies above the predicted interval. That failed prediction is evidence against assigning Z to this gap, not evidence that the whole periodic arrangement must be discarded.	5
Total Question 4			5
05.1	Table P follows one consistent measured quantity, atomic weight. However, its unlike chemical properties in the same columns weaken the proposed repeating pattern. Table Q groups elements by similar chemical properties even when this requires changing strict weight order. Leaving gaps is a risk because undiscovered elements might not exist or might not match the predicted properties. Discovering elements with the predicted properties would support Table Q's gaps and grouping. Later measurements of atomic number would explain the order objectively and could resolve the weight-order anomalies.	Judge each arrangement by both its rule and its ability to explain chemical patterns. Table Q takes a testable risk by predicting missing elements, while Table P avoids gaps at the cost of poor chemical groupings. New-element discoveries and proton-number measurements provide independent tests of the competing arrangements.	6
Total Question 5			6

4.1.2.3 · Metals and non-metals

Tier 1 · Easy

Mark scheme for 4.1.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	T is a metal. It forms T²⁺ ions.	Loss of electrons produces a positive ion, which is the defining chemical behaviour of a metal. Losing two negatively charged electrons leaves charge 2+.	2
Total Question 1			2
02.1	E is a non-metal. It gains an electron rather than losing electrons to form a positive ion.	Metals characteristically lose electrons and form positive ions. Gaining an electron is characteristic of a non-metal.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Magnesium is on the left of the table and has two outer electrons. It can lose those electrons to form Mg²⁺, so it is a metal. Sulfur is towards the right and has six outer electrons. Sulfur does not form positive ions in its typical reactions, so it is a non-metal.	Link location to outer-shell behaviour. Magnesium reaches a stable arrangement by losing its two outer electrons and therefore forms a positive ion. Sulfur lies in the non-metal region and does not show positive-ion formation.	4
Total Question 1			4
02.1	Some non-metals have atypical physical properties, so one physical property is not decisive. Metals react to form positive ions. Non-metals do not form positive ions in their typical reactions.	Treat graphite as an exception to the usual non-conducting pattern. The specification's chemical distinction is whether atoms form positive ions, so classification should not rely on conductivity alone.	3
Total Question 2			3
03.1	A typical metal conducts electricity, whereas a typical non-metal does not. A typical metal is malleable, whereas a solid non-metal is usually brittle. A metal reacts to form positive ions, whereas a non-metal does not form positive ions. Metals are mainly towards the left and bottom of the periodic table; non-metals are towards the right and top.	Make a direct contrast for each requested feature. Chemical classification is tied to positive-ion formation, while position and physical properties provide supporting patterns.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	A is a metal: it is lustrous, malleable and conducts electricity. Formation of A³⁺ confirms positive-ion behaviour. A is expected on the left or towards the lower part of the table. B is a non-metal: it is brittle and non-conducting. B is expected towards the upper-right part of the table.	Use several independent observations. A shows the physical pattern of a metal and the decisive chemical evidence of positive-ion formation. B lacks the characteristic metallic properties. Map the classifications to the broad metal and non-metal regions of the table.	6
Total Question 1			6
02.1	C should be classified as a non-metal because it does not form positive ions. Its conductivity is an exception to the usual non-metal pattern and is not sufficient to classify it as a metal. D should be classified as a metal because it forms positive ions. The poor conductivity observation conflicts with the typical metal pattern but may depend on the form or conditions tested. Chemical ion formation gives the decisive classification in the specification.	Weigh the evidence instead of counting observations. Physical properties are typical patterns with exceptions, whereas positive-ion formation is the stated chemical distinction between metals and non-metals.	5
Total Question 2			5
03.1	J is likely to be a metal. J is likely to lose its one outer electron and form a 1+ ion. K is likely to be a non-metal and does not form positive ions. J is likely to conduct electricity and be malleable. Solid K is likely to be a poor conductor and brittle compared with J.	Use periodic position and outer-shell structure to predict chemical behaviour, then apply the characteristic physical-property contrast between metals and non-metals.	5
Total Question 3			5
04.1	Three oxide ions give a total charge of 6−. The two E ions must therefore provide 6+ in total, so each is E³⁺. Each E atom loses three electrons to form the ion. E is a metal because metals form positive ions. E is expected on the left or towards the centre of the periodic table rather than the upper right.	Use electrical neutrality to reverse-engineer the unknown ion charge. Three O²⁻ ions contribute 6−, so two E ions must each be 3+. Formation by electron loss is characteristic of a metal and supports a left-or-central periodic-table position.	5
Total Question 4			5
05.1	The conductor rule is false because graphite conducts electricity but carbon is a non-metal. Conductivity is therefore useful evidence but is not sufficient on its own. The hardness rule is false because Group 1 metals are soft and have relatively low melting points for metals. The negative-ion rule is false because Group 0 non-metals are unreactive and do not normally gain electrons. Metal and non-metal properties are typical patterns with exceptions, not universal single-property tests. Classification should combine periodic-table position with several physical and chemical properties, especially whether positive ions form in reactions.	Attack each universal rule with the named counterexample. Graphite defeats a conductivity-only test, Group 1 defeats an all-metals-are-hard claim, and Group 0 defeats an all-non-metals-form-ions claim. A robust classification uses multiple consistent lines of evidence.	6
Total Question 5			6

4.1.2.4 · Group 0

Tier 1 · Easy

Mark scheme for 4.1.2.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Neon has a full outer shell of eight electrons. This is a stable electron arrangement, so neon has little tendency to react.	Inspect the outer shell in 2,8. Because it is complete, neon does not need to gain, lose or share electrons to obtain a stable arrangement.	2
Total Question 1			2
02.1	They exist as single atoms rather than bonded molecules. They are monatomic.	Group 0 elements do not normally bond to one another, so each gas consists of separate atoms; this is described as monatomic.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	−108 °C. Xenon is below krypton in Group 0 and has a greater relative atomic mass. Boiling point increases down Group 0.	Continue the stated group trend. Xenon must have a boiling point higher than krypton's −153 °C; only −108 °C satisfies that direction.	3
Total Question 1			3
02.1	neon < argon < xenon Relative atomic mass increases down Group 0. Boiling point increases as relative atomic mass increases down the group.	Locate the elements from higher to lower in Group 0. The specification trend is increasing boiling point down the group as the atoms become heavier.	3
Total Question 2			3
03.1	The first shell can hold a maximum of two electrons. Helium's only shell is therefore full. This is a stable electron arrangement, so helium is unreactive.	Apply shell capacity rather than treating eight as a rule for every shell. Helium's first energy level is complete with two electrons, giving the same stability principle as the full outer shells of the other noble gases.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.2.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The gas is argon. Its 18 electrons give the structure 2,8,8. Its full outer shell is stable, making it unreactive and unlikely to form molecules. Argon is below neon and has greater relative atomic mass, so its boiling point is higher.	Count the electrons to obtain atomic number $2+8+8=18$ , which identifies argon. Connect the full outer shell to unreactive monatomic particles, then apply the increasing boiling-point trend down Group 0.	5
Total Question 1			5
02.1	Neon is a gas because −155 °C is above its boiling point. Argon is a gas because −155 °C is above its boiling point. Krypton is a liquid because −155 °C lies between its melting and boiling points. Both melting point and boiling point increase from neon to argon to krypton. This is the increasing-temperature trend down Group 0 as relative atomic mass increases.	Compare −155 °C with each melting and boiling point. Above the boiling point means gas; between melting and boiling points means liquid. The progressively less negative transition temperatures show both values increasing down the group.	5
Total Question 2			5
03.1	Argon has a full outer shell of electrons. The stable arrangement makes argon unreactive. It therefore does not react with the hot metal filament. Argon atoms do not easily form bonds with each other, so the gas is monatomic.	Link the filled outer shell to both observations: chemical inertness protects the filament, and the lack of a drive to bond means argon exists as separate atoms.	4
Total Question 3			4
04.1	Helium is collected first because it has the lowest boiling point, −269 °C. Argon is collected next at about −186 °C. Xenon is collected last because it has the highest boiling point, −108 °C. Fractional distillation separates the substances using their different boiling points. The atoms are not chemically changed or combined, so each monatomic Group 0 element is recovered unchanged.	Order the elements from the most negative to the least negative boiling point. On warming, the component with the lowest boiling point vaporises first. Because the process only changes physical state and separates existing atoms, it does not create new substances.	5
Total Question 4			5
05.1	Boiling point increases down Group 0, and radon's value of −62 °C extends this physical trend beyond xenon at −108 °C. The claim is incorrect because greater mass, density and boiling point do not by themselves show greater chemical reactivity. A radon atom has a full outer electron shell. Its stable outer-shell arrangement means that electron transfer or sharing is not favoured. Radon is consequently very unreactive; its position at the bottom of the boiling-point trend does not overturn the Group 0 reactivity explanation.	Read the six values in group order to establish the increase in boiling point through radon. Then separate the physical evidence from the chemical claim. Chemical reactivity is explained by the full outer shell, which remains present in radon despite its greater mass and density.	5
Total Question 5			5

4.1.2.5 · Group 1

Tier 1 · Easy

Mark scheme for 4.1.2.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Sodium hydroxide and hydrogen. Suitable observation: fizzing, movement on the surface, melting into a ball, or eventual disappearance.	Apply the general alkali-metal reaction: metal + water → metal hydroxide + hydrogen. Hydrogen production causes effervescence; sodium also moves and may melt because the reaction releases heat.	3
Total Question 1			3
02.1	lithium oxide	A Group 1 metal reacting with oxygen forms the metal oxide, so lithium forms lithium oxide.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.1.2.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	2Na + Cl₂ → 2NaCl Both sodium and potassium have one outer electron. Each loses one electron to form a 1+ ion, while chlorine forms a 1− ion, giving a one-to-one ratio.	Chlorine is diatomic, so make two NaCl units and then balance sodium with a coefficient of 2. The common single outer electron explains why either Group 1 metal forms a 1+ ion and hence a 1:1 chloride formula.	4
Total Question 1			4
02.1	2Li + 2H₂O → 2LiOH + H₂ Suitable observation: fizzing, movement on the surface, or gradual disappearance.	Use two lithium atoms and two water molecules to make two LiOH units, leaving two hydrogen atoms for H₂. Hydrogen production causes fizzing; the floating metal also moves and is used up.	4
Total Question 2			4
03.1	A is lithium. B is sodium. C is potassium. Reactivity increases from lithium to sodium to potassium, going down Group 1.	Match the observations to the increasing vigour of the first three alkali metals. Lithium reacts least vigorously, sodium melts and moves more rapidly, and potassium can ignite with a lilac flame.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.2.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Lithium < sodium < potassium in vigour. The outer electron is farther from the nucleus down the group. There are more inner shells and greater shielding. The nuclear attraction to the outer electron is weaker, so it is lost more easily. Rubidium would react even more vigorously than potassium.	Use the known downward increase in Group 1 reactivity. Explain it through distance and shielding rather than electron number: both reduce the attraction holding the single outer electron. Extrapolate the same pattern to rubidium below potassium.	6
Total Question 1			6
02.1	Caesium reacts more quickly or vigorously than lithium. Caesium's outer electron is farther from the nucleus and more shielded. The weaker attraction allows the outer electron to be lost more easily. Both trials produce the same final amount of hydrogen. The equation shows that two atoms of either Group 1 metal produce one hydrogen molecule, and equal metal particle numbers were used.	Use atomic structure to compare rates: the caesium outer electron is less strongly attracted and is lost more readily. Keep rate separate from yield. The same 2:1 metal-to-hydrogen ratio applies to both metals, so equal atom numbers give equal final hydrogen amounts when water is in excess.	5
Total Question 2			5
03.1	The comparison is not valid because more than one variable changes between trials. Use freshly cut pieces with the same exposed surface area. Use the same initial water temperature for all three trials. Use the same volume of water and the same apparatus. Measure reaction speed consistently, for example by timing a fixed observation from the moment the metal touches the water. The improved comparison should show lithium slowest, sodium intermediate and potassium fastest.	A fair comparison changes only the identity of the Group 1 metal. Control size, temperature, water volume and measurement, then apply the established increase in reactivity down the group.	6
Total Question 3			6
04.1	The oxide formula is M₂O because two M⁺ ions balance one O²⁻ ion. The chloride formula is MCl. The hydroxide formula is MOH. Each M atom loses one outer-shell electron to form M⁺. All Group 1 metals have one outer-shell electron, so they form compounds with the same charge ratios.	Balance the total positive and negative charges in each compound. One M⁺ balances each 1− ion, while two are needed for O²⁻. The repeated +1 charge follows from the shared one-electron outer-shell structure of Group 1 atoms.	5
Total Question 4			5
05.1	A is sodium and B is potassium. The prediction is incorrect: potassium reacts faster with water than sodium. B's outer electron is farther from the nucleus and has more shielding from inner electrons. The weaker attraction makes B's outer electron easier to lose. Both reactions form hydrogen and the corresponding metal hydroxide: sodium hydroxide or potassium hydroxide.	Use electron totals to identify the metals, then evaluate the causal claim rather than accepting only its distance statement. The extra occupied shell in potassium increases distance and shielding, weakening attraction to the outer electron and increasing reactivity. Group membership keeps the product pattern the same.	5
Total Question 5			5

4.1.2.6 · Group 7

Tier 1 · Easy

Mark scheme for 4.1.2.6 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Their atoms each have seven electrons in the outer shell. The elements form diatomic molecules made from pairs of atoms.	Use one atomic feature and one molecular feature: seven outer electrons fixes the group, while the elemental halogens exist as X₂ molecules.	2
Total Question 1			2
02.1	Bromine: liquid. Iodine: solid.	Recall the room-temperature progression down Group 7: bromine is a liquid and iodine is a solid.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.2.6 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Potassium chloride and bromine form. Cl₂ + 2KBr → 2KCl + Br₂ Chlorine is above bromine in Group 7 and is more reactive. Chlorine therefore displaces bromine from bromide ions.	Compare group positions to select the more reactive halogen. Chlorine displaces bromine, so swap the halogens in the salt. Balance the two atoms in each diatomic molecule by using two formula units of each potassium halide.	5
Total Question 1			5
02.1	Astatine should have a higher boiling point than iodine. Astatine should be less reactive than iodine. Its outer shell is farther from the nucleus and experiences more shielding. The weaker attraction makes gaining an electron more difficult.	Extrapolate both group trends: boiling point rises but reactivity falls down Group 7. Explain the reaction trend through the reduced nuclear attraction for an incoming electron.	4
Total Question 2			4
03.1	A chlorine molecule has formula Cl₂. Each chlorine atom gains one electron when it forms a chloride ion. The chloride ion has charge 1−, written Cl⁻. Bromine also has seven outer-shell electrons, so it undergoes similar reactions.	Halogens exist as pairs of atoms. One gained electron completes the outer shell and produces a 1− ion; the shared seven-electron outer structure explains similar reactions within Group 7.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.1.2.6 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	No reaction occurs with sodium chloride because bromine is less reactive than chlorine. Bromine displaces iodine from sodium iodide. Br₂ + 2NaI → 2NaBr + I₂ Reactivity decreases down Group 7, so bromine is more reactive than iodine but less reactive than chlorine. Down the group, increased distance and shielding make attraction for an incoming electron weaker.	Place the halogens in reactivity order chlorine > bromine > iodine. Bromine cannot displace chloride but can displace iodide. Form sodium bromide and iodine, then balance the diatomic halogens and sodium salts. Relate the trend to the increasing difficulty of gaining an electron down the group.	6
Total Question 1			6
02.1	P is chlorine. Q is bromine. R is iodine. Increasing boiling point: chlorine < bromine < iodine. A more reactive halogen displaces a less reactive halogen from a halide solution. The results therefore give the reactivity order P > Q > R. Reactivity decreases down Group 7 because greater distance and shielding weaken attraction for an incoming electron.	Translate each displacement into a pairwise reactivity comparison. P is above both others, Q is between them and R is lowest, matching chlorine, bromine and iodine. The boiling-point order runs in the opposite direction down the group.	6
Total Question 2			6
03.1	Magnesium bromide has formula MgBr₂. Magnesium bromide has ionic bonding. Hydrogen bromide has formula HBr. Hydrogen bromide has covalent bonding. Magnesium is a metal and bromine is a non-metal, whereas hydrogen and bromine are both non-metals.	A Group 7 element forms an ionic compound with a metal and a covalent compound with another non-metal. Balance Mg²⁺ with two Br⁻ ions to obtain MgBr₂.	5
Total Question 3			5
04.1	Chlorine is a gas at 50 °C because this is above its boiling point. Bromine is a liquid at 50 °C because the temperature lies between its melting and boiling points. Iodine is a solid at 50 °C because this is below its melting point. Bromine is the required liquid halogen. Bromine does not react with potassium chloride because it is less reactive than chlorine. Bromine reacts with potassium iodide because it is more reactive than iodine and displaces it.	Compare 50 °C with both phase-change temperatures for each halogen. Bromine is the only liquid, so apply the Group 7 reactivity order chlorine greater than bromine greater than iodine to its two displacement tests.	6
Total Question 4			6
05.1	U contains potassium bromide. No reaction with bromine rules out iodide because bromine would displace iodine. A reaction with chlorine rules out chloride because chlorine cannot displace itself. Chlorine is more reactive than bromine, so it displaces bromine from potassium bromide. Cl₂ + 2KBr → 2KCl + Br₂.	Use each result as a constraint. The bromine test leaves chloride or bromide possible, while the chlorine displacement leaves bromide or iodide possible. Their only common possibility is bromide, and the reactivity order gives the balanced replacement equation.	5
Total Question 5			5

4.1.3.1 · Comparison with Group 1 elements (chemistry only)

Tier 1 · Easy

Mark scheme for 4.1.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Iron is denser. Iron is stronger or harder. Iron has a higher melting point. Iron is less reactive with water, oxygen or halogens.	Treat iron as a named transition metal and sodium as a Group 1 metal. Any three correct contrasts from density, strength, hardness, melting point or reactivity earn the marks.	3
Total Question 1			3
02.1	Chromium. Copper.	Chromium and copper are named transition elements. Potassium is in Group 1 and calcium is in Group 2.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.1.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Nickel is a transition metal whereas potassium is Group 1. Nickel has a much higher melting point. Nickel is stronger and harder, so it resists deformation. Nickel is much less reactive with water.	Translate each design condition into a property, then apply the transition-metal versus Group 1 comparison. Nickel meets the high-temperature, mechanical-strength and low-reactivity requirements; potassium does not.	4
Total Question 1			4
02.1	Sodium is a Group 1 metal and reacts rapidly with water. Sodium also reacts readily with oxygen in air. Oil keeps water and oxygen away from the sodium. Iron is a transition metal and is much less reactive with water and oxygen, so it does not need the same storage.	Link the storage condition to chemical reactivity. Sodium must be isolated from air and moisture because Group 1 metals react readily; iron is much less reactive under the same conditions.	4
Total Question 2			4
03.1	Density of chromium = 35.7 ÷ 5.0 = 7.14 g/cm³. Chromium is much denser than sodium. This supports the general pattern that transition elements have higher densities than Group 1 elements.	Divide the sample mass by its volume and retain the stated density unit. Then compare the numerical values and link the result to the specified transition-metal versus Group 1 pattern.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	U is a Group 1 metal. Its low melting point, low density, softness and vigorous water reaction support this. V is a transition metal. Its high melting point, high density, hardness and lower reactivity support this. A valid example of V is chromium, manganese, iron, cobalt, nickel or copper.	Compare the whole pattern rather than relying on one value. U matches the soft, low-density, low-melting and highly reactive Group 1 pattern. V shows the opposing transition-metal pattern. Finish with one of the named exemplars required by AQA.	6
Total Question 1			6
02.1	The conclusion is plausible but density alone is insufficient. High melting point and low reactivity with water also match the typical transition-metal pattern. Evidence that X is hard or strong would strengthen the classification. Evidence that X is much less reactive with oxygen or halogens than a Group 1 metal would strengthen the comparison. Comparison with a named transition element such as iron would provide a useful reference.	Do not classify from one property because the trends are general rather than definitions. Combine physical and chemical evidence, then seek characteristic transition chemistry or comparison with a specified example.	5
Total Question 2			5
03.1	Potassium is more reactive with chlorine than copper. The evidence is consistent with potassium being a Group 1 metal and copper being a transition element. Potassium reacts vigorously with water, whereas copper does not react with cold water. Copper is stronger than potassium. Copper is harder than potassium.	Use the halogen reaction as direct evidence for the reactivity contrast. Then apply the specified general differences: transition elements are less reactive, stronger and harder than Group 1 metals.	5
Total Question 3			5
04.1	Density of A = 24.0 ÷ 30.0 = 0.800 g/cm³. Density of B = 21.0 ÷ 2.50 = 8.40 g/cm³. A is consistent with a Group 1 metal because it has low density, low melting point and high reactivity with water. B is consistent with a transition metal because it is dense, has a high melting point and is less reactive with water. The claim is incorrect because density depends on both mass and volume, not mass alone. Although the A sample is heavier, its much larger volume makes it less dense than B.	Calculate each density from its own mass and volume before comparing the materials. Then combine the density result with melting-point and water-reaction evidence. The raw sample masses are misleading because the sample volumes differ by a factor of 12.	6
Total Question 4			6
05.1	Use freshly exposed samples with the same dimensions or exposed surface area for the water test. Use the same volume and initial temperature of water in identical apparatus. Measure reactivity consistently, for example by timing the same observable change. Apply the same measured force with the same tool to equal-shaped samples for the hardness test. Compare the depth of indentation or deformation as the hardness measurement. The Group 1 metal should react faster with water and deform more, while the transition metal should react less and resist deformation.	Design each comparison so that metal identity is the main changed variable. Equal surface area and water conditions make reaction times comparable; equal force and sample shape make indentation comparable. The combined chemical and physical pattern is stronger than either observation alone.	6
Total Question 5			6

4.1.3.2 · Typical properties (chemistry only)

Tier 1 · Easy

Mark scheme for 4.1.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	They can form ions with different charges. They form coloured compounds. They or their compounds can act as catalysts.	Recall the three specification headings for typical transition chemistry: variable ion charge, coloured compounds and catalytic activity.	3
Total Question 1			3
02.1	Transition elements form coloured compounds.	The blue colour is evidence for the typical property that transition elements form coloured compounds.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.1.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Iron is 2+ in FeCl₂. Iron is 3+ in FeCl₃. Iron forms ions with different charges. Its different compounds can have different colours.	A neutral formula must have total charge zero. Two chloride ions contribute 2−, so Fe is 2+ in FeCl₂; three contribute 3−, so Fe is 3+ in FeCl₃. The formulas and observations show both variable charge and coloured compounds.	4
Total Question 1			4
02.1	FeO is iron(II) oxide. Fe₂O₃ is iron(III) oxide. Iron forms ions with different charges. The iron ion is 2+ in FeO and 3+ in Fe₂O₃.	In FeO, one Fe ion balances one O²⁻, so iron is 2+. In Fe₂O₃, three oxide ions total 6−, shared between two iron ions, so each is 3+. Roman numerals state these ion charges.	4
Total Question 2			4
03.1	The pink cobalt compound demonstrates that transition elements form coloured compounds. Fe²⁺ and Fe³⁺ demonstrate that a transition element can form ions with different charges. The nickel compound demonstrates that transition-element compounds can be useful catalysts.	Map each observation directly to one of the three specified general properties: coloured compounds, variable ion charges and catalytic activity.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.1.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The copper compound is the more effective catalyst. Its trial is 4 times faster than the uncatalysed trial. A catalyst increases reaction rate. A catalyst is not used up overall or remains chemically unchanged at the end. Transition elements and their compounds commonly show catalytic activity.	For equal reaction amounts, rate is proportional to $1/t$ . The copper trial has the shortest time, so it is fastest. The rate factor is $(1/60)/(1/240)=240/60=4$ . Link the result to the typical catalytic property and state that a catalyst speeds the reaction without being consumed overall.	6
Total Question 1			6
02.1	Both solids are associated with faster reactions, so both initially appear to increase rate. A is not supported as a catalyst because it is not recovered unchanged in mass and may have been consumed. B is supported as a catalyst because it gives the shortest reaction time. B is recovered with its original mass, consistent with not being used up overall. The colours show that the solids are coloured compounds but do not by themselves prove catalytic activity. Further chemical testing would be needed to show that B is chemically unchanged, not just unchanged in mass.	Separate the two catalyst criteria: faster reaction and no overall consumption. A meets only the rate observation. B meets the timing and mass evidence, although equal mass alone cannot prove identical composition. Colour is a separate typical property and is not evidence of catalysis by itself.	6
Total Question 2			6
03.1	The claim is incorrect because both completed trials produce 36 g of product. The larger mass after 40 s shows that product forms faster when the copper compound is present. The copper compound therefore increases the rate of reaction, not the final amount of product. Its recovery unchanged supports its identification as a catalyst because it is not used up overall. This exemplifies the typical catalytic behaviour of transition-element compounds.	Separate rate evidence at a fixed time from final-yield evidence at completion. The early masses differ, but the equal final masses refute the yield claim; unchanged recovery supplies the catalyst evidence.	5
Total Question 3			5
04.1	Q has the stronger evidence for being a transition element. QCl₂ contains Q²⁺ because two Cl⁻ ions balance it. QCl₃ contains Q³⁺, so Q forms ions with different charges. The blue and green compounds show the typical formation of coloured compounds. The unchanged Q oxide that increases rate shows catalytic behaviour. P's single 1+ charge and colourless compounds do not provide these characteristic lines of evidence.	Translate the chloride formulae into ion charges, then keep the three evidence chains separate: variable charge, coloured compounds and catalysis. Q satisfies all three characteristic patterns, whereas the information about P does not.	6
Total Question 4			6
05.1	Run a control without the compound under otherwise identical conditions. Measure rate consistently in both trials, rather than relying only on the finishing time. Recover the compound after the reaction. Wash and dry it before comparing its final mass with its initial mass. Reuse the recovered compound and check that it still increases the reaction rate. Compare completed trials to show that the compound changes the rate rather than being consumed to create extra final product.	Colour alone is not catalyst evidence. A controlled rate comparison establishes the speed change, while recovery after washing and drying tests whether the compound is used up overall. Successful reuse and unchanged completed yield strengthen the catalyst interpretation over the alternatives.	6
Total Question 5			6