← Topic 8
⚡ Challenge Paper Preparation

Challenge Prep: The Periodic Table

IGCSE Chemistry 0620 — Topic 8

Topic 8 looks like the easiest topic on the paper — a table, two groups, a few colours. Then the challenge paper asks you to explain why reactivity increases down Group I but decreases down Group VII, and half the candidates in the room write the same explanation for both. This guide hunts every trap in the topic: the reversed trend, the "more shells" non-explanation that scores zero, group confused with period, displacement predicted backwards, noble gases that are supposedly unreactive "because they are gases", transition metals treated as if they were sodium, and the missing 2 in Cl₂. Get these right and Topic 8 becomes the most predictable marks on the paper.

⚠️ Common Traps & Misconceptions

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Twelve traps that cost students marks on Topic 8 questions. Every one of them appears on challenge papers regularly.

⚠️ TRAP
Trap 1: Reversing the reactivity trend between Group I and Group VII
The Trap"Reactivity increases down every group, so iodine is more reactive than chlorine." Or the mirror error: "reactivity decreases down Group I, so lithium is the most violent with water." Roughly one candidate in three gets one of the two groups backwards, and many write the same explanation for both.
The TruthReactivity INCREASES down Group I but DECREASES down Group VII. The physical cause is identical — going down, the atom is bigger, the outer shell is further from the nucleus and there is more shielding, so the attraction is weaker. The consequence flips because the two groups want opposite things: a Group I atom needs to LOSE an electron (weak attraction helps → more reactive) while a Group VII atom needs to GAIN one (weak attraction hinders → less reactive).
Why It MattersThis single distinction is worth 3–4 marks on almost every Topic 8 challenge paper, and getting it backwards poisons everything downstream: your displacement predictions, your data-table extrapolation and your prediction for an unknown element all invert together.
Example Question"Explain why potassium is more reactive than lithium, but iodine is less reactive than chlorine. Refer to electronic structure in both cases. [5]"
⚠️ TRAP
Trap 2: Explaining a trend with "more shells" and nothing else
The Trap"Potassium is more reactive because it has more shells." "Iodine is less reactive because it is bigger." Both sentences are true and both score zero on a 3-mark explain question.
The Truth"More shells" is an observation, not an explanation. The mark scheme wants the causal chain: (1) more occupied shells so the outer electron is further from the nucleus; (2) more shielding by the inner shells; (3) therefore the electrostatic attraction between nucleus and outer electron is weaker; (4) so the electron is lost more easily (Group I) or gained less easily (Group VII). Three or four separate marking points, three or four separate clauses.
Why It MattersCambridge marks these questions point by point. A candidate who understands the chemistry perfectly but compresses it into one clause typically scores 1 out of 3. Writing the words distance, shielding and attraction explicitly is the difference between a grade C and a grade A answer.
Example Question"Explain, in terms of electronic structure, why caesium is more reactive than sodium. [3]"
⚠️ TRAP
Trap 3: Confusing group number with period number
The Trap"The configuration 2,8,3 means Group 3, Period 3 — no wait, Group 2, Period 8?" Candidates who have not fixed the rule guess, and about half guess wrong. A related version: "an element in Period 4 has four outer electrons."
The TruthGroup number = number of OUTER-SHELL electrons = the last number in the configuration. Period number = number of OCCUPIED SHELLS = how many numbers there are. So 2,8,3 has three numbers (Period 3) ending in 3 (Group III) ⇒ aluminium. And 2,8,8,2 has four numbers (Period 4) ending in 2 (Group II) ⇒ calcium.
Why It MattersAlmost every Topic 8 question begins by placing an element. Get the position wrong and every subsequent part — ion charge, formula, metal or non-metal, oxide acidity — is wrong too. It is the highest-leverage single fact in the topic.
Example Question"An atom has 19 protons. State its group and period, and give the formula of its oxide. [3]"
⚠️ TRAP
Trap 4: Predicting displacement the wrong way round
The Trap"Iodine will displace bromine from potassium bromide because iodine is bigger and heavier." "Bromine displaces chlorine from sodium chloride." Both invert the rule, and both usually follow from Trap 1.
The TruthA MORE reactive halogen displaces a LESS reactive one from a solution of its halide — and reactivity decreases down Group VII, so a halogen can only displace one BELOW it. Chlorine displaces bromine and iodine; bromine displaces only iodine; iodine displaces nothing. Cl2 + 2KBr → 2KCl + Br2 works; I2 + KBr does nothing at all.
Why It MattersDisplacement is set almost every year, often as a results table you must complete. One inverted rule turns a 5-mark table into 0. It also underpins the industrial extraction of bromine from seawater, a favourite application question.
Example Question"Complete the table by writing 'reaction' or 'no reaction' for each combination of halogen and halide solution, and give the colour of any solution that changes. [5]"
⚠️ TRAP
Trap 5: "Noble gases are unreactive because they are gases"
The Trap"They are unreactive because they are gases." "Because they are very light." "Because they have no electrons in the outer shell." All three appear every session and all three score zero.
The TruthNoble gases are unreactive because they have a FULL OUTER ELECTRON SHELL (8 electrons; helium has 2). A full shell is a stable arrangement, so the atom has no tendency to lose, gain or share electrons — and that is what a chemical reaction requires. Being a gas is a consequence of being monatomic with weak forces between atoms, not the cause of anything. Chlorine is also a gas and is one of the most reactive elements on the syllabus.
Why It MattersThis is a guaranteed 2-mark question and one of the easiest marks in the whole paper — but only if you name the full outer shell. The examiner is checking whether you can distinguish a cause from a correlation.
Example Question"Explain, in terms of electronic structure, why argon does not form compounds. [2]"
⚠️ TRAP
Trap 6: Treating transition metals as if they were Group I
The Trap"Iron reacts violently with cold water to give iron hydroxide and hydrogen." "Copper should be stored under oil." "Iron is soft and has a low melting point because it is a metal." "Nickel chloride is white."
The TruthTransition elements are almost the opposite of Group I in every property: high density (Cu 8.9 vs Na 0.97), high melting point (Fe 1538 °C vs Na 98 °C), hard rather than soft, much less reactive (iron rusts slowly; copper does not react with water at all), variable oxidation number (Fe2+ and Fe3+) and coloured compounds (CuSO4 blue, FeSO4 green). They are also used as catalysts, which no Group I metal ever is.
Why It MattersThe Group I versus transition comparison is the standard 4–6 mark question in 8.4, and it is marked as paired statements. Writing "transition metals are metals so they react with water" collapses the entire distinction the question is testing.
Example Question"Give three differences between the properties of potassium and those of copper, and support each with data or an example. [6]"
⚠️ TRAP
Trap 7: Forgetting that halogens are diatomic in equations
The Trap"Na + Cl → NaCl." "Cl + KBr → KCl + Br." "Cl2 + KBr → KCl2 + Br." Also the reverse error: writing the ion as Cl2− or "2Cl−" when a single chloride is meant.
The TruthThe element is always diatomic: F2, Cl2, Br2, I2 — and so are H2, O2 and N2. The ion is always single: Cl−, Br−, I−. Because one Cl2 molecule provides two chlorine atoms, the balancing follows automatically: 2Na + Cl2 → 2NaCl and Cl2 + 2KBr → 2KCl + Br2.
Why It MattersEquation marks are all-or-nothing. A displacement question typically awards 2 marks for the equation — correct formulae and correct balancing — and dropping the subscript 2 loses both, even though the chemistry in your head was right.
Example Question"Write a balanced symbol equation, and an ionic equation, for the reaction between chlorine and aqueous potassium iodide. [3]"
⚠️ TRAP
Trap 8: Saying Group I atoms have "more electrons to lose"
The Trap"Potassium is more reactive than sodium because it has more electrons, so more can be lost." Or, for halogens: "chlorine is more reactive because it has more electrons to gain."
The TruthEvery Group I atom loses exactly one electron; every Group VII atom gains exactly one. The number is fixed by the group and never varies. Reactivity is about how easily that single electron moves, which depends on distance and shielding — not about how many electrons are available.
Why It MattersThis answer signals a fundamental misunderstanding of why groups exist, so examiners give it no credit at all even when the conclusion ("potassium is more reactive") happens to be right. Marks come from the reasoning, not the verdict.
Example Question"State the number of electrons lost by a rubidium atom when it reacts, and explain why rubidium reacts more vigorously than potassium. [3]"
⚠️ TRAP
Trap 9: Explaining halogen melting points with covalent bonds
The Trap"Iodine has a higher melting point than chlorine because its covalent bonds are stronger." Or: "because the atoms are held together more tightly."
The TruthMelting a molecular substance separates whole molecules from one another; the covalent bonds inside each I2 molecule are not broken at all. The correct chain is: down the group the molecules get larger, so the intermolecular forces between molecules are stronger, so more energy is needed to separate them and the melting and boiling points rise — taking the elements from gas (Cl2) to liquid (Br2) to solid (I2).
Why It MattersPhysical and chemical trends in Group VII have different causes and are marked separately: intermolecular forces for melting points, electron gain for reactivity. Using one explanation for both loses the marks for the other, and the two even point in opposite directions.
Example Question"Explain why iodine is a solid at room temperature whereas chlorine is a gas, and state whether iodine is more or less reactive than chlorine. [4]"
⚠️ TRAP
Trap 10: Getting halogen colours and states imprecise
The Trap"Iodine is purple." "Bromine is a brown gas." "Chlorine is a green liquid." "Bromine is orange." Each is close enough to feel right and wrong enough to lose the mark.
The TruthLearn all three exactly: chlorine — pale yellow-green GAS; bromine — red-brown LIQUID; iodine — grey-black SOLID that sublimes to a purple vapour. Solid iodine is grey-black and shiny; the purple belongs to the vapour and to solutions in organic solvents, while iodine in water is pale brown. In displacement reactions, displaced bromine turns a solution orange and displaced iodine turns it brown.
Why It MattersThese are 1-mark recall items that also serve as the evidence in longer observation questions. If you cannot say what colour the solution turned, you cannot prove which halogen was displaced, and you lose the deduction marks as well as the recall mark.
Example Question"State the colour and physical state of bromine at room temperature, and describe what is seen when chlorine water is added to aqueous potassium bromide. [3]"
⚠️ TRAP
Trap 11: Confusing the element with its ion
The Trap"Chlorine is added to drinking water and it is poisonous, so table salt must be dangerous." "Sodium is in your food." "Iodine is added to salt." Also in formulae: writing Cl2 inside NaCl, or "the sodium in seawater will react with the water".
The TruthAn atom that has lost or gained an electron is a different species with different properties. Sodium metal (Na) explodes in water; the sodium ion (Na+) is a stable component of table salt and of your blood. Chlorine (Cl2) is a toxic green gas; the chloride ion (Cl−) is harmless. That is why reactive Group I metals are never found uncombined in nature, and why "the sodium in seawater" is already an ion and cannot react further.
Why It MattersApplication questions on water treatment, fertilisers, batteries and nutrition all hinge on this distinction, and examiners write the questions specifically to test it. Naming the species precisely — "chloride ions, Cl−", not "chlorine" — is usually worth a mark on its own.
Example Question"Explain why sodium is never found as the free metal in nature, but sodium compounds are abundant in seawater. [2]"
⚠️ TRAP
Trap 12: Assuming every trend is smooth, and ignoring given data
The Trap"Density increases perfectly down Group I, so potassium must be denser than sodium." "All Group I metals float." "The melting point drops by the same amount each time, so caesium melts at 28 °C exactly." Candidates recite the rule instead of reading the table printed right in front of them.
The TruthDensity down Group I is not perfectly regular: Li 0.53, Na 0.97, K 0.86, Rb 1.53, Cs 1.88 — potassium is the anomaly, and examiners include it deliberately. The general direction is upward, so Li, Na and K float on water while Rb and Cs sink. Melting points fall by a shrinking amount each step (83, then 35, then 24), so extrapolate by the pattern of the gaps rather than by a constant difference, and give a range if you are unsure.
Why It MattersData-response questions award marks for using the data supplied. If the table shows the potassium anomaly and you contradict it from memory, you lose the mark even though your general rule is fine. Always quote a number from the stem in your answer.
Example Question"Use the data in the table to predict the density of rubidium, and state with a reason whether rubidium would float on water. [3]"

🧩 Multi-Step Reasoning Walkthroughs

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Six challenging questions broken down step by step. Try each step yourself before revealing the next.

Walkthrough 1 — The Two Opposite Trends in One AnswerExplain, in terms of electronic structure, why potassium is more reactive than lithium but iodine is less reactive than chlorine. Your answer must make clear why the same change in atomic structure produces opposite effects. [6]
1

Lose or gain — decide this first

A Group I atom has one outer electron and reacts by LOSING it to reach a full shell underneath. A Group VII atom has seven outer electrons and reacts by GAINING one. Write these two sentences down before anything else — the entire answer hinges on them, and most candidates who go wrong go wrong here.

2

One physical change, applying to both groups

Going down any group, each element has one more occupied electron shell than the one above. Therefore: the atomic radius increases, the outer shell is further from the nucleus, and there is more shielding by the inner shells of electrons. The net result in both groups is the same: the attraction between the nucleus and the outer-shell region is weaker.

3

Weaker attraction makes losing EASIER

Potassium (2,8,8,1) has three shielding shells to lithium's one, and its outer electron sits much further out. A weaker hold on that electron means it is lost more easily, so potassium is more reactive. Evidence: lithium fizzes steadily and stays solid; potassium ignites with a lilac flame.

4

Weaker attraction makes gaining HARDER

An iodine atom is much larger than a chlorine atom, so the outer shell — the place where an incoming electron must go — is further from the nucleus and better shielded. The nucleus therefore attracts an incoming electron less strongly, so the electron is gained less easily and iodine is less reactive. Evidence: chlorine displaces iodine from potassium iodide, but iodine cannot displace chlorine.

5

The sentence that earns the top band

"The same change — a bigger, better-shielded atom with a weaker attraction for its outer electrons — makes it easier to lose an electron and harder to gain one. Because Group I reacts by losing and Group VII by gaining, the reactivity trends run in opposite directions." Examiners specifically look for a candidate who links the two halves rather than answering them as unrelated questions.

Final AnswerGroup I atoms react by losing their single outer electron; Group VII atoms react by gaining one [1]. Going down either group, each atom has one more occupied shell, so the outer shell is further from the nucleus and is shielded by more inner shells [1], making the attraction between the nucleus and the outer electrons weaker [1]. In Group I this means the outer electron is lost more easily, so potassium is more reactive than lithium [1]. In Group VII it means an incoming electron is attracted less strongly and gained less easily, so iodine is less reactive than chlorine [1]. The same structural change therefore produces opposite reactivity trends [1].
Examiner's NoteThis is the flagship Topic 8 question and the mark scheme is unusually generous to candidates who structure their answer. Marks are lost in three predictable ways: writing "more shells" without distance/shielding/attraction; explaining Group VII with "loses an electron more easily" (copied straight from the Group I answer); and never stating the contrast at all. Note that nuclear charge increases down both groups — if you mention it, you must say that distance and shielding outweigh it, or you will argue yourself into the wrong conclusion.
Walkthrough 2 — A Full Displacement Results TableThree colourless solutions, potassium chloride, potassium bromide and potassium iodide, are each treated with chlorine water, bromine water and iodine solution. (a) Complete the nine-cell table of results. (b) Write equations for the reactions that occur. (c) Explain what the pattern proves. (d) Suggest what would be seen if astatine solution were added to potassium iodide. [9]
1

Cl2 > Br2 > I2

Reactivity decreases down Group VII, so chlorine is the most reactive and iodine the least. Write this order at the top of your answer. Every one of the nine cells now follows mechanically from a single rule: a halogen reacts only with the halide of an element below it.

2

Reactions only above the diagonal

Chlorine water: KCl no reaction; KBr reaction — turns orange; KI reaction — turns brown. Bromine water: KCl no reaction (chlorine is above it); KBr no reaction (same element); KI reaction — turns brown. Iodine solution: no reaction with any of the three, because nothing on the list is below iodine. Three reactions out of nine, all in the top-right corner.

3

Every halogen carries its subscript 2

Cl2 + 2KBr → 2KCl + Br2
Cl2 + 2KI → 2KCl + I2
Br2 + 2KI → 2KBr + I2
The 2 in front of the potassium halide is forced by the diatomic molecule: one Cl2 produces two Cl−, so two K+ partners are needed. Ionically, cancelling the spectator potassium: Cl2 + 2Br− → 2Cl− + Br2.

4

The table IS the reactivity order

Chlorine displaced both of the others, bromine displaced only iodine, and iodine displaced nothing. That establishes experimentally that reactivity decreases down Group VII. In electron terms, the halogen that wins is the one that attracts an electron most strongly: chlorine takes an electron from the bromide ion, so chlorine is reduced and is the stronger oxidising agent, while bromide is oxidised.

5

No reaction — and say why

Astatine is below iodine, so it is less reactive than iodine and cannot displace it: no reaction, no colour change. The trap here is to reason "astatine is the biggest and heaviest so it must be the most powerful" — that is the Group I instinct applied to Group VII. Size makes a halogen weaker, not stronger.

Final Answer(a) Chlorine reacts with KBr (orange) and KI (brown); bromine reacts with KI only (brown); iodine reacts with none [3]
(b) Cl2 + 2KBr → 2KCl + Br2; Cl2 + 2KI → 2KCl + I2; Br2 + 2KI → 2KBr + I2 [3]
(c) A more reactive halogen displaces a less reactive one from its halide, so the order is Cl2 > Br2 > I2, proving that reactivity decreases down Group VII [2]
(d) No reaction — astatine is below iodine and therefore less reactive, so it cannot displace iodide [1]
Examiner's NoteTwo subtleties separate the top answers. First, in the "no reaction" cells the mixture is still coloured, because the bromine or iodine solution you added had a colour of its own — candidates who write "turns orange" for bromine water plus KCl have described the reagent, not a reaction. Second, examiners accept "orange" or "red-brown" for displaced bromine and "brown" for displaced iodine, but not "dark" or "changes colour". In (d) the mark is for the reason, not the verdict; a bare "no reaction" often scores nothing.
Walkthrough 3 — Deducing an Element from CluesElement Q has an atom containing 20 protons. Element R is a diatomic gas whose atoms have the configuration 2,8,7. (a) Give the group and period of each. (b) Predict the type of bonding in the compound of Q and R, and give its formula. (c) Predict the pH of the solution formed when the oxide of Q dissolves in water, with a reason. (d) State how the reactivity of R compares with the element directly below it, and explain. [9]
1

Protons → electrons → shells

Q has 20 protons and therefore 20 electrons: 2, then 8, then 8, leaving 2 ⇒ 2,8,8,2. Four numbers means Period 4; the last number 2 means Group II. That is calcium. R is given as 2,8,7: three numbers ⇒ Period 3, ending in 7 ⇒ Group VII. That is chlorine, and the clue "diatomic gas" confirms it.

2

Metal + non-metal ⇒ ionic

Q is on the far left ⇒ a metal that loses electrons. R is on the far right ⇒ a non-metal that gains them. Electrons transfer, so the bonding is ionic. Q loses two electrons to form Q2+; R gains one to form R−. Balancing the charges needs two R ions per Q ion: QR2, that is CaCl2.

3

Metal oxide ⇒ basic ⇒ alkaline solution

Q is a metal, and metal oxides are basic. A soluble basic oxide dissolves to give an alkaline solution, so predict a pH of about 12–13. (Calcium oxide gives limewater, roughly pH 12.) Say the reason explicitly: "because Q is a metal on the left of the Periodic Table, so its oxide is basic."

4

Careful — this is Group VII, not Group I

The element below R (bromine) is less reactive than R, because reactivity decreases down Group VII. Reason: the bromine atom is larger, so the outer shell where an incoming electron must go is further from the nucleus and there is more shielding, so the attraction for the incoming electron is weaker and it is gained less easily. Evidence: R (chlorine) displaces bromine from potassium bromide.

Final Answer(a) Q: 2,8,8,2 — Group II, Period 4; R: Group VII, Period 3 [2]
(b) Ionic bonding, because a metal reacts with a non-metal and electrons are transferred [1]; Q2+ and R− give the formula QR2 [2]
(c) About pH 12–13 (alkaline), because Q is a metal and metal oxides are basic, dissolving to give an alkaline solution [2]
(d) The element below R is less reactive [1], because its atom is larger with more shielding, so the attraction for an incoming electron is weaker and the electron is gained less easily [1]
Examiner's NoteMulti-part deduction questions like this are the standard Extended structure, and they are designed so that an early error propagates. Candidates who place Q in "Group 4, Period 2" go on to predict a covalent compound and an acidic oxide, losing five marks from one slip. Always write the configuration out, count the numbers, and read the last one. In (d), examiners see a large number of Group I explanations pasted into a Group VII question — the giveaway phrase is "lost more easily", which cannot be right for an element that gains electrons.
Walkthrough 4 — Data-Response on Group IA table gives melting points (Li 181, Na 98, K 63 °C) and densities (Li 0.53, Na 0.97, K 0.86 g/cm3). (a) Describe the trend in melting point and predict the value for rubidium. [2] (b) Comment on the density data and predict whether caesium (density 1.88) would float. [3] (c) Write an equation for the reaction of rubidium with water and describe two observations. [3] (d) Explain why the reaction of rubidium is more vigorous than that of sodium. [3]
1

Direction plus magnitude

The melting point decreases down the group, and the decreases are getting smaller: 181 → 98 is a fall of 83, but 98 → 63 is a fall of only 35. Extrapolating a further fall of roughly 20–25 gives about 40 °C for rubidium (true value 39). Quote at least one number from the table — data-response marks are for using the data, not reciting the rule.

2

Do not pretend the pattern is smooth

The density generally increases down the group, but the trend is not regular: potassium (0.86) is less dense than sodium (0.97). Say so explicitly — examiners award a mark for spotting it. Caesium at 1.88 g/cm3 is greater than the density of water (1.00), so caesium would sink, unlike Li, Na and K which all float.

3

2M + 2H2O → 2MOH + H2

2Rb(s) + 2H2O(l) → 2RbOH(aq) + H2(g). Observations must be things you could see: the metal ignites immediately / the reaction is explosive; there is rapid effervescence; the metal disappears; and universal indicator turns purple because a soluble hydroxide forms. "Hydrogen is produced" is a deduction, not an observation — pair it with "bubbles of gas".

4

Distance, shielding, attraction

A rubidium atom has more occupied shells than a sodium atom, so its single outer electron is further from the nucleus [1] and is shielded by more inner shells of electrons [1]; the attraction between the nucleus and that electron is therefore weaker, so it is lost more easily [1]. Note that both atoms lose exactly one electron — never write "more electrons to lose".

Final Answer(a) Melting point decreases down the group, by a diminishing amount (83 °C then 35 °C) [1]; predicted rubidium value about 40 °C [1]
(b) Density generally increases down the group but the trend is irregular — potassium (0.86) is less dense than sodium (0.97) [2]; caesium at 1.88 g/cm3 is denser than water so it would sink [1]
(c) 2Rb(s) + 2H2O(l) → 2RbOH(aq) + H2(g) [1]; observations: the metal ignites / the reaction is explosive, and there is rapid effervescence as the metal disappears; the solution turns universal indicator purple [2]
(d) The outer electron in rubidium is further from the nucleus [1] and more shielded by inner shells [1], so the attraction is weaker and it is lost more easily [1]
Examiner's NotePart (b) is the discriminator. The syllabus states that density increases down Group I, but the printed data contradicts it at one point, and examiners want to see whether you read the table or recited the rule. Candidates who write "density increases regularly, so potassium is denser than sodium" lose two marks for contradicting data on the page in front of them. In (a), a prediction with no number scores nothing, and a number wildly outside the pattern (say 5 °C or 90 °C) scores nothing either — examiners allow a sensible range.
Walkthrough 5 — Transition Elements versus Group IA student is given two unlabelled metals, X and Y. X is soft, floats on water and reacts violently with it; its chloride is a white solid dissolving to a colourless solution. Y is hard and dense, does not react with cold water, forms two different chlorides and its solutions are coloured. (a) Identify the block each metal belongs to. [2] (b) Give the names and formulae of Y's two chlorides if Y is iron. [2] (c) Explain why Y's compounds need a Roman numeral in the name but X's do not. [2] (d) Suggest one industrial use for Y that X could never fulfil, with a reason. [2]
1

Every clue points the same way

X: soft, low density (floats), very reactive with water, white/colourless compounds, one chloride ⇒ a Group I alkali metal. Y: hard, dense, unreactive with cold water, coloured compounds, two chlorides (variable oxidation number) ⇒ a transition element, in the block between Groups II and III.

2

Match the numeral to the charge

Iron forms Fe2+ and Fe3+, and chloride is always Cl−. So the compounds are iron(II) chloride, FeCl2 (pale green solution) and iron(III) chloride, FeCl3 (yellow-brown solution). The Roman numeral gives the oxidation number of the metal, not the number of chlorides — a distinction examiners test directly.

3

Ambiguity is the whole point

Y shows variable oxidation number, so the name "iron chloride" would be ambiguous — FeCl2 and FeCl3 are genuinely different compounds with different colours and formulae. The numeral removes the ambiguity. X (a Group I metal) forms only a 1+ ion, so there is nothing to distinguish and no numeral is needed. Nobody writes "sodium(I) chloride".

4

Match the property to the job

Best answers: iron as the catalyst in the Haber process — transition elements catalyse reactions and are chemically unchanged, whereas Group I metals are not catalysts and would react with everything in the reactor. Or structural use in bridges and girders — iron is hard and melts at 1538 °C, while sodium is knife-soft, melts at 98 °C and would react with rain. Always give the property and why X fails it.

Final Answer(a) X is a Group I (alkali) metal; Y is a transition element [2]
(b) Iron(II) chloride, FeCl2 and iron(III) chloride, FeCl3 [2]
(c) Y has variable oxidation number so the name would otherwise be ambiguous [1]; X forms only a 1+ ion, so no numeral is needed [1]
(d) Y can be used as a catalyst (iron in the Haber process) because transition elements catalyse reactions and are unchanged at the end [1]; X could not, because Group I metals are not catalysts and are far too reactive — they would react with the water, air and reactants present [1]
Examiner's NotePart (d) separates the top band because it demands that you connect a property to a use and explain the failure of the alternative. A bare "iron is used in bridges" scores one mark at best. Also watch part (c): a surprising number of candidates say the numeral shows "how many chlorines there are". It does not — it is the oxidation number of the metal, which is why copper(II) oxide is CuO with only one oxygen.
Walkthrough 6 — Noble Gases, Uses and the Diatomic Contrast(a) Explain, in terms of electronic structure, why the noble gases are unreactive and monatomic, whereas chlorine is diatomic and very reactive. [4] (b) Explain why helium rather than hydrogen is used in balloons. [2] (c) Explain why argon is used in filament lamps and in arc welding. [3] (d) Nitrogen is also used to protect food from oxidation. Suggest why a museum might still prefer argon. [2]
1

Two consequences from one cause

A noble gas atom has a full outer electron shell (He 2; Ne 2,8; Ar 2,8,8). It therefore has no tendency to lose, gain or share electrons, which is what a chemical reaction requires ⇒ unreactive. And because it does not even need to share electrons with another atom of its own kind, it has no reason to bond at all ⇒ it exists as single atoms, that is monatomic.

2

Seven is not eight

A chlorine atom (2,8,7) is one electron short of a full shell. Two chlorine atoms therefore share a pair of electrons so that each achieves eight ⇒ the element is diatomic, Cl2. That same one-electron deficit makes chlorine eager to take an electron from anything else, so it is very reactive. The contrast is exact: a full shell means no bonding and no reactions; a nearly full shell means both.

3

Density AND safety

Helium has a very low density, so it provides lift [1], and it is unreactive / non-flammable, so it cannot burn or explode [1]. Hydrogen is actually lighter and cheaper, so density alone does not explain the choice — the safety point is the one the question is really after. The Hindenburg disaster of 1937 is the standard illustration.

4

Same property, two applications

Filament lamp: at over 2000 °C a tungsten filament would react with oxygen and burn away in air; argon is unreactive so it will not attack the filament, and the bulb lasts far longer. Arc welding: molten metal reacts with oxygen and nitrogen to form brittle oxides and nitrides that weaken the joint; a stream of argon blankets the weld, excluding air while taking no part itself. Argon is chosen over the other noble gases because it makes up about 0.9% of air and is therefore cheap.

5

Why argon beats nitrogen for the fussiest jobs

Nitrogen is unreactive because its N≡N triple bond is very hard to break — but it is not incapable of reacting. At high temperatures it forms nitrides with lithium and magnesium, and nitrogen oxides in an engine. Argon needs no such caveat: it has a full outer shell, so there is nothing to break and no reaction is possible under any conditions a museum or a welder will meet.

Final Answer(a) Noble gases have a full outer shell [1] so no tendency to lose, gain or share electrons, hence unreactive and existing as single atoms (monatomic) [1]. Chlorine has seven outer electrons, one short of a full shell [1], so two atoms share a pair of electrons to form Cl2, and the same deficit makes it very reactive [1]
(b) Helium has a low density so it gives lift [1] and is non-flammable / unreactive, unlike hydrogen which burns explosively [1]
(c) Argon is unreactive [1]; in a lamp it stops the hot filament reacting with oxygen and burning away [1]; in welding it blankets the molten metal, excluding oxygen and nitrogen so no brittle oxides or nitrides form [1]
(d) Nitrogen is unreactive only because its triple bond is hard to break and it can react under some conditions, whereas argon has a full outer shell and cannot react at all [2]
Examiner's NotePart (a) is worth four marks and is regularly answered with one sentence: "they have full outer shells so they are unreactive." That earns two at most — the question also asks about monatomic and about chlorine, and each needs its own explanation. In (b) the single most common error is claiming helium is lighter than hydrogen; it is not, and examiners take a dim view of an invented fact used to justify a correct conclusion. Part (d) is a stretch-and-challenge item: the distinction between kinetically unreactive (nitrogen) and genuinely inert (argon) is exactly the kind of nuance that separates the top grades.

🔍 Spot the Difference

▼

Pairs of questions that look nearly identical but have different answers. Spot the key distinction.

Question A
Is caesium more reactive than sodium?
Yes. Group I reactivity increases down the group. Caesium's outer electron is further from the nucleus and better shielded, so it is lost more easily.
Question B
Is iodine more reactive than chlorine?
No. Group VII reactivity decreases down the group. Iodine's outer shell is further from the nucleus and better shielded, so an electron is gained less easily.
Key DifferenceSame structural change, opposite outcome. A weaker attraction between nucleus and outer shell makes an electron easier to lose but harder to gain. Group I loses, Group VII gains — so the trends run in opposite directions. Before answering any trend question, ask yourself first: is this element trying to lose or to gain?
Question A
An atom is 2,8,3. Which group?
Group III. Read the last number — three outer-shell electrons. It is aluminium.
Question B
An atom is 2,8,3. Which period?
Period 3. Count how many numbers there are — three occupied shells. Same element, different question.
Key DifferenceGroup = the value of the last number (outer electrons). Period = the count of the numbers (occupied shells). They coincide by accident here, which is exactly why examiners choose configurations like 2,8,8,2 — four numbers ending in 2, so Period 4, Group II. Say both words out loud as you answer.
Question A
Does chlorine react with potassium bromide solution?
Yes. Chlorine is above bromine, so it is more reactive and displaces it. The solution turns orange. Cl2 + 2KBr → 2KCl + Br2.
Question B
Does bromine react with potassium chloride solution?
No. Bromine is below chlorine and less reactive, so it cannot displace it. Nothing happens — the mixture simply stays the orange colour of the bromine water added.
Key DifferenceDisplacement runs downwards only: a halogen displaces the halide of an element below it. Note the observational trap in B — the tube is coloured, but only because the reagent was coloured to begin with. Compare with the original colour, not with water.
Question A
Why does reactivity change down Group VII?
Because the atom is larger and more shielded, so the nucleus attracts an incoming electron less strongly and it is gained less easily. This is about electrons.
Question B
Why does the melting point change down Group VII?
Because the molecules are larger, so the intermolecular forces between them are stronger and more energy is needed to separate them. This is about forces between molecules.
Key DifferenceTwo trends, two completely different causes — and they point in opposite directions (reactivity falls, melting point rises). Using the electron explanation for melting points, or intermolecular forces for reactivity, loses the mark even though both sentences are individually true.
Question A
Is chlorine, Cl2, dangerous?
Yes. A toxic pale yellow-green gas that bleaches damp litmus and was used as a chemical weapon.
Question B
Is the chloride ion, Cl−, dangerous?
No. It is a stable ion with a full outer shell, present in table salt, seawater and your own blood plasma.
Key DifferenceGaining one electron produces a completely different species with completely different properties. The same applies to sodium: the metal explodes in water, the Na+ ion is a nutrient. Whenever an application question mentions salt, brine, fertiliser or drinking water, name the ion explicitly — that precision is usually a mark.
Question A
Does potassium need a Roman numeral in its compound names?
No. Potassium forms only K+, so "potassium chloride" is unambiguous. Group I metals have one fixed oxidation state.
Question B
Does copper need a Roman numeral?
Yes. Copper forms Cu+ and Cu2+, so "copper oxide" is ambiguous: Cu2O is red and CuO is black. Variable oxidation number demands the numeral.
Key DifferenceThe Roman numeral is a flag for variable oxidation number, a transition-element property. It gives the charge on the metal, not the number of the other atoms — copper(II) oxide is CuO, with a single oxygen.
Question A
Why is argon unreactive?
Because it has a full outer shell (2,8,8), so there is no tendency to lose, gain or share electrons. Nothing needs to be broken because nothing is bonded.
Question B
Why is nitrogen unreactive?
Because N2 molecules are held by a very strong triple bond that takes a lot of energy to break. Nitrogen can react — with lithium, with magnesium, in car engines.
Key DifferenceArgon is inert (no bonds, no reactions possible); nitrogen is merely unreactive (bonds that are hard to break). Same practical outcome, entirely different reason — and challenge papers ask you to distinguish them, usually in a food-packaging or museum-conservation context.
Question A
Does sodium react with cold water?
Yes, violently. It melts into a ball, darts about and fizzes: 2Na + 2H2O → 2NaOH + H2. The solution becomes strongly alkaline.
Question B
Does iron react with cold water?
Essentially no. Iron rusts slowly over days and only when both water and oxygen are present. There is no fizzing, no hydrogen, no alkali.
Key DifferenceTransition elements are far less reactive than Group I metals. This is why iron builds bridges and sodium is stored under oil, and why "iron reacts with water to give iron hydroxide and hydrogen" is one of the most frequently penalised sentences in 8.4.
Question A
How many electrons does a potassium atom lose when it reacts?
Exactly one — the same as lithium, sodium, rubidium and caesium. Every Group I atom loses one and forms a 1+ ion.
Question B
Then why is potassium more reactive than lithium?
Not because it loses more, but because it loses that one electron more easily — greater distance from the nucleus and more shielding mean a weaker attraction.
Key DifferenceReactivity is about how easily, never how many. "More electrons to lose" is the single most penalised phrase in 8.2, and it also gives away that the candidate has not understood what a group is.
Question A
What colour is solid iodine?
Grey-black and shiny. In water it gives a pale brown solution, and displaced iodine turns a solution brown.
Question B
What colour is iodine vapour?
Purple / violet, seen when solid iodine is gently warmed and sublimes without melting.
Key DifferenceThe famous purple belongs to the vapour, not the solid. Writing "iodine is purple" when asked for the appearance of the solid loses the mark, and writing "the solution turned purple" in a displacement question loses another — displaced iodine in aqueous solution is brown.

🔗 Periodic Table Concept Map

▼

Click each node to see how the subtopics connect.

⭐ CORE FRAMEWORK 1
Electronic structure is the engine behind every square of the table
Reading Position from Configuration ▶
Group number = number of outer-shell electrons = the last number in the configuration.
Period number = number of occupied shells = how many numbers there are.
Order of the table = increasing proton number (not atomic mass — that was Mendeleev's table).
Worked example: 2,8,8,2 → four numbers so Period 4; last number 2 so Group II → calcium.
Position Predicts Charge, Formula and Bonding ▶
Groups I, II, III lose electrons → charge = group number, positive (Na+, Mg2+, Al3+).
Groups V, VI, VII gain electrons → charge = 8 − group number, negative (N3−, O2−, Cl−).
Group IV usually shares → covalent (CO2, SiO2, GeCl4).
Metal + non-metal → ionic; non-metal + non-metal → covalent. Position alone tells you which — a direct link to Topic 3.
Metals, Non-metals and Their Oxides ▶
The staircase from boron down to astatine divides metals (left) from non-metals (right). The six elements on it — B, Si, Ge, As, Sb, Te — are metalloids.
Metal oxides are basic (Na2O, CaO → alkaline solutions); non-metal oxides are acidic (SO2, CO2). A pH measurement therefore locates an element on the table — straight into Topic 7.
Metallic character increases down a group and decreases across a period. Group IV runs carbon (non-metal) → silicon/germanium (metalloid) → tin/lead (metal).
Reactive metals (Group I, Group II, aluminium) are extracted by electrolysis, because carbon cannot displace them — a link to Topic 5 and Topic 12.
Predicting an Unknown Element ▶
Step 1: locate it — group gives outer electrons, period gives shells.
Step 2: its properties lie between those of the elements immediately above and below.
Step 3: quote the trend as your reason — a prediction without a reason usually scores half.
Mendeleev's proof: he left gaps and predicted the properties of gallium, scandium and germanium before they were discovered.
⭐ CORE FRAMEWORK 2
One cause, two opposite trends: Group I versus Group VII
The Shared Cause ▶
Going down either group: one more occupied shell → larger atomic radius → outer shell further from the nucleus.
More inner shells → more shielding of the nuclear charge.
Net result in both groups: the attraction between nucleus and outer shell is weaker.
Nuclear charge does increase too, but distance and shielding outweigh it — say so if you mention it, or you will argue the wrong way.
Group I: Losing an Electron ▶
One outer electron → reacts by losing it → forms a 1+ ion.
Weaker attraction ⇒ electron lost more easily ⇒ reactivity INCREASES down the group.
Physical trends: melting point decreases, softness increases, density generally increases (K is an anomaly at 0.86).
Key reaction: 2M + 2H2O → 2MOH + H2; the hydroxide is soluble so the solution is alkaline, pH 13–14.
Group VII: Gaining an Electron ▶
Seven outer electrons → reacts by gaining one → forms a 1− halide ion.
Weaker attraction ⇒ electron gained less easily ⇒ reactivity DECREASES down the group.
Physical trends (a different cause — intermolecular forces): melting and boiling points increase, colour darkens: yellow-green gas → red-brown liquid → grey-black solid.
Elements are diatomic: Cl2, Br2, I2. The ion is single: Cl−.
Displacement: the Trend Made Visible ▶
Rule: a more reactive halogen displaces a less reactive one from a solution of its halide — so a halogen displaces only what is below it.
Cl2 + 2KBr → 2KCl + Br2 (orange) · Cl2 + 2KI → 2KCl + I2 (brown) · Br2 + 2KI → 2KBr + I2 (brown).
Redox view: Cl2 + 2e− → 2Cl− means chlorine is reduced and is the oxidising agent; the halide is oxidised. Oxidising power decreases down the group — a link to Topic 6.
Industrial use: bromine is extracted from seawater by bubbling chlorine through brine, which is this reaction at scale.
⭐ CORE FRAMEWORK 3
The two blocks that break the pattern: transition elements and noble gases
Five Transition Element Properties ▶
High density (Fe 7.9, Cu 8.9 g/cm3) and high melting point (Fe 1538 °C, Cu 1085 °C).
Variable oxidation number — Fe2+/Fe3+, Cu+/Cu2+; hence Roman numerals in the names.
Coloured compounds — CuSO4 blue, FeSO4 green, FeCl3 yellow-brown, KMnO4 purple.
Catalysts — Fe (Haber), V2O5 (Contact), Ni (hydrogenation), MnO2 (decomposing H2O2).
Contrast with Group I ▶
Hardness and strength: transition metals hard and tough; Group I soft enough to cut with a knife.
Reactivity: transition metals react slowly or not at all with water; Group I reacts violently with cold water.
Compounds: transition compounds coloured with variable oxidation states; Group I compounds white, colourless in solution, always +1.
Storage and use: transition metals sit on an open shelf and build bridges and catalysts; Group I metals live under oil and are used only as ions.
Noble Gases: Full Shells ▶
Full outer shell — He 2, Ne 2,8, Ar 2,8,8 ⇒ no tendency to lose, gain or share electrons ⇒ unreactive.
Monatomic — single atoms, He and Ar, never He2. Every other gaseous element on the syllabus is diatomic.
Very low boiling points — only weak forces between separate atoms; helium liquefies at 4 K.
Never say "unreactive because they are gases". Chlorine is a gas and is one of the most reactive elements on the syllabus.
Uses: Inertness Plus One More Property ▶
Helium in balloons — unreactive/non-flammable and low density (hydrogen is lighter but explodes).
Argon in filament lamps — unreactive, so the hot tungsten cannot oxidise and burn away; also cheap, being 0.9% of air.
Argon in arc welding — blankets the molten metal, excluding oxygen and nitrogen so no brittle oxides or nitrides form.
Neon in signs — unreactive and glows red-orange under an electrical discharge. Two reasons, two marks, every time.

❌ "Why Is This Wrong?" Exercises

▼

Spot the error in each student's answer. Think before revealing.

Exercise 1: "Explain why iodine is less reactive than chlorine. [3]"
Student's Answer"Iodine is bigger, so its outer electron is further from the nucleus and is lost more easily, which makes it less reactive."
The FlawThe student has pasted the Group I explanation into a Group VII question. Halogens do not lose electrons at all — they gain one. And the internal logic collapses: losing an electron more easily would make an element more reactive, not less, so the answer contradicts its own conclusion.
Correct Answer"An iodine atom has more occupied electron shells than a chlorine atom, so its outer shell is further from the nucleus and there is more shielding by the inner shells [1]. The nucleus therefore attracts an incoming electron less strongly [1], so the electron is gained less easily and iodine is less reactive [1]."
Key RuleBefore writing any Group VII explanation, say to yourself: halogens GAIN. If the word "lost" appears in your Group VII answer, you have copied the wrong template.
Exercise 2: "Explain why caesium is more reactive than sodium. [3]"
Student's Answer"Because caesium has more shells than sodium."
The FlawTrue but empty. "More shells" is a description of the difference, not an explanation of the consequence. The examiner has asked why more shells matters, and the student has simply restated the observation. This typically scores zero out of three.
Correct Answer"Caesium has more occupied shells, so its outer electron is further from the nucleus [1] and is shielded by more inner shells of electrons [1]. The electrostatic attraction between the nucleus and the outer electron is therefore weaker, so the electron is lost more easily and caesium is more reactive [1]."
Key RuleThree marks means three clauses. Count them as you write: distance, shielding, weaker attraction → lost more easily. Never let a trend explanation be a single sentence.
Exercise 3: "An atom has the configuration 2,8,8,1. State its group and period. [2]"
Student's Answer"Group 4, Period 1."
The FlawThe two answers have been swapped. The student has used the count of the numbers (four) as the group and the last number (one) as the period — exactly the wrong way round.
Correct Answer"Group I (one outer-shell electron) [1], Period 4 (four occupied shells) [1]." The element is potassium.
Key RuleGroup = the last number. Period = how many numbers. Write the words in full — "Group I because there is 1 outer electron; Period 4 because there are 4 shells" — and the marks are secure even if you momentarily doubt yourself.
Exercise 4: "Predict what happens when iodine solution is added to potassium bromide solution. [2]"
Student's Answer"The iodine displaces the bromine because iodine is bigger and heavier, and the solution turns orange."
The FlawThe displacement rule has been inverted. "Bigger and heavier" is not a measure of reactivity in Group VII — in fact a larger halogen is less reactive, because it attracts an incoming electron less strongly. Iodine is below bromine, so it cannot displace it.
Correct Answer"No reaction [1]. Iodine is below bromine in Group VII, so it is less reactive and cannot displace bromine from its halide; a halogen can only displace one that is below it in the group [1]. There is no colour change beyond the brown of the iodine solution added."
Key RuleWrite the reactivity order Cl2 > Br2 > I2 at the top of your answer before predicting anything. Then every cell of a displacement table follows from one rule instead of from instinct.
Exercise 5: "Explain why the noble gases do not form compounds. [2]"
Student's Answer"Because they are gases, and gases are too light and spread out to react with anything."
The FlawBeing a gas has nothing to do with reactivity. Chlorine, oxygen and fluorine are all gases and all extremely reactive; fluorine is the most reactive element there is. The student has mistaken a correlation for a cause, and has said nothing about electrons.
Correct Answer"Noble gas atoms have a full outer electron shell — eight electrons, or two for helium [1]. They therefore have no tendency to lose, gain or share electrons, which is what forming a compound requires, so they are unreactive [1]."
Key RuleEvery "explain why unreactive" answer in Group VIII must contain the words full outer shell. Being monatomic and being a gas are consequences of that full shell, not causes of the inertness.
Exercise 6: "Compare the properties of sodium and iron. [4]"
Student's Answer"Both are metals, so both are shiny, both conduct electricity, both react with water to give a hydroxide and hydrogen, and both form white compounds."
The FlawTwo factual errors and a misread question. Iron does not react with cold water to give hydrogen — it rusts slowly and only when oxygen is also present. Iron compounds are coloured, not white: FeSO4 green, FeCl3 yellow-brown. And "compare" asks for differences, not a list of things all metals share.
Correct Answer"Iron has a much higher melting point (1538 °C against 98 °C) and is much denser (7.9 against 0.97 g/cm3, so sodium floats on water and iron sinks) [1]. Iron is much less reactive — sodium reacts violently with cold water while iron only rusts slowly [1]. Iron shows variable oxidation number (Fe2+ and Fe3+) whereas sodium is always +1 [1]. Iron forms coloured compounds and is used as a catalyst; sodium compounds are white and sodium is not a catalyst [1]."
Key Rule"Compare" means give both sides of each difference, ideally with data. Shared properties of all metals earn nothing, because they do not discriminate between the two blocks.
Exercise 7: "Write an equation for the reaction between chlorine and potassium bromide solution. [2]"
Student's Answer"Cl + KBr → KCl + Br"
The FlawBoth halogens have lost their subscript 2. The elements chlorine and bromine exist as diatomic molecules, Cl2 and Br2, never as lone atoms. The equation as written is balanced but chemically meaningless, and scores nothing.
Correct Answer"Cl2(aq) + 2KBr(aq) → 2KCl(aq) + Br2(aq) [2]. The 2 in front of KBr and KCl is forced by the diatomic molecule: one Cl2 supplies two chlorine atoms, each needing its own K+ partner. Ionically: Cl2 + 2Br− → 2Cl− + Br2."
Key RuleLearn the seven diatomic elements — H2, N2, O2, F2, Cl2, Br2, I2. Write the formulae first, then balance. Balancing an equation with wrong formulae guarantees a wrong equation.
Exercise 8: "Explain why potassium reacts more vigorously with water than lithium. [3]"
Student's Answer"Potassium has 19 electrons and lithium only has 3, so potassium has far more electrons available to give away in the reaction."
The FlawBoth atoms lose exactly one electron — the single outer-shell electron. The inner electrons are never involved; removing them would leave an unstable ion and cost enormous energy. The student has answered a "how many" question when the examiner asked a "how easily" question.
Correct Answer"Both atoms lose one outer electron, but in potassium that electron is in a shell further from the nucleus [1] and is shielded by more inner shells [1], so the attraction holding it is weaker and it is lost more easily, making potassium more reactive [1]."
Key RuleThe number of electrons lost is fixed by the group, not by the size of the atom. If your explanation would give a different answer for potassium and sodium about how many electrons move, it is wrong.
Exercise 9: "Explain why iodine has a higher melting point than chlorine. [2]"
Student's Answer"Because iodine is less reactive, so its atoms hold on to each other more strongly and the covalent bonds are harder to break."
The FlawTwo errors. First, reactivity and melting point are unrelated trends with different causes — one is about electron gain, the other about forces between molecules. Second, melting a molecular solid does not break covalent bonds; the I2 molecules survive intact and merely move apart.
Correct Answer"Iodine molecules are larger than chlorine molecules, so the intermolecular forces between the molecules are stronger [1], and more energy is needed to separate them, giving a higher melting point [1]."
Key RuleWhenever you explain the melting point of a simple molecular substance, the phrase must be "forces between molecules". Saying "bonds" without saying which kind is the fastest way to lose the mark — a link straight back to Topic 3.
Exercise 10: "Suggest why argon is used in filament light bulbs. [2]"
Student's Answer"Because argon conducts electricity well and helps the filament to glow more brightly."
The FlawArgon does not conduct the working current and does not make the light brighter — the current passes through the tungsten filament. The student has invented a mechanism instead of using the one property the noble gases actually have.
Correct Answer"Argon is unreactive because it has a full outer shell [1], so it does not react with the white-hot tungsten filament; it also excludes oxygen, which would otherwise oxidise the filament and cause it to burn away, so the bulb lasts much longer [1]."
Key RuleFor any noble gas use, the answer is "because it is unreactive" plus one job-specific property. If you find yourself inventing a chemical action for a noble gas, stop — the entire point of these elements is that they do nothing.
Exercise 11: "The table shows densities: Li 0.53, Na 0.97, K 0.86 g/cm3. Describe the trend. [2]"
Student's Answer"The density increases steadily down the group, so potassium is denser than sodium."
The FlawThe final clause contradicts the data printed in the question: potassium is 0.86 and sodium is 0.97, so potassium is the less dense of the two. The student has recited the general rule instead of reading the table, which is the exact behaviour a data-response question is designed to catch.
Correct Answer"The density generally increases down the group, from 0.53 for lithium to 0.86 for potassium [1], but the trend is not regular — potassium (0.86) is slightly less dense than sodium (0.97) [1]. All three are less dense than water, so all three float."
Key RuleIn a data question, quote at least one number from the stem and check your general statement against every row before writing it. Examiners include anomalies deliberately, and spotting one is usually worth its own mark.
Exercise 12: "Sodium chloride is used in food. Explain why this is safe when sodium reacts violently with water and chlorine is a toxic gas. [3]"
Student's Answer"Because there is only a small amount of each in the salt, and they cancel each other out when mixed."
The Flaw"Cancelling out" is not chemistry, and quantity is irrelevant — a small amount of sodium metal is still dangerous. The student has missed the entire point: sodium chloride does not contain sodium metal and chlorine gas at all.
Correct Answer"In sodium chloride the elements are present as ions, not as atoms or molecules [1]. Each sodium atom has lost its outer electron to form Na+ and each chlorine atom has gained one to form Cl−, so both now have full outer shells and are stable [1]. An ion has completely different properties from the element it came from, so sodium chloride is a stable, unreactive ionic compound and is safe to eat [1]."
Key RuleElement ≠ ion. This distinction underpins every application question in Topic 8, from drinking water to fertiliser to lithium batteries. Name the species precisely — "chloride ions, Cl−" — and the mark follows.

✍️ Ultra-Detailed Practice Questions

▼

Ten Cambridge-style challenge questions. Write your answer, then reveal the model answer with mark scheme and examiner's notes.

Question 1
[8 marks]
Meera in Pune is studying two groups of the Periodic Table. (a) Explain, in terms of electronic structure, why reactivity increases down Group I. [3] (b) Explain why reactivity decreases down Group VII. [3] (c) State clearly why the same change in atomic structure produces opposite trends in the two groups. [2]
Model Answer(a) Going down Group I each atom has one more occupied electron shell, so the single outer electron is further from the nucleus [1] and is shielded by more inner shells of electrons [1]. The attraction between the nucleus and the outer electron is therefore weaker, so it is lost more easily and the element is more reactive [1]
(b) Going down Group VII each atom is likewise larger with more shielding, so the outer shell where an incoming electron must go is further from the nucleus [1] and the attraction for an incoming electron is weaker [1], so the electron is gained less easily and the element is less reactive [1]
(c) A Group I atom reacts by losing an electron while a Group VII atom reacts by gaining one [1]. A weaker attraction makes losing easier but gaining harder, so the identical structural change produces opposite reactivity trends [1]
Examiner's NotesThis question is deliberately built so that a candidate who has memorised one explanation writes it twice and scores about three marks out of eight. The word "lost" must appear in (a) and the word "gained" in (b) — examiners scan for exactly that. Part (c) is the top-band discriminator and is answered well by fewer than a quarter of candidates: it needs the link, not a third repetition of distance and shielding. Candidates who mention increasing nuclear charge must add that distance and shielding outweigh it, otherwise the argument points the wrong way and the final mark is withheld.
Question 2
[9 marks]
A technician sets up nine test tubes, adding chlorine water, bromine water and iodine solution to aqueous potassium chloride, potassium bromide and potassium iodide. (a) State which three combinations react, and give the colour change in each. [4] (b) Write a balanced symbol equation and an ionic equation for the reaction of chlorine with potassium iodide. [3] (c) Explain what the overall pattern demonstrates about Group VII. [2]
Model Answer(a) Chlorine + potassium bromide: colourless to orange / red-brown as bromine is displaced [1]. Chlorine + potassium iodide: colourless to brown as iodine is displaced [1]. Bromine + potassium iodide: turns brown as iodine is displaced [1]. All other combinations show no reaction [1]
(b) Cl2(aq) + 2KI(aq) → 2KCl(aq) + I2(aq) — correct formulae [1], correctly balanced [1]
Ionic: Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq), the potassium ions being spectators [1]
(c) A more reactive halogen displaces a less reactive one from a solution of its halide, and chlorine displaced both while iodine displaced neither [1]. The order is therefore Cl2 > Br2 > I2, showing that reactivity decreases down Group VII [1]
Examiner's NotesColour marks are awarded strictly: displaced bromine gives orange or red-brown, displaced iodine gives brown, and "purple" for aqueous iodine is not accepted (that is the vapour). Many candidates describe a colour change in the no-reaction tubes because the added reagent was itself coloured — state the change from colourless to a new colour and this cannot happen. In (b) the ionic equation is often written as Cl2 + I− → Cl− + I2, which is not balanced for charge or atoms; check both. Part (c) needs the general rule and the conclusion about the group — one without the other scores one mark.
Question 3
[8 marks]
The table gives melting points for Group I: Li 181 °C, Na 98 °C, K 63 °C, and densities Li 0.53, Na 0.97, K 0.86 g/cm3. (a) Describe the melting point trend and predict a value for caesium, two places below potassium. [2] (b) Comment critically on the density data, and predict whether caesium (1.88 g/cm3) would float on water. [3] (c) Write a balanced equation with state symbols for caesium reacting with water, and give two observations. [3]
Model Answer(a) The melting point decreases down the group, and the size of each decrease gets smaller (83 °C then 35 °C) [1]; a prediction of about 25–35 °C for caesium is acceptable (true value 29 °C) [1]
(b) The density generally increases down the group [1] but the trend is irregular — potassium (0.86 g/cm3) is less dense than sodium (0.97 g/cm3) [1]. Caesium at 1.88 g/cm3 is denser than water (1.00), so it would sink [1]
(c) 2Cs(s) + 2H2O(l) → 2CsOH(aq) + H2(g) [1]; observations: the reaction is explosive / the metal ignites immediately [1]; there is very rapid effervescence and the metal disappears, and universal indicator turns purple because an alkaline solution is formed [1]
Examiner's NotesPart (b) is where the marks are won and lost. The syllabus statement is that density increases down Group I, but the data in front of the candidate contradicts it at one point, and the examiner wants to know which one you trust. Answers that assert a smooth increase lose two marks. In (a) a prediction with no number scores nothing; so does a number outside a sensible range. In (c), watch the balancing — the 2Cs and 2H2O are forced by the diatomic H2 — and remember that observations must be visible: "hydrogen is produced" is a deduction and needs "bubbles of gas" alongside it.
Question 4
[7 marks]
Element J has 17 protons. Element L has the electronic configuration 2,8,8,2. (a) Give the group and period of each. [2] (b) Predict the type of bonding and the formula of the compound of J and L. [3] (c) Predict, with a reason, the approximate pH of the solution formed when the oxide of L is added to water. [2]
Model Answer(a) J has 17 electrons, arranged 2,8,7 ⇒ Group VII, Period 3 [1]. L is 2,8,8,2 ⇒ Group II, Period 4 [1]
(b) L is a metal (left of the table) and J is a non-metal (right), so electrons are transferred and the bonding is ionic [1]. L loses two electrons to give L2+ and J gains one to give J− [1], so two J ions are needed per L ion: LJ2 [1]
(c) About pH 12–13, that is alkaline [1], because L is a metal and metal oxides are basic, dissolving in water to give an alkaline solution [1]
Examiner's NotesThis is a chain question, so an error in (a) propagates through the whole answer — though examiners normally allow error carried forward if the later reasoning is sound. Two frequent slips: writing the formula as J2L (the subscript attached to the wrong symbol) and giving pH 7 in (c) on the grounds that the oxide "just dissolves". A neutral prediction earns nothing; the marks are for recognising that metal oxides are basic. Note also that the question says "approximate pH" — a range or a single sensible value both score, but "alkaline" with no number risks losing the first mark.
Question 5
[8 marks]
A student is given samples of potassium and of copper. (a) Give four differences between their physical or chemical properties, quoting data or an example for each. [4] (b) Explain why copper compounds are named with a Roman numeral but potassium compounds are not. [2] (c) Copper is used as the wiring in a house and iron as the girders. Suggest why no Group I metal could be used for either purpose. [2]
Model Answer(a) Any four, each with support: copper is much denser (8.9 against 0.86 g/cm3, so potassium floats on water) [1]; copper has a much higher melting point (1085 °C against 63 °C) [1]; copper is hard while potassium can be cut with a knife [1]; copper is far less reactive — it does not react with cold water, whereas potassium ignites [1]; copper shows variable oxidation number (Cu+, Cu2+) while potassium is always +1; copper compounds are coloured (CuSO4 blue) while potassium compounds are white [max 4]
(b) Copper has variable oxidation number, so "copper oxide" would be ambiguous — Cu2O and CuO are different compounds [1]. Potassium forms only K+, so there is nothing to distinguish and no numeral is needed [1]
(c) Group I metals are far too reactive — they react violently with water and with air, so wiring or girders would corrode away and produce flammable hydrogen [1]; they are also too soft and too low-melting (potassium melts at 63 °C) to carry a structural load or survive a hot environment [1]
Examiner's NotesIn (a) the instruction "quoting data or an example" is doing real work: unsupported statements such as "copper is denser" typically score half the available marks. Candidates who list four similarities (both metals, both conduct, both shiny) score zero, because none of them discriminates. Part (c) rewards two distinct reasons — reactivity and mechanical properties — and answers that give reactivity twice in different words earn one mark. A neat extra in (b): noting that this is why sodium chloride is never written "sodium(I) chloride" shows genuine understanding.
Question 6
[7 marks]
(a) Explain, in terms of electronic structure, why the noble gases are unreactive and exist as single atoms. [3] (b) Explain why helium is used in weather balloons rather than hydrogen. [2] (c) A food company packs crisps under nitrogen; a museum stores a bronze artefact under argon. Suggest why the museum does not use nitrogen. [2]
Model Answer(a) A noble gas atom has a full outer electron shell — eight electrons, or two in helium [1]. It therefore has no tendency to lose, gain or share electrons, which is what a chemical reaction requires, so it is unreactive [1]. Because it does not need to share electrons even with another atom of its own kind, it does not form molecules and exists as single atoms, that is monatomic [1]
(b) Helium has a very low density, so the balloon rises [1], and it is unreactive and non-flammable, unlike hydrogen which burns explosively in air [1]
(c) Nitrogen is unreactive only because the N≡N triple bond is very strong and hard to break, and it can react under some conditions (for example forming nitrides with certain metals) [1]. Argon has a full outer shell and therefore cannot react at all, which matters for an irreplaceable artefact stored for decades [1]
Examiner's NotesPart (a) carries three marks and is regularly answered in one sentence about full outer shells, which caps the score at two — the monatomic point needs its own explanation. In (b) the classic error is asserting that helium is lighter than hydrogen; it is not, and inventing a fact to support a correct conclusion is penalised. Part (c) is a genuine stretch item testing the distinction between kinetically unreactive and genuinely inert; candidates who simply write "argon is more unreactive" without a reason score one at most. Cost is a legitimate secondary point (nitrogen is cheaper, which is why food packing uses it) and is credited if the chemistry is also there.
Question 7
[8 marks]
Bromine is extracted commercially at the Dead Sea by bubbling chlorine gas through concentrated brine. (a) Explain why this works, and write the ionic equation. [3] (b) Identify what is oxidised and what is reduced, with reasons. [2] (c) A student suggests using iodine instead of chlorine because iodine is cheaper to transport as a solid. Evaluate this suggestion. [2] (d) State one safety precaution needed when handling bromine, with a reason. [1]
Model Answer(a) Brine contains bromine as bromide ions, Br−, not as the element [1]. Chlorine is above bromine in Group VII and therefore more reactive, so it displaces bromine from its halide [1]. Ionic equation: Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq) [1]
(b) The bromide ions are oxidised, because each loses an electron to become part of a Br2 molecule [1]. The chlorine is reduced, because each atom gains an electron to become Cl−; chlorine is therefore the oxidising agent [1]
(c) The suggestion will not work [1]: iodine is below bromine in Group VII and therefore less reactive, so it cannot displace bromide ions — no reaction would occur, whatever the transport cost [1]
(d) Work in a fume cupboard (or wear gloves and eye protection), because bromine is toxic and corrosive and its vapour is harmful [1]
Examiner's NotesPart (a) has a hidden mark that many candidates miss: recognising that seawater contains bromide ions rather than bromine molecules, so a chemical reaction is required rather than a physical separation. In (b), "chlorine is reduced" alone is not enough — the mark requires the electron statement. Part (c) tests whether you can reject an economically attractive but chemically impossible idea; answers that weigh up cost without addressing reactivity score nothing. In (d) the reason must be specific: "it is dangerous" does not earn the mark, whereas "toxic vapour" does.
Question 8
[7 marks]
Element T lies directly below astatine in Group VII and has never been isolated in a weighable quantity. (a) Predict its physical state and colour at room temperature, with a reason. [2] (b) Predict the formula of its compound with calcium, explaining your reasoning. [2] (c) Predict, with a reason, whether T would displace iodine from potassium iodide solution. [2] (d) Explain how chemists can be confident about T's group even though no chemistry has been done with it. [1]
Model Answer(a) A black (very dark) solid with a melting point higher than that of astatine [1], because going down Group VII the molecules get larger so intermolecular forces get stronger, raising melting and boiling points, and the colour becomes progressively darker [1]
(b) Group VII means seven outer electrons, so T gains one to form T−; calcium is Group II and forms Ca2+ [1]. Two T− are needed to balance one Ca2+, giving CaT2 [1]
(c) No [1] — T is below iodine, so it is less reactive than iodine and cannot displace it from its halide [1]
(d) The Periodic Table is ordered by proton number, which is measured directly when the atom is made, so its position — and hence its outer-shell electron count — follows without any chemistry being performed [1]
Examiner's NotesPrediction questions are marked on the reason at least as much as the prediction. In (a) the melting-point reason must be intermolecular forces, not "the bonds get stronger" and not reactivity. In (c) the trap is powerful: a heavier, larger element intuitively feels more powerful, and a substantial minority answer "yes". Part (d) is a synoptic mark rewarding candidates who understand what the table is ordered by — answers citing atomic mass do not score, since that was Mendeleev's criterion, not the modern one.
Question 9
[8 marks]
A student writes: "Iron is a Group I metal because it is a metal that forms positive ions. Like sodium it reacts violently with cold water to give a hydroxide and hydrogen, its compounds are white, and it should be stored under oil. Iron chloride has the formula FeCl." (a) Identify and correct four errors. [4] (b) State the block iron actually belongs to and give three properties that place it there. [3] (c) Explain why iron can be used as a catalyst in the Haber process but sodium could not. [1]
Model Answer(a) Error 1: forming positive ions is common to all metals and does not place iron in Group I — iron lies in the transition block between Groups II and III [1]. Error 2: iron does not react with cold water to give hydrogen; it rusts slowly, and only when both water and oxygen are present [1]. Error 3: iron compounds are coloured, not white — iron(II) sulfate is pale green and iron(III) chloride yellow-brown [1]. Error 4: iron is far too unreactive to need oil, and the formula must be FeCl2 or FeCl3 depending on the oxidation state; "FeCl" would require Fe+, which iron does not form [1]
(b) Iron is a transition element [1]. Any three: high melting point (1538 °C); high density (7.9 g/cm3); variable oxidation number (Fe2+ and Fe3+); coloured compounds; catalytic behaviour [2]
(c) Transition elements act as catalysts and are chemically unchanged at the end, whereas sodium is far too reactive — it would react with the water vapour, air and reactants in the plant rather than catalysing anything [1]
Examiner's NotesError-correction questions award the mark only when the error and its correction are given as a pair — simply writing "the third sentence is wrong" scores nothing. This particular stem is built from four of the most common misconceptions in 8.4, and the FeCl formula error rewards candidates who realise that the Roman numeral is not decoration but a statement about charge. In (b), listing "it is a metal" or "it conducts electricity" does not place iron anywhere, because every metal does that; the marks are for discriminating properties.
Question 10
[10 marks]
A researcher in London is given four unlabelled samples. P is a soft metal stored under oil that floats on water and reacts to give a solution of pH 14. Q is a red-brown liquid that turns aqueous potassium iodide brown. R is a hard, dense metal forming two chlorides, one green and one yellow-brown in solution. S is a colourless gas that forms no compounds and exists as single atoms. (a) Identify the group or block of each and name a possible element for each. [4] (b) Write an equation for the reaction of P with water and one for the reaction of Q with potassium iodide. [3] (c) Explain, in terms of electronic structure, why P and S behave so differently. [2] (d) State one property of R that neither P nor S possesses, and give an industrial use that depends on it. [1]
Model Answer(a) P: Group I — soft, low density (floats), violent with water, alkaline product; e.g. sodium or potassium [1]. Q: Group VII — a red-brown liquid that displaces iodine; it must be bromine [1]. R: transition element — hard, dense, two chlorides (variable oxidation number), coloured solutions; e.g. iron [1]. S: Group VIII / 0 — unreactive and monatomic; e.g. argon or helium [1]
(b) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) [2 — products [1], balancing and state symbols [1]]
Br2(aq) + 2KI(aq) → 2KBr(aq) + I2(aq) [1]
(c) P has one electron in its outer shell, which is easily lost to give a stable full shell underneath, so it is very reactive [1]. S has a full outer shell, so it has no tendency to lose, gain or share electrons and is unreactive — and, needing no bonds, it exists as single atoms [1]
(d) R shows catalytic behaviour (or variable oxidation number, or coloured compounds) — for example iron is the catalyst in the Haber process for making ammonia [1]
Examiner's NotesThis is a synoptic identification question of the kind that closes a Topic 8 paper, and it rewards systematic reading of the clues rather than guesswork. Q is fully determined: red-brown liquid can only be bromine, and the displacement of iodine confirms it sits above iodine. R is determined by the two chlorides — a single clue that rules out every main-group metal at once. In (b) state symbols carry their own mark and (aq) for the displaced halogen is expected. Part (c) must contrast one outer electron with a full outer shell; candidates who describe only P, or only S, score one. In (d) the property and the use must match — naming "high density" and then citing the Haber process gains nothing.
IGCSE Chemistry 0620 — Topic 8 Challenge Prep Guide. Work through every section, and revisit the twelve traps the night before your exam — especially Trap 1.