The Periodic Table

• Four questions, four answers, one method. Where is it? → proton number. What are its electrons doing? → period and group. What will it do? → outer shell. How sure can I be? → only as sure as the data in front of you.
• If the question says explain and your answer contains no electrons, you have almost certainly answered a describe question instead.
• Eight of these nine errors are single words: mass for proton number, purple for grey-black, always for often, no for full. Precision of vocabulary is most of the work in this topic.
• Count the shells to get the period. Count the outer electrons to get the group. Both readings come from the same electronic configuration, which comes from the proton number.
• Before predicting anything, ask one question: does what I have been given actually fix the outer shell? A proton number does. A configuration does. A group does. A period, on its own, does not.
• Direction, property, evidence, qualification. Then ask: am I inside the data or outside it? Inside → a number is acceptable. Outside → a direction and a range.
• Period number \(=\) how many entries in the configuration. Group number \(=\) the last entry. For Group I the last entry is always \(1\), so the ion is always \(+1\).
• Free halogen \(=\) two atoms: \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\), \(\mathrm{I_2}\). Halide ion in a compound \(=\) one atom with one extra electron: \(\mathrm{Cl^{-}}\), \(\mathrm{Br^{-}}\), \(\mathrm{I^{-}}\). Two different things with two different formulae.

• Describe the Periodic Table as an arrangement of elements in periods (horizontal rows) and groups (vertical columns), in order of increasing proton number. Core 8.1.1
• Describe the change from metallic to non-metallic character across a period. Core 8.1.2
• Describe lithium, sodium and potassium as relatively soft metals. Core 8.2.1
• Describe the three trends down Group I: melting point decreases, density generally increases, reactivity increases. Core 8.2.1
• Describe chlorine, bromine and iodine as diatomic non-metals, and state their appearance at r.t.p. — a pale yellow-green gas, a red-brown liquid and a grey-black solid. Core 8.3.1 Core 8.3.2
• Describe the two trends down Group VII: density increases, reactivity decreases. Core 8.3.1
• Describe the displacement reactions of halogens with halide ions. Core 8.3.3
• Describe the transition elements as metals that have high densities and high melting points, form coloured compounds, and often act as catalysts — as elements and in compounds. Core 8.4.1
• Describe transition elements as having ions with variable oxidation numbers, including iron(II) and iron(III). Supplement 8.4.2
• Describe the Group VIII noble gases as unreactive, monatomic gases. Core 8.5.1
• Explain the similarities in chemical properties of elements in the same group in terms of their outer-shell electronic configuration. Core 8.1.4
• Explain a halogen displacement reaction in terms of the relative reactivity of the two halogens and the transfer of electrons. Core 8.3.3
• Explain the unreactivity of the noble gases in terms of their full outer electron shells. Core 8.5.1
• Identify the relationship between the group number and the charge of the ion an element forms, within the main-group pattern. Core 8.1.3
• Identify trends in a group you have not studied, from data you are given in the question — a table, a graph or a short paragraph. Supplement 8.1.6
• Predict the properties of an element from its position in the Periodic Table. Core 8.1.5
• Predict properties of other elements in Group I from information supplied about lithium, sodium and potassium. Core 8.2.2
• Predict properties of other elements in Group VII from information supplied about chlorine, bromine and iodine. Core 8.3.4

Why it matters. The noble gases are the reference point for the whole of bonding. Every ion in this chapter forms in order to reach the electron arrangement of a noble gas. Understanding why that arrangement is the destination is the same as understanding why Group I loses one electron and Group VII gains one — three sections of this chapter collapse into a single idea here.

• 1 · Rows and columns swapped Defect Calling a vertical column a period, or a horizontal row a group. Fix Period is the horizontal row — think of a sentence running left to right and ending in a full stop. Group is the vertical column — a family standing in a line. Period number = number of occupied shells; group number = number of outer electrons.
• 2 · “Arranged in order of increasing atomic mass” Defect Writing that the elements are ordered by relative atomic mass. Fix They are ordered by increasing proton number. The two orders agree most of the time, which is why the error survives, but they are not the same statement and only one of them is the syllabus wording. Tellurium sits before iodine despite having the larger relative atomic mass.
• 3 · “Group number = ion charge” applied everywhere Defect Deducing a \(+4\) ion for carbon, or a \(+7\) ion for chlorine, by reading the group number as the charge. Fix The relationship holds for Groups I, II and III (charge \(=\) group number, positive) and for Groups V, VI and VII (charge \(=\) group number \(-\,8\), negative). Group IV and the transition elements do not follow it, and Group VIII normally forms no ions at all.
• 4 · Treating the density trend as exact Defect Writing that density increases down Group I as though every step were guaranteed. Fix The syllabus calls all three Group I trends general trends, and density is where you can see why. Density generally increases, and the data has a wobble in it — potassium is slightly less dense than sodium. Keep the word “generally” and the statement is safe; drop it and the statement is falsifiable from a data table the examiner may well print.
• 5 · “Iodine is a purple solid” Defect Replacing iodine's actual appearance with the colour of its vapour or of its solution. Fix At r.t.p. solid iodine is grey-black, with a slight metallic sheen. Purple is what you see when it is heated and sublimes; brown is what you see in aqueous solution. Three different observations, three different conditions — the syllabus asks for the first.
• 6 · Halogen and halide used interchangeably Defect Writing “the chlorine in sodium chloride”, or writing \(\mathrm{Cl}\) for the element and \(\mathrm{Cl_2^{-}}\) for the ion. Fix The halogen is the free element, a diatomic molecule: \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\), \(\mathrm{I_2}\). The halide is the single negative ion inside a compound: \(\mathrm{Cl^{-}}\), \(\mathrm{Br^{-}}\), \(\mathrm{I^{-}}\). They have different colours, different reactivity and different formulae.
• 7 · Displacement run the wrong way Defect Predicting that iodine displaces chloride, usually justified by “iodine is bigger”. Fix Only a more reactive halogen displaces a less reactive halide, and reactivity decreases down Group VII, so the order is \(\mathrm{Cl_2 > Br_2 > I_2}\). Iodine is the least reactive of the three and displaces nothing from the other two. “No reaction” is a complete and correct answer.
• 8 · “All transition metal compounds are coloured” Defect Turning a typical property into a universal law, and then using it to reject a correct identification. Fix Transition elements form coloured compounds and are often catalysts. Neither statement is “always”. Titanium(IV) oxide, the white pigment in paint, and scandium oxide are both white solids. One property alone never classifies an element — look for the cluster.
• 9 · “Noble gases are unreactive because they have no electrons” Defect Confusing a full outer shell with an empty one, or writing \(\mathrm{Ar_2}\) by analogy with \(\mathrm{Cl_2}\). Fix Argon has \(18\) electrons, arranged \(2,8,8\). It is unreactive because the outer shell is full, so there is no advantage in losing, gaining or sharing. A full shell also means no bonding to another argon atom: noble gases are monatomic, written \(\mathrm{He}\), \(\mathrm{Ne}\), \(\mathrm{Ar}\).
• Error 1 · “The groups are the horizontal rows.” Defect Row and column swapped. Chemistry A group is a chemical family, and family resemblance comes from a shared outer-shell arrangement, which is what the columns share. The rows share only the number of shells. Repair Period \(=\) row. Group \(=\) column. Transfer Sodium and potassium are in the same what? Group.
• Error 2 · “The elements are arranged in order of increasing relative atomic mass.” Defect Wrong ordering quantity. Chemistry Proton number defines the element and fixes the electron arrangement; mass does not. The two orders disagree in a few places, and the table follows proton number every time. Repair Replace “relative atomic mass” with “proton number”. Transfer Tellurium has a greater relative atomic mass than iodine. Which comes first in the table? Tellurium, because \(52 < 53\).
• Error 3 · “The period number tells you the number of outer-shell electrons.” Defect The two readings of a configuration have been exchanged. Chemistry In \(2,8,7\) the number of entries \((3)\) is the count of occupied shells, which is the period; the last entry \((7)\) is the count of outer electrons, which is the group. Repair Period \(=\) how many numbers. Group \(=\) the last number. Transfer \(2,8,8,2\): period and group? Period 4, Group II.
• Error 4 · “Carbon is in Group IV, so it forms \(\mathrm{C^{4+}}\) ions.” Defect The group-to-charge rule applied outside its range. Chemistry Four outer electrons is equally far from losing all of them and from gaining four. Group IV elements normally share electrons instead, forming covalent bonds as in \(\mathrm{CO_2}\) and \(\mathrm{CH_4}\). Repair The rule covers Groups I–III and V–VII. Group IV is an exception, and so are the transition elements. Transfer What ion does silicon \((2,8,4)\) form in its simple compounds? None — it shares.
• Error 5 · “Iron is in the eighth column, so its ion is \(\mathrm{Fe^{8+}}\).” Defect Main-group reasoning applied to a transition element. Chemistry The group-number-to-charge rule covers Groups I–III and V–VII only. The transition elements sit outside it entirely, so no charge can be deduced from a column count. Repair For a transition element, read the charge from the name or the formula of the compound — never from a column count. Transfer What is the charge on the copper ion in copper(II) sulfate? \(+2\). Supplement — Extended candidates Supplement 8.4.2 Why The reason no single charge follows from the position is that transition elements have ions with variable oxidation numbers. Iron forms \(\mathrm{Fe^{2+}}\) and \(\mathrm{Fe^{3+}}\), so there is no one answer for the question to have. A Core candidate needs the rule's limit; an Extended candidate needs the reason as well.
• Error 6 · “Group VIII elements form ions with a charge of \(8+\).” Defect The pattern continued into a group where it does not apply. Chemistry Ions form in order to reach a full outer shell. A noble gas already has one, so there is no reason to lose or gain anything. Group VIII normally forms no ions at all. Repair Group VIII: no ion. Write it as an atom — \(\mathrm{Ar}\), \(\mathrm{Ne}\). Transfer What ion does neon form? None.
• Error 7 · “Density increases down Group I.” Defect A missing qualifier turns a true statement into a false one. Chemistry The data is not perfectly regular: potassium \((0.86\;\mathrm{g/cm^3})\) is slightly less dense than sodium \((0.97\;\mathrm{g/cm^3})\). The syllabus calls all three Group I trends general trends, and density is the one where you can see why. Repair Insert one word: density generally increases down Group I. Transfer Which of the three trends most needs the word “generally”, and why? Density — the potassium value dips below sodium's.
• Error 8 · “Melting point increases down Group I.” Defect Direction reversed. Chemistry The data runs \(181\,{}^\circ\mathrm{C}\), \(98\,{}^\circ\mathrm{C}\), \(63\,{}^\circ\mathrm{C}\) for lithium, sodium and potassium — falling at every step. Repair Melting point decreases down Group I. Transfer Rubidium is below potassium. Higher or lower melting point? Lower.
• Error 9 · “Reactivity decreases down Group I, like Group VII.” Defect One group's trend copied onto the other. Chemistry The two run in opposite senses. Group I reactivity increases down the group; Group VII reactivity decreases. Repair Learn them as a contrasting pair, never separately. Transfer Which is more reactive, potassium or lithium? Potassium. Chlorine or iodine? Chlorine.
• Error 10 · “Sodium cannot be a proper metal because you can cut it with a knife.” Defect One physical property treated as disqualifying. Chemistry Sodium is shiny when freshly cut, conducts heat and electricity, and forms a positive ion — all metallic properties. The syllabus says relatively soft, meaning softer than typical metals, not that it is not one. Repair Keep the word relatively, and remember it is a comparison with metals in general. Transfer Name two metallic properties lithium still has. Shiny when cut; conducts electricity.
• Error 11 · “Iodine is a purple solid.” Defect The colour of the vapour given as the colour of the solid. Chemistry At r.t.p. iodine is a grey-black solid with a slight metallic sheen. Purple is the vapour, seen on warming; brown is the aqueous solution. Repair Answer the conditions the question named. “At r.t.p.” means the solid. Transfer Colour of solid iodine? Grey-black. Colour of aqueous iodine? Brown.
• Error 12 · “Chlorine is a green liquid; bromine is a brown gas.” Defect States shuffled between elements. Chemistry At r.t.p. the three states run gas, liquid, solid going down the group: chlorine is a pale yellow-green gas, bromine a red-brown liquid, iodine a grey-black solid. Repair Fix the order gas → liquid → solid to the order Cl → Br → I, and the rest follows. Transfer Which halogen is a liquid at r.t.p.? Bromine.
• Error 13 · Writing \(\mathrm{Cl + KBr \rightarrow KCl + Br}\) Defect Free halogens written as single atoms. Chemistry Halogen atoms have seven outer electrons and pair up to complete their shells, so the elements are diatomic: \(\mathrm{Cl_2}\), \(\mathrm{Br_2}\), \(\mathrm{I_2}\). Repair \(\mathrm{Cl_2 + 2KBr \rightarrow 2KCl + Br_2}\). Note that the subscript forces the \(2\) in front of \(\mathrm{KBr}\). Transfer Formula of the element bromine? \(\mathrm{Br_2}\). Of the bromide ion? \(\mathrm{Br^{-}}\).
• Error 14 · “Potassium bromide solution is red-brown.” Defect The colour of the element attributed to its ion. Chemistry Halide ions in solution are colourless. Red-brown is the free element bromine. This matters because the colourless starting solution is what makes a displacement visible. Repair Halide solution \(=\) colourless. Halogen in solution \(=\) coloured. Transfer What colour is aqueous potassium iodide? Colourless.
• Error 15 · “Iodine displaces chloride because iodine atoms are bigger.” Defect Displacement run backwards, on an irrelevant reason. Chemistry Displacement needs the added halogen to be the more reactive, and reactivity decreases down Group VII. Iodine is the least reactive of the three and displaces neither of the others. Atomic size is not a Topic 8 explanation. Repair Compare positions first: only a halogen above the halide can displace it. Transfer Does bromine displace chloride? No. Does bromine displace iodide? Yes.
• Error 16 · Writing \(\mathrm{Cl_2 + Br^{-} \rightarrow Cl^{-} + Br_2}\) Defect Ionic equation unbalanced in both atoms and charge. Chemistry \(\mathrm{Cl_2}\) supplies two chlorine atoms, so two bromide ions are needed and two chloride ions are produced. Charge: \(-1\) on the left against \(-1\) on the right looks fine, but the atoms do not balance — Br is \(1\) left, \(2\) right. Repair \(\mathrm{Cl_2(aq) + 2Br^{-}(aq) \rightarrow 2Cl^{-}(aq) + Br_2(aq)}\). Atoms: \(2\) and \(2\) both ways. Charge: \(-2\) and \(-2\). Transfer Balance \(\mathrm{Br_2 + \_\,I^{-} \rightarrow \_\,Br^{-} + I_2}\). Both blanks are \(2\).
• Error 17 · “All transition metal compounds are coloured, so a white compound rules it out.” Defect A typical property hardened into a universal law, then used backwards. Chemistry Transition elements form coloured compounds and are often catalysts. Neither is “always”: titanium(IV) oxide is the white pigment in paint. And a property that is typical of a class cannot be used to exclude a member on one counter-example. Repair Classify on the cluster: high melting point, high density, coloured compounds, catalysis. (Extended candidates add variable oxidation number as a fifth.) Transfer A metal melts at \(1668\,{}^\circ\mathrm{C}\), has density \(4.51\;\mathrm{g/cm^3}\), and forms a white oxide. Transition element? Probably yes — two strong signals, and one white compound does not rule it out.
• Error 18 · “Noble gases are unreactive because they have no electrons — and argon is \(\mathrm{Ar_2}\).” Defect Two errors: a full shell mistaken for an empty one, and a diatomic formula for a monatomic element. Chemistry Argon has \(18\) electrons, arranged \(2,8,8\). Its outer shell is full, which is why there is no tendency to lose, gain or share — and why it does not bond even to another argon atom. Repair “Full outer shell”, and write \(\mathrm{Ar}\), never \(\mathrm{Ar_2}\). Transfer How many electrons does argon have altogether, and how many in its outer shell? \(18\) altogether, \(8\) in the outer shell.

• Two things worth being clear about. First, the practical paper is taken by both routes — experimental skill is not an Extended-only requirement, and the Paper 5 / Paper 6 choice is made by your centre, not by your tier. Second, Extended is not a separate syllabus: an Extended paper can ask you anything from the Core content as well as the Supplement content. In this chapter that matters, because thirteen of the fifteen statements are Core — most of an Extended paper's Topic 8 marks are Core marks.
• A question style worth recognising. Questions on this topic often use letters — element \(\mathrm{A}\), element \(\mathrm{Q}\) — rather than real symbols, so that you cannot answer from memory. Treat that as a gift rather than a trap: when a proton number, a configuration or a group is supplied, the answer is derivable from it — and when none of those is supplied, saying what is missing is the answer, as cases G and H in the position studio show. If you find yourself trying to remember which element \(\mathrm{Q}\) “really is”, you have left the method.
• What the two incomplete cases are teaching. Real questions do sometimes give you less than you need, and recognising that is the answer. “The period tells me there are four shells, but not the group, so I cannot predict the ion” is a complete answer. Inventing the rest is not.
• The Extended habit this is training. On a familiar group, memory and data agree and you barely notice which one you used. On an unfamiliar group they cannot agree, because there is no memory to consult — so any trend you state has to be visible in the printed numbers, and any property that shows no trend has to be reported as showing no trend. “There is no simple trend in this property” is a complete answer, and it is one that a candidate answering from memory can never give.
• The one-line self-check. Before you commit an answer, read it back and ask whether a reader who could not see the table would know (1) which way you were going, (2) what changed, and (3) what evidence you used. If any of the three is missing, the sentence is not finished.
• Describing the appearance. If asked what is seen when a piece of sodium is cut, the observation has two halves: a shiny, silvery surface that quickly becomes dull. Both halves matter. Saying only “shiny” misses the observation the demonstration was arranged to show; saying only “dull grey” describes the sample before it was cut.
• A sentence worth memorising. “Going down the group, [property] [increases / decreases / generally increases], from [first value with unit] to [last value with unit], so [element] will be [above / below] [nearest value], probably in the region of [range].” Fill in the brackets and you have written a complete prediction, every time.
• Three separate retrieval directions. Practise this three ways round, not one: given the name, produce the formula, the state and the colour; given the colour, produce the element; given the state, produce the element. A student who has only rehearsed name → colour is stuck the moment a question runs the other way, which a multiple-choice stem can easily do.
• The diagonal is worth noticing. Three of the six non-reactions are the “same element” cases along the diagonal, and they are non-reactions for a different reason from the other three. If a question asks why nothing happened, say which of the two reasons applies: either the added halogen was less reactive, or it was the same element.
• A three-part observation sentence that always works. “The [starting colour] solution of [named halide] turns [final colour], because [named halogen] has been displaced and is present as [formula, aqueous].” If you cannot fill in the first blank, the question has not given you enough to answer — and saying so is better than a guess.
• Two directions, one comparison. Whenever a question asks about two halogens, there is only ever one fact in play: which is higher in the group. From it come the reactivity order, the direction of any displacement, the density comparison, and often the state. Find that one fact first and the rest of the answer writes itself.
• What a three-property answer needs. For “give three properties that show \(\mathrm{Y}\) is a transition element”, the answer needs three distinct properties each tied to the data you were given — not to three ways of saying “it is dense”. Scan the four Core properties and pick the ones the question has actually supplied evidence for; Extended candidates have a fifth to scan as well.
• How to use the ladder. Level 1 should become automatic — if you have to think, keep drilling it. Level 2 is where the rules of the topic sit. Level 3 cannot be memorised at all; it has to be practised on information you have not seen before, which is exactly why every Level 3 prompt supplies its own data.
• Marking yourself honestly. Award a mark only where your answer contains the underlined idea, not merely something adjacent to it. “It goes brown” is not “the colourless solution turns brown”. “Density increases” is not “density generally increases down the group”. The gap between those pairs is where the marks in this topic actually live.

• Topic 8 is unusual: it is one of the few topics where nearly every mark can be earned from information printed on the paper in front of you. You are given a Periodic Table. Questions frequently supply a small data table for a group you have never studied. The skill being tested is not recall of a hundred elements — it is disciplined reading.
• Cambridge IGCSE Chemistry 0620 has two routes. You are entered for one of them, and which one decides both your theory papers and the grades available to you. The practical component is the same choice on both routes.
• Two things worth being clear about. First, the practical paper is taken by both routes — experimental skill is not an Extended-only requirement, and the Paper 5 / Paper 6 choice is made by your centre, not by your tier. Second, Extended is not a separate syllabus: an Extended paper can ask you anything from the Core content as well as the Supplement content. In this chapter that matters, because thirteen of the fifteen statements are Core — most of an Extended paper's Topic 8 marks are Core marks.
• Short, factual and unforgiving. Typical stems: which element is in Group VII and Period 3; which of four ions has the configuration \(2,8\); which halogen displaces bromide but not chloride; which property is not typical of a transition element. There is no partial credit, so precision on the three r.t.p. appearances and the two Group VII trends pays directly.
• Where the explaining happens. Expect: deduce group and period from a configuration; explain why two elements are in the same group; complete an ionic equation for a displacement; state and explain a trend from a supplied table; list properties that identify an element as a transition metal. The explanations wanted here are electron-based: an answer that describes without naming the outer shell has not answered the explain command word.
• Colour changes on mixing halogen solutions with halide solutions, and the identification of an unknown halogen or halide from a small set of test results. The examiner is checking that you separate what was seen from what it means, and that you name the starting colour as well as the final one.