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Question 1 -- Building the Periodic Table
Total: 12 marks
A class in Cape Town is studying how the Periodic Table came to take its modern form, and how the position of an element can be used to predict its chemistry.
(a)[4]
(i) State the property by which Mendeleev ordered the elements, and the property used to order them today. [1]
(ii) Explain why the modern order is the better one. [1]
(iii) Explain why Mendeleev deliberately left gaps in his table. [1]
(iv) State what all the elements in a single group have in common. [1]
Model Answer -- 1(a)
(i) Mendeleev used relative atomic mass; the modern table uses proton number (atomic number) [1]
(ii) Ordering by mass puts a few pairs in the wrong place — argon has a greater relative atomic mass than potassium, yet argon must come first if it is to sit with the noble gases; proton number never produces such an anomaly [1]
(iii) He left a gap wherever the next known element did not fit the pattern of a column, predicting that an undiscovered element would fill it, which kept chemically similar elements together [1]
(iv) They all have the same number of electrons in the outer shell, which is why they react in similar ways [1]
⚠ If you missed marks here: Saying the modern table is better "because it is more modern" earns nothing; you need a specific example of a pair that mass ordering gets wrong. In (iv), "they have similar properties" restates the question — the mark is for the reason, which is the number of outer-shell electrons.
(b)[4]
Consider Period 3, running from sodium to argon.
(i) Describe how the number of outer-shell electrons changes across the period. [1]
(ii) Describe how the metallic character of the elements changes across the period. [1]
(iii) Describe how the acid–base character of the oxides changes across the period. [1]
(iv) Name the Period 3 element whose oxide is amphoteric. [1]
Model Answer -- 1(b)
(i) The number of outer-shell electrons increases by one for each element, from one in sodium to eight in argon [1]
(ii) The elements change from metals on the left, through the borderline elements, to non-metals on the right [1]
(iii) The oxides change from basic (Na₂O, MgO), through amphoteric, to acidic (P₄O₁₀, SO₃) [1]
(iv) Aluminium; Al₂O₃ reacts with acids to give aluminium salts and with hot concentrated alkali to give aluminates [1]
⚠ If you missed marks here: A very common error is to say that the number of shells increases across a period; it is the number of electrons in the outermost shell that changes, while the number of occupied shells stays at three. In (iii) both ends of the change are needed for the mark.
(c)[4]
Element R forms an ion R³⁺ and a chloride of formula RCl₃. Element T forms an ion T²⁻.
(i) State the group of R. [1]
(ii) State the group of T. [1]
(iii) Give the formula of the compound formed between R and T. [1]
(iv) State which of R and T forms the acidic oxide, and explain why. [1]
Model Answer -- 1(c)
(i) R is in Group III, because it loses three outer-shell electrons to form a 3⁺ ion and combines with three chloride ions [1]
(ii) T is in Group VI, because it gains two electrons to complete its outer shell, so it must start with six [1]
(iii) R₂T₃ — two 3⁺ ions balance three 2⁻ ions, giving a total charge of zero [1]
(iv) T forms the acidic oxide, because T is a non-metal and non-metal oxides that dissolve in water give acidic solutions; R is a metal and its oxide is basic or amphoteric [1]
⚠ If you missed marks here: For (ii), remember that a negative ion tells you how many electrons were gained, so the group number is 8 minus the size of the charge. In (iii), writing R₃T₂ means the charges have been crossed over the wrong way; always check that the total positive charge cancels the total negative charge.
Question 2 -- The Alkali Metals in Data and in Reaction
Total: 12 marks
A student in Delhi collects published data for four Group I metals and then carries out a quantitative experiment with potassium.
metal
melting point / °C
density / g cm⁻³
time for a 0.1 g piece to disappear in cold water / s
lithium
181
0.53
95
sodium
98
0.97
22
potassium
63
0.86
6
rubidium
39
1.53
less than 1
(a)[4]
(i) Describe two clear trends shown by the data as the group is descended. [2]
(ii) Predict the melting point and the density of caesium, the next metal down the group. [1]
(iii) Potassium is stored under oil, but caesium is supplied in a sealed glass tube filled with argon. Suggest why. [1]
Model Answer -- 2(a)
(i) The melting point falls steadily down the group, from 181 °C for lithium to 39 °C for rubidium [1]
(i) The reaction with water becomes faster and more vigorous down the group, the time falling from 95 s to under 1 s [1]
(ii) Caesium should melt a little below rubidium, at about 29 °C, and should be denser than rubidium, at about 1.9 g cm⁻³ [1]
(iii) Caesium is so reactive that it slowly attacks even the hydrocarbon oil, whereas argon has a complete outer shell and reacts with nothing at all, so it gives complete protection [1]
⚠ If you missed marks here: Note that density does not fall smoothly — potassium is less dense than sodium — so quoting "density decreases down the group" as a trend loses the mark. Trends must be supported by figures from the table to be convincing.
(b)[5]
The student reacts 0.78 g of potassium (Aₓ = 39) with an excess of cold water and collects the gas produced.
(i) Write the balanced symbol equation for the reaction, including state symbols. [2]
(ii) Calculate the volume of gas collected at room temperature and pressure. One mole of any gas occupies 24 dm³ at r.t.p. [2]
(iii) State the approximate pH of the solution left behind. [1]
Model Answer -- 2(b)
(i) 2K + 2H₂O → 2KOH + H₂ [1]
(i) With state symbols: 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g) [1]
(ii) Moles of potassium = 0.78 ÷ 39 = 0.020 mol, and the 2 : 1 ratio gives 0.010 mol of hydrogen [1]
(ii) Volume = 0.010 × 24 = 0.24 dm³, which is 240 cm³ [1]
(iii) The solution is potassium hydroxide, a strong alkali, so the pH is about 13 to 14 [1]
⚠ If you missed marks here: The single most costly error is using a 1 : 1 ratio, which doubles the volume to 480 cm³. Also check the units: the gas volume comes out in dm³ when you multiply by 24, so it must be multiplied by 1000 if the answer is wanted in cm³.
(c)[3]
Sodium also burns brightly when lowered into a gas jar of chlorine.
(i) Write the balanced symbol equation for this reaction. [1]
(ii) Name the type of bonding in the product and explain how it forms in terms of electrons. [1]
(iii) State one safety precaution needed for this demonstration and give a reason. [1]
Model Answer -- 2(c)
(i) 2Na + Cl₂ → 2NaCl [1]
(ii) Ionic bonding: each sodium atom transfers its single outer electron to a chlorine atom, giving Na⁺ and Cl⁻ ions that attract one another strongly [1]
(iii) Carry out the demonstration in a fume cupboard, or behind a safety screen with the class at a distance, because chlorine is toxic and the reaction is vigorous and very bright [1]
⚠ If you missed marks here: Writing Na + Cl → NaCl loses the mark because chlorine exists as Cl₂ molecules, so the equation must be balanced with two sodium atoms. In (ii), saying "electrons are shared" describes covalent bonding and is the wrong mechanism for a metal reacting with a non-metal.
Question 3 -- Halogen Atoms, Ions and Reactivity
Total: 12 marks
The halogens take their name from the Greek for "salt former". This question looks at how the structure of a halogen atom explains both the salts it forms and the way its reactivity changes down the group.
(a)[4]
(i) Write the electronic configuration of a fluorine atom and of a chlorine atom. [2]
(ii) Describe how a chlorine atom becomes a chloride ion, and give the electronic configuration and the charge of that ion. [2]
Model Answer -- 3(a)
(i) Fluorine is 2,7 [1]
(i) Chlorine is 2,8,7 [1]
(ii) The atom gains one electron into its outer shell, completing it [1]
(ii) The chloride ion has the configuration 2,8,8 and a charge of 1⁻, because it now has one more electron than it has protons [1]
⚠ If you missed marks here: Do not write that the chlorine atom loses seven electrons; gaining one is far easier and is what actually happens. The charge must be explained by comparing the numbers of protons and electrons, not simply asserted.
(b)[4]
The halogens react with hydrogen under the following conditions: fluorine explodes even in the dark; chlorine explodes in bright sunlight; bromine reacts only when heated with a catalyst; iodine reacts slowly and the reaction is reversible.
(i) State what these observations show about the trend in reactivity down Group VII. [2]
(ii) Explain this trend in terms of atomic structure. [2]
Model Answer -- 3(b)
(i) Increasingly harsh conditions are needed as the group is descended — darkness, then sunlight, then heat and a catalyst, then only a slow and incomplete change [1]
(i) This shows that reactivity decreases steadily down Group VII [1]
(ii) Down the group each element has one more occupied shell, so the atomic radius increases and the outer shell is further from the nucleus, with more inner shells shielding it [1]
(ii) An incoming electron is therefore attracted less strongly, so it is gained less readily and the element is less reactive [1]
⚠ If you missed marks here: Part (i) asks what the observations show, so you must refer to the conditions themselves rather than simply stating the trend. In (ii), a very common error is to write that the outer electron is "lost less easily" — that is the Group I argument applied to the wrong group.
(c)[4]
(i) Chlorine is bubbled into water containing a few drops of litmus solution. Describe what is seen. [1]
(ii) State one large-scale use of chlorine that depends on this behaviour. [1]
(iii) Give one hazard associated with chlorine and a precaution that reduces the risk. [1]
(iv) The word halogen means "salt former". Give the name and formula of the salt formed when a halogen reacts directly with sodium. [1]
Model Answer -- 3(c)
(i) The litmus first turns red, because the solution is acidic, and is then bleached to colourless [1]
(ii) Chlorine is added in very small amounts to drinking water and to swimming pools, where it kills bacteria and sterilises the water [1]
(iii) Chlorine is toxic and attacks the lungs, so it must be used in a fume cupboard or in a well-ventilated room and only in small quantities [1]
⚠ If you missed marks here: In (i) both stages are needed — many candidates mention only the bleaching and miss the initial colour change to red. In (iii) a hazard on its own is not enough; the mark requires a matching precaution.
Question 4 -- Displacement, Redox and Calculation
Total: 12 marks
A laboratory technician in Auckland prepares three test tubes so that a class can see for themselves which halogens displace which.
(a)[3]
Predict, with a reason in each case, what would be seen in each of the following mixtures.
(i) Bromine water added to aqueous potassium iodide. [1]
(ii) Chlorine water added to aqueous sodium chloride. [1]
(iii) Iodine solution added to aqueous potassium bromide. [1]
Model Answer -- 4(a)
(i) The solution turns dark brown, because bromine is above iodine in the group and displaces it from the iodide [1]
(ii) No change, because the halogen added and the halide present are the same element, so there is nothing to displace [1]
(iii) No change, because iodine is below bromine and a halogen cannot displace one that lies above it [1]
⚠ If you missed marks here: Part (ii) catches candidates who look only at the names and assume that mixing a halogen with a halide must produce something; check whether it is the same element first. Every mark here needs both the observation and the reason.
(b)[5]
Consider the reaction between bromine and aqueous potassium iodide.
(i) Write the ionic equation, including state symbols. [2]
(ii) Identify the species that is oxidised and the species that is reduced, justifying each choice. [2]
(iii) Define oxidation in terms of electrons. [1]
Model Answer -- 4(b)
(i) Br₂ + 2I⁻ → 2Br⁻ + I₂ [1]
(i) With state symbols: Br₂(aq) + 2I⁻(aq) → 2Br⁻(aq) + I₂(aq); the potassium ions are spectators and are left out [1]
(ii) The iodide ions are oxidised, because each loses one electron as it becomes part of an iodine molecule [1]
(ii) Bromine is reduced, because each bromine atom gains an electron to become a bromide ion [1]
(iii) Oxidation is the loss of electrons; reduction is the gain of electrons [1]
⚠ If you missed marks here: Marks are frequently lost for leaving the potassium ions in the ionic equation, and for writing "iodine is oxidised" when it is the iodide ion that loses the electron. Remember OIL RIG and always state which species loses and which gains.
(c)[4]
1.60 g of bromine, Br₂ (Mₓ = 160), is added to an excess of aqueous potassium iodide.
(i) Calculate the number of moles of bromine used. [1]
(ii) State the number of moles of iodine formed. [1]
(iii) Calculate the mass of iodine formed. The Mₓ of I₂ is 254. [1]
(iv) Explain why adding more potassium iodide would not increase the mass of iodine obtained. [1]
(ii) The equation shows a 1 : 1 ratio between bromine and iodine, so 0.0100 mol of iodine is formed [1]
(iii) Mass = 0.0100 × 254 = 2.54 g [1]
(iv) The potassium iodide is already in excess, so the bromine is the limiting reactant; once all the bromine has reacted no further iodine can be produced however much iodide is present [1]
⚠ If you missed marks here: The 2 in front of the iodide ion in the equation refers to the ions, not to the iodine molecules, so the bromine to iodine ratio really is 1 : 1. In (iv) the key phrase is "limiting reactant" — saying only that "there is already enough" does not explain why.
Question 5 -- Group VIII: The Noble Gases
Total: 10 marks
Argon makes up nearly one per cent of the air, yet it went unnoticed until 1894 because it takes no part in the reactions used to remove the other gases.
(a)[4]
(i) State where the noble gases are found in the Periodic Table and describe the outer electron shell of a noble gas atom. [2]
(ii) State two other physical or chemical properties shared by all the noble gases. [2]
Model Answer -- 5(a)
(i) They form Group VIII, the last column on the far right of the Periodic Table [1]
(i) Each atom has a complete outer shell — eight electrons in every case except helium, which has two [1]
(ii) They are monatomic, existing as separate single atoms rather than as molecules [1]
(ii) They are colourless gases at room temperature with very low boiling points, and they are extremely unreactive [1]
⚠ If you missed marks here: Helium is the exception that examiners look for: it has a full shell of two, not eight, so an answer that says "all have eight outer electrons" is not quite right. Do not offer "unreactive" twice as two separate properties.
(b)[3]
Name the noble gas used in each of the following applications and give a reason for the choice in each case.
(i) Filling airships and party balloons. [1]
(ii) Providing a protective atmosphere around a hot metal weld. [1]
(iii) Making brightly glowing red-orange display tubes. [1]
Model Answer -- 5(b)
(i) Helium, because it has a very low density so the balloon rises, and unlike hydrogen it is unreactive and cannot catch fire [1]
(ii) Argon, because it forms a completely inert blanket that keeps oxygen away from the hot metal so that the weld does not oxidise [1]
(iii) Neon, because it emits a bright red-orange glow when a high voltage is passed through it at low pressure [1]
⚠ If you missed marks here: Each mark needs the gas and the reason together. A frequent slip is to say that helium is used "because it is lighter than air and burns well" — the whole advantage over hydrogen is that it does not burn at all.
(c)[3]
A student writes: "The noble gases are unreactive because their atoms are too heavy to move about and meet other atoms."
(i) Give one piece of evidence that shows this explanation cannot be correct. [1]
(ii) Give the correct explanation for their lack of reactivity. [1]
(iii) Xenon can be made to react with fluorine to form XeF₄. State what this shows about the claim that noble gases never react. [1]
Model Answer -- 5(c)
(i) Helium is one of the lightest substances known, and its atoms move faster than those of almost any other gas, yet it is completely unreactive; so mass cannot be the reason [1]
(ii) Each atom has a complete outer electron shell, so there is no tendency to lose, gain or share electrons, and every kind of chemical bonding requires one of those three processes [1]
(iii) It shows that unreactivity is very great but not absolute: in the largest atoms the outer electrons are far from the nucleus and heavily shielded, so an extremely reactive element such as fluorine can force them into bonding under harsh conditions [1]
⚠ If you missed marks here: Part (i) asks for evidence against the student's idea, so a general statement about full shells does not earn it; you must give a counter-example, and helium is the clearest one. In (iii), avoid concluding that the noble gases are "really quite reactive" — the point is that the rule has rare exceptions among the heaviest members.
Question 6 -- Transition Elements in Use
Total: 12 marks
The central block of the Periodic Table contains the metals on which most engineering and much of the chemical industry depends. This question compares them with the metals of Group I.
(a)[4]
State four characteristic properties of the transition elements, giving a named example to illustrate each one. [4]
Model Answer -- 6(a)
High density — iron has a density of 7.87 g cm⁻³ and copper 8.96 g cm⁻³, both far above that of any Group I metal [1]
High melting point — copper melts at 1085 °C and iron at 1538 °C [1]
Variable oxidation number — iron forms Fe²⁺ and Fe³⁺, and copper forms Cu⁺ and Cu²⁺ [1]
Coloured compounds and catalytic activity — copper(II) sulfate solution is blue, and iron catalyses the Haber process while nickel catalyses the hydrogenation of vegetable oils [1]
⚠ If you missed marks here: The question asks for a named example with each property, so a bare list of four properties can score only partial credit. Avoid offering "conducts electricity" or "is shiny", since these belong to all metals and are not characteristic of the transition block.
(b)[4]
Compare a typical transition element with a typical Group I metal under each of the following headings. [4]
(i) Hardness. [1]
(ii) Melting point. [1]
(iii) Reaction with cold water. [1]
(iv) Charge on the ions formed. [1]
Model Answer -- 6(b)
(i) Transition metals are hard and strong; Group I metals are soft enough to be cut with a knife [1]
(ii) Transition metals melt at high temperatures, typically above 1000 °C; Group I metals melt below 200 °C and caesium melts at only 29 °C [1]
(iii) Transition metals react slowly or not at all with cold water; every Group I metal reacts vigorously, giving hydrogen and an alkaline hydroxide [1]
(iv) Transition metals form ions of more than one charge, such as 2⁺ and 3⁺; a Group I metal forms only a 1⁺ ion [1]
⚠ If you missed marks here: Both halves of every comparison are needed — writing "transition metals are hard" without saying that Group I metals are soft leaves the comparison incomplete and the mark is not awarded.
(c)[4]
Two solids are supplied. Solid X is white and dissolves in water to give a colourless solution. Solid Y is blue-green and dissolves to give a blue solution; the metal in Y also forms a second, colourless chloride in which its oxidation number is +1.
(i) State which solid contains a transition metal, giving two pieces of evidence. [2]
(ii) Suggest which group the metal in X is most likely to belong to, with a reason. [1]
(iii) Describe one further test that would support your identification of Y, and give the expected result. [1]
Model Answer -- 6(c)
(i) Y contains the transition metal, because its compound is coloured and gives a coloured solution, which is characteristic of the transition block [1]
(i) Y also shows variable oxidation number, since the same metal forms a +1 chloride as well as the +2 compound; a main-group metal has only one oxidation number [1]
(ii) X is most likely a Group I (or Group II) metal compound, because its solid is white and its solution colourless, which is typical of main-group metal compounds [1]
(iii) Add aqueous sodium hydroxide to the solution of Y: a pale blue precipitate of the metal hydroxide forms, whereas a Group I compound would give no precipitate at all [1]
⚠ If you missed marks here: Part (i) asks for two pieces of evidence, so colour alone earns only one mark; the second oxidation number is the other clue given in the stem. In (iii) the expected result must be stated, not just the test.
Question 7 -- Deducing Position from Formulae
Total: 10 marks
The table gives information about five elements, labelled G, H, J, K and L. The letters are not the chemical symbols of the elements.
element
proton number
formula of chloride
formula of oxide
melting point / °C
G
3
GCl
G₂O
181
H
12
HCl₂
HO
649
J
16
JCl₂
JO₂
115
K
20
KCl₂
KO
842
L
10
no chloride
no oxide
−249
(a)[4]
State the group of the Periodic Table to which each of G, H, J and L belongs, giving evidence from the table in each case. [4]
Model Answer -- 7(a)
G is in Group I: the formulae GCl and G₂O show a 1⁺ ion, and proton number 3 gives the configuration 2,1 [1]
H is in Group II: HCl₂ and HO show a 2⁺ ion, and proton number 12 gives 2,8,2 [1]
J is in Group VI: proton number 16 gives 2,8,6, and JO₂ is the acidic oxide of a non-metal with a low melting point of 115 °C [1]
L is in Group VIII: proton number 10 gives the full outer shell 2,8, it forms neither a chloride nor an oxide, and its melting point is extremely low [1]
⚠ If you missed marks here: J is the one that catches candidates out: the chloride JCl₂ looks like a Group II formula, so use the proton number to get the configuration before deciding. The absence of any compounds at all is decisive evidence for a noble gas.
(b)[3]
Elements H and K belong to the same group.
(i) State which of them is the more reactive. [1]
(ii) Explain your answer in terms of atomic structure. [1]
(iii) State which of the two oxides gives the more alkaline solution in water. [1]
Model Answer -- 7(b)
(i) K is the more reactive; its proton number of 20 places it one period lower in the group than H [1]
(ii) K has an extra occupied shell, so its outer electrons are further from the nucleus and are shielded by more inner shells; the attraction on them is weaker so they are lost more easily [1]
(iii) The oxide of K gives the more alkaline solution, because the more reactive metal forms the more strongly basic oxide and its hydroxide is more soluble [1]
⚠ If you missed marks here: Note that the higher melting point of K is not evidence of greater reactivity; melting point and chemical reactivity are separate properties. The reasoning must go through shells, distance and shielding.
(c)[3]
A sixth element, M, has proton number 4.
(i) State the group and period of M. [1]
(ii) Predict the formula of the chloride of M. [1]
(iii) Predict how the reactivity of M compares with that of H, and explain why. [1]
Model Answer -- 7(c)
(i) The configuration of M is 2,2, so M is in Group II and Period 2 [1]
(ii) MCl₂, because M forms a 2⁺ ion which needs two chloride ions to balance it [1]
(iii) M is less reactive than H, because M lies above H in Group II; its outer electrons are closer to the nucleus and less shielded, so they are held more tightly and lost less easily [1]
⚠ If you missed marks here: Work out the electron configuration first — a proton number of 4 gives 2,2, so both the group and the period follow immediately. In (iii) the reasoning must be the mirror image of the argument used for going down the group.
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