IGCSE Chemistry Paper 4 (Theory / Extended)

Topic 8: The Periodic Table -- Challenge Paper

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0620

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

Question 1 -- Identifying Four Elements from Combined Evidence

Total: 12 marks

A research student in Zurich is given four sealed samples labelled only P, Q, R and S, together with the following notes made by a previous worker.

sample	observations recorded
P	soft enough to cut with a knife; floats on water and reacts at once, giving a gas and a solution of pH 14; burns with a lilac flame
Q	dense dark red-brown liquid; turns a colourless iodide solution dark brown, but has no effect on a colourless chloride solution
R	density 8.9 g cm⁻³; melts at 1085 °C; forms one chloride that dissolves to a blue solution and a second chloride that is white
S	a gas of separate single atoms; boils at −186 °C; forms no compounds with any reagent tried

(a) [4]

State the group, or region, of the Periodic Table to which each of P, Q, R and S belongs, and give the single strongest piece of evidence for each. [4]

Model Answer -- 1(a)

P is a Group I metal: only the alkali metals are soft enough to cut, float on water and react immediately to give a strongly alkaline solution; the lilac flame identifies it as potassium [1]

Q is a Group VII halogen: the only common element that is a dark red-brown liquid is bromine, and it displaces iodine but not chlorine, which fixes its position between them [1]

R is a transition element: the high density and high melting point rule out a main-group metal, and two chlorides with different colours show variable oxidation number [1]

S is a Group VIII noble gas: it is monatomic and forms no compounds at all, which no other kind of element does [1]

⚠ If you missed marks here: The question asks for the strongest single piece of evidence, so quoting a property shared by many elements, such as "R is dense", is weak; the two differently coloured chlorides are what really identify a transition metal. Watch that for Q it is the pair of results together, not either one alone, that pins down its position.

1 mark -- P Group I soft floats pH 14 lilac flame potassium 1 mark -- Q Group VII bromine red brown liquid displaces iodide not chloride 1 mark -- R transition high density melting point two chlorides variable oxidation 1 mark -- S noble gas monatomic forms no compounds Group VIII

(b) [4]

(i) Give the formula of the compound formed between P and Q, name the type of bonding, and explain how the bonding forms. [2]

(ii) Write the balanced symbol equation for the reaction of Q with aqueous sodium iodide, and name the type of reaction. [2]

Model Answer -- 1(b)

(i) The formula is PQ, since a 1⁺ ion and a 1⁻ ion combine in a 1 : 1 ratio; using potassium and bromine the compound is KBr [1]

(i) The bonding is ionic: the single outer electron of P is transferred to Q, completing Q's outer shell, and the oppositely charged ions then attract one another strongly [1]

(ii) Br₂ + 2NaI → 2NaBr + I₂ [1]

(ii) It is a displacement reaction, and also a redox reaction, since iodide is oxidised while bromine is reduced [1]

⚠ If you missed marks here: Identifying P and Q from part (a) first makes the equation much easier; candidates who leave everything in letters often forget that Q exists as Q₂ molecules and so write an unbalanced equation. "Neutralisation" is a common wrong answer for the reaction type — no acid or base is involved anywhere here.

1 mark -- formula PQ one to one KBr 1 mark -- ionic bonding electron transferred outer shell completed attraction 1 mark -- Br2 2NaI 2NaBr I2 balanced 1 mark -- displacement redox iodide oxidised bromine reduced

(i) Explain why R can form two different chlorides but P can form only one. [2]

(ii) Predict what would be observed if S were bubbled through a solution of the compound of P and Q, and explain your prediction. [1]

(iii) State which of P, Q and R would be most likely to act as a catalyst in an industrial process, and why. [1]

Model Answer -- 1(c)

(i) R is a transition element, and such elements can form ions of more than one charge, so more than one chloride is possible — here one with the metal in the +1 state and one in the +2 state [1]

(i) P has only one electron in its outer shell, so it can lose only that one electron; removing a second would mean breaking into a complete inner shell, which needs far more energy than any reaction can supply [1]

(ii) Nothing at all would be seen: S has a complete outer shell, so it has no tendency to lose, gain or share electrons and simply bubbles through unchanged [1]

(iii) R, because transition elements and their compounds are the most widely used industrial catalysts, for example iron in the Haber process and nickel in hydrogenation [1]

⚠ If you missed marks here: In (i) the second mark is for explaining why P is limited, not just for saying that it is; the argument about breaking into a full inner shell is what examiners look for. In (ii), "no reaction" alone is worth nothing without the reason.

1 mark -- R transition ions of more than one charge two chlorides 1 mark -- P one outer electron second removal breaks full inner shell too much energy 1 mark -- nothing seen full outer shell no tendency lose gain share unchanged 1 mark -- R catalyst transition elements Haber nickel hydrogenation

Question 2 -- A Quantitative Group I Investigation

Total: 12 marks

A student in Melbourne drops a 0.10 g piece of each of three alkali metals into water and collects the hydrogen produced in an inverted measuring cylinder held over the reaction. Her results are shown below. (Aₓ: Li = 7, Na = 23, K = 39; one mole of gas occupies 24 dm³ at r.t.p.)

metal	mass used / g	volume of gas collected / cm³
lithium	0.10	168
sodium	0.10	51
potassium	0.10	19

(a) [4]

(i) Lithium is the least reactive of the three metals, yet it produced by far the largest volume of gas. Explain why. [2]

(ii) Calculate the volume of hydrogen that 0.10 g of sodium should produce, and comment on how well the measured value agrees. [2]

Model Answer -- 2(a)

(i) The volume of gas depends on the number of moles of metal reacting, not on how violent the reaction looks; reactivity governs the rate, not the yield [1]

(i) Lithium has by far the smallest relative atomic mass, so 0.10 g of it contains about 5.6 times as many atoms as 0.10 g of potassium, and each atom releases the same share of hydrogen [1]

(ii) Moles of sodium = 0.10 ÷ 23 = 4.35 × 10⁻³ mol; the 2 : 1 ratio gives 2.17 × 10⁻³ mol of hydrogen [1]

(ii) Volume = 2.17 × 10⁻³ × 24 = 0.052 dm³, or about 52 cm³, which agrees very closely with the measured 51 cm³ [1]

⚠ If you missed marks here: The trap in (i) is assuming that the more vigorous reaction must give more product; separate rate from amount in your mind. In (ii), forgetting the 2 : 1 ratio doubles the answer, and the comment on agreement is worth a mark in its own right, so do not leave it out.

1 mark -- volume depends on moles not vigour reactivity governs rate 1 mark -- lithium smallest relative atomic mass more atoms in same mass 1 mark -- moles sodium 0.00435 moles hydrogen 0.00217 two to one 1 mark -- volume 52 cm3 agrees closely with measured 51

(b) [5]

The measured value for potassium, 19 cm³, is well below the value calculated from its mass.

(i) Suggest two reasons connected with the reaction itself that would make the collected volume too low. [2]

(ii) State one variable the student needed to control, and explain why. [1]

(iii) Suggest an improvement to the apparatus and explain how it would make the results more reliable. [2]

Model Answer -- 2(b)

(i) The reaction with potassium is so fast that much of the hydrogen escapes before the measuring cylinder can be positioned over it [1]

(i) The heat released ignites some of the hydrogen, which burns to water instead of being collected; the metal may also skate out from under the cylinder [1]

(ii) The temperature of the water must be kept the same for each metal, because gas volume depends on temperature and a warmer sample would read high; equally the mass of metal and the volume of water should be fixed [1]

(iii) Use a sealed conical flask fitted with a bung and delivery tube leading to a gas syringe, so that no gas can escape at any stage [1]

(iii) A gas syringe also reads to the nearest 0.5 cm³ rather than to the nearest 1 or 2 cm³, and repeating each run and taking a mean would reduce the effect of random error [1]

⚠ If you missed marks here: "Human error" and "the equipment was inaccurate" are never credited; every suggestion must identify a specific mechanism by which gas was lost or a reading was distorted. In (iii) the improvement and its effect are marked separately, so state both.

1 mark -- reaction too fast gas escapes before cylinder positioned 1 mark -- hydrogen ignites burns metal skates away from cylinder 1 mark -- control temperature same mass volume water gas volume depends 1 mark -- sealed flask bung delivery tube gas syringe no escape 1 mark -- syringe finer scale repeat runs mean reduces random error

(i) Calculate the volume of hydrogen that 0.10 g of rubidium (Aₓ = 85.5) would produce if all of it reacted and all the gas were collected. [2]

(ii) State one additional hazard that would arise with rubidium and the precaution it makes necessary. [1]

Model Answer -- 2(c)

(i) Moles of rubidium = 0.10 ÷ 85.5 = 1.17 × 10⁻³ mol, so moles of hydrogen = 5.85 × 10⁻⁴ mol [1]

(i) Volume = 5.85 × 10⁻⁴ × 24 = 0.014 dm³, which is about 14 cm³ [1]

(ii) Rubidium reacts explosively with water and the hydrogen ignites at once, so the experiment could only be done behind a safety screen with a very small piece — in practice it is not performed in schools at all [1]

⚠ If you missed marks here: Notice that the calculated volume keeps falling as the group is descended, purely because the relative atomic mass keeps rising; this is a good check that your arithmetic follows the pattern of the earlier results. The hazard mark requires a matching precaution, not just the word "explosive".

1 mark -- moles rubidium 0.00117 moles hydrogen 0.000585 1 mark -- volume about 14 cm3 1 mark -- explosive hydrogen ignites safety screen tiny piece not done in schools

Question 3 -- Predicting the Properties of Francium and Astatine

Total: 12 marks

Francium and astatine sit at the bottom of Group I and Group VII respectively. Both are intensely radioactive and only a few thousand atoms of either exist on Earth at any moment, so almost everything known about them has been predicted rather than measured.

Group I	melting point / °C	density / g cm⁻³	Group VII	melting point / °C	colour
lithium	181	0.53	chlorine	−101	pale yellow-green
sodium	98	0.97	bromine	−7	red-brown
potassium	63	0.86	iodine	114	grey-black
rubidium	39	1.53	astatine	?	?
caesium	29	1.93

(a) [4]

Use the Group I data to predict, with a reason in each case, the following properties of francium.

(i) Its melting point. [1]

(ii) Its density. [1]

(iii) How its atomic radius compares with that of caesium. [1]

(iv) What would be seen if a small piece were dropped into water, and the pH of the resulting solution. [1]

Model Answer -- 3(a)

(i) Melting point below 29 °C, perhaps 25 °C or a little lower, since the values fall steadily down the group and the differences are getting smaller [1]

(ii) Density above 1.93 g cm⁻³, perhaps about 2.4 g cm⁻³, since density rises down the group apart from the small dip at potassium [1]

(iii) Its atomic radius would be the largest in the group, because francium has one more occupied electron shell than caesium [1]

(iv) The reaction would be explosively violent, the metal vanishing instantly with the hydrogen igniting; the solution would be francium hydroxide, FrOH, with a pH of 14 [1]

⚠ If you missed marks here: Predictions must be numerical where the question implies a value, and each one must follow the direction of the trend in the table. Note the potassium density anomaly: quoting the trend as "always increases" is not quite right, and the best answers acknowledge it.

1 mark -- melting point below 29 falls down group differences smaller 1 mark -- density above 1.93 rises down group 1 mark -- largest atomic radius one more occupied shell than caesium 1 mark -- explosive hydrogen ignites francium hydroxide FrOH pH 14

(b) [4]

Now use the Group VII data to predict, with a reason in each case, the following properties of astatine.

(i) Its colour. [1]

(ii) Its melting point and physical state at room temperature. [1]

(iii) Its position in the reactivity order of the halogens. [1]

(iv) What would be observed if astatine solution were added to aqueous sodium iodide. [1]

Model Answer -- 3(b)

(i) Black, or so dark as to appear black, because the colours deepen steadily from pale yellow-green through red-brown to grey-black [1]

(ii) A solid at room temperature melting somewhere above 114 °C, probably near 300 °C, since the melting points rise down the group and the gaps are widening [1]

(iii) It would be the least reactive halogen of all, because reactivity decreases down Group VII [1]

(iv) No change would be seen: astatine lies below iodine, so it cannot displace iodine from an iodide solution [1]

⚠ If you missed marks here: The two groups run in opposite directions for reactivity, and candidates who have just answered part (a) frequently carry the Group I pattern across and predict that astatine is the most reactive halogen. Check the direction of each trend separately before you write.

1 mark -- black very dark colours deepen down group 1 mark -- solid melting point above 114 near 300 gaps widening 1 mark -- least reactive halogen reactivity decreases down group 1 mark -- no change below iodine cannot displace iodide

(i) Explain how one single idea about atomic structure accounts for reactivity increasing down Group I but decreasing down Group VII. [2]

(ii) A student predicts that francium and astatine would combine to give FrAt, "the most ionic compound that could ever exist". Comment on whether the formula is justified, and give one reason why the compound could never be prepared in a weighable amount. [2]

Model Answer -- 3(c)

(i) In both groups, going down means one more occupied shell, so the outer shell is further from the nucleus and is screened by more inner shells, and the net attraction between the nucleus and the outer region falls [1]

(i) A Group I atom reacts by losing an electron, which that weaker attraction makes easier, so reactivity rises; a Group VII atom reacts by gaining an electron, which the same weaker attraction makes harder, so reactivity falls [1]

(ii) The formula FrAt is justified: francium would form a 1⁺ ion and astatine a 1⁻ ion, so they combine in a 1 : 1 ratio, and the pairing of the most easily oxidised metal with a halogen would indeed give highly ionic bonding [1]

(ii) Neither element has a stable isotope; both are intensely radioactive with half-lives of minutes or less and exist only as a handful of atoms at a time, so no weighable sample of either element, let alone of the compound, could ever be assembled [1]

⚠ If you missed marks here: Part (i) is asking you to unify two trends under one cause, so an answer that simply describes each trend separately will not score. In (ii), be careful to separate the chemistry, which is sound, from the practical objection, which is about radioactivity and scarcity rather than about bonding.

1 mark -- one more shell further from nucleus more shielding attraction falls 1 mark -- losing electron easier Group I gaining harder Group VII opposite trends 1 mark -- FrAt justified 1 plus and 1 minus ions one to one highly ionic 1 mark -- both radioactive short half lives few atoms no weighable sample

Question 4 -- Designing and Evaluating a Displacement Investigation

Total: 12 marks

A student in Toronto sets out to place chlorine, bromine and iodine in order of reactivity by adding each halogen solution to each of three potassium halide solutions. One of her nine results does not fit the pattern.

added to →	KCl(aq)	KBr(aq)	KI(aq)
chlorine water	no change	turns orange	turns dark brown
bromine water	no change	no change	turns dark brown
iodine solution	no change	turns pale orange	no change

(a) [3]

(i) State which result is anomalous, and give the result that should have been obtained. [1]

(ii) Explain why the recorded result cannot be correct. [1]

(iii) Suggest one practical cause of the anomalous result. [1]

Model Answer -- 4(a)

(i) The anomalous result is iodine solution with potassium bromide, recorded as turning pale orange; it should have shown no change [1]

(ii) An orange colour would mean that bromine had been set free, but iodine lies below bromine in Group VII and is the less reactive of the two, so it cannot take electrons from bromide ions [1]

(iii) A dropping pipette used earlier for chlorine or bromine water may not have been rinsed, contaminating the tube; alternatively the pale orange colour of the iodine solution itself may have been mistaken for a change [1]

⚠ If you missed marks here: Identifying the odd result is only the first mark; the explanation must refer to the relative positions of the two halogens in the group. Suggested causes must be specific and practical — "she made a mistake" is not creditworthy.

1 mark -- iodine with potassium bromide anomalous should be no change 1 mark -- orange means bromine freed iodine below bromine less reactive cannot 1 mark -- contaminated pipette not rinsed or iodine own colour mistaken

(b) [5]

(i) Describe how the investigation should be carried out so that the results are valid, including two variables that must be controlled. [3]

(ii) Give two safety precautions needed and a reason for each. [2]

Model Answer -- 4(b)

(i) Place 2 cm³ of each halide solution in a separate labelled test tube, add 2 cm³ of the halogen solution with a clean pipette for each transfer, stopper and shake, then record the colour against a white background [1]

(i) Set up a control tube containing the halide solution alone, so that the original colour is known and any change can be judged reliably [1]

(i) Controlled variables: the concentration and the volume of every solution must be the same, and all tubes must be at the same temperature [1]

(ii) Wear eye protection and work in a fume cupboard, because chlorine and bromine vapours are toxic and attack the eyes and lungs [1]

(ii) Use only small volumes, keep the tubes stoppered and pour waste into the labelled halogen residues container, because the solutions are harmful and staining and should not enter the sink [1]

⚠ If you missed marks here: Given the anomaly in part (a), a clean pipette for every transfer and a control tube are exactly the details the examiner is hoping to see. Safety marks are only awarded when the precaution is tied to a specific hazard, so "be careful" scores nothing.

1 mark -- equal volumes labelled tubes clean pipette shake record against white 1 mark -- control tube halide alone original colour comparison 1 mark -- same concentration volume temperature controlled variables 1 mark -- eye protection fume cupboard chlorine bromine vapours toxic 1 mark -- small volumes stoppered halogen residues container harmful

(i) Using only the valid results, place the three halogens in order of reactivity, most reactive first. [1]

(ii) Write the ionic equation for the reaction between chlorine and aqueous potassium bromide, and identify the oxidising agent, justifying your choice. [2]

(iii) The student proposes extending the method to place astatine in the order. Explain why this would not work in practice. [1]

Model Answer -- 4(c)

(i) Chlorine, then bromine, then iodine [1]

(ii) Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) [1]

(ii) Chlorine is the oxidising agent, because it takes electrons from the bromide ions and so causes their oxidation, while chlorine itself gains electrons and is reduced [1]

(iii) Astatine has no stable isotope and exists only as a few thousand atoms worldwide at any moment; no solution of a weighable amount could be made, and even if it could, the colour change from so few atoms would be invisible [1]

⚠ If you missed marks here: The oxidising agent is the species that is itself reduced, which trips up many candidates — work out which species gains electrons first, then name that one as the oxidising agent. Remember to exclude the anomalous result when deducing the order.

1 mark -- order chlorine bromine iodine most reactive first 1 mark -- ionic equation chlorine bromide chloride bromine state symbols 1 mark -- chlorine oxidising agent takes electrons itself reduced 1 mark -- astatine no stable isotope few atoms no solution colour invisible

Question 5 -- Deducing Variable Oxidation Number

Total: 10 marks

An analyst in Pune is given two chlorides of the same metal M and determines the percentage of M by mass in each. Chloride A contains 44.1 % M and chloride B contains 34.4 % M. Take Aₓ(M) = 56 and Aₓ(Cl) = 35.5.

(a) [4]

(i) Show by calculation that the formula of A is MCl₂. [2]

(ii) Show by calculation that the formula of B is MCl₃. [1]

(iii) State what these two formulae tell you about the position of M in the Periodic Table. [1]

Model Answer -- 5(a)

(i) In 100 g of A there are 44.1 g of M, which is 44.1 ÷ 56 = 0.788 mol, and 55.9 g of chlorine, which is 55.9 ÷ 35.5 = 1.575 mol [1]

(i) Dividing by the smaller value gives a ratio of 1 : 2.00, so the formula of A is MCl₂ [1]

(ii) In 100 g of B there are 34.4 ÷ 56 = 0.614 mol of M and 65.6 ÷ 35.5 = 1.848 mol of chlorine, a ratio of 1 : 3.01, so B is MCl₃ [1]

(iii) M forms ions of two different charges, 2⁺ and 3⁺, so it shows variable oxidation number and must be a transition element from the central block [1]

⚠ If you missed marks here: Remember that the chlorine percentage is 100 minus the percentage of M — forgetting this step is the commonest error. Always divide both mole values by the smaller one; comparing the raw moles directly does not give the formula.

1 mark -- 44.1 divided 56 gives 0.788 and 55.9 divided 35.5 gives 1.575 moles 1 mark -- ratio one to two formula MCl2 1 mark -- 0.614 and 1.848 ratio one to three formula MCl3 1 mark -- two charges 2 plus 3 plus variable oxidation number transition central block

(b) [3]

(i) Predict two further properties, other than variable oxidation number, that M would be expected to show. [2]

(ii) Name the element M. [1]

Model Answer -- 5(b)

(i) Its compounds would be coloured — here a pale green solution for the 2⁺ state and an orange-brown one for the 3⁺ state [1]

(i) It would have a high density and a high melting point, and it or its compounds could act as a catalyst [1]

(ii) M is iron, since Aₓ = 56 and it forms FeCl₂ and FeCl₃ [1]

⚠ If you missed marks here: The question rules out variable oxidation number, so repeating it in another form earns nothing. Note how the two colours predicted in (i) match the two chlorides in the stem — using the data given always strengthens an answer.

1 mark -- coloured compounds pale green orange brown solutions 1 mark -- high density high melting point catalyst 1 mark -- M is iron FeCl2 FeCl3

A student comments: "Sodium ought to form NaCl and also NaCl₂, if only enough energy were supplied."

Evaluate this comment, explaining what is right and what is wrong about it, and state what the comparison shows about the difference between Group I metals and transition metals. [3]

Model Answer -- 5(c)

NaCl₂ is not formed: sodium has only one electron in its outer shell, so only one electron is available to be lost in ordinary chemistry [1]

Removing a second electron would mean taking it from the complete shell beneath, which is much closer to the nucleus and far more strongly held; the energy needed is far greater than any chemical reaction could release, so the compound would never be stable [1]

This is exactly why Group I metals show only the 1⁺ state while transition metals such as M show two or more: in a transition element the electrons available for bonding lie in shells of very similar energy, so removing an extra one costs comparatively little [1]

⚠ If you missed marks here: An answer that just says "sodium is always 1⁺" is a statement, not an evaluation. The mark scheme rewards the energy argument — why a second electron is so much harder to remove — and the explicit contrast with the transition block.

1 mark -- NaCl2 not formed sodium only one outer electron available 1 mark -- second electron from complete inner shell closer strongly held energy too great 1 mark -- Group I only 1 plus transition two or more similar energy shells

Question 6 -- The Discovery and Behaviour of the Noble Gases

Total: 12 marks

In 1894 Lord Rayleigh found that nitrogen separated from air was consistently denser than nitrogen prepared by decomposing a nitrogen compound. The two densities, measured at the same temperature and pressure, were 1.2572 g dm⁻³ and 1.2505 g dm⁻³ respectively. Working with William Ramsay, he traced the difference to a new element.

(a) [4]

(i) Explain what the difference in density suggested about the gas obtained from air. [2]

(ii) Outline how the new gas could be separated from a sample of air and how the workers could tell that it was a genuinely new element. [2]

Model Answer -- 6(a)

(i) The gas from a compound is pure nitrogen, so the sample from air, being denser, must contain nitrogen mixed with something else that is denser than nitrogen [1]

(i) The difference of 0.0067 g dm⁻³ is about 0.5 %, which is far larger than the uncertainty of the measurements, so it could not be dismissed as experimental error [1]

(ii) Remove the oxygen, carbon dioxide and water vapour chemically, then pass the remaining gas repeatedly over hot magnesium, which combines with nitrogen to form solid magnesium nitride; a small residue of gas is left behind [1]

(ii) The residue reacted with nothing they tried, and its emission spectrum showed lines that matched no known element, so it had to be a new one [1]

⚠ If you missed marks here: The second mark in (i) is for recognising that the size of the discrepancy matters — a difference in the fourth decimal place would have proved nothing. In (ii), "they tested it" is too vague; describe the removal of nitrogen and the evidence from the spectrum.

1 mark -- gas from compound pure nitrogen air sample contains denser component 1 mark -- difference 0.0067 about 0.5 percent larger than experimental uncertainty 1 mark -- remove oxygen carbon dioxide water pass over hot magnesium nitride residue 1 mark -- residue reacted with nothing new spectrum lines unknown element

(b) [4]

(i) Explain, in terms of particles, why argon gas is denser than nitrogen gas at the same temperature and pressure. [1]

(ii) Explain why argon exists as separate atoms whereas nitrogen exists as N₂ molecules. [1]

(iii) The boiling points of the noble gases are helium −269 °C, neon −246 °C, argon −186 °C and krypton −152 °C. Describe and explain this trend. [2]

Model Answer -- 6(b)

(i) Equal volumes of gases contain equal numbers of particles, and an argon atom has a mass of 40 compared with 28 for an N₂ molecule, so the same volume of argon weighs more [1]

(ii) Argon already has a complete outer shell of eight electrons, so bonding to another atom would gain it nothing; a nitrogen atom has five outer electrons and completes its shell by sharing three pairs with a second nitrogen atom [1]

(iii) The boiling point rises steadily down the group, from −269 °C to −152 °C — remember that a less negative value is a higher temperature [1]

(iii) The atoms become larger down the group and the attractive forces between them become stronger, so more energy is needed to separate them into a gas [1]

⚠ If you missed marks here: Negative numbers are the classic trap in (iii): many candidates read the figures as falling and describe the trend backwards. In (i) the answer must compare the mass of one argon atom with that of a whole N₂ molecule, not with a single nitrogen atom.

1 mark -- equal volumes equal numbers particles argon 40 nitrogen molecule 28 1 mark -- argon complete outer shell nothing gained nitrogen shares three pairs 1 mark -- boiling point rises down group less negative is higher 1 mark -- larger atoms stronger attraction more energy to separate

(i) Titanium welded in an atmosphere of nitrogen becomes brittle, whereas titanium welded under argon does not. Explain this difference. [2]

(ii) Evaluate the statement "the noble gases never form compounds". [2]

Model Answer -- 6(c)

(i) Nitrogen is unreactive at room temperature but not at welding temperatures, where it combines with hot titanium to form a hard, brittle nitride within the metal [1]

(i) Argon has a complete outer shell and reacts with nothing at any temperature, so it forms a genuinely inert blanket that keeps both oxygen and nitrogen away from the weld [1]

(ii) The statement is very nearly true and is a good working rule: the noble gases take part in essentially no ordinary chemistry, and helium, neon and argon form no compounds at all [1]

(ii) It is not strictly true, however, because krypton and xenon do form fluorides such as XeF₄ under extreme conditions; in these larger atoms the outer electrons lie further from the nucleus and are heavily shielded, so a partner as reactive as fluorine can force them into bonding [1]

⚠ If you missed marks here: An evaluation needs both sides: say what is right about the statement before explaining the exception. Simply writing "false, xenon forms XeF₄" misses the mark for recognising that the rule holds for the lighter members and for almost all conditions.

1 mark -- nitrogen reacts hot titanium brittle nitride formed 1 mark -- argon complete shell reacts with nothing inert blanket keeps oxygen away 1 mark -- statement nearly true good working rule helium neon argon no compounds 1 mark -- krypton xenon fluorides XeF4 outer electrons shielded fluorine forces bonding

Question 7 -- Placing an Unfamiliar Element

Total: 10 marks

Element Z has proton number 31 and lies in Period 4. A sample supplied to a laboratory in Lagos is found to be a soft, shiny solid that conducts electricity and melts in the hand at about 30 °C. It forms a single chloride, ZCl₃, and a single oxide, Z₂O₃; the oxide dissolves both in dilute hydrochloric acid and in hot sodium hydroxide solution. All its compounds are colourless.

(a) [4]

(i) Deduce the group of Z, giving two independent pieces of evidence. [2]

(ii) State whether Z is a metal or a non-metal, and justify your answer using two pieces of evidence. [2]

Model Answer -- 7(a)

(i) Z is in Group III, because the formulae ZCl₃ and Z₂O₃ both show that Z forms a 3⁺ ion, so it must have three outer-shell electrons to lose [1]

(i) The proton number of 31 gives the configuration 2,8,18,3, which again shows three electrons in the outer shell and places Z in Period 4 [1]

(ii) Z is a metal: it is shiny and it conducts electricity, and its oxide reacts with dilute acid to give a salt, which is basic behaviour [1]

(ii) The oxide also dissolves in hot alkali, so it is amphoteric; this places Z close to the stepped line between metals and non-metals, exactly where a Group III metal in the middle of the table would be expected [1]

⚠ If you missed marks here: The very low melting point tempts candidates to call Z a non-metal, but conductivity and the reaction of its oxide with acid settle the matter; melting point alone never decides whether an element is metallic. Amphoteric behaviour is a clue about position, not evidence against being a metal.

1 mark -- Group III ZCl3 Z2O3 forms 3 plus ion three outer electrons 1 mark -- proton number 31 configuration 2,8,18,3 Period 4 1 mark -- metal shiny conducts electricity oxide reacts with acid basic 1 mark -- oxide dissolves in alkali amphoteric near stepped line borderline

(b) [3]

Aluminium is also in Group III.

(i) State one similarity you would expect between the compounds of Z and those of aluminium. [1]

(ii) Predict which of the two elements is the more reactive, and explain why. [1]

(iii) Aluminium melts at 660 °C but Z melts at 30 °C. Comment on what this shows about using group trends to predict physical properties. [1]

Model Answer -- 7(b)

(i) Both have three outer-shell electrons, so their compounds have matching formulae — AlCl₃ and ZCl₃, Al₂O₃ and Z₂O₃ — and both oxides are amphoteric [1]

(ii) Z should be the more reactive metal, because it lies below aluminium in the group, so its outer electrons are further from the nucleus and shielded by more inner shells and are therefore lost more easily [1]

(iii) Chemical reactivity and melting point do not have to follow the same pattern; melting point depends on the strength of the metallic bonding in the solid, so a group trend that is reliable for chemical behaviour may not carry across to physical properties [1]

⚠ If you missed marks here: Part (iii) is asking for a comment about the method of prediction itself, not another fact about Z; the mark is for recognising that different properties depend on different things and that trends must be checked property by property.

1 mark -- three outer electrons same formulae AlCl3 ZCl3 amphoteric oxides 1 mark -- Z more reactive below aluminium electrons further shielded lost easily 1 mark -- reactivity and melting point different patterns metallic bonding trends checked separately

(i) Predict the formula of the compound formed between Z and a Group VI element T. [1]

(ii) Predict the formula of the compound formed between Z and a Group VII element X. [1]

(iii) Z lies in Period 4, where the transition elements are also found. Explain why Z cannot itself be a transition element. [1]

Model Answer -- 7(c)

(i) Z₂T₃, because two 3⁺ ions balance three 2⁻ ions [1]

(ii) ZX₃, because one 3⁺ ion needs three 1⁻ ions to balance it [1]

(iii) Z forms only one chloride and one oxide, so it has a single oxidation number, and all its compounds are colourless; both facts rule out the transition block, which lies between Groups II and III and is defined by variable oxidation number and coloured compounds [1]

⚠ If you missed marks here: Being in Period 4 is not enough to make an element a transition metal — the central block occupies only part of that period. The strongest answers quote both the single oxidation number and the absence of colour.

1 mark -- formula Z2T3 two 3 plus balance three 2 minus 1 mark -- formula ZX3 one 3 plus needs three 1 minus 1 mark -- single oxidation number colourless compounds outside central block

Self-Assessment

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