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A-Level · Chemistry

AC6 The Periodic Table and Inorganic Chemistry

AQA 7402/7405/7408 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question is about trends across Period 3, from sodium (Na) to argon (Ar).

(a)State how atomic radius changes from sodium to chlorine across Period 3.(1)

(b)Explain why atomic radius decreases across Period 3 from sodium to chlorine.(2)

(c)The first ionisation energy generally increases across Period 3, but there is a decrease between magnesium and aluminium. Explain this decrease.(3)

(d)There is also a decrease in first ionisation energy between phosphorus and sulfur. Explain this decrease.(2)

(Total for Question 1 is 8 marks)

The table below gives approximate melting points for four Period 3 elements.
Na: 371 K Mg: 923 K Al: 933 K Si: 1687 K

(a)Explain, in terms of structure and bonding, why the melting point increases from sodium to magnesium to aluminium.(3)

(b)Silicon has a much higher melting point than aluminium. Explain why.(2)

(c)The melting point drops sharply between silicon and phosphorus (P4 melts at 317 K). Explain why.(2)

(d)Predict, with a reason, how the melting points of P4, S8 and Cl2 compare with one another.(3)

(Total for Question 2 is 10 marks)

This question is about the oxides of Period 3 elements and their reactions with water.

(a)Write an equation for the reaction of sodium oxide, Na2O, with water, and state the approximate pH of the resulting solution.(2)

(b)Write an equation for the reaction of phosphorus(V) oxide, P4O10, with water, and state the approximate pH of the resulting solution.(2)

(c)Aluminium oxide, Al2O3, does not react with water, but it reacts with both dilute hydrochloric acid and dilute sodium hydroxide solution. Write an equation for each of these two reactions and state the term used to describe an oxide that behaves in this way.(4)

(d)Using your answers to parts (a) to (c), deduce and explain the overall pattern in the acid-base character of Period 3 oxides as you move from sodium oxide to phosphorus(V) oxide.(3)

(Total for Question 3 is 11 marks)

A student reacts small samples of magnesium ribbon and calcium metal, in turn, with cold water.

(a)Write a balanced equation, including state symbols, for the reaction of calcium with cold water.(2)

(b)Describe the observations you would expect when comparing magnesium and calcium each added to cold water.(2)

(c)Explain why reactivity with water increases down Group 2 from magnesium to calcium.(3)

(d)Magnesium does not react with cold water but does react with steam. Write a balanced equation for the reaction of magnesium with steam, and explain why the product formed is different from the product formed when calcium reacts with cold water.(3)

(Total for Question 4 is 10 marks)

This question is about solubility trends of Group 2 compounds and a required practical test for sulfate ions.

(a)State how the solubility of Group 2 hydroxides changes down the group, and use this to explain why magnesium hydroxide, rather than barium hydroxide, is used as the active ingredient in indigestion remedies.(2)

(b)State how the solubility of Group 2 sulfates changes down the group, and explain why barium sulfate, despite being an insoluble compound of a toxic metal, can be used safely as a 'barium meal' in medical X-ray imaging.(2)

(c)Describe how you would carry out a chemical test on a solution to confirm the presence of sulfate ions, SO4²-, including the reagents used, the order of addition, and the expected positive result.(3)

(d)A student adds excess barium chloride solution to 25.0 cm3 of 0.150 mol dm-3 magnesium sulfate solution. Calculate the mass, in g, of barium sulfate precipitate formed. (Mr of BaSO4 = 233.4)(3)

(Total for Question 5 is 10 marks)

This question uses practical techniques from the required practical on acid-base titrations. A student wants to determine the mass of magnesium hydroxide, Mg(OH)2, in an indigestion tablet. The tablet is crushed and reacted with 50.0 cm3 of 0.500 mol dm-3 hydrochloric acid, an amount known to be in excess.
Equation: Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
The excess (unreacted) acid is then titrated against 0.200 mol dm-3 sodium hydroxide solution, requiring a mean titre of 24.60 cm3.
(Mr of Mg(OH)2 = 58.3)

(a)Calculate the initial amount, in mol, of HCl added to the crushed tablet.(1)

(b)Calculate the amount, in mol, of NaOH used in the titration, and hence the amount, in mol, of HCl that was left unreacted (in excess) after reacting with the tablet.(2)

(c)Calculate the amount, in mol, of HCl that reacted with the Mg(OH)2 in the tablet, and hence the amount, in mol, of Mg(OH)2 present.(2)

(d)Calculate the mass, in g, of Mg(OH)2 in the tablet.(2)

(e)The tablet packaging states that each tablet contains 600 mg of Mg(OH)2. Evaluate whether the student's experimental result in part (d) supports this claim.(2)

(Total for Question 6 is 9 marks)

This question is about physical property trends of the Group 7 halogens: fluorine, chlorine, bromine and iodine.

(a)State the trend in boiling point down Group 7, and state the physical state at room temperature of chlorine, bromine and iodine.(2)

(b)Explain the trend in boiling point down Group 7.(3)

(c)Explain the trend in electronegativity down Group 7.(2)

(d)Fluorine is considerably more reactive and electronegative than would be predicted simply by extrapolating the Group 7 trend from chlorine. Suggest, with a reason, why fluorine's properties deviate from this trend.(2)

(Total for Question 7 is 9 marks)

This question is about halogen displacement and disproportionation reactions.

(a)Chlorine water is added to a colourless solution of potassium bromide. Write an ionic equation for the reaction that occurs and state the observation.(3)

(b)Explain, in terms of oxidising power, why chlorine displaces bromide ions from solution but iodine does not displace chloride ions from solution.(2)

(c)Chlorine gas is bubbled into cold, dilute sodium hydroxide solution. Write a balanced equation for this reaction and state one commercial use of the products formed.(4)

(d)Explain what is meant by the term 'disproportionation', using oxidation numbers to show the changes in oxidation state of chlorine in the reaction in part (c).(3)

(Total for Question 8 is 12 marks)

This question uses practical techniques from the required practical qualitative test for halide ions. A student is given a colourless solution and asked to identify which halide ion it contains. The student adds a few drops of dilute nitric acid, followed by silver nitrate solution, AgNO3(aq), and then tests the solubility of any precipitate formed in dilute and concentrated ammonia solution.

(a)State the purpose of adding dilute nitric acid to the solution before the silver nitrate solution.(1)

(b)Complete the following, stating the colour of the precipitate formed with silver nitrate solution, and its solubility in ammonia, for chloride ions, bromide ions and iodide ions.(3)

(c)Write an ionic equation for the reaction between silver ions and bromide ions.(1)

(d)A student carries out this test on an unknown solution, X, and observes a cream precipitate that dissolves only when concentrated ammonia solution is added (it does not dissolve in dilute ammonia). Identify the halide ion present in X and justify your answer.(2)

(e)Explain why this silver nitrate/ammonia test alone would not distinguish between a solution containing only chloride ions and a solution containing a mixture of chloride and bromide ions.(2)

(Total for Question 9 is 9 marks)

This question is about the reactions of solid sodium halides with concentrated sulfuric acid.

(a)Write an equation for the reaction of solid sodium chloride with concentrated sulfuric acid, and state the observation made.(2)

(b)With solid sodium bromide, the HBr initially formed is partly further oxidised by concentrated sulfuric acid, forming bromine and sulfur dioxide. Write an equation for this further reaction, and state the oxidation state of sulfur in H2SO4 and in SO2.(3)

(c)With solid sodium iodide, concentrated sulfuric acid is reduced much further, all the way to hydrogen sulfide, H2S. Write the overall equation for this reaction, and calculate the oxidation state of sulfur in H2SO4 and in H2S.(4)

(d)Explain why the reducing power of the halide ions increases down Group 7, from chloride to iodide.(3)

(Total for Question 10 is 12 marks)

A solid compound X is known to be a Group 2 metal halide, MX2, where M is one of Mg, Ca or Ba, and X is one of Cl, Br or I. Describe a series of chemical tests, with reagents, conditions and expected observations, that a student could use to identify both the metal cation and the halide anion present in solid compound X. Your answer should include the results expected for at least two possible cations and at least two possible anions.

(Total for Question 11 is 6 marks)

Group 2 carbonates decompose on heating to form the metal oxide and carbon dioxide gas.
(Molar volume of a gas at RTP = 24 dm3 mol-1; Mr of MgCO3 = 84.3)

(a)Write a balanced equation for the thermal decomposition of magnesium carbonate, MgCO3.(1)

(b)A 5.05 g sample of magnesium carbonate is heated until it decomposes completely. Calculate the volume, in dm3, of carbon dioxide gas produced, measured at room temperature and pressure (RTP).(3)

(c)Barium carbonate, BaCO3, requires a much higher temperature than magnesium carbonate to decompose completely. Explain this difference in terms of the polarising effect of the metal cation on the carbonate ion.(3)

(d)Predict, with a reason, whether strontium carbonate, SrCO3, would need a higher or lower temperature than magnesium carbonate to decompose completely.(2)

(Total for Question 12 is 9 marks)

Mark scheme · AC6 The Periodic Table and Inorganic Chemistry

Question 1

(a) B1 atomic radius decreases (across the period) oe
(a) Answer: Atomic radius decreases from Na to Cl
(b) B1 nuclear charge (number of protons) increases across the period
(b) B1 number of shells (and shielding from inner electrons) stays the same, so the increased nuclear charge pulls the outer electrons in more strongly, reducing the radius oe
(b) Answer: Nuclear charge increases across the period while the number of shells (shielding) stays constant, so outer electrons are pulled closer to the nucleus, decreasing atomic radius
(c) B1 in Mg the outer electron removed is from a full 3s subshell, but in Al the outer electron removed is from the 3p subshell oe
(c) B1 the 3p electron in Al is at a (slightly) higher energy than the 3s electrons and/or is shielded by the 3s electrons oe
(c) B1 so less energy is needed to remove the 3p electron from Al than to remove a 3s electron from Mg, giving a lower first ionisation energy despite Al having one more proton
(c) Answer: Aluminium's outer electron is removed from the higher-energy 3p subshell (shielded slightly by the 3s electrons), so it requires less energy to remove than a 3s electron from magnesium, even though Al has a greater nuclear charge
(d) B1 in P the 3p subshell is half-filled (3p3, one electron in each of three orbitals), which is an extra-stable arrangement oe
(d) B1 in S the fourth 3p electron must pair up with an existing electron in the same orbital; the repulsion between the two paired electrons makes it easier to remove one, lowering the first ionisation energy oe
(d) Answer: Phosphorus has a stable, half-filled 3p3 configuration; sulfur's fourth 3p electron must pair in an already-occupied orbital, and the extra electron-electron repulsion makes this electron easier to remove

Question 2

(a) B1 Na, Mg and Al are all giant metallic structures held together by metallic bonding oe
(a) B1 going from Na to Mg to Al, each atom contributes one more delocalised electron (1, then 2, then 3) to the 'sea' of electrons, and the ionic charge on the metal ion also increases (+1, +2, +3)
(a) B1 stronger electrostatic attraction between the metal cations and the increased number/density of delocalised electrons requires more energy to overcome, raising the melting point oe
(a) Answer: Na, Mg and Al are giant metallic lattices; going across the series the ionic charge and the number of delocalised electrons per atom both increase, strengthening the metallic bonding and so raising the melting point
(b) B1 silicon has a giant covalent (macromolecular) structure, in which each Si atom is joined to four others by strong covalent bonds oe
(b) B1 melting requires breaking many strong covalent bonds throughout the lattice, which needs much more energy than overcoming the (weaker, delocalised) metallic bonding in aluminium oe
(b) Answer: Silicon has a giant covalent lattice held together by strong covalent bonds in all directions; melting means breaking these strong bonds, which needs far more energy than overcoming the metallic bonding in aluminium
(c) B1 phosphorus (as P4) exists as small, simple covalent molecules, unlike the giant covalent lattice of silicon oe
(c) B1 only weak London (van der Waals) forces act between P4 molecules, and these require little energy to overcome, so the melting point is much lower oe
(c) Answer: P4 is a simple molecular substance; only weak London forces exist between the P4 molecules, which need much less energy to overcome than the strong covalent bonds in giant covalent silicon
(d) B1 predicted order: S8 has the highest melting point, then P4, then Cl2 the lowest (accept correct order stated either way round with reasoning)
(d) B1 melting point depends on the strength of London (van der Waals) forces between molecules, which increases with the number of electrons in the molecule / with relative molecular mass oe
(d) B1 S8 has the most electrons (largest Mr), then P4, then Cl2 has the fewest, so this order matches the strength of the London forces between molecules oe
(d) Answer: S8 (highest) > P4 > Cl2 (lowest); melting point rises with the number of electrons per molecule (London forces), and S8 has the most electrons per molecule, followed by P4, then Cl2

Question 3

(a) B1 Na2O + H2O -> 2NaOH cao (state symbols not required)
(a) B1 pH approximately 13-14 (accept 'strongly alkaline')
(a) Answer: Na2O + H2O -> 2NaOH; pH approximately 13-14
(b) B1 P4O10 + 6H2O -> 4H3PO4 cao
(b) B1 pH approximately 0-1 (accept 'strongly acidic')
(b) Answer: P4O10 + 6H2O -> 4H3PO4; pH approximately 0-1
(c) B1 Al2O3 + 6HCl -> 2AlCl3 + 3H2O cao
(c) B1 Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4 (sodium aluminate) oe (accept 2NaOH + Al2O3 -> 2NaAlO2 + H2O)
(c) B1 term: amphoteric
(c) B1 amphoteric means the oxide can react as both a base (with acid) and an acid (with alkali) oe
(c) Answer: Al2O3 + 6HCl -> 2AlCl3 + 3H2O; Al2O3 + 2NaOH + 3H2O -> 2NaAl(OH)4; amphoteric (reacts with both acids and bases)
(d) B1 the oxides change from strongly basic (Na2O, MgO) through amphoteric (Al2O3) to acidic then strongly acidic (SiO2 weakly/not reactive, P4O10 strongly acidic) across the period oe
(d) B1 this matches a change in bonding from ionic (Na2O, MgO, metal oxides) to covalent (SiO2, P4O10, non-metal oxides), with Al2O3 intermediate/having mixed ionic-covalent character oe
(d) B1 ionic oxides release OH- ions (or accept H+ readily) making solutions basic, while covalent oxides of non-metals react with water to form acidic solutions (releasing H+) oe
(d) Answer: Across Period 3 the oxides go from strongly basic (Na2O, MgO) to amphoteric (Al2O3) to acidic (SiO2, weakly) to strongly acidic (P4O10), reflecting the change in bonding from ionic (metal oxides) to covalent (non-metal oxides)

Question 4

(a) M1 correct formulae: Ca, H2O, Ca(OH)2, H2
(a) A1 Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g) fully balanced with correct state symbols cao
(a) Answer: Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)
(b) B1 magnesium reacts very slowly (if at all) with cold water, with few/no visible bubbles over a reasonable timescale oe
(b) B1 calcium reacts noticeably/steadily with cold water, producing a stream of bubbles (hydrogen gas) and forming a cloudy white suspension/precipitate of calcium hydroxide oe
(b) Answer: Magnesium shows almost no reaction with cold water; calcium reacts steadily, fizzing (releasing hydrogen gas) and forming a cloudy white suspension of calcium hydroxide
(c) B1 atomic radius increases down the group (extra electron shell for each successive element) oe
(c) B1 shielding by inner shells increases, which more than compensates for the increase in nuclear charge oe
(c) B1 so the outer two electrons are less strongly attracted to the nucleus / (first and second) ionisation energies decrease down the group, so the atoms lose their outer electrons more easily, making the metal more reactive oe
(c) Answer: Atomic radius and shielding increase down the group, outweighing the increase in nuclear charge; this lowers ionisation energies down the group, so the outer electrons are lost more easily, increasing reactivity
(d) B1 Mg(s) + H2O(g) -> MgO(s) + H2(g) cao
(d) B1 magnesium hydroxide, Mg(OH)2, is (much less soluble/almost insoluble) and unstable at the high temperature of steam, so it decomposes to the oxide oe
(d) B1 at the higher temperature of steam magnesium reacts fast enough to form the oxide directly, whereas at room temperature with cold water the reaction is far too slow to be observed oe
(d) Answer: Mg(s) + H2O(g) -> MgO(s) + H2(g); the reaction with steam occurs at a much higher temperature, forming the oxide (as any hydroxide formed is not stable/is far less soluble than Ca(OH)2), whereas cold water reacts too slowly with Mg to be observed

Question 5

(a) B1 solubility of Group 2 hydroxides increases down the group oe
(a) B1 Mg(OH)2 is only sparingly soluble, so it forms a mild suspension that safely neutralises small amounts of excess stomach acid without the solution becoming too concentrated/strongly alkaline, unlike the more soluble Ba(OH)2 oe
(a) Answer: Solubility of Group 2 hydroxides increases down the group; Mg(OH)2 is only sparingly soluble, giving a mild, safely-controlled antacid, whereas the more soluble Ba(OH)2 would give a dangerously concentrated alkaline solution
(b) B1 solubility of Group 2 sulfates decreases down the group oe
(b) B1 BaSO4 is so insoluble that essentially no Ba2+ ions are released into the body/bloodstream, so it is non-toxic even though soluble barium compounds are poisonous oe
(b) Answer: Solubility of Group 2 sulfates decreases down the group; BaSO4 is so insoluble that virtually no Ba2+ ions enter the bloodstream, making it safe despite the toxicity of soluble barium salts
(c) B1 add a few drops of dilute hydrochloric acid to the test solution first oe
(c) B1 then add barium chloride solution (BaCl2(aq)) oe
(c) B1 a (white) precipitate forms if sulfate ions are present (BaSO4); the dilute HCl added first removes/prevents a false positive from carbonate or sulfite ions, which would also give a white precipitate with BaCl2 but which dissolves in/reacts away with the acid oe
(c) Answer: Add dilute HCl to the solution first (to remove any carbonate/sulfite ions), then add BaCl2(aq); a white precipitate (BaSO4) confirms sulfate ions are present
(d) M1 n(MgSO4) = (25.0/1000) x 0.150 = 3.75 x 10^-3 mol
(d) M1 n(BaSO4) = n(MgSO4) = 3.75 x 10^-3 mol (1:1 ratio, BaCl2 in excess)
(d) A1 mass = 3.75e-3 x 233.4 = 0.875 g cao (awrt 0.875 g)
(d) Answer: 0.875 g

Question 6

(a) B1 0.0250 mol cao
(a) Answer: 0.0250 mol
(b) M1 n(NaOH) = 0.200 x (24.60/1000) = 4.92 x 10^-3 mol
(b) A1 n(excess HCl) = n(NaOH) = 4.92 x 10^-3 mol cao (1:1 ratio, NaOH + HCl -> NaCl + H2O)
(b) Answer: 4.92 x 10^-3 mol
(c) M1 n(HCl reacted with tablet) = 0.0250 - 4.92e-3 = 0.02008 mol ft from (a) and (b)
(c) A1 n(Mg(OH)2) = 0.02008/2 = 0.01004 mol cao (awrt 0.0100-0.0101 mol) ft
(c) Answer: 0.01004 mol
(d) M1 mass = 0.01004 x 58.3 ft from (c)
(d) A1 0.585 g cao (awrt 0.585 g) ft
(d) Answer: 0.585 g
(e) M1 percentage difference = (0.600 - 0.585)/0.600 x 100 = 2.5% ft from (d)
(e) A1 valid conclusion: the result (0.585 g) is close to the stated 0.600 g, within a small (about 2.5%) experimental difference, so it broadly supports the manufacturer's claim (allowing for rounding of atomic masses and experimental/titration error) oe ft
(e) Answer: The experimental mass (0.585 g) is about 2.5% below the stated 0.600 g, which is a small difference consistent with normal experimental error, so the result broadly supports the label claim

Question 7

(a) B1 boiling point increases down the group oe
(a) B1 chlorine is a gas, bromine is a liquid, iodine is a solid (all three needed) at room temperature
(a) Answer: Boiling point increases down the group; Cl2 is a gas, Br2 is a liquid, and I2 is a solid at room temperature
(b) B1 the halogens exist as simple diatomic molecules held together (between molecules) only by London (van der Waals/induced dipole-dipole) forces oe
(b) B1 the number of electrons per molecule increases down the group (F2 to I2), increasing the size of instantaneous/induced dipoles oe
(b) B1 this increases the strength of the London forces between molecules, so more energy is needed to separate the molecules, raising the boiling point oe
(b) Answer: Down the group the number of electrons per molecule increases, strengthening the London (van der Waals) forces between molecules, so more energy is needed to boil the substance, raising the boiling point
(c) B1 electronegativity decreases down the group oe
(c) B1 atomic radius (and shielding) increases down the group, so a bonding pair of electrons is less strongly attracted to the nucleus of the larger atom oe
(c) Answer: Electronegativity decreases down the group; increasing atomic radius and shielding mean the nucleus attracts a shared (bonding) pair of electrons less strongly
(d) B1 the fluorine atom is exceptionally small, so its outer electrons are extremely close to (and strongly attracted by) the nucleus, giving an unusually high charge density oe
(d) B1 this makes the F-F bond in F2 relatively weak (due to strong lone pair-lone pair repulsion between the very close atoms), and makes F- ions/F atoms attract electrons unusually strongly, so fluorine is more reactive/electronegative than a simple trend would predict oe
(d) Answer: Fluorine's very small atomic radius gives it an unusually high charge density; strong lone pair repulsion weakens the F-F bond and its outer electrons attract a bonding pair very strongly, making fluorine more reactive and electronegative than the group trend predicts

Question 8

(a) M1 correct species: Cl2, Br-, Cl-, Br2
(a) A1 Cl2 + 2Br- -> 2Cl- + Br2 fully balanced cao
(a) B1 solution turns (from colourless) to orange oe
(a) Answer: Cl2 + 2Br- -> 2Cl- + Br2; the solution turns orange
(b) B1 oxidising power (ability to gain electrons/be reduced) decreases down Group 7, so chlorine is a stronger oxidising agent than bromine (and iodine) oe
(b) B1 chlorine, as the stronger oxidising agent, can oxidise (remove electrons from) Br-; but iodine is a weaker oxidising agent than chlorine, so it cannot oxidise Cl- (remove electrons from the already more strongly held chloride ion) oe
(b) Answer: Oxidising power decreases down Group 7, so Cl2 (a stronger oxidising agent) can oxidise Br- to Br2, but I2 (a weaker oxidising agent than Cl2) cannot oxidise Cl- to Cl2
(c) M1 correct species: Cl2, NaOH, NaCl, NaOCl (or NaClO), H2O
(c) A1 Cl2 + 2NaOH -> NaCl + NaOCl + H2O fully balanced cao
(c) B1 commercial use: household bleach / disinfectant oe
(c) B1 this is an example of a disproportionation reaction
(c) Answer: Cl2 + 2NaOH -> NaCl + NaOCl + H2O; the mixture (containing sodium chlorate(I), NaOCl) is used as household bleach/disinfectant; this is a disproportionation reaction
(d) B1 disproportionation is a reaction in which the same element is simultaneously oxidised and reduced oe
(d) B1 chlorine in Cl2 has oxidation state 0, and is reduced to -1 in Cl- (NaCl) ft from (c)
(d) B1 chlorine in Cl2 (oxidation state 0) is also oxidised to +1 in ClO- (NaOCl) ft from (c)
(d) Answer: Disproportionation is when the same element is simultaneously oxidised and reduced; here Cl (0 in Cl2) is reduced to -1 (in Cl-) and oxidised to +1 (in ClO-) in the same reaction

Question 9

(a) B1 to remove/react away any carbonate or hydroxide ions present, which would otherwise also form a precipitate with Ag+ and give a false-positive/misleading result oe
(a) Answer: It removes carbonate or hydroxide ions that would otherwise also form a precipitate with silver ions, giving a false result
(b) B1 chloride: white precipitate, dissolves in dilute ammonia solution
(b) B1 bromide: cream precipitate, dissolves only in concentrated ammonia solution
(b) B1 iodide: yellow precipitate, insoluble in ammonia (even concentrated)
(b) Answer: Cl-: white precipitate, dissolves in dilute NH3; Br-: cream precipitate, dissolves only in concentrated NH3; I-: yellow precipitate, insoluble even in concentrated NH3
(c) B1 Ag+ + Br- -> AgBr cao
(c) Answer: Ag+(aq) + Br-(aq) -> AgBr(s)
(d) B1 bromide ions, Br-
(d) B1 justification: a cream precipitate that is insoluble in dilute ammonia but dissolves in concentrated ammonia is characteristic of AgBr (unlike AgCl, which dissolves in dilute ammonia, or AgI, which does not dissolve even in concentrated ammonia) oe
(d) Answer: Bromide ions (Br-); the observations (cream precipitate, insoluble in dilute ammonia but soluble in concentrated ammonia) uniquely match AgBr
(e) B1 both AgCl and AgBr would be present as a mixed precipitate, and the overall colour/behaviour may be dominated by or hard to distinguish from a pure chloride result oe
(e) B1 in particular, both AgCl and AgBr dissolve in (a suitable concentration of) ammonia, and small amounts of a second halide can be masked by the excess of the first, so the test cannot reliably confirm the absence of a second halide ion oe
(e) Answer: A mixture of AgCl and AgBr precipitates would form together, and since both dissolve in ammonia (AgCl in dilute, AgBr only in concentrated), a small amount of a second halide can be masked by an excess of the other, so the simple test cannot reliably distinguish a pure chloride solution from a chloride/bromide mixture

Question 10

(a) B1 NaCl + H2SO4 -> NaHSO4 + HCl cao
(a) B1 steamy/misty white fumes (of HCl gas) observed oe
(a) Answer: NaCl + H2SO4 -> NaHSO4 + HCl; steamy white fumes of HCl gas are observed
(b) B1 2HBr + H2SO4 -> Br2 + SO2 + 2H2O cao
(b) B1 sulfur in H2SO4 = +6
(b) B1 sulfur in SO2 = +4
(b) Answer: 2HBr + H2SO4 -> Br2 + SO2 + 2H2O; S in H2SO4 = +6, S in SO2 = +4
(c) M1 8HI + H2SO4 -> 4I2 + H2S + 4H2O cao
(c) A1 sulfur in H2SO4 = +6
(c) A1 sulfur in H2S = -2
(c) B1 this is a change of 8 in oxidation state (from +6 to -2), showing iodide is a much stronger reducing agent than bromide (which only reduces S to +4 in SO2) oe
(c) Answer: 8HI + H2SO4 -> 4I2 + H2S + 4H2O; S in H2SO4 = +6, S in H2S = -2
(d) B1 reducing power depends on how readily the halide ion loses an electron (is oxidised) oe
(d) B1 ionic radius increases down the group (Cl- < Br- < I-), so the outer electron to be lost is further from the nucleus and more shielded oe
(d) B1 this electron is less strongly attracted/held, so it is lost more easily down the group, meaning I- loses an electron more readily than Cl- and is therefore a stronger reducing agent oe
(d) Answer: Ionic radius increases down the group, so the outer electron of I- is further from the nucleus and more shielded than in Cl-; it is therefore lost more easily, making I- a stronger reducing agent than Cl-

Question 11

Level 3 (5-6): A clear, logically ordered method is given for both the cation and the anion test, with correct reagents and conditions throughout, and correct, distinguishing observations given for at least two possible metal cations and at least two possible halide anions.
Level 2 (3-4): A mostly correct method for identifying the cation and/or the anion is given, with some correct observations, but with minor omissions (such as a missing acidification step, or observations given for only one ion of a pair) or a small inaccuracy.
Level 1 (1-2): Some relevant but isolated or incomplete points made, such as naming a test without stating reagents/conditions, or stating an observation without a valid method; limited coverage of cations or anions.
Level 0 (0): No relevant chemistry given, or the response is entirely incorrect or irrelevant.
Indicative content:
- Dissolve a small sample of solid X in distilled water to make a solution for testing.
- Flame test for the cation: dip a clean nichrome (or platinum) wire into concentrated hydrochloric acid, then into a sample of the solid or solution, and hold it in a blue Bunsen flame.
- Observe the flame colour: magnesium gives no distinctive flame colour (or is not usually flame tested); calcium gives a brick-red/orange-red flame; barium gives a green (apple-green) flame.
- Anion (halide) test: add a few drops of dilute nitric acid to a fresh sample of the solution (to remove any carbonate/hydroxide interference), then add silver nitrate solution, AgNO3(aq).
- Observe the precipitate: a white precipitate that dissolves in dilute ammonia solution indicates chloride ions (AgCl); a cream precipitate that dissolves only in concentrated ammonia solution indicates bromide ions (AgBr); a yellow precipitate that is insoluble even in concentrated ammonia solution indicates iodide ions (AgI).
- Test the solubility of any precipitate in dilute ammonia solution, then in concentrated ammonia solution, to distinguish between the three halides.
- An alternative or confirmatory anion test is to add a small sample of the solid to concentrated sulfuric acid: chloride gives only steamy white fumes (HCl); bromide gives steamy fumes plus orange bromine vapour/fumes (from further redox with H2SO4); iodide gives purple/black solid or vapour (iodine) together with the toxic, rotten-egg-smelling gas hydrogen sulfide, from more extensive reduction of the sulfuric acid.
- Combining the flame test result (identifying the metal cation) with the silver nitrate/ammonia (or concentrated sulfuric acid) result (identifying the halide anion) allows the full identity of compound X to be deduced.

Question 12

(a) B1 MgCO3 -> MgO + CO2 cao
(a) Answer: MgCO3 -> MgO + CO2
(b) M1 n(MgCO3) = 5.05/84.3 = 0.0599 mol
(b) M1 n(CO2) = n(MgCO3) = 0.0599 mol (1:1 ratio from equation in (a)) ft
(b) A1 volume = 0.0599 x 24 = 1.44 dm3 cao (awrt 1.44 dm3) ft
(b) Answer: 1.44 dm3
(c) B1 Mg2+ is a smaller, more highly charge-dense ion than Ba2+ (both have the same 2+ charge, but Mg2+ has a much smaller ionic radius) oe
(c) B1 the smaller, more charge-dense Mg2+ ion polarises (distorts) the electron cloud of the neighbouring carbonate ion more strongly than the larger Ba2+ ion does oe
(c) B1 this greater distortion weakens a C-O bond within the carbonate ion more in MgCO3, so less energy (a lower temperature) is needed to break it down to the oxide and CO2, whereas BaCO3 needs a much higher temperature oe
(c) Answer: Mg2+ is smaller and more charge-dense than Ba2+, so it polarises the carbonate ion's electron cloud more strongly, weakening a C-O bond more and allowing MgCO3 to decompose at a lower temperature than BaCO3
(d) B1 SrCO3 would need a higher temperature than MgCO3 to decompose oe
(d) B1 Sr2+ is larger than Mg2+ (lower down Group 2), so it is less polarising/has a lower charge density, causing less distortion of the carbonate ion and so requiring more energy (a higher temperature) to decompose oe
(d) Answer: SrCO3 needs a higher temperature than MgCO3; Sr2+ is larger and less polarising than Mg2+, so it distorts the carbonate ion less, making SrCO3 more thermally stable

The Periodic Table and Inorganic Chemistry - Worksheets, Questions and Revision