The electron configuration of copper makes it a useful metal.

Determine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.

The electron configuration of copper makes it a useful metal.

Explain why a copper(II) solution is blue, using section 17 of the data booklet.

The electron configuration of copper makes it a useful metal.

Copper plating can be used to improve the conductivity of an object.

State, giving your reason, at which electrode the object being electroplated should be placed.

$«E=\frac{745 000 \mathrm{J} {\mathrm{mol}}^{-1}}{6.02\times {10}^{23} {\mathrm{mol}}^{-1}}=»1.24\times {10}^{-18} \mathrm{J}$ ✔

$«E=h\nu »$
$«1.24\times {10}^{-18} \mathrm{J}=6.63\times {10}^{-34} \mathrm{J} \mathrm{s}\times \mathrm{\nu }»$
$\mathrm{\nu }=1.87\times {10}^{15} «{\mathrm{s}}^{-1}/\mathrm{Hz}»$ ✔

Award [2] for correct final answer.
Award [1] for 1.12 × 10³⁹ «Hz».

orange light is absorbed «and the complementary colour is observed» ✔

Any TWO from:
partially filled d-orbitals ✔
«ligands/water cause» d-orbitals «to» split ✔
light is absorbed as electrons move to a higher energy orbital «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to «orange» light in the visible region of the spectrum ✔

cathode/negative «electrode» AND ${\mathrm{Cu}}^{2+}$ reduced «at that electrode» ✔

Accept cathode/negative «electrode» AND copper forms «at that electrode».

Determining the frequency of a photon that will cause the first ionization of copper was the most challenging question on the exam. Many could not do it all, although some came up with the answer that came from using the result that would arise from the ionization energy in J/mole (and frequently kJ/mole) rather than J/atom.

Many students were able to fully explain why solutions containing Cu²⁺ appear blue, however the misconception between absorption and emission spectra is still quite evident.

Surprisingly not that well answered. Most students identified the cathode as the electrode where electroplating occurs but few could adequately justify why.

Formulate an equation for the oxidation of nickel(II) sulfide to nickel(II) oxide.

The nickel obtained from another ore, nickeliferous limonite, is contaminated with iron. Both nickel and iron react with carbon monoxide gas to form gaseous complexes, tetracarbonylnickel, $\text{Ni(CO}{\text{)}}_{\text{4}}\text{(g)}$, and pentacarbonyliron, $\text{Fe(CO}{\text{)}}_{\text{5}}\text{(g)}$. Suggest why the nickel can be separated from the iron successfully using carbon monoxide.

Calculate the standard entropy change, $\mathrm{\Delta }{S}^{\theta }$, of the reaction, in $\text{J}{\text{K}}^{-1}$, using the values given.

Calculate a value for $\mathrm{\Delta }{H}^{\theta }$ in kJ.

Use your answers to (c)(i) and (c)(ii), to determine the temperature, in °C, at which the decomposition of liquid tetracarbonylnickel to nickel and carbon monoxide becomes favourable.

(If you did not get answers to (c)(i) and (c)(ii), use $-500\text{ J}{\text{K}}^{-1}$ and $-200\text{ kJ}$ respectively but these are not the correct answers.)

Suggest why experiments involving tetracarbonylnickel are very hazardous.

$\text{2NiS(s)}+\text{3}{\text{O}}_{\text{2}}\text{(g)}\to \text{2NiO(s)}+\text{2S}{\text{O}}_{\text{2}}\text{(g)}$

[1 mark]

formation of «gaseous» pentacarbonyliron is slower
OR
«gaseous» complexes form at different rates
OR
gases have different rates of diffusion «due to difference in masses»
OR
difference in thermal stability of «gaseous» complexes
OR
difference in boiling points of «gaseous» complexes
OR
difference in solubility of «gaseous» complexes
OR
difference in surface affinity «onto solid absorbent»
OR
difference in chemical properties of «gaseous» complexes

Accept any other valid answer.

[1 mark]

$\sum S_{\text{RHS}}^{\theta }=313.4\ll \text{J}{\text{K}}^{-1}\gg$
AND
$\sum S_{\text{LHS}}^{\theta }=\ll (4\times 197.6)+29.9\text{ J}{\text{K}}^{-1}=\gg 820.3\ll \text{J}{\text{K}}^{-1}\gg$

$\mathrm{\Delta }{S}^{\theta }\ll =\sum S_{\text{RHS}}^{\theta }-\sum S_{\text{LHS}}^{\theta }=313.4-820.3\gg =-506.9\ll \text{J}{\text{K}}^{-1}\gg$

Award [2] for correct final answer.

[2 marks]

$\mathrm{\Delta }{H}^{\theta }\ll =-633.0-4\times (-110.5)\gg =-191\ll kJ\gg$

[1 mark]

«when» $\mathrm{\Delta }G=0$ «forward and backward reactions are equally favourable»

«when $\mathrm{\Delta }G=0$, $\text{T}=\frac{\mathrm{\Delta }H}{\mathrm{\Delta }S}$», $\text{T}=\ll \frac{191\text{ kJ}}{0.5069\text{ kJ}{\text{K}}^{-1}}=\gg 377\ll \text{K}\gg$

«temperature =» 104 «°C»

Award [3] for correct final answer. Use of –500 J K^–1 and –200 kJ gives 127 °C.

Award [2 max] for T < 104 «°C».

Accept ΔG < 0 and T > 104 «°C».

[3 marks]

CO is toxic/poisonous
OR
Ni(CO)₄ decomposition deposits nickel in the lungs
OR
tetracarbonylnickel is toxic/poisonous
OR
tetracarbonylnickel is highly flammable «auto-ignition temperature of 60 °C»

[1 mark]

5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)₂•2H₂O, was dissolved in water to make 1.00 dm³ of solution. 25.0 cm³ samples of this solution were titrated against a 0.100 mol dm^-3 solution of sodium hydroxide using a suitable indicator.

(COOH)₂ (aq) + 2NaOH (aq) → (COONa)₂ (aq) + 2H₂O (l)

The mean value of the titre was 14.0 cm³.

(i) Suggest a suitable indicator for this titration. Use section 22 of the data booklet.

(ii) Calculate the amount, in mol, of NaOH in 14.0 cm³ of 0.100 mol dm^-3 solution.

(iii) Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm³ sample.

(iv) Determine the percentage purity of the hydrated ethanedioic acid sample.

Draw the Lewis (electron dot) structure of the ethanedioate ion, ^–OOCCOO^–.

Outline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.

Explain how ethanedioate ions act as ligands.

i
phenolphthalein
OR
phenol red

ii
«n(NaOH) = $\left(\frac{14.0}{1000}\right)$ dm³ × 0.100 mol dm^-3 =» 1.40 × 10^-3 «mol»

iii
«$\frac{1}{2}$ × 1.40 × 10^-3 =» 7.00 × 10^-4 «mol»

iv
ALTERNATIVE 1:
«mass of pure hydrated ethanedioic acid in each titration = 7.00 × 10^-4 mol × 126.08 g mol^-1 =» 0.0883 / 8.83 × 10^-2 «g»

mass of sample in each titration = «$\frac{25}{1000}$ × 5.00 g =» 0.125 «g»

«% purity = $\frac{0.0883\mathrm{g}}{0.125\mathrm{g}}$ × 100 =» 70.6 «%»

ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm³ solution = 7.00 × 10^-4 × $\frac{1000}{25}$=» 2.80 × 10^-2 «mol»

«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10^-2 mol × 126.08 g mol^-1 =» 3.53 «g»

«% purity = $\frac{3.53\mathrm{g}}{5.00\mathrm{g}}$ × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = $\frac{5.00\mathrm{g}}{126.08\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}$ = 0.03966 «mol»

actual amount of hydrated ethanedioic acid = «7.00 × 10^-4 × $\frac{1000}{25}$ =» 2.80 × 10^-2 «mol»

«% purity = $\frac{2.80\times {10}^{-2}}{0.03966}$ × 100 =» 70.6 «%»

Award suitable part marks for alternative methods.
Award [3] for correct final answer.
Award [2 max] for 50.4 % if anhydrous ethanedioic acid assumed.

Accept single negative charges on two O atoms singly bonded to C.
Do not accept resonance structures.
Allow any combination of dots/crosses or lines to represent electron pairs.

electrons delocalized «across the O–C–O system»
OR
resonance occurs

Accept delocalized π-bond(s).
No ECF from (d).

122 «pm» < C–O < 143 «pm»

Accept any answer in range 123 «pm» to 142 «pm».
Accept “bond intermediate between single and double bond” or “bond order 1.5”.

coordinate/dative/covalent bond from O to «transition» metal «ion»
OR
acts as a Lewis base/nucleophile

can occupy two positions
OR
provide two electron pairs from different «O» atoms
OR
form two «coordinate/dative/covalent» bonds «with the metal ion»
OR
chelate «metal/ion»

Describe the bonding in metals.

Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:

Calculate the relative atomic mass of titanium to two decimal places.

State the number of protons, neutrons and electrons in the ${}_{\text{22}}^{\text{48}}\text{Ti}$ atom.

State the full electron configuration of the ${}_{\text{22}}^{\text{48}}\text{T}{\text{i}}^{2+}$ ion.

Suggest why the melting point of vanadium is higher than that of titanium.

Sketch a graph of the first six successive ionization energies of vanadium on the axes provided.

Explain why an aluminium-titanium alloy is harder than pure aluminium.

Describe, in terms of the electrons involved, how the bond between a ligand and a central metal ion is formed.

Outline why transition metals form coloured compounds.

State the type of bonding in potassium chloride which melts at 1043 K.

A chloride of titanium, $\text{TiC}{\text{l}}_{\text{4}}$, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.

Formulate an equation for this reaction.

Suggest one disadvantage of using this smoke in an enclosed space.

electrostatic attraction

between «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons

Accept “mobile electrons”.

Do not accept “metal atoms/nuclei”.

[2 marks]

$\frac{(46\times 7.98)\text{ + }(47\times 7.32)\text{ + }(48\times 73.99)\text{ + }(49\times 5.46)\text{ + }(50\times 5.25)}{100}=47.93$

Answer must have two decimal places with a value from 47.90 to 48.00.

Award [2] for correct final answer.

Award [0] for 47.87 (data booklet value).

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

$\text{1}{\text{s}}^{\text{2}}\text{2}{\text{s}}^{\text{2}}\text{2}{\text{p}}^{\text{6}}\text{3}{\text{s}}^{\text{2}}\text{3}{\text{p}}^{\text{6}}\text{3}{\text{d}}^{\text{2}}$

[1 mark]

vanadium has smaller ionic radius «leading to stronger metallic bonding»

Accept vanadium has «one» more valence electron«s» «leading to stronger metallic bonding».

Accept “atomic” for “ionic”.

[1 mark]

regular increase for first five AND sharp increase to the 6th

A log graph is acceptable.

Accept log plot on given axes (without amendment of y-axis).

Award mark if gradient of 5 to 6 is greater than “best fit line” of 1 to 5.

[1 mark]

titanium atoms/ions distort the regular arrangement of atoms/ions

OR

titanium atoms/ions are a different size to aluminium «atoms/ions»

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

pair of electrons provided by the ligand

Do not accept “dative” or “coordinate bonding” alone.

[1 mark]

partially filled d-orbitals

«ligands cause» d-orbitals «to» split

light is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted

energy gap corresponds to light in the visible region of the spectrum

colour observed is the complementary colour

[4 marks]

ionic

OR

«electrostatic» attraction between oppositely charged ions

[1 mark]

«simple» molecular structure

OR

weak«er» intermolecular bonds

OR

weak«er» bonds between molecules

Accept specific examples of weak bonds such as London/dispersion and van der Waals.

Do not accept “covalent”.

[1 mark]

$\text{TiC}{\text{l}}_{\text{4}}\text{(l)}+\text{2}{\text{H}}_{\text{2}}\text{O(l)}\to \text{Ti}{\text{O}}_{\text{2}}\text{(s)}+\text{4HCl(aq)}$ correct products
correct balancing

Accept ionic equation.

Award M2 if products are HCl and a compound of Ti and O.

[2 marks]

HCl causes breathing/respiratory problems

OR

HCl is an irritant

OR

HCl is toxic

OR

HCl has acidic vapour

OR

HCl is corrosive

Accept TiO₂ causes breathing

problems/is an irritant.

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

Identify one organic functional group that can react with acidified K₂Cr₂O₇(aq).

Corrosion of iron is similar to the processes that occur in a voltaic cell. The initial steps involve the following half-equations:

Fe²⁺(aq) + 2e^– $\rightleftharpoons$ Fe(s)

$\frac{1}{2}$O₂(g) + H₂O(l) + 2e^– $\rightleftharpoons$ 2OH^–(aq)

Calculate E ^θ, in V, for the spontaneous reaction using section 24 of the data booklet.

Calculate the Gibbs free energy, ΔG ^θ, in kJ, which is released by the corrosion of 1 mole of iron. Use section 1 of the data booklet.

Explain why iron forms many different coloured complex ions.

Zinc is used to galvanize iron pipes, forming a protective coating. Outline how this process prevents corrosion of the iron pipes.

hydroxyl/OH
OR
aldehyde/CHO

Accept “hydroxy/alcohol” for “hydroxyl”.

Accept amino/amine/NH₂.

[1 mark]

«E ^θ =» +0.85 «V»

Accept 0.85 V.

[1 mark]

ΔG ^θ «= –nFE ^θ» = –2 «mol e^–» x 96500 «C mol^–1» x 0.85 «V»

«ΔG ^θ =» –164 «kJ»

Accept “«+»164 «kJ»” as question states energy released.

Award [1 max] for “+” or “–” 82 «kJ».

Do not accept answer in J.

[2 marks]

incompletely filled d-orbitals

colour depends upon the energy difference between the split d-orbitals

variable/multiple/different oxidation states

different «nature/identity of» ligands

different number of ligands

[3 marks]

Zn/zinc is a stronger reducing agent than Fe/iron
OR
Zn/zinc is oxidized instead of Fe/iron
OR
Zn/zinc is the sacrificial anode

Accept “Zn is more reactive than Fe”.

Accept “Zn oxide layer limits further corrosion”.

Do not accept “Zn layer limits further corrosion”.

[1 mark]

The stable isotope of rhenium contains 110 neutrons.

State the nuclear symbol notation ${}_{\text{Z}}^{\text{A}}\text{X}$ for this isotope.

Suggest the basis of these predictions.

A scientist wants to investigate the catalytic properties of a thin layer of rhenium metal on a graphite surface.

Describe an electrochemical process to produce a layer of rhenium on graphite.

Predict two other chemical properties you would expect rhenium to have, given its position in the periodic table.

Describe how the relative reactivity of rhenium, compared to silver, zinc, and copper, can be established using pieces of rhenium and solutions of these metal sulfates.

State the name of this compound, applying IUPAC rules.

Calculate the percentage, by mass, of rhenium in ReCl₃.

Suggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 6 of the data booklet.

Deduce the coefficients required to complete the half-equation.

ReO₄⁻ (aq) + ____H⁺ (aq) + ____e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + ____H₂O (l)        E^θ = +0.36 V

Predict, giving a reason, whether the reduction of ReO₄⁻ to [Re(OH)₂]²⁺ would oxidize Fe²⁺ to Fe³⁺ in aqueous solution. Use section 24 of the data booklet.

${}_{\text{75}}^{\text{185}}\text{Re}$    [✔]

gap in the periodic table
OR
element with atomic number «75» unknown
OR
break/irregularity in periodic trends     [✔]

«periodic table shows» regular/periodic trends «in properties»      [✔]

electrolyze «a solution of /molten» rhenium salt/Reⁿ⁺     [✔]

graphite as cathode/negative electrode
OR
rhenium forms at cathode/negative electrode     [✔]

Note: Accept “using rhenium anode” for M1.

Any two of:
variable oxidation states     [✔]

forms complex ions/compounds     [✔]

coloured compounds/ions     [✔]

«para»magnetic compounds/ions     [✔]

Note: Accept other valid responses related to its chemical metallic properties.

Do not accept “catalytic properties”.

place «pieces of» Re into each solution    [✔]

if Re reacts/is coated with metal, that metal is less reactive «than Re»    [✔]

Note: Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.

rhenium(III) chloride
OR
rhenium trichloride    [✔]

«M_r ReCl₃ = 186.21 + (3 × 35.45) =» 292.56    [✔]
«100 × $\frac{186.21}{292.56}$ =» 63.648 «%»   [✔]

same group as Mn «which forms MnO₄^-»
OR
in group 7/has 7 valence electrons, so its «highest» oxidation state is +7    [✔]

ReO₄⁻ (aq) + 6H⁺ (aq) + 3e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + 2H₂O (l)    [✔]

no AND ReO₄⁻ is a weaker oxidizing agent than Fe³⁺
OR
no AND Fe³⁺ is a stronger oxidizing agent than ReO₄⁻
OR
no AND Fe²⁺ is a weaker reducing agent than [Re(OH)₂]²⁺
OR
no AND [Re(OH)₂]²⁺ is a stronger reducing agent than Fe²⁺
OR
no AND cell emf would be negative/–0.41 V     [✔]

It was expected that this question would be answered correctly by all HL candidates. However, many confused the A-Z positions or calculated very unusual numbers for A, sometimes even with decimals.

This is a NOS question which required some reflection on the full meaning of the periodic table and the wealth of information contained in it. But very few candidates understood that they were being asked to explain periodicity and the concept behind the periodic table, which they actually apply all the time. Some were able to explain the “gap” idea and other based predictions on properties of nearby elements instead of thinking of periodic trends. A fair number of students listed properties of transition metals in general.

Generally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.

Generally well answered though some students suggested physical properties rather than chemical ones.

Many candidates chose to set up voltaic cells and in other cases failed to explain the actual experimental set up of Re being placed in solutions of other metal salts or the reaction they could expect to see.

Almost all candidates were able to name the compound according to IUPAC.

Most candidates were able to answer this stoichiometric question correctly.

This should have been a relatively easy question but many candidates sometimes failed to see the connection with Mn or the amount of electrons in its outer shell.

Surprisingly, a great number of students were unable to balance this simple half-equation that was given to them to avoid difficulties in recall of reactants/products.

Many students understood that the oxidation of Fe²⁺ was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for E^Θ often failing to realise that the oxidation of Fe²⁺ would have the inverse sign to the reduction reaction.

State the relationship between NH₄⁺ and NH₃ in terms of the Brønsted–Lowry theory.

Determine the concentration, in mol dm^–3, of the solution formed when 900.0 dm³ of NH₃(g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm³ of solution. Use sections 1 and 2 of the data booklet.

Calculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.

Calculate the concentration, in mol dm^–3, of ammonia molecules in the solution with pH = 9.3. Use section 21 of the data booklet.

An aqueous solution containing high concentrations of both NH₃ and NH₄⁺ acts as an acid-base buffer solution as a result of the equilibrium:

NH₃ (aq) + H⁺ (aq) $\rightleftharpoons$ NH₄⁺ (aq)

Referring to this equilibrium, outline why adding a small volume of strong acid would leave the pH of the buffer solution almost unchanged.

Magnesium salts form slightly acidic solutions owing to equilibria such as:

Mg²⁺(aq) + H₂O (l) $\rightleftharpoons$ Mg(OH)⁺(aq) + H⁺(aq)

Comment on the role of Mg²⁺ in forming the Mg(OH)⁺ ion, in acid-base terms.

Mg(OH)⁺ is a complex ion, but Mg is not regarded as a transition metal. Contrast Mg with manganese, Mn, in terms of one characteristic chemical property of transition metals, other than complex ion formation.

conjugate «acid and base» ✔

amount of ammonia $⟨⟨=\frac{P.V}{R.T}=\frac{100.0 kPa\times 900.0 d{m}^3}{8.31 J K^{-1}mo{l}^{-1}\times 300.0 K}⟩⟩ = 36.1 «\mathrm{mol}»$ ✔

concentration $⟨⟨=\frac{n}{V}=\frac{36.1}{2.00}⟩⟩=18.1 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

$\left[{\mathrm{OH}}^{-}\right] ⟨⟨=\frac{{\mathrm{K}}_{\mathrm{W}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-9.3}}={10}^{-4.7}⟩⟩=2.0 \times {10}^{-5}⟨⟨\mathrm{mol} {\mathrm{dm}}^{-3}⟩⟩$  ✔

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}/\frac{{10}^{-4.7}\times {10}^{-4.7}}{\left[N{H}_3\right]}⟨⟨={10}^{-4.75}⟩⟩$ ✔

$\left[{\mathrm{NH}}_3\right]=⟨⟨=\frac{{10}^{-9.4}}{{10}^{-4.75}}={10}^{-4.65}⟩⟩=2.24\times {10}^{-5}«\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Accept other methods of carrying out the calculation.

Award [2] for correct answer.

equilibrium shifts to right/H⁺ reacts with NH₃ ✔

«as large excess» ratio [NH₃]:[NH₄⁺] «and hence pH» almost unchanged ✔

Accept “strong acid/H⁺ converted to a weak acid/NH₄⁺ «and hence pH almost unchanged».

Lewis acid ✔

accepts «a lone» electron pair «from the hydroxide ion» ✔

Do not accept electron acceptor without mention of electron pair.

ALTERNATIVE 1

Property: variable oxidation state ✔

Comparison: Mn compounds can exist in different valencies/oxidation states AND Mg has a valency/oxidation state of +2 in all its compounds ✔

Accept valency.
Accept for second statement “Mg «always» has the same oxidation state”.

ALTERNATIVE 2

Property: coloured ions/compounds/complexes ✔

Comparison: Mn ions/compounds/complexes coloured AND Mg ions/compounds white/«as solids»/colourless «in aqueous solution» ✔

Accept Mn forms coloured ions/compounds/complexes and Mg does not.

ALTERNATIVE 3

Property: catalytic activity ✔

Comparison: «many» Mn compounds act as catalysts AND Mg compounds do not «generally» catalyse reactions ✔


For any property accept a correct specific example, for example manganate(VII) is purple.
Do not accept differences in atomic structure, such as partially filled d sub-levels, but award ECF for a correct discussion.

Well done; However, instead of identifying the conjugate acid-base relationship, some simply identified these as Brønsted–Lowry base and acid.

Good performance. Some teachers suggested the question had an error in units, but this was not the case. The question had to be solved, first by using the data provided for application of gas law to determine the number of moles of gas. Next, given volume of solution, V = 2.00 dm³, determine its concentration.

Concentration of [OH^˗] was asked for but some calculated [H₃O⁺] instead. On the whole, question was done well.

Mediocre performance. Since a mark was given for the K_b expression, that mark could also be scored for the Henderson Hasselbalch (HH) equation, provided it is specific to the equilibrium reaction. Unfortunately, there was poor understanding of the application of the equation in most cases. Students should be strongly encouraged to use the HH equation only when a buffer is involved. Appropriate K_a or K_b expressions should be used when buffer solutions are not involved.

Mediocre performance. One mark was scored for suggesting equilibrium shifts to right or H⁺ reacts with NH₃. However, some made reference to ammonia being a strong base or no reference to the strong acid, H⁺ being converted to a weak acid, NH₄⁺.

Mediocre performance; although some Mg²⁺ was identified as a Lewis acid, the reasoning given was that it accepts an electron, rather than an electron pair or references were made to Bronsted-Lowry theory.

Deduce the ratio of Fe²⁺:Fe³⁺ in Fe₃O₄.

State the type of spectroscopy that could be used to determine their relative abundances.

State the number of protons, neutrons and electrons in each species.

Iron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.

Determine the specific heat capacity of iron, in J g⁻¹K⁻¹. Use section 1 of the data booklet.

A voltaic cell is set up between the Fe²⁺(aq) | Fe (s) and Fe³⁺ (aq) | Fe²⁺ (aq) half-cells.

Deduce the equation and the cell potential of the spontaneous reaction. Use section 24 of the data booklet.

The figure shows an apparatus that could be used to electroplate iron with zinc. Label the figure with the required substances.

Outline why, unlike typical transition metals, zinc compounds are not coloured.

Transition metals like iron can form complex ions. Discuss the bonding between transition metals and their ligands in terms of acid-base theory.

1:2 ✔

Accept 2 Fe³⁺: 1 Fe²⁺
Do not accept 2:1 only

mass «spectroscopy»/MS ✔

Award [1 max] for 4 correct values.

specific heat capacity « = $\frac{q}{m\times ∆T}/\frac{1000 \mathrm{J}}{50 \mathrm{g}\times 44 \mathrm{K}}$» = 0.45 «J g⁻¹K⁻¹» ✔

Equation:
2Fe³⁺(aq) + Fe(s) → 3Fe²⁺(aq) ✔

Cell potential:
«+0.77 V − (−0.45 V) = +»1.22 «V» ✔

Do not accept reverse reaction or equilibrium arrow.

Do not accept negative value for M2.

left electrode/anode labelled zinc/Zn AND right electrode/cathode labelled iron/Fe ✔

electrolyte labelled as «aqueous» zinc salt/Zn²⁺ ✔

Accept an inert conductor for the anode.

Accept specific zinc salts such as ZnSO₄.

« Zn²⁺» has a full d-shell
OR
does not form « ions with» an incomplete d-shell ✔

Do not accept “Zn is not a transition metal”.

Do not accept zinc atoms for zinc ions.

ligands donate pairs of electrons to metal ions
OR
forms coordinate covalent/dative bond✔

ligands are Lewis bases
AND
metal «ions» are Lewis acids ✔

Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.

Suggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.

The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.

Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl.

KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq)

Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution.

State the equilibrium constant expression, K_c.

Predict, with a reason, the effect on the equilibrium constant, K_c, when the temperature is increased.

Determine an approximate order of magnitude for K_c, using sections 1 and 2 of the data booklet. Assume ΔG^Θ for the forward reaction is approximately +50 kJ at 298 K.

Suggest one reason why urea is a solid and ammonia a gas at room temperature.

Sketch two different hydrogen bonding interactions between ammonia and water.

The combustion of urea produces water, carbon dioxide and nitrogen.

Formulate a balanced equation for the reaction.

Calculate the maximum volume of CO₂, in cm³, produced at STP by the combustion of 0.600 g of urea, using sections 2 and 6 of the data booklet.

Describe the bond formation when urea acts as a ligand in a transition metal complex ion.

The C–N bonds in urea are shorter than might be expected for a single C–N bond. Suggest, in terms of electrons, how this could occur.

The mass spectrum of urea is shown below.

Identify the species responsible for the peaks at m/z = 60 and 44.

The IR spectrum of urea is shown below.

Identify the bonds causing the absorptions at 3450 cm⁻¹ and 1700 cm⁻¹ using section 26 of the data booklet.

Predict the number of signals in the ¹H NMR spectrum of urea.

Predict the splitting pattern of the ¹H NMR spectrum of urea.

Outline why TMS (tetramethylsilane) may be added to the sample to carry out ¹H NMR spectroscopy and why it is particularly suited to this role.

molar mass of urea «4 $\times$ 1.01 + 2 $\times$ 14.01 + 12.01 + 16.00» = 60.07 «g mol^_-1»

«% nitrogen = $\frac{\text{2}\times \text{14.01}}{\text{60.07}}$ $\times$ 100 =» 46.65 «%»

Award [2] for correct final answer.

Award [1 max] for final answer not to two decimal places.

[2 marks]

«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»

OR

«cost» increases AND inefficient/too much/about half mass not nitrogen

Accept other reasonable explanations.

Do not accept answers referring to safety/explosions.

[1 mark]

Note: Urea’s structure is more complex than that predicted from VSEPR theory.

[3 marks]

n(KNCO) «= 0.0500 dm³ $\times$ 0.100 mol dm^–3» = 5.00 $\times$ 10^–3 «mol»

«mass of urea = 5.00 $\times$ 10^–3 mol $\times$ 60.07 g mol^–1» = 0.300 «g»

Award [2] for correct final answer.

[2 marks]

$K_{\text{c}}=\frac{[{({\text{H}}_2\text{N})}_2\text{CO}]\times [{\text{H}}_2\text{O}]}{{[\text{N}{\text{H}}_3]}^2\times [\text{C}{\text{O}}_2]}$

[1 mark]

«K_c» decreases AND reaction is exothermic

OR

«K_c» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

ln K « = $\frac{-\mathrm{\Delta }{G}^{\mathrm{\Theta }}}{RT}=\frac{-50\times {10}^3\text{ J}}{8.31\text{ J }{\text{K}}^{-1}\text{ mo}{\text{l}}^{-1}\times 298\text{ K}}$ » = –20

«K_c =» 2 $\times$ 10^–9

OR

1.69 $\times$ 10^–9

OR

10^–9

Accept range of 20-20.2 for M1.

Award [2] for correct final answer.

[2 marks]

Any one of:

urea has greater molar mass

urea has greater electron density/greater London/dispersion

urea has more hydrogen bonding

urea is more polar/has greater dipole moment

Accept “urea has larger size/greater van der Waals forces”.

Do not accept “urea has greater intermolecular forces/IMF”.

[1 mark]

Award [1] for each correct interaction.

If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.

Penalize solid line to represent H-bonding only once.

[2 marks]

2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g)

correct coefficients on LHS

correct coefficients on RHS

Accept (H₂N)₂CO(s) + $\frac{3}{2}$O₂(g) → 2H₂O(l) + CO₂(g) + N₂(g).

Accept any correct ratio.

[2 marks]

«V = $\frac{\text{0.600 g}}{\text{60.07 g mo}{\text{l}}^{-1}}$ $\times$ 22700 cm³ mol^–1 =» 227 «cm³»

[1 mark]

lone/non-bonding electron pairs «on nitrogen/oxygen/ligand» given to/shared with metal ion

co-ordinate/dative/covalent bonds

[2 marks]

lone pairs on nitrogen atoms can be donated to/shared with C–N bond

OR

C–N bond partial double bond character

OR

delocalization «of electrons occurs across molecule»

OR

slight positive charge on C due to C=O polarity reduces C–N bond length

[1 mark]

60: CON₂H₄⁺

44: CONH₂⁺

Accept “molecular ion”.

[2 marks]

3450 cm^–1: N–H

1700 cm^–1: C=O

Do not accept “O–H” for 3450 cm^–1.

[2 marks]

1

[2 marks]

singlet

Accept “no splitting”.

[1 mark]

acts as internal standard

OR

acts as reference point

one strong signal

OR

12 H atoms in same environment

OR

signal is well away from other absorptions

Accept “inert” or “readily removed” or “non-toxic” for M1.

[2 marks]

Hydrogen spectral data give the frequency of 3.28 × 10¹⁵ s⁻¹ for its convergence limit.

Calculate the ionization energy, in J, for a single atom of hydrogen using sections 1 and 2 of the data booklet.

Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.

Deduce any change in the colour of the electrolyte during electrolysis.

Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.

Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.

IE «= ΔE = hν = 6.63 × 10^–34 J s × 3.28 × 10¹⁵ s^–1» = 2.17 × 10^–18 «J»

[1 mark]

«$\lambda =\frac{C}{\text{v}}=\frac{3.00\times {10}^8\text{ m}{\text{s}}^{-1}}{3.28\times {10}^{15}{\text{s}}^{-1}}=$» 9.15 × 10^–8 «m»

[1 mark]

no change «in colour»

Do not accept “solution around cathode will become paler and solution around the anode will become darker”.

[1 mark]

oxygen/O₂

Accept “carbon dioxide/CO2”.

[1 mark]

Transition metals:

«contain» d and s orbitals «which are close in energy»

OR

«successive» ionization energies increase gradually

Alkali metals:

second electron removed from «much» lower energy level

OR

removal of second electron requires large increase in ionization energy

[2 marks]

State the electron configuration of a bromine atom.

Sketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.

Draw two Lewis (electron dot) structures for BrO₃⁻.

Determine the preferred Lewis structure based on the formal charge on the bromine atom, giving your reasons.

Predict, using the VSEPR theory, the geometry of the BrO₃⁻ ion and the O−Br−O bond angles.

Bromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.

Deduce the half-equation for this reduction reaction.

Bromate(V) ions oxidize iron(II) ions, Fe²⁺, to iron(III) ions, Fe³⁺.

Deduce the equation for this redox reaction.

Calculate the standard Gibbs free energy change, ΔG^Θ, in J, of the redox reaction in (ii), using sections 1 and 24 of the data booklet.

E^Θ (BrO₃⁻ / Br⁻) = +1.44 V

State and explain the magnetic property of iron(II) and iron(III) ions.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

OR

[Ar] 4s² 3d¹⁰ 4p⁵ ✔

Accept 3d before 4s.

Accept double-headed arrows.

Structure I - follows octet rule:

Structure II - does not follow octet rule:

Accept dots, crosses or lines to represent electron pairs.

«structure I» formal charge on Br = +2

OR

«structure II» formal charge on Br = 0/+1 ✔

structure II is preferred AND it produces formal charge closer to 0 ✔

Ignore any reference to formal charge on oxygen.

Geometry:
trigonal/pyramidal ✔

Reason:
three bonds AND one lone pair
OR
four electron domains ✔

O−Br−O angle:
107° ✔

Accept “charge centres” for “electron domains”.

Accept answers in the range 104–109°.

BrO₃⁻ (aq) + 6e⁻ + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l)

correct reactants and products ✔

balanced equation ✔

Accept reversible arrows.

BrO₃⁻ (aq) + 6Fe²⁺ (aq) + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l) + 6Fe³⁺ (aq) ✔

E^Θ_reaction = «+1.44 V – 0.77 V =» 0.67 «V» ✔

ΔG^Θ = «–nFE^Θ_reaction = – 6 × 96500 C mol^–1 × 0.67 V =» –3.9 × 10⁵ «J» ✔

both are paramagnetic ✔

«both» contain unpaired electrons ✔

Accept orbital diagrams for both ions showing unpaired electrons.

Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.

State the shape of the complex ion.

Deduce the charge on the complex ion and the oxidation state of cobalt.

Describe, in terms of acid-base theories, the type of reaction that takes place between the cobalt ion and water to form the complex ion.

Any three of:

Group 1:
atomic/ionic radius increases

smaller charge density

OR

force of attraction between metal ions and delocalised electrons decreases

Do not accept discussion of attraction between valence electrons and nucleus for M2.

Accept “weaker metallic bonds” for M2.

Group 17:
number of electrons/surface area/molar mass increase

London/dispersion/van der Waals’/vdw forces increase

Accept “atomic mass” for “molar mass”.

[Max 3 Marks]

«distorted» octahedral

Accept “square bipyramid”.

Charge on complex ion: 1+/+
Oxidation state of cobalt: +2

Lewis «acid-base reaction»

H2O: electron/e^– pair donor

OR

Co²⁺: electron/e^– pair acceptor

State why NH₃ is a Lewis base.

Calculate the pH of a 1.00 × 10⁻² mol dm⁻³ aqueous solution of ammonia.

pK_b = 4.75 at 298 K.

Justify whether a 1.0 dm³ solution made from 0.10 mol NH³ and 0.20 mol HCl will form a buffer solution.

Sketch the shape of one sigma ($\text{σ}$) and one pi ($\mathrm{\pi }$) bond.

Identify the number of sigma and pi bonds in HCN.

State the hybridization of the carbon atom in HCN.

Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.

Explain why transition metal cyanide complexes are coloured.

donates «lone/non-bonding» pair of electrons ✔

Kb = 10^-4.75 /1.78 x 10^-5
OR
Kb = $\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[{\mathrm{NH}}_3\right]}$ ✔

[OH^–] = « $\sqrt{\left(1.00\times {10}^{-2}\times {10}^{-4.75}\right)}$ =» 4.22 × 10^–4 «(mol dm^–3)» ✔

pOH« = –log₁₀ (4.22 × 10^–4)» = 3.37
AND
pH = «14 – 3.37» = 10.6

OR

[H⁺]« =$\frac{1.00\times {10}^{-14}}{4.22\times {10}^{-4}}$» = 2.37 × 10^–11
AND
pH« = –log₁₀ 2.37 × 10^–11» = 10.6 ✔

Award [3] for correct final answer.

no AND is not a weak acid conjugate base system

OR

no AND weak base «totally» neutralized/ weak base is not in excess

OR

no AND will not neutralize small amount of acid ✔

Accept “no AND contains 0.10 mol NH₄Cl + 0.10 mol HCl”.

Sigma ($\sigma$):

Pi ($\pi$):

Accept overlapping p-orbital(s) with both lobes of equal size/shape.

Shaded areas are not required in either diagram.

Sigma ($\text{σ}$): 2 AND Pi ($\mathrm{\pi }$): 2 ✔

sp ✔

HCN has stronger dipole–dipole attraction ✔

Do not accept reference to H-bonds.

Any three from:

partially filled d-orbitals ✔

«CN- causes» d-orbitals «to» split ✔

light is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔

energy gap corresponds to light in the visible region of the spectrum ✔

Do not accept “colour observed is the complementary colour” for M4.

The main error was the omission of lone electron "pair", though there was also a worrying amount of very confused answers for a very basic chemistry concept where 40% provided very incorrect answers.

Rather surprisingly, many students got full marks for this multi-step calculation; others went straight to the pH/pKa acid/base equation so lost at least one of the marks: students often seem less prepared for base calculations, as opposed to acid calculations.

Poorly answered revealing little understanding of buffering mechanisms, which is admittedly a difficult topic.

This proved to be the most challenging question (10%). It was a good question, where candidates had to explain a huge difference in boiling point of two covalent compounds, requiring solid understanding of change of state where breaking bonds cannot be involved). Yet most considered the triple bonds in HCN as the cause, suggesting covalent bonds break when substance boil, which is very worrying. Others considered H-bonds which at least is an intermolecular force, but shows they are not too familiar with the conditions necessary for H-bonding.

This question appears frequently in exams but with slightly different approaches. In general candidates ignore the specific question and give the same answers, failing in this case to describe why complexes are coloured rather than what colour is seen. These answers appear to reveal that many candidates don't really understand this phenomenon, but learn the answer by heart and make mistakes when repeating it, for example, stating that the ‘d-orbitals of the ligands were split’- an obvious misconception. The average mark was 1.6/3, with a MS providing 4 ideas that would merit a mark

Explain why Si has a smaller atomic radius than Al.

Explain why the first ionization energy of sulfur is lower than that of phosphorus.

State the condensed electron configurations for Cr and Cr3⁺.

Describe metallic bonding and how it contributes to electrical conductivity.

Deduce, giving a reason, which complex ion [Cr(CN)₆]³⁻ or [Cr(OH)₆]³⁻ absorbs higher energy light. Use section 15 of the data booklet.

[Cr(OH)₆]³⁻ forms a green solution. Estimate a wavelength of light absorbed by this complex, using section 17 of the data booklet.

Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF₄, and sulfur dichloride, SCl₂.

Suggest, giving reasons, the relative volatilities of SCl₂ and H₂O.

nuclear charge/number of protons/Z/Z_eff increases «causing a stronger pull on the outer electrons» ✓

same number of shells/«outer» energy level/shielding ✓

P has «three» unpaired electrons in 3p sub-level AND S has one full 3p orbital «and two 3p orbitals with unpaired electrons»
OR
P: [Ne]3s²3p_x¹3p_y¹3p_z¹ AND S: [Ne]3s²3p_x²3p_y¹3p_z¹ ✓

Accept orbital diagrams for 3p sub-level for M1. Ignore other orbitals or sub-levels.

repulsion between paired electrons in sulfur «and therefore easier to remove» ✓

Accept “removing electron from S gives more stable half-filled sub-level" for M2.

Cr:
[Ar] 4s¹3d⁵ ✓

Cr³⁺:
[Ar] 3d³ ✓

Accept “[Ar] 3d⁵4s¹”.

Accept “[Ar] 3d³4s⁰”.

Award [1 max] for two correct full electron configurations “1s²2s²2p⁶3s²3p⁶4s¹3d⁵ AND 1s²2s²2p⁶3s²3p⁶3d³”.

Award [1 max] for 4s¹3d⁵ AND 3d³.

electrostatic attraction ✓

between «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓

mobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓

Do not accept “nuclei” for “cations/positive ions” in M2.

Accept “mobile/free” for “delocalized” electrons in M2.

Accept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.

[Cr(CN)₆]³⁻ AND CN⁻/ligand causes larger splitting «in d-orbitals compared to OH⁻»
OR
[Cr(CN)₆]³⁻ AND CN⁻/ligand associated with a higher Δ/«crystal field» splitting energy/energy difference «in the spectrochemical series compared to OH⁻ » ✓

Accept “[Cr(CN)₆]³⁻ AND «CN⁻» strong field ligand”.

any value or range between 647 and 700 nm ✓

SF₄/SCl₂ structure does not have to be 3-D for mark.

Penalize missing lone pairs of electrons on halogens once only.

Accept any combination of dots, lines or crosses for bonds/lone pairs.

Accept “non-linear” for SCl₂ molecular geometry.

Award [1] for two correct electron domain geometries, e.g. trigonal bipyramidal for SF₄ and tetrahedral for SCl₂.

H₂O forms hydrogen bonding «while SCl₂ does not» ✓

SCl₂ «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓

Alternative 1:
H₂O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓

Alternative 2:
SCl₂ less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓

Ignore reference to Van der Waals.

Accept “SCl₂ has «much» larger molar mass/electron density” for M2.

State the electron configuration of the Cu⁺ ion.

Copper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.

4HCl (g) + O₂ (g) → 2Cl₂ (g) + 2H₂O (g)

Calculate the standard enthalpy change, ΔH^θ, in kJ, for this reaction, using section 12 of the data booklet.

The diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.

Annotate both charts to show the activation energy for the catalysed reaction, using the label E_{a (cat)}.

Explain how the catalyst increases the rate of the reaction.

Solid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl₂•$x$H₂O.

A student heated a sample of hydrated copper(II) chloride, in order to determine the value of $x$. The following results were obtained:

Mass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g

Determine the value of $x$.

State how current is conducted through the wires and through the electrolyte.

Wires: 

Electrolyte:

Write the half-equation for the formation of gas bubbles at electrode 1.

Bubbles of gas were also observed at another electrode. Identify the electrode and the gas.

Electrode number (on diagram):

Name of gas: 

Deduce the half-equation for the formation of the gas identified in (c)(iii).

Determine the enthalpy of solution of copper(II) chloride, using data from sections 18 and 20 of the data booklet.

The enthalpy of hydration of the copper(II) ion is −2161 kJ mol⁻¹.

Calculate the cell potential at 298 K for the disproportionation reaction, in V, using section 24 of the data booklet.

Comment on the spontaneity of the disproportionation reaction at 298 K.

Calculate the standard Gibbs free energy change, ΔG^θ, to two significant figures, for the disproportionation at 298 K. Use your answer from (e)(i) and sections 1 and 2 of the data booklet.

Suggest, giving a reason, whether the entropy of the system increases or decreases during the disproportionation.

Deduce, giving a reason, the sign of the standard enthalpy change, ΔH^θ, for the disproportionation reaction at 298 K.

Predict, giving a reason, the effect of increasing temperature on the stability of copper(I) chloride solution.

Describe how the blue colour is produced in the Cu(II) solution. Refer to section 17 of the data booklet.

Deduce why the Cu(I) solution is colourless.

When excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH₃)₄(H₂O)₂]²⁺, forms.

State the molecular geometry of this complex ion, and the bond angles within it.

Molecular geometry:

Bond angles: 

Examine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl₄]²⁻.

[Ar] 3d¹⁰
OR
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ ✔

ΔH^θ = ΣΔH^θ_f (products) − ΣΔH^θ_f (reactants) ✔
ΔH^θ = 2(−241.8 «kJ mol⁻¹») − 4(−92.3 «kJ mol⁻¹») = −114.4 «kJ» ✔

NOTE: Award [2] for correct final answer.

E_{a (cat)} to the left of E_a ✔                        

peak lower AND E_{a (cat)} smaller ✔

«catalyst provides an» alternative pathway ✔

«with» lower E_a
OR
higher proportion of/more particles with «kinetic» E ≥ E_a(cat) «than E_a» ✔

mass of H₂O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl₂ = «17.917 g – 16.221 g =» 1.696 «g» ✔

moles of H₂O = «$\frac{0.443 \text{g}}{18.02 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$= » 0.0126 «mol» ✔

«water : copper(II) chloride = 1.95 : 1»

«$x$ =» 2 ✔

NOTE: Accept «$x$ =» 1.95.

NOTE: Award [3] for correct final answer.

Wires:
«delocalized» electrons «flow» ✔

Electrolyte:
«mobile» ions «flow» ✔

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → $\frac{1}{2}$Cl₂ (g) + e⁻ ✔

NOTE: Accept e for e⁻.

«electrode» 3 AND oxygen/O₂ ✔

NOTE: Accept chlorine/Cl₂.

2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– ✔

NOTE: Accept 2Cl^– (aq) → Cl₂ (g) + 2e^–.
Accept 4OH⁻ → 2H₂O + O₂ + 4e⁻

enthalpy of solution = lattice enthalpy + enthalpies of hydration «of Cu²⁺ and Cl⁻» ✔

«+2824 kJ mol^–1 − 2161 kJ mol^–1 − 2(359 kJ mol^–1) =» −55 «kJ mol^–1» ✔

NOTE: Accept enthalpy cycle.
Award [2] for correct final answer.

E^θ = «+0.52 – 0.15 = +» 0.37 «V» ✔

spontaneous AND E^θ positive ✔

ΔG^θ = «−nFE = −1 mol × 96 500 C Mol^–1 × 0.37 V=» −36 000 J/−36 kJ ✔

NOTE: Accept “−18 kJ mol^–1 «per mole of Cu⁺»”.

Do not accept values of n other than 1.

Apply SF in this question.

Accept J/kJ or J mol⁻¹/kJ mol⁻¹ for units.

2 mol (aq) → 1 mol (aq) AND decreases ✔

NOTE: Accept “solid formed from aqueous solution AND decreases”.
Do not accept 2 mol → 1 mol without (aq).

ΔG^θ < 0 AND ΔS^θ < 0 AND ΔH^θ < 0
OR
ΔG^θ + TΔS^θ < 0 AND ΔH^θ < 0 ✔

TΔS more negative «reducing spontaneity» AND stability increases ✔

NOTE: Accept calculation showing non-spontaneity at 433 K.

«ligands cause» d-orbitals «to» split ✔

light absorbed as electrons transit to higher energy level «in d–d transitions»
OR
light absorbed as electrons promoted ✔

energy gap corresponds to «orange» light in visible region of spectrum ✔

colour observed is complementary ✔

full «3»d sub-level/orbitals
OR
no d–d transition possible «and therefore no colour» ✔

octahedral AND 90° «180° for axial» ✔

NOTE: Accept square-based bi-pyramid.

Any two of:
ligand/chloride ion Lewis base AND donates e-pair ✔
not Brønsted–Lowry base AND does not accept proton/H⁺ ✔
Lewis definition extends/broader than Brønsted–Lowry definition ✔

Suggest what can be concluded about the gold atom from this experiment.

Subsequent experiments showed electrons existing in energy levels occupying various orbital shapes.

Sketch diagrams of 1s, 2s and 2p.

State the electron configuration of copper.

Copper is a transition metal that forms different coloured complexes. A complex [Cu(H₂O)₆]²⁺(aq) changes colour when excess Cl⁻(aq) is added.

Explain the cause of this colour change, using sections 3 and 15 from the data booklet.

Most ⁴He²⁺ passing straight through:

most of the atom is empty space
OR
the space between nuclei is much larger than ⁴He²⁺ particles
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔

Very few ⁴He²⁺ deviating largely from their path:

nucleus/centre is positive «and repels ⁴He²⁺ particles»
OR
nucleus/centre is «more» dense/heavy «than ⁴He²⁺ particles and deflects them»
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔

Do not accept the same reason for both M1 and M2.

Accept “most of the atom is an electron cloud” for M1.

Do not accept only “nucleus repels ⁴He²⁺ particles” for M2.

1s AND 2s as spheres ✔

one or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation (p_x, p_y p_z) ✔

1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰
OR
[Ar] 4s¹3d¹⁰ ✔

Accept configuration with 3d before 4s.

chloride is lower in the spectrochemical series ✔

«ligand cause» decreased/lesser splitting «in d-orbitals compared to H₂O» ✔

frequency/energy of light absorbed is decreased
OR
wavelength of light absorbed is increased ✔

Accept ·chloride a weaker ligand than water/produces a smaller energy difference than water for M1.

Award [2 max] for mentioning splitting of orbitals is changed AND frequency/ wavelength/energy of light absorbed
are different/changed without mentioning correct decrease or increase.

Deduce the full electron configuration of Fe²⁺.

Explain why, when ligands bond to the iron ion causing the d-orbitals to split, the complex is coloured.

State the nuclear symbol notation, ${}_{\text{Z}}^{\text{A}}\text{X}$, for iron-54.

Mass spectrometry analysis of a sample of iron gave the following results:

Calculate the relative atomic mass, A_r, of this sample of iron to two decimal places.

An iron nail and a copper nail are inserted into a lemon.

Explain why a potential is detected when the nails are connected through a voltmeter.

Calculate the standard electrode potential, in V, when the Fe²⁺ (aq) | Fe (s) and Cu²⁺ (aq) | Cu (s) standard half-cells are connected at 298 K. Use section 24 of the data booklet.

Calculate ΔG^θ, in kJ, for the spontaneous reaction in (f)(i), using sections 1 and 2 of the data booklet.

Calculate a value for the equilibrium constant, K_c, at 298 K, giving your answer to two significant figures. Use your answer to (f)(ii) and section 1 of the data booklet. 

(If you did not obtain an answer to (f)(ii), use −140 kJ mol⁻¹, but this is not the correct value.)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶   [✔]

«frequency/wavelength of visible» light absorbed by electrons moving between d levels/orbitals    [✔]

colour due to remaining frequencies
OR
complementary colour transmitted    [✔]

${}_{\text{26}}^{\text{54}}\text{Fe}$     [✔]

«A_r =» 54 × 0.0584 + 56 × 0.9168 + 57 × 0.0217 + 58 × 0.0031

OR

«A_r =» 55.9111    [✔]

«A_r =» 55.91    [✔]

Note: Award [2] for correct final answer

Do not accept data booklet value (55.85).

lemon juice is the electrolyte
OR
lemon juice allows flow of ions
OR
each nail/metal forms a half-cell with the lemon juice    [✔]

Any one of:
iron is higher than copper in the activity series
OR
each half-cell/metal has a different redox/electrode potential     [✔]

iron is oxidized
OR
Fe → Fe²⁺ + 2e⁻
OR
Fe → Fe³⁺ + 3e⁻
OR
iron is anode/negative electrode of cell   [✔]

copper is cathode/positive electrode of cell
OR
reduction occurs at the cathode
OR
2H⁺ + 2e⁻ → H₂   [✔]

electrons flow from iron to copper   [✔]

«E^θ = +0.34 V −(−0.45 V) = +»0.79 «V»   [✔]

«ΔG^θ = −nFE^θ = −2mol × 96 500 C mol⁻¹ × $\frac{0.79\text{ J }{\text{C}}^{-1}}{1000}$ =» −152 «kJ»    [✔]

Note: Accept range 150−153 kJ.

«lnK_c = $-\frac{\mathrm{\Delta }{G}^{\theta }}{RT}=-\frac{-152\times {10}^3\text{ Jmo}{\text{l}}^{-1}}{8.31\text{J}{\text{K}}^{-1}\text{mo}{\text{l}}^{-1}\times 298\text{K}}$ =» 61.38    [✔]

K = 4.5 × 10²⁶    [✔]

Note: Accept answers in range 2.0 × 10²⁶ to 5.5 × 10²⁶.

Do not award M2 if answer not given to two significant figures.

If −140 kJmol⁻¹ used, answer is 3.6 × 10²⁴.

Done fairly well with common mistakes leaving in the 4s² electrons as part of Fe²⁺ electron configuration, or writing 4s¹ 3d⁵

This was poorly answered and showed a clear misconception and misunderstanding of the concepts. Most of the candidates failed to explain why the complex is coloured and based their answers on the emission of light energy when electrons fall back to ground state and not on light absorption by electrons moving between the split d-orbitals and complementary colour transmitted of certain frequencies.

Many candidates wrote the nuclear notation for iron as Z over A.

This question on average atomic mass was the best answered question on the exam. A few candidates did not write the answer to two decimal places as per instructions.

Very few candidates scored M1 regarding the lemon juice role as electrolyte. Some earned M2 but a lot of answers were too vague, such as ‘electrons move through the circuit’, etc.

Only 50 % of candidates earned this relatively easy mark on calculate EMF from 2 half-cell electrode potentials.

Average performance; typical errors were using the incorrect value for n, the number of electrons, or not using consistent units and making a factor of 1000 error in the final answer.

This question was left blank by quite a few candidates. Common errors included not using correct units, or more often, calculation error in converting ln K_c into K_c value.