Estimate the H−N−H bond angle in methanamine using VSEPR theory.

State the electron domain geometry around the nitrogen atom and its hybridization in methanamine.

Ammonia reacts reversibly with water.
$$\text{N}{\text{H}}_{\text{3}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{NH}}_{\text{4}}^{+}\text{(aq)}+\text{O}{\text{H}}^{-}\text{(aq)}$$
Explain the effect of adding ${\text{H}}^{+}\text{(aq)}$ ions on the position of the equilibrium.

Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second H⁺ is so small it can be neglected.)

$${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq)}+{\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{+}\text{(aq)}+\text{O}{\text{H}}^{-}\text{(aq)}$$

$$\text{p}{K}_{\text{b}}\text{ (hydrazine)}=5.77$$

Calculate the pH of a $0.0100\text{ mol}\text{d}{\text{m}}^{-3}$ solution of hydrazine.

Suggest a suitable indicator for the titration of hydrazine solution with dilute sulfuric acid using section 22 of the data booklet.

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

Determine the enthalpy change of reaction, $\mathrm{\Delta }H$, in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.

$${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}\to {\text{N}}_{\text{2}}\text{(g)}+\text{2}{\text{H}}_{\text{2}}\text{(g)}$$

The standard enthalpy of formation of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}$ is $+50.6\text{ kJ}\text{mo}{\text{l}}^{-1}$. Calculate the enthalpy of vaporization, $\mathrm{\Delta }{H}_{\text{vap}}$, of hydrazine in $\text{kJ}\text{mo}{\text{l}}^{-1}$. $${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}$$ (If you did not get an answer to (f), use $-85\text{ kJ}$ but this is not the correct answer.)

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from $\text{1000 d}{\text{m}}^{\text{3}}$ of the sample.

Calculate the volume, in $\text{d}{\text{m}}^{\text{3}}$, of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = $\text{24.8 d}{\text{m}}^{\text{3}}$ at SATP.)

107°

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

tetrahedral

sp³

No ECF allowed.

[2 marks]

removes/reacts with $\text{O}{\text{H}}^{-}$

moves to the right/products «to replace $\text{O}{\text{H}}^{-}$ ions»

Accept ionic equation for M1.

[2 marks]

K_b = 10^–5.77 / 1.698 x 10^–6
OR
$K_{\text{b}}=\frac{\left[{\text{N}}_{\text{2}}{\text{H}}_5^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]}{\left[{\text{N}}_{\text{2}}{\text{H}}_4\right]}$

 [OH^–]² «= 1.698 × 10^–6 × 0.0100» = 1.698 × 10^–8

OR

[OH^–] «$=\sqrt{1.698\times {10}^{-8}}$» = 1.303 × 10^–4 «mol dm^–3»

pH «$=-\text{lo}{\text{g}}_{10}\frac{1\times {10}^{-14}}{1.3\times {10}^{-4}}$» = 10.1

Award [3] for correct final answer.

Give appropriate credit for other methods containing errors that do not yield correct final answer.

[3 marks]

methyl red

OR

bromocresol green

OR

bromophenol blue

OR

methyl orange

[1 mark]

bubbles

OR

gas

OR

magnesium disappears

${\text{2NH}}_{\text{4}}^{+}\text{(aq)}+\text{Mg(s)}\to \text{M}{\text{g}}^{\text{2}+}\text{(aq)}+\text{2N}{\text{H}}_{\text{3}}\text{(aq)}+{\text{H}}_{\text{2}}\text{(g)}$

Do not accept “hydrogen” without reference to observed changes.

Accept "smell of ammonia".

Accept 2H⁺(aq) + Mg(s) $\to$ Mg²⁺(aq) + H₂(g)

Equation must be ionic.

[2 marks]

bonds broken:

E(N–N) + 4E(N–H)

OR

$158\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg +4\times 391\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg /1722\ll \text{kJ}\gg$

bonds formed:

E(N$\equiv$N) + 2E(H–H)

OR

$945\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg +2\times 436\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg /1817\ll \text{kJ}\gg$

$\ll \mathrm{\Delta }H=\text{bonds broken}-\text{bonds formed}=1722-1817=\gg -95\ll \text{kJ}\gg$

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

OR

$\mathrm{\Delta }{H}_{\text{vap}}=-50.6\text{ kJ}\text{mo}{\text{l}}^{-1}-\text{(}-95\text{ kJ}\text{mo}{\text{l}}^{-1}\text{)}$

$\ll \mathrm{\Delta }{H}_{vap}=\gg +44\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg$

Award [2] for correct final answer. Award [1 max] for –44 «kJ mol^–1».

Award [2] for:

ΔH_vap = –50.6 kJ mol^–1 – (–85 J mol^–1) = ＋34 «kJ mol^–1».

Award [1 max] for –34 «kJ mol^–1».

[2 marks]

total mass of oxygen $\ll =8.0\times {10}^{-3}\text{ g}\text{d}{\text{m}}^{-3}\times 1000\text{ d}{\text{m}}^3\gg =8.0\ll \text{g}\gg$

$\text{n(}{\text{O}}_{\text{2}}\text{) }\ll =\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=\gg 0.25\ll \text{mol}\gg$

OR

$\text{n(}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{)}=\text{n(}{\text{O}}_{\text{2}}\text{)}$

$\ll \text{mass of hydrazine}=0.25\text{ mol}\times 32.06\text{ g}\text{mo}{\text{l}}^{-1}=\gg 8.0\ll \text{g}\gg$

Award [3] for correct final answer.

[3 marks]

$\ll \text{n(}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{)}=\text{n(}{\text{O}}_{\text{2}}\text{)}=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=\gg 0.25\ll \text{mol}\gg$

$\ll \text{volume of nitrogen}=0.25\text{ mol}\times 24.8\text{ d}{\text{m}}^3\text{mo}{\text{l}}^{-1}\gg =6.2\ll \text{d}{\text{m}}^3\gg$

Award [1] for correct final answer.

[1 mark]

State the full electronic configuration of Fe²⁺.

Explain why there is a large increase from the 8th to the 9th ionization energy of iron.

Calculate the oxidation state of sulfur in iron(II) disulfide, FeS₂.

Describe the bonding in iron, Fe (s).

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ ✔

Any two of:

IE₉: electron in lower energy level
OR
IE₉: more stable/full electron level ✔

IE₉: electron closer to nucleus
OR
IE₉: electron more tightly held by nucleus ✔

IE₉: less shielding by «complete» inner levels ✔

–1 ✔

Accept “– I”.

electrostatic attraction/hold between «lattice of» positive ions/cations AND delocalized «valence» electrons ✔

Mostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s² 3d⁴.

Despite some confusion regarding which sub-level the electrons were being removed from, many candidates were able to make at least one valid point, commonly in terms of lower energy/ full sub level/closer to nucleus.

This was an easy question, yet 30% of the candidates were unable to work it out; some wrote the oxidation state in the conventionally incorrect format, 1- and lost the mark.

Most candidates knew the bonding in Fe is metallic but some did not “describe” it or missed the type of attraction, a minor mistake; others referred to nuclei or protons instead of cations/positive ions. In some cases, candidates referred too ionic bonding, probably still thinking of FeS₂ (not reading the question well). Overall, only 30% answered satisfactorily.

Outline how a catalyst increases the rate of reaction.

Explain why an increase in temperature increases the rate of reaction.

Discuss, referring to intermolecular forces present, the relative volatility of propanone and propan-2-ol.

The diagram shows an unlabelled voltaic cell for the reaction

${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Ni}\left(\mathrm{s}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Pb}\left(\mathrm{s}\right)$

Label the diagram with the species in the equation.

Calculate the standard cell potential, in $\mathrm{V}$, for the cell at $298 \mathrm{K}$. Use section 24 of the data booklet

Calculate the standard free energy change, $∆{\mathrm{G}}^{⦵}$, in $\mathbf{kJ}$, for the cell using sections 1 and 2 of the data booklet.

Suggest a metal that could replace nickel in a new half-cell and reverse the electron flow. Use section 25 of the data booklet.

Describe the bonding in metals.

Nickel alloys are used in aircraft gas turbines. Suggest a physical property altered by the addition of another metal to nickel.

provides an alternative pathway/mechanism AND lower E_a ✔

Accept description of how catalyst lowers E_a (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”).

more/greater proportion of molecules with E E_a ✔

greater frequency/probability/chance of collisions «between the molecules»
OR
more collision per unit of time/second ✔

hydrogen bonding/bonds «and dipole–dipole and London/dispersion forces are present in» propan-2-ol ✔

dipole–dipole «and London/dispersion are present in» propanone ✔

propan-2-ol less volatile AND hydrogen bonding/bonds stronger «than dipole–dipole »
OR
propan-2-ol less volatile AND «sum of all» intermolecular forces stronger ✔

$«-0.13 \mathrm{V}-(-0.26 \mathrm{V})=+»0.13«\mathrm{V}»$ ✔

$«{\mathrm{\Delta G}}^{\mathrm{\Theta }}=-{\mathrm{nFE}}^{\mathrm{\Theta }}=-2\times 96 500\times \frac{0.13}{1000}=»-25 «\mathrm{k} \mathrm{J}»$ ✔

$\mathrm{Bi}/\mathrm{Cu}/\mathrm{Ag}/\mathrm{Pd}/\mathrm{Hg}/\mathrm{Pt}/\mathrm{Au}$ ✔

Accept $Sb$ OR $As$.

electrostatic attraction ✔

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

Accept “mobile/free electrons”.

Any of:

malleability/hardness
OR
«tensile» strength/ductility
OR
density
OR
thermal/electrical conductivity
OR
melting point
OR
thermal expansion ✔

Do not accept corrosion/reactivity or any chemical property.

Accept other specific physical properties.

Although fairly well done some candidates did not mention that providing an alternate pathway to the reaction was how the activation energy was lowered and hence did not gain the mark.

Almost all candidates earned at least 1 mark for the effect of temperature on rate. Some missed increase in collision frequency, others the idea that more particles reached the required activation energy.

The average mark was 1.9/3. Almost all candidates could recognize hydrogen bonding in alcohol but many missed the dipole-dipole attraction in propanone. There was also some confusion on the term volatility, with some thinking stronger IMF meant higher volatility.

A surprising number of No Response for a question where candidates simply had to label a diagram with the species in the equation. Some candidates had the idea but did not use the species for electrolytic cell, e.g., Pb(SO₄) instead of Pb²⁺(aq).

80% of candidates could correctly calculate a cell potential by using a reduction table and a balanced redox reaction. 

This was similar to 2f(ii) where many could apply the formula for Gibbs free energy change, ΔG^ө, correctly however some did not get the units correct.

80% could correctly pick a metal to reverse the electron flow, however some candidates thought a more reactive, rather than a less reactive metal than nickel would reverse the electron flow.

Most candidates were aware that metallic bonding involved a "sea of electrons", but were unsure about surrounding what and could not identify that it was electrostatic attraction holding the metal together.

Almost all candidates could correctly identify a physical property of a metal which might be altered when alloying.

Draw the Lewis structures of oxygen, O₂, and ozone, O₃.

Outline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.

Reason: 

Length:

Predict the bond angle in the ozone molecule.

Discuss how the different bond strengths between the oxygen atoms in O₂ and O₃ in the ozone layer affect radiation reaching the Earth’s surface.

Identify the steps which absorb ultraviolet light.

Determine, showing your working, the wavelength, in m, of ultraviolet light absorbed by a single molecule in one of these steps. Use sections 1, 2 and 11 of the data booklet.

Ozone depletion is catalysed by nitrogen monoxide, NO, which is produced in aircraft and motor vehicle engines, and has the following Lewis structure.

Show how nitrogen monoxide catalyses the decomposition of ozone, including equations in your answer.

NOTES: Coordinate bond may be represented by an arrow.

Do not accept delocalized structure for ozone.

resonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔

NOTE: Accept any length between these two values.

any value from 110°–119° ✔

«bond» in O₂ stronger than in O₃ ✔

ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔

NOTE: Accept ozone «layer» absorbs a range of frequencies.

steps 1 AND 3 ✔

ALTERNATIVE 1:
for oxygen:

$E=«\frac{498 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 8.27\times {10}^{-19} «\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{8.27\times {10}^{-19} \text{J}}=» 2.40\times {10}^{-7} «\text{m}»$✔

ALTERNATIVE 2:
for ozone:

similar calculation using 200 < bond enthalpy < 400 for ozone, such as

$E=«\frac{300 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 4.98\times {10}^{-19}«\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{4.98\times {10}^{-19} \text{J}}=» 3.99\times {10}^{-7} «\text{m}»$✔

NOTE: Award [2] for correct final answer.

•NO + O₃ → •NO₂ + O₂ ✔

•NO₂ + O₃ → •NO + 2O₂ ✔

NOTE: Accept •NO₂ → •NO + •O AND •O + O₃ → 2O₂ for M2.

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]

Outline why metals, like iron, can conduct electricity.

Justify why sulfur is classified as a non-metal by giving two of its chemical properties.

Sketch the first eight successive ionisation energies of sulfur.

Describe the bonding in this type of solid.

State a technique that could be used to determine the crystal structure of the solid compound.

State the full electron configuration of the sulfide ion.

Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.

Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

Explain why the addition of small amounts of carbon to iron makes the metal harder.

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

forms covalent/bonds compounds «with other non-metals» ✔

forms anions «rather than cations» ✔

behaves as an oxidizing agent «rather than a reducing agent» ✔

Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.

two regions of small increases AND a large increase between them✔

large increase from 6th to 7th ✔

Accept line/curve showing these trends.

electrostatic attraction ✔

between oppositely charged ions/between Fe²⁺ and S²⁻ ✔

X-ray crystallography ✔

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔

Accept 2,8 (for O^2–) and 2,8,8 (for S^2–)

allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔

Accept other valid answers.

4FeS(s) + 7O₂(g) → 2Fe₂O₃(s) + 4SO₂(g) ✔

Accept any correct ratio.

+6
OR
−2 to +4 ✔

Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2- to 4+.

sulfur dioxide/SO₂ causes acid rain ✔

Accept sulfur dioxide/SO₂/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.

disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔

prevents layers/atoms sliding over each other ✔

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]

Carbon dioxide can be represented by at least two resonance structures, I and II.

Calculate the formal charge on each oxygen atom in the two structures.

Deduce, giving a reason, the more likely structure.

Absorption of UV light in the ozone layer causes the dissociation of oxygen and ozone.

Identify, in terms of bonding, the molecule that requires a longer wavelength to dissociate.

Carbon and silicon are elements in group 14.

Explain why CO₂ is a gas but SiO₂ is a solid at room temperature.

Award [1] for any two correctly filled cells.

[2 marks]

structure I AND no formal charges

OR

structure I AND no charge transfer «between atoms»

[1 mark]

O₃ has bond between single and double bond AND O₂ has double bond

OR

O₃ has bond order of 1.5 AND O₂ has bond order of 2

OR

bond in O₃ is weaker/longer than in O₂

O₃ requires longer wavelength

M1: Do not accept “ozone has one single and one double bond”.

[2 marks]

CO₂ «non-polar» «weak» London/dispersion forces/instantaneous induced dipole-induced dipole forces between molecules

SiO₂ network/lattice/3D/giant «covalent» structure

M1: The concept of “between” is essential.

[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.

Describe the nature of ionic bonding.

Describe how the relative atomic mass of a sample of calcium could be determined from its mass spectrum.

When calcium compounds are introduced into a gas flame a red colour is seen; sodium compounds give a yellow flame. Outline the source of the colours and why they are different.

Suggest two reasons why solid calcium has a greater density than solid potassium.

Outline why solid calcium is a good conductor of electricity.

Sketch a graph of the first six ionization energies of calcium.

Calcium carbide reacts with water to form ethyne and calcium hydroxide.

CaC₂(s) + H₂O(l) → C₂H₂(g) + Ca(OH)₂(aq)

Estimate the pH of the resultant solution.

Describe how sigma (σ) and pi ($\pi$) bonds are formed.

Deduce the number of σ and $\pi$ bonds in a molecule of ethyne.

electrostatic attraction AND oppositely charged ions

[1 mark]

multiply relative intensity by «m/z» value of isotope

OR

find the frequency of each isotope

sum of the values of products/multiplication «from each isotope»

OR

find/calculate the weighted average

Award [1 max] for stating “m/z values of isotopes AND relative abundance/intensity” but not stating these need to be multiplied.

[2 marks]

«promoted» electrons fall back to lower energy level

energy difference between levels is different

Accept “Na and Ca have different nuclear charge” for M2.

[2 marks]

Any two of:

stronger metallic bonding

smaller ionic/atomic radius

two electrons per atom are delocalized

OR

greater ionic charge

greater atomic mass

Do not accept just “heavier” or “more massive” without reference to atomic mass.

[2 marks]

delocalized/mobile electrons «free to move»

[1 mark]

general increase

only one discontinuity between “IE2” and “IE3”

[2 marks]

pH > 7

Accept any specific pH value or range of values above 7 and below 14.

[1 mark]

sigma (σ):

overlap «of atomic orbitals» along the axial/internuclear axis

OR

head-on/end-to-end overlap «of atomic orbitals»

pi ($\pi$):

overlap «of p-orbitals» above and below the internuclear axis

OR

sideways overlap «of p-orbitals»

Award marks for suitable diagrams.

[2 marks]

sigma (σ): 3

AND

pi ($\pi$): 2

[1 mark]

State the name of Compound B, applying International Union of Pure and Applied Chemistry (IUPAC) rules.

Compound A and Compound B are both liquids at room temperature and pressure. Identify the strongest intermolecular force between molecules of Compound A.

State the number of $\text{σ}$ (sigma) and $\mathrm{\pi }$ (pi) bonds in Compound A.

Deduce the hybridization of the central carbon atom in Compound A.

Identify the isomer of Compound B that exists as optical isomers (enantiomers).

Draw the structural formula of the alkene required.

Explain why the reaction produces more (CH₃)₃COH than (CH₃)₂CHCH₂OH.

Deduce the structural formula of the repeating unit of the polymer formed from this alkene.

Deduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).

Identify the type of reaction.

Outline the requirements for a collision between reactants to yield products.

Explain the mechanism of the reaction using curly arrows to represent the movement of electron pairs.

The polarity of the carbon–halogen bond, C–X, facilitates attack by HO^–.

Outline, giving a reason, how the bond polarity changes going down group 17.

2-methylpropan-2-ol /2-methyl-2-propanol ✔

Accept methylpropan-2-ol/ methyl-2-propanol.

Do not accept 2-methylpropanol.

dipole-dipole ✔

Do not accept van der Waals’ forces.

$\text{σ}$: 9
AND
$\mathrm{\pi }$: 1 ✔

sp² ✔

butan-2-ol/CH₃CH(OH)C₂H₅ ✔

carbocation formed from (CH₃)₃COH is more stable / (CH₃)₃C⁺ is more stable than (CH₃)₂CHCH₂⁺ ✔

«because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e^- density
OR
«greater number of alkyl groups» are more electron releasing
OR
«greater number of alkyl groups creates» greater inductive/+I effect ✔

Do not award any marks for simply quoting Markovnikov’s rule.

Do not penalize missing brackets or n.

Do not award mark if continuation bonds are not shown.

no change «in colour/appearance/solution» ✔

«nucleophilic» substitution
OR
SN2 ✔

Accept “hydrolysis”.

Accept SN1

energy/E ≥ activation energy/E_a ✔

correct orientation «of reacting particles»
OR
correct geometry «of reacting particles» ✔

curly arrow going from lone pair/negative charge on O in ^-OH to C ✔

curly arrow showing I leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

Accept OH^- with or without the lone pair.

Do not allow curly arrows originating on H, rather than the -, in OH^-.

Accept curly arrows in the transition state.

Do not penalize if HO and I are not at 180°.

Do not award M3 if OH–C bond is represented.

Award [2 max] if S_N1 mechanism shown.

decreases/less polar AND electronegativity «of the halogen» decreases ✔

Accept “decreases” AND a correct comparison of the electronegativity of two halogens.

Accept “decreases” AND “attraction for valence electrons decreases”.

Naming the organic compound using IUPAC rules was generally done well.

Mediocre performance in stating the number of σ (sigma) and π (pi) bonds in propanone; the common answer was 3 σ and 1 π instead of 9 σ and 1 π, suggesting the three C-H σ bonds in each of the two methyl groups were ignored.

sp² hybridization of the central carbon atom in the ketone was very done well.

Mediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH₃CH(OH)C₂H₅ as the isomer that exists as an optical isomer.

Good performance; some had a H and CH₃ group on each C atom across double bond instead of having two H atoms on one C and two CH₃ groups on the other.

Poor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.

Mediocre performance; deducing structural formula of repeating unit of the polymer was challenging in which continuation bonds were sometimes missing, or structure included a double bond or one of the CH₃ group was missing.

Mediocre performance; deducing whether the tertiary alcohol could be oxidized solicited mixed responses ranging from the correct one, namely no change (in colour, appearance or solution), to tertiary alcohol will be reduced, or oxidized, or colour will change will occur, and such.

Excellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.

Good performance. For the requirements for a collision between reactants to yield products, some suggested necessary, sufficient or enough energy or even enough activation energy instead of energy/E ≥ activation energy/E_a.

Mechanism for SN2 not done well. Often the negative charge on OH was missing, the curly arrow was not going from lone pair/negative charge on O in -OH to C, or the curly arrow showing I leaving placed incorrectly and specially the negative charge was missing in the transition state. Formation of a carbocation intermediate indicating SN1 mechanism could score a maximum of 2 marks.

Good performance on how the polarity of C-X bond changes going down group 17.

Outline why ethanoic acid is classified as a weak acid.

A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.

Cl₂ (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H₂O (l)

Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.

Draw a Lewis (electron dot) structure of chloramine.

State the hybridization of the nitrogen atom in chloramine.

Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.

Molecular geometry:

H–N–H bond angle:

State the type of bond formed when chloramine is protonated.

Sketch a graph of pH against volume of hydrochloric acid added to ammonia solution, showing how you would determine the pK_a of the ammonium ion.

Suggest a suitable indicator for the titration, using section 22 of the data booklet.

Explain, using two equations, how an equimolar solution of ammonia and ammonium ions acts as a buffer solution when small amounts of acid or base are added.

partial dissociation «in aqueous solution»    [✔]

ethanoic acid/vinegar reacts with NaOH    [✔]

moves equilibrium to left/reactant side    [✔]

releases Cl₂ (g)/chlorine gas
OR
Cl₂ (g)/chlorine gas is toxic    [✔]

Note: Accept “ethanoic acid produces H⁺ ions”

Accept “ethanoic acid/vinegar reacts with NaOCl”.

Do not accept “2CH₃COOH + NaOCl + NaCl → 2CH₃COONa + Cl₂ + H₂O” as it does not refer to equilibrium.

Accept suitable molecular or ionic equations for M1 and M3.

    [✔]

Note: Accept any combination of dots/crosses or lines to represent electron pairs.

sp³    [✔]

Molecular geometry:
«trigonal» pyramidal   [✔]

H–N–H bond angle:
107°    [✔]

Note: Accept angles in the range of 100–109.

covalent/dative/coordinate    [✔]

correct shape of graph AND vertical drop at Vn    [✔]

pK_a = pH at $\frac{\text{Vn}}{2}$/half neutralization/half equivalence    [✔]

Note: M1: must show buffer region at pH > 7 and equivalence point at pH < 7. Graph must start below pH = 14.

methyl orange
OR
bromophenol blue
OR
bromocresol green
OR
methyl red    [✔]

NH₃ (aq) + H⁺ (aq) → NH₄ + (aq)    [✔]

NH₄ + (aq) + OH⁻ (aq) → NH₃ (aq) + H₂O(l)    [✔]

Note: Accept reaction arrows or equilibrium signs in both equations.

Award [1 max], based on two correct reverse equations but not clearly showing reacting with acid or base but rather dissociation.

Majority of candidates understood weak acids do not fully dissociate.

The average score was 1 out 3. Many could not suggest why it is dangerous to mix chlorine with vinegar. Most students gained at least one mark for stating that “chlorine gas will be produced” but couldn’t link it to equilibrium ideas.

Most candidates correctly drew the Lewis structure of chloramine. Some left off lone pair electrons.

Mostly correct with a surprising number stating sp or sp² hybridization.

Generally well done with some candidates misinterpreting the bond angle from the stated geometry.

“Ionic bond”, “hydrogen bond” and “intermolecular forces” were some common answers.

Quite poorly done with many candidates not indicating a vertical drop but rather a weak acid/weak base curve. Some did not have the correct location for the equivalence point.

Generally well done although a number of candidates chose bromothymol blue as a suitable indicator for weak base with a strong acid.

Nearly 30 % of candidates did not attempt to answer this question about buffer equations. It was also poorly answered because equations were not used to explain buffer action or the dissociation equations for the base and acid were given rather than their reactions with H⁺ or OH^- .

Discuss the bonding in the resonance structures of ozone.

Deduce one resonance structure of ozone and the corresponding formal charges on each oxygen atom.

The first six ionization energies, in kJ mol^–1, of an element are given below.

Explain the large increase in ionization energy from IE₃ to IE₄.

At a given time, the concentration of NO₂(g) and N₂O₄(g) were 0.52 and $0.10\text{ mol}\text{d}{\text{m}}^{-3}$ respectively.

Deduce, showing your reasoning, if the forward or the reverse reaction is favoured at this time.

Comment on the value of ΔG when the reaction quotient equals the equilibrium constant, Q = K.

lone pair on p orbital «of O atom» overlaps/delocalizes with pi electrons «from double bond»

both O–O bonds have equal bond length
OR
both O–O bonds have same/1.5 bond order
OR
both O–O are intermediate between O–O AND O=O 

both O–O bonds have equal bond energy

Accept “p/pi/$\pi$ electrons are delocalized/not localized”.

[3 marks]

ALTERNATIVE 1:

FC: –1 AND +1 AND 0

ALTERNATIVE 2:

FC: 0 AND +1 AND –1

Accept any combination of lines, dots or crosses to represent electrons.

Do not accept structure that represents 1.5 bonds.

Do not penalize missing lone pairs if already penalized in 3(b).

If resonance structure is incorrect, no ECF.

Any one of the structures with correct formal charges for [2 max].

[2 marks]

Any two of:
IE₄: electron in lower/inner shell/energy level
OR
IE₄: more stable/full electron shell

IE₄: electron closer to nucleus
OR
IE₄: electron more tightly held by nucleus

IE₄: less shielding by complete inner shells

Accept “increase in effective nuclear charge” for M2.

[2 marks]

«Q_c = $\frac{0.10}{{0.52}^2}$ =» 0.37
reaction proceeds to the left/NO₂(g) «until Q = K_c»
OR
reverse reaction «favoured»

Do not award M2 without a calculation for M1 but remember to apply ECF.

[2 marks]

ΔG = 0
reaction at equilibrium
OR
rate of forward and reverse reaction is the same
OR
constant macroscopic properties

[2 marks]

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

Write an equation for the complete combustion of ethyne.

Deduce the Lewis (electron dot) structure of ethyne.

Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C₂H₆.

Identify the type of interaction that must be overcome when liquid ethyne vaporizes.

State the name of product B, applying IUPAC rules.

Determine the enthalpy change for the reaction, in kJ, to produce A using section 11 of the data booklet.

The enthalpy change for the reaction to produce B is −213 kJ.

Predict, giving a reason, which product is the most stable.

The IR spectrum and low resolution ¹H NMR spectrum of the actual product formed are shown.

Deduce whether the product is A or B, using evidence from these spectra together with sections 26 and 27 of the  data booklet.

Identity of product:

One piece of evidence from IR:

One piece of evidence from ¹H NMR:

Deduce the splitting pattern you would expect for the signals in a high resolution ¹H NMR spectrum.

2.3 ppm:

9.8 ppm:

Suggest the reagents and conditions required to ensure a good yield of product B.

Reagents: 

Conditions:

Deduce the average oxidation state of carbon in product B.

Explain why product B is water soluble.

C₂H₂ (g) + 2.5O₂ (g) → 2CO₂ (g) + H₂O (l)

OR

2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (l)    [✔]

    [✔]

Note: Accept any valid combination of lines, dots and crosses.

«ethyne» shorter AND a greater number of shared/bonding electrons

OR

«ethyne» shorter AND stronger bond     [✔]

London/dispersion/instantaneous dipole-induced dipole forces    [✔]

ethanal    [✔]

«sum of bond enthalpies of reactants =» 2(C—H)+C ≡ C + 2(O—H)
OR
2 × 414 «kJ mol^-1» + 839 «kJ mol^-1» + 2 × 463 «kJ mol^-1»
OR
2593 «kJ»    [✔]

«sum of bond enthalpies of A =» 3(C—H) + C=C + C—O + O—H
OR
3 × 414 «kJ mol^-1» + 614 «kJ mol^-1» + 358 «kJ mol^-1» + 463 «kJ mol^-1»
OR
2677 «kJ»     [✔]
«enthalpy of reaction = 2593 kJ – 2677 kJ» = –84 «kJ»     [✔]

Note: Award [3] for correct final answer.

B AND it has a more negative/lower enthalpy/«potential» energy

OR

B AND more exothermic «enthalpy of reaction from same starting point»     [✔]

Identity of product: «B»

IR spectrum:
1700–1750 «cm^–1 band» AND carbonyl/CO group present
OR
no «band at» 1620–1680 «cm^–1» AND absence of double bond/C=C
OR
no «broad band at» 3200–3600 «cm^–1 » AND absence of hydroxyl/OH group    [✔]

¹H NMR spectrum:
«only» two signals AND A would have three
OR
«signal at» 9.4–10.0 «ppm» AND «H atom/proton of» aldehyde/–CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of alkyl/CH next to» aldehyde/CHO present
OR
«signal at» 2.2–2.7 «ppm» AND «H atom/proton of» RCOCH_2- present
OR
no «signal at» 4.5–6.0 «ppm» AND absence of «H atom/proton next to» double bond/C=C ✔

Note: Accept a specific value or range of wavenumbers and chemical shifts.

Accept “two signals with areas 1:3”.

2.3 ppm: doublet    [✔]

9.8 ppm: quartet    [✔]

Reagents:
acidified/H⁺ AND «potassium» dichromate«(VI)»/K₂Cr₂O₇/Cr₂O₇^2-    [✔]

Conditions:
distil «the product before further oxidation»       [✔]

Note: Accept “«acidified potassium» manganate(VII)/KMnO₄/MnO₄^-/permanganate”.

Accept “H₂SO₄” or “H₃PO₄” for “H⁺”.

Accept “more dilute dichromate(VI)/manganate(VII)” or “excess ethanol”.

Award M1 if correct reagents given under “Conditions”.

–1     [✔]

Any three of:

has an oxygen/O atom with a lone pair      [✔]

that can form hydrogen bonds/H-bonds «with water molecules»     [✔]

hydrocarbon chain is short «so does not disrupt many H-bonds with water molecules»     [✔]

«large permanent» dipole-dipole interactions with water      [✔]

All candidates were able to write the correct reactants/products for combustion of ethyne, but a few failed to balance correctly.

Most drew correct Lewis structures for ethyne, though some drew ethene.

Surprisingly very few explained the difference in bond length/strength looking at electrons shared and just gave the shorter/triple or longer/single bond answer.

Good to see that most candidates identified the specific IMF correctly.

Most candidates gave the correct IUPAC name.

Candidates were able to calculate the ΔH of the given reaction correctly; a few inverted the calculations or made mathematical errors.

Generally well done, most common error was stating that the enthalpy change was “larger” without the indication that it was an exothermic change or the sign.

Interpretation of spectra was very good and the few candidates that lost marks with ¹H NMR data rather than IR, for example simply mentioning two signals for B. However, most candidates that attempted this question got full marks.

The stronger candidates were able to predict the splitting pattern correctly, others inverted the answer, but many others repeated the information for protons with the given chemical shift, which is unexpected since wording was straightforward.

Candidates seemed to be confused by the prompts, reagent and conditions, so often included the acid among conditions. Careless errors were common such as the wrong charge on the dichromate ion. Few candidates suggest permanganate as an option.

Most candidates were able to calculate oxidation state of carbon in B.

Candidates did not understand that they must mention the IMF responsible for the solubility. Most candidates explained the polarity of the aldehyde and water but did not mention that this results in permanent dipole-dipole interactions; many did mention H-bonding. The mention of the lone pair on O atom and short hydrocarbon chain were very rare.

Outline, giving a reason, the effect of a catalyst on a reaction.

On the axes, sketch Maxwell–Boltzmann energy distribution curves for the reacting species at two temperatures T₁ and T₂, where T₂ > T₁.

Explain the effect of increasing temperature on the yield of SO₃.

Draw the Lewis structure of SO₃.

Explain the electron domain geometry of SO₃.

State the product formed from the reaction of SO₃ with water.

State the meaning of a strong Brønsted–Lowry acid.

increases rate AND lower E_a ✔

provides alternative pathway «with lower E_a»
OR
more/larger fraction of molecules have the «lower» E_a ✔

Accept description of how catalyst lowers E_a for M2 (e.g. “reactants adsorb on surface «of catalyst»”, “reactant bonds weaken «when adsorbed»”, “helps favorable orientation of molecules”).

both axes correctly labelled ✔

peak of T₂ curve lower AND to the right of T₁ curve ✔

lines begin at origin AND correct shape of curves AND T₂ must finish above T₁ ✔

Accept “probability «density» / number of particles / N / fraction” on y-axis.

Accept “kinetic E/KE/E_k” but not just “Energy/E” on x-axis.

decrease AND equilibrium shifts left / favours reverse reaction ✔

«forward reaction is» exothermic / ΔH is negative ✔

✔

Note:

Accept any of the above structures as formal charge is not being assessed.

three electron domains «attached to the central atom» ✔

repel/as far away as possible /120° «apart» ✔

sulfuric acid/H₂SO₄ ✔

Accept “disulfuric acid/H₂S₂O₇”.

fully ionizes/dissociates ✔

proton/H⁺ «donor »✔

Overall well answered though some answers were directed to explain the specific example rather than the simple and standard definition of the effect of a catalyst.

Few got the 3 marks for this standard question (average mark 1.7), the most common error being incomplete/incorrect labelling of axes, curves beginning above 0 on y-axis or inverted curves.

Many candidates got one mark at least, sometimes failing to state the effect on the production of SO₃ though they knew this quite obviously. This failure to read the question properly also resulted in an incorrect prediction based exclusively on kinetics instead of using the information provided to guide their answers.

Drawing the Lewis structure of SO₃ proved to be challenging, with lots of incorrect shapes, lone pair on S, etc.; accepting all resonant structures allowed many candidates to get the mark which was fair considering no formal charge estimation was required.

Most were focussed on the shape itself instead of explaining what led them to suggest that shape; number of electron domains allowed most candidates to get one mark and eventually a mention of bond angles resulted in only 35% getting both marks. In general, students were not able to provide clear explanations for the shape (not a language issue) but rather were happy to state the molecular geometry which they knew, but wasn't what was actually required for the mark.

6(d)(i)-(ii): These simple questions could be expected to be answered by all HL candidates. However 20% of the candidates suggested hydroxides or hydrogen as products of an aqueous dissolution of sulphur oxide. In the case of the definition of a strong Brønsted-Lowry acid, only 50% got both marks, often failing to define "strong" but in other cases defining them as bases even.

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.

Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.

Write an equation for the reaction of white phosphorus (P₄) with chlorine gas to form phosphorus trichloride (PCl₃).

Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl₃.

Outline the reason why PCl₅ is a non-polar molecule, while PCl₄F is polar.

Calculate the standard enthalpy change (ΔH^⦵) for the forward reaction in kJ mol⁻¹.

ΔH^⦵_f PCl₃(g) = −306.4 kJ mol⁻¹

ΔH^⦵_f PCl₅(g) = −398.9 kJ mol⁻¹

Calculate the entropy change, ΔS, in J K⁻¹mol⁻¹, for this reaction.

Chemistry 2e, Chpt. 21 Nuclear Chemistry, Appendix G: Standard Thermodynamic Properties for Selected Substances https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for- selectedsubstances# page_667adccf-f900-4d86-a13d-409c014086ea © 1999-2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. (CC BY 4.0) https://creativecommons.org/licenses/by/4.0/.

Calculate the Gibbs free energy change (ΔG), in kJ mol⁻¹, for this reaction at 25 °C. Use section 1 of the data booklet.

If you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol⁻¹ and −150.5 J mol⁻¹K⁻¹ respectively, but these are not the correct answers.

Determine the equilibrium constant, K, for this reaction at 25 °C, referring to section 1 of the data booklet.

If you did not obtain an answer in (c)(iii), use ΔG = –43.5 kJ mol⁻¹, but this is not the correct answer.

State the equilibrium constant expression, K_c, for this reaction.

State, with a reason, the effect of an increase in temperature on the position of this equilibrium.

Accept any diagram with each P joined to the other three.
Accept any combination of dots, crosses and lines.

P₄(s) + 6Cl₂(g) → 4PCl₃(l) ✔

Electron domain geometry: tetrahedral ✔

Molecular geometry: trigonal pyramidal ✔

Bond angle: 100«°» ✔

Accept any value or range within the range 91−108«°» for M3.

PCl₅ is non-polar:

symmetrical
OR
dipoles cancel ✔

PCl₄F is polar:

P–Cl has a different bond polarity than P–F ✔

non-symmetrical «dipoles»
OR
dipoles do not cancel ✔

Accept F more electronegative than/different electronegativity to Cl for M2.

«−398.9 kJ mol⁻¹ − (−306.4 kJ mol⁻¹) =» −92.5 «kJ mol⁻¹» ✔

«ΔS = 364.5 J K^–1mol^–1 – (311.7 J K^–1mol^–1 + 223.0 J K^–1mol^–1)=» –170.2 «J K^–1mol^–1» ✔

«ΔS =» –0.1702 «kJ mol^–1K^–1»
OR
298 «K» ✔

«ΔG = –92.5 kJ mol^–1 – (298 K × –0.1702 kJ mol^–1K^–1) =» –41.8 «kJ mol^–1» ✔

Award [2] for correct final answer.

If –87.6 and -150.5 are used then –42.8.

«ΔG = –41.8 kJ mol^–1 = $-\frac{8.31 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}}{1000}$ × 298 K × lnK»
OR
«ΔG = –41800 J mol^–1 = –8.31 J mol^–1K^–1 × 298 K × lnK»

«lnK = =» 16.9 ✔

«K = e^16.9 =» 2.19 × 10⁷ ✔

Award [2] for correct final answer.

Accept range of 1.80 × 10⁶–2.60 × 10⁷.

If –43.5 is used then 4.25 × 10⁷.

«K_c =» $\frac{\left[{\mathrm{PCl}}_5\right]}{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}$ ✔

«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔

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 ✔

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»

Outline what is measured by BOD.

A student dissolved 0.1240 ± 0.0001 g of Na₂S₂O₃ to make 1000.0 ± 0.4 cm³ of solution to use in the Winkler Method.

Determine the percentage uncertainty in the molar concentration.

Calculate the amount, in moles of Na₂S₂O₃ used in the titration.

Deduce the mole ratio of O₂ consumed in step I to S₂O₃²⁻ used in step III.

Calculate the concentration of dissolved oxygen, in mol dm⁻³, in the sample.

The three steps of the Winkler Method are redox reactions.

Deduce the reduction half-equation for step II.

Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.

«amount of» oxygen used to decompose the organic matter in water ✔

«$\frac{0.0001 \mathrm{g}}{0.1240 \mathrm{g}}\times 100\%=$» 0.08 «%»
OR
«$\frac{0.4 {\mathrm{cm}}^3}{1000.0 {\mathrm{cm}}^3}\times 100\%=$» 0.04 «%» ✔

«0.08 % + 0.04 % =» 0.12/0.1 «%» ✔

Award [2] for correct final answer.

Accept fractional uncertainties for M1, i.e., 0.0008 OR 0.0004.

«$\frac{37.50 {\mathrm{cm}}^3}{1000}$× 5.000 × 10⁻⁴mol dm⁻³ =» 1.875 × 10⁻⁵«mol» ✔

1:4 ✔

Accept “4 mol S₂O₃^2– :1 mol O₂“, but not just 4:1.

«$1.875\times {10}^{-5} \mathrm{mol}\times \frac{1}{4}=$» 4.688 × 10⁻⁶«mol» ✔

«$\frac{4.688\times {10}^{-6} \mathrm{mol}}{{\displaystyle \frac{25.00 {\mathrm{cm}}^3}{1000}}}=$» 1.875 × 10⁻⁴«mol dm⁻³» ✔

Award [2] for correct final answer.

MnO₂(s) + 2e⁻ + 4H⁺(aq) → Mn²⁺(aq) + 2H₂O (l) ✔

rate of reaction of oxygen with impurities depends on temperature
OR
rate at which bacteria/organisms grow/respire depends on temperature ✔

Draw the Lewis (electron dot) structures of PF₃ and PF₅ and use the VSEPR theory to deduce the molecular geometry of each species including bond angles.

Predict whether the molecules PF₃ and PF₅ are polar or non-polar.

State the type of hybridization shown by the phosphorus atom in PF₃.

Accept any combination of dots, crosses and lines.

Penalize missing lone pairs once only.

Do not apply ECF for molecular geometry.

Accept values in the range 95–109 for PF₃.

PF₃ polar AND PF₅ non-polar

Apply ECF from part (a) molecular geometry.

sp³

Identify the wavenumber of one peak in the IR spectrum of benzoic acid, using section 26 of the data booklet.

Identify the spectroscopic technique that is used to measure the bond lengths in solid benzoic acid.

Outline one piece of physical evidence for the structure of the benzene ring.

Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.

Outline why both C to O bonds in the conjugate base are the same length and suggest a value for them. Use section 10 of the data booklet.

The pH of an aqueous solution of benzoic acid at 298 K is 2.95. Determine the concentration of hydroxide ions in the solution, using section 2 of the data booklet.

Formulate the equation for the complete combustion of benzoic acid in oxygen using only integer coefficients.

The combustion reaction in (f)(ii) can also be classed as redox. Identify the atom that is oxidized and the atom that is reduced.

Suggest how benzoic acid, M_r = 122.13, forms an apparent dimer, M_r = 244.26, when dissolved in a non-polar solvent such as hexane.

State the reagent used to convert benzoic acid to phenylmethanol (benzyl alcohol), C₆H₅CH₂OH.

Any wavenumber in the following ranges:
2500−3000 «cm⁻¹» [✔]
1700−1750 «cm⁻¹» [✔]
2850−3090 «cm⁻¹» [✔]

X-ray «crystallography/spectroscopy» [✔]

Any one of:

«regular» hexagon

OR

all «H–C–C/C-C-C» angles equal/120º [✔]
all C–C bond lengths equal/intermediate between double and single

OR

bond order 1.5 [✔]

     [✔]

Note: Accept Kekulé structures.
Negative sign must be shown in correct position.

electrons delocalized «across the O–C–O system»

OR

resonance occurs [✔]

122 «pm» < C–O < 143 «pm» [✔]

Note: Accept “delocalized π-bond”.
Accept “bond intermediate between single and double bond” or “bond order 1.5” for M1.
Accept any answer in range 123 to 142 pm.

ALTERNATIVE 1:
[H⁺] «= 10^−2.95» = 1.122 × 10⁻³ «mol dm⁻³» [✔]

«[OH⁻] = $\frac{1.00\times {10}^{-14}\text{ mo}{\text{l}}^2\text{ d}{\text{m}}^{-6}}{1.22\times {10}^{-3}\text{ mol d}{\text{m}}^{-3}}$ =» 8.91 × 10⁻¹² «mol dm⁻³» [✔]

ALTERNATIVE 2:
pOH = «14 − 2.95 =» 11.05 [✔]
«[OH⁻] = 10^−11.05 =» 8.91 × 10⁻¹² «mol dm⁻³» [✔]

Note: Award [2] for correct final answer.
Accept other methods.

2C₆H₅COOH (s) + 15O₂ (g) → 14CO₂ (g) + 6H₂O (l)
correct products    [✔]
correct balancing    [✔]

Oxidized:

C/carbon «in C₆H₅COOH»

AND

Reduced:

O/oxygen «in O₂»      [✔]

«intermolecular» hydrogen bonding    [✔]

Note: Accept diagram showing hydrogen bonding.

lithium aluminium hydride/LiAlH₄    [✔]

Most candidates could identify a wavenumber or range of wavenumbers in the IR spectrum of benzoic acid.

Less than half the candidates identified x-ray crystallography as a technique used to measure bond lengths. There were many stating IR spectroscopy and quite a few random guesses.

Again less than half the candidates could accurately give a physical piece of evidence for the structure of benzene. Many missed the mark by not being specific, stating ‘all bonds in benzene with same length’ rather than ‘all C-C bonds in benzene have the same length’.

Very poorly answered with only 1 in 5 getting this question correct. Many did not show all the bonds and all the atoms or either forgot or misplaced the negative sign on the conjugate base.

This question was a challenge. Candidates were not able to explain the intermediate bond length and the majority suggested the value of either the bond length of C to O single bond or double bond.

Generally well done with a few calculating the pOH rather than the concentration of hydroxide ion asked for.

Most earned at least one mark by correctly stating the products of the reaction.

Another question where not reading correctly was a concern. Instead of identifying the atom that is oxidized and the atom that is reduced, answers included formulas of molecules or the atoms were reversed for the redox processes.

The other question where only 10 % of the candidates earned a mark. Few identified hydrogen bonding as the reason for carboxylic acids forming dimers. There were many G2 forms stating that the use of the word “dimer” is not in the syllabus, however the candidates were given that a dimer has double the molar mass and the majority seemed to understand that the two molecules joined together somehow but could not identify hydrogen bonding as the cause.

Very few candidates answered this part correctly and scored the mark. Common answers were H₂SO₄, HCl & Sn, H₂O₂. In general, strongest candidates gained the mark.

Outline two differences between the bonding of carbon atoms in C₆₀ and diamond.

Explain why C₆₀ and diamond sublime at different temperatures and pressures.

State two features showing that propane and butane are members of the same homologous series.

Describe a test and the expected result to indicate the presence of carbon–carbon double bonds.

Draw the full structural formula of (Z)-but-2-ene.

Write the equation for the reaction between but-2-ene and hydrogen bromide.

State the type of reaction.

Suggest two differences in the ¹H NMR of but-2-ene and the organic product from (d)(ii).

Predict, giving a reason, the major product of reaction between but-1-ene and steam.

Explain the mechanism of the reaction between 1-bromopropane, CH₃CH₂CH₂Br, and aqueous sodium hydroxide, NaOH (aq), using curly arrows to represent the movement of electron pairs.

Deduce the splitting pattern in the ¹H NMR spectrum for 1-bromopropane.

Calculate the enthalpy change of the reaction, ΔH, using section 11 of the data booklet.

Draw and label an enthalpy level diagram for this reaction.

Any two of:

C₆₀ fullerene: bonded to 3 C AND diamond: bonded to 4 C ✔

C₆₀ fullerene: delocalized/resonance AND diamond: not delocalized / no resonance ✔

C₆₀ fullerene: sp² AND diamond: sp³ ✔

C₆₀ fullerene: bond angles between 109–120° AND diamond: 109° ✔

Accept "bonds in fullerene are shorter/stronger/have higher bond order OR bonds in diamond longer/weaker/have lower bond order".

diamond giant/network covalent AND sublimes at higher temperature ✔

C₆₀ molecular/London/dispersion/intermolecular «forces» ✔

Accept “diamond has strong covalent bonds AND require more energy to break «than intermolecular forces»” for M1.

same general formula / C_nH_2n+2 ✔

differ by CH₂/common structural unit ✔

Accept "similar chemical properties".

Accept “gradation/gradual change in physical properties”.

ALTERNATIVE 1:

Test:

add bromine «water»/Br₂ (aq) ✔

Result:

«orange/brown/yellow» to colourless/decolourised ✔

Do not accept “clear” for M2.

ALTERNATIVE 2:

Test:

add «acidified» KMnO₄ ✔

Result:

«purple» to colourless/decolourised/brown ✔

Accept “colour change” for M2.

ALTERNATIVE 3:

Test:

add iodine /${\text{I}}_2$ ✔

Result:

«brown» to colourless/decolourised ✔

Accept

CH₃CH=CHCH₃ + HBr (g) → CH₃CH₂CHBrCH₃

Correct reactants ✔

Correct products  ✔

Accept molecular formulas for both reactants and product

«electrophilic» addition/EA ✔

Do not accept nucleophilic or free radical addition.

ALTERNATIVE 1: Any two of:

but-2-ene: 2 signals AND product: 4 signals ✔

but-2-ene: «area ratio» 3:1/6:2 AND product: «area ratio» 3:3:2:1 ✔

product: «has signal at» 3.5-4.4 ppm «and but-2-ene: does not» ✔

but-2-ene: «has signal at» 4.5-6.0 ppm «and product: does not» ✔

ALTERNATIVE 2:

but-2-ene: doublet AND quartet/multiplet/4 ✔

product: doublet AND triplet AND quintet/5/multiplet AND sextet/6/multiplet ✔

Accept “product «has signal at» 1.3–1.4 ppm «and but-2-ene: does not»”.

CH₃CH₂CH(OH)CH₃ ✔

«secondary» carbocation/CH₃CH₂CH⁺CH₃ more stable ✔

Do not accept “Markovnikov’s rule” without reference to carbocation stability.

curly arrow going from lone pair/negative charge on O in HO^– to C ✔

curly arrow showing Br breaking ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

formation of organic product CH₃CH₂CH₂OH AND Br– ✔

Do not allow curly arrow originating on H in HO^–.

Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition
state.

Do not penalize if HO and Br are not at 180° to each other.

Award [3 max] for S_N1 mechanism.

triplet/3 AND multiplet/6 AND triplet/3 ✔

bond breaking: C–H + Cl–Cl / 414 «kJ mol^–1» + 242 «kJ mol^–1»/656 «kJ»
OR
bond breaking: 4C–H + Cl–Cl / 4 × 414 «kJ mol^–1» + 242 «kJ mol^–1» / 1898 «kJ» ✔

bond forming: «C–Cl + H–Cl / 324 kJ mol^–1 + 431 kJ mol^–1» / 755 «kJ»
OR
bond forming: «3C–H + C–Cl + H–Cl / 3 × 414 «kJ mol^–1» + 324 «kJ mol^–1» + 431 kJ mol^–1» / 1997 «kJ» ✔

«ΔH = bond breaking – bond forming = 656 kJ – 755 kJ» = –99 «kJ» ✔

Award [3] for correct final answer.

Award [2 max] for 99 «kJ».

reactants at higher enthalpy than products ✔

ΔH/-99 «kJ» labelled on arrow from reactants to products
OR
activation energy/E_a labelled on arrow from reactant to top of energy profile ✔

Accept a double headed arrow between reactants and products labelled as ΔH for M2.

A challenging question, requiring accurate knowledge of the bonding in these allotropes (some referred to graphite, clearly the most familiar allotrope). The most frequent (correct) answer was the difference in number of bonded C atoms and hybridisation in second place. However, only 30% got a mark.

Again, this was a struggle between intermolecular forces and covalent bonds and this proved to be even harder than (a)(i) with only 25% of candidates getting full marks. The distinction between giant covalent/covalent network in diamond and molecular in C60 and hence resultant sublimation points, was rarely explained. There were many general and vague answers given, as well as commonly (incorrectly) stating that intermolecular forces are present in diamond. As another example of insufficient attention to the question itself, many candidates failed to say which would sublime at a higher temperature and so missed even one mark.

This easy question was quite well answered; same/similar physical properties and empirical formula were common errors.

Candidates misinterpreted the question and mentioned CH3⁺, i.e., the lost fragment; the other very common error was -COOH which shows a complete lack of understanding of MS considering the question is about butane so O should never appear.

Well answered by most, but some basic chemistry was missing when reporting results, perhaps as a result of little practical work due to COVID. A significant number suggested IR spectrometry, very likely because the question followed one on H NMR spectroscopy, thus revealing a failure to read the question properly (which asks for a test). Some teachers felt that adding "chemical" would have avoided some confusion.

Most were able to draw this isomer correctly, though a noticeable number of students included the Z as an atom in the structural formula, showing they were completely unfamiliar with E/Z notation.

Well done in general and most candidates wrote correct reagents, eventually losing a mark when considering H₂ to be a product alongside 2-bromobutane.

Very well answered, some mentioned halogenation which is a different reaction.

A considerable number of students (40%) got at least 1 mark here, but marks were low (average mark 0.9/2). Common errors were predicting 3 peaks, rather than 4 for 2 -bromobutane and vague / unspecific answers, such as ‘different shifts’ or ‘different intensities’. It is surprising that more did not use H NMR data from the booklet; they were not directed to the section as is generally done in this type of question to allow for more general answers regarding all information that can be obtained from an H NMR spectrum.

Product was correctly predicted by many, but most used Markovnikov's Rule to justify this, failing to mention the stability of the secondary carbocation, i.e., the chemistry behind the rule.

As usual, good to excellent candidates (47.5%) were able to get 3/4 marks for this mechanism, while most lost marks for carelessness in drawing arrows and bond connectivity, issues with the lone pair or negative charge on the nucleophile, no negative charge on transition state, or incorrect haloalkane. The average mark was thus 1.9/4.

Another of the very poorly answered questions where most candidates (90%) failed to predict 3 peaks and when they did, considered there would be a quartet instead of multiplet/sextet; other candidates seemed to have no idea at all. This is strange because the compound is relatively simple and while some teachers considered that predicting a sextet may be beyond the current curriculum or just too difficult, they could refer to a multiplet; a quartet is clearly incorrect.

Only the very weak candidates were unable to calculate the enthalpy change correctly, eventually missing 1 mark for inverted calculations.

Most candidates drew correct energy profiles, consistent with the sign of the energy change calculated in the previous question. And again, only very weak candidate failed to get at least 1 mark for correct profiles.

Draw a Lewis (electron dot) structure for ozone.

Discuss the relative length of the two O−O bonds in ozone.

Explain why there are frequencies of UV light that will dissociate O₃ but not O₂.

Explain, using equations, how the presence of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ results in a chain reaction that decreases the concentration of ozone in the stratosphere.

✔

Accept any combination of lines, dots or crosses to represent electrons.

Do not accept structures that represent 1.5 bonds.

both equal ✔

delocalization/resonance ✔

Accept bond length between 121 and 148 pm/ that of single O−O bond and double O=O bond for M1.

bond in O₃ is weaker
OR
O₃ bond order 1.5/< 2 ✔

Do not accept bond in O₃ is longer for M1.

lower frequency/longer wavelength «UV light» has enough energy to break the O–O bond in O₃ «but not that in O₂» ✔

Accept “lower frequency/longer wavelength «UV light» has lower energy”.

${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{(g) →∙CClF}}_{\text{2}}\text{(g) Cl}\bullet \text{(g)}$ ✔

${\text{Cl•(g)+O}}_{\text{3}}{\text{(g)→O}}_{\text{2}}\text{(g)+ClO•(g)}$
AND
${\text{ClO∙(g)+O}}_{\text{3}}{\text{(g)→2O}}_{\text{2}}\text{(g)+Cl}\bullet \text{(g)}$ ✔

Do not penalize missing radical.

Accept:for M2:
${\text{Cl∙(g) + O}}_{\text{3}}{\text{(g) → O}}_{\text{2}}\text{(g) + ClO}\bullet \text{(g)}$
AND
${\text{ClO∙(g) + O(g) → O}}_{\text{2}}\text{(g) + Cl}\bullet \text{(g)}$

Predict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR

State the type of hybridization shown by the central carbon atom in molecule B.

State the number of sigma ($\mathrm{\sigma }$) and pi ($\mathrm{\pi }$) bonds around the central carbon atom in molecule B.

The IR spectrum of one of the compounds is shown:

COBLENTZ SOCIETY. Collection © 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Deduce, giving a reason, the compound producing this spectrum.

Compound A and B are isomers. Draw two other structural isomers with the formula ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$.

The equilibrium constant, $K_{\mathrm{c}}$, for the conversion of A to B is $1.0\times {10}^8$ in water at $298 \mathrm{K}$.

Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.

Calculate the standard Gibbs free energy change, $∆G^{⦵}$, in $\mathbf{kJ}\mathbf{ }{\mathbf{mol}}^{\mathbf{–}\mathbf{1}}$, for the reaction (A to B) at $298 \mathrm{K}$. Use sections 1 and 2 of the data booklet.

Propanone can be synthesized in two steps from propene. Suggest the synthetic route including all the necessary reactants and steps.

Propanone can be synthesized in two steps from propene.

Suggest why propanal is a minor product obtained from the synthetic route in (g)(i).

Electron domain geometry: tetrahedral ✔

Molecular geometry: bent/V-shaped ✔

${\mathrm{sp}}^2$ ✔

$\mathrm{\sigma }$-bonds: $3$
AND
$\mathrm{\pi }$-bonds: $1$ ✔

B AND $\mathrm{C}=\mathrm{O}$ absorption/$1750«{\mathrm{cm}}^{-1}»$ 
OR
B AND absence of $\mathrm{O}–\mathrm{H} /3200-3600«{\mathrm{cm}}^{-1} \mathrm{absorption}»$✔

Accept any value between $\mathit{1700}\mathit{-}\mathit{1750}\mathit{ }c{m}^{\mathit{-}\mathit{1}}$.

Accept any two $C_3{H}_{6}O$ isomers except for propanone and propen-2-ol:

✔✔

Penalize missing hydrogens in displayed structural formulas once only.

$\mathrm{B}$ AND $K_{\mathrm{c}}$ is greater than $1$/large ✔

$«\mathrm{\Delta }{G}^{\mathrm{\Theta }}=-RT\ln K=0.00831 \mathrm{kJ} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1} (298 \mathrm{K}) (\ln 1.0\times {10}^8)=»$
$-46«\mathrm{kJ} {\mathrm{mol}}^{-1}»$ ✔