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]

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

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]

Several compounds can be synthesized from but-2-ene. Draw the structure of the final product for each of the following chemical reactions.

Determine the change in enthalpy, ΔH, for the combustion of but-2-ene, using section 11 of the data booklet. 

CH₃CH=CHCH₃(g) + 6O₂ (g) → 4CO₂(g) + 4H₂O (g)

State the hybridization of the carbon I and II atoms in but-2-ene.

Draw diagrams to show how sigma (σ) and pi (π) bonds are formed between atoms.

Sketch the mechanism for the reaction of 2-methylbut-2-ene with hydrogen bromide using curly arrows.

Explain why the major organic product is 2-bromo-2-methylbutane and not 2-bromo-3-methylbutane.

Deduce two features of this molecule that can be obtained from the mass spectrum. Use section 28 of the data booklet.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.

Identify the bond responsible for the absorption at A in the infrared spectrum. Use section 26 of the data booklet.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce
on behalf of the United States of America. All rights reserved.

Deduce the identity of the unknown compound using the previous information, the ¹H NMR spectrum and section 27 of the data booklet.

SDBS, National Institute of Advanced Industrial Science and Technology (AIST).

Draw the stereoisomers of butan-2-ol using wedge-dash type representations.

Outline how two enantiomers can be distinguished using a polarimeter.

Penalize missing hydrogens in displayed structural formulas once only.

Accept condensed structural formulas: CH₃CH(OH)CH₂CH₃ / CH₃CH₂CH₂CH₃ or skeletal structures.

Bonds broken:
2(C–C) + 1(C=C) + 8(C–H) + 6O=O / 2(346) + 1(614) + 8(414) + 6(498) / 7606 «kJ» ✓

Bonds formed:
8(C=O) + 8(O–H) / 8(804) + 8(463) / 10 136 «kJ» ✓

Enthalpy change:
«Bonds broken – Bonds formed = 7606 kJ – 10 136 kJ =» –2530 «kJ» ✓

Award [3] for correct final answer.

Award [2 max] for «+» 2530 «kJ».

Sigma (σ):

Accept any diagram showing end to end/direct overlap of atomic/hybridized orbitals and electron density concentrated between nuclei.

Pi (π):

Accept any diagram showing sideways overlap of unhybridized p/atomic orbitals and electron density above and below plane of bond axis.

Alternative 1

Penalize incorrect bond e.g., -CH-H₃C or –CH₃C only once in the paper.

Alternative 2

curly arrow going from C=C to H of HBr AND curly arrow showing Br leaving ✓

representation of carbocation ✓

curly arrow going from lone pair/negative charge on Br⁻ to C⁺ ✓

«2-bromo-2-methylbutane involves» formation of more stable «tertiary» carbocation/intermediate
OR
«2-bromo-3-methylbutane involves» formation of less stable «secondary» carbocation/intermediate ✓

«intermediate» more stable due to «increased positive» inductive/electron-releasing effect of extra –R/alkyl group/–CH₃/methyl ✓

Do not award marks for quoting Markovnikov’s rule without any explanation.

m/z 58:
molar/«relative» molecular mass/weight/Mr «is 58 g mol⁻¹/58» ✓

m/z 43:
«loses» methyl/CH₃ «fragment»
OR
COCH₃⁺ «fragment» ✓

Do not penalize missing charge on the fragments.

Accept molecular ion «peak»/ CH₃COCH₃⁺/C₃H₆O⁺.

Accept any C₂H₃O⁺ fragment/ CH₃CH₂CH₂⁺/C₃H₇⁺.

C=O ✓

Accept carbonyl/C=C.

Information deduced from ¹H NMR:

«one signal indicates» one hydrogen environment/symmetrical structure
OR
«chemical shift of 2.2 indicates» H on C next to carbonyl ✓

Compound:

propanone/CH₃COCH₃ ✓

Accept “one type of hydrogen”.

Accept .

enantiomers rotate «plane of» plane-polarized light ✓

equal degrees/angles/amounts AND opposite directions/rotation ✓

Accept “optical isomers” for “enantiomers”.

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 ✔

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

${\mathrm{H}}_2\mathrm{O}$/water «and ${\mathrm{H}}^{+}$» ✔

${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right){\mathrm{CH}}_3$/propan-2-ol ✔

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$/«potassium» dichromate(VI) AND ${\mathrm{H}}^{+}$
OR
${\mathrm{KMnO}}_4$/«acidified potassium» manganate(VII) ✔

Accept $H_{\mathit{3}}{O}^{\mathit{+}}$.

primary carbocation «intermediate forms»
OR
minor product «of the water addition would be» propan-1-ol
OR
anti-Markovnikov addition of water ✔

primary alcohol/propan-1-ol oxidizes to an aldehyde/propanal ✔

The majority of students got at least one of electron domain geometry or molecular geometry correct.

The vast majority of students could identify the hybridization around a central carbon atom.

The vast majority of students could identify BOTH sigma and pi bonds in a molecule.

Many candidates identified B having C = O and a peak at 1750.

A surprising number of candidates drew propanone here as an option, either failing to read the question or perhaps finding the structural formulae provided difficult to understand.

Most candidates identified B, the product, as being in greater concentration at equilibrium however some lost the mark because they did not include a reason.

Most candidates could apply the formula for Gibbs free energy change, ΔG^Θ, correctly however some did not get the units correct.

The mean mark was ⅔ for the required synthetic route. Some candidates failed to identify water as a reagent in the hydration reaction, or note that dichromate ion oxidation requires acidic conditions. This was also the question with most No Response.

This question regarding the formation of a minor product was not well answered. Many candidates struggled to explain the formation of propan-1-ol and to then oxidize it to propanal.

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 equation for the reaction of each substance with water.

Draw a diagram showing the delocalization of electrons in the conjugate base of butanoic acid.

Deduce the average oxidation state of carbon in butanoic acid.

A 0.250 mol dm⁻³ aqueous solution of butanoic acid has a concentration of hydrogen ions, [H⁺], of 0.00192 mol dm⁻³. Calculate the concentration of hydroxide ions, [OH⁻], in the solution at 298 K.

Determine the pH of a 0.250 mol dm⁻³ aqueous solution of ethylamine at 298 K, using section 21 of the data booklet.

Sketch the pH curve for the titration of 25.0 cm³ of ethylamine aqueous solution with 50.0 cm³ of butanoic acid aqueous solution of equal concentration. No calculations are required.

Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.

State a suitable reagent for the reduction of butanoic acid.

Deduce the product of the complete reduction reaction in (e)(i).

Butanoic acid:
CH₃CH₂CH₂COOH (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂CH₂COO⁻ (aq) + H₃O⁺ (aq) ✔

Ethylamine:
CH₃CH₂NH₂ (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂NH₃⁺ (aq) + OH⁻ (aq) ✔

Diagram showing:
dotted line along O–C–O AND negative charge

Accept correct diagrams with pi clouds.

–1 ✔

«$\frac{1.00\times {10}^{-14}\text{mo}{\text{l}}^2\text{d}{\text{m}}^{-6}}{0.00192\text{mol}\text{d}{\text{m}}^{-3}}$» = 5.21 × 10^–12 «mol dm^–3» ✔

«pK_b = 3.35, K_b = 10^–3.35 = 4.5 × 10^–4»

«C₂H₅NH₂ + H₂O $\rightleftharpoons$ C₂H₅NH₃⁺ + OH^–»

K_b = $\frac{\left[\text{O}{\text{H}}^{-}-\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{\text{H}}_{\text{2}}\right]}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{\left[\text{O}{\text{H}}^{-}\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{0.250}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{x^2}{0.250}$ ✔

« x = [OH^–] =» 0.011 «mol dm^–3» ✔

«pH = –log$\frac{1.00\times {10}^{-14}}{0.011}=$» 12.04

OR

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

Award [3] for correct final answer.

decreasing pH curve ✔

pH close to 7 (6–8) at volume of 25 cm³ butanoic acid ✔

weak acid/base shape with no flat «strong acid/base» parts on the curve ✔

Any two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔

Accept “butanoic acid forms dimers”

Accept “butanoic acid has larger M_r/hydrocarbon chain/number of electrons” for M2.

Accept “butanoic acid has larger «permanent» dipole/more polar” for M3.

lithium aluminium hydride/LiAlH₄ ✔

butan-1-ol/1-butanol/CH₃CH₂CH₂CH₂OH ✔

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

Formulate two equations to show how nitrogen(II) oxide, NO, catalyses the destruction of ozone.

Suggest why the loss of ozone is an international environmental concern.

$\text{NO}\bullet \text{(g)}+{\text{O}}_{\text{3}}\text{(g)}\to \text{N}{\text{O}}_{\text{2}}\bullet \text{(g)}+{\text{O}}_{\text{2}}\text{(g)}$

$\text{N}{\text{O}}_{\text{2}}\bullet \text{(g)}+\text{O}\bullet \text{(g)}\to \text{NO}\bullet \text{(g)}+{\text{O}}_{\text{2}}\text{(g)}$
OR
$\text{N}{\text{O}}_{\text{2}}\bullet \text{(g)}+{\text{O}}_{\text{3}}\text{(g)}\to \text{NO}\bullet \text{(g)}+\text{2}{\text{O}}_{\text{2}}\text{(g)}$

Allow representation of radicals without $\bullet$ if consistent throughout.

[2 marks]

«loss of ozone» allows UV radiation to penetrate atmosphere/reach earth

UV radiation causes skin cancer
OR
UV radiation causes tissue damage

[2 marks]

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

Deduce a Lewis (electron dot) structure of the nitric acid molecule, HNO₃, that obeys the octet rule, showing any non-zero formal charges on the atoms.

Explain the relative lengths of the three bonds between N and O in nitric acid.

State a technique used to determine the length of the bonds between N and O in solid HNO₃.

Write an equation for the reaction between the acids to produce the electrophile, NO₂⁺.

Draw the structural formula of the carbocation intermediate produced when this electrophile attacks benzene.

Deduce the number of signals that you would expect in the ¹H NMR spectrum of nitrobenzene and the relative areas of these.

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

bonds and non-bonding pairs correct ✔

formal charges correct ✔

Accept dots, crosses or lines to represent electron pairs.

Do not accept resonance structures with delocalised bonds/electrons.

Accept + and – sign respectively.

Do not accept a bond between nitrogen and hydrogen.

For an incorrect Lewis structure, allow ECF for non-zero formal charges.

Any three of:

two N-O same length/order ✔
delocalization/resonance ✔

N-OH longer «than N-O»
OR
N-OH bond order 1 AND N-O bond order 1½ ✔

Award [2 max] if bond strength, rather than bond length discussed.

Accept N-O between single and double bond AND N-OH single bond.

X-ray crystallography ✔

HNO₃ + 2H₂SO₄ $\rightleftharpoons$ NO₂⁺ + H₃O⁺ + 2HSO₄^- ✔

Accept “HNO₃ + H₂SO₄ $\rightleftharpoons$ NO₂⁺ + H₂O + HSO₄^-”.

Accept “HNO₃ + H₂SO₄ $\rightleftharpoons$ H₂NO₃⁺ + HSO₄^-” AND “H₂NO₃⁺ $\rightleftharpoons$ NO₂⁺ + H₂O”.

Accept single arrows instead of equilibrium signs.

Accept any of the five structures.

Do not accept structures missing the positive charge.

Number of signals: three/3 ✔

Relative areas: 2 : 2 : 1 ✔

Drawing arrows in the boxes to represent the electron configuration of a nitrogen atom was done extremely well.

Drawing the Lewis structure of HNO₃ was performed extremely poorly with structures that included H bonded to N, no double bond or a combination of single, double and even a triple bond or incorrect structures with dotted lines to reflect resonance. Many did not calculate non-zero formal charges.

Poorly done; some explained relative bond strengths between N and O in HNO₃, not relative lengths; others included generic answers such as triple bond is shortest, double bond is longer, single longest.

A majority could not state the technique to determine length of bonds; answers included NMR, IR, and such instead of X-ray crystallography.

Many had difficulties writing the balanced equation(s) for the formation of the nitronium ion.

Again, many had difficulty drawing the structural formula of the carbocation intermediate produced in the reaction.

Deducing the number of signals in the 1H NMR spectrum of nitrobenzene, which depend on the number of different hydrogen environments, was done poorly. Also, instead of relative areas, the common answer included chemical shift (ppm) values.

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 why ozone in the stratosphere is important.

Dinitrogen monoxide in the stratosphere is converted to nitrogen monoxide, NO (g).

Write two equations to show how NO (g) catalyses the decomposition of ozone.

State one analytical technique that could be used to determine the ratio of ¹⁴N : ¹⁵N.

A sample of gas was enriched to contain 2 % by mass of ¹⁵N with the remainder being ¹⁴N.

Calculate the relative molecular mass of the resulting N₂O.

Predict, giving two reasons, how the first ionization energy of ¹⁵N compares with that of ¹⁴N.

Explain why the first ionization energy of nitrogen is greater than both carbon and oxygen.

Nitrogen and carbon:

Nitrogen and oxygen:

State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.

State, giving a reason, the shape of the dinitrogen monoxide molecule.

Deduce the hybridization of the central nitrogen atom in the molecule.

absorbs UV/ultraviolet light «of longer wavelength than absorbed by O₂»     [✔]

NO (g) + O₃ (g) → NO₂ (g) + O₂ (g)       [✔]
NO₂ (g) + O₃ (g) → NO (g) + 2O₂ (g)     [✔]

Note: Ignore radical signs.

Accept equilibrium arrows.

Award [1 max] for NO₂ (g) + O (g) → NO (g) + O₂ (g).

mass spectrometry/MS     [✔]

« $\frac{(98\times 14)+(2\times 15)}{100}$ =» 14.02    [✔]

«M_r = (14.02 × 2) + 16.00 =» 44.04    [✔]

Any two:

same AND have same nuclear charge /number of protons/Z^_eff      [✔]

same AND neutrons do not affect attraction/ionization energy/Z^_eff
OR
same AND neutrons have no charge       [✔]

same AND same attraction for «outer» electrons     [✔]

same AND have same electronic configuration/shielding     [✔]

Note: Accept “almost the same”.

“Same” only needs to be stated once.

Nitrogen and carbon:

N has greater nuclear charge/«one» more proton «and electrons both lost from singly filled p-orbitals»    [✔]

Nitrogen and oxygen:

O has a doubly filled «p-»orbital
OR
N has only singly occupied «p-»orbitals     [✔]

Note: Accept “greater e– ^- e^- repulsion in O” or “lower e– ^- e^- repulsion in N”.

Accept box annotation of electrons for M2.

delocalization

OR

delocalized π-electrons    [✔]

Note: Accept “resonance”.

linear AND 2 electron domains

OR

linear AND 2 regions of electron density    [✔]

Note: Accept “two bonds AND no lone pairs” for reason.

sp     [✔]

Candidates sometimes failed to identify how ozone works in chemical terms, referring to protects/deflects, i.e., the consequence rather than the mechanism.

Many candidates recalled the first equation for NO catalyzed decomposition of ozone only. Some considered other radical species.

All candidates, with very few exceptions, answered this correctly.

Most candidates were able to calculate the accurate mass of N₂O, though quite a few candidates just calculated the mass of N and didn’t apply it to N₂O, losing an accessible mark.

Many students realized that neutrons had no charge and could not affect IE significantly, but many others struggled a lot with this question since they considered that ¹⁵N would have a higher IE because they considered the greater mass of the nucleus would result in an increase of attraction of the electrons.

Mixed responses here; the explanation of higher IE for N with respect to C was less well explained, though it should have been the easiest. It was good to see that most candidates could explain the difference in IE of N and O, either mentioning paired/unpaired electrons or drawing box diagrams.

Most candidates identified resonance for this given Lewis representation.

Though quite a number of candidates suggested a linear shape correctly, they often failed to give a complete correct explanation, just mentioning the absence of lone pairs but not two bonds, instead of referring to electron domains.

Hybridisation of the N atom was correct in most cases.

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.

Distinguish between a sigma and pi bond.

Identify the hybridization of carbon in ethane, ethene and ethyne.

State, giving a reason, if but-1-ene exhibits cis-trans isomerism.

State the type of reaction which occurs between but-1-ene and hydrogen iodide at room temperature.

Explain the mechanism of the reaction between but-1-ene with hydrogen iodide, using curly arrows to represent the movement of electron pairs.

State, giving a reason, if the product of this reaction exhibits stereoisomerism.

Deduce the rate expression for this reaction.

Deduce the units of the rate constant.

Determine the initial rate of reaction in experiment 4.

Deduce, with a reason, the mechanism of the reaction between 2-chloropentane and sodium hydroxide.

Discuss the reason benzene is more reactive with an electrophile than a nucleophile.

Sigma (σ) bond:

overlap «of atomic orbitals» along the axial / intermolecular axis / electron density is between nuclei
OR
head-on/end-to-end overlap «of atomic orbitals» ✔

Pi (π) bond:

overlap «of p-orbitals» above and below the internuclear axis/electron density above and below internuclear axis
OR
sideways overlap «of p-orbitals» ✔

Accept a suitable diagram.

All 3 required for mark.

no AND 2 groups on a carbon «in the double bond» are the same/hydrogen «atoms»

OR

no AND molecule produced by rearranging atoms bonded on a carbon «in the double bond» is the same as the original ✔

«electrophilic» addition ✔

Do not allow nucleophilic addition.

curly arrow going from C=C to H of HI AND curly arrow showing I leaving ✔

representation of carbocation ✔

curly arrow going from lone pair/negative charge on I^– to C⁺ ✔

2-iodobutane formed ✔

Penalize incorrect bond, e.g. –CH–H₃C or –CH₃C once only.

yes AND has a carbon attached to four different groups
OR
yes AND it contains a chiral carbon ✔

Accept yes AND mirror image of molecule different to original/non-superimposable on original.

«rate =» k[NaOH][C₅H₁₁Cl] ✔

mol^–1dm³s^–1 ✔

ALTERNATIVE 1:

«k = » 1.25 «mol^–1dm³s^–1» ✔

«rate = 1.25 mol^–1dm³s^–1 × 0.60 mol dm^–3 × 0.25 mol dm^–3»

1.9 x 10^–1 «mol dm^–3s^–1» ✔

ALTERNATIVE 2:

«[NaOH] exp. 4 is 3 × exp. 1»

«[C₅H₁₁Cl] exp. 4 is 2.5 × exp. 1»

«exp. 4 will be » 7.5× faster ✔

1.9 x 10^–1 «mol dm^–3s^–1» ✔

Award [2] for correct final answer.

S_N2 AND rate depends on both OH^– and 2-chloropentane ✔

Accept E2 AND rate depends on both OH^– and 2-chloropentane.

delocalized electrons/pi bonds «around the ring»
OR
molecule has a region of high electron density/negative charge ✔

electrophiles are attracted/positively charged AND nucleophiles repelled/negatively charged ✔

Do not accept just “nucleophiles less attracted” for M2.

Accept “benzene AND nucleophiles are both electron rich” for “repels nucleophiles”.

(i) Draw diagrams to show how sigma (σ) and pi (π) bonds are formed between atoms.

(ii) State the number of sigma (σ) and pi (π) bonds in propane and propene.

Construct the mechanism of the formation of 2-bromopropane from hydrogen bromide and propene using curly arrows to denote the movement of electrons.

i

ii

Award [1] for two or three correct answers.
Award [2] for all four correct.

curly arrow going from C=C to H of HBr and curly arrow showing Br leaving
representation of carbocation
curly arrow going from lone pair/negative charge on Br^– to C⁺

Award [2 max] for formation of 1-bromopropane.

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.

Deduce what information can be obtained from the ¹H NMR spectrum.

Identify the functional group that shows stretching at 1710 cm^–1 in the infrared spectrum of this compound using section 26 of the data booklet and the ¹H NMR.

Suggest the structural formula of this compound.

Bromine was added to hexane, hex-1-ene and benzene. Identify the compound(s) which will react with bromine in a well-lit laboratory.

Deduce the structural formula of the main organic product when hex-1-ene reacts with hydrogen bromide.

State the reagents and the name of the mechanism for the nitration of benzene.

Outline, in terms of the bonding present, why the reaction conditions of halogenation are different for alkanes and benzene.

Below are two isomers, A and B, with the molecular formula C₄H₉Br.

Explain the mechanism of the nucleophilic substitution reaction with NaOH(aq) for the isomer that reacts almost exclusively by an S_N2 mechanism using curly arrows to represent the movement of electron pairs.

Number of hydrogen environments: 3

Ratio of hydrogen environments: 2:3:9

Splitting patterns: «all» singlets

Accept any equivalent ratios such as 9:3:2.

Accept “no splitting”.

[3 marks]

carbonyl
OR
C=O

Accept “ketone” but not “aldehyde”.

[1 mark]

Accept (CH₃)₃CCH₂COCH₃.

Award [1] for any aldehyde or ketone with C₇H₁₄O structural formula.

[2 marks]

hexane AND hex-1-ene

Accept “benzene AND hexane AND hex-1-ene”.

[1 mark]

CH₃CH₂CH₂CH₂CHBrCH₃

Accept displayed formula but not molecular formula.

[1 mark]

Reagents: «concentrated» sulfuric acid AND «concentrated» nitric acid

Name of mechanism: electrophilic substitution

[2 marks]

benzene has «delocalized» $\pi$ bonds «that are susceptible to electrophile attack» AND alkanes do not

Do not accept “benzene has single and double bonds”.

[1 mark]

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

curly arrow showing Br 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 in OH^–.

Accept curly arrows in the transition state.

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

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

Award [2 max] if wrong isomer is used.

[3 marks]

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.

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]

Deduce the Lewis (electron dot) structure and molecular geometry and the bond angles of PCl₃.

Lewis structure:

Molecular geometry:

trigonal/triangular pyramidal

Bond angles:

< 109.5°

Penalize missing lone pairs once only between this question and 4(b)(ii).

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

Do not apply ECF.

Do not accept answer equal to or less than 90°.

Literature value is 100.1°.

[3 marks]

State the full electron configuration of the chlorine atom.

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

The mass spectrum of chlorine is shown.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Outline the reason for the two peaks at $m/z=35$ and $37$.

Explain the presence and relative abundance of the peak at $m/z=74$.

Calculate the amount, in $\mathrm{mol}$, of manganese(IV) oxide added.

Determine the limiting reactant, showing your calculations.

Determine the excess amount, in $\mathrm{mol}$, of the other reactant.

Calculate the volume of chlorine, in ${\mathrm{dm}}^3$, produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

State the oxidation state of manganese in ${\mathrm{MnO}}_2$ and ${\mathrm{MnCl}}_2$.

Deduce, referring to oxidation states, whether ${\mathrm{MnO}}_2$ is an oxidizing or reducing agent.

Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.

State the formula of the conjugate base of hypochlorous acid.

Calculate the concentration of ${\mathrm{H}}^{+} \left(\mathrm{aq}\right)$ in a $\mathrm{HClO} \left(\mathrm{aq}\right)$ solution with a $\mathrm{pH}=3.61$.

State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.

Predict, giving a reason, whether ethane or chloroethane is more reactive.

Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, $\mathrm{NaOH} \left(\mathrm{aq}\right)$, using curly arrows to represent the movement of electron pairs.

Ethoxyethane (diethyl ether) can be used as a solvent for this conversion.
Draw the structural formula of ethoxyethane

Deduce the number of signals and chemical shifts with splitting patterns in the ¹H NMR spectrum of ethoxyethane. Use section 27 of the data booklet.

Calculate the percentage by mass of chlorine in ${\mathrm{CCl}}_2{\mathrm{F}}_2$.

Comment on how international cooperation has contributed to the lowering of $\mathrm{CFC}$ emissions responsible for ozone depletion.

$\mathrm{CFC}$s produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^5$ ✔

Do not accept condensed electron configuration.

${\mathrm{Cl}}^{-}$ AND more «electron–electron» repulsion ✔

Accept $C{l}^{\mathit{-}}$ AND has an extra electron.

$\mathrm{Cl}$ has a greater nuclear charge/number of protons/${\mathrm{Z}}_{\mathrm{eff}}$ «causing a stronger pull on the outer electrons» ✔

same number of shells
OR
same «outer» energy level
OR
similar shielding ✔

«two major» isotopes «of atomic mass $35$ and $37$» ✔

«diatomic» molecule composed of «two» chlorine-37 atoms ✔

chlorine-37 is the least abundant «isotope»
OR
low probability of two ${}_{{}_{37}}\mathrm{Cl}$ «isotopes» occurring in a molecule ✔

$«\frac{2.67 \mathrm{g}}{86.94 \mathrm{g} {\mathrm{mol}}^{-1}}=»0.0307 «\mathrm{mol}»$ ✔

$«{\mathrm{n}}_{\mathrm{HCl}}=2.00 \mathrm{mol} {\mathrm{dm}}^{-3}\times 0.2000 {\mathrm{dm}}^3»=0.400 \mathrm{mol}$✔

$«\frac{0.400}{4}=» 0.100 \mathrm{mol}$ AND ${\mathrm{MnO}}_2$ is the limiting reactant ✔

Accept other valid methods of determining the limiting reactant in M2.

$«0.0307 \mathrm{mol}\times 4=0.123 \mathrm{mol}»$

$«0.400 \mathrm{mol}–0.123 \mathrm{mol}=»0.277 «\mathrm{mol}»$ ✔

$«0.0307 \mathrm{mol}\times 22.7 {\mathrm{dm}}^3 {\mathrm{mol}}^{-1}=» 0.697 «{\mathrm{dm}}^3»$ ✔

Accept methods employing $pV=nRT$.

${\mathrm{MnO}}_2: +4$ ✔

${\mathrm{MnCl}}_2: +2$ ✔

oxidizing agent AND oxidation state of $\mathrm{Mn}$ changes from $+4$ to $+2$/decreases ✔

partially dissociates/ionizes «in water» ✔

${\mathrm{ClO}}^{-}$ ✔

$«\left[{\mathrm{H}}^{+}\right]={10}^{-3.61}=» 2.5\times {10}^{-4} «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

«free radical» substitution/${\mathrm{S}}_{\mathrm{R}}$ ✔

Do not accept electrophilic or nucleophilic substitution.

chloroethane AND C–Cl bond is weaker/$324 \mathrm{kJ} {\mathrm{mol}}^{-1}$ than C–H bond/$414 \mathrm{kJ} {\mathrm{mol}}^{-1}$
OR
chloroethane AND contains a polar bond ✔

Accept “chloroethane AND polar”.

curly arrow going from lone pair/negative charge on $\mathrm{O}$ in ⁻OH to $\mathrm{C}$ ✔

curly arrow showing $\mathrm{Cl}$ leaving ✔

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

Accept $O{H}^{\mathit{-}}$ with or without the lone pair.

Do not accept curly arrows originating on $H$ in $O{H}^{-}$.

Accept curly arrows in the transition state.

Do not penalize if $HO$ and $Cl$ are not at 180°.

Do not award M3 if $OH-C$ bond is represented. 

 / ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3$ ✔

Accept $\mathit{(}C{H}_{\mathit{3}}C{H}_{\mathit{2}}{\mathit{)}}_{\mathit{2}}O$.

2 «signals» ✔

0.9−1.0 AND triplet ✔

3.3−3.7 AND quartet ✔

Accept any values in the ranges.

Award [1] for two correct chemical shifts or two correct splitting patterns.

$«M\left({\mathrm{CCl}}_2{\mathrm{F}}_2\right) =» 120.91 «\mathrm{g} {\mathrm{mol}}^{-1}»$ ✔

$\frac{2\times 35.45 \mathrm{g} {\mathrm{mol}}^{-1}}{120.91 \mathrm{g} {\mathrm{mol}}^{-1}}\times 100\%=» 58.64 «\%»$ ✔

Award [2] for correct final answer.

Any of:

research «collaboration» for alternative technologies «to replace $\mathrm{CFC}$s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔

Do not accept just “collaboration”.

Do not accept any reference to $CFC$ as greenhouse gas or product of fossil fuel combustion.

Accept reference to specific measures, such as agreement on banning use/manufacture of $CFC$s.

${\mathrm{O}}_3+\mathrm{Cl}·\to \mathrm{O}2+\mathrm{ClO}·$ ✔

$\mathrm{ClO}·+\mathrm{O}·\to {\mathrm{O}}_2+\mathrm{Cl}·$
OR
$\mathrm{ClO}·+{\mathrm{O}}_3\to \mathrm{Cl}·+2{\mathrm{O}}_2$ ✔

Penalize missing/incorrect radical dot (∙) once only.

Well answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.

Most candidates could correctly explain the relative sizes of chlorine atom and chloride ion.

Fairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.

More than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.

Not well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.

Surprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.

Most candidates could find the volume of gas produced in a reaction under standard conditions.

More than 90% could identify the oxidation number of manganese in both MnO₂ and MnCl₂.

Most candidates stated that MnO₂ is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.

Another well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water. 

Roughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.

Vast majority of candidates could calculate the concentration of H⁺ (aq) in a HClO (aq) solution with a pH =3.61.

Many identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.

The underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.

Few earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.

Approximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.

A poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.

Another good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition. 

Somewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.

Few could construct the propagation equations showing how CFCs affect ozone, and many lost marks by failing to identify ClO· as a radical.

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» ✔