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Thinka Jan 2026 Cambridge International A Level-Style Mock — Chemistry (YCH11)

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An original Thinka practice paper modelled on the structure and difficulty of the Jan 2026 Cambridge International A Level Chemistry (YCH11) paper. Not affiliated with or reproduced from Cambridge.

Unit 1: Section A

Answer all multiple-choice questions. Select the single best answer from A to D.

20 PastPaper.question · 20 PastPaper.marks

PastPaper.question 1 · multiple-choice

1 PastPaper.marks

An element $ \text{X} $ has the following successive ionisation energies in $ \text{kJ mol}^{-1} $:
$ \text{IE}_1 = 578 $
$ \text{IE}_2 = 1817 $
$ \text{IE}_3 = 2745 $
$ \text{IE}_4 = 11578 $
$ \text{IE}_5 = 14831 $

What is the formula of the oxide of $ \text{X} $?

A.$ \text{XO} $
B.$ \text{X}_2\text{O} $
C.$ \text{X}_2\text{O}_3 $
D.$ \text{XO}_2 $

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PastPaper.workedSolution

There is a very large increase between the third and fourth ionisation energies (from 2745 to 11578 $ \text{kJ mol}^{-1} $). This indicates that the fourth electron is being removed from a inner shell, closer to the nucleus and less shielded. Thus, element $ \text{X} $ has three valence electrons and forms a stable $ \text{X}^{3+} $ ion. Oxygen forms a $ \text{O}^{2-} $ oxide ion. Combining these in a neutral compound gives the formula $ \text{X}_2\text{O}_3 $.

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1 mark: Correct answer is C.
0 marks: Any other option selected.

PastPaper.question 2 · multiple-choice

1 PastPaper.marks

Which of the following represents the correct electronic configuration of a copper(I) ion, $ \text{Cu}^+ $?

A.$ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 4s^1 $
B.$ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} $
C.$ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2 $
D.$ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 4s^2 $

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PastPaper.workedSolution

A neutral copper atom has the anomalous electronic configuration $ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1 $. When transition metals form positive ions, they lose electrons from the outer $ 4s $ subshell before the $ 3d $ subshell. Therefore, removing one electron to form $ \text{Cu}^+ $ results in the configuration $ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} $, which is represented by option B.

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1 mark: Correct answer is B.
0 marks: Any other option selected.

PastPaper.question 3 · multiple-choice

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When $ 10\text{ cm}^3 $ of a gaseous hydrocarbon is completely combusted in $ 70\text{ cm}^3 $ of oxygen (an excess), the final gas volume at room temperature and pressure is $ 50\text{ cm}^3 $. Passing this gas through aqueous sodium hydroxide reduces the volume to $ 20\text{ cm}^3 $.

(All gas volumes are measured under the same conditions of temperature and pressure, where water is a liquid).

What is the molecular formula of the hydrocarbon?

A.$ \text{C}_3\text{H}_6 $
B.$ \text{C}_3\text{H}_8 $
C.$ \text{C}_4\text{H}_{10} $
D.$ \text{C}_2\text{H}_6 $

PastPaper.showAnswers

PastPaper.workedSolution

Passing the final mixture through aqueous NaOH absorbs $ \text{CO}_2 $. The decrease in volume is $ 50 - 20 = 30\text{ cm}^3 $, which represents the volume of $ \text{CO}_2 $ produced. Since $ 10\text{ cm}^3 $ of hydrocarbon produces $ 30\text{ cm}^3 $ of $ \text{CO}_2 $, the hydrocarbon has 3 carbon atoms ($ x = 3 $).
The remaining $ 20\text{ cm}^3 $ is the unreacted excess oxygen, meaning the volume of oxygen reacted was $ 70 - 20 = 50\text{ cm}^3 $.
The ratio of hydrocarbon reacted to oxygen reacted is $ 10 : 50 = 1 : 5 $.
For the combustion equation $ \text{C}_3\text{H}_y + 5\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} $, balancing oxygen atoms gives: $ 10 = 6 + \frac{y}{2} \Rightarrow \frac{y}{2} = 4 \Rightarrow y = 8 $. Therefore, the molecular formula is $ \text{C}_3\text{H}_8 $.

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1 mark: Correct answer is B.
0 marks: Any other option selected.

PastPaper.question 4 · multiple-choice

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Which of the following correctly describes the number of bonding pairs of electrons, the number of lone pairs of electrons around the central atom, and the molecular shape of chlorine trifluoride, $ \text{ClF}_3 $?

A.3 bonding pairs, 2 lone pairs, T-shaped
B.3 bonding pairs, 1 lone pair, Trigonal pyramidal
C.3 bonding pairs, 2 lone pairs, Trigonal planar
D.3 bonding pairs, 0 lone pairs, Trigonal planar

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PastPaper.workedSolution

The central chlorine atom in $ \text{ClF}_3 $ has 7 valence electrons. It forms three single covalent bonds with three fluorine atoms (using 3 electrons), leaving 4 non-bonding electrons. These 4 electrons form 2 lone pairs. With 3 bonding pairs and 2 lone pairs, the electron-pair geometry is trigonal bipyramidal, but the molecular shape is T-shaped to minimise lone pair-lone pair repulsions.

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1 mark: Correct answer is A.
0 marks: Any other option selected.

PastPaper.question 5 · multiple-choice

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Which of the following alkenes can exist as a pair of $ E $/$ Z $ isomers?

A.2-methylbut-2-ene
B.hex-1-ene
C.3-methylpent-2-ene
D.2,3-dimethylbut-2-ene

PastPaper.showAnswers

PastPaper.workedSolution

For $ E $/$ Z $ isomerism to exist, each carbon atom of the $ \text{C}=\text{C} $ double bond must be attached to two different groups.
- In 2-methylbut-2-ene, one carbon is bonded to two identical methyl groups (no isomerism).
- In hex-1-ene, the terminal carbon is bonded to two hydrogen atoms (no isomerism).
- In 3-methylpent-2-ene ($ \text{CH}_3\text{-CH}=\text{C(CH}_3)\text{-CH}_2\text{-CH}_3 $), C2 is bonded to H and $ \text{-CH}_3 $ (different), and C3 is bonded to $ \text{-CH}_3 $ and $ \text{-CH}_2\text{CH}_3 $ (different). Thus, it has $ E $/$ Z $ isomers.
- In 2,3-dimethylbut-2-ene, both carbons have identical methyl groups (no isomerism).

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1 mark: Correct answer is C.
0 marks: Any other option selected.

PastPaper.question 6 · multiple-choice

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Which of the following equations represents a termination step in the free-radical chlorination of methane that produces a hydrocarbon byproduct?

A.$ \text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2 $
B.$ \text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl} $
C.$ \text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6 $
D.$ \text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl} $

PastPaper.showAnswers

PastPaper.workedSolution

In the free-radical chlorination of methane, termination steps involve the combination of any two free radicals. When two methyl radicals ($ \text{CH}_3^\bullet $) combine, they form ethane ($ \text{C}_2\text{H}_6 $), which is a hydrocarbon byproduct (or impurity) in the reaction mixture.

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1 mark: Correct answer is C.
0 marks: Any other option selected.

PastPaper.question 7 · multiple-choice

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Copper can be extracted by heating copper(II) oxide with carbon:

$ 2\text{CuO(s)} + \text{C(s)} \rightarrow 2\text{Cu(s)} + \text{CO}_2\text{(g)} $

What is the percentage atom economy by mass for the production of copper in this reaction?

(Relative atomic masses, $ A_r $: $ \text{C} = 12.0 $, $ \text{O} = 16.0 $, $ \text{Cu} = 63.5 $)

A.37.1%
B.74.3%
C.79.9%
D.84.1%

PastPaper.showAnswers

PastPaper.workedSolution

Atom economy $ = \frac{\text{mass of desired product}}{\text{total mass of all reactants}} \times 100 $
Desired product is $ 2\text{Cu} $: $ 2 \times 63.5 = 127.0\text{ g mol}^{-1} $.
Reactants are $ 2\text{CuO} + \text{C} $: $ 2 \times (63.5 + 16.0) + 12.0 = 2(79.5) + 12.0 = 159.0 + 12.0 = 171.0\text{ g mol}^{-1} $.
Atom economy $ = \frac{127.0}{171.0} \times 100 \approx 74.3\% $.

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1 mark: Correct answer is B.
0 marks: Any other option selected.

PastPaper.question 8 · multiple-choice

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Which of the following molecules contains polar bonds but is overall a non-polar molecule?

A.$ \text{NH}_3 $
B.$ \text{SF}_6 $
C.$ \text{CHCl}_3 $
D.$ \text{H}_2\text{S} $

PastPaper.showAnswers

PastPaper.workedSolution

In sulfur hexafluoride ($ \text{SF}_6 $), the S-F bonds are highly polar because fluorine is much more electronegative than sulfur. However, because the molecule has a perfectly symmetrical octahedral shape, the individual bond dipoles cancel each other out completely, resulting in a net dipole moment of zero (non-polar molecule).

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1 mark: Correct answer is B.
0 marks: Any other option selected.

PastPaper.question 9 · multiple-choice

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An element in Period 3 of the Periodic Table has successive ionization energies as shown below:

$\text{IE}_1 = 578\text{ kJ mol}^{-1}$
$\text{IE}_2 = 1817\text{ kJ mol}^{-1}$
$\text{IE}_3 = 2745\text{ kJ mol}^{-1}$
$\text{IE}_4 = 11577\text{ kJ mol}^{-1}$
$\text{IE}_5 = 14842\text{ kJ mol}^{-1}$

Identify the element.

A.Sodium
B.Magnesium
C.Aluminium
D.Silicon

PastPaper.showAnswers

PastPaper.workedSolution

The successive ionization energies show a very large increase between the third and fourth ionization energies (from $2745\text{ kJ mol}^{-1}$ to $11577\text{ kJ mol}^{-1}$). This indicates that the fourth electron is removed from an inner quantum shell that is closer to the nucleus and less shielded, meaning the element has three valence electrons. In Period 3, the element with three valence electrons is aluminium.

PastPaper.markingScheme

[1] C - Aluminium. Large jump between third and fourth ionization energies indicates three valence electrons.

PastPaper.question 10 · multiple-choice

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A $0.440\text{ g}$ sample of an unknown gas occupies a volume of $249\text{ cm}^3$ at a pressure of $100\text{ kPa}$ and a temperature of $300\text{ K}$. What is the molar mass of the gas in $\text{g mol}^{-1}$?

[Gas constant, $R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}$]

A.16.0
B.28.0
C.44.0
D.64.0

PastPaper.showAnswers

PastPaper.workedSolution

Using the ideal gas equation:
$PV = nRT$

Convert units to SI:
$P = 100 \times 10^3\text{ Pa}$
$V = 249 \times 10^{-6}\text{ m}^3$
$T = 300\text{ K}$

Calculate the number of moles, $n$:
$n = \frac{PV}{RT} = \frac{100 \times 10^3 \times 249 \times 10^{-6}}{8.31 \times 300} = \frac{24.9}{2493} \approx 0.009988\text{ mol}$

Calculate the molar mass, $M$:
$M = \frac{\text{mass}}{n} = \frac{0.440}{0.009988} \approx 44.05\text{ g mol}^{-1}$

This is closest to $44.0\text{ g mol}^{-1}$.

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[1] C - 44.0. Correct application of ideal gas equation to find molar mass.

PastPaper.question 11 · multiple-choice

1 PastPaper.marks

Which of the following species has a molecular shape that is trigonal pyramidal?

A.$\text{BF}_3$
B.$\text{H}_3\text{O}^+$
C.$\text{NH}_4^+$
D.$\text{CO}_3^{2-}$

PastPaper.showAnswers

PastPaper.workedSolution

In $\text{H}_3\text{O}^+$, the central oxygen atom has three bonding pairs of electrons and one lone pair of electrons (since oxygen has 6 valence electrons, losing one for the positive charge leaves 5, three of which are used in single covalent bonds to hydrogen). According to VSEPR theory, a species with three bonding pairs and one lone pair has a trigonal pyramidal shape. $\text{BF}_3$ and $\text{CO}_3^{2-}$ are trigonal planar, while $\text{NH}_4^+$ is tetrahedral.

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[1] B - $\text{H}_3\text{O}^+$. Identify that the species has 3 bonding pairs and 1 lone pair around the central atom.

PastPaper.question 12 · multiple-choice

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What is the systematic IUPAC name for the major organic product formed when 2-methylpropane reacts with a limited amount of chlorine in the presence of ultraviolet light, where substitution occurs at the tertiary carbon?

A.1-chloro-2-methylpropane
B.2-chloro-2-methylpropane
C.2-chlorobutane
D.1-chloro-3-methylpropane

PastPaper.showAnswers

PastPaper.workedSolution

2-methylpropane has the formula $\text{(CH}_3\text{)}_3\text{CH}$. The central carbon (carbon-2) is a tertiary carbon because it is bonded to three other methyl groups. Replacing the hydrogen atom on this tertiary carbon with a chlorine atom yields $\text{(CH}_3\text{)}_3\text{CCl}$. The longest carbon chain contains three carbon atoms, with a chlorine atom and a methyl group both attached to carbon-2. Therefore, the systematic IUPAC name is 2-chloro-2-methylpropane.

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[1] B - 2-chloro-2-methylpropane. Correctly identify the product of substitution at the tertiary carbon.

PastPaper.question 13 · multiple-choice

1 PastPaper.marks

Which of the following alkenes can exhibit stereoisomerism (E/Z isomerism)?

A.2-methylbut-2-ene
B.propene
C.pent-1-ene
D.hex-3-ene

PastPaper.showAnswers

PastPaper.workedSolution

For an alkene to exhibit E/Z isomerism, each carbon atom of the C=C double bond must be bonded to two different atoms or groups of atoms.
- In 2-methylbut-2-ene, one C=C carbon is bonded to two methyl groups.
- In propene and pent-1-ene, the terminal C=C carbon is bonded to two hydrogen atoms.
- In hex-3-ene ($\text{CH}_3\text{CH}_2\text{CH=CHCH}_2\text{CH}_3$), each of the double-bonded carbons is bonded to one hydrogen atom ($\text{-H}$) and one ethyl group ($\text{-CH}_2\text{CH}_3$), allowing for distinct E and Z isomers.

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[1] D - hex-3-ene. Correct identification of the alkene meeting the structural requirements for E/Z isomerism.

PastPaper.question 14 · multiple-choice

1 PastPaper.marks

What is the electronic configuration of a $\text{Fe}^{3+}$ ion in its ground state?

A.$[\text{Ar}] 4s^2 3d^3$
B.$[\text{Ar}] 3d^5$
C.$[\text{Ar}] 4s^1 3d^4$
D.$[\text{Ar}] 4s^2 3d^5$

PastPaper.showAnswers

PastPaper.workedSolution

The electronic configuration of a neutral iron atom ($\text{Fe}$, $Z=26$) is $[\text{Ar}] 4s^2 3d^6$. When first row d-block elements form cations, electrons are lost from the $4s$ subshell first, before any $3d$ electrons. To form a $\text{Fe}^{3+}$ ion, three electrons are removed: two from the $4s$ orbital and one from the $3d$ orbital, leaving the configuration as $[\text{Ar}] 3d^5$.

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[1] B - $[\text{Ar}] 3d^5$. Award mark for correct removal of electrons from 4s then 3d orbitals.

PastPaper.question 15 · multiple-choice

1 PastPaper.marks

Calcium oxide is manufactured by the thermal decomposition of calcium carbonate:

$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$

What is the percentage atom economy for the production of calcium oxide in this reaction?

[Molar masses: $\text{CaCO}_3 = 100.1\text{ g mol}^{-1}$; $\text{CaO} = 56.1\text{ g mol}^{-1}$; $\text{CO}_2 = 44.0\text{ g mol}^{-1}$]

A.44.0%
B.56.0%
C.78.4%
D.100%

PastPaper.showAnswers

PastPaper.workedSolution

Atom economy is calculated using the following formula:
$\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\%$

Here, the desired product is $\text{CaO}$ and the reactant is $\text{CaCO}_3$.

$\text{Atom Economy} = \frac{56.1}{100.1} \times 100\% \approx 56.0\%$

PastPaper.markingScheme

[1] B - 56.0%. Correct calculation using molar masses of calcium oxide and calcium carbonate.

PastPaper.question 16 · multiple-choice

1 PastPaper.marks

Which of the following molecules has a permanent net dipole moment (is polar)?

A.$\text{CF}_4$
B.$\text{SF}_6$
C.$\text{PCl}_3$
D.$\text{CO}_2$

PastPaper.showAnswers

PastPaper.workedSolution

A molecule has a permanent net dipole moment if its individual bond dipoles do not cancel each other out due to symmetry.
- $\text{CF}_4$ is tetrahedral and highly symmetrical, so dipoles cancel out (non-polar).
- $\text{SF}_6$ is octahedral and highly symmetrical, so dipoles cancel out (non-polar).
- $\text{CO}_2$ is linear and symmetrical, so dipoles cancel out (non-polar).
- $\text{PCl}_3$ is trigonal pyramidal with a lone pair on the phosphorus atom. The asymmetrical shape means the polar $\text{P-Cl}$ bonds do not cancel out, resulting in a net dipole moment.

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[1] C - $\text{PCl}_3$. Correctly identify that the trigonal pyramidal geometry of phosphorus trichloride results in a net dipole.

PastPaper.question 17 · multiple_choice

1 PastPaper.marks

The successive ionization energies of a Period 3 element, $X$, are shown in the table below.

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\text{Ionization energy number} & 1\text{st} & 2\text{nd} & 3\text{rd} & 4\text{th} & 5\text{th} & 6\text{th} \\
\hline
\text{Ionization energy / kJ mol}^{-1} & 1012 & 1903 & 2912 & 4957 & 6274 & 21268 \\
\hline
\end{array}
\]

Which of the following is element $X$?

A.Silicon
B.Phosphorus
C.Sulfur
D.Chlorine

PastPaper.showAnswers

PastPaper.workedSolution

There is a massive jump between the 5th and the 6th ionization energies (from 6274 to 21268 $\text{kJ mol}^{-1}$). This indicates that the 6th electron is removed from an inner quantum shell, which is closer to the nucleus and experiences significantly less shielding. Therefore, element $X$ has 5 valence electrons and belongs to Group 15 of the Periodic Table.

In Period 3, the Group 15 element is phosphorus.

- Silicon is in Group 14 (4 valence electrons; large jump between 4th and 5th IE).
- Sulfur is in Group 16 (6 valence electrons; large jump between 6th and 7th IE).
- Chlorine is in Group 17 (7 valence electrons; large jump between 7th and 8th IE).

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[1 mark] B - Phosphorus
- Award 1 mark for correct selection.
- No partial marks.

PastPaper.question 18 · multiple_choice

1 PastPaper.marks

A sample of $2.12\text{ g}$ of a metal carbonate, $\text{M}_2\text{CO}_3$, reacted completely with excess dilute hydrochloric acid to form carbon dioxide gas.

\[ \text{M}_2\text{CO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{MCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \]

At room temperature and pressure (r.t.p.), $480\text{ cm}^3$ of carbon dioxide was collected.

What is the identity of the metal $\text{M}$?

[Molar volume of gas at r.t.p. = $24.0\text{ dm}^3\text{ mol}^{-1}$]

A.Lithium
B.Sodium
C.Potassium
D.Rubidium

PastPaper.showAnswers

PastPaper.workedSolution

1. Calculate the moles of $\text{CO}_2$ produced:
\[ n(\text{CO}_2) = \frac{480\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.020\text{ mol} \]

2. Determine the moles of $\text{M}_2\text{CO}_3$ reacted:
Since the stoichiometry ratio of $\text{M}_2\text{CO}_3 : \text{CO}_2$ is $1:1$:
\[ n(\text{M}_2\text{CO}_3) = 0.020\text{ mol} \]

3. Calculate the molar mass ($M_r$) of $\text{M}_2\text{CO}_3$:
\[ M_r(\text{M}_2\text{CO}_3) = \frac{\text{mass}}{n} = \frac{2.12\text{ g}}{0.020\text{ mol}} = 106\text{ g mol}^{-1} \]

4. Determine the relative atomic mass ($A_r$) of $\text{M}$:
\[ 2 \times A_r(\text{M}) + 12.0 + (3 \times 16.0) = 106 \]
\[ 2 \times A_r(\text{M}) + 60.0 = 106 \]
\[ 2 \times A_r(\text{M}) = 46.0 \]
\[ A_r(\text{M}) = 23.0 \]

Comparing this to values in the Periodic Table, the metal with $A_r \approx 23.0$ is sodium ($\text{Na}$).

PastPaper.markingScheme

[1 mark] B - Sodium
- Award 1 mark for correct calculation and identification.
- No partial marks.

PastPaper.question 19 · multiple_choice

1 PastPaper.marks

What is the number of bonding pairs of electrons, the number of lone pairs of electrons around the central atom, and the shape of the tetrachloroiodate(I) ion, $\text{ICl}_4^-$?

A.4 bonding pairs, 0 lone pairs, tetrahedral
B.4 bonding pairs, 1 lone pair, see-saw
C.4 bonding pairs, 2 lone pairs, square planar
D.6 bonding pairs, 0 lone pairs, octahedral

PastPaper.showAnswers

PastPaper.workedSolution

1. The central atom is iodine, which is in Group 17 and has 7 valence electrons.
2. The negative charge of $-1$ adds 1 extra electron, giving a total of 8 valence electrons on the central iodine atom.
3. Iodine forms 4 single covalent bonds with the four chlorine atoms, using 4 of its valence electrons to form 4 bonding pairs.
4. The remaining $8 - 4 = 4$ valence electrons form 2 lone pairs on the iodine atom.
5. With 4 bonding pairs and 2 lone pairs (a total of 6 electron pairs), the electron-pair geometry is octahedral. To minimize lone-pair to lone-pair repulsion, the two lone pairs are positioned opposite each other, resulting in a square planar molecular shape.

PastPaper.markingScheme

[1 mark] C - 4 bonding pairs, 2 lone pairs, square planar
- Award 1 mark for the correct combination.
- No partial marks.

PastPaper.question 20 · multiple_choice

1 PastPaper.marks

Which of the following alkenes can exist as a pair of geometric ($E$/$Z$) isomers?

A.2-methylbut-2-ene
B.2-methylpent-1-ene
C.3-methylpent-2-ene
D.2,3-dimethylbut-2-ene

PastPaper.showAnswers

PastPaper.workedSolution

For geometric ($E$/$Z$) isomerism to occur, both carbon atoms involved in the double bond must be bonded to two different groups.

- **A: 2-methylbut-2-ene** ($\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3$). The left-hand double-bonded carbon is bonded to two identical methyl groups, so it cannot show geometric isomerism.
- **B: 2-methylpent-1-ene** ($\text{CH}_2=\text{C}(\text{CH}_3)(\text{CH}_2\text{CH}_2\text{CH}_3)$). The left-hand double-bonded carbon is bonded to two identical hydrogen atoms, so it cannot show geometric isomerism.
- **C: 3-methylpent-2-ene** ($\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)(\text{CH}_2\text{CH}_3)$). The left-hand double-bonded carbon is bonded to $\text{-H}$ and $\text{-CH}_3$ (different). The right-hand double-bonded carbon is bonded to $\text{-CH}_3$ and $\text{-CH}_2\text{CH}_3$ (different). Thus, it can exist as $E$ and $Z$ isomers.
- **D: 2,3-dimethylbut-2-ene** ($\text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)-\text{CH}_3$). Both double-bonded carbons are bonded to two identical methyl groups, so it cannot show geometric isomerism.

PastPaper.markingScheme

[1 mark] C - 3-methylpent-2-ene
- Award 1 mark for correct identification of the isomerizing alkene.
- No partial marks.

Unit 1: Section B

Answer all structured questions in the spaces provided.

*(a)(i)*
- **M1**: Correct skeletal drawing of E-but-2-ene and Z-but-2-ene (1)
- **M2**: Correctly labeled E and Z isomers (1)
*(a)(ii)*
- **M1**: Restricted rotation around the C=C double bond (due to pi-bond) (1)
- **M2**: For E-Z isomerism, each carbon of the double bond must be attached to two different groups (1)
- **M3**: One carbon in but-1-ene has two identical hydrogen atoms attached, so it cannot show E-Z isomerism (1)
*(b)(i)*
- **M1**: Curly arrow from the C=C double bond of propene to H of H-Br AND dipole on H-Br correctly shown (Hδ+ - Brδ-) (1)
- **M2**: Curly arrow from H-Br bond to Br (1)
- **M3**: Correct drawing of secondary carbocation intermediate with a positive charge on the middle carbon AND Br- with a lone pair (1)
- **M4**: Curly arrow from the lone pair of Br- to the positive carbon (1)
*(b)(ii)*
- **M1**: Name major product: 2-bromopropane (1)
- **M2**: Identify that the secondary carbocation is more stable than the primary carbocation (1)
- **M3**: Explain that the secondary carbocation has two electron-releasing alkyl groups which stabilize the positive charge (positive inductive effect) (1)

Unit 2: Section A

Answer all multiple-choice questions.

12 PastPaper.question · 12 PastPaper.marks

PastPaper.question 1 · Multiple Choice

1 PastPaper.marks

Which of the following describes and explains the trend in thermal stability of Group 2 carbonates down the group from magnesium carbonate to barium carbonate?

A.Thermal stability increases because the cationic radius increases, leading to less polarisation of the carbonate ion.
B.Thermal stability decreases because the cationic radius increases, leading to more polarisation of the carbonate ion.
C.Thermal stability increases because the charge density of the cation increases, leading to less polarisation of the carbonate ion.
D.Thermal stability decreases because the lattice energy of the carbonate increases.

PastPaper.showAnswers

PastPaper.workedSolution

Down Group 2 from magnesium to barium, the ionic radius of the $M^{2+}$ cation increases while its charge remains constant at $+2$. This causes a decrease in charge density of the cation down the group. As a result, the larger cations have less polarising power and cause less distortion (polarisation) of the large carbonate ion's electron cloud. Therefore, more thermal energy is required to decompose the carbonate, meaning thermal stability increases down the group.

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1 mark for correct option (A).

PastPaper.question 2 · Multiple Choice

1 PastPaper.marks

Under identical reaction conditions, which of the following halogenoalkanes reacts fastest when heated with aqueous silver nitrate in ethanol?

A.1-chlorobutane
B.1-bromobutane
C.2-bromobutane
D.2-iodobutane

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PastPaper.workedSolution

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. The $\text{C-I}$ bond is weaker than the $\text{C-Br}$ and $\text{C-Cl}$ bonds, so iodoalkanes react faster than bromoalkanes and chloroalkanes. Additionally, secondary halogenoalkanes (like 2-iodobutane) react faster than primary halogenoalkanes (like 1-iodobutane, 1-bromobutane, or 1-chlorobutane) under these conditions due to the greater stability of the carbocation intermediate or easier nucleophilic substitution pathways. Therefore, 2-iodobutane reacts the fastest.

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1 mark for correct option (D).

PastPaper.question 3 · Multiple Choice

1 PastPaper.marks

Which of the following organic compounds has the highest boiling temperature?

A.Propane-1,2-diol
B.Propan-1-ol
C.Propan-2-ol
D.Butan-1-ol

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PastPaper.workedSolution

Boiling temperature depends on the strength of intermolecular forces. All of these compounds can form hydrogen bonds, which are the strongest intermolecular forces. Propane-1,2-diol has two hydroxyl ($-\text{OH}$) groups per molecule, allowing it to form extensive hydrogen-bonded networks. Propan-1-ol, propan-2-ol, and butan-1-ol only have one $-\text{OH}$ group per molecule. The additional hydrogen bonding in propane-1,2-diol significantly increases the energy required to separate the molecules, resulting in the highest boiling temperature.

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1 mark for correct option (A).

PastPaper.question 4 · Multiple Choice

1 PastPaper.marks

Using the standard enthalpy changes of combustion ($\Delta_c H^\ominus$) below, what is the standard enthalpy change of formation ($\Delta_f H^\ominus$) of propane, $\text{C}_3\text{H}_8\text{(g)}$? $\Delta_c H^\ominus[\text{C(s)}] = -393.5\text{ kJ mol}^{-1}$; $\Delta_c H^\ominus[\text{H}_2\text{(g)}] = -285.8\text{ kJ mol}^{-1}$; $\Delta_c H^\ominus[\text{C}_3\text{H}_8\text{(g)}] = -2219.2\text{ kJ mol}^{-1}$

A.$-104.5\text{ kJ mol}^{-1}$
B.$-1539.9\text{ kJ mol}^{-1}$
C.$+104.5\text{ kJ mol}^{-1}$
D.$+2539.9\text{ kJ mol}^{-1}$

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PastPaper.workedSolution

The formation equation for propane is: $3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}$. Using Hess's law with enthalpy changes of combustion: $\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$. Therefore, $\Delta_f H^\ominus = [3 \times (-393.5) + 4 \times (-285.8)] - (-2219.2) = [-1180.5 - 1143.2] + 2219.2 = -2323.7 + 2219.2 = -104.5\text{ kJ mol}^{-1}$.

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1 mark for correct option (A).

PastPaper.question 5 · Multiple Choice

1 PastPaper.marks

Which of the following substances is the yellow solid formed when solid sodium iodide reacts with concentrated sulfuric acid?

A.Sulfur dioxide
B.Sulfur
C.Hydrogen sulfide
D.Iodine

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PastPaper.workedSolution

In the reaction of solid sodium iodide with concentrated sulfuric acid, iodide ions are oxidized to iodine (which is a grey-black solid or a purple vapour). The sulfuric acid is reduced to several products including sulfur dioxide (colourless, choking gas), sulfur (yellow solid), and hydrogen sulfide (gas with a rotten-egg odour). The yellow solid observed is sulfur (S).

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1 mark for correct option (B).

PastPaper.question 6 · Multiple Choice

1 PastPaper.marks

How does the addition of a catalyst affect the Maxwell-Boltzmann distribution of molecular energies for a gas-phase reaction at constant temperature?

A.The peak of the distribution curve shifts to the right and becomes lower.
B.The peak of the distribution curve shifts to the left and becomes higher.
C.The shape of the distribution curve remains unchanged, but the activation energy barrier shifts to a lower value.
D.The shape of the distribution curve remains unchanged, but the activation energy barrier shifts to a higher value.

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PastPaper.workedSolution

A catalyst provides an alternative reaction pathway with a lower activation energy, but it does not alter the temperature or the kinetic energy of the reacting molecules. Therefore, the shape of the Maxwell-Boltzmann distribution curve itself remains completely unchanged. However, the activation energy line ($E_a$) is shifted to a lower energy value (to the left), meaning a larger fraction of molecules have energy greater than or equal to the activation energy.

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1 mark for correct option (C).

PastPaper.question 7 · Multiple Choice

1 PastPaper.marks

An organic compound has the molecular formula $\text{C}_3\text{H}_6\text{O}_2$. Its infrared (IR) spectrum shows a very broad, strong absorption band in the range $2500\text{--}3300\text{ cm}^{-1}$ and a sharp, strong absorption band at approximately $1715\text{ cm}^{-1}$. Which compound is consistent with this information?

A.Ethyl formate
B.Methyl acetate
C.Propanoic acid
D.Propan-1-ol

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PastPaper.workedSolution

The broad, strong absorption band in the region $2500\text{--}3300\text{ cm}^{-1}$ is characteristic of the O-H stretching vibration in carboxylic acids. The sharp, strong band at $1715\text{ cm}^{-1}$ corresponds to the C=O stretch of a carbonyl group. Together, these indicate the presence of a carboxylic acid functional group (-COOH). Propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) is a carboxylic acid with the molecular formula $\text{C}_3\text{H}_6\text{O}_2$, making it the correct option. Ethyl formate and methyl acetate are esters (no O-H stretch), and propan-1-ol is an alcohol ($\text{C}_3\text{H}_8\text{O}$, O-H stretch is in the range $3200\text{--}3650\text{ cm}^{-1}$ and there is no C=O stretch).

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1 mark for correct option (C).

PastPaper.question 8 · Multiple Choice

1 PastPaper.marks

When chlorine gas is reacted with cold, dilute aqueous sodium hydroxide, what are the oxidation states of chlorine in the products formed?

A.$-1$ and $+1$
B.$-1$ and $+5$
C.0 and $-1$
D.$+1$ and $+5$

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PastPaper.workedSolution

When chlorine gas reacts with cold, dilute sodium hydroxide, it undergoes disproportionation: $\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}$. In sodium chloride ($\text{NaCl}$), the oxidation state of chlorine is $-1$. In sodium chlorate(I) ($\text{NaClO}$), the oxidation state of chlorine is $+1$. (Reacting chlorine with hot, concentrated sodium hydroxide yields $\text{NaCl}$ and $\text{NaClO}_3$, where chlorine is in the $-1$ and $+5$ oxidation states respectively).

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1 mark for correct option (A).

PastPaper.question 9 · multiple-choice

1 PastPaper.marks

Which of the following anhydrous Group 2 nitrates decomposes at the highest temperature?

A.Magnesium nitrate
B.Calcium nitrate
C.Strontium nitrate
D.Barium nitrate

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PastPaper.workedSolution

Thermal stability of Group 2 nitrates increases down the group. As the group is descended, the ionic radius of the $M^{2+}$ cation increases, while its charge remains $+2$. This causes the charge density of the cation to decrease, meaning it has a lower polarizing effect on the nitrate anion. Consequently, the nitrogen-oxygen bond in the anion is weakened to a lesser extent, making the nitrate more thermally stable and requiring a higher temperature for decomposition. Therefore, barium nitrate is the most thermally stable.

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1 mark for correct choice D. Reject other choices as thermal stability increases down the group.

PastPaper.question 10 · multiple-choice

1 PastPaper.marks

Under identical reaction conditions, which of the following halogenoalkanes reacts most rapidly with aqueous silver nitrate in ethanol?

A.1-chlorobutane
B.1-bromobutane
C.2-bromo-2-methylpropane
D.2-chloro-2-methylpropane

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PastPaper.workedSolution

The rate of hydrolysis of halogenoalkanes depends on two factors: the strength of the carbon-halogen bond and the mechanism. The C-Br bond is weaker than the C-Cl bond, so C-Br bonds break more easily. Tertiary halogenoalkanes (2-bromo-2-methylpropane and 2-chloro-2-methylpropane) react via the $S_{\text{N}}1$ mechanism, which is much faster than the $S_{\text{N}}2$ mechanism of primary halogenoalkanes because the tertiary carbocation intermediate is highly stable. Thus, 2-bromo-2-methylpropane has both a tertiary structure and a weaker carbon-halogen bond, making it react the most rapidly.

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1 mark for correct choice C. Reject tertiary chloroalkane (D) as C-Cl is stronger than C-Br. Reject primary bromoalkane (B) as it reacts via the slower SN2 mechanism.

PastPaper.question 11 · multiple-choice

1 PastPaper.marks

Which of the following compounds has the highest boiling temperature?

A.Propan-1-ol
B.Propanal
C.Propanone
D.Methoxyethane

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PastPaper.workedSolution

Propan-1-ol is an alcohol and contains a highly polar $\text{O-H}$ group, which enables it to form intermolecular hydrogen bonds. Hydrogen bonds are the strongest type of intermolecular force. Propanal, propanone, and methoxyethane do not have $\text{O-H}$ groups and can only form permanent dipole-dipole forces and London forces. Therefore, propan-1-ol requires the most energy to overcome its intermolecular forces, resulting in the highest boiling temperature.

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1 mark for correct choice A. Reject other options because they lack hydrogen bonding.

PastPaper.question 12 · multiple-choice

1 PastPaper.marks

Consider the following standard enthalpy changes of combustion:

$\Delta_{\text{c}}H^{\ominus}[\text{C}_2\text{H}_4(\text{g})] = -1411\text{ kJ mol}^{-1}$

$\Delta_{\text{c}}H^{\ominus}[\text{H}_2(\text{g})] = -286\text{ kJ mol}^{-1}$

$\Delta_{\text{c}}H^{\ominus}[\text{C}_2\text{H}_6(\text{g})] = -1560\text{ kJ mol}^{-1}$

What is the standard enthalpy change of hydrogenation of ethene, $\Delta_{\text{r}}H^{\ominus}$, for the reaction shown below?

$\text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_6(\text{g})$

A.$-137\text{ kJ mol}^{-1}$
B.$+137\text{ kJ mol}^{-1}$
C.$-3257\text{ kJ mol}^{-1}$
D.$+3257\text{ kJ mol}^{-1}$

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PastPaper.workedSolution

According to Hess's Law, using standard enthalpy changes of combustion: $\Delta_{\text{r}}H^{\ominus} = \sum \Delta_{\text{c}}H^{\ominus}(\text{reactants}) - \sum \Delta_{\text{c}}H^{\ominus}(\text{products})$. Substituting the given values: $\Delta_{\text{r}}H^{\ominus} = [(-1411) + (-286)] - [-1560] = -1697 + 1560 = -137\text{ kJ mol}^{-1}$.

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1 mark for correct choice A. Reject B (incorrect sign). Reject C and D (incorrect combination of enthalpy values).

Unit 2: Section B

Answer all structured inorganic and energetics questions.

Part (a) [4 marks]
- MP1: Correct labels on both axes: y-axis as 'Number of molecules'/'Fraction of molecules', x-axis as 'Energy'/'Kinetic Energy'. (1)
- MP2: Correct shape of the curve: starts at origin, single peak, does not touch/cross the x-axis at the high-energy tail. (1)
- MP3: $E_{\text{a}}$ correctly positioned on the x-axis. (1)
- MP4: Shading of the area to the right of $E_{\text{a}}$ under the curve. (1)

Part (b)(i) [3 marks]
- MP1: $E_{\text{cat}}$ marked to the left of $E_{\text{a}}$ on the x-axis. (1)
- MP2: Explains that the catalyst provides an alternative pathway with a lower activation energy. (1)
- MP3: Explains that more molecules have energy $\ge E_{\text{cat}}$ / greater area under curve is shaded, leading to more successful collisions per unit time. (1)

Part (b)(ii) [3 marks]
- MP1: Curve at higher temperature $T_2$ has a lower peak shifted to the right. (1)
- MP2: Curve at $T_2$ is higher at high energies (crosses $T_1$ curve once and remains above it at the tail). (1)
- MP3: Explains that a larger fraction of molecules now have energy $\ge E_{\text{a}}$, resulting in a higher frequency of successful collisions / more successful collisions per unit time. (1)

Unit 2: Section C

Answer all structured organic synthesis and classification questions.

1 PastPaper.question · 20 PastPaper.marks

PastPaper.question 1 · Organic Synthesis & Structural Identification

20 PastPaper.marks

This question is about Compound A, an organic intermediate used widely in the design of synthetic fragrances and flavorings. Compound A has the molecular formula $\text{C}_5\text{H}_{10}\text{O}$.

(a) The infrared (IR) spectrum of Compound A contains a broad, strong absorption peak in the range $3200-3600\text{ cm}^{-1}$ and a sharp, weaker absorption peak at $1650\text{ cm}^{-1}$.

Identify the bonds and functional groups responsible for both of these IR absorption peaks. [2]
In the mass spectrum of Compound A, a prominent fragment peak is observed at $m/z = 71$. Deduce the formula of the species responsible for this peak, including its charge. [1]
Compound A is a branched-chain primary alcohol that does not show E/Z (geometric) isomerism. Draw the skeletal formula of Compound A. [2]

(b) Compound A undergoes several standard organic reactions.

When Compound A is reacted with phosphorus(V) chloride, $\text{PCl}_5$, a vigorous reaction occurs. State one observation for this reaction and draw the skeletal formula of the organic product. [2]
Under different conditions, Compound A can be oxidized using acidified potassium dichromate(VI), $\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+$. Describe the experimental setup and technique required to prepare and obtain a pure sample of the aldehyde, 3-methylbut-2-enal, rather than the carboxylic acid. Explain your answer in terms of intermolecular forces. [3]
Draw the skeletal formula of the carboxylic acid formed when Compound A is heated under reflux with excess acidified potassium dichromate(VI). [1]
Describe the color change observed when bromine in an organic solvent is added to Compound A, and draw the skeletal formula of the organic product formed. [3]

(c) An alternative "green" synthesis of Compound A involves the acid-catalyzed addition of 2-methylpropene to methanal (formaldehyde):

$\text{C}_4\text{H}_8 + \text{HCHO} \rightarrow \text{C}_5\text{H}_{10}\text{O}$

Calculate the atom economy of this reaction. Show your working. [2]
Suggest two other principles of green chemistry, other than atom economy, that make this catalytic pathway preferable to a traditional multi-step halogenation-substitution route. [2]
In a laboratory preparation, $11.2\text{ g}$ of 2-methylpropene ($M_{\text{r}} = 56.0$) was reacted with excess methanal ($M_{\text{r}} = 30.0$) to produce $12.9\text{ g}$ of Compound A ($M_{\text{r}} = 86.0$). Calculate the percentage yield of Compound A, giving your answer to three significant figures. [2]

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Part (a)

(i) The broad peak at $3200-3600\text{ cm}^{-1}$ is due to the $\text{O-H}$ bond stretch of an alcohol [1]. The peak at $1650\text{ cm}^{-1}$ is due to the $\text{C=C}$ bond stretch of an alkene [1].
(ii) The molecular mass of Compound A ($\text{C}_5\text{H}_{10}\text{O}$) is $86$. The loss of a methyl radical ($\cdot\text{CH}_3$, mass = $15$) yields a fragment with $m/z = 71$. The species is $\text{[C}_4\text{H}_7\text{O]}^+$ [1]. (Charge must be shown; accept structurally written cations like $\text{[(CH}_3\text{)C=CH-CH}_2\text{OH]}^{+}$ or similar).
(iii) To prevent E/Z isomerism, one of the doubly-bonded carbons must be attached to two identical groups. In a branched 5-carbon skeleton containing a primary alcohol, this corresponds to 3-methylbut-2-en-1-ol (where C3 is attached to two methyl groups):

Award [2] for correct skeletal structure showing the double bond at C2-C3, branch at C3, and primary alcohol group. Award [1] if structural/displayed formula drawn correctly instead of skeletal, or if a branched isomer is drawn that incorrectly exhibits E/Z isomerism.

Part (b)

(i) Observation: Misty white / steamy fumes [1].
Skeletal formula: Same skeleton as A but with chlorine instead of the hydroxyl group (1-chloro-3-methylbut-2-ene) [1].
(ii) Set up the apparatus for distillation [1] (or distill off the aldehyde as it forms).
The aldehyde lacks an $\text{O-H}$ bond and cannot form intermolecular hydrogen bonds (only dipole-dipole forces), meaning it has a lower boiling point than both the starting alcohol and the carboxylic acid product [1].
Immediate distillation removes the aldehyde from the reaction mixture, preventing further oxidation to the carboxylic acid [1].
(iii) Heating under reflux with excess oxidising agent oxidises the primary alcohol group to a carboxylic acid. Skeletal formula of 3-methylbut-2-enoic acid: same carbon skeleton with a $\text{=O}$ and $\text{-OH}$ on the terminal carbon [1].
(iv) Color change: Orange / brown / yellow to colorless (decolorized) [1] (Reject: 'discolored' or 'clear').
Skeletal formula of addition product (2,3-dibromo-3-methylbutan-1-ol): 2 bromine atoms added across the C=C double bond, single bond remaining, alcohol group intact [2] (Award [1] if carbon skeleton is correct but bromine atoms or alcohol are incorrectly placed).

Part (c)

(i) In addition reactions where only one product is formed, all atoms of the reactants are converted into the desired product. Therefore, the atom economy is 100% [1].
Working: $\text{Atom economy} = \frac{M_{\text{r}}(\text{desired product})}{\sum M_{\text{r}}(\text{reactants})} \times 100 = \frac{86.0}{56.0 + 30.0} \times 100 = 100\%$ [1].
(ii) Any two principles of Green Chemistry from:
- Use of a catalyst (increases reaction rate, allows lower reaction temperatures/pressures) [1].
- Prevention of waste (no toxic byproducts like $\text{HCl}$ are produced) [1].
- Safer reagents/solvents (avoids hazardous/chlorinated reagents) [1].
- Energy efficiency (reduces energy demands of chemical processes) [1].
(iii) $\text{Moles of 2-methylpropene} = \frac{11.2}{56.0} = 0.200\text{ mol}$ [1].
Since stoichiometry is 1:1, theoretical moles of Compound A = $0.200\text{ mol}$.
\[ \text{Theoretical yield of A} = 0.200\text{ mol} \times 86.0\text{ g mol}^{-1} = 17.2\text{ g} \]
\[ \text{Percentage yield} = \frac{12.9}{17.2} \times 100\% = 75.0\% \]
[1] (Accept 75% or 75.0%).

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(a)(i) [2 marks]
- 1 mark: O-H bond stretch (alcohol) at 3200-3600 cm^-1. (Accept 'hydroxyl group')
- 1 mark: C=C bond stretch (alkene) at 1650 cm^-1. (Reject: C=O)

(a)(ii) [1 mark]
- 1 mark: [C4H7O]+ or [(CH3)2C=CHCH2]+ (Charge is essential. Reject neutral formula).

(a)(iii) [2 marks]
- 2 marks: Correct skeletal formula of 3-methylbut-2-en-1-ol.
- 1 mark: Correct atoms/groups but drawn as structural/displayed formula, OR skeletal formula of a branched 5-carbon isomer that has E/Z isomerism (e.g., 2-methylbut-2-en-1-ol).

(b)(i) [2 marks]
- 1 mark: Misty white fumes / steamy fumes. (Accept: cloudy gas. Reject: white smoke).
- 1 mark: Correct skeletal formula of 1-chloro-3-methylbut-2-ene.

(b)(ii) [3 marks]
- 1 mark: Distillation setup / distill off product as it forms.
- 1 mark: Aldehyde has lower boiling point / cannot form hydrogen bonds (only dipole-dipole).
- 1 mark: Distillation prevents further contact with oxidizing agent / prevents further oxidation.

(b)(iii) [1 mark]
- 1 mark: Correct skeletal formula of 3-methylbut-2-enoic acid.

(b)(iv) [3 marks]
- 1 mark: Orange/yellow/brown to colorless. (Reject: clear).
- 2 marks: Correct skeletal formula of 2,3-dibromo-3-methylbutan-1-ol. (1 mark if bromine atoms placed on incorrect carbons of the skeleton).

(c)(i) [2 marks]
- 1 mark: Showing sum of reactant masses (56.0 + 30.0) or noting addition reaction with single product.
- 1 mark: 100%.

(c)(ii) [2 marks]
- 1 mark each for any two valid green chemistry principles listed in the solution.

(c)(iii) [2 marks]
- 1 mark: Correct calculation of moles of reactant (0.200 mol).
- 1 mark: Correct percentage yield of 75.0% (allow 75%).

PastPaper.section Unit 3

Answer all structured questions testing qualitative tests and experimental volumetric skills.

- Part (a): [3 marks]
- 1 mark for identifying bromide ($\text{Br}^-$).
- 1 mark for justifying with cream precipitate and its solubility in concentrated ammonia.
- 1 mark for stating that nitric acid reacts with and removes carbonate/sulfite impurities.
- Part (b): [5 marks]
- 1 mark for identifying $\text{Ca}^{2+}$.
- 1 mark for linking $\text{Ca}^{2+}$ to the brick-red flame.
- 1 mark for linking $\text{Ca}^{2+}$ to the white precipitate with sulfuric acid.
- 1 mark for identifying $\text{NH}_4^+$.
- 1 mark for linking $\text{NH}_4^+$ to the alkaline gas evolved with sodium hydroxide.
- Part (c): [2.5 marks]
- 1.5 marks for correct formula of reactants and products.
- 1 mark for correct state symbols: $(aq)$ for ions, $(s)$ for calcium sulfate.
- Part (d): [2 marks]
- 1 mark for reagent (concentrated hydrochloric acid / hydrogen chloride gas).
- 1 mark for correct observation (dense white fumes).

Unit 4: Section A

Answer all multiple-choice physical chemistry questions.

20 PastPaper.question · 20 PastPaper.marks

PastPaper.question 1 · Multiple Choice

1 PastPaper.marks

The rate equation for the reaction between peroxodisulfate(VI) ions and iodide ions is:

$$\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$$

Which of the following changes will increase the value of the rate constant, $k$?

A.Increasing the concentration of \(\text{I}^-\right) ions.
B.Adding a catalyst to the reaction mixture.
C.Increasing the concentration of $\text{S}_2\text{O}_8^{2-}$ ions.
D.Carrying out the reaction in a flask with a larger volume.

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PastPaper.workedSolution

The rate constant $k$ is temperature-dependent and is affected by the addition of a catalyst, which provides an alternative pathway with a lower activation energy, thus increasing $k$. It is independent of the concentrations of the reactants or the volume of the reaction vessel.

PastPaper.markingScheme

[1 mark] B is the correct answer.
- Reject A and C because changing reactant concentrations changes the rate of reaction but not the rate constant.
- Reject D because changing the volume changes concentration but not the rate constant.

PastPaper.question 2 · Multiple Choice

1 PastPaper.marks

For a certain reaction, $\Delta H^{\ominus} = -114.4\text{ kJ mol}^{-1}$ at a temperature of $25\text{ }^{\circ}\text{C}$.

What is the standard entropy change of the surroundings, $\Delta S_{\text{surroundings}}^{\ominus}$, at this temperature?

A.$-384\text{ J K}^{-1}\text{ mol}^{-1}$
B.$-0.384\text{ J K}^{-1}\text{ mol}^{-1}$
C.$+384\text{ J K}^{-1}\text{ mol}^{-1}$
D.$+0.384\text{ J K}^{-1}\text{ mol}^{-1}$

PastPaper.showAnswers

PastPaper.workedSolution

Use the equation:
$$\Delta S_{\text{surroundings}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T}$$

Convert temperature to Kelvin:
$$T = 25 + 273 = 298\text{ K}$$

Convert enthalpy change to $\text{J mol}^{-1}$:
$$\Delta H^{\ominus} = -114.4 \times 10^3\text{ J mol}^{-1} = -114400\text{ J mol}^{-1}$$

Calculate entropy change:
$$\Delta S_{\text{surroundings}}^{\ominus} = -\frac{-114400}{298} = +383.89\text{ J K}^{-1}\text{ mol}^{-1} \approx +384\text{ J K}^{-1}\text{ mol}^{-1}$$

PastPaper.markingScheme

[1 mark] C is the correct answer.
- Award [1] for calculating $\approx +384\text{ J K}^{-1}\text{ mol}^{-1}$.
- Reject A (incorrect sign).
- Reject B and D (incorrect unit conversion, left in $\text{kJ K}^{-1}\text{ mol}^{-1}$).

PastPaper.question 3 · Multiple Choice

1 PastPaper.marks

The following equilibrium is established at temperature $T$:

$$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$$

At equilibrium in a closed vessel of fixed volume, the mole fractions of the gases are:
$\chi(\text{SO}_2) = 0.20$, $\chi(\text{O}_2) = 0.30$, $\chi(\text{SO}_3) = 0.50$.
The total pressure of the system is $2.0\text{ atm}$.

What is the value of the equilibrium constant, $K_p$, in $\text{atm}^{-1}$?

A.$2.08$
B.$4.17$
C.$10.4$
D.$20.8$

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PastPaper.workedSolution

First, calculate the equilibrium partial pressures of each gas:
$$p(\text{SO}_2) = 0.20 \times 2.0 = 0.40\text{ atm}$$
$$p(\text{O}_2) = 0.30 \times 2.0 = 0.60\text{ atm}$$
$$p(\text{SO}_3) = 0.50 \times 2.0 = 1.00\text{ atm}$$

Next, write the expression for $K_p$:
$$K_p = \frac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2 \times p(\text{O}_2)}$$

Substitute the partial pressures into the expression:
$$K_p = \frac{(1.00)^2}{(0.40)^2 \times (0.60)} = \frac{1.00}{0.16 \times 0.60} = \frac{1.00}{0.096} \approx 10.4\text{ atm}^{-1}$$

This corresponds to Option C.

PastPaper.markingScheme

[1 mark] C is the correct answer.
- Award [1] for calculating $10.4$.
- Reject A (calculated as $\frac{0.50^2}{0.20^2 \times 0.30}$ without multiplying by total pressure).
- Reject B (incorrect math or incorrect power dependence).
- Reject D (incorrectly multiplying the final answer by total pressure).

PastPaper.question 4 · Multiple Choice

1 PastPaper.marks

A weak monobasic acid, $\text{HA}$, has a $\text{p}K_a$ of $4.82$ at $298\text{ K}$.

What is the $\text{pH}$ of a $0.050\text{ mol dm}^{-3}$ solution of $\text{HA}$ at this temperature?

A.$2.41$
B.$3.06$
C.$4.82$
D.$6.12$

PastPaper.showAnswers

PastPaper.workedSolution

Calculate $K_a$:
$$K_a = 10^{-\text{p}K_a} = 10^{-4.82} = 1.514 \times 10^{-5}\text{ mol dm}^{-3}$$

Using the weak acid approximation:
$$[\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]}$$
$$[\text{H}^+] = \sqrt{(1.514 \times 10^{-5}) \times 0.050} = \sqrt{7.57 \times 10^{-7}} = 8.70 \times 10^{-4}\text{ mol dm}^{-3}$$

Calculate $\text{pH}$:
$$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(8.70 \times 10^{-4}) \approx 3.06$$

PastPaper.markingScheme

[1 mark] B is the correct answer.
- Reject A (this is simply $\frac{1}{2} \text{p}K_a$).
- Reject C (this is the $\text{p}K_a$ value itself).
- Reject D (calculated by adding $0.65$ instead of subtracting).

PastPaper.question 5 · Multiple Choice

1 PastPaper.marks

The rate of reaction between two reactants, $\text{X}$ and $\text{Y}$, was studied. The following results were obtained:

- When the concentration of $\text{X}$ was doubled, while keeping the concentration of $\text{Y}$ constant, the rate of reaction quadrupled.
- When the concentration of $\text{Y}$ was halved, while keeping the concentration of $\text{X}$ constant, the rate of reaction decreased by a factor of eight.

What is the overall order of the reaction?

A.$2$
B.$3$
C.$4$
D.$5$

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PastPaper.workedSolution

Determine individual reactant orders:
1. Since doubling $[\text{X}]$ causes the rate to quadruple ($2^2 = 4$), the order with respect to $\text{X}$ is $2$.
2. Since halving $[\text{Y}]$ causes the rate to decrease by a factor of eight ($(0.5)^3 = 0.125$), the order with respect to $\text{Y}$ is $3$.

Calculate the overall order of reaction:
$$\text{Overall order} = 2 + 3 = 5$$

PastPaper.markingScheme

[1 mark] D is the correct answer.
- Reject A because this assumes first order with respect to both reactants.
- Reject B because this assumes order of 1 for $\text{X}$ and 2 for $\text{Y}$.
- Reject C because this assumes order of 2 for both reactants.

PastPaper.question 6 · Multiple Choice

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For the decomposition of calcium carbonate:

$$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$$

The standard enthalpy change, $\Delta H^{\ominus}$, is $+178\text{ kJ mol}^{-1}$ and the standard entropy change of the system, $\Delta S_{\text{system}}^{\ominus}$, is $+160\text{ J K}^{-1}\text{ mol}^{-1}$.

Assuming these values do not change with temperature, above what temperature does this reaction become thermodynamically feasible?

A.$111\text{ K}$
B.$288\text{ K}$
C.$1113\text{ K}$
D.$2848\text{ K}$

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PastPaper.workedSolution

A reaction is thermodynamically feasible when the standard Gibbs free energy change is negative:
$$\Delta G^{\ominus} < 0$$
$$\Delta H^{\ominus} - T\Delta S_{\text{system}}^{\ominus} < 0$$
$$T > \frac{\Delta H^{\ominus}}{\Delta S_{\text{system}}^{\ominus}}$$

Convert $\Delta H^{\ominus}$ to $\text{J mol}^{-1}$:
$$\Delta H^{\ominus} = 178 \times 10^3\text{ J mol}^{-1}$$

Substitute the values:
$$T > \frac{178000}{160} = 1112.5\text{ K}$$

Therefore, the reaction becomes feasible above $1113\text{ K}$.

PastPaper.markingScheme

[1 mark] C is the correct answer.
- Award [1] for calculating $1113\text{ K}$.
- Reject A (incorrect calculation by not converting kJ to J).
- Reject B or D (incorrect mathematical operations).

PastPaper.question 7 · Multiple Choice

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The industrial synthesis of methanol is represented by the equilibrium:

$$\text{CO}(\text{g}) + 2\text{H}_2(\text{g}) \rightleftharpoons \text{CH}_3\text{OH}(\text{g}) \quad \Delta H^{\ominus} = -91\text{ kJ mol}^{-1}$$

Which of the following statements correctly describes the effect of increasing the temperature of this reaction at constant pressure?

A.The rate of the forward reaction decreases, and the value of $K_c$ increases.
B.The rate of both the forward and reverse reactions increases, and the value of $K_c$ decreases.
C.The position of equilibrium shifts to the right, and the value of $K_c$ increases.
D.The position of equilibrium shifts to the left, and the value of $K_c$ remains constant.

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PastPaper.workedSolution

Increasing the temperature increases the kinetic energy of the particles, leading to more frequent and successful collisions, thereby increasing the rate of both the forward and reverse reactions. Because the forward reaction is exothermic, according to Le Chatelier's principle, an increase in temperature shifts the equilibrium position to the left (the endothermic direction). As the equilibrium shifts to the left, the concentration of the product decreases while the concentrations of reactants increase, which causes the value of the equilibrium constant $K_c$ to decrease.

PastPaper.markingScheme

[1 mark] B is the correct answer.
- Reject A because increasing the temperature always increases the rate of reactions.
- Reject C because the equilibrium shifts to the left for an exothermic reaction, decreasing $K_c$.
- Reject D because $K_c$ changes with temperature.

PastPaper.question 8 · Multiple Choice

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Four different aqueous solutions, each of concentration $0.10\text{ mol dm}^{-3}$, are tested with a pH meter at $298\text{ K}$.

Which of the following lists the solutions in order of increasing $\text{pH}$ (from lowest $\text{pH}$ to highest $\text{pH}$)?

A.$\text{HCl(aq)} < \text{CH}_3\text{COOH(aq)} < \text{CH}_3\text{COONa(aq)} < \text{NH}_3\text{(aq)}$
B.$\text{CH}_3\text{COOH(aq)} < \text{HCl(aq)} < \text{NH}_3\text{(aq)} < \text{CH}_3\text{COONa(aq)}$
C.$\text{HCl(aq)} < \text{CH}_3\text{COOH(aq)} < \text{NH}_3\text{(aq)} < \text{CH}_3\text{COONa(aq)}$
D.$\text{CH}_3\text{COOH(aq)} < \text{HCl(aq)} < \text{CH}_3\text{COONa(aq)} < \text{NH}_3\text{(aq)}$

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PastPaper.workedSolution

Let's determine the relative $\text{pH}$ values of each solution at $0.10\text{ mol dm}^{-3}$:

1. $\text{HCl(aq)}$: Strong acid. It fully dissociates to release a high concentration of $\text{H}^+$, so $\text{pH} = -\log_{10}(0.10) = 1.0$. (Lowest pH)
2. $\text{CH}_3\text{COOH(aq)}$: Weak acid. It only partially dissociates, so $[\text{H}^+] < 0.10\text{ mol dm}^{-3}$, resulting in a moderately low pH (around 2.9).
3. $\text{CH}_3\text{COONa(aq)}$: Salt of a weak acid and a strong base. The acetate ion acts as a weak conjugate base and undergoes hydrolysis with water:
$$\text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq})$$
This produces a weakly alkaline solution (pH around 8.9).
4. $\text{NH}_3\text{(aq)}$: Weak base. It reacts with water to produce hydroxide ions:
$$\text{NH}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{NH}_4^+(\text{aq}) + \text{OH}^-(\text{aq})$$
At $0.10\text{ mol dm}^{-3}$, this produces a higher concentration of $\text{OH}^-\right) than acetate salt hydrolysis, resulting in a higher pH (around 11.1). (Highest pH)

Therefore, the correct increasing order of $\text{pH}\) is:
$$\text{HCl(aq)} < \text{CH}_3\text{COOH(aq)} < \text{CH}_3\text{COONa(aq)} < \text{NH}_3\text{(aq)}$$

PastPaper.markingScheme

[1 mark] A is the correct answer.
- Reject B, C, and D because they put weak acid before strong acid, or mistake the weak base as less basic than the salt.

PastPaper.question 9 · Multiple Choice

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Consider the reaction: $2\text{A(aq)} + \text{B(aq)} \rightarrow \text{C(aq)}$. The following initial rates were obtained at a constant temperature:
- Run 1: $[\text{A}] = 0.10 \text{ mol dm}^{-3}$, $[\text{B}] = 0.10 \text{ mol dm}^{-3}$, Initial Rate = $1.2 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}$
- Run 2: $[\text{A}] = 0.20 \text{ mol dm}^{-3}$, $[\text{B}] = 0.10 \text{ mol dm}^{-3}$, Initial Rate = $4.8 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}$
- Run 3: $[\text{A}] = 0.20 \text{ mol dm}^{-3}$, $[\text{B}] = 0.20 \text{ mol dm}^{-3}$, Initial Rate = $9.6 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}$
What is the overall order of the reaction and the units of the rate constant, $k$?

A.Overall order = 2, Units of $k$ = $\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$
B.Overall order = 3, Units of $k$ = $\text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}$
C.Overall order = 3, Units of $k$ = $\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$
D.Overall order = 4, Units of $k$ = $\text{dm}^9 \text{ mol}^{-3} \text{ s}^{-1}$

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PastPaper.workedSolution

From Run 1 to Run 2, [A] doubles while [B] is constant, and the rate increases by a factor of 4 (from $1.2 \times 10^{-3}$ to $4.8 \times 10^{-3}$). Therefore, the reaction is second order with respect to A.
From Run 2 to Run 3, [B] doubles while [A] is constant, and the rate increases by a factor of 2 (from $4.8 \times 10^{-3}$ to $9.6 \times 10^{-3}$). Therefore, the reaction is first order with respect to B.
The overall order is $2 + 1 = 3$.
The rate equation is: $\text{Rate} = k[\text{A}]^2[\text{B}]$.
Units of $k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}$.

PastPaper.markingScheme

[1] B - Overall order = 3, Units of $k$ = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1}.

PastPaper.question 10 · Multiple Choice

1 PastPaper.marks

A plot of $\ln(k)$ against $\frac{1}{T}$ (where $T$ is temperature in Kelvin) for a particular reaction gives a straight line with a gradient of $-1.25 \times 10^4 \text{ K}$. What is the activation energy, $E_a$, of this reaction? (Gas constant, $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$)

A.$+104 \text{ kJ mol}^{-1}$
B.$-104 \text{ kJ mol}^{-1}$
C.$+1.50 \text{ kJ mol}^{-1}$
D.$+104 \times 10^3 \text{ kJ mol}^{-1}$

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PastPaper.workedSolution

According to the Arrhenius equation: $\ln(k) = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln(A)$.
The gradient of a plot of $\ln(k)$ against $\frac{1}{T}$ is equal to $-\frac{E_a}{R}$.
$\text{gradient} = -\frac{E_a}{R} = -1.25 \times 10^4 \text{ K}$.
$E_a = 1.25 \times 10^4 \times 8.31 = 103875 \text{ J mol}^{-1} = +104 \text{ kJ mol}^{-1}$.

PastPaper.markingScheme

[1] A - +104 \text{ kJ mol}^{-1}.

PastPaper.question 11 · Multiple Choice

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Consider the reaction for the Haber process: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$.
Using the standard entropy values below, what is the standard entropy change of the system, $\Delta S^{\ominus}_{\text{system}}$, in $\text{J K}^{-1} \text{ mol}^{-1}$?
- $S^{\ominus}[\text{N}_2\text{(g)}] = 191.6 \text{ J K}^{-1} \text{ mol}^{-1}$
- $S^{\ominus}[\text{H}_2\text{(g)}] = 130.6 \text{ J K}^{-1} \text{ mol}^{-1}$
- $S^{\ominus}[\text{NH}_3\text{(g)}] = 192.3 \text{ J K}^{-1} \text{ mol}^{-1}$

A.$-129.9 \text{ J K}^{-1} \text{ mol}^{-1}$
B.$-198.8 \text{ J K}^{-1} \text{ mol}^{-1}$
C.$+198.8 \text{ J K}^{-1} \text{ mol}^{-1}$
D.$-383.4 \text{ J K}^{-1} \text{ mol}^{-1}$

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PastPaper.workedSolution

The entropy change of the system is calculated as: $\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}_{\text{products}} - \sum S^{\ominus}_{\text{reactants}}$.
$\Delta S^{\ominus}_{\text{system}} = (2 \times S^{\ominus}[\text{NH}_3\text{(g)}]) - (S^{\ominus}[\text{N}_2\text{(g)}] + 3 \times S^{\ominus}[\text{H}_2\text{(g)}])$.
$\Delta S^{\ominus}_{\text{system}} = (2 \times 192.3) - (191.6 + 3 \times 130.6)$.
$\Delta S^{\ominus}_{\text{system}} = 384.6 - (191.6 + 391.8) = 384.6 - 583.4 = -198.8 \text{ J K}^{-1} \text{ mol}^{-1}$.

PastPaper.markingScheme

[1] B - -198.8 \text{ J K}^{-1} \text{ mol}^{-1}.

PastPaper.question 12 · Multiple Choice

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For a chemical reaction, $\Delta H^{\ominus} = -45.0 \text{ kJ mol}^{-1}$ and $\Delta S^{\ominus}_{\text{system}} = -125 \text{ J K}^{-1} \text{ mol}^{-1}$. At what temperature does this reaction cease to be thermodynamically feasible?

A.Above $360 \text{ K}$
B.Below $360 \text{ K}$
C.Above $0.36 \text{ K}$
D.Below $0.36 \text{ K}$

PastPaper.showAnswers

PastPaper.workedSolution

A reaction is thermodynamically feasible when $\Delta G^{\ominus} \le 0$.
$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}_{\text{system}}$.
Setting $\Delta G^{\ominus} = 0$:
$T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}_{\text{system}}} = \frac{-45000 \text{ J mol}^{-1}}{-125 \text{ J K}^{-1} \text{ mol}^{-1}} = 360 \text{ K}$.
Since both $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ are negative, the reaction is feasible at lower temperatures (where the exothermic term dominates) and becomes non-feasible at higher temperatures where the entropy term $-T\Delta S^{\ominus}$ becomes positive and larger in magnitude than $\Delta H^{\ominus}$.
Therefore, the reaction ceases to be feasible above $360 \text{ K}$.

PastPaper.markingScheme

[1] A - Above 360 K.

PastPaper.question 13 · Multiple Choice

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In an equilibrium mixture of three gases, $\text{X}$, $\text{Y}$, and $\text{Z}$, the reaction is represented by: $\text{X(g)} + 2\text{Y(g)} \rightleftharpoons 2\text{Z(g)}$.
At a certain temperature, the equilibrium partial pressures are:
- $p(\text{X}) = 0.20 \text{ atm}$
- $p(\text{Y}) = 0.40 \text{ atm}$
- $p(\text{Z}) = 0.80 \text{ atm}$
What is the numerical value of the equilibrium constant, $K_p$, at this temperature, and its units?

A.$20 \text{ atm}^{-1}$
B.$10 \text{ atm}^{-1}$
C.$20 \text{ atm}$
D.$0.05 \text{ atm}$

PastPaper.showAnswers

PastPaper.workedSolution

The expression for the equilibrium constant is: $K_p = \frac{p(\text{Z})^2}{p(\text{X}) \cdot p(\text{Y})^2}$.
Substituting the equilibrium partial pressures:
$K_p = \frac{0.80^2}{0.20 \times 0.40^2} = \frac{0.64}{0.20 \times 0.16} = \frac{0.64}{0.032} = 20$.
The units of $K_p$ are: $\frac{\text{atm}^2}{\text{atm} \cdot \text{atm}^2} = \text{atm}^{-1}$.

PastPaper.markingScheme

[1] A - 20 atm^{-1}.

PastPaper.question 14 · Multiple Choice

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The gas-phase reaction below is endothermic: $2\text{SO}_3\text{(g)} \rightleftharpoons 2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \quad \Delta H > 0$.
If the temperature of the system is increased at constant volume, what happens to the value of the equilibrium constant, $K_c$, and the yield of $\text{O}_2\text{(g)}$?

A.$K_c$ increases, yield of $\text{O}_2$ increases
B.$K_c$ decreases, yield of $\text{O}_2$ decreases
C.$K_c$ stays constant, yield of $\text{O}_2$ increases
D.$K_c$ increases, yield of $\text{O}_2$ decreases

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PastPaper.workedSolution

Since the forward reaction is endothermic ($\Delta H > 0$), according to Le Chatelier's principle, an increase in temperature shifts the equilibrium position to the right (the endothermic direction) to absorb the added thermal energy. This increases the concentration of products ($\text{SO}_2$ and $\text{O}_2$) and decreases the concentration of reactants ($\text{SO}_3$). Therefore, the equilibrium constant, $K_c$, which is the ratio of product concentrations to reactant concentrations, must increase, and the yield of $\text{O}_2$ also increases.

PastPaper.markingScheme

[1] A - K_c increases, yield of O_2 increases.

PastPaper.question 15 · Multiple Choice

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A solution of a weak monoprotic acid, $\text{HA}$, has a concentration of $0.050 \text{ mol dm}^{-3}$. The acid dissociation constant, $K_a$, of $\text{HA}$ is $1.8 \times 10^{-5} \text{ mol dm}^{-3}$ at $298 \text{ K}$. What is the pH of this solution at $298 \text{ K}$?

A.$3.02$
B.$2.04$
C.$4.74$
D.$1.30$

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PastPaper.workedSolution

For a weak monoprotic acid, we can use the approximation:
$[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$.
$[\text{H}^+] = \sqrt{1.8 \times 10^{-5} \times 0.050} = \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4} \text{ mol dm}^{-3}$.
$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(9.49 \times 10^{-4}) = 3.02$.

PastPaper.markingScheme

[1] A - 3.02.

PastPaper.question 16 · Multiple Choice

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A buffer solution is prepared by mixing $50.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ propanoic acid ($K_a = 1.30 \times 10^{-5} \text{ mol dm}^{-3}$) with $50.0 \text{ cm}^3$ of $0.050 \text{ mol dm}^{-3}$ sodium propanoate solution. What is the pH of the resulting buffer solution?

A.$4.59$
B.$4.89$
C.$5.19$
D.$3.59$

PastPaper.showAnswers

PastPaper.workedSolution

Using the Henderson-Hasselbalch equation:
$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$.
First, calculate the moles of propanoic acid (HA) and propanoate ions ($\text{A}^-$):
$n(\text{HA}) = 0.0500 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.0050 \text{ mol}$.
$n(\text{A}^-) = 0.0500 \text{ dm}^3 \times 0.050 \text{ mol dm}^{-3} = 0.0025 \text{ mol}$.
Now, calculate $\text{p}K_a$:
$\text{p}K_a = -\log_{10}(1.30 \times 10^{-5}) = 4.89$.
Substitute the values into the equation:
$\text{pH} = 4.89 + \log_{10}\left(\frac{0.0025}{0.0050}\right) = 4.89 + \log_{10}(0.5) = 4.89 - 0.30 = 4.59$.

PastPaper.markingScheme

[1] A - 4.59.

PastPaper.question 17 · Multiple Choice

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The reaction between nitrogen monoxide and hydrogen is represented by the equation: \[2\text{NO}(g) + 2\text{H}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g)\] The rate equation is: \[\text{rate} = k[\text{NO}]^2[\text{H}_2]\] In an experiment, the concentration of $\text{NO}$ is doubled and the concentration of $\text{H}_2$ is halved. By what factor does the initial rate of reaction change?

A.It is unchanged
B.It increases by a factor of 2
C.It increases by a factor of 4
D.It increases by a factor of 8

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PastPaper.workedSolution

Let the initial rate be $\text{rate}_1 = k[\text{NO}]^2[\text{H}_2]$. When $[\text{NO}]$ is doubled to $2[\text{NO}]$ and $[\text{H}_2]$ is halved to $0.5[\text{H}_2]$, the new rate is: \[\text{rate}_2 = k(2[\text{NO}])^2(0.5[\text{H}_2]) = k \times 4[\text{NO}]^2 \times 0.5[\text{H}_2] = 2k[\text{NO}]^2[\text{H}_2] = 2\text{rate}_1\] Therefore, the rate increases by a factor of 2.

PastPaper.markingScheme

• B is the correct answer (1 mark). • A is incorrect because it assumes no overall change. • C is incorrect because it forgets to halve the rate due to hydrogen. • D is incorrect because it assumes both are doubled or incorrect application of the squared term.

PastPaper.question 18 · Multiple Choice

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For a reaction at $298\text{ K}$, the standard enthalpy change is $\Delta H^{\ominus} = -115\text{ kJ mol}^{-1}$ and the standard entropy change of the system is $\Delta S_{\text{system}}^{\ominus} = -185\text{ J K}^{-1}\text{ mol}^{-1}$. What is the total entropy change, $\Delta S_{\text{total}}^{\ominus}$, for this reaction at $298\text{ K}$?

A.-571 J K^{-1} mol^{-1}
B.-201 J K^{-1} mol^{-1}
C.+201 J K^{-1} mol^{-1}
D.+571 J K^{-1} mol^{-1}

PastPaper.showAnswers

PastPaper.workedSolution

First, calculate the entropy change of the surroundings: \[\Delta S_{\text{surroundings}}^{\ominus} = -\frac{\Delta H^{\ominus}}{T} = -\frac{-115 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = +385.9\text{ J K}^{-1}\text{ mol}^{-1}\] Next, calculate the total entropy change: \[\Delta S_{\text{total}}^{\ominus} = \Delta S_{\text{system}}^{\ominus} + \Delta S_{\text{surroundings}}^{\ominus} = -185 + 385.9 = +200.9\text{ J K}^{-1}\text{ mol}^{-1} \approx +201\text{ J K}^{-1}\text{ mol}^{-1}\]

PastPaper.markingScheme

• C is the correct answer (1 mark). • A is incorrect because it uses an incorrect sign for the surroundings entropy change. • B is incorrect because of an incorrect subtraction. • D is incorrect because of combining incorrect signs for both terms.

PastPaper.question 19 · Multiple Choice

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A buffer solution is prepared containing $0.10\text{ mol dm}^{-3}$ of a weak acid, $\text{HA}$, and $0.20\text{ mol dm}^{-3}$ of its conjugate base, $\text{A}^-$. The acid dissociation constant, $K_a$, of $\text{HA}$ is $2.0 \times 10^{-5}\text{ mol dm}^{-3}$. What is the pH of this buffer solution?

A.4.40
B.4.70
C.5.00
D.5.40

PastPaper.showAnswers

PastPaper.workedSolution

The pH of a buffer solution can be calculated using: \[\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\] First, calculate $\text{p}K_a$: \[\text{p}K_a = -\log_{10}(2.0 \times 10^{-5}) = 4.70\] Then, substitute the concentrations into the equation: \[\text{pH} = 4.70 + \log_{10}\left(\frac{0.20}{0.10}\right) = 4.70 + 0.30 = 5.00\]

PastPaper.markingScheme

• C is the correct answer (1 mark). • A is incorrect because the ratio was inverted in the calculation. • B is incorrect because it is the value of $\text{p}K_a$ without applying the ratio correction. • D is incorrect because of an incorrect buffer formula.

PastPaper.question 20 · Multiple Choice

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Which carbonyl compound reacts with hydrogen cyanide, $\text{HCN}$, to form a product that exists as a racemic mixture of optical isomers?

A.Propanal
B.Propanone
C.Methanal
D.Pentan-3-one

PastPaper.showAnswers

PastPaper.workedSolution

Nucleophilic addition of $\text{CN}^-$ to a carbonyl group creates a chiral center if the resulting carbon is bonded to four different groups. Propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) reacts to form 2-hydroxybutanenitrile, $\text{CH}_3\text{CH}_2\text{CH(OH)CN}$. The central carbon has four different groups attached ($-\text{H}$, $-\text{OH}$, $-\text{CN}$, $-\text{CH}_2\text{CH}_3$), making it chiral. Because the carbonyl group is planar, attack from above and below is equally likely, yielding a 50:50 racemic mixture. The other options yield products with at least two identical groups on the reaction center carbon and are therefore achiral.

PastPaper.markingScheme

• A is the correct answer (1 mark). • B, C and D are incorrect because their reaction products do not contain a chiral carbon atom, meaning they cannot exhibit optical isomerism.

Unit 4: Section B

Answer all structured physical chemistry questions.

(a) [Total: 4 marks]
- 1 Mark: Correctly deduces first-order for A.
- 1 Mark: Explains reasoning by comparing Experiment 1 and 2.
- 1 Mark: Correctly deduces second-order for B.
- 1 Mark: Explains reasoning by comparing Experiment 1 and 3.

(b) [Total: 1 mark]
- 1 Mark: Correct rate equation matching part (a) ($\text{Rate} = k[\text{A}][\text{B}]^2$).

(c) [Total: 3.4 marks]
- 1 Mark: Correct rearrangement of rate equation to solve for $k$.
- 1.4 Marks: Correct calculated value of $0.24$.
- 1 Mark: Correct units of $\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}$.

(d) [Total: 2 marks]
- 1 Mark: Correct substitution of new concentrations.
- 1 Mark: Correct final value of $1.44 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}$ (accept standard form or decimal).

Unit 4: Section C

Answer all questions on rates of organic reaction systems.

1 PastPaper.question · 18 PastPaper.marks

PastPaper.question 1 · Organic Spectroscopy & Reaction Kinetics

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Answer all parts of the question. (a) The rate of hydrolysis of a chiral halogenoalkane A, $\text{C}_4\text{H}_9\text{Br}$, by aqueous sodium hydroxide, $\text{NaOH(aq)}$, was investigated at a constant temperature. The following experimental data were obtained: - Experiment 1: $[\text{C}_4\text{H}_9\text{Br}] = 0.050\text{ mol dm}^{-3}$; $[\text{OH}^-] = 0.10\text{ mol dm}^{-3}$; Initial rate = $3.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ - Experiment 2: $[\text{C}_4\text{H}_9\text{Br}] = 0.100\text{ mol dm}^{-3}$; $[\text{OH}^-] = 0.10\text{ mol dm}^{-3}$; Initial rate = $7.00 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ - Experiment 3: $[\text{C}_4\text{H}_9\text{Br}] = 0.100\text{ mol dm}^{-3}$; $[\text{OH}^-] = 0.20\text{ mol dm}^{-3}$; Initial rate = $7.00 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}$ (i) Deduce the order of reaction with respect to $\text{C}_4\text{H}_9\text{Br}$ and with respect to $\text{OH}^-$, explaining your reasoning. (ii) Write the rate equation for this reaction. (iii) Calculate the value of the rate constant, $k$, stating its units. (b) A single enantiomer of the chiral halogenoalkane A (2-bromobutane) is hydrolysed under these conditions to form alcohol B ($\text{C}_4\text{H}_{10}\text{O}$). State the mechanism ($\text{S}_\text{N}1$ or $\text{S}_\text{N}2$) of this reaction, and explain the effect of this mechanism on the optical activity of the product alcohol B. (c) Alcohol B is oxidized by heating under reflux with acidified potassium dichromate(VI) to form compound C. Describe how Infrared (IR) spectroscopy can be used to monitor the progress of this reaction and confirm when the oxidation is complete. (d) The $^1\text{H}$ NMR spectrum of the purified organic product C contains three peaks: - A triplet at $\delta = 1.0\text{ ppm}$ (integration 3H) - A singlet at $\delta = 2.1\text{ ppm}$ (integration 3H) - A quartet at $\delta = 2.4\text{ ppm}$ (integration 2H) Explain how this $^1\text{H}$ NMR spectrum confirms that product C is butanone.

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(a)(i) Order wrt $\text{C}_4\text{H}_9\text{Br}$ is 1 because comparing Experiments 1 and 2, when $[\text{C}_4\text{H}_9\text{Br}]$ doubles and $[\text{OH}^-]$ is held constant, the rate doubles. Order wrt $[\text{OH}^-]$ is 0 because comparing Experiments 2 and 3, when $[\text{OH}^-]$ doubles and $[\text{C}_4\text{H}_9\text{Br}]$ is held constant, the rate remains unchanged. (ii) $\text{Rate} = k[\text{C}_4\text{H}_9\text{Br}]$. (iii) Using Exp 2: $k = \frac{\text{Rate}}{[\text{C}_4\text{H}_9\text{Br}]} = \frac{7.00 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}}{0.100\text{ mol dm}^{-3}} = 7.00 \times 10^{-4}\text{ s}^{-1}$. (b) The reaction proceeds via an $\text{S}_\text{N}1$ mechanism. In the slow step, the $\text{C-Br}$ bond breaks heterolytically to form a planar carbocation intermediate. The nucleophile ($\text{OH}^-$) can attack this planar intermediate with equal probability from either side, resulting in an equimolar (racemic) mixture of both enantiomers of alcohol B. Therefore, the product B is optically inactive due to the cancellation of the optical rotations. (c) The reaction is monitored by the disappearance of the broad $\text{O-H}$ stretching absorption of the alcohol (B) at $3200-3600\text{ cm}^{-1}$ and the appearance of the sharp $\text{C=O}$ carbonyl stretching absorption of the ketone (C) at $1675-1750\text{ cm}^{-1}$. The reaction is complete when the $\text{O-H}$ absorption is entirely absent from the spectrum. (d) The structure of butanone is $\text{CH}_3\text{COCH}_2\text{CH}_3$. The triplet at $\delta = 1.0\text{ ppm}$ corresponds to the $\text{-CH}_3$ protons split by the 2 adjacent protons of the $\text{-CH}_2-$. The quartet at $\delta = 2.4\text{ ppm}$ corresponds to the $\text{-CH}_2-$ protons split by the 3 adjacent protons of the $\text{-CH}_3$. The singlet at $\delta = 2.1\text{ ppm}$ corresponds to the isolated $\text{-CH}_3$ protons directly attached to the carbonyl group (no neighboring protons). The 3:3:2 integration ratio matches the proton environments.

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(a)(i) 1 mark for correct order wrt C4H9Br with reasoning. 1 mark for correct order wrt OH^- with reasoning. 1 mark for overall clear logical deduction. (ii) 1 mark for Rate = k[C4H9Br]. (iii) 1 mark for value 7.00 x 10^-4. 1 mark for unit s^-1. (b) 1 mark for stating SN1. 1 mark for explaining that the slow step forms a carbocation. 1 mark for describing the carbocation intermediate as planar. 1 mark for stating the nucleophile attacks with equal probability from either side. 1 mark for stating that this forms a racemic mixture which is optically inactive. (c) 1 mark for loss of broad O-H absorption band in range 3200-3600 cm^-1. 1 mark for gain of sharp C=O absorption band in range 1675-1750 cm^-1. 1 mark for stating the reaction is complete when the O-H band is completely absent. (d) 1 mark for matching the triplet at 1.0 ppm to CH3 adjacent to CH2. 1 mark for matching the quartet at 2.4 ppm to CH2 adjacent to CH3. 1 mark for matching the singlet at 2.1 ppm to CH3 adjacent to C=O. 1 mark for linking the integration values to the total protons/environments in butanone.

Unit 5: Section A

Answer all transition metal and nitrogen-containing organic compound questions.

12 PastPaper.question · 12 PastPaper.marks

PastPaper.question 1 · multiple_choice

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Which of the following electronic configurations represents the ground state of a $\text{Cr}^{3+}$ ion?

A.$[\text{Ar}] 4\text{s}^2 3\text{d}^1$
B.$[\text{Ar}] 3\text{d}^3$
C.$[\text{Ar}] 4\text{s}^1 3\text{d}^2$
D.$[\text{Ar}] 3\text{d}^4$

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The electronic configuration of a neutral chromium atom, $\text{Cr}$ (atomic number 24), is $[\text{Ar}] 4\text{s}^1 3\text{d}^5$. When forming the $\text{Cr}^{3+}$ ion, three electrons are removed. The single electron in the $4\text{s}$ orbital is lost first, followed by two electrons from the $3\text{d}$ subshell. This leaves three electrons in the $3\text{d}$ subshell, resulting in the ground state configuration of $[\text{Ar}] 3\text{d}^3$.

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B is correct (1 mark).

PastPaper.question 2 · multiple_choice

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Which of the following is the correct order of increasing basicity (weakest base to strongest base) in aqueous solution for the compounds: phenylamine (I), ammonia (II), ethylamine (III), and diethylamine (IV)?

A.I < II < III < IV
B.I < II < IV < III
C.II < I < III < IV
D.IV < III < II < I

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Phenylamine (I) is the weakest base because the lone pair of electrons on the nitrogen atom is partially delocalised into the benzene ring, making it less available to accept a proton. Ammonia (II) is stronger than phenylamine as there is no delocalisation. Ethylamine (III) is stronger than ammonia due to the positive inductive effect of the ethyl group. Diethylamine (IV) is the strongest because it has two electron-releasing ethyl groups, further increasing the electron density on the nitrogen atom. Thus, the correct order is I < II < III < IV.

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A is correct (1 mark).

PastPaper.question 3 · multiple_choice

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When $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ reacts with $\text{1,2-diaminoethane}$ (en), a ligand substitution reaction occurs: $[\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 3\text{en}(\text{aq}) \rightarrow [\text{Co}(\text{en})_3]^{2+}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})$. What are the signs of the enthalpy change, $\Delta H$, and the entropy change of the system, $\Delta S_{\text{system}}$, for this reaction?

A.$\Delta H \gg 0$ (highly endothermic) and $\Delta S_{\text{system}} > 0$
B.$\Delta H \ll 0$ (highly exothermic) and $\Delta S_{\text{system}} < 0$
C.$\Delta H \approx 0$ and $\Delta S_{\text{system}} > 0$
D.$\Delta H \approx 0$ and $\Delta S_{\text{system}} < 0$

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Because six $\text{Co-N}$ coordinate bonds are formed while six $\text{Co-O}$ coordinate bonds are broken, and these bonds have very similar strengths, the enthalpy change $\Delta H$ is close to zero. The reaction converts 4 reactant particles into 7 product particles in solution, which significantly increases the disorder and results in a positive entropy change of the system, $\Delta S_{\text{system}} > 0$.

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C is correct (1 mark).

PastPaper.question 4 · multiple_choice

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Which of the following represents the major organic species present in a solution of alanine, $\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}$, at $\text{pH } 1$?

A.$\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-$
B.$\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-$
C.$\text{CH}_3\text{CH}(\text{NH}_2)\text{COOH}$
D.$\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}$

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At a highly acidic pH of 1, the high concentration of hydrogen ions causes both basic sites of the amino acid to be protonated. The carboxylate group is protonated to form a carboxylic acid group ($\text{-COOH}$), and the amine group is protonated to form an ammonium group ($\text{-NH}_3^+$). Therefore, the major species is $\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}$.

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D is correct (1 mark).

PastPaper.question 5 · multiple_choice

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A solution containing $\text{VO}_2^+$ ions is reduced using zinc and hydrochloric acid. Which of the following shows the correct sequence of colours observed as vanadium is reduced from its $+5$ to its $+2$ oxidation state?

A.Yellow $\rightarrow$ Blue $\rightarrow$ Green $\rightarrow$ Violet
B.Yellow $\rightarrow$ Violet $\rightarrow$ Blue $\rightarrow$ Green
C.Blue $\rightarrow$ Yellow $\rightarrow$ Green $\rightarrow$ Violet
D.Violet $\rightarrow$ Green $\rightarrow$ Blue $\rightarrow$ Yellow

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The reduction steps of vanadium are: $\text{VO}_2^+$ ($+5$, yellow) is reduced to $\text{VO}^{2+}$ ($+4$, blue), then to $\text{V}^{3+}$ ($+3$, green), and finally to $\text{V}^{2+}$ ($+2$, violet). The sequence is Yellow to Blue to Green to Violet.

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A is correct (1 mark).

PastPaper.question 6 · multiple_choice

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Which of the following reaction processes produces an amide as the primary organic product?

A.Propanenitrile reacting with $\text{LiAlH}_4$ in dry ether
B.Propanoic acid reacting with ammonia followed by heating
C.Bromopropane reacting with excess ammonia in ethanol under pressure
D.Nitrobenzene reacting with tin and concentrated hydrochloric acid

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Reacting a carboxylic acid (propanoic acid) with ammonia initially forms an ammonium salt (ammonium propanoate). Heating this salt dry causes it to undergo thermal dehydration, producing an amide (propanamide). The other processes produce primary amines (propylamine in A and C, phenylamine in D).

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B is correct (1 mark).

PastPaper.question 7 · multiple_choice

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Which of the following statements correctly explains the origin of colour in aqueous transition metal complex ions?

A.d-orbitals split in energy; electrons emit specific frequencies of visible light as they transition to a lower energy d-orbital, and this emitted light is seen.
B.Electrons transition from s-orbitals to d-orbitals, absorbing UV light, which causes the solution to fluoresce in the visible region.
C.d-orbitals split in energy; electrons absorb specific frequencies of visible light to transition to a higher energy d-orbital, and the complementary colour is transmitted.
D.Ligands absorb specific wavelengths of light, causing ligand-to-metal charge transfer that results in the emission of the complementary colour.

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When ligands coordinate to a transition metal ion, the d-orbitals split into different energy levels. When visible light shines on the complex, an electron absorbs a specific wavelength of visible light to transition to a higher energy d-orbital. The remaining unabsorbed wavelengths are transmitted, and we see the complementary colour.

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C is correct (1 mark).

PastPaper.question 8 · multiple_choice

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Which pair of monomers can undergo condensation polymerisation to form a polyamide?

A.$\text{Benzene-1,4-diol}$ and $\text{benzene-1,4-dicarboxylic acid}$
B.$\text{Ethane-1,2-diol}$ and $\text{hexane-1,6-dioic acid}$
C.$\text{Phenylethene}$ and $\text{buta-1,3-diene}$
D.$\text{Benzene-1,4-dicarboxylic acid}$ and $\text{benzene-1,4-diamine}$

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Polyamides are formed by the reaction of diamines with dicarboxylic acids (or diacyl chlorides) with the elimination of water (or HCl). Benzene-1,4-dicarboxylic acid and benzene-1,4-diamine react to form Kevlar, which is a polyamide. The other choices form polyesters (A, B) or addition polymers (C).

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Unit 5: Section B

Answer all transition metal complexes and arenes questions.

(a)(i) M1: Both reactant ions are negatively charged [1]. M2: Electrostatic repulsion leads to high activation energy [1]. (a)(ii) M3: Equation showing oxidation of Fe2+ to Fe3+ by S2O82- [1]. M4: Equation showing reduction of Fe3+ back to Fe2+ by I- [1]. (b)(i) M5: Defines autocatalytic: a product acts as the catalyst [1]. (b)(ii) M6: Identifies Mn2+ as the catalyst [1]. (b)(iii) M7: Graph sketch with correctly labeled axes showing correct sigmoidal shape (initially slow, then steep, then plateaus near the x-axis) [1]. M8: Explains initial slow rate is due to lack of Mn2+ catalyst [1]. M9: Explains subsequent rapid rate is due to accumulation of Mn2+ catalyst [1]. M10: Explains final slow rate is due to reactant depletion [1].

Unit 5: Section C

Answer all organic synthesis pathway and catalytic cycle questions.

1 PastPaper.question · 20 PastPaper.marks

PastPaper.question 1 · Organic Synthesis & Coordination Chemistry

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This question is about the synthesis of a bidentate organic ligand, its coordination chemistry with copper(II) ions, and the catalytic behavior of transition metal ions.

Part (a) Synthesis of 2-aminophenol

A research chemist designs a pathway to synthesize the bidentate ligand, 2-aminophenol, starting from phenol.

(i) Phenol is reacted with dilute nitric acid at room temperature to form a mixture of 2-nitrophenol and 4-nitrophenol. Write the equation for the formation of 2-nitrophenol from phenol, using structural or molecular formulae for the organic compounds, and state the type of reaction mechanism. (3 marks)

(ii) Explain why 2-nitrophenol and 4-nitrophenol can be separated by steam distillation, making reference to the types of intermolecular forces present in each isomer. (3 marks)

(iii) State the reagents and conditions required to reduce 2-nitrophenol to 2-aminophenol. (2 marks)

Part (b) Coordination Chemistry of 2-aminophenol

Under basic conditions, the phenolic $\text{-OH}$ group of 2-aminophenol is deprotonated to form the 2-aminophenoxide ion, which acts as a bidentate ligand. This ligand reacts with aqueous copper(II) ions to form a neutral, square-planar coordination complex with the molecular formula $[\text{Cu}(\text{C}_6\text{H}_6\text{NO})_2]$.

(i) Draw the structure of the deprotonated 2-aminophenoxide ligand, clearly showing the lone pairs on the donor atoms that form coordinate bonds to the metal center. (2 marks)

(ii) Draw the structure of one of the geometric isomers of the neutral square-planar complex $[\text{Cu}(\text{C}_6\text{H}_6\text{NO})_2]$, clearly showing the coordinate bonds. (2 marks)

(iii) Explain why transition metal complexes, such as $[\text{Cu}(\text{C}_6\text{H}_6\text{NO})_2]$, are colored. (2 marks)

Part (c) Transition Metals as Catalysts

Transition metals and their compounds are widely used as catalysts. A classic example is the catalysis of the reaction between peroxodisulfate ions, $\text{S}_2\text{O}_8^{2-}$, and iodide ions, $\text{I}^-$, by aqueous iron(II) ions, $\text{Fe}^{2+}$.

(i) Write two ionic equations to show how $\text{Fe}^{2+}$ acts as a catalyst in this reaction, and explain why the uncatalyzed reaction has a high activation energy. (4 marks)

(ii) State the type of catalysis shown by $\text{Fe}^{2+}$ in this reaction, and explain how a catalyst increases the rate of reaction. (2 marks)

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Part (a)

(i) Equation:
$\text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_4(\text{OH})\text{NO}_2 + \text{H}_2\text{O}$
Mechanism: Electrophilic substitution.

(ii) 2-nitrophenol exhibits intramolecular hydrogen bonding (between the $\text{-OH}$ group and the adjacent $\text{-NO}_2$ group on the same molecule). In contrast, 4-nitrophenol exhibits intermolecular hydrogen bonding (between the $\text{-OH}$ of one molecule and the $\text{-NO}_2$ of another). Consequently, 2-nitrophenol is more volatile (has a lower boiling point) and is easily vaporized during steam distillation, whereas 4-nitrophenol has stronger intermolecular forces and remains behind.

(iii) Reagents: Tin ($\text{Sn}$) and concentrated hydrochloric acid ($\text{HCl}$) heated under reflux. Followed by addition of an alkali, such as sodium hydroxide ($\text{NaOH}$), to liberate the free amine (2-aminophenol).

Part (b)

(i) The deprotonated 2-aminophenoxide ligand has a benzene ring with an $\text{-O}^-$[oxygen atom with a negative charge and lone pairs] and an $\text{-NH}_2$[nitrogen atom with a lone pair] on the adjacent carbon atom. The donor atoms are the phenoxide oxygen ($\text{O}^-$) and the amine nitrogen ($\text{N}$), both showing a lone pair.

(ii) The complex $[\text{Cu}(\text{C}_6\text{H}_6\text{NO})_2]$ is square planar. Copper(II) is the central metal ion. Two bidentate ligands coordinate via their $\text{O}$ and $\text{N}$ atoms. In the cis-isomer, the two $\text{O}$ atoms are adjacent to each other and the two $\text{N}$ atoms are adjacent. In the trans-isomer, the two $\text{O}$ atoms are opposite each other, as are the two $\text{N}$ atoms. Coordinate bonds are shown with arrows pointing from $\text{O}$ and $\text{N}$ to $\text{Cu}$.

(iii) The presence of ligands causes the d-orbitals of the copper(II) ion to split into two energy levels with a small energy gap ($\Delta E$). An electron is excited from a lower d-orbital to a higher d-orbital (a d-d transition) by absorbing a photon of visible light (where $\Delta E = h\nu$). The complementary color of the light that is not absorbed is seen.

Part (c)

(i) Step 1: $\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}$
Step 2: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$
The uncatalyzed reaction is between two negatively charged ions ($\text{S}_2\text{O}_8^{2-}$ and $\text{I}^-$). Since like charges repel each other, the collision has a very high activation energy. The catalyzed route involves collisions of opposite charges (negative and positive), which have a much lower activation energy.

(ii) Homogeneous catalysis. A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy, so a larger fraction of colliding particles have energy greater than or equal to the activation energy.

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Part (a) [Total: 8 marks]

(i) [3 marks]
• M1: Correct reactants and products ($\text{C}_6\text{H}_5\text{OH}$ and $\text{HNO}_3$ to $\text{C}_6\text{H}_4(\text{OH})\text{NO}_2$ and $\text{H}_2\text{O}$) [1 mark]
• M2: Correctly balanced equation [1 mark]
• M3: Electrophilic substitution [1 mark]

(ii) [3 marks]
• M1: Identifies intramolecular hydrogen bonding in 2-nitrophenol [1 mark]
• M2: Identifies intermolecular hydrogen bonding in 4-nitrophenol [1 mark]
• M3: Connects this to volatility / boiling point differences (2-nitrophenol is more volatile / has weaker intermolecular forces, so distillable) [1 mark]

(iii) [2 marks]
• M1: Tin ($\text{Sn}$) and concentrated hydrochloric acid ($\text{HCl}$) (and heat/reflux) [1 mark]
• M2: Sodium hydroxide ($\text{NaOH}$) / alkali to liberate the amine [1 mark] (Reject: LiAlH4 or NaBH4 as these do not typically reduce nitroarenes in the Pearson Edexcel specification; Sn/HCl is the standard expected reagent)

Part (b) [Total: 6 marks]

(i) [2 marks]
• M1: Correct structural drawing of 2-aminophenoxide (benzene ring with adjacent $\text{-O}^-$/$\text{-O}$ and $\text{-NH}_2$) [1 mark]
• M2: Showing lone pair on the oxygen and nitrogen donor atoms [1 mark]

(ii) [2 marks]
• M1: Square planar representation around $\text{Cu}$ with four coordination sites [1 mark]
• M2: Correct coordinate arrows from oxygen and nitrogen of two ligands to $\text{Cu}$, forming a neutral complex [1 mark]

(iii) [2 marks]
• M1: Ligands cause d-orbitals to split into different energy levels [1 mark]
• M2: d-d transition / promotion of electrons from lower to higher d-orbitals absorbs light in the visible range, transmitting/reflecting the complementary color [1 mark]

Part (c) [Total: 6 marks]

(i) [4 marks]
• M1: $\text{S}_2\text{O}_8^{2-} + 2\text{Fe}^{2+} \rightarrow 2\text{SO}_4^{2-} + 2\text{Fe}^{3+}$ [1 mark]
• M2: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ [1 mark]
• M3: Uncatalyzed reaction involves repulsion between negatively charged ions ($\text{S}_2\text{O}_8^{2-}$ and $\text{I}^-$) [1 mark]
• M4: Catalyzed steps involve oppositely charged ions colliding, lowering the activation energy [1 mark]

(ii) [2 marks]
• M1: Homogeneous catalysis [1 mark]
• M2: Provides an alternative reaction pathway with a lower activation energy [1 mark]

PastPaper.section Unit 6

Answer all questions testing experimental design, purification, and organic preparation.

(a)
- M1: Equation: $ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- $ (or equivalent). (1)
- M2: Formula of electrophile: $ \text{NO}_2^+ $. (1)

(b)
- M1: To prevent further nitration / dinitration / multiple nitrations. (1)
- M2: Identify that methyl 3,5-dinitrobenzoate would be formed as an impurity. (0.5)

(c)
- M1: To precipitate the product AND safely dilute the concentrated acids. (1)

(d) (i)
- M1: Highly soluble in hot ethanol but insoluble in cold ethanol (whereas it is insoluble in water at all temperatures). (1)
(ii)
- M1: Wash with a small portion of ice-cold ethanol. (1)
- M2: Dry in a desiccator / warm oven (below melting point) / between filter papers. (1)

(e) (i)
- M1: Calculation of mass of methyl benzoate: $ 4.0 \times 1.09 = 4.36 \text{ g} $. (1)
- M2: Moles of methyl benzoate = $ \frac{4.36}{136.0} = 0.0321 \text{ mol} $ (accept $ 0.032 \text{ mol} $). (1)

(e) (ii)
- M1: Calculation of $ M_r $ of methyl 3-nitrobenzoate as $ 181.0 \text{ g mol}^{-1} $. (1)
- M2: Theoretical mass of product = $ 0.03206 \times 181.0 = 5.80 \text{ g} $. (1)
- M3: Percentage yield = $ \frac{3.65}{5.80} \times 100 = 62.9\% $ (accept range $ 62.8\% - 63.0\% $, must be 3 sig figs). (1)

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