2023 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

The equation for the thermal decomposition of sodium hydrogencarbonate is: 2\(\text{NaHCO}_3\)(s) \(\rightarrow\) \(\text{Na}_2\text{CO}_3\)(s) + \(\text{CO}_2\)(g) + \(\text{H}_2\text{O\)(l). On cooling to r.t.p., the water vapour condenses into liquid water, so only carbon dioxide remains in the gas phase. 0.10 mol of \(\text{NaHCO}_3\) produces 0.050 mol of \(\text{CO}_2\) gas. The volume of 0.050 mol of gas at r.t.p. is 0.050 mol \(\times\) 24.0 dm\(^3\) mol\(^{-1}\) = 1.20 dm\(^3\).

Award 1 mark for the correct option A. Method: Identify the balanced equation and molar ratio of NaHCO3 to CO2 (2:1). Account for the condensation of water at r.t.p., leaving only CO2 gas. Calculate volume using molar volume of 24.0 dm3/mol.

The balanced chemical equation for the neutralisation is: \(\text{H}_2\text{X}\) + 2\(\text{NaOH}\) \(\rightarrow\) \(\text{Na}_2\text{X}\) + 2\(\text{H}_2\text{O}\). Number of moles of \(\text{NaOH}\) reacted = 0.0375 dm\(^3\) \(\times\) 0.100 mol dm\(^{-3}\) = 0.00375 mol. Since 1 mole of \(\text{H}_2\text{X}\) reacts with 2 moles of \(\text{NaOH}\), the number of moles of \(\text{H}_2\text{X}\) is 0.00375 mol / 2 = 0.001875 mol. The concentration of \(\text{H}_2\text{X}\) = 0.001875 mol / 0.0250 dm\(^3\) = 0.0750 mol dm\(^{-3}\).

Award 1 mark for the correct option B. Method: Calculate moles of NaOH. Use the stoichiometry of a diprotic acid (1:2 ratio) to find moles of H2X. Divide by the volume of H2X solution to find concentration.

Down Group 2, the size (ionic radius) of the metal cation increases while keeping the same +2 charge. This decreases the charge density of the cation. A cation with lower charge density has a weaker polarising effect on the electron cloud of the nitrate anion, meaning it is less able to weaken the N-O bonds in the nitrate ion. Therefore, more thermal energy is needed to decompose the nitrates down the group, so thermal stability increases.

Award 1 mark for the correct option A. Explanation: Down Group 2, cationic radius increases, charge density decreases, polarising power decreases, and thermal stability increases.

The solubility of Group 2 hydroxides increases down the group, so magnesium hydroxide, \(\text{Mg(OH)}_2\), is insoluble/sparingly soluble and precipitates as a white solid when aqueous sodium hydroxide is added. Barium hydroxide, \(\text{Ba(OH)}_2\), is highly soluble and remains in solution. Conversely, the solubility of Group 2 sulfates decreases down the group, so sodium sulfate would precipitate barium ions as \(\text{BaSO}_4\) while magnesium ions would remain dissolved.

Award 1 mark for the correct option A. Key knowledge: solubility trends of Group 2 hydroxides (increases down the group) and sulfates (decreases down the group).

Magnesium chloride, \(\text{MgCl}_2\), undergoes slight hydrolysis in water due to the relatively high charge density of the magnesium ion, resulting in a weakly acidic solution of pH \(\approx\) 6.5. Phosphorus chlorides (such as \(\text{PCl}_3\) or \(\text{PCl}_5\)) react vigorously and completely with water to produce strong acids (like \(\text{HCl}\)), giving a highly acidic solution of pH \(\approx\) 1-2. Sodium chloride forms a neutral solution (pH \(\approx\) 7), and aluminium chloride undergoes significant hydrolysis to form a solution with pH \(\approx\) 3.

Award 1 mark for the correct option A. Assessment of Period 3 chloride hydrolysis: MgCl2 is weakly acidic (pH 6.5), AlCl3 is moderately acidic (pH 3), PCl3/PCl5 are strongly acidic (pH 2 or less), NaCl is neutral (pH 7).

Silicon (P) has a giant molecular/covalent structure, which gives it a very high melting point, and all of its valence electrons are held localized in covalent bonds, so it does not conduct electricity in the solid state. Aluminium (Q) is a metal with a high melting point and a sea of delocalised electrons, making it an excellent electrical conductor. Chlorine (R) is a simple molecular element that exists as diatomic \(\text{Cl}_2\) molecules held by weak London dispersion forces, giving it a low melting point.

Award 1 mark for the correct option A. Candidates must relate bonding and structure to the physical properties of Period 3 elements.

Oxidative cleavage of alkenes with hot, concentrated, acidified potassium manganate(VII) breaks the carbon-carbon double bond. If a double-bonded carbon is attached to two alkyl groups, it is oxidised to a ketone. 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\), is symmetrical and has two double-bonded carbons, each bonded to two methyl groups. Cleaving this bond produces two identical molecules of propanone, \(\text{CH}_3\text{COCH}_3\), which is a ketone. Hence, propanone is the sole organic product.

Award 1 mark for the correct option A. Method: Identify the oxidative cleavage products of each alkene option. Only 2,3-dimethylbut-2-ene yields propanone as the sole organic product.

Propane, \(\text{CH}_3\text{CH}_2\text{CH}_3\), is a non-polar hydrocarbon whose molecules are held together only by weak instantaneous dipole-induced dipole (London) forces, giving it the lowest boiling point. Methoxymethane, \(\text{CH}_3\text{OCH}_3\), is polar and experiences permanent dipole-dipole forces in addition to London forces, resulting in a higher boiling point than propane. Ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\), contains a highly polar O-H bond and forms strong hydrogen bonds between its molecules, giving it the highest boiling point of the three.

Award 1 mark for the correct option A. Method: Compare the strength of the intermolecular forces (London dispersion vs permanent dipole-dipole vs hydrogen bonding) in compounds of comparable molecular mass.

1. Calculate the number of moles of \(\text{CO}_2\) gas collected:
\(n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol}\).

2. From the stoichiometric equation, the mole ratio of \(M\text{CO}_3\) to \(\text{CO}_2\) is \(1:1\):
\(n(M\text{CO}_3) = 0.010\text{ mol}\).

3. Calculate the molar mass (\(M_r\)) of \(M\text{CO}_3\):
\(M_r(M\text{CO}_3) = \frac{1.97\text{ g}}{0.010\text{ mol}} = 197\text{ g mol}^{-1}\).

4. Calculate the relative atomic mass (\(A_r\)) of the metal \(M\):
\(A_r(M) = M_r(M\text{CO}_3) - [A_r(\text{C}) + 3 \times A_r(\text{O})] = 197 - [12.0 + (3 \times 16.0)] = 197 - 60.0 = 137\).

5. Identify the Group 2 element with \(A_r \approx 137\), which is Barium (\(\text{Ba}\)).

[1 mark] C is correct.
- Award 1 mark for the correct calculation of moles, relative molecular mass, and identifying the metal as Barium.

Let's analyze each statement:
- Option A is incorrect: The solubility of Group 2 hydroxides increases down the group.
- Option B is incorrect: The thermal stability of Group 2 nitrates increases down the group because the cation size increases, reducing its charge density and polarizing power on the nitrate anion.
- Option C is correct: The solubility of Group 2 sulfates decreases down the group due to the hydration enthalpy decreasing more rapidly than the lattice enthalpy.
- Option D is incorrect: The reactivity of Group 2 metals with oxygen increases down the group as first and second ionisation energies decrease.

[1 mark] C is correct.
- Award 1 mark for correctly identifying the trend in solubility of Group 2 sulfates.

The oxide \(X_2\text{O}_3\) is amphoteric because it reacts with both acids (dilute hydrochloric acid) and bases (hot aqueous sodium hydroxide).
In Period 3, the only amphoteric oxide is aluminium oxide, \(\text{Al}_2\text{O}_3\).
Aluminium (\(\text{Al}\)) is the element \(X\), and its proton (atomic) number is 13.

[1 mark] B is correct.
- Award 1 mark for identifying the amphoteric oxide as aluminium oxide and determining the proton number of aluminium is 13.

Hot, concentrated, acidified \(\text{KMnO}_4\) cleaves the carbon-carbon double bond (\(\text{C}=\text{C}\)) of alkenes:
- If a double-bonded carbon is bonded to two alkyl groups, it is oxidized to a ketone.
- If a double-bonded carbon is bonded to two hydrogen atoms (\(=\text{CH}_2\)), it is oxidized to carbon dioxide and water.

Since the products are propanone (a ketone with two methyl groups) and carbon dioxide, the reactant alkene must have one double-bonded carbon with two methyl groups, and the other double-bonded carbon with two hydrogens.
This corresponds to 2-methylpropene, \((\text{CH}_3)_2\text{C}=\text{CH}_2\).

[1 mark] C is correct.
- Award 1 mark for analyzing the oxidative cleavage products and identifying the correct alkene.

All four compounds have similar relative molecular masses (around 58-60) and thus similar instantaneous dipole-induced dipole forces (London dispersion forces):
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (propan-1-ol) has a highly polar \(\text{O-H}\) bond and can form strong intermolecular hydrogen bonds.
- \(\text{CH}_3\text{OCH}_2\text{CH}_3\) (methoxyethane) and \(\text{CH}_3\text{CH}_2\text{CHO}\) (propanal) are polar molecules but only exhibit permanent dipole-permanent dipole forces, which are weaker than hydrogen bonds.
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) (butane) is non-polar and only exhibits instantaneous dipole-induced dipole forces.

Since hydrogen bonding is the strongest type of intermolecular force among these, propan-1-ol (option a) has the highest boiling point.

[1 mark] A is correct.
- Award 1 mark for identifying that hydrogen bonding in propan-1-ol leads to the highest boiling point.

1. Determine the empirical formula by calculating the molar ratio of the elements:
- Moles of \(\text{C} = \frac{40.0}{12.0} = 3.33\text{ mol}\)
- Moles of \(\text{H} = \frac{6.7}{1.0} = 6.7\text{ mol}\)
- Moles of \(\text{O} = \frac{53.3}{16.0} = 3.33\text{ mol}\)

2. Divide each by the smallest value (3.33):
- \(\text{C} = 1.0\)
- \(\text{H} \approx 2.0\)
- \(\text{O} = 1.0\)

Therefore, the empirical formula is \(\text{CH}_2\text{O}\).

3. Calculate the empirical formula mass:
- Mass of \(\text{CH}_2\text{O} = 12.0 + (2 \times 1.0) + 16.0 = 30.0\text{ g mol}^{-1}\).

4. Determine the molecular formula multiplier:
- Multiplier = \(\frac{\text{Relative molecular mass}}{\text{Empirical formula mass}} = \frac{180}{30.0} = 6\).

5. Multiply the empirical formula by 6 to obtain the molecular formula:
- Molecular formula = \(\text{C}_6\text{H}_{12}\text{O}_6\).

[1 mark] D is correct.
- Award 1 mark for finding the correct empirical formula and using the molar mass to determine the molecular formula.

1. Calculate the moles of \(\text{H}_2\text{SO}_4\) used:
\(n(\text{H}_2\text{SO}_4) = 0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 0.00250\text{ mol}\).

2. Calculate the moles of \(\text{H}^+\) ions provided by sulfuric acid (since it is diprotic):
\(n(\text{H}^+) = 2 \times 0.00250 = 0.00500\text{ mol}\).

3. Calculate the moles of \(X(\text{OH})_n\) reacted:
\(n(X(\text{OH})_n) = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00250\text{ mol}\).

4. For a complete neutralization reaction:
\(\text{moles of H}^+ = \text{moles of OH}^-\)
\(0.00500\text{ mol} = n \times n(X(\text{OH})_n)\)
\(0.00500 = n \times 0.00250\)
\(n = 2\).

Therefore, the formula is \(X(\text{OH})_2\), and the value of \(n\) is 2.

[1 mark] B is correct.
- Award 1 mark for calculating the correct moles of hydrogen ions and hydroxide ions, and finding \(n = 2\).

Let's check the physical properties and bonding types of Period 3 elements:
- Aluminum (\(\text{Al}\)) is a metal with a giant metallic lattice containing highly mobile delocalized electrons, hence it has a very high electrical conductivity. (Matches \(P\))
- Silicon (\(\text{Si}\)) is a metalloid with a giant covalent structure. It acts as a semiconductor with moderate electrical conductivity. (Matches \(Q\))
- Phosphorus (\(\text{P}\)) is a non-metal with simple molecular structure (\(\text{P}_4\)). Since all outer shell valence electrons are localized in covalent bonds, it is an insulator and does not conduct electricity. (Matches \(R\))

Therefore, the correct row is Option A.

[1 mark] A is correct.
- Award 1 mark for correctly identifying the matching elements based on their electrical conductivity and bonding type.

First, calculate the number of moles of carbon dioxide gas evolved:
\[ n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol} \]

Using the balanced equation:
\[ M\text{CO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow M\text{Cl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \]

The mole ratio of \(M\text{CO}_3\) to \(\text{CO}_2\) is 1:1, so:
\[ n(M\text{CO}_3) = 0.010\text{ mol} \]

Now, calculate the molar mass of \(M\text{CO}_3\):
\[ M_{\text{r}}(M\text{CO}_3) = \frac{1.00\text{ g}}{0.010\text{ mol}} = 100\text{ g mol}^{-1} \]

Since \(M_{\text{r}}(M\text{CO}_3) = A_{\text{r}}(M) + A_{\text{r}}(\text{C}) + 3 \times A_{\text{r}}(\text{O})\):
\[ A_{\text{r}}(M) + 12.0 + 3(16.0) = 100 \]
\[ A_{\text{r}}(M) + 60.0 = 100 \]
\[ A_{\text{r}}(M) = 40.0 \]

This corresponds to Calcium (\(A_{\text{r}} = 40.1\)).

Award 1 mark for the correct calculation leading to option A.
- Reject other choices as they have different atomic masses.

Down Group 2, the atomic and cationic radii of the metal elements increase. As the radius of the \(M^{2+}\) cation increases, its charge density decreases, which significantly reduces its polarising power. Consequently, the larger cation polarises the large, electron-rich nitrate anion (\(\text{NO}_3^-\)) to a lesser extent. Since the covalent character in the nitrate ion is weakened less, more thermal energy is required to decompose the nitrate, making the compound more stable to heat. Therefore, option B is correct.

Award 1 mark for explaining that thermal stability increases down the group because larger cations have lower polarising power (Option B).
- Reject other options: A incorrectly states they become less stable; C refers to changing anion size; D incorrectly claims electronegativity increases.

- \(\text{Na}_2\text{O}\) is a basic oxide; it reacts with water to form an alkaline solution (\(\text{NaOH}\)) and does not react with \(\text{NaOH}\).
- \(\text{Al}_2\text{O}_3\) is amphoteric but is completely insoluble in water and does not react with it.
- \(\text{SiO}_2\) is an acidic oxide but is giant covalent and insoluble in water; it only reacts with concentrated base at high temperatures.
- \(\text{P}_4\text{O}_{10}\) is an acidic molecular oxide. It reacts vigorously with water to form phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)), which is an acidic solution. Being an acidic oxide, it also reacts readily with aqueous sodium hydroxide to form a salt and water.

Award 1 mark for identifying phosphorus(V) oxide (Option D).
- Reject options A, B, and C as they do not meet both criteria of reacting with water to form an acidic solution and reacting with aqueous sodium hydroxide.

1. Reaction of 2-methylbut-2-ene, \(\text{(CH}_3\text{)}_2\text{C=CHCH}_3\), with cold, dilute, acidified \(\text{KMnO}_4\) results in mild oxidation to form a diol. The carbon-carbon double bond is converted to a single bond, and an -OH group is added to each double-bonded carbon. This gives the diol \(\text{(CH}_3\text{)}_2\text{C(OH)CH(OH)CH}_3\) (product \(X\)).

2. Reaction with hot, concentrated, acidified \(\text{KMnO}_4\) causes complete cleavage of the carbon-carbon double bond:
- The carbon atom bonded to two methyl groups, \(\text{(CH}_3\text{)}_2\text{C=}\), is oxidized to a ketone: propanone, \(\text{CH}_3\text{COCH}_3\) (product \(Y\)).
- The carbon atom bonded to one methyl group and one hydrogen atom, \(\text{=CHCH}_3\), is oxidized to a carboxylic acid: ethanoic acid, \(\text{CH}_3\text{COOH}\) (product \(Z\)).

Thus, option A is correct.

Award 1 mark for identifying the correct structures of X, Y and Z (Option A).
- Reject other options since B contains an aldehyde (which would be oxidized), C uses the wrong starting alkene, and D contains the wrong cleavage fragments.

All four compounds have 4 carbon atoms and similar relative molecular masses, but different functional groups and shapes:
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (butan-1-ol) is a primary alcohol with a linear chain. It contains highly polar O-H bonds, allowing it to form strong intermolecular hydrogen bonds. Its linear shape also maximizes surface-to-surface contact, maximizing instantaneous dipole-induced dipole (London dispersion) forces.
- \(\text{(CH}_3\text{)}_3\text{COH}\) (2-methylpropan-2-ol) is a tertiary alcohol. Although it can form hydrogen bonds, its highly spherical, branched shape reduces surface contact area, leading to weaker London dispersion forces and a lower boiling point than butan-1-ol.
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\) (butanal) is an aldehyde. It has permanent dipole-dipole attractions but cannot form intermolecular hydrogen bonds.
- \(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3\) (ethoxyethane) is an ether. It is weakly polar and has only dipole-dipole and dispersion forces, with no intermolecular hydrogen bonding.

Therefore, butan-1-ol has the highest boiling point.

Award 1 mark for choosing Option A.
- Reject other options because they lack hydrogen bonding or have lower intermolecular surface contact due to branching.

Using the ideal gas equation:
\[ pV = nRT = \frac{m}{M_{\text{r}}}RT \]

Rearranging to solve for \(M_{\text{r}}\):
\[ M_{\text{r}} = \frac{mRT}{pV} \]

Substitute the given values into the equation, ensuring all SI units are correct:
- Mass, \(m = 0.184\text{ g}\)
- Temperature, \(T = 127 + 273 = 400\text{ K}\)
- Volume, \(V = 74.0\text{ cm}^3 = 74.0 \times 10^{-6}\text{ m}^3\)
- Pressure, \(p = 1.01 \times 10^5\text{ Pa}\)
- Gas constant, \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

\[ M_{\text{r}} = \frac{0.184 \times 8.31 \times 400}{1.01 \times 10^5 \times 74.0 \times 10^{-6}} \]
\[ M_{\text{r}} = \frac{611.616}{7.474} \approx 81.83 \]

Rounding to the nearest whole number gives 82.

Award 1 mark for correct calculations leading to Option B.
- Reject other options due to temperature conversion errors (e.g. using 127 K instead of 400 K) or unit errors.

Down Group 2, the solubility trends are as follows:
1. The solubility of hydroxides increases down the group. Therefore, \(\text{Mg(OH)}_2\) is insoluble (forming a white precipitate), while the hydroxides of the elements further down the group are more soluble and do not readily precipitate.
2. The solubility of sulfates decreases down the group. Therefore, \(\text{MgSO}_4\) is highly soluble (forming no precipitate with dilute sulfuric acid), while sulfates of heavier Group 2 metals (such as \(\text{SrSO}_4\) and \(\text{BaSO}_4\)) are highly insoluble and form distinct white precipitates.

Since \(X^{2+}\) forms a precipitate with hydroxide ions but no precipitate with sulfate ions, the metal \(X\) must be magnesium.

Award 1 mark for choosing Option A.
- Reject options B, C, and D because they would form a precipitate with sulfuric acid and/or not form a precipitate with sodium hydroxide.

For an alkene to exhibit cis-trans isomerism, each carbon of the double bond must be attached to two different groups.

- In 2-methylbut-2-ene, \(\text{(CH}_3\text{)}_2\text{C=CHCH}_3\), carbon-2 is bonded to two identical methyl groups, so it cannot show cis-trans isomerism.
- In 2,3-dimethylbut-2-ene, \(\text{(CH}_3\text{)}_2\text{C=C(CH}_3\text{)}_2\), both double-bonded carbons are bonded to two identical methyl groups.
- In 3-methylpent-2-ene, \(\text{CH}_3\text{CH=C(CH}_3\text{)CH}_2\text{CH}_3\):
- Carbon-2 is bonded to -H and -\(\text{CH}_3\) (two different groups).
- Carbon-3 is bonded to -\(\text{CH}_3\) and -\(\text{CH}_2\text{CH}_3\) (two different groups).
Therefore, 3-methylpent-2-ene exists as cis and trans isomers.
- In 2-methylpent-1-ene, \(\text{CH}_2\text{=C(CH}_3\text{)CH}_2\text{CH}_2\text{CH}_3\), carbon-1 is bonded to two identical hydrogen atoms.

Award 1 mark for Option C.
- Reject other options because at least one of the double-bonded carbon atoms is bonded to two identical groups.

Calculate the number of moles of hydrogen gas collected: \(n(\text{H}_2) = \frac{960 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.040 \text{ mol}\). Since \(\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}\), the number of moles of magnesium in the alloy is also 0.040 mol. Calculate the mass of magnesium: \(m(\text{Mg}) = 0.040 \text{ mol} \times 24.3 \text{ g mol}^{-1} = 0.972 \text{ g}\). Calculate the mass of copper in the alloy: \(m(\text{Cu}) = 1.60 \text{ g} - 0.972 \text{ g} = 0.628 \text{ g}\). Calculate the percentage by mass of copper: \(\%\text{Cu} = \frac{0.628 \text{ g}}{1.60 \text{ g}} \times 100\% = 39.25\% \approx 39.3\%\).

Award 1 mark for the correct calculation of 39.3% (Option A). Accept alternative valid rounding. Reject 60.8% (which is the percentage of magnesium).

Down Group 2, the cationic radius of the \(M^{2+}\) ions increases. Consequently, the charge density and polarising power of the cation decrease. The cation polarises the carbonate anion's \(\text{C-O}\) bond less effectively, making the carbonate compound more stable to thermal decomposition. Hence, thermal stability increases down the group.

Award 1 mark for identifying that thermal stability increases down the group because the larger cation causes less polarisation of the carbonate anion (Option A).

Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), reacts with water to form phosphoric(V) acid, which is strongly acidic (pH 1-2). Aluminium oxide, \(\text{Al}_2\text{O}_3\), is insoluble in water and forms a neutral mixture (pH 7). Magnesium oxide, \(\text{MgO}\), is sparingly soluble and reacts to form a weakly alkaline solution (pH 9). Sodium oxide, \(\text{Na}_2\text{O}\), reacts vigorously to form strongly alkaline sodium hydroxide (pH 13-14). The order of increasing pH is therefore \(\text{P}_4\text{O}_{10} < \text{Al}_2\text{O}_3 < \text{MgO} < \text{Na}_2\text{O}\).

Award 1 mark for the correct order from most acidic to most alkaline (Option A).

Alkenes react with cold, dilute, alkaline \(\text{KMnO}_4\) to form diols via mild oxidation. Let us analyze the diols formed by each option: but-1-ene forms butane-1,2-diol which contains a chiral carbon (carbon-2) and thus exhibits optical isomerism; cis-but-2-ene and trans-but-2-ene form butane-2,3-diol which contains chiral carbons (carbons-2 and 3) and exhibits optical isomerism; 2-methylpropene forms 2-methylpropane-1,2-diol, \((\text{CH}_3)_2\text{C(OH)CH}_2\text{OH}\). In this diol, neither carbon-1 nor carbon-2 is a chiral center (carbon-2 has two identical methyl groups attached, and carbon-1 has two hydrogen atoms attached). Therefore, 2-methylpropane-1,2-diol has no optical isomers.

Award 1 mark for selecting 2-methylpropene as the only alkene whose diol product has no chiral centers (Option D).

All four options have similar relative molecular masses (around 58-60). Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) contains a hydrogen atom directly bonded to a highly electronegative oxygen atom, allowing it to form strong intermolecular hydrogen bonds. Propanone and propanal only experience permanent dipole-dipole forces, while butane only experiences weak London dispersion forces. Due to the significantly higher energy required to break hydrogen bonds, propan-1-ol has the highest boiling point.

Award 1 mark for identifying propan-1-ol as the substance with the highest boiling point due to hydrogen bonding (Option A).

Calculate the initial moles of \(\text{H}^+\): \(n(\text{H}^+) = 0.0250 \text{ dm}^3 \times 0.200 \text{ mol dm}^{-3} = 5.00 \times 10^{-3} \text{ mol}\). Calculate the moles of \(\text{OH}^-\): \(n(\text{OH}^-) = 0.0150 \text{ dm}^3 \times 0.300 \text{ mol dm}^{-3} = 4.50 \times 10^{-3} \text{ mol}\). The reaction is 1:1, so the excess moles of \(\text{H}^+\) is \(5.00 \times 10^{-3} - 4.50 \times 10^{-3} = 5.00 \times 10^{-4} \text{ mol}\). The total volume of the mixture is \(25.0 + 15.0 = 40.0 \text{ cm}^3 = 0.0400 \text{ dm}^3\). The concentration of remaining hydrogen ions is \(\frac{5.00 \times 10^{-4} \text{ mol}}{0.0400 \text{ dm}^3} = 0.0125 \text{ mol dm}^{-3}\).

Award 1 mark for the correct calculation of the remaining hydrogen ion concentration (Option A).

The decomposition reaction is: \(\text{MCO}_3\text{(s)} \rightarrow \text{MO(s)} + \text{CO}_2\text{(g)}\). Calculate the moles of \(\text{CO}_2\) gas: \(n(\text{CO}_2) = \frac{1.20 \text{ dm}^3}{24.0 \text{ dm}^3 \text{ mol}^{-1}} = 0.050 \text{ mol}\). This means there was 0.050 mol of \(\text{MCO}_3\). Determine the molar mass of \(\text{MCO}_3\): \(M_r = \frac{4.215 \text{ g}}{0.050 \text{ mol}} = 84.3 \text{ g mol}^{-1}\). Since \(M_r(\text{MCO}_3) = A_r(\text{M}) + 12.0 + 3(16.0) = 84.3\), we have \(A_r(\text{M}) + 60.0 = 84.3\), which gives \(A_r(\text{M}) = 24.3\). This corresponds to magnesium.

Award 1 mark for correctly identifying the metal as magnesium (Option A) based on the calculated molar mass.

Chloride P must be magnesium chloride (\(\text{MgCl}_2\)), which undergoes slight hydrolysis in water due to the moderate polarising power of the \(\text{Mg}^{2+}\) cation, producing a weakly acidic solution of pH around 6.5. Sodium chloride dissolves to form a completely neutral solution of pH 7.0. Chloride Q must be silicon(IV) chloride (\(\text{SiCl}_4\)), which reacts violently and completely with water to yield \(\text{SiO}_2\text{.H}_2\text{O}\) and HCl gas (highly acidic, pH 1-2, white fumes). This matches the pair described in Option B.

Award 1 mark for identifying P as magnesium chloride and Q as silicon(IV) chloride (Option B).

Step 1: Calculate the moles of \(\text{CO}_2\) gas collected.
\(\text{Moles of CO}_2 = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.010\text{ mol}\)

Step 2: Use the stoichiometric equation for the decomposition of Group 2 carbonates:
\(\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})\)
The mole ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is \(1:1\). Therefore, moles of \(\text{MCO}_3 = 0.010\text{ mol}\).

Step 3: Calculate the relative molecular mass (\(M_r\)) of \(\text{MCO}_3\).
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.97\text{ g}}{0.010\text{ mol}} = 197\text{ g mol}^{-1}\)

Step 4: Determine the relative atomic mass (\(A_r\)) of metal \(\text{M}\).
\(A_r(\text{M}) + 12.0 + (3 \times 16.0) = 197\)
\(A_r(\text{M}) + 60.0 = 197 \implies A_r(\text{M}) = 137\)

Looking at the periodic table, \(A_r \approx 137.3\) corresponds to Barium (\(\text{Ba}\)).

1 mark for the correct calculation identifying the metal as Barium (D). No partial marks.

A is incorrect: The solubility of Group 2 hydroxides increases down the group, so barium hydroxide is more soluble than calcium hydroxide.
B is incorrect: Thermal stability of carbonates increases down the group, so magnesium carbonate decomposes at a lower temperature than calcium carbonate.
C is correct: The solubility of Group 2 sulfates decreases down the group, meaning barium sulfate is significantly less soluble in water than magnesium sulfate.
D is incorrect: Group 2 nitrates decompose on heating to form the metal oxide, nitrogen dioxide, and oxygen gas, not the nitrite.

1 mark for the correct option (C). No partial marks.

The chloride of element \(X\) is neutral (\(\text{pH} \approx 7\)), which corresponds to sodium chloride, \(\text{NaCl}\). Thus, element \(X\) is sodium (\(\text{Na}\)).
The chloride of element \(Y\) undergoes violent hydrolysis in water to produce an acidic solution and a white precipitate of silicon dioxide: \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\). This matches silicon (\(\text{Si}\)).

1 mark for identifying element X as sodium and element Y as silicon (B).

When an alkene is cleaved by hot, concentrated, acidified \(\text{KMnO}_4\), the double bond is completely broken.
- If a carbon atom in the double bond is attached to two alkyl groups, it forms a ketone.
- Since propanone, \((\text{CH}_3)_2\text{C}=\text{O}\), is the ONLY organic product, the alkene must be symmetrical and contain the \((\text{CH}_3)_2\text{C}=\) group on both sides of the double bond.
- The structure of alkene \(W\) is therefore \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\), which is 2,3-dimethylbut-2-ene.

1 mark for the correct IUPAC name of the alkene (B).

Ethanol contains a hydrogen atom directly bonded to a highly electronegative oxygen atom (\(\text{O}-\text{H}\)), which enables ethanol molecules to form strong intermolecular hydrogen bonds. Dimethyl ether contains polar \(\text{C}-\text{O}\) bonds but lacks hydrogen atoms bonded directly to nitrogen, oxygen, or fluorine; thus, it cannot form hydrogen bonds with itself. The stronger intermolecular hydrogen bonds in ethanol require significantly more thermal energy to overcome, leading to a much higher boiling point.

1 mark for identifying the presence of intermolecular hydrogen bonds in ethanol as the primary reason (B).

Step 1: Calculate the mass and moles of Carbon.
\(\text{Mass of C} = 0.220\text{ g} \times \frac{12.0}{44.0} = 0.060\text{ g}\)
\(\text{Moles of C} = \frac{0.060\text{ g}}{12.0\text{ g mol}^{-1}} = 0.005\text{ mol}\)

Step 2: Calculate the mass and moles of Hydrogen.
\(\text{Mass of H} = 0.090\text{ g} \times \frac{2.0}{18.0} = 0.010\text{ g}\)
\(\text{Moles of H} = \frac{0.010\text{ g}}{1.0\text{ g mol}^{-1}} = 0.010\text{ mol}\)

Step 3: Calculate the mass and moles of Oxygen.
\(\text{Mass of O} = 0.150\text{ g} - 0.060\text{ g} - 0.010\text{ g} = 0.080\text{ g}\)
\(\text{Moles of O} = \frac{0.080\text{ g}}{16.0\text{ g mol}^{-1}} = 0.005\text{ mol}\)

Step 4: Find the simplest whole-number ratio.
\(\text{Ratio C : H : O} = 0.005 : 0.010 : 0.005 = 1 : 2 : 1\)
Therefore, the empirical formula is \(\text{CH}_2\text{O}\).

1 mark for the correct empirical formula derivation (B).

Reactivity of Group 2 oxides with water increases down the group, so barium oxide reacts more vigorously with water than magnesium oxide.
Furthermore, the solubility of Group 2 hydroxides increases down the group. Barium hydroxide, \(\text{Ba(OH)}_2\), is highly soluble and dissociates to release a high concentration of \(\text{OH}^-\) ions, resulting in a strongly alkaline pH (~13). Magnesium hydroxide, \(\text{Mg(OH)}_2\), is only sparingly soluble, meaning fewer \(\text{OH}^-\) ions are in solution, resulting in a much lower pH (~9-10). Thus, the mixture containing barium oxide has a higher pH.

1 mark for identifying both the higher vigor of reaction and the higher pH for barium oxide (A).

For an alkene to exhibit cis-trans (E-Z) isomerism, each carbon of the double bond must be bonded to two different groups.
- A: 2-methylbut-2-ene is \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\). One of the double-bonded carbons has two identical methyl groups, so it cannot have stereoisomers.
- B: hex-1-ene is \(\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3\). The terminal double-bonded carbon has two identical hydrogen atoms, so it cannot have stereoisomers.
- C: 3-methylpent-2-ene is \(\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3\). Carbon 2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups). Carbon 3 is bonded to \(-\text{CH}_3\) and \(-\text{CH}_2\text{CH}_3\) (two different groups). Therefore, it exhibits stereoisomerism.
- D: 2-methylpent-2-ene is \((\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3\). Carbon 2 has two identical methyl groups, so no stereoisomerism is possible.

1 mark for the correct identification of the stereoisomeric alkene (C).

(a) The balanced ionic equation for the reaction is:
\(\text{MnO}_4^-\text{(aq)} + 5\text{Fe}^{2+}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow \text{Mn}^{2+}\text{(aq)} + 5\text{Fe}^{3+}\text{(aq)} + 4\text{H}_2\text{O(l)}\)

(b) Moles of \(\text{MnO}_4^-\right.\):
\(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{18.50}{1000}\text{ dm}^3 = 3.70 \times 10^{-4}\text{ mol}\)

(c) Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\):
According to the stoichiometry, \(1\text{ mol}\) of \(\text{MnO}_4^-\right.\) reacts with \(5\text{ mol}\) of \(\text{Fe}^{2+}\).
\(n(\text{Fe}^{2+}) = 5 \times 3.70 \times 10^{-4}\text{ mol} = 1.85 \times 10^{-3}\text{ mol}\)

(d) Moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\) (the original sample):
\(n(\text{Fe}^{2+})_{\text{total}} = 1.85 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 1.85 \times 10^{-2}\text{ mol}\)

Mass of anhydrous \(\text{FeSO}_4\):
\(m = 1.85 \times 10^{-2}\text{ mol} \times 151.9\text{ g mol}^{-1} = 2.81\text{ g}\) (to 3 s.f.)

(e)(i) Molar mass of \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\):
If \(1.85 \times 10^{-2}\text{ mol}\) has a mass of \(5.14\text{ g}\):
\(M_r = \frac{5.14\text{ g}}{1.85 \times 10^{-2}\text{ mol}} = 277.8\text{ g mol}^{-1}\) (or 278)

(ii) Determine \(x\):
Mass of water of crystallisation per mole of hydrated salt:
\(x \times M_r(\text{H}_2\text{O}) = 277.8 - 151.9 = 125.9\text{ g mol}^{-1}\)
\(x = \frac{125.9}{18.0} = 6.99 \approx 7\)
Therefore, \(x = 7\).

Part (a): [2 marks]
- 1 mark for correct reactants and products: \(\text{MnO}_4^-\text{(aq)} + \text{Fe}^{2+}\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{Mn}^{2+}\text{(aq)} + \text{Fe}^{3+}\text{(aq)} + \text{H}_2\text{O(l)}\)
- 1 mark for correct stoichiometry: 1, 5, 8 \(\rightarrow\) 1, 5, 4

Part (b): [1 mark]
- 1 mark for \(3.70 \times 10^{-4}\text{ mol}\) (accept \(0.00037\))

Part (c): [1 mark]
- 1 mark for multiplying moles of manganate by 5 to get \(1.85 \times 10^{-3}\text{ mol}\) (allow error carried forward (ECF) from (b))

Part (d): [2 marks]
- 1 mark for calculating moles in \(250.0\text{ cm}^3\) (\(1.85 \times 10^{-2}\text{ mol}\))
- 1 mark for multiplying by 151.9 to get \(2.81\text{ g}\) (or \(2.810\text{ g}\)) (allow ECF)

Part (e)(i): [3 marks]
- 1 mark for equating moles of hydrated salt to total moles of \(\text{Fe}^{2+}\) (\(1.85 \times 10^{-2}\text{ mol}\))
- 1 mark for expression: \(M_r = \text{mass} / \text{moles}\)
- 1 mark for calculated \(M_r = 277.8\text{ g mol}^{-1}\) (accept range 277-278) (allow ECF)

Part (e)(ii): [3 marks]
- 1 mark for subtracting 151.9 from their calculated \(M_r\) (e.g., \(277.8 - 151.9 = 125.9\))
- 1 mark for dividing by 18.0
- 1 mark for rounding to nearest whole number: \(x = 7\) (allow ECF)

(a) Down Group 2, the thermal stability of the nitrates increases. This is because:
- The ionic radius of the Group 2 cation increases down the group while keeping the same \(+2\) charge.
- The charge density of the cation decreases, which makes it less polarising.
- Consequently, the cation exerts a weaker polarising effect on the electron cloud of the nitrate anion, causing less distortion/weakening of the \(\text{N}-\text{O}\) covalent bonds.
- Therefore, more thermal energy is required to decompose the nitrate down the group.

(b) \(2\text{Ca(NO}_3)_2\text{(s)} \rightarrow 2\text{CaO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\)

(c)(i) The solubility of the Group 2 hydroxides increases down the group.
- Both lattice energy and hydration energy of the hydroxides decrease down the group.
- However, because the hydroxide ion is relatively small, the lattice energy decreases less rapidly than the hydration energy of the cations (meaning the enthalpy change of solution becomes more exothermic / less endothermic down the group).

(ii) The pH of the resulting solution increases down the group, because the increasing solubility of the hydroxides leads to a higher concentration of \(\text{OH}^-\text{(aq)}\) ions in solution.

(d) \(\text{Mg(OH)}_2\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{H}_2\text{O(l)}\)

Part (a): [4 marks]
- 1 mark for stating thermal stability increases down the group.
- 1 mark for stating that ionic radius of the cation increases / charge density decreases.
- 1 mark for stating that the polarising power of the cation decreases.
- 1 mark for stating that there is less distortion/polarisation of the nitrate anion (or \(\text{N}-\text{O}\) bond is weakened less).

Part (b): [2 marks]
- 1 mark for correct formulae of products: \(\text{CaO}\), \(\text{NO}_2\), and \(\text{O}_2\).
- 1 mark for correct balancing and state symbols: \(2\text{Ca(NO}_3)_2\text{(s)} \rightarrow 2\text{CaO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\)

Part (c)(i): [3 marks]
- 1 mark for stating solubility increases down the group.
- 1 mark for stating that both lattice energy and hydration energy decrease down the group.
- 1 mark for explaining that the lattice energy decreases slower than hydration energy / enthalpy change of solution becomes more exothermic/less endothermic down the group.

Part (c)(ii): [1 mark]
- 1 mark for stating that pH increases down the group because \([\text{OH}^-]\) increases.

Part (d): [2 marks]
- 1 mark for correct species: \(\text{Mg(OH)}_2\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{H}_2\text{O(l)}\)
- 1 mark for correct state symbols.

(a)(i) Sodium burns vigorously with a bright yellow flame to form a white solid (sodium oxide).
Equation: \(4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}\)
(Accept \(2\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow \text{Na}_2\text{O}_2\text{(s)}\) with observation of yellow flame).

(ii) Phosphorus burns with a very bright white flame and produces thick white smoke/clouds of phosphorus(V) oxide.
Equation: \(\text{P}_4\text{(s)} + 5\text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}\)

(b)(i) Equation: \(\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\)
- Approximate pH of the resulting solution: \(1\) or \(2\) (highly acidic).

(ii) Silicon dioxide has a giant covalent structure. The strong \(\text{Si}-\text{O}\) covalent bonds throughout the giant lattice require too much energy to break, so it does not react with or dissolve in water.
- It reacts with hot, concentrated sodium hydroxide because it is an acidic oxide, neutralising the strong base to form a silicate salt.
- Behavior shown: acidic oxide.

(c)(i) \(\text{Al}_2\text{O}_3\text{(s)} + 6\text{H}^+\text{(aq)} \rightarrow 2\text{Al}^{3+}\text{(aq)} + 3\text{H}_2\text{O(l)}\)
(ii) \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2[\text{Al(OH)}_4]^-\text{(aq)}\)
(Accept: \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} \rightarrow 2\text{AlO}_2^-\text{(aq)} + \text{H}_2\text{O(l)}\))

Part (a)(i): [2 marks]
- 1 mark for observation: yellow flame / white solid.
- 1 mark for equation: \(4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}\) (or for peroxide: \(2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2\)).

Part (a)(ii): [2 marks]
- 1 mark for observation: white flame / white smoke/solid.
- 1 mark for equation: \(\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}\) (or \(2\text{P}_2\text{O}_5\)).

Part (b)(i): [2 marks]
- 1 mark for equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\).
- 1 mark for pH: 1–2.

Part (b)(ii): [3 marks]
- 1 mark for explaining giant covalent structure of \(\text{SiO}_2\) with strong covalent bonds that cannot be broken by water.
- 1 mark for explaining reaction with NaOH because it is an acidic oxide.
- 1 mark for identifying behavior as acidic.

Part (c)(i): [1 mark]
- 1 mark for correct ionic equation: \(\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}\)

Part (c)(ii): [2 marks]
- 1 mark for correct reactants and products: \(\text{Al}_2\text{O}_3 + 2\text{OH}^- \rightarrow 2[\text{Al(OH)}_4]^-\right.\) or \(2\text{AlO}_2^-\right.\)
- 1 mark for correct balancing: \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\right.\) (or \(\text{Al}_2\text{O}_3 + 2\text{OH}^- \rightarrow 2\text{AlO}_2^- + \text{H}_2\text{O}\))

(a)(i) Skeletal structures of cis-pent-2-ene and trans-pent-2-ene:
- cis-pent-2-ene: A bent backbone of 5 carbons with the double bond between C2 and C3, both alkyl groups (methyl and ethyl) are on the same side of the double bond.
- trans-pent-2-ene: A zig-zag backbone of 5 carbons with the double bond between C2 and C3, with the methyl and ethyl groups on opposite sides of the double bond.

(ii) cis-trans isomerism arises due to restricted rotation about the \(\text{C}=\text{C}\) double bond (due to the presence of the \(\pi\) bond).
- In pent-2-ene, each of the carbon atoms in the double bond is attached to two different groups (C2 is attached to \(-\text{H}\) and \(-\text{CH}_3\); C3 is attached to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\)).
- In 2-methylbut-2-ene, one of the double-bonded carbons is attached to two identical methyl (\(-\text{CH}_3\)) groups, so swapping them does not produce a different isomer.

(b)(i) Mechanism:
- Step 1: Curly arrow starts from the \(\text{C}=\text{C}\) double bond of 2-methylbut-1-ene to the hydrogen atom of \(\text{H}^{\delta+}-\text{Br}^{\delta-}\).
- Curly arrow from the \(\text{H}-\text{Br}\) bond to the bromine atom.
- Intermediate: Draw the tertiary carbocation \(\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\) and a bromide ion, \(:\text{Br}^-\).
- Step 2: Curly arrow from the lone pair on \(:\text{Br}^-\) to the positively charged carbon of the carbocation.
- Final Product: 2-bromo-2-methylbutane.

(ii) The major product is formed via the more stable carbocation intermediate.
- The reaction to form 2-bromo-2-methylbutane goes via a tertiary (\(3^\circ\)) carbocation, \(\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\).
- The reaction to form 1-bromo-2-methylbutane goes via a primary (\(1^\circ\)) carbocation, \(^+ \text{CH}_2-\text{CH(CH}_3)\text{CH}_2\text{CH}_3\).
- Tertiary carbocations are more stable than primary carbocations due to the greater electron-donating inductive effect of three alkyl groups compared to one.

(c) When pent-2-ene polymerises, the double bond opens up. The main polymer chain is formed by the two carbons of the double bond:
\(-\text{[CH(CH}_3)-\text{CH(C}_2\text{H}_5)-\text{CH(CH}_3)-\text{CH(C}_2\text{H}_5)\text{]-}\)
(Two repeating units shown with continuation bonds at the ends).

Part (a)(i): [2 marks]
- 1 mark for correct skeletal structure of cis-pent-2-ene, clearly labelled.
- 1 mark for correct skeletal structure of trans-pent-2-ene, clearly labelled.

Part (a)(ii): [2 marks]
- 1 mark for stating that there is restricted rotation around the \(\text{C}=\text{C}\) double bond.
- 1 mark for explaining that pent-2-ene has two different groups on each double-bonded carbon, whereas 2-methylbut-2-ene has two identical (methyl) groups on one of the double-bonded carbons.

Part (b)(i): [4 marks]
- 1 mark for correct dipoles on \(\text{H}-\text{Br}\) and curly arrow from double bond to \(\text{H}\).
- 1 mark for curly arrow breaking the \(\text{H}-\text{Br}\) bond to form \(\text{Br}^-\).
- 1 mark for drawing the correct tertiary carbocation intermediate structure and \(\text{Br}^-\).
- 1 mark for curly arrow from lone pair on \(\text{Br}^-\) to the carbocation carbon to form 2-bromo-2-methylbutane.

Part (b)(ii): [2 marks]
- 1 mark for identifying that the major product forms via a tertiary (\(3^\circ\)) carbocation while the minor product forms via a primary (\(1^\circ\)) carbocation.
- 1 mark for stating that tertiary carbocations are more stable than primary carbocations because of the inductive effect of the three electron-donating alkyl groups.

Part (c): [2 marks]
- 1 mark for showing the correct single backbone unit with side-chains \(-\text{CH}_3\) and \(-\text{CH}_2\text{CH}_3\) (or \(-\text{C}_2\text{H}_5\)).
- 1 mark for drawing exactly two repeating units correctly connected with open continuation bonds at both ends.

(a) Electronegativity is the power / ability of an atom in a covalent bond to attract the shared pair of electrons towards itself.

(b)
- In \(\text{CO}_2\), the carbon atom has no lone pairs and forms two double bonds, resulting in a linear molecular shape (bond angle \(180^\circ\)).
- Due to the symmetrical linear shape, the dipoles from the polar \(\text{C}=\text{O}\) bonds act in opposite directions and cancel each other out, leaving no net dipole moment.
- In \(\text{SO}_2\), the sulfur atom has a lone pair of electrons in addition to the double bonds, resulting in a bent/non-linear molecular shape (bond angle approximately \(119^\circ\)).
- Because the shape is asymmetrical, the dipoles of the polar \(\text{S}=\text{O}\) bonds do not cancel, resulting in a net dipole moment and making the molecule polar.

(c)(i)
- Fluorine is highly electronegative, creating a highly polar \(\text{H}-\text{F}\) bond.
- This allows \(\text{HF}\) molecules to form hydrogen bonds between them.
- Hydrogen bonds are significantly stronger than the intermolecular forces (permanent dipole-dipole and London forces) present in \(\text{HCl}\), and therefore require much more energy to overcome.

(ii)
- From \(\text{HCl}\) to \(\text{HI}\), the size of the halogen atom increases, meaning the total number of electrons in the molecule increases.
- This increases the strength of the instantaneous dipole-induced dipole (London / dispersion) forces between the molecules.
- This increase in London forces outweighs the decrease in permanent dipole-dipole forces, so more energy is required to overcome the intermolecular attractions down the group, raising the boiling point.

Part (a): [2 marks]
- 1 mark for "power/ability of an atom to attract the shared pair of electrons".
- 1 mark for specifying "in a covalent bond".

Part (b): [4 marks]
- 1 mark for stating that \(\text{CO}_2\) is linear.
- 1 mark for explaining that the polar bonds in \(\text{CO}_2\) cancel each other out (symmetrical dipoles).
- 1 mark for stating that \(\text{SO}_2\) is non-linear / bent (due to lone pair on S).
- 1 mark for explaining that the dipoles do not cancel out (non-symmetrical dipoles).

Part (c)(i): [3 marks]
- 1 mark for stating that \(\text{HF}\) has hydrogen bonding.
- 1 mark for explaining that fluorine is highly electronegative (making \(\text{HF}\) highly polar).
- 1 mark for stating that hydrogen bonds are stronger than the permanent dipole-dipole / London forces in \(\text{HCl}\) (require more energy to break).

Part (c)(ii): [3 marks]
- 1 mark for stating that from \(\text{HCl}\) to \(\text{HI}\), the number of electrons in the molecules increases.
- 1 mark for stating that instantaneous dipole-induced dipole (London/dispersion) forces become stronger.
- 1 mark for stating that more energy is needed to break these stronger London forces (which outweighs the decrease in permanent dipole-dipole forces).

1. Assuming a mean volume of FB 2 = 24.15 cm^3.
2. Moles of MnO4^- = (24.15 / 1000) * 0.0200 = 4.83 * 10^-4 mol.
3. Moles of H2O2 = 2.5 * 4.83 * 10^-4 = 1.21 * 10^-3 mol.
4. Concentration of H2O2 = 1.21 * 10^-3 / 0.0250 = 0.0483 mol dm^-3.

Table of results (3 marks): Headings and units present; all burette readings to 0.05 cm^3; concordant titres within 0.10 cm^3. Accuracy marks (4 marks): based on proximity to supervisor value. Calculations (4.33 marks): Mean titre selected correctly (1 mark), moles of MnO4^- correct to 3 sig figs (1.33 marks), moles and concentration of H2O2 calculated correctly with stoichiometric ratio 5:2 (2 marks). Percentage error of pipette (2 marks): calculation of error for 25.0 cm^3 pipette with uncertainty of 0.06 cm^3 ((0.06/25.0)*100 = 0.24%).

1. Assuming a Delta T of 20.9 degrees C from the graph extrapolation.
2. Heat energy q = 25.0 * 4.18 * 20.9 = 2184.15 J = 2.18 kJ.
3. Moles of CuSO4 = 0.0250 * 0.400 = 0.0100 mol.
4. Delta H = -2.18 / 0.0100 = -218 kJ mol^-1.

Recording temperatures (2 marks): All temperatures recorded to nearest 0.5 degrees C or 0.1 degrees C. Graph (4 marks): Linear scales, points plotted correctly, straight lines drawn for cooling and heating stages, extrapolation to t=4 mins shown. Temperature rise (1.33 marks): Delta T determined correctly from extrapolation. Calculations (4 marks): Heat energy q calculated with correct units (2 marks), moles of CuSO4 and negative sign for exothermic reaction Delta H in kJ mol^-1 (2 marks). Evaluation (2 marks): Suggestion of a practical improvement (e.g., use of a lid or vacuum flask to minimize heat loss).

Observations:
FB 6: NaOH(aq) - no change; NH3(aq) - no change; H2SO4(aq) - white precipitate; AgNO3(aq) - white precipitate soluble in NH3(aq).
FB 7: NaOH(aq) - white precipitate insoluble in excess; NH3(aq) - white precipitate insoluble in excess; H2SO4(aq) - no change; AgNO3(aq) - no change.
FB 8: NaOH(aq) - no change; NH3(aq) - no change; H2SO4(aq) - no change; AgNO3(aq) - yellow precipitate insoluble in NH3(aq).

Identity:
FB 6 contains Ba2+ and Cl^- (Barium chloride).
FB 7 contains Mg2+ and NO3^- (Magnesium nitrate).
FB 8 contains Na+ and I^- (Sodium iodide).

Observations table (5 marks): 1 mark for each correct set of observations for FB 6, FB 7, FB 8. Identification (4.33 marks): 2.33 marks for identifying Ba2+, Mg2+, and Na+ with supporting evidence; 2 marks for identifying Cl^-, NO3^-, and I^- with supporting evidence. Ionic equations (4 marks): 2 marks for Ba^2+(aq) + SO4^2-(aq) -> BaSO4(s) with state symbols; 2 marks for Ag^+(aq) + I^-(aq) -> AgI(s) or Ag^+(aq) + Cl^-(aq) -> AgCl(s) with state symbols.