2023 Edexcel IAS-Level Chemistry (XCH11) Practice Paper with Answers

The successive ionization energies show a massive increase between the third and fourth ionization energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner shell, which experiences less shielding. Therefore, there are three electrons in the outer shell of element \(X\), placing it in Group 13.

Correct answer receives 1 mark.

Using the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT}\). Converting units: \(p = 1.01 \times 10^5 \text{ Pa}\), \(V = 2.45 \times 10^{-4} \text{ m}^3\), \(T = 298 \text{ K}\). Therefore, \(n = \frac{1.01 \times 10^5 \times 2.45 \times 10^{-4}}{8.31 \times 298} \approx 0.0100 \text{ mol}\). Thus, \(M_r = \frac{\text{mass}}{n} = \frac{0.581}{0.0100} = 58.1\).

Correct answer receives 1 mark.

\(\text{H}_3\text{O}^+\) has three bonding pairs and one lone pair of electrons on the central oxygen atom. This gives a trigonal pyramidal shape, which is non-planar. All other choices are planar.

Correct answer receives 1 mark.

A propagation step involves a reaction between a radical and a molecule to produce a different radical and a different molecule. The reaction \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) fits this definition. Option A is initiation, while options B and D are termination steps.

Correct answer receives 1 mark.

For \(E\)/\(Z\) isomerism, each carbon of the double bond must be attached to two different groups. In hex-3-ene, each double-bonded carbon is attached to a hydrogen atom and an ethyl group. In all other options, at least one of the double-bonded carbons is attached to two identical groups.

Correct answer receives 1 mark.

The electronic configuration of a neutral copper atom is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1\). When forming the \(\text{Cu}^+\) ion, the electron in the \(4\text{s}\) orbital is lost first, leaving the filled \(3\text{d}\) subshell: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10}\).

Correct answer receives 1 mark.

Calculate moles per 100g: \(\text{C}: 52.2 / 12 = 4.35\), \(\text{H}: 13.0 / 1 = 13.0\), \(\text{O}: 34.8 / 16 = 2.175\). Dividing each by \(2.175\) yields the mole ratio: \(\text{C} = 2\), \(\text{H} = 6\), \(\text{O} = 1\). Thus, the empirical formula is \(\text{C}_2\text{H}_6\text{O}\).

Correct answer receives 1 mark.

Covalent character is maximized when polarization of the anion is greatest. This is favored by a highly polarizing cation (small ionic radius: \(\text{Li}^+\)) and a highly polarizable anion (large ionic radius: \(\text{I}^-\)). Hence, \(\text{LiI}\) has the greatest covalent character.

Correct answer receives 1 mark.

The successive ionization energies show a massive increase between the 5th and 6th ionization energies (from 6270 to 21270 kJ mol^-1). This indicates that the first five electrons are in the outer shell, and the 6th electron is removed from a shell closer to the nucleus. Therefore, the element has 5 valence electrons and belongs to Group 15. In Period 3, this element is phosphorus, which forms the hydride PH3.

Correct answer is B. 1 mark for identifying the jump between the 5th and 6th ionization energies, identifying phosphorus as the element, and selecting PH3.

Let the hydrocarbon be C_xH_y. The volume of CO2 produced is absorbed by NaOH, so V(CO2) = 40 cm^3. Since 20 cm^3 of hydrocarbon was burned, x = 40 / 20 = 2. The decrease in volume is given by V(hydrocarbon) + V(O2 reacted) - V(CO2 produced) = 30 cm^3. Thus, 20 + 20(x + y/4) - 40 = 30. Substituting x = 2 gives 20 + 40 + 5y - 40 = 30, which simplifies to 20 + 5y = 30, so y = 2. The molecular formula is C2H2.

Correct answer is C. 1 mark for determining that the hydrocarbon contains 2 carbons and 2 hydrogens, leading to the molecular formula C2H2.

BF3 has a trigonal planar shape because the central boron atom has three bonding pairs of electrons and no lone pairs in its outer shell. This results in a symmetrical geometry with bond angles of exactly 120 degrees.

Correct answer is C. 1 mark for identifying that BF3 has a trigonal planar geometry with a bond angle of 120 degrees.

A propagation step must have a radical and a non-radical as reactants, and produce a new radical and a new non-radical as products. The reaction of ethane with a chlorine radical to form an ethyl radical and hydrogen chloride represents this process.

Correct answer is C. 1 mark for selecting the correct propagation equation.

The reaction of 2-methylbut-2-ene with HBr proceeds via electrophilic addition. Protonation of the double bond yields a tertiary carbocation (at C2) as the more stable intermediate compared to the secondary carbocation (at C3), due to the greater electron-releasing inductive effect of the three alkyl groups. Attack of the bromide ion on the tertiary carbocation produces 2-bromo-2-methylbutane.

Correct answer is B. 1 mark for identifying 2-bromo-2-methylbutane as the major product formed via the tertiary carbocation intermediate.

An atom of manganese has the configuration [Ar] 3d^5 4s^2. When forming the Mn^2+ ion, the two 4s electrons are lost first, resulting in the configuration [Ar] 3d^5.

Correct answer is B. 1 mark for identifying Mn^2+ as having the d^5 configuration.

Mass of water lost = 4.93 - 2.41 = 2.52 g. Molar amount of anhydrous MgSO4 = 2.41 / 120.4 = 0.0200 mol. Molar amount of water lost = 2.52 / 18.0 = 0.140 mol. The molar ratio of H2O to MgSO4 is 0.140 / 0.0200 = 7. Thus, x = 7.

Correct answer is C. 1 mark for the calculation of the mole ratio and obtaining x = 7.

Bond polarity is determined by the difference in electronegativity between the two bonded atoms. The electronegativities are F (4.0), O (3.4), N (3.0), C (2.5), and Cl (3.0). The electronegativity difference for C-F is 1.5, which is greater than that of C-Cl (0.5), N-F (1.0), and O-F (0.6). Thus, the C-F bond is the most polar.

Correct answer is A. 1 mark for selecting the C-F bond as having the largest electronegativity difference.

To identify the element, we look for the largest relative jump in successive ionization energies, which indicates the removal of an electron from a shell closer to the nucleus (a new inner shell).

- 1st to 2nd: \(1903 - 1012 = 891\text{ kJ mol}^{-1}\)
- 2nd to 3rd: \(2912 - 1903 = 1009\text{ kJ mol}^{-1}\)
- 3rd to 4th: \(4957 - 2912 = 2045\text{ kJ mol}^{-1}\)
- 4th to 5th: \(6274 - 4957 = 1317\text{ kJ mol}^{-1}\)
- 5th to 6th: \(21269 - 6274 = 14995\text{ kJ mol}^{-1}\) (a very large increase, more than three-fold)

This extremely large jump between the 5th and 6th ionization energies shows that the 6th electron is removed from a shell closer to the nucleus. Therefore, the element has 5 valence electrons in its outer shell.

In Period 3, the element with 5 valence electrons (Group 15 / Group 5) is phosphorus.

1 mark for the correct option B.
- Reject A, C, D.

1. Write the balanced equation for the reaction:
\(\text{MgCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\)

2. Calculate the number of moles of \(\text{HCl}\):
\(n(\text{HCl}) = \text{volume in dm}^3 \times \text{concentration}\)
\(n(\text{HCl}) = \frac{50.0}{1000} \times 2.00 = 0.100\text{ mol}\)

3. Determine the moles of \(\text{CO}_2\) produced:
The molar ratio of \(\text{HCl} : \text{CO}_2\) is \(2 : 1\).
Since \(\text{MgCO}_3\) is in excess, \(\text{HCl}\) is the limiting reactant.
\(n(\text{CO}_2) = \frac{0.100}{2} = 0.0500\text{ mol}\)

4. Calculate the volume of \(\text{CO}_2\) gas at rtp:
\(V(\text{CO}_2) = n(\text{CO}_2) \times 24.0\text{ dm}^3\text{ mol}^{-1}\)
\(V(\text{CO}_2) = 0.0500 \times 24.0 = 1.20\text{ dm}^3\)

Therefore, the correct option is B.

1 mark for the correct option B.
- Reject A (calculated with incorrect mole ratio or concentration error)
- Reject C (forgot to divide moles of HCl by 2)
- Reject D (multiplied moles of HCl by 2)

- \(\text{BCl}_3\): Boron has 3 valence electrons and forms 3 bonding pairs with chlorine atoms, with no lone pairs. According to electron-pair repulsion theory, 3 bonding pairs repel equally to minimize repulsion, resulting in a **trigonal planar** shape with bond angles of \(120^\circ\).
- \(\text{NH}_4^+\): Nitrogen in the ammonium ion has 4 bonding pairs and no lone pairs around it. These 4 bonding pairs repel equally, resulting in a **tetrahedral** shape with bond angles of \(109.5^\circ\).

Thus, the correct option is B.

1 mark for the correct option B.
- Reject A, C, D.

1. **Determine the priority at C2:**
- The two groups attached to C2 are a methyl group (\(\text{-CH}_3\)) and a hydrogen atom (\(\text{-H}\)).
- Carbon (atomic number 6) has a higher priority than hydrogen (atomic number 1).
- High priority group at C2 = **methyl group**.

2. **Determine the priority at C3:**
- The two groups attached to C3 are an ethyl group (\(\text{-CH}_2\text{CH}_3\)) and a propyl group (\(\text{-CH}_2\text{CH}_2\text{CH}_3\)).
- First atom of both groups is Carbon (\(\text{-CH}_2-\)).
- For ethyl, the second carbon is bonded to (H, H, H).
- For propyl, the second carbon is bonded to (C, H, H).
- Since Carbon has a higher atomic number than Hydrogen, the propyl group has a higher priority than the ethyl group.
- High priority group at C3 = **propyl group**.

3. **Determine E/Z assignment:**
- The question states that the methyl group (high priority at C2) and the propyl group (high priority at C3) are on **opposite sides** of the double bond.
- Since the two highest-priority groups are on opposite sides, the stereoisomer is designated as **(E)**.

4. **Determine IUPAC name:**
- The longest carbon chain containing the double bond has 6 carbons, making the parent chain **hex-2-ene**.
- There is an ethyl substituent at carbon 3.
- Thus, the name is **(E)-3-ethylhex-2-ene**.

Therefore, the correct option is A.

1 mark for the correct option A.
- Reject B (incorrect Z assignment)
- Reject C and D (incorrect parent chain selection, violating the longest chain rule containing the double bond)

(a) The large jump between the 3rd and 4th ionization energies indicates that the fourth electron is being removed from an inner quantum shell. Therefore, the element has 3 valence electrons and is in Group 13 (Group 3). Since it is in Period 3, the element is aluminium (Al). (b) The electronic configuration of an aluminium atom is \(1s^2 2s^2 2p^6 3s^2 3p^1\). Removing two electrons gives the \(Al^{2+}\) ion: \(1s^2 2s^2 2p^6 3s^1\). (c) The fourth electron is removed from the second principal quantum shell (2p subshell), which is closer to the nucleus and experiences significantly less shielding than the third shell, resulting in a much stronger electrostatic attraction from the nucleus.

M1: Identifies element X as aluminium / Al (1) M2: Gives correct electronic configuration of \(Al^{2+}\) as \(1s^2 2s^2 2p^6 3s^1\) (1) M3: Explains that the fourth electron is removed from an inner shell / shell closer to the nucleus (2nd shell instead of 3rd shell) (1) M4: States that this inner shell experiences less shielding / stronger nuclear attraction (1)

First, calculate the moles of carbon in the CO2 produced: \(n(CO_2) = \frac{2.64}{44.0} = 0.0600\text{ mol}\), which means \(n(C) = 0.0600\text{ mol}\). The mass of C is \(0.0600 \times 12.0 = 0.720\text{ g}\). Next, calculate the moles of hydrogen in the H2O produced: \(n(H_2O) = \frac{1.62}{18.0} = 0.0900\text{ mol}\), which means \(n(H) = 2 \times 0.0900 = 0.180\text{ mol}\). The mass of H is \(0.180 \times 1.0 = 0.180\text{ g}\). Find the mass of oxygen in compound Y by subtraction: \(\text{Mass of O} = 1.38 - (0.720 + 0.180) = 0.480\text{ g}\). Convert mass of O to moles: \(n(O) = \frac{0.480}{16.0} = 0.0300\text{ mol}\). Find the simplest whole-number molar ratio: C : H : O = 0.0600 : 0.180 : 0.0300 = 2 : 6 : 1. The empirical formula of Y is \(C_2H_6O\).

M1: Calculates moles/mass of C (0.0600 mol / 0.720 g) AND moles/mass of H (0.180 mol / 0.180 g) (1) M2: Calculates mass of O by subtraction (0.480 g) and converts to moles of O (0.0300 mol) (1) M3: Divides by the smallest mole value to find the ratio C : H : O = 2 : 6 : 1 (1) M4: Deduces the correct empirical formula \(C_2H_6O\) (1)

Ammonia, \(NH_3\), has 3 bonding pairs of electrons and 1 lone pair around the central nitrogen atom. The repulsion from the lone pair is stronger than that from bonding pairs, pushing the N-H bonds closer together to form a trigonal pyramidal shape with a bond angle of \(107^\circ\) (accept \(106.5^\circ - 107.5^\circ\)). The ammonium ion, \(NH_4^+\), has 4 bonding pairs of electrons and no lone pairs around the central nitrogen atom. These 4 bonding pairs repel each other equally, resulting in a regular tetrahedral shape with a bond angle of \(109.5^\circ\).

M1: States the shape and bond angle of \(NH_3\) as trigonal pyramidal and \(107^\circ\) (1) M2: States the shape and bond angle of \(NH_4^+\) as tetrahedral and \(109.5^\circ\) (1) M3: Explains that \(NH_3\) contains 3 bonding pairs and 1 lone pair, whereas \(NH_4^+\) contains 4 bonding pairs (and no lone pairs) (1) M4: Explains that lone pairs repel more strongly than bonding pairs, which decreases the bond angle in \(NH_3\) (1)

(a) Initiation: Under UV light, the Cl-Cl bond undergoes homolytic fission: \(Cl_2 \xrightarrow{UV} 2Cl^\bullet\). Propagation 1: A chlorine radical reacts with methane: \(CH_4 + Cl^\bullet \rightarrow ^\bullet CH_3 + HCl\). Propagation 2: The methyl radical reacts with a chlorine molecule: \(^\bullet CH_3 + Cl_2 \rightarrow CH_3Cl + Cl^\bullet\). (b) Ethane is formed during a termination step when two methyl radicals collide and combine: \(^\bullet CH_3 + ^\bullet CH_3 \rightarrow C_2H_6\).

M1: Correct equation for initiation step with UV indicated (1) M2: Correct equation for propagation step 1 showing radical on C in the methyl radical (1) M3: Correct equation for propagation step 2 showing regeneration of chlorine radical (1) M4: Explains that ethane is formed via the termination step where two methyl radicals combine (including equation \(2^\bullet CH_3 \rightarrow C_2H_6\)) (1)

(a) The major product is 2-bromopropane. (b) The mechanism is electrophilic addition: Propene \(CH_3CH=CH_2\) reacts with \(H^{\delta+}-Br^{\delta-}\). A curly arrow goes from the double bond to the H atom, and another curly arrow goes from the H-Br bond to the Br atom. This forms a secondary carbocation intermediate \(CH_3CH^+CH_3\) and a bromide ion \(Br^-\). A curly arrow then goes from the lone pair on the bromide ion to the positively charged carbon. (c) This is the major product because the secondary carbocation intermediate is more stable than the alternative primary carbocation intermediate \(CH_3CH_2CH_2^+\). This is due to the electron-donating inductive effect of two methyl groups, which disperses the positive charge more effectively than one propyl group.

M1: Identifies 2-bromopropane as the major product (1) M2: Draws curly arrow from double bond to H and from H-Br bond to Br, showing correct dipoles (1) M3: Draws the structure of the secondary carbocation intermediate and curly arrow from the lone pair of the bromide ion to the C+ (1) M4: Explains that the secondary carbocation intermediate is more stable than the primary carbocation due to the electron-donating inductive effect of two alkyl groups (1)

Convert all values to SI units: \(P = 101\text{ kPa} = 101000\text{ Pa}\), \(T = 97.0 + 273.15 = 370.15\text{ K}\) (or \(370\text{ K}\)), \(V = 78.5\text{ cm}^3 = 78.5 \times 10^{-6}\text{ m}^3\). Use the ideal gas equation: \(PV = nRT \Rightarrow n = \frac{PV}{RT}\). \(n = \frac{101000 \times 78.5 \times 10^{-6}}{8.31 \times 370.15} = \frac{7.9285}{3075.95} = 2.5775 \times 10^{-3}\text{ mol}\). Find the molar mass: \(M = \frac{\text{mass}}{n} = \frac{0.231}{2.5775 \times 10^{-3}} = 89.62\text{ g mol}^{-1}\). Rounding to 3 significant figures gives 89.6.

M1: Converts temperature to Kelvin (370 K or 370.15 K) AND volume to \(\text{m}^3\) (\(7.85 \times 10^{-5}\text{ m}^3\)) (1) M2: Rearranges ideal gas equation correctly to make n or M the subject (1) M3: Calculates moles correctly as \(2.58 \times 10^{-3}\text{ mol}\) (1) M4: Calculates molar mass as 89.6 (accept 89.5 to 89.8) to 3 significant figures (1)

Sodium has metallic bonding, consisting of a giant metallic lattice of positive sodium ions surrounded by a sea of delocalised valence electrons. These delocalised electrons are free to move and carry charge through the structure in both the solid and liquid states. Sodium chloride has a giant ionic lattice consisting of alternating sodium ions (\(Na^+\)) and chloride ions (\(Cl^-\)) held together by strong electrostatic attractions. In the solid state, these ions are held in fixed positions within the lattice and cannot move, so solid sodium chloride does not conduct electricity. When molten or in aqueous solution, the giant lattice is broken down, and the ions are free to move and carry the electric current.

M1: States that sodium has metallic bonding with delocalised electrons (1) M2: Explains that these delocalised electrons are free to move and carry charge in both solid and liquid states (1) M3: States that sodium chloride has a giant ionic lattice with ions in fixed positions in the solid, so it cannot conduct (1) M4: Explains that when molten or aqueous, the ionic bonds are broken/loosened, allowing the ions to move and carry charge (1)

Stereoisomerism occurs when molecules have the same molecular and structural formula but a different arrangement of their atoms in three-dimensional space. In alkenes, stereoisomerism (specifically E/Z isomerism) arises due to the restricted rotation about the carbon-carbon double bond, caused by the overlap of p-orbitals forming the pi bond. For stereoisomerism to occur, both carbon atoms involved in the C=C double bond must be attached to two different atoms or groups of atoms. In but-2-ene, each double-bonded carbon is attached to a hydrogen atom and a methyl group, so it exists as E-but-2-ene and Z-but-2-ene. In but-1-ene, the first carbon (C1) is bonded to two identical hydrogen atoms, meaning it cannot show stereoisomerism.

M1: Defines stereoisomerism as molecules with the same structural formula but different arrangement of atoms in space (1) M2: Identifies that restricted rotation around the C=C double bond (due to the pi bond) is required (1) M3: Explains that each carbon of the C=C bond must be attached to two different groups (1) M4: States that but-1-ene fails this condition because C1 has two identical hydrogen atoms, while but-2-ene has two different groups on both carbons (1)

1. Calculate the mass of water lost:
\( 4.76\text{ g} - 2.60\text{ g} = 2.16\text{ g} \)

2. Calculate the molar mass of anhydrous cobalt(II) chloride, \( \text{CoCl}_2 \):
\( M_{\text{r}}(\text{CoCl}_2) = 58.9 + (2 \times 35.5) = 129.9\text{ g mol}^{-1} \)

3. Calculate the number of moles of \( \text{CoCl}_2 \):
\( n(\text{CoCl}_2) = \frac{2.60\text{ g}}{129.9\text{ g mol}^{-1}} = 0.0200\text{ mol} \)

4. Calculate the number of moles of water, \( \text{H}_2\text{O} \):
\( n(\text{H}_2\text{O}) = \frac{2.16\text{ g}}{18.0\text{ g mol}^{-1}} = 0.120\text{ mol} \)

5. Determine the simplest molar ratio:
\( \text{Ratio} = \frac{0.120}{0.0200} = 6.00 \)

Therefore, \( x = 6 \).

- Mark 1: Calculating the mass of water lost (\( 2.16\text{ g} \)) and molar mass of \( \text{CoCl}_2 \) (\( 129.9\text{ g mol}^{-1} \)).
- Mark 2: Calculating moles of \( \text{CoCl}_2 \) as \( 0.0200\text{ mol} \) (accept 0.02).
- Mark 3: Calculating moles of \( \text{H}_2\text{O} \) as \( 0.120\text{ mol} \) (accept 0.12).
- Mark 4: Deducing the mole ratio of 1 : 6 and stating \( x = 6 \).

1. Around the central boron atom in \( \text{BF}_3 \), there are three bonding pairs of electrons and zero lone pairs. To minimize repulsion, these electron pairs arrange themselves as far apart as possible, resulting in a trigonal planar shape with a bond angle of \( 120^\circ \).

2. Around the central nitrogen atom in \( \text{NF}_3 \), there are three bonding pairs of electrons and one lone pair. The total of four electron pairs arrange themselves in a tetrahedral arrangement, but because we only see the atoms, the molecular shape is trigonal pyramidal.

3. Lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, which pushes the \( \text{N-F} \) bonds closer together, reducing the bond angle from the tetrahedral \( 109.5^\circ \) to approximately \( 107^\circ \).

- Mark 1: Stating \( \text{BF}_3 \) shape is trigonal planar and bond angle is \( 120^\circ \) with explanation of 3 bonding pairs and 0 lone pairs.
- Mark 2: Stating \( \text{NF}_3 \) shape is trigonal pyramidal and bond angle is \( 107^\circ \) (accept \( 106^\circ - 108^\circ \)).
- Mark 3: Stating \( \text{NF}_3 \) has 3 bonding pairs and 1 lone pair around the N atom.
- Mark 4: Explaining that lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, which reduces the bond angle.

(a) The initiation step involves the homolytic fission of the bromine-bromine bond by UV light to form two bromine radicals:
\( \text{Br}_2 \rightarrow 2\text{Br}^\bullet \)

(b) The two propagation steps are:
1. \( \text{C}_2\text{H}_6 + \text{Br}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HBr} \)
2. \( \text{C}_2\text{H}_5^\bullet + \text{Br}_2 \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{Br}^\bullet \)

(c) A trace of butane is formed in a termination step when two ethyl radicals combine:
\( 2\text{C}_2\text{H}_5^\bullet \rightarrow \text{C}_4\text{H}_{10} \)

- Mark 1: Correct initiation equation showing \( \text{Br}_2 \rightarrow 2\text{Br}^\bullet \) (dots indicating unpaired electrons are required).
- Mark 2: Correct first propagation step forming \( \text{C}_2\text{H}_5^\bullet \) and \( \text{HBr} \).
- Mark 3: Correct second propagation step forming \( \text{C}_2\text{H}_5\text{Br} \) and regeneration of \( \text{Br}^\bullet \).
- Mark 4: Correct termination equation showing combination of two ethyl radicals to form butane.

1. The addition of \( \text{HBr} \) to propene proceeds via a carbocation intermediate.

2. The reaction to form 2-bromopropane goes via a secondary carbocation intermediate (\( \text{CH}_3\text{C}^+\text{HCH}_3 \)), whereas the reaction to form 1-bromopropane goes via a primary carbocation intermediate (\( \text{CH}_3\text{CH}_2\text{C}^+\text{H}_2 \)).

3. Secondary carbocations are more stable than primary carbocations.

4. This is because the two electron-releasing (inductive effect) alkyl (methyl) groups in the secondary carbocation help to spread and stabilize the positive charge more effectively than the single ethyl group in the primary carbocation.

- Mark 1: Identifying that the reaction proceeds via a carbocation intermediate.
- Mark 2: Correctly identifying the secondary carbocation intermediate for 2-bromopropane and the primary carbocation intermediate for 1-bromopropane.
- Mark 3: Stating that the secondary carbocation is more stable than the primary carbocation.
- Mark 4: Explaining the stability in terms of the positive inductive effect / electron-donating effect of two alkyl groups versus one.

(a) Element X is aluminium (\( \text{Al} \)).
Justification: There is a large increase/jump between the third and fourth ionization energies (from \( 2745 \) to \( 11578\text{ kJ mol}^{-1} \)). This indicates that the fourth electron is removed from an inner shell closer to the nucleus, meaning X has 3 valence electrons and is in Group 3 (Group 13).

(b) The equation for the third ionization energy of aluminium is:
\( \text{Al}^{2+}\text{(g)} \rightarrow \text{Al}^{3+}\text{(g)} + \text{e}^- \)

(c) The electronic configuration of Mg is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 \), and Al is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^1 \). The outer electron of aluminium is in a \( 3\text{p} \) subshell, which is higher in energy and further from the nucleus than the \( 3\text{s} \) subshell of magnesium. This \( 3\text{p} \) electron also experiences additional shielding from the \( 3\text{s}^2 \) electrons, making it easier to remove.

- Mark 1: Identifying X as aluminium (\( \text{Al} \)) with the justification of a large jump between the 3rd and 4th ionization energies.
- Mark 2: Equation for the 3rd ionization energy of \( \text{Al} \) with correct state symbols: \( \text{Al}^{2+}\text{(g)} \rightarrow \text{Al}^{3+}\text{(g)} + \text{e}^- \) (accept \( \text{X} \) in place of \( \text{Al} \)).
- Mark 3: Stating that the outer electron of Al is in a \( 3\text{p} \) orbital/subshell, whereas the outer electron of Mg is in a \( 3\text{s} \) orbital/subshell.
- Mark 4: Stating that the \( 3\text{p} \) orbital is higher in energy / further from the nucleus and experiences shielding by the inner \( 3\text{s} \) electrons.

(a) The equation for the standard enthalpy of formation of pentane is:
\( 5\text{C(s)} + 6\text{H}_2\text{(g)} \rightarrow \text{C}_5\text{H}_{12}\text{(l)} \)

(b) Using the Hess's Law cycle where both reactants and products are combusted to form \( 5\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \):
\( \Delta_{\text{f}} H^\ominus = \sum \Delta_{\text{c}} H^\ominus(\text{reactants}) - \sum \Delta_{\text{c}} H^\ominus(\text{products}) \)

\( \Delta_{\text{f}} H^\ominus = [5 \times \Delta_{\text{c}} H^\ominus(\text{C(s)}) + 6 \times \Delta_{\text{c}} H^\ominus(\text{H}_2\text{(g)})] - \Delta_{\text{c}} H^\ominus(\text{C}_5\text{H}_{12}\text{(l)}) \)

\( \Delta_{\text{f}} H^\ominus = [5 \times (-394) + 6 \times (-286)] - (-3509) \)

\( \Delta_{\text{f}} H^\ominus = [-1970 - 1716] + 3509 \)

\( \Delta_{\text{f}} H^\ominus = -3686 + 3509 = -177\text{ kJ mol}^{-1} \)

- Mark 1: Correct equation for standard enthalpy of formation of pentane including state symbols: \( 5\text{C(s)} + 6\text{H}_2\text{(g)} \rightarrow \text{C}_5\text{H}_{12}\text{(l)} \).
- Mark 2: Designing a valid Hess's Law cycle or expressing the enthalpy change relation: \( \Delta_{\text{f}} H = 5 \times \Delta_{\text{c}} H(\text{C}) + 6 \times \Delta_{\text{c}} H(\text{H}_2) - \Delta_{\text{c}} H(\text{pentane}) \).
- Mark 3: Correctly substituting the values: \( [5 \times (-394) + 6 \times (-286)] - (-3509) \).
- Mark 4: Calculating final answer of \( -177\text{ kJ mol}^{-1} \) (must have a minus sign and units).

The equation for the formation of butane is: \(4\text{C(s)} + 5\text{H}_2\text{(g)} \rightarrow \text{C}_4\text{H}_{10}\text{(g)}\). Using a Hess's cycle with enthalpies of combustion: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\). Therefore, \(\Delta_f H^\theta = [4 \times (-393.5)] + [5 \times (-285.8)] - (-2876.5) = -1574.0 - 1429.0 + 2876.5 = -126.5\text{ kJ mol}^{-1}\).

1 mark: Correct calculation of standard enthalpy of formation showing working and correct sign (-126.5).

Propan-1-ol (option C) has the highest boiling temperature because it contains an \(-\text{OH}\) group which allows it to form strong hydrogen bonds between its molecules. The other molecules only have weaker London forces (options A and B) or permanent dipole-dipole forces (option D), which require less thermal energy to overcome.

1 mark: Correct compound selected (C).

As you descend Group 2, the ionic radius of the \(M^{2+}\) cation increases while the charge remains the same. This decreases the charge density and therefore the polarizing power of the cation. Consequently, the nitrate ion is polarized (distorted) less, making the compound more thermally stable.

1 mark: Correctly identifies that the thermal stability increases down the group because the larger cation has less polarizing power, polarizing the nitrate ion less (A).

Iodide ions are strong reducing agents and reduce the sulfur in sulfuric acid (oxidation state +6) to sulfur dioxide (\(\text{SO}_2\), oxidation state +4), sulfur (\(\text{S}\), oxidation state 0), and hydrogen sulfide (\(\text{H}_2\text{S}\), oxidation state -2). These are all reduction products of sulfuric acid. Iodine (\(\text{I}_2\)) is an oxidation product of iodide, and hydrogen iodide (\(\text{HI}\)) is formed in an acid-base reaction without any change in oxidation states.

1 mark: Correctly identifies the list containing only reduction products of sulfuric acid (A).

At a higher temperature, the average kinetic energy of the molecules increases, shifting the peak of the Maxwell-Boltzmann distribution curve to the right. Since the total number of molecules remains constant, the area under the curve must remain constant, causing the peak of the curve to become lower.

1 mark: Correctly identifies that the peak shifts to the right and is lower, and the area remains constant (A).

The rate of hydrolysis is determined by the strength of the carbon-halogen bond. The \(\text{C}-\text{I}\) bond is the weakest (lowest bond enthalpy) among the halogenoalkanes, so 1-iodobutane reacts the fastest. Bond polarity would suggest 1-chlorobutane is fastest, but bond enthalpy is the dominant factor here.

1 mark: Correct halogenoalkane and correct explanation (A).

The peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of broad peaks in the region \(3200 - 3750\text{ cm}^{-1}\) (alcohol \(\text{O}-\text{H}\)) and \(2500 - 3300\text{ cm}^{-1}\) (carboxylic acid \(\text{O}-\text{H}\)) indicates that there are no hydroxyl groups. Thus, the compound must be a ketone (propanone).

1 mark: Correctly identifies propanone as the only compound containing a carbonyl group but no hydroxyl group (C).

Only a change in temperature can change the value of the equilibrium constant, \(K_c\). Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature shifts the equilibrium position to the right to absorb the added thermal energy, which increases the concentrations of products and decreases the concentration of reactant, thereby increasing \(K_c\).

1 mark: Correctly identifies that increasing the temperature of an endothermic reaction increases \(K_c\) (A).

The equation for the standard enthalpy of formation of propan-1-ol is:
\(3\text{C}(s, \text{graphite}) + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_3\text{H}_7\text{OH}(l)\)

Using Hess's Law and standard enthalpies of combustion:
\(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\)
\(\Delta_f H^\ominus = [3 \times (-394) + 4 \times (-286)] - [-2021]\)
\(\Delta_f H^\ominus = [-1182 - 1144] + 2021\)
\(\Delta_f H^\ominus = -2326 + 2021 = -305\text{ kJ mol}^{-1}\)

1 mark for the correct answer (A).

Propan-1-ol contains a highly polar \(\text{O}-\text{H}\) bond which enables hydrogen bonding between its molecules. Hydrogen bonding is significantly stronger than the permanent dipole-dipole forces present in propanal and 1-chloropropane, and the London forces in butane. Since they all have similar numbers of electrons and molar masses, the stronger intermolecular forces in propan-1-ol result in the highest boiling temperature.

1 mark for the correct answer (A).

Going down Group 2, the cationic radius increases, so the magnesium ion is smaller than the barium ion. As a result, the \(\text{Mg}^{2+}\) ion has a higher charge density and is more polarising. It polarises the large electron cloud of the carbonate ion more strongly, weakening the covalent carbon-oxygen bonds within the carbonate group, which makes thermal decomposition occur at a lower temperature.

1 mark for the correct answer (A).

The reaction of solid \(\text{KI}\) with concentrated \(\text{H}_2\text{SO}_4\) initially produces \(\text{HI}\) and \(\text{KHSO}_4\) via an acid-base (non-redox) reaction. Due to the strong reducing power of the iodide ion, the \(\text{HI}\) is further oxidised to \(\text{I}_2\), while the sulfur in \(\text{H}_2\text{SO}_4\) is reduced to \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\). Therefore, \(\text{I}_2\), \(\text{H}_2\text{S}\) and \(\text{SO}_2\) are all redox products, whereas \(\text{KHSO}_4\) and \(\text{HI}\) are acid-base products.

1 mark for the correct answer (B).

As the temperature increases, molecules gain kinetic energy, shifting the peak of the distribution to the right (higher energy). Because the total number of molecules remains constant, the area under the curve is unchanged. Therefore, to accommodate the wider spread of energies, the peak height must decrease.

1 mark for the correct answer (A).

Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium position to the right (in the exothermic direction) to oppose the decrease in temperature. This increases the concentration of \(\text{SO}_3\) at equilibrium and decreases the concentrations of the reactants, resulting in a larger value of \(K_c\). Although increasing pressure increases the yield of \(\text{SO}_3\), it does not change \(K_c\) because only temperature changes the equilibrium constant.

1 mark for the correct answer (A).

The rate of hydrolysis of halogenoalkanes increases down Group 7 because the carbon-halogen bond strength decreases (the C-I bond is the weakest and easiest to break). Therefore, 1-iodobutane reacts the fastest. The iodide ions (\(\text{I}^-\)) released react with silver ions (\(\text{Ag}^+\)) to form a yellow precipitate of silver iodide (\(\text{AgI}\)).

1 mark for the correct answer (A).

The infrared spectrum of compound \(\mathbf{Y}\) indicates the presence of a carboxylic acid, characterised by the broad \(\text{O}-\text{H}\) absorption (\(2500 - 3300\text{ cm}^{-1}\)) and the carbonyl \(\text{C}=\text{O}\) absorption (\(1710\text{ cm}^{-1}\)). A carboxylic acid is formed by the complete oxidation of a primary alcohol under reflux. Among the options, only propan-1-ol is a primary alcohol. Propan-2-ol is secondary (oxidises to a ketone), 2-methylpropan-2-ol is tertiary (resistant to oxidation), and propanone is already a ketone.

1 mark for the correct answer (A).

The equation for the formation of propan-1-ol is: \(3\text{C(s)} + 4\text{H}_2\text{(g)} + 0.5\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}\). Using Hess's Law with standard combustion enthalpies: \(\Delta H^\ominus_\text{f} = \sum \Delta H^\ominus_\text{c}(\text{reactants}) - \sum \Delta H^\ominus_\text{c}(\text{products}) = [3 \times (-394) + 4 \times (-286)] - [-2021] = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\).

1 mark: Correctly identifies and calculates the enthalpy change of formation as -305 kJ mol^-1.

The rate of hydrolysis of halogenoalkanes is determined by the strength (bond enthalpy) of the carbon-halogen bond, not the bond polarity. Since the bond enthalpy decreases down Group 7 (C-F > C-Cl > C-Br > C-I), the C-I bond is the weakest and requires the least energy to break, resulting in 1-iodobutane reacting the fastest.

1 mark: Identifies 1-iodobutane as the fastest-reacting species due to having the weakest carbon-halogen bond.

The initial reaction between NaBr and concentrated sulfuric acid produces HBr, observed as steamy acidic fumes. Bromide ions are sufficiently strong reducing agents to reduce sulfuric acid to sulfur dioxide (SO2), and are themselves oxidized to bromine (Br2), which is observed as brown fumes. Bromide is not strong enough to reduce sulfur further to sulfur (yellow solid) or hydrogen sulfide (rotten egg smell).

1 mark: Correctly identifies the observations (steamy acidic fumes and brown vapour) and the matching reduction product (SO2).

Propan-1-ol has hydrogen bonding which is the strongest type of intermolecular force, giving it the highest boiling temperature. Propanal experiences permanent dipole-dipole forces which are stronger than the London forces present in butane. Butane only has London forces, giving it the lowest boiling temperature. Therefore, the decreasing order is propan-1-ol > propanal > butane.

1 mark: Correctly identifies the order of decreasing boiling temperatures based on the strength of intermolecular forces.

Both butane-1-ol and butane-1,4-diol experience London forces, permanent dipole-dipole forces, and hydrogen bonding. However, butane-1,4-diol contains two hydroxyl (\(-\text{OH}\)) groups, whereas butane-1-ol contains only one. Consequently, butane-1,4-diol can form more (up to twice as many) hydrogen bonds per molecule than butane-1-ol. This extensive network of hydrogen bonding creates much stronger intermolecular attractions, requiring significantly more kinetic (thermal) energy to separate the molecules and transition from liquid to gas.

M1: Identifies that both molecules experience hydrogen bonding as their dominant intermolecular force.
M2: Explains that butane-1,4-diol has two hydroxyl groups per molecule whereas butane-1-ol has only one, allowing butane-1,4-diol to form more/twice as many hydrogen bonds per molecule.
M3: Relates this directly to the requirement of more energy to overcome these stronger intermolecular forces, leading to a higher boiling temperature.

This is a nucleophilic substitution reaction, specifically following the \(\text{S}_\text{N}2\) mechanism because 1-iodobutane is a primary halogenoalkane. In the rate-determining step, the nucleophile (the hydroxide ion, \(\text{OH}^-\)) attacks the electron-deficient carbon atom. This is shown by a curly arrow from a lone pair of electrons on the oxygen of the \(\text{OH}^-\) ion to the carbon attached to the iodine (\(\text{C}^{\delta+}\)). Simultaneously, the carbon-iodine bond breaks heterolytically, represented by a curly arrow from the C–I bond to the iodine atom.

M1: Correctly identifies the reaction type and mechanism as nucleophilic substitution (or \(\text{S}_\text{N}2\)).
M2: Describes the curly arrow starting from the lone pair of the oxygen on the hydroxide ion to the carbon bonded to iodine (\(\text{C}^{\delta+}\)).
M3: Describes the curly arrow starting from the C–I bond to the iodine atom, representing heterolytic bond fission.

The equation for the formation of ethyl ethanoate is:
\(4\text{C}(\text{s}) + 4\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3(\text{l})\)

According to Hess's Law, using combustion data:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)

\(\Delta_f H^\theta = [4 \times \Delta_c H^\theta(\text{C}) + 4 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{ethyl ethanoate})\)
\(\Delta_f H^\theta = [4 \times (-394) + 4 \times (-286)] - (-2238)\)
\(\Delta_f H^\theta = [-1576 - 1144] + 2238\)
\(\Delta_f H^\theta = -2720 + 2238 = -482\text{ kJ mol}^{-1}\)

M1: Writes the balanced equation for the formation of ethyl ethanoate, or shows a correct thermodynamic cycle linking formation with combustion products.
M2: Correctly sets up the Hess's Law calculation: \([4 \times (-394) + 4 \times (-286)] - (-2238)\).
M3: Calculates the correct final value of \(-482\text{ kJ mol}^{-1}\) with the negative sign and appropriate unit.

The thermal stability of Group 2 carbonates increases down the group. As you descend Group 2, the ionic radius of the \(\text{M}^{2+}\) cation increases while its charge remains constant at \(2+\). This leads to a decrease in charge density of the cation. Consequently, the larger cation has a weaker polarizing effect on the carbonate ion (\(\text{CO}_3^{2-}\)), causing less distortion of its electron cloud. Because the carbon-oxygen bonds in the carbonate ion are weakened to a lesser extent, more thermal energy is needed to decompose the carbonate into the metal oxide and carbon dioxide.

M1: States that thermal stability increases down the group and identifies that the ionic radius of the cation increases while charge remains constant (causing a decrease in charge density).
M2: Explains that the cation has a weaker polarizing effect (or causes less distortion) on the carbonate ion's electron cloud.
M3: Concludes that the C–O bonds are weakened less, meaning more energy/higher temperatures are required to break them.

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (\(E_{\text{cat}} < E_a\)). In a Maxwell-Boltzmann distribution, the x-axis represents kinetic energy, and the activation energy is marked as a threshold. When a catalyst is introduced, the threshold shifts to the left (to a lower energy value). As a result, the area under the curve to the right of this threshold increases, meaning a significantly greater fraction of the reacting molecules possess energy equal to or greater than the activation energy. This leads to a higher frequency of successful collisions per unit time.

M1: Mentions that a catalyst provides an alternative pathway with a lower activation energy.
M2: Relates this to the Maxwell-Boltzmann distribution, stating that the activation energy threshold shifts to the left.
M3: Explains that a larger area under the curve (greater fraction of molecules) now has energy \(\ge\) the catalysed activation energy, leading to more frequent successful collisions.

The strong, sharp absorption band at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group (\(\text{C=O}\) stretch). The lack of any broad absorption band in the region \(3200-3600\text{ cm}^{-1}\) confirms the absence of a hydroxyl group (\(\text{O-H}\) stretch), meaning **X** cannot be an alcohol. Given the molecular formula \(\text{C}_3\text{H}_6\text{O}\), which has one double bond equivalent, and containing a carbonyl group with no hydroxyl group, **X** must be the ketone propanone (or the aldehyde propanal; however, a ketone's carbonyl stretch is typically closer to \(1715\text{ cm}^{-1}\)).

M1: Identifies the carbonyl group (\(\text{C=O}\)) from the peak at \(1715\text{ cm}^{-1}\).
M2: Identifies that the absence of a broad band at \(3200-3600\text{ cm}^{-1}\) indicates no \(\text{O-H}\) group is present (rules out alcohols).
M3: Deduces the IUPAC name as propanone (accept propanal) based on the formula \(\text{C}_3\text{H}_6\text{O}\) and the functional groups identified.

The balanced ionic equation for the reaction of chlorine with cold, dilute sodium hydroxide is:
\(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\)

To show this is a disproportionation reaction, we examine the oxidation numbers of chlorine:
- In reactant \(\text{Cl}_2\), the oxidation state of chlorine is \(0\).
- In the chloride product (\(\text{Cl}^-\)), the oxidation state of chlorine is \(-1\) (decrease in oxidation number; reduction).
- In the chlorate(I) product (\(\text{ClO}^-\)), the oxidation state of chlorine is \(+1\) (increase in oxidation number; oxidation).

Since the same chlorine species is simultaneously oxidized and reduced, the reaction is a disproportionation.

M1: Provides the correct balanced ionic equation: \(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\).
M2: States correct oxidation numbers: \(0\) for reactant \(\text{Cl}_2\), \(-1\) for product \(\text{Cl}^-\), and \(+1\) for product \(\text{ClO}^-\).
M3: Defines disproportionation clearly as the simultaneous oxidation (from \(0\) to \(+1\)) and reduction (from \(0\) to \(-1\)) of the same element in a single reaction.

1. Calculate the heat energy released (\(q\)):
\(q = m \times c \times \Delta T\)
\(m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\)
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 10.5\text{ }^\circ\text{C} = 2194.5\text{ J} = 2.1945\text{ kJ}\)

2. Calculate the amount of moles of copper(II) sulfate (\(n\)) reacting:
\(n = c \times V = 0.200\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0100\text{ mol}\)

3. Calculate the enthalpy change (\(\Delta H\)):
\(\Delta H = -\frac{q}{n} = -\frac{2.1945\text{ kJ}}{0.0100\text{ mol}} = -219.45\text{ kJ mol}^{-1}\)

Rounding to 3 significant figures, \(\Delta H = -219\text{ kJ mol}^{-1}\).

M1: Correctly calculates heat transferred: \(q = 50.0 \times 4.18 \times 10.5 = 2194.5\text{ J}\) (or \(2.19\text{ kJ}\)).
M2: Correctly calculates moles of limiting reactant: \(n = 0.200 \times 0.0500 = 0.0100\text{ mol}\).
M3: Divides heat by moles and applies the correct negative sign for an exothermic reaction to obtain \(-219\text{ kJ mol}^{-1}\) (allow \(-220\text{ kJ mol}^{-1}\) if rounded values were used at earlier stages; negative sign is required).

To calculate the enthalpy change of the reaction, we use the formula:
\(\Delta H = \sum \text{bond enthalpies of reactants} - \sum \text{bond enthalpies of products}\)

1. Calculate the total energy absorbed to break bonds in the reactants:
- 1 mol of \(\text{C}=\text{C}\) bonds = \(1 \times 611 = 611\text{ kJ}\)
- 1 mol of \(\text{C}-\text{C}\) bonds = \(1 \times 347 = 347\text{ kJ}\)
- 6 mol of \(\text{C}-\text{H}\) bonds = \(6 \times 413 = 2478\text{ kJ}\)
- 2 mol of \(\text{O}-\text{H}\) bonds = \(2 \times 463 = 926\text{ kJ}\)
Total energy absorbed = \(611 + 347 + 2478 + 926 = 4362\text{ kJ mol}^{-1}\)

2. Calculate the total energy released when bonds are formed in the products (propan-1-ol):
- 2 mol of \(\text{C}-\text{C}\) bonds = \(2 \times 347 = 694\text{ kJ}\)
- 7 mol of \(\text{C}-\text{H}\) bonds = \(7 \times 413 = 2891\text{ kJ}\)
- 1 mol of \(\text{C}-\text{O}\) bonds = \(1 \times 358 = 358\text{ kJ}\)
- 1 mol of \(\text{O}-\text{H}\) bonds = \(1 \times 463 = 463\text{ kJ}\)
Total energy released = \(694 + 2891 + 358 + 463 = 4406\text{ kJ mol}^{-1}\)

3. Calculate \(\Delta H\):
\(\Delta H = 4362 - 4406 = -44\text{ kJ mol}^{-1}\)

- 1 mark: Calculation of total bond enthalpies of reactants broken = 4362 kJ/mol (or calculation of net bonds broken: 1xC=C + 1xO-H = 1074 kJ/mol).
- 1 mark: Calculation of total bond enthalpies of products formed = 4406 kJ/mol (or calculation of net bonds formed: 1xC-C + 1xC-H + 1xC-O = 1118 kJ/mol).
- 1 mark: Evaluation of \(\Delta H = -44\text{ kJ mol}^{-1}\) (must include the negative sign).

1. Although the chlorine atom is more electronegative than bromine, making the \(\text{C}-\text{Cl}\) bond more polar than the \(\text{C}-\text{Br}\) bond, polarity is not the deciding factor for the rate of nucleophilic substitution here.
2. The rate depends primarily on the strength of the carbon-halogen bond. The \(\text{C}-\text{Br}\) bond is longer and weaker (has a lower bond enthalpy) than the \(\text{C}-\text{Cl}\) bond because bromine has a larger atomic radius.
3. Because the \(\text{C}-\text{Br}\) bond is weaker, it requires less activation energy to break during the rate-determining step, leading to a faster rate of hydrolysis.

- 1 mark: Identifies that the \(\text{C}-\text{Cl}\) bond is more polar than the \(\text{C}-\text{Br}\) bond (or chlorine is more electronegative than bromine).
- 1 mark: States that the \(\text{C}-\text{Br}\) bond is weaker / has a lower bond enthalpy than the \(\text{C}-\text{Cl}\) bond.
- 1 mark: Explains that bond strength/enthalpy is the dominant factor determining the reaction rate (or lowering activation energy), meaning the C-Br bond breaks more easily.

1. Down Group 2, the ionic radius of the metal cation increases, while the overall ionic charge remains \(2+\).
2. This results in a decrease in the charge density of the cation down the group.
3. The smaller magnesium ion has a higher charge density and thus polarizes (distorts) the electron cloud of the carbonate ion more severely than the larger barium ion. This polarization weakens the \(\text{C}-\text{O}\) bond within the carbonate ion, making it easier to decompose upon heating. Therefore, barium carbonate is much more thermally stable and requires more heat energy to decompose than magnesium carbonate.

- 1 mark: States that thermal stability of Group 2 carbonates increases down the group.
- 1 mark: Explains that the cation size/ionic radius increases and charge density of the cation decreases.
- 1 mark: Explains that larger cations cause less polarization/distortion of the carbonate ion, meaning more energy is needed to break the carbon-oxygen bond.

1. Both compounds have the same molecular formula and therefore a similar number of electrons, meaning their London (dispersion) forces are of comparable strength.
2. Propan-1-ol contains a highly polar \(\text{O}-\text{H}\) bond, allowing it to form intermolecular hydrogen bonds between its molecules.
3. Methoxyethane does not contain an \(\text{O}-\text{H}\) bond, so it cannot form hydrogen bonds; instead, it only experiences weaker London forces and permanent dipole-dipole forces. Hydrogen bonds are much stronger and require significantly more thermal energy to overcome, resulting in a much higher boiling point for propan-1-ol.

- 1 mark: Identifies that propan-1-ol can form hydrogen bonds (due to the polar \(\text{O}-\text{H}\) bond).
- 1 mark: Identifies that methoxyethane can only form London forces and permanent dipole-dipole forces (cannot form hydrogen bonds).
- 1 mark: States that hydrogen bonds are stronger than London forces / permanent dipole-dipole forces and require more energy to overcome.

A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy (\(E_a\)). On a Maxwell-Boltzmann distribution curve, the activation energy line for the catalyzed reaction (\(E_{\text{cat}}\)) is shifted to the left (closer to the origin) compared to the uncatalyzed activation energy line (\(E_a\)). Consequently, the area under the curve to the right of the activation energy line—which represents the fraction of molecules with energy greater than or equal to the activation energy—increases significantly. This leads to a greater frequency of successful collisions.

- 1 mark: States that a catalyst provides an alternative pathway with a lower activation energy (\(E_a\)).
- 1 mark: Describes that on a Maxwell-Boltzmann curve, the activation energy line is shifted to the left (lower energy).
- 1 mark: Explains that a larger fraction/proportion of molecules now possess energy equal to or exceeding the new activation energy (\(\ge E_{\text{cat}}\)), leading to more frequent successful collisions.

1. Identification from IR spectrum:
The sharp, strong peak at \(1715\text{ cm}^{-1}\) corresponds to a carbonyl group (\(\text{C}=\text{O}\)). The lack of a broad absorption peak between \(3200-3670\text{ cm}^{-1}\) indicates that there is no alcohol (\(\text{O}-\text{H}\)) group present, ruling out propan-1-ol.

2. Identification from mass spectrum:
The molecular ion peak at \(m/z = 58\) corresponds to the molecular mass of \(\text{C}_3\text{H}_6\text{O}\) (which can be either propanone or propanal). The fragment at \(m/z = 43\) is due to the loss of a methyl group (\(\text{CH}_3\), mass = 15) to form the acylium ion \([\text{CH}_3\text{CO}]^+\) (mass = 43), which is highly stable and characteristic of propanone: \([\text{CH}_3\text{COCH}_3]^+\cdot \rightarrow [\text{CH}_3\text{CO}]^+ + \cdot\text{CH}_3\). Therefore, **X** is propanone.

- 1 mark: Deduces that **X** is propanone.
- 1 mark: Uses IR data to justify the presence of \(\text{C}=\text{O}\) (1715 cm\(^{-1}\)) and the absence of an \(\text{O}-\text{H}\) group (3200-3670 cm\(^{-1}\)).
- 1 mark: Identifies the fragment at \(m/z = 43\) as the acylium ion \([\text{CH}_3\text{CO}]^+\) (formed by loss of \(\text{CH}_3\) from propanone).

1. The purple vapour is iodine (\(\text{I}_2\)), which is formed by the oxidation of iodide ions (\(\text{I}^-\)).
2. The yellow solid is elemental sulfur (\(\text{S}\)), produced from the deep reduction of sulfuric acid.
3. The choking gas is sulfur dioxide (\(\text{SO}_2\)).
4. For the reduction of sulfuric acid (\(\text{H}_2\text{SO}_4\)) to sulfur (\(\text{S}\)), the oxidation state of sulfur changes from +6 to 0. This requires a 6-electron transfer.

The balanced reduction half-equation is:
\(\text{H}_2\text{SO}_4 + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{S} + 4\text{H}_2\text{O}\)
(or ionic equivalent: \(\text{SO}_4^{2-} + 8\text{H}^+ + 6\text{e}^- \rightarrow \text{S} + 4\text{H}_2\text{O}\))

- 1 mark: Correctly identifies all three physical products: purple vapour = \(\text{I}_2\), yellow solid = \(\text{S}\), choking gas = \(\text{SO}_2\) (accept hydrogen sulfide, \(\text{H}_2\text{S}\), as a choking gas/rotten egg gas).
- 1 mark: Correctly balances the species in the half-equation (accept molecular \(\text{H}_2\text{SO}_4\) or ionic \(\text{SO}_4^{2-}\)).
- 1 mark: Correctly balances the charges with electrons (6e\(^-\)) in the reduction half-equation.

1. Determine mass of solution:
\(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)
Since density is \(1.00\text{ g cm}^{-3}\), \(m = 100.0\text{ g}\).

2. Determine temperature change:
\(\Delta T = 27.9 - 21.2 = 6.7\text{ K}\) (or \(6.7^\circ\text{C}\))

3. Calculate heat released (\(q\)):
\(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.8006\text{ kJ}\)

4. Calculate moles of reaction (water formed):
\(n(\text{H}_2\text{O}) = n(\text{H}^+) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)

5. Calculate molar enthalpy of neutralization (\(\Delta H_{\text{neutralization}}\)):
\(\Delta H = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.012\text{ kJ mol}^{-1}\)

To three significant figures, the enthalpy change is \(-56.0\text{ kJ mol}^{-1}\).

- 1 mark: Calculation of heat transferred: \(q = 2.80\text{ kJ}\) (or 2800.6 J).
- 1 mark: Calculation of moles of water produced: \(n = 0.0500\text{ mol}\).
- 1 mark: Final value of \(-56.0\text{ kJ mol}^{-1}\) (must have negative sign, accept values in the range of -56.0 to -56.01).

1. Butane molecules are held together only by weak London forces (instantaneous dipole-induced dipole forces).
2. Propan-1-ol contains a highly polar -OH group, allowing its molecules to form hydrogen bonds in addition to London forces.
3. Hydrogen bonds are significantly stronger than London forces, which means much more thermal energy is needed to separate the propan-1-ol molecules, resulting in a higher boiling temperature.

M1 (1 mark): State that butane has only London forces (accept: dispersion forces / instantaneous dipole-induced dipole forces).
M2 (1 mark): State that propan-1-ol has hydrogen bonding (due to the -OH group).
M3 (1 mark): State that hydrogen bonds are stronger than London forces and require more energy to overcome/break.

1. 1-iodobutane reacts / hydrolyses faster (forming a yellow precipitate of silver iodide more quickly than the white precipitate of silver chloride forms from 1-chlorobutane).
2. The C-I bond is weaker than the C-Cl bond (it has a lower bond enthalpy).
3. Therefore, less energy is required to break the C-I bond, allowing hydrolysis to occur at a faster rate, despite the C-Cl bond being more polar.

M1 (1 mark): Identify that 1-iodobutane hydrolyses faster (or forms a precipitate faster).
M2 (1 mark): State that the C-I bond is weaker than the C-Cl bond / has a lower bond enthalpy (reject explanation based on electronegativity/polarity).
M3 (1 mark): Relate bond weakness to reactivity, stating that less energy is needed to break the C-I bond.

Effect: The calculated concentration of the sodium hydroxide solution will be higher than the actual value. [1 mark] Explanation: The water remaining in the burette dilutes the hydrochloric acid, decreasing its actual concentration. [1 mark] Therefore, a larger volume of the diluted acid is required to neutralise the NaOH. Because the calculation assumes the acid concentration is exactly 0.100 mol dm^-3, this larger volume leads to an overestimation of the moles of acid used, which in turn leads to a higher calculated concentration of NaOH. [0.77 mark]

Award 1 mark for stating that the calculated concentration is higher than actual. Award 1 mark for explaining that the acid is diluted by the water in the burette. Award 0.77 marks for linking the dilution to a larger volume of acid needed and therefore a higher calculated concentration of NaOH.

Assumption 1: The density of the reaction mixture is assumed to be 1.00 g cm^-3, so that the mass of 50.0 cm^3 of solution is taken to be 50.0 g. [1 mark] Assumption 2: The specific heat capacity of the mixture is assumed to be 4.18 J g^-1 K^-1, which is the value for pure water. [1 mark] Assumption 3: The heat capacity of the calorimeter or container is negligible. [0.77 mark for any of these two, or alternative correct assumptions up to 2.77 marks total]

Award 1 mark for assuming the density of the mixture is 1.00 g cm^-3 (or same as water). Award 1 mark for assuming the specific heat capacity is 4.18 J g^-1 K^-1 (or same as water). Award 0.77 marks for stating that the heat capacity of the container/calorimeter is negligible.

Test to identify the layer: Add a small volume of water (or water-soluble dye) to the mixture. The layer that increases in size is the aqueous layer, identifying the other as the organic layer. Alternatively, check the density: 1-bromobutane is denser than water and will form the lower layer. [1.38 marks] Separation: Transfer the mixture to a separating funnel, allow the layers to settle, open the tap to carefully run off the lower organic layer into a separate flask, and close the tap before the upper aqueous layer passes through. [1.39 marks]

Award 1.38 marks for describing a valid test (such as adding water and observing which layer increases, or checking density). Award 1.39 marks for describing the correct use of a separating funnel to run off and collect the layers separately.

Step 1: Acidify the solution with dilute nitric acid, then add aqueous silver nitrate. Observation: Chloride ions form a white precipitate of silver chloride, bromide ions form a cream precipitate of silver bromide, and iodide ions form a yellow precipitate of silver iodide. [1 mark] Step 2: Add dilute aqueous ammonia to the precipitates. Observation: The white precipitate (silver chloride) dissolves, but the cream and yellow precipitates do not. [1 mark] Step 3: Add concentrated aqueous ammonia to the remaining precipitates. Observation: The cream precipitate (silver bromide) dissolves, while the yellow precipitate (silver iodide) remains insoluble. [0.77 mark]

Award 1 mark for adding nitric acid and silver nitrate with correct initial precipitate colours. Award 1 mark for testing with dilute ammonia and noting that only silver chloride dissolves. Award 0.77 marks for testing with concentrated ammonia and noting silver bromide dissolves but silver iodide remains insoluble.

Role of ethanol: Ethanol acts as a common solvent allowing the polar water/silver nitrate and the non-polar halogenoalkane to mix and react in a single phase. [1 mark] Trend: The rate of reaction increases in the order 1-chlorobutane < 1-bromobutane < 1-iodobutane (precipitate forms fastest with 1-iodobutane). [1 mark] Explanation: The carbon-halogen bond strength (bond enthalpy) decreases down the group (C-Cl > C-Br > C-I). The C-I bond requires the least energy to break, leading to the fastest rate of hydrolysis. [0.77 mark]

Award 1 mark for stating that ethanol acts as a mutual/common solvent. Award 1 mark for the correct trend in rate (1-iodobutane is fastest / rate increases down Group 7). Award 0.77 marks for explaining that carbon-halogen bond enthalpy decreases down the group (C-I is the weakest bond).

Reason 1: Heat is lost to the surrounding air, the copper calorimeter, or the draft shields, rather than being transferred entirely to the water. [1 mark] Reason 2: Incomplete combustion of methanol occurs, producing carbon monoxide or soot instead of carbon dioxide, which releases less energy per mole. [1 mark] Other acceptable answers: Evaporation of methanol from the wick before or after weighing [0.77 mark] or reactions occurring under non-standard conditions [0.77 mark]. Total marks capped at 2.77.

Award 1 mark for identifying heat loss to the surroundings/calorimeter. Award 1 mark for identifying incomplete combustion of methanol. Award 0.77 marks for identifying fuel evaporation or non-standard conditions.

Part (i): Dissolving the solid in a beaker first ensures all the solid is completely dissolved before transfer, preventing undissolved particles from getting trapped in the narrow neck of the volumetric flask where they would be difficult to dissolve by swirling. [1.38 marks] Part (ii): Ensuring the bottom of the meniscus is aligned with the graduation mark at eye level prevents parallax error and ensures the final volume is exactly 250 cm^3. [1.39 marks]

Award 1.38 marks for explaining that dissolving in a beaker first ensures complete dissolution/prevents solid trapping in the neck. Award 1.39 marks for explaining that aligning at eye level avoids parallax error and ensures exact volume.

Procedure: Dip a nichrome (or platinum) wire into concentrated hydrochloric acid and heat it in a hot Bunsen flame until no colour is seen to ensure it is clean. [1 mark] Dip the clean wire back into concentrated hydrochloric acid, touch the solid sample to adhere some to the wire, and place it in the blue Bunsen flame. [1 mark] Observation: Calcium ions produce a brick-red (or orange-red) flame. [0.77 mark]

Award 1 mark for cleaning the wire with concentrated HCl and heating. Award 1 mark for dipping the wire in acid, picking up the solid, and placing in a blue flame. Award 0.77 marks for the correct brick-red flame color.

First, calculate the amount in moles of sodium carbonate: \(n = \frac{2.65\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0250\text{ mol}\). Next, convert the volume of the solution from cubic centimetres to cubic decimetres: \(V = \frac{250.0\text{ cm}^3}{1000} = 0.250\text{ dm}^3\). Finally, calculate the concentration: \(C = \frac{0.0250\text{ mol}}{0.250\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\).

Award 1 mark for correct calculation of moles (0.0250 mol). Award 1 mark for dividing by the correct volume in dm3 (0.250 dm3). Award 0.77 marks for the correct concentration to 3 significant figures with appropriate units (0.100 mol dm^-3).

First, calculate the temperature change: \(\Delta T = 26.1 - 19.5 = 6.6^\circ\text{C}\) (which is equivalent to \(6.6\text{ K}\)). Second, calculate the total mass of the mixture: \(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\), which weighs \(100.0\text{ g}\) assuming a density of \(1.00\text{ g cm}^{-3}\). Finally, apply the formula \(q = m c \Delta T\): \(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.6\text{ K} = 2758.8\text{ J}\). Rounding to 3 significant figures gives \(2760\text{ J}\).

Award 1 mark for identifying the correct total mass of 100.0 g and temperature change of 6.6 K. Award 1 mark for the correct substitution of these values into the energy equation. Award 0.77 marks for the correct final answer rounded to 3 significant figures (2760 J).

The aqueous sodium hydrogencarbonate is added to neutralise any residual acid impurities (such as unreacted hydrogen bromide or sulfuric acid catalyst) present in the crude organic layer. This neutralisation reaction produces carbon dioxide gas as a byproduct. Inverting the separating funnel and opening the tap periodically is required to release the built-up gas pressure, preventing the stopper from being forced out.

Award 1.38 marks for stating that the sodium hydrogencarbonate neutralises residual acid impurities. Award 1.39 marks for explaining that carbon dioxide gas is generated and venting releases the pressure build-up.

The formation of a cream precipitate when treated with aqueous silver nitrate confirms the presence of bromide ions (\(\text{Br}^-\)). The simplest ionic equation representing the precipitation reaction is: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\).

Award 1.00 mark for correctly identifying the halide ion as bromide. Award 1.77 marks for the correct, balanced ionic equation including correct state symbols.

Heating to constant mass (by heating, cooling, weighing, and repeating the process) ensures that the thermal decomposition is complete and that all water of crystallisation has been fully evaporated from the hydrated salt, leaving only anhydrous copper(II) sulfate.

Award 2.77 marks for stating that heating to constant mass ensures all water of crystallisation has been completely removed or evaporated.

First, calculate the temperature change: \(\Delta T = 24.6 - 18.4 = 6.2^\circ\text{C}\). Because two temperature readings are taken (initial and final), the absolute uncertainty in the temperature change is doubled: \(2 \times 0.1^\circ\text{C} = 0.2^\circ\text{C}\). Finally, calculate the percentage uncertainty: \(\text{Percentage uncertainty} = \frac{0.2}{6.2} \times 100\% = 3.2258\%\), which rounds to \(3.2\%\) to 2 significant figures.

Award 1 mark for calculating the temperature difference of 6.2. Award 1 mark for doubling the reading uncertainty to 0.2. Award 0.77 marks for the correct calculation and rounding to 3.2%.

Instead of heating under reflux, the student must set up the apparatus for distillation (with immediate collection). Ethanal has a lower boiling point than ethanol and water because it cannot form hydrogen bonds. By using immediate distillation, the ethanal vaporises and is distilled off as soon as it is formed, removing it from contact with the remaining oxidising agent in the reaction flask and preventing further oxidation to ethanoic acid.

Award 1.00 mark for identifying distillation (or immediate distillation). Award 1.77 marks for explaining that ethanal has a lower boiling point and is removed from contact with the oxidising agent, preventing its further oxidation to ethanoic acid.

Chlorine is a stronger oxidising agent than iodine and displaces iodide ions from potassium iodide to form aqueous iodine: \(\text{Cl}_2(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\). When cyclohexane is added, the non-polar iodine molecules dissolve preferentially in the cyclohexane layer. Since cyclohexane is less dense than water, it forms the upper layer. Dissolved molecular iodine (\(\text{I}_2\)) in hydrocarbon solvents exhibits a distinct purple or violet colour.

Award 1.38 marks for stating the colour of the upper organic layer is purple (or violet). Award 1.39 marks for identifying the chemical species responsible as molecular iodine or I2.

1. Determine the temperature change (\(\Delta T\)):
\(\Delta T = 35.5 - 19.5 = 16.0\ ^\circ\text{C}\)

2. Determine the total uncertainty for two temperature readings (initial and maximum):
\(\text{Total uncertainty} = 2 \times 0.5\ ^\circ\text{C} = 1.0\ ^\circ\text{C}\)

3. Calculate the percentage uncertainty:
\(\text{Percentage uncertainty} = \frac{1.0}{16.0} \times 100\% = 6.25\%\)

- **Award 1.38 marks** for calculating the temperature change of \(16.0\ ^\circ\text{C}\) and doubling the uncertainty to \(1.0\ ^\circ\text{C}\).
- **Award 1.39 marks** for the final percentage uncertainty of \(6.25\%\) (accept 6.3%).
- Reject calculations using only a single uncertainty value (e.g., \(0.5 / 16.0 \times 100 = 3.13\%\)), which can receive a maximum of 1.38 marks.

Halogenoalkanes are organic compounds that are insoluble in water because they cannot form hydrogen bonds with water molecules. Aqueous silver nitrate is a polar aqueous solution. Without a co-solvent, the reaction mixture would form two separate layers (immiscible), which would significantly hinder the reaction rate. Ethanol has both a polar hydroxyl group and a non-polar ethyl group, allowing it to dissolve both the halogenoalkanes and the aqueous reactants. It acts as a mutual solvent to ensure the reactants are in a single homogeneous phase.

- **Award 1.38 marks** for identifying that halogenoalkanes are insoluble in water or immiscible with the aqueous silver nitrate solution.
- **Award 1.39 marks** for explaining that ethanol acts as a mutual/common solvent (or dissolves both reactants) to allow them to mix/react together in a single phase.
- Accept 'cosolvent' or 'solubilising agent' for ethanol.
- Reject statements claiming that ethanol acts as a catalyst or a nucleophile.