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There is a very large 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 a lower energy level closer to the nucleus, meaning the element has three valence electrons and forms a \(3+\) ion (\(X^{3+}\)). Since oxide ions are \(\text{O}^{2-}\), the formula of its oxide is \(X_2\text{O}_3\).

1 mark for identifying the large increase between the 3rd and 4th ionization energies, deducing a 3+ charge on the ion, and writing the correct formula X2O3.

First, convert the volume to \(\text{dm}^3\): \(96.0\text{ cm}^3 = 0.0960\text{ dm}^3\). Next, calculate the amount in moles of the gas: \(n = 0.0960 / 24.0 = 0.00400\text{ mol}\). Calculate the molar mass of the diatomic gas: \(M = 0.112 / 0.00400 = 28.0\text{ g mol}^{-1}\). Since the element is diatomic, the relative molecular mass is 28.0, which corresponds to nitrogen (\(\text{N}_2\)).

1 mark for calculating the moles of gas, finding the molar mass of 28.0 g/mol, and correctly identifying nitrogen as the diatomic gas.

The oxygen atom in \(\text{H}_3\text{O}^+\) has 5 valence electrons (6 minus 1 for the positive charge). Three electrons are shared in single covalent bonds with hydrogen atoms, leaving one lone pair. With three bonding pairs and one lone pair, the electron geometry is tetrahedral, resulting in a trigonal pyramidal molecular shape. The bond angle is reduced from the tetrahedral angle of \(109.5^\circ\) to approximately \(107^\circ\) due to greater repulsion from the lone pair.

1 mark for identifying the three bonding pairs and one lone pair, deducing the trigonal pyramidal shape, and selecting the correct bond angle of 107 degrees.

There are three structural isomers of \(\text{C}_5\text{H}_{12}\): pentane, 2-methylbutane, and 2,2-dimethylpropane. Pentane can form 1-chloropentane, 2-chloropentane, and 3-chloropentane (3 isomers). 2-methylbutane can form 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, and 1-chloro-3-methylbutane (4 isomers). 2,2-dimethylpropane can only form 1-chloro-2,2-dimethylpropane (1 isomer). Therefore, only 1 structural isomer of \(\text{C}_5\text{H}_{12}\) (pentane) yields exactly three structural isomers of monochloroalkane.

1 mark for evaluating the number of monochloroalkanes formed by each structural isomer of C5H12 and determining that only pentane meets the condition.

But-1-ene does not exhibit E/Z isomerism because one of the doubly bonded carbons is attached to two identical hydrogen atoms. Upon electrophilic addition with hydrogen bromide, the major product is 2-bromobutane according to Markovnikov's rule (via the more stable secondary carbocation). 2-bromobutane contains a chiral carbon (C-2 is bonded to hydrogen, methyl, ethyl, and bromine groups).

1 mark for identifying that but-1-ene has no E/Z isomerism and forms the chiral major product 2-bromobutane.

Molar mass of desired product (\(\text{C}_4\text{H}_9\text{Cl}\)) = \(4 \times 12.0 + 9 \times 1.0 + 35.5 = 92.5\text{ g mol}^{-1}\). Total mass of reactants (\(\text{C}_4\text{H}_{10} + \text{Cl}_2\)) = \((4 \times 12.0 + 10 \times 1.0) + (2 \times 35.5) = 58.0 + 71.0 = 129.0\text{ g mol}^{-1}\). Atom economy = \((92.5 / 129.0) \times 100\% \approx 71.7\%\).

1 mark for calculating the mass of the desired product and total mass of reactants, and computing the correct percentage of 71.7%.

First, find the relative abundance of \(^{22}\text{Ne}\): \(100\% - 90.48\% - 0.27\% = 9.25\%\). Then calculate the weighted average of the isotopic masses: \(A_r = [(20 \times 90.48) + (21 \times 0.27) + (22 \times 9.25)] / 100 = [1809.6 + 5.67 + 203.5] / 100 = 2018.77 / 100 = 20.1877\), which rounds to 20.19.

1 mark for determining the abundance of the third isotope and calculating the relative atomic mass as 20.19.

Phosphorus has the electron configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^3\), with one electron in each of the three 3p orbitals. Sulfur has the configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^4\), where one of the 3p orbitals contains a pair of electrons. The mutual repulsion between these two paired electrons (spin-pair repulsion) in the same orbital makes it easier to remove one of them, resulting in a lower first ionization energy for sulfur compared to phosphorus.

1 mark for selecting the correct explanation based on spin-pair repulsion in the paired 3p orbital of sulfur.

The successive ionization energies show a very large increase (jump) between the third and fourth ionization energies (from 2745 to 11577 kJ mol\(^{-1}\)). This indicates that the fourth electron is removed from an inner shell, which is closer to the nucleus and experiences significantly less shielding. Therefore, element \(Y\) has three valence electrons and forms a stable \(Y^{3+}\) ion. To form a neutral chloride, three chloride ions (\(\text{Cl}^-\)) are needed, giving the formula \(Y\text{Cl}_3\).

1 mark: Correctly identifies \(Y\text{Cl}_3\) (C) based on the ionization energy jump.

The atomic number of iron (\(\text{Fe}\)) is 26, so a neutral iron atom has the electronic configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\). When transition metal atoms form positive ions, they lose their \(4s\) electrons before their \(3d\) electrons. Therefore, the \(\text{Fe}^{2+}\) ion loses both \(4s\) electrons, resulting in the ground state configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\).

1 mark: Correctly identifies configuration A.

First, calculate the moles of \(\text{MgCO}_3\) reacted:

\(\text{moles of MgCO}_3 = \frac{4.215\text{ g}}{84.3\text{ g mol}^{-1}} = 0.050\text{ mol}\).

The equation for the reaction is:

\(\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\).

The molar ratio between \(\text{MgCO}_3\) and \(\text{CO}_2\) is 1:1, so 0.050 mol of \(\text{CO}_2\) is produced.

Now, calculate the volume of gas at RTP:

\(\text{Volume} = 0.050\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\).

1 mark: Correctly calculates 1.20 dm\(^3\) (A).

First, calculate the total molecular mass of the reactants:

\(\text{Mass of reactants} = M_r(\text{Fe}_2\text{O}_3) + 3 \times M_r(\text{CO})\)

\(= (2 \times 55.8 + 3 \times 16.0) + 3 \times (12.0 + 16.0)\)

\(= 159.6 + 84.0 = 243.6\text{ g mol}^{-1}\).

Next, calculate the mass of the desired product (\(2\text{Fe}\)):

\(\text{Mass of desired product} = 2 \times 55.8 = 111.6\text{ g mol}^{-1}\).

Finally, calculate the percentage atom economy:

\(\text{Atom Economy} = \frac{111.6}{243.6} \times 100\% \approx 45.8\%\).

1 mark: Correct calculation of atom economy to 45.8% (A).

In \(\text{BF}_3\), the boron atom has three valence electrons and forms three single covalent bonds with fluorine atoms, leaving no lone pairs on the central boron atom. This results in a trigonal planar geometry with bond angles of exactly 120\(^\circ\) to minimize repulsion between the three bonding electron pairs.

1 mark: Correctly identifies \(\text{BF}_3\) (B).

Covalent character in an ionic compound arises from polarization of the anion by the cation. According to Fajans' rules, polarization is maximized when the cation is small with a high charge density (having high polarizing power) and the anion is large (highly polarizable). Among the choices, \(\text{Mg}^{2+}\) is smaller than \(\text{Ba}^{2+}\) and has a higher charge density. The iodide ion (\(\text{I}^-\)) is larger than the chloride ion (\(\text{Cl}^-\)) and is more easily polarized. Therefore, \(\text{MgI}_2\) has the most covalent character.

1 mark: Correctly identifies \(\text{MgI}_2\) (A).

In a propagation step of a free-radical reaction, a free radical reacts with a stable molecule to produce a new free radical and a new stable molecule. The equation \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) is a propagation step because a chlorine radical reacts with a methane molecule to yield a methyl radical and hydrogen chloride. Option A is initiation, while options C and D are termination steps.

1 mark: Correctly identifies the propagation step (B).

During the electrophilic addition of \(\text{HBr}\) to propene, the electrophile (\(\text{H}^+\)) adds first. Addition can form either a primary carbocation (\(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\)) or a secondary carbocation (\(\text{CH}_3\text{C}^+\text{HCH}_3\)). The secondary carbocation is more stable than the primary carbocation because it has two electron-donating alkyl groups (methyl groups) attached to the positively charged carbon. These groups release electron density via the inductive effect, dispersing the positive charge and stabilizing the intermediate. Thus, the major product is formed via this more stable secondary carbocation.

1 mark: Correctly identifies the explanation involving secondary carbocation stability (B).

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

2. Calculate the number of moles of \(\text{CO}_2\) gas produced:
\(\text{Moles of CO}_2 = \frac{\text{Volume in dm}^3}{24.0} = \frac{0.300\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0125\text{ mol}\)

3. Determine the moles of \(\text{NaHCO}_3\) required from the \(1:1\) stoichiometry:
\(\text{Moles of NaHCO}_3 = 0.0125\text{ mol}\)

4. Calculate the required mass of \(\text{NaHCO}_3\):
\(\text{Mass} = \text{moles} \times \text{molar mass} = 0.0125\text{ mol} \times 84.0\text{ g mol}^{-1} = 1.05\text{ g}\)

1 mark: Correct answer A.

Incorrect Distractors:
- B: Correctly calculates moles of gas but incorrectly assumes a 2:1 ratio (as in thermal decomposition).
- C: Uses an incorrect 1:2 ratio.
- D: Fails to convert \(300\text{ cm}^3\) to \(\text{dm}^3\) correctly.

1. Analyze the successive ionization energies to find the largest relative increase:
- From 1st to 2nd: \(1817 - 578 = 1239\text{ kJ mol}^{-1}\)
- From 2nd to 3rd: \(2745 - 1817 = 928\text{ kJ mol}^{-1}\)
- From 3rd to 4th: \(11578 - 2745 = 8833\text{ kJ mol}^{-1}\) (a very large jump)

2. The massive jump between the 3rd and 4th ionization energies indicates that the fourth electron is removed from an inner quantum shell. Therefore, element \(X\) has 3 valence electrons and belongs to Group 3 (Group 13).

3. In Group 3, the element forms \(X^{3+}\) ions. Combined with the oxide ion (\(\text{O}^{2-}\)), the formula of its oxide is \(X_2\text{O}_3\).

1 mark: Correct answer C.

Incorrect Distractors:
- A: Formula for a Group 1 element oxide.
- B: Formula for a Group 2 element oxide.
- D: Formula for a Group 4 element oxide.

- \(\text{CO}_2\) has 2 bonding regions (double bonds) and 0 lone pairs on the central carbon atom, giving a linear shape (bond angle \(180^\circ\)).
- \(\text{BeF}_2\) has 2 bonding pairs and 0 lone pairs on the central beryllium atom, giving a linear shape (bond angle \(180^\circ\)).
- \(\text{H}_2\text{S}\) has 2 single bonding pairs and 2 lone pairs on the central sulfur atom. Repulsion between these 4 electron pairs results in a bent/non-linear molecular shape (bond angle approximately \(104.5^\circ\)).
- \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs on the central boron atom, giving a trigonal planar shape (bond angle \(120^\circ\)).

1 mark: Correct answer C.

- The reaction of an alkene with a hydrogen halide is an electrophilic addition reaction.
- Adding \(\text{H}^+\) to but-1-ene can form a primary carbocation (minor path) or a secondary carbocation (major path).
- A secondary carbocation is more stable than a primary carbocation due to the greater electron-releasing inductive effect of the two alkyl groups compared to one.
- Therefore, the secondary carbocation intermediate is preferred, yielding 2-bromobutane as the major product.
- Statement D is correct.

1 mark: Correct answer D.

Incorrect Distractors:
- A: 1-bromobutane is the minor product.
- B: The major product is formed via the more stable secondary carbocation.
- C: The reaction mechanism is electrophilic addition, not nucleophilic.

(a) Relative isotopic mass is defined as the mass of an individual atom of an isotope relative to 1/12th of the mass of a carbon-12 atom.

(b) Let \(x\) be the abundance of \(^{63}\text{Cu}\). The abundance of \(^{65}\text{Cu}\) is therefore \(100 - x\).
\(63.00x + 65.00(100 - x) = 63.55 \times 100\)
\(63x + 6500 - 65x = 6355\)
\(-2x = -145\)
\(x = 72.5\%\) (abundance of \(^{63}\text{Cu}\))
\(100 - x = 27.5\%\) (abundance of \(^{65}\text{Cu}\))

(c)(i) Copper has an anomalous electronic configuration to achieve a stable full d-subshell: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (or \([\text{Ar}] 3d^{10} 4s^1\)).
(c)(ii) When forming a transition metal ion, electrons are lost from the 4s orbital first, then 3d: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\) (or \([\text{Ar}] 3d^9\)).

(d) Copper has 29 protons in its nucleus, while potassium has 19. Both have their outer electrons in the 4th shell, meaning shielding is similar. The much larger nuclear charge in copper exerts a stronger electrostatic attraction on the outer electron, pulling it closer and requiring more energy to remove.

(e) The 11th ionization energy removes the last electron from the 3rd shell. The 12th ionization energy removes an electron from the 2nd shell (specifically, the \(2p\) subshell), which is much closer to the nucleus and experiences far less shielding, resulting in a significantly larger energy requirement (the jump).

(f) Copper belongs to the d-block because its highest energy occupied subshell is a d-subshell.

(g) Periodicity is the repeating trend of physical and chemical properties of elements across periods in the Periodic Table.

Detailed Marks:
(a)
- 1 Mark: Mass of an atom of an isotope.
- 1 Mark: Compared to 1/12th of the mass of an atom of carbon-12.

(b)
- 1 Mark: Setting up a correct algebraic expression: e.g., \(63x + 65(100-x) = 6355\).
- 1 Mark: Correct percentage of \(^{63}\text{Cu} = 72.5\%\).
- 1 Mark: Correct percentage of \(^{65}\text{Cu} = 27.5\%\).

(c)(i)
- 1 Mark: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (accept noble gas core: \([\text{Ar}] 3d^{10} 4s^1\)).

(c)(ii)
- 1 Mark: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^9\) (accept noble gas core: \([\text{Ar}] 3d^9\)).

(d)
- 1 Mark: Copper has a greater nuclear charge / more protons (29 protons vs 19 protons for K).
- 1 Mark: Outer electrons in both are in the 4th shell / similar shielding.
- 1 Mark: Stronger electrostatic attraction between the nucleus and the outer electron in Cu (requires more energy to remove).

(e)
- 1 Mark: The 12th electron is removed from an inner shell / second shell (or shell closer to the nucleus).
- 1 Mark: This shell experiences significantly less shielding from the nucleus (or has stronger electrostatic attraction).

(f)
- 1 Mark: d-block.
- 1 Mark: Highest energy occupied subshell is a d-subshell (or outer d-subshell is being filled).

(g)
- 1 Mark: Repeating pattern/trend of properties across a period of the Periodic Table.

(a) The ionic equation is the reaction of metal cations, \(\text{M}^{2+}(aq)\), with carbonate anions, \(\text{CO}_3^{2-}(aq)\), to form the insoluble solid metal carbonate, \(\text{MCO}_3(s)\):
\(\text{M}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{MCO}_3(s)\).

(b) The precipitate must be washed with distilled water to remove any soluble ions/impurities (such as sodium ions and nitrate ions) adhering to the surface of the precipitate, which would otherwise increase the measured mass.

(c) The precipitate is dried to remove any water, ensuring the measured mass represents only the dry solid and does not include the mass of water.

(d) Calculation of relative atomic mass of \(\text{M}\):
The total mass of \(\text{M(NO}_3)_2 \cdot 6\text{H}_2\text{O}\) used is 5.126 g in \(250.0\text{ cm}^3\).
Therefore, the mass of \(\text{M(NO}_3)_2 \cdot 6\text{H}_2\text{O}\) in the \(25.0\text{ cm}^3\) portion is:
\(5.126 \times \frac{25.0}{250.0} = 0.5126\text{ g}\).

Let the relative atomic mass of \(\text{M}\) be \(A_r\).

Molar mass of \(\text{M(NO}_3)_2 \cdot 6\text{H}_2\text{O} = A_r + 2(14.0 + 3 \times 16.0) + 6 \times 18.0 = A_r + 124.0 + 108.0 = A_r + 232.0\text{ g mol}^{-1}\).
Moles of hydrated salt in \(25.0\text{ cm}^3 = \frac{0.5126}{A_r + 232.0}\).

Molar mass of \(\text{MCO}_3 = A_r + 12.0 + 3 \times 16.0 = A_r + 60.0\text{ g mol}^{-1}\).
Moles of precipitate \(\text{MCO}_3 = \frac{0.169}{A_r + 60.0}\).

Since 1 mole of \(\text{M(NO}_3)_2 \cdot 6\text{H}_2\text{O}\) yields 1 mole of \(\text{MCO}_3\), these moles are equal:
\(\frac{0.5126}{A_r + 232.0} = \frac{0.169}{A_r + 60.0}\)
\(0.5126(A_r + 60.0) = 0.169(A_r + 232.0)\)
\(0.5126 A_r + 30.756 = 0.169 A_r + 39.208\)
\(0.3436 A_r = 8.452\)
\(A_r = \frac{8.452}{0.3436} = 24.598 \approx 24.6\).

(e) If some metal carbonate remains dissolved, the obtained mass of the precipitate (0.169 g) is lower than it should be. Looking at the equation:
\(\frac{\text{Mass of Hydrated Salt}}{\text{Mass of Precipitate}} = \frac{A_r + 232.0}{A_r + 60.0}\)
If the mass of the precipitate is smaller, the ratio on the left is larger, which algebraically results in a smaller calculated value for \(A_r\).

(f) Moles of \(\text{Mg(NO}_3)_2 = \frac{2.97\text{ g}}{148.3\text{ g mol}^{-1}} = 0.0200\text{ mol}\).
According to the balanced equation, \(2\text{ moles of Mg(NO}_3)_2\) produce \(5\text{ moles of gas}\) (\(4\text{ NO}_2 + 1\text{ O}_2\)).
Total moles of gas produced = \(0.0200 \times \frac{5}{2} = 0.0500\text{ mol}\).
Total volume of gas = \(0.0500\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3\).

Detailed Marks:
(a)
- 1 Mark: Correct formula of reactants and products: \(\text{M}^{2+}(aq) + \text{CO}_3^{2-}(aq) \rightarrow \text{MCO}_3(s)\).
- 1 Mark: Correct state symbols.

(b)
- 1 Mark: To remove soluble impurities / unreacted ions (e.g. sodium nitrate). Reject 'remove water'.

(c)
- 1 Mark: To ensure that only the mass of the dry carbonate is measured (otherwise mass would be artificially high due to remaining water).

(d)
- 1 Mark: Calculate mass of hydrated salt in \(25.0\text{ cm}^3\) portion = 0.5126 g.
- 1 Mark: Express molar mass of hydrated salt as \(A_r + 232\) AND molar mass of carbonate as \(A_r + 60\).
- 1 Mark: Set up equation: \(\frac{0.5126}{A_r + 232} = \frac{0.169}{A_r + 60}\).
- 1 Mark: Expand and rearrange correctly: \(0.3436 A_r = 8.452\).
- 1 Mark: Final calculation of \(A_r = 24.6\) (accept range 24.5 to 24.6).

(e)
- 1 Mark: Explains that the recorded mass of precipitate is too low.
- 1 Mark: Link this to a lower calculated value of \(A_r\).

(f)
- 1 Mark: Calculate moles of \(\text{Mg(NO}_3)_2 = 0.0200\text{ mol}\).
- 1 Mark: Identifies the 2:5 molar ratio of reactant to gas (or states that \(0.0200\text{ mol}\) produces \(0.0400\text{ mol of NO}_2\) and \(0.0100\text{ mol of O}_2\)).
- 1 Mark: Calculate total moles of gas = \(0.0500\text{ mol}\).
- 1 Mark: Calculate volume of gas = \(1.20\text{ dm}^3\).

(a) The central nitrogen atom shares one pair of electrons with each of the three chlorine atoms. Nitrogen has 5 valence electrons, so it has 1 lone pair remaining. Each chlorine atom has 7 valence electrons, so after sharing 1 electron with nitrogen, each chlorine has 3 lone pairs remaining.

(b) The shape is trigonal pyramidal. The bond angle is \(107^\circ\) (accept \(106^\circ - 108^\circ\)). The nitrogen atom has four electron pairs (3 bonding pairs and 1 lone pair) in its outer shell. These pairs repel each other to get as far apart as possible. Since lone-pair to bonding-pair repulsion is greater than bonding-pair to bonding-pair repulsion, the bond angle is compressed from the tetrahedral angle of \(109.5^\circ\) to \(107^\circ\).

(c)(i) Chlorine has a higher electronegativity than nitrogen, causing the bonding pair of electrons in each \(\text{N}-\text{Cl}\) bond to be pulled closer to the chlorine atom, creating a dipole (\(\text{N}^{\delta+}-\text{Cl}^{\delta-}\)).
(c)(ii) The molecule is polar. Because the shape is asymmetric (trigonal pyramidal), the dipoles of the three \(\text{N}-\text{Cl}\) bonds do not cancel out, resulting in a net dipole moment.

(d)(i) The shape of \(\text{PCl}_5\) is trigonal bipyramidal. The bond angles are \(120^\circ\) (equatorial-equatorial) and \(90^\circ\) (axial-equatorial).
(d)(ii) Phosphorus is in Period 3 and has access to vacant d-orbitals in its valence shell, allowing it to accommodate 10 electrons (expand its octet) to form 5 covalent bonds. Nitrogen is in Period 2 and has no d-orbitals, meaning its valence shell is capped at 8 electrons, preventing it from forming 5 covalent bonds.

(e) Graphite is a good conductor of electricity, while diamond is an insulator. In graphite, each carbon atom is bonded to three others in hexagonal layers, leaving one delocalized electron per carbon atom. These electrons are free to drift throughout the structure and carry electrical charge. In diamond, each carbon is bonded to four others in a 3D giant tetrahedral structure; all valence electrons are locked in covalent bonds and cannot move.

Detailed Marks:
(a)
- 1 Mark: Correctly showing 3 shared pairs between N and 3 Cl atoms.
- 1 Mark: Correct number of non-bonding electrons (1 lone pair on N and 3 lone pairs on each Cl).

(b)
- 1 Mark: Trigonal pyramidal shape AND bond angle in range \(106^\circ - 108^\circ\).
- 1 Mark: 4 regions of electron density (3 bonding pairs and 1 lone pair) repel to minimize repulsion.
- 1 Mark: Lone pairs repel more than bonding pairs.

(c)(i)
- 1 Mark: Chlorine is more electronegative than nitrogen (leading to unequal electron sharing).

(c)(ii)
- 1 Mark: The molecule is asymmetric / trigonal pyramidal.
- 1 Mark: The bond dipoles do not cancel / there is a net overall dipole.

(d)(i)
- 1 Mark: Trigonal bipyramidal.
- 1 Mark: Bond angles of \(90^\circ\) AND \(120^\circ\) (both needed).

(d)(ii)
- 1 Mark: Phosphorus can expand its octet because it has vacant 3d orbitals.
- 1 Mark: Nitrogen is in Period 2 / has no d-orbitals and cannot expand its octet.

(e)
- 1 Mark: Correctly states graphite conducts while diamond does not.
- 1 Mark: Graphite: each carbon bonded to 3 others, resulting in delocalized electrons that are free to move.
- 1 Mark: Diamond: each carbon bonded to 4 others, all valence electrons are localized in covalent bonds.

(a)(i) Stereoisomers are molecules with the same structural formula but a different arrangement of their atoms in space.

(a)(ii)
Skeletal structure of E-pent-2-ene: A zigzag chain of 5 carbons with the double bond between carbons 2 and 3. The methyl group on C2 and the ethyl group on C3 are trans (on opposite sides of the double bond).
Skeletal structure of Z-pent-2-ene: The methyl group on C2 and the ethyl group on C3 are cis (on the same side of the double bond).

(a)(iii)
1. There is restricted rotation around the carbon-carbon double bond (\(\text{C}=\text{C}\)) due to the presence of the \(\pi\)-bond.
2. Each carbon atom of the double bond is attached to two different groups: Carbon-2 is bonded to \(\text{-H}\) and \(\text{-CH}_3\); Carbon-3 is bonded to \(\text{-H}\) and \(\text{-CH}_2\text{CH}_3\).
3. According to Cahn-Ingold-Prelog (CIP) priority rules, on Carbon-2, \(\text{-CH}_3\) has higher priority than \(\text{-H}\) (higher atomic number of C vs H). On Carbon-3, \(\text{-CH}_2\text{CH}_3\) has higher priority than \(\text{-H}\). When the two high-priority groups are on opposite sides of the double bond, it is the E-isomer; on the same side, it is the Z-isomer.

(b)(i)
The structure of 2-methylbut-2-ene is \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\).
When \(\text{HBr}\) adds across the double bond:
- The major product is 2-bromo-2-methylbutane (bromine adds to the more substituted carbon, C2).
- The minor product is 2-bromo-3-methylbutane (bromine adds to C3).

(b)(ii)
Mechanism:
1. Polar \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) approaches the double bond of 2-methylbut-2-ene.
2. A curly arrow starts from the \(\text{C}=\text{C}\) double bond and points to the hydrogen atom of \(\text{H}-\text{Br}\).
3. A concurrent curly arrow starts from the \(\text{H}-\text{Br}\) bond and points to the bromine atom.
4. This results in the formation of a tertiary carbocation intermediate, \(\text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\), and a bromide ion, \(\text{Br}^-\).
5. A curly arrow starts from a lone pair on the \(\text{Br}^-\) ion and points to the positively charged carbon atom of the carbocation to form the major product.

(b)(iii)
The addition of \(\text{H}^+\) to C3 forms a tertiary carbocation intermediate (\(\text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\)), whereas addition of \(\text{H}^+\) to C2 forms a secondary carbocation intermediate (\(\text{(CH}_3)_2\text{CH}-\text{C}^+\text{HCH}_3\)). The tertiary carbocation is more stable than the secondary carbocation because it has three electron-releasing alkyl groups (compared to two) that stabilize the positive charge by the inductive effect. The more stable intermediate is formed more rapidly, leading to the major product.

Detailed Marks:
(a)(i)
- 1 Mark: Molecules/compounds with the same structural formula.
- 1 Mark: Different arrangement of atoms in space.

(a)(ii)
- 1 Mark: Correct skeletal structure of E-pent-2-ene.
- 1 Mark: Correct skeletal structure of Z-pent-2-ene.

(a)(iii)
- 1 Mark: Restricted rotation around the \(\text{C}=\text{C}\) double bond (or because of the \(\pi\)-bond).
- 1 Mark: Each of the carbon atoms in the double bond is bonded to two different groups.
- 1 Mark: Application of CIP rules: \(\text{-CH}_3\) has higher priority than \(\text{-H}\) and \(\text{-CH}_2\text{CH}_3\) has higher priority than \(\text{-H}\).

(b)(i)
- 1 Mark: Major product: 2-bromo-2-methylbutane.
- 1 Mark: Minor product: 2-bromo-3-methylbutane (accept 3-bromo-2-methylbutane).

(b)(ii)
- 1 Mark: Correct dipole shown on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) AND curly arrow from the double bond to the \(\text{H}\).
- 1 Mark: Curly arrow from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\).
- 1 Mark: Correct structure of the tertiary carbocation intermediate AND \(\text{Br}^-\).
- 1 Mark: Curly arrow from a lone pair on the \(\text{Br}^-\) to the positive carbon of the intermediate.

(b)(iii)
- 1 Mark: Major product proceeds via a tertiary carbocation, while minor product proceeds via a secondary carbocation.
- 1 Mark: Tertiary carbocations are more stable (due to the electron-releasing inductive effect of 3 alkyl groups).

Hydrogen fluoride, HF, has hydrogen bonding between its molecules, which is significantly stronger than the intermolecular forces in the other hydrogen halides. Therefore, HF has the highest boiling temperature. For HCl, HBr, and HI, the intermolecular forces are dominated by London (dispersion) forces. Since the number of electrons in the molecules increases from HCl to HBr to HI, the strength of the London forces increases, leading to higher boiling temperatures in the order: \( \text{HCl} < \text{HBr} < \text{HI} \). Combining these trends gives the overall order of increasing boiling temperatures: \( \text{HCl} < \text{HBr} < \text{HI} < \text{HF} \).

1 mark: Correct option chosen (B).

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. Down Group 7, the bond length increases and the bond enthalpy of the C-X bond decreases: C-Cl (\( 346\text{ kJ mol}^{-1} \)) > C-Br (\( 290\text{ kJ mol}^{-1} \)) > C-I (\( 228\text{ kJ mol}^{-1} \)). The weaker the bond, the more easily it is broken during nucleophilic substitution. Therefore, 1-iodobutane reacts the fastest to produce silver iodide precipitate, followed by 1-bromobutane, and 1-chlorobutane is the slowest.

1 mark: Correct option chosen (B).

As you descend Group 2, the ionic radius of the \( \text{M}^{2+} \) cation increases while its charge remains constant. This results in a decrease in the charge density of the cation down the group. A lower charge density means the cation has less polarising power and is less able to polarise/distort the carbonate ion's electron cloud. Since the carbonate ion is less distorted, more thermal energy is required to decompose it, so the thermal stability of the carbonates increases down the group.

1 mark: Correct option chosen (A).

Using Hess's Law and a cycle based on combustion products:
\( \text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)} \)
\( \Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products}) \)
\( \Delta_f H^\ominus = \Delta_c H^\ominus[\text{C(s)}] + 2 \times \Delta_c H^\ominus[\text{H}_2\text{(g)}] - \Delta_c H^\ominus[\text{CH}_4\text{(g)}] \)
\( \Delta_f H^\ominus = -393.5 + 2(-285.8) - (-890.3) \)
\( \Delta_f H^\ominus = -393.5 - 571.6 + 890.3 = -74.8\text{ kJ mol}^{-1} \).

1 mark: Correct option chosen (A).

The Maxwell-Boltzmann distribution curve represents the distribution of kinetic energies of the molecules, which depends solely on the temperature. Therefore, adding a catalyst does not change the shape or position of the curve. Instead, a catalyst provides an alternative reaction pathway with a lower activation energy, effectively shifting the activation energy threshold (\( E_a \)) to the left (a lower value) on the energy axis. This allows a larger fraction of molecules to have sufficient energy to react successfully.

1 mark: Correct option chosen (C).

In concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)), sulfur has an oxidation state of +6. Solid potassium iodide is a very strong reducing agent and can reduce sulfur to several species: \( \text{SO}_2 \) (oxidation state +4), \( \text{S} \) (oxidation state 0), and \( \text{H}_2\text{S} \) (oxidation state -2). All of these represent reduced states of sulfur. Iodine (\( \text{I}_2 \)) is formed by the oxidation of iodide ions. Hydrogen iodide (\( \text{HI} \)) and potassium hydrogensulfate (\( \text{KHSO}_4 \)) are products of an acid-base reaction, not redox.

1 mark: Correct option chosen (B).

The strong, sharp absorption band at \( 1715\text{ cm}^{-1} \) indicates the presence of a carbonyl group (\( \text{C=O} \)). The absence of a broad band in the \( 3200\text{ - }3750\text{ cm}^{-1} \) range shows that there is no hydroxyl group (\( \text{O-H} \)). Therefore, the compound is a carbonyl compound (a ketone or aldehyde) and not an alcohol or a carboxylic acid. Propanone, \( \text{CH}_3\text{COCH}_3 \), is a ketone with the molecular formula \( \text{C}_3\text{H}_6\text{O} \) containing only a carbonyl functional group, which perfectly fits this spectroscopic data.

1 mark: Correct option chosen (C).

The forward reaction is exothermic (\( \Delta H = -197\text{ kJ mol}^{-1} \)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the direction that releases heat (the exothermic direction, which is the forward direction). This shifts the equilibrium to the right, increasing the yield of \( \text{SO}_3\text{(g)} \). Increasing the temperature would decrease the yield. Decreasing pressure shifts the equilibrium to the side with more gaseous moles (the left), decreasing the yield. Adding a catalyst increases the rate of both forward and reverse reactions equally, leaving the equilibrium yield unchanged.

1 mark: Correct option chosen (D).

The boiling temperature depends on the strength of the intermolecular forces. All four options are halogenoalkanes which exhibit London forces and permanent dipole-dipole forces. 1-iodobutane has the largest number of electrons and the largest surface area compared to the other options. This leads to the strongest London forces, requiring the most thermal energy to overcome, and thus results in the highest boiling temperature.

[1] Correct option identified (C). Reject all other options.

Thermal stability of Group 2 carbonates increases down the group. Going down Group 2, the cationic radius increases, so the charge density of the metal cation decreases. Consequently, the metal cation has a weaker polarizing effect on the carbonate ion's electron cloud, distorting it less and making the C-O bond harder to break. Therefore, barium carbonate is the most thermally stable and requires the highest temperature to decompose.

[1] Correct option identified (D). Reject other options.

The rate of precipitate formation depends on the rate of carbon-halogen bond fission. C-I bonds are weaker than C-Cl bonds, so iodoalkanes react faster than chloroalkanes. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster than primary halogenoalkanes because they react via the \(S_N1\) mechanism involving a highly stable tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane reacts the fastest.

[1] Correct option identified (D). Reject other options.

The equation for the combustion of ethane is: \(C_2H_6(g) + 3.5O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\). Using Hess's Law: \(\Delta H_{\text{c}}^\ominus = \sum \Delta H_{\text{f}}^\ominus(\text{products}) - \sum \Delta H_{\text{f}}^\ominus(\text{reactants})\). This gives \(\Delta H_{\text{c}}^\ominus = [2 \times (-393.5) + 3 \times (-285.8)] - [-84.7] = [-787.0 - 857.4] + 84.7 = -1644.4 + 84.7 = -1559.7 \text{ kJ mol}^{-1}\).

[1] Correct option identified (B). Reject other options.

A catalyst provides an alternative reaction pathway with a lower activation energy. It does not alter the actual kinetic energies of the molecules, so the shape of the Maxwell-Boltzmann distribution curve (including the peak position and total area) remains unchanged. However, the activation energy threshold (\(E_{\text{a}}\) value) shifts to the left (a lower energy value) on the horizontal axis, meaning a larger fraction of molecules have energy greater than or equal to the activation energy.

[1] Correct option identified (C). Reject other options.

In the reaction of sodium bromide with concentrated sulfuric acid, hydrogen bromide (HBr) is initially produced in an acid-base reaction (not a redox reaction). Bromide ions are strong enough reducing agents to reduce concentrated sulfuric acid to sulfur dioxide (\(SO_2\)), in which the oxidation state of sulfur decreases from +6 to +4. Bromide is oxidized to bromine (\(Br_2\)), which is the oxidation product. Bromide ions are not strong enough reducing agents to reduce sulfur to hydrogen sulfide (\(H_2S\)), which only occurs with iodide ions.

[1] Correct option identified (C). Reject other options.

The empirical formula mass of \(C_2H_4O\) is \(2(12.0) + 4(1.0) + 16.0 = 44.0\). Given the molecular ion peak is at \(m/z = 88\), the molecular formula must be twice the empirical formula, which is \(C_4H_8O_2\). Among the options: Butan-1-ol is \(C_4H_{10}O\) (\(M_{\text{r}} = 74\)), Butanoic acid is \(C_4H_8O_2\) (\(M_{\text{r}} = 88\)), Diethyl ether is \(C_4H_{10}O\) (\(M_{\text{r}} = 74\)), and Propan-1-ol is \(C_3H_8O\) (\(M_{\text{r}} = 60\)). Therefore, the compound must be butanoic acid.

[1] Correct option identified (B). Reject other options.

Enthalpy change of reaction using bond enthalpies is calculated as: \(\Delta H = \sum \text{Bond enthalpies of bonds broken} - \sum \text{Bond enthalpies of bonds formed}\). Bonds broken: 1 \(\times\) C-H bond = 413, 1 \(\times\) Cl-Cl bond = 243. Total bonds broken = \(413 + 243 = 656 \text{ kJ mol}^{-1}\). Bonds formed: 1 \(\times\) C-Cl bond = 346, 1 \(\times\) H-Cl bond = 432. Total bonds formed = \(346 + 432 = 778 \text{ kJ mol}^{-1}\). \(\Delta H = 656 - 778 = -122 \text{ kJ mol}^{-1}\).

[1] Correct option identified (A). Reject other options.

Propanone (\( \text{CH}_3\text{COCH}_3 \)) contains a highly polar carbonyl group (\( \text{C}=\text{O} \)), giving it a permanent dipole. The lone pairs on the highly electronegative oxygen atom can accept hydrogen bonds from the highly positive hydrogen atoms (\( \text{H}^{\delta+} \)) of water molecules. However, since propanone contains no hydrogen atoms directly bonded to a highly electronegative atom (N, O, or F), it cannot form hydrogen bonds with other propanone molecules. Propan-1-ol can form hydrogen bonds with itself. Butane is non-polar and cannot form hydrogen bonds. Chloromethane has a permanent dipole but does not form hydrogen bonds with water as chlorine is not sufficiently electronegative.

[1] Correct option B (1 mark).

Thermal stability of Group 2 nitrates increases down the group. As you go down the group from Mg to Ba, the cationic radius of the \( \text{M}^{2+} \) ion increases while its charge remains constant. This leads to a decrease in charge density and polarizing power. Consequently, the larger barium ion (\( \text{Ba}^{2+} \)) polarizes and distorts the electron cloud of the nitrate anion less than the smaller cations, making the nitrate N-O bonds more stable and requiring the highest temperature to decompose.

[1] Correct option D (1 mark).

The rate of hydrolysis of halogenoalkanes is determined by two main factors: the strength of the carbon-halogen bond and the stability of the intermediate carbocation. The C-I bond is the longest and weakest (lowest bond enthalpy) compared to C-Br and C-Cl bonds, meaning it breaks most easily. Furthermore, 2-iodo-2-methylpropane is a tertiary halogenoalkane, which reacts via the \( \text{S}_\text{N}1 \) mechanism. The tertiary carbocation intermediate is highly stabilized by the inductive effect of three alkyl groups, making this pathway significantly faster than those of primary or secondary halogenoalkanes.

[1] Correct option D (1 mark).

At a higher temperature, the average kinetic energy of the molecules increases. This shifts the Maxwell-Boltzmann distribution curve to the right (higher energy) and flattens it (the peak becomes lower in height) to keep the total area under the curve constant (as the number of particles is constant). The activation energy (\( E_a \)) is a characteristic constant of a given chemical reaction and is independent of temperature; only the addition of a catalyst can change the activation energy of the reaction pathway.

[1] Correct option B (1 mark).

**Part (a)(i)**
When 2-bromo-2-methylpropane is heated with aqueous KOH, the hydroxide ion acts as a nucleophile, replacing the bromine atom. The product is 2-methylpropan-2-ol, \((CH_3)_3COH\). The reaction type is nucleophilic substitution.

**Part (a)(ii)**
When heated with ethanolic KOH, elimination takes place. The hydroxide ion acts as a base, abstracting a proton from an adjacent carbon atom. The product is methylpropene, \((CH_3)_2C=CH_2\). The reaction type is elimination, and the hydroxide ion acts as a base.

**Part (b)(i)**
- Step 1: The C-Br bond breaks heterolytically to form a tertiary carbocation intermediate and a bromide ion: \((CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-\). Draw a curly arrow starting from the C-Br bond pointing towards the Br atom. Include a dipole showing \(\delta+\) on the central carbon and \(\delta-\) on the bromine.
- Step 2: The hydroxide ion attacks the carbocation: \((CH_3)_3C^+ + OH^- \rightarrow (CH_3)_3COH\). Draw a curly arrow from the lone pair on the oxygen of the \(OH^-\)
ion to the positively charged carbon of the carbocation.

**Part (b)(ii)**
2-bromo-2-methylpropane is a tertiary halogenoalkane. It reacts via \(S_N1\) because the tertiary carbocation intermediate is highly stable due to the positive inductive effect of three electron-releasing methyl groups. Furthermore, steric hindrance by the bulky methyl groups prevents direct nucleophilic attack from the back, making an \(S_N2\) pathway highly unfavorable.

**Part (c)(i)**
- Reagent: Acidified potassium dichromate(VI) / \(K_2Cr_2O_7 + H_2SO_4\)
- Condition: Heat under reflux (to ensure complete oxidation to the carboxylic acid)

**Part (c)(ii)**
- Alcohol Y (butan-1-ol) shows a characteristic broad absorption due to the O-H (alcohol) bond in the range \(3200-3600\text{ cm}^{-1}\) and lacks any C=O absorption.
- Carboxylic acid Z (butanoic acid) shows a very broad absorption due to the O-H (acid) bond in the range \(2500-3300\text{ cm}^{-1}\) and a strong, sharp absorption due to the C=O bond in the range \(1675-1750\text{ cm}^{-1}\).

**Part (a)(i)**
- Award 1 mark for correct identification of the product: 2-methylpropan-2-ol (allow name or correct structural formula).
- Award 1 mark for stating reaction type: nucleophilic substitution.

**Part (a)(ii)**
- Award 1 mark for correct identification of the product: methylpropene (allow name or correct structural formula).
- Award 1 mark for stating reaction type: elimination.
- Award 1 mark for stating the role of the hydroxide ion: base (accept proton acceptor).

**Part (b)(i)**
- Award 1 mark for a correct curly arrow from the C-Br bond to the Br atom, showing the heterolytic fission.
- Award 1 mark for drawing the correct structure of the carbocation intermediate, showing the positive charge on the central carbon atom: \((CH_3)_3C^+\).
- Award 1 mark for drawing a correct curly arrow from a lone pair on the oxygen of the \(OH^-\)
ion to the positively charged carbon of the intermediate.
- Award 1 mark for showing correct partial charges on the C-Br bond: \(\text{C}^{\delta+}\text{-Br}^{\delta-}\).

**Part (b)(ii)**
- Award 1 mark for explaining that the tertiary carbocation intermediate is stable (due to electron-releasing methyl groups) OR steric hindrance of the methyl groups prevents back-side attack.

**Part (c)(i)**
- Award 1 mark for identifying the correct reagent: acidified potassium dichromate(VI) / \(K_2Cr_2O_7\) and \(H_2SO_4\) (allow other suitable oxidizing agents such as acidified sodium dichromate(VI)). Reject potassium manganate(VII) if no mention of acid.
- Award 1 mark for specifying the correct condition: heating under reflux (must mention reflux to achieve carboxylic acid).

**Part (c)(ii)**
- Award 1 mark for identifying the O-H bond in alcohol Y at \(3200-3600\text{ cm}^{-1}\) (or lack of C=O).
- Award 1 mark for identifying the O-H bond in carboxylic acid Z at \(2500-3300\text{ cm}^{-1}\) and the C=O bond at \(1675-1750\text{ cm}^{-1}\).

**Part (a)(i)**
Magnesium carbonate decomposes on heating to yield solid magnesium oxide and carbon dioxide gas:
\[MgCO_3(s) \rightarrow MgO(s) + CO_2(g)\]

**Part (a)(ii)**
- Trend: The thermal stability of Group 2 carbonates increases down the group from magnesium to barium (they require higher temperatures to decompose).
- Explanation:
- The ionic radius / size of the Group 2 cation increases down the group.
- The charge on the cation remains constant at \(2+\).
- Therefore, the charge density of the cation decreases down the group.
- The smaller magnesium ion has a higher charge density and is more polarising, so it distorts the electron cloud of the carbonate ion (\(CO_3^{2-}\)) more strongly.
- This weakening of the carbon-oxygen bonds in the carbonate ion makes it easier to break on heating. Barium, with its larger cation, has a much weaker polarising effect, meaning more heat energy is required to decompose its carbonate.

**Part (b)(i)**
- Misty fumes: Hydrogen chloride gas / \(HCl(g)\).
- Equation:
\[NaCl(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HCl(g)\]
(Note: The reaction forming sodium sulfate, \(2NaCl(s) + H_2SO_4(l) \rightarrow Na_2SO_4(s) + 2HCl(g)\), is also acceptable if balanced correctly, though sodium hydrogensulfate is the major product under standard laboratory conditions.)

**Part (b)(ii)**
- Purple vapor: Iodine / \(I_2\)
- Yellow solid: Sulfur / \(S\)
- Gas with a smell of rotten eggs: Hydrogen sulfide / \(H_2S\)
- Explanation:
- Iodide ions (\(I^-\)) are much stronger reducing agents than chloride ions (\(Cl^-\)).
- Chloride ions are not strong enough to reduce the sulfur in sulfuric acid (which remains at an oxidation state of +6). Iodide ions, being larger with more shielding, lose their outer shell electrons much more easily and reduce sulfur from +6 in \(H_2SO_4\) to 0 in \(S\) and -2 in \(H_2S\).

**Part (a)(i)**
- Award 1 mark for correct reactant and product formulas: \(MgCO_3 \rightarrow MgO + CO_2\).
- Award 1 mark for correct state symbols: \((s) \rightarrow (s) + (g)\).

**Part (a)(ii)**
- Award 1 mark for stating that thermal stability increases down the group.
- Award 1 mark for stating that the ionic radius/size of the cation increases down the group (whilst charge remains \(2+\)).
- Award 1 mark for stating that the charge density of the cation decreases down the group.
- Award 1 mark for explaining that the polarising power of the cation decreases (or that the smaller magnesium ion has a higher polarising power).
- Award 1 mark for explaining that this polarising action distorts / weakens the carbon-oxygen bond (C-O) in the carbonate ion, making decomposition easier.

**Part (b)(i)**
- Award 1 mark for identifying the misty fumes as hydrogen chloride / \(HCl\) (reject hydrochloric acid).
- Award 1 mark for a balanced equation: \(NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl\) (or balanced equation forming \(Na_2SO_4\)).

**Part (b)(ii)**
- Award 1 mark for identifying all three observations correctly: Purple vapor = iodine (\(I_2\)); Yellow solid = sulfur (\(S\)); Gas with smell of rotten eggs = hydrogen sulfide (\(H_2S\)). (Award 1 mark for any two correct, 2 marks for all three).
- Award 1 mark for stating that iodide ions are stronger reducing agents than chloride ions.
- Award 1 mark for explaining that iodide ions are larger / have more electron shells / more shielding, so their outer electrons are lost more easily.

**Part (a)(i)**
- Expression:
\[K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\]
- Units:
\[\text{Units} = \frac{\text{mol dm}^{-3}}{\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2} = \frac{1}{\text{mol}^2\text{ dm}^{-6}} = \text{dm}^6\text{ mol}^{-2}\]

**Part (a)(ii)**
First, calculate the equilibrium concentrations by dividing the number of moles by the volume (\(2.0\text{ dm}^3\)):
- \([CO] = \frac{0.15\text{ mol}}{2.0\text{ dm}^3} = 0.075\text{ mol dm}^{-3}\)
- \([H_2] = \frac{0.24\text{ mol}}{2.0\text{ dm}^3} = 0.12\text{ mol dm}^{-3}\)
- \([CH_3OH] = \frac{0.38\text{ mol}}{2.0\text{ dm}^3} = 0.19\text{ mol dm}^{-3}\)

Now substitute these values into the \(K_c\) expression:
\[K_c = \frac{0.19}{0.075 \times (0.12)^2} = \frac{0.19}{0.075 \times 0.0144} = \frac{0.19}{0.00108} \approx 175.93\text{ dm}^6\text{ mol}^{-2}\]

Rounding to two or three significant figures gives \(180\) or \(176\text{ dm}^6\text{ mol}^{-2}\).

**Part (a)(iii)**
- Effect: The equilibrium yield of methanol will decrease.
- Reason: The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature will shift the equilibrium position to the left (in the endothermic direction) to absorb the added heat.

**Part (a)(iv)**
- Effect: No change.
- Reason: The value of the equilibrium constant \(K_c\) is only affected by changes in temperature. Changes in pressure do not alter \(K_c\) (even though they alter the position of equilibrium).

**Part (b)(i)**
- Axes: The y-axis should be labelled \(\text{Number of molecules}\) (or \(\text{Fraction of molecules with energy } E\)) and the x-axis should be labelled \(\text{Energy}\) (or \(\text{Kinetic energy}\)).
- Curve shape: Starts at the origin, rises to a peak, and then decreases asymmetrically towards the x-axis, never quite touching it (asymptotic).
- Labels: \(E_{cat}\) should be positioned to the left of \(E_a\) on the x-axis, showing that the activation energy with a catalyst is lower than without a catalyst.

**Part (b)(ii)**
- A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{cat} < E_a\)).
- Therefore, a significantly larger fraction of molecules possess kinetic energy greater than or equal to \(E_{cat}\) (as shown by the larger shaded area under the curve to the right of \(E_{cat}\)). This results in a higher frequency of successful collisions, thus increasing the rate of reaction.

**Part (a)(i)**
- Award 1 mark for the correct expression: \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\) (square brackets are mandatory; reject round brackets).
- Award 1 mark for correct units: \(dm^6 mol^{-2}\) (allow \(mol^{-2} dm^6\)).

**Part (a)(ii)**
- Award 1 mark for calculating the three equilibrium concentrations correctly:
- \([CO] = 0.075\text{ mol dm}^{-3}\)
- \([H_2] = 0.12\text{ mol dm}^{-3}\)
- \([CH_3OH] = 0.19\text{ mol dm}^{-3}\)
- Award 1 mark for correct substitution into their \(K_c\) expression: \(K_c = \frac{0.19}{0.075 \times (0.12)^2}\).
- Award 1 mark for correct final calculated value of 176 (or 180) to 2 or 3 significant figures. (Ignore incorrect units if already marked in part (i)).

**Part (a)(iii)**
- Award 1 mark for stating that the yield of methanol decreases.
- Award 1 mark for stating that the forward reaction is exothermic, so the equilibrium shifts to the left / in the endothermic direction to oppose the temperature rise.

**Part (a)(iv)**
- Award 1 mark for stating that there is no change in the value of \(K_c\).
- Award 1 mark for explaining that \(K_c\) is constant at a constant temperature / is only affected by temperature.

**Part (b)(i)**
- Award 1 mark for correct labels on axes: y-axis = Number/fraction of molecules; x-axis = Energy.
- Award 1 mark for drawing a correct asymmetric curve starting at the origin and asymptotic to the x-axis.
- Award 1 mark for marking \(E_a\) and \(E_{cat}\) on the x-axis, with \(E_{cat}\) to the left of \(E_a\).

**Part (b)(ii)**
- Award 1 mark for explaining that a catalyst provides an alternative pathway with a lower activation energy.
- Award 1 mark for stating that more molecules have energy \(\ge E_{cat}\) (or referring to the larger shaded area under the curve), leading to more successful collisions per unit time.

(a) Formation of hydrogen bromide: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (or \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\)). Excess concentrated sulfuric acid must be avoided because concentrated sulfuric acid is a strong oxidizing agent that can oxidize the bromide ions (or \(\text{HBr}\)) to elemental bromine (\(\text{Br}_2\)), reducing the yield of the desired reactant.

(b)(i) The purpose of washing with aqueous sodium hydrogencarbonate is to neutralize any remaining acid impurities (such as sulfuric acid or hydrogen bromide).

(b)(ii) Safety precaution: Release the pressure in the separating funnel frequently by inverting the funnel and carefully opening the tap. Explanation: The neutralization reaction produces carbon dioxide (\(\text{CO}_2\)) gas, which causes a rapid build-up of pressure inside the closed funnel.

(c)(i) Suitable drying agents: anhydrous calcium chloride (\(\text{CaCl}_2\)), anhydrous magnesium sulfate (\(\text{MgSO}_4\)), or anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)).

(c)(ii) The organic liquid changes from cloudy/milky to clear/transparent.

(d) Yield calculation:
1. Calculate moles of reactant (butan-1-ol): \(n(\text{butan-1-ol}) = \frac{10.0\text{ g}}{74.1\text{ g mol}^{-1}} = 0.13495\text{ mol}\)
2. Theoretical moles of 1-bromobutane = \(0.13495\text{ mol}\) (since the stoichiometry is 1:1)
3. Theoretical mass of 1-bromobutane: \(0.13495\text{ mol} \times 137.0\text{ g mol}^{-1} = 18.488\text{ g}\)
4. Percentage yield: \(\text{Percentage Yield} = \frac{11.5\text{ g}}{18.488\text{ g}} \times 100\% = 62.202\%\). To 3 significant figures, this is \(62.2\%\).

(e) Mechanism (\(S_N2\)):
1. The hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile. A curly arrow starts from a lone pair of electrons on the oxygen atom of the \(\text{OH}^-\) ion and points to the carbon atom of the C-Br bond.
2. Dipoles must be shown on the polar C-Br bond: \(\text{C}^{\delta+}-\text{Br}^{\delta-}\).
3. A second curly arrow starts from the C-Br bond and points to the bromine atom.
4. This is a single-step concerted mechanism (or via a transition state with partial bonds to OH and Br) yielding butan-1-ol and a bromide ion (\(\text{Br}^-\)).

(f) Infrared spectroscopy comparison:
- Butan-1-ol contains an O-H (alcohol) bond, which shows a broad, strong absorption band in the range \(3200 - 3750\text{ cm}^{-1}\), and a C-O bond, which shows an absorption band in the range \(1000 - 1300\text{ cm}^{-1}\).
- 1-bromobutane does not contain O-H or C-O bonds, but instead contains a C-Br bond, which shows an absorption band in the range \(500 - 600\text{ cm}^{-1}\).
- Complete conversion is confirmed by the complete disappearance of the broad absorption band at \(3200 - 3750\text{ cm}^{-1}\) and the appearance of the C-Br absorption band at \(500 - 600\text{ cm}^{-1}\).

(a) [2 marks]
- 1 mark: Correct balanced equation for the formation of \(\text{HBr}\). (Accept: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) or \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\)).
- 1 mark: Correct explanation that concentrated sulfuric acid is an oxidizing agent that oxidizes hydrogen bromide / bromide ions to bromine.

(b)(i) [1 mark]
- 1 mark: State that it neutralizes / reacts with remaining acid (accept sulfuric acid / hydrogen bromide / acid impurities). Reject: 'to remove impurities' without specifying acid neutralization.

(b)(ii) [2 marks]
- 1 mark: Invert and open the tap regularly / vent the funnel to release gas / pressure.
- 1 mark: State that carbon dioxide (\(\text{CO}_2\)) gas is produced causing pressure build-up.

(c)(i) [1 mark]
- 1 mark: Suggest anhydrous \(\text{CaCl}_2\) / \(\text{MgSO}_4\) / \(\text{Na}_2\text{SO}_4\). Reject: anhydrous copper(II) sulfate or cobalt(II) chloride.

(c)(ii) [1 mark]
- 1 mark: Liquid changes from cloudy to clear / transparent.

(d) [4 marks]
- 1 mark: Calculation of moles of butan-1-ol = \(0.135\text{ mol}\) (allow 0.13495).
- 1 mark: Calculation of theoretical mass of 1-bromobutane = \(18.5\text{ g}\) (allow 18.488).
- 1 mark: Correct percentage yield calculation setup (Actual mass / Theoretical mass x 100).
- 1 mark: Final value of \(62.2\%\) (must be rounded correctly to 3 significant figures). Correct answer with no working scores 4 marks.

(e) [4 marks]
- 1 mark: Curly arrow starting from a lone pair on the oxygen of \(\text{OH}^-\) to the \(\text{C}\) atom attached to the halogen.
- 1 mark: Dipole correctly labeled on the \(\text{C-Br}\) bond (\(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\)).
- 1 mark: Curly arrow starting from the \(\text{C-Br}\) bond to the \(\text{Br}\) atom.
- 1 mark: Correct organic product (butan-1-ol) and leaving group (\(\text{Br}^-\)) shown. (Max 3 marks if transition state is drawn incorrectly, but full marks can be given for a correctly shown transition state with dashed partial bonds and correct charge overall).

(f) [3 marks]
- 1 mark: Mention the absence of the broad \(\text{O-H}\) alcohol absorption band in the region \(3200 - 3750\text{ cm}^{-1}\) (or absence of C-O stretch in region \(1000 - 1300\text{ cm}^{-1}\)).
- 1 mark: Mention the presence of the \(\text{C-Br}\) stretch in the region \(500 - 600\text{ cm}^{-1}\).
- 1 mark: Correctly link these changes to complete conversion of butan-1-ol to 1-bromobutane.

(a) 1. Weigh 1.45 g of the solid using a weighing bottle, then weigh the bottle after emptying to record the exact mass transferred by difference. 2. Dissolve the solid in a beaker containing less than 200 cm³ of deionised water, stirring with a glass rod. 3. Transfer the solution quantitatively to a 250 cm³ volumetric flask using a funnel. 4. Rinse the beaker, glass rod, and funnel with deionised water and add all washings to the flask. 5. Make up to the graduation mark with deionised water, adding the last drops with a pipette until the bottom of the meniscus is on the mark. Stopper and invert several times to mix.

(b) Titrations 1, 2, and 3 are concordant because they are within \(\pm 0.10 \text{ cm}^3\) of each other. The rough titration is discarded. Mean titre = \(\frac{20.35 + 20.25 + 20.30}{3} = 20.30 \text{ cm}^3\).

(c) Moles of \(\text{NaOH} = 0.0800 \times \frac{20.30}{1000} = 1.624 \times 10^{-3} \text{ mol}\).
Moles of \(\text{H}_2\text{A}\) in 25.0 cm³ = \(\frac{1.624 \times 10^{-3}}{2} = 8.12 \times 10^{-4} \text{ mol}\).
Moles of \(\text{H}_2\text{A}\) in 250.0 cm³ = \(8.12 \times 10^{-4} \times 10 = 8.12 \times 10^{-3} \text{ mol}\).
Molar mass of \(\text{H}_2\text{A} = \frac{1.45 \text{ g}}{8.12 \times 10^{-3} \text{ mol}} = 178.57 \text{ g mol}^{-1}\), which rounds to 179 \(\text{g mol}^{-1}\) (3 sig figs).

(d) Percentage uncertainty = \(\frac{0.15}{250.0} \times 100\% = 0.060\%\).

(a) Max [5 marks]:
- Weigh by difference OR weigh in a weighing bottle and record mass [1]
- Dissolve solid in a beaker with deionised water and stir with a glass rod [1]
- Quantitative transfer to volumetric flask including washings of beaker and rod [1]
- Fill to graduation mark such that the bottom of the meniscus is on the line [1]
- Stopper and invert flask several times [1]

(b) [2 marks]:
- Identifies concordant titres (1, 2, and 3) OR states they are within 0.10 cm³ [1]
- Correct calculation of mean titre = 20.30 cm³ [1]

(c) [4.5 marks]:
- Moles of NaOH = \(1.624 \times 10^{-3}\) [1]
- Moles of H2A in 25 cm³ = \(8.12 \times 10^{-4}\) (moles of NaOH / 2) [1]
- Moles of H2A in 250 cm³ = \(8.12 \times 10^{-3}\) (moles in 25 cm³ x 10) [1]
- Molar mass calculation: \(1.45 / 8.12 \times 10^{-3} = 178.57\) [1]
- Give answer to 3 sig figs (179) with correct units (g/mol) [0.5]

(d) [1 mark]:
- Percentage uncertainty = 0.060% [1]

(a) Addition and mixing of the reactants takes time, so the reaction is not instantaneous and the temperature at exactly 4 minutes cannot be measured accurately during active mixing.

(b) 1. Plot a graph of temperature against time. 2. Draw a horizontal line of best fit through the initial points (0 to 3 minutes) and a straight cooling line of best fit back-extrapolated from the cooling points (5 to 10 minutes) to 4 minutes. 3. Find the vertical temperature difference between these two lines of best fit at the 4th minute (time of mixing).

(c) \(\Delta T = 31.0 - 19.5 = 11.5 \text{ °C}\).

(d) \(q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 11.5 \text{ K} = 2403.5 \text{ J}\).

(e) Moles of \(\text{CuSO}_4 = 0.200 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.0100 \text{ mol}\).
\(\Delta H = -\frac{q}{\text{moles}} = -\frac{2403.5 \text{ J}}{0.0100 \text{ mol}} = -240350 \text{ J mol}^{-1} = -240.35 \text{ kJ mol}^{-1}\).
To 3 significant figures, this is \(-240 \text{ kJ mol}^{-1}\).

(f) The reaction may not have gone to absolute completion, OR the heat capacity of the polystyrene cup/thermometer was neglected.

(a) [1 mark]:
- Reaction is not instantaneous / mixing takes time [1]

(b) [3 marks]:
- Plot temperature against time [1]
- Draw best-fit line through points 0-3 min AND back-extrapolate the cooling line from points 5-10 min to 4 min [1]
- Measure vertical distance between the two lines at 4 minutes [1]

(c) [1 mark]:
- \(\Delta T = 11.5 \text{ °C}\) [1]

(d) [2 marks]:
- use of \(q = m c \Delta T\) with mass = 50.0 g [1]
- \(q = 2403.5 \text{ J}\) (or \(2.40 \text{ kJ}\)) [1]

(e) [4.5 marks]:
- Moles of CuSO4 = 0.0100 mol [1]
- \(\Delta H = \frac{q}{\text{moles}}\) division [1]
- Negative sign included [1]
- Correct value 240 (or 240.4) [1]
- Correct unit \(\text{kJ mol}^{-1}\) and 3 sig figs (\(-240\)) [0.5]

(f) [1 mark]:
- Incomplete reaction OR zinc was oxidized/coated in copper stopping reaction OR heat capacity of cup ignored [1] (Do not accept heat loss)

(a) i) Clean a nichrome/platinum wire by dipping it into concentrated hydrochloric acid and placing it in a hot Bunsen flame until no colour is seen. Then dip the clean wire back into concentrated HCl, touch it to the solid sample, and place it into a non-luminous Bunsen flame. ii) The brick-red flame indicates the presence of calcium ions, \(\text{Ca}^{2+}\).

(b) i) A cream precipitate with silver nitrate indicates the presence of bromide ions, \(\text{Br}^-\). The ionic equation is: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\). ii) Adding dilute sulfuric acid to calcium bromide produces a white precipitate of calcium sulfate, \(\text{CaSO}_4\).

(c) i) Solid calcium bromide reacts with concentrated sulfuric acid to produce: 1. Steamy/misty fumes (due to \(\text{HBr}\)). 2. Orange/brown fumes (due to \(\text{Br}_2\) vapour). 3. A choking gas (due to \(\text{SO}_2\)). (Any two are sufficient). ii) Bromide ions are stronger reducing agents than chloride ions, so they can reduce concentrated sulfuric acid to sulfur dioxide, whereas chloride ions cannot reduce sulfuric acid.

(a) [4 marks total]:
- i) Dip nichrome/platinum wire in conc. HCl and heat in flame to clean [1]
- i) Dip wire in conc. HCl, touch sample, place in non-luminous (blue) flame [1]
- ii) Calcium / \(\text{Ca}^{2+}\) [1]

(b) [5 marks total]:
- i) Bromide / \(\text{Br}^-\)[1]
- i) \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\) equation [1], correct state symbols [1]
- ii) Observation: White precipitate / solid [1]
- ii) Name: Calcium sulfate [1]

(c) [3.5 marks total]:
- i) Any two from: Orange/brown fumes [1], steamy/misty fumes [1], choking gas [1] (Max 2 marks, plus 0.5 bonus if all observations are completely correct)
- ii) Bromide is a stronger reducing agent than chloride [1]

(a) Reflux prevents the loss of volatile organic reactants (butan-1-ol) and products (1-bromobutane) by condensing them back into the reaction flask, allowing them reaction time to reach completion without boiling dry.

(b) \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (or \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\)).

(c) i) Add the crude mixture and aqueous sodium hydrogencarbonate solution to the separating funnel. Stopper and shake gently. Invert the funnel and open the tap regularly to release pressure built up by carbon dioxide gas. Allow the layers to separate, then run off the lower organic layer (which is denser than water). ii) Dry with anhydrous calcium chloride (or anhydrous magnesium sulfate). The organic layer is dry when it changes from cloudy to clear.

(d) Transfer the dried liquid to a distillation flask and distil. Collect the fraction boiling over a narrow range around the boiling point of 1-bromobutane (101–104 °C). A sharp boiling point confirms the purity of the sample.

(e) Moles of butan-1-ol = \(\frac{7.41 \text{ g}}{74.1 \text{ g mol}^{-1}} = 0.100 \text{ mol}\).
Theoretical yield of 1-bromobutane = \(0.100 \text{ mol} \times 137.0 \text{ g mol}^{-1} = 13.70 \text{ g}\).
Percentage yield = \(\frac{8.22 \text{ g}}{13.70 \text{ g}} \times 100\% = 60.0\%\).

(a) [2 marks]:
- Prevents loss of volatile components / reactants / products [1]
- Condenses vapour back into flask to allow reaction to go to completion [1]

(b) [1 mark]:
- Correctly balanced equation for HBr formation [1]

(c) [5.5 marks total]:
- i) Add mixture and NaHCO3 solution to separating funnel and shake [1]
- i) Release pressure / vent CO2 by opening tap while inverted [1]
- i) Run off lower organic layer / separate layers [1.5]
- ii) Drying agent: Anhydrous calcium chloride / magnesium sulfate / sodium sulfate [1]
- ii) Appearance changes from cloudy to clear [1]

(d) [3 marks]:
- Perform simple distillation [1]
- Collect fraction boiling at 101-104 °C (boiling point of 1-bromobutane) [1]
- Pure compound boils over a narrow range [1]

(e) [1 mark]:
- Yield = 60.0% [1]