2025 Edexcel IAL Chemistry (YCH11) 模擬試題連答案詳解

When excess aqueous ammonia is added to hexaaquacopper(II) ions, a ligand substitution reaction occurs where four water ligands are replaced by four ammonia ligands. This forms the diaquatetraamminecopper(II) complex, \([Cu(NH_3)_4(H_2O)_2]^{2+}\), which is deep blue in color. The equation for this reaction is: \([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\).

1 mark: Correct identification of both the formula and the deep blue color (Option A).

Ethylamine is the most basic because the electron-donating ethyl group increases electron density on the nitrogen atom, making its lone pair more available to accept a proton. Ammonia has no alkyl groups, making it less basic than ethylamine. Phenylamine is the least basic because the lone pair of electrons on the nitrogen atom is partially delocalised into the \(\pi\)-system of the benzene ring, making it much less available to accept a proton.

1 mark: Correctly ranking ethylamine as the most basic and phenylamine as the least basic (Option A).

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen (C-X) bond. Since the C-I bond is the longest and weakest (has the lowest bond enthalpy), it is broken most easily. Thus, 1-iodobutane reacts the fastest to form a precipitate of silver iodide. The high bond polarity of C-Cl does not make it react faster because bond enthalpy is the dominant factor determining the rate of nucleophilic substitution here.

1 mark: Correctly identifying that 1-iodobutane is the fastest due to lowest bond enthalpy (Option C).

The central chlorine atom in \(\text{ClF}_3\) has 7 valence electrons, plus 3 electrons from the 3 fluorine-chlorine single bonds, giving 10 electrons in its outer shell (5 electron pairs). There are 3 bonding pairs and 2 lone pairs. According to VSEPR theory, the 5 pairs arrange themselves in a trigonal bipyramidal base, with the 2 lone pairs occupying equatorial positions to minimize repulsion. This results in a T-shaped molecular geometry. Due to strong lone pair-bonding pair repulsions, the equatorial-axial-equatorial angles are compressed from \(90^\circ\) to approximately \(87.5^\circ\).

1 mark: Correct shape (T-shaped) and compressed bond angle (Option C).

The equation for the thermal decomposition of magnesium nitrate is: \(2\text{Mg(NO}_3)_2(s) \rightarrow 2\text{MgO}(s) + 4\text{NO}_2(g) + \text{O}_2(g)\). From this equation, \(2\text{ moles}\) of anhydrous \(\text{Mg(NO}_3)_2\) produce \(5\text{ moles}\) of gas (\(4\text{ NO}_2 + 1\text{ O}_2\)). Therefore, \(0.10\text{ mol}\) of \(\text{Mg(NO}_3)_2\) produces: \(0.10 \times \frac{5}{2} = 0.25\text{ mol}\) of gas. The total volume of gas is: \(0.25\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 6.0\text{ dm}^3\).

1 mark: Correct stoichiometry calculation and volume (Option C).

The atomic number of iron (Fe) is 26. Its neutral ground-state electronic configuration is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2\) (or \([Ar] 3\text{d}^6 4\text{s}^2\)). When forming transition metal cations, electrons are lost from the \(4\text{s}\) subshell first, and then from the \(3\text{d}\) subshell. To form the \(\text{Fe}^{3+}\) ion, 3 electrons are removed: 2 from \(4\text{s}\) and 1 from \(3\text{d}\). This leaves the configuration as \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5\).

1 mark: Correct electronic configuration (Option A).

Alanine is an amino acid. In strongly acidic conditions (such as pH 1), there is a high concentration of \(\text{H}^+\) ions. This causes both the basic amine group to be protonated to \(\text{-NH}_3^+\) and the acidic carboxylate group to remain fully protonated as \(\text{-COOH}\). The resulting species is a cation with a 1+ charge: \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\).

1 mark: Correctly identifying the protonated cationic form of alanine at low pH (Option B).

The sharp, strong peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C=O}\)). The absence of a broad peak at \(3200\text{--}3600\text{ cm}^{-1}\) means there is no alcohol (\(\text{O-H}\)) group. Propanone has the formula \(\text{C}_3\text{H}_6\text{O}\) and contains a carbonyl group but no hydroxyl group, which perfectly fits this spectrum. Propan-1-ol has molecular formula \(\text{C}_3\text{H}_8\text{O}\) and would show an \(\text{O-H}\) stretch but no \(\text{C=O}\). Propanoic acid has molecular formula \(\text{C}_3\text{H}_6\text{O}_2\). Methyl vinyl ether does not contain a carbonyl group.

1 mark: Correct identification of propanone as the compound fitting the IR details (Option C).

Cobalt has an atomic number of 27, so its neutral ground state electronic configuration is \( [\text{Ar}] 3\text{d}^7 4\text{s}^2 \). When forming a 2+ ion, the transition metal loses its two 4s electrons first, resulting in the ground state configuration of \( [\text{Ar}] 3\text{d}^7 \).

Correct answer (A) gives 1 mark. Any other response or no response gives 0 marks.

Ethylamine is a primary aliphatic amine. Its alkyl group is electron-releasing via the inductive effect, which increases the electron density on the nitrogen atom and makes its lone pair more available to accept a proton. In ammonia, there are no alkyl groups to release electrons. In phenylamine, the lone pair on the nitrogen atom is delocalized into the benzene pi system, making it less available. In ethanamide, the lone pair is strongly delocalized onto the carbonyl oxygen atom, rendering it almost neutral.

Correct answer (C) gives 1 mark. Any other response or no response gives 0 marks.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. Down Group 7, the bond length increases and the bond enthalpy (bond strength) decreases: \( \text{C-Cl} > \text{C-Br} > \text{C-I} \). Since the C-I bond is the weakest, it is broken most easily, resulting in the fastest rate of reaction and precipitate formation. The C-Cl bond is the strongest and reacts the slowest.

Correct answer (A) gives 1 mark. Any other response or no response gives 0 marks.

Water (\( \text{H}_2\text{O} \)) has two bonding pairs and two lone pairs on the central oxygen atom. According to VSEPR theory, the four electron domains arrange themselves tetrahedrally to minimize repulsion, but the presence of two lone pairs repels the bonding pairs more strongly, reducing the standard tetrahedral angle from \( 109.5^\circ \) to \( 104.5^\circ \). Carbon dioxide is linear (\( 180^\circ \)), sulfur dioxide is bent with a bond angle of around \( 119^\circ \) due to having one lone pair, and methane is tetrahedral (\( 109.5^\circ \)).

Correct answer (B) gives 1 mark. Any other response or no response gives 0 marks.

The chemical equation for the reaction is: \( \text{CaCO}_3\text{(s)} + 2\text{HCl}\text{(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O}\text{(l)} + \text{CO}_2\text{(g)} \). First, calculate the amount in moles of \( \text{CaCO}_3 \): \( n = \frac{5.00}{100.1} = 0.0500\text{ mol} \). According to the 1:1 stoichiometry, the amount of \( \text{CO}_2 \) produced is also \( 0.0500\text{ mol} \). The volume of \( \text{CO}_2 \) at rtp is: \( V = 0.0500\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 1.20\text{ dm}^3 \).

Correct answer (A) gives 1 mark. Any other response or no response gives 0 marks.

The oxidation states of vanadium and their corresponding solution colors are: +5 is yellow (\( \text{VO}_2^+ \)), +4 is blue (\( \text{VO}^{2+} \)), +3 is green (\( \text{V}^{3+} \)), and +2 is violet (\( \text{V}^{2+} \)). Thus, the correct sequence of colors observed during the reduction process is Yellow to Blue to Green to Violet.

Correct answer (A) gives 1 mark. Any other response or no response gives 0 marks.

In a strongly alkaline solution, there is a high concentration of hydroxide ions which deprotonate the amino acid. The amine group exists in its neutral, unprotonated form (\( \text{-NH}_2 \)) and the carboxylic acid group exists in its deprotonated conjugate base form (\( \text{-COO}^- \)). Therefore, the entire species is anionic: \( \text{CH}_3\text{CH(NH}_2)\text{COO}^- \).

Correct answer (A) gives 1 mark. Any other response or no response gives 0 marks.

The molecular formula \( \text{C}_3\text{H}_6\text{O} \) corresponds to a compound with one degree of unsaturation (either a double bond or a ring). The strong, sharp peak at \( 1715\text{ cm}^{-1} \) indicates the presence of a carbonyl group (\( \text{C}=\text{O} \)). The absence of a broad band in the \( 3200\text{--}3600\text{ cm}^{-1} \) range indicates that there is no hydroxyl (\( \text{O-H} \)) group, which rules out propan-2-ol. Propanoic acid contains an \( \text{O-H} \) group and has the molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \). Methyl vinyl ether has the correct molecular formula but lacks a carbonyl group. Therefore, the compound is propanone.

Correct answer (B) gives 1 mark. Any other response or no response gives 0 marks.

The addition of excess aqueous ammonia to copper(II) ions results in the formation of \([Cu(NH_3)_4(H_2O)_2]^{2+}\). In this complex, the coordination number remains 6 as four water molecules are substituted by four ammonia molecules, leaving two water molecules in the axial positions of the distorted octahedral structure.

Award 1 mark for the correct answer B. Reject all other options.

Aliphatic amines are stronger bases than ammonia due to the positive inductive effect of alkyl groups, which increase the electron density on the nitrogen atom, making its lone pair more available for donation to a proton. Diethylamine, being a secondary amine with two ethyl groups, has a stronger inductive effect than the primary amine ethylamine, making it the strongest base among the options. Phenylamine is the weakest base because the lone pair of electrons on the nitrogen is delocalised into the benzene ring.

Award 1 mark for the correct answer D. Reject all other options.

The broad absorption band at \(3350\text{ cm}^{-1}\) in the infrared spectrum indicates the presence of an alcohol \(O-H\) group. The compound \(X\) has the formula \(C_4H_{10}O\) (molecular mass = 74). In the mass spectrum, 2-methylpropan-2-ol, \((CH_3)_3COH\), fragments by losing a methyl group (mass 15) to form the very stable tertiary carbocation-like fragment \([(CH_3)_2COH]^+\). This corresponds to a major peak at \(m/z = 59\). Since it has no \(-CH_2OH\) or \(-CH(OH)CH_3\) groups, it does not produce significant peaks at \(m/z = 31\) or \(m/z = 45\), unlike the other options.

Award 1 mark for the correct answer D. Reject all other options.

The amide ion, \(NH_2^-\), has two bonding pairs of electrons and two lone pairs of electrons around the central nitrogen atom, giving a tetrahedral arrangement of electron pairs. Due to the extra repulsion from the two lone pairs, the bond angle is reduced from the perfect tetrahedral angle of \(109.5^\circ\) to approximately \(104.5^\circ\). \(CO_2\) and \(NO_2^+\) are linear (\(180^\circ\)), and \(SO_2\) has only one lone pair on the sulfur atom, resulting in a bent shape with a bond angle of approximately \(119^\circ\).

Award 1 mark for the correct answer B. Reject all other options.

First, find the moles of \(H_2\): \(n(H_2) = 1.44 / 24.0 = 0.060\text{ mol}\). According to the stoichiometric equation, \(2\text{ moles of Al}\) produce \(3\text{ moles of H}_2\). Thus, \(n(\text{Al}) = 0.060 \times 2 / 3 = 0.040\text{ mol}\). The mass of pure aluminium is \(0.040 \times 27.0 = 1.08\text{ g}\). The percentage purity of the aluminium sample is \((1.08 / 1.35) \times 100\% = 80.0\%\).

Award 1 mark for the correct answer C. Reject all other options.

The electronic configuration of \(\text{Co}^{2+}\) is \([Ar] 3d^7\). The seven d-electrons occupy five d-orbitals as two pairs and three unpaired electrons. The electronic configuration of \(\text{Cr}^{3+}\) is \([Ar] 3d^3\), which contains exactly three unpaired d-electrons. Therefore, \(\text{Cr}^{3+}\) has the same number of unpaired d-electrons. Both \(\text{Fe}^{3+}\) and \(\text{Mn}^{2+}\) have the \([Ar] 3d^5\) configuration with five unpaired electrons, and \(\text{Ni}^{2+}\) has the \([Ar] 3d^8\) configuration with two unpaired electrons.

Award 1 mark for the correct answer C. Reject all other options.

At a pH much higher than the isoelectric point (pH 12.0 > 6.0), the solution is strongly alkaline and contains a high concentration of hydroxide ions. Under these conditions, the amino acid acts as an acid and loses protons. Both the carboxylic acid group and the protonated amine group are deprotonated to form \(\text{-COO}^-\) and \(\text{-NH}_2\) respectively, making the overall ion anionic with the structure \(\text{CH}_3\text{CH(NH}_2)\text{COO}^-\).

Award 1 mark for the correct answer D. Reject all other options.

1-Bromobutane is a primary halogenoalkane and reacts via the \(S_N2\) mechanism. This is a single-step concerted process with a bimolecular rate-determining step, so its rate is dependent on both the concentrations of the halogenoalkane and the hydroxide nucleophile. 2-Bromo-2-methylpropane is a tertiary halogenoalkane and reacts via the \(S_N1\) mechanism, which has a unimolecular rate-determining step (formation of a carbocation) and does not depend on the concentration of the hydroxide ion. Tertiary halogenoalkanes hydrolyse much faster than primary halogenoalkanes due to the high stability of the intermediate tertiary carbocation.

Award 1 mark for the correct answer D. Reject all other options.

The electronic configurations of the neutral transition metals and their respective ions are:
- \(Mn\): \([Ar]\, 3d^5 4s^2\). For \(Mn^{2+}\), two electrons are removed from the outer \(4s\) subshell, leaving \([Ar]\, 3d^5\).
- \(Fe\): \([Ar]\, 3d^6 4s^2\). For \(Fe^{2+}\), two electrons are removed from the outer \(4s\) subshell, leaving \([Ar]\, 3d^6\).
- \(Cr\): \([Ar]\, 3d^5 4s^1\). For \(Cr^{3+}\), one electron from the \(4s\) and two from the \(3d\) subshells are removed, leaving \([Ar]\, 3d^3\).
- \(Co\): \([Ar]\, 3d^7 4s^2\). For \(Co^{3+}\), two electrons from the \(4s\) and one from the \(3d\) subshells are removed, leaving \([Ar]\, 3d^6\).

1 mark for selecting the correct option (A).

When excess concentrated hydrochloric acid is added to aqueous \(Cu^{2+}\), ligand substitution occurs. Due to the larger size of the chloride ligand compared to water, only four chloride ligands can fit around the central copper ion to avoid steric hindrance. This results in the formation of a tetrahedral complex with the formula \([CuCl_4]^{2-}\). The coordination number changes from 6 to 4, and the geometry changes from octahedral to tetrahedral.

1 mark for selecting the correct option (B).

In aqueous solution, the basicity of amines is determined by the availability of the lone pair of electrons on the nitrogen atom to accept a proton. Aliphatic alkyl groups are electron-donating by the inductive effect, which increases the electron density on the nitrogen atom, making aliphatic amines stronger bases than ammonia. Diethylamine (a secondary aliphatic amine) has two electron-donating ethyl groups, making it a stronger base than ethylamine (a primary aliphatic amine). Phenylamine is a much weaker base because the lone pair of electrons on the nitrogen atom is delocalised into the aromatic \(\pi\)-system of the benzene ring, making it much less available to accept a proton.

1 mark for selecting the correct option (C).

In a highly acidic solution (\(pH = 1.0\)), there is a very high concentration of \(H^+\) ions. These protons associate with both basic and acidic groups on the amino acid. The carboxylate group is protonated to form carboxylic acid (\(-COOH\)), and the amine group is protonated to form an ammonium ion (\(-NH_3^+\)). Thus, the predominant species in solution is the cation \(CH_3CH(NH_3^+)COOH\).

1 mark for selecting the correct option (C).

Primary halogenoalkanes (such as \(CH_3CH_2CH_2CH_2Br\) and \((CH_3)_2CHCH_2Br\)) react almost exclusively via an \(S_N2\) mechanism due to steric accessibility and the instability of primary carbocations. Secondary halogenoalkanes (such as \(CH_3CH_2CH(Br)CH_3\)) react via a mixture of \(S_N1\) and \(S_N2\). Tertiary halogenoalkanes (such as \((CH_3)_3CBr\)) react primarily via an \(S_N1\) mechanism because the bulky methyl groups hinder nucleophilic attack from the backside, and the tertiary carbocation intermediate formed is highly stable due to the positive inductive electron-donating effects of three surrounding methyl groups.

1 mark for selecting the correct option (D).

Bromine exists naturally as two major stable isotopes, \(^{79}Br\) and \(^{81}Br\), in an approximate abundance ratio of \(1:1\). Therefore, any molecular ion or fragment that contains exactly one bromine atom will exhibit two distinct isotopic peaks separated by 2 mass units (representing \(M\) containing \(^{79}Br\) and \(M+2\) containing \(^{81}Br\)) with approximately equal relative heights (a ratio of \(1:1\)).

1 mark for selecting the correct option (B).

To find the molecular geometry:
- For \(SF_4\), sulfur has 6 valence electrons and forms 4 single covalent bonds with fluorine, leaving 1 lone pair. This gives 5 electron pairs around the central atom, leading to a trigonal bipyramidal electron-pair geometry. The lone pair occupies an equatorial position to minimise repulsion, which gives a see-saw molecular geometry.
- \(XeF_4\) has 4 bonding pairs and 2 lone pairs, which gives a square planar shape.
- \(SiF_4\) has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape.
- \(SF_6\) has 6 bonding pairs and 0 lone pairs, giving an octahedral shape.

1 mark for selecting the correct option (A).

We use the ideal gas equation:
\(pV = nRT \implies V = \frac{nRT}{p}\)

Convert units to SI:
- Pressure, \(p = 150\text{ kPa} = 150 \times 10^3\text{ Pa}\)
- Amount of substance, \(n = 0.200\text{ mol}\)
- Temperature, \(T = 323\text{ K}\)

Calculate volume in \(\text{m}^3\):
\(V = \frac{0.200 \times 8.31 \times 323}{150 \times 10^3} = 0.00357884\text{ m}^3\)

Convert \(\text{m}^3\) to \(\text{dm}^3\) (multiply by 1000):
\(V = 0.00357884 \times 1000 = 3.58\text{ dm}^3\) (to 3 significant figures).

1 mark for selecting the correct option (A).

Chromium (Z = 24) has a ground-state electronic configuration of \( [\text{Ar}] 4s^1 3d^5 \). When it forms a \( \text{Cr}^{2+} \) ion, it loses two electrons (the 4s electron first, then one of the 3d electrons), leaving it with the configuration \( [\text{Ar}] 3d^4 \).

1 mark: Correct identification of the ground state electronic configuration of the transition metal ion.

Ethylamine is more basic than ammonia because the electron-donating ethyl group increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton. Phenylamine is less basic than ammonia because the lone pair of electrons on the nitrogen atom is delocalised into the benzene \(\pi\)-system, making it less available to accept a proton.

1 mark: Correct order of basic strength based on amine structure and electron density of the nitrogen lone pair.

Tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react predominantly via the \( \text{S}_{\text{N}}1 \) mechanism, which is much faster than the nucleophilic substitution of primary halogenoalkanes. Furthermore, the C-I bond is weaker than the C-Cl bond (lower bond enthalpy), meaning iodoalkanes hydrolyse much faster than chloroalkanes.

1 mark: Correct identification of both the fastest reacting halogenoalkane and the correct mechanistic pathway.

Xenon tetrafluoride, \( \text{XeF}_4 \), has six electron pairs around the central xenon atom (four bonding pairs and two lone pairs). The geometry based on octahedral arrangement with the two lone pairs opposite each other minimizes repulsion, resulting in a square planar molecular shape.

1 mark: Correct identification of the square planar molecule based on VSEPR theory.

The balanced equation for the complete combustion of ethane is: \( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \). Amount of ethane = \( \frac{3.0\text{ g}}{30.0\text{ g mol}^{-1}} = 0.10\text{ mol} \). Using the stoichiometry of the reaction, the amount of \( \text{O}_2 \) required = \( 0.10\text{ mol} \times \frac{7}{2} = 0.35\text{ mol} \). Volume of \( \text{O}_2 \) = \( 0.35\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 8.4\text{ dm}^3 \).

1 mark: Correct calculation of the volume of oxygen.

Concentrated HCl provides chloride ligands which undergo ligand exchange with hexaaquacopper(II) ions: \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O} \). Because chloride ions are larger and carry a negative charge, only four can coordinate around the copper ion due to electrostatic and steric repulsion, resulting in a tetrahedral geometry.

1 mark: Correctly identifies both the formula and the tetrahedral geometry of the tetrachlorocuprate(II) ion.

At a highly alkaline pH of 12 (well above the isoelectric point of 6.0), the high concentration of hydroxide ions deprotonates both the carboxylic acid group (forming \( \text{COO}^- \)) and any protonated amine groups (leaving the neutral \( \text{NH}_2 \) group). Therefore, the predominant species is the anion.

1 mark: Correctly identifies the ionic form of alanine at high pH.

For a reaction to be feasible under standard conditions, \( E^\ominus_{\text{cell}} \) must be positive. For \( \text{Fe}^{3+} \) to oxidize \( \text{I}^- \): \( E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 - 0.54 = +0.23\text{ V} \) (feasible). For \( \text{Fe}^{3+} \) to oxidize \( \text{Br}^- \): \( E^\ominus_{\text{cell}} = 0.77 - 1.09 = -0.32\text{ V} \) (not feasible).

1 mark: Correctly applies standard electrode potentials to determine feasibility of redox reactions.

Adding excess chloride ions (\(Cl^-\)) to a solution containing pale blue hexaaquacopper(II) ions, \([Cu(H_2O)_6]^{2+}\), results in a ligand substitution reaction. The tetrahedral tetrachlorocuprate(II) ion, \([CuCl_4]^{2-}\), is formed. This complex has a characteristic yellow color. When mixed with remaining unreacted blue aquo complex, the solution appears green or yellow-green.

Award 1 mark for selecting correct option B.
- Incorrect options A, C, and D do not represent the correct yellow tetrahedral complex.

Iron (\(Fe\)) has an atomic number of 26. Its neutral ground state electron configuration is \([Ar] 3d^6 4s^2\). To form the \(Fe^{3+}\) ion, three electrons are removed. The two electrons in the highest energy occupied orbital, the \(4s\) subshell, are removed first, followed by one electron from the \(3d\) subshell. This leaves a configuration of \([Ar] 3d^5\).

Award 1 mark for the correct option A.
- Option B is incorrect as it represents \(Fe^{5+}\).
- Option C is incorrect as it represents \(Fe^{4+}\) configuration under an incorrect removal sequence.
- Option D represents the \(Fe^{2+}\) configuration.

Phenylamine is the weakest base because the lone pair on the nitrogen atom is partially delocalized into the benzene \(\pi\)-system, making it less available to accept a proton. Ammonia is a stronger base than phenylamine, but weaker than aliphatic amines because the alkyl groups in aliphatic amines (like ethylamine) are electron-donating, which increases the electron density on the nitrogen atom (positive inductive effect). Diethylamine is a secondary aliphatic amine and is a stronger base than the primary ethylamine because it has two electron-donating ethyl groups. Therefore, the order of increasing basicity is phenylamine < ammonia < ethylamine < diethylamine.

Award 1 mark for selecting correct option A.
- Reject other options where the order of delocalization and positive inductive effects is incorrect.

At a pH of 12 (highly alkaline), which is significantly above the isoelectric point of alanine (6.0), the concentration of hydroxide ions is high. These basic conditions promote deprotonation of both functional groups. The carboxylic acid group is deprotonated to form the carboxylate ion, \(-COO^-\), and the protonated amino group (\(-NH_3^+\)) loses a proton to become a neutral amino group (\(-NH_2\)). Thus, the overall charge is negative, represented by the species \(H_2NCH(CH_3)COO^-\).

Award 1 mark for the correct option A.
- Option B is the zwitterionic form which is dominant near the pI (pH 6.0).
- Option C is the cationic form dominant at very low pH (highly acidic).
- Option D represents an uncharged non-zwitterion form which does not exist in significant quantities in solution.

Naturally occurring bromine consists of two stable isotopes, \(^{79}Br\) and \(^{81}Br\), in an approximate 1:1 abundance ratio. This leads to characteristic twin molecular ion peaks (\(M\) and \(M+2\)) of equal intensity in the mass spectrum of any monobrominated organic compound. Chlorine would produce an \(M\) and \(M+2\) peak in a 3:1 ratio due to \(^{35}Cl\) and \(^{37}Cl\). Iodine and fluorine are monoisotopic and would produce only a single main molecular ion peak.

Award 1 mark for the correct option B.
- Reject option A because chlorine has a 3:1 isotope ratio.
- Reject options C and D because fluorine and iodine are monoisotopic.

Sulfur tetrafluoride, \(SF_4\), has a central sulfur atom with 6 valence electrons, bonding with 4 fluorine atoms (using 4 electrons) and leaving 2 non-bonding electrons as 1 lone pair. This gives 5 electron pairs in total, which arrange themselves in a trigonal bipyramidal electron-pair geometry. To minimize electron repulsion, the lone pair occupies an equatorial position, giving rise to a see-saw molecular shape. \(XeF_4\) is square planar (4 bonding pairs, 2 lone pairs), and both \(SiF_4\) and \(NH_4^+\) are tetrahedral (4 bonding pairs, 0 lone pairs).

Award 1 mark for the correct option A.
- Reject options B, C, and D as their geometries are square planar and tetrahedral respectively.

The chemical equation for the reaction is:
\(CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)\)

First, calculate the amount of calcium carbonate:
\(n(CaCO_3) = \frac{10.0\text{ g}}{100.1\text{ g mol}^{-1}} \approx 0.0999\text{ mol}\)

According to the stoichiometry of the reaction, 1 mole of \(CaCO_3\) produces 1 mole of \(CO_2\).
\(n(CO_2) \approx 0.0999\text{ mol}\)

Now, calculate the volume of gas at RTP:
\(V(CO_2) = 0.0999\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} \approx 2.40\text{ dm}^3\)

Award 1 mark for the correct option A.
- Option B is incorrect (corresponds to an incorrect stoichiometry ratio of 1:2).
- Option C is incorrect (corresponds to using 12.0 as molar volume).
- Option D is incorrect (corresponds to an arithmetic/decimal place error).

For a reaction to be thermodynamically feasible, \(E^\circ_{\text{cell}}\) must be positive. The half-cell with the more positive \(E^\circ\) value will undergo reduction (acting as the cathode):
\(\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}\)

The other half-cell will undergo oxidation (acting as the anode):
\(2\text{I}^-(aq) \rightarrow \text{I}_2(aq) + 2e^- \quad E^\circ = -0.54\text{ V}\)

The standard cell potential is:
\(E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\)

Award 1 mark for selecting correct option A.
- Option B is incorrect as it has a negative sign.
- Options C and D are incorrect as they result from adding the absolute potential values.

When excess concentrated hydrochloric acid is added to aqueous copper(II) ions, a ligand exchange reaction occurs. The larger chloride ligands replace the water ligands. Due to steric hindrance and electrostatic repulsion between the negative chloride ligands, only four chloride ligands coordinate to the copper ion, forming the tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\), which has a tetrahedral shape.

1 mark: Correctly identifies \([\text{CuCl}_4]^{2-}\) and its tetrahedral shape.

Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalised into the \(\pi\)-system of the benzene ring, making it less available to accept a proton. Ammonia is stronger than phenylamine but weaker than alkylamines. Ethylamine has one electron-releasing ethyl group which increases the electron density on the nitrogen atom (positive inductive effect). Diethylamine has two electron-releasing ethyl groups, making the lone pair on nitrogen even more available to accept a proton, making it the strongest base among the choices.

1 mark: Correctly identifies the order of basicity from weakest to strongest.

The presence of two molecular ion peaks of approximately equal intensity (1:1 ratio) separated by 2 mass units (\(m/z = 136\) and \(138\)) indicates the presence of a single bromine atom, which exists as two stable isotopes, \(^{79}\text{Br}\) and \(^{81}\text{Br}\), in roughly equal abundance. The molecular formula for bromobutane is \(\text{C}_4\text{H}_9\text{Br}\). Calculating the molecular mass with \(^{79}\text{Br}\): \((4 \times 12.0) + (9 \times 1.0) + 79.0 = 48.0 + 9.0 + 79.0 = 136.0\). With \(^{81}\text{Br}\), the molecular mass is \(138.0\). Thus, 1-bromobutane is the correct compound.

1 mark: Identifies 1-bromobutane based on isotope abundance and relative molecular mass calculation.

The central chlorine atom in \(\text{ClF}_3\) has 7 valence electrons. It forms 3 single covalent bonds with fluorine atoms, leaving 4 non-bonding electrons (2 lone pairs). According to VSEPR theory, these 5 electron pairs arrange themselves in a trigonal bipyramidal geometry to minimize repulsion. To minimize lone pair-lone pair repulsion, the two lone pairs occupy equatorial positions, resulting in a T-shaped molecular geometry. The strong repulsion from the lone pairs compresses the axial-equatorial bond angles from \(90^\circ\) to approximately \(87.5^\circ\).

1 mark: Correctly identifies the T-shaped geometry and the correct compressed bond angle.

Using the ideal gas equation: \(pV = nRT\). First, convert all quantities to SI units: \(p = 100 \text{ kPa} = 100 \times 10^3 \text{ Pa}\), \(V = 2.40 \text{ dm}^3 = 2.40 \times 10^{-3} \text{ m}^3\), \(T = 100^\circ\text{C} = 100 + 273 = 373 \text{ K}\). Calculate the number of moles, \(n\): \(n = pV / RT = (100 \times 10^3 \times 2.40 \times 10^{-3}) / (8.31 \times 373) \approx 0.07743 \text{ mol}\). Calculate the mass of \(\text{CO}_2\): \(\text{Mass} = n \times M_r = 0.07743 \times 44.0 \approx 3.41 \text{ g}\).

1 mark: Correctly calculates mass using ideal gas equation conversion and molar mass.

This is homogeneous catalysis because the reactants (aqueous peroxodisulfate and iodide ions) and the catalyst (aqueous iron(II) ions) are all in the same phase (aqueous). The uncatalysed reaction is slow due to the high activation energy required for the collision of two negatively charged ions. The catalyst works by providing an alternative pathway with lower activation energy. The first step involves the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) by \(\text{S}_2\text{O}_8^{2-}\): \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\). The \(\text{Fe}^{3+}\) ions then oxidize the \(\text{I}^-\) ions back to \(\text{I}_2\), regenerating the \(\text{Fe}^{2+}\) catalyst.

1 mark: Correctly identifies homogeneous catalysis definition and the correct initial step of the reaction mechanism.

Alanine is an amino acid. In strongly acidic solutions (such as pH 1), the concentration of hydrogen ions, \(\text{H}^+\), is very high. The carboxylate group is protonated to form \(-\text{COOH}\), and the amine group is also protonated to form \(-\text{NH}_3^+\). Therefore, the overall species has a positive charge, making \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\) the predominant species.

1 mark: Correctly identifies the fully protonated cationic form of the amino acid at low pH.

Lattice energy depends on ionic charge and ionic radius. The greater the charges on the ions and the smaller their ionic radii, the stronger the electrostatic attraction between them, resulting in a more exothermic lattice energy. Aluminium oxide, \(\text{Al}_2\text{O}_3\), contains \(\text{Al}^{3+}\) and \(\text{O}^{2-}\) ions, which have the highest charges among the choices. In comparison, \(\text{NaCl}\) contains \(1+\) and \(1-\), \(\text{MgO}\) contains \(2+\) and \(2-\), and \(\text{MgF}_2\) contains \(2+\) and \(1-\). Hence, \(\text{Al}_2\text{O}_3\) has the most exothermic lattice energy.

1 mark: Selects Aluminium oxide based on the combination of highest charges and small ionic sizes.

The atomic number of nickel is 28, so a neutral nickel atom has the electronic configuration \([\text{Ar}] 3\text{d}^8 4\text{s}^2\). When forming a \(\text{Ni}^{2+}\) ion, two electrons are lost from the outer 4s subshell first. This leaves the configuration as \([\text{Ar}] 3\text{d}^8\).

1 mark: Correctly identifies the configuration as option A.

Addition of excess concentrated hydrochloric acid replaces the water ligands with larger chloride ligands to form the tetrachlorocuprate(II) ion, \([\text{Cu}\text{Cl}_4]^{2-}\). Due to steric hindrance and repulsion between the charged chloride ligands, the coordination number decreases from 6 to 4, adopting a tetrahedral geometry with a characteristic yellow-green color.

1 mark: Correctly identifies the tetrahedral geometry, coordination number of 4, and yellow-green color.

Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalized into the \(\pi\)-system of the benzene ring, making it less available to accept a proton. Ammonia is stronger than phenylamine but weaker than ethylamine. Ethylamine is the strongest base because the ethyl group is electron-releasing via the inductive effect, which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton. Therefore, the order of increasing basicity is phenylamine < ammonia < ethylamine.

1 mark: Correctly identifies option C.

Under acidic conditions, the carboxylic acid group of alanine reacts with methanol to form an ester group (\(-\text{COOCH}_3\)). Simultaneously, the basic amine group (\(-\text{NH}_2\)) is protonated by the strong acid catalyst to form an ammonium cation (\(-\text{NH}_3^+\)). Therefore, the resulting organic cation is \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOCH}_3\).

1 mark: Correctly identifies the protonated ester product as B.

The rate of hydrolysis of halogenoalkanes depends on the strength (bond enthalpy) of the carbon-halogen bond rather than its polarity. The C-I bond has the lowest bond enthalpy (weakest bond) among the C-X bonds, so it is broken most easily. Consequently, 1-iodobutane reacts the fastest, forming a yellow precipitate of silver iodide rapidly.

1 mark: Correctly identifies 1-iodobutane as the fastest reactant due to having the weakest carbon-halogen bond (Option B).

In \(\text{NH}_4^+\), there are 4 bonding pairs and 0 lone pairs of electrons around the nitrogen atom, giving a tetrahedral shape with a bond angle of \(109.5^\circ\). In \(\text{NH}_3\), there are 3 bonding pairs and 1 lone pair, resulting in a trigonal pyramidal shape; the lone pair-bonding pair repulsion reduces the bond angle to approximately \(107^\circ\). In \(\text{NH}_2^-\), there are 2 bonding pairs and 2 lone pairs, causing even greater repulsion and reducing the bond angle further to about \(104.5^\circ\). Thus, the correct order of decreasing bond angle is \(\text{NH}_4^+ > \text{NH}_3 > \text{NH}_2^-\).

1 mark: Correctly identifies the order of decreasing bond angles as option A.

According to the balanced equation, 2 moles of \(\text{Ca}(\text{NO}_3)_2\) produce 5 moles of gaseous products (\(4\text{ mol of NO}_2\) and \(1\text{ mol of O}_2\)). Therefore, the total number of moles of gas produced from 0.050 mol of \(\text{Ca}(\text{NO}_3)_2\) is \(0.050 \times \frac{5}{2} = 0.125\text{ mol}\). The total volume of gas is \(0.125\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 3.0\text{ dm}^3\).

1 mark: Correctly calculates the volume of gas as 3.0 dm3 (Option C).

The half-equations are: Reduction: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\quad E^\ominus = +0.77\text{ V}\), Oxidation: \(2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{e}^-\quad E^\ominus = +0.54\text{ V}\). The standard cell potential is calculated using \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.77 - (+0.54) = +0.23\text{ V}\). Since \(E^\ominus_{\text{cell}}\pi\) is positive, the reaction is thermodynamically feasible under standard conditions.

1 mark: Correctly calculates +0.23 V and determines the reaction is feasible (Option B).

1. Find the mass of oxygen in the oxide:
\(\text{Mass of O} = 2.39\text{ g} - 2.07\text{ g} = 0.32\text{ g}\)

2. Calculate the moles of lead and oxygen atoms:
\(\text{Moles of Pb} = \frac{2.07\text{ g}}{207.2\text{ g mol}^{-1}} = 0.010\text{ mol}\)
\(\text{Moles of O} = \frac{0.32\text{ g}}{16.0\text{ g mol}^{-1}} = 0.020\text{ mol}\)

3. Determine the simplest whole-number ratio:
\(\text{Pb} : \text{O} = 0.010 : 0.020 = 1 : 2\)

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

[1] B is the correct option.
Accept: B
Reject: All other options

The arrangement of electron pairs (electron geometry) depends on the total number of bonding and lone pairs on the central atom:
- \(\text{PCl}_4^+\): Phosphorus has 5 valence electrons. Subtracting 1 for the positive charge gives 4 valence electrons. These form 4 single bonds, meaning there are 4 bonding pairs and 0 lone pairs (tetrahedral arrangement).
- \(\text{SF}_4\): Sulfur has 6 valence electrons, forming 4 single bonds and leaving 1 lone pair (5 electron pairs in total, which gives a trigonal bipyramidal arrangement with a seesaw molecular geometry).
- \(\text{H}_3\text{O}^+\): Oxygen has 6 valence electrons. Subtracting 1 for the positive charge gives 5 valence electrons, which form 3 single bonds and leave 1 lone pair (4 electron pairs in total, tetrahedral arrangement with trigonal pyramidal molecular geometry).
- \(\text{NH}_2^-\): Nitrogen has 5 valence electrons. Adding 1 for the negative charge gives 6 valence electrons, which form 2 single bonds and leave 2 lone pairs (4 electron pairs in total, tetrahedral arrangement with bent molecular geometry).

[1] B is the correct option.
Accept: B
Reject: All other options

Reaction with aqueous silver nitrate involves the nucleophilic substitution (hydrolysis) of the halogenoalkane.
- Bromoalkanes react faster than chloroalkanes because the \(\text{C-Br}\) bond is weaker than the \(\text{C-Cl}\) bond, making it easier to break.
- Tertiary halogenoalkanes (such as 2-bromo-2-methylpropane) react much faster than secondary and primary halogenoalkanes because they undergo hydrolysis via the \(\text{S}_\text{N}1\) mechanism, which proceeds via a highly stable tertiary carbocation intermediate.

[1] D is the correct option.
Accept: D
Reject: All other options

- The broad, strong absorption band in the region \(2500 - 3300\text{ cm}^{-1}\) is characteristic of an \(\text{O-H}\) stretch of a carboxylic acid (alcohols usually absorb at higher frequencies, \(3200 - 3650\text{ cm}^{-1}\), and form a narrower band).
- The sharp, strong absorption band at \(1715\text{ cm}^{-1}\) is characteristic of a \(\text{C=O}\) stretch.
- This indicates the presence of a carboxylic acid group (\(\text{COOH}\)).
- Among the options, only \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\) (butanoic acid) is a carboxylic acid and matches the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\).

[1] C is the correct option.
Accept: C
Reject: All other options

- Iron (\(\text{Fe}\), atomic number 26) has the electronic configuration \([\text{Ar}] 3d^6 4s^2\). When forming the \(\text{Fe}^{3+}\) ion, it loses 3 electrons (two from \(4s\) and one from \(3d\)), giving \([\text{Ar}] 3d^5\).
- Manganese (\(\text{Mn}\), atomic number 25) has the electronic configuration \([\text{Ar}] 3d^5 4s^2\). When forming the \(\text{Mn}^{2+}\) ion, it loses 2 electrons from the \(4s\) orbital, giving \([\text{Ar}] 3d^5\).
- Therefore, \(\text{Mn}^{2+}\) and \(\text{Fe}^{3+}\) have the identical electronic configuration.

[1] B is the correct option.
Accept: B
Reject: All other options

The reaction that takes place is a ligand exchange reaction:
\([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)
- Concentrated hydrochloric acid provides a high concentration of chloride ions (\(\text{Cl}^-\)), which displace the water ligands.
- Due to the larger size and greater electrostatic repulsion between chloride ligands, only four chloride ligands coordinate around the copper ion, forming a tetrahedral complex \([\text{CuCl}_4]^{2-}\).
- The copper ion remains in the +2 oxidation state.

[1] C is the correct option.
Accept: C
Reject: All other options

- Aliphatic amines are stronger bases than ammonia because the alkyl groups are electron-donating (+I inductive effect), which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton.
- Secondary aliphatic amines, such as diethylamine \((\text{CH}_3\text{CH}_2)_2\text{NH}\), have two electron-donating ethyl groups, making the nitrogen lone pair more available than in primary aliphatic amines like ethylamine.
- Phenylamine is a much weaker base because the nitrogen lone pair is delocalized into the benzene \(\pi\)-system, making it less available for protonation.

[1] D is the correct option.
Accept: D
Reject: All other options

- At the isoelectric point (\(\text{pH } 6.0\)), alanine exists as a neutral zwitterion, \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\).
- In highly acidic conditions (\(\text{pH } 2.0\)), the concentration of hydrogen ions is high. Both the basic carboxylate group (\(\text{COO}^-\)) and the amine group are protonated.
- The carboxylate group is protonated to form a neutral carboxylic acid group (\(\text{COOH}\)), while the amine group remains protonated as a positively charged ammonium group (\(\text{NH}_3^+\)).
- Therefore, the predominant species is the cation \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).

[1] B is the correct option.
Accept: B
Reject: All other options

The electronic configuration of a neutral copper atom is \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\). To form the \(\text{Cu}^+\) ion, one electron is lost from the 4s orbital first, which is the highest energy occupied orbital. This leaves the electronic configuration of \(\text{Cu}^+\) as \([\text{Ar}] 3\text{d}^{10}\).

1 mark: Correct option B selected. Reject all other options.

The addition of excess concentrated hydrochloric acid (containing chloride ions, \(\text{Cl}^-\)) to hexaaquacopper(II) ions, \([\text{Cu(H}_2\text{O)}_6]^{2+}\), results in a ligand substitution reaction. Due to the larger size and greater electrostatic repulsion of the chloride ligands, only four chloride ligands can coordinate, forming the tetrahedral tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\).

1 mark: Correct option B selected. Reject all other options.

Phenylamine is the weakest base because the nitrogen lone pair is delocalised into the benzene \(\pi\)-ring system, reducing its availability to accept a proton. Ammonia is stronger than phenylamine but weaker than alkylamines. Ethylamine is stronger due to the electron-releasing (positive inductive) effect of the ethyl group, which increases electron density on the nitrogen. Diethylamine has two electron-releasing ethyl groups, making it even more basic than ethylamine.

1 mark: Correct option A selected. Reject all other options.

The rate of hydrolysis of halogenoalkanes is determined by the bond enthalpy of the carbon-halogen bond. The C-I bond is the longest and weakest (has the lowest bond enthalpy), so it is broken most easily. Therefore, 1-iodobutane reacts the fastest.

1 mark: Correct option B selected. Reject all other options.

The ammonium ion, \(\text{NH}_4^+\), has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape with a bond angle of \(109.5^\circ\). Ammonia, \(\text{NH}_3\), has 3 bonding pairs and 1 lone pair, resulting in a trigonal pyramidal shape with a bond angle of \(107^\circ\). The amide ion, \(\text{NH}_2^-\), has 2 bonding pairs and 2 lone pairs, resulting in a bent shape with a bond angle of \(104.5^\circ\). Since lone pairs repel more than bonding pairs, the bond angle decreases as the number of lone pairs on the nitrogen atom increases.

1 mark: Correct option A selected. Reject all other options.

First, write the balanced equation: \(2\text{Ca(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{CaO(s)}\). Calculate the moles of Ca: \(n(\text{Ca}) = \frac{2.0\text{ g}}{40.0\text{ g mol}^{-1}} = 0.050\text{ mol}\). According to the stoichiometry, \(n(\text{O}_2) = \frac{1}{2} \times n(\text{Ca}) = 0.025\text{ mol}\). Calculate the volume of \(\text{O}_2\) gas: \(V = 0.025\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.60\text{ dm}^3\).

1 mark: Correct option A selected. Reject all other options.

At a pH below its isoelectric point (acidic conditions, \(\text{pH } 2.0 < 6.0\)), the basic amino group of alanine is protonated to form \(-\text{NH}_3^+\), and the carboxylate group remains in its protonated neutral form, \(-\text{COOH}\). Thus, the predominant species is the cation \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\).

1 mark: Correct option C selected. Reject all other options.

The absorption at \(1740\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). The absence of a broad absorption in the \(2500\text{--}3300\text{ cm}^{-1}\) range rules out an \(\text{O}-\text{H}\) bond of a carboxylic acid. Thus, the compound must be an ester rather than a carboxylic acid. Propanoic acid contains an acid \(\text{O}-\text{H}\) group, whereas methyl ethanoate is an ester with no \(\text{O}-\text{H}\) group. Therefore, the compound is methyl ethanoate.

1 mark: Correct option B selected. Reject all other options.

(a) The balanced ionic equation for the reaction is:
\([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)

(b) The coordination number of cobalt changes from 6 in \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) to 4 in \([\text{CoCl}_4]^{2-}\). Consequently, the geometry of the complex changes from octahedral to tetrahedral. This occurs because the chloride ion (\(\text{Cl}^-\)) is larger than the water molecule (\(\text{H}_2\text{O}\)), and its negative charge results in greater electrostatic repulsion, meaning only four chloride ligands can be accommodated around the central cobalt(II) ion compared to six water molecules.

Part (a):
- 1 Mark: Correct reactants and products with correct formula and stoichiometry.
- 1 Mark: Correct state symbols and equilibrium sign.

Part (b):
- 1 Mark: Identifies coordination number change from 6 to 4 AND shape change from octahedral to tetrahedral.
- 1 Mark: Explains that chloride ligands are larger (and/or cause greater repulsion) than water, limiting the number of ligands that can coordinate.

(a) At pH 1 (highly acidic conditions), the amino group is protonated: \(\text{C}_6\text{H}_5\text{CH}_2\text{CH}(\text{NH}_3^+)\text{COOH}\).
At pH 13 (highly alkaline conditions), the carboxylic acid group is deprotonated: \(\text{C}_6\text{H}_5\text{CH}_2\text{CH}(\text{NH}_2)\text{COO}^-\).

(b) Phenylalanine exists as a zwitterion (\(\text{C}_6\text{H}_5\text{CH}_2\text{CH}(\text{NH}_3^+)\text{COO}^-\)) in the solid state. This creates strong electrostatic attractions (ionic bonding behavior) between the positively charged ammonium group (\(-\text{NH}_3^+\)) of one molecule and the negatively charged carboxylate group (\(-\text{COO}^-\)) of an adjacent molecule. These strong electrostatic forces require a very large amount of energy to overcome, resulting in a much higher melting temperature compared to simple molecular organic substances of comparable size that only experience London forces or hydrogen bonding.

Part (a):
- 1 Mark: Correct structure of phenylalanine at pH 1 (protonated amine, neutral carboxyl).
- 1 Mark: Correct structure of phenylalanine at pH 13 (neutral amine, deprotonated carboxyl).

Part (b):
- 1 Mark: Explains that solid phenylalanine exists as a zwitterion with both negative (carboxylate) and positive (ammonium) charges.
- 1 Mark: Explains that there are strong electrostatic forces of attraction / ionic bonds between adjacent zwitterions, requiring substantial energy to break.

(a) An \(\text{S}_{\text{N}}2\) mechanism proceeds via a single transition state where the nucleophile (\(\text{OH}^-\)) attacks the carbon atom from the opposite side relative to the leaving group (\(\text{Br}^-\)). This results in complete inversion of configuration at the chiral carbon (Walden inversion). Since only one enantiomer of the reactant is used, a single enantiomer of the product butan-2-ol is formed, and the product mixture remains optically active.

(b) If the reaction proceeds via an \(\text{S}_{\text{N}}1\) mechanism, the first step is the loss of the bromide ion to form a planar carbocation intermediate. The nucleophile (\(\text{OH}^-\)) can then attack the planar carbocation with equal probability from either side. This results in the formation of equal amounts of both enantiomers (a racemic mixture). Consequently, the optical activity of the product is lost, and the mixture is optically inactive because the opposing rotations of plane-polarised light by the two enantiomers cancel out.

Part (a):
- 1 Mark: States that inversion of configuration occurs.
- 1 Mark: Explains that the nucleophile attacks from the opposite side to the leaving group (backside attack).

Part (b):
- 1 Mark: Explains that S_N1 involves a planar carbocation intermediate which can be attacked equally from either side.
- 1 Mark: Concludes that a racemic mixture is formed, which is optically inactive because the optical activities cancel.

(a) Silicon dioxide (\(\text{SiO}_2\)) exists as a giant covalent (macromolecular) structure where each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement, and each oxygen atom is bonded to two silicon atoms. Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) has a simple molecular structure consisting of discrete \(\text{P}_4\text{O}_{10}\) molecules held together by weak intermolecular forces, although covalent bonding exists within each molecule.

(b) To melt silicon dioxide, a large number of strong covalent bonds throughout the giant lattice must be broken, which requires a substantial amount of thermal energy. In contrast, melting phosphorus(V) oxide does not involve breaking any covalent bonds; it only requires overcoming the relatively weak intermolecular forces (such as London dispersion forces and dipole-dipole attractions) between individual molecules, which requires much less energy. Thus, silicon dioxide has a significantly higher melting temperature.

Part (a):
- 1 Mark: States that SiO2 is giant covalent (macromolecular) with covalent bonding throughout.
- 1 Mark: States that P4O10 is a simple molecular structure with covalent molecules.

Part (b):
- 1 Mark: Explains that melting SiO2 requires breaking strong covalent bonds throughout the lattice (requiring high energy).
- 1 Mark: Explains that melting P4O10 only requires overcoming weak intermolecular forces between molecules (requiring low energy).

(a) Step 1: Calculate the mass of water of crystallisation lost.
Mass of \(\text{H}_2\text{O} = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\).

Step 2: Calculate the number of moles of anhydrous \(\text{CuSO}_4\) and \(\text{H}_2\text{O}\).
Moles of \(\text{CuSO}_4 = \frac{3.19\text{ g}}{159.6\text{ g mol}^{-1}} = 0.0200\text{ mol}\).
Moles of \(\text{H}_2\text{O} = \frac{1.80\text{ g}}{18.0\text{ g mol}^{-1}} = 0.100\text{ mol}\).

Step 3: Determine the ratio \(x\).
\(x = \frac{\text{Moles of } \text{H}_2\text{O}}{\text{Moles of } \text{CuSO}_4} = \frac{0.100}{0.0200} = 5\).
Therefore, the formula is \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) and \(x = 5\).

(b) Heating to constant mass (by heating, cooling, weighing, and repeating until the mass remains unchanged) ensures that the thermal decomposition/dehydration is complete and that all the water of crystallisation has been driven off. If any water remained, the measured mass of anhydrous salt would be too high, leading to an underestimation of \(x\).

Part (a):
- 1 Mark: Calculates the mass of water lost as 1.80 g and the moles of both substances correctly (0.0200 mol and 0.100 mol).
- 1 Mark: Correctly finds the molar ratio of 1 : 5, yielding x = 5.

Part (b):
- 1 Mark: Explains that constant mass indicates the reaction is complete (all water of crystallisation has been removed).
- 1 Mark: Notes that incomplete dehydration would lead to inaccurate masses (overestimation of the remaining solid mass) and an incorrect value for x.

(a) For the oxidation of iodide ions by iron(III) ions:
Oxidation: \(2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2e^- \quad E^\ominus = +0.54\text{ V}\)
Reduction: \(\text{Fe}^{3+}(\text{aq}) + e^- \rightarrow \text{Fe}^{2+}(\text{aq}) \quad E^\ominus = +0.77\text{ V}\)

To construct the overall cell, double the reduction half-equation and combine:
\(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

Calculating the cell potential:
\(E_{\text{cell}}^\ominus = E_{\text{reduction}}^\ominus - E_{\text{oxidation}}^\ominus = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\).
Since \(E_{\text{cell}}^\ominus\) is positive, the reaction is thermodynamically feasible under standard conditions.

(b) For the oxidation of chloride ions by iron(III) ions:
The oxidation half-reaction is: \(2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{aq}) + 2e^- \quad E^\ominus = +1.36\text{ V}\).
\(E_{\text{cell}}^\ominus = E_{\text{reduction}}^\ominus - E_{\text{oxidation}}^\ominus = +0.77\text{ V} - (+1.36\text{ V}) = -0.59\text{ V}\).
Since \(E_{\text{cell}}^\ominus\) is negative, the reaction is not thermodynamically feasible under standard conditions.

Part (a):
- 1 Mark: Correctly balanced overall ionic equation (with correct state symbols or implied).
- 1 Mark: Correct calculation of E_cell = +0.23 V and states that it is feasible because the value is positive.

Part (b):
- 1 Mark: Correct calculation of E_cell = -0.59 V for the iron(III)/chloride system.
- 1 Mark: Correctly concludes that it is not feasible because the standard cell potential is negative.

(a) The entropy of the system increases (positive \(\Delta S_{\text{system}}^\ominus\)) because a solid reactant decomposes to produce a solid and a gas (carbon dioxide). Gas molecules are highly disordered and have much higher entropy than solids because they are free to move randomly and have many more possible ways of arranging energy (higher translational disorder).

(b) For a reaction to be thermodynamically feasible, the Gibbs free energy change must be less than or equal to zero (\(\Delta G^\ominus \le 0\)).
We use the Gibbs free energy equation: \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S_{\text{system}}^\ominus\).

At the point of feasibility, let \(\Delta G^\ominus = 0\):
\(0 = \Delta H^\ominus - T\Delta S_{\text{system}}^\ominus\)
\(T = \frac{\Delta H^\ominus}{\Delta S_{\text{system}}^\ominus}\)

Convert \(\Delta S_{\text{system}}^\ominus\) into \(\text{kJ K}^{-1}\text{ mol}^{-1}\):
\(\Delta S_{\text{system}}^\ominus = \frac{160}{1000} = 0.160\text{ kJ K}^{-1}\text{ mol}^{-1}\).

Substitute values into the expression for temperature:
\(T = \frac{178\text{ kJ mol}^{-1}}{0.160\text{ kJ K}^{-1}\text{ mol}^{-1}} = 1112.5\text{ K}\).

Thus, the reaction becomes thermodynamically feasible at temperatures equal to or greater than \(1113\text{ K}\) (to 3 significant figures, or \(1112.5\text{ K}\)).

Part (a):
- 1 Mark: Explains that there is an increase in disorder because a gas (CO2) is formed from a solid reactant (CaCO3).
- 1 Mark: Mentions that gases have higher entropy / random motion / more microstates than solids.

Part (b):
- 1 Mark: Converts entropy to kJ K^-1 mol^-1 (0.160) and uses Gibbs free energy equation set to zero.
- 1 Mark: Calculates T = 1112.5 K or 1113 K.

(a) Concentrated sulfuric acid is a catalyst and a stronger acid than nitric acid. It protonates nitric acid, facilitating the elimination of water to generate the electrophile, the nitronium ion (\(\text{NO}_2^+\)).
Chemical equations:
\(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-\)
\(\text{H}_2\text{NO}_3^+ \rightarrow \text{NO}_2^+ + \text{H}_2\text{O}\)
Or overall:
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)

(b) The mechanism proceeds as follows:
1. A curly arrow is drawn from the delocalised \(\pi\) ring of benzene to the nitrogen atom of the \(\text{NO}_2^+\)\ electrophile.
2. This forms a positively charged intermediate (arenium ion) with a broken ring (represented as a horseshoe opened towards the carbon bonded to the nitro group and hydrogen, with a \(+\) charge inside the horseshoe).
3. A curly arrow is drawn from the \(\text{C-H}\) bond on that carbon back into the system of the ring to restore the aromatic delocalisation, releasing an \(\text{H}^+\) ion (which is subsequently picked up by \(\text{HSO}_4^-\)\ to regenerate the sulfuric acid catalyst).

Part (a):
- 1 Mark: States that sulfuric acid acts as a catalyst / proton donor (or stronger acid than nitric acid) to generate the nitronium ion (NO2+).
- 1 Mark: Writes a balanced equation showing the generation of the NO2+ electrophile.

Part (b):
- 1 Mark: Draws a curly arrow from the benzene ring to the NO2+ ion, and represents the intermediate correctly with a partial ring and a positive charge.
- 1 Mark: Draws a curly arrow from the C-H bond of the sp3 carbon in the intermediate into the ring, releasing H+ and reforming the benzene ring.

(a) In \([Cu(H_2O)_6]^{2+}\), the copper ion is coordinated to 6 water molecules, so its coordination number is 6. In \([CuCl_4]^{2-}\), it is coordinated to 4 chloride ions, so its coordination number is 4. (b) In an octahedral complex, the five 3d orbitals are split into two groups of different energy levels by the electric field of the ligands. An electron in a lower energy d-orbital absorbs a photon of light in the visible range and is promoted to a higher energy d-orbital (d-d transition). The remaining wavelengths of light are transmitted, and the complementary colour of the absorbed light is observed.

1 Mark for part (a): Correctly identifies coordination numbers as 6 for \([Cu(H_2O)_6]^{2+}\) and 4 for \([CuCl_4]^{2-}\). 1 Mark for part (b): Explains that the d-orbitals split into two different energy levels in the presence of ligands. 1 Mark for part (b): Explains that an electron absorbs light in the visible spectrum to promote from a lower energy level to a higher energy level (d-d transition). 1.5357 Marks for part (b): Mentions that the remaining light is transmitted, giving the complementary colour.

(a) At low pH (highly acidic, pH = 1), both basic amine groups (\(-NH_2\)) are protonated to \(-NH_3^+\). The carboxylic acid group remains as \(-COOH\). (b) In its solid state, lysine exists as a zwitterion, possessing both positive (\(-NH_3^+\)) and negative (\(-COO^-\)) charges. This leads to strong electrostatic forces of attraction (ionic bonding) between different zwitterions in a giant ionic-like lattice. These strong attractions require significant energy to overcome. Conversely, hexylamine cannot form zwitterions; its intermolecular forces are hydrogen bonds and London forces, which are significantly weaker than the electrostatic forces in lysine.

1.5357 Marks for part (a): Describing or drawing the structure with both amine groups protonated as \(-NH_3^+\) and the carboxylic acid group remaining as \(-COOH\). 1 Mark for part (b): Mentioning that lysine exists as a zwitterion in its solid state. 1 Mark for part (b): Explaining that there are strong electrostatic attractions (ionic bonds) between zwitterions in a lattice. 1 Mark for part (b): Explaining that hexylamine only has hydrogen bonding and London forces which are much weaker, requiring less energy to break.

(a) In the \(S_N2\) transition state for 1-bromobutane with \(OH^-\), the central carbon is bonded to a propyl group, two hydrogen atoms, and is partially bonded to the incoming hydroxide (\(HO^{\delta-}\)) and the departing bromide (\(Br^{\delta-}\)) ion. The central carbon has a trigonal bipyramidal arrangement with a negative charge. (b) In tertiary halogenoalkanes, three bulky alkyl groups surround the carbon-halogen bond, causing severe steric hindrance which prevents the nucleophile from attacking the carbon atom from the back (precluding \(S_N2\)). However, they undergo \(S_N1\) because the tertiary carbocation intermediate formed by heterolytic fission of the C-X bond is stabilized by the positive inductive effect of three electron-donating alkyl groups, which disperse the positive charge.

1.5357 Marks for part (a): Describing transition state correctly with: pentacoordinate carbon; partial bonds to OH and Br; correct partial negative charges on both oxygen and bromine. 1.5357 Marks for part (b): Explaining steric hindrance in tertiary halogenoalkanes preventing nucleophilic backside attack. 1.5 Marks for part (b): Explaining the stability of the tertiary carbocation intermediate due to the electron-donating inductive effect of three alkyl groups.

(a) In the chlorate(III) ion (\(ClO_2^-\)), the central chlorine atom has 7 valence electrons, plus 1 from the negative charge, giving 8. It forms two bonding regions with oxygen, leaving two lone pairs of electrons. Total electron pairs around Cl = 4 (2 bonding pairs/regions, 2 lone pairs). The electron geometry is tetrahedral, but the molecular shape is bent / V-shaped. Due to the greater repulsion of lone pairs compared to bonding pairs, the bond angle is reduced to approximately \(104.5^\circ\). (b) In carbon dioxide (\(CO_2\)), the central carbon atom has 4 valence electrons and forms two double bonds with two oxygen atoms. There are 2 bonding regions (double bonds) and 0 lone pairs around the central carbon atom. To minimize repulsion, these two bonding regions position themselves as far apart as possible, resulting in a linear molecular shape with a bond angle of exactly \(180^\circ\).

1 Mark for part (a): Correctly identifies that the central Cl atom has 2 bonding regions/pairs and 2 lone pairs. 1 Mark for part (a): Correctly states the shape is bent / V-shaped and predicts a bond angle between \(103^\circ\) and \(106^\circ\). 1 Mark for part (b): Correctly identifies that the central C atom has 2 bonding regions and 0 lone pairs. 1.5357 Marks for part (b): Correctly states the shape of \(CO_2\) is linear and the bond angle is \(180^\circ\), explaining that bonding regions repel each other to a position of minimum repulsion.

(a) Mass of \(C = 2.64 \times (12.0 / 44.0) = 0.720\text{ g}\). Mass of \(H = 1.08 \times (2.0 / 18.0) = 0.120\text{ g}\). Mass of \(O = 1.48 - (0.720 + 0.120) = 0.640\text{ g}\). Moles of \(C = 0.720 / 12.0 = 0.060\text{ mol}\). Moles of \(H = 0.120 / 1.0 = 0.120\text{ mol}\). Moles of \(O = 0.640 / 16.0 = 0.040\text{ mol}\). Dividing by the smallest value (0.040) gives a ratio of \(C : H : O = 1.5 : 3 : 1\), which scales to the empirical formula \(C_3H_6O_2\). (b) Using \(pV = nRT\), \(n = (101 \times 10^3 \times 207 \times 10^{-6}) / (8.31 \times 373) = 0.00675\text{ mol}\). Molar mass \(M = 0.500 / 0.00675 = 74.0\text{ g mol}^{-1}\). Since the empirical formula mass of \(C_3H_6O_2\) is 74.0, the molecular formula is also \(C_3H_6O_2\).

1 Mark: Correct calculation of masses of C and H (0.720 g and 0.120 g respectively), and thus mass of O (0.640 g). 1 Mark: Correct determination of mole ratios to give the empirical formula \(C_3H_6O_2\). 1.5357 Marks: Using \(pV = nRT\) correctly with conversion of units to find the number of moles of vapour, \(n \approx 0.00675\text{ mol}\). 1 Mark: Calculating the molar mass \(M \approx 74.0\text{ g mol}^{-1}\) and showing that the molecular formula is \(C_3H_6O_2\).

(a) Combining the half-equations \(Fe^{2+} \rightarrow Fe^{3+} + e^-\) and \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) gives: \(5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O\). (b) Moles of \(MnO_4^-\) used: \(n(MnO_4^-) = 0.0200 \times (22.40 / 1000) = 4.48 \times 10^{-4}\text{ mol}\). Moles of \(Fe^{2+}\) in \(25.00\text{ cm}^3\) portion: \(n(Fe^{2+}) = 5 \times 4.48 \times 10^{-4} = 2.24 \times 10^{-3}\text{ mol}\). Moles of \(Fe^{2+}\) in the total \(250.0\text{ cm}^3\) flask: \(n(Fe^{2+})_{\text{total}} = 2.24 \times 10^{-2}\text{ mol}\). Mass of \(Fe^{2+}\) in sample: \(2.24 \times 10^{-2} \times 55.8 = 1.25\text{ g}\). Percentage by mass = \((1.25 / 2.50) \times 100 = 50.0\%\).

1 Mark for part (a): Correct balanced ionic equation. 1 Mark for part (b): Calculates moles of \(MnO_4^-\) used as \(4.48 \times 10^{-4}\text{ mol}\). 1 Mark for part (b): Calculates moles of \(Fe^{2+}\) in \(250.0\text{ cm}^3\) as \(2.24 \times 10^{-2}\text{ mol}\). 1.5357 Marks for part (b): Calculates mass of \(Fe^{2+}\) as \(1.25\text{ g}\) and percentage by mass as \(50.0\%\).

The order of basicity is: ethylamine > ammonia > phenylamine. A base acts as a proton acceptor via the lone pair of electrons on the nitrogen atom. In ethylamine, the alkyl (ethyl) group is electron-donating due to the positive inductive effect, which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton. Ammonia has no such groups and acts as the baseline. In phenylamine, the lone pair on the nitrogen atom overlaps with and becomes delocalized into the \(\pi\)-system of the benzene ring, significantly reducing the electron density on the nitrogen and making the lone pair much less available to accept a proton.

1 Mark: Correctly ranks the basicity as ethylamine > ammonia > phenylamine. 1.5357 Marks: Explains that basicity depends on the availability of the nitrogen lone pair to accept a proton (form a dative covalent bond). 1 Mark: Explains that in ethylamine, the electron-donating ethyl group increases electron density on the nitrogen atom. 1 Mark: Explains that in phenylamine, the nitrogen lone pair delocalizes into the benzene \(\pi\)-ring, decreasing its availability.

(a) Bromoethane consists of the ethyl group combined with a bromine atom. The two molecular ion peaks represent: \([C_2H_5^{79}Br]^+\) at \(m/z = 108\) and \([C_2H_5^{81}Br]^+\) at \(m/z = 110\). Because the isotopes \(^{79}Br\) and \(^{81}Br\) exist in a nearly \(1:1\) ratio in nature, there is an equal probability of a bromoethane molecule containing either isotope, resulting in equal peak intensities. (b) The peak at \(m/z = 29\) corresponds to the ethyl carbocation fragment, which is formed by the loss of a bromine radical from the molecular ion: \([C_2H_5Br]^{+\bullet} \rightarrow [C_2H_5]^+ + Br^\bullet\). The formula of this fragment ion is \([C_2H_5]^+\) (or \(CH_3CH_2^+\)).

1 Mark for part (a): Identifies the formulas of both molecular ions correctly: \([C_2H_5^{79}Br]^+\) and \([C_2H_5^{81}Br]^+\). 1.5357 Marks for part (a): Explains that the nearly equal abundance of the two bromine isotopes leads to equal probability and hence equal peak intensities. 2 Marks for part (b): Correctly identifies the fragment as the ethyl cation, \([C_2H_5]^+\) or \(CH_3CH_2^+\) (1 mark for the formula, 1 mark for the positive charge; reject without charge).

(a) The reaction of hexaaquacopper(II) with excess chloride ions produces the tetrachlorocuprate(II) complex ion:
\([Cu(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)

(b) The starting complex, \)[Cu(H_2O)_6]^{2+}\), has an octahedral geometry with a coordination number of 6. The product complex, \([CuCl_4]^{2-}\), has a tetrahedral geometry with a coordination number of 4. The coordination number decreases because chloride ions are significantly larger than water molecules, leading to steric crowding and greater repulsion between ligands if 6 chloride ions were to bind.

(c) The process is a ligand exchange/substitution reaction. The chloride ion behaves as a monodentate ligand (or Lewis base) by donating a lone pair of electrons to the copper ion.

M1: Correct balanced ionic equation, including states and charges: \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CuCl_4]^{2-} + 6H_2O\) (accept reversible arrow). [1 mark]
M2: Identifies the change in geometry from octahedral to tetrahedral. [1 mark]
M3: Explains that chloride ligands are larger than water molecules (or experience greater steric hindrance/repulsion), causing a reduction in coordination number. [1 mark]
M4: Correctly identifies the reaction type as ligand substitution/exchange and the chloride ion as a ligand/Lewis base/nucleophile. [1 mark]

1. Calculate the number of moles of manganate(VII) ions used:
\(n(MnO_4^-) = C \times V = 0.0200 \text{ mol dm}^{-3} \times \frac{50.0}{1000} \text{ dm}^3 = 1.00 \times 10^{-3} \text{ mol}\)

2. Calculate the number of moles of electrons gained by the manganate(VII) ions:
Since each \(MnO_4^-\) ion gains 5 electrons:
\(n(e^-) = 5 \times n(MnO_4^-) = 5 \times 1.00 \times 10^{-3} \text{ mol} = 5.00 \times 10^{-3} \text{ mol}\)

3. Calculate the number of moles of vanadium ions in the titrated sample:
\(n(V) = 0.100 \text{ mol dm}^{-3} \times \frac{25.0}{1000} \text{ dm}^3 = 2.50 \times 10^{-3} \text{ mol}\)

4. Determine the ratio of moles of electrons lost per mole of vanadium:
\(\text{Ratio} = \frac{n(e^-)}{n(V)} = \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 2\)
This means each vanadium ion loses 2 electrons during oxidation to the +5 state.

5. Calculate the initial oxidation state of vanadium:
\(\text{Initial state} = +5 - 2 = +3\).

M1: Calculates the amount in moles of \(MnO_4^-\) used as \(1.00 \times 10^{-3}\) mol. [1 mark]
M2: Calculates the amount in moles of electrons gained by \(MnO_4^-\) as \(5.00 \times 10^{-3}\) mol. [1 mark]
M3: Calculates the amount in moles of vanadium ions as \(2.50 \times 10^{-3}\) mol. [1 mark]
M4: Deduces the electron-to-vanadium molar ratio of 2 and states that the initial oxidation state is +3. [1 mark]

(a) The reduction of nitrobenzene to phenylamine is performed in two steps:
1. Heating nitrobenzene under reflux with tin (\(Sn\)) and concentrated hydrochloric acid (\(HCl\)) to form the phenylammonium ion.
2. Reacting the mixture with aqueous sodium hydroxide (\(NaOH\)) to release phenylamine from its conjugate acid.

(b) Basicity is determined by the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(H^+\)):
- In phenylamine, the p-orbital containing the lone pair of electrons on the nitrogen atom overlaps with the delocalised \(\pi\)-system of the benzene ring. This delocalisation reduces the electron density on the nitrogen, making the lone pair less available for protonation.
- In ethylamine, the ethyl group is alkyl and acts as an electron-donating group due to the positive inductive effect. This increases the electron density on the nitrogen atom, making the lone pair more available to coordinate with a proton.

M1: Identifies tin (\(Sn\)) and concentrated hydrochloric acid (\(HCl\)), followed by sodium hydroxide (\(NaOH\)) treatment. [1 mark]
M2: Identifies the condition of heating under reflux. [1 mark]
M3: Explains that the lone pair on the nitrogen atom in phenylamine is delocalised into the benzene ring's \(\pi\)-system, reducing its availability to accept a proton. [1 mark]
M4: Explains that the ethyl group in ethylamine has a positive inductive effect (or is electron-donating), increasing the nitrogen lone pair's availability. [1 mark]

(a) In the zwitterion form of serine, the carboxylic acid group loses a proton to become a carboxylate group, and the amine group accepts a proton to become an ammonium group:
\(H_3N^+-CH(CH_2OH)-COO^-\).

(b) (i) Under acidic conditions (dilute \(HCl\)), the carboxylate group is protonated back to a carboxylic acid, leaving the amine group protonated:
\(H_3N^+-CH(CH_2OH)-COOH\).
(ii) Under alkaline conditions (dilute \(NaOH\)), the ammonium group is deprotonated back to an amine, leaving the carboxylate group negatively charged:
\(H_2N-CH(CH_2OH)-COO^-\).

(c) Organic molecules of comparable molecular mass are usually held together by relatively weak intermolecular forces (such as London dispersion forces or hydrogen bonding). However, crystalline amino acids exist as zwitterions held together by a network of strong electrostatic forces of attraction (ionic bonds) between the oppositely charged groups (\(-NH_3^+\) and \(-COO^-\)) in the lattice, requiring much higher thermal energy to melt.

M1: Correct zwitterionic structure of serine showing both positive charge on N and negative charge on O. [1 mark]
M2: Correct cationic structure in acidic solution: \(H_3N^+-CH(CH_2OH)-COOH\). [1 mark]
M3: Correct anionic structure in alkaline solution: \(H_2N-CH(CH_2OH)-COO^-\). [1 mark]
M4: Explains that amino acids exist as zwitterions held together by strong electrostatic attractions (or ionic bonds) between oppositely charged ions, which require substantial energy to overcome. [1 mark]

(a) The balanced equation for the hydrolysis of 2-bromo-2-methylpropane with water is:
\((CH_3)_3CBr + H_2O \rightarrow (CH_3)_3COH + H^+ + Br^-\)
(Accept: \((CH_3)_3CBr + H_2O \rightarrow (CH_3)_3COH + HBr\))

(b) 2-bromo-2-methylpropane hydrolyses much faster than 1-bromobutane because:
- 2-bromo-2-methylpropane is a tertiary halogenoalkane and hydrolyses via the \(S_N1\) mechanism. The rate-determining step involves the heterolytic fission of the \(C-Br\) bond to form a tertiary carbocation, \((CH_3)_3C^+\). This carbocation is highly stable because the three electron-donating methyl groups release charge density towards the positive carbon (positive inductive effect), lowering the activation energy for the step.
- 1-bromobutane is a primary halogenoalkane and hydrolyses via the \(S_N2\) mechanism. It cannot form a stable carbocation because it only has one alkyl group to disperse the positive charge. The \(S_N2\) transition state is more crowded and has a higher activation energy, making the overall reaction slower.

M1: Correct balanced structural chemical equation for the hydrolysis of 2-bromo-2-methylpropane. [1 mark]
M2: Correctly identifies that 2-bromo-2-methylpropane reacts faster. [1 mark]
M3: Identifies that 2-bromo-2-methylpropane reacts via the \(S_N1\) mechanism involving a carbocation, while 1-bromobutane reacts via \(S_N2\). [1 mark]
M4: Explains that the tertiary carbocation is highly stable due to the positive inductive effect of three alkyl groups, lowering the activation energy of the rate-determining step. [1 mark]

(a) The strong, sharp absorption at 1715 cm\(^{-1}\) corresponds to the stretching vibration of a carbonyl group, \(C=O\). Since there is no absorption in the 3200–3600 cm\(^{-1}\) range, there is no hydroxyl (\(O-H\)) group, ruling out alcohols. Therefore, **X** must be a carbonyl compound: either a ketone (butanone) or an aldehyde (butanal).

(b) In mass spectrometry, molecules fragment under electron ionization. A major peak at \(m/z = 43\) can correspond to:
1. The acylium ion, \([CH_3CO]^+\) (mass = 12 + 3 + 12 + 16 = 43), formed by \(\alpha\)-cleavage of butanone.
2. The propyl carbocation, \([C_3H_7]^+\) (or \([CH_3CH_2CH_2]^+\), mass = 36 + 7 = 43), formed by loss of the formyl or acyl group from either butanal or butanone.

M1: Identifies the 1715 cm\(^{-1}\) band as \(C=O\) (carbonyl) and notes the absence of an \(O-H\) stretch at 3200-3600 cm\(^{-1}\). [1 mark]
M2: Concludes that **X** is a ketone or aldehyde. [1 mark]
M3: Identifies \([CH_3CO]^+\) including the positive charge. [1 mark]
M4: Identifies \([C_3H_7]^+\) (or \([CH_3CH_2CH_2]^+\)) including the positive charge. [1 mark]

(a) In ammonia, \(NH_3\), the central nitrogen atom has 5 valence electrons and forms 3 single covalent bonds with hydrogen atoms, leaving 1 lone pair. The 4 electron pairs locate themselves in a tetrahedral arrangement to minimize electron repulsion.
Because lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the three N-H bonding pairs are pushed closer together, compressing the bond angle from 109.5\(^\circ\) to approximately 107\(^\circ\). The geometry is trigonal pyramidal.

(b) In the ammonium ion, \(NH_4^+\), the central nitrogen atom has 4 bonding pairs and 0 lone pairs. The 4 bonding pairs arrange themselves symmetrically in space to minimize repulsion, yielding equal repulsion between all pairs. The shape is perfectly tetrahedral, and the bond angle is 109.5\(^\circ\).

M1: For \(NH_3\): Predicts shape is trigonal pyramidal AND identifies 3 bonding pairs and 1 lone pair. [1 mark]
M2: For \(NH_3\): Predicts bond angle is ~107\(^\circ\) AND explains that lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion. [1 mark]
M3: For \(NH_4^+\): Predicts shape is tetrahedral AND identifies 4 bonding pairs and 0 lone pairs. [1 mark]
M4: For \(NH_4^+\): Predicts bond angle is 109.5\(^\circ\) AND explains that all bonding pairs repel equally. [1 mark]

1. State the ideal gas equation and rearrange for molar mass (\(M\)):
\(PV = nRT = \frac{m}{M}RT \implies M = \frac{mRT}{PV}\)

2. Convert all quantities to SI units:
- Mass, \(m = 0.322\) g
- Pressure, \(P = 101 \text{ kPa} = 101 \times 10^3 \text{ Pa}\) (or \(\text{N m}^{-2}\))
- Volume, \(V = 112 \text{ cm}^3 = 112 \times 10^{-6} \text{ m}^3\)
- Temperature, \(T = 98.0 + 273.15 = 371.15\) K (or \(371\) K)

3. Calculate the moles of gas, \(n\):
\(n = \frac{PV}{RT} = \frac{(101 \times 10^3 \text{ Pa}) \times (112 \times 10^{-6} \text{ m}^3)}{8.31 \text{ J K}^{-1}\text{ mol}^{-1} \times 371.15 \text{ K}}\)
\(n = \frac{11.312}{3084.26} = 3.6676 \times 10^{-3} \text{ mol}\)

4. Calculate the molar mass, \(M\):
\(M = \frac{m}{n} = \frac{0.322 \text{ g}}{3.6676 \times 10^{-3} \text{ mol}} = 87.795 \approx 87.8 \text{ g mol}^{-1}\)

(If using \(T = 371\) K: \(n = 3.669 \times 10^{-3}\) mol, \(M = 87.8 \text{ g mol}^{-1}\)).

M1: Converts temperature to Kelvin (371.15 K or 371 K) AND pressure to Pascal (\(1.01 \times 10^5\) Pa) AND volume to \(\text{m}^3\) (\(1.12 \times 10^{-4} \text{ m}^3\)). [1 mark]
M2: Rearranges the ideal gas equation to express \(n\) or \(M\) as the subject: \(n = \frac{PV}{RT}\) or \(M = \frac{mRT}{PV}\). [1 mark]
M3: Calculates the number of moles correctly: \(n = 3.67 \times 10^{-3}\) mol (accept range \(3.66 \times 10^{-3}\) to \(3.68 \times 10^{-3}\) mol). [1 mark]
M4: Calculates the molar mass as 87.8 (accept 87.5 to 88.1 to 3 significant figures) with units of \(\text{g mol}^{-1}\). [1 mark]

When concentrated hydrochloric acid is added to aqueous copper(II) ions, a ligand substitution reaction takes place:
\([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\) (or with a single arrow).

The tetrachlorocuprate(II) ion, \([CuCl_4]^{2-}\), has a tetrahedral shape with a coordination number of 4. This change in shape (from octahedral to tetrahedral) and coordination number (from 6 to 4) occurs because:
1. Chloride ions (\(Cl^-\)) are larger than water molecules (\(H_2O\)).
2. There is greater electrostatic repulsion between the negatively charged chloride ligands compared to the neutral polar water ligands.

Thus, fewer chloride ligands can fit around the central copper(II) ion, resulting in a change from a 6-coordinate octahedral complex to a 4-coordinate tetrahedral complex.

M1 (1 mark): Correct and balanced equation: \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\) (state symbols are not required but must be correct if included; accept single arrow).
M2 (1 mark): Identifies the change in coordination number from 6 to 4 and the change in geometry from octahedral to tetrahedral.
M3 (1 mark): Explains that chloride ligands are larger than water ligands OR there is greater repulsion between negatively charged chloride ligands compared to neutral polar water molecules.
M4 (1.5357 marks): Completes explanation by linking size/charge repulsion to the steric/electrostatic limit of only fitting 4 chloride ligands around the central copper ion.

(i) The end-point is reached when the first excess of \(MnO_4^-\) added does not get reduced, which results in a persistent pale pink color. The solution changes from colorless (or very pale green/yellow of \(Fe^{2+}/Fe^{3+}\)) to pale pink.

(ii) Half-reactions:
\(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)
\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)
Combining them gives:
\(MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\)

(iii) Calculation:
Amount of \(MnO_4^-\) used:
\(n(MnO_4^-) = 0.0200 \times \frac{22.45}{1000} = 4.49 \times 10^{-4}\text{ mol}\)
From the stoichiometry of the equation, \(n(Fe^{2+}) = 5 \times n(MnO_4^-)\):
\(n(Fe^{2+}) = 5 \times 4.49 \times 10^{-4} = 2.245 \times 10^{-3}\text{ mol}\) (in \(25.0\text{ cm}^3\))

Concentration of \(Fe^{2+}\) in \(\text{mol dm}^{-3}\):
\(c(Fe^{2+}) = \frac{2.245 \times 10^{-3}}{0.0250} = 0.0898\text{ mol dm}^{-3}\)

Concentration of \(Fe^{2+}\) in \(\text{g dm}^{-3}\):
\(c(Fe^{2+}) = 0.0898\text{ mol dm}^{-3} \times 55.8\text{ g mol}^{-1} = 5.01084\text{ g dm}^{-3} \approx 5.01\text{ g dm}^{-3}\) (to 3 significant figures).

M1 (1 mark): (i) Colorless (or pale green) to pale pink (reject purple, reject self-indicating without color description).
M2 (1 mark): (ii) Correct balanced ionic equation: \(MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O\).
M3 (1 mark): (iii) Correct calculation of moles of \(MnO_4^-\) and moles of \(Fe^{2+}\) (using \(5:1\) ratio): \(n(Fe^{2+}) = 2.245 \times 10^{-3}\text{ mol}\).
M4 (1.5357 marks): Correct calculation of final concentration of \(Fe^{2+}\) in \(\text{g dm}^{-3}\) to 3 significant figures: \(5.01\text{ g dm}^{-3}\). (Allow TE for incorrect ratio/equation).

(i) The zwitterion structure of alanine is \(H_3N^+-CH(CH_3)-COO^-\) where both positive and negative charges reside on the same molecule.
Alanine exists as a zwitterion in the solid state. Consequently, it exhibits strong electrostatic attractions (ionic bonding) between the negatively charged carboxylate group (\(-COO^-\)) of one molecule and the positively charged ammonium group (\(-NH_3^+\)) of an adjacent molecule. This forms a 3D ionic lattice. These strong ionic interactions require a substantial amount of thermal energy to overcome, resulting in a very high melting temperature compared to covalent molecules of similar molecular mass which only exhibit weak intermolecular forces (such as hydrogen bonds or London dispersion forces).

(ii) In strongly acidic conditions (pH 1), the amine group is protonated, and the carboxylate group is also protonated:
Structure: \(CH_3CH(NH_3^+)COOH\).

In strongly alkaline conditions (pH 12), the ammonium group is deprotonated, and the carboxyl group remains deprotonated:
Structure: \(CH_3CH(NH_2)COO^-\).

M1 (1 mark): Correctly draws the zwitterion structure of alanine: \(H_3N^+-CH(CH_3)-COO^-\) (charges must be shown on the nitrogen and oxygen atoms respectively).
M2 (1 mark): Explains high melting point by referencing the strong electrostatic attractions / ionic bonds between oppositely charged parts of different zwitterions in a 3D lattice, requiring high thermal energy to break.
M3 (1 mark): Correct structural formula at pH 1: \(CH_3CH(NH_3^+)COOH\) (accept skeletal/displayed; must show positive charge on nitrogen/ammonium).
M4 (1.5357 marks): Correct structural formula at pH 12: \(CH_3CH(NH_2)COO^-\). (Must show negative charge on oxygen).

(i) Preparing ethylamine from bromoethane:
Reagent: Concentrated ammonia (\(NH_3\)) dissolved in ethanol (ethanolic ammonia).
Conditions: Heat in a sealed tube (under pressure).

An excess of ammonia is crucial because the primary amine product (ethylamine) is itself a nucleophile and can react with unreacted bromoethane to form diethylamine (a secondary amine), triethylamine, and quaternary ammonium salts. Excess ammonia ensures that bromoethane is much more likely to collide with ammonia than with ethylamine, maximizing the yield of the primary amine.

(ii) Comparing basic strength:
Basicity is a measure of the ability of the nitrogen lone pair to accept a proton (\(H^+\)).
- **Ethylamine** is a stronger base than ammonia. The ethyl group is an electron-donating alkyl group (+I inductive effect). This increases the electron density on the nitrogen atom, making its lone pair more readily available to accept a proton.
- **Ammonia** is of intermediate basicity.
- **Phenylamine** is a much weaker base than ammonia. The lone pair of electrons on the nitrogen atom overlaps with the delocalized \(\pi\)-electron cloud of the benzene ring. This delocalization decreases the electron density on the nitrogen atom, making the lone pair much less available to accept a proton.

M1 (1 mark): Reagent and conditions: (concentrated) ammonia in ethanol / ethanolic ammonia AND heat in a sealed tube / under pressure (Do not accept reflux unless sealed/pressure is specified).
M2 (1 mark): Explanation of excess: To prevent further substitution / reaction of the product (ethylamine) with bromoethane / to maximize the yield of the primary amine.
M3 (1 mark): Relative basicity order: Ethylamine > ammonia > phenylamine.
M4 (1.5357 marks): Explains that the ethyl group is electron-donating (+I effect), which increases lone pair availability, whereas the phenyl ring delocalizes the lone pair on nitrogen into the aromatic system, reducing its availability to accept a proton.

(i) Under \(S_N1\) conditions, the reaction proceeds in two steps:
1. **First step (slow):** The carbon-bromine bond in 2-bromo-2-methylpropane breaks heterolytically.
\((CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-\)
- Curly arrow starts from the C-Br bond and goes to the Br atom.
- Intermediate: tertiary carbocation, \((CH_3)_3C^+\) (planar around the central carbon).

2. **Second step (fast):** The hydroxide ion (\(OH^-\)) acts as a nucleophile and attacks the carbocation.
- Curly arrow starts from a lone pair on the oxygen of \(OH^-\)_ and points to the positive carbon atom of the carbocation.
- Product: \((CH_3)_3COH\) (2-methylpropan-2-ol).

(ii) Tertiary halogenoalkanes undergo \(S_N1\) rather than \(S_N2\) due to:
1. **Stability of the carbocation:** The tertiary carbocation intermediate has three electron-donating alkyl groups (methyl groups) attached to the positive carbon. These groups push electron density toward the positive carbon via the inductive effect (+I effect), stabilizing the positive charge.
2. **Steric Hindrance:** In an \(S_N2\) mechanism, the nucleophile must attack the carbon atom from the side opposite to the leaving group (back-side attack). In a tertiary halogenoalkane, the bulky methyl groups block the approach of the nucleophile, preventing the formation of the \(S_N2\) transition state.

M1 (1 mark): Step 1 of mechanism showing correct curly arrow from C-Br bond to Br, forming carbocation and \(Br^-\).
M2 (1 mark): Step 2 of mechanism showing correct structure of carbocation intermediate and curly arrow from the lone pair of \(OH^-\) to the positive carbon atom.
M3 (1 mark): Explains that the tertiary carbocation is stabilized by the electron-donating (+I inductive) effect of three alkyl/methyl groups.
M4 (1.5357 marks): Explains that steric hindrance / bulkiness of the three methyl groups prevents the back-side attack of the nucleophile, making the \(S_N2\) pathway highly unfavorable.

(i) Dot-and-cross diagram details:
- Central atom is Cl.
- There are three single covalent bonds between Cl and F. Each Cl-F bond consists of one electron from Cl (dot) and one from F (cross).
- Cl has 2 lone pairs (4 electrons shown as dots) in its outer shell.
- Each F atom has 3 lone pairs (6 electrons shown as crosses) in its outer shell.

(ii) Explanation using VSEPR:
- The central chlorine atom has 5 regions of electron density (3 bonding pairs and 2 lone pairs).
- To minimize repulsion, these 5 electron pairs adopt a trigonal bipyramidal parent geometry.
- Lone pairs occupy the equatorial positions to minimize \(90^\circ\) repulsions.
- This leaves the 3 bonding pairs in a **T-shaped** geometry.
- Order of repulsion: lone pair - lone pair > lone pair - bond pair > bond pair - bond pair.
- The two equatorial lone pairs repel the axial Cl-F bonding pairs, compressing the bond angle from the ideal \(90^\circ\) to approximately \(87.5^\circ\) (accept any value less than \(90^\circ\), typically \(86^\circ - 88^\circ\)).

M1 (1 mark): Correct dot-and-cross diagram showing 3 bonding pairs and 2 lone pairs on the central chlorine atom (F outer shells must also show 8 electrons total).
M2 (1 mark): Identifies the shape as **T-shaped** and states the bond angle is less than \(90^\circ\) (e.g., \(86^\circ - 88^\circ\)).
M3 (1 mark): Explains that the central chlorine atom has 5 electron pairs (3 bonding, 2 lone pairs) which arrange in a trigonal bipyramidal geometry to minimize repulsion.
M4 (1.5357 marks): Explains that lone pairs repel more than bonding pairs (lp-lp > lp-bp > bp-bp) and occupy equatorial positions, which compresses the axial-equatorial angle below \(90^\circ\).

(i) Empirical formula calculation:
- Molar mass of \(CO_2 = 44.0\text{ g mol}^{-1}\)
- Molar mass of \(H_2O = 18.0\text{ g mol}^{-1}\)

Mass of Carbon in \(3.52\text{ g}\) of \(CO_2\):
\(m(C) = 3.52 \times \frac{12.0}{44.0} = 0.960\text{ g}\)

Mass of Hydrogen in \(1.80\text{ g}\) of \(H_2O\):
\(m(H) = 1.80 \times \frac{2.0}{18.0} = 0.200\text{ g}\)

Mass of Oxygen in the sample of **X**:
\(m(O) = 1.48 - m(C) - m(H) = 1.48 - 0.960 - 0.200 = 0.320\text{ g}\)

Convert masses to moles:
- \(n(C) = \frac{0.960}{12.0} = 0.080\text{ mol}\)
- \(n(H) = \frac{0.200}{1.0} = 0.200\text{ mol}\)
- \(n(O) = \frac{0.320}{16.0} = 0.020\text{ mol}\)

Divide by the smallest value (\(0.020\)) to find simple ratio:
- Ratio \(C = 4\), \(H = 10\), \(O = 1\).
Thus, the empirical formula is \(C_4H_{10}O\).

(ii) Molecular formula:
The empirical formula mass of \(C_4H_{10}O = (4 \times 12.0) + (10 \times 1.0) + 16.0 = 74.0\text{ g mol}^{-1}\).
Since the relative molecular mass is \(74.0\), the molecular formula is identical to the empirical formula: \(C_4H_{10}O\).

(iii) Balanced chemical equation:
\(C_4H_{10}O + 6O_2 \rightarrow 4CO_2 + 5H_2O\)

M1 (1 mark): Correctly calculates the mass of C (\(0.960\text{ g}\)) and H (\(0.200\text{ g}\)) from the given combustion product masses.
M2 (1 mark): Calculates the mass of O by subtraction (\(0.320\text{ g}\)) and converts all masses to moles: \(C = 0.080\), \(H = 0.200\), \(O = 0.020\).
M3 (1 mark): Divides by the smallest mole value to find the simplest whole number ratio (\(4:10:1\)), concluding that the empirical formula is \(C_4H_{10}O\).
M4 (1.5357 marks): Confirms molecular formula is \(C_4H_{10}O\) based on given \(M_r = 74.0\) and writes the fully balanced combustion equation: \(C_4H_{10}O + 6O_2 \rightarrow 4CO_2 + 5H_2O\).

Titanium(III) has the electron configuration \([Ar] 3d^1\).

1. **Splitting of d-orbitals:** In the octahedral field of six water ligands, the five degenerate 3d orbitals of the \(Ti^{3+}\) ion split into two sets of non-degenerate energy levels: three orbitals of lower energy and two of higher energy.

2. **Excitation / Absorption:** The single 3d electron occupies one of the lower energy d-orbitals. When white light is shone on the solution, the electron absorbs a photon of light corresponding to the energy difference (\(\Delta E = h\nu\)) between the split d-orbitals. This causes the electron to be promoted (excited) from the lower energy level to the higher energy level (this is a **d-d transition**).

3. **Color observed:** The energy absorbed (\(\Delta E\)) corresponds to a specific wavelength of light in the visible spectrum (specifically in the yellow-green region). The remaining wavelengths of light are not absorbed; they are transmitted (or reflected). The combination of these transmitted wavelengths represents the complementary color to yellow-green, which is perceived by the eye as **purple**.

M1 (1 mark): Explains that d-orbitals split into two non-degenerate energy levels in the presence of ligands / octahedral environment.
M2 (1 mark): States that an electron absorbs light energy (a photon) and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition).
M3 (1 mark): Relates the energy absorbed to the gap between the split d-orbitals using \(\Delta E = h\nu\) or \(\Delta E = \frac{hc}{\lambda}\).
M4 (1.5357 marks): Explains that the color observed (purple) is due to the transmitted/remaining wavelengths of light that are complementary to the absorbed color (yellow-green).

1. The solution turns from pale blue to a yellow-green colour (often described as green due to the mixture of the blue reactant and yellow product).
2. The reaction is a ligand substitution: \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\).
3. The coordination number decreases from 6 to 4 because the chloride ligands are larger than water molecules and carry a negative charge, leading to greater steric hindrance and electrostatic repulsion between ligands.
4. The molecular geometry changes from octahedral (6-coordinate) to tetrahedral (4-coordinate).

Award 4.5357 marks as follows:
- 1 mark: Stating the correct final colour of the solution (green / yellow-green).
- 1.5357 marks: Providing the correct balanced equilibrium equation including charges.
- 1 mark: Explaining the change in coordination number from 6 to 4 and stating the change in shape from octahedral to tetrahedral.
- 1 mark: Explaining that chloride ligands are larger / have greater mutual repulsion, which prevents 6-coordination.

1. At pH 2.0, which is highly acidic and well below the isoelectric point, the excess hydrogen ions protonate the amine group (\(-NH_2\)) to form \(-NH_3^+\) while the carboxyl group remains protonated as \(-COOH\). The predominant species is a cation: \(CH_3CH(NH_3^+)COOH\). During electrophoresis, this positively charged species migrates towards the negative electrode (cathode).
2. At pH 10.0, which is highly alkaline and well above the isoelectric point, the excess hydroxide ions deprotonate the carboxyl group (\(-COOH\)) to form \(-COO^-\) and deprotonate any protonated amine. The predominant species is an anion: \(CH_3CH(NH_2)COO^-\). During electrophoresis, this negatively charged species migrates towards the positive electrode (anode).

Award 4.5357 marks as follows:
- 1.5357 marks: Correct structure of the cation at pH 2.0 and correctly identifying that it migrates to the cathode.
- 1.5 marks: Correct structure of the anion at pH 10.0 and correctly identifying that it migrates to the anode.
- 1.5 marks: Correctly explaining that low pH causes protonation of amine and high pH causes deprotonation of carboxylic acid group.

1. The molecular formula \(C_3H_6O\) indicates one double bond equivalent (unsaturated ring or pi bond).
2. The strong absorption at \(1715\text{ cm}^{-1}\) in the IR spectrum is characteristic of a carbonyl group (C=O bond).
3. The absence of a broad absorption band above \(3200\text{ cm}^{-1}\) indicates that there is no O-H bond, which rules out any alcohol isomers.
4. Since compound Y does not react with Fehling's solution, it is not oxidized, ruling out the aldehyde isomer (propanal). Ketones do not react with Fehling's solution.
5. Therefore, the compound must be the only possible ketone with three carbons, which is propanone, \(CH_3COCH_3\).

Award 4.5357 marks as follows:
- 1 mark: Correctly identifying compound Y as propanone and writing its structural formula \(CH_3COCH_3\).
- 1 mark: Identifying that the absorption at \(1715\text{ cm}^{-1}\) indicates a carbonyl (C=O) group.
- 1.5357 marks: Explaining that the lack of reaction with Fehling's solution rules out an aldehyde (propanal) and confirms the ketone functional group.
- 1 mark: Explaining that the absence of a broad band above \(3200\text{ cm}^{-1}\) rules out O-H groups / alcohols.

1. The central chlorine atom is in Group 7 and has 7 valence electrons. It forms three single covalent bonds with three fluorine atoms, using 3 electrons, which leaves 4 non-bonding valence electrons (2 lone pairs).
2. The total number of electron pairs around the central atom is 5 (3 bonding pairs and 2 lone pairs), which corresponds to a trigonal bipyramidal electron-pair arrangement.
3. To minimize electron-pair repulsion, the 2 lone pairs occupy the equatorial positions where they experience less repulsion (at \(120^\circ\) to each other).
4. This leaves the 3 bonding pairs to occupy the remaining two axial positions and one equatorial position, giving the molecule a T-shaped geometry.
5. Due to the greater repulsive force of the lone pairs compared to the bonding pairs, the axial fluorine atoms are pushed slightly towards the equatorial fluorine atom, reducing the bond angle from \(90^\circ\) to approximately \(87.5^\circ\).

Award 4.5357 marks as follows:
- 1 mark: Stating that chlorine has 3 bonding pairs and 2 lone pairs (5 electron pairs in total).
- 1.5357 marks: Explaining that lone pairs occupy the equatorial positions to minimize repulsion in a trigonal bipyramidal arrangement.
- 1 mark: Identifying the final molecular shape as T-shaped.
- 1 mark: Stating the approximate bond angle of \(87.5^\circ\) (or less than \(90^\circ\)) and attributing it to lone pairs repelling more strongly than bonding pairs.

1. Calculate the moles of carbon in the sample:
\(n(CO_2) = \frac{\text{mass}}{\text{Molar mass}} = \frac{6.60\text{ g}}{44.0\text{ g mol}^{-1}} = 0.150\text{ mol}\).
Since there is 1 carbon atom per \(CO_2\) molecule, \(n(C) = 0.150\text{ mol}\).
Mass of carbon = \(0.150\text{ mol} \times 12.0\text{ g mol}^{-1} = 1.80\text{ g}\).

2. Calculate the moles of hydrogen in the sample:
\(n(H_2O) = \frac{2.70\text{ g}}{18.0\text{ g mol}^{-1}} = 0.150\text{ mol}\).
Since there are 2 hydrogen atoms per \(H_2O\) molecule, \(n(H) = 2 \times 0.150\text{ mol} = 0.300\text{ mol}\).
Mass of hydrogen = \(0.300\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.30\text{ g}\).

3. Confirm that the total mass matches the hydrocarbon mass:
\(1.80\text{ g (C)} + 0.30\text{ g (H)} = 2.10\text{ g}\), which matches the sample mass.

4. Find the simplest ratio of carbon to hydrogen:
Ratio \(C : H = 0.150 : 0.300 = 1 : 2\).

5. The empirical formula is \(CH_2\).

Award 4.5357 marks as follows:
- 1.5357 marks: Correct calculation of moles of Carbon (\(0.150\text{ mol}\)) from the given mass of \(CO_2\).
- 1.5 marks: Correct calculation of moles of Hydrogen (\(0.300\text{ mol}\)) from the given mass of \(H_2O\) (must account for the factor of 2).
- 1.5 marks: Correctly determining the simplest integer ratio as 1:2 and stating the empirical formula as \(CH_2\).

1. Write down the reduction half-equation:
\(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\)
2. Write down the oxidation half-equation:
\(Fe^{2+} \rightarrow Fe^{3+} + e^-\)
3. Multiply the oxidation half-equation by 6 to balance the number of electrons:
\(6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-\)
4. Add both half-equations together to cancel the electrons:
\(Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O\)
5. The starting dichromate(VI) solution is orange, and the resulting chromium(III) solution is green, so the observed colour change is orange to green.

Award 4.5357 marks as follows:
- 1 mark: Stating or deriving the correct reduction half-equation.
- 2.5357 marks: Writing the fully balanced overall ionic equation with all correct species and state symbols (if included).
- 1 mark: Stating the correct colour change from orange to green.

1. Basicity depends on the ease with which the lone pair of electrons on the nitrogen atom can be donated to a proton (\(H^+\)).
2. Ethylamine is more basic than ammonia because the ethyl group is electron-donating due to the positive inductive effect (+I). This increases the electron density around the nitrogen atom, making the lone pair more available for coordinate bonding to a proton.
3. Phenylamine is less basic than ammonia because the lone pair of electrons on the nitrogen atom overlaps with the delocalized pi-electron system of the benzene ring. This delocalization reduces the electron density on the nitrogen, making the lone pair much less available to accept a proton.
4. Therefore, the order of increasing basicity is phenylamine < ammonia < ethylamine.

Award 4.5357 marks as follows:
- 1 mark: Correct order of basicity (phenylamine < ammonia < ethylamine).
- 1.5357 marks: Explaining that ethylamine is more basic than ammonia due to the positive inductive effect of the ethyl group enhancing nitrogen's lone pair availability.
- 2 marks: Explaining that phenylamine is less basic due to the delocalization of the nitrogen lone pair of electrons into the benzene ring's pi-cloud, reducing its availability.

1. Thermal stability of Group 2 carbonates increases down the group from \(MgCO_3\) to \(BaCO_3\).
2. Down the group, the ionic radius of the Group 2 cation (\(M^{2+}\)) increases, while its charge remains constant at \(2+\).
3. Consequently, the charge density of the cation decreases from \(Mg^{2+}\) to \(Ba^{2+}\).
4. A cation with higher charge density (like \(Mg^{2+}\)) is more polarizing. It distorts the electron cloud of the nearby carbonate ion (\(CO_3^{2-}\)) more strongly, weakening the carbon-oxygen bonds within the carbonate ion.
5. Since the carbon-oxygen bonds are already weakened by polarization in \(MgCO_3\), it requires less thermal energy to decompose into magnesium oxide and carbon dioxide compared to \(BaCO_3\).

Award 4.5357 marks as follows:
- 1 mark: Stating that thermal stability increases down the group.
- 1 mark: Identifying that the cationic radius increases down the group, while the charge remains the same.
- 1.5357 marks: Explaining that the charge density and polarizing power of the cation decrease down the group.
- 1 mark: Connecting lower polarizing power to less distortion/weakening of the C-O bonds in the carbonate ion, which requires more heat to decompose.

1. Catalysis by \(\text{Fe}^{2+}\) involves a homogeneous catalytic cycle where transition metal ions can readily change oxidation states.
2. First step: \(\text{Fe}^{2+}\) reduces the peroxodisulfate(VI) ion:
\(2\text{Fe}^{2+}\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}\)
3. Second step: The newly formed \(\text{Fe}^{3+}\) oxidises the iodide ions to iodine, regenerating the \(\text{Fe}^{2+}\) catalyst:
\(2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}\)
4. The uncatalysed reaction involves collision between two anions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)). Due to electrostatic repulsion between like-charged ions, the collision is highly unfavourable, resulting in a high activation energy. By using a catalyst, both steps involve collisions between oppositely charged ions (positive iron ions and negative reactant ions), which have much lower activation energies.

- **Mark 1**: Explanation of electrostatic repulsion between negative reactant ions causing a high activation energy in the uncatalysed reaction.
- **Mark 2**: Correct balanced equation for the reduction of peroxodisulfate(VI) by iron(II) ions.
- **Mark 3**: Correct balanced equation for the oxidation of iodide ions by iron(III) ions.
- **Mark 4**: Statement that the iron(II) catalyst is regenerated in the cycle, making the overall pathway more kinetically viable.

1. Basicity of nitrogen-containing compounds depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.
2. In **ethylamine**, the ethyl group is an alkyl group that releases electron density (+I inductive effect) towards the nitrogen atom. This increases the electron density on the nitrogen, making its lone pair more available to accept a proton than in ammonia.
3. **Ammonia** has no substituents other than hydrogen, serving as a baseline comparison.
4. In **phenylamine**, the lone pair on the nitrogen atom is adjacent to the benzene ring and overlaps with the \(\pi\)-system, becoming delocalised. This significantly reduces the electron density on the nitrogen atom, making the lone pair far less available to accept a proton.
5. The relative basicity order is therefore: ethylamine > ammonia > phenylamine.

- **Mark 1**: Correct order of basicity (ethylamine > ammonia > phenylamine).
- **Mark 2**: Correctly identifies that basicity is due to the ability of the lone pair on the nitrogen atom to accept a proton.
- **Mark 3**: Explains that the alkyl group in ethylamine is electron-donating (+I effect), which increases the availability of the lone pair.
- **Mark 4**: Explains that the lone pair in phenylamine is delocalised into the benzene ring's \(\pi\)-system, reducing its availability.

1. The hydrolysis of a halogenoalkane is a nucleophilic substitution reaction: \(\text{R-X} + \text{OH}^- \rightarrow \text{R-OH} + \text{X}^-\).
2. **Bond Polarity**: The electronegativity of the halogens decreases in the order \(\text{Cl} > \text{Br} > \text{I}\). Therefore, the C-Cl bond is the most polar and has the largest partial positive charge on the carbon atom, which would make it most susceptible to nucleophilic attack. If polarity were dominant, 1-chlorobutane would be the fastest.
3. **Bond Enthalpy**: The bond strength decreases in the order \(\text{C-Cl} > \text{C-Br} > \text{C-I}\) because the halogen atomic size increases, resulting in less effective orbital overlap with carbon. Thus, the C-I bond is the weakest and easiest to break.
4. **Dominant Factor**: Experimentally, 1-iodobutane hydrolyses the fastest, which demonstrates that bond enthalpy is the dominant factor in determining the rate of nucleophilic substitution.

- **Mark 1**: Correct order of rates of hydrolysis (1-iodobutane > 1-bromobutane > 1-chlorobutane).
- **Mark 2**: Discussion of bond polarity (C-Cl is most polar because Cl is the most electronegative) and how this would theoretically affect nucleophilic attack.
- **Mark 3**: Discussion of bond enthalpy (C-I is the weakest bond due to poorer orbital overlap from the larger iodine atom).
- **Mark 4**: Identification of bond enthalpy as the dominant factor and linking the weakest bond to the highest reaction rate.

1. In both \(\text{NH}_3\) and \(\text{NH}_4^+\), the central nitrogen atom has four pairs of valence shell electrons (coordination number of 4).
2. In ammonia (\(\text{NH}_3\)):
- There are 3 bonding pairs and 1 lone pair.
- The shape is **trigonal pyramidal**.
- The bond angle is compressed to **\(107^\circ\)** (accept \(107^\circ\) - \(108^\circ\)) because lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion.
3. In the ammonium ion (\(\text{NH}_4^+\)):
- There are 4 bonding pairs and 0 lone pairs.
- The shape is **tetrahedral**.
- The bond angle is the ideal tetrahedral angle of **\(109.5^\circ\)** because all four bonding electron pairs repel each other equally.

- **Mark 1**: Correct shapes for both species (trigonal pyramidal for \(\text{NH}_3\) and tetrahedral for \(\text{NH}_4^+\)) and correct bond angles (\(107^\circ\) and \(109.5^\circ\)).
- **Mark 2**: Identifies that both nitrogen atoms are surrounded by four pairs of electrons.
- **Mark 3**: States that lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion.
- **Mark 4**: Connects the absence of a lone pair in \(\text{NH}_4^+\) to equal repulsion and symmetric tetrahedral geometry.

1. Calculate mass of water of crystallisation lost:
\(\text{Mass of water} = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}\)
2. Calculate moles of water lost:
\(n(\text{H}_2\text{O}) = \frac{1.80\text{ g}}{18.0\text{ g/mol}} = 0.100\text{ mol}\)
3. Calculate molar mass of anhydrous copper(II) sulfate, \(\text{CuSO}_4\):
\(M(\text{CuSO}_4) = 63.5 + 32.1 + (4 \times 16.0) = 159.6\text{ g/mol}\)
4. Calculate moles of anhydrous \(\text{CuSO}_4\):
\(n(\text{CuSO}_4) = \frac{3.19\text{ g}}{159.6\text{ g/mol}} = 0.0200\text{ mol}\)
5. Find simple whole number ratio of \(\text{H}_2\text{O}\) to \(\text{CuSO}_4\):
\(\text{Ratio} = \frac{0.100}{0.0200} = 5\)
6. Therefore, \(x = 5\), and the chemical formula is \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\).

- **Mark 1**: Calculates mass of water lost (\(1.80\text{ g}\)) and converts it to moles (\(0.100\text{ mol}\)).
- **Mark 2**: Correctly calculates the molar mass of anhydrous \(\text{CuSO}_4\) as \(159.6\text{ g/mol}\).
- **Mark 3**: Correctly calculates the moles of anhydrous \(\text{CuSO}_4\) as \(0.0200\text{ mol}\).
- **Mark 4**: Shows the mole ratio is \(5:1\) and states the correct formula \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\).

1. **Origin of Colour**:
- Ligands field splits the five d-orbitals of the transition metal ion into two non-degenerate levels with an energy gap, \(\Delta E\).
- Electrons absorb a quantum of visible light corresponding to \(\Delta E = h\nu\) and are excited from the lower level to the higher level (d-d transition).
- The complementary wavelengths of light that are not absorbed are transmitted or reflected, which we perceive as colour (pale blue for hydrated copper(II) ions).
2. **Ligand Exchange and Colour Change**:
- When excess ammonia is added, a ligand exchange reaction takes place:
\([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\text{(aq)} + 4\text{NH}_3\text{(aq)} \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)}\)
- Ammonia ligand creates a different crystal field strength compared to water, changing the d-orbital splitting energy gap (\(\Delta E\)).
- As a result, a different wavelength of visible light is absorbed, and the observed complementary colour shifts to a deep blue.

- **Mark 1**: Explanation of d-orbital splitting into two energy levels by ligands.
- **Mark 2**: Linking of electron transition (d-d transition) to the absorption of visible light, with the complementary colour being seen.
- **Mark 3**: Identification of the ligand exchange reaction resulting in the formation of \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).
- **Mark 4**: Explaining that different ligands change the d-orbital energy gap (\(\Delta E\)), shifting the absorbed wavelength and changing the color.

1. Amino acids join via a condensation reaction where water is eliminated to form a peptide (amide) link.
2. Since two different amino acids are reacting, two combinations are possible:
- **Alanyl-glycine**: The alanine residue provides the free N-terminus (amine group) and the glycine residue provides the free C-terminus (carboxylic acid group).
Structure: \(\text{H}_2\text{N-CH(CH}_3\text{)-CONH-CH}_2\text{-COOH}\)
- **Glycyl-alanine**: The glycine residue provides the free N-terminus and the alanine residue provides the free C-terminus.
Structure: \(\text{H}_2\text{N-CH}_2\text{-CONH-CH(CH}_3\text{)-COOH}\)
3. The peptide link is the central covalent covalent bond block: \(-\text{C}(=\text{O})-\text{N}(\text{H})-\).

- **Mark 1**: Correct structural formula of Alanyl-glycine.
- **Mark 2**: Correct structural formula of Glycyl-alanine.
- **Mark 3**: Description or drawing that correctly identifies the peptide (amide) linkage as \(-\text{CONH}-\).
- **Mark 4**: Explicitly states that peptide bond formation is a condensation reaction that releases a water molecule.

1. **IR Analysis**:
- The strong absorption at \(1715\text{ cm}^{-1}\) indicates a carbonyl group, \(\text{C=O}\).
- The absence of any broad absorption band above \(3200\text{ cm}^{-1}\) proves there is no alcohol or acid \(\text{O-H}\) group present.
2. **Isomer Identification**:
- For the molecular formula \(\text{C}_3\text{H}_6\text{O}\), the only two possibilities containing a carbonyl group and no other double bonds are **propanal** (an aldehyde) and **propanone** (a ketone).
3. **Mass Spec Analysis**:
- The molecular ion peak \(M^+\) of \(\text{C}_3\text{H}_6\text{O}\) is at \(m/z = 58\).
- A major fragment peak at \(m/z = 43\) represents a loss of 15 mass units (\(58 - 15 = 43\)), which corresponds to the loss of a methyl group (\(\cdot\text{CH}_3\)).
- Propanone (\(\text{CH}_3\text{COCH}_3\)) undergoes easy single cleavage next to the carbonyl to give the highly stable resonance-stabilised acylium ion, \([\text{CH}_3\text{CO}]^+\), which has a mass of 43.
- In contrast, propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) would fragment to yield \([\text{CH}_3\text{CH}_2]^+\) (\(m/z = 29\)) or \([\text{CHO}]^+\) (\(m/z = 29\)). Therefore, the compound must be propanone.

- **Mark 1**: Associates the peak at \(1715\text{ cm}^{-1}\) with the presence of a carbonyl group (\(\text{C=O}\)).
- **Mark 2**: Uses the absence of peak above \(3200\text{ cm}^{-1}\) to rule out the presence of any hydroxyl/alcohol group (\(\text{O-H}\)).
- **Mark 3**: Deduces that the fragment at \(m/z = 43\) represents the acylium ion \([\text{CH}_3\text{CO}]^+\), resulting from a loss of a methyl group.
- **Mark 4**: Correctly identifies the compound as propanone (by name or formula) and rules out propanal based on mass spectrometry.

First, compare the relative sizes of the chloride and water ligands: chloride ions are larger and have greater mutual repulsion. Therefore, only four chloride ligands can fit around the central cobalt(II) ion compared to six smaller water molecules. Second, write the equation for the ligand substitution reaction: \([Co(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CoCl_4]^{2-} + 6H_2O\). Calculate the change in the number of particles: 5 reactant particles produce 7 product particles. An increase in the number of particles in solution increases the disorder of the system, resulting in a positive entropy change.

Mark 1: Mentions that chloride ligands are larger than water ligands (or have greater mutual repulsion).
Mark 2: Explains that steric hindrance / lack of space around the cobalt ion prevents six chloride ligands from coordinating.
Mark 3: Writes or states the reaction shows 5 reactant particles forming 7 product particles.
Mark 4: Connects the increase in the number of particles to an increase in disorder / positive entropy change.
Mark 4.5357: Expresses ideas clearly using correct chemical terminology throughout.

Basicity is a measure of the ability of a species to donate a lone pair of electrons to a proton. In ethylamine, the ethyl group has a positive inductive effect (+I) and releases electron density towards the nitrogen atom, increasing the availability of its lone pair. In ammonia, there is no inductive effect from alkyl groups. In phenylamine, the lone pair on the nitrogen atom overlaps with the delocalised \(\pi\)-system of the benzene ring, reducing the electron density on the nitrogen and making the lone pair much less available to accept a proton.

Mark 1: Correct order of basicity: ethylamine > ammonia > phenylamine.
Mark 2: Explains that the ethyl group in ethylamine is electron-donating, increasing electron density on the nitrogen atom.
Mark 3: Explains that the lone pair in phenylamine is delocalised into the benzene ring.
Mark 4: Relates these electronic effects to the relative availability of the lone pair on the nitrogen atom to accept a proton.
Mark 4.5357: Consistently links structure to basic strength in a logical sequence.

In the \(S_N1\) mechanism of 2-bromobutane: the C-Br bond breaks first to form a planar carbocation intermediate. Nucleophilic attack by \(OH^-\)\ can occur with equal probability from above or below the plane, leading to a 50:50 mixture of two enantiomers (racemisation). In the \(S_N2\) mechanism of 2-bromooctane: the reaction is concerted and proceeds via a transition state. The nucleophile attacks from the side opposite the leaving group (backside attack). This forces the other three groups to invert (like an umbrella), yielding a single enantiomer and thus preserving optical activity.

Mark 1: Identifies that 2-bromobutane undergoes \(S_N1\) involving a planar carbocation intermediate.
Mark 2: Explains that attack on the planar carbocation is equally likely from either side, resulting in a racemic mixture / optical inactivity.
Mark 3: Identifies that 2-bromooctane undergoes \(S_N2\) involving a single concerted transition state.
Mark 4: Explains that backside attack by the nucleophile causes inversion of configuration, retaining optical activity.
Mark 4.5357: Correctly uses precise terminology ('planar carbocation', 'backside attack', 'inversion of configuration').

For \(XeF_4\): Xenon has 8 valence electrons, plus 4 from fluorine, giving 12 electrons (6 pairs). There are 4 bonding pairs and 2 lone pairs. The octahedral arrangement of 6 electron pairs is adopted, but to minimise lone-pair/lone-pair repulsion, the two lone pairs sit opposite each other (axial positions), giving a square planar molecular geometry with bond angles of \(90^{\circ}\). For \(SF_4\): Sulfur has 6 valence electrons, plus 4 from fluorine, giving 10 electrons (5 pairs). There are 4 bonding pairs and 1 lone pair. The trigonal bipyramidal arrangement of 5 electron pairs is adopted. The lone pair occupies an equatorial position to minimise repulsion, resulting in a see-saw molecular geometry with equatorial bond angles of approximately \(102^{\circ}\) (or \(< 120^{\circ}\)) and axial-equatorial angles of approximately \(87^{\circ}\) (or \(< 90^{\circ}\)).

Mark 1: Correctly determines the number of bonding pairs and lone pairs for both: \(XeF_4\) (4 BP, 2 LP) and \(SF_4\) (4 BP, 1 LP).
Mark 2: Predicts square planar shape and bond angle of \(90^{\circ}\) for \(XeF_4\).
Mark 3: Predicts see-saw (or distorted tetrahedral) shape and bond angles of \(<120^{\circ}\) / \(<90^{\circ}\) for \(SF_4\).
Mark 4: Explains that lone pairs repel more than bonding pairs, and explains the spatial arrangement chosen to minimise repulsion (lone pairs opposite in \(XeF_4\), equatorial in \(SF_4\)).
Mark 4.5357: Clear logical explanation linking electron pair geometry to molecular shape for both compounds.

Mass of water lost = \(3.92\text{ g} - 2.84\text{ g} = 1.08\text{ g}\).
Molar mass of \(H_2O = 18.0\text{ g mol}^{-1}\).
Moles of \(H_2O = 1.08 / 18.0 = 0.060\text{ mol}\).
Molar mass of anhydrous \((NH_4)_2Fe(SO_4)_2\) = \(2 \times [14.0 + (4 \times 1.0)] + 55.8 + 2 \times [32.1 + (4 \times 16.0)] = 284.0\text{ g mol}^{-1}\).
Moles of anhydrous salt = \(2.84 / 284.0 = 0.010\text{ mol}\).
Ratio of moles of water to anhydrous salt = \(0.060 / 0.010 = 6\).
Therefore, \(x = 6\).

Mark 1: Calculates the mass of water lost (\(1.08\text{ g}\)) and correctly calculates the molar mass of anhydrous \((NH_4)_2Fe(SO_4)_2\) (\(284.0\text{ g mol}^{-1}\)).
Mark 2: Calculates the moles of water (\(0.060\text{ mol}\)) and the moles of anhydrous salt (\(0.010\text{ mol}\)).
Mark 3: Determines the simplest molar ratio by dividing both values by the smaller number of moles (\(0.060 / 0.010 = 6\)).
Mark 4: States final answer as \(x = 6\).
Mark 4.5357: Shows complete working with correct units and appropriate significant figures throughout.

Without a catalyst, the two negative reactant ions (\(S_2O_8^{2-}\) and \(I^-\)) repel each other, meaning collisions between them rarely have enough energy to overcome the high activation energy barrier. Iron(II) ions act as a homogeneous catalyst by providing an alternative pathway with a lower activation energy, involving species of opposite charges. First, \(Fe^{2+}\) is oxidised to \(Fe^{3+}\) by \(S_2O_8^{2-}\): \(2Fe^{2+}(aq) + S_2O_8^{2-}(aq) \rightarrow 2Fe^{3+}(aq) + 2SO_4^{2-}(aq)\). Second, \(Fe^{3+}\) is reduced back to \(Fe^{2+}\) by oxidising \(I^-\): \(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\). The catalyst is regenerated at the end of the reaction.

Mark 1: Explains that repulsion between like-charged anions leads to a high activation energy for the uncatalysed reaction.
Mark 2: Explains that the catalyst provides an alternative pathway of lower activation energy.
Mark 3: Writes balanced equation for the oxidation of \(Fe^{2+}\) by peroxodisulfate.
Mark 4: Writes balanced equation for the reduction of \(Fe^{3+}\) by iodide.
Mark 4.5357: Notes that the catalyst \(Fe^{2+}\) is regenerated, demonstrating the definition of a catalyst.

At highly acidic pH (pH 1.0), both the amine group and the carboxylate group are protonated, giving a net positive charge: \(H_3N^+-CH(CH_3)-COOH\). At the isoelectric point (pH 6.0), the molecule exists as a neutral zwitterion: \(H_3N^+-CH(CH_3)-COO^-\). At highly alkaline pH (pH 12.0), both groups lose protons, yielding a net negative charge: \(H_2N-CH(CH_3)-COO^-\). During electrophoresis, charged particles migrate towards the electrode of opposite charge. Since alanine has a net negative charge at pH 12.0, it will migrate towards the positive electrode (the anode).

Mark 1: Draws/describes correct protonated structure at pH 1.0.
Mark 2: Draws/describes correct zwitterion structure at pH 6.0.
Mark 3: Draws/describes correct deprotonated structure at pH 12.0.
Mark 4: Predicts correct direction of migration (towards the positive electrode / anode).
Mark 4.5357: Explains that migration occurs because the negative ion is attracted to the positive electrode.

1) Infrared: The absence of a broad peak at \(3200-3600\text{ cm}^{-1}\) rules out an alcohol (no O-H bond). The strong, sharp peak at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (C=O). 2) Chemical test: Since **X** does not react with Tollens' reagent, it cannot be an aldehyde. It must be a ketone. 3) Identification: The only ketone with the formula \(C_4H_8O\) is butanone, \(CH_3COCH_2CH_3\). 4) Mass spectrum: The molecular ion peak is at \(m/z = 72\), which corresponds to \(C_4H_8O^+\). Fragmentation of butanone occurs via cleavage of the C-C bond adjacent to the carbonyl group. Cleavage of the ethyl group (mass 29) leaves behind the acylium ion \([CH_3CO]^+\) with \(m/z = 43\) (specifically, \(72 - 29 = 43\)).

Mark 1: Identifies the carbonyl group (C=O) from the peak at \(1715\text{ cm}^{-1}\) and rules out alcohol (O-H) due to the lack of peak at \(3200-3600\text{ cm}^{-1}\).
Mark 2: Deduces that **X** is a ketone (and not an aldehyde) because it does not react with Tollens' reagent.
Mark 3: Concludes that **X** is butanone (and provides the name or structure).
Mark 4: Identifies the fragment at \(m/z = 43\) as the acylium ion \([CH_3CO]^+\) (positive charge must be included).
Mark 4.5357: Explains how the fragment \(m/z = 43\) arises from the loss of an ethyl radical (\(\cdot CH_2CH_3\)) from the molecular ion.

First reaction: \([Cu(H_2O)_6]^{2+}(aq) + 4Cl^{-}(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)\). The starting complex \([Cu(H_2O)_6]^{2+}\) is octahedral and pale blue. The product complex \([CuCl_4]^{2-}\) is tetrahedral and yellow-green. The second reaction is the reverse process upon dilution with water: \([CuCl_4]^{2-}(aq) + 6H_2O(l) \rightleftharpoons [Cu(H_2O)_6]^{2+}(aq) + 4Cl^{-}(aq)\). The coordinate number decreases from 6 to 4 because \(Cl^{-}\) ligands are larger than \(H_2O\) ligands and experience greater steric repulsion around the central \(Cu^{2+}\) ion. Color changes because different ligands split the d-orbitals by different amounts (energy gap \(\Delta E\) changes). The wavelength of absorbed light is given by \(E = \frac{hc}{\lambda}\). When d-orbital splitting changes, the absorbed wavelength changes, and the complementary color observed also changes.

1 mark for correct balanced equation for the formation of the tetrachloro complex. 1 mark for correct balanced equation for the dilution back to the hexaaqua complex. 1 mark for identifying the geometries of both complexes (octahedral and tetrahedral). 1 mark for explaining the change in coordination number due to the larger size/steric hindrance of the chloride ligands. 1 mark for explaining d-orbital splitting by ligands. 1.625 marks for linking the change in splitting energy to the change in wavelength of light absorbed and the observed complementary color.

1. Hydrolysis: Reflux the peptide sample with \(6\text{ mol dm}^{-3}\) \(HCl\) for approximately 24 hours to break the peptide bonds. 2. Chromatography setup: Apply a small spot of the concentrated hydrolysate alongside reference spots of pure phenylalanine and alanine on a pencil baseline on chromatography paper. 3. Development: Suspend the paper in a chromatography tank containing a suitable solvent system (e.g., butan-1-ol, ethanoic acid, water). Allow the solvent front to run near the top of the paper, then mark the solvent front. 4. Visualization: Amino acids are colorless, so spray the dried chromatogram with ninhydrin reagent and heat in an oven to reveal purple/brown spots. 5. Analysis: Calculate the Rf value for each spot using \(R_f = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}\). Compare these \(R_f\) values with the \(R_f\) values of the reference amino acids under identical experimental conditions to identify them.

1 mark for specifying reflux with \(6\text{ mol dm}^{-3}\) \(HCl\). 1 mark for chromatography setup details (pencil line, spotting hydrolysate and reference samples). 1 mark for running the solvent and marking the solvent front. 1 mark for spraying with ninhydrin and heating to visualize spots. 1 mark for correct formula of \(R_f\) value. 1.625 marks for explaining the comparison of experimental \(R_f\) values with reference standards to confirm identity of the amino acids.

1. IR interpretation: The broad absorption at \(3350\text{ cm}^{-1}\) is typical of an O-H stretching vibration in alcohols. The peak at \(1050\text{ cm}^{-1}\) corresponds to the C-O stretch. 2. Chemical test interpretation: Tertiary alcohols do not undergo oxidation by acidified potassium dichromate(VI) because they lack a hydrogen atom on the carbon bearing the -OH group, hence the solution remains orange. 3. Structural deduction: The molecular formula \(C_4H_{10}O\) has isomers. The tertiary alcohol isomer is 2-methylpropan-2-ol, which has the formula \((CH_3)_3COH\). 4. Mass spectrometry fragmentation: The molecular ion of 2-methylpropan-2-ol is \([C_4H_{10}O]^{+}\) with \(m/z = 74\). Loss of a methyl radical (\(CH_3\), mass 15) via cleavage of a C-C bond adjacent to the oxygen atom yields the stable tertiary carbocation \([(CH_3)_2COH]^{+}\), which has an \(m/z\) of 59 (\(74 - 15 = 59\)).

1 mark for identifying the O-H group from \(3350\text{ cm}^{-1}\). 1 mark for deducing that the orange color remaining means it is a tertiary alcohol. 1 mark for drawing/identifying 2-methylpropan-2-ol as the only tertiary alcohol. 1 mark for stating its correct IUPAC name. 1 mark for identifying the mass of the molecular ion as 74 and explaining that the peak at 59 is due to the loss of a methyl radical (M-15). 1.625 marks for identifying the formula of the fragment ion as \([(CH_3)_2COH]^{+}\) with a positive charge.

Sodium chloride (\(NaCl\)) consists of a giant ionic lattice of \(Na^{+}\) and \(Cl^{-}\) ions held together by strong electrostatic forces of attraction in all directions (ionic bonding). Silicon(IV) oxide (\(SiO_2\)) consists of a giant covalent tetrahedral network where each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. Both have high melting temperatures because a large amount of thermal energy is required to overcome the strong forces of attraction (ionic attractions in \(NaCl\) and covalent bonds in \(SiO_2\)). Solid \(NaCl\) does not conduct electricity because the ions are fixed in the lattice; however, when molten or dissolved in water, the lattice breaks down, and the ions are free to move and carry charge. \(SiO_2\) does not conduct electricity in any state because the valence electrons are tightly held in localized covalent bonds, and there are no free-moving ions or delocalized electrons to act as charge carriers.

1 mark for describing NaCl as a giant ionic lattice with strong ionic bonds. 1 mark for describing SiO2 as a giant covalent structure with strong covalent bonds. 1 mark for explaining that both have high melting points due to the large amount of energy needed to break these strong bonds/attractions. 1 mark for explaining that solid NaCl does not conduct electricity but molten/aqueous NaCl does because ions become mobile. 1 mark for explaining that SiO2 does not conduct electricity because electrons are localized in covalent bonds and there are no mobile charge carriers. 1.625 marks for structured comparative language highlighting differences and similarities clearly.

1. Write the ionic equation: \(MnO_4^{-}(aq) + 5Fe^{2+}(aq) + 8H^{+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)\). 2. Moles of \(MnO_4^{-}\) used: \(n(MnO_4^{-}) = 0.0100\text{ mol dm}^{-3} \times 0.00800\text{ dm}^3 = 8.00 \times 10^{-5}\text{ mol}\). 3. From stoichiometry, \(n(Fe^{2+})\) in \(25.00\text{ cm}^3\) aliquot = \(5 \times n(MnO_4^{-}) = 5 \times 8.00 \times 10^{-5} = 4.00 \times 10^{-4}\text{ mol}\). 4. Total \(n(Fe^{2+})\) in \(250.0\text{ cm}^3\) = \(4.00 \times 10^{-4} \times 10 = 4.00 \times 10^{-3}\text{ mol}\). This is equal to the moles of anhydrous salt in the \(1.568\text{ g}\) sample. 5. Molar mass of anhydrous \((NH_4)_2Fe(SO_4)_2\): \(2 \times (14.0 + 4.0) + 55.8 + 2 \times (32.1 + 4 \times 16.0) = 284.0\text{ g mol}^{-1}\). 6. Mass of anhydrous \((NH_4)_2Fe(SO_4)_2\) = \(4.00 \times 10^{-3}\text{ mol} \times 284.0\text{ g mol}^{-1} = 1.136\text{ g}\). 7. Mass of water of crystallisation = \(1.568\text{ g} - 1.136\text{ g} = 0.432\text{ g}\). 8. Moles of water of crystallisation = \(0.432\text{ g} / 18.0\text{ g mol}^{-1} = 0.0240\text{ mol}\). 9. Calculate x: \(x = n(H_2O) / n(Fe^{2+}) = 0.0240 / (4.00 \times 10^{-3}) = 6\).

1 mark for the correct balanced ionic equation. 1 mark for calculating the moles of MnO4- and multiplying by 5 to find moles of Fe2+ in the aliquot. 1 mark for scaling up to the 250.0 cm3 volumetric flask. 1 mark for calculating the molar mass of the anhydrous salt (284.0 g mol-1) and finding its mass (1.136 g). 1 mark for finding the mass of water (0.432 g) and moles of water (0.0240 mol). 1.625 marks for dividing the moles of water by the moles of anhydrous salt to find x = 6 with clear logical presentation.

Without a catalyst, the reaction \(S_2O_8^{2-}(aq) + 2I^{-}(aq) \rightarrow 2SO_4^{2-}(aq) + I_2(aq)\) involves collision between two negatively charged anions. The strong electrostatic repulsion between like-charged ions results in a very high activation energy (\(E_a\)), meaning very few collisions have energy greater than or equal to \(E_a\), making the reaction slow at room temperature. Iron(II) ions act as a homogenous catalyst: Step 1: \(S_2O_8^{2-}(aq) + 2Fe^{2+}(aq) \rightarrow 2SO_4^{2-}(aq) + 2Fe^{3+}(aq)\). Step 2: \(2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\). Both steps have lower activation energy than the uncatalysed route. Transition metals are highly effective homogenous catalysts because they have variable oxidation states that differ in stability by relatively small energy gaps, allowing them to readily transition between oxidation states to act as electron carriers.

1 mark for explaining that the uncatalysed reaction is slow because both reactants are negatively charged and repel each other, resulting in high activation energy. 1 mark for the first catalytic equation. 1 mark for the second catalytic equation. 1 mark for explaining that catalyst steps involve ions of opposite charge, lowering activation energy. 1 mark for explaining that transition metals have variable oxidation states. 1.625 marks for linking variable oxidation states to the ability of transition metals to act as intermediate electron transfer agents, facilitating the overall reaction.

Route A (Direct substitution): Bromoethane reacts with concentrated ethanolic ammonia under pressure in a sealed tube: \(CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_2 + HBr\). However, ethylamine is itself a nucleophile due to the lone pair on the nitrogen atom and will react further with remaining bromoethane to form diethylamine, triethylamine, and finally tetraethylammonium bromide. This leads to a mixture of products and a low yield of the primary amine. Route B (Nitrile reduction): To make ethylamine, chloromethane or bromomethane is reacted with potassium cyanide (\(KCN\)) in aqueous ethanol under reflux to produce acetonitrile (\(CH_3CN\)) via nucleophilic substitution (\(S_N2\)). Acetonitrile is then reduced using \(LiAlH_4\) in dry ether, or hydrogen gas with a nickel catalyst, to form ethylamine: \(CH_3CN + 4[H] \rightarrow CH_3CH_2NH_2\). The nitrile route yields exclusively the primary amine, preventing further alkylation, making it the highly preferred synthetic route for purity. However, a major disadvantage of this route is the high toxicity of potassium cyanide, which requires stringent safety precautions.

1 mark for detailing conditions of direct substitution (heating with ethanolic ammonia under pressure). 1 mark for explaining further alkylation in direct substitution, leading to a mixture of products. 1 mark for stating conditions of the nitrile route: KCN in aqueous ethanol under reflux, followed by reduction using LiAlH4 in dry ether or H2/Ni. 1 mark for explaining that the nitrile reduction route produces exclusively the primary amine. 1 mark for discussing safety issues (toxicity of cyanide vs high pressure of ammonia). 1.625 marks for justifying the choice of the nitrile route based on higher yield/easier purification of the primary amine product.

Equation: \(C_6H_{11}OH \rightarrow C_6H_{10} + H_2O\). Concentrated phosphoric(V) acid is used as a dehydrating catalyst. It is preferred over concentrated sulfuric acid because \(H_2SO_4\) is a strong oxidising agent that can oxidise cyclohexanol to carbon/by-products and is reduced to toxic sulfur dioxide gas, lowering the yield of cyclohexene. Preparation & Distillation: Heat cyclohexanol and concentrated phosphoric(V) acid in a round-bottom flask with anti-bumping granules. The cyclohexene co-distils with water and is collected in a cooled receiver flask. Washing: Transfer the crude distillate to a separating funnel, add saturated sodium chloride solution (or sodium hydrogencarbonate). Shake, release pressure, and allow layers to separate. Discard the lower aqueous layer. Drying: Transfer the organic layer to a conical flask, add a spatula of anhydrous calcium chloride as a drying agent, and swirl until the liquid becomes clear. Purification: Decant the dried liquid into a clean flask and perform fractional distillation, collecting the pure cyclohexene fraction at its boiling point of approximately \(83^\circ\text{C}\).

1 mark for the correct balanced equation/structural representation of the dehydration. 1 mark for explaining that phosphoric acid is less oxidising than sulfuric acid. 1 mark for describing the initial distillation setup during heating of the reaction mixture. 1 mark for describing washing in a separating funnel and discarding the aqueous layer. 1 mark for specifying a suitable anhydrous drying agent (e.g. CaCl2 or MgSO4) and indicating the mixture must turn clear. 1.625 marks for describing final purification by fractional distillation and collecting the product at its specific boiling temperature (approx 83 degrees C).

Step 1: Write the ionic equation for the reduction: Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) -> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l). Step 2: Write the precipitation equation: Cr3+(aq) + 3OH-(aq) -> Cr(OH)3(s). Step 3: Calculate the moles of dichromate(VI) ions: Moles of Cr2O72- = 0.0250 dm3 * 0.0450 mol dm-3 = 1.125 * 10^-3 mol. Step 4: Use the reacting ratio (1 mol of Cr2O72- reacts with 6 mol of Fe2+) to find the moles of Fe2+ required: Moles of Fe2+ = 6 * 1.125 * 10^-3 mol = 6.75 * 10^-3 mol. Step 5: Calculate the volume of FeSO4 solution needed: Volume = Moles / Concentration = (6.75 * 10^-3 mol) / 0.150 mol dm-3 = 0.0450 dm3 = 45.0 cm3. Step 6: Explain the d-orbital splitting: In an isolated chromium(III) ion, the five 3d orbitals are degenerate. When six water ligands approach octahedrally, the electrostatic repulsion splits the 3d orbitals into two sets of different energy levels: a lower-energy triplet (t2g) and a higher-energy doublet (eg). Chromium(III) has a d3 electronic configuration, meaning three electrons occupy the lower-energy d-orbitals singly with parallel spins, minimizing repulsion.

Award 1 mark for the correct balanced reduction equation: Cr2O72- + 14H+ + 6Fe2+ -> 2Cr3+ + 6Fe3+ + 7H2O. Award 1 mark for the correct precipitation equation: Cr3+ + 3OH- -> Cr(OH)3. Award 1 mark for calculating the moles of dichromate(VI) as 1.125 * 10^-3 mol. Award 1 mark for using the 1:6 ratio to find moles of Fe2+ as 6.75 * 10^-3 mol. Award 1 mark for calculating the volume as 45.0 cm3 (allow ecf from incorrect moles). Award 1.625 marks for explaining d-orbital splitting: mentioning degeneracy loss, splitting into lower and higher energy sets in an octahedral field, and the d3 electron configuration occupying the lower three orbitals singly.

The order of basicity from most basic to least basic is: ethylamine > phenylamine > hexanamide. Basicity depends on the availability of the nitrogen lone pair to accept a proton (H+). In ethylamine, the alkyl (ethyl) group is electron-releasing due to the positive inductive effect (+I). This increases the electron density on the nitrogen atom, making the lone pair highly available to accept a proton. In phenylamine, the lone pair of electrons on the nitrogen atom is adjacent to the delocalized pi-system of the benzene ring and becomes partially delocalized into the ring. This significantly reduces the electron density on the nitrogen atom, making the lone pair much less available to accept a proton. In hexanamide, the nitrogen atom is bonded directly to a highly electronegative carbonyl group (C=O). The strongly electron-withdrawing nature of the carbonyl group (via resonance) delocalizes the nitrogen lone pair extensively into the C-O pi-system. This makes the lone pair extremely unavailable, rendering hexanamide virtually neutral/non-basic.

Award 1.625 marks for stating the correct order of basicity: ethylamine > phenylamine > hexanamide. Award 2 marks for explaining ethylamine: mentioning the positive inductive effect of the ethyl group, which increases electron density on the nitrogen and makes the lone pair more available. Award 1.5 marks for explaining phenylamine: describing how the nitrogen lone pair delocalizes into the benzene ring's pi-system, reducing its availability. Award 1.5 marks for explaining hexanamide: describing the electron-withdrawing effect of the carbonyl group, which strongly delocalizes the nitrogen lone pair and makes it non-basic.

The halogenoalkane A is 2-bromo-2-methylpropane. SN1 Mechanism: Step 1 (slow, rate-determining): Heterolytic fission of the C-Br bond occurs as the bromide ion leaves, forming a tertiary carbocation intermediate: (CH3)3C+. This is represented by a curly arrow from the C-Br bond to the Br atom. Step 2 (fast): The hydroxide nucleophile attacks the positively charged carbon of the carbocation. This is represented by a curly arrow from a lone pair on the oxygen of OH- to the positive carbon, yielding 2-methylpropan-2-ol. Why SN1 is favored: Tertiary carbocations are highly stabilized by the positive inductive (+I) effect of three electron-releasing methyl groups, which spread and minimize the positive charge density on the carbon. Additionally, steric hindrance around the tertiary carbon prevents the nucleophile from attacking from the backside, making the transition state of an SN2 mechanism energetically unfavorable. NMR identification: 2-bromo-2-methylpropane is highly symmetrical and contains 9 chemically equivalent protons, which will produce a single singlet peak in its 1H NMR spectrum. In contrast, 1-bromobutane has 4 distinct proton environments and would produce 4 separate peaks with complex spin-spin splitting patterns (e.g., triplet, sextet, quintet, triplet) and different integration values.

Award 2 marks for the correct curly arrow mechanism of SN1: showing heterolytic fission of C-Br with correct arrows, and the nucleophilic attack of OH- on the carbocation. Award 1 mark for drawing/describing the correct structure of the (CH3)3C+ carbocation. Award 1.625 marks for explaining that tertiary halogenoalkanes favor SN1 due to inductive stabilization of the carbocation by three methyl groups, and steric hindrance blocking the SN2 pathway. Award 2 marks for the NMR comparison: identifying that 2-bromo-2-methylpropane yields a single singlet peak (9H) due to equivalent environments, while 1-bromobutane yields four distinct splitting patterns due to four different environments.

Carbon dioxide (CO2) has a simple molecular structure with strong covalent double bonds between carbon and oxygen within each molecule. However, the intermolecular forces between non-polar CO2 molecules are weak London dispersion forces, which require very little thermal energy to overcome, explaining why CO2 is a gas at room temperature. Silicon dioxide (SiO2) has a giant covalent (macromolecular) structure. Each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement, and each oxygen atom is bonded to two silicon atoms. Melting SiO2 requires breaking many strong, localized covalent bonds throughout the entire giant lattice, which requires an immense amount of thermal energy, resulting in a very high melting point and a hard structure. Sulfur dioxide (SO2) has a simple molecular structure. The central sulfur atom has 6 valence electrons. It forms two double bonds (or one double and one coordinate bond) with two oxygen atoms and retains one lone pair of electrons (3 electron domains in total). According to VSEPR theory, these three domains of electron density repel each other to minimize repulsion, adopting a trigonal planar electron-pair geometry. The presence of the lone pair, which repels bonding pairs more strongly than bonding pairs repel each other, compresses the O-S-O angle to approximately 119 degrees, resulting in a bent (V-shaped) molecular geometry.

Award 1.5 marks for describing CO2: simple molecular structure with weak London dispersion forces between molecules requiring minimal energy to overcome. Award 2 marks for describing SiO2: giant covalent macromolecular structure with strong covalent bonds in a tetrahedral network that require significant energy to break. Award 2 marks for explaining SO2: describing 3 electron domains around the sulfur (two bonding pairs/regions, one lone pair), VSEPR theory state of minimizing repulsion, and lone pair-bonding pair repulsion compressing the bond angle to approximately 119 degrees (bent shape). Award 1.125 marks for coherent comparison of the physical states based on intermolecular vs covalent intramolecular network bonding.

Step 1: Write the balanced chemical equation: Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). Step 2: Calculate the moles of HCl used in the titration: Moles of HCl = Volume * Concentration = 0.02450 dm3 * 0.100 mol dm-3 = 2.450 * 10^-3 mol. Step 3: Use the stoichiometric ratio from the balanced equation (1 mol Na2CO3 reacts with 2 mol HCl) to find the moles of Na2CO3 in the 25.0 cm3 portion: Moles of Na2CO3 in 25.0 cm3 = (2.450 * 10^-3) / 2 = 1.225 * 10^-3 mol. Step 4: Calculate the moles of Na2CO3 in the total 250 cm3 solution: Moles of Na2CO3 in 250 cm3 = 1.225 * 10^-3 mol * 10 = 1.225 * 10^-2 mol. Step 5: Calculate the molar mass of the hydrated sodium carbonate Na2CO3.xH2O: Molar mass (M) = mass / moles = 3.50 g / 1.225 * 10^-2 mol = 285.71 g mol-1. Step 6: Find the mass contribution of xH2O: Molar mass of anhydrous Na2CO3 = (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g mol-1. Mass of xH2O = 285.71 - 105.99 = 179.72 g mol-1. Step 7: Calculate x: x = 179.72 / 18.02 = 9.97, which rounds to the nearest whole integer, 10. Thus, the formula is Na2CO3.10H2O. Step 8: Calculate the percentage uncertainty of the pipette: Percentage uncertainty = (Tolerance / Measured Volume) * 100 = (0.06 / 25.0) * 100 = 0.24%.

Award 1 mark for writing the correct balanced equation: Na2CO3 + 2HCl -> 2NaCl + H2O + CO2. Award 1 mark for calculating the moles of HCl (2.450 * 10^-3 mol) and the moles of Na2CO3 in 25.0 cm3 (1.225 * 10^-3 mol). Award 1 mark for scaling to 250 cm3 (1.225 * 10^-2 mol). Award 1 mark for determining the molar mass of the hydrated compound as approximately 285.7 g mol-1. Award 1 mark for correctly solving for x = 10 (accept values between 9.9 and 10.1). Award 1.625 marks for calculating the correct percentage uncertainty of the pipette: (0.06 / 25.0) * 100 = 0.24%.

The overall uncatalyzed reaction is: S2O82-(aq) + 2I-(aq) -> 2SO42-(aq) + I2(aq). This reaction is thermodynamically favorable but extremely slow because both reactant ions (S2O82- and I-) are negatively charged. The electrostatic repulsion between like-charged ions creates a high activation energy barrier. When iron(II) ions are added, they provide an alternative reaction pathway with lower activation energy. The catalysis occurs in two steps. Step 1: The negatively charged peroxodisulfate ions oxidize the positive iron(II) ions to iron(III) ions: S2O82-(aq) + 2Fe2+(aq) -> 2SO42-(aq) + 2Fe3+(aq). This step involves oppositely charged ions attracting each other, lowering the activation energy. Step 2: The newly formed iron(III) ions then oxidize the iodide ions, regenerating the iron(II) catalyst: 2Fe3+(aq) + 2I-(aq) -> 2Fe2+(aq) + I2(aq). This step also involves oppositely charged ions, ensuring a low activation energy. Transition metals such as iron are uniquely suited for this role because they have variable oxidation states (specifically +2 and +3) with close energy levels, allowing them to readily donate and accept electrons during chemical cycles.

Award 1.5 marks for explaining the high activation energy of the uncatalyzed reaction due to the electrostatic repulsion between two negatively charged reactant ions (S2O82- and I-). Award 1.5 marks for writing the first step of the catalyzed pathway: S2O82- + 2Fe2+ -> 2SO42- + 2Fe3+. Award 1.5 marks for writing the second step of the catalyzed pathway: 2Fe3+ + 2I- -> 2Fe2+ + I2. Award 2.125 marks for explaining the role of transition metals: mentioning that variable oxidation states (+2 and +3 for iron) allow them to act as effective electron-transfer intermediaries, and explaining how attracting opposite charges lowers the activation energy of both steps.

A zwitterion is an internally neutral molecule that contains both a positive and a negative charge (dipolar ion) with a net charge of zero. Lysine contains two basic amine (-NH2) groups (one alpha-amino and one side-chain epsilon-amino) and one acidic carboxylic acid (-COOH) group. At low pH (pH 1), the high concentration of H+ protons causes all basic groups to be protonated, while the acidic group remains protonated. The resulting structure is H3N+-CH2-CH2-CH2-CH2-CH(NH3+)-COOH, giving a net charge of +2. At high pH (pH 12), the low concentration of H+ protons causes all groups to lose protons. The amine groups remain neutral (-NH2) and the carboxylic acid group is deprotonated, yielding H2N-CH2-CH2-CH2-CH2-CH(NH2)-COO-, giving a net charge of -1. At its isoelectric point (pH 9.7), lysine exists primarily as a species where the carboxylic acid group is deprotonated (-COO-) and one of the amino groups is protonated (-NH3+), resulting in a net overall charge of zero. To separate the mixture at pH 6.0 using paper electrophoresis: 1) Alanine (isoelectric point 6.0) has a net charge of zero at pH 6.0, so it remains stationary at the starting line. 2) Aspartic acid (isoelectric point 3.0) is in an environment above its isoelectric point, meaning it loses H+ ions to become negatively charged (anion). It will migrate toward the positive anode (+). 3) Lysine (isoelectric point 9.7) is in an environment below its isoelectric point, meaning it gains H+ ions to become positively charged (cation). It will migrate toward the negative cathode (-). This difference in migration direction separates the three amino acids clearly.

Award 1 mark for the correct definition of a zwitterion: a dipolar ion with a net charge of zero. Award 1 mark for describing the structure of lysine at pH 1: both amine groups protonated to -NH3+, carboxyl group as -COOH (net charge +2). Award 1 mark for describing the structure of lysine at pH 12: both amine groups as -NH2, carboxyl group deprotonated to -COO- (net charge -1). Award 1 mark for describing the isoelectric point behavior (net charge of zero). Award 2.625 marks for the electrophoresis explanation: stating that at pH 6.0, alanine does not move because it has a net zero charge; aspartic acid is negatively charged and moves to the anode; lysine is positively charged and moves to the cathode.

The balanced chemical equation for the substitution reaction is: [Cu(H2O)6]2+(aq) + 3en(aq) -> [Cu(en)3]2+(aq) + 6H2O(l). In both [Cu(H2O)6]2+ and [Cu(en)3]2+, the coordination number of the copper(II) ion is 6. This is because water is a monodentate ligand (donating one lone pair) and 1,2-diaminoethane is a bidentate ligand (donating two lone pairs from its two nitrogen atoms). The high stability of the product complex, known as the chelate effect, is explained by thermodynamic principles. Enthalpy change (dH): During the reaction, six Cu-O coordinate bonds are broken, and six Cu-N coordinate bonds are formed. Because the bond strengths are very similar, the enthalpy change of the reaction (dH) is close to zero (or slightly exothermic). Entropy change (dS): The reaction starts with 4 reactant particles in solution (one complex ion and three 1,2-diaminoethane molecules) and produces 7 product particles (one complex ion and six water molecules). This representing a large increase in the disorder of the system, so the system entropy change (dS) is highly positive. Gibbs Free Energy (dG): According to the equation dG = dH - TdS, because dH is small and dS is highly positive, dG is highly negative. This makes the reaction extremely feasible and thermodynamically driven by entropy, resulting in a highly stable chelated product complex.

Award 1.5 marks for the correct balanced equation: [Cu(H2O)6]2+ + 3en -> [Cu(en)3]2+ + 6H2O. Award 1 mark for correctly identifying the coordination number of both complexes as 6. Award 1.5 marks for describing the enthalpy change: stating that breaking six Cu-O bonds and forming six Cu-N bonds results in an enthalpy change close to zero. Award 1.5 marks for describing the entropy change: explaining that the increase from 4 reactant particles to 7 product particles leads to a highly positive system entropy change. Award 1.125 marks for linking these factors to the Gibbs free energy equation (dG = dH - TdS), showing that a highly negative dG drives the reaction forward and accounts for the high stability of the chelate complex.