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First, construct an ICE table (Initial, Change, Equilibrium) in terms of concentrations (since volume is 1.00 dm\(^3\), moles equal concentration):

- \( [\text{A}]_{\text{initial}} = 2.00 \text{ mol dm}^{-3} \)
- \( [\text{B}]_{\text{initial}} = 1.00 \text{ mol dm}^{-3} \)
- \( [\text{C}]_{\text{initial}} = 0.00 \text{ mol dm}^{-3} \)

At equilibrium, \( [\text{C}] = 0.80 \text{ mol dm}^{-3} \). Since \( 2 \text{ moles of C} \) are produced from \( 2 \text{ moles of A} \) and \( 1 \text{ mole of B} \):
- Change in \( [\text{C}] = +0.80 \text{ mol dm}^{-3} \)
- Change in \( [\text{A}] = -0.80 \text{ mol dm}^{-3} \implies [\text{A}]_{\text{eq}} = 2.00 - 0.80 = 1.20 \text{ mol dm}^{-3} \)
- Change in \( [\text{B}] = -0.40 \text{ mol dm}^{-3} \implies [\text{B}]_{\text{eq}} = 1.00 - 0.40 = 0.60 \text{ mol dm}^{-3} \)

Now, calculate \( K_c \):
\( K_c = \frac{[\text{C}]^2}{[\text{A}]^2[\text{B}]} = \frac{(0.80)^2}{(1.20)^2 \times 0.60} = \frac{0.64}{1.44 \times 0.60} = \frac{0.64}{0.864} \approx 0.74 \text{ dm}^3 \text{ mol}^{-1} \).

1 mark for the correct calculation of equilibrium concentrations and substituting into the correct Kc expression to obtain 0.74.

The electronic configuration of a gaseous cobalt atom is \( [\text{Ar}] 3\text{d}^7 4\text{s}^2 \). When it forms the \( \text{Co}^{2+} \) ion, the two 4s electrons are lost first, giving the configuration \( [\text{Ar}] 3\text{d}^7 \).
According to Hund's rule, the 7 electrons occupy the five d-orbitals as follows: two orbitals contain paired electrons, and three orbitals contain single, unpaired electrons. Thus, \( \text{Co}^{2+} \) has 3 unpaired d-electrons.

Let us check the other options:
- \( \text{Fe}^{2+} \) has configuration \( [\text{Ar}] 3\text{d}^6 \), which has 4 unpaired d-electrons.
- \( \text{Mn}^{2+} \) has configuration \( [\text{Ar}] 3\text{d}^5 \), which has 5 unpaired d-electrons.
- \( \text{Cr}^{3+} \) has configuration \( [\text{Ar}] 3\text{d}^3 \), which has 3 unpaired d-electrons.
- \( \text{Ni}^{2+} \) has configuration \( [\text{Ar}] 3\text{d}^8 \), which has 2 unpaired d-electrons.

Therefore, \( \text{Cr}^{3+} \) has the same number of unpaired d-electrons as \( \text{Co}^{2+} \).

1 mark for correctly identifying that both Co2+ and Cr3+ contain 3 unpaired d-electrons.

First, locate the possible structural isomers formed by substituting one hydrogen atom on 2-methylbutane, \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3 \):

1. Substitution on C1 (or the methyl on C2): produces 1-chloro-2-methylbutane, \( \text{CH}_2\text{ClCH}(\text{CH}_3)\text{CH}_2\text{CH}_3 \). C2 is now a chiral center (bonded to \( -\text{H} \), \( -\text{CH}_2\text{Cl} \), \( -\text{CH}_3 \), and \( -\text{CH}_2\text{CH}_3 \)). This exists as a pair of optical isomers (2 isomers).
2. Substitution on C2: produces 2-chloro-2-methylbutane, \( \text{CH}_3\text{CCl}(\text{CH}_3)\text{CH}_2\text{CH}_3 \). C2 is not chiral because it has two identical methyl groups attached. (1 isomer).
3. Substitution on C3: produces 2-chloro-3-methylbutane, \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CHClCH}_3 \). C3 is a chiral center (bonded to \( -\text{H} \), \( -\text{Cl} \), \( -\text{CH}_3 \), and \( -\text{CH}(\text{CH}_3)_2 \)). This exists as a pair of optical isomers (2 isomers).
4. Substitution on C4: produces 1-chloro-3-methylbutane, \( \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{Cl} \). No chiral centers exist here. (1 isomer).

Total number of monochloroalkane isomers = 2 + 1 + 2 + 1 = 6.

1 mark for identifying all four structural monochloroalkane products and correctly accounting for optical isomerism at the chiral carbons to arrive at a total of 6.

1. Shaking with aqueous NaOH removes carbon dioxide gas. The decrease in volume is \( 60.0 - 20.0 = 40.0 \text{ cm}^3 \), which is the volume of \( \text{CO}_2 \) produced.
2. The remaining gas (20.0 cm\(^3\)) is unreacted excess oxygen.
3. The volume of oxygen that reacted is \( 80.0 - 20.0 = 60.0 \text{ cm}^3 \).
4. Using Avogadro's hypothesis, the mole ratio of \( \text{C}_x\text{H}_y : \text{CO}_2 \) is \( 10.0 : 40.0 = 1 : 4 \), so \( x = 4 \).
5. The combustion equation is \( \text{C}_4\text{H}_y + (4 + y/4)\text{O}_2 \rightarrow 4\text{CO}_2 + y/2 \text{H}_2\text{O} \).
6. The volume ratio of hydrocarbon to reacted \( \text{O}_2 \) is \( 10.0 : 60.0 = 1 : 6 \).
7. Therefore, \( 4 + y/4 = 6 \implies y/4 = 2 \implies y = 8 \).
8. The hydrocarbon is \( \text{C}_4\text{H}_8 \).

1 mark for calculating the volume of carbon dioxide produced and volume of oxygen reacted, and using gas stoichiometry to determine the molecular formula of C4H8.

In highly basic conditions (such as pH 12), the amino acid exists in its anionic form. The carboxylic acid group (\( -\text{COOH} \)) loses a proton to become \( -\text{COO}^- \), and the amino group remains in its neutral deprotonated form (\( -\text{NH}_2 \)). Thus, the predominant species is \( \text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^- \).

1 mark for identifying the correct form of the carboxylic acid and amine functional groups of alanine at high pH.

According to Le Chatelier's principle, for an exothermic forward reaction, decreasing the temperature shifts the position of equilibrium to the right, which increases the yield of \( \text{SO}_3 \). Since the equilibrium constant \( K_p \) is only dependent on temperature, and the shift is to the right (products side), the value of \( K_p \) also increases.

Changing the volume of the container changes the yield but has no effect on \( K_p \). Adding a catalyst speeds up the rate of both forward and reverse reactions equally, having no effect on either yield or \( K_p \).

1 mark for identifying that only a decrease in temperature will increase both the equilibrium constant (Kp) and the yield for an exothermic reaction.

Optical isomerism in octahedral complexes occurs when the complex lacks a plane of symmetry, resulting in non-superimposable mirror images. This typically happens in octahedral complexes with bidentate ligands, such as ethylenediamine ('en').

- cis-\( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \) lacks a plane of symmetry and has non-superimposable mirror images, so it exists as a pair of optical isomers.
- trans-\( [\text{Co}(\text{en})_2\text{Cl}_2]^+ \) is symmetrical and has a plane of symmetry, so it does not show optical isomerism.
- \( [\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} \) and trans-\( [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ \) are highly symmetrical and do not exhibit optical isomerism.

1 mark for recognizing that the cis-isomer of an octahedral complex containing two bidentate ligands lacks a plane of symmetry and exhibits optical isomerism.

The cracking reaction can be represented by the balanced chemical equation:
\( \text{C}_8\text{H}_{18} \rightarrow \text{C}_3\text{H}_6 \text{ (propene)} + \text{C}_2\text{H}_4 \text{ (ethene)} + \text{Y} \)

To find the formula of \( Y \), subtract the carbon and hydrogen atoms of the alkenes from the starting alkane:
- Number of Carbon atoms in \( Y = 8 - 3 - 2 = 3 \)
- Number of Hydrogen atoms in \( Y = 18 - 6 - 4 = 8 \)

So the molecular formula of alkane \( Y \) is \( \text{C}_3\text{H}_8 \), which is propane.

1 mark for applying conservation of mass (atom balance) to deduce the molecular formula of the missing alkane as propane.

To find the equilibrium constant, we must first determine the equilibrium concentrations of all species.

1. Find the change in moles of each species:
- Initial moles: \(X = 2.0\text{ mol}\), \(Y = 1.5\text{ mol}\), \(Z = 0\text{ mol}\).
- At equilibrium, there are \(0.80\text{ mol}\) of \(Z\), so the change in moles of \(Z\) is \(+0.80\text{ mol}\).
- Using the stoichiometry of the reaction, \(2X + Y \rightleftharpoons 2Z\):
- Moles of \(X\) reacted: \(0.80\text{ mol}\)
- Moles of \(Y\) reacted: \(0.40\text{ mol}\)

2. Calculate the equilibrium moles:
- \(X = 2.0 - 0.80 = 1.2\text{ mol}\)
- \(Y = 1.5 - 0.40 = 1.1\text{ mol}\)
- \(Z = 0.80\text{ mol}\)

3. Calculate the equilibrium concentrations (Volume \(V = 2.0\text{ dm}^3\)):
- \([X] = \frac{1.2\text{ mol}}{2.0\text{ dm}^3} = 0.60\text{ mol dm}^{-3}\)
- \([Y] = \frac{1.1\text{ mol}}{2.0\text{ dm}^3} = 0.55\text{ mol dm}^{-3}\)
- \([Z] = \frac{0.80\text{ mol}}{2.0\text{ dm}^3} = 0.40\text{ mol dm}^{-3}\)

4. Calculate \(K_c\):
\(K_c = \frac{[Z]^2}{[X]^2 [Y]} = \frac{(0.40)^2}{(0.60)^2 \times 0.55} = \frac{0.16}{0.36 \times 0.55} = \frac{0.16}{0.198} \approx 0.81\text{ dm}^3\text{ mol}^{-1}\).

1 mark for the correct option A.
- Award 1 mark for calculating the equilibrium concentrations as [X] = 0.60, [Y] = 0.55, [Z] = 0.40 mol dm^-3, substituting correctly into the Kc expression, and obtaining 0.81.

A: When an excess of concentrated hydrochloric acid is added to pink octahedral aqueous cobalt(II) ions, \([Co(H_2O)_6]^{2+}\) (coordination number 6), the chloride ligands replace the water ligands to form blue tetrahedral \([CoCl_4]^{2-}\) (coordination number 4).
B: Aqueous copper(II) ions, \([Cu(H_2O)_6]^{2+}\) (coordination number 6), react with excess ammonia to form the octahedral complex \([Cu(NH_3)_4(H_2O)_2]^{2+}\), which retains a coordination number of 6.
C: Aqueous iron(III) ions undergo a precipitation reaction with sodium hydroxide to form a neutral, insoluble hydroxide precipitate, not a soluble 4-coordinate complex ion.
D: Aqueous chromium(III) ions undergo reaction with excess sodium hydroxide to form a 6-coordinate hexahydroxochromate(III) complex, \([Cr(OH)_6]^{3-}\), retaining a coordination number of 6.

1 mark for option A. Successfully identifying that the reaction of cobalt(II) with excess concentrated hydrochloric acid leads to a decrease in coordination number from 6 to 4.

Propagation steps must involve a radical reactant and produce a radical product.
- Option A: \(C_2H_5\cdot + Cl_2 \rightarrow C_2H_5Cl + Cl\cdot\) is a propagation step, but it leads to the formation of the monochlorinated alkane, monochloroethane.
- Option B: \(C_2H_5Cl + Cl\cdot \rightarrow C_2H_4Cl\cdot + HCl\) is a propagation step where a chlorine radical abstracts a hydrogen atom from monochloroethane, producing a chloroethyl radical. In a subsequent propagation step, this radical reacts with molecular chlorine to form dichloroethane. Thus, this step directly initiates the pathway to a dichlorinated alkane.
- Option C: \(C_2H_4Cl\cdot + C_2H_5\cdot \rightarrow C_4H_9Cl\) involves two radicals reacting to form a single neutral product, which represents a termination step.
- Option D: This reaction is not a pathway involved in the free-radical chlorination of ethane.

1 mark for option B. Identifying the correct propagation step that leads to secondary substitution.

1. Liquid water has negligible volume at room temperature, so the remaining gases are unreacted \(O_2\) and produced \(CO_2\).
2. Shaking with \(NaOH(aq)\) absorbs \(CO_2\). The volume decrease is:
\(\text{Volume of } CO_2 = 85.0 - 55.0 = 30.0\text{ cm}^3\).
3. Since \(10.0\text{ cm}^3\) of \(C_xH_y\) produced \(30.0\text{ cm}^3\) of \(CO_2\):
\(x = \frac{30.0}{10.0} = 3\).
4. The gas left after \(NaOH\) treatment is unreacted \(O_2 = 55.0\text{ cm}^3\).
5. Since \(100.0\text{ cm}^3\) of \(O_2\) was added originally, the volume of reacted \(O_2\) is:
\(100.0 - 55.0 = 45.0\text{ cm}^3\).
6. The combustion equation is:
\(C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O\)
Therefore:
\(x + y/4 = \frac{45.0}{10.0} = 4.5\)
7. Substitute \(x = 3\):
\(3 + y/4 = 4.5 \Rightarrow y/4 = 1.5 \Rightarrow y = 6\).
8. The hydrocarbon is \(C_3H_6\).

1 mark for option B. Method: Determine x = 3 from the carbon dioxide volume change, and y = 6 from the oxygen gas volume reacted.

At a highly acidic pH of 1, there is a very high concentration of \(H^+\) ions in solution.
- The basic amino group (\(-NH_2\)) gains a proton to become protonated as \(-NH_3^+\).
- The acidic carboxylate group (\(-COO^-\)) gains a proton to become \(-COOH\).
- Therefore, the predominant species in solution at \(\text{pH } 1\) is the cationic form, \(CH_3CH(NH_3^+)COOH\).

1 mark for option A. Identifying that both functional groups are protonated under highly acidic conditions.

- Position of equilibrium: Since the forward reaction is exothermic, increasing the temperature shifts the equilibrium in the endothermic direction (to the left) according to Le Chatelier's principle.
- Rate of forward reaction: An increase in temperature increases the average kinetic energy of the reactant molecules. This leads to more frequent collisions and a larger fraction of molecules having energy greater than the activation energy. Thus, the rate of the forward reaction increases.
- Equilibrium constant (\(K_p\)): Since the equilibrium position shifts to the left, the ratio of product partial pressures to reactant partial pressures decreases, which results in a decrease in the value of \(K_p\).

1 mark for option A. Correctly identifying that the equilibrium position shifts left, the forward rate increases, and Kp decreases when temperature is raised in an exothermic reaction.

In transition metal complexes, the presence of ligands causes the five d-orbitals of the transition metal ion to split into two sets of orbitals with different energy levels. When visible light falls on the complex, an electron in a lower energy d-orbital absorbs a photon of specific frequency and is promoted to a higher energy d-orbital (d-to-d transition). The complementary colour to the absorbed light is transmitted or reflected, which is the colour observed.

1 mark for option A. Correctly describing d-orbital splitting, absorption of light for promotion, and the observation of the complementary transmitted colour.

Let the formula of the unknown alkene \(Y\) be \(C_nH_{2n}\) and the unknown alkane \(Z\) be \(C_mH_{2m+2}\).
The balanced chemical equation is:
\(C_{14}H_{30} \rightarrow 2C_3H_6 + C_nH_{2n} + C_mH_{2m+2}\)

Equating carbon atoms:
\(14 = 2(3) + n + m \Rightarrow 14 = 6 + n + m \Rightarrow n + m = 8\)

Equating hydrogen atoms:
\(30 = 2(6) + 2n + 2m + 2 \Rightarrow 30 = 12 + 2(n + m) + 2 \Rightarrow 30 = 14 + 2(n + m) \Rightarrow n + m = 8\)

Therefore, the sum of carbon atoms in alkene \(Y\) and alkane \(Z\) must be exactly 8.
- Option A: ethene (2) + pentane (5) = 7 carbons. (Incorrect)
- Option B: but-1-ene (4) + butane (4) = 8 carbons. (Correct)
- Option C: propene (3) + hexane (6) = 9 carbons. (Incorrect)
- Option D: pent-1-ene (5) + hexane (6) = 11 carbons. (Incorrect)

1 mark for option B. Setting up a balanced equation to determine the total number of carbon atoms in the products Y and Z, then verifying which option fits this relationship.

The partial pressure of each gas is calculated by multiplying its mole fraction by the total pressure: \(p(N_2) = 0.20 \times 5.0 \times 10^5 = 1.0 \times 10^5\text{ Pa}\), \(p(H_2) = 0.60 \times 5.0 \times 10^5 = 3.0 \times 10^5\text{ Pa}\), and \(p(NH_3) = 0.20 \times 5.0 \times 10^5 = 1.0 \times 10^5\text{ Pa}\). The expression for \(K_p\) is: \(K_p = \frac{p(NH_3)^2}{p(N_2) \times p(H_2)^3}\). Substituting the values: \(K_p = \frac{(1.0 \times 10^5)^2}{(1.0 \times 10^5) \times (3.0 \times 10^5)^3} = \frac{1.0 \times 10^{10}}{2.7 \times 10^{21}} = 3.7 \times 10^{-12}\text{ Pa}^{-2}\).

Award 1 mark for selecting A. Award 0 marks for any other option.

An increase in temperature increases the rate of both the forward and the reverse reactions because a larger fraction of molecules have kinetic energy greater than the activation energy for both directions. Since the forward reaction is exothermic, Le Chatelier's principle states that an increase in temperature shifts the equilibrium position in the endothermic direction (to the left). This decreases the concentration of products relative to reactants, meaning the value of the equilibrium constant \(K_p\) decreases.

Award 1 mark for selecting C. Award 0 marks for any other option.

The electronic configuration of a gaseous manganese atom (Mn) is \([Ar] 3d^5 4s^2\). When forming a transition metal ion, the 4s electrons are lost first. To form \(Mn^{3+}\), three electrons are removed: two from the 4s orbital and one from the 3d orbital, giving the configuration \([Ar] 3d^4\). For \(Fe^{3+}\), the configuration is \([Ar] 3d^5\). For \(Cr^{3+}\), it is \([Ar] 3d^3\). For \(V^{3+}\), it is \([Ar] 3d^2\).

Award 1 mark for selecting A. Award 0 marks for any other option.

The addition of ammonia leads to a ligand exchange reaction, forming \([Cu(NH_3)_4(H_2O)_2]^{2+}\). Ammonia is a stronger field ligand than water, which causes a larger splitting of the d-orbitals (greater value of \(\Delta E\)). Consequently, the complex absorbs light of a higher frequency (shorter wavelength, in the yellow-green region), making the transmitted complementary color a deeper blue.

Award 1 mark for selecting C. Award 0 marks for any other option.

The structure of 2-methylbutane is \(CH_3-CH(CH_3)-CH_2-CH_3\). There are four distinct carbon environments where a hydrogen atom can be replaced by a chlorine atom: 1) On either of the two equivalent methyl groups at position 1, giving 1-chloro-2-methylbutane. 2) On the CH carbon at position 2, giving 2-chloro-2-methylbutane. 3) On the \(CH_2\) carbon at position 3, giving 2-chloro-3-methylbutane. 4) On the terminal methyl carbon at position 4, giving 1-chloro-3-methylbutane. This yields exactly 4 structural isomers.

Award 1 mark for selecting B. Award 0 marks for any other option.

The general formula for an alkane is \(C_nH_{2n+2}\). Using the relative atomic masses (C = 12, H = 1), the molecular mass is \(12n + (2n + 2) = 14n + 2 = 72\), which gives \(14n = 70\) and \(n = 5\). Thus, the alkane is pentane, \(C_5H_{12}\). The balanced combustion equation is: \(C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O\). Therefore, 1 mole of pentane requires exactly 8 moles of \(O_2\) for complete combustion.

Award 1 mark for selecting C. Award 0 marks for any other option.

Upon cooling from \(120^\circ\text{C}\) to room temperature, water vapor condenses to liquid of negligible volume, causing a contraction of \(95 - 55 = 40\text{ cm}^3\). This corresponds to the volume of gaseous water produced. Since \(10\text{ cm}^3\) of hydrocarbon was used, each molecule of hydrocarbon must contain 8 hydrogen atoms (as \(10\text{ cm}^3 \times y/2 = 40\text{ cm}^3 \implies y = 8\)). Treatment with NaOH absorbs \(CO_2\), resulting in a contraction of \(55 - 25 = 30\text{ cm}^3\). This corresponds to the volume of \(CO_2\) produced. Thus, each molecule of hydrocarbon contains 3 carbon atoms (as \(10\text{ cm}^3 \times x = 30\text{ cm}^3 \implies x = 3\)). The molecular formula is \(C_3H_8\).

Award 1 mark for selecting C. Award 0 marks for any other option.

At highly alkaline pH values (pH 12), the concentration of hydroxide ions is high. Both acidic and basic functional groups will exist in their deprotonated forms. The carboxylic acid group is deprotonated to form the carboxylate ion, \(-COO^-\), and the protonated amine group (if present as an ammonium ion) loses a proton to form the neutral amine group, \(-NH_2\). This results in the anionic species \(CH_3CH(NH_2)COO^-\).

Award 1 mark for selecting C. Award 0 marks for any other option.

First, calculate the equilibrium moles of each species using stoichiometry: Initial moles: X = 2.00 mol, Y = 2.00 mol, Z = 0 mol. Since 1.00 mol of Z is formed at equilibrium, the change in Z is +1.00 mol. According to the balanced equation, the changes are: X = -1.00 mol, Y = -0.50 mol. Equilibrium moles: X = 2.00 - 1.00 = 1.00 mol, Y = 2.00 - 0.50 = 1.50 mol, Z = 1.00 mol. Next, calculate equilibrium concentrations by dividing by the volume of 2.00 dm\(^{3}\): \([\text{X}] = 1.00 / 2.00 = 0.50\text{ mol dm}^{-3}\), \([\text{Y}] = 1.50 / 2.00 = 0.75\text{ mol dm}^{-3}\), \([\text{Z}] = 1.00 / 2.00 = 0.50\text{ mol dm}^{-3}\). Now, substitute these concentrations into the equilibrium constant expression: \(K_c = \frac{[\text{Z}]^2}{[\text{X}]^2 [\text{Y}]} = \frac{(0.50)^2}{(0.50)^2 \times 0.75} = \frac{1}{0.75} = 1.33\text{ dm}^3\text{mol}^{-1}\).

Award 1 mark for the correct calculation of Kc value (1.33 dm3 mol-1). Method: Correctly find equilibrium concentrations [X]=0.50, [Y]=0.75, [Z]=0.50, and substitute into the Kc expression to obtain 1.33.

Let us determine the d-orbital electron configurations for each transition metal ion in its ground state: For A: \(\text{Cr}^{3+}\) is \(3d^3\) (3 unpaired electrons); \(\text{Fe}^{3+}\) is \(3d^5\) (5 unpaired electrons). For B: \(\text{Ni}^{2+}\) is \(3d^8\) (2 unpaired electrons in its three paired and two unpaired orbitals); \(\text{Ti}^{2+}\) is \(3d^2\) (2 unpaired electrons). Both have exactly 2 unpaired electrons. For C: \(\text{V}^{3+}\) is \(3d^2\) (2 unpaired electrons); \(\text{Mn}^{2+}\) is \(3d^5\) (5 unpaired electrons). For D: \(\text{Cu}^{2+}\) is \(3d^9\) (1 unpaired electron); \(\text{Zn}^{2+}\) is \(3d^{10}\) (0 unpaired electrons).

Award 1 mark for identifying the correct pair Ni2+ and Ti2+ (both having exactly 2 unpaired d-electrons).

The target reaction is: \([\text{CuCl}_4]^{2-}(\text{aq}) + 4\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightleftharpoons [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq})\). This reaction can be written by reversing equation (1) and adding it to equation (2). Therefore, the equilibrium constant for this reaction is: \(K_c = \frac{K_{\text{stab2}}}{K_{\text{stab1}}} = \frac{1.0 \times 10^{13}}{4.0 \times 10^5} = 2.5 \times 10^7\).

Award 1 mark for the correct calculation showing that Kc is the quotient of the two stability constants (2.5 x 10^7).

The structure of 2-methylbutane is \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\). Monochlorination can substitute a hydrogen on different carbon positions: 1) Substitution at C1 (or the methyl on C2) yields 1-chloro-2-methylbutane. 2) Substitution at C2 yields 2-chloro-2-methylbutane. 3) Substitution at C3 yields 2-chloro-3-methylbutane. 4) Substitution at C4 yields 1-chloro-3-methylbutane. This results in exactly 4 distinct structural isomers.

Award 1 mark for identifying that 2-methylbutane has 4 unique carbon positions for monosubstitution, yielding 4 structural isomers.

1) Aqueous NaOH absorbs carbon dioxide. The volume reduction is equal to the volume of \(\text{CO}_2\) produced: \(V(\text{CO}_2) = 55 - 25 = 30\text{ cm}^3\). Since 10 \(\text{cm}^3\) of \(\text{C}_x\text{H}_y\) yields 30 \(\text{cm}^3\) of \(\text{CO}_2\), we find \(x = 3\). 2) The remaining 25 \(\text{cm}^3\) of gas is unreacted \(\text{O}_2\). Therefore, volume of \(\text{O}_2\) reacted = \(70 - 25 = 45\text{ cm}^3\). 3) The combustion stoichiometry for 1 mole of \(\text{C}_3\text{H}_y\) is: \(\text{C}_3\text{H}_y + (3 + y/4)\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O}\). The mole ratio of hydrocarbon to reacted \(\text{O}_2\) is 10 : 45 = 1 : 4.5. Thus, \(3 + y/4 = 4.5 \implies y/4 = 1.5 \implies y = 6\). The hydrocarbon is \(\text{C}_3\text{H}_6\).

Award 1 mark for identifying the hydrocarbon as C3H6 by showing that 30 cm3 of CO2 is produced (giving x=3) and 45 cm3 of O2 is consumed (giving y=6).

In a highly alkaline solution (pH 13), both of the acidic carboxylic acid groups (\(-\text{COOH}\)) are fully deprotonated to form carboxylate anions (\(-\text{COO}^-\)). The basic amine group (\(-\text{NH}_2\)) remains unprotonated as a neutral amine (\(-\text{NH}_2\)). Thus, the resulting major species is the dicarboxylate anion, \(\text{^{-}OOC-CH}_2\text{-CH(NH}_2)\text{-COO}^-\).

Award 1 mark for selecting option C, which correctly shows both carboxylic groups deprotonated and the amino group unprotonated at pH 13.

The value of the equilibrium constant, \(K_p\), is dependent solely on the temperature. Catalysts, pressure changes, or concentration/partial pressure changes of reactants or products do not change the value of \(K_p\). For an exothermic forward reaction (\(\Delta H < 0\)), Le Chatelier's principle dictates that decreasing the temperature will shift the equilibrium position to the right, increasing the proportion of products relative to reactants, which increases the value of \(K_p\).

Award 1 mark for identifying that only a temperature decrease will increase the value of Kp for an exothermic reaction.

1) Calculate the molar mass of \(\text{Na}_2\text{CO}_3\): \(M_r = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\). 2) Calculate the amount in moles of \(\text{Na}_2\text{CO}_3\): \(\text{moles} = 2.65 / 106.0 = 0.0250\text{ mol}\). 3) Each formula unit of \(\text{Na}_2\text{CO}_3\) contains 3 ions (two \(\text{Na}^+\) ions and one \(\text{CO}_3^{2-}\) ion). 4) Calculate the total moles of ions: \(0.0250 \times 3 = 0.0750\text{ mol}\). 5) Calculate the total number of ions: \(\text{Number of ions} = 0.0750 \times 6.02 \times 10^{23} = 4.52 \times 10^{22}\).

Award 1 mark for the correct calculation of the number of ions (4.52 x 10^22) by converting mass to moles, multiplying by 3 (number of ions per formula unit), and then multiplying by Avogadro's constant.

To find the equilibrium constant \(K_c\), first determine the equilibrium moles of each species using an ICE (Initial, Change, Equilibrium) table:

- **Initial moles:**
\(\text{A} = 2.0\text{ mol}\), \(\text{B} = 2.0\text{ mol}\), \(\text{C} = 0\text{ mol}\)

- **Change in moles:**
Since \(0.80\text{ mol}\) of \(\text{C}\) is formed, the change in \(\text{C}\) is \(+0.80\text{ mol}\).
By stoichiometry, the change in \(\text{A}\) is \(-0.80\text{ mol}\) and the change in \(\text{B}\) is \(-0.40\text{ mol}\).

- **Equilibrium moles:**
\(\text{A} = 2.0 - 0.80 = 1.20\text{ mol}\)
\(\text{B} = 2.0 - 0.40 = 1.60\text{ mol}\)
\(\text{C} = 0.80\text{ mol}\)

- **Equilibrium concentrations** (in a volume of \(10.0\text{ dm}^3\)):
\([\text{A}] = \frac{1.20\text{ mol}}{10.0\text{ dm}^3} = 0.120\text{ mol dm}^{-3}\)
\([\text{B}] = \frac{1.60\text{ mol}}{10.0\text{ dm}^3} = 0.160\text{ mol dm}^{-3}\)
\([\text{C}] = \frac{0.80\text{ mol}}{10.0\text{ dm}^3} = 0.080\text{ mol dm}^{-3}\)

- **Equilibrium expression \(K_c\):**
\(K_c = \frac{[\text{C}]^2}{[\text{A}]^2[\text{B}]} = \frac{(0.080)^2}{(0.120)^2 \times 0.160} = \frac{0.0064}{0.0144 \times 0.160} = 2.78\text{ dm}^3\text{ mol}^{-1}\)

Award 1 mark for the correct calculation of equilibrium concentrations and substituting into the \(K_c\) expression to obtain \(2.78\text{ dm}^3\text{ mol}^{-1}\). Reject values that do not account for the volume or stoichiometric ratios.

According to Le Chatelier's principle:
1. **Temperature increase:** The forward reaction is exothermic (\(\Delta H = -88\text{ kJ mol}^{-1}\)). Increasing the temperature will shift the position of equilibrium in the endothermic direction (to the left) to absorb the added heat, resulting in a **decrease** in the yield of \(\text{PCl}_5\).
2. **Pressure increase:** There are 2 moles of gas on the reactant side and only 1 mole of gas on the product side. Increasing the total pressure shifts the position of equilibrium to the side with fewer gas molecules (to the right) to reduce the pressure, resulting in an **increase** in the yield of \(\text{PCl}_5\).

Award 1 mark for identifying that increasing temperature decreases the yield and increasing pressure increases the yield.

Let's find the number of unpaired d-electrons for each transition metal ion:
- \(\text{Fe}^{3+}\): Iron's ground-state electronic configuration is \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). Removing three electrons gives \([\text{Ar}] 3\text{d}^5\). By Hund's rule, all 5 d-orbitals are singly occupied, giving **5** unpaired electrons.
- \(\text{Co}^{2+}\): Cobalt is \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). Removing two electrons gives \([\text{Ar}] 3\text{d}^7\). There are 2 fully paired d-orbitals and **3** unpaired electrons.
- \(\text{Ni}^{2+}\): Nickel is \([\text{Ar}] 3\text{d}^8 4\text{s}^2\). Removing two electrons gives \([\text{Ar}] 3\text{d}^8\). There are 3 fully paired d-orbitals and **2** unpaired electrons.
- \(\text{Cu}^{2+}\): Copper is \([\text{Ar}] 3\text{d}^{10} 4\text{s}^1\). Removing two electrons gives \([\text{Ar}] 3\text{d}^9\). There are 4 fully paired d-orbitals and **1** unpaired electron.

Award 1 mark for the correct selection of \(\text{Fe}^{3+}\) (option A) based on its \(3\text{d}^5\) electronic configuration.

Transition metal complexes are coloured because the presence of ligands causes the degenerate d-orbitals of the transition metal ion to split into two groups of non-degenerate energy levels. When visible light falls on the complex, an electron is promoted from a lower d-orbital to a higher d-orbital by absorbing light of a specific frequency corresponding to the energy gap (\(\Delta E = h\nu\)). The colour we observe is the complementary colour of the absorbed frequency.

Award 1 mark for identifying the splitting of d-orbitals by ligands and the absorption of specific light frequencies during electron promotion as the source of colour.

Let's draw and list all structural isomers of dichlorobutane (based on a straight-chain carbon skeleton):
1. **1,1-dichlorobutane:** \(\text{CHCl}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3\)
2. **1,2-dichlorobutane:** \(\text{CH}_2\text{Cl}-\text{CHCl}-\text{CH}_2-\text{CH}_3\)
3. **1,3-dichlorobutane:** \(\text{CH}_2\text{Cl}-\text{CH}_2-\text{CHCl}-\text{CH}_3\)
4. **1,4-dichlorobutane:** \(\text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Cl}\)
5. **2,2-dichlorobutane:** \(\text{CH}_3-\text{CCl}_2-\text{CH}_2-\text{CH}_3\)
6. **2,3-dichlorobutane:** \(\text{CH}_3-\text{CHCl}-\text{CHCl}-\text{CH}_3\)

Any other naming, such as 2,4-dichlorobutane, represents a duplicate of a previously numbered isomer (1,3-dichlorobutane in this case) due to symmetry. Thus, there are exactly 6 structural isomers.

Award 1 mark for identifying that exactly 6 structural isomers of dichlorobutane can be drawn.

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

2. From the stoichiometry of the equation, 1 mole of \(\text{MCO}_3\) produces 1 mole of \(\text{CO}_2\):
\(n(\text{MCO}_3) = n(\text{CO}_2) = 0.010\text{ mol}\)

3. Calculate the molar mass (\(M_{\text{r}}\)) of \(\text{MCO}_3\):
\(M_{\text{r}}(\text{MCO}_3) = \frac{1.48\text{ g}}{0.010\text{ mol}} = 148\text{ g mol}^{-1}\)

4. Find the relative atomic mass (\(A_{\text{r}}\)) of metal \(\text{M}\):
\(A_{\text{r}}(\text{M}) = M_{\text{r}}(\text{MCO}_3) - M_{\text{r}}(\text{CO}_3^{2-})\)
\(A_{\text{r}}(\text{M}) = 148 - (12.0 + 3 \times 16.0) = 148 - 60.0 = 88.0\)

5. Comparing this to the Group 2 metals in the periodic table, strontium (Sr) has an \(A_{\text{r}}\) of 87.6, which is the closest.

Award 1 mark for the correct calculation steps yielding \(A_{\text{r}} = 88.0\) and identifying strontium as the correct Group 2 metal.

1. Find the molar mass of ethanol, \(\text{C}_2\text{H}_5\text{OH}\):
\(M_{\text{r}}(\text{C}_2\text{H}_5\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\text{ g mol}^{-1}\)

2. Calculate the number of moles of ethanol in the sample:
\(n(\text{ethanol}) = \frac{4.60\text{ g}}{46.0\text{ g mol}^{-1}} = 0.100\text{ mol}\)

3. Each molecule of ethanol contains 6 hydrogen atoms. Therefore, 0.100 mol of ethanol contains:
\(0.100 \times 6 = 0.600\text{ mol of H atoms}\)

4. Calculate the number of H atoms using Avogadro's constant (\(L = 6.02 \times 10^{23}\text{ mol}^{-1}\)):
\(N(\text{H}) = 0.600 \times 6.02 \times 10^{23} = 3.61 \times 10^{23}\)

Award 1 mark for calculating the correct number of hydrogen atoms as \(3.61 \times 10^{23}\). Reject answers that ignore the presence of 6 H atoms in each ethanol molecule.

At a highly alkaline pH of 12:
- The carboxylic acid group (\(-\text{COOH}\)) acts as an acid, losing its proton to become a negatively charged carboxylate ion: \(-\text{COO}^-\).
- The basic amine group remains unprotonated as a neutral amine group: \(-\text{NH}_2\).

Thus, the predominant species in solution at pH 12 is the anionic form of the amino acid, represented by \(\text{CH}_3\text{CH(NH}_2\text{)COO}^-\).

Award 1 mark for identifying the correct deprotonated form of both functional groups at pH 12, yielding the carboxylate anion and neutral amine group.

(a) A dynamic equilibrium is a state in which:
- The rate of the forward reaction equals the rate of the reverse reaction. [1]
- The concentrations of reactants and products remain constant in a closed system. [1]

(b)(i) Expression for \(K_c\):
\[K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}\] [1]
Units: None / no units [1]

(b)(ii) From the stoichiometric equation, 1 mole of ethanoic acid reacts with 1 mole of ethanol to produce 1 mole of ethyl ethanoate and 1 mole of water.
Given equilibrium amount of ethyl ethanoate = 0.75 mol.
- Equilibrium amount of water formed = 0.75 mol [1]
- Equilibrium amount of ethanoic acid = 1.00 - 0.75 = 0.25 mol [1]
- Equilibrium amount of ethanol = 1.50 - 0.75 = 0.75 mol [1]

(b)(iii) \(K_c = \frac{0.75 \times 0.75}{0.25 \times 0.75} = 3.00\) [2]
(Since the volume terms cancel out, the mole values can be substituted directly into the equilibrium expression).

(c)(i) The position of equilibrium shifts to the left (reactant side). [1]
This is because the reverse reaction is endothermic and absorbs the added heat energy to counteract the increase in temperature. [1]

(c)(ii) The value of \(K_c\) decreases. [1]
Since the position of equilibrium shifts to the left, the concentration of products decreases and the concentration of reactants increases, resulting in a smaller value of \(K_c\). [1]

(d) The catalyst increases the rate of both the forward and reverse reactions equally. [1]
It does this by providing an alternative reaction pathway with a lower activation energy, so the position of equilibrium is reached faster but remains unchanged. [1]

- (a) Rate of forward reaction = rate of reverse reaction [1]; concentrations of reactants and products remain constant [1].
- (b)(i) Correct expression for \(K_c\) [1]; 'No units' stated [1].
- (b)(ii) Moles of water = 0.75 mol [1]; Moles of ethanoic acid = 0.25 mol [1]; Moles of ethanol = 0.75 mol [1].
- (b)(iii) Show correct substitution of moles into the expression [1]; final answer 3.00 (or 3) [1].
- (c)(i) Shift to the left / reactants [1]; because the reverse reaction is endothermic / absorbs heat [1].
- (c)(ii) \(K_c\) decreases [1]; because concentration of products decreases relative to reactants [1].
- (d) Rate of forward and reverse reactions are increased equally [1]; because it lowers activation energy for both pathways by the same amount [1].

(a)(i) Relative formula mass is the weighted average mass of one formula unit of a compound compared to 1/12th of the mass of an atom of carbon-12. [2]

(a)(ii) Mass of \(CuSO_4\) = 3.19 g
\(M_r(CuSO_4) = 63.5 + 32.1 + (16.0 \times 4) = 159.6\)
Moles of \(CuSO_4 = \frac{3.19}{159.6} = 0.0200\) mol [1]
Mass of water lost = 4.99 - 3.19 = 1.80 g
Moles of water = \frac{1.80}{18.0} = 0.100\) mol [1]
Ratio \(x = \frac{0.100}{0.0200} = 5\)
Therefore, \(x = 5\). [1]

(b)(i) Moles of \(N_2H_4 = \frac{64.0}{32.0} = 2.00\) mol [1]
From the equation, 1 mole of \(N_2H_4\) produces 1 mole of \(N_2\).
So, moles of \(N_2 = 2.00\) mol [1]
Volume of \(N_2\) gas = \(2.00 \times 24.0 = 48.0 \text{ dm}^3\) [1]

(b)(ii) Total moles of gas produced per mole of \(N_2H_4\) is 5 mol (1 mol of \(N_2\) + 4 mol of \(H_2O\)).
So, total moles of gas from 2.00 mol of \(N_2H_4 = 2.00 \times 5 = 10.0\) mol [1]
Total volume of gas = \(10.0 \times 24.0 = 240 \text{ dm}^3\) [1]

(c)(i) Moles of \(MnO_4^- = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\) mol [2]

(c)(ii) Moles of \(Fe^{2+} = 5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3}\) mol [1]
Volume of \(Fe^{2+}\) solution = \(25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3\) [1]
Concentration of \(Fe^{2+} = \frac{2.25 \times 10^{-3}}{0.0250} = 0.0900 \text{ mol dm}^{-3}\) [1]

- (a)(i) Weighted average mass of one formula unit [1] compared to 1/12th of the mass of carbon-12 [1].
- (a)(ii) Correct moles of \(CuSO_4\) (0.0200) [1]; correct moles of water (0.100) [1]; correct ratio giving \(x = 5\) [1].
- (b)(i) Moles of hydrazine = 2.00 mol [1]; 1:1 mole ratio shown or implied [1]; Volume of \(N_2 = 48.0 \text{ dm}^3\) [1].
- (b)(ii) Total moles of gas = 10.0 mol [1]; Total volume = 240 \text{ dm}^3\) [1]. (Allow error carried forward from (b)(i)).
- (c)(i) Moles of manganate(VII) = \(4.50 \times 10^{-4}\) mol (award 1 mark for incorrect division by 1000, 2 marks for correct answer) [2].
- (c)(ii) Moles of \(Fe^{2+}\) = \(2.25 \times 10^{-3}\) mol [1]; volume conversion to \(\text{dm}^3\) [1]; final concentration = 0.0900 \(\text{mol dm}^{-3}\) [1].

(a)(i) A compound containing hydrogen and carbon atoms only. [1]

(a)(ii) Contains only single C-C bonds / no multiple bonds. [1]

(b)(i) \(C_6H_{14} \rightarrow C_4H_{10} + C_2H_4\) [1]

(b)(ii) High temperature [1] and high pressure. [1]

(c)(i) Free-radical substitution [1]

(c)(ii) Homolytic fission [1]

(c)(iii)
- Initiation step:
\(Cl_2 \xrightarrow{UV} 2Cl\bullet\) [1]
- Propagation steps:
\(C_2H_6 + Cl\bullet \rightarrow \bullet C_2H_5 + HCl\) [1]
\(\bullet C_2H_5 + Cl_2 \rightarrow C_2H_5Cl + Cl\bullet\) [1]
- Termination step producing butane:
\(2\bullet C_2H_5 \rightarrow C_4H_{10}\) [1]

(d)(i) The skeletal formula of 2-methylpropane is a central carbon bonded to three methyl groups (represented as a 'Y' shape). [1]

(d)(ii) Butane has a higher boiling point than 2-methylpropane. [1]
Butane is a straight-chain isomer and has a larger surface area / contact area than the branched 2-methylpropane. [1]
Consequently, butane has stronger instantaneous dipole-induced dipole (London dispersion) forces, which require more thermal energy to overcome. [1]

- (a)(i) Compound consisting of carbon and hydrogen only [1].
- (a)(ii) Contains only single C-C bonds [1].
- (b)(i) Correct equation: \(C_6H_{14} \rightarrow C_4H_{10} + C_2H_4\) [1].
- (b)(ii) High temperature [1]; high pressure [1].
- (c)(i) Free-radical substitution [1].
- (c)(ii) Homolytic fission [1].
- (c)(iii) Initiation equation: \(Cl_2 \rightarrow 2Cl\bullet\) [1]; First propagation equation [1]; Second propagation equation [1]; Termination equation to butane [1].
- (d)(i) Correct skeletal formula of 2-methylpropane [1].
- (d)(ii) Butane has a higher boiling point [1]; butane has a larger surface area [1]; butane has stronger London dispersion forces [1].

(a)(i) The 3d and 4s subshells in transition metals are very close in energy. [1]
Consequently, different numbers of electrons from both the 3d and 4s subshells can be lost or shared during chemical bonding without requiring a very high amount of energy. [1]

(a)(ii)
- Ti: \(3d^2 4s^2\) (or \(4s^2 3d^2\)) [1]
- \(Ti^{3+}\): \(3d^1\) [1]

(b)(i) \([CuCl_4]^{2-}\) [1]

(b)(ii)
- Geometry: octahedral to tetrahedral [1]
- Coordination number: decreases from 6 to 4 [1]

(b)(iii) \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CuCl_4]^{2-} + 6H_2O\) [1]

(c)(i) *cis*-*trans* isomerism (or geometrical isomerism) [1]

(c)(ii)
- Cis-isomer: Two Cl ligands are adjacent to each other (at a 90° angle). [1]
- Trans-isomer: Two Cl ligands are opposite to each other (at a 180° angle). [1]
(Drawings should clearly show octahedral geometry with wedge/dash or standard grid lines).

(d)(i)
- Ligands approach the transition metal ion and cause the degenerate d-orbitals to split into two sets of different energy levels. [1]
- Electrons occupy the lower energy d-orbitals. [1]
- An electron absorbs energy in the form of visible light and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition). [1]
- The complementary color of the absorbed wavelength of light is transmitted or reflected, which is the color we observe. [1]

- (a)(i) Energy levels of 3d and 4s orbitals are close [1]; similar energy allows both 3d and 4s electrons to be lost / used in bonding [1].
- (a)(ii) Ti: \(3d^2 4s^2\) [1]; \(Ti^{3+}\): \(3d^1\) [1].
- (b)(i) \([CuCl_4]^{2-}\) [1].
- (b)(ii) Geometry changes from octahedral to tetrahedral [1]; coordination number changes from 6 to 4 [1].
- (b)(iii) \([Cu(H_2O)_6]^{2+} + 4Cl^- \rightarrow [CuCl_4]^{2-} + 6H_2O\) [1].
- (c)(i) *cis*-*trans* / geometrical isomerism [1].
- (c)(ii) Diagram of trans-isomer (Cl ligands opposite) [1]; Diagram of cis-isomer (Cl ligands adjacent) [1].
- (d)(i) Split of d-orbitals into two different energy levels by ligands [1]; absorption of light / photon in the visible range [1]; promotion of d-electron from lower to higher energy level (d-d transition) [1]; complementary color is observed [1].

(a) \(K_c = \frac{[\text{NO}]^2[\text{Cl}_2]}{[\text{NOCl}]^2}\). (b) Initial moles: \(\text{NOCl} = 0.800\), \(\text{NO} = 0\), \(\text{Cl}_2 = 0\). At equilibrium, \(n(\text{Cl}_2) = 0.240\) mol. By stoichiometry: \(n(\text{NO}) = 2 \times 0.240 = 0.480\) mol, and \(n(\text{NOCl}) = 0.800 - (2 \times 0.240) = 0.320\) mol. Volume = 2.00 dm\(^3\). Equilibrium concentrations: (i) \([\text{NOCl}] = 0.320 / 2.00 = 0.160\) mol dm\(^{-3}\), (ii) \([\text{NO}] = 0.480 / 2.00 = 0.240\) mol dm\(^{-3}\), (iii) \([\text{Cl}_2] = 0.240 / 2.00 = 0.120\) mol dm\(^{-3}\). (c) \(K_c = \frac{(0.240)^2 \times 0.120}{(0.160)^2} = 0.270\) mol dm\(^{-3}\). (d) Total moles of gas at equilibrium, \(n_{\text{total}} = 0.320 + 0.480 + 0.240 = 1.040\) mol. Using \(pV = nRT\): \(p = \frac{1.040 \times 8.31 \times 500}{2.00 \times 10^{-3}} = 2.16 \times 10^6\) Pa. (e) If the temperature is increased, the equilibrium position shifts to the right (the endothermic direction) to absorb the added heat, thus increasing the concentrations of products and decreasing the concentration of reactant. Consequently, the value of \(K_c\) increases.

(a) [1 mark] Correct expression for \(K_c\). (b) [3 marks] [1 mark] for each correct concentration: \([\text{NOCl}] = 0.160\) mol dm\(^{-3}\), \([\text{NO}] = 0.240\) mol dm\(^{-3}\), \([\text{Cl}_2] = 0.120\) mol dm\(^{-3}\). (c) [3 marks] [1 mark] for correct substitution, [1 mark] for value 0.270, [1 mark] for units mol dm\(^{-3}\). (d) [2 marks] [1 mark] for calculating total moles of gas = 1.040 mol, [1 mark] for correct calculation of pressure as \(2.16 \times 10^6\) Pa. (e) [2 marks] [1 mark] for stating that \(K_c\) increases, [1 mark] for explaining that the endothermic direction is favored with temperature increase.

(a) Partial pressure is the pressure that a gas in a mixture would exert if it alone occupied the entire volume of the container at the same temperature. (b) Let initial moles of \(\text{N}_2\text{O}_4 = 1\). At equilibrium: \(n(\text{N}_2\text{O}_4) = 1 - \alpha\), \(n(\text{NO}_2) = 2\alpha\). Total moles = \(1 - \alpha + 2\alpha = 1 + \alpha\). Mole fractions: \(\chi(\text{N}_2\text{O}_4) = \frac{1-\alpha}{1+\alpha}\), \(\chi(\text{NO}_2) = \frac{2\alpha}{1+\alpha}\). Partial pressures: \(p(\text{N}_2\text{O}_4) = \frac{1-\alpha}{1+\alpha} P\), \(p(\text{NO}_2) = \frac{2\alpha}{1+\alpha} P\). \(K_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{(\frac{2\alpha}{1+\alpha} P)^2}{\frac{1-\alpha}{1+\alpha} P} = \frac{4\alpha^2 P}{1-\alpha^2}\). (c) \(P = 1.20 \times 10^5\) Pa, \(\alpha = 0.450\). Total moles factor \(1+\alpha = 1.450\). \(p(\text{N}_2\text{O}_4) = \frac{1-0.450}{1.450} \times 1.20 \times 10^5 = 4.55 \times 10^4\) Pa. \(p(\text{NO}_2) = \frac{2 \times 0.450}{1.450} \times 1.20 \times 10^5 = 7.45 \times 10^4\) Pa. (d) \(K_p = \frac{(7.45 \times 10^4)^2}{4.55 \times 10^4} = 1.22 \times 10^5\) Pa (or using the formula in (b): \(K_p = \frac{4 \times (0.450)^2 \times 1.20 \times 10^5}{1 - 0.450^2} = 1.22 \times 10^5\) Pa). (e) If total pressure is increased, the equilibrium shifts to the side with fewer gas moles (the left side) to oppose the increase in pressure. Consequently, the degree of dissociation \(\alpha\) decreases.

(a) [1 mark] Correct definition of partial pressure. (b) [3 marks] [1 mark] for expressing equilibrium moles, [1 mark] for mole fractions, [1 mark] for correct final expression \(K_p = \frac{4\alpha^2 P}{1-\alpha^2}\). (c) [3 marks] [1 mark] for calculating total moles factor 1.450, [1 mark] for \(p(\text{N}_2\text{O}_4) = 4.55 \times 10^4\) Pa, [1 mark] for \(p(\text{NO}_2) = 7.45 \times 10^4\) Pa. (d) [2 marks] [1 mark] for value \(1.22 \times 10^5\) (accept range 1.21-1.23 x 10^5), [1 mark] for unit Pa. (e) [2 marks] [1 mark] for stating \(\alpha\) decreases, [1 mark] for explanation (shifts to the side with fewer gas molecules/moles).

(a) The electronic configuration of Ni is \([\text{Ar}] 3d^8 4s^2\). Thus, the configuration of the \(\text{Ni}^{2+}\) ion is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^8\). (b)(i) The diagram must show a central \(\text{Ni}\) ion coordinated to six water molecules in an octahedral arrangement (using wedges and dashed bonds to show 3D geometry). Bond angles of 90\(^\circ\) and 180\(^\circ\) must be indicated. The coordinate bonds must point from the oxygen lone pair of the water molecules to the nickel ion. (b)(ii) Water ligands split the degenerate 3d orbitals of \(\text{Ni}^{2+}\) into two non-degenerate sub-levels with an energy gap, \(\Delta E\). Electrons in the lower energy d-orbitals absorb a photon of visible light and are promoted to the higher energy d-orbitals (d-d transition). The color green is transmitted/reflected, which is the complementary color of the light absorbed. (c)(i) \([\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{NiCl}_4]^{2-} + 6\text{H}_2\text{O}\). (c)(ii) Chloride ligands are larger than water molecules and are negatively charged. This leads to greater steric hindrance and electrostatic repulsion between ligands around the nickel central ion, meaning only 4 chloride ligands can fit around \(\text{Ni}^{2+}\) compared to 6 neutral water molecules.

(a) [1 mark] for terminating with \(3d^8\) (or \(3d^8 4s^0\)). (b)(i) [3 marks] [1 mark] for octahedral shape with wedges and dashes, [1 mark] for correct bond angles (90 and 180 degrees), [1 mark] for coordinate bonds pointing from O to Ni. (b)(ii) [3 marks] [1 mark] for d-orbital splitting due to ligands, [1 mark] for excitation/promotion of d-electrons absorbing a photon of visible light (d-d transition), [1 mark] for light not absorbed being transmitted as green (complementary color). (c)(i) [2 marks] [1 mark] for correct reactants and products, [1 mark] for correct balancing. (c)(ii) [2 marks] [1 mark] for mentioning that chloride ligands are larger/bulkier than water, [1 mark] for mentioning increased steric hindrance or charge repulsion.

(a) The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion in solution from its constituent ions or molecules (often replacing solvent/water ligands). (b)(i) \([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O}\). (b)(ii) \(K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4}\). (c)(i) \(K_{\text{stab}} = 10^{13.1} = 1.26 \times 10^{13}\) dm\(^{12}\) mol\(^{-4}\). (c)(ii) \([\text{Cu}(\text{EDTA})]^{2-}\) is much more stable because it has a larger \(K_{\text{stab}}\) value. The reaction involves the substitution of four monodentate ammonia ligands (and two water ligands) by a hexadentate EDTA ligand. This results in a larger number of free molecules on the product side (e.g., releasing 6 free monodentate molecules), which causes a significant increase in disorder (system entropy, \(\Delta S^\ominus > 0\)). Consequently, the Gibbs free energy change, \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), becomes highly negative, driving the reaction forward. (d)(i) A bidentate ligand is a species that donates two lone pairs of electrons to a central metal ion to form two coordinate bonds. (d)(ii) The drawing must show an octahedral copper center with three loops representing the bidentate 1,2-diaminoethane (en) ligand, showing 3D wedges and dashes indicating a chiral (non-superimposable) mirror-image isomer.

(a) [2 marks] [1 mark] for equilibrium constant for the formation of a complex, [1 mark] for starting from its constituent ions/molecules in solvent. (b)(i) [1 mark] Correct balanced equation. (b)(ii) [1 mark] Correct expression for \(K_{\text{stab}}\). (c)(i) [1 mark] \(1.26 \times 10^{13}\) (accept units or no units). (c)(ii) [3 marks] [1 mark] for identifying \([\text{Cu}(\text{EDTA})]^{2-}\) is more stable, [1 mark] for stating that there is an increase in the number of particles (entropy of system increases, \(\Delta S > 0\)), [1 mark] for linking positive \(\Delta S\) to a more negative \(\Delta G\) driving stability. (d)(i) [1 mark] Correct definition of bidentate. (d)(ii) [2 marks] [1 mark] for correct octahedral coordination around Cu, [1 mark] for correct 3D representation of one optical isomer using 'en' loops.

(a) \(\text{C}_4\text{H}_{10} + \text{Cl}_2 \rightarrow \text{C}_4\text{H}_9\text{Cl} + \text{HCl}\). (b)(i) Initiation: \(\text{Cl}_2 \xrightarrow{h\nu} 2\text{Cl}^\bullet\). Propagation 1: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2\dot{\text{C}}\text{HCH}_3 + \text{HCl}\). Propagation 2: \(\text{CH}_3\text{CH}_2\dot{\text{C}}\text{HCH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3 + \text{Cl}^\bullet\). (b)(ii) Termination: \(2\text{CH}_3\text{CH}_2\dot{\text{C}}\text{HCH}_3 \rightarrow \text{C}_8\text{H}_{18}\) (specifically 3,4-dimethylhexane) or simply \(2\text{C}_4\text{H}_9^\bullet \rightarrow \text{C}_8\text{H}_{18}\). (c)(i) The secondary butyl radical (\(\text{CH}_3\text{CH}_2\dot{\text{C}}\text{HCH}_3\)) is more stable than the primary butyl radical (\(\text{CH}_3\text{CH}_2\text{CH}_2\dot{\text{C}}\text{H}_2\)) because it has two electron-donating alkyl groups attached to the radical carbon instead of one. The electron-donating inductive effect of the alkyl groups stabilizes the electron-deficient radical carbon, lowering the activation energy for its formation. (c)(ii) Optical isomerism. 2-chlorobutane has a chiral carbon (carbon-2) bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{Cl}\). Draw two non-superimposable tetrahedral mirror images. (d) Use a large excess of butane relative to chlorine, which ensures that chlorine radicals are far more likely to collide with butane molecules than with monochlorobutane molecules.

(a) [1 mark] Correct overall monochlorination equation. (b)(i) [3 marks] [1 mark] for initiation step, [1 mark] for propagation step 1 (radical on C2), [1 mark] for propagation step 2. (b)(ii) [1 mark] Correct termination step yielding a C8 compound (e.g., \(\text{C}_8\text{H}_{18}\)). (c)(i) [2 marks] [1 mark] for identifying the secondary radical is more stable than the primary radical, [1 mark] for explaining stability via the electron-donating inductive effect of two alkyl groups. (c)(ii) [3 marks] [1 mark] for identifying optical isomerism, [2 marks] for drawing two clear, non-superimposable 3D mirror images of 2-chlorobutane. (d) [1 mark] State that a large excess of butane (or limiting chlorine) should be used.

(a) \(\text{C}_7\text{H}_{16}(l) + 11\text{O}_2(g) \rightarrow 7\text{CO}_2(g) + 8\text{H}_2\text{O}(l)\). (b)(i) Carbon monoxide is toxic because it binds irreversibly to hemoglobin in red blood cells, forming carboxyhemoglobin. This prevents hemoglobin from transporting oxygen to vital body organs and tissues. (b)(ii) Catalytic converters contain platinum, palladium, or rhodium catalysts. Carbon monoxide and nitrogen monoxide adsorb onto the catalytic surface, where they react to form non-toxic carbon dioxide and nitrogen gases. Equation: \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\). (c)(i) \(\text{C}_7\text{H}_{16} \rightarrow \text{C}_3\text{H}_8 + \text{C}_4\text{H}_8\) (butene). (c)(ii) Catalyst: Zeolite (or aluminosilicates). Conditions: High temperature (approx. 500\(^\circ\)C) and slight pressure. (d) Moles of heptane burned = \(15.0 / 100.2 = 0.1497\) mol. From the equation in (a), 1 mole of heptane produces 7 moles of \(\text{CO}_2\). Moles of \(\text{CO}_2\) produced = \(7 \times 0.1497 = 1.048\) mol. Volume of \(\text{CO}_2\) at r.t.p. = \(1.048 \times 24.0 = 25.15\) dm\(^3\) (approx. 25.1 dm\(^3\) or 25.2 dm\(^3\)).

(a) [1 mark] Correctly balanced equation. (b)(i) [1 mark] Irreversible binding to hemoglobin, reducing oxygen transport. (b)(ii) [3 marks] [1 mark] for catalyst name (Pt/Pd/Rh), [1 mark] for describing adsorption and reaction to form \(\text{CO}_2\) and \(\text{N}_2\), [1 mark] for correct equation: \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\). (c)(i) [1 mark] Correct equation: \(\text{C}_7\text{H}_{16} \rightarrow \text{C}_3\text{H}_8 + \text{C}_4\text{H}_8\). (c)(ii) [2 marks] [1 mark] for zeolite/aluminosilicate catalyst, [1 mark] for high temperature (approx 500\(^\circ\)C). (d) [3 marks] [1 mark] for calculating moles of heptane = 0.150 (or 0.1497) mol, [1 mark] for moles of \(\text{CO}_2\) = 1.05 (or 1.048) mol, [1 mark] for correct final volume = 25.1 or 25.2 dm\(^3\).

(a)(i) Mass of Carbon: \(2.200 \times \frac{12.0}{44.0} = 0.600\) g. Mass of Hydrogen: \(0.900 \times \frac{2.0}{18.0} = 0.100\) g. (a)(ii) Mass of Oxygen in sample: \(1.500 - (0.600 + 0.100) = 0.800\) g. Moles of C = \(0.600 / 12.0 = 0.050\) mol. Moles of H = \(0.100 / 1.0 = 0.100\) mol. Moles of O = \(0.800 / 16.0 = 0.050\) mol. Dividing by the smallest value (0.050) gives the molar ratio C : H : O = 1 : 2 : 1. The empirical formula of **X** is \(\text{CH}_2\text{O}\). (b)(i) The molecular ion peak \(M^+\) corresponds to the unfragmented ionized molecule, and its \(m/z\) value represents the relative molecular mass of the compound. (b)(ii) The empirical formula mass of \(\text{CH}_2\text{O} = 30.0\). Since the \(M^+\) peak is at \(m/z = 90.0\), the ratio is \(90.0 / 30.0 = 3\). Thus, the molecular formula is \(\text{C}_3\text{H}_6\text{O}_3\). (c) Reaction with \(\text{Na}_2\text{CO}_3\) yielding \(\text{CO}_2\) gas confirms the presence of a carboxylic acid group (\(-\text{COOH}\)). Reaction with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) yielding a product that does not react with Tollens' reagent confirms the presence of a secondary alcohol group (\(-\text{CH(OH)}-\)) that oxidized to a ketone, which cannot be further oxidized. A primary alcohol would oxidize to an aldehyde and then a carboxylic acid, which would also not react with Tollens' but wouldn't match the molecular formula with one carboxyl group. Thus, **X** has both a carboxylic acid group and a secondary alcohol group. Based on \(\text{C}_3\text{H}_6\text{O}_3\), the structural formula of **X** is \(\text{CH}_3\text{CH(OH)COOH}\) (2-hydroxypropanoic acid / lactic acid).

(a)(i) [2 marks] [1 mark] for C mass = 0.600 g, [1 mark] for H mass = 0.100 g. (a)(ii) [3 marks] [1 mark] for O mass = 0.800 g, [1 mark] for calculating the moles of C, H, and O, [1 mark] for the empirical formula \(\text{CH}_2\text{O}\). (b)(i) [1 mark] State that \(m/z\) of \(M^+\) gives the relative molecular mass (\(M_r\)). (b)(ii) [1 mark] Correct molecular formula \(\text{C}_3\text{H}_6\text{O}_3\). (c) [4 marks] [1 mark] for identifying the carboxylic acid group and linking to carbonate reaction, [1 mark] for identifying the secondary alcohol group and linking to ketone formation (no Tollens' reaction), [1 mark] for explaining why the secondary alcohol fits, [1 mark] for the correct structural formula \(\text{CH}_3\text{CH(OH)COOH}\).

(a) A zwitterion is a dipolar ion containing both a positive and a negative charge, but is overall neutral. The zwitterionic structure of alanine is \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\). (b)(i) At pH 2.0 (highly acidic), the amine group is protonated: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\). (b)(ii) At pH 12.0 (highly alkaline), the carboxylic acid group is deprotonated: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\). (c)(i) The dipeptide structure is \(\text{CH}_3\text{CH}(\text{NH}_2)\text{CONHCH}(\text{CH}_3)\text{COOH}\). The peptide bond (\(-\text{CO-NH}-\)) must be circled. (c)(ii) Condensation reaction (or nucleophilic substitution/elimination). (d) In the solid state, alanine exists as a zwitterion, where oppositely charged groups (\(-\text{NH}_3^+\) and \(-\text{COO}^-\)) form strong electrostatic attractions (ionic bonds) in a 3D lattice. In contrast, lactic acid exists as neutral molecules, which are held together only by hydrogen bonding and weaker van der Waals forces. Because ionic bonds are significantly stronger than hydrogen bonds, much more thermal energy is required to break the forces holding the lattice together in alanine than in lactic acid.

(a) [2 marks] [1 mark] for definition of zwitterion, [1 mark] for structure \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\). (b)(i) [1 mark] Correct protonated structure. (b)(ii) [1 mark] Correct deprotonated structure. (c)(i) [2 marks] [1 mark] for correct dipeptide backbone, [1 mark] for circling the peptide/amide bond. (c)(ii) [1 mark] Condensation reaction. (d) [4 marks] [1 mark] for identifying alanine exists as a zwitterion with ionic bonding, [1 mark] for identifying lactic acid exists as neutral molecules with hydrogen bonding, [1 mark] for stating that ionic bonds are much stronger than hydrogen bonds, [1 mark] for concluding that more energy is needed to melt alanine.

(a)(i) \([Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftharpoons [Ni(NH_3)_6]^{2+} + 6H_2O\)

(a)(ii) Non-bonding d-orbitals on the metal ion are split into two energy levels by the electric field of the ligands. An electron in a lower d-orbital absorbs a photon of visible light and is promoted to a higher d-orbital (d-d transition). The energy gap, \(\Delta E\), is related to the frequency of light absorbed by \(\Delta E = h\nu\). Since \(NH_3\) is a stronger-field ligand than \(H_2O\) (higher in the spectrochemical series), it causes a larger splitting of the d-orbitals (larger \(\Delta E\)). Consequently, a different wavelength of light is absorbed by \([Ni(NH_3)_6]^{2+}\) compared to \([Ni(H_2O)_6]^{2+}\), resulting in different complementary transmitted colours (changing from green to blue).

(b)(i) Stability constant, \(K_{stab}\), is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.

(b)(ii) The ligand exchange reaction is: \([Ni(NH_3)_6]^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+} + 6NH_3\).
The left-hand side of the reaction contains 4 reactant particles, while the right-hand side contains 7 product particles. This increase in the total number of free particles in solution leads to a significant increase in disorder, resulting in a highly positive entropy change (\(\Delta S^\theta > 0\)). Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), a positive \(\Delta S^\theta\) makes \(\Delta G^\theta\) much more negative, driving the equilibrium far to the right and making the chelated complex exceptionally stable.

(c) The expression for the stability constant of \([Ni(en)_3]^{2+}\) is:
\(K_{stab} = \frac{[[Ni(en)_3]^{2+}]}{[[Ni(H_2O)_6]^{2+}][en]^3}\)
Substitute the given values into the expression:
\(2.0 \times 10^{18} = \frac{0.025}{[[Ni(H_2O)_6]^{2+}] \times (0.050)^3}\)
Calculate the denominator term:
\((0.050)^3 = 1.25 \times 10^{-4}\)
Rearrange the equation for the concentration of the aqua complex:
\([[Ni(H_2O)_6]^{2+}] = \frac{0.025}{2.0 \times 10^{18} \times 1.25 \times 10^{-4}}\)
\([[Ni(H_2O)_6]^{2+}] = \frac{0.025}{2.50 \times 10^{14}} = 1.0 \times 10^{-16} \text{ mol dm}^{-3}\)

(a)(i) [1 mark]
- Correctly written balanced equation with state symbols or correct formulas: \([Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftharpoons [Ni(NH_3)_6]^{2+} + 6H_2O\).

(a)(ii) [3 marks]
- 1 mark: Mention that d-orbitals are split into two energy levels by ligands, and promotion of electrons (d-d transition) absorbs light/photons.
- 1 mark: State that \(NH_3\) is a stronger ligand/higher in the spectrochemical series, causing a larger energy gap (\(\Delta E\)) than \(H_2O\).
- 1 mark: Connect the larger \(\Delta E\) to the absorption of a shorter wavelength / higher frequency of light, meaning a different complementary color (blue instead of green) is transmitted.

(b)(i) [2 marks]
- 1 mark: Mentions it is the equilibrium constant.
- 1 mark: Specifies it is for the formation of a complex ion from its constituent ligands and metal ion in solution.

(b)(ii) [2 marks]
- 1 mark: Explains that there is an increase in the number of particles (from 4 to 7 molecules) during ligand substitution, leading to a positive entropy change (\(\Delta S > 0\)).
- 1 mark: Connects the positive entropy change to making \(\Delta G\) more negative (the chelate effect), which increases the thermodynamic stability of the complex.

(c) [3 marks]
- 1 mark: Correctly writes the expression for \(K_{stab}\) of \([Ni(en)_3]^{2+}\).
- 1 mark: Correct substitution of values into the expression: \(2.0 \times 10^{18} = \frac{0.025}{[[Ni(H_2O)_6]^{2+}] \times (0.050)^3}\).
- 1 mark: Correct calculated final value of \(1.0 \times 10^{-16}\) with correct units (\(\text{mol dm}^{-3}\)).

**(a)** Shaking methodology and safety:
- Shake the stoppered funnel vigorously for several minutes, then allow the layers to settle, and repeat to ensure solute partitions fully and dynamic equilibrium is reached.
- Safety precaution: Invert the funnel and open the tap regularly during shaking to release pressure buildup from vaporizing solvent.

**(b)** Separation and phase identification:
- Run off the bottom layer through the tap, closing it exactly when the boundary interface reaches the tap. Pour out the top layer from the top of the funnel to prevent contamination.
- Identification test: Add a few drops of water to a small sample of one of the layers. If the mixture remains a single, uniform phase, that layer is the aqueous phase. Alternatively, add a few drops of water to the separating funnel and observe which layer increases in volume.

**(c)** Data processing:
Because equal volumes of both layers (\(10.0\text{ cm}^3\)) are titrated against the same concentration of \(\text{NaOH}\) (\(0.100\text{ mol dm}^{-3}\)), the partition coefficient can be calculated directly from the ratio of the titre volumes:
\[K_{pc} = \frac{V_{\text{NaOH(organic)}}}{V_{\text{NaOH(aqueous)}}}\]
- **Run 1:** \(37.40 / 12.50 = 2.99\)
- **Run 2:** \(45.35 / 15.10 = 3.00\)
- **Run 3:** \(32.10 / 8.40 = 3.82\)
- **Run 4:** \(54.60 / 18.20 = 3.00\)
- **Run 5:** \(59.90 / 20.00 = 3.00\)

- **Anomalous run:** Run 3 (value of 3.82 is significantly higher than the rest).
- **Plausible experimental error for Run 3:** The student may have overshot the endpoint of the organic layer titration, or used an organic sample of volume greater than \(10.0\text{ cm}^3\) due to a pipette calibration/usage error, or failed to shake the mixture long enough to reach equilibrium (leaving excess acid in the organic layer).
- **Mean \(K_{pc}\):** \((2.99 + 3.00 + 3.00 + 3.00) / 4 = 3.00\) (to 3 sig fig).

**(d)** Uncertainty calculations:
- **Pipette:** \(\text{Percentage uncertainty} = \frac{0.04}{10.0} \times 100\% = 0.40\%\)
- **Burette:** Since a titre volume requires two separate readings (initial and final), the absolute uncertainty is doubled: \(2 \times 0.05 = \pm 0.10\text{ cm}^3\).

\[\text{Percentage uncertainty} = \frac{0.10}{12.50} \times 100\% = 0.80\%\]
- **Comparison:** The burette measurement contributes more to the experimental error because its percentage uncertainty (\(0.80\%\)) is twice that of the pipette (\(0.40\%\)).

**(a) [3 marks total]**
- 1 mark: Shake vigorously and allow to settle repeatedly to ensure equilibrium is established.
- 1 mark: Invert and open the tap of the separating funnel to vent/release pressure.
- 1 mark: Wear safety goggles/gloves and perform venting away from self/others (or in a fume cupboard) due to organic solvent vapors.

**(b) [3 marks total]**
- 1 mark: Run the lower layer out of the bottom tap, stopping exactly at the interface, and pour the upper layer out of the top.
- 1 mark: Add a small volume of water to a sample of one layer.
- 1 mark: State that if the mixture remains single-phase / mixes completely, it is the aqueous layer (or if it forms two layers, it is the organic layer).

**(c) [5 marks total]**
- 1 mark: Correctly calculates all 5 \(K_{pc}\) values to 3 sig figs (2.99, 3.00, 3.82, 3.00, 3.00).
- 1 mark: Identifies Run 3 as the anomalous run.
- 1 mark: Explains the anomaly with a logical experimental error (e.g., overshoot of organic endpoint, or aqueous layer volume too low due to air bubble in pipette).
- 1 mark: Calculates the mean \(K_{pc}\) as 3.00.
- 1 mark: Gives final answer to 3 significant figures with appropriate working.

**(d) [4 marks total]**
- 1 mark: Correctly calculates pipette percentage uncertainty as 0.40%.
- 1 mark: Recognizes that a burette titre involves two readings, doubling the absolute uncertainty to \(\pm 0.10\text{ cm}^3\).
- 1 mark: Correctly calculates burette percentage uncertainty as 0.80%.
- 1 mark: Concludes that the burette measurement contributes more to the experimental error.

**(a)** Colorimetry principles:
- Copper(II) complexes with amine ligands are typically deep blue/violet and absorb light in the complementary orange/red region of the visible spectrum.
- A red filter (wavelength around \(600-650\text{ nm}\)) should be used because it corresponds to the region of maximum absorbance of the complex, maximizing the sensitivity of the absorbance measurements relative to concentration.

**(b)** Calculations for the table:
- For Tube 3: \(x = 4.0 / 10.0 = 0.40\).
\(A_{\text{corrected}} = 0.448 - (1 - 0.40) \times 0.080 = 0.448 - 0.048 = 0.400\).
- For Tube 5: \(x = 6.5 / 10.0 = 0.65\).
\(A_{\text{corrected}} = 0.678 - (1 - 0.65) \times 0.080 = 0.678 - 0.028 = 0.650\).
- For Tube 6: \(x = 7.0 / 10.0 = 0.70\).
\(A_{\text{corrected}} = 0.624 - (1 - 0.70) \times 0.080 = 0.624 - 0.024 = 0.600\).
- For Tube 7: \(x = 8.0 / 10.0 = 0.80\).
\(A_{\text{corrected}} = 0.416 - (1 - 0.80) \times 0.080 = 0.416 - 0.016 = 0.400\).

Completed rows:
- Tube 3: \(x = 0.40\), \(A_{\text{corrected}} = 0.400\)
- Tube 5: \(x = 0.65\), \(A_{\text{corrected}} = 0.650\)
- Tube 6: \(x = 0.70\), \(A_{\text{corrected}} = 0.600\)
- Tube 7: \(x = 0.80\), \(A_{\text{corrected}} = 0.400\)

**(c)** Constructing the plot and finding \(n\):
- Plot \(A_{\text{corrected}}\) on the y-axis against \(x\) (mole fraction of \(\text{en}\)) on the x-axis.
- Draw two best-fit straight lines: one through the rising points (where \(\text{en}\) is limiting) and one through the falling points (where \(\text{Cu}^{2+}\) is limiting).
- Determine the coordinates of the point where these two lines intersect. The x-value at this intersection is \(x_{\text{max}}\), representing the stoichiometric ratio.
- Calculation: \(x_{\text{max}} = 0.67 = 2/3\).
\[x = \frac{n}{1 + n} \implies 0.67 = \frac{n}{1 + n} \implies n = \frac{0.67}{1 - 0.67} = \frac{0.67}{0.33} \approx 2\]
- Therefore, the formula of the complex is \([\text{Cu}(\text{en})_2]^{2+}\).

**(d)** Controlling variables:
- According to the Beer-Lambert law, absorbance is directly proportional to concentration. To ensure that changes in absorbance are solely due to shifts in the position of the equilibrium and stoichiometry (rather than dilution effects), the total volume must be kept constant.
- This ensures that the total concentration of the active metal and ligand species combined (\([\text{Cu}^{2+}] + [\text{en}]\)) remains constant (at \(0.100\text{ mol dm}^{-3}\)) across all trials, which is a fundamental requirement of Job's method.

**(e)** Deviation and Stability Constant:
- Deviation occurs because the formation reaction is a reversible equilibrium. At the stoichiometric point, the complex is partially dissociated back into free \(\text{Cu}^{2+}\) and \(\text{en}\), reducing the concentration of the complex relative to the theoretical maximum (extrapolated point where dissociation is assumed to be zero).
- To find \(K_{\text{stab}}\): The extrapolated intersection absorbance yields the concentration of complex if \(100\%\) conversion occurred. By comparing the actual absorbance to this maximum, the equilibrium concentration of \([\text{Cu}(\text{en})_2]^{2+}\) is determined. The concentrations of unreacted \(\text{Cu}^{2+}\) and \(\text{en}\) can then be calculated by mass balance, and substituted into:
\[K_{\text{stab}} = \frac{[[\text{Cu}(\text{en})_2]^{2+}]}{[\text{Cu}^{2+}][\text{en}]^2}\]

**(a) [2 marks total]**
- 1 mark: State that the copper-ethylenediamine complex is colored and absorbs visible light (while the ligand does not).
- 1 mark: Explain that a red/orange filter (around 600-650 nm) is used because it is complementary to the blue/violet complex, maximizing absorbance sensitivity.

**(b) [3 marks total]**
- 1 mark: Correctly calculates all four mole fractions (x) to 2 decimal places (0.40, 0.65, 0.70, 0.80).
- 2 marks: Correctly calculates corrected absorbance (A_corrected) to 3 decimal places (Tube 3: 0.400, Tube 5: 0.650, Tube 6: 0.600, Tube 7: 0.400). Deduct 1 mark for any arithmetic error or rounding error.

**(c) [4 marks total]**
- 1 mark: Plots corrected absorbance on the y-axis against mole fraction x on the x-axis.
- 1 mark: Describes drawing two best-fit intersecting straight lines through the points.
- 1 mark: Identifies the intersection point as representing the stoichiometry of the complex.
- 1 mark: Shows working using the formula \(n = x / (1 - x)\) to find \(n = 2\) and states the formula of the complex as \([\text{Cu}(\text{en})_2]^{2+}\).

**(d) [3 marks total]**
- 1 mark: States that keeping volume constant ensures that the total concentration of metal + ligand remains constant.
- 1 mark: Mentions that this controls the variable of dilution.
- 1 mark: Connects concentration to absorbance using the Beer-Lambert law (absorbance is directly proportional to concentration).

**(e) [3 marks total]**
- 1 mark: States that the complex is in equilibrium and undergoes partial dissociation at the stoichiometric point.
- 1 mark: Explains that the extrapolated line intersection gives the concentration of complex at 100% conversion (no dissociation).
- 1 mark: Explains that the ratio of actual to extrapolated absorbance allows calculation of equilibrium concentrations of all species to evaluate \(K_{\text{stab}}\).