2023 IB DP Chemistry Practice Paper with Answers

Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), is a non-metal oxide that reacts with water to produce phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), which is a moderately strong acid. Sodium oxide and magnesium oxide are basic metal oxides that produce alkaline solutions, whereas silicon dioxide is a giant covalent structure that is insoluble in water and has no effect on pH.

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

The carbonate ion, \(\text{CO}_3^{2-}\), has three electron domains around the central carbon atom and no lone pairs. According to VSEPR theory, this results in a trigonal planar geometry with bond angles of exactly 120 degrees due to resonance. Ammonia, \(\text{NH}_3\), and the hydronium ion, \(\text{H}_3\text{O}^+\), both have tetrahedral electron domain geometries with one lone pair, leading to a trigonal pyramidal molecular geometry with bond angles of approximately 107 degrees. Oxygen difluoride, \(\text{OF}_2\), is bent with a bond angle of approximately 103 degrees.

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

The structural formula of propanenitrile is \(\text{H}_3\text{C}-\text{CH}_2-\text{C}\equiv\text{N}\). Single covalent bonds consist of 1 \(\sigma\) bond, while triple covalent bonds consist of 1 \(\sigma\) and 2 \(\pi\) bonds. Counting the bonds: five C-H single bonds (5 \(\sigma\)), two C-C single bonds (2 \(\sigma\)), and one C-N triple bond (1 \(\sigma\) and 2 \(\pi\)). Total \(\sigma\) bonds = 5 + 2 + 1 = 8. Total \(\pi\) bonds = 2.

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

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). \(\text{H}_2\text{PO}_4^-\), acting as a Bronsted-Lowry acid, donates a proton to form its conjugate base, \(\text{HPO}_4^{2-}\).

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

In pure water, the concentration of hydrogen ions is equal to the concentration of hydroxide ions, so \([\text{H}^+] = [\text{OH}^-]\). Since \(K_w = [\text{H}^+][\text{OH}^-] = [\text{H}^+]^2\), we find that \([\text{H}^+] = \sqrt{K_w} = \sqrt{4.0 \times 10^{-14}} = 2.0 \times 10^{-7} \text{ mol dm}^{-3}\).

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

3-methylbutan-1-ol has the molecular structure \((\text{CH}_3)_2\text{CH}-\text{CH}_2-\text{CH}_2\text{OH}\). The hydroxyl group (-\text{OH}) is attached to a carbon atom that is bonded to only one other carbon atom, making it a primary alcohol. The molecule contains 5 carbon atoms in total, with 4 carbon-carbon single bonds connecting them.

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

Ethyl propanoate has the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\) and belongs to the ester homologous series. Carboxylic acids are functional group isomers of esters. Pentanoic acid has the same molecular formula (\(\text{C}_5\text{H}_{10}\text{O}_2\)) but belongs to a different homologous series (carboxylic acids). Methyl butanoate and propyl ethanoate are esters (chain/positional isomers), while pentan-3-one is a ketone (\(\text{C}_5\text{H}_{10}\text{O}\)).

Award 1 mark for the correct choice A. Reject all other options.

Calculating the oxidation states of the transition metals: In \([\text{Fe}(\text{H}_2\text{O})_6]^{3+}\), Fe is +3. In \(\text{Cr}_2\text{O}_7^{2-}\), \(2(x) + 7(-2) = -2\) leads to \(Cr = +6\). In \(\text{MnO}_4^-\), \(x + 4(-2) = -1\) leads to \(Mn = +7\). In \([\text{Co}(\text{NH}_3)_6]^{3+}\), Co is +3. The highest oxidation state is +7 for manganese in permanganate.

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

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, dihydrogen phosphate (\(\text{H}_2\text{PO}_4^-\)) acts as a Br\u00f8nsted-Lowry acid by donating a proton to become hydrogen phosphate (\(\text{HPO}_4^{2-}\)), its conjugate base. Thus, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) form a conjugate acid-base pair.

Award [1] for correct option B. Any other option receives [0] marks.

Let the volume of each solution be \(V\text{ dm}^3\). The total volume of the mixture is \(2V\text{ dm}^3\). Calculate the initial moles of \(\text{H}^+\) and \(\text{OH}^-\): Moles of \(\text{H}^+\) = \(0.10 \times V\). Moles of \(\text{OH}^-\) = \(0.08 \times V\). Reaction: \(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\). Since \(\text{OH}^-\) is the limiting reactant, the excess moles of \(\text{H}^+\) = \(0.10V - 0.08V = 0.02V\). The concentration of excess \(\text{H}^+\) in the mixture is \([\text{H}^+] = \frac{0.02V}{2V} = 0.01\text{ mol dm}^{-3}\). Finally, \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(10^{-2}) = 2.0\).

Award [1] for correct option B. Any other option receives [0] marks.

The carbonate ion (\(\text{CO}_3^{2-}\)) has three bonding electron domains and no lone pairs around the central carbon atom, leading to a trigonal planar geometry. Due to resonance, the pi electrons are delocalized equally over all three carbon-oxygen bonds, resulting in three identical bond angles of exactly \(120^\circ\). Ozone (\(\text{O}_3\)) and sulfur dioxide (\(\text{SO}_2\)) have bent molecular structures with a lone pair on the central atom, reducing their bond angles to less than \(120^\circ\). Ammonia (\(\text{NH}_3\)) has a trigonal pyramidal molecular geometry with bond angles of approximately \(107^\circ\).

Award [1] for correct option B. Any other option receives [0] marks.

Bond length decreases as bond order increases. Let us determine the nitrogen-oxygen bond orders: 1. \(\text{NO}^+\) has 10 valence electrons, giving a triple bond between N and O (bond order = 3.0). 2. \(\text{NO}_2^-\) has 18 valence electrons. Its resonance structures distribute 3 bonding pairs over 2 positions, giving a bond order of 1.5. 3. \(\text{NO}_3^-\) has 24 valence electrons. Resonance distributes 4 bonding pairs over 3 positions, giving a bond order of 1.33. Since higher bond order means shorter bond length, the order of increasing bond length (shortest to longest) is \(\text{NO}^+ < \text{NO}_2^- < \text{NO}_3^-\).

Award [1] for correct option A. Any other option receives [0] marks.

First, calculate the Index of Hydrogen Deficiency (IHD) for \(\text{C}_4\text{H}_8\text{O}_2\): \(\text{IHD} = \frac{2(4) + 2 - 8}{2} = 1\). This indicates the molecule contains either one double bond or one ring. For acyclic isomers, there must be exactly one double bond. I. A carboxylic acid (e.g., butanoic acid) contains one \(\text{C}=\text{O}\) bond (IHD = 1) and two oxygen atoms. This is possible. II. An ester (e.g., ethyl ethanoate) contains one \(\text{C}=\text{O}\) bond (IHD = 1) and two oxygen atoms. This is possible. III. A molecule containing both a ketone group and a hydroxyl group (e.g., 3-hydroxybutan-2-one) contains one \(\text{C}=\text{O}\) bond (IHD = 1) and two oxygen atoms in total. This is also possible. Therefore, all three options are possible.

Award [1] for correct option D. Any other option receives [0] marks.

Amines are classified as primary, secondary, or tertiary based on the number of alkyl or aryl groups directly bonded to the nitrogen atom. In propylamine, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\), only one carbon is attached to the nitrogen (primary). In tert-butylamine, \((\text{CH}_3)_3\text{CNH}_2\), only one carbon is directly attached to the nitrogen (primary, even though that carbon is tertiary). In \((\text{CH}_3)_2\text{CHNHCH}_3\), there are two carbon atoms attached to the nitrogen (secondary). In \((\text{CH}_3)_2\text{NCH}_2\text{CH}_3\), there are three carbon atoms directly bonded to the nitrogen atom (two methyls and one ethyl), making it a tertiary amine.

Award [1] for correct option D. Any other option receives [0] marks.

Across Period 3, the bonding structures of oxides change from giant ionic (\(\text{Na}_2\text{O}\), \(\text{MgO}\)) to giant covalent (\(\text{SiO}_2\)), and then to molecular covalent (\(\text{P}_4\text{O}_{10}\), \(\text{SO}_3\)). Thus, the overall structural transition is from giant ionic to molecular covalent. Concurrently, the chemical behavior of the oxides changes from basic (\(\text{Na}_2\text{O}\)) to amphoteric (\(\text{Al}_2\text{O}_3\)), and then to acidic (\(\text{SiO}_2\), \(\text{P}_4\text{O}_{10}\), \(\text{SO}_3\)). Thus, the overall trend in acid-base character is from basic to acidic.

Award [1] for correct option A. Any other option receives [0] marks.

The standard cell potential of a voltaic cell is calculated as: \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}}\), where the cathode undergoes reduction (more positive \(E^\theta\)) and the anode undergoes oxidation (more negative \(E^\theta\)). To obtain the maximum possible potential, we couple the half-cell with the most positive standard reduction potential as the cathode and the most negative as the anode. This corresponds to the silver cathode and the zinc anode: \(E^\theta_{\text{cell}} = +0.80\text{ V} - (-0.76\text{ V}) = +1.56\text{ V}\). Other combinations produce smaller potentials (\(\text{Zn}/\text{Cu}\) yields \(+1.10\text{ V}\), \(\text{Cu}/\text{Ag}\) yields \(+0.46\text{ V}\)).

Award [1] for correct option B. Any other option receives [0] marks.

An amphiprotic species can act as both a Brønsted–Lowry acid (by donating a proton, \(\text{H}^+\)) and a Brønsted–Lowry base (by accepting a proton). For \(\text{HCO}_3^-\): Acid form is \(\text{CO}_3^{2-}\) and base form is \(\text{H}_2\text{CO}_3\), so it is amphiprotic. For \(\text{H}_2\text{PO}_4^-\): Acid form is \(\text{HPO}_4^{2-}\) and base form is \(\text{H}_3\text{PO}_4\), so it is amphiprotic. For \(\text{CO}_3^{2-}\): It can act as a base to form \(\text{HCO}_3^-\), but it has no proton to donate, so it cannot act as an acid. Therefore, only I and II can act as both.

[1] Award for correct identification of amphiprotic species (I and II only).

To find the number of \(\sigma\) and \(\pi\) bonds, we can draw the structural formula: \(\text{H}_2\text{C}=\text{CH}-\text{C}\equiv\text{N}\). Single bonds contain 1 \(\sigma\) bond, double bonds contain 1 \(\sigma\) and 1 \(\pi\) bond, and triple bonds contain 1 \(\sigma\) and 2 \(\pi\) bonds. Here, we have three \(\text{C}-\text{H}\) single bonds (3 \(\sigma\)), one \(\text{C}=\text{C}\) double bond (1 \(\sigma\) and 1 \(\pi\)), one \(\text{C}-\text{C}\) single bond (1 \(\sigma\)), and one \(\text{C}\equiv\text{N}\) triple bond (1 \(\sigma\) and 2 \(\pi\)). Total \(\sigma\) = 3 + 1 + 1 + 1 = 6. Total \(\pi\) = 1 + 2 = 3. Therefore, there are 6 \(\sigma\) and 3 \(\pi\) bonds.

[1] Award for correct identification of 6 \(\sigma\) and 3 \(\pi\) bonds.

Amines are classified as primary, secondary, or tertiary based on the number of alkyl or aryl groups attached directly to the nitrogen atom. In \((\text{CH}_3)_3\text{CNH}_2\), the nitrogen is bonded to only one carbon atom, making it a primary amine. In \((\text{CH}_3)_2\text{CHNHCH}_3\), the nitrogen is bonded to two carbon atoms, making it a secondary amine. In \((\text{CH}_3)_3\text{N}\), the nitrogen is bonded to three carbon atoms (three methyl groups), making it a tertiary amine. In \((\text{CH}_3)_2\text{CHCH}_2\text{NH}_2\), the nitrogen is bonded to only one carbon atom, making it a primary amine.

[1] Award for identifying \((\text{CH}_3)_3\text{N}\) as the tertiary amine.

Let's analyze the species: \(\text{Cl}^-\), \(\text{S}^{2-}\), and \(\text{P}^{3-}\) are isoelectronic (each has 18 electrons). As nuclear charge decreases from chlorine (17 protons) to sulfur (16 protons) to phosphorus (15 protons), the nucleus exerts less pull on the electron cloud, so the radius increases: \(\text{Cl}^- < \text{S}^{2-} < \text{P}^{3-}\). In contrast, \(\text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+}\) (decreasing radius). For cations down a group, the correct increasing order is \(\text{Li}^+ < \text{Na}^+ < \text{K}^+\). For the isoelectronic series in option D, the correct increasing order is \(\text{Na}^+ < \text{F}^- < \text{O}^{2-}\).

[1] Award for identifying the correct increasing order of ionic radius for isoelectronic species (Option B).

Let's calculate the oxidation state of sulfur (S) in each species: In \(\text{S}_2\text{O}_3^{2-}\), \(2(\text{S}) + 3(-2) = -2 \implies \text{S} = +2\). In \(\text{S}_4\text{O}_6^{2-}\), \(4(\text{S}) + 6(-2) = -2 \implies \text{S} = +2.5\). In \(\text{SO}_3^{2-}\), \((\text{S}) + 3(-2) = -2 \implies \text{S} = +4\). In \(\text{SO}_2\text{Cl}_2\), oxygen is more electronegative than sulfur (\(-2\)) and chlorine is also more electronegative than sulfur (\(-1\)), so \((\text{S}) + 2(-2) + 2(-1) = 0 \implies \text{S} = +6\). Thus, the highest oxidation state of sulfur is in \(\text{SO}_2\text{Cl}_2\).

[1] Award for selecting the correct species with the highest oxidation state (+6).

For any pure water sample, \([\text{H}^+] = [\text{OH}^-]\) because autoionization produces equal amounts of hydrogen and hydroxide ions. Since the concentrations are equal, the water is neutral. Since \(K_w = [\text{H}^+][\text{OH}^-] = [\text{H}^+]^2 = 5.5 \times 10^{-14}\), we get \([\text{H}^+] = \sqrt{5.5 \times 10^{-14}} \approx 2.35 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}\). Since this concentration is greater than \(1.0 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}\), the pH is less than 7.00 (approx 6.63). Therefore, the pH is less than 7.00, but the water remains neutral.

[1] Award for identifying that pH is less than 7.00 and the water is neutral.

The length of a carbon-carbon bond depends on its bond order: \(\text{C}_2\text{H}_6\) (ethane) has a single covalent bond (bond order = 1), which is the longest. \(\text{C}_6\text{H}_6\) (benzene) has intermediate bonds due to resonance delocalization (bond order = 1.5). \(\text{C}_2\text{H}_4\) (ethene) has a double covalent bond (bond order = 2). \(\text{C}_2\text{H}_2\) (ethyne) has a triple covalent bond (bond order = 3), which is the shortest. Therefore, the decreasing order of bond length is \(\text{C}_2\text{H}_6 > \text{C}_6\text{H}_6 > \text{C}_2\text{H}_4 > \text{C}_2\text{H}_2\).

[1] Award for identifying the correct decreasing order of bond length.

Let's analyze the structural formula: 1. \(-\text{CO}-\) positioned between two carbon atoms represents a ketone group. 2. \(-\text{CH(OH)}-\) contains a hydroxyl group, representing an alcohol. 3. \(-\text{COOCH}_3\) is a carbonyl group bonded to an oxygen which is bonded to a methyl group, representing an ester. Thus, the functional groups are ketone, alcohol, and ester.

[1] Award for identifying Ketone, Alcohol, Ester.

A conjugate acid-base pair consists of two species that differ by exactly one proton, \(\text{H}^+\). In the given reaction, \(\text{H}_2\text{PO}_4^-\) acts as an acid by donating a proton to form its conjugate base, \(\text{HPO}_4^{2-}\). Therefore, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) represent a conjugate acid-base pair. Option a is incorrect as they are from different conjugate systems. Option b also lists species from different pairs, and option d lists two reactants.

Award 1 mark for the correct option (C). No marks are awarded for any of the other options.

First, calculate the initial amount in moles of \(\text{H}^+\) ions: \(n(\text{H}^+) = 0.0500\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 5.0 \times 10^{-3}\text{ mol}\). Next, calculate the initial amount in moles of \(\text{OH}^-\) ions: \(n(\text{OH}^-) = 0.0500\text{ dm}^3 \times 0.08\text{ mol dm}^{-3} = 4.0 \times 10^{-3}\text{ mol}\). Since \(\text{H}^+\) reacts with \(\text{OH}^-\) in a 1:1 ratio, the excess moles of \(\text{H}^+\) is \(5.0 \times 10^{-3}\text{ mol} - 4.0 \times 10^{-3}\text{ mol} = 1.0 \times 10^{-3}\text{ mol}\). The total volume of the mixture is \(50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3 = 0.100\text{ dm}^3\). The concentration of the excess \(\text{H}^+\) is \([\text{H}^+] = 1.0 \times 10^{-3}\text{ mol} / 0.100\text{ dm}^3 = 1.0 \times 10^{-2}\text{ mol dm}^{-3}\). Finally, \(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-2}) = 2.0\).

Award 1 mark for the correct option (B). No marks are awarded for other options.

Bond length decreases as the bond order increases. Let's determine the nitrogen-oxygen bond order in each species: In \(\text{NO}_2^-\), the resonance structures show a bond order of 1.5. In \(\text{NO}_3^-\), the resonance structures show a bond order of 1.33. \(\text{NO}^+\) is isoelectronic with \(\text{N}_2\) and contains a triple bond, giving a bond order of 3. \(\text{NO}_2^+\) is a linear molecule containing two double bonds, giving a bond order of 2. Since \(\text{NO}^+\) has the highest bond order of 3, it has the shortest nitrogen-to-oxygen bond length.

Award 1 mark for the correct option (C). No marks are awarded for other options.

A non-polar molecule with polar bonds must have a highly symmetrical geometry where the individual bond dipole moments cancel out. \(\text{PCl}_3\) is trigonal pyramidal and polar. \(\text{SF}_4\) is seesaw-shaped and polar. \(\text{CH}_2\text{Cl}_2\) is an asymmetrical tetrahedral molecule and polar. \(\text{SO}_3\) is trigonal planar and highly symmetrical; the individual polar sulfur-oxygen double bonds cancel each other's dipoles completely, making the molecule non-polar.

Award 1 mark for the correct option (C). No marks are awarded for other options.

An alcohol is classified as tertiary if the carbon atom bonded to the hydroxyl (\(-\text{OH}\)) group is also directly bonded to three other carbon atoms. In 2-methylbutan-2-ol, the second carbon is bonded to the hydroxyl group, two methyl groups, and one ethyl group. This carbon is bonded to three other carbon atoms, making it a tertiary alcohol. In contrast, 3-methylbutan-2-ol is a secondary alcohol, while 2,2-dimethylpropan-1-ol and 2-methylbutan-1-ol are primary alcohols.

Award 1 mark for the correct option (A). No marks are awarded for other options.

Analyzing the molecular formula: the \(-\text{OH}\) group on the second carbon represents an alcohol. The \(-\text{COO}-\) group represents an ester. The \(-\text{CH}=\text{CH}_2\) group represents an alkene. Therefore, the functional groups present are alcohol, ester, and alkene.

Award 1 mark for the correct option (B). No marks are awarded for other options.

Electronegativity is the ability of an atom to attract a bonding pair of electrons. Down Group 17, atomic radius and shielding increase, meaning the nucleus has a weaker attraction for bonding electrons, so electronegativity decreases. Across Period 3 (from \(\text{Na}\) to \(\text{Cl}\)), the effective nuclear charge increases while inner-shell shielding remains relatively constant, so the nucleus attracts bonding electrons more strongly, meaning electronegativity increases. Thus, electronegativity perfectly fits this trend.

Award 1 mark for the correct option (D). No marks are awarded for other options.

We calculate the oxidation state of sulfur in each species. In \(\text{S}_2\text{O}_3^{2-}\), \(2x + 3(-2) = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\). In \(\text{SO}_3^{2-}\), \(x + 3(-2) = -2 \Rightarrow x = +4\). In \(\text{H}_2\text{SO}_3\), \(2(+1) + x + 3(-2) = 0 \Rightarrow x = +4\). In \(\text{SO}_2\text{Cl}_2\), since oxygen and chlorine are more electronegative than sulfur, O is \(-2\) and Cl is \(-1\), leading to \(x + 2(-2) + 2(-1) = 0 \Rightarrow x = +6\). Sulfur has its highest oxidation state of \(+6\) in \(\text{SO}_2\text{Cl}_2\).

Award 1 mark for the correct option (D). No marks are awarded for other options.

The initial pH is \(2.0\), which corresponds to a hydrogen ion concentration of \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.0}\text{ mol dm}^{-3} = 0.01\text{ mol dm}^{-3}\). Diluting a volume of \(10.0\text{ cm}^3\) to \(1000.0\text{ cm}^3\) represents a dilution factor of \(\frac{1000.0}{10.0} = 100\). Thus, the concentration of hydrogen ions in the diluted solution is \([\text{H}^+]_{\text{diluted}} = \frac{0.01\text{ mol dm}^{-3}}{100} = 0.0001\text{ mol dm}^{-3} = 10^{-4.0}\text{ mol dm}^{-3}\). The pH of this diluted solution is \(-\log_{10}(10^{-4.0}) = 4.0\).

Award [1] for the correct option C.
Award [0] for other options.
- Option A incorrect: indicates an increase in concentration.
- Option B incorrect: assumes a 10-fold dilution.
- Option D incorrect: assumes a 1000-fold dilution.

According to the Brønsted-Lowry theory, a conjugate acid-base pair consists of two species that differ from each other by a single proton (\(\text{H}^+\)). In this reaction, \(\text{HPO}_4^{2-}\) acts as a Brønsted-Lowry base by accepting a proton to form its conjugate acid, \(\text{H}_2\text{PO}_4^-\). Therefore, \(\text{HPO}_4^{2-}\) and \(\text{H}_2\text{PO}_4^-\)

Award [1] for the correct option D.
Award [0] for other options because they do not represent two species differing by exactly one proton.

The hydronium ion, \(\text{H}_3\text{O}^+\), has three bonding pairs of electrons and one lone pair on the central oxygen atom (a total of 4 electron domains, tetrahedral electron domain geometry). The presence of the lone pair results in a trigonal pyramidal molecular shape. \(\text{BF}_3\) and \(\text{SO}_3\) both have trigonal planar geometries (3 electron domains, 0 lone pairs). \(\text{ClF}_3\) has 5 electron domains (3 bonding pairs, 2 lone pairs) and a T-shaped geometry.

Award [1] for the correct option B.
Award [0] for any other option.

Bond length is inversely proportional to bond order. Ethyne (\(\text{C}_2\text{H}_2\)) contains a carbon-carbon triple bond (bond order 3) and has the shortest bond length. Ethene (\(\text{C}_2\text{H}_4\)) contains a carbon-carbon double bond (bond order 2). Benzene (\(\text{C}_6\text{H}_6\)) contains a delocalized π system where the carbon-carbon bonds have a bond order of approximately 1.5. Ethane (\(\text{C}_2\text{H}_6\)) contains a carbon-carbon single bond (bond order 1) and has the longest bond length. Therefore, the correct order of increasing bond length is: \(\text{C}_2\text{H}_2 < \text{C}_2\text{H}_4 < \text{C}_6\text{H}_6 < \text{C}_2\text{H}_6\).

Award [1] for the correct option B.
Award [0] for other options.

Aspirin contains: (1) a benzene ring, which classifies it as an arene; (2) a carboxylic acid group (\(-\text{COOH}\)), which classifies it as a carboxylic acid; and (3) an ester linkage (\(-\text{OCOCH}_3\)), which classifies it as an ester. There are no ketone, alcohol, aldehyde, or phenol groups in its structure.

Award [1] for the correct option B.
Award [0] if any other option is selected.

In a tertiary halogenoalkane, the halogen-bearing carbon is directly bonded to three other carbon atoms. In 2-chloro-2-methylpropane, the C-2 carbon atom (holding the chlorine) is bonded to three methyl groups (three other carbon atoms). Therefore, it is a tertiary halogenoalkane. 2-chlorobutane and 3-chloropentane are secondary halogenoalkanes. 1-chloro-2,2-dimethylpropane is a primary halogenoalkane because the carbon bonded to chlorine is connected to only one other carbon atom (the quaternary carbon).

Award [1] for the correct option A.
Award [0] for other options.

Phosphorus has a outer electron configuration of \(3\text{s}^2 3\text{p}^3\), meaning each of the three \(3\text{p}\) orbitals is singly occupied (half-filled stability). Sulfur has a configuration of \(3\text{s}^2 3\text{p}^4\), meaning one of the \(3\text{p}\) orbitals contains a pair of electrons. The electrostatic repulsion between these two paired electrons in the same orbital (spin-pair repulsion) makes it easier to remove one electron from sulfur than from phosphorus, resulting in sulfur having a lower first ionization energy.

Award [1] for the correct option C.
Award [0] for other options.
- Option A is incorrect because sulfur has a smaller atomic radius than phosphorus due to higher effective nuclear charge.
- Option B is incorrect as effective nuclear charge works to increase ionization energy, not decrease it.
- Option D is incorrect as both electrons are removed from the same principal energy level (\(n=3\)).

By applying the rules for assigning oxidation numbers:
- In \(\underline{\text{I}}\text{O}_4^-\): \(x + 4(-2) = -1 \Rightarrow x = +7\).
- In \(\underline{\text{Cr}}_2\text{O}_7^{2-}\): \(2y + 7(-2) = -2 \Rightarrow 2y = +12 \Rightarrow y = +6\).
- In \(\underline{\text{P}}\text{O}_4^{3-}\): \(z + 4(-2) = -3 \Rightarrow z = +5\).
- In \(\underline{\text{S}}\text{O}_3^{2-}\): \(w + 3(-2) = -2 \Rightarrow w = +4\).
Thus, the iodine atom in the periodate ion, \(\text{IO}_4^-\), has the highest oxidation state (\(+7\)).

Award [1] for the correct option A.
Award [0] for any other option.

(a) \(K_a = \frac{[\text{H}^+][\text{Lac}^-]}{[\text{HLac}]}\) (b) \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.43} = 3.72 \times 10^{-3}\text{ mol dm}^{-3}\). Assuming \([\text{H}^+] = [\text{Lac}^-]\), and \([\text{HLac}]_{\text{eq}} = 0.150 - 3.72 \times 10^{-3} = 0.146\text{ mol dm}^{-3}\) (or keeping \([\text{HLac}] \approx 0.150\text{ mol dm}^{-3}\)). If using \(0.146\text{ mol dm}^{-3}\): \(K_a = \frac{(3.72 \times 10^{-3})^2}{0.146} = 9.48 \times 10^{-5}\text{ mol dm}^{-3}\). If using \(0.150\text{ mol dm}^{-3}\): \(K_a = \frac{(3.72 \times 10^{-3})^2}{0.150} = 9.23 \times 10^{-5}\text{ mol dm}^{-3}\). (c)(i) \(\text{HLac(aq)} + \text{NaOH(aq)} \rightarrow \text{NaLac(aq)} + \text{H}_2\text{O(l)}\) (or ionic: \(\text{HLac(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{Lac}^-\text{(aq)} + \text{H}_2\text{O(l)}\)). (ii) Initial moles: \(n(\text{HLac}) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 7.50 \times 10^{-3}\text{ mol}\); \(n(\text{NaOH}) = 0.0250\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\). Remaining moles after reaction: \(n(\text{HLac}) = 7.50 \times 10^{-3} - 3.00 \times 10^{-3} = 4.50 \times 10^{-3}\text{ mol}\); \(n(\text{Lac}^-) = 3.00 \times 10^{-3}\text{ mol}\). Total volume = \(75.0\text{ cm}^3 = 0.0750\text{ dm}^3\). Final concentrations: \([\text{HLac}] = \frac{4.50 \times 10^{-3}\text{ mol}}{0.0750\text{ dm}^3} = 0.0600\text{ mol dm}^{-3}\); \([\text{Lac}^-] = \frac{3.00 \times 10^{-3}\text{ mol}}{0.0750\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\). (iii) Using Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log\frac{[\text{Lac}^-]}{[\text{HLac}]}\). For \(K_a = 9.48 \times 10^{-5}\), \(\text{p}K_a = 4.02\); \(\text{pH} = 4.02 + \log\left(\frac{0.0400}{0.0600}\right) = 3.84\text{ or } 3.85\). For \(K_a = 9.23 \times 10^{-5}\), \(\text{p}K_a = 4.03\); \(\text{pH} = 4.03 + \log\left(\frac{0.0400}{0.0600}\right) = 3.85\text{ or } 3.86\).

(a) 1 mark for correct expression. (b) 1 mark for calculating \([\text{H}^+] = 3.72 \times 10^{-3}\text{ mol dm}^{-3}\); 1 mark for correct setting up of \(K_a\) equation; 1 mark for final calculated value of \(K_a\) around \(9.2 \times 10^{-5}\) to \(9.5 \times 10^{-5}\). (c)(i) 1 mark for correct balanced chemical equation. (ii) 1 mark for calculating correct initial moles; 1 mark for calculating remaining moles of \(\text{HLac}\) and formed moles of \(\text{Lac}^-\); 1 mark for dividing by total volume to find \([\text{HLac}] = 0.0600\text{ mol dm}^{-3}\) and \([\text{Lac}^-] = 0.0400\text{ mol dm}^{-3}\). (iii) 1 mark for \(\text{p}K_a\) calculation; 1 mark for final \(\text{pH}\) value of \(3.84 - 3.86\) (allow ECF from previous parts).

(a)(i) A Brønsted-Lowry base is a proton (\(\text{H}^+\)) acceptor. (ii) \(\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}\). Here, \(\text{NH}_3\) is the base and \(\text{NH}_4^+\) is its conjugate acid; \(\text{H}_2\text{O}\) is the acid and \(\text{OH}^-\) is its conjugate base. (b) Ammonium chloride dissociates fully to form \(\text{NH}_4^+\text{(aq)}\) and \(\text{Cl}^-\text{(aq)}\). The ammonium ion, \(\text{NH}_4^+\), is a weak acid (conjugate acid of the weak base ammonia) and undergoes hydrolysis in water: \(\text{NH}_4^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}\) (or \(\text{NH}_4^+\text{(aq)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}^+\text{(aq)}\)). The production of hydronium ions (\(\text{H}_3\text{O}^+\)) makes the solution acidic. (c)(i) The pH curve starts at a weak base pH (around \(11\)), drops gradually through a buffer region, has an equivalence point at pH \(< 7\) (around \(5\)) at a volume of \(25.0\text{ cm}^3\) of acid, and levels off at a strong acid pH (around \(1\)). (ii) Methyl orange is a suitable indicator. The equivalence point pH for a weak base-strong acid titration is acidic (approx. pH \(5\)), which lies closest to the rapid pH change region and matches the color change interval of methyl orange (\(3.1 - 4.4\)).

(a)(i) 1 mark for 'proton/H+ acceptor'. (ii) 1 mark for balanced equilibrium equation; 1 mark for correctly identifying both conjugate pairs. (b) 1 mark for correct hydrolysis equation (reversible arrows not strictly penalized, but preferred); 1 mark for explaining that hydronium/hydrogen ions are generated. (c)(i) 1 mark for starting pH between 10-12 and ending pH around 1-2; 1 mark for showing equivalence point at exactly \(25.0\text{ cm}^3\); 1 mark for drawing the equivalence pH inflection point below 7 (between 4 and 6). (ii) 1 mark for selecting methyl orange; 1 mark for justifying that the equivalence point pH is acidic/less than 7, which fits the range of methyl orange.

(a)(i) For \(\text{XeF}_2\): Xenon (\(\text{Xe}\)) is single-bonded to two fluorine (\(\text{F}\)) atoms, with three lone pairs on the central \(\text{Xe}\) atom (total of 10 electrons around \(\text{Xe}\)), and three lone pairs on each \(\text{F}\) atom. For \(\text{XeF}_4\): \(\text{Xe}\) is single-bonded to four \(\text{F}\) atoms, with two lone pairs on the central \(\text{Xe}\) atom (total of 12 electrons around \(\text{Xe}\)), and three lone pairs on each \(\text{F}\) atom. (ii) \(\text{XeF}_2\): 5 electron domains around \(\text{Xe}\) (2 bonding, 3 lone pairs) gives a linear molecular geometry. \(\text{XeF}_4\): 6 electron domains around \(\text{Xe}\) (4 bonding, 2 lone pairs) gives a square planar molecular geometry. (iii) Both molecules are non-polar. In \(\text{XeF}_2\), the linear arrangement is highly symmetrical, so the polar bonds point in opposite directions and cancel out. In \(\text{XeF}_4\), the square planar arrangement is also symmetrical, meaning the four polar Xe-F bonds cancel each other out. (b)(i) Structure 1 (octet rule obeyed): Sulfur is single-bonded to all four oxygen atoms. All four oxygens have three lone pairs (formal charge of \(-1\) each), and sulfur has no lone pairs (formal charge of \(+2\)). Structure 2 (minimized formal charges): Sulfur has two double bonds to two oxygen atoms and two single bonds to two oxygen atoms. Double-bonded oxygens have two lone pairs (formal charge of \(0\)), single-bonded oxygens have three lone pairs (formal charge of \(-1\)), and sulfur has six bonding pairs and no lone pairs (formal charge of \(0\)). (ii) For Structure 1: \(\text{FC(S)} = 6 - 0 - 4 = +2\). For Structure 2: \(\text{FC(S)} = 6 - 0 - 6 = 0\).

(a)(i) 1 mark for correct \(\text{XeF}_2\) Lewis structure with all lone pairs; 1 mark for correct \(\text{XeF}_4\) Lewis structure with all lone pairs. (ii) 1 mark for linear (\(\text{XeF}_2\)); 1 mark for square planar (\(\text{XeF}_4\)). (iii) 1 mark for stating both are non-polar; 1 mark for explaining that symmetrical geometries cause bond dipole moments to cancel. (b)(i) 1 mark for octet-rule compliant structure (four single bonds, S has \(+2\) formal charge); 1 mark for expanded octet structure (two single bonds, two double bonds, S has \(0\) formal charge). (ii) 1 mark for calculating S formal charge of \(+2\) in octet structure; 1 mark for calculating S formal charge of \(0\) in minimized structure.

(a)(i) The carbonate ion has 24 valence electrons. In each of the three resonance structures, the central carbon is double-bonded to one oxygen (which has two lone pairs and a formal charge of 0) and single-bonded to two oxygens (each having three lone pairs and a formal charge of -1). The double bond rotates among all three oxygen positions. (ii) In the carbonate ion, the 4 bonding electron pairs are shared equally among 3 carbon-oxygen bonds, giving a bond order of \(\frac{4}{3} = 1.33\). In carbon dioxide (\(\text{CO}_2\)), carbon forms two double bonds with two oxygen atoms, so the carbon-oxygen bond order is \(2\). (iii) Experimental evidence shows that all three carbon-oxygen bond lengths in the carbonate ion are identical (and intermediate in length/strength between a single C-O bond and a double C=O bond). (b)(i) A \(\sigma\) bond is formed by the head-on (axial) overlap of \(sp^2\) hybrid orbitals between the two carbon atoms. A \(\pi\) bond is formed by the sideways (lateral) overlap of the unhybridized \(p\) orbitals of the two carbon atoms, which lie perpendicular to the plane of the molecule. (ii) The carbon atoms in ethene are \(sp^2\) hybridized. The H-C-H bond angle is approximately \(120^\circ\) (accept \(117^\circ - 120^\circ\)).

(a)(i) 2 marks for drawing three structures with correct bonds and lone pairs; 1 mark for correct formal charges on oxygen atoms (or indicating the overall 2- charge on each resonance structure). (ii) 1 mark for bond order of 1.33 in carbonate; 1 mark for bond order of 2 in \(\text{CO}_2\). (iii) 1 mark for stating that all carbon-oxygen bonds are of equal length/strength (intermediate between single and double). (b)(i) 1 mark for describing head-on overlap of orbitals for \(\sigma\) bond; 1 mark for describing sideways/lateral overlap of p orbitals for \(\pi\) bond. (ii) 1 mark for \(sp^2\) hybridization; 1 mark for bond angle of \(120^\circ\) (accept range \(117^\circ - 120^\circ\)).

(a) Three possible acyclic structural isomers of \(\text{C}_4\text{H}_8\text{O}\) and their functional groups: 1. Butanal (aldehyde group): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\); 2. Butanone (ketone/carbonyl group): \(\text{CH}_3\text{COCH}_2\text{CH}_3\); 3. But-3-en-1-ol (alkene and alcohol/hydroxyl groups): \(\text{CH}_2\text{=CH-CH}_2\text{CH}_2\text{OH}\) (or ethoxyethene / other unsaturated ethers/alcohols). (b)(i) Butanal skeletal: a four-carbon zig-zag chain with a double-bonded oxygen at carbon 1. Butanone skeletal: a four-carbon zig-zag chain with a double-bonded oxygen at carbon 2. (ii) Butane is non-polar and experiences only weak London dispersion forces between its molecules. Butanal and butanone are polar due to the highly polar carbonyl group (\(\text{C=O}\)), so they additionally experience stronger dipole-dipole forces, resulting in higher boiling points than butane. Butan-1-ol has a hydroxyl group (\(\text{-OH}\)) and is capable of forming intermolecular hydrogen bonds, which are significantly stronger than dipole-dipole forces, giving it the highest boiling point of all. (c)(i) Acidified potassium dichromate(VI) (\(\text{H}^+ / \text{Cr}_2\text{O}_7^{2-}\)), heated and distilled off immediately. (ii) Acidified potassium dichromate(VI) (\(\text{H}^+ / \text{Cr}_2\text{O}_7^{2-}\)), heated under reflux.

(a) 1 mark for each correct full structural formula and matching functional group name (up to 3 marks). (b)(i) 1 mark for correct skeletal structure of butanal; 1 mark for correct skeletal structure of butanone. (ii) 1 mark for identifying London dispersion forces in butane; 1 mark for identifying dipole-dipole forces in butanal/butanone; 1 mark for identifying hydrogen bonding in butan-1-ol. (c)(i) 1 mark for acidified potassium dichromate(VI) (or acidified potassium manganate(VII)) AND distillation. (ii) 1 mark for acidified potassium dichromate(VI) AND reflux.

(a)(i) 1-chlorobutane is a primary halogenoalkane. 2-chloro-2-methylpropane is a tertiary halogenoalkane. (ii) 1-chlorobutane reacts primarily via the \(\text{S}_\text{N}2\) mechanism. 2-chloro-2-methylpropane reacts primarily via the \(\text{S}_\text{N}1\) mechanism. (b) The \(\text{S}_\text{N}2\) mechanism involves: 1. A curly arrow starting from a lone pair on the oxygen of \(\text{OH}^-\)\ to the carbon atom attached to chlorine. 2. A curly arrow showing the cleavage of the C-Cl bond starting from the C-Cl bond line to the chlorine atom. 3. Draw a transition state: carbon is surrounded by five groups with dotted lines representing partial bonds to the incoming hydroxyl group and the leaving chlorine group. The overall charge of the transition state must be shown as negative, inside brackets (i.e. \([\dots]^-\)). 4. Draw the final products: butan-1-ol and a chloride ion (\(\text{Cl}^-\)). (c) Tertiary halogenoalkanes have three bulky alkyl groups attached to the carbon bonded to the halogen. This causes significant steric hindrance, which blocks the backside attack of the nucleophile, preventing the \(\text{S}_\text{N}2\) mechanism. However, when the leaving group departs, a stable tertiary carbocation is formed, stabilized by the positive inductive electron-donating effect of the three alkyl groups, which highly favors the two-step \(\text{S}_\text{N}1\) mechanism.

(a)(i) 1 mark for classifying 1-chlorobutane as primary; 1 mark for 2-chloro-2-methylpropane as tertiary. (ii) 1 mark for \(\text{S}_\text{N}2\) for 1-chlorobutane; 1 mark for \(\text{S}_\text{N}1\) for 2-chloro-2-methylpropane. (b) 1 mark for curly arrow from hydroxide lone pair to the carbon; 1 mark for curly arrow from the C-Cl bond to chlorine; 1 mark for drawing the correct transition state structure (with partial bonds and overall negative charge); 1 mark for correct products (butan-1-ol and \(\text{Cl}^-\)). (c) 1 mark for explaining that steric hindrance prevents backside attack in the \(\text{S}_\text{N}2\) pathway; 1 mark for explaining that the tertiary carbocation intermediate is stabilized by the positive inductive effect of three alkyl groups in the \(\text{S}_\text{N}1\) pathway.

(a)(i) Electronegativity increases across Period 3 (from sodium to chlorine). This is because nuclear charge increases (more protons in the nucleus) while shielding remains approximately constant (electrons are added to the same main energy level), creating a stronger electrostatic attraction between the nucleus and shared bonding electrons. (ii) Sodium and magnesium have low electronegativities, resulting in a large electronegativity difference with oxygen, which leads to ionic bonding in \(\text{Na}_2\text{O}\) and \(\text{MgO}\). Non-metals on the right of Period 3 have higher electronegativities, leading to a small electronegativity difference with oxygen, and thus covalent bonding in oxides like \(\text{P}_4\text{O}_{10}\) and \(\text{SO}_3\). (b)(i) Sodium oxide reaction with water: \(\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}\) (or ionic: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{Na}^+ + 2\text{OH}^-\)). Phosphorus(V) oxide reaction with water: \(\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\). (ii) Aluminium oxide is amphoteric (reacts with both acids and bases). Reaction with acid (e.g., \(\text{H}^+\)): \(\text{Al}_2\text{O}_3\text{(s)} + 6\text{H}^+\text{(aq)} \rightarrow 2\text{Al}^{3+}\text{(aq)} + 3\text{H}_2\text{O(l)}\). Reaction with base (e.g., \(\text{OH}^-\)): \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2[\text{Al(OH)}_4]^-\text{(aq)}\) (or \(\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} \rightarrow 2\text{AlO}_2^-\text{(aq)} + \text{H}_2\text{O(l)}\)).

(a)(i) 1 mark for stating that electronegativity increases; 1 mark for explaining via increased nuclear charge and constant shielding. (ii) 1 mark for linking a large electronegativity difference to ionic character; 1 mark for linking a small electronegativity difference to covalent character. (b)(i) 1 mark for balanced equation of \(\text{Na}_2\text{O}\) with water; 1 mark for balanced equation of \(\text{P}_4\text{O}_{10}\) with water. (ii) 1 mark for identifying \(\text{Al}_2\text{O}_3\) as amphoteric; 1 mark for correct equation showing reaction with acid (\(\text{H}^+\)); 2 marks for correct equation showing reaction with base (\(\text{OH}^-\)) (1 mark for reactants/products, 1 mark for balancing).

(a)(i) Oxidation occurs at the anode: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\). Reduction occurs at the cathode: \(\text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)}\). (ii) Overall cell reaction: \(\text{Zn(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Ag(s)}\). Standard cell notation: \(\text{Zn(s)} \mid \text{Zn}^{2+}\text{(aq)} \parallel \text{Ag}^+\text{(aq)} \mid \text{Ag(s)}\). (iii) Electrons flow through the external circuit from the zinc electrode (anode) to the silver electrode (cathode). In the salt bridge, potassium cations (\(\text{K}^+\)) migrate towards the silver half-cell (cathode) to balance the loss of positive charge, and nitrate anions (\(\text{NO}_3^-\)) migrate towards the zinc half-cell (anode) to balance the gain of positive charge. (b)(i) \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (-0.76\text{ V}) = +1.56\text{ V}\). (ii) \(\Delta G^\theta = -nFE^\theta_{\text{cell}}\). Here, \(n = 2\) moles of electrons. \(\Delta G^\theta = -2 \times 96\,485\text{ C mol}^{-1} \times 1.56\text{ V} = -301\,033.2\text{ J mol}^{-1} = -301\text{ kJ mol}^{-1}\).

(a)(i) 1 mark for zinc oxidation half-equation; 1 mark for silver reduction half-equation; 1 mark for identifying zinc as the anode and silver as the cathode. (ii) 1 mark for correct overall equation; 1 mark for correct cell notation showing phase boundaries and salt bridge. (iii) 1 mark for stating electron flow is from Zn to Ag; 1 mark for stating that cations migrate to the silver half-cell and anions to the zinc half-cell. (b)(i) 1 mark for calculating \(E^\theta_{\text{cell}} = +1.56\text{ V}\). (ii) 1 mark for setting up equation \(\Delta G^\theta = -nFE^\theta\) using \(n = 2\); 1 mark for correct calculation of \(-301\text{ kJ mol}^{-1}\) (must have negative sign and correct units).

(a) At the anode (oxidation): \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-\). At the cathode (reduction): \(\text{Fe}^{3+}(\text{aq}) + e^- \rightarrow \text{Fe}^{2+}(\text{aq})\). Overall balanced cell equation: \(\text{Zn(s)} + 2\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq})\). Standard cell potential: \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (-0.76\text{ V}) = +1.53\text{ V}\). (b) Electrons flow through the external wire from the zinc electrode (anode, where oxidation occurs) to the platinum electrode (cathode, where reduction occurs). To maintain charge neutrality, \(\text{Na}^+\) cations from the salt bridge migrate into the cathode compartment to balance the loss of positive charge concentration (as \(\text{Fe}^{3+}\) is reduced to \(\text{Fe}^{2+}\)), while \(\text{NO}_3^-\) anions from the salt bridge migrate into the anode compartment to balance the accumulation of positive charge as \(\text{Zn}^{2+}\) ions are produced. (c) Charge passed: \(Q = I \times t = 1.50\text{ A} \times (45.0 \times 60\text{ s}) = 4050\text{ C}\). Moles of electrons: \(n(e^-) = \frac{Q}{F} = \frac{4050\text{ C}}{96500\text{ C mol}^{-1}} = 0.04197\text{ mol}\). The reduction equation is \(\text{M}^{3+}(\text{aq}) + 3e^- \rightarrow \text{M(s)}\). Moles of metal \(\text{M}\) deposited: \(n(\text{M}) = \frac{n(e^-)}{3} = \frac{0.04197}{3} = 0.01399\text{ mol}\). Molar mass of \(\text{M} = \frac{\text{mass}}{n} = \frac{0.725\text{ g}}{0.01399\text{ mol}} = 51.8\text{ g mol}^{-1}\). The metal with a molar mass closest to this value is Chromium (\(\text{Cr}\), molar mass \(52.00\text{ g mol}^{-1}\)).

Part (a) [4 marks]: - Correct anode half-equation: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-\) [1 mark]. - Correct cathode half-equation: \(\text{Fe}^{3+}(\text{aq}) + e^- \rightarrow \text{Fe}^{2+}(\text{aq})\) [1 mark]. - Correct overall cell equation: \(\text{Zn(s)} + 2\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq})\) [1 mark]. - Correct calculation of cell potential: \(E^\theta_{\text{cell}} = +1.53\text{ V}\) [1 mark]. Part (b) [3 marks]: - Direction of electrons: from the zinc electrode to the platinum electrode / anode to cathode through the wire [1 mark]. - Cations (\(\text{Na}^+\)) migrate to the cathode/iron half-cell [1 mark]. - Anions (\(\text{NO}_3^-\)) migrate to the anode/zinc half-cell [1 mark]. Part (c) [3 marks]: - Calculates charge \(Q = 4050\text{ C}\) and moles of electrons \(n(e^-) = 0.0420\text{ mol}\) [1 mark]. - Correctly uses the 1:3 stoichiometry to find moles of metal \(n(\text{M}) = 0.0140\text{ mol}\) [1 mark]. - Calculates molar mass of \(51.8\text{ g mol}^{-1}\) (accept range 51.7 to 52.0) and identifies the metal as Chromium / \(\text{Cr}\) [1 mark].

(a) The independent variable is time (\(t\)), and the dependent variable is the volume of oxygen gas evolved (\(V\)).

(b) Average rate during the first \(30\text{ s}\) is calculated as:
\[\text{Average rate} = \frac{\Delta V}{\Delta t} = \frac{36.0\text{ cm}^3 - 0.0\text{ cm}^3}{30\text{ s} - 0\text{ s}} = 1.20\text{ cm}^3\text{ s}^{-1}\]

(c) As the reaction proceeds, the concentration of the reactant, hydrogen peroxide (\(\text{H}_2\text{O}_2\)), decreases because it is being consumed. According to collision theory, a lower concentration means fewer reactant particles per unit volume, which reduces the frequency of successful collisions between reactant particles per unit time, thereby decreasing the rate of reaction.

(d) The instantaneous rate at \(t = 20\text{ s}\) is equal to the gradient of the tangent line at that point:
\[\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{45.0\text{ cm}^3 - 9.0\text{ cm}^3}{40\text{ s} - 0\text{ s}} = \frac{36.0\text{ cm}^3}{40\text{ s}} = 0.90\text{ cm}^3\text{ s}^{-1}\]

(e) Using a syringe with a larger graduation uncertainty (\(\pm 0.5\text{ cm}^3\) instead of \(\pm 0.1\text{ cm}^3\)) increases the random uncertainty of the measured volume.

(a) Award [1] if both variables are correctly identified:
- Independent variable: time [0.5]
- Dependent variable: volume (of \(\text{O}_2\) gas) [0.5]

(b) Award [1] for correct calculation of average rate:
- \(\frac{36.0}{30} = 1.20\text{ cm}^3\text{ s}^{-1}\) [1]

(c) Award [2] for a complete explanation:
- Reactant/\(\text{H}_2\text{O}_2\) concentration decreases OR reactant particles are consumed [1]
- Fewer collisions per unit time / lower collision frequency [1]

(d) Award [2] for correct instantaneous rate calculation and units:
- Correct calculation of gradient: \(\frac{45.0 - 9.0}{40 - 0} = 0.90\) [1]
- Correct units: \(\text{cm}^3\text{ s}^{-1}\) (or \(\text{cm}^3/\text{s}\)) [1]

(e) Award [1] for:
- Increases the random uncertainty / precision decreases [1]

(a) Calculate the heat change, \(q\):
\[\Delta T = 29.00\text{ }^\circ\text{C} - 21.00\text{ }^\circ\text{C} = 8.00\text{ }^\circ\text{C} = 8.00\text{ K}\]
\[q = m \cdot c \cdot \Delta T = 105.00\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 8.00\text{ K} = 3511.2\text{ J} \approx 3510\text{ J} \text{ (or } 3.51\text{ kJ)}\]

(b) The process is exothermic because the temperature of the reaction mixture increased (\(\Delta T > 0\)), indicating that heat was released to the surroundings.

(c) Calculate the moles of \(\text{CaCl}_2\):
\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00\text{ g}}{110.98\text{ g mol}^{-1}} = 0.045053\text{ mol} \approx 0.0451\text{ mol}\]

(d) Calculate the enthalpy change of solution, \(\Delta H_{\text{sol}}\):
\[\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{3.5112\text{ kJ}}{0.045053\text{ mol}} = -77.9\text{ kJ mol}^{-1}\]
(Since the reaction is exothermic, \(\Delta H\) must be negative. The answer is given to 3 significant figures.)

(e) Two sources of experimental error that could explain why the experimental value is less exothermic (i.e., less negative) than the literature value of \(-81.3\text{ kJ mol}^{-1}\):
1. Heat loss to the surroundings (e.g., to the air or the polystyrene cup), resulting in a lower maximum temperature recorded.
2. Incomplete dissolution of the anhydrous calcium chloride solid, meaning less heat was generated than expected for the mass used.

(a) Award [2] for correct calculation:
- \(\Delta T = 8.00\text{ K}\) [1]
- \(q = 105.00 \times 4.18 \times 8.00 = 3510\text{ J}\) (or \(3.51\text{ kJ}\)) [1]

(b) Award [1] for:
- Exothermic AND because the temperature increased [1]

(c) Award [1] for:
- \(n = \frac{5.00}{110.98} = 0.0451\text{ mol}\) [1]

(d) Award [2] for:
- \(\Delta H_{\text{sol}} = -77.9\text{ kJ mol}^{-1}\) (must have negative sign and be to 3 significant figures) [2]
- Award [1] if magnitude is correct (77.9) but sign is positive OR if there is a rounding/sig fig error with correct sign.

(e) Award [2] for any two of the following:
- Heat loss to the surroundings/cup [1]
- Incomplete dissolution of \(\text{CaCl}_2\) [1]
- The heat capacity of the cup was neglected [1]
- Assumed the specific heat capacity of the solution is the same as pure water, but it may differ [1]

(a) Butanoic acid contains an \(\text{O}-\text{H}\) bond (carboxylic acid), which shows a broad absorption band in the range of \(2500{-}3000\text{ cm}^{-1}\). This is absent in ethyl ethanoate (which only has \(\text{C}-\text{H}\) stretching near \(2850{-}3100\text{ cm}^{-1}\) and no \(\text{O}-\text{H}\) group).

(b) Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) has three distinct proton environments, giving three signals in its \({}^1\text{H}\text{ NMR}\) spectrum:
- A singlet (3H) for the \(\text{CH}_3\) adjacent to the carbonyl group (\(\text{CH}_3\text{CO}-\)), because there are no adjacent protons.
- A quartet (2H) for the \(\text{CH}_2\) protons (\(-\text{OCH}_2\text{CH}_3\)), because they are adjacent to a methyl group (3 adjacent protons, splitting = \(3+1 = 4\)).
- A triplet (3H) for the methyl protons (\(-\text{OCH}_2\text{CH}_3\)), because they are adjacent to a \(\text{CH}_2\) group (2 adjacent protons, splitting = \(2+1 = 3\)).

In contrast, butanoic acid (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)) has four distinct proton environments, producing four signals. Thus, the number of signals (3 vs 4) and the specific splitting patterns (singlet, quartet, triplet for ethyl ethanoate) uniquely identify ethyl ethanoate.

M1: Broad \(\text{O}-\text{H}\) absorption in the range of \(2500\text{ to }3000\text{ cm}^{-1}\) (accept any range within these limits, e.g., \(2500{-}3300\text{ cm}^{-1}\)). [1]
M2: State that ethyl ethanoate has 3 signals/peaks (while butanoic acid has 4 signals). [1]
M3: Identify the singlet signal in ethyl ethanoate and attribute it to \(\text{CH}_3\text{CO}-\) protons / protons with no adjacent hydrogen atoms. [1]
M4: Identify the quartet and triplet splitting pattern (ethyl group / \(\text{O}-\text{CH}_2-\text{CH}_3\)). [1]
M5: Correctly apply the \(n+1\) rule to explain the quartet (split by 3 adjacent protons) OR the triplet (split by 2 adjacent protons). [1]

(a) Using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\)
\(7.40 = 6.36 + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)\)
\(\log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 7.40 - 6.36 = 1.04\)
\(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{1.04} \approx 11.0\) (or 11)

(b) When excess lactic acid releases \(\text{H}^+\) ions into the bloodstream, the conjugate base, hydrogencarbonate (\(\text{HCO}_3^-\)), reacts with the added hydronium ions to form weak, largely undissociated carbonic acid:
\(\text{HCO}_3^-\text{(aq)} + \text{H}^+\text{(aq)} \rightleftharpoons \text{H}_2\text{CO}_3\text{(aq)}\)
This shifts the equilibrium to the right, neutralizing the added acid and preventing a significant decrease in pH.

M1: Recall or correctly set up the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\). [1]
M2: Correctly calculate the logarithmic difference: \(\log\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 1.04\). [1]
M3: Correctly solve for the ratio: \(11\) or \(11.0\) (accept range \(10.9\) to \(11.0\)). [1]
M4: Write a balanced equation showing \(\text{HCO}_3^-\) reacting with \(\text{H}^+\): \(\text{HCO}_3^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{H}_2\text{CO}_3\text{(aq)}\). [1]
M5: Explain that this reaction removes free \(\text{H}^+\) ions from solution, thereby minimizing the pH change. [1]

(a) Molecules that form liquid crystals typically have:
- Long, rod-like or disc-like rigid shapes (often containing multiple benzene rings to prevent rotation/flexing).
- Polar groups (e.g., \(\text{C}\equiv\text{N}\), \(\text{COOR}\)) which create strong dipole-dipole interactions, maintaining alignment while allowing flow.
- Alkyl chains at the ends to provide flexibility and prevent crystallization.

(b) Distinguishing between nematic and smectic phases:
- Nematic phase: Molecules have translational freedom (they can move past one another like a liquid) but have directional order (on average, they point in the same direction along a common axis).
- Smectic phase: Molecules not only point in the same direction (directional order) but are also arranged in parallel layers or planes (translational/positional order).
- Relative order: Smectic phases are more ordered than nematic phases.

M1: State that molecules must be rigid/rod-shaped/planar (containing benzene rings / biphenyl groups). [1]
M2: State that they must contain polar groups (for dipole-dipole attractions) OR contain flexible hydrocarbon tails (to prevent easy crystallization). [1]
M3: Describe the nematic phase: molecules have parallel alignment / point in the same direction BUT are randomly distributed/lack layer structure. [1]
M4: Describe the smectic phase: molecules are parallel AND arranged in distinct layers. [1]
M5: Explicitly state that the smectic phase has a higher degree of order / is more ordered than the nematic phase. [1]

(a) At the anode, methanol is oxidized to carbon dioxide:
\(\text{CH}_3\text{OH}\text{(aq)} + \text{H}_2\text{O}\text{(l)} \rightarrow \text{CO}_2\text{(g)} + 6\text{H}^+\text{(aq)} + 6\text{e}^-
\)
At the cathode, oxygen is reduced to water:
\(\text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}\text{(l)}
\)

(b) Multiplying the anode reaction by 2 and the cathode reaction by 3 to balance electrons yields the overall cell reaction:
\(2\text{CH}_3\text{OH}\text{(aq)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O}\text{(l)}
\)

(c) Advantages of methanol over hydrogen:
- Methanol is a liquid at room temperature and pressure, making it much easier and safer to store, transport, and distribute than high-pressure hydrogen gas.
- Methanol has a higher volumetric energy density than hydrogen gas, meaning more energy can be stored in a smaller volume.

M1: Correct anode half-equation: \(\text{CH}_3\text{OH} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + 6\text{H}^+ + 6\text{e}^-\). [1]
M2: Correct cathode half-equation: \(\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}\). [1]
M3: Correct overall equation: \(2\text{CH}_3\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 4\text{H}_2\text{O}\) (or fractional coefficients equivalent). [1]
M4: State one valid advantage (e.g., liquid at room temp / easier storage / safer transport / utilizes existing liquid fuel infrastructure). [1]
M5: State a second distinct advantage (e.g., higher volumetric energy density / simpler tank design / less leakage risk). [1]

(a) Heterogeneous catalysts provide a surface for reactant molecules to adsorb. Transition metals are effective because:
- They have partially filled d-orbitals that can form weak coordinate covalent bonds with reactant molecules, facilitating adsorption and weakening the bonds within the reactants.
- They have variable oxidation states, allowing them to gain or lose electrons easily, thereby providing low-activation-energy pathways for redox processes.

(b) For the Haber process:
(i) Catalyst: Iron (\(\text{Fe}\)) (or finely divided iron). [Alternatively, for the Contact process: Vanadium(V) oxide (\(\text{V}_2\text{O}_5\))].
(ii) Steps:
- Adsorption: Nitrogen (\(\text{N}_2\)) and hydrogen (\(\text{H}_2\)) gas molecules are adsorbed onto the active sites on the surface of the iron catalyst, weakening the \(\text{N}\equiv\text{N}\) and \(\text{H}-\text{H}\) bonds.
- Reaction: The adsorbed atoms react on the surface to form ammonia (\(\text{NH}_3\)) molecules.
- Desorption: The newly formed ammonia molecules desorb from the catalyst surface, freeing the active sites for further reactants.

M1: State that transition metals have vacant/partially filled d-orbitals that allow reactants to adsorb/bond to the surface. [1]
M2: State that they exhibit variable oxidation states, allowing them to act as temporary electron reservoirs / facilitate electron transfer. [1]
M3: Correctly identify the catalyst for a named process: Iron/\(\text{Fe}\) for the Haber process OR Vanadium(V) oxide/\(\text{V}_2\text{O}_5\) for the Contact process. [1]
M4: Describe adsorption: reactants bind to the active sites on the catalyst surface, which weakens their bonds. [1]
M5: Describe reaction and desorption: chemical reaction occurs on the surface, and the products desorb/leave the catalyst surface. [1]

(a) Kevlar is formed from 1,4-diaminobenzene (benzene-1,4-diamine) and benzene-1,4-dicarboxylic acid (or benzene-1,4-dioyl dichloride).
The structural formulas are:
- Monomer 1: \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2\) (benzene-1,4-diamine)
- Monomer 2: \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) (benzene-1,4-dicarboxylic acid) or \(\text{ClOC}-\text{C}_6\text{H}_4-\text{COCl}\) (benzene-1,4-dioyl dichloride)

(b) Kevlar is formed via a condensation polymerization reaction, eliminating water (\(\text{H}_2\text{O}\)) [or hydrogen chloride (\(\text{HCl}\)) if the diacyl chloride is used].

(c) Kevlar's high tensile strength arises because:
- The linear, rigid polymer chains (containing planar benzene rings) align parallel to one another.
- Strong hydrogen bonds form between the polar amide groups (\(\text{C}=\text{O}\) of one chain and \(\text{N}-\text{H}\) of an adjacent chain), holding the sheets of chains tightly together.
- Additionally, London dispersion forces (and \(\pi\)-\(\pi\) stacking) between the aromatic benzene rings stabilize the layered sheets.

M1: Correct structural formula of the diamine monomer: \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2\) (or drawn equivalent). [1]
M2: Correct structural formula of the dicarboxylic acid / diacyl chloride monomer: \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) or \(\text{ClOC}-\text{C}_6\text{H}_4-\text{COCl}\) (or drawn equivalent). [1]
M3: Identify the reaction type as condensation polymerization AND the small molecule as \(\text{H}_2\text{O}\) (or \(\text{HCl}\)). [1]
M4: State that the polymer chains are linear and rigid, allowing them to pack closely/align parallel. [1]
M5: Identify extensive hydrogen bonding between \(\text{C}=\text{O}\) and \(\text{N}-\text{H}\) groups of adjacent chains as the key force holding the chains together. [1]