2025 IB DP Chemistry Practice Paper with Answers

All three species have 4 electron domains around the central nitrogen atom, giving a basic tetrahedral arrangement of electron domains. In \( \text{NH}_4^+ \), there are 4 bonding pairs and 0 lone pairs, resulting in a regular tetrahedral shape with a bond angle of \( 109.5^\circ \). In \( \text{NH}_3 \), there are 3 bonding pairs and 1 lone pair; the lone pair-bonding pair repulsion reduces the bond angle to approximately \( 107^\circ \). In \( \text{NH}_2^- \), there are 2 bonding pairs and 2 lone pairs; the stronger lone pair-lone pair repulsion reduces the bond angle further to approximately \( 104.5^\circ \). Therefore, the correct order of increasing bond angle is \( \text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+ \).

Award [1] for correct option A. No other options are accepted.

First, analyze the structural formula of acrylic acid: \( \text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})-\text{OH} \). For the \( \sigma \) bonds, there are three \( \text{C}-\text{H} \) single bonds, one \( \text{C}=\text{C} \) double bond (1 \( \sigma \)), one \( \text{C}-\text{C} \) single bond, one \( \text{C}=\text{O} \) double bond (1 \( \sigma \)), one \( \text{C}-\text{O} \) single bond, and one \( \text{O}-\text{H} \) single bond, summing up to 8 \( \sigma \) bonds. For the \( \pi \) bonds, there is one in the \( \text{C}=\text{C} \) double bond and one in the \( \text{C}=\text{O} \) double bond, making 2 \( \pi \) bonds in total. The carbonyl carbon is bonded to three other atoms with no lone pairs, representing 3 electron domains, which corresponds to \( \text{sp}^2 \) hybridization.

Award [1] for correct option B. No other options are accepted.

To determine the relative bond lengths, we look at the carbon-oxygen bond order in each species: in carbonate (\( \text{CO}_3^{2-} \)), the three carbon-oxygen bonds are equivalent due to resonance, giving a bond order of 1.33; in carbon dioxide (\( \text{CO}_2 \)), each bond is a double bond with a bond order of 2.0; in carbon monoxide (\( \text{CO} \)), there is a triple bond with a bond order of 3.0. Because bond length is inversely related to bond order, the order of decreasing bond length is \( \text{CO}_3^{2-} > \text{CO}_2 > \text{CO} \).

Award [1] for correct option A. No other options are accepted.

\( \text{XeF}_4 \) has 6 electron domains on the central xenon atom (4 bonding pairs and 2 lone pairs). The lone pairs occupy axial positions opposite to each other, while the four fluorine atoms lie in a square planar arrangement around the central atom. The individual polar Xe-F bond dipoles cancel each other out completely, yielding a net dipole moment of zero. All other options are asymmetrical (see-saw, T-shaped, or bent) and have non-zero net dipole moments.

Award [1] for correct option B. No other options are accepted.

Atom economy measures the percentage of starting reactant materials that end up in the desired final product. Since this is an addition reaction with only one product (methyl methacrylate), all of the atoms present in the reactants are incorporated into the desired product. Therefore, no by-products or waste are generated, yielding an atom economy of exactly 100%.

Award [1] for correct option D. No other options are accepted.

The delivered volume is calculated as the difference between the final and initial readings: \( V_{\text{delivered}} = V_{\text{final}} - V_{\text{initial}} \). If both readings are systematically higher by the same value (\( +0.15\text{ cm}^3 \)), the offset cancels out: \( V_{\text{measured}} = (V_{\text{true, final}} + 0.15) - (V_{\text{true, initial}} + 0.15) = V_{\text{true, final}} - V_{\text{true, initial}} \). Thus, the calculated volume delivered is accurate, and the calculated concentration will be unaffected.

Award [1] for correct option C. No other options are accepted.

First, calculate the percentage uncertainty in the mass: \( \frac{0.1}{50.0} \times 100\% = 0.2\% \). Next, calculate the percentage uncertainty in the temperature change: \( \frac{0.2}{8.0} \times 100\% = 2.5\% \). For quantities combined by multiplication, the total percentage uncertainty is the sum of the individual percentage uncertainties: \( 0.2\% + 2.5\% = 2.7\% \).

Award [1] for correct option C. No other options are accepted.

One of the 12 principles of Green Chemistry is 'Design for Degradation' (Principle 10), which states that chemical products should be designed so that at the end of their function they break down into innocuous degradation products. Using large excesses of reactants (Option A) creates waste; increasing the use of solvents (Option B) contradicts minimizing auxiliaries; and using fossil fuels (Option D) contradicts using renewable feedstocks.

Award [1] for correct option C. No other options are accepted.

Sulfur tetrafluoride (\(SF_4\)) has 34 valence electrons. The sulfur atom forms 4 single covalent bonds and has 1 non-bonding lone pair, resulting in 5 electron domains. This gives a trigonal bipyramidal electron domain geometry and a seesaw molecular geometry. \(XeF_4\) has 6 electron domains (4 bonding, 2 non-bonding) resulting in a square planar geometry. \(CF_4\) and \(BF_4^-\) both have 4 electron domains with no lone pairs, resulting in a tetrahedral geometry.

Award 1 mark for A. Reject other options based on their molecular geometries (XeF4 is square planar, CF4 and BF4- are tetrahedral).

Bond length is inversely proportional to bond order. The bond order in carbon monoxide (\(CO\)) is 3 (triple bond). The bond order in carbon dioxide (\(CO_2\)) is 2 (double bond). The bond order in the carbonate ion (\(CO_3^{2-}\)) is 1.33 due to resonance. Therefore, the order of increasing bond length is \(CO < CO_2 < CO_3^{2-}\).

Award 1 mark for A. Award 0 marks for any other option. Higher bond order corresponds to shorter bond length.

The major resonance structures of \(N_2O\) are \(N \equiv N - O\) and \(N = N = O\). In both structures, the central nitrogen atom forms 4 covalent bonds and has no lone pairs. The formal charge is calculated as: Formal Charge = (valence electrons) - (non-bonding electrons) - 0.5 * (bonding electrons) = 5 - 0 - 4 = +1.

Award 1 mark for C. Correct calculation of formal charge for the central nitrogen in either resonance contributor.

Prop-2-enal has the structure \(H_2C=CH-C(=O)H\). There are 4 C-H single bonds, 1 C-C single bond, 1 C=C double bond, and 1 C=O double bond. This gives a total of 7 sigma bonds (4 from C-H, 1 from C-C, 1 from C=C, and 1 from C=O) and 2 pi bonds (1 from C=C and 1 from C=O). All 3 carbon atoms have 3 electron domains, which means they are all \(sp^2\) hybridized.

Award 1 mark for D. Correct identification of 7 sigma bonds, 2 pi bonds, and 3 sp2 hybridized carbon atoms.

If heat is lost to the surroundings, the maximum temperature recorded during the reaction will be lower than the theoretical value. This results in a smaller temperature change (\(\Delta T\)), which leads to a lower calculated heat of reaction (\(q = mc\Delta T\)) and thus an enthalpy change that is less exothermic (closer to zero) than the literature value.

Award 1 mark for C. Heat loss systematically reduces the measured temperature change, leading to a less exothermic calculated value. Reject other options as they do not lead to a less exothermic value in this systematic manner.

Atom economy is calculated as (mass of desired product / total mass of reactants) * 100%. Desired product mass = \(2 \times 55.85 = 111.70\text{ g mol}^{-1}\). Total mass of reactants = \(Fe_2O_3 + 3CO = (2 \times 55.85 + 3 \times 16.00) + 3 \times (12.01 + 16.00) = 159.70 + 84.03 = 243.73\text{ g mol}^{-1}\). Atom economy = \((111.70 / 243.73) \times 100\% = 45.83\%\), which rounds to 45.8%.

Award 1 mark for B. Correct calculation of desired mass (111.70 g/mol) and total reactant mass (243.73 g/mol) and applying the atom economy equation.

The percentage uncertainty in the mass measurement is \((0.01 / 1.50) \times 100\% = 0.67\%\). The percentage uncertainty in the volume measurement is \((0.10 / 25.00) \times 100\% = 0.40\%\). When quantities are multiplied or divided, their percentage uncertainties are added: \(0.67\% + 0.40\% = 1.07\%\).

Award 1 mark for D. Sum of percentage uncertainties of the mass (0.67%) and volume (0.40%) equals 1.07%.

Oxygen (\(O_2\)) has a double bond between the oxygen atoms (bond order = 2). Ozone (\(O_3\)) undergoes resonance, with each oxygen-oxygen bond having a bond order of 1.5. A lower bond order results in a weaker and longer bond. Therefore, the oxygen-oxygen bond in \(O_3\) is weaker and longer than the bond in \(O_2\).

Award 1 mark for B. Recognizes the resonance-hybrid bond order of 1.5 in ozone compared to 2 in molecular oxygen, and correctly correlates bond order with bond strength and length.

To find the most stable resonance structure, we assign formal charges to minimize charge separation and place negative formal charges on the more electronegative element. Dinitrogen monoxide can be represented by resonance structures:
1. \(\text{N} \equiv \text{N} - \text{O}\)
- Terminal N: 5 valence electrons - 2 non-bonding electrons - 3 bonding electrons = 0
- Central N: 5 valence electrons - 0 non-bonding electrons - 4 bonding electrons = +1
- Oxygen: 6 valence electrons - 6 non-bonding electrons - 1 bonding electron = -1

2. \(\text{N} = \text{N} = \text{O}\)
- Terminal N: 5 - 4 - 2 = -1
- Central N: 5 - 0 - 4 = +1
- Oxygen: 6 - 4 - 2 = 0

Since oxygen is more electronegative (3.4) than nitrogen (3.0), the structure with the negative formal charge on the oxygen atom (Structure 1) is more stable and contributes most significantly to the hybrid.

[1 mark] Selects A. Accept if correct reasoning about formal charge distribution and electronegativity is applied.

To determine the hybridization of each carbon atom, we count the number of electron domains around it:
- Left carbon: 3 electron domains (two single bonds to H, one double bond to C). This corresponds to \(\text{sp}^2\) hybridization.
- Central carbon: 2 electron domains (two double bonds to adjacent C atoms). This corresponds to \(\text{sp}\) hybridization.
- Right carbon: 3 electron domains (two single bonds to H, one double bond to C). This corresponds to \(\text{sp}^2\) hybridization.

[1 mark] Selects A. Identify electron domains around each carbon to deduce hybridization correctly.

The percentage atom economy is calculated as:

\(\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100\%\)

1. Molar mass of reactants:
- \(\text{C}_3\text{H}_8 = 3(12) + 8(1) = 44\text{ g mol}^{-1}\)
- \(\text{Br}_2 = 2(80) = 160\text{ g mol}^{-1}\)
- Sum of reactants = \(44 + 160 = 204\text{ g mol}^{-1}\)

2. Molar mass of desired product (1-bromopropane):
- \(\text{C}_3\text{H}_7\text{Br} = 3(12) + 7(1) + 80 = 123\text{ g mol}^{-1}\)

3. Calculation:
- \(\text{Atom Economy} = \frac{123}{204} \times 100\% \approx 60.3\%\)

[1 mark] Selects A. Correctly calculates molar mass of reactants and products and applies the atom economy formula.

The heat released in the reaction (\(q\)) is proportional to the number of moles of reacting species. In the second experiment, the volume of each reactant is doubled, so the amount of reaction (and thus the total heat energy released, \(q'\)) is doubled: \(q' = 2q\).

However, the total volume of the reaction mixture also doubles (from \(100.0\text{ cm}^3\) to \(200.0\text{ cm}^3\)), meaning the mass of solution to be heated (\(m'\)) is doubled: \(m' = 2m\).

Using the calorimetric equation:
\(\Delta T' = \frac{q'}{m' \cdot c} = \frac{2q}{2m \cdot c} = \Delta T\).

Therefore, the temperature rise remains the same (\(\Delta T\)).

[1 mark] Selects B. Correctly reasons that both heat produced and mass of solution double, keeping temperature change constant.

Bond length is inversely proportional to bond order. We determine the nitrogen-to-oxygen bond orders in the species:
- \(\text{NO}^+\): Isoelectronic with \(\text{N}_2\); contains a triple bond (bond order = 3.0).
- \(\text{NO}_2^+\): Linear geometry \([\text{O}=\text{N}=\text{O}]^+\); contains double bonds (bond order = 2.0).
- \(\text{NO}_2^-\): Trigonal planar electron domain geometry with one lone pair on nitrogen; resonance structures average to a bond order of 1.5.
- \(\text{NO}_3^-\): Trigonal planar geometry with resonance structures averaging to a bond order of 1.33.

Since \(\text{NO}^+\) has the highest bond order (3.0), it has the shortest bond length.

[1 mark] Selects A. Deduces bond orders and links higher bond order to shorter bond length.

A molecule has a non-zero dipole moment if its dipoles do not cancel out due to symmetry:
- \(\text{SF}_6\): Octahedral geometry; highly symmetrical, polar bonds cancel out perfectly (non-polar).
- \(\text{XeF}_4\): Square planar geometry (6 electron domains: 4 bonding pairs and 2 lone pairs in trans positions); polar bonds cancel out (non-polar).
- \(\text{SF}_4\): Seesaw geometry (5 electron domains: 4 bonding pairs and 1 equatorial lone pair); the asymmetry results in net dipoles not cancelling (polar).
- \(\text{PF}_5\): Trigonal bipyramidal geometry; symmetrical structure, polar bonds cancel out (non-polar).

[1 mark] Selects C. Identifies molecular geometries and determines which molecular structure lacks complete symmetry.

Let's analyze each error:
- I. Heat loss to the surroundings means less heat is transferred to the water, giving a smaller temperature rise and a less exothermic experimental value. (Contributes)
- II. Incomplete combustion of ethanol releases less energy per mole of ethanol than complete combustion, resulting in a lower temperature rise and a less exothermic experimental value. (Contributes)
- III. Evaporation of ethanol after extinguishing the flame but before final weighing increases the measured mass change (the apparent mass of fuel consumed). Since \(\Delta H_c = -\frac{q}{n}\), a falsely inflated value for \(n\) (moles of ethanol) reduces the calculated magnitude of \(\Delta H_c\), making it less exothermic. (Contributes)

[1 mark] Selects D. Understands how systematic and operational errors impact calculated calorimetric values.

Let the mass of both zinc and sulfur be \(m\) grams.
- Moles of zinc present = \(\frac{m}{65.38}\)
- Moles of sulfur present = \(\frac{m}{32.06}\)

Since \(65.38 > 32.06\), the number of moles of zinc is smaller than the number of moles of sulfur (\(n_{\text{Zn}} < n_{\text{S}}\)). Because they react in a 1:1 molar ratio according to the balanced equation, zinc will be completely consumed first and is therefore the limiting reactant.

[1 mark] Selects B. Compares molar amounts from equal masses to identify the limiting reactant.

To determine the most stable Lewis structure, we calculate the formal charge (FC) for each atom where \(\text{FC} = \text{valence electrons} - \text{lone pair electrons} - \frac{1}{2}(\text{bonding electrons})\).

For Structure A (\(:\text{N} \equiv \text{N} - \ddot{\text{O}}:\)):
- Terminal nitrogen: \(5 - 2 - 3 = 0\)
- Central nitrogen: \(5 - 0 - 4 = +1\)
- Terminal oxygen: \(6 - 6 - 1 = -1\)

For Structure B (\(\ddot{\text{N}} = \text{N} = \ddot{\text{O}}\)):
- Terminal nitrogen: \(5 - 4 - 2 = -1\)
- Central nitrogen: \(5 - 0 - 4 = +1\)
- Terminal oxygen: \(6 - 4 - 2 = 0\)

Both structures have a net charge of zero and a central nitrogen with a \(+1\) formal charge. Since oxygen is more electronegative than nitrogen, the structure that places the negative formal charge on the oxygen atom (Structure A) is more stable and is the major resonance contributor.

Award [1] for the correct option (A).
- Reject other choices as they either state incorrect formal charges, choose the less stable structure, or apply incorrect reasoning regarding electronegativity.

Let's analyze the shapes and bond angles of the given molecules:
- \(\text{CH}_4\): Tetrahedral shape, all bonding pairs, bond angle is \(109.5^\circ\).
- \(\text{NH}_3\): Trigonal pyramidal shape, 1 lone pair, bond angle is approximately \(107.8^\circ\).
- \(\text{H}_2\text{O}\): Bent shape, 2 lone pairs, bond angle is approximately \(104.5^\circ\).
- \(\text{PH}_3\): Trigonal pyramidal shape. Phosphorus is a Period 3 element with a lower electronegativity and a larger atomic size than nitrogen. The P-H bonding pairs are further from the central atom, reducing bonding pair-bonding pair repulsion. Consequently, the lone pair repels the bonds much more strongly, compressing the bond angle down to approximately \(93.5^\circ\).

Award [1] for the correct option (D).
- Reject choices A, B, and C as their bond angles are significantly larger than that of \(\text{PH}_3\).

Let's break down the structure of methyl cyanoacrylate:

1. Identify the single, double, and triple bonds:
- \(\text{CH}_2=\text{C}\): contains two \(\text{C}-\text{H}\) single bonds (\(2\sigma\)) and one \(\text{C}=\text{C}\) double bond (\(1\sigma, 1\pi\)).
- Central C bonded to \(-\text{CN}\): one \(\text{C}-\text{C}\) single bond (\(1\sigma\)).
- Cyano group (\(-\text{C}\equiv\text{N}\)): one \(\text{C}\equiv\text{N}\) triple bond (\(1\sigma, 2\pi\)).
- Central C bonded to carbonyl C of \(-\text{COOCH}_3\): one \(\text{C}-\text{C}\) single bond (\(1\sigma\)).
- Carbonyl group (\(\text{C}=\text{O}\)): one \(\text{C}=\text{O}\) double bond (\(1\sigma, 1\pi\)).
- Carbonyl C bonded to ester oxygen: one \(\text{C}-\text{O}\) single bond (\(1\sigma\)).
- Ester oxygen bonded to methyl carbon: one \(\text{O}-\text{C}\) single bond (\(1\sigma\)).
- Methyl group (\(-\text{CH}_3\)): three \(\text{C}-\text{H}\) single bonds (\(3\sigma\)).

2. Sum them up:
- Sigma (\(\sigma\)) bonds: \(2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 3 = 12\)
- Pi (\(\pi\)) bonds: \(1\,(\text{from C=C}) + 2\,(\text{from } \text{C}\equiv\text{N}) + 1\,(\text{from C=O}) = 4\)

Award [1] for the correct option (B).
- Reject choices with incorrect counts of sigma or pi bonds due to missing the triple bond in the nitrile group or treating double/triple bonds incorrectly.

Using the ideal gas equation: \(PV = nRT\)

First, convert units to SI standard:
- \(P = 100.6\text{ kPa} = 1.006 \times 10^5\text{ Pa}\)
- \(V = 250.0\text{ cm}^3 = 2.500 \times 10^{-4}\text{ m}^3\)
- \(T = 98.0 + 273.15 = 371.15\text{ K}\)

Calculate the number of moles (\(n\)):
\(n = \frac{PV}{RT} = \frac{1.006 \times 10^5 \times 2.500 \times 10^{-4}}{8.31 \times 371.15} = \frac{25.15}{3084.26} \approx 0.008154\text{ mol}\)

Calculate the molar mass (\(M\)):
\(M = \frac{\text{mass}}{n} = \frac{0.812\text{ g}}{0.008154\text{ mol}} \approx 99.6\text{ g mol}^{-1}\)

Award [1] for the correct option (C).
- Option A corresponds to not converting Celsius to Kelvin (\(T = 98.0\)).
- Option B corresponds to using room temperature (\(298.15\text{ K}\)).
- Option D represents another calculation error.

1. Calculate the heat absorbed by the water (\(q\)):
\(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 18.2\text{ K} = 7607.6\text{ J} = 7.608\text{ kJ}\)

2. Calculate the amount of methanol combusted (\(n\)):
\(n = \frac{\text{mass}}{\text{Molar mass}} = \frac{0.480\text{ g}}{32.04\text{ g mol}^{-1}} = 0.01498\text{ mol}\)

3. Calculate the experimental enthalpy of combustion (\(\Delta H_c\)):
\(\Delta H_c = -\frac{q}{n} = -\frac{7.608\text{ kJ}}{0.01498\text{ mol}} \approx -508\text{ kJ mol}^{-1}\) (exothermic, hence negative sign).

Award [1] for the correct option (B).
- Option A corresponds to using the mass of methanol (\(0.480\text{ g}\)) instead of the mass of water in the heat capacity equation.
- Option C is incorrect due to a positive sign (combustion is always exothermic).
- Option D corresponds to the absolute heat value without dividing by the moles of fuel correctly.

1. Write the balanced chemical equation:
\(\text{H}_2\text{SO}_4\text{ (aq)} + 2\text{NaOH (aq)} \rightarrow \text{Na}_2\text{SO}_4\text{ (aq)} + 2\text{H}_2\text{O (l)}\)

2. Calculate moles of \(\text{H}_2\text{SO}_4\):
\(n(\text{H}_2\text{SO}_4) = 0.150\text{ mol dm}^{-3} \times 0.01640\text{ dm}^3 = 0.00246\text{ mol}\)

3. Determine moles of \(\text{NaOH}\) using the mole ratio (\(1:2\)):
\(n(\text{NaOH}) = 2 \times 0.00246\text{ mol} = 0.00492\text{ mol}\)

4. Calculate the concentration of \(\text{NaOH}\):
\([\text{NaOH}] = \frac{0.00492\text{ mol}}{0.0200\text{ dm}^3} = 0.246\text{ mol dm}^{-3}\)

Award [1] for the correct option (C).
- Option A represents a \(1:1\) mole ratio mistake (\(0.123\text{ mol dm}^{-3}\)).
- Option D represents an error where the mole ratio was incorrectly doubled or applied upside down.

- Option A: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{CH}_2\text{CHO}\) contains a primary amine group (\(-\text{NH}_2\)) and an aldehyde group (\(-\text{CHO}\)).
- Option B: \(\text{CH}_3\text{NHCH}_2\text{COCH}_3\) contains a secondary amine group (\(-\text{NH}-\)) and a ketone group (\(-\text{CO}-\)) separated by a carbon, which act as individual groups.
- Option C: \(\text{CH}_3\text{CONHCH}_2\text{CH}_3\) is an amide (specifically N-ethylethanamide). The carbonyl and amine groups are directly bonded, forming a single amide functional group.
- Option D: \(\text{CH}_3\text{NHCOOCH}_3\) is a carbamate (ester and amide features together).

Award [1] for the correct option (B).
- Reject choice C, as amides are distinct from separate ketones and amines.

These species form an isoelectronic series (all have \(18\) electrons and the electronic configuration of argon, \(2, 8, 8\)):
- \(\text{S}^{2-}\) (\(16\) protons, \(18\) electrons)
- \(\text{Cl}^-\n\) (\(17\) protons, \(18\) electrons)
- \(\text{K}^+\n\) (\(19\) protons, \(18\) electrons)
- \(\text{Ca}^{2+}\) (\(20\) protons, \(18\) electrons)

In an isoelectronic series, as nuclear charge (number of protons) increases, the nucleus pulls the same number of electrons more strongly, resulting in a smaller radius. Therefore, the ionic radius decreases in the order: \(\text{S}^{2-} > \text{Cl}^- > \text{K}^+ > \text{Ca}^{2+}\).

Award [1] for the correct option (A).
- Option B is the opposite order (increasing ionic radius).
- Options C and D are incorrect orders for their respective isoelectronic series.

Each species has 4 electron domains around the central nitrogen atom. According to VSEPR theory, the bond angles depend on the number of non-bonding pairs (lone pairs):
- \( \text{NH}_4^+ \) has 0 lone pairs, giving a tetrahedral geometry with a bond angle of \( 109.5^\circ \).
- \( \text{NH}_3 \) has 1 lone pair, giving a trigonal pyramidal geometry with a bond angle of \( 107^\circ \).
- \( \text{NH}_2^- \) has 2 lone pairs, giving a bent geometry with a bond angle of \( 104.5^\circ \).

Since lone pairs repel more strongly than bonding pairs, increasing the number of lone pairs decreases the bond angle. Therefore, the order of increasing bond angle is: \( \text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+ \).

[1 mark] for identifying option B as the correct increasing order of bond angles.
[0 marks] for any other option.

The carbonate ion (\( \text{CO}_3^{2-} \)) exhibits resonance. The \( \pi \) electrons are delocalized across all three C–O bonds, which means all three bonds are identical in length and strength. The bond order is calculated as:

\( \text{Bond Order} = \frac{\text{Total number of bonding pairs in resonance structures}}{\text{Number of resonance positions}} = \frac{4}{3} \approx 1.33 \).

Additionally, the carbon atom is \( \text{sp}^2 \) hybridized (having 3 electron domains), and the ion contains 3 \( \sigma \) bonds and 1 delocalized \( \pi \) bond.

[1 mark] for choosing C.
[0 marks] for incorrect options.

The Lewis structure of ozone can be drawn as \( \text{O}=\text{O}-\text{O} \) with formal charges. The central oxygen atom forms 3 bonds (one double bond and one single bond) and has 1 lone pair (2 non-bonding electrons).

Using the formal charge formula:
\( \text{FC} = V - N - \frac{1}{2}B \)
where:
- \( V = 6 \) (valence electrons of free oxygen atom)
- \( N = 2 \) (non-bonding electrons)
- \( B = 6 \) (bonding electrons, from 3 shared pairs)

\( \text{FC} = 6 - 2 - \frac{1}{2}(6) = +1 \).

[1 mark] for choosing C.
[0 marks] for incorrect options.

Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)) contains a highly polar \( \text{O}-\text{H} \) bond, enabling the molecules to form strong intermolecular hydrogen bonds. Ethanal (\( \text{CH}_3\text{CHO} \)), methoxymethane (\( \text{CH}_3\text{OCH}_3 \)), and fluoroethane (\( \text{CH}_3\text{CH}_2\text{F} \)) lack an H atom bonded directly to a highly electronegative atom (N, O, or F) and cannot form hydrogen bonds. They only experience weaker dipole-dipole forces and London dispersion forces, resulting in lower boiling points.

[1 mark] for identifying option A as correct.
[0 marks] for other options.

Atom economy is defined as:

\( \text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\% \)

- In Route 1, ethanol is the sole product of the reaction. Therefore, 100% of the reactant atoms are converted into the desired product, giving an atom economy of exactly 100%.
- In Route 2, carbon dioxide (\( \text{CO}_2 \)) is formed as a by-product alongside ethanol, meaning some of the starting mass is wasted, resulting in an atom economy of less than 100% (approx. 51.1%).
- Renewable status and mass conservation do not dictate the atom economy value.

[1 mark] for selecting A as the correct option.
[0 marks] for selecting B, C, or D.

Let us evaluate the effect of each factor on the calculated enthalpy of combustion (\( \Delta H_c = -q / n \), where \( q = m c \Delta T \)):
- **I (Heat loss)**: Heat lost to the surroundings means the temperature rise (\( \Delta T \)) recorded is lower than expected. Thus, the calculated \( q \) is too low, making \( \Delta H_c \) less exothermic.
- **II (Incomplete combustion)**: Incomplete combustion releases less energy per mole than complete combustion, which reduces the final temperature of the water. This also decreases \( \Delta T \) and results in a less exothermic calculated value.
- **III (Evaporation of methanol)**: If methanol evaporates before the final mass of the burner is recorded, the measured mass change of the burner will be greater than the mass actually combusted. This overestimates the number of moles burned (\( n \)). Since \( \Delta H_c \propto 1/n \), an artificially high \( n \) reduces the magnitude of the calculated \( \Delta H_c \), making it less exothermic.

Therefore, all three factors (I, II, and III) contribute to a less exothermic experimental value.

[1 mark] for selecting option D.
[0 marks] for other options.

An ideal gas is assumed to have zero intermolecular forces and negligible molecular volume. Real gases deviate from this behavior, but approach it under conditions of **high temperature** and **low pressure**:
- High temperature (\( 500\text{ K} \)) gives gas molecules high kinetic energy, enabling them to overcome intermolecular attractions.
- Low pressure (\( 50\text{ kPa} \)) means the molecules are spread far apart, meaning the volume occupied by the gas molecules themselves is negligible compared to the volume of the container.

Therefore, the gas behaves most ideally at \( 500\text{ K} \) and \( 50\text{ kPa} \).

[1 mark] for selecting D.
[0 marks] for incorrect options.

In phosgene (\( \text{COCl}_2 \)), the central carbon atom forms a double bond with the oxygen atom and single bonds with two chlorine atoms. This corresponds to 3 electron domains around the carbon atom and no non-bonding pairs on it. To minimize electron domain repulsion, the domains adopt a trigonal planar arrangement. The molecular geometry is therefore trigonal planar, with bond angles close to \( 120^\circ \).

[1 mark] for identifying trigonal planar and \( 120^\circ \) (option A).
[0 marks] for any other response.

For (a): Mass of solution m = 50.0 + 50.0 = 100.0 g. Temperature change \(\Delta T = 27.6 - 21.3 = 6.3\text{ K}\). Heat released q = m * c * \(\Delta T\) = 100.0 g * 4.18 J g-1 K-1 * 6.3 K = 2633.4 J = 2.63 kJ. For (b): Moles of reactants reacting n = c * V = 1.00 mol dm-3 * 0.0500 dm3 = 0.0500 mol. \(\Delta H_{\text{neu}}\) = -q / n = -2.6334 kJ / 0.0500 mol = -52.7 kJ mol-1. For (c): Percentage error = |(-57.2 - (-52.7)) / -57.2| * 100% = 7.9% (or 7.87% using unrounded value 7.93%). For (d): Heat loss to the surroundings is a key systematic error. Another error is assuming the specific heat capacity of the solution is exactly that of pure water, or neglecting the heat absorbed by the polystyrene cup itself. This heat loss can be minimized by adding a lid, using double-nested polystyrene cups for extra insulation, or by plotting a temperature-time cooling curve and extrapolating back to the time of mixing to find the theoretical maximum temperature.

Part (a): [2 marks total] - Award 1 mark for correct calculation of temperature change (6.3 K) and mass (100.0 g). Award 1 mark for correct calculation of heat released in kJ (2.63 kJ, accept 2.6 kJ). Part (b): [3 marks total] - Award 1 mark for calculating moles of reactants (0.0500 mol). Award 1 mark for dividing heat by moles. Award 1 mark for correct value with negative sign (-52.7 kJ mol-1, accept -52.6 to -53.0 depending on rounding). Part (c): [2.66 marks total] - Award 1 mark for correct percentage error formula. Award 1.66 marks for correct final percentage error (7.9% or 7.87%, accept error propagated from b). Part (d): [4 marks total] - Award 1 mark for each valid systematic error (e.g., heat lost to surroundings, heat absorbed by calorimeter, density/specific heat capacity assumptions) up to 2 marks. Award 2 marks for a clear, valid method to minimize one of the identified errors (e.g., using a temperature vs time graph to extrapolate the maximum temperature corrected for cooling, or adding a lid/insulating jacket).

For (a): Comparing Exp 1 and Exp 2: [Y] is held constant. When [X] is doubled (from 0.10 to 0.20), the rate doubles (from 1.2 * 10^-3 to 2.4 * 10^-3). Therefore, the order with respect to X is 1. Comparing Exp 1 and Exp 3: [X] is held constant. When [Y] is doubled (from 0.10 to 0.20), the rate increases by a factor of 4 (from 1.2 * 10^-3 to 4.8 * 10^-3). Therefore, the order with respect to Y is 2. For (b): The rate expression is Rate = k[X][Y]^2. To find k, use data from Exp 1: 1.2 * 10^-3 = k(0.10)(0.10)^2 => 1.2 * 10^-3 = k(1.0 * 10^-3) => k = 1.2. Units of k for a third-order reaction are dm6 mol-2 s-1. For (c): The rate constant is highly temperature-dependent. According to collision theory, an increase in temperature increases the average kinetic energy of the particles. Consequently, a significantly higher fraction of colliding particles have energy equal to or greater than the activation energy (E >= Ea), leading to more successful collisions per unit time and a larger rate constant k.

Part (a): [4 marks total] - Award 1 mark for order wrt X is 1. Award 1 mark for showing that rate doubles when [X] doubles. Award 1 mark for order wrt Y is 2. Award 1 mark for showing that rate quadruples when [Y] doubles. Part (b): [3.66 marks total] - Award 1 mark for the correct rate expression: Rate = k[X][Y]^2. Award 1.66 marks for the correct numerical calculation of k = 1.2 (accept values calculated from other runs). Award 1 mark for correct units: dm6 mol-2 s-1 (or dm^6 mol^-2 s^-1). Part (c): [4 marks total] - Award 1 mark for stating that rate constant k varies with temperature. Award 1 mark for stating that higher temperature increases the average kinetic energy of particles. Award 1 mark for stating that a greater fraction of molecules have energy greater than or equal to the activation energy (E >= Ea). Award 1 mark for relating this to an increased frequency of successful collisions, leading to a higher k.

For (a): Ethane is a non-polar hydrocarbon, so its dominant intermolecular force is London dispersion forces. Fluoromethane has a polar C-F bond and is asymmetrical, so its dominant force is dipole-dipole attractions. Methanol contains a highly electronegative oxygen bonded to hydrogen (O-H), so its dominant intermolecular force is hydrogen bonding. For (b): Both ethane (Mr = 30.08 g mol-1) and fluoromethane (Mr = 34.04 g mol-1) have similar electron clouds and London dispersion forces. However, fluoromethane is polar, introducing permanent dipole-dipole attractions. These extra attractions require more thermal energy to overcome, resulting in a higher boiling point. For (c): Methanol is capable of forming intermolecular hydrogen bonds, which are much stronger than dipole-dipole or London dispersion forces. Breaking these strong hydrogen bonds requires a high amount of energy, which significantly elevates its boiling point compared to the other compounds. For (d): Methanol is the most soluble in water. Water is a highly polar solvent capable of hydrogen bonding. Methanol can form strong hydrogen bonds with water molecules via its hydroxyl (-OH) group, making it highly miscible.

Part (a): [3 marks total] - Award 1 mark for identifying London dispersion forces (or dispersion/van der Waals forces) for ethane. Award 1 mark for identifying dipole-dipole forces for fluoromethane. Award 1 mark for identifying hydrogen bonding for methanol. Part (b): [3 marks total] - Award 1 mark for stating that both have similar molar masses/electron counts (similar strength of dispersion forces). Award 1 mark for noting that fluoromethane is polar while ethane is non-polar. Award 1 mark for stating that dipole-dipole forces in fluoromethane require more energy to overcome than dispersion forces. Part (c): [3 marks total] - Award 1 mark for identifying that methanol undergoes hydrogen bonding. Award 1 mark for noting that hydrogen bonds are significantly stronger than dipole-dipole and dispersion forces. Award 1 mark for connecting stronger forces to the need for more energy to separate the molecules. Part (d): [2.66 marks total] - Award 1 mark for predicting methanol. Award 1.66 marks for explaining that methanol forms hydrogen bonds with water molecules, facilitating dissolution.

(a) For \(\text{SO}_4^{2-}\) with an expanded octet, the central sulfur atom has 6 valence electrons and is bonded to four oxygen atoms. To minimize formal charges, sulfur forms two double bonds (\(\text{S}=\text{O}\)) and two single bonds (\(\text{S}-\text{O}^{-}\)) with the oxygen atoms. Formal Charge (FC) of S = \(6 - 0 - 6 = 0\). FC of double-bonded O = \(6 - 4 - 2 = 0\). FC of single-bonded O = \(6 - 6 - 1 = -1\). For \(\text{SO}_2\), the central sulfur atom has one lone pair, one single bond to oxygen, and one double bond to oxygen (exhibiting resonance with a bond order of 1.5).

(b) \(\text{CO}_2\) consists of small, non-polar linear molecules. The intermolecular forces between \(\text{CO}_2\) molecules are weak London dispersion forces, requiring very little energy to overcome, resulting in a low melting point. \(\text{SiO}_2\) is a giant covalent/network structure where each silicon atom is covalently bonded to four oxygen atoms in a tetrahedral arrangement. Breaking this giant lattice requires a huge amount of thermal energy to break strong covalent bonds, leading to a high melting point. Neither substance conducts electricity because all valence electrons are localized within covalent bonds and there are no mobile ions or delocalized electrons.

(c) Hybridization is the mixing of atomic orbitals (such as s and p) on a central atom to produce a set of equivalent hybrid orbitals of intermediate energy. In ethene (\(\text{C}_2\text{H}_4\)), each carbon atom has three electron domains, giving \(sp^2\) hybridization. In ethyne (\(\text{C}_2\text{H}_2\)), each carbon atom has two electron domains, giving \(sp\) hybridization. \(sp\) hybrid orbitals have 50% s-character, whereas \(sp^2\) orbitals have 33.3% s-character. The greater s-character in \(sp\) hybrids means the bonding electrons are held closer to the carbon nucleus, making the carbon-carbon triple bond in ethyne stronger and shorter than the double bond in ethene.

(d) Both \(\text{NH}_3\) and \(\text{H}_2\text{O}\) have four electron domains around the central atom, giving them a tetrahedral electron domain geometry. \(\text{NH}_3\) has three bonding pairs and one lone pair, resulting in a trigonal pyramidal molecular geometry with a bond angle of approximately \(107^\circ\). \(\text{H}_2\text{O}\) has two bonding pairs and two lone pairs, resulting in a bent/V-shaped molecular geometry with a bond angle of approximately \(104.5^\circ\). According to VSEPR theory, the order of electron pair repulsion is lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. The presence of two lone pairs in water causes greater repulsion and compresses the bonding pairs more tightly than the single lone pair in ammonia, resulting in a smaller bond angle.

(a) [6 marks]
- 1 mark for correct Lewis structure of expanded octet \(\text{SO}_4^{2-}\) (S with 12 valence electrons, two double bonds, two single bonds, brackets with 2- charge).
- 1 mark for correct Lewis structure of \(\text{SO}_2\) (S with one lone pair, one double bond, one single bond or resonance structure equivalent).
- 1 mark for calculating correct formal charge on S (0).
- 1 mark for calculating correct formal charges on oxygen atoms (0 for double-bonded, -1 for single-bonded).
- 1 mark for stating that the sum of the formal charges equals the overall charge of the ion (-2).
- 1 mark for explaining that the expanded octet structure is preferred because it minimizes formal charges across the molecule.

(b) [6 marks]
- 1 mark for describing \(\text{CO}_2\) as a simple molecular structure with weak intermolecular (London dispersion) forces.
- 1 mark for describing \(\text{SiO}_2\) as a giant covalent network structure with strong covalent bonds throughout.
- 1 mark for linking \(\text{CO}_2\)'s structure to its low melting/sublimation point (requires little energy to overcome weak intermolecular forces).
- 1 mark for linking \(\text{SiO}_2\)'s structure to its high melting point (requires large amounts of energy to break strong covalent bonds in the giant lattice).
- 1 mark for stating that neither conducts electricity.
- 1 mark for explaining that both lack mobile charge carriers (no free-moving ions or delocalized electrons) because all valence electrons are localized in covalent bonds.

(c) [6 marks]
- 1 mark for defining hybridization (mixing of atomic orbitals of different energy to form new hybrid orbitals of equal energy).
- 1 mark for identifying ethene carbon as \(sp^2\) hybridized.
- 1 mark for identifying ethyne carbon as \(sp\) hybridized.
- 1 mark for stating ethyne has a triple bond while ethene has a double bond.
- 1 mark for explaining that \(sp\) orbitals have higher s-character than \(sp^2\) orbitals and hold electrons closer to the nucleus.
- 1 mark for concluding that the triple bond in ethyne is shorter and stronger than the double bond in ethene.

(d) [4.5 marks]
- 1 mark for identifying that both have 4 electron domains and tetrahedral domain geometry.
- 1 mark for identifying \(\text{NH}_3\) as trigonal pyramidal with a bond angle of \(107^\circ\) (accept \(106^\circ - 108^\circ\)).
- 1 mark for identifying \(\text{H}_2\text{O}\) as bent with a bond angle of \(104.5^\circ\) (accept \(104^\circ - 105^\circ\)).
- 1 mark for stating the relative repulsion strengths: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair.
- 0.5 marks for stating that the two lone pairs in water exert more repulsive force, compressing the H-O-H bond angle more than the single lone pair in ammonia.

(a) Moles of salicylic acid = \(\frac{5.00\text{ g}}{138.12\text{ g mol}^{-1}} = 0.0362\text{ mol}\).
Moles of acetic anhydride = \(\frac{6.00\text{ g}}{102.09\text{ g mol}^{-1}} = 0.0588\text{ mol}\).
Since the reaction stoichiometry is 1:1, salicylic acid is the limiting reactant (0.0362 mol < 0.0588 mol).
Theoretical moles of aspirin produced = 0.0362 mol.
Theoretical mass of aspirin = \(0.0362\text{ mol} \times 180.15\text{ g mol}^{-1} = 6.52\text{ g}\).

(b) Percentage yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{4.85\text{ g}}{6.52\text{ g}} \times 100 = 74.4\%\).
Reasons for yield < 100%: 1. Some aspirin remains dissolved in the cold solvent and does not crystallize out during filtration. 2. Loss of product during transfer between glassware (mechanical loss) or during the washing step.

(c) Atom economy is a measure of the proportion of starting materials that are converted into the useful/desired product. \(\text{Atom economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100 = \frac{180.15}{138.12 + 102.09} \times 100 = 75.0\%\). Difference: Percentage yield measures the practical efficiency of a chemical process (how much of the reactant actually converted to product), whereas atom economy measures the theoretical efficiency of the chemical reaction design (how many atoms are wasted as byproducts in the balanced equation).

(d) Principles demonstrated: 1. Use of catalytic reagents (acid catalyst) to speed up the reaction without being consumed, saving energy. 2. Relatively high atom economy (75%) meaning most atoms end up in the desired product. Improvements: 1. Substitute hazardous organic solvents or catalysts with more environmentally benign alternatives (e.g., solid-acid catalysts). 2. Recycle or find a secondary commercial use for the ethanoic acid byproduct instead of disposing of it as waste.

(e) Recrystallization procedure: 1. Dissolve the crude aspirin in the minimum volume of hot solvent (e.g., ethanol or water). 2. Filter the hot solution to remove any insoluble impurities. 3. Allow the hot filtrate to cool slowly to room temperature, then place it in an ice bath to maximize crystallization of pure aspirin. 4. Filter the crystals under vacuum using a Buchner funnel, wash with a small portion of ice-cold solvent, and dry. Chemical Principle: The desired product is highly soluble in the hot solvent but significantly less soluble in the cold solvent, causing it to crystallize out of solution upon cooling. Impurities remain in the solution because they are present in much smaller amounts and remain below their saturation limit.

(a) [6 marks]
- 1 mark for calculating moles of salicylic acid (0.0362 mol).
- 1 mark for calculating moles of acetic anhydride (0.0588 mol).
- 1 mark for identifying salicylic acid as the limiting reactant with a valid comparison.
- 1 mark for stating the mole ratio of reactant to product is 1:1, hence 0.0362 mol of aspirin is expected.
- 1 mark for using the correct molar mass of aspirin (180.15 g/mol).
- 1 mark for the correct final theoretical mass of 6.52 g (accept 6.51 - 6.53 g depending on intermediate rounding).

(b) [4.5 marks]
- 1.5 marks for calculating percentage yield: \(\frac{4.85}{6.52} \times 100 = 74.4\%\) (accept 74.3 - 74.5% based on previous value).
- 1.5 marks for first experimental reason (e.g., some product remains dissolved in cold solvent; mechanical transfer loss).
- 1.5 marks for second experimental reason (e.g., incomplete reaction; side reactions).

(c) [4 marks]
- 1 mark for definition of atom economy (molecular mass of desired product divided by sum of molecular masses of all reactants, expressed as a percentage).
- 1 mark for calculating atom economy: \(\frac{180.15}{240.21} \times 100 = 75.0\%\).
- 2 marks for explaining the difference: yield is a measure of reaction completeness/practical efficiency, while atom economy is a measure of theoretical synthetic design and inherent waste generation.

(d) [4 marks]
- 1 mark for each of two green chemistry principles demonstrated (e.g., use of catalyst; high atom economy; moderate temperature).
- 1 mark for each of two green chemistry improvements (e.g., using a greener solvent; recycling the ethanoic acid byproduct; using a less hazardous catalyst).

(e) [4 marks]
- 1 mark for dissolving crude solid in the minimum volume of hot solvent.
- 1 mark for cooling slowly to allow pure crystals to form.
- 1 mark for filtering under vacuum/washing with cold solvent and drying.
- 1 mark for explaining the solubility principle (the desired product crystallizes out upon cooling because its solubility decreases significantly, while impurities remain dissolved because of their low concentration).

(a) Hydrazine (\(\text{N}_2\text{H}_4\)) has a single bond between the two nitrogen atoms (\(\text{H}_2\text{N}-\text{NH}_2\)), with each N atom bonded to two H atoms and possessing one lone pair. Around each N atom, there are 4 electron domains, giving a tetrahedral domain geometry and a trigonal pyramidal molecular geometry with a bond angle of approximately \(107^\circ\). Dinitrogen difluoride (\(\text{N}_2\text{F}_2\)) has a double bond between the nitrogen atoms (\(\text{F}-\text{N}=\text{N}-\text{F}\)), with each N atom bonded to one F atom and possessing one lone pair. Around each N atom, there are 3 electron domains, giving a trigonal planar domain geometry and a bent molecular geometry with a bond angle of approximately \(120^\circ\).

(b) \(\text{N}_2\) (bp = \(77\text{ K}\)) is a non-polar homonuclear diatomic molecule. The only intermolecular forces acting between \(\text{N}_2\) molecules are weak London dispersion forces. \(\text{N}_2\text{F}_2\) (bp = \(162\text{ K}\)) contains polar \(\text{N}-\text{F}\) bonds. It exhibits permanent dipole-dipole attractions in addition to London dispersion forces, which require more thermal energy to break. Hydrazine (\(\text{N}_2\text{H}_4\), bp = \(387\text{ K}\)) has highly polar \(\text{N}-\text{H}\) bonds with nitrogen being highly electronegative. This allows hydrazine to form strong hydrogen bonds, which are the strongest type of intermolecular force and require the most thermal energy to overcome.

(c) Fluorine is more electronegative than oxygen, so the hydrogen bonds in \(\text{HF}\) are individually stronger than those in \(\text{H}_2\text{O}\). However, each water molecule can form an average of four hydrogen bonds because it has two hydrogen atoms and two lone pairs on the oxygen atom. In contrast, \(\text{HF}\) has only one hydrogen atom and three lone pairs per molecule, meaning it is limited to an average of only two hydrogen bonds per molecule. The extensive 3D hydrogen-bonded network in water requires significantly more total energy to disrupt than the chains in \(\text{HF}\), resulting in water's higher boiling point. [Diagram shows two water molecules with partial charges \(\delta^+\) on H and \(\delta^-\) on O, with a dotted line between a lone pair on O of one molecule and H of another].

(d) In graphite, each carbon atom is \(sp^2\) hybridized and covalently bonded to three other carbon atoms in hexagonal planar sheets. The fourth valence electron of each carbon atom occupies a p-orbital and is delocalized within the layers. These mobile delocalized electrons allow graphite to conduct electricity. The sheets are held together by weak London dispersion forces, allowing them to slide easily over one another, making graphite soft and a good lubricant. In diamond, each carbon atom is \(sp^3\) hybridized and covalently bonded tetrahedrally to four other carbon atoms in a giant 3D network. All valence electrons are localized in strong single covalent bonds, meaning there are no mobile electrons to conduct electricity (insulator). The rigid, highly directional 3D covalent framework makes diamond extremely hard.

(a) [6 marks]
- 1 mark for correct Lewis structure of \(\text{N}_2\text{H}_4\) (showing correct single bonds and lone pairs).
- 1 mark for correct Lewis structure of \(\text{N}_2\text{F}_2\) (showing double bond between N atoms and correct lone pairs).
- 1 mark for trigonal pyramidal molecular geometry for N in \(\text{N}_2\text{H}_4\).
- 1 mark for bent molecular geometry for N in \(\text{N}_2\text{F}_2\).
- 1 mark for bond angle of \(107^\circ\) (accept \(105^\circ - 109^\circ\)) in \(\text{N}_2\text{H}_4\).
- 1 mark for bond angle of \(120^\circ\) (accept \(115^\circ - 120^\circ\)) in \(\text{N}_2\text{F}_2\).

(b) [6.5 marks]
- 1 mark for identifying London dispersion forces as dominant in \(\text{N}_2\).
- 1 mark for identifying dipole-dipole forces as dominant in \(\text{N}_2\text{F}_2\) (due to polar \(\text{N}-\text{F}\) bonds and molecular asymmetry in the cis-isomer/overall).
- 1.5 marks for identifying hydrogen bonding in \(\text{N}_2\text{H}_4\).
- 1 mark for ranking the strength of intermolecular forces: hydrogen bonding > dipole-dipole > London dispersion forces.
- 1 mark for linking stronger intermolecular forces directly to the higher thermal energy required to overcome them (hence higher boiling points).
- 1 mark for correct correlation with the numerical values given.

(c) [5 marks]
- 1 mark for stating that fluorine is more electronegative than oxygen, making individual \(\text{H}-\text{F}\) hydrogen bonds stronger than \(\text{H}-\text{O}\) hydrogen bonds.
- 1 mark for explaining that \(\text{H}_2\text{O}\) can form an average of 4 hydrogen bonds per molecule because it has 2 H atoms and 2 lone pairs on O.
- 1 mark for explaining that \(\text{HF}\) is limited to an average of 2 hydrogen bonds per molecule due to having only 1 H atom.
- 1 mark for concluding that the greater number of hydrogen bonds in water creates a more extensive network requiring more energy to break.
- 1 mark for drawing a correct, labeled diagram showing H-bonding (dotted/dashed line, partial charges \(\delta^+\) on H and \(\delta^-\) on O, and lone pairs clearly shown on O).

(d) [5 marks]
- 1 mark for describing graphite as having \(sp^2\) hybridized carbons in layers with delocalized electrons.
- 1 mark for explaining that delocalized electrons are mobile and can carry charge, making graphite conductive.
- 1 mark for explaining that weak dispersion forces between layers allow them to slide over each other, making it a good lubricant.
- 1 mark for describing diamond as a giant covalent network with \(sp^3\) hybridized carbons in a tetrahedral 3D structure.
- 1 mark for stating that all electrons in diamond are localized in strong covalent bonds, meaning no mobile charge carriers (insulator) and high resistance to deformation (hardness).

(a) Moles of \(\text{NaHCO}_3\) = \(\frac{2.10\text{ g}}{84.01\text{ g mol}^{-1}} = 0.0250\text{ mol}\).
From the balanced chemical equation, the mole ratio of \(\text{NaHCO}_3\) to \(\text{Na}_2\text{CO}_3\) is 2:1.
Therefore, moles of \(\text{Na}_2\text{CO}_3\) produced = \(\frac{0.0250\text{ mol}}{2} = 0.0125\text{ mol}\).
Theoretical mass of the solid residue (\(\text{Na}_2\text{CO}_3\)) = \(0.0125\text{ mol} \times 105.99\text{ g mol}^{-1} = 1.32\text{ g}\).

(b) The ideal gas equation is \(PV = nRT\).
Rearranging for volume: \(V = \frac{nRT}{P}\).
Substituting the values into the equation:
\(V = \frac{0.0125\text{ mol} \times 8.314\text{ J K}^{-1}\text{ mol}^{-1} \times 298.15\text{ K}}{1.013 \times 10^5\text{ Pa}} = 3.059 \times 10^{-4}\text{ m}^3\).
To convert from \(\text{m}^3\) to \(\text{dm}^3\), multiply by 1000:
\(V = 3.059 \times 10^{-4} \times 1000 = 0.306\text{ dm}^3\).

(c) Three systematic errors that can cause a lower volume than expected:
1. Gas leaks in the connections or joints of the gas syringe setup (some gas escapes to the surroundings before being measured).
2. Collection of gas over water (carbon dioxide is moderately soluble in water and some gas dissolves, decreasing the volume of gas collected).
3. Incomplete decomposition of the sample (if the sample was not heated to a constant mass, less gas was generated than theoretical calculations assume).
4. High friction in the gas syringe plunger (the resistance forces the gas to remain compressed under a pressure slightly higher than atmospheric pressure, resulting in a lower measured volume).

(d) The two assumptions of the kinetic molecular theory that fail for real gases are:
1. The volume occupied by the gas particles is negligible compared to the total volume of the container.
2. There are no intermolecular attractive forces between the gas particles.
Carbon dioxide behaves most like an ideal gas under conditions of high temperature and low pressure. At high temperatures, the particles have high kinetic energy, which easily overcomes any weak intermolecular attractions between the molecules. At low pressures, the gas particles are spread far apart, meaning the volume occupied by the gas molecules themselves is negligible compared to the massive volume of the empty space between them.

(a) [5 marks]
- 1 mark for calculating moles of \(\text{NaHCO}_3\) (0.0250 mol).
- 1 mark for identifying the 2:1 molar ratio between \(\text{NaHCO}_3\) and \(\text{Na}_2\text{CO}_3\).
- 1 mark for calculating the moles of \(\text{Na}_2\text{CO}_3\) produced (0.0125 mol).
- 1 mark for multiplying the moles of residue by the molar mass of \(\text{Na}_2\text{CO}_3\) (105.99 g/mol).
- 1 mark for the correct final answer with units (1.32 g).

(b) [4.5 marks]
- 1 mark for stating the ideal gas equation: \(PV = nRT\).
- 1 mark for rearranging the equation to \(V = \frac{nRT}{P}\) and substituting correct values with consistent SI units.
- 1.5 marks for calculating the volume in \(\text{m}^3\) (\(3.06 \times 10^{-4}\text{ m}^3\)).
- 1 mark for converting the volume correctly to \(\text{dm}^3\) (\(0.306\text{ dm}^3\)).

(c) [6 marks]
- 1 mark for each of three valid systematic errors (e.g., gas leaks; gas dissolving in water; incomplete heating/reaction; friction in syringe).
- 1 mark for explaining each error's effect (e.g., leaks allow gas to escape to the atmosphere so less reaches the syringe; dissolution in water removes gas from the gas phase; incomplete reaction produces fewer moles of gas; friction creates resistance keeping the pressure inside slightly higher and volume lower).

(d) [7 marks]
- 1 mark for stating that kinetic molecular theory assumes gas particles have negligible volume.
- 1 mark for stating that it assumes no intermolecular forces.
- 1 mark for explaining that at high pressures/low temperatures, these assumptions fail because particles are closer together and move slower.
- 1 mark for identifying that \(\text{CO}_2\) behaves most like an ideal gas at high temperature.
- 1 mark for identifying that it behaves most like an ideal gas at low pressure.
- 1 mark for explaining that high temperature overcomes intermolecular attractions due to high kinetic energy.
- 1 mark for explaining that low pressure keeps particles far apart, making the volume of the particles negligible relative to the container.