2025 IB DP Chemistry 模擬試題連答案詳解

To form a transition metal ion, electrons are lost first from the \(4\text{s}\) subshell, followed by the \(3\text{d}\) subshell. Therefore, first-row transition metal ions generally do not have electrons in the \(4\text{s}\) subshell (i.e., they are \(4\text{s}^0\)). Let us evaluate the options:

- \([\text{Ar}] 4\text{s}^2 3\text{d}^1\) represents a neutral scandium (\(\text{Sc}\)) atom.
- \([\text{Ar}] 4\text{s}^2 3\text{d}^3\) represents a neutral vanadium (\(\text{V}\)) atom.
- \([\text{Ar}] 3\text{d}^5\) represents an iron(III) ion (\(\text{Fe}^{3+}\)), since a neutral \(\text{Fe}\) atom has the configuration \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). Losing three electrons (two from the \(4\text{s}\) subshell and one from the \(3\text{d}\) subshell) yields \([\text{Ar}] 3\text{d}^5\).
- \([\text{Ar}] 4\text{s}^1 3\text{d}^5\) represents a neutral chromium (\(\text{Cr}\)) atom.

Therefore, option C is correct.

[1 mark] for choosing C. Award 0 marks for any other option.

According to the combined gas law:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

Rearranging to solve for the final pressure \(P_2\):

\(P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}\)

Substitute the given values into the equation:

\(P_2 = \frac{100\text{ kPa} \times 2.00\text{ dm}^3 \times 450\text{ K}}{1.00\text{ dm}^3 \times 300\text{ K}}\)

\(P_2 = \frac{90000}{300} = 300\text{ kPa}\)

Therefore, the new pressure is \(300\text{ kPa}\).

[1 mark] for choosing B. Award 0 marks for any other option.

First, calculate the empirical formula mass of \(\text{C}_2\text{H}_4\text{O}\):

\(\text{Empirical mass} = (2 \times 12.01) + (4 \times 1.01) + 16.00 = 24.02 + 4.04 + 16.00 = 44.06\text{ g mol}^{-1}\)

Next, determine the molar mass (\(M_{\text{r}}\)) of the compound using the given mass and amount in moles:

\(M_{\text{r}} = \frac{\text{mass}}{\text{moles}} = \frac{8.8\text{ g}}{0.10\text{ mol}} = 88\text{ g mol}^{-1}\)

Determine the ratio of the molecular mass to the empirical formula mass:

\(\text{Ratio} = \frac{88\text{ g mol}^{-1}}{44.06\text{ g mol}^{-1}} \approx 2\)

Multiply the subscripts in the empirical formula by this ratio (2):

\(\text{Molecular formula} = (\text{C}_2\text{H}_4\text{O}) \times 2 = \text{C}_4\text{H}_8\text{O}_2\)

Therefore, the molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).

[1 mark] for choosing B. Award 0 marks for any other option.

A molecule's molecular geometry differs from its electron domain geometry if there is at least one non-bonding lone pair of electrons on the central atom.

- \(\text{CF}_4\): Carbon has 4 valence electrons, all bonded to fluorine atoms. It has 4 bonding electron domains, so both geometries are tetrahedral.
- \(\text{BF}_3\): Boron has 3 valence electrons, all bonded to fluorine atoms. It has 3 bonding electron domains, so both geometries are trigonal planar.
- \(\text{PF}_3\): Phosphorus has 5 valence electrons. Three are shared in covalent bonds with fluorine atoms, leaving one non-bonding lone pair. With 4 total electron domains (3 bonding, 1 non-bonding), its electron domain geometry is tetrahedral, but its molecular geometry is trigonal pyramidal. Because they are different, this is the correct answer.
- \(\text{SF}_6\): Sulfur has 6 valence electrons, all bonded to fluorine atoms. It has 6 bonding electron domains, so both geometries are octahedral.

[1 mark] for choosing C. Award 0 marks for any other option.

1. Find the total mass of the reaction mixture:
\(\text{Total volume} = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)
\(\text{Mass } (m) = 100.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 100.0\text{ g}\)

2. Calculate the heat energy released (\(q\)):
\(q = m \cdot c \cdot \Delta T\)
\(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.0\text{ K} = 2508\text{ J} = 2.508\text{ kJ}\)

3. Calculate the amount of moles of water formed during neutralization:
\(n = C \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)

4. Calculate the enthalpy change of neutralization (\(\Delta H_{\text{neu}}\)):
\(\Delta H_{\text{neu}} = -\frac{q}{n} = -\frac{2.508\text{ kJ}}{0.0500\text{ mol}} = -50.16\text{ kJ mol}^{-1} \approx -50.2\text{ kJ mol}^{-1}\)

Therefore, the enthalpy change of neutralization is \(-50.2\text{ kJ mol}^{-1}\).

[1 mark] for choosing B. Award 0 marks for any other option.

When the temperature of a gas is increased, the average kinetic energy of the particles increases. Consequently:

1. The peak of the Maxwell-Boltzmann curve (which represents the most probable kinetic energy) shifts to the right (higher energy).
2. Because the total number of particles remains constant, the total area under the curve must remain constant. To accommodate the wider distribution of energies at higher temperatures, the height of the peak must decrease (become lower).

Therefore, the peak shifts to the right and becomes lower.

[1 mark] for choosing B. Award 0 marks for any other option.

To identify the spontaneous reaction and the cell potential:

1. The half-cell with the more positive reduction potential undergoes reduction at the cathode. Therefore, copper(II) ions are reduced:
\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})\) (Cathode, \(E^\ominus = +0.34\text{ V}\))

2. The other half-cell undergoes oxidation at the anode. Therefore, iron metal is oxidized:
\(\text{Fe}(\text{s}) \rightarrow \text{Fe}^{2+}(\text{aq}) + 2\text{e}^-\) (Anode, \(E^\ominus = -0.44\text{ V}\))

3. Combining these half-reactions gives the spontaneous net cell reaction:
\(\text{Fe}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Fe}^{2+}(\text{aq}) + \text{Cu}(\text{s})\)

4. Calculate the standard cell potential:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}\)

Therefore, option A is correct.

[1 mark] for choosing A. Award 0 marks for any other option.

An alcohol is classified as tertiary (\(3^\circ\)) if the carbon atom bonded directly to the hydroxyl group (\(-\text{OH}\)) is itself bonded to three other carbon atoms.

- **2-methylbutan-2-ol**: The carbon skeleton has a 4-carbon chain (butane). At carbon-2, there is both a methyl group and a hydroxyl group. The C2 atom is bonded directly to three other carbon atoms (C1, C3, and the methyl group carbon). This is a tertiary alcohol.
- **3-methylbutan-2-ol**: The hydroxyl group is on carbon-2, which is bonded only to carbon-1 and carbon-3. This is a secondary alcohol.
- **2-methylbutan-1-ol**: The hydroxyl group is on carbon-1, which is bonded only to carbon-2. This is a primary alcohol.
- **pentan-3-ol**: The hydroxyl group is on carbon-3, which is bonded to carbon-2 and carbon-4. This is a secondary alcohol.

[1 mark] for choosing A. Award 0 marks for any other option.

Using the combined gas law: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\). Convert the temperatures to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). Rearrange the equation to solve for \(P_2\): \(P_2 = \frac{P_1 V_1 T_2}{V_2 T_1} = \frac{100\text{ kPa} \times 2.0\text{ dm}^3 \times 600\text{ K}}{1.0\text{ dm}^3 \times 300\text{ K}} = 400\text{ kPa}\).

Award 1 mark for the correct option C.

\(\text{PF}_3\) is trigonal pyramidal with a lone pair of electrons on the phosphorus atom, causing an asymmetric charge distribution that results in a net dipole moment. The other molecules (trigonal planar \(\text{BF}_3\), tetrahedral \(\text{CF}_4\), and octahedral \(\text{SF}_6\)) are highly symmetrical, so their individual bond dipoles cancel out completely.

Award 1 mark for the correct option B.

For a spontaneous cell reaction, the half-cell with the more positive standard reduction potential undergoes reduction at the cathode: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq})\) with \(E^\theta_{\text{red}} = +0.77\text{ V}\). The other half-cell undergoes oxidation at the anode: \(\text{Sn}^{2+}(\text{aq}) \rightarrow \text{Sn}^{4+}(\text{aq}) + 2\text{e}^-\) with \(E^\theta_{\text{ox}} = +0.15\text{ V}\). The overall cell potential is: \(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = +0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}\).

Award 1 mark for the correct option A.

Let the initial rate be \(\text{Rate}_1 = k[\text{A}]^2[\text{B}]\). Substituting the new concentrations into the rate expression gives: \(\text{Rate}_2 = k\left(\frac{1}{2}[\text{A}]\right)^2(2[\text{B}]) = k \cdot \frac{1}{4}[\text{A}]^2 \cdot 2[\text{B}] = \frac{1}{2}k[\text{A}]^2[\text{B}] = \frac{1}{2}\text{Rate}_1\). Therefore, the initial rate decreases by a factor of 2.

Award 1 mark for the correct option A.

The equilibrium constant expression is: \(K_c = \frac{[\text{Y}][\text{Z}]}{[\text{X}]^2}\). Substituting the known values gives: \(4.0 = \frac{2.0 \times 2.0}{[\text{X}]^2}\), which simplifies to \(4.0 = \frac{4.0}{[\text{X}]^2}\). Therefore, \([\text{X}]^2 = 1.0\), so \([\text{X}] = 1.0\text{ mol dm}^{-3}\).

Award 1 mark for the correct option A.

The atomic number of iron is 26, so a neutral iron atom has the configuration \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). Transition metals lose electrons from their outermost \(\text{s}\) orbital first when forming ions. Thus, forming the \(\text{Fe}^{2+}\) ion involves the loss of two \(4\text{s}\) electrons, leaving the configuration \([\text{Ar}] 3\text{d}^6\).

Award 1 mark for the correct option A.

First, calculate the total molar mass of urea: \(12.01 + 16.00 + (2 \times 14.01) + (4 \times 1.01) = 60.07\text{ g mol}^{-1}\). The mass contribution of nitrogen in one mole of urea is \(2 \times 14.01 = 28.02\text{ g}\). Thus, the percentage by mass of nitrogen is: \(\frac{28.02}{60.07} \times 100\% \approx 46.7\%\).

Award 1 mark for the correct option B.

The total volume is \(100.0\text{ cm}^3\), corresponding to a mass of \(100.0\text{ g}\). The heat energy released is: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.8424\text{ kJ}\). The amount of acid or base reacting is: \(n = c \times V = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). The enthalpy of neutralization is: \(\Delta H = -\frac{q}{n} = -\frac{2.8424\text{ kJ}}{0.0500\text{ mol}} = -56.8\text{ kJ mol}^{-1}\), which rounds to \(-57\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A.

First, identify the element using the proton number. An atomic number of 34 (34 protons) corresponds to selenium (\(\text{Se}\)). Next, calculate the mass number as the sum of protons and neutrons: \(34 + 45 = 79\), giving the isotope symbol \(^{79}_{34}\text{Se}\). Finally, determine the charge of its most stable monoatomic ion: selenium is in Group 16 of the periodic table and has 6 valence electrons, meaning it gains 2 electrons to complete its valence octet, forming a \(2-\)\ ion (\(\text{Se}^{2-}\)). Therefore, the correct complete symbol is \(^{79}_{34}\text{Se}^{2-}\).

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

Manganese (\(\text{Mn}\)) has an atomic number of 25, with a ground-state neutral configuration of \([\text{Ar}] 4\text{s}^2 3\text{d}^5\). When transition metals form cations, they lose electrons from the outer \(4\text{s}\) subshell first. Thus, losing two electrons to form the \(\text{Mn}^{2+}\) ion removes the two \(4\text{s}\) electrons, leaving the configuration \([\text{Ar}] 3\text{d}^5\). In contrast, \(\text{Fe}^{2+}\) is \([\text{Ar}] 3\text{d}^6\), \(\text{Cr}\) is \([\text{Ar}] 4\text{s}^1 3\text{d}^5\), and \(\text{V}^{2+}\) is \([\text{Ar}] 3\text{d}^3\).

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

Lattice enthalpy depends on the charges of the ions and their ionic radii, in accordance with Coulomb's law. Magnesium oxide (\(\text{MgO}\)) consists of \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) ions, which have a high charge product of 4. Although calcium oxide (\(\text{CaO}\)) also has a charge product of 4, the ionic radius of the \(\text{Mg}^{2+}\) ion is smaller than that of the \(\text{Ca}^{2+}\) ion because it has fewer electron shells. Therefore, the interionic distance in \(\text{MgO}\) is smaller, resulting in stronger electrostatic forces of attraction and a more exothermic lattice enthalpy than \(\text{NaCl}\), \(\text{MgF}_2\), or \(\text{CaO}\).

Award [1] for the correct choice C. Award [0] for any other choice.

In \(\text{XeF}_4\), the central xenon atom has 8 valence electrons. It forms 4 covalent bonds with fluorine atoms and retains 2 lone pairs, resulting in 6 electron domains (octahedral domain geometry). To minimize electron-pair repulsion, the lone pairs occupy opposite axial positions, resulting in a square planar molecular geometry. \(\text{CF}_4\) and \(\text{BF}_4^-\) are tetrahedral (4 bonding pairs, 0 lone pairs), while \(\text{SF}_4\) is see-saw shaped (4 bonding pairs, 1 lone pair).

Award [1] for the correct choice D. Award [0] for any other choice.

Across Period 3, the number of protons in the nucleus increases (increasing the nuclear charge) while the shielding from inner electron shells remains relatively constant because electrons are added to the same main energy level (\(n=3\)). This increases the effective nuclear charge felt by outer valence electrons, pulling them closer to the nucleus (decreasing the atomic radius) and holding them more tightly, which generally increases the energy required to remove the first electron.

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

For a redox reaction to be spontaneous under standard conditions, the cell potential must be positive (\(E^\theta_{\text{cell}} > 0\)). The cell potential is calculated using \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). For the reaction in option B, iron is oxidized to \(\text{Fe}^{2+}\) (\(E^\theta_{\text{oxidation}} = -0.44\text{ V}\)) and copper(II) ions are reduced to copper (\(E^\theta_{\text{reduction}} = +0.34\text{ V}\)). This gives \(E^\theta_{\text{cell}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}\), which is positive and therefore spontaneous. All other options yield negative cell potentials.

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

1. Find mass of the solution: \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). 2. Find temperature change: \(\Delta T = 26.8^\circ\text{C} - 20.0^\circ\text{C} = 6.8\text{ K}\). 3. Calculate heat energy released: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 2842.4\text{ J} = 2.8424\text{ kJ}\). 4. Calculate moles of reactant: \(n(\text{HCl}) = n(\text{NaOH}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). 5. Calculate enthalpy of neutralization per mole of water formed: \(\Delta H = -\frac{q}{n} = -\frac{2.8424\text{ kJ}}{0.0500\text{ mol}} = -56.8\text{ kJ mol}^{-1}\). The value is negative because the temperature rose, indicating an exothermic process.

Award [1] for the correct choice C. Award [0] for any other choice.

Assuming a \(100.0\text{ g}\) sample of the oxide: 1. Calculate the mass of each element: \(70.0\text{ g}\) of \(\text{Fe}\) and \(30.0\text{ g}\) of \(\text{O}\). 2. Determine the moles of each element: \(n(\text{Fe}) = \frac{70.0\text{ g}}{55.85\text{ g mol}^{-1}} \approx 1.253\text{ mol}\) and \(n(\text{O}) = \frac{30.0\text{ g}}{16.00\text{ g mol}^{-1}} = 1.875\text{ mol}\). 3. Find the simplest whole-number molar ratio: divide each by the smallest number of moles: \(\text{Fe} = \frac{1.253}{1.253} = 1\) and \(\text{O} = \frac{1.875}{1.253} \approx 1.5\). Multiplying both by 2 to convert to integers gives a ratio of \(2:3\). The empirical formula is therefore \(\text{Fe}_2\text{O}_3\).

Award [1] for the correct choice D. Award [0] for any other choice.

First, determine the number of protons in the ion. Since the ion has a \( 2- \) charge and contains 18 electrons:
\( \text{Number of protons} = \text{Number of electrons} + \text{charge} = 18 + (-2) = 16 \).

Next, use the mass number to find the number of neutrons:
\( \text{Mass number} = \text{protons} + \text{neutrons} \)
\( 34 = 16 + \text{neutrons} \)
\( \text{Neutrons} = 34 - 16 = 18 \).

[1 mark] for correct identification of 18 neutrons. Award 0 marks for any other response.

Use the combined gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).
Convert temperatures from Celsius to Kelvin:
\( T_1 = 27 + 273 = 300 \text{ K} \)
\( T_2 = 127 + 273 = 400 \text{ K} \)

Rearrange to solve for \( V_2 \):
\( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} \)
\( V_2 = 2.0 \text{ dm}^3 \times \frac{100 \text{ kPa}}{200 \text{ kPa}} \times \frac{400 \text{ K}}{300 \text{ K}} \)
\( V_2 = 2.0 \times 0.5 \times \frac{4}{3} = 1.33 \text{ dm}^3 \).

[1 mark] for the correct volume of 1.33.

\( \text{NH}_3 \) (ammonia) has three bonding pairs and one lone pair around the central nitrogen atom. Due to the extra repulsion from the lone pair, the bond angles are reduced from the tetrahedral angle of \( 109.5^\circ \) to approximately \( 107^\circ \).
- \( \text{BF}_3 \) is trigonal planar (\( 120^\circ \)).
- \( \text{CH}_4 \) is tetrahedral (\( 109.5^\circ \)).
- \( \text{H}_2\text{O} \) is bent with two lone pairs (\( 104.5^\circ \)).

[1 mark] for NH3 as the correct answer. Other options incorrect.

A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. This allows a greater fraction of collisions to have energy greater than or equal to the activation energy, increasing the rate of successful collisions. It does not alter the average kinetic energy of the molecules (which is only changed by temperature), does not significantly change collision frequency, and does not affect the position of equilibrium.

[1 mark] for identifying that it provides an alternative pathway with lower activation energy.

The heat change of the surroundings is calculated using \( q = m c \Delta T \).
- \( m = 50.0 \text{ g} \)
- \( c = 4.18 \text{ J g}^{-1} \text{K}^{-1} \)
- \( \Delta T = T_{\text{final}} - T_{\text{initial}} = 15.0^\circ\text{C} - 20.0^\circ\text{C} = -5.0^\circ\text{C} \) (or \( -5.0 \text{ K} \)).

Therefore, the correct expression is \( 50.0 \times 4.18 \times (-5.0) \).

[1 mark] for the correct calculation expression with the correct negative temperature change.

Let the oxidation state of sulfur be \( x \). The oxidation state of oxygen is typically \( -2 \). The sum of oxidation states in the polyatomic ion must equal its overall charge of \( -2 \):
\( 2(x) + 3(-2) = -2 \)
\( 2x - 6 = -2 \)
\( 2x = +4 \)
\( x = +2 \).

[1 mark] for correct calculation of +2.

(a)
First, calculate the temperature change:
\(\Delta T = 31.8^\circ\text{C} - 21.2^\circ\text{C} = 10.6^\circ\text{C} = 10.6\text{ K}\)

Next, calculate the heat energy absorbed:
\(q = m \cdot c \cdot \Delta T\)
\(q = 50.00\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.6\text{ K} = 2215.4\text{ J}\)
\(q = 2.2154\text{ kJ} \approx 2.22\text{ kJ}\)

(b)
Calculate the amount (in moles) of \(\text{CaCl}_2\):
\(n(\text{CaCl}_2) = \frac{3.00\text{ g}}{110.98\text{ g mol}^{-1}} = 0.02703\text{ mol}\)

Calculate the enthalpy change of solution:
\(\Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{2.2154\text{ kJ}}{0.02703\text{ mol}} = -81.96\text{ kJ mol}^{-1}\)

To the appropriate number of significant figures (3 significant figures based on the temperature change and the mass of calcium chloride):
\(\Delta H_{\text{sol}} = -82.0\text{ kJ mol}^{-1}\)

(c)
Absolute uncertainty in \(\Delta T\):
\(\Delta T = (31.8 \pm 0.1) - (21.2 \pm 0.1) = 10.6 \pm 0.2^\circ\text{C}\)

Percentage uncertainty in \(\Delta T\):
\(\frac{0.2^\circ\text{C}}{10.6^\circ\text{C}} \times 100\% = 1.89\% \approx 1.9\%\)

(d)
Systematic error: Heat loss to the surroundings / calorimeter absorbing heat.
Effect on \(\Delta H_{\text{sol}}\): The maximum temperature reached is lower than theoretical, resulting in a smaller temperature change (\(\Delta T\)). Thus, the calculated \(\Delta H_{\text{sol}}\) value is less exothermic (smaller absolute value).

(a) [2 marks]
- \(\Delta T = 10.6\text{ K}\) [1 mark]
- \(q = 2.22\text{ kJ}\) (accept \(2.215\text{ kJ}\) or \(2.22\text{ kJ}\); award full marks for correct final value) [1 mark]

(b) [3 marks]
- \(n(\text{CaCl}_2) = 0.0270\text{ mol}\) [1 mark]
- Negative sign included and correct calculation of \(\Delta H_{\text{sol}} = -82.0\text{ kJ mol}^{-1}\) [1 mark]
- Expressing the final answer to 3 significant figures [1 mark]

(c) [2 marks]
- Correct absolute uncertainty of \(\pm 0.2^\circ\text{C}\) [1 mark]
- Correct calculation of percentage uncertainty to \(1.9\%\) (accept \(1.89\%\)) [1 mark]

(d) [1.33 marks]
- Identification of a valid systematic error (e.g., heat loss to surroundings / heat absorbed by the beaker) AND explanation that this lowers \(\Delta T\), making \(\Delta H_{\text{sol}}\) less exothermic / less negative [1.33 marks].

(a) The volume remains constant because the reaction has reached completion. Magnesium is the limiting reactant and has been completely consumed.

(b)
Average rate = \(\frac{\text{Change in volume}}{\text{Change in time}} = \frac{34.0\text{ cm}^3 - 0.0\text{ cm}^3}{30\text{ s} - 0\text{ s}} = 1.13\text{ cm}^3\text{ s}^{-1}\) (or \(1.1\text{ cm}^3\text{ s}^{-1}\)).

(c)
To find the instantaneous rate at \(t = 20\text{ s}\):
1. Draw a tangent line to the curve at the point where \(t = 20\text{ s}\).
2. Calculate the gradient (slope) of the tangent line using \(\frac{\Delta y}{\Delta x}\).

(d)
Using magnesium powder increases the total surface area of the reactant. This increases the frequency of successful collisions, resulting in a faster initial rate of reaction.
1. The curve will be steeper at the start (higher initial gradient).
2. The curve will reach the plateau (horizontal line) sooner than \(50\text{ s}\).
3. It will level off at the exact same volume (\(42.0\text{ cm}^3\)) because the mass of magnesium is identical, producing the same maximum quantity of hydrogen gas.

(a) [1.33 marks]
- Magnesium has been completely consumed / limiting reactant used up / reaction completed [1.33 marks].

(b) [2 marks]
- Calculation: \(\frac{34.0}{30} = 1.13\text{ cm}^3\text{ s}^{-1}\) (accept \(1.1\text{ cm}^3\text{ s}^{-1}\)) [1 mark]
- Correct unit: \(\text{cm}^3\text{ s}^{-1}\) or \(\text{cm}^3 / \text{s}\) [1 mark]

(c) [2 marks]
- Draw a tangent to the curve at \(t = 20\text{ s}\) [1 mark]
- Calculate the gradient/slope of the tangent [1 mark]

(d) [3 marks]
- Steeper initial curve / steeper initial slope [1 mark]
- Reaches completion / horizontal plateau earlier [1 mark]
- Levels off at the same final volume (\(42.0\text{ cm}^3\)) [1 mark]

(a)
- Mass of anhydrous \(\text{CuSO}_4 = 25.95\text{ g} - 24.35\text{ g} = 1.60\text{ g}\)
- Mass of water lost = \(26.85\text{ g} - 25.95\text{ g} = 0.90\text{ g}\)

(b)
- Moles of anhydrous \(\text{CuSO}_4 = \frac{1.60\text{ g}}{159.62\text{ g mol}^{-1}} = 0.01002\text{ mol}\)
- Moles of water lost = \frac{0.90\text{ g}}{18.02\text{ g mol}^{-1}} = 0.04994\text{ mol}\)
- Ratio \(x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.04994}{0.01002} = 4.98 \approx 5\)
- Formula: \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)

(c)
Heating to constant mass involves repeatedly heating, cooling, and weighing the sample until two consecutive mass measurements are identical. This guarantees that all water of crystallization has been driven off, ensuring a complete dehydration reaction.

(d)
Decomposition of \(\text{CuSO}_4\) releases gaseous sulfur trioxide (\(\text{SO}_3\)) which escapes the crucible. This extra loss of mass would be incorrectly calculated as extra lost water. Therefore, the calculated mass of water lost would be higher than actual, resulting in an overestimated (higher) value of \(x\).

(a) [2 marks]
- Mass of anhydrous salt = \(1.60\text{ g}\) [1 mark]
- Mass of water lost = \(0.90\text{ g}\) [1 mark]

(b) [3.33 marks]
- \(n(\text{CuSO}_4) = 0.0100\text{ mol}\) AND \(n(\text{H}_2\text{O}) = 0.0500\text{ mol}\) [1.33 marks]
- Mole ratio calculation: \(x = \frac{0.0500}{0.0100} = 5\) [1 mark]
- Correct chemical formula: \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) [1 mark]

(c) [1.5 marks]
- Explanation: Ensures all water of crystallization is fully evaporated/removed so that the mass of anhydrous salt is accurate [1.5 marks].

(d) [1.5 marks]
- Calculated \(x\) is too high/overestimated [0.5 marks]
- Explanation: Gas (\(\text{SO}_3\)) escapes, increasing the apparent mass of water lost, which increases the calculated moles of water [1.0 mark].

(a) (i)
\( M_r(\text{AgCl}) = 107.87 + 35.45 = 143.32\text{ g mol}^{-1} \)
\( n(\text{AgCl}) = \frac{6.963\text{ g}}{143.32\text{ g mol}^{-1}} = 0.04858\text{ mol} \)

(ii)
Since each mole of \( \text{MCl}_2 \) produces two moles of \( \text{AgCl} \):
\( n(\text{MCl}_2) = \frac{0.04858}{2} = 0.02429\text{ mol} \)

\( M_r(\text{MCl}_2) = \frac{3.148\text{ g}}{0.02429\text{ mol}} = 129.6\text{ g mol}^{-1} \)

\( M_r(\text{M}) = 129.6 - 2(35.45) = 58.7\text{ g mol}^{-1} \)

The metal \( \text{M} \) is nickel (\( \text{Ni} \)) (molar mass \( 58.69\text{ g mol}^{-1} \)).

(b)
Mass of anhydrous \( \text{NiCl}_2 \) remaining = \( 5.000\text{ g} - 2.271\text{ g} = 2.729\text{ g} \)
\( n(\text{NiCl}_2) = \frac{2.729\text{ g}}{129.59\text{ g mol}^{-1}} = 0.02106\text{ mol} \)

Mass of water lost = \( 2.271\text{ g} \)
\( n(\text{H}_2\text{O}) = \frac{2.271\text{ g}}{18.02\text{ g mol}^{-1}} = 0.1260\text{ mol} \)

Ratio \( x = \frac{n(\text{H}_2\text{O})}{n(\text{NiCl}_2)} = \frac{0.1260}{0.02106} = 5.98 \approx 6 \).

Therefore, \( x = 6 \).

(a) (i)
Award [1] for calculating molar mass of AgCl as \( 143.32\text{ g mol}^{-1} \).
Award [1] for calculating moles of AgCl as \( 0.04858\text{ mol} \) (accept 0.0486 mol).

(ii)
Award [1] for finding moles of \( \text{MCl}_2 = 0.02429\text{ mol} \).
Award [1] for calculating molar mass of metal as \( 58.7\text{ g mol}^{-1} \) (allow 58.6 - 59.0).
Award [1] for identifying the metal as Nickel/Ni (or Cobalt/Co).

(b)
Award [1] for calculating mass of anhydrous salt (\( 2.729\text{ g} \)) AND its moles (\( 0.02106\text{ mol} \)).
Award [1] for calculating moles of water (\( 0.1260\text{ mol} \)).
Award [1.33] for calculating the ratio \( 5.98 \) and rounding to the nearest whole integer (\( x = 6 \)).

(a) Standard enthalpy change of solution (\( \Delta H^\theta_{\text{sol}} \)) is the enthalpy change when one mole of an ionic solute dissolves in excess solvent to form an infinitely dilute solution under standard conditions.

(b) \( q = m c \Delta T \)
\( m = 100.0\text{ g} \) (since density of water is \( 1.00\text{ g cm}^{-3} \))
\( \Delta T = 29.1 - 21.3 = 7.8\\ ^\circ\text{C} \)
\( q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\\ ^\circ\text{C}^{-1} \times 7.8\\ ^\circ\text{C} = 3260.4\text{ J} = 3.26\text{ kJ} \) (or \( 3.3\text{ kJ} \))

(c) Moles of \( \text{CaCl}_2 = \frac{4.25\text{ g}}{110.98\text{ g mol}^{-1}} = 0.0383\text{ mol} \)
Since the temperature of the water increased, the reaction is exothermic, so \( \Delta H \) is negative.
\( \Delta H_{\text{sol}} = -\frac{q}{n} = -\frac{3.26\text{ kJ}}{0.0383\text{ mol}} = -85.1\text{ kJ mol}^{-1} \) (accept range \( -84.5 \) to \( -85.2 \) depending on rounding of \( q \)).

(d) Assumption: No heat loss to the surroundings / calorimeter has zero heat capacity.
Effect: Heat loss results in a smaller measured temperature rise (\( \Delta T \)). Consequently, the calculated heat energy (\( q \)) is lower, making the calculated \( \Delta H_{\text{sol}} \) less exothermic (less negative/more positive) than the actual value.

(a)
Award [1.33] for a complete definition mentioning \"one mole of solute\", \"excess solvent/infinite dilution\", and \"standard conditions\". Award [1] if one of these is missing.

(b)
Award [1] for calculating \( \Delta T = 7.8\\ ^\circ\text{C} \).
Award [1] for \( q = 3.26\text{ kJ} \) (or \( 3.3\text{ kJ} \)).

(c)
Award [1] for calculating moles of \( \text{CaCl}_2 = 0.0383\text{ mol} \).
Award [1] for dividing \( q \) by moles.
Award [1] for negative sign and value \( -85.1\text{ kJ mol}^{-1} \) (accept range \( -84.5 \) to \( -85.2 \)).

(d)
Award [1] for identifying assumption (e.g., \"no heat loss to the surroundings\" or \"heat capacity of the cup is negligible\").
Award [1] for explaining the effect (e.g., \"makes \( \Delta H \) less negative/less exothermic because the recorded temperature change is smaller than it should be\").

(a) The \( \text{Fe}^{3+}/\text{Fe}^{2+} \) half-reaction has the more positive potential, so reduction occurs here:
\( 2\text{Fe}^{3+}(\text{aq}) + 2\text{e}^- \rightarrow 2\text{Fe}^{2+}(\text{aq}) \)

Oxidation occurs at the \( \text{Sn}^{4+}/\text{Sn}^{2+} \) half-cell:
\( \text{Sn}^{2+}(\text{aq}) \rightarrow \text{Sn}^{4+}(\text{aq}) + 2\text{e}^- \)

Overall reaction:
\( 2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq}) \)

\( E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V} \)

(b) The reducing agent is \( \text{Sn}^{2+}(\text{aq}) \) (or tin(II) ions) because it donates electrons and is oxidized.

(c) Acidified dichromate(VI) oxidizes \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \):
Oxidation: \( 6\text{Fe}^{2+}(\text{aq}) \rightarrow 6\text{Fe}^{3+}(\text{aq}) + 6\text{e}^- \)
Reduction: \( \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) \)

Overall balanced ionic equation:
\( \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) + 6\text{Fe}^{3+}(\text{aq}) \)

\( E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = 1.33\text{ V} - 0.77\text{ V} = +0.56\text{ V} \)

(d) The color change is from orange (due to \( \text{Cr}_2\text{O}_7^{2-} \)) to green (due to \( \text{Cr}^{3+} \)).

(a)
Award [1] for the correct overall equation.
Award [1] for correct state symbols in the equation.
Award [1] for \( E^\theta_{\text{cell}} = +0.62\text{ V} \).

(b)
Award [1] for \( \text{Sn}^{2+}(\text{aq}) \) (do not accept Sn or \( \text{Sn}^{4+} \)).

(c)
Award [1] for writing the correct oxidation half-equation or recognizing the 1:6 ratio.
Award [1.33] for the fully balanced overall ionic equation including state symbols.
Award [1] for \( E^\theta_{\text{cell}} = +0.56\text{ V} \).

(d)
Award [1] for \"orange to green\" (both colors required in correct order).

(a)
- Lewis structure of \( \text{CO}_2 \): Oxygen is double-bonded to carbon on both sides. Carbon has no lone pairs, and each oxygen has two lone pairs: \( \text{:O=C=O:} \)

- Lewis structures of \( \text{CO}_3^{2-} \): Carbon has three electron domains. Resonance structures must show one double bond \( \text{C=O} \) and two single bonds \( \text{C-O} \) with formal negative charges on the singly bonded oxygens. Three resonance structures should be shown separated by double-headed arrows, or a resonance hybrid drawn with dashed lines:
Resonance structures: \( [\text{O=C(O^-)_2}] \leftrightarrow [\text{^-O-C(=O)-O^-}] \leftrightarrow [\text{(^-O)_2C=O}] \).

(b)
- In methanol, \( \text{CH}_3\text{OH} \), the carbon-oxygen bond is a single covalent bond with a bond order of 1.
- In the carbonate ion, \( \text{CO}_3^{2-} \), the pi electrons are delocalized over three carbon-oxygen bonds due to resonance, giving a bond order of \( 1.33 \) (or \( 4/3 \)).
- In carbon dioxide, \( \text{CO}_2 \), there are two localized double bonds, giving a bond order of 2.

Since a higher bond order results in a stronger attraction between the nuclei and bonding electrons, the bond length decreases as bond order increases.
Order of decreasing bond length (longest to shortest): \( \text{CH}_3\text{OH} > \text{CO}_3^{2-} > \text{CO}_2 \).

(c)
(i) \( \text{CO}_2 \): molecular geometry is **linear**, bond angle is **\( 180^\circ \)**.
(ii) \( \text{CO}_3^{2-} \): molecular geometry is **trigonal planar**, bond angle is **\( 120^\circ \)**.

(a)
Award [1] for correct Lewis structure of \( \text{CO}_2 \) including all lone pairs.
Award [2] for correct Lewis structures of \( \text{CO}_3^{2-} \) showing either three resonance structures with double-headed arrows and overall charge, or a correct resonance hybrid with fractional bonds and delocalized charge.

(b)
Award [1] for stating the correct order of decreasing bond length: \( \text{CH}_3\text{OH} > \text{CO}_3^{2-} > \text{CO}_2 \).
Award [1] for stating correct bond orders (\( \text{CH}_3\text{OH} = 1 \), \( \text{CO}_3^{2-} = 1.33 \), \( \text{CO}_2 = 2 \)).
Award [1.33] for explaining that higher bond order means greater electron density between nuclei, stronger electrostatic attraction, and thus shorter bonds.

(c)
(i) Award [1] for linear AND \( 180^\circ \).
(ii) Award [1] for trigonal planar AND \( 120^\circ \).

(a) When an atom absorbs energy, electrons are promoted/excited from lower energy levels (ground state) to higher energy levels (excited states). These excited states are unstable, so the electrons cascade/fall back down to lower energy levels. As they relax, they emit energy in the form of electromagnetic radiation (photons). Because energy levels are quantized/discrete, the transitions involve specific amounts of energy (\( \Delta E = h\nu \)), resulting in lines of specific frequencies/wavelengths in the emission spectrum.

(b) (i) The transition is from \( n = \infty \rightarrow n = 1 \).

(ii) Transitions in the Lyman series all end at the ground state, \( n = 1 \). Since the energy gap between \( n = 1 \) and any higher level is very large, the emitted photons have high energy and high frequency, falling into the ultraviolet region. In contrast, the Balmer series transitions end at \( n = 2 \). The energy gaps to \( n = 2 \) are much smaller, corresponding to lower-energy, lower-frequency radiation in the visible spectrum.

(c)
(i) \( \text{Cu}: 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1 \) (Note: The anomalous configuration is due to the extra stability of a fully-filled 3d subshell).

(ii) \( \text{Cu}^{2+}: 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9 \) (Note: Transition metals lose their outer s electrons first, before losing d electrons).

(a)
Award [1] for mentioning absorption of energy/promotion of electrons to higher energy levels.
Award [1] for explaining that electrons fall back to lower levels, emitting photons/electromagnetic radiation.
Award [1] for connecting discrete energy levels (quantization) to specific lines/wavelengths/frequencies of light.

(b)
(i) Award [1] for \( n = \infty \rightarrow n = 1 \) (accept \"from infinity to one\").
(ii) Award [1] for stating that Lyman series ends at \( n=1 \) and Balmer at \( n=2 \).
Award [1] for explaining that the energy difference to \( n=1 \) is much larger than to \( n=2 \), resulting in higher energy/frequency UV radiation versus visible light.

(c)
(i) Award [1.33] for the full configuration: \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1 \). Do not accept abbreviated [Ar] notation as full configuration was requested. Do not award marks for \( 3\text{d}^9 4\text{s}^2 \).
(ii) Award [1] for \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9 \).

(a) Isomers of \( \text{C}_4\text{H}_{10}\text{O} \) belonging to different homologous series can be an alcohol and an ether.
- Isomer 1: Butan-1-ol, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \). Functional group: hydroxyl group / alcohol.
- Isomer 2: Ethoxyethane, \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \). Functional group: ether.
(Other combinations like butan-2-ol and methoxypropane are acceptable).

(b)
(i) Since both belong to the same homologous series (alcohols) and **A** is oxidized to a carboxylic acid (which turns litmus red), **A** must be a primary alcohol. Therefore, **A** is butan-1-ol (structure: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \)) or 2-methylpropan-1-ol.
Since **B** is resistant to oxidation, **B** must be a tertiary alcohol. Therefore, **B** is 2-methylpropan-2-ol (structure: \( (\text{CH}_3)_3\text{COH} \)).

(ii) **Y** is the product of full oxidation of butan-1-ol (Isomer **A**), which is butanoic acid (structure: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \). The functional group is the carboxyl group (or carboxylic acid).

(c) Butan-1-ol has a much higher boiling point than ethoxyethane. Butan-1-ol molecules can form strong intermolecular hydrogen bonds due to the presence of the polar \( \text{-O-H} \) group. Ethoxyethane molecules cannot form hydrogen bonds with each other (as they lack an \( \text{H} \) atom bonded directly to a highly electronegative \( \text{O} \) atom) and only experience weaker dipole-dipole and London dispersion forces, which require less energy to overcome.

(a)
Award [1] each for two correct structural formulas (one alcohol, one ether) [2 max].
Award [1] each for correct corresponding IUPAC names and functional group identification [2 max].

(b)
(i)
Award [1] for deducing the structure/name of **A** as butan-1-ol (or 2-methylpropan-1-ol).
Award [1] for deducing the structure/name of **B** as 2-methylpropan-2-ol.

(ii)
Award [1] for the IUPAC name: butanoic acid (or 2-methylpropanoic acid depending on A).
Award [0.33] for functional group: carboxyl group / carboxylic acid.

(c)
Award [1] for explaining that butan-1-ol has hydrogen bonding (which is stronger), whereas ethoxyethane only has dipole-dipole/dispersion forces (which are weaker and require less energy to break).