2025 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

Element X is silicon (\text{Si}). Silicon forms a giant covalent lattice with strong covalent bonds throughout, giving it the highest melting point in Period 3. Element Y has a very high third ionisation energy compared to its second, indicating that the third electron is removed from a shell closer to the nucleus (an inner shell). This means Y has two valence electrons, placing it in Group 2. The Group 2 element in Period 3 is magnesium (\text{Mg}).

1 mark for identifying both silicon as the highest melting point element and magnesium as the element with the large increase between the second and third ionisation energies.

According to Hess's Law, the enthalpy change of formation of propane, represented by the equation 3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}, can be calculated from the enthalpies of combustion of the reactants and products: \Delta H_{\text{f}}^{\ominus} = 3 \times \Delta H_{\text{c}}^{\ominus}[\text{C(graphite)}] + 4 \times \Delta H_{\text{c}}^{\ominus}[\text{H}_2\text{(g)}] - \Delta H_{\text{c}}^{\ominus}[\text{C}_3\text{H}_8\text{(g)}] = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106 \text{ kJ mol}^{-1}.

1 mark for the correct calculation showing -106 kJ mol^{-1}.

We use the ideal gas equation: pV = nRT = (m / M_{\text{r}})RT. Rearranging for M_{\text{r}} gives: M_{\text{r}} = mRT / (pV). Convert units to SI: m = 0.130 g, p = 101 \times 10^3 Pa, V = 45.4 \times 10^{-6} m^3, T = 100 + 273 = 373 K. Thus, M_{\text{r}} = (0.130 \times 8.31 \times 373) / (101 \times 10^3 \times 45.4 \times 10^{-6}) = 402.95 / 4.5854 = 87.9, which rounds to 88.

1 mark for the correct conversion of all quantities to SI units and calculation of M_r as 88.

4-chlorohex-2-ene has two stereogenic centers: 1. The C=C double bond at carbon-2 has different groups attached to each carbon atom (carbon-2 is bonded to -H and -CH3; carbon-3 is bonded to -H and -CH(Cl)CH2CH3), meaning it exhibits cis-trans (E/Z) isomerism. 2. Carbon-4 is a chiral center because it is bonded to four different groups: -H, -Cl, -CH2CH3, and -CH=CHCH3. The total number of stereoisomers is 2^n, where n is the number of stereogenic centers. Here, n = 2, so 2^2 = 4 stereoisomers exist (cis-R, cis-S, trans-R, trans-S).

1 mark for identifying both the double bond and the chiral center, leading to 4 stereoisomers.

The rate of hydrolysis of halogenoalkanes is determined by the strength (bond enthalpy) of the carbon-halogen bond, not by the bond polarity. Down Group 17, the bond length increases and the bond enthalpy decreases: C–Cl > C–Br > C–I. Since the C–I bond is the weakest, it is broken most easily, resulting in 1-iodobutane undergoing nucleophilic substitution at the fastest rate.

1 mark for identifying that bond enthalpy determines the rate and that the C-I bond has the lowest bond enthalpy.

Reduction is defined as a decrease in oxidation number. In KMnO4, potassium (K) is +1 and oxygen (O) is -2. Thus, the oxidation number of manganese (Mn) is +7. In MnSO4, because the sulfate ion (SO4^{2-}) has a charge of -2, the oxidation number of manganese (Mn) is +2. The oxidation number of Mn decreases from +7 to +2, which is reduction. Meanwhile, carbon in H2C2O4 is oxidized from +3 to +4 in CO2.

1 mark for identifying that manganese is reduced and its oxidation number changes from +7 to +2.

The forward reaction is exothermic (\Delta H < 0). According to Le Chatelier's principle, decreasing the temperature shifts the position of equilibrium to the right to release heat, thereby increasing the yield of SO3(g). Because the equilibrium constant K_c is only affected by temperature, and this temperature decrease shifts the equilibrium to the right, the concentration of products increases relative to the reactants, resulting in an increased value of K_c. Increasing pressure also increases the yield of SO3(g), but it does not change the value of K_c.

1 mark for identifying that decreasing the temperature increases both the yield of products and the value of K_c.

Let us determine the electronic configuration of each ion: 1. Cr atoms have configuration [Ar] 3d^5 4s^1. Loss of three electrons gives Cr^{3+} as [Ar] 3d^3. 2. Fe atoms have configuration [Ar] 3d^6 4s^2. Loss of three electrons (two from the 4s orbital and one from the 3d orbital) gives Fe^{3+} as [Ar] 3d^5. 3. Co atoms have configuration [Ar] 3d^7 4s^2. Loss of three electrons gives Co^{3+} as [Ar] 3d^6. 4. Ni atoms have configuration [Ar] 3d^8 4s^2. Loss of two electrons gives Ni^{2+} as [Ar] 3d^8. Therefore, Fe^{3+} is the correct ion.

1 mark for identifying Fe^{3+} as having the [Ar] 3d^5 configuration.

To calculate the relative atomic mass (\(A_r\)) of the element, we take the weighted average of the relative isotopic masses:

\(A_r(M) = \frac{(46.00 \times 8.00) + (48.00 \times 74.00) + (50.00 \times 18.00)}{100}\)

\(A_r(M) = \frac{368 + 3552 + 900}{100}\)

\(A_r(M) = \frac{4820}{100} = 48.20\)

1 mark for the correct calculation of the weighted average leading to 48.20 (accuracy mark).

Using the ideal gas equation: \(pV = nRT = \frac{m}{M_r}RT\)

Rearranging to solve for \(M_r\):
\(M_r = \frac{mRT}{pV}\)

Convert units to SI units:
- Temperature, \(T = 97 + 273 = 370\text{ K}\)
- Volume, \(V = 85.0 \times 10^{-6}\text{ m}^3\)
- Pressure, \(p = 1.01 \times 10^5\text{ Pa}\)
- Mass, \(m = 0.282\text{ g}\)

Substitute the values:
\(M_r = \frac{0.282 \times 8.31 \times 370}{1.01 \times 10^5 \times 85.0 \times 10^{-6}}\)

\(M_r = \frac{867.09}{8.585} \approx 101\text{ g mol}^{-1}\)

1 mark for the correct conversion of units and calculation of molar mass as 101.

1. Calculate heat energy released, \(q\):
\(q = mc\Delta T\)
- Total mass of solution, \(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\)
- Temperature change, \(\Delta T = 28.2 - 21.5 = 6.7\text{ K}\)
\(q = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J} = 2.8006\text{ kJ}\)

2. Calculate moles of water formed:
\(n(\text{H}_2\text{O}) = n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\)

3. Calculate enthalpy change of neutralization per mole of water formed:
\(\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.8006\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\)

1 mark for calculating the heat change using total mass (100 g) and dividing by moles of water formed (0.0500 mol) with a negative sign indicating an exothermic reaction.

By systematically placing three chlorine atoms on a three-carbon chain, we find the following structural isomers:
1. 1,1,1-trichloropropane (\(\text{CH}_3\text{CH}_2\text{CCl}_3\))
2. 1,1,2-trichloropropane (\(\text{CH}_3\text{CHClCHCl}_2\))
3. 1,1,3-trichloropropane (\(\text{CH}_2\text{ClCH}_2\text{CHCl}_2\))
4. 1,2,2-trichloropropane (\(\text{CH}_3\text{CCl}_2\text{CH}_2\text{Cl}\))
5. 1,2,3-trichloropropane (\(\text{CH}_2\text{ClCHClCH}_2\text{Cl}\))

No other structural variations exist (e.g., 2,2,3- is identical to 1,2,2-). Thus, there are exactly 5 structural isomers.

1 mark for systematically identifying all 5 unique structural isomers of trichloropropane.

Down Group 2, the cationic radius of the \(M^{2+}\) metal ion increases while the charge remains constant. This results in a lower charge density and weaker polarizing power of the cation. Consequently, the electron cloud of the nitrate anion is polarized (distorted) to a lesser extent, making the nitrate ion more stable to thermal decomposition (higher temperatures are required to decompose it).

1 mark for identifying that stability increases down the group and linking it to the decreasing charge density/polarizing power of the cation.

1. Write the reduction half-equation for dichromate(VI):
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)

2. Write the oxidation half-equation for hydrogen peroxide:
\(\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-\)

3. Multiply the oxidation half-equation by 3 to balance the electrons transferred (6 electrons):
\(3\text{H}_2\text{O}_2 \rightarrow 3\text{O}_2 + 6\text{H}^+ + 6\text{e}^-\)

4. Combine and simplify the overall ionic equation:
\(\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{O}_2 \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{O}_2\)

Thus, the mole ratio of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{H}_2\text{O}_2\) is \(1 : 3\).

1 mark for balancing the redox electron transfer and finding the correct stoichiometry of 1 : 3.

1. Compound \(P\) has an \(\text{O-H}\) group (broad peak at \(3200\text{–}3600\text{ cm}^{-1}\)) and no \(\text{C=O}\) group (no peak at \(1640\text{–}1750\text{ cm}^{-1}\)). Therefore, \(P\) is an alcohol.
2. Compound \(Q\) has a \(\text{C=O}\) group (peak at \(1640\text{–}1750\text{ cm}^{-1}\)) but has no alcohol \(\text{O-H}\) or carboxylic acid \(\text{O-H}\) remaining (no absorption in either \(3200\text{–}3600\text{ cm}^{-1}\) or \(2500\text{–}3000\text{ cm}^{-1}\)). This indicates \(Q\) is a ketone.
3. Since \(Q\) is a ketone and is formed by oxidation, \(P\) must have been a secondary alcohol.

1 mark for analyzing the IR absorption peaks to conclude that P is a secondary alcohol and Q is a ketone.

1. Ground state configuration of cobalt atom (atomic number 27): \([\text{Ar}] 3\text{d}^7 4\text{s}^2\).
2. When forming the \(\text{Co}^{2+}\) ion, two electrons are removed from the highest energy occupied principal quantum shell first, which is the \(4\text{s}\) subshell. Thus, \(\text{Co}^{2+}\) has the configuration \([\text{Ar}] 3\text{d}^7\).
3. By Hund's rule, the 7 electrons in the five \(3\text{d}\) orbitals pair up as follows: five orbitals get one electron each (unpaired), and then two of those orbitals receive a second electron (paired). This leaves exactly 3 unpaired electrons.

1 mark for correctly identifying that 4s electrons are removed first and applying Hund's rule to find 3 unpaired electrons in the 3d subshell.

Let the percentage abundance of \( ^{24}X \) be \( a \) and the percentage abundance of \( ^{26}X \) be \( b \).
Since the sum of the abundances of all isotopes is \( 100\% \):
\( a + 10.0 + b = 100 \implies a = 90.0 - b \)

Using the formula for relative atomic mass (\( A_r \)):
\( A_r = \frac{24a + 25(10.0) + 26b}{100} = 24.32 \)

Substitute \( a = 90.0 - b \) into the equation:
\( 24(90.0 - b) + 250 + 26b = 2432 \)
\( 2160 - 24b + 250 + 26b = 2432 \)
\( 2410 + 2b = 2432 \)
\( 2b = 22 \implies b = 11.0\% \)

Thus, the percentage abundance of \( ^{26}X \) is \( 11.0\% \).

1 mark for the correct calculation showing the relative abundance of \( ^{26}X \) is \( 11.0\% \).

Using the ideal gas equation: \( pV = nRT = \frac{m}{M_r} RT \)
Rearranging for \( M_r \):
\( M_r = \frac{mRT}{pV} \)

Convert all quantities to SI units:
- Temperature, \( T = 100 + 273 = 373\text{ K} \)
- Volume, \( V = 83.1\text{ cm}^3 = 83.1 \times 10^{-6}\text{ m}^3 \)
- Mass, \( m = 0.236\text{ g} \)
- Pressure, \( p = 1.00 \times 10^5\text{ Pa} \)

Substitute the values:
\( M_r = \frac{0.236 \times 8.31 \times 373}{1.00 \times 10^5 \times 83.1 \times 10^{-6}} \)
\( M_r = \frac{0.236 \times 8.31 \times 373}{8.31} \)
\( M_r = 0.236 \times 373 = 88.0\text{ g mol}^{-1} \)

1 mark for the correct calculation showing \( M_r = 88.0 \).

The equation for the formation of propane is:
\( 3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)} \)

Using Hess's Law with standard enthalpy changes of combustion:
\( \Delta H^\theta_f = \sum \Delta H^\theta_c (\text{reactants}) - \sum \Delta H^\theta_c (\text{products}) \)
\( \Delta H^\theta_f = 3 \times \Delta H^\theta_c [\text{C(graphite)}] + 4 \times \Delta H^\theta_c [\text{H}_2\text{(g)}] - \Delta H^\theta_c [\text{C}_3\text{H}_8\text{(g)}] \)
\( \Delta H^\theta_f = 3(-394) + 4(-286) - (-2220) \)
\( \Delta H^\theta_f = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1} \)

1 mark for the correct cycle application and final value of \( -106\text{ kJ mol}^{-1} \).

First, find the equilibrium moles of all species:
- Initial moles: \( n(\text{N}_2) = 2.00 \), \( n(\text{H}_2) = 6.00 \), \( n(\text{NH}_3) = 0 \)
- Equilibrium moles of \( \text{NH}_3 \) is \( 2.00 \), meaning a change of \( +2.00\text{ mol} \).
- According to stoichiometry, change in \( n(\text{N}_2) = -1.00\text{ mol} \) and change in \( n(\text{H}_2) = -3.00\text{ mol} \).
- Equilibrium moles: \( n(\text{N}_2) = 2.00 - 1.00 = 1.00\text{ mol} \), \( n(\text{H}_2) = 6.00 - 3.00 = 3.00\text{ mol} \).

Since the volume is \( 1.00\text{ dm}^3 \), concentrations equal moles:
\( [\text{N}_2] = 1.00\text{ mol dm}^{-3} \)
\( [\text{H}_2] = 3.00\text{ mol dm}^{-3} \)
\( [\text{NH}_3] = 2.00\text{ mol dm}^{-3} \)

\( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{2.00^2}{1.00 \times 3.00^3} = \frac{4.00}{27.0} \approx 0.148\text{ dm}^6\text{ mol}^{-2} \)

1 mark for calculating correct equilibrium concentrations and substituting into \( K_c \) expression to yield \( 0.148\text{ dm}^6\text{ mol}^{-2} \).

As you descend Group 2, the solubility of sulfates decreases. This is because both hydration enthalpy (\( \Delta H_\text{hyd} \)) and lattice energy (\( \Delta H_\text{lat} \)) become less exothermic, but \( \Delta H_\text{hyd} \) of the cation decreases much more rapidly than the lattice energy due to the large size of the sulfate anion. Hence, the dissolution process becomes less energetically favorable down the group.

1 mark for identifying the correct trend and scientific explanation.

Esters with the formula \( \text{C}_4\text{H}_8\text{O}_2 \) can be systematically listed:
1. Propyl methanoate (\( \text{HCOOCH}_2\text{CH}_2\text{CH}_3 \))
2. Isopropyl methanoate (\( \text{HCOOCH(CH}_3)_2 \))
3. Ethyl ethanoate (\( \text{CH}_3\text{COOCH}_2\text{CH}_3 \))
4. Methyl propanoate (\( \text{CH}_3\text{CH}_2\text{COOCH}_3 \))

None of these isomers contain a chiral carbon atom, so there are no optical isomers. Therefore, there are exactly 4 isomeric esters in total.

1 mark for the correct number of isomeric esters (4).

The rate of hydrolysis depends on:
1. The strength of the C–Halogen bond: C–I is the weakest bond, so iodoalkanes hydrolyze the fastest.
2. The class of halogenoalkane: tertiary halogenoalkanes react much faster than secondary or primary because they proceed via the stable tertiary carbocation in the \( \text{S}_\text{N}1 \) mechanism.

Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) is hydrolyzed the fastest and produces a precipitate the most rapidly.

1 mark for identifying that 2-iodo-2-methylpropane reacts fastest.

A strong, broad absorption band in the region \( 2500\text{--}3000\text{ cm}^{-1} \) is characteristic of the O–H stretching vibration in carboxylic acids (alcohols have a very broad O–H stretch in the range \( 3200\text{--}3600\text{ cm}^{-1} \)).
A strong, sharp absorption band at approximately \( 1710\text{ cm}^{-1} \) is characteristic of the C=O stretching vibration in carbonyl groups.
Combining both of these features indicates the presence of a carboxylic acid.

1 mark for identifying carboxylic acid as the correct functional group.

Since the empirical formula is \(\text{CH}_2\text{Cl}\) (formula mass = 49.5) and the molecular ion peaks are around 98–102, the molecular formula must be \(\text{C}_2\text{H}_4\text{Cl}_2\). This means each molecule contains exactly two chlorine atoms.

The probability of each isotopic combination is:
- Both \(^{35}\text{Cl}\) atoms (giving \(m/z = 98\)): \(0.75 \times 0.75 = 0.5625\)
- One \(^{35}\text{Cl}\) and one \(^{37}\text{Cl}\) (giving \(m/z = 100\)): \(2 \times 0.75 \times 0.25 = 0.3750\)
- Both \(^{37}\text{Cl}\) atoms (giving \(m/z = 102\)): \(0.25 \times 0.25 = 0.0625\)

Dividing these probabilities by the lowest value (0.0625) gives the ratio:
\(0.5625 / 0.0625 = 9\)
\(0.3750 / 0.0625 = 6\)
\(0.0625 / 0.0625 = 1\)

Thus, the peak ratio is 9 : 6 : 1.

1 mark for calculating the correct probability ratio for the three peak heights based on a binomial distribution for two chlorine atoms and choosing option C.

Using the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT}\)

First, convert all values into standard SI units:
- \(p = 1.00 \times 10^5\text{ Pa}\)
- \(V = 77.5\text{ cm}^3 = 77.5 \times 10^{-6}\text{ m}^3\)
- \(T = 100^\circ\text{C} = 100 + 273 = 373\text{ K}\)

Now, calculate the number of moles, \(n\):
\(n = \frac{(1.00 \times 10^5\text{ Pa}) \times (77.5 \times 10^{-6}\text{ m}^3)}{(8.31\text{ J K}^{-1}\text{ mol}^{-1}) \times (373\text{ K})} = \frac{7.75}{3099.63} \approx 0.00250\text{ mol}\)

Finally, calculate the relative molecular mass, \(M_r\):
\(M_r = \frac{\text{mass}}{n} = \frac{0.185\text{ g}}{0.00250\text{ mol}} = 74.0\)

1 mark for correct unit conversions, calculation of moles using pV = nRT, and final calculation of Mr to select option C.

The equation representing the standard enthalpy change of formation of dimethyl ether is:
\(2\text{C (s)} + 3\text{H}_2\text{ (g)} + \frac{1}{2}\text{O}_2\text{ (g)} \to \text{CH}_3\text{OCH}_3\text{ (g)}\)

By Hess's Law, utilizing the standard enthalpies of combustion:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus (\text{reactants}) - \sum \Delta H_c^\ominus (\text{products})\)

Substitute the values from the table:
\(\Delta H_f^\ominus = [2 \times \Delta H_c^\ominus(\text{C}) + 3 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{CH}_3\text{OCH}_3)\)
\(\Delta H_f^\ominus = [2(-394) + 3(-286)] - (-1460)\)
\(\Delta H_f^\ominus = [-788 - 858] + 1460\)
\(\Delta H_f^\ominus = -1646 + 1460 = -186\text{ kJ mol}^{-1}\)

1 mark for establishing the stoichiometric coefficients in the formation equation, constructing the Hess's Law calculation, and obtaining -186 kJ/mol.

All three elements exist as simple molecular structures held together by weak instantaneous dipole-induced dipole (London dispersion) forces. The strength of these intermolecular forces depends on the number of electrons per molecule.
- \(\text{S}_8\) has \(8 \times 16 = 128\) electrons.
- \(\text{P}_4\) has \(4 \times 15 = 60\) electrons.
- \(\text{Cl}_2\) has \(2 \times 17 = 34\) electrons.

Since \(\text{S}_8\) has the largest number of electrons, its intermolecular forces are the strongest, giving it the highest melting point. Since \(\text{Cl}_2\) has the fewest electrons, it has the weakest intermolecular forces and the lowest melting point. Therefore, the order of decreasing melting points is \(\text{S}_8 > \text{P}_4 > \text{Cl}_2\).

1 mark for relating the melting point order to the strength of van der Waals forces which depend on the number of electrons in the molecules.

Reaction of Group 2 hydroxides with dilute sulfuric acid yields the corresponding metal sulfate and water:
\(\text{M(OH)}_2\text{ (s)} + \text{H}_2\text{SO}_4\text{ (aq)} \to \text{MSO}_4 + 2\text{H}_2\text{O}\)

The solubility of Group 2 sulfates decreases down the group:
- \(\text{MgSO}_4\) is highly soluble, so it forms a clear, colourless solution.
- \(\text{CaSO}_4\) is sparingly soluble, forming a precipitate / cloudy mixture.
- \(\text{BaSO}_4\) is highly insoluble, forming a thick white precipitate.

Therefore, only the test tube containing magnesium hydroxide will form a clear, colourless solution.

1 mark for recalling the trend of decreasing solubility of Group 2 sulfates down the group to identify that only magnesium sulfate is soluble.

First, let's list all possible structural isomers of \(\text{C}_4\text{H}_8\text{Cl}_2\) with a butane (linear) skeleton and check for chirality:
1. 1,1-dichlorobutane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHCl}_2\) (no chiral carbon)
2. 1,2-dichlorobutane: \(\text{CH}_3\text{CH}_2\text{*CH(Cl)CH}_2\text{Cl}\) (carbon 2 is chiral)
3. 1,3-dichlorobutane: \(\text{CH}_3\text{*CH(Cl)CH}_2\text{CH}_2\text{Cl}\) (carbon 3 is chiral)
4. 1,4-dichlorobutane: \(\text{ClCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\) (no chiral carbon)
5. 2,2-dichlorobutane: \(\text{CH}_3\text{CH}_2\text{CCl}_2\text{CH}_3\) (no chiral carbon)
6. 2,3-dichlorobutane: \(\text{CH}_3\text{*CH(Cl)*CH(Cl)CH}_3\) (carbons 2 and 3 are chiral)

Now, let's look at the methylpropane (branched) skeleton:
7. 1,1-dichloro-2-methylpropane: \(\text{(CH}_3)_2\text{CHCHCl}_2\) (no chiral carbon)
8. 1,2-dichloro-2-methylpropane: \(\text{(CH}_3)_2\text{C(Cl)CH}_2\text{Cl}\) (no chiral carbon)
9. 1,3-dichloro-2-methylpropane: \(\text{CH}_3\text{CH(CH}_2\text{Cl})_2\) (no chiral carbon)

Only structural isomers 2, 3, and 6 contain at least one chiral carbon atom. Therefore, there are exactly 3 structural isomers containing a chiral carbon atom.

1 mark for systematically analyzing the structural isomers of C4H8Cl2, identifying which ones contain a chiral carbon atom, and correctly choosing 3 (option B).

The rate of hydrolysis of halogenoalkanes depends on:
1. The strength of the carbon-halogen bond: The \(\text{C-I}\) bond is the weakest (lowest bond enthalpy) compared to \(\text{C-Br}\) and \(\text{C-Cl}\), meaning iodoalkanes react faster.
2. The classification of the halogenoalkane: Tertiary halogenoalkanes hydrolyse much faster than primary halogenoalkanes because they react via the \(S_N1\) mechanism, which goes through a relatively stable tertiary carbocation intermediate.

Thus, 2-iodo-2-methylpropane (a tertiary iodoalkane) will hydrolyse the fastest and produce a silver iodide precipitate in the shortest time.

1 mark for identifying that both the weakest C-I bond and the tertiary SN1 mechanism make 2-iodo-2-methylpropane react fastest.

Analysis of the IR spectrum of the product:
- A very broad absorption band in the range \(2500-3000\text{ cm}^{-1}\) is characteristic of the \(\text{O-H}\) stretch of a carboxylic acid.
- A strong, sharp peak at \(1710\text{ cm}^{-1}\) is characteristic of the \(\text{C=O}\) stretch of a carbonyl group.

Thus, the product formed is a carboxylic acid. Heating a primary alcohol under reflux with excess acidified potassium dichromate(VI) results in complete oxidation to a carboxylic acid. Since propan-1-ol is a primary alcohol, it will be oxidised to propanoic acid, which produces this spectrum.

Propan-2-ol (a secondary alcohol) would oxidise to propanone (a ketone), which lacks the carboxylic acid \(\text{O-H}\) stretch. Propanone and ethyl methanoate do not oxidise to carboxylic acids under these conditions.

1 mark for identifying that the IR peaks correspond to a carboxylic acid, and that a primary alcohol (propan-1-ol) is required to form a carboxylic acid under reflux.

The ground-state electron configuration of a phosphorus atom (Z = 15) is \(1s^2 2s^2 2p^6 3s^2 3p^3\). The three electrons in the 3p subshell occupy separate orbitals singly (according to Hund's rule), giving 3 unpaired electrons. The ground-state electron configuration of a cobalt atom (Z = 27) is \([Ar] 3d^7 4s^2\). To form the \(Co^{2+}\) ion, the two 4s electrons are lost, leaving the configuration as \([Ar] 3d^7\). In the 3d subshell, the five d-orbitals are occupied by 7 electrons: two orbitals are doubly occupied, and three orbitals are singly occupied, resulting in 3 unpaired electrons. Thus, both species have 3 unpaired electrons.

Award 1 mark for the correct selection of option B.

Using the ideal gas equation: \(pV = nRT\). Convert all quantities to SI units: \(p = 1.01 \times 10^5\ \text{Pa}\), \(V = 62.2\ \text{cm}^3 = 62.2 \times 10^{-6}\ \text{m}^3\), and \(T = 100 + 273 = 373\ \text{K}\). Calculate the number of moles, \(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 62.2 \times 10^{-6}}{8.31 \times 373} \approx 0.002027\ \text{mol}\). Now calculate the relative molecular mass: \(M_r = \frac{\text{mass}}{n} = \frac{0.150\ \text{g}}{0.002027\ \text{mol}} \approx 74.0\).

Award 1 mark for the correct calculation and selection of option C.

The equation for the formation of methanol is: \(C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)\). Using Hess's law and the enthalpy changes of combustion: \(\Delta H_f^\ominus = [\Delta H_c^\ominus(C) + 2 \times \Delta H_c^\ominus(H_2)] - \Delta H_c^\ominus(CH_3OH) = [-393.5 + 2(-285.8)] - (-726.0) = -965.1 + 726.0 = -239.1\ \text{kJ\ mol}^{-1}\).

Award 1 mark for the correct application of Hess's Law and calculation leading to option A.

We determine the stoichiometry by balancing the electron transfer. The oxidation half-equation is \(Sn^{2+} \rightarrow Sn^{4+} + 2e^-\). The reduction half-equation is \(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\). To balance the electron transfer, we multiply the oxidation half-equation by 3, giving 6 electrons lost. The overall ionic equation shows 1 mole of \(Cr_2O_7^{2-}\) reacting with 3 moles of \(Sn^{2+}\).

Award 1 mark for the correct balanced redox stoichiometry leading to option C.

Magnesium chloride, \(MgCl_2\) (where X is magnesium), dissolves in water to form a slightly acidic solution with a pH of around 6.5 due to the slight polarising power of the hydrated \(Mg^{2+}\) ion. Silicon tetrachloride, \(SiCl_4\) (where Y is silicon), reacts vigorously with water (hydrolyses) to produce silicon dioxide and hydrogen chloride gas, resulting in a strongly acidic solution of pH 2. Sodium chloride (X = sodium) forms a completely neutral solution of pH 7, which does not match the pH of 6.5.

Award 1 mark for correctly identifying the behaviors of Period 3 chlorides in water and selecting option B.

Magnesium carbonate decomposes on heating to produce magnesium oxide and carbon dioxide gas, \(CO_2\), which turns limewater cloudy. Magnesium nitrate decomposes on heating to produce magnesium oxide, oxygen gas, and nitrogen dioxide gas, \(NO_2\), which is a brown gas. Since both observations (limewater turning cloudy and brown gas) are made, a mixture of magnesium carbonate and magnesium nitrate is consistent with these results.

Award 1 mark for identifying the decomposition products of Group 2 carbonates and nitrates, leading to option A.

The structure of 4-chlorohex-2-ene is \(CH_3-CH=CH-CH(Cl)-CH_2-CH_3\). First, the carbon-carbon double bond has different groups attached to each carbon of the double bond (\(-H\) and \(-CH_3\) on carbon-2; \(-H\) and \(-CH(Cl)CH_2CH_3\) on carbon-3), allowing for cis/trans (geometric) isomerism (2 configurations). Second, the carbon atom at position 4 is a chiral centre because it is bonded to four different groups: \(-H\), \(-Cl\), \(-CH_2CH_3\), and \(-CH=CHCH_3\), allowing for optical isomerism (2 configurations). The total number of stereoisomers is \(2^1 \times 2^1 = 4\) (cis-R, cis-S, trans-R, and trans-S).

Award 1 mark for identifying both geometric and optical isomerism components and calculating 4 stereoisomers (Option C).

The strong absorption peak at \(1715\ \text{cm}^{-1}\) indicates the presence of a carbonyl group (\(C=O\)) in product Y. The lack of any broad absorption peaks in the ranges \(2500-3000\ \text{cm}^{-1}\) (carboxylic acid \(O-H\)) and \(3200-3600\ \text{cm}^{-1}\) (alcohol \(O-H\)) indicates that Y is a ketone and that no unreacted alcohol remains. Ketones are produced by the oxidation of secondary alcohols. Propan-2-ol is a secondary alcohol and is oxidised to propanone (a ketone). Propan-1-ol and propanal are oxidised to propanoic acid (showing a carboxylic acid \(O-H\) peak). 2-Methylpropan-2-ol is a tertiary alcohol and does not oxidise.

Award 1 mark for analyzing the IR data to identify a ketone product and matching it to a secondary alcohol reactant (Option B).

(a)(i) Sodium reacts vigorously with cold water to form sodium hydroxide and hydrogen gas: \( 2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g) \). The resulting solution is strongly alkaline, hence pH is 13-14.
(ii) Magnesium reacts extremely slowly with cold water, but reacts rapidly when heated with steam: \( \text{Mg}(s) + \text{H}_2\text{O}(g) \rightarrow \text{MgO}(s) + \text{H}_2(g) \). Observation: burns with a bright white flame to produce a white solid.

(b)(i) Sodium oxide is a basic oxide that reacts readily with water to produce a colorless solution of sodium hydroxide: \( \text{Na}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) \). The resulting pH is 13-14.
(ii) Aluminium oxide is amphoteric. It does not react with or dissolve in water because of its high lattice energy, which arises from the strong electrostatic attraction between highly charged \( \text{Al}^{3+} \) and \( \text{O}^{2-} \) ions.
(iii) Sulfur trioxide is an acidic oxide that reacts violently with water to form sulfuric acid: \( \text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq) \). This forms a strongly acidic solution with a pH of 1-2.

(c)(i) Phosphorus trichloride reacts rapidly/violently with water, producing misty white fumes of hydrogen chloride gas and a strongly acidic solution containing phosphorous acid: \( \text{PCl}_3(l) + 3\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{PO}_3(aq) + 3\text{HCl}(aq) \). The pH of the solution is 1-2.
(ii) This reaction is classified as hydrolysis because water molecules break down the covalent bonds in \( \text{PCl}_3 \).

(a)(i) [1] mark for the correct balanced equation: \( 2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 \) (state symbols not required but must be correct if used). [1] mark for stating pH is 13 or 14.
(ii) [1] mark for correct description (heating magnesium in steam/hot water) and stating observation (bright white light/flame or white ash/solid formed). [1] mark for correct balanced equation: \( \text{Mg} + \text{H}_2\text{O} \rightarrow \text{MgO} + \text{H}_2 \).

(b)(i) [1] mark for correct balanced equation: \( \text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \) and observation (dissolves to form a colorless solution). [1] mark for stating pH is 13 or 14.
(ii) [1] mark for explaining that it does not dissolve due to high lattice energy / very strong ionic bonds. [1] mark for stating that it is amphoteric.
(iii) [1] mark for correct balanced equation: \( \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \). [1] mark for stating pH is 1 or 2.

(c)(i) [1] mark for describing the reaction (misty fumes/heat evolved). [1] mark for correct balanced equation: \( \text{PCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_3 + 3\text{HCl} \). [1] mark for stating pH is 1 or 2.
(ii) [1.5] marks for stating 'hydrolysis' (accept 'nucleophilic substitution' only if hydrolysis is also mentioned; otherwise reject).

(a) The expression for the gas equilibrium constant \( K_p \) is:
\[ K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times p(\text{H}_2)^2} \]
Units: \( \frac{\text{kPa}}{\text{kPa} \times \text{kPa}^2} = \text{kPa}^{-2} \) (or \( \text{Pa}^{-2} \)).

(b)(i) Write down the ICE (Initial, Change, Equilibrium) table:
- Initial moles: \( n(\text{CO}) = 1.00 \), \( n(\text{H}_2) = 2.00 \), \( n(\text{CH}_3\text{OH}) = 0 \).
- Equilibrium mole of \( \text{CH}_3\text{OH} = 0.40 \text{ mol} \), so change \( = +0.40 \text{ mol} \).
- Moles of \( \text{CO} \) reacted = \( 0.40 \text{ mol} \). Equilibrium \( n(\text{CO}) = 1.00 - 0.40 = 0.60 \text{ mol} \).
- Moles of \( \text{H}_2 \) reacted = \( 2 \times 0.40 = 0.80 \text{ mol} \). Equilibrium \( n(\text{H}_2) = 2.00 - 0.80 = 1.20 \text{ mol} \).

(ii) Total number of moles at equilibrium = \( 0.60 + 1.20 + 0.40 = 2.20 \text{ mol} \).
- Mole fraction of \( \text{CO} \), \( x(\text{CO}) = \frac{0.60}{2.20} = 0.2727 \approx 0.273 \)
- Mole fraction of \( \text{H}_2 \), \( x(\text{H}_2) = \frac{1.20}{2.20} = 0.5455 \approx 0.545 \)
- Mole fraction of \( \text{CH}_3\text{OH} \), \( x(\text{CH}_3\text{OH}) = \frac{0.40}{2.20} = 0.1818 \approx 0.182 \)

(iii) Calculate partial pressures (\( p = x \times P_{\text{total}} \)):
- \( p(\text{CO}) = 0.2727 \times 500 \text{ kPa} = 136.36 \text{ kPa} \)
- \( p(\text{H}_2) = 0.5455 \times 500 \text{ kPa} = 272.73 \text{ kPa} \)
- \( p(\text{CH}_3\text{OH}) = 0.1818 \times 500 \text{ kPa} = 90.91 \text{ kPa} \)

Substitute into the \( K_p \) expression:
\[ K_p = \frac{90.91}{136.36 \times (272.73)^2} = \frac{90.91}{136.36 \times 74381.65} = \frac{90.91}{10142681.8} = 8.96 \times 10^{-6} \text{ kPa}^{-2} \]
(If calculated using Pa, total pressure \( = 5.00 \times 10^5 \text{ Pa} \), \( K_p = 8.96 \times 10^{-12} \text{ Pa}^{-2} \)).

(c)(i) Since the forward reaction is exothermic (\( \Delta H = -91 \text{ kJ mol}^{-1} \)), increasing the temperature shifts the position of equilibrium to the left (reactants side) to absorb the excess heat. This causes the value of \( K_p \) to decrease.
(ii) Adding a catalyst increases the rate of both forward and reverse reactions equally. Therefore, there is no effect on the position of equilibrium, and the value of \( K_p \) remains unchanged.

(a) [1.5] marks for correct expression: \( K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \cdot p(\text{H}_2)^2} \) (reject square brackets representing concentrations). [1] mark for correct units (\( \text{kPa}^{-2} \) or \( \text{Pa}^{-2} \) or \( \text{atm}^{-2} \)).

(b)(i) [1] mark for equilibrium moles of \( \text{CO} = 0.60 \text{ mol} \). [1] mark for equilibrium moles of \( \text{H}_2 = 1.20 \text{ mol} \).
(ii) [1] mark for calculating total moles = \( 2.20 \text{ mol} \). [2] marks for all three correct mole fractions to 3 significant figures (deduct 1 mark for any incorrect fraction or rounding error).
(iii) [1] mark for calculating all correct partial pressures. [2] marks for substitution and correct evaluation of the number (8.96 \( \times 10^{-6} \) or 9.0 \( \times 10^{-6} \)). [1] mark for correct units matching the calculation (e.g. \( \text{kPa}^{-2} \)).

(c)(i) [0.5] mark for stating equilibrium shifts left. [1] mark for explaining that the reaction is exothermic, hence \( K_p \) decreases.
(ii) [1.5] marks for stating that the catalyst has no effect on either the equilibrium position or \( K_p \) because it increases the rate of both forward and reverse reactions equally.

(a)(i) Structural isomerism occurs when two or more compounds share the same molecular formula but have different structural arrangements of atoms (different structural formulas).
(ii) The four structural isomers of \( \text{C}_4\text{H}_9\text{Cl} \) are:
1. **1-Chlorobutane**: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \) (Skeletal: straight 4-carbon chain with a Cl on carbon-1).
2. **2-Chlorobutane**: \( \text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3 \) (Skeletal: straight 4-carbon chain with a Cl on carbon-2).
3. **1-Chloro-2-methylpropane**: \( (\text{CH}_3)_2\text{CHCH}_2\text{Cl} \) (Skeletal: 3-carbon chain with a methyl branch on carbon-2, Cl on carbon-1).
4. **2-Chloro-2-methylpropane**: \( (\text{CH}_3)_3\text{C-Cl} \) (Skeletal: 3-carbon chain with a methyl branch and a Cl on carbon-2).

(b)(i) The isomer is **2-chlorobutane**. It exhibits **optical isomerism** (enantiomerism).
(ii) Carbon-2 in 2-chlorobutane is a **chiral carbon** (asymmetric carbon atom) because it is bonded to four different groups: a hydrogen atom (\( -\text{H} \)), a chlorine atom (\( -\text{Cl} \)), a methyl group (\( -\text{CH}_3 \)), and an ethyl group (\( -\text{CH}_2\text{CH}_3 \)).
(iii) The 3D drawings must show a central carbon bonded tetrahedrally (using wedged, dashed, and normal lines) to the four different groups, forming non-superimposable mirror images. The mirror plane should be drawn or clearly implied.

(c) The isomer that yields three different alkenes upon elimination is **2-chlorobutane**.
Elimination can remove a hydrogen atom from either carbon-1 or carbon-3:
- Removal from carbon-1 yields: **but-1-ene**.
- Removal from carbon-3 yields: **but-2-ene**, which exists as two stereoisomers: **cis-but-2-ene** (or (Z)-but-2-ene) and **trans-but-2-ene** (or (E)-but-2-ene).
Therefore, the three alkene products are but-1-ene, cis-but-2-ene, and trans-but-2-ene.

(a)(i) [1] mark for stating "same molecular formula". [1] mark for "different structural formula/arrangement of atoms".
(ii) [4] marks for drawing the correct skeletal structures of the four isomers (1 mark each). [2] marks for the correct IUPAC names matching the drawings (0.5 marks each).

(b)(i) [1] mark for identifying "2-chlorobutane". [0.5] mark for stating "optical isomerism" (accept enantiomerism).
(ii) [1] mark for explaining that it contains a chiral carbon (or carbon attached to 4 different groups: \( -\text{H} \), \( -\text{Cl} \), \( -\text{CH}_3 \), \( -\text{CH}_2\text{CH}_3 \)).
(iii) [1.5] marks for drawing the tetrahedral 3D structure of one enantiomer with correct wedge/dash notation. [1] mark for drawing the second enantiomer as a clear, non-superimposable mirror image of the first.

(c) [0.5] mark for identifying 2-chlorobutane. [1] mark for listing all three IUPAC names correctly (but-1-ene, cis-but-2-ene, and trans-but-2-ene; accept Z/E nomenclature).

(a)(i) Disproportionation is a redox reaction in which the same element is simultaneously oxidized (its oxidation number increases) and reduced (its oxidation number decreases).
(ii) Reaction of chlorine with cold, aqueous sodium hydroxide:
\[ \text{Cl}_2(g) + 2\text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{NaClO}(aq) + \text{H}_2\text{O}(l) \]
Oxidation states of chlorine:
- Reactant \( \text{Cl}_2 \): 0
- Product \( \text{NaCl} \) (chloride): -1
- Product \( \text{NaClO} \) (chlorate(I)): +1

(iii) Reaction of chlorine with hot, aqueous sodium hydroxide:
\[ 3\text{Cl}_2(g) + 6\text{NaOH}(aq) \rightarrow 5\text{NaCl}(aq) + \text{NaClO}_3(aq) + 3\text{H}_2\text{O}(l) \]
Oxidation states of chlorine:
- Reactant \( \text{Cl}_2 \): 0
- Product \( \text{NaCl} \) (chloride): -1
- Product \( \text{NaClO}_3 \) (chlorate(V)): +5

(b)(i)
- **Oxidation half-equation**: Iron(II) is oxidized to Iron(III):
\[ \text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^- \]
- **Reduction half-equation**: Dichromate(VI) is reduced to Chromium(III) in acid:
\[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \]

(ii) Multiply the oxidation half-equation by 6 to balance the electrons, then combine:
\[ 6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^- \]
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
Overall ionic equation:
\[ \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l) \]

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

(a)(i) [1] mark for stating that the same element is oxidised and reduced simultaneously. [1] mark for mentioning the increase and decrease in oxidation numbers.
(ii) [1.5] marks for the correct balanced equation: \( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \) (or ionic equivalent). [2] marks for all three correct oxidation states: 0, -1, and +1 (deduct 1 mark if one is incorrect; 0 marks if more than one is incorrect).
(iii) [1.5] marks for the correct balanced equation: \( 3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + \text{H}_2\text{O} \). [2] marks for all three correct oxidation states: 0, -1, and +5 (deduct 1 mark if one is incorrect; 0 marks if more than one is incorrect).

(b)(i) [1] mark for the correct oxidation half-equation: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \). [2] marks for the correct reduction half-equation: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \) (1 mark for correct formulas of species, 1 mark for correct balancing with electrons).
(ii) [1.5] marks for the correct combined overall equation: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \).
(iii) [1] mark for the correct color change: Orange to green (both colors must be correct).