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Question 1

Topic: 9.2

(a) Complete Table 1.1 using relevant information from the Periodic Table.

(b) State and explain the difference in the ionic radius of \(Al^{3+}\) compared to \(Mg^{2+}\).

(c) Draw a labelled diagram to show the structure and bonding in sodium.

(d) Fig. 1.1 shows the variation in melting point of some Period 3 elements in their standard states at room temperature and pressure.

(i) Explain why Si has a high melting point.
(ii) Complete Fig. 1.1 to show the variation in the melting points of the elements P, S and Cl.
(e) Two Period 3 elements react with an excess of oxygen at room pressure.
(i) Complete Table 1.2.

(ii) The solutions made in column 3 of Table 1.2 are mixed together. Name the type of reaction that occurs.
(iii) Write an equation to describe the reaction between P4O10 and an excess of water.
(f) Aluminium hydroxide is amphoteric.
(i) Explain what is meant by amphoteric.
(ii) Write an equation to describe the reaction that occurs when aluminium hydroxide, Al(OH)₃, reacts with NaOH(aq).

▶️ Answer/Explanation
Solution

(a)

Explanation: The table is completed using the Periodic Table. Na (sodium) forms a \(Na^+\) ion, Mg (magnesium) forms \(Mg^{2+}\), and Al (aluminium) forms \(Al^{3+}\). The ionic radii decrease across the period due to increasing nuclear charge.

(b) \(Al^{3+}\) is smaller than \(Mg^{2+}\).

Explanation: \(Al^{3+}\) has a greater nuclear charge (\(+13\)) compared to \(Mg^{2+}\) (\(+12\)), attracting the remaining electrons more strongly. Both ions have the same number of electron shells, but the higher charge on \(Al^{3+}\) reduces its ionic radius.

(c)

Explanation: Sodium has a metallic lattice structure with \(Na^+\) ions in a sea of delocalized electrons. The diagram shows the closely packed ions and free-moving electrons responsible for conductivity.

(d)(i) Silicon has a giant covalent structure.

Explanation: Silicon forms a tetrahedral network of strong covalent bonds. A large amount of energy is required to break these bonds, resulting in a high melting point.

(d)(ii)

Explanation: P, S, and Cl are simple molecular substances with weak van der Waals forces. Their melting points decrease as the size of the molecules decreases (P₄ > S₈ > Cl₂).

(e)(i)

Explanation: The table is completed as follows: Na₂O (sodium oxide) forms NaOH (aq), and P₄O₁₀ (phosphorus pentoxide) forms H₃PO₄ (aq).

(e)(ii) Acid-base (neutralization) reaction.

Explanation: NaOH (a base) reacts with H₃PO₄ (an acid) to form salt and water, which is a neutralization reaction.

(e)(iii) \(P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4\)

Explanation: Phosphorus pentoxide reacts vigorously with water to form phosphoric acid, releasing heat in the process.

(f)(i) Amphoteric substances react with both acids and bases.

Explanation: Aluminium hydroxide can act as an acid (reacting with NaOH) or a base (reacting with HCl), making it amphoteric.

(f)(ii) \(Al(OH)_3 + NaOH \rightarrow NaAl(OH)_4\)

Explanation: Al(OH)₃ dissolves in NaOH to form sodium aluminate, demonstrating its acidic behavior.

Question 2

Topic: 5.1

Separate samples of Na₂CO₃ and NaHCO₃ react with HCl(aq) to produce the same products, as shown in Table 2.1.

Table 2.1

(a) Complete the reaction pathway diagram in Fig. 2.1 for reaction 2. Label the diagram to show the enthalpy change, \(∆H_2\), and the activation energy, \(E_A\).

Fig. 2.1

(b) The value for ∆H1 is determined by experiment using the following method.
• 50.0 cm³ of \(2.00moldm^{–3}\) HCl(aq) is added to a polystyrene cup.
• The initial temperature of the acid is recorded as 19.6°C.
• 0.0400mol of Na₂CO₃ is added and the mixture is stirred.
• All the solid Na₂CO₃ disappears and a colourless solution is produced.
The maximum temperature recorded during the reaction is 26.2°C.
(i) Describe one other observation that shows the reaction is complete.

(ii) Calculate the value of ∆H1 in kJmol⁻¹. Assume the specific heat capacity of the reaction mixture is the same as for water and no heat is lost to the surroundings. Show your working.

(iii) Thermal decomposition occurs when NaHCO₃ is heated.
reaction 3: 2NaHCO₃ → Na₂CO₃+ H₂O + CO₂
Calculate the enthalpy change for reaction 3, \(∆H_r\), using the data in Table 2.1 and the value of \(∆H_1\) calculated in (b)(ii). (If you were unable to calculate a value for \(∆H_1\) in (b)(ii), assume the enthalpy change is \(–38.4kJmol^{–1}\). This is not the correct value.)

(c) Z is a salt that contains a Period 4 element from Group 2. When Z is heated brown gas forms. Identify the formula of Z and use it to write an equation for the reaction.

▶️ Answer/Explanation
Solution

(a)

Answer to Fig. 2.1

Explanation: The reaction pathway diagram shows the enthalpy change (\(∆H_2\)) as the difference between the energy levels of reactants and products. The activation energy (\(E_A\)) is the energy barrier from reactants to the transition state.

(b)(i) No more bubbles of gas are observed.

Explanation: The reaction produces CO₂ gas, so the cessation of bubbling indicates completion.

(b)(ii) \(∆H_1 = -33.0 \, \text{kJmol}^{-1}\)

Explanation: Using \(q = mc∆T\), the heat change is calculated as \(50.0 \, \text{g} \times 4.18 \, \text{Jg}^{-1}\text{K}^{-1} \times (26.2 – 19.6) \, \text{K} = 1.38 \, \text{kJ}\). For \(0.0400 \, \text{mol}\) of Na₂CO₃, \(∆H_1 = -1.38 \, \text{kJ} / 0.0400 \, \text{mol} = -33.0 \, \text{kJmol}^{-1}\).

(b)(iii) \(∆H_r = +91.4 \, \text{kJmol}^{-1}\)

Explanation: Using Hess’s Law, \(∆H_r = 2∆H_2 – ∆H_1 = 2(-58.3) – (-33.0) = +91.4 \, \text{kJmol}^{-1}\).

(c) Ca(NO₃)₂ → CaO + 2NO₂ + ½O₂

Explanation: The brown gas is NO₂, indicating thermal decomposition of calcium nitrate. The balanced equation shows the formation of CaO, NO₂, and O₂.

Question 3

Topic: 7.1

(a) Describe what is meant by dynamic equilibrium.

(b) Reaction 4 describes the reversible reaction between yellow Fe³⁺ (aq) and colourless SCN⁻ (aq) to produce red FeSCN²⁺ (aq).
reaction 4 Fe³⁺(aq) + SCN⁻(aq) \( \rightleftharpoons \) FeSCN²⁺(aq)
yellow colourless red
An equilibrium mixture contains Fe³⁺(aq), SCN⁻ (aq) and FeSCN²⁺ (aq). A few colourless crystals of soluble KSCN(s) are added. The mixture is then left until it reaches equilibrium again. The temperature of both equilibrium mixtures is the same.
(i) Deduce the changes that occur, if any, in the equilibrium mixture after KSCN(s) is added compared to the original equilibrium mixture.
• change in appearance
• change in relative concentration of Fe³⁺(aq)
• change in value of the equilibrium constant, \(K_c\).

(ii) The expression for the equilibrium constant, \(K_c\), for reaction 4 is shown

\(5.00 × 10^{–5}mol\) of Fe³⁺(aq) and \(5.00 × 10^{–5}mol\) of SCN⁻ (aq) are added together and allowed to reach equilibrium. The total volume of the mixture is 25.0 cm³. At equilibrium the concentration of FeSCN²⁺(aq) is \(4.23 × 10^{–4}moldm^{–3}\). Calculate the equilibrium constant, \(K_c\), for reaction 4. Include the units in your answer.

(c) Determine the full electronic configuration of Fe³⁺.
(d) SCN⁻ (aq) is colourless. Complete the dot-and-cross diagram in Fig. 3.1 to show the arrangement of outer electrons in an SCN⁻ ion.

▶️ Answer/Explanation
Solution

(a) Dynamic equilibrium occurs when the forward and reverse reaction rates are equal, and the concentrations of reactants and products remain constant over time. This is a state where no macroscopic changes are observed, but molecules continue to react at the molecular level.

Explanation: In dynamic equilibrium, the system is not static—reactants and products are continuously interconverting, but their concentrations do not change because the rates of formation and decomposition are balanced.

(b)(i)

  • Change in appearance: The mixture becomes darker red due to increased formation of FeSCN²⁺.
  • Change in [Fe³⁺(aq)]: The concentration of Fe³⁺ decreases as more reacts with the added SCN⁻.
  • Change in \(K_c\): No change, as \(K_c\) is temperature-dependent and the temperature remains constant.

Explanation: Adding KSCN increases [SCN⁻], shifting the equilibrium to the right (Le Chatelier’s Principle), producing more FeSCN²⁺ and consuming Fe³⁺. \(K_c\) remains unchanged because temperature is constant.

(b)(ii) \(K_c = 1380 \, \text{dm}^3 \text{mol}^{-1}\)

Explanation: First, calculate initial concentrations: \([Fe³⁺]_0 = [SCN⁻]_0 = \frac{5.00 \times 10^{-5} \text{mol}}{0.025 \text{dm}^3} = 2.00 \times 10^{-3} \text{mol dm}^{-3}\). At equilibrium, \([FeSCN²⁺] = 4.23 \times 10^{-4} \text{mol dm}^{-3}\). Using the ICE table, equilibrium concentrations are \([Fe³⁺] = [SCN⁻] = 1.577 \times 10^{-3} \text{mol dm}^{-3}\). Substituting into \(K_c = \frac{[FeSCN²⁺]}{[Fe³⁺][SCN⁻]}\), we get \(K_c = 1380 \, \text{dm}^3 \text{mol}^{-1}\).

(c) \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\)

Explanation: Fe has atomic number 26. Fe³⁺ loses 3 electrons (2 from 4s and 1 from 3d), resulting in the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5\).

(d)

Explanation: The SCN⁻ ion has a lone pair on the sulfur (S), carbon (C), and nitrogen (N) atoms, with covalent bonds between S-C and C-N. The negative charge is delocalized across the ion.

Question 4

Topic: 15.1

CH₃(CH₂)₅CHBrCH₃ exists as a pair of stereoisomers.

(a) Draw the three-dimensional structures of the two stereoisomers of CH₃(CH₂)₅CHBrCH₃. R can be used to represent CH₃(CH₂)₅.

Stereoisomers Template

(b) A sample of CH₃(CH₂)₅CHBrCH₃ reacts with NaOH to make CH₃(CH₂)₅CH(OH)CH₃ in an \(S_N1\) mechanism. Complete Fig. 4.1 to show the mechanism for the reaction of CH₃(CH₂)₅CHBrCH₃ and NaOH. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Reaction Mechanism Template

(c) Separate samples of CH₃(CH₂)₅CHBrCH₃, CH₃(CH₂)₅CH(OH)CH₃, and CH₃(CH₂)₅CHCH₂ are tested with different reagents. Complete Table 4.1. If no reaction occurs, write × in the relevant box.

Table 4.1

(d) CH₃(CH₂)₅CHBrCH₃ is heated with D to produce three different molecules, E, F and G.

Reaction Scheme

(i) Name the type of reaction.
(ii) Identify D and the conditions used.

(e) (i) Both σ and π bonds are present in a molecule of E as a result of different types of hybridisation in the carbon atoms. Complete Table 4.2 to show the number of carbon atoms with each type of hybridisation in a molecule of E.

Table 4.2

(ii) Describe the essential feature of an unbranched hydrocarbon that causes its molecules to show stereoisomerism. Explain how this feature leads to stereoisomerism.

▶️ Answer/Explanation
Solution

(a)

Stereoisomers of CH₃(CH₂)₅CHBrCH₃

Explanation: The two stereoisomers are enantiomers (non-superimposable mirror images) due to the chiral center at the carbon bearing the Br atom. The 3D structures show the R and S configurations with R representing CH₃(CH₂)₅.

(b)

SN1 Mechanism

Explanation: The \(S_N1\) mechanism involves two steps: (1) Formation of a carbocation intermediate after Br⁻ leaves (shown with curly arrow), and (2) Nucleophilic attack by OH⁻ on the planar carbocation, resulting in CH₃(CH₂)₅CH(OH)CH₃.

(c) Table 4.1 Answers: CH₃(CH₂)₅CHBrCH₃ reacts with NaOH (✓), CH₃(CH₂)₅CH(OH)CH₃ reacts with acidified K₂Cr₂O₇ (✓), and CH₃(CH₂)₅CHCH₂ reacts with Br₂ (✓). All other combinations are ×.

(d)
(i) Elimination reaction
(ii) D = KOH (alcoholic), Conditions = Heat
Explanation: Heating with alcoholic KOH causes elimination of HBr, producing alkene (E), alkene (F), and possibly a minor product (G).

(e)
(i) Table 4.2: sp³ hybridised carbons = 5, sp² hybridised carbons = 2
(ii) Presence of a C=C double bond (π bond) restricts rotation, allowing different spatial arrangements when each double-bonded carbon has two different groups attached.

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