CAPE Chemistry Unit 1 — Paper 2 Solutions

The Lyman Series. All the transitions shown end at n = 1 (the first / ground energy level), which defines the Lyman series. (Other series for reference: Balmer ends at n = 2, Paschen ends at n = 3.)

The s orbital is spherical — symmetrical about the nucleus in all directions. The p orbitals (p_x, p_y, p_z) are dumbbell-shaped with two lobes on either side of the nucleus, each oriented along one of the three Cartesian axes.

Hybridization is the mixing or combining of atomic orbitals to form new hybrid orbitals that are usually of lower energy and more stable. The hybrid orbitals usually have different shapes, energies and orientations than the original atomic orbitals. In the case of carbon, mixing one 2s and three 2p atomic orbitals forms four sp³ hybrid orbitals.

Resonance is a phenomenon used to describe electrons being delocalised across multiple atoms. Resonance is observed in a molecule or ion that can be represented by two or more Lewis (resonance) structures that differ in the position of electrons and not in the arrangement of the atoms.

1. Electron pairs around the central atom repel each other; the shape of the molecule is determined by these repulsions between bonding pairs and lone pairs.
2. The electron pairs arrange themselves as far apart as possible in order to minimise repulsion.
3. Bonding pairs and lone pairs occupy space around the central atom, but lone pairs exert greater repulsion than bonding pairs — they distort bond angles and alter the ideal shape (multiple bonds count as a single electron domain).

BeCl₂ has 2 bonding pairs and 0 lone pairs on the central Be atom. The two electron domains repel and arrange as far apart as possible — at 180°.

Shape: linear. Bond angle: 180°.    Cl–Be–Cl

CH₃⁻ has 3 bonding pairs and 1 lone pair on the central C atom (sp³ hybridised, four electron domains). The lone pair compresses the bonding pairs slightly below the ideal tetrahedral angle.

Shape: trigonal pyramidal. Bond angle: ≈ 107° (slightly less than 109.5° due to the lone-pair–bond-pair repulsion).

The lone pair sits above the carbon; three C–H bonds point down and out, giving a trigonal pyramid:

Use the equation: 2 MnO₄⁻(aq) + 5 NO₂⁻(aq) + 6 H⁺(aq) → Mn²⁺(aq) + 5 NO₃⁻(aq) + 3 H₂O(l).

(i) Calculate the number of moles of KMnO₄ used in the titration. (1 mark)

n = c × V = 0.02 × (27.50 ÷ 1000) = 0.000550 mol = 5.50 × 10⁻⁴ mol

(ii) Deduce the number of moles of KNO₂ in the titre volume. (1 mark)

From the balanced equation, MnO₄⁻ : NO₂⁻ = 2 : 5 = 1 : 2.5

n(KNO₂) in 25.0 cm³ = 0.000550 × 2.5 = 0.001375 mol

(iii) Calculate the moles of KNO₂ present in the residue. (1 mark)

The 25.0 cm³ aliquot is 1/40 of the 1 dm³ residue solution, so:

n(KNO₂) in 1000 cm³ = 0.001375 × (1000 ÷ 25) = 0.0550 mol

(iv) Calculate the mass of KNO₂ present in the residue. (2 marks)

M(KNO₂) = 39.10 + 14.01 + (2 × 16.00) = 85.10 g mol⁻¹

mass = n × M = 0.0550 × 85.10 = 4.68 g

(v) If 6.20 g KNO₃ were completely decomposed, determine the mass of KNO₂ formed. (3 marks)

M(KNO₃) = 39.10 + 14.01 + (3 × 16.00) = 101.11 g mol⁻¹

n(KNO₃) = 6.20 ÷ 101.11 = 0.0613 mol

From 2 KNO₃ → 2 KNO₂ + O₂, the ratio KNO₃ : KNO₂ = 1 : 1, so n(KNO₂) = 0.0613 mol.

mass = 0.0613 × 85.10 = 5.22 g

(vi) Using results from (f)(iv) and (v), calculate the percentage of KNO₂ formed. (2 marks)

% KNO₂ = (actual mass formed ÷ mass formed if completely decomposed) × 100

= (4.68 ÷ 5.22) × 100 = ≈ 89.6 %

(vii) Outline the steps taken in the laboratory to prepare 1 dm³ of the 0.02 mol dm⁻³ potassium manganate(VII) solution. (5 marks)

1. Calculate the mass needed. n = c × V = 0.02 × 1 = 0.02 mol; M(KMnO₄) = 158.04 g mol⁻¹; mass = n × M = 0.02 × 158.04 = 3.16 g.
2. Using an electronic balance, accurately weigh out 3.16 g of KMnO₄ solid into a clean weighing boat.
3. Transfer the KMnO₄ into a beaker and add about 200 mL of distilled water; stir with a glass rod until it dissolves completely.
4. Transfer the solution into a 1 dm³ volumetric flask using a funnel; rinse the beaker and funnel into the volumetric flask with distilled water (so all of the KMnO₄ reaches the flask).
5. Add distilled water until the bottom of the meniscus sits exactly on the 1 dm³ calibration mark, taking care to read at eye level.
6. Stopper the flask and invert it several times to mix thoroughly.

A dynamic equilibrium refers to the state in a closed system where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations (or partial pressures) of both reactants and products remain constant. The reactions are still occurring in both directions; there is simply no net change.

(i) Write an expression for the equilibrium constant of the reverse reaction. (1 mark)

K_reverse = [A][B] ⁄ [C][D]  = 1 ⁄ K_forward

(ii) State the numerical value of the equilibrium constant for the reverse reaction. (1 mark)

K_reverse = 1 ⁄ 4.0 = 0.25

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (concentration, pressure or temperature), the equilibrium will shift in the direction that counteracts the change and re-establishes equilibrium.

Equilibrium A: H₂O(g) + C(s) ⇌ H₂(g) + CO(g)    ΔH = +131 kJ mol⁻¹

Equilibrium B: 2 CrO₄²⁻(aq) + 2 H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)

(i) Increasing the pressure on Equilibrium A. (2 marks)

Counting gaseous moles: LHS = 1 (H₂O); RHS = 2 (H₂ + CO). Solid carbon is ignored. Increasing pressure favours the side with fewer gas moles, so the equilibrium shifts to the LEFT (backward). The volume / concentration of gaseous H₂ and CO decreases.

(ii) Decreasing the temperature on Equilibrium A. (2 marks)

The forward reaction is endothermic (ΔH > 0). Decreasing temperature favours the exothermic direction, i.e. the reverse reaction. The equilibrium shifts to the LEFT and the yields of H₂ and CO decrease.

(iii) Increasing [H⁺(aq)] on Equilibrium B. (2 marks)

Adding H⁺ increases the concentration of a reactant; the system responds by consuming H⁺, so the equilibrium shifts to the RIGHT. [Cr₂O₇²⁻] increases (the solution turns from yellow towards orange).

A buffer solution is one that resists changes in pH when small amounts of acid or alkali (or water on dilution) are added.

Use the Henderson–Hasselbalch equation:

pH = pK_a + log₁₀([A⁻] ⁄ [HA])

pK_a = −log(7.4 × 10⁻⁵) = 4.131

log₁₀(0.5 ⁄ 0.2) = log₁₀(2.5) = 0.398

pH = 4.131 + 0.398 = ≈ 4.53

1. Measure out 250 cm³ of 0.2 mol dm⁻³ HA using a measuring cylinder, then transfer it to a beaker.
2. Calculate the mass of NaA needed for [NaA] = 0.5 mol dm⁻³ in 250 cm³: n = 0.5 × 0.250 = 0.125 mol → mass = 0.125 × M(NaA). Weigh out the NaA on an electronic balance.
3. Dissolve the solid NaA in the 250 cm³ of HA solution from step 1, stirring with a glass rod until fully dissolved.
4. Place a small portion of the buffer solution into three separate test tubes and measure the initial pH of each (using a pH meter or a narrow-range universal indicator).
5. Test 1 (acid challenge): add a small aliquot of dilute HCl to one tube and re-measure the pH — a true buffer shows only a small change in pH.
6. Test 2 (alkali challenge): add a small aliquot of dilute NaOH to a second tube and re-measure the pH — again, only a small change is expected.
7. Test 3 (control): the third tube is the untouched control. Comparing the three confirms the solution is acting as a buffer.

The mixture supplies a weak base (NH₃) and its conjugate acid (NH₄⁺, from (NH₄)₂SO₄):

(NH₄)₂SO₄(aq) → 2 NH₄⁺(aq) + SO₄²⁻(aq)    (eq 1)

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)    (eq 2)

If acid (H⁺) is added: NH₃ reacts with H⁺ to form NH₄⁺:

NH₃(aq) + H⁺(aq) → NH₄⁺(aq)

The added H⁺ is consumed, so [H⁺] (and therefore pH) is barely changed.

If alkali (OH⁻) is added: NH₄⁺ reacts with OH⁻ to re-form NH₃ and water:

NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l)

The added OH⁻ is consumed, so pH is barely changed.

Half-life (t_½) is the time taken for the concentration of a reactant to fall to half of its initial value. For a first-order reaction, t_½ is constant — independent of starting concentration.

A catalyst provides an alternative reaction pathway with a lower activation energy, E_a(catalysed). On the Boltzmann curve this lowers the activation-energy line, so a much larger fraction of molecules have energy ≥ E_a at the same temperature, and successful collisions per unit time increase. The rate of reaction increases.

Although both sodium and aluminium are in Period 3, they are in different groups and as we move from left to right across the period, the group number increases. Sodium is in Group 1 with 1 valence electron, hence an oxidation number of +1. Aluminium is in Group 3 with 3 valence electrons, hence an oxidation number of +3. The maximum positive oxidation number across the period therefore matches the group number.

• Magnesium chloride, MgCl₂: ionic bond (large electronegativity difference).
• Silicon chloride, SiCl₄: covalent bond.
• Phosphorus(III) chloride, PCl₃: covalent bond.

Silicon chloride is hydrolysed by water to form an acidic solution, releasing fumes of hydrogen chloride gas:

SiCl₄(l) + 4 H₂O(l) → SiO₂(s) + 4 HCl(aq)

The reaction is vigorous: white fumes of HCl are seen above the mixture (HCl(g) + moist air → mist), and a white precipitate of silicon(IV) oxide / silica forms in the solution.

(i) Define the term 'ligand'. (2 marks)

A ligand is an ion or a molecule with one or more lone pairs of electrons that can be donated to a transition-element ion to form a coordinate (dative) bond, producing a complex.

(ii) Write the formula of the complex formed. (1 mark)

If X has a charge of n+ and is bonded to six OH⁻ ligands (each contributing −1):

[X(OH)₆]⁽ⁿ⁻⁶⁾  — for the typical CAPE-set example with n = 3 (e.g. Al³⁺ or Cr³⁺) the complex is [X(OH)₆]³⁻.

(iii) State the shape of the complex formed. (1 mark)

Octahedral.

(iv) Using a suitable diagram, illustrate your answer in (b)(iii). (2 marks)

(v) State what is meant by the term 'coordination number', in reference to a complex ion. (2 marks)

The coordination number is the number of coordinate (dative) bonds that ligands form with the central transition-metal ion in the complex. For [X(OH)₆]³⁻, the coordination number is 6.

The light-blue [Cu(H₂O)₆]²⁺ ions are progressively replaced by chloride ligands, passing through green intermediates to a yellow solution of [CuCl₄]²⁻:

[Cu(H₂O)₆]²⁺  (light blue)  → [Cu(H₂O)₅Cl]⁺  (green)  → [Cu(H₂O)₄Cl₂]  (dark yellow)  → [Cu(H₂O)₃Cl₃]⁻  (dark yellow)  → [CuCl₄]²⁻  (yellow / orange)

The solution changes colour from light blue → green → yellow / orange.

The overall ligand-substitution equilibrium can be written as:

[Cu(H₂O)₆]²⁺(aq) + 4 Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq) + 6 H₂O(l)

Concentrated HCl drives the equilibrium far to the right because the chloride complex is the more stable product. Hence:

[CuCl₄]²⁻ has the higher stability constant K_stab.

(i) Describe the changes which occur at the end-point of this titration. (2 marks)

During the titration, each addition of KMnO₄ is decolourised by the Fe²⁺ in the flask. At the end-point, all the Fe²⁺ has been oxidised, so the next drop of KMnO₄ is no longer decolourised — a permanent pale pink (faint purple) colour persists in the flask. KMnO₄ is its own indicator.

(ii) Write the formula and state the colour of the species formed from the oxidation of Fe²⁺ ions. (2 marks)

Fe²⁺ is oxidised to Fe³⁺, which gives a reddish-brown / yellow-brown colour in solution.

Half-equation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻

Overall: MnO₄⁻(aq) + 8 H⁺(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq)