WJEC A-Level Chemistry AS Unit 1, June 2025: Question 11

✅ Model Answer

NH₃ forms hydrogen bonds between molecules, whereas PH₃ only has weaker van der Waals (induced dipole-dipole) forces.

Hydrogen bonds are significantly stronger than van der Waals forces, requiring more energy to overcome in NH₃.

💡 Key Knowledge

• Hydrogen bonding occurs when H is bonded to a small, highly electronegative atom with a lone pair (N, O, or F).
• Nitrogen is electronegative enough to form hydrogen bonds; phosphorus has a lower electronegativity, so PH₃ cannot form intermolecular hydrogen bonds.
• Boiling breaks intermolecular forces, not covalent bonds!

🧠 Exam Technique

• Name both forces explicitly: You must name the force in NH₃ (hydrogen bonding) AND the force in PH₃ (van der Waals forces).
• Energy comparison: Always explicitly connect force strength to the energy needed to overcome them.

❌ Common Errors

• Stating that covalent bonds are broken when boiling hydrides (automatic loss of marks).
• Failing to mention that forces exist between molecules.
• Assuming PH₃ can form hydrogen bonds simply because it contains hydrogen and a Group 5 element.

✅ Model Answer

AsH₃ has more electrons than PH₃ (As has atomic number 33 vs P with 15).

Therefore, AsH₃ has stronger / more van der Waals forces between molecules, requiring more thermal energy to overcome.

💡 Key Knowledge

• Both PH₃ and AsH₃ have van der Waals forces as their primary intermolecular attraction.
• As you descend a group, atomic size and total electron count increase.
• More electrons = larger electron cloud = greater polarisability = stronger temporary induced dipoles.

🧠 Exam Technique

Always cite electron count rather than "molecular mass" or "size". WJEC mark schemes specifically look for the phrase "more electrons" leading to "stronger van der Waals forces".

❌ Common Errors

• Writing "AsH₃ is heavier" without referring to the number of electrons.
• Claiming AsH₃ forms dipole-dipole or hydrogen bonds.

✅ Model Answer

• Electron count: Phosphorus has 5 outer electrons, forming 3 bonding pairs and 1 lone pair of electrons.
• Repulsion principle: Electron pairs repel and adopt positions of minimum repulsion (spread as far apart as possible).
• Molecular shape: Trigonal pyramidal (or pyramidal).
• Bond angle: 107° (or any value less than 109° / less than tetrahedral) because lone pair–bonding pair repulsion is greater than bonding pair–bonding pair repulsion.

💡 Key Knowledge & Diagramming

• Tetrahedral base: 4 total pairs give a tetrahedral electron geometry (approx. 109.5°).
• Lone pair effect: Each lone pair compresses bond angles by approximately 2°–2.5°.
• Drawing guidance: Central P with one lone pair (two dots or a lobe) on top; three H atoms below with one standard line, one wedged bond (pointing forward), and one dashed bond (receding into the page).

🧠 4-Step Answer Framework

1. State total electron pairs: 3 BP + 1 LP .
2. State the fundamental rule: electron pairs repel as far apart as possible.
3. Name the 3D shape based solely on atom positions: pyramidal.
4. State bond angle ( 107° ) and justify via lone pair repulsion.

❌ Common Errors

• Calling the shape "trigonal planar" (confusing it with BF₃, which has 3 BP and 0 LP).
• Giving an angle of 120° (planar angle) or exactly 109.5°.
• Forgetting to state that lone pairs repel more than bonding pairs.

Part (c)(i) • Silver Nitrate Test (2 Marks)

Claim: Yellow precipitate means only iodide is present.

Judgment: Incorrect.

Justification: Although the yellow precipitate confirms the presence of silver iodide (AgI), any white precipitate (AgCl) or cream precipitate (AgBr) formed simultaneously would be masked / obscured by the intense yellow colour and cannot be seen clearly.

Part (c)(ii) • Bromine Water Test (2 Marks)

Claim: No colour change means the solution only contains bromide.

Judgment: Disagree.

Justification: While no change confirms iodide is absent (as Br₂ would oxidise I⁻ to brown I₂), chloride ions could also be present because bromine is not a strong enough oxidising agent to displace/oxidise chloride ions (no reaction occurs with Cl⁻ either).

💡 Halogen Displacement Rules

Oxidising ability decreases down Group 7:

Cl₂ > Br₂ > I₂

• Br₂ oxidises I⁻ → I₂ (orange/brown colour forms).
• Br₂ cannot oxidise Cl⁻ (solution remains orange Br₂; no reaction).
• Therefore, no colour change means I⁻ is absent, but tells you nothing about whether Cl⁻ is present!

🧠 Exam Technique: "Evaluate the Claim"

• Always start with an unambiguous Agree / Disagree or Correct / Incorrect.
• Credit in these questions relies on considering what is hidden:
  • In (c)(i): darker precipitates hide paler ones.
  • In (c)(ii): unreactive species produce no observable change.

📐 Step-by-Step Oxidation State Tracking

1. Chlorine:

• In elemental Cl₂: oxidation state = 0
• In HCl: oxidation state = −1
• Change: 0 → −1 (gain of electrons → REDUCED)

2. Oxygen:

• In H₂O: oxidation state = −2
• In elemental O₂: oxidation state = 0
• Change: −2 → 0 (loss of electrons → OXIDISED)

✅ Model Answer

The reaction is a redox reaction because both oxidation and reduction occur:

• Oxygen is oxidised: It loses electrons, and its oxidation state increases from −2 in H₂O to 0 in O₂.
• Chlorine is reduced: It gains electrons, and its oxidation state decreases from 0 in Cl₂ to −1 in HCl.

🧠 Examiner Tip

To secure full marks on 2-mark redox questions, you must specify:

1. Which element is oxidised AND its initial and final oxidation states (or mention of electron loss).
2. Which element is reduced AND its initial and final oxidation states (or mention of electron gain).

❌ Common Errors

• Stating "water is oxidised" instead of naming the specific element: oxygen.
• Confusing signs, e.g. writing +2 instead of −2 for oxygen in water.
• Assuming this is a disproportionation reaction (it is NOT: chlorine only decreases its oxidation state; oxygen increases its state).