AQA A-Level Chemistry Paper 1, 2023: Question 1

✅ Correct Answer

A (central) metal atom or ion surrounded by ligands (bonded via coordinate/dative covalent bonds).

💡 Key Knowledge

• A complex consists of a central metal species (atom or positive ion) coordinated to species with lone pairs (ligands).
• Alternative acceptable definition: A species in which the number of coordinate bonds exceeds the oxidation state of the metal.

✅ Correct Answer

Name of shape: Pyramidal (or trigonal pyramidal / tetrahedral)

Drawing: Central P atom with 3 single bonds to F atoms (using wedges/dashes or standard bonds showing 3D geometry) plus 1 lone pair drawn on the phosphorus atom.

🧠 Exam Technique

• Count valence electrons: Phosphorus has 5 valence electrons. It forms 3 single covalent bonds with fluorine (3 bonding pairs), leaving 2 electrons (1 lone pair).
• 4 total electron pairs = tetrahedral electron geometry; 3 bonding pairs + 1 lone pair = trigonal pyramidal molecular shape (~107° bond angle).
• The prompt specifically says: "Include any lone pairs of electrons that influence the shape." Forgetting the lone pair on P loses the drawing mark!

❌ Common Errors

• Naming the shape "trigonal planar" — confusing PF₃ (Group 15, 1 lone pair) with BF₃ (Group 13, 0 lone pairs).
• Drawing lone pairs on the fluorine atoms instead of on phosphorus (F lone pairs do not determine the central shape).

✅ Correct Answer

PF₃ is a neutral molecule (has no charge) AND the overall complex is neutral (has no charge).

🧠 Exam Technique: Two-Part Reasoning

To score this mark, you must state both halves of the equation:

• 1. The charge on the ligand (PF₃ = 0)
• 2. The charge on the entire complex (Cr(PF₃)₆ = 0)
• Therefore: Oxidation state of Cr = 0 - (6 × 0) = 0.

❌ Common Errors

• Only stating that "PF₃ is neutral" without mentioning that the complex itself has no charge.
• Trying to explain using electronegativity differences — the mark scheme explicitly states: "Ignore electronegativity".

01.4: Oxidation State of Cr in [Cr(NH₃)₄Cl₂]Cl [1 mark]

Answer: +3 (or 3)

01.5: Bond Between N and H in Ammonia [1 mark]

Answer: Covalent (bond)

✅ Correct Answer: Isomer Drawings

Draw two octahedral complexes with square brackets and a + charge on each:

✅ Type of Isomerism

Cis-trans isomerism
(also accept: geometric isomerism or E-Z isomerism)

🧠 Exam Technique: Securing All 3 Marks

• Mark 1: Both 3D octahedral isomers drawn with correct connectivity (Cr in center, 4 NH₃, 2 Cl). Cis has the two Cl ligands at 90°; trans has them at 180° directly opposite each other.
• Mark 2: Square brackets with the overall + charge shown on the complex ions. (Notice the question specifies "the complex ion", so omit the chloride counter-ion).
• Mark 3: Correct naming of the specific isomerism type.

❌ Common Errors

• Writing just "stereoisomerism" — the mark scheme explicitly states "Ignore stereoisomerism" because it is too broad; you must specify cis-trans or geometric.
• Omitting the + charge outside the square brackets.
• Drawing optical isomers (enantiomers) — MA₄B₂ octahedral complexes do NOT show optical isomerism, only cis-trans!

✅ Correct Equation

CrCl₃ + 5 H₂O → [Cr(H₂O)₅Cl]Cl₂

Alternative acceptable ionic product:
[Cr(H₂O)₅Cl]²⁺ + 2 Cl⁻

💡 How to Deduce the Formula

• Look at stoichiometry: 5 H₂O molecules are reacting with 1 CrCl₃.
• Chromium(III) almost always has a coordination number of 6 (octahedral).
• 5 coordination sites are taken by the 5 H₂O molecules; the 6th site must be filled by 1 chloride ligand inside the complex sphere: [Cr(H₂O)₅Cl] .
• The remaining 2 chlorine atoms stay outside the coordination sphere as counter-ions: Cl₂ .