Edexcel A-Level Chemistry Paper 1, June 2025: Question 6

✅ Correct Answers

(a)(i): D — Cr: [Ar] 3d⁵4s¹  |  Cr³⁺: [Ar] 3d³ [1 Mark]

(a)(ii): Chromium is a transition metal because it forms at least one stable ion with an incompletely filled (partially filled) d-subshell (or d-orbitals). [1 Mark]

💡 Key Knowledge

• Chromium is an anomaly: an electron is promoted from 4s to 3d to minimise electron repulsion and provide a symmetrical, stable half-filled 3d⁵ subshell.
• When forming ions, transition metals always lose their 4s electrons before 3d electrons. For Cr³⁺, 1 electron is lost from 4s and 2 from 3d, leaving 3d³.
• IUPAC transition metal definition: an element whose atom has a partially filled d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

❌ Common Errors & Examiner Traps

• "d-shell" rejected: Examiners explicitly state: "Ignore d-shell". You must write d-subshell or d-orbital(s).
• Forgetting "stable": The definition refers to stable ions.
• Writing [Ar] 3d⁴4s² for ground-state Cr: this is the standard Aufbau expectation, but Cr and Cu are exceptions.

✅ Correct Answers

(b)(i) 3D Structure: [2 Marks]

• Mark 1: Central Cr surrounded by 6 water ligands using clear 3D representation: 2 bonds in the plane (straight lines), 2 pointing forward (wedges), and 2 pointing backward (hatched/dashed lines).
• Mark 2: Dative covalent bonds clearly originating from the oxygen atom of each H₂O ligand to the central Cr (e.g. drawn with arrows pointing towards Cr: Cr ← OH₂ , or bonds terminating specifically at the lone pair on the O atom).

(b)(ii) Shape: Octahedral [1 Mark]

🧠 Examiner's Drawing Guide

To secure full marks on diagram (b)(i):

• Draw central Cr.
• Draw one vertical bond straight up to OH₂ and one straight down to OH₂.
• On the left: one wedged bond pointing forward to OH₂, one dashed bond pointing back to OH₂.
• On the right: one wedged bond to OH₂, one dashed bond to OH₂.
• Crucial: Always write the formula as H₂O—Cr or Cr—OH₂ so the bond line touches the O, never the H!

❌ Common Errors

• Writing "octagonal" or "octagon" instead of octahedral — automatically scored 0 marks!
• Drawing the bond touching a hydrogen atom ( Cr—H₂O ). Oxygen donates the lone pair, so bonding must explicitly link to O.
• Flat cross / planar drawings: 3D wedges and dashes are strictly required when the prompt specifies "3-dimensional shape".

✅ Correct Answers: Part (c) [3 Marks]

Reaction as a base (with acid, H⁺): [1 Mark]
[Cr(OH)₃(H₂O)₃](s) + 3H⁺(aq) → [Cr(H₂O)₆]³⁺(aq)

Reaction as an acid (with base, OH⁻): [1 Mark]
[Cr(OH)₃(H₂O)₃](s) + 3OH⁻(aq) → [Cr(OH)₆]³⁻(aq) + 3H₂O(l)
(Also accepted: reacting with 1 OH⁻ to give [Cr(OH)₄(H₂O)₂]⁻)

State symbols: All state symbols correct across both equations [1 Mark].

💡 Understanding Amphoteric Hydroxides

• Amphoteric: Capable of reacting both as an acid (donating protons / reacting with OH⁻) and as a base (accepting protons / reacting with H⁺).
• [Cr(OH)₃(H₂O)₃] is an uncharged grey-green precipitate (hence state symbol (s)).
• With H⁺, the three OH⁻ ligands are protonated back into H₂O molecules, reforming the green/violet hexaaqua ion in solution: (aq).
• With OH⁻, protons are pulled off water ligands, forming soluble hydroxo complexes such as deep green [Cr(OH)₆]³⁻(aq).

🧠 Exam Technique: State Symbols

The third mark is solely dependent on getting all state symbols correct:

• Neutral complex precipitate = (s)
• Aqueous ions = (aq)
• Water formed in product = (l)

Missing a single state symbol or writing (aq) for the precipitate forfeits this mark completely.

✅ 4-Mark Perfect Response for (e)(i)

1. Ligand approach splits d-orbitals: Ligands approach the Cr³⁺ ion, causing the five degenerate d-orbitals to split into two sets of different energy levels (ΔE). [1 Mark]
2. Promotion of electron: An electron absorbs energy and is promoted from a lower-energy d-orbital to a higher-energy d-orbital (d–d transition). [1 Mark]
3. Light absorption: The energy absorbed corresponds to the energy of photons of visible light (ΔE = hν). [1 Mark]
4. Transmission of colour: The remaining unabsorbed wavelengths are transmitted / reflected, so the complementary colour is observed. [1 Mark]

❌ Fatal Errors in Colour Questions

• "d-orbitals split" vs "d-orbital splits": Never say a single d-orbital splits. The set of d-orbitals (or the d-subshell) splits.
• Emission vs Absorption: Transition metal complex colour is due to absorption of light! Students who write "light is emitted when electrons fall back down" (confusing it with flame tests) automatically lose Mark 3 and Mark 4!
• Failing to specify visible light.

✅ Part (e)(ii): Explaining Different Colours [1 Mark]

Reason: The complexes have different ligands (H₂O vs Cl⁻) and different coordination numbers / shapes (octahedral 6-coordinate vs tetrahedral 4-coordinate).

This results in a different energy gap (ΔE) between the split d-orbitals, meaning a different frequency / wavelength of visible light is absorbed.

💡 Factors Affecting ΔE and Colour

ΔE depends on four factors:

• 1. Nature of the ligand (spectrochemical series)
• 2. Coordination number and geometry (octahedral vs tetrahedral)
• 3. Oxidation state of the metal ion
• 4. Identity of the central metal ion

📐 Step-by-Step Calculation: Part (f)(i) [2 Marks]

Step 1: Identify oxidation and reduction half-cells

• H₂O₂/OH⁻ has more positive E^⦵ (+0.88 V) → Reduction (forward direction).
• CrO₄²⁻/Cr(OH)₃ has more negative E^⦵ (-0.13 V) → Oxidation (reverse direction).

Step 2: Equalise electron transfer (LCM of 3e⁻ and 2e⁻ = 6e⁻)

Multiply oxidation half-reaction by 2:
2Cr(OH)₃ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O + 6e⁻

Multiply reduction half-reaction by 3:
3H₂O₂ + 6e⁻ → 6OH⁻ [1 Mark]

Step 3: Combine and cancel common species

Left: 2Cr(OH)₃ + 10OH⁻ + 3H₂O₂
Right: 2CrO₄²⁻ + 8H₂O + 6OH⁻
Cancel 6 OH⁻ from both sides:

Overall Equation:
2Cr(OH)₃ + 3H₂O₂ + 4OH⁻ → 2CrO₄²⁻ + 8H₂O [1 Mark]

📐 Part (f)(ii): E^⦵_cell & Feasibility [2 Marks]

Step 1: Calculate E^⦵_cell

E^⦵_cell = E^⦵(reduction) − E^⦵(oxidation)

E^⦵_cell = (+0.88 V) − (−0.13 V) = +1.01 V

[1 Mark awarded for +1.01 V]

Step 2: Feasibility statement

Because E^⦵_cell is positive (> 0), the reaction is thermodynamically feasible. [1 Mark]

💡 Thermodynamic Relationships: Why Option A is Correct

The two key equations linking cell potential to thermodynamic quantities are:

ΔS_total = n × F × E^⦵_cell

ln K = (n × F × E^⦵_cell) / (R × T)

• E^⦵_cell is directly proportional to ΔS_total (making B and C incorrect).
• E^⦵_cell is directly proportional to ln K, NOT to K itself (making D incorrect).