Transition Metal Complexes, Ligand Substitution, Chelate Effect, Colorimetry & Catalysis
A transition metal is strictly defined as a d-block element that forms at least one stable ion with a partially filled (incomplete) d-subshell of electrons (1 to 9 d-electrons), which excludes Scandium (Sc³⁺ is 3d⁰) and Zinc (Zn²⁺ is 3d¹⁰). Because their 3d and 4s energy levels lie close together, transition metals exhibit variable oxidation states, catalytic activity, and complex ion formation via coordinate (dative covalent) bonding from ligands. When ligands bond to a central transition metal ion, electrostatic repulsion splits the five degenerate 3d orbitals into two energy levels separated by an energy gap ΔE = hν = hc/λ, allowing d-electrons to absorb specific visible wavelengths and transmit the complementary colour. Examiners test complex ion stereoisomerism (cisplatin), the entropy-driven chelate effect with EDTA⁴⁻, heterogeneous vs homogeneous catalysis (including autocatalysis), and aqueous metal ion precipitations.
- Transition Metal Definition & Why Sc and Zn Are Excluded
- A transition metal is an element that forms at least one stable ion with a partially filled d-subshell (d¹ to d⁹). Scandium only forms Sc³⁺ ([Ar] 3d⁰, empty 3d subshell) and Zinc only forms Zn²⁺ ([Ar] 3d¹⁰, completely full 3d subshell); therefore, neither Sc nor Zn is a transition metal, and their compounds are white/colourless because d-d electron transitions cannot occur.
- Complex Ions, Coordination Number & Ligand Types
- A complex ion consists of a central metal ion surrounded by ligands bonded via coordinate (dative covalent) bonds, where each ligand donates at least one lone pair of electrons. Coordination number = total number of coordinate bonds (not number of ligands) to the central metal ion. Monodentate ligands (1 lone pair donated): H₂O:, :NH₃, :Cl⁻, :CN⁻. Bidentate ligands (2 lone pairs donated): ethane-1,2-diamine (en, H₂NCH₂CH₂NH₂) and ethanedioate (C₂O₄²⁻). Multidentate/Hexadentate ligand (6 lone pairs): EDTA⁴⁻.
- Shapes & Stereoisomerism of Complex Ions
- 1. Coordination number 6 (small ligands like H₂O, NH₃, en, EDTA⁴⁻, e.g. [Cu(H₂O)₆]²⁺): Octahedral, 90° (exhibits cis-trans and optical isomerism in [M(AA)₂X₂] or [M(AA)₃]). 2. Coordination number 4 (large Cl⁻ ligands that cannot fit 6 around the ion, e.g. [CuCl₄]²⁻, [CoCl₄]²⁻): Tetrahedral, 109.5°. 3. Coordination number 4 in d⁸ complexes (Pt²⁺, Ni²⁺, e.g. cisplatin [Pt(NH₃)₂Cl₂]): Square planar, 90° (cis-isomer has identical ligands at 90°; trans-isomer at 180°). 4. Coordination number 2 (e.g. Tollens' reagent [Ag(NH₃)₂]⁺): Linear, 180°.
Transition Metal Complexes, Ligand Substitution, Chelate Effect, Colorimetry & Catalysis: 3D Molecular, Lattice, DNA & Vector Explorer
Drag directly on the 3D model to rotate it through any angle, or switch tabs to inspect VSEPR bond angles, crystal lattices, DNA base pairing, and 3D space diagonals.
CH₄ / NH₄⁺ — Bond Angle: 109.5°
- Bonding Pairs:
- 4
- Lone Pairs on Central Atom:
- 0
- Molecule Polarity:
- Non-polar
4 bonding pairs and 0 lone pairs repel equally in 3D space to the vertices of a regular tetrahedron (109.5°).
Key Concepts & Exam Definitions (6)
- Transition Metal Definition & Why Sc and Zn Are Excluded
- A transition metal is an element that forms at least one stable ion with a partially filled d-subshell (d¹ to d⁹). Scandium only forms Sc³⁺ ([Ar] 3d⁰, empty 3d subshell) and Zinc only forms Zn²⁺ ([Ar] 3d¹⁰, completely full 3d subshell); therefore, neither Sc nor Zn is a transition metal, and their compounds are white/colourless because d-d electron transitions cannot occur.
- Complex Ions, Coordination Number & Ligand Types
- A complex ion consists of a central metal ion surrounded by ligands bonded via coordinate (dative covalent) bonds, where each ligand donates at least one lone pair of electrons. Coordination number = total number of coordinate bonds (not number of ligands) to the central metal ion. Monodentate ligands (1 lone pair donated): H₂O:, :NH₃, :Cl⁻, :CN⁻. Bidentate ligands (2 lone pairs donated): ethane-1,2-diamine (en, H₂NCH₂CH₂NH₂) and ethanedioate (C₂O₄²⁻). Multidentate/Hexadentate ligand (6 lone pairs): EDTA⁴⁻.
- Shapes & Stereoisomerism of Complex Ions
- 1. Coordination number 6 (small ligands like H₂O, NH₃, en, EDTA⁴⁻, e.g. [Cu(H₂O)₆]²⁺): Octahedral, 90° (exhibits cis-trans and optical isomerism in [M(AA)₂X₂] or [M(AA)₃]). 2. Coordination number 4 (large Cl⁻ ligands that cannot fit 6 around the ion, e.g. [CuCl₄]²⁻, [CoCl₄]²⁻): Tetrahedral, 109.5°. 3. Coordination number 4 in d⁸ complexes (Pt²⁺, Ni²⁺, e.g. cisplatin [Pt(NH₃)₂Cl₂]): Square planar, 90° (cis-isomer has identical ligands at 90°; trans-isomer at 180°). 4. Coordination number 2 (e.g. Tollens' reagent [Ag(NH₃)₂]⁺): Linear, 180°.
- Origin of Colour in Transition Metal Complexes (ΔE = hν = hc/λ) & Colorimetry
- In an isolated gaseous ion, all five 3d orbitals are degenerate (same energy). When ligands form coordinate bonds, electrostatic repulsion splits the 3d orbitals into a lower-energy set and a higher-energy set separated by an energy gap ΔE. Visible light shines on the complex; electrons in the lower 3d orbitals absorb light of frequency ν where ΔE = hν = hc / λ and are excited to the higher 3d orbitals, while the remaining unabsorbed wavelengths are transmitted (giving the complementary colour). ΔE depends on: 1. Central metal ion identity, 2. Oxidation state, 3. Ligand identity, and 4. Coordination number/shape. Colorimetry uses a complementary filter and a calibration curve of absorbance vs concentration (Beer-Lambert law).
- Ligand Substitution & The Chelate Effect (Entropy Driven)
- Replacing monodentate ligands with a bidentate or multidentate ligand (e.g. [Cu(H₂O)₆]²⁺(aq) + EDTA⁴⁻(aq) → [Cu(EDTA)]²⁻(aq) + 6H₂O(l)) is thermodynamically very favourable (the chelate effect): 1. Because 6 Cu-O bonds are broken and 6 similar Cu-O/Cu-N bonds are formed, ΔH ≈ 0. 2. However, 2 moles of reactant particles produce 7 moles of product particles in solution (2 → 7), causing a huge increase in disorder and a large positive entropy change (ΔS ≫ 0). 3. Therefore, ΔG = ΔH - TΔS is strongly negative (ΔG ≪ 0), driving the substitution to completion.
- Heterogeneous, Homogeneous & Autocatalysis
- 1. Heterogeneous catalysis (catalyst in different phase from reactants, e.g. V₂O₅(s) in Contact process: SO₂ + V₂O₅ → SO₃ + V₂O₄, then V₂O₄ + ½O₂ → V₂O₅; or Fe(s) in Haber process): reactants adsorb onto active sites, weakening covalent bonds, react, and desorb (poisoned by impurities like sulfur/lead blocking active sites). 2. Homogeneous catalysis (same aqueous phase via variable oxidation states): Fe²⁺/Fe³⁺ catalysing S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq) (uncatalysed reaction is slow because two negative ions repel; Fe²⁺ is oxidised by S₂O₈²⁻ to Fe³⁺, and Fe³⁺ is reduced by I⁻ back to Fe²⁺, so each step attracts oppositely charged ions). 3. Autocatalysis: Mn²⁺ product catalyses the reaction between MnO₄⁻ and C₂O₄²⁻.
Quick-Fire Flashcards (6)
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Why are Scandium (Sc) and Zinc (Zn) classified as d-block elements but not transition metals, and why are their aqueous ions colourless?
Explain why [Cu(H₂O)₆]²⁺ is blue, and state the four factors that change the colour (ΔE) of a transition metal complex.
Write the ionic equation and state the colour and geometry change when excess concentrated HCl is added to aqueous [Cu(H₂O)₆]²⁺.
Explain in terms of ΔH, ΔS, and ΔG why multidentate ligands like EDTA⁴⁻ replace monodentate H₂O ligands so readily (the chelate effect).
Why is the uncatalysed reaction between peroxodisulfate ions (S₂O₈²⁻) and iodide ions (I⁻) very slow, and how do Fe²⁺ ions catalyse it?
What is autocatalysis in the titration of ethanedioate ions (C₂O₄²⁻) with acidified manganate(VII) ions (MnO₄⁻)?
Exam-Style Questions & Mark Schemes (4)
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- Q1
Explain why aqueous solutions containing [Cu(H₂O)₆]²⁺ ions are coloured, whereas aqueous solutions containing [Cu(H₂O)₂]⁺ or [Zn(H₂O)₆]²⁺ ions are colourless. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. In [Cu(H₂O)₆]²⁺, the Cu²⁺ ion has a partially filled 3d⁹ subshell, and the coordinate bonds from the six H₂O ligands split the five 3d orbitals into two different energy levels separated by an energy gap ΔE [1 mark]. 2. When white visible light passes through the solution, electrons in the lower-energy 3d orbitals absorb light of a specific frequency/wavelength corresponding to ΔE = hν = hc/λ and are excited (promoted) to the higher-energy 3d orbitals [1 mark]. 3. The remaining unabsorbed wavelengths of visible light are transmitted, giving the solution its complementary blue colour [1 mark]. 4. In [Cu(H₂O)₂]⁺ (Cu⁺ is 3d¹⁰) and [Zn(H₂O)₆]²⁺ (Zn²⁺ is 3d¹⁰), the 3d subshell is completely full (10 electrons) [1 mark]. 5. Therefore, no empty space exists in the upper 3d orbitals for d-d electron promotion to occur, so no visible light is absorbed and the solutions appear colourless [1 mark].
- Q2
Write balanced ionic equations and describe the observations when: (a) aqueous sodium hydroxide is added dropwise and then in excess to aqueous iron(II) sulfate, and (b) aqueous ammonia is added dropwise and then in excess to aqueous copper(II) sulfate. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Adding NaOH(aq) to [Fe(H₂O)₆]²⁺(aq) (pale green solution) produces a green gelatinous precipitate (which darkens/turns rusty brown on standing in air due to oxidation to Fe(III)) that does not dissolve in excess NaOH [1 mark]. 2. Equation for (a) (deprotonation / acid-base reaction): [Fe(H₂O)₆]²⁺(aq) + 2OH⁻(aq) → [Fe(H₂O)₄(OH)₂](s) + 2H₂O(l) [1 mark]. 3. (b) Adding NH₃(aq) dropwise to [Cu(H₂O)₆]²⁺(aq) (pale blue solution) first forms a pale blue precipitate of [Cu(H₂O)₄(OH)₂](s) by deprotonation: [Cu(H₂O)₆]²⁺(aq) + 2NH₃(aq) → [Cu(H₂O)₄(OH)₂](s) + 2NH₄⁺(aq) [1 mark]. 4. Adding excess NH₃(aq) causes partial ligand substitution, dissolving the pale blue precipitate to form a deep royal-blue solution [1 mark]. 5. Equation for excess NH₃: [Cu(H₂O)₄(OH)₂](s) + 4NH₃(aq) → [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2OH⁻(aq) + 2H₂O(l) (or overall from [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O) [1 mark].
- Q3
Explain with the aid of equations how solid vanadium(V) oxide (V₂O₅) acts as a heterogeneous catalyst in the Contact process (2SO₂(g) + O₂(g) ⇌ 2SO₃(g)), and explain why heterogeneous catalysts are supported on a ceramic honeycomb mesh. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. V₂O₅ is in a different physical state (solid) from the gaseous reactants SO₂ and O₂, and uses vanadium's variable oxidation states (+5 and +4) to provide an alternative pathway of lower activation energy [1 mark]. 2. Step 1 (Vanadium reduced from +5 to +4): V₂O₅(s) + SO₂(g) → V₂O₄(s) + SO₃(g) [1 mark]. 3. Step 2 (Vanadium re-oxidised from +4 back to +5 by oxygen, regenerating the catalyst): V₂O₄(s) + ½O₂(g) → V₂O₅(s) [1 mark]. 4. Coating a thin layer of V₂O₅ onto an inert ceramic honeycomb support maximises the surface area of exposed active sites while minimising the mass and cost of expensive vanadium oxide required [1 mark].
- Q4
Explain why aqueous [Al(H₂O)₆]³⁺ has a pH of ~3 (more acidic than [Fe(H₂O)₆]²⁺, which has a pH of ~6), and explain why adding aqueous sodium carbonate (Na₂CO₃) to [Al(H₂O)₆]³⁺ produces effervescence and a white precipitate, whereas adding Na₂CO₃ to [Fe(H₂O)₆]²⁺ produces a green precipitate with no gas. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. The Al³⁺ ion has a higher charge (+3 vs +2) and a smaller ionic radius than Fe²⁺, giving Al³⁺ a much higher charge density (polarising power) [1 mark]. 2. Consequently, Al³⁺ withdraws electron density much more strongly from the O-H covalent bonds of the coordinated H₂O ligands, weakening the O-H bonds so they dissociate more readily to release H⁺(aq) ions: [Al(H₂O)₆]³⁺(aq) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H⁺(aq) [1 mark]. 3. Because [Al(H₂O)₆]³⁺ is sufficiently acidic, the released H⁺ ions protonate carbonate ions (CO₃²⁻ + 2H⁺ → H₂O + CO₂(g)), shifting the hydrolysis equilibrium all the way to the neutral hydroxide precipitate [1 mark]. 4. Equation for Al³⁺ + CO₃²⁻: 2[Al(H₂O)₆]³⁺(aq) + 3CO₃²⁻(aq) → 2[Al(H₂O)₃(OH)₃](s) + 3CO₂(g) + 3H₂O(l) (white precipitate + bubbles of CO₂) [1 mark]. 5. Because [Fe(H₂O)₆]²⁺ is not acidic enough to protonate CO₃²⁻ to CO₂, it undergoes simple precipitation of insoluble iron(II) carbonate with no gas: [Fe(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → FeCO₃(s) + 6H₂O(l) [1 mark].
Common Exam Questions (4)
What does 'amphoteric' mean for metal hydroxides like [Al(H₂O)₃(OH)₃](s) or [Cr(H₂O)₃(OH)₃](s)?
An amphoteric hydroxide can act as both a Brønsted-Lowry base (reacting with added acid H⁺ to dissolve back into [M(H₂O)₆]³⁺) and a Brønsted-Lowry acid (donating H⁺ from coordinated H₂O ligands to excess strong alkali OH⁻ to dissolve into a soluble tetrahydroxo or hexahydroxo complex like [Al(OH)₄]⁻ or [Cr(OH)₆]³⁻).
What is the difference between a deprotonation (acid-base) reaction and a ligand substitution reaction when NH₃(aq) is added to [Cu(H₂O)₆]²⁺?
When small amounts of NH₃ are added, NH₃ acts as a Brønsted-Lowry base and pulls H⁺ ions off two H₂O ligands (deprotonation) to precipitate [Cu(H₂O)₄(OH)₂](s) + 2NH₄⁺; the coordination number (6) and Cu-O bonds do not change. Only when excess NH₃ is added does NH₃ act as a ligand, replacing 4 ligands in a ligand substitution reaction to form [Cu(NH₃)₄(H₂O)₂]²⁺.
Why must a manganate(VII) (KMnO₄) redox titration be acidified with dilute sulfuric acid (H₂SO₄) rather than HCl or HNO₃?
MnO₄⁻ requires a high concentration of H⁺(aq) to be reduced completely to colourless Mn²⁺(aq) (instead of brown solid MnO₂). Hydrochloric acid (HCl) cannot be used because MnO₄⁻ is a strong enough oxidising agent to oxidise Cl⁻ ions to Cl₂ gas (giving a false high titre), and nitric acid (HNO₃) is itself an oxidising agent.
How does haem in haemoglobin illustrate complex ion chemistry and carbon monoxide poisoning?
Haem is an octahedral Fe²⁺ complex with four planar coordinate bonds to a porphyrin ring (multidentate ligand), one coordinate bond to the globin protein (histidine), and one reversible coordinate site for O₂ (or H₂O). Carbon monoxide (:CO) is a strong ligand that forms a much stronger coordinate bond to Fe²⁺ than O₂, irreversibly substituting O₂ in a ligand substitution reaction.
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