1. Transition Metal Definition & Characteristic Properties
- Scandium (Sc): Electronic configuration is [Ar] 3d1 4s2. It forms only one stable ion, Sc3+, which has the configuration [Ar] 3d0. Because the 3d sub-level is completely empty, scandium does not satisfy the definition.
- Zinc (Zn): Electronic configuration is [Ar] 3d10 4s2. It forms only the Zn2+ ion, with configuration [Ar] 3d10. Because the 3d sub-level is completely full, zinc does not satisfy the definition.
Four Characteristic Properties of Transition Metals
- Variable oxidation states: e.g. Fe exists as Fe2+ and Fe3+; Mn exists from +2 to +7. (The 3d and 4s electrons have very similar energies, allowing differing numbers of electrons to be lost).
- Formation of complex ions: Readily coordinate with lone-pair donor ligands.
- Formation of coloured compounds: Absorption of visible light due to d-d electronic transitions.
- Catalytic activity: Act as heterogeneous and homogeneous catalysts due to variable oxidation states and vacant d orbitals.
2. Complex Ions, Ligands & Coordination Number
- Complex Ion: A central transition metal ion surrounded by coordinately bonded ligands.
- Ligand: An atom, ion, or molecule that possesses at least one lone pair of electrons and donates it to a central metal ion to form a coordinate (dative covalent) bond.
- Coordination Number: The total number of coordinate bonds formed between the central metal ion and its surrounding ligands.
Classification of Ligands by Denticity
| Denticity | Definition | Representative Examples | Typical Coordination Number & Complex |
|---|---|---|---|
| Monodentate | Donates ONE lone pair per ligand molecule/ion. | :H2O, :NH3, :Cl-, :CN-, :OH- | Coord 6: [Fe(H2O)6]2+, [Cu(H2O)6]2+
Coord 4: [CuCl4]2- (Cl- is large and charged) |
| Bidentate | Donates TWO lone pairs from different atoms in the same molecule. | 1,2-diaminoethane (en, H2NCH2CH2NH2)
Ethanedioate ion (C2O4 2-) |
Coord 6: [Fe(C2O4)3]3-, [Cr(en)3]3+ |
| Multidentate | Donates THREE OR MORE lone pairs from different atoms in the same molecule. | EDTA4- (hexadentate: donates 6 lone pairs from 4 carboxylate oxygens and 2 amine nitrogens) | Coord 6: [Cu(EDTA)]2-, [Fe(EDTA)]- |
3. The Chelate Effect & Thermodynamic Stability
When monodentate ligands (such as H2O or NH3) are replaced by bidentate or multidentate ligands (such as EDTA4- or 1,2-diaminoethane), the resulting complex is vastly more stable. This phenomenon is known as the chelate effect.
Thermodynamic Justification of the Chelate Effect
Consider the ligand substitution of hexaaquacopper(II) by EDTA4-:
[Cu(H2O)6]2+(aq) + EDTA4-(aq) -> [Cu(EDTA)]2-(aq) + 6H2O(l)
- Enthalpy change (delta H): Metal-ligand coordinate bond strengths are very similar (Cu-O and Cu-N bonds broken vs formed), so delta H is approximately zero.
- Entropy change (delta S): On the left side of the equation, there are 2 particles in solution (one complex ion and one EDTA4-). On the right side, there are 7 particles (one chelated complex ion and six released water molecules). The increase in independent particles in solution creates substantial disorder, producing a large positive entropy change (delta S > 0).
- Gibbs free energy (delta G): Since delta G = delta H - T * delta S, the large positive delta S term makes delta G highly negative, driving the reaction decisively forward and rendering the chelate complex exceptionally stable.
4. Geometries of Transition Metal Complexes
- Octahedral (Coordination Number 6, 90 degree bond angles): Formed with small, uncharged ligands like H2O and NH3. Examples: [Fe(H2O)6]2+, [Cu(H2O)6]2+, [Fe(CN)6]3-, [Cr(NH3)6]3+.
- Tetrahedral (Coordination Number 4, 109.5 degree bond angles): Formed when ligands are large and charged, such as chloride ions (Cl-). Steric hindrance and electrostatic repulsions prevent 6 chloride ions from packing around the metal. Examples: [CuCl4]2- (yellow-green), [CoCl4]2- (deep blue).
- Square Planar (Coordination Number 4, 90 degree bond angles): Formed by platinum(II) and nickel(II) complexes. The most famous example is the anticancer chemotherapy drug cisplatin, [Pt(NH3)2Cl2].
- Linear (Coordination Number 2, 180 degree bond angles): Formed by silver(I) complexes, notably Tollens' reagent: [Ag(NH3)2]+.
5. D-Orbital Splitting Diagram
The diagram below displays how the five degenerate 3d orbitals split into two distinct energy levels under the electrostatic influence of octahedral ligands:
6. Origin of Color in Complex Ions
The distinctive, vibrant colors of transition metal complexes arise from electronic transitions between split d-orbitals:
- In an isolated gaseous transition metal atom or ion, all five 3d orbitals are degenerate (identical in energy).
- When ligands approach the metal ion along the Cartesian axes (in an octahedral complex), the lone pairs on the ligands electrostatically repel electrons in the d-orbitals that point directly along the axes (dx2-y2 and dz2) more strongly than those pointing between the axes (dxy, dyz, dxz).
- This unequal repulsion splits the 3d sub-level into two non-degenerate energy levels separated by an energy gap, delta E.
- When white light shines on the complex, an electron absorbs a specific frequency of visible light and is promoted from the lower d-level to the higher d-level. The energy absorbed is governed by Planck's equation:
delta E = h * nu = (h * c) / lambda - The frequencies of light that are NOT absorbed are transmitted or reflected through the solution. The human eye perceives the complementary color to the absorbed wavelength (e.g. absorbing red-orange light at ~600 nm transmits blue light, making hydrated copper(II) appear blue).
For a d-d transition to occur, there must be an electron in a lower d-orbital capable of promotion, AND a vacancy in a higher d-orbital to accept it.
- Sc3+ ([Ar] 3d0) has no 3d electrons to promote: colorless.
- Zn2+ ([Ar] 3d10) has all 3d orbitals completely filled; there is no vacant orbital to receive an excited electron: colorless.
7. Catalytic Mechanisms
1. Heterogeneous Catalysis
The catalyst is in a different physical phase from the reactants (usually a solid catalyst with gaseous reactants). Reactant molecules adsorb onto active sites on the solid catalyst surface, weakening bonds and lowering activation energy. Products then desorb.
- Haber Process: Solid iron (Fe) catalyses N2(g) + 3H2(g) <=> 2NH3(g).
- Contact Process: Vanadium(V) oxide (V2O5) catalyses SO2(g) + 1/2 O2(g) <=> SO3(g).
Step 1: SO2 + V2O5 -> SO3 + V2O4 (V reduced from +5 to +4)
Step 2: V2O4 + 1/2 O2 -> V2O5 (V oxidised back to +5)
2. Homogeneous Catalysis
The catalyst is in the same physical phase as the reactants (typically both in aqueous solution). The transition metal ion functions by shuttling between variable oxidation states.
Peroxodisulfate-Iodide Reaction Catalysed by Fe2+
The uncatalysed reaction S2O8 2- + 2I- -> 2SO4 2- + I2 is extremely slow because both reactants are negatively charged, creating high electrostatic repulsion (high activation energy). Fe2+ ions provide a low-activation alternative pathway by reacting with one anion at a time:
- Step 1: S2O8 2- + 2Fe2+ -> 2SO4 2- + 2Fe3+ (Fe2+ oxidised to Fe3+)
- Step 2: 2Fe3+ + 2I- -> 2Fe2+ + I2 (Fe3+ reduced back to Fe2+)
Autocatalysis: Oxidation of Ethanedioic Acid by Manganate(VII)
2MnO4- + 16H+ + 5C2O4 2- -> 2Mn2+ + 8H2O + 10CO2. The reaction starts very slowly at room temperature due to repulsion between negative MnO4- and C2O4 2- ions. However, as the reaction proceeds, product Mn2+ ions accumulate and act as an autocatalyst, dramatically accelerating the reaction rate.
8. Worked Calculations
Calculate the energy gap (delta E) between the split d-orbitals in J and in kJ mol-1.
(Planck constant h = 6.63 x 10^-34 J s; speed of light c = 3.00 x 10^8 m s-1; Avogadro constant L = 6.022 x 10^23 mol-1).
Step 1: Calculate delta E for a single photon:
delta E = (h * c) / lambda
delta E = (6.63 x 10^-34 * 3.00 x 10^8) / (5.00 x 10^-7) = 3.978 x 10^-19 J
Step 2: Convert to energy per mole (kJ mol-1):
delta E_molar = delta E * L = (3.978 x 10^-19 J) * (6.022 x 10^23 mol-1) = 2.396 x 10^5 J mol-1
delta E_molar = 2.396 x 10^5 / 1000 = 240 kJ mol-1
Step 1: Write the balanced equation:
[Ni(H2O)6]2+(aq) + 3en(aq) <=> [Ni(en)3]2+(aq) + 6H2O(l)
Step 2: Explain using thermodynamics:
4 particles on the reactant side (1 complex + 3 en) form 7 particles on the product side (1 complex + 6 water molecules). The substantial increase in particles in solution produces a large positive entropy change (delta S > 0). Since delta H is close to zero, delta G = delta H - T * delta S becomes strongly negative, making the equilibrium position lie almost entirely to the right.
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: Which of the following elements is classified as a d-block element but NOT a transition metal?
Show Answer & Explanation
Correct Answer: D
Zinc forms only the Zn2+ ion, which has a full 3d10 sub-level. A transition metal must form at least one stable ion with a partially filled d sub-level (3d1 to 3d9).
Question 2: Why does the ligand exchange of [Cr(H2O)6]3+ with EDTA4- proceed with a high equilibrium constant?
Show Answer & Explanation
Correct Answer: B
This is the chelate effect: 1 [Cr(H2O)6]3+ + 1 EDTA4- -> 1 [Cr(EDTA)]- + 6 H2O. An increase from 2 to 7 independent particles creates significant disorder, driving the reaction via positive delta S.
Question 3: What is the coordination number and shape of the complex ion [CuCl4]2-?
Show Answer & Explanation
Correct Answer: B
Chloride ligands (Cl-) are relatively large and carry a negative charge. Due to steric crowding and electrostatic repulsion, only four chloride ligands can fit around copper, forming a tetrahedral complex.
Question 4: In the reaction between peroxodisulfate ions and iodide ions, why can Fe2+ ions act as an effective homogeneous catalyst?
Show Answer & Explanation
Correct Answer: B
Fe2+ readily oxidises to Fe3+ by reducing S2O8 2-, and Fe3+ subsequently oxidises I- to I2 while regenerating Fe2+. Alternating between Fe2+ and Fe3+ bypasses the high-repulsion barrier between like-charged anions.