d- and f-Block Elements · Lesson 6 of 12
Complexes, Catalysts and Useful Metal Materials
“Connect ligand binding, surface chemistry and crystal structure with industrial applications.”
• Identify ligands, charge, oxidation state and coordination number in representative complexes. • Explain heterogeneous and redox catalysis without changing equilibrium. • Distinguish interstitial compounds from substitutional alloys. • Relate material and catalytic uses to chemical properties.
Ligands create new chemical environments
When ammonia is added to a suitable copper(II) solution, the familiar pale-blue environment can change to an intensely blue ammine complex. Molecules or ions with donor electron pairs bind to a metal centre; they are called ligands. Binding changes electron energies, colour, redox stability and sometimes solubility. Small, relatively highly charged metal ions attract donor groups strongly, and suitable metal orbitals participate in the bonding. Describing this solely as “empty d orbitals accept pairs” is too narrow: metal–ligand bonding can involve s, p and d character and covalent interactions.
| Complex | Metal state and d count | Ligands and coordination number |
|---|---|---|
| [Fe(CN)₆]³⁻ | Fe(III), d⁵ | Six monodentate cyanides; CN=6 |
| [Fe(CN)₆]⁴⁻ | Fe(II), d⁶ | Six monodentate cyanides; CN=6 |
| [Cu(NH₃)₄]²⁺ | Cu(II), d⁹ | Four N donors in the simplified formula; CN=4 |
| [PtCl₄]²⁻ | Pt(II), d⁸ | Four monodentate chlorides; CN=4 |
Coordination number counts directly attached donor atoms, not simply the number of ligand molecules when a ligand has several donor atoms. In aqueous copper ammine solutions, additional axial water interactions may be present, so the compact [Cu(NH₃)₄]²⁺ formula is not a complete claim that water has disappeared from every coordination environment. For [PtCl₄]²⁻ a square-planar structure is familiar, but a coordination number of four alone could also describe a tetrahedral arrangement in another metal–ligand system.
Problem
Explain the metal-state difference between [Fe(CN)₆]³⁻ and [Fe(CN)₆]⁴⁻.
- 1.Each cyanide ligand has formal charge −1. For the first ion, x−6=−3 gives Fe(III). For the second, x−6=−4 gives Fe(II).
- 2.Both contain six donor atoms and may have the same coordination-number description, while differing by one electron and one net charge.
- 3.Ligand strength can also affect spin pairing. Oxidation state, coordination number and magnetic behaviour are related but distinct properties.
A catalyst offers a different path
A catalyst increases reaction rate by providing a pathway with a lower effective activation barrier. It participates in intermediate steps and is regenerated overall. It does not change the standard Gibbs energy between the same reactants and products, the equilibrium constant or the stoichiometric material balance. “Regenerated” does not mean “never reacts.” Transition metals are useful because they can bind reactants and often interconvert between oxidation states.
On finely divided nickel, hydrogen and an unsaturated organic molecule adsorb at the surface. Binding concentrates reactants near one another and can weaken bonds before a different sequence produces hydrogenated material and releases it from the surface. More exposed surface can increase accessible catalytic sites, but excessive binding could also hinder release. Iron in ammonia synthesis and V₂O₅-based catalysts in sulfur dioxide oxidation illustrate that the catalyst must suit the specific reaction and operating conditions.
| Process | Catalyst or metal system | Property put to work |
|---|---|---|
| N₂+3H₂⇌2NH₃ | Iron-based catalyst | Surface activation of strongly bonded reactants |
| 2SO₂+O₂⇌2SO₃ | V₂O₅-based catalyst | Accessible vanadium redox chemistry |
| Hydrogenation of unsaturated fats | Finely divided Ni | Adsorption and hydrogen activation |
| Ethene to ethanal, Wacker oxidation | Pd(II)/Cu(II) chloride system | Coordination, oxidation and catalyst regeneration |
| Ethene polymerisation | Ti chloride/organoaluminium systems | Controlled coordination and chain growth |
| Alkyne cyclotrimerisation to aromatic products | Suitable Ni complexes | Substrate coordination and bond reorganisation |
Problem
How can Fe³⁺/Fe²⁺ catalyse oxidation of iodide by peroxodisulfate?
- 1.In one step, 2Fe³⁺+2I⁻→2Fe²⁺+I₂. Iron(III) oxidises iodide and becomes iron(II).
- 2.In another, 2Fe²⁺+S₂O₈²⁻→2Fe³⁺+2SO₄²⁻. Peroxodisulfate reoxidises the iron(II).
- 3.Add and cancel both iron forms: 2I⁻+S₂O₈²⁻→I₂+2SO₄²⁻. Iron is absent from the net reaction but present in the pathway.
- 4.Check charge: −4 on each side. Regeneration permits turnover; these equations describe a catalytic redox cycle without claiming each written overall step is a proven single elementary collision.
Small atoms can fit between metal atoms
A metal lattice contains spaces between its atoms. Small H, C or N atoms can occupy some of these spaces while much of the metal framework remains intact. Such interstitial compounds can retain metallic conductivity while becoming harder and more refractory. Occupancy need not correspond to a fixed ratio of small atoms to metal atoms, so compositions such as VH₀.₅₆ or TiH₁.₇ are possible. Other examples include TiC and manganese nitrides.
An interstitial formula is not a reliable basis for assigning a simple integral ionic oxidation state to every metal atom. These solids combine metallic and covalent contributions and can have non-stoichiometry. Many are hard, high-melting and chemically resistant, but “all interstitial compounds are inert under every condition” would be too strong; hydrides, for example, can exchange hydrogen under suitable conditions.
Alloys modify a metal lattice in a different way
An alloy is a metallic material containing two or more elements, at least one of which is a metal. Similar-sized metal atoms can substitute for one another in a lattice. A radius difference of roughly 15% or less is a useful criterion favouring extensive substitutional solid solubility, not a guarantee: electronic structure and crystal form also matter. Alloys need not all be uniform random solutions; ordered phases and multiple phases are possible. Carbon-containing steels also illustrate that alloying can include interstitial contributions.
| Material or compound | Composition or feature | Use linked to a property |
|---|---|---|
| Steel and stainless steel | Fe with controlled C and alloying metals such as Cr, Ni, Mn, V, Mo or W | Strength, hardness or corrosion resistance |
| Brass | Cu–Zn alloy | Useful mechanical and working properties |
| Bronze | Cu–Sn alloy | Harder, useful engineering material |
| Mischmetal | Typically about 95% rare-earth metals and about 5% Fe with minor constituents | Lighter flints and alloy additions |
| TiO₂ | Stable light-scattering white oxide | Pigment |
| MnO₂ | Redox-active oxide | Dry-cell chemistry |
| Zn, Ni and Cd | Appropriate electrode chemistry | Battery systems |
| AgBr | Light-sensitive silver halide | Photographic recording |
Applications should be connected to the form actually used. White pigment is predominantly titanium dioxide, TiO₂, not the lower oxide TiO. Silver bromide is a compound with photochemical properties, whereas silver metal is conductive. Alloy additions improve specific properties through structure and phase behaviour, not merely because “transition metals are strong.” Some uses are historical or specialised, such as particular coin alloys or photographic films; the underlying chemistry remains useful even when products change.
Quick Quiz
Quiz
What is the Fe state in [Fe(CN)₆]⁴⁻?
Explanation: Six cyanides contribute −6, so Fe must contribute +2.
What does a catalyst change for a fixed net reaction?
Explanation: A catalyst affects rate through a different pathway and is regenerated.
Why can VH₀.₅₆ have a nonintegral formula?
Explanation: A bulk average composition can reflect incomplete interstitial occupancy.
Which is the white titanium pigment?
Explanation: Titanium dioxide is the widely used white light-scattering oxide.
Which substrate gives ethanal in the familiar Wacker process?
Explanation: Pd-based Wacker chemistry oxidises ethene to ethanal.
Practice Problems
- Distinguish oxidation state, coordination number and ligand denticity using [Fe(CN)₆]³⁻.
- Add the two iron catalytic equations and verify the net balance.
- Explain why powdering Ni can increase its catalytic activity.
- Compare substitutional alloying and interstitial compound formation.
- Why is a 15% radius match insufficient to guarantee a homogeneous alloy?
- Connect three applications to three different properties.
Fe is +3 because x−6=−3. The coordination number is six because six donor atoms attach to Fe. Each cyanide is monodentate in this representation. These numbers answer different questions and should not be substituted for one another.
Key Takeaways
• Ligands alter a metal centre’s chemical and electronic environment. • Metal state and coordination number require separate calculations. • Catalysts react along a pathway and are regenerated overall. • Catalysts change rates while preserving the equilibrium of a specified net reaction. • Interstitial occupancy can produce hard conducting solids with variable composition. • Alloys can be substitutional, interstitial, ordered or multiphase materials. • Useful applications follow specific properties of the actual metal, compound or alloy.