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Unit IV · The d- and f-block elements
9 min readConcept sheet2-mark "give reasons" question; 3-mark question with the Fe³⁺ catalysis equations; 1-mark MCQ or assertion-reason

Catalytic properties, interstitial compounds and alloys

Transition metals are good catalysts because of variable oxidation states and surface bonding. Small atoms in their lattices give hard interstitial compounds; similar sizes give alloys.

Why good catalystsVariable oxidation states, ability to form complexes, and a surface that can bond to reactants using 3d and 4s electrons
Key catalystsV₂O₅ (Contact process), finely divided Fe (Haber process), Ni (hydrogenation of oils)
Interstitial compoundsSmall H, C or N atoms trapped in metal lattices, e.g. TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇; usually non-stoichiometric
Alloy ruleMetals whose atomic radii are within about 15% of each other form alloys (brass Cu-Zn, bronze Cu-Sn, steels)

Why transition metals are good catalysts

Transition metals and their compounds are well-known catalysts. There are two reasons. First, they can change oxidation state easily, so they can take electrons from one reactant and pass them to another. Second, solid catalysts bond to reactant molecules at their surface, using their 3d and 4s electrons. This brings reactants together on the surface, raising their concentration there, and weakens the bonds in the reacting molecules. Both effects lower the activation energy.

Transition metal catalysts in industry

ProcessCatalystReaction
Contact process (sulphuric acid)V₂O₅2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Haber process (ammonia)Finely divided ironN₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Hydrogenation of vegetable oilsNickelUnsaturated oil + H₂ → saturated fat (vanaspati)

How Fe³⁺ catalyses the iodide-persulphate reaction

Overall: 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻. This is slow on its own, because two negative ions repel each other.

  1. 1
    2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
    Fe³⁺ oxidises iodide and is reduced to Fe²⁺.
  2. 2
    2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻
    Persulphate re-oxidises Fe²⁺ back to Fe³⁺.
  3. 3★
    Add the steps: 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻
    Fe³⁺ is given back unchanged, so it is a catalyst. Each step joins oppositely charged ions, so it is fast.

★ key step

Interstitial compounds

  • They form when small atoms like H, C or N are trapped in the empty spaces (interstices) of a metal's crystal lattice.
  • Their formulas do not follow simple whole-number ratios (non-stoichiometric), and their bonding is not simply ionic or covalent: TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇.
  • They have higher melting points than the pure metals.
  • They are very hard; some borides are nearly as hard as diamond.
  • They keep metallic conductivity and are chemically inert.
  • Steel and cast iron are hard because a little carbon sits in the holes of the iron lattice.

Alloy formation

An alloy is made by mixing metals together. In many alloys the atoms of one metal simply take random places among the atoms of the other, giving a uniform (homogeneous) solid solution. This works when the atomic radii differ by no more than about 15%. Transition metals have similar radii, so they form alloys readily. Such alloys are hard and often have high melting points. Cr, V, W, Mo and Mn are added to iron to make different steels, including stainless steel. Alloys of transition and non-transition metals, such as brass (copper and zinc) and bronze (copper and tin), are also very useful.

In real life

Vanaspati is made by hydrogenating vegetable oil over a nickel catalyst, which is why it is solid at room temperature. Your stainless steel tiffin box and plates are an alloy of iron with chromium and nickel.

Exam angle

2 marks: “Why do transition metals and their compounds act as catalysts? Give two examples.” Variable oxidation states and surface bonding (using 3d and 4s electrons) lower the activation energy. V₂O₅ in the Contact process; finely divided Fe in the Haber process.

2 marks: “What are interstitial compounds? State two of their properties.” Compounds formed when small atoms (H, C, N) are trapped in metal lattices, e.g. TiC. They are very hard and have higher melting points than the pure metal; they keep metallic conductivity.

3 marks: “Explain, with equations, how Fe³⁺ ions catalyse the reaction between iodide and persulphate ions.” 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂; 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻. Net: 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻; Fe³⁺ is regenerated, and each step joins oppositely charged ions.

Remember it

Catalysts: V for Vitriol (V₂O₅, sulphuric acid), Iron for ammonia, Nickel for vanaspati.