Catalytic and interstitial properties

Why transition metals catalyse reactions. Two independent mechanisms, and a given metal may use either or both.

• Variable oxidation states (homogeneous catalysis). A transition-metal ion can flip between two stable, close-in-energy oxidation states, $\text{M}^{n+} \rightleftharpoons \text{M}^{(n+1)+}$, acting as an electron shuttle. It gets consumed forming an intermediate in one step and is regenerated in a later step, so the net amount is unchanged at the end — the signature of a true catalyst. This is only possible because transition metals commonly have two or more oxidation states of comparable stability; an alkali-metal ion, with essentially one accessible oxidation state, cannot do it.
• Surface adsorption (heterogeneous catalysis). A transition-metal surface offers a large area with vacant or partly filled d-orbitals. Reactant gas molecules bind (adsorb) to these sites; the bonding weakens bonds inside the adsorbed molecule and holds two reactants in close, favourable orientation, lowering the activation energy for them to react. The metal itself is not consumed — it is a physical surface effect, so finely dividing the metal (more surface area) makes it a better catalyst.

Why transition metals trap small atoms — interstitial compounds.

• Transition metals crystallise in close-packed metallic lattices, but "close-packed" still leaves small octahedral and tetrahedral gaps (interstitial sites) between the metal atoms.
• Small non-metal atoms — H, C, N, B — are little enough to occupy these gaps without pushing the metal atoms apart or destroying the metallic lattice. The result is an interstitial compound: $\text{Fe}_3\text{C}$ (steel/cementite), $\text{TiN}$, $\text{WC}$ (tungsten carbide) are the standard examples.
• The trapped atoms act like pins between the metal layers, making it mechanically harder for one layer to slide over another. Interstitial compounds are consequently harder and more brittle than the pure metal and have very high melting points, while the delocalised electron sea of the original lattice survives largely intact, so they remain metallic conductors.

Why transition metals mix with each other — alloys. Atomic radii change only slowly across a transition series (the "filling" electrons go into an inner d subshell, largely shielding the outer size), so metals within or near the same series are similar enough in size that one can substitute for another in the lattice without straining it. This gives substitutional alloys — stainless steel (Fe–Cr–Ni), brass (Cu–Zn) — with mechanical or chemical properties distinct from any parent metal. Alloying is size-matching substitution; it is a different mechanism from interstitial trapping, and the two are easy to conflate on an exam.

where: $\text{M}^{n+}$ is the catalyst's starting oxidation state, step 1 is oxidation or reduction of the catalyst by one reactant, and step 2 restores $\text{M}^{n+}$ by reaction with the second reactant — the net catalyst amount cancels between the two steps, which is the algebraic signature that distinguishes a true catalytic cycle from a one-off reagent.

The whole topic reduces to one geometric fact and one electronic fact about the 3d series, and every named property follows from one or the other.

Step 1. Transition metals have partly filled d-orbitals close in energy across two adjacent oxidation states — the electronic fact. This is why $\text{Fe}^{2+}/\text{Fe}^{3+}$ and similar pairs are accessible with modest energy input: a $\text{d}^n$ ion can lose or gain one electron without a large jump in stability, unlike $\text{Na}^+$, whose next state would mean breaking a stable noble-gas core.

Step 2. Because two oxidation states are both stable and interconvertible, the metal ion acts as an electron relay — oxidised by one reactant, reduced back by another — so it appears unchanged overall but has broken the net reaction into two easier steps (homogeneous catalysis). On a metal surface, the same partly filled d-orbitals bind gas molecules, weakening their internal bonds and holding reactants in a favourable geometry (heterogeneous catalysis) — a different route to the same outcome, a lower effective activation energy.

Step 3. Separately, the geometric fact: metal atoms in a crystal lattice leave interstitial gaps too small for another metal atom but large enough for H, C, N or B. Occupying these gaps does not disturb the metallic bonding — it adds mechanical resistance to layers sliding past each other, which is exactly what "hardness" means here, while the delocalised electrons (and hence conductivity) stay intact.

Step 4. Because atomic radius changes only gradually across a transition series, a metal atom is often close enough in size to a neighbouring transition metal to occupy its lattice site directly — substitution, not insertion — giving an alloy rather than an interstitial compound. The single geometric test: small enough to fit in a gap, or similar enough in size to substitute.

Problem. Explain, with the specific reaction, how $\text{Fe}^{3+}$ catalyses the reaction between iodide ions and persulphate ions, and then explain why steel (Fe with a little carbon) is harder than pure iron yet still conducts electricity like a metal.

Step 1. The uncatalysed reaction is $2\text{I}^- + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2\text{SO}_4^{2-}$. It is slow because two negative ions must approach closely, and electrostatic repulsion raises the activation energy.

Step 2. Add $\text{Fe}^{3+}$. It first oxidises iodide: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ — a reaction between oppositely signed species, so it proceeds far more easily than ion-ion repulsion allows.

Step 3. The $\text{Fe}^{2+}$ produced is then reoxidised by persulphate: $2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}$. Adding steps 2 and 3 reproduces the original overall equation exactly, with $\text{Fe}^{3+}/\text{Fe}^{2+}$ cancelling to zero net change — confirming it is a genuine catalyst, not a consumed reagent.

Step 4. For steel: pure iron has a close-packed lattice with small interstitial gaps. Carbon atoms are small enough to occupy some of these gaps, forming the interstitial compound $\text{Fe}_3\text{C}$ dispersed through the iron. The trapped carbon pins the iron layers, resisting slip under stress, so steel is harder and more brittle than pure iron. Because carbon occupies gaps rather than replacing Fe atoms, the original metallic bonding and delocalised electron sea survive largely unchanged, so steel still conducts electricity and has a high melting point like a metal.

Answer: Fe$^{3+}$ catalyses the reaction by cycling $\text{Fe}^{3+} \rightleftharpoons \text{Fe}^{2+}$ (net catalyst unchanged), and steel is harder than pure iron because carbon pins the interstitial gaps of the Fe lattice while metallic bonding, and hence conductivity, is retained.

Competitive stretch

Classify $\text{TiN}$ vs stainless steel in one line: $\text{TiN}$ is interstitial (small N atoms fill gaps in the Ti lattice, giving cutting-tool hardness); stainless steel is a substitutional alloy (Cr, Ni atoms of comparable radius replace Fe atoms, giving corrosion resistance). Same test both times — gap-filling or site-substitution.

• Confusing interstitial compounds (small non-metal atoms in lattice gaps, e.g. Fe–C) with alloys (substitution of one similarly sized metal atom for another, e.g. Fe–Cr–Ni) — different atom type, different mechanism.
• Assuming any transition metal catalyses any reaction — catalytic activity depends on having accessible oxidation states (homogeneous) or a surface that adsorbs the specific reactants (heterogeneous); it is not a universal property.
• Treating interstitial compounds as ionic or covalent solids — they retain essentially metallic bonding and conductivity; only hardness, brittleness and melting point rise sharply.
• Forgetting that alloying needs atomic radii to be reasonably close — even between two transition metals, a large size mismatch prevents good substitutional mixing.

Ask the tutor: want more worked examples of transition-metal catalysis, such as Ni in alkene hydrogenation or MnO2 decomposing hydrogen peroxide?