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Ask why transition metals make good catalysts and most students say "variable oxidation states" and stop. That earns the first mark. The rest come from showing how a changing oxidation state lets a catalyst take part in a reaction and come back unchanged, and from knowing which metal does which industrial job. This post covers NCERT's explanation, the examples to know, and a worked electrode-potential check of an iron-catalysed reaction.
Where is this in the syllabus?
"Catalytic behaviour" of the first-row transition elements is in the d- and f-block unit of the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus. The JEE (Advanced) 2026 brochure lists "catalytic properties" under d-block elements, and homogeneous and heterogeneous catalysis under chemical kinetics. The matching text is section 4.3.12 of the NCERT Class 12 chapter The d- and f-Block Elements (reprint 2026-27). Check your exam's current documents.
Why are transition metals good catalysts?
NCERT gives two reasons: these metals can adopt multiple oxidation states, and they can form complexes. It then describes two ways this helps.
At a solid surface. Reactant molecules form bonds with atoms on the catalyst surface; first-row metals use their 3d and 4s electrons for this. That raises the concentration of reactants at the surface and weakens bonds in the reacting molecules, lowering the activation energy.
By changing oxidation state. An ion oxidises one reactant and is reduced, then the other reactant oxidises it back. Over one cycle it is unchanged.
The kinetics chapter adds the limits: a catalyst does not alter ΔG of a reaction, and it speeds up the forward and backward reactions to the same extent, so the equilibrium constant stays the same and equilibrium is simply reached sooner.
Which industrial examples should you know?
NCERT names these:
Vanadium(V) oxide, V2O5: oxidation of SO2 in the Contact process for sulphuric acid.
Finely divided iron: the Haber process for ammonia from N2 and H2. Fritz Haber received the 1918 Nobel Prize in Chemistry "for the synthesis of ammonia from its elements".
Nickel: catalytic hydrogenation, including hydrogenation of fats.
TiCl4 with Al(CH3)3: Ziegler catalysts for making polythene. Our post on Ziegler and Natta tells that story.
The Contact process shows a surface catalyst changing and recovering. As Chemguide explains, V2O5 oxidises SO2 to SO3 and is reduced to vanadium(IV) oxide, which oxygen then re-oxidises:
SO2 + V2O5 → SO3 + V2O4
V2O4 + ½O2 → V2O5
Add them and the vanadium oxides cancel, leaving SO2 + ½O2 → SO3.
How does iron catalyse the iodide-persulphate reaction?
NCERT's example is 2I⁻ + S2O8²⁻ → I2 + 2SO4²⁻, catalysed by iron(III). Chemguide notes the uncatalysed reaction is very slow and explains why: it needs a collision between two negative ions, which repel. With iron present, it runs in two steps, each between oppositely charged ions:
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I2
2Fe²⁺ + S2O8²⁻ → 2Fe³⁺ + 2SO4²⁻
Add the steps and the iron cancels. Chemguide adds that either iron(II) or iron(III) can be used, since each step regenerates the other.
Worked example: are both steps feasible?
Use standard reduction potentials from the NCERT Electrochemistry chapter: Fe³⁺/Fe²⁺ = +0.77 V and I2/I⁻ = +0.54 V. The same chapter gives 1.96 V for the S2O8²⁻/SO4²⁻ couple. E°cell = E°(couple reduced) − E°(couple oxidised).
Step 1 (Fe³⁺ reduced, I⁻ oxidised): 0.77 − 0.54 = +0.23 V
Step 2 (S2O8²⁻ reduced, Fe²⁺ oxidised): 1.96 − 0.77 = +1.19 V
Overall (S2O8²⁻ reduced, I⁻ oxidised): 1.96 − 0.54 = +1.42 V
Both steps are feasible. Now use ΔG° = −nFE° with n = 2 and F = 96487 C mol⁻¹:
Step 1: −2 × 96487 × 0.23 = −44,384 J, about −44.4 kJ mol⁻¹
Step 2: −2 × 96487 × 1.19 = −229,639 J, about −229.6 kJ mol⁻¹
Sum: −274.0 kJ mol⁻¹, equal to −2 × 96487 × 1.42 = −274,023 J for the overall reaction.
The catalyst leaves the overall ΔG° unchanged; it supplies a new path made of feasible steps. Under standard conditions, both steps have a positive E° only if the catalyst's couple lies between 0.54 V and 1.96 V, which is why the iron couple at 0.77 V fits.
Which mistakes cost marks?
Writing only "variable oxidation states" in a longer answer. Add the mechanism: reduced by one reactant, re-oxidised by the other.
Putting the catalyst in the overall equation. Fe³⁺ appears only in the steps.
Saying a catalyst shifts equilibrium. It does not; see our post on collision theory and catalysts.
Mixing up the pairs: iron for Haber, V2O5 for Contact, nickel for hydrogenation.
Sign errors in E°cell. Subtract the potential of the couple being oxidised.
A practice routine
Write NCERT's two reasons, then a three-line explanation of each.
From memory, pair each catalyst with its process.
Write both steps of the iron and vanadium cycles, add them, and cancel the catalyst.
Redo the E° check without notes, then revise oxidation states of transition metals.
If an activation energy barrier is hard to picture, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read.
Tonight, write the two steps of the iron cycle from memory and check that they add up to the overall equation.
