Read the passage carefully and answer the questions.
The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron, and nickel (in Catalytic Hydrogenation) are some of the examples. Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst (first row transition metals utilise 3d and 4s electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (the activation energy is lowering). Also because the transition metal ions can change their oxidation states, they become more effective as catalysts.
Reaction between iodide and persulphate ions takes place as follows:
\( \mathrm{2I^{-}+S_2O_8^{2-}\rightarrow I_2+2SO}_4^{2-} \)
Which of the following catalyses the above reaction?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\mathrm{Fe}^{3+}\)
A transition metal ion that can switch between two oxidation states can catalyse a redox reaction: it takes electrons from one reactant and passes them on to the other. The reaction between iodide and persulphate ions is slow, because both reactants are negative ions that repel each other. Iron(III) catalyses it in two fast steps.
First, \( \ce{Fe^3+} \) oxidises iodide to iodine and is reduced to \( \ce{Fe^2+} \). Then persulphate oxidises \( \ce{Fe^2+} \) back to \( \ce{Fe^3+} \). The two steps add up to the reaction in the question, and the \( \ce{Fe^3+} \) ion is given back unchanged. \( \ce{Fe^2+} \) is only the intermediate formed in the first step.
NCERT (2026-27 edition):
Catalytic Properties
