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?

$\mathrm{Fe}^{2+}$

The correct answer is Option (1) → $\mathrm{Fe}^{2+}$

Fe²⁺ catalyses the reaction between iodide and persulphate ions.

Reasoning:
The reaction

2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻

is very slow in the absence of a catalyst. Transition metal ions like Fe²⁺ speed it up by providing an alternative pathway involving change in oxidation states.

The catalysis proceeds through two steps:

Step 1:
Fe²⁺ + S₂O₈²⁻ → Fe³⁺ + SO₄²⁻ + SO₄•⁻
(Fe²⁺ is oxidised to Fe³⁺)

Step 2:
Fe³⁺ + I⁻ → Fe²⁺ + ½ I₂
(Fe³⁺ is reduced back to Fe²⁺)

Thus, Fe²⁺ is regenerated and acts as a true catalyst by cycling between Fe²⁺ and Fe³⁺ oxidation states.

Why other options are incorrect:

• Fe³⁺ is not the effective catalyst here; the catalytic cycle is initiated by Fe²⁺.

• Mn²⁺ and Sn²⁺ do not efficiently catalyse this specific redox reaction.

 Therefore, Fe²⁺ is the correct answer.