Redox reaction where iodide reduces X²⁺ to metal X Steps:

• Identify the redox process: X²⁺ gains 2 electrons to form X(s), so X²⁺ is reduced; I⁻ loses electrons to form I₂, so I⁻ is oxidized.
• Determine spontaneity condition: For the reaction to occur, standard reduction potential E°(X²⁺/X) > E°(I₂/I⁻) = +0.54 V.
• Classify metals: Group II metals have E° << +0.54 V (e.g., Mg²⁺/Mg = -2.37 V), so their ions cannot be reduced by I⁻.
• Conclude metal type: Only some transition metals have E° > +0.54 V (e.g., Pd²⁺/Pd = +0.99 V), allowing reduction to metal.

Why D is correct:

• Transition metals like Pd have E°(X²⁺/X) > +0.54 V, enabling I⁻ to reduce X²⁺ per the Nernst equation for positive cell potential.

Why the others are wrong:

• A: Group II metals have negative E° values, preventing reduction by I⁻.
• B: Group II ions cannot be reduced by I⁻ due to unfavorable potentials.
• C: While X is a transition metal, the key process is reduction of X²⁺, not oxidation.

Final answer: D