JEE Advanced 2025 Paper 1 Q03 Chemistry Inorganic Chemistry d-Block Elements Easy

JEE Advanced 2025 Paper 1 · Q03 · d-Block Elements

One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is:

  1. A. $I_2$
  2. B. $IO_3^-$
  3. C. $IO_4^-$
  4. D. $IO_2^-$
Reveal answer + step-by-step solution

Correct answer:B

Solution

In neutral or slightly alkaline medium, $MnO_4^-$ oxidises $I^-$ past $I_2$: the standard outcome in neutral (weakly basic) conditions is iodate, $IO_3^-$ ($Mn$ goes to $MnO_2$). Reaction: $2 MnO_4^- + I^- + H_2O \to 2 MnO_2 + IO_3^- + 2 OH^-$ (balanced). In strongly acidic medium $I^-$ would go to $I_2$; only in neutral/slightly basic conditions is $IO_3^-$ favoured. Answer: (B) $IO_3^-$.

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