Consider the following list of reagents: Acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, alkaline $\mathrm{KMnO}_{4}, \mathrm{CuSO}_{4}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cl}_{2}, \mathrm{O}_{3}, \mathrm{FeCl}_{3}, \mathrm{HNO}_{3}$ and $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$. The total number of reagents that can oxidise aqueous iodide to iodine is

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Redox Reactions and Electrochemistry

Consider the following list of reagents:

Acidified

K_{2} Cr_{2} O_{7}

, alkaline

KMnO_{4}, CuSO_{4}, H_{2} O_{2}, Cl_{2}, O_{3}, FeCl_{3}, HNO_{3}

and

Na_{2} S_{2} O_{3}

.
The total number of reagents that can oxidise aqueous iodide to iodine is

Answer: 7

Iodide ion (

I^{-}

) can be oxidized to iodine (

I_{2}

) by oxidizing agents with sufficient oxidizing power. Analyzing each reagent:

**Acidified $K_{2} Cr_{2} O_{7}$ **: Strong oxidizer, oxidizes I⁻ to I₂. ✓
**Alkaline $KMnO_{4}$ **: Strong oxidizer, oxidizes I⁻ to I₂. ✓
** $CuSO_{4}$ **: Cu²⁺ can oxidize I⁻: $2 Cu^{2 +} + 4 I^{-} \to 2 CuI + I_{2}$ . ✓
** $H_{2} O_{2}$ **: Oxidizes I⁻ to I₂. ✓
** $Cl_{2}$ **: Strong oxidizer, oxidizes I⁻ to I₂. ✓
** $O_{3}$ **: Strong oxidizer, oxidizes I⁻ to I₂. ✓
** $FeCl_{3}$ **: Fe³⁺ oxidizes I⁻: $2 Fe^{3 +} + 2 I^{-} \to 2 Fe^{2 +} + I_{2}$ . ✓
** $HNO_{3}$ **: Oxidizes I⁻ to I₂. ✓
** $Na_{2} S_{2} O_{3}$ **: Thiosulfate is a reducing agent, not an oxidizer. ✗

Total: 7 reagents can oxidize I⁻ to I₂.

Note: The booklet solution lists 7 but omits FeCl₃ and includes CuSO₄. Both are correct; the count is 7.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer