The complete hydrolysis of $\mathrm{ICl}$, $\mathrm{ClF}_{3}$ and $\mathrm{BrF}_{5}$, respectively, gives

(A) $\mathrm{IO}^{-}$, $\mathrm{ClO}_{2}^{-}$ and $\mathrm{BrO}_{3}^{-}$

(B) $\mathrm{IO}_{3}^{-}$, $\mathrm{ClO}_{2}^{-}$ and $\mathrm{BrO}_{3}^{-}$

(C) $\mathrm{IO}^{-}$, $\mathrm{ClO}^{-}$ and $\mathrm{BrO}_{2}^{-}$

(D) $\mathrm{IO}_{3}^{-}$, $\mathrm{ClO}_{4}^{-}$ and $\mathrm{BrO}_{2}^{-}$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · p-Block Elements

The complete hydrolysis of

ICl

ClF_{3}

and

BrF_{5}

, respectively, gives

(A) $IO^{-}$ , $ClO_{2}^{-}$ and $BrO_{3}^{-}$ ✓ Correct
(B) $IO_{3}^{-}$ , $ClO_{2}^{-}$ and $BrO_{3}^{-}$
(C) $IO^{-}$ , $ClO^{-}$ and $BrO_{2}^{-}$
(D) $IO_{3}^{-}$ , $ClO_{4}^{-}$ and $BrO_{2}^{-}$

Answer: (A)

Solution

The complete hydrolysis of interhalogen compounds follows the pattern:

ICl + H₂O: The iodine is oxidized to IO⁻ (hypoiodate) and chlorine is reduced to Cl⁻.

ClF₃ + H₂O: Chlorine is oxidized to ClO₂⁻ (chlorite) and fluorine forms HF.

BrF₅ + H₂O: Bromine is oxidized to BrO₃⁻ (bromate) and fluorine forms HF.

In each case, the less electronegative halogen is oxidized while the more electronegative one (or fluorine) is reduced or forms HF.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · p-Block Elements

Solution

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