JEE Advanced 2025 Paper 2 Q02 Chemistry Inorganic Chemistry p-Block Elements (Groups 15-18) Medium

JEE Advanced 2025 Paper 2 · Q02 · p-Block Elements (Groups 15-18)

The complete hydrolysis of $ICl$, $ClF_3$ and $BrF_5$, respectively, gives:

  1. A. $IO^-$, $ClO_2^-$ and $BrO_3^-$
  2. B. $IO_3^-$, $ClO_2^-$ and $BrO_3^-$
  3. C. $IO^-$, $ClO^-$ and $BrO_2^-$
  4. D. $IO_3^-$, $ClO_4^-$ and $BrO_2^-$
Reveal answer + step-by-step solution

Correct answer:A

Solution

In the hydrolysis of an interhalogen compound, the smaller-electronegativity halogen ends up as the central atom of an oxyanion of the same oxidation state. $ICl + H_2O \to HCl + HIO$; iodine is in +1 state, giving $IO^-$ (hypoiodite). $ClF_3 + 2 H_2O \to 3 HF + HClO_2$; chlorine is in +3 state, giving $ClO_2^-$ (chlorite). $BrF_5 + 3 H_2O \to 5 HF + HBrO_3$; bromine is in +5 state, giving $BrO_3^-$ (bromate). Hence the products are $IO^-$, $ClO_2^-$, $BrO_3^-$ — option (A).

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