Q29 / 150 · MHT CET 2025 (21 Apr Shift 2) · Chemistry

Identify correct statement for the reaction. $\mathrm{I}_2+\mathrm{KClO}_3 \xrightarrow{\Delta} \mathrm{ICl}+\mathrm{KIO}_3$

  • A
    $\mathrm{I}_2$ is an oxidising agent
  • B
    $\mathrm{KClO}_3$ is Reducing agent
  • Correct answer
    C
    Oxidation number of 'Cl' decreases by 6
  • D
    Oxidation number of I increases by 2

Solution

Oxidation state analysis: In $\mathrm{I}_2$, iodine is elemental with oxidation state 0. For $\mathrm{KClO}_3$, oxygen is $-2$ and potassium is $+1$, requiring chlorine at $+5$ to balance: $+1 + x + 3(-2) = 0$ gives $x = +5$.

In product $\mathrm{ICl}$, chlorine ($-1$) implies iodine at $+1$. In $\mathrm{KIO}_3$, potassium $+1$ and oxygen $-2$ require iodine at $+5$: $+1 + z + 3(-2) = 0$ gives $z = +5$.

Iodine is oxidized from 0 to $+1$ (increase of 1) and $+5$ (increase of 5), making $\mathrm{I}_2$ the reducing agent. Chlorine is reduced from $+5$ to $-1$ (decrease of $+5 - (-1) = 6$), making $\mathrm{KClO}_3$ the oxidizing agent.

The chlorine oxidation number decreases by 6, corresponding to option C.

Hence, the correct answer is option: 'C'.

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