Statement-1 : I_2 rightarrow IO_3^- + I^-; Th | vedclass

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(C) In Statement-$1$,the oxidation state of $I$ in $I_2$ is $0$. In $IO_3^-$,the oxidation state of $I$ is $+5$ (calculated as $x + 3(-2) = -1 \Righta

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Both Statement-$1$ & $2$ are correct.

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(C) In Statement-$1$,the oxidation state of $I$ in $I_2$ is $0$. In $IO_3^-$,the oxidation state of $I$ is $+5$ (calculated as $x + 3(-2) = -1 \Rightarrow x = +5$). In $I^-$,the oxidation state of $I$ is $-1$. Since the same element $(I)$ is simultaneously oxidized ($0$ to $+5$) and reduced ($0$ to $-1$),this is a disproportionation reaction. Thus,Statement-$1$ is correct.In Statement-$2$,the oxidation state of Iodine $(I)$ ranges from $-1$ (in $I^-$) to $+7$ (in $IO_4^-$). Thus,Statement-$2$ is also correct.

Statement-$1$ : $I_2 \rightarrow IO_3^- + I^-$; This is a disproportionation reaction.
Statement-$2$ : Oxidation number of $I$ can vary from $-1$ to $+7$.

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Statement-$1$ : $I_2 \rightarrow IO_3^- + I^-$; This is a disproportionation reaction.Statement-$2$ : Oxidation number of $I$ can vary from $-1$ to $+7$.