The oxidation states of iodine in HIO_4,H_3IO | vedclass

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Question

(C) To calculate the oxidation state of iodine $(x)$ in these compounds,we use the rule that the sum of oxidation states in a neutral molecule is $0$.

Vedclass · Accepted Answer

$+ 7, + 7, + 7$

Vedclass · Answer

(C) To calculate the oxidation state of iodine $(x)$ in these compounds,we use the rule that the sum of oxidation states in a neutral molecule is $0$. The oxidation state of $H$ is $+ 1$ and $O$ is $- 2$.For $HIO_4$: $1 + x + 4(- 2) = 0 \implies 1 + x - 8 = 0 \implies x = + 7$.For $H_3IO_5$: $3(1) + x + 5(- 2) = 0 \implies 3 + x - 10 = 0 \implies x = + 7$.For $H_5IO_6$: $5(1) + x + 6(- 2) = 0 \implies 5 + x - 12 = 0 \implies x = + 7$.Thus,the oxidation states of iodine in all three compounds are $+ 7, + 7, + 7$.

Column-$I$	Column-$II$
$A$. Positively charged ion	$1$. $+7$
$B$. Oxidation number of all neutral atoms	$2$. $-1$
$C$. Oxidation number of hydrogen $(H^+)$	$3$. $+1$
$D$. Oxidation number of fluorine in $NaF$	$4$. $0$
$E$. Negatively charged ion	$5$. Cation
	$6$. Anion

The oxidation states of iodine in $HIO_4$,$H_3IO_5$,and $H_5IO_6$ are respectively:

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