# What are the oxidation states of the iodine in HIO_4 and HIO_3 ?

Jan 2, 2015

Iodine has an oxidation number of +7 in $H I {O}_{4}$ and an oxidation number of +5 in $H I {O}_{3}$.

In both cases, iodine's oxidation number can be determined using $H$ and $O$'s known oxidation numbers of +1 and -2, respectively.

So, for the first compound

${H}^{+ 1} {I}^{x} {O}_{4}^{- 2}$

The sum of each individual atom's oxidation number must equal the molecule's overall charge. Since $H I {O}_{4}$ is a neutral molecule, we get

$1 \cdot \left(+ 1\right) + 1 \cdot x + 4 \cdot \left(- 2\right) = 0 \implies x = 8 - 1 = 7$

So iodine has a +7 oxidation number in $H I {O}_{4}$.

The exact same technique applies for $H I {O}_{3}$:

${H}^{+ 1} {I}^{x} {O}_{3}^{- 2}$, so

$1 \cdot \left(+ 1\right) + x + 3 \cdot \left(- 2\right) = 0 \implies x = 6 - 1 = 5$

Iodine's oxidation number is +5 in $H I {O}_{3}$.