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

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Answer 1 · 2015-01-02T22:35:23

Iodine has an oxidation number of +7 in $HIO_4$ and an oxidation number of +5 in $HIO_3$ .

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 $HIO_4$ is a neutral molecule, we get

$1 * (+1) + 1 * x + 4* (-2) = 0 => x= 8 -1 =7$

So iodine has a +7 oxidation number in $HIO_4$ .

The exact same technique applies for $HIO_3$ :

$H^(+1)I^(x)O_3^(-2)$ , so

$1*(+1) + x + 3 *(-2) =0=>x = 6-1 = 5$

Iodine's oxidation number is +5 in $HIO_3$ .

What are the oxidation states of the iodine in HIO_4HIO_4 and HIO_3HIO_3 ?