Question

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

Answer 1

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`.