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

Iodine has an 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 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##.