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QUESTION

# 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.