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

Question

31046 views around the world

You can reuse this answer
Creative Commons License

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#.