KIO4 is an example of polyatomic ion.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion.
The charge on this polyatomic ion is 0.
To get the oxidation number of I, we will let x as the oxidation number of I.
We can get the oxidation number of Iodine by solving like this;
KIO4 : (+1) + x + 4(-2) = 0
(+1) + x - 8 = 0
x - 7 = 0
x = +7
Therefore, the oxidation number of iodine atom in KIO4 is +7.