What is the oxidation number of iodine in KIO4? (1) + 1 (3) + 7 (2) - 1 (4) - 7

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.