Question

1. what is the oxidation number of I in the poly atomic anion perioddate IO^4- a. +7 B. -1 C. 0 D. +5

Answer 1

Step 1: Explanation

Rule for assigning oxidation number

1. The oxidation number of an element in its free (uncombined) state is zero ---- for example, Al(s) or Zn(s). This is also true for elements found in nature as diatomic (two-atom) elements
2. The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion --- for example: Na⁺ ------> Oxidation number =+1
3. The sum of all oxidation numbers in a neutral compound is zero.
4. The sum of all oxidation numbers in a polyatomic  ion is equal to the charge on the ion.
5. The oxidation number of oxygen in a compound is usually –2. If, however, the oxygen is in a class of compounds called peroxides (for example, hydrogen peroxide), then the oxygen has an oxidation number of –1. If the oxygen is bonded to fluorine, then oxidation number is +1.
6. The oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), then the oxidation state of hydrogen is –1.
7. The oxidation number of fluorine is always –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they are in combination with an oxygen or fluorine.

Step 2: Solution

(a) IO₄^- ------> oxidation number of I = +7

IO₄^-    has a charge of -1 and we just discussed that Oxygen will usually have an oxidation number of -2.

if the Oxidation number of the Iodine ion be X

then , X + 4(-2) = -1 => X =+7

So, the oxidation number of I in the IO₄^- ion is +7. ---------> option(a)