Question

Oxidation number of iodine in, $$I{O_3}^{-}$$, $$I{O_4}^{-}$$, $$KI$$ and $$I_2$$ respectively is
A. -2, -5, -1, 0.
B. +5, +7, -1, 0.
C. +2, +5, +1, 0.
D. -1, +1, 0, +1.

Answer 1

Hint: Iodine belongs to the Group VIIA (Halogens) having the electronic configuration $$5s^{2}5p^5$$. In compounds of iodine (where known), the most common oxidation numbers of iodine are: 7, 5, 3, 1, and -1.

Complete step by step solution:

Let’s know some common rules of finding the oxidation number:
Oxidation number of an atom is zero in a substance that contains atoms of only one element for example $$S_8$$.
Oxidation number of simple ions is equal to the charge on the ion. For example, the oxidation number of $$C{l}^{-}$$ ion is -1.
Oxidation number of hydrogen is +1 when it is combined with a nonmetal
Oxidation number of hydrogen is -1 when it is combined with a metal.
Group IA forms compounds in which the metal atom has an oxidation number of +1. For example, $$L{i}_{3}N$$.
Group IIA form compounds in which the metal atom has an oxidation number +2. For example, $$CaC{O}_3$$.
Oxygen usually has an oxidation number of -2. For example, $$H_2{O}_2$$.
Group VIIA often form compounds in which the nonmetal has an oxidation number -1. For example, $$ZnB{r}_2$$.
The total charge on the compound or ion is taken on the right-hand side.
Now in the given question we have to find the Oxidation number of Iodine in $$I{O_3}^{-}$$.
Let's take the oxidation number of iodine as x, we know the oxidation number for oxygen is -2. So, by solving
x + 3 × (-2) = -1
x + (-6) = -1
x = +5
Oxidation number of iodine in $$I{O_3}^{-}$$ is +5.

Similarly, in $$I{O_4}^{-}$$
x + 4 × (-2) = -1
x + (-8) = -1
x = +7
Oxidation state of iodine in $$I{O_4}^{-}$$ is +7.

In KI
+1 + x = 0
x = -1
Oxidation number of iodine in KI is -1.

In $$I_2$$
2x = 0
x = 0
Oxidation number of Iodine in $$I_2$$ is 0.

Therefore, from the above statements we can conclude that the correct option is (b).

Note: Balancing a redox reaction first requires identifying the oxidation numbers that are there in the net ionic equation, then, breaking the equation into half reactions and finally, adding the electrons to the reactions.