Question

Find the oxidation number of $I$ in $\mathop {KIO}\nolimits_3$

Answer 1

Oxidation number is defined as the total number of electrons than an atom either gains or loses in order to form a chemical bond with the atom and form a compound.

Complete step by step answer:
Iodine belongs from the $7A$ group and it is a non-metal and having oxidation state $- 1$ i.e., it always needs 1 electron to complete its octet and come in stable state. For example, if we consider $\mathop {BrCl}\nolimits_2$ in this bromine having oxidation state of $+ 2$ because $Cl$ is more electronegative. For calculating the oxidation number, we take all the charges on the right-hand side and equal to zero.
The oxidation state is the number of electrons that are lost, gained, or shared by a certain atom with other atoms in the same molecule
Here, we will focus on all the fixed components and then work out the oxidation state of $I$. The net charge on the molecule is 0. Therefore, all the oxidation states will add up to 0 after they are multiplied by the number of molecules.
$({\text{oxidation state of}} K) + ({\text{oxidation state of I}}) + (3 \times {\text{oxidation state of O}}) = 0$
Let the oxidation state of $I$ be $x$
We know that almost all elements from group 1 have an oxidation state of $+ 1$. This rule never changes. Thus, we will assume that the oxidation state of $K$ is $+ 1$. And the oxidation state of $O$ is $- 2$.
Now plugging in the numbers in the equation given above, we get,
$[ + 1] + [x] + [3 \times ( - 2)] = 0$
$\Rightarrow 1 + x - 6 = 0$
$\Rightarrow x - 5 - 0$
$\Rightarrow x = + 5$
So, the oxidation state of $I$ is $+ 5$
Thus, the oxidation state of $I$ $+ 5$ is when the oxidation states of $K$ and $O$ are $+ 1$ and $- 2$ respectively.
Note: Always check what the net charge on the given ion or molecule is, this will change the number present on the R.H.S. of the equation.
For example: If we consider $\mathop {SO}\nolimits_4^{2 - }$ then to calculate the oxidation state of $S$, we need to consider the equation.
$({\text{oxidation state of S}}) + (4 \times {\text{oxidation state of O}}) = - 2$
Let the oxidation state of $S$ be $x$
Then, $[x] + [4 \times ( - 2)] = - 2$
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
Here the oxidation state of S will be +6. If we do not pay attention to the net charge, the answer will turn out to be +8, which is not possible