Question

What is the oxidation of all iodine atoms in $K{{I}_{3}}$ ?

Answer 1

We know that Oxidation number or oxidation state of an element can be defined as the degree of oxidation of an element in a given compound. In simpler terms, it can be understood as the number of the electrons gained or lost by an atom while forming a compound.

Complete answer:
This results in forming a net charge over this element, which is referred to as the oxidation state. Depending on the number of electrons present in the valence shell of an atom, an element may exhibit a single or in some cases multiple oxidation states, depending on the atoms they are combining with. Now, forming the equation for calculating the oxidation states of iodine and potassium.
Let us assume the oxidation number of l is x.
In $KI_3$, the oxidation number of K is +1.
$1\left( +1 \right)\text{ }+\text{ }3\left( x \right)\text{ }=\text{ }0$
$\Rightarrow +1\text{ }+3x\text{ }=\text{ }0$
$\Rightarrow 3x\text{ }=\text{ }-1$
$\Rightarrow x\text{ }=\text{ }-1/3$
The cumulative number of electrons that an atom either gains in or loses to form a chemical bond with another atom, this is known as oxidation number. An oxidation number that represents the capacity to gain, transfer or disperse electrons is given for each atom that participates in an oxidation reduction reaction.

Note:
Remember that the oxidation number of all halogens will always be equal to $\left( -1 \right).$This is because of the number of valence electrons present in them. They have only $7$ valence electrons, which means that they are one electron short of completing their octets. Hence, all halogens generally just prefer to accept one electron and hence their oxidation state is $\left( -1 \right)$.