Question

What is the oxidation state of chlorine in $Cl{O_2}$ ?

Answer 1

Hint: The oxidation state is often referred to as the oxidation number. The number of electrons that every atom of an element gains, loses, or shares in order to establish a chemical bond with another atom to create a compound or molecule is known as the oxidation number.
To put it another way, an atom's oxidation number is the charge that appears on it when it forms ionic bonds with other heteroatoms.

Complete answer:
So, to get the oxidation of chlorine in the given compound $Cl{O_2}$
We will do some calculations as follows,
Let us assume that the oxidation number of the chlorine in the compound =x.
As we know that the values of the oxygen atom found in the given compound are,
The oxidation number of the oxygen \[O = - 2\]
And for we have already assumed the oxidation number of the chlorine =x
And we know that the total sum of the oxidation numbers of the atoms of a compound should be zero.
So, as per the formula for the given compound $Cl{O_2}$
We put the oxidation state and equate it equal to zero.
$x + 2( - 2) = 0$
$ \Rightarrow x - 4 = 0$
$ \Rightarrow x = 4$
So, we had calculated the oxidation state of chlorine in $Cl{O_2}$ is 4.

Note:
The oxidation state or oxidation number of some atoms, molecules, or ions is constant:
Because the net charge on a neutral atom or molecule is zero, the overall oxidation state of the atom or molecule is zero.
The oxidation state of charged ions is the same as the net charge of the ion. For example, all alkali metal ions have an oxidation state of +1, all alkaline earth metal ions have an oxidation state of +2, and the boron family has an oxidation state of +3. Hydrogen has an oxidation state of +1 in proton ( ${H^ + }$ ) and +1 in hydride.