Question

What is the oxidation state of each individual carbon atom in ${{C}_{2}}{{O}_{4}}^{2-}$?

Answer 1

Oxidation state is also known as oxidation number. It is defined as the total number of electrons that an atom either gains or loses to form a chemical bond. Each oxygen atom will have a $-2$ oxidation state because no oxygen is involved in peroxide linkage.

Complete answer: Oxidation states are typically represented by integers which may be positive, zero or negative. In some cases, the average oxidation state of an element is a fraction. For example in case of $F{{e}_{3}}{{O}_{4}}$ (magnetite) the oxidation state of iron is $\dfrac{8}{3}$.
Oxidation state shows the total number of electrons which have been removed from an element or added to an element to get to its present state. Oxidation involves an increase in oxidation state and reduction involves a decrease in oxidation state.
Carbon atoms show $+1,+2,+3,+4$ oxidation states whereas oxygen atoms commonly show $-2$ oxidation states. Here in case of ${{C}_{2}}{{O}_{4}}^{2-}$ known as oxalate ion all the oxygen atom shows $-2$ oxidation state.
${{C}_{2}}{{O}_{4}}^{2-}$ is a polyatomic ion with a $-2$ charge on it. As the chemical structure of this ion is symmetrical , both carbon are equivalent.
Now we have to calculate the oxidation state of the carbon atom, so let us assume the oxidation state of carbon is $x$.
Now, using formula we calculate the oxidation state as:
$2x+(-2)\times 4=-2$
$2x-8=-2$
$2x=+6$
$x=+3$
Therefore, each carbon shows $+3$ oxidation state.

Note:
The oxidation number of oxygen atoms is always $-2$ except in $H_2O$, $O{{F}_{2}}$. In ${{H}_{2}}{{O}_{2}}$ oxidation state of oxygen atom is $-1$ and in $O{{F}_{2}}$ oxidation state of oxygen atom is $+2$. The highest oxidation state is $+9$ in $IrO_{4}^{+}$ cation and lowest oxidation state is $-5$ in $A{{l}_{3}}BC$.