Question

Give the formula of a compound in which carbon exhibits an oxidation state of:
A.+4
B.+2
C.-4

Answer 1

This question requires the knowledge of the concept of oxidation states of an element in different compounds. An element can exhibit different oxidation states in different compounds depending upon the formula of the compound and the structure of that compound. Some complex structured compounds exhibit exceptional oxidation states of various elements.

Complete answer: The oxidation state of an atom (also known as the oxidation number) in a chemical compound gives information about how many electrons it has lost and so describes the atom's degree of oxidation. The hypothetical charge that an atom would have if all of its links to other atoms were totally ionic in nature can be characterized as the oxidation state of that atom.
The oxidation state of carbon in $C{O_2}$ is +4, for example, since the hypothetical charge borne by the carbon atom if both carbon-oxygen double bonds were entirely ionic would be +4. (each oxygen atom would hold a charge of -2 since oxygen is more electronegative than carbon).
The oxidation state of an atom is not considered the atom's true charge. When the oxidation state of a substance increases, it is referred to as oxidation; when the oxidation state decreases, it is referred to as reduction. For carbon in $C{H_4}$, the oxidation state is - 4.
So, carbon has +4 oxidation state in $C{O_2}$, +2 oxidation state in $CO$ and -4 oxidation state in $C{H_4}$.

Note:
The phrases oxidation state and oxidation number are commonly interchanged. They're the numbers that describe how many electrons are lost in an atom. The oxidation state should ideally be zero, positive, or negative. When the oxidation numbers of a compound's atoms are added together, the algebraic sum must be zero. When the oxidation numbers of the atoms in a polyatomic ion are added together, the algebraic sum must equal the charge on the ion.