Question

Oxidation states of carbon atoms in diamond and graphite are.
A. $ + 2, + 4$
B. $ + 4, + 2$
C. $ - 4,4$
D. Zero, zero

Answer 1

Hint: We know that the oxidation number of a molecule or ion is defined as the number of charges it would carry if electrons are completely transferred . Oxidation states are represented by integers which may be positive ,zero, or negative. In some cases, oxidation number is represented by fraction which is the average oxidation state of an element for example $\dfrac{8}{3}$ for iron in magnetite $F{e_3}{O_4}$.
If there are increases in the oxidation number of elements ,it is known as oxidation and a decrease in oxidation state is known as a reduction . In this type of reaction there is involvement of formal transfer of electrons. Oxidation state is zero for pure elements.

Complete step by step answer:
There are general rules for assigning oxidation number:
$1.$ The oxidation number of an element in its free states is zero.
$2.$ The oxidation number of a monatomic ion is the same charge on the ion, for example : $N{a^{ + 1}}$: O.N=1
$3.$ H shows +1 oxidation with non metals and -1 oxidation with metals.
$4.$ O always shows -2 oxidation number and exceptional in case of peroxide where it shows -1 oxidation state.

Hence, Graphite and diamond by itself is elemental carbon and the oxidation state of an element in free state is zero . So in graphite and diamond, carbon is in zero oxidation state.
So, the correct answer is “Option D”.

Note: Graphite by itself is elemental carbon and has an oxidation number of 0. However, graphite can form intercalation compounds with metal cations or nonmetal or complex anions. Formulation can include: $K{C_{8,}}K{C_{24,}}Li{C_6},Ca{C_6},{C_{24}}\left( {HS{O_4}} \right),{C_8}\left( {As{F_6}} \right),{C_{12}}Br,$ amongst others. The average oxidation state per $C$ atom is respectively$ - \dfrac{1}{8}, - \dfrac{1}{{24}}, - \dfrac{1}{6}, - \dfrac{1}{3}, + \dfrac{1}{{24}}, + \dfrac{1}{8}, + \dfrac{1}{{12}}$ in these examples.