# How many oxidation states does carbon adopt?

Sep 7, 2016

How many oxidation states does carbon exhibit? Probably about 9.

#### Explanation:

As a Group IV element, carbon has a large redox manifold. ${C}^{- I V}$ to ${C}^{I V}$. Carbon is slightly more electronegative than hydrogen.

I will try to provide examples for each one.

${C}^{- I V} :$ $C {H}_{4}$

${C}^{- I I I} :$ ${H}_{3} C - C {H}_{3}$

${C}^{- I I} :$ ${H}_{3} {C}^{- I I I} - {C}^{- I I} {H}_{2} - {C}^{- I I I} {H}_{3}$

${C}^{- I} :$ ${\left({H}_{3} C\right)}_{3} {C}^{- I} H$, or $\text{benzene}$

${C}^{0} :$ $\text{Soot or bbq charcoal}$ or $C {H}_{2} {X}_{2}$

${C}^{I} :$ ${\left({H}_{3} C\right)}_{3} C - X$

${C}^{+ I I} :$ ""^(-)C-=O^+, (H_3C)_2CX_2

${C}^{+ I I I} :$ ${X}_{3} C - C {X}_{3}$

${C}^{+ I V} :$ $C {O}_{2}$, $C {X}_{4}$

Note that for the purposes of oxidation state assignment, when we break a $C - C$ bond, the electrons are assume to be shared equally, i.e. $C - C \rightarrow 2 \times \cdot C$. Also note that oxidation states are a FORMALISM, they represent an ad hoc way of counting electrons and not a physical reality.