# In which molecule does carbon express its LOWEST oxidation number: "methane, ethane, carbon monoxide, carbon dioxide?"

Jun 1, 2017

$\text{Methane......}$

#### Explanation:

In methane we have.............. ${\stackrel{- I V}{\text{CH}}}_{4}$.

For carbon dioxide, ${\stackrel{+ I V}{\text{CO}}}_{2}$.

In each case, the sum of the oxidation numbers equals the charge on the species, and of course we may have a neutral molecule.

And thus $\stackrel{+ I I}{\text{CO}}$, and $\stackrel{- I I}{C} {H}_{3} O H$.

As always, to assign oxidation state, we distribute the charge to the MOST electronegative atom. When we assign oxidation numbers to carbon compounds (which is an unusual exercise), we assume that the $\text{carbon-carbon}$ bond is broken, the electrons are shared between the carbon radicals, i.e.

${H}_{3} C - C {H}_{3} \rightarrow {H}_{3} C \cdot + \cdot C {H}_{3}$, i.e. $\stackrel{- I I I}{C} {H}_{3}$.

And thus we speak of carbon oxidation:

$C {H}_{4} , C {H}_{3} O H , {H}_{2} C \left(= O\right) , H C \left(= O\right) O H , C {O}_{2} :$

stackrel(-IV)C;stackrel(-II)C;stackrel(0)C;stackrel(+II)C; stackrel(+IV)C......................

And for ethane, ${H}_{3} C - C {H}_{3} \rightarrow 2 \times \dot{C} {H}_{3}$, that is $\stackrel{- I I I}{C}$.