Question

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

Answer 1

`"Methane......"`

Explanation:

In methane we have.............. `stackrel(-IV)"CH"_4`.

For carbon dioxide, `stackrel(+IV)"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(+II)"CO"`, and `stackrel(-II)CH_3OH`.

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 `"carbon-carbon"` bond is broken, the electrons are shared between the carbon radicals, i.e.

`H_3C-CH_3 rarr H_3C* +*CH_3`, i.e. `stackrel(-III)CH_3`.

And thus we speak of carbon oxidation:

`CH_4, CH_3OH, H_2C(=O), HC(=O)OH, CO_2:`

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

And for ethane, `H_3C-CH_3rarr2xxdotCH_3`, that is `stackrel(-III)C`.