Question

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

Answer 1

The degree of oxidation (loss of electrons) of an atom in a chemical compound is described by the oxidation state, often known as the oxidation number. The oxidation state, which can be positive, negative, or zero, is the hypothetical charge that an atom would have if all of its connections to other atoms were entirely ionic, with no covalent component. When it comes to genuine bonds, this is never the case.

Complete answer:
The oxidation state of an atom has nothing to do with its "real" formal charge or any other atomic characteristic. This is especially true at high oxidation states, when the ionisation energy necessary to generate a multiply positive ion exceeds the energies available in chemical processes. Furthermore, depending on the electronegativity scale employed in their computation, the oxidation states of atoms in a particular molecule may differ. As a result, an atom's oxidation state in a compound is merely a formalism. It is, nevertheless, crucial to comprehending inorganic compound nomenclature standards.
We always allocate the charge to the most electronegative atom when assigning oxidation state. When we assign oxidation numbers to carbon compounds (which is an uncommon task), we presume that the carbon-carbon link has been broken, that the electrons have been divided amongst the carbon radicals, and that the carbon-carbon connection has been broken.
METHANE $C{{H}_{4}}$
The oxidation state of hydrogen = $\text{4 x }\left( \text{+ 1} \right)\text{ = 4}$
Since it Is a neutral atom
$x+\text{ }4=0$
Hence the oxidation state of carbon is $x=-4$
ETHANE ${{C}_{2}}{{H}_{6}}$
The oxidation state of hydrogen = $\text{6 x }\left( \text{+ 1} \right)\text{ = 6}$
Since it Is a neutral atom
$2x+\text{ 6}=0$
Hence the oxidation state of carbon is $x=-3$
Carbon monoxide $CO$
The oxidation state of oxygen = $\text{1 x }\left( \text{-2} \right)\text{ = -2}$
Since it Is a neutral atom
$x-\text{ 2}=0$
Hence the oxidation state of carbon is $x=+2$
Carbon dioxide $C{{O}_{2}}$
The oxidation state of oxygen = $\text{2 x }\left( \text{-2} \right)\text{ = -4}$
Since it Is a neutral atom
$x-\text{ 4}=0$
Hence the oxidation state of carbon is $x=+4$
Hence methane is the correct answer.

Note:
Antoine Lavoisier used the term "oxidation" to describe the interaction of a material with oxygen. Much later, it was discovered that the material loses electrons when oxidised, and the definition was expanded to encompass any process in which electrons are lost, regardless of whether oxygen is present. Oxidation is the process of increasing an atom's oxidation state through a chemical reaction; reduction is the process of decreasing an atom's oxidation state. The formal transfer of electrons is involved in such reactions: a net gain of electrons is a reduction, and a net loss of electrons is an oxidation. The oxidation state of pure elements is zero.