# Why is the oxidation state of Carbon (-4)?

Based on my understanding, atoms try to achieve the electronic configuration of the nearest noble gas, by losing or gaining electrons. For example, Sodium is closest to Neon, so loses an electron so the outer shell is full (2nd shell is full). This doesn't include the transition metals or the lanthanides or the actinides, which I realize follow something different because of their orbitals.

In the case of Carbon, it is equally distant from Neon as it is from Helium. There's the octet rule, but why exactly does it prefer Neon's configuration over Helium?

• Valency don't have a sign. Do you mean oxidation state? – Ivan Neretin Feb 9 '18 at 13:15
• And -4 Is far from the only possible oxidation state for carbon. We can get any whole number from -4 to +4 inclusive, with some fractions thrown in on the side (e.g. cyclopropenyl ion =+(1/3)). – Oscar Lanzi Feb 9 '18 at 15:19

Oxidation state is a hypothetical number assigned to each atom in a molecular structure (refer this) which assumes all bonds to be purely ionic. So it doesn't mean that that the carbon atoms in $\ce{C_2H_2}$ having an oxidation state of $-1$ have gained $1$ electron each to result in 5 electrons in the valence shell.
If you meant the $\ce{C^{4-}}$ (carbide) ion, carbon actually prefers this because removing 4 electrons is difficult as it is quite endoergic as it requires nearly 14000 kJ/mol (refer this) to form a $\ce{C^{4+}}$ basically due to its small size, which means that electrons are closer to the nucleus and we can remove it only at the expense of high energy. Thus, $\ce{C^{4-}}$ is comparatively easy to form as it is easier to add electrons into the valence shell.
• $\ce{C^{4-}}$ is impossible to form. No atomic ion with 2 or more negative charges is bound. Only when imagining a compound as purely ionic does $\ce{C^{4-}}$ come to mind. – DavePhD Feb 9 '18 at 15:28