-2
$\begingroup$

In the formation of NaCl, Na gets oxidised because it loses an electron, while Cl gets reduced because it gains an electron.

But how does this concept work in the case of covalent compounds, especially those compounds that have non-polar bonds?

For example, according to my textbook, in the formation of methane Hydrogen is getting oxidised from state 0 to 1, while Carbon is getting reduced from state 0 to -4.

But how is hydrogen losing an electron here? Doesn't it gain an electron due to sharing? C-H bonds are non polar so we cannot say that carbon pulls hydrogen's electrons towards it either right?

$\endgroup$
16
  • $\begingroup$ In these cases of nonpolar covalent bonding, oxidation and reduction is formal, similarly as oxidation states/numbers. If a bond is easily broken, like N-Cl, ON can be formally determined from ON of products $\endgroup$
    – Poutnik
    Nov 3, 2022 at 9:15
  • $\begingroup$ @Poutnik What do we mean by oxidation and reduction is formal? What does formal here? $\endgroup$ Nov 3, 2022 at 10:12
  • $\begingroup$ There is no gaining nor loosing electrons in $\ce{C(s) + 2 H2(g) -> CH4(g}$). There is speculative gaining and loosing electrons in thought heterolytic $\ce{H3C-H -> H3C- + p}$ bond breaking, on which the formal oxidation number is based on. $\endgroup$
    – Poutnik
    Nov 3, 2022 at 11:50
  • $\begingroup$ @Poutnik Oh, Ok. Thank you so much.. So in throry, if we take an isolated ch4 atom then on breaking one of its bonds we will end up with a methyl anion and a hydrogen cation? has this been experimentally checked too? And pardon me for asking a slightly silly question but umm, is it possible to just break one bond while leaving rest of the three untouched? $\endgroup$ Nov 3, 2022 at 13:03
  • $\begingroup$ @Poutnik another doubt that I have is, that according to this speculation, if we keep on breaking the bonds, then we keep on creating more H cations or in other words protons. But I have been taught that CH4 disassociates into H2O and CO2 but here how will protons form H20? $\endgroup$ Nov 3, 2022 at 13:11

1 Answer 1

1
$\begingroup$

Carbon is slightly more "electronegative" than hydrogen, so by convention C in CH4 is -4 and each hydrogen is +1, even though the bonds are considered covalent.

Differences in electronegativity are important in understanding polar molecules. More reading here.

With CO2, the carbon oxidation state is +4, because oxygen has a stronger affinity for electrons than carbon.

$\endgroup$
2
  • $\begingroup$ So the definition that oxidation is the loss of electrons is wrong in the case of CH4 right? The only reason H gets oxidised is because its oxidation state increases from 0 to1. And the reason we write its oxidation state as 1 is that carbon is more electronegative than hydrogen. Nothing about electrons right? $\endgroup$ Nov 4, 2022 at 5:53
  • $\begingroup$ @Obinna we can observe that the net charge of elemental H, C, and molecular CH4 remain 0, and the bonding is "covalent", or (more or less) sharing the electrons to fill vacant orbitals. $\endgroup$ Nov 4, 2022 at 7:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.