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My teacher said that even though interstitial compounds are neither ionic nor covalent, we still define an oxidation state for them. The example she gave was $\ce{TiC}$. She said the carbon is in $-4$ oxidation state. My thought was that there is no chemical bonding in interstitial compounds, and oxidation state is defined as the hypothetical charge on an atom if all its bonds were ionic, and since there are no bonds here, all atoms would be in their elemental state, and oxidation states would be zero. Which one is it?

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    $\begingroup$ In my opinion nobody can determine the oxidation state of such interstitial compounds. $\endgroup$
    – Maurice
    Aug 3, 2023 at 19:59
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    $\begingroup$ " there is no chemical bonding"? Of course there is! It can be considered metallic, but that's irrelevant. C atoms don't bind with each other here, so if you reaally want to ascribe ox. states they are just like you said $\endgroup$
    – Mithoron
    Aug 3, 2023 at 21:00
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    $\begingroup$ From a different perspective perhaps, if the Gibbs free energy changes (decreases) upon introduction of the interstitials then there is 'binding' of some kind going on. This is certainly the case for TiC, and way too many to list other transition metal carbides, oxides, nitrides, borides, ... $\endgroup$
    – Jon Custer
    Aug 3, 2023 at 21:40
  • $\begingroup$ Also, interstitial compounds are sometimes non-stoichiometric. So, it's hard to determine their oxidation state. $\endgroup$ Aug 5, 2023 at 3:43

2 Answers 2

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In the case of titanium carbide $\ce{TiC}$, there is a slight reaction with water to form methane, consistent with an oxidation state assignment $\ce{Ti^{IV}C^{-IV}}$, and even Group 5 carbides with the 1:1 stoichiometry (like niobium carbide $\ce{NbC}$) give methane as well.

From Avgustinik et al[1]:

Contrary to the generally held view that carbides of the transition metals of groups IV and V are inert with respect to water, the authors show on the example of titanium carbide that the carbides are decomposed by water, although the extent of attack is not more than 5–30 atomic layers, i. e., not more than 100–150 Å.

The hydrolysis of TiC proceeds according to the reaction: $\ce{TiC + xH2O -> CH4 + TiO2·xH2O}.$ [sic; the coefficient on water should be $x+2$]

Methane is the principal constituent of the gaseous hydrolysis products, and consequently it is possible to regard carbides of the transition metals of groups IV and V as methane derivatives. Hydrogen evolution in both carbides of group V and nonstoichiometric carbides of group IV is believed to take place as a result of the presence of free electrons in the carbides.

The "free electrons" in the reference apparently refer to the possibility that stoichiometric Group V carbides may contain metal(V) and free electrons (e.g. $\ce{Nb^{V}C^{-IV}e^-}$) or a Group 4 carbide might be carbon deficient and thus not have all valence electrons bound to carbon (e.g. $\ce{Ti^{IV}C_{1-x}^{-IV}(e^-)_{4x}}$). The overall story is that at least for methanides, the distinction between "ionic/saline" and "metallic/interstitial" carbides is actually not sharp.

Reference

  1. A. I. Avgustinik; G. V. Drozdetskaya; S. S. Ordan'yan (1967). "Reaction of titanium carbide with water". Powder Metallurgy and Metal Ceramics. 6 (6): 470–473. doi:10.1007/BF00780135. S2CID 134209836
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Although $\ce{TiC}$ cannot be strictly described only by covalent or ionic bonds, it does include both combined. The hardness of $\ce{TiC}$ is much greater than that of Ti and C in the same configuration, which simply means that they interact to form bonds, contrary to the argument where "there is no chemical bonding" that suggests a process like an intercalation of Li between two graphene layers in Li-ion batteries, where there are no such bonds (covalent or ionic).

Impurities tend to localize orbitals, as this affects the translational symmetry, the ionic effect is more effective. But this explanation is not intuitive and satisfying. There are articles on the subject giving an explanation in the case of $\ce{TiC}$ as a combination of a metallic, a partial ionic and covalent bonds. As they interact, the charge fluctuates and there is an oxidation state.

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  • $\begingroup$ Oxidation state doesn't actually have anything to do with charge, or electronegativity. It's an arbitrary parameter. Also, even in things like intercalated Li, there is a lot of chemical bonding - more of a metallic kind again. $\endgroup$
    – Mithoron
    Aug 4, 2023 at 18:08
  • $\begingroup$ @Mithoron This is a very clumsy criticism, if you confirm that you have a very good understanding of electronic structure and quantum mechanics, I will expand my answer or respond to your comment. $\endgroup$
    – M06-2x
    Aug 4, 2023 at 19:38

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