5
$\begingroup$

To get the oxidation state of carbon (denoted $x$) in octane, $\ce{C8H18}$, I used the fact that hydrogen has an oxidation state of $+1$:

$$8x + 18\cdot 1 = 0 \qquad \Longrightarrow \qquad x = -\frac{18}{8} = -2.25$$

But in that way, I get an oxidation number that is fractional, or not an integer. Is that normal?

$\endgroup$
  • $\begingroup$ We hardly ever get fractional oxidation states, you know. We see them only in things like Fe3O4 (iron), alkali metal superoxides (oxygen), YBa2Cu3O7 high temperature superconductor (copper), aromatic ions (carbon), trihydrogen cation in interstellar space (hydrogen), ... . $\endgroup$ – Oscar Lanzi Mar 8 '18 at 10:59
  • 1
    $\begingroup$ Also you shouldn't just start from the formula, you don't know anything about connectivities for example. In this case it's pretty obvious but there are also compounds with more than one carbon-carbon bond and those are usually not counted when you try to find oxidation states. But answering your question. There are examples where this is postulated but then it is often dependend on the system. Like with non-innocent ligands you can distribute electron density over your system then sometimes weird oxidation states are reported although this is not true. $\endgroup$ – Justanotherchemist Mar 8 '18 at 11:08
  • $\begingroup$ I also read about an example with a solid state cyanide of cobalt where they had a fractional oxidation state. I never overlooked that system that well but there are things like Mg(I) for example where you have a Mg-Mg bond. If you force it enough you may start doing redox-chemistry somewhere completely else. And then it would suddenly end up without a fractional oxidation state again. $\endgroup$ – Justanotherchemist Mar 8 '18 at 11:11
19
$\begingroup$

The "oxidation number" is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons "owned" by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.

Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.

It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.

$\endgroup$
9
$\begingroup$

In addition to bob's answer, it's worth noting that oxidation number assumes that bonding is happening mostly ionically, which isn't really the case with alkanes where the electronegativities of Carbon and Hydrogen are very similar.

$\endgroup$
4
$\begingroup$

IUPAC recognizes that there are fractional oxidation states, but asks that you avoid writing them.

IR 4.6.1 says not to write an oxidation state "where it is not feasible or reasonable to define" because:

This avoids the use of fractional oxidation states.

Examples:

  1. $\ce{O2-}$

  2. $\ce{Fe4S4^3+}$

See also IR-5.4.2.2 which says "oxidation numbers are no longer recommended when naming homopolyatomic ions" because "ions such as pentabismuth(4+) (see Section IR-5.3.2.3) and dioxide(1—) (see Section IR-5.3.3.3), with fractional formal oxidation numbers, could not be named at all"

However, in the particular example in the question, not all the carbons are equivalent. Some will have 3 hydrogens, some 2 hydrogens, and depending upon the isomer, perhaps some with 1 or 0 hydrogens. So there is no fractional oxidation state in this example.

$\endgroup$
0
$\begingroup$

Oxidation state is a bit of a formalism but fractional states are rare but possible. For instance, Ag2F has silver in +1/2. There is a fair amount of metal-metal bonding so it's not like you can think of just having a mix of +1 and 0 atoms. (As you would have +2/+3 in a ferrite.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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