Ask A Scientist Chemistry Archive

name         Greg
status       other
grade        other
location     CA

Question -   Although I can easily balance the reaction of paraffin
C_22H_46 and oxygen by inspection, I am attempting to balance this redox
reaction using the oxidation number method. How does one assign oxidation
numbers to the carbon and hydrogen in a paraffin? It does not seem to be
possible to obtain a electrically neutral paraffin molecule using integer
oxidation numbers.
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   The point you raise is larger than your specific question. The use of
"valence" or "oxidation numbers" is a METHOD, not a law of nature. The
OBJECTIVE is to obtain mass/charge/ etc. balances in chemical reactions
required by various conservation laws. There are situations in which a
particular method is useful, and other cases where the method breaks down,
or is not useful. In which case, the ineffective method is abandoned and a
different method must is used. This distinction is frequently not made in
introductory chemistry because the teaching examples used are precisely
those where the method is applicable. "Oxidation number" in many, if not
most organic chemical reactions involving covalent rather than ionic
bonding is not useful, and does not reflect reality. Another example where
the "oxidation number" method fails is in non-stiochiometric compositions.
These compositions are frequently encountered in minerals, in metallic
alloys, and in cage (clathrate) compounds. All these are very common in
the real world of chemistry. Here even the concept of the law of combining
proportions (being in the ratio of small whole numbers) does not reflect
the compositional reality, so one could end up with strange looking
formula like: Fe1.8As0.3S1.7O0.2.
That is a hypothetical case, but it could express the analytical results. It
is the method that is limited, not the chemical analysis.

You are very right, "oxidation number" becomes nonsensical in the
combustion reactions of hydrocarbons.

Vince Calder
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C & H are regarded as having very similar degrees of electronegativity,
and  C-H bonds have very low electric polarization.
So why assume there is electron transfer, or oxidation number?
The system of oxidation numbers implicitly treats all bonds as 
partly-ionic bonds.
C-H bonds are purely covalent.  Almost exactly equal sharing of each 
other's electrons.
The oxidation number of each element was 0 before formation of the paraffin,
and it remains zero after forming paraffin.  The C's are 0 (zero), 
and the H's are 0.

Then when the paraffin reacts with oxygen, CO2 and H2O are formed.
Oxygen is highly electronegative compared with C or H,
and CO2 and H2O both have some polar character in each bond. 
(admittedly in CO2 it isn't much)
So now there is a nominal partial electron transfer
which we count up as non-zero oxidation numbers.
Because each oxygen likes to be -2 and the product molecules are neutral (0),
the C's become +4, and the H's become +1.

Considering C & H equal leads to a few weird results in other molecules.
The C is considered variable, and any H bonded to C is considered 0.
Formaldahyde (O=C)H2 would have O=-2, H=0, C=+2.
Formic acid (O=C)HOH would have C:+3.
     The H on the C would be zero,
    and the H in the OH would be +1 because it is on an 
 electronegative Oxygen.

Oxidation numbers are an approximation system
for guessing the bond-count preferences of elements in unfamiliar molecules
from the bond-count preferences already observed in more common molecules.
No such set of rules works right for every possible situation.
Maybe you would come a little closer if you had about two capacity numbers,
one for ionic or polar bonds, one for covalent.
But that sounds like work,  and it has not evolved yet.

Jim Swenson
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