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Topic: doubt on oxidation number  (Read 15832 times)

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Offline Borek

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Re: doubt on oxidation number
« Reply #15 on: September 15, 2006, 12:31:11 PM »
ou are wanting me to say “It is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.” (taken directly from Wikipedia). And you want to pounce on me about the hypothetical part about it not being a real “matter property”. However, by doing it for both sides it does show you the actual loss and gain of electrons, and in that case it is your precious “matter property”.

Charge transfer occurs between molecules, oxidation numbers try to attach charge to individual atoms in these molecules. That's simply wrong, charge is delocalized. Charge transfer in the redox reactions is much better described by half reactions - they show what is happening (where the real charge moves), without the need of using "hypothetical charge".

My site at

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

will only repeat what I am writing here, for the second opinion look here:

http://www.av8n.com/physics/balance-charge-atom.htm#sec-oxnum

While I not necesarilly agree with John Denker's idea of balancing equations using the method he proposes, I completely agree with his remarks on oxidation numbers.
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Offline enahs

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Re: doubt on oxidation number
« Reply #16 on: September 15, 2006, 01:21:17 PM »
You are now just arguing semantics.

You are in fact not making any valid point, other then to backwardly say I do not know what I am talking about.

They are a simple method to recognize when electron structure of atoms change. Nobody said it is always right, and nobody said that electrons where not delocalized. It is just a simple method to recognize what has changed, because that is very important to Chemist.

Offline Borek

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Re: doubt on oxidation number
« Reply #17 on: September 15, 2006, 02:34:48 PM »
say I do not know what I am talking about.

No, you are just repeating what you have been taught during your Chem101. Try to think out of the box.

Thing is, too many people believe that ON precisely describe what is really happening, that when permanganate gets reduced to manganate it is manganese that gets an electron. It is not a manganese atom that gets anything, it is whole ion that changes it charge.

To be honest I was taught the same thing and there was a time when I put more meaning into ON then they deserve. I have just later learnt that I was wrong :)
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Offline enahs

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Re: doubt on oxidation number
« Reply #18 on: September 15, 2006, 04:03:49 PM »
Quote

No, you are just repeating what you have been taught during your Chem101. Try to think out of the box.
No I am not. I am giving reason in a High School Chemistry forum why it is useful that someone at this level could under stand. Everything you learn in high school, and 99% bachelors level college chemistry is in fact wrong, just simplifications to make life easier and explain things and understand things.  Point of fact I am in a Masters program for chemistry.

Quote
Thing is, too many people believe that ON precisely describe what is really happening, that when permanganate gets reduced to manganate it is manganese that gets an electron. It is not a manganese atom that gets anything, it is whole ion that changes it charge.

And for every example you can give for a high school level or B.S. level where it is wrong I can give you an example where it is more or less correct.


My only comment was to you because you basically say for all purposes “they suck and are useless”. They are not useless (but they do suck…..), I never made any claims about them representing the actual electron distribution or charge, etc. I said they where a useful shortcut in explaining and figuring out various things, that is all. If they are really that bad then why does every chemistry book written in the past 60+ years mention them?

It is also very nice of you to be willing to point out the truth to help people and to further their education, but you are not saying that in general, you are directing it directly at me as if I am saying something I am not.

Offline xstrae

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Re: doubt on oxidation number
« Reply #19 on: September 16, 2006, 10:40:28 PM »
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This is what I would have done : x + 1(3) + -2(2) + 1(1) = 0 ; x = 0
How do I figure out that the O is in the peroxide state?


You hit the right nail on the head.
Having known nothing about structure of a compound you can calculate oxidation numbers nearby at your pleasure taking into account only two rules (1.ON for elements=0, and sum of all ONS is eqaul to 0 for compounds or is equal to charge of ion), and counting taken and lost electrons you can solve any redox equation. This is an aproach which has nothing common with quantum theory but always efficient. Moreover you use math on the level of primary school.

For compound CH4O2 formal ON for C is 0 (using +1 as ON for H and -2 as ON fot O). But when you know its structure CH3OOH ON for C is -2.
Decomposition of CH4O2 leads us to the reaction
CH4O2 = CH2O + H2O
which is non-redox reaction for ON of C=0 or a redox reaction (disproportionation) for ON of C=-2. In the later case you can learn something on a mechamism (chemistry) of this reaction (which Borek put in question). Note, this approach is for slightly more advanced chemists.

ok thanks :)

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