Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: xstrae on September 13, 2006, 12:29:27 PM
-
Hi, I have been asked to find the oxidation state of:
C in CH3COOH
2x - 2(2) + 1(4) = 0 ; x = 0
is that correct?
-
OK
http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure
Scroll down to the end of the page. Oxidation numbers in organic compunds are crazy.
-
konichiwa2 ,you did it correctly.
Contrary to Borek, I think oxidation numbers problem in organic chemistry is a quite simple one.
You can calculate a mean oxidation number for for a specific kind of atoms in a molecule, or oxidation numbers for each atom.
For sepatate atoms:
CH3 ON for C is -3 [x+3=0]
COOH ON for C is +3 [x-2(2)+1=0]
And mean is (-3+3)/2=0
C-C bonds do not contribute to ON in this method. Why?
Other example
CH3-CH2-COOH ON(for C) [ 3x+6-4=0] => x=-2/3
For separate atoms: -3 -2 and +3 respectively for C1, C2 and C3 (from the left)
mean from above numbers is (-3-2+3)/3=-2/3
Try to count oxidation numbers for all atoms in CH3OOH (peroxide)
Use separate atom method
Just for checking)
(C-2, H+1, O-1)
-
Contrary to Borek, I think oxidation numbers problem in organic chemistry is a quite simple one.
You can calculate a mean oxidation number for for a specific kind of atoms in a molecule, or oxidation numbers for each atom.
I have never stated they are difficult, I just can't see any rationale behind.
Do they reflect any real matter property? Any real property of atoms in a molecule? Do they have any use other than accounting when doing redox reactions, redox reactions that can be done without introducing non-existing concepts?
If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.
-
ok thanks
Try to count oxidation numbers for all atoms in CH3OOH (peroxide) Use separate atom method
I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0
-
I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0
Then how would you have connected all the other atoms without breaking valencey and octet/duet rules?
If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.
Since chemical reactions are defined as transfer of electrons it is very important.
Important in real synthesis (where you just do not remember a bunch of list of chemicals that react with each other). Determining mechanisms of reactions by following the electrons, etc etc. It is not art for arts sake, it does however have very limited to use at lower levels.
-
If not, why do the students have to learn them? And learn how to assign them to every atom in organic molecules? For me that's a waste of time, art for the art's sake.
Since chemical reactions are defined as transfer of electrons it is very important.
Important in real synthesis (where you just do not remember a bunch of list of chemicals that react with each other). Determining mechanisms of reactions by following the electrons, etc etc. It is not art for arts sake, it does however have very limited to use at lower levels.
Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?
-
I got that when I read that it is a peroxide. But how do i determine whether O in a given compound is a peroxide? If you hadnt told me it is a peroxide, i think I would have ended up with ON of C = 0
Then how would you have connected all the other atoms without breaking valencey and octet/duet rules?
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?
-
[
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.
-
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.
When you know structure of the compound you know it is peroxide, that's enough to suspect what is going on during decomposition. I rest my case.
-
Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?
It is not limited to organic chemistry and synthesis.
However, the easiest example to point to (because I am lazy) is electrochemistry (making batteries).
-
Can you show examples of organic reactions where ON help determine reaction mechanism? Can you elaborate on this "etc, etc" part?
It is not limited to organic chemistry and synthesis.
However, the easiest example to point to (because I am lazy) is electrochemistry (making batteries).
You are mistaking balancing redox reaction with determining reaction mechanism.
But let's start from the beginning. Define oxidation number.
-
You are mistaking balancing redox reaction with determining reaction mechanism.
No I am not. I am just trying to illustrate the uses of oxidation numbers without writing a paper or listing 20 reactions.
I do not understand what your problem with them is. As a chemist you want to understand what is happening in a chemical reaction. Oxidations numbers are useful for that. Knowing what gained and lost electrons is very important. You might see no use for it to get a good grade on the test, but just by the definition of the word it is clearly important in understanding what is taking place in a chemical reaction.
-
I do not understand what your problem with them is. As a chemist you want to understand what is happening in a chemical reaction. Oxidations numbers are useful for that. Knowing what gained and lost electrons is very important. You might see no use for it to get a good grade on the test, but just by the definition of the word it is clearly important in understanding what is taking place in a chemical reaction.
For the oxidation numbers to be usefull they have to reflect some matter property. They don't. Please provide definition of oxidation number.
-
For the oxidation numbers to be usefull they have to reflect some matter property. They don't. Please provide definition of oxidation number.
You 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”.
-
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.
-
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.
-
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 :)
-
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.
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.
-
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 :)