Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Jules18 on December 07, 2009, 11:17:17 PM
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hey guys, I'm just wondering how to figure out the formal oxidation number of an atom in a compound when you have the lewis structure?
(apparently it's totally different than formal charge)
~Julie~
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It's pretty easy actually... just have to follow a set of rules.
http://en.wikipedia.org/wiki/Oxidation_state
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Thanks so much.
Wikipedia gives two different ways of doing it, depending on whether you have a lewis structure or not, but when I use both ways on the same compound (XeF8) I get two different values. Does that happen a lot? Do you just assume the one based on the lewis structure is right?
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What values are you getting and how are you getting them? You should be getting the same value regardless.
I believe (and I don't remember all the oxidation state rules... it's been a while) that halogens are always -1. So each fluorine has the oxidation state of -1. Since XeF8 (are you sure it's not XeF6?) is overall neutral then Xe has to be +8. What are you getting?
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XeF8 doesn't exist
Xe can't have +VIII oxidation state.
It might've been XeF6
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re Savy2020: Yeah I don't think XeF8 exists either... that's why I asked if the person was sure. Either way though the point of how to calculate oxidation numbers is the same.
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Of course whichever method from whereever you use you do get the same oxidation state
@Jules18
How do you get two different oxidation states. Could you please explain...
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XeF8 doesn't exist
Xe can't have +VIII oxidation state.
It might've been XeF6
nope, it's definitely XeF8. The question gives me a Lewis structure and it shows 8 fluorines. I don't know if the compound could actually exist, but maybe whoever made the question doesn't care and just wants to test if we know how to figure out oxidation
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Well then the answer is +VIII.
Wikipedia gives two different ways of doing it, depending on whether you have a lewis structure or not, but when I use both ways on the same compound (XeF8) I get two different values. Does that happen a lot? Do you just assume the one based on the lewis structure is right?
What were the two diff values you got?
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What values are you getting and how are you getting them? You should be getting the same value regardless.
I believe (and I don't remember all the oxidation state rules... it's been a while) that halogens are always -1. So each fluorine has the oxidation state of -1. Since XeF8 (are you sure it's not XeF6?) is overall neutral then Xe has to be +8. What are you getting?
When I did it that way, I did get +8. But when I used the other method listed in wikipedia under "From a Lewis structure" http://en.wikipedia.org/wiki/Oxidation_state , I got +6.
(Xe had 8 single bonds with fluorine, which is more electronegative, and two unbonded electrons.)
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From where do you get the 2 unbonded electrons??
Xe has only 8 valence electrons.
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you're right, it wouldn't really have 2 unbonded electrons, but that's the diagram the question gave me. so i guess the question isn't very realistic and was just testing my method for figuring out formal oxidation, and I'm confused for no reason.
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Stick with the rules method, even though you have a lewis structure. I used the rules method all through gen chem (even with a lewis structure present) and it always worked just fine.
The "short cut" with the lewis structure in wikipedia is used in organic chemistry...it's a fine shortcut but I'm not sure that the examples used on wikipedia are calculated correctly.