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Topic: Formal Oxidation Number  (Read 17245 times)

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Offline Jules18

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Formal Oxidation Number
« on: December 07, 2009, 11:17:17 PM »
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~

Offline stewie griffin

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Re: Formal Oxidation Number
« Reply #1 on: December 08, 2009, 09:43:38 AM »
It's pretty easy actually... just have to follow a set of rules.
http://en.wikipedia.org/wiki/Oxidation_state

Offline Jules18

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Re: Formal Oxidation Number
« Reply #2 on: December 08, 2009, 07:30:57 PM »
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?
« Last Edit: December 08, 2009, 07:47:16 PM by Jules18 »

Offline stewie griffin

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Re: Formal Oxidation Number
« Reply #3 on: December 09, 2009, 08:26:59 AM »
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?

Offline savy2020

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Re: Formal Oxidation Number
« Reply #4 on: December 09, 2009, 11:11:03 AM »
XeF8 doesn't exist
Xe can't have +VIII oxidation state.
It might've been XeF6
:-) SKS

Offline stewie griffin

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Re: Formal Oxidation Number
« Reply #5 on: December 09, 2009, 11:15:16 AM »
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.

Offline savy2020

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Re: Formal Oxidation Number
« Reply #6 on: December 09, 2009, 11:23:36 AM »
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...
:-) SKS

Offline Jules18

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Re: Formal Oxidation Number
« Reply #7 on: December 09, 2009, 04:03:07 PM »
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

Offline savy2020

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Re: Formal Oxidation Number
« Reply #8 on: December 09, 2009, 04:06:23 PM »
Well then the answer is +VIII.
Quote
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?
:-) SKS

Offline Jules18

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Re: Formal Oxidation Number
« Reply #9 on: December 09, 2009, 04:09:51 PM »
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.)

Offline savy2020

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Re: Formal Oxidation Number
« Reply #10 on: December 09, 2009, 04:12:39 PM »
From where do you get the 2 unbonded electrons??
Xe has only 8 valence electrons.
:-) SKS

Offline Jules18

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Re: Formal Oxidation Number
« Reply #11 on: December 09, 2009, 04:39:36 PM »
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. 

Offline stewie griffin

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Re: Formal Oxidation Number
« Reply #12 on: December 09, 2009, 05:32:11 PM »
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.

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