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Author Topic: Oxidation states/geometries of N2O3 and N205  (Read 14374 times)

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spirochete

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Oxidation states/geometries of N2O3 and N205
« on: May 13, 2009, 03:19:44 PM »

I can't figure out how to rationalize the oxidation states of these two nitrogen oxides from their Lewis structures.  All I have are molecular formulas.  I attached my drawings as a word file.  I am aware of the charges/lone pairs I omitted them for simplicity.

N2O3 is actually planar with bent bonds to all oxygens.  It's also got a +3 oxidation state.  I've drawn N203 with one nitrogen having a lone pair and two bonds to oxygen, and the central bond to the other carbon.  The other nitrogen looks like a nitro group.  This molecule has 5 total bonds to oxygen.  With two nitrogen atoms shouldn't this be a +2.5 oxidation state?

Also shouldn't the nitrogen on the left be SP hybridized due to resonance, making the left side linear?  In reality it's not but I can't see why, unless it's just the instability of that resonance structure.

Similar deal with dinitrogen pentoxide, which is actuality has a +5 oxidation state.  Here we have 2 nitrogens with 8 total bonds to oxygen, so why not a +4 oxidation state?
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Re: Oxidation states/geometries of N2O3 and N205
« Reply #1 on: May 13, 2009, 07:27:15 PM »

The Lewis structures are not completed - some electrons are missing.
For N2O3 omit bond between nitrogen atoms for oxidation state calculations.
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spirochete

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Re: Oxidation states/geometries of N2O3 and N205
« Reply #2 on: May 14, 2009, 03:28:13 AM »

I intentionally left out the lone pairs and charges.  Also I did leave out the N-N bonds when calculating oxidation state.  N2O3 has 5 total bonds to oxygen and the N2O5 has 8 total bonds.  Most or all are delocalized through resonance but that that shouldn't matter.
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BluRay

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Re: Oxidation states/geometries of N2O3 and N205
« Reply #3 on: May 14, 2009, 08:40:37 AM »

Referring to your pictures, you have:
1. in N2O3, attributing all the bond electrons  between N and O to the oxygen, one N (the one at left) is left with 1 lone pair (that you didn't draw) + 1 electron from the bond with the other N, so 3 electrons as total --> oxidation number = + 2  (5-3); the other N atom is left only with the electron that comes from the bond with the previous N, so ox. number = + 4  (5-1). The average is +3: [(+2) + (+4)]/2 = +3 which is nitrogen's ox. numb. in that compound.
2. in N2O5, after attributing the bond electrons to the most electronegative element (oxygen), both N atoms are left with no electrons, so their ox. numb. = +5  (5-0).
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spirochete

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Re: Oxidation states/geometries of N2O3 and N205
« Reply #4 on: May 16, 2009, 07:15:51 AM »

oh I see now thanks.  The major thing is that I was forgetting to include the formal positive charge on nitrogen when calculating oxidation state.
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hahahanna

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Re: Oxidation states/geometries of N2O3 and N205
« Reply #5 on: July 05, 2009, 02:03:17 PM »

Hi!

I'm stuck with my N2O3 compound.  ???
I have draw the Lewisstructur and I have understand that the geometry is planar, but what is the VSEPR for it and what is the space geometry for it? And what is the sigma and pi bonds in this compound?

I will be really greateful if someone can answer my questions and explain it for me.


The Lewisstructure:


O-N-N-O2
Double bond to the oxygens and one electron pair on the one nitrogen.
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Re: Oxidation states/geometries of N2O3 and N205
« Reply #6 on: July 05, 2009, 10:00:47 PM »

The Lewis structure for N2O3 is OK
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