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### Topic: Lewis Structure for N2O5  (Read 38015 times)

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#### jumper1127

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##### Lewis Structure for N2O5
« on: October 01, 2008, 10:11:25 PM »
http://en.wikipedia.org/wiki/Image:N2O5.svg

(Sorry, I can not link to the picture, so just put URL here.)

Now we are learning Lewis Structure and Formal Charges. I encounter a problem when I was asked to draw the Lewis structure of N2O5. The problem tells me that there are no N-N or O-O bonds, and the structure should be symmetrical.

Formal Charge=valence electrons-nonbonding electrons-bongding electrons/2

According to the formal charge rule, a structure should be most stable without any formal charges. So I draw a Lewis Structure different from the one which is correct (See picture). I used N=O bonds to connect all the oxygens at four corners (sorry I can not show my work, since I can't upload here). And the oxygen at four corners have another 2 lone pairs of electrons. Calculate their formal charges, I got 6-4-4/2=0. And the middle part of the structure is the same as the picture shows, which is N-O-N. Nitrogen has 5 valence electrons, so if it has 5 single bonds and no lone electron, it has no formal charge. And the oxygen in the middle, which has two single bonds and two lone pairs of electrons, the formal charge is also 0. Every one is happy, and the electrons in total is 40, which corresponds to 6*5+5*2=40.

However, the correct structure has formal charges. Why we use this structure instead of the one I thought was right? I'm so confused.

#### enahs

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##### Re: Lewis Structure for N2O5
« Reply #1 on: October 01, 2008, 10:25:18 PM »
The structure you link is not correct....nor is any Lewis structure if you want to get very specific.

Lewis structures are just nice easy ways of drawing things. They are very useful, but not always a good representation. The major flaw in them is saying electrons occur in a specific bond, which is not the case. Electrons are free to move, and have probabilities of being in the bond. To account for this in Lewis structures, there is resonance. In other words, I can write 4 other completley valid structures for that molecule according formal charges and such....which one is correct?
http://en.wikipedia.org/wiki/Resonance_(chemistry)

If you look at the picture you linked with the theory of resonance, you can see all the bonds according to Lewis structure and resonance will be equal.

#### jumper1127

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##### Re: Lewis Structure for N2O5
« Reply #2 on: October 01, 2008, 11:34:04 PM »
Yes, I know what you mean, and I know there are four resonance structures here. But why my thought was wrong? My structure has no formal charge, which is better than the correct structure. I can't understand.

#### nj_bartel

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##### Re: Lewis Structure for N2O5
« Reply #3 on: October 02, 2008, 01:29:53 AM »
Just going to ignore the octet rule and give nitrogen 10 valence electrons?

#### jumper1127

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##### Re: Lewis Structure for N2O5
« Reply #4 on: October 02, 2008, 02:09:12 PM »
Quote
There are many exceptions of Octet Rule. P can have 5 bonds, why not N? They are in the same column.

Yes, I think this is a problem. P can have 5 bonds, since it can have a d orbital if necessary. But N can't, because it can not have a d orbital. Is that right?
There is no exception to the octet rule because it is absolutely valid only for period 2 elemets of the periodic table
« Last Edit: October 03, 2008, 01:13:03 AM by AWK »

#### Astrokel

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##### Re: Lewis Structure for N2O5
« Reply #5 on: October 02, 2008, 02:54:33 PM »
yes you are right  period III elements have vacant d orbitals allowing expansion of octet structure.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

#### jumper1127

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##### Re: Lewis Structure for N2O5
« Reply #6 on: October 02, 2008, 10:02:51 PM »
Thanks.

Another question. Draw the Lewis Structure of NO+.

I used double bond to connect two atoms. O has 4 lone electrons, and N has 2 lone electrons, with a +. No formal charge.

But the correct is triple bond between O and N. N has 2 lone electrons, and O has 2, with a +. This structure also has no formal charge.

Why the positive sign should be on O instead of N? O has a larger electronegativity than N, it should attract more electrons.

#### nj_bartel

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##### Re: Lewis Structure for N2O5
« Reply #7 on: October 02, 2008, 10:14:01 PM »
In order to satisfy the octet of both, oxygen must give up one of its lone pairs to bond.  Although nitrogen would more favorably give up its lone pair to bond due to electronegativity, that would violate oxygen's octet.

#### jumper1127

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##### Re: Lewis Structure for N2O5
« Reply #8 on: October 02, 2008, 10:18:37 PM »
Oh yeah, Octet rule! I forgot to consider it again. Damn it.

But I think my structure does not violate oxygen's octet rule. Instead, nitrogen's octet rule is violated because in my structure, nitrogen just has 6 electrons around it.

Thank you.

#### nj_bartel

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##### Re: Lewis Structure for N2O5
« Reply #9 on: October 02, 2008, 10:22:42 PM »
I'm sorry, I might have said something wrong.  Let me reevaluate real quick.

#### nj_bartel

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##### Re: Lewis Structure for N2O5
« Reply #10 on: October 02, 2008, 10:29:54 PM »
Yea, the structure I was thinking of would have 2 of nitrogen's lone electrons bonding to two of oxygen's lone electrons, and oxygen donating its lone pair to bond.  However, this would leave a free electron on nitrogen if I'm not mistaken.  Gonna leave it for someone else to answer sorry.

#### jumper1127

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##### Re: Lewis Structure for N2O5
« Reply #11 on: October 02, 2008, 10:50:46 PM »
How about N contributes 3 electrons and O also contributes 3 electrons to build the triple bond between them? Because we have a positive charge here, and if O loses 1 electron to have this positive charge, it will have 3 lone single electrons and a pair of lone electrons. So both N and O has 3 lone single electrons now, and they are happy to build a triple bond with these electrons. And besides bonding electrons, N has a pair of lone electrons, and O has a pair of lone electrons and a positive charge. This structure fulfills the octet rule, and has no formal charge. So it's the best one.