Hybridizing oxygen and nitrogen orbitals

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I was thinking about how you'd hybridize oxygen and nitrogen orbitals (should be pretty straightforward), but I was getting confused. It's clear to me for carbon binding to 3 atoms with a double bond on one (say C in benzene ring: electron from 2s gets promoted to 2p, then you have 3 sp2 orbitals and a 2p orbital), but nitrogen and oxygen (in say the NO2 of nitrobenzene) throw me off. I feel like it should be simple, but I haven't been able to find an explanation on how to go about it.

With nitrogen, I can have 3 sp2 orbitals (one for C, and one for each O), which looks ok, but I'm not positive. For oxygen single bound to nitrogen (has a negative charge to begin with doesn't it? or how can it not), by promoting one electron from 2s to the last 2p, all you have is a single 2s orbital which is in the direction of nitrogen (or it could not promote and be a 2p orbital..somewhere). However, for oxygen double bound to nitrogen (neutral charge I believe), you have, after promoting an electron from 2s to 2p, a 2s orbital and a 2p orbital hybridize to make 2 sp orbitals. That can't be right though, because there needs to be one hybridized orbital to bind with nitrogen, and then a non-hybridized orbital (like a 2p orbital) for the pi bond/delocalized electron ring (cloud) from resonance structures.

I'm pretty sure everything I said about oxygen there is nonsense though, so does anyone know what the issue is?
 
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DrDu
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Hybridization was mainly introduced to describe the tetravalency of carbon, e.g. in CH_4. In nitrogen and oxygen compounds, it is often not necessary to use hybridized orbitals, especially since the promotion energy in e.g. oxygen is much higher than in carbon. For the N in the nitro-group assumption of sp2 hybridization is ok. But for the oxygens bound to it, I would describe both the sigma and the pi bonding in terms of pure p orbitals. The s-orbital remains doubly occupied and mostly inert in that process.
 
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Hybridization was mainly introduced to describe the tetravalency of carbon, e.g. in CH_4. In nitrogen and oxygen compounds, it is often not necessary to use hybridized orbitals, especially since the promotion energy in e.g. oxygen is much higher than in carbon. For the N in the nitro-group assumption of sp2 hybridization is ok. But for the oxygens bound to it, I would describe both the sigma and the pi bonding in terms of pure p orbitals. The s-orbital remains doubly occupied and mostly inert in that process.
So do I not have to differentiate between the oxygen with a double bond and the oxygen with a single bond (not looking at the combined resonance structure)? The oxygen with a single bond only has one p orbital to make a sigma bond, while the oxygen with a double bond has a p orbital to make a sigma bond and another p orbital to make a pi bond. But how how can it make a pi bond without something else to make the pi bond with (since nitrogen doesn't have anything available, and the other O only has a p orbital to bond with N)? Does it only make the pi bond when you look at resonance structures?
 
  • #4
DrDu
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In each of the resonance structures one oxygen forms a pi bond with the nitrogen and the other one has a lone electron pair (a doubly filled p orbital). Both atoms have the same number of orbitals. The sigma bond is formed by e.g. the p_x orbitals. The p_y is always filled with two electrons ( a lone pair). The p_z contains one electron on the oxygen with the double bond and this electron is paired with the electron in the p_z on the nitrogen. On the ohter atom, the p_z contains two electrons which are paired with each other. In the second resonance structure the roles of atom 1 and atom 2 are interchanged.
 
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In each of the resonance structures one oxygen forms a pi bond with the nitrogen and the other one has a lone electron pair (a doubly filled p orbital). Both atoms have the same number of orbitals. The sigma bond is formed by e.g. the p_x orbitals. The p_y is always filled with two electrons ( a lone pair). The p_z contains one electron on the oxygen with the double bond and this electron is paired with the electron in the p_z on the nitrogen. On the ohter atom, the p_z contains two electrons which are paired with each other. In the second resonance structure the roles of atom 1 and atom 2 are interchanged.
"Both atoms have the same number of orbitals." Didn't you also say that the double bond oxygen has two p orbitals (one in z-direction forming pi bond with nitrogen, and one in x-direction forming sigma bond with nitrogen), and that the single bond oxygen just has the p_x orbital forming sigma bond with nitrogen (and the p_y and p_z are just two lone pairs). So the atoms don't have the same number of orbitals? Since the first has two (the double bond O), and the other has one (the single bond O).

Also, how does the double bond oxygen form a pi bond with nitrogen?

This is what I thought nitrogen's valence orbital diagram looked like: http://i.imgur.com/ocsdl.png

However, how can it form the pi bond without another orbital (a p orbital) to do so? It only has enough combine the 2s and two of the 2p to make three sp2 that are needed to bind nitrogen to Carbon, and the two Oxygens.
 
  • #6
DrDu
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First I my count of orbitals doesn't depend on how they are occupied while yours apparently does.
Secondly, take in mind that in each resonance structure, nitrogen has a positive charge while one of the oxygens (the one which does not form a pi bond) has a negative charge.
So in your diagramm, there should only be 4 electrons on nitrogen, not 5.
 
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http://www.goiit.com/posts/list/organic-chemistry-orbital-structure-of-the-nitro-group-in-nitroalkane-984143.htm#1180277 [Broken]

Is this at all right? It's really contradicting what you said :s
 
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  • #8
DrDu
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That's somehow a mixture of valence bond and molecular orbital pictures. Especially organic chemists tend to treat heteroatoms like carbon atoms. As I told you, using hybrid orbitals on oxygen will lead to a description which, although not strictly wrong, will lead to inferior values e.g. for binding energy and other molecular characteristics.
If you want to gain a deeper understanding, I recommend you to have a look at Linus Paulings book: The Nature of the Chemical Bond and the Structure of Molecules and Crystals: Introduction to Modern Structural Chemistry.
Although some concepts are somewhat outdated, he at least explains very well the reason for considering hybrid orbitals and treats nicely the bonding in all kinds of substances, covering certainly also your nitro group example.
 

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