When is nitrogen sp2 hybridized

The valence-bond concept of orbital hybridization can be extrapolated to other atoms including nitrogen, oxygen, phosphorus, and sulfur. In other compounds, covalent bonds that are formed can be described using hybrid orbitals. The nitrogen is sp 3 hybridized which means that it has four sp 3 hybrid orbitals. Two of the sp 3 hybridized orbitals overlap with s orbitals from hydrogens to form the two N-H sigma bonds. One of the sp3 hybridized orbitals overlap with an sp 3 hybridized orbital from carbon to form the C-N sigma bond.

The lone pair electrons on the nitrogen are contained in the last sp 3 hybridized orbital. Due to the sp3 hybridization the nitrogen has a tetrahedral geometry. The oxygen is sp 3 hybridized which means that it has four sp 3 hybrid orbitals. One of the sp 3 hybridized orbitals overlap with s orbitals from a hydrogen to form the O-H signma bonds. One of the sp3 hybridized orbitals overlap with an sp 3 hybridized orbital from carbon to form the C-O sigma bond.

Both the sets of lone pair electrons on the oxygen are contained in the remaining sp 3 hybridized orbital. Due to the sp3 hybridization the oxygen has a tetrahedral geometry.

What if there was some compensating effect whereby a lone pair unhybridized p-orbital was actually more stable than if it was in a hybridized orbital? This turns out to be the case in many situations where the lone pair is adjacent to a pi bond! The most common and important example is that of amides , which constitute the linkages between amino acids.

The nitrogen in amides is planar sp 2 , not trigonal pyramidal sp 3 , as proven by x-ray crystallography. Why is trigonal planar geometry favoured here? Better orbital overlap of the p orbital with the pi bond vs. The drawing below tries to show how a change in hybridization from sp 3 to sp 2 brings the p-orbital closer to the adjoining p-orbitals of the pi bond, allowing for better orbital overlap. Better orbital overlap allows for stronger pi-bonding between the nitrogen lone pair and the carbonyl p-orbital, which results in an overall lowering of energy.

Two important consequences of this interaction are restricted rotation in amides, as well as the fact that acid reacts with amides on the oxygen, not the nitrogen lone pair! The oxygen in esters and enols is also also sp 2 hybridized, as is the nitrogen in enamines and countless other examples. For this reason, the most basic site of pyrrole is not the nitrogen lone pair, but on the carbon C-2! Another example where the actual hybridization differs from what we might expect from the shortcut is in cases with geometric constraints.

For instance in the phenyl cation below, the indicated carbon is attached two two atoms and zero lone pairs. In fact, the geometry around the atom is much closer to sp 2. A quote passed on to me from Matt seems appropriate:. Ask Matt about scheduling an online tutoring session here.

Third row elements like phosphorus and sulfur can exceed an octet of electrons by incorporating d-orbitals in the hybrid. An amine connected to three different substituents R 1 R 2 and R 3 should be chiral, since it has in total 4 different substituents including the lone pair. However, all early attempts to prepare enantiomerically pure amines met with failure. It was later found that amines undergo inversion at room temperature, like an umbrella being forced inside-out by a strong wind.

In the transition state for inversion the nitrogen is trigonal planar. One can thus calculate the difference in energy between the sp 3 and sp 2 geometries by measuring the activation barrier for this process notably by the work of Kurt Mislow RIP ]. Note 3 :A fun counter-example might be Coelenterazine. One would not expect both nitrogen atoms to be sp 2 hybridized, because that would lead to a cyclic, flat, conjugated system with 8 pi electrons : in other words, antiaromatic.

This is incorrect, and was proven wrong years ago. This way, every atom follows the octet rule in each resonance structure. Yes and no. It is more to the point of what orbitals are involved in the overall bonding scheme. What is the hybridization of the carbonyl carbon there?

Is it sp2? Is it sp2-? Is it somewhere in between? What about the hybridization in di-central rhenium complexes with quaternary bond? You can observe similar deshielding effects in, say, azulene for the protons on the 7-membered ring. NOTE : this image is purposefully incorrect! Instead, it stays unhybridized and holds the lone pair because that way, electron delocalization is allowed due to symmetry compatibility, and provides energetic stability aromaticity.

Ultimately we found that the reasons why the lone pair in pyridine is outside the ring and in pyrrole it is inside the ring are:. Why is the nitrogen's lone pair part of sp2 orbital in pyridine but part of p orbital in pyrrole? Why so? I've read half a dosen discriptions online, but am clearly missing something. Truong-Son N. Jul 12, Below, I actually go into how I know this. Basically: hatE is an identity operation that returns the same molecule back without doing anything defined for the purpose of completeness.

For both molecules, this is the only rotation axis they have. This operation returns the molecule reflected through the xz plane coplanar with the ring. This operation returns the molecule reflected through the yz plane perpendicular to the ring.