Chemistry - What is the hybridization of the nitrogen in the azide ion?

Solution 1:

Here are the resonance structures we can draw for the azide anion.

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  • Resonance structure A has an octet on each nitrogen and just one unit of formal charge on each nitrogen
  • Resonance structures B and C also have an octet around each nitrogen, but there is a rather large formal charge of -2 on one of the nitrogens. Such a large charge on a small atom like nitrogen makes structures B and C less favorable and therefore we would expect them to count less to the "true" representation of the azide anion, than resonance structure A
  • Resonance structures D and E do not have an octet around the singly-bound, terminal nitrogen, these resonance structures will count even less. In fact, since the azide ion is linear and these two resonance structures are bent, we can say that they don't really contribute to the true picture of the azide ion.

So if we consider only resonance structures A and B (C is equivalent to B) we can see that the central nitrogen is $\ce{sp}$ hybridized in both cases (like in allene for A and like in acetylene for B). The $\ce{sp}$ hybridization of the central nitrogen is consistent with the linear structure of the azide ion.

In A, the terminal nitrogens may both be $\ce{sp^2}$ hybridized; after all the drawing looks similar to the double bond in ethylene. However, there are other options. For example, both nitrogens could be $\ce{sp}$ hybridized, with an $\ce{sp}$ single bond, one lone pair in a $\ce{sp}$ orbital and the other lone pair in a $\ce{p}$ orbital (just like the 2 lone pairs on oxygen in water). Without more data there is no way to tell for sure, but in an introductory class $\ce{sp^2}$ hybridization would be a reasonable answer.

In B, the terminal nitrogen with the triple bond appears to be $\ce{sp}$ hybridized just like in acetylene. The hybridization of the other terminal nitrogen in resonance structure B is anybody's guess; there are many possibilities and since there is only one ligand attached (the central nitrogen), we don't know what direction the 2 electron lone pairs are pointing in. This nitrogen might be unhybridized, $\ce{sp}$ hybridized, $\ce{sp^2}$ hybridized - again, we just don't know. Since the true hybridization of the terminal nitrogen is an equal mixture of these two hybridizations (e.g. a blend of B and C), we really don't know how the terminal nitrogens in B and C are hybridized.

In an introductory class, the answer would be that the central nitrogen is $\ce{sp}$ hybridized, and, since resonance structure A is the major resonance contributor, we might say that the terminal nitrogen is $\ce{sp^2}$ hybridized (but, from the above discussion, we know that is an oversimplification).

Solution 2:

I have said it a couple of times already on our site,[1] but I'm not getting tired of saying it again:

  1. Hybridisation always follows the (spatial) geometry of the molecule, never the other way around. As such it is a way to interpret bonding, not to predict it. Following from that, knowing the molecular structure may always give you the ability to suggest a hybridisation scheme.
  2. As a rule of thumb, terminal atoms (excluding hydrogen) have almost always (maximally) approximately sp hybridised orbitals. While other hybridisation schemes can be applied, they are usually not a good representation.[2]
  3. As another rule of thumb, the lesser the mixing between s and p orbitals, the (probably) more correct the bonding will be described.[3]
  4. Any scheme using d-orbitals or higher (in main group chemistry) is likely to be outdated and needs a more refined approach to bonding.

Using the above it is fairly simple to derive the hybridisation scheme of all the atoms. Since the molecule is linear, the central nitrogen is constrained to have at most sp orbitals. With the rule of thumb, the terminal nitrogen also only have sp orbitals. These orbitals will form the σ bonds and the σ lone pairs. The remaining two p orbitals at every nitrogen will form the π bonds. If you further think about it, the molecule is isoelectronic to carbon dioxide. I have written about the electronic structure before: π Bonding in Carbon Dioxide.

Because of the high symmetry of the molecule, D∞h, automatic interpretation (with natural bond orbital theory) of this bonding situation is very difficult. A natural resonance theory analysis on the DF-BP86/def2-QZVPP level of theory gives us the following structures: $$\ce{ \underset{24\%}{[N-N#N]-} <-> \underset{49\%}{[N=N=N]^-} <-> \underset{24\%}{[N#N-N]-} }$$

Unfortunately the generated hybrid orbitals don't look much like we would expect them to look, because they are also constrained to symmetry. (Occupied orbitals in blue/orange.) You can see that the central nitrogen basically has two lone pairs, which have a very low occupancy (1.29 el.), while the (antibonding) π bond is between the the terminal nitrogen. A linear combination of these two would lead to the expected bonding picture. The virtual combinations of the π bond (red/yellow) on the other hand have a quite high occupancy (0.71 el.). This is often an indicator that a hybridisation scheme is not going to work very well. Resonance effects are not described very well with hybrid orbitals.

natural bond orbitals of azide ion


  1. See for example:
    What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride?
    What is the hybridization of chlorine in vinyl chloride?
    Hybridisation of terminal nitrogen in diazomethane
  2. It is very important to understand that atoms are never hybridised. (Even though organic chemists tend to use this jargon - even in textbooks.) The orbitals at an atom can be hybridised. Hybridisation is just one way of looking at the bonding situation, it is one model. As such, this interpretation is correct, whether or not your teacher might agree with it. There is no way to probe an hybridisation scheme, there are only instances where one explanation fits the experimental observations better than another. For example see below.
  3. One most popular example of misapplying hybridisation schemes is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially means that the lone pairs are equivalent. This interpretation is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. For more on this I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124. DOI: 10.1021/ed064p124).