# Determining the structure of N2O using hybridization

I have learnt how to determine the structure of molecules where only the central atom is hybridised like $\ce{ClF3}$, $\ce{C2H2}$, $\ce{PCl5}$, but in $\ce{N2O}$ it seems as though both nitrogen and oxygen have hybrid orbitals. How do I find the hybridization of oxygen and nitrogen in $\ce{N2O}$ and finally determine its structure?

Research Effort: I watched some lectures on hybridization, but all of them included cases in which only the central atom had hybrid orbitals. I also read the topic in my book but this issue wasn't addressed there too.

• I don't see how the hybridsation model could yield a structure prediction. In my understanding hybridisation can be used a post Lewis/VSEPR justification of a structure prediction. In that line you 1. draw a proper Lewis representation (possibly even some mesomeric forms) 2. apply VSEPR and 3. then explain it using hybridisation. – Raphael J.F. Berger Jul 29 '17 at 10:38

Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.

If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.

$$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$

From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.

So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.

After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).

• I am/ was a bit hesitant to up-vote this, but given the two wrong answers, this is much better. Okay, back to nitpicking: The only correct way to use hybridisation is to use it post-geometry determination (so it's not only practically incorrect). In principle there is no reason to assume any other external field than local linear coordination at terminal atoms. Constrained to this sp hybridisation is the max you can get. – Martin - マーチン Nov 27 '17 at 8:25

To do this, you will need to follow the steps mentioed by R_Berger in the comments. I will use $\ce{CO2}$ as an example.

1. Draw a Lewis diagram.

1. Look at bonding & lone pairs on each atom.

Carbon has 2 double bonds which means 2 $\times \ \sigma$ bonds and 2 $\times \ \pi$ bonds. Therefore, it has sp hybridization (2 $\times$ sp orbitals and 2 $\times$ p orbitals)

Each oxygen has 1 double bond and 2 lone-pairs which means 1 $\times \ \pi$ bond and 3 $\times \ \sigma$ bonds (actually 1 bond and 2 orbitals with lone pairs). Therefore they are sp2 hybridization. (3 $\times$ sp2 orbitals and 1 $\times$ p orbital)

1. Determine the shape

This can be done directly from the Lewis diagram using VSEPR theory. However, we only need to consider the central atom as that's where the bond angles will be! (You need two bonds to measure an angle...).

So for carbon dioxide, we have got sp hybridization at the central atom (or two areas of electron density with no lone pairs), so the shape will be linear.

Aside: You probably were already fine with the determining hybridization part. You just need to use the same rules on outer atoms!

• Oxygen does not have sp² hybridisation. Terminal atoms have at most sp hybrid orbitals. Also note that atoms are never hybridised, only orbitals are. It is perfectly valid, to have a variety of hybrid orbitals at one atom. Just because there is one sp² orbital, it does not mean that there need to be two more. (Have a look here for more on carbon dioxide.) – Martin - マーチン Jul 31 '17 at 6:21
• That link is not one I've found before so I've just spent some time reading it. A few notes (please correct me if needed)... if I understand this, the method I described above (and is used in a majority of places if I search) is outdated? Where do the lone pairs reside? From your link, I see the central argument is that the electron density map is a cylinder which could not happen with my sp<sup>2</sup>, but does fit an sp hybridisation. Do you have any links for those maps? (Aside: my knowledge is from teaching IB level Chemistry, so much of the advanced theories has gone past me!) – Michael Roberts Jul 31 '17 at 6:48
• After reading around the topics more, I believe that my answer is correct AND Martin's answer is also correct! I have answered the question using Valence Bond theory (where the hybridisation sp² describes what happens on the oxygen atom itself) whlie Martin used Molecular Orbtial theory (where the sp hydridisation is between the carbon's s and oxygen's p orbitals) (See here) – Michael Roberts Jul 31 '17 at 7:20
• I am sorry, but you have not answered the question with Valence Bond theory, which is essentially an equivalent theory to molecular orbitals. The sp² orbitals for such atoms used to be part of a VB description, but (at the latest) with the formulation of Bent's rule that basically became obsolete. Mathematically this would still work, physically it simply does not make any sense. For the simple case of water it can be proven, that the lone pairs are not equivalent; this can be extended to more complex molecules. Side remark: MOT does not use or need hybridisation. – Martin - マーチン Jul 31 '17 at 9:13
• Thank you for the above. I'm now reading about Bent's rule... there's a lot to take in / relearn here. I've mainly been involved at secondary/ high school level where the older hybridisations are still considered as the correct answer. Once I've done more reading I'll look at updating my answer. – Michael Roberts Jul 31 '17 at 11:38

Based on hybridisation theory, the hybridisation states of the atoms are $\mathrm{sp}$, $\mathrm{sp}$ and $\mathrm{sp^3}$, for the atoms N, N and O, which are connected to each other in this order:

Conventionally, as most textbooks suggest, hybridisation states are derived from electronic geometry of the atoms (i.e. how the electron density regions are arranged around the atoms). For 2 electron density regions, it is sp-hybridised. For 3 electron density regions, it is $\mathrm{sp^2}$-hybridised. For 4 electron density regions, it is $\mathrm{sp^3}$-hybridised etc.

From the Lewis structure I have drawn, we can derive the results as such.