# Sp5 hybridization in cyclopropane?

I have never before heard/read about something as a $sp^5$ hybridization. Today, Henry Rzepa's blog post made me aware of the existance of such a bonding system. That made me search a little bit and I found an entry in a german chemistry forum, where this question was also asked ... they answered it with a mathematical construction$^\ast$:

Cyclopropane has the following angles:

$\angle \ce{HCH}=118^\circ~\text{resp.}~\gg 120^\circ$
$\angle \ce{CCC}~\text{with bent bonds:}~60 + 2 \cdot 21 = 102^\circ$

The orbitals towards the protons are $sp^2$ because of the $120^\circ$ angles.
The orbitals towards the carbons originate in the following relation:

$$1 + a \cos~\alpha = 0$$ ... where $\alpha$ is the bond angle and $a$ is the p-amount in sp$^a$ for the orbitals, which make up the angle.

This means for the orbitals, which span the 102 degree angle: $$1 + a \cos 102^\circ = 0$$ $$1 + a \cdot (-0.20) = 0$$ $$a = \frac{-1}{-0.20} = 5$$ $$\Rightarrow \text{sp}^5 \text{-orbitals}$$

Test:

• In a single sp$^a$ orbital, the s-amount is: $\frac{1}{1+a}$, because $1+a$ equals the sum of all amounts of s and p
• In a single sp$^a$ orbital, the p-amount is: $\frac{a}{1+a}$
• For s:

• In the orbitals that are oriented towards the protons, the s-amount is $\frac{1}{1+2} = \frac{1}{3}$
• In the orbitals that are oriented towards the carbons, the s-amount is $\frac{1}{1+5} = \frac{1}{6}$
• Addition of all s-amounts at a single carbon with all four bond orbitals yield: $\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=1$, which is correct, because there is only one single s-orbital at every carbon atom.
• For p:
• In the orbitals that are oriented towards the protons, the p-amount is $\frac{2}{1+2} = \frac{2}{3}$
• In the orbitals that are oriented towards the carbons, the p-amount is $\frac{5}{1+5}=\frac{5}{6}$
• Addition of all p-amounts at a single carbon with four bond orbitals yield: $\frac{2}{3}+\frac{2}{3}+\frac{5}{6}+\frac{5}{6}$, which is correct, because there are 3 p-orbitals at every carbon atom.

This means, that the bent bonds with $21^\circ$ from the $\ce{C-C}$-bond are spanned by sp$^5$ orbitals.

So math-magically this seems to make sense, but is there another explanation that might base more on chemical intuition or "real" chemical concepts?

A fast calculation ($\omega$B97X-D/def2-TZVPP) and a subsequent analysis of the isosurface of the Laplacian of the electron density, showed at least the expected "nonlinear", slightly curved bond between the carbon atoms.

Whoever might want to see the electron localization function (ELF), which also shows the bent bonds quite good:

$^\ast$ While I tried to translate it to my best, some errors might have been introduced by this . . . please correct me, where I'm wrong.

• en.wikipedia.org/wiki/Isovalent_hybridization – Mithoron Sep 13 '15 at 13:27
• Hybridisation isn't real - it's always only mathematical concept - you're talking about it's more complicated version, which is rather incompatible with normal version. – Mithoron Sep 13 '15 at 13:30
• Coulson's theorem is the key here. It can be used to interconvert bond angles and hybridization indices. See this earlier answer where it is applied to cyclopropane and ethylene. As pointed out in this answer, the bent-bond analysis (Pauling) is equivalent to the more common pi bond treatment (Hückel). – ron Sep 13 '15 at 13:46
• The notation is just a short form. It just states, that there is much more p character in the bonds than in the traditional ones, which we usually use. $$\mathrm{sp}^n\equiv\mathrm{s}^{\frac{1}{n+1}}\mathrm{p}^{\frac{n}{n+1}}$$ So this proof is just restating the original definition. Coulson's theorem (as stated in this proof) only works for equivalent hybrid orbitals, the more general approach is covered by Bent's rule (=observation). – Martin - マーチン Sep 13 '15 at 15:12
• Using hybridization for more complicated examples seems like embedding circular orbits on circular orbits to explain retrograde planetary motion in a geocentric model. If you just admit that the model is flawed and use a better heliocentric model, the complication goes away. – Zhe Feb 15 '18 at 22:53

Until this post, $\mathrm{sp^1,~sp^2~\text{and}~sp^3}$ meant for me, that the hybrid orbitals would consist from one s orbital and one, two or three p orbitals. I thought of it being so, as this seems to be what the most textbooks suggest, as can be seen in the following image:

One 2s orbital and three 2p orbitals hybridize to four $\mathrm{sp^3}$ orbitals.

This scheme seems to work but it also seems to be a working simplification as can be seen by my question about $\mathrm{sp^5}$ hybrid orbitals, which simply cannot consist from one s orbital and five p orbitals, as there are only three p orbitals in reach.

Then what does this nomenclature actually mean?

As Martin commented, it is more like the following: $$\mathrm{sp^n \equiv s^\frac{1}{n+1}p^\frac{n}{n+1}}$$ or to say it slightly different, a $\mathrm{sp^n}$-bond has $\frac{100}{n+1}~\%$ s-amount and $\frac{100\,n}{n+1}~\%$ p-amount.

\begin{array}{lll} \hline \text{bond} & \text{s-amount} & \text{p-amount}\\ \hline \mathrm{sp} & 50~\% & 50~\%\\ \mathrm{sp^2} & 33~\% & 67~\%\\ \mathrm{sp^3} & 25~\% & 75~\%\\ \mathrm{sp^4} & 20~\% & 80~\%\\ \mathrm{sp^5} & 17~\% & 83~\%\\ \ldots & \ldots & \ldots \\ \mathrm{sp^n} & \frac{100}{n+1}~\% & \frac{100\;n}{n+1}~\%\\ \hline \end{array}

• But this cannot work. If we make 3 sp5 orbitals we are only using about 50% of the s-orbital and 2.5 p-orbitals. So there would be left over s and p orbitals. We get a total of 7 (partial) orbitals, how do we distribute our 6 electrons here? It also doesn't work out if we make 6 sp5 orbitals (and how do we distribute 6 electrons and make 3 bonds with 6 orbitals?) – DSVA Feb 15 '18 at 20:58
• @DSVA A slightly different approach to the sp5 bonds can be found in this answer of mine. Hybridisation is a simple concept, obviously when you form linear combinations of the orbitals while maintaining the total number of orbitals, and the orthonormalisation. Also note that in the given description, every carbon utilises $\ce{2\times sp^5}$ orbitals for two C-C bonds, and $\ce{2\times sp^2}$ for two C-H bonds. – Martin - マーチン Feb 16 '18 at 9:03

I think the simplest way to go at this is to start with the thought that we have no clue about the hybridization of the orbitals used in the C-C bonds in cyclopropane. On the other hand, the H-C-H on any of the carbons is 'opened' up, to almost $\pu{120^{\circ}}$ (it's not quite 120, but we're going with the big picture here). So, gross assumption here - those C-H bonds are each using $\mathbf{sp^2}$ hybrid orbitals (or 33% s, 67% p, twice). We still have overall one s and 3 p orbitals to work with at each carbon, so to get carried away a bit,

$$\pu{3.00 p - 0.67 p - 0.67 p = 1.66 p is left.}$$

For the amount of s remaining,

$$\pu{1.00 s - 0.33 s - 0.33 s = 0.33 s is left.}$$

Since we have two bonds (the (C-C)'s) left over to describe, divide what's left by two; that's

$$\pu{0.83 p (or 83\%) and 0.17 s (or 17\%).}$$

That's 5 times as much p as s (and like Answer 1 reads), $\mathbf{sp^5}$. And yes, $\mathrm{sp^5}$ corresponds to about $\pu{101.6^{\circ}}$ angles between the orbitals, so cyclopropane in this model still has bent bonds.