organic chemistry - 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: $\frac{a}{1+a}$


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  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:

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