Cyclopropane has the complying with angles:
$angle ceHCH=118^circ~ extresp.~gg 120^circ$ $angle ceCCC~ extvia bent bonds:~60 + 2 cdot 21 = 102^circ$
The orbitals towards the prolots are $sp^2$ bereason of the $120^circ$ angles. The orbitals towards the carbons originate in the following relation:
$$1 + a cos~alpha = 0$$ ... wbelow $alpha$ is the bond angle and $a$ is the p-amount in sp$^a$ for the orbitals, which consist of the angle.
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This suggests for the orbitals, which expectancy the 102 level angle: $$1 + a cos 102^circ = 0$$ $$1 + a cdot (-0.20) = 0$$ $$a = frac-1-0.20 = 5$$ $$Rightarrow extsp^5 ext-orbitals$$
Test:In a single sp$^a$ orbital, the s-amount is: $frac11+a$, because $1+a$ equates to the sum of all quantities of s and p In a solitary sp$^a$ orbital, the p-amount is: $fraca1+a$
For s:In the orbitals that are oriented in the direction of the proloads, the s-amount is $frac11+2 = frac13$ In the orbitals that are oriented towards the carbons, the s-amount is $frac11+5 = frac16$ Addition of all s-quantities at a single carbon with all four bond orbitals yield: $frac13+frac13+frac16+frac16=1$, which is correct, bereason tbelow is only one single s-orbital at eincredibly carbon atom. For p: In the orbitals that are oriented in the direction of the prolots, the p-amount is $frac21+2 = frac23$ In the orbitals that are oriented towards the carbons, the p-amount is $frac51+5=frac56$ Addition of all p-amounts at a solitary carbon via four bond orbitals yield: $frac23+frac23+frac56+frac56$, which is correct, because tright here are 3 p-orbitals at eincredibly carbon atom.
This implies, that the bent bonds with $21^circ$ from the $ceC-C$-bond are spanned by sp$^5$ orbitals.
So math-magically this appears to make sense, however is there one more explanation that might base more on chemical intuition or "real" chemical concepts?
A quick calculation ($omega$B97X-D/def2-TZVPP) and also a succeeding evaluation of the isosurface of the Laplacian of the electron density, proved at leastern the intended "nonlinear", slightly curved bond in between the carbon atoms.
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Whoever before might want to watch the electron localization feature (ELF), which additionally reflects the bent bonds quite good:
$^ast$ While I tried to translate it to my best, some errors could have been introduced by this . . . please correct me, where I"m wrong.