I have read some answers about the $\mathrm{sp^5}$ hybridisation of carbon atoms in the $\ce{C-C}$ bond in cyclopropane as well as the $\sigma$-bond resonance in methylcyclopropyl cation.

Overall, the answers say that the $\ce{H-C}$ bond is due to overlap of $\mathrm{sp^2}$ orbital of carbon and $\mathrm{1s}$ orbital of hydrogen, while the bent bonds of the ring are formed by overlap of $\mathrm{sp^5}$ hybrid orbitals.

But, overall true separation of orbitals in such cases is not defined i.e. we cannot say that some orbital is completely $\mathrm{sp^2}$ hybridised.

Similarly, if the loss of a proton occurs from cyclopropane then the following:

Source: chemthes.com

In such an ion, will the hybridization change and $\mathrm{s}$ and $\mathrm{p}$ characters of the orbitals be redistributed?

My assumptions for this are:

  • In the $\mathrm{sp^2}$ bond the lone pair due to loss of proton will pose increased steric hinderance as it occupies greater volume. This will cause increased repulsion and decreases stability. Moreover, shouldn't the $\mathrm{s}$-character increase to accommodate the increased negative charge?

  • Accordingly, wouldn't the $\mathrm{sp^5}$ bonds will now have greater $\mathrm{p}$-character to ease even greater strain produced?

Then, what will be the change in the hybridisation in the overall molecule on formation of this ion?


  • $\begingroup$ Initially, I thought about an accidental claim of an carbon atom orbital thought to be hybridized $sp^5$ (with a five) instead of $sp^3$ (with a three), because the hybridizations of carbon atom orbitals taught in school are $sp$, $sp^2$, and $sp^3$. Because the body of your question repeats the claim of $sp^5$ more than once, may you either indicate a reference for this unusual concept, or consider to revert (back) to $sp^3$ (because, the cyclopropyl anion still is a cyclopropane)? $\endgroup$ – Buttonwood Aug 1 at 9:10
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    $\begingroup$ I tried to check your claims, it seems that, C-C bond (without anion) came out to be $\ce{sp^{3.5}}$ and the anion ones are $\ce{sp^{2.6}}$ towards, $\ce{sp^{4.06}}$ away and the lone pair was in $\ce{sp^{1.64}}$. so i think there wont be sterric hinderance,due to the s, but p character decreased against your claim $\endgroup$ – user96208 Aug 1 at 9:26
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    $\begingroup$ Michael B. Smith- March's advanced organic chemistry_ reactions and mechanisms and structure-wiley 2020 $\endgroup$ – user96208 Aug 1 at 15:07
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    $\begingroup$ I was reading it, I couldnt find it though....Also, I maybe $5\%$ wrong about aromatic nature...I was just going through the huckel MO, and found that there is a small break in the orbitals plotted at -16.266eV (in the negative sign of wave region) (It might be due to a computational fault) $\endgroup$ – user96208 Aug 1 at 15:50
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    $\begingroup$ I don't even know what you are saying! $\endgroup$ – Zhe Aug 1 at 19:17


Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated. The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$

where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

angle described in Coulson's Theorem

See these earlier answers for interesting examples where the theorem is applied:

The Question

The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem

$$1 + \lambda^2 \cos(114^\circ) = 0$$

where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation

$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$

(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.

Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$ Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.

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  • $\begingroup$ I don't get the leap that s character will be removed from the bonds. I would assume the lone pair to be in a p like orbital, increasing s character in the bonds. Could you clarify that please. $\endgroup$ – Martin - マーチン Aug 3 at 12:45
  • $\begingroup$ @Martin-マーチン Just like in ammonia where the molecule is approximately tetrahedral with the lone pair in an approximately $\ce{sp^3}$ orbital; this being a lower energy geometry than the molecule being planar with the lone pair in a p orbital. $\endgroup$ – ron Aug 3 at 13:03
  • $\begingroup$ Yes, you're right, I somehow got my reasoning the wrong way around. $\endgroup$ – Martin - マーチン Aug 3 at 16:27

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