I am learning about how to calculate s-orbital character. While doing so, I came across a formula
$$\cos\theta = \frac{s}{s-1}.$$
For cycloproprane, I used $\theta = 60^\circ,$ and got $s = -1.$
Question which I want to solve
Find out the stability of the following carbanion:
I found Does negative sign of s-orbital exist?, but failed to understand it. If that answers my questions, please tell.
The formula only works when the orbital is directed along the bond line between two atoms. Since no orbitals can give you a $60^o$ bond angle, the orbitals are directed outside the triangle, forming a “banana” bond and the formula is being used inappropriately. So the negative sign in your Question #1 is not acceptable. We must do something different.
Let’s look at the H-C-H bond angle. The H-C-H bond angle in cyclopropane is $115^o$ Ref 1. Using that angle, and assuming that the orbitals are directed exactly between the atoms, s = 29.7% for the C-H bond, which is broken to form the carbanion. The acidity of cyclopropane is also indicated by its pKa: 46.
Cyclopropane is interesting because you might think, starting with atomic orbitals, that such a structure would not form, since the smallest angle you could imagine for a carbon atom would be $90^o$; how could you possibly get $60^o$?
Well, that's where your imagination comes into play. Don't go to negative s-bonds (that's too imaginative). Consider how the simple (i.e., positive) addition of individual atomic orbitals could lead to something like the bonding that you do observe:
With simple $sp^3$ orbitals, you can point one pair from each carbon just outside the ring (there will still be considerable overlap: Figure b) and each carbon will be attached to 2 hydrogens above and below the ring with the other two $sp^3$ orbitals. These ring bonds are called banana bonds - not directly in line with the atoms, but bonding nevertheless. Figure a) shows a more complicated bond scheme where the carbon is hybridized $sp^2p$ with the $p$ obital in the plane of the carbons: you get two bonds and one antibond (the whites overlap, the reds overlap, and the red-white antibonds). But then you get three $sp^2$ orbitals poking right into the ring to form a 3-atom bond - again, the bonding is not in line with the atoms. (The hydrogens are bound with $sp^2$ bonds, above and below the ring. You might use the equation to determine the s and p character of these C-H bonds if you know the H-C-H angle (Ref 1).
A very interesting (11 slides) presentation about the bonding is made at Ref 2.
Ref 1: https://cyclopropanelasalle.weebly.com/structure.html
But the upshot is that all (most?) equations have limited ranges of usefulness. Whenever some experimenter makes something that goes beyond, a new equation is needed.
Cyclohexane, at the other end of your question, has a pKa of 52 (very non-acidic!), and a probable H-C-H bond angle of $109.5^o$, and s-character in the C-H bond of 25% (note that the molecule has a chair structure and is not planar).
Other pKa’s of interest are:
acetylene, pKa =15, C-H bond is $sp$, 50% s
ethylene, pKa = 44, C-H bond is $sp^2$, 33% s
methane, pKa = 48, C-H bond is $sp^3$, 25% s
(interesting that cyclohexane is similarly hybridized but even less acidic)
Perhaps the best answer to the stability question is that the smaller rings are more acidic, i.e., have more stable anions because the acidic proton has more s-character in the more strained ring.
Question #2 discussed sign in your link. “Sign” (+/-) in the equation for s-character is a measure of amount: how much s-character, in %, is in the bond. But when describing orbitals, we use + and - as indicators of the polarity of the wavefunction, not the amount of s or p character. Plusses combine in such a way that electrons flow from one orbital to another without hindrance, in a bond. A - polarity next to a + requires a node: no electrons there at the node, therefore no bonding. So s-orbitals are designated with one sign (either + or -, depending on whether it is bonding with an adjacent atom; higher levels have internal nodes, but that is no concern for us here). p-Orbitals have a node at the nucleus which is indicated by calling one lobe + and the other lobe -. Bonding molecular orbitals have the same sign of the wavefunction adjacent; in the picture, red-red and blue-blue (in (c)) are bonding; red-blue (in (a) and (b)) is antibonding because the wavefunction splits between the atoms.
