How specific is the exception of negative charge location (orbital vs atom) for determining stability of conjugate base?

Question

Chapter 3 Acid-base reactions in Klein's Organic Chemistry As a Second Language explains an exception to the typical order of important factors for determining how stable a negative charge on a conjugate base is [1, p. 61]:

This example illustrates that a negative charge on an $\mathrm{sp}$ hybridized carbon atom is more stable than a negative charge on an $\mathrm{sp^3}$ hybridized nitrogen atom. For this reason, $\ce{NH2-}$ can be used as a base to deprotonate a triple bond.

How specific is this exception? Does it only hold true for nitrogen? Or could an oxygen atom be $\mathrm{sp^3}$ or $\mathrm{sp^2}$ hybridized and still be less stable than the structure on the left despite its higher electronegativity?

Reference

1. Klein, D. R. Organic Chemistry As a Second Language: First Semester Topics, 4th ed.; Wiley: Hoboken, 2016. ISBN 978-1-119-11066-8.

Answer 1

The enhanced electronegativity that comes from having carbon with $\mathrm{sp}$-hybridized orbitals is enough to allow terminal acetylenes to protonate nitrogen-anion bases like $\ce{NH2^-}$, but (in the absence of additional enhancements such as conjugation) not an oxyanion base like $\ce{OH^-}$. In effect carbon with $\mathrm{sp}$-hybridized orbitals lies between ordinary nitrogen and oxygen atoms.

We can get the same effect with $\mathrm{sp^3}$-hybridized orbitals on carbon in a cyclopentadiene ring, as with the parent compound $\ce{C5H6}$ which is an even stronger acid than acetylene. The acidity comes from the ring becoming aromatic when it is deprotonated and negatively charged; as a result cyclopentadiene is weakly deprotonated even by highly concentrated sodium hydroxide solutions, as well as being strongly deprotonated by amide ions. When you study aromaticity you should see this effect, as well as the opposite effect of cyclopropenyl compounds acting as Lewis bases to make that ring aromatic.