Stability of Carbanions For comparing stability of carbanion, we use the fact that &quot;negative charge is stable on more electronegative atom.&quot; So, The stability order is, {electronegative order: $\mathrm{sp \gt sp^2 \gt  sp^3}$} and therefore, $\ce{HC#C^- \gt H2C=C^-H \gt CH3-C^-H2}$.
My doubt is that, we generally say, carbanion is stable when there is less electron density on $\ce{C^-}$.
In $\ce{HC#C^-}$, $\ce{C}$ is $\mathrm{sp}$ hybridized that means it is more electronegative, so it will pull electrons more as compared to other two. So, electron density order will be reverse of above order. And, $\ce{HC#C^-}$ will be least stable.
By, my thought, answer is completely opposite.
Please help me to get out of this doubt and please correct me if I said something wrong.
$\mathrm{sp}$ hybridized carbanions are more stable as the lone pair of electrons is placed in an orbital consisting of high s-character. In general, electrons are more stable in s orbitals rather than the p orbitals of the same shell. This can be understood in a simple way, by thinking that the s orbitals are closer to the nucleus, and can also be verified by the n+l rule.
Now, when we say $\mathrm {sp}$ hybridized species are more electronegative than $sp^2$ hybridized species, we mean that the $\mathrm{sp}$ orbital are more accomplished in holding an electron within them when compared to their $\mathrm{sp}^2$ counterparts. As electrons are more stable in an $\mathrm{sp}$ orbital, electrons are pulled more towards them rather than the $\mathrm{sp}^2$ orbitals, which makes us say that $\mathrm{sp}$ orbitals are more electronegative than $\mathrm{sp}^2$ orbitals.