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For comparing stability of carbanion, we use the fact that "negative charge is stable on more electronegative atom." 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.

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  • $\begingroup$ Read this. $\endgroup$ – Mathew Mahindaratne Mar 22 at 7:51
  • $\begingroup$ I don't understand your doubt. In term of Electronegativity you have to interpret it as the CH is pulling from the C(-). $\endgroup$ – Alchimista Mar 22 at 10:24
  • $\begingroup$ s-Orbitals are lower in energy than p-orbitals. The same reasoning applies to hybrid orbitals. This may help. chemistry.stackexchange.com/questions/124896/… $\endgroup$ – user55119 Apr 20 at 15:17
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$\mathrm{sp}$ 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.

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  • $\begingroup$ Can the downvoter take a moment to describe what is wrong with this answer? I am always open for feedback. :) $\endgroup$ – Nisarg Bhavsar May 7 at 4:20
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The stability works on the following principle:

More spread out the electron density, more stable is the compound.

As you can see, sp hybridized carbon atom pulls the electron density to a larger extent, thereby, effectively spreading out the charge.

Can you also think of why this is the reverse for cations?

In the case of carbocations, the sp hybridized carbon atom pulls electron density causing even more positive density on the carbocation.

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    $\begingroup$ Would the downvoter care to explain the reason? $\endgroup$ – DatBoi Mar 22 at 6:59
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    $\begingroup$ Yeah, it seems like someone has mass downvoted all answers to this question. My answer has also been downvoted without stating a reason. $\endgroup$ – Nisarg Bhavsar Mar 22 at 8:08
  • $\begingroup$ @DatBoi and @ Nisarg Bhavsar. Thanks for your reply. But ☹️ I think you didn't understand my doubt. My doubt is that, In HC---C- , C- is sp hybridized so it is more electronegative. So due to high electron pulling capacity C - will have more electron density . Since , more electron density on Carbanion decreases it's stability . So, HC ---C- will be least stable. This is my thought. 😢please help. 😭I am getting totally opposite stability order $\endgroup$ – The Explorer Mar 22 at 17:04
  • $\begingroup$ @The Explorer You are thinking about it in the opposite sense. Assume if we had acetylene and ethene. Which of them would be more reactive towards a base?? Obviously acetylene because $\ce{#}$ is more electronegative than a $\ce{=}$. What you need to understand is that the $\ce{#}$ will pull the electrons from the carbon and not the other way around. $\endgroup$ – Nisarg Bhavsar Mar 22 at 17:18
  • $\begingroup$ @TheExplorer Stop! I suggest you rather ask yourself, what does it even mean for a hybridised orbital to pull away electron density.? The answer you will get will be that in a sp orbital, electron density is closer to the nucleus and thus the least instability. $\endgroup$ – Rishi May 26 at 13:31

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