I am not an expert in chemistry, but these are the few facts I know from Standard XII chemistry books (these facts could be wrong or incomplete or modified by now).

Facts I know -

  • Ethane has 'sp$^3$' hybridisation, Ethene has 'sp$^2$', and Ethyne 'sp'.

  • The % of 's' orbital present follows in each of the hybridisation follows the order sp > sp$^2$ > sp$^3$.

  • Since 's' orbital has greater penetration (to the nucleus) depth, the H atom in Ethyne comes the closest to the Carbon atom nucleus than in Ethene and Ethane, hence C-H bond length of Ethyne is the lowest, followed by Ethene and then Ethane.

  • On the other hand, since 's' orbital has greater penetration depth, 'sp' hybridised orbitals has the most Electronegativity, followed by 'sp$^2$' and then 'sp$^3$'. Because of this, Ethyne is most acidic, since the carbanion (HC $\equiv$ C $^-$) is more stable than (H$_2$C = CH$^-$) and (H$_3$C $-$ CH$_2$$^-$). Ethyne can release $H^+$ more easily than Ethene and Ethane, hence Ethyne is most acidic, followed by Ethene and then Ethane.

  • Following a similar reasoning of the nature of 's' orbital, one finds that the C-H bond energy of Eythne is the highest, followed by Ethene and then Ethane.

Question -

How can Ethyne be the most acidic and yet have the shortest C-H bond length. Isn't there a general rule that "the shorter the bond length, the stronger the bond strength, and so the weaker the acid"? Shouldn't shorter X-H bond lengths (X is some atom) mean that X and H has greater bond strength and so it does not dissociate much into $X^-$ and $H^+$ ions, thereby not producing much $H^+$ ions?

Is there something wrong with the Facts that I stated or is the statement of my Question not true in general?