# How to prove that chloroform is more acidic than fluoroform? [duplicate]

How to prove that chloroform is more acidic than fluoroform?

Using the concept that the inductive effect is dominant over the resonance (mesomeric effect) in the case of halide substituents: $\require{mediawiki-texvc}$

1. Since both have single hydrogen, let's compare stability of carbanions $\ce{CF3-}$ and $\ce{CCl3-}$.
2. Since fluorine shows more negative inductive effect than chlorine, $\ce{CF3-}$ should have been more stable than $\ce{CCl3-}$, so some greater cause must be acting here.
3. In $\ce{CCl3-}$ carbon has non bonded electrons in 2p orbital and chlorine has an empty 3d orbital, but since the energy difference is very high, $\ce{p\pi-d\pi}$ backbonding is barely possible. Worst case scenario, even if it's possible, then giving preference to backbonding will violate our initial rule (that inductive > mesomeric).
4. Bond lengths, $\ce{C-Cl}=1.75\AA, \ce{C-F}=1.35\AA$, but since chlorine is bigger than fluorine, they both will have approximately same electronic $\ce{p-\pi}$ repulsion (of lone pair of halogen and at anionic centre).

Which of my reasoning are wrong? How do I prove the fact that chloroform is more acidic than fluoroform?

Data: $\mathrm pK_\mathrm a (\ce{CHBr3})=13.7$ and $\mathrm pK_\mathrm a (\ce{CHCl3})=15.5$ and $\mathrm pK_\mathrm a (\ce{CHF3})=28$ (source - page 2)

(1) $\ce{CF3-}$ is highly unstable due to huge repulsions between the negative charge on carbon and the lone pair of fluorine.
(2)While in $\ce{CCl3-}$ chlorine has an empty d orbital thus the negative will be more dispersed into the this empty orbital hence the repulsions will be quite less as compared to $\ce{CF3-}$.
• I would $+1$ a self-answer, but I cannot $+1$ anything that assumes the usefulness of d-orbitals on chlorine. – Jan Dec 12 '15 at 23:47