How to prove that chloroform is more acidic than fluoroform? 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: $\require{mediawiki-texvc}$


Since both have single hydrogen, let's compare stability of carbanions $\ce{CF3-}$ and $\ce{CCl3-}$. 
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.   
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).
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)
My Views (I'm not entirely sure about these reasons)

(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 $\ce{CF3-}$.