Why is trichloromethane more acidic than trifluoromethane? In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 78 under the topic "Back Bonding" it is given:


  $\ce{CHCl3}$ is more acidic than $\ce{CHF3}$ which is explained as follows:
  
  $$
\begin{align}
\ce{CHCl3 &amp;&lt;=&gt; H+ + ^-CCl3}\\
\ce{CHF3 &amp;&lt;=&gt; H+ + ^-CF3}
\end{align}
$$
  
  The lone pair on the $\ce{C}$ atom gets delocalised through 2pπ–3dπ bonding in $\ce{^-CCl3}$ which is not possible in the case of $\ce{^-CF3}.$


My doubt is, fluorine being the most electronegative element exerts a greater negative inductive effect (compared to that of chlorine) which stabilises the conjugate base formed. So I feel trifluoromethane must be more acidic than trichloromethane. But the reverse order is observed. 

Please explain why is this so. Should we want to prefer the conclusion from back bonding over inductive effect?
We know that fluorine ($\ce{F}$) is more electronegative than chlorine ($\ce{Cl}$). Therefore, -I effect (Inductive Effect) of $\ce{F}$ would be more and trifluoromethane ($\ce{CHF3}$) would be more acidic.

But this is not the case. As -M (Mesomeric Effect) effect also plays an important role in determining the acidity. We also know that -M is considered over -I (Inductive Effect).

Now, If we compare the conjugate base of both the given molecules, $\ce{Cl3C-}$ and $\ce{F3C-}$, Chlorine has vacant $\mathrm{3d}$ orbitals whereas $\ce{F}$ does not have any vacant $\mathrm{d}$ orbitals (in fact, it does not even have $\mathrm{d}$-orbitals). Therefore, $\ce{Cl}$ can exert -M effect due to availability of vacant $\mathrm{d}$ orbital but $\ce{F}$ cannot exert the -M effect. Generally, -M effect stabilises the negative charge and increases acidic strength. Therefore, trichloromethane ($\ce{CHCl3}$) is more acidic than trifluoromethane ($\ce{CHF3}$).