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This question already has an answer here:

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:

$$ \begin{align} \ce{CHCl3 &<=> H+ + ^-CCl3}\\ \ce{CHF3 &<=> H+ + ^-CF3} \end{align} $$

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?

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marked as duplicate by Jan organic-chemistry Aug 24 at 17:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Related: How to prove that chloroform is more acidic than fluoroform? $\endgroup$ – Loong Aug 24 at 13:22
  • $\begingroup$ @Loong, Thanks for sharing the link. The question is similar to that of mine but the accepted answer's author is not confident about the answer. But his reasoning is understandable. I thought of starting a bounty for that question but that is marked duplicate addressing a completely different question. Anyway thanks for spending time in helping me. Hoping for a better answer. $\endgroup$ – Intellex Aug 24 at 13:27
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    $\begingroup$ Also related: Which molecule is more acidic: CHF3 or CHBr3? I think that is more believable than some ideas of d-orbital involvement (which has fallen out of favour over the last few decades). But I also wonder what has been written in the primary literature. $\endgroup$ – orthocresol Aug 24 at 14:10
  • $\begingroup$ @orthocresol, Thank you for the link. Initially, I thought that was a different question but it cleared my doubts. $\endgroup$ – Intellex Aug 24 at 14:32
  • $\begingroup$ Since OP mentioned the question answered theirs, I have no regrets dupehammering. $\endgroup$ – Jan Aug 24 at 17:20
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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 ($\ce{CHF3}$).

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