The bond angle $\ce{\angle(F-C-F)}$ of $\ce{CHF_3 > CH_2F_2}$ whereas that of $\ce{\angle(Cl-C-Cl)}$ is $\ce{CH_2Cl_2 > CHCl_3}$ why is this the case? Some explanations I read suggest the Bent's rule but I am not able to understand the idea well.


1 Answer 1


According to NIST CCCBDB, the $\angle(\ce{F-C-F})$ bond angles of $\ce{CH2F2}$ and $\ce{CHF3}$ are ${108.421 ^\circ}$ and $108.099 ^\circ$ respectively as calculated using CCSD=FULL/aug-cc-pVTZ.

Similarly, the $\angle(\ce{Cl-C-Cl})$ bond angles of $\ce{CH2Cl2}$ and $\ce{CHCl3}$ are ${112.329 ^\circ}$ and $109.992 ^\circ$ respectively as calculated using CCSD/aug-cc-pVTZ.

Both show a similar trend for bond angles unlike as mentioned in the question.

Now, this trend can be explained using Bent's rule which states:

Atomic s character concentrates in orbitals directed toward electropositive substituents

As the number of electronegative substituents increases, the p-character in each bonding orbital between $\ce{C-Cl}$ increases. This increase in p character leads to a decrease of the bond angle as observed here.

There is a similar decrease for $\ce{C-F}$ as seen above. The reason for the chloromethanes having a higher bond angle is due to steric effects. Chlorine has an A-value of 0.43 where fluorine has an A-value of 0.15. Therefore, the steric effect is more pronounced in the chloromethanes.

  • $\begingroup$ The book mentions $108^\circ6'$ for CHF3 and $108^\circ3'$ for CH2F2 whereas it is $111^\circ3'$ for CHCL3 and $111^\circ8'$ for CH2CL2 which is just so off that given on the website. Not sure where they might have taken the data from $\endgroup$
    – Ashish
    Jun 9, 2021 at 7:44
  • 1
    $\begingroup$ @Ashish Convert the degrees into decimal or vice versa and you will find that both are nothing but equal. $\endgroup$ Jun 9, 2021 at 11:53

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