# Why is the bond angle of trifluoromethane higher than that of difluoromethane?

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.

• Where did you see this result? Jun 9 at 7:08
• "Wiley Physical Chemistry for JEE" Jun 9 at 7:24
• @Ashish This may help. chemistry.stackexchange.com/questions/152367/… Jun 9 at 7:36

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.

• 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 Jun 9 at 7:44
• @Ashish Convert the degrees into decimal or vice versa and you will find that both are nothing but equal. Jun 9 at 11:53