Identify the bonding in both molecules

• In $BF_3$, boron is $sp^2$ hybridised and forms three $\sigma$ bonds with fluorine.
• Boron has only 6 electrons around it in $BF_3$, so it is electron-deficient.
• Each fluorine has lone pairs, and one of these lone pairs can overlap with the empty unhybridised $2p$ orbital on boron.

Hence, in $BF_3$ there is back bonding: $F \rightarrow B$ involving $p\pi - p\pi$ overlap.

Effect of back bonding in $BF_3$

This $p\pi - p\pi$ back donation gives partial double-bond character to the $B-F$ bonds. Therefore:

• bond length decreases,
• bond strength increases,
• bond dissociation energy becomes higher.

So the unusually high bond dissociation energy of $B-F$ in $BF_3$ is due to this extra $\pi$-bonding contribution.

Now consider $CF_4$

• In $CF_4$, carbon is $sp^3$ hybridised and already has a complete octet.
• There is no vacant low-energy $2p$ orbital on carbon available to accept lone-pair donation from fluorine.
• Therefore, effective $p\pi - p\pi$ back bonding like that in $BF_3$ is not possible.

Thus, the $C-F$ bonds in $CF_4$ are essentially only strong $\sigma$ bonds, without the additional strengthening due to back bonding.

Evaluate the options

• A: Incorrect. The main reason is not a stronger $\sigma$ bond in $BF_3$; rather it is the extra $p\pi-p\pi$ interaction.
• B: Correct. $BF_3$ has significant $p\pi-p\pi$ interaction, while $CF_4$ does not.
• C: Incorrect. This states the opposite of the true situation.
• D: Incorrect. Boron is not smaller than carbon in a way that explains this trend, and size is not the key reason here.

Final answer

$\boxed{\text{B}}$

Comparison with stored correct answer

Stored correct answer = B.

My derived answer also = B. Hence, they agree.