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The bond dissociation energy of B−F in BF3 is 646 kJ mol−1, whereas that of C−F in CF4 is 515 kJ mol−1. The correct reason for higher B−F bond dissociation energy as compared to that of C−F bond is
- stronger σ bond between B and F in BF3 as compared to that between C and F in CF4
- significant pπ−pπ interaction between B and F in BF3 whereas there is no such interaction between C and F in CF4
- lower degree of pπ−pπ interaction between B and F in BF3 than that between C and F in CF4
- smaller size of B−atom as compared to that of C−atom
Correct answer: significant pπ−pπ interaction between B and F in BF3 whereas there is no such interaction between C and F in CF4
Solution
Boron has a vacant 2p orbital, so fluorine lone pairs donate into it giving significant pi(p-p) back-bonding in BF3, which strengthens (shortens) the B-F bond. Carbon in CF4 has no vacant p orbital for such interaction, so C-F is weaker. Hence the correct reason is significant p-pi-p-pi interaction in BF3.
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