 Solution

The expected order is B-F > C-F > N-F.

B-F has a large favorable polar/ionic contribution because electron-deficient boron and fluorine have a large electronegativity difference. Fluorine lone-pair donation into boron's empty 2p orbital can additionally strengthen bonding in planar BF${}_3$. C-F is a strong, polar second-row sigma bond, but tetrahedral CF${}_4$ has no equivalent empty carbon 2p acceptor for that pi donation.

N-F has less favorable bond polarity and substantial repulsion between nitrogen and fluorine lone pairs, making it weaker. Invoking occupied boron d orbitals is unnecessary and inappropriate for this second-row comparison.