= Solution
The expected order is \b[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.
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