Key idea: Simple electronegativity arguments predict $\ce{BF3}$ should be the stronger Lewis acid (since F withdraws more electron density from B, leaving it more electron-poor). Experimentally, however, the opposite trend is observed: $\ce{BCl3} > \ce{BF3}$ in Lewis acidity toward most bases.

Explanation via π-donation: Boron in $\ce{BX3}$ is $sp^2$ hybridized with an empty $p$ orbital perpendicular to the molecular plane. The halogen lone pairs can donate into this empty orbital, forming a partial $\pi$ bond that reduces the electron deficiency at boron.

• Fluorine's $2p$ lone pairs match well in size and energy with boron's empty $2p$ orbital, giving strong π-donation from F to B. This significantly quenches the Lewis acidity of boron in $\ce{BF3}$.
• Chlorine's $3p$ lone pairs are larger and more diffuse, giving poorer orbital overlap with boron's $2p$ orbital, so π-donation is weaker in $\ce{BCl3}$.

Because $\ce{BF3}$ enjoys more effective π back-donation, its boron center is less electron-deficient than expected from electronegativity alone, making $\ce{BF3}$ the weaker Lewis acid of the two — despite F being more electronegative. Thus the correct explanation is the π-donation effect, not raw electronegativity.

Answer: (B)