Q.Why does boron trifluoride behave as a Lewis acid?

Group 13 trivalent compounds such as $BF_3$ and $BCl_3$ have only six electrons around the central atom — two short of a stable octet — making them "electron-deficient" and therefore strong Lewis acids: they readily accept an electron pair from a donor (a Lewis base) to complete their octet. $BF_3$ reacts eagerly with ammonia, forming the adduct $F_3B \leftarrow NH_3$, converting boron from planar $sp^2$ geometry to tetrahedral $sp^3$ geometry in the adduct. This electron deficiency also explains the differing hydrolysis behaviour of $BCl_3$ versus $CCl_4$: $BCl_3$ hydrolyses instantly in water (boron's empty orbital readily accepts a lone pair from water's oxygen), while $CCl_4$ resists hydrolysis entirely, because carbon has no accessible d orbital or empty low-energy orbital to accept the incoming lone pair, and its coordination number is already saturated at 4 with no room to expand. Similarly, $SiCl_4$, unlike $CCl_4$, DOES hydrolyse, because silicon's 3d orbitals give it somewhere to accept the donated lone pair before losing HCl stepwise. Electron deficiency also accounts for observed geometry/bond-length effects: in $BF_3$, partial $\pi$-donation from fluorine's lone pairs into boron's empty p orbital shortens and strengthens the B–F bond (130 pm) relative to $BF_4^-$ (143 pm), where boron is already a full-octet, $sp^3$-hybridised, non-electron-deficient tetrahedral ion with no such back-donation. $BCl_3$'s zero net dipole moment, despite each individual B–Cl bond being polar, follows from the planar, symmetric $sp^2$ (trigonal) geometry forced by boron's electron deficiency, in which the three bond-dipole vectors cancel exactly.