Shown here, is the molecule BF3 represented in three different ways. The ball-and-stick model is drawn on the lower left. Boron is shown in purple and the fluorine atoms are shown in green. On the right, different views of the empty 2p-orbital belonging to boron are shown. With three bonds to fluorine (sp2 hybridization), and no lone pairs, there remains one 2p-orbital that is not hybridized and empty. Thus, it wants a lone pair of electrons to give it an octet of electrons. Also, fluorine is highly electronegative, withdrawing electron density from the boron atom. This is represented in the electrostatic potential model at the upper-left, with flourine atoms in red (partial negative charge) and boron in blue (partial positive charge). For all these reasons, the molecule BF3 is a good acceptor of electrons and therefore a good Lewis acid.
In this picture, an acid/base reaction is shown. NH3 has a lone pair of electrons. This is represented by the red area on the electrostatic potential model. The H+ is obviously positively charged as shown by its blue color. The lone pair of electrons on the NH3 molecule will donate these pairs of electrons to the H+, so here the NH3 is acting as a Lewis base as well as a Bronsted-Lowry base, and the H+ is acting a a Lewis acid and well as a Bronsted-Lowry acid. Notice that by convention, the arrow points from the electron donor to the electron acceptor. The bottom picture shows H+ bound to the NH3 molecule forming NH4+. Please note that the molecule is now sp3 hybridized and has a formal charge of +1.
In this diagram, NH3 again acts as a Lewis base. BF3 acts as a Lewis acid when it accepts the lone pair of electrons that NH3 donates. This reaction fills BF3's empty 2p-orbital, and now boron is sp3 hybridized when previously (as BF3) it was sp2 hybridized.