BF3 molecule is planar but NH3 pyramidal. Explain

B in \(\mathrm{BF_3}\) is sp² hybridised with no lone pair, so \(\mathrm{BF_3}\) is trigonal planar; N in \(\mathrm{NH_3}\) is sp³ hybridised with one lone pair, so \(\mathrm{NH_3}\) is pyramidal.

\(\mathrm{BF_3}\): B is \(1s^{2}2s^{2}2p^{1}\). In the excited state (\(2s^{1}2p_x^{1}2p_y^{1}\)) the 2s and two 2p orbitals mix to form three sp² orbitals in one plane at 120°. Each overlaps with a 2p orbital of an F atom. B has three bond pairs and no lone pair, so all four atoms lie in one plane, with F–B–F angles of 120°.

\(\mathrm{NH_3}\): N is \(1s^{2}2s^{2}2p^{3}\). Its four orbitals form four sp³ hybrid orbitals directed tetrahedrally; three form N–H bonds and one holds a lone pair. The lone pair occupies one corner of the tetrahedron, so the atoms form a pyramid with N at the top. The extra repulsion of the lone pair reduces the H–N–H angle to about 107°.

So the difference comes from the type of hybridisation and the presence of a lone pair on N.