Question

What is the hybridization state of B in $B{{F}_{3}}$and $BF_{4}^{-}$?
$B{{F}_{3}}+{{F}^{-}}\to BF_{4}^{-}$
(A)$s{{p}^{2}},s{{p}^{3}}$
(B)$s{{p}^{3}},s{{p}^{3}}$
(C)$s{{p}^{2}},s{{p}^{2}}$
(D)$s{{p}^{3}},s{{p}^{3}}d$

Answer 1

Hint In hybridization atomic orbitals fuse to form new hybridized orbitals. If four bond pairs are present around the central atom and no lone pair is formed then the shape is tetrahedral. For tetrahedral shape, four substituents are located at the corners of the central atom. For tetrahedral hybridization state is $s{{p}^{3}}$ and for planar molecules we have $s{{p}^{2}}$ hybridized state.

Complete Step By Step Solution:
$B{{F}_{3}}$ has an empty 2p orbital and fluorine atom has one lone pair of electrons, so basically for three fluorine atoms there are three lone pairs of electrons in 2p orbital. In $B{{F}_{3}}$ molecule the central bromine atom has $s{{p}^{2}}$ hybridization due to which an equilateral triangle is formed with three fluorine atoms at the corner. This results in Trigonal Planar geometry with bond angle of ${{120}^{o}}$.
In $BF_{4}^{-}$ four bonds are present around the central atom boron i.e. four fluorine atoms are present at the corners of the boron atom which results in the formation of Tetrahedral geometry. There are 3 valence electrons in boron as it has atomic no. 5 and each of the four fluorides contributes one electron to each covalent bond formed and the overall negative charge of the molecule contributes another electron, so on overall we have 3+4+1=8 valence electrons which form 4 electron pairs. So, the geometry is Tetrahedral.

So the correct choice is (A).

Note: Study the concept of molecular geometry properly with all the structures and the changes in the bond angles on the presence of lone pairs. If in tetrahedral geometry one lone pair is present along with the three bond pairs then it changes to trigonal pyramidal. Molecular geometry determines the position of each atom along with the respective bond angles and bond lengths.