Question

The geometry and the type of hybrid orbital present about the central atom in \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\] is
A) Linear, \[sp\]
B) Trigonal planar, \[s{p^2}\]
C) Tetrahedral, \[s{p^3}\]
D) Pyramidal, \[s{p^3}\]

Answer 1

Hint: To find out the geometry of \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\]molecule, first we have to find out the hybridization of the boron atom. To calculate geometry, we count the electron groups surrounding the boron atom.

Complete step by step solution:We know, boron atoms have three valence electrons. So, it can form three numbers of covalent bonds. In \[{\rm{B}}{{\rm{F}}_{\rm{3}}}\]molecule, three fluorine atoms are bonded to the boron atom. Therefore, the count of electron groups is 3, the hybridization of the molecule is \[s{p^3}\] and therefore, the molecular geometry is of trigonal planar.
If in a molecule, the central atom is surrounded by two electron groups, then the hybridization of the compound is \[sp\] . And an \[sp\]hybridized molecule is linear in shape.
In an \[s{p^3}\] hybridized molecule, four electron groups surround the centrally placed atom of the molecule. And the shape of the molecule is tetrahedral in nature.

Therefore, option B is right.

Note: If a molecule is \[s{p^3}\]hybridized and has no lone pair, then its geometry and molecular shape is same, that is, tetrahedral. For example, in methane (\[{\rm{C}}{{\rm{H}}_{\rm{4}}}\]), geometry and shape is same, that is, tetrahedral because there is no lone pair. But if the molecule is \[s{p^3}\]hybridized and has three bond pairs and a lone pair , then the molecule geometry is tetrahedral but the molecular shape is pyramidal. For example, ammonia (\[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] ) has tetrahedral geometry and pyramidal shape because of the presence of a lone pair at the nitrogen atom.