Question

What is the shape of \[B{F_4}^ - \] ?
A.Tetrahedral
B.Pyramidal
C.Trigonal Planar
D.Bent

Answer 1

Hint: The shape of a molecule can be determined using VSEPR Theory from the number of valence electrons on the central atom.

Complete step by step answer:
Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) can be used to determine the shape of molecules. Shape of a molecule is determined in VSEPR Theory by counting the valence electrons on the central atom. The central atom is usually the least electronegative atom in the molecule.
First let us find the central atom on \[B{F_4}^ - \] . Among Boron and Fluorine, Boron is least electronegative. Hence Boron is the central atom in \[B{F_4}^ - \] .
Secondly, count the number of valence electrons on Boron. Boron is a Group $13$ element. Hence the number of valence electrons on Boron is three. Since there is a negative charge present on the molecule, we count this electron also. Hence the total number of electrons available for bonding is four. We have four fluorine atoms in the molecule. Each fluorine atom contains an unpaired electron on one of the p-orbitals. Hence in order to satisfy all the valence electrons we need four orbitals. Therefore according to VSEPR Theory, the four orbitals on Boron hybridise to give four identical $s{p^3}$ hybrid orbitals. Each $s{p^3}$ orbital contains one electron. The p-orbital on Fluorine which contains one electron will combine with the $s{p^3}$ hybridised orbital of Boron to form a $\sigma $ bond, giving a total of four $\sigma $ bonds.
$s{p^3}$ hybridisation results in a tetrahedral geometry. Since all the valence electrons on Boron participate in bonding, there will be no lone-pair electrons on Boron. Hence the shape of \[B{F_4}^ - \] will be tetrahedral.
Hence option A is correct.

Note:
The $s{p^3}$ hybridised orbitals adapts tetrahedral shape in order to minimize repulsion. The geometry with minimum repulsion for $s{p^3}$ orbital is tetrahedral. Shape will be different if lone-pair electrons are present.