Question

Shape of \[BF_4^ - \] is:
A. Tetrahedral
B. Pyramidal
C. Trigonal planar
D. Bent

Answer 1

Hint: The number of outer electrons can be used to determine the shape of a molecule using VSEPR theory.

Complete Step by Step Solution:
The shape of molecules can be determined using the Valance Shell Electron Pair Repulsion (VSEPR) Theory. In VSEPR theory, the shape of a molecule is determined by counting the valence or outermost electrons on the central atom. The centre atom is usually the molecule's least electronegative atom.
Let us start with the centre atom of \[BF_4^ - \]. Boron has a lower electronegativity than fluorine. As a result, the central atom in \[BF_4^ - \] is the boron.
Count how many valence electrons there are in Boron. Boron belongs to the Group \[13\] of the periodic table. Boron has three valence electrons. We count electron of charge as well because the molecule has a negative charge. As a result, there are four electrons accessible for bonding. The molecule has four fluorine atoms. On one of the p-orbitals of each fluorine atom is an unpaired electron. As a result, four orbitals are required to satisfy all of the valence electrons. According to VSEPR theory, the four Boron orbitals hybridise to produce four identical \[s{p^3}\] hybrid orbitals. One electron resides in each \[s{p^3}\]orbital. Fluorine's p-orbital, which contains one electron, will combine with Boron's \[s{p^3}\]hybridised orbital to form a bond, resulting in a total of four bonds.
Tetrahedral geometry is produced by \[s{p^3}\] hybridization. On Boron, there are no lone pairs of electrons since all of the valence electrons are involved in bonding. As a result, \[BF_4^ - \] will have a tetrahedral shape.
As a result, the correct answer is option A tetrahedral.

Note: To reduce repulsion, the \[s{p^3}\]hybridised orbitals adopt a tetrahedral form. Tetrahedral geometry has the least amount of repulsion for the \[s{p^3}\] orbital. If lone-pair electrons are present, the shape will change.