Question

According to the VSEPR theory, which shape is possible for a molecule with the molecular formula of $ A{B_3} $ (Where the number of total electron groups is unstated)?

Answer 1

Hint: Depending upon the number of atoms that are involved in bonding between a central metal atom and other atoms the hybridization will be based. The molecule with the molecular formula of $ A{B_3} $ has trigonal planar, trigonal pyramidal, and T-shape.

Complete Step By Step Answer:
VSEPR theory stands for Valence shell electron pair repulsion theory which determines the hybridization based on the valence electrons of the central metal atom and bonding between central metal atom and other atoms.
The molecular formula of $ A{B_3} $ will have the molecular formulas of $ B{F_3},N{H_3},Cl{F_3} $
In $ B{F_3} $ , Boron is the central metal atom and is involved in $ s{p^2} $ hybridization. The shape of $ B{F_3} $ is trigonal planar.
In $ N{H_3} $ , Nitrogen is the central metal atom and is involved in $ s{p^3} $ hybridization. The shape of $ N{H_3} $ is tetrahedral and nitrogen consisting of one lone pair of electrons.
In $ Cl{F_3} $ , Chlorine is the central metal atom and is involved in $ s{p^3}d $ hybridization. The shape of $ Cl{F_3} $ is T-shape and chlorine consisting of two lone pairs of electrons.
Thus, according to the VSEPR theory trigonal planar, tetrahedral, and T-shape were possible for a molecule with the molecular formula of $ A{B_3} $ .

Note:
The number of valence electrons of the central metal atom and the electrons that were involved in bonding and lone pair of electrons decide the hybridization. $ A{B_3} $ molecule has the central metal atom of A, and there are three atoms of B. Though the number of atoms is the same, the lone pair of electrons decides the hybridization and shape of the molecule.