Question

How many atom(s) of $B{F_2}^{-}$ lie in the same plane?

Answer 1

Hint: The concept of intermixing of different orbitals of atoms and then redistribution of energy to given equivalent energy of new orbitals, identical shape, and symmetric orientation in space is known as hybridization. As a result of hybridization, new orbitals are named hybrid orbitals, and bonds formed between hybrid orbitals are hybrid bonds.

Complete step by step solution:
Hybridization of $B{F_2}^{-}$:
The central atom in $B{F_2}^{-}$ is the boron atom which is involved in the hybridization.
The electron configuration of boron = $1s^{2}2s^{2}2p^1$ where boron atom with three outermost electrons in the ground state. One born electron is unpaired in p-orbital and one lone pair in s-orbital in the ground state.
During the formation of the compound$B{F_2}^{-}$, one s-orbital and two $2p$ orbitals hybridize overlapped with these hybridized $s{p}^2$ orbitals then bonds are formed. According to the $s{p}^2$ hybridization, the structure of $B{F_2}^{-}$ will is a trigonal planar. But due to lone pair-bond pair repulsions, then the structure distorted to bent shape.
Based on the valence bond electron pair repulsion theory, the shape of $B{F_2}^{-}$ is the bent shape and the F-B-F bond angle is ${120}^o$, which means all three atoms $B{F_2}^{-}$ are in the same plane.

Note: The characteristics of hybridization are the number of hybrid orbitals formed is equal to the number of atomic orbitals that are undergoing hybridization and these hybrid orbitals are equivalent in energy with shape. The stability of hybrid orbitals more than atomic orbitals and these hybrid orbitals have different orientations in space.