Question

Explain hybridization of central atom in: $Br{F_5}$

Answer 1

For solving this question first we need to understand the concepts of hybridization. Hybridization which is also known as orbital hybridization is the concept of mixing atomic orbitals with new hybrid orbitals.

Complete step by step answer:

The concept of hybridization is based on the difference in energies, shapes, and sizes when compared with the component of atomic orbitals. Thus, this concept is suitable for the pairing of electrons to form chemical bonds according to the valence bond theory.
The different types of hybridization are $s{p^3},{\text{ }}s{p^2},{\text{ }}sp,\,{\text{ }}ds{p^2},{\text{ }}s{p^3}d,{\text{ }}s{p^3}{d^2},{\text{ }}s{p^3}{d^3}$
Now, in the case of $Br{F_5}$, hybridization can be determined by the number of lone pairs around iodine and the number of sigma bonds formed between Br and F. For determining the hybridization of $Br{F_5}$ first we need to understand that when 1s, 3p, 2d-orbitals of same element mix and recast to form hybrid orbitals of same energy with an equal orientation in space is known as $s{p^3}{d^2}$ hybridization.
Now, according to the question, we know that the central atom iodine has 7 valence electrons out of which there are 5 electrons that form 5 sigma bonds with F atoms. Also, there are 2 electrons that form 1 lone pair made from the static number 6. Thus, it further implies that the hybridization of the central atom possesses $s{p^3}{d^2}$ hybridization having one 4s, three 4p and two 4-d orbitals. Therefore, $Br{F_5}$ has a geometrical shape of an octahedral and the shape is square pyramidal.

Note:
When we refer to hybridization Bent rule is an important concept. Bent rule is defined and explained as the relationship between the orbital hybridization of central atoms in the molecules and the electronegativity of substituent.