Question

Explain hybridization of the central atom in: $I{F_3}$

Answer 1

Hint: In chemistry, orbital hybridisation (or hybridization) is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds in valence bond theory.

Complete Step by step answer: The hybridization of ​ $I{F_3}$ can be determined by the number of lone pairs around iodine and the number of sigma bonds formed between I and F. Since the iodine has 7 valence electrons out of which 3 electrons form sigma bond with F atoms and left with forming 2 lone pair. Hence this shows that the steric number is 5.
If you draw out the Lewis dot structure you should know that fluorine can only have 8 electrons due to the octet rule. Iodine can have more electrons because it's past the 3rd period and has empty spots available in its d orbitals.
Each one of those atoms has 7 valence electrons so you should have 28 electrons in your lewis dot structure. Since 3 of the electrons on the iodine atom need to be shared with the flourishes, you only have 4 electrons left which form 2 lone pairs.
You have 2 lone pairs and 3 bonds which means you need a hybrid orbital that can house 10 electrons at the same energy level. That results in the $s{p^3}d$ orbital mentioned at the beginning.
Therefore the hybridization is $s{p^3}d$ and the geometry is trigonal planar and the shape is T-shaped where the 2 F atoms are axial and one is at equatorial position.

Note: The T-shaped molecular structure of unstable IF3 has been characterized for the first time by X-ray analysis. In the solid state the iodine atom is pentagonal-planar coordinated, as a result of two weak bonds bridging through fluorine atoms of neighboring molecules.