In IF3 (iodine trifluoride), the iodine atom (I) forms three covalent bonds with fluorine atoms (F). To determine the hybridization of the central iodine atom, we can use the valence shell electron pair repulsion (VSEPR) theory.
In IF3, the iodine atom has five valence electrons (group 7) and each fluorine atom has seven valence electrons (group 17). The total number of valence electrons in IF3 is:
Iodine (I) = 5 valence electrons Fluorine (F) = 3 valence electrons (since there are three fluorine atoms)
Total = 5 + (3 x 7) = 26 valence electrons
To distribute these electrons, we first place a single bond between the iodine atom and each fluorine atom, which accounts for 6 electrons (3 pairs). This leaves us with 20 electrons to distribute.
Next, we place lone pairs on the iodine atom to satisfy the octet rule. Since iodine can accommodate more than eight electrons due to its d-orbitals, it can expand its octet. In IF3, two lone pairs are placed on the iodine atom, accounting for 4 electrons (2 pairs). This leaves us with 16 electrons.
Finally, we distribute the remaining electrons as lone pairs on the fluorine atoms. Each fluorine atom will have one lone pair, accounting for 6 electrons (3 pairs). This exhausts all 16 electrons.
The electron geometry of IF3 is trigonal bipyramidal because there are five electron pairs around the central iodine atom (three bonding pairs and two lone pairs). However, the molecular geometry is T-shaped because the two lone pairs create repulsion, causing the fluorine atoms to push closer together.
Regarding the hybridization of the iodine atom, it undergoes sp3d hybridization. This means that one s orbital, three p orbitals, and one d orbital from the iodine atom participate in hybridization to form five sp3d hybrid orbitals. These hybrid orbitals then overlap with the fluorine's p orbitals to form three sigma bonds in the molecule.