Hybridization of I in IF5 is?
The Correct Answer and Explanation is :
The hybridization of iodine (I) in IF₅ is sp³d².
Explanation:
Step 1: Valence Electrons
Iodine is in group 17 of the periodic table and has 7 valence electrons. Fluorine, also in group 17, has 7 valence electrons but forms a single bond with iodine, using 1 electron from each atom.
Since there are 5 fluorine atoms in IF₅, each forms a single bond with iodine, accounting for 5 of iodine’s 7 valence electrons.
Step 2: Lone Pairs
After forming 5 bonds with fluorine, iodine has 2 electrons (1 pair) left. These will remain as a lone pair on iodine.
Step 3: Electron Geometry
To determine the hybridization, we use the concept of electron domains around the iodine atom. In IF₅, iodine has:
- 5 bond pairs (due to the 5 bonds with fluorine atoms),
- 1 lone pair (as discussed).
That gives a total of 6 electron domains. For an atom with 6 electron domains, the corresponding hybridization is sp³d².
Step 4: Shape and Bond Angles
- The molecular geometry is square pyramidal, as derived from the octahedral electron geometry but with one position occupied by a lone pair.
- In this arrangement, the lone pair will take an equatorial position, minimizing electron-electron repulsion with the bonding pairs.
- The bond angles around the iodine atom are slightly less than 90° due to the lone pair’s repulsion.
Orbitals in sp³d² Hybridization
In sp³d² hybridization, the orbitals involved are:
- 1 s orbital,
- 3 p orbitals,
- 2 d orbitals.
These combine to form 6 hybrid orbitals, accommodating 5 fluorine atoms and 1 lone pair on iodine.
Therefore, iodine in IF₅ is sp³d² hybridized, forming a square pyramidal structure.