What is the hybridization of the central iodine atom in I3 -?
The Correct Answer and Explanation is:
The hybridization of the central iodine atom in the triiodide ion (I₃⁻) is sp³d.
In the triiodide ion, I₃⁻, three iodine atoms are bonded together in a linear arrangement. To determine the hybridization of the central iodine atom, consider the number of electron domains (regions of electron density) around it. The central iodine forms two sigma bonds with the terminal iodine atoms and possesses three lone pairs of electrons. This results in a total of five electron domains.
According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, five electron domains correspond to a trigonal bipyramidal electron geometry. The hybrid orbitals that accommodate this geometry involve one s orbital, three p orbitals, and one d orbital, giving rise to sp³d hybridization.
Despite the trigonal bipyramidal electron geometry, the molecular shape is linear. This occurs because the three lone pairs occupy the equatorial positions in the trigonal bipyramidal geometry to minimize electron repulsion, while the two bonding pairs occupy the axial positions. As a result, the observable shape of the ion is linear, consistent with a bond angle of approximately 180°.
The central iodine atom thus uses sp³d hybrid orbitals to accommodate the five regions of electron density. The two hybrid orbitals in the axial positions form sigma bonds with the terminal iodine atoms, while the other three hybrid orbitals accommodate the lone pairs.
This hybridization allows for the correct geometry, bonding, and electron distribution as observed in the I₃⁻ ion. Additionally, I₃⁻ is often encountered in solutions containing iodide (I⁻) and iodine (I₂), where the linear triiodide ion is formed as a result of the equilibrium:
I₂ + I⁻ ⇌ I₃⁻
The sp³d hybridization of the central iodine is critical for this linear bonding and the overall stability of the ion.
