Understanding Hybridisation of I₃⁻: Triiodide Ion

Triiodide Ion (I₃⁻) is a type of polyhalogen ion, and even though it has lone pairs, it is a linear molecule. It’s an example of sp³d hybridisation (which decides the molecule’s geometry) in inorganic chemistry.

Let us understand how hybridisation happens in I₃⁻. Read on to learn how it leads to its bonding and molecular shape.

What is the Hybridisation of I₃⁻?

The triiodide ion (I₃⁻) consists of three iodine atoms. The central iodine atom is bonded to two outer iodines through single covalent bonds. In addition to these bonds, the central iodine also carries three lone pairs of electrons.

According to the VSEPR theory, the central atom has five electron domains (2 bond pairs + 3 lone pairs). This leads to sp³d hybridisation.

Using the Hybridisation Formula for I₃⁻

We can determine the hybridisation of the triiodide ion using the simple formula:

formula

For I₃⁻:

Valence electrons of iodine = 7
Number of monovalent atoms (outer iodines) = 2
Negative charge = 1
Positive charge = 0

Substituting:

formula

A hybridisation number of 5 corresponds to sp³d hybridisation.

Breakdown of I₃⁻ Hybridisation

Type: Inorganic polyatomic ion
Electron geometry: Trigonal bipyramidal (from 5 hybrid orbitals)
Molecular shape: Linear (3 lone pairs occupy equatorial positions, pushing the two bonds to opposite axial positions)

Electronic Configuration of Iodine

Atomic number of iodine: 53
Ground state: [Kr] 4d¹⁰ 5s² 5p⁵
In I₃⁻, the central iodine does not need to promote electrons for bonding in the usual sense. This bonding is explained by the formation of an extended 3-center 4-electron bond.
For the purposes of VSEPR/hybridisation description, we model it as sp³d hybridisation (mixing one s orbital, three p orbitals, and one d orbital) to accommodate 5 electron domains.

Formation of Hybrid Orbitals

Orbitals involved: 1s + 3p + 1d → 5 sp³d hybrid orbitals
Arrangement: Trigonal bipyramidal
Distribution:

2 orbitals → σ bonds with the outer iodines
3 orbitals → lone pairs

Bonding in Triiodide Ion

Bond pairs: 2 (I–I) σ bonds
Lone pairs: 3 on the central iodine
Hybridisation type: sp³d
Bond angle: 180°
Geometry: Linear

Molecular Geometry of Triiodide Ion

Details At A Glance

Property	Details
Molecule	Triiodide Ion (I₃⁻)
Hybridisation	sp³d
Geometry	Linear
Bond angle	180°
Bonding	2 σ bonds (I–I), 3 lone pairs
Unhybridised Orbitals	0
Iodine valency satisfied?	Yes, by forming 2 bonds with each iodine atom and having 3 LP

Formal Charge in I₃⁻

To determine if the Lewis structure of I₃⁻ is stable, we calculate the formal charge on each atom using the formula:

Formal charge = Valence electrons − (Lone pair electrons + ½ × Bonding electrons)

Step-by-step for each atom:

Central Iodine (I):

Valence electrons: 7
Lone pairs: 3 (6 electrons, 2 in each pair)
Bonding electrons: 4 (4 electrons from two sigma bonds with I)

Formal charge = 7 − (6 + ½ × 4) = 7 − (6 + 2) = −1

Terminal Iodine (I) – each

Valence electrons: 7
Lone pairs: 3 (6 electrons, 2 in each pair)
Bonding electrons: 2 (2 electrons from one sigma bond with central I)

Formal charge = 7 − (6 + ½ × 2) = 0

Thus, the total charge on I₃⁻ is −1, confirming that it matches its chemical formula.

Summing Up

The central iodine in I₃⁻ bonds with other iodines (2 σ) and also has lone pairs. sp³d hybridisation leads to a trigonal bipyramidal, but due to lone pair repulsion, they occupy the equatorial positions, giving it a linear molecular shape (the bond pairs in axial position).