Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct shape and $\text{I}-\text{I}-\text{I}$ bond angles respectively in ${\text{I}}_3^{-}$, ion are

Decoding the shape of a polyatomic ion can feel like solving a microscopic puzzle. But with the right tools—specifically VSEPR theory—it becomes a logical and highly satisfying process. Let's embark on a journey to uncover the exact 3D geometry of the triiodide ion, ${\text{I}}_3^{-}$.

Every molecular geometry problem begins with identifying the central atom. In the case of ${\text{I}}_3^{-}$, one iodine atom sits in the center, flanked by two others.

To figure out its 3D shape, we need to use VSEPR theory (Valence Shell Electron Pair Repulsion). The master key here is the Steric Number, which is simply the sum of the bond pairs and lone pairs around the central atom.

Let's count the electrons. Our central iodine belongs to Group 17 of the periodic table, giving it $7$ valence electrons.

But wait, there's a negative charge on the ion! That negative sign means the system has gained one extra electron. Adding this to our initial count, we have a total of $8$ valence electrons on the central iodine.

Now, this central iodine is attached to two other iodine atoms. Each single bond consumes one electron from the central atom. Therefore, we clearly have $2$ bond pairs.

We started with $8$ valence electrons and used $2$ for bonding. That leaves us with $6$ non-bonding electrons. Since electrons pair up, we divide by $2$, giving us exactly $3$ lone pairs.

Let's calculate the steric number.

A steric number of $5$ means the central atom requires five hybrid orbitals, which corresponds to $s{p}^{3}d$ hybridization.

For $s{p}^{3}d$ hybridization, the foundational electron geometry is trigonal bipyramidal. Imagine an equator with three positions spaced at $120^{\circ }$, and an axis with two positions pointing straight up and down.

This is where mistakes often happen. Where do we place the $3$ lone pairs?

According to Bent's rule and VSEPR theory, lone pairs demand more spatial volume than bonding pairs. To minimize electrostatic repulsion, they must occupy the equatorial positions.

By placing all three lone pairs on the equator, the two bonded iodine atoms are forced into the axial positions. This creates a perfectly linear shape.

Since the three iodine atoms are aligned in a perfectly straight line, the $\text{I}-\text{I}-\text{I}$ bond angle is exactly $180^{\circ }$.

And that is our final answer! The elegance of molecular geometry lies in how perfectly these repulsive forces balance out to create stable, symmetrical structures.