1. Identify the central atom and count electron pairs

In $I_3^-$, the central iodine is bonded to two terminal iodine atoms.

Total valence electrons: $3 \times 7 + 1 = 22$

The central iodine forms two bond pairs with the two terminal iodines.

2. Find lone pairs on the central atom

Using the Lewis structure, after completing octets of terminal iodines, the central iodine is left with 3 lone pairs.

So, around the central iodine, there are:

• $2$ bond pairs
• $3$ lone pairs

Thus, total electron pairs around the central atom $= 5$.

3. Determine electron-pair geometry using VSEPR

For $5$ electron pairs, the electron-pair geometry is trigonal bipyramidal.

In trigonal bipyramidal arrangement, lone pairs prefer equatorial positions because they experience less repulsion there.

So the $3$ lone pairs occupy all three equatorial positions, leaving the two bonded iodine atoms in the two axial positions.

4. Determine molecular shape

Since the two bonded atoms are opposite each other in axial positions, the molecular shape is: $\text{Linear}$

5. Determine bond angle

The angle between the two axial bonds is: $180^\circ$

Hence, the correct shape and bond angle in $I_3^-$ are: $\boxed{\text{Linear; }180^\circ}$

6. Check options

• A: Linear; $180^\circ$ ✅
• B: T-shaped; $180^\circ$ and $90^\circ$ ❌
• C: Trigonal planar; $120^\circ$ ❌
• D: Distorted trigonal planar; $135^\circ$ and $90^\circ$ ❌

Therefore, the correct option is A.