Count total valence electrons in $I_3^-$:

• Each iodine atom has $7$ valence electrons.
• For $3$ iodine atoms: $3 \times 7 = 21$
• The negative charge adds $1$ extra electron.

Hence, total valence electrons: $21 + 1 = 22$

Choose the central atom:

In $I_3^-$, one iodine is the central atom and the other two are terminal atoms.

Form two single bonds between central iodine and the two terminal iodines:

• Two $I-I$ bonds use $2 \times 2 = 4$ electrons.

Remaining electrons: $22 - 4 = 18$

Complete octets of terminal iodine atoms:

• Each terminal iodine needs $6$ more electrons as lone-pair electrons after forming one bond.
• For two terminal iodines: $2 \times 6 = 12$ electrons.

Remaining electrons: $18 - 12 = 6$

Place remaining electrons on central iodine:

• The remaining $6$ electrons stay on the central iodine.
• $6$ electrons correspond to: $\frac{6}{2} = 3 \text{ lone pairs}$

Answer:

The number of lone pairs on the central iodine atom in $I_3^-$ is: $\boxed{3}$