1. In an interhalogen compound $AB_3$, the central atom $A$ is a halogen.

2. A halogen has $7$ valence electrons.

3. In $AB_3$, atom $A$ forms $3$ single bonds with three $B$ atoms.

   • Each bond uses one electron from $A$.
   • So, $3$ of the $7$ valence electrons of $A$ are used in bonding.

4. Remaining electrons on $A$: $7 - 3 = 4$ These $4$ electrons form: $\frac{4}{2} = 2$ lone pairs.

5. This also matches the known VSEPR shape:

   • $3$ bond pairs + $2$ lone pairs $\Rightarrow$ steric number $5$
   • Electron geometry: trigonal bipyramidal
   • Molecular shape: T-shaped

Therefore, the number of lone pairs on $A$ is $2$