AB$$_3$$ is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is _________ (Integer answer)

We are told that the molecule is an inter-halogen of the type $$\mathrm{AB_3}$$ and that its observed geometry is T-shaped. In any VSEPR (Valence Shell Electron Pair Repulsion) discussion we first count total electron pairs around the central atom $$\mathrm A$$, because the geometry is fixed by the steric number (total regions of electron density = bond pairs + lone pairs).

Each halogen atom has seven valence electrons. Therefore the central halogen $$\mathrm A$$ also possesses $$7$$ valence electrons. We now account for the three $$\mathrm{A{-}B}$$ single bonds.

Every single (sigma) bond is a bond pair involving one electron from $$\mathrm A$$ and one from $$\mathrm B$$. So the three $$\mathrm{A{-}B}$$ bonds consume $$3$$ of the $$7$$ valence electrons of $$\mathrm A$$:

$$7\;(\text{valence electrons of A}) \;-\;3\;(\text{electrons used in three } \sigma\text{ bonds}) \;=\;4$$

The $$4$$ electrons that remain on $$\mathrm A$$ must stay as non-bonding electrons. Because two electrons make one lone pair, we have

$$\dfrac{4\; \text{non-bonding electrons}}{2\; \text{electrons per lone pair}} \;=\;2\; \text{lone pairs}.$$

Hence the electron-pair distribution around $$\mathrm A$$ is

$$3 \text{ bond pairs } + 2 \text{ lone pairs } = 5 \text{ regions of electron density}.$$

According to VSEPR theory, five regions adopt a trigonal-bipyramidal arrangement. When two of those five positions are occupied by lone pairs (preferentially the equatorial sites, because 90° repulsions are minimized), the three bond pairs occupy the remaining positions and the molecular shape observed is T-shaped, exactly as stated in the question. This matches the description $$\mathrm{AB_3E_2}$$, where $$\mathrm E$$ denotes a lone pair.

So, the number of lone pairs on the central atom $$\mathrm A$$ is clearly

$$2.$$

Hence, the correct answer is Option C.