Number of lone pair (s) of electrons on central atom and the shape of BrF$$_3$$ molecule respectively, are:

We need to find the number of lone pairs on the central atom and the shape of BrF₃.

Bromine (Br) has 7 valence electrons, and each F atom contributes 1 electron. Total electron pairs around Br = $$\frac{7 + 3 \times 1}{2}$$. Br forms 3 bonds with F atoms, using 3 electrons, leaving 7 − 3 = 4 electrons that form 2 lone pairs. Thus, there are 3 bond pairs and 2 lone pairs, for a total of 5 electron pairs, corresponding to a trigonal bipyramidal electron geometry.

In this arrangement, the 2 lone pairs occupy equatorial positions to minimize repulsion, while the 3 F atoms occupy the 2 axial and 1 equatorial positions. This gives a bent T-shape molecular geometry.

Number of lone pairs on central atom = 2, Shape = bent T-shape. The answer is Option C: 2, bent T-shape.