Find the total valence electrons around the central atom

In $BrF_3$, the central atom is bromine ($Br$).

• Bromine belongs to group 17, so it has $7$ valence electrons.
• Each fluorine also has $7$ valence electrons, but for shape determination we focus on electron pairs around the central atom.

Determine how many bond pairs and lone pairs are on bromine

Bromine forms $3$ single bonds with $3$ fluorine atoms.

So, out of bromine’s $7$ valence electrons:

• $3$ electrons are used in forming $3$ bond pairs
• Remaining electrons on bromine $= 7 - 3 = 4$ electrons

These $4$ electrons form: $\frac{4}{2} = 2$ lone pairs.

Hence, the central atom has 2 lone pairs.

Use VSEPR theory to determine the shape

Total electron pairs around bromine:

• Bond pairs $= 3$
• Lone pairs $= 2$

Thus, steric number $= 5$.

For steric number $5$, the electron pair geometry is trigonal bipyramidal.

In trigonal bipyramidal arrangement, lone pairs prefer equatorial positions to minimize repulsion.

So with $2$ lone pairs in equatorial positions, the molecular shape formed by the three bonded fluorine atoms becomes T-shaped.

Match with the given options

• A: $0$, triangular planar $\,\rightarrow$ incorrect
• B: $1$, pyramidal $\,\rightarrow$ incorrect
• C: $2$, bent T-shape $\,\rightarrow$ correct in intent for lone pairs $=2$ and shape $=T$-shape
• D: $1$, bent T-shape $\,\rightarrow$ incorrect

The standard description is 2 lone pairs, T-shaped.

Final answer

$\boxed{\text{C}}$

because $BrF_3$ has 2 lone pairs on the central atom and a T-shaped molecular geometry.