(A) The steric number of the central Bromine atom is $=5$ bond pairs$+$one lone pair $=6$. Hence the shape is square pyramidal structure and the hybridisation is ${\mathrm{sp}}^3{\mathrm{d}}^2$.

(B) ${\mathrm{H}}_2\mathrm{O}$ molecular geometry is bent because it has two bound atoms and two lone pairs on oxygen, the central atom. The two lone pairs on the oxygen atom push the two hydrogen atoms down further away from the oxygen atom and closer to each other, creating a bent shape and $104.5$ degrees angle between the two $\mathrm{H}$ atoms.

(C) The molecular geometry of ${\mathrm{ClF}}_3$ is said to be T-shaped. ${\mathrm{ClF}}_3$ acquires such a shape because two lone pairs take up equatorial positions as they require more space and greater repulsions. They are arranged in a trigonal bipyramidal shape with $3$ bonds and $2$ lone pairs.

(D)  ${\mathrm{SF}}_4$ has an ${\mathrm{sp}}^3\mathrm{d}$ type of hybridisation. In a molecule of ${\mathrm{SF}}_4$, each $\mathrm{S}$ atom is bonded to four fluorine atoms.