Consider the given list of molecules: XeF_2 , ClF_3 , SF_4 , PCl_5...

The dipole moment of a molecule depends on its geometry and the presence of lone pairs.

$\text{XeF}_2$: The central $\text{Xe}$ atom has $3$ lone pairs and $2$ bond pairs ($sp^3d$ hybridization). It has a linear shape with lone pairs at equatorial positions. The bond dipoles cancel out, so $\mu = 0$.

$\text{ClF}_3$: The central $\text{Cl}$ atom has $2$ lone pairs and $3$ bond pairs ($sp^3d$ hybridization). It has a T-shape. The bond dipoles do not cancel out, so $\mu \neq 0$.

$\text{SF}_4$: The central $\text{S}$ atom has $1$ lone pair and $4$ bond pairs ($sp^3d$ hybridization). It has a see-saw shape. The bond dipoles do not cancel out, so $\mu \neq 0$.

$\text{PCl}_5$: The central $\text{P}$ atom has $0$ lone pairs and $5$ bond pairs ($sp^3d$ hybridization). It has a regular trigonal bipyramidal geometry. All bond dipoles cancel out, so $\mu = 0$.

$\text{BrF}_5$: The central $\text{Br}$ atom has $1$ lone pair and $5$ bond pairs ($sp^3d^2$ hybridization). It has a square pyramidal shape. The axial bond dipole is not canceled, so $\mu \neq 0$.

$\text{XeF}_4$: The central $\text{Xe}$ atom has $2$ lone pairs and $4$ bond pairs ($sp^3d^2$ hybridization). It has a square planar shape with lone pairs at opposite axial positions. The bond dipoles cancel out, so $\mu = 0$.

Therefore, the molecules possessing a permanent dipole moment are $\text{ClF}_3$, $\text{SF}_4$, and $\text{BrF}_5$. There are $3$ such molecules.

Answer: $3$