$$sp^3d^2$$ hybridization is not displayed by

To decide whether the central atom in a molecule or ion uses $$sp^3d^2$$ hybrid orbitals, we first recall a basic rule from valence-shell electron-pair repulsion (VSEPR) theory.

The steric number of the central atom is defined as

$$\text{steric number}= \text{number of atoms directly bonded} + \text{number of lone pairs on the central atom}.$$

Once this number is known, the hybridization is fixed by the following correspondence:

$$ \begin{aligned} \text{steric number } 2 &\;\rightarrow\; sp,\\ \text{steric number } 3 &\;\rightarrow\; sp^2,\\ \text{steric number } 4 &\;\rightarrow\; sp^3,\\ \text{steric number } 5 &\;\rightarrow\; sp^3d,\\ \text{steric number } 6 &\;\rightarrow\; sp^3d^2.\\ \end{aligned} $$

Now we examine every given species one by one.

For $$SF_6$$:

The sulfur atom is bonded to six fluorine atoms and possesses no lone pair. Therefore

$$\text{steric number}=6+0=6.$$

A steric number of $$6$$ demands $$sp^3d^2$$ hybridization. So $$SF_6$$ indeed shows $$sp^3d^2$$ hybridization.

For $$BrF_5$$:

The bromine atom is bonded to five fluorine atoms and has one lone pair. Hence

$$\text{steric number}=5+1=6.$$

Again the steric number is $$6$$, and consequently bromine uses $$sp^3d^2$$ hybrid orbitals in $$BrF_5$$.

For $$PF_5$$:

The phosphorus atom is bonded to five fluorine atoms and carries no lone pair. So

$$\text{steric number}=5+0=5.$$

A steric number of $$5$$ matches $$sp^3d$$ hybridization, not $$sp^3d^2$$. Therefore $$PF_5$$ does not employ $$sp^3d^2$$ hybridization.

For $$[CrF_6]^{3-}$$:

The chromium ion $$Cr^{3+}$$ lies at the center of an octahedral coordination environment with six fluoride ligands and no lone pair. Thus

$$\text{steric number}=6+0=6.$$

An octahedral arrangement (steric number $$6$$) once more requires $$sp^3d^2$$ hybridization when an outer-orbital (high-spin) set of orbitals is used with weak-field fluoride ligands. Hence $$[CrF_6]^{3-}$$ is also $$sp^3d^2$$.

Among all the options, only $$PF_5$$ has a steric number less than $$6$$ and therefore lacks $$sp^3d^2$$ hybridization.

Hence, the correct answer is Option C.