1. We need the species whose hybrid orbitals use only one $d$-orbital of the central atom.

   Let us examine the hybridization of the central atom in each option.

2. Option A: $XeF_4$

   • Shape: square planar
   • Hybridization traditionally assigned: $sp^3d^2$
   • This involves two $d$-orbitals.

   So, $XeF_4$ does not satisfy the condition.

3. Option B: $[Ni(CN)_4]^{2-}$

   • Oxidation state of Ni: \Rightarrow x = +2$$
   • So Ni is $Ni^{2+}$ with configuration: $[Ar]3d^8$
   • $CN^-$ is a strong field ligand, so electrons pair up.
   • For square planar complex, hybridization is: $dsp^2$
   • This uses one $d$-orbital, one $s$, and two $p$ orbitals.

   Hence, this option does satisfy the condition.

4. Option C: $[CrF_6]^{3-}$

   • Oxidation state of Cr: \Rightarrow x = +3$$
   • So Cr is $Cr^{3+}$ with configuration: $[Ar]3d^3$
   • $F^-$ is a weak field ligand.
   • Octahedral complex formed with outer orbital hybridization: $sp^3d^2$
   • This uses two $d$-orbitals.

   So, this option does not satisfy the condition.

5. Option D: $BrF_5$

   • Shape: square pyramidal
   • Hybridization traditionally assigned: $sp^3d^2$
   • This involves two $d$-orbitals.

   So, this option does not satisfy the condition.

6. Conclusion

   Only $[Ni(CN)_4]^{2-}$ has hybridization $dsp^2$, involving exactly one $d$-orbital of the central atom.

$\boxed{[Ni(CN)_4]^{2-}}$