To determine which species has a regular tetrahedral structure, we need to analyze the hybridization and VSEPR theory for each option.

A regular tetrahedral structure is observed when the central atom undergoes sp³ hybridization and is surrounded by four identical atoms or groups, with no lone pairs on the central atom. The bond angles are approximately 109.5°.

In ${\mathrm{BF}}_4^{-}$, the central atom is Boron (B). Boron has 3 valence electrons. When it forms four bonds with four fluorine atoms and carries a negative charge, it effectively has 4 electron pairs. The hybridization of B in ${\mathrm{BF}}_4^{-}$ is sp³. Since there are no lone pairs on the central boron atom and four bond pairs, its geometry is regular tetrahedral.

• A) ${\mathrm{XeF}}_4$: Xenon (Xe) has 8 valence electrons. It forms 4 bonds with fluorine atoms and has 2 lone pairs (8 - 4 = 4 electrons, so 2 lone pairs). The steric number is 4 (bond pairs) + 2 (lone pairs) = 6. This corresponds to sp³d² hybridization, and the geometry is square planar due to the two lone pairs occupying axial positions.
• B) ${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$: In this complex, Nickel (Ni) is in the +2 oxidation state. Ni²⁺ has a d⁸ configuration. Cyanide (CN^-) is a strong field ligand. In the presence of strong field ligands, the d electrons pair up, leading to dsp² hybridization. This results in a square planar geometry.
• D) ${\mathrm{SF}}_4$: Sulfur (S) has 6 valence electrons. It forms 4 bonds with fluorine atoms and has 1 lone pair (6 - 4 = 2 electrons, so 1 lone pair). The steric number is 4 (bond pairs) + 1 (lone pair) = 5. This corresponds to sp³d hybridization, and the geometry is see-saw (a distorted tetrahedron) due to the presence of the lone pair.

Correct Answer: (C)