To determine if a molecule or ion has a regular tetrahedral geometry, we need to consider its hybridization and the number of lone pairs on the central atom. A regular tetrahedral geometry requires the central atom to be sp³ hybridized and have no lone pairs.

Step 1: Analyze the central atom and its hybridization for each option.

• A) ${\mathrm{SF}}_4$: Sulfur (S) is the central atom. It has 6 valence electrons. It forms 4 single bonds with fluorine atoms. This leaves 6 - 4 = 2 electrons, which form 1 lone pair. The steric number is 4 (bonds) + 1 (lone pair) = 5. This corresponds to sp³d hybridization. The geometry is see-saw, which is a distorted tetrahedron.
• B) ${\mathrm{BF}}_4^{-}$: Boron (B) is the central atom. It has 3 valence electrons. The negative charge adds 1 electron, so it has 3 + 1 = 4 electrons available for bonding. It forms 4 single bonds with fluorine atoms. There are no lone pairs (4 - 4 = 0). The steric number is 4 (bonds) + 0 (lone pairs) = 4. This corresponds to sp³ hybridization. With no lone pairs and sp³ hybridization, the geometry is regular tetrahedral.
• C) ${\mathrm{XeF}}_4$: Xenon (Xe) is the central atom. It has 8 valence electrons. It forms 4 single bonds with fluorine atoms. This leaves 8 - 4 = 4 electrons, which form 2 lone pairs. The steric number is 4 (bonds) + 2 (lone pairs) = 6. This corresponds to sp³d² hybridization. The geometry is square planar.
• D) ${\mathrm{ClF}}_3$: Chlorine (Cl) is the central atom. It has 7 valence electrons. It forms 3 single bonds with fluorine atoms. This leaves 7 - 3 = 4 electrons, which form 2 lone pairs. The steric number is 3 (bonds) + 2 (lone pairs) = 5. This corresponds to sp³d hybridization. The geometry is T-shaped.

Step 2: Identify the option with regular tetrahedral geometry.

Only ${\mathrm{BF}}_4^{-}$ has sp³ hybridization and no lone pairs on the central atom, resulting in a regular tetrahedral geometry.

Option Analysis:

• A) ${\mathrm{SF}}_4$: See-saw geometry due to one lone pair.
• B) ${\mathrm{BF}}_4^{-}$: Regular tetrahedral geometry due to sp³ hybridization and no lone pairs.
• C) ${\mathrm{XeF}}_4$: Square planar geometry due to two lone pairs.
• D) ${\mathrm{ClF}}_3$: T-shaped geometry due to two lone pairs.

Correct Answer: (B)