Which one of the following has the regular tetrahedral structure?

What Makes a Geometry a "Regular" Tetrahedron?

A molecule or ion has a regular tetrahedral structure if its central atom is bonded to four identical surrounding groups with zero lone pairs on the central atom. The presence of lone pairs causes asymmetric electron repulsion, distorting the ideal bond angles away from $$109.5^\circ$$.

Structure and Geometry Analysis:

• Option A: $$\text{XeF}_4$$ (Xenon Tetrafluoride)

  • Xenon ($$\text{Xe}$$) has 8 valence electrons. It forms 4 single bonds with fluorine atoms, leaving 4 non-bonding electrons ($$2 \text{ lone pairs}$$).
  • Steric Number: $$4 \text{ bonds} + 2 \text{ lone pairs} = 6$$ ($$sp^3d^2$$ hybridization).
  • Geometry: Square Planar.



• Option B: $$[\text{Ni(CN)}_4]^{2-}$$ (Tetracyanidonickelate(II))

  • Nickel is in the $$+2$$ oxidation state with a $$d^8$$ electronic configuration. Because $$\text{CN}^-$$ is a powerful strong field ligand, it forces the $$3d$$ electrons to pair up, leaving one inner $$3d$$ orbital vacant.
  • Hybridization: $$dsp^2$$.
  • Geometry: Square Planar.



• Option C: $$\text{BF}_4^-$$ (Tetrafluoroborate Anion)

  • Boron ($$\text{B}$$) normally has 3 valence electrons, but the negative charge adds 1 more, giving it 4 valence electrons. It uses all 4 electrons to form 4 covalent bonds with fluorine atoms, leaving zero lone pairs on the boron atom.
  • Steric Number: $$4 \text{ bonds} + 0 \text{ lone pairs} = 4$$ ($$sp^3$$ hybridization).
  • Geometry: Regular Tetrahedral (all bond angles are perfectly symmetric at $$109.5^\circ$$).



• Option D: $$\text{SF}_4$$ (Sulfur Tetrafluoride)

  • Sulfur ($$\text{S}$$) has 6 valence electrons. It forms 4 single bonds with fluorine atoms, leaving 2 non-bonding electrons ($$1 \text{ lone pair}$$).
  • Steric Number: $$4 \text{ bonds} + 1 \text{ lone pair} = 5$$ ($$sp^3d$$ hybridization).
  • Geometry: See-saw (distorted due to the equatorial lone pair).

Conclusion:

Only the tetrafluoroborate ion ($$\text{BF}_4^-$$) features $$sp^3$$ hybridization with no lone pairs on the central atom to disrupt its structural symmetry.

Answer: Option C — $$\text{BF}_4^-$$