Which of the following has the square planar structure?

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

  • Xenon ($$\text{Xe}$$) belongs to the noble gases and has 8 valence electrons. It forms 4 single covalent bonds with fluorine atoms, leaving 4 non-bonding electrons.
  • Steric Number: $$4 \text{ bonding pairs} + 2 \text{ lone pairs} = 6$$ ($$sp^3d^2$$ hybridization).
  • Geometry: The 6 electron pairs arrange octahedrally. To minimize repulsion, the 2 lone pairs occupy opposite axial positions, forcing the 4 fluorine atoms into a perfectly symmetric square planar configuration.



• Option B: $$\text{NH}_4^+$$ (Ammonium Ion)

  • Nitrogen normally has 5 valence electrons, but the positive charge leaves it with 4. It forms 4 single bonds with hydrogen atoms, leaving 0 lone pairs.
  • Steric Number: $$4 \text{ bonding pairs} + 0 \text{ lone pairs} = 4$$ ($$sp^3$$ hybridization).
  • Geometry: Tetrahedral.



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

  • Boron has 3 valence electrons, and the negative charge adds 1 more. It forms 4 single bonds with fluorine atoms, leaving 0 lone pairs.
  • Steric Number: $$4 \text{ bonding pairs} + 0 \text{ lone pairs} = 4$$ ($$sp^3$$ hybridization).
  • Geometry: Tetrahedral.



• Option D: $$\text{CCl}_4$$ (Carbon Tetrachloride)

  • Carbon has 4 valence electrons and forms 4 single bonds with chlorine atoms, leaving 0 lone pairs.
  • Steric Number: $$4 \text{ bonding pairs} + 0 \text{ lone pairs} = 4$$ ($$sp^3$$ hybridization).
  • Geometry: Tetrahedral.

Conclusion:

While the other three options possess an $$sp^3$$ tetrahedral geometry, only $$\text{XeF}_4$$ has the $$sp^3d^2$$ electronic distribution that yields a square planar molecular geometry.

Answer: Option A — $$\text{XeF}_4$$