To determine the geometry of a molecule, we need to find the steric number (number of bond pairs + lone pairs) around the central atom and apply VSEPR theory.

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the geometry of molecules based on minimizing the repulsion between electron pairs (both bonding and non-bonding) in the valence shell of the central atom. The number of electron domains (bond pairs + lone pairs) determines the electron geometry, which in turn influences the molecular geometry.

For ${\mathrm{BF}}_3$:

• Central atom: Boron (B)
• Valence electrons of B: 3
• Number of F atoms bonded: 3
• Number of bond pairs: 3
• Number of lone pairs on B: $(3-3\times 1)/2=0$
• Steric number: $3+0=3$
• Electron geometry: Trigonal planar
• Molecular geometry: Trigonal planar
• Hybridization: ${\mathrm{sp}}^2$

Option Analysis:

• A) ${\mathrm{IF}}_3$: Central atom is Iodine (I). I has 7 valence electrons. It forms 3 bonds with F atoms. Number of lone pairs = $(7-3\times 1)/2=2$. Steric number = $3\text{ (bond pairs)}+2\text{ (lone pairs)}=5$. Electron geometry is trigonal bipyramidal, but due to 2 lone pairs, the molecular geometry is T-shaped.
• B) ${\mathrm{PCl}}_3$: Central atom is Phosphorus (P). P has 5 valence electrons. It forms 3 bonds with Cl atoms. Number of lone pairs = $(5-3\times 1)/2=1$. Steric number = $3\text{ (bond pairs)}+1\text{ (lone pair)}=4$. Electron geometry is tetrahedral, but due to 1 lone pair, the molecular geometry is trigonal pyramidal.
• C) ${\mathrm{NH}}_3$: Central atom is Nitrogen (N). N has 5 valence electrons. It forms 3 bonds with H atoms. Number of lone pairs = $(5-3\times 1)/2=1$. Steric number = $3\text{ (bond pairs)}+1\text{ (lone pair)}=4$. Electron geometry is tetrahedral, but due to 1 lone pair, the molecular geometry is trigonal pyramidal.
• D) ${\mathrm{BF}}_3$: As explained above, Boron has 3 valence electrons and forms 3 bonds with F atoms, with no lone pairs. Steric number is 3, leading to a trigonal planar geometry.

Correct Answer: (D)