Concept: Molecular geometry is determined by the number of electron domains (bonding pairs and lone pairs) around the central atom, according to VSEPR theory. A trigonal planar geometry arises when there are three electron domains and all three are bonding pairs, with no lone pairs on the central atom.

Why (D) is correct:

Let's analyze the central atom and its electron domains for each molecule:

• A) ${\mathrm{IF}}_3$: Iodine (I) 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. So, the central atom has 3 bonding pairs + 2 lone pairs = 5 electron domains. According to VSEPR theory, 5 electron domains lead to a trigonal bipyramidal electron geometry. With 2 lone pairs, the molecular geometry is T-shaped.
• B) ${\mathrm{PCl}}_3$: Phosphorus (P) is the central atom. It has 5 valence electrons. It forms 3 single bonds with chlorine atoms. This leaves 5 - 3 = 2 electrons, which form 1 lone pair. So, the central atom has 3 bonding pairs + 1 lone pair = 4 electron domains. According to VSEPR theory, 4 electron domains lead to a tetrahedral electron geometry. With 1 lone pair, the molecular geometry is trigonal pyramidal.
• C) ${\mathrm{NH}}_3$: Nitrogen (N) is the central atom. It has 5 valence electrons. It forms 3 single bonds with hydrogen atoms. This leaves 5 - 3 = 2 electrons, which form 1 lone pair. So, the central atom has 3 bonding pairs + 1 lone pair = 4 electron domains. According to VSEPR theory, 4 electron domains lead to a tetrahedral electron geometry. With 1 lone pair, the molecular geometry is trigonal pyramidal.
• D) ${\mathrm{BF}}_3$: Boron (B) is the central atom. It has 3 valence electrons. It forms 3 single bonds with fluorine atoms. This leaves 3 - 3 = 0 electrons, meaning no lone pairs on the central atom. So, the central atom has 3 bonding pairs + 0 lone pairs = 3 electron domains. According to VSEPR theory, 3 electron domains with no lone pairs lead to a trigonal planar molecular geometry. The bond angles are approximately 120°.

Option Analysis:

• A) ${\mathrm{IF}}_3$: T-shaped (due to 3 bonding pairs and 2 lone pairs).
• B) ${\mathrm{PCl}}_3$: Trigonal pyramidal (due to 3 bonding pairs and 1 lone pair).
• C) ${\mathrm{NH}}_3$: Trigonal pyramidal (due to 3 bonding pairs and 1 lone pair).
• D) ${\mathrm{BF}}_3$: Trigonal planar (due to 3 bonding pairs and 0 lone pairs).

Correct Answer: (D)