Concept: The bond angle of $109^{\circ }28^{\prime }$ is characteristic of a perfect tetrahedral geometry, which arises when the central atom undergoes sp³ hybridization and has no lone pairs of electrons.

Why (A) is correct:

• For $\left[{\mathrm{NH}}_4{}^{+}\right]$: The central atom is Nitrogen (N). Nitrogen has 5 valence electrons. In ${\mathrm{NH}}_4{}^{+}$, it forms 4 bonds with hydrogen atoms and has a +1 charge. This means N contributes 4 electrons to bonding (5 - 1 = 4). All 4 valence electrons are involved in forming 4 single bonds. Thus, there are 4 bond pairs and 0 lone pairs. The steric number is 4, leading to sp³ hybridization and a perfect tetrahedral geometry with a bond angle of $109^{\circ }28^{\prime }$.
• For $\left[{\mathrm{BF}}_4^{-}\right]$: The central atom is Boron (B). Boron has 3 valence electrons. In ${\mathrm{BF}}_4^{-}$, it forms 4 bonds with fluorine atoms and has a -1 charge. This means B effectively has 4 electrons for bonding (3 + 1 = 4). All 4 valence electrons are involved in forming 4 single bonds. Thus, there are 4 bond pairs and 0 lone pairs. The steric number is 4, leading to sp³ hybridization and a perfect tetrahedral geometry with a bond angle of $109^{\circ }28^{\prime }$.

Option Analysis:

• B) $\left[{\mathrm{NH}}_4{}^{+}\right],\left[{\mathrm{BF}}_3\right]$: $\left[{\mathrm{NH}}_4{}^{+}\right]$ has a tetrahedral geometry. However, ${\mathrm{BF}}_3$ has sp² hybridization with 3 bond pairs and 0 lone pairs, resulting in a trigonal planar geometry with a bond angle of $120^{\circ }$.
• C) ${\mathrm{NH}}_3,\left[{\mathrm{BF}}_4^{-}\right]$: $\left[{\mathrm{BF}}_4^{-}\right]$ has a tetrahedral geometry. However, ${\mathrm{NH}}_3$ has sp³ hybridization with 3 bond pairs and 1 lone pair, resulting in a trigonal pyramidal geometry with a bond angle of approximately $107^{\circ }$ (due to lone pair-bond pair repulsion).
• D) $\left[{\mathrm{NH}}_3\right],\left[{\mathrm{BF}}_3\right]$: Neither ${\mathrm{NH}}_3$ (trigonal pyramidal, $\approx 107^{\circ }$) nor ${\mathrm{BF}}_3$ (trigonal planar, $120^{\circ }$) has a bond angle of $109^{\circ }28^{\prime }$.

Correct Answer: (A)