To determine the lowest bond angle, we need to analyze the geometry and hybridization of each molecule/ion, considering the presence of lone pairs.

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the geometry of molecules based on minimizing repulsion between electron pairs (both bonding and non-bonding) around the central atom. Lone pairs exert more repulsion than bonding pairs, leading to a reduction in bond angles.

• A) ${\mathrm{NH}}_3$ (Ammonia): The central atom is Nitrogen (N). It has 3 bond pairs (with H atoms) and 1 lone pair. The steric number is 4 (3 bond pairs + 1 lone pair), indicating ${\mathrm{sp}}^3$ hybridization. Due to the presence of one lone pair, the geometry is trigonal pyramidal, and the bond angle is approximately $107^{\circ }$.
• B) ${\mathrm{BeF}}_2$ (Beryllium fluoride): The central atom is Beryllium (Be). It has 2 bond pairs (with F atoms) and 0 lone pairs. The steric number is 2, indicating $\mathrm{sp}$ hybridization. The geometry is linear, and the bond angle is $180^{\circ }$.
• C) ${\mathrm{H}}_3{\mathrm{O}}^{+}$ (Hydronium ion): The central atom is Oxygen (O). It has 3 bond pairs (with H atoms) and 1 lone pair. The steric number is 4, indicating ${\mathrm{sp}}^3$ hybridization. Similar to ammonia, due to one lone pair, the geometry is trigonal pyramidal, and the bond angle is approximately $107^{\circ }$.
• D) ${\mathrm{CH}}_4$ (Methane): The central atom is Carbon (C). It has 4 bond pairs (with H atoms) and 0 lone pairs. The steric number is 4, indicating ${\mathrm{sp}}^3$ hybridization. The geometry is tetrahedral, and the bond angle is $109.5^{\circ }$.

Comparing the bond angles: ${\mathrm{BeF}}_2$ ($180^{\circ }$), ${\mathrm{CH}}_4$ ($109.5^{\circ }$), ${\mathrm{NH}}_3$ ($107^{\circ }$), ${\mathrm{H}}_3{\mathrm{O}}^{+}$ ($107^{\circ }$).

Both ${\mathrm{NH}}_3$ and ${\mathrm{H}}_3{\mathrm{O}}^{+}$ have bond angles of approximately $107^{\circ }$. However, typically, the bond angle in ${\mathrm{NH}}_3$ is slightly less than that in ${\mathrm{H}}_3{\mathrm{O}}^{+}$ due to the higher electronegativity of nitrogen compared to oxygen in the context of the positive charge on oxygen in ${\mathrm{H}}_3{\mathrm{O}}^{+}$ which pulls electron density away from the central atom, slightly reducing lone pair-bond pair repulsion compared to ${\mathrm{NH}}_3$. However, for the purpose of this question, both are close, and the key is that they are significantly lower than ${\mathrm{CH}}_4$ and ${\mathrm{BeF}}_2$.

Given the options, ${\mathrm{NH}}_3$ has one of the lowest bond angles among the choices.

Correct Answer: (A)