To determine the correct order of bond angles, we need to analyze the geometry and hybridization of each molecule.

• ${\mathrm{H}}_2\mathrm{S}$ (Hydrogen Sulfide): The central atom is Sulfur (S). Sulfur has 6 valence electrons. Two are used to form bonds with two Hydrogen atoms, leaving two lone pairs. The steric number is 4 (2 bond pairs + 2 lone pairs), indicating ${\mathrm{sp}}^3$ hybridization. The geometry is bent or V-shaped. Due to the presence of two lone pairs and the larger size of S compared to O (in H2O), the bond angle is less than the ideal tetrahedral angle (109.5°). The bond angle in ${\mathrm{H}}_2\mathrm{S}$ is approximately 92.1°.
• ${\mathrm{NH}}_3$ (Ammonia): The central atom is Nitrogen (N). Nitrogen has 5 valence electrons. Three are used to form bonds with three Hydrogen atoms, leaving one lone pair. The steric number is 4 (3 bond pairs + 1 lone pair), indicating ${\mathrm{sp}}^3$ hybridization. The geometry is trigonal pyramidal. The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, which compresses the bond angle from the ideal tetrahedral angle (109.5°). The bond angle in ${\mathrm{NH}}_3$ is approximately 107°.
• ${\mathrm{SiH}}_4$ (Silane): The central atom is Silicon (Si). Silicon has 4 valence electrons. All four are used to form bonds with four Hydrogen atoms. There are no lone pairs. The steric number is 4 (4 bond pairs + 0 lone pairs), indicating ${\mathrm{sp}}^3$ hybridization. The geometry is tetrahedral. With no lone pairs, the bond angle is the ideal tetrahedral angle, 109.5°.
• ${\mathrm{BF}}_3$ (Boron Trifluoride): The central atom is Boron (B). Boron has 3 valence electrons. All three are used to form bonds with three Fluorine atoms. There are no lone pairs. The steric number is 3 (3 bond pairs + 0 lone pairs), indicating ${\mathrm{sp}}^2$ hybridization. The geometry is trigonal planar. With no lone pairs, the bond angle is the ideal trigonal planar angle, 120°.

• ${\mathrm{H}}_2\mathrm{S}\approx 92.1^{\circ }$
• ${\mathrm{NH}}_3\approx 107^{\circ }$
• ${\mathrm{SiH}}_4\approx 109.5^{\circ }$
• ${\mathrm{BF}}_3\approx 120^{\circ }$

${\mathrm{H}}_2\mathrm{S}<{\mathrm{NH}}_3<{\mathrm{SiH}}_4<{\mathrm{BF}}_3$

• A) ${\mathrm{H}}_2\mathrm{S}<{\mathrm{NH}}_3<{\mathrm{SiH}}_4<{\mathrm{BF}}_3$: This order matches our calculated bond angles.
• B) ${\mathrm{NH}}_3<{\mathrm{H}}_2\mathrm{S}<{\mathrm{SiH}}_4<{\mathrm{BF}}_3$: Incorrect, as ${\mathrm{H}}_2\mathrm{S}$ has a smaller bond angle than ${\mathrm{NH}}_3$.
• C) ${\mathrm{H}}_2\mathrm{S}<{\mathrm{SiH}}_4<{\mathrm{NH}}_3<{\mathrm{BF}}_3$: Incorrect, as ${\mathrm{NH}}_3$ has a smaller bond angle than ${\mathrm{SiH}}_4$.
• D) ${\mathrm{H}}_2\mathrm{S}<{\mathrm{NH}}_3<{\mathrm{BF}}_3<{\mathrm{SiH}}_4$: Incorrect, as ${\mathrm{SiH}}_4$ has a smaller bond angle than ${\mathrm{BF}}_3$.

Correct Answer: (A)