Identify the molecular shapes and approximate bond angles

We compare the bond angles in the given molecules:

• $H_2S$
• $NH_3$
• $BF_3$
• $SiH_4$

$H_2S$

• Central atom: S
• Shape: bent
• Sulfur has two lone pairs and two bond pairs.
• Since sulfur is larger and shows very little hybridization, the $H-S-H$ bond angle is quite small.
• Approximate bond angle: $\angle HSH \approx 92^\circ$

$NH_3$

• Central atom: N
• Shape: trigonal pyramidal
• Nitrogen has one lone pair and three bond pairs.
• Tetrahedral electron pair geometry gives ideal angle $109.5^\circ$, but lone pair-bond pair repulsion compresses it.
• Approximate bond angle: $\angle HNH \approx 107^\circ$

$SiH_4$

• Central atom: Si
• Shape: tetrahedral
• No lone pairs on Si.
• Ideal tetrahedral bond angle: $\angle HSiH = 109.5^\circ$

$BF_3$

• Central atom: B
• Shape: trigonal planar
• No lone pairs on B.
• Ideal trigonal planar bond angle: $\angle FBF = 120^\circ$

Arrange in increasing order

Using the approximate values: $H_2S \approx 92^\circ < NH_3 \approx 107^\circ < SiH_4 = 109.5^\circ < BF_3 = 120^\circ$

Therefore, the correct order is: $H_2S < NH_3 < SiH_4 < BF_3$

Check options

• A: $H_2S < SiH_4 < NH_3 < BF_3$ ❌ because $NH_3 < SiH_4$
• B: $H_2S < NH_3 < BF_3 < SiH_4$ ❌ because $BF_3 > SiH_4$
• C: $H_2S < NH_3 < SiH_4 < BF_3$ ✅
• D: $NH_3 < H_2S < SiH_4 < BF_3$ ❌ because $H_2S$ has the smallest angle

Final answer

The correct option is C.