The correct order of bond angles (smallest first) in $$H_2S, NH_3, BF_3$$ and $$SiH_4$$ is

• 1. $$\text{H}_2\text{S}$$ (Hydrogen Sulfide)

  Sulfur belongs to the 3rd period. According to Drago's Rule, when the central atom belongs to the 3rd period or higher and is bonded to less electronegative elements like hydrogen, the hybridization of the central atom is negligible. The bonding occurs almost purely via orthogonal $$3p$$ orbitals. Consequently, the bond angle is exceptionally close to $$90^\circ$$ (experimentally, it is $$\sim 92^\circ$$), making it the smallest in the group.




• 2. $$\text{NH}_3$$ (Ammonia)

  The nitrogen atom undergoes $$sp^3$$ hybridization, yielding a steric number of 4 ($$3 \text{ bonding pairs} + 1 \text{ lone pair}$$). While an ideal tetrahedral geometry possesses a bond angle of $$109.5^\circ$$, the strong lone pair-bonding pair (lp-bp) repulsions compress the internal $$\text{H--N--H}$$ bond angle down to $$\sim 107^\circ$$.




• 3. $$\text{SiH}_4$$ (Silane)

  The silicon atom is $$sp^3$$ hybridized and bonded to four hydrogen atoms with zero lone pairs. Because the electronic configuration around the central atom is perfectly symmetric, it forms a regular tetrahedral geometry with no lone-pair distortions. The internal $$\text{H--Si--H}$$ bond angle stands exactly at the ideal value of $$109.5^\circ$$.




• 4. $$\text{BF}_3$$ (Boron Trifluoride)

  The boron atom features $$sp^2$$ hybridization with three bond pairs and zero lone pairs. This configuration produces a perfectly symmetrical trigonal planar geometry where the internal $$\text{F--B--F}$$ bond angles are maximize at exactly $$120^\circ$$, representing the largest angle in this selection.

Arranging the molecules by their bond angles from smallest to largest gives:

$$\text{H}_2\text{S } (\sim 92^\circ) < \text{NH}_3\text{ } (\sim 107^\circ) < \text{SiH}_4\text{ } (109.5^\circ) < \text{BF}_3\text{ } (120^\circ)$$

Correct Option: C — $$\text{H}_2\text{S} < \text{NH}_3 < \text{SiH}_4 < \text{BF}_3$$