Question

The correct order of bond angles of H$_2$S, NH$_3$, BF$_3$, and SiH$_4$ is

• A. H$_2$S<NH$_3$<SiH$_4$<BF$_3$
• B. NH$_3$<H$_2$S<SiH$_4$<BF$_3$
• C. H$_2$S<SiH$_4$<NH$_3$<BF$_3$
• D. H$_2$S<NH$_3$<BF$_3$<SiH$_4$

Hint:

Lone pairs reduce bond angles due to greater repulsion compared to bonding pairs.

Answer 1

The Correct Option is A

- H$_2$S: Bent molecule, large lone pair repulsion, smallest bond angle ($\approx 92^\circ$)
- NH$_3$: Trigonal pyramidal with one lone pair, bond angle $\approx 107^\circ$
- SiH$_4$: Tetrahedral, bond angle $\approx 109.5^\circ$
- BF$_3$: Trigonal planar, bond angle exactly $120^\circ$
Hence, the order: H$_2$S<NH$_3$<SiH$_4$<BF$_3$