Question

The correct increasing order for bond angles among \( \text{BF}_3, \, \text{PF}_3, \, \text{and} \, \text{CF}_3 \) is:

• A. \( \text{PF}_3 \, < \, \text{BF}_3 \, < \, \text{CF}_3 \)
• B. \( \text{BF}_3 \, < \, \text{PF}_3 \, < \, \text{CF}_3 \)
• C. \( \text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3 \)
• D. \( \text{BF}_3 \, = \, \text{PF}_3 \, < \, \text{CF}_3 \)

Answer 1

The Correct Option is C

BF$_3$: Planar structure with 120$^\circ$ bond angles ($sp^2$ hybridization).
PF$_3$: Tetrahedral geometry distorted by lone pair on phosphorus, bond angle $<$ 109.5$^\circ$.
CF$_3$: Tetrahedral geometry with strong electron-withdrawing fluorine atoms, bond angle $\sim$ 104$^\circ$.
The order of bond angles is CF$_3$ $<$ PF$_3$ $<$ BF$_3$.