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

Answer: (C)

$\text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3$

Answer 2

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$_3$.