Determine the shapes and bond angles of each molecule

(i) $\mathrm{BF}_3$

• Central atom: $\mathrm{B}$
• Valence electrons on $\mathrm{B}$ = 3
• It forms 3 bond pairs and has no lone pair.
• According to VSEPR, geometry is trigonal planar.

Therefore, $\angle FBF = 120^\circ$

(ii) $\mathrm{PF}_3$

• Central atom: $\mathrm{P}$
• Valence electrons on $\mathrm{P}$ = 5
• It forms 3 bond pairs and has 1 lone pair.
• Electron pair geometry: tetrahedral
• Molecular shape: trigonal pyramidal

Due to lone pair-bond pair repulsion, bond angle becomes less than tetrahedral angle $109.5^\circ$. For $\mathrm{PF}_3$, the bond angle is about $\angle FPF \approx 97^\circ$

(iii) $\mathrm{ClF}_3$

• Central atom: $\mathrm{Cl}$
• Valence electrons on $\mathrm{Cl}$ = 7
• It forms 3 bond pairs and has 2 lone pairs.
• Total electron pairs = 5, so trigonal bipyramidal electron pair arrangement.
• The two lone pairs occupy equatorial positions.
• Molecular shape becomes T-shaped.

In a T-shaped molecule, the relevant $\angle FClF$ is slightly less than $90^\circ$ (and one angle is close to $180^\circ$). The smaller bond angle is taken for comparison here: $\angle FClF < 90^\circ \quad (\text{about } 87.5^\circ)$

Compare the bond angles

We have: $\mathrm{ClF}_3 \approx 87.5^\circ$ $\mathrm{PF}_3 \approx 97^\circ$ $\mathrm{BF}_3 = 120^\circ$

Hence, the increasing order is $\mathrm{ClF}_3 < \mathrm{PF}_3 < \mathrm{BF}_3$

Match with the given options

This corresponds to: Option C