Determine the geometry and bond angle of each species

We compare the bond angles using VSEPR theory.

(i) $BF_3$

• Central atom: $B$
• Valence electrons on $B$: 3
• It forms 3 bond pairs and has 0 lone pairs.
• Geometry: trigonal planar
• Bond angle: $\angle FBF = 120^\circ$

(ii) $NH_3$

• Central atom: $N$
• Valence electrons on $N$: 5
• It forms 3 bond pairs and has 1 lone pair.
• Electron pair geometry: tetrahedral
• Molecular shape: trigonal pyramidal
• Lone pair-bond pair repulsion compresses the ideal tetrahedral angle $109.5^\circ$.
• Bond angle: $\angle HNH \approx 107^\circ$

(iii) $PF_3$

• Central atom: $P$
• Valence electrons on $P$: 5
• It forms 3 bond pairs and has 1 lone pair.
• Shape: trigonal pyramidal
• Since $F$ is highly electronegative, bond pair electrons are drawn away from $P$, so bond pair-bond pair repulsion decreases and the bond angle becomes smaller than in $NH_3$.
• Thus: $\angle FPF < \angle HNH$
• Approximate bond angle: $\angle FPF \approx 97^\circ$

(iv) $I_3^-$

• Total valence electrons: $3\times 7 + 1 = 22$
• Central iodine has 5 electron pairs around it: 2 bond pairs and 3 lone pairs.
• Electron pair geometry: trigonal bipyramidal
• The 3 lone pairs occupy equatorial positions.
• Molecular shape: linear
• Bond angle: $\angle I-I-I = 180^\circ$

Arrange in decreasing order of bond angle

From the above values: $I_3^- = 180^\circ$ $BF_3 = 120^\circ$ $NH_3 \approx 107^\circ$ $PF_3 \approx 97^\circ$

Therefore, the decreasing order is: $I_3^- > BF_3 > NH_3 > PF_3$

Match with the given options

This corresponds to Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

Hence, they agree.