Concept: Bond angles are influenced by the electronegativity of the surrounding atoms and the presence of lone pairs. More electronegative atoms tend to pull electron density away from the central atom, reducing the repulsion between bonding pairs and thus decreasing the bond angle. Conversely, less electronegative atoms or groups allow electron density to remain closer to the central atom, increasing repulsion and potentially increasing the bond angle, or if they are bulky, they can increase steric repulsion.

Why (B) is correct:

In the ${\mathrm{FBrO}}_3$ molecule, Bromine (Br) is the central atom. It is bonded to one Fluorine (F) atom and three Oxygen (O) atoms. The molecule has a trigonal pyramidal geometry around the central Br atom, with one lone pair and four bond pairs (one Br-F and three Br-O bonds). However, the question asks to compare two specific bond angles: $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}$ and $\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$.

1. Electronegativity: Fluorine (4.0) is significantly more electronegative than Oxygen (3.5). When Fluorine is bonded to Bromine, it pulls electron density from the Br-F bond pair more strongly towards itself compared to Oxygen pulling electron density from the Br-O bond pair.

2. Effect on Bond Angle: The electron density in the Br-F bond is shifted further away from the central Bromine atom. This reduces the repulsion between the Br-F bond pair and the Br-O bond pairs. Consequently, the $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}$ bond angle will be smaller.

3. In contrast, the electron density in the Br-O bonds is closer to the central Bromine atom (as Oxygen is less electronegative than Fluorine). This leads to greater repulsion between the Br-O bond pairs, resulting in a larger $\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$ bond angle.

Therefore, $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}<\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$.

Option Analysis:

• A) $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}>\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$: This is incorrect because the higher electronegativity of F reduces the bond angle involving F.
• B) $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}<\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$: This is correct due to the greater electronegativity of Fluorine compared to Oxygen, which pulls electron density away from the central Br atom, reducing the repulsion and thus the bond angle involving F.
• C) $\mathrm{F}-\widehat{\mathrm{Br}}-\mathrm{O}=\mathrm{O}-\widehat{\mathrm{Br}}-\mathrm{O}$: This is incorrect as the different electronegativities of F and O will lead to different bond angles.
• D) None of these: This is incorrect as option B correctly describes the relationship.

Correct Answer: (B)