Identify the molecule(X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO₃$HNO_{3}$, H₂SO₄$H_{2}SO_{4}$, NF₃$NF_{3}$ and O₃$O_{3}$. Choose the correct bond angle made by the central atom of the molecule (X).

Core Logic

Draw Lewis structures for all given molecules and sum the total lone pairs on all atoms. HNO₃$HNO_3$: 7 lone pairs total. H₂SO₄$H_2SO_4$: 8 lone pairs total. O₃$O_3$: 6 lone pairs total. NF₃$NF_3$: Nitrogen has 1 lone pair, and each of the three fluorine atoms has 3 lone pairs. Total = 1 + (3 × 3) = 10$1 + (3 \times 3) = 10$ lone pairs.

Step 1: Identifying Molecule X

Molecule X is NF₃$NF_3$ due to possessing the maximum number of lone pairs (10).

Step 2: Bond Angle Analysis

In NF₃$NF_3$, Nitrogen is sp³$sp^3$ hybridized. Due to the high electronegativity of Fluorine, the bond pair electron density shifts towards fluorine. This reduces bond pair-bond pair repulsion around the central nitrogen compared to ammonia (NH₃$NH_3$). As a result, the lone pair compresses the F-N-F bond angle more severely than in NH₃$NH_3$ (107^°$107^\circ$). The resulting F-N-F bond angle is approximately 102^°$102^\circ$.

Pattern Recognition

Electronegativity rules: If the surrounding atoms are more electronegative than the central atom, bond angles decrease because bonding electrons are pulled away from the central atom, allowing the lone pair to expand further and crush the angle.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The p-Block Elements