Question

Identify the molecule (X) with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among HNO₃, H₂SO₄, NF₃, and O₃. Choose the correct bond angle made by the central atom of the molecule (X).

• A. 116°
• B. 120°
• C. 107°
• D. 102°

Hint:

In NF₃, the bond angle (102°) is smaller than in NH₃ (107°) because the highly electronegative F atoms pull the bonding electrons away from the central atom, reducing bond-pair bond-pair repulsion.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To identify molecule (X), we must count the total number of lone pairs on all atoms in the Lewis structures of the given molecules. Then, we determine the geometry and bond angle of that specific molecule.

Step 2: Key Formula or Approach:
1. Total lone pairs = (Sum of lone pairs on central atom and all peripheral atoms).
2. VSEPR theory determines the bond angle.
Step 3: Detailed Explanation:
- HNO₃: Total 8 lone pairs (1 on N, 2 on each double-bonded O, 3 on single-bonded O).
- H₂SO₄: Total 8 lone pairs (2 on each double-bonded O, 2 on each -OH oxygen).
- O₃: Total 6 lone pairs.
- NF₃: Total 10 lone pairs (1 on N and 3 on each of the 3 F atoms).
Thus, (X) is NF₃.
In NF₃, Nitrogen has 3 bond pairs and 1 lone pair ($sp^3$ hybridized). Due to high electronegativity of Fluorine, the lone pair-bond pair repulsion is significant, and the bond pairs are pulled closer to F, reducing the F-N-F angle significantly below the tetrahedral 109.5°. The experimental value is approximately 102°.
Step 4: Final Answer:
The molecule (X) is NF₃ and the bond angle is 102°.