Question

Identify the pair of molecules in which the hybridization of the central atom is \( sp^2 \) with bent geometry

• A. \( \text{H}_2\text{O}, \text{SO}_2 \)
• B. \( \text{SO}_2, \text{O}_3 \)
• C. \( \text{H}_2\text{O}, \text{O}_3 \)
• D. \( \text{N}_2\text{O}, \text{H}_2\text{O} \)

Hint:

Formula for steric number: \( \frac{1}{2}(V + M - C + A) \). \( \text{SO}_2 \): \( \frac{1}{2}(6 + 0) = 3 \Rightarrow sp^2 \). \( \text{O}_3 \): Central O has 6 valence. Acts like \( \text{SO}_2 \). \( sp^2 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We need to find molecules with: 1. \( sp^2 \) hybridization (Steric Number = 3). 2. Bent geometry (meaning at least one lone pair).
Step 3: Detailed Explanation:

Analyze each molecule: 1. \( \text{H}_2\text{O} \): Oxygen has 6 valence e\(^-\). 2 bonds, 2 lone pairs. Steric No = 4. Hybridization \( sp^3 \). Bent shape. (Incorrect hybridization). 2. \( \text{SO}_2 \): Sulphur has 6 valence e\(^-\). Forms 2 double bonds (superficially), but sigma bonds = 2. Lone pair = 1. Total domains = 2 sigma + 1 lone pair = 3. Hybridization: \( sp^2 \). Shape: Bent (V-shape). (Matches criteria). 3. \( \text{O}_3 \) (Ozone): Central O has 6 valence e\(^-\). Forms double bond with one O, coordinate bond (single sigma) with other. Sigma bonds = 2. Lone pairs = 1. Total domains = 3. Hybridization: \( sp^2 \). Shape: Bent. (Matches criteria). 4. \( \text{N}_2\text{O} \): Linear. \( sp \) hybridization. Pair matching criteria: \( \text{SO}_2 \) and \( \text{O}_3 \).
Step 4: Final Answer:

The pair is \( \text{SO}_2, \text{O}_3 \).