Question

The percentage s-character of the hybrid orbitals in\( No^+_2\), \(NO^-_3\) and\( NH^+_4\) respectively are

• A. 25%, 50%, 33.3%
• B. 33.3%, 50%, 25%
• C. 33.3%, 25%, 50%
• D. 50%, 33.3%, 25%

Answer 1

The Correct Option is D

To determine the percentage of s-character in the hybrid orbitals, we must analyze the type of hybridization for each molecule.
1. For \( \text{NO}_2^+ \): \( \text{NO}_2^+ \) has a linear geometry with nitrogen having 2 bonding pairs and no lone pairs of electrons, implying \( sp \) hybridization. In \( sp \) hybridization, the s-character is 50%.
2. For \( \text{NO}_3^- \): \( \text{NO}_3^- \) has a trigonal planar geometry with nitrogen having 3 bonding pairs and no lone pairs, implying \( sp^2 \) hybridization. In \( sp^2 \) hybridization, the s-character is 33.3%.
3. For \( \text{NH}_4^+ \): \( \text{NH}_4^+ \) has a tetrahedral geometry with nitrogen having 4 bonding pairs and no lone pairs, implying \( sp^3 \) hybridization. In \( sp^3 \) hybridization, the s-character is 25%.
Thus, the percentage of s-character in the hybrid orbitals of nitrogen in \( \text{NO}_2^+ \), \( \text{NO}_3^- \), and \( \text{NH}_4^+ \) are 50%, 33.3%, and 25%, respectively.

The correct option is (D): 50%, 33.3%, 25%