Question

Which of the following species have same bond order? I. \( \mathrm{H_2} \)
II. \( \mathrm{B_2} \)
III. \( \mathrm{O_2^{2-}} \)
IV. \( \mathrm{Be_2} \)
V. \( \mathrm{N_2} \)
Correct option is:

• A. I and II
• B. I, II and III
• C. I, II and IV
• D. I, IV and V

Hint:

Always remember: \( \mathrm{O_2^{2-}} \) has bond order 1 (like single bond).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Bond order is calculated based on Molecular Orbital (MO) Theory. It indicates the stability of a molecule.
Step 3: Detailed Explanation:
1. \(\text{H}_{2}\) (\(2e^{-}\)): \((\sigma 1s)^{2}\).
\[ BO = \frac{1}{2}(2 - 0) = 1 \]
2. \(\text{B}_{2}\) (\(10e^{-}\)): \((\sigma 1s)^{2} (\sigma^* 1s)^{2} (\sigma 2s)^{2} (\sigma^* 2s)^{2} (\pi 2p_x)^{1} (\pi 2p_y)^{1}\).
\[ BO = \frac{1}{2}(6 - 4) = 1 \]
3. \(\text{O}_{2}^{2-}\) (\(18e^{-}\)): Peroxide ion has configuration like \(F_{2}\).
\[ BO = \frac{1}{2}(10 - 8) = 1 \]
4. \(\text{Be}_{2}\) (\(8e^{-}\)): \((\sigma 1s)^{2} (\sigma^* 1s)^{2} (\sigma 2s)^{2} (\sigma^* 2s)^{2}\).
\[ BO = \frac{1}{2}(4 - 4) = 0 \]
5. \(\text{N}_{2}\) (\(14e^{-}\)):
\[ BO = \frac{1}{2}(10 - 4) = 3 \]
Thus, species I, II, and III all have a bond order of 1.
Step 4: Final Answer:
The correct option is (B).