Question

Identify the pair in which the difference in bond order value is maximum.

• A. \( O_2^-, O_2^+ \)
• B. \( O_2^{2-}, O_2^{2+} \)
• C. \( O_2, O_2^{2+} \)
• D. \( O_2^+, O_2^{2+} \)

Hint:

Bond order decreases with the addition of electrons to antibonding orbitals.
- Higher bond order implies stronger and shorter bonds.

Answer 1

The Correct Option is B

The bond order of an oxygen molecule and its ions can be calculated using: \[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \] where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons. 1. Bond Order Calculations: - \( O_2 \) (neutral) = 2.0 - \( O_2^+ \) = 2.5 - \( O_2^- \) = 1.5 - \( O_2^{2+} \) = 3.0 - \( O_2^{2-} \) = 1.0 2. Difference in Bond Order: - \( O_2^{2-} \) vs \( O_2^{2+} \) \[ \Delta B.O = 3.0 - 1.0 = 2.0 \] - This is the maximum difference among the given pairs. Thus, the correct answer is \(\boxed{O_2^{2-}, O_2^{2+}}\).