Question

The sum of bond order of \(O_2\), \(O_2^-\), \(O_2^{2-}\) and \(O_2^+\) is equal to

• A. 5
• B. 4
• C. 6
• D. 9

Hint:

A useful shortcut for bond orders of O₂ and its ions: O₂ has a bond order of 2. Adding an electron (to an antibonding orbital) decreases the bond order by 0.5. Removing an electron (from an antibonding orbital) increases it by 0.5. O₂⁺: 2 + 0.5 = 2.5 O₂: 2.0 O₂⁻: 2 - 0.5 = 1.5 O₂²⁻: 2 - 1.0 = 1.0

Answer 1

The Correct Option is D

Step 1: Calculate Bond Orders: - \( O_2 \) (16 electrons): Bond Order = 2.0 - \( O_2^+ \) (15 electrons): Bond Order = 2.5 (Removed antibonding electron) - \( O_2^- \) (17 electrons): Bond Order = 1.5 (Added antibonding electron) - \( O_2^{2+} \) (14 electrons, like \( N_2 \)): Bond Order = 3.0
Step 2: Summation: Sum = \( 2.5 + 1.5 + 2.0 + 3.0 = 9.0 \)