Question:

In which of the following sets, the sum of bond orders of the three species is maximum?

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Higher bond order corresponds to greater stability and stronger bonds.
Updated On: Mar 24, 2025
  • \( B_2, CN^-, O_2^{2-} \)
  • \( O_2, F_2, O_2^+ \)
  • \( O_2, N_2, O_2^+ \)
  • \( C_2, O_2, He_2^{2+} \)
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The Correct Option is C

Solution and Explanation

Step 1: Determining Bond Orders Using Molecular Orbital Theory \[ BO = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \] For the given species: - \( O_2 \) → \( BO = 2 \) - \( N_2 \) → \( BO = 3 \) - \( O_2^+ \) → \( BO = 2.5 \)
Step 2: Calculating Total Bond Order For option (3): \[ BO_{\text{O}_2} + BO_{\text{N}_2} + BO_{\text{O}_2^+} = 2 + 3 + 2.5 = 7.5 \] This is the highest among all the options.
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