Question

The correct bond order in the following species is :

• A. ${O_2^+ < O_2^- < O_2^{2+} }$
• B. ${O_2^- < O_2^+ < O_{2}^{2+} }$
• C. ${O_2^{2+} < O_2^+ < O_2^- }$
• D. ${O_2^{2+} < O_2^- < O_2^+ }$

Answer 1

The Correct Option is B

To find the correct order of bond order for various oxygen species, we need to use the Molecular Orbital Theory (MOT). The bond order is calculated using the formula: 

\(BO = \frac{(N_b - N_a)}{2}\)

where \(N_b\) is the number of electrons in bonding molecular orbitals and \(N_a\) is the number of electrons in antibonding molecular orbitals.

1. For \(O_2^+\):
   • Electron configuration: \(\sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1\)
   • Bonding electrons \((N_b) = 10\)
   • Antibonding electrons \((N_a) = 5\)
   • Bond order: \(BO = \frac{10 - 5}{2} = 2.5\)
2. For \(O_2^-\\):
   • Electron configuration: \(\sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^2\)
   • Bonding electrons \((N_b) = 10\)
   • Antibonding electrons \((N_a) = 6\)
   • Bond order: \(BO = \frac{10 - 6}{2} = 2.0\)
3. For \(O_2^{2+}\):
   • Electron configuration: \(\sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^0\)
   • Bonding electrons \((N_b) = 10\)
   • Antibonding electrons \((N_a) = 4\)
   • Bond order: \(BO = \frac{10 - 4}{2} = 3.0\)

Therefore, the correct order of bond order is: \({O_2^- < O_2^+ < O_{2}^{2+}}\)