Question

Among the following species, which one has the highest bond order according to Molecular Orbital Theory?

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

Hint:

Removing electrons from antibonding orbitals increases bond order, while adding electrons into antibonding orbitals decreases bond order.

Answer 1

The Correct Option is B

Concept: According to Molecular Orbital Theory (MOT), bond order is given by: \[ \text{Bond Order}=\frac{N_b-N_a}{2} \] where:
• \(N_b\) = Number of electrons in bonding molecular orbitals
• \(N_a\) = Number of electrons in antibonding molecular orbitals A higher bond order indicates a stronger and shorter bond.

Step 1: Bond order of \(O_2\). For oxygen molecule: \[ BO=\frac{10-6}{2}=2 \] Thus, \[ BO(O_2)=2 \]

Step 2: Bond order of \(O_2^+\). Removal of one electron occurs from antibonding \(\pi^*\) orbital. Antibonding electrons decrease by one. \[ BO=\frac{10-5}{2}=2.5 \]

Step 3: Bond order of \(O_2^-\). One electron is added into antibonding orbital. \[ BO=\frac{10-7}{2}=1.5 \]

Step 4: Bond order of \(O_2^{2-}\). Two electrons are added into antibonding orbital. \[ BO=\frac{10-8}{2}=1 \]

Step 5: Compare all values. \[ O_2^+ (2.5) > O_2 (2) > O_2^- (1.5) > O_2^{2-}(1) \] Hence the highest bond order belongs to: \[ O_2^+ \] Therefore option (B) is correct.