To determine the bond order of O\u2082\u207b, we first need to write its molecular orbital (MO) electronic configuration.

Step 1: Determine the total number of electrons.
An oxygen atom has 8 electrons. So, O\u2082 has $2\times 8=16$ electrons. For O\u2082\u207b, there is an additional electron, making the total number of electrons $16+1=17$.

Step 2: Write the molecular orbital configuration.
The MO configuration for molecules with more than 14 electrons is:
${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2)({\pi }_{2p_x}^{\ast 2}={\pi }_{2p_y}^{\ast 1})$
Filling 17 electrons: $2+2+2+2+2+4+3=17$

Step 3: Calculate the bond order.
The bond order (BO) is calculated using the formula:
$BO=\frac{1}{2}(N_b-N_a)$
Where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.

From the MO configuration:
$N_b=2({\sigma }_{1s})+2({\sigma }_{2s})+2({\sigma }_{2p_z})+4({\pi }_{2p_x},{\pi }_{2p_y})=10$
$N_a=2({\sigma }_{1s}^{\ast })+2({\sigma }_{2s}^{\ast })+3({\pi }_{2p_x}^{\ast },{\pi }_{2p_y}^{\ast })=7$

Therefore, $BO=\frac{1}{2}(10-7)=\frac{1}{2}(3)=1.5$

Option Analysis:

• A) 0.5: This would imply a very weak bond, typically found in highly unstable species. Incorrect.
• B) 1.5: This is the correct bond order for O\u2082\u207b, indicating a bond strength between a single and a double bond.
• C) 3.5: This bond order is too high; it would imply a very strong bond, stronger than a triple bond, which is not possible for O\u2082\u207b. Incorrect.
• D) 2.5: This bond order is for species like N\u2082\u207a or O\u2082\u207a, not O\u2082\u207b. Incorrect.

Correct Answer: (B)