To determine the bond order of the O\u2082 molecule, 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 molecule has $2\times 8=16$ electrons.

Step 2: Write the molecular orbital configuration.
For molecules with more than 14 electrons (like O\u2082), the order of MOs is:
${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2({\pi }_{2p_x}^2={\pi }_{2p_y}^2)({\pi }_{2p_x}^{\ast 1}={\pi }_{2p_y}^{\ast 1})$

Step 3: Calculate the bond order.
Bond order $=\frac{1}{2}(N_b-N_a)$, 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 configuration:
$N_b=2({\sigma }_{1s})+2({\sigma }_{2s})+2({\sigma }_{2p_z})+2({\pi }_{2p_x})+2({\pi }_{2p_y})=10$
$N_a=2({\sigma }_{1s}^{\ast })+2({\sigma }_{2s}^{\ast })+1({\pi }_{2p_x}^{\ast })+1({\pi }_{2p_y}^{\ast })=6$

Bond order $=\frac{1}{2}(10-6)=\frac{1}{2}(4)=2$

Option Analysis:

• A) 1.5: This would imply a bond order between a single and a double bond, which is incorrect for O\u2082.
• B) 2: This is the correct bond order for O\u2082, indicating a double bond.
• C) 1: This would imply a single bond, which is incorrect for O\u2082.
• D) 3: This would imply a triple bond, which is incorrect for O\u2082.

Correct Answer: (B)