Question

The bond order of \( O_2 \) molecule and its magnetic property are respectively:

• A. 3, paramagnetic
• B. 2, paramagnetic
• C. 3, diamagnetic
• D. 1.5, paramagnetic
• E. 2, diamagnetic

Hint:

In molecular orbital theory, bond order is used to predict the stability of a molecule. A higher bond order indicates greater stability. Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired.

Answer 1

The Correct Option is B

Step 1: The molecular orbital configuration of \( O_2 \) is given by:

\[ \sigma_{1s}^2, \, \sigma_{1s}^*{}^2, \, \sigma_{2s}^2, \, \sigma_{2s}^*{}^2, \, \sigma_{2p_z}^2, \, \pi_{2p_x}^1, \, \pi_{2p_y}^1. \] There are 12 electrons in the valence shell, and the molecular orbitals fill according to the Aufbau principle and Hund's rule.

Step 2: The bond order of a molecule is calculated using the formula:

\[ \text{Bond order} = \frac{1}{2} \left( \text{number of bonding electrons} - \text{number of antibonding electrons} \right). \] For \( O_2 \), the number of bonding electrons is 8, and the number of antibonding electrons is 4. Thus,

\[ \text{Bond order} = \frac{1}{2} \left( 8 - 4 \right) = 2. \] Step 3: Since there are unpaired electrons in the molecular orbitals (\( \pi_{2p_x}^1, \pi_{2p_y}^1 \)), the molecule exhibits paramagnetism.

Thus, the bond order of \( O_2 \) is 2, and it is paramagnetic.

Hence, the correct answer is option (B).