Kerala KEAM 5 June 2024 Solved Paper

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