COMEDK 2025 Morning Solved Paper

Concept:The stability of a diatomic molecule or ion is measured by its bond order, and magnetic character depends on the number of unpaired electrons. The correct pair must show an increase in bond order on forming the monovalent ion while keeping the same number of unpaired electrons.Explanation:Using Molecular Orbital theory, we examine each option.For $\mathrm{O}_2 \to \mathrm{O}_2^+$:$\mathrm{O}_2$ valence MO configuration: $\sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2, \pi_{2p_x}^{*1}, \pi_{2p_y}^{*1}$.Bond order = $\frac{8-4}{2}=2$; two unpaired electrons → paramagnetic.$\mathrm{O}_2^+$ removes one electron from a $\pi^*$ orbital.Bonding electrons remain 8; antibonding = $2(\sigma_{2s}^*) + 1(\pi^*) = 3$.Bond order = $\frac{8-3}{2}=2.5$ (higher than 2, so stability increases).One unpaired electron remains → still paramagnetic (magnetic character unchanged).Now check the other options:$\mathrm{C}_2 / \mathrm{C}_2^+$: bond order drops from 2 to 1.5, magnetic changes from diamagnetic to paramagnetic.$\mathrm{B}_2 / \mathrm{B}_2^+$: bond order drops from 1 to 0.5, both paramagnetic but stability decreases.$\mathrm{N}_2 / \mathrm{N}_2^+$: bond order drops from 3 to 2.5, magnetic changes from diamagnetic to paramagnetic.Only $\mathrm{O}_2 / \mathrm{O}_2^+$ satisfies both conditions.Answer:B. $\mathrm{O}_2 / \mathrm{O}_2^+$