COMEDK 2025 Afternoon Solved Paper

Concept:The stability of a diatomic anion depends on whether the added electron occupies a bonding or antibonding molecular orbital, which changes the bond order.Explanation:For $\mathrm{C}_2$ (8 valence electrons), the MO configuration is:$\sigma_{2s}^2$ $\sigma_{2s}^*2$ $\pi_{2p_x}^2$ $\pi_{2p_y}^2$.Bond order = $\frac{6 - 2}{2} = 2$.Adding one electron to form $\mathrm{C}_2^-$ (9 valence electrons) places it in $\sigma_{2p_z}$, a bonding orbital.New bond order = $\frac{7 - 2}{2} = 2.5$ → increased bond order and greater stability.For $\mathrm{N}_2$ (10 valence electrons), bond order = 3.The extra electron in $\mathrm{N}_2^-$ goes into $\pi^*2p$ (antibonding), lowering bond order to 2.5 → less stable.For $\mathrm{O}_2$ (12 valence electrons), bond order = 2.$\mathrm{O}_2^-$ adds electron to $\pi^*2p$, bond order becomes 1.5 → less stable.For $\mathrm{F}_2$ (14 valence electrons), bond order = 1.$\mathrm{F}_2^-$ adds electron to $\pi^*2p$, bond order becomes 0.5 → less stable.Only $\mathrm{C}_2$ has an extra electron entering a bonding orbital, increasing bond order and stability.Answer:Option A, $\mathrm{C}_2$.