A quick shortcut for 14-electron systems (like $\text{N}_2$ or $\text{O}_2^{2+}$): they always have a bond order of 3.0 and are diamagnetic.
Each addition or removal of an electron changes the bond order by 0.5.
Step 1: Recall the tools needed for matching. Every entry in List-I is a diatomic species whose bond order and magnetic behaviour follow from its molecular orbital electron configuration, using $\text{Bond Order} = \dfrac{N_b-N_a}{2}$ and the rule that any unpaired electron in an MO makes the species paramagnetic. Step 2: Work from a known anchor. Take $\text{C}_2$, whose configuration $\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^4$ gives $N_b=8$, $N_a=4$, so its bond order is 2 with all electrons paired, making it diamagnetic. This anchors one entry and lets the rest be checked against List-II by elimination rather than derived fully from scratch. Step 3: Cross-check the remaining species. For each remaining diatomic ion or molecule, count total electrons, fill the MO diagram in the correct order, keeping in mind the crossover of $\sigma_{2p}$ and $\pi_{2p}$ levels for lighter diatomics, read off $N_b$ and $N_a$, and note whether the highest filled level leaves unpaired electrons. Step 4: Line up the pairs against the options. Only one lettered arrangement is consistent with every bond order and magnetic nature worked out this way. Final answer: Option 4, A-III, B-I, C-II, D-IV.
