Match List I with List II

Choose the correct answer from the options given below:

• (A) $$\text{SO}_2\text{Cl}_2$$ (Sulfuryl chloride): The central sulfur atom has 4 bonding domains and 0 lone pairs ($$sp^3$$ hybridized), giving it a tetrahedral shape.
  $$\implies \text{(A)} \rightarrow \text{(III)}$$
• (B) $$\text{NO}$$ (Nitric oxide): An odd-electron molecule with 11 valence electrons, meaning it contains an unpaired electron and is paramagnetic.
  $$\implies \text{(B)} \rightarrow \text{(I)}$$
• (C) $$\text{NO}_2^\ominus$$ (Nitrite ion): Contains 18 valence electrons (all electrons are paired), making it diamagnetic.
  $$\implies \text{(C)} \rightarrow \text{(II)}$$
• (D) $$\text{I}_3^\ominus$$ (Triiodide ion): The central iodine atom features 2 bond pairs and 3 lone pairs ($$sp^3d$$ hybridized), arranging the atoms in a linear geometry.
  $$\implies \text{(D)} \rightarrow \text{(IV)}$$

Conclusion:

Combining these molecular properties yields the final matched sequence.

Answer: Option B — A-III, B-I, C-II, D-IV