Match List I with List II Choose the correct answer from the options given below:
- A-II, B-III, C-I, D-IV
- A-III, B-I, C-II, D-IV
- A-IV, B-I, C-III, D-II
- A-III, B-IV, C-II, D-I
Answer
(2) A-III, B-I, C-II, D-IV
Solution
(A): SO2Cl2 — S: 4 σ bonds, no lp (sp3) ⇒ (A) is III (tetrahedral)
(B): NO — 15 electrons (odd) ⇒ one unpaired e- ⇒ (B) is I (paramagnetic)
(C): NO2- — 24 valence electrons, all paired ⇒ (C) is II (diamagnetic)
(D): I3- — 2 bp + 3 lp (sp3d, lp equatorial) ⇒ (D) is IV (linear)
∴ A-III, B-I, C-II, D-IV
Answer: (2)