Match List I with List II
| List I Complex | List II Hybridisation, Spin state |
| A. [CoF6]3– | I. d2sp3, low spin |
| B. [NiCl4]2– | II. sp3d2, high spin |
| C. [Co(NH3)6]3+ | III. sp3, high spin |
| D. [Ni(CO)4] | IV. sp3, no unpaired electron |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-III, C-I, D-IV
In valence bond theory, a weak field ligand leaves the d electrons unpaired, so the metal uses its outer nd orbitals (sp3d2, high spin, outer orbital complex). A strong field ligand forces the d electrons to pair, freeing inner (n−1)d orbitals (d2sp3, low spin, inner orbital complex). Four-coordinate complexes of weak field ligands are usually tetrahedral (sp3).
A. [CoF6]3– — This matches II. sp3d2, high spin. \( \ce{Co^{3+}} \) is 3d6 (Co: [Ar]3d74s2, minus three electrons); \( \ce{F^{-}} \) is a weak field ligand, so the four unpaired 3d electrons stay unpaired and 4s, 4p, 4d orbitals are used.
B. [NiCl4]2– — This matches III. sp3, high spin. \( \ce{Ni^{2+}} \) is 3d8; \( \ce{Cl^{-}} \) is weak field, so the two unpaired electrons remain and the complex is tetrahedral.
C. \( \ce{[Co(NH3)6]^{3+}} \) — This matches I. d2sp3, low spin. \( \ce{Co^{3+}} \) is 3d6; \( \ce{NH3} \) pairs all six electrons into three 3d orbitals, leaving two 3d orbitals for bonding (diamagnetic).
D. \( \ce{[Ni(CO)4]} \) — This matches IV. sp3, no unpaired electron. Ni(0) is 3d84s2; CO pushes the 4s electrons into 3d, giving 3d10, and the empty 4s and 4p orbitals form tetrahedral sp3 bonds.
Shortcut: The pair C-I appears in Options (1) and (2), and A-II then leaves only Option (2).