Match List-I with List-II List-I Complex List-II hybridisation (A) [C…

The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

In valence bond theory the metal ion uses hybrid orbitals to accept the ligand electron pairs. A strong-field ligand (\( \ce{NH3} \), \( \ce{CN^{-}} \)) pairs up the d electrons and frees inner d orbitals (inner orbital complex); a weak-field ligand (\( \ce{F^{-}} \), \( \ce{Cl^{-}} \)) does not, so outer orbitals are used. Six pairs need six orbitals (d²sp³ or sp³d², octahedral); four pairs need four (dsp², square planar, or sp³, tetrahedral).

• (A) \([CoF_6]^{3-}\) — This matches (IV) sp³d². Co (Z = 27) is [Ar]3d⁷4s², so \( \ce{Co^{3+}} \) is 3d⁶, but \( \ce{F^{-}} \) is weak, so 3d: [↑↓][↑][↑][↑][↑] (4 unpaired); no 3d orbital is free and the 4s, 4p and two 4d orbitals are used: sp³d² (outer orbital, high spin).

• (B) \([Co(NH_3)_6]^{3+}\) — This matches (III) d²sp³. \( \ce{Co^{3+}} \) is again 3d⁶ (the two 4s electrons and one 3d electron removed), and \( \ce{NH3} \) pairs these six electrons in three orbitals, 3d: [↑↓][↑↓][↑↓][ ][ ], leaving two 3d orbitals empty for d²sp³ (inner orbital, diamagnetic).

• (C) \([NiCl_4]^{2-}\) — This matches (I) sp³. Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸, 3d: [↑↓][↑↓][↑↓][↑][↑] (2 unpaired). \( \ce{Cl^{-}} \) is weak, so the 4s and three 4p orbitals are used: sp³, tetrahedral, paramagnetic.

• (D) \([Ni(CN)_4]^{2-}\) — This matches (II) dsp². \( \ce{Ni^{2+}} \) is 3d⁸, but strong \( \ce{CN^{-}} \) pairs the two odd electrons: 3d: [↑↓][↑↓][↑↓][↑↓][ ]. One 3d orbital is empty, so dsp² hybridisation, square planar, diamagnetic.

Shortcut: The pair (A)–(IV) appears only in Option (4).

Note: The same/similar question has been asked in different years by NTA as per the details given below: