Match List I with List II List I List II A. [Co(NH_3)_6]^3+ I. sp^3 B…

The correct answer is Option (1) → A-III, B-IV, 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. \([Co(NH_3)_6]^{3+}\) — This matches III. d²sp³. Co (Z = 27) is [Ar]3d⁷4s²; removing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. \( \ce{NH3} \) pairs these six electrons in three orbitals, 3d: [↑↓][↑↓][↑↓][ ][ ], leaving two 3d orbitals empty for d²sp³ (inner orbital, diamagnetic).

• B. \([CoF_6]^{3-}\) — This matches IV. sp³d². \( \ce{Co^{3+}} \) is again 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).

• 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: A-III appears in Options (1) and (2); of these only Option (1) has C-I.

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