Match the complexes given in List-I with oxidation state and the type…

The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

The oxidation state of the metal comes from a charge balance on the complex. By valence bond theory, a complex that uses the inner (n−1)d orbitals for hybridisation (d²sp³, dsp²) is an inner orbital complex; one that uses the outer ns, np (and nd) orbitals (sp³, sp³d²) is an outer orbital complex. Strong field ligands such as \( \ce{CN^{-}} \) (and \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) pair up the d electrons and free inner d orbitals; weak field ligands such as \( \ce{F^{-}} \) and \( \ce{Cl^{-}} \) do not.

• (A) \([Co(NH_3)_6]^{3+}\) — This matches (IV) +3 and Inner orbital complex. \( \ce{NH3} \) is neutral, so Co is +3. Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron gives \( \ce{Co^{3+}} \): 3d⁶. The six electrons pair up in three 3d orbitals, the two free 3d orbitals take part in d²sp³ hybridisation, and the complex is diamagnetic.

• (B) \([CoF_6]^{3-}\) — This matches (III) +3 and Outer orbital complex. x + 6(−1) = −3 gives x = +3, again \( \ce{Co^{3+}} \): 3d⁶. \( \ce{F^{-}} \) is a weak field ligand, so the four unpaired electrons stay unpaired and the 4s, 4p and 4d orbitals are used: sp³d², paramagnetic.

• (C) \([NiCl_4]^{2-}\) — This matches (II) +2 and Outer orbital complex. x + 4(−1) = −2 gives x = +2. Ni (Z = 28) is [Ar]3d⁸4s²; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d⁸. \( \ce{Cl^{-}} \) cannot pair the two unpaired electrons, so the outer 4s and 4p orbitals are used: sp³, tetrahedral, paramagnetic.

• (D) \([Ni(CN)_4]^{2-}\) — This matches (I) +2 and Inner orbital complex. x + 4(−1) = −2 gives \( \ce{Ni^{2+}} \): 3d⁸. \( \ce{CN^{-}} \) is a strong field ligand; the two unpaired electrons pair up, one 3d orbital is freed and dsp² hybridisation gives a square planar, diamagnetic complex.

Shortcut: (A)-(IV) appears only in Option (2).