Match List-I with List-IIList-ICOORDINATION COMPOUNDList-IIOxidation…

The correct answer is Option (3) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)

For each complex, find the oxidation state of the metal from the charges, the coordination number from the donor atoms (a bidentate ligand such as \( \ce{C2O4^{2-}} \) or en counts twice), and the d electron count. In an octahedral field the d orbitals split into t_2g (lower) and e_g (higher). A strong field ligand pairs the electrons in t_2g (low spin); a weak field ligand puts electrons into e_g before pairing (high spin). For d¹–d³ ions all electrons go into t_2g either way.

• (A) \( \ce{K3[Co(C2O4)3]} \) — This matches (II) O.S.= +3, C.N.= 6, t_2g⁶e_g⁰. Cobalt is +3 (x − 6 = −3), three bidentate oxalates give C.N. 6, Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron gives Co³⁺: 3d⁶, which is low spin with oxalate, so all six electrons pair in t_2g.

• (B) cis-\( \ce{[CrCl2(en)2]Cl} \) — This matches (I) O.S.= +3, C.N.= 6, t_2g³. Chromium is +3 (x − 2 = +1), two \( \ce{Cl^{-}} \) and two bidentate en give C.N. 6, Cr (Z = 24) is [Ar]3d⁵4s¹; losing the 4s electron and two 3d electrons gives Cr³⁺: 3d³, and these three electrons occupy t_2g singly.

• (C) \( \ce{K3[Fe(CN)6]} \) — This matches (IV) O.S.= +3, C.N.= 6, t_2g⁵e_g⁰. Iron is +3 (x − 6 = −3). Fe (Z = 26) is [Ar]3d⁶4s²; losing the two 4s electrons and one 3d electron gives Fe³⁺: 3d⁵, and the strong field CN⁻ ligands keep all five electrons in t_2g (low spin, one unpaired electron).

• (D) \( \ce{[Mn(H2O)6]SO4} \) — This matches (III) O.S.= +2, C.N.= 6, t_2g³e_g². Manganese is +2 (the cation balances \( \ce{SO4^{2-}} \)). Mn (Z = 25) is [Ar]3d⁵4s²; losing the two 4s electrons gives Mn²⁺: 3d⁵, and weak field H₂O gives the high spin arrangement with five unpaired electrons.

Shortcut: (D) is the only complex with the metal in the +2 state, so (D)–(III); this pair appears only in Option (3).