Match the LIST-I with LIST-II.LIST-I(Coordination Complex)LIST-II(Cha…

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

The d electron count of the metal ion and the strength of the ligand field decide the arrangement of electrons and the shape of a complex. In an octahedral field, the d orbitals split into t_2g (lower) and e_g (higher). A strong field ligand (\( \ce{CN^{-}} \), CO, \( \ce{NH3} \) with \( \ce{Co^{3+}} \)) makes the splitting larger than the pairing energy, so the electrons pair up in t_2g. A weak field ligand (\( \ce{F^{-}} \)) gives a small splitting, so the electrons occupy e_g before pairing.

• (A) \( \ce{[Co(NH3)6]Cl3} \) — This matches (IV) t_2g⁶. \( \ce{NH3} \) is neutral and three \( \ce{Cl^{-}} \) ions are outside the bracket, so cobalt is +3. Co (Z = 27) is [Ar]3d⁷4s²; losing the two 4s electrons and one 3d electron leaves 3d⁶, and \( \ce{NH3} \) acts as a strong field ligand with \( \ce{Co^{3+}} \), so all six electrons pair in t_2g (low spin, d²sp³, diamagnetic).

• (B) \( \ce{[Ni(CO)4]} \) — This matches (III) Tetrahedral. CO is neutral, so nickel is in the zero oxidation state: Ni (Z = 28) is [Ar]3d⁸4s². CO is a strong field ligand, so the 4s electrons pair into 3d, giving 3d¹⁰. The empty 4s and 4p orbitals give sp³ hybridisation, and the complex is tetrahedral and diamagnetic.

• (C) \( \ce{[CoF6]Br3} \) — This matches (II) t_2g⁴e_g². The complex ion \( \ce{[CoF6]^{3-}} \) has cobalt in the +3 state (x + 6(−1) = −3), so it is Co³⁺, 3d⁶ as worked out in (A); as printed, [CoF₆]Br₃ would need cobalt in the +9 state, so the item is read as the complex ion, and \( \ce{F^{-}} \) is a weak field ligand. The electrons go into e_g before pairing, giving the high spin arrangement t_2g⁴e_g² with four unpaired electrons (sp³d², paramagnetic).

• (D) \( \ce{[Ni(CN)4]Cl2} \) — This matches (I) Square Planar. The complex ion \( \ce{[Ni(CN)4]^{2-}} \) has nickel in the +2 state (x + 4(−1) = −2); losing the two 4s electrons of Ni ([Ar]3d⁸4s²) gives Ni²⁺, 3d⁸ (as printed, [Ni(CN)₄]Cl₂ would need nickel in the +6 state). The strong field \( \ce{CN^{-}} \) ligands pair the two unpaired electrons, leaving one 3d orbital empty. This gives dsp² hybridisation, so the complex is square planar and diamagnetic.

Shortcut: The pair A–IV appears in Options (3) and (4), and B–III then leaves only Option (3).

Note: It seems that there is a mistake in the original NTA paper, which prints [CoF₆]Br₃ (item C) and [Ni(CN)₄]Cl₂ (item D) when it should read K₃[CoF₆] and K₂[Ni(CN)₄].