Match the LIST-I with LIST-II.LIST-I ComplexLIST-II PropertiesA. [Mn(…

The correct answer is Option (1) → A-III, B-IV, C-I, D-II

A four-coordinate complex is either tetrahedral (sp³) or square planar (dsp²). With weak field ligands (\( \ce{Cl^{-}} \), \( \ce{Br^{-}} \)) or a d¹⁰ metal, the inner d orbitals are not free and the complex is tetrahedral. A d⁸ ion with a strong field, or a d⁸ ion of the 4d/5d series (where the splitting is always large), pairs its electrons, frees one d orbital and becomes square planar.

• A. \( \ce{[Mn(Br)4]^{2-}} \) — This matches III. Tetrahedral, sp³ and 5 unpaired electrons. x + 4(−1) = −2 gives \( \ce{Mn^{2+}} \); Mn (Z = 25) [Ar]3d⁵4s² loses the two 4s electrons: 3d⁵. \( \ce{Br^{-}} \) is weak field, so all five electrons stay unpaired.

• B. \( \ce{[Ni(CO)4]} \) — This matches IV. Tetrahedral, sp³ and 0 unpaired electrons. CO is neutral, so Ni is in the 0 state; Ni (Z = 28) is [Ar]3d⁸4s². The strong field CO pushes the 4s electrons into 3d, giving 3d¹⁰, all paired; 4s and 4p orbitals form sp³ hybrids.

• C. \( \ce{[Pt(Cl)4]^{2-}} \) — This matches I. Square planar, dsp² and 0 unpaired electrons. x + 4(−1) = −2 gives \( \ce{Pt^{2+}} \); Pt (Z = 78) [Xe]4f¹⁴5d⁹6s¹ loses the 6s electron and one 5d electron: 5d⁸. For a 5d metal the splitting is large even with \( \ce{Cl^{-}} \), so the electrons pair and the complex is square planar.

• D. \( \ce{[Ni(Cl)4]^{2-}} \) — This matches II. Tetrahedral, sp³ and 2 unpaired electrons. x + 4(−1) = −2 gives \( \ce{Ni^{2+}} \); Ni (Z = 28) [Ar]3d⁸4s² loses the two 4s electrons: 3d⁸. \( \ce{Cl^{-}} \) is weak field, so no pairing occurs; two electrons remain unpaired and the complex is tetrahedral.

Shortcut: The pair A–III appears in Options (1) and (4), and B–IV then leaves only Option (1).