Match the complexes given in List-I with the geometry given in List-I…

The correct answer is Option (4) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)

List-I Complex	List-II Geometry
(A) $[NiCl_4]^{2-}$	(I) Tetrahedral, paramagnetic
(B) $[Ni(CN)_4]^{2-}$	(II) Square planer, diamagnetic
(C) $[COF_6]^{3-}$	(IV) Octahedral, paramagnetic
(D) $[Co(NH_3)_6]^{3+}$	(III) Octahedral, diamagnetic

To determine the correct matching, we analyze each complex based on:

• Oxidation state of the metal
• Strength of ligand (strong field or weak field)
• Hybridisation and geometry
• Magnetic behavior (presence of unpaired electrons)

(A) [NiCl₄]²⁻

Oxidation state of Ni:

x + 4(−1) = −2

x = +2

Electronic configuration of Ni²⁺:

Ni = [Ar] 3d⁸ 4s²

Ni²⁺ = 3d⁸

Cl⁻ is a weak field ligand, so pairing does not occur.

Hybridisation becomes sp³, giving tetrahedral geometry.

Because electrons remain unpaired, the complex is paramagnetic.

Thus:

(A) → Tetrahedral, paramagnetic

(A) – (I)

(B) [Ni(CN)₄]²⁻

Oxidation state of Ni:

Ni²⁺ → 3d⁸

CN⁻ is a strong field ligand, which causes pairing of electrons.

Hybridisation becomes dsp², giving square planar geometry.

All electrons become paired, so the complex is diamagnetic.

Thus:

(B) → Square planar, diamagnetic

(B) – (II)

(C) [CoF₆]³⁻

Oxidation state of Co:

x + 6(−1) = −3

x = +3

Co³⁺ configuration:

Co = [Ar] 3d⁷ 4s²

Co³⁺ = 3d⁶

F⁻ is a weak field ligand, so no electron pairing occurs.

The complex forms outer orbital hybridisation (sp³d²), producing an octahedral complex with unpaired electrons.

Thus the complex is paramagnetic.

(C) → Octahedral, paramagnetic

(C) – (IV)

(D) [Co(NH₃)₆]³⁺

Oxidation state:

Co³⁺ → 3d⁶

NH₃ is a strong field ligand with Co³⁺, causing pairing of electrons.

Hybridisation becomes d²sp³, producing an octahedral complex.

All electrons become paired, so the complex is diamagnetic.

Thus:

(D) → Octahedral, diamagnetic

(D) – (III)