Which of the following complexes are square planar?

(A) $[Ni(CN)_4]^{2-}$
(B) $[NiCl_4]^{2-}$
(C) $[PtCl_4]^{2-}$
(D) $[PdCl_4]^{2-}$

Choose the correct answer from the options given below:

(A), (C) and (D) only

The correct answer is Option (4) → (A), (C) and (D) only

Square planar geometry is commonly shown by d⁸ metal ions when strong field ligands are present.

Let us analyze each complex.

Option (A) [Ni(CN)₄]²⁻

Nickel oxidation state:

Ni + 4(CN⁻) = -2

Ni = +2

Ni²⁺ → 3d⁸ configuration

CN⁻ is a strong field ligand.

Strong field ligands cause pairing of electrons and lead to dsp² hybridisation.

dsp² hybridisation gives square planar geometry.

Hence, this complex is square planar.

Option (B) [NiCl₄]²⁻

Ni²⁺ again has 3d⁸ configuration.

But Cl⁻ is a weak field ligand.

Weak field ligands do not cause pairing of electrons.

Hence, hybridisation becomes sp³.

sp³ leads to tetrahedral geometry.

Therefore, this complex is tetrahedral, not square planar.

Option (C) [PtCl₄]²⁻

Pt²⁺ → 5d⁸ configuration

For 5d metals like platinum, even weak field ligands can cause pairing due to large crystal field

Thus, dsp² hybridisation occurs.

This leads to square planar geometry.

Hence, this complex is square planar.

Option (D) [PdCl₄]²⁻

Pd²⁺ → 4d⁸ configuration

Like platinum, palladium also belongs to heavier transition metals.

Large crystal field splitting favors electron pairing.

Thus, dsp² hybridisation occurs.

Hence, square planar geometry is formed.