Explanation

We calculate unpaired electrons in each complex one by one.

1. [Ni(CO)₄]

Oxidation state of Ni = 0

Electronic configuration of Ni (0):

$$ [Ar]\,3d^8\,4s^2 $$

CO is a strong field ligand, causing pairing of electrons.

Hybridization = sp³, tetrahedral, all electrons paired.

Unpaired electrons = 0

2. [NiCl₄]²⁻

Oxidation state of Ni = +2

Ni²⁺ configuration:

$$ [Ar]\,3d^8 $$

Cl⁻ is a weak field ligand, no pairing occurs.

Geometry = tetrahedral (sp³)

d⁸ tetrahedral → 2 unpaired electrons

3. [PtCl₂(NH₃)₂]

Oxidation state of Pt = +2

Pt²⁺ configuration:

$$ 5d^8 $$

Square planar complex → dsp² hybridization

All electrons paired.

Unpaired electrons = 0

4. [Ni(CN)₄]²⁻

Oxidation state of Ni = +2 → d⁸

CN⁻ is a strong field ligand.

Square planar, dsp² hybridization.

All electrons paired.

Unpaired electrons = 0

5. [Pt(CN)₄]²⁻

Oxidation state of Pt = +2 → d⁸

Strong field ligand, square planar.

All electrons paired.

Unpaired electrons = 0

Total unpaired electrons:

$$ 0 + 2 + 0 + 0 + 0 = \boxed{2} $$

Related JEE Main Chemistry Questions