Number of unpaired electrons in the square planar $[Pt(CN)_4]^{2-}$ are:

Zero unpaired electron

The correct answer is Option (2) → Zero unpaired electron

Core Concept:

Strong field ligands cause pairing of electrons.

Explanation:

First find oxidation state of Pt.

Let oxidation state = x

x + 4(−1) = −2

x − 4 = −2

x = +2

So:

Pt²⁺

Electronic configuration of Pt:

Pt → [Xe] 4f¹⁴ 5d⁹ 6s¹

Pt²⁺ → remove 6s¹ and one 5d electron

Pt²⁺ → 5d⁸

CN⁻ is a strong field ligand → causes pairing.

Square planar complexes of d⁸ metals are low spin.

All electrons pair up.

Hence: 

No unpaired electrons.

Option 1: Incorrect — pairing occurs.

Option 2: Correct — all electrons paired.

Option 3: Incorrect — no unpaired electron remains.

Option 4: Incorrect — strong field prevents this.