The number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion are:

Question

Accepted Answer

The question asks for the number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion. The correct answer is Option 1, which states that there are 0 unpaired electrons. Let's explore why this is the case and why the other options are incorrect.

### Explanation for Option 1 (Correct Answer):

1. **Electronic Configuration of Platinum (Pt):**
   - The atomic number of platinum is 78, and its ground state electronic configuration is [Xe] 4f¹⁴ 5d⁹ 6s¹.
   - In the complex $[Pt(CN)_{4}]^{2-}$, platinum is in the +2 oxidation state. This means it loses two electrons, resulting in the configuration [Xe] 4f¹⁴ 5d⁸.

2. **Nature of the Ligand:**
   - Cyanide (CN⁻) is a strong field ligand and causes a large splitting of the d-orbitals in the crystal field theory. This leads to the pairing of electrons in the lower energy orbitals.

3. **Square Planar Geometry:**
   - The $[Pt(CN)_{4}]^{2-}$ complex adopts a square planar geometry, which is typical for d⁸ metal ions with strong field ligands.
   - In a square planar field, the d-orbitals split into different energy levels. The order of filling is such that all electrons pair up in the lower energy orbitals, leaving no unpaired electrons.

4. **Resulting Electron Configuration:**
   - In the square planar arrangement, the 5d orbitals are split into different energy levels, and the electrons fill these levels in a way that minimizes unpaired electrons.
   - For $[Pt(CN)_{4}]^{2-}$, all eight d-electrons are paired, resulting in 0 unpaired electrons.

### Explanation for Why Other Options are Incorrect:

- **Option 2 (1 unpaired electron):**
  - If there were 1 unpaired electron, it would imply that one of the d-orbitals is singly occupied. However, due to the strong field nature of CN⁻ and the square planar geometry, all electrons are paired.

- **Option 3 (2 unpaired electrons):**
  - Having 2 unpaired electrons would suggest that two d-orbitals are singly occupied. This is not possible in a square planar complex with a strong field ligand like CN⁻, which causes all electrons to pair.

- **Option 4 (3 unpaired electrons):**
  - This would mean three d-orbitals are singly occupied, which is highly unlikely in a square planar complex with strong field ligands. The electron pairing energy is overcome by the crystal field splitting energy, leading to no unpaired electrons.

In conclusion, due to the strong field nature of the CN⁻ ligand and the square planar geometry of the $[Pt(CN)_{4}]^{2-}$ complex, all d-electrons are paired, resulting in 0 unpaired electrons. Therefore, Option 1 is the correct answer.