Which of the following compounds is paramagnetic ?

Options

A. `[Ni(CO)_(4)]`
B. `K_(2)[Ni(CN)_(4)]`
C. `K_(3)[CoF_(6)]`
D. `[Cr(NH_(3))_(6)]Cl_(3)`

Answer

To determine which of the given compounds is paramagnetic, we need to analyze each compound for the presence of unpaired electrons. Paramagnetic substances have unpaired electrons, while diamagnetic substances have all electrons paired.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are:
- Nickel tetracarbonyl (Ni(CO)₄)
- Potassium tetra cyano nickel (K₂Ni(CN)₄)
- Potassium hexafluorocobaltate (K₃CoF₆)
- Chromium hexamine chloride (Cr(NH₃)₆Cl₃)

2. **Analyze Nickel Tetracarbonyl (Ni(CO)₄)**:
- Nickel in this complex has an oxidation state of 0.
- The electronic configuration of Ni in the ground state is [Ar] 3d⁸ 4s².
- CO is a strong field ligand, which causes pairing of electrons.
- After hybridization (sp³), the configuration becomes 3d¹⁰, with no unpaired electrons.
- **Conclusion**: Ni(CO)₄ is **diamagnetic**.

3. **Analyze Potassium Tetra Cyano Nickel (K₂Ni(CN)₄)**:
- Nickel here has an oxidation state of +2.
- The electronic configuration for Ni²⁺ is 3d⁸.
- CN⁻ is also a strong field ligand, causing pairing of electrons.
- The 3d configuration after pairing is fully filled (3d⁸ → 3d¹⁰).
- **Conclusion**: K₂Ni(CN)₄ is **diamagnetic**.

4. **Analyze Potassium Hexafluorocobaltate (K₃CoF₆)**:
- Cobalt in this complex has an oxidation state of +3.
- The electronic configuration for Co³⁺ is 3d⁶.
- F⁻ is a weak field ligand, which does not cause pairing.
- The 3d configuration has 4 unpaired electrons (3d⁶).
- **Conclusion**: K₃CoF₆ is **paramagnetic**.

5. **Analyze Chromium Hexamine Chloride (Cr(NH₃)₆Cl₃)**:
- Chromium here has an oxidation state of +3.
- The electronic configuration for Cr³⁺ is 3d³.
- NH₃ is a neutral ligand and does not cause pairing.
- The 3d configuration has 3 unpaired electrons (3d³).
- **Conclusion**: Cr(NH₃)₆Cl₃ is **paramagnetic**.

### Final Answer:
The compounds that are paramagnetic are:
- K₃CoF₆ (Option C)
- Cr(NH₃)₆Cl₃ (Option D)