Which of the following complexes is diamagnetic

Options

A. `Ni(CO)_4`
B. `Ni(Cl_4)^(2-)`
C. `Ni(Br)_4^(2-)`
D. `NiCl_2. 4H_2O`

Explanation

To determine which of the given complexes is diamagnetic, we need to analyze the electronic configuration of the metal ions in each complex and the nature of the ligands involved. Here’s a step-by-step solution:

### Step 1: Identify the Complexes
We need to identify the complexes provided in the question. In this case, we have:
1. Nickel complex with carbon (Ni(CO)₄)
2. Nickel(II) complex with chlorine (NiCl₂)
3. Nickel(II) complex with fluorine (NiF₂)
4. Nickel(II) complex with water (Ni(H₂O)₂)

### Step 2: Determine the Oxidation State and Electron Configuration
- **For Ni(CO)₄**:
- Carbon (CO) is a neutral ligand, so the oxidation state of nickel is 0.
- The electron configuration of Ni (atomic number 28) is [Ar] 3d⁸ 4s². In Ni(0), it will have 10 electrons in the 3d and 4s orbitals.

- **For NiCl₂**:
- Chlorine is a -1 ligand, so the oxidation state of nickel is +2.
- The electron configuration for Ni²⁺ is [Ar] 3d⁸ (losing 2 electrons from 4s).

- **For NiF₂**:
- Fluorine is also a -1 ligand, so the oxidation state of nickel is +2.
- The electron configuration for Ni²⁺ is the same as above, [Ar] 3d⁸.

- **For Ni(H₂O)₂**:
- Water is a neutral ligand, so the oxidation state of nickel is +2.
- The electron configuration for Ni²⁺ remains [Ar] 3d⁸.

### Step 3: Analyze the Ligands
- **CO**: A strong field ligand that causes pairing of electrons.
- **Cl⁻**: A weak field ligand that does not cause pairing.
- **F⁻**: A weak field ligand that does not cause pairing.
- **H₂O**: A weak field ligand that does not cause pairing.

### Step 4: Determine Magnetic Properties
- **Ni(CO)₄**: Since CO is a strong field ligand, it will cause the electrons to pair up. Therefore, there will be no unpaired electrons, making it **diamagnetic**.

- **NiCl₂, NiF₂, Ni(H₂O)₂**: In these complexes, since Cl⁻, F⁻, and H₂O are weak field ligands, they do not cause electron pairing. Thus, Ni²⁺ in these complexes will have unpaired electrons, making them **paramagnetic**.

### Conclusion
From the analysis, the only complex that is diamagnetic is **Ni(CO)₄**.

### Final Answer
**The complex that is diamagnetic is Ni(CO)₄.**

---