Amongst `Ni(CO)_(4),[Ni(CN)_(4)]^(2-)` and `[NiCl_(4)]^(2-)`

Options

A. `[Ni(CO)_4], [NiCl_4]^(2-)` are diamagnetic and `[Ni(CN)_4]^(2-)` is paramagnetic.
B. `[NiCl_4]^(2-) , [Ni(CN)_4]^(2-) ` are diamagnetic and `[Ni(CO)_4] ` is a paramagnetic .
C. `[Ni(CO)_4], [Ni(CN)_4]^(2-)` are diamagnetic and `[NiCl_4]^(2-)` is paramagnetic .
D. `[Ni(CO)_4]` is diamagnetic and `[NiCl_4],[Ni(CN)_4]^(2-)` are paramagnetic .

Answer

To determine the magnetic nature of the coordination compounds `Ni(CO)_(4)`, `[Ni(CN)_(4)]^(2-)`, and `[NiCl_(4)]^(2-)`, we will analyze each compound step by step.

### Step 1: Analyze `Ni(CO)_(4)`

1. **Determine the oxidation state of Ni**:
- Carbonyl (CO) is a neutral ligand, so the oxidation state of Ni in `Ni(CO)_(4)` is 0.

2. **Electronic configuration of Ni**:
- The ground state electronic configuration of Ni is `3d^8 4s^2`.

3. **Effect of the ligand**:
- CO is a strong field ligand, which causes pairing of electrons.
- In the presence of CO, the 3d electrons will pair up, leading to the configuration: `3d^{10}` (all electrons paired).

4. **Magnetic nature**:
- Since all electrons are paired, `Ni(CO)_(4)` is **diamagnetic**.

### Step 2: Analyze `[Ni(CN)_(4)]^(2-)`

1. **Determine the oxidation state of Ni**:
- Cyanide (CN) is a -1 charge ligand. For the complex `[Ni(CN)_(4)]^(2-)`, the oxidation state of Ni can be calculated as follows:
- Let the oxidation state of Ni be x.
- x + 4*(-1) = -2 → x - 4 = -2 → x = +2.

2. **Electronic configuration of Ni in +2 state**:
- The electronic configuration of Ni in the +2 state is `3d^8`.

3. **Effect of the ligand**:
- CN is also a strong field ligand, which leads to low spin complexes and pairing of electrons.
- The configuration will be `3d^{10}` (all electrons paired).

4. **Magnetic nature**:
- Since all electrons are paired, `[Ni(CN)_(4)]^(2-)` is **diamagnetic**.

### Step 3: Analyze `[NiCl_(4)]^(2-)`

1. **Determine the oxidation state of Ni**:
- Chloride (Cl) is a -1 charge ligand. For the complex `[NiCl_(4)]^(2-)`, the oxidation state of Ni can be calculated as follows:
- Let the oxidation state of Ni be x.
- x + 4*(-1) = -2 → x - 4 = -2 → x = +2.

2. **Electronic configuration of Ni in +2 state**:
- The electronic configuration of Ni in the +2 state is `3d^8`.

3. **Effect of the ligand**:
- Cl is a weak field ligand, which leads to high spin complexes and no pairing of electrons.
- The configuration will remain as `3d^8`, resulting in 2 unpaired electrons.

4. **Magnetic nature**:
- Since there are unpaired electrons, `[NiCl_(4)]^(2-)` is **paramagnetic**.

### Conclusion

- `Ni(CO)_(4)` is **diamagnetic**.
- `[Ni(CN)_(4)]^(2-)` is **diamagnetic**.
- `[NiCl_(4)]^(2-)` is **paramagnetic**.

Thus, the correct statement regarding the magnetic nature of these complexes is that the first two complexes are diamagnetic and the last one is paramagnetic.