There are four complexes of Ni. Select the complexes/es which will be attracted by magnetic field :
(I) `[Ni(CN)_(4)]^(2-)` (II) `[NiCl_(4)]^(2-)` (III) `[Ni(CO)_(4)]` (IV)`[Ni(NH_(3))_(6)]^(2+)`

Options

A. I only
B. II only
C. II, III and IV
D. II and IV

Answer

To determine which of the given nickel complexes are attracted by a magnetic field, we need to identify whether they are paramagnetic (having unpaired electrons) or diamagnetic (having all paired electrons).

### Step-by-Step Solution:

1. **Identify the Oxidation State of Nickel in Each Complex:**
- (I) `[Ni(CN)_(4)]^(2-)`: The CN ligand is a strong field ligand. The overall charge is -2, so Ni must be in the +2 oxidation state.
- (II) `[NiCl_(4)]^(2-)`: The Cl ligand is a weak field ligand. Again, the overall charge is -2, so Ni is in the +2 oxidation state.
- (III) `[Ni(CO)_(4)]`: CO is a strong field ligand and is neutral. Therefore, Ni is in the 0 oxidation state.
- (IV) `[Ni(NH_(3))_(6)]^(2+)`: NH3 is a moderate field ligand. The overall charge is +2, indicating Ni is in the +2 oxidation state.

2. **Determine the Electron Configuration of Nickel:**
- Nickel has an atomic number of 28, so its electron configuration is `[Ar] 3d^8 4s^2`.
- In the +2 oxidation state, Ni loses two electrons from the 4s orbital, resulting in the configuration: `3d^8`.

3. **Analyze Each Complex for Unpaired Electrons:**
- (I) `[Ni(CN)_(4)]^(2-)`: CN is a strong field ligand, which causes pairing of electrons. Thus, `3d^8` becomes `3d^8` with no unpaired electrons. This complex is **diamagnetic**.
- (II) `[NiCl_(4)]^(2-)`: Cl is a weak field ligand, which does not cause pairing. Therefore, `3d^8` retains unpaired electrons. This complex is **paramagnetic** and will be attracted to a magnetic field.
- (III) `[Ni(CO)_(4)]`: CO is a very strong field ligand, which causes all electrons to pair. Thus, `3d^8` becomes `3d^{10}` with no unpaired electrons. This complex is **diamagnetic**.
- (IV) `[Ni(NH_(3))_(6)]^(2+)`: NH3 is a moderate field ligand. In an octahedral field, the `3d^8` configuration can have unpaired electrons. Typically, this results in 2 unpaired electrons. This complex is **paramagnetic** and will be attracted to a magnetic field.

4. **Conclusion:**
- The complexes that will be attracted by a magnetic field are:
- (II) `[NiCl_(4)]^(2-)`
- (IV) `[Ni(NH_(3))_(6)]^(2+)`

### Final Answer:
The complexes that will be attracted by a magnetic field are (II) `[NiCl_(4)]^(2-)` and (IV) `[Ni(NH_(3))_(6)]^(2+)`.