Transition metals and many of their compounds show paramagnetic behaviour where there are unpaired electron or electrons. The magnetic moment arises from the spin and orbital motions in ions or molecule. Magnetic moment of n unpaired electrons is given as
`mu=sqrt(n(n+2))` Bohr magneton
Magnetic moment increases as the number of unpaired electrons increases.
Q. Magnetic moment of `[Ni(CN)_4]^(2-)` is zero but that of `[Ni(H_2O)_4]^(2+)` is `2.83BM`. is because of :

Options

A. `CN^(ɵ)` is a strong ligand making two unpaired electrons in `Ni^(2+)` to pair up, while in `[Ni(H_2O)_4]^(2+)` two electrons remain unpaired as `H_2O` is a weak ligand.
B. Different oxidation state of `Ni` in two complexes
C. Both (a) and (b)
D. none of the above

Answer

To solve the question regarding the magnetic moments of the complexes `[Ni(CN)_4]^(2-)` and `[Ni(H_2O)_4]^(2+)`, we need to analyze the electronic configuration of nickel in both complexes and the nature of the ligands involved.

### Step-by-Step Solution:

1. **Determine the Oxidation State of Nickel in Both Complexes:**
- For `[Ni(CN)_4]^(2-)`:
- The cyanide ion (CN^-) has a charge of -1.
- Since there are four cyanide ions, the total negative charge contributed by CN^- is -4.
- The overall charge of the complex is -2.
- Let the oxidation state of Ni be \( x \).
- The equation becomes: \( x + (-4) = -2 \)
- Solving for \( x \): \( x = +2 \).

- For `[Ni(H_2O)_4]^(2+)`:
- Water (H2O) is a neutral ligand, so it does not contribute to the charge.
- The overall charge of the complex is +2.
- Let the oxidation state of Ni be \( y \).
- The equation becomes: \( y = +2 \).

2. **Determine the Electronic Configuration of Ni in Both Complexes:**
- Nickel has an atomic number of 28, so its ground state electronic configuration is \( [Ar] 3d^8 4s^2 \).
- In the +2 oxidation state, the configuration will be \( 3d^8 \) (since two electrons are removed, typically from the 4s orbital first).

3. **Analyze the Ligands:**
- **Cyanide (CN^-)**: This is a strong field ligand, which causes pairing of electrons in the d-orbitals.
- **Water (H2O)**: This is a weak field ligand, which does not cause pairing of electrons.

4. **Electron Configuration in the Complexes:**
- For `[Ni(CN)_4]^(2-)`:
- The strong field ligand CN^- causes the 3d electrons to pair up. Thus, the 3d orbitals will be filled as follows:
- \( 3d: \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \) (8 electrons, all paired).
- Therefore, there are **0 unpaired electrons**.

- For `[Ni(H_2O)_4]^(2+)`:
- The weak field ligand H2O does not cause pairing. Thus, the filling will be:
- \( 3d: \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \) (8 electrons, with 2 unpaired).
- Therefore, there are **2 unpaired electrons**.

5. **Calculate the Magnetic Moment:**
- The formula for magnetic moment \( \mu \) is given by:
\[ \mu = \sqrt{n(n+2)} \text{ Bohr magneton} \]
- For `[Ni(CN)_4]^(2-)` (0 unpaired electrons):
\[ \mu = \sqrt{0(0+2)} = 0 \text{ Bohr magneton} \]
- For `[Ni(H_2O)_4]^(2+)` (2 unpaired electrons):
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \text{ Bohr magneton} \]

### Conclusion:
The magnetic moment of `[Ni(CN)_4]^(2-)` is zero because all electrons are paired due to the strong field ligand CN^-, while the magnetic moment of `[Ni(H_2O)_4]^(2+)` is 2.83 BM due to the presence of 2 unpaired electrons caused by the weak field ligand H2O.