Valence bond theory successfully explains the magnetic behaviour of complexes. The substances which contains unpaired electrons. and paramagnetic character increases as the mumber of unpaired electrons increases. Magnetic moment of a complex can be determined experimentally and by using formula `sqrt(n(n+2))` and we can determine the number of unpaired electrons in it. This information is important in writing the electronic structure of complex which in turm is also useful in deciding the geometry of complex.
The magnetic moments (spin only) of `[NiCl_(4)]^(2-) and [Ni(CN)_(4)]^(2-)` are:

Options

A. 0,0
B. 2.82 B.M., 2.82 B.M.
C. 0,2.82 B.M.
D. 2.82 B.M., 0

Explanation

To solve the problem regarding the magnetic moments of the complexes \([NiCl_4]^{2-}\) and \([Ni(CN)_4]^{2-}\), we will follow these steps:

### Step 1: Determine the oxidation state of Nickel in both complexes.
- For \([NiCl_4]^{2-}\):
- Let the oxidation state of Ni be \(x\).
- The charge contributed by 4 Cl ligands (each with a charge of -1) is \(-4\).
- The overall charge of the complex is \(-2\).
- Therefore, the equation is:
\[ x + (-4) = -2 \implies x = +2 \]

- For \([Ni(CN)_4]^{2-}\):
- Similarly, let the oxidation state of Ni be \(x\).
- The charge contributed by 4 CN ligands (each with a charge of -1) is \(-4\).
- The overall charge of the complex is \(-2\).
- Therefore, the equation is:
\[ x + (-4) = -2 \implies x = +2 \]

### Step 2: Write the electronic configuration of Ni in the +2 oxidation state.
- Nickel (Ni) has an atomic number of 28. The electronic configuration of neutral Ni is \([Ar] 3d^8 4s^2\).
- In the +2 oxidation state, Ni loses two electrons, typically from the 4s orbital first:
- Thus, the electronic configuration for \([Ni]^{2+}\) is \([Ar] 3d^8\).

### Step 3: Determine the nature of the ligands and their effect on electron pairing.
- **For \([NiCl_4]^{2-}\)**:
- Chloride (Cl) is a weak field ligand, which means it does not cause strong pairing of electrons.
- According to Hund's rule, the electrons will fill the d-orbitals singly before pairing.
- The filling of the \(3d\) orbitals will yield 2 unpaired electrons.

- **For \([Ni(CN)_4]^{2-}\)**:
- Cyanide (CN) is a strong field ligand, which causes pairing of electrons in the lower energy orbitals.
- Therefore, all 8 electrons will pair up in the \(3d\) orbitals, resulting in 0 unpaired electrons.

### Step 4: Calculate the magnetic moments using the formula.
- The formula for magnetic moment (\(\mu\)) is given by:
\[ \mu = \sqrt{n(n+2)} \]
where \(n\) is the number of unpaired electrons.

- **For \([NiCl_4]^{2-}\)**:
- Here, \(n = 2\) (2 unpaired electrons).
- Thus,
\[ \mu = \sqrt{2(2+2)} = \sqrt{2 \times 4} = \sqrt{8} = 2.83 \text{ Bohr Magneton} \]

- **For \([Ni(CN)_4]^{2-}\)**:
- Here, \(n = 0\) (0 unpaired electrons).
- Thus,
\[ \mu = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ Bohr Magneton} \]

### Summary of Results:
- The magnetic moment of \([NiCl_4]^{2-}\) is \(2.83 \text{ Bohr Magneton}\).
- The magnetic moment of \([Ni(CN)_4]^{2-}\) is \(0 \text{ Bohr Magneton}\).